IIT JEE Physics (1978 to 2018) Topic-wise Complete Solutions for IIT JEE with H C Verma Harish Chandra Verma Bharati Bhawan

Table of contents :
Mechanics......Page 11
Units and Measurements......Page 13
Rest and Motion: Kinematics......Page 19
Newton's Laws of Motion......Page 29
Friction......Page 35
Circular Motion......Page 46
Work and Energy......Page 51
Centre of Mass, Linear Momentum, Collision......Page 63
Rotational Mechanics......Page 84
Gravitation......Page 129
Simple Harmonic Motion......Page 140
Fluid Mechanics......Page 158
Some Mechanical Properties of Matter......Page 175
Waves......Page 187
Wave Motion and Waves on a String......Page 189
Sound Waves......Page 202
Light Waves......Page 223
Optics......Page 241
Geometrical Optics......Page 243
Optical Instruments......Page 283
Dispersion and Spectra......Page 285
Photometry......Page 288
Thermodynamics......Page 289
Heat and Temperature......Page 291
Kinetic Theory of Gases......Page 300
Calorimetry......Page 307
Laws of Thermodynamics......Page 312
Specific Heat Capacities of Gases......Page 319
Heat Transfer......Page 339
Electromagnetism......Page 355
Electric Field and Potential......Page 357
Gauss's Law......Page 375
Capacitors......Page 390
Electric Current in Conductors......Page 400
Thermal and Chemical Effects of Electric Current......Page 424
Magnetic Field......Page 428
Magnetic Field due to a Current......Page 447
Permanent Magnets......Page 462
Electromagnetic Induction......Page 468
Alternating Current......Page 493
Electromagnetic Waves......Page 500
Modern Physics......Page 501
Electric Current through Gases......Page 503
Photoelectric Effect and Wave-Particle Duality......Page 505
Bohr's Model and Physics of the Atom......Page 515
X-rays......Page 531
Semiconductors and Semiconductor Devices......Page 536
The Nucleus......Page 539
Mechanics......Page 559
Waves......Page 562
Optics......Page 564
Heat and Thermodynamics......Page 565
Electricity and Magnetism......Page 566
Modern Physics......Page 569

Citation preview

IIT JEE PHYSICS (1978–2018: 41 Years) Topic-wise Complete Solutions

Combined Volume Mechanics, Waves and Optics Heat, Electromagnetism and Modern Physics

Jitender Singh Shraddhesh Chaturvedi

PsiPhiETC 2018

ii

Published by PsiPhiETC 116, Nakshatra Colony, Balapur Hyderbad 500005, Telangana, India.

IIT JEE Physics c 2018 by Authors Copyright All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the authors. Request for permission to make copies of any part of the work should be mailed to: 116, Nakshatra Colony, Balapur, Hyderbad 500005, Telangana, India. The authors have taken care in preparation of this book, but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information contained herein. Typeset in TEX. Third Edition, 2018

2

Visit us at: www.concepts-of-physics.com

We dedicate this book to the hundreds of anonymous professors at IITs who formulated the challenging problems for IIT-JEE. The book is a showcase of their creation.

v

Foreword Physics starts with observing the nature. The systematic observation results in simple rules which unlock the doors to the nature’s mystery. Having learned a handful of simple rules, we can combine them logically to obtain more complicated rules and gain an insight into the way this world works. The skill, to apply the theoretical knowledge to solve any practical problem, comes with regular practice of solving problems. The aim of the present collection of problems and solutions is to develop this skill. IIT JEE questions had been a challenge and a center of attraction for a big section of students at intermediate and college level. Independent of their occurrence as an evaluation tool, they have good potential to open up thinking threads in mind. Jitender Singh and Shraddhesh Chaturvedi have used these questions to come up with a teaching material that can benefit students. The explanations accompanying the problems could bring conceptual clarity and develop the skills to approach any unseen problem, step by step. These problems are arranged in a chapter sequence that is used in my book Concepts of Physics. Thus a student using both the books will find it as an additional asset. Both Jitender Singh and Shraddhesh Chaturvedi have actually been my students at IIT, Kanpur. Jitender Singh has been closely associated with me since long. It gives me immense pleasure to see that my own students are furthering the cause of Physics education. I wish them every success in this work and expect much more contribution from them in future! Dr. H C Verma Professor of Physics IIT Kanpur

vii

Preface This book provides a comprehensive collection of IIT JEE problems and their solutions. We have tried to keep our explanations simple so that any reader, with basic knowledge of intermediate physics, can understand them on his/her own without any external assistance. It can be, therefore, used for self-study. To us, every problem in this book, is a valuable resource to unravel a deeper understanding of the underlying physical concepts. The time required to solve a problem is immaterial as far as Physics is concerned. We believe that getting the right answer is often not as important as the process followed to arrive at it. The emphasis in this text remains on the correct understanding of the principles of Physics and on their application to find the solution of the problems. If a student seriously attempts all the problems in this book, he/she will naturally develop the ability to analyze and solve complex problems in a simple and logical manner using a few, well-understood principles. For the convenience of the students, we have arranged the problems according to the standard intermediate physics textbook. Some problems might be based on the concepts explained in multiple chapters. These questions are placed in a later chapter so that the student can try to solve them by using the concept(s) from multiple chapters. This book can, thus, easily complement your favorite text book as an advanced problem book. The IIT JEE problems fall into one of the nine categories: (i) MCQ with single correct answer (ii) MCQ with one or more correct answers (iii) Paragraph based (iv) Assertion Reasoning based (v) Matrix matching (vi) True False type (vii) Fill in the blanks (viii) Integer Type, and (ix) Subjective. Each chapter has sections according to these categories. In each section, the questions are arranged in the descending order of year of appearance in IIT JEE. Detailed solutions are given for each problem. We advise you to solve each problem yourself. You may cover the solution with a piece of paper to focus your attention to the problem. If you can’t solve a problem, you can always look at the solution later. However, trying it first will help you identify the critical points in the problems, which in turn, will accelerate the learning process. Furthermore, it is advised that even if you think that you know the answer to a problem, you should turn to its solution and check it out, just to make sure you get all the critical points. This book has a companion website, www.concepts-of-physics.com. The site will host latest version of the errata list and other useful material. We would be glad to hear from you for any suggestions on the improvement of the book. We have tried our best to keep the errors to a minimum. However, they might still remain! So, if you find any conceptual errors or typographical errors, howsoever small and insignificant, please inform us so that it can be corrected in the later editions. We believe, only a collaborative effort from the readers and the authors can make this book absolutely error-free, so please contribute. Many friends and colleagues have contributed greatly to the quality of this book. First and foremost, we thank Dr. H. C. Verma, who was the inspiring force behind this project. Our close friends and classmates from IIT Kanpur, Deepak Sharma, Chandrashekhar Kumar and Akash Anand stood beside us throughout this work. This work would not have been possible without the constant support of our wives Reena and Nandini and children Akshaj, Viraj and Maitreyi. Jitender Singh, [email protected] Shraddhesh Chaturvedi, [email protected]

Contents

I

Mechanics . . . . . . . . . . . . . . . . . . . . . . . .

1

22 Calorimetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297

1

Units and Measurements . . . . . . . . . . . . . . . . . . .

3

23 Laws of Thermodynamics . . . . . . . . . . . . . . . . . . 302

2

Rest and Motion: Kinematics . . . . . . . . . . . . . . .

9

24 Specific Heat Capacities of Gases . . . . . . . . . . . 309

3

Newton’s Laws of Motion . . . . . . . . . . . . . . . . . .

19

25 Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329

4

Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25

5

Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . .

36

6

Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . .

41

7

Centre of Mass, Linear Momentum, Collision .

53

8

Rotational Mechanics . . . . . . . . . . . . . . . . . . . . . .

74

9

Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

V Electromagnetism . . . . . . . . . . . . . . . . . 345 26 Electric Field and Potential . . . . . . . . . . . . . . . . 347 27 Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 28 Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 29 Electric Current in Conductors . . . . . . . . . . . . . 390 10 Simple Harmonic Motion . . . . . . . . . . . . . . . . . . 130

30 Thermal and Chemical Effects of Electric Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414

11 Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 148

31 Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 418

12 Some Mechanical Properties of Matter . . . . . . . 165

32 Magnetic Field due to a Current . . . . . . . . . . . . 437

II Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

33 Permanent Magnets . . . . . . . . . . . . . . . . . . . . . . . 452

13 Wave Motion and Waves on a String . . . . . . . . 179

34 Electromagnetic Induction . . . . . . . . . . . . . . . . . 458

14 Sound Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

35 Alternating Current . . . . . . . . . . . . . . . . . . . . . . . 483

15 Light Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

36 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . 490

III Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

VI Modern Physics . . . . . . . . . . . . . . . . . . . 491

16 Geometrical Optics . . . . . . . . . . . . . . . . . . . . . . . . 233

37 Electric Current through Gases . . . . . . . . . . . . . 493

17 Optical Instruments . . . . . . . . . . . . . . . . . . . . . . . 273

38 Photoelectric Effect and Wave-Particle Duality 495

18 Dispersion and Spectra . . . . . . . . . . . . . . . . . . . . 275

39 Bohr’s Model and Physics of the Atom . . . . . . 505

19 Photometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

40 X-rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 41 Semiconductors and Semiconductor Devices . . 526

IV Thermodynamics . . . . . . . . . . . . . . . . . . 279

42 The Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529

20 Heat and Temperature . . . . . . . . . . . . . . . . . . . . . 281

A Quick Reference Formulae . . . . . . . . . . . . . . . . . . 549

21 Kinetic Theory of Gases . . . . . . . . . . . . . . . . . . . 290 ix

Part I

Mechanics

θ L

m ~v

M

1

Chapter 1 Units and Measurements

One Option Correct

scale (m is counted beyond MSR). The value measured by this calipers is

Q 1. There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1 ) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2 ) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1 and C2 , respectively are (2016) 2

3

X = MSR + x = MSR + mxm − vxv .

(1)

In calipers C1 , MSR1 = 2.8 cm, m1 = 7 and v1 = 7 and in calipers C2 , MSR2 = 2.8 cm, m2 = 8 and v2 = 7. Substitute these values in equation (1) to get X1 = MSR1 + m1 xm1 − v1 xv1 = 2.8 + 7(0.1) − 7(0.09) = 2.87 cm. X2 = MSR2 + m2 xm2 − v2 xv2

4

= 2.8 + 8(0.1) − 7(0.11) = 2.83 cm.

C1 0 2

5

10

3

The Vernier calipers are generally of type C1 having m = v and least count LC = xm −xv . For these calipers, equation (1) gives X = MSR + v × LC. Ans. B

4

C2 0

(A) 2.85 and 2.82 (C) 2.87 and 2.86

5

10

Q 2. The diameter of a cylinder is measured using a Vernier calipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is (2013) (A) 5.112 cm (B) 5.124 cm (C) 5.136 cm (D) 5.148 cm

(B) 2.87 and 2.83 (D) 2.87 and 2.87

Sol. In both calipers C1 and C2 , 1 cm is divided into 10 equal divisions on the main scale. Thus, 1 division on the main scale is equal to xm1 = xm2 = 1 cm/10 = 0.1 cm. In calipers C1 , 10 equal divisions on the Vernier scale are equal to 9 main scale divisions. Thus, 1 division on the Vernier scale of C1 is equal to xv1 = 9xm1 /10 = 0.09 cm. In calipers C2 , 10 equal divisions on the Vernier scale are equal to 11 main scale divisions. Thus, 1 division on the Vernier scale of C2 is equal to xv2 = 11xm2 /10 = 0.11 cm. MSR1

1 MSD = 5.15 cm − 5.10 cm = 0.05 cm, 2.45 cm = 0.049 cm. 1 VSD = 50

7xm1 x1

2

Sol. From the given data, a main scale division (MSD) and a vernier scale division (VSD) are

3

4

C1 0

5 7xv1

X1 MSR2

LC = 1 MSD − 1 VSD = 0.001 cm.

8xm2 x2

2

The least count (LC) of the given Vernier calipers is

10

3

For the given measurement, main scale reading (MSR) is 5.10 cm and the vernier scale reading (VSR) is 24. Hence, diameter D of the cylinder is

4

C2 0 X2

5

10

D = MSR + VSR × LC

7xv2

= 5.10 + 24 × 0.001 = 5.124 cm. Let main scale reading be MSR and v th division of the Vernier scale coincides with mth division of the main

Ans. B 3

4

Part I. Mechanics

Q 3. The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has relative error of 2%, the percentage error in density is (2011)

(A) 0.9 % (B) 2.4 % (C) 3.1 % (D) 4.2 % Sol. The density of a ball of mass m and diameter D is given by
(1) ρ = 6m/ πD3 . Differentiate equation (1) to get 6 dρ = π



D3 dm − 3mD2 dD D6

 .

(2)

Divide equation (2) by (1) to get

Q 4. A Vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier calipers, the least count is (2010) (A) 0.02 mm (B) 0.05 mm (C) 0.1 mm (D) 0.2 mm Sol. For the given Vernier calipers, a main scale division (MSD) is 1 MSD = 1 mm. Since 20 vernier scale divisions (VSD) are equal to 16 MSD, we get 1 VSD = 16/20 = 0.8 mm. The least count (LC) is given by LC = 1 MSD − 1 VSD = 1.0 − 0.8 = 0.2 mm. Ans. D Q 5. In a screw gauge, the zero of main scale coincides with fifth division of circular scale as shown in the figure (i). The circular scale of a screw gauge has 50 divisions and its pitch is 0.5 mm. The diameter of the ball being measured in the figure (ii) is (2006) 0

dρ dm dD = −3 . ρ m D

(3)

In error analysis, measured mass mmeasured and actual mass mactual are related by

10 5 0

fig. (i)

0

30 25 20

fig. (ii)

(A) 1.2 mm (B) 1.25 mm (C) 2.20 mm (D) 2.25 mm Sol. One division on main scale is p = 0.5 mm (pitch) and one division on circular scale (Least Count) is

mactual = mmeasured ± ∆m, where ∆m is a small positive number representing measurement error. Let ∆m and ∆D be the measurement errors (both positive) in m and D. From equation (3), dρ is maximum when dm = ∆m and dD = −∆D. Thus, the error in ρ is ∆ρ ∆m ∆D = +3 . ρ m D

(4)

The least count (LC) of a screw gauge, with pitch p and total number of divisions on the circular scale N , is given by

LC = p/N = 0.5/50 = 0.01 mm. From the figure (i), zero error corresponding to five circular scale divisions is e = 5 × LC = 5 × 0.01 = 0.05 mm. From the figure (ii), measurement by screw gauge corresponding to two main scale divisions and 25 circular scale divisions is Dmeasured = 2 × p + 25 × LC = 2 × 0.5 + 25 × 0.01 = 1.25 mm. The diameter of the ball is equal to its measured value minus zero error of the screw gauge i.e.,

LC = p/N = 0.5/50 = 0.01 mm. LC is error in measurement of the ball diameter D i.e., ∆D = 0.01 mm. The measured diameter for the given main scale reading (MSR) and circular scale reading (n) is D = MSR + (n) LC = 2.5 + (20)0.01 = 2.70 mm. 2 Given ∆m m = 100 = 0.02. Substitute the values in equation (4) to get

∆ρ 0.01 = 0.02 + 3 × = 0.031 = 3.1%. ρ 2.70 Ans. C

Dactual = Dmeasured − e = 1.25 − 0.05 = 1.20 mm. Ans. A Q 6. A student performs an experiment for determi  4π 2 l nation of g = T 2 . The error in pendulum length l (≈ 1 m) is ∆l. To measure time period T , the student takes total time of n oscillations with a stop watch of least count ∆Tlc and (s)he commits a human error of ∆Th = 0.1 s. The amplitude of oscillation is A. The measurement of g is most accurate for (2006) (A) ∆l = 5 mm, ∆Tlc = 0.2 s, n = 10, A = 5 mm (B) ∆l = 5 mm, ∆Tlc = 0.2 s, n = 20, A = 5 mm (C) ∆l = 5 mm, ∆Tlc = 0.1 s, n = 20, A = 1 mm (D) ∆l = 1 mm, ∆Tlc = 0.1 s, n = 50, A = 1 mm

Chapter 1. Units and Measurements

5

Sol. The time period of given pendulum is p T = 2π l/g ≈ 2 s,

One or More Option(s) Correct

where we used l = 1 m and g = 9.8 m/s2 . Differentiate and simplify the formula for g to get ∆g ∆l ∆T = +2 . g l T

(1)

Let t be the total time of n oscillations. The error in measurement of t is due to least count of stop watch (∆Tlc ) and human error (∆Th ) i.e., ∆t = ∆Tlc + ∆Th . Thus, measurement error in time period is ∆Tlc ∆Th ∆t = + . ∆T = n n n Substitute ∆T in equation (1) to get ∆l ∆Tlc + ∆Th ∆g = +2 . g l nT

(2)

In equation (2), substitute l = 1 m, T = 2 s, ∆Th = 0.1 s and values of ∆l, ∆Tlc and n from given options. The measurement of g is most accurate for ∆l = 1 mm, ∆T = 0.1 s and n = 50. Relative error in this case is ∆g/g = 0.005. Ans. (D) Q 7. A wire has a mass m = 0.3 ± 0.003 g, radius r = 0.5 ± 0.005 mm and length l = 6 ± 0.06 cm. The maximum percentage error in the measurement of its density is (2004) (A) 1 (B) 2 (C) 3 (D) 4 Sol. The density of a wire of mass m, length l, and radius r is given by ρ=

m m = 2 . V πr l

(1)

Differentiate equation (1) and divide it by ρ to get ∆m ∆r ∆l ∆ρ + . = + 2 (2) ρ max m r l From the given data, m = 0.3 g, ∆m = 0.003 g, r = 0.5 mm, ∆r = 0.005 mm, l = 6 cm, and ∆l = 0.06 cm. Substitute the values in equation (2) to get ∆ρ 0.003 2 × 0.005 0.06 = + + = 0.04 = 4%. ρ 0.3 0.5 6 Ans. D Q 8. The edge of a cube is measured to be 1.2 × 10−2 m. Its volume should be recorded as (2003) (A) 1.7 × 10−6 m3 (B) 1.73 × 10−6 m3 −6 3 (C) 1.70 × 10 m (D) 1.728 × 10−6 m3 Sol. The volume V = (1.2 × 10−2 )3 = 1.728 × 10−6 m3 shall be recorded as 1.7 × 10−6 m3 upto two significant figures. Ans. A

Q 9. In an experiment to determine the acceleration due to gravity g, the formula usedqfor the time period of a periodic motion is T = 2π 7(R−r) . The val5g ues of R and r are measured to be (60 ± 1) mm and (10 ± 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true? (2016)

(A) (B) (C) (D)

The The The The

error error error error

in in in in

the the the the

measurement of r is 10%. measurement of T is 3.57%. measurement of T is 2%. determined value of g is 11%.

Sol. The relative percentage error in the measurement of r is given by er =

∆r 1 × 100 = × 100 = 10%. r 10

The mean value of time period is given by P5

0.52 + 0.56 + 0.57 + 0.54 + 0.59 1 Ti = 5 5 = 0.556 s ≈ 0.56 s,

T¯ =

where we have truncated at 2 significant figures. The absolute error and the relative percentage error in the measurement of T are given by P5

|Ti − T¯| 5 0.04 + 0 + 0.01 + 0.02 + 0.03 = = 0.02. 5 ∆T 0.02 × 100 = 3.57%. eT = ¯ × 100 = 0.56 T

∆T =

1

Given equation for time period can be written as g=

28π 2 R − r . 5 T2

Differentiate to get the relative error in g as ∆g ∆(R − r) ∆T ∆R + ∆r ∆T = +2 = +2 g R−r T R−r T 1+1 0.02 = +2 = 0.11 = 11%. 60 − 10 0.56 It q is interesting to note that the time period T = 2π 7(R−r) is same as the time period of a small spher5g ical ball of mass m and radius r when it rolls without slipping on a rough concave surface of large radius R. We encourage you to derive it. Ans. A, B, D

6

Part I. Mechanics

Sol. In given Vernier callipers, each 1 cm is equally divided into 8 main scale divisions (MSD). Thus, 1 MSD = 1/8 = 0.125 cm. Further, 4 main scale divisions coincide with 5 Vernier scale divisions (VSD) i.e., 4 MSD = 5 VSD. Thus, 1 VSD = 4/5 MSD = 0.8 × 0.125 = 0.1 cm. The least count of the Vernier callipers is given by LC = 1 MSD − 1 VSD = 0.125 − 0.1 = 0.025 cm. In screw gauge, let l be the distance between two adjacent divisions on the linear scale. The pitch p of the screw gauge is the distance travelled on the linear scale when it makes one complete rotation. Since circular scale moves by two divisions on the linear scale when it makes one complete rotation, we get p = 2l. The least count of the screw gauge is defined as ratio of the pitch to the number of divisions on the circular scale (n) i.e., lc = p/n = 2l/100 = l/50.

Sol. From the given expression, d = λ/(2 sin θ). Differentiate to get ∆d = −(λ/2) csc θ cot θ ∆θ. The absolute error in d is |∆d| = (λ/2) csc θ cot θ |∆θ|. The figure shows that csc θ cot θ (and hence |∆d|) decreases as θ increases from 0◦ to 90◦ . csc θ cot θ

(2015)

(A) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm. (B) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm. (C) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm. (D) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

θ in the range 0 to 90◦ . The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0◦ , (2013) (A) the absolute error in d remains constant. (B) the absolute error in d increases. (C) the fractional error in d remains constant. (D) the fractional error in d decreases.

0◦

θ

90◦

The fractional error in d is −(λ/2) csc θ cot θ∆θ ∆d = = − cot θ∆θ. d λ/(2 sin θ)

cot θ

Q 10. Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then,

0◦

θ

90◦

The figure shows that cot θ (and hence ∆d/d) decreases as θ increases from 0◦ to 90◦ . Ans. D

(1) Integer Type

If p = 2 LC = 2(0.025) = 0.05 cm, then l = p/2 = 0.025 cm. Substitute l in equation (1) to get the least count of the screw gauge lc = 0.025/50 = 5 × 10−4 cm = 0.005 mm. If l = 2 LC = 2(0.025) = 0.05 cm then equation (1) gives lc = 0.05/50 = 1 × 10−3 cm = 0.01 mm. Ans. B, C Q 11. Using the expression 2d sin θ = λ, one calculates the value of d by measuring the corresponding angles

Q 12. The energy of a system as a function of time t is given as E(t) = A2 exp(−αt), where α = 0.2 s−1 . The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t = 5 s is . . . . . . . (2015) Sol. Differentiate the expression E(t) = A2 e−αt to get dE = 2Ae−αt dA − A2 αe−αt dt.

(1)

Divide equation (1) by E(t) and simplify to get dA dA dt dE =2 − αdt = 2 − αt . E A A t

(2)

Chapter 1. Units and Measurements

7

The error in measurement of a parameter x is generally defined by xactual = xmeasured ± ∆x,

Since 10 vernier scale divisions (VSD) are equal to 9 MSD, we get 1 VSD = 9/10 = 0.9 mm.

where ∆x is a small positive number representing measurement error. Let ∆A and ∆t be the measurement errors (both positives) in A and t. From equation (2), dE is maximum when dA = ∆A and dt = −∆t. Thus, the percentage relative error in E(t) is given by ∆A ∆t ∆E =2 + αt E A t = 2(1.25%) + (0.2)(5)(1.5%) = 4%.

The least count (LC) is given by LC = 1 MSD − 1 VSD = 1.0 − 0.9 = 0.1 mm. Given, the main scale reading (MSR) is 10 and the vernier scale reading (VSR) is one. The measured value of the edge is given by a = MSR × MSD + VSR × LC = 10 × 1 + 1 × 0.1 = 10.1 mm.

We encourage you to derive equation (2) by taking logarithm on both sides of E(t) = A2 e−αt and then differentiating it. Ans. 4 Q 13. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density ρ of the fog, intensity (power/area) S of the light from the signal and its frequency f . The engineer finds that d is proportional to S 1/n . The value of n is . . . . . . . (2014) Sol. From the given information, d = k ρa S b f c ,

(1)

where k is some dimensionless proportionality constant and a, b, c are unknown parameters. Substitute dimensions of physical quantities in equation (1) to get [L1 ] = [ML−3 ]a [MT−3 ]b [T−1 ]c .

(2)

Equate the exponents of M and L in equation (2) to get a + b = 0,

(3)

1 = −3a.

(4)

m/V = 2.736/1.03 = 2.6563 = 2.66 g/cm3 , (after rounding off to three significant digits). Ans. 2.66 g/cm3 Q 15. The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area (in cm2 ) of the wire in appropriate number of significant figures. (2004) Sol. The distance moved on the linear scale when circular scale makes one complete rotation is p = 1 mm (pitch). The number of divisions on the circular scale is N = 100. Thus, one division on the circular scale is LC = p/N = 1/100 = 0.01 mm. The linear scale reading (LSR) is 1 mm and the circular scale reading (CSR) is 47. Thus, the diameter of the wire is d = LSR + CSR × LC = 1 + 47 × 0.01 = 1.47 mm = 0.147 cm.

Solve equations (3) and (4) to get b = 1/3. Ans. 3 Descriptive Q 14. The edge of a cube is measured using a Vernier calipers (9 divisions of the main scale are equal to 10 divisions of Vernier scale and 1 main scale division is 1 mm). The main scale division reading is 10 and first division of Vernier scale was found to be coinciding with the main scale. The mass of the cube is 2.736 g. Calculate the density in g/cm3 upto correct significant figures. (2005) Sol. From the given data, a main scale division (MSD) is 1 MSD = 1 mm.

The measurement of a has three significant digits. The volume of the cube is V = a3 = 1.03 cm3 and the density is

The curved surface area of the cylinder is A = πdl = 3.14(0.147)(5.6) = 2.5848 cm2 . Rounding off to two significant digits gives A = 2.6 cm2 . Ans. 2.6 cm2 Q 16. N divisions on the main scale of a Vernier calipers coincide with (N + 1) divisions on its Vernier scale. If each division on the main scale is of a units, determine the least count of instrument. (2003) Sol. Given, a main scale division (MSD) of the Vernier calipers is 1 MSD = a.

8

Part I. Mechanics

Since (N + 1) vernier scale divisions (VSD) are equal to N main scale divisions, we get 1 VSD =

N Na MSD = . N +1 N +1

The least count is given by LC = 1 MSD − 1 VSD =

a . N +1 Ans.

a N +1

Chapter 2 Rest and Motion: Kinematics

One Option Correct

The sound of impact travels upwards with a velocity v = 300 m/s. The time taken by the sound to travel a distance L is t2 = L/v = 0.067 s. Thus, time interval between dropping the stone and receiving the sound of impact is

Q 1. Consider an expanding sphere of instantaneous radius r whose total mass remains constant. The expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate of fractional is constant. The velocity v change in density ρ1 dρ dt of any point on the surface of the expanding sphere is proportional to (2017) (A) r (B) 1/r (C) r3 (D) r2/3

T = t1 + t2 =

1 δT = 2

are constants and

dρ dt

r

2 δL δL δL + = gL v L

1 2

s

2L L + g v

! ,

which gives δL = L

Differentiate w.r.t. time t to get  1/3 dr 1 3m 1 dρ r 1 dρ =− =− . dt 3 4πρ ρ dt 3 ρ dt 1 dρ ρ dt

2L/g + L/v.

Differentiate to get

Sol. The velocity of any point, on the surface of the expanding sphere of instantaneous radius r, is radially outwards. Its magnitude is given by v = dr/dt. The density of the sphere of mass m is given by ρ = m/( 43 πr3 ) which gives,  1/3 3m r= . 4πρ

Note that m and

p

δT 1 2

q

2L g

+

0.01

=

L v

1 2

q

2(20) 10

+

≈ 0.01 = 1%. 20 300

Ans. (A) Q 3. The given graph shows the variation of velocity (v) with displacement (x). Which one of the following graph correctly represents the variation of acceleration (a) with displacement (2005)

is negative. Ans. (A)

Q 2. A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δT = 0.01 s and he measures the depth of the well to be L = 20 m. Take the acceleration due to gravity g = 10 m/s2 and the velocity of sound is 300 m/s. Then the fractional error in the measurement, δL/L, is closest to (2017) (A) 1% (B) 5% (C) 3% (D) 0.2%

v v0

O

(A)

Sol. The stone is dropped from rest and falls under a constant acceleration g = 10 m/s2 before it hits the bottom of the well at depth L = 20 m. The time taken by the pstone to reach the bottom of the well is given by t1 = 2L/g = 2 s.

a

O

(C)

•

(B) x

a

O

x

a

O

(D) x

x0

x

a

O

x

v L

Sol. Given graph is a straight line with equation

g

v = v0 − 9

v0 x. x0

10

Part I. Mechanics

Differentiate the velocity v w.r.t. time t to get the acceleration

resistance, its velocity v varies with height h above the ground as (2000) (A)

dv dv dx dv = =v dt dx dt dx  v0 v0 v2 v2 =− v0 − x = − 0 + 02 x. x0 x0 x0 x0

v

v

(B)

a=

O

Thus, a-x graph is a straight line with negative intercept on a axis and a positive slope. Ans. A Q 4. A particle starts from rest. Its acceleration a (in m/s2 ) versus time t (in s) is as shown in the figure. The maximum speed of the particle will be (2004)

v

O

h

v

(D)

O

h

h

Sol. The equation v 2 − u2 = 2gh gives velocity of the ball at a height h during its downward journey as p ~vdown = − 2g(d − h) ˆ.

a(m/s2 ) 10

O

(C)

O

h

t(s)

11

y ˆ

(A) 110 m/s (B) 55 m/s (C) 550 m/s (D) 660 m/s

d

Sol. The acceleration a is decreasing but positive from t = 0 to t = 11 s. Thus, speed is maximum at t = 11 s and is given by the area under t-a graph i.e., vmax =

1 2

× 11 × 10 = 55 m/s.

We encourage you to solve R t this problem5 analytically. 10 Hint: a = 10 − 11 t, v = 0 a dt = 10t − 11 t2 , and v is maximum at t = 11 s. Ans. B Q 5. A small block slides without friction down an inclined plane starting from rest. Let sn be the distance sn travelled from t = n − 1 to t = n. Then sn+1 is (2004) 2n−1 2n+1 2n−1 2n (A) 2n (B) 2n−1 (C) 2n+1 (D) 2n+1 Sol. The downward acceleration of the block on an inclined plane with inclination angle θ is given by a = g sin θ. Initial velocity of the block is u = 0. The distance travelled in the first n seconds is given by s(n) = un + 21 g sin θ n2 =

1 2

g sin θ n2 .

Thus, the distances travelled in the n seconds are

th

which is positive and becomes zero at h = d/2. Ans. A Q 7. In 1.0 s, a particle goes from point A to point B, moving in a semicircle of radius 1.0 m (see figure). The magnitude of the average velocity is (1999)

1m •

and the (n+1)

B

th

(A) 3.14 m/s (B) 2.0 m/s (C) 1.0 m/s (D) zero

1 2

sn+1 = s(n + 1) − s(n) = (2n + 1) g sin θ. sn sn+1

x

This velocity remains negative till the ball strikes the ground. After bouncing from the ground the ball rises to a√height d/2. Thus the speed just after bouncing is u = gd. The velocity of the ball during upward motion at a height h is p ~vup = gd − 2gh ˆ,

A

sn = s(n) − s(n − 1) = (2n − 1) 12 g sin θ,

Divide equation (1) by (2) to get

d 2

=

(1) (2)

2n−1 2n+1 .

Ans. C Q 6. A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air

Sol. The average velocity is given by ~v = ∆~r/∆t, where ∆~r is the change in the displacement vector and ∆t is time interval. The magnitude of the displacement vector changes by |∆~r| = |AB| = 2 m in time interval ∆t = 1 s. Thus, |~v | = |∆~r|/∆t = 2/1 = 2 m/s. Ans. B

Chapter 2. Rest and Motion: Kinematics Q 8. A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t = 0 along the horizontal string AB, with the speed v. Friction between the bead and the string may be neglected. Let tP and tQ be the respective times taken by P and Q to reach the point B. Then, (1993) Q

A

B

11 Let ~vb and ~vr be the velocities of the boat and the river current w.r.t. the ground. The velocity of the boat in still water is equal to the relative velocity of the boat w.r.t. water i.e., ~vb/r = ~vb − ~vr , which gives |~vb/r |2 = |~vb |2 + |~vr |2 .

C

(A) tP < tQ (C) tP > tQ

(B) tP = tQ length of arc ACB P = length (D) ttQ of chord AB

Sol. Let l be the length of the chord AB. The time taken by the particle Q and the particle P to reach the point B are tQ = l/v, and tP = l/¯ vh , where v¯h is the average horizontal velocity of the particle P. A

Q

B N

P

N

•

•

mg

mg

C

The forces acting on the particle P are its weight mg and normal reaction N . The weight is always in downward vertical direction and hence cannot affect horizontal velocity vh . However, N has a horizontal component that accelerates particle P from A to C and retards it from C to B. By symmetry, P will have same velocity on points symmetrically located about C. At B, velocity is again vh and at every point between A and B it is greater than vh . Thus, the average horizontal velocity of P is more than its starting value i.e., v¯h > v. Hence, tP < tQ . Ans. A Q 9. A boat, which has a speed of 5 km/h in still water, crosses a river of width 1 km along the shortest possible path in 15 min. The velocity of the river water (in km/h ) is (1988) √ (A) 1 (B) 3 (C) 4 (D) 41

~vr ~vb/r

~vb

d = 1km

Sol. The boat crosses the river by the shortest path if it moves perpendicular to the river current.

(1)

Given, speed of the boat in still water |~vb/r | = 5 km/h and the speed of the boat relative to the ground |~vb | =

P

(∵ ~vb ⊥ ~vr ).

1 d = = 4 km/h. t 15/60

Substitute |~vb | and |~vb/r | in equation (1) to get |~vr | = 3 km/h. Ans. B Q 10. A river is flowing from west to east at a speed of 5 m/min. A man on the south bank of the river, capable of swimming at 10 m/min in still water, wants to swim across the river in the shortest time. He should swim in a direction (1983) (A) due north (B) 30◦ east of north (C) 30◦ west of north (D) 60◦ east of north Sol. Let ~vm and ~vr be velocities of the man and the river current w.r.t. the ground. The velocity of the man in still water is equal to the relative velocity of the man w.r.t. water i.e., ~vm/r = ~vm − ~vr . Q R ~vr T ~vm/r

α P

θ

~vm N E

Let ~vm/r and ~vm make angle α and θ with the north direction. Let d be the width of the river. The time taken by the man to cross the river is given by width of the river component of man velocity along north d = |~vm | cos θ d . = |~vm/r | cos α

t=

(1)

(∵ PQ = t|~vm | cos θ = t|~vm/r | cos α). From equation (1), t is minimum when denominator is maximum. Given, |~vm/r | = 10 m/min, a constant. Thus, t become minimum when cos α = 1 i.e., α = 0. Hence, the man takes the shortest time when he swims perpendicular to the river velocity i.e., towards north. Note that man will reach the point T and not Q. Ans. A

12

Part I. Mechanics

Q 11. In the arrangement shown in the figure the end P and Q of an unstretchable string move downwards with uniform speed U . Pulleys A and B are fixed. Mass M moves upwards with a speed (1982) (A) 2U cos θ (B) U/ cos θ (C) 2U/ cos θ (D) U cos θ

N −~vi

~v f −

45◦

~v i

Sol. In the right angled triangle AOM, let side AO = x, OM = y, and MA = r. Pythagoras theorem gives x2 + y 2 = r 2 .

(1) A

O

P

M

The average acceleration of the particle is given by

Thus, the magnitude of the average acceleration is 0.707 m/s2 and its direction is north-west. Ans. C

Q

y dy/dt = r dr/dt.

(2)

In triangle AOM, y/r = cos θ. As the string is unstretchable, the rate of decrease of r (side AM) is equal to the rate at which P is pulled down i.e., dr/dt = U. Substitute in equation (2) to get the upward speed of M as V = dy/dt = (dr/dt)r/y = U r/y = U/ cos θ. Note that V increases when string is pulled down and becomes infinite at θ = 90◦ . ~ be the upward velocity of the mass Aliter: Let V M . The tangential velocity of any point on the string should be a constant, U (because string is unstretchable). Consider a point of string which is in contact ~ along the string direcwith M . The component of V tion MA is V cos θ. Hence, V cos θ = U . Ans. B Q 12. A particle is moving eastwards with a velocity of 5 m/s. In 10 s the velocity changes to 5 m/s northwards. The average acceleration in this time is (1982) (A) zero. (B) √12 m/s2 towards north-east. √1 2

E

~vi

∆~v ~vf − ~vi 5n ˆ − 5 eˆ = = ∆t tf − ti 10     √ n ˆ − eˆ n ˆ − eˆ √ √ = 0.5 2 = 0.707 . 2 2

When the ends P and Q of the string are pulled down, r and y decrease but x remains constant. Differentiate equation (1) w.r.t. time to get

(C)

O

~aavg =

B

θ

θ

~vf

One or More Option(s) Correct Q 13. A particle of mass m moves on the x-axis as follows: it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times (0 < t < 1). If α denotes the instantaneous acceleration of the particle, then, (1993) (A) α cannot remain positive for all t in the interval 0 ≤ t ≤ 1. (B) |α| cannot exceed 2 at any point in its path. (C) |α| must be ≥ 4 at some point or points in its path. (D) α must change sign during the motion, but no other assertion can be made with the information given. Sol. The positions of the particle at t = 0 and at t = 1 s are x(0) = 0 m,

x(1) = 1 m.

The particle is at rest at t = 0 and comes to rest at t = 1 s. Thus, initial and final velocities of the particle are v(0) = 0,

v(1) = 0.

(1)

The velocity v(1) of a particle having instantaneous acceleration α is given by Z v(1) = v(0) +

1

α dt.

(2)

0

α

m/s2 towards north-west.

(D) 1/2 m/s2 towards north. Sol. Let eˆ and n ˆ be the unit vectors along east and north directions, respectively. Initially, velocity of the particle is ~vi = 5 ˆe m/s. The particle velocity after t = 10 s is ~vf = 5 n ˆ m/s.

1

t(s)

R1 The equations (1) and (2) give, 0 αdt = 0. Thus, area under the α-t curve is zero. If α = 0 for all 0 < t < 1 then area becomes zero but this also makes x(1) = 0, which is not true. If α > 0 for all 0 < t < 1 then

Chapter 2. Rest and Motion: Kinematics

13 x

area under the curve is positive and if α < 0 for all 0 < t < 1 then area under the curve is negative. Thus, α must change sign at some time in R 1 0 < t < 1. Only conclusions that can be made are 0 αdt = 0, α 6= 0, and α must change sign (at least once) in 0 < t < 1.

0 •O2

O2 •

x d2 α α1 0 −α2

0 •O1

•

θ1 d1 tc

1

θ2

t

Lens

v(m/s) 2

tc> 12 tc< 12

0

O1

tc

1

• I01

• I02

Retina • I1,I2

t

If α change sign only once (say at t = tc ) and α = α1 in 0 < t < tc and α = −α2 in tc < t < 1, where α1 and α2 are some positive constants, then condition x(1) = 1 m is violated if α < 4 at all points. We encourage you to explore area under v-t graph to show that α ≥ 4 at some point(s). Hint: The area under v-t graph, 12 × 1 × vmax = 1, gives vmax = 2 m/s. Now, vmax = 2 = α1 tc = α2 (1 − tc ). If tc < 21 than α1 > 4 otherwise α2 > 4. Ans. A, C

The explanation has to do with how human mind/eye perceive the motion. Take a simple situation of two point objects O1 and O2 (see figure). The images I1 and I2 formed on the retina of eye are point images. As the object moves, the image on the retina also moves. The distance moved by the image on the retina is perceived as motion/velocity. Since eye is a lens with distance of retina/screen w.r.t. lens almost fixed, the angular displacement of the image is perceived as velocity. For same relative velocity of objects O1 and O2 , angular displacement of nearby object (θ1 in the figure) is more than the angular displacement of the far-off object (θ2 ). Hence nearby objects appear to move faster. Ans. B True False Type

Q 14. Statement 1: For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary. Statement 2: If the observer and the object are ~1 and V ~2 respectively with refmoving at velocities V erence to a laboratory frame, the velocity of the object ~2 − V ~1 . with respect to the observer is V (2008) (A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. (B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true. Sol. Both, statement 1 and statement 2 are true but statement 2 is not a correct explanation of statement 1. The relative velocity of the nearby as well as the far-off objects is same.

Q 15. A projectile fired from the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path. (1984) Sol. Let the projectile be fired with an initial velocity v at an angle θ from the horizontal. y v

P •

v cos θ

v sin θ

Assertion Reasoning Type

θ O v cos θ

x

The horizontal component of velocity, v cos θ, remains constant throughout the trajectory because there is no acceleration in the horizontal direction. The vertical component of velocity starts decreasing from the initial value v sin θ to zero at the topmost point P and then again increases to v sin θ when projectile hits the ground. Thus, the speed of the projectile at the top, v cos θ, is minimum. We encourage you to use conservation of mechanical energy to arrive at the same conclusion. Ans. T

14

Part I. Mechanics

Q 16. Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion with the same speed. [Neglect air resistance.] (1983) Sol. Apply v 2 − u2 = 2as (or the conservation of mechanical energy) to show that speeds of the balls at the projection point (in the downward journey) are equal to their initial projection speeds. Ans. T Fill in the Blank Type Q 17. The trajectory of a projectile in a vertical plane is y = ax−bx2 , where a, b are constants, and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is . . . . . . and the angle of projection from the horizontal is . . . . . . (1997) Sol. Trajectory equation is y = ax − bx2 . The slope of the trajectory is zero at maximum height i.e., dy = a − 2bx = 0. dx

(1)

Solve equation (1) to get x = a/(2b). Substitute it in the given trajectory equation to get the maximum height a  a 2 a2 ymax = a −b = . 2b 2b 4b The slope of trajectory at the projection point is given by dy tan θ = = a − 2b(0) = a. dx x=0 Thus, the angle of projection is θ = tan−1 a. 2 Ans. a4b , tan−1 a Q 18. Four persons K, L, M, N are initially at the four corners of a square of side d. Each person now moves with a uniform speed v in such a way that K always moves directly towards L, L directly towards M, M directly towards N and N directly towards K. The four persons will meet at a time . . . . . . (1984)

Consider a small time interval ∆t at the beginning of journey. In this time interval, all of them travel a distance v∆t in the direction of person they are facing i.e., person K moves a distance v∆t along the line KL, person L moves a distance v∆t along the line LM and so on. In the next time interval, K starts moving in the direction of new position of L, L starts moving in the direction of new position of M and so on. Through this process, all persons finally meet at the centre of the square O. The symmetry plays a key role in this problem. Four persons always remain at the corners of a square. The sides of this square decrease and rotate continuously but its centre remains fixed at O. Now, concentrate on the motion of K. The magnitude of its velocity is v and the direction of its velocity at any instant is along the side K0 L0 of the square formed by the four persons at that instant (see figure). Thus, its velocity vector always makes 45◦ angle with the line joining current position of K and centre of the square O. Hence, the √ component of its velocity towards O is v cos 45◦ = v/ 2 (a constant). The net displacement of K from the initial position K to√the final posi# » tion O is vector KO with magnitude d/ 2. Thus, the time taken by K to reach O is the ratio of ‘displacement’ and √‘component of velocity along displacement’ d/ 2 √ = d. i.e., t = v/ v 2 Aliter: The velocity of K is v along KL. The velocity of L is zero along KL (because it is perpendicular to the line KL). Thus, the separation KL decreases at a rate v. Since this rate is constant, the time taken to reduce the separation from d to zero is d/v. Ans. d/v Q 19. A particle moves in a circle of radius R. In half the period of revolution its displacement is . . . . . . and distance covered is . . . . . . (1983) Sol. Let the particle travels from the point P to the point Q in half of the time period.

R P•

•

O

•

Q

Sol. The trajectory followed by the four persons is very interesting (see figure). N

M0

M

N0

O

v

L0

Integer Type

v √ 2

K

The distance travelled by the particle is half of the circle perimeter i.e., πR. The displacement of the particle is a vector from the initial point to the final point # » i.e., PQ. The magnitude of the displacement vector is # » |PQ| = 2R. Ans. 2R, πR

v

K0

v d

L

Q 20. A ball is projected from the ground at an angle of 45◦ with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon

Chapter 2. Rest and Motion: Kinematics

15

hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30◦ with the horizontal surface. The maximum height it reaches after the bounce, in meters, is . . . . . . . (2018) Sol. The maximum height of the ball projected from the ground with an initial speed u at an angle θ with the horizontal is given by H=

u2 sin2 θ . 2g

(1)

y

Sol. Let us analyse the problem in the laboratory frame. Let v0 be the speed of the rocket when balls are thrown, say at time t = 0. In the laboratory frame, the speed of a ball thrown from the left end is vl = v0 +0.3 and that of a ball thrown from the right end is vr = v0 −0.2. Since the rocket is accelerating towards the right with an acceleration a = 2 m/s2 , its velocity keeps on increasing and the left face of the rocket collides with a ball thrown from the left end before this ball collides with the ball thrown from the right end. This collision takes place at a time t when the distance travelled by the rocket is equal to the distance travelled by the ball thrown from the left i.e.,

u H

v0 t + 12 (2)t2 = (v0 + 0.3)t.

u0

θ

θ0

O

P

H0

x Q

By conservation of energy, the kinetic energy of the ball just before it hits the ground is equal to its initial kinetic energy (K = mu2 /2). The ball loses half of its kinetic energy upon hitting the ground. Thus, kinetic energy of the ball just after hitting the ground is K0 =

1 1 1 1 2 mu0 = K = · mu2 , 2 2 2 2

(2)

2

which gives u0 = u2 /2. The velocity ~u 0 makes an angle θ0 = 30◦ with the horizontal. Apply equation (1) to get the maximum height after hitting the ground

Solve to get t = 0.3 s. The displacement of left ball relative to the left face  of rocket at time t is  (v0 + 0.3)t − (v0 t + 21 (2)t2 ) which attains a maximum value of 2.25 cm at time t = 0.15 s. Type of collision (elastic/inelastic) between the rocket and the ball is not given in the question. Since the relative distance of the ball thrown from the left w.r.t. rocket’s left face is negligible in comparison to the chamber length, we assume that the ball and the left face are located approximately at the same distance from the right face. Thus, we need to find the time at which a ball thrown from the right end collides with the left face of the rocket. The right ball collides with the left face when their displacements are equal i.e.,

2

u0 sin2 θ0 u2 sin2 θ0 = · 2g 2 2g 2 0 1 2gH sin θ H sin2 θ0 = · = · 2 2 sin θ 2g 2 sin2 θ (120) sin2 30◦ = 30 m. = 2 sin2 45◦

H0 =

(using (2)) (using (1))

We encourage you to find the range OQ of the ball when it hits the ground for the second time. Ans. 30 Q 21. A rocket is moving in a gravity free space with a constant acceleration of 2 m/s2 along +x direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from the left end of the chamber in +x direction with a speed of 0.3 m/s relative to the rocket. At the same time, another ball is thrown in −x direction with a speed of 0.2 m/s from its right end relative to the rocket. The time in seconds when the two balls hit each other is . . . . . . . (2014) 0.3 m/s 0.2 m/s 4m

v0 t + 12 (2)t2 = 4 + (v0 − 0.2)t. Solve to get t = 1.9 s ≈ 2 s. Ans. 2 Q 22. Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30◦ and 60◦ with respect to the horizontal respectively as shown in √ the figure. The speed of A is 100 3 m/s. At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = t0 , A just escapes being hit by B, t0 in seconds is . . . . . . . (2014)

A

30◦

B 60◦

a = 2 m/s2 x

~A and V ~B be the velocity vectors of airplane Sol. Let V A and B in a frame attached to the ground. The figure ~A , V ~B , and V ~B/A , the velocity of B relative to A. shows V

16

Part I. Mechanics

The expressions for these three vectors are u sin 60◦

√ √ ~A = 100 3 cos 30◦ ˆı + 100 3 sin 30◦ ˆ V √ = 150 ˆı + 50 3 ˆ, ~B = VB cos 60◦ ˆı + VB sin 60◦ ˆ V √ VB 3VB = ˆı + ˆ, 2 2 ~B/A = V ~B − V ~A V     √ VB VB − 150 ˆı + 3 − 50 ˆ. = 2 2

y

u

x 60◦ u0 + u cos 60◦

(1)

The time of √ flight is given by t = 2uy /g = 2u sin 60◦ /g = 3 s. The distance travelled by the train and the ball in time t are xtrain = u0 t + 21 at2 and xball = (u0 + u cos 60◦ )t. Since the boy moves forward by 1.15 m, we have

y

xball = xtrain + 1.15, (u0 + u cos 60◦ )t = u0 t + 12 at2 + 1.15.

~B/A V

√ Substitute u = 10 m/s and t = 3 s in equation (1) and then solve to get a = 5 m/s2 . We encourage you to solve this problem in a frame attached to the train. Ans. 5

~B V

60◦

(1)

~A V 30◦ x

O

Descriptive The observer in A sees B moving with a constant velocity perpendicular to the line of motion of A i.e., ~B/A ⊥ V ~A . Thus, V ~B/A · V ~A = 150(VB − 200) = 0. V

(2)

Solve equation (2) to get VB = 200 m/s. Substitute √ ~B/A = −50ˆı + 50 3ˆ VB in equation (1) to get V  and ~ |VB/A | = 100 m/s. The time taken to travel a relative distance of 500 m with a relative speed of 100 m/s is t0 = 500/100 = 5 s. Ans. 5 Q 23. A train is moving along straight line with a constant acceleration a. A boy standing in the train throws a ball forward with a speed of 10 m/s at an angle 60◦ from horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train in m/s2 is . . . . . . . (2011) Sol. Consider the motion in a frame attached to the ground. Let train’s velocity at the time of projection of ball be u0 . The ball is projected with a velocity u = 10 m/s (relative to the train) at an angle 60◦ from the horizontal. Thus the horizontal and vertical component of the initial velocity of the ball are

Q 24. On a frictionless horizontal surface, assumed to be x-y plane, a small trolley A is moving along a straight line parallel√to the
y-axis (see figure) with a constant velocity of 3 − 1 m/s. At a particular instant when the line OA makes an angle of 45◦ with the x-axis, a ball is thrown along the surface from the origin O. Its velocity makes an angle φ with the x-axis and it hits the trolley. (2002) y A

45◦

x

O

(a) The motion of the ball is observed from the frame of the trolley. Calculate the angle θ made by the velocity vector of the ball with the x-axis in this frame. (b) Find the speed of the ball with respect to the surface, if φ = 4θ/3. Sol. The velocities of the trolley and the ball w.r.t. surface are

ux = u0 + u cos 60◦ ,

√ ~vA = ( 3 − 1) ˆ, and

uy = u sin 60◦ .

~vB = vB cos φ ˆı + vB sin φ ˆ.

Chapter 2. Rest and Motion: Kinematics y

17

•D

~vA •A ~vB a φ •

45◦

x

•

a

B

C

The distance travelled by the ball along x and y axes are vB cos φ t = a,

(1)

vB sin φ t = a + AD = a + vA t.

(2)

Q 26. A large heavy box is sliding without friction down a smooth plane of inclination θ. From a point P on the bottom of the box, a particle is projected inside the box. The initial speed of the particle with respect to the box is u and the direction of projection makes an angle α with the bottom as shown in the figure. (1998) (a) Find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands. [Assume that the particle does not hit any other surface of the box. Neglect air resistance.] (b) If the horizontal displacement of the particle as seen by an observer on the ground is zero, find the speed of the box with respect to the ground at the instant when the particle was projected.

•

α

Eliminate a from equations (1) and (2) to get

Q

•

P

vB sin φ − vA = vB cos φ.

θ

(3)

The velocity of the ball in a frame attached to the trolley is equal to the relative velocity of B w.r.t. A i.e., ~vB/A = ~vB − ~vA = vB cos φ ˆı + (vB sin φ − vA ) ˆ.

Sol. Let x-y frame be attached to the ground as shown in the figure. Since box moves without friction on a smooth plane, its acceleration is ax,b = −g sin θ,

ay,b = 0.

The angle θ made by ~vB/A with x-axis is given by tan θ =

(vB/A )y vB sin φ − vA = 1. = (vB/A )x vB cos φ

u

(4)

Solve equation (4) to get θ = 45◦ . Thus, to an observer in trolley, the ball will appear to be coming towards √ him. Substitute φ = 4θ/3 = 4×45/3 = 60◦ and vA = ( 3−1) in equation (3) to get vB = 2 m/s. Ans. (a) 45◦ (b) 2 m/s Q 25. An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0 the plank starts moving along the +x direction with an acceleration 1.5 m/s2 . At the same instant a stone is projected from the origin with a velocity ~u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45◦ to the horizontal. All the motions are in x-y plane. Find ~u and the time after which the stone hits the object. [Take g = 10 m/s2 .] (2000) y(m)

x θ

Let ux,b = −u0,b and uy,b = 0 be the components of the box velocity when the particle is projected. The acceleration of the particle is given by ax,p = −g sin θ,

The components of particle’s initial velocity (relative to the box) are ux,p/b = u cos α,

uy,p/b = u sin α.

The relative velocity, ux,p/b = ux,p − ux,b , gives ux,p = u cos α − u0,b and similarly uy,p = u sin α. Let the particle lands after time t. The distance travelled by the box and the position coordinates of the particle after time t are given by

yp = 0 = u sin αt − 3.0

ay,p = −g cos θ.

xp = (u cos α − u0,b )t −

P ~ u

O

α

θ

xb = −u0,b t − 12 g sin θ t2 ,

A 1.25

y

in gs

x(m)

Ans. (3.75 ˆı + 6.25 ˆ) m/s, 1 s

(1)

2 1 2 g sin θt , 2

1 2 g cos θ t

.

(2) (3)

The equation (3) gives t=

2u sin α . g cos θ

(4)

18

Part I. Mechanics

Subtract equation (1) from (2) and substitute the value of t to get u2 sin 2α . g cos θ

(t1 − t2 )2 − (t1 − t2 ) = 0.

The particle lands at the starting point if xp = 0. Substitute xp = 0 in equation (2) and replace t by equation (4) to get u0,b = u cos α − 21 g sin θ t =

u cos(α + θ) . cos θ

We encourage you to solve this problem in a frame attached to the box. 2 u cos(α+θ) sin 2α Ans. (a) ug cos θ (b) cos θ Q 27. Two guns situated on the top of a hill of√height 10 m fire one shot each with the same speed 5 3 m/s at some interval of time. One gun fires horizontally and other fires upwards at an angle 60◦ with the horizontal. The shots collide in air at point P. Find, [Take g = 10 m/s2 .] (1996) (a) the time interval between the firings. (b) the coordinates of the point P. Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectory in x-y plane. Sol. Let √ gun 1 fires horizontally at time t1 with speed u = 5 3 m/s and gun 2 fires at time t2 with the same speed at an angle of 60◦ with the horizontal. y

60◦

u

y0 = 10m

Sol. Let the body is released from A, it strikes the inclined plane at B and finally reaches the ground at C. A •

v

◦ B

H h

C ◦

The direction of velocity of the body becomes horizontal after striking at B. Thus, the time taken by the body to travel from B to C is p tBC = 2h/g.

1 x

Q 28. A body falling freely from a given height H hits an inclined plane in its path at a height h. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of (h/H) the body will take maximum time to reach the ground? (1986)

The time taken by the body to travel from A to B is given by p tAB = 2(H − h)/g.

u 2

Q

Solve to get t1 − t2 = 1 s or t1 − t2 = 0. If t1 = t2 then shots collide only at Q, a trivial solution. Thus, t1 − t√ 2 = 1 s which gives t − t1 = 1 s, t − t2 = 2 s, x = 5 3 m and y = 5 m. √ Ans. (a) 1 s (b) (5 3 m, 5 m)

H −h

xp − xb = u cos αt =

Substitute y0 = 10 m, g = 10 m/s2 , t − t1 = t1 − t2 , t − t2 = 2(t1 − t2 ), uy,1 = 0 and uy,2 = 7.5 m/s to get

P (x, y) y

O

x

Let two shots collide at time t on the point P (x, y). The horizontal and vertical components of velocity of √ 3 m/s, the shot from gun 1 and gun 2 are u = 5 x,1 √ uy,1 = 0, ux,2 = u cos 60◦ = 2.5 3 m/s and uy,2 = u sin 60◦ = 7.5 m/s. As horizontal distances covered by the two shots are equal, we get x = ux,1 (t − t1 ) = ux,2 (t − t2 ). Substitute values to get t = 2t1 − t2 . Similarly, equate y coordinates at time t to get y = y0 + uy,1 (t − t1 ) − 12 g(t − t1 )2 = y0 + uy,2 (t − t2 ) − 12 g(t − t2 )2 .

Thus, the total time taken by the body to reach the ground is p p t = tAB + tBC = 2(H − h)/g + 2h/g. The value of h for maximizing t is given by dt/dh = 0 i.e., # r " 2 −1 1 dt p = +√ = 0. dh g h (H − h) Solve to get h = H/2. Ans. 1/2

Chapter 3 Newton’s Laws of Motion

One Option Correct

Q 2. Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance a from the center P (see figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F . As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x, is (2007)

Q 1. A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is (2009) a 2a a a (B) 2gk (C) gk (D) 4gk (A) gk

F

Sol. Let us analyse the problem in the laboratory frame. Let the bead be at position (x0 , y0 ). The forces acting on the bead are its weight mg and the normal reaction N .

m• a

(A)

y N sin θ

•m P

(C)

(B) (D)

F √ x 2m √a2 −x2 a2 −x2 F 2m x

=

N

F √ a 2m a2 −x2 F x 2m a

a

•

Sol. Let T be the tension in the string and A be the acceleration of the masses.

θ N cos θ

mg x

F

O

Resolve N in the horizontal and the vertical directions. Apply Newton’s second law to get N sin θ = mg,

(1)

N cos θ = ma.

(2)

m•

T

P

a x

•m

F = 2T sin θ, (3)

mA = T cos θ.

Now, tan θ is the slope of the normal to the parabola y = kx2 at (x0 , y0 ) i.e.,

Eliminate T to get

1 tan θ = − = −1/(2kx0 ). dy/dx

T

Resolve T in the horizontal and the vertical directions and apply Newton’s second law to get

Divide equation (1) by (2) to get tan θ = g/a.

a θ x

A= (4)

F F x √ cot θ = . 2 2m 2m a − x2 Ans. B

Eliminate tan θ from equations (3) and (4) to get Q 3. A particle moves in the x-y plane under the influence of a force such that its linear momentum is p~(t) = A (ˆı cos kt − ˆsin kt), where A and k are constants. The angle between the force and the momentum is (2007) ◦ ◦ ◦ ◦ (A) 0 (B) 30 (C) 45 (D) 90

x0 = −a/(2gk). We encourage you to solve this problem in a frame attached to the wire by using a pseudo force −ma ˆı. Ans. B 19

20

Part I. Mechanics

Sol. Given, linear momentum of the particle is p~ = A(ˆı cos kt − ˆsin kt). By Newton’s second law, the force on the particle is F~ = d~ p/dt = A(−ˆı k sin kt − ˆk cos kt). The cosine of the angle between p~ and F~ is given by cos θ =

p~ · F~ = − sin kt cos kt + sin kt cos kt = 0, |~ p||F~ |

which gives an upward acceleration of a = mg/(2m) = g/2 just after the string is cut. The downward force on the block of mass m is mg which gives it a downward acceleration g. Ans. A Q 5. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by (2001)

m

and hence θ = 90◦ . Ans. D Q 4. System shown in the figure is in equilibrium and at rest. The string is massless and inextensible and spring is massless. The acceleration of the mass 2m and m just after the string is cut will be (2006)

√ (A) p2M g (C) (M + m)2 + m2 g

√ (B) p2mg (D) (M + m)2 + M 2 g

m

g/2 upwards, g downwards g upwards, g/2 downwards g upwards, 2g downwards 2g upwards, g downwards

R

Sol. In equilibrium, let T be the tension in the string and x be the elongation of the spring having a spring constant k. The forces on the block of mass m are its weight mg downwards and tension T upwards. The forces on the block of mass 2m are its weight 2mg downwards, tension T downwards, and spring force kx upwards.

T m

T

mg

T mg T

Ans. D

2mg

Apply Newton’s second law on two blocks to get kx = 2mg + T = 3mg.

T

Total force on the pulley is zero. Thus, reaction by the clamp is equal and opposite to the resultant of mg and two tension forces i.e., p p R = T 2 + (T + mg)2 = g M 2 + (M + m)2 .

kx

2m

T + mg

Sol. Let T be the tension in the string. Forces on the block of mass M are its weight M g (downwards) and the tension T (upwards). In equilibrium, T = M g. Forces on the pulley are its weight mg (downwards), tension T (downwards), tension T (leftwards) and the reaction from the clamp R.

2m

(A) (B) (C) (D)

M

Q 6. The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be (2001)

T = mg, (1)

When the string is cut (T = 0), the upward force on the block of mass 2m is F = kx − 2mg = 3mg − 2mg = mg,

θ θ √ m

2m

(A) 0◦ (B) 30◦ (C) 45◦ (D) 60◦

m

Chapter 3. Newton’s Laws of Motion

21

Sol. Let T be the tension in the string. The forces on the block of mass m are its weight mg (downwards) and the tension T (upward). In equilibrium, T = mg.

(1) T

T θ

=

θ

T cos θ

T sin θ √

frame in which this particle is at ‘rest’ or in ‘uniform motion’ is called an inertial frame. A frame having uniform motion w.r.t. another inertial frame is also inertial. An accelerating or rotating frame is non-inertial. Thus, the frame attached to the earth is non-inertial because this frame is rotating due to spinning of the earth about its axis as well as due to revolution of the earth around the sun. We encourage you to find the angular velocity of this frame due to spinning and revolution of the earth. Hint: ωspin = 15◦ /h, ωrevolution = 0.04◦ /h. Ans. B, D

2mg

True False Type

√

The forces on the block of mass 2m are its weight 2mg and the tension from two strings (along the strings). Resolve T in the horizontal and the vertical directions. Equilibrium condition gives √ 2mg = 2T cos θ. (2) √

Eliminate T from equations (1) and (2) to get θ = 45◦ . Ans. C Q 7. A ship of mass 3 × 107 kg initially at rest, is pulled by a force of 5 × 104 N through a distance of 3 m. Assuming that the resistance due to water is negligible, the speed of the ship is (1980) (A) 1.5 m/s (B) 60 m/s (C) 0.1 m/s (D) 5 m/s Sol. By Newton’s second law, acceleration of the ship of mass m when pulled by a force F is a = F/m. Initial speed of the ship is u = 0 and the distance travelled is s = 3 m. Apply v 2 − u2 = 2as to get the speed of the ship after travelling a distance s, p √ v = 2as = 2(F/m)s q = 2((5 × 104 )/(3 × 107 ))3 = 0.1 m/s.

Q 9. Two identical trains are moving on rails along the equator on the earth in opposite directions with the same speed. They will exert the same pressure on the rails. (1985) Sol. This problem is analysed in an inertial frame. The earth rotates about its axis in an anticlockwise direction when looking from the north pole. Let ω be the angular velocity of the earth and r be its radius. The rotation of the earth causes each point on the equator to move with a velocity ve = ωr w.r.t. an inertial frame. Let speed of the train w.r.t. the earth be v. Thus, when looking from this inertial frame, the speed of the train, moving in an anticlockwise direction, is va = ve + v = ωr + v and that of the train, moving in a clockwise direction, is vc = ve − v = ωr − v. Na a

va

•

mg r

ω

•

mg vc

Ans. C

•

c

Nc

One or More Option(s) Correct Q 8. A reference frame attached to the earth (1986) (A) is an inertial frame by definition. (B) cannot be inertial frame because the earth is revolving around the sun. (C) is an inertial frame because Newton’s laws are applicable in this frame. (D) cannot be an inertial frame because the earth is rotating about its own axis. Sol. The inertial frame is defined by Newton’s first law of motion. Suppose there is a particle with no physical force (gravitational, electromagnetic, weak, or nuclear) acting on it. By Newton’s first law, this particle is at ‘rest’ or in ‘uniform motion’. But ‘rest’ or ‘uniform motion’ are defined w.r.t. a reference frame. The reference

The forces on the trains are gravitational attraction mg towards the centre of the earth and the normal reaction N . The net force on the train provides the centripetal acceleration for the circular motion in the inertial frame. Apply Newton’s second law to the two trains to get Na = mg − mva2 /r = mg − m(ωr + v)2 /r, Nc = mg −

mvc2 /r

2

= mg − m(ωr − v) /r.

(1) (2)

From equations (1) and (2), Nc > Na i.e., train moving in the clockwise direction exerts more pressure on the rails. We encourage you to analyse this problem in a rotating frame attached to the earth (a noninertial frame). Note that adding a centrifugal force

22

Part I. Mechanics

F~Centrifugal = −m~ ω × (~ ω × ~r) as pseudo force is not sufficient. When the body has a non-zero velocity ~v in a rotating frame then F~Coriolis = −2m~ ω × ~v is another pseodo force that needs to be added along with the centrifugal force. Ans. F Q 10. The pulley arrangements of figure (a) and (b) are identical. The mass of the rope is negligible. In figure (a) the mass m is lifted up by attaching a mass 2m to the other end of the rope. In figure (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2mg. The acceleration of m is the same in both cases. (1984)

Descriptive Q 11. A particle of mass 10−2 kg is moving along the positive x-axis under the influence of force F (x) = − 2xk2 , where k = 10−2 N m2 . At time t = 0 it is at x = 1.0 m and its velocity v = 0. (1998) (a) Find its velocity when it reaches x = 0.5 m. (b) Find the time at which it reaches x = 0.25 m. Sol. The acceleration of given particle is dv dv dx dv F k . = =v = =− dt dx dt dx m 2mx2 Integrate v

Z m

2m

(a)

0

F=2mg

m

k v dv = − 2m

r

(b)

Sol. Consider the case given in the figure (a). The forces on the blocks are their weights and the tension Ta in the rope. The blocks move in opposite directions with equal accelerations Aa (because the rope is inextensible).

mg

2m

2mg (a)

dx , x2

k 1−x , m x

(1)

(minus sign indicates the correct direction of v). F (x) x = 0.25m

x = 0.5m

t =?

v =?

x = 1m v =t= 0

Substitute x = 0.5 m in equation (1) to get v = −1 m/s. Write equation (1) as

Tb

m Ta

1

x

to get v=−

Ta

Z

m

Tb

dx v= =− dt

mg

r

k 1−x , m x

F = 2mg

and integrate

(b)

Z Apply Newton’s second law on the two blocks to

− 1

get Ta − mg = mAa ,

(1)

2mg − Ta = 2mAa .

(2)

Eliminate Ta from equations (1) and (2) to get Aa = g/3. Now consider the case given in the figure (b). The tension in the rope is equal to the applied force i.e., Tb = 2mg (because the rope is of negligible mass). Apply Newton’s second law to the block of mass m to get its acceleration Ab as Ab = (Tb − mg)/m = (2mg − mg)/m = g. Ans. F

0.25

r

m x dx = k 1−x

Z

t

dt, 0

to get t = 1.48 s. Ans. (a) −1.0 m/s (b) 1.48 s Q 12. A smooth semicircular wire track of radius R is fixed in a vertical plane (see figure). One end of massless spring of natural length 3R/4 is attached to the lowest point O of the wire track. A small ring of mass m which can slide on the track is attached to the other end of the spring. The ring is held stationary at point P such that the spring makes an angle 60◦ with the vertical. The spring constant is k = mg/R. Consider the instant when the ring is released (a) draw the free body diagram of the ring, and (b) determine the tangential acceleration of the ring and the normal reaction. (1996)

Chapter 3. Newton’s Laws of Motion

23 (a) Find the tension at the midpoint of the lower wire. (b) Find the tension at the midpoint of the upper wire. S

C 2.9 kg

◦P 60◦ O•

1.9 kg

Sol. The forces on the ring are its weight mg, normal reaction N , and the spring force F .

Sol. Let m1 = 1.9 kg and m2 = 2.9 kg be the masses of the two blocks, l = 1 m be the length of each wire and λ = 0.2 kg/m be the mass per unit length of the lower wire. Consider the mass m1 and the lower half of the lower wire as a system.

C N F

S ◦P 60◦

O•

at

m2

T1

mg

Consider the triangle COP when the ring is just released. The length CO = CP = R which gives ∠COP = ∠CPO = 60◦ . Thus, triangle COP is an equilateral triangle with length OP = R. The extension of the spring is x = OP − 34 R = 14 R and hence the spring force is mg R mg F = kx = = . R 4 4

(1)

Resolve F and mg along the tangential and the normal directions. Since there is no acceleration in the normal direction, forces are balanced along it i.e., N + F cos 60◦ = mg cos 60◦ .

(2)

Eliminate F from equations (1) and (2) to get N = 3mg/8. The net force in the tangential direction is √ 5 3 ◦ ◦ Ft = F sin 60 + mg sin 60 = mg. (3) 8 Newton’s second law gives the tangential acceleration √ Ft 5 3 as at = m = 8 g. √ Ans. (b) 5 8 3 g, 3mg 8 Q 13. Two blocks of mass 2.9 kg and 1.9 kg are suspended from a rigid support S by two inextensible wires each of length 1 m (see figure). The upper wire has negligible mass and the lower wire has a uniform mass of 0.2 kg/m. The whole system of blocks, wire and support have an upward acceleration of 0.2 m/s2 . [Take g = 9.8 m/s2 .] (1989)

l 2

mw m1

a

m1

(m1 + mw )g

The mass of the lower half of the wire is mw = λ(l/2) = 0.1 kg. Let T1 be the tension at the mid point of the lower wire. The forces on the system are tension T1 (upwards) and the system weight (m1 + mw )g (downwards). The system is moving up with an acceleration a = 0.2 m/s2 . Apply Newton’s second law on the system to get T1 − (m1 + mw )g = (m1 + mw )a. Substitute values to get T1 = (m1 + mw )(g + a) = 20 N. S

T2 l 2

m2

m2 m0w

m1

a

m1 (m1 +m2 +m0w )g

24

Part I. Mechanics

Now, consider the mass m1 , the mass m2 , lower wire of mass m0w = λl = 0.2 kg, and the lower half of the upper wire together as a system. Let T2 be the tension at the mid point of the upper wire. The forces on the system are tension T2 (upwards) and the system weight (m1 + m2 + m0w )g (downwards). The system is moving up with an acceleration a = 0.2 m/s2 . Apply Newton’s second law on the system to get T2 − (m1 + m2 + m0w )g = (m1 + m2 + m0w )a. Substitute the values to get T2 = (m1 + m2 + m0w )(g + a) = 50 N. Ans. (a) 20 N (b) 50 N Q 14. A uniform rope of length L and mass M lying on a smooth table is pulled by a constant force F . What is the tension in the rope at a distance l from the end where the force is applied? (1978) Sol. Apply Newton’s second law on the rope to get its acceleration a = F/M . Consider a cross-section of the rope at a distance l from the end where the force is applied. Let T be the tension at this cross-section. Let A and B be the portions of the rope towards the left and the right of this cross-section. Mass of A is mA = M (L−l)/L, only force on it is tension T (towards the right), and acceleration is a = F/M . L F

L−l A

l T T

B

F

Apply Newton’s second law on A to get T = mA a = F (L − l)/L. Ans. F 1 −

l L




Chapter 4 Friction

We encourage you to show that if P < P1 then the block starts sliding down and if P > P2 then block starts moving up. Note that P1 > 0 because tan θ > µ. Ans. A

One Option Correct Q 1. A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is µ and tan θ > µ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg(sin θ − µ cos θ) to P2 = mg(sin θ +µ cos θ), the frictional force f versus P graph will look like (2010)

Q 2. What is the maximum value of the force F such that the block shown in the arrangement does not move? [Take g = 10 m/s2 .] (2003) F 60◦

√

3 kg

µ=

1 √ 2 3

m

(A) 20 N (B) 10 N (C) 12 N (D) 15 N

P

Sol. The forces acting of the block are applied force F , weight mg, normal reaction N , and the frictional force f as shown in the figure.

θ

(A) f

(B) f P2 P1

(C)

P

P1

(D)

f P1

N

P

f P1

P

P2

P2

P2

F cos 60◦

f

P

mg

Sol. The forces acting on the block are its weight mg, normal reaction N , applied force P and frictional force f .

F sin 60◦

Resolve F in the horizontal and the vertical directions and apply Newton’s second law to get

N P

N = F sin 60◦ + mg,

f

mg θ

nθ

si

θ = mg

f = F cos 60 .

The force F becomes maximum when the friction force f attains its maximum value i.e., f = µN.

(3)

Eliminate f and N from equations (1)–(3) to get

0 = P + f − mg sin θ,

Fmax =

which gives f = −P + mg sin θ.

µmg = 20 N. cos 60◦ − µ sin 60◦ Ans. A

(1)

Q 3. An insect crawls up a hemispherical surface very slowly (see figure). The coefficient of friction between the surface and the insect is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle α with the vertical, the maximum possible value of α is given by (2001)

This is a straight line with slope −1. Substitute the values of P1 and P2 in equation (1) to get the frictional force at these points i.e., and

(2)

mg cos θ

Resolve mg along and normal to the inclined plane and apply Newton’s second law to get

f1 = µmg cos θ,

(1)

◦

f2 = −µmg cos θ. 25

26

Part I. Mechanics Sol. Lubrication reduces the non-conservative frictional forces. This increases the efficiency of the machine. Ans. B

α •

(A) cot α = 3 (C) sec α = 3

(B) tan α = 3 (D) cosec α = 3

Sol. The forces on the insect are its weight mg, normal reaction N , and the frictional force f . N α

α =

mg

Sol. The forces on the block of mass m = 2 kg are its weight mg, normal reaction N , and the frictional force f .

•

co

sα

f

mg

sin

Q 6. A block of mass 2 kg rests on a rough inclined plane making an angle of 30◦ with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on√the block is (1980) (A) 9.8 N √ (B) 0.7 × 9.8 × 3 N (C) 9.8 × 3 N (D) 0.7 × 9.8 N

α N

mg f

Resolve mg along and perpendicular to the normal. Apply equilibrium condition to get

Q 4. A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is (1994) (A) 2.5 N (B) 0.98 N (C) 4.9 N (D) 0.49 N Sol. The forces on the block are applied force F = 5 N, normal reaction N , weight mg and the frictional force f . f

F N

mg

◦

The angle α attains its maximum value when the frictional force reaches its maximum limit of fmax = µN . Substitute in equation (1) and then divide by equation (2) to get tan α = µ = 1/3. Thus, cot α = 3. Ans. A

30◦

30

(2)

30◦ =

os

N = mg cos α.

m

gc

(1)

30

m

f = mg sin α,

◦

in gs

The net force on the block is zero because it is at rest. Resolve mg in the directions parallel and perpendicular to the inclined plane. Apply Newton’s second law in these directions to get N = mg cos 30◦ = (2)(9.8)(0.866) = 16.97 N, f = mg sin 30◦ = (2)(9.8)(0.5) = 9.8 N. Note that f is less than fmax = µN = (0.7)(16.97) = 11.88 N. Ans. A One or More Option(s) Correct Q 7. A small block of mass 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure. The block remains stationary if [Take g = 10 m/s2 .] (2012) Q

mg

In equilibrium, N = F = 5 N and f = mg = 0.1×9.8 = 0.98 N. Note that f is less than its maximum possible value of fmax = µN = 0.5 × 5 = 2.5 N. Ans. B Q 5. If a machine is lubricated with oil, (1980) (A) the mechanical advantage of the machine increases. (B) the mechanical efficiency of the machine increases. (C) both its mechanical advantage and mechanical efficiency increases. (D) its efficiency increases, but its mechanical advantage decreases.

1N θ O

P

(A) θ = 45◦ (B) θ > 45◦ and frictional force acts on the block towards P (C) θ > 45◦ and frictional force acts on the block towards Q (D) θ < 45◦ and frictional force acts on the block towards Q

Chapter 4. Friction

27

Sol. The forces acting on the block are F = 1 N towards the left, weight mg = 0.1 × 10 = 1 N downwards, normal force N , and the frictional force f . Resolve F and mg along and perpendicular to the inclined plane. f

In this case, the body will just start moving when the horizontal component of Fpull is equal to or greater than the maximum value of friction force i.e., Fpull cos θ = fpull = µNpull .

(1)

Since there is no acceleration in the vertical direction Fc θ os

Npull = mg − Fpull sin θ.

N

(2)

=

Eliminate Npull from equations (1) and (2) to get F

Fpull =

µmg . cos θ + µ sin θ

mg

θ θ in sθ F s g co m

Npush

sin

=

θ

mg

Fpush θ

When θ = 45◦ , the net force that brings the block down is fpush

1 1 Fd = mg sin θ − F cos θ = √ − √ = 0. 2 2

mg

Thus, the block is stationary if θ = 45◦ . When θ > 45◦ , the force Fd > 0, and hence the block has a tendency to move down. Thus, the frictional force f acts on the block upwards i.e., towards Q. Ans. A, C Assertion Reasoning Type Q 8. Statement 1: It is easier to pull a heavy object than to push it on a level ground. Statement 2: The magnitude of frictional force depends on the nature of the two surfaces in contact. (2008)

(A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. (B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true. Sol. Both, statement 1 and statement 2, are true but statement 2 is not a correct explanation of statement 1. The forces acting on the body in pull case are shown in the figure. Npull Fpull

The forces acting on the body in push case are shown in the figure. In this case, Fpush cos θ = fpush = µNpush , Npush = mg + Fpush sin θ, µmg Fpush = . cos θ − µ sin θ Note that Fpull = Fpush = µmg if θ = 0. Ans. B Q 9. Statement 1: A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table. Statement 2: For every action there is an equal and opposite reaction. (2007) (A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. (B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true. Sol. Both statements are true but statement 2 is not a correct explanation for statement 1. Generally, statement 1 is attributed to Newton’s first law but this is not entirely correct. We encourage you to repeat this experiment by pulling the cloth slowly. The outcome of experiment depends on acceleration acloth of the cloth. N

θ f fpull

acloth mg

mg

28

Part I. Mechanics

The forces acting on the dish are its weight mg, normal reaction N , and the frictional force f . Maximum value of the frictional force is fmax = µN = µmg.

are stationary if θ ≤ θr otherwise they are moving. Consider the limiting case, θ = θr , when the blocks are at rest and f = µN2 .

By Newton’s second law, acceleration of the dish is adish = f /m. If acloth is small then the cloth and the dish move together i.e., adish = acloth . Maximum value of acloth for the cloth and the dish to move together is given by

(1)

Apply Newton’s second law to m1 , R = m1 g sin θ,

(2)

N1 = m1 g cos θ,

(3)

and to m2 ,

acloth = adish = fmax /m = µg. Beyond this limit, acloth > adish = µg and hence the cloth comes out leaving the dish on table. Ans. B Matrix or Matching Type

f = R + m2 g sin θ,

(4)

N2 = m2 g cos θ.

(5)

Eliminate R, N2 , and f from equations (1)–(5) to get −1

Q 10. A block of mass m1 = 1 kg and another block of mass m2 = 2 kg are placed together on an inclined plane with angle of inclination θ (see figure). Various values of θ are given in Column I. The coefficient of friction between the block m1 and the plane is always zero. The coefficient of static and dynamic friction between the block m2 and the plane are equal to µ = 0.3. In Column II expressions for the friction on block m2 are given. Match the correct expression of the friction in Column II with the angles given in Column I. The acceleration due to gravity is denoted by g. [Given, tan(5.5◦ ) ≈ 0.1, tan(11.5◦ ) ≈ 0.2, tan(16.5◦ ) ≈ 0.3].

θr = tan



µm2 m1 + m2



−1

= tan



0.3 × 2 1+2



= tan−1 (0.2) = 11.5◦ . Thus, for θ = 5◦ and θ = 10◦ , blocks are at rest with frictional force f = R + m2 g sin θ = (m1 + m2 )g sin θ. For θ = 15◦ and θ = 20◦ , blocks are moving with frictional force f = µN2 = µm2 g cos θ,

(limiting value).

(2014)

Ans. P7→2, Q7→2, R7→3, S7→3 Column I (P) (Q) (R) (S)

θ θ θ θ

Column II

◦

=5 = 10◦ = 15◦ = 20◦

(1) (2) (3) (4)

m2 g sin θ (m1 + m2 )g sin θ µm2 g cos θ µ(m1 + m2 )g cos θ

Sol. The forces on the block of mass m1 are its weight m1 g, normal reaction from the inclined plane N1 , and the reaction from the second block R. N1

R

N2

f m1 g

True False Type Q 11. When a person walks on a rough surface, the frictional force exerted by the surface on the person is opposite to the direction of his motion. (1981) Sol. To walk in the forward direction, the person pushes his foot backward. Thus, the foot has a tendency to move backward against the rough surface. To oppose this movement, the friction force on the foot acts in the forward direction. Note that the frictional force is the only horizontal force on the person. Thus, a person can accelerate in forward direction only if the frictional force is forward. Ans. F

R m2 g θ

Similarly, forces on the block of mass m2 are m2 g, N2 , R, and the frictional force f . If θ is slowly increased, f starts increasing and attains its maximum value f = µN2 at θ = θr (angle of repose). The blocks

Fill in the Blank Type Q 12. A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5 m/s2 , the frictional force acting on the block is . . . . . . N. (1984)

29

Sol. The acceleration of the block is equal to the acceleration of the truck i.e., a = 5 m/s2 . Only horizontal force acting on the block is frictional force f . N

a = 25 m/s2

Chapter 4. Friction

θ

f a

mg

cos θ = 4/5, sin θ = 3/5

f1

N1

Sol. Suppose block just slides up in case 1 and just slides down in case 2. The forces acting on the block are its weight (mg), normal reaction (R), applied force (F1 in case 1 and F2 in case 2 ) and frictional force (f1 in case 1 and f2 in case 2 ), as shown in the figure. R1

f1

F1

R2

mg

f2

F2

mg

45◦

45◦

Case (1)

Case (2)

The equilibrium conditions in case 1 and case 2

in θ

f2

ma θ mg ma co sθ =

Q 13. A block is moving on an inclined plane making an angle 45◦ with the horizontal and the coefficient of friction is µ. The force required to just push it up the inclined plane is three times the force required to just prevent it from sliding down. If we define N = 10µ then N is . . . . . . . (2011)

as

Integer Type

Sol. Let us solve the problem in a frame attached to the disc. This frame is accelerating towards the left with an acceleration a = 25 m/s2 . We can apply Newton’s second law in this non-inertial frame if we apply a psedoforce ma acting towards the right.

m

Apply Newton’s second law to get f = ma = (1)(5) = 5 N. Note that the limiting value of frictional force is fmax = µN = µmg = 0.6(1)(9.8) = 5.88 N. Ans. 5

N2

The forces acting on the block are pseudo force ma towards the right, its weight mg into the paper, normal reaction on the bottom surface of the block N1 coming out of the paper, frictional force f1 = µN1 corresponding to N1 , normal reaction on the side surfaces of the block N2 and frictional force f2 = µN2 corresponding to N2 (see figure). The constraint that the block moves in a horizontal plane along the groove gives N1 = mg, N2 = ma sin θ.

give R1 = mg cos 45◦ , ◦

R2 = mg cos 45 ,

F1 = mg sin 45◦ + f1 ,

Apply Newton’s second law along the groove to get

◦

F2 = mg sin 45 − f2 .

Using F1 = 3F2 , f1 = µR1 , and f2 = µR2 , we get µ = 0.5. Thus, N = 5. Ans. 5 Descriptive Q 14. A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 kg is placed as shown in the figure. The coefficient of friction between the block and all surfaces of groove in contact is µ = 2/5. The disc has an acceleration of 25 m/s2 . Find the acceleration of the block with respect to disc. (2006)

ma cos θ − f1 − f2 = mar . Substitute the values of the parameters to get ar = 10 m/s2 . Ans. 10 m/s2 Q 15. Two blocks A and B of equal masses are released from an inclined plane of inclination 45◦ at t = 0. Both the blocks are initially at rest. The coefficient of kinetic friction between the block A and the inclined plane is 0.2 while it is 0.3 for the block B. Initially the block A is √ 2 m behind the block B. When and where their front faces will come in a line. [Take g = 10 m/s2 .] (2004)

30

Part I. Mechanics √ 2m

(a) Draw a free body diagram of mass M , clearly showing all the forces. (b) Let the magnitude of the force of friction between m1 and M be f1 and that between m2 and ground be f2 . For a particular force F it is found that f1 = 2f2 . Find f1 and f2 . Write equations of motion of all the masses. Find F , tension in the string and acceleration of the masses.

B

A

B A 45◦

Sol. The forces acting on the blocks are weight mg, normal reaction N and the frictional force f = µN . N

co g

f1

θ

mg

N1

T

n

=

si

θ

g

m

m

sθ

f

Sol. Let T be the tension in the string, N be the reaction force between M and ground, N1 be the reaction force between m1 and M , f1 be the friction force on m1 , and f2 be the friction force on m2 . The free body diagrams of m1 , m2 , and M are shown in the figure.

T

θ

T

Newton’s second law gives

F Mg

N = mg cos θ,

T

ma = mg sin θ − f = mg sin θ − µN

N

= mg sin θ − µmg cos θ.

N2

Thus, acceleration of the two blocks are √ aA = g sin θ − µA g cos θ = 8/ 2, √ aB = g sin θ − µB g cos θ = 7/ 2,

m2

Q 16. In the figure masses m1 , m2 and M are 20 kg, 5 kg and 50 kg respectively. The coefficient of friction between M and ground is zero. The coefficient of friction between m1 and M and between m2 and ground is 0.3. The pulleys and the strings are massless. The string is perfectly horizontal between P1 and m1 and also between P2 and m2 . The string is perfectly vertical between P1 and P2 . An external horizontal force F is applied to the mass M . [Take g = 10 m/s2 .] (2000)

P2 m2

m1

M

T

T

m1

f2

where µA = 0.2, µB = 0.3 and θ = 45◦ . Let front faces comes in a line at time t. The distances travelled by A and B in time t are √ SA = 12 aA t2 = (4/ 2)t2 , √ SB = 12 aB t2 = (3.5/ 2)t2 . √ √ Using SA − SB = 2, we get t = 2 s and S√ A = 8 2 m. Ans. SA = 8 2 m, 2 s

P1

N1

F

f1 m2 g

m1 g

Apply Newton’s second law on m1 , m2 , and M in the vertical direction to get N1 = m1 g,

N2 = m2 g,

N = N1 + M g.

Limiting (maximum) values of the frictional forces are f1,max = µN1 = µm1 g = 0.3 × 20 × 10 = 60 N, f2,max = µN2 = µm2 g = 0.3 × 5 × 10 = 15 N. Let a1 , a2 , and A be the rightward accelerations of m1 , m2 , and M w.r.t. the ground. As string is inextensible, the accelerations of m1 and m2 are equal i.e., a1 = a2 = a. Apply Newton’s second law on m1 , m2 , and M in the horizontal direction to get T − f2 = m2 a2 = m2 a,

(1)

f1 − T = m1 a1 = m1 a,

(2)

F − f1 = M A.

(3)

The magnitude of F decides the values of the frictional forces f1 and f2 and the motion of the blocks. There are two possible cases (i) m1 does not slide over M (i.e., A = a), and (ii) m1 slides over M (i.e., a 6= A) and hence f1 = f1,max .

Chapter 4. Friction

31

Consider the horizontal motion of the blocks in case (i). In this case, the external forces on the system of m1 , m2 , and M are F and f2 . The system starts moving only when F > f2,max .

Beyond F = 150 N, the assumption that m1 does not slide over M is invalid and case (ii) comes into play. In this case, the acceleration a remains at its maximum value of 1.8 m/s2 . Using equation (3), the acceleration A varies with F as

f2 (N)

A = (F − f1,max )/M = (F − 60)/50.

15

O

F(N)

15

The force f2 adjusts itself such that it is equal to F till it reaches the maximum value of f2,max i.e., ( F, if F ≤ 15 N; f2 = (4) 15 N, otherwise. Eliminate T and A from equations (1)–(3) to get the frictional force f1 =

f2 M + (m1 + m2 )F . m1 + m2 + M

(5)

f1 (N)

From equations (5) and (6), the condition f1 = 2f2 occurs when f2 = f2,max = 15 N and f1 = 2f2,max = 30 N. Substitute the values of f1 and f2 to get a = 0.6 m/s2 , F = 60 N, and T = 18 N. Ans. (b) f1 = 30 N, f2 = 15 N, F = 60 N, T = 18 N, a = 35 m/s2 Q 17. Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in the figure. The wedge is inclined at 45◦ to the horizontal on both sides. The coefficient of friction between block A and the wedge is 2/3 and that between block B and the wedge is 1/3. If the blocks A and B are released from rest, find, (1997)

60

150

F(N)

From equations (4) and (5), the f1 is equal to F till F ≤ 15 N, then it increases linearly with F with a slope m1 +m2 m1 +m2 +M till it reaches a value of f1,max = 60 N, and then remains constant at its limiting value (see figure). Mathematically,   if F ≤ 15 N; F, 15M +(m1 +m2 )F f1 = , if 15 N < F ≤ 150 N; m1 +m2 +M   60 N, otherwise. (6) Note that f1 = 60 N when F = 150 N. a(m/s2 ) A a

1.8 0.6 O 15 60

150

45◦

(a) the acceleration of A. (b) tension in the string. (c) the magnitude and direction of friction force acting on A. Sol. Let block B is accelerating down the plane with an acceleration a. The string is inextensible which makes block A to move up the plane with the same acceleration a. The tension T in the string remains same throughout the string as it is massless and the pulley is frictionless. The forces on the two blocks are shown in the figure. NA

F(N)

B

A

15 O 15

T

NB

T

Eliminate T from equations (1) and (2) to get the acceleration a = (f1 − f2 )/(m1 + m2 ).

(7)

Substitute f2 and f1 from equations (5) and (6) in equation (7) to get   if F < 15 N; 0, F −15 F −15 a = m1 +m2 +M = 75 , if 15 N < F ≤ 150 N;   1.8 otherwise.

B

A

fB fA 45◦

a mg

2mg

45◦

Resolve forces along and normal to the plane. Apply Newton’s second law in normal direction to get √ NA = mg cos 45◦ = mg/ 2, (1) √ NB = 2mg cos 45◦ = 2 mg. (2)

32

Part I. Mechanics

Apply Newton’s second law along the plane to get √ (3) T − mg/ 2 − fA = ma, √ 2 mg − T − fB = 2ma. (4)

Sol. Let the force F be applied at an angle θ to the horizontal plane. The forces acting on the block of mass m are its weight mg, normal reaction N , frictional force f , and the applied force F . N

Eliminate T from equations (3) and (4) to get

F

F sin θ

 √ 1  mg/ 2 − fA − fB . 3m

(5)

When blocks move relative to the plane then frictional forces attain their maximum values which are given by √ √ fA = fA,max = µA NA = (2/3)(mg/ 2) = 2mg/3, √ √ fB = fB,max = µB NB = (1/3)( 2 mg) = 2mg/3. Substitute these values in equation (5) to get a = √ −g/9 2. The negative sign shows that our assumption of block B moving down the plane (and hence assumed directions of fA and fB ) was wrong. If we assume block B to move up and rewrite equations (3) and (4), we get  √ 1  a= −mg/ 2 − fA − fB , 3m again a negative quantity. Thus, blocks are neither moving up or down i.e., a = 0. Substitute a = 0 in equations (3) and (4) to get √ fA = T − mg/ 2, √ fB = 2mg/ 2 − T.

=

a=

θ

f

F cos θ

mg

Resolve F~ in the horizontal and the vertical directions. The net force in the vertical direction is zero (because there is no acceleration in the vertical direction) i.e., N + F sin θ = mg.

(1)

The block will just start moving when the horizontal component of the applied force is equal to the limiting value of the frictional force i.e., F cos θ = fmax = µN.

(2)

Eliminate N from equations (1) and (2) and simplify to get F =

µmg . cos θ + µ sin θ

(3)

The applied force is minimum when dF/dθ = 0 i.e., The minimum √ and the maximum values of fA and fB are zero and 2 mg/3. The variation of fA and fB with tension T is shown in the figure.

which gives θ = tan−1 µ. Substitute θ in equation (3) to get

f 2mg √ 2

−T √ T − mg 2

fB O

µmg dF =− (− sin θ + µ cos θ) = 0, dθ (cos θ + µ sin θ)2

mg √ 2

fA 2mg √ 2

T

It can be seen that fB attains its maximum value earlier than fA attains√its maximum. Substitute fB = √ √ 2 mg/3, to get T = 2 2√mg/3 and fA = 2 mg/3. √ , downwards Ans. (a) 0 (b) 2 3 2 mg (c) 3mg 2 Q 18. A block of mass m rests on a horizontal floor with which it has a coefficient of static friction µ. It is desired to make the body move by applying the minimum possible force F . Find the magnitude of F and the direction in which it has to be applied. (1987)

µmg Fmin = p . 1 + µ2 We encourage you to solve this problem if the block rests on an inclined plane. Ans. √µmg 2 , tan−1 µ 1+µ

Q 19. Masses M1 , M2 and M3 are connected by strings of negligible mass which passes over massless and frictionless pulleys P1 and P2 as shown in figure. The masses move such that the portion of the string between P1 and P2 is parallel to the inclined plane and the portion of the string between P2 and M3 is horizontal. The masses M2 and M3 are 4.0 kg each and the coefficient of kinetic friction between the masses and the surface is 0.25. The inclined plane makes an angle of 37◦ with the horizontal. If the mass M1 moves downwards with a uniform velocity, find, [Take g = 9.8 m/s2 , sin 37◦ ≈ 3/5.] (1981)

Chapter 4. Friction

33 of m1 and m2 with the inclined plane are 0.75 and 0.25 respectively. Assuming the string to be taut, find (a) the common acceleration of two masses, and (b) the tension in the string. [Take sin 37◦ = 0.6, cos 37◦ = 0.8, g = 9.8 m/s2 .] (1979)

P1

M 2

M1

P2

37◦

M3

m

1

(a) the mass of M1 . (b) the tension in the horizontal portion of the string.

m

2

Sol. The masses M1 , M2 , and M3 are connected by the inextensible strings. Since M1 moves downwards with a uniform velocity, all the three masses will moves with the same uniform velocity because they are connected by the inextensible strings. Thus, accelerations of M1 , M2 , and M3 are zero. By Newton’s second law, the net force on each mass is zero. Let T1 be the tension in the string that connects M1 and M2 and T2 be the tension in the string that connects M2 and M3 . The normal reactions on M2 and M3 are N2 and N3 , and the frictional forces on these blocks are f2 and f3 . The free body diagrams of the three masses are shown in the figure. T1

N2

T1

37◦

Sol. The forces acting on the block of mass m1 = 4 kg are its weight m1 g, normal reaction N1 , frictional force f1 , and the string tension T . Similarly, forces on the block of mass m2 = 2 kg are its weight m2 g, normal reaction N2 , frictional force f2 , and the string tension T. N1

f1

M M1 37

M1 g

T2 T2

◦

1

37◦

N3

2

m

T

N2

T f2

m

2

m1 g m2 g 37◦

M3

f2

f3

M2 g

M3 g

Resolve M2 g along and normal to the inclined plane. The frictional force on the mass M2 is f2 = µN2 and that on the mass M3 is f3 = µN3 because these blocks are sliding on the rough surfaces with the coefficient of kinetic friction µ = 0.25. Apply Newton’s second law on the three blocks to get T1 = M1 g,

(1)

N2 = M2 g cos 37◦ ,

(2)

T1 = T2 + µN2 + M2 g sin 37◦ , (∵ f2 = µN2 ) (3) N3 = M3 g,

(4)

T2 = µN3 .

(∵ f3 = µN3 ). (5)

Solve equations (1)–(5) to get T2 = µM3 g = (0.25)(4)(9.8) = 9.8 N, ◦

◦

M1 = µM3 + µM2 cos 37 + M2 sin 37

= (0.25)(4) + (0.25)(4)(4/5) + (4)(3/5) = 21/5 kg. Ans. (a) 4.2 kg (b) 9.8 N Q 20. Two blocks connected by a massless string slides down an inclined plane having an angle of inclination of 37◦ . The masses of the two blocks are m1 = 4 kg and m2 = 2 kg respectively and the coefficients of friction

Resolve m1 g and m2 g in directions parallel and perpendicular to the incline. Apply Newton’s second law on m1 and m2 in a direction perpendicular to the incline to get N1 = m1 g cos 37◦ ,

(1)

◦

N2 = m2 g cos 37 .

(2)

The frictional forces on the sliding blocks m1 and m2 are equal to their limiting values f1 = µ1 N1 = µ1 m1 g cos 37◦ , ◦

f2 = µ2 N2 = µ2 m2 g cos 37 ,

(3) (4)

where µ1 = 0.75 and µ2 = 0.25 are the coefficients of friction for m1 and m2 . Let blocks are accelerating down with an equal acceleration a (note that the string remains taut). Apply Newton’s second law on m1 and m2 in a direction parallel to the inclined plane to get m1 g sin 37◦ + T − f1 = m1 a, ◦

m2 g sin 37 − T − f2 = m2 a.

(5) (6)

Add equations (5) and (6) and substitute f1 and f2 from equations (3) and (4) to get µ1 m1 + µ2 m2 g cos 37◦ , m1 + m2 (0.75)(4) + (0.25)(2) = (9.8)(0.6) − (9.8)(0.8) 4+2 = 1.3 m/s2 .

a = g sin 37◦ −

34

Part I. Mechanics

Now, multiply equation (5) by m2 and equation (6) by m1 and then subtract to get, (µ1 − µ2 )m1 m2 g cos 37◦ , m1 + m2 (0.75 − 0.25)(4)(2) = (9.8)(0.8) = 5.2 N. 4+2

T =

Ans. (a) 1.3 m/s2 (b) 5.2 N Q 21. In the figure, the blocks A, B and C have masses 3 kg, 4 kg and 8 kg respectively. The coefficient of sliding friction between any two surfaces is 0.25. A is held at rest by a massless rigid rod fixed to the wall, while B and C are connected by a light flexible cord passing around a fixed frictionless pulley. Find the force F necessary to drag C along the horizontal surface to the left at a constant speed. Assume that the arrangement shown in the figure i.e., B on C and A on B, is maintained throughout. [Take g = 10 m/s2 .] (1978)

NA = mA g,

(1)

NB = NA + m B g = (mA + mB )g,

B

(using (1))

(2)

(using (2))

(3)

NC = NB + mC g = (mA + mB + mC )g.

A

F

The free body diagrams of the blocks A, B, and C are shown in the figure. The forces on the block A are its weight mA g, normal reaction NA , frictional force f1 , and reaction from the rod RA . The forces on the block B are mB g, normal reaction NB on the lower surface, normal reaction NA on the upper surface, tension T from the string, frictional force f1 on the upper surface and frictional force f2 on the lower surface. The forces on the block C are mC g, normal reaction NC on the lower surface, normal reaction NB on the upper surface, tension T from the string, frictional force f2 on the upper surface and frictional force f3 on the lower surface. Newton’s second law on the block A, B, and C gives

The frictional forces in the sliding motion are equal to their limiting values i.e.,

C

f1 = µNA Sol. The masses of the blocks are mA = 3 kg, mB = 4 kg, and mC = 8 kg. The coefficient of sliding friction between any two surfaces is µ = 0.25. Let the frictional force between the block A and the block B be f1 , between the block B and the block C be f2 , and between the block C and the ground be f3 . The applied force F slides the block C along the horizontal surface to the left at a constant speed. The frictional force f3 on the block C resists this movement by acting in the rightward direction. Since the string is inextensible and the pulley is fixed, the block B starts moving towards the right with a speed equal to that of C. The frictional force f2 on B resists this motion by acting in the leftward direction. The frictional force f2 on the block C acts in the rightward direction (Newton’s third law). The block A is fixed and the block B moves towards the right relative to the block A. The frictional force f1 on the block B acts in the leftward direction to oppose this motion. The force f1 on the block A acts in the rightward direction. NA RA

A

f1

mA g NB

= µmA g, = µ(mA + mB )g,

B

= µ(mA + mB + mC )g.

Q 22. A block of mass 2 kg slides on an inclined plane which makes an angle of 30◦ with the horizontal. The coefficient √ √ of friction between the block and the surface is 3/ 2. What force along the plane should be applied to the block so that it moves (a) down, and (b) up, without any acceleration? [Take g = 10 m/s2 .] (1978) Sol. Let F be the force required to move the block downwards with a constant velocity. The forces on the block are its weight mg, normal reaction N , and frictional force f . The frictional force in the sliding motion is equal to its limiting value f = fmax = µN . N f ◦

in

30 =

mg

◦

30◦

30

30

m

gs

os gc

NB

(6)

Ans. 80 N

m

mC g

(using (3))

F = T + f2 + f3 = µ(4mA + 3mB + mC )g = 80 N.

F f2 T f3

(5)

T = f1 + f2 = µmA g + µ(mA + mB )g = 25 N.

T

C

(using (2))

Apply Netwton’s second law on the block B and the block C to get the force F required to drag the block C at constant speed,

mB g NA NC F

(4)

f3 = µNC

f1 f2

(using (1))

f2 = µNB

Chapter 4. Friction

35

Resolve mg in the directions parallel and perpendicular to the incline. Apply Newton’s second law to get N = mg cos 30◦ ,

(1)

F = f − mg sin 30◦ = µN − mg sin 30◦ (∵ f = µN ) = mg(µ cos 30◦ − sin 30◦ ) (using (1)) ! √ √ 3 3 1 − = 11.21 N. = (2)(10) √ 2 2 2 Let F 0 be the force required to move the block upwards with a constant velocity. The direction of frictional force f is downwards and its magnitude is f = fmax = µN . The normal reaction N is given by equation (1). N F0 ◦

30

30 =

◦

30

mg

os

f

gc

30◦

in

m

m

gs

Apply Newton’s second law in the direction parallel to the incline to get F 0 = f + mg sin 30◦ = µN + mg sin 30◦ = mg(µ cos 30◦ + sin 30◦ ) ! √ √ 3 3 1 + = 31.21 N. = (2)(10) √ 2 2 2 Ans. (a) 11.21 N (b) 31.21 N

Chapter 5 Circular Motion

From equation (3), normal reaction N is positive (radially outward) if θ < cos−1 (2/3) and negative (radially inward) if θ > cos−1 (2/3). By Newton’s third law, force applied by the bead on the wire is N 0 = −N , which is radially inwards initially and radially outwards later. Ans. D

One Option Correct Q 1. A wire, which passes through the hole in a small bead, is bent in the form of a quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is

Q 2. Consider a disc rotating in the horizontal plane with constant angular speed ω about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed 1/8 rotation, (ii) their range is less than half the disc radius, and (iii) ω is constant throughout. Then, (2012)

(2014) A

90◦

B

(A) always radially outwards. (B) always radially inwards. (C) radially outwards initially and radially inwards later. (D) radially inwards initially and radially outwards later.

•

R

ω

y •

Sol. Let v be speed of the bead when it makes an angle θ with the vertical direction.

x

O

Q

•

A N •

P

r θ

mg

(A) P lands in the shaded region and Q in the unshaded region. (B) P lands in the unshaded region and Q in the shaded region. (C) Both P and Q land in the unshaded region. (D) Both P and Q land in the shaded region.

B

The forces acting on the bead are its weight mg and normal reaction N . Since all the forces acting on the bead are conservative, the mechanical energy of the bead is conserved i.e., mgr(1 − cos θ) =

2 1 2 mv .

Sol. Let radius of the disc be r0 and time period of rotation be T = 2π/ω. Let the pebble Q be at the point (0, y0 ) at the time of projection. For projectile motion, let r be the range and t be the time of flight. It is given that t < T /8 and r < r0 /2. We analyse the projectile motion in a frame attached to the ground. If the disc is not rotating then pebbles projected from P and Q land at P1 and Q1 travelling a distance r in the direction of R. The rotation of the disc provides additional initial velocity to the pebbles given by ∆~uQ = −y0 ω ˆı and

(1)

The radially inward components of force provide centripetal acceleration to the bead i.e., mg cos θ − N = mv 2 /r.

(2)

Eliminate v 2 from equations (1) and (2) to get N = mg(3 cos θ − 2).

(3) 36

Chapter 5. Circular Motion

37

∆~uP = r0 ω ˆı. These velocities are in the horizontal plane and are perpendicular to the plane of projection. The time of flight t is not affected by these velocities as it depends only on vertical component of initial velocity. Also, distance travelled in the direction of projection is not affected as it depends on t and velocity component in the direction of projection. However, these additional velocities affect distance travelled in the direction perpendicular to the plane of projection. These introduce additional displacements given by ∆~rQ = −y0 ωt ˆı and ∆~rP = r0 ωt ˆı and pebbles land at Q01 and P10 . Note that range QQ01 and PP01 are increased due to additional velocities (QQ01 > QQ1 , PP01 > PP1 ).

l •

(A) 9 (B) 18 (C) 27 (D) 36 Sol. This is an example of conical pendulum. Let string makes an angle θ with the vertical and angular velocity of the particle be ω. The forces acting on the particle are its weight mg and tension T in the string.

R

T cos θ

θ ω

Q01 y0 ωt Q 1

=

l T

θ

•

T sin θ

r θ Q

y

y0

mg x

O θ0 ωt r0

P01

P1 r0 ωt r P

r0 ω t

T sin θ = mω 2 r = mω 2 (l sin θ).

From triangle OQ1 Q01 , −1

θ = tan



y0 ωt y0 + r



≤ tan−1 (ωt) (∵ tan−1 x is an increasing function) ≤ ωt.

Resolve T along the horizontal and the vertical directions. As the particle moves in a horizontal plane, net vertical force on it is zero. Net horizontal force on it provides the necessary centripetal acceleration for circular motion. Apply Newton’s second law in the horizontal direction to get

p This equation gives ω = T /(ml). The angular velocity ω attains maximum value when tension T in the string is maximum. Substitute the values to get r r Tmax 324 ωmax = = = 36 rad/s. ml 0.5 × 0.5 Ans. D

(∵ tan−1 x ≤ x; for all x).

The angle θ is less than the angular displacement of disc in time t. Hence, landing point Q01 lies in unshaded region. The distance travelled by P in time interval t is r0 ωt along the arc of the disc. The angle subtended by this arc at O is ωt. From the figure, θ0 > ωt. Hence, landing point P10 lies in unshaded region. Ans. C Q 3. A ball of mass (m) 0.5 kg is attached to the end of a string having length (l) 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of the ball (in rad/s) is (2011)

Q 4. A bob of mass m is suspended by a massless string of length l. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies, (2008) B •

θ

l

•

• A

(A) θ = π4 (C) π2 < θ
R where ρ0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed V as a function of distance r (0 < r < ∞) from the centre of the system is represented by (2008) (A)

V

(B)

V

r

r

R

(C)

R

V

(D)

V

P •

r

3R

(A) (C)

2GM 7R GM 4R

√
4 2−5

R

Sol. The gravitational force provides the centripetal acceleration to the particle of mass m i.e.,

4R

√
(B) − 2GM 5 7R √4 2 −
(D) 2GM 2−1 5R

GM m/r2 = mV 2 /r.

1/2

V = (4πGρ0 /3)

r.

For r > R, M = ρ0 (4πR3 /3). Substitute it in equation (1) to get
1/2 −1/2 V = 4πGρ0 R3 /3 r .

P •

2

R √ 16

+ 2

r

r

3R •

4R

(1)

For r ≤ R, M = ρ0 (4πr3 /3) is the mass inside a sphere of radius r. Substitute it in equation (1) to get

Sol. We need to find gravitational potential energy of a unit mass placed at the point P.

4R

r

R

4R

O

Ans. C Q 6. A double star system consists of two stars A and B which have time periods TA and TB , radius RA and RB and mass MA and MB . Choose the correct option. (2006)

The surface mass density of the annular disc is σ = M/(16πR2 − 9πR2 ) = M/(7πR2 ).

(A) (B) (C) (D)

If TA > TB then RA > RB If TA > TB then MA > MB 3/2 TA /TB = (RA /RB ) TA = TB

Chapter 9. Gravitation

121

Sol. Let the distance between A and B be r and C be the centre of mass of the double star system. By definition of the centre of mass, the distances of A and B from C are given by rA =

MB r , MA + MB

rB =

MA r . MA + MB

Q 8. A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth surface (Re = 6400 km) will approximately be (2002) (A) 12 h (B) 1 h (C) 2 h (D) 4 h Sol. Time period of a geostationary satellite is T36000 = 24 h. Keplar’s third law, T 2 ∝ a3 , gives the time period of a satellite in an orbit close to the earth surface as T6400 = 24 × (6400/36000)

MA • A

MB • B

C •

r

rA

rB

2 GMA MB /r2 = MA vA /rA .

The time period of A is given by

Q 9. A simple pendulum has a time period T1 when on the earth’s surface and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The √ value of T2 /T1 is (2001) (A) 1 (B) 2 (C) 4 (D) 2 Sol. The time period of a simple pendulum of length l is given by p T = 2π l/g. (1) The acceleration due to gravity on the earth’s surface is

2πrA 2π r3/2 . =p vA G(MA + MB )

g1 = GM /R2 ,

Similarly, TB =

≈ 1.8 h.

Since time period increases with a, time period of a satellite moving few hundred kilometers above the earth surface is ≈ 2 h. Ans. C

The stars A and B move in a plane containing C with orbits defined by the initial conditions. For simplicity, let orbits of A and B be circular. The gravitational force provides the centripetal acceleration i.e.,

TA =

3/2

(2)

and that at a height R is 2πrB 2π r3/2 . =p vB G(MA + MB )

g2 = GM /(R + R)2 = GM /(4R2 ).

Ans. D Q 7. If W1 , W2 and W3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively (see figure) in the gravitational field of a point mass m. Find the correct relation between W1 , W2 and W3 . (2003) •

B

2

1

3 •

A

(A) W1 > W2 > W3 (C) W1 < W2 < W3

(B) W1 = W2 = W3 (D) W2 > W1 > W3

Sol. The gravitational force is conservative and hence the work done by a gravitational force is independent of the path and depends only on the end points. Ans. B

(3)

The equations (1)–(3) give s r g1 GM/R2 T2 = = = 2. T1 g2 GM/(4R2 ) Ans. D Q 10. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth, (1998) (A) the acceleration of S is always directed towards centre of the earth. (B) the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant. (C) the total mechanical energy of S varies periodically with time. (D) the linear momentum of S remains constant in magnitude. Sol. The gravitational force on the satellite S is towards the centre of the earth (central force) i.e., GM m rˆ. F~ = − r2

122

Part I. Mechanics ~v

S•

Sol. Let the gravitational force on the planet by the star is

~ r

~ r F

F = kR−5/2 ,

• E

By Newton’s second law, acceleration is also towards the centre of the earth. The torque on S about the centre of the earth is ~τ = ~r × F~ = ~0 because ~r and F~ are anti-parallel. Thus, the angular momentum, ~ = m~r × ~v , of S is conserved i.e., both magnitude and L ~ remain constant. The angular momendirection of L tum is coming out of the paper for the orbital motion shown in the figure. Gravitational force is conservative so energy of S is conserved. The magnitude of linear momentum of S remains constant in the circular orbit but not in the elliptical orbit. It is maximum in the elliptical orbit when S is closest to the earth. Ans. A Q 11. An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy E0 . Its potential energy is (1997) (A) −E0 (B) 1.5E0 (C) 2E0 (D) E0 Sol. The kinetic energy K and potential energy U of a satellite in circular orbit are related by K = −U/2. The total energy is E0 = U + K = U + (−U/2) = U/2. Thus, U = 2E0 . Ans. C Q 12. If the distance between the earth and the sun were half of its present value, the number of days in a year would have been (1996) (A) 64.5 (B) 129 (C) 182.5 (D) 730 Sol. According to Kepler’s third law, square of the time period is proportional to the cube of the distance (semimajor axis for the elliptical orbit) i.e.,

where k is some constant and R is the distance between the planet and the star. This force provides the centripetal acceleration to the planet of mass m moving with a velocity v in the circular orbit of radius R, mv 2 /R = kR−5/2 , which gives 1/2  k . v= mR3/2 The time period of revolution is the time taken to complete one revolution i.e., 2πR T = = v



4π 2 mR7/2 k

1/2 ,

which gives T 2 ∝ R7/2 . We encourage you to generalize the result for F = kR−n . Hint: T 2 ∝ Rn+1 . Ans. B Q 14. If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth, is (1983) (A) 21 mgR (B) 2mgR (C) mgR (D) 14 mgR Sol. The acceleration due to gravity on the earth’s surface is given by g = GM/R2 ,

(1)

where M is the mass of the earth and R is its radius. The potential energies of mass m on the earth’s surface and at a height R above the earth’s surface are U1 = −GM m/R, U2 = −GM m/(R + R) = −GM m/(2R).

T 2 ∝ r3 . Let the distance between the earth and the sun be r1 . Time period of the orbital motion of the earth is T1 = 365 days. When distance is halved, r2 = r1 /2, the time period becomes T2 = T1 (r2 /r1 )3/2 = 365(1/2)3/2 = 129 days. Ans. B Q 13. Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T . If the gravitational force of attraction between the planet and the star is proportional to R−5/2 , then, (1989) (A) T 2 ∝ R2 (B) T 2 ∝ R7/2 (C) T 2 ∝ R3/2 (D) T 2 ∝ R3.75

The increase in potential energy of mass m when it is raised from the earth’s surface to a height R above the earth’s surface is ∆U = U2 − U1 = −GM m/(2R) − (−GM m/R) = GM m/(2R) = (GM/R2 )(mR/2) = mgR/2,

(using (1)).

Note that the increase in the potential energy is not mgR. Why? Ans. A Q 15. If the radius of the earth were to shrink by one per cent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would (1981) (A) decrease (B) remain unchanged (C) increase (D) be zero

Chapter 9. Gravitation

123

Sol. The acceleration due to gravity on the earth surface is given by g = GM/R2 .

(1)

Thus, g increases when radius R is decreased. Differentiate equation (1) and simplify to get percentage increase in the acceleration due to gravity, ∆g/g = −2(∆R/R) = −2(−1) = 2%. Ans. C One or More Option(s) Correct Q 16. Two bodies, each of mass M , are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement(s) is (are) (2013) (A) The minimal initial velocity of the mass m to escape q the gravitational field of the two bodies is 4 GM L . (B) The minimal initial velocity of the mass m to escape q the gravitational field of the two bodies is

2 GM L . (C) The minimal initial velocity of the mass m to escape q the gravitational field of the two bodies is 2GM L .

(D) The energy of the mass m remains constant. Sol. The gravitational potential energy and kinetic energy of the particle at the point O and at the point P are given by UO = − KO =

2GM m , L

2GM m UP = − √ . L2 + x2 1 KP = mv 2 . 2

1 mu2 , 2

√ L2 +

x2

P x u M•

•

L

O

L

•M

Gravitational force is conservative so total energy of the particle is conserved i.e., KO + UO = KP + UP . Substitute these to get   1 1 2GM m 2GM m 2 2 mv = mu − +√ . (1) 2 2 L L2 + x2 The particle will escape if quantity inside the bracket m in equation (1) is positive i.e., 12 mu2 ≥ 2GM . The L equality sign gives minimal initial velocity to escape p umin = 2 GM /L.

Aliter: The gravitational binding energy of the particle at O is BE = −UO . The particle will escape if KO ≥ BE i.e., KO ≥ −UO . Ans. B, D Q 17. Two spherical planets P and Q have the same uniform density ρ, masses MP and MQ , and surface areas A and 4A, respectively. A spherical planet R has uniform density ρ and its mass is (MP + MQ ). The escape velocities from the planets P, Q and R, are VP , VQ and VR , respectively. Then, (2012) (A) VQ > VR > VP (B) VR > VQ > VP (C) VR /VP = 3 (D) VP /VQ = 1/2 Sol. Let rP , rQ , and rR be the radii of the planet P, Q, and R, respectively. The fact that the area of Q is four times the area of P and densities of all three are same, gives √ 3 rQ = 2rP , MQ = 8MP , MR = 9MP , rR = 9rP . Substitute these values to get the escape velocities from the planet P, Q, and R as r 2GMP VP = , r s P 2GMQ VQ = = 2VP , rQ r √ 2GMR 3 VR = = 9VP . rR Ans. B, D Q 18. The magnitudes of the gravitational field at distance r1 and r2 from the centre of a uniform sphere of radius R and mass M are F1 and F2 respectively, then, (1994)

=

(B)

F1 F2 F1 F2

(C)

F1 F2

=

(D)

F1 F2

=

(A)

=

r1 r2 r22 r12 r13 r23 r12 r22

if r1 < R and r2 < R if r1 > R and r2 > R if r1 < R and r2 < R if r1 < R and r2 < R

Sol. The gravitational field due to a uniform solid sphere of radius R and mass M at a distance r from its centre is given by ( GM R3 r, if r ≤ R; F = GM if r > R. r2 , Ans. A, B Q 19. A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of coordinates. Two spheres of equal radii 1 unit, with their centres at A(−2, 0, 0) and B(2, 0, 0) respectively, are taken out of the solid sphere leaving behind spherical cavities as shown in the figure. Then, (1993)

124

Part I. Mechanics y

The gravitational potential by the object C at the point P (x, y, z) is given by VC = V − VA − VB

A•

x

B

•

O

z

(A) the gravitational field due to this object at the origin is zero. (B) the gravitational field at the point B(2, 0, 0) is zero. (C) the gravitational potential is the same at all points on circle y 2 + z 2 = 36. (D) the gravitational potential is the same at all points on circle y 2 + z 2 = 4. Sol. The mass of a solid sphere of radius R = 4 units is M = 43 π(4)3 ρ = 64m0 , where m0 = 4πρ/3. The masses of sphere A and B of radius r = 1 unit is m = 4 3 3 π(1) ρ = m0 . The solid sphere is a combination of three bodies, sphere A, sphere B, and remaining part C. The gravitational field by the solid sphere at its centre O is zero i.e., ~ ~ A (O) + E ~ B (O) + E ~ C (O) = ~0. E(O) =E Thus, gravitational field at O by the remaining part C is ~ C (O) = E(O) ~ ~ A (O) − E ~ B (O) E −E Gm0 Gm0 = ~0 − 2 (−ˆı) − 2 ˆı = ~0. 2 2 Note the direction of gravitational field. The gravitational field by the solid sphere at (2,0,0) is ~ 0, 0) = − GM x ˆı = − G(64m0 )(2) ˆı E(2, R3 43 = −2Gm0 ˆı. Thus, the gravitational field at (2,0,0) by the remaining part C is ~ C (2, 0, 0) = E(2, ~ 0, 0) − E ~ A (2, 0, 0) − E ~ B (2, 0, 0) E Gm0 = −2Gm0 ˆı − 2 (−ˆı) − ~0 4 31 = − Gm0 ˆı. 16 y 2 + z 2 = 36 y P (x, y, z)

Gm0 G(64m0 ) −p (x + 2)2 + y 2 + z 2 x2 + y 2 + z 2 Gm0 . −p (x − 2)2 + y 2 + z 2

=p

The curve y 2 + z 2 = 36 is a cylinder of radius 6 units with its axis along x. The potential is not constant on this cylinder as x varies along its length although y 2 + z 2 is constant. However, potential is a constant on any circle on this cylinder (each circle is cross-section of a cylinder at fixed x) so any circle on this cylinder is an equi-potential surface. Same argument is valid for y 2 + z 2 = 4. Ans. A, C, D Assertion Reasoning Type Q 20. Statement 1: An astronaut in an orbiting space station above the Earth experiences weightlessness. Statement 2: An object moving around the Earth under the influence of Earth’s gravitational force is in a state of free-fall. (2008) (A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. (B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true. Sol. The weight of a mass m, when measured by a weighing balance, is equal to the reaction force R acting on the balance. For a stationary mass m on the earth’s surface, Newton’s second law gives R = mg. If m is placed in a lift moving with a downward acceleration a then ma = mg − R, which gives R = mg − ma. If a = g then R = 0 and it is called a condition of free-fall. For an astronaut in orbiting space station, mv 2 /r = GM m/r2 − R. Since mv 2 /r = GM m/r2 , we get R = 0 and astronaut experiences weightlessness. Ans. A Matrix or Matching Type

6 A z

O B

x

Q 21. A planet of mass M , has two natural satellites with masses m1 and m2 . The radii of their circular orbits are R1 and R2 respectively. Ignore the gravitational force between the satellites. Define v1 , L1 , K1 and T1 to be, respectively, the orbital speed, angular

Chapter 9. Gravitation

125

momentum, kinetic energy and time period of revolution of satellite 1; and v2 , L2 , K2 and T2 to be the corresponding quantities of satellite 2. Given m1 /m2 = 2 and R1 /R2 = 1/4, match the ratios in Column-I to the numbers in Column-II. (2018) Column I (P) (Q) (R) (S)

Column II

v1 /v2 L1 /L2 K1 /K2 T1 /T2

(1) (2) (3) (4)

1/8 1 2 8

Sol. In a circular orbit, gravitational force provides the necessary centripetal acceleration to the satellite i.e., p GM m mv 2 = which gives orbital speed v = GM/R. 2 R R Thus, the ratio of the orbital speeds of two satellites is r √ R2 v1 = = 4 = 2. v2 R1 The angular momentum of the satellite moving in a circular orbit is given by L = mvR. Thus, the ratio of the angular momentum of two satellites is L1 m 1 v1 R 1 = = (2)(2)(1/4) = 1. L2 m 2 v2 R 2 The ratio of the kinetic energies is given by K1 m1 m1 v12 /2 = = K2 m2 v22 /2 m2



v1 v2

2

= (2)(2)2 = 8.

Fill in the Blank Type Q 23. A particle is projected vertically upwards from the surface of earth (radius R) with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of earth is . . . . . . (1997) Sol. The expressions for escape velocity and minimum kinetic energy for a particle to escape from a planet of radius R and mass M are p Kmin = 12 mve2 = GM m/R. ve = 2GM/R, Initial potential energy and kinetic energy of the projectile are Ui = −

GM m , R

Ki =

1 GM m Kmin = . 2 2R

At the maximum height h, the velocity of the projectile is zero. The potential and kinetic energy of the projectile at the maximum height are Uf = −

GM m , R+h

Kf = 0.

Use conservation of energy, Ui + Ki = Uf + Kf , to get h = R. Ans. R Q 24. The ratio of earth’s orbital angular momentum (about the sun) to its mass is 4.4 × 1015 m2 /s. The area enclosed by earth’s orbit is approximately . . . . . . m2 . (1997)

The time period of a satellite moving in a circular orbit is given by T = 2πR/v. Thus, the ratio of the time periods of two satellites is 1 2πR1 /v1 R1 v2 1 1 T1 = = = · = . T2 2πR2 /v2 R2 v1 4 2 8 Ans. P7→(3), Q7→(2), R7→(4), S7→(1) True False Type Q 22. It is possible to put an artificial satellite into orbit in such a way that it will always remain directly over New Delhi. (1984) Sol. A satellite can remain stationary above a point lying on the equatorial plane (geostationary). Since New Delhi does not lie on the equatorial plane it is not possible to have a geostationary satellite over New Delhi. Note that the time period of a geostationary satellite is 24 h and its orbit lies in the equatorial plane. A satellite with a time period of 24 h that may or may not lie on an equatorial plane is called geosynchronous. All satellites in Indian Regional Navigation Satellite System (IRNSS) are geosynchronous and some of them are geostationary. Ans. F

Sol. The areal velocity of a planet of mass m and angular momentum L in an elliptical orbit is given by dA/dt = L/(2m). Integrate from t = 0 to t = T to get Z T L LT A = dA = dt = 2m 0 2m LT = = (4.4 × 1015 )(365 × 24 × 60 × 60/2) m2 ≈ 6.94 × 1022 m2 . Z

Note that in a circular orbit of radius r, L mvr r 2πr r πr2 A = =v = = = . 2m 2m 2 T 2 T T Ans. 6.94 × 1022 Q 25. The masses and radii of the Earth and the Moon are M1 , R1 and M2 , R2 respectively. Their centres are at a distance d apart. The minimum speed with which a particle of mass m should be projected from a point midway between the two centres so as to escape to infinity is . . . . . . (1988)

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Part I. Mechanics

Sol. The gravitational potential energy of the particle of mass m located at the midpoint of line joining the centres of the earth and the moon is

g 0 = g0 − ω 2 R,

GM1 m GM2 m − . U =− d/2 d/2 M1

where g0 = 9.8 m/s2 is a constant, ω is the angular velocity of the earth and R = 6400 km = 6.4 × 106 m is the radius of the earth. Substitute the p values to g0 /R = get ω at which g 0 becomes zero i.e., ω = 1.24 × 10−3 rad/s. Ans. 1.24 × 10−3

M2

R1

R2

m • P

•

d/2

•

d/2

Integer Type

The particle will escape to infinity if its total mechanical energy at the point P is greater than or equal to zero, i.e., K +U =

1 GM1 m GM2 m mv 2 − − ≥ 0. 2 d/2 d/2

(1)

From equation (1), the minimum projection speed at which the particle escapes to infinity is p vmin = 2 G(M1 + M2 )/d. q Ans. v = 2 G d (M1 + M2 ) Q 26. A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth where R is the radius of the earth. The time period of another satellite at a height of 2.5R from the surface of the earth is . . . . . . h. (1987) Sol. For the geostationary satellite, distance from the centre of the earth is R1 = R + 6R = 7R and the time period is T1 = 24 h. The distance of another satellite from the centre of the earth is R2 = R + 2.5R = 3.5R. Apply Kepler’s third law to get the time period of this satellite as T2 = (R2 /R1 )

3/2

Sol. The effective acceleration due to gravity at the equator is given by

T1 = (3.5/7)3/2 24 = 8.48 h. Ans. 8.48

Q 27. According to Kepler’s second law, the radius vector to a planet from the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of . . . . . . (1985) Sol. Kepler’s second law is a consequence of the conservation of angular momentum. Ans. angular momentum Q 28. The numerical value of the angular velocity of rotation of the earth should be . . . . . . rad/s in order to make the effective acceleration due to gravity at the equator equal to zero. (1984)

Q 29. A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r = 3l
from M . The M , the tension in the rod is zero for m = k 288 value of k is . . . . . . . (2015) M

m •

•

m •

r

l

Sol. The acceleration ~a of the point masses are equal because they are connected by a massless rigid rod. ~a M •

f1

~a f3 f2

• m f2

r

• m

l

Consider the situation when tension in the rod is zero. The gravitational forces on the two point masses are shown in the figure. The forces f1 = GM m/r2 and f3 = GM m/(r + l)2 are due to the attraction by the larger mass M . The force f2 = Gmm/l2 is due to mutual attraction between the two point masses. Apply Newton’s second law on the two point masses to get GM m Gmm − = ma, r2 l2 GM m Gmm + = ma. 2 (r + l) l2

(1) (2)

Eliminate a from equations (1) and (2) and then substitute r = 3l to get m = 7M/288. Ans. 7

Chapter 9. Gravitation

127

Q 30. A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet’s gravity is 1/4th of its value at the surface of the planet. √ If the escape velocity from the planet is vesc = v N , then the value of N is . . . . . . . [ignore energy loss due to atmosphere.] (2015)

Descriptive

Sol. The acceleration due to gravity at the surface of a spherical planet of radius R and mass M is given by g0 = GM/R2 , where G is the universal gravitational constant. Let h be the maximum height attained by the bullet of mass m when fired vertically upwards with a velocity v. The acceleration due to gravity at a height h above the surface of the planet is given by g = GM/(R+ h)2 . Given, g = g0 /4 i.e.,

Sol. The potential energy of a particle of mass m placed in the gravitational field of a uniform solid sphere of mass M and radius R is given by

Q 32. There is a crater of depth R/100 on the surface of moon (radius R). A projectile is fired vertically upward from the crater with velocity equal to the escape velocity v from the surface of the moon. Find the maximum height attained by the projectile. (2003)

U =−


GM m 3R2 − r2 , 3 2R

where r < R. We encourage you to derive this result.

GM /(R + h)2 = 41 (GM /R2 ).

h •

Solve to get h = R. At the maximum height, kinetic energy of the bullet becomes zero. Apply conservation of energy between the initial and the final (maximum height) positions of the bullet to get

R 100 99R 100

•

−GM m/R + 12 mv 2 = −GM m/(R + h) + 0. Substitute h = R and solve to get v 2 = GM/R. The escape velocity from the planet is given by √ p √ vesc = 2(GM/R) = 2v 2 = v 2. Ans. 2 Q 31. Gravitational acceleration on the surface of a √ planet is 116 g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km/s, the escape velocity on the surface of the planet (in km/s) will be . . . . . . . (2010) Sol. The acceleration due to gravity on the surface of a planet of mass Mp and radius Rp is given by gp = GMp /Rp2 .

The escape velocity from the surface is p ve = 2GM/R, and kinetic energy of the projectile of mass m at the point of projection is Ki = 12 mve2 = GM m/R. The total energy of projectile at the point of projection is Ti = Ui + Ki i GM m GM m h 2 2 . =− 3R − (99R/100) + 2R3 R

(1) (4/3)πRp3 ρp .

The mass is related to density ρp by Mp = Substitute in equation (1) and simplify to get Rp = 3gp /(4πGρp ).

At the maximum height h, kinetic energy, potential energy, and total energy of the projectile are Kf = 0,

The escape velocity from the surface of the planet is given by q p vp = 2GMp /Rp = 2gp Rp q = 3gp2 /(2πGρp ). (2) Substitute values in equation (2) to get s s gp2 /ge2 vp 6/121 3 = = = , ve ρp /ρe 2/3 11 which gives vp = 3 km/s. Ans. 3

Uf = −

GM m , R+h

Tf = −

GM m . R+h

The conservation of energy, Ti = Tf , gives h = 99.5R. Ans. 99.5R Q 33. Distance between the centres of two stars is 10a. The masses of the stars are M and 16M and their radii a and 2a respectively. A body of mass m is fired straight from the surface of the larger star towards the surface of the smaller star. What should be its minimum initial speed to reach the surface of the smaller star? Obtain the expression in terms of G, M and a. (1996) Sol. Let the gravitational attraction by the two stars is equal and opposite at the point P.

128

Part I. Mechanics

M

Given vo = 12 ve . Substitute vo from equation (2) and ve from equation (3) to get h = R = 6400 km. When satellite is stopped in the orbit, its kinetic and potential energy become

10a r2

r1 P •

16M

r

Ui = −

Ki = 0, 1 mv 2 min 2

U

GM m . 2R

On reaching the earth’s surface, kinetic and potential energies are The mass m should be projected with a velocity such that it just reaches P, beyond which it is automatically attracted by the mass M . At P, G(16M )m GM m = , r12 r22

1 16GM m GM m 16GM m GM m mv 2 − − =− − . 2 min 2a 10a − 2a r2 r1 Substitute r1 = 2a and r2 = 8a to get p vmin = 32 5GM /a. We encourage you to analyse variation of potential energy (U ) in region between two stars. See the role of Umax at P. You can imagine that particle is moving on this potential energy curve. q Ans.

3 2

5GM a

Q 34. An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. (1990) (a) Determine the height of the satellite above the earth’s surface. (b) If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, find the speed with which it hits the surface of the earth. Sol. Consider a satellite of mass m in a circular orbit of radius R+h. In a circular orbit, gravitational attraction provides the centripetal acceleration i.e., GM m = . (R + h)2 R+h

1 mv 2 , 2

Uf = −

GM m . R

Conservation of energy, Ui + Ki = Uf + Kf , gives

which gives r2 = 4r1 . Using, r1 + r2 = 10a, we get r1 = 2a and r2 = 8a. Let vmin be the minimum velocity required for mass m to reach P. The conservation of energy gives

mvo2

Kf =

(1)

The equation (1) gives the orbital speed of the satellite as p vo = GM /(R + h). (2) The escape velocity from the surface of earth is given by p ve = 2GM /R. (3)

p p GM /R = gR q = (9.8)(6400 × 103 ) = 7.9 × 103 m/s.

v=

Ans. (a) 6400 km (b) 7.9 km/s Q 35. Three particles, each of mass m, are situated at the vertices of an equilateral triangle of side length a. The only force acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation a. Find the initial velocity that should be given to each particle and also the time period of the circular motion. (1988) Sol. Apply law of cosines in triangle OBC to get the length of a side √ r = a/ 3. A • =

=

F

F O

•

r B•

30

120◦

◦

r 30◦

a

•C

The gravitational forces on the particle at A due to the particles at B and C are of the magnitude |F~ | = Gm2 /a2 . Resolve F~ along and perpendicular to AO. The resultant force on the particle at A is towards the centre O and its magnitude is √ Fr = |F~ | cos 30◦ + F~ cos 30◦ = 3Gm2 /a2 . By symmetry, the resultant forces on the particles at B and C are also towards the centre O with the same

Chapter 9. Gravitation

129

magnitude Fr . The force Fr provides the necessary centripetal force for the particle to move in a circle of radius r with constant speed v i.e., √ √ 3Gm2 /a2 = mv 2 /r = 3mv 2 /a, p which gives v = GM/a. The time period in a uniform circular motion is given by r √ 2πa/ 3 2πr a3 =p = 2π . T = v 3Gm GM/a q q a3 Ans. v = GM , T = 2π a 3Gm Q 36. Two satellites S1 and S2 revolves around a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 h and 8 h respectively. The radius of the orbit of S1 is 104 km. When S2 is closest to S1 , find, (1986) (a) the speed of S2 relative to S1 . (b) the angular speed of S2 as actually observed by an astronaut in S1 . Sol. For satellite S1 , period of revolution is T1 = 1 h and orbital radius is r1 = 104 km. The period of revolution for satellite S2 is T2 = 8 h. Apply Kepler’s third law to get the orbital radius of S2 as 2/3

r2 = (T2 /T1 )

r1 = 4 × 104 km. v2

v1

S2 • r2 S1

• •

r1

The orbital speeds of the satellites S1 and S2 are given by v1 = 2πr1 /T1 = (2π × 104 )/1 = 2π × 104 km/h, v2 = 2πr2 /T2 = (2π × 4 × 104 )/8 = π × 104 km/h. The satellites S1 and S2 are closest to each other if they lie on the same radial vector (see figure). The speed of S2 relative to S1 at the closest distance is given by v2/1 = v2 − v1 = −π × 104 km/h. The angular speed of S2 as actually observed by an astronaut in S1 is ω=

v2/1 π × 104 π =− = − rad/h, 4 4 r2 − r1 3 4 × 10 − 10

where negative sign indicates the clockwise direction of rotation. We encourage you to find the number of revolutions by S1 and S2 between the successive events of the closest distance. Ans. (a) −π × 104 km/h (b) 3 × 10−4 rad/s

Chapter 10 Simple Harmonic Motion

One Option Correct

Aliter: The problem can be solved graphically using vector addition. The displacement x1 (t) can be represented by a vector ~v1 = A∠0 and displacement x2 (t) by ~v2 = A∠(2π/3) (see figure). The resultant of these two is ~v = A∠(π/3). The resultant of the three is zero if displacement x3 (t) is −~v = A∠(4π/3).

Q 1. A small block is connected to one end of a massless spring of unstretched length 4.9 m. The other end of the spring (see figure) is fixed. The system lies on horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes SHM with angular frequency ω = π/3 rad/s. Simultaneously, at t = 0, a small pebble is projected with speed v from point P at an angle of 45◦ as shown in the figure. Point P is at a distance of 10 m from O. If the pebble hits the block at t = 1 s, the value of v is [Take g = 10 m/s2 .]

~v2 2π 3

~v1 4π 3

(2012) z

v 45◦

O

~v

−~v •

x

P

Ans. B

10 m

√ (A) √50 m/s (C) 52 m/s

Q 3. The x-t graph of a particle undergoing SHM is shown in the figure. The acceleration of the particle at t = 4/3 s is (2009)

√ (B) √51 m/s (D) 53 m/s

Sol. Since pebble hits the block after 1 s, its time of flight is 1 s i.e.,

x(cm) 1 0

2v sin θ = 1. g

4

8

12

t(s)

-1 √

√ Substitute θ and g to get v = 50 m/s. We encourage you to show that both pebble and the block are at x = 2 2θ 5 m after 1 s. Hint: xpebble = 10 − v sin and xblock = g 4.9 + 0.2 cos ωt. Ans. A

(A) (C)

3 2 2 32 π cm/s 2 π 2 32 cm/s

2

2 (B) −π 32√ cm/s (D) − 323 π 2 cm/s2

Sol. The general equation for SHM is x = A sin(ωt + δ).

(1)

Q 2. A point mass is subjected to two simultaneous sinusoidal displacements in x
direction, x1 (t) = A sin(ωt) and x2 (t) = A sin ωt + 2π 3 . Adding a third sinusoidal displacement x3 (t) = B sin(ωt + φ) brings the mass to complete (2011) √ rest. The value of B and φ are (A) √2A, 3π/4 (B) A, 4π/3 (C) 3A, 5π/6 (D) A, π/3

From the given graph, the amplitude is A = 1 cm. At t = 0, the displacement is zero i.e., x = 1 sin δ = 0, which gives, δ = 0. Also, one oscillation is completed at t = 8 s, which gives ω = 2π/T = 2π/8 = π/4. Substitute A, δ, and ω in equation (1) to get

Sol. Adding x3 (t) brings the mass to complete rest. Thus, x1 (t) + x2 (t) + x3 (t) = 0, which gives

Differentiate equation (2) twice w.r.t. time to get the particle acceleration

x = sin(πt/4).

B sin(ωt + φ) = −A [sin(ωt) + sin (ωt + 2π/3)]

(2)

d2 x/dt2 = −(π 2 /16) sin(πt/4).

= −A [2 sin (ωt + π/3) cos(−π/3)]

At√t = 4/3 s, the acceleration is −(π 2 /16) sin(π/3) = − 3π 2 /32 cm/s2 . Ans. D

= −A sin (ωt + π/3) = A sin (ωt + 4π/3) 130

Chapter 10. Simple Harmonic Motion

131

Q 4. The mass M shown in the figure oscillates in SHM with amplitude A. The amplitude of the point P is (2009) k1

k2

•

M

P

(A)

k1 A k2

(B)

k2 A k1

(C)

k1 A k1 +k2

(D)

(M L2 /12)

a1 a2 •

P

M

k2

d2 θ = −kL2 θ/2. dt2

Simplify to get

k2 A k1 +k2

Sol. When M is at the maximum displacement A, the spring with spring constant k1 has maximum elongation, say a1 . The elongation of the other spring is a2 = A − a1 . The spring forces on a small element at P are k1 a1 towards the left and k2 (A − a1 ) towards the right.

k1

The moment of inertia I of the rod about the oscillation axis is M L2 /12. The angular acceleration of the rod is α = −d2 θ/dt2 , (anticlockwise direction is considered as positive). Using τ = Iα, we get

The mass of the element is zero because the spring is massless. Hence, by Newton’s second law, resultant force on this element is zero. Thus, k1 a1 = k2 (A − a1 ), 2A . which gives a1 = kk1 +k 2 Ans. D Q 5. A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is (2009)

d2 θ = −(6k/M )θ = −ω 2 θ. dt2 This p equation represents a SHM with angular p frequency ω 1 6k/M . = 2π ω = 6k/M and frequency ν = 2π Ans. C Q 6. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table. The least count for length is 0.1 cm and least count for time is 0.1 s. Student Length of pendulum (cm) I II III

No. of oscil- Time for (n) Time lations oscillations period (n) (s) (s)

64.0 64.0 20.0

8 4 4

128.0 64.0 36.0

16.0 16.0 9.0

If EI , EII  and EIII are the percentage errors in g i.e.,  ∆g g × 100 for students I, II and III, respectively, (2008)

(A) EI = 0 (C) EI = EII

(B) EI is minimum (D) EII is maximum

Sol. The time period of a simple pendulum is related to its length l and acceleration due to gravity g by p (1) T = 2π l/g.

•

If n oscillations takes time t then time period is T = t/n. Substitute it in equation (1) to get (A)

1 2π

q

2k M

(B)

1 2π

q

k M

(C)

1 2π

q

6k M

(D)

1 2π

q

24k M

Sol. When rod is displaced by a small angle θ, both the springs get stretched by (L/2) sin θ ≈ (L/2)θ causing a restoring force kLθ/2 on the rod by each spring. The restoring torque τ on the rod about the pivot point O is in anti-clockwise direction and is given by (kLθ/2)(L/2) + (kLθ/2)(L/2) = kL2 θ/2.

g = 4π 2 ln2 /t2 .

(2)

Differentiate and simplify equation (2) to get ∆g/g = ∆l/l + 2∆t/t.

(3)

Given, ∆l = 0.1 cm and ∆t = 0.1 s. Substitute the values in equation (3) for the given cases to get EI = (0.1/64.0 + 2 × 0.1/128) × 100 = 0.31%, EII = (0.1/64.0 + 2 × 0.1/64) × 100 = 0.46%, EIII = (0.1/20.0 + 2 × 0.1/36) × 100 = 1.05%.

L 2

θ •O

L 2

It appears that increasing l and n will keep on improving the accuracy of g measurement. Can you explain why this is not true in a real pendulum? Ans. B

132

Part I. Mechanics

Q 7. A simple pendulum has time period T1 . The point of suspension is now moved upwards according to the relation y = kt2 , (k = 1 m/s2 ) where y is the vertical displacement. The time period now becomes T2 . The ratio of T1 2 /T2 2 is [Take g = 10 m/s2 .] (2005) (A) 6/5 (B) 5/6 (C) 1 (D) 4/5 Sol. Let us analyse the problem in an inertial frame (G) attached to the ground. In this frame, point of suspension O has an upward acceleration given by ~as/g = ay ˆ = d2 y/dt2 ˆ = 2k ˆ = 2 m/s2 ˆ.

Use sin θ ≈ θ in equation (3) to get the time period 2π 1 T = = ω 2π

s

l . g + ay

In the first case, ay = 0 and in the second case ay = 2 m/s2 . Thus, g + ay T12 10 + 2 6 = = = . T22 g 10 5 We encourage you to analyse the problem in the frame S. q Hint: Use the psuedo force −m~a to get T = l 2π |~g−~ a| .

y ~a

Ans. A

x

O

Q 8. A block P of mass m is placed on a horizontal frictionless plane. A second block of same mass m is placed on it and is connected to a spring of spring constant k. Then two blocks are pulled by a distance A. Block Q oscillates without slipping. What is the maximum value of frictional force between the two blocks? (2004)

θ l

τ cos θ =

τ

τ sin θ

•

θ

~ab/s

mg k Q

Consider another frame (S ) attached to the point O i.e., frame S has an acceleration ~as/g w.r.t. frame G. The bob executes SHM as seen from S, i.e., θ = θ0 sin ωt,

P

(A) kA/2 (B) kA (C) µs mg (D) zero

where ω = 2π/T is the angular frequency, l is the pendulum’s length, θ0 is the amplitude, and θ is angle from the vertical. The acceleration of the bob in frame S is tangential with magnitude |~ab/s | = l d2 θ/dt2 = lω 2 θ. Resolve ~ab/s along the x and y directions to get ~ab/s = −lω θ(cos θ ˆı + sin θ ˆ). The relative acceleration, ~ab/s = ~ab/g − ~as/g , gives acceleration of the bob in frame G as ~ab/g = ~ab/s + ~as/g = −lω 2 θ cos θ ˆı − (lω 2 θ sin θ − ay ) ˆ. The forces on the bob are its weight mg and tension τ . Newton’s second law applied in frame G gives τ sin θ = mlω 2 θ cos θ,

(1) 2

τ cos θ − mg = may − mlω θ sin θ.

Sol. There is no friction between the block P and the horizontal plane. Also the block Q oscillates without slipping. Thus, two blocks move together in SHM with amplitude A. The maximum acceleration amax of the blocks occurs at the extreme positions and is given by Newton’s second law, (2m)amax = kA.

2

(1)

The frictional force f is the only horizontal force on the block P. The force f becomes maximum when acceleration of P attains its maximum value given by equation (1). Thus, fmax = mamax = kA/2. Ans. A Q 9. For a particle executing SHM, the displacement x is given by x = A cos ωt. Identify the graph which represents the variation of potential energy (PE) as a function of time t and displacement x, (2003) PE I

PE

II

(2)

Eliminate τ from equations (1) and (2) to get (g + ay ) sin θ = ω 2 lθ.

µs

III t

(3)

(A) I, III (B) II, IV (C) II, III (D) I, IV

IV

x

Chapter 10. Simple Harmonic Motion

133

Sol. In SHM, the potential energy varies with displacement x as PE = 12 kx2 (a parabola). The variation with time t is given by PE = 21 kA2 cos2 ωt. Ans. A Q 10. An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is (2002) 2M g Mg Mg g (B) (C) (D) (A) 4M k k k 2k

Q 12. The period of oscillation of simple pendulum of length L suspended from the roof of the vehicle which moves without friction, down an inclined plane of inclination α, (2000) q q is given by L L (A) 2π g cos α (B) 2π g sin α q q L (C) 2π Lg (D) 2π g tan α Sol. The forces acting on the vehicle of mass M are its weight M g and normal reaction N from the plane.

Sol. Initially, the mass M is stationary and the spring is unstretched. Thus, the kinetic energy of the mass as well as the elastic potential energy of the spring are zero.

α θ T

θ • α mg

k

x0

M x α

Let x be the maximum extension of the spring. At the maximum extension, mass is stationary but the spring is stretched by x. The loss in gravitational potential energy of the mass is equal to the gain in elastic energy of spring i.e., M gx =

2 1 2 kx ,

which gives x = 2M g/k. We encourage you to show that in equilibrium x0 = M g/k, and when disturbed, it starts SHM about equilibrium position with an amplitude M g/k. Ans. B Q 11. A particle executes SHM between x = −A and x = +A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2 , then, (2001) (A) T1 < T2 (B) T1 > T2 (C) T1 = T2 (D) T1 = 2T2 Sol. The displacement of a particle undergoing SHM with amplitude A and time period T is given by x = A sin (2πt/T ) . The displacement becomes A/2 at time t1 given by A/2 = A sin (2πt1 /T ) . Solve to get t1 = T /12. The displacement becomes A at time t2 , A = A sin (2πt2 /T ) . Solve to get t2 = T /4. Thus, T1 = t1 = T /12, and T2 = t2 − t1 = T /6 = 2T1 . Ans. A

Resolve M g along and normal to the inclined plane and apply Newton’s second law to get N = M g cos α,

M a0 = M g sin α.

Thus, vehicle accelerates with the downward acceleration a0 = g sin α. Let us find the mean position (rest position of the bob in the frame attached to vehicle) of the bob of mass m. In the mean position, vehicle and bob move downwards with the same acceleration a0 = g sin α. Suppose pendulum’s string makes an angle α + θ with the vertical in the mean position (see figure). The forces on the bob are its weight mg and tension T . Resolve mg and T along and normal to the plane and apply Newton’s second law to get mg cos α = T cos θ,

(1)

mg sin α + T sin θ = ma0 = mg sin α.

(2)

Solve equations (1) and (2) to get θ = 0 and T = mg cos α. When bob is disturbed from the mean position, it starts SHM with effective acceleration due to gravity geff = g cos α and time period p p τ = 2π L/geff = 2π L/(g cos α). We encourage you to find θ and T by solving the problem in the frame attached to vehicle. Also explore the case when vehicle moves with friction. In general, if a vehicle containing pendulum accelerates with p an acceleration ~a0 , then ~geff = ~g − ~a0 and τ = 2π L/|~geff |. Ans. A Q 13. A particle free to move along the h x-axis i has 2 potential energy given by U (x) = k 1 − e−x for

134

Part I. Mechanics

−∞ ≤ x ≤ +∞, where k is a positive constant of appropriate dimensions. Then, (1999) (A) at points away from the origin, the particle is in unstable equilibrium. (B) for any finite non-zero value of x, there is a force directed away from the origin. (C) if its total mechanical energy is k/2, it has its minimum kinetic energy at origin. (D) for small displacement from x = 0, the motion is simple harmonic.  2 Sol. The potential energy U (x) = k 1 − e−x , varies with x as shown in the figure.

When displacement is equal to a (amplitude), the velocity v = 0, kinetic energy K = 21 mv 2 = 0, potential energy U = ka3 , and the total energy U +K = ka3 . Let v be the velocity of the particle when displacement is x (0 < x < a). At this instant, K = 21 mv 2 and U = kx3 . The conservation of energy, 2 1 2 mv

+ kx3 = ka3 ,

gives dx = v= dt

r

2k 3 (a − x3 ). m

(1)

By symmetry, time taken from x = 0 to x = a is T /2. Integrate equation (1) from x = 0 to x = a to get r Z a Z T /2 m dx T √ dt = = . (2) 3 2 2k a − x3 0 0

U k x

The potential energy has a local minimum dU dx =
2 d U 0, dx2 > 0 at stable equilibrium and local maximum
dU d2 U dx = 0, dx2 < 0 at unstable equilibrium. It is given that potential energy has a stable equilibrium at x = 0. When particle is displaced from the origin, the force F = −dU /dx brings it back. This force points towards the origin. There is no unstable equilibrium. Total mechanical energy, U + K = k/2, gives kinetic energy of the particle as K = k/2 − U , which is maximum at x = 0. The force on a particle of mass m placed near x = 0 is given by   2 dU x2 F =− = −2kxe−x = −2kx 1 − + ··· dx 2!

Either carry out the integration in equation (2) by using x = a sin2/3√ θ or use the dimensional analysis to show that T ∝ 1/ a. Ans. A Q 15. Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that of N is (1988) p p (A) k1 /k2 (B) k2 /k1 (C) k2 /k1 (D) k1 /k2 Sol. The two blocks undergo SHM in the vertical direction about their mean positions.

≈ −2kx. k1

The force F is proportional to x and points towards x = 0. Thus, p particle executes SHM with angular frequency ω = 2k/m. Ans. D Q 14. A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is U (x) = k|x|3 , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is √ (A) proportional to 1/ a √ (C) proportional to a

(1998)

(B) independent of a (D) proportional to a3/2

Sol. The potential energy U = k|x|3 versus x is plotted in the figure.

k2 2A1 m

x0 2A2 m

Let the amplitudes of oscillations be A1 and A2 and the angular frequencies be ω1 and ω2 . The angular frequencies of oscillations are given by p p ω1 = k1 /m, ω2 = k2 /m. The velocities are maximum at the mean positions and are given by v1,max = A1 ω1 ,

v2,max = A2 ω2 .

U ka3

−a

O

a

x

p The condition v1,max = v2,max gives A1 /A2 = k2 /k1 . We encourage you to find the distance between the two mean positions (x0 ). Note that the elongation in stiff spring (k1 > k2 in the figure) is less and its frequency is more than the other spring. Ans. B

Chapter 10. Simple Harmonic Motion

135

Q 16. A particle executes SHM with a frequency f . The frequency with which its kinetic energy oscillates is (1987) (A) f /2 (B) f (C) 2f (D) 4f Sol. The displacement of a particle undergoing SHM varies with time t as x = A sin 2πf t, where A is the amplitude and f is the frequency. The velocity of the particle is v = dx/dt = 2πAf cos 2πf t.

The final time period of oscillation (T = 2π/ω) in both the cases is same because ω10 = ω20 . For SHM in horizontal plane, mean position x0 is equal to the natural length of the spring. Thus, the mean position x0 remains same in both the cases. k v

M x0

x

before T

after

t m

k

K

v10

M

T

x0

t

The kinetic energy of a particle of mass m moving with velocity v is K = 21 mv 2 = 21 m(2πf )2 A2 cos2 (2πf t) = mπ 2 f 2 A2 (1 + cos 4πf t). Thus, the kinetic energy varies with a frequency 2f as shown in the figure. Ans. C One or More Option(s) Correct Q 17. A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x0 . Consider two cases: (1) when the block is at x0 ; and (2) when the block is at x = x0 + A. In both the cases, a particle with mass m (< M ) is softly placed on the block after which they stick to each other. Which of the following statement(s) is(are) true about the motion after the mass m is placed on the mass M ?

Let the initial amplitude of SHM be A, final amplitude in case 1 be A01 , and final amplitude in case 2 be A02 . In case 1, the speed of the block of mass M at the mean position just before the particle of mass m is softly placed over it is v = ωA. Let the speed just after mass m is placed over it be v10 . The linear momentum of the system along the direction of motion is conserved because there is no external force on it in the direction of motion. Apply conservation of linear momentum, M v = (M + m)v10 , to get v10 = M v/(M + m) = M ωA/(M + m).

From equation (1), the instantaneous speed at the mean position is decreased i.e., v10 < v. The speed at the mean position v10 is related to the amplitude A01 by v10 = A01 ω10 . Substitute for v10 and ω10 to get A01 =

p r k/M MA v10 M p = . =A ω10 M + m k/(M + m) M +m (2)

(2016)

(A) The amplitude of oscillation in the first case q M changes by a factor of m+M , whereas in the second case it remains unchanged. (B) The final time period of oscillation in both the cases is same. (C) The total energy decreases in both the cases. (D) The instantaneous speed at x0 of the combined masses decreases in both the cases. Sol. The angular frequency of the spring mass system depends on the spring constant and the attached mass. Initial angular frequency of the system (ω), final angular frequency of the system in case 1 (ω10 ), and final angular frequency of the system in case 2 (ω20 ) are given by p ω = k/M , p ω10 = k/(M + m), p ω20 = k/(M + m).

(1)

Let T be the initial total energy, T10 be the final total energy in case 1, and T20 be the final total energy in case 2. The final energy in case 1 is given by T10 =

1 M M 1 02 kA1 = kA2 =T . 2 2 M +m M +m (using (2)) k

V =0 M

x0

A

k

V0 =0 m

before after

M x0

A

136

Part I. Mechanics

In case 2, the particle of mass m is placed over the block when it is at the extreme position x = x0 + A. Thus, velocity of the block just before m is placed over it is zero. The conservation of linear momentum gives that the velocity of the combined masses just after m is placed over the block is also zero. Hence the position x = x0 + A remains as the extreme position for SHM of combined masses. Thus, the amplitude of the SHM remains unchanged in case 2 i.e., A02 = A. The total energy of the system also remains unchanged in case 2 2 i.e., T20 = 21 kA02 = 12 kA2 = T . The instantaneous speed at the mean position in case 2 is given by r r k M 0 0 0 v2 = ω2 A2 = A= ωA M +m M +m M = v. M +m Ans. A, B, D Q 18. Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω1 and ω2 and have total energies E1 and E2 , respectively. The variations of their momenta p and position x are shown in the figures. If a/b = n2 and a/r = n, then the correct equation(s) is (are) (2015)

Q 19. A particle of mass m is attached to one end of a massless spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at t = 0 with an initial velocity u0 . When the speed of the particle is 0.5u0 , it collides elastically with a rigid wall. After this collision, (2013) (A) the speed of the particle when it returns to its equilibrium position is u0 . (B) the time at which the particle passes throughpthe equilibrium position for the first time is t = π m k. (C) the time at which the maximum compression of the pm spring occurs is t = 4π 3 k. (D) the time at which the particle passes through the equilibrium position for the second time is t = pm 5π 3 k. Sol. At t = 0, the particle is at equilibrium position Q with maximum velocity u0 . From Q to R, its motion is like SHM with speed u = u0 cos ωt, where ω = 2π/T , and T = 2π

p

p

u0

b x a

P

x

Q

m/k. 0.5u0

R

r

Energy= E1

(A) E1 ω1 = E2 ω2 (C) ω1 ω2 = n2

p

At R, it undergoes elastic collision with the wall and the direction of its velocity get reversed. From R to Q and then to P its motion is like SHM. At P, spring has maximum compression. Let tQR is the time taken to move from Q to R. The speed at R is

Energy= E2

(B) ω2 /ω1 = n2 (D) E1 /ω1 = E2 /ω2

Sol. In harmonic oscillator, total energy is equal to the potential energy at the extreme positions (maximum displacement) and it is also equal to the kinetic energy at the central position (zero displacement). The potential energy at a displacement x is given by U = 12 mω 2 x2 and kinetic energy at linear momentum p is given by K = p2 /(2m). For the harmonic oscillator having elliptical x-p graph, U1,max = E1 = 12 mω12 a2 ,

(1)

K1,max = E1 = b2 /(2m).

(2)

For the harmonic oscillator having circular x-p graph, U2,max = E2 = 12 mω22 R2 , 2

K2,max = E2 = R /(2m).

(3) (4)

Use equations (1)–(4) and the relations a/b = n2 and a/R = n to get ω2 /ω1 = n2 and E1 /ω1 = E2 /ω2 . Ans. B, D

u = 0.5u0 = u0 cos(ωtQR ). Solve to get tQR

T π = = 6 3

r

m . k

Particle p m passes Q for the first time after time 2tQR = 2π 3 k and its speed at this instant is u0 . The time to reach P is given by r T T 7π m 2tQR + tQP = + = . 3 4 6 k The time to arrive at Q for the second time is r T 5π m T . 2tQR + 2tQP = + = 3 2 3 k Ans. A, D

Chapter 10. Simple Harmonic Motion

137

Q 20. A metal rod of length L and mass m is pivoted at one end. A thin disc of mass M and radius R (< L) is attached at its centre to the free end of rod. Consider the two ways the disc is attached. (Case A) The disc is not free to rotate about its centre, and (Case B ) the disc is free to rotate about its centre. The roddisc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is (are) true? (2011)

In Case B, the rod undergoes pure rotation about O and the disc has both translation and rotation. The kinetic energy of the rod is f 2 KB,rod = 12 IO,rod ωB =

2 2 11 2 3 mL ωB

2 = 16 mL2 ωB ,

and the kinetic energy of the disc is f 2 KB,disc = 21 M vcm + 12 Icm ω 2

= 12 M (LωB )2 + 2 = 12 M L2 ωB +

2 2 11 2 2MR ω 2 2 1 4MR ω ,

f where ω is the angular velocity of the disc. Using, KB = f f f f KB,rod + KB,disc and equating KA to KB , we get

(A) restoring torque in case A = restoring case B. (B) restoring torque in case A < restoring case B. (C) Angular frequency of case A > Angular of case B. (D) Angular frequency of case A < Angular of case B.

torque in torque in frequency frequency

Sol. Consider motion of the centre of mass C located at a distance rC = MM+m/2 +m L from the pivot point O. O

rC

L /2

•

θ •

C•

L mg

Mg

2 2 ωB = ωA +


M R2 3 2 · ωA − ω2 . 2 2 2 mL + 3M R

Now, consider the motion of the disc alone. When released, the disc was at rest with its angular velocity ω = 0 and the angular momentum about its centre of mass L = Icm ω = 0. There is no external torque on the disc about its centre of mass because the disc is free to rotate and its weight pass through the centre of mass. Thus, angular momentum of the disc about its centre of mass is conserved i.e., angular momentum and angular velocity of the disc always remains zero. Substitute ω = 0 in equation (1) to get ωB > ωA . Ans. A, D Q 21. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 s for this and records 40 s for 20 oscillations. For this observation, which of the following statement(s) is (are) true?

•

The weight of disc-rod system, (M + m)g, acts at C giving an anticlockwise restoring torque about O, τ = (M + m)grC sin θ = (M + m/2) gL sin θ, where θ is angular displacement. Since forces remain the same, restoring torques are equal in both, case A and case B. Since system is released from the same position, initial potential energy of the system is equal in both the cases i.e., UAi = UBi . Initial kinetic energy is i i zero in both the cases i.e., KA = KB = 0. The potential energy at the lowest point is also zero i.e., UAf = UBf = 0. f i The energy conservation, UAi + KA = UAf + KA and f f f f i i UB + KB = UB + KB , gives KA = KB . In Case A, the disc is rigidly fixed to the rod and the disc-rod system has pure rotational motion about an axis passing through O. Thus,
2 f 2 = 12 13 mL2 + 12 M R2 + M L2 ωA KA = 21 IO ωA
2 = 16 mL2 + 12 M L2 + 14 M R2 ωA .

(1)

(2010)

(A) (B) (C) (D)

Error ∆T in measuring time period T is 0.05 s. Error ∆T in measuring time period T is 1 s. Percentage error in the determination of g is 5%. Percentage error in the determination of g is 2.5%.

Sol. Let t = 40 s be the time taken for n = 20 oscillations. The time period is T = t/n = 2 s. The error in T is given by ∆T = ∆t/n, where ∆t = 1 s is the error in t (least count of the watch). Thus, ∆T = 0.05 s. The acceleration due to gravity is related to the pendulum’s length l and time period T by g = 4π 2 l/T 2 . Differentiate and simplify to get ∆g/g = ∆l/l + 2∆T /T, (note that the errors from both the parameters are added together). Substitute ∆l = 0 (since l = 1 m exactly), ∆T = 0.05 s, and T = 2 s to get ∆g/g = 0.05 = 5%. Ans. A, C

138

Part I. Mechanics

Q 22. Function x = A sin2 ωt + B cos2 ωt + C sin ωt cos ωt represents SHM, (2006) (A) For any value of A, B and C (except √ C = 0) (B) If A = −B, C = 2B, amplitude = |B 2| (C) If A = B, C = 0 (D) If A = B, C = 2B, amplitude = |B| Sol. Using trigonometric identities, we get x = A sin2 ωt + B cos2 ωt + C sin ωt cos ωt = = =

A B 2 (1 − cos 2ωt) + 2 (1 + cos 2ωt) A+B + C2 sin 2ωt + B−A 2 2 cos 2ωt A+B + D sin(2ωt + δ), 2

+

C 2

sin 2ωt

where D = 21 (C 2 +B 2 +A2 −2AB)1/2 and tan δ = B−A C . √ If A = −B and C = 2B then D = 2B and δ = π/4. If A = B and C = 0 then x = B, a constant. If A = B and C = 2B then D = B and δ = 0. Ans. A, B, D Q 23. The coordinates of a particle moving in a plane are given by x = a cos pt and y = b sin pt where a, b (< a) and p are positive constants of appropriate dimensions. Then, (1999) (A) the path of the particle is an ellipse. (B) the velocity and acceleration of the particle are norπ mal to each other at t = 2p . (C) the acceleration of the particle is always directed towards a focus. (D) the distance travelled by the particle in time interπ val t = 0 to t = 2p is a. Sol. Given x = a cos pt and y = b sin pt. Eliminate t to 2 2 get the path, xa2 + yb2 = 1, an ellipse. y ~v1 ~a1 ~a0

Q 24. Three simple harmonic motions in the same direction having the same amplitude and same period are superimposed. If each differ in phase from the next by 45◦ , then, (1999) √
(A) the resultant amplitude is 1 + 2 a. (B) the phase of the resultant motion relative to the first is 90◦ . (C) the energy √
associated with the resulting motion is 3 + 2 2 times the energy associated with any single motion. (D) the resulting motion is not simple harmonic. Sol. By superposition principle, resultant displacement of the given SHMs is y = y1 + y2 + y3 = a sin ωt + a sin(ωt + 45◦ ) + a sin(ωt + 90◦ ) = a [(sin ωt + sin(ωt + 90◦ )) + sin(ωt + 45◦ )] √ = a( 2 + 1) sin(ωt + 45◦ ). 0 The √ resultant motion is0 a SHM with amplitude a = ( 2 + 1)a, frequency ω = ω, and phase difference of 45◦ relative to the first SHM. The total energy of the resultant SHM is given by

√ 2 2 E = 21 ma0 ω 0 = (3 + 2 2) 21 ma2 ω 2 .

~v0 b

At time t = 0, ~r0 = a ˆı, ~v0 = bp ˆ, and ~a0 = −p2 a ˆı. At π time t = 2p , ~r1 = b ˆ, ~v1 = −ap ˆı, and ~a1 = −p2 b ˆ. The velocity ~v1 and acceleration ~a1 are normal to each other. The acceleration ~a = −p2~r is always directed towards the origin. The distance travelled from ~r0 to ~r1 is one quarter of the perimeter of the ellipse. We encourage you to analyze this motion as the superposition of two perpendicular SHMs of different amplitudes. Ans. A, B, C

x

Ans. A, C

a

Let ~r, ~v , and ~a be the position, velocity, and acceleration vectors of the particle. Writing position in the vector form ~r = x ˆı + y ˆ = a cos pt ˆı + b sin pt ˆ. Differentiate ~r w.r.t. time t to get velocity ~v and then differentiate ~v w.r.t. time t to get acceleration ~a, ~v = −ap sin pt ˆı + bp cos pt ˆ, ~a = −ap2 cos pt ˆı − bp2 sin pt ˆ = −p2 (a cos pt ˆı + b sin pt ˆ) = −p2~r.

Q 25. A linear harmonic oscillator of force constant 2 × 106 N/m and amplitude 0.01 m has a total mechanical energy of 160 J. Its (1989) (A) maximum potential energy is 100 J. (B) maximum kinetic energy is 100 J. (C) maximum potential energy is 160 J. (D) maximum potential energy is zero. Sol. The potential energy U (x), kinetic energy K(x), and total energy E of a linear harmonic oscillator are related by U (x) + K(x) = E.

(1)

Chapter 10. Simple Harmonic Motion E

139 Q 26. The phase space diagram for a ball thrown vertically up from ground is (B) Momentum (A) Momentum

U (x)

Position

(C)

x −A

Position

60 U0

U

Umax

Kmax

K

160

O

(D)

Momentum

Momentum

A

Position

The potential energy of a linear harmonic oscillator of force constant k at a displacement x is given by U (x) = U0 +

2 1 2 kx ,

(2)

where U0 is a constant. The kinetic energy of the oscillator becomes zero at the maximum displacement (x = ±A). Substitute K(x) = 0, x = A = 0.01 m, and k = 2 × 106 N/m in equations (1) and (2) and simplify to get U0 = E − 12 kA2 = 160 − 21 (2 × 106 )(0.01)2 = 60 J.

(3)

Use equations (1)–(3) to get the minimum and maximum potential and kinetic energies as x=0: x = ±A :

Umin = 60 J,

Kmax = 100 J;

Umax = 160 J,

Kmin = 0. Ans. B, C

Position

Sol. Let the ball of mass m be thrown up with an initial velocity u. Its velocity v and displacement x are related by v 2 − u2 = −2gx, where g is the acceleration due to gravity. The momentum (p = mv) is given by p2 = m2 u2 − 2m2 gx, which gives p p = ± m2 u2 − 2m2 gx. At x = 0, the momentum is mu when the ball starts going up and it becomes −mu when the ball comes back. At the maximum height, x = u2 /(2g), the momentum becomes zero. Ans. D Q 27. The phase space diagram for SHM is a circle centered at the origin. In the figure, two circles represent the same oscillator but for different initial conditions, and E1 and E2 are total mechanical energies, respectively. Then,

Paragraph Type

Momentum E1

Paragraph for Questions 26-28

Momentum

Phase space diagrams are useful tools in analysing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) versus p(t) curve in this plane. The arrow on curve indicates time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upward (or to right) is positive and downwards (or towards left) is negative. (2011)

Position

E2 2a a

√ (A) E1 = 2E2 (C) E1 = 4E2

Position

(B) E1 = 2E2 (D) E1 = 16E2

Sol. In SHM, the energy of an oscillator of mass m oscillating with an angular frequency ω and amplitude A is E = 21 mω 2 A2 . From the given figure, E1 /E2 = A21 /A22 = a2 /(4a2 ) = 1/4. Ans. C Q 28. Consider the spring mass system, with mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is

140

Part I. Mechanics V (x)

(A) Momentum

(B)

Momentum

Position

(C)

Position

(D)

Momentum

E O

Momentum

Position

Position

Sol. The position x and momentum p of a spring-mass system undergoing SHM with an angular frequency ω and amplitude A are given by x = A cos ωt, p = m dx/dt = −mA sin ωt. In the first quarter of the cycle, position is positive and momentum is negative.

V0 x0

x

From the given figure, V (x) ≥ 0 for all x. If E ≤ 0 for some x then kinetic energy K(x) = E − V (x) ≤ 0 for those x and hence the motion of the particle is not allowed for those x. On the other hand, if E ≥ V0 then K(x) = E − V (x) ≥ 0 for all x and particle is allowed to move for all x, including infinity (in this case the particle will escape to infinity). Thus, for the particle to have a periodic motion, V0 > E > 0. In this case, particle is allowed at points where K(x) = E − V (x) = E − αx4 ≥ 0 i.e., |x| ≤ (E/α)1/4 .

x

Ans. C t

O

Q 30. For periodic motion of small amplitude A, the time period T of this p mparticle is pproportional p αto p 1 α 1 (A) A m α (B) A α (C) A m (D) A m

p

When system is submerged in the water, the damping causes continuous reduction of the amplitude. Ans. B Paragraph for Questions 29-31 When a particle of mass m moves on the x-axis in a potential of the form V (x) = kx2 , it performs SHM. The p corresponding time period is proportional to m/k, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V (x) = αx4 (α > 0) for |x| near the origin and becomes a constant equal to V0 for |x| ≥ x0 (see figure). (2010)

Q 31. The acceleration of this particle for |x| > |x0 | is V0 (A) proportional to mx (B) proportional to V0 q0 V0 (C) proportional to mx0 (D) zero Sol. For |x| > |x0 |, potential energy is a constant i.e., V (x) = V0 . The force on the particle in this region is given by F = −dV (x)/dx = 0. Newton’s second law gives acceleration a = F/m = 0. Aliter: The kinetic energy of the particle in this region is constant i.e., K = E − V (x) = E − V0 .

V (x)

V0 x0

x

Q 29. If the total energy of the particle is E, it will have periodic motion only if (A) E < 0 (B) E > 0 (C) V0 > E > 0 (D) E > V0 Sol. The kinetic energy of the particle cannot be negative. The total energy E is the sum of the kinetic energy K(x) and potential energy V (x) i.e., E = K(x) + V (x).

Sol. The dimensions of A is [L], m is [M] and α is V (x)/x4 = [ML−2 T−2 ]. From, the dimensional analyp sis, the time period is proportional to A1 m/α. Ans. B

(1)

Hence, particle moves with a constant speed, v = 2K/m. The acceleration of a particle moving along xaxis with a constant constant speed is zero (a = dv/dt = 0). Ans. D Paragraph for Questions 32-34 A uniform thin cylindrical disc of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disc symmetrically on either side at a distance d from its centre. The axle is massless and both the springs

Chapter 10. Simple Harmonic Motion

141

and the axle are in horizontal plane. The unstretched length of each spring is L. The disc is initially at its equilibrium position with its centre of mass at a distance L from the wall. The disc rolls without slipping ~ = V0ˆı. The coefficient of friction is µ. with velocity V (2008)

Q 33. The centre of mass of the disc undergoes SHM with q angular frequency ωq equal to q q k M

(A)

(B)

2k M

(C)

2k 3M

4k 3M

(D)

Sol. At a displacement x, the net external force is −4kx/3. Apply Newton’s second law to get
M d2 x/dt2 = − 4k 3 x, which can be written as
4k d2 x/dt2 = − 3M x = −ω 2 x.

y

This equation q represents a SHM with an angular fre4k . quency ω = 3M Ans. D

2d d

d

V0

R

Q 34. The maximum value of V0 for which the disc will roll without is q q q slippingq

x

(A) µg

Q 32. The net external force acting on the disc when its centre of mass is at displacement x with respect to its equilibrium position is (A) −kx (B) −2kx (C) − 32 kx (D) − 43 kx Sol. The forces acting on the disc are spring force 2kx, frictional force f , weight M g and normal reaction N . y

N

a

M 2k

(B) µg

(C) µg

3M k

(D) µg

5M 2k

Sol. The disc will stop ‘rolling without slipping’ if frictional force is not adequate. The maximum value of frictional force occurs when spring elongation is maximum i.e., fmax = µN = µM g

(∵ N = M g)

= 2kxmax /3,

(∵ f = 2kx/3).

which gives x

2kx

M k

xmax = 3µM g/(2k). The energy is conserved in ‘rolling without slipping’ i.e., Ki + Ui = Kf + Uf (we encourage you to find why energy is conserved in-spite of frictional force). In SHM, Ki = 0 when the displacement is maximum and Uf = 0 when the velocity is maximum. The total kinetic energy comprises of translation and rotation and is given by

α

f Mg

There is no acceleration in the vertical direction and hence N = M g. Let a and α be the linear and the angular acceleration as shown in the figure. As disc is rolling without slipping, a and α are related by a = αR.

(1)

Apply Newton’s second law in the horizontal direction to get 2kx − f = M a.

(2)

The torque τ about the axle passing through the centre of mass is related to α by τ = Iα, where I is the moment of inertia of the disc about its axis of rotation. Substitute τ = f R and I = M R2 /2 in τ = Iα to get f R = αM R2 /2.

(3)

Eliminate a and α from equations (1)–(3) to get f = 2kx/3. The net external force is given by Fext = −2kx + f = −2kx + 2kx/3 = −4kx/3. Ans. D

Kf = 21 M V02 + 12 IΩ2 = 12 M V02 +

1 2

2 1 2MR




2

(V0 /R) = 34 M V02 ,

where angular velocity Ω = V0 /R as the disc is rolling without slipping. The total potential energy of the two springs is  2
g . Ui = 2 21 kx2max = k 3µM 2k The conservation of energy, Ui + Ki = Uf + Kf , gives p V0 = µg 3M/k. Ans. C Matrix or Matching Type Q 35. A particle of unit mass is moving along the xaxis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in Column I (a and U0 are constants). Match the potential energies in Column I to the corresponding statement(s) in Column II. (2015)

142

Part I. Mechanics

Column I h
2 i2 (A) U1 (x) = U20 1 − xa (B) U2 (x) = U20


x 2 a

U0 2


x 2 a

(C) U3 (x) =

(D) U4 (x) = U20

h

x a

h exp −

−

1 3

Column II


x 2 a

i

(r) The force acting on the particle is zero at x = −a.


i x 3

(s) The particle experiences an attractive force towards x = 0 in the region |x| < a. (t) The particle with total energy U0 can oscillate 4 about the point x = −a.

a

Sol. The force F (x) on a particle having a potential energy U (x) is given by F (x) = −dU/dx. In case (A), the potential energy and the force on the particle are  2 U0 x2 U1 (x) = 1− 2 , 2 a   2U0 x x2 F1 (x) = 1− 2 . (1) a a

−a

O F1 (x)

a

x

O

a

x

From equation (1), the force F1 (x) = 0 at the points x = −a, x = 0, and x = a. The variation of the potential energy and the force on the particle with respect to x are shown in the figure. The potential energy has a local maximum at x = 0 with value U1 (x = 0) = U0 /2. Thus, the particle experiences a repulsive force at/near x = 0. The particle with total energy U0 /4 (less than U0 /2) can oscillate about the point x = −a (if x < 0 initially) or x = +a (if x > 0 initially). In case (B), the potential energy and the force on the particle are U0 x2 , U2 (x) = 2 a2 U0 x F2 (x) = − 2 . a

−a

−a

O F2 (x)

a

x

O

a

x

From equation (2), the force F2 (x) = 0 at the point x = 0. The potential energy has a minimum at x = 0. Thus, the particle experiences an attractive force at/near x = 0 in the region |x| < a. The particle particle can oscillate about the point x = 0 only. Note that the given potential energy corresponds to SHM about x = 0. We encourage you to find the properties of this SHM if the particle is released from x = a. In case (C), the potential energy and the force on the particle are U0 x2 −x2 /a2 e , 2 a2   2 2 U0 x x2 F3 (x) = − 2 1 − 2 e−x /a . a a

U3 (x) =

−a

O F3 (x)

a

x

O

a

x

From equation (3), the force F3 (x) = 0 at the point x = −a, x = 0, and x = a. The variation of the potential energy and the force on the particle w.r.t. x are shown in the figure. The potential energy has a minimum at x = 0 and maxima at x = −a and x = a. Thus, the particle experiences an attractive force at/near x = 0 in the region |x| < a. The particle can oscillate about the point x = 0 only. In case (D), the potential energy and the force on the particle are   U0 x 1 x3 U4 (x) = − , 2 a 3 a3   U0 x2 F4 (x) = − 1− 2 . (4) 2a a

U0 4

−a

U4 (x) O

U0 3

a

x

a

x

F4 (x)

(2)

(3)

U3 (x)

−a

U1 (x)

−a

U2 (x)

(p) The force acting on the particle is zero at x = a. (q) The force acting on the particle is zero at x = 0.

−a

O

Chapter 10. Simple Harmonic Motion

143

From equation (4), the force F4 (x) = 0 at the point x = −a and x = a. The potential energy has a local minimum at x = −a and a local maximum at x = a. Thus, the particle experiences an attractive force at/near x = −a and a repulsive force at/near x = a. The potential energy at x = a is U4 (x = a) = U0 /3. The particle with total energy U0 /4 (less than U0 /3) can oscillate about the point x = −a. You can visualize the attractive and repulsive nature of the forces by rolling a particle on a surface that has the shape of the potential energy curve. The forces are repulsive at the top of the hill and attractive at the bottom of the valley. The motion can be periodic if particle is confined in a potential well (valley). Ans. A7→(p, q, r, t), B7→(q, s), C7→(p,q,r, s), D7→(p, r, t) Q 36. Column I gives a list of possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphs given in Column II. (2008) Column I

through the origin if a = 0 and y0 6= 0, and a parabola if a 6= 0.

The range of a projectile thrown with an initial velocity u making an angle θ with the horizontal is given by

R = u2 sin 2θ/g.

Hence, u-R curve is a parabola.

Column II

(A) Potential energy of a simple pendulum (y axis) as a function of displacement (x axis).

(p)

(B) Displacement (y axis) as a function of time (x axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive x direction.

(q)

(C) Range of a projectile (y axis) as a function of its velocity (x axis) when projected at a fixed angle.

(r)

(D) The square of time period (y axis) of a simple pendulum as a function of its length (x axis).

(s)

The square of time period of a simple pendulum of length l is given by

y x O y x O

T 2 = l/(4π 2 g). y x O y x O

Sol. The potential energy of a simple pendulum of length l and mass m is

This has a linear relationship with l and this line passes through the origin.

Ans. A7→p, B7→(q,r,s), C7→s, D7→q

U = mgl(1 − cos θ) ≈ mglθ2 /2, where we have assumed the angular displacement θ to be small. This is the equation of a parabola symmetric about θ = 0. The displacement of a particle moving with a constant acceleration is y = y0 + v0 t + 12 at2 . This is the equation of a straight line passing through the origin if a = 0 and y0 = 0, straight line not passing

Q 37. Column I describes some situations in which a small object moves. Column II describes some characteristics of these motions. Match the situations in Column I with the characteristics in Column II. (2007)

144

Part I. Mechanics Column I

Column II

(A) The object moves on the x-axis under a conservative force in such a way that its speed and √ position satisfy v = c1 c2 − x2 , where c1 and c2 are positive constants. (B) The object moves on the x-axis in such a way that its velocity and its displacement from the origin satisfy v = −kx, where k is a positive constant. (C) The object is attached to one end of a massless spring of a given spring constant. The other end of the spring is attached to the ceiling of an elevator. Initially everything is at rest. The elevator starts going upwards with a constant acceleration a. The motion of the object is observed from the elevator during the period it maintains this acceleration. (D) The object is projected from the earth’s surface vertically p upwards with a speed 2 GMe /Re , where, Me is the mass of the earth and Re is the radius of the earth. Neglect forces from objects other than the earth.

(p) The object executes a SHM.

(q) The object does not change its direction.

Fill in the Blank Type (r) The kinetic energy of the object keeps on decreasing.

Q 38. An object of mass 0.2 kg executes simple harmonic oscillations along the x-axis with frequency of 25/π Hz. At the position x = 0.04, the object has kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillations is. . . . . . m. (1994) Sol. Total energy of a particle of mass m in SHM with amplitude A and frequency ν is given by T = 21 kA2 = 12 mω 2 A2 = 2π 2 mν 2 A2 .

(1)

The conservation of energy gives (s) The object can change its direction only once.

T = K + U = 0.5 + 0.4 = 0.9 J.

(2)

Solve equations (1) and (2) to get  A=

T 2π 2 mν 2

1/2

 =

0.9 2π 2 (0.2)(25/π)2

1/2

= 0.06 m.

√ Sol. In case (A), v = dx/dt = c1 c2 − x2 . Differentiate to get d2 x c1 dx = √ (−2x) = −c21 x. dt2 dt 2 c2 − x2 The equation (1) represents a SHM. In case (B), v = dx/dt = −kx. Integrate, Z t Z x dx = −k dt, x0 x 0 to get x = x0 e−kt . The kinetic energy is 2 1 2 mv

As shown in the figure, the object will not change its direction and its kinetic energy will keep on decreasing. Case (C) is a spring-mass system. It will have SHM with mean position x0 . The forces acting on the mass at the mean position are upward spring force kx0 and downward weight mg. Apply Newton’s second law to get, ma = kx0 − mg, which gives x0 = m(a + g)/k. In case (D), the object is projected with escape velocity. It will not change direction and its kinetic energy will keep on decreasing. Ans. A7→p, B7→(q, r), C7→p, D7→(q, r)

= 12 mk 2 x2 = 12 mk 2 x0 e−2kt = K0 e−2kt . x0 K0 x K t

(1)

Ans. 0.06 Q 39. Two simple harmonic motions are represented by the equations y1
= 10 sin (3πt + π/4) and y2 = √ 5 sin 3πt + 3 cos 3πt . Their amplitudes are in the ratio of . . . . . . (1986) Sol. The amplitude of SHM y1 = 10 sin (3πt + π/4) is 10. Rewrite the displacement of second SHM as   √ y2 = 5 sin 3πt + 3 cos 3πt ! √ 1 3 = 10 sin 3πt + cos 3πt 2 2 = 10 sin(3πt + 60◦ ). Thus, the amplitudes of both the SHM are 10. We encourage you to show that y2 is a superposition of two SHMs with a phase difference of 90◦ between them. Ans. 1 : 1

Chapter 10. Simple Harmonic Motion

145

Integer Type

Descriptive

Q 40. A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0 N/m and the mass of the block is 2.0 kg. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass 1.0 kg moving with a speed of 2.0 m/s collides elastically with the first block. The collision is such that the 2.0 kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is . . . . . . . (2018)

Q 41. A mass m is undergoing SHM in the vertical direction about the mean position y0 with amplitude A and angular frequency ω. At a distance y from the mean position, the mass detaches from the spring. Assume that the spring contracts and does not obstruct the motion of m. Find the distance y (measured from the mean position) such that the height h attained by the block is maximum. [Aω 2 > g]. (2005)

1 kg

2 m/s

2 kg

y

Sol. Let the spring constant be k = 2 N/m and masses of two blocks be m1 = 1 kg and m2 = 2 kg. Before collision, the speed of the block of mass m1 is v1 = 2 m/s and that of the block of mass m2 is v2 = 0 (at rest). After collision, let the speed of the block of mass m1 be v10 and that of the block of mass m2 be v20 . v1

v10

v20

m1

m2

m

Sol. The mass m will reach the maximum height if it detaches from the spring while it is moving up and is above the mean position.

m m1

m2

A y

y0 Before Collision

After Collision

Consider the two blocks together as a system. In horizontal direction, the external force on the system of two blocks is zero (floor is frictionless and spring force is negligibly small). Thus, the linear momentum of the system in the horizontal direction is conserved i.e., 2 = −v10 + 2v20 .

(1)

The kinetic energy of the system is conserved in elastic collision i.e., 2

2 1 2 mv

= 21 kA2 − 12 ky 2 ,

which gives

m1 v1 + m2 v2 = −m1 v10 + m2 v20 ,

2 2 1 1 2 m1 v1 + 2 m2 v2 2 2 4 = v10 + 2v20 .

Let the block detaches when it is at a distance y above the mean position y0 . The kinetic energy of the mass at the time of detachment is

2

= 12 m1 v10 + 12 m2 v20 , (2)

Solve equations (1) and (2) to get v10 = 2/3 m/s and v20 = 4/3 m/s. After collision, the block of p p mass m2 undergoes SHM with time period T = 2π m2 /k = 2π 2/2 = 2π s. The spring returns to its unstretched position after time t = T /2 = π s. In this time interval, the block of mass m2 returns to its original position but the block of mass m1 keeps moving leftward with a constant speed v10 . Thus, the separation between the two blocks at this instant is equal to the distance traveled by the block of mass m1 in time t i.e., x = v10 t = (2/3)(π) = 2.09 m. Can you show that coefficient of restitution (e) is one in elastic collision? Now, use e = 1 in place of equation (2) to get the result. Ans. 2.09

v 2 = ω 2 (A2 − y 2 ), where ω 2 = k/m. The height of a projectile thrown vertically up with a velocity v from an initial height y is given by h=y+

v2 2g

=y+

2 ω2 2g (A

− y 2 ).

To find y for which h is maximum, we shall have dh/dy = 0, which gives y = g/ω 2 < A (since Aω 2 > g). Ans. g/ω 2 Q 42. Two identical balls A and B, each of mass 0.1 kg, are attached to two identical massless springs. The spring-mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in the figure. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius 0.06 m. Each spring has a natural length of 0.06π m and spring constant 0.1 N/m. Initially, both the balls are displaced by an angle θ = π/6 rad with respect to the diameter PQ of the circle (see figure) and released from rest. (1993)

146

Part I. Mechanics

A P

B

6m 0.0 π

π 6

6

Q

Thus, the initial potential energy of the system is Ui = U1,i + U2,i = 4kR2 θ2 and the initial kinetic energy is Ki = K1,i + K2,i = 0. In the final position (balls at P and Q), each spring is in its natural unstretched state giving Uf = U1,f + U2,f = 0. Final kinetic energy of the system is Kf = K1,f + K2,f = 21 mv 2 + 12 mv 2 = mv 2 ,

(a) Calculate the frequency of oscillation of ball B. (b) Find the speed of ball A when A and B are at the two ends of the diameter PQ. (c) What is the total energy of the system?

where v is the velocity of each ball. The conservation of energy, Ui + Ki = Uf + Kf , gives v = 2Rθ

Sol. Given, mass of each ball m = 0.1 kg, spring constant of each spring k = 0.1 N/m, natural length of each spring l = 0.06π m and radius of the circle R = 0.06 m. The length l is equal to the half of the circle perimeter i.e., l = πR. Thus, the equilibrium positions of the ball A and the ball B are at P and Q, respectively.

B θ

R θ

s

P

F Q

Let θ be the angular displacement of each ball from its equilibrium position. In this condition, one of the spring is compressed by length Rθ + Rθ = 2Rθ and other is extended by the same length. The forces on the ball B by the two springs are in the same direction providing tangential restoring force

k/m

= 2(0.06)(π/6)

p

0.1/0.1 = 0.0628 m/s.

Total energy of the system is E = Ui = Kf = 4kR2 θ2 = 4(0.1)(0.06)2 (π/6)2 = 3.9 × 10−4 J.

at

F1 A F2

p

Ans. (a)

1 π

Hz (b) 0.0628 m/s (c) 3.9 × 10−4 J

Q 43. Two light springs of force constant k1 and k2 and a block of mass m are in one line AB on a smooth horizontal table such that one end of each spring is fixed on rigid supports and the other end is free as shown in the figure. The distance CD between the free ends of the springs is 60 cm. If the block moves along AB with a velocity 120 cm/s in between the springs, calculate the period of oscillation of the block. [Take k1 = 1.8 N/m, k2 = 3.2 N/m, m = 200 g.] (1985) 60 cm k1

m •

F = F1 + F2 = k(2Rθ) + k(2Rθ) = 4kRθ.

A

C

k2 v D

B

The tangential acceleration of the ball B is given by at =

d2 s d2 (Rθ) d2 θ = =R 2. 2 2 dt dt dt

Apply Newton’s second law to get mR d2 θ/dt2 = −4kRθ, which can be written as d2 θ/dt2 = − (4k/m) θ = −ω 2 θ. This equation represents a SHM with frequency ν=

1 p 1 p 1 ω = 4k/m = 4(0.1)/0.1 = Hz. 2π 2π 2π π

In the initial position, θ = π/6, and each ball is at rest. The potential energy of each spring is U1,i = U2,i = 12 k(2Rθ)2 = 2kR2 θ2 .

Sol. The block travels a distance d = 60 cm between the points C and D with a constant speed v = 120 cm/s. Thus, the time taken by the block in moving from C to D and from D to C are tCD = tDC = d/v = 60/120 = 0.5 s.

(1)

After striking the spring of spring constant k2 = 3.2 N/m at D, the block moves as if it is doing SHM with the point D as the mean position. The time taken by the block in traveling from D to the point of maximum compression of the spring and then back to D is equal to the half of time period of SHM i.e., r r 2π m 0.2 T2 t2 = = = 3.14 = 0.785 s, (2) 2 2 k2 3.2 where m = 200 g is mass of the block. Similarly, the time taken by the block in traveling from C to the point

Chapter 10. Simple Harmonic Motion of maximum compression of the spring and then back to C is r r 0.2 T1 2π m t1 = = = 3.14 = 1.047 s. (3) 2 2 k1 1.8 The equations (1)–(3) give the period of oscillation of the block as T = tCD + t2 + tDC + t1 = 0.5 + 0.785 + 0.5 + 1.047 = 2.832 s. We encourage you to find the maximum compression of the two springs. Ans. 2.832 s Q 44. Two masses m1 and m2 are suspended together by a massless spring of spring constant k (see figure). When the masses are in equilibrium, m1 is removed without disturbing the system. Find the angular frequency and the amplitude of oscillation of m2 . (1981)

m2 m1

Sol. Let the natural length of the spring of spring constant k be l. Let x1 be the elongation of the spring when both masses m1 and m2 are suspended from it and x2 be the elongation when only mass m2 is suspended from it. l x2

m 2 Q x1 P m2 m1

In equilibrium, weight of the suspended mass is balanced by the spring force. Apply equilibrium condition to get x2 = m2 g/k, x1 = (m1 + m2 )g/k. When the mass m1 is slowly removed, the mass m2 undergoes SHM around its equilibrium position. The amplitude of oscillation is equal to the initial displacement of m2 i.e., distance of m2 from its equilibrium position when the mass m1 is just removed. Thus, the amplitude of SHM is A = PQ = x1 − x2 = (m1 + m2 )g/k − m2 g/k = m1 g/k. p The frequency of SHM is given by ωq = k/m2 . Ans. ω = mk2 , A = mk1 g

147 Q 45. A mass m attached to a spring oscillates with a period of 2 s. If the mass is increased by 2 kg the period increases by 1 s. Find the initial mass m assuming Hooke’s law is obeyed. (1979) Sol. The time period of a spring-mass system of p mass m and the spring constant k is given by T = 2π m/k. Substitute the values for the two cases to get p (1) 2 = 2π m/k, p 3 = 2π (m + 2)/k, (2) Divide equation (1) by (2) and simplify to get m = 1.6 kg. Ans. 1.6 kg

Chapter 11 Fluid Mechanics

One Option Correct

2R

Q 1. A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in water in half-submerged state. If ρc is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is (2012) (A) more than half-filled if ρc is less than 0.5. (B) more than half-filled if ρc is more than 1.0. (C) half-filled if ρc is more than 0.5. (D) less than half-filled if ρc is less than 0.5.

B A h

C D

(A) (B) (C) (D)

Sol. For cylindrical shell, let inner radius be r1 , outer radius r2 , height h, and material density ρ.

|2P0 Rh + πR2 ρgh − 2RT | |2P0 Rh + Rρgh2 − 2RT | |P0 πR2 + Rρgh2 − 2RT | |P0 πR2 + Rρgh2 + 2RT |

Sol. The forces acting on one side of the section ABCD by water on the other side of this section are caused by hydrostatic pressure and surface tension. To calculate the force due to hydrostatic pressure, consider a small strip of thickness dx at a depth x.

r2 r1 h

2R x

A

h 2

B x

h

Let the cylinder be filled with water of density ρw upto a height x. Given ρ/ρw = ρc . The cylinder is in half submerged state i.e., height h/2 of the cylinder is inside the water. In equilibrium condition, weight of the cylinder (including the water inside it) is equal to the upthrust on it i.e.,

C

D

The pressure at the depth x is P = P0 + ρgx and the force on the strip is dF = P (2Rdx) = 2R(P0 + ρgx)dx. Integrate dF from x = 0 to x = h to get the total force due to hydrostatic pressure on section ABCD Z h F = 2R(P0 + ρgx)dx = 2RP0 h + Rρgh2 .

(πr22 − πr12 )h ρg + πr12 x ρw g = πr22 (h/2) ρw g. Simplify to get  2  r x = h 22 (0.5 − ρc ) + ρc r  1  r22 − r12 = h 0.5 + (0.5 − ρc ) . r12

dx

0

The force due to surface tension acts only on the surface so it will act on side AB and have magnitude 2T R. This force will try to pull the side AB towards other side of the surface i.e., it is opposite to the force due to hydrostatic pressure. Thus, the magnitude of the desired force is |2RP0 h + Rρgh2 − 2T R|. Ans. B

(1)

The equation (1) gives x > 0.5h if ρc < 0.5 (∵ r2 > r1 ). Ans. A Q 2. Water is filled up to a height h in a beaker of radius R as shown in the figure. The density of water is ρ, the surface tension of water is T and the atmospheric pressure is P0 . Consider a vertical section ABCD of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude (2007)

Q 3. Water is filled in a beaker upto a height of 3 m. An orifice (hole) is at a height of 52.5 cm from the bottom of beaker (see figure). The ratio of cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is [Take g = 10 m/s2 .] (2005) 148

Chapter 11. Fluid Mechanics

149

52.5 cm

3m

Let l changes to l0 when coin falls into the water. In equilibrium, weight of the block is balanced by the buoyancy force i.e., ρb Vb g = Ab l0 ρw g.

(2)

Subtract equation (2) from (1) to get 2

(A) 50 (m/s) (C) 51 (m/s)2

(B) 50.5 (m/s) (D) 52 (m/s)2

2

Ab (l − l0 ) = ρc Vc /ρw .

Sol. Let A1 be the cross-sectional area of the beaker and v1 be the flow velocity at a height h1 = 3 m. Let A2 be the cross-sectional area of the orifice and v2 be the flow velocity at the orifice located at a height h2 = 0.525 m. Using continuity equation on an imaginary streamline from top of the beaker to the orifice, A1 v1 = A2 v2 , we get v1 = (A2 /A1 )v2 .

(1)

Bernoulli’s equation, P0 + ρgh1 + 12 ρv12 = P0 + ρgh2 + 1 2 2 ρv2 , gives v22 = v12 + 2g(h1 − h2 ).

(2)

Eliminate v1 from equations (1) and (2) to get v22 =

2(10)(3 − 0.525) 2g(h1 − h2 ) = = 50 (m/s)2 . 1 − (A2 /A1 )2 1 − 0.01 Ans. A

Q 4. A wooden block, with a coin placed on its top, floats in water as shown in the figure. The distance l and h are shown there. After some time the coin falls into the water. Then, (2002)

(3)

The equation (3) gives l0 < l. Total volume of the coin, water, and immersed portion of the block after the coin falls into the water is Aw h0 = Vc + Vw + Ab l0 = Vc + (Aw h − Ab l) + Ab l0 = Aw h + Vc − Ab (l − l0 ) = Aw h + Vc − ρc Vc /ρw   = Aw h − Vρcwρc 1 − ρρwc .

(4)

The equation (4) gives h0 < h because ρw < ρc . We encourage you to use the above analysis in formulating a qualitative solution. Ans. D Q 5. A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and mass M . It is suspended by a string in a liquid of density ρ, where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is (2001)

h Coin ρ l h

(A) (B) (C) (D)

2R

(A) M g (C) M g + πR2 hρg

l decreases and h increases. l increases and h decreases. Both l and h increase. Both l and h decrease.

Sol. Let ρc , ρb , and ρw be the densities and Vc , Vb , and Vw be the volumes of the coin, block, and water. Let Ab and Aw be the cross-section areas of the block and the water container. The volume of water displaced when the block (with coin above it) is placed in water is Ab l. The volume of the water in the beaker is Vw = Aw h − Ab l. In equilibrium, weights of the block and coin are balanced by the buoyancy force i.e., ρc Vc g + ρb Vb g = Ab lρw g.

(1)

(B) M g − V ρg (D) ρg(V + πR2 h)

Sol. The upthrust (Fupthrust ) by the liquid on the cylinder is equal to the weight of the liquid displaced by the cylinder i.e., Fupthrust = ρV g. The Fupthrust is the resultant force applied by the liquid on the cylinder. It is resultant of the upward force F1 on the bottom surface and the downward force F2 = πR2 ρhg on the top surface. Thus, Fupthrust = F1 − F2 , which gives F1 = ρgV + πR2 ρhg. Aliter: Let H be the total height of the cylinder. The volume of the cylinder after removing the hemisphere is V = πR2 H − 32 πR3 .

150

Part I. Mechanics Q 7. A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of gravity. Then, the pressure in the compartment is (1999) (A) same everywhere (B) lower in front side (C) lower in rear side (D) lower in upper side

h

dF Rcos θ

H θ

dθ

Sol. Let A be the cross-section area and ρ be the gas density. Consider a small element of width dx at a distance x.

R

Consider a small element of the hemisphere with angle from the vertical varying from θ to θ + dθ. The radius of this element is R sin θ, its width is Rdθ and its area is 2πR2 sin θdθ. The pressure at the depth of this element is (h + H − R cos θ)ρg and radial force due to this pressure is dF = 2πR2 ρg(h + H − R cos θ) sin θdθ. Resolve dF in the horizontal and the vertical directions. By symmetry, horizontal component of force integrates to zero. The vertical component of the force is

p

p+dp ~a

x

dx

Let pressure at the distance x be p and that at distance x + dx be p + dp. The net force on the element due to pressure difference is dF = [pA − (p + dp)A] = −Adp, (rightwards). This force provides acceleration a to the element. Apply Newton’s second law on the element,

dFV = dF cos θ = 2πR2 ρg(h + H − R cos θ) sin θ cos θdθ.

−Adp = ma = (ρAdx)a,

Integrate dFV from θ = 0◦ to θ = 90◦ to get FV = πR2 ρg(h + H) − 32 πρgR3 = πR2 ρgh + ρV g. We encourage you to find the tension in the string. Ans. D Q 6. A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal to (2000)

(A)

√L 2π

(B) 2πL (C) L (D)

L 2π

Sol. The areas of the square hole of side L and the circular hole of radius R are As = L2 ,

Ac = πR2 .

√ The velocity of efflux at a depth h is given by v = 2gh. Thus, the velocities of efflux at the square and circular hole are p p p vc = 2g(4y) = 2 2gy. vs = 2gy,

to get dp/dx = −ρa. Negative sign indicates that the pressure decreases with x i.e., as we go from the left to the right. Ans. B Q 8. Water from a tap emerges vertically downwards with an initial speed of 1.0 m/s. The cross-sectional area of tap is 10−4 m2 . Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of stream 0.15 m below the tap is (1998) (A) 5.0 × 10−4 m2 (B) 1.0 × 10−4 m2 (C) 5.0 × 10−5 m2 (D) 2.0 × 10−5 m2 Sol. Let v1 = 1.0 m/s, A1 = 10−4 m2 , h = 0.15 m, v2 be the velocity at the depth h, and A2 be the crosssectional area of stream at the depth h. A1 v1 h

The volume flow rate for both the holes is same i.e., A2

As v s = Ac v c .

(1)

Substitute area √ and velocity of efflux in equation (1) to get R = L/ 2π. √ We encourage you to derive the velocity of efflux, v = 2gh, by using Bernoulli’s equation. Ans. A

v2

The continuity equation, A1 v1 = A2 v2 , gives A2 = A1 v1 /v2 .

(1)

Chapter 11. Fluid Mechanics

151

Bernoulli’s equation, p1 + ρgh1 + 12 ρv12 = p2 + ρgh2 + 12 ρv22 , with p1 = p2 gives v22 = v12 + 2g(h1 − h2 ) = v12 + 2gh.

(2)

Substitute v2 from equation (2) into equation (1) to get A2 = p

A1 v 1 v12 + 2gh

=p

(10−4 )(1) (1)2 + 2(10)(0.15)

= 5.0 × 10−5 m2 . Ans. C Q 9. A homogeneous solid cylinder of length L and cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure P0 . Then, density D of solid is given by

Q 10. A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density ρ at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is (1990)  1/2  1/2 k−Aρg k+Aρg 1 1 (B) 2π (A) 2π M M    1/2 1/2 2 k+ρgL k+Aρg 1 1 (C) 2π (D) M 2π Aρg Sol. The volume of the cylinder submerged in the liquid in equilibrium condition is 21 LA. Let x0 be the spring extension in the equilibrium condition. The forces acting on the cylinder are its weight M g, upthrust 12 LAρg and spring force kx0 .

(1995)

d

2d

(A)

5 4d

(B)

4 5d

(C) 4d (D)

d 5

In equilibrium, net force on the cylinder is zero i.e.,

Sol. Let A be the topmost point of the cylinder, B be the point on the interface, and C be the lowest point of the cylinder. P0 A

h 3L 4 L 4

d

kx0 + 12 LAρg = M g. Let the cylinder is displaced further down by a distance x. Now, the forces acting
on the cylinder are its weight M g, upthrust L2 + x Aρg and spring force k(x0 + x). Net downward force on the cylinder is

B C

F = M g − k(x0 + x) −

2d

L 2


+ x Aρg


= M g − kx0 − 12 LAρg − (k + Aρg)x Let the depth of point A be h. The pressures at A, B and C are PA = P0 + hdg, L 4 (2d)g = PA

Newton’s second law gives d2 x F k + Aρg = =− x = −ω 2 x. 2 dt M M

PB = PA + 34 Ldg, PC = PB +

= −(k + Aρg)x.

+ 54 Ldg.

The forces on the cylinder are its weight W = D(LA/5)g (downwards), hydrostatic force at the top surface FA = PA (A/5) (downwards), and hydrostatic force at the bottom surface FC = PC (A/5) (upwards). In equilibrium, FA + W = FC i.e., PA A/5 + D(LA/5)g = (PA + 54 Ldg)A/5. Simplify to get D = 5d/4. Ans. A

This equation represents a SHM with frequency ν = q k+Aρg 1 ω 2π = 2π M . Ans. B Q 11. A vessel contains oil (density = 0.8 g/cm3 ) over mercury (density = 13.6 g/cm3 ). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere (in g/cm3 ) is (1988) (A) 3.3 (B) 6.4 (C) 7.2 (D) 12.8

152

Part I. Mechanics

Sol. Let V be the volume of the sphere and ρ be its density. Half of the sphere is in mercury of density ρHg = 13.6 g/cm3 and another half is in oil of density ρoil = 0.8 g/cm3 . The forces on the sphere are its weight V ρg (downwards), buoyant force due to mercury V ρHg g/2 (upwards), and the buoyant force due to oil V ρoil g/2 (upwards) as shown in the figure. V 2

ρoil g ρoil

(A) (B) (C) (D)

ρ ρHg V 2

Q 13. A body floats in a liquid contained in a beaker. The whole system as shown in the figure, falls freely under gravity. The upthrust on the body is (1982)

ρHg g V ρg

The sphere floats if the net force on it is zero i.e.,

zero. equal to the weight of the liquid displaced. equal to the weight of the body in air. equal to the weight of the immersed portion of the body.

Sol. The forces acting on the body are its weight mg and the upthrust F .

V ρg = V ρHg g/2 + V ρoil g/2. F

Substitute the values to get ρ = ρHg /2 + ρoil /2 = 7.2 g/cm3 . Ans. C Q 12. A U-tube of uniform cross-section is partially filled with a liquid I. Another liquid II which does not mix with liquid I is poured into one side. It is found that the liquid levels of the two sides of the tube are same, while the level of liquid I has risen by 2 cm. If the specific gravity of the liquid I is 1.1, the specific gravity of liquid II must be (1983)

g

mg

The downward acceleration of the body in free fall is g. Apply Newton’s second law of motion to get mg − F = ma = mg,

i.e.,

F = 0. Ans. A

II

Q 14. A vessel containing water is given a constant acceleration a towards the right along a straight horizontal path. Which of the following diagrams represent the surface of the liquid? (1981)

I

(A) (A) 1.12 (B) 1.1 (C) 1.05 (D) 1.0 Sol. If the level of liquid I rises by height h in the right limb then it must fall by the same height in the left limb (because volume of the liquid is constant and U-tube is of uniform cross-section).

(D)

None

a

I B

N cos θ

I

h A

(C)

a

Sol. The surface of the liquid will have a shape as shown in the figure.

h h II

(B) a

Thus, the final configuration of the two liquids will be as shown in the figure. Let atmospheric pressure be p0 . The hydrostatic pressure at the point A and at the point B should be equal because these points are at the same level in liquid I. Hence, p0 + (2h)ρII g = p0 + (2h)ρI g, i.e., ρII = ρI = 1.1. Ans. B

N = θ

a N sin θ

mg

θ

Consider a small element of mass m on the surface. The forces acting on this element are its weight mg and normal reaction N . Note that the reaction force

Chapter 11. Fluid Mechanics

153

cannot have a component tangential to the liquid surface (otherwise element will start moving tangentially). Let liquid surface makes an angle θ with the horizontal. Resolve N in the horizontal and the vertical directions. Apply Newton’s second law on the element in the horizontal and the vertical directions to get N cos θ = mg,

(1)

N sin θ = ma.

(2)

Eliminate N from equations (1) and (2) to get θ = tan−1 (a/g). Ans. C One or More Option(s) Correct Q 15. A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P (r) is the pressure at r(r < R), then the correct option(s) is (are) (2015) (r=3R/4) 63 = (A) P (r = 0) = 0 (B) P P (r=2R/3) 80 (C)

P (r=3R/5) P (r=2R/5)

=

16 21

(D)

P (r=R/2) P (r=R/3)

=

20 27

Sol. The acceleration due to gravity at a radial distance r (r < R) from the centre of a sphere of constant mass density ρ is given by G( 43 πr3 )ρ r2

4πρG r. 3 Consider a spherical shell of radius r and thickness dr. g=

=

dP

dr

P

P

+

A r O

Thus, P (0) = 23 πGρ2 R2 . The ratio of pressures at the desired points are 63 P (3R/4) R2 − (3R/4)2 = = 2 , 2 P (2R/3) R − (2R/3) 80 16 R2 − (3R/5)2 P (3R/5) = = 2 , 2 P (2R/5) R − (2R/5) 21 P (R/2) 27 R2 − (R/2)2 = = 2 . 2 P (R/3) R − (R/3) 32 Ans. B, C Q 16. A solid sphere of radius R and density ρ is attached to one end of a massless spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3ρ. The complete arrangement is placed in a liquid of density 2ρ and is allowed to reach equilibrium. The correct statement(s) is (are) (2013) 3 ρg . (A) the net elongation of the spring is 4πR 3k 3

ρg (B) the net elongation of the spring is 8πR . 3k (C) the light sphere is partially submerged. (D) the light sphere is completely submerged.

Sol. Let the height of the container is large enough. Consider the two spheres and the mass-less spring as a system. The external forces on this system are (i) Weight of the heavier sphere in downward direction (ii) Weight of lighter sphere in the downward direction, (iii) Upthrust of the liquid on both the spheres. Here, we have not considered the spring force as it is internal to the system. ρ

R

2ρ 3ρ

Let P and P + dP be the fluid pressures inside and outside the shell. Consider a small element of area A and mass dm = ρAdr on the shell. The forces on this element are gravitational force (dm g) directed inwards, the force due to pressure of the fluid outside the shell ((P + dP )A) directed inwards, and the force due to pressure of the fluid inside the shell (P A) directed outwards. In equilibrium, net force on the element is zero i.e., (ρA dr) 43 πρGr + (P + dP )A = P A, or dP = − 43 πGρ2 r dr. Note that the pressure decreases with increase in r. The pressure is zero at the surface of the sphere i.e., P = 0 at r = R. Integrate to get Z P (r) Z r rdr dP = P (r) = − 43 πGρ2 R

0

= 23 πGρ2 (R2 − r2 ).

The heavier sphere (density 3ρ) will move downwards pulling the lighter sphere (density ρ) along with it. Let volume V of the lighter sphere is submerged in liquid. In equilibrium, the net force on the system is zero i.e., 3 4 3 πR (3ρ)g

+ 34 πR3 (ρg) = 43 πR3 (2ρ)g + V (2ρ)g.

Solve to get V = 43 πR3 . Thus, lighter sphere will be completely inside the liquid in the equilibrium. Let the elongation of the spring be x in the equilibrium condition. As the two sphere system is in equilibrium, two spheres are also in equilibrium separately. Newton’s second law on both the spheres give 3 3 4 4 3 πR (ρ)g + kx = 3 πR (2ρ)g 3 3 4 4 3 πR (2ρ)g + kx = 3 πR (3ρ)g.

154

Part I. Mechanics

We get same equation for both the spheres. Solve to get x = 4πR3 ρg/(3k). Ans. A, D Q 17. Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a string. They are fully immersed in a fluid of density dF . They get arranged into equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if (2011)

Sol. Let us assume that the balances A and B are not biased. The mass of the block is m = 2 kg and mass of the beaker (with liquid) is M = 5 kg. When placed in the liquid, an upward buoyant force acts on the block. This will reduce the reading of the balance A. By Newton’s third law, an equal and opposite force acts on the liquid. This force will increase the reading of the balance B. Ans. B, C Paragraph Type

A

Paragraph for Questions 19-20

B

(A) dA < dF (C) dA > dF

(B) dB > dF (D) dA + dB = 2dF

Sol. Let V be the volume of both the spheres. The forces acting on the sphere A are its weight (dA V g, downwards), buoyant force (dF V g, upwards), and tension in the string (T , downwards). In equilibrium, T + dA V g = dF V g.

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere. (2014)

(1)

The forces acting on the sphere B are its weight (dB V g, downwards), buoyant force (dF V g, upwards), and tension in the string (T , upwards). In equilibrium, (2)

Q 19. If the piston is pushed at a speed of 5 mm/s, the air comes out of the nozzle with a speed of (A) 0.1 m/s (B) 1 m/s (C) 2 m/s (D) 8 m/s

Use T > 0 in equations (1) and (2) to get dA < dF and dB > dF . Ans. A, B, D

Sol. The speed of air inside the piston (spray gun) is equal to the speed of the piston i.e., v1 = 5 mm/s. The area of the piston is A1 = πr12 and that of the nozzle is A2 = πr22 , where r1 = 20 mm and r2 = 1 mm. The continuity equation, A1 v1 = A2 v2 , gives speed of the air at the nozzle as

T + dF V g = dB V g. Eliminate T from equations (1) and (2) to get dA + dB = 2dF .

Q 18. The spring balance A reads 2 kg with a block m suspended from it. A balance B reads 5 kg when a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in the figure. In this situation, (1985) A

m B

(A) the balance A will read more than 2 kg. (B) the balance B will read more than 5 kg. (C) the balance A will read less than 2 kg and B will read more than 5 kg. (D) the balance A and B will read 2 kg and 5 kg, respectively.

v2 =

A1 v 1 π(20)2 (5) = 2000 mm/s = 2 m/s. = A2 π(1)2 Ans. C

Q 20. If the density of air is ρa and that of the liquid is ρl , then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to q q √ ρa ρl (B) ρ ρ (C) (A) a l ρl ρa (D) ρl Sol. Let the speed of the piston be va,1 . The continuity 1 equation gives speed of air at the nozzle, va,2 = A A2 va,1 , a constant for fixed va,1 . The air expands when it comes out to the atmosphere (pressure P0 ) and finally its speed reduces to zero. P2 P0

P0

Chapter 11. Fluid Mechanics

155

Let pressure of the air at the nozzle be P2 . Bernoulli’s equation for the air gives 2 = P0 . P2 + 12 ρa va,2

(1)

The liquid rises due to pressure difference (P0 − P2 ). Apply Bernoulli’s equation for the liquid in the vertical tube (assume height to be negligibly small) to get P0 = 12 ρl vl2 + P2 .

Q 22. Let the cylinder is prevented from moving up, by applying a force and water level is further decreased. Then, height of water level (h2 in figure) for which the cylinder remains in original position without application of force is 4r h ρ

(2) 2r

Eliminate (P0 − P2 ) from equations (1) and (2) to get 2 1 2 ρ l vl

2 = 21 ρa va,2 ,

h2

(A) 4h/9 (B) 5h/9 (C) h (D) 2h/3

p which gives, vl = va,2 ρa /ρl . If the cross-section area of the tube is A pthen the volume flow rate of the liquid is Avl = Ava,2 ρa /ρl . Ans. A Paragraph for Questions 21-23 A wooden cylinder of diameter 4r, height h and density ρ/3 is kept on a hole of diameter 2r of a tank, filled with liquid of density ρ as shown in the figure. (2006)

Sol. In this case, forces acting on the cylinder are F1 = 4πr2 p0 on its upper surface (downwards), F2 = 3πr2 (p0 + ρgh2 ) on its lower surface exposed to water (upwards), F3 = πr2 p0 on its lower surface exposed to air (upwards) and its weight W = 43 πr2 ρgh (downwards). In equilibrium, F1 + W = F2 + F3 , i.e., 4πr2 p0 + 34 πr2 ρgh = 3πr2 (p0 + ρgh2 ) + πr2 p0 , which gives h2 = 4h/9. Ans. A

4r

h1 h ρ

2r

Q 21. Now, level of the liquid starts decreasing slowly. When the level of liquid is at a height h1 above the cylinder, the block just starts moving up. Then value of h1 is (A) 2h/3 (B) 5h/4 (C) 5h/3 (D) 5h/2 Sol. Let p0 be the atmospheric pressure. When cylinder just starts moving up, the forces acting on it are: F1 = π(2r)2 (p0 + ρgh1 ) on its upper surface, F2 = 3πr2 (p0 + ρg(h1 + h)) on its lower surface exposed to water, F3 = πr2 p0 on its lower surface exposed to air, and its weight W = π(2r)2 h(ρ/3)g.

Sol. If height h2 is further reduced then upward forces will decrease. The cylinder will not move up and remains at its original position. As calculated earlier, if 5h 4h 3 ≤ h1 and h2 ≥ 9 , cylinder tries to move up. Beyond these limits, the upward forces are not sufficient to move it up. Ans. A Assertion Reasoning Type

F1 h1 ρ/3

Q 23. If height h2 of water level is further decreased, then, (A) cylinder will not move up and remains at its original position. (B) for h2 = h/3, cylinder again starts moving up. (C) for h2 = h/4, cylinder again starts moving up. (D) for h2 = h/5, cylinder again starts moving up.

h ρ

F2 W F3

In equilibrium, F1 + W = F2 + F3 , i.e., 4πr2 (p0 +ρgh1 ) + 43 πr2 ρgh = 3πr2 (p0 +ρgh1 +ρgh) + πr2 p0 , which gives h1 = 5h/3. Ans. C

Q 24. Statement 1: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down. Statement 2: In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant. (2008) (A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. (B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true.

156

Part I. Mechanics

Sol. The statement 2 is same as the continuity equation, Av = constant, where A is the cross-section area and v is the fluid velocity. When pipe is held vertically up, the energy conservation makes v to decrease as water goes up and hence A increases by continuity equation. It is the other way around when the pipe is held vertically down. Ans. A Matrix or Matching Type Q 25. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, state of the lift’s motion is given in Column I and the distance where the water jet hits the floor of the lift is given in Column II. Match the statements from Column I with those in Column II.

True False Type Q 26. A block of ice with a lead shot embedded in it is floating on water contained in a vessel. The temperature of the system is maintained at 0 ◦ C as the ice melts. When the ice melts completely the level of water in the vessel rises. (1986) Sol. To begin with, let there is no lead shot in the ice block. The ice block, of density ρi , cross-section area Ai and height hi , is floating in the water of density ρw and cross-section area Aw . ρi hi h hw

Ai Aw

ρw

(2014)

Column I

Column II

(P) Lift is accelerating vertically up. (Q) Lift is accelerating vertically down with an acceleration less than the gravitational acceleration. (R) Lift is moving vertically up with constant speed. (S) Lift is falling freely.

(1) d = 1.2 m (2) d > 1.2 m

(3) d < 1.2 m (4) No water leaks out of the jar.

Sol. The velocity of the efflux depends on the hydrostatic pressure difference and is given by p v = 2geff H.

H h d

After coming out of the jar, water follows the projectile motion. Time taken in traversing a vertical distance h is p t = 2h/geff . The horizontal distance covered in this time is p √ p d = vt = 2geff H 2h/geff = 2 Hh, which is independent of geff . In the case of free fall, hydrostatic pressure difference is zero (geff = 0) and hence the water does not come out. Ans. P7→1, Q7→1, R7→1, S7→4

Let the height h of the ice block is inside the water and the height of the water level is hw . The weight of the block is balanced by the buoyant force, ρi (Ai hi )g = ρw (Ai h)g, i.e., h/hi = ρi /ρw .

(1)

Thus, fraction of the ice inside the water is equal to the ratio of the densities of ice and water. When the ice melts, density changes from ρi to ρw (at constant temperature). The mass conservation, ρi Ai hi = ρw Vw , gives the increase in the volume of the water Vw = ρi Ai hi /ρw . Thus, the height h0w of the water level after the melting of ice is given by Aw h0w = (Aw hw − Ai h) + ρi Ai hi /ρw = Aw h, where we have used equation (1) to replace ρi /ρw by h/hi . Thus, the height of the water level does not change when ice melts. Now, consider the case where lead shot of volume Vl and density ρl is embedded into the ice block. The volume of water displaced by the ice block and lead shot increase to balance the weight of the lead shot i.e., ∆Vw = Vl ρl /ρw . When the ice melts, additional increase in the volume of the water due to the lead shot is ∆Vw0 = Vl . Since ρl > ρw , we get ∆Vw0 < ∆Vw . Hence, height of the water level decreases when the ice block melts. Ans. F Fill in the Blank Type Q 27. A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where crosssectional area is 10 cm2 , the water velocity is 1 m/s and the pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is 5 cm2 , is . . . . . . Pa. [Density of water = 103 kg/m3 .] (1994)

Chapter 11. Fluid Mechanics

157

Sol. The continuity equation, A1 v1 = A2 v2 , gives

Descriptive Q 29. A U-shaped tube contains a liquid of density ρ and it is rotated about the line as shown in the figure. Find the difference in height H of the liquid column. [Assume diameter of the tube d  L.] (2005)

v2 = A1 v1 /A2 = (10)(1)/5 = 2 m/s. Bernoulli’s equation,

ω

p1 + 12 ρv12 + ρgh1 = p2 + 12 ρv22 + ρgh2 , with h1 = h2 gives

H

p2 = p1 + 12 ρ(v12 − v22 ) = 2000 + 21 (1000)(12 − 22 ) = 500 Pa.

L

Ans. 500 Sol. Consider a small element of thickness dx at a distance x from the axis of rotation.

Integer Type

ω

Sol. Let the height of the vessel be H0 = 500 mm, cross-section area be A, and the final height of the water column be h = 200 mm. Let P be the pressure in the vessel (above the water surface) when water stops coming out of the orifice. The water stops coming out when the hydrostatic pressure at the orifice is equal to the atmospheric pressure P0 i.e., P + hρg = P0 .

(1)

Initially, the pressure of air in the vessel is P0 and it decreases to P due to increase in the volume from V0 = (H0 − H)A to V = (H0 − h)A. Appply Boyle’s law on the air inside the vessel, P0 V0 = P V , to get P0 (H0 − H)A = P (H0 − h)A.

(2)

Eliminate P from equations (1) and (2) and simplify to get hρg(H0 − h) P0 (0.2)(1000)(10)(0.5 − 0.2) = = 0.006 m. 105

H −h=

Ans. 6

dx pA

(p+dp)A L

Let p and p + dp be the pressures on the left and the right side of this element. The resultant force acting on this element is 2

2

dF = (p + dp) πd4 − p πd4 =

πd2 4 dp.

This force provides the centripetal acceleration to the element. Using Newton’s second law, π(d2 /4)dp = ρπ(d2 /4)dxω 2 x, we get dp = ρω 2 xdx. Integrate from x = 0 to x = L to get the pressure difference between the two ends Z L p= ρω 2 xdx = ρω 2 L2 /2. 0

This pressure is equal to the hydrostatic pressure by a column of height H i.e., ρω 2 L2 /2 = ρgH, which gives H = ω 2 L2 /(2g). Ans. H =

ω 2 L2 2g

Q 30. Consider a horizontally oriented syringe containing water located at a height of 1.25 m above the ground. The diameter of the plunger is 8 mm and the diameter of the nozzle is 2 mm. The plunger is pushed with a constant speed of 0.25 m/s. Find the horizontal range of water stream on the ground. (2004) 2 mm 1.25 m

(2009)

x

8 mm

Q 28. A cylindrical vessel of height 500 mm has in orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice. [Take atmospheric pressure = 1.0 × 105 N/m2 , density of water = 1000 kg/m3 and g = 10 m/s2 . Neglect any effect of surface tension.]

158

Part I. Mechanics

Sol. Apply continuity equation, A1 v1 = A2 v2 , on the two ends of the syringe with A1 = π(d1 /2)2 = 16π mm2 , v1 = 0.25 m/s and A2 = π(d2 /2)2 = π mm2 , to get v2 = 4 m/s. After coming out of the syringe, water falls like a projectile thrown horizontally with a speed v2 = 4 m/s from a heightph = 1.25 m. The time taken by the projectile is t = 2h/g = 0.5 s and the horizontal distance travelled in this time is v2 t = 2 m. Ans. 2 m Q 31. A solid sphere of radius R is floating in a liquid of density ρ with half of its volume submerged. If the sphere is slightly pushed and released, it starts performing SHM. Find the frequency of these oscillations. (2004)

Sol. In the floating condition, buoyancy force is equal and opposite to the weight of the sphere i.e., mg =

1 2

· 43 πR3 ρg,

which gives m = 23 πR3 ρ. FB

x

mg

Let the sphere be displaced downwards by a small distance x. If x  R then additional volume of liquid displaced is approximately equal to the volume of a cylinder of radius R and height x. Thus, the total volume of the displaced liquid is V = 23 πR3 + πR2 x, and the buoyancy force on the sphere is FB = V ρg = 23 πR3 ρg + πR2 xρg. Apply Newton’s second law to get m

d2 x = mg − 23 πR3 ρg − πR2 xρg = −πR2 xρg, dt2

which can be written as d2 x 3g x = −ω 2 x. = − 2R dt2 This equation represents a SHM with frequency ν = p ω 1 = 3g/(2R). 2π 2π q 3g 1 Ans. 2π 2R Q 32. A uniform solid cylinder of density 0.8 g/cm3 floats in equilibrium in a combination of two non-mixing liquids A and B with its axis vertical. The densities of the liquids A and B are 0.7 g/cm3 and 1.2 g/cm3 , respectively. The height of liquid A is hA = 1.2 cm. The length of the part of the cylinder immersed in liquid B is hB = 0.8 cm. (2002)

Air

h

A

hA

B

hB

(a) Find the total force exerted by liquid A on the cylinder. (b) Find h, the length of the part of the cylinder in air. (c) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is then released. Find the acceleration of the cylinder immediately after it is released. Sol. Let A be the cross-section area of the cylinder, ρC = 1.2 g/cm3 be the density of the cylinder, ρA = 0.7 g/cm3 be the density of liquid A, and ρB = 1.2 g/cm3 be the density of liquid B. We can consider cylinder to be made up of a stack of thin cylindrical discs. By symmetry, net force on the curved surfaces of each disc is zero. Hence, the net force on the cylinder by the liquid A is zero. Similarly, the net force on the curved surface of cylinder by the liquid B is zero. Let p0 be the atmospheric pressure. The hydrostatic pressure at the bottom of the cylinder and the upward force on the cylinder due to this pressure are p = p0 + ρA ghA + ρB ghB , FB = pA = (p0 + ρA ghA + ρB ghB )A. Other forces acting on the cylinder are its weight mg and the force FT due to pressure on the top surface which are given by mg = ρC gA(h + hA + hB ), FT = p0 A. In equilibrium, FT + mg = FB , which gives h = hA = 1.2



ρA ρC 0.7 0.8

   − 1 + hB ρρB − 1 C

− 1 + 0.8 1.2 0.8 − 1 = 0.25 cm.

The forces on the cylinder, when its top face is just below the liquid A, are weight mg, force due to pressure at the top face FT0 = p0 A = FT , and the force due to pressure at the bottom face FB0 = (p0 + ρA ghA + ρB g(h + hB ))A = FB + ρB ghA.

Chapter 11. Fluid Mechanics

159

Net upward force on the cylinder and its acceleration are F = FB0 − mg − FT0 = ρB ghA + (FB − mg − FT ) = ρB ghA, F ρB hg a= = m ρc (h + hA + hB ) (1.2)(0.25)g g = = . 0.8(0.25 + 1.2 + 0.8) 6 Ans. (a) zero (b) 0.25 cm (c) g/6 Q 33. A wooden stick of length L, radius R and density ρ has a small metal piece of mass m (of negligible volume) attached to its one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in equilibrium in a liquid of density σ (> ρ). (1999) Sol. Let A be the lowest point of the stick where mass m is attached. The mass of the stick is M = πR2 Lρ and its centre of mass C is at a distance AC = L/2 from the point A. mw g

Q 34. A non-viscous liquid of constant density 1000 kg/m3 flows in streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross-section of the tube at two points P and Q at heights of 2 m and 5 m are respectively 4 × 10−3 m2 and 8 × 10−3 m2 . The velocity of the liquid at point P is 1 m/s. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q. (1997) Q P 2m

5m

Sol. Let A1 = 4 × 10−3 m2 , v1 = 1 m/s, h1 = 2 m, and p1 be the pressure at the end P. Corresponding parameters at the end Q are A2 = 8 × 10−3 m2 , v2 , h2 = 5 m, and p2 . The continuity equation between P and Q gives

θ

D

The equation (3) is quadratic in m with two solutions √
√ √ for m. Solve this to get mmin = πR2 L ρ σ − ρ . We encourage you to show that the centre of mass of the stick-mass system, say C 0 , lies at B in this limiting case. If m < mmin then C 0 lies above B and the equilibrium becomes unstable. √
√ √ Ans. πR2 L ρ σ − ρ

C B Mg A

v2 = A1 v1 /A2 = 0.5 m/s.

mg

Bernoulli’s equation between P and Q gives

Let the length AD = l of the stick be inside the liquid. The weight of the displaced liquid is mw g = πR2 lσg and the centre of mass of the displaced liquid, B, is at a distance AB = l/2 from A. Let the stick makes an angle θ from the horizontal. The forces acting on the stick are its weight M g acting downwards at C, weight of the attached mass mg acting downwards at A, and the force of buoyancy mw g acting upwards at B. In equilibrium, net force on the stick is zero i.e.,

p1 + 21 ρv12 + ρgh1 = p2 + 12 ρv22 + ρgh2 , p1 − p2 = ρg(h2 − h1 ) + 21 ρ(v22 − v12 ). The work done per unit volume by the pressure as the fluid flows from the end P to the end Q is Wp = p1 − p2 = ρg(h2 − h1 ) + 12 ρ(v22 − v12 ) = (1000)(9.8)(5 − 2) + 12 (1000)(0.52 − 12 ) = 29025 J/m3 . The work done per unit volume by the gravity is

M g + mg = mw g,

Wh = ρg(h1 − h2 ) = −29400 J/m3 .

which gives

Ans. 29025 J/m3 , −29400 J/m3 ,

πR2 (σl − ρL) = m.

(1)

Also, net torque on the stick about its centre of mass C is zero, i.e., mg(L/2) cos θ = mw gBC cos θ = πR2 lσg cos θ(L − l)/2.

(2)

Eliminate l from equations (1) and (2) to get m2 + 2(πR2 Lρ)m + (πR2 Lρ)2 (1 − σ/ρ).

(3)

Q 35. A large open top container of negligible mass and uniform cross-sectional area A has a small hole of crosssectional area A/100 in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density ρ and mass m0 . Assuming that the liquid starts flowing out horizontally through the hole at t = 0. Calculate, (1997) (a) the acceleration of the container. (b) velocity of efflux when 75% of the liquid has drained out.

160

Part I. Mechanics

Sol. The height of the container is given by H = VA = m0 /ρ 0 = m A ρA . The velocity of efflux when container is completely filled with the liquid is p p v = 2gH = 2m0 g/(ρA).

Sol. Let A be the cross-sectional area of the rod. The mass of the rod is m = d1 AL and its weight is mg = d1 ALg. F

A θ L 2

H A 100

mg P

v

F

The mass of the liquid coming out of the hole in one second and the momentum carried by this liquid in one second are ∆m = ρv(A/100) = ρvA/100, ∆p = ∆m v = ρAv 2 /100. The rate of change of momentum of the container is equal to the force acting on the container i.e., F = ∆p/∆t = ∆p/1 = ρAv 2 /100. Newton’s second law gives acceleration of the container as 1 ρA 2m0 g g 1 ρAv F = = . = m0 m0 100 m0 100 ρA 50

The height of the liquid, when 75% of the liquid is drained out is, H 0 = H/4 and the velocity of efflux at that instant is p p p v 0 = 2gH 0 = gH/2 = m0 g/(2ρA). Ans. (a)

g 50

(b)

q

gm0 2Aρ

Q 36. A thin rod of length L and uniform cross-section is pivoted at its lowest point P inside a stationary homogeneous and non-viscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through P. The density d1 of the material of the rod is smaller than the density d2 of the liquid. The rod is displaced by small angle θ from its equilibrium position and then released. Show that the motion of the rod is simple harmonic and determine its angular frequency in terms of the given parameters. (1996)

d1 P

d2

F = d2 ALg. Let θ be the angular displacement at time t. The torque on the rod about the fixed point P is τ = mg L2 sin θ − F L2 sin θ = 12 gAL2 (d1 − d2 ) sin θ   (∵ sin θ ≈ θ). ≈ − 12 gAL2 (d2 − d1 ) θ. (1) The rod undergoes SHM because the restoring torque is proportional to the angular displacement. The torque about a fixed point P is related to the angular acceleration α by mL2 d2 θ d1 AL3 d2 θ = . (2) 3 dt2 3 dt2 where I is the moment of inertia of the rod about the axis of rotation passing through the fixed point P. Eliminate τ from equations (1) and (2) to get   3g(d2 − d1 ) d2 θ =− θ = −ω 2 θ. dt2 2d1 L τ = Iα =

2

a=

The buoyant force on the rod is equal to the weight of the displaced liquid i.e.,

This equation represents a SHM with an angular frequency s 3(d2 − d1 )g ω= . 2d1 L r   g 3 d2 −d1 Ans. 2 d1 L Q 37. A container of large uniform cross-sectional area A resting on a horizontal surface, holds two immiscible, non-viscous and incompressible liquids of densities d and 2d, each of height H/2 as shown in the figure. The lower density liquid is open to the atmosphere having pressure p0 . (1995)

H 2

d

H 2

2d

h x

Chapter 11. Fluid Mechanics

161

(a) A homogeneous solid cylinder of length L (L < H/2), cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquidliquid interface with length L/4 in the denser liquid. Determine, (i) the density of the solid. (ii) the total pressure at the bottom of the container. (b) The cylinder is removed and the original arrangement is restored. A tiny hole of area s (s  A) is punched on the vertical side of the container at a height h (h < H/2). Determine, (i) the initial speed of efflux of the liquid at the hole. (ii) the horizontal distance x travelled by the liquid initially. (iii) the height hm at which the hole should be punched so that the liquid travels the maximum distance xm initially. Also calculate xm . [Neglect the air resistance in these calculations.] Sol. Let point A be at the top of the cylinder, point B be on the interface and point C be at the bottom of the cylinder. Let the depth of A be h. p0 A

h

d

3L 4 L 4

B C

2d

The pressures at A, B and C are pA = p0 + hdg, pB = pA + ( 3L 4 )dg, pC = pB + ( L4 )(2d)g = pA + 54 Ldg. The forces on the cylinder are its weight W = D(LA/5)g (downwards), hydrostatic force at the top surface FA = pA (A/5) (downwards), and hydrostatic force at the bottom surface FC = pC (A/5) (upwards). In equilibrium, FA + W = FC i.e., pA A/5 + DLgA/5 = (pA + 54 Ldg)A/5, which gives D = 5d/4. Let p be the pressure at the bottom of the container. The forces on the liquid-cylinder system are pA at the bottom surface (upwards), p0 A at the top surface (downwards), weight of the upper liquid 12 HAdg (downwards), weight of the lower liquid HAdg (down1 wards), and weight of the cylinder L( A5 )( 5d 4 )g = 4 LAdg (downwards). In equilibrium, pA = p0 A + 12 HAdg + HAdg + 14 LAdg, which gives p = p0 + 14 (L + 6H)dg.

H 2

d

H 2

2d

h x

Apply Bernoulli’s equation between the interface surface and the hole to get

p0 + H2 dg + H2 − h (2d)g + 0 = p0 + 12 (2d)v 2 , which gives the velocity of efflux, p v = (3H/4 − h)2g. The time taken by the water to fall by a vertical distance h is p t = 2h/g. The horizontal distance covered in time t is given by p x = vt = h(3H − 4h). The distance x is maximum when dx 1 = p (3H − 8h) = 0, dh 2 h(3H − 4h) which gives hm = 3H/8 and maximum value of x as xm = 3H/4. dg(6H+L) Ans. (a) (i) 5d 4 (ii) p = p0 + 4 p p 3H (b) (i) (3H − 4h) g2 (ii) h(3H − 4h) (iii) 3H , 8 4 Q 38. A ball of density d is dropped on to a horizontal solid surface. It bounces elastically from the surface and returns to its original position in time t1 . Next, the ball is released and it falls through the same height before striking the surface of density dL . (1992) (a) If d < dL , obtain an expression (in terms of d, t1 and dL ) for the time t2 the ball takes to come back to the position from which it was released. (b) Is the motion of the ball simple harmonic? (c) If d = dL , how does the speed of the ball depend on its depth inside the liquid? Neglect all frictional and other dissipative forces. Assume the depth of the liquid to be large. Sol. Let V be the volume of the ball of material density d and mass m = V d. The ball comes back to its original position after an elastic collision with the horizontal surface. • h F • mg

dL

162

Part I. Mechanics

Let the ball is dropped from a height h. The time taken by the ball in its downward journey is p t0 = 2h/g, (1) and the speed of the ball just before the collision with the surface is p v0 = 2hg. (2) The speed of the ball just after the collision is v0 (because of elastic collision), and the time taken in the upward journey is t0 . The equations (1) and (2) and the condition t1 = 2t0 gives

(a) Find the angle θ that the radius to the interface makes with the vertical in equilibrium position. (b) If the whole liquid column is given a small displacement from its equilibrium position, show that the resulting oscillations are simple harmonic. Find the time period of these oscillations. Sol. The liquids are non-viscous, incompressible and immiscible which allows us to consider them as rigid bodies for small angular motion about the equilibrium position. Consider two liquids together as a system. Let V be the volume of each liquid and m1 = 1.5ρV and m2 = ρV be their masses. Let A and B be the centres of mass of the two liquids.

v0 = 21 gt1 . The forces acting on the ball inside the liquid are its weight mg = V dg and the upthrust F = V dL g. Since d < dL , net force is upwards and Newton’s second law gives upward acceleration as
a = (F − mg)/m = dLd−d g.

R O 45

t=

C

θ

A

This retardation reduces the speed from v0 to zero in a time

+

Dθ

B m2 g

m1 g

t1 d v0 = . a 2(dL − d)

Then, the upward acceleration increases the speed from zero to v0 in a time t. The ball comes out of the liquid with speed v0 and continues its upward journey to the original position. Total time taken by the ball is t2 = tdown + tup = ( t21 + t) + (t +

t1 2)

= t1 + 2t

= t1 dL /(dL − d). The motion of the ball is periodic but not simple harmonic because the restoring force is not proportional to the displacement. If d = dL then a = 0 and the ball descends with a constant velocity v0 = 12 gt1 . L Ans. (a) dtL1 d−d (b) no (c) remains same Q 39. Two non-viscous, incompressible and immiscible liquids of densities ρ and 1.5ρ are poured into the two limbs of a circular tube of radius R and small crosssection kept fixed in a vertical plane as shown in the figure. Each liquid occupies one-fourth the circumference of the tube. (1991)

By geometry, ∠AOC = 45◦ −θ and ∠BOD = 45◦ + θ. The forces on the system are weights m1 g and m2 g acting at A and B respectively and normal reaction from the circular tube. In equilibrium, resultant force and the resultant torque on the system about O are zero. The torque on the system about the fixed point O due to normal reaction is zero as it passes through O. Thus, torque on the system about the fixed point O is m1 g AC − m2 g BD = m1 gR sin(45◦ − θ) − m2 gR sin(45◦ + θ) = 0, which gives sin(45◦ + θ) m1 3 = = . sin(45◦ − θ) m2 2 Solve to get θ = tan−1 (1/5). Let the liquids be slightly displaced by a small angle φ in the anticlockwise direction. The torque on the system is τ = m1 gR sin(45◦ −θ−φ)−m2 gR sin(45◦ +θ+φ)

R O θ

= 1.5ρgV R [sin(45◦ −θ) cos φ−cos(45◦ −θ) sin φ] − ρgV R [sin(45◦ +θ) cos φ+cos(45◦ +θ) sin φ] . Using small angle approximation sin φ ≈ φ, cos φ ≈ 1, and the result from the previous part, we get τ = −ρgV R [cos(45◦ + θ) + 1.5 cos(45◦ − θ)] φ.

Chapter 11. Fluid Mechanics

163

The negative sign indicates the restoring nature of the torque and it is proportional to the angular displacement φ. Thus, system executes SHM. The moment of inertia of the system about an axis passing through O is I = m1 R2 + m2 R2 = 2.5ρV R2 . 2

Using I ddtφ2 = τ , we get d2 φ ρgV R =− [cos(45◦ + θ) + 1.5 cos(45◦ − θ)] φ dt2 I = −ω 2 φ, which gives the time period of SHM as T = 2π/ω s = 2π

ρgV s

= 2π

R[cos(45◦

2.5ρV R2 + θ) + 1.5 cos(45◦ − θ)]

1.38R . g

The forces acting on the plank are its weight mg acting at plank’s centre of mass C, buoyancy force FB acting at the centre of mass of the displaced water P, and reaction R acting on the hinge point O. From geometry, perpendicular distances of the point C and the point P from a vertical line passing through O are rC = 2l sin θ and rP = h2 tan θ, respectively. Also, the length of the plank inside the water is h sec θ. The weight of the plank and the buoyancy force on it are mg = ρAlg, FB = σA(h sec θ)g. In equilibrium, net torque on the plank about the fixed point O is zero. The torque of the reaction force R about the hinge point O is zero because R passes through O. Thus, the net torque on the plank about O is τ = FB (h/2) tan θ − mg(l/2) sin θ
σh2 sec θ tan θ − ρl2 sin θ = 0, = Ag 2 which gives

Ans. (a) tan

−1

1 5




(b) 2π

q

1.38R g

Q 40. A wooden plank of length 1 m and uniform crosssection is hinged at one end to the bottom of a tank as shown in the figure. The tank is filled with water upto a height 0.5 m. The specific gravity of the plank is 0.5. Find the angle θ that the plank makes with the vertical in the equilibrium position. [Exclude the case θ = 0◦ .] (1984)

cos2 θ = (h2 /l2 ) (σ/ρ) = (1/4) (2) = 1/2. Solve to get θ = 45◦ . Ans. 45◦ Q 41. A gas bubble, from an explosion under water, oscillates with a period T proportional to pa db E c , where p is the static pressure, d is the density of water and E is the total energy of the explosion. Find the values of a, b and c. (1981) Sol. The dimensions of time period T is [T], pressure p is [ML−1 T−2 ], density d is [ML−3 ], and energy E is [ML2 T−2 ]. The given equation is T = kpa db E c ,

where k is a dimensionless constant. The dimensions of parameters on both sides of equation (1) should be same. Write dimensions of various physical quantities in equation (1) to get

θ

Sol. The length of the plank is l = 1 m and height of the water level is h = 0.5 m. Let A be the cross-section area of the plank, ρ be the density of the plank and σ be the density of the water. Given, ρ = 0.5σ. h 2

a

b

[M0 L0 T1 ] = [ML−1 T−2 ] [ML−3 ] [ML2 T−2 ] =[Ma+b+c L−a−3b+2c T−2a−2c ].

c

(2)

Equate exponents of M, L, and T on the two sides of equation (2) to get tan θ FB

h

P θ O

(1)

C

mg l 2

sin θ

a + b + c = 0,

(3)

− a − 3b + 2c = 0,

(4)

− 2a − 2c = 1.

(5)

Solve equations (3)–(5) to get a = −5/6, b = 1/2, and c = 1/3. Ans. a = − 65 , b = 12 , c = 13

164

Part I. Mechanics

Q 42. Two identical cylindrical vessels with their bases at the same level, each contain a liquid of density ρ. The height of the liquid in one vessel is h1 and in the other is h2 . The area of either base is A. What is the work done by gravity in equalising the levels when the two vessels are connected? (1981) Sol. Let h be the final height of the liquid when cylinders are connected.

Q 44. A column of mercury of length 10 cm is contained in the middle of a horizontal tube of length 1 m which is closed at both the ends. The two equal lengths contain air at standard atmospheric pressure of 0.76 m of mercury. The tube is now turned to vertical position. By what distance will the column of mercury be displaced? Assume temperature to be constant. (1978) Sol. Let A and B be the portions of the tube on the left and the right sides of the mercury column. In horizontal condition, both A and B have equal lengths l0 = 45 cm and equal pressures p0 = 0.76 m of Hg.

h1

h = (h1 + h2 )/2.

Ui = (ρAh1 )g(h1 /2) + (ρAh2 )g(h2 /2) =

2 1 2 ρAg(h1

+

h22 ),

Uf = (ρAh)g(h/2) + (ρAh)g(h/2) = ρAgh2 . The work done by the gravity is W = Ui − Uf = 12 ρAg h21 + h22 − 2h2
= 12 ρAg h21 + h22 − 12 (h1 + h2 )2 , = =

2 1 4 ρAg(h1 1 4 ρAg(h1

+

h22


(using (1))

− 2h1 h2 ) ρAg 4

(h1 − h2 )

B

p0

p0

pA

45cm

45cm

A

Let mercury column is dips down by a distance x when tube is turned to the vertical position. Lengths of A and B change to lA = (45 − x) cm and lB = (45 + x) cm. Let the pressures in A and B change to pA and pB . As temperature is constant, apply Boyle’s law, pi Vi = pf Vf , on A and B to get p0 (45A) = pA (45 − x)A,

(1)

p0 (45A) = pB (45 + x)A,

(2)

where A is the cross-sectional area of the tube. In the vertical position, the pressure at the lower face of mercury (pA ) is equal to the sum of the pressure at the upper face (pB ) and the hydrostatic pressure due to mercury column of height 10 cm, i.e.,

Sol. Let the density of the stones be ρ and that of the water be σ (< ρ). Let the volume of the stones be V . In floating condition, weight of the stones is balanced by the buoyant force V1 = V (ρ/σ),

(3)

Eliminate pA and pB from equations (1)–(3) to get

2

Q 43. A boat floating in a water tank is carrying a number of large stones. If the stones are unloaded into water, then water level will rises up? (1979)

i.e.,

A

pA = pB + 10ρg.

− h2 )2 . Ans.

ρV g = σV1 g

10cm

(1)

Note that the decrease in height in the left cylinder is equal to the increase in height in the right cylinder. The work done by the gravity is equal to the change in the potential energy of the liquid. The potential energy of a liquid column of height h and mass m is equal to mg(h/2). Thus, the initial and the final potential energies of the liquid are

45+x

Total volume of the liquid in two cylinders remains same whether they are connected or not i.e., Ah1 + Ah2 = 2Ah, which gives

pB 10cm

h2

45−x

B

h

(1)

where V1 is the volume of displaced water. When the stones are thrown into the water, volume of the displaced water is V2 = V . Since ρ > σ, equation (1) gives V1 > V2 i.e., volume of displaced water in floating condition is more. Thus, level of water in the floating condition should be higher. Ans. F

x=

45(10ρg) 45(10ρg) = = 2.96 cm. 2p0 2(76ρg) Ans. 2.96 cm

Chapter 12 Some Mechanical Properties of Matter

Eliminate (P0 − P ) from equations (2) and (3) and substitute R from equation (1) to get

One Option Correct Q 1. A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height h, where the radius of its cross-section is b. If the surface tension of water is S, its density is ρ, and its contact angle with glass is θ, the value of h will be (where g is acceleration due to gravity) (2014)

h=


2S 2S = cos θ + α2 . ρgR bρg Ans. D

Q 2. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is (2013) (A) 0.25 (B) 0.5 (C) 2 (D) 4

h

Sol. Young’s modulus of the wire material is defined as Y = (A) (C)

2S bρg 2S bρg

cos(θ − α) cos(θ − α/2)

(B) (D)

2S bρg 2S bρg

cos(θ + α) cos(θ + α/2)

∆l = R θ+ α 2

P0

h

2 ∆lthin lthin /rthin L/R2 = = 2. = 2 ∆lthick lthick /rthick 2L/(4R2 )

α 2

Ans. C

By geometry, this radius makes an angle θ + α2 with the horizontal where
cos θ + α2 = b/R. (1)

Q  3. In the  determination of Young’s modulus 4M Lg Y = πld2 by using Searle’s method, a wire of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of wire is observed. Quantities d and l are measured using screw gauge and micrometer, respectively. They have same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement,

Let P0 be the atmospheric pressure and P1 be the pressure just below the meniscus. Excess pressure on the concave side of meniscus of radius R is given by P0 − P1 = 2S/R.

(2)

The hydrostatic pressure gives P0 − P1 = hρg.

Fl . Y πr2

Young’s modulus Y is same for both the wires (same material) and so is the applied force F . Thus, the ratio of the elongation in the given wires is

b θ

(1)

where F is the tensile force, r is the radius of the wire, l is the length of the wire, and ∆l is the elongation of the wire due to force F . The equation (1) gives elongation of the wire as

Sol. Let R be the radius of the meniscus formed with a contact angle θ.

P0 P1

F/(πr2 ) Fl stress = = , strain ∆l/l ∆l πr2

(2012)

(A) due to the error in the measurements of d and l are the same.

(3) 165

166

Part I. Mechanics

(B) due to the error in the measurement of d is twice that due to the error in the measurement of l. (C) due to the error in the measurement of l is twice that due to the error in the measurement of d. (D) due to the error in the measurement of d is four times that due to the error in measurement of l. Sol. Differentiate the expression Y = divide it by Y to get

4M Lg πld2

and then

∆Y /Y = ∆l/l + 2 ∆d/d.

(1)

From the given data, the least counts of screw gauge and micrometer are pitch divided by the number of divisions on the circular scale i.e., 0.5/100 = 0.005 m. Hence, ∆d = ∆l = 0.005 m. The equation (1) gives the error contribution of d in the measurement of Y as ed = 2 ∆d/d = 2(0.005)/0.5 = 0.02, and the error contribution due to the error in l as el = ∆l/l = 0.005/0.25 = 0.02.

Q 5. A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ±0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ±0.01 mm. Take g = 9.8 m/s2 (exact). The Young’s modulus obtained from the reading is (2007) (A) (2.0 ± 0.3) × 1011 N/m2 (B) (2.0 ± 0.2) × 1011 N/m2 (C) (2.0 ± 0.1) × 1011 N/m2 (D) (2.0 ± 0.05) × 1011 N/m2 Sol. Young’s modulus of a material is given by

Ans. A

Y =

Q 4. A glass tube of uniform internal radius r has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius r. End 2 has subhemispherical soap bubble as shown in the figure. Just after opening the valve, (2008)

2

The radius r2 of the sub-hemispherical bubble is more than the radius r1 of the hemispherical bubble i.e., r2 > r1 . Hence, from equations (1) and (2), P1 > P2 and air moves from end 1 to end 2. As air moves out from end 1, the volume of soap bubble at this end decreases. Ans. B

F/A 4F L stress = = , strain l/L πd2 l

(1)

where F is the applied force, L is the length of the wire, l is the extension in the length of the wire and d is the diameter of the wire. From the given data, F = mg = 9.8 N, L = 2.0 m, l = 0.8 × 10−3 m, and d = 0.4 × 10−3 m. Substitute the values in equation (1) to get Y = 1.95 × 1011 ≈ 2.0 × 1011 N/m2 . Differentiate equation (1) and divide it by Y to get the error in Y ∆Y /Y = 2 ∆d/d + ∆l/l.

1

Substitute ∆d = 0.01 mm and ∆l = 0.05 mm to get (A) air from end 1 flows towards end 2. No change in the volume of the soap bubbles. (B) air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases. (C) no change occurs. (D) air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases. Sol. The excess pressure inside a soap bubble of radius r and surface tension T is given by ∆p = P − P0 = 4T /r.

∆Y = Y (2 × 0.01/0.4 + 0.05/0.8) = Y (0.1125) ≈ 0.2 × 1011 N/m2 . Ans. B Q 6. The pressure of a medium is changed from 1.01 × 105 Pa to 1.165 × 105 Pa and change in volume is 10% keeping temperature constant at 20 ◦ C. The bulk modulus of the medium is (2005) (A) 204.8 × 105 Pa (B) 102.4 × 105 Pa (C) 51.2 × 105 Pa (D) 1.55 × 105 Pa Sol. The bulk modulus B of a fluid at pressure p and volume V is given by

r

1 1 dV =− . B V dp

P0 P

Substitute the values to get Thus, the pressures inside the end 1 and end 2 of the tube are P1 = P0 + 4T /r1 ,

(1)

P2 = P0 + 4T /r2 .

(2)

B=−

∆p (1.01 × 105 − 1.165 × 105 ) =− ∆V /V 0.1

= 1.55 × 105 Pa. Ans. D

Chapter 12. Some Mechanical Properties of Matter Q 7. The adjacent graph shows the extension (∆l) of a wire of length 1 m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is 10−6 m2 , calculate, from the graph, the Young’s modulus of the material of the wire. (2003) ∆l(×10−4

m)

4 3

20 40 60 80

(A) 2 × 10 N/m (C) 3 × 1012 N/m2

W (N) −11

k(l1 − l) F1 L =L+ . AY AY /L

Let the mass m is displaced by a small distance x from its equilibrium position. Let L2 and l2 be the new lengths of the wire and the spring i.e.,

(B) 2 × 10 N/m (D) 2 × 1013 N/m2

L2 = L +

2

W/A = ∆l/L . −6 2

Sol. Young’s modulus is defined as Y Given, the cross-section area of the wire A = 10 m and the length of the wire L = 1 m. From the given graph, elongation of the wire is ∆l = 10−4 m at a load of W = 20 N. Substitute the values to get Y =

L1 = L +

20/10−6 = 2 × 1011 N/m2 . 10−4 /1 Ans. A

Q 8. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant k. A mass m hangs freely from the free end of the spring. The area of cross-section and the Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period (1993) q T equal to pm m(Y A+kL) (B) 2π (A) 2π k q q Y Ak mY A mL (C) 2π (D) 2π kL YA Sol. Let L and l be the natural lengths of the wire and the spring. Let the lengths become L1 and l1 when mass m hangs from the spring. The forces on mass m are its weight mg and the spring force F1 = k(l1 − l). L

l

L1

l1

(1)

The upward force on the mass m is F2 = k(l2 − l) which is equal to the downward force on the wire. Thus,

1

2

mass m, which is F1 . The expression for Young’s mod1 /A ulus, Y = (L1F−L)/L , gives

L2 + l2 = L1 + l1 + x.

2

11

167

L2

l2

x

k(l2 − l) . AY /L

Substitute l1 , L1 , and L2 in equation (1) and simplify to get −1  1 1 + x. k(l2 − l) − mg = k AY /L Thus, the restoring force on the mass m is  −1 1 1 + x = keq x, F2 − mg = k AY /L where keq is the equivalent spring constant. The time period is given by r r m m(kL + AY ) . T = 2π = 2π keq kAY Note that problem is equivalent to two springs of spring constants AY /L and k connected in series. Ans. B Q 9. A highly rigid cubical block A of small mass M and side L is fixed rigidly on to another cubical block B of the same dimensions and of low modulus of rigidity η such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of A. After the force is withdrawn, block A executes small oscillations, the time period of which is given (1992) q by √ Mη (A) 2π M ηL (B) 2π q q L M (C) 2π MηL (D) 2π ηL Sol. The shape of the upper cube is not distorted as it is rigid. Consider a time when tangential strain in the lower cube is θ.

In equilibrium, F1 = mg, which gives •

l1 = l + mg/k. Since the spring is massless, the forces on the two ends of the spring are equal and opposite. Thus, the downward force on the wire is equal to the upward force on

x

A F

F L

θ

B

168

Part I. Mechanics

The modulus of rigidity, η = F/A θ , gives the rightward force on the top surface of the lower cube as F = ηAθ = ηL2 (x/L) = ηLx, where x is the linear displacement of the top surface of the lower cube. Since the upper cube is not distorted, the linear displacement of the centre of mass of the upper cube is also x. Also, by Newton’s third law, the restoring force on the upper cube is equal and opposite to F . Apply Newton’s second law on the upper cube to get 2

which gives d2 x/dt2 = − (ηL/M ) x = −ω 2 x. This equation represents a SHM with time period T = q M 2π ω = 2π ηL . Ans. D Q 10. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied? (1981) (A) Length = 50 cm, diameter = 0.5 mm (B) Length = 100 cm, diameter = 1 mm (C) Length = 200 cm, diameter = 2 mm (D) Length = 300 cm, diameter = 3 mm /A 4T L Sol. Young’s modulus is defined as Y = Tl/L = πD 2l , where T is the tension in the wire, D is its diameter, L is its length and l is its elongation. Thus, elongations of the given wires are

4T πY 4T lb = πY 4T lc = πY 4T ld = πY

La Da2 Lb Db2 Lc Dc2 Ld Dd2

4T πY 4T = πY 4T = πY 4T = πY =

Sol. Consider the portion of water that rises in the capillary tube. In equilibrium, the upward force due to surface tension, F = 2πrσ cos θ, is balanced by the weight of the water, W = πr2 hρg, i.e., 2πrσ cos θ = πr2 hρg which gives h=

2

M d x/dt = −F = −ηLx,

la =

(B) For a given material of the capillary tube, h is independent of σ. (C) If this experiment is performed in a lift going up with a constant acceleration, then h decreases. (D) h is proportional to contact angle θ.

50 4T = 200 0.52 πY 100 4T = 100 2 1 πY 200 4T = 50 22 πY 300 4T ≈ 33 32 πY Ans. A

One or More Option(s) Correct Q 11. A uniform capillary tube of inner radius r is dipped vertically into a beaker filled with water. The water rises to a height h in the capillary tube above the water surface in the beaker. The surface tension of water is σ. The angle of contact between water and the wall of the capillary tube is θ. Ignore the mass of water in the meniscus. Which of the following statements is (are) true? (2018) (A) For a given material of the capillary tube, h decreases with increase in r.

2σ cos θ . rρg

(1)

From equation (1), h decreases with increase in r, it depends on σ and it is proportional to cos θ. The contact angle θ depends on the material of the capillary tube. In a lift going up with a constant acceleration a, the force due to surface tension remains same but weight of the water increases i.e., Wlift = πr2 hρ(g + a). Thus, the height of the water rise in a capillary becomes h=

2σ cos θ 2σ cos θ = , rρ(g + a) rρgeff

where geff is the effective g in the lift. Hence, h decrease if the experiment is performed in a lift going up with a constant acceleration. Can you guess value of h if the experiment is performed in a lift falling freely? Ans. (A), (C) Q 12. Consider a thin square plate floating on a viscous liquid in large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity u0 . Which of the following statements is (are) true? (2018) (A) The resistive force of liquid on the plate is inversely proportional to h. (B) The resistive force of liquid on the plate is independent of the area of the plate. (C) The tangential (shear) stress on the floor of the tank increases with u0 . (D) The tangential (shear) stress on the plate varies linearly with the viscosity η of the liquid. Sol. In a viscous liquid, there is no relative motion between the solid surface and the liquid layer in contact with it i.e., the liquid layer moves with the speed of the solid surface. Thus, the liquid layer in contact with the plate moves with a speed u0 and the liquid layer in contact with the floor of the tank is stationary. The velocity gradient is given by dv/dz ≈ (u0 − 0)/h = u0 /h, where we have used the fact that h is very small in comparison to the tank width. Consider a liquid layer which is in contact with the plate. The viscous drag

Chapter 12. Some Mechanical Properties of Matter

169

force (backward) on this layer, due to the layer just below it, is

L1

σ1 V g P

Fon top layer = ηA(dv/dz) = ηA(u0 /h).

ρ1 V g

T

σ2 V g

T

u0

Q h

L2

Similarly, forces on the sphere Q are ρ2 V g, T , and σ1 V g. In equilibrium, the net force on the spheres P and Q are separately zero i.e.,

u0

Fon

plate Fon top layer

The top layer moves with a constant speed u0 . Thus, the plate should apply an equal but opposite force (forward) on the top layer to balance viscous drag on it. In turn, by Newton’s third law, the top layer should apply equal but opposite force (backward) force on the plate. Thus, the resistive force of liquid on the plate is Fon plate = Fon top layer = ηA(u0 /h). The tangential (shear) stress on the plate is given by σon plate = Fon plate /A = η(u0 /h). Apply similar argument on liquid layer in contact with the floor of the tank. The resistive force of liquid on the floor of the tank is Fon floor = ηA(u0 /h) and the tangential (shear) stress on the floor of the tank is σon floor = Fon floor /A = η(u0 /h). Ans. (A), (C), (D) Q 13. Two spheres P and Q of equal radii have densities ρ1 and ρ2 , respectively. The spheres are connected by a massless string and placed in liquids L1 and L2 of densities σ1 and σ2 and viscosities η1 and η2 , respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being taut (see figure). If sphere P alone in L2 has terminal velocity ~vP and Q alone in L1 has terminal velocity ~vQ , then

T + ρ1 V g = σ1 V g,

(1)

T + σ2 V g = ρ2 V g.

(2)

The tension T > 0 because the string is taut. Thus, equation (1) gives ρ1 < σ1 and equation (2) gives ρ2 > σ2 . Eliminate T from equations (1) and (2) to get σ1 − ρ1 = ρ2 − σ2 .

L1 P L2 Q

|~ vP | η2 (B) |~ vQ | = η1 (D) ~vP · ~vQ < 0

Sol. Let V = 43 πr3 be the volume of the spheres P and Q of equal radii r. The forces acting on the sphere P are its weight ρ1 V g, tension from the string T , and the buoyancy force σ1 V g.

(3)

Now, consider the situation when the sphere P moves in liquid L2 and the sphere Q moves in liquid L1 . These spheres will attain the terminal velocities ~vP and ~vQ after some time. The direction of the velocity (upwards or downwards) will depend on the density of the sphere in comparison to the density of the liquid. Let us consider the case when ρ1 > σ2 . In this case, the velocity of the sphere P is downwards. From equation (3), if ρ1 > σ2 then ρ2 < σ1 . If the density of a sphere is less than the density of the liquid in which it is immersed, it will move up. Thus, the velocity of the sphere Q is upwards i.e., the directions of ~vP and ~vQ are opposite. L1

σ1 V g

~vQ

Q 6πη1 rvQ ρ2 V g σ2 V g

6πη2 rvP P L2

(2015)

|~ vP | η1 (A) |~ vQ | = η2 (C) ~vP · ~vQ > 0

ρ2 V g

ρ1 V g

~vP

Hence, ~vP · ~vQ < 0. The forces on the sphere P are its weight ρ1 V g, buoyancy force σ2 V g, and viscous drag 6πη2 rvP (see figure). Similarly, the forces on the sphere Q are ρ2 V g, σ1 V g and 6πη1 rvQ . Net forces on the spheres are zero when they move with terminal velocities i.e., 6πη2 rvP + ρ1 V g = σ2 V g,

(4)

6πη1 rvQ + σ2 V g = ρ1 V g.

(5)

Simplify equations (4) and (5) to get 2r2 (ρ1 − σ2 ) , 9η2 2r2 (σ1 − ρ2 ) vQ = . 9η1

vP =

(6) (7)

170

Part I. Mechanics

Divide equation (6) by (7) to get |~vP | η 1 ρ1 − σ2 η1 = = . |~vQ | η2 σ2 − ρ2 η2

(using (3)).

We encourage you to show the results for ρ1 < σ2 . Ans. A, D

Strain

Q 14. In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is (are) (2015)

P

When liquid medicine of density ρ is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper. (2010) Q 15. If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r  R) is 2 2 (A) 2πrT (B) 2πRT (C) 2πrR T (D) 2πRr T Sol. The force due to surface tension T acts on the drop-dropper interface (a circle). The direction of this force is tangential to the surface of the spherical drop.

Q

Dropper

Stress

interface

(A) (B) (C) (D)

P has more tensile strength than Q. P is more ductile than Q. P is more brittle than Q. The Young’s modulus of P is more than that of Q.

dF θ r R θ

r

Sol. The tensile strength of a material is generally defined as the stress at which the material breaks.

Strain

P

Q

Stress SQ SP

From the figure, P has more strength than Q (because SP > SQ ). The material Q is brittle because its stress-strain curve does not show any yield. The material P is ductile as its stress-strain curve clearly show yielding (the part beyond the small kink). Thus, P is more ductile than Q. In the elastic region (straight line portion), the slope of P is more than that of Q i.e.,  

strain stress



stress strain



 > P

 < P

strain stress



stress strain



, i.e., Q

. Q

Thus, Young’s modulus of P is less than that of Q. Ans. A, B Paragraph Type Paragraph for Questions 15-17

Consider a small element of length dl on the dropdropper interface. The force on this element is dF = T dl, and it makes an angle θ from the horizontal. Resolve dF in the horizontal and the vertical directions to get, dFh = T dl cos θ and dFv = T dl sin θ. By symmetry, the force dFh on two diametrically opposite elements on circular interface is equal and opposite. R Thus, total force in the horizontal direction, Fh = dFh = 0. Integrate the vertical component over the circular interface to get Fv = T (2πr) sin θ = 2πrT (r/R) = 2πr2 T /R. Ans. C Q 16. If r = 5 × 10−4 m, ρ = 103 kg/m3 , g = 10 m/s2 , T = 0.11 N/m, the radius of the drop when it detaches from the dropper is approximately (A) 1.4 × 10−3 m (B) 3.3 × 10−3 m (C) 2.0 × 10−3 m (D) 4.1 × 10−3 m Sol. The drop will detach from the dropper when its weight is equal to the vertical force due to surface tension i.e., (4/3)πR3 ρg = 2πr2 T /R. Solve to get  1/4 R = 3r2 T /(2ρg) ≈ 1.4 × 10−3 m. Ans. A

Chapter 12. Some Mechanical Properties of Matter

171

Q 17. After the drop detaches, its surface energy is (A) 1.4 × 10−6 J (B) 2.7 × 10−6 J (C) 5.4 × 10−6 J (D) 8.1 × 10−6 J

The equilibrium condition on piston-mass system gives

Sol. The surface energy of the spherical drop of radius R is given by U = T (4πR2 ) = (0.11)4π(1.4 × 10−3 )2 = 2.7 × 10−6 J. Ans. B

∆pA = M g.

(1)

If the volume V of a body changes by ∆V when pressure applied on it is changed by ∆p then its bulk modulus is 1 ∆V 1 . (2) = K V ∆p The volume of a sphere of radius R is V = 43 πR3 . Differentiate it to get

Fill in the Blank Type Q 18. A uniform rod of length L and density ρ is being pulled along a smooth floor with a horizontal acceleration α (see figure). The magnitude of the stress at the transverse cross-section through the mid-point of the rod is . . . . . . (1993) L

∆V ∆R =3 . V R

(3)

Use equations (1)–(3) to get ∆R/R = M g/(3AK). Ans. M g/(3AK) Q 20. A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y . If the wire is stretched by an amount x, the work done is . . . . . .

α

(1987)

Sol. Let T be the tension at the middle point of the rod. This tension accelerates left half of the rod with acceleration α. The mass of the left half is m = L2 Aρ, where A is the cross-sectional area of the rod. L 2

α T

By Newton’s second law, the tension T and stress σ are given by

Sol. The work done by the applied force is equal to the elastic potential energy of the strained body i.e., W =U =

1 2

× stress × strain × volume.

In the given case, strain = x/L and the stress = Y × strain = Y x/L. The volume of the wire is AL. Substitute these to get   1 1 YA W = (x/L)(Y x/L)(AL) = x2 . 2 2 L Note that the effective spring constant of the wire is Y A/L.
Ans. 21 YLA x2

T = mα = 12 LAρα, σ = T /A = 12 Lρα. Ans.

1 2 Lρα

Q 19. A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid. When a mass M is placed on the piston to compress the liquid the fractional change in the radius of the sphere, ∆R/R, is . . . . . . (1988) Sol. Let ∆p be the increase in the liquid pressure when mass M is placed over the piston. The forces on the piston-mass system are its weight M g (downwards) and the force due to water ∆pA (upwards). M A V

Integer Type Q 21. A steel wire of diameter 0.5 mm and Young’s modulus 2 × 1011 N/m2 carries a load of mass M . The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is . . . . . . . [Take g = 10 m/s2 and π = 3.2.] (2018) Sol. This problem is based on Searle’s experiment. The mains scale (fixed on reference wire) and the vernier scale (fixed on test wire) are used to measure the extension of the test wire under load.

172

Part I. Mechanics

The test wire of length l = 1 m is made of steel of Young’s modulus Y = 2 × 1011 N/m2 and its crosssection area is A = πD2 /4 = (3.2)(0.5 × 10−3 )2 /4 = 2 × 10−7 m2 . The extension in wire due to increase in load F = mg = (1.2)(10) = 12 N is given by (12)/(2 × 10 ) F/A = Y /l (2 × 1011 )/(1)

= 0.3 × 10−3 m = 0.3 mm. The least count of main scale is 1 mm i.e., 1 MSD = 1 mm. Since 10 VSD correspond to 9 MSD, we get 1 VSD = (9/10)MSD = 0.9 mm. The least count of the vernier scale is given by

Reference Wire

Test Wire

0

Main Scale

Vernier Scale

LC = 1 MSD − 1 VSD = 1 − 0.9 = 0.1 mm.

0

10

10 M

 K=

∆U +1 4πSR2

3 .

Substitute ∆U = 10−3 J, S = 0.1 4π N/m and R = 10−2 m to get K = (100 + 1)3 ≈ 106 . Ans. 6

−7

∆l =

Eliminate r from equations (1) and (2) to get

Q 23. Consider two solid spheres P and Q each of density 8 g/cm3 and diameters 1 cm and 0.5 cm, respectively. Sphere P is dropped into a liquid of density 0.8 g/cm3 and viscosity η = 3 poiseulles. Sphere Q is dropped into a liquid of density 1.6 g/cm3 and viscosity η = 2 poiseulles. The ratio of the terminal velocities of P and Q is . . . . . . . (2016) Sol. The terminal velocity of a sphere of radius r and density ρ, immersed in a liquid of density σ and viscosity η, is given by v=

0

0

2 (ρ − σ)r2 g . 9 η

Substitute the values of given parameters to get vP (ρP − σP ) rP2 ηQ = 2 η vQ (ρQ − σQ ) rQ P

10

10

=

m M

(8 − 0.8) (1/2)2 (2) = 3. (8 − 1.6) (0.5/2)2 (3) Ans. 3

Initially, zero of vernier scale coincides with the zero of main scale. Due to extension of the wire, the vernier scale shifts by ∆l = 0.3 mm. Hence, 3rd division (∆l/LC = 3) of the vernier scale will coincide with a main scale division. Ans. 3 Q 22. A drop of liquid of radius R = 10−2 m having surface tension S = 0.1 4π N/m divides itself into K identical drops. In this process the total change in the surface energy ∆U = 10−3 J. If K = 10α the the value of α is . . . . . . . (2017) Sol. Let a spherical drop of liquid of radius R divides itself into K spherical drops, each of radius r. The density of liquid does not change in this process. Hence, conservation of mass gives 3 4 3 πR

= 43 πr3 K,

i.e., r = RK −1/3 .

(1)

The surface energy of a drop of radius R is Ui = 4πR2 S and total surface energy of K drops of radius r is Uf = 4πr2 KS. Note that the surface energy increases when a bigger drop is divided into multiple smaller drops. Total change in surface energy is ∆U = Uf − Ui = 4πS(Kr2 − R2 ).

(2)

Q 24. During Searle’s experiment, zero of the Vernier scale lies between 3.20 × 10−2 m and 3.25 × 10−2 m of the main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 × 10−2 m and 3.25 × 10−2 m of the main scale but now 45th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 × 10−7 m2 . The least count of the Vernier scale is 1.0 × 10−5 m. The maximum percentage error in the Young’s modulus of the wire is . . . . . . . (2014) Sol. The difference between the two measurements by Vernier scale gives elongation of the wire caused by the additional load of 2 kg. In the first measurement, main scale reading is MSR = 3.20 × 10−2 m and Vernier scale reading is VSR = 20. The least count of Vernier scale is LC = 1 × 10−5 m. Thus, the first measurement by Vernier scale is L1 = MSR + VSR × LC = 3.20 × 10−2 + 20(1 × 10−5 ) = 3.220 × 10−2 m.

Chapter 12. Some Mechanical Properties of Matter In the second measurement, MSR = 3.20 × 10−2 m and VSR = 45. Thus, the second measurement by Vernier scale is L2 = 3.20 × 10−2 + 45(1 × 10−5 ) = 3.245 × 10−2 m. The elongation of the wire due to force F = 2g is l = L2 − L1 = 0.025 × 10−2 m. The maximum error in the measurement of l is ∆l = LC = 1 × 10−5 m. Young’s modulus is given by Y = FL lA . The maximum percentage error in the measurement of Y is −5

∆Y ∆l 1 × 10 ×100 = 4%. ×100 = ×100 = Y l 0.025 × 10−2 We encourage you to compute the values of Y and ∆Y . Hint: Y = 2 × 1011 N/m2 and ∆Y = 0.08 × 1011 N/m2 . Ans. 4 Q 25. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its crosssectional area is 4.9 × 10−7 m2 . If the mass is pulled a little in the vertically downward direction and released, it performs SHM of angular frequency 140 rad/s. If the Young’s modulus of the material of the wire is n × 109 N/m2 , the value of n is . . . . . . . (2010) Sol. When mass m is pulled by a force F , the wire elongation x, length l, cross-sectional area A, and Young’s modulus of wire material Y are related by Y = F/A x/l i.e.,

173 Q 26. Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m2 . The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soapwater used to make bubbles is 0.04 N/m. Find the ratio nB /nA , where nA and nB are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity.] (2009) Sol. Let rA = 2 cm be the radius of the bubble A, rB = 4 cm be the radius of the bubble B, P0 = 8 N/m2 be the chamber pressure and T = 0.04 N/m be the surface tension. Let PA and PB be the pressure inside the bubble A and bubble B, respectively. The volumes of the two bubbles are 3 , VA = 43 πrA 3 . VB = 43 πrB

The excess pressure inside a soap bubble of radius r is given by 4T /r. Thus, PA = P0 + 4T /rA , PB = P0 + 4T /rB . Since the temperatures of bubble A and bubble B are equal, the ideal gas equation for the air inside the soap bubbles gives PA VA /nA = PB VB /nB . Substitute PA , VA , PB , and VB to get 3 nB (P0 + 4T /rB ) rB = 3 nA (P0 + 4T /rA ) rA

=

(8 + 4(0.04)/0.04) (0.04)3 = 6. (8 + 4(0.04)/0.02) (0.02)3 Ans. 6

F = (Y A/l) x. Descriptive Q 27. In a Searle’s experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count of 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young’s modulus of the material of wire from these data. (2004)

l

x

Fr

F

Sol. Young’s modulus is given by The restoring force by the wire is equal but opposite to F i.e., Fr = −F . Apply Newton’s second law to get

Y =

F/A 4F L 4(50)(110.0) = = 2 l/L πd l 3.14(0.050)2 (0.125)

= 2.24 × 107 N/cm2 = 2.24 × 1011 N/m2 . 2

2

(1)

2

m d x/dt = −(Y A/l) x = −ω x. This equation p represents SHM with an angular frequency ω = Y A/(lm). Substitute the values to get Y = ω 2 lm/A = 4 × 109 N/m2 . Ans. 4

Differentiate the expression for Y and simplify it to get ∆Y ∆L ∆d ∆l = +2 + Y L d l 0.1 2 × 0.001 0.001 = + + = 0.049. 110.0 0.050 0.125

(2)

174

Part I. Mechanics

From equations (1) and (2), ∆Y = 1.09 × 1010 N/m2 . Ans. 1.09 × 1010 N/m2 Q 28. A small sphere falls from rest in a viscous liquid. Due to friction, heat is produced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity. (2004) Sol. The forces acting on the sphere are its weight 4 4 3 3 3 πr ρg downwards, buoyancy force 3 πr σg upwards, and viscous force 6πηrv upwards. The sphere attains the terminal velocity vt when the resultant force on it is zero i.e., 3 4 3 πr ρg

= 43 πr3 σg + 6πηrvt .

Solve to get the terminal velocity vt =

2r2 (ρ − σ)g . 9η

which gives T =

λlg λga = . 2l sin θ 2y

where we have assumed sin θ ≈ θ ≈ tan θ = y/a. Ans. λga 2y Q 30. A liquid of density 900 kg/m3 is filled in a cylindrical tank of upper radius 0.9 m and lower radius 0.3 m. A capillary tube of length l is attached at the bottom of the tank as shown in the figure. The capillary has outer radius 0.002 m and inner radius a. When pressure p is applied at the top of the tank volume flow rate of the liquid is 8 × 10−6 m3 /s and if capillary tube is detached, the liquid comes out from the tank with a velocity of 10 m/s. Determine the coefficient of viscosity of the liquid. [Given : πa2 = 10−6 m2 and a2 /l = 2 × 10−6 m.] (2003)

The rate of heat generation is equal to the rate of work done by the viscous force which, in turn, is equal to its power. Thus,

0.9m

8π(ρ − σ)2 g 2 r5 dQ = (6πηrvt )vt = . dt 27η Ans.

dQ dt

Q 29. A container of width 2a is filled with a liquid. A thin wire of mass per unit length λ is gently placed over the liquid surface in the middle of the surface as shown in the figure. As a result, the liquid surface is depressed by a distance y ( a). Determine the surface tension of the liquid. (2004) y

•

2a

Sol. Let l be the length of the wire and m = λl be its mass. The forces acting on the wire are its weight mg = λlg and the force due to surface tension F = T l, where T is surface tension. y

F

F θ

y

0.3m

l

Sol. Let r1 = 0.9 m be the radius at the top, v1 be the velocity at the top, r2 = 0.3 m be the radius at the bottom, and v2 = 10 m/s be the velocity at the bottom (when capillary tube is detached). The continuity equation, πr12 v1 = πr22 v2 , gives v1 = 10/9 m/s. Let p0 be the atmospheric pressure. Apply Bernoulli’s equation between the top and bottom p0 + p + ρgH + 12 ρv22 = p0 + 0 + 12 ρv12 , to get p + ρgH = 12 ρ(v22 − v12 ) =

4 9

× 105 N/m2 .

The pressure difference across the capillary is

θ

a

a

mg

Resolve F in the horizontal and the vertical directions and apply Newton’s second law in the vertical direction to get 2F sin θ = mg,

H

∝ r5

∆p = (p0 + p + ρgH) − p0 = p + ρgH. By Poiseuille’s equation, volume flow rate in a capillary is given by Q=

π∆pa4 ∆p(πa2 )(a2 /l) = . 8ηl 8η

Substitute the values to get η =

1 720

N s/m2 . 1 Ans. 720 N s/m2

Chapter 12. Some Mechanical Properties of Matter Q 31. A soap bubble is being blown at the end of a very narrow tube of radius b. Air (density ρ) moves with a velocity v inside the tube and comes to rest inside the bubble. The surface tension of the soap solution is T . After some time the bubble, having grown to a radius r, separates from the tube. Find the value of r. Assume that r  b so that you can consider the air to be falling normally on the bubble’s surface. (2003) Bubble Ring of radius b

~v

175 the force acting on the bubble is equal and opposite to the force acting on the tube i.e., Fbubble,left = 4πb2 T /r. The bubble detaches from the tube when Fbubble,right = Fbubble,left , which gives, r = 4T /(ρv 2 ). 4T Ans. ρv 2 Q 32. A point mass m is suspended at the end of massless wire of length l and cross-sectional area A. If Y is the Young’s modulus of elasticity of the material of the wire, obtain the expression for the frequency of the SHM along the vertical line. (1978) Sol. When mass m is pulled by a force F , the wire elongation x, length l, cross-sectional area A, and Young’s modulus of wire material Y are related by Y = F/A x/l i.e.,

Blown Air

F = (Y A/l) x. Sol. The air strikes the inner surface of the bubble and imparts a momentum to it. The mass of the air striking in time ∆t is ∆m = ρ · πb2 · v∆t. The momentum transferred to the bubble in time ∆t is

l

∆p = ∆mv = πb2 ρv 2 ∆t.

Fr

x

m F

θ T dl

θ

O

The restoring force by the wire is equal but opposite to F i.e., Fr = −F . Apply Newton’s second law to get d2 x Y A/l =− x = −ω 2 x. dt2 m

Newton’s second law gives the force on the bubble as Fbubble,right = ∆p/∆t = πb2 ρv 2 . The surface tension produces the force on the tube at the end connected to the bubble (see figure). The force on a small element dl is dFtube = T dl. Resolve this force in the parallel and perpendicular directions to the tube surface to get dFtube,⊥ = T dl cos θ and dFtube,k = T sin θdl ≈ (T b/r) dl. By symmetry, the total force on the tube in the perpendicular direction is zero. Total force on the tube in the parallel direction is Z Ftube,k =

2πb

(T b/r) dl = 2πb2 T /r.

0

The soap bubble has two surfaces and hence the total force on the tube is 4πb2 T /r. By Newton’s third law,

This equation represents SHM with a frequency ν = q Y A/l 1 ω 2π = 2π m . Note that the equivalent spring constant of the wire is Y A/l. q Ans. ν =

1 2π

YA ml

Part II

Waves

S2

d S1

D

177

Chapter 13 Wave Motion and Waves on a String

One Option Correct

A

Q 1. A transverse sinusoidal wave moves along a string in the positive x direction at a speed of 10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snap-shot of the wave is shown in the figure. The velocity of point P when its displacement is 5 cm is (2008)

B

(A) (C)

3π ˆ m/s √50 3π ı m/s 50 ˆ

Sol. Let Tl and Tr be the tensions in the left and the right strings. The forces acting on the rod BD are Tl and Tr upward and mg downwards. In equilibrium, the resultant force and the resultant torque on the system about P are zero i.e.,

P

√

(B) (D)

3π − √50 ˆ m/s 3π − 50 ˆı m/s

Tl + Tr = mg,

(1)

Tl x − Tr (l − x) = 0.

(2)

Solve equations (1) and (2) to get

Sol. The general equation of a wave travelling in the positive x direction is y = A sin(kx − ωt + δ).

D

(A) l/5 (B) l/4 (C) 4l/5 (D) 3l/4

x

√

P m

y •

C x

Tl = mg(l − x)/l,

(3)

Tr = mgx/l.

(4)

(1) The fundamental frequencies in the left and the right strings are given by

For the given wave, δ = 0, A = 0.1 m, k = 2π/λ = 2π/0.5 = 4π, and ω = 2πν = 2πv/λ = 2π 0.1/0.5 = 0.4π. The displacement of the point P is y = 0.05 m. Substitute these values in equation (1) to get

1p Tl /µ, 2y 1p νr = Tr /µ, 2y νl =

0.05 = 0.10 sin(4πx − 0.4πt).

(5) (6)

where, y is the length of the string and µ is the mass per unit length of the string. Given, νl = 2νr . Substitute νl and νr from equations (5) and (6) to get Tl = 4Tr . Substitute Tl and Tr from equations (3) and (4) to get mg(l − x)/l = 4 mgx/l. Solve to get x = l/5. Ans. A

Solutions of this equation are (4πx − 0.4πt) = π/6 and (4πx − 0.4πt) = 5π/6. The point P corresponds to (4πx − 0.4πt) = 5π/6. Differentiate equation (1) to get the velocity of P, dy/dt = −Aω cos(kx − ωt + δ) = −0.10 (0.4π) cos(4πx − 0.4πt) √ = −0.04π cos(5π/6) = 3π/50 m/s. The direction of velocity is along the y-axis or ˆ. Ans. A

Q 3. A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is 1500 m/s and in air it is 300 m/s. The frequency of sound recorded by an observer who is standing in air is (2004) (A) 200 Hz (B) 3000 Hz (C) 120 Hz (D) 600 Hz

Q 2. A massless rod BD is suspended by two identical massless strings AB and CD of equal length. A block of mass m is suspended from point P such that BP is equal to x. If the fundamental frequency of the left wire is twice the fundamental frequency of right wire, then the value of x is (2006)

Sol. The frequency is a characteristic of the source and does not change when sound is transmitted from one medium to another. Since both source and observer are stationary, the frequency of sound recorded by the observer is equal to the source frequency, which is 600 Hz. Ans. D 179

180

Part II. Waves

Q 4. A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is (2002) (A) 25 kg (B) 5 kg (C) 12.5 kg (D) 1/25 kg Sol. The velocity of wave in a string under tension T and mass per unit length µ is p p v = T /µ = M g/µ.

The energy of this element in the case of y1 is dE1 = 12 dma21 ω12 = 21 (m/L)A2 ω 2 sin2 (πx/L)dx, (1) and, in the case of y2 is, dE2 = 21 dma22 ω22 = 2(m/L)A2 ω 2 sin2 (2πx/L)dx. (2) Integrate equations (1) and (2) from x = 0 to x = L to get E1 = 41 mA2 ω 2 , E2 = mA2 ω 2 .

λ1 /2

We encourage you to show that the energy of nth normal mode, yn = A sin(nπx/L) sin nωt, is 41 mn2 A2 ω 2 . Ans. C

l

λ2 /2

The frequency ν of waves in the two cases is same because the string resonates with the same tuning fork. In the first case, 5 v1 5 1p 5 = l = λ1 = 9g/µ, (1) 2 2 ν 2ν and, in the second case, 3 3 v2 3 1p l = λ2 = = M g/µ. (2) 2 2 ν 2ν Divide equation (2) by (1) to get M = 25 kg. Ans. A Q 5. The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment the displacement of the wire is y1 = A sin (πx/L) sin ωt and energy is E1 and in
other experiment its displacement is y2 = A sin 2πx/L sin 2ωt and energy is E2 . Then, (2001)

(A) E2 = E1 (C) E2 = 4E1

(B) E2 = 2E1 (D) E2 = 16E1

Sol. Let mass of the string be m and its mass per unit length be m/L. Consider a small element of length dx at a distance x from x = 0. This element undergoes SHM with amplitude a1 = A sin πx/L and angular frequency ω1 = ω in the case of y1 and amplitude a2 = A sin 2πx/L and angular frequency ω2 = 2ω in the case of y2 . x

dx A

x=0

x=L A

Q 6. Two pulses in a stretched string, whose centres are initially 8 cm apart, are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 s the total energy of the pulses will be (2001)

8cm

(A) (B) (C) (D)

zero. purely kinetic. purely potential. partly kinetic and partly potential.

Sol. Phase difference between the given pulses is 180◦ . The distance traveled by each pulse in 2 s is 4 cm. Thus, after two seconds both pulses will interfere destructively with resultant displacement zero. The potential energy, 1 2 2 kx , also becomes zero. By conservation of energy, total energy of the pulses will be purely kinetic. Ans. B Q 7. Two vibrating strings of the same material but of lengths L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency ν1 and the other with frequency ν2 . The ratio ν1 /ν2 is given by (2000) (A) 2 (B) 4 (C) 8 (D) 1 Sol. Let the lengths and radii of the two strings be l1 = L, r1 = 2r, l2 = 2L, and r2 = r, respectively. Let ρ be the density of the material. The mass and mass per unit length of the first string are m1 = ρ(πr12 l1 ) = 4πρr2 L, µ1 = m1 /l1 = 4πρr2 ,

(1)

Chapter 13. Wave Motion and Waves on a String

181

and that of the second string are m2 = ρ(πr22 l2 ) = 2πρr2 L, µ2 = m2 /l2 = πρr2 .

(2)

In fundamental mode, l = λ/2 = v/(2ν) which gives s v 1 T ν= = . 2l 2l µ Substitute µ from equations (1) and (2) to get s s s 1 1 T1 1 T T ν1 = = , = 2l1 µ1 2L 4πρr2 4L πρr2 s s s T2 T 1 T 1 1 = . ν2 = = 2 2l2 µ2 2(2L) πρr 4L πρr2 Ans. D −2

Q 8. A string of length 0.4 m and mass 10 kg is tightly clamped at its ends. The tension in the string is 1.6 N. Identical wave pulses are produced at one end at equal intervals of time ∆t. The minimum value of ∆t, which allows constructive interference between successive pulses is (1998) (A) 0.05 s (B) 0.10 s (C) 0.20 s (D) 0.40 s Sol. The velocity of sound in a string of mass per unit length µ and tension T is given by s s T 1.6 v= = = 8 m/s. −2 µ 10 /0.4 l A

B

Q 10. The extension in a string, obeying Hooke’s law, is x. The speed of transverse wave in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of transverse wave will be (1996) (A) 1.22v (B) 0.61v (C) 1.50v (D) 0.75v Sol. The velocity of transverse wave in a string under tension T and mass per unit length µ is given by p (1) v = T /µ. The tension in the string is related to its extension x by Hooke’s law T = kx, where k is a constant. The tension is kx in the first case and 1.5kx in the second case. Substitute T = 1.5kx in equation (1) to get velocity in the second case p √ p v 0 = 1.5kx/µ = 1.5 kx/µ = 1.22v. We have assumed x to be much smaller than the string length to keep µ constant. Ans. A Q 11. An object of specific gravity ρ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is 300 Hz. The object is immersed in water, so that one half of its volume is submerged. The new fundamental frequency (in Hz) is (1995)

(A) (C)



1

2 300 2ρ−1  2ρ  2ρ 300 2ρ−1

(B) (D)

1



2 2ρ 300 2ρ−1   300 2ρ−1 2ρ

Sol. Let V be the volume of the object of specific gravity ρ and density ρρw , where ρw is the density of the water. The mass of the object is m = V ρρw , its weight is mg = V ρρw g and tension in the string is T = mg = V ρρw g.

The phase difference between the two pulses should be integral multiple of 2π for constructive interference. The reflection at fixed end introduces an additional phase difference of π. Thus, if a pulse undergoes two reflections before superimposing with a pulse just produced then their interference is constructive. The minimum time required for a wave to go from the end A to end B, get reflected at B, come back to end A, get reflected at A and then get superimposed with a wave pulse just produced is tmin = 2l/v = 2(0.4)/8 = 0.1 s. Ans. B Q 9. A travelling wave in a stretched string is described by the equation: y = A sin(kx − ωt). The maximum particle velocity is (1997) (A) Aω (B) ω/k (C) dω/dk (D) x/ω dy dt

= −Aω cos(kx − ωt), atSol. The particle speed, tains its maximum value Aω when | cos(kx − ωt)| = 1. Ans. A

The fundamental frequency of a string of length l, mass per unit length µ and tension T is s 1 T ν= . (1) 2l µ Let FB be the upward buoyant force on the object when half of its volume is immersed in water. By Archimedes principle, the buoyant force is equal to the weight of the displaced liquid i.e., FB = (V /2)ρw g. After immersion in water, the tension in the string is reduced to T 0 = mg − FB  = V ρρw g − (V /2)ρw g =   2ρ − 1 = T. 2ρ

2ρ − 1 2ρ

 V ρρw g (2)

182

Part II. Waves

The equations (1) and (2) give the new fundamental frequency as ν0 =

1 2l

s

s r r T 0 1 T 2ρ − 1 2ρ − 1 = = 300 Hz. µ 2l µ 2ρ 2ρ Ans. A

Q 12. The displacement y of a particle executing periodic motion is given by y = 4 cos2 (t/2) sin(1000t). This expression may be considered to be a result of the superposition of . . . . . . independent harmonic motions.

One or More Option(s) Correct Q 15. A block M hangs vertically at the bottom end of a uniform rope of constant mass per unit length. The top end of the rope is attached to a fixed rigid support at O. A transverse wave pulse (Pulse 1) of wavelength λ0 is produced at point O on the rope. The pulse takes time TOA to reach point A. If the pulse of wavelength λ0 is produced at point A (Pulse 2) without disturbing the position of M it takes time TAO to reach point O. Which of the following option(s) is(are) correct? (2017) O

Pulse 1

(1992)

(A) two (B) three (C) four (D) five Sol. Given displacement can be written as y = 4 cos2 (t/2) sin(1000t) = 2(1 + cos t) sin(1000t)

Pulse 2

= 2 sin(1000t) + 2 sin(1000t) cos(t)

A M

= 2 sin(1000t) + sin(999t) + sin(1001t). Thus, y is a superposition of three independent harmonic motions of angular frequencies 999 rad/s, 1000 rad/s, and 1001 rad/s. Ans. B Q 13. A wave represented by the equation y = a cos(kx − ωt) is superimposed with another wave to form a stationary wave such that point x = 0 is a node. The equation for the other wave is (1988) (A) a sin(kx + ωt) (B) −a cos(kx − ωt) (C) −a cos(kx + ωt) (D) −a sin(kx − ωt) Sol. A stationary wave is formed by the superposition of two identical waves travelling in the opposite directions. The point x = 0 is a node if displacement of the resultant wave is zero at this point at all times. Thus, the wave superimposed on y = a cos(kx − ωt) is y 0 = −a cos(kx + ωt). Ans. C

(A) The time TAO = TOA . (B) The wavelength of Pulse 1 becomes longer when it reaches point A. (C) The velocity of any pulse along the rope is independent of its frequency and wavelength. (D) The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rope. Sol. Let m be the mass of the block M, l be the length of the rope and µ be the mass per unit length of the rope. Consider a point P at a distance x from O (see figure). Pulse 1 O

x

Pulse 1

x •P

l−x Pulse 2

Q 14. A transverse wave is described by the equation y = y0 sin (2π (f t − x/λ)). The maximum particle velocity is equal to four times the wave velocity if (1984) (A) λ = πy0 /4 (B) λ = πy0 /2 (C) λ = πy0 (D) λ = 2πy0 Sol. It can be seen that the wave y = y0 sin (2π (f t − x/λ)) has the frequency f and wavelength λ. Thus, the wave velocity is v = f λ. The particle velocity for the given wave is vp = ∂y/∂t = y0 (2πf ) cos (2π (f t − x/λ)) . Thus, the maximum particle velocity is vp,max = 2πf y0 . The condition, vp,max = 4v, gives λ = πy0 /2. Ans. B

A M

Pulse 2

The tension in the rope at P balances weight of the block M and weight of the rope segment AP i.e., T (x) = mg + µ(l − x)g. The speed of pulse at point P is given by p p v(x) = T (x)/µ = g(m/µ + l − x). (1) The speed of pulse is a characteristic of the medium i.e., it depends on tension and mass per unit length. The frequency of the pulse, ν, is characteristic of the source of pulse generation. The wavelength relates these two by λ(x) = v(x)/ν. From equation (1), the speed of pulse 1 decreases as it travels from O (x = 0) to A (x = l). Thus, the

Chapter 13. Wave Motion and Waves on a String

183

wavelength of pulse 1 become shorter when it reaches A. Also, speed of pulse 1 and pulse 2 are same at each point because speed is a characteristic of the medium. Hence, speed of the two pulses are same at the mid point (x = l/2). However, since pulse 1 is moving in +x direction and pulse 2 is moving in −x direction, their velocities (which is a vector quantity) at the mid point are not same. Since the speed of two pulses are same atR each dx , point, these pulses will take same time, T = v(x) to travel from O to A or from A to O i.e., TOA = TAO . Let us extend this problem beyond the answer. Use equation (1) p at points O and A p to get the speed of pulse i.e., vO = g(m/µ + l) and vA = gm/µ. Since wavelength of pulse 1 at O and wavelength of pulse 2 at A are equal to λ0 , the frequencies of two pulses are given by p p g(m/µ + l) gm/µ vA vO = , and ν2 = = . ν1 = λ0 λ0 λ0 λ0 The wavelengths of pulse 1 and pulse 2 at point P are given by λ1 (x) = v(x)/ν1 and λ2 (x) = v(x)/ν2 . The change in wavelengths of two pulses as they travel from one end to another end can be seen in the figure. The time taken by pulse 1 to travel from O to A is Z

l

TOA = 0

dx = v(x)

l

Z

dx p , g(m/µ + l − x)

0

0

3m

x

The fundamental frequency is given by ν0 =

v v 100 25 = = = Hz. λ 4l 4(3) 3

Thus, the waveform will satisfy equations (1) and (2), and the permissible frequencies will be odd multiples of ν0 . Ans. A, C, D Q 17. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation,     y(x, t) = (0.01 m) sin (62.8 m−1 )x cos (628 s−1 )t Assuming π = 3.14, the correct statement(s) is (are) (2013)

(A) The number of nodes is 5. (B) The length of the string is 0.25 m. (C) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m. (D) The fundamental frequency is 100 Hz. Sol. The general expression for a standing wave on a string fixed at both ends is

and time taken by pulse 2 to travel from A to O is y = 2A sin kx cos ωt. Z

0

TAO = l

dx = −v(x)

Z 0

l

dx

p . g(m/µ + l − x)

(1) L

Ans. (A), (C) P

Q 16. One end of a taut string of length 3 m along the x axis is fixed at x = 0. The speed of the waves in the string is 100 m/s. The other end of the string is vibrating in the y direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are) (2014) 50πt cos (A) y(x, t) = A sin πx 6 3 100πt (B) y(x, t) = A sin πx 3 cos 3 250πt (C) y(x, t) = A sin 5πx 6 cos 3 (D) y(x, t) = A sin 5πx 2 cos 250πt Sol. The displacement of a stationary wave is given by   2πx y(x, t) = A sin cos(2πνt). λ The boundary conditions give node at x = 0 and antinode at x = 3 m i.e., y(0, t) = 0,

(1)

y(3, t) = ±A.

(2)

In the fifth harmonic, kL = 5π and ω = 5ω0 = 5 × 2πν0 , where L is the length of the string and ν0 is its fundamental frequency. The fifth harmonic is shown in the figure. It has six nodes. Compare the given expression,     y(x, t) = (0.01 m) sin (62.8 m−1 )x cos (628 s−1 )t , with equation (1) to get 2A = 0.01 m, k = 62.8 m−1 , and ω = 628 s−1 . Use these values to get maximum displacement of midpoint (antinode at P ) as 0.01 m, L = 0.25 m, and ν0 = 20 Hz. Ans. B, C Q 18. Standing waves can be produced (1999) (A) on a string clamped at both ends. (B) on a string clamped at one end and free at the other. (C) when incident wave gets reflected from a wall. (D) when two identical waves with a phase difference of π are moving in the same direction.

184

Part II. Waves

Sol. Standing waves are produced when two waves of same frequency and same amplitude travel through a medium in the opposite directions. Ans. A, B, C Q 19. In a wave motion y = a sin(kx − ωt), y can represent (1999) (A) electric field (B) magnetic field (C) displacement (D) pressure Sol. The y = a sin(kx − ωt) represents a progressive wave traveling along the x direction. The parameter y is electric or/and magnetic field in electromagnetic waves, displacement of a particle in transverse waves on string, and pressure in the case of longitudinal sound waves. Ans. A, B, C, D 0.8 Q 20. y(x, t) = (4x+5t) 2 +5 represents a moving pulse where x and y are in metre and t is in second. Then,

(1999)

(A) (B) (C) (D)

pulse is moving in positive x direction. in 2 s it will travel a distance of 2.5 m. its maximum displacement is 0.16 m. it is a symmetric pulse.

Sol. The displacement of a point on the wave is given by y = a sin (2πf t − 2πx/λ) . The velocity of the wave is v = f λ and velocity of a particle on the wave at fixed x is u = ∂y/∂t = 2πaf cos (2πf t − 2πx/λ) . Maximum value of u is umax = 2πaf . Substitute a = 10−3 m, v = 10 m/s and umax = v/10 = 1 m/s to get 3 −2 m. f = 10 2π Hz and λ = 2π × 10 Ans. A, C Q 22. The (x, y) coordinates of the corners of a square plate are (0, 0), (L, 0), (L, L), and (0, L). The edges of the plate are clamped and transverse standing waves are set-up in it. If u(x, y) denotes the displacement of the plate at the point (x, y) at some instant of time, the possible expression(s) for u is (are) [Where a is a positive constant.] (1998) πy πy πx (A) a cos 2L cos 2L (B) a sin πx sin L L 2πy πy (C) a sin πx (D) a cos 2πx L sin L L sin L

Sol. The pulse at t = 0, 0.8 , Y (x, 0) = 16x2 + 5 is shown in the figure. Y 0.16m v

Q 21. A transverse sinusoidal wave of amplitude a, wavelength λ and frequency f is travelling on a stretched string. The maximum speed of any point on the string is v/10, where v is the speed of propagation of the wave. If a = 10−3 m and v = 10 m/s, then λ and f are given by (1998) (A) λ = 2π × 10−2 m (B) λ = 10−3 m 3 (D) f = 104 Hz (C) f = 10 2π Hz

•

Sol. The plate is shown in the figure. y

−x

t=0

x

y (0, L) C

x

It is a symmetric pulse as Y (−x, 0) = Y (x, 0) for all x. The amplitude of the pulse is Y (x, 0)max = 0.16 m. Let us consider a point on the pulse having a displacement y (a constant). Since this point lies on the pulse, we have 0.8 y= , (1) (4x + 5t)2 + 5 for all x and t. As the time progresses, the value of x changes such that equation (1) remains valid. In other words, x varies with t, i.e., point moves with time. Differentiate equation (1) w.r.t. time t (note that y is constant) to get

O (0, 0)

B (L, L)

A x (L, 0)

The four edges of the plate are OA, AB, BC, and CO. Since edges are clamped, the displacement u(x, y) of the plate on all the four edges should be zero. These are called boundary conditions on the transverse standing waves set-up in the plate. The four edges are given by OA : {0 ≤ x ≤ L, y = 0}, AB : {x = L, 0 ≤ y ≤ L},

(2)

BC : {0 ≤ x ≤ L, y = L}, CO : {x = 0, 0 ≤ y ≤ L}.

Thus, point moves with a speed v = −1.25 m/s. Negative sign indicates a movement towards the −x direction. The argument is valid for any point on the pulse and hence entire pulse moves towards the left with speed v = −1.25 m/s. In two seconds, pulse travels a distance of vt = 1.25 × 2 = 2.5 m. Ans. B, C, D

The displacement becomes zero on all the four edges in the following two cases πy πx sin , u(x, y) = a sin L L πx 2πy u(x, y) = a sin sin . L L Ans. B, C

dx/dt = −1.25 m/s.

Chapter 13. Wave Motion and Waves on a String

185

Q 23. A wave disturbance in a medium is described by, y(x, t) = 0.02 cos (50πt + π/2) cos(10πx), where x and y are in metre and t is in second. (A) A node occurs at x = 0.15 m (B) An antinode occurs at x = 0.3 m (C) The speed of wave is 5 m/s (D) The wavelength of wave is 0.2 m

(1995)

Sol. Given wave disturbance, y(x, t) = 0.02 cos (50πt + π/2) cos(10πx),

(1)

Q 25. The displacement of particles in a string stretched in the x direction is represented by y. Among the following expressions for y, those describing wave motion is (are) (1987) (A) cos kx sin ωt (B) k 2 x2 − ω 2 t2
(C) cos2 (kx + ωt) (D) cos k 2 x2 − ω 2 t2 Sol. For a wave travelling along x-axis, the displacement should be a function of x ± vt, where v is the wave velocity. The displacement y = cos kx sin ωt can be rewritten as

is of the form, y(x, t) = A cos (ωt + π/2) cos(kx).

(2)

Compare equations (1) and (2) to get A = 0.02, ω = 50π, and k = 10π. The frequency is ν = ω/(2π) = 25 Hz and the wavelength is λ = 2π/k = 0.2 m. The speed of the wave is v = νλ = 5 m/s. The nodes (zero displacement) occur when cos(kx) = 0 i.e., at xn = (2n − 1)π/(2k),

(3)

where n = 1, 2, . . .. Substitute the values in equation (3) to get x1 = 0.05 m, x2 = 0.15 m, x2 = 0.25 m etc. The anti-nodes (maximum displacement) occur when cos(kx) = ±1 i.e., at x0n = nπ/k,

y = cos kx sin ωt     kx + ωt 1 kx − ωt 1 − sin = sin 2 2 2 2       1 ω  ω  k 1 k = sin x + t − sin x− t , 2 2 k 2 2 k

(4)

where n = 0, 1, 2, . . .. Substitute the values in equation (4) to get x00 = 0 m, x01 = 0.1 m, x02 = 0.2 m, x03 = 0.3 m etc. Ans. A, B, C, D Q 24. A wave is represented by the equation: y = A sin (10πx + 15πt + π/3), where x is in metre and t is in second. The expression represents (1990) (A) a wave travelling in positive x direction with a velocity 1.5 m/s. (B) a wave travelling in negative x direction with a velocity 1.5 m/s. (C) a wave travelling in negative x direction with a wavelength 0.2 m. (D) a wave travelling in positive x direction with a wavelength 0.2 m. Sol. Compare the given equation, y = A sin (10πx + 15πt + π/3) , with the general equation of the wave, y = A sin(kx + ωt + φ), to get λ = 2π/k = 2π/(10π) = 0.2 m and ν = ω/2π = 15π/(2π) = 7.5 Hz. The speed of the wave is v = νλ = 7.5 × 0.2 = 1.5 m/s. The wave is travelling in the negative x direction. Ans. B, C

which represents the superposition of two waves travelling in the opposite directions with a velocity ω/k. The displacement y = cos2 (kx + ωt) can be written as h  ω i 1 1 y = cos2 (kx + ωt) = + cos 2k x + t , 2 2 k which represents a wave travelling along −x direction with a velocity ω/k. The displacements y = k 2 x2 −ω 2 t2 and y = cos(k 2 x2 −ω 2 t2 ) cannot be written as a function of x ± vt. Aliter: The differential equation for the wave motion is ∂2y ∂2y = c , ∂x2 ∂t2

(1)

where c is a constant, ∂y/∂x is the partial derivative of y w.r.t. x and ∂y/∂t is the partial derivative of y w.r.t. time t. We encourage you to check the given displacements with equation (1). Ans. A, C Q 26. The dimensions of the quantities in one (or more) of the following pairs are the same. Identify the pair(s), (1986)

(A) (B) (C) (D)

torque and work angular momentum and work energy and Young’s modulus light year and wavelength

Sol. The dimensions of the torque and the work are [ML2 T−2 ]. The dimesions of light year and wavelength are [L]. We encourage you to find the dimensions of other quantities. Ans. A, D Q 27. A wave equation which gives the displacement along the y direction is given by: y = 10−4 sin(60t + 2x) where x and y are in metre and t is in second. This represents a wave (1982)

186

Part II. Waves

(A) travelling with a velocity of 30 m/s in the negative x direction. (B) of wavelength π m. (C) of frequency 30/π Hz. (D) of amplitude 10−4 m .

1 Sol. The disturbance at t = 0 is y = (1+x) 2 and at 1 t = 2 is y = 1+(x−1)2 . The statement “shape of the disturbance does not change with time” is not correct which can be seen from the figure.

y

Sol. Compare y = 10−4 sin(60t + 2x) with the general equation of a progressive wave y = A sin(kx + ωt),

t=0

1.0 • 0.5 •

(1)

to get amplitude A = 10−4 m, k = 2 m−1 , and ω = 60 rad/s. The wavelength of the given wave is λ = 2π/k = 2π/2 = π m, and the frequency is ν = ω/(2π) = 60/(2π) = 30/π Hz. The velocity of the wave is v = νλ = ω/k = 60/2 = 30 m/s. The wave in equation (1) is travelling in the negative x direction. Ans. A, B, C, D

−1

y = A sin (2πνt + 2πx/λ) , where A = 2.5 × 10−5 m, ν = 25 Hz, and λ = v/ν = 300/25 = 12 m. Let x1 and x2 be the two points separated by 6 m i.e., x2 = x1 + 6. The phase at x1 is φ1 = 2πνt + 2πx1 /λ,

(1)

x

t=0

t=2

0

1

x

We believe that there is a typographical error in the question. We encourage you to solve this problem 1 by taking disturbance at t = 0 as y = 1+(1+x) 2 instead 1 of y = (1+x)2 , other things remains the same. Hint: See figure. Ans. 1 Q 30. In a sonometer wire, the tension is maintained by suspending a 50.7 kg mass from the free end of the wire. The suspended mass has a volume of 0.0075 m3 . The fundamental frequency of vibration of the wire is 260 Hz. If the suspended mass is completely submerged in water, the fundamental frequency will become . . . . . . Hz. (1987) Sol. The tension in the wire when mass m = 50.7 kg is suspended from it is T = mg.

and the phase at x2 is φ2 = 2πνt + 2πx2 /λ = 2πνt + 2π(x1 + 6)/λ.

1

y

−1

Sol. Equation of the given progressive wave is

0

In this case, we can comment on the speed with which maximum of the disturbance travels. At t = 0, the maximum is at x = −1 m and at t = 2 s maximum is at x = 1 m. Thus, maximum moves from x = −1 m to 1−(−1) = x = 1 m in 2 s, giving its speed as v = ∆x ∆t = 2 1 m/s. We encourage you to find the speed of the point y = 1.

Fill in the Blank Type Q 28. A plane progressive wave of frequency 25 Hz, amplitude 2.5 × 10−5 m and initial phase zero, propagates along the negative x direction with a velocity of 300 m/s. At any instant, the phase difference between the oscillations at two points 6 m apart along the line of propagation is . . . . . . and the corresponding amplitude difference is . . . . . . m. (1997)

t=2

(2)

Subtract equation (1) from (2) to get the phase difference between x1 and x2 as φ2 − φ1 = π. The amplitude does not change in a plane progressive wave. Ans. π, zero Q 29. The amplitude of a wave disturbance travelling 1 in the positive x direction is given by y = (1+x) 2 at time 1 t = 0 and by y = 1+(x−1)2 at t = 2 s, where x and y are in metres. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is . . . . . . m/s. (1990)

When the suspended mass of volume V = 0.0075 m3 is submerged in water of density ρ = 1000 kg/m3 , the tension in the wire reduces due to the buoyant force on the submerged mass. The tension in the wire now becomes T 0 = mg − FB = mg − ρV g. The fundamental frequencies of sonometer wire of length l, mass per unit length µ, and under the tensions T and T 0 are given by p p 1 1 T /µ = 2l mg/µ, (1) ν = 2l p p 1 1 ν 0 = 2l T 0 /µ = 2l (mg − ρV g)/µ. (2)

Chapter 13. Wave Motion and Waves on a String

187

Divide equation (2) by (1) and substitute the values to get r

m − ρV ν m

r

50.7 − 1000(0.0075) 260 = 240 Hz. 50.7

ν0 = =

Q 33. Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, π/3, 2π/3 and π. When they are superposed, the intensity of the resulting wave is nI0 . The value of n is . . . . . . . (2015)

Ans. 240 Q 31. Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particles have maximum amplitude of vibration is . . . . . . m. [Speed of sound = 330 m/s.] (1984) Sol. The wall acts as a rigid boundary and reflects the incident wave. Thus, there is a displacement node at the wall (similar to a pipe closed at one end). A

N

A

N

l = λ/4

The incident wave and the reflected waves superimpose to give stationary waves. The displacement of the stationary wave is maximum at the antinodes. Minimum distance of the antinode from the wall is l=

Integer Type

Sol. The intensity of a wave is proportional to the square of its amplitude i.e., I0 = cA2 , where c is a constant. The amplitudes of four harmonic waves are equal as their intensities are equal. Let these waves be travelling along the x direction with wave vector k and angular frequency ω. The resultant displacement of these waves is given by y = y1 + y2 + y3 + y4 = A sin(ωt − kx + 0) + A sin(ωt − kx + π/3) + A sin(ωt − kx + 2π/3)+A sin(ωt − kx + π) = A sin(ωt − kx + π/3) + A sin(ωt − kx + 2π/3) = 2A sin(ωt − kx + π/2) cos(π/6) √ = 3A cos(ωt − kx). √ The amplitude of the resultant wave is Ar = 3A and its intensity is Ir = cA2r = 3cA2 = 3I0 . Note that y1 and y4 are out of phase and interfere destructively. The displacement y2 and y3 have a phase difference of δ = π/3. Thus, we can arrive at the resultant intensity by using the formula Ir = I0 + I0 + 2

p p I0 I0 cos δ

= 2I0 + 2I0 cos(π/3) = 3I0 .

λ v 330 = = = 0.125 m. 4 4ν 4(660)

Aliter: The figure show the four harmonic waves on the phase diagram.

Ans. 0.125 Q 32. A travelling wave has the frequency ν and the particle displacement amplitude A. For the wave the particle velocity amplitude is . . . . . . and the particle acceleration amplitude is . . . . . . (1983)

Ar

y3

y2 2π 3

Sol. The displacement of a travelling wave of frequency ν and particle displacement amplitude A is given by y4

π 3

y1

y = A cos(2πx/λ − 2πνt).

(2)

The magnitude of each vector is A. The displacements y1 and y4 are out of phase. The resultant of these two vectors is zero. The angle between y2 and y3 is π/3. Use the law of vector addition to get the resultant of y2 √ and y3 as Ar = 3A. Ans. 3

From equations (1) and (2), vp,max = 2πνA and ap,max = 4π 2 ν 2 A. Ans. 2πνA, 4π 2 ν 2 A

Q 34. When two progressive waves y1 = 4 sin(2x − 6t) and y2 = 3 sin(2x−6t− π2 ) are superimposed, the amplitude of the resultant wave is . . . . . . . (2010)

Differentiate w.r.t. time t to get the particle velocity and differentiate it again to get the particle acceleration i.e., vp = dy/dt = 2πνA sin(2πx/λ − 2πνt), 2

2

2 2

ap = d y/dt = −4π ν A cos(2πx/λ − 2πνt).

(1)

188

Part II. Waves

Sol. The resultant wave is given by y = y1 + y2 = 4 sin(2x − 6t) + 3 sin(2x − 6t − π/2) = 4 sin(2x − 6t) − 3 cos(2x − 6t)  p 4 2 2 = 4 +3 √ sin(2x − 6t) 2 4 + 32  3 −√ cos(2x − 6t) 42 + 32   = 5 cos δ sin(2x − 6t) − sin δ cos(2x − 6t) = 5 sin(2x − 6t − δ),

which attains a maximum value of ω 2 A. Given, ωA = 3 m/s and ω 2 A = 90 m/s2 . Solve to get ω = 30 rad/s and A = 0.1 m. The wave velocity v = νλ = ω/k = 20 m/s gives k = ω/v = 3/2 m−1 . The wave travelling towards the +x direction is y = 0.1 sin (3x/2 − 30t + φ) , and that travelling towards the −x direction is y = 0.1 sin (3x/2 + 30t + φ) . Ans. y = 0.1 sin 30t ± 23 x + φ

−1

where δ = cos (4/5). We encourage you to find the resultant by vector addition method. Hint: y1 and y2 have a phase difference of π/2. Ans. 5 Q 35. A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string. (2009)




Q 37. A string of mass per unit length µ is clamped at both ends such that one end of the string is at x = 0 and the other is at x = l. When string vibrates in fundamental mode, amplitude of the mid-point O of the string is a, and tension in the string is T . Find the total oscillation energy stored in the string. (2003) Sol. In the fundamental mode, l = λ/2. x

Sol. The speed of sound in a string having tension T = −3 0.5 N and mass per unit p length µ = m/l = 10 /0.2 = −3 5 × 10 kg/m is v = T /µ = 10 m/s.

dx a

x=0

x=l

20 cm

N

N

N

N

N

The velocity of sound in a string p with tension T and mass per unit length µ is v = T /µ. Thus, the frequency of vibration is

λ/2

ν= The wavelength of wave with frequency 100 Hz is λ = v/ν = 10/100 = 0.1 m = 10 cm. The distance between the two successive nodes is λ/2 = 5 cm (see figure). Ans. 5 Descriptive Q 36. A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration of 3 m/s and 90 m/s2 respectively. Velocity of the wave is 20 m/s. Find the waveform. (2005) Sol. The equation of a progressive wave travelling in the +x direction is y = A sin(kx − ωt + φ) = A sin (2πx/λ − 2πνt + φ) . Differentiate w.r.t. time to get the particle velocity dy/dt = −Aω cos(kx − ωt + φ), which attains a maximum value of ωA. Differentiate again w.r.t. time to get the particle acceleration d2 y/dt2 = −Aω 2 sin(kx − ωt + φ),

v 1p T /µ. = λ 2l

The amplitude in the middle (i.e., at x = l/2) is a and amplitude at a distance x from x = 0 is A = a sin(2πx/λ) = a sin(πx/l). Consider a small element of length dx at a distance x from x = 0. This element of mass dm = µdx undergoes SHM with frequency ν and amplitude A. The energy of this element is dE = =

1 2

dm ω 2 A2 =

1 2

µdx · 4π 2 ν 2 · a2 sin2 (πx/l)

π 2 a2 T sin2 (πx/l) dx. 2l2

Integrate dE from x = 0 to x = l to get E = π 2 a2 T /(4l). Ans. π 2 a2 T /(4l) Q 38. A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has length 4.8 m and mass 0.06 kg. QR has length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N. A sinusoidal wave pulse of amplitude 3.5 cm is sent along the wire PQ from the end P. No power is dissipated during the propagation of the wave pulse. Calculate, (1999)

Chapter 13. Wave Motion and Waves on a String

189 x=0

(a) the time taken by the wave pulse to reach the other end R. (b) the amplitude of the reflected and transmitted wave pulse after the incident wave pulse crosses the joint Q. Sol. The speed of sound in a wire p with mass per unit length µ and tension T is v = T /µ. Substitute the values to get the speed of sound in wire PQ, s s T1 80 v1 = = = 80 m/s, µ1 0.06/4.8 and that in the wire QR, s 80 v2 = = 32 m/s. 0.2/2.56

l = 1m

The free ends of the rod are anti-nodes, clamped point is a node, and there are two nodes on either sides of clamped point. From the figure, length l = 1 m = 5λ/2, which gives λ = 0.4 m. The speed of the wave in the rod is given by s r Y 2 × 1011 v= = = 5000 m/s, ρ 8000 and its frequency is ν = v/λ = 12500 Hz. The equation of a stationary wave with node at x = 0 is given by

Time taken by the pulse to reach R is tPQ + tQR = 4.8/80 + 2.56/32 = 0.14 s. The wave is reflected and transmitted at Q. Let the incident, reflected, and the transmitted waves be,

y = A sin (2πx/λ) sin(2πνt),

(1)

where A = 2 × 10−6 m is the amplitude at the antinode. Substitute the values of parameters in equation (1) to get y = 2 × 10−6 sin(5πx) sin(25000πt).

yi = Ai sin(ωt − ki x), yr = Ar sin(ωt + kr x),

(2)

Substitute x = 0.02 m in equation (2) to get the equation of motion of a particle at x = 0.02 m,

yt = At sin(ωt − kt x), where Ai = 3.5 cm is the amplitude of the incident wave and ki = ω/v1 is the wave vector of the incident wave. The incident wave and the reflected wave are in the same medium which gives k1 = ki = kr = ω/v1 . The transmitted wave is in an another medium which gives k2 = kt = ω/v2 . The superposed wave in PQ is y1 = yi +yr and wave in QR is y2 = yt . At the boundary, say x = 0, the displacement and its slope are continuous dy2 1 i.e., y1 |x=0 = y2 |x=0 and dy dx x=0 = dx x=0 . These conditions give, Ai + Ar = At ,

(1)

k1 (Ai − Ar ) = k2 At .

(2) v2 −v1 v2 +v1 Ai

and Solve equations (1) and (2) to get Ar = 2v2 At = v2 +v1 Ai . Substitute Ai , v1 , and v2 , to get, Ar = −1.5 cm and At = 2 cm. Ans. (a) 0.14 s (b) Ar = 1.5 cm, At = 2.0 cm Q 39. A metallic rod of length 1 m is rigidly clamped at its mid point. Longitudinal stationary waves are set-up in the rod in such a way that there are two nodes on either side of the mid-point. The amplitude of an antinode is 2 × 10−6 m. Write the equation of motion at a point 2 cm from the mid-point and those of the constituent waves in the rod. [Young’s modulus of the material of the rod = 2 × 1011 N/m2 , density = 8000 kg/m3 .] (1994) Sol. The nodes and the anti-nodes are shown in the figure.

y = 2 × 10−6 sin(0.1π) sin(25000πt). The equations of constituent progressive waves are y1 = (A/2) sin (2πνt − 2πx/λ) = 10−6 sin(25000πt − 5πx), y2 = (A/2) sin (2πνt + 2πx/λ) = 10−6 sin(25000πt + 5πx). 10

−6

Ans. y = 2 × 10−6 sin(0.1π) sin(25000πt), y1 = sin(25000πt − 5πx), y2 = 10−6 sin(25000πt + 5πx)

Q 40. The displacement of the medium in a sound wave is given by the equation yi = A cos(ax + bt), where A, a and b are positive constants. The wave is reflected by an obstacle situated at x = 0. The intensity of the reflected wave is 0.64 times that of the incident wave. (1991)

(a) What are the wavelength and frequency of incident wave? (b) Write the equation for the reflected wave. (c) In the resultant wave formed after reflection, find the maximum and minimum values of the particle speeds in the medium. (d) Express the resultant wave as a superposition of a standing wave and a travelling wave. What are the positions of the antinodes of the standing wave? What is the direction of propagation of travelling wave?

190

Part II. Waves

Sol. Compare yi = A cos(ax + bt) with the standard wave equation y = A cos(kx + ωt) to get λ=

2π 2π = , k a

ν=

ω b = . 2π 2π

y A

z12 = z1 + z2 = A cos(kx − ωt) + A cos(kx + ωt)

yi

= 2A cos kx cos ωt.

x

O

Sol. The standing waves are formed by superposition of two similar waves travelling in the opposite directions. This condition is satisfied by waves z1 = A cos(kx − ωt) and z2 = A cos(kx + ωt). The displacement of the resultant standing wave is given by

yr 0.8A

A12

This wave is travelling in the negative x direction. The reflected wave moves in the positive x direction with the same λ and ν. The reflection from a denser medium (obstacle) introduces an additional phase difference of π. The amplitude of the reflected wave is 0.8A as its intensity (I ∝ A2 ) is 0.64 times the intensity of the incident wave. Thus, the equations of reflected wave yr and the resultant waves y are yr = 0.8A cos(ax − bt + π) = −0.8A cos(ax − bt), y = yi + yr = A cos(ax + bt) − 0.8A cos(ax − bt).

(1)

Differentiate equation (1) w.r.t. time to get the particle speed vp = dy/dt = −Ab sin(ax + bt) − 0.8Ab sin(ax − bt). Maximum particle speed is 1.8Ab at the points satisfying sin(ax + bt) = ±1 and sin(ax − bt) = ±1. Minimum particle speed is zero at the points satisfying sin(ax + bt) = 0 and sin(ax − bt) = 0. Rewrite the resultant wave of equation (1) as y = 0.8A [cos(ax + bt) − cos(ax − bt)] + 0.2A cos(ax + bt) = −1.6A sin ax sin bt + 0.2A cos(ax + bt).

(2)

The second part of equation (2) is a travelling wave of amplitude 0.2A moving in the −x direction. The first part is a stationary wave. The anti-nodes of the stationary waves i are points where sin ax = ±1 i.e., at h (−1)n π x= n+ 2 a. Ans. (a) 2π/a, b/(2π) (b) −0.8A cos(ax − bt) (c) 1.8Ab, zero (d) y = −1.6A sin ax h i sin bt+0.2A cos(ax+bt). Antinodes at x = n+

(−1)n 2

π a.

Negative x direction.

Q 41. The following equations represent transverse waves: z1 = A cos(kx − ωt), z2 = A cos(kx + ωt), z3 = A cos(ky − ωt). Identify the combination(s) of the waves which will produce (a) standing wave(s), (b) a wave travelling in the direction making an angle of 45◦ with the positive x and positive y-axes. In each case, find the position at which the resultant intensity is always zero. (1987)

x I12 x

The amplitude of resultant wave varies with x as A12 = 2A cos kx. The intensity of a wave is proportional to the square of its amplitude. Thus, intensity becomes zero wherever the amplitude A12 = 2A cos kx = 0 i.e., at the points π x = (2n + 1) 2k , where n = 0, ±1, ±2, . . . .

The wave z1 = A cos(kx − ωt) travels along the positive x direction and z3 = A cos(ky − ωt) travels along the positive y direction. The resultant wave is z13 = z1 + z3 = A cos(kx − ωt) + A cos(ky − ωt)     k(x + y) k(x − y) cos − ωt . = 2A cos 2 2 We encourage you to prove analytically or otherwise that the resultant wave makes an angle of 45◦ with the positive x and positive y axes.h The amplitude of the i k(x−y) resultant wave is A13 = 2A cos . The intensity 2 becomes zero wherever amplitude A13 = 0 i.e., at the points x − y = (2n + 1) πk , where n = 0, ±1, ±2, . . . . Ans. (a) z1 and z2 ; π x = (2n + 1) 2k where n = 0, ±1, ±2, . . . (b) z1 and z3 ; x − y = (2n + 1) πk where n = 0, ±1, ±2, . . . Q 42. The vibration of a string of length 60 cm fixed at both
ends are represented by the equation y = 4 sin πx 15 cos(96πt), where x and y are in cm and t in second. (1985) (a) What is the maximum displacement of a point at x = 5 cm? (b) Where are the nodes located along the string? (c) What is the velocity of the particle at x = 7.5 cm at t = 0.25 s. (d) Write down the equations of the component waves whose superposition gives above wave.

Chapter 13. Wave Motion and Waves on a String

191

Sol. The displacement of a point on the string at position x and time t is given by y(x, t) = 4 sin (πx/15) cos(96πt),

(1)

where x and y are in cm and t is in second. Substitute x = 5 cm in equation (1) to get the displacement of this point as y(x = 5, t) = 4 sin (π/3) cos(96πt) √ = 2 3 cos(96πt).

Sol. The rope has mass M = 12 kg, length l = 12 m, and mass per unit length µ = M/l = 1 kg/m. A block of mass m = 2 kg is hanging from the lower end. Note that the tension in the rope varies along its length because the rope is not massless. Consider a point P located at a height x above the lowest point of the rope. Mass of the portion below P is Mx = µx = xM/l. The tension Tx in the rope at P balances the weight of the rope below P and the weight of the block of mass m i.e.,

(2) Tx = mg + Mx g = mg + M (x/l)g.

The displacement at x = 5 cm becomes maximum when √ cos(96πt) = 1 and the maximum displacement is 2 3 cm.

Thus, the tension in the rope varies linearly with x with its values T0 = mg at the bottom and Tl = (m + M )g at the top.

y O •

0

•

•

•

•

15

30

45

60

x Tx

The nodes are points at which the displacement remains zero for all values of t. From equation (1), the nodes are the points that lie between the two ends of the string (i.e., 0 ≤ x ≤ 60 cm) and satisfy

P

l x

T0 mg

sin(πx/15) = 0.

(3)

Solve equation (3) to get x = 0 cm, 15 cm, 30 cm, 45 cm, 60 cm. The particle velocity is given by vp (x, t) = ∂y/∂t = −384π sin (πx/15) sin (96πt) .

(4)

Substitute x = 7.5 cm and t = 0.25 s in equation (4) to get vp = 0 cm/s. Use trigonometric identity to rewrite equation (1) as y(x, t) = 2 (2 sin (πx/15) cos(96πt)) = 2 sin (πx/15+96πt)+2 sin (πx/15−96πt) = y1 + y2 .

(5)

Thus, the wave given by equation (5) is a superposition of two waves, y1 and y2 . We encourage you to find the amplitude, frequency, velocity and direction for each of these waves. √ Ans. (a) 2 3 cm (b) x =
πx 0, 15 cm, 30 cm . . . (c) zero
(d) y1 = 2 sin 15 − 96πt πx and y2 = 2 sin 15 + 96πt Q 43. A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.06 m is produced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope? (1984)

The velocity of a wave in a string having mass per unit length µ and tension Tx is given by s s r Tx (m + M x/l)g g(lm + xM ) = = . vx = µ M/l M Note that the velocity of the wave p varies along length of the rope p with its values v0 = lgm/M at the bottom and vl = gl(m + M )/M at the top. The frequency of the wave does not vary as it travels in a medium. Given, wavelength of the wave at the bottom is λ0 = 0.06 m. Thus, frequency of the given wave is p glm/M v0 ν= = . λ0 λ0 Hence, the variation of the wavelength with x is given by p g(lm + xM )/M vx p λx = = λ0 ν glm/M r lm + xM λ0 . = lm p Substitute the values to get λl = (8/2) 0.06 = 0.12 m. The figure shows the variation of λx along the rope. Ans. 0.12 m

Chapter 14 Sound Waves

One Option Correct

where, v(= 330 m/s) is the velocity of sound, us (= +36 km/h = +10 m/s) is the velocity of source (siren) and uo = +10 m/s is the velocity of observer (driver). Substitute these values to get v + uo v + uo v ν0 = ν0 = 8.5 kHz. ν2 = v v − us v − us

Q 1. A student is performing the experiment of resonance column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38 ◦ C at which the speed of sound is 336 m/s. The zero of meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is (2012) (A) 14.0 cm (B) 15.2 cm (C) 16.4 cm (D) 17.6 cm

Aliter: The building is like a fixed mirror (a reflector). The driver looks into the mirror and finds a source of frequency ν0 moving towards him with the same speed (10 m/s). Thus, using Doppler’s effect, we o get, ν2 = v+u v−us ν0 . Ans. A

Sol. Let l1 be the length of the air column and d be the end correction. First resonance will occur when v 336 λ = = 16.4 cm. l1 + d = = 4 4ν 4(512)

Q 3. A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the string is 50 N and the speed of sound is 320 m/s, the mass of the string is (2010) (A) 5 grams (B) 10 grams (C) 20 grams (D) 40 grams

(1)

The end correction is related to the tube radius r by d = 0.6r = 0.6(2) = 1.2 cm. Substitute d in equation (1) to get l1 = 16.4 − 1.2 = 15.2 cm. Ans. B

Sol. The fundamental mode in a pipe closed at one end and the second harmonic in a string are shown in the figure.

Q 2. A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/h towards a tall building which reflect the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is (2011) (A) 8.50 kHz (B) 8.25 kHz (C) 7.75 kHz (D) 7.50 kHz

ν1 ν1 ν2

The source of ν1 is a moving siren and observer is a stationary building. For ν2 , source is a stationary building and observer is a moving driver. Change in frequency due to Doppler’s effect is given by ν1 =

v ν0 , v − us

ν2 =

λp /4

λs

where vp = 320 m/s is the velocity of sound in the pipe and Lp = 0.8 m is length of the pipe. For string of mass m, length Ls and having tension T , velocity of sound in the string is given by p r T /(m/Ls ) vs T νs = = = λs Ls mLs s 50 10 = =√ . (2) m(0.5) m

ν0

vo

Ls

It can be seen that λp /4 = Lp and λs = Ls . For the pipe closed at one end, vp 320 vp = = = 100 Hz, (1) νp = λp 4Lp 4(0.8)

Sol. Let ν0 be the frequency of sound emitted by siren, ν1 be apparent frequency incident on building (which is the same as the frequency reflected by it), and ν2 be the frequency heard by the driver. vs

Lp

At resonance, νp = νs . Substitute νp and νs from equations (1) and (2) to get m = 0.01 kg = 10 g. Ans. B

v + uo ν1 , v 192

Chapter 14. Sound Waves

193

Q 4. A vibrating string of certain length l under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces to 2 per second. Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork (in Hz) is (2008) (A) 344 (B) 336 (C) 117.3 (D) 109.3 Sol. The third harmonic in a pipe closed at one end occurs at a frequency ν = 3v/(4L) = 3 × 340/(4 × 0.75) = 340 Hz. Since string and closed pipe are in resonance, the frequency of string is ν = 340 Hz. The tuning fork of frequency n produces 4 beats with the string of frequency ν. Thus, n − ν = 4 or ν − n = 4 i.e., n = 344 Hz or n = 336 Hz. ν:

l2

The figure shows the variation of particle displacement along the pipe. We encourage you to draw the variation of excess pressure. Ans. A Q 6. A tuning fork of 512 Hz is used to produce resonance in a resonance tube experiment. The level of water at first resonance is 30.7 cm and at second resonance is 63.2 cm. The error in calculating velocity of sound is [Speed of sound = 330 m/s.] (2005) (A) 204.1 cm/s (B) 110 cm/s (C) 58 cm/s (D) 280 cm/s Sol. Let d be the end correction. The conditions for the first and the second resonances are l1 + d = λ/4, l2 + d = 3λ/4.

+4

336 −2

ν : 334

l2 + d

n:

l1 + d

340 −4

0

l1

344 +2

−2

+2

338

342

346

The frequency of string increases to ν 0 with increase √ in tension T as ν ∝ T . Now tuning fork produces 2 beats. Thus, n − ν 0 = 2 or ν 0 − n = 2. The frequency n = 336 Hz gives ν 0 = 334 Hz or ν 0 = 338 Hz, which is not possible since ν 0 > ν. The frequency n = 344 Hz gives ν 0 = 342 Hz or ν 0 = 346 Hz, which is possible. Ans. A

Subtract to get λ = 2(l2 − l1 ). The measured speed of sound is

Q 5. In the experiment to determine the speed of sound using a resonance column, (2007) (A) prongs of the tuning fork are kept in a vertical plane. (B) prongs of the tuning fork are kept in a horizontal plane. (C) in one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air. (D) in one of the two resonances observed, the length of the resonating air column is close to half of the wavelength of sound in air.

Q 7. An open pipe is in resonance in second harmonic with frequency f1 . Now one end of the tube is closed and frequency is increased to f2 such that the resonance again occurs in nth harmonic. Choose the correct option. (2005) (A) n = 3, f2 = 43 f1 (B) n = 3, f2 = 54 f1 (C) n = 5, f2 = 54 f1 (D) n = 5, f2 = 43 f1

v = νλ = 2ν(l2 − l1 ) = 2(512)(0.632 − 0.307) = 332.8 m/s. Thus, the measurement error is 332.8−330.0 = 2.8 m/s. Ans. D

Sol. The second harmonic in an open pipe occurs when λo = l. Thus, v v f1 = = . λo l λo /2

Sol. The sound waves are longitudinal in nature. The prongs of a tuning fork are kept in the vertical plane to produce the longitudinal waves moving towards the resonance columns (placed downwards). The first resonance occurs at l1 = λ/4 and the second resonance occurs at l2 = 3λ/4.

l

λc

194

Part II. Waves

The nth harmonic in a pipe closed at one end occurs when nλc /4 = l (here n is odd). Thus, f2 =

v nv n = = f1 . λc 4l 4 5 4 f1

is one possible Ans. C

Q 8. A pipe of length l1 , closed at one end is kept in a chamber of gas of density ρ1 . A second pipe open at both ends is placed in a second chamber of gas of density ρ2 . The compressibility of both the gases is equal. Calculate the length of the second pipe if frequency of first overtone in both the cases (2004) q is equal. q ρ1 ρ2 4l1 4l1 (A) l1 /3 (B) 4l1 /3 (C) 3 ρ2 (D) 3 ρ1 Sol. The first overtones in a pipe closed at one end and in a pipe open at both ends are as shown. l1 ρ1 N

A

N

A

and the frequency of the stationary siren heard by the motor-cyclist is 330 + v v + vo νs = × 165. v − vs 330 − 0

νs0 =

(2)

Since the motor-cyclist does not observe the beats, νc0 = νs0 . Substitute νc0 and νs0 from equations (1) and (2) to get v = 22 m/s. Ans. B Q 10. In an experiment of the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is 0.1 m. When this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. Calculate the end correction. (2003) (A) 0.012 m (B) 0.025 m (C) 0.05 m (D) 0.024 m

d1 + e = λ/4,

N

(1)

where d1 is the length of the air column, e is the end correction and λ is the wavelength. The next resonance (with the first overtone) occurs when

ρ2 N

(1)

Sol. The first resonance (with fundamental frequency) occurs when

l2

A

v + vo 330 − v νc = × 176, v − vs 330 − 22

νc0 =

Since f2 > f1 , n = 5 and f2 = solution.

A

Sol. By Doppler’s effect, frequency of the police car horn heard by the motor-cyclist is

A

In the closed pipe, 34 λ1 = l1 and in the open pipe, λ2 = l2 . The expressions for frequency of sound in closed and open pipe are

d2 + e = 3λ/4.

(2)

Eliminate λ from equations (1) and (2) to get e = (d2 − 3d1 )/2 = (0.35 − 3 × 0.1)/2 = 0.025 m.

p B/ρ1 v1 ν1 = , = λ1 4l1 /3 p B/ρ2 v2 ν2 = = , λ2 l2

Ans. B

where Bqis the bulk modulus. Condition ν1 = ν2 gives l2 = 4l31 ρρ12 . Ans. C Q 9. A police car moving at 22 m/s sounds a horn at 176 Hz while chasing a motor-cyclist. Both, police car and motor-cyclist, are moving towards a stationary siren of frequency 165 Hz. What is the speed of the motorcycle if it is given that the motor-cyclist does not observe beats? [Speed of sound = 330 m/s.] (2003) Police car (176 Hz)

22 m/s

Motorcycle

v

Stationarysiren (165 Hz)

(A) 33 m/s (B) 22 m/s (C) zero (D) 11 m/s

Q 11. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz, while the train approaches the siren. During his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that of train A is (2002) (A) 242/252 (B) 2 (C) 5/6 (D) 11/6 Sol. Change in frequency due to Doppler’s effect is given by ν0 =

v + vo ν, v − vs

(1)

where vo is velocity of the observer considered positive when moving towards the source, and vs is the velocity of source considered positive when moving towards the observer. Substitute values in equation (1) to get 5.5 =

v + vA (5), v

(2)

Chapter 14. Sound Waves

195

in the first case and 6=

v + vB (5), v

(3)

in the second case. Eliminate v from equations (2) and (3) to get vB /vA = 2. Ans. B Q 12. Two monatomic ideal gases 1 and 2 of molecular masses m1 and m2 respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to that in the gas 2 is given by q (2000) q m1 m2 m1 m2 (B) (C) (D) (A) m2 m1 m2 m1 Sol. The speed of sound in a gas with molecular mass M and kept at temperature T is given by p (1) v = γRT /M . For the given monatomic gases, γ1 = γ2 = 5/3 and temperature T1 = T2 . Substitute in equation (1) to get p p v1 /v2 = M2 /M1 = m2 /m1 . Ans. B Q 13. A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1 . If the train’s speed is reduced to 17 m/s, the frequency registered is f2 . If the speed of sound is 340 m/s then the ratio f1 /f2 is (2000) (A) 18/19 (B) 1/2 (C) 2 (D) 19/18 Sol. Change in frequency due to Doppler’s effect is given by f0 =

v + vo f, v − vs

where v = 340 m/s is the speed of sound in the medium, vo is the speed of the observer w.r.t. medium, considered positive when moving towards the source, and vs is the speed of source w.r.t. medium, considered positive when moving towards the observer. In the first case vo = 0 and vs = 34 m/s gives 340 + 0 340 f1 = f= f. 340 − 34 306

(1)

In the second case vo = 0 and vs = 17 m/s gives f2 =

340 + 0 340 f= f. 340 − 17 323

(2)

Divide equation (1) by (2) to get f1 /f2 = 323/306 = 19/18. Ans. D Q 14. The ratio of the speed of sound in nitrogen gas to that (1999) p in heliumpgas, at 300 √K is √ (A) 2/7 (B) 1/7 (C) 3/5 (D) 6/5

Sol. The speed of sound in a gas is given by p v = γRT /M .

(1)

The nitrogen is a diatomic gas with γN2 = 7/5 and MN2 = 28. The helium is a monatomic gas with γHe = 5/3 and MHe = 4. Substitute these values in equation (1) to get s √ γN2 MHe 3 vN2 = = . vHe γHe MN2 5 Ans. C Q 15. In hydrogen spectrum the wavelength of Hα line is 656 nm; whereas in the spectrum of a distant galaxy Hα line wavelength is 706 nm. Estimated speed of galaxy with respect to earth is (1999) (A) 2 × 108 m/s (B) 2 × 107 m/s (C) 2 × 106 m/s (D) 2 × 105 m/s Sol. The wavelength of Hα line is increased from λ1 = 656 nm to λ2 = 706 nm i.e., line shifts towards the red color of spectrum (red shift). The red shift occurs due to receding source (galaxy). If a source recedes with velocity v then the red shift is approximately ∆λ = λ2 − λ1 ≈

λ1 v , c

(1)

where c = 3 × 108 m/s is the speed of light. Substitute the values in equation (1) to get v = 0.23 × 108 m/s. We encourage you to derive equation (1) from the o Doppler-effect equation f2 = v+v v−vs f1 . Ans. B Q 16. A given quantity of an ideal gas is at pressure p and absolute temperature T . The isothermal bulk modulus of the gas is (1998) (A) 23 p (B) p (C) 32 p (D) 2p Sol. Isothermal bulk modulus is the ratio of volumetric stress to volumetric strain at constant temperature i.e., dp dp = −V B=− . (1) dV /V dV T Differentiate the ideal gas equation, pV = nRT , at constant T to get dp p =− . (2) dV T V Substitute dp/dV from equation (2) into equation (1) to get B = p. Ans. B Q 17. A whistle giving out 450 Hz approaches a stationary observer at a speed of 33 m/s. The frequency heard by the observer is [Speed of sound = 330 m/s.] (1997)

(A) 409 Hz (B) 429 Hz (C) 517 Hz (D) 500 Hz

196

Part II. Waves

Sol. Change in frequency due to Doppler’s effect is given by ν=

330 + 0 v + vo ν0 = × 450 = 500 Hz. v − vs 330 − 33

Q 20. A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is opened at both ends the fundamental frequency that can be excited is (in Hz) (1986)

Ans. D Q 18. An open pipe is suddenly closed at one end with the result that the frequency of the third harmonic of the closed pipe is found to be higher by 100 Hz than the fundamental frequency of the open pipe. The fundamental frequency (in Hz) of the open pipe is (1996) (A) 200 (B) 300 (C) 240 (D) 480

(A) 1024 (B) 512 (C) 256 (D) 128 Sol. Let l be the length of the tube. The wavelength of fundamental mode for the tube closed at one end is λc = 4l and that for the tube open at both ends is λo = 2l.

Sol. Length of the pipe is same in the both cases. Fundamental frequency of the open pipe is fo = v/(2l),

l = λc /4

(1)

and frequency of the third harmonic of the pipe closed at one end is fc = 3v/(4l).

(2)

Given, fc = fo + 100. Use equations (1) and (2) to get fo = 200 Hz. Ans. A Q 19. An organ pipe P1 , closed at one end and vibrating in its first harmonic, and another pipe P2 , open at both ends and vibrating in its third harmonic, are in resonance with a given tuning fork. The ratio of the length of P1 to that of P2 is (1988) (A) 8/3 (B) 3/8 (C) 1/6 (D) 1/3 Sol. The first harmonic in a pipe closed at one end and the third harmonic in a pipe opened at both ends are shown in the figure. For these harmonics, the pipe lengths l1 and l2 and wavelengths λ1 and λ2 are related by l1 = λ1 /4,

(1)

l2 = 3λ2 /2.

(2) l1 = λ1 /4

l = λo /2

Thus, fundamental frequency for the two cases are nc = v/λc = v/(4l) = 512 Hz,

(1)

no = vλo = v/(2l),

(2)

where v is the speed of sound in the tube. Solve equations (1) and (2) to get no = 1024 Hz. Ans. A Q 21. A cylindrical tube, open at both ends, has a fundamental frequency f in air. The tube is dipped vertically in water so that half of its length is in water. The fundamental frequency of the air column is now (1981) (A) f /2 (B) 3f /4 (C) f (D) 2f Sol. The fundamental modes in the pipe opened at both ends and the pipe closed at one end are shown in the figure. A

A

l/2 l

N

N

P1 A

N

A

l2 = 3λ2 /2 P2 A

N

A

N

A

N

A

Since these harmonics are in resonance with the same tuning fork, the frequencies of these harmonics are equal. Thus, the wavelengths are equal for these harmonics (because speed of sound is same in both the pipes) i.e., λ1 = λ2 . Substitute in equations (1) and (2) to get l1 /l2 = 1/6. Ans. C

In the pipe opened at both ends, l = λo /2, and the fundamental frequency is fo = v/λo = v/(2l).

(1)

In the pipe closed at one end, l/2 = λc /4, and the fundamental frequency is fc = v/λc = v/(2l).

(2)

From equations (1) and (2), fc = fo = f . Ans. C

Chapter 14. Sound Waves

197

Q 22. In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500 Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7 cm and 83.9 cm. Which of the following statements is (are) true? (2018) (A) The speed of sound determined from this experiment is 332 m/s. (B) The end correction in this experiment is 0.9 cm. (C) The wavelength of the sound wave is 66.4 cm. (D) The resonance at 50.7 cm corresponds to the fundamental harmonic. Sol. Let the two successive resonances correspond to (2n + 1)th and (2(n + 1) + 1)th harmonics of the resonance tube, where n = 0, 1, 2, . . . The conditions for these resonances are

respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let ν(t) represent the beat frequency measured by a person sitting in the car at time t. Let νP , νQ and νR be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is 330 m/s. Which of the following statement(s) is(are) true regarding the sound heard by the person? (2016) (A) νP + νR = 2νQ . (B) The rate of change in beat frequency is maximum when the car passes through Q. (C) The plot below represents schematically the variation of beat frequency with time. ν(t) P Q

νQ

l1 + d = (2n + 1)λ/4,

(1)

l2 + d = (2n + 3)λ/4,

(2)

where l1 = 50.7 cm and l2 = 83.9 cm are the lengths of the air column for successive resonances, d is the end correction and λ is the wavelength of sound. Subtract equation (1) from equation (2) to get λ = 66.4 cm = 0.664 m. The speed of sound in air is given by v = νλ = (500)(0.664) = 332 m/s.

R

(D) The plot below represents schematically the variation of beat frequency with time. ν(t) P νQ

Q

Q 23. Two loudspeakers M and N are located 20 m apart and emit sound at frequencies 118 Hz and 121 Hz,

t

Sol. The frequencies of the sources M and N are νM = 118 Hz and νN = 121 Hz. The distances MQ = QN = d = 10 m and PQ = QR = D = 1800 m (see figure). The speed of sound in air is v = 330 m/s and speed of the car is u = 60 km/hr = 50/3 m/s. The car will reach Q at time tQ = D/u = 108 s and it will reach R at time tR = 2D/u = 216 s.

θ

θ os

u S1 u sin θ

P ut

u sin θ θ

Q D−ut

D = 1800 m

d = 10 m

uc

d = 10 m

(νN = 121 Hz) N

=

l2 + d

l1 + d

R

In a pipe closed at one end, the resonance occurs when length of the air column is an odd multiple of λ/4 i.e., at lengths λ/4 = 16.6 cm, 3λ/4 = 49.8 cm, 5λ/4 = 83 cm etc. Thus, the resonance at column length l1 is third harmonic (n = 1) and resonance at column length l2 is fifth harmonic. Substitute n = 1 in equation (1) to get the end correction d = −0.9 cm. We are not comfortable with negative value of end correction. The effective length (distance between node at closed end and anti-node at open end) from experimental data is found to be greater than physical length of the air column. Thus, the end correction should be positive. Its value for a resonance tube of radius r is d = 0.6r. Ans. (A), (B), (C)

t

M (νM = 118 Hz)

S2 u

=

One or More Option(s) Correct

u R sθ

co

D = 1800 m

Consider the time t (≤ tQ ) when car is at S1 between P and Q. The distance travelled by the car in time t is PS1 = ut. At this instant, the lines S1 N and S1 M both make angle θ with the velocity vector ~u. The component of observer (person sitting in the car) velocity towards

198

Part II. Waves

the sources N and M is uo = u cos θ. The sources N and M are at rest i.e., us = 0. Apply Doppler’s effect equation to get frequencies of the sources N and M heard by the observer as 0 νN

v + uo v + u cos θ = νN = νN v − us v ! D − ut u νN , = 1+ p v d2 + (D − ut)2

which gives νP +νR = 2νQ . Differentiate equations (1)– (2) w.r.t. time t to get rate of change of beat frequency dν(t) u2 d2 . = −(νN − νM ) dt v (d2 + (D − ut)2 )3/2 ν(t) (Hz)

3.15

v + uo v + u cos θ νM = νM v − us v ! D − ut u νM . = 1+ p v d2 + (D − ut)2

3.0

0 νM =

The beat frequency heard by the observer at time t (≤ tQ ) is 0 0 ν(t) = νN − νM

=

D − ut u 1+ p v d2 + (D − ut)2

! (νN − νM ). (1)

Now, consider the time t (≥ tQ ) when car is at S2 between Q and R. The distance travelled by the car in time t is PS2 = ut. At this instant, the lines S2 N and S2 M both make angle (180◦ − θ) with the velocity vector ~u. The component of observer velocity towards the sources N and M is uo = −u cos θ. Apply Doppler’s effect equation to get v + uo v − u cos θ νN = νN v − us v ! u ut − D νN , = 1− p v d2 + (ut − D)2

0 νN =

v + uo v − u cos θ νM νM = v − us v ! u ut − D = 1− p νM . v d2 + (ut − D)2

0 νM =

The beat frequency heard by the observer at time t (≥ tQ ) is 0 0 ν(t) = νN − νM

=

1−

ut − D u p v d2 + (ut − D)2

! (νN − νM ). (2)

Substitute t = 0 and t = tQ = D/u in equation (1) to get νP and νQ and substitute t = tR = 2D/u in equation (2) to get νR i.e.,   D u (νN − νM ), νP = ν(t = 0) = 1 + √ v d2 + D 2 νQ = ν(t = D/u) = (νN − νM ),   u D (νN − νM ), νR = ν(t = 2D/u) = 1 − √ v d2 + D2

(3)

2.85 t (s) 0

108

216

From equation (3), the slope is negative and its magnitude is maximum when t = D/u = tQ (denominator is minimum). Thus the rate of change of beat frequency is maximum when car passes through Q. The figure shows that beat frequency is equal to 3.15 Hz at P, it reduces slowly till the car reaches close to Q, at Q the beat frequency reduces sharply, and then it reduces slowly to 2.85 Hz when the car reaches R. Ans. A, B, C Q 24. A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s−1 . He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs √ is (0.350 ± 0.005) m, the gas 167RT = 640 J1/2 mol−1/2 , in the tube is [Given: √ 1/2 −1/2 140RT = 590 J mol . The molar masses M q in 10 grams are given in the options. Take the value of M for each gas as givenqthere.] (2014)
7 (A) Neon M = 20, 10 20 = 10 q
3 (B) Nitrogen M = 28, 10 28 = 5 q
9 (C) Oxygen M = 32, 10 32 = 16 q
17 (D) Argon M = 36, 10 36 = 32 Sol. The speed of sound in a gas with molecular mass M , ratio of specific heat γ, and temperature T is given by r v=

γRT . M

The minimum height of air column for the resonance to occur is r λ v 1 γRT l= = = . (1) 4 4ν 4ν M The ratio of specific heat is γm = 5/3 = 1.67 for monatomic gases and γd = 7/5 = 1.4 for diatomic gases.

Chapter 14. Sound Waves

199

Substitute these values in equation (1) to get r √ r 1 1.67RT 167RT 10 lNe = = 4(244) 20 × 10−3 4(244) 20 640 7 = = 0.459 m, 4(244) 10 r √ 140RT 10 590 3 lN2 = = = 0.363 m, 4(244) 28 4(244) 5 r √ 140RT 10 590 9 lO2 = = = 0.340 m, 4(244) 32 4(244) 16 r √ 167RT 10 640 17 lAr = = = 0.348 m. 4(244) 36 4(244) 32

which is greater than f1 (assuming u > w). In the case (ii), uo is +(u + w) and us is +(u − w). Substitute these values in equation (1) to get   v + (u + w) f2 = f1 , v − (u − w) which is again greater than f1 . We encourage you to show that equation (1) can be generalized to   v + w + uo f1 , f2 = v + w − us

Thus, only lAr lies in the specified range of (0.350 ± 0.005) m. Ans. D Q 25. Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency f1 . An observer in the other vehicle hears the frequency of the whistle to be f2 . The speed of sound in still air is v. The correct statement(s) is (are) (2013) (A) If the wind blows from the observer to the source, f2 > f1 . (B) If the wind blows from the source to the observer, f2 > f1 . (C) If the wind blows from the observer to the source, f2 < f1 . (D) If the wind blows from the source to the observer, f2 < f1 . Sol. Doppler’s effect gives   v + uo f2 = f1 . v − us

(1)

Let the wind speed be w and wind moves towards the source in case (i) and towards the observer in case (ii) (see figure). f1 S

u

w

u

O f2

Case (i) f1 S

u

w

u

O f2

Case (ii)

In the case (i), uo , the speed of observer w.r.t. medium considered positive when it moves towards the source is +(u − w) and us , the speed of source w.r.t. medium considered positive when it moves towards observer is +(u + w). Substitute these values in equation (1) to get   v + (u − w) f2 = f1 , v − (u + w)

where w is positive if the wind blows in the direction of sound and negative if it blows opposite to the direction of sound. Ans. A, B Q 26. A person blows into open-end of a long pipe. As a result, a high pressure pulse of air travels down the pipe. When this pulse reaches the other end of pipe, (2012)

(A) a high pressure pulse starts travelling up the pipe, if the other end of pipe is open. (B) a low pressure pulse starts travelling up the pipe, if the other end of pipe is open. (C) a low pressure pulse starts travelling up the pipe, if the other end of pipe is closed. (D) a high pressure pulse starts travelling up the pipe if the other end of pipe is closed. Sol. The high pressure pulse is a compression pulse. When this gets reflected from a rigid boundary (closed pipe), the reflected pressure wave has the same phase as the incident wave i.e., a compression pulse is reflected as a compression pulse. In case of open pipe, there is a phase change of 180◦ when it is reflected by an open end i.e., a compression pulse is reflected as a rarefaction pulse. Ans. B, D Q 27. A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then, (2009) (A) the intensity of the sound heard at the first resonance was more than that at the second resonance. (B) the prongs of the tuning fork were kept in a horizontal plane above the resonance tube. (C) the amplitude of vibration of the ends of the prongs is typically around 1 cm. (D) the length of the air-column at the first resonance was somewhat shorter than 1/4 th of the wavelength of the sound in air. Sol. Let prongs of a tuning fork lie in the x-z plane.

200

Part II. Waves Total intensity over the spherical surface is same as P , a constant. We encourage you to find the intensity variation for the cylindrical waves. Ans. A, C, D

z x S0

S

The longitudinal waves produced by the tuning fork move in a line perpendicular to its symmetry axis S-S 0 . Thus, the prongs of the tuning fork are kept in a vertical plane above the resonance tube. Let l1 and l2 be the length of air column for first and second resonances and d be the end correction. The first and second resonances occur when l1 + d = λ/4,

(1)

l2 + d = 3λ/4.

(2)

From equation (1), l1 is somewhat shorter than λ/4. The intensity of the sound heard decreases as the length of the air-column increases. Ans. A, D Q 28. As a wave propagates, (1999) (A) the wave intensity remains constant for a plane wave. (B) the wave intensity decreases as the inverse of the distance from the source for a spherical wave. (C) the wave intensity decreases as the inverse square of the distance from the source for a spherical wave. (D) total intensity of the spherical wave over spherical surface centred at the source remains constant at all times. Sol. Let P be the power of the source. The intensity is energy crossing per unit area per unit time i.e., I = P/A. In the case of a plane waves, area of the wavefront remains constant as the wave propagates and hence its intensity remains constant. Plane Wave

In the case of a spherical waves, the area of the wavefront increases as A = 4πr2 , where r is the distance from the source. Spherical Wave

Q 29. A sound wave of frequency f travels horizontally to the right. It is reflected from a large vertical plane surface moving to left with a speed v. The speed of sound in medium is c. (1995) (A) The number of waves striking the surface per sec. ond is f (c+v) c (B) The wavelength of reflected wave is fc(c−v) (c+v) . (C) The frequency of the reflected wave is f (c+v) c−v . (D) The number of beats heard by a stationary listener vf to the left of the reflecting surface is c−v . Sol. Doppler’s effect gives frequency of the wave striking the vertical plane as f1 =

v f

I = P/(4πr2 ).

f1 f2

This frequency is same as the number of waves striking the plane per second. Now, the wall acts as a source for reflected waves. The frequency and the wavelength of the reflected wave are c + vo c+v c f1 = f, f1 = c − vs c−v c−v c c(c − v) λ2 = . = f2 f (c + v) f2 =

Number of beats heard by the stationary observer is 2v ∆f = f2 − f = c−v f. Ans. A, B, C Q 30. Two identical straight wires are stretched so as to produce 6 beats/s when vibrating simultaneously. On changing the tension slightly in one of them, the beat frequency remains unchanged. Denoting by T1 and T2 the higher and lower initial tension in the strings, then it could be said that while making the above changes in tension (1991) (A) T2 was decreased (B) T2 was increased (C) T1 was decreased (D) T1 was increased Sol. Frequencies in the two stretched wires having mass per unit length µ, length l, and tension T1 and T2 are given by ν1 =

Thus, intensity of a spherical wave varies as

c+v c + vo f= f. c − vs c

np T1 /µ, 2l

ν2 = ν1

T1 6 Hz T2

ν2

np T2 /µ. 2l

Chapter 14. Sound Waves

201

The condition T1 > T2 gives ν1 > ν2 . The wires produce 6 beats/s if ν1 = ν2 + 6.

(1) ν1

T1

ν2

T2 6 Hz

T10

ν10

If T1 is changed to T10 then frequency ν1 changes to ν10 . The beat frequency remains the same only if ν10 = ν2 − 6.

Q 32. An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 264 Hz, if the length of the column is [Speed of sound = 330 m/s.] (1985) (A) 31.25 cm (B) 62.50 cm (C) 93.75 cm (D) 125 cm Sol. Natural frequencies of a pipe of length l, closed at one end, are given by, ν = nv/(4l), where v = 330 m/s is the speed of sound and n is an odd integer. Thus, the possible length of pipe (closed at one end) that can resonate with a tuning fork of frequency ν = 264 Hz are

(2)

From equations (1) and (2), ν10 = ν1 − 12, which gives T10 < T1 . T20 6 Hz

ν1

T2

ν2

n(330) nv = = 0.3125n m, 4ν 4(264)

which gives l1 = 31.25 cm, l3 = 93.75 cm, l5 = 156.25 cm etc. Ans. A,C

ν20

T1

ln =

Paragraph Type

T20

If T2 is changed to then the frequency ν2 changes to ν20 . The beat frequency remains the same only if ν20 = ν1 + 6.

(3)

From equations (1) and (3), ν20 = ν2 + 12 which gives T20 > T2 . Ans. B, C Q 31. Velocity of sound in air is 320 m/s. A pipe closed at one end has a length of 1 m. Neglecting end corrections, the air column in the pipe can resonate for sound of frequency (1989) (A) 80 Hz (B) 240 Hz (C) 320 Hz (D) 400 Hz

Paragraph for Questions 33-35 Two trains A and B are moving with speed 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of train A blows a whistle. Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f2 = 1120 Hz, as shown in the figure. The spread in the frequency (highest frequency − lowest frequency) is thus 320 Hz. The speed of sound in air is 340 m/s. (2007) Intensity

Sol. The boundary conditions give an antinode at the open-end and a node at the closed-end. l = λo /4 N

f1

freq.

A

l = 3λ/4 N

A

l = 5λ/4 N

A

The fundamental mode occurs with wavelength λo /4 = l = 1 m which gives fundamental frequency as νo =

f2

v v 320 = = = 80 Hz. λo 4l 4(1)

Only odd harmonics are allowed in a pipe closed at one end (see figure). Thus, allowed frequencies are odd multiples of fundamental frequency i.e., 80 Hz, 240 Hz, 400 Hz, etc. Ans. A, B, D

Q 33. The speed of sound of the whistle is (A) 340 m/s for passengers in A and 310 m/s for passengers in B. (B) 360 m/s for passengers in A and 310 m/s for passengers in B. (C) 310 m/s for passengers in A and 360 m/s for passengers in B. (D) 340 m/s for passengers in both the trains. Sol. The speed of sound in air is vS = 340 m/s. The sound travels towards the passengers in A as well as the passengers in B. 20 m/s P 340 m/s

30 m/s A

P 340 m/s

B

202

Part II. Waves

For the passengers in train A, velocity of the sound relative to the passenger is ~vS/A = ~vS − ~vA = (−340ˆı) − (20ˆı) = −360 m/s ˆı.

Two plane harmonic sound waves are expressed by the equations: y1 = A cos (0.5πx − 100πt) and y2 = A cos (0.46πx − 92πt). [All parameters are in MKS units.] (2006)

Note that ~vS = −340 m/s ˆı as engine is at the front end of train A. For the passengers in train B, velocity of the sound relative to the passenger is

Q 36. How many times does an observer hear maximum intensity in one second? (A) 4 (B) 10 (C) 6 (D) 8

~vS/B = ~vS − ~vB = (340ˆı) − (30ˆı) = 310 m/s ˆı.

Sol. The equation of a progressive wave moving in the +x direction is

Ans. B

Intensity

Intensity

Q 34. The distribution of the sound intensity of the whistle as observed by the passengers in train A is best represented by (A) (B)

freq.

f1

(D)

f1

f2

freq.

f2

freq.

Intensity

f2

Intensity

f1

(C)

f1

f2

freq.

Sol. Let f10 and f20 be the frequencies heard by the passenger in train A. Doppler’s effect gives f10 =

v + vo f1 , v − vs

(1)

where v = 340 m/s is the speed of sound in the medium, vo = 20 m/s is the velocity of the observer w.r.t. medium, considered positive when moving towards the source; vs = −20 m/s is the velocity of source w.r.t., medium considered positive when moving towards the observer. Substitute these values in equation (1) to get f10 = f1 . Similarly, f20 = f2 . If the velocity of the observer and the source is equal then the frequency heard by the observer is equal to the source frequency. Ans. A Q 35. The spread of frequency as observed by the passengers in train B is (A) 310 Hz (B) 330 Hz (C) 350 Hz (D) 290 Hz Sol. Let f10 and f20 be the frequencies heard by the passenger in train B. Use the Doppler-effect equation with v = 340 m/s, vo = −30 m/s, and vs = 20 m/s to get

y = A cos (2πx/λ − 2πνt) , where λ is the wavelength and ν is the frequency. Compare with the given equations to get, λ1 = 4 m, ν1 = 50 Hz, λ2 = 2/0.46 m and ν2 = 46 Hz. The observer hears an intensity variation due to the phenomenon of beats and number of beats heard in one second is ∆ν = ν1 − ν2 = 50 − 46 = 4 Hz. Ans. A Q 37. What is the speed of the sound? (A) 200 m/s (B) 180 m/s (C) 192 m/s (D) 96 m/s Sol. The speed of sound is v = νλ = ν1 λ1 = ν2 λ2 = 200 m/s. Ans. A Q 38. At x = 0 how many times the amplitude of y1 + y2 is zero in one second? (A) 192 (B) 48 (C) 100 (D) 96 Sol. At x = 0, the displacement of the superposed wave is y = y1 + y2 = 2A cos(96πt) cos(4πt). The displacement y becomes zero when either of cos(96πt) or cos(4πt) becomes zero. The cosine has two zeros in every interval of 2π. Thus the number of zeros of cos(96πt) when argument varies from 0 to 96π are 96π 2π × 2 = 96. Similarly, the number of zeros of cos(4πt) 4π × 2 = 4. As when argument varies from 0 to 4π are 2π these two functions do not overlap, their product gives us 96 + 4 = 100 zeros in one second. cos(4πt)

340 − 30 310 f10 = f1 = × 800 = 775 Hz, 340 − 20 320 310 f20 = × 1120 = 1085 Hz. 320 Thus, the spread of frequency is f20 − f10 = 1085 − 775 = 310 Hz. Ans. A Paragraph for Questions 36-38

cos(96πt)

t(s) 0

1

We encourage you to show that cos(4πt) and cos(96πt) do not have a common zero for any t. Hint: 96πt = (2n + 1)π/2 and 4πt = (2m + 1)π/2 do not have any solution for n, m ∈ [0, 1, 2, . . .]. Ans. C

Chapter 14. Sound Waves

203

Matrix or Matching Type Q 39. Column I shows four systems, each of the same length L for producing standing waves. The lowest possible natural frequency of the system is called its fundamental frequency, whose wavelength is denoted by λf . Match each system with statement given in column II describing the nature and wavelength of standing waves, (2011)

Column I (A) Pipe close at one end. 0

(p) Longitudinal waves

L

(B) Pipe open at both ends. 0

(q) Transverse waves

L

(C) Stretched wire clamped at both ends. 0

(r) λf = L

L

(D) Stretched wire clamped at both ends and at mid point. 0

Column II

L 2

(s) λf = 2L

L

(t) λf = 4L Sol. The waves in a pipe (open or closed) are longitudinal and that in a stretched wire are transverse.

Sol. Let ν0 be the frequency of the source S moving with a speed us towards the stationary observer O. ν0

ν0 S•

us

O• ν0

ν0

ν0

The observer will hear the sound directly from the source and after reflection from the wall. Apply the Doppler-effect equation to get ν0 =

v v + uo ν0 = ν0 , v − us v − us

where ν 0 is the frequency heard by the observer when the sound directly comes to him. The apparent frequency of the sound striking the wall is also equal to ν 0 because the wall is stationary. After reflection, the wall acts as a stationary source of frequency ν 0 . The frequency heard by the observer after reflection from the wall is also equal to ν 0 because both the source (wall) and the observer are stationary. Thus, the observer will not hear beats. Aliter: The wall is a reflector similar to a plane mirror. Thus, the image of the source moves towards the observer with speed vs . Hence, the observer hear sound from two sources of frequency ν0 coming towards him from the opposite directions with equal speeds. Thus, he will not hear beats. Ans. F Q 41. The ratio of the velocity of sound in hydrogen gas (γ = 7/5) to thatpin helium gas (γ = 5/3) at the same temperature is 21/5. (1983) Sol. The speed of sound in a gas with molecular mass M , ratio of specific heats γ, and temperature T is given by p v = γRT /M .

The relationship between wavelength corresponding to fundamental mode (λf ) and length (L) for pipe closed at one end is λf /4 = L; for pipe open at both ends is λf /2 = L; for wire clamped at both ends is λf /2 = L; for wire clamped at both ends and at the middle is λf = L (see figure). Ans. A7→(p,t), B7→(p,s), C7→(q,s), D7→(q,r)

For hydrogen, γH2 = 7/5 and MH2 = 2 and for helium γHe = 5/3 and MHe = 4. The ratio of velocities of sound in hydrogen and helium gases at the same temperature is s p γH2 RT /MH2 γH2 MHe vH = p = vHe γHe MH2 γHe RT /MHe r r 21 4 42 = · = . 25 2 25 Ans. F

True False Type

Fill in the Blank Type

Q 40. A source of sound with frequency 256 Hz is moving with a velocity v towards a wall and an observer is stationary between the source and the wall. When the observer is between the source and the wall he will hear beats. (1985)

Q 42. A bus is moving towards a huge wall with a velocity of 5 m/s. The driver sounds a horn of frequency 200 Hz. The frequency of the beats heard by a passenger of the bus will be . . . . . . Hz. [Speed of sound = 342 m/s.] (1994)

204

Part II. Waves

Sol. Given, velocity of the bus u = 5 m/s, velocity of the sound v = 342 m/s, and frequency of the horn f = 200 Hz. f f1

vs vo f

f1 2

Doppler’s effect gives frequency of the wave striking the wall as f1 =

v + vo v f= f. v − vs v−u

(1)

The wall acts as a stationary source of frequency f1 . The frequency of reflected waves heard by the passenger is f2 =

v+u v + vo f1 . f1 = v − vs v

Integer Type Q 44. Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 m/s and the man behind walks at a speed 2.0 m/s. A third man is standing at a height 12 m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of sound in air is 330 m/s. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is . . . . . . . (2018) Sol. From similar triangles OSP and OSQ, ∠P = ∠Q = θ. From given lengths, OP = 13 m and cos θ = 5/13. O

(2)

•

Substitute f1 from equation (1) into equation (2) and use the given values to get f2 =

342 + 5 v+u f= × 200 ≈ 206 Hz. v−u 342 − 5

12m

Thus, the beat frequency heard by the passenger is f2 − f = 6 Hz. Ans. 6 •

Q 43. A cylindrical resonance tube open at both ends has fundamental frequency f in air. Half of the length of the tube is dipped vertically in water. The fundamental frequency to the air column now is . . . . . . (1992) Sol. The boundary conditions give an antinode at the open-end and a node at the closed-end. The fundamental modes in the two cases are shown in the figure.

P

θ

5m 2m/s

5m S

θ • Q θ

1m/s

Let the speeds of two men be uP = 2 m/s and uQ = 1 m/s. The component of velocity of source P in the direction of observer O is us = uP cos θ (approaching O). By Doppler’s effect equation, the frequency of P heard by the stationary observer (uo = 0) is v + uo v+0 νP = νP v − us v − uP cos θ 330 × 1430 . = 330 − 2 cos θ

νP0 = l/2 l

The component of velocity of source Q in the direction of observer O is us = −uQ cos θ (moving away from O). The frequency of Q heard by the observer O is In an open pipe, λo /2 = l,

fo = v/λo = v/(2l).

and in the pipe closed at one end, λc /4 = l/2,

fc = v/λc = v/(2l).

Thus, the fundamental frequency is equal in both the cases. Ans. f

v + uo v+0 νQ = νQ v − us v − (−uQ cos θ) 330 × 1430 = . 330 + cos θ

0 νQ =

The beat frequency heard by the observer is   330 × 1430 330 × 1430 0 0 ∆ν = νP − νQ = − 330 − 2 cos θ 330 + cos θ 330 × 1430 × 3(5/13) ≈ = 5 Hz. 330 × 330

Chapter 14. Sound Waves

205

Let the observer start his clock at this instant i.e., he sets t = 0 when two men are at the positions given in this problem. Can you help him to find the expression for beat frequency as a function of time t? Check your results when (i) P reaches the point S and (ii) t → ∞. Ans. 5 Q 45. A stationary source emits sound of frequency f0 = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 m/s. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? [Given that the speed of sound in air is 330 m/s and the car reflects the sound at the frequency it has received.]. (2017) Sol. Doppler’s effect gives the frequency of the wave striking the large car as f1 =

c+v c + vo f= f, c − vs c

where c = 330 m/s is speed of sound, vo = v = 2 m/s is speed of the observer (approaching large car), vs = 0 is speed of the source (stationary) and f = 492 Hz is the frequency of emitted sound.

where v is the velocity of sound (note the sign convention). For the reflected waves, these cars act as sources of frequency f1 and f2 moving with velocities v1 and v2 towards the stationary observer. Thus, the values of frequency of the reflected waves are given by v − vo v f1 = f1 , and v − vs v − v1 v v − vo f2 = f2 . f20 = v − vs v − v2 f10 =

Substitute the values of f1 and f2 to get f10 =

v + v1 v + v2 f0 , and f20 = f0 . v − v1 v − v2

Using f10 − f20 = (1.2/100)f0 , we get v + v1 v + v2 − = 0.012, i.e., v − v1 v − v2 2v(v1 − v2 ) = 0.012. (v − v1 )(v − v2 ) Now, since v1  v, v2  v, we can approximate v−v1 ≈ v and v − v2 ≈ v. After these simplifications, we get, v1 − v2 = 0.012v/2 = 7.12 ≈ 7 km/h. Ans. 7

v f

f1

Descriptive

f2

f1

Q 47. An observer standing on a railway crossing receives frequency of 2.2 kHz and 1.8 kHz when the train approaches and recedes from the observer. Find the velocity of the train. [Speed of sound = 300 m/s.] (2005)

Now, the large car acts as a source for reflected waves. The frequency of the reflected waves is same as the frequency of incident waves. The frequency received by the stationary observer is given by f2 =

c + vo c+v c f1 = f, f1 = c − vs c−v c−v

where vo = 0 (stationary observer) and vs = 2 m/s (approaching source). Number of beats heard by the 2v stationary observer is ∆f = f2 − f = c−v f = 6 Hz. Ans. 6 Q 46. A stationary source is emitting sound at a fixed frequency f0 , which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0 . What is the difference in the speeds of the cars (in km/h) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 m/s. (2010) Sol. Let the two cars be moving with velocities v1 and v2 towards a stationary source emitting frequency f0 . Doppler’s effect gives frequency of the sound incident on these cars as v − v0 v + v1 f1 = f0 = f0 , and v − vs v v − v0 v + v2 f2 = f0 = f0 , v − vs v

Sol. Let u be the velocity of the train and ν0 be the frequency emitted by it. Change in frequency due to Doppler’s effect is given by ν=

v + uo ν0 . v − us

(1)

Substitute ν = 2.2 kHz, v = 300 m/s, uo = 0, and us = u in equation (1) for the approaching train to get 2.2 kHz =

300 ν0 . 300 − u

(2)

Substitute ν = 1.8 kHz, v = 300 m/s, uo = 0, and us = −u in equation (1) for the receding train to get 1.8 kHz =

300 ν0 . 300 + u

(3)

Eliminate ν0 from equations (2) and (3) to get u = 30 m/s. Ans. 30 m/s Q 48. In a resonance tube experiment to determine the speed of sound in air, a pipe of diameter 5 cm is used. The air column in pipe resonates with a tuning fork of frequency 480 Hz when the minimum length of the air column is 16 cm. Find the speed of sound in air at room temperature. (2003)

206

Part II. Waves

Sol. The end correction in resonance tube of radius r is e = 0.6r. The first resonance occurs when d + e = λ/4 = v/(4ν), which gives v = 4ν(d + 0.6r) = 4(480)(0.16 + 0.6 × 0.025) = 336 m/s.

that in a pipe closed at both ends is λ0B /2 = l. The ratio of fundamental frequencies in this case is s ν0A vA /λ0A vA γA MB = = = ν0B vB /λ0B vB γB M A r r 25 189 9 3 = = . = 21 400 16 4 Ans. (a)

Ans. 336 m/s Q 49. Two narrow cylindrical pipes A and B have the same length. Pipe A is open at both ends and is filled with a monatomic gas of molar mass MA . Pipe B is open at one end and closed at the other end, and is filled with a diatomic gas of molar mass MB . Both gases are at the same temperature. (2002) (a) If the frequency of the second harmonic of pipe A is equal to the frequency of the third harmonic of the fundamental mode in pipe B, determine the value of MA /MB . (b) Now the open end of the pipe B is closed (so that the pipe is closed at both ends). Find the ratio of the fundamental frequency in pipe A to that in pipe B. Sol. Let l be the length of pipe A and pipe B. The conditions for second harmonic in an open pipe A and third harmonic in a closed pipe B are λA = l,

(1)

3λB /4 = l.

(2) A B l

which gives MA 16 5/3 400 = = . MB 9 7/5 189 The wavelength corresponding to fundamental frequency in a pipe opened at both ends is λ0A /2 = l and

(b)

3 4

Q 50. A boat is travelling in a river with a speed of 10 m/s along the stream flowing with a speed of 2 m/s. From this boat a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is 14.45 mm. Assume the attenuation of sound in water and air is negligible. (2001) (a) What will be the frequency detected by a receiver kept inside the river downstream? (b) The transmitter and the receiver are now pulled up into air. The air is blowing with a speed of 5 m/s in the direction opposite to the river stream. Determine the frequency of the sound detected by the receiver? [Temperature of the air and water = 20◦ C, Density of river water = 103 kg/m3 , Bulk modulus of the water = 2.088 × 109 Pa, R = 8.31 J/mol K, Mean molecular mass of air = 28.8 × 10−3 kg/mol, Cp /CV of air = 1.4.] Sol. The speed of sound in water is s s β 2.088 × 109 = = 1445 m/s, vw = ρ 103 and that in air is r r γRT 1.4 × 8.31 × 293 va = = = 344 m/s. M 28.8 × 10−3

The velocities of sound in pipe A and pipe B are given by p vA = γA RT /MA , (3) p vB = γB RT /MB , (4) where γA = 5/3 (monatomic) and γB = 7/5 (diatomic). The frequency of sound in the two cases is same i.e., vA /λA = vB /λB . Substitute λA , λB , vA , and vB from equations (1)–(4) to get p p γA /MA = (3/4) γB /MB ,

400 189

Case (a): Vw=2m/s Vs=10m/s

Vo = 0

Case (b): Va=5m/s

The frequency of the source ν=

vw 1445 = = 105 Hz, λw 14.45 × 10−3

is independent of the medium in which it is placed. Change in frequency due to Doppler’s effect is given by ν0 =

v + vo ν, v − vs

(1)

where v is the speed of sound in the medium, vo is the velocity of the observer w.r.t. medium considered

Chapter 14. Sound Waves

207

positive when moving towards the source, and vs is the velocity of source w.r.t. medium considered positive when moving towards the observer. In case (a), v = 1445 m/s. The velocity of observer w.r.t. medium ~o/w = V ~o − V ~w = 0ˆı − 2ˆı = −2 m/s ˆı (to(water) is V wards left i.e., towards the source), and hence vo = 2 m/s. Similarly, velocity of the source w.r.t. water is ~s/w = V ~s − V ~w = 10ˆı − 2ˆı = 8 m/s ˆı (towards right i.e., V towards the observer) and hence vs = 8 m/s. Substitute the values in equation (1) to get νw =

1445 + 2 × 105 = 1.0069 × 105 Hz. 1445 − 8

πr2 v = πR2 V,

In case (b), v = 344 m/s. The velocity of observer ~o/a = V ~o − V ~a = 0ˆı − (−5ˆı) = w.r.t. medium (air) is V 5 m/s ˆı (towards right i.e., away from the source), and hence vo = −5 m/s. Similarly, velocity of the source ~s/a = V ~s − V ~a = 10ˆı − (−5ˆı) = 15 m/s ˆı w.r.t. air is V (towards right i.e., towards the observer) and hence vs = 15 m/s. Substitute the values in equation (1) to get νa =

344 − 5 × 105 = 1.0304 × 105 Hz. 344 − 15 5

5

Ans. (a) 1.0069 × 10 Hz (b) 1.0304 × 10 Hz Q 51. A 3.6 m long pipe resonates with a source of frequency 212.5 Hz when water level is at certain height in the pipe. Find the heights of water level (from the bottom of the pipe) at which resonances occur. Neglect end correction. Now the pipe is filled to height H(≈ 3.6 m). A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as a function of H. If the radii of the pipe and the hole are 2 × 10−2 m and 1 × 10−3 m respectively, calculate the time interval between the occurrence of first two resonances. [Speed of sound in air is 340 m/s and g = 10 m/s2 .] (2000) Sol. The wavelength of the source is λ = v/ν = 340/212.5 = 1.6 m. 2R l1 l2

H

where n = 1, 2, 3, . . .. Thus, resonances occur when l1 = λ/4 = 0.4 m, l2 = 3λ/4 = 1.2 m, l3 = 5λ/4 = 2.0 m, l4 = 7λ/4 = 2.8 m, and l5 = 9λ/4 = 3.6 m. Corresponding heights of the water level are Hn = 3.6− ln i.e., H1 = 3.2 m, H2 = 2.4 m, H3 = 1.6 m, H4 = 0.8 m and H5 = 0 m. Let R = 2 × 10−2 m be the radius of the pipe and r = 1 × 10−3 m be the radius of the hole. Let V be the velocity of the water at the height H (top) and v be the velocity of the water coming out of the hole. The continuity equation gives

H1 H2

(1)

and Bernoulli’s equation between the top and bottom points gives 2 1 2 ρV

+ ρgH + P0 = 21 ρv 2 + P0 ,

(2)

where P0 is the atmospheric pressure. Eliminate v from equations (1) and (2) to get r √ 2gr4 √ −2 V = H = 1.11 × 10 H. R4 − r 4 The velocity V is same as the rate of decrease in height H with time i.e., V =−

√ dH = 1.11 × 10−2 H. dt

(3)

The height decreases from H1 = 3.2 m to H2 = 2.4 m between the first two resonances. Integrate equation (3), Z

H2

−

H

−1/2

dH = 1.1 × 10

−2

Z dt,

H1

to get √ √ 2( H1 − H2 ) t= = 43.5 s. 1.1 × 10−2 √ Ans. 3.2 m, 2.4 m, 1.6 m, 0.8 m, −2 − dH = 1.11 × 10 H, 43.5 s dt Q 52. The air column in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed of sound in air is 330 m/s. End corrections may be neglected. Let p0 denote the mean pressure at any point in the pipe and ∆p0 the maximum amplitude of pressure variation. (1998)

The resonances occur when length l of the air column in pipe (closed at one end) is an odd multiple of λ/4 (odd harmonics) i.e., ln = (2n − 1)λ/4,

(a) Find the length L of air column. (b) What is the amplitude of pressure variation at the middle of the column? (c) What are the maximum and minimum pressures at the open end of the pipe? (d) What are the maximum and minimum pressures at the close end of the pipe?

208

Part II. Waves

Sol. The wavelength of the wave is λ = v/ν = 330/440 = 3/4 m. l/2

0 A

N

A

x

l N

A

N

In a second overtone, length of the pipe is related to wavelength by l = 5λ/4 = 15/16 m. Let the pressure variation in the pipe is given by ∆p = ∆p0 sin (2πx/λ + φ) .

(1)

The open end of the pipe is a displacement antinode and a pressure node. Thus, ∆p = 0 at x = 0 which gives φ = 0. Thus, equation (1) becomes ∆p = ∆p0 sin (2πx/λ) .

Q 53. A band playing music at a frequency f is moving towards a wall at a speed vb . A motorist is following the band with a speed vm . If v is the speed of sound, obtain an expression for the beat frequency heard by the motorist. (1997) Sol. The sound waves reach the motorist directly with a frequency f1 and after reflection from the wall with a frequency f2 .

•

f vm

•

f f2

vb f0 f0

Doppler’s effect gives f1 =

v + vo v + vm v + vm f= f= f. v − vs v − (−vb ) v + vb

The frequency of the sound incident at the wall is v+0 v f = f= f, v − vb v − vb 0

and that heard by the motorist (after reflection) is f2 =

We encourage you to find f2 by considering the source as an image of the band in the wall. m) Ans. 2vbv(v+v f 2 −v 2 b

Q 54. The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the lengths of the pipes. [ Speed of sound in air is 330 m/s. ] (1997) Sol. The first overtones in a pipe closed at one end and in a pipe open at both ends are as shown. lc

(2)

Substitute λ = 3/4 m and x√= L/2 = 15/32 m in equation (2) to get ∆p = ∆p0 / 2 at the mid point of the pipe. The open end is a pressure node with ∆p = 0 giving pmax = p0 + ∆p = p0 and pmin = p0 − ∆p = p0 . The closed end is a pressure antinode with ∆p = ∆p0 giving pmax = p0 + ∆p0 and pmin = p0 − ∆p0 . ∆p √0 Ans. (a) 15 16 m (b) ± 2 (c) pmax = pmin = p0 (d) p0 + ∆p0 , p0 − ∆p0

f1

Thus, the beat frequency heard by the motorist is given by   v + vm v + vm 2vb (v + vm ) f. f2 − f1 = − f= v − vb v + vb v 2 − vb2

v + vm 0 v + vm f = f. v+0 v − vb

A

N

A

N

lo

A

N

A

N

A

The fundamental frequency ν1c of a closed organ pipe is related to its length lc by lc =

λ1c v 330 = = 0.75 m. = 4 4ν1c 4(110)

The frequency of first overtone of the closed organ pipe (i.e., its third harmonic) is ν3c = 3ν1c = 3(110) = 330 Hz. The first overtone of an open organ pipe (i.e., its second harmonic ν2o ) beats with the first overtone of a closed organ pipe (i.e., its third harmonic ν3c ) with a beat frequency of ∆ν = 2.2 Hz. Hence, ν2o = ν3c ± ∆ν = (330 ± 2.2) Hz. The frequency of first overtone of the open organ pipe is related to its length lo by lo = λ2o =

v 330 = = 1 ∓ 0.0067 m. ν2o 330 ± 2.2 Ans. lo = 1 ± 0.0067 m, lc = 0.75 m

Q 55. A whistle emitting a sound of frequency 440 Hz is tied to a string of 1.5 m length and rotated with an angular velocity of 20 rad/s in the horizontal plane. Calculate the range of frequencies heard by an observer stationed at a large distance from the whistle. [Speed of sound = 330 m/s.] (1996)

Chapter 14. Sound Waves

209

Sol. Let the whistle of frequency ν0 = 440 Hz be rotating in the anticlockwise direction with a speed vs = rω = 1.5 × 20 = 30 m/s. vs

A •

O

B vs

By Doppler’s effect, frequency heard by the observer O is minimum when the whistle is at A (source receding with a velocity vs ) and maximum when the whistle is at B (source approaching with a velocity vs ). The minimum and maximum frequencies heard by the observer are given by 330 × 440 v ν0 = = 403.3 Hz, v + vs 330 + 30 330 × 440 v = 484 Hz. ν0 = = v − vs 330 − 30

νmin = νmax

Ans. 403.3 Hz to 484 Hz Q 56. Two radio stations broadcast their programs at the same amplitude A and at slightly different frequencies ω1 and ω2 respectively, where ω1 − ω2 = 103 Hz. A detector receives the signals from the two stations simultaneously. It can only detect signals of intensity ≥ 2A2 . (1993) (a) Find the time interval between successive maxima of the intensity of the signal received by the detector. (b) Find the time for which the detector remains idle in each cycle of the intensity of the signal. Sol. Without any loss of generality, we can assume detector to be placed at x = 0. The waves at the detector are

Amplitude of
the superposed wave is A0 = 2 2 2A cos ω1 −ω t and its frequency is ω0 = ω1 +ω . In2 2 tensity of the superposed wave   ω1 − ω2 t I = A20 = 4A2 cos2 2 = 2A2 [1 + cos ((ω1 − ω2 )t)] , varies with a frequency (ω1 − ω2 ), as shown in the figure. The time interval between the successive maxima of intensity is T =

2π 2π = 3 = 6.28 × 10−3 s. ω1 − ω2 10

Consider the cycle from t = 0 to t = T . In this cycle, I < 2A2 from t1 to t2 where t1 and t2 are the solutions of cos ((ω1 − ω2 )t) = 0.

(1)

The solutions of equation (1) are (ω1 − ω2 )t1 = π/2 and (ω1 − ω2 )t2 = 3π/2. Substitute these values to get π = 3.14 × 10−3 s. Note that this problem t2 −t1 = ω1 −ω 2 represents the beat phenomenon. Ans. (a) 6.28 × 10−3 s (b) 3.14 × 10−3 s Q 57. A source of sound is moving along a circular path of radius 3 m with an angular velocity of 10 rad/s. A sound detector located far away from the source is executing linear SHM along the line BD (see figure) with an amplitude BC = CD = 6 m. The frequency of oscillation of the detector is 5/π Hz. The source is at the point A when the detector is at the point B. If the source emits a continuous sound wave of frequency 340 Hz, find the maximum and the minimum frequencies recorded by the detector. [Speed of sound = 340 m/s.] (1990) 6m 3m

•

A

6m

•

•

•

B

C

D

y1 = A1 sin(ω1 t − k1 x) = A sin(ω1 t), y2 = A2 sin(ω2 t − k2 x) = A sin(ω2 t). Sol. The speed of the source is us = ωs r = 10 × 3 = 30 m/s. The maximum speed of the detector performing SHM with amplitude A = 6 m is

I 4A2 2A2 0 t1

t2 T

t

ud = ωd A = 2πνd A = 2π ·

vd

The superposed wave at the detector is y = y1 + y2 = A [sin(ω1 t) + sin(ω2 t)]     ω1 − ω2 ω1 + ω2 = 2A cos t sin t 2 2 = A0 sin(ω0 t).

5 · 6 = 60 m/s. π

A

B

C

B

C

D

us us ud A

D

210

Part II. Waves

The detector attains its maximum speed at the mid point C. By Doppler’s effect, the frequency recorded by the detector is maximum when the detector moves towards the approaching source and the frequency is minimum when the detector moves away from the receding source (see figure). Note that the positions of source and detector, shown in the figure, use the fact that the angular frequencies of the source and the detector are equal. The maximum and minimum frequencies recorded by the detector are given by 340 + 60 v + vo f= × 340 = 438.7 Hz, v − vs 340 − 30 v + vo 340 + (−60) = f= × 340 = 257.3 Hz. v − vs 340 − (−30)

fmax = fmin

uo = 40 km/h. The velocity of sound is v = 1200 km/h. Change in frequency due to Doppler’s effect is given by     v + uo 1200 + 40 ν1 = ν1 = 580 = 599.33 Hz. v − us 1200 − 0 Now, consider the sound reflected from the hill. The source of frequency ν1 is the stationary hill and the observer is moving train. When looking from the frame attached to the medium, the source (hill) is moving towards the observer (train) with a velocity us = 40 km/h and the observer is stationary i.e., uo = 0. Change in frequency due to Doppler’s effect is given by     v + uo 1200 ν2 = ν0 = 599.33 = 620 Hz. v − us 1200 − 40

Ans. 438.7 Hz, 257.3 Hz

1 km

Q 58. A train approaching a hill at a speed of 40 km/h sounds a whistle of frequency 580 Hz when it is at a distance of 1 km from the hill. A wind with a speed of 40 km/h is blowing in the direction of motion of the train. Find, [Velocity of sound = 1200 km/h.] (1988) (a) the frequency of the whistle as heard by an observer on the hill. (b) the distance from the hill at which the echo from the hill is heard by the driver and its frequency. Sol. Let ν0 = 580 Hz be the frequency of the sound emitted by the train, ν1 be the apparent frequency incident on the hill (which is same as the frequency reflected by it), and ν2 be the frequency of echo heard by the driver.

ν1

u •

u

ν1 ν2

ˆ

B •

x

Let the train be at position A when it blows the whistle and at position B when an echo is heard by the driver. The distance of A from the hill is 1 km. Let the distance of B from the hill be x. The time time taken by the sound to travel from A to the hill and back to B is equal to the time taken by the train to travel from A to B i.e., 1 x 1−x + = . 1200 1200 40 Solve to get x = 29/31 km. Ans. (a) 599.33 Hz (b) 29/31 km, 620 Hz

~ um/g

ν0 •

•

A

ˆı

The source of frequency ν0 is moving towards a stationary observer at the hill. The velocities of the source, the observer, and the medium (air) w.r.t. the ground are

Q 59. Two tuning forks with natural frequencies of 340 Hz each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards him at the same speed. The observer hears beats of frequency 3 Hz. Find the speed of the tuning fork. [Speed of sound = 340 m/s.] (1986) Sol. Let the tuning fork A is moving towards the observer O with a speed u and the tuning fork B is moving away from the observer with the same speed u. νB

~us/g = 40 km/h ˆı, ~uo/g = ~0, ~um/g = 40 km/h ˆı. A •

Thus, the velocity of the source and the observer w.r.t. the medium are ~us/m = ~us/g − ~um/g = ~0,

~uo/m = ~uo/g − ~um/g

= −40 km/h ˆı. Hence, when looking from a frame attached to the medium, the source is stationary i.e., us = 0 and the observer is moving towards the source with a velocity

ν

u

ν

•

•

O

B

u

νA

Apply the Doppler-effect equation to get the apparent frequencies of the tuning forks A and B as heard by the observer, νA =

v ν, v−u

νB =

v ν, v+u

Chapter 14. Sound Waves

211

where ν = 340 Hz is the frequencies of the tuning forks and v = 340 m/s is the speed of sound. The beat frequency heard by the observer is νA − νA =

Sol. The first overtone in the string and fundamental mode in the pipe closed at one end are given in the figure.

2(340)u 2uv ν= 340 = 3 Hz. 2 −u (340)2 − u2

v2

2uv v 2 −u2

ls = λs

2u v

Solve to get u = 1.5 m/s. Note that ≈ if u  v. Ans. 1.5 m/s Q 60. A sonometer wire under tension of 64 N vibrating in its fundamental mode is in resonance with a vibrating tuning fork. The vibrating portion of the sonometer wire has a length of 10 cm and mass of 1 g. The vibrating tuning fork is now moved away from the vibrating wire with a constant speed and an observer standing near the sonometer hears one beat per second. Calculate the speed with which the tuning fork is moved, if the speed of sound in air is 300 m/s. (1983) Sol. The sonometer wire has a length l = 10 cm, mass m = 1 g and the tension T = 64 N. The fundamental frequency of this wire is given by p p T /µ T l/m v = = ν0 = 2l 2l 2l p (64)(0.1)/0.001 = 400 Hz. = 2(0.1) The tuning fork is in resonance with the wire of frequency 400 Hz. Thus, the frequency of the tuning fork is ν = 400 Hz. Let the tuning fork is moving away from the observer with a speed u. Apply the Doppler-effect equation to get the frequency of the tuning fork as heard by the observer, v + uo (300)(400) 300 + 0 ν = 400 = . ν= v − us 300 − (−u) 300 + u 0

The observer hears one beat per second. Thus, the difference in the fundamental frequency of the sonometer and the apparent frequency of the tuning fork is one hertz i.e., ν0 − ν 0 = 400 −

(300)(400) = 1 Hz. 300 + u

lp = λp /4

The length of the string is ls = 25 cm and its mass is ms = 2.5 g. The frequency of the first overtone in the string is given by p r T /(ms /ls ) T vs νs = = = , (1) λs ls m s ls where T is the tension and vs is the speed of sound in the string. The fundamental frequency of the pipe of length ls = 40 cm is given by νp =

v 320 v = = = 200 Hz. λp 4lp 4(0.40)

The beat frequency heard by the observer is 8 Hz i.e., νs − νp = 8 Hz,

or,

νp − νs = 8 Hz.

From equation (1), the frequency νs decreases with a decrease in tension T . Since the beat frequency decreases with a decrease in tension, νs − νp = 8 Hz, correctly represents the relation between νs and νp . Substitute the values from equations (1) and (2) to get T = ms ls (8 + 200)2 = (2.5 × 10−3 ) (0.25) (208)2 = 27.04 N. Ans. 27.04 N Q 62. A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity of 5 m/s. How many beats per second will be heard by the observer on source itself if sound travels at a speed of 330 m/s? (1981)

Sol. Let ν0 = 256 Hz be the frequency of sound emitted by the source, ν1 be the apparent frequency incident on the wall (which is same as the frequency reflected by it), and ν2 be the frequency heard by the observer on source.

Solve to get u = 0.75 m/s.

vs

ν0 ν1

•

Ans. 0.75 m/s Q 61. A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats/s are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s find the tension in the string. (1982)

(2)

ν1

vo ν2

For calculation of ν1 , the source is moving and the observer is stationary (wall). For ν2 , the source is stationary (wall) and the observer is moving. Change in frequency due to Doppler’s effect is given by ν1 =

v ν0 , v − us

ν2 =

v + uo ν1 , v

212

Part II. Waves

where, v = 330 m/s is the velocity of sound, us = +5 m/s is the velocity of source and uo = +5 m/s is the velocity of observer. Substitute these values to get v + uo v v + uo ν0 = ν0 v v − us v − us 330 + 5 256 = 263.88 Hz. = 330 − 5

ν2 =

The number of beats heard per second by the observer is ∆ν = ν2 − ν0 = 263.88 − 256 = 7.88 Hz. Ans. 7.88 Hz

Chapter 15 Light Waves

One Option Correct

Q 4. In a Young’s double slit experiment, bi-chromatic light of wavelength 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1 m. The minimum distance between two successive regions of complete darkness is (2004) (A) 4 mm (B) 5.6 mm (C) 14 mm (D) 28 mm

Q 1. In the Young’s double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is (2013) (A) (2n + 1) λ2 (B) (2n + 1) λ4 λ (C) (2n + 1) λ8 (D) (2n + 1) 16

Sol. The positions of minima in Young’s double slit experiment are given by

Sol. The intensity variation in Young’s double slit experiment with identical slits is given by I = 4I0 cos2 (δ/2),

(1)

x = (2n − 1)

where I0 is the intensity due to a single slit and δ is the phase difference. The equation (1) gives the peak intensity as 4I0 . For points having half the peak intensity,

λD , 2d

where n ∈ {1, 2, 3, . . .}, d = 0.1 mm, D = 1 m and λ is the wavelength of light. The condition for minima of λ1 = 400 nm to occur at the same places as the minima of λ2 = 560 nm is

cos2 (δ/2) = 1/2. Solve to get the phase difference δ = (2n + 1) π2 . The λ path difference is ∆x = 2π δ = (2n + 1) λ4 . Ans. B

(2n1 − 1)

Q 2. Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe width recorded are βG , βR , and βB , respectively. Then, (2012) (A) βG > βB > βR (B) βB > βG > βR (C) βR > βB > βG (D) βR > βG > βB

λ1 D λ2 D = (2n2 − 1) , 2d 2d

(1)

where n1 and n2 are positive integers. Substitute the values in equation (1) to get 5n1 = 7n2 − 1.

(2)

The equation (2) is satisfied for (n1 , n2 ) ∈ {(4, 3), (11, 8), . . .}. The minimum distance between the two successive region of complete darkness is

Sol. The fringe width β is related to the wavelength λ by, β = λD/d. In the visible spectrum λB < λG < λR (VIBGYOR), and hence βR > βG > βB . Ans. D

∆x = ((2n01 − 1) − (2n1 − 1))

Q 3. In Young’s double slit experiment intensity at a point is one fourth of the maximum intensity. Angular position of this (2005)
point is
λ (A) sin−1 λd
(B) sin−1 2d
λ λ (D) sin−1 4d (C) sin−1 3d

λ1 D 2d

= ((2 × 11 − 1) − (2 × 4 − 1))

400 × 10−9 × 1 2 × 0.1 × 10−3

= 28 mm. We encourage you to analyse the figure given below to see how the intensities Iλ1 , Iλ2 and Iλ1 + Iλ2 vary with x.

Sol. The intensity variation in Young’s double slit experiment is given by I = Imax cos2 (δ/2),

I

where δ is the phase difference between the interfering waves. The condition I = Imax /4 gives δ = 2π/3. The path difference at an angular position θ is ∆x = d sin θ =

Iλ1+Iλ2 Iλ1 Iλ2

λ δ. 2π

Substitute δ = 2π/3 to get θ = sin−1

0 λ 3d




14

42

x(mm)

. Ans. C

Ans. D 213

214

Part II. Waves

Q 5. In the adjacent diagram CP represents a wavefront and AO and BP, the corresponding two rays. Find the condition of θ for constructive interference at P between the ray BP and the reflected ray OP. (2003) O C

R

Sol. When the two waves of intensities I1 and I2 having a phase difference δ interfere then the resultant intensity is given by p I = I1 + I2 + 2 I1 I2 cos δ. Substitute the given values to get √ IA = I + 4I + 2 4I 2 cos(π/2) = 5I, √ IB = I + 4I + 2 4I 2 cos(π) = I.

θ θ d

A

(1) (2)

P

Subtract equation (2) from (1) to get IA − IB = 4I. Ans. B

B

(A) cos θ = 3λ 2d (C) sec θ − cos θ =

λ d

λ (B) cos θ = 4d (D) sec θ − cos θ =

4λ d

Sol. The path difference between the two waves is CO + OP. From triangle OCP, CO = OP cos 2θ and from triangle ORP, OP = d/ cos θ. Simplify to get the path difference CO + OP =

d cos 2θ d + = 2d cos θ. cos θ cos θ

The reflection from the plane mirror introduces an additional phase difference of π. Thus, for constructive interference, path difference should be an odd multiple of λ/2 which gives cos θ = (2n − 1)λ/(4d), where n = 1, 2, 3, . . .. Ans. B Q 6. In the ideal double-slit experiment, when a glassplate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is (2002) (A) 2λ (B) 2λ/3 (C) λ/3 (D) λ Sol. The path difference introduced by the glass plate should be integral multiple of λ for constructive interference to occur at the position of central maximum. Thus, (µ − 1)t = nλ, where n = 1, 2, . . .. Substitute n = 1 and µ = 1.5 to get the minimum thickness tmin =

nmin λ 1×λ = = 2λ. µ−1 1.5 − 1 Ans. A

Q 7. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between resultant intensities at A and B is (2001) (A) 2I (B) 4I (C) 5I (D) 7I

Q 8. In a Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of the light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by (2001) (A) 12 (B) 18 (C) 24 (D) 30 Sol. The fringe width in Young’s double slit experiment is given by β = λD/d. The length of a segment with 12 fringes of fringe width β1 = 600D/d is y = 12β1 . The number of fringes of fringe width β2 = 400D/d which can be accommodated in y are y/β2 = 18. Ans. B Q 9. In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern, (2000) (A) the intensities of both maxima and the minima increase. (B) the intensity of the maxima increases and the minima has zero intensity. (C) the intensity of maxima decreases and that of minima increases. (D) the intensity of maxima decreases and the minima has zero intensity. Sol. Let the intensity of light passing through the two slits be I1 and I2 . The intensity at a point having a phase difference δ is given by p I = I1 + I2 + 2 I1 I2 cos δ. Thus, the intensity at maxima (δ = 0) and minima (δ = π) are p Imax = I1 + I2 + 2 2I1 I2 , p Imin = I1 + I2 − 2 I1 I2 . When the slits are of same width, I1 = I0 , I2 = I0 , Imax = 4I0 and Imin = 0. When the width of one √ of the slit is doubled, I1 = I0 , I√ 2 = 2I0 , Imax = (3 + 2 2)I0 = 5.83I0 and Imin = (3 − 2 2)I0 = 0.17I0 . We encourage you to show that both Imax and Imin increase if I2 = xI0 for some x > 1. Ans. A

Chapter 15. Light Waves

215

Q 10. Yellow light is used in a single slit diffraction experiment with slit width of 0.6 mm. If yellow light is replaced by X-rays, then the observed pattern will reveal (1999) (A) that the central maximum is narrower. (B) more number of fringes. (C) less number of fringes. (D) no diffraction pattern. Sol. Let b = 0.6 mm be the slit width. The wavelength of yellow light is 550 × 10−9 m whereas that of X-rays is of the order of 10−10 m. The first minimum of diffraction pattern occurs at an angle θ given by b sin θ = λ. Substitute the values of b and λ to get θyellow = 0.05◦ and θX-ray = 0.00001◦ . Thus, the central maximum is much narrower for X-rays. Note that θX-ray is so small that X-ray diffraction pattern is hardly observed. The wavelength should be of the order of b to see the diffraction pattern. Ans. A Q 11. A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat plate as shown. The observed interference fringes from this combination shall be (1999)

Sol. Let AC be the slit with midpoint B and the first minimum of diffraction pattern occurs at point P. Consider a small portion of the slit above A and an equal and similar portion above B. C

P

B

O

A

The waves emanating from these two portions superimpose at P to produce zero intensity. Thus, phase difference between waves coming from A and B is π. In this way, for each portion of slit in AB there is an equal and similar portion in BC to produce the minimum at P. The phase difference between the waves coming from B and C is also π. Hence, phase difference between the waves coming from A and C is 2π. Ans. D Q 13. A narrow slit of width 1 mm is illuminated by monochromatic light of wavelength 600 nm. The distance between the first minima on the either side of a screen at a distance of 2 m is (1994) (A) 1.2 cm (B) 1.2 mm (C) 2.4 cm (D) 2.4 mm Sol. The first minimum of a diffraction pattern by a single slit is obtained at a position given by λ = d sin θ ≈ d (y/D).

(A) (B) (C) (D)

y

straight. circular. equally spaced. having fringe spacing which increases as we go outwards.

Sol. The interference fringes are the locus of those points which have a constant phase difference between the waves refracted by the cylindrical slice and reflected by the plane surface. These are straight lines parallel to the axis of cylinder. The concept is similar to Newton’s ring, where spherical slice produces circular fringes.

d

θ y

D

The distance between the first minimum on either side of the screen is 2λD 2(600 × 10−9 )(2) = d 1 × 10−3 −3 = 2.4 × 10 m = 2.4 mm.

2y =

Ans. D Q 14. Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are (1988)

Ans. A Q 12. A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is (1998) (A) zero (B) π/2 (C) π (D) 2π

(A) 5I and I (C) 9I and I

(B) 5I and 3I (D) 9I and 3I

Sol. When two coherent waves of the intensities I1 and I2 interfere with each other, the resultant intensity is given by p Ir = I1 + I2 + 2 I1 I2 cos δ, (1)

216

Part II. Waves

where δ is the phase difference between the interfering waves. From equation (1), the maximum and the minimum intensities are p Imax = I1 + I2 + 2 I1 I2 , (when δ = 0, 2π, . . .), (2) p Imin = I1 + I2 − 2 I1 I2 , (when δ = π, 3π, . . .). (3)

P1

∆θ

S1 d

Substitute I1 = I and I2 = 4I in equations (2) and (3) to get Imax = 9I and Imin = I. Ans. C Q 15. In Young’s double slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is (1981) (A) unchanged (B) halved (C) doubled (D) quadrupled Sol. In Young’s double slit experiment, fringe width is given by β = λD/d, where d is the separation between the slits and D is the distance of the screen from the slits. When d is reduced to d0 = d/2 and D is increased to D0 = 2D, the fringe width become

P2

S2

(A) The angular separation between two consecutive bright spots decreases as we move from P1 to P2 along the first quadrant. (B) A dark spot will be formed at the point P2 . (C) The total number of fringes produced between P1 and P2 in the first quadrant is close to 3000. (D) At P2 the order of the fringe will be maximum. Sol. The path difference between the waves from S1 and S2 meeting at the point P2 is ∆x = S1 P2 − S2 P2 = d = 1.8 mm = 3000λ, where λ = 600 nm is given wavelength. The path difference at P2 is integral multiple of λ. Thus, constructive interference takes place at P2 (bright spot).

β 0 = λD0 /d0 = λ(2D)/(d/2) = 4λD/d = 4β. Ans. D

P1

Q 16. When a ray of light enters a glass slab from air, P

(1980)

(A) (B) (C) (D)

its wavelength decreases. its wavelength increases. its frequency increases. neither its wavelength nor its frequency changes.

Sol. The frequency ν of light does not change when light travels from one medium to another. The refractive index of a medium is defined as the ratio of speed of light in vacuum to the speed of light in medium i.e., n = c/v. The wavelength in a medium is given by λ = v/ν = c/(nν) = λ0 /ν, where λ0 = c/ν is the wavelength in vacuum. The refractive index of glass is n = 1.5. Thus, the wavelength of the light reduces from λ0 to λ0 /1.5 when it enters a glass slab from air. Ans. A One or More Option(s) Correct Q 17. Two coherent monochromatic point sources S1 and S2 of wavelength λ = 600 nm are placed symmetrically on either side of the centre of the circle as shown. The sources are separated by a distance d = 1.8 mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is ∆θ. Which of the following option(s) is(are) correct? (2017)

Q θ S1

d

S2

P2

The path difference at P1 is ∆x = P1 S1 − P2 S2 = 0. Thus, a bright spot occurs at P1 also. The path difference increases from zero to 3000λ as we move from P1 to P2 in the first quadrant. There is a maximum wherever path difference is integral multiple of λ i.e., wherever ∆x = λ, 2λ, . . . , 2999λ. Thus, there are 2999 maxima between P1 and P2 (excluding maxima at P1 and P2 ). Hence, total number of fringes between P1 and P2 is close to 3000. Consider a general point P at an angle θ (see figure). The path difference between the waves from S1 and S2 , meeting at the point P , is ∆x = S1 P − S2 P = S1 Q ≈ d sin θ.

(1)

Let nth maximum occurs at θ and (n + 1)th maximum occurs at θ0 = (θ + ∆θ). Note that θ increases with increase in n. Substitute these conditions in equation (1) to get ∆xn+1 − ∆xn = λ = d (sin θ0 − sin θ) = d∆θ cos θ,

Chapter 15. Light Waves

217

where, we have used small angle approximations sin ∆θ ≈ ∆θ and cos ∆θ ≈ 1. Thus, the angular separation between two consecutive maxima, ∆θ = λ/(d cos θ), increases as we move from P1 to P2 in the first quadrant (see figure). We encourage you to find expression for intensity (on the circumference) as a function of θ and deduce these results from that expression. Ans. (C), (D)

z •

O

•

S1

S2

Screen

Q 18. While conducting the Young’s double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S1 , S2 ) emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3 m from the mid-point of S1 S2 , as shown schematically in the figure. The distance between the sources is d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining S1 S2 . Which of the following is(are) true of the intensity pattern on the screen? (2016)

y

x

d D

(A) Straight bright and dark bands parallel to the xaxis. (B) The region very close to the point O will be dark. (C) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction. (D) Semi circular bright and dark bands centered at point O. Sol. The waves from the sources S1 and S2 interfere at the origin O. The path difference between these waves is ∆ = S1 O − S2 O = d = 0.6003 mm = 2001(λ/2), where λ = 600 nm. The path difference at O is odd integral multiple of λ/2. Thus there will be dark fringe at O. z P (x, z)

at O. It can be proved by taking a point P (x, z) on the screen. The path difference at P is given by ∆ = S1 P − S2 P q q = x2 + z 2 + (D+ d2 )2 − x2 + z 2 + (D− d2 )2 . Simplify to get (you need to square it twice) x2 + z 2 =

∆2 D 2 d2 d2 2 + − D = r2 , − ∆2 4 4

which is an equation of a circle with centre O and radius r. Note that ∆ is constant at a fringe. Since the screen covers upper half of the x-z plane, the student will see semi-circular bright and dark bands centred at point O. Ans. B, D Q 19. A light source, which emits two wavelengths λ1 = 400 nm and λ2 = 600 nm, is used in Young’s double slit experiment. If recorded fringe widths for λ1 and λ2 are β1 and β2 and the number of fringes for them within a distance y on one side of the central maximum are m1 and m2 , respectively, then (2014) (A) β2 > β1 (B) m1 > m2 (C) From the central maximum, 3rd maximum of λ2 overlaps with 5th minimum of λ1 (D) The angular separation of fringes of λ1 is greater than λ2 Sol. The fringe width in Young’s double slit experiment is given by β = λD/d. Thus, the ratio of fringe width for the two wavelengths is β1 /β2 = λ1 /λ2 = 400/600 = 2/3. I Iλ1

Iλ2

y

The number of fringes in a distance y are m = y/β. Hence, the ratio of number of fringes of two wavelengths in a given distance y is m1 /m2 = β2 /β1 = 3/2. The third maximum for λ2 and fifth minimum of λ1 will occur at y2 = 3β2 = 3 × 600 D/d = 1800 D/d, y1 = (5 − 1/2) β1 = 4.5 × 400 D/d = 1800 D/d.

x O

It can be easily seen that the path difference is same for all points on the screen which lie on a circle centered

We encourage you to show that the second maximum of λ2 and third maximum of λ1 occur at the same place (see figure). Ans. A, B, C

218

Part II. Waves

Q 20. In a Young’s double slit experiment, the separation between the two slits is d and the wavelength of the light is λ. The intensity of light falling on the slit 1 is four times the intensity of light falling on slit 2. Choose the correct option(s), (2008) (A) If d = λ, the screen will contain only one maximum. (B) If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screen. (C) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase. (D) If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase. Sol. The path difference at a point having an angle θ is given by ∆x = d sin θ. For the first maximum (other than central maximum), the path difference will be equal to λ i.e., sin θ1 = λ/d. If d = λ then θ1 = 90◦ which is not possible for a screen placed at some finite distance. Hence, the screen will have only one (central) maximum. If λ < d < 2λ then 1/2 < sin θ1 < 1. Thus, we get at least one more maximum besides the central one. Let the intensity of slit 1 and slit 2 be I1 = 4I0 and I2 = I0 . The intensity at a point having a phase difference δ is given by p I = I1 + I2 + 2 I1 I2 cos δ. Thus, the maximum and minimum intensities are q Imax = 4I0 + I0 + 2 4I02 = 9I0 , q Imin = 4I0 + I0 − 2 4I02 = I0 . If the intensity of the slit 1 is reduced to I0 then the resultant intensity becomes 4I0 cos2 (δ/2) which gives Imax = 4I0 and Imin = 0. If the intensity of slit 2 is increased to 4I0 then the resultant intensity becomes 16I0 cos2 (δ/2) which gives Imax = 16I0 and Imin = 0. Ans. A, B Q 21. In an interference arrangement similar to Young’s double slit experiment, the slits S1 and S2 are illuminated with coherent microwave sources, each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by a distance d = 150.0 m. The intensity I(θ) is measured as a function of θ, where θ is defined as shown. If I0 is the maximum intensity, then I(θ) for 0 ≤ θ ≤ 90◦ is given by (1995)

S1

d 2

θ

d 2

(A) (B) (C) (D)

S2

I(θ) = I0 /2 for θ = 30◦ I(θ) = I0 /4 for θ = 90◦ I(θ) = I0 for θ = 0◦ I(θ) is constant for all θ

Sol. The intensity variation in Young’s double slit experiment is given by I = I0 cos2 (δ/2),

(1)

where I0 is the maximum intensity and δ is the phase difference between the interfering waves. The phase difference is given by 2πνd 2π d sin θ = sin θ λ c 2π(106 )(150) sin θ = π sin θ. = 3 × 108

δ=

(2)

Substitute δ from equation (2) into equation (1) to get I(θ) = I0 cos

2



π sin θ 2

 .

Substitute the given values of θ to get π I0 = , 4 2 ◦ 2 π = 0, and I(90 ) = I0 cos 2 ◦ I(0 ) = I0 .

I(30◦ ) = I0 cos2

Ans. A, C Q 22. White light is used to illuminate the two slits in a Young’s double slit experiment. The separation between the slits is b and the screen is at a distance d ( b) from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing. Some of these missing wavelengths are (1984) 2 2 2 b2 (A) λ = bd (B) λ = 2bd (C) λ = 3d (D) λ = 2b 3d Sol. Let the point P be located on the screen directly in front of the slit S1 . S1 b

P b 2

θ

S2 d

Chapter 15. Light Waves

219

The distance of the point P from the central maximum is y = b/2. The path difference between the waves interfering at P is given by ∆x = S2 P − S1 P = b sin θ ≈ b tan θ = b(b/2)/d = b2 /(2d). The wavelengths which undergo destructive interference at P will be missing at P. The wavelength λ undergoes destructive interference at P if the path difference at P is equal to an odd integral multiple of λ/2 i.e., n(λ/2) = ∆x = b2 /(2d),

for n = 1, 3, 5, . . ..

Substitute n = 1, 3, 5, . . . to get the missing wavelengths 2 b2 b2 , λ5 = 5d etc. λ1 = bd , λ3 = 3d Ans. A, C Q 23. In the Young’s double slit experiment, the interference pattern is found to have an intensity ratio between the bright and dark fringes as 9. This implies that (1982) (A) the intensities at the screen due to the two slits are 5 units and 4 units respectively. (B) the intensities at the screen due to the two slits are 4 units and 1 units respectively. (C) the amplitude ratio is 3. (D) the amplitude ratio is 2. Sol. Let I1 and I2 be the intensities of light coming out of two slits. The ratio of maximum and minimum intensities at the screen is given by √
2 √ I1 + I2 Imax = √ √
2 = 9. Imin I1 − I2 Simplify to get I1 /I2 = 4. The intensities are propor2 tional to the square of the amplitudes i.e., √I ∝ √ A . Thus, the ratio of the amplitude is A1 /A2 = I1 / I2 = 2. Ans. B, D Paragraph Type Paragraph for Questions 24-25 Most materials have refractive index, n > 1. So, when a light ray from air enters a naturally occurring θ1 n2 material, then by Snell’s law, sin sin θ2 = n1 , it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, √ n = vc = ± r µr , where c is the speed of electromagnetic waves in vacuum, v is its speed in medium, r and µr are the relative permittivity and permeability of the medium respectively. In normal materials, both r and µr are positive, implying positive n for the medium. When both r and µr are negative, one must choose

the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behavior, without violating any physical laws. Since n is negative, it results in a change in direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials. (2012)

Q 24. For light incident from air on a meta-material, the appropriate ray diagram is (B) (A) θ1 θ1 Air

Air

Meta-Material

Meta-Material

θ2

θ2

(C)

(D)

θ1

θ1

Air

Air

Meta-Material

Meta-Material

θ2

θ2

Sol. Since n < 0, by Snell’s law sin θ2 = sinnθ1 < 0. We encourage you to demonstrate this using Huygens principle. Ans. C Q 25. Choose the correct statement, (A) The speed of light in meta material is v = c|n|. c (B) The speed of light in meta material is v = |n| . (C) The speed of light in meta material is v = c. (D) The wavelength of the light in meta-material (λm ) is given by λm = λair |n|, where λair is the wavelength of the light in air. c Sol. The speed of light in a meta-material is |n| (modulus is taken because n < 0 for a meta-material). Ans. B

Paragraph for Questions 26-28 The figure shows a surface XY separating two transparent media, medium-1 and medium-2. The lines ab and cd represent wavefronts of a light wave travelling in medium-1 and incident on XY. The lines ef and gh represents wavefronts of the light wave in medium-2 after refraction. (2007) b

d medium-1

X

c

a

f

h

Y medium-2

e

g

Q 26. Light travels as a (A) parallel beam in each medium. (B) convergent beam in each medium. (C) divergent beam in each medium.

220

Part II. Waves

(D) divergent beam in one medium and convergent beam in other medium. Sol. The wavefronts in medium 1 are parallel to each other. Hence incident beam consists of parallel light rays. Same is true for refracted beam in the medium 2. Ans. A Q 27. The phases of the light wave at c, d, e and f are φc , φd , φe and φf respectively. It is given that φc 6= φf . (A) φc cannot be equal to φd . (B) φd can be equal to φe . (C) (φd − φf ) is equal to (φc − φe ). (D) (φd − φc ) is not equal to (φf − φe ). Sol. By definition, the phase at all points on a wavefront is equal. Thus, φc = φd and φe = φf . Hence, (φd − φf ) = (φc − φe ). Also, since φc 6= φf , we get (φc = φd ) 6= (φe = φf ). Ans. C Q 28. Speed of light is (A) the same in medium-1 and medium-2. (B) larger in medium-1 than in medium-2. (C) larger in medium-2 than in medium-1. (D) different at b and d. Sol. The ray of light travels normal to the wavefront. The ray bends towards the normal while going from medium 1 to medium 2. b

Column I P2 P1 P0

(A) S2 S1

(B) (µ − 1)t = λ/4

Column II (p) δ(P0 ) = 0

(q) δ(P1 ) = 0 P2 P1 P0

S2 S1

(C) (µ − 1)t = λ/2

(r) I(P1 ) = 0 P2 P1 P0

S2 S1

(D) (µ − 1)t = 3λ/4

(s) I(P0 ) > I(P1 )

P2 P1 P0

S2 S1

(t) I(P2 ) > I(P1 ) Sol. The phase difference δ for a given path difference ∆d is given by δ = ∆d (2π/λ).

(1)

In case (A), equation (1) gives the phase differences at P0 , P1 , and P2 as δ(P0 ) = (S1 P0 − S2 P0 )2π/λ = 0,

d a

δ(P1 ) = (S1 P1 − S2 P1 )2π/λ = π/2,

c

medium-1 medium-2

f h

e g

δ(P2 ) = (S1 P2 − S2 P2 )2π/λ = 2π/3. The intensity at a point having a phase difference δ is given by I = 4I0 cos2 (δ/2),

Hence, by Snell’s law, medium 2 is denser than medium 1 i.e., µ2 > µ1 . Thus, v2 = c/µ2 < c/µ1 = v1 i.e., speed of light is larger in medium 1. Ans. B Matrix or Matching Type Q 29. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits S1 and S2 . In each of these cases S1 P0 = S2 P0 , S1 P1 − S2 P1 = λ/4 and S1 P2 − S2 P2 = λ/3, where λ is the wavelength of the light used. In the case B, C, and D, a transparent sheet of refractive index µ and thickness t is pasted on slit S2 . The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by δ(P ) and the intensity by I(P ). Match each situation given in Column I with the statement(s) in Column II valid for that situation. (2009)

(2)

where I0 is the intensity at each slit. Substitute δ in equation (2) to get the intensities at P0 , P1 , and P2 as I(P0 ) = 4I0 cos2 (δ(P0 )/2) = 4I0 , I(P1 ) = 4I0 cos2 (π/4) = 2I0 , I(P2 ) = 4I0 cos2 (π/3) = I0 . The transparent sheet of refractive index µ and thickness t introduces an additional path difference of ∆d = (µ − 1)t.

(3)

In case (B), the equations (1) and (3) give the phase differences at P0 , P1 , and P2 as, δ(P0 ) = [S1 P0 − (S2 P0 + (µ − 1)t)] 2π/λ = (−λ/4)2π/λ = −π/2, δ(P1 ) = [S1 P1 − (S2 P1 + (µ − 1)t)] 2π/λ = 0, δ(P2 ) = [S1 P2 − (S2 P2 + (µ − 1)t)] 2π/λ = π/6,

Chapter 15. Light Waves

221

and equation (2) gives intensities at these points as

Air

60◦

I(P0 ) = 4I0 cos2 (−π/4) = 2I0 ,

r

2

I(P1 ) = 4I0 cos (0) = 4I0 ,

Water

2

I(P2 ) = 4I0 cos (π/12) < 4I0 . Similarly, for case (C), δ(P0 ) = −π,

δ(P1 ) = −π/2,

δ(P2 ) = −π/3,

I(P0 ) = 0,

I(P1 ) = 2I0 ,

I(P2 ) = 3I0 .

For case (D), δ(P0 ) = −3π/2, I(P0 ) = 2I0 ,

δ(P1 ) = −π,

I(P1 ) = 0,

δ(P2 ) = −5π/6,

and Fill in the Blank Type

I(P2 ) = 4I0 cos2 (−5π/12) > 0. Ans. A7→(p,s), B7→q, C7→t, D7→(r,s,t) True False Type Q 30. In a Young’s double slit experiment performed with a source of white light, only black and white fringes are observed. (1987) Sol. The fringe width in Young’s double slit experiment with slit separation d and screen distance D is given by β = λD/d. Thus, fringe width is different for different colours (λ). Hence, minima and maxima of different colours will occur at different places on the screen. However, all the colours will have a common central maximum (white). Ans. F Q 31. Two slits in a Young’s double slit experiment are illuminated by two different sodium lamps emitting light of the same wavelength. No interference pattern will be observed on the screen. (1984) Sol. The beams coming from the two slits must be coherent to observe the interference pattern. The beams from the two different sodium lamps are not coherent. Ans. T Q 32. A plane wave of sound travelling in air is incident upon a plane water surface. The angle of incidence is 60◦ . Assuming Snell’s law to be valid for sound waves, it follows that the sound wave will be refracted into water away from the normal. (1984) Sol. The refractive index of a medium is the ratio of the speed of light in vacuum to the speed of the light in the medium i.e., n = c/v. Snell’s law, n1 sin i = n2 sin r, for refraction from medium of refractive index n1 = c/v1 to the medium of refractive index n2 = c/v2 can be written as v2 sin i = v1 sin r.

The speed of sound in water is more than the speed of sound in air i.e., v2 > v1 . Thus, from equation (1), r > i i.e., sound wave is bent away from the normal. Note that water acts as a rarer medium and air act as a denser medium for the sound waves. We encourage you to substitute the values of v1 = 330 m/s, v2 = 1440 m/s and i = 60◦ in equation (1) and find out the possible reasons for inconsistency. Ans. T

(1)

Q 33. A slit of width d is placed in front of a lens of focal length 0.5 m and is illuminated normally with light of wavelength 5.89 × 10−7 m. The first diffraction minima on either side of the central diffraction maximum are separated by 2 × 10−3 m. The width d of the slit is . . . . . . m. (1997) Sol. The diffraction pattern is observed at the focal plane of the lens. Hence, screen is placed at a distance D = f = 0.5 m. The first minimum on either side is symmetrically distributed about the central maximum. Thus, the distance of first minimum from the central maxumum is y = (2 × 10−3 )/2 = 10−3 m. The condition for the first minimum is d sin θ ≈ d tan θ = d (y/D) = λ. Substitute the values to get, d=

(5.89 × 10−7 )(0.5) λD = 2.945 × 10−4 m. = y 10−3 Ans. 2.945 × 10−4

Q 34. A monochromatic beam of light of wavelength 6000 ˚ A in vacuum enters a medium of refractive index 1.5. In the medium its wavelength is . . . . . . and its frequency is . . . . . . (1997) Sol. The speed of light in vacuum is c = 3 × 108 m/s. The wavelength of given light in vacuum is λ0 = 6000 ˚ A and its frequency in vacuum is ν0 =

3 × 108 c = = 5 × 1014 Hz. λ0 6000 × 10−10

The refractive index (n) of a medium is the ratio of speed of light in vacuum (c) to the speed of light in medium (v). Thus, the speed of light in the medium of refractive index n = 1.5 is v = c/n = (3 × 108 )/1.5 = 2 × 108 m/s.

222

Part II. Waves

The frequency of light does not change when it travel from one medium to another. Hence, the wavelength of light in the medium is λ=

c/n c/ν0 λ0 6000 v = = = = = 4000 ˚ A. ν ν0 n n 1.5 Ans. 4000 ˚ A, 5 × 1014 Hz

Q 35. A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at a distance 9 m and 25 m respectively from the source. The ratio of amplitudes of the waves at P and Q is . . . . . . (1989) Sol. A point source emits spherical waves.

rP S

Sol. The velocity of light in a medium of refractive index n = 1.5 is v = c/n = 3 × 108 /1.5 = 2 × 108 m/s. The wavelength of light of frequency ν = 5 × 1014 Hz in the medium is given by λ=

2 × 108 v = = 4 × 10−7 m. ν 5 × 1014 Ans. 2 × 108 m/s, 4 × 10−7 m

Integer Type Q 38. A Young’s double slit interference arrangement with slits S1 and S2 is immersed in water (refractive index = 4/3) as shown in the figure. The positions of maxima on the surface of water are given by x2 = p2 m2 λ2 − d2 , where λ is the wavelength of light in air (refractive index = 1), 2d is the separation between the slits and m is an integer. The value of p is . . . . . . .

rQ P

(2015) Q S1 d

The intensity of a spherical wave varies with the distance from the source (r) as I = I0 /r2 , where I0 is a constant. The intensity is proportional to the square of the wave amplitude A, i.e., I = A2 . Eliminate √ I, to get the amplitude of the spherical wave as, A = I0 /r. Thus, the ratio of the amplitudes at points P and Q located at distances rP = 9 m and rQ = 25 m is

Air

x

d

Water

S2

Sol. The waves emanating from the two slits interfere at the point P.

√ AP I0 /rP rQ 25 . =√ = = AQ rP 9 I0 /rQ

S1 d

x

d

Ans. 25/9 Q 36. In Young’s double slit experiment, the two slits act as coherent sources of equal amplitude A and wavelength λ. In another experiment with the same set-up the two slits are sources of equal amplitude A and wavelength λ, but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is . . . . . . (1986) Sol. Since the two slits act as two sources of equal amplitude (A), the intensities of light from the two slits are equal, say I. In the case of coherent sources, constructive interference takes place at the mid-point of the screen with intensity I1 = 4I0 . In the case of incoherent sources, there is no interference and intensities simply add up to give I2 = 2I. Ans. 2 Q 37. A light wave of frequency 5 × 1014 Hz enters a medium of refractive index 1.5. In the medium the velocity of light wave is . . . . . . and its wavelength is . . . . . . (1983)

P

Air

Water

S2

The waves from slit S1 travel a distance S1 P in the medium of refractive index µ1 = 1 (air). The √ optical path length for these waves is l1 = µ1 S1 P = x2 + d2 . The waves from the slit S2 travel a distance S2 P in the medium of refractive index µ2 = 4/3 (water). The optical √ path length for these waves is l2 = µ2 S2 P = (4/3) x2 + d2 . The path difference between the waves interfering at P is p ∆ = l2 − l1 = 13 x2 + d2 . (1) For maxima, the path difference is equal to an integral multiple of the wavelength λ i.e., ∆ = mλ,

for m = 1, 2, 3, . . ..

(2)

Eliminate ∆ from equations (1)–(2) and simplify to get x2 = 9m2 λ2 − d2 . Ans. 3

Chapter 15. Light Waves

223

Descriptive Q 39. In a Young’s double slit experiment, two wavelengths of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maxima coincide again? Take D/d = 103 . Symbols have their usual meaning. (2004) Sol. In Young’s double slit experiment, the path difference at a distance x from the centre is given by ∆ = d sin θ = d (x/D). The interference pattern attains maxima when the path difference is an integral multiple of wavelength i.e., d (x/D) = nλ, where n is a positive integer. Two wavelengths, λ1 = 500 nm and λ2 = 700 nm, have their maxima at the same point (same x) if n1 λ1 D/d = n2 λ2 D/d, which gives 5n1 = 7n2 . The solutions of the equation 5n1 = 7n2 are (n1 , n2 ) = { (7, 5), (14, 10), . . .}. Minimum distance of such a maxima is x = n1 λ1 (D/d) = 7(500 × 10−9 )103 m = 3.5 mm. Ans. 3.5 mm Q 40. A point source S emitting light of wavelength 600 nm is placed at a very small height h above a flat reflecting surface AB (see figure). The intensity of the reflected light is 36% of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it. (2002) Screen

D •S

h A

A2 + A22 − 2A1 A2 (A1 − A2 )2 Imin = 12 = 2 Imax (A1 + A2 ) A1 + A22 + 2A1 A2 √ I1 + I2 − 2 I1 I2 1 √ = = . 16 I1 + I2 + 2 I1 I2 The path difference between the two waves at P is ∆ = 2h + λ/2 (the reflected wave travels an additional distance 2h and undergoes an additional phase change of π due to reflection from the denser medium). Moving the mirror by a distance x introduces a path difference of 2x. To get two consecutive maxima at P, the path difference caused by the mirror movement should be equal to 2xmin = λ. Thus, the minimum distance through which the mirror is to be moved is xmin = λ/2 = 600/2 = 300 nm. You may visualize the two sources as S and its mirror image S 0 . Ans. (a) circular (b) 1/16 (c) 300 nm Q 41. A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line perpendicular to the plane AB and passing through O, the middle point of S1 and S2 . A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in the figure. Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. Now, a liquid is poured into the vessel and filled upto OQ. The central bright fringe is found to be at Q. Calculate the refractive index of the liquid. (2001)

B

(a) What is the shape of the interference fringes on the screen? (b) Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point P (shown in the figure). (c) If the intensity at point P corresponds to a maximum, calculate the minimum distance through which the reflecting surface AB should be shifted so that the intensity at P again becomes maximum.

A

O

Q

S2

S B 2m

Sol. The wavefront emanating from a point source S is spherical. The plane screen intersects the spherical wavefronts in circles. Similarly AB reflects the spherical wavefront as spherical wavefront which intersects the

D

S1 P 40 cm

P

screen in circle. Thus, the interference pattern on the screen will have circular fringes. Let I1 = A21 = I0 be the intensity of light directly coming to the screen from S and I2 = A22 = 0.36I0 be the intensity of light coming to the screen after reflection. These two light waves interfere on the screen producing interference pattern with Amin = A1 − A2 and Amax = A1 + A2 . The ratio of minimum to maximum intensities is

C 10 cm

Sol. Let the central maximum be observed at R.

224

Part II. Waves 1

R S1 P

d

µ = 1.0

y2 θ

α

y1

2

t

Q

µ2 = 1.8 µ1 = 1.5

S2

S D1

D2

Thus, the path difference between the two waves interfering at R is zero. The path difference between the two waves coming through slit S1 and S2 is given by

Ray 1 undergoes reflection from a denser medium µ2 = 1.8 causing a phase change of φ1 = π. Ray 2 undergoes reflection from a rarer medium µ1 = 1.5 without causing a phase change. Ray 2 also travels an additional path 2t in a medium of refractive index µ2 , causing an additional optical path difference ∆x2 = 2µ2 t. Thus, the total path difference between ray 1 and ray 2 is

∆x = SS1 R − SS2 R

∆x = ∆x2 − φ1 (λ/2π) = 2µ2 t − λ/2.

= SS1 + S1 R − SS2 − S2 R = (SS1 − SS2 ) − (S2 R − S1 R) = d sin α − d sin θ

The constructive interference takes place when the total path difference is an integral multiple of wavelength λ,

≈ d tan α − d tan θ

∆x = 2µ2 t − λ/2 = nλ, i.e.,

= d(y1 /D1 − y2 /D2 ).

3.6t = (n + 1/2) λ,

Substitute ∆x = 0 to get y2 = y1 D2 /D1 = 40 × 10/200 = 2 cm. Note that ∆x = 0 is equivalent to θ = α i.e., SS1 RS2 S is a parallelogram. The central maximum is shifted to Q when a liquid of refractive index µ is filled upto point Q. The wave S1 Q travels in air but S2 Q travels in liquid causing a path difference S2 Q − S1 Q = (µ − 1)D2 . The path difference between waves interfering at Q is ∆x0 = SS1 Q − SS2 Q = (SS1 − SS2 ) − (S2 Q − S1 Q) = d sin α − (µ − 1)D2 ≈ d tan α − (µ − 1)D2 = dy1 /D1 − (µ − 1)D2 .

where n = 0, 1, 2, . . .. The minimum value of t occurs for n = 0. Substitute the values in equation (1) to get 648 λ = 90 nm. = 4µ2 4 × 1.8
Ans. 3.6t = n + 12 λ with n = 0, 1, 2, . . ., = 90 nm

tmin =

tmin

Q 43. Young’s double slit experiment is done in a medium of refractive index 4/3. A light of 600 nm wavelength is falling on the slits having 0.45 mm separation. The lower slit S2 is covered by thin glass sheet of thickness 10.4 µm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in the figure. (1999)

For central maximum at Q, ∆x0 = 0, which gives µ=1+

(1)

y

0.08 × 40 y1 d =1+ = 1.0016. D1 D2 10 × 200

S1 S•

O

Ans. 2 cm above point Q on side CD, µ = 1.0016 Q 42. A glass plate of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8. Light of wavelength λ travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. Write the condition for their constructive interference. If λ = 648 nm, obtain the least value of t for which the rays interfere constructively. (2000) Sol. We consider only the near normal incidence.

S2

(a) Find the location of central maximum (bright fringe with zero path difference) on the y-axis. (b) Find the light intensity at point O relative to the maximum fringe intensity. (c) Now, if 600 nm light is replaced by white light of range 400 nm to 700 nm, find the wavelengths of the light that form maxima exactly at point O. [All wavelengths in the problem are for the given medium of refractive index 4/3. Ignore dispersion.]

Chapter 15. Light Waves

225 y

d=1 mm

Sol. Given d = 0.45 mm, D = 1.5 m, λ = 600 nm, t = 10.4 µm, µg = 1.5, and µw = 4/3. Let the central maximum occurs at P, a distance y below O. 30◦ y S1

x

T S•

d

O

D = 1.0 m

y

θ P S2 D

The path difference at P without the glass sheet is given by ∆x1 = S1 P − S2 P = d sin θ = yd/D. The glass sheet of thickness t, refractive index µg , placed in a medium of refractive index µw introduces a path difference

(a) If the incident beam falls normally on the double slit apparatus, find the y-coordinates of all the interference minima on the screen. (b) If the incident beam makes an angle of 30◦ with the x-axis (as in the dotted arrow shown in figure), find the y-coordinates of the first minima on either side of the central maximum. Sol. The wavelength is λ = 0.5 mm, distance between the slits S1 and S2 is d = 1.0 mm and the distance of the screen is D = 1.0 m. S1

∆x2 = (µg /µw − 1) t.

θ

d

Since the path difference at P is zero, we get ∆x1 = ∆x2 . Solve to get

P θ

O

nθ

i

s S2 d

D

y = (D/d) (µg /µw − 1) t = 4.33 mm. At point O, the path difference is introduced by the glass plate only. The phase difference between the waves interfering at O is δ = (2π/λ) ∆x2 = (2π/λ) (µg /µw − 1) t = 13π/3. If I0 is the intensity of each slit then the intensity at O is given by IO = 4I0 cos2 (δ/2) = Imax cos2 (13π/6) = (3/4)Imax , where Imax = 4I0 is the intensity of central maximum. The point O will have a maximum when path difference is an integral multiple of wavelength i.e., ∆x2 = nλn , where n = 1, 2, 3, . . .. Thus the wavelengths that can give maximum at O are λ1 = 1300 nm, λ2 = 650 nm, λ3 = 433.33 nm, λ4 = 325 nm, . . . . The wavelengths λ2 and λ3 lie in the range 400 nm to 700 nm. Ans. (a) −4.33 mm (b) I = 43 Imax (c) 650 nm, 433.33 nm Q 44. A coherent parallel beam of microwaves of wavelength λ = 0.5 mm falls on a Young’s double slit apparatus. The separation between the slits is 1.0 mm. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of 1.0 m from it as shown in the figure. (1998)

Consider the case when beam falls normally on the apparatus. The path difference between the two waves interfering at P is given by ∆x = S2 P − S1 P ≈ d sin θ. For minimum intensity at P (destructive interference), the path difference should be an odd multiple of λ/2 i.e., ∆x = d sin θ = (2n − 1) (λ/2) , where n = 1, 2, 3, . . .. Substitute the values of d, λ, and n, to get sin θ1 = 1/4 for n = 1 and sin θ2 = 3/4 for n = 2. Other values of n give sin θ > 1, which is inconsistent. Thus, the distances of these minima from O are √ y1 = D tan θ1 = 1/ 15 m, and √ y2 = D tan θ2 = 3/ 7 m. By symmetry, we also get the minima on the other side of O at √ √ y10 = −1/ 15 m, y20 = −3/ 7 m. S1 α

d

P

sin

θ

θ

O

α

nθ

i

S2

ds

D

226

Part II. Waves

Now, consider the case where incident beam makes an angle α = 30◦ with the x axis. The geometry is shown in the figure. The wave incident on slit S1 travels an extra path as compared to the wave incident on slit S2 . This additional path difference is given by

The phase difference corresponding to ∆x is δ = (2π/λ) ∆x = (2π/λ) (µ2 − µ1 ) t.

The intensity at a point with a phase difference δ is given by

∆x0 = d sin α = d sin 30◦ = 0.5 mm. Thus, the total path difference between the waves interfering at P is

I(δ) = Imax cos2 (δ/2) .

cos2 (δ/2) = 3/4.

For a central maximum, ∆x = d sin θ0 − d sin α = 0, √ which gives θ0 = α = 30◦ and y0 = D tan θ0 = 1/ 3 m. First minimum will occur when ∆x = d sin θ1 − d sin α = λ/2, which √ gives sin θ1 = 3/4 and hence y1 = D tan θ1 = 3/ 7 m. The minimum on the other side will occur when ∆x = d sin θ10 − d sin α = −λ/2, √ which gives y10 = 1/ 15√m. √ √ Ans. (a) ±1/ 15 m, ±3/ 7 m (b) 1/ 15 m, √ 3/ 7 m Q 45. In a Young’s double slit experiment, the upper slit is covered by a thin glass plate of refractive index 1.4, while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400 ˚ A. It is found that the point P on the screen, where the central maximum (n = 0) falls before the glass plates were inserted, now has 3/4 the original intensity. It is further observed that what used to be the fifth maximum earlier lies below the point P while the sixth minima lies above P. Calculate the thickness of glass plate. [Absorption of light by glass plate may be neglected.] (1997) Sol. Let t be the thickness of the plates of refractive indices µ1 = 1.4 and µ2 = 1.7. µ1 6th min P 5th max

S2 µ2

The path difference introduced by the plates between the two waves interfering at P is ∆x = ∆x2 − ∆x1 = (µ2 − 1)t − (µ1 − 1)t = (µ2 − µ1 )t.

(2)

Substitute I(δ) = (3/4)Imax in equation (2) to get

∆x = d sin θ − d sin α.

S1

(1)

(3)

Since what used to be the 5th maximum earlier lies below P and what used to be the 6th minimum earlier lies above P (see figure), the path difference at P should lie between 5λ and 5λ + λ2 i.e., 5λ < ∆x < 5λ + λ2 or 10π < δ < 11π. cos2

δ 2

3 4

π

3π

5π

7π

5th max P 6th min δ 31π 11π 9π 3

The equation (3) along with the condition 10π < δ < 11π give δ = 10π+ π3 = 31π 3 . Substitute δ in equation (1) to get λδ 2π(µ2 − µ1 ) (5400 × 10−10 ) (31π/3) = = 9.3 × 10−6 m. 2π(1.7 − 1.4)

t=

Ans. 9.3 µm Q 46. In Young’s double slit experiment, the source is red light of wavelength 7 × 10−7 m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by 10−3 m to the position previously occupied by the 5th bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength 5 × 10−7 m, the central fringe shifts to a position initially occupied by the 6th bright fringe due to red light. Find the refractive index of glass for green light. Also estimate the change in fringe width due to the change in wavelength. (1997) Sol. The path difference introduced by a thin plate of refractive index µ and thickness t is ∆x = (µ − 1)t.

(1)

The plate shifts the central maximum to a position previously occupied by 5th bright fringe i.e., ∆x = 5λred . Substitute in equation (1) to get t=

5 × 7 × 10−7 5λred = = 7 × 10−6 m. µ−1 1.5 − 1

Chapter 15. Light Waves

227

With green light, the path difference introduced by the thin plate is ∆x0 = (µgreen − 1)t, where µgreen is the refractive index of the glass plate for green light. The plate shifts the central maximum of green light to a position previously occupied by the 6th bright fringe of red light i.e., ∆x0 = 6λred . Substitute the values to get 6 (7 × 10−7 ) 6λred +1= + 1 = 1.6. t 7 × 10−6 Five fringes of red light occupy a distance of 10−3 m. Thus, the fringe width for red light is µgreen =

10−3 = 2 × 10−4 m. 5 The fringe width β is related to wavelength λ by β = λD/d. Thus, βred =

λgreen βred βgreen = λred 5 × 10−7 = × 2 × 10−4 = 1.43 × 10−4 m. 7 × 10−7 The change in fringe width is ∆β = βgreen − βred = −0.57 × 10

−4

m.

Ans. 7 × 10−6 m, 1.6, −0.57 × 10−4 m Q 47. A double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and distance between the plane of slits and screen is 1.33 m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300 ˚ A. (1996) (a) Calculate the fringe width. (b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minimum at the axis. Sol. Young’s double slit apparatus with d = 1 mm and D = 1.33 m is immersed in a liquid of refractive index µ = 1.33. The wavelength of light in the given liquid is λ0 = λ/µ = 6300/1.33 = 4737 ˚ A. The fringe width is given by β=

Ans. (a) 0.63 mm (b) 1.579 µm Q 48. Angular width of central maximum in the Fraunhoffer diffraction pattern of the slit is measured. The slit is illuminated by light of wavelength 6000 ˚ A. When the slit is illuminated by light of another wavelength, the angular width decreased by 30%. Calculate the wavelength of this light. The same decrease in the angular width of central maximum is obtained when the original apparatus is immersed in a liquid. Find refractive index of the liquid. (1996) Sol. Let the slit of width b is used to observe Fraunhoffer diffraction. First minimum is observed at an angular displacement θ given by b sin θ = λ, where λ is the wavelength of the light. For small angles, sin θ ≈ θ, and angular width of central maximum is given by β = 2θ = 2λ/b. When the wavelength is changed from λ = 6000 ˚ A to λ0 , the angular width is decreased by 30%, i.e., β 0 = 2λ0 /b = (1 − 0.30)β = (1 − 0.30) (2 × 6000/b) , which gives λ0 = 0.7 × 6000 = 4200 ˚ A. The wavelength of light in liquid of refractive index µ is given by λ00 = λ/µ. Since β is decreased by the same amount, we have λ00 = λ0 , which gives µ = λ/λ0 = 6000/4200 = 1.43. Ans. 4200 ˚ A, 1.43 Q 49. A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in the figure. Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects 25 per cent of the light incident on it and transmits the remaining. Find the ratio of the minimum and maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate.

λ0 D λD (6300 × 10−10 ) 1.33 = = d µ d 1.33 (1 × 10−3 )

(1990) I2

= 0.63 × 10−3 = 0.63 mm. One of the slit is covered by a glass sheet of thickness t and refractive index µ0 = 1.53. The path difference introduced by the glass sheet is

I1

∆x = (µ0 /µ − 1) t = (1.53/1.33 − 1) t = 0.15t. (1)

I

Now, for the first minimum to shift to the position of central maximum, ∆x = λ0 /2. Substitute it in equation (1) to get 4737 × 10−10 λ0 t= = 0.15 × 2 0.15 × 2 −6 = 1.579 × 10 = 1.579 µm.

1

2

Sol. Let I be the intensity of the incident beam. The intensity of the beam reflected back by the first plate is I1 = I/4.

228

Part II. Waves

The intensity of transmitted beam is 34 I. This beam undergoes reflection on other plate to give intensity of 3 I. Finally, three-fourth intensity reflected beam as 16 of this beam is transmitted by the first plate to give intensity of beam coming back as

1.5. The path and phase difference caused by this film are ∆x = (µ − 1)t = (1.5 − 1)2000 = 1000 ˚ A, 2π 2π π δ= ∆x = 1000 = . λ 6000 3

I2 = 9I/64.

Two waves of powers PA and PB and a phase difference of δ = π/3 interfere to give the resultant power p P = PA + PB + 2 PA PB cos δ = 7 × 10−6 W. (1)

3 I 64 9 I 64 3 I 16 1 I 4

Note that equation √ (1) is a generalized intensity formula I = I1 + I2 + 2 I1 I2 cos δ when the slits/apertures are of different sizes. Ans. 7 × 10−6 W

3 I 4

I

The beams with intensities I1 and I2 interfere to give ratio of minimum to maximum intensities as √ √ 1 ( I1 − I2 )2 Imin √ √ = . = Imax 49 ( I1 + I2 )2 Ans. 1/49 Q 50. In a modified Young’s double slit experiment, a monochromatic uniform and parallel beam of light of wavelength 6000 ˚ A and intensity 10/π W/m2 is incident normally on two apertures A and B of radii 0.001 m and 0.002 m respectively. A perfectly transparent film of thickness 2000 ˚ A and refractive index 1.5 for the wavelength of 6000 ˚ A is placed in front of aperture A (see figure). Calculate the power (in W) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot. (1989)

Q 51. A beam of light consisting of two wavelengths, 6500 ˚ A and 5200 ˚ A is used to obtain interference fringe in a Young’s double slit experiment. (1985) (a) Find the distance of the third bright fringe on the screen from the central maximum of wavelength 6500 ˚ A. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm. Sol. Given, the distance between the slits is d = 2 mm and the distance of the screen from the slits is D = 120 cm. The fringe width for wavelength λ1 = 6500 ˚ A is given by λ1 D (6500 × 10−10 )(1.20) = d 2 × 10−3 −4 = 3.90 × 10 m.

β1 =

A

The distance of the third bright fringe from the central maximum for wavelength λ1 = 6500 ˚ A is equal to the width of three fringes i.e.,

F B

y3 = 3β1 = 3(3.90 × 10−4 ) = 1.17 × 10−3 m = 1.17 mm.

Sol. The powers received by the aperture A and B are 2 PA,i = I(πrA ) = (10/π)(π × (0.001)2 ) = 10−5 W, 2 PB,i = I(πrB ) = (10/π)(π

2

−5

× (0.002) ) = 4×10

I

Iλ2

Iλ1

W.

Only 10% of the power received by the apertures goes in the original direction. Thus, the powers coming out of the two apertures in the original direction are PA = 0.1 PA,i = 10−6 W, PB = 0.1 PB,i = 4 × 10−6 W. The light from aperture A passes through a transparent film of thickness t = 2000 ˚ A and refractive index µ =

y

Let nth 1 maximum of wavelength λ1 coincides with maximum of wavelength λ2 at a point P on the screen. Thus, the distance between the central maximum and the point P contains n1 fringes of λ1 and n2 fringes of λ2 i.e., nnd 2

yn1 = n1 λ1 D/d = n2 λ2 D/d,

(1)

Chapter 15. Light Waves

229

which gives n1 /n2 = λ2 /λ1 = 5200/6500 = 4/5. Thus, the minimum integral values of n1 and n2 are 4 and 5, respectively. Substitute these in equation (1) to get the minimum distance from the central maximum where bright fringes from both the wavelengths coincide λ1 D = 4β1 = 4(3.90 × 10−4 ) d = 1.56 × 10−3 m = 1.56 mm.

y = n1

Ans. (a) 1.17 mm (b) 1.56 mm Q 52. In Young’s double slit experiment using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 µm is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in experiment. (1983) Sol. Let d be the separation between the slits and D be the distance of the screen from the slits. The path difference at a point P on the screen located at a distance y from the central maximum is given by ∆x = yd/D. The condition for the nth maximum on the screen is ∆x = nλ, which gives the position of the nth maximum as yn = nλD/d. A mica sheet of refractive index µ = 1.6 and thickness t = 1.964 µm introduces an additional path difference (n − 1)t. Thus, the total path difference between the waves interfering at P becomes ∆x = yd/D + (µ − 1)t. The condition for nth maximum on the screen ∆x = nλ gives the new position of the nth maximum as yn0 = [nλ − (µ − 1)t] D/d. The fringes shift towards the central maximum by a distance ∆y = yn − yn0 = (µ − 1)tD/d.

(1)

The fringe width for the light of wavelength λ is given by β = λD/d. When the distance of the screen from the slits is doubled, fringe width becomes β 0 = 2λD/d.

(2)

Given, β 0 = ∆y. Substitute the values from equations (1) and (2) to get λ = (µ − 1)t/2 = (1.6 − 1)(1.964 × 10−6 )/2 = 5892 ˚ A. Ans. 5892 ˚ A

Part III

Optics

A

i

231

δr

Chapter 16 Geometrical Optics

One Option Correct I02

(−50, 0)

I2 (0, 0)

m

y

c 50

Q 1. A small object is placed 50 cm to the left of a thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle θ = 30◦ to the axis of the lens, as shown in the figure.

θ θ

x

x

f = 30 cm 50 cm θ

x R

(0, 0)

=

(−50, 0)

0

10 cm

50 cm

What happens to the image if mirror is tilted by an angle θ? Consider a ray along the principal axis of the lens. If mirror is not tilted then this ray starts from the object, refracts through the lens without deviation, incident normally on the mirror and finally comes to the image I2 after reflection from the mirror. If the mirror is tilted by an angle θ then this ray is incident on the mirror at an angle of incidence θ and reflected by the mirror with an angle of reflection θ and finally forms a new image at I20 (see figure). Thus, the axis on which image lies makes an angle 2θ = 60◦ with the principal axis of the lens. Also, the image distance will remain same upto the first order of approximation (image distance will remain exactly same in case of plane mirror). Thus, (x, y) coordinates of the image are

√ (50 + 50 3, −50)

If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) at which the image is formed are (2016) √ (A) (0, 0) √ (B) (50 − 25 3,√25) (C) (25, 25 3) (D) (125/3, 25/ 3) Sol. The object distance for the lens of focal length f = 30 cm is u = −50 cm. Apply lens formula, v1 − u1 = f1 , to get the image distance v = 75 cm. Thus, image I1 by the lens is formed 25 cm behind the mirror.

I2

O

50 cm

x = 50 − 50 cos 60◦ = 25 cm, √ y = 50 sin 60◦ = 25 3 cm.

I1 x

50 cm

25 cm

Aliter: The image I1 acts as a virtual object for the tilted mirror. The object distance (along the √ principal axis of the mirror) is u = 25 cos 30◦ = 25 3/2 and height of the object is h = 25 sin 30◦ = 25/2. The √mirror formula gives the image distance v = √ −50 3/(4 − 3). The√magnification by the mirror is m = −v/u = 4/(4 − √ 3). Thus, the image height 0is VI02 = mh = 50/(4− 3). In right angled triangle PVI2 , √ angle θ = 30◦ and distance PI02 = 100/(4 − 3) cm.

Let us consider the case when mirror is not tilted. The image I1 will act as a virtual object with u = 25 cm for a convex mirror of focal length fm = R/2 = 50 cm. Apply the mirror formula, v1 + u1 = f1m , to get the image distance v = −50 cm. Thus, image I2 by the mirror is formed 50 cm left of the mirror. 233

234

Part III. Optics I20 (x, y) 50 √ 4− 3

Solve equations (1) and (2) to get µl = √

100 √ 4− 3

V √ 50 √3 4− 3

θ θ

Ans. C

x P

(0, 0)

µb r 2.72(11.54/2) = 1.36. =p 2 2 r +h (11.54/2)2 + 102

25 2

√ 25 3 2

Q 3. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is 2/3 times the wavelength in free space. The radius of the curved surface of the lens is (2013) (A) 1 m (B) 2 m (C) 3 m (D) 6 m

50 cm

The x and y coordinates of the image I02 are given by   100 1 √ cos 60◦ = 50 1 − √ x = 50 − 4− 3 4− 3 = 28 cm, √ 100 50 3 √ sin 60◦ = √ = 38 cm. y= 4− 3 4− 3

f = uv/(u − v) = +6 m. The refractive index of the lens material is given by,

Ans. C Q 2. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is (2014) Liquid

S

Sol. The magnification of the given lens is, m = v/u = −1/3. Note that m is negative because image is real and inverted. The image distance is v = +8 m. Substitute v to get u = −24 m. Substitute u and v in the lens formula, 1/f = 1/v − 1/u, to get,

Block

(A) 1.21 (B) 1.30 (C) 1.36 (D) 1.42 Sol. The rays stop coming to liquid side if the angle of incidence at the block-liquid interface is greater than the critical angle θc .

µ = c/v = λspace /λmedium = 3/2. The lens maker’s formula for the plano-convex lens gives,   1 1 1 = (µ − 1) − , f ∞ −R where R is the desired radius. Substitute the values of µ and f to get R = 3 m. Ans. C Q 4. A√ ray
of light travelling in the direction 1 3  ˆ is incident on a plane mirror. After reˆ ı + 2 √
flection, it travels along 21 ˆı − 3 ˆ . The angle of incidence is (2013) (A) 30◦ (B) 45◦ (C) 60◦ (D) 75◦ √
√
Sol. Let vˆi = 12 ˆı + 3ˆ  and vˆr = 12 ˆı − 3ˆ  be the unit vectors along the incident and the reflected rays. From Snell’s law, angle of incidence (say θ) is equal to the angle of reflection. Thus, angle between −ˆ vi and vˆr is 2θ. y

r θc

O

h

x

θc vˆi

Let µb = 2.72 be the refractive index of the block and µl be the refractive index of the liquid. The critical angle is given by sin θc = µl /µb . From geometry, p sin θc = r/ r2 + h2 .

(1)

(2)

θ

vˆr

Using, dot products of vectors, we get cos(2θ) = −ˆ vi · vˆr = 1/2. Hence, θ = 30◦ . You can see that vˆi = cos 60◦ ˆı + sin 60◦ ˆ and vˆr = cos(−60◦ ) ˆı + sin(−60◦ ) ˆ. Draw these vectors to get desired result. Ans. A

Chapter 16. Geometrical Optics

235

Q 5. A biconvex lens is formed with two thin planoconvex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R = 14 cm. For this biconvex lens, for an object distance of 40 cm, the image distance will be (2012) n = 1.5

n = 1.2

Thus, R = 1 and T = 0 for all θ ≥ θc . The energy conservation (assuming no absorption) gives R + T = 1,

for all θ.

These two arguments are sufficient to get the answer. You might be thinking that R = 0 and T = 1 at θ = 0 (normal incidence) but detailed analysis shows that 2 n1 − n2 = 4%, and n1 + n2 4n1 n2 T = = 96%, (n1 + n2 )2 

R=

R = 14 cm

when light travels from the glass to air at normal incidence. Ans. C

(A) −280 cm (B) 40 cm (C) 21.5 cm (D) 13.3 cm Sol. Let f1 and f2 be the focal lengths of the left and the right plano-convex lenses. Using lens maker’s formula, 1/f = (µ − 1) [1/R1 − 1/R2 ] , we get, 1/f1 = (1.5 − 1) [1/14 − 1/∞] = 1/28,

(1)

1/f2 = (1.2 − 1) [1/∞ − 1/(−14)] = 1/70.

(2)

The thin lenses are in contact with each other. Thus, focal length of the complete system is given by 1/f = 1/f1 + 1/f2 .

(3)

Substitute f1 and f2 from equations (1) and (2) into equation (3) to get f = 20 cm. Substitute f = 20 cm and u = −40 cm in the lens formula, 1/f = 1/v − 1/u, to get the image distance v = 40 cm. Ans. B Q 6. A light ray travelling in glass medium is incident on glass-air interface at an angle of incidence θ. The reflected (R) and transmitted (T) intensities, both as a function of θ are plotted. The correct sketch is (2011) (B) T R

0◦

θ

90◦

R θ

90◦

0◦

gives v1 = 30 cm. 30 cm

10 cm

•

•

•

•

O

I2

I3

I1

20 cm

20 cm

R θ

90◦

T R θ

90◦

Sol. The light undergoes total internal reflection when it travels from a dense medium (glass, n1 = 1.5) to a rare medium (air, n2 = 1) and its angle of incidence (θ) is greater than the critical angle, θc = sin−1 (n2 /n1 ) = 41.8◦ .

1 1 1 − = , v1 −30 15

16 cm

(D) 100% T

Sol. The lens formula, 1/v1 − 1/u1 = 1/f1 , applied for refraction at the lens,

T

0◦ Intensity

Intensity

(C) 100%

0◦

100% Intensity

Intensity

(A) 100%

Q 7. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is (2010) (A) virtual and at a distance of 16 cm from the mirror. (B) real and at a distance of 16 cm from the mirror. (C) virtual and at a distance of 20 cm from the mirror. (D) real and at a distance of 20 cm from the mirror.

The image of the object O is formed at I1 , 20 cm right of the mirror. Before converging at I1 , the rays are reflected back by the plane mirror. The mirror formula, 1/v2 + 1/u2 = 1/f2 , applied for reflection at the plane mirror, 1 1 1 + = , v2 20 ∞ gives v2 = −20 cm (the I1 acts as a virtual object for this reflection). The image of I1 is formed at I2 , 10 cm left of the lens. Before converging at I2 , the rays undergo second refraction at the lens. The lens formula,

236

Part III. Optics

1/v3 − 1/u3 = 1/f3 , applied for the second refraction at the lens, 1 1 1 − = , v3 −10 −15 gives v3 = −6 cm (note the negative sign for f3 as ray is incident from the left side). The final image I3 is 16 cm left of the mirror. Since rays actually converge at this point, image is real. Note that O is placed at a distance 2f from the lens and hence I1 shall be at 2f . Also, image and the object distances from the plane mirror are equal. Ans. B Q 8. A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as [Take g = 10 m/s2 .] (2009) (A) 9 m/s (B) 12 m/s (C) 16 m/s (D) 21.33 m/s Sol. Let the ball O be at a height x above the water surface. Let the fish sees the image I of the ball at a height x0 above the water surface. •I

Q 9. A light beam is traveling from Region I to Region IV (see figure). The refractive index in Region I, II, III and IV are n0 , n20 , n60 and n80 , respectively. The angle of incidence θ for which the beam just misses entering region IV is (2008) Region I n0

Region II

Region III

Region IV

n0 2

n0 6

n0 8

θ 0

(A) sin−1 (C) sin−1

0.2 m

(B) sin−1 (D) sin−1

3 4
1 4




0.6 m 1 8
1 3




Sol. The path of the beam is shown in the figure. I

II

III

n0

n0 2

θ2

IV

θ2 θ1 θ1 θ

n0 6

n0 8

The beam will just miss entering region IV if the angle of incidence at III-IV interface is equal to the critical angle i.e.,

•O

x0

sin θ2 =

x

The formula for refraction at the spherical surface, µ2 /v − µ1 /u = (µ2 − µ1 )/R, 0

relates x and x . Substitute µ2 = 4/3, µ1 = 1, R = ∞, and u = −x, to get v = −x0 = −4x/3 i.e., x0 = (4/3)x. Thus, dx0 /dt = (4/3)dx/dt.

(1)

The ball falls by a distance s = 20 − 12.8 = 7.2 m when fish sees it. Thus, actual velocity of the ball when it is 12.8 m above the water surface is p p dx/dt = 2gs = 2(10)(7.2) = 12 m/s. (2) Use equations (1) and (2) to get apparent velocity of the ball as seen by the fish as dx0 /dt = (4/3)12 = 16 m/s. Ans. C

n0 /8 3 = . n0 /6 4

The angle of incidence at III-IV interface is equal to the angle of refraction at II-III interface. Using, Snell’s law for refraction at II-III interface, we get      n0 /6 3 1 1 sin θ1 = sin θ2 = = . n0 /2 4 3 4 Using similar argument for I-II interface, we get      n0 /2 1 1 1 sin θ = sin θ1 = = . n0 4 2 8 Ans. B Q 10. A ray of light traveling in water is incident on its surface open to air. The angle of incidence is θ, which is less than the critical angle. Then there will be (2007) (A) only a reflected ray and no refracted ray. (B) only a refracted ray and no reflected ray. (C) a reflected ray and a refracted ray and the angle between them would be less than 180◦ − 2θ. (D) a reflected ray and a refracted ray and the angle between them would be greater than 180◦ − 2θ. Sol. The angle of incidence θ is less than the critical angle. Thus, the incident rays will undergo reflection as well as refraction.

Chapter 16. Geometrical Optics

237 In given case, when eye is shifted towards left, the image appears to the right of the object pin. Thus, image lies between the eye and the object pin. Hence, the object is near to the pole P in comparison to its image. This is possible when the object is placed between f and 2f i.e., f < x < 2f . Ans. B

r air α water θ θ

By the laws of reflection, angle of incidence is equal to the angle of reflection. Also, by Snell’s law, sin θ/sin r = 1/µw < 1. Thus, sin θ < sin r and hence θ < r. The angle between the refracted and the reflected rays is α = 180◦ − (r + θ) < 180◦ − 2θ

Q 12. A point object is placed at a distance of 20 cm from a thin plano-convex lens of focal length 15 cm. The plane surface of the lens is now silvered. The image created by the system is at (2006)

(∵ r > θ).

•

Ans. C Q 11. In an experiment to determine the focal length (f ) of a concave mirror by the u-v method, a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then, (2007) (A) x < f (B) f < x < 2f (C) x = 2f (D) x > 2f Sol. The concave mirror forms the inverted image only if object is placed beyond the focus i.e., x > f . Also, if x = 2f then the object and the image coincides with each other and the student cannot observe parallax. In general, the concept of parallax is memorized by the students without understanding it. Let A and B be two points in space. Place your eye at E on the line AB. E1

20 cm

(A) (B) (C) (D)

60 cm 60 cm 12 cm 12 cm

to to to to

the the the the

left of the system right of the system left of the system right of the system

Sol. Substitute u = −20 cm and f = 15 cm in the lens formula, 1/v − 1/u = 1/f,

to get the image distance v = 60 cm. The image I1 by the lens is formed 60 cm to the right of the system. Before converging, the rays are reflected by the plane mirror i.e., this image act as an object for the plane mirror. The image I2 by the plane mirror is formed 60 cm towards left of the system.

•

40 cm

E• E2

(1)

•

A

20 cm

60 cm

•

B

•

If you move eye in one direction then the point A, which is near to the eye, appears to move in the opposite direction. On the other hand, the point B, which is far from the eye, appears to move in the direction of eye movement. Move the eye leftward from the point E to the point E1 . The line joining the eye to the far point B moves from EB to E1 B and the near point A appears to move rightward relative to this line. The line joining the eye to the near point A move from EA to E1 A and the far point B appears to move rightward relative to this line. We encourage you to experience this phenomenon by placing forefingers of two hands in front of their eyes and then moving the head to change eye’s position. If the points A and B coincides with each other then parallax is not observed i.e., parallax between A and B is removed.

I2

•

I3

I1

12 cm

The reflected rays, while going back, are refracted again by the lens i.e., image by the mirror acts as an object for the lens. This time, rays are travelling from the right to the left and hence the first focus of the lens is relevant. Thus, u = −60 cm and f = −15 cm. Substitute these values in equation (1) to get v = −12 cm i.e., the image I3 is formed 12 cm to the left of the system. Ans. C Q 13. A biconvex lens of focal length f forms a circular image of radius r of sun in focal plane. Then which option is correct? (2006) (A) πr2 ∝ f (B) πr2 ∝ f 2

238

Part III. Optics

(C) If lower half part is covered by black sheet, then area of the image is equal to πr2 /2. (D) If f is doubled, intensity will increase. Sol. The image of a distant object formed by a biconvex lens is inverted and lies at the focal plane.

v (cm) 11

10 45◦

(A) 5 ± 0.1 (C) 0.5 ± 0.1

-10

9 -9

(B) 5 ± 0.05 (D) 0.5 ± 0.05

Sol. Substitute u = −10 cm and v = 10 cm in the lens formula, 1/f = 1/v − 1/u, to get f = 5.0 cm. Differentiate and simplify to get ∆f /f 2 = ∆v/v 2 + ∆u/u2 . Substitute u = −10 cm, v = 10 cm, ∆u = 0.1 cm, ∆v = 0.1 cm, and f = 5.0 cm to get ∆f = 0.05 cm. We encourage you to get the answer by differentiating f = uv/(u − v) also. Ans. B Q 15. A convex lens is in contact with concave lens. The magnitude of the ratio of their focal length is 3/2. Their equivalent focal length is 30 cm. What are their individual focal lengths? (2005) (A) −75 cm, 50 cm (B) −10 cm, 15 cm (C) 75 cm, 50 cm (D) −15 cm, 10 cm

25 cm

15 cm 33.25 cm

I• •

O

(A) 10 cm (B) 15 cm (C) 20 cm (D) 25 cm Sol. The ray emanating from the object O undergoes (i) refraction at the water-air interface, (ii) reflection at the mirror, and (iii) refraction at the air-water interface. The ray diagram is shown in the figure. The point O is the object for the first refraction and I1 is the corresponding image. For this refraction, substitute µ2 = 1, µ1 = 1.33, u = −33.25 cm, and R = ∞ in the general equation for refraction at the spherical surface, µ2 /v − µ1 /u = (µ2 − µ1 )/R,

(1)

to get v = −33.25/1.33 = −25 cm. The image I1 is formed at the same place as the final image I (see figure). The point I1 will act as an object for reflection at mirror with object distance u = −15 − 25 = −40 cm.

• I2

25 cm

Q 14. The graph between object distance u and the image distance v for a lens is as shown. The focal length of the lens is (2006)

µ = 1.33

15 cm

The sun is an extended circular object. The rays at bottom of the image come from the top of the sun and rays at the top of the image come from the bottom of the sun (see figure). The angle subtended by the sun on the lens (2α) is equal to the angle subtended by its image on the lens. If α is small then r = αf . Thus, πr2 ∝ f 2 . Ans. B

Q 16. A container is filled with water (µ = 1.33) upto a height of 33.25 cm. A concave mirror is placed 15 cm above the water level and the image of an object placed at the bottom is formed 25 cm below the water level. The focal length of the mirror is (2005)

33.25 cm

f

u (cm) -11

1/F = 1/f1 + 1/f2 . Substitute f1 = +f , f2 = −3f /2, and F = 30 cm to get f = 10 cm. Thus, f1 = 10 cm and f2 = −15 cm. Ans. D

r

α

Sol. Let the focal length of the convex lens be f1 = +f and the focal length of the concave lens be f2 = −3f /2. The equivalent focal length (F ) of the two thin lenses in contact is given by

• I,I1 •

O

For the second refraction, the image I is at distance v = −25 cm. Substitute µ2 = 1.33, µ1 = 1, v = −25, and R = ∞ in equation (1) to get the object distance u = −25/1.33 = −18.8 cm. This object is shown as point I2 in the figure. The point I2 is the image formed after reflection by mirror. Thus, the image distance for

Chapter 16. Geometrical Optics

239

reflection at mirror is v = −15 − 18.8 = −33.8 cm. The mirror formula,

Sol. The convex lens forms image (I1 ) at its focus.

1/f = 1/v + 1/u, with u = −40 cm and v = −33.8 cm gives f = −18.32 cm. We encourage you to use the concept of the apparent depth and the real depth to find the image and/or object distance for the two refractions. Ans. C Q 17. A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of the sphere is (2004) (A) 2 cm (B) 4 cm (C) 6 cm (D) 12 cm Sol. The ray is refracted at the glass-air interface.

I1

26 cm

This image acts as an object for the concave lens. The lens formula, 1/v − 1/u = 1/f, with u = 4 cm and f = −20 cm gives v = 5 cm for the image I2 . The magnification by concave lens is m = v/u = 1.25 and hence size of I2 is 1.25 × 2 = 2.5 cm. Ans. B

Substitute µ1 = 1.5, µ2 = 1, u = −6 cm, and R = −6 cm in the general equation for refraction at the spherical surface,

Air r

r

Glass

Q 18. A ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation which of the following is true? (2004)

R S

PQ is horizontal QR is horizontal RS is horizontal Either PQ or RS is horizontal

Sol. The ray inside the prism (QR) is parallel to the base (horizontal) of the prism for the minimum deviation. Ans. B Q 19. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image. (2003) (A) 1.25 cm (B) 2.5 cm (C) 1.05 cm (D) 2 cm

(A)

4 3

sin i (B)

1 sin i

i

(C) 4/3 (D) 1

Sol. Using Snell’s law for glass to water refraction, sin i/ sin r = µw /µg . For water to air refraction, sin r/ sin 90◦ = 1/µw (note that r is the critical angle). Solve to get µg = 1/ sin i. Ans. B Q 21. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle 30◦ at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is (2002) √ 2 3m B 0.2 m

to get v = −6 cm. Thus, the image is formed at the centre of the sphere. In the ray diagram, the rays emanating from the object hits the surface of the sphere along the normal and hence are transmitted un-deviated. Ans. C

P

µw = 34

Water

µ2 /v − µ1 /u = (µ2 − µ1 )/R,

(A) (B) (C) (D)

4 cm

Q 20. A ray of light is incident at the glass-water surface at an angle i. It emerges finally parallel to the surface of water, then the value of glass refractive index µg would be (2003)

O,I

Q

I2

30◦ A

(A) 28 (B) 30 (C) 32 (D) 34 Sol. The angle of incidence is equal to the angle of reflection. Thus, ray will strike the mirror B after the first reflection at mirror A at a distance √ x1 = 0.2 tan 30◦ = 0.2/ 3 m.

240

Part III. Optics

After the first reflection, the ray will undergo two reflections (first √ at mirror A and second at mirror B) at every 2x1 = 0.4/ 2√m. Thus, √ the total √ number of reflections in distance 2 3 are 2 3/(0.2/ 3) = 30. Ans. B

Obs. C i i D

Q 22. Which of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams, (2002) (B) (A) R1

R2

(C)

B

h

∞

R

O

E 2h

(D) R

3h

r A

R

∞

R

Sol. The dispersion occurs due to variation of the focal length f with the refractive index µ i.e., if df /dµ 6= 0. The focal length of a spherical lens of refractive index µ and radius of curvature R1 and R2 is given by the lens maker’s formula, 1/f = (µ − 1) [1/R1 − 1/R2 ] . Substitute the values of R1 and R2 to get the focal lengths for the four given cases as,

Using geometry, √ in triangle ABC, side AB = 2h, BC = 2h, AC = 2 2h, and √ (2) sin i = AB/AC = 1/ 2, ◦ which gives angle i as √ 45 . In triangle ODE, side OE = h, DE = 2h, OD = 5h and √ sin r = OE/OD = 1/ 5. (3)

The equations (1)–(3) give µ =

p

= (µ − 1) [1/R1 + 1/R2 ] ,

Q 24. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will suffer (2001)

1/fB = (µ − 1) [1/R − 1/∞] = (µ − 1)/R, 1/fC = (µ − 1) [1/R − 1/R] = 0, 1/fD = (µ − 1) [1/(−R) − 1/∞] = −(µ − 1)/R. As fC is independent of µ, system C does not exhibit dispersion. Ans. C Q 23. An observer can see through a pin-hole at the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius is h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is (2002)

3h

h 2h 5 2

(B)

q

5 2

(C)

q

3 2

(D)

3 2

Sol. By Snell’s law of refraction, sin i/ sin r = µ.

Q P

(A) (B) (C) (D)

R

greater deviation. no deviation. same deviation as before. total internal reflection.

Sol. At the minimum deviation, the ray inside P is parallel to its base. This ray will remain parallel to the base as it goes from P to Q and then to R (because prisms are symmetrical and are made of same material). The refraction on the right face of R is similar to the refraction at the right face of P. Hence, ray will suffer same deviation as before. In other words, deviations from P and Q are equal and opposite and R produces same deviation as P. Ans. C

Obs.

(A)

5/2. Ans. B

1/fA = (µ − 1) [1/R1 − 1/(−R2 )]

(1)

Q 25. A ray of light passes through four transparent media with refractive index µ1 , µ2 , µ3 and µ4 as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have (2001)

Chapter 16. Geometrical Optics µ1

241 This condition is satisfied when angle i2 is equal to or greater than the critical angle i.e.,

µ3 µ4 D

µ2

C

B

i2,min = sin−1 (n2 /n1 ) .

A

(1)

Snell’s law for refraction at the face AB gives (A) µ1 = µ2 (B) µ2 = µ3 (C) µ3 = µ4 (D) µ4 = µ1

sin α/ sin r1 = n1 /n2 .

(2)

Sol. Snell’s law for refraction at the interface 1-2 gives µ1 sin i12 = µ2 sin r12 .

(1)

For refraction at the interface 2-3, angle of incidence i23 = r12 . Snell’s law for this refraction gives µ2 sin r12 = µ3 sin r23 .

(2)

Similarly for refraction at interface 3-4, µ3 sin r23 = µ4 sin r34 .

(3)

The equations (1)–(3) gives

Thus, the maximum value of α occurs when r1 is maximum. In right-angled triangle PQR, r1 = 90◦ − i2 , which gives r1,max = 90◦ − i2,min .

(3)

The equations (1)–(3) give   −1 n1 sin(r1,max ) αmax = sin n2    −1 n1 −1 n2 = sin cos sin . n2 n1

µ1 sin i12 = µ4 sin r34 .

Ans. A

Thus, the condition i12 = r34 is possible only when µ1 = µ4 . Note that after passing through the multiple parallel slabs the ray emerges parallel to the incident ray if the initial and the final mediums have same refractive index. Ans. D

Q 27. A point source of light S, placed at a distance L in front of the centre of a plane mirror of width d, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. The greatest distance over which he can see image of the light source in the mirror is (2000)

Q 26. A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n2 (n1 > n2 ). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence αmax , such that the ray comes out only from the other surface CD, is given by (2000) A ax

n2

m

α

B

L 2L

(A) d/2 (B) d (C) 2d (D) 3d

D n1

S

d

Sol. The man can see the source between the points B and D.

C D

(A) sin−1

h



i

n1 cos sin−1 nn21 h n2  i (B) sin−1 n1 cos sin−1 n12   (C) sin−1 nn21   (D) sin−1 nn21

C

θ A

R

D

i2 r2 r1

α

P B

Q

θ L

Sol. The ray comes out from the face CD only if it does not escape through the face AD. Thus, the ray should undergo total internal reflection at the face AD. A

•S

d

n1 n2 C

E

G

F 2L

B

At the point B, he receives rays reflected by the lowest point A of the mirror. At the point D, he receives rays reflected by the top-most point C of the mirror. The angle of incidence is equal to the angle of reflection. Using geometrical similarity in triangle ASE and AGB, length BG = 2EF = 2ES = 2 × d/2 = d. By symmetry, length BD = 2BG + AC = 3d. Ans. D

242

Part III. Optics

Q 28. A diverging beam of light from a source S having divergence angle α falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and its refractive index is n, then the divergence angle of the emergent beam is (2000) S α i

i n

Sol. The angle of refraction at the upper surface is equal to the angle of incidence at the lower surface. S α rr i

Ans. A

t

(A) zero (B) α (C) sin−1 (1/n) (D) 2 sin−1 (1/n)

i

Sol. The focal length of a concave lens of radius of curvature R and refractive index µ2 = 1.5 placed in a medium of refractive index µ1 = 1.75 is given by the lens maker’s formula    µ2 1 1 1 = −1 − f µ1 R1 R2    1.5 1 1 1 = −1 − = . 1.75 −R R 3.5R

i n

t

Q 31. A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to (1998) (A) 5R (B) 3R (C) 2R (D) 1.5R Sol. The ray diagram is shown in the figure.

i

µ1 = 1.0

Snell’s law can be used to show that the angle of emergence is equal to angle of incidence i.e., glass slab only introduces a lateral shift to the incident ray. Thus, divergence angle α does not change as shown. Ans. B Q 29. A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids L1 or L2 having refractive indices n1 and n2 , respectively (n2 > n1 > 1). The lens will diverge parallel beam of light if it is filled with (2000) (A) air and placed in air. (B) air and immersed in L1 . (C) L1 and immersed in L2 . (D) L2 and immersed in L1 . Sol. Let the double concave lens of radius of curvature R and refractive index n0 is placed in a medium of refractive index n. The lens maker’s formula gives its focal length f as  0    0   1 n 1 1 n −n −2 = −1 − = . f n R1 R2 n R Thus, the focal length is negative if n0 > n. Hence, the double concave lens behaves as a diverging lens if the refractive index of the lens material is greater than the refractive index of the medium in which it is placed. Ans. D Q 30. A concave lens of glass, refractive index 1.5, has both surfaces of the same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a (1999) (A) convergent lens of focal length 3.5R (B) convergent lens of focal length 3.0R (C) divergent lens of focal length 3.5R (D) divergent lens of focal length 3.0R

•

P

µ2 = 1.5 •

Q

O x

x

The object distance u = −|PO| = −x and the image distance v = |OQ| = x are related by the equation for refraction at the spherical surface, µ2 /v − µ1 /u = (µ2 − µ1 )/R. Substitute the values to get x = 5R. Ans. A Q 32. A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let O be the pole of the mirror and C its center of curvature. A point object is placed at C. It has a real image, also located at C. If the mirror is now filled with water, the image will be (1998) (A) real and will remain at C. (B) real and located at a point between C and ∞. (C) virtual and located at a point between C and O. (D) real and located at a point between C and O. Sol. The ray emanating from C is normal to the concave mirror. C

S P

R

Q O

Chapter 16. Geometrical Optics If there is no water then ray CP is reflected by the mirror and reflected ray traces its path forming a real image at C. When the mirror is filled with water, refraction at P bends the ray towards the normal, reflection at Q is such that QR and QP make equal angle with the normal QC, refraction at R causes RS to bend away from the normal forming a real image at S. Ans. D Q 33. A real image of a distant object is formed by a plano-convex lens on its principal axis. Spherical aberration (1998) (A) is absent. (B) is smaller if the curved surface of the lens faces the object. (C) is smaller if the plane surface of the lens faces the object. (D) is the same whichever side of the lens faces the object. Sol. The spherical aberration is less if the total deviation of the rays is distributed over two surfaces of the lens. Consider a plano-convex lens forming image of a distant object. If the plane surface faces the incident rays, the spherical aberration is much larger because deviation of incident parallel rays by plane surface is zero and total deviation is by the single curved surface. On the other hand, if the curved surface faces parallel incident rays then total deviation is distributed over both the surfaces.

243 Sol. The images formed by convex mirror and concave lens are virtual and cannot be obtained on the screen. Thus, only a concave mirror or a convex lens can be used. Consider the case when we place a concave mirror.

x3 x2 x1

Let u = −x1 , v = −x2 , and f = −x3 , where x1 , x2 , and x3 are positive. The magnification is given by m = −v/u = −x2 /x1 . For a diminished image, |m| < 1, which gives x2 < x1 . The distance between the object and the image is d = x1 − x2 . The mirror formula, 1/v + 1/u = 1/f , gives x3 =

x1 x2 x1 (x1 − d) . = x1 + x2 2x1 − d

Thus, if we take x1 = 2d then we get a real image at x2 = d with magnification m = −1/2 by using a concave mirror of focal length f = −x3 = −2d/3. However, there is a practical problem in the use of a concave mirror. The screen obstructs the rays going to the mirror. Now, consider the case when we place a convex lens. x3

We encourage you to analyse the situation when object is near to the lens. Ans. B Q 34. An eye specialist prescribes spectacles having combination of convex lens of focal length 40 cm in contact with a concave lens of focal length 25 cm. The power of thin lens combination in diopters is (1997) (A) +1.5 (B) −1.5 (C) +6.67 (D) −6.67 Sol. The equivalent focal length f of combination of two thin lenses in contact is given by 1/f = 1/f1 + 1/f2 . Substitute f1 = 40 cm and f2 = −25 cm (concave lens) to get 1/f = −3/200 cm−1 = −3/2 m−1 . Thus, the power of the combination in diopter is 1/f = −1.5. Ans. B Q 35. A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by placing (1995) (A) a concave mirror of suitable focal length. (B) a convex mirror of suitable focal length. (C) a convex lens of focal length less than 0.25 m. (D) a concave lens of suitable focal length.

x1

x2

Let u = −x1 , v = x2 , and f = x3 , where x1 , x2 , and x3 are positive. The magnification is given by m = v/u = −x2 /x1 . For diminished image, |m| < 1, which gives x2 < x1 . The distance between the object and the image is d = x1 + x2 . The lens formula, 1/v − 1/u = 1/f , gives x3 =

x1 x2 x1 (d − x1 ) = . x1 + x2 d

The condition x2 < x1 and x1 + x2 = d gives x1 > d/2 which sets an upper limit on the focal length fmax = x3,max = d/4. We encourage you to prove analytically or graphically that x3,max = d/4 which occurs at x1 = d/2. Ans. C Q 36. An isosceles prism of angle 120◦ has a refractive index 1.44. Two parallel rays of monochromatic light enter the prism parallel to each other in air as shown. The rays emerging from the opposite face (1995)

244

Part III. Optics

120◦

rays

separation ∆ as shown in the figure. Taking the origin of coordinates, O, at the centre of the first lens, the x and y-coordinates of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by (1993) y

∆

(A) (B) (C) (D)

O

are parallel to each other. are diverging. make an angle 2[sin−1 (0.72)−30◦ ] with each other. make an angle 2 sin−1 (0.72) with each other.

Sol. The triangle ABC is isosceles with ∠B = 120◦ and ∠A = ∠C = 30◦ . From geometry, angle of incidence is i = 30◦ . Snell’s law gives angle of refraction as r = sin−1 (n sin i) = sin−1 (1.44 sin 30◦ )

d

(A) x = (B) x = (C) x = (D) x =

= sin−1 (0.72).

x

f1 f2 f1 +f2 , y = ∆ f1 (f2 +d) ∆2 f1 +f2 −d , y = f1 +f2 f1 f2 +d(f1 −d) ∆(f1 −d) f1 +f2 −d , y = f1 +f2 −d f1 f2 +d(f1 −d) f1 +f2 −d , y = 0

Sol. The lens L1 converges parallel rays at its focus to form an image I1 on the right side of lens L2 (since d < f1 ). This image acts as an object for lens L2 .

C

30◦ i

r

L2 L1

δ

B i

r

δ

y

I1

A

The ray diagram is shown in the figure. The emergent rays are converging with angle 2δ between them. The normal makes an angle of 30◦ with the incident ray which gives δ = r − 30◦ . Thus angle between the emergent rays is 2δ = 2[sin−1 (0.72) − 30◦ ]. Ans. C Q 37. Spherical aberration in a thin lens can be reduced by (1994) (A) using a monochromatic light. (B) using a doublet combination. (C) using a circular annular mask over the lens. (D) increasing the size of the lens. Sol. The spherical aberration occurs due to convergence of the paraxial and the marginal rays at different places on the axis of the lens. This can be reduced by either stopping the paraxial or the marginal rays by using circular annular mask (also called stop) over the lens. Ans. C Q 38. Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d (where d < f1 , d < f2 ) and their centres are displaced by a vertical

I2

∆ O x

30◦

d

f1−d

Using lens formula, 1/v − 1/u = 1/f , with u = f1 − d and f = f2 we get v=

f2 (f1 − d) . f2 + f1 − d

Thus, x coordinate of the image I2 is x=d+v =

f1 f2 + d(f1 − d) . f1 + f2 − d

The magnification by lens L2 is m=

f2 v = . u f2 + f1 − d

The image I1 is at a distance ∆ below the optic axis of L2 . Thus, image I2 is at a distance m∆ below the optic axis of L2 . Hence, y coordinate of the image I2 is y = ∆ − m∆ =

∆(f1 − d) . f1 + f2 − d Ans. C

Chapter 16. Geometrical Optics

245

Q 39. A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. The size of the image is approximately equal to, (1988) 1/2  1/2  f u−f (B) b u−f (A) b f  2 2  f (C) b u−f (D) b u−f f Sol. Let the centre of the object of size b be at a distance u from the pole of the mirror. u + b/2

1

r0

(A) sin−1 (tan r) (C) sin−1 (tan r0 )

(B) sin−1 (cot i) (D) tan−1 (sin i)

Sol. The laws of reflection gives r = i. Since angle between the reflected and the refracted ray is 90◦ , we get

u − b/2 2

i r

u

r0 = 90◦ − r = 90◦ − i,

(∵ r = i).

The object distances for the end 1 and the end 2 are n1

u1 = −(u − b/2),

u2 = −(u + b/2). n2

The focal length of the concave mirror is −f . Substitute in the mirror formula to get the distances for the images of the end 1 and the end 2, 1 1 1 = , + v1 −(u − b/2) −f 1 1 1 = . + v2 −(u + b/2) −f

f (u − b/2) f (u + b/2) − f − (u − b/2) f − (u + b/2)  2 bf 2 f = ≈b , (f − u)2 − b2 /4 f −u

n2 sin i sin i = = tan i. = n1 sin r0 sin(90◦ − i) The critical angle is given by ic = sin−1

v1 − v2 =

n2 = sin−1 (tan i) = sin−1 (tan r). n1 Ans. A

because b is very small. Aliter: Differentiate the mirror formula (1)

to get ∆v/v 2 + ∆u/u2 = 0.

r0

Apply Snell’s law for refraction from the medium of refractive index n1 to the medium of refractive index n2 to get

Note that b, u, and f are positive distances. The length of the image is the distance between the images of the end 1 and the end 2 and is given by

1/v + 1/u = 1/f,

i r

(2)

Here, the object length is b = |∆u| and the image length is l = |∆v|. Eliminate v from equations (1) and (2) to get l = bf 2 /(u − f )2 . Ans. D Q 40. A ray of light from denser medium strikes a rarer medium at an angle of incidence i (see figure). The reflected and refracted rays make an angle of 90◦ with each other. The angles of reflection and refraction are r and r0 . The critical angle is (1983)

Q 41. A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. The power of the combination is (1982) (A) −1.5D (B) −6.5D (C) +6.5D (D) +6.67D Sol. The convex length of focal length f1 = 40 cm is in contact with a concave lens of focal length f2 = −25 cm. The power of the combination is given by 1 1 1 1 3 1 = + = + =− cm−1 f f1 f2 40 −25 200 = −3/2 m−1 = −1.5 D.

P =

Ans. A Q 42. A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face AB is totally reflected to reach the face BC if (1981)

246

Part III. Optics B

A θ

∞

C

(A) sin θ > 8/9 (C) sin θ < 2/3

Sol. Consider the point F, the focus of the concave mirror. All rays emanating from F will be reflected back parallel to the principal axis PF. Thus, virtual image F0 of the point F is formed at ∞. N0 •

M0

(B) 2/3 < sin θ < 8/9 (D) None of these

Sol. The ray is incident normally on the face AB and enters into the prism undeviated. It strikes the face AC at Q. P

B

θ

A

M •

N •

x ∞ F0

•

L0

•

P fˆ

L

45◦ •

N1

F x

fˆ/2 fˆ

θ Q C

In right angled triangle APQ, ∠Q = 90◦ − θ. Thus, angle of incidence at Q is θ. The critical angle for the total internal reflection from the glass (refractive index n1 = 1.5) to water (refractive index n2 = 4/3) is,   8 −1 n2 −1 −1 4/3 ic = sin = sin . = sin n1 1.5 9 Thus, the ray undergoes total internal reflection at the face AC to reach the face BC if θ > ic = sin−1 (8/9) i.e., sin θ > 8/9. Ans. A One or More Option(s) Correct Q 43. A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f , as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire? (These figures are not to the scale.) (2018)

45◦ f /2

(A)

f

(B)

∞

α > 45◦ α

(C)

(D) α 0◦ < α < 45◦

∞

Consider the point L at an object distance u = −|f |/2. For convenience, we use symbol fˆ for |f | i.e, fˆ = |f |. Use mirror formula, 1/v + 1/u = 1/f , with f = −fˆ to get the image distance v = fˆ. Thus, the virtual image L0 of the point L is formed on the left side of the mirror at a distance fˆ. The image of a point on LF will lie on the line L0 F0 . Now, consider the point M. Its object distance is u = −fˆ/2. Thus, the virtual image M0 of the point M is also formed at a distance fˆ from the mirror. The distance of M0 above the principal axis is given by hM 0 = −(v/u)(LM) = −(−2)(fˆ/2) = fˆ (magnification by a spherical mirror). Now consider a point N on the line segment FM. Let N1 be the projection of N on the principal axis PF such that FN1 = x. The object distance is u = −(fˆ−x). The mirror formula gives the image distance v = fˆ(fˆ − x)/x. The distance of the image N0 above the principal axis is given by hN 0 = −(v/u)(N1 N) = (fˆ/x)(x) = fˆ (it is independent of x). Since N is an arbitrary point on FM, the image of all points on FM will be formed at a distance fˆ above the optic axis. There is a subtle issue in this problem. Can we use paraxial rays approximation (mirror formula, magnification formula etc.) for the given length LM = f /2? Ans. (D) Q 44. For an isosceles prism of angle A and refractive index µ, it is found that the angle of minimum deviation δm = A. Which of the following option(s) is(are) correct? (2017) (A) For the angle of incidence i1 = A, the ray inside the prism is parallel to the base of the prism. (B) At minimum deviation, the incident angle i1 and the refracting angle r1 at the first refracting surface are related by r1 = (i1 /2). (C) For this prism, the emergent ray at the second surface will be tangential to the surface when the

Chapter 16. Geometrical Optics

247

angle hof incidence at the first isurface is i1 = q −1 A 2 sin A 4 cos 2 − 1 − cos A . sin (D) For this prism, the refractive index µ and
the angle of prism A are related as A = 12 cos−1 µ2 . Sol. Apply Snell’s law for refraction at P and Q to get sin i1 = µ sin r1 , and µ sin r2 = sin i2 .

(1)

(2016)

(2)

(A) The refractive index of the lens is 2.5. (B) The radius of curvature of the convex surface is 45 cm. (C) The faint image is erect and real. (D) The focal length of the lens is 20 cm.

Since ∠AQV = ∠ARV = 90◦ , we get A = r1 + r2 ,

(3)

and geometry gives angle of deviation δ = (i1 + i2 ) − (r1 + r2 ).

(4)

A

A i1

δ

Q

r1 r2

R

Sol. For refraction by the plano-convex lens, object distance is u = −30 cm. The size of the image is double the size of the object i.e., magnification m = v/u = ±2, where positive sign is for erect image and negative sign is for inverted image. Thus, the image distance is v = 60 cm (inverted image) or v = −60 cm (erect image). Apply lens formula, f1 = v1 − u1 , to get f = 20 cm (inverted image) and f = 60 cm (erect image).

i2

V

P

Q 45. A plano-convex lens is made of material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is(are) true?

f = 20cm

S

µ B

F C

At minimum deviation, the angle of emergence is equal to the angle of incidence i.e., i2 = i1 . Substitute i2 = i1 in equations (1) and (2) to get r2 = r1 . Substitute r2 = r1 in equation (3) to get r1 = A/2. Also, it is given that angle of minimum deviation δm = A. Substitute δ = A, i2 = i1 and r2 = r1 = A/2 in equation (4) to get i1 = A. As ∠B = ∠C (because prism is isosceles) and ∠BQR = ∠CRQ (because r2 = r1 ), the line QR is parallel to the base BC. Substitute the values of i1 and r1 in equation (1) to get A = 2 cos−1 (µ/2)

i.e.,

µ = 2 cos(A/2).

(5)

The emergent ray is tangential to the prism surface, if i2 = 90◦ . Substitute in equation (2) to get the critical angle r2 = sin−1 (1/µ). Substitute r2 in equation (3) to get r1 = A − sin−1 (1/µ). Substitute the expressions for r1 and µ (equation (5)) in equation (1) and simplify to get    1 i1 = sin−1 µ sin A − sin−1 µ     1 = sin−1 µ sin A cos sin−1 − cos A µ " # r A −1 2 = sin sin A 4 cos − 1 − cos A . 2 Note that cos(sin−1 x) =

√

I

O

1 − x2 . Ans. (A), (B), (C)

f = 60cm

O

I

F

Let R be the radius of the convex surface of the plano-convex lens which is made of material of refractive index n. Apply lens-maker’s formula to get its focal length f as   1 1 1 n−1 = (n − 1) − . (1) = f R ∞ R

O

I F

C

The faint image is formed due to reflection from the convex surface, which acts as a convex mirror. The focal length of this mirror is fm = R/2, object distance is u = −30 cm and image distance is v = 10 cm (the image by convex mirror is erect, virtual, and is formed towards the right of the pole). Apply the mirror formula, f1m = 1 1 v + u , to get fm = 15 cm. Thus, the radius of curvature of the convex surface is R = 2fm = 30 cm. Substitute R = 30 cm in equation (1) to get n = 2.5 for f = 20 cm (inverted image) and n = 1.5 for f = 60 cm (erect image). Ans. A, D

248

Part III. Optics

Q 46. A transparent slab of thickness d has a refractive index n(z) that increases with z. Here z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices n1 and n2 (> n1 ), as shown in the figure. A ray of light is incident with angle θi from medium 1 and emerges in the medium 2 with refraction angle θf with a lateral displacement l. Which of the following statement(s) is(are) true? (2016) n1 = const

θi

1

n(z)

d

z l

n2 = const

2 θf

By Snell’s law, n1 sin i1 = n sin r1 and n sin i2 = n2 sin r2 . By geometry, i2 = r1 and the lateral displacement is given by l=d

sin r1 dn1 sin i1 . =q cos r1 n2 − n21 sin2 i1 Ans. A, C, D

Q 47. A parallel beam of light is incident from air at an angle α on the side PQ of√a right angled triangular prism of refractive index n = 2. Light undergoes total internal reflection in the prism at the face PR when α has a minimum value of 45◦ . The angle θ of the prism is (2016) P

(A) (B) (C) (D)

n1 sin θi = n2 sin θf n1 sin θi = (n2 − n1 ) sin θf l is independent of n2 l is dependent of n(z)

θ

Sol. We can approximate the given transparent slab (of varying refractive index) by a large number of parallel slabs of small and equal thicknesses but having different refractive indices. As thickness of each slab is small, we can assume refractive index within a slab as uniform.

α

n=

Q n1 = const

θi

1

√

2

R

(A) 15◦ (B) 22.5◦ (C) 30◦ (D) 45◦

d

z l

n2 = const

2 θf

Sol. The ray diagram when a ray strikes the face PR at the critical angle is shown in the figure. This condition occurs when angle of incidence at the face PQ is minimum i.e., α = 45◦ .

When a ray passes through a stack of parallel slabs then its angle of emergence θf depends on the angle of incidence θi and the refractive indices (n1 and n2 ) of the media surrounding the stack i.e.,

P

θ

sin θf = (n1 /n2 ) sin θi . We encourage you to prove this for a stack of two or three parallel slabs of different refractive indices. Snell’s law for the medium of varying refractive index can be written as n(z) sin θz = constant at all points on the ray path. The lateral displacement of a ray by a parallel slab depends on the angle of incidence θi , refractive index n1 , slab thickness d and the refractive index of the slab n (it is independent of n2 ). Consider the parallel slab shown in the figure. n1 n n2

i1

α A

Q

r ic n=

B √

2

R

Apply Snell’s law, sin α/ sin r = n, at interface PQ to get the angle of refraction,  
◦ r = sin−1 sinn α = sin−1 sin√45 = 30◦ . 2

1 r1

i2

d

l

2 r2

The critical angle for total internal reflection at the prism-air interface PR is given by  
ic = sin−1 n1 = sin−1 √12 = 45◦ .

Chapter 16. Geometrical Optics

249 n1

In triangle PAB, θ + (90◦ + r) + (90◦ − ic ) = 180◦ . Substitute the values of r and ic and solve to get θ = 15◦ . Ans. A Q 48. Two identical glass rods S1 and S2 (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surface at a distance d as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod S1 on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside S2 . The distance d is (2015) S1 P •

d

(A) 60 cm (B) 70 cm (C) 80 cm (D) 90 cm Sol. Apply the formula for the refraction at the spher1 ical surface of S1 , µv2 − µu1 = µ2 −µ R , to get 1.0 1.5 1 − 1.5 − = . v (−50) (−10) Solve to get the image distance v = 50 cm (point Q in the figure). 50cm 20cm S1 Q

•

50cm

(B) |f1 | = 2.8R (D) |f2 | = 1.4R

Sol. Consider the parallel rays traversing from air to glass. These rays undergo first refraction at the airfilm interface. The object and the image distances for refraction at a spherical surface are related by µ2 /v − µ1 /u = (µ2 − µ1 )/r.

(1)

Substitute µ2 = n1 = 1.4, µ1 = 1, u = −∞, and r = R in equation (1) to get the image distance

This image acts as an object for the second refraction at the film-glass interface. For this refraction, µ2 = n2 = 1.5, µ1 = n1 = 1.4, u = 3.5R, v = f1 , and r = R. Substitute these values in equation (1) to get f1 = 3R. Now, consider the parallel rays traversing from the glass to air. These rays undergo first refraction at the glass-film interface. For this refraction, µ2 = n1 = 1.4, µ1 = n2 = 1.5, u = −∞, and r = −R (note the sign convention). Substitute these values in equation (1) to get the image distance v = n1 R/(n2 − n1 ) = 14R. This image acts as an object for the second refraction at the film-air interface. For this refraction, µ2 = 1, µ1 = n1 = 1.4, u = 14R, r = −R, and v = f2 . Substitute values in equation (1) to get f2 = 2R (this focus is towards the left of the interface). Ans. A, C

S2 P

(A) |f1 | = 3R (C) |f2 | = 2R

v = n1 R/(n1 − 1) = 3.5R.

S2

50cm

n2

Air

•

d

The image Q acts as an object for refraction at the spherical surface of S2 . The object distance is u = −(d − 50). The image distance is v = ∞ (because rays are parallel in S2 ). Apply the formula for refraction at the spherical surface of S2 , to get

Q 50. A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60◦ (see figure). √ If the refractive index of the material of the prism is 3, which of the following is (are) correct? (2010) B

O

1 1.5 − 1 1.5 − = . ∞ −(d − 50) (10)

60◦

60◦ P

Solve to get d = 70 cm.

C 135◦

Ans. B Q 49. A transparent thin film of uniform thickness and refractive index n1 = 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index n2 = 1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f1 from the film, while rays of light traversing from glass to air get focused at distance f2 from the film. Then, (2014)

90◦ A

75◦ D

(A) The ray gets totally internally reflected at face CD. (B) The ray comes out through face AD. (C) The angle between the incident ray and the emergent ray is 90◦ . (D) The angle between the incident ray and the emergent ray is 120◦ .

250

Part III. Optics

Sol. Consider the refraction at the face AB. Snell’s law, √ sin i/ sin r = sin 60◦ / sin r = 3, gives r = 30◦ . B 60◦ 60◦ 30◦

P

135◦

C

45◦ 45◦ 30◦ 90◦ A

60◦

R

Q

75◦ D

The geometry in BCQP gives ∠BP Q = 30◦ + 90◦ = 120◦ , ∠CQP = 360◦ −(135◦ +60◦ +120◦ ) = 45◦ . Thus, the angle of incidence at Q is i1 = 45◦ . The critical angle is √ √ given by √ for prism to air refraction sin ic = 1/ 3. Since sin i1 = 1/ 2 > 1/ 3, we get i1 > ic i.e., the angle of incidence is greater than the critical angle. Thus, the ray undergoes total internal reflection at Q. The laws of reflection gives r1 = i1 = 45◦ . In triangle QRD, ∠QRD = 60◦ and hence the angle of incidence at R is i2 = 30◦ . Apply Snell’s law for refraction at R to get r2 = 60◦ . Ans. A, B, C Q 51. A student performed the experiment for determination of focal length of a concave mirror by using u-v method using an optical bench of length 1.5 m. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are: (42, 56), (48, 48), (60, 40), (66, 33) and (78, 39). The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is (are) (2009) (A) (42, 56) (B) (48, 48) (C) (66, 33) (D) (78, 39) Sol. Using mirror formula, 1/v + 1/u = 1/f , calculate v for given u of five measurements. The (u, v) pairs with calculated values of v are (42, 56), (48, 48), (60, 40), (66, 37.7) and (78, 34.7). The absolute error in v for the last two measurements is 37.7 − 33 = 4.7 cm and 39 − 34.7 = 4.3 cm, which is more than maximum error in v (0.2 cm). Ans. C, D Q 52. A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence 45◦ . The ray undergoes total internal reflection. If n is the refractive index of the medium with respect to air, select the possible value(s) of n from the following, (1998) (A) 1.3 (B) 1.4 (C) 1.5 (D) 1.6

Sol. The critical angle ic is related to the refractive index n by sin ic = 1/n. The ray undergoes total internal reflection if angle of incidence is greater than or equal to the critical angle. Thus, ic ≤ 45◦ and sin ic ≤ sin 45◦ ◦ (because sin x is an increasing function √ for 0 < x < 90 ). ◦ Hence, n = 1/sin ic ≥ 1/sin 45 = 2. Ans. C, D Q 53. Which of the following form(s) a virtual and erect image for all positions of the object? (1996) (A) Convex lens (B) Concave lens (C) Convex mirror (D) Concave mirror Sol. The concave lens and convex mirror form virtual and erect images for all positions of the object. We encourage you to verify this by their ray diagrams and/or by analysing lens and mirror formula. Ans. B, C Q 54. A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen, (1986) (A) half of the image will disappear. (B) complete image will be formed. (C) intensity of the image will increase. (D) intensity of the image will decrease. Sol. The image formed by the lens will be complete even if upper half of the lens is covered. The intensity of the image decreases because energy at each image point will be coming from lower part only. We encourage you to draw ray diagram and show that image is complete. Ans. B, D Paragraph Type Paragraph for Questions 55-56 Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2 . The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1 . The numerical aperture (NA) of the structure is defined as sin im . (2015) n1 > n2 n2

Cladding

Air i

θ

Core

n1

√ Q 55. For two structures namely S1 with n1 = 45/4 and n2 = 3/2, and S2 with n1 = 8/5 and n2 = 7/5 and taking refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is (are)

Chapter 16. Geometrical Optics

251

(A) NA of S1 immersed in water is the same as that of S√ 2 immersed in a liquid of refractive index 16/3 15. (B) NA √ of S1 immersed in liquid of refractive index 6/ 15 is the same as that of S2 immersed in water. (C) NA of S1 placed in air is the same as that √ of S2 immersed in liquid of refractive index 4/ 15. (D) NA of S1 placed in air is the same as that of S2 placed in water. Sol. Let the whole structure be placed in the medium of refractive index n0 . From geometry, if the angle of incidence at Q is θ then the angle of refraction at P is 90◦ − θ. n0 P i

Q θ

n2

In case (C), substitute n0 = 1 in√equation (4) to get NA1 = 3/4 and substitute n0 = 4/ 15 equation (5) to get NA2 = 3/4. In case (D), substitute n0 = 1 in equation (4) to get NA1 = 3/4 √ and substitute n0 = 4/3 equation (5) to get NA2 = 3 15/20. Ans. A, C Q 56. If two structures of same cross-sectional area, but different numerical apertures NA1 and NA2 (NA2 < NA1 ) are joined longitudinally, the numerical aperture of the combined structure is NA1 NA2 (B) NA1 + NA2 (C) NA1 (D) NA2 (A) NA 1 +NA2 Sol. It is not difficult to see that the angle of emergence from the given structure is equal to the angle of incidence to the structure.

n1 n0

90◦ −θ

P i

The ray will undergo total internal reflection at Q if the angle of incidence at Q is greater than or equal to the critical angle i.e., θ ≥ θc = sin−1 (n2 /n1 ).

(1)

Apply Snell’s law for refraction at P to get n0 sin i = n1 sin(90◦ − θ) = n1 cos θ p = n1 1 − sin2 θ.

(2)

From equation (2), the angle of incidence is maximum (i = im ) when θ is minimum i.e., when θ = θc (from equation (1)). Thus, the numerical aperture is given by p p n21 − n22 n1 1 − sin2 θc NA = sin im = = (3) n0 n0 √ Substitute n1 = 45/4 and n2 = 3/2 in equation (3) to get the numerical aperture for the structure S1 as p 45/16 − 9/4 3 NA1 = = . (4) n0 4n0 Similarly, substitute n1 = 8/4 and n2 = 7/5 in equation (3) to get the numerical aperture for the structure S2 as p √ 64/25 − 49/25 15 NA2 = = . (5) n0 5n0 In case (A), substitute n0 = 4/3 in equation √ (4) to get NA1 = 9/16 and substitute n0 = 16/3 15 in equation (5) to get NA2 = 9/16. √ In case (B), √ substitute n0 = 6/ 15 in equation (4) to get NA1 = 15/8 and √ substitute n0 = 4/3 in equation (5) to get NA2 = 3 15/20.

Q θ

n2 n1 i

90◦ −θ

Let im,1 and im,2 be the maximum values of the incidence angles for the rays to undergo total internal reflection in S1 and S2 , when they are isolated. The ray will undergo total internal reflection in both the structures connected in series if the angle of incidence is less than or equal to the minimum of im,1 and im,2 . Thus, the numerical aperture of the combined structure is governed by the structure having lower numerical aperture. Thus, desired numerical aperture is NA2 because NA2 < NA1 . We encourage you to analyse the cases (i) angle of incidence between im,1 and im,2 and (ii) angle of incidence greater than im,1 and im,2 . Ans. D Assertion Reasoning Type Q 57. Statement 1: The formula connecting u, v and f for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature. Statement 2: Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces. (2007)

(A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. (B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true. Sol. The mirror formula, 1/u + 1/v = 1/f , is valid for paraxial rays i.e., rays close to the principal axis. The paraxial ray condition is true for the mirrors having very small size in comparison to their radii of curvature. The restriction comes from the small angle approximations used in deriving the formula and has nothing to do with the validity of the laws of reflection.

252

Part III. Optics Ans. C

Matrix or Matching Type Q 58. Four combination of two thin lenses are given in Column I. The radius of curvature of all curved surfaces is r and refractive index of all the lenses is 1.5. Match lens combination in Column I with their focal length in Column II. (2014)

rectangular block at normal incidence. Depending upon the relationships between µ1 , µ2 and µ3 , it takes one of the four possible paths ‘ef’, ‘eg’, ‘eh’, or ‘ei’. Match the paths in Column I with conditions of refractive indices in Column II. (2013) f e

45◦

g h

Column I (P)

Column II

µ1 i

(1) 2r

µ2

(Q)

(2) r/2

(R)

(3) −r

(S)

(4) r

Sol. The lens maker’s formula gives focal length of the bi-convex lens as   1 1 1 = (µ − 1) − f1 R1 R2   1 1 1 = (1.5 − 1) − = . (1) r −r r The focal length of a plano-convex (also convex-plane) lens is given by   1 1 1 1 − (2) = (1.5 − 1) = , f2 ∞ −r 2r and that of a plano-concave (also concave-plano) lens is given by   1 1 1 1 =− . = (1.5 − 1) − (3) f3 ∞ r 2r Use equations (1)–(3) to get the effective focal length in case P, Q, R, and S as 1/fP = 1/f1 + 1/f1 = 2/f1 = 2/r, 1/fQ = 1/f2 + 1/f2 = 2/f2 = 1/r, 1/fR = 1/f3 + 1/f3 = 2/f3 = −1/r, 1/fS = 1/f1 + 1/f3 = 1/r + 1/(−2r) = 1/(2r). Ans. P7→2, Q7→4, R7→3, S7→1 Q 59. A right angled prism of refractive index µ1 is placed in a rectangular block of refractive index µ2 , which is surrounded by a medium of refractive index µ3 , as shown in the figure. A ray of light ‘e’ enters the

Column I (p) (q) (r) (s)

e→f e→g e→h e→i

(1) (2) (3) (4)

µ3

Column II √ µ1 > 2 µ2 µ2 > µ1 and µ2 > µ3 µ1 = µ2 √ µ2 < µ1 < 2 µ2 and µ2 > µ3

Sol. For the path ‘ef’, the ray bends towards the normal while going from the medium of refractive index µ1 to the medium of refractive index µ2 . This is possible when µ2 > µ1 . This ray bends away from the normal while going from the medium of refractive index µ2 to the medium of refractive index µ3 . This is possible when µ2 < µ3 . For the path ‘eg’, the ray does not bend while going from the medium of refractive index µ1 to the medium of refractive index µ2 and hence µ1 = µ2 . For path ‘eh’, the ray bends away from the normal while going from the medium of refractive index µ1 to the medium of refractive index µ2 which is possible when µ1 > µ2 . Also, this ray does not undergo total internal reflection while going from medium of refractive index µ1 to the medium of refractive index µ2 . Thus angle of incidence (45◦ ) is less than the critical√angle i.e., sin i < sin ic = µ2 /µ1 which gives µ1 < 2 µ2 . This ray bends away from the normal when going from the medium of refractive index µ2 to the medium of refractive index µ3 and hence µ2 < µ3 . For the path ‘ei’, the ray undergoes the total internal reflection while going from the medium of refractive index µ1 to the medium of refractive index µ2 . Hence the angle of incidence is greater than the critical angle √ i.e. sin i > sin ic = µ2 /µ1 which gives µ1 > 2 µ2 . Ans. p7→2, q7→3, r7→4, s7→1 Q 60. Two transparent media of refractive indices µ1 and µ3 have a solid lens shaped transparent material of refractive index µ2 between them as shown in figures in Column II. A ray traversing these media is also shown in the figures. In Column I different relationship between µ1 , µ2 and µ3 are given. Match them to the ray diagrams shown in Column II. (2010)

Chapter 16. Geometrical Optics Column I

253 Column II

(A) µ1 < µ2

(p) µ3

(B) µ1 > µ2

(q) µ3

µ2

µ2

µ1

µ1

F, image at infinity when S is placed at F, and the virtual and the magnified image when S is placed between F and O. The convex mirror produces the virtual and the diminished image. The lens formula, 1/v − 1/u = 1/f , gives image distance v and magnification m as v = uf /(u + f ),

(C) µ2 = µ3

(r) µ3

µ1 µ2

(D) µ2 > µ3

(s) µ3

µ2

(t) µ3

µ1

µ1 µ2

Sol. If µ1 < µ2 then the ray bends towards the normal after refraction at µ1 -µ2 interface. Incident rays parallel to the optic axis bend towards the optic axis by the convex lens and away from the optic axis by the concave lens. If µ1 > µ2 then the ray bends away from the normal. If µ2 = µ3 then the ray goes straight without bending at the µ2 -µ3 interface. If µ2 > µ3 then the ray bends away from the normal. We encourage you to use these facts to draw the ray diagrams for each case. Ans. A7→(p,r), B7→(q,s,t), C7→(p,r,t), D7→(q,s) Q 61. An optical component and an object S placed along its optic axis are given in Column I. The distance between the object and the component can be varied. The properties of the images are given in Column II. Match all the properties of images from Column II with the appropriate components given in Column I. (2008) Column I (A) (B)

S•

S•

m = v/u = f /(u + f ).

The image is real if v > 0 and virtual if v < 0. The image is erect if m > 0 and inverted if m < 0. The image is magnified if |m| > 1 and diminished in |m| < 1. If S is placed between ∞ and 2F (i.e., −∞ < u < −2f ) then the image is real (v > 0), inverted (m < 0), and diminished (|m| < 1). If S is placed between 2F and F (i.e., −2f < u < −f ) then the image is real, inverted, and magnified. If S is placed at F (u = −f ) then the image is formed at infinity (v = +∞). If S is placed between F and O (i.e., −f < u < 0) then the image is virtual, erect, and magnified. We encourage you to draw a ray-diagram for each case. Ans. A7→(p,q,r,s), B7→q, C7→(p,q,r,s), D7→(p,q,r,s) True False Type Q 62. A parallel beam of white light falls on a combination of a concave and a convex lens, both of the same material. Their focal lengths are 15 cm and 30 cm respectively for the mean wavelength in white light. On the same side of the lens system, one sees coloured patterns with violet colour at the outer edge. (1988) Sol. The effective focal length of two thin lenses of focal length f1 and f2 placed in contact with each other is given by 1/fe = 1/f1 + 1/f2 . Substitute f1 = −15 cm (concave) and f2 = 30 cm (convex) to get fe = −30 cm. Thus, the combination will act as a concave lens of focal length −30 cm. violet red

Column II (p) Real image (q) Virtual image fv

(C)

S•

(r) Magnified image

(D)

S•

(s) Image at infinity

Sol. Let O be the optic centre and F be the first focus. The concave mirror produces the real and diminished image if the object S is placed between ∞ and 2F, real and magnified image when S is placed between 2F and

fr

The refractive index of the material for the violet colour is more than that of the red colour. Thus, by lens maker’s formula, the focal length for the violet colour is less than that of the red colour (fv < fr ). The parallel rays, after passing through the concave lens, appear to be diverging from the focus. The ray diagram clearly shows violet colour pattern on the outer edge. Ans. T

254

Part III. Optics

Q 63. A convex lens of focal length 1 m and concave lens of focal length 0.25 m are kept 0.75 m apart. A parallel beam of light first passes through the convex lens, then through the concave lens and comes to a focus 0.5 m away from the concave lens. (1983)

Sol. The ray incident normally on the face AB goes undeviated inside the prism. This ray strikes the face BC at an angle of incidence i. B

Sol. The focal length of the convex lens is f1 = 1 m and that of the concave lens is f2 = −0.25 m. These lenses are separated by a distance d = 0.75 m. The parallel beam of light converges at the effective focus of the system. The effective focal length of the system is given by

Q

i

r

C

In triangle BPQ, ∠BQP = 180◦ − (90◦ + 30◦ ) = 60◦ .

Thus, fe = ∞ i.e., parallel beams remain parallel after passing through the system. We encourage you to show the result by a ray diagram. Ans. F Q 64. A beam of white light passing through a hollow prism gives no spectrum. (1983) Sol. The prism disperses white light because (i) refractive index of prism material is different for different colors (wavelength) (ii) angle of deviation is different for different colors (as it depends on the refractive index). The air inside the hollow prism has the same refractive index for all colors. Thus, angle of deviation is same for all the colors. The walls of the hollow prism are thin parallel slabs which introduce lateral shifts only. Ans. T Fill in the Blank Type Q 65. Two thin lenses, when in contact, produce a combination of power +10 D. When they are 0.25 m apart, the power reduces to +6 D. The focal length of the lenses are . . . . . . m and . . . . . . m. (1997) Sol. The effective focal length of two thin lenses of focal length f1 and f2 in contact is (1)

and that for lenses separated by a distance d is 1/F 0 = 1/f1 + 1/f2 − d/(f1 f2 ).

P

A

1 1 d 1 = + − fe f1 f2 f1 f2 1 1 0.75 = + − = 0. 1 −0.25 (1)(−0.25)

1/F = 1/f1 + 1/f2 ,

30◦

Thus, i = 90◦ −∠BQP = 30◦ . Substitute in Snell’s law, sin i √1 , to get r = 45◦ . The angle of deviation i.e., sin r = 2 the angle between the emergent ray and the incident ray is r − i = 15◦ . Ans. 15◦ Q 67. A ray of light undergoes deviation of 30◦ when √ incident on an equilateral prism of refractive index 2. The angle made by the ray inside the prism with the base of the prism is . . . . . . ◦ . (1992) Sol. The prism angle A, angle of minimum deviation δm , and the refractive index µ of the prism material are related by m sin A+δ 2 . sin A2 √ Substitute µ = 2 and A = 60◦ (equilateral prism) to get δm = 30◦ . Thus, the given ray undergoes minimum deviation. At minimum deviation, the ray inside the prism is parallel to its base. Ans. zero

µ=

Q 68. A slab of material of refractive index 2, shown in the figure, has a curved surface APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given in the figure. An object O is placed at a distance 15 cm from the pole P as shown. The distance of the final image of O from P, as viewed from left is . . . . . . (1991)

(2)

Substitute 1/F = 10 m−1 in equation (1), 1/F 0 = 6 m−1 and d = 0.25 m in equation (2) and then solve to get f1 = 1/8 m and f2 = 1/2 m. Ans. 0.125, 0.5 Q 66. A ray of light is incident normally on one of the faces of a prism of apex angle 30◦ and refractive index √ 2. The angle of deviation of the ray is . . . . . . ◦ . (1997)

n1 = 1

A C n2 = 2 O •

P

E

15 cm B 20 cm

D

n3= 43

Chapter 16. Geometrical Optics

255

Sol. The rays emanating from O will undergo refraction at the spherical surface APB. Using µ2 µ1 µ2 − µ1 − = , v u R with µ2 = 1, µ1 = 2, u = 15 cm and R = 10 cm, we get v = 30 cm. A

C

•

P

•

O

D 30 cm

The virtual image is formed at 30 cm towards the right of P. Note that the virtual image is not affected by the presence of water on the right side of CD. Ans. 30 cm to the right of P. Virtual Image. Q 69. A thin rod of length f /3 is placed along the optic axis of a concave mirror of focal length f such that its image, which is real and elongated, just touches the rod. The magnification is . . . . . . (1991) Sol. Let F be the focus and C be the centre of curvature of the concave mirror.

Q

C P f 3

(3)

Q 71. A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a distance d between them. If a parallel beam of light falling on A leaves B as parallel beam, then d is equal to . . . . . . cm. (1985) Sol. Focal length of the convex lens A is fA = 20 cm and that of the concave lens B is fB = −5 cm (note the sign convention). The effective focal length of the combination of two lenses separated by a distance d is (1)

A parallel beam of light falling on A leave B as a parallel beam if effective focal length is infinite i.e., F = ∞. Substitute values in equation (1) and solve to get d = 15 cm. Ans. 15

f 2f

The image by the concave mirror is real and elongated if the object is placed between F and C. The image of an object placed at C is formed at C. Thus, one end of the rod shall be at C as its image touches it. The image of the end P is formed at Q. Using mirror’s formula, v10 + u10 = f10 , with u0 = −(2f − f /3) = −5f /3 and f 0 = −f , we get v 0 = −5f /2. The magnification is m=

(2)

Divide equation (2) by (3) and solve to get the focal length of the lens when it is placed in water i.e., f = 60 cm. Ans. 60

1 1 1 d = + − . F fA fB fA fB

O

F

where R1 and R2 are the radii of the two surfaces. Apply equation (1) to the cases when lens of refractive index µ2 = 1.5 is placed in air (µ1 = 1) and placed in water (µ1 = 4/3), to get    1.5 1 1 1 = −1 − , 15 1 R1 R2    1 1.5 1 1 = −1 − . f 4/3 R1 R2

I

B 15 cm

Sol. The lens maker’s formula gives the focal length of a spherical lens of refractive index µ2 placed in a medium of refractive µ1 as    1 µ2 1 1 = −1 − , (1) f µ1 R1 R2

QC 5f /2 − 2f = = 1.5. CP f /3

Integer Type Q 72. Sunlight of intensity 1.3 kW/m2 is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kW/m2 , at a distance 22 cm from the lens on the other side is . . . . . . . (2018)

Ans. 1.5 Q 70. A thin lens of refractive index 1.5 has a focal length of 15 cm in air. When the lens is placed in a medium of refractive index 4/3, its focal length will become . . . . . . cm. (1987)

Sol. The sun rays are parallel to each other and get focussed by the convex lens at its focus F (at 20 cm away from the lens). Let r2 be the radius of the circle located at 22 cm from the lens such that this circle contains all the rays that pass through F. By geometry of similar triangles, r1 /r2 = 20/2 = 10.

256

Part III. Optics rk+1

k+1

nk+1

r1 F •

r2

rk

k

ik

nk nk−1

k−1 20cm

2cm

ik-1

The lens converges the energy in a circular area A1 = πr12 to a circular area A2 = πr22 . Let I1 and I2 be the light intensities at these circles. Equating the energy passing through these circles per unit time, I1 A1 = I2 A2 , we get πr2 I1 A1 I 1 = 12 = I2 = I2 πr2



r1 r2

1.6 sin 30◦ = (1.6 − 0.1m) sin 90◦ ,

2

2

I1 2

= (10) × 1.3 = 130 kW/m . You might have used a convex lens to make a hole in the paper by focusing sunlight on it. Can you now explain why paper should be placed close to the focus? It is the concentrated light intensity at the paper that burn it! Ans. 130 Q 73. A monochromatic light is travelling in a medium of refractive index n = 1.6. It enters a stack of glass layers from the bottom side at an angle θ = 30◦ . The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as nm = n − m∆n, where nm is the refractive index of the mth slab and ∆n = 0.1 (see the figure). The ray is refracted out parallel to the interface between the (m − 1)th and mth slabs from the right side of the stack. What is the value of m? (2017) m m−1

n − m∆n

which gives m = 8. We encourage you to find the speed of light in mth layer. It exceeds the speed of light in vacuum, which is not allowed by Einstein’s theory of relativity. The phrase “To err is human; to forgive, divine” is valid for IIT professors too :). Ans. 8 Q 74. A monochromatic beam of light is incident at 60◦ on one face of an equilateral prism of refractive index n and emerges from the opposite face making √ an angle θ(n) with the normal (see figure). For n = 3 the value of θ is 60◦ and dθ/dn = m. The value of m is . . . . . . . (2015)

60◦

θ

Sol. Let r1 be the angle of refraction at the face AP and r2 be the angle of incidence at the face AQ of the prism.

n − (m − 1)∆n

≈

For a ray entering first layer, angle of incidence i0 = θ = 30◦ and refractive index n0 = n = 1.6. For a ray entering mth layer, angle of refraction is rm = 90◦ and refractive index nm = n − m∆n = 1.6 − 0.1m. Substitute values in equation (1) to get

≈ A 3

n − 3∆n

2

n − 2∆n

1

n − ∆n

60◦ 60◦

r1 r2

n

θ

O

θ

n0 n

Sol. Let ik−1 be the angle of incidence and rk be the angle of refraction for a ray entering k th glass layer of refractive index nk (see figure). Since glass layers are parallel to each other, the angle of refraction at an interface is equal to the angle of incidence for refraction at the next interface i.e., ik = rk . Apply Snell’s law for refraction at successive layers to get n0 sin i0 = n1 sin r1 = · · · = nm sin rm .

Q

P

(1)

From triangle OPQ, r1 + r2 = 60◦ .

(∵ ∠POQ = 120◦ .)

(1)

Let n0 be the refractive index of the medium in which prism is placed. Apply Snell’s law for refraction at P and Q to get n0 sin 60◦ = n sin r1 ,

(2)

n sin r2 = n0 sin θ.

(3)

Chapter 16. Geometrical Optics

257

√ Given, θ = 60◦ for n = 3. Substitute these values in equations (1)–(3) and solve to get n0 = 1, r1 = 30◦ , and r2 = 30◦ . Now, eliminate r2 from equations (1) and (3) to get n sin(60◦ − r1 ) n0 n (sin 60◦ cos r1 − cos 60◦ sin r1 ) = n0 √ s n2 1 n 3 n2 − 2 sin2 r1 − = sin r1 2 2 n0 n0 2 n0 √ s √ 3 n2 3 3 = − − , (using (2)). 2 n20 4 4

Solve to get the image position v = 20 cm (right side of the lens) and magnification by the lens mlens = v/u = 20/(−20) = −1. The magnification by the set-up is given by m1 = mmirror mlens = (−2)(−1) = 2.

sin θ =

Differentiate to get √  −1/2 3 n n2 dθ 3 . cos θ = − dn 2 n20 n20 4 Substitute n = 2.

√

3, n0 = 1, and θ = 60◦ to get dθ/dn = Ans. 2

Q 75. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification m1 . When the set-up is kept in a medium of refractive index 7/6, the magnification becomes m2 . The magnitude |m2 /m1 | is . . . . . . . (2015)

15 cm 30 cm

20 cm

20 cm

50 cm

Now, consider the set-up placed in a medium of refractive index µ01 = 7/6. The focal length of the mirror does not change. Thus, the distance of the image formed by the mirror and its magnification does not change. The focal length of the lens changes. The refractive index of the lens material is µ2 = 1.5. Apply lens maker’s formula to get the new focal length of the lens    µ2 − µ01 1 1 1 = − f0 µ01 R1 R2   1 µ2 − µ01 µ1 µ2 − µ1 1 − = µ0 µ2 − µ1 µ1 R1 R2  1 0   µ2 − µ1 µ1 1 = µ2 − µ1 µ0 f   1  1.5 − 7/6 2 1 1 = = . (1) 1.5 − 1 7/6 10 35 From equation (1), focal length of the lens is f 0 = 35/2 cm. Apply lens formula, 1/v − 1/u = 1/f 0 , to get 1 1 1 − = . v (−20) 35/2

15 cm

50 cm

Sol. Consider the set-up placed in air of refractive index µ1 = 1. Apply the mirror formula, 1/v+1/u = 1/f , to get 1 1 1 + = . v 15 10 Solve to get the image position v = 30 cm (right side of the mirror) and the magnification by the mirror mmirror = −v/u = 30/15 = −2. This image acts as an object for the lens with object distance u = −(50 − 30) = 20 cm. Apply the lens formula, 1/v − 1/u = 1/f , to get 1 1 1 − = . v (−20) 10

Solve to get the image distance v = 140 cm and the magnification by the lens m0lens = v/u = 140/(−20) = −7. The magnification by the set-up is given by m2 = m0mirror m0lens = (−2)(−7) = 14. Thus, the ratio of the magnification in two cases is m2 /m1 = 14/2 = 7. Ans. 7 Q 76. Water (with refractive index 4/3) in a tank is 18 cm deep. Oil of refractive index 7/4 lies on water making a convex surface of radius of curvature R = 6 cm as shown in the figure. Consider oil to act as thin lens. An object S is placed 24 cm above water surface. The location of its image is at x cm above the bottom of tank. Then x is . . . . . . . (2011) S •

µ = 1.0 R = 6 cm µ = 7/4 µ = 4/3

258

Part III. Optics

Sol. The equation for refraction at the spherical interface is µ2 /v − µ1 /u = (µ2 − µ1 )/R.

•

•

O1

O2

•

(1) 25 m

24 cm

•S

18 cm

•

I2 I1

50 7 m

50 m

25 3 m

µ = 1.0 µ = 7/4

µ = 4/3

For air-oil interface, µ2 = 7/4, µ1 = 1.0, u = −24 cm, and R = +6 cm. Substitute in equation (1) to get the image distance v = +21 cm. This image acts as a source for refraction at oil-water interface. For this refraction, µ2 = 4/3, µ1 = 7/4, u = +21 cm, and R = ∞. Substitute values in equation (1) to get v = +16 cm. Thus, height of the image from bottom of water surface is x = 18 − 16 = 2 cm. Ans. 2 Q 77. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image m25 changes from m25 to m50 . The ratio m is . . . . . . . 50

Given, v1 = 25/3 m for image I1 and v2 = 50/7 m for image I2 . Substitute in mirror formula, 1/v + 1/u = 1/f , to get the object distance u1 = −50 m and u2 = −25 m (negative sign indicates that the object lies on left side of the mirror). Thus, the distance travelled by the object in 30 s is 25 m which gives a speed of 3 km/h. Ans. 3 Q 79. A large slab (µ = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R? (2010)

Sol. The ray will not emerge from the slab if the incident angle is more than the critical angle θc . The critical angle is given by sin θc = 1/µ = 3/5 which gives q tan θc = sin θc / 1 − sin2 θc = 3/4. R

R

(2010) θc

8 cm

Sol. The lens formula, 1/v − 1/u = 1/f , for the two locations of the object gives

θc •

1 1 1 − = , v1 −25 20 1 1 1 = . − v2 −50 20

S

(1) (2)

The equation (1) gives v1 = 100 cm and equation (2) gives v2 = 100/3 cm. Thus magnification in the first case is m25 = v1 /u1 = −4 and that in the second case is m50 = v2 /u2 = −2/3. Hence, m25 /m50 = 6. Ans. 6 Q 78. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is ob50 served to move from 25 3 m to 7 m in 30 seconds. What is the speed (in km/h) of the object? (2010) Sol. The focal length of the convex mirror of radius of curvature R is given by f = R/2 = 20/2 = 10 m. Let I1 and I2 be the image formed at t = 0 s and t = 30 s, respectively and corresponding object positions are O1 and O2 .

From the given figure, R = 8 tan θc = 6 cm. Ans. 6 Descriptive Q 80. A ray of √ light is incident on a prism ABC of refractive index 3 as shown in the figure. (2005) B

D

60◦

60◦ 60◦

A

C

E

(a) Find the angle of incidence for which the deviation of light ray by the prism ABC is minimum. (b) By what angle the second identical prism must be rotated, so that the final ray suffers net minimum deviation.

Chapter 16. Geometrical Optics

259

Sol. At minimum deviation, ray inside the prism is parallel to the base which gives r = A/2 = 30◦ . B,D

E

A i

r

A

r

Q 82. Figure shows √ an irregular block of material of refractive index 2. A ray of light strikes the face AB as shown in the figure. After refraction it is incident on a spherical surface CD of radius of curvature 0.4 m and enters a medium of refractive index 1.514 to meet PQ at E. Find the distance OE upto two places of decimal.

i

(2004) B

C

C

45◦

√

Snell’s law, sin i/ sin r = µ = 3, gives i = √ sin−1 ( 3/2) = 60◦ . The prism DCE should be rotated about C in the anticlockwise direction through 60◦ so that the final emergent ray is parallel to the incident ray and the angle of deviation is zero. Ans. (a) 60◦ (b) 60◦

P

µ=1

µ=

√

O E 2

Q

µ = 1.514

60◦ A

D

Sol. Snell’s law for refraction at the face AB gives Q 81. AB and CD are two slabs. The medium between the slabs has refractive index 2. Find the minimum angle of incidence at Q, so that the ray is totally reflected by the two slabs. (2005) A

Q

√ µ= 2

B

√ µ= 3

D

Sol. The critical angle for a ray travelling from medium of refractive index µ1 to a medium of refractive index µ2 (< µ1 ) is ic = sin−1 (µ2 /µ1 ). Thus, the critical angle at the upper slab is √ ic,upper = sin−1 ( 2/2) = 45◦ , and that at the lower slab is √ ic,lower = sin−1 ( 3/2) = 60◦ .

√ Substitute µ2 = 1.514, µ1 = 2, u = −∞, and R = 0.4 m to get v = OE = 6.06 m. Ans. 6.06 m Q 83. An object is approaching a thin convex lens of focal length 0.3 m with a speed of 0.01 m/s. Find the magnitudes of the rate of change of position and rate of change of lateral magnification, of image when the object is at a distance of 0.4 m from the lens. (2004) Sol. The lens formula, 1/v − 1/u = 1/f , gives v=

uf . u+f

(1)

Differentiate equation (1) w.r.t. time to get the rate of change of image position

Q 60◦

√ Substitute i = 45◦ , µ2 = 2, and µ1 = 1 to get r = 30◦ . The refracted ray is parallel to the optic axis and is again refracted by the spherical surface. The equation for refraction at the spherical surface is µ2 /v − µ1 /u = (µ2 − µ1 )/R.

µ=2 P C

sin i/sin r = µ2 /µ1 .

60◦

The angle of incidence at the upper slab shall be greater than or equal to 45◦ for a ray to be totally reflected by this slab. In case of reflection, angle of incidence is equal to the angle of reflection. By geometry, angle of reflection at the upper slab is equal to the angle of incidence at the lower slab. But, the angle of incidence at the lower slab shall be greater than or equal to 60◦ for the ray to be totally reflected by this slab. Thus, the minimum angle of incidence at Q for the ray to be totally reflected by both slabs is 60◦ . Ans. 60◦

dv f2 du = , dt (u + f )2 dt

(2)

f and differentiate magnification m = uv = u+f w.r.t. time to get the rate of change of lateral magnification

dm f du =− . dt (u + f )2 dt

(3)

Substitute u = −0.4 m and f = 0.3 m in equation (1) to get v = 1.2 m. Substitute du/dt = 0.01 m/s in equations (2) and (3) to get dv/dt = 0.09 m/s and dm/dt = −0.3 s−1 . Ans. 0.09 m/s, −0.3 s−1

260

Part III. Optics

Q 84. In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is R. Determine the focal length of this system for µ1 < µ2 < µ3 . (2003)

µ1

µ3

A

R

60◦

µ2 B

µ1 µ2 − µ1 µ2 − = . v1 −∞ +R

(1)

The image formed by this refraction acts as an object for the refraction at the right surface i.e., u = v1 for refraction at the right surface. Thus,

√

3 C

The ray undergoes multiple reflections in coated film of refractive index µ = 2.2 and thickness d. The path difference between any consecutive transmitted waves is 2µd. These waves interfere constructively (maximum intensity) when the path difference is an integral multiple of wavelength i.e., 2µd = nλ, where n = 1, 2, 3, . . .. The minimum value of d occurs at n = 1 and its value is dmin = λ/(2µ) = 6600/(2 × 2.2) = 1500 ˚ A. Ans. (a) zero (b) 1500 ˚ A

µ3 µ2 µ3 − µ2 . − = v2 v1 +R

(2)

3R Solve equations (1) and (2) to get v2 = µµ3 −µ (focal 1 length). It is interesting to note that the focal length is independent of µ2 . 3R Ans. µµ3 −µ 1

Q 85. A prism of refracting angle 30◦ is coated with a thin film of transparent material of refractive index 2.2 on face AC as shown in the figure. A light of wavelength 6600 ˚ A is incident on face AB such that angle of incidence is 60◦ . Find, (a) the angle of emergence and (b) the minimum value of thickness of the coated film on the face AC for which the light emerging from the face has maximum intensity.√[Given refractive index of the material of the prism is 3.] (2003) A µ = 2.2 30◦

60◦ µ= B

30◦ S

µ=

Sol. The rays parallel to the optic axis converge at the focus. The relation between u and v for refraction at a spherical surface is, µ2 /v − µ1 /u = (µ2 − µ1 )/R. For refraction of the parallel rays at the left surface,

µ = 2.2

30◦

√

3 C

Sol. Snell’s law, sin i/ sin r = √ µ, for refraction on the face AB gives sin r = sin 60◦ / 3 = 1/2 i.e., r = 30◦ . In triangle ARS, the angle at S is 90◦ . Thus, ray falls normally on face AC and emerges normally to AC i.e., angle of emergence is 0◦ .

Q 86. A thin convex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object is placed 15 cm above the lens on its principal axis, the object coincides with its own image. On repeating with another liquid, the object and the image again coincide at a distance of 25 cm from the lens. Calculate the refractive index of the liquid. (2001)

Sol. The image coincides with the object if ray retraces its path after reflection from the plane mirror. This condition is satisfied if rays incident on the mirror are along its normal i.e., rays are parallel to the optic axis of the lens. The rays become parallel to the optic axis when the object is placed at the focus. Let R be the radius of curvature of the bi-convex lens having refractive index µg = 3/2. The focal length of the lens is given by   1 1 1 = (µg − 1) − f1 R1 R2    3 1 1 1 = −1 − = . (1) 2 R −R R The water (µw = 4/3) between the bi-convex lens and the mirror acts as a plano-concave lens of focal length f2 given by    1 4 1 1 1 −1 − =− . (2) = f2 3 −R ∞ 3R

Chapter 16. Geometrical Optics

261

Use equations (1) and (2) to get the effective focal length of the combined system as 1 1 1 1 2 1 = + = − = . F f1 f2 R 3R 3R

= −(µ − 1)/R,

B0

1/F = 1/f1 +

1/f20

(4)

= (2 − µ)/R.

(5)

The image coincides with the object when the effective focal length is equal to the object distance i.e., F 0 = 25 cm. Substitute F 0 = 25 cm and R = 10 cm in equation (5) to get µ = 1.6. Ans. 1.6 Q 87. A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axis PQ and RS parallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens. If A0 B0 is the image after refraction from the lens and the reflection from the mirror, find the distance A0 B0 from the pole of the mirror and obtain its magnification. Also locate positions of A0 and B 0 with respect to the optic axis RS. (2000) A Q

P 0.6 cm R 30 cm

B

A0

and the effective focal length of the combined system becomes 0

B1

(3)

The image coincides with the object when the effective focal length is equal to the object distance i.e., F = 15 cm. Substitute F = 15 cm in equation (3) to get R = 2F/3 = 10 cm. In the second case, liquid of refractive index µ acts as a lens of focal length 1/f20

A

B S 20 cm

Sol. The rays from object AB undergo refraction by the lens and then reflection by the mirror. The lens formula, 1/v1 − 1/u1 = 1/f1 , with u1 = −20 cm and f1 = 15 cm gives v1 = 60 cm. The image by the lens, say A1 B1 , is formed at a distance 60 cm from lens towards its left (see figure). The magnification by the lens is m1 = v1 /u1 = −3, i.e., size of the inverted image A1 B1 is three times the size of object AB. The point B1 lies on the optic axis of the lens and point A1 at a distance |A1 B1 | = 3|AB| = 3 × 1.2 = 3.6 cm below the optic axis of the lens.

15 cm A1 30 cm 60 cm

20 cm

The image A1 B1 acts as a virtual object for the mirror. Using mirror formula, 1/v2 + 1/u2 = 1/f2 , with u2 = 30 cm and f2 = −15 cm, we get v2 = −15 cm. The image by the mirror, say A0 B0 , is located 15 cm right of the mirror. The magnification by the mirror is m2 = −v2 /u2 = 1/2, i.e., mirror does not invert the image A1 B1 but reduces its size by half. Thus, the final image is inverted and is of size |A0 B0 | = 21 |A1 B1 | = 1.8 cm with total magnification m1 m2 = −3/2. The point B1 lies on the optic axis of the lens i.e., at yB1 = 0.6 cm above the optic axis of the mirror. The image of B1 will lie at yB 0 = m2 yB1 = 21 × 0.6 = 0.3 cm i.e., 0.3 cm above the optic axis of the mirror. The point A1 lies 3.6 cm below the optic axis of the lens i.e., at yA1 = −3.6 + 0.6 = −3 cm from the optic axis of the mirror. The image of A1 will lie at yA0 = m2 yA1 = 1 2 × (−3) = −1.5 cm i.e., 1.5 cm below the optic axis of the mirror. Ans. 15 cm, −3/2 Q 88. The x-y plane is the boundary between two transparent√media. Medium 1 with z ≥ 0 has a refractive index 2 and medium 2 with z < 0 has refractive √ index √3. A ray √ of light in medium 1, given by vector ˆ is incident on the plane of ~ = 6 3 ˆı + 8 3 ˆ − 10 k, A separation. Find the unit vector in the direction of the refracted ray in medium 2. (1999) √ √ ˆ is going ~ = 6 3 ˆı +8 3 ˆ−10 k, Sol. The incident ray, A from the medium 1 to the medium 2. Normal to the plane separating the two media (x-y plane) and coming ˆ The angle of incidence, i, ~1 = k. out to medium 1 is N ~ is the angle between N1 and the reverse of incident ray ~ Use dot product of two vectors to get i.e., −A. !   ~ ·N ~1 (−A) 1 −1 −1 i = cos = cos = 60◦ . ~ ~ 2 |(−A)| |N1 | ~1 N ~ A

z

i

n1 r

~2 N

~ B

n2

262

Part III. Optics Apply Snell’s law, n1 sin i = n2 sin r, to get √ √ !   n1 2 3 −1 −1 √ r = sin = 45◦ . sin i = sin n2 3 2

~ = xˆı + yˆ Let B  + z kˆ be the refracted ray. The angle ~ and the normal of refraction is the angle between B ˆ ~ N2 = −k. The dot product gives cos 45◦ =

~ ·N ~2 1 B −z =√ . =p ~ |N ~ 2| 2 x2 + y 2 + z 2 |B|

This equation gives z 2 = x2 + y 2 and z ≤ 0. By Snell’s law, incident ray, refracted ray, and normal lie in the same plane. The cross product of two vectors in this plane gives √ to the plane i.e., nor√ normal ~×N ~ 1 = 8 3ˆı − 6 3ˆ  (coming out of the mal ~n = A ~ lies in this plane, the angle between B ~ paper). Since B and ~n is 90◦ i.e., √ √ ~ · ~n = 8 3x − 6 3y = 0, B

For refraction at the plane surface, substitute u = −mR and radius as ∞ in equation (1) to get v = −1.5mR. The image I is formed at a distance 1.5mR towards left of the plane surface and at a distance 1.5mR + R towards left of the curved surface (rays are paraxial i.e., very close to the optic axis). This image acts as a virtual object for refraction at the curved surface. Since emergent ray is parallel to the optic axis, the final image is formed at infinity. Substitute u = −(1.5m + 1)R and v = ∞ in equation (1) to get m = 4/3. Ans. 4/3 Q 90. A prism of refractive index n1 and another prism of refractive index n2 are stuck together with gap as shown in the figure. The angles of the prism are as shown. The n1 and n2 depends on λ, the wavelength 4 and n2 = of light, according to: n1 = 1.20 + 10.8×10 λ2 4 , where λ is in nm. (1998) 1.45 + 1.80×10 λ2 D C

4 2 3 x. Thus, z = 5 − 3 x (negative

2

2

2

16 2 9 x

which gives y = x +y = x + = 25 2 x gives z = root is taken because 9 ˆ ~ = x (3ˆı + 4ˆ  − 5k) z ≤ 0). Substitute y and z to get B 3 ~ and a unit vector in direction of B as ~ 1 ˆ ˆ= B = √ B (3ˆı + 4ˆ  − 5k). ~ 5 2 |B| 1 √

Ans.

5 2



3ˆı + 4ˆ  − 5kˆ



Q 89. A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a distance mR from it. Find the value of m for which a ray from P will emerge parallel to the table as shown in the figure. (1999)

70◦ n2 20◦ n1

60◦

40◦

A

B

(a) Calculate the wavelength λ0 for which rays incident at any angle on the interface BC pass through without bending at that interface. (b) For light of wavelength λ0 , find the angle of incidence i on the face AC such that the deviation produced by the combination of prisms is minimum. Sol. The ray incident at any angle on the surface BC passes undeviated only when n1 = n2 i.e., 1.20 +

10.8 × 104 1.80 × 104 = 1.45 + . 2 λ0 λ20 O

P

60◦ D

mR

R

C

70◦ n2

Sol. The relation between u and v for refraction at a spherical surface is given by

20◦ n1 40◦ B

(1)

P mR 1.5mR

r1 60◦

A

µ2 µ1 µ2 − µ1 − = . v u R

I

i1

R

Solve to get λ0 = 600 nm. At λ0 , the given system is a part of the equilateral prism ABO of refractive index n1 = n2 = 1.5 (see figure). The deviation is minimum when r1 = ∠O/2 = 30◦ , where r1 is the angle of refraction at the face AC. Snell’s
law, sin i1 = n1 sin r1 = 1.5 × 12 = 43 , gives i1 = sin−1 34 .
Ans. (a) 600 nm (b) sin−1 43

Chapter 16. Geometrical Optics

263

Q 91. A thin equi-convex lens of glass of refractive index µ = 3/2 and of focal length 0.3 m in air is sealed into an opening at one end of a tank filled with water (µ = 4/3). On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in the figure. The separation between the lens and the mirror is 0.8 m. A small object is placed outside the tank in front of lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the object formed by the system. (1997) 0.9 m

0.8 m

acts as a virtual object for the mirror and its real image I3 is formed at a distance 0.4 m in front of the mirror. The image I3 acts as a real object for the refraction at water glass interface 4/3 3/2 − 4/3 3/2 − = , v −0.4 0.3 (see sign conventions carefully for a ray going from right to left), which gives v = −0.54 m i.e., image I4 is formed at a distance 0.54 m right of the lens. This image acts as an object for glass-air refraction 1 3/2 1 − 3/2 − = , v −0.54 −0.3 which gives v = −0.9 m i.e., final image I5 is at a distance of 0.9 m from the lens towards its right. Ans. 0.9 m right of the lens.

•

Q 92. A right angled prism (45◦ -90◦ -45◦ ) of refractive index n has a plane of refractive index n1 (n1 < n) cemented to its diagonal face. The assembly is in air. The ray is incident on AB. (1996)

Sol. The lens maker’s formula,   1 1 1 = (µ − 1) − , f R1 R2

A

n1

with f = 0.3 m, µ = 32 , R1 = R, and R2 = −R gives R = 0.3 m. The image and the object distances for refraction at the spherical surface are related by B

µ1 µ2 − µ1 µ2 − = . v u R 0.9 m

•

n

(a) Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle. (b) Assuming n = 1.352, calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated.

0.8 m

I4 ••

I3

C

•

I2 •

I1 •

I5

Sol. From geometry, ∠D = 180◦ − ∠A = 135◦ ,

(1)

◦

∠D = 180 − r1 − i2 . For refraction at the air-glass interface of the lens 3/2 1 3/2 − 1 − = , v −0.9 0.3

The equations (1) and (2) give r1 = 45◦ − i2 .

(3) A

which gives v = 2.7 m i.e., the image I1 is formed at a distance of 2.7 m towards right of the lens. The image I1 acts as a virtual object for refraction at the glasswater interface of the lens. For this refraction

45◦ i2

which gives v = 1.2 m i.e., the image I2 is formed at a distance of 1.2 m towards right of the lens. The image I2 lies on right of the mirror at a distance of 0.4 m. This

r2

r1 D

i1

4/3 3/2 4/3 − 3/2 − = , v 2.7 −0.3

(2)

45◦ B

C

The critical angle for refraction at the face AC is θc = sin−1 (n1 /n). If the ray strikes the face AC with

264

Part III. Optics 1.8 m

critical angle then i2 = θc . Apply Snell’s law for refraction at the face AB to get

d

I

sin i1 = n sin r1 = n sin(45◦ − i2 ) = n sin(45◦ − θc ) ◦

I0

(using (3).) (∵ i2 = θc .) −1

O


(n1 /n)

= n sin 45 − sin 
= n sin 45◦ cos sin−1 (n1 /n)
 − cos 45◦ sin sin−1 (n1 /n) q  1 =√ n2 − n21 − n1 , 2 p 
 which gives i1 = sin−1 √12 n2 − n21 − n1 . The refracted ray passes through the face AC undeviated if i2 = 0◦ (normal incidence) or n1 = n (but, n1 < n). Thus, for normal incidence at AC

d0

Similarly, for the lower half part of the lens u = −d0 , v = 1.8 − d0 , and 1 1.8 1 1 − = . = f 1.8 − d0 −d0 1.8d0 − d0 2

(2)

Eliminate f from equations (1) and (2) to get d0 = 1.8 − d,

(3)

0

(d = d is not allowed because the two halves are separated). The magnification by the upper half of the lens is v 1.8 − d m= = , (4) u −d and that by the lower half of the lens is

Substitute in Snell’s law to get i1 = sin−1 (n sin r1 ) = sin−1 (1.352 sin 45◦ ) = sin−1 (0.956) = 73◦ . Ans. (a) sin−1

h

√1 2

p

n2 − n21 − n1

i

m0 = (b) 73◦

Q 93. A thin plano-convex lens of focal length f is split into two halves. One of the halves is shifted along the optical axis. The separation between object and image planes is 1.8 m. The magnification of the image formed by one of the half lens is 2. Find the focal length of the lens and separation between the halves. Draw the ray diagram for image formation. (1996)

O

1.8 m

Sol. The focal length of a lens does not change when it is cut into two halves i.e., focal length of each half is f . Also, the object and the image positions are same for both the halves. Let upper half part of the lens is placed at a distance d from the object plane and the lower half part of the lens is placed at a distance d0 from the object plane. Thus, for the upper half part of the lens, u = −d and v = 1.8 − d. The lens formula gives 1 1 1 1 1 1.8 = − = − = . f v u 1.8 − d −d 1.8d − d2

1.8 − d0 . −d0

(5)

If |m| = 2 then equation (4) gives d = 0.6 m. Substitute in other equations to get d0 = 1.2 m, f = 0.4 m, m = −2, and m0 = −1/2. If |m0 | = 2 then equation (5) gives d0 = 0.6 m. Substitute in other equations to get d = 1.2 m, f = 0.4 m, m = −1/2, and m0 = −2. The ray diagram when d = 0.6 m is shown in the figure. Here, we have placed the object such that the rays passing through one half of the lens do not pass through another half of the lens. The situation is more complicated if rays pass through both the parts of the lens. Ans. 0.4 m, 0.6 m Q 94. A ray of light travelling in air is incident at grazing angle (incident angle = 90◦ ) on a long rectangular slab of a transparent medium of thickness t = 1.0 m. The point of incidence is the origin A(0, 0). The medium has a variable index of refraction n(y) given  1/2 by n(y) = ky 3/2 + 1 , where k = 1.0 m−3/2 . The refractive index of air is 1.0. (1995) y Air t = 1m

r1 = 45◦ − i2 = 45◦ − 0◦ = 45◦ .

P (x1 , y1 )

medium •

B(x, y) θ

(1)

A(0, 0)

x Air

Chapter 16. Geometrical Optics

265

(a) Obtain a relation between the slope of the trajectory of the ray at a point B(x, y) in the medium and the incident angle at that point. (b) Obtain an equation for the trajectory y(x) of the ray in the medium. (c) Determine the coordinates (x1 , y1 ) of the point P, where the ray intersects the upper surface of the slab-air boundary. (d) Indicate the path of the ray subsequently. Sol. The refractive index n(y) varies along the y axis. Hence, boundary separating the two mediums with different refractive indices is parallel to the x axis and its normal is parallel to the y axis. The incident angle i is as shown in the figure. y P (x1 , y1 ) • B(x, y)

t

i θ

x

Substitute y = 1 m in equation (4) to get coordinate of the point P as (4 m, 1 m). The angle of incidence at point P is given by i = sin−1 p

1 y 3/2

+1

= sin−1 √

1 13/2

+1

= 45◦ .

The refractive index of the medium at P is p p √ n1 = n(y) = y 3/2 + 1 = 13/2 + 1 = 2, and the refractive index of the air at P is n2 = 1. Using Snell’s law at P, n1 sin i = n2 sin r, we get, r = 90◦ i.e., the emergent ray is parallel to the incident ray. Ans. (a) slope = cot i (b) 4y 1/4 = x (c) (4 m, 1 m) (d) the ray will emerge parallel to the incident ray Q 95. An image Y is formed of a point object X by a lens whose optic axis AB is as shown in the figure. Draw a ray diagram to locate the lens and its focus. If the image Y of the object X is formed by a concave mirror (having the same optic axis as AB) instead of lens, draw another ray diagram to locate the mirror and its focus. Write down the steps of construction of the ray diagrams. (1994)

A(0, 0) X •

r i

B

A

B •

Y

From geometry, the angle between the x axis and the incident ray at the point (x, y) is θ = 90◦ − i, and the slope of the trajectory is dy = tan θ = tan(90◦ − i) = cot i. dx

(1)

Now, consider a point B(x, y) very close to A such that a ray from A to B can be approximated by a straight line and refractive index of the medium at A is approximately equal to that of B which is n(y). Thus, the angle of incidence at B is equal to the angle of refraction at A i.e., r = i. Now, for refraction at A, n1 = 1, i1 = 90◦ , n2 = n(y), and r2 = r = i. Snell’s law applied at A(0, 0) gives sin i =

1 1 =p . 3/2 n(y) y +1

(2)

Sol. The inverted image is formed by a convex lens. X •

Z F

A

O

B •

Y

The intersection of XY and AB is the optical centre O of the lens. A line passing through O and perpendicular to AB represents the lens. Draw XZ parallel to AB from the object to the lens. The intersection of ZY and AB is the focus F. For a concave mirror, draw YY0 perpendicular to AB such that Y and Y 0 are equidistant from AB. Y0 •

Use equation (2) in equation (1) to get

X •

dy = cot i = y 3/4 . dx

(3)

Integrate equation (3), Z y Z x y −3/4 dy = dx, 0

O

•

F

•

B

C

•

Y

0

to get the equation of the trajectory as 4y 1/4 = x.

A

(4)

Extend Y0 X such that it intersects AB at O, pole of the mirror. The intersection of XY and AB, say C, is the centre of curvature of the mirror and the mid-point of OC is focus as f = R/2, for a mirror. Ans. See solution

266

Part III. Optics

Q 96. In given figure, S is a monochromatic point source emitting light of wavelength λ = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 mm. The distance along the axis from S to L1 and L2 is 0.15 m while that from L1 and L2 to O is 1.30 m. The screen at O is normal to SO. (1993) Screen L1

three fringe widths i.e., OA = 3β = 3

λD (500 × 10−9 )(1) = 10−3 m. = 3× d 1.5 × 10−3

Reducing the gap between L1 and L2 reduces y and y1 thereby decreases d. Thus, β and OA increase. Ans. (a) 1 mm (b) increase Q 97. Light is incident at an angle α on one planar end of a transparent cylindrical rod of refractive index n. Determine the least value of n so that the light entering the rod does not emerge from the curved surface of the rod irrespective of the value of α. (1992)

A S

0.5 mm

O α

L2 0.15 m

1.30 m

(a) If the third intensity maximum occurs at the point A on the screen, find the distance OA. (b) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OA increase, decrease, or remain the same. Sol. The images formed by L1 and L2 act as the light sources to produce the interference pattern on the screen. Let us find the position of the image formed by L1 . Using lens formula,

Sol. The ray will not emerge out of the curved surface if the angle of incidence at B is greater than or equal to the critical angle i.e., i0 ≥ θc = sin−1 (1/n) . B i0 α

A

r

n

By geometry, r = 90◦ − i0 ,

1/v − 1/u = 1/f,

(1)

(2)

and by Snell’s law, with u = −0.15 m and f = 0.1 m we get v = 0.3 m. The magnification by the lens is given by m = v/u = −2. The object S is placed at a distance y = 0.25 mm below the optic axis of L1 . Thus image I1 is formed at a distance y1 = |m|y = 0.5 mm above (because m is negative) the optic axis of L1 . Thus, the distance of I1 from the line SO is y + y1 = 0.25 + 0.5 = 0.75 mm. By symmetry, image formed by L2 is at a distance 0.75 mm below the line SO. Thus the distance between the two images is d = 0.75 + 0.75 = 1.5 mm and the distance of the screen from the images is D = 1.3 − v = 1.3 − 0.3 = 1 m. L1

A I1 y1

y

O

S

L2 0.15 m

I2 1.3 m

The waves from I1 and I2 interfere to give third maxima at A (see figure). The distance OA contains

sin α = n sin r.

(3)

Eliminate r from equations (2) and (3) to get sin α = n cos(i0 ). Substitute i0 in inequality (1) to get √ sin α 1 n2 − 1 i0 = cos−1 ≥ sin−1 = cos−1 , (4) n n n The inequality (4) gives p n ≥ 1 + sin2 α.

(5)

Note that cos−1 θ1 ≥ cos−1 θ2 =⇒ θ1 ≤ θ2 for angles lying in the first quadrant. Thus, by inequality (5), the ray will not√ emerge out√for any incident angles, 0 < α < 90◦ , if n ≥ 1 + 12 = 2. √ Ans. 2 Q 98. A parallel beam of light travelling in water (refractive index = 4/3) is refracted by a spherical air bubble of radius 2 mm situated in water. Assuming the light rays to be paraxial, (1988) (a) Find the position of the image due to refraction at the first surface and the position of the final image. (b) Draw a ray diagram showing the positions of both the images.

Chapter 16. Geometrical Optics

267

Sol. Let O be the centre of the air bubble of radius R = 2 mm. The ray first undergoes refraction at P and then refraction at Q. For refraction at P, the ray bends away from the normal as it travels from a dense medium (µ1 = 4/3) to a rare medium (µ2 = 1). The object is at infinity (u = −∞) and the image distance is given by the formula for refraction at the spherical surface µ1 µ2 − µ1 µ2 − = , i.e., v u R

Q P I2 P 0

A ir R

O

S

By geometry, the angle of incidence at R is ir = ∠A and that at S is is = ∠B. Note that the angle of incidence is equal to the angle of reflection. The condition ∠A ≥ ∠B gives ir ≥ is . Thus, if condition for the total internal reflection is satisfied at S (i.e., is ≥ sin−1 n1 ) then it is also satisfied at R. Hence, the desired condition is

Q0

B ≥ sin−1 (1/n),

5 mm

(∵ is = B);

1/n ≤ sin B, (∵ sin x is increasing for x ∈ [0◦ , 90◦ ]);

2 mm 2 mm

For refraction at Q, the ray bends towards normal as it travels from a rare medium (µ1 = 1) to a dense medium (µ2 = 4/3). The image I1 acts as a virtual object with object distance (u ≈ −(6 + 4) = −10 mm). The image distance is given by µ2 µ1 µ2 − µ1 4/3 1 4/3 − 1 − = , i.e., − = , v u R v −10 −2 which gives v = −5 mm. Note that OQ ≈ OQ0 for paraxial rays. Thus, image I2 is formed at a distance of 1 mm towards the left of the sphere. The ray diagram and position of two images are shown in the figure. Note that the rays are paraxial and the point P is about 2 mm left of O and point Q is about 2 mm right of O. Ans. (a) −6 mm, −5 mm Q 99. A right angled prism is to be made by selecting a proper material and the angles A and B (B ≤ A), as shown in the figure. It is desired that a ray of light incident on the face AB emerges parallel to the incident direction after two internal reflections. (1987) A

is C

i.e., 6 mm

B

1 4/3 1 − 4/3 − = , v −∞ 2

which gives v = −6 mm. Thus, image I1 is formed at a distance of 6 mm towards left of the sphere (see figure). Note that OP ≈ OP0 because rays are paraxial.

I1

Sol. Let the incident ray undergoes the first total internal reflection at the point R and the second total internal reflection at the point S.

B

i.e.,

n ≥ 1/ sin B,

i.e.,

nmin = 1/ sin Bmax , √ nmin = 2,

i.e.,

(∵ Bmax = 45◦ as A + B = 90◦ ). If B = 30◦ then is = 30◦ and ir = 60◦ . The critical angle for n = 5/3 is ic = sin−1 (3/5) = 37◦ . Thus, the total internal reflection will take place at R but not at S. √ Ans. (a) 2 (b) No Q 100. Monochromatic light is incident on a plane interface AB between two media of refractive indices n1 and n2 (n2 > n1 ) at an angle of incidence θ as shown in the figure. The angle θ is infinitesimally greater than the critical angle for the two media so that total internal reflection takes place. Now if a transparent slab DEFG of uniform thickness and of refractive index n3 is introduced on the interface (see figure), show that for any value of n3 all light will ultimately be reflected back again into medium II. Consider separately the cases (a) n3 < n1 and (b) n3 > n1 . (1986) medium I (n1 ) D

E medium III (n3 )

A G

F θ

C

(a) What should be the minimum refractive index n for this to be possible? (b) For n = 5/3, is it possible to achieve this with the angle B equal to 30◦ ?

B

medium II (n2 )

Sol. The critical angle for a ray travelling from the medium II of refractive index n2 to the medium I of refractive index n1 is given by θc(II,I) = sin−1 (n1 /n2 ).

(1)

268

Part III. Optics

Angle θ is infinitesimally greater than θc(II,I) .

Q 101. A plano-convex lens has a thickness of 4 cm. When placed on a horizontal table, with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face of the lens is found to be 25/8 cm. Find the focal length of the lens. [Assume thickness to be negligible while finding its focal length.] (1984)

medium I (n1 ) D

E medium III (n3 )

A G

F θ

B

medium II (n2 )

Now, medium III of refractive index n3 is introduced between the medium II and the medium I. Consider the case where n3 < n1 . The critical angle for a ray travelling from the medium II to the medium III is given by n3 n1 n3 = sin θc(II,III) = n2 n1 n2 n3 n1 = sin θc(II,I) , (∵ θc(II,I) = sin−1 ) n1 n2 < sin θc(II,I) , (∵ n3 < n1 ), < sin θ,

(∵ θ > θc(II,I) ).

n1

n3 A G n2

E r

r F

B

θ

Apply Snell’s law to get n2 sin r = sin θ n3 n1 n2 = sin θ n3 n1 sin θ n1 = sin θc(III,I) , (∵ θc(III,I) = sin−1 ) n1 /n2 n3 sin θ n1 = sin θc(III,I) , (∵ θc(II,I) = sin−1 ) sin θc(II,I) n2 > sin θc(III,I) ,

(∵ θ > θc(II,I) ).

3 cm I1

4 cm

O Case (1)

(2)

The equation (2) shows that the condition of total internal reflection is satisfied for a ray going from the medium II to the medium III. Consider the case where n3 > n1 . In this case, θc(II,III) > θc(II,I) which gives θ < θc(II,III) (since θ is infinitesimally greater than θc(II,I) ). Thus, the ray going from the medium II to the medium III is not totally reflected. The ray is refracted to the medium III. The angle of refraction (r) at the interface of medium II and III is equal to the angle of incidence at the interface of medium III and I (because slab DEFG is of uniform thickness). D

Sol. Let refractive index of the lens material be n and the radius of its curved surface be R. In case (1), the ray emanate from the object O and undergoes refraction at the plane surface. The image of O is formed at I1 (see figure).

(3)

The equation (3) shows that the condition of total internal reflection is satisfied for a ray going from the medium III to the medium I. Ans. See solution

I2

25 8

cm

4 cm

O Case (2)

The object distance is u = −4 cm and image distance is v = −3 cm (note that the distances are measured from the plane refracting surface). Apply formula for the refraction at the spherical surface (a plane surface is a spherical surface with radius ∞), µ2 µ1 µ2 − µ1 1 n 1−n − = , i.e., − = = 0, v u R −3 −4 ∞ which gives n = 4/3. Note that n is the ratio of the actual depth to the apparent depth. In case (2), the ray undergoes refraction at the spherical surface of radius R. The image of O is formed at I2 . The object distance is u = −4 cm and image distance is v = −25/8 cm. Apply formula for the refraction at the spherical surface to get 4/3 1 − 4/3 1 − = , −25/8 −4 −R which gives R = 25 cm. Now, use lens maker’s formula to get focal length of the plano-convex lens   1 1 1 = (n − 1) − f R1 R2    4 1 1 1 = −1 − = . 3 ∞ −25 75 Thus, the focal length of the given lens is f = 75 cm. Ans. 75 cm

Chapter 16. Geometrical Optics

269

Q 102. An object is placed 21 cm in front of a concave mirror of radius of curvature 20 cm. A glass slab of thickness 3 cm and refractive index 1.5 is placed close to the mirror in the space between the object and the mirror. Find the position of the final image formed. The distance of the nearer surface of the slab from the mirror is 1.0 cm. (1980)

Sol. The silvered side (convex side) of the concavoconvex lens will act like a concave mirror of radius of curvature R = 20 cm. C O,I

Sol. The ray emanating from the object O will undergo refraction at two faces of the glass slab of thickness t = 3 cm and refractive index n = 1.5.

O

O0

1 cm 3 cm 21 cm

The apparent shift of the object is given by 0

OO = t − t/n = 3 − 3/1.5 = 1 cm. The apparent object O0 acts as the virtual object for the concave mirror. The focal length of concave mirror is f = R/2 = −10 cm and object distance is u = −(21 − OO0 ) = −20 cm. Substitute values in the mirror formula, 1/v + 1/u = 1/f , 1 1 1 + = , v −20 −10 to get v = −20 cm. Note that the image coincides with the object because object was placed at the centre of curvature of the mirror. After reflection from the mirror, the ray again undergo refraction at the glass slab, with an apparent shift of 1 cm in opposite direction. Thus, the final image will occur at a distance of 21 cm from the concave mirror. Note that the final image coincides with the object and the ray retraces its path after reflection from the mirror. Ans. 21 cm Q 103. The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface. (1981)

The image I is formed at the same place as the object O if the rays from the object, after refraction at the concave surface, fall normally on the silvered surface and retrace their path. Thus, the image formed by refraction at the concave surface should lie at the centre of curvature (C ) of the silvered convex surface. The formula for refraction at the spherical surface is µ1 µ2 − µ1 µ2 − = . v u R

(1)

Substitute µ2 = 1.5, µ1 = 1, v = −20 cm, and R = −60 cm in equation (1) to get 1.5 1 1.5 − 1 − = , −20 u −60

i.e.,

u = −15 cm.

When the concave part is filled with water, the rays emanating from the object undergo refraction at the water surface, refraction at the concave surface, reflection at the silvered surface and then retrace their path. C O,I

The image formed after refraction at the concave surface must be at C as earlier. Substitute µ2 = 1.5, µ1 = 4/3, v = −20 cm, and R = −60 cm in equation (1) to get the object distance for this refraction 4/3 1.5 − 4/3 1.5 − = , i.e., u = −18.46 cm. −20 u −60 The image after refraction at the water surface will act as the object of refraction at the concave surface. Substitute µ2 = 4/3, µ1 = 1, v = −18.66 cm, and R = ∞ in equation (1) to get the object distance for the refraction at water surface i.e.,

(a) Where should a pin be placed on the optic axis such that its image is formed at the same place? (b) If the concave part is filled with water of refractive index 4/3, find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin.

1 4/3 − 1 4/3 − = , −18.46 u ∞

i.e.,

u = −13.84 cm.

Thus, the object and the image in this case are at u = −13.84 cm. Hence, the pin is to be moved downward by a distance 15 − 13.84 = 1.16 cm. Ans. (a) 15 cm (b) 1.16 cm (downwards)

270

Part III. Optics

Q 104. A rectangular block of glass is placed on a printed page lying on a horizontal surface. Find the minimum value of the refractive index of glass for which the letters on the page are not visible from any of the vertical faces of the block. (1980) Sol. Consider the ray emanating from the point P on the printed page and going to the vertical wall RT of the rectangular block. S

T

i

There is a thin air film between the printed page and the rectangular block of refractive index n. The ray from P undergoes refraction at Q. After refraction, the ray travels in the block before undergoing second refraction at R. If ray undergoes total internal refraction at R then an observer on the right of wall RT cannot see the point P. In right angled triangle QRS, r is the angle of refraction at Q and 90◦ − r is the angle of incidence at R. The condition for total internal refraction at R is n ≥ sec r.

i.e.,

(1)

From inequality (1), n is minimum when r is maximum. The angle of refraction at Q is maximum when angle of incidence is i = 90◦ (grazing incidence). Apply Snell’s law for refraction at Q to get sin r = 1/n,

sec r = √

i.e.,

(∵ µ = µ1 ).

The focal length f is negative if µ1 < µ2 . Ans. µ1 < µ2 Q 106. The radius of curvature of the convex face of a plano-convex lens is 12 cm and its refractive index µ = 1.5. (1979)

r

P

90◦ − r ≥ sin−1 (1/n),

= 2(µ1 − µ2 )/R,

R

90−r

Q

In figure (b), parallel rays are diverged by the lens. Thus, lens acts as a concave lens with a negative focal length. Apply lens makers formula to get   1 1 1 = (µ − µ2 ) − f R −R

n n2

−1

.

(2)

From equations (1)–(2), minimum value of the refrac√ tive index is n ≥ 2. √ Ans. 2 Q 105. What is the relation between the refractive indices µ1 and µ2 ? If the behaviour of the light rays is as shown in the figure. (1979)

(a) Find the focal length of the lens. (b) The plane face of the lens is now silvered. At what distance from the lens will parallel rays incident on the convex surface converge? (c) Sketch the ray diagram to locate the image, when a point object is placed on the axis 20 cm from the lens. (d) Calculate the image distance when the object is placed as in (c) above. Sol. Apply lens maker’s formula to get the focal length of the plano-convex lens of radius of curvature R = 12 cm and refractive index µ = 1.5,   1 1 1 = (µ − 1) − f R1 R2   1 1 1 − . = (1.5 − 1) = 12 ∞ 24 Thus, the focal length of the plano-convex lens is f = 24 cm.

I2

I3

I1

12 cm µ1

µ

µ1

µ2

µ

µ2

Sol. In figure (a), parallel rays are not deviated by the lens. This is possible if refractive index of the lens and its surroundings are equal i.e., µ = µ1 .

µ1

µ

(a)

µ1

µ2

µ

(b)

µ2

36 cm

36 cm

The plane surface of the lens is silvered. The parallel rays are first refracted at the convex surface, gets reflected from the silvered plane surface, and finally gets refracted at the convex surface. The object distance for the parallel rays is u = −∞. For the first refraction at the convex surface, apply formula for refraction at the spherical surface µ2 µ1 µ2 − µ1 − = , v u R 1.5 1 1.5 − 1 − = , v −∞ 12

Chapter 16. Geometrical Optics

271

to get v = 36 cm. Thus, image I1 after first refraction is formed at a distance 36 cm from the lens (towards right). The image I1 acts as an object for the plane mirror. The plane mirror forms image I2 of the object I1 at a distance 36 cm towards its left. The image I2 acts as an object for the second refraction from the convex surface with object distance u = −36 cm. Apply formula for refraction at the spherical surface

A

30◦

30◦

P

Q

r1

60◦

1.5 1 − 1.5 1 − = , v −36 12 B

to get v = −12 cm. Note that the values of µ1 and µ2 are reversed. Thus, the parallel rays converge at a distance 12 cm towards the left of the lens. So the system acts as a concave mirror of focal length 12 cm.

I

Substitute values in equation (1) to get i2 = 0◦ i.e., the ray strikes the second face at the normal incidence. In triangle APQ, ∠Q = 90◦ and ∠A = 30◦ , so ∠P = 60◦ . Hence, angle of refraction at P is r1 = 90◦ − ∠P = 30◦ . Apply Snell’s law at P to get refractive index of the prism µ=

O

C

√ sin 60◦ sin i1 = = 3. ◦ sin r1 sin 30 Ans. µ =

20 cm 30 cm

When the object is placed at a distance 20 cm from the lens, we can use the above procedure to get position of the final image. However, we can use the result of previous part i.e., system acts as a concave mirror of focal length 12 cm, to get the position of image. Apply mirror formula, 1 1 1 + = , v u f 1 1 1 + = , v −20 −12

√

3

Q 108. A pin is placed 10 cm in front of a convex lens of focal length 20 cm and made of a material of refractive index 1.5. The convex surface of the lens farther away from the pin is silvered. Determine the position of the final image. Is the image real or virtual? (1978) Sol. Apply the lens maker’s formula to the convex lens of focal length f = 20 cm and refractive index µ = 1.5,   1 1 1 = (1.5 − 1) − , 20 R −R to get the radius of curvature R = 20 cm. The pin is placed at a distance u = −10 cm from the lens. The convex surface away from the pin is silvered. The silvered surface will act as a concave mirror of focal length f = R/2 = 10 cm.

to get v = −30 cm. Ans. (a) 24 cm (b) 12 cm (d) v = −30 cm Q 107. A ray of light is incident at an angle of 60◦ on one face of a prism which has an angle of 30◦ . The ray emerging out of the prism makes an angle of 30◦ with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of the prism.

I1 , I2

O, I3

10 cm 20 cm

(1978)

Sol. Given, the prism angle A = 30◦ , the deviation angle δ = 30◦ , and the incidence angle i1 = 60◦ . These angles are related by δ = (i1 + i2 ) − A, where i2 is angle of incidence at the second face.

(1)

The rays from the object are first refracted at the convex surface, get reflected from the silvered surface, and finally get refracted at the first convex surface. For the first refraction at the convex surface, apply formula for refraction at the spherical surface, 1.5 1 1.5 − 1 − = , v −10 20

272

Part III. Optics

to get v = −20 cm. Thus, image I1 after the first refraction is formed at a distance 20 cm from the lens (towards left). The image I1 acts as an object for the concave mirror of focal length f = −10 cm. Apply mirror formula, 1 1 1 + = , v −20 −10 to get v = −20 cm. The image I2 of the object I1 coincides with the object because it is placed at the centre of curvature of the concave mirror. The image I2 acts as an object for the second refraction at the first convex surface with object distance u = −20 cm. Apply formula for refraction at the first spherical surface, 1.5 1 − 1.5 1 − = , v −20 20 to get v = −10 cm. Thus, a real image is formed at the location of the pin. Ans. 10 cm, real

Chapter 17 Optical Instruments

One Option Correct

One or More Option(s) Correct

Q 1. In a compound microscope, the intermediate image is (2000) (A) virtual, erect and magnified (B) real, erect and magnified (C) real, inverted and magnified (D) virtual, erect and reduced

Q 3. A planet is observed by an astronomical refracting telescope having an objective of focal length 16 m and an eyepiece of focal length 2 cm, (1992) (A) the distance between the objective and the eyepiece is 16.02 m. (B) the angular magnification of the planet is −800. (C) the image of the planet is inverted. (D) the objective is larger than the eyepiece.

Sol. We encourage you to draw a ray diagram to show real, inverted, and magnified nature of the intermediate image. Ans. C

Sol. In the normal adjustment of a telescope, image by the objective is formed at the focal plane of the eyepiece.

Q 2. The focal lengths of the objective and the eyepiece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eyepiece is 15.0 cm. The final image formed by the eyepiece is at infinity. The two lenses are thin. The distance (in cm) of the object and the image produced by the objective, measured from the objective lens, are respectively, (1995) (A) 2.4 and 12.0 (B) 2.4 and 15.0 (C) 2.0 and 12.0 (D) 2.0 and 3.0

fo

Thus, the length of the telescope is L = fo + fe = 16 + 0.02 = 16.02 m. An inverted final image by the eyepiece is formed at infinity. The magnification of the telescope is m = − ffoe = − 1600 = −800. In telescope, 2 the aperture size of objective is larger than that of the eyepiece to get more light from the distant stars. Ans. A, B, C, D

Sol. In the normal adjustment of a compound microscope, the image produced by the objective falls on the focal plane of the eyepiece so the final image is formed at infinity. Eyepiece

Objective

Q 4. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length fo of the objective and the focal length fe of the eyepiece are (1989) (A) fo = 45 cm, fe = −9 cm (B) fo = 50 cm, fe = 10 cm (C) fo = 7.2 cm, fe = 5 cm (D) fo = 30 cm, fe = 6 cm

∞

O u

v

fe

fe d

Thus, image formed by the objective is at a distance d − fe = 15.0 − 3.0 = 12.0 cm from the objective. Using lens formula, 1/v − 1/u = 1/fo , with fo = 2.0 cm and v = 12.0 cm, we get

Sol. In the normal adjustment of a telescope, image by the objective lens is formed at the focal plane of the eyepiece. Thus, length of the telescope is

vfo 2 × 12 u= = = −2.4 cm. fo − v 2 − 12 Ans. A

fo + fe = L = 36 cm. 273

(1)

274

Part III. Optics fo

fe

The magnitude of angular magnification is |m| = |−fo /fe | = 5 which allows two possibilities

where µ is refractive index of the medium between the object and the objective, θ is the angle subtended by a radius of the objective lens on the object and λ is wavelength used. The resolving power of electron microscope is much larger than that of the optical microscope because de-Broglie wavelength of the electron is much smaller than the wavelength of visible light. Ans. smaller

fo /fe = −5,

(2)

Descriptive

fo /fe = 5.

(3)

Q 7. A telescope has an objective of focal length 50 cm and an eyepiece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away from the objective. Calculate (a) the separation between the objective and the eyepiece, and (b) the magnification produced by the telescope. (1980)

The equations (1) and (2) give fo = 30 cm and fe = 6 cm. The equations (1) and (3) give fo = 45 cm and fe = −9 cm. We encourage you to draw the ray diagrams for both the cases and comment on the type of image (inverted/erect). Ans. A, D Matrix or Matching Type Q 5. Column I gives some properties of a simple telescope used to view distant objects having eyepiece and objective lens of focal lengths fe and fo . Match these properties with parameters given in Column II. (2006) Column I

Column II

(A) Intensity of light received by lens (B) Angular magnification

(p) radius of aperture (R) (q) dispersion of lens (r) focal length fo , fe (s) spherical aberration

(C) Length of telescope (D) Sharpness of image

Sol. The intensity of light falling on a lens is proportional to its aperture area πR2 . In normal adjustment, the angular magnification (m = − ffoe ) and length of the telescope (L = fo + fe ) depends on both fo and fe . The sharpness of image depends on the radius of aperture, dispersion of lens and the spherical aberration. We encourage you to find expressions for m and L when the telescope is adjusted such that the final image is formed at the near point of the eye. Ans. A7→p, B7→r, C7→r, D7→(p,q,s) Fill in the Blank Type Q 6. The resolving power of electron microscope is higher than that of an optical microscope because the wavelength of electron is . . . . . . than wavelength of the visible light. (1992) Sol. The resolving power of a microscope is given by R=

2µ sin θ , λ

Sol. The focal length of the objective is fo = 50 cm and the object distance is u = −200 cm. Substitute values in lens formula, v1 − u1 = f1 , 1 1 1 − = , v −200 50 to get the image distance v = 200/3 cm. The manification produced by the objective is mo = v/u = (200/3)/(−200) = −1/3. The image formed by the objective acts as an object for eyepiece. The focal length of the eyepiece is fe = 5 cm. The image by the eyepiece is formed at the distance of distinct vision i.e., v = −25 cm. Substitute values in lens formula, 1 1 1 − = , −25 u 5 to get the object distance u = −25/6 cm. The magnification produced by the eyepiece is me = v/u = (−25)/(−25/6) = 6. The length of the telescope is L = 200/3 + 25/6 = 70.73 cm. The magnification produced by the telescope is m = mo me = (−1/3)(6) = −2. We encourage you to show the image formed by the objective and the the eyepiece by a ray diagram. Ans. (a)70.73 cm (b) m = −2

Chapter 18 Dispersion and Spectra

One Option Correct

where A is a small positive constant. The wavelength increases in order VIBGYOR and hence refractive index decreases in this order. Hence, the critical angle, ic = sin−1 (1/µ), increases in order VIBGYOR. The critical angles for Yellow, Orange, and Red are greater than the critical angle for Green and hence these rays do not undergo total internal reflection. Ans. A

Q 1. Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60◦ ). In the position of minimum deviation, the angle of refraction will be (2008) (A) 30◦ for both the colours. (B) greater for the violet colour. (C) greater for the red colour. (D) equal but not 30◦ for both the colours.

Q 3. A thin prism P1 with angle 4◦ and made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of the prism P2 is (1990) (A) 5.33◦ (B) 4◦ (C) 3◦ (D) 2.6◦

Sol. At the angle of minimum deviation (δm ), angle of incidence is equal to the angle of emergence, the angle of refraction (r) is equal to half of the prism angle (A) and ray inside the prism is parallel to the prism base. Further, refractive index is given by µ=

m sin A+δ 2 , A sin 2

Sol. The deviation by a thin prism of refractive index µ and angle A is δ = (µ − 1)A.

(1)

and the angle of incidence by i=

A1

A + δm . 2

A2

(2)

From equations (1) and (2), δm and i depend on µ (colours). However, for the given prism, r = A/2 = 30◦ is independent of µ. We encourage you to find δm and i for red and violet colours if µred = 1.514 and µviolet = 1.523. Ans. A

In dispersion without deviation by two prisms, total deviation angle is zero i.e., (µ1 − 1)A1 + (µ2 − 1)A2 = 0. Substitute values and solve to get

Q 2. White light is incident on the interface of glass and air as shown in the figure. If green light is just totally internally reflected then the emerging ray in air contains (2004) Air Glass

A2 = −

µ1 − 1 1.54 − 1 A1 = − × 4 = −3◦ . µ2 − 1 1.72 − 1

Negative sign indicates that two prisms are placed with their angles inverted with respect to each other (see figure). Ans. C

Green

White

(A) (B) (C) (D)

Q 4. A beam of light consisting of red, green and blue colours is incident on a right-angled prism. The refractive indices of the material of the prism for the above red, green, and blue wavelengths are 1.39, 1.44 and 1.47, respectively. The prism will (1989)

yellow, orange, red violet, indigo, blue all colours all colours except green

Sol. The refractive index µ of a material increases with decrease in the wavelength λ of light and is given by Cauchy’s formula, µ = µ0 +

A , λ2

45◦

275

276

Part III. Optics and that for the flint glass is

(A) separate the red colour from the green and blue colours. (B) separate the blue colour from the red and green colours. (C) separate all the three colours from one another. (D) not separate even partially any colour from the other two colours.

Let refracting angle of crown and flint glass prisms be A and A0 , respectively. The deviation by crown glass prism is

Sol. The critical angle for the color of refractive index µc is given by

and that by flint glass prism is

−1

θc = sin

(1/µc ) .

µ0y = (µ0b + µ0r )/2 = (1.77 + 1.73)/2 = 1.75.

δ = (µy − 1)A = (1.5 − 1)6 = 3◦ ,

δ 0 = (µ0y − 1)A0 = (1.75 − 1)A0 = 0.75A0 . The total deviation is zero, i.e., δ + δ 0 = 3◦ + 0.75A0 = 0◦ ,

i

r 45◦

red

blue, green

which gives A0 = −4◦ . The negative sign indicates inverted orientation of the flint glass prism. Net dispersion of the system is given by δnet = (δb − δr ) + (δb0 − δr0 ) = [(µb − 1)A − (µr − 1)A]

This color undergoes total internal reflection if its angle of incidence is greater than the critical angle i.e., i > θc or sin i > sin θc = 1/µc (because sin x is increasing function for 0◦ ≤ x ≤ 90◦ ). The angle of incidence for all colors for glass to air refraction is i = 45◦ . Substitute the values to get ◦

sin 45 = ◦

sin 45 = sin 45◦ =

1 1.41 1 1.41 1 1.41

< > >

1 1.39 1 1.44 1 1.47

= 1/µred = 1/µgreen = 1/µblue .

Hence, red light is not totally reflected but green and blue colors are totally reflected (see figure). We encourage you to find the angle of reflection and refraction for these colors. Ans. A

+ [(µ0b − 1)A0 − (µ0r − 1)A0 ] = (µb − µr )A + (µ0b − µ0r )A0 = (1.51 − 1.49)6◦ + (1.77 − 1.73)(−4◦ ) = −0.04◦ . Ans. (a) 4◦ (b) −0.04◦ Q 6. Two parallel beams of light P and Q (separation d) containing radiations of wavelengths 4000 ˚ A and 5000 ˚ A (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in the figure. The refractive index of the prism as a function of wavelength is given by the relation, µ(λ) = 1.20 + λb2 , where λ is in ˚ A and b is a positive constant. The value of b is such that the condition for total internal reflection at the face AC is just satisfied for one wavelength and is not satisfied for the other.

Descriptive Q 5. The refractive indices of the crown glass for blue and red light are 1.51 and 1.49 respectively and those of the flint glass are 1.77 and 1.73 respectively. An isosceles prism of angle 6◦ is made of crown glass. A beam of white light is incident at a small angle on this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incident light. (2001) (a) Determine the angle of the flint glass prism. (b) Calculate the net dispersion of the combined system. Sol. The deviation by a prism having small refracting angle A is given by δ = (µ − 1)A. The refractive index for the crown glass is, µy = (µb + µr )/2 = (1.51 + 1.49)/2 = 1.50,

(1991) A P

sin θ = 0.8 θ

d Q B

C

(a) Find the value of b. (b) Find the deviation of the beams transmitted through the face AC. (c) A convergent lens is used to bring these transmitted beams into focus. If the intensities of the upper and the lower beams immediately after transmission from the face AC, are 4I and I respectively, find the resultant intensity at the focus.

Chapter 18. Dispersion and Spectra

277

Sol. The refractive index, µ(λ) = 1.20 + b/λ2 ,

(1)

is more for a small λ = 4000 ˚ A. Thus, the critical angle, θc = sin−1 (1/µ), is less for λ = 4000 ˚ A and this wavelength undergoes the total internal reflection at the face AC. The critical angle for 4000 ˚ A is θc = i1 . A

∆x = µ0 QS − PR = PS(µ0 sin i − sin r) = 0. sin θ = 0.8

Hence, two beams interfere constructively giving a maxima of intensity p p √ √ Imax = ( I1 + I2 )2 = ( 4I + I)2 = 9I.

θ i1

d

δ1 r2 i2

δ2

B

C

By geometry, i1 = i2 = θ. Thus, µ = 1/ sin θ = 1/0.8. Substitute λ = 4000 ˚ A and µ = 1/0.8 into equation (1) to get 2 b = 8 × 105 ˚ A .

The deviation angle for this wavelength is δ1 = 90◦ − i1 = 90◦ − θ = 90◦ − sin−1 0.8 = 37◦ . 2 ˚ and b = 8 × 105 ˚ Substitute λ = 5000 A A into equation (1) to get refractive index for 5000 ˚ A as

µ0 = 1.20 + 8 × 105 /(5000)2 = 1.232. Apply Snell’s law, sin i2 /sin r2 = 1/µ0 , to get r2 = sin−1 (µ0 sin i2 ) = sin−1 (µ0 sin θ) = sin−1 (1.232 × 0.8) = 80.26◦ . The deviation angle for this wavelength is δ2 = r2 − i2 = r2 − θ = 80.26 − sin−1 0.8 = 27.13◦ . A θ P r i d

R Q B

to travel of distance QS in the medium of refractive index µ0 and distance PR in air. Hence, optical path difference is ∆x = µ0 QS − PR. Now, in triangle PQS, ∠S = (90◦ − i) which gives QS = PS cos(90◦ − i) = PS sin i. In triangle PRS, ∠P = (90◦ − r) which gives PR = PS cos(90◦ − r) = PS sin r. By Snell’s law µ0 sin i = sin r. Thus,

S C

Two beams of 5000 ˚ A come out of the face AC. These beams interfere at the focus of the lens. The intensity depends on the phase difference between these beams. From the given figure, two beams have zero path difference upto PQ and there is no change in path difference beyond RS. Thus, path difference arises due

2 Ans. (a) b = 8 × 105 ˚ A (b) δ1 = 37◦ , δ2 = 27.13◦ (c) 9I

Chapter 19 Photometry

True False Type

Sol. The intensity at a point is defined as the energy per unit time per unit area. The intensity of a point source of radiant power F at a distance r is given by

Q 1. The intensity of light at a distance r from the axis of a long cylindrical source is inversely proportional to r.

I = F/(4πr2 ).

(1981)

Sol. Let the cylindrical source emits energy E per unit length per unit time uniformly in all radial directions. Let Q be a point at a radial distance r from the cylindrical source. Consider a cylinder of radius r and length l coaxial to the source.

C

•

m

1.5 60◦ P

60◦

3m

Q r •

1.5 m

A•

B S

l

Note that the radiant power F is uniformly distributed over a spherical shell of radius r. Substitute values to get intensities at P due to the sources A and B,

The energy emitted by length l of the source strikes normal to the curved surface of the cylinder. The area of the curved surface is 2πrl and energy emitted by length l in unit time is El. Thus, incident energy per unit area per unit time (intensity) at the point Q is I = El/(2πrl) = E/(2πr). Ans. T

FA 90 = = 0.796 W/m2 , 2 4πrAP 4 (3.14) (3)2 FB 180 = = = 6.369 W/m2 , 2 4πrBP 4 (3.14) (1.5)2

IA,P = IB,P

Descriptive

IC,P = 20 W/m2 ,

Q 2. Screen S is illuminated by two point sources A and B. Another source C sends a parallel beam of light towards point P on the screen (see figure). Line AP is normal to the screen and the lines AP, BP and CP are in one plane. The distances AP, BP and CP are 3 m, 1.5 m and 1.5 m. The radiant powers of sources A and B are 90 W and 180 W respectively. The beam from C is of intensity 20 W/m2 . Calculate intensity at P on the screen. (1982)

(Given).

The normal to the screen makes an angle 0◦ with AP but makes an angle 60◦ with BP and CP. Consider a unit area on the screen at the point P. The energy received by this unit area in one second is IA,P cos 0◦ from the source A, IB,P cos 60◦ from the source B and IC,P cos 60◦ from the source C. Thus, the total energy received per unit time per unit area at P is IP = IA,P cos 0◦ + IB,P cos 60◦ + IC,P cos 60◦ = 0.796 + 6.369(1/2) + 20(1/2) = 13.98 W/m2 .

C

•

Ans. 13.98 W/m2 60◦ A•

P

60◦ •

B S

278

Part IV

Thermodynamics

P TL

TH

Ad

Iso the rm ia ba t

W O

V

279

Chapter 20 Heat and Temperature

One Option Correct

(A)

Q 1. A real gas behaves like an ideal gas if its (A) pressure and temperature are both high. (B) pressure and temperature are both low. (C) pressure is high and temperature is low. (D) pressure is low and temperature is high.

β

(B)

β

(2010) p

(C)

Sol. In an ideal gas, the average force of attraction between the molecules and volume of the molecules (in comparison to volume of the gas) are negligibly small. These conditions are satisfied for a real gas when pressure is low and temperature is high. Ans. D

(1) (2)

p

V = nRT /p.

(1)

Differentiate equation (1) w.r.t. pressure p at constant temperature T to get dV nRT =− 2 . dp p

(2)

Use equations (1) and (2) to get  p   nRT  1 1 dV β=− =− − 2 = . V dp nRT p p Thus, β-p curve is a rectangular hyperbola. Note that β is defined as compressibility of the gas. Ans. A Q 4. When a block of iron floats in mercury at 0 ◦ C, fraction k1 of its volume is submerged, while at the temperature 60 ◦ C, a fraction k2 is seen to be submerged. If the coefficient of volume expansion of iron is γFe and that of mercury is γHg , then the ratio k1 /k2 can be expressed as (2001) 1+60γHg 1+60γFe 1−60γFe 1+60γFe (A) 1+60γ (B) (C) (D) 1+60γHg 1−60γHg 1+60γFe Hg

Since the length of each rod is increased by the same amount, we get l10 − l1 = l20 − l2 .

(D)

Sol. The ideal gas equation, pV = nRT , gives

Sol. After thermal expansion, length of the aluminium and the steel rod become

l20 = l2 (1 + αs t).

p β

p

Q 2. Two rods, one of aluminium and the other made of steel, having initial lengths l1 and l2 are connected together to form a single rod of length l1 + l2 . The coefficients of linear expansion for aluminium and steel are αa and αs , respectively. If the length of each rod increases by the same amount when their temperature 1 are raised by t◦ C, then find the ratio l1 l+l , (2003) 2 αa αs αa αs (A) αa (B) αs (C) αa +αs (D) αa +αs

l10 = l1 (1 + αa t),

β

(3)

Sol. Let VFe,0 , ρFe,0 , VHg,0 and ρHg,0 be the volumes and densities of iron and mercury blocks at 0 ◦ C. At 60 ◦ C, these quantities become

Substitute l10 and l20 from equations (1) and (2) into equation (3) to get, l1 αa = l2 αs . Hence, l1 l1 αs = = . l1 + l2 l1 + (l1 αa /αs ) αa + αs

VFe,60 = VFe,0 (1 + 60γFe ), ρFe,60 = ρFe,0 /(1 + 60γFe ), VHg,60 = VHg,0 (1 + 60γHg ),

Ans. C

ρHg,60 = ρHg,0 /(1 + 60γHg ). Q 3. Which of the following graphs correctly represent the variation of β = − V1 dV dp with p for an ideal gas at constant temperature? (2002)

Note that density varies with temperature due to volume change caused by thermal expansion (ρ ∝ V1 ). 281

282

Part IV. Thermodynamics

In hydrostatic equilibrium, weight of the iron block is balanced by the buoyancy force on it. Thus, ρFe,0 VFe,0 g = ρHg,0 k1 VF e,0 g,

(1)

ρFe,60 VFe,60 g = ρHg,60 k2 VF e,60 g.

(2)

l Y1 , α 1

F1

Y2 , α 2

F2

Divide equation (1) by (2) to get k1 ρFe,0 /ρHg,0 ρFe,0 /ρFe,60 1 + 60γFe = = = . k2 ρFe,60 /ρHg,60 ρHg,0 /ρHg,60 1 + 60γHg Ans. A Q 5. An ideal gas is initially at temperature T and volume V . Its volume is increased by ∆V due to an increase in temperature ∆T , pressure remaining constant. The quantity δ = V1 ∆V ∆T varies with temperature as (2000) δ

δ

(B)

(A)

T

T + ∆T

T

δ

T

T

T + ∆T

T

T + ∆T

δ

(C)

(D)

T

T + ∆T

T

T

Sol. The ideal gas equation, pV = nRT , gives p∆V = nR∆T at constant pressure p. Thus,  p   nR  1 ∆V 1 δ= = = . V ∆T nRT p T Hence, δ-T graph is a rectangular hyperbola. Note that δ is the coefficient of volume expansion of the gas at constant pressure. Ans. C Q 6. Two rods of different materials having coefficients of thermal expansion α1 , α2 and Young’s modulii Y1 , Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If α1 : α2 = 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to (1989)

(A) 2 : 3 (B) 1 : 1 (C) 3 : 2 (D) 4 : 9

But, the lengths of the rods cannot increase because they are fixed between two rigid walls. To oppose thermal expansion, the wall pushes the rod by reaction forces F1 and F2 as shown in the figure. The stresses due to these forces (called thermal stress) are σ1 = F1 /A and σ2 = F2 /A. These stresses compress the rods with compressions given by ∆l10 = lσ1 /Y1 ,

(3)

∆l20

(4)

= lσ2 /Y2 .

Given, thermal stresses are equal in both the rods i.e., σ1 = σ2 . Also, there is no bending of the rods when expansion is equal to compression i.e., ∆l10 = ∆l1 and ∆l20 = ∆l2 . Substitute in equations (1)–(4) and simplify to get Y1 /Y2 = α2 /α1 = 3/2. Note that the thermal stresses in the two rods are σ1 = α1 Y1 ∆T and σ2 = α2 Y2 ∆T . Ans. C Q 7. A metal ball immersed in alcohol weighs W1 at 0 ◦ C and W2 at 50 ◦ C. The coefficient of cubical expansion of the metal is less than that of the alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that (1980) (A) W1 > W2 (B) W1 = W2 (C) W1 < W2 (D) All of these Sol. Let ρ1 and V1 be the density and the volume of metal ball at 0 ◦ C and ρ2 and V2 be the corresponding values at 50 ◦ C. Let σ1 and σ2 be the densities of alcohol at 0 ◦ C and 50 ◦ C, respectively. The forces acting on the ball are its weight mg and the buoyant force F , acting opposite to each other. The weighing machine measures the net force on the ball. Thus, the weights measured at 0 ◦ C and 50 ◦ C are W1 = mg − F1 = mg − σ1 V1 g,

(1)

W2 = mg − F2 = mg − σ2 V2 g = mg − σ1 (1 − 50γa ) V1 (1 + 50γm )g = mg − σ1 V1 g [1 − 50(γa − γm )] (∵ γa γm ≈ 0)

Sol. Let the lengths of the rods before heating be l and their cross-section areas be A. On heating, the temperature of the rods is increased by ∆T . The length of each rod increases due to thermal expansion and is given by ∆l1 = lα1 ∆T,

(1)

∆l2 = lα2 ∆T.

(2)

= mg − σ1 V1 g + 50σ1 V1 g(γa − γm ) = W1 + 50σ1 V1 g(γa − γm ) > W1 .

(using (1)) (∵ γa > γm ).

Note that the volume of the metal at 50 ◦ C is V2 = V1 (1 + 50γm ) and the density of the alcohol at 50 ◦ C is σ2 = σ1 (1 − 50γa ). Ans. C

Chapter 20. Heat and Temperature

283

One or More Option(s) Correct

True False Type

Q 8. A bimetallic strip is formed out of two identical strips - one of copper and the other of brass. The coefficients of linear expansion of the two metals are αC and αB . On heating, the temperature of the strip goes up by ∆T and the strip bends to form an arc of radius of curvature R. Then, R is (1999) (A) proportional to ∆T . (B) inversely proportional to ∆T . (C) proportional to |αB − αC |. (D) inversely proportional to |αB − αC |.

Q 9. Water in a closed tube (see figure) is heated with one arm vertically placed above a lamp. Water will begin to circulate along the tube in counter-clockwise direction. (1983)

Sol. Let d and l0 be the widths and the initial lengths of the two strips. The lengths of the strips after heating are

Sol. The density of water at A decreases when it is heated up. The lighter water at A moves up which sets a clockwise current in the tube. Ans. F

lC = l0 (1 + αC ∆T ),

αC

B

Q 10. A barometer made of a very narrow tube (see figure) is placed at normal temperature and pressure. The coefficient of volume expansion of mercury is 0.00018 /◦ C and that of the tube is negligible. The temperature of mercury in the barometer is now raised by 1 ◦ C but the temperature of the atmosphere does not change. Then, the mercury height in the tube remains unchanged. (1983)

lB = l0 (1 + αB ∆T ).

θ

A

αB

R

Vacuum d

The strips bend as shown in the figure. From figure, the lengths of the two strips are lB = (R + d)θ,

(1)

lC = Rθ.

(2)

Hg

Divide equation (1) by (2) to get R+d lB 1 + αB ∆T = = R lC 1 + αC ∆T = (1 + αB ∆T )(1 + αC ∆T )−1 ≈ (1 + αB ∆T )(1 − αC ∆T ) ≈ 1 + (αB − αC )∆T,

(3)

where we have neglected the higher order terms. Simplify equation (3) to get R=

d . (αB − αC )∆T

(4)

The bending shown in the figure assumes αB > αC (which is generally the case for brass and copper). If αB < αC , bending of strip will be in the opposite direction. To take care of both the cases, equation (4) is written as d . R= |αB − αC | ∆T

Sol. The volume of mercury increases from V0 to V = V0 (1 + γ∆T ) when its temperature is raised by ∆T . Thus, the density of mercury changes to m m m ρ= = = (1 + γ∆T )−1 V V0 (1 + γ∆T ) V0 ≈ ρ0 (1 − γ∆T ). (1) Let h0 be the initial height of mercury column and h be the height when its temperature is increased by ∆T = 1 ◦ C. The atmospheric pressure (p0 ) remains constant and pressure in the vacuum section of the tube is zero. Thus, the hydrostatic pressure due to mercury column is equal to the atmospheric pressure in both the cases i.e., p0 = h0 ρ0 g = hρg = hρ0 (1 − γ∆T )g, which gives h=

Ans. B, D

(using (1))

h0 h0 = ≈ 1.00018h0 . 1 − γ∆T 1 − 0.00018 Ans. F

284

Part IV. Thermodynamics

Q 11. The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressure p1 and p2 are as shown in the figure. It follows from the graphs that p1 is greater than p2 . (1982) V

p1

Q 13. The
equation of state of a real gas is given by p + Va2 (V − b) = RT , where p, V and T are pressure, volume and temperature respectively and R is the universal gas constant. The dimensions of the constant a in this equation is . . . . . . (1997) Sol. The dimensions of [pV 2 ] = [ML5 T-2 ].

a V2

and p are same. Thus, [a] = Ans. [ML5 T-2 ]

p2

Q 14. A piece of metal floats on mercury. The coefficient of volume expansion of the metal and mercury are γ1 and γ2 , respectively. If the temperatures of both mercury and the metal are increased by an amount ∆T , the fraction of the volume of the metal submerged in mercury changes by the factor . . . . . . (1991)

T

Sol. The ideal gas equation gives V = (nR/p) T. Thus V -T graph is a straight line with slope nR/p. From the given graph, slope of line with pressure p1 is more than the slope of line with pressure p2 i.e., nR/p1 > nR/p2 , which gives p2 > p1 . Ans. F Fill in the Blank Type

x = Vi /V = ρ1 /ρ2 .

Q 12. A ring shaped tube contains two ideal gases with equal masses and relative molar masses M1 = 32 and M2 = 28. The gases are separated by one fixed partition and another movable stopper S which can move freely without friction inside the ring. The angle α as shown in the figure is . . . . . . ◦ . (1997) M1

M2

The densities of metal and mercury decrease due increase in volume by thermal expansion. The densities become ρ2 ρ1 , and ρ02 = . ρ01 = 1 + γ1 ∆T 1 + γ2 ∆T

x0 =

ρ01 ρ1 1 + γ2 ∆T . = 0 ρ2 ρ2 1 + γ1 ∆T

(2)

The equations (1) and (2) give the fractional change in the fraction of volume inside mercury as

S

Sol. In equilibrium, the pressure on the two sides of stopper S is equal because S is free to move i.e., p1 = p2 . Ideal gas equation on two sides gives m2 p 2 V2 = RT2 . M2

Divide and substitute p1 = p2 , T1 = T2 , and m1 = m2 to get

x0 1 + γ2 ∆T = . x 1 + γ1 ∆T Ans.

1+γ2 ∆T 1+γ1 ∆T

Q 15. During an experiment, an ideal gas is found to obey an additional law p2 V = constant. The gas is initially at a temperature T and volume V . When it expands to a volume 2V, the temperature becomes . . . . . . (1987)

M2 28 V1 . = = V2 M1 32

(1)

If A is the cross-sectional area of the tube and r is radius of the ring then V1 A(2π − α)r 2π − α = = . V2 Aαr α Solve equations (1) and (2) to get α =

(1)

In floating condition, fraction of the volume inside mercury is

α

m1 p1 V1 = RT1 , M1

Sol. Let V be the total volume of the metal, Vi be the volume of the metal immersed in mercury, ρ1 be the density of the metal, and ρ2 be the density of mercury. The metal float on mercury if its weight, V ρ1 g, is balanced by the upthrust, Vi ρ2 g, i.e., V ρ1 g = Vi ρ2 g. This gives the fraction of volume inside mercury as

(2)

Sol. Let the initial and final pressure, volume, and temperature of the gas be (pi , Vi , Ti ), and (pf , Vf , Tf ). Apply ideal gas equation at initial and final states to get pi Vi /Ti = pf Vf /Tf .

(1)

Apply p2 V = constant at initial and final states to get 8 15

× 2π = 192◦ . Ans. 192

p2i Vi = p2f Vf .

(2)

Chapter 20. Heat and Temperature

285

Sol. Young’s modulus is given by Y = where F = mg is the applied weight, A = πr is the crosssectional area of the wire, L is the length of the wire, and ∆Lm is the elongation in the wire due to longitudinal stress. Thus, we can write ∆Lm =

mgL mg/(πr2 ) = 2 . Y /L πr Y

The thermal contraction of the wire when its temperature is reduced is ∆Lt = Lα∆T, where α is the coefficient of linear expansion and ∆T is the change in temperature. In the given problem, ∆Lm = ∆Lt , which gives πr2 Y α∆T π(10−3 )2 (1011 )(10−5 )(10) = g 10 = π ≈ 3.

(3)

= ρl /(1 + γl ∆T ).

Now, the volume of the displaced liquid is a0 2 h and the buoyancy force is a0 2 hρ0l g. In equilibrium, equate weight of the cube to the buoyancy force to get a3 ρs g = a0 2 hgρ0l = a2 (1 + αs ∆T )2 hg [ρl /(1 + γl ∆T )] ,

(4)

where we have used equations (2) and (3) to eliminate a0 and ρ0l . Divide equation (1) by (4) to get 1 + γl ∆T = (1 + αs ∆T )2 ≈ 1 + 2αs ∆T, which gives γl = 2αs . Ans. γl = 2αs Q 18. The apparatus shown in the figure consists of four glass columns connected by horizontal sections. The height of two central columns B and C are 49 cm each. The two outer columns A and D are open to the atmosphere. A and C are maintained at a temperature of 95 ◦ C while the column B and D are maintained at 5 ◦ C. The height of the liquid in A and D measured from the base line are 52.8 cm and 51 cm respectively. Determine the linear coefficient of thermal expansion of the liquid. (1997) A

D B

C 5◦ C

F/A ∆Lm /L , 2

(2)

ρ0l

95◦ C

Q 16. Steel wire of length L at 40 ◦ C is suspended from ceiling and then a mass m is hung from its free end. The wire is cooled down from 40 ◦ C to 30 ◦ C to regain its original length L. The coefficient of linear thermal expansion of the steel is 10−5 /◦ C, Young’s modulus for steel is 1011 N/m2 and radius of the wire is 1 mm. Assume L  diameter of the wire. Then the value of m (in kg) is nearly . . . . . . . (2011)

a0 = a(1 + αs ∆T ),

5◦ C

Integer Type

When the temperature is increased by ∆T , edge length of the cube increases to a0 and density of the liquid decreases to ρ0l , which are given by

95◦ C

Substitute√Vi = V and Vf = 2V in equation (2) to 2pf . Now, substitute Vi = V , Vf = 2V , get p√ i = pi = √2pf , and Ti = T in equation (1) and solve to get Tf = 2T . √ Ans. 2T

m=

Sol. The volume and density of a liquid varies with the temperature as

Ans. 3 VT = V0 (1 + γT ), Descriptive Q 17. A cube of coefficient of linear expansion αs is floating in a bath containing a liquid of coefficient of volume expansion γl . When the temperature is raised by ∆T , the depth up to which the cube is submerged in the liquid remains the same. Find the relation between αs and γl showing all the steps. (2004) Sol. Let cube has the material density ρs and edge length a. Also, let its height h be inside the liquid of density ρl . The forces acting on the cube are its weight a3 ρs g and the buoyancy force a2 hρl g. In floating condition, a3 ρs g = a2 hρl g.

(1)

and

ρT = ρ0 /(1 + γT ),

where γ is the coefficient of volume expansion. Let hA = 52.8 cm be the height of column A, hD = 51.0 cm be the height of column D, and h = 49 cm be the height of columns B and C. The pressures at the top of column B and column C are equal i.e., p0 + hA ρ95 g − hρ5 g = p0 + hD ρ5 g − hρ95 g, which simplifies to hD + h ρ95 = . ρ5 hA + h

(1)

Also, from the expression of density, ρ95 ρ0 1 + 5γ 1 + 5γ = · = . ρ5 1 + 95γ ρ0 1 + 95γ

(2)

286

Part IV. Thermodynamics

Divide equation (1) by (2) and simplify to get

at the same pressure p. When the tube is held at an angle 60◦ with the vertical direction, the length of the air column above and below mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure p in centimetre of mercury. [The temperature of the system is kept at 30 ◦ C.] (1986)

hA − hD 95hD + 90h − 5hA 52.8 − 51 = = 2 × 10−4 /◦ C. 95(51) + 90(49) − 5(52.8)

γ=

The coefficient of linear expansion is α = γ/3 = 6.7 × 10−5 /◦ C. Ans. 6.7 × 10−5 /◦ C Q 19. A thin rod of negligible mass and area of crosssection 4 × 10−6 m2 , suspended vertically from one end, has a length of 0.5 m at 100 ◦ C. The rod is cooled to 0 ◦ C, but prevented from contracting by attaching a mass at the lower end. Find (a) the mass, and (b) the energy stored in the rod. [Given, for the rod, Young’s modulus = 1011 N/m2 , coefficient of linear expansion = 10−5 /K and g = 10 m/s2 .] (1997) Sol. The length of the rod at 100 ◦ C is l = l100 = 0.5 m and at 0 ◦ C it is

Sol. Let x be the length of the air columns on two sides of mercury when the tube is horizontal. The liquid mercury can be considered as incompressible in comparison to the air located on its two sides. Thus, the length of mercury column remains as 5 cm when the tube is tilted at 30◦ from the horizontal. x

5cm

x

p

p

cm

46 m 5c cm 5 . 44 p1 30◦

p2

l0 = l100 (1 + α∆T ) = 0.5(1 + 10−5 (0 − 100)) = 0.4995 m.

Since the length of the tube is constant, we get

Thus, decrease in length, due to thermal contraction, is given by ∆l1 = l100 − l0 = 0.5 − 0.4995 = 0.5 × 10−3 m.

(1)

The increase in length due to elongation by mass M is ∆l2 =

M gl M g/A = . Y /l YA

(2)

Since the rod is not allowed to contract, the decrease in length due to contraction is equal to the increase in length due to elongation i.e., ∆l1 = ∆l2 . Substitute ∆l1 from equation (1) and ∆l2 from equation (2) to get Y ∆l1 A lg (1011 )(0.5 × 10−3 )(4 × 10−6 ) = = 40 kg. (0.5)10

M=

The elastic potential energy stored in the elongated rod is given by 1 AY U= (∆l)2 2 l (4 × 10−6 )(1011 ) = × (0.5 × 10−3 )2 = 0.1 J. 2 × 0.5 Ans. (a) 40 kg (b) 0.1 J Q 20. A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal ends containing air

2x + 5 = 44.5 + 5 + 46,

(1)

which gives x = 45.25 cm. Let A be the cross-sectional area of the tube. The temperature of the gas remains constant at 30 ◦ C during the tilting process. Thus,the process is isothermal and hence pV = constant for both the chambers of the gas. Let p be the initial pressure and p1 and p2 be the final pressures inside the chambers (see figure). Apply pV = constant for both the chambers to get p(xA) = p1 (44.5A),

(2)

p(xA) = p2 (46A).

(3)

Now, consider the two sides of mercury column. The pressure at two ends of mercury column are p1 and p2 . The height difference between two ends of mercury column is 5 sin 30◦ = 2.5 cm (note that the tube is thin). Thus, the difference in p1 and p2 is equal to the hydrostatic pressure caused by the height difference of 2.5 cm i.e., p1 = p2 + 2.5. (pressures are in cm of Hg).

(4)

Eliminate p1 , p2 , and x from equations (1)–(4) to get p = 75.4 cm of Hg. Ans. 75.4 cm of Hg Q 21. Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0 ◦ C and pressure 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62 ◦ C. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible. (1985)

Chapter 20. Heat and Temperature

287

Sol. Let V be the volume of each bulb. Initial pressure of each bulb is pi = 76 mm of Hg and initial temperature is Ti = 0 ◦ C = 273 K. Apply ideal gas equation to get the number of moles of the gas in each bulb ni =

∆l2 =

pi V . RTi

nf,1 =

pf V , RTf,1

T /A Tl T (1) = = Y /l YA (2 × 1011 ) (10−6 )

= 5 × 10−6 T.

The pressures in both the bulbs are equal because they are connected by a narrow tube. Let pf be the common final pressure of the two bulbs. Final temperature of the bulb placed in ice is Tf,1 = 0 ◦ C = 273 K and that of the bulb placed in water bath is Tf,2 = 62 ◦ C = 335 K. Apply ideal gas equation to get the final number of moles of the gas in two bulbs nf,2 =

pf V . RTf,2

By conservation of mass, the number of moles of the gas in two bulbs remains constant i.e., 2ni = nf,1 + nf,2 , which gives 2

tensed). The elongation of the wire due to tension T is given by

pf V pf V pi V = + . RTi RTf,1 RTf,2

Solve to get the common final pressure in the bulbs 2pi Tf,1 Tf,2 Ti (Tf,1 + Tf,2 ) 2(76)(273)(335) = = 83.75 mm of Hg. 273(273 + 335)

pf =

In equilibrium, the magnitude of the tension is such that the elongation by the tension is equal to the thermal contraction. Equate ∆l1 = ∆l2 to get T = 48.4 N. In fundamental mode of vibration, wavelength of the wave is related to length of the wire by λ = 2l = 2(1) = 2 m. The mass per unit length of the given wire is µ = m/l = 0.1/1 = 0.1 kg/m. The wave velocity in a wire of mass per unit length µ under tension T is given by p p v = T /µ = 48.4/0.1 = 22 m/s. The frequency of the fundamental mode of vibration is ν = v/λ = 22/2 = 11 Hz. Ans. 11 Hz Q 23. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M . The piston and the cylinder have equal crosssectional area A. Atmospheric pressure is p0 and when the piston is in equilibrium, the volume of the gas is V . The piston is now displaced slightly from its equilibrium position. Assuming that the system is completely isolated from its surroundings, show that the piston executes SHM and find the frequency of oscillation. (1981)

Ans. 83.75 cm of Hg Q 22. A steel wire of length 1 m, mass 0.1 kg and uniform cross-sectional area 10−6 m2 is rigidly fixed at both ends. The temperature of the wire is lowered by 20 ◦ C. If transverse waves are set-up by plucking the string in the middle, calculate the frequency of the fundamental mode of vibration. [Given Ysteel = 2 × 1011 N/m2 , αsteel = 1.21 × 10−5 /◦ C.] (1984) Sol. Given wire has length l = 1 m, mass m = 0.1 kg, and the cross-sectional area A = 10−6 m2 . For steel, Young’s modulus is Y = 2 × 1011 N/m2 and the coefficient of linear expansion is α = 1.21 × 10−5 /◦ C. The decrease in temperature of the wire is ∆T = 20 ◦ C. Thus, the decrease in the length of the wire due to thermal contraction is

Sol. Let p be the pressure inside the cylinder of volume V at equilibrium position of the piston. The forces acting on the piston are its weight M g (downwards), force due to the outer atmosphere p0 A (downwards), and force due to the inside gas pA (upwards). In equilibrium, the net force on the piston is zero i.e., pA = p0 A + M g.

Let the piston is displaced down by a small distance x. p0 x p

∆l1 = lα∆T = (1) (1.21 × 10−5 ) (20) = 2.42 × 10−4 m. The wire is not allowed to contract because it is rigidly fixed at both the ends. The thermal contraction is opposed by elongation due to tension T in the wire (because thermal contraction causes the wire to become

(1)

The gas inside the cylinder is compressed to a volume V 0 = V − Ax.

(2)

The compression is an adiabatic process because the system is completely isolated from its surroundings.

288

Part IV. Thermodynamics γ

Apply adiabatic equation, pV γ = p0 V 0 , to get the pressure of the compressed gas  γ V 0 p =p V0  γ V =p . (using (2)). (3) V − Ax The net force on the piston under this condition is F = (p0 A + M g) − p0 A  γ V , (using (3)) = (p0 A + M g) − pA V − Ax " #  −γ Ax = (p0 A + M g) 1 − 1 − , V (using (1))    Ax , ≈ (p0 A + M g) 1 − 1 + γ V (∵ Ax  V ) 2

γA (p0 + M g/A) =− x. V

(4)

From equation (4), the restoring force on the piston is proportional to the displacement. Thus, the piston undergoes SHM. Apply Newton’s second law on the piston to get d2 x γA(M g + p0 A) =− x = −ω 2 x. dt2 MV ω The frequency of oscillation is ν = 2π = q γ(p0 +M g/A) A . 2π MV q γ(p0 +M g/A) A Ans. 2π MV

Q 24. A composite rod is made by joining a copper rod, end to end, with a second rod of different material but of the same cross-section. At 25 ◦ C, the composite rod is 1 m in length, of which the length of the copper rod is 30 cm. At 125 ◦ C the length of the composite rod increases by 1.91 mm. When the composite rod is not allowed to expand by holding it between two rigid walls, it is found that the length of the two constituents do not change with rise in temperature. Find the Young’s modulus and the coefficient of linear expansion of the second rod. [For copper, coefficient of linear expansion = 1.7 × 10−5 /◦ C, Young’s modulus = 1.3 × 1011 N/m2 .] (1979) Sol. Let l1 = 30 cm be the length of the copper rod at 25 ◦ C and α1 = 1.7 × 10−5 /◦ C be its coefficient of linear expansion. Let α2 be the coefficient of linear expansion for the other rod of length l2 = 100 − 30 = 70 cm at 25 ◦ C. Cu 30 cm

The increase in the length of the rod when its temperature rises from 25 ◦ C to 125 ◦ C is ∆l = ∆l1 + ∆l2 = l1 α1 ∆T + l2 α2 ∆T, 0.191 = (30)(1.7 × 10−5 )(100) + (70)α2 (100). Solve to get α2 = 2 × 10−5 /◦ C. Let A be the cross-sectional area of the rods, Y1 = 1.3 × 1011 N/m2 be Young’s modulus of copper and Y2 be Young’s modulus of the other rod. When the composite rod is not allowed to expand by holding it between two rigid walls, compression forces are developed in the rods to oppose thermal expansion. The compressive forces in two rods are equal (otherwise, it will deform at the joint). In equilibrium, the length contraction due to the compressive force is equal to the thermal expansion i.e., ∆l1 = F l1 /(AY1 ) = l1 α1 ∆T,

(1)

∆l2 = F l2 /(AY2 ) = l2 α2 ∆T.

(2)

Divide equation (1) by (2) and simplify to get (1.3 × 1011 )(1.7 × 10−5 ) Y1 α1 = α2 2 × 10−5 11 = 1.105 × 10 N/m2 .

Y2 =

Ans. 1.105 × 1011 N/m2 , 2 × 10−5 /◦ C Q 25. A copper wire is held at the two ends by rigid supports. At 30 ◦ C, the wire is just taut, with negligible tension. Find the speed of transverse waves in this wire at 10 ◦ C. [Given, for copper: Young’s modulus = 1.3 × 1011 N/m2 , coefficient of linear expansion (1979) = 1.7 × 10−5 ◦ C−1 , density = 9 × 103 kg/m3 .] Sol. Let l and A be the length and the cross-sectional area of the wire. The density of the wire is ρ = 9 × 103 kg/m3 , mass of the wire is m = ρlA, and the mass per unit length is µ = m/l = ρA. The decrease in the length of the wire when its temperature is reduced by ∆T = 30 ◦ C − 10 ◦ C = 20 ◦ C is given by ∆l = lα∆T,

(1)

where α = 1.7 × 10−5 ◦ C−1 is the coefficient of linear expansion. Since, the wire is held at its two ends by rigid supports, the tension in the wire increases to oppose the reduction in its length. The tension required to elongate a wire of Young’s modulus Y = 1.3 × 1011 N/m2 by ∆l is given by T = Y A∆l/l = Y Aα∆T.

(using (1)).

The speed of the transverse waves in a wire of mass per unit length µ under tension T is given by p p p v = T /µ = Y Aα∆T /(ρA) = Y α∆T /ρ q = (1.3 × 1011 )(1.7 × 10−5 )(20)/(9 × 103 ) = 70.08 m/s.

70 cm

Ans. 70.08 m/s

Chapter 20. Heat and Temperature

289

Q 26. A sinker of weight W0 has an apparent weight W1 when placed in a liquid at a temperature T1 and W2 when weighed in the same liquid at temperature T2 . The coefficient of cubical expansion of the material of the sinker is β. What is the coefficient of volume expansion of the liquid? (1978) Sol. Let ρ0 , ρ1 , and ρ2 be the densities of the sinker at temperatures T0 , T1 , and T2 , respectively. The volumes of the sinker at these temperatures are V0 , V1 , and V2 and the corresponding densities of the liquid at these temperatures are σ0 , σ1 , and σ2 . The apparent weight of the sinker placed in a liquid is equal to its actual weight minus the buoyant force. Thus, the weights at temperature T1 and T2 are W1 = W0 − V1 σ1 g = W0 − (V1 ρ1 g)(σ1 /ρ1 ) = W0 − W0 (σ1 /ρ1 ),

(1)

W2 = W0 − V2 σ2 g = W0 − (V2 ρ2 g)(σ2 /ρ2 ) = W0 − W0 (σ2 /ρ2 ),

(2)

where we have used the fact that the mass of the sinker is same at all temperatures i.e., V0 ρ0 = V1 ρ1 = V2 ρ2 and sinker is completely submerged into the liquid. Divide equation (1) by (2) and simplify to get W0 − W1 (σ1 /σ2 ) = . (ρ1 /ρ2 ) W0 − W2

(3)

The densities of the sinker and the liquid at temperatures T1 and T2 are related by ρ1 /ρ2 = 1 + β(T2 − T1 ), σ1 /σ2 = 1 + γ(T2 − T1 ), where γ is the coefficient of volume expansion of the liquid and β is the coefficient of cubical expansion of the material of the sinker. Substitute in equation (3) and simplify to get   W0 − W1 W2 − W1 γ=β . + W0 − W2 (W0 − W2 )(T2 − T1 )   W2 −W1 0 −W1 Ans. β W W0 −W2 + (W0 −W2 )(T2 −T1 )

Chapter 21 Kinetic Theory of Gases

One Option Correct

Q 3. A mixture of 2 mol of helium gas (atomic mass=4 u) and 1 mol of argon gas (atomic mass=40 u) is kept at 300 K in a container. The ratio of rms speed vrms (helium) (2012) vrms (argon) is (A) 0.32 (B) 0.45 (C) 2.24 (D) 3.16

Q 1. Two non-reactive monatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature is 4 : 3. The ratio of their densities is (2013) (A) 1 : 4 (B) 1 : 2 (C) 6 : 9 (D) 8 : 9

Sol. The rms speed of a gas kept at an absolute temperature T is given by p vrms = 3kT /m,

Sol. The ideal gas equation, pV = nRT, can be written as   ρV RT, pV = M

where m is the mass of a gas molecule and k is Boltzmann constant. Thus, r p 3kT /(4u) vrms (helium) 40 =p = = 3.16. vrms (argon) 4 3kT /(40u)

where ρ is density and M is molecular mass. The molecular mass is equal to the atomic mass for monatomic gas. Thus, for given two gases

Ans. D

ρ1 p 1 M1 8 p1 M1 /(RT ) 2 4 = = = × = . ρ2 p2 M2 /(RT ) p 2 M2 3 3 9

Q 4. A gas mixture consists of 2 mol of oxygen and 4 mol of argon at temperature T . Neglecting all vibrational modes, the total internal energy of the system is (1999) (A) 4RT (B) 15RT (C) 9RT (D) 11RT

Ans. D Q 2. Two moles of ideal helium gas are in rubber balloon at 30 ◦ C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35 ◦ C. The amount of heat required in raising the temperature is nearly [Take R = 8.31 J/mol K.] (2012) (A) 62 J (B) 104 J (C) 124 J (D) 208 J

Sol. The internal energy of n moles of an ideal gas at temperature T , with each molecule having f degrees of freedom, is given by f U = n RT. 2

Sol. The pressure inside the balloon is equal to the constant ambient pressure because balloon is fully expandable. Thus, the given process is isobaric, so heat exchanged in the process is given by ∆Q = nCp ∆T.

(1)

Oxygen is a diatomic gas with fO2 = 5 (three translational, two rotational) and argon is a monatomic gas with fAr = 3 (three translational). Substitute the values in equation (1) to get the total internal energy of the system as

(1)

Helium is a monatomic gas with three degrees of freedom (f = 3). Thus, the specific heats at constant volume (Cv ) and constant pressure (Cp ) for helium are given by

5 3 U = UO2 + UAr = 2 × RT + 4 × RT = 11RT. 2 2 Ans. D Q 5. A vessel contains a mixture of 1 mol of O2 and 2 mol of N2 at 300 K. The ratio of the average rotational kinetic energy per O2 molecule to that of per N2 molecule is (1998) (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) depends on the moment of inertia of the molecules.

Cv = (f /2)R = (3/2)R, Cp = Cv + R = (3/2)R + R = (5/2)R. Substitute the values in equation (1) to get ∆Q = (2)(5/2)(8.31)(35 − 30) = 207.75 ≈ 208 J. Ans. D 290

Chapter 21. Kinetic Theory of Gases Sol. Average kinetic energy per molecule of a gas at temperature T is K=

f kT, 2

where f is degrees of freedom of each molecule and k is Boltzmann constant. Both O2 and N2 are diatomic gases with two rotational degrees of freedom. Thus, average rotational kinetic energy per molecule for these gases are KO2 = KN2 = kT . Ans. A Q 6. The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21 × 10−21 J and 484 m/s, respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour) (1997) (A) 12.42 × 10−21 J, 968 m/s (B) 8.78 × 10−21 J, 684 m/s (C) 6.21 × 10−21 J, 968 m/s (D) 12.42 × 10−21 J, 684 m/s Sol. A molecule of mass m in a gas kept at temperature T has average translational kinetic energy E = 32 kT and p rms speed vrms = 3kT /m. Thus, √ E600 = 2E300 = 12.42 × 10−21 J and vrms,600 = 2vrms,300 = 1.414 × 484 = 684 m/s. Ans. D Q 7. The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules (in eV) at the same temperature is (1997) (A) 0.0015 (B) 0.003 (C) 0.048 (D) 0.768 Sol. Average translational kinetic energy, E = 32 kT , depends only on T . Ans. C Q 8. A vessel contains 1 mol of O2 gas (molar mass 32) at a temperature T . The pressure of the gas is p. An identical vessel containing one mole of the gas (molar mass 4) at a temperature 2T has a pressure of (1997) (A) p/8 (B) p (C) 2p (D) 8p Sol. Ideal gas equation, pV = nRT , for the gas in 1st vessel is pV = RT and for the gas in 2nd vessel is p0 V 0 = R(2T ). Substitute V 0 = V to get p0 = 2p. Ans. C Q 9. Three closed vessels A, B and C are at the same temperature T and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only O2 , B only N2 , and C a mixture of equal quantities of O2 and N2 . If the average speed of the O2 molecules in vessel A is v1 , that of N2 molecules in vessel B is v2 , the average speed of the O2 molecules in vessel C is [M is the mass of an O2 molecule.] (1992) p √ 2 (A) v1 +v (B) v1 (C) v1 v2 (D) 3kT /M 2

291 Sol. The average speed of an ideal gas molecule of p molecular mass M , at temperature T , is v¯ = 8RT /(πM ). Since M and T are same for O2 gas in vessel A and vessel C, the average speed of O2 in vessel C and A are equal (v1 ). Ans. B Q 10. At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s. The gas is (1984) (A) H2 (B) F2 (C) O2 (D) Cl2 Sol. The rms speed of the gas molecules, having molecular mass M and temperature T , is given by p vrms = 3RT /M . Substitute vrms = 1930 m/s, T = 298 K (room temperature), and R = 8.314 J/(mol K) to get M = 2.0 × 10−3 kg/mol = 2.0 g/mol. Thus, the given diatomic gas is H2 . Ans. A One or More Option(s) Correct Q 11. A flat plate is moving normal to its plane through a gas under the action of a constant force F . The gas is kept at a very low pressure. The speed of the plate v is much less than the average speed u of the gas molecules. Which of the following option(s) is(are) true? (2017) (A) At a later time the external force F balances the resistive force. (B) The plate will continue to move with constant nonzero acceleration, at all times. (C) The resistive force experienced by the plate is proportional to v. (D) The pressure difference between the leading and trailing faces of the plate is proportional to uv. Sol. Let M be the mass of the plate and A be the area of its leading/trailing face. Let ρ be the density of the gas, m be the mass of a gas molecule and u be the average speed of the gas molecules. The plate is moving with a speed v ( u) under the influence of a constant force F . trailing Ft

F leading Fl

trailing Ft

F leading Fl

v m

u

u

v m

Before Collision

ut

m m

ul

After Collision

Consider collision of a gas molecule on the leading face of the plate. It is easy to analyse this collision in a frame attached to the plate. In this frame, the gas molecules strikes the leading face with a speed (u + v), undergoes the elastic collision at the plate, and then

292

Part IV. Thermodynamics

reflects back with the same speed (u + v). Thus, the speed of the gas molecule after the collision is ul = u+2v in the laboratory frame. We encourage you to use the relative velocity formula, ~um/p = ~um/l −~up/l , to get this result (here, in the subscripts, m is molecule, p is plate and l is laboratory). The increase in the linear momentum of the gas molecule due to collision at the leading face in laboratory frame is ∆pl = m(u + 2v) − (−mu) = 2m(u + v). A molecule, located at a distance (u+v) or less from the leading face and moving towards it, will hit the leading face of the plate within next one second. For simplicity, we assume the random motion of the molecules to be in one dimension (normal to the plate) so that half of the molecules are moving towards the plate and another half are moving away from it. Thus, the number of molecules that strike the leading face in one second, Nl , is equal to one half of the number of molecules lying in the volume A(u + v) i.e., Nl =

1 ρA(u+v) . 2 m

Thus, increase in linear momentum of the gas in one second (which is equal to the force by the plate on the gas) is Nl ∆pl = ρA(u + v)2 . By Newton’s third law, the gas apply equal and opposite force on the leading face of the plate. Thus, the force on the leading face of the plate due to collision by the gas molecules, is Fl = ρA(u + v)2 . The pressure on the leading side of the plate is Pl = Fl /A = ρ(u + v)2 . Similar analysis on the trailing face of the plate gives ∆pt = m(u − 2v) − (−mu) = 2m(u − v), Nt =

1 ρA(u−v) , 2 m 2

Ft = ρA(u − v) , Pt = Ft /A = ρ(u − v)2 . The pressure difference between the leading and trailing faces, Pl − Pt = ρ(u + v)2 − ρ(u − v)2 = 4ρuv, is proportional to uv. The resistive force on the plate, Fr = Fl − Ft = 4ρAuv, is proportional to v. The net force on the plate is Fn = F − Fr = F − 4ρAuv.

Newton’s second law gives acceleration of the plate a=

Fn F − 4ρAuv = . M M

The acceleration a is positive (provided F > 4ρAuv at initial time t = 0). Hence, speed v of the plate increases with time. As v increases, the resistive force Fr also increases. This continues till the net force Fn on the plate becomes zero. The speed of the plate when net force becomes zero is vt = F/(4ρAu) (this is called terminal velocity). Henceforth,the plate continues to move with constant velocity vt (zero acceleration). We encourage you to find the expressions for v and a as a function of time t (with suitable initial conditions). Draw the graphs of time versus v, a, and Fr . Also analyse the motion if v > vt at t = 0. Ans. (A), (C), (D) Q 12. A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium in equilibrium at temperature T . Assuming the gases are ideal, the correct statement(s) is (are) (2015) (A) The average energy per mole of the gas mixture is 2RT . (B) The ratio of speed of p sound in the gas mixture to that in helium gas is 6/5. (C) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2. (D) The ratio of the rms speed of √ helium atoms to that of hydrogen molecules is 1/ 2. Sol. The internal energy of one mole of an ideal gas at temperature T is given by U = f2 RT , where f is the degrees of freedom of the gas molecule. The degrees of freedom for hydrogen (diatomic) and helium (monatomic) gases are fH2 = 5 and fHe = 3, respectively. Thus, UH2 = 52 RT and UHe = 32 RT . The total internal energy of the gas mixture is Utotal = UH2 + UHe = 25 RT + 32 RT = 4RT. The mixture contains two moles of the gases. The internal energy per mole of the mixture is Umix = Utotal /2 = 2RT . The specific heat at constant volume is given by Cv = dU/dT . Thus, the specific heats at constant volume for helium and the mixture are Cv,He = dUHe /dT = 23 R, and Cv,mix = dUmix /dT = 2R. The specific heats at constant pressure, Cp = Cv + R, for these gases are Cp,He = Cv,He + R = 52 R, Cp,mix = Cv,mix + R = 3R.

and

Chapter 21. Kinetic Theory of Gases

293

The ratio of specific heats, γ = Cp /Cv , are γHe = 5/3 and γmix = 3/2. The speed ofp sound, in a gas of molecular mass M , is given by vs = γRT /M . The molecular mass of the gas mixture is nH2 MH2 + nHe MHe nH2 + nHe (1)(2) + (1)(4) = = 3 g/mol, 1+1

Mmix =

where nH2 = 1 and nHe = 1 are the number of moles of hydrogen and helium in the gas mixture. The ratio of the speeds of sound in the gas mixture and helium is vs,mix vs,He

s p γmix RT /Mmix γmix MHe = p = γHe Mmix γHe RT /MHe s r (3/2) (4) 6 = = . (5/3) (3) 5

The prms speed of the atoms/molecules is given by vrms = 3RT /M . The ratio of the rms speed of helium atoms to that of hydrogen molecules is vrms,He vrms,H2

r r p r 3RT /MHe MH2 2 1 = = . =p = MHe 4 2 3RT /MH2 Ans. A, B, D

Q 13. Let v¯, vrms and vp respectively denote the mean speed, root mean square speed and most probable speed of the molecules in an ideal monatomic gas at absolute temperature T . The mass of the molecule is m. Then, (A) (B) (C) (D)

√ (1998) no molecule can have speed greater than √ 2vrms . no molecule can have speed less than vp / 2. vp < v¯ < vrms . the average kinetic energy of a molecule is 34 mvp2 .

Note that the speed distribution is asymmetric about vp . It expands more towards the right of vp , which makes v¯ > vp (see figure). The average kinetic energy of monatomic gas molecule is given by K = 32 kT = 43 mvp2 . Ans. C, D Q 14. The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the rms velocity of the gas molecules is v, at 480 K it becomes (1996) (A) 4v (B) 2v (C) v/2 (D) v/4 Sol. The rms speed of the gas molecules at temperature T is given by p vrms = 3kT /m, where m is the mass of a gas molecule and k is Boltzmann constant. Thus, the rms speed of gas molecules at temperature T = 120 K and T 0 = 480 K are p p v = 3k(120)/m, v 0 = 3k(480)/m = 2v. Ans. B Q 15. From the following statements concerning ideal gas at any given temperature T , select the correct one(s), (1995) (A) The coefficient of volume expansion at constant pressure is the same for all ideal gases. (B) The average translational kinetic energy per molecule of oxygen gas is 3 kT , k being Boltzmann constant. (C) The mean-free path of molecules increases with decrease in the pressure. (D) In a gaseous mixture, the average translational kinetic energy of the molecules of each component is different.

vp =

p

2kT /m,

v¯ =

p

8kT /(πm),

Sol. Differentiate ideal gas equation, pV = nRT , to get pdV + V dp = nRdT . This reduces to pdV = nRdT at constant pressure (dp = 0). The coefficient of volume 1 expansion at constant pressure, γ = V1 dV dT = T , is the same for all ideal gases. The average translational kinetic energy per molecule of ideal gas is 32 kT . At a given temperature, the decrease in pressure leads to increase in volume. Thus, the mean free path increases with decrease in pressure. Ans. A, C

vrms =

p

3kT /m.

Assertion Reasoning Type

Sol. The molecules in a gas travel with different speeds. The number of molecules having speed between v and v + dv varies with v as shown in the figure. This variation is given by the Maxwell’s speed distribution law. This law gives expressions for the most probable speed (vp ), average speed v¯, and rms speed vrms as

n

vp v¯ vrms

v

Q 16. Statement 1: The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume. Statement 2: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision. (2007)

294

Part IV. Thermodynamics

(A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. (B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true. Sol. The total kinetic energy of n moles of an ideal gas at temperature T is given by U=

Fill in the Blank Type Q 19. A gas thermometer is used as a standard thermometer for measurement of temperature. When the gas container of the thermometer is immersed in water at its triple point 273.16 K, the pressure in the gas thermometer reads 3.0 × 104 N/m2 . When the gas container of the same thermometer is immersed in another system, the gas pressure reads 3.5 × 104 N/m2 . The temperature of this system is therefore . . . . . . ◦ C. (1997)

f nRT, 2

(1)

where f is the degrees of freedom. There are three degrees of freedom (f = 3) for translational motion. The ideal gas equation gives P V = nRT.

(2)

From equations (1) and (2), U = 1.5P V . Ans. B

Sol. The volume and number of moles of the gas in a gas thermometer remain constant. The ideal gas equation gives p1 V1 = n1 RT1 ,

p2 V2 = n2 RT2 ,

for the two measurements. Substitute V1 = V2 and n1 = n2 to get (3.5 × 104 )(273.16) p2 T1 = p1 3.0 × 104 = 318.68 K = 45.53 ◦ C.

T2 = True False Type Q 17. The rms speed of oxygen molecules (O2 ) at a certain temperature T (degree absolute) is v. If the temperature is doubled and oxygen gas dissociates into atomic oxygen, the rms speed remains unchanged. (1987)

Sol. The rms speed of a gas at temperature T is given by

Ans. 45.53 ◦ C Q 20. The variation of temperature of a material as heat is given to it at a constant rate is as shown in the figure. The material is in solid state at the point O. The state of the material at the point P is . . . . . . (1985) C

r vrms =

T

3RT , M

(1)

r 0 vrms

=

3RT 0 = M0

s

r 3R(2T ) 3RT =2 = 2vrms . M/2 M Ans. F

Q 18. Two different gases at the same temperature have equal rms velocities. (1982) Sol. The rms velocity of a gas, with molecular mass m at temperature T , is given by vrms =

p

3kT /m.

Thus, the two gases of different molecular masses will have different vrms at the same temperature. Ans. F

D

P B

O

where M is the molecular mass of the gas. When temperature is raised to T 0 = 2T , oxygen molecule dissociates into two atoms with atomic mass M 0 = M/2. Substitute in equation (1) to get

A

Heat added

Sol. The material is in solid state from O to A. It transform to a mixture of solid and liquid from point A to B. The heat added from A to B is used to convert the material from solid to liquid state (latent heat of fusion). It does not change its temperature. The material is in liquid state from B to C. From C to D, the material is a mixture of liquid and gas. The heat given from C to D is used to convert the material from liquid to gas (latent heat of vaporization). We encourage you to relate the slopes of the graph with the specific heats of the material. Ans. partly solid and partly liquid Descriptive Q 21. An insulated box containing a monatomic gas of molar mass M moving with a speed v0 is suddenly stopped. Find the increment in gas temperature as a result of stopping the box. (2003)

Chapter 21. Kinetic Theory of Gases

295

Sol. Let m be the total mass of the gas containing n = m/M moles. The decrease in kinetic energy of the gas is ∆K = 12 mv02 .

(1)

This energy increases internal energy of the gas. The increase in the internal energy of the gas is   m 3R ∆T. (2) ∆U = nCv ∆T = M 2 The energy conservation gives, ∆K = ∆U . Substitute the value of ∆K and ∆U from equations (1) and (2) to get, ∆T = M v02 /(3R). M v2 Ans. 3R0 Q 22. A cubical box of side 1 m contains helium gas (atomic weight 4) at a pressure of 100 N/m2 . During an observation time of 1 s, an atom travelling with the rms speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without any collision with other atoms. Take, R = 25 3 J/mol K and k = 1.38 × 10−23 J/K. (2002) (a) Evaluate the temperature of the gas. (b) Evaluate the average kinetic energy per atom. (c) Evaluate the total mass of helium gas in the box. Sol. The atoms travel a distance of 2 m between successive collisions to a wall. The number of collisions per second to a particular wall is 500. Thus, the distance traveled by an atom moving with rms speed in a time interval of 1 s is 500 × 2 = 1000 m. Hence, rms speed of the atom is vrms = 1000 m/s. The vrms is related to temperature T by 3 1 1 M 2 1 M 2 2 kT = mvrms = v , v = 2 2 2 NA rms 2 R/k rms

(1)

where, m is the mass of an atom, M = 4 g is molar mass, and NA = R/k is Avogadro number. Substitute the values in equation (1) to get T =

2 M vrms (4 × 10−3 )(1000)2 = = 160 K. 3R 3 × (25/3)

The average kinetic energy per atom for a gas at temperature T = 160 K is given by 3 3 K = kT = (1.38 × 10−23 )(160) 2 2 = 3.312 × 10−21 J. The ideal gas equation, pV = nRT = gives mgas =

mgas M RT ,

pV M (100)(1)(4) = = 0.3 g. RT (25/3)(160)

Ans. (a) 160 K (b) 3.312 × 10−21 J (c) 0.3 g

Q 23. An ideal gas has a specific heat at constant pressure Cp = 52 R. The gas is kept in a closed vessel of volume 0.0083 m3 , at a temperature of 300 K and pressure of 1.6 × 106 N/m2 . An amount of 2.49 × 104 J of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas. (1987) Sol. Given, initial pressure pi = 1.6 × 106 N/m2 , volume Vi = 0.0083 m3 , and temperature Ti = 300 K. Final volume of the gas is Vf = 0.0083 m3 as gas is kept in a closed vessel. The work done by the gas is zero because volume of the gas does not change. Substitute ∆W = 0 in the first law of thermodynamics, ∆Q = ∆U + ∆W , to get ∆Q = ∆U = nCv ∆T, (∵ ∆U = nCv ∆T ) pi Vi (Cp − R)(Tf − Ti ). = RTi (∵ pV = nRT, Cp − Cv = R) Solve for Tf and substitute the values to get Tf = Ti +

RTi ∆Q 2Ti ∆Q = Ti + pi Vi (Cp − R) 3pi Vi

= 300 +

2(300)(2.49 × 104 ) = 675 K. 3(1.6 × 106 )(0.0083)

The number of moles of the gas does not change between the initial and the final states i.e., n = pi Vi /Ti = pf Vf /Tf . Use this to get the final pressure pf =

pi Tf (1.6 × 106 )(675) pi Vi Tf = = Vf Ti Ti 300

= 3.6 × 106 N/m2 . Ans. 675 K, 3.6 × 106 N/m2 Q 24. One gram mole of oxygen at 27 ◦ C and one atmospheric pressure is enclosed in a vessel. (a) Assuming the molecules to be moving with vrms , find the number of collisions per second which the molecules makes with one square metre area of vessel. (b) The vessel is next thermally insulated and moved with constant speed v0 . It is then suddenly stopped. The process results in a rise of the temperature of the gas by 1 ◦ C. Calculate the speed v0 . (1983) Sol. The rms speed of the oxygen gas (molecular mass M = 32 g/mol) at temperature T = 27 ◦ C = 300 K is given by s r 3RT 3(8.314)(300) vrms = = = 483.56 m/s. M 32 × 10−3 The pressure of the gas is p0 = 1.013 × 105 N/m2 . Thus, the force acting on the wall of area A = 1 m2 is F = p0 A = 1.013 × 105 N. Let n molecules of the oxygen gas hit the wall per square metre per second.

296

Part IV. Thermodynamics

The molecules collide the wall with a speed vrms and recoil with the same speed. Thus, the momentum transferred to one square metre of the wall in one second is ∆p = n(2mvrms ), where m is the mass of each molecule. The momentum transferred per second is equal to the force i.e., F = 2nmvrms . Substitute values to get F F N0 F = = 2mvrms 2(M/N0 )vrms 2M vrms (1.013 × 105 ) (6.023 × 1023 ) = 1.97 × 1027 . = 2(32 × 10−3 ) (483.56)

n=

The mass of n = 1 mol of the oxygen gas is mg = nM = 32 × 10−3 kg. The kinetic energy of the gas contained in an insulated vessel moving with speed v0 is used to increase the temperature of the diatomic gas by ∆T i.e., 2 1 2 m g v0

= ∆U = nCv ∆T = n 52 R∆T,

which gives s v0 =

5nR∆T = mg

s

5(1)(8.314)(1) = 36 m/s. 32 × 10−3

Ans. (a) 1.96 × 1027 /s (b) 36 m/s

Chapter 22 Calorimetry

One Option Correct

Q 2. Water of volume 2 litre in a container is heated with a coil of 1 kW at 27 ◦ C. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27 ◦ C to 77 ◦ C? [Specific heat of water is 4.2 kJ/(kg◦ C).] (2005) (A) 8 min 20 s (B) 6 min 2 s (C) 7 min (D) 14 min

Q 1. A water cooler of storage capacity 120 litres can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30 ◦ C and the entire stored 120 litres of water is initially cooled to 10 ◦ C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is [Specific heat of water is 4.2 kJ kg−1 K−1 and the density of water is 1000 kg/m3 .] (2016)

Sol. The heater coil gives energy at a rate of 1000 J/s, out of which 160 J/s is dissipated through the lid. Thus, energy for heating the water is available at a rate of P = 1000 − 160 = 840 J/s.

The heat energy required to raise the temperature of mass m of water from T1 to T2 is

Device

Cooler

(1)

Hot

Q = mS(T2 − T1 ),

(2)

where S is the specific heat of the water. If t is the time required to raise temperature of mass m from T1 to T2 then, by energy conservation,

Cold

(A) 1600 (B) 2067 (C) 2533 (D) 3933 P t = Q.

Sol. Let Pd = 3 kW be the power generated by the device. Let the water temperature increases from T to T + ∆T in a time interval ∆t. In this time interval, the energy generated by the device is Pd ∆t, the energy pumped out by the cooler is P ∆t and the energy used to increase the temperature of water is mS∆T . The energy of the system is conserved because it is thermally insulated i.e.,

(3)

The density of water is 1000 kg/m3 . Thus, the mass of 2 litre of water is m = 2 kg. Substitute P and Q from equations (1) and (2) into equation (3) to get (2)(4.2 × 103 )(77 − 27) mS(T2 − T1 ) = P 840 = 500 s = 8 min 20 s.

t=

Pd ∆t = P ∆t + mS∆T, Ans. A

which gives P = Pd −

mS∆T ρV S∆T = Pd − . ∆t ∆t

Q 3. One calorie is defined as the amount of heat required to raise temperature of 1 g of water by 1 ◦ C in a certain interval of temperature and at certain pressure. The temperature interval and pressure is (2005) (A) from 14.5 ◦ C to 15.5 ◦ C at 760 mm of Hg (B) from 98.5 ◦ C to 99.5 ◦ C at 760 mm of Hg (C) from 13.5 ◦ C to 14.5 ◦ C at 76 mm of Hg (D) from 3.5 ◦ C to 4.5 ◦ C at 76 mm of Hg

(1)

The minimum value of P occurs when ∆T attains its maximum allowed value i.e., ∆T = 30 ◦ C − 10 ◦ C = 20 ◦ C. Substitute the values of various parameters in equation (1) to get P = 3 × 103 −

(1000)(120 × 10−3 )(4.2 × 103 )(20) 3 × 3600

Sol. One calorie is defined as the heat required to raise the temperature of 1 g of water from 14.5 ◦ C to 15.5 ◦ C at atmospheric pressure. Ans. A

= 2067 W. Ans. B 297

298

Part IV. Thermodynamics

Time

(B)

Temp.

(A)

Temp.

Temp.

(2000)

Time

Sol. Initially, the temperature of liquid oxygen will increase linearly with a slope given by ∆T 1 ∆Q = , ∆t mSL ∆t

(1)

(C)

Heat supplied

(D)

Heat supplied

where m is mass and SL is specific heat of liquid oxygen. This will continue till there is a phase change from liquid state to gaseous state at constant temperature. During the phase change process, liquid oxygen gets converted to gaseous oxygen without any change in temperature. Hereafter, the temperature will again increase linearly with a slope given by 1 ∆Q ∆T = . ∆t mSG ∆t

Heat supplied Temp.

Time

Q 6. A block of ice at −10 ◦ C is slowly heated and converted to steam at 100 ◦ C. Which of the following curves represents the phenomenon qualitatively?

Time

(D)

Temp.

(C)

Hence, the final mass of water in the container is mw + m = 6 kg. We encourage you to prove that the final state of matter cannot be (i) only water or (ii) only ice. Hint: These states violate energy conservation. Ans. B

Temp.

(B)

Temp.

(A)

Temp.

Q 4. Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which of the following graphs represent the variation of temperature with time? (2004)

(2)

Note that ∆Q/∆t in equations (1) and (2) are equal (rate of heating is constant). The slopes of T -t curve are different because specific heats of liquid and gaseous states are different (SL 6= SG ). Ans. C Q 5. 2 kg of ice at −20 ◦ C is mixed with 5 kg of water at 20 ◦ C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are 1 kcal/(kg◦ C) and 0.5 kcal/(kg◦ C) while the latent heat of fusion of ice is 80 kcal/kg. (2003) (A) 7 kg (B) 6 kg (C) 4 kg (D) 2 kg Sol. The heat energy absorbed, when 2 kg of ice at −20 ◦ C is heated to ice at 0 ◦ C, Qice = mi Si ∆T = (2)(0.5)(20) = 20 kcal. The heat energy released, when 5 kg of water at 20 ◦ C is cooled to water at 0 ◦ C,

Heat supplied

Sol. Initially, temperature of the ice increases from −10 ◦ C to 0 ◦ C. The heat supplied and temperature increase are related by ∆Q = mSice ∆T . Thus, the slope of the T -Q curve is ∆T 1 = . ∆Q mSice At 0 ◦ C, the ice melts into water by absorbing mLfusion amount of heat. The temperature remains constant during the melting process. Then, temperature of water starts increasing with a slope 1 ∆T = . ∆Q mSwater At 100 ◦ C, the water vaporizes by absorbing mLvaporization amount of heat. The temperature remains constant during the vaporization process. Ans. A Q 7. Steam at 100 ◦ C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15 ◦ C till the temperature of the calorimeter and its contents rises to 80 ◦ C. The mass of the steam condensed (in kg) is (1986) (A) 0.130 (B) 0.065 (C) 0.260 (D) 0.135 Sol. The heat required to raise the temperature of the water of mass mw = 1.1 kg and calorimeter of water equivalent mc = 0.02 kg from initial temperature Ti = 15 ◦ C to final temperature 80 ◦ C is given by

Qwater = mw Sw ∆T = (5)(1)(20) = 100 kcal. Thus, the energy available to melt the ice is Q = Qwater − Qice = 100 − 20 = 80 kcal, and the mass of ice melted by this energy is m = Q/Lfusion = 80/80 = 1 kg.

Q1 = (mw + mc )S(Tf − Ti ) = (1.1 + 0.02)(1000)(80 − 15) = 72800 cal. The heat released in the condensation of mass ms of steam at 100 ◦ C is Q2 = ms L = ms (540000) cal.

Chapter 22. Calorimetry

299

The energy conservation, Q1 = Q2 , gives ms = 72800/540000 = 0.135 kg = 135 g. We encourage you to find ms , if the final temperature of the condensed steam is reduced to 80 ◦ C. Hint: ms = 130 g. Ans. D One or More Option(s) Correct Q 8. The figure shows variation of specific heat capacity (C) of a solid as a function of temperature (T ). The temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, which of the following statement(s) is (are) correct to a reasonable approximation. (2013)

Fill in the Blank Type Q 9. A substance of mass M kg requires a power input of P watts to remain in the molten state at its melting point. When the power source is turned off, the sample completely solidifies in time t seconds. The latent heat of fusion of the substance is . . . . . . (1992) Sol. The substance remains in the molten state at its melting point if the rate of heat loss is equal to the rate of heat gain, which is P . When source is switched off, the loss continues at the same rate. Thus, the heat lost in time t is Q1 = P t. The heat energy released when mass M is transformed from molten state to solid state is Q2 = M L. The energy conservation, Q1 = Q2 , gives L = P t/M . Ans. PMt J/kg Q 10. 300 g of water at 25 ◦ C is added to 100 g of ice at 0 ◦ C. The final temperature of the mixture is . . . . . . ◦ C.

C

(1989) 100 200 300 400 500

T (K)

(A) the rate at which heat is absorbed in the range 0 − 100 K varies linearly with temperature T . (B) heat absorbed in increasing the temperature from 0 − 100 K is less than the heat required for increasing the temperature from 400 − 500 K. (C) there is no change in the rate of heat absorption in the range 400 − 500 K. (D) the rate of heat absorption increases in the range 200 − 300 K. Sol. The specific heat capacity (C) of a material relates rise in temperature of the material to the heat absorbed by it, ∆Q = mC∆T . Thus, the rate of absorption of heat by a material is given by dQ dT = mC . dt dt The temperature is increased at a constant rate i.e., dT /dt is a constant. Hence, rate of absorption of heat is proportional to the specific heat. From the figure, the heat capacity (and hence the rate of absorption of heat) increase in the temperature range 0 K to 100 K but the variation is not linear. The rate of heat absorption increases in the temperature range 200 K to 300 K but becomes constant in the temperature range 400 K to 500 K. The heat absorbed in increasing the temperature from T1 to T2 , Z T2 Q= mC dT, T1

is proportional to the area under C-T curve. This area is less for temperature range 0 K - 100 K as compared to temperature range 400 K - 500 K. Ans. B, C, D

◦

Sol. When Mw = 300 g mass of water at 25 C is mixed with Mi = 100 g mass of ice at 0 ◦ C, possible final state of the system are (a) A mixture of water (mass Mw + m) and ice (mass Mi − m) at 0 ◦ C. (b) Mass Mw + Mi of water at some temperature T , 0 ◦ C < T < 25 ◦ C. Following changes take place to reach the state (a), (1) Mass Mw of water cools down from 25 ◦ C to 0 ◦ C releasing energy Qw = Mw S∆T = 300(1)(25) = 7500 cal. (2) Mass m of ice transforms from solid state to liquid state absorbing energy Qi = mL = 80m cal. Conservation of energy, Qw = Qi , gives m = 7500/80 = 93.75 g. Thus, the mixture contains 393.75 g of water and 6.25 g of ice at 0 ◦ C. We encourage you to show that the state (b) is not allowed. Hint: energy conservation, Mw S(25−T ) = Mi L+Mi (T −0), gives negative T. Ans. 0 ◦ C Integer Type Q 11. A piece of ice of mass m grams is at −5 ◦ C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 g of ice has melted. Assuming there is no other heat exchange in the process, the value of m is . . . . . . . [Given: ice heat capacity = 2100 J/(kg◦ C) and latent heat = 3.36 × 105 J/kg.] (2010)

Sol. In the final state, ice-water mixture is in equilibrium. Thus, the temperature of m grams of ice is raised from −5 ◦ C to 0 ◦ C. The heat absorbed in this process is Q1 = mS∆T.

(1)

300

Part IV. Thermodynamics

The state of m1 = 1 g of ice is changed from solid to liquid. The heat absorbed in the melting process is

Q (cal) 40500 36000

Q2 = m1 L.

(2)

(e)

32000

(d)

27000 (b)

The heat supplied is Q = 420 J. By energy conservation, Q = Q1 + Q2 . Substitute Q1 and Q2 from equations (1) and (2) to get

Qrel (c) Qabs 4500

420 − (10−3 )(3.36 × 105 ) Q − m1 L = m= S∆T (2100)(5)

(a) 253 273

= 8 × 10−3 kg = 8 g. Ans. 8 Descriptive Q 12. In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Find the final temperature of the mixture (in Kelvin). [Given, Lfusion = 80 cal/g = 336 J/g, Lvaporization = 540 cal/g = 2268 J/g, Sice = 2100 J/kg K = 0.5 cal/(g K), Swater = 4200 J/kg K = 1 cal/(g K).] (2006)

Sol. Possible final states of the system are (i) steam-water mixture at 373 K. (ii) water at temperature Tw . (iii) water-ice mixture at 273 K. (iv) ice at temperature Ti . Let us analyse the transformation to state (iii). To reach state (iii), the ice and the steam undergo following transformations, (a) ice at 253 K to ice at 273 K by absorbing heat Qa = mice Sice ∆T = (450)(0.5)(20) = 4500 cal.

373

T

By energy conservation, absorbed energy should be equal to the released energy. This condition is satisfied at T = 273 K. Analytically, energy conservation, Qa + Qb = Qc + Qd + Qe , gives y = 9x − 55/8. This equation admits multiple feasible solutions for x ∈ [0, 1] and y ∈ [0, 1]. These solutions lie on the line segment joining (0.763, 0) and (0.875, 1). We encourage you to show that the energy conservation does not allow state (i), (ii) and (iv) described above. Ans. 273 K Q 13. An ice cube of mass 0.1 kg at 0 ◦ C is placed in an isolated container which is at 227 ◦ C. The specific heat S of the container varies with temperature T according to the empirical relation S = A + BT , where A = 100 cal/(kg K) and B = 2 × 10−2 cal/(kg K2 ). If the final temperature of the container is 27 ◦ C, determine the mass of the container. [Latent heat of fusion for water = 8 × 104 cal/kg, specific heat of water = 103 cal/(kg K).] (2001) Sol. Let m be the mass of the container. The heat energy released by the container when its temperature decreases from T to T − dT is dQr = mS[(T − dT ) − (T )] = −mSdT = −m(A + BT )dT.

(b) fraction x of mice from ice at 273 K to water at 273 by absorbing heat Qb = xmice Lfusion = x(450)(80) = 36000x cal. (c) steam at 373 K to water at 373 K by releasing heat

Integrate dQr to get the energy released by the container when its temperature decreases from Ti = 273 + 227 = 500 K to Tf = 273 + 27 = 300 K, Z Tf Qr = − m(A + BT )dT Ti

Qc = msteam Lvaporization = (50)(540) = 27000 cal. (d) water at 373 K to water at 273 K by releasing heat Qd = msteam Swater ∆T = (50)(1)(100) = 5000 cal. (e) fraction y of msteam from water at 273 K to ice at 273 K by releasing heat Qe = ymsteam Lfusion = y(50)(80) = 4000y cal.

1 = −mA(Tf − Ti ) − mB(Tf2 − Ti2 ) 2 = −(m)(100)(300 − 500) 1 − (m)(2 × 10−2 )(3002 − 5002 ) 2 = 2.16 × 104 m. The temperature of ice/water increases from T0 = 273 K to Tf = 300 K. Heat absorbed by the ice cube is Qa = mi L + mi Si (Tf − T0 )

The figure shows the variation of absorbed energy (Qabs = Qa + Qb ) and released energy (Qrel = Qc + Qd + Qe ) with temperature.

= (0.1)(8 × 104 ) + (0.1)(103 )(300 − 273) = 1.07 × 104 cal.

Chapter 22. Calorimetry

301

By energy conservation, Qr = Qa . Substitute values to get m = 0.495 kg. Ans. 0.495 kg Q 14. A 5 m long cylindrical steel wire with radius 2 × 10−3 m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring losses. [For the steel wire: Young’s modulus = 2.1 × 1011 Pa, density = 7860 kg/m3 , specific heat = 420 J/kg K.] (2001) Sol. The length and radius of the wire are L = 5 m and r = 2 × 10−3 m. For the wire material, density is ρ = 7860 kg/m3 , specific heat is S = 420 J/kg K, and Young’s modulus is Y = 2.1 × 1011 Pa. The mass of the bob is M = 100 kg. The force acting on the wire is equal to the weight of the bob i.e., F = M g. The cross section area of the wire is A = πr2 . Thus, the longitudinal stress acting on the wire is Stress =

Mg F = . A πr2

(1)

The definition of Young modulus, Y = stress/strain, and equation (1) gives strain in the wire as strain =

l Mg = 2 . L πr Y

(2)

Thus, elastic potential energy of the stretched wire can be expressed as U=

1 1 M 2 g2 L stress × strain × volume = . 2 2 πr2 Y

(3)

When the wire is snapped, stored elastic energy is released and temperature of the wire increases. The heat energy required to raise the temperature of a mass m by an amount ∆T is Q = mS∆T = (πr2 Lρ)S∆T.

(4)

The conservation of energy gives U = Q. Substitute U and Q from equations (3) and (4) to get ∆T = =

1 M 2 g2 2 π 2 r4 ρSY 2(3.14)2 (2

In this process, mass m of steam at temperature Ts = 100 ◦ C is transformed to water at temperature Ts by releasing the heat, Qr,1 = mL.

(1)

Then, mass m of transformed water at temperature Ts is cooled to a temperature T2 by releasing heat Qr,2 = mSw (Ts − T2 ).

(2)

The mass mw of water is heated from temperature T1 to temperature T2 by absorbing heat Qa = mw Sw (T2 − T1 ).

(3)

By energy conservation, Qr,1 + Qr,2 = Qa . Using equations (1)–(3), we get (100)(1)(90 − 24) mw Sw (T2 − T1 ) = L + Sw (Ts − T2 ) 540 + (1)(100 − 90) = 12 g.

m=

Ans. 12 g Q 16. A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is 27 ◦ C. [For lead, melting point = 327 ◦ C, specific heat = 0.03 cal/g-◦ C, latent heat of fusion = 6 cal/g and J = 4.2 J/cal.] (1981) Sol. Let v be the velocity of the bullet of mass m kilogram. The bullet loses 25% of its kinetic energy to the obstacle. Thus, the energy used to melt the bullet is Q = 0.75( 12 mv 2 ). In the melting process, (a) the temperature of the bullet is raised from Ti = 27 ◦ C to melting point Tf = 327 ◦ C by absorbing heat Qa = mS∆T = m(4.2)(0.03 × 103 )(327 − 27) = 37.8 × 103 m, and, (b) the bullet is transformed from solid to liquid at its melting point by absorbing heat

×

(100)2 (9.8)2 (7860)(420)(2.1 × 1011 )

10−3 )4

= 4.39 × 10−3 K. Ans. 4.39 × 10−3 K Q 15. The temperature of 100 g of water is to be raised from 24 ◦ C to 90 ◦ C by adding steam to it. Calculate the mass of the steam required for this purpose. (1996) Sol. Let m be the mass of the steam required to raise the temperature of mass mw = 100 g of water from temperature T1 = 24 ◦ C to temperature T2 = 90 ◦ C.

Qb = mL = m(4.2)(6 × 103 ) = 25.2 × 103 m. Apply conservation of energy, Q = Qa + Qb , to get r r 2(Qa + Qb ) 2(37.8m + 25.2m)103 v= = 0.75m 0.75m = 409.8 m/s. Ans. 409.8 m/s

Chapter 23 Laws of Thermodynamics

One Option Correct

Q 3. The figure shows the p-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then, (2009)

Q 1. In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas, (2001) (A) the temperature will decrease. (B) the volume will increase. (C) the pressure will remain constant. (D) the temperature will increase.

p A

3 2 D

Sol. First law of thermodynamics, dQ = dU + dW , gives dU < 0. For an ideal gas, internal energy decreases due to decrease in temperature. Ans. A

B

1

0

C 1

2

3

V

(A) the process during the path A → B is isothermal. (B) heat flows out of the gas during the path B → C → D. (C) work done during the path A → B → C is zero. (D) positive work is done by the gas in the cycle ABCDA.

One or More Option(s) Correct Q 2. One mole of a monatomic ideal gas undergoes a cyclic process as shown in the figure (where V is the volume and T is the temperature). Which of the statements below is (are) true? (2018) T

Sol. The process A → B lies on a circle having centre (2, 2) and radius 1 unit. The equation of the circle is given by

II

I

III

(p − 2)2 + (V − 2)2 = 1. IV V

(A) (B) (C) (D)

The ideal gas equation is pV = nRT . The product pV is not constant in the process A → B and hence the process is not isothermal. In the process B → C → D, the product pV decreases from 6 units to a value less than 2 units. Thus, the temperature of the gas reduces in this process. Hence, internal energy of the gas decreases i.e., ∆U < 0. Also, work done by the gas is negative, ∆W < 0 (since volume decreases). Substitute in the first law of thermodynamics, ∆Q = ∆U + ∆W , to get ∆Q < 0. The negative sign of ∆Q indicates heat flows out of the gas. The work done by the gas is given by area under the p-V diagram. It is positive for the process A → B (volume increases) and negative for the process B → C (volume decreases). The magnitude of area under AB is more than area under BC. Note that,

Proecess I is an isochoric process. In process II, gas absorbs heat. In process IV, gas releases heat. Process I and III are not isobaric.

Sol. The volume is constant in an isochoric process. It is parallel to T axis on a T -V diagram. The process II is isothermal expansion as temperature is constant and volume is increased. The change in internal energy of a gas in an isothermal process is zero i.e., ∆U = 0. The work done byRthe gas in an expansion process is positive i.e., ∆W = pdV > 0. By first law of thermodynamics, ∆Q = ∆U + ∆W = ∆W > 0, i.e., heat is absorbed by the gas. The process IV is isothermal compression as temperature is constant and volume is decreased. Thus, ∆U = 0, ∆W < 0 and ∆Q = ∆W < 0 i.e., heat is released by the gas. The pressure is constant in an isobaric process. It is a straight line, T = (p/nR)V , on a T -V diagram. Can you relate the given cycle with Carnot cycle? Ans. (B), (C), (D)

WAB = 2 + π/2, WBC = −(1 + π/2), WABC = WAB + WBC = 1. Thus, the work is done by the gas in the process A → B → C as WABC > 0. 302

Chapter 23. Laws of Thermodynamics

303

Using the same argument for the cycle ABCDA, positive work is done by the gas in this cycle. This work is equal to the area enclosed by the curve ABCDA. Ans. B, D Q 4. During the melting of slab of ice at 273 K at atmospheric pressure, (1998) (A) positive work is done by the ice-water system on the atmosphere. (B) positive work is done on the ice-water system by the atmosphere. (C) the internal energy of the ice-water system increases. (D) the internal energy of the ice-water system decreases. Sol. The volume of ice-water system decreases from Vi to Vf when ice melts at 273 K. Thus, the work done by the ice-water system on the atmosphere, ∆W = p(Vf − Vi ) < 0,

Q 6. While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the from   piston  the top is  p0 πR2 −M g 2p0 πR2 2L (A) πR2 p0 +M g 2L (B) 2    πR p02  p0 πR2 +M g p0 πR (C) 2L (D) πR2 p0 −M g 2L πR2 p0 Sol. Let x be the distance of the piston from the top and p be the pressure inside the cylinder. Apply Boyle’s law, p0 (πR2 2L) = p(πR2 x), to get p = 2p0 L/x.

(1)

The forces acting on the piston are its weight M g, force due to inside gas pπR2 , and force due to outside atmosphere p0 πR2 .

(1)

Mg

is negative and the work done on the ice water-system by the atmosphere is positive. The heat absorbed by the system in melting mass m of ice, ∆Q = mL > 0,

p0 πR2

In equilibrium condition, M g + pπR2 = p0 πR2 .

(2)

is positive, where L is the latent heat of fusion. Using equations (1) and (2) in the first law of thermodynamics, ∆Q = ∆U + ∆W , we get,

pπR2

x

(2)

Substitute p from equation (1) into equation (2) and solve to get x=

∆U = Uf − Ui = ∆Q − ∆W > 0.

2p0 πR2 L . p0 πR2 − M g Ans. D

Ans. B, C Paragraph Type Paragraph for Questions 5-7 A fixed thermally conducting cylinder has a radius R and height L0 . The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface as shown in the figure. The atmospheric pressure is p0 . (2007)

Q 7. The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is ρ. In equilibrium, the height H of the water column in the cylinder satisfies

2R

L0

L

H L0

Piston

Q 5. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be Mg p0 Mg (A) p0 (B) p20 (C) p20 + πR 2 (D) 2 − πR2 Sol. Since hole at the top is open, the pressure inside the cylinder is equal to the pressure outside it, which is p0 . Ans. A

(A) (B) (C) (D)

ρg(L0 − H)2 + p0 (L0 − H) + L0 p0 ρg(L0 − H)2 − p0 (L0 − H) − L0 p0 ρg(L0 − H)2 + p0 (L0 − H) − L0 p0 ρg(L0 − H)2 − p0 (L0 − H) + L0 p0

=0 =0 =0 =0

Sol. Let pressure inside the cylinder be p. Initial pressure and volume of the cylinder are p0 and πR2 L0 . Apply Boyle’s law, p0 πR2 L0 = p πR2 (L0 − H), to get the pressure of air inside the cylinder p = p0 L0 /(L0 − H).

(1)

304

Part IV. Thermodynamics

The hydrostatic pressure at the height H of the water column in the cylinder is given by ph = p0 + ρg(L0 − H).

(2)

In equilibrium, the inside pressure is equal to the hydrostatic pressure i.e., p = ph . Substitute p and ph from equations (1) and (2) to get ρg(L0 − H)2 + p0 (L0 − H) − p0 L0 = 0. Ans. C

Integer Type Q 9. A thermodynamic system is taken from an initial state i with internal energy Ui = 100 J to the final state f along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along the path af, ib and bf are Waf = 200 J, Wib = 50 J and Wbf = 100 J respectively. The heat supplied to the system along the path iaf, ib and bf are Qiaf , Qib and Qbf respectively. If the internal energy of the system in the state b is Ub = 200 J and Qiaf = 500 J, the ratio Qbf /Qib is . . . . . . . (2014) p

Matrix or Matching Type Q 8. Match the following for the given process,

(2006)

p 30

i

Process Process Process Process

V

M K

L

10

20

Column I (A) (B) (C) (D)

b

J

20 10

f

a

J →K K→L L→M M →J

V (m3 )

Column II (p) (q) (r) (s)

W >0 Q 0. The temperature TL > TK because volume VL > VK . Thus, ∆U = UL − UK > 0. Hence, ∆Q = ∆U + ∆W > 0. Process L → M is isochoric with ∆W = 0. The temperature TM > TL because pressure pM > pL . Thus, ∆U = UM − UL > 0. Hence, ∆Q = ∆U + ∆W > 0. In the process M → J, pJ VJ = 300 and pM VM = 400. The temperature TJ < TM because pJ VJ < pM VM . Thus, ∆U = UJ − UM < 0. The work done by the gas is negative because of decrease in its volume i.e., ∆W < 0. The first law of thermodynamics gives ∆Q = ∆U + ∆W < 0. Ans. A7→q, B7→(p,s), C7→s, D7→(q,r)

Sol. In a thermodynamics process, the heat supplied to the system, the increase in internal energy of the system, and the work done by the system are related by the first law of thermodynamics, ∆Q = ∆U + ∆W . The first law of thermodynamics for the process iaf gives Qiaf = Uiaf + Wiaf = (Uf − Ui ) + (Wia + Waf ). (1) Substitute Qiaf = 500 J, Ui = 100 J, Wia = 0 (constant volume), and Waf = 200 J in equation (1) to get Uf = 400 J. In the process ib, Qib = Uib + Wib = (Ub − Ui ) + Wib .

(2)

Substitute Ub = 200 J, Ui = 100 J, and Wib = 50 J in equation (2) to get Qib = 150 J. In the process bf, Qbf = Ubf + Wbf = (Uf − Ub ) + Wbf .

(3)

Substitute Uf = 400 J, Ub = 200 J and Wbf = 100 J in equation (3) to get Qbf = 300 J. Thus, Qbf /Qib = 300/150 = 2. Ans. 2 Descriptive Q 10. A metal of mass 1 kg at constant atmospheric pressure and at initial temperature 20 ◦ C is given a heat of 20000 J. Find (a) change in temperature (b) work done and (c) change in internal energy. [Given: Specific heat = 400 J/kg ◦ C, coefficient of cubical expansion γ = 9 × 10−5 /◦ C, density ρ = 9000 kg/m3 , atmospheric pressure = 105 N/m2 .] (2005)

Chapter 23. Laws of Thermodynamics

305

Sol. The heat given to the metal is related to rise in its temperature by ∆Q = mS∆T.

WBC = nRTB ln(VC /VB )

Substitute ∆Q = 20000 J, m = 1 kg, and S = 400 J/kg ◦ C to get ∆T = 50 ◦ C. Initial volume of metal is V0 = m/ρ = 1/9000 m3 . The change in volume due to thermal expansion is given by ∆V = V0 γ∆T = (1/9000)(9 × 10 −7

= 5 × 10

−5

)(50)

m .

5

∆W = p0 ∆V = (10 )(5 × 10

−7

) = 0.05 J.

The first law of thermodynamics, ∆Q = ∆U + ∆W , gives ∆U = ∆Q − ∆W = 20000 − 0.05 = 19999.95 J. Ans. (a) 50 ◦ C (b) 0.05 J (c) 19999.95 J Q 11. A monatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure. The volume ratio are VB /VA = 2 and VD /VA = 4. If the temperature TA at A is 27 ◦ C, calculate, [Give answer in terms of the gas constant R.] (2001) V VD D

VB

and change in the internal energy in this process is ∆UBC = 0. The first law of thermodynamics gives QBC = ∆UBC + WBC = 831.6R.

QCD = nCv ∆T = nCv (TD − TC ) = nCv (TA − TB ) = −900R. The work done by the gas in the isothermal process DA is WDA = nRTA ln(VA /VD ) = 2 × R × 300 ln(1/4) = −831.6R, and change in the internal energy in this process is ∆UDA = 0. The first law of thermodynamics gives QDA = ∆UDA + WDA = −831.6R. In a complete cycle, ∆U = 0. Apply the first law of thermodynamics for a complete cycle, ∆Q = ∆U +∆W , to get ∆W = ∆Q = QAB + QBC + QCD + QDA

C

= 1500R + 831.6R − 900R − 831.6R = 600R. Ans. (a) 600 K (b) 1500R, 831.6R, −900R, −831.6R (c) 600R

B A TA

= (2)(R)(600) ln 2 = 831.6R,

The process CD is isochoric so the absorbed heat is given by

3

The work done in changing the volume by ∆V against the atmospheric pressure p0 is given by

VA O

The work done by the gas in the isothermal process BC is

TB

T

(a) the temperature of the gas at point B. (b) heat absorbed or released by the gas in each process. (c) the total work done by the gas during the complete cycle. Sol. The straight line AB passes through the origin O. Thus, V /T is a constant on line AB (isobaric process). Using, VA /TA = VB /TB , we get TB = (VB /VA )TA = 2(273 + 27) = 600 K. For a monatomic ideal gas, Cv = 32 R and Cp = 52 R. The heat absorbed in the isobaric process AB is QAB = nCp (TB − TA ) = (2)(5R/2)(600 − 300) = 1500R.

Q 12. Two moles of an ideal monatomic gas is taken through a cycle ABCA as shown in the p-T diagram. During the process AB, pressure and temperature of the gas vary such that pT = constant. If T1 = 300 K, calculate, [Give answers in terms of the gas constant R.] (2000) p 2p1

B

p1

C

A T1

2T1

T

(a) the work done on the gas in the process AB. (b) the heat absorbed or released by the gas in each of the processes.

306

Part IV. Thermodynamics

Sol. Differentiate the ideal gas equation, pV = nRT , to get pdV + V dp = nRdT.

(1)

In the process AB, pT = constant. Differentiate this to get pdT + T dp = 0.

(2)

Eliminate dp from equations (1) and (2) to get pdV = 2nRdT . The work done by the gas in the process AB is Z Z TB WAB = pdV = 2nRdT = 2nR(TB − TA )

Sol. The p-V diagrams of the given processes are shown in the figure. Given, TA = 300 K, γ = 1.4, n = 1, VA /VB = 16, and VC /VB = 2. For the adiabatic compression AB, TA VAγ−1 = TB VBγ−1 , which gives TB = (VA /VB )γ−1 TA = (16)1.4−1 × 300 = 909 K. p B

C

TA

= 2(2)R(300 − 600) = −1200R.

D

(3)

A

The work done on the gas in this process is 1200R. The change in the internal energy of n moles of an ideal gas, with f degrees of freedom, is given by ∆U = n(f /2)R∆T . The degrees of freedom for a monatomic gas is three. Thus, for the process AB, (4)

Use equations (3) and (4) in the first law of thermodynamics to get heat absorbed in the process AB as QAB = ∆UAB + WAB = −900R − 1200R = −2100R. The heat released in this process is 2100R. The process BC is isobaric. The work done by the gas, increase in the internal energy of the gas, and heat absorbed by the gas are given by WBC = pB (VC − VB ) = nR(TC − TB ) = 600R, f ∆UBC = n R(TC − TB ) = 900R, 2 QBC = ∆UBC + WBC = 1500R. You may use QBC = nCp ∆T to get the heat absorbed. The process CA is isothermal with ∆UCA = 0. The work done by the gas and the heat absorbed in this process are given by WCA = nRTC ln (VA /VC ) = nRTC ln (pC /pA ) = 2R(600) ln 2 = 831.6R,

V

In the isobaric process BC, VC /TC = VB /TB , which gives

In the adiabatic expansion CD, TC VCγ−1 = TD VDγ−1 , which gives TD = (VC /VD )γ−1 TC = (2/16)1.4−1 × 1818 = 791.4 K. The efficiency of a cycle is defined as the ratio of net work done by the gas to the heat absorbed. For the adiabatic processes, QAB = QCD = 0. The specific heats for a diatomic gas are Cv = 5R/2 and Cp = 7R/2. The heats absorbed in the isobaric process BC and isochoric process DA are QBC = nCp (TC − TB ) = 1(7/2)(8.314)(1818 − 909) = 26451 J, QDA = nCv (TA − TD ) = 1(5/2)(8.314)(300 − 791.4) = −10214 J. The negative sign in QDA indicates that the heat is released. Thus, the heat absorbed in the cycle ABCD is QAbsorbed = QBC = 26451 J. The work done by the gas in the given processes are

QCA = ∆UCA + WCA = 831.6R. QBC

16V0

TC = (VC /VB )TB = 2 × 909 = 1818 K.

∆UAB = n(f /2)R(TB − TA ) = 2(3/2)R(300 − 600) = −900R.

V0 2V0

Ans. (a) 1200R (b) QAB = −2100R, = 1500R, QCA = 831.6R

Q 13. One mole of a diatomic gas (γ = 1.4) is taken through a cyclic process starting from A. The process A → B is an adiabatic compression, B → C is isobaric expansion, C → D an adiabatic expansion and D → A is isochoric. The volume ratio are VA /VB = 16 and VC /VB = 2 and the temperature at A is TA = 300 K. Calculate the temperature of the gas at the points B and D and find the efficiency of the cycle. (1997)

nR(TA − TB ) γ−1 (1)(8.314)(300 − 909) = −12658 J, = 1.4 − 1 nR(TC − TD ) = γ−1 (1)(8.314)(1818 − 791.4) = = 21338 J, 1.4 − 1 = pC VC − pB VB = nR(TC − TB ) = 7557 J,

WAB =

WCD

WBC

WDA = 0.

Chapter 23. Laws of Thermodynamics Thus, the net work done in the cycle ABCDA is WABCD = WAB + WBC + WCD + WDA = −12658 + 7557 + 21338 + 0 = 16237 J, and the efficiency of the cycle is η=

−QReleased We encourage you to show that η = QAbsorbed × QAbsorbed 100. Ans. TB = 909 K, TD = 791.4 K, η = 61.4%

Q 14. A sample of 2 kg monatomic helium (assume ideal) is taken through the process ABC and another sample of 2 kg of the same gas is taken through the process ADC (see figure). Given molecular mass of helium = 4. (1997)

5

A 10

D 20

V (m3 )

Sol. The number of moles of helium is n = m/M = (2 × 103 )/4 = 500. The ideal gas equation, pV = nRT , at the point A, B, C, and D gives TA = TB = TC = TD =

= 2.25 × 106 J. WABC = WAB + WBC = 0 + pB (VC − VB ) = 106 J. The first law of thermodynamics gives the heat absorbed in the process as QABC = ∆UABC + WABC = 3.25 × 106 J. Similarly, for the process ADC, increase in the internal energy, work done by the gas, and heat absorbed in the process are given by ∆UADC = 2.25 × 106 J, WADC = WAD + WDC = pA (VB − VA ) + 0 = 0.5 × 106 J,

Ans. (a) TA = 120.28 K, TB = 240.56 K, TC = 481.11 K, TD = 240.56 K (b) No (c) QABC = 3.25 × 106 J, QADC = 2.75 × 106 J

C

(a) What is the temperature of helium in each of the states A, B, C and D? (b) Is there any way of telling afterwards which sample of helium went through the process ABC and which went through the process ADC? Write Yes or No. (c) How much is the heat involved in the process ABC and ADC?

pA VA nR pB VB nR pC VC nR pD VD nR

∆UABC = UC − UA = (3/2)nR(TC − TA )

QADC = ∆UADC + WADC = 2.75 × 106 J.

p(×104 N/m2 ) B

f = 3 is the degrees of freedom. Thus, the change in internal energy in the process ABC is

The work done by the gas in the process ABC is

WABCD 16237 × 100 = × 100 = 61.4%. QAbsorbed 26451

10

307

(5 × 104 )(10) = = 120.28 K, (500)(8.314) (10 × 104 )(10) = = 240.56 K, (500)(8.314) (10 × 104 )(20) = = 481.11 K, (500)(8.314) (5 × 104 )(20) = = 240.56 K. (500)(8.314)

The parameters like pressure, volume, and temperature are the state functions i.e., they only describe the state of the gas at a point and not the path taken by the gas to reach that point. Hence, it is not possible to tell afterwards which sample went through ABC or ADC. The internal energy of n moles of a monatomic gas at temperature T is given by, U = (f /2)nRT , where

Q 15. An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are Q1 = 5960 J, Q2 = −5585 J, Q3 = −2980 J and Q4 = 3645 J, respectively. The corresponding quantities of work involved are W1 = 2200 J, W2 = −825 J, W3 = −1100 J and W4 respectively. (1994) (a) Find the value of W4 . (b) What is the efficiency of the cycle? Sol. The change in the internal energy of an ideal gas in a cyclic process is zero i.e., ∆U = 0. Apply the first law of thermodynamics, ∆Q = ∆U + ∆W , to get Q1 + Q2 + Q3 + Q4 = W1 + W2 + W3 + W4 , which gives W4 = Q1 + Q2 + Q3 + Q4 − W1 − W2 − W3 = 5960 + (−5585) + (−2980) + 3645 − 2200 − (−825) − (−1100) = 765 J. The efficiency of a cycle is the ratio of total work done to the heat absorbed. The heat absorbed in a process is taken as positive. Thus, Q1 and Q4 denote that the heat is absorbed whereas Q2 and Q3 denote that the heat is released. The efficiency of the cycle is given by W1 + W2 + W3 + W4 Q1 + Q4 2200 + (−825) + (−1100) + 765 = 5960 + 3645 = 0.108 = 10.8 %.

η=

308

Part IV. Thermodynamics Ans. (a) 765 J (b) 10.8 %

Q 16. Two moles of helium gas undergo a cyclic process as shown in the figure. Assuming the gas to be ideal, calculate the following quantities in this process. (1992) p(atm.) 2

A

1 D 300

B

C 400

T (◦ K)

(a) The net change in the heat energy. (b) The net work done. (c) The net change in internal energy. Sol. Helium is a monatomic gas with Cp = 25 R. The heat absorbed in the isobaric processes AB and CD are QAB = nCp (TB − TA ) = (2)(5/2)(8.314)(400 − 300) = 4157 J, QCD = nCp (TD − TC ) = (2)(5/2)(8.314)(300 − 400) = −4157 J. In the isothermal processes BC and DA, ∆UBC = 0 and ∆UDA = 0. By the first law of thermodynamics, ∆Q = ∆U +∆W , the heats absorbed in these processes are given by QBC = WBC = nRTB ln (pB /pC ) = 2(8.314)(400) ln(2/1) = 4610 J, QDA = WDA = nRTD ln (pD /pA ) = 2(8.314)(300) ln(1/2) = −3457 J. Thus, the net heat exchange in the cycle ABCDA is Qnet = QAB + QBC + QCD + QDA = 4157 + 4610 − 4157 − 3457 = 1152 J. The change in the internal energy of the system in a cycle is zero i.e., ∆Unet = 0. The first law of thermodynamics gives the net work done by the gas as Wnet = Qnet − ∆Unet = 1152 J. Ans. (a) 1152 J (b) 1152 J (c) zero

Chapter 24 Specific Heat Capacities of Gases

One Option Correct

final state (pf , Vf , Tf ) is given by ∆U = Uf − Ui = f2 (nRTf − nRTi )

Q 1. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure pi = 105 Pa and volume Vi = 10−3 m3 changes to a final state pf = (1/32) × 105 Pa and Vf = 8 × 10−3 m3 in an adiabatic quasi-static process, such that p3 V 5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at pi followed by an isochoric (isovolumetric) process at volume Vf . The amount of heat supplied to the system in the two-step process is approximately (2016) (A) 112 J (B) 294 J (C) 588 J (D) 813 J

= =

The change in internal energy ∆U is same in the two processes because it is a state function. Apply first law of thermodynamics, ∆Q = ∆U + W , to get heat supplied to the gas as ∆Q = ∆U + W = −112.5 + 700 = 587.5 J. Ans. C Q 2. 5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. Taking the initial temperature to be T1 , the work done in the process is (2011) 9 (A) 89 RT1 (B) 32 RT1 (C) 15 RT (D) RT 1 1 8 2 Sol. For an ideal gas, pV = nRT . In an adiabatic process, T V γ−1 = constant, and the work done by the gas is given by p1 V1 − p2 V2 nRT1 − nRT2 = γ−1 γ−1 "  γ−1 # V1 nRT1 1− , = γ−1 V2

W =

where V1 = 5.6 litre and V2 = 0.7 litre are initial and final volume of the gas. Now, 5.6 litre at STP gives n = 5.6/22.4 = 1/4 mol. Helium, a monatomic gas, has γ = 5/3. Substitute these values to get W = − 98 RT1 . Negative sign indicates that the work is done on the gas. Ans. A

1 2 f

Vi




= 105 (8 × 10−3 − 10−3 ) = 700 J.

Substitute γ = 5/3 to get f = 3 (given gas is monatomic).

pf

× 105 × 8 × 10−3 − 105 × 10−3

W = pi (Vf − Vi )

dU f = R, dT 2 f +2 Cp = Cv + R = R, 2 Cp f +2 γ= = . Cv f

i

1 32

Negative sign of ∆U indicates that the internal energy of final state is less than that of the initial state. The work done by the gas in the isochoric process is zero and in the isobaric process is

Cv =

pi

(pf Vf − pi Vi )

= −112.5 J.

Sol. The equation, p3 V 5 = constant, of given adiabatic process can be written as pV 5/3 = constant. Compare this with equation for adiabatic process, pV γ = constant, to get ratio of the specific heats γ = 5/3. Let f be the degree of freedom of the molecules of the gas. The internal energy of one mole gas at temperature T is given by U = f2 RT . The specific heats at constant volume Cv , specific heat at constant pressure Cp , and the ratio of the specific heats are given by

p

f 2 3 2

Vf

V

Q 3. An ideal gas is expanding such that P T 2 = constant. The coefficient of volume expansion of the gas is (2008) (A) 1/T (B) 2/T (C) 3/T (D) 4/T

The increase in internal energy of n moles of the gas when it goes from the initial state (pi , Vi , Ti ) to the 309

310

Part IV. Thermodynamics

Sol. The coefficient of volume expansion is defined as (A) p A

1 dV γ= . V dT

(B) p A

(1)

The ideal gas equation, P V = nRT , gives P = nRT /V . Substitute P in P T 2 = constant to get V = kT 3 ,

C V

B

(C) p A

C

B

V

(D) p A

(2)

for some constant k. Differentiate equation (2) with respect to T and substitute in equation (1) to get γ = 3/T . Ans. C Q 4. An ideal gas expands isothermally from a volume V1 to V2 and then compressed to original volume V1 adiabatically. Initial pressure is p1 and final pressure is p3 . The total work done is W . Then, (2004) (A) p3 > p1 , W > 0 (B) p3 < p1 , W < 0 (C) p3 > p1 , W < 0 (D) p3 = p1 , W = 0 Sol. On a p-V diagram, magnitude of slope of an adiabatic process is greater than that of an isothermal process (adiabatic process is steeper). This fact is applied at the point B to draw the isothermal expansion process AB and the adiabatic compression process BC. p

B

C V

C

B

V

Sol. The process AB is isothermal and BC is isobaric. The product pV is constant in an isothermal process and hence AB is a rectangular hyperbola on a p-V diagram. For the isobaric process BC, TC > TB , and hence VC > VB . The analysis till this point gives (A) as a possible option. Note: The question gives additional incorrect information that AC is adiabatic. The adiaγ batic process on a p-T diagram is given by p = kT γ−1 , γ p dp (k > 0, is constant) and hence slope dT = γ−1 T > 0 for any γ > 1. Ans. A Q 6. An ideal gas is taken through the cycle ABCA, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is (2002)

C

p3

V (m3 )

p1

2

A

C

B

B V1

V2

V

1

A p(N/m2 )

10

Note that the slope is negative in both the cases because it is defined as the tangent (tan) of an angle from the positive x-axis measured in the anti-clockwise direction. From the graph p3 > p1 . The work done by the gas is the area under the p-V curve and it is considered positive for expansion and negative for contraction. Thus, WAB is positive and WBC is negative and |WBC | > |WAB |. Thus, total work done W = WAB + WBC < 0. We encourage you to find the expressions for p3 and W in terms of the given parameγ−1 ters. Hint: p3 = p1 (V2 /V1 ) , WAB = p1 V1 ln VV21 , and
p1 V1 γ−1 WBC = γ−1 1 − (V2 /V1 ) . Ans. C Q 5. The p-T diagram for an ideal gas is shown in the figure, where AC is an adiabatic process. Corresponding p-V diagram is given by (2003) p A

(A) −5 J (B) −10 J (C) −15 J (D) −20 J Sol. The work done by the gas in the isochoric process BC is WBC = 0 and in the isobaric process AB is WAB = p(VB − VA ) = 10(2 − 1) = 10 J. In the cycle ABCA, heat supplied to the gas is QABCA = 5 J and change in the internal energy of the gas is ∆U = 0. Apply the first law of thermodynamics QABCA = ∆U + WABCA = ∆U + (WAB + WBC + WCA ), to get, WCA = −5 J. Ans. A Q 7. p-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to (2001) p

B

1

C T

2

V

Chapter 24. Specific Heat Capacities of Gases

(A) He and O2 (C) He and Ar

311 a temperature T2 by releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1 /T2 is given by

(B) O2 and He (D) O2 and N2

Sol. The slope of an adiabatic process, pV γ constant, is given by

=

p dp = −γ . dV V Thus, the slope at a given point (V, p) is proportional to γ. Comparing slopes at the common point of the given curves, we get, γ1 < γ2 . Thus, γ1 (= 75 ) corresponds to diatomic O2 and γ2 (= 53 ) to monatomic He. Ans. B Q 8. Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2 if purely isobaric and W3 if purely adiabatic, then, (2000) (A) W2 > W1 > W3 (B) W2 > W3 > W1 (C) W1 > W2 > W3 (D) W1 > W3 > W2 Sol. Let p1 , V1 , and T1 be the initial pressure, volume, and temperature of the gas. For an ideal gas, p1 V1 = nRT1 . The work done by the gas in the isothermal, isobaric, and adiabatic processes are given by   V2 W1 = p1 V1 ln , V1 W2 = p1 (V2 − V1 ), "

W3 =

p1 V1 p1 V1 − p2 V2 = 1− γ−1 γ−1



V1 V2

γ−1 # ,

(2000) 2/3

(A) (L1 /L2 ) (C) L2 /L1

(B) L1 /L2 2/3 (D) (L2 /L1 )

Sol. Let A be the cross-sectional area of the cylinder, V1 = AL1 be its initial volume, and V2 = AL2 be its final volume. The adiabatic equation, pV γ = constant, can be written as T V γ−1 = constant by using ideal gas equation pV = nRT . Thus, T1 V1γ−1 = T2 V2γ−1 .

(1)

For a monatomic gas, γ = Cp /Cv = 5/3. Substitute γ, V1 , and V2 in equation (1) to get 5

T1 /T2 = (V2 /V1 ) 3 −1 = (L2 /L1 )2/3 . Ans. D Q 10. Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is (1998) (A) 30 K (B) 18 K (C) 50 K (D) 42 K Sol. The piston in the cylinder A is free to move. Hence pressure of the gas is constant and the heat is given to it at constant pressure i.e., QA = nCp ∆TA .

where p2 and V2 are the final pressure and volume. If a relationship among W1 , W2 , and W3 holds in general then it should also hold good for any special case, say V2 = 2V1 and γ = 2. In this case, W1 = 0.693p1 V1 , W2 = p1 V1 , and W3 = 0.5p1 V1 , which gives W2 > W1 > W3 . Aliter: The isothermal, isobaric, and adiabatic processes starting at the same initial point (p1 , V1 ) and with same final volume V2 are drawn in the figure. p p1

2

The piston of the cylinder B is fixed. Hence the volume of the gas is constant and the heat is given at constant volume i.e., QB = nCv ∆TB .

(2)

The ratio of specific heats for a diatomic gas is Cp /Cv = 7/5. The heat given to the two gases is equal, QA = QB . Substitute QA and QB from equations (1) and (2), to get ∆TB = (Cp /Cv ) ∆TA = (7/5) 30 = 42 K.

1

Ans. D

3 V1

(1)

V2

V

The work done in a process is equal to the area under the curve on a p-V diagram. It is clear from the diagram that W2 > W1 > W3 . Ans. A Q 9. A monatomic ideal gas, initially at temperature T1 , is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to

Q 11. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V . The mass of the gas in A is mA and that in B is mB . The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V . The changes in the pressure in A and B are found to be ∆p and 1.5∆p, respectively. Then, (1998)

(A) 4mA = 9mB (C) 3mA = 2mB

(B) 2mA = 3mB (D) 9mA = 4mB

312

Part IV. Thermodynamics

Sol. Let M be the molecular mass of the gas. Number of moles of the gas in two containers are nA = mA /M and nB = mB /M . The product pV is constant in an isothermal process. Thus, for the two containers, pA VA = p0A VA0 ,

pB VB = p0B VB0 .

Substitute VA = VB = V and VA0 = VB0 = 2V to get pA − p0A = 21 pA = ∆p,

pB − p0B = 12 pB = 1.5∆p.

Divide these to get pA /pB = 2/3. The ideal gas equations for the two containers, pA V = (mA /M )RT,

pB V = (mB /M )RT,

give mA /mB = pA /pB = 2/3. Ans. C Q 12. When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is (1990) (A) 2/5 (B) 3/5 (C) 3/7 (D) 5/7 Sol. The change in the internal energy of an ideal gas, when its temperature changes by ∆T , is given by ∆U = nCv ∆T.

(1)

The heat supplied at constant pressure, to increase the gas temperature by ∆T , is given by ∆Q = nCp ∆T.

(2)

Divide equation (1) by (2) to get ∆U Cv 5 1 = = = , ∆Q Cp γ 7 where γ = 7/5 is the ratio of specific heats for a diatomic gas. Ans. D Q 13. If one mole of a monatomic gas (γ = 5/3) is mixed with one mole of a diatomic gas (γ = 7/5), the value of γ for the mixture is (1988) (A) 1.40 (B) 1.50 (C) 1.53 (D) 3.07 Sol. The internal energy of one mole of an ideal gas at temperature T is f U = RT = Cv T, 2

Q 14. 70 cal of heat is required to raise the temperature of 2 mol of an ideal diatomic gas at constant pressure from 30 ◦ C to 35 ◦ C. The amount of heat required to raise the temperature of the same gas through the same range (30 ◦ C to 35 ◦ C) at constant volume is (1985) (A) 30 cal (B) 50 cal (C) 70 cal (D) 90 cal Sol. The heats required to raise the temperature of n moles of an ideal gas by an amount ∆T at constant volume and at constant pressure are ∆Qv = nCv ∆T,

(1)

∆Qp = nCp ∆T.

(2)

Divide equation (1) by (2) to get ∆Qv =

∆Qp 70 ∆Qp = = 50 cal, = Cp /Cv γ 7/5

where γ = 7/5 is the ratio of specific heats for the diatomic gas. We encourage you to calculate Cp by substituting the values in equation (2) and then use Cv = (Cp − R) in equation (1) to get ∆Qv . Ans. B Q 15. An ideal monatomic gas is taken round the cycle ABCDA as shown in the p-V diagram (see figure). The work done during the cycle is (1983) p 2p

p

(1)

where f is the degrees of freedom and Cv is the molar specific heat at constant volume. The degrees of freedom for a monatomic gas is three and that for a diatomic gas is five. Thus, from equation (1), the internal energy of a monatomic gas is Um = 3RT /2 and that of a diatomic gas is Ud = 5RT /2. The mixture contains the two moles of gases with the total internal energy Umix = Um + Ud = 23 RT + 52 RT = 4RT.

Thus, the internal energy of one mole of the mixture is Umix /2 = 2RT . Hence, the molar specific heat at constant volume is Cv,mix = 2R. The molar specific heat at constant pressure is Cp,mix = Cv + R = 3R. The ratio of specific heats for the mixture is γmix = Cp,mix /Cv,mix = 3/2. We encourage you to derive the expression for γmix when n1 moles of an ideal gas with f1 degrees of freedom are mixed with n2 moles of another ideal gas with f2 degrees of freedom. Hint: γmix = n1 (f1 +2)+n2 (f2 +2) . n1 f1 +n2 f2 Ans. B

B

C

A

D

V

(A) pV (B) 2pV (C)

1 2 pV

2V

V

(D) zero

Sol. The work done by a gas in a cyclic process is equal to the area enclosed by its p-V diagram. The enclosed area for the given process is W = pV . We encourage you to find the work done in the processes AB, BC, CD, and DA, individually. Ans. A

Chapter 24. Specific Heat Capacities of Gases

313

One or More Option(s) Correct Q 16. An ideal monatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T1 , pressure p1 and volume V1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2 , pressure p2 and volume V2 . During the process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement(s) is (are) (2015)

Use results from equations (2) and (3) to get the energy stored in the spring
Uspring = 12 kx2 = 12 (kx)x = 21 21 p1 A x = 14 p1 (xA) = 14 p1 V1 .

(4)

The internal energy of n moles of an ideal gas at temperature T is given by U = f2 nRT , where f is the degrees of freedom (f = 3 for the monatomic gases). Thus, the change in the internal energy of the gas is given by ∆U = U2 − U1 = f2 nRT2 − f2 nRT1 = 32 nR(3T1 ) − 32 nRT1

(∵ T2 = 3T1 , f = 3)

= 3nRT1 = 3p1 V1 .

(A) If V2 = 2V1 and T2 = 3T1 , then the energy stored in the spring is 41 p1 V1 . (B) If V2 = 2V1 and T2 = 3T1 , then the change in internal energy is 3p1 V1 . (C) If V2 = 3V1 and T2 = 4T1 , then the work done by the gas is 37 p1 V1 . (D) If V2 = 3V1 and T2 = 4T1 , then the heat supplied to the gas is 17 6 p1 V1 . Sol. We assume that the spring side of the cylinder is open to the atmosphere. p2 A

p0 A

In the cases (C) and (D), V2 = 3V1 and T2 = 4T1 . Substitute these in equation (1) to get p2 = 34 p1 . Substitute p2 = 43 p1 and p0 = p1 in equation (2) to get kx = 13 p1 A. The volume V2 = 3V1 = V1 + xA, gives xA = 2V1 . Substitute these in equation (4) to get
Uspring = 12 kx2 = 12 (kx)x = 21 31 p1 A x = 16 p1 (xA) = 13 p1 V1 . The gas does work against the spring and the atmospheric pressure p0 = p1 . The work done by the gas against the spring force is equal to the stored potential energy of the spring i.e., Wspring = Uspring = 13 p1 V1 . The work done by the gas against the atmospheric pressure is Watm = p0 (V2 − V1 ) = p1 (3V1 − V1 ) = 2p1 V1 .

kx

Thus, the total work done by the gas is

x

Let the atmospheric pressure be p0 , spring constant be k, and cross-sectional area of the cylinder be A. Initially, the spring is in its relaxed state. Thus, the pressure on two sides of the piston is equal i.e., p0 = p1 . In case (A) and (B), V2 = 2V1 and T2 = 3T1 . Apply the ideal gas equation between the initial and final states to get p1 V1 /T1 = p2 V2 /T2 = p2 (2V1 )/(3T1 ).

W = Wspring + Watm = 31 p1 V1 + 2p1 V1 = 73 p1 V1 . The change in the internal energy of the gas is ∆U = 23 nR(4T1 ) − 32 nRT1 = 92 nRT1 = 92 p1 V1 . Apply the first law of thermodynamics to get the heat supplied to the gas ∆Q = ∆U + W = 92 p1 V1 + 73 p1 V1 =

(1)

Solve equation (1) to get p2 = 32 p1 . The piston comes to the equilibrium state when spring is compressed by a distance x. The forces acting on the piston are force due to gas in the cylinder (p2 A), force due to outside atmosphere (p0 A), and the spring force (kx). In equilibrium, net force on the piston is zero i.e.,

Q 17. One mole of an ideal gas in initial state A undergoes a cyclic process ABCA as shown in the figure. Its pressure at A is p0 . Choose the correct option(s) from the following, (2010) V B

(2)

Substitute p2 = 32 p1 and p0 = p1 in equation (2) to get kx = 12 p1 A. The volume V2 = 2V1 = V1 + xA, gives xA = V1 .

(3)

41 6 p1 V1 .

Ans. A, B, C

4V0

p2 A = p0 A + kx.

(∵ nRT1 = p1 V1 ).

V0

C

A

T0

T

314

(A) (B) (C) (D)

Part IV. Thermodynamics

Internal energies at A and B are the same. Work done by the gas in process AB is p0 V0 ln 4. Pressure at C is p0 /4. Temperature at C is T0 /4.

Sol. The internal energy of one mole of an ideal gas at temperature T is given by U = 3RT /2. The process AB is isothermal i.e., TA = TB = T0 , which makes UA = UB . The work done in the isothermal process AB is WAB = nRT ln(VB /VA ) = nRT0 ln(4V0 /V0 ) = p0 V0 ln 4. Assume that the line BC pass through the origin. In the process BC, the slope V /T is constant. Thus, BC is an isobaric process (as V /T = nR/p, by the ideal gas equation). Thus, pC = pB = RTB /VB

Q 19. An ideal gas is taken from the state A (p0 , V0 ) to the state B ( p20 , 2V0 ) along a straight line path in the p-V diagram. Select the correct statement(s) from the following, (1993) (A) The work done by the gas in the process A to B exceeds the work that would be done if the system were taken from A to B along an isotherm. (B) In the T -V diagram, the path AB becomes a part of parabola. (C) In the p-T diagram, the path AB becomes a part of hyperbola. (D) In going from A to B, the temperature T of the gas first increases to a maximum value and then decreases. Sol. The equation of a straight line through the points A(p0 , V0 ) and B( p20 , 2V0 ) is given by p=−

p0 3 V + p0 . 2V0 2

(1)

p

= RT0 /(4V0 ) = (RT0 /V0 )/4 = p0 /4. Apply the ideal gas equation for the states A and C to get

p0

A

p0 2

(p0 /4)V0 pC VC T0 = T0 = T0 /4. TC = p0 V0 p0 V0

B V0

Ans. A, B, C, D Q 18. Cv and Cp denotes the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then, (2009) (A) Cp − Cv is larger for a diatomic ideal gas than for a monatomic ideal gas. (B) Cp + Cv is larger for a diatomic ideal gas than for a monatomic ideal gas. (C) Cp /Cv is larger for a diatomic ideal gas than for a monatomic ideal gas. (D) Cp · Cv is larger for a diatomic ideal gas than for a monatomic ideal gas. Sol. The molar heat capacities of an ideal gas, with f degrees of freedom, are given by Cv = R(f /2),

(1)

Cp = Cv + R = R(f + 2)/2.

(2)

Use equations (1) and (2) to get Cp + Cv = (1 + f )R, Cp 2 =1+ , Cv f f (2 + f ) 2 Cp · Cv = R . 4 The value of f for the monatomic and diatomic gases are 3 and 5. Ans. B, D

2V0

V

The isothermal process from A to B is a rectangular hyperbola (pV = constant). The two processes are shown in the figure. The work done by the gas is the area under the p-V graph, which is more for the straight line than for the rectangular hyperbola. The ideal gas equation, pV = nRT , gives p = nRT /V . Substitute it in equation (1) to get "  # 2 T0 V V T =− , −3 2 V0 V0 where, T0 = p0 V0 /(nR). Thus, T -V curve is a parabola as shown in the figure. Similarly, substitute V = nRT /p in equation (1) to get # "  2 p 3 p T = −2T0 − . p0 2 p0 T T0

A V0

B 2V0

V

Hence, T -p curve is also a parabola. The figure shows that the temperature of the gas first increases and then decreases. We encourage you to find TA , TB , and Tmax in terms of p0 and V0 . Ans. A, B, D

Chapter 24. Specific Heat Capacities of Gases

315

Q 20. For an ideal gas, (1989) (A) the change in internal energy in a constant pressure process from temperature T1 to T2 is equal to nCV (T2 −T1 ), where CV is the molar heat capacity at constant volume and n the number of moles of the gas. (B) the change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process. (C) the internal energy does not change in an isothermal process. (D) no heat is added or removed in an adiabatic process. Sol. In an isobaric process, the work done by the gas is Z V2 ∆W = pdV V1

Z

T2

dT = nR(T2 − T1 ),

= nR

(1)

T1

Column 1 (W1→2 )

Column 2

Column 3 p 1

(I)

p2 V2 −p1 V1 γ−1

2

(i) Isothermal (P) V p

(II) pV1 −pV2

(ii) Isochoric

1

(Q)

2 V p

(III) 0

(iii) Isobaric

1

2

(R) V p

(IV) −nRT ln VV12

1

(iv) Adiabatic (S) 2 V

and the heat absorbed by the gas is ∆Q = nCp (T2 − T1 ).

(2)

By first law of thermodynamics ∆Q = ∆U + ∆W.

(3)

Use equations (1)–(3) to get the change in the internal energy ∆U = ∆Q − ∆W = nCp (T2 − T1 ) − nR(T2 − T1 ) = n(Cp − R)(T2 − T1 ) = nCV (T2 − T1 ). In an adiabatic process, there is no heat exchange between the system and the surroundings i.e., ∆Q = 0. Substitute it in the first law of thermodynamics, equation (3), to get ∆U = −∆W which gives |∆U | = |∆W |. The internal energy of a gas depends on its temperature. Since the temperature is constant in an isothermal process, internal energy does not change. By definition, no heat is added or removed in an adiabatic process. Ans. A, B, C, D Paragraph Type Paragraph for Questions 21-23 An ideal gas is undergoing a cyclic thermodynamics process in different ways as shown in corresponding p-V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations used in thermodynamic processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n. (2017)

Q 21. Which one of the following options correctly represents a thermodynamic process that is used as a correction in the determination of the speed of sound in an ideal gas? (A) IV, ii, R (B) I, ii, Q (C) I, iv, Q (D) III, iv, R Sol. Adiabatic process is used as a correction in the determination of the speed of sound in an ideal gas. The p-V diagram of this process is shown in (Q). The work done on the gas in an adiabatic process is given 2 −p1 V1 by W1→2 = p2 Vγ−1 . Ans. (C) Q 22. Which of the following options is the only correct representation of a process in which ∆U = ∆Q−p∆V ? (A) II, iii, S (B) II, iii, P (C) III, iii, P (D) II, iv, R Sol. Compare the given relation, ∆U = ∆Q − p∆V , with the first law of thermodynamics, ∆Q = ∆U +∆W , to get work done by the gas as ∆W = p∆V . This is the work done in an isobaric process. The p-V diagram of this process is shown in (P). The work done in this R process is W1→2 = pdV = pV1 − pV2 . Ans. (B) Q 23. Which of the following options is the correct combination? (A) II, iv, P (B) III, ii, S (C) II, iv, R (D) IV, ii, S Sol. The work done in an isochoric process is zero. The p-V diagram of this process is shown in (S). We encourage you to identify the p-V diagram and work done in an isothermal process. Ans. (B)

316

Part IV. Thermodynamics Paragraph for Questions 24-25

In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are Cv = 23 R, Cp = 52 R, and those for an ideal diatomic gas are Cv = 25 R, Cp = 72 R.

By energy conservation, Ql = Qu . Substitute Ql and Qu from equations (1) and (2) to get Tf = 525 K. The work done by the gas in an isobaric process is given by W = nR(Tf − Ti ). Substitute the values to get the work done by the gas in the lower compartment as Wl = 2R(525 − 700) = −350R and that by the gas in the upper compartment as Wu = 2R(525 − 400) = 250R. Hence, the total work done by the gases is Wl + Wu = −350R + 250R = −100R. Ans. D

(2014)

Paragraph for Questions 26-28

Q 24. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be (A) 550 K (B) 525 K (C) 513 K (D) 490 K Sol. When the partition is fixed, lower compartment is at constant volume and upper compartment is at constant pressure (because the piston is freely movable). Let the equilibrium temperature be T . The heat lost by the lower compartment is Ql = nCv ∆T = 2 (3R/2) (700 − T ). The heat gained by the upper compartment is Qu = nCp ∆T = 2 (7R/2) (T − 400). Since the container and piston are made of perfectly insulating material, energy conservation Ql = Qu , gives T = 490 K. Ans. D Q 25. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be (A) 250R (B) 200R (C) 100R (D) −100R Sol. In this case, the lower and upper compartments exchange heat at constant pressure. Let the equilibrium temperature be Tf . The heat lost by the lower compartment is Ql = nCp ∆T = 2 (5R/2) (700 − Tf ),

(1)

and the heat absorbed by the upper compartment is Qu = nCp ∆T = 2 (7R/2) (Tf − 400).

(2)

A small spherical monatomic ideal gas bubble (γ = 5/3) is trapped inside a liquid of density ρl (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is T0 , the height of the liquid is H and the atmospheric pressure is p0 . [Neglect surface tension.] (2008) p0

H y

Q 26. As the bubble moves upwards, besides the buoyancy force the following forces are acting on it, (A) Only the force of gravity. (B) The force due to gravity and the force due to the pressure of the liquid. (C) The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquid. (D) The force due to gravity and the force due to viscosity of the liquid. Sol. The forces acting on the bubble are force due to gravity, viscous force opposite to the direction of motion and force due to pressure of the liquid. The force due to pressure of the liquid is same as buoyancy. Ans. D Q 27. When the gas bubble is at a height y from the bottom, its temperature is 2/5  2/5  p0 +ρl g(H−y) l gH (B) T (A) T0 pp00+ρ 0 +ρl gy p0 +ρl gH  3/5  3/5 p0 +ρl g(H−y) p0 +ρl gH (D) T0 (C) T0 p0 +ρl gy p0 +ρl gH Sol. The bubble does not exchange heat with its surroundings. Hence, it undergoes adiabatic expansion

Chapter 24. Specific Heat Capacities of Gases

317

given by pV γ = constant. Also, the gas in the bubble obeys ideal gas equation pV = nRT . Eliminate V to get p1−γ T γ = constant. Thus, T2γ , T1γ = p1−γ p1−γ 2 1

(1)

where p1 = p0 + ρl gH is the hydrostatic pressure at the bottom, T1 = T0 , and p2 = p0 + ρl g(H − y) is the hydrostatic pressure at a height y. Substitute the values in equation (1) and simplify to get  T2 = T0

p0 + ρl g(H − y) p0 + ρl gH

2/5

0 +ρl gH) (A) ρl nRgT0 (p (p0 +ρl gy)7/5

(D)

ρl nRgT0 (p0 +ρl gH)2/5 [p0 +ρl g(H−y)]3/5 3/5 0 +ρl gH) ρl nRgT0 (p (p0 +ρl gy)8/5 ρl nRgT0 (p0 +ρl gH)3/5 [p0 +ρl g(H−y)]2/5

Sol. The buoyancy force at a height y is equal to the weight of the liquid displaced by the bubble i.e., FB = ρl V g,

(1)

where V is the volume of the bubble. The ideal gas equation gives V = nRT2 /p2 .

(2)

Eliminate V from equations (1) and (2) and substitute for p2 and T2 to get the buoyancy force FB =

(P) In process I

(1) Work done by the gas is zero (2) Temperature of the gas remained unchanged (3) No heat is exchanged between the gas and its surroundings (4) Work done by the gas is 6p0 V0

(Q) In process II (R) In process III

(S) In process IV

.

Q 28. The buoyancy force acting on the gas bubble is [Here R is the universal gas constant.]

(C)

Column II

2/5

Ans. B

(B)

Column I

Sol. The process II is isobaric (constant pressure) and process III is isochoric (constant volume). The work done by the gas in isobaric expansion process II is ∆W = p∆V = 3p0 (3V0 − V0 ) = 6p0 V0 . The work done by the gas in isochoric process III is zero. The slope of an adiabatic process is greater than the slope of an isothermal process at the point of intersection. Thus, process I is adiabatic and process IV is isothermal. Hence, no heat is exchanged between the gas and its surroundings in process I and temperature of the gas remained unchanged in process IV. For a monatonic ideal gas, find the ratio of slope of an adiabatic process to the slope of an isothermal process shown on a p-V diagram. Ans. P7→(3), Q7→(4), R7→(1), S7→(2) Q 30. One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E and E → F → H → E as shown in the p-V diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. Match the paths in Column I with the magnitudes of the work done in Column II. (2013)

ρl nRgT0 . (p0 + ρl gH)2/5 [p0 + ρl g(H − y)]3/5

p 32p0

F

Ans. B Matrix or Matching Type

p0

Q 29. One mole of a monatomic ideal gas undergoes four thermodynamics processes as shown schematically in p-V diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in Column I with the corresponding statements in Column II. (2018) p

G

H

V0

32V0

Column I (P) (Q) (R) (S)

G→ E G→ H F→ H F→ G

V

Column II (1) (2) (3) (4)

160p0 V0 ln 2 36p0 V0 24p0 V0 31p0 V0

II

3p0

p0

E

I

V0

IV

III

3V0

V

Sol. The process G → E is isobaric i.e., at constant pressure p0 . The work done by the gas in this process is given by Z WGE = pdV = p0 (VE − VG ) = −31p0 V0 .

318

Part IV. Thermodynamics

The process G → H is also isobaric and the work done by the gas is

The equation (1) gives the change in the internal energy for a process i → f ,

WGH = p0 (VH − VG ). The magnitude |WGH | is less than the magnitude |WGE | (since VE < VH ). Only valid answer among the given choices is WGH = −24p0 V0 . This also gives us VH = 8V0 . On a p-V diagram, magnitude of the slope of an adiabatic process is more than that of an isothermal process (adiabatic process is steeper). Comparing the slopes of F → G and F → H at the point F, we conclude that F → G is isothermal and F → H is adiabatic. The work done by the gas in an isothernal process is given by WFG = nRT ln (VG /VF ). The ideal gas equation gives nRT = 32p0 V0 . Substitute the values of VG , VF , and nRT to get

∆Ui→f = Uf − Ui = (3/2)(pf Vf − pi Vi ).

(2)

If ∆Ui→f > 0 then the internal energy of the gas increases else it decreases. Substitute the values of p and V for the various processes to get

∆UA→B = −9pV,

∆UB→C = −3pV,

∆UC→D = 9pV,

∆UD→A = 0.

WFG = 32p0 V0 ln (32V0 /V0 ) = 160p0 V0 ln 2. The work done by the gas in an adiabatic process F → H is given by WFH =

pF VF − pH VH 32p0 V0 − p0 8V0 = = 36p0 V0 , γ−1 5/3 − 1

where γ = 5/3 for a monatomic gas. Ans. P7→4, Q7→3, R7→2, S7→1 Q 31. One mole of monatomic gas is taken through a cycle ABCDA as shown in the p-V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I. (2011)

3p 1p

A

C

O 1V

D 3V

9V

Column I (A) A → B (B) B → C (C) C → D (D) D → A

WA→B = −6pV,

WB→C = 0,

WC→D = 8pV,

WD→A < 0.

The WD→A is negative because the volume of the gas decreases in this process. The first law of thermodynamics, ∆Q = ∆U + ∆W , gives the heat absorbed in a process. The heat is absorbed (gained) by the gas if ∆Q > 0 and released (lost) by the gas if ∆Q < 0. Using the change in the internal energy and the work done by the gas in various processes, we get

p B

The work done by the gas is given by the area under the p-V diagram. The work done by the gas is positive if its volume increases and negative otherwise. From the given p-V diagram, we get

V

Column II (p) Internal energy decreases. (q) Internal energy increases. (r) Heat is lost. (s) Heat is gained. (t) Work is done on the gas.

Sol. The internal energy of one mole of an ideal monatomic gas is U = (3/2)RT = (3/2)pV.

(1)

QA→B = −15pV,

QB→C = −3pV,

QC→D = 17pV,

QA→B < 0.

Ans. A7→(p,r,t), B7→(p,r), C7→(q,s), D7→(r,t)

Q 32. Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. (2008)

Chapter 24. Specific Heat Capacities of Gases

319 we get

Column I

Column II

(A) An insulated container has two chambers separated by a valve. Chamber I contains an ideal gas and the chamber II has vacuum. The valve is opened.

(p) Temperature of the gas decreases.

I

= (1/2)nRT1 , = −(3/4)nRT1 , ∆Q = ∆U + ∆W = −(3/4)nRT1 + (1/2)nRT1

Ideal gas Vacuum

= −(1/4)nRT1 . (q) Temperature of the gas increases or remains constant. (r) The gas loses heat.

2V1

(s) The gas gains heat.

V

Sol. The process (A) is a free expansion of the gas. The pressure against which the gas expands is zero (p = 0 due to vacuum). Thus, the work done by the gas in this process is ∆W = p(V2 − V1 ) = 0. The insulated container prohibits the heat exchange with the surroundings i.e., ∆Q = 0. Apply the first law of thermodynamics, ∆Q = ∆U + ∆W , to get ∆U = 0. Thus, the temperature of the gas remains constant. The internal energy of n moles of an ideal monatomic gas (three degrees of freedom), at temperature T , is given by U = (f /2)nRT = (3/2)nRT.

(1)

The work done by an ideal gas in the process pV x = k (a constant) is given by Z

V2

∆W =

pdV = V1

p1 V1 − p2 V2 . x−1

(2)

T2 = (2)−1/3 T1 ,

∆U = (3/2)nR(T2 − T1 ) h i = (3/2)nRT1 (2)−1/3 − 1 , h i ∆Q = ∆U + ∆W = (3/2)nRT1 1 − (2)−1/3 . Since ∆Q > 0, the gas absorbs heat in this process. In the process (D), the product p2 V2 is greater than p1 V1 . Thus, T2 > T1 (by the ideal gas equation). Hence, ∆U = U2 −U1 > 0. The work by the gas in an expansion process is positive i.e., ∆W > 0. By the first law of thermodynamics, ∆Q = ∆U + ∆W > 0 and hence the gas absorbs heat. Ans. A7→q, B7→(p,r), C7→(p,s), D7→(q,s) True False Type Q 33. At a given temperature, the specific heat of a gas at a constant pressure is always greater than its specific heat at constant volume. (1987) Sol. For a gas, the specific heat at constant pressure (Cp ) and the specific heat at constant volume (Cv ) are related by Cp = Cv + R. Ans. T Q 34. The curves A and B in the figure shows p-V graphs for an isothermal and adiabatic process for an ideal gas. The isothermal process is represented by the curve A. (1985) p

Eliminate p from pV x = k and pV = nRT to get T V x−1 = k/(nR).

Since ∆Q < 0, the gas loses heat in this process. In the process (C), V2 = 2V1 and x = 4/3. Using equations (1)–(3) and the first law of thermodynamics, we get

h i ∆W = 3nR(T1 − T2 ) = 3nRT1 1 − (2)−1/3 ,

p V1

∆W = (p1 V1 − p2 V2 ) = nR(T1 − T2 ) ∆U = U2 − U1 = (3/2)nRT2 − (3/2)nRT1

II

(B) An ideal monatomic gas expands to twice its original volume such that its pressure p ∝ V12 , where V is the volume of the gas. (C) An ideal monatomic gas expands to twice its original volume such that its 1 , where pressure p ∝ V 4/3 V is its volume. (D) An ideal monatomic gas expands such that its pressure p and volume V follows the behaviour shown in the figure.

T2 = T1 (V1 /V2 ) = T1 /2,

(3)

A B

In the process (B), V2 = 2V1 and x = 2. Using equations (1)–(3) and the first law of thermodynamics,

V

320

Part IV. Thermodynamics

Sol. On a p-V graph, the slopes of an isothermal process (pV = constant) and a adiabatic process (pV γ = constant) are given by

Q 37. One mole of a monatomic ideal gas is mixed with one mole of a diatomic ideal gas. The molar specific heat of the mixture at constant volume is . . . . . . (1984)

(dp/dV )isothermal = −p/V, (dp/dV )adiabatic = −γp/V. Thus, the slope of an adiabatic curve is γ times the slope of the isothermal curve (at common point of these processes on p-V graph). Since γ > 1, adiabatic curve is steeper than isothermal curve. Ans. T Fill in the Blank Type Q 35. An ideal gas with pressure p, volume V and temperature T is expanded isothermally to a volume 2V and a final pressure pi . If the same gas is expanded adiabatically to a volume 2V , the final pressure is pa . The ratio of the specific heats of the gas is 1.67. The ratio pa /pi is . . . . . . (1994) Sol. The product pV remains constant in an isothermal process and pV γ remains constant in an adiabatic process. p isothermal

pi pa adiabatic

2V

Thus, the heat required to raise the temperature of (nm + nd ) = 2 moles of the mixture by ∆T (at constant volume) is Qmix = Qm + Qd = 23 R∆T + 25 R∆T = 4R∆T = (nm + nd )Cv,mix ∆T = 2Cv,mix ∆T. This equation gives the specific heat of the mixture at constant volume as Cv,mix = 2R. Ans. 2R Integer Type Q 38. One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 J mol−1 K−1 , the decrease in its internal energy, in Joule, is . . . . . . .

p

V

Sol. The degrees of freedom for a monatomic gas is fm = 3 and that for a diatomic gas is fd = 5. The heats required to raise the temperature of nm = 1 mole of a monatomic and nd = 1 mole of a diatomic gas by ∆T (at constant volume) are
∆Qm = ∆Um = nm f2m R ∆T = (1) 23 R ∆T,
∆Qd = ∆Ud = nd f2d R ∆T = (1) 52 R ∆T.

V

(2018)

Substitute the initial and final values for the isothermal process, pV = pi (2V ), to get pi = p/2. Substitute these in the adiabatic equation, pV γ = pa (2V )γ , to get pa = p/2γ . Thus, pa /pi = 21−γ = 21−1.67 = 0.628. The figure shows these processes on a p-V diagram. Ans. 0.628 Q 36. A container of volume 1 m3 is divided into two equal parts by a partition. One part has an ideal gas at 300 K and the other part is vacuum. The whole system is thermally isolated from the surroundings. When the partition is removed, the gas expands to occupy the whole volume. Its temperature will now be . . . . . . (1993)

Sol. The given process is a free expansion of an ideal gas as the other side is vacuum. The pressure against which the gas expands is zero i.e., p = 0. Thus, R no work is done by the gas in this process i.e., ∆W = pdV = 0. Also, ∆Q = 0 because the system is thermally isolated from the surroundings. The first law of thermodynamics, ∆Q = ∆U + ∆W , gives ∆U = 0. Thus, the temperature remains fixed at 300 K. Ans. 300 K

Sol. The degrees of freedom of a monatomic gas is f = 3. Its molar specific heats are Cv = f2 RT and Cp = Cv + R = (f +2) 2 RT . The ratio of specific heats is given by γ = Cp /Cv = (f + 2)/f = 5/3. In an adiabatic process, T V γ−1 = constant. Hence, T1 V1γ−1 = T2 V2γ−1 , which gives γ−1

T2 = T1 (V1 /V2 )

= 100(1/8)5/3−1 = 25 K,

where we used T1 = 100 K and V2 = 8V1 . The change in internal energy of one mole of the ideal gas is f f f RT2 − RT1 = R(T2 − T1 ) 2 2 2 = (3/2)(8)(25 − 100) = −900 J.

∆U = U2 − U1 =

Thus, the decrease in internal energy of the gas is 900 J. Ans. 900 Q 39. A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of the gas is Ti (in Kelvin) and the final temperature is aTi , the value of a is . . . . . . . (2010)

Chapter 24. Specific Heat Capacities of Gases Sol. The diatomic gas has five degrees of freedom i.e., f = 5. Thus, the internal energy per mole, specific heat at constant volume, specific heat at constant pressure, and the ratio of specific heats for a diatomic gas, are given by U = (f /2)RT = (5/2)RT,

Cv = dU/dT = 5R/2,

Cp = Cv + R = 7R/2,

γ = Cp /Cv = 7/5.

321 Sol. The process A → B → C is shown in the figure on a p-V diagram. The process AB is adiabatic and the process BC is isochoric. p C p2 p1

In an adiabatic process Ti Viγ−1 = Tf Vfγ−1 , which gives γ−1

Tf = Ti (Vi /Vf )

= Ti (32)

2/5

B A V2

V1

V

= 4Ti . Ans. 4

Descriptive Q 40. The piston cylinder arrangement shown in the figure contains a diatomic gas at temperature 300 K. The cross-sectional area of the cylinder is 1 m2 . Initially the height of the piston above the base of the cylinder is 1 m. The temperature is now raised to 400 K at constant pressure. Find the new height of the piston above the base of the cylinder. If the piston is now brought back to its original height without any heat loss, find the new equilibrium temperature of the gas. You can leave the answer in fraction. (2004)

1m

Sol. Initial volume of the gas in the cylinder is V1 = 1 m3 . Using the ideal gas equation (at constant pressure), the volume of the gas, when its temperature is raised from T1 = 300 K to T2 = 400 K, is given by

The ratio of the specific heats for a monatomic ideal gas is γ = Cp /Cv = 5/3. The work done by the gas in a adiabatic process is given by WAB =

i p1 V1 h γ−1 1 − (V1 /V2 ) γ−1 h i 3 2/3 = − p1 V1 (V1 /V2 ) − 1 . 2

WAB =

Work done by the gas in an isochoric process is zero i.e., WBC = 0. Total work done by the gas is WABC = WAB + WBC . Heat transfer to the gas in an adiabatic process is zero i.e., QAB = 0 and that in the isochoric process is QBC = Q. Thus, QABC = QAB + QBC = Q. The first law of thermodynamics, QABC = UABC + WABC , gives h i 3 2/3 UABC = Q + p1 V1 (V1 /V2 ) − 1 . 2

(2)

Ideal gas equation, pV = nRT , at point A gives T1 = p1 V1 /(2R). The internal energy of n moles of an ideal gas having f degrees of freedom (f = 3 for a monatomic gas) and kept at temperature T is U = (f /2)nRT . Thus, the change in the internal energy from A to C is f UABC = n R(TC − T1 ) = 3R(TC − T1 ). 2

Ans. 4/3 m, 448.8 K Q 41. Two moles of an ideal monatomic gas initially at pressure p1 and volume V1 undergo an adiabatic compression until its volume is V2 . Then the gas is given heat Q at constant volume V2 . [Give answer in terms of p1 , V1 , V2 , Q and R.] (1999) (a) Sketch the complete process on a p-V diagram. (b) Find the total work done by the gas, the total change in its internal energy and the final temperature of the gas.

(1)

In an adiabatic process, p1 V1γ = p2 V2γ . Thus, the presγ sure p2 = p1 (V1 /V2 ) . Substitute p2 in equation (1) to get

V2 = V1 (T2 /T1 ) = 1(400/300) = 4/3 m3 . Hence, the new height of the piston above the base is 4/3 m. The process to bring back the piston to its original height is adiabatic (no heat loss) and thus, T2 V2γ−1 = T3 V3γ−1 . Final volume of the gas is V3 = 1 m3 . For a diatomic gas, γ = 7/5. Substitute the values to get  γ−1  7/5−1 V2 4/3 T3 = T2 = (400) = 448.8 K. V3 1

p1 V1 − p2 V2 . γ−1

(3)

Eliminate UABC from equations (2) and (3) and substitute for T1 to get the temperature TC =

Q p1 V1 + 3R 2R

(b)

− 23 p1 V1

Q 3R

+

p1 V 1 2R

h



V1 V2

2/3 .

Ans. solution i h (a) See i 2 3 3 (V1 /V2 ) − 1 , 2 p1 V1 (V1 /V2 ) − 1 + Q, 2 3

2

(V1 /V2 ) 3

322

Part IV. Thermodynamics

Q 42. One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in the figure. Calculate, (1998)

T B

3T0 2T0

C

T0

A

p 3p0 p0

V0

B

A V0

C 2V0

V

(a) the work done by the gas. (b) the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB. (c) the net heat absorbed by the gas in the path BC. (d) the maximum temperature attained by the gas during the cycle. Sol. The work done by the gas is the area under the curve on a p-V diagram, which is positive if the volume of the gas increases. Thus, from the given figure, WAB = 0, WBC = p0 (2V0 − V0 ) + 21 (2V0 − V0 )(3p0 − p0 ) = 2p0 V0 , WCA = (V0 − 2V0 )p0 = −p0 V0 . Hence, the work done by the gas in the cycle ABCA is ∆W = WAB + WBC + WCA = p0 V0 .

(1)

Note that the work done by the gas in the clockwise cyclic process is equal to the enclosed area on a p-V diagram. The specific heats for a monatomic gas are Cv = 3R/2 and Cp = 5R/2. The heats absorbed by the gas in the isobaric process CA and the isochoric process AB are

= (5R/2) (p0 V0 /R − 2p0 V0 /R) = −(5/2)p0 V0 ,

(2)

QAB = nCv ∆T = (3R/2)(Tf − Ti ) = (3R/2) (3p0 V0 /R − p0 V0 /R) = 3p0 V0 .

(3)

V

The temperature, T = pV /(nR) = pV /R, is maximum when the product pV attains its maximum value. In the process AB, pV is maximum at B with maximum value 3p0 V0 and in the process CA, pV is maximum at C with the maximum value 2p0 V0 . The process BC is a straight line (on a p-V diagram) passing through B(V0 , 3p0 ) and C(2V0 , p0 ). The equation of this line is p = − (2p0 /V0 ) V + 5p0 . The temperature at a point (p, V ) on this line is given by T = pV /R   = (p0 V0 /R) −2(V /V0 )2 + 5(V /V0 ) .

(4)

The temperature attains its maximum value when dT /dV = 0. Differentiate equation (4) and solve dT /dV = 0 to get V = 5V0 /4. Substitute V in equation (4) to get Tmax = 25p0 V0 /(8R). The figure shows the T -V diagrams of given processes. Ans. (a) p0 V0 (b) 5p0 V0 /2, 3p0 V0 (c) p0 V0 /2 (d) 25p0 V0 /(8R) Q 43. At 27 ◦ C two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate, (1996) (a) the final temperature of the gas. (b) change in its internal energy. (c) the work done by the gas during this process. Sol. In an adiabatic process, T V γ−1 = constant i.e., T1 V1γ−1 = T2 V2γ−1 . Substitute, T1 = 273 + 27 = 300 K, V1 = V , V2 = 2V , and γ = 5/3 (monatomic gas) to get T2 = T1 (V1 /V2 )

QCA = nCp ∆T = (1)(5R/2)(Tf − Ti )

2V0

γ−1

= 300(1/2)5/3−1 = 189 K.

The internal energy of n moles of a monatomic gas (degrees of freedom f = 3) at temperature T is given by U = (f /2)nRT = (3/2)nRT . Thus, the change in the internal energy in the given process is ∆U = U2 − U1 = (3/2)nR(T2 − T1 ) = (3/2)(2)(8.314)(189 − 300) = −2768 J.

In a cyclic process, the gas returns to its initial state, and hence ∆U = 0. The first law of thermodynamics, ∆Q = ∆U + ∆W , reduces to ∆Q = ∆W for a cyclic process. Thus, QAB + QBC + QCA = ∆W . Substitute the values from equations (1)–(3) to get QBC = 12 p0 V0 .

There is no heat transfer in the adiabatic process i.e., ∆Q = 0. The first law of thermodynamics, ∆Q = ∆U + ∆W , gives work done by the gas as ∆W = −∆U = 2768 J. Ans. (a) 189 K (b) −2768 J (c) 2768 J

Chapter 24. Specific Heat Capacities of Gases

323

Q 44. A gaseous mixture enclosed in a vessel of volume V consists of one mole of gas A with γ = Cp /CV = 5/3 and another gas B with γ = 7/5 at a certain temperature T . The molecular weights of the gases A and B are 4 and 32, respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation pV 19/13 = constant, in adiabatic processes. (1995) (a) Find the number of moles of the gas B in the gaseous mixture. (b) Compute the speed of sound in the gaseous mixture at 300 K. (c) If T is raised by 1 K from 300 K, find the percentage change in the speed of sound in the gaseous mixture. (d) The mixture is compressed adiabatically to 1/5 th of its initial volume V . Find the change in its adiabatic compressibility in terms of the given quantities.

Differentiate equation (2) with respect to T and simplify to get ∆v 1 ∆T 1 = = = 0.00167 = 0.167%. v 2 T 2 × 300 The bulk modulus is defined as dp dp B=− = −V . dV /V dV

(3)

Differentiate pV γm = constant, to get dp/dV = −γm p/V for an adiabatic process. Substitute it in equation (3) to get B = γm p. Thus, adiabatic compressibility is β = 1/B = 1/(γm p). Let β = 1/(γm p) and β 0 = 1/(γm p0 ) be the compressibilities when the volumes are V and V 0 = V /5. The γ adiabatic equation, pV γm = p0 V 0 m , gives p0 = 5γm p. Also, p = nm RT /V . Substitute the values to get

UA = nA Cv,A T,

where

Cv,A = R/(γA − 1),


V (1 − 5−γm ) 5−γm − 1 = − γm p γm nm RT
V 1 − 5−19/13 =− (19/13)(1 + 2)(8.314)(300)

UB = nB Cv,B T,

where

Cv,B = R/(γB − 1),

= −8.27 × 10−5 V.

Sol. Eliminate Cp from the relations Cp − Cv = R and γ = Cp /Cv to get, Cv = R/(γ − 1). The internal energies of the gases A and B are

The molar specific heat of the mixture, at the constant volume is Cv,m =

Q 45. A closed container of volume 0.02 m3 contains a mixture of neon and argon gases, at a temperature of 27 ◦ C and pressure of 1 × 105 N/m2 . The total mass of the mixture is 28 g. If the molar masses of neon and argon are 20 g/mol and 40 g/mol respectively, find the masses of the individual gases in the container assuming them to be ideal [R = 8.314 J/mol K]. (1994) Sol. Let mn and ma be the masses of neon and argon. Given, total mass of the mixture is

(1)

The adiabatic equation for the mixture is, pV 19/13 = constant, which gives γm = 19/13. Substitute nA = 1, γA = 5/3, γB = 7/5, and γm = 19/13 in equation (1) to get nB = 2. Molecular mass of the mixture is

mn + ma = 28.

(1)

The number of moles of neon are nn = mn /20 and that of argon are na = ma /40. Apply the ideal gas equation, pV = nRT , to get the partial pressure of neon and argon as pn = nn RT /V,

pa = na RT /V.

Dalton’s law of partial pressure gives pn + pa = p = 1 × 105 N/m2 . Substitute the values to get

Mtotal n A M A + n B MB = nm nA + nB 1(4) + 2(32) = = 22.67 g/mol, 1+2

Mm =

and the speed of sound in the mixture is r γm RT v= Mm s (19/13)(8.314)(300) = 401 m/s. = 22.67 × 10−3

1

Ans. (a) 2 mol (b) 401 m/s (c) 0.167% (d) −8.27 × 10−5 V

and the internal energy of the mixture is   nB nA + RT. U = UA + UB = γA − 1 γB − 1

1 dU nA + nB dT   nB 1 nA + R = nA + nB γA − 1 γB − 1 R = . γm − 1

∆β = β 0 − β =

2mn + ma = 20pV /(RT ). Solve equations (1) and (2) to get, 20pV 20(1 × 105 )(0.02) − 28 = − 28 RT (8.314)(300) = 4.074 g,

mn = (2)

ma = 28 − mn = 23.926 g. Ans. mn = 4.074 g, ma = 23.926 g

(2)

324

Part IV. Thermodynamics

Q 46. One mole of a monatomic ideal gas is taken through the cycle shown in the figure. A → B is adiabatic expansion, B → C is cooling at constant volume, C → D is adiabatic compression, and D → A is heating at constant volume. The pressure and temperature at A, B, etc., are denoted by pA , TA , pB , TB etc., respectively. Given that TA = 1000 K, pB = 32 pA and pC = 13 pA , calculate the following quantities, [Given: 2/5 (2/3) = 0.85] (1993) p A B

D

C V

(a) The work done by the gas in the process A → B. (b) The heat lost by the gas in the process B → C. (c) The temperature TD . Sol. For a monatomic gas, γ = 5/3. Given n = 1, TA = 1000 K, pB = 2pA /3, and pC = pA /3. In the adiabatic process A → B, p1−γ TAγ = p1−γ TBγ , A B which gives 

 γ1 −1

  35 −1 3 TA = TA 2

pA pB  2/5 2 (1000) = 850 K. = 3

TB =

The work done by the gas in the adiabatic process A → B is WAB

nR(TA − TB ) (1)(8.314)(1000 − 850) = = γ−1 5/3 − 1 = 1870 J.

In the adiabatic process C → D and the isochoric process D → A, p1−γ TDγ = p1−γ TCγ , D C

(1)

pD /TD = pA /TA .

(2)

Solve equations (1) and (2) to get 1−γ pC TCγ TA1−γ pA  −2/3 1 (425)5/3 (1000)−2/3 = 500 K. = 3 

TD =

Ans. (a) 1870 J (b) −5300 J (c) 500 K
Q 47. Three moles of an ideal gas Cp = 72 R at pressure pA and temperature TA is isothermally expanded to twice its initial volume. It is then compressed at constant pressure to its original volume. Finally, gas is compressed at constant volume to its original pressure pA . (1991) (a) Sketch p-V and p-T diagrams for the complete process. (b) Calculate the net work done by the gas, and net heat supplied to the gas during the complete process. Sol. On a p-V diagram, isothermal process (AB) is a rectangular hyperbola, isobaric process (BC) is a straight line parallel to V axis, and isochoric process is a straight line parallel to p axis. Using the ideal gas equation at A and B, pA VA = pB VB = pB (2VA ), we get pB = pA /2. The p-V diagram of the given processes is as shown in the figure. p pA pA 2

A

C VA

In the isochoric process B → C, pB /TB = pC /TC , which gives     pC pA /3 TC = TB = 850 = 425 K. pB 2pA /3 The work done in an isochoric pressure is zero i.e., WBC = 0. Using the first law of thermodynamics, QBC = UBC + WBC , we get QBC = UBC = UC − UB = (f /2)R(TC − TB ) = (3/2)(8.314)(425 − 850) = −5300 J, where f = 3 is the degrees of freedom for a monatomic gas. Negative sign indicates the heat loss by the gas.

p pA pA 2

B 2VA

V

A

C TA 2

B TA

T

On p-T diagram, isothermal process AB is a straight line parallel to p axis, isobaric process BC is a straight line parallel to T axis, and isochoric process CA is a straight line. The ideal gas equation at B and C, VB /TB = VC /TC , gives TC = VC TB /VB = VA TA /(2VA ) = TA /2.

Chapter 24. Specific Heat Capacities of Gases

325

WAB = nRTA ln (VB /VA ) = 3RTA ln 2 = 2.08RTA , WBC = pB (VC − VB ) = (pA /2)(VA − 2VA ) = −(1/2)pA VA = −(1/2)nRTA = −(3/2)RTA = −1.5RTA , WCA = 0, Wnet = WAB + WBC + WCA = 2.08RTA − 1.5RTA = 0.58RTA . The change in the internal energy of an ideal gas in a cyclic process is zero i.e., ∆U = 0. Thus, by the first law of thermodynamics, Qnet = ∆U + Wnet = 0.58RTA . Ans. (a) See solution (b) 0.58RTA , 0.58RTA Q 48. An ideal gas having initial pressure p, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T /2. (1990)

of 2.4 × 10−3 m3 and the spring is in its relaxed (unstretched, uncompressed) state. The gas is heated by a small electric heater until the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied (in Joules) by the heater. [The force constant of the spring is 8000 N/m, and the atmospheric pressure 1.0 × 105 Nm−2 . The cylinder and the piston are thermally insulated. The piston is massless and there is no friction between the piston and the cylinder. Neglect heat loss through the lead wires of the heater. The heat capacity of the heater coil is negligible. Assume the spring to be massless.] (1989) open atmosphere Heater

Work done by the gas in the isothermal process AB, in the isobaric process BC, in the isochoric process CA, and in the entire cycle are given by

Sol. Initially, the spring is in its relaxed state and there is no friction between the piston and the cylinder. Thus, pressure inside the cylinder is equal to the atmospheric pressure i.e., pi = po = 1 × 105 N/m2 . x

γ−1

γ = 1 + ln 2/ln 5.66 = 1.4.

(1)

An ideal gas, with f degrees of freedom, has Cv = f2 RT , Cp = Cv + R, and γ = Cp /Cv = 1 + 2/f.

(2)

Solve equations (1) and (2) to get f = 5. γ In an adiabatic process, pV γ = p0 V 0 , which gives the final pressure of the gas γ

p0 = p (V /V 0 ) = p/(5.66)1.4 , and the work done by the gas in an adiabatic expansion is pV − p0 V 0 γ−1 pV − (p/(5.66)1.4 )(5.66V ) = = 1.25pV. 1.4 − 1

W =

Ans. (a) 5 (b) 1.25pV Q 49. An ideal monatomic gas is confined in a cylinder by a spring loaded piston of cross-section 8.0 × 10−3 m2 . Initially, the gas is at 300 K and occupies a volume

Heater

(a) How many degrees of freedom do gas molecules have? (b) Obtain the work done by the gas during the expansion as a function of the initial pressure p and volume V. Sol. In an adiabatic process, T V γ−1 = T 0 V 0 . Substitute T 0 = T /2, V 0 = 5.66V to get (5.66)γ−1 = 2 which gives

rigid support

pf A

p0 A kx

On heating, the piston moves by a distance x = 0.1 m which causes the spring to get compressed by the same amount. Thus, the forces acting on the piston of area A = 8.0 × 10−3 m2 from outside are p0 A (due to open atmosphere) and kx (due to spring). These forces are balanced by the force of the gas at pressure pf i.e., p0 A + kx = pf A, which gives kx (8000)(0.1) = 1 × 105 + A 8.0 × 10−3 5 2 = 2 × 10 N/m .

pf = p0 +

Initial volume of the gas is Vi = 2.4 × 10−3 m3 . Its volume increases due to movement of the piston. Thus, the final volume of the gas is Vf = Vi + xA = 2.4 × 10−3 + (0.1)(8.0 × 10−3 ) = 3.2 × 10−3 m3 . Apply the ideal gas equation, pi Vi /Ti = pf Vf /Tf , to get pf Vf Tf (2 × 105 ) (3.2 × 10−3 ) (300) = pi Vi (1 × 105 ) (2.4 × 10−3 ) = 800 K.

Tf =

The gas does work against the atmospheric pressure and the spring. The work done by the gas on the spring is equal to the potential energy of compressed spring i.e., Wspring = 12 kx2 = 21 (8000)(0.1)2 = 40 J.

(1)

326

Part IV. Thermodynamics

The work done by the gas on the open atmosphere is Watm = p0 (Vf − Vi ) = 1 × 105 (3.2 × 10−3 − 2.4 × 10−3 ) = 80 J.

The temperature at C is TC = TA = 300 K. In an adiabatic process T V γ−1 = constant, which gives 

(2)

From equations (1) and (2), the total work done by the gas is,

VC = VB

pC =

= 600 J. Apply the first law of thermodynamics to get the heat supplied to the gas as

1  γ−1

 = 40

600 300

1  5/3−1

= 113 litre.

The ideal gas equation, pV = nRT , gives the final pressure as

W = Wspring + Watm = 40 + 80 = 120 J. The increase in the potential energy of n moles of a monatomic gas (degrees of freedom, f = 3), when its temperature is raised from Ti to Tf , is    f pi Vi f ∆U = n R(Tf − Ti ) = R(Tf − Ti ) 2 RTi 2    1 × 105 × 2.4 × 10−3 3 = (800 − 300) 300 2

TB TC

nRTC 2(8.314)(300) = 0.44 × 105 N/m2 . = VC 113 × 10−3

The work done by the gas in the isobaric process AB is WAB = pA (VB − VA ) = (2.49 × 105 ) (20 × 10−3 ) = 4980 J. In an adiabatic process, there is no heat exchange between the system and the surroundings i.e., ∆Q = 0. Apply the first law of thermodynamics, ∆Q = ∆U +W , to get the work done by the gas in the adiabatic process BC,

∆Q = W + ∆U = 120 + 600 = 720 J. Ans. 800 K, 720 J Q 50. Two moles of helium gas (γ = 5/3) are initially at temperature 27 ◦ C and occupy a volume of 20 litre. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value. (1988)

(a) Sketch the process on a p-V diagram. (b) What are the final volume and pressure of the gas? (c) What is the work done by the gas? Sol. Let A be the initial state of the gas, AB be an isobaric process, and BC be an adiabatic process. p(×105 )N/m2 2.49

A

B

C

0.44 20 40

V (l)

113

Initial volume of the gas is VA = 20 litre, initial temperature is TA = 300 K, and the number of moles is n = 2. The ideal gas equation, pV = nRT , gives initial pressure as pA =

nRTA 2(8.314)(300) = = 2.49 × 105 N/m2 . VA 20 × 10−3

The volume at B is VB = 2VA = 40 litre. Since AB is an isobaric process, pB = pA = 2.49 × 105 N/m2 and TB = TA (VB /VA ) = 600 K.

WBC = −∆U = −(UC − UB )   = − n f2 RTC − n f2 RTB = n f2 R(TB − TC )
= 2 32 (8.314)(600 − 300) = 7483 J, where f = 3 is the degrees of freedom for the monatomic helium. Total work done by the gas is W = WAB + WBC = 4980 + 7483 = 12463 J. Ans. (a) See solution (b) 113 litre, 0.44 × 105 N/m2 , (c) 12463 J Q 51. The rectangular box shown in the figure has a partition which can slide without friction along the length of the box. Initially each of the two chambers of the box has one mole of a monatomic ideal gas (γ = 5/3) at a pressure p0 , volume V0 and temperature T0 . The chamber on the left is slowly heated by an electric heater. The walls of the box and the partition are thermally insulated. Heat loss through the lead wires of the heaters is negligible. The gas in the left chamber expands pushing the partition until the final pressure in both chambers become 243p0 /32. Determine (a) the final temperature of the gas in each chamber, and (b) the work done by the gas in the right chamber. (1984)

Chapter 24. Specific Heat Capacities of Gases

327

Sol. Let V1 and V2 be the final volumes and T1 and T2 be the final temperatures of the left and the right chambers. Total volume of the box remains constant i.e., V1 + V2 = 2V0 .

(1)

Initially, each chamber has n = 1 mole of a monatomic ideal gas (γ = 5/3). The number of moles of the gas in each chamber remains one as the gas is not leaked from one chamber to another. Apply ideal gas equation at initial and final states of the two chambers to get p0 V0 /T0 = p1 V1 /T1 = (243p0 /32)V1 /T1 ,

(2)

p0 V0 /T0 = p2 V2 /T2 = (243p0 /32)V2 /T2 ,

(3)

where p1 = p2 = 243p0 /32 is the common final pressure of the two chambers. Eliminate V1 and V2 from equations (1)–(3) to get T1 + T2 = (243/16)T0 .

(4)

There is no heat exchange between the right chamber and its surroundings. Thus, the gas in the right chamber undergoes adiabatic compression. Apply the adiabatic equation, p1−γ T γ = constant, at initial and final states of the right chamber to get p1−γ T0γ = p1−γ T2γ = (243p0 /32)1−γ T2γ , 0 2

γ=

Cv + R 3R/2 + R 5 Cp = = = . Cv Cv 3R/2 3

In an adiabatic process pi Viγ = pf Vfγ . Substitute the values to get the final pressure of the gas pf = pi (Vi /Vf )γ = 105 (6/2)5/3 = 6.24 × 105 N/m2 . The work done by a perfect gas in an adiabatic process is given by pi Vi − pf Vf γ−1 (105 ) (6 × 10−3 ) − (6.24 × 105 ) (2 × 10−3 ) = 5/3 − 1 = −972 J.

W =

The minus sign indicates that the work is done on the gas. Ans. −972 J Q 53. A cyclic process ABCA shown in the V -T diagram is performed with a constant mass of an ideal gas. Show the same process on a p-V diagram. [In the figure, line AB passes through origin.] (1981) V

which gives  T2 = T0  = T0

Sol. Given, initial pressure pi = 105 N/m2 , initial volume Vi = 6 litre = 6 × 10−3 m3 , final volume Vf = 2 litre = 2 × 10−3 m3 , and molar specific heat at constant volume Cv = 3R/2. The ratio of specific heats is

32 243

(1−γ)/γ

2431/5 321/5

 = T0

2

32 243

V2

−2/5

9 = T0 . 4

V1

(5)

Solve equations (4) and (5) to get T1 = 207 16 T0 . The gas in the right chamber undergoes adiabatic compression i.e., ∆Q = 0. Apply the first law of thermodynamics, ∆Q = ∆U + ∆W , to get the work done by the gas in right chamber as ∆W = −∆U = −(U2 − U0 ) = U0 − U2 = n f2 RT0 − n f2 RT2 = n f2 R(T0 − T2 )
= (1) 32 R T0 − 94 T0 = − 15 8 RT0 ,

C

B

A T1

T2

T

Sol. The process AB in the V -T diagram is represented by a straight line passing through the origin. The slope of this straight line is (V2 −V1 )/(T2 −T1 ) and its equation is   V2 − V1 V = T. (1) T2 − T1 V V2

where f = 3 is the degrees of freedom of a monatomic gas. Ans. (a) T1 = 12.94T0 , T2 = 2.25T0 (b) −1.875RT0 Q 52. Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are 105 N/m2 and 6 litre respectively. The final volume of the gas is 2 litre. Molar specific heat of the gas at constant volume is 3R/2. (1982)

C

V1

A

O

T1

B

T2

T

The ideal gas equation gives T = pV /(nR). Substitute it in equation (1) to get   T2 − T1 p= nR = p1 = const. (2) V2 − V1

328

Part IV. Thermodynamics p p1

A

p2 O

B

C V1

V2

V

Thus, the process AB is an isobaric process at constant pressure p1 given by equation (2). The process BC is an isochoric process at constant volume V2 . The pressure at B is p1 . Apply ideal gas equation to get the pressure at C, p2 = (T1 /T2 )p1 . The process CA is an isothermal process at constant temperature T1 . The ideal gas equation, pV = nRT1 , gives a rectangular hyperbola on p-V diagram for constant temperature. The pressure at C is p2 and that at A is p1 . Ans. See solution

Chapter 25 Heat Transfer

One Option Correct

x

Q 1. The end Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 10 ◦ C. Now the end P is maintained at 10 ◦ C, while the end S is heated and maintained at 400 ◦ C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2 × 10−5 K−1 , the change in length of the wire PQ is (2016) (A) 0.78 mm (B) 0.90 mm (C) 1.56 mm (D) 2.34 mm

2κ

Q,R

10◦ C

dl = α∆T dx = α(130x + 10 − 10)dx = 130αxdx.

∆x = 1m

κ

Integrate from x = 0 m to x = 1 m to get increase in length as

S

1 x2 l= 130αxdx = 130α 2 0 0 1 = 130(1.2 × 10−5 ) = 0.78 mm. 2 Z

400◦ C

T

∆x = 1m

In steady state, the rate of heat flow from the end S to R is equal to the rate of heat flow from the end Q to P because the system is insulated from the surrounding. Thus,

T (x) 400◦ C

1m

2m



Q 2. Parallel rays of light of intensity I = 912 Wm−2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant σ = 5.7 × 10−8 Wm−2 K−4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to (2014) (A) 330 K (B) 660 K (C) 990 K (D) 1550 K

which gives T = 140 ◦ C.

0m

1

Detailed analysis of heat transfer reveals that temd2 T perature obeys the differential equation c12 ∂T ∂t = dx2 , where c is a constant. In steady state, temperature is d2 T independent of time i.e., ∂T ∂t = 0 which gives dx2 = 0. Integrate it twice to get T (x) = mx + c. Ans. A

dQSR dQQP = , i.e., dt dt 2κA(T − 10) κA(400 − T ) = , ∆x ∆x

140◦ C 10◦ C

dx

Now, consider a small element dx of the wire PQ located at a distance x from the end P. Initial temperature of this element was 10 ◦ C and final temperature (in steady state) is T (x) = 130x + 10. The coefficient of linear thermal expansion of PQ is 1.2 × 10−5 K−1 . The increase in length of this element due to thermal expansion is given by

Sol. Let steady state temperature of the junction (Q,R) be T . P

Q

P

x

Let P be the origin and x-axis points towards Q. In steady state, the temperature on the wire varies linearly with the distance x i.e., T (x) = mx + c, where m is the slope and c is the intercept on T axis. In wire PQ, the temperature T = 10 ◦ C at x = 0 m and the temperature T = 140 ◦ C at x = 1 m. Substitute these values in T (x) = mx + c to get c = 10 ◦ C and m = 130 ◦ C/m. Thus, steady state temperature on the wire PQ is given by

Sol. Let the steady state temperature of the sphere be T . The sphere gains heat due to incident parallel beam of intensity I and cross-sectional area πr2 . The sphere losses heat due to radiation to the surroundings.

r I

T (x) = 130x + 10. 329

330

Part IV. Thermodynamics The rates of heat gain and heat loss by the sphere

are dQin /dt = πr2 I, dQout /dt = 4πr2 σ(T 4 − (300)4 ). In steady state, rate of heat gain is equal to the rate of heat loss i.e., dQin /dt = dQout /dt. Substitute the values to get  1/4 T = I/(4σ) + (300)4  1/4 ≈ 330 K. = 912/(4 × 5.7 × 10−8 ) + (300)4 Ans. A Q 3. Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity κ and the other 2κ. The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuration II is (2013)

2T

Sol. Let Th and Tl be the temperatures of hot and cold ends and Tm be the temperature at the middle point of configuration I. The rate of heat flow is equal in both the blocks of configuration I because the blocks are connected in series, i.e., ∆QI /∆tI = κA(Th − Tm )/x (1)

Solve equation (1) to get Tm = (Th + 2Tl )/3. Thus, equation (1) becomes

3T

Stefan-Boltzmann law gives the rate of heat loss and heat gain by the middle plate as dQout /dt = σAT04 + σAT04 ,

(2)

Q 5. A body with area A and emissivity e = 0.6 is kept inside a spherical black body. Total heat radiated by the body at temperature T is (2005) (A) 0.6σeAT 4 (B) 0.8σeAT 4 (C) 1.0σeAT 4 (D) 0.4σeAT 4 Sol. By Stefan-Boltzmann law, energy radiated per unit time by a body of surface area A, emissivity e, and absolute temperature T is eσAT 4 . Ans. C Q 6. Variation of radiant energy emitted by the sun, filament of tungsten lamp, and welding arc as a function of its wavelength is shown in the figure. Which of the following option gives the correct match? (2005)

In the configuration II, total rate of heat flow is the sum of heat flows through the two blocks because the blocks are joined in parallel i.e.,

Eλ T3 T2 T1

∆QII /∆tII = κA(Th − Tl )/x + 2κA(Th − Tl )/x = 3κA(Th − Tl )/x.

T0

The steady state condition, dQout /dt = dQin /dt, gives 1/4 T0 = (97/2) T . Ans. C

(A) 2.0 s (B) 3.0 s (C) 4.5 s (D) 6.0 s

∆QI /∆tI = 2κA(Th − Tl )/(3x).

Sol. Let the steady state temperature of the middle plate be T0 . In the steady state, the heat radiated per unit time by the middle plate is equal to the heat received per unit time by it. The middle plate radiates heat from both the surfaces and receives the heat radiated by the first and the third plates.

dQin /dt = σA(2T )4 + σA(3T )4 .

Configuration II Configuration I 2κ κ κ 2κ x

= 2κA(Tm − Tl )/x.

Q 4. Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very large thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T , respectively. The temperature of the middle (i.e. second) plate under steady state condition is (2012)

97 1/4 65 1/4 T (B) 4 T (A) 2
1/4 1/4 (C) 97 T (D) (97) T 2

λ

(3)

Divide equation (2) by (3). Substitute ∆QI = ∆QII and ∆tI = 9 s to get ∆tII = 2 s. Ans. A

(A) (B) (C) (D)

Sun-T1 , Sun-T2 , Sun-T3 , Sun-T1 ,

tungsten tungsten tungsten tungsten

filament-T2 , filament-T1 , filament-T2 , filament-T3 ,

welding welding welding welding

arc-T3 arc-T3 arc-T1 arc-T2

Chapter 25. Heat Transfer

331

Sol. Wien’s displacement law is λmax T = b. The knowledge of λmax or T of the given sources will help in getting the answer. Typical temperature of tungsten filament in lamp is 3000 K. The surface temperature of the sun is 5800 K. The temperature of welding arc varies from 6000 K to 30000 K. Ans. B

which gives

Q 7. In which of the following process, convection does not take place primarily? (2005) (A) Sea and land breeze. (B) Boiling of water. (C) Warming of glass bulb through filament. (D) Circulation of air around furnace.

Q 10. Two identical conducting rods are first connected independently to two vessels, one containing water at 100 ◦ C and the other containing ice at 0 ◦ C. In the second case, the rods are joined end to end and connected to the same vessels. Let q1 and q2 grams per second be the rate of melting of ice in the two cases, respectively. The ratio q1 /q2 is (2004) (A) 1/2 (B) 2 (C) 4 (D) 1/4

Sol. The glass bulb heats up due to radiations from the filament. Convection is the heat transfer by the movement of the medium. Ans. C αZ

− kθ , p is pressure, Z is Q 8. In the relation p = α βe distance, k is Boltzmann constant and θ is temperature. The dimensional formula of β is (2004) (A) [M0 L2 T0 ] (B) [ML2 T1 ] (C) [ML0 T−1 ] (D) [M0 L2 T−1 ]

Sol. The argument of exponent, αZ kθ , is dimensionless. Thus dimensions of α are same as dimensions of kθ Z = ML2 T−2 −2 = MLT (since kθ is energy). The dimensions L MLT−2 0 2 0 of β are same as that of αp = ML −1 T−2 = [M L T ]. Ans. A Q 9. Three discs, A, B and C having radii 2 m, 4 m and 6 m, respectively are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm, respectively. The power radiated by them are QA , QB and QC . Then, (2004) (A) QA is maximum (B) QB is maximum (C) QC is maximum (D) QA = QB = QC Sol. The temperature of a black body and the wavelength corresponding to maximum intensity are related by Wien’s displacement law, λm T = b.

(1)

The power radiated by a black body at temperature T , with a surface area A and emissivity e is given by Stefan’s law, Q = σeAT 4 .

(2)

Thin disc of radius r emits from both the surfaces. Thus, A = 2πr2 . Eliminate T from equations (1) and (2) and substitute for A to get
Q = 2πσeb4 r2 /λ4m ,

QA : QB : QC =

2 r2 r2 rA : B : C 4 4 λA λB λ4m

= 0.049 : 0.0625 : 0.057. Ans. B

Sol. The rate of heat conduction through a material, with conductivity κ, cross-section area A, length ∆x, and temperature difference between the two ends ∆T , is given by ∆Q/∆t = κA∆T /∆x. 0◦ C

100◦ C l

In the case (1), two rods are connected in parallel. The rate of heat transfer through each rod is, ∆Q/∆t = κA(100/l). Thus, the rate of heat transfer to the ice is ∆Q1 /∆t = 2(∆Q/∆t) = κA(200/l). 0◦ C

(1)

100◦ C 2l

In the case (2), two identical rods are connected in series, with effective length 2l. The rate of heat transfer to the ice is given by ∆Q2 /∆t = κA(100/2l).

(2)

The heat transferred to the ice is used to melt it. The rate of melting is q = ∆m/∆t = (1/L)∆Q/∆t, where L is the latent heat of fusion. Use equations (1) and (2) to get the ratio of rate of melting in two cases ∆Q2 1 i.e., q1 /q2 = ∆Q ∆t / ∆t = 4. Ans. C Q 11. The temperature (T ) versus time (t) graphs of two bodies X and Y with equal surface areas are shown in the figure. If the emissivity and the absorptivity of X and Y are Ex , Ey and ax , ay , respectively, then, (2003)

332

Part IV. Thermodynamics Sol. Let T be the temperature of the junction. The rate of heat flow to the junction is

T

dQin /dt = κA(90 − T )/l + κA(90 − T )/l

y x

(A) (B) (C) (D)

Ex Ex Ex Ex

> Ey < Ey > Ey < Ey

and and and and

ax ax ax ax

= 2κA(90 − T )/l,

t

(1)

and the rate of heat flow out of the junction is

< ay > ay > ay < ay

dQout /dt = κA(T − 0)/l = κAT /l.

Sol. The rate of cooling for x is greater than that of y. The cooling occurs due to the heat loss by radiation. By Stefan’s law, the rate of heat loss is dQ/dt = σEA(T 4 − T04 ), where E is emissivity. Hence, Ex > Ey . Since good emitters are good absorbers, the absorptivity ax > ay . Ans. C Q 12. An ideal black-body at room temperature is thrown into a furnace. It is observed that, (2002) (A) initially it is the darkest body and at later times the brightest. (B) it is the darkest body at all times. (C) it cannot be distinguished at all times. (D) initially it is the darkest body and at later times it cannot be distinguished. Sol. Let T be the temperature of the furnace and T0 be the room temperature. When a black body at temperature T0 is thrown into the furnace it emits radiation energy at a rate ebody σAT04 = σAT04 (since ebody = 1). On the other hand, a portion of the furnace having same area A emits the radiation energy at a rate efurnace σAT 4 . Generally, σAT04 < efurnace σAT 4 , (because T0  T ), so the black body appears as the darkest body. The temperature of the black body rises till it reaches T . At this temperature, the black body emits the radiation energy at a rate σAT 4 whereas the furnace emits the radiation energy at a rate efurnace σAT 4 . Since efurnace < 1, the rate of radiation energy emitted by the black body is more than that by the furnace so the black body appears to be the brightest. Ans. A Q 13. Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0 ◦ C and 90 ◦ C respectively. The temperature of junction of the three rods will be (2001) 90◦ C

By the energy conservation, dQin /dt = dQout /dt. Use equations (1) and (2) to get T = 60 ◦ C. Ans. B Q 14. The plots of intensity versus wavelength for three black bodies at temperatures T1 , T2 and T3 , respectively are as shown in the figure. Their temperatures are such that (2000)

I T1

T3

T2

λ

(A) T1 > T2 > T3 (C) T2 > T3 > T1

(B) T1 > T3 > T2 (D) T3 > T2 > T1

Sol. Wien’s displacement law relates the wavelength at the maximum intensity to the temperature of the body, λm T = b. In the given cases, λm,1 < λm,3 < λm,2 and hence T1 > T3 > T2 . Ans. B Q 15. A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1 , between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3 . [The Wien constant, b = 2.88 × 106 nm K]. Then, (1998) (A) U1 = 0 (B) U3 = 0 (C) U1 > U2 (D) U2 > U1 Sol. Wien’s displacement law gives λm =

b 2.88 × 106 = = 1000 nm. T 2880

Thus the radiated energy Eλ attains its maximum at λm = 1000 nm. The variation of Eλ with λ is shown in the figure. Eλ

0◦ C U3 1499 1500

U2 999 1000

(A) 45 ◦ C (B) 60 ◦ C (C) 30 ◦ C (D) 20 ◦ C

U1 499 500

90◦ C

(2)

λ (nm)

Chapter 25. Heat Transfer

333

The area under the curve for a given wavelength range gives the radiated energy in that wavelength range i.e., Uλ = Eλ ∆λ. The figure clearly shows that U1 < U3 < U2 . Ans. D Q 16. A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated (in W) would be (1997) (A) 225 (B) 450 (C) 900 (D) 1800 Sol. The radiated power of a black body at temperature T , with surface area A and emissivity e, is given by dQ/dt = σeAT 4 = 450 W.

(1)

Initially, A = 4πr2 and temperature is T . Finally, A0 = 4π(r/2)2 = A/4 and T 0 = 2T . Substitute these values to get dQ0 /dt = σeA0 T 04 = 4σeAT 4 .

(2)

Divide equation (2) by (1) to get dQ0 /dt = 1800 W. Ans. D Q 17. The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the north star has the maximum value at 350 nm. If these stars behaves like black bodies, then the ratio of the surface temperature of the sun and the north star is (1997) (A) 1.46 (B) 0.69 (C) 1.21 (D) 0.83 Sol. Wien’s displacement law relates the wavelength at the maximum intensity to the temperature of the body, λm T = b. Thus, λm,star 350 Tsun = 0.69. = = Tstar λm,sun 510 Ans. B Q 18. Two metallic spheres S1 and S2 are made of the same material and have got identical surface finish. The mass of S1 is thrice that of S2 . Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of S1 to that of S2 is (1995) √
1/3 1 1 1 (A) 3 (B) √3 (C) 3 (D) 3 Sol. Let r1 and r2 be the radii of S1 and S2 and ρ be the density of the material. The masses of S1 and S2 are m1 = (4/3)πr13 ρ,

m2 = (4/3)πr23 ρ.

Since m1 = 3m2 , we get r1 = 31/3 r2 .

Let T be the common temperature of the two spheres and T0 be the room temperature. The rates of radiation heat loss by S1 and S2 are dQ1 /dt = 4πσer12 (T 4 − T04 ), dQ2 /dt =

4πσer22 (T 4

−

(1)

T04 ).

(2)

The temperatures of the spheres reduce due to radiation heat losses. The rates of temperature change for S1 and S2 are given by dQ1 /dt = −m1 S(dT1 /dt),

(3)

dQ2 /dt = −m2 S(dT2 /dt).

(4)

Eliminate dQ1 /dt and dQ2 /dt from equations (1)–(4) to get dT1 /dt = −3σe(T 4 − T04 )/(r1 ρS), 4

dT2 /dt = −3σe(T −

T04 )/(r2 ρS).

(5) (6)

Divide equation (5) by (6) to get the ratio of rate of cooling as  1/3 1 r2 dT1 /dt = = . dT2 /dt r1 3 Ans. D Q 19. Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC, right angled at B. The points A and B are √ maintained at temperatures T and 2T , respectively. In the steady state, the temperature of the point C is Tc . Assuming that only heat conduction takes place, Tc /T is (1995) 3 1 (C) √3 √12−1 (D) √2+1 (A) 2 √12−1 (B) √2+1 ( ) ( ) √ Sol. Let TA = T , TB = 2T , and the steady state temperature of C is TC . Since TB > TA , heat flows from B to A. If TC < T then heat flows to C from A as well as from √ B leading to a non-steady state. Similarly, if TC > 2T then the heat flows from C to A as well as to B leading to a√non-steady state. Thus, in the steady state T < TC < 2T and the heat flow from B to C is equal to heat flow from C to A. TC √

A T

2x

x

C x

√

B 2T

The rates of heat flow from B to C and from C to A are given by √ dQBC κA(TB − TC ) κA( 2T − TC ) = = , (1) dt x x dQCA κA(TC − TA ) κA(TC − T ) √ √ = = . (2) dt 2x 2x

334

Part IV. Thermodynamics

In the steady state, dQBC /dt = dQCA /dt. Use equations (1) and (2) to get √ √ √ TC = (TA + 2TB )/(1 + 2) = 3T /(1 + 2). Ans. B Q 20. A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of material of thermal conductivity K2 . The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is (1988) K1 +3K2 3K1 +K2 K1 K2 (D) (A) K1 + K2 (B) K1 +K2 (C) 4 4 Sol. Let ∆T = (Th − Tl ) be the temperature difference between the two ends and x be the length of the cylinders. 2R R

Th

Tl

K1 K2 x

The rate of heat flow through the cylinder of radius R and thermal conductivity K1 is dQ1 /dt = K1 A1 ∆T /x = K1 (πR2 )∆T /x. The rate of heat flow through the cylindrical shell of inner radius R, outer radius 2R, and thermal conductivity K2 is dQ2 /dt = K2 A2 ∆T /x = K2 (4πR2 − πR2 )∆T /x 2

= 3K2 πR ∆T /x.

Sol. According to Wien’s displacement law (λm T = b), if absolute temperature T of the body is increased then wavelength λm corresponding to the peak intensity in the spectrum of electromagnetic radiation emitted by the body decreases i.e., peak of the spectrum shifts to shorter wavelengths. The rate of radiation energy emitted by a body of surface area A, emissivity e, and kept at an absolute temperature T is given by Stefan’s law dQe /dt = σeAT 4 , where σ is Stefan’s constant. If this body is kept at a surrounding temperature T0 then rate of radiation energy absorbed by the body is dQa /dt = σeAT04 . Thus, the net rate of energy loss by the body is dQi /dt = dQe /dt − dQa /dt = σeA(T 4 − T04 ).

The rate of heat flow through the system from one end to the other end is dQ/dt = dQ1 /dt + dQ2 /dt = (K1 + 3K2 )(πR2 )∆T /x
2 = K1 +3K (4πR2 )∆T /x 4 = Keff Aeff ∆T /x.

the value of σT04 = 460 W/m2 (where σ is the StefanBoltzmann constant). Which of the following option(s) is(are) correct? (2017) (A) If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths. (B) If the surrounding temperature reduces by a small amount ∆T0  T0 , then to maintain the same body temperature the same (living) human being needs to radiate ∆W = 4σT03 ∆T0 more energy per unit time. (C) The amount of energy radiated by the body in 1 s is close to 60 Joules. (D) Reducing the exposed surface area of the body (e.g., by curling up) allows human to maintain the same body temperature while reducing the energy lost by radiation.

(1)

From equation (1), the effective thermal conductivity is Keff = (K1 + 3K2 )/4. Ans. C One or More Option(s) Correct Q 21. A human body has a surface area of approximately 1 m2 . The normal body temperature is 10 K above the surrounding room temperature T0 . Take the room temperature to be T0 = 300 K. For T0 = 300 K,

If surrounding temperature is reduced by ∆T0 ( T0 ) then net rate of energy lost by the body becomes h i dQf 4 = σeA T 4 − (T0 − ∆T0 ) dt "  4 # ∆T0 4 4 = σeA T − T0 1 − T0    ∆T0 ≈ σeA T 4 − T04 1 − 4 T0 dQi = + 4σeAT03 ∆T0 . dt Thus, decrease in the surrounding temperature increases the rate of energy loss by 4σeAT03 ∆T0 . To maintain the body temperature, the human being needs to increase the rate of energy generation (through chemical reactions in the body) by 4σeAT03 ∆T0 to balance the increase in rate of energy loss.

Chapter 25. Heat Transfer

335

The human beings can also maintain the body temperature (without increasing the rate of energy generation) by reducing the surface area (e.g., by curling up) of their body. If surrounding temperature is reduced by ∆T0 and surface area of the body is reduced by ∆A ( A) then net rate of energy loss by the body becomes h i dQf 4 = σe(A − ∆A) T 4 − (T0 − ∆T0 ) dt dQi + 4σeAT03 ∆T0 − σe∆A(T 4 − T04 ). ≈ dt

where x, y, and z are constants to be determined. Equate the exponents of M, L, and T on the two sides of equation (1) to get

∆T0 4AT03 ∆T0 ≈A , T 4 − T04 ∆T

where, the body temperature is T = T0 + ∆T . We assume body to be black i.e., e = 1. The amount of energy emitted by the body in one second is 4

dQe /dt = σeAT = σeA(T0 + ∆T )

(2)

0 = 2x + y + 3z,

(3)

0 = −x − y − 2z.

(4)

Solve equations (2)–(4)√to√get√x = 1/2, y = 1/2, and z = −1/2. Thus, M ∝ h c/ G. Similarly, the dimensions of length can be written as

Thus, the rate of energy loss by radiation remains same (dQf /dt = dQi /dt) if reduction in surface area is ∆A =

1 = x − z,

[L]1 = [h]x [c]y [G]z   = Mx−z L2x+y+3z T−x−y−2z .

Equate the exponents of M, L, and T on the two sides of equation (5) to get 0 = x − z,

(6)

1 = 2x + y + 3z,

(7)

0 = −x − y − 2z.

(8)

4

Solve equations (6)–(8) to get x = 1/2, y = −3/2, and z = 1/2. Thus, L ∝ h1/2 G1/2 c−3/2 . Ans. A, C, D

≈ eAσT04 (1 + 4∆T /T0 ) = (1)(1)(460)(1 + 4(10)/300) = 521 J. The amount of energy lost by the body in one second is
dQi /dt = σeA(T 4 − T04 ) = σeA T0 + ∆T )4 − T04 ≈ eAσT04 (4∆T /T0 ) = 61 J. The question is little ambiguous because reader may interpret ‘energy radiated by the body’ as ‘energy radiated by the body through emission’ or ‘net energy radiated by the body’. Also, emissivity of the body is not given. Ans. (D) Q 22. Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M . Then the correct option(s) is (are) (2015) √ √ (A) M ∝√ c (B) M ∝√ G (C) L ∝ h (D) L ∝ G Sol. The formula E = hν gives the dimensions of Planck’s constant as [h] = [M1 L2 T−1 ] and the formula F = Gm1 m2 /r2 gives the dimensions of universal gravitational constant as [G] = [M−1 L3 T−2 ]. The dimensions of speed of light is [c] = [L1 T−1 ]. The dimensions of mass can be written in terms of h, c, and G as

Q 23. A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then in steady state, (2011) 0 1L

heat

x

y

A 2K

3L

(A) (B) (C) (D)

heat flow through A and E slabs are same. heat flow through slab E is minimum. temperature difference across slab E is minimum. heat flow through C = heat flow through B + heat flow through D.

Sol. Let T1 , T2 , T3 and T4 be the temperatures at slab interfaces as shown in the figure. 0

1L

5L 6L B 3K

A 2K 3L

(1)

E 6K

C 4K D 5K

4L

z

[M] = [h] [c] [G]  x  1 −1 y  −1 3 −2 z = M1 L2 T−1 L T M L T  x−z 2x+y+3z −x−y−2z  = M L T ,

5L 6L B 3K

1L

1L 1

(5)

4L T1 T2

E C 4K

6K

D 5K T3 T4

336

Part IV. Thermodynamics

The rate of heat flow across a slab, with thermal conductivity κ, width w, thickness l, length x, and temperature difference ∆T , is given by dQ/dt = κlw (∆T /x) . Thus, the heat flow rates through the given slabs are dQA (2K)(4Lw)(T1 − T2 ) = = 8Kw(T1 − T2 ), dt L (1) dQB (3K)(Lw)(T2 − T3 ) 3 = = Kw(T2 − T3 ), dt 4L 4 (2) (4K)(2Lw)(T2 − T3 ) dQC = = 2Kw(T2 − T3 ), dt 4L (3) dQD (5K)(Lw)(T2 − T3 ) 5 = = Kw(T2 − T3 ), dt 4L 4 (4) (6K)(4Lw)(T3 − T4 ) dQE = = 24Kw(T3 − T4 ). dt L (5) In the steady state, heat flow into the system is equal to the heat flow out of the system i.e., dQA /dt = dQE /dt. Use equations (1) and (5) to get T1 − T2 = 3(T3 − T4 ).

(6)

Similarly, the steady state condition for the interface at temperature T2 is dQA /dt = dQB /dt + dQC /dt + dQD /dt, which gives (by using equations (1)–(4)) 2(T1 − T2 ) = T2 − T3 .

(7)

Use equations (6) and (7) to get T3 − T4 = (1/3)(T1 − T2 ) = (1/6)(T2 − T3 ). Also, equations (2)–(4) give dQC /dt = dQB /dt + dQD /dt. Ans. A, C, D Q 24. In a dark room with ambient temperature T0 , a black body is kept at temperature T. Keeping the temperature of the black body constant (at T ), sun rays are allowed to fall on the black body through a hole in the roof of the dark room. Assuming that there is no change in the ambient temperature of the room, which of the following statement(s) is (are) correct? (2006)

(A) The quantity of radiation absorbed by the black body in unit time will increase. (B) Since emissivity = absorptivity, hence the quantity of radiation emitted by black body in unit time will increase. (C) Black body radiates more energy in unit time in the visible spectrum. (D) The reflected energy in unit time by the black body remains same. Sol. When sun rays fall on a black body, a portion of incident photons are absorbed and the remaining are reflected. So, the net quantity of radiations absorbed by the black body in unit time increases. At the same time, since the emissivity and the absorptivity are equal, the quantity of radiations emitted by the black body in unit time also increases. The wavelength λmax at which the maximum radiation takes place depends on the black body temperature T and is given by Wien’s displacement law λmax T = b. Ans. A, B Q 25. Two bodies A and B have thermal emissivities of 0.01 and 0.81, respectively. The outer surface area of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength λB corresponding to maximum spectral radiance in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiance in the radiation from A by 1.00 µm. If the temperature of A is 5802 K, (1994)

(A) (B) (C) (D)

the temperature of B is 1934 K. λB = 1.5 µm. the temperature of B is 11604 K. the temperature of B is 2901 K.

Sol. Let eA = 0.01, eB = 0.81, and TA = 5802 K. Stefan’s law gives the radiant powers of the two bodies as dQA /dt = σAeA TA4 , dQB /dt = σAeB TB4 . Solve dQA /dt = dQB /dt to get 1/4

TB = (eA /eB )

T

× 5802

= 1934 K. Wien’s displacement law, λm T = b, gives λB = (TA /TB )λA = (5802/1934)λA = 3λA .

(1)

Also, since λB > λA and |λB − λA | = 1 µm, we get λB − λA = 1 µm.

T0

1/4

TA = (0.01/0.81)

(2)

Solve equations (1) and (2) to get λA = 0.5 µm and λB = 1.5 µm. Ans. A, B

Chapter 25. Heat Transfer

337

Matrix or Matching Type Q 26. Match the physical quantities given in List I to their dimensions in List II. (2013) Column I (P) (Q) (R) (S)

dQ1 /dt 4πσe (1)2 (4000)4 = 1. = dQ2 /dt 4πσe (4)2 (2000)4

Column II

Boltzmann constant Coefficient of viscosity Planck constant Thermal conductivity

(1) (2) (3) (4)

where σ is Stefan’s constant, e is emissivity, and A is surface area. Substitute the given values of radii and temperatures in equation (1) to get

[ML2 T−1 ] [ML−1 T−1 ] [MLT−3 K−1 ] [ML2 T−2 K−1 ]

Ans. F Fill in the Blank Type

Sol. A particle at temperature T has the thermal energy E = kT (k is Boltzmann constant). A photon of light with frequency ν has the energy E = hν (h is Planck constant). The viscous force exerted by a layer of area A on the fluid having a velocity gradient dv dz is dv given by F = −ηA dz (η is the coefficient of viscosity). The rate of heat flow through a slab of area A is related dQ dT to the temperature gradient, dT dx , by dt = κA dx (κ is the thermal conductivity). Ans. P7→4, Q7→2, R7→1, S7→3 Q 27. Column I gives some devices and Column II gives some processes on which the functioning of these devices depend. Match the devices in Column I with the processes in Column II. (2007)

Q 29. The earth receives 1400 W/m2 of solar power. If all the solar energy falling on a lens of area 0.2 m2 is focused onto a block of ice of mass 280 g, the time taken to melt the ice will be . . . . . . minutes. [ Latent heat of fusion of ice = 3.3 × 105 J/kg. ] (1997) Sol. The solar energy received per second at the lens of area 0.2 m2 is P = (1400)(0.2) = 280 J/s. The heat energy required to melt the ice block is Q = mL = (0.280)(3.3 × 105 ) = 9.24 × 104 J. Thus, time required to melt the ice is t=

9.24 × 104 Q = = 330 s = 5.5 minutes. P 280 Ans. 5.5

(B) Steam engine (C) Incandescent lamp (D) Electric fuse

(p) Radiation from a hot body (q) Energy conversion (r) Melting (s) Thermal expansion of solids

Sol. The bimetallic strip works on the differential thermal expansion of two metals. Steam engine converts the heat energy into the mechanical energy. Incandescent lamp radiates light when it becomes hot. The electric fuse breaks the circuit by melting if the current exceeds the safety limits. Ans. A7→s, B7→q, C7→p, D7→r True False Type Q 28. Two spheres of the same material have radii 1 m and 4 m, temperature 4000 K and 2000 K respectively. The energy radiated per second by the first sphere is greater than that by the second. (1988) Sol. By Stefan’s law, the energy radiated per second by a sphere of radius r at temperature T is given by dQ/dt = σeAT 4 = 4πσer2 T 4 ,

(1)

Q 30. Two metal cubes A and B of the same size are arranged as shown in the figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of A and B are 300 W/m ◦ C and 200 W/m ◦ C, respectively. After steady state is reached the temperature T of the interface will be . . . . . . (1996)

A

B

0◦ C

(A) Bimetallic strip

Column II

100◦ C

Column I

T

Sol. Let a be the side length of the cubes. The rates of heat flow through the cube A and cube B are given by dQA KA a2 (100 − T ) = = KA a(100 − T ), dt a dQB KB a2 (T − 0) = = KB aT. dt a

(1) (2)

In the steady state, dQA /dt = dQB /dt. Use equations (1) and (2) to get the temperature of the interface, T = 100KA /(KA + KB ) = 100(300)/(300 + 200) = 60 ◦ C. Ans. 60 ◦ C

338

Part IV. Thermodynamics

Q 31. A solid copper sphere (density ρ and specific heat c) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. The time required for the temperature of the sphere to drop to 100 K is . . . . . . (1991) Sol. Stefan’s law gives the rate of energy radiated by the sphere as dQ/dt = σAT 4 = 4πσr2 T 4 .

(1)

Q 33. The earth receives at its surface radiation from the sun at the rate of 1400 W/m2 . The distance of the centre of the sun from the earth is 1.5 × 1011 m and the radius of the sun is 7 × 108 m. Treating the sun as a black body, it follows from the above data that its surface temperature is . . . . . . K. (1989) Sol. Given, radius of the sun r = 7 × 108 m, distance of the earth from the centre of the sun R = 1.5 × 1011 m, and the intensity at the earth surface I = 1400 W/m2 .

The temperature of the sphere reduces due to energy loss. The rate of energy loss is related to the rate of temperature decrease by

R

dQ/dt = −mc (dT /dt)

r

3

= −(4/3)πr ρc (dT /dt) .

S

(2)

Eliminate dQ/dt from equations (1) and (2) to get ρrc (dT /dt) = −3σT 4 .

(3) Let the absolute temperature of the sun’s surface be T . By Stefan’s law, the rate of energy radiated by the sun is

Integrate equation (3), Z Tf Z t −4 ρrc T dT = −3σ dt, Ti

dQ/dt = σAT 4 = σ(4πr2 )T 4 .

0

(1)

to get Thus, rate of energy received per unit area at the earth surface is

" # 1 ρrc 1 − 3 t= 9σ Tf3 Ti   1 ρrc 1 − = 2003 9(5.67 × 10−8 ) 1003 = 1.71ρrc.

I=

dQ/dt σr2 T 4 = . 4πR2 R2

(2)

Substitute the values in equation (2) to get

Ans. 1.71ρrc

IR2 T = σr2 

"

1/4 =


2 #1/4 (1400) 1.5 × 1011
2
5.67 × 10−8 7 × 108

Q 32. A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. If the temperature difference between the outer and inner surface of the shell is not to exceed T , the thickness of the shell should not be less than . . . . . . (1991)

Q 34. Planck’s constant has dimensions . . . . . .

Sol. The heat produced per unit time by the source is P . This heat is distributed over the inner surface of the spherical shell having an area A = 4πR2 . Let x be the thickness of the shell. The rate of heat transfer due to conduction is

Sol. Plank’s constant is h = E/ν, where E is energy and ν is frequency. The dimensions of energy are ML2 T−2 and that of frequency are T−1 . Thus, Plank’s constant has dimensions ML2 T−1 . Ans. [ML2 T-1 ]

dQ/dt = KAT /x,

(1)

where T is the temperature difference between the inner and outer surfaces of the shell. For the temperature T to be constant, the rate of heat incident on the inner surface of the shell should be equal to the rate of heat transfer through conduction i.e., P = dQ/dt. Use equation (1) to get x = KAT /P = 4πR2 KT /P. Ans.

4πR2 KT P

= 5803 K. Ans. 5803 (1985)

Integer Type Q 35. Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T1 = 300 K and T2 = 100 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1 and K2 respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, the K1 /K2 =. . . . . . . (2018)

Chapter 25. Heat Transfer

339 It is given that x1 = 1. Subtract equation (1) from (2) and simplify to get
x2 = x1 + log2 σeA(3040)4 /P0
− log2 σeA(760)4 /P0   (3040)4 = 1 + 4 log2 4 = 9. = 1 + log2 (760)4

Insulating Material

T1

K1

K2

L

L

T2

Sol. The temperature of the source is T1 = 300 K and that of the sink is T2 = 100 K. The temperature at the junction of two cylinders is T = 200 K. The radius of the bigger cylinder, r2 , is twice the radius of the smaller cylinder, r1 i.e., r2 = 2r1 . The rate of heat conduction through a material, with conductivity K, cross-section area A, length ∆x, and temperature difference between the two ends ∆T , is given by ∆Q/∆t = KA∆T /∆x. Thus, the rates of heat conduction through the two cylinders are ∆Q1 /∆t = K1 (πr12 )(T1 − T )/L1 , ∆Q2 /∆t = K2 (πr22 )(T − T2 )/L2 . There is no heat loss from the cylinders because they are covered with an insulating material. Also, there is no heat accumulation in steady state. Thus, ∆Q1 /∆t = ∆Q2 /∆t, which gives r2 (T − T2 )/L2 (2r1 )2 (200 − 100)/L K1 = 22 = = 4. K2 r1 (T1 − T )/L1 r12 (300 − 200)/L Ans. 4 Q 36. A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P ) by the metal. The sensor has a scale that displays log2 (P/P0 ), where P0 is a constant. When the metal surface is at a temperature of 487 ◦ C, the sensor shows a value of 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767 ◦ C? (2016) Sol. The power radiated by a metal surface is given by P = σeAT 4 , where σ is Stefan’s constant, e is emissivity of the metal, A is surface area and T is the absolute temperature of the metal. Thus, the powers radiated at temperatures T1 = 487 ◦ C = 487 + 273 = 760 K and T2 = 2767 ◦ C = 3040 K are 4

P1 = σeA(760) ,

Ans. 9 Q 37. Two spherical stars A and B emits blackbody radiation. The radius of A is 400 times that of B and A emits 104 times the power emitted from B. The ratio λA /λB of their wavelengths λA and λB at which the peaks occur in their respective radiation curves is . . . . . . . (2015) Sol. Let TA and TB be the temperatures and RA and RB be the radii of the stars A and B. Stefan’s law gives the power radiated by the two stars as 2 PA = (dQ/dt)A = σ 4πRA TA4 ,

PB = (dQ/dt)B = σ

2 4πRB

(1)

TB4 .

(2)

Given RA = 400RB and PA = 104 PB . Divide equation (2) by (1) to get  1/4  1/2 TB PB RA = = 2. (3) TA PA RB Apply Wien’s displacement law to get λA /λB = TB /TA = 2. Ans. 2 Q 38. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1 and T2 , respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B? (2010)

Sol. Stefan-Boltzmann law gives the energy radiated per unit time by a spherical black body of area A = 4πr2 and temperature T as E = σAT 4 = 4πσr2 T 4 .

(1)

Wien’s displacement law relates the temperature of the black body to the wavelength at maximum intensity by λm T = b.

(2)

Eliminate T from equations (1) and (2) to get E = 4πσb4 (r2 /λ4 ),

4

P2 = σeA(3040) .

The sensor’s measurement for these radiated powers are
x1 = log2 (P1 /P0 ) = log2 σeA(760)4 /P0 , (1)
4 x2 = log2 (P2 /P0 ) = log2 σeA(3040) /P0 . (2)

which gives E1 /E2 = (r1 /r2 )2 (λ2 /λ1 )4 = (6/18)2 (1500/500)4 = 9. Ans. 9

340

Part IV. Thermodynamics

Q 39. A metal rod AB of length 10x has its one end A in ice at 0 ◦ C and the other end B in water at 100 ◦ C. If a point P on the rod is maintained at 400 ◦ C, then it is found that equal amounts of water and ice evaporates and melts per unit time. The latent heat of evaporation of water is 540 cal/g and latent heat of melting of ice is 80 cal/g. If the point P is at a distance of λx from the ice end A, find the value of λ. [Neglect any heat loss to the surroundings.] (2009) Sol. The situation is shown in the figure. 0◦ C

P

KA(T1 − T2 )/L = σeA(T24 − Ts4 ).

(1)

Now, T2 = Ts + ∆T gives T24 = Ts4 (1 + ∆T /Ts )4 ≈ Ts4 (1 + 4∆T /Ts ) or T24 − Ts4 = 4Ts3 ∆T. Substitute T2 = Ts + ∆T and T24 − Ts4 = 4Ts3 ∆T in equation (1) and simplify to get

400◦ C 100◦ C

A

Sol. The heat gain through conduction by the end exposed to the surroundings is equal to the radiation heat loss by it i.e.,

B

λx

∆T =

10x

Let A be the cross-sectional area of the rod and k be the thermal conductivity of the rod’s material. The rates of heat flow from the point P towards the end A and the end B, are given by dQA /dt = kA(TP − TA )/(λx) = 400kA/(λx), dQB /dt = kA(TP − TB )/(10x − λx) = 300kA/(10x − λx). Let the rate of melting of ice be dmi /dt and the rate of evaporation of water be dmw /dt. The rates of heat required for the ice and the water are dQA /dt = (dmi /dt)Li ,

K(T1 − Ts ) . 4eσLTs3 + K Ans.

K 4eσLTs3 +K

Q 41. The top of an insulated cylindrical container is covered by a disc having radiation emissivity 0.6, thermal conductivity 0.167 W m−1 K−1 and thickness 1 cm. The temperature is maintained by circulating oil as shown in the figure. The temperature of the upper surface of disc is 127 ◦ C and temperature of the surrounding is 27 ◦ C. Find (a) the rate of radiation loss to the surroundings by unit area of the disc and (b) the temperature of the circulating oil. Neglect the heat loss −8 due to convection. [Given σ = 17 W m−2 K−4 .] 3 × 10 (2003) out

dQB /dt = (dmw /dt)Lw , where Li = 80 cal/g and Lw = 540 cal/g are the latent heats of fusion and evaporation, respectively. Use dmi /dt = dmw /dt and simplify to get,     400 10 − λ 80 · = , λ 300 540 which gives, λ = 9. Ans. 9 Descriptive Q 40. One end of a rod of length L and cross-sectional area A is kept in a furnace of temperature T1 . The other end of the rod is kept at temperature T2 . The thermal conductivity of the material of the rod is K and emissivity of the rod is e. It is given that T2 = Ts + ∆T , where ∆T  Ts , Ts being the temperature of the surroundings. If ∆T ∝ (T1 − Ts ), find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is T2 . (2004) Insulated Furnace T1

Rod L Insulated

Ts T2

in

Sol. The temperature of the radiating surface is T = 273 + 127 = 400 K and that of the surroundings is T0 = 273 + 27 = 300 K. The rate of heat loss by radiation to the surroundings per unit area, is given by
1 dQrad = eσ T 4 − T04 A dt  
17 −8 = 0.6 × 10 4004 − 3004 3 = 595 W/m2 .

(1)

The temperature can be maintained when the rate of heat loss due to radiation is equal to the rate of heat gain (from oil) due to conduction. The rate of heat gain by conduction per unit area, is given by 1 dQcon K(Toil − T ) 0.167(Toil − 400) = = . (2) A dt x 0.01 Equate the right hand sides of equations (1) and (2) to get Toil = 435.6 K = 162.6◦ C. Ans. (a) 595 W/m2 (b) 162.6 ◦ C

Chapter 25. Heat Transfer

341

Q 42. A solid body X of heat capacity C is kept in an atmosphere whose temperature is TA = 300 K. At time t = 0, the temperature of X is T0 = 400 K. It cools according to Newton’s law of cooling. At time t1 its temperature is found to be 350 K. At this time (t1 ) the body X is connected to a large body Y at atmospheric temperature TA through a conducting rod of length L, cross-sectional area A and thermal conductivity K. The heat capacity of Y is so large that any variation in its temperature may be neglected. The cross-sectional area A of the connecting rod is small compared to the surface area of X. Find the temperature of X at time t = 3t1 . (1998)

From equations (6) and (7), dT /dt = − [ln 2/t1 + KA/(LC)] (T − TA ) Integrate equation (8) from t = t1 to t = 3t1 ,   Z 3t1 Z T2 dT ln 2 KA =− + dt, t1 LC T1 T − TA t1 to get  ln

(1)

Integrate equation (1) from t = 0 to t = t1 , Z t1 Z T1 dT = −k dt, 0 T0 T − TA to get

(2)

The rate of heat loss is related to the rate of temperature change by (3)

Substitute dT /dt from equation (1) and k from equation (2) to get dQr /dt = Ck(T − TA ) = C(T − TA ) ln 2/t1 .

T2 = 300 + 12.5e−

. 

300 + 12.5e−

(6)

which is related to the rate of temperature change of X by (7)

2KAt1 CL



K

Sol. Given, room temperature Tr = 27 ◦ C, atmosphere temperature Ta = 0 ◦ C, glass thickness xg = 0.01 m, air-space thickness xa = 0.05 m, area A = 1 m2 , thermal conductivity of the glass Kg = 0.8 W m−1 K−1 , and thermal conductivity of the air Ka = 0.08 W m−1 K−1 . Let the room-side glass-air interface temperature be T1 and the out-side glass-air interface temperature be T2 . Tr

T1

T2

0.01m

0.05m

Ta

0.01m

The rate of heat flow through the room-side glass is dQg1 /dt = Kg A(Tr − T1 )/xg ,

= C(T − TA )ln 2/t1 + KA(T − TA )/L

(9)

Q 43. A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state, the room glass interface and the glass-outdoor interface are at constant temperature of 27 ◦ C and 0 ◦ C respectively. Calculate the rate of heat flow through the window pane. Also find the temperatures of the other interfaces. Given thermal conductivities of glass and air as 0.8 and 0.08 W m−1 K−1 , respectively. (1997)

(5)

dQt /dt = dQr /dt + dQc /dt

dQt /dt = −CdT /dt.

2KAt1 . LC

(4)

The equations (4) and (5) give the total rate of heat loss by X as

= C [ln 2/t1 + KA/(LC)] (T − TA ),

2KAt1 LC

Ans.

In the second case, heat is lost to the atmosphere by radiation and to the body Y by conduction. The rate of heat loss by X due to conduction is dQc /dt = KA(T − TA )/L.

= −2 ln 2 −

Atmosphere

dQr /dt = −mS (dT /dt) = −C (dT /dt) .



Room

 T1 − TA T0 − TA   350 − 300 = − ln = ln 2. 400 − 300 

kt1 = − ln

T2 − TA T1 − TA

Substitute the values in equation (9) and simplify to get

Sol. In the first case, the body X loses the thermal energy due to radiation. Newton’s law of cooling for a body at temperature T kept in an ambient temperature TA gives the rate of cooling dT /dt = −k(T − TA ).

(8)

(1)

and through the air-space is dQa /dt = Ka A(T1 − T2 )/xa ,

(2)

and through the out-side glass is dQg2 /dt = Kg A(T2 − Ta )/xg .

(3)

342

Part IV. Thermodynamics

In the steady state, dQg1 /dt = dQa /dt. Use equations (1) and (2) to get

Integrate equation (3), Z

Kg xa (Tr − T1 ) = Ka xg (T1 − T2 ).

(4)

Also, dQg2 /dt = dQa /dt gives

T1

dT κA = T0 − T hmS

Z

t

dt, 0

to get

Tr − T1 = T2 − Ta .

(5)

  hmS T0 − T1 ln κA T0 − T2   (0.4)(0.4)(600) 400 − 300 = ln 10(0.04) 400 − 350

t=

Solve equations (4) and (5) to get T1 =

T2

(Kg xa + Ka xg )Tr + Ka xg Ta = 26.48 ◦ C, 2Ka xg + Kg xa

= 166 s.

T2 = 0.52 ◦ C. Substitute these values in equation (2) to get the rate of heat flow dQa Ka A(T1 − T2 ) = dt xa 0.08(1)(26.48 − 0.52) = 41.5 W. = 0.05 Ans. 41.5 W, 26.48 ◦ C, 0.52 ◦ C Q 44. A cylindrical block of length 0.4 m and area of cross-section 0.04 m2 is placed coaxially on a thin metal disc of mass 0.4 kg and of the same cross-section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If the thermal conductivity of the material of the cylinder is 10 W m−1 K−1 and the specific heat capacity of the material of the disc is 600 J kg−1 K−1 , how long will it take for the temperature of the disc to increase to 350 K? Assume, for purpose of calculation, the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder.

Ans. 166 s Q 45. An electric heater is used in a room of total wall area 137 m2 to maintain a temperature of +20 ◦ C inside it, when the outside temperature is −10 ◦ C. The walls have three different layers. The innermost layer is of wood of thickness 2.5 cm, the middle layer is of cement of thickness 1.0 cm and the outermost layer is of brick of thickness 25.0 cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and ceiling. The thermal conductivities of wood, cement and brick are 0.125, 1.5 and 1.0 W m−1 ◦ C−1 , respectively. (1986) Sol. The thickness of the wood, the cement, and the brick layers are x1 = 2.5 cm, x2 = 1 cm, and x3 = 25 cm and thermal conductivities of these layers are K1 = 0.125 W m−1 ◦ C−1 , K2 = 1.5 W m−1 ◦ C−1 , and K3 = 1 W m−1 ◦ C−1 (see figure). Total wall area is A = 137 m2 , inside temperature is 20 ◦ C, and outside temperature is −10 ◦ C. Let T1 and T2 be the temperatures at the interface of wood-cemnet and cement-brick layers, respectively.

(1992) 20◦ C

Sol. Let T be the temperature of the disc at time t. The rate of heat inflow to the disc is dQ/dt = κA(T0 − T )/h.

-10◦ C

T1 T2

K1

K2

K3

x1

x2

x3

(1)

T0

The rates of heat flow through the layers are given dQ dt

by

h

dQ1 /dt = K1 A(20 − T1 )/x1 ,

T

dQ2 /dt = K2 A(T1 − T2 )/x2 , This heat is used to increase the temperature of the disc i.e., dQ/dt = mS (dT /dt) .

(2)

Eliminate dQ/dt from equations (1) and (2) to get κA dT = (T0 − T ). dt hmS

(3)

dQ3 /dt = K3 A(T2 + 10)/x3 . In the steady state, the rate of heat flow is equal in all the layers i.e., dQ/dt = dQ1 /dt = dQ2 /dt = dQ3 /dt, which gives K1 (20 − T1 )/x1 = K2 (T1 − T2 )/x2 ,

(1)

K2 (T1 − T2 )/x2 = K3 (T2 + 10)/x3 .

(2)

Chapter 25. Heat Transfer

343

Solve equations (1) and (2) for T1 and T2 and substitute the values to get the rate of heat loss through the wall as 20 − (−10) + Kx22A + Kx33A 30 = 0.025 0.01 + 0.125(137) (1.5)(137) +

dQ = dt

x1 K1 A

0.25 (1)(137)

Sol. The heater of resistance R = 20 Ω, operated at voltage V = 200 V, produces heat at a rate

= 9000 W. To maintain the room temperature at 20 ◦ C, the rate of heat produced in the room should be equal to the rate of heat loss through the walls. Thus, the power of the electric heater should be 9000 W. Aliter: The thermal resistances of the three layers are R1 = x1 /(K1 A), R2 = x2 /(K2 A) and R3 = x3 /(K3 A). These thermal resistances are connected in series giving the effective thermal resistance of the wall as x2 x3 x1 + + . Reff = R1 + R2 + R3 = K1 A K2 A K3 A The rate of heat flow through a wall having temperature difference ∆T = 20 − (−10) = 30 ◦ C and thermal resistance Reff is given by dQ ∆T = = dt Reff

30 x1 K1 A

+

x2 K2 A

+

x3 K3 A

= 9000 W.

Q 46. A solid sphere of copper of radius R and a hollow sphere of the same material of inner radius r and outer radius R are heated to the same temperature and allowed to cool in the same environment. Which of them starts cooling faster? (1982) Sol. Both the spheres have same surface area A, same emissivity e, same temperature T , and same ambient temperature T0 . The net rate of heat radiation by both the spheres is equal to (1)

The rate of heat radiation is equal to the rate of heat loss due to decrease in temperature i.e., dQ/dt = mS(−dT /dt),

(2)

where m is the mass of the sphere and S is the specific heat. From equations (1) and (2), the rate of cooling is given by −

σeA(T 4 − T04 ) dT = . dt mS

P = V 2 /R = (200)2 /20 = 2000 W. To maintain the room temperature at Tr = 20 ◦ C, the rate of heat produced by the heater should be equal to the rate of heat loss through the window. The rate of heat loss through the window of area A = 1 m2 , thickness x = 0.2 cm, thermal conductivity K = 0.2 cal m−1 s−1◦ C−1 , and ambient temperature Ta is dQ/dt = KA(Tr − Ta )/x. Equate P = dQ/dt and simplify to get (2000)(0.2 × 10−2 ) Px = 20 − KA (0.2 × 4.2)(1) ◦ = 15.24 C.

Ta = Tr −

Ans. 15.24 ◦ C Ans. 9000 W

dQ/dt = σeA(T 4 − T04 ).

Q 47. A room is maintained at 20 ◦ C by a heater of resistance 20 Ω connected to 200 V mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area 1 m2 and thickness 0.2 cm. Calculate the temperature outside. Thermal conductivity of glass is 0.2 cal m−1 s−1◦ C−1 and mechanical equivalent of heat is 4.2 J/cal. (1978)

(3)

Since mass of the hollow sphere is less than that of the solid sphere, the rate of cooling of the hollow sphere is more than that of the solid sphere. Ans. Hollow sphere

Part V

Electromagnetism

A

R I

l

~v

L

B x

345

Chapter 26 Electric Field and Potential x0

One Option Correct Q 1. Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It is found to move at 45◦ to the vertical just after release. Then X is nearly (2012) (A) 10−5 V (B) 10−7 V (C) 10−9 V (D) 10−10 V

x qE

k(x+x0 )

x

O0

O

At the new mean position O0 , the block is in equilibrium due to electrostatic and spring forces i.e., qE = kx0 ,

Sol. The electric field in the region between the two plates is given by E = X/d.

(1)

which gives x0 = qE/k. At O0 , the spring is compressed by a distance x0 . Let the spring be further compressed by a distance x. Apply Newton’s second law at this position to get

d qE

m d2 x/dt2 = −k(x + x0 ) + qE = −kx,

(2)

=

45◦

where we have used qE from equation (1). The equation (2) represents a SHM with frequency ν0 . Note that the net force on the block is zero at the mean position. We encourage you to draw analogy of this problem with a vertically hanging spring mass system. Ans. A

mg

The proton moves at 45◦ to the vertical if the acceleration (resultant force) is in this direction. The resultant of electric force qE and gravitational force mg makes an angle of 45◦ with the vertical if qE = mg i.e., q(X/d) = mg. Thus,

Q 3. Consider a system of three charges 3q , 3q and − 2q 3 placed at points A, B and C, respectively, as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60◦ . (2008)

mgd (1.67 × 10−27 )(9.8)(1 × 10−2 ) = q 1.6 × 10−19 −9 ≈ 10 V.

X=

y B

Ans. C

•

Q 2. A wooden block performs SHM on a frictionless surface with frequency ν0 . The block contains a charge ~ is +q on its surface. If now, a uniform electric field E switched on as shown, then SHM of the block will be

C•

O

x

60◦ •

(2011)

A

(A) of the same frequency and with shifted mean position. (B) of the same frequency and with same mean position. (C) of changed frequency and with shifted mean position. (D) of changed frequency with same mean position.

(A) The electric field at point O is 8πq0 R2 directed along the negative x-axis. (B) The potential energy of the system is zero. (C) The magnitude of the force between the charge C q2 and B is 54π 2. 0R q (D) The potential at point O is 12π . 0R

Sol. Let m be theqmass of the block, k be the spring k 1 ~ constant, ν0 = 2π m (frequency of SHM when E is switched off), and O be the mean position.

Sol. The charges at A, B, and C are qA = q/3, qB = q/3, and qC = −2q/3. The electric fields at O due to qA and qB are equal in magnitude but opposite in 347

348

Part V. Electromagnetism

direction. Thus, the resultant electric field at O is only due to charge qC and is given by ~O = − E

q ˆı. 6π0 R2

The triangle ABC is right-angled with ∠A = 60◦ , ∠C √ = 90◦ , and rAB = 2R. Thus, rAC = R and rBC = 3R. The potential energy for the given charge distribution is   1 qA qB qA qC qB qC U= + + 4π0 rAB rAC rBC   2 2 1 q 2q 2q 2 6= 0. = − − √ 4π0 18R 9R 9 3R

path. We encourage you to show that work done in taking a unit charge from A to B is zero even if both the charges are positive. Ans. C Q 5. Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at O is double the electric field when only one positive charge of same magnitude is placed at R. Which of the following arrangements of charge is possible for P, Q, R, S, T and U respectively, (2004) Q

P

The magnitude of force between qC and qB is FBC =

1 qB qC q2 . = 2 4π0 rBC 54π0 R2

U

•

The potential at O is V =

O

T

1 (qA /R + qB /R + qC /R) = 0. 4π0

(A) +, −, +, −, −, + (C) +, +, −, +, −, − Ans. C

Q 4. Positive and negative
point charges
of equal magnitude are kept at 0, 0, a2 and 0, 0, − a2 , respectively. The work done by the electric field when another positive point charge is moved from (−a, 0, 0) to (0, a, 0) is (2007) (A) positive (B) negative (C) zero (D) depends on the path connecting the initial and final positions. Sol. The charge configuration is shown in√the figure. The point A(−a, 0, 0) is at a distance rA = 5a/2 from both the charges. Also, the point B(0, a, 0) is at a dis√ tance rB = 5a/2 from both the charges. y (0, a, 0) • B −q• (0, 0, − a2 ) (−a, 0, 0) • x A q • (0, 0, a ) 2 z

The potentials at the point A and B are given by

R

S

(B) +, −, +, −, +, − (D) −, +, +, −, +, −

Sol. The given condition is met if the charge at U is negative, charge at R is positive and field at O due to P, Q, S and T is zero. +

−

−

−

+

−

+

+

−

+

+

−

This is possible if the line joining the two charges and passing through O has charges of same sign on its two ends. Two such possibilities are shown in the figure. Ans. D Q 6. A metallic shell has a point charge q kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of force? (2003) (A) (B)

(C)

(D)

1 q 1 q − = 0, 4π0 rA 4π0 rA 1 q 1 q VB = − = 0. 4π0 rB 4π0 rB

VA =

Since VA = VB , the work done in taking a unit charge from A to B is zero. The electrostatic forces are conservative and work done by them do not depend on the

Sol. The electric field inside the conductor is zero. The field lines are normal to the equipotential surface of the

Chapter 26. Electric Field and Potential

349

conductor. Note that field outside the metallic shell is same as the field of a point charge kept at its centre. Ans. D Q 7. Two equal point charges are fixed at x = −a and x = +a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to (2002)

(A) x (B) x2 (C) x3 (D) 1/x Sol. Let O be the origin and O0 be a point to the right of O at a distance x. q •

(−a, 0)

Q • O O0 x

q

The potentials at the point B and C relative to the point A are given by Z Z 1 ~ · d~rx = VA − VB = VA − E Edx = VA − Ex, 0 Z ~ · d~ry = VA . ~ ⊥ d~ry ). VC = VA − E (∵ E ~ but does not Note that the potential decreases along E ~ change in a direction perpendicular to E. Ans. B Q 9. Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in (2001) (B) (A)

•

• •

The potentials at O and O0 due to charges at (−a, 0) and (a, 0) are q q q + = , VO = 4π0 a 4π0 a 2π0 a q q VO0 = + 4π0 (a + x) 4π0 (a − x)   q a = . 2π0 a2 − x2

~ = E ˆı. Sol. The uniform electric field in the region is E Let d~rx = dx ˆı and d~ry = dy ˆ be the small displacement vectors along x and y-axes.

B A

1

x(cm)

• • •

Sol. The electric field lines emanate from a positive charge. They do not intersect and do not form closed loops in electrostatics.

Q 8. A uniform electric field pointing in positive x direction exists in a region. Let A be the origin, B be the point on the x-axis at x = +1 cm, and C be the point on the y-axis at y = +1 cm. Then the potentials at the points A, B and C satisfy (2001) (A) VA < VB (B) VA > VB (C) VA < VC (D) VA > VC

~ E

(D) •

•

Ans. B

y(cm)

•

•

(C) •

The potential energy of charge Q placed in a potential V is QV . Thus, the change in potential energy of charge Q when it is displaced by a small distance x is   qQ 1 a ∆U = QVO0 − QVO = − 2π0 a2 − x2 a 2 2 qQ qQ x x = ≈ . (for x  a). 2π0 a(a2 − x2 ) 2π0 a3

1 C

•

•

(a, 0)

q •

q•

•q

•

E 6= 0

By symmetry, the field is zero at the centroid. The fields at the middle point of each side are non-zero. The direction of electric field along the perpendicular bisector is as shown in the figure. Ans. C Q 10. The dimensions of 12 0 E 2 (where 0 is permittivity of free space and E is electric field) is (2000) (A) [MLT−1 ] (B) [ML2 T−2 ] (C) [ML−1 T−2 ] (D) [ML2 T−1 ] Sol. The energy density (energy per unit volume) in a region, with electric field E, is given by 12 0 E 2 . Thus, the dimensions of 21 0 E 2 are same as the dimensions of the energy density which are [ML2 T−2 ]/[L3 ] = [ML−1 T−2 ]. Ans. C Q 11. Three charges Q, +q and +q are placed at the vertices of a right angled isosceles triangle as shown in the figure. The net electrostatic energy of the configuration is zero, if Q is equal to (2000)

350

Part V. Electromagnetism (A) the electric field E at all points on the x-axis has the same direction. (B) work has to be done in bringing a test charge from ∞ to the origin. (C) electric field at all point on y-axis is along x-axis. (D) the dipole moment is 2qd along the x-axis.

Q

+q

+q a

(A)

−q √ 1+ 2

(B)

−2q √ 2+ 2

~ on x axis is along −ˆı for x < Sol. The electric field E −d, along +ˆı for −d < x < d, and along −ˆı for x > d.

(C) −2q (D) +q

Sol. The electrostatic energy of charges q1 and q2 , sepq1 q2 arated by a distance r, is given by U = 4π . Electro0r static energy of the given configuration is   1 Qq qq Qq = 0. U= + +√ 4π0 a a 2a

y E =

2E cos θ =

E −q

+q

θ (−d, 0) •

Ans. B Q 12. A charge +q is fixed at each of the points x = x0 , x = 3x0 , x = 5x0 , · · · , ∞ on the x-axis and a charge −q is fixed at each of the points x = 2x0 , x = 4x0 , x = 6x0 , · · · , ∞. Here x0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/(4π0 r). Then, the potential at the origin due to the above system of charges is (1998)

(A) zero (B)

q 8π0 x0 ln 2

(C) infinite (D)

q ln 2 4π0 x0

Sol. The potential at the origin due to the given system of charges is   q q 1 1 1 1 − + − + ... = V = ln 2. 4π0 x0 1 2 3 4 4π0 x0 Ans. D Q 13. An electron of mass me , initially at rest, moves through a certain distance in a uniform electric field in time t1 . A proton of mass mp , also initially at rest, takes time t2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio t2 /t1 is nearly equal to (1997) 1/2 1/2 (A) 1 (B) (mp /me ) (C) (me /mp ) (D) 1836 Sol. The magnitude of electric force on the electron and the proton is equal i.e., Fe = Fp = qE. The acceleration and distance travelled by the electron and proton are ae =

qE , me

ap =

qE ; mp

xe =

1 2 qEt21 ae t1 = , 2 2me

xp =

1 2 qEt22 ap t2 = . 2 2mp

Equate xe = xp to get t2 /t1 = (mp /me )1/2 . Ans. B Q 14. Two point charges +q and −q are held fixed at (−d, 0) and (d, 0) respectively of a x-y coordinate system. Then, (1995)

•

O

x

(d, 0)

The potential at the origin O is zero and hence no work is done in bringing a test charge from ∞ to O. The electric field at any point on the y axis is along ˆı as shown in the figure. The dipole moment of the configuration is p~ = −2qd ˆı. Ans. C Q 15. Two identical thin rings, each of radius R, are coaxially placed a distance R apart. If Q1 and Q2 are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that√of the other is (1992) 1 −Q2 ) √ (A) zero (B) ( 2−1)q(Q 2(4π R) √

(C)

2q(Q1 +Q2 ) 4π0 R

(D)

0 √ ( 2+1)q(Q1 +Q2 ) √ 2(4π0 R)

Sol. Let S and T be the centres of two rings carrying charges Q1 and Q2 , respectively. Q1

Q2 √ 2R

Solve to get Q =

−2q √ . 2+ 2

S

R

R T

The distance of the centre from any point on the √ other ring is 2R. The potentials at the points S and T due to the two rings are   1 Q1 Q2 VS = +√ , 4π0 R 2R   1 Q2 Q1 VT = +√ . 4π0 R 2R Thus, the work done in taking a charge q from T to S is √ q(Q1 − Q2 )( 2 − 1) √ W = q(VS − VT ) = . 4 2π0 R Ans. B

Chapter 26. Electric Field and Potential

351

Q 16. A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to (1987) (A) −Q/2 (B) −Q/4 (C) +Q/4 (D) +Q/2

force on charge Q is along −x direction and is given by

Sol. Let the separation between the two particles of charges Q be 2a. Coulomb’s forces on the charge q due to the other two charges are equal and opposite. Hence, charge q is always in equilibrium irrespective of its sign and magnitude.

The net force on charge Q is towards the mean position O but it is not proportional to the distance from the mean position. Thus, the motion of the charge Q is not SHM. The charge Q starts moving from (2a, 0) towards O, crosses the origin O and moves upto the point (−2a, 0), changes the direction of motion at (−2a, 0) and repeats the journey to the starting point. We encourage you to find the time period of oscillation. Ans. D

• Q

FQ

Fq

• Q

•

q a

Q2 /(16π0 a2 ) = −Qq/(4π0 a2 ), which gives q = −Q/4. Ans. B Q 17. Two equal negative charges −q are fixed at points (0, −a) and (0, a) on y-axis. A positive charge Q is released from rest at the point (2a, 0) on the xaxis. The charge Q will (1984) (A) execute SHM about the origin. (B) move to the origin and remain at rest. (C) move to infinity. (D) execute oscillatory but not SHM.

−q •

F

θ

F cos θ

θ

•Q

F sin θ

F −q

Q 19. An alpha particle of energy 5 MeV is scattered through 180◦ by a fixed uranium nucleus. The distance of closest approach is of the order of (1981) (A) 1 ˚ A (B) 10−10 cm (C) 10−12 cm (D) 10−15 cm Sol. Initially, kinetic energy of the alpha particle is K0 = 5 MeV and its potential energy is U0 = 0 (because it is far away from the nucleus). The charge of the uranium nucleus is Q = Ze = 92e and charge on the alpha particle is q = 2e. Let O be the centre of the uranium nucleus. The alpha particle starts moving towards O and is scattered by an angle 180◦ at P. +Ze O

P

=

F cos θ

a •

Sol. The potential inside a hollow conducting sphere is constant and its value is equal to the potential at the surface. Thus, the potential at the centre is 10 V. Note that the electric field inside the hollow conducting sphere is zero. Ans. B

d

=

O x

Q 18. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. The potential at the centre of the sphere is (1983) (A) zero. (B) 10 V. (C) same as at a point 5 cm away from the surface. (D) same as at a point 25 cm away from the surface.

F sin θ

Sol. Let the charge Q be located at a distance x from the origin O. The electrostatic attraction forces on charge Q due to charge −q located at (0, a) and charge −q located at (0, −a) are equal in magnitude and their directions are as shown in the figure.

y

2Qqx . 4π0 (x2 + a2 )3/2

a

Coulomb’s force on a charge Q due to another charge Q is repulsive in nature and has magnitude FQ = Q2 /(16π0 a2 ). For the charge Q to be in equilibrium, Coulomb’s force on it due to charge q should be attractive and of magnitude FQ i.e.,

a

Fnet = 2F cos θ =

The magnitude of electrostatic forces is Qq F = . 4π0 (a2 + x2 ) Resolve the forces along the x and y directions. The components along the y directions cancel out. The net

The distance of the closest approach is OP = d. The kinetic energy of the alpha particle at P is KP = 0 (since its velocity is zero) and its potential energy is UP = 2Ze2 /(4π0 d). Apply conservation of energy, K0 + U0 = KP + UP , to get d=

(9 × 109 ) (2) (92) (1.6 × 10−19 )2 2Ze2 = 4π0 K0 (5 × 106 ) (1.6 × 10−19 )

= 5.3 × 10−14 m. Ans. C

352

Part V. Electromagnetism

One or More Option(s) Correct

Ans. A, B, C

Q 20. A length-scale (l) depends on the permittivity () of a dielectric material, Boltzmann constant (kB ), the absolute temperature (T ), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for l is(are) dimensionally correct? (2016)

(A) l = (C) l =

q q

nq 2 kB T q2 n2/3 kB T

(B) l =

q

kB T nq 2

(D) l =

q

q2 n1/3 kB T

Sol. The dimensions of thermal energy kB T is ML2 T−2 . From Coulomb’s law, F = q1 q2 /(4πr2 ), the dimensions of q 2 / is ML3 T−2 . The dimensions of number per unit volume n is L−3 . Substitute these q dimensions in given expressions to get dimensions of knqB2T q q2 and n1/3 as L. kB T Ans. B, D Q 21. Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the q 1 figure. Given that K = 4π 2 , which of the following 0 L statement(s) is (are) correct? (2012) L

F +q

E −q

P A +2q

S

B +q

(A) (B) (C) (D)

T O R

D −2q

C −q

The electric field at O is 6K along OD. The potential at O is zero. The potential at all points on the line PR is same. The potential at all points on the line ST is same.

Q 22. Under the influence of the Coulomb field of charge +Q, a charge −q is moving around it in an elliptical orbit. Find out the correct statement(s). (2009) (A) The angular momentum of the charge −q is constant. (B) The linear momentum of the charge −q is constant. (C) The angular velocity of the charge −q is constant. (D) The linear speed of the charge −q is constant. Sol. The torque on a charge −q due to Coulomb force 1 Qq ~ ˆ is ~τ = ~r × F~ = ~0. Since ~τ = dL/dt = ~0, F~ = − 4π 2 r 0 r ~ is a constant. In elliptical orbit, angular momentum L r varies and hence to keep L = mω 2p r constant, ω must vary with r. The speed v = ωr = Lr/m also varies with r. By Netwon’s second law, F~ = d~ p/dt 6= ~0 and hence p~ cannot be a constant. We encourage you to draw analogy between this problem and the planetary motion (Kepler’s laws). Ans. A Q 23. A positively charged thin metal ring of radius R is fixed in the x-y plane with its centre at the origin O. A negatively charged particle P is released from rest at the point (0, 0, z0 ) where z0 > 0. Then the motion of P is (1998) (A) periodic for all values of z0 satisfying 0 < z0 < ∞. (B) simple harmonic for all values of z0 satisfying 0 < z0 ≤ R. (C) approximately simple harmonic provided z0  R. (D) such that P crosses O and continues to move along the negative z axis towards z = −∞. Sol. Let Q be the charge on a ring of radius R and centre O. z

Sol. The electric field at O due to the charges at A and D is 4K along OD, due to the charges at B and E is 2K along OE and due to the charges at C and F is 2K along OC. For the given geometry, resultant of these fields is 6K along OD. The potential at O is X 1 qi 1 X VO = = qi = 0. 4π0 L 4π0 L For any point on PR, we have pairs of equal and opposite charges at the same distance making the potential at any point on PR zero. It may be seen that potential at points on OS is positive and that on OT is negative. We encourage you to show that the potential on ST (at a distance x from O, taken positive towards the right) is  q 2 2 √ V (x) = −√ 4π0 L2 + x2 + xL L2 + x2 − xL  4x − 2 . L − x2

• P(0, 0, z0 )

~ F

R

O

y

x

The electric field at a point P (0, 0, z0 ) due to the ring is ~ = E

Qz0 ˆ k, 4π0 (R2 + z02 )3/2

and the force on the charge −q placed at P is F~ = −

Qqz0 ˆ k. 4π0 (R2 + z02 )3/2

This force accelerates the charge towards O (see figure). When particle crosses O and moves to the other side of

Chapter 26. Electric Field and Potential

353

the ring, force direction is again towards O. Thus, the particle executes a periodic motion about O. Qqz0 ˆ If z0  R, the force becomes F~ ≈ − 4π 3 k, which 0R is proportional to the displacement z0 and is always towards O. In this q case, the particle execute SHM with frequency ω =

Qq 4π0 mR3 .

Ans. A, C Q 24. A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in the figure as (1996) 1 2 3 4

(A) 1 (B) 2 (C) 3 (D) 4 Sol. In electrostatics (i.e., when charges are not moving or the charge density does not vary with the time), electric field inside a conductor is zero. The field lines are normal to the surface and never enter inside a conductor. Ans. D

Q 26. Estimate the wavelength at which plasma reflection will occur for a metal having the density of electron N = 4 × 1027 m−3 . Take 0 = 10−11 and m = 10−30 , where these quantities are in proper SI units. (A) 800 nm (B) 600 nm (C) 300 nm (D) 200 nm Sol. The plasma reflection occurs at the q frequency e2 ω = ωp . Thus, λ = 2πc/ωp = 2πc/ N m0 , where c = 3 × 108 m/s and e = 1.6 × 10−19 C. Substitute the values to get λ = 589 nm ≈ 600 nm. Ans. B Matrix or Matching Type Q 27. Four charges q1 , q2 , q3 and q4 of same magnitude are fixed along the x axis at x = −2a, −a, +a, and +2a, respectively. A positive charge q is placed on the positive y axis at a distance b > 0. Four options of the sign of these charges are given in Column I. The direction of the forces on the charge q is given in Column II. Match Column I with Column II. (2014) (0, b) q

q1

Paragraph Type

q2

(−2a,0) (−a,0)

q3

q4

(a,0) (2a,0)

Paragraph for Questions 25-26 A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let N be number density of free electrons, each of mass m. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field become zero, the electrons begin to oscillate about positive ions with natural frequency ωp , which is called plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency ω, where a part of energy is absorbed and a part of it is reflected. As ω approaches ωp , all the free electrons are set to resonate together and all the energy is reflected. This is the explanation for high reflectivity of metals. (2011) Q 25. Taking the electronic charge as e and permittivity as 0 , use dimensional analysis to determine correct expression for ωp . q q p m0 p m0 Ne N e2 (B) (C) (A) m0 Ne m0 (D) N e2

Column I

Column II

(P) q1 , q2 , q3 , q4 all positive (Q) q1 , q2 positive; q3 , q4 negative (R) q1 , q4 positive; q2 , q3 negative (S) q1 , q3 positive; q2 , q4 negative

(3) +y (4) −y

Sol. Let |q1 | = |q2 | = |q3 | = |q4 | = q 0 . The forces on q by q1 , q2 , q3 , and q4 in all the four cases (P, Q, R, S) are shown in the figure. (0, b) q

q1

q2

(−2a,0) (−a,0)

q3

q4

(a,0) (2a,0)

2

1 q Sol. Using Coulomb’s law, F = 4π 2 , the dimensions 0 r 2 3 −2 of e /0 is given by [ML T ]. The number density q N e2 −3 has dimensions [L ]. This gives dimensions of N m0 q e2 as T−1 . Hence ωp = N m0 . Ans. C

(1) +x (2) −x

P

Q

R

S

354

Part V. Electromagnetism

Coulomb’s law gives the magnitude of the force by q1 and q4 as 0

F1,4 =

qq , 4π0 (b2 + 4a2 )

and that by q2 and q3 as F2,3 =

Thus, F1,4 < F2,3 . See the figure for the directions of forces in four cases. Resolve the forces in x and y directions and compare the magnitudes to get the answer. Ans. P7→3, Q7→1, R7→4, S7→2 Q 28. Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II. (2007)

(A) GMe Ms , where G is universal gravitational constant, Me mass of the earth and Ms mass of the Sun. (B) 3RT M , where R is universal gas constant, T absolute temperature and M molar mass. 2 (C) q2FB 2 , where F is force, q charge and B magnetic field. e (D) GM Re , where G is universal gravitational constant, Me mass of the earth and Re radius of the earth.

Q 29. A ring of radius R carries a uniformly distributed charge +Q. A point charge −q is placed on the axis of the ring at a distance 2R from the centre of the ring and released from rest. The particle executes a SHM along the axis of the ring. (1988) Sol. The electric field due to a uniformly charged ring of radius R and charge Q at a point P on its axis (see figure) is given by

qq 0 . 4π0 (b2 + a2 )

Column I

True False Type

Qx 1 ˆı. 2 4π0 (R + x2 )3/2 +Q

√ x2

y +

R

x R2

−q x

O

P

Thus, force on a negative charge (−q) placed at a distance x on the axis of the ring is

Column II (p) volt coulomb metre

1 Qqx F~ (x) = − ˆı. 4π0 (R2 + x2 )3/2

(1)

The force is restoring in nature but it is not proportional to x. Thus, the motion of the particle is not SHM but periodic if x is large (x = 2R). We encourage you to show, for x  R, that equation (1) reduces to   F~ (x) = − Q/(4π0 R3 ) x ˆı,

(q) kg m3 s−2

(r) m2 s−2

(s) farad kg−1

~ E(x) =

2

volt

Sol. From Newton’s law of gravitation, F = GMe Ms /r2 , units of GMe Ms are same as that of F r2 which are kg m3 s−2 . Other units of F r2 are energymetre. One of the units of energy are coulomb-volt since U = qV . Thus, volt-coulomb-metre are also the units of F r2 and GMe Ms . The internal energy of an ideal gas is 23 nRT = 3 2 (m/M )RT , which gives the units of 3RT /M as energykg−1 i.e., m2 s−2 . The energy stored in a capacitor, U = 21 CV 2 , gives the units of energy as farad-volt2 . Thus, another units of 3RT /M are farad-volt2 kg−1 . The force on a current carrying wire in a magnetic field, F = ilB = (q/t)lB, gives units of F 2 /(q 2 B 2 ) as m2 s−2 which are same as farad-volt2 kg−1 . The gravitational potential energy, U = −GMe m/Re , gives the units of GMe /Re as energykg−1 which are same as m2 s−2 and farad-volt2 kg−1 . Ans. A7→(p,q), B7→(r,s), C7→(r,s), D7→(r,s)

and the particle p executes SHM with an angular frequency ω = Q/(4π0 mR3 ). Ans. F Q 30. An electric line of force in the x-y plane is given by the equation x2 + y 2 = 1. A particle with unit positive charge, initially at rest at the point x = 1, y = 0 in the x-y plane, will move along the circular line of force. (1988) Sol. The electric force on a positive unit charge placed at a point P is along the tangent to electric lines of force at P. The path of the particle depends on the initial conditions (position, velocity) and acceleration. In the given case, initial position is P0 (1, 0), initial velocity is zero, and initial acceleration is ~a = qE/m ˆ. Thus, the particle starts moving in ˆ direction. y

P1 •

•

O

•

1 P0

x

Chapter 26. Electric Field and Potential Let the particle moves to a new position P1 in a small time interval ∆t. The position of the particle after next ∆t time interval will depend on the velocity and acceleration at P1 , which cannot be deduced from the given information. Therefore, it cannot be concluded that the particle moves along the given line of force. Ans. F Q 31. Two protons A and B are placed in between the two plates of a parallel plate capacitor charged to a potential difference V as shown in the figure. The forces on the two protons are identical. (1986) +

−

+

−

+ + +

•

B •

A

− −

V

355 Q 34. The work done in carrying a point charge from one point to another in an electrostatic field depends on the path along which the point charge is carried. (1981)

Sol. The electrostatic field/force is conservative in nature. The work done by a conservative force is independent of the path and depends only on the end points. Note that the work done by a conservative force in a closed path is always zero. Ans. F Fill in the Blank Type Q 35. The electric potential V at any point x, y, z (all in metre) in space is given by V = 4x2 volt. The electric field at the point (1 m, 0 m, 2 m) is . . . . . . V/m. (1992)

−

Sol. The electric field is given by Sol. The magnitude of electric field between the two parallel plates of a capacitor with potential difference V is E = V /d, where d is the separation between the plates. Hence, forces qE on two protons (of charge q) are same irrespective of their locations within the capacitor. Ans. T Q 32. Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge Q coulomb and the other an equal negative charge. Their masses after charging are different. (1983) Sol. Let M0 be the mass of the neutral sphere, m be the mass of an electron and −e be the charge on an electron. A sphere is given the positive charge Q by taking away n = Q/e electrons from it. Thus, the mass of the positively charged sphere is

~ = − ∂V ˆı − ∂V ˆ − ∂V kˆ = −8x ˆı. E ∂x ∂y ∂z ~ at the point (1 m, 0 m, 2 m) Substitute x = 1 to get E ~ = −8 ˆı. i.e., E Ans. −8ˆı Q 36. Five point charges, each of value +q coulomb, are placed on five vertices of a regular hexagon of side L metre. The magnitude of the force on the point charge of value −q coulomb placed at the centre of the hexagon is . . . . . . newton. (1992) q

Mneg = M0 + mn = M0 + mQ/e.

q

Sol. The high energy X-rays cause ejection of the electrons from the metal ball (photoelectric effect). Thus, the ball gets positively charged. The positively charged ball is deflected in the direction of electric field. Ans. T

q

Sol. The forces on −q due to charges at 1 and 4 are equal and opposite. q 5

We encourage you to find the charges on the spheres if the difference in their masses is 1 µg. Ans. T Q 33. A small metal ball is suspended in a uniform electric field with the help of an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the direction of the field. (1983)

q

−q

Mpos = M0 − mn = M0 − mQ/e. Another sphere is given the negative charge −Q by putting n = Q/e electrons on it. Thus, the mass of the negatively charged sphere is

q

6

q 4 3 q

−q 1 q

2 q

Also, the forces on −q due to charges at 2 and 5 are equal and opposite. Thus, the net force on −q is 2 1 q due to charge at 3 and its magnitude is F = 4π 2 = 0 L 2

9 × 109 Lq 2 . 2

Ans. 9 × 109 Lq 2

356

Part V. Electromagnetism

Q 37. A point charge q moves from point P to point S along the path PQRS (see figure) in a uniform electric field E pointing parallel to the positive direction of the x-axis. The coordinates of points P, Q, R and S are (a, b, 0), (2a, 0, 0), (a, −b, 0), (0, 0, 0) respectively. The work done by the field in the above process is given by the expression . . . . . . (1989)

Sol. In the state of weightlessness, the net force on the charges are only due to Coulomb repulsion. The repulsion causes the two charges to move farthest away from each other. In this position, the angle between the two strings is 180◦ and separation between the two charges is r = 2L. Q

y

F

P

~ E

S

x

Q R

Sol. The work done by the conservative forces (electrostatic, gravitational, etc.) is independent of the path i.e., it depends only on the initial and final points. Thus, the work done by the field along path P → Q → R → S is same as the work done along the path P → T → S.

L T

L O

Q 39. Figure shows lines of constant potential in a region in which an electric field is present. The values of the potential of each line is also shown. Of the points A, B and C, the magnitude of the electric field is the greatest at the point . . . . . . (1984)

A

~ E P

B

b a

x

Q

T

50 V 40 V C 30 V 20 V 10 V

R

The work done by the field in moving a charge q along the path P → T → S is given by Z

S

Z

S

~ · d~l E

qdV = −q

W = P

P T

Z

~ · d~l − q E

= −q P Z 0

= −q

Z

S

~ · d~l E

T

Z

(E ˆı) · (−dy ˆ) − q b

Z

0

(E ˆı) · (−dx ˆı) a

Sol. The electric field is related to electric potential by E = −dV /dx. The potential difference between the successive lines of constant potential is ∆V = 10 V. The perpendicular distances between successives lines at the point A and at the point C are almost equal but this distance is smaller at the point B i.e., ∆xA = ∆xC > ∆xB . Hence, |EB | = ∆V /∆xB > ∆V /∆xA = |EA | = |EC |. Ans. B

0

dx = −qEa.

= qE

F

The forces acting on each charge are Coulomb force F = Q2 /(4π0 r2 ) = Q2 /(16π0 L2 ) and tension T . In equilibrium, T = Q2 /(16π0 L2 ). Q2 Ans. 180◦ , 16π 2 0L

y

S

Q

T

a

Note that the potentials at the point S and P are related by VP = VS − Ea. Aliter: The force on the charge is F~ = qE ˆı and ~ = ~rS − ~rP = −a ˆı − b ˆ. Thus, its displacement is S ~ ~ W = F · S = (qE ˆı) · (−a ˆı − b ˆ) = −qEa. Ans. −qEa Q 38. Two small balls having equal positive charges Q (coulomb) on each are suspended by two insulating strings of equal length L (metre) from a hook fixed to a stand. The whole set-up is taken in a satellite into space where there is no gravity (state of weightlessness). The angle between the strings is . . . . . . and the tension in each string is . . . . . . newton. (1986)

Integer Type Q 40. A particle, of mass 10−3 kg and charge 1.0 C, is initially at rest. At time t = 0, the particle comes ~ under the influence of an electric field E(t) = E0 sin ωt ˆı, where E0 = 1.0 N/C and ω = 103 rad/s. Consider the effect of only the electric force on the particle. Then the maximum speed, in m/s, attained by the particle at subsequent times is . . . . . . . (2018) Sol. The force on a particle of mass m and charge ~ is given by F~ = q E. ~ q placed in an electric field E By Newton’s second law, acceleration of the particle is given by ~a(t) =

~ F~ qE qE0 = = sin ωt ˆı. m m m

(1)

Chapter 26. Electric Field and Potential

357

The particle starts from rest i.e., initial velocity ~v0 = ~0. Integrate equation (1) to get velocity of the particle Z ~v (t) = ~v0 +

t

qE0 ~a dt = m

t

Z

sin ωt dt ˆı   qE0 2qE0 ωt = (1 − cos ωt) ˆı = sin2 ˆı. (2) mω mω 2 0

0

By symmetry, the magnitudes of force on the other three charges are same as the magnitude of force on charge at B. The directions of these forces are as shown in the figure. In equilibrium, the force due to surface tension (on line BC) is equal and opposite to xcomponent of electrostatic force acting on the charges placed at point B and C i.e., γa = 2F cos 45◦ =

a(t) t

which gives r √  1/3 1  2 + 1/ 2 q 2 /γ a= 3 . 4π0

v(t) t

From equation (2), the particle attains the maximum speed when sin(ωt/2) = 1 i.e., vmax =

Let the particle starts from the origin. Can you find the expression for its position x(t) and plot it on x-t graph? Don’t get surprised by seeing the particle drifting away from the origin under the influence of sinusoidal force. What is the average speed of the particle? Ans. 2 Q 41. Four point charges, each of +q, are rigidly fixed at the four corners of a square planar soap film of side a. The surface tension of soap film is γ. The system of charges and planar film are in equilibrium and a =  1/N k q 2 /γ where k is a constant. Then N is . . . . . . . (2011)

Sol. The force on the line BC due to the surface tension is γa.

Descriptive

(2005)

Sol. The volume of liquid in a bubble of radius a and thickness t is Vb = 43 π((a + t)3 − a3 ) ≈ 4πa2 t (since t  a). The volume of liquid in the droplet of radius r is Vd = 43 πr3 . Equate Vb = Vd to get r = (3a2 t)1/3 . The potential on a spherical shell of radius a and charge q is V = q/(4π0 a). Thus, charge on the bubble having potential V is q = 4π0 aV . The charge conservation gives charge on the droplet as q = 4π0 aV . The potential on the droplet is given by Vd =

 a 1/3 q 4π0 aV = = V. 4π0 r 3t (3a2 t)1/3

F

γa

D


a 1/3 3t

Q 43. Eight point charges are placed at the corners of a cube of edge a as shown in the figure. Find the work done in disassembling this system of charges. (2003)

y x

Ans. V

B 45◦

A

F

Ans. 3

Q 42. A conducting bubble of radius a and thickness t (t  a) has potential V . Now the bubble collapses into a droplet. Find the potential of the droplet.

2 (1) (1) 2qE0 = = 2 m/s. mω (10−3 ) (103 )

F

√ i 1 q2 h 2 + 1/ 2 , 4π0 a2

a

F

The electrostatic forces on charge at B due to charges at A, D, and C are along AB, DB, and CB, respectively. Thus, the total force on the charge at B due to other three charges is  2   1 q q2 q 2 ˆı + ˆ √ F~ = ˆ ı +  ˆ + 4π0 a2 a2 2a2 2     2 √ 1 q 1 ˆı + ˆ √ = 2+ . 4π0 a2 2 2

−q

+q

◦ C 45

+q

−q −q +q

+q −q

Sol. In the given system, there are 8 C2 = 28 pairs of charges. The charge pairs on cube edges (twelve edges of length a each) have unlike charges. The charge pairs on face-diagonals (six faces, √ two diagonals per face, twelve face-diagonal of length 2a each) have like charges.

358

Part V. Electromagnetism

The charge pairs on √ main-diagonals (four main-diagonal each with length 3a) have unlike charges. Thus, the potential energy of the system is

Fe •B

vB

TB mg

  1 q2 5.824 q 2 q2 q2 U= =− −12 +12 √ −4 √ . 4π0 a 4π0 a 2a 3a

•q

r

The work done to disassemble the system is

vA

•

W = −U =

A

5.824 q 2 . 4π0 a Ans.

2 5.824 q 4π0 a

Q 44. A positive point charge q is fixed at origin. A dipole with a dipole moment p~ is placed along the x-axis far away from the origin with p~ pointing along positive x-axis. Find (a) the kinetic energy of the dipole when it reaches a distance r from the origin and (b) force experienced by the charge q at this moment. (2003) Sol. Total energy of a dipole p~ = p ˆı when it is far away from the charge q, is zero. Now, this dipole is placed in the electric field of charge q. The electric field of charge q and the potential energy of the dipole are given by ~q = E

1 q ˆı, 4π0 r2

~q = − U = −~ p·E

qp . 4π0 r2

The conservation of energy, K + U = 0, gives dipole kinetic energy as K = −U = qp/(4π0 r2 ). The electric field at the origin by the dipole and force on the charge q are ~ p = 2p ˆı, E 4π0 r3

~ p = 2pq ˆı. F~q = q E 4π0 r3

We encourage you to find the force on p~ by differentiating its potential energy, Fp = − dU dr . Apply Newton’s third law to find the force on q. Ans. (a) 4πqp0 r2 (b) 2πpq0 r3 ˆı Q 45. A small ball of mass 2 × 10−3 kg having a charge of 1 µC is suspended by a string of length 0.8 m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball, so that it can make complete revolution. (2001) Sol. Let q = 1 µC be the charge and m = 2 × 10−3 kg be the mass of a particle moving in a circle of radius r = 0.8 m. The forces acting on the particle are its weight mg downward, tension T radially inward, and 2 1 q electrostatic force Fe = 4π 2 radially outward. Let 0 r vA and vB be its velocities at the lowest point A and the highest point B.

At B, the radially inward force provides the centripetal acceleration i.e., 2 mg + TB − Fe = mvB /r.

(1)

The electrostatic potential energies of the particle at A and B are same. Using the conservation of mechanical energy between A and B, we get 2 1 2 mvA

2 = 12 mvB + 2mgr.

(2)

Eliminate vB from equations (1) and (2) to get 2 vA = 5rg/m + rTB /m − rFe /m.

(3)

Velocity vA is minimum when TB = 0 (since TB ≥ 0). Substitute TB = 0 in equation (3) to get the minimum value of vA as  vA =

5rg −

rFe m

1/2

 =

5gr −

1 q2 4π0 mr

1/2 .

Substitute the values of q, m, r, and g to get vA = 5.86 m/s. We encourage you to analyse the problem if Fe > mg. Ans. 5.86 m/s Q 46. Four point charges +8 µC,−1 µC, and p µC, −1 p +8 µC are fixed at the points − 27/2 m, − 3/2 m, p p + 3/2 m and + 27/2 m respectively on the y-axis. A particle of mass 6 × 10−4 kg and charge +0.1 µC moves along the x direction. Its speed at x = +∞ is v0 . Find the least value of v0 for which the particle will cross the origin. Also find the kinetic energy of the particle h at the origin. Assume i that space is gravity free. 9 1 2 2 (2000) 4π0 = 9 × 10 Nm /C . Sol. Given q = 1 µC = 10−6 C, Q = 8 µC = 8 × 10−6 p C, q0 = 0.1 µC = 10−7 C, m = 6 × 10−4 kg and a = 3/2 m. Consider a point P at a distance x from the origin.

Chapter 26. Electric Field and Potential y 3a • +Q a • −q q0

P

•

•

v0 m

x

−a • −q

359 Sol. Consider a ring of radius r and width dr. The charge on the ring is dq = 2πrσdr, where σ is the surface charge density of the disc. The potential due to the ring at a point P located at a height h on the axis of the disc is 1 2πσrdr √ . dV = 4π0 r2 + h2 Q P

−3a • +Q V Hh V0 x0

O r

x U

The potential at P due to given charge distribution

2mga

is V (x) =

  1 2Q 2q √ −√ . 4π0 x2 + 9a2 x2 + a 2

√

3mga √ a/ 3

h

The potential varies with x and attains its maximum at x0 (see figure). The V (x) becomes maximum when   2x dV (x) Q q =− − 2 dx 4π0 (x2 + 9a2 )3/2 (x + a2 )3/2 = 0. (1)

Integrate dV from r = 0 to r = a to get the potential due to the complete disc i σ hp 2 V = a + h2 − h . 20

Substitute in equation (1) and solve to get p Q = 8q p x0 = 5/3 a = 5/2 m. The potential at x0 is V0 = V (x0 ) = 2.7 × 104 V. The kinetic energy of charge q0 at x = ∞ should be sufficient to cross the potential barrier V0 , p 2 1 =⇒ v0 = 2q0 V0 /m = 3 m/s. 2 mv0 = q0 V0 ,

For the particle to just reach O, decrease in its gravitational potential energy should be equal to the increase in its electrostatic potential energy i.e.,

The potential energy of q0 at the origin is   1 2Qq0 2qq0 U= − = 2.4 × 10−3 J. 4π0 3a a Let K be the kinetic energy of q0 at the origin. The conservation of energy, 21 mv02 = K + U , gives K = 12 mv02 − U = 3 × 10−4 J. Ans. 3 m/s, 3 × 10−4 J Q 47. A non-conducting disc of radius a and uniform positive surface charge density σ is placed on the ground with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disc from a height H with zero initial velocity. The particle has q/m = 40 g/σ. (1999) (a) Find the value of H if the particle just reaches the disc. (b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position.

Thus, the potentials at O and Q are given by  σa σ p 2 VO = a + H2 − H . , VQ = 20 20

mgH = q(VO − VQ ) i p qσ h = a + H − a2 + H 2 . 20 Substitute q/m = 40 g/σ to get p a2 + H 2 = a + H/2, which gives, H = 4a/3. Total potential energy (U ) of the particle at point P is the sum of its gravitational and electrostatic potential energies i.e.,  qσ p 2 a + h2 − h U = mgh + 2 h p 0 i = mg 2 a2 + h2 − h . The potential energy attains extremum at the equilibrium position i.e.,   dU 2h = mg √ − 1 = 0, dh a2 + h2 √ √ which gives hmin = a/ 3 and Umin = 3mga. The figure shows the variation of U with height h. Note that the equilibrium is stable (i.e., d2 U/dh2 > 0).√ Ans. (a) 4a/3 (b) a/ 3

360

Part V. Electromagnetism

Q 48. A circular ring of radius R with uniform positive charge density λ per unit length is located in the y-z plane with its centre at the origin O. A particle of mass m and √ positive charge q is projected from the point P (R 3, 0, 0) on the positive x-axis directly towards O, with an initial speed v. Find the smallest (non-zero) value of the speed v such that the particle does not return to P. (1993) Sol. The charge on the ring is Q = 2πRλ. The potential at a point (x, 0, 0) on the axis of the ring is given by V (x) =

Q 1 √ . 2 4π0 R + x2

Thus, √ the potentials at the point O(0, 0, 0) and P (R 3, 0, 0) are 1 Q VO = , 4π0 R

1 Q VP = . 4π0 2R

y

Sol. The electric field inside and outside of a uniformly charged sphere having a charge Q and radius R is given by ( Qr if r ≤ R; 3, E(r) = 4πQ0 R , if r > R. 2 4π0 r The energy density (energy per unit volume) in space, having an electric field E, is given by 12 0 E 2 . Let us calculate the electrostatic potential energy stored inside and outside the sphere separately. Take a spherical shell of radius r and thickness dr inside the sphere. The potential energy of this shell is dU1 =

Q2 1 r4 dr. 0 E 2 (4πr2 dr) = 2 8π0 R6

Integrate dU1 from r = 0 to r = R to get Z R Q2 Q2 4 U1 = r dr = . 8π0 R6 0 40π0 R Similarly, the potential energy stored in a spherical shell of radius r and thickness dr outside the sphere is

v O

•

x

dU2 =

P

z

V

•

O

x

P

The potential increases monotonically from P to O and attains its maximum value at O. The particle will not come back to P if it just crosses O. The energy required by a particle of charge q to reach the point O from the point P is q(VO − VP ). Thus, 1 1 qQ qλ mv 2 = q(VO − VP ) = = , 2 4π0 2R 40 p which gives v = qλ/(20 m). p Ans. qλ/(20 m) Q 49. Answer the following questions, (a) A charge Q is uniformly distributed over a spherical volume of radius R. Obtain an expression for the energy of the system. (b) What will be the corresponding expression for the energy needed to completely disassemble the planet earth against the gravitational pull among its constituent particles? [Assume the earth to be sphere of uniform mass density. Calculate this energy, given the product of the mass and the radius of the earth to be 2.5 × 1031 kg m.] (c) If the same charge Q as in part (a) is given to a spherical conductor of the same radius R, what will be the energy of the system? (1992)

1 Q2 dr 0 E 2 (4πr2 dr) = . 2 8π0 r2

Integrate from r = R to r = ∞ to get the potential energy stored outside the sphere as Z ∞ Q2 Q2 dr U2 = = . 8π0 R r2 8π0 R 2

3Q Total potential energy is U = U1 + U2 = 20π . 0R In case of a spherical conductor, the electric field inside the conductor is zero giving U1 = 0. The field and energy outside the conductor are

E=

Q , 4π0 r2

U = U2 =

Q2 . 8π0 R

We encourage you to compare this with the potential energy of a spherical capacitor of radius R and charge Q. Many results from electrostatics can be directly used in gravitation by replacing charge Q by mass M and constant 1/(4π0 ) by universal gravitational constant G. Thus, gravitational potential energy of a 2 (negasphere of mass M and radius R is U = − 35 GM R tive because gravitational forces are always attractive). Hence, energy required to completely disassemble the sphere is E = −U = =

3 GM 3 3 GM 2 M R = gM R = 2 5 R 5 R 5

3 (9.8)(2.5 × 1031 ) = 1.5 × 1032 J. 5 Ans. (a)

(c)

Q2 8π0 R

2 3 Q 20 π0 R

(b)

3 GM 2 5 R ,

1.5 × 1032 J

Chapter 26. Electric Field and Potential

361

Q 50. Two fixed charges −2Q and Q are located at the points with coordinates (−3a, 0) and (+3a, 0) respectively in the x-y plane. (1991) (a) Show that all points in the x-y plane where the electric potential due to the two charges is zero, lie on a circle. Find its radius and the location of its centre. (b) Give the expression V (x) at a general point on the x-axis and sketch the function V (x) on the whole x-axis. (c) If a particle of charge +q starts from rest at the centre of the circle, show by a short quantitative argument that the particle eventually crosses the circle. Find its speed when it does so. Sol. The net electric potential at the point P (x, y) due to the charge −2Q located at (−3a, 0) and the charge Q located at (3a, 0) is given by " # 1 2 Q p −p . V = 4π0 (x − 3a)2 + y 2 (x + 3a)2 + y 2

from a higher potential to a lower potential. By conservation of energy, decrease in the potential energy is equal to increase in kinetic energy i.e., r 1 qQ Qq 2 mv = , which gives v = . 2 16π0 a 8π0 ma

(b) Vx =

Q 4π0



1 |3a−x|

−

2 |3a+x|



(c)

Ans. q (a) 4a, (5a, 0) Qq 8π0 ma

Q 51. A point particle of mass M attached to one end of a massless rigid non-conducting rod of length L. Another point particle of the same mass is attached to the other end of the rod. The two particles carry charges +q and −q respectively. This arrangement is held in a region of a uniform electric field E such that the rod makes a small angle θ (say about 5◦ ) with the field direction as shown in the figure. Find the expression for the minimum time needed for the rod to become parallel to the field after it is set free. (1989) • +q

The potential is zero when

1 √ (x−3a)2 +y 2 2 2

2 =√ , (x+3a)2 +y 2 2

θ

which simplifies to (x − 5a) + y = (4a) . This is the equation of a circle of radius 4a and centre (5a, 0).

−q •

y, V

−2Q −3a

Sol. Coulomb’s force qE acts on each charge as shown in the figure.

Q a

3a

5a

9a

x +q M• •

The potential on the x axis is given by   Q 1 −2 V (x) = + . 4π0 |x + 3a| |x − 3a|

L 2 sin θ

qE

(1)

The modulus function is defined as |x| = −x if x < 0 and |x| = x if x ≥ 0. Using definition of modulus function, we can write equation (1) as    Q 2 1  + , if x ≤ −3a;  4π 3a+x 3a−x   0  Q −2 1 V (x) = 4π0 x+3a + 3a−x , if −3a < x ≤ 3a; (2)      −2 1  Q 4π0 x+3a + x−3a , if x > 3a. It can be seen from equation (2) that V → −∞ as x → −3a and V → ∞ as x → 3a. The potential is zero at x = a and at x = 9a (see figure). The potential at the centre of circle (x = 5a) is   −2 Q 1 Q + = , V = 4π0 8a 2a 16π0 a which has a positive value. The potential at the circumference of the circle is zero. A positive charge moves

θ

qE

L 2 sin θ

O

~ E

•M −q

The torque of these forces about the point O is in clockwise direction and is given by τ = qE(L/2) sin θ + qE(L/2) sin θ = qEL sin θ.

(1)

The torque τ is related to the angular acceleration α = −d2 θ/dt2 (note that the torque is restoring in nature) by Iα = τ i.e., −I

d2 θ = qEL sin θ ≈ qELθ, dt2

(∵ sin θ ≈ θ),

(2)

where I is the moment of inertia of the rod (along with the two particles of mass M each) about an axis passing through O. As the rod is massless, the total moment of inertia is given by I = M (L/2)2 + M (L/2)2 = 12 M L2 . Substitute it in equation (2) to get d2 θ 2qE =− θ = −ω 2 θ. 2 dt ML

362

Part V. Electromagnetism

This equation q represents a SHM with an angular frequency ω = 2qE M L and the time period s 2π ML T = = 2π . ω 2qE The minimum time for the rod to become parallel to the field is tmin = T /4. q ML Ans. π2 2qE Q 52. Three particles, each of mass 1 g and carrying a charge q, are suspended from a common point by insulated massless strings, each 100 cm long. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side length 3 cm, calculate the charge q on each particle. [Take g = 10 m/s2 .] (1988) Sol. Let the three charged particles, each of charge q, be located at the three corners of an equilateral triangle of side a = 3 cm in the horizontal plane. The distance of the charges from the centroid O is r. Apply law of cosines in triangle OBC to get √ √ r = a/ 3 = 3 cm.

a

30◦ r

A •

x

30◦ 30◦

=

O

•

=

y a

a

•

C

Consider the electrostatic force on the charge at A due to the charges at B and C. The magnitude of forces due to each charge is |F~ | =

q2 , 4π0 a2

and their directions are as shown in the figure. Resolve F~ along and perpendicular to AO. The resultant electrostatic force on charge at A is given by √ 2 3q ◦ ~ ˆı. Fe = 2F cos 30 ˆı = 4π0 a2 P •

T cos θ

θ l

T =

z •

r

O x T sin θ

θ •

A

mg

1/2 4π0 a2 mg tan θ √ q= 3 1/2  2 (0.03) (10−3 )(10)(0.017) = 9 × 109 (1.73) 

Ans. 3.17 × 10−9 C

~ F

F sin 30◦

30◦ r

Divide equation (1) by (2) and simplify to get

~ F 2F cos 30◦

•

The forces on the charge at A are Fe , tension T , and weight mg. Resolve T in the horizontal and the vertical directions. The equilibrium condition on charge at A gives √ 2 3q , (1) T sin θ = Fe = 4π0 a2 T cos θ = mg. (2)

= 3.17 × 10−9 C.

F sin 30◦

B

The charges are connected with strings of length l = 100 cm. Let the other end of each string is hanging from the point P, which is vertically above O (see figure). In triangle OAP, √ r 3 tan θ = √ =√ = 0.017. l2 − r 2 1002 − 3

Fe

Q 53. Three point charges q, 2q and 8q are to be placed on a 9 cm long straight line. Find the positions where the charges should be placed such that the potential energy of this system is minimum. In this situation, what is the electric field at the position of the charge q due to the other two charges? (1987) Sol. The potential energy of two charges q1 and q2 , separated by a distance r, is given by U = q1 q2 /(4π0 r). For the potential energy of the given system to be minimum, the charges of the larger magnitude should be placed at the extreme positions. 2q

q

•

•

x

8q • 9−x

Let the charge q be placed at a distance x from the charge 2q. The total potential energy of the given system is   (2q)(8q) (2q)(q) (q)(8q) 1 + + U= 4π0 9 x 9−x   2 q 8 1 4 = + + . 2π0 9 x 9 − x The value of x for which potential energy is minimum is given by dU/dx = 0 i.e.,   dU q2 1 4 = − 2+ = 0. dx 2π0 x (9 − x)2

Chapter 26. Electric Field and Potential Solve to get x = 3. Note that another solution, x = −9, is not acceptable. The electric field at the position of q due to the other two charges is   8q 2q ~ = 1 − ˆı E 4π0 x2 (9 − x)2   q 8 2 = − ˆı = ~0. 4π0 (3)2 (9 − 3)2 Note that the force on the charge q is given by F~ = ~ = −dU/dx ˆı. The charge q is placed at the position qE of a stable equilibrium. Ans. (a) 2q and 8q at ends, q at 3 cm from 2q (b) zero Q 54. Two fixed, equal, positive charges, each of magnitude q = 5 × 10−5 C are located at points A and B separated by a distance 6 m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge, when reaches the point C at a distance of 4 m from O, has a kinetic energy of 4 J. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C. (1985) A

• +q

D

•

−q •C

O • +q

Sol. Let the farthest point D be at a distance rOD from O. The velocity of negative charge at D is vd = 0 (because the charge changes its direction of motion at √ D). From geometry, rAC = rBC = 32 + 42 = 5 m. A

• +q

3m •

rOD

O 3m

4m

−q •C

• +q

B

The electrostatic potential energies of the negative charge placed at C and D are   q2 1 1 UC = − + 4π0 rAC rBC 2(9 × 109 )(5 × 10−5 )2 = −9 J, 5   q2 1 1 45 =− + =− J. 4π0 rAD rBD rAD =− UD

mechanical energy, KC + UC = KD + UD , to get rAD = 45/5 = 9 m. Pythagoras theorem in triangle AOD gives q p √ 2 − r2 rOD = rAD 92 − 32 = 72 = 8.48 m. AO = Ans. From O 8.48 m Q 55. A thin fixed ring of radius 1 m has a positive charge 1 × 10−5 C uniformly distributed over it. A particle of mass 0.9 g and having a negative charge of 1 × 10−6 C is placed on the axis at a distance of 1 cm from the centre of the ring. Show that the motion of the negatively charged particle is approximately simple harmonic. Calculate time period of oscillations. (1982) Sol. A positive charge Q = 10−5 C is uniformly distributed over a ring of radius r = 1 m. The negative charge on the particle is q = 10−6 C and its mass is m = 0.9 g. +Q

√ r2

r O

x

+

x2

−q

x

Let the particle be placed at a distance x from the centre of the ring. Apply Coulomb’s law to show that the electrostatic force acting on the particle is Qqx ˆı 4π0 (r2 + x2 )3/2 Qqx =− ˆı 4π0 r3 (1 + x2 /r2 )3/2 Qq ≈− x ˆı. 4π0 r3 [∵ (1 + x2 /r2 )3/2 ≈ 1 for x  r].

F~ = −

B

D

363

The kinetic energy of the negative charge at C is KC = 4 J and at D is KD = 12 mvd2 = 0. Apply conservation of

The restoring force F~ is proportional to the displacement from the centre of motion. Thus, the particle executes SHM. Apply Newton’s second law on the particle to get its equation of motion d2 x Qq =− x = −ω 2 x, dt2 4π0 mr3 where ω is the angular frequency of SHM. The time period of oscillations is given by s 2π 4π0 mr3 = 2π T = ω Qq s (0.9 × 10−3 ) (1)3 = 2(3.14) = 0.628 s. (9 × 109 ) (10−5 ) (10−6 ) Note that the motion of the particle is periodic but not SHM, if x 6 r. Ans. 0.628 s

364

Part V. Electromagnetism

Q 56. A pendulum bob of mass 80 mg and carrying a charge of 2 × 10−8 C is at rest in a horizontal uniform electric field of 20000 V/m. Find the tension in the thread of the pendulum and the angle it makes with the vertical. [Take g = 9.8 m/s2 .] (1979) Sol. The ball of mass m = 80 mg and charge q = 2 × 10−8 C is placed in a horizontal uniform electric field E = 20000 V/m. Let T be the tension in the thread and θ be the angle it makes with the vertical.

T cos θ

θ T θ •

A

qE

mg

60◦

30◦

B

C

(a)

(i) The angle α. (ii) The tension in the cord. (iii) The normal reaction on the beads. (b) If the cord is now cut what are the values of the charges for which the beads continue to remain stationary?

Resolve T in the horizontal and the vertical directions. In equilibrium, net force on the ball is zero i.e., T sin θ = qE,

Fe

mg B

qE (2 × 10−8 )(20000) θ = tan−1 = tan−1 = 27◦ , mg (80 × 10−6 )(9.8) p T = (qE)2 + (mg)2 q = (2 × 10−8 × 20000)2 + (80 × 10−6 × 9.8)2

N2

α

α

T

(2)

Solve equations (1) and (2) to get

A

N1

(1)

T cos θ = mg.

= 8.8 × 10

q2 l

Sol. The forces acting on the charge q1 are its weight mg, normal reaction N1 , string tension T and Coulomb’s force Fe = q1 q2 /(4π0 l2 ). Similarly, forces on the charge q2 are its weight mg, normal reaction N2 , string tension T and Coulomb’s force Fe = q1 q2 /(4π0 l2 ).

T sin θ

−4

α

q1

30◦

90−α

l

T mg

α

60◦

N1 + Fe sin α = mg cos 30◦ + T sin α, Fe cos α + mg sin 30 = T cos α,

Ans. 8.8 × 10−4 N, 27◦ Q 57. A charged particle is free to move in an electric field. It will always move along an electric line of force.

C

The forces are shown in the figure. In triangle ABC, ∠A = 90◦ . Resolve the forces on q1 and q2 in the directions parallel and perpendicular to the sides of the frame. Balance the forces on q1 to get ◦

N.

Fe

(1) (2)

and on q2 to get N2 + Fe cos α = mg cos 60◦ + T cos α, ◦

Fe sin α + mg sin 60 = T sin α.

(3) (4)

(1979)

Sol. The electric force on a charge particle is tangential to the electric lines of forces. However, if the initial velocity of the particle makes an angle with the direction of force, the particle moves in a curved path e.g., particle moves perpendicular to the force in the uniform circular motion. Ans. F Q 58. A rigid insulated wire frame in the form of a right angled triangle ABC, is set in a vertical plane as shown in the figure. Two beads of equal masses m each and carrying charges q1 and q2 are connected by a cord of length l and can slide without friction on the wires. Considering the case when the beads are stationary determine, (1978)

Solve equations (2) and (4) to get α = 60◦ and T = mg + Fe . Substitute √ these values in equations (1) and (3) to get N1 = 3mg and N2 = mg. The tension becomes zero when the cord is cut. Substitute T = 0 to get Fe = −mg i.e., q1 q2 = −4π0 l2 mg. √ 1 q1 q2 Ans. (a) (i) 60◦ (ii) 4π + mg (iii) 3mg, mg 2 0 l (b) q1 q2 = −4π0 mgl2

Chapter 27 Gauss’s Law

Sol. By using Gauss’s law, Ein = 0, Eout = Q 4π0 r . Vin =

Q 1. Charges q, 2q and 4q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii r/2, r and 2r respectively, as shown in the figure. If magnitude of the electric fields at point P at a distance r from the centre of spheres 1, 2 and 3 are E1 , E2 and E3 respectively, then (2014) 4q P• r q

P• 2q r

•

P• r

•

r 2

z (a, 0, a)

(a, a, a)

•

r y

2r (0, 0, 0)

(A) E1 > E2 > E3 (C) E2 > E1 > E3

q , 4π0 r2

E2 =

(A) 2E0 a2 (B)

4q q (r3 ) = , 3 (2r) 2

E3 =

2q . 4π0 r2

q0 q = . 2 4π0 r 8π0 r2

0

(C)

R

~ |E|

0

r V

R

r

(B)

~ |E|

0 ~ (D) |E|

0

(2)

Use equations (1) and (2) to get ˆ = E0 a2 . φ = (E0 ˆı) · a2 (ˆı − k) Ans. C Q 4. A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure). The force F is proportional to (2010)

V

R

E0 a 2 2

ˆ = a2 (ˆı − k). ˆ ~ = (a ˆ) × (a ˆı + ak) S

Q 2. Consider a thin spherical shell of radius R with its centre at origin, carrying uniform positive surface charge density. The variation of the magnitude of the ~ electric field |E(r)| and the electric potential V (r) with distance r from the centre, is best represented by the graph? (2012) V

2E0 a2 (C) E0 a2 (D)

The area of the shaded region is the cross product of vectors representing the two sides i.e.,

Ans. C

~ (A) |E|

√

~ due to an electric field Sol. The flux through an area S ~ is given by E I ~ · dS ~ φ= E I ~ ~=E ~ · S. ~ (∵ E ~ is a constant.) = E · dS (1)

For the sphere 3, only the charge inside the spherical volume of radius r contributes to the electric field E3 . This charge and the electric field E3 are q0 =

(0, a, 0)

x

(B) E3 > E1 > E2 (D) E3 > E2 > E1

Sol. The magnitude of the electric field E1 and E2 due to the spheres 1 and 2 at a distance r from their centres are E1 =

Q 4π0 r 2

and R ~ ~ · dr, Vout = Since Ein = 0 and V (r) = − E we get constant. Using the fact that V (r) is continuous at r = R, we get, Vin = 4πQ0 R . Ans. D ~ = E0 ˆı, where E0 is a Q 3. Consider an electric field E constant. The flux through the shaded region (as shown in the figure) due to this field is (2011)

One Option Correct

r F

F

V

R

(A)

r

365

1 2 2 0 σ R

(B)

1 2 0 σ R

(C)

1 σ2 0 R

(D)

1 σ2 0 R 2

366

Part V. Electromagnetism

Sol. Let a charge carrying conductor of area d~s and surface charge density σ is placed in an electric field ~ 1 on one side and E ~ 2 on the other side. having value E

E~1

E~2 σ

d~s

Q2 + Q1 . Thus, the charge density on the outer surface of the middle shell is σ2 = (Q1 + Q2 )/(16πR2 ). Same argument applies for the outermost shell. Induced charge on the inner surface is Q3i = −Q2o = −(Q1 + Q2 ). The charge on the outer surface is Q3o = Q3 − Q3i = Q3 + Q1 + Q2 . The charge density on this surface is σ3 = (Q1 + Q2 + Q3 )/(36πR2 ).

Detailed analysis shows that electric force on the ~ average = σ ds (E ~1 + E ~ 2 )/2. In the conductor is σ ds E given problem, take a small area element d~s (radially outward) on the shell. The field inside the shell is ~ average | = zero and outside the shell is σ/0 . Thus, |E σ/(20 ) is in radially outward direction. The force on σ2 d~s. Integration of d~s over this element is dF~ = 2 0 the hemisphere gives the projected area πR2 . Thus, 2 2 |F~ | = π2 σ R . 0 Aliter: Use dimensional analysis to get the correct option. Ans. A Q 5. Three concentric metallic spherical shells of radii R, 2R and 3R, are given charges Q1 , Q2 and Q3 , respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1 : Q2 : Q3 , is (2009) (A) 1 : 2 : 3 (B) 1 : 3 : 5 (C) 1 : 4 : 9 (D) 1 : 8 : 18 Sol. The charge Q1 on the innermost shell is uniformly distributed on its outer surface giving a charge density σ1 = Q1 /(4πR2 ). + + + − − − + + − + 3R + − + + − − − − + + + − − +− −+ 2R + + − + −+ +− + −

R

+−

−+ + − + − − − + + − − + + + − − − + + +

− +

+

− +

Take a spherical Gaussian surface passing through the inside of middle shell. The flux through this surface is zero because shell is conducting and hence field inside the shell is zero. Thus, by Gauss’s law, charge enclosed within this Gaussian surface is zero. Hence, induced charge on the inner surface of the middle shell is Q2i = −Q1 . Total charge on the middle shell is the sum of charges on its outer surface Q2o and inner surface Q2i i.e., Q2 = Q2i + Q2o , which gives Q2o = Q2 − (−Q1 ) =

Equate σ1 = σ2 = σ3 and simplify to get Q1 : Q2 : Q3 = 1 : 3 : 5. Ans. B Q 6. A disc of radius a/4 having a uniformly distributed charge 6 C is placed in the x-y plane with its center at (−a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8 C is placed on the x-axis from x = a/4 to x = 5a/4. The two point charges −7 C and 3 C are placed at (a/4, −a/4, 0) and (−3a/4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x = ±a/2, y = ±a/2, z = ±a/2. The electric flux through the cubical surface is (2009) y

3C •

x • −7C

(A) − 2C 0 (B)

2C 0

(C)

10C 0

(D)

12C 0

Sol. By Gauss’s law, flux through a closed surface enclosing a charge qenc is qenc /0 . The charge enclosed within the given cubical surface consists of charge on half of the disc, charge on one-fourth of the rod, and the point charge −7 C. Charge on half of the disc is 3 C and that on one-fourth of the rod is 2 C. Thus, qenc = 3 C + 2 C + (−7 C) = −2 C. Hence the desired flux is −2C/0 . Ans. A Q 7. A spherical portion has been removed from a solid sphere having a charge distribution uniformly in its volume as shown in the figure. The electric field inside the emptied space is (2007)

Chapter 27. Gauss’s Law

(A) zero everywhere (C) non-uniform

367

(B) non-zero and uniform (D) zero only at its center

Sol. Let ρ be the charge density of the bigger sphere and O be its centre.

O ~ r

~ r1

O0 ~ r2 P

Removing a spherical portion from the bigger sphere is equivalent to adding a spherical portion of opposite charge density (−ρ). The electric field at a point in the cavity is equal to superposition of the fields by bigger sphere with charge density ρ and field by the smaller sphere with charge density −ρ. Using Gauss’s law, field at an inside point P by the bigger sphere is ~ 1 = ρ~r1 , E 30 and field by the smaller sphere is ~ 2 = − ρ~r2 . E 30 The resultant field at P is ~ =E ~1 + E ~ 2 = ρ(~r1 − ~r2 ) = ρ~r . E 30 30 Hence field in the cavity is non-zero and uniform as ~r remains same for all points inside the cavity. Ans. B Q 8. Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then, (2007) (A) negative and distributed uniformly over the surface of the sphere. (B) negative and appears only at the point on the sphere closest to the point charge. (C) negative and distributed non-uniformly over the entire surface of the sphere. (D) zero. Sol. The positive point charge induces negative charge in its vicinity but an equal amount of positive charge is induced in the remaining portion of the sphere. The net charge on the sphere remains zero. Ans. D Q 9. A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. (2007)

(A) A potential difference appears between the two cylinders when a charge density is given to the inner cylinder. (B) A potential difference appears between the two cylinders when a charge density is given to the outer cylinder. (C) No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders. (D) No potential difference appears between the two cylinders when same charge density is given to both the cylinders. Sol. Let the radii of the inner and outer cylinders be r1 and r2 , respectively. The problem has cylindrical symmetry and Gauss’s law can be applied to find the electric field E. r2 r1

l

In the case (A), surface charge density σ is given to the inner cylinder. Take a cylinder of radius r (r1 < r < r2 ) and length l as Gaussian surface. Apply Gauss’s law to get the electric field E(2πrl) =

qenc σ(2πr1 l) = , 0 0

i.e.,

E=

σr1 . 0 r

Potential difference between the two cylinders is given by   Z r2 σr1 r2 ∆V = − Edr = − ln .  r1 0 r1 In the case (B), charge density σ is given to the outer cylinder. Apply Gauss’s law to get E = 0 (because qenc = 0) and hence ∆V = 0. In the case (C), uniform linear charge density λ is kept along the common axis of cylinders. Apply Gauss’s law as in case (A) to get the electric field E(2πrl) = λl/0 ,

i.e.,

E = λ/(2π0 r).

Potential difference between the two cylinders is   Z r2 r2 λ ∆V = − Edr = − ln . 2π0 r1 r1 In the case (D) charge density σ is given to both the cylinders. Using the argument similar to the case (A), r2 1 we get, ∆V = − σr 0 ln( r1 ). Ans. A Q 10. Which of the following sets have different dimensions, (2005) (A) Pressure, Young’s modulus, Stress (B) EMF, Potential difference, Electric potential (C) Heat, Work done, Energy (D) Dipole moment, Electric flux, Electric field

368

Part V. Electromagnetism

Sol. Dipole moment of two equal and opposite charges ~ passing separated by distance a is qa. The flux of field E ~ is E ~ · S. ~ through area S Ans. D Q 11. Three infinitely long charge sheets are placed parallel to x-y plane as shown in the figure. The electric field at point P is (2005) z σ −2σ

z = 3a P z=0

−σ

z = −a

•

(A)

2σ ˆ 0 k

ˆ (B) − 2σ 0 k (C)

4σ ˆ 0 k

Sol. The potential at infinity is V∞ = 0. The potential at the centre of the ring is given by Z dq V0 = 4π0 r q (9 × 109 )(1.11 × 10−10 ) = = = 2 V. 4π0 r 0.5 By the definition of potential, Z l=0 ~ = V0 − V∞ = 2 V. ~ · dl −E l=∞

x

Ans. A

ˆ (D) − 4σ 0 k

Sol. The electric field at point P is given by ˆ = − 2σ k. ˆ ~ = −σ kˆ + −2σ kˆ + σ (−k) E 20 20 20 0 Ans. B Q 12. Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be due to (2004)

Q 14. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V . If the shell is given a charge of −3Q, the new potential difference between the same two surfaces is (1989) (A) V (B) 2V (C) 4V (D) −2V Sol. The charge on a solid conducting sphere is distributed uniformly over its surface. Thus, the potential due to a solid conducting sphere of radius R and charge Q is equal to the potential due to a hollow conducting shell of same radius and charge. r2

B • q2

A r1

• +q1 • −q1

(A) q2 (C) all the charges

(B) only the positive charges (D) +q1 and −q1

Sol. Gauss’s law gives the electric flux φ through a surface S by Z ~ · dS ~ = qenc . φ= E (1) 0 S In the given case, φ = 0 because qenc = q1 + (−q1 ) = 0. ~ in equation (1) is the net However, the electric field E electric field by all the charges. It is interesting to note ~ on Gaussian surface adjusts that if q2 is moved Rthen E ~ ~ remains zero. in such a way that E · dS Ans. C Q 13. A non-conducting ring of radius 0.5 m carries a charge of 1.11 × 10−10 C distributed non-uniformly on its circumference producing an electric field E everyR l=0 ~ ~ · dl, where in space. The value of the integral l=∞ −E (l = 0 being centre of the ring) in Volt is (1997) (A) +2 (B) −1 (C) −2 (D) zero

The potential at a distance r from the centre of a solid conducting sphere (or hollow conducting shell) is given by ( Q , if r ≤ R; 0R V (r) = 4π (1) Q otherwise. 4π0 r , Let r1 and r2 be the radii of the sphere and the shell. In the case (1), charge on the sphere is qsphere = Q and that on the shell is qshell = 0. Use equation (1) to get the potentials at point A, point B, and potential difference between these points as qshell Q qsphere + = , 4π0 r1 4π0 r2 4π0 r1 qsphere qshell Q VB = + = , 4π0 r2 4π0 r2 4π0 r2 Q Q ∆VAB = VA − VB = − 4π0 r1 4π0 r2   Q 1 1 = − . 4π0 r1 r2 VA =

(2)

Chapter 27. Gauss’s Law

369

In the case (2), charge on the sphere is qsphere = Q and that on the shell is qshell = −3Q. The potentials at the point A, point B, and potential difference between these points are qshell Q 3Q qsphere + = − 4π0 r1 4π0 r2 4π0 r1 4π0 r2   1 3 Q − , = 4π0 r1 r2 qsphere qshell Q 3Q VB0 = + = − 4π0 r2 4π0 r2 4π0 r2 4π0 r2 2Q =− , 4π0 r2   Q 1 1 0 0 0 ∆VAB = VA − VB = − . 4π0 r1 r2

Gaussian surface is the charge on the line segment PQ i.e., √ qenc = PQ λ = 3R λ, where p we used law of cosines in triangle √ OPQ to get PQ = R2 + R2 − 2(R)(R) cos 120◦ = 3R. The electric flux through the shell (Gaussian surface) is given by Gauss’s law √ φ = qenc /0 = 3R λ/0 .

VA0 =

(3)

From equations (2) and (3), the potential difference between A and B does not change even though potentials at both A and B are changed. Note that the potential difference ∆VAB depends on the charge on the inner sphere only. Ans. A

The electric field at a point due to the infinitely long line charge is radial i.e., perpendicular to the line PQ. The non-conducting spherical shell does not affect or alter the electric field. Thus, electric field at a point lying on the surface of the shell is perpendicular to the z-axis i.e., its z component is zero. The figure shows electric field on the surface of the shell for a positive line charge. Note that the magnitude of electric field is inversely proportional to the distance of the point from the line charge. z λ •

P

One or More Option(s) Correct •

Q 15. An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density λ. It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 120◦ at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is 0 . Which of the following statements is (are) true? (2018)

•

R 120◦

•

•

x

O

•

•

Q

•

z λ P

You are encouraged to find the electric field at a point (x, y, z) lying on the surface of the shell. Hint: If line PQ passes through x = −R/2 and y = 0 then

R 120◦

O

~ E(x, y, z) =

λ (x + R/2)ˆı + yˆ  . 2π0 (x + R/2)2 + y 2 Ans. (A), (B)

Q

√ (A) The electric flux through the shell is 3Rλ/0 . (B) The z-component of the electric field is zero at all the points on the surface of the shell.√ (C) The electric flux through the shell is 2Rλ/0 . (D) The electric field is normal to the surface of the shell at all points. Sol. Let Gaussian surface be same as the nonconducting spherical shell. The charge enclosed by this

Q 16. A point charge +q is placed just outside an imaginary hemispherical surface of radius r as shown in the figure. Which of the following statement(s) is(are) correct? (2017) +q

r

370

Part V. Electromagnetism

(A) The electric flux passing through the  curved surface  of the hemisphere is − 2q0 1 −

√1 2

.

(B) The component of the electric field normal to the flat surface is constant over the surface. (C) Total electric flux through the curved and the flat surfaces is q/0 . (D) The circumference of the flat surface is an equipotential. Sol. The electric field lines emanate uniformly from the point charge +q in radially outward directions. Since charge +q is placed just outside the hemispherical surface, any electric field line that enters the closed hemispherical surface (through the curved surface) will exit it from (a) the curved surface or (b) the flat surface. Thus, the net flux through the closed hemispherical surface is zero i.e., φcurved + φflat = 0 which gives

The solid angle subtended by the flat surface at the location of point charge is   Ω = 2π(1 − cos α) = 2π(1 − cos 45◦ ) = 2π 1 − √12 . Thus, Gauss’s law gives the flux through the flat surface     q Ω q 1 φflat = . = 1− √ 0 4π 20 2 Consider a point P on the flat surface at a distance x from the point O (see figure). The magnitude of electric field at P is given by E=

1 q . 2 4π0 (r + x2 )

The component of electric field normal to the flat surface,

φcurved = −φflat . Note that an electric field line that enters the curved surface and exits through the curved surface does not contribute to φcurved . On the other hand, an electric field line that enters the curved surface and exits through the flat surface contributes equally to φcurved and φflat (but with opposite signs). r sin φ

E⊥ = E cos θ =

varies with x. It is not constant. By symmetry, any point on the circumference of the flat surface is equidistant from the point charge. Thus, circumference is an equipotential with its potential given by

rd φ

V = φ α

r

1 qr , 4π0 (r2 + x2 )3/2

1 q √ . 4π0 2 r

dφ

We encourage you to find the flux through the flat surface by integrating Z Z ~ ~ E · dS = E⊥ dS. φflat =

C

The solid angle subtended by a cone of half angle α at its vertex is 2π(1 − cos α). To obtain this result, consider a solid cone of slant height r such that its vertex C is located at the centre of a sphere of radius r. The solid angle subtended by the cone is equal to the area of spherical cap divided by r2 i.e., Z α 1 Ω= 2 (2πr sin φ) (rdφ) = 2π(1 − cos α), r 0 where we divided spherical cap into small rings of radius r sin φ and thickness rdφ to get its area. +q

disc

disc

Hint: Take a ring of radius x and thickness dx to integrate over the disc. Ans. (A), (D) Q 17. The figure depicts two situations in which two infinitely long static line charges of constant positive line charge density λ are kept parallel to each other. In their resulting electric field, point charge q and −q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statements(s) is (are) (2015)

45◦ √

θ √

r

2r

λ

P r

O

λ

λ

λ

r 2 + x2

x

Ek

θ E⊥ E ~

+q

x

−q

x

Chapter 27. Gauss’s Law

371

r2

(A) Both charges execute SHM. (B) Both charges will continue moving in the direction of their displacement. (C) Charge +q executes SHM while charge −q continues moving in the direction of its displacement. (D) Charge −q executes SHM while charge +q continues moving in the direction of its displacement.

λ

λ d+x

d−x +q

x

a O

~ is uniform, its magnitude is independent of r2 (A) E but its direction depends on ~r. ~ is uniform, its magnitude depends on r2 and its (B) E direction depends on ~r. ~ is uniform, its magnitude is independent of a but (C) E its direction depends on ~a. ~ is uniform and both its magnitude and direction (D) E depend on ~a. Sol. Let ρ be the charge density of the spherical charge distribution of radius r1 centred at the origin O. A spherical cavity of radius r2 centred at P with distance OP = a = r1 − r2 is made in the spherical charge distribution.

~a

Let the charge +q be displaced by a small distance x (x  d) from its equilibrium position (see figure). The net electric field at the new location of +q due to the two line charges is ~ = E

r1

r2

Q

~r −

Sol. Let the separation between the two line charges be 2d. The electric field by a positive line charge is radially outwards and its magnitude at a perpendicular distance d is given by E = λ/(2π0 d). At the mid point, the electric fields due to the two line charges are equal and opposite to each other. Thus, the net force on the charge placed at the mid point is zero.

P

P ~ a

~ r O

λ λ λ ˆı − ˆı ≈ − x ˆı. 2π0 (d + x) 2π0 (d − x) π0 d2

r1

The force on the charge +q is ~ = − λq x ˆı. F~ = q E π0 d2 The force on +q is proportional to the displacement x and is directed towards the equilibrium position (restoring force). Thus, the charge +q executes SHM about its equilibrium position. We encourage you to find the time period of this SHM. The force on the charge −q when displaced by small distance x is ~ = F~ = −q E

λq x ˆı. π0 d2

This force is directed away from the equilibrium position. Hence, the charge −q continues to move in the direction of its displacement. Ans. C Q 18. Consider a uniform spherical charge distribution of radius r1 centered at the origin O. In this distribution, a spherical cavity of radius r2 , centered at P with distance OP = a = r1 − r2 (see figure) is made. If the ~ r), then electric field inside the cavity at position ~r is E(~ the correct statement(s) is (are) (2015)

The sphere with cavity is equivalent to a sphere of uniform charge density −ρ and radius r2 centred at P embedded in the original sphere. Thus, the electric field at a point Q in the cavity is superposition of (i) electric field at Q due to the sphere of charge density ~ 1 ), and (ii) electric ρ and radius r1 centred at O (say E field at Q due to the sphere of charge density −ρ and ~ 2 ). Let ~a, ~r, and ~r − ~a be radius r2 centred at P (say E the vectors as shown in the figure. The electric fields ~ 1, E ~ 2 , and their superposition E ~ 12 are given by E 1 43 π|~r|3 ρ ρ ~r, rˆ = 4π0 |~r|2 30 ~ 2 = − ρ (~r − ~a), E 30 ~ ~ ~ 2 = ρ ~a. E12 = E1 + E 30

~1 = E

Thus, the electric field at a point within the cavity is uniform and its magnitude and direction both depend on ~a. We encourage you to draw the electric lines of forces inside and outside the cavity. Ans. D

372

Part V. Electromagnetism

Q 19. Let E1 (r), E2 (r) and E3 (r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density λ, and an infinite plane with uniform surface charge density σ. If E1 (r0 ) = E2 (r0 ) = E3 (r0 ) at a given distance r0 , then (2014) λ (A) Q = 4πσr02 (B) r0 = 2πσ (C) E1 ( r20 ) = 2E2 ( r20 ) (D) E2 ( r20 ) = 4E3 ( r20 ) Sol. The electric fields at a distance r by the given charge distributions are E1 (r) =

Q , 4π0 r2

E2 (r) =

λ , 2π0 r

E3 (r) =

σ . 20

Equate E1 (r0 ) = E2 (r0 ) to get Q = 2λr0 , E1 (r0 ) = E3 (r0 ) to get Q = 2πσr02 and, E2 (r0 ) = E3 (r0 ) to get λ r0 = πσ . Use these equalities to get the fields at r = r0 /2, E1

E2

  Q 2λr0 λ = = 2 π0 r02 π0 r02 2π0 (r0 /2)
r0 = 2E2 2 ,

λ πσr0 σ r0 r0 2 = π r = π r =  = 2E3 2 . 0 0 0 0 0 r0 2




=

Ans. C Q 20. Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge densities +ρ and −ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region, (2013)

Consider a point P in the overlapping region. Let ~r1 is the position vector of P w.r.t. C1 and rˆ1 is the unit vector along ~r1 . Similarly, ~r2 is the position vector of P w.r.t. C2 and rˆ2 is the unit vector along ~r2 . The electric field at P is given by 1 43 π|~r2 |3 ρ 1 43 π|~r1 |3 ρ r ˆ − rˆ2 1 4π0 |~r1 |2 4π0 |~r2 |2 ρ ρ = (~r1 − ~r2 ) = ~r. 30 30

~P = E

~ P is a constant in magnitude and The electric field E has same direction (from C1 to C2 ) at all points in the overlapping region. Since the electric field is non zero, the potential is not constant. Ans. C, D Q 21. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities ρ1 and ρ2 respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, along the line joining the centres of the sphere, is zero. The ratio ρ1 /ρ2 can be (2013) (A) −4 (B) −32/25 (C) 32/25 (D) 4 Sol. The electric field at Q,  4 3 4 3 3 πR ρ1 3 πR ρ2 ~Q = 1 ˆ ı − ˆ ı = ~0, E 4π0 (2R)2 R2 gives ρ1 /ρ2 = 4. 2R ρ2 R ρ1 P•

2R

ρ

Q•

•

x

•

2R

−ρ

R2

R1

The electric field at P,  4 3 4 3 1 3 πR ρ1 3 π(2R) ρ2 ~ EP = − ˆı + ˆı = ~0, 4π0 (2R)2 (5R)2 gives ρ1 /ρ2 = −32/25. Ans. B, D

(A) (B) (C) (D)

the the the the

electrostatic electrostatic electrostatic electrostatic

field is zero. potential is constant. field is constant in magnitude. field has same direction.

Q 22. A cubical region of side a has it centre at the origin. It encloses three fixed point charges −q at (0, −a/4, 0) +3q at (0, 0, 0) and −q at (0, a/4, 0). Choose the correct option(s). (2012)

Sol. Let C1 and C2 be the centres of the two spheres and ~r be position vector from C1 to C2 .

~ r1

R1

a

−q

ρ C1

z

~ r

P

−ρ

•

~ r2 C2 R2 x

• 3q

−q •

y

Chapter 27. Gauss’s Law

373

(A) The net electric flux crossing the plane x = +a/2 is equal to the net electric flux crossing the plane x = −a/2. (B) The net electric flux crossing the plane y = +a/2 is more than the net electric flux crossing the plane y = −a/2. (C) The net electric flux crossing the entire region is q/0 . (D) The net electric flux crossing the plane z = +a/2 is equal to the net electric flux crossing the plane x = +a/2. Sol. From the reflection symmetry of the given charge distribution about y-z plane, the electric flux crossing the plane x = +a/2 is equal to the electric flux crossing the plane x = −a/2. Also, by the reflection symmetry about x-z plane, the electric flux crossing the plane y = +a/2 is equal to the electric flux crossing the plane y = −a/2. By the rotational symmetry about y-axis, the electric flux crossing the plane x = +a/2 is equal to the electric flux crossing the plane z = a/2. From Gauss’s law, the net electric flux crossing the entire region is qenc /0 = q/0 . Ans. A, C, D Q 23. A spherical metal shell A of radius RA and a solid metal sphere of radius RB (< RA ) are kept far apart and each is given a charge +Q. Now they are connected by a thin metal wire. Then, (2011) inside = 0 (B) QA > QB (A) EA on surface on surface B A < EB =R (D) EA (C) σσB RA Sol. When connected by a wire, charges on A and B are redistributed until potential on both becomes equal. After the charge redistribution, A and B are not influenced by each other because they are far apart. Thus, inside = 0 as field inside a conducting shell is zero. EA

Q 24. Which of the following statement(s) is (are) correct? (2011) (A) If electric field due to a point charge varies as r−2.5 instead of r−2 , then the Gauss law will still be valid. (B) The Gauss law can be used to calculate the field distribution around an electric dipole. (C) If the electric field between two point charges is zero somewhere, then the sign of two charges is the same. (D) The work done by the external force in moving a unit positive charge from point A at potential VA to point B at potential VB is VB − VA . Sol. Gauss’s law is based on 1/r2 (inverse square) na~ due to a point charge has ture of electric field. Let E k −2.5 ~ r dependence i.e., E = r2.5 rˆ. Consider a sphere of radius r as Gaussian surface. The direction of electric field is radial and its magnitude is same at all points of Gaussian surface. Thus, the flux through Gaussian surface is I ~ = k × (4πr2 ) = 4πk ~ · dS √ . φ= E r2.5 r The flux varies with r and hence Gauss’s law is not valid (φ 6= qenc /0 ). Gauss’s law is useful for calculating the electric field in problems with some underlying symmetry e.g., spherical, cylindrical, or plane charge distribution etc. The electric field between two point charges becomes zero only when both charges have the same sign. Last option is true by the definition of electric potential. Ans. C, D Q 25. A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that (2010)

QA QB RA •

RB •

The charge on a metal sphere B resides on its surface. Equating the potential on surface of A and B, QA QB 4π0 RA = 4π0 RB , gives QA > QB (since RA > RB ). The ratio of surface charge densities is 2 σA QA /(4πRA ) QA /(4π0 RA ) RB RB = = , 2 = Q /(4π R ) R σB QB /4πRB RA B 0 B A

and the ratio of surface electric fields is surface 2 EA QA /(4π0 RA ) RB = < 1. 2 ) = R surface QB /(4π0 RB EB A

Ans. A, B, C, D

Q1

•

•

Q2

(A) |Q1 | > |Q2 |. (B) |Q1 | < |Q2 |. (C) at a finite distance to the left of Q1 the electric field is zero. (D) at a finite distance to the right of Q2 the electric field is zero. Sol. From the direction of electric field lines, Q1 is positive and Q2 is negative. The density of electric field lines (which is an indication of flux) is more around Q1 in comparison to Q2 . In other words, the flux φ1 through a small sphere containing Q1 is more than the

374

Part V. Electromagnetism

flux φ2 through a similar sphere containing Q2 . From Gauss’s law, flux φ = qenc /0 . Thus, φ1 > φ2 implies |Q1 | > |Q2 |. The electric field at a distance x towards the right of Q2 is given by ~ = |E|

1 Q2 1 Q1 − , 4π0 (d + x)2 4π0 x2

where d is the separation between Q1 and Q2 . Since ~ becomes zero for some finite x. We enQ1 > Q2 , |E| courage you to show that there are two such points given by √ √ Q2 − Q1 Q2 Q2 + Q1 Q2 d, x2 = d. x1 = Q1 − Q2 Q1 − Q2 Also, show that the electric field is non-zero at all finite distances towards the left of Q1 . Draw the electric field lines in the entire region. Ans. A, D Q 26. For spherical symmetrical charge distribution, variation of electric potential, V , with distance from centre, r, is given in the figure. Which of the following option(s) is (are) correct? (2006) q 4π0 R0

V q 4π0 r

r

R0

(A) (B) (C) (D)

Total charge within 2R0 is q. Total electrostatic energy for r ≤ R0 is zero. At r = R0 electric field is discontinuous. There will be no charge anywhere except at r = R0 .

Sol. Use relation, E = − dV dr , to get the electric field from given potential ( q if r ≤ R0 , V (r) = 4πq 0 R0 ; if r > R0 , 4π0 r ( 0 if r ≤ R0 , E(r) = q if r > R0 . 4π0 r 2 Consider a spherical shell of radius 2R0 as Gaussian surface. Using Gauss’s law for this surface, we get qenc = 0 φ = 0 ·

q · 16πR02 = q. 16π0 R02

The electrostatic energy density for r ≤ R0 is = 0 (since E = 0) and hence the total electric energy stored in this region is zero. The electric field is discontinuous at r = R0 . This can be seen by taking left and right limits 1 2 2 0 E

lim E(r) = 0,

r→R0−

lim E(r) =

r→R0+

q . 4π0 R02

Since electric field is continuous at all points except at r = R0 , there is no charge distribution except at r = R0 . We encourage you to realize that the given potential is same as the potential due to a spherical shell of radius R0 having a charge q. Ans. A, B, C, D Q 27. An elliptical cavity is carved within a perfect conductor. A positive charge q is placed at the centre of the cavity. The points A and B are on the cavity surface as shown in the figure. Then, (1999) A • •

q

B

•

(A) electric field near A in the cavity = electric field near B in the cavity. (B) charge density at A = charge density at B. (C) potential at A = potential at B. (D) total electric field flux through the surface of the cavity is q/0 . Sol. The distance between the charge q and point A is less than the distance between the charge q and point B. The inverse square dependence of electric field, E ∝ r12 , gives EA > EB . The electric field in the vicinity of conductor with a surface charge density σ is given by E = σ/0 . Thus, EA = σA /0 and EB = σB /0 . The relation EA > EB gives σA > σB . Under electrostatic conditions, the field inside the conductor is zero and potential is same at all points on the conductor. This gives VA = VB . Gauss’s law gives flux crossing the cavity as φ = qenc /0 = q/0 . Ans. C, D Q 28. A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere, at a distance r from its centre, (1998) (A) increases as r increases for r < R. (B) decreases as r increases for 0 < r < ∞. (C) decreases as r increases for R < r < ∞. (D) is discontinuous at r = R. Sol. Let q be the charge on the sphere. Gauss’s law gives the field inside and outside the sphere as qr q Einside = , Eoutside = . 4π0 R3 4π0 r2 E(r)

r O

R

The variation of E with r is as shown in the figure. Ans. A, C

Chapter 27. Gauss’s Law

375

Paragraph Type

Assertion Reasoning Type

Paragraph for Questions 29-31 The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density ρ(r) (charge per unit volume) is dependent only on the radial distance r from the centre of the nucleus as shown in the figure. The electric field is only along the radial direction. (2008)

(B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

d

a

(C) Statement 1 is true, statement 2 is false.

r R

(D) Statement 1 is false, statement 2 is true.

Q 29. The electric field at r = R is (A) independent of a. (B) directly proportional to a. (C) directly proportional to a2 . (D) inversely proportional to a. Sol. The problem has spherical symmetry. Let us consider a spherical shell of radius R as Gaussian surface. Using Gauss’s law, E(4πR2 ) = qenc /0 . But qenc = Ze is independent of a. Hence, E is independent of a. Ans. A Q 30. For a = 0, the value of d (maximum value of ρ as shown in the figure) is 3Ze 3Ze 4Ze Ze (A) 4πR 3 (B) πR3 (C) 3πR3 (D) 3πR3 Sol. If a = 0 then the expression for volume charge density in the region 0 < r < R is ρ(r) = d(1 − r/R) which is a straight line passing through (0, d) and (R, 0). Consider a spherical shell of radius r and thickness dr. The charge inside this shell is ρ(r)4πr2 dr. Total charge inside a sphere of radius R is Z

R

ρ(r)4πr2 dr = 4πd

0

Statement 2: The electrical potential of a sphere of radius R with charge Q uniformly distributed on the (2008) surface is given by 4πQ0 R . (A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

ρ(r)

qenc =

Q 32. Statement 1: For practical purposes, the earth is used as reference at zero potential in electrical circuits.

Z

R

(1 − r/R)r2 dr

0

Sol. The potential at a point is generally defined with respect to a reference of zero potential at infinity. The potential on the surface of a sphere of radius R with charge Q uniformly distributed on it is 4πQ0 R when reference point is at infinity. The earth has a negative charge density of approximately σe = 1 nC/m2 giving us Qe = 4πRe2 σe ≈ 0.5 × 106 C and Ve ≈ 0.7 × 109 V, where Re = 6378 km is the radius of the earth. But for all practical purposes, the earth is used as a reference with zero potential. The reasons are: (i) Due to large size of the earth, its potential does not change even if a small amount of charge is given to the earth or taken away from it. Also, all conductors on the earth which are not given any external charge, are very nearly at the potential Ve . The choice of reference is quite arbitrary and is largely governed by convenience. (ii) The earth is a good conductor. It is easily accessible for electrical circuits geographically distributed over it and hence it is used as a common reference with zero potential.

πdR3 = . 3

Ans. B

Substitute qenc = Ze to get d = 3Ze/(πR3 ). Ans. B Q 31. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies, (A) a = 0 (B) a = R/2 (C) a = R (D) a = 2R/3 Sol. The electric field inside a uniformly charged sphere is proportional to r. The volume charge density inside a uniformly charged sphere is constant and hence a = R. We encourage you to find qenc in terms R a of a, R, and d and then analyse. Hint: qenc = 0 d 4πr2 dr + R R d(R−r) πd 2 3 2 2 3 R−a 4πr dr = 3 (R + a R + aR + a ). a Ans. C

Matrix or Matching Type

Q 33. The electric field E is measured at a point P (0, 0, d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. Column-I contains different relations between E and d. Column-II describes different electric charge distributions, along with their locations. Match the functions in Column-I with the related charge distribution in Column-II. (2018)

376

Part V. Electromagnetism Column I

(P) E is independent of d (Q) E ∝ 1/d

(R) E ∝ 1/d2

(S) E ∝ 1/d3

Column II (1) A point charge q at the origin. (2) A small dipole with point charge q at (0, 0, l) and −q at (0, 0, −l). Take 2l  d. (3) An infinite line charge coincident with the x-axis, with uniform linear charge density λ. (4) Two infinite wires carrying uniform linear charge density parallel to the xaxis. The one along (y = 0, z = l) has a charge density +λ and the one along (y = 0, z = −l) has a charge density −λ. Take 2l  d. (5) Infinite plane charge coincident with the x-y plane with uniform surface charge density.

Sol. The electric field at point P (0, 0, d) due to a point charge q at the origin is given by ˆ ~ 1 = 1 q k. E 4π0 d2

Note that power of d has increased by one when configuration is changed from point charge (mono-pole) to dipole. Similar was the case when configuration is changed from single line charge to two line charges (of opposite charge density). Think over it! Ans. P7→(5), Q7→(3), R7→(1,4), S7→(2) Integer Type Q 34. An infinitely long uniform charge distribution of charge per unit length λ lies parallel to the y-axis in √ the y-z plane at z = 23 a (see figure). If the magnitude of the flux of the electric field through the rectangular surface ABCD lying in the x-y plane with its centre at the origin is λL/(n0 ) (0 = permittivity of free space), then the value of n is . . . . . . . (2015) z

1 2p ˆ k. 4π0 d3

The electric field at point P (0, 0, d) due to an infinite line charge coincident with the x-axis and having a uniform linear charge density λ is given by ~3 = E

λ ˆ k. 2π0 d

Two infinite wires carry a uniform linear charge density parallel to x-axis. The one along (y = 0, z = l) has a charge density +λ and the one along (y = 0, z = −l) has a charge density −λ. The electric field at point P (0, 0, d) due to this configuration is given by ~4 = E

λ λl ˆ λ kˆ − kˆ ≈ k. 2π0 (d − l) 2π0 (d + l) π0 d2

The electric field at point P (0, 0, d), due to an infinite plane charge coincident with the x-y plane, and having a uniform surface charge density σ, is given by ˆ ~ 5 = σ k. E 20

C

a A

√ 3 a 2

y B

x

Sol. The configuration is shown in the figure. Cross-section of Gaussian Surface

The electric field at point P (0, 0, d) due to a small dipole with point charge q at (0, 0, l) and −q at (0, 0, −l) ˆ is i.e., field due to a dipole of dipole moment p~ = 2lq k, given by ~2 = E

L

D

Line Charge

30◦ √3a 2

P

a

Q

Rectangular Surface

√ Consider a cylindrical Gaussian surface of radius a 3/2 and length L. The charge enclosed by Gaussian surface is qenc = λL. Apply Gauss’s law to get the total flux through Gaussian surface φ = qenc /0 = λL/0 . The flux through the rectangular surface is equal to the flux through the cylindrical section with chord PQ (see figure). From geometry, the chord PQ subtends an angle α = 60◦ at the axis of the cylinder. Thus, the flux through the rectangular surface is φrectangle = φ(α/360) = (λL/0 )(60/360) = λL/(60 ). We encourage you to find the R flux through the rectan~ · dS. ~ gular surface by using φ = E Ans. 6

Chapter 27. Gauss’s Law

377

Q 35. An infinitely long solid cylinder of radius R has a uniform charge density ρ. It has a spherical cavity of radius R/2 with its centre on the axis of cylinder, as shown in the figure. The magnitude of electric field at point P, which is at a distance 2R from the axis of the 23ρR . The value of cylinder, is given by the expression 16k 0 k is . . . . . . . (2012) z

R R 2

P

y

Ans. 6 Q 36. A solid sphere of radius R has a charge Q distributed in its volume with a charge density ρ = κra , where κ and a are constants and r is the distance from its centre. If the electric field at r = R/2 is 1/8 times that at r = R, find the value of a. (2009) Sol. The direction of electric field E(r) is radial and its magnitude is same at all points on the spherical shell of radius r due to spherical symmetry. Consider a spherical Gaussian surface of radius r (< R). Gauss’s law gives E(r)4πr2 = qenc /0 .

2R x

R dr

Sol. Given configuration is equivalent to an infinitely long solid cylinder of radius R having a uniform charge density ρ and a solid sphere of radius R/2 having a uniform charge density −ρ. By the principle of super~ = E ~ cyl + E ~ sph . position, the electric field at P is E ~ sph and E ~ cyl point towards the By symmetry, both E y-direction. z

R 2

P ~ sph E

E(r) =

2R R

P

y ~ cyl E

For sphere, consider a shell of radius 2R as Gaussian surface. Apply Gauss’s law, (4/3) π (R/2)3 (−ρ) , 0

~ sph = −ρR/(960 ) ˆ. For cylinder, consider a to get E cylinder of radius 2R and length l as Gaussian surface. Apply Gauss’s law, E cyl · 2π(2R)l =

The ratio,

E(R/2) E(R)

=

1 2a+1

= 18 , gives a = 2. Ans. 2

Descriptive

x

E sph · 4π(2R)2 =

qenc κra+1 = . 4π0 r2 0 (a + 3)

This equation gives the electric field at r = R/2 and r = R as κ(R/2)a+1 κ(R)a+1 E(R/2) = , E(R) = . 0 (a + 3) 0 (a + 3)

z

l

To determine qenc , consider a spherical shell of radius r and thickness dr. The charge on this shell is ρ4πr2 dr. Integrate from 0 to r to get the charge enclosed within a sphere of radius r i.e., Z r ra+3 qenc = κra 4πr2 dr = 4πκ . a+3 0 Substitute qenc in the expression of Gauss’s law to get

y

2R

x

r

Q 37. Two large parallel metallic plates S1 and S2 carrying surface charge densities σ1 and σ2 respectively (σ1 > σ2 ) are placed at a distance d apart in vacuum. Find the work done by the electric field in moving a point charge q a distance a (a < d) from S1 towards S2 along a line making an angle π/4 with the normal to the plates. (2004) σ1

π R2 l(ρ) , 0

~ cyl = ρR/(40 ) ˆ. Thus, the resultant electric to get E field at P is given by ~ =E ~ cyl + E ~ sph = ρR ˆ − ρR ˆ = 23ρR ˆ. E 40 960 960

σ2

B a 45◦ A

d

378

Part V. Electromagnetism

Sol. The electric field near a large metallic plate having a charge density σ is σ/0 . The electric field by two plates in the region of interest is opposite to each other, giving the net electric field ~ = σ1 − σ2 ˆı, E 0 where ˆı is the unit vector normal to the plates. The ~ force on a charge particle in an electric field is F~ = q E ~ ~ and work done by F in displacing charge q by S is ~= W = F~ · S

q(σ1 − σ2 ) qa(σ1 − σ2 ) √ a cos 45◦ = . 0 20 Ans.

(σ1 −σ2 )qa √ 20

Q 38. A conducting sphere S1 of radius r is attached to an insulating handle. Another conducting sphere S2 of radius R is mounted on an insulating stand. S2 is initially uncharged. S1 is given a charge Q, brought into contact with S2 and removed. S1 is recharged such that the charge on it is again Q and it is again brought into contact with S2 and removed. This procedure is repeated n times. (1998) (a) Find the electrostatic energy of S2 after n such contacts with S1 . (b) What is the limiting value of this energy as n → ∞? Sol. The potential on the surface of a conducting sphere S1 of radius r and charge Q is V = Q/(4π0 r). When S1 is brought in contact to a uncharged conducting sphere S2 of radius R, a charge q1 is transferred from S1 to S2 . The charge transfer continues till potential on both the spheres become equal   q1 R Q − q1 = , =⇒ q1 = Q . 4π0 r 4π0 R R+r The sphere S1 is recharged to Q before the second contact. Let q2 be the charge transferred from S1 to S2 in the second contact. The charge transfer stops when potential on both the spheres become equal  2 q1 + q2 R Q − q2 = , =⇒ q2 = Q . 4π0 r 4π0 R R+r Similarly, if q3 is the charge transfer in the third contact
3 q1 +q2 +q3 R 3 then Q−q and 4π0 r = 4π0 R , which gives q3 = Q R+r th this process continues
till the n contact with a charge n R . Total charge on S2 after n transfer qn = Q R+r contacts is Q0 = q1 + q2 + · · · + qn "   2  n # R R R =Q + +· · ·+ R+r R+r R+r   n  QR R = 1− . r R+r (sum of a geometric series.)

The capacitance of spherical conductor of radius R is C = 4π0 R and hence electrostatic energy of S2 is   n 2 2 R Q0 Q2 R 1− Un = = . 2C 8π0 r2 R+r As n → ∞, sum of geometric progression becomes Q0∞ = QR r and hence the limiting value of electrostatic energy is U∞ =

Q0∞2 Q2 R . = 2C 8π0 r2

We encourage you to evaluate U∞ = limn→∞ Un to cross verify this solution. h
n i 2 Q2 R R Ans. (a) Un = 8π 1 − 2 R+r 0r (b) U∞ =

Q2 R 8π0 r 2

Q 39. Two isolated metallic solid spheres of radii R and 2R are charged such that both of these have same charge density σ. The spheres are located far away from each other and connected by a thin conducting wire. Find the new charge density on the bigger sphere. (1996) Sol. Initially, charge on the sphere of radius R is q1 = 4πR2 σ and that on the sphere of radius 2R is q2 = 4π(2R)2 σ = 16πR2 σ. Let q10 and q20 be the charges on two spheres after they are connected by a thin conducting wire. Charge conservation gives q10 + q20 = q1 + q2 = 20πR2 σ.

(1)

The charge transfer through the conducting wire stops when the potentials on surface of two spheres are equal 1 q10 1 q20 = . 4π0 R 4π0 2R

(2)

Solve equations (1) and (2) to get the charges and charge densities 20 2 πR σ, 3 5 q10 σ10 = = σ, 2 4πR 3

40 2 πR σ, 3 q20 5 σ20 = = σ. 2 4π(2R) 6

q10 =

q20 =

Ans.

5 6σ

Q 40. Three concentric spherical metallic shells A, B and C of radii a, b and c (a < b < c) have surface charge densities σ, −σ and σ, respectively. (1990) (a) Find the potential of the three shells A, B and C. (b) If the shells A and C are at the same potential, obtain the relation between the radii a, b and c. Sol. Three concentric shells are shown in the figure. σ −σ σ C B A

c

a b

Chapter 27. Gauss’s Law

379

The charges on the three shells are qA = 4πa2 σ, qB = −4πb2 σ and qC = 4πc2 σ. The potential due to a conducting shell of radius R and charge q, at a radial distance r, is given by ( q , if r ≤ R; V (r) = 4πq 0 R if r > R. 4π0 r , Thus, the potentials on the three shells are σ qB qC i 1 h qA = + + [a − b + c] , 4π0 a b c 0   σ a2 qB qC i 1 h qA = + + −b+c , VB = 4π0 b b c 0 b   1 h qA σ a2 qB qC i b2 VC = = + + − +c . 4π0 c c c 0 c c VA =

2

2

If VA = VC then a − b + c = ac − bc + c which gives a + b = c. Ans. (a) VA = σ0 (a − b + c), VB =  2  2 2 2  σ a VC = σ0 a −bc +c (b) a + b = c 0 b −b+c , Q 41. A charge Q is distributed over two concentric hollow spheres of radii r and R(> r) such that the surface charge densities are equal. Find the potential at the common centre. (1981) Sol. Let the charge q be distributed uniformly over the hollow sphere of radius r and remaining charge Q − q be distributed uniformly on the hollow sphere of radius R. The charge densities of two spheres are equal q Q−q = 2 4πr 4πR2

i.e.,

q=

Qr2 . R2 + r 2

Q−q q O

r R

The potential at O due to both the spheres is   1 q Q−q Q(R + r) VO = + = . 4π0 r R 4π0 (R2 + r2 ) Ans.

Q(R+r) 4π0 (R2 +r 2 )

Chapter 28 Capacitors

One Option Correct

Let the final (after switch is turned to position 2) charges on the two capacitors be Qf1 and Qf2 . The charge conservation gives

Q 1. In the given circuit, a charge of +80 µC is given to the upper plate of the 4 µF capacitor. Then in the steady state, the charge on the upper plate of the 3 µF capacitor is (2012)

Qf1 + Qf2 = Qi .

(1)

The capacitors C1 and C2 are connected in parallel. Thus, the voltages across both the capacitors are equal i.e.,

+80µC 4µF

Qf2 Qf1 = . C1 C2 2µF

3µF

(2)

Solve equations (1) and (2) to get the charges 1 C1 Qi = Qi , C1 + C2 5 C2 4 = Qi = Qi , C1 + C2 5

Qf1 = (A) +32 µC (B) +40 µC (C) +48 µC (D) +80 µC

Qf2

Sol. Let the charges on 3 µF and 2 µF capacitors be q and q 0 respectively. Charge on lower plate of 4 µF capacitor is −80 µC. Lower plate of 4 µF capacitor and upper plates of 2 µF and 3 µF capacitors form an isolated system. Hence, net charge on this system must be zero i.e., q + q 0 − 80 µC = 0.

and the electrostatic potential energies   Qf1 2 1 Qi 2 1 Uf1 = = Ui , = 2C1 25 2C1 25 Uf2 =

(1)

The energy loss is Ui − (Uf1 + Uf2 ) = 0.80Ui i.e., 80%. Can you explain where is this energy lost? Ans. D

The potentials (V = q/C) across these two capacitors are equal which gives q/3 = q 0 /2.

Q 3. Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2 . The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is (2002)

(B) 14 C V12 + V22 (A) 14 C V12 − V22 2 2 (C) 14 C (V1 − V2 ) (D) 41 C (V1 + V2 )

(2)

Eliminate q 0 from equations (1) and (2) to get q = 48 µC. Ans. C Q 2. A 2 µF capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is (2011) 1 ◦ V

Sol. Initially, the total electrostatic energy stored in the two capacitors is Ui = 12 CV12 + 12 CV22 . Initial charges on the two capacitors are q1 = CV1 and q2 = CV2 . Let q10 and q20 be the charges on the capacitors when same terminals of the two capacitors are connected to each other.

2 ◦ ◦S 2µF

Qf2 2 4 Ui . = 2C2 25

8µF

+ −

(A) 0 % (B) 20 % (C) 75 % (D) 80 % Sol. Let C1 = 2 µF and C2 = 8 µF be the capacitances of given capacitors. Initially, charge on the capacitor C1 Q2 is Qi = C1 V and its electrostatic energy is Ui = 2Ci1 .

+ −

380

Chapter 28. Capacitors

381

The charge conservation gives q10 + q20 = q1 + q2 .

(1)

0

The potentials V across the two capacitors are equal, i.e., q10 /C = q20 /C.

(2)

Solve equations (1) and (2) to get q10 = q20 = (q1 + q2 )/2,

V 0 = (V1 + V2 )/2.

and

Thus, the final electrostatic energy stored in the two capacitors is

Use the energy conservation, UA + UB = UA0 + UB0 , to get q 0 = 0 or q 0 = 2q. We encourage you to interpret the two solutions of q 0 . Ans. A Q 5. A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K1 , K2 and K3 as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor then its dielectric constant K is given by (2000) A/2

Uf = 12 CV 2 + 12 CV 2 = 14 C(V12 + V22 + 2V1 V2 ),

A/2 •

K1

and loss in energy is Ui − Uf = 14 C(V1 − V2 )2 . Ans. C

K2

d K3

Q 4. Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is (2001) +

−

+

−

+

−

+

−

+

−

S

A

B

(A) zero (B) q/2 (C) q (D) 2q Sol. Initially, charge on the right plate of capacitor A is −q and that on the left plate of capacitor B is zero. The electric field inside the two capacitors and energy stored in them are q/A −q/A q EA = − = , 20 20 A0 EB = 0; 1 q2 d 0 EA 2 (Ad) = , 2 20 A UB = 0. UA =

Let the charge q 0 moves from the right plate of A to the left plate of B when the switch S is closed. By charge conservation, charge remaining on the right plate of A is −q − q 0 and charge on the left plate of B is q 0 . The electric field inside the two capacitors and energy stored in them are   q q0 q/A (−q − q 0 )/A 0 − = 1+ , EA = 20 20 A0 2q q 0 /A 0 EB = , 20  2 1 q2 q0 0 2 (Ad) = 1+ , UA0 = 0 EA 2 20 2q 2

UB0 =

q0 d . 80 A

•

1 (A) K 1 (B) K 1 (C) K (D) K

q

d 2

= = = =

1 K1

+

1 K2

1 K1 +K2 K1 K2 K1 +K2 K1 K3 K1 +K3

1 2K3 1 2K3

+

+ + 2K3 K3 + KK22+K 3

Sol. The capacitance of a parallel plate capacitor of plate area A, plate separation d, and dielectric constant K is given by C = Kd0 A . Given configuration can be considered as a combination of three capacitors having capacitances 0 A K1 0 (A/2) = K1 = K1 C0 , d/2 d K2 0 (A/2) C2 = = K2 C0 , d/2 K 3 0 A C3 = = 2K3 C0 . d/2 C1 =

where C0 = 0 A/d. The capacitor C1 and C2 are connected in parallel with their effective capacitance C12 = C1 + C2 = (K1 + K2 )C0 . The C12 is combined in series with C3 giving effective capacitance of the three capacitors as C123 =

C12 C3 (K1 + K2 )2K3 K0 A = C0 = . C12 + C3 K1 + K2 + 2K3 d

Simplify to get

1 K

=

1 2K3

+

1 K1 +K2 .

Ans. B Q 6. Two identical metal plates are given positive charges q1 and q2 (< q1 ) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is (1999) q1 +q2 q1 +q2 q1 −q2 q1 −q2 (A) 2C (B) C (C) C (D) 2C

382

Part V. Electromagnetism

Sol. Let A be the area of the plates and d be the separation between them. q1

q2 ~1 E

~2 E

y x

Sol. Initially, charge on the capacitor of capacitance C is q1 = CV and that on the capacitor of capacitance 2C is q2 = (2C)(2V ) = 4CV . Let q10 and q20 be the charges on the two capacitors when the positive terminal of one is connected to the negative terminal of the other (see figure). C

+ +

d

−

−

+ +

−

The electric field, in the region between the plates, due to the left plate is

−

2C

~ 1 = σ1 ˆı = q1 ˆı, E 20 20 A

The charge conservation gives q10 + q20 = −q1 + q2 = 3CV.

and due to the right plate is

(1)

Also, the potentials across the two capacitors are equal

~ 2 = − σ2 ˆı = − q2 ˆı. E 20 20 A

q10 /C = q20 /(2C).

~ and potential difference between The resultant field E the two plates are

(2)

Solve equations (1) and (2) to get q10 = CV and q20 = 2CV . Final energy of the configuration is

~ =E ~1 + E ~ 2 = q1 − q2 ˆı, E 20 A q1 − q2 q1 − q2 V = Ed = = . 2(0 A/d) 2C

2

U=

2

q10 q0 3 + 2 = CV 2 . 2C 2(2C) 2 Ans. B

Ans. D Q 7. For the circuit shown, which of the following statements is true? (1999) S1

V1 = 30V C1 = 2µF

(A) (B) (C) (D)

with with with with

S1 S3 S1 S1

S3

V2 = 20V

S2

Q 9. Seven capacitors each of capacitance 2 µF are connected in a configuration to obtain an effective capacitance 10 11 µF. Which of the following combination will achieve the desired result? (1990) (A) (B)

C2 = 3µF

closed, V1 = 15 V, V2 = 20 V. closed, V1 = V2 = 25 V. and S2 closed, V1 = V2 = 0. and S3 closed, V1 = 30 V, V2 = 20 V.

Sol. The charge on the left capacitor is Q1 = C1 V1 = 60 µC and that on the right capacitor is Q2 = C2 V2 = 60 µC. The potential or charge on the capacitor does not change in any case. Ans. D Q 8. A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is (1995) 2 (A) zero (B) 32 CV 2 (C) 25 (D) 92 CV 2 6 CV

(C)

(D)

Sol. In option A, equivalent capacitance of five capacitors connected in parallel is C1 = 10 µF and that of two capacitors connected in series is C2 = 1 µF. The C1 and C2 are connected in series so equivalent capacitance is C2 = 10 Ceq = CC11+C 11 µF. 2 Ans. A One or More Option(s) Correct Q 10. A parallel plate capacitor having plates of area s and plate separation d, has capacitance C1 in air. When two dielectrics of different relative permittivities (1 = 2 and 2 = 4) are introduced between the two plates as shown in the figure, the capacitance becomes C2 . The ratio C2 /C1 is (2015)

Chapter 28. Capacitors

383 d/2 2

+ 1

E1

Q2

E2

s/2

s/2

−

d

(A) (C)

(A) 6/5 (B) 5/3 (C) 7/5 (D) 7/3 Sol. The capacitance of an air filled parallel plate capacitor having plates of area s and plate separation d is given by C1 = 0 s/d. Consider the case when two dielectrics of relative permittivities (1 = 2 and 2 = 4) are introduced between the two plates as shown in the figure. We can think of this configuration to be made up of three parts: (a) Capacitor of plate area s/2, plate separation d, and filled with a dielectric of relative permittivity 1 = 2 (lower half). The capacitance of this part is Ca = s 1 0 s/2 d = 0 d = C 1 . (b) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity 1 = 2 (left upper half). The capacitance of this s/2 part is Cb = 1 0 d/2 = 20 ds = 2C1 . (c) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity 2 = 4 (right upper half). The capacitance of this s/2 part is Cc = 2 0 d/2 = 40 ds = 4C1 . The capacitors Cb and Cc are connected in series with equivalent capacitance Cbc =

Q1

E1 E2 Q1 Q2

=1 3 =K

(B) (D)

Q 11. A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is C1 . When the capacitor is charged, the plate area covered by the dielectric gets charge Q1 and the rest of the area gets charge Q2 . The electric field in the dielectric is E1 and that in the other portion is E2 . Choose the correct option/options, ignoring edge effects. (2014)

1 K 2+K K

K0 (A/3) K0 A = , d 3d 20 A 0 (2A/3) = . C2 = d 3d

C1 =

These two portions are connected in parallel. Hence, 0A the total capacitance is C = C1 + C2 = (2+K) . 3d In parallel connection, potential between the plates V /d 1 is equal which gives E E2 = V /d = 1. The ratio of the charges on the two portions is

Q1 Q2

=

C1 V C2 V

=K 2 . Ans. A, D

Q 12. In the circuit shown in figure, there are two parallel plate capacitors each of capacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and then released. The switch S2 is then pressed to charge the capacitor C2 . After some time, S2 is released and then S3 is pressed. After some time, (2013) S1

The capacitor Cbc is connected in parallel with Ca . The equivalent capacitance is

We encourage you to find the capacitance when positive and negative plates (terminals) are placed at the top and bottom in the given configuration (instead of the left and right sides). Ans. D

= =

Sol. Let A be the total plate area of the capacitor and d be the separation between the plates. The plate area for the portion with dielectric is A1 = A/3 and that without dielectric is A2 = 2A/3. The capacitances of the portion with dielectric and without dielectric are

Cb Cc 4 (2C1 )(4C1 ) = C1 . = Cb + Cc (2C1 ) + (4C1 ) 3

4 7 C2 = Cabc = Ca k Cbc = C1 + C1 = C1 . 3 3

E1 E2 C C1

2V0

(A) (B) (C) (D)

the the the the

charge charge charge charge

S2 C1

on on on on

the the the the

upper upper upper upper

S3 C2

plate plate plate plate

of of of of

V0

C1 C1 C2 C2

is is is is

2CV0 CV0 0 −CV0

Sol. Pressing switch S1 charges C1 to a charge Q = C1 V = 2CV0 with its upper plate (attached to positive terminal) having charge +2CV0 . Releasing switch S1 and pressing switch S2 equally distributes the charge on C1 and C2 (identical capacitors). Thus, the charge on the upper plate of C1 and C2 is +CV0 . When switch S2 is released and switch S3 is pressed, charge on the upper plate of C1 remains +CV0 but charge on C2 changes. The battery attached to C2 charges it to charge CV0 with upper plate (attached to negative terminal) having charge −CV0 . Ans. B, D

384

Part V. Electromagnetism

Q 13. A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x = 0 and positive plate is at x = 3d. The slab is equidistant from the plates. The capacitor is given some charge. As x goes from 0 to 3d, (1998) (A) the magnitude of the electric field remains the same. (B) the direction of the electric field remains the same. (C) the electric potential increases continuously. (D) the electric potential increases at first, then decreases and again increases. Sol. Let σ > 0 be the charge density of the plate at x = 3d. − − − −

+ +

−

+

−

+

−

−

+ + + +

0

d

2d

3d

Q 15. A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system in question, in the process of inserting the slab, then, (1991) (B) Q = 0 KAV (A) Q = 0 AV d d 2   V 1 (C) E = Kd (D) W = 0 AV 1− K 2d Sol. Initially, the capacitance of the capacitor, charge on its plates, and electrostatic energy stored in it are

x

The electric field in the various regions inside the capacitor is given by  σ  if 0 < x ≤ d; − 0 ˆı, σ ~ E(x) = − 0 K ˆı, if d < x ≤ 2d;   σ if 2d < x ≤ 3d. − 0 ˆı, R The electric potential is given by V (x) = − E dx. If the plate at x = 0 is assumed to be at zero potential then the potential in the various regions is  σx  if 0 < x ≤ d;  0 , σd V (x) = 0 + σ(x−d) , if d < x ≤ 2d; 0 K   σd σ(x−2d) σd , if 2d < x ≤ 3d. 0 + 0 K + 0 Ans. B, C ~ in the annular Q 14. The magnitude of electric field E region of a charged cylindrical capacitor, (1996) (A) is same throughout. (B) is higher near the outer cylinder than near the inner cylinder. (C) varies as 1/r where r is the distance from the axis. (D) varies as 1/r2 where r is the distance from the axis. Sol. Consider a cylindrical Gaussian surface of length l and radius r. A

the inner cylinder then the charge enclosed by Gaussian H ~ · dS ~ = qenc /0 , surface is qenc = λl. Gauss’s law, E gives E = 2πλ0 r . Ans. C

B

~ is radial and is of By symmetry, electric field E same magnitude at all points at a radial distance r. The fluxes through the end surfaces A and B of the cylinder are H zero. Hence, the flux through Gaussian surface is ~ · dS ~ = 2πrlE. If λ is the charge per unit length on E

C = 0 A/d, Q = CV = 0 AV /d, U = 21 CV 2 = 0 AV 2 /(2d). The charge on the capacitor plates does not change if the dielectric slab is inserted after the battery disconnection. Finally, the capacitance, charge, and stored energy are C 0 = K0 A/d, Q0 = Q = 0 AV /d, 2

U 0 = Q0 /(2C 0 ) = 0 AV 2 /(2dK). The energy conservation gives work done on the system as   0 AV 2 1 W = U0 − U = − 1− . 2d K The potential difference across the capacitor plates becomes V 0 = Q0 /C 0 = V /K and electric field between the plates is E 0 = V 0 /d = V /(Kd). Ans. A, C, D Q 16. A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles, (1987) (A) the charge on the capacitor increases. (B) the voltage across the plates increases. (C) the capacitance increases. (D) the electrostatic energy stored in the capacitor increases. Sol. The capacitance of a parallel plate area A and plate separation d increasing the plate separation from itance decreases to C 0 = 0 A/d0
Q0 (B) V > V0 (C) E > E0 (D) U > U0 Sol. The potential across the capacitor is equal to the battery voltage if the battery is not removed i.e., V = V0 . The capacitance of the capacitor increases from C0 to C = KC0 when a dielectric slab of dielectric constant K is introduced to fill the gap between the plates. The charge on the capacitor plates become Q = CV = (KC0 )V0 = KQ0 > Q0 (because K > 1). The electric field between the plates separated by a distance d is given by E = V /d = V0 /d = E0 . The electrostatic energy stored in the capacitor is given by
U = 21 CV 2 = 12 (KC0 )V02 = K 12 C0 V02 = KU0 > U0 .

(∵ K > 1) Ans. A, D

Paragraph Type Paragraph for Questions 18-19 Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r  h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0 and the top plate at −V0 . Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. [Ignore gravity.] (2016)

A

HV

Q 18. Which one of the following statements is correct? (A) The balls will stick to the top plate and remain there. (B) The balls will bounce back to the bottom plate carrying the same charge they went up with. (C) The balls will bounce back to the bottom plate carrying the opposite charge they went up with. (D) The balls will execute simple harmonic motion between the two plates. Sol. The distance between the two plates is h. The potential of the bottom plate is V0 and that of the top plates is −V0 . The electric field between the plates is E = 2V0 /h (directed upwards). The radius of each ball is r ( h). Let m be the mass and C be the capacitance of each ball. −V0 −q F

~ E h

F +q +V0

When the ball touches the bottom plate, it gets a positive charge q = CV0 (we assume that the charge transfer is instantaneous). This positively charged ball experiences an upward force, F = qE = 2CV02 /h, which accelerates the ball upwards. Since the force is constant, the ball cannot do SHM (for SHM, the force should be proportional to the displacement and directed towards the centre). When the ball hits the top plate, it transfers the positive charge to the plate and gets negative charge q = −CV0 . This negatively charged ball again experience a force F = qE (downward) and starts accelerating downwards. Thus, the ball keeps moving between the bottom and the top plates carrying a charge +q upwards and −q downwards. Ans. C Q 19. The average current in the steady state registered by the ammeter in the circuit will be (A) zero (B) proportional to the potential V0 1/2 (C) proportional to V0 (D) proportional to V02 Sol. The average current when a charge q moves from the bottom plate to the top plate in time T is i = q/T .

386

Part V. Electromagnetism

The ball moves with a uniform acceleration a = F/m = 2CV02 /(mh). The distance travelled in time T is r 1 2CV02 2 h m 1 h = aT 2 = T , i.e., T = . 2 2 mh V0 C

Q 21. Five identical capacitor plates, each of area A, are arranged such that adjacent plates are at a distance d apart, the plates are connected to a source of emf V as shown in the figure. The charge on plate 1 is . . . . . . and on plate 4 is . . . . . . (1984)

The average current carried by each ball is r C C 2 q = V . i= T h m 0 1 2 3 4 5

Ans. D Fill in the Blank Type Q 20. Two parallel plate capacitors of capacitance C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the region between the plates of capacitor C is completely filled with a material of dielectric constant K. The potential difference across the capacitors now becomes . . . . . . (1988)

Sol. Plate numbers 1, 3, and 5 are connected to the positive terminal of the battery of emf V and plate numbers 2 and 4 are connected to the negative terminal of the battery. The equivalent circuit is shown in the figure. 1 2

Sol. Let the two capacitors are connected to the battery as shown in the figure.

3 2 3 4

C

5 4

q1 −

+ q2

V 2C KC q10 q20 2C

The charges on the plates connected to the positive terminal of the battery of potential V are q1 = CV and q2 = 2CV . Let q10 and q20 be the charges on these plates after the battery is disconnected and the dielectric is inserted. The capacitance of the first capacitor becomes KC when a material of dielectric constant K is inserted into it. By charge conservation, total charge on the plates connected to positive terminal remains constant, once the battery is disconnected (isolated system) i.e., q10 + q20 = q1 + q2 = 3CV.

(1)

The potential across the two capacitors after insertion of dielectric is given by V 0 = q10 /(KC), 0

V =

q20 /(2C).

(2) (3)

Eliminate q10 and q20 from equations (1)–(3) to get V 0 = 3V /(K + 2). Ans. 3V /(K + 2)

There are four capacitors, each of capacity C = 0 A/d, connected in parallel. The charge on each capacitor is q = 0 AV /d. Since plate 1 is connected to positive terminal of the battery and is not part of any other capacitor except first, the charge on this plate is q1 = 0 AV /d. The plate 4 is common to two capacitors (formed by the plates 3 and 4 and the plates 5 and 4) each having charge q = 0 AV /d. Thus, charge on this plate is q4 = −20 AV /d (negative because this plate is connected to negative terminal of the battery). We encourage you to find the charges on each side (surface) of all the plates. 20 AV Ans. 0 AV d , − d Integer Type Q 22. Three identical capacitors C1 , C2 and C3 have a capacitance of 1.0 µF each and they are uncharged initially. They are connected in a circuit as shown in the figure and C1 is then filled completely with a dielectric material of relative permittivity r . The cell electromotive force (emf ) is V0 = 8 V. First the switch S1 is closed while the switch S2 is kept open. When the capacitor C3 is fully charged, S1 is opened and S2 is closed simultaneously. When all the capacitors reach equilibrium, the charge on C3 is found to be 5 µF. The value of r = . . . . . . . (2018)

Chapter 28. Capacitors

387 C

S2 V0

C1

S1

C2

Sol. When the switch S1 is closed and the switch S2 is kept open, the capacitor C3 (1 µF) gets fully charged by a battery of emf V0 (8 V). On complete charging, the charge on the capacitor C3 becomes q3 = C3 V0 = (1)(8) = 8 µC (i.e., charge +8 µC on the plate connected to the positive terminal of the battery and −8 µC on the plate connected to the negative terminal of the battery). +3µF

V0

+8µC −8µC

−3µF

C1 +5µF

C3 +3µF −3µF

−5µF

2µF

+ 100V A

C3

3µF

Sol. Initial charges on the capacitor A and B are given by qA0 = CA VA = 3 × 100 = 300 µC, qB0 = CB VB = 2 × 180 = 360 µC. Let qA , qB , and qC be the final charges on the three capacitors. The charge on each plate is as shown in the figure. +qC −qC

C2

C

+qA −qA

A

B

−qB +qB

By charge conservation, total charge on the upper plate of A and the left plate of C is equal to the initial charge on the upper plate of A i.e., qA + qC = qA0 = 300 µC.

(1)

Similarly, total charge on the upper plate of B and the right plate of C is equal to the initial charge on the upper plate of B i.e., qB + qC = qB0 = 360 µC.

(2)

The net potential in the loop is zero. Moving clockwise starting with A, we get qA qC qB qC qB qA − + = 0. (3) − + = 0, i.e., CA CC CB 3 2 2 Solve equations (1)–(3) to get qA = 90 µC,

q1 q2 q3 + = , C10 C2 C3 3 µC 3 µC 5 µC + = . r (1 µF) 1 µF 1 µF

− 180V B

(a) the final charge on the three capacitors. (b) the amount of electrostatic energy stored in the system before and after completion of the circuit.

C3

When the switch S1 is opened and the switch S2 is closed, the positive charge starts flowing from the upper plate of C3 to the upper plate of C1 . Similarly, the negative charge starts flowing from the lower plate of C3 to the lower plate of C2 . An equal but opposite charge is induced on the other plates of C1 and C2 . In equilibrium, charge on the upper plate of C3 is +5 µC and charge on the lower plate of C3 is −5 µC. By conservation of charge, charge on the upper plate of C1 is +3 µC and charge on the lower plate of C2 is −3 µC. The potential difference across a capacitor of capacitance C having a charge q is given by V = q/C. When capacitor C1 is filled with a dielectric of relative permittivity r , its capacitance increases from C1 to C10 = r C1 . The potential difference across C3 is equal to the potential difference across C1 and C2 together i.e.,

2µF

qB = 150 µC, and qC = 210 µC.

The stored electrostatic energy in the initial and final configurations are qA0 2 qB0 2 + 2CA 2CB (300 × 10−6 )2 (360 × 10−6 )2 = + −6 2(3 × 10 ) 2(2 × 10−6 )

Ui =

Solve to get r = 1.5. Ans. 1.5 Descriptive Q 23. Two capacitors A and B with capacities 3 µF and 2 µF are charged to a potential difference of 100 V and 180 V respectively. The plates of the capacitors are connected as shown in the figure with one wire of each capacitor free. The upper plate of A is positive and that of B is negative. An uncharged 2 µF capacitor C with lead wires falls on the free ends to complete the circuit. Calculate, (1997)

= 47.4 × 10−3 J, qA 2 qB 2 qC 2 + + 2CA 2CB 2CC −6 2 (150×10−6 )2 (210×10−6 )2 (90×10 ) = + + 2(3×10−6 ) 2(2×10−6 ) 2(2×10−6 )

Uf =

= 18.0 × 10−3 J. Ans. (a) qA = 90 µC, qB = 150 µC, qC = 210 µC (b) Ui = 47.4 mJ, Uf = 18 mJ

388

Part V. Electromagnetism

Q 24. The capacitance of a parallel plate capacitor with plate area A and separation d, is C. The space between the plates is filled with two wedges of dielectric constant K1 and K2 respectively (see figure). Find the capacitance of the resulting capacitor. (1996)

Sol. Given, side of the square plate l = 1 m, separation between the plates d = 0.01 m, dielectric constant of oil K = 11 and emf of the battery V = 500 V. Let at a time instant t, length x of the plate be in oil and remaining length (l − x) be in air.

A

V K2 d

C1

v

K1

x

C2 d

Sol. Let l be the length and b be the width of the capacitor plate of area A = lb. Consider a small element of length dx and width b at a distance x from the left edge.

This configuration is equivalent to two capacitors of capacitance C1 = 0 l(l − x)/d,

x θ

A B

K1

connected in parallel giving equivalent capacitance

K2

C = C1 + C2 =

d

C

This element consists of two capacitors, AB and BC, connected in series. The plate area for capacitors is bdx, plate separation for AB is d1 = x tan θ = xd/l, and plate separation for BC is d2 = d−d1 = (1 − x)d/l. The capacitances of AB and BC are C1 =

0 K1 b dx , xd/l

and C2 = K0 lx/d

dx

and

C2 =

0 K2 b dx . (l − x)d/l

The capacitor AB and BC are connected in series giving effective capacitance C1 C2 0 lb K1 K2 dx = C1 + C2 d K1 l + (K2 − K1 )x C0 K1 K2 dx = , K1 l + (K2 − K1 )x

0 l [l + (K − 1)x] . d

The charge on the capacitor is q = CV and current through the circuit is   dq dC 0 l(K − 1)v i= =V =V , dt dt d where v = dx/dt is the speed at which the plates are being lowered. Substitute the values to get   (8.85 × 10−12 )(1)(11 − 1)(0.001) i = 500 0.01 = 4.425 × 10−9 A. Ans. 4.425 × 10−9 A

dC =

where C0 = 0 A/d. Integrate dC from x = 0 to x = l to get the capacitance Z l C0 K1 K2 K2 C0 K1 K2 dx C= = ln . K l + (K − K )x K K1 1 2 1 2 − K1 0

Q 26. Two parallel plate capacitors A and B have the same separation d = 8.85 × 10−4 m between the plates. The plate areas of A and B are 0.04 m2 and 0.02 m2 , respectively. A slab of dielectric constant (relative permittivity) K = 9 has dimensions such that it can exactly fill the space between the plates of capacitor B. (1993)

We encourage you to use this result to find C when K2 = K1 . K2 2 Ans. 0dA KK21−K ln K K1 1 Q 25. Two square metal plates of side 1 m are kept 0.01 m apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of emf 500 V. The plates are then lowered vertically into the oil at a speed of 0.001 m/s. Calculate the current drawn from the battery during the process. [Dielectric constant of oil = 11, 0 = 8.85 × 10−12 C2 N−1 m−2 .] (1994)

B

A

110V (i)

(ii)

A

B

(iii)

(a) The dielectric slab is placed inside A as shown in the figure (i). A is then charged to a potential difference of 110 V. Calculate the capacitance of A and the energy stored in it. (b) The battery is disconnected and then the dielectric slab is removed from A. Find the work done by the external agency in removing the slab from A.

Chapter 28. Capacitors (c) The same dielectric slab is now placed inside B, filling it completely. The two capacitors A and B are then connected as shown in the figure (iii). Calculate the energy stored in the system. Sol. Let plate area of the capacitor B is A = 0.02 m2 . The area of dielectric slab is A = 0.02 m2 and the plate area of the capacitor A is 2A = 0.04 m2 . The dielectric slab fills half portion of the capacitor A. The capacitor A with dielectric slab inside it is equivalent to two capacitors of capacitance 0 A/d and K0 A/d connected in parallel. Thus, the equivalent capacitance of A with dielectric slab is (1 + K)0 A 0 A K0 A + = d d d (1 + 9)(8.85 × 10−12 )(0.02) = = 2 × 10−9 F. 8.85 × 10−4

CA =

The charge on capacitor A and electrostatic energy stored in it are qA = CA V = (2 × 10−9 )(110) = 2.2 × 10−7 C, UA = 21 CA V 2 = = 1.21 × 10

1 2 −5

389 Q 27. The figure shows two identical parallel plate capacitors connected to a battery with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric. (1983) S V

A

× (2 × 10−9 )(110)2

S

J. A

and the stored energy after the removal of dielectric is 2

0 qA (2.2 × 10−7 )2 = = 6.05 × 10−5 J. 0 2CA 2(0.4 × 10−9 )

The stored energy increases because the work is done by the external source. By energy conservation, the work done by the external source is W = UA0 − UA = 4.84 × 10−5 J. The capacitance of the capacitor B with dielectric slab is CB = K0 A/d = 1.8 × 10−9 F. The capacitors A and B are connected in parallel giving equivalent capacitance 0 C = CA + CB = 0.4 × 10−9 + 1.8 × 10−9

= 2.2 × 10−9 F. Total charge on the equivalent capacitor is q = qA and the stored energy is q2 (2.2 × 10−7 )2 U= = = 1.1 × 10−5 J. 2C 2(2.2 × 10−9 ) Ans. (a) 2 × 10−9 F, 1.21 × 10−5 J J (c) 1.1 × 10−5 J

C

B

C

S V

0 CA = 0 (2A)/d = 0.4 × 10−9 F,

(b) 4.84 × 10

C

U = UA + UB = CV 2 .

V

−5

B

Sol. Let CA = C and CB = C be the initial capacitances of the capacitors A and B, respectively. Initially, potential difference across both the capacitors is V and electrostatic energies stored in two capacitors are UA = 12 CA V 2 = 12 CV 2 and UB = 12 CB V 2 = 12 CV 2 . Thus, the total energy stored in two capacitors is

The charge on A remains same after the dielectric 0 = qA . The capacitance of A after is removed i.e., qA the removal of dielectric is

UA0 =

C

A

0 CA B

0 CB

When a dielectric of dielectric constant K = 3 is introduced in the two capacitors, their capacitances be0 0 come CA = KCA = 3C and CB = KCB = 3C. The potential difference across capacitor A does not change when the switch is opened. Thus, the electrostatic energy stored in capacitor A is 0 UA0 = 21 CA V 2 = 12 (3C)V 2 = 32 CV 2 .

The charge on capacitor B does not change when the switch is opened i.e., charge remains equal to its initial value of Q = V /CA = V /C. Thus, the electrostatic energy stored in capacitor B is 2

UB0 =

Q2 (V /C)2 1 Q0 = = = CV 2 . 0 2CB 2(3C) 6C 6

Total energy stored in the two capacitors after the introduction of dielectric is U 0 = UA0 + UB0 = 32 CV 2 + 16 CV 2 = 53 CV 2 . Hence, the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric is Ui /Uf = 3/5. Ans. 3/5

Chapter 29 Electric Current in Conductors

One Option Correct

length l is given by Z ~ = j(r, t)(2πrl). ~j(r, t) · dA I=

Q 1. An infinite line charge of uniform electric charge density λ lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity  and electrical conductivity σ. The electrical conduction in the material follows Ohm’s law. Which one of the following graphs best describes the subsequent variation of the magnitude of the current density j(t) at any point in the material? (2016) (A)

j(t)

(B) t

(0,0)

(C)

j(t)

(D)

By conservation of charge, the current I is equal to the rate of decrease of charge q on axial line segment of length l i.e., I=−

j(t)

(0,0)

(3)

surface

dλ(t) dq = −l . dt dt

(4)

From equations (1)–(4) σλ(t) dλ(t) =− . dt 

t

j(t)

Integrate with initial condition λ(t = 0) = λ0 to get σt

t

(0,0)

(0,0)

λ(t) = λ0 exp e−  ,

t

Substitute λ(t) in equation (2) to get

Sol. Let the charge per unit length on the axis be λ(t) at a time t. The electric field due to this line charge at a radial distance r is given by ~ t) = λ(t) rˆ. E(r, 2πr

~j(r, t) = σλ0 e− σt  . 2πr Note that current density varies as 1/r with radial distance r. It decreases exponentially with time and becomes zero as t → ∞. We encourage you to show that equation (2) is equivalent to the popular form of Ohm’s law, V = IR. Ans. C

(1)

R ~ E(r) l

r

Q 2. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Ω, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is (2014)

~j(r)

The free charges inside the conductor start moving radially due to the presence of electric field. The current density at a point is defined as the current flowing across a unit area placed perpendicular to the direction of current flow. The current density at a point is related to the electric field at that point by Ohm’s law i.e., ~ t), ~j(r, t) = σ E(r,

R

(2)

90Ω

40cm

where σ is the electrical conductivity of the conductor. The current through a cylindrical shell of radius r and

(A) 60 ± 0.15 Ω (C) 60 ± 0.25 Ω 390

(B) 135 ± 0.56 Ω (D) 135 ± 0.23 Ω

Chapter 29. Electric Current in Conductors

391

Sol. Let λ be the resistance per unit length (in Ω/cm) of the potentiometer wire. Total length of the wire is 100 cm and null point is obtained at x = 40 cm. The resistances of four branches of Wheatstone bridge are, R1 = R, R2 = 90 Ω, R3 = λx, and R4 = λ(100 − x).

resistance per unit length (in Ω/cm). Since Wheatstone bridge is balanced at null condition, X P = , 10 Q

=⇒

X=

10P 10(53λ) = = 10.6 Ω. Q 50λ Ans. B

R2

R1

Q 4. Consider a thin square sheet of side L and thickness t, made of material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is (2010)

↑ G R4

R3

V

Wheatstone bridge is balanced if R1 R3 = , R2 R4

x R = . 90 100 − x

or

t

(1) L

Solve equation (1) to get R = 60 Ω. The least count of scale gives error in measurement of x, i.e., ∆x = 0.1 cm. To find error in measurement of R, differentiate equation (1) and simplify to get ∆x ∆x ∆R = + . R x 100 − x

(2)

Substitute the values in equation (2) and solve to get ∆R = 0.25 Ω. Ans. C Q 3. A metre bridge is set-up as shown, to determine an unknown resistance X using a standard 10 Ω resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of X is (2011) X

10 Ω

N

A

B

(A) 10.2 Ω (B) 10.6 Ω (C) 10.8 Ω (D) 11.1 Ω Sol. Let N be the null point on the wire. Given AN = 52 cm and NB = 100 − 52 = 48 cm. X A

10Ω G

B Q

P N

Let A0 and B 0 represent the two end points with end corrections i.e., A0 N = 52 + 1 = 53 cm and NB0 = 48 + 2 = 50 cm. The resistance of branch A0 N is P = 53λ and that of branch N B 0 is Q = 50λ, where λ is the

(A) (B) (C) (D)

directly proportional to L. directly proportional to t. independent of L. independent of t.

Sol. The sheet is of square shape with thickness t, width w = L, and length l = L. The resistance between the two opposite faces is given by R=

ρl ρl ρL ρ = = = . A wt Lt t Ans. C

Q 5. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R100 , R60 and R40 , respectively, the relation between these resistances is (2010) 1 (A) R100 = R140 + R160 (B) R100 = R40 + R60 1 (C) R100 > R60 > R40 (D) R100 > R160 > R140 Sol. The power of a bulb having resistance R and operating at voltage V is given by P = V 2 /R. Let the three bulbs operate at temperatures T100 , T60 and T40 above room temperature. The resistances and powers of three bulbs at operating temperatures are given by 0 R100 = R100 (1 + αT100 ),

0 V 2 /R100 = 100,

0 R60 = R60 (1 + αT60 ),

0 V 2 /R60 = 60,

0 R40 = R40 (1 + αT40 ),

0 V 2 /R40 = 40,

where α is the thermal coefficient of resistance. Elimi0 0 0 nate R100 , R60 , and R40 to get 1 100 = 2 (1 + αT100 ), R100 V 1 60 = 2 (1 + αT60 ), R60 V 1 40 = 2 (1 + αT40 ). R40 V

392

Part V. Electromagnetism

We expect the higher power bulb to have a higher temperature i.e., T100 > T60 > T40 . This gives us, 1 1 1 1 1 1 R100 > R60 > R40 and R100 6= R60 + R40 . We encourage you to measure the resistance of a bulb (by using a multimeter etc.) at room temperature, calculate resistance at operating temperature, and get some estimates of α and operating temperature. Ans. D

1Ω 1Ω 1Ω

3V

1Ω

1Ω

1Ω

1Ω 1Ω

1Ω

1Ω

(R1 )

Q 6. To verify Ohm’s law, a student is provided with a test resister RT , a high resistance R1 , and a small resistance R2 , two identical galvanometers G1 and G2 , and a variable voltage source V . The correct circuit to carry out the experiment is (2010)

3V

(R2 )

1Ω

1Ω

1Ω

1Ω

3V

G1

(A) R2

1Ω G2

RT

R1 (R3 )

V

(A) P1 > P2 > P3 (C) P2 > P1 > P3

G1

(B) R1 G2

RT

R2

V

(C)

G1

R1 RT

Sol. In configuration (R1 ), the resistances form a balanced Wheatstone bridge. Effective resistance of this configuration is R1 = 2Ω k 2Ω = 1 Ω. The effective resistance of the configuration (R2 ) is R2 = (2Ω k 2Ω) k 1Ω = 1Ω k 1Ω = 0.5 Ω. The effective resistance of the configuration (R3 ) is R3 = 2 Ω. Thus, the powers dissipated in the three configurations are

G2

P1 = V 2 /R1 = 9 W,

R2

P2 = V 2 /R2 = 18 W,

G1

and

2

P3 = V /R3 = 4.5 W.

V

(D)

(B) P1 > P3 > P2 (D) P3 > P2 > P1

R2

Ans. C G2

RT

R1 V

Sol. To verify Ohm’s law, we need to measure the voltage across the test resistance RT and current passing through it. The voltage can be measured by connecting a high resistance R1 in series with galvanometer. This combination becomes a voltmeter and should be connected in parallel to RT . The current can be measured by connecting a low resistance R2 (shunt) in parallel with galvanometer. This combination becomes an ammeter and should be connected in series to measure the current through RT . Ans. C Q 7. Figure shows three resistor configurations (R1 ), (R2 ), and (R3 ) each connected to 3 V battery. If the power dissipated by the configuration (R1 ), (R2 ) and (R3 ) is P1 , P2 and P3 , respectively, then (2008)

Q 8. A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K = 2. The level of liquid is d/3 initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time t is (2008) C d

(A) (C)

60 R 5d+3V t 60 R 5d−3V t

(B) (D)

d 3

R

(15d+9V t)0 R 2d2 −3dV t−9V 2 t2 (15d−9V t)0 R 2d2 +3dV t−9V 2 t2

Sol. At a time t, thicknesses of the upper part (without dielectric) and the lower part (with dielectric) are given by du = 2d/3 + V t,

dl = d/3 − V t.

Chapter 29. Electric Current in Conductors

393

The capacitances of the upper and the lower part at time t are

3µF X 6µF S

0 A 0 = , du 2d/3 + V t K0 A 20 Cl = = . dl d/3 − V t

Cu =

3Ω

6Ω

Y 9V

Capacitors Cu and Cl are connected in series and hence the effective capacitance and time constant are given by

Sol. Let q1 and V1 be the charge and voltage for C1 = 3 µF capacitor and q2 and V2 be the charge and voltage for C2 = 6 µF capacitor. When switch S is open, two capacitors are connected in series and hence charge on the two is equal

60 Cu Cl = , Ceff = Cu + Cl 5d + 3V t 60 R τ = RCeff = . 5d + 3V t Ans. A Q 9. A resistance of 2 Ω is connected across one gap of a meter-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2 Ω, is connected across the other gap. When the resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is (2007) (A) 3 Ω (B) 4 Ω (C) 5 Ω (D) 6 Ω Sol. Let X be the unknown resistance. The null point is obtained when Wheatstone bridge is balanced i.e., ρl1 /A l1 X 2 = ρl2 /A = l2 , where l1 and l2 are as shown in the figure.

X

2Ω

l1

l2

2Ω

(A) 0 (B) 54 µC (C) 27 µC (D) 81 µC

X

C1 V1 = C2 V2 .

(1)

Also, sum of voltages across the two capacitors is equal to the battery voltage V1 + V2 = 9 V.

(2)

Solve equations (1) and (2) to get V1 = 6 V, V2 = 3 V, q1 = 18 µC, and q2 = 18 µC. Total charge on the plates connected to X is −q1 + q2 = 0 because they form an isolated circuit. When the switch is closed, the voltage V10 across C1 is equal to the voltage across R1 = 3 Ω and voltage V20 across C2 is equal to the voltage across R2 = 6 Ω. Effective resistance of the circuit is R1 +R2 = 9 Ω which gives a current I = 1 A flowing through R1 and R2 . Thus, the voltages and charges on the capacitors are V10 = IR1 = 3 V,

V20 = IR2 = 6 V,

q10 = C1 V10 = 9 µC,

q20 = C2 V20 = 36 µC.

Total charge on plates connected to X is −q10 + q20 = −9 + 36 = 27 µC, which is flown from Y to X. Ans. C Q 11. Find the time constant for the given R-C circuits in correct order (in µs), [Given R1 = 1 Ω, R2 = 2 Ω, C1 = 4 µF, C2 = 2 µF.] (2006)

l10

l20

V R1

V V

C2

R1

C1

R2

C2

Since X > 2 Ω, we get l1 > l2 . Also, l1 + l2 = 100 cm. When the resistances are interchanged, the null point shifts by 20 cm. As X > 2 Ω, the null point will shift towards the left i.e., l10 = l1 − 20 and l20 = l2 + 20. l0 2 The balance condition gives X = l10 = ll12 −20 +20 . Solve to 2 get X = 3 Ω. Ans. A

(A) 18, 4, 8/9 (C) 4, 18, 8/9

Q 10. A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is (2007)

Sol. The time constant of a R-C circuit is given by τ = Re Ce , where Re is the equivalent resistance and Ce is the equivalent capacitance. The equivalent resistance,

C1 R2

C1

R1 R2

C2

(B) 18, 8/9, 4 (D) 4, 8/9, 18

394

Part V. Electromagnetism

equivalent capacitance, and time constants for the given circuits are:

Q 13. Find out the value of current through 2 Ω resistance for the given circuit. (2005)

1 : Re = R1 + R2 = 3 Ω, 10V

Ce = C1 + C2 = 6 µF,

10Ω

5Ω

20V

2Ω

τ1 = 18 µs. R1 R2 2 2 : Re = = Ω, R1 + R2 3 C1 C2 8 Ce = = µF, C1 + C2 6 8 τ2 = µs. 9 3 : Re = 2/3 Ω,

(A) 5 A (B) 2 A (C) zero (D) 4 A Sol. Kirchhoff’s junction law prohibits current flow through 2 Ω resistance. Ans. C

Ce = 6 µF, τ3 = 4 µs. Ans. B Q 12. A 4 µF capacitor and a 2.5 MΩ resistance are in series with 12 V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given ln 2 = 0.693] (2005) (A) 13.86 s (B) 6.93 s (C) 7 s (D) 14 s

Q 14. A moving coil galvanometer of resistance 100 Ω is used as an ammeter using a resistance 0.1 Ω. The maximum deflection current in the galvanometer is 100 µA. Find the current in the circuit, so that the ammeter shows maximum deflection. (2005) (A) 100.1 mA (B) 1000.1 mA (C) 10.01 mA (D) 1.01 mA Sol. A galvanometer of resistance G is converted to an ammeter by connecting a small shunt resistance S in parallel. i A

Sol. The circuit for charging the capacitor is shown in the figure.

ig

G

B i

i − ig S

4µF

2.5MΩ

Kirchhoff’s loop law gives ig G − (i − ig )S = 0,

12V

Here, V = 12 V is the battery voltage, R = 2.5 MΩ is the series resistance, and C = 4 µF is the capacitance. The potential across the capacitor at time t is h i t VC = V 1 − e− RC , (1)

=⇒

i = ig (G + S)/S.

The maximum deflection current of galvanometer sets upper limit on the current measured by this ammeter. Substitute the values to get i = ig (G + S)/S = (100 × 10−6 ) ((100 + 0.1)/0.1)) = 100.1 mA.

and the potential across the resistor is

Ans. A

VR = V − VC .

(2)

Using equations (1) and (2), and the condition VC = 3VR , we get

Q 15. Six equal resistances are connected between points P, Q and R as shown in the figure. Then, the net resistance will be maximum between (2004) P

t

e− RC = 1/4. Take logarithm on both sides and simplify to get t = 2RC ln 2 = 2 (2.5 × 106 ) (4 × 10−6 ) (0.693)

Q

R

= 13.86 s. Ans. A

(A) P and Q (C) P and R

(B) Q and R (D) any two points

Chapter 29. Electric Current in Conductors

395

Sol. The equivalent resistances between P and Q, between Q and R, and between P and R are given by

Sol. The unknown resistance X is connected between the points A and D.

RPQ = R k (R/2 + R/3) = R k 5R/6

B

R(5R/6) 5 = = R, R + 5R/6 11 RQR RPR

D

X V A

4 R, = (R/2) k (R + R/3) = 11 3 = (R/3) k (R + R/2) = R. 11

C1

B1 G

Ans. A Q 16. A capacitor is charged using an external battery with a resistance x in series. The dashed line shows the variation of ln I with respect to time. If the resistance is changed to 2x, the new graph will be (2004)

Let the resistance of branch BC be P , resistance of branch CD be Q, and resistance of rheostat branch AB be R. Wheatstone bridge is balanced when P/Q = R/X which gives X = QR/P . Ans. C Q 18. Which of the following set-up can be used to verify Ohm’s law? (2003) A (B) (A)

S R

ln(I)

C

A

V

V

Q P t

A

(C)

V

(D)

(A) P (B) Q (C) R (D) S Sol. The charge q on a capacitor of capacitance C while it is being charged by a battery of voltage V and connected in series to a resistor R is given by i h t q = CV 1 − e− RC . Differentiate to get the current t dq V = e− RC . dt R Take logarithm on both sides to get     V 1 ln I = ln − t. R RC

Q 17. For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between (2004) D

(A) B and C (C) A and D

R1

R2 G

A x

B

C

(A) x (B) x/4 (C) 4x (D) 2x Sol. If the radius of the wire AB is doubled then its resistance (R = ρl/A) becomes one fourth and current (i = V /R) becomes four times. However, the potential drop per unit length, V /l, remains the same. Ans. A Q 20. The three resistances of equal value are arranged in the different combination shown below. Arrange them in increasing order of power dissipation. (2003)

A B1

Sol. The verification of Ohm’s law (V = IR) requires the measurements of current through and voltage across the variable resistance. Ans. B

(1)

The equation (1) represents a straight line between t
and ln I with an intercept ln VR on ln I axis and slope 1 − RC . When R is changed from x to 2x, the intercept

V decreases from ln Vx to ln 2x and slope increases 1 1 to − 2xC . from − xC Ans. B

C

A

Q 19. In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of galvanometer is x, what should be its value if the radius of the wire AB is doubled? (2003)

I=

B

V

C1

(B) C and D (D) B1 and C1

i i (I)

(II)

396

Part V. Electromagnetism i

Thus, effective resistance of the circuit is Re = 2Rr (4R k 4R) k 2r = R+r . We encourage you to exploit the symmetry to check that i3 = 0. Ans. A

i

(III)

(IV)

(A) III < II < IV < I (C) I < IV < III < II

(B) II < III < IV < I (D) I < III < II < IV

Sol. The equivalent resistances and power dissipation of the given combination are given by RI = R + R + R = 3R,

PI = i2 RI = 3 i2 R,

RII = (R + R) k R = 23 R,

PII = i2 RII =

RIII = RIV =

R k R k R = 13 R, (R k R) + R = 32 R,

PIII = PIV =

B1

2 2 3 i R, 2 i RIII = 13 i2 R, i2 RIV = 32 i2 R.

Q 21. The effective resistance between points P and Q of the electrical circuit shown in the figure is (2002) 2R

250V

R1 = V 2 /100,

r

r

(A)

2Rr R+r

(B)

8R(R+r) 3R+r

2R

(C) 2r + 4R (D)

5R 2

+ 2r

Sol. The given circuit is a good example of symmetry. When looking from P, branch PAB looks same as the branch PFE and hence the same current i1 should flow through these branches. 2R

A

B i3

i1 i2

r

i1 E

2R

2

W2 = i2 R2 ≈ 23 (250/V ) ,

It is interesting to note that B1 (100 W) is dimmer than B2 (60 W) which in turn is dimmer than B3 (60 W). Ans. D

2R 2R

2

W1 = i2 R1 ≈ 14 (250/V ) ,

2

Q

i3

Substitute the values of i, R1 , and R2 to get the output powers

W3 = (250)2 /R3 = 60 (250/V ) .

r O

R3 = V 2 /60.

and C

2R

P

F

2R

R2 = V 2 /60,

In the given configuration, the current through B1 and B2 is     250 100 × 60 250 75 250 = 2 = 2 . i= R1 + R2 V 100 + 60 V 2

Q

2R 2R

(B) W1 > W2 > W3 (D) W1 < W2 < W3

Sol. Let the given power ratings be defined at an operating voltage V . The resistances of the three bulbs are given by

2R P

B2 B3

(A) W1 > W2 = W3 (C) W1 < W2 = W3

Ans. A

2R

Q 22. A 100 W bulb B1 and two 60 W bulbs B2 and B3 , are connected to a 250 V source, as shown in the figure. Now W1 , W2 and W3 are the output powers of the bulb B1 , B2 and B3 respectively. Then, (2002)

D

Similarly, when looking from O, branches OB and OE look identical and same current i3 should flow through them. Branches through P and Q are also identical. Apply Kirchhoff’s law in loop ABOPA and BCQOB to get

Q 23. In the given circuit, it is observed that the current I is independent of the value of the resistance R6 . Then the resistance values must satisfy (2001) R5 I

R1

R3 R6

2Ri1 + 2Ri3 − i2 r = 0,

(1)

2R(i1 − i3 ) − r(i2 + 2i3 ) − 2Ri3 = 0.

(2)

Solve equations (1) and (2) to get i3 = 0 i.e., no current flows in the branch OB and OE. Hence, branch OB and OE can be removed from the circuit without affecting the effective resistance.

R2

(A) R1 R2 R5 = R3 R4 R6 1 (B) R15 + R16 = R1 +R + 2 (C) R1 R4 = R2 R3 (D) R1 R3 = R2 R4

1 R3 +R4

R4

Chapter 29. Electric Current in Conductors

397

Sol. Effective resistance of the circuit depends on R6 unless Wheatstone bridge formed by R1 , R2 , R3 , R4 and R6 is balanced. Thus, the current becomes independent of R6 only when the bridge is balanced i.e., R1 /R2 = R3 /R4 . It might be tedious but worth to find the current I in terms of R1 , R2 , R3 , R4 , R5 and R6 and explicitly show the independence of I in the balanced condition. Hint: Use Kirchhoff’s law in the three loops and solve for I. Ans. C Q 24. In the given circuit, with steady current, the potential difference across the capacitor must be (2001) V

R

V

C

2V

2R

Q

P S

↑ G

R

V

(A) IR = IG (B) IP = IG (C) IQ = IG (D) IQ = IR Sol. Let G be the resistance of the galvanometer. Consider the case when switch S is open. In this case, resistances R and G are connected in series giving IR = IG . Also, the resistances P and Q are connected in series giving IP = IQ . B IP A

C

R

(A) V (B) V /2 (C) V /3 (D) 2V /3

D

Sol. In the steady state, a capacitor acts like an open circuit element (see figure).

V

2V

A B i

IG

↑ G

Kirchhoff’s law in the loop ABCA and the loop ADCA (through battery) gives

R

V

Q

S IR

V

IQ

P

2R

IP = IQ = V /(P + Q),

(1)

IR = IG = V /(R + G).

(2)

Now consider the case when the switch S is closed. 0 0 Let the currents in P , Q, R, and G be IP0 , IQ , IR and 0 IG . The current through the galvanometer does not change when S is closed i.e.,

Apply Kirchhoff’s loop law to get 2V − 2iR − iR − V = 0,

0 IG = IG ,

which gives the current i = V /(3R). Traversing from B to A through the upper branch gives the potential of A w.r.t. potential of B as VA = VB − iR − V + V = VB − V /3. Ans. C Q 25. A quantity X is given by 0 L ∆V ∆t , where 0 is the permittivity of free space, L is a length, ∆V is a potential difference and ∆t is a time interval. The dimensional formula for X is the same as that of (2001) (A) resistance (B) charge (C) voltage (D) current Sol. The formula for potential, V = 4πq 0 r , gives the     dimensions of 0 L Vt = qt , which are same as the dimensions of current. Ans. D Q 26. In the circuit shown P 6= R, the reading of galvanometer is same with switch S open or closed. Then, (1999)

(3)

and Kirchhoff’s law applied to the loop ADCA gives 0 0 IR R + IG G = V.

(4)

0 0 The equations (2)–(4) give IR = IG . Thus, no current flows through the branch BD and Wheatstone bridge is balanced i.e.,

P/Q = R/G.

(5)

The equations (1), (2), and (5) give IP = IG (R/P ), IQ = IG (R/P ), and IQ = IR (R/P ). Since P 6= R, we get IP 6= IG , IQ 6= IG , and IQ 6= IR . Ans. A Q 27. In the circuit shown in the figure, the current through (1998) 3Ω 9V

2Ω

8Ω 2Ω

2Ω 8Ω

2Ω

2Ω

4Ω

398

(A) (B) (C) (D)

Part V. Electromagnetism

the the the the

3Ω 3Ω 4Ω 4Ω

resistor resistor resistor resistor

is is is is

drift velocity vd is given by i = neAvd . Thus, both E and vd are inversely proportional to A. Ans. D

0.50 A 0.25 A 0.50 A 0.25 A

Sol. Let R1 and R2 be the effective resistances of the circuits to the right of CD and AB, respectively (see figure). These resistances are given by R1 = 2 + 4 + 2 = 8 Ω and R2 = 2 + (8 k R1 ) + 2 = 2 + (8 k 8) + 2 = 8 Ω. 3Ω i 9V

A 2Ω i1

8Ω 2Ω

B

2Ω

D

R

R

R

E, 4Ω

C 2Ω i2 8Ω

Q 30. A battery of internal resistance 4 Ω is connected to the network resistances as shown in the figure. In order that the maximum power can be delivered to the network, the value of R (in Ω) should be (1995)

6R R

R

4R

4Ω

(A) 4/9 (B) 2 (C) 8/3 (D) 18 2Ω

Sol. Given circuit forms a balanced Wheatstone bridge. Effective resistance of the complete circuit is R = 3+(8 k R2 )+2 = 3+(8 k 8)+2 = 9 Ω. Thus the current in 3 Ω resistor is i = V /R = 9/9 = 1 A. Current i is equally divided at node A giving i1 = 0.5 A. Current i1 is further divided into two equal parts at node C giving i2 = 0.25 A, the current through 4 Ω resistor. Ans. D Q 28. A parallel combination of 0.1 MΩ resistor and a 10 µF capacitor is connected across a 1.5 V source of negligible resistance. The time required for the capacitor to get charged upto 0.75 V is approximately (in second) (1997) (A) infinite (B) loge 2 (C) log10 2 (D) zero Sol. Since the capacitor plates are directly connected to a battery of zero internal resistance, the time required for the capacitor to get charged upto 0.75 V (or to any potential from 0 V to 1.5 V) is approximately zero. 0.1MΩ

C R

2R

A

6R 2R

B 4R

D E, 4Ω

Thus, the resistance 6R in branch CD can be removed. Equivalent resistance between the nodes A and B is Re = (R + 2R) k (2R + 4R) = 3R k 6R = 2R. The E current through the circuit is i = ReE+r = 2R+4 and the power delivered to the network is P = i 2 Re =

E2 R . 2 (R + 2)2

The power attains its maxima when E 2 (2 − R) dP = = 0, dR 2 (R + 2)3 which gives R = 2 Ω. Note that the power delivered by the battery is maximum when the load resistance (Re ) is equal to the internal resistance of the battery. Ans. B

10µF

1.5V

Ans. D Q 29. A steady current flows in a metallic conductor of non-uniform cross-section. The quantity/quantities constant along the length of the conductor is (are)

Q 31. The current i in the circuit (see figure) is

(1983)

i 2V

30Ω

30Ω

(1997)

(A) (B) (C) (D)

current, electric field and drift speed drift speed only current and drift speed current only

Sol. In the steady state, charge cannot accumulate in a conductor. Thus, a constant current i = dq/dt flows along the length of conductor. The current density at a point having cross-section area A is given by J = i/A. The current density is related to electric field by J = σE, where conductivity σ is a material property. The

30Ω

(A) 1/45 A (B) 1/15 A (C) 1/10 A (D) 1/5 A Sol. The effective resistance of the given circuit is Re = (30 + 30) k 30 = 60 k 30 30 × 60 = = 20 Ω. 30 + 60 Thus, current i = V /Re = 2/20 = 1/10 A. Ans. C

Chapter 29. Electric Current in Conductors One or More Option(s) Correct Q 32. Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance of the galvanometers Rc < R/2, which of the following statement(s) about any one of the galvanometers is(are) true? (2016) (A) The maximum voltage range is obtained when all the components are connected in series. (B) The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer. (C) The maximum current range is obtained when all the components are connected in parallel. (D) The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors. Sol. Let ig be the current that gives full deflection of the galvanometer. When all components are connected in series (see figure), effective resistance of the circuit is Re = 2R + 2Rc and maximum current allowed in the circuit is ig . A

R

Rc

R ig

Rc

399 Let i and ig be the currents through the resistors and the galvanometers. By Kirchhoff’s law, iR = ig Rc , which gives i = ig Rc /R. The current between A and B is IAB = 2i + 2ig = 2ig (1 + Rc /R). Consider the case when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors.

A

i

R

i

R

0 IAB

ig

Rc

Rc

B

0 IAB = ig + 2i = ig + 2ig (2Rc /R)

= 2ig (1 + Rc /R) − (1 − 2Rc /R)ig = IAB − (1 − 2Rc /R)ig < IAB

(∵ Rc < R/2). Ans. B, C

B

Thus, the voltage between A and B is VAB = ig Re = 2ig (R + Rc ).

Q 33. In the circuit shown in the figure, the key is pressed at time t = 0. Which of the following statement(s) is(are) true? (2016) 40µF

Consider the case when two resistors and one galvanometer are connected in series and the second galvanometer is connected in parallel.

− V + 50kΩ

25kΩ

20µF

A A

R

R

ig R c

Key

B

ig R c

The maximum current through each galvanometer is ig and the maximum current through the resistors is 2ig . Apply Kirchhoff’s law to get the voltage between A and B as 0 VAB = 2ig R + 2ig R + ig Rc

= 2ig (R + Rc ) + 2ig (R − Rc /2) = VAB + 2ig (R − Rc /2) > VAB

(∵ Rc < R/2).

Consider the case when all four components are connected in parallel.

A

IAB

i

R

i

R

ig

Rc

ig

Rc

5V

(A) The voltmeter displays −5 V as soon as the key is pressed, and displays +5 V after a long time. (B) The voltmeter will display 0 V at time t = ln 2 seconds. (C) The current in the ammeter becomes 1/e of the initial value after 1 second. (D) The current in the ammeter becomes zero after a long time. Sol. The impedance (effective resistance) of the capacitors is zero immediately after the key is pressed (t → 0+ ). Thus, capacitors will act as short circuit elements (see figure). Q

B

50kΩ A

− V +

25kΩ

P 5V

400

Part V. Electromagnetism

Thus, potential at the node Q is VQ = 5 V and potential at the node P is VP = 0 V. The voltmeter will display V = VP − VQ = 0 − 5 = −5 V. The impedance of the capacitors is infinite in steady state (t → ∞) and they act as open circuit elements (see figure).

Q 34. In an aluminium (Al) bar of square cross-section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 2.7 × 10−8 Ω m and 1.0 × 10−7 Ω m, respectively. The electrical resistance between the two faces P and Q of the composite bar is (2015)

Q 25kΩ

Q Al

50kΩ

Fe

P

A

50 m m

− V +

2mm P

5V

7mm

We assume voltmeter to be ideal with infinite resistance. Thus, no current flows through the circuit and hence ammeter will read zero. Also, potential at the node Q is VQ = 0 V and potential at the node P is VP = 5 V. The voltmeter will display V = VP − VQ = 5 − 0 = 5 V. The capacitor C1 = 40 µF is connected across a V = 5 V battery by a series resistance R1 = 25 kΩ. Similarly, the capacitor C2 = 20 µF is connected across the same battery by a series resistance R2 = 50 kΩ. We can remove voltmeter from the circuit for current calculation as it has infinite resistance. This circuit is used to charge both the capacitors. 40µF

5V

i2 = (V /R2 )e

− R tC 2 2

= 125e−t µA.

1875 49

µΩ (D)

2475 132

µΩ

(1 × 10−7 ) (5 × 10−2 ) ρFe lFe = = 1250 µΩ. AFe 4 × 10−6

ρAl lAl (2.7 × 10−8 ) (5 × 10−2 ) = 30 µΩ. = AAl 4.5 × 10−5

Req = RFe k RAl =

The currents through the capacitors at time t (charging of the capacitor) are given by t

µΩ (C)

The iron and aluminium bars are connected in parallel between P and Q. Thus, the equivalent resistance of the composite bar is

20µF

i1 = (V /R1 )e− R1 C1 = 200e−t µA,

1875 64

For the aluminium bar, AAl = (7 × 10−3 )2 − (2 × 10−3 )2 = 4.5 × 10−5 m2 , lAl = 5 × 10−2 m, and ρAl = 2.7 × 10−8 Ω m. The resistance of the aluminium bar is RAl =

P 50kΩ

µΩ (B)

RFe =

Q

i2

2475 64

Sol. The resistance of a conductor of length l, crosssectional area A, and material resistivity ρ is given by R = ρl/A. For iron bar, length is lFe = 5 × 10−2 m, cross-sectional area is AFe = (2×10−3 )2 = 4 × 10−6 m2 , and resistivity is ρFe = 1.0 × 10−7 Ω m. The resistance of the iron bar is

25kΩ

i1 A

(A)

=

RFe RAl (1250)(30) = RFe + RAl 1250 + 30

1875 µΩ. 64

(1) (2)

From equations (1) and (2), the current through the ammeter at time t = 0 s is i0 = i1 + i2 = 325 µA. The current through the ammeter at time t = 1 s is i = i1 + i2 = 200e−1 + 125e−1 = 325/e = i0 /e. The potentials at points P and Q at a time t are given by VP = V − i2 R2 = 5(1 − e−t ),

(3)

VQ = 0 + i2 R1 = 5e−t .

(4)

From equations (3) and (4), the voltmeter at time t = ln 2 will display V = Vp − VQ = 5 − 10e− ln 2 = 0 V. Ans. A, B, C, D

Ans. B Q 35. Two ideal batteries of emf V1 and V2 and three resistances R1 , R2 and R3 are connected as shown in the figure. The current in resistance R2 would be zero if (2014) V1

R1 R2 V2 R3

(A) (B) (C) (D)

V1 = V2 and R1 = R2 = R3 V1 = V2 and R1 = 2R2 = R3 V1 = 2V2 and 2R1 = 2R2 = R3 2V1 = V2 and 2R1 = R2 = R3

Chapter 29. Electric Current in Conductors

401

Sol. Let i1 and i2 be the currents as shown in the figure. i1

A

B

V1

R1 R2

D

We claim, by symmetry, that i = i0 because circuit looks symmetrical when we see it from either node A or B. This can be shown by applying Kirchhoff’s loop law in loops APQA and SBTS which gives

C

i1 −i2

V2 F

i1 R1 + (i1 − i2 )R2 = V1 ,

(1)

i2 R3 − (i1 − i2 )R2 = V2 .

(2)

Multiply equation (1) by R3 and (2) by R1 and then subtract to get the current through R2 as V 1 R3 − V 2 R1 (i1 − i2 ) = . R1 R3 + R2 R3 − R1 R2 The current through R2 becomes zero when V1 R3 = V 2 R1 . Ans. A, B, D Q 36. For the resistance network shown in the figure, choose the correct option(s), (2012) S

2Ω

2Ω

2Ω 1Ω

i = i0 = (4i1 + i2 )/7.

Again, by using symmetry, currents in branch PQ and ST are zero. This can be shown by applying Kirchhoff’s loop law in PSTQP which gives 2i2 + (i2 − i0 ) − 4(i1 − i2 ) − (i − i2 ) = 0.

(4)

Substitute i and i0 from equation (3) into equation (4) and simplify to get i1 = 23 i2 , and i = i2 . Now, apply Kirchhoff’s loop law in APSBA (through battery) to get, 6i2 = 12. Solve to get, i2 = i = i0 = 2 A and i1 = 3 A. Taking the path QAPS, the potential at S is VS = VQ + 4(i1 − i) − 2i − 2i2 = VQ − 4 V. It may be noted that the problem can be solved by using Kirchhoff’s laws only without utilizing symmetry arguments. The readers are encouraged to put the conditions which break the symmetry and solve the problem. Ans. A, B, C, D Q 37. For the circuit shown in the figure,

I

4Ω 4Ω

Q

(3)

(2009)

1Ω

4Ω i1

(2)

Solve equations (1) and (2) to get

Apply Kirchhoff’s law in the loop ABCDA and CEFDC to get

P i2

(1)

2i0 − 4(i1 − i0 ) − (i2 − i0 ) = 0.

E

i2

R3

2i + (i − i2 ) − 4(i1 − i) = 0,

2 kΩ R1

24 V

T

6 kΩ R2 RL 1.5 kΩ 12V

(A) (B) (C) (D)

The current through PQ is zero. i1 = 3 A The potential at S is less than that at Q. i2 = 2 A

Sol. Let i be the current leaving node A through branch AP and i0 be the current entering node B through branch SB. Kirchhoff’s junction law gives the currents in other branches. P 2Ω

i2 2Ω

S

i2 −i0

i−i2

i1 −i

i1 4Ω

4Ω Q i1 −i2 12V

T

Sol. The effective resistance of R2 and RL combination is

= 6 × 1.5/(6 + 1.5) = 1.2 kΩ.

2Ω

1Ω

1Ω

the current I through the battery is 7.5 mA. the potential difference across RL is 18 V. ratio of powers dissipated in R1 and R2 is 3. if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9.

RR2 kRL = R2 RL /(R2 + RL ) i0

i A

(A) (B) (C) (D)

B 4Ω i1 −i0

The resistance RR2 kRL is connected in series with R1 . Thus, the total effective resistance of the circuit is Reff = R1 + RR2 kRL = 2 + 1.2 = 3.2 kΩ. The current through the battery is I = V /Reff = 24/3.2 = 7.5 mA. The potentials across R1 , R2 , and

402

Part V. Electromagnetism

RL and the corresponding power dissipations are VR1 = IR1 = 7.5 × 2 = 15 V, PR1 = VR21 /R1 = 152 /(2 × 103 ), VR2 = V − VR1 = 24 − 15 = 9 V, PR2 = VR22 /R2 = 92 /(6 × 103 ),

Vmax ≈ 10 V. The microammeter can be used to measure higher current from A to B by connecting a parallel resistance S. Maximum current
from A to B that can be measured is imax = ig 1 + G S . For S = 1 Ω,   100 ≈ 5×10−3 A = 5 mA. imax = 50×10−6 1 + 1

VRL = VR2 = 9 V,

Ans. B, C

PRL = VR2L /RL = 92 /(1.5 × 103 ). Thus, the ratio, PR1 /PR2 = 25/3. When R1 and R2 are interchanged, the effective resistance becomes 0 Reff = R2 + RR1 kRL

= 6 + (2 × 1.5)/(2 + 1.5) = 48/7 kΩ. The current through the battery is I 0 = 24/(48/7) = 3.5 mA. The potentials across R1 , R2 , and RL and the corresponding power dissipations are VR0 2 = I 0 R2 = 3.5 × 6 = 21 V, 2

PR0 2 = VR0 2 /R2 = 212 /(6 × 103 ),

Q 39. Capacitor C1 of capacitance 1 µF and capacitor C2 of capacitance 2 µF are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t = 0. (1989) (A) The current in each of the two discharging circuits is zero at t = 0. (B) The currents in the two discharging circuits at t = 0 are equal but not zero. (C) The currents in the two discharging circuits at t = 0 are unequal. (D) Capacitor C1 loses 50% of its initial charge sooner than C2 loses 50% of its initial charge. Sol. Let C1 = 1 µF, C2 = 2 µF, and V be the emf of the battery. Initial charges on the two capacitors are q0,1 = C1 V and q0,2 = C2 V .

VR0 1 = V − VR0 2 = 24 − 21 = 3 V, 2

PR0 1 = VR0 1 /R1 = 32 /(2 × 103 ), VR0 L = VR0 1 = 3 V, PR0 L =

2 VR0 L/RL

C q

= 32 /(1.5 × 103 ).

i

Thus, the ratio, PR0 L /PRL = 32 /92 = 1/9. Ans. A, D Q 38. A microammeter has a resistance of 100 Ω and full scale range of 50 µA. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combination(s), (1991) (A) 50 V range with 10 kΩ resistance in series. (B) 10 V range with 200 kΩ resistance in series. (C) 5 mA range with 1 Ω resistance in parallel. (D) 10 mA range with 1 Ω resistance in parallel. Sol. The microammeter has a resistance G = 100 Ω and the full scale range ig = 50 µA. It can be used to measure the voltage across A and B by connecting a series resistance R. R

G

A ig

↑

B

i−ig ig

G

Consider the discharging of capacitors at time t. Let the charge on the capacitor be q, its potential is q/C, and current through it is i = −dq/dt (negative sign because the charge q is decreasing with time). Apply Kirchhoff’s loop law to get 2q/C − iR = 0,

i.e.,

dq 1 + q = 0. dt RC

(1)

Integrate equation (1) to get t

q = q0 e− RC ,

(2)

where q0 is the initial charge on the capacitor. Substitute initial value of charges in equation (2) to get the charges at time t on the two capacitors as t

t

t

t

q1 = q0,1 e− RC1 = V C1 e− RC1 , q2 = q0,2 e− RC2 = V C2 e− RC2 .

S

A i

R

↑

B

Maximum voltage across A and B that can be measured is Vmax = ig (G + R). For R = 10 kΩ, Vmax = 50 × 10−6 (100 + 10000) ≈ 0.5 V and for R = 200 kΩ,

Differentiate to get the currents (i = −dq/dt) through the two capacitors as t

i1 = (V /R) e− RC1 , t

i2 = (V /R) e− RC2 .

Chapter 29. Electric Current in Conductors

403

Thus, the initial currents through the two capacitors are i0,1 = i0,2 = V /R. Note that the initial current depends only on the initial potential of the capacitor and the discharging resistance. From equation (2), the charge on the capacitor becomes half of its initial value at time T = RC ln 2, which gives T1 = RC1 ln 2, T2 = RC2 ln 2, and T1 = T2 /2. Ans. B, D Paragraph Type Paragraph for Questions 40-41 Consider a simple RC circuit shown in figure 1. Process 1: In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V0 (i.e., charging continues for time T  RC). In the process some dissipation (ED ) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is EC . Process 2: In a different process the voltage is first set to V0 /3 and maintained for a charging time T  RC. Then the voltage is raised to 2V0 /3 without discharging the capacitor and again maintained for a time T  RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in process 1. These two processes are depicted in figure 2. (2017)

S V

+ −

R C

2V0 3 V0 3

Process 1

Process 2

T  RC

T Figure 1

2T

t

Figure 2

Q 40. In process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are related by (A) EC = ED ln 2 (B) EC = ED (C) EC = 2ED (D) EC = ED /2 Sol. In process 1, when the capacitor is fully charged (i.e., T  RC), charge on the capacitor is Q = CV0 and potential across it is VC = V0 . The electrostatic energy stored in the capacitor is EC = 21 CVC2 = 12 CV02 .

(1)

In the charging process, the charge Q flows through the battery at potential VB = V0 (battery emf ). Thus, the work done by the battery is EB = QVB = (CV0 )V0 = CV02 .

EB = EC + ED .

(3)

Use equations (1)–(3) to get EC = ED . Ans. (B) Q 41. In process 2, total energy dissipated across the resistance ED is (A) ED = 61 CV02 (B) ED = 32 CV02 2 (C) ED = 3CV0 (D) ED = 21 CV02 Sol. Let us divide the process 2 in three parts (1) battery set to potential VB1 = 13 V0 (2) battery set to potential VB2 = 23 V0 and (3) battery set to potential VB3 = V0 . In part (1), potential across the capacitor is VC1 = 1 V , charge on the capacitor is QC1 = CVC1 = 13 CV0 0 3 and the charge flowing through the battery is QB1 = QC1 = 13 CV0 . Thus, the work done by the battery is

EB1 = QB1 VB1 = 31 CV0 V30 = 91 CV02 . In part (2), potential across the capacitor is VC2 = charge on the capacitor is QC2 = CVC2 = 23 CV0 and the charge flowing through the battery is QB2 = QC2 − QC1 = 13 CV0 (additional charge supplied to the capacitor). Thus, the work done by the battery is

EB2 = QB2 VB2 = 31 CV0 23 V0 = 29 CV02 . 2 3 V0 ,

In part (3), potential across the capacitor is VC3 = V0 , charge on the capacitor is QC3 = CVC3 = CV0 and charge flowing through the battery is QB3 = QC3 − QC2 = 13 CV0 . Thus, the work done by the battery is
EB3 = QB3 VB3 = 31 CV0 V0 = 13 CV02 .

V V0

The energy supplied (EB ) by the battery is stored in the capacitor (EC ) and dissipated by the resistor (ED ). By conservation of energy

(2)

Thus, the total work done by the battery is EB = EB1 + EB2 + EB3 = 23 CV02 . The total electrostatic energy stored in the capacitor is EC = 12 CV02 . The energy conservation gives total energy dissipated in the resistor R as ED = EB − EC = 16 CV02 . Ans. (A) Assertion Reasoning Type Q 42. Statement 1: In a metre bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance. Statement 2: Resistance of a metal increase with increase in temperature. (2008) (A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. (B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true.

404

Part V. Electromagnetism

Sol. Let the unknown resistance be X and standard resistance be R. X

R

l1

Fill in the Blank Type Q 45. The dimensions of electrical conductivity is ...... (1997) Sol. The current density J is proportional to the electric field E with proportionality constant σ i.e., J = σE. J     dI 2   L A2  Thus, [σ] = E = = VI/A /d = P A = ML2 T−3 L2 [M−1 L−3 T3 A2 ], where P = V I is power. Ans. [M-1 L-3 T3 A2 ]

l2

The null point is obtained when Wheatstone bridge ρl1 /A l1 is balanced. Thus, X R = ρl2 /A = l2 . The resistance of a metal increases with temperature. Thus X increases when its temperature is increased. To obtain the null point at the same location (same l1 and l2 ), R should be increased as ∆R = ll21 ∆X. Ans. D

Q 46. In the given circuit, each battery is 5 V and has an internal resistance of 0.2 Ω. The reading in the ideal voltmeter V is . . . . . . V. (1997)

True False Type

Sol. Let E = 5 V and r = 0.2 Ω. Apply Kirchhoff’s loop law, 8E − 8ir = 0, to get i = Er = 25 A. The voltmeter reads V = E − ir = 0 V. Ans. zero

Q 43. The current-voltage graphs for a given metallic wire at two different temperatures T1 and T2 are shown in the figure. The temperature T2 is greater than T1 . (1985)

V

Q 47. The equivalent resistance between points A and B of the given circuit is . . . . . . (1997)

I T1 A

2R

2R

R

B

T2

V

Sol. Let R1 and R2 be the resistances of the metallic wire at temperature T1 and T2 , respectively. Ohm’s law, V = IR, gives the slope of the I-V graph as dI/dV = 1/R. In the given graph, slope at T1 is greater than the slope at T2 i.e., 1/R1 > 1/R2 or R1 < R2 . The resistance of the metallic wire varies with the temperature as R1 = R0 (1 + α(T1 − T0 ),

(1)

R2 = R0 (1 + α(T2 − T0 ),

(2)

where α > 0 is the thermal coefficient of resistance and R0 is the resistance at temperature T0 . Substitute R1 and R2 from equations (1) and (2) into the inequality R1 < R2 to get T1 < T2 . Ans. T Q 44. Electrons in a conductor have no motion in the absence of a potential difference across it. (1982) Sol. The electrons inside the conductors have random motion. In the absence of potential difference across a conductor, the motion of the electrons is such that their drift velocity is zero. Ans. F

Sol. All the three resistors are connected in parallel. The equivalent resistance is Req = (2R k 2R) k R = R k R = R/2. Ans. R/2 Integer Type Q 48. In the following circuit, the current through the resistor R(= 2 Ω) is I Amperes. The value of I is . . . . . . . (2015) R(= 2Ω)

1Ω 8Ω 2Ω

6Ω 6.5V

2Ω 4Ω 10Ω

12Ω

4Ω

Sol. Consider the resistors that join the nodes B, C, D, and E. These resistors form a balanced Wheatstone bridge between the nodes B and D. Thus, 8 Ω resistor in the branch CE can be removed without affecting the circuit (see the right figure below). Effective resistance between the nodes B and D is RBD = (1 Ω + 2 Ω) k (2 Ω + 4 Ω) = (3 Ω) k (6 Ω) = 2 Ω.

Chapter 29. Electric Current in Conductors A

R(= 2Ω) B

1Ω

C

A

R(= 2Ω) B

405

1Ω

C

8Ω 6Ω

2Ω

2Ω

E

6.5V

D

10Ω

12Ω

2Ω 4Ω

F

4Ω

10Ω

12Ω

G

2Ω

E

6.5V

4Ω

F

6Ω

D

Q 50. At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V? [Take ln 5 = 1.6, ln 3 = 1.1]. (2010)

4Ω

2µF

2MΩ

G

A

The equivalent circuit is shown in the left figure below. In this circuit, consider the resistors which join the nodes B, D, F, and G. These resistors form a balanced Wheatstone bridge between B and G. Thus, 10 Ω resistor between D and F can be removed without affecting the circuit. Effective resistance between B and G is (middle figure below) RBG = (6 Ω + 12 Ω) k (2 Ω + 4 Ω) = (18 Ω) k (6 Ω) = 4.5 Ω. Thus, the effective resistance of the entire circuit (right figure below) is Reff = 2 Ω + 4.5 Ω = 6.5 Ω. The current through the resistor R = 2 Ω is I = V /Reff = 6.5/6.5 = 1 A.

B 2MΩ

2µF

Sol. The resistors are connected in parallel which gives the equivalent resistance R = 2 k 2 = 1 MΩ. The capacitors connected in parallel give the equivalent capacitance C = 2 k 2 = 4 µF. The equivalent circuit is shown in the figure. V0 10V

A

R(= 2Ω)B

A

6Ω

R(= 2Ω)B

2Ω

6.5V

A

6Ω

12Ω

10Ω

6.5V

D

F

4Ω

4.5Ω

D

12Ω

G

4Ω

G

G

Ans. 1 Q 49. Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB (in volts) is . . . . . . . (2011) 6V

1Ω

A

B 2Ω

3V

A

C 4µF

B

2Ω

6.5V F

R(= 2Ω)B

R 1MΩ

The voltage across the equivalent capacitor is same as the voltage across the individual capacitors (parallel combination). Thus, we need to find time t at which the voltage across C become 4 V in the equivalent circuit (charging of a capacitor). The voltage across C at time t is h i V = V0 1 − e−t/(RC) , which simplifies to   V0 . t = RC ln V0 − V Substitute V0 = 10 V, V = 4 V, R = 1 × 106 Ω and C = 4 × 10−6 F to get t = 4 ln(5/3) = 4(ln 5 − ln 3) = 2 s. Ans. 2

Sol. Let I be the current in the loop. Descriptive 6V

I

1Ω

A

B

3V

Q 51. R1 , R2 , R3 are different values of resistor R and A, B, C are the null points obtained corresponding to R1 , R2 and R3 , respectively. For which resistor, the value of X will be most accurate and why? (2005)

2Ω

Kirchhoff’s loop law, 6 − 3 − 2I − 1I = 0, gives I = 1 A. Now, going from B to A along the upper branch, we get, VB − 1 + 6 = VA , which gives VA − VB = 5 V. Ans. 5

X

R G

A

B

C

406

Part V. Electromagnetism

Sol. Let l be the total length of the wire. Let x be the length of the wire towards the left of the null point. Wheatstone bridge is balanced when X = Rx/(l − x). Differentiate and simplify to get the relative error in measurement of X, l∆x ∆X = , X x(l − x)

(1)

Aliter: Differentiate q = q0 (1 − e−αt ) to get the current through the capacitor ic = αq0 e−αt . Substitute t = 0 to get the initial current ic0 = αq0 . The capacitor behaves like a short circuit when the switch is connected at t = 0 (see figure). Thus, the initial current through the capacitor is ic0 = V /R1 . R1

R1

i

i R2

where ∆x is the measurement error in x. The value of ∆X in equation (1) attains its minimum value when the denominator is maximum i.e., x = l/2. Ans. R2 Q 52. At t = 0, switch S is closed (see figure). The charge on the capacitor is varying with time as q = q0 (1 − e−αt ). Obtain the value of q0 and α in the given circuit parameters. (2005) R1 S R2

C V

Sol. Let the charge on the capacitor be q, current through it be ic = dq/dt, and potential across it be Vc = q/C at a time t. i R1 ic C

V t=∞

t=0

In the steady state (t → ∞), the capacitor behaves like an open circuit element. The charge on capacitor in the steady state is q0 and the potential across it is VC = q0 /C. The current through R2 is i = V /(R1 + R2 ) and the potential across it is VR2 = iR2 = V R2 /(R1 + R2 ). R2 . Substitute The condition VC = VR2 gives q0 = RCV 1 +R2 R1 +R2 q0 in V /R1 = αq0 to get α = CR1 R2 . There is a semi-quantitative argument to explain the initial and steady state behaviour of capacitor. The capacitive reactance is given by XC = 1/(ωC). When switch is connected, there is sudden change w.r.t. time i.e., ω → ∞. Thus, XC = limω→∞ 1/(ωC) = 0 (closed). In the steady state, there is no change w.r.t. time i.e., ω → 0. Thus, XC = limω→0 1/(ωC) = ∞ (open). R2 R1 +R2 Ans. q0 = RCV , α = CR 1 +R2 1 R2 Q 53. Draw the circuit for experimental verification of Ohm’s law using a source of variable DC voltage, a main resistance of 100 Ω, two galvanometers and two resistances of values 106 Ω and 10−3 Ω, respectively. Clearly show the positions of the voltmeter and the ammeter.

i − ic

V

R2

V

R2

(2004)

The capacitor C and resistor R2 are connected in parallel and hence the potential difference across these two is equal VC = (i − ic )R2 .

(1)

(2)

Eliminate i from equations (1) and (2) and substitute VC = q/C and iC = dq/dt to get dq (R1 + R2 ) V + q= . dt CR1 R2 R1

106 Ω

G

100Ω G

Kirchhoff’s loop law gives V = iR1 + (i − ic )R2 .

Sol. Verification of Ohm’s law requires the measurement of the voltage and current.

(3)

Integrate equation (3) with initial condition q = 0 at t = 0 to get i R +R2 CV R2 h t − 1 q= 1 − e C(R1 R2 ) . R1 + R2

10−3 Ω

The galvanometer can be converted to a voltmeter by connecting a very high resistance (106 Ω) in series and to an ammeter by connecting a very low resistance (10−3 Ω) in parallel, as shown. Ans. See solution. Q 54. Show by diagram, how can we use a rheostat as the potential divider? (2003) Sol. The rheostat is a resistance coil with two ends A and B and a slider C.

Chapter 29. Electric Current in Conductors

407

RL C

A

S

B

V

To use rheostat as a potential divider, connect A and B across the battery (V0 ) and B and C across the load resistance (RL ). As slider C moves from A to B, the potential across RL changes from V0 to zero. Ans. See solution. Q 55. A thin uniform wire AB of length 1 m, an unknown resistance X and a resistance of 12 Ω are connected by thick conducting strips, as shown in the figure. A battery and galvanometer (with a sliding jockey connected to it) are also available. Connections are to be made to measure the unknown resistance X using the principle of Wheatstone bridge. Answer the following questions. (2002)

X B

C

B

(a) the charge q on the capacitor at time t. (b) the current in AB at time t. What is its limiting value as t → ∞? Sol. Consider a time instant t. Let i be current through the battery, q be charge on the capacitor C, i1 = dq/dt be the current through C, and VC = q/C be potential across C (see figure for polarity). i

A i1 i−i1

V

R

R R

+ −

C

B

The Kirchhoff’s loop law gives,

12Ω C

R

R R

V0

A

A

D

(a) Are there positive and negative terminals on the galvanometer? (b) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points. (c) After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the value of the resistance X. Sol. The galvanometer does not have positive or negative terminals. The circuit diagram to measure the unknown resistance X is given in the figure.

V − (i − i1 )R − iR = 0,

(1)

V − i1 R − q/C − iR = 0.

(2)

Eliminate i from equation (1) and equation (2) to get i1 = dq/dt,   2 V dq + q= , (3) dt 3RC 3R and then integrate to get i 2t CV h 1 − e− 3RC . 2

q=

(4)

Use equation (1) and equation (4) to get current through the capacitor and the branch AB, dq V − 2t = e 3RC dt 3R   i1 V 1 − 2t V 3RC − = 1− e = i − i1 = 2R 2 2R 3

i1 = X A

N

B

12Ω C

iAB

D

G

Let R be the total resistance of the potentiometer wire of length 100 cm and N be the null point. The resistance of branch AN is RAN = 60(R/100) = 0.6R and of branch BN is RBN = 0.4R. The balancing condition of Wheatstone bridge, RAN /RBN = 12/X, gives X = 8 Ω. Ans. (a) No (b) See solution (c) 8 Ω Q 56. In the circuit shown in figure, the battery is an ideal one, with emf V. The capacitor is initially uncharged. The switch S is closed at time t = 0. Find, (1998)

(5)

Taking the limits t → ∞ in equation (5), we get steady state current in AB as V /(2R). The readers may note that capacitor becomes an open circuit element in steady state. h i 2t 2t V V − 3RC V Ans. (a) CV 1 − e− 3RC (b) 2R − 6R e , 2R 2 Q 57. Find the emf (V ) and the internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs V1 and V2 and internal resistances r1 and r2 respectively, with polarities as shown in the figure. (1997)

408

Part V. Electromagnetism V2

r2 A

−

+

r1

B

V1

Sol. The emf of the equivalent battery is V = VA − VB and its internal resistance r is the resistance between A and B, when A and B are not connected to external resistance. Let i be the current in the loop when A and B are open. r2 A

i

V2 −

+

r1

τ = RC = B

d K0 A K0 5(8.85 × 10−12 ) = = σA d σ 7.4 × 10−12

= 5.98 s. The charge on the capacitor and the current through the circuit at time t are

V1

q = q0 e−t/τ ,

Kirchhoff’s loop law

i = dq/dt = −(q0 /τ )e−t/τ ,

and

where q0 = 8.85 × 10−6 C is the initial charge. The magnitude of the current at t = 12 s is

V1 − ir1 − ir2 + V2 = 0, 2 gives the current i = Vr11 +V +r2 . Traversing from B to A along the lower branch relates the potential at A and B by

VA = VB + V1 − ir1 = VB + V1 −

(V1 + V2 )r1 , r1 + r2

which gives the emf of the equivalent battery as V = VA − VB =

The capacitance of parallel plate capacitor with plate area A, separation d, and dielectric constant K is K0 A . (1) C= d The conductivity is related to the current density J and electric field E by J = σE. The electric field and the potential difference V are related by V = Ed. Thus, the resistance R can be written as V Ed d V = = = . (2) R= I JA σEA σA The equations (1) and (2) give the time constant of the circuit as

i=

8.85 × 10−6 −12/5.98 e = 0.198 × 10−6 A. 5.98 Ans. 0.198 µA

Q 59. An electrical circuit is shown in the figure. Calculate the potential difference across the resistor of 400 Ω as will be measured by the voltmeter V of resistance 400 Ω either by applying Kirchhoff’s rules or otherwise. (1996)

V1 r2 − V2 r1 . r1 + r2

V 400Ω

The internal resistance of the battery is r = r1 k r2 = r1 r2 r1 +r2 . Ans. V =

V1 r2 −V2 r1 r1 +r2 ,

r=

r1 r2 r1 +r2

Q 58. A leaky parallel plate capacitor is filled completely with a material having electrical conductivity σ = 7.4 × 10−12 Ω−1 m−1 and dielectric constant K = 5. If the charge on the capacitor at instant t = 0 is q = 8.85 µC, then calculate the leakage current at the instant t = 12 s. (1997)

100Ω

100Ω

I2

100Ω I1 I

10V

Sol. The resistance of voltmeter is 400 Ω as shown in the figure.

Sol. The charge in a leaky capacitor decreases (leaks) due to presence of in-built resistance. Equivalent circuit of the leaky capacitor is an R-C circuit as shown in the figure.

400Ω I3 100Ω

C

R

200Ω

I2 A

100Ω B

200Ω C

100Ω I1 I

400Ω

10V

D

Chapter 29. Electric Current in Conductors

409

The equivalent resistance of the two 400 Ω resistors connected in parallel is 200 Ω. Now, it can be seen that the given
circuit is a balanced Wheatstone RAB RAC = R bridge R with no current flowing through BD CD 100 Ω resistor connecting B and C. The resistance of upper branch connecting A and D is R1 = 100 + (400 k 400) = 100 + 200 = 300 Ω and that of lower branch connecting A and D is R2 = 100 + 200 = 300 Ω. Thus, the equivalent resistance between A and D is R = R1 k R2 = 150 Ω. Thus, the currents I, I1 , and I2 are given by I = V /R = 10/150 = 1/15 A,

Solve equation (1) to get i1 +i3 = E1 /R3 = 6/4 = 1.5 A and then substitute in equation (2) to get E2 + E3 − (i1 + i3 )R3 2 + 3 − (1.5)4 = R2 + R4 2+3 = −1/5 A.

i3 =

The potential difference across the capacitor C is same as the potential difference across e and f which in turn is equal to the potential difference across e and d i.e., V = Ve − Vd = E2 − i3 R2 = 2 − (−0.2)(2) = 2.4 V. The energy stored in the capacitor is

I1 = I2 = I/2 = 1/30 A.

E = 12 CV 2 = 12 (5 × 10−6 ) (2.4)2 = 1.44 × 10−5 J.

The current through 400 Ω resistor and voltage across it are

Q 61. An infinite ladder network of resistances is constructed with 1 Ω and 2 Ω resistances (see figure). The 6 V battery between A and B has negligible internal resistance. (1987)

I3 = I2 /2 = 1/60 A, V400 = 400I3 = 400(1/60) = 20/3 V. Ans.

20 3

V

Q 60. In the given circuit: E1 = 3E2 = 2E3 = 6 V and R1 = 2R4 = 6 Ω, R3 = 2R2 = 4 Ω, C = 5 µF. Find the current in R3 and the energy stored in the capacitor. (1988)

C R2

E2

R3

R4

E3

Sol. In the steady state, the capacitor will act as an open circuit element as shown in the figure. R1

f

E1

i1 E2 a

R2 d

R3

1Ω

6V

2Ω

1Ω

1Ω

2Ω

2Ω

B

Sol. Let the effective resistance between A and B be R. The ladder network consists of infinite number of units, where each unit consists of two resistances of values R1 = 1 Ω and R2 = 2 Ω. The effective resistance of the ladder will not change by removal of one unit, say the unit close to the battery. A

g

i

i1

i3

e

A

(a) Show that the effective resistance between A and B is 2 Ω. (b) What is the current that passes through the 2 Ω resistance nearest to the battery?

E1

R1

Ans. 1.5 A, 1.44 × 10−5 J

6V

c

1Ω

C i − i1 i1 2Ω

R

i3 E3

R4

B

b

Let i1 be the current flowing through the battery E1 and i3 be the current flowing through the battery E3 . Kirchhoff’s junction law at junctions c and d gives the current through R3 as i1 + i3 and current through R2 as i3 . Apply Kirchhoff’s loop law in the loops fgcdf and abcdea to get E1 − (i1 + i3 )R3 = 0,

(1)

E3 − i3 R4 − (i1 + i3 )R3 − i3 R2 + E2 = 0.

(2)

Thus, the effective resistance between A and B is equal to the resistance of the circuit shown in the figure i.e., R = R1 + (R2 k R) R2 R 2R = R1 + =1+ . R2 + R 2+R Solve to get R = 2 Ω. The effective resistance between A and B is R = 2 Ω. Thus, the current through the battery of emf E =

410

Part V. Electromagnetism

6 V is i = E/R = 6/2 = 3 A. Since R2 = R = 2 Ω, current i is equally divided at the node C giving i1 = i/2 = 1.5 A. Ans. (b) 1.5 A Q 62. A part of circuit in steady state along with the currents flowing in the branches, the values of resistances etc., is shown in the figure. Calculate the energy stored in the capacitor C(4 µF). (1986)

3Ω 2A

5Ω

C G

A 3Ω

2Ω

2Ω 2V

1Ω 2Ω

1V i1 D

Sol. In the steady state, no current flows through the capacitor as it acts as an open circuit element. Apply Kirchhoff’s first law at the junctions a and b to get the current flowing out of these junctions as 3 A.

1A 4V

2A 3Ω

3A a

5Ω

4µF 3V

1Ω 2A

3Ω

B

4Ω

1A

3Ω

H

D

1Ω

1Ω 2A

2Ω

F

Sol. The circuit is shown in the figure.

3Ω 4µF

3V

B

(a) the potential difference between B and D. (b) the potential difference across the terminals of each of the cells G and H.

1A 4V

E A

b 3A

1Ω

i1 − i2 i2

3V 3Ω

C

Apply Kirchhoff’s loop law in the loop BADB and DCBD to get 2 − 2i1 − i1 − 1 − 2(i1 − i2 ) = 0,

(1)

3 − 3i2 − i2 − 1 + 2(i1 − i2 ) = 0.

(2)

Solve equations (1) and (2) to get i1 = 5/13 A and i2 = 6/13 A. The potential difference between B and D is

1Ω 2Ω

1V

4Ω

VB − VD = −2(i1 − i2 ) = −2(5/13 − 6/13) = 2/13 V.

1A

Now, traversing from a to b through 5 Ω, 1 Ω, and 2 Ω resistances, we get

The potential difference between the cells of batteries G and H are VG = 3 − 3i2 = 3 − 3(6/13) = 21/13 V, VH = 1 + (1)i2 = 1 + 6/13 = 19/13 V.

Va − (3)(5) − (3)(1) + (3)(2) = Vb . Thus, the potential difference across the capacitor is Vab = Va − Vb = 12 V. The energy stored in the capacitor is given by 2 = 12 (4 × 10−6 )(12)2 = 2.88 × 10−4 J. E = 21 CVab

Ans. 0.288 mJ Q 63. In the circuit shown in figure E, F, G, H are cells of emf 2, 1, 3 and 1 V, respectively and their internal resistances are 2, 1, 3 and 1 Ω, respectively. Calculate, (1984)

Ans. (a)

2 13

V (b)

21 13

V,

19 13

V

Q 64. Two resistors, 400 Ω and 800 Ω are connected in series with a 6 V battery. It is desired to measure the current in the circuit. An ammeter of 10 Ω resistance is used for this purpose. What will be the reading in the ammeter? Similarly, if a voltmeter of 1000 Ω resistance is used to measure the potential difference across the 400 Ω resistor, what will be the reading in the voltmeter? (1982) Sol. The ammeter of resistance 10 Ω is connected in series to measure current in the circuit.

Chapter 29. Electric Current in Conductors 400Ω

411 i1

800Ω

2Ω

A 10Ω

B i − i1

i

3Ω

A 4Ω

i 6V

2.8Ω

6V

Effective resistance of the circuit is RA = 400 + 800 + 10 = 1210 Ω. The current through the circuit is i = V /RA = 6/1210 = 4.96 × 10 D

1000Ω

V

−3

A.

C

i1 A

400Ω i − i1

The effective resistance of the circuit is

800Ω B

i

R = (2 k 3) + 2.8 =

2×3 + 2.8 = 4 Ω. 2+3

The current i through the circuit is i = V /R = 6/4 = 1.5 Ω. The current i gets divided into two branches at the node A. Apply Kirchhoff’s loop law in the loop containing 2 Ω and 3 Ω resistors to get 2i1 = 3(i − i1 ) = 3(1.5 − i1 ),

6V

The voltmeter of resistance 1000 Ω is connected in parallel to 400 Ω resistor to measure the voltage across it. Effective resistance of the circuit is RV = (400 k 1000) + 800 400 × 1000 = + 800 = 1085.7 Ω. 400 + 1000 The current through the battery is i = V /RV = 6/1085.7 = 5.53 × 10−3 A. The current i gets divided into two branches at the node A. Apply Kirchhoff’s loop law in ABCDA to get 400(i − i1 ) = 1000i1 i.e., i1 = 2i/7 = 1.58 × 10−3 A. The potential difference across the voltmeter is V = i1 (1000) = 1.58 V. Note that ideal resistances of the ammeter and the voltmeter are zero and infinite, respectively. Ans. 4.96 mA, 1.58 V Q 65. Calculate the steady state current in the 2 Ω resistor shown in the circuit (see figure). The internal resistance of the battery is negligible and the capacitance of the condenser is 0.2 µF. (1982)

which gives i1 = 0.9 A. Ans. 0.9 A Q 66. A steady current passes through a cylindrical conductor. Is there an electric field inside the conductor? (1982) Sol. The electric field inside a conductor is zero under the electrostatic conditions i.e., when charges are not moving. When a steady current is set through a cylinder, the charges are in motion (not static). In this case, electric field inside the conductor is not zero. In fact, it is the electric field inside the conductor that provides the drift velocity for the flow of charges. Ans. yes Q 67. Find the potential difference between the points A and B and between the points B and C in the steady state (see figure). (1981) 3µF F

B

3µF

1µF G 1µF E

D 1µF

10Ω

2Ω A 3Ω C

4Ω

6V

2.8Ω

Sol. In the steady state, the capacitor acts as an open circuit element. The equivalent circuit in the steady state is shown in the figure.

20Ω

C 100V

Sol. In the steady state, the capacitors act as open circuit elements, breaking all paths for flow of current through the battery. There is no current through 10 Ω and 20 Ω resistors i.e., no potential drop across these resistors. So these resistors can be replaced by conducting wires. Equivalent circuit for the steady state is shown in the figure.

412

Part V. Electromagnetism 3µF F

B

3µF

1µF G 1µF

D

A

E 1µF

A

C

1Ω

i1

3V

1Ω

i2

2V

i3

C 1Ω

100V 6µF F

B

1V

2µF G

There is no current through R = 1 Ω. Apply Kirchhoff’s junction law at node C to get

1µF A

C 100V

i1 + i2 + i3 = 0.

Further, two 3 µF capacitors between B and D are connected in parallel giving their effective capacitance 6 µF. Similarly, two 1 µF capacitors between B and E are connected in parallel giving their effective capacitance 2 µF. The equivalent circuit is shown in the figure. Let CAB = 6 µF and VAB be potential across this capacitor. The charge on CAB is qAB = CAB VAB . Similarly, let CBC = 2 µF and VBC be the potential difference across this capacitor. The charge on CBC is qBC = CBC VBC . The charges on CAB and CBC are equal because they are connected in series i.e., qAB = qBC , which gives 3VAB = VBC .

(1)

Apply Kirchhoff’s loop law in the upper and lower loops to get 3 − i1 + i2 − 2 = 0,

(2)

2 − i2 + i3 − 1 = 0.

(3)

Solve equations (1)–(3) to get i1 = 1 A, i2 = 0 A, and i3 = −1 A. Since i2 = 0 A and current through R is zero, VAB = VCB = E2 = 2 V. The circuit after shorting r2 = 1 Ω and connecting point A to B is shown in the figure. E

1Ω

(1) C

Total potential difference across CAB and CBC is equal to the battery voltage, i.e., VAB + VBC = 100.

(2)

Solve equations (1) and (2) to get VAB = 25 V and VBC = 75 V. Ans. VAB = 25 V, VBC = 75 V Q 68. In the circuit shown in figure E1 = 3 V, E2 = 2 V, E3 = 1 V and R = r1 = r2 = r3 = 1 Ω. (1981)

A

B

1Ω

R

r1

E1

r2

E2

r3

B

E3

(a) Find the potential difference between the points A and B and the currents through each branch. (b) If r2 is short circuited and the point A is connected to point B, find the currents through E1 , E2 , E3 and the resistor R.

i1

3V

i2

2V

B

i3 H

1Ω

G

1V

i1 + i2 + i 3 D

1Ω

A

Apply Kirchhoff’s loop law in loops ABCDA, FECBF, and BCHGB to get 2 − (i1 + i2 + i3 ) = 0,

(4)

3 − i1 − 2 = 0,

(5)

2 + i3 − 1 = 0.

(6)

Solve equations (4)–(6) to get i1 = 1 A, i2 = 2 A, and i3 = −1 A. The current through the resistor R is i1 + i2 + i3 = 1 + 2 − 1 = 2 A. Ans. (a) 2 V, 1 A, 0, −1 A (b) 1 A, 2 A, −1 A, 2A Q 69. All resistances in the figure are in Ω. Find the effective resistance between the points A and B. (1980) 3

3 3

3 6 6

B

6 3

3

Sol. Let i1 , i2 , and i3 be the currents through the batteries E1 = 3 V, E2 = 2 V, E3 = 1 V, respectively.

F

A

Chapter 29. Electric Current in Conductors

413

Sol. In the leftmost loop, two 3 Ω resistors are connected in parallel to one 6 Ω resistor. Effective resistance of this loop is R = (3 + 3) k 6 = 6 k 6 = 3 Ω. Repeat the same process for the next two loops to get the effective resistance of the three loops (from left) as 3 Ω. Thus, the resistance across AB consists of two 3 Ω resistors connected in series and one 3 Ω resistor connected in parallel, giving the effective resistance

circuit has two branches, APB and AQB, each of resistance 2R, connected in parallel. Thus, the effective resistance between A and B is RAB = (2R) k (2R) = R. Ans. R Q 72. A copper wire is stretched to make it 0.1% longer. What is the percentage change in its resistance?

RAB = (3 + 3) k 3 = 6 k 3 = 2 Ω.

(1978)

Ans. 2 Ω Q 70. A 25 W and a 100 W bulb are joined in series and connected to the mains. Which bulb will glow brighter? (1979)

Sol. Let the powers of the bulbs, P1 = 25 W and P2 = 100 W, be specified at a voltage V . The resistances of the bulbs are R1 = V 2 /P1 and R2 = V 2 /P2 . When connected in series, the currents through the bulbs are equal, say i. The powers consumed by the bulbs, connected in the series, are P10 = i2 R1 = i2 V 2 /P1 = P20

2

2

2

= i R2 = i V /P2 =

1 2 2 25 i V , 1 2 2 100 i V .

(1) (2)

From equations (1) and (2), P10 = 4P20 . Hence, the bulb with 25 W power will glow brighter. Ans. 25 W Q 71. If each of the resistances in the network shown in the figure is R, what is the resistance between the terminals A and B ? (1978)

A

B

Sol. Given circuit is redrawn in the figure after moving out the terminals A and B from inside the triangle. P R

R A

B

R R

R Q

It is a balanced Wheatstone bridge. Hence, resistance in the branch PQ can be removed without affecting the effective resistance of the circuit. Hence, the

Sol. Let A be the cross-sectional area of the wire of length l. When the wire is stretched, its volume V = Al, and resistivity ρ remain constant. The resistance of the wire is given by R = ρl/A = ρl2 /(Al) = (ρ/V )l2 . Differentiate and simplify to get ∆R/R = 2 ∆l/l = 2 (0.1%) = 0.2%. Ans. 0.2% increase

Chapter 30 Thermal and Chemical Effects of Electric Current

One Option Correct

Sol. The heat generated by the current i = 1 A flowing through a resistance R = 100 Ω in time interval t = 5 min is given by

Q 1. Two bars of radius r and 2r are kept in contact as shown. An electric current I is passed through the bars. Which one of following is correct? (2006) l/2

There is no heat loss as the container is thermally insulated. The work done by the gas, ∆W , is zero because the container is rigid. The first law of thermodynamics, ∆Q = ∆U + ∆W , gives the change in internal energy ∆U = 30 kJ. Ans. D

l/2

2r

I

∆Q = i2 Rt = (1)2 (100)(5 × 60) = 30 kJ.

r C

A

B

(A) Heat produced in bar BC is 4 times the heat produced in bar AB. (B) Electric field in both halves is equal. (C) Current density across AB is double that across BC. (D) Potential difference across AB is 4 times that of across BC.

Q 3. A wire of length L and three identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by ∆T in time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount ∆T in the same time. The value of N is (2001) (A) 4 (B) 6 (C) 8 (D) 9

Sol. Let ρ be the resistivity of the material. The resistances of the bar AB and BC are given by RAB =

ρ(l/2) ρl = , 2 4πr 8πr2

RBC =

Sol. Let R and m be the resistance and mass of the wire of length L. Let E be the emf of each cell. The potential difference across the wire when it is connected to three cells in series is 3E. The heat produced in time interval t is given by

ρ(l/2) ρl = . 2 πr 2πr2

The currents flowing through both the bars are equal to I as they are connected in series. The heats produced per unit time in two bars are 2

PAB = I 2 RAB =

I ρl , 8πr2

Q1 = (3E)2 t/R = 9E 2 t/R.

2

PBC = I 2 RBC =

I ρl . 2πr2

This heat increases the temperature of the wire. If S is the specific heat of the material of the wire then rise in temperature ∆T is given by

The potential drop V , electric field E, and the current density J for the two bars are given by VAB EAB JAB

Iρl = IRAB = , 8πr2 Iρ VAB = = , l/2 4πr2 I = , 4πr2

VBC EBC JBC

Q1 = 9E 2 t/R = mS∆T.

Iρl = IRBC = ; 2πr2 VBC Iρ = = 2; l/2 πr I = 2. πr

(1)

In the second case, potential difference across the wire is N E, resistance is 2R, mass is 2m, and rise in temperature is same as ∆T . Thus, Q2 = N 2 E 2 t/(2R) = (2m)S∆T.

Ans. A

(2)

Solve equations (1) and (2) to get N = 6. Ans. B

Q 2. Ideal gas is contained in a thermally insulated and rigid container and it is heated through a resistance of 100 Ω by passing a current of 1 A. Change in internal energy of the gas after 5 min will be (2005) (A) zero (B) 10 kJ (C) 20 kJ (D) 30 kJ

Q 4. In the circuit shown in figure the heat produced in the 5 Ω resistor due to the current flowing through it is 10 cal/s. The heat generated in the 4 Ω resistor is (1981) 414

Chapter 30. Thermal and Chemical Effects of Electric Current 4Ω

415

r in a small segment of length l ( l0 ). The resistance of ρl ρl this small segment increases from πr 2 to πr 2 (∵ r < r0 ), 0 where ρ is the resistivity of the tungsten.

6Ω

l0

5Ω r0

r

(A) 1 cal/s (B) 2 cal/s (C) 3 cal/s (D) 4 cal/s Sol. Let i1 and i2 be the currents through the 5 Ω and 4 Ω resistors, respectively. The heats produced in the 5 Ω and 4 Ω resistors due to the current flowing through them are P1 = 5i21 = 10 cal/s, P2 =

(1)

4i22 .

(2) i2

4Ω

6Ω B

A

C i1



i2 i1

2

4 P1 = 5

The heat generated in this small segment in a time interval t, i2 t ρl/(πr2 ), is more than the heat generated in other segment of same length but of radius r0 . This heat increases temperature of this segment which further increases its resistance through (i) temperature dependence of resistance (ii) increase in rate of evaporation. Thus, the temperature T of this segment is higher than the temperature T0 of other segment. A black body at higher temperature emits more radiation at higher band of frequencies (see figure).

5Ω

Apply Kirchhoff’s loop law in ABCA to get, 10i2 − 5i1 = 0 i.e., i2 /i1 = 1/2. Divide equation (2) by (1) and substitute i2 /i1 = 1/2 to get 4 P2 = 5

l

 2 1 (10) = 2 cal/s. 2

We encourage you to find i2 from the given data. Hint: i1 = 2.89 A. Ans. B One or More Option(s) Correct Q 5. An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true? (2016) (A) The temperature distribution over the filament is uniform. (B) The resistance over small sections of the filament decreases with time. (C) The filament emits more light at higher band of frequencies before it breaks up. (D) The filament consumes less electrical power towards the end of the life of the bulb. Sol. The filament breaks up at random location due to non-uniform evaporation of tungsten from the filament. The non-uniform evaporation is caused by the non-uniform temperature distribution over the filament. Let l0 be the length and r0 be the radius of the filament. Let radius of the filament reduces from r0 to

Iλ

T

T0

λ

Towards the end of life of the bulb, effective resistance Re of the filament increases (due to decrease in radius caused by evaporation). Thus, the electrical power consumed by the filament, V 2 /Re , decreases. Ans. C, D Q 6. Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K? (2014) (A) 4 if wires are in parallel (B) 2 if wires are in series (C) 1 if wires are in series (D) 0.5 if wires are in parallel Sol. The resistance of the wire of length L and diameter d is R1 = 4ρL πd2 and of length L and diameter 2d is R2 = ρL R1 πd2 = 4 . When the two wires of resistances R2 are connected in series, the effective resistance is Rs = 2R2 = R1 /2.

(1)

and when they are connected in parallel, the effective resistance is Rp = R2 /2 = R1 /8.

(2)

416

Part V. Electromagnetism

Let V be the applied voltage, m = 0.5 kg be the mass of the water, and S be the specific heat of the water. Initially, the heat produced by a resistance R1 in time t1 = 4 min is V 2 t1 /R1 . This heat is used to raise the temperature of the water by ∆T = 40 K. Thus, V 2 t1 /R1 = mS∆T.

Sol. Let E be the emf and r = 1 Ω be the internal resistance of each battery. The series and parallel combination of two batteries are shown in the figure. E

(3) E

Let ts and tp be the time taken to raise the temperature of same amount of water by ∆T when the resistances are connected in series and parallel. Thus,

E

1Ω

1Ω

R 1Ω

E

1Ω

V 2 ts /Rs = mS∆T.

(4)

R

V 2 tp /Rp = mS∆T.

(5)

When the two batteries are connected in series, effective emf becomes 2E and effective internal resistance is 2r. Thus, the current through R is i1 = 2E/(2r + R) and the heat produced in R is given by  2 2E 2 J1 = i1 R = R. 2r + R

Divide equation (4) by (3) and use the equation (1) Rs t1 = 12 × 4 = 2 min. Similarly, divide to get ts = R 1 equation (5) by (3) and use the equation (2) to get R tp = Rp1 t1 = 18 × 4 = 0.5 min. Ans. B, D Fill in the Blank Type Q 7. An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb delivers 500 W is . . . . . . Ω. (1987) Sol. The resistance of the bulb of power rating P = 500 W at applied voltage 100 V is Rb = V 2 /P = (100)2 /500 = 20 Ω. Rb

R

When the batteries are connected in parallel, the effective emf is E and the effective internal resistance is r/2. Thus, the current through R is i2 = E/(r/2 + R) and the heat produced in it is given by  2 E 2 J2 = i2 R = R. r/2 + R Substitute J1 and J2 in the given relation, J1 = 2.25J2 , and solve to get R = 4 Ω. We encourage you to find the effective emf and the effective internal resistance for series and parallel combination by using Kirchhoff’s law. Ans. 4

i

Descriptive 200V

The effective resistance of the circuit when resistance R is connected in series to the bulb is Re = Rb +R. The current through the circuit when connected to 200 V supply is i = V /Re = 200/(Rb + R). The power deliverd by the bulb in this configuration is 500 W if P = i 2 Rb =



200 Rb + R

2 Rb = 500.

Substitute Rb = 20 Ω and solve to get R = 20 Ω. Ans. 20 Integer Type Q 8. When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1 . When the same batteries are connected in parallel across R, the rate is J2 . If J1 = 2.25J2 then the value of R (in Ω) is . . . . . . . (2010)

Q 9. A copper wire having cross-sectional area of 0.5 mm2 and a length of 0.1 m is initially at 25 ◦ C and is thermally insulated from the surroundings. If a current of 1.0 A is set up in this wire, (a) find the time in which the wire will start melting. The change of resistance with the temperature of the wire may be neglected. (b) What will this time be, if the length of the wire is doubled? [For Copper, Melting point = 1075 ◦ C, Specific resistance = 1.6 × 10−8 Ω m, Density = 9 × 103 kg/m3 , Specific heat = 9 × 10−2 cal/(kg ◦ C).] (1979) Sol. A current i = 1.0 A is flowing through a copper wire of cross-sectional area A = 0.5 mm2 and length l = 0.1 m. The wire is thermally insulated from the surroundings. Hence, the heat produced in the wire is used to raise its temperature to the melting point Tm = 1075 ◦ C i.e., i2 Rt = mS∆T,

(1)

where R = ρl/A is the resistance (ρ = 1.6 × 10−8 Ω m is resistivity), t is the time required for melting to start,

Chapter 30. Thermal and Chemical Effects of Electric Current m = DAl is the mass of the wire (D = 9 × 103 kg/m3 is mass density), S = 9 × 10−2 cal/(kg ◦ C) is specific heat, and ∆T = Tm − T0 = 1075 − 25 = 1050 ◦ C is the rise in temperature. Substitute the values in equation (1) to get (DAl)S∆T DA2 S∆T mS∆T = 2 = , 2 i R i (ρl/A) i2 ρ (9 × 103 )(0.5 × 10−6 )2 (9 × 10−2 × 4.18)(1050) = (1)2 (1.6 × 10−8 ) = 55.55 s. (2)

t=

From equation (2), the time t is independent of the wire length. Thus, the time required for the melting to start remains equal to 55.55 s when the length of the wire is doubled. Ans. (a) 55.55 s (b) 55.55 s Q 10. A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of 10 Ω and a resistance R, to a 100 V mains as shown in the figure. What will be the value of R so that the heater operates with a power of 62.5 W? (1978) 10Ω B

Heater

C

R 100V

Sol. The heater power is P = 1000 W at an applied voltage of V = 100 V. The resistance of the heater is Rh = V 2 /P = (100)2 /1000 = 10 Ω. The potential difference across the heater for it to operate at P 0 = 62.5 W is given by p p √ Vh = P 0 Rh = (62.5)(10) = 625 = 25 V. The current through the heater is ih = Vh /10 = 2.5 A. ih

10Ω i

Heater

ir R 100V

The potential difference across 10 Ω resistance connected in series to the heater is V = 100 − Vh = 100 − 25 = 75 V and current through it is i = V /10 = 7.5 A. The current flowing through the resistance R connected in parallel to the heater is ir = i−ih = 7.5−2.5 = 5 A. The potential difference across this resistor is 25 V. Hence, R = 25/5 = 5 Ω. Ans. 5 Ω

417

Chapter 31 Magnetic Field

y

One Option Correct B0

Q 1. A thin flexible wire of length L is connected to two adjacent fixed points and carries current I in the clockwise direction as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is (2010)

0

(A) z

(C)

(C)

IBL 2π

(D)

(B) x

a 2a 3a

3a

x

(D) a 2a 3a

z 0

z 0

IBL π

2a

-B0

0

(A) IBL (B)

a

x

a 2a 3a

x

z 0

a 2a 3a

x

IBL 4π

Sol. The magnetic force on a positive charge q, when it enters the region a < x < 2a, is given by

Sol. Consider a small arc of circular loop subtending an angle 2δ at its centre.

ˆ ~ = qv0ˆı × B0 ˆ = qv0 B0 k. F~m = q~v × B F T cos δ

δ

=

=

δ

δ

δ

T sin δ

T sin δ

T

This force provides the centripetal acceleration for the particle to move along a circle of radius r in x-z plane i.e., qv0 B0 = mv 2 /r, which gives r = mv0 /(qB0 ).

T cos δ T

z

O

The length of the arc is Lδ/π. The forces acting on the arc are tension T on both ends in the tangential direction and magnetic force F in a radially outward direction. The magnetic force is F = IB(Lδ/π). Under the equilibrium condition,

2a

3a

x

If r < a then particle will never come out of region a < x < 2a. Let r > a and particle comes out of this region. The direction of magnetic field reverses in region 2a < x < 3a which causes magnetic force to reverse its direction. The particle changes its trajectory to a circle having equal radius and same tangent at x = 2a (see figure). Ans. A

2T sin δ ≈ 2T δ = IB(Lδ/π), which gives T =

a

ILB 2π .

Ans. C ~ = B0 ˆ exists in the Q 2. A magnetic field vector B ~ region a < x < 2a and vector B = −B0 ˆ, in the region 2a < x < 3a, where B0 is a positive constant. A positive point charge moving with a velocity vector ~v = v0ˆı, where v0 is a positive constant, enters the magnetic field at x = a. The trajectory of the charge in this region can be like (2007)

Q 3. An electron moving with a speed u along the positive x-axis at y = 0 enters a region of uniform magnetic ~ = −B0 kˆ which exists to the right of y-axis. The field B electron exits from the region after sometime with the speed v at coordinate y, then (2004) 418

Chapter 31. Magnetic Field

419

y e- u

x

•

(A) v > u, y < 0 (C) v > u, y > 0

Sol. Let the initial velocity of the particle be ~v = vˆı. ~+ The force acting on the particle is given by F~ = q E ~ The forces in four given cases are q~v × B. ˆ F~(A) = −qvc ˆ + qvb k, F~(B) = qa ˆı − qvc ˆ, ˆ F~(C) = −qvb ˆ + qvc k,

(B) v = u, y > 0 (D) v = u, y < 0

~ is always Sol. The magnetic force, F~B = q~v × B, perpendicular to the displacement vector making work done by F~B zero. By work-energy theorem, the kinetic energy of electron does not change and hence v = u. ˆ = −euB0 ˆ, is ~ = −euˆı × (−B0 k) Now, F~B = q~v × B towards the negative y-axis which makes final position y < 0. The electron travels in a circular path with radius mu/(eB0 ). Ans. D Q 4. A conducting loop carrying a current I is placed in a uniform magnetic field pointing into the plane of the paper as shown. The loop will have a tendency to (2003) y ~ B ⊗

ˆ F~(D) = qa ˆı − qvc ˆ + qvb k. Given motion is possible in case (B) only. Ans. B Q 6. A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a region containing a uniform magnetic field B directed along the negative z direction, extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is (2002)

(A)

qbB m

(B)

q(b−a)B m

(C)

qaB m

(D)

q(b+a)B 2m

Sol. The magnetic force, qvB, provides the necessary centripetal acceleration for the circular path of radius r, i.e., qvB = mv 2 /r. The particle will just enter the region x > b if tangent to the path at A is along the y axis.

x I 2r

(A) (B) (C) (D)

contract expand move towards +ve x axis move towards −ve x axis

A y x

v (b − a)

Sol. Consider a small element of length d~l. The force ~ on d~l is radially outward and thus the loop (Id~l × B) has a tendency to expand. Ans. B Q 5. For a positively charged particle moving in a x-y plane initially along the x-axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond P. The curved path is shown in the x-y plane and is found to be non-circular. Which of the following combination is possible? [Here a, b and c are non-zero positive constants.] (2003)

This condition gives r = (b − a). Eliminate r to get v = q(b−a)B . If particle velocity is greater than q(b−a)B m m then the particle will enter region x > b otherwise it will come back. Ans. B Q 7. Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB respectively and the trajectories are as shown in the figure. Then, (2001)

y A B O

(A) (C)

~ = 0, E ~ E = 0,

~ = bˆ B  + ckˆ ~ B = cˆ  + bkˆ

P

(B) (D)

x

~ = aˆı, B ~ = ckˆ + bˆı E ~ ~ = ckˆ + bˆ E = aˆı, B 

(A) (B) (C) (D)

mA vA < mB vB mA vA > mB vB mA < mB and vA < vB mA = mB and vA = vB

420

Part V. Electromagnetism

Sol. The magnetic force provides the centripetal accelA vA eration i.e., qvB = mv 2 /r, which gives rA = mqB mB vB and rB = qB . In figure, rA > rB , and hence mA vA > mB vB . Ans. B Q 8. An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along +z direction, then (2000) (A) positive ions deflect towards +y direction and negative ions towards −y direction. (B) all ions deflect towards +y direction. (C) all ions deflect towards −y direction. (D) positive ions deflect towards −y direction and negative ions towards +y direction. Sol. Initially, velocity is zero so magnetic force F~B = ~ = ~0. q~v × B

FE FB

v

• −q

z

Q 11. In the formula X = 3Y Z 2 , X and Z have dimensions of capacitance and magnetic induction, respectively. The dimensions of Y in MKSQ system is (1995) (A) [M−3 L−1 T3 Q4 ] (B) [M−3 L−2 T4 Q4 ] (C) [M−2 L−2 T4 Q4 ] (D) [M−3 L−2 T4 Q] Sol. The dimensions of capacitance are C = 2

Q QV F vQ

= =

2

[Q ] [ML2 T−2 ] [MLT−2 ] [LT−1 Q]

+q •

Q V

=

and of magnetic induction are B = = [MT−1 Q−1 ]. [X] [Z 2 ]

=

[C] [B 2 ]

Thus, the dimen=

[M−1 L−2 T2 Q2 ] [M2 T−2 Q−2 ]

=

Ans. B

FB v x

~ B

2

have used K = 21 mv 2 = (mv) 2m . Using, md = 2mp , mα = 4m√ p , qd = qp , qα = 2qp and Kp = Kd = Kα , we get, rd = 2 rp and rα = rp . Ans. A

sions of Y are [Y ] = [M−3 L−2 T4 Q4 ].

y FE

Sol. The magnetic force on a charge q, moving with a velocity v in a normal magnetic field B, is qvB. It provides the centripetal acceleration to√the particle i.e., 2 2mK which gives r = mv qvB = mv r qB = qB , where we

~ E

~ on the charge +q is The electrostatic force F~E = q E in +x direction, provides it a velocity towards +x. Now, magnetic force starts acting in −y direction causing the charge to deflect towards −y. The electrostatic force on the charge −q is in −x direction, provides it a velocity towards −x. The magnetic force on it is in −y direction deflecting it towards −y. Ans. C Q 9. A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a

Q 12. Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of the mass of X to that of Y is (1988) 1/2 (A) (R1 /R2 ) (B) R2 /R1 2 (C) (R1 /R2 ) (D) R1 /R2 Sol. The kinetic energy of a particle of mass m and charge q, after being accelerated through a potential difference V , is given by qV = 12 mv 2 .

(1)

The magnetic force qvB provides the centripetal acceleration for motion in a circle of radius R i.e., qvB = mv 2 /R.

(2)

(1999)

(A) straight line (B) circle (C) helix (D) cycloid ~ accelerates the particle Sol. The electrostatic force q E ~ Thus, the velocity ~v is parallel to in the direction of E. ~ which in turn is parallel to magnetic field B. ~ MagE ~ = ~0 and the particle netic force on the particle is q~v × B ~ moves along a straight line parallel to E. Ans. A Q 10. A proton, a deuteron and an α-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp , rd and rα denote respectively the radii of the trajectories of these particles, then (1997) (A) rα = rp < rd (B) rα > rd > rp (C) rα = rd > rp (D) rp = rd = rα

Eliminate v from equations (1) and (2) to get m = qB 2 R2 /V.

(3)

The particles X and Y are of equal charge q, accelerated through the same potential V , and are placed in the same magnetic field B. Thus, from equation (3), the ratio of the mass of X and Y is mX /mY = (R1 /R2 )2 . Ans. C Q 13. A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic ~ 0 such that B ~ 0 is perpendicular to the plane of field B the loop. The magnetic force acting on the loop is (1983)

~ 0 (B) 2πirB ~ 0 (C) zero (D) πirB ~0 (A) irB

Chapter 31. Magnetic Field

421

Sol. The force, on a current carrying conductor of ~ is given by F~ = i d~l× B. ~ length d~l in a magnetic field B, Consider the two small elements, each of length d~l, at two diametrically opposite points P and Q.

mass, speed, and linear momentum of the charged particle then radius r of the circular path is given by

i

When the charged particle enters the magnetic field ~ on it is along at point P1 , the magnetic force, q~v × B, +y direction. The centre of the circular path, C0 , will lie on the y-axis (at y > −R) because the direction of force at P1 is along +y-direction (velocity direction at P1 is tangential to the circular path). The particle will re-enter the region 1 if radius of the circular path is less than OP2 i.e., r < 23 R. Substitute r from equation (1) into this inequality to get p B > 23 qR . Thus, if magnetic field is more than a threshold value then the particle completes a semicircular path in region 2. The position of the centre of the semicircular path C0 and its radius C0 E0 are governed by the magnetic field. The point of re-entry into the region 1, E0 , lies at a distance r above the centre C0 i.e.,

~ dl

~0 B ⊗

Q

~ F

~ F

•

r

r

O

P

~ dl

The forces on these elements are equal and opposite giving net force zero. Similarly, forces on the other diametrically opposite pair of elements cancel out. Hence, net force on the loop is zero. Ans. C

r=

p mv = . qB qB

One or More Option(s) Correct

2mv . qB

P1 E0 = 2r = Q 14. A uniform magnetic field B exists in the region between x = 0 and x = 3R/2 (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge +q and momentum p directed along xaxis enters region 2 from region 1 at point P1 (y = −R). Which of the following option(s) is(are) correct? (2017)

(1)

For fixed q, v and B, the distance P1 E0 is directly proportional to the mass m of the particle. Region 1

y

Region 2

E1 Region 1

y

Region 2

B

Region 3 E0

B

O

Region 3

P2

x 5 R 8

+q • P1 (y = −R) 3R/2

(A) When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between √ point P1 and the farthest point from y-axis is p/ 2. 8 p (B) For B = 13 qR , the particle will enter region 3 through the point P2 on x-axis. p (C) For B > 23 qR , the particle will re-enter region 1. (D) For a fixed B, particle of same charge q and same velocity v, the distance between the point P1 and the point of re-entry into region 1 is inversely proportional to the mass of the particle. Sol. In a uniform magnetic field B, a charged particle travels in a circular path when it is projected perpendicular to direction of the field. If q, m, v, and p are charge,

p ~f

C2 C1 C0 O

13 R 8

R 2 3 R 2

D

P2

x

R

+q P1

p ~i

The particle re-enters the region 1 through the longest path if r = 32 R. In this case, the coordinates of the centre C1 are x = 0 and y = r − R = 12 R. The coordinates of the farthest point from the y-axis, D, are x = 32 R and y = R2 . The linear momentum of the particle at P1 is p~i = pˆı and that at D is p~f = pˆ . Since magnetic force acts perpendicular to the direction of linear momentum, it only changes the direction of linear momentum and not its magnitude. In other words, the magnitude of the linear momentum remains constant because work done by the magnetic force is zero. The change in linear momentum is

422

Part V. Electromagnetism

∆~ p = p~i − p~f = pˆı − pˆ . The √ magnitude of change in linear momentum is |∆~ p| = 2p. 8 p Substitute B = 13 qR in equation (1) to get r = 13 3 8 R, which is more than 2 R. Thus, the particle will enter the region 3. In this case, the coordinates of the centre C2 are x = 0 and y = r − R = 58 R. Apply Pythagoras theorem in right angled triangle C2 OP2 to get C2 P2 = 13 8 R, which is equal to the radius of circular path. Thus, the point P2 lies on the circular path and the particle enters the region 3 through P2 . Ans. (B), (C) Q 15. A conductor (shown in the figure) carrying constant current I is kept in the x-y plane in a uniform ~ If F is the magnitude of the total magnetic field B. magnetic force acting on the conductor, then the correct statement(s) is (are) (2015)

I L

R

R π/4

Sol. The path followed by the particle is a circular arc PQ with velocities ~u1 and ~u2 along the tangents at P and Q. y

y R π/6

Q 16. A particle of mass m and positive charge q, moving with a constant velocity ~u1 = 4ˆı m/s, enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the √ other side
after 10 milliseconds with a velocity ~u2 = 2 3ˆı + ˆ m/s. The correct statement(s) is (are) (2013) (A) The direction of the magnetic field is −z direction. (B) The direction of the magnetic field is +z direction. (C) The magnitude of the magnetic field is 50πm 3q units. (D) The magnitude of the magnetic field is 100πm 3q units.

x

⊗B ~

O θ

R

L ~ u2

(A) (B) (C) (D)

If If If If

~ B ~ B ~ B ~ B

is is is is

along along along along

zˆ, F x ˆ, F yˆ, F zˆ, F

∝ (L + R) =0 ∝ (L + R) =0

q P

a e c I a

b L

x

π/4 d

R

y

R

R π/6

√ The ~u2 = 2( 3 ˆı + ˆ) gives θ = π/6. For this ~ is in −z direction so that motion, magnetic field (B) ~ provides the necessary centripetal acceleration. q ~v × B 2 From Newton’s second law, qvB = mv r , which gives ω = vr = qB m , where v is velocity, ω is angular velocity and r is radius of the circular path. The angular displacement in time t is θ = ωt = qB m t. Substitute for θ πm and t to get B = 50 . 3 q Ans. A, C

g

f R

x ~ u1 L

Sol. The force on a conducting element of length d~l, ~ is given by carrying a current I in a magnetic field B, ~ If the field is uniform, the total force on dF~ = I d~l × B. the conductor is given by Z g  Z g → ~ =I ~ = I ag ~ (1) F~ = Id~l × B d~l × B ×B. a

θ

L

~ is taken out of the integral sign beNote that B cause it is constant. The vector ag ~ = 2(L + R) x ˆ. The magnetic forces on the conductor in the given cases are Case (A):

F~ = I(2(L + R) x ˆ) × (B zˆ)

Case (B):

= −2IB(L + R) yˆ, F~ = I(2(L + R) x ˆ) × (B x ˆ) = ~0,

Case (C):

F~ = I(2(L + R) x ˆ) × (B yˆ) = 2IB(L + R) zˆ.

We encourage you to show the above results by integrating equation (1) for the given cases. Ans. A, B, C

Q 17. Consider the motion of a positive charge in a region where there are simultaneous uniform electric and ~ = E0 ˆ and B ~ = B0 ˆ. At time t = 0, magnetic fields E this charge has velocity v in the x-y plane making an angle θ with the x axis. Which of the following option(s) is (are) correct for t > 0? (2012) (A) If θ = 0, the charge moves in a circular path in the x-z plane. (B) If θ = 0, the charge undergoes helical motion with constant pitch along y axis. (C) If θ = 10◦ , the charge undergoes helical motion with its pitch increasing with time, along the y axis. (D) If θ = 90◦ , the charge undergoes linear but accelerated motion along y axis. Sol. Let ~v = v cos θ ˆı + v sin θ ˆ be the velocity of the particle.

Chapter 31. Magnetic Field y

423 Q 19. A particle of mass m and charge q, moving with velocity v enters Region II to the boundary as shown in the figure. Region II has a uniform magnetic field B perpendicular to the plane of the paper. The length of the region II is l. Choose the correct choice(s). (2008)

B0 E0

=

v sin θ

v

θ v cos θ

x

~ The total force on the charged particle, due to E ~ is given by and B,

Region I

Region II

Region III

v

~ + qE ~ F~ = q~v × B = q(v cos θ ˆı + v sin θ ˆ) × B0 ˆ + qE0 ˆ = qvB0 cos θ kˆ + qE0 ˆ. ~ gives rise to The first component of force (due to B) circular motion about y-axis and the second component ~ causes accelerated linear motion along y(due to E) axis. When 0◦ ≤ θ < 90◦ , the particle undergoes helical motion with its pitch increasing with time. If θ = 90◦ , the magnetic force is zero and the charge undergoes linear accelerated motion along y-axis. We encourage θ , you to show that the radius of the helix is r = mvqBcos 0 2πm time period is T = B0 , and pitch of the helix at time qE0 2 0 t is p = v sin θ T + 12 qE m T + m T t. Ans. C, D Q 18. An electron and a proton are moving on straight parallel paths with same velocity. They enter into semiinfinite region of uniform magnetic field perpendicular to their velocity. Which of the following statement(s) is (are) true? (2011) (A) They will never come out of magnetic field region. (B) They will come out traveling along parallel paths. (C) They will come out at the same time. (D) They will come out at different times.

2rp

Sol. In the region of magnetic field, both particles travel in a circular path (see figure) with magnetic force providing the necessary centripetal acceleration i.e., qvB = mv 2 /r.

2re

p e

m v

p The radius of proton path is rp = qB and of elecme v tron path is re = qB . The times spent by the two πr πm particle in the magnetic field are tp = v p = qBp πre πme and te = v = qB . Since mp ≈ 1840me , we get tp ≈ 1840te . Ans. B, D

l

(A) The particle enters region III only if its velocity v > qlB/m. (B) The particle enters region III only if its velocity v < qlB/m. (C) Path length of the particle in Region II is maximum when v = qlB/m. (D) Time spent in Region II is same for any velocity v as long as the particle returns to Region I. Sol. The particle undergoes circular motion in the region II.

I

•

II

III

v l

The magnetic force provides the centripetal acceleration i.e., qvB = mv 2 /r which gives r = mv/(qB). The particle will enter the region III only if radius r > l which gives v > qlB/m. As seen from the figure, path length is maximum when r = l i.e., v = qlB/m. The time spent in the region II, t = πr/v = πm/(qB), is independent of v. Ans. A, C, D Q 20. An infinitely long wire carrying current I1 passes through O and is perpendicular to the plane of paper. Another current carrying loop ABCD lies in plane of paper as shown in the figure. Which of the following statement(s) is (are) correct? (2006)

424

Part V. Electromagnetism (B) O2+ will be deflected most. (C) He+ and O2+ will be deflected equally. (D) all will be deflected equally.

C B I1 O

I2 O0

Sol. The magnetic force provides the centripetal acceleration to a charge q, moving with speed v, in a circular path of radius r i.e., qvB = mv 2 /r. The radius r is related to kinetic energy K by √ r = mv/qB = 2mK/(qB).

A D

(A) net force on the loop is zero. (B) net torque on the loop is zero. (C) loop will rotate clockwise about axis OO0 when seen from O. (D) loop will rotate anticlockwise OO0 when seen from O. ~ by the current I1 is circumSol. The magnetic field B ~ is ferential. For each element d~l on the branch AB, B parallel to the current direction, so the force F~AB =

Z

B

~ = ~0. I2 d~l × B

A

~ is anti-parallel to d~l. Thus, the For the branch CD, B total force on branch CD is F~CD = ~0.

dl B dl

I1 O

O0

B

The particle, moving in minimum radius path, undergoes maximum deflection. Thus, the deflection of H+ is maximum and of He+ and O2+ are equal. Ans. A, C Q 22. A particle of charge +q and mass m moving under the influence of a uniform electric field Eˆı and uniform magnetic field B kˆ follows a trajectory from P to Q as shown in figure. The velocities at P and Q are vˆı and −2vˆ . Which of the following statement(s) is (are) correct? (1991)

C

B B

Substitute the values of m and q for the given particles to get p √ 2(1)K 2K rH+ = = , (1)B B p √ 2(4)K 2K rHe+ = =2 , (1)B B p √ 2(16)K 2K rO2+ = =2 . (2)B B

dl B

A dl

y

D

P

~ is perpendicular to d~l, and For the branch BC, B ~ is coming out of the paper, so F~BC = FBC , d~l × B where FBC is the magnitude of force. For the branch ~ is going into the paper, so F~DA = FDA ⊗, DA, d~l × B where FDA is the magnitude of force. By symmetry, F~BC = −F~DA . Thus, the net force acting on ABCDA is zero. However, there is non-zero clockwise (when looking from O) torque that rotates the loop in clockwise direction. We encourage you to show that   Z r2 r2 µ0 I1 I2 ln . FBC = I2 B(r)dr = 2π r1 r1 Ans. A, C Q 21. H+ , He+ and O2+ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of H+ , He+ and O2+ are 1 u, 4 u and 16 u respectively. Then, (1994) (A) H+ will be deflected most.

~ E v

B ~

a Q

x

2a 2v

(A) E =

3 4

h

i 2

mv qa

.

(B) Rate h ofi work done by the electric field at P is 3 mv 3 . 4 a (C) Rate of work done by the electric field at P is zero. (D) Rate of work done by both the fields at Q is zero. Sol. The electric and magnetic forces on the charged ~ The particle are given by F~e = qE ˆı and F~m = q~v × B. work done by the electric force in making a displacement ~r = ~rQ − ~rP = 2a ˆı − a ˆ is We = F~e · ~r = qE ˆı · (2a ˆı − a ˆ) = 2qaE.

(1)

The work done by the magnetic force is zero. By work-energy theorem, the work done on the particle is

Chapter 31. Magnetic Field

425

equal to the change in its kinetic energy i.e.,

Paragraph Type Paragraph for Questions 24-26

We = KQ − KP = 12 m(2v)2 − 12 mv 2 = 32 mv 2 .

(2)

Solve equations (1) and (2) to get E = 3mv 2 /(4qa). The rate of work done (power) by the magnetic force is zero. The rates of work done by the electric force at the points P and Q are

A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0) with a given ~ and a uniform velocity ~v . A uniform electric field E ~ magnetic field B exists everywhere. The velocity ~v , elec~ and magnetic field B ~ are given in columns tric field E 1, 2 and 3, respectively. The quantities E0 and B0 are positive in magnitude. (2017)

F~e · ~vP = (qE ˆı) · (v ˆı) = qEv = 3mv 3 /(4a), F~e · ~vQ = (qE ˆı) · (−2v ˆ) = 0.

Column 1

Ans. A, B, D Q 23. A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields, respectively. Then, the region of space may have (1985)

(B) E = 0, B 6= 0 (D) E = 6 0, B 6= 0

(A) E = 0, B = 0 (C) E = 6 0, B = 0

Sol. The force on a proton moving with a velocity ~v , ~ and magnetic field B, ~ is in a region of electric field E given by ~ + q~v × B. ~ F~ = F~e + F~m = q E

(1)

If both E and B are zero then there is no force acting on the proton. Thus, acceleration of the proton is zero. The proton keeps moving without any change in its velocity. ~ = ~0 and B ~ k ~v then equation (1) gives F~ = ~0. If E The proton keeps moving with the constant velocity. ~ 6= ~0 and B ~ = ~0 then F~ = q E ~ 6= ~0. Thus, If E acceleration of the particle is non-zero i.e., velocity of the proton changes with time.

E0 Electron, ~v = 2 B x ˆ 0 E0 Electron, ~v = B0 yˆ Proton, ~v = 0 E0 Proton, ~v = 2 B x ˆ 0

Q 24. In which case straight line along the moves along −ˆ y )? (A) IV, ii, S (B) (C) III, ii, R (D)

(i) (ii) (iii) (iv)

E0 zˆ −E0 yˆ −E0 x ˆ E0 x ˆ

Column 3 ~ (B) (P) (Q) (R) (S)

−B0 x ˆ B0 x ˆ B0 yˆ B0 zˆ

would the particle move in a negative direction of y-axis (i.e., II, iii, Q III, ii, P

Sol. Lorentz force on a charged particle having a charge q, moving with a velocity ~v in a region of uniform elec~ and uniform magnetic field B, ~ is given by tric field E ~ + q~v × B. ~ F~net = F~E + F~B = q E

(1)

The particle will move along −ˆ y axis if (i) initial velocity ~v is along −ˆ y and net force is either zero or directed along −ˆ y (ii) initial velocity ~v is zero and net force is directed along −ˆ y . A proton with initial velocity ~v = ~0 ~ = −E0 yˆ and magnetic field placed in an electric field E ~ = B0 yˆ will experience a net force F~net = −qE0 yˆ. It B moves in a straight line along −ˆ y. Ans. (C) Q 25. In which case will the particle move in a straight line with constant velocity? (A) II, iii, S (B) III, iii, P (C) IV, i, S (D) III, ii, R

~ E ~e F ~v ~ B

(I) (II) (III) (IV)

Column 2 ~ (E)

~m F

~ 6= ~0 and B ~ 6= ~0, there are non-zero forces If E ~ and F~m = q~v × B. ~ However, if directions and F~e = q E ~ and B ~ are adjusted in such a way that magnitudes of E ~ = −q~v × B, ~ the net force on the proton becomes zero qE (see figure). The proton keeps moving with a constant velocity. Ans. A, B, D

Sol. The particle will move in a straight line with constant velocity (zero acceleration) if net force on it is ~ = −~v × B. ~ An zero. Net Lorentz force is zero if E E0 electron with initial velocity ~v = B0 yˆ will move in a straight line with constant velocity when placed in an ~ = −E0 x ~ = B0 zˆ. electric field E ˆ and magnetic field B Ans. (A) Q 26. In which case will the particle describe a helical path with axis along the positive z direction? (A) II, ii, R (B) III, iii, P (C) IV, i, S (D) IV, ii, R

426

Part V. Electromagnetism

Sol. The particle will describe a helical path with axis along +z direction if (i) component of net force in x-y plane provides the centripetal acceleration and (ii) the particle has a non-zero velocity and/or non-zero force along +z direction. A proton with initial velocity E0 x ˆ will move in a helical path with axis along ~v = 2 B 0 ~ = E0 zˆ +z direction when placed in an electric field E ~ and magnetic field B = B0 zˆ. We encourage you to find the radius and pitch of the helix as a function of time. Ans. (C)

~ = Bˆ magnetic field B . Thus, the magnetic force on the moving electrons is ~ = (−e)(−vd ˆı) × (B ˆ) F~m = q~vd × B IB ˆ k. = evd B kˆ = nwd

(1)

l w

S I

d P

K

R

y I z

M

x

Q

Paragraph for Questions 27-28 In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are l, w and d, respectively. A uniform magnetic field ~ is applied on the strip along the positive y-direction. B Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.

Note that an electron has a negative charge q = −e. From equation (1), the electrons move towards the face PQRS which leads to a negative potential at this face. This potential establishes an electric field in a direction from K to M. If V is the potential difference ˆ ~ = (V /w) k. between these faces then electric field is E The electric force on the electrons is ˆ ~ = (−e) V kˆ = − eV k. F~e = q E w w

The accumulation of electrons on face PQRS continues till the net force on the electrons become zero (equilibrium condition). Use equations (1) and (2) to get the potential difference in the equilibrium condition

(2015)

V = l w

S I

d

K

R

P

IB . ned

(3)

y I z

M

(2)

x

Q

From equation (3), V depends on d and not on w. Thus, if d1 = d2 then V2 = V1 and if d1 = 2d2 then V2 = 2V1 . Ans. A, D

Q 27. Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1 and w2 and thicknesses are d1 and d2 , respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is (are) (A) If w1 = w2 and d1 = 2d2 , then V2 = 2V1 . (B) If w1 = w2 and d1 = 2d2 , then V2 = V1 . (C) If w1 = 2w2 and d1 = d2 , then V2 = 2V1 . (D) If w1 = 2w2 and d1 = d2 , then V2 = V1 .

Q 28. Consider two different metallic strips (1 and 2) of same dimensions (length l, width w and thickness d) with carrier densities n1 and n2 , respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in magnetic field B2 , both along positive y-directions. Then V1 and V2 are the potential differences developed between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips, the correct options(s) is (are) (A) If B1 = B2 and n1 = 2n2 , then V2 = 2V1 . (B) If B1 = B2 and n1 = 2n2 , then V2 = V1 . (C) If B1 = 2B2 and n1 = n2 , then V2 = 0.5V1 . (D) If B1 = 2B2 and n1 = n2 , then V2 = V1 .

Sol. The charge carriers in a metallic strip are electrons. The electrons move in a direction opposite to the current flow with a drift velocity given by

Sol. The ratio of potential difference in two strips is given by

vd =

I I = , neA newd

where n is the number density of the electrons (number of electrons per unit volume), e is the electronic charge, and A = wd is the cross-sectional area of the strip. The electrons move with velocity ~vd = −vd ˆı in a uniform

V2 B2 n1 = . V1 B1 n2 Substitute B1 = B2 and n1 = 2n2 to get V2 = 2V1 . Substitute B1 = 2B2 and n1 = n2 to get V2 = 0.5V1 . You may recall that the given paragraph is an explanation of Hall effect. Ans. A, C

Chapter 31. Magnetic Field

427

Matrix or Matching Type Q 29. Column I gives three physical quantities. Select the appropriate units for the choices given in Column II. Some of the physical quantities may have more than one choice. (1990) Column I (A) Capacitance (B) Inductance (C) Magnetic Induction

Column II

Sol. The magnetic force on a charged particle, moving in the magnetic field, is perpendicular to its ve~ The work done by the magnetic locity, F~ = q~v × B. force in a small time interval dt is dW = F~ · d~s = ~ · (~v dt) = 0. Thus, the work done by the mag(q~v × B) netic force on the charged particle is zero and its energy remains constant. Ans. T

(p) ohm-second (q) coulomb2 (joule)-1 (r) coulomb (volt)-1

Q 32. A charged particle enters a region of uniform magnetic field at an angle of 85◦ to the magnetic lines of force. The path of the particle is a circle. (1983)

(s) newton (ampere metre)-1 (t) volt-second (ampere)-1

Sol. The charged particle follows a circular path if its velocity is perpendicular to the magnetic field. In the given case, the charged particle will follow a helical path. We encourage you to derive the expressions for pitch and radius of the helix. Ans. F

2

Q Sol. The energy stored in a capacitor, U = 2C , gives 2 the unit of the capacitance C as coulomb (joule)-1 . Also, Q = CV , gives the unit of C as coulomb (volt)-1 . The time constant of L-R circuit, τ = L/R, gives the unit of inductance L as ohm-second. The induced voltdi age across an inductor, e = −L dt , gives the unit of L as -1 volt-second (ampere) . The force on a current carrying conductor in the magnetic field, F = ilB, gives the unit of magnetic field B as newton (ampere metre)-1 . Ans. A7→(q,r), B7→(p,t), C7→s

Q 33. No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field. (1981) Sol. Let the magnetic field B be coming out of the paper and current i, in rectangular loop of length l and width w, flows in the counter-clockwise direction. FCD D

True False Type Q 30. An electron and a proton are moving with the same kinetic energy along the same direction. When they pass through a uniform magnetic field perpendicular to the direction of their motion, they describe circular paths of the same radius. (1985) Sol. The magnetic force on a charge q, moving with the velocity v in a perpendicular magnetic field B, provides the centripetal acceleration for circular motion i.e., qvB = mv 2 /r, which gives the radius of the circle as √ mv 2mK r= = , qB qB where K = 21 mv 2 is the kinetic energy of the particle. Thus, the radius of the circular paths of the proton and the electron, are in the ratio rp /re =

q

C y

mp /me 6= 1. Ans. F

Q 31. There is no change in the energy of a charged particle moving in magnetic field although a magnetic force is acting on it. (1983)

x

FAD

A

FBC

B FAB

The magnetic forces on the four branches of the loop are F~AB = −ilB ˆ, F~BC = iwB ˆı,

F~CD = ilB ˆ, F~DA = −iwB ˆı.

Net force on the loop is F~AB + F~BC + F~CD + F~DA = ~0. Ans. T Fill in the Blank Type Q 34. A uniform magnetic field with a slit system as shown in the figure is to be used as momentum filter for high-energy charged particles. With a field B Tesla, it is found that the filter transmits α-particles each of energy 5.3 MeV. The magnetic field is increased to 2.3B Tesla and deuterons are passed into the filter. The energy of each deuteron transmitted by the filter is . . . . . . MeV. (1997)

428

Part V. Electromagnetism z

D I

C

E Source

Detector

Sol. The magnetic force provides the centripetal acceleration to a charge q moving with speed v in a circular path of radius r i.e., qvB = mv 2 /r. The radius r is related to the kinetic energy K of the particle by √ r = mv/(qB) = 2mK/(qB). Thus, radii of the circular paths for the α particle and the deuteron are p p rα = 2mα Kα /(qα Bα ), rd = 2md Kd /(qd Bd ).

B x

F y

O A

Sol. The current in branch AB (EF) is parallel (antiparallel) to the field direction so forces on these branches are zero. The magnetic forces on the branches BC and DE are equal in magnitude but opposite in direction and hence cancel out. The force on branch CD is given by ˆ ~ ×B ~ = I(L ˆı) × (B ˆ) = ILB k. F~CD = I L Ans. ILB, +ve z

The slit acts as a filter for both the particles if rα = rd , which gives  q 2  B 2  m   1 2  2.3 2  4  d d α Kd = Kα = 5.3 qα Bα md 2 1 2 = 14.0185 MeV.

Q 37. A neutron, a proton, an electron and an alpha particle enter a region of constant magnetic field with equal velocities. The magnetic field is along the inward normal to the plane of the paper. The tracks of the particles are labeled in the figure. The electron follows track . . . . . . and the alpha particle follows track . . . . . .

Ans. 14.0185

(1984)

Q 35. A metallic block carrying current I is subjected ~ as shown in the figto a uniform magnetic induction B ure. The moving charges experience a force F~ given by . . . . . . which results in the lowering of the potential of the face . . . . . . . Assume the speed of the charges to be v. (1996) y

~ B

E F

A C

z

G B

H x I

D

Sol. The charge carriers in a metal are free electrons. The charge on each electron is q = −e = −1.6 × 10−19 C. The electrons move opposite to the current direction I with a velocity ~v = −v ˆı. The magnetic force on an electron is given by ˆ ~ = −e(−v ˆı) × (B ˆ) = evB k. F~ = q~v × B Thus, the electrons (negative charge) move towards the z-axis i.e., face ABCD. Accumulation of negative charges on the face ABCD lowers its potential. ˆ ABDC Ans. evB k, Q 36. A wire ABCDEF (with each side of length L) bent as shown in the figure and carrying a current I is placed in a uniform magnetic induction B parallel to the positive y direction. The force experienced by the wire is . . . . . . in the . . . . . . direction. (1990)

C

B

A D

Sol. The magnetic force on a charge q, moving with a ~ is given by F~m = q~v ×B. ~ velocity ~v in a magnetic field B, The magnetic force on neutron (neutral) is zero and it follows the path C. In the given magnetic field, the magnetic force on the positive charges is towards the left and on the negative charges is towards the right. Thus, the electron (negative charge) follows the path D. The magnetic force provides the centripetal acceleration for circular motion of the charged particle i.e., mv 2 /r = qvB, which gives r = mv/(qB).

(1)

The mass of alpha particle is four times the mass of proton, mα = 4mp , and the charge on alpha particle is two times the charge on proton, qα = 2qp . Substitute these values in equation (1) to get rα mα v/(qα B) = = 2. rp mp v/(qp B) Thus, alpha particle follows the path B and proton follow the path A. Note that except C the given paths are of circular shapes with different radii. Ans. D, B

Chapter 31. Magnetic Field

429

Integer Type

The time taken by the particle to travel from P to

Q 38. In the x-y plane, the region y > 0 has a uniform magnetic field B1 kˆ and the region y < 0 has another ˆ A positively charged partiuniform magnetic field B2 k. cle is projected from the origin along the positive y-axis with speed v0 = π m/s at t = 0, as shown in the figure. Neglect gravity in this problem. Let t = T be the time when the particle crosses the x-axis from below for the first time. If B2 = 4B1 , the average speed of the particle, in m/s, along the x-axis in the time interval T is . . . . . . . (2018) y

Q is T2 = πr2 /v0 = πm/(qB2 ). At Q, the particle crosses x-axis from below for the first time. Thus, T = T1 + T2 . The average speed of the particle along the x-axis in the time interval T is given by R T1 R T1 +T2 RT v dt + vx2 dt v dt x1 x 0 T1 = v= 0 T T + T2  1  2mv0 1 1 q B1 + B2 2r1 + 2r2   = = πm 1 1 T1 + T2 + q B1 B2

B1

=

v0 = π m/s x

•

B2

Sol. The charged particle enters the magnetic field region at O. Its velocity is perpendicular to the field direction. The magnetic force on the particle, qv0 B1 , provides centripetal acceleration to move on a circular path from O to P. The radius of the circular path is given by r1 = mv0 /(qB1 ). y

Q 39. A proton and an alpha particle, after being accelerated through same potential difference, enter uniform magnetic field, the direction of which is perpendicular to their velocities. Find the ratio of radii of the circular paths of the two particles. (2004)

r1 v0 Q • O1

•

O2 •

•

P

x

r2 v0

B2

The time taken by the particle to travel from O to P is T1 = πr1 /v0 = πm/(qB1 ). At P, the particle enters the magnetic field B2 with its velocity perpendicular to the field direction. The magnetic force qv0 B2 provides the centripetal acceleration to move in a circular path from P to Q. The radius of the circular path is r2 = mv0 /(qB2 ).

where vx is the magnitude of the x-component of the particle velocity. Can you find the average velocity of the particle in the time interval T ? You can see that particle is effectively moving along x axis. Suppose you injected the particle in the magnetic field at O and then went for a walk. You came back after time t ( T ). Where will you find the particle? On another day, you forget to carry the watch. After coming back from the walk you found the particle on the x axis at a distance x from O. Can you find the time? Ans. 2 Descriptive

B1

O•

2v0 2(π) = = 2 m/s, π π

Sol. Let a particle of charge q and mass m is accelerated through a potential V . In this process, the electrical potential energy qV is converted to the kinetic energy of the particle i.e., qV = 12 mv 2 . Thus, the vep locity of the accelerated particle is v = 2qV /m. The magnetic force provides the centripetal acceleration for the charge to move in a circular path of radius r i.e., qvB = mv 2 /r. The radius r is s mv 1 2mV r= = . qB B q Use this equation to get the ratio of the radii of circular paths of the proton and the alpha particle as r r rp mp qα 1 2 1 = = · =√ . rα mα qp 4 1 2 √ Ans. 1/ 2

430

Part V. Electromagnetism

Q 40. A ring of radius R having uniformly distributed charge Q is mounted on a rod suspended by two identical strings. The tension in strings in equilibrium is T0 . Now a vertical magnetic field is switched on and ring is rotated at constant angular velocity ω. Find the maximum ω with which the ring can be rotated if the strings (2003) can withstand a maximum tension of 3T2 0 .

the vertically upward direction as the z-axis. A uniform ˆ exists in the region. The ~ = B0 (3ˆı +4k) magnetic field B loop is held in the x-y plane and a current I is passed through it. The loop is now released and is found to stay in the horizontal position in equilibrium. (2002) z

D Q

P T0

T0

a S

ω0

x

~ B

Sol. Initially, forces acting on the rod-ring system are weight mg and tension 2T0 . In equilibrium, mg = 2T0 . The ring takes time T = 2π/ω to complete one revolution, setting a current i and magnetic moment µ, given by i=

Q Qω = , T 2π

µ = iA =

1 ωQR2 . 2

D

T1

y

R b

(a) What is the direction of the current I in PQ? (b) Find the magnetic force on the arm RS. (c) Find the expression for I in terms of B0 , a, b and m. Sol. The forces acting on the loop are its weight mg, magnetic force F~B , and the reaction F~R at the hinge axis. In equilibrium, the resultant torque about the hinge point is zero. The torque about the hinge point by the force F~R is zero because it passes through the hinge point. The torque of weight mg about the hinge point is 
 . (1) ~τmg = a2 ˆı × −mg kˆ = 21 mgaˆ

T2

Let the current I in the loop be flowing in anticlockwise direction. The magnetic moment of the loop ˆ Torque on the loop by the magnetic force is µ ~ = abI k. ~ FB is

ω0

~ B

The direction of µ is towards the right. The torque ~ = 1 ωBQR2 (anticlockwise). on the loop is τ = |~ µ × B| 2 To balance this torque, tensions in the strings change to new values T1 and T2 such that T1 > T2 . In equilibrium, the resultant force and resultant torque about centre of the disc are zero i.e., T1 + T2 = mg, D D 1 T1 − T2 = ωBQR2 . 2 2 2

ˆ × (3B0 ˆı + 4B0 k) ˆ ~ = (abI k) ~τB = µ ~ ×B = 3abIB0 ˆ.

(2)

In equilibrium, resultant torque on the loop, ~τmg + ~τB , is zero. Using equations (1) and (2) we get ~τmg + ~τB = (mga/2 + 3abIB0 )ˆ  = ~0.

(1) (2)

Eliminate T2 from equations (1) and (2) and use mg = 2T0 to get T1 = T0 + ωBQR2 /(2D).

mg . The negative sign in I Solve to get I = − 6bB 0 indicates the clockwise direction of the current i.e., curmg flows from P to Q. rent 6bB 0 Magnetic force on the arm RS is given by

ˆ F~B,RS = (−Ib ˆ) × (3B0 ˆı + 4B0 k) ˆ = 1 mg(−4 ˆı + 3 k). 6

Maximum possible value of T1 is 3T0 /2 which gives DT0 ωmax = BQR 2. DT0 Ans. BQR 2 Q 41. A rectangular loop PQRS made from a uniform wire has length a, width b and mass m. It is free to rotate about the arm PQ, which remains hinged along a horizontal line taken as the y-axis (see figure). Take

Aliter: The force on the loop by Bz lies in the x-y plane. The torque of this force does not have any contribution in balancing ~τmg . The magnetic forces on PS and QR due to Bx are zero. The torque about the hinge point due to magnetic force on PQ is zero. Thus, the torque of magnetic force on RS (due to Bx ) balances ~τmg . This force, say F , shall be vertically up and hence

Chapter 31. Magnetic Field

431

current I in RS is from R to S. Also, F = IbBx = 3bIB0 . The torque balancing gives F a = mga/2 i.e., mg . F = mg/2. Solve to get I = 6bB 0
Ans. (a) P to Q (b) IbB0 3kˆ − 4ˆı (c) mg

sides parallel to x and y-axes. Each side of the frame is of mass M and length L. (1998) y

6bB0

(a) Find the value of L if the particle emerges from the region of magnetic field with its final velocity at an angle 30◦ to its initial velocity. (b) Find the final velocity of the particle and the time spent by it in the magnetic field, if the magnetic field now expands to 2.1L. Sol. Let the particle enters the region of magnetic field at the point P. The magnetic force when the particle just enters this region is given by ˆ = qv0 B0 ˆ. F~ = qv0ˆı × (−B0 k) This force is perpendicular to the velocity and moves the particle in a circular path. The centre of the circle, O, lies on the y axis at a distance r from P. The magnetic force provides the centripetal acceleration for the circular motion i.e., qv0 B0 = mv02 /r. The geometry gives, L/r = sin θ = sin 30◦ = 1/2. Solve to get r = 2L mv0 . and L = 2B 0q

I0

S

Q 42. The region between x = 0 and x = L is filled ˆ A particle of mass with uniform magnetic field −B0 k. m, positive charge q and velocity v0ˆı travels along x-axis and enters the region of the magnetic field. Neglect the gravity throughout the question. (1999)

R

~ B 45◦

x

O Q

P

(a) What is the torque ~τ acting on the frame due to the magnetic field? (b) Find the angle by which the frame rotates under the action of this torque in a short interval of time ∆t and the axis about which this rotation occurs (∆t is so short that any variation in the torque during this interval may be neglected). [Given: the moment of inertia of the frame about an axis through its centre perpendicular to its plane is 43 M L2 .] Sol. The magnitude of magnetic moment of the loop is I0 L2 and its direction is coming out of the paper (anticlockwise current). Thus, the magnetic moment is ˆ The magnetic field at the location of the µ ~ = I0 L2 k. loop is given by B ~ = B cos 45◦ ˆı + B sin 45◦ ˆ = √ B (ˆı + ˆ). 2 y ~ τ S

y

R

v0 x

L

P ~ ⊗B

O θ

The torque on the loop is given by

r

L2 B ~ = I0√ ~τ = µ ~ ×B (−ˆı + ˆ). 2

v0 L P

v0 L 2.1L

Q

L

θ x

If the magnetic field expands to the distance 2.1L then the particle completes a semicircle. The particle velocity when it comes out of magnetic field region is ~v = −v0 ˆı. Time spent by the particle in the field is πr 2πL πm v0 = v0 = B0 q . mv0 πm Ans. (a) 2B (b) −v0ˆı, B 0q 0q ~ is directed Q 43. A uniform constant magnetic field B ◦ at an angle of 45 to the x-axis in x-y plane. PQRS is rigid square wire frame carrying a steady current I0 , with its centre at the origin O. At time t = 0, the frame is at rest in the position shown in the figure with its

This torque will rotate the loop about axis SQ (see figure). Let ISQ be the moment of inertia of the loop about axis SQ. The symmetry gives ISQ = IPR . Using perpendicular axis theorem, ISQ + IPR = Izz = 34 M L2 , we get ISQ = 32 M L2 . The direction of angular acceleration of the loop is anticlockwise when looking from S to Q and its magnitude is given by α=

|~τ | I0 L2 B 3I0 B = = . ISQ 2M L2 /3 2M

The angle by which the frame rotates in time ∆t is θ=

1 3I0 B α(∆t)2 = (∆t)2 . 2 4M Ans. (a)

I0√ L2 B 2

(ˆ  − ˆı) (b)

3 I0 B 2 4 M (∆t)

432

Part V. Electromagnetism

Q 44. A particle of mass m and charge q is moving in a region where uniform constant electric and magnetic ~ and B ~ are present. E ~ and B ~ are parallel to each fields E other. At time t = 0, the velocity ~v0 of the particle is ~ (assume that its speed is always perpendicular to E  c, the speed of light in vacuum). Find the velocity ~v of the particle at time t. You must express your answer ~ and B ~ and their in terms of t, q, m, the vector ~v0 , E magnitude v0 , E and B. (1998) Sol. Let the coordinate system be as shown in the fig~ = E ˆ, B ~ = B ˆ, and ~v0 = v0 ˆı. Let ure. Here E ~v = vx ˆı + vy ˆ + vz kˆ be the velocity of the particle of charge q and mass m at time t. y ~ E

where we have used the initial conditions vz = 0 and dvz /dt = 0 to get C = 0. The velocity of the particle is given by qBt qEt qBt ˆ ˆı + ˆ + v0 sin k m m m !   ~ qt ~ qBt ~v0 × B qBt . ~v0 + E + sin = cos m m m B

~v = v0 cos

We encourage you to visualize the particle motion. The projected motion in x-z plane is circular with ω = qB/m. The particle moves in a helix with increasing pitch.      ~ qt ~ ~ v0 ×B ~v0 + m E + sin qBt Ans. cos qBt m m B Q 45. A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20 A, is in equilibrium at a height of 0.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady current of 30 A, as shown in the figure. Show that when AB is slightly depressed, it executes SHM. Find the period of oscillations. (1994)

~ B

x ~v0 z

A

Lorentz force, acting on the particle at time t, is given by ~ + q~v × B ~ = qE ˆ + qB(vx kˆ − vz ˆı) F~ = q E ˆ = −qBvz ˆı + qE ˆ + qBvx k. Newton’s second law gives the components of acceleration as dvx /dt = −(qB/m)vz = −ωvz ,

(1)

dvy /dt = qE/m,

(2)

dvz /dt = (qB/m)vx = ωvx ,

(3)

where ω = qB/m is a constant. Integrate equation (2) with the initial condition vy = 0 at t = 0 to get vy = qEt/m. The equations for vx and vz are coupled together. We use a trick to get vx and vz . Differentiate equation (1) and substitute dvz /dt from equation (3) to get d2 vx /dt2 = −ω (dvz /dt) = −ω 2 vx .

(4)

The equation (4) is similar to SHM equation and its solution is given by vx = A sin ωt + B cos ωt = v0 cos ωt,

(5)

where we have used the initial conditions vx = v0 and dvx /dt = 0 to get B = v0 and A = 0. Substitute vx from equation (5) into equation (3) to get dvz /dt = ωv0 cos ωt. Integrate to get vz = v0 sin ωt + C = v0 sin ωt,

B

C

D

Sol. Let m be the mass per unit length of the wire AB. The magnetic force between the wires is repulsive 0 i1 i2 and its magnitude per unit length is Fm = µ2πd . In µ0 i1 i2 equilibrium, Fm = mg i.e., 2πd = mg. Fm A

x

B

i2

d

mg C

i1

D

Let AB is slightly displaced by x such that its distance from CD becomes (d + x). The net force on the unit length of the wire AB is given by µ0 i1 i2 µ0 i1 i2  x −1 F = − mg = 1+ − mg 2π(d + x) 2πd d µ0 i1 i2  x ≈ 1− − mg 2πd d µ i i  µ i i 0 1 2 0 1 2 = − mg − x 2πd 2πd2 µ0 i1 i2 =− x. 2πd2 The restoring force is proportional to x and hence the wire executes SHM. Apply Newton’s second law, 2 0 i1 i2 m ddt2x = F = − µ2πd 2 x, to get d2 x µ0 i1 i2 g =− x = − x = −ω 2 x, 2 2 dt 2πd m d

Chapter 31. Magnetic Field

433

p where angular frequency ω = g/d p p and the time period T = 2π/ω = 2π d/g = 2 × 3.14 0.01/9.8 = 0.2 s. Ans. 0.2 s Q 46. An electron gun G emits electrons of energy 2 keV travelling in the positive x direction. The electrons are required to hit the spot S where GS = 0.1 m, and the line GS makes an angle of 60◦ with the x-axis ~ as shown in the figure. A uniform magnetic field B parallel to GS exists in the region outside the electron gun. Find the minimum value of B needed to make the electrons hit S. (1993)

is minimum at n = 1. The minimum value of the magnetic field is 2πmv cos θ dq (2 × 3.14)(9.1 × 10−31 )(2.65 × 107 )(0.5) = (0.1)(1.6 × 10−19 )

Bmin =

= 4.73 × 10−3 T. Ans. 4.73 × 10−3 T Q 47. A wire loop carrying a current I is placed in the x-y plane as shown in the figure. (1991)

S

y

M

x a 120◦

~ B I 60◦

P

x

v

G

~v

N

Sol. The speed of the electron is given by s r 2K 2 × 2 × 103 × 1.6 × 10−19 v= = m 9.1 × 10−31 = 2.65 × 107 m/s. The path of the electron is helical because initial velocity makes an angle θ = 60◦ with the magnetic field. The velocity component perpendicular to the field is v⊥ = v sin θ. The magnetic force gives the centripetal mv 2 acceleration i.e., qv⊥ B = r ⊥ . The time period of the circular motion is T =

2πm 2πr . = v⊥ qB

(a) If a particle with charge +q and mass m is placed at the centre P and given a velocity ~v along NP (see figure), find its instantaneous acceleration. ~ = (b) If an external uniform magnetic induction field B Bˆı is applied, find the force and the torque acting on the loop due to this field. Sol. The field at the point P by the arc MN is ˆ ~ 1 = µ0 Iα kˆ = µ0 I(2π/3) kˆ = µ0 I k, B 4πa 4πa 6a and by the chord NM is ~ 2 = − µ0 I [cos θ1 − cos θ2 ] kˆ B 4πd µ0 I =− [cos 30◦ − cos 150◦ ] kˆ 4π(a cos 60◦ ) √ 3µ0 I ˆ =− k, 2πa

y S

and the resultant field at P is z

G

x

ˆ ~ =B ~1 + B ~ 2 = − 0.11µ0 I k. B a

The velocity component parallel to the field is v|| = v cos θ. The pitch of the helix is given by p = v|| T =

x 30 a

2πm v cos θ. qB

60 d 60 P

I

The distance between G and S is d = 0.1 m. The electron will hit S if it completes integral number of rotation before reaching S i.e., the distance d = np, where cos θ n = 1, 2, . . .. Substitute p to get B = 2πnmv which dq

y

M

30 N

~v

434

Part V. Electromagnetism

The velocity ~v of the charge particle, the magnetic force F~ on it, and its acceleration ~a are given by √ i vh ˆı + 3 ˆ , ~v = v cos 60◦ ˆı + v sin 60◦ ˆ = 2  √  0.11µ Iqv 0 ~ = F~ = q~v × B − 3 ˆı + ˆ , 2a  ~ F 0.11µ0 Iqv  √ − 3 ˆı + ˆ . ~a = = m 2ma The current carrying chord NM in a uniform ex~ = Bˆı experiences a force given ternal magnetic field B by √ ˆ F~1 = −|NM| BI kˆ = − 3aBI k. It can be seen that the arc MN experiences an equal but opposite force. Thus, the net force on the loop is zero. The area of the loop, its magnetic moment, and the torque acting on it, are given by   √ 1 1 A = a2 α − |MN|d = a2 π/3 − 3/4 = 0.61a2 , 2 2 ˆ µ ~ = IA kˆ = 0.61Ia2 k, ˆ × (B0 ˆı) = 0.61Ia2 B ˆ. ~ = (0.61Ia2 k) ~τ = µ ~ ×B Ans. (a) (b) zero, 0.61Ia2 B ˆ

0.11µ0 Iqv 2am

√
ˆ − 3ˆı

Q 48. A beam of protons with velocity 4 × 105 m/s enters a uniform magnetic field of 0.3 T at an angle of 60◦ to the magnetic field. Find the radius of the helical path taken by the protons beam. Also find the pitch of the helix (which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of rotation). (1986) Sol. Let the magnetic field be directed towards the x~ = B ˆı = 0.3 ˆı T. Let the proton enters the axis i.e., B magnetic field region at the origin O. Its velocity lies in the x-y plane and makes an angle θ = 60◦ with the x-axis. y

v sin 60◦

p

The magnetic force on the proton is perpendicular to the velocity ~v and is directed towards the −z axis. The work done by this force is zero. Hence, by work-energy theorem, the speed of the proton does not change. The component of velocity along x-axis, vk = v cos θ, does not change (because there is no force along x-axis). The component of velocity in y-z plane changes direction but its magnitude remains constant at v⊥ = v sin θ. The magnetic force provides the centripetal acceleration for motion in a circle of radius r (in y-z plane) i.e., qvB sin θ =

2 mv 2 sin2 θ mv⊥ = , r r

which gives r=

(1.67 × 10−27 )(4 × 105 ) sin 60◦ mv sin θ = qB (1.6 × 10−19 )(0.3)

= 1.2 × 10−2 m. Besides its circular motion, the proton moves along the x-axis with a constant speed vk = v cos θ. The superposition of circular motion in y-z plane and linear motion along x-axis gives a helical motion (see figure). The distance moved by the proton along x-axis while it completes one revolution in y-z plane is called pitch and is given by 2πr 2πmv sin θ/(qB) = (v cos θ) v⊥ v sin θ 2πmv cos θ = qB 2(3.14)(1.67 × 10−27 )(4 × 105 ) cos 60◦ = (1.6 × 10−19 )(0.3)

p = vk

= 4.37 × 10−2 m. Ans. 1.2 × 10−2 m, 4.37 × 10−2 m

2r

60◦ v cos 60◦

ˆ ~ = −qvB sin θ k. F~ = q~v × B

Q 49. A particle of mass m = 1.6 × 10−27 kg and charge q = 1.6 × 10−19 C enters a region of uniform magnetic field of strength 1 T along the direction shown in figure. The speed of the particle is 107 m/s. (1984)

~v

O

1.6 × 10−19 C, when it enters the magnetic field region, is given by

x ~ B

z

θ F E

Thus, the initial velocity of the proton is ~v = v cos θ ˆı + v sin θ ˆ, where v = 4 × 105 m/s is the speed of the proton. The magnetic force on the proton of charge q =

45◦

(a) The magnetic field is directed along the inward normal to the plane of paper. The particle leaves the region of the field at the point F. Find the distance EF and the angle θ.

Chapter 31. Magnetic Field

435

(b) If the direction of the field is along the outward normal to the plane of paper, find the time spent by the particle in the region of the magnetic field after entering it at E. Sol. The magnetic force on a charge q, moving with ~ is given by a velocity ~v inside a magnetic field B, ~ Since force is perpendicular to ~v , the F~ = q~v × B. speed of the charge does not change inside the magnetic field. The charge moves in a circular path of radius r. The centripetal acceleration for the circular motion is provided by the magnetic force i.e., mv 2 /r = qvB.

(1) θ F r P

O

by the particle in the magnetic field is 3 2πr 3 2π (mv/(qB)) 3πm · = · = 4 v 4 v 2qB −27 3 (3.14) (1.6 × 10 ) = 4.71 × 10−8 s. = 2 (1.6 × 10−19 ) (1)

t=

Ans. (a) 0.14 m, 45◦ (b) 4.71 × 10−8 s Q 50. A particle of mass 1 × 10−26 kg and charge +1.6 × 10−19 C traveling with a velocity 1.28 × 106 m/s in the +x direction enters a region in which a uniform electric field E and a uniform magnetic field of induction B are present such that Ex = Ey = 0, Ez = −102.4 kV/m and Bx = Bz = 0, By = 8 × 10−2 T. The particle enters this region at the origin at time t = 0. Determine the location (x, y, z) of the particle at t = 5 × 10−6 s. If the electric field is switched off at this instant (with the magnetic field still present), what will be the position of the particle at t = 7.45 × 10−6 s?

r

(1982) E 45◦

The velocity of the particle is along the tangent to the circle. Thus, normal to the velocities at E and F intersect at the centre of the circle, O (see figure). Draw OP as perpendicular to EF. In triangle OPE, ∠E = 90◦ − 45◦ = 45◦ and in triangle OPF, ∠F = 90◦ − θ. Since sides OE = OF = r, we get ∠F = ∠E, which gives θ = 45◦ . The distance EF is given by EF = 2r cos 45◦   mv =2 cos 45◦ (using (1)) qB √ 2 (1.6 × 10−27 ) (107 ) = = 0.14 m. (1.6 × 10−19 ) (1) Note that the charge travels one quadrant of the circle inside the magnetic field.

45◦

Sol. The particle of mass m = 1 × 10 kg, charge q = +1.6 × 10−19 C, and velocity ~v = 1.28 × 106 m/s ˆı enters the electric and the magnetic field region at the origin O. y ~ B ~ E ~E F O• ~B F

~v

P •

x

Q•

z

~ = The electric field in the region is E ~ = 8 × 10−2 T ˆ. −102.4 kV/m kˆ and magnetic field is B The forces on the particle due to electric and magnetic fields, as the particle enters at the point O, are given by ˆ ~ = (1.6 × 10−19 ) (−102.4 × 103 k) F~E = q E ˆ N, = −1.6384 × 10−14 k

(1)

~ F~B = q~v × B

E r

P θ r

−26

O

F

When the direction of the magnetic field is reversed, the direction of magnetic force also gets reversed. The particle follows a circular path as shown in the figure. The radius of the circle does not change because magnitude of magnetic force remains the same. It is not difficult to see that charge travels three quadrant of a circle in the magnetic field. Thus, time spent

= (1.6 × 10−19 ) (1.28 × 106 ˆı) × (8 × 10−2 ˆ) ˆ N. = 1.6384 × 10−14 k (2) From equations (1) and (2), the electric and magnetic forces on the particle are equal and opposite. Thus, the net force on the particle is zero. The particle continues moving with velocity ~v along +x axis. Distance travelled by the particle in time t1 = 5 × 10−6 s is x = OP = vt1 = (1.28 × 106 )(5 × 10−6 ) = 6.4 m. Thus, the coordinates of the particle after time t1 = 5 × 10−6 s are P (6.4 m, 0, 0). When the particle is at the point P, electric field is switched-off. Thus, force on the particle is due to

436

Part V. Electromagnetism

magnetic field only. The magnitude of the force is FB = 1.6384 × 10−14 N and its direction is always perpendicular to the velocity. The particle starts moving in a circle of radius r with constant velocity v. The centripetal acceleration for uniform circular motion is provided by the magnetic force i.e., mv 2 /r = FB , which gives r=

(10−26 ) (1.28 × 106 )2 mv 2 = = 1 m. FB 1.6384 × 10−14

We are interested in the particle position at time t2 = 7.45 × 10−6 s. The angular displacement of the particle in time t = t2 − t1 = 2.45 × 10−6 s is θ = ωt = (v/r)t = (1.28 × 106 /1)(2.45 × 10−6 ) = 3.14 rad = 180◦ . Thus, the particle completes a semicircle in x-z plane in time t = 7.45 × 10−6 s to reach the point Q (see figure). The coordinates of point Q are (6.4 m, 0, 2 m). Ans. (6.4 m, 0, 0), (6.4 m, 0, 2 m) Q 51. A potential difference of 600 V is applied across the plates of a parallel plate condenser. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates, with a velocity of 2 × 106 m/s moves un-deflected between the plates. Find the magnitude and direction of the magnetic field in the region between the condenser plates. [Charge of the electron = 1.6 × 10−9 C, Neglect the edge effects.] (1981) 600 V + + + + + +

-

Sol. The direction of the electric field between the plates is from the positive plate to the negative plate. The magnitude of the electric field is E = V /d, where V is the potential difference and d is the separation between the plates. The electrostatic force on the electron (negative charge) is towards the left and its magnitude is FE = qE. An electron projected vertically up with a velocity of v = 2 × 106 m/s moves un-deflected only if the net force on it is zero. Thus, magnetic force, ~ should be equal and opposite to F~E (see F~B = q~v × B, figure). 600 V ~ E + ~ ⊗B + + ~v + + ~ ~B F F + E

-

The magnetic force F~B on the moving electron (negative charge) is towards right if the direction of magnetic field is into the paper. Equate FE = FB to get the magnitude of the magnetic field FE qE V = = qv qv dv 600 = = 0.1 T. (3 × 10−3 )(2 × 106 )

B=

Ans. 0.1 T, perpendicular to paper inwards Q 52. Give MKS units for (a) Young’s modulus, (b) magnetic induction, and (c) power of a lens. (1980) Sol. The MKS units of Young’s modulus is N/m2 , magnetic induction is Tesla, and power of a lens is m−1 . Ans. (a) N/m2 (b) Tesla (c) m−1

Chapter 32 Magnetic Field due to a Current

One Option Correct

It is easy to see that magnitic field at O due to any ~ AB . Hence the net straight wire segment is equal to B magnetic field at O is

Q 1. A symmetric star shaped conducting wire loop is carrying a steady current i as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the center of the loop is (2017)

√  ~ = 12B ~ AB = µ0 i 6 B 3−1 . 4πa Ans. (A) Q 2. An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the ~ as a function of the radial distance magnetic field, |B| r from the axis is best represented by (2012)

i

4a

~ (A) |B|

(A) (C)

µ0 i 4πa 6 µ0 i 4πa 3

√
√3 − 1
3−1

(B) (D)

µ0 i 4πa 6 µ0 i 4πa 3

~ (B) |B|

√

r


3+ √1
2− 3

R/2

r

R

~ (C) |B|

R/2

R

R/2

R

~ (D) |B|

Sol. The configuration is shown in the figure. r

A

R/2

Sol. The magnetic field of a cylindrical conductor is in circumferential direction due to cylindrical symmetry. The magnitude of magnetic field is same at all points lying on a circle of radius r. We use Ampere’s law, H ~ = µ0 Ienc , to find the magnetic field. Consider a ~ · dl B loop of radius r as shown in the figure.

i ◦ B 30

C

60◦

P D

r

R

O

E ~ B r R 2

By symmetry, ∠AOC = 360◦ /6 = 60◦ , ∠OPB = 90 and side OA = 4a/2 = 2a. Since AB = BC, we get ∠AOB = ∠COB = 60◦ /2 = 30◦ . In triangle OAP, ∠OAP = 30◦ and in triangle OBP, ∠OBP = 60◦ . Also, the perpendicular distance OP = OA cos 60◦ = a. The magnetic field at O due to straight wire segment AB is coming out of the paper and its magnitude is  µ0 i µ0 i 1 √ BAB = (cos 30◦ − cos 60◦ ) = 3−1 . 4πa 4πa 2 ◦

R

Apply Ampere’s law on this loop to get B = The values of Ienc for different regions are

Ienc =

  0,

I(πr 2 −πR2 /4) ,  πR2 −πR2 /4

 437

I,

if r < R/2; if R/2 ≤ r < R; otherwise.

µ0 Ienc 2πr .

438

Part V. Electromagnetism

and the corresponding magnetic field is

B=

  0,

2µ0 I 3πr   µ0 I 2πr ,

2

r −

R2 4




Sol. The position vector of the point P is ~r = x ˆı + y ˆ.

if r < R/2; , if R/2 ≤ r < R; otherwise.

y y

Ans. D

~ r

Q 3. A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in x-y plane and a steady current I flows through the wire. The z component of magnetic field at the centre of spiral is (2011)

µ0 N I b 2(b−a) ln a
µ0 N I b 2b ln a




The perpendicular distance of P from the wire is p ˆ x2 + y 2 and unit vector along ~r is rˆ = √x ˆı+y . 2 2 x +y

(B) (D)

) ~ = µ0 I (−kˆ × rˆ) = µ0 I (yˆı − xˆ . B 2πr 2π x2 + y 2 Ans. A

µ0 N I b+a 2(b−a) ln b−a
µ0 N I b+a 2b ln b−a




Sol. Consider a small circular element of thickness dr at a distance r from the centre of spiral.

Q 5. A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the centre is (2001) (A)

r

dr

a

µ0 N I b

(B)

2µ0 N I a

(C)

µ0 N I 2(b−a)

ln ab (D)

µ0 I N 2(b−a)

ln ab

Sol. See solution of similar problem on page 438. Ans. C

b N The number of turns in this element is b−a dr and N the current flowing through this element is i = b−a drI (current I is flowing through each turn). The magnetic field at the centre of spiral due to this element is given by

dB =

x

x b

(A) (C)

~ B

The magnetic field by a current carrying wire, having a ˆ is given current I along the negative z direction (−k), by

a

I

x

I

r=

y

P (x, y)

Q 6. A non-planar loop of conducting wire carrying a current I is placed as shown in the figure. Each of the straight sections of the loop is of length 2a. The magnetic field due to this loop at the point P (a, 0, a) points in the direction (2001)

z

µ0 i µ0 N I dr = . 2r 2(b − a) r

Integrate dB from r = a to r = b to get   Z b µ0 N I dr µ0 N I b B= = ln . 2(b − a) r 2(b − a) a a

y

x

Ans. A Q 4. A long straight wire along the z-axis carries a current I in the negative z direction. The magnetic field ~ at a point having coordinate (x, y) on the z = 0 vector B plane is (2002) µ0 I(yˆ ı−xˆ ) µ0 I(xˆ ı+yˆ ) (A) 2π(x2 +y2 ) (B) 2π(x2 +y2 ) (C)

µ0 I(xˆ −yˆ ı) 2π(x2 +y 2 )

(D)

µ0 I(xˆ ı−yˆ ) 2π(x2 +y 2 )

(A) (C)

√1 − ˆ + kˆ 2  √1 ˆ ı + ˆ + kˆ 3





(B) (D)

√1 − ˆ + 3  √1 ˆ ı + kˆ 2



 kˆ + ˆı

Sol. Consider the six straight sections as shown in the figure.

Chapter 32. Magnetic Field due to a Current z

Q 7. An infinitely long conductor PQR is bent to form a right angle as shown in the figure. A current I flows through PQR. The magnetic field due to this current at the point M is B1 . Now, another infinitely long straight conductor QS is connected at Q, so that current is I/2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now B2 . The ratio B1 /B2 is given by (2000)

F

5 P(a, 0, a)

G 4 y 6

E D

3 2

H A θ2

√

I √

2a

C x

B

1

439

M 3a

2a

−∞ P

P

θ1

The distance of point P from the mid-point √ √ of each section is 2a and from two end-points is √ 3a (see figure). The √ cosine of two angles are cos θ1 = 1/ 3 and cos θ2 = −1/ 3. The magnitude of magnetic field at P, due to each section, is given by B1 =

µ0 I µ I √ (cos θ1 − cos θ2 ) = √ 0 . 2π 2a 6πa

The direction of magnetic field at P by each section is given by cross product of the direction of current carrying segment and unit vector from the middle point of section to P (ˆi× rˆ). The direction of current carrying segment and unit vector from middle point to P for the six sections are

R −∞

(A) 1/2 (B) 1 (C) 2/3 (D) 2 Sol. The magnetic field at the point P (see figure) by a current carrying wire is given by B=

µ0 I (cos θ1 − cos θ2 ). 4πd θ2 I ⊗

d

rˆ1 =

√1 (ˆ + 2

ˆi2 = −ˆ ,

rˆ2 =

√1 (−ˆ ı+ 2

ˆ k);

ˆi3 = −ˆı,

rˆ3 =

√1 (−ˆ + 2

ˆ k);

ˆ ˆi4 = k,

rˆ4 =

√1 (ˆ ı− 2

ˆ);

ˆi5 = −ˆ ,

rˆ5 =

√1 (ˆ ı− 2

ˆ k);

ˆ ˆi6 = −k,

rˆ6 =

√1 (ˆ ı+ 2

ˆ).

P

θ1

ˆ k);

ˆi1 = ˆı,

∞ S

I Q

In the first case, the field at the point M by parts PQ, QR, and the total field are given by µ0 I µ0 I (cos 0◦ − cos 90◦ ) = , 4πd 4πd µ0 I BQR = (cos 180◦ − cos 180◦ ) = 0, 4πd µ0 I B1 = BPQ + BQR = . 4πd BPQ =

The net field at P is given by  ~ = √µ0 I ˆi1 × rˆ1 + ˆi2 × rˆ2 + ˆi3 × rˆ3 + ˆi4 × rˆ4 B 6πa  +ˆi5 × rˆ5 + ˆi6 × rˆ6  µ0 I  =√ ˆı + kˆ . 3πa Aliter: There is an elegant way to find the direction ~ Given current loop can be considered as superpoof B. sition of two loops, ABCDA and EFGHE, as shown in the figure. The magnetic field at P due to ABCDA is along the z-axis and due to EFGHE is along the x-axis. Also, the magnitudes of field due to two loops are equal. Thus, the direction of resultant magnetic field is along ˆ √1 (ˆ ı + k). 2

Ans. D

In the second case, the field at M by parts PQ and QR remains same as in the first case. The field by part QS and the total field are given by µ0 (I/2) µ0 I (cos 90◦ − cos 180◦ ) = , 4πd 8πd 3 µ0 I B2 = BPQ + BQR + BQS = . 2 4πd BQS =

Ans. C Q 8. Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field B along the line XX 0 is given by (2000)

440

Part V. Electromagnetism

(A)

(B) X0

X

~v y

X0

X

−I

I d

d

d

(C)

(D) X0

X d

d

d

~ B d

d X0

X

d

Sol. Let one of the wire is located at x = −d and the other at x = d.

x

X0

x = −d

~ = − 2µ0 I ˆ = − 2µ0 I ˆ. The field This field is B 2πd/2 πd is parallel to the velocity ~v of the charge. Thus, F~ = ~ = ~0. q~v × B Ans. D Q 10. A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the centre. The value of the magnetic induction at the centre due to the current in the ring is (1995) (A) proportional to (180◦ − θ) (B) inversely proportional to r (C) zero, only if θ = 180◦ (D) zero for all values of θ

y

X

x=d

The current is coming out of the paper. The magnetic field by two wires is circumferential as shown in the figure. The magnitude of magnetic field at a perpendicular distance r from a straight wire is given by 0I B = µ2πr . Thus, the field at a distance x from the centre of the given wires is given by µ0 I µ0 I + 2π(x + d) 2π(x − d)   µ0 I x = . π x2 − d2

Sol. The circumference of the circle is 2πr and resisR . The resistance of the tance per unit length is 2πr Rθ smaller arc AB is R1 = 2π and the current flowing through it is i1 = RV1 = 2πV Rθ . A V

B=

0I B < 0 for x < −d, and B ≈ µπx if x  −d B > 0 for −d < x < 0 B = 0 at x = 0 B < 0 for 0 < x < d 0I if x  d B > 0 for x > d, and B ≈ µπx lim − B = −∞, lim + B = ∞, lim− B = −∞,

x→−d

x→−d

i1

x→d

lim+ B = ∞.

x→d

Ans. B Q 9. Two very long straight parallel wires carry steady currents I and −I respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity ~v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is (1998) Iqv 0 Iqv 0 Iqv (B) µ0πd (C) 2µπd (D) zero (A) µ2πd Sol. The magnetic field at the mid point by the two wires is in the same direction.

θ

O

i2

B

(1)

We encourage you to show the following by using equation (1), (a) (b) (c) (d) (e) (f)

x

Similarly, resistance of the larger arc is R2 = flowing through it is i2 = RV2 = of magnetic field at O by the smaller arc AB is coming out of the paper and its magnitude is µ0 i1 θ µ0 V B1 = = . 4πr 2Rr The direction of magnetic field at O by the larger arc is into the paper and its magnitude is R(2π−θ) and the current 2π 2πV . The direction R(2π−θ)

µ0 V µ0 i2 (2π − θ) = . 4πr 2Rr Thus, B1 and B2 are equal in magnitude but opposite in direction. Hence the resultant magnetic field at O is zero. Ans. D B2 =

Q 11. A current I flows along the length of an infinitely long, straight, thin-walled pipe. Then the magnetic field (1993)

(A) at all points inside the pipe is the same, but not zero. (B) at any point inside the pipe is zero. (C) is zero only on the axis of the pipe. (D) is different at different points inside the pipe.

Chapter 32. Magnetic Field due to a Current

441

Sol. By symmetry, the magnetic field inside the pipe is circumferential. Take a circular loop of radius r with centre on the axis of the pipe. By symmetry, the magnitude of magnetic field is same throughout the loop.

Fqr

I

q

Fpq

Q 12. Two thin long parallel wires separated by a distance b are carrying a current i ampere each. The magnitude of the force per unit length exerted by one wire on the other is (1986) µ0 i2 µ0 i µ0 i µ0 i2 (A) b2 (B) 2πb (C) 2πb (D) 2πb2 Sol. The magnetic field at a distance b from a long wire carrying a current i is circumferential in direction and has magnitude B = µ0 i/(2πb). Second wire is placed perpendicular to the direction of magnetic field of the first wire. The magnetic force on unit length of second wire carrying a current i in a perpendicular magnetic field B is given by f = iB = µ0 i2 /(2πb). Ans. B

Frs p

d

H ~ · d~l = B(2πr) = µ0 Ienc . By Ampere’s law, B Since Ienc = 0, we get B = 0 for all r. Thus, B = 0 inside the pipe. We encourage you to explain why there cannot be a uniform magnetic field along the axis, similar to the field of a solenoid. Ans. B

r i b

s Fsp a

The force on a current carrying conductor of length ~ is given by l and current i, placed in a magnetic field B, ~ The direction of forces on the four branches F~ = i~l× B. of the square loop are as shown in the figure. By symmetry, the force on branch qr is equal and opposite to the force on branch sp i.e., Fqr = Fsp . The branch pq is closer to the long wire as compared to the branch sr. Thus, the magnetic field at the branch pq is more than the magnetic field at the branch sr. Hence, force on the branch pq is more than the force on the branch rs i.e., Fpq > Frs . Thus, net force on the loop is towards the long wire and the loop starts moving towards it. The torque on the loop about its centre of mass is zero. We encourage you to find the expressions for Fpq , Frs , and Fqr in terms of parameters given in the figure. Hint: µ0 Iib µ0 Iib , Frs = , 2πd 2π(d + a) µ0 Ii  a = ln 1 + . 2π d

Fpq = Fqr

Ans. C Q 13. A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If steady current I is established in the wire as shown in the figure, the loop will (1985) i I

(A) (B) (C) (D)

rotate about an axis parallel to the wire. move away from the wire. move towards the wire. remain stationary.

Sol. The magnetic field at the location of the square loop due to the long wire is directed into the paper and is given by ~ = µ0 I ⊗, B 2πr where r is the perpendicular distance from the long wire.

One or More Option(s) Correct Q 14. Two infinitely long straight wires lie in the x-y plane along the line x = ±R. The wire located at x = +R carries a constant current i1 and the wire located at x = −R carries a constant current i2 . A circular √ loop of radius R is suspended with its centre at (0, 0, 3R) and in a plane parallel to the x-y plane. This loop carries a constant current i in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the +ˆ  direction. Which of the following statements regarding the magnetic field ~ is (are) true? B (2018) ~ (A) If i1 = i2 , then B cannot be equal to zero at the origin (0, 0, 0). ~ can be equal to zero (B) If i1 > 0 and i2 < 0, then B at the origin (0, 0, 0). ~ can be equal to zero (C) If i1 < 0 and i2 > 0, then B at the origin (0, 0, 0). (D) If i1 = i2 , then the z-component of the
magnetic 0i field at the centre of the loop is − µ2R . Sol. The configuration is shown in the figure.

442

Part V. Electromagnetism ~ 1 at the point C due to the The magnetic field B infinitely long straight wire located at x = +R is perpendicular to the line AC (see figure). It lies in the x-z plane and makes an angle 30◦ with the x axis. Resolve ~ 1 along the x and z directions to get B

z i y

C• √

3R i2

i1

•

•

B

~ 1 (C) = B

R

x

•

O

R

A

Similarly, the magnetic field at C due to the infinitely long straight wire located at x = −R is ~ 2 (C) = B

The magnetic field at the origin O (0, 0, 0) due to the infinitely long straight wire located at x = +R is ˆ ~ 1 (O) = µ0 i1 k, B 2πR

ˆ ~ 2 (O) = µ0 i2 (−k). B 2πR The magnetic field at O due to the circular loop is µ0 iR2 ˆ = µ0 i (−k). ˆ ~ 3 (O) = √ B (−k) 16R 2(R2 + ( 3R)2 )3/2 The resultant magnetic field at the origin (0, 0, 0) is ~ ~ 1 (O) + B ~ 2 (O) + B ~ 3 (O) B(O) =B µ0 ˆ = (8(i1 − i2 ) − πi) k. 16πR ~ 1 (O) = −B ~ 2 (O) and B(O) ~ Thus, if i1 = i2 then B = ~ B3 (O) 6= ~0. ~ 1 (O) and B ~ 2 (O) If i1 > 0 and i2 < 0 then both B ˆ ˆ ~ are along k but B3 (O) is along −k. Thus, resultant field ~ B(O) can be zero. ~ 1 (O), B ~ 2 (O) and B ~ 3 (O) If i1 < 0 and i2 > 0 then B ˆ ~ are all along −k. Thus, resultant field B(O) cannot be zero. We encourage you to relate these results with the feasible solutions of 8(i1 − i2 ) − πi = 0 for i > 0. Now, let us find the magnetic field at the centre of the loop i.e., point C in the figure. z ~1 B i C•

i2 60◦ ⊗ B R

 µ0 i2  cos 30◦ ˆı − sin 30◦ kˆ . 2π(2R)

The magnetic field at C due to the circular loop is ˆ ~ 3 (C) = µ0 i (−k). B 2R

and due to the infinitely long straight wire located at x = −R is

2R

 µ0 i1  cos 30◦ ˆı + sin 30◦ kˆ . 2π(2R)

~3 B •

O

30◦ 30◦ ~2 B 2R 60◦ i1 ⊗ R A

x

The resultant magnetic field at C is ~ ~ 1 (C) + B ~ 2 (C) + B ~ 3 (C) B(C) =B   √ µ0 3(i1 + i2 ) ˆı + (i1 − i2 − 4πi) kˆ . = 8πR ~ 1 (C) and B ~ 2 (C) are If i1 = i2 then magnitudes of B equal and their z-components are equal in magnitude but opposite in direction. Thus, z-component of the magnetic field at C
is due to the circular loop only and 0i its value is − µ2R . We encourage you to find the expression for magnetic field at a general point (0, 0, z). Ans. (A), (B), (D) Q 15. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. The cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is (are) (2013) (A) In the region 0 < r < R, the magnetic field is nonzero. (B) In the region R < r < 2R, the magnetic field is along the common axis. (C) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r centered on the axis. (D) In the region r > 2R, the magnetic field is nonzero. ~ cyl be the magnetic field due to the hollow Sol. Let B ~ sol be the magnetic cylindrical conductor of radius R, B ~ net be the field due to the solenoid of radius 2R, and B ~ cyl is tangenresultant of these two. The direction of B ˆ and its magnitude remains the same at all tial (along θ) points lying on a circle of radius r.

Chapter 32. Magnetic Field due to a Current

443

θˆ

rˆ r

R

x

2R

H ~ · d~l = µ0 Ienc , gives the magnetic Ampere’s law, B field due to the cylindrical conductor as ( ~ if r < R; ~ cyl = 0, (1) B µ0 I ˆ 2πr θ, otherwise. The field due to the solenoid is given by ( ~ sol = µ0 nI zˆ, if r < 2R; B ~0, otherwise.

Q 17. When d ≈ a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case, (A) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ a. (B) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ a. (C) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ 1.2a. (D) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ 1.2a. Sol. Let X be the point located on the axis of the loop at a height h above it. Let Bloop be the magnetic field at X due to the loop of radius a and carrying a current I. The direction of Bloop is coming out of the paper and its magnitude is

(2) Bloop =

The equations (1) and (2) give the net field as   if r < R; µ0 nI zˆ, ~ net = µ0 nI zˆ + µ0 I θ, ˆ B if R < r < 2R; 2πr   µ0 I ˆ θ, otherwise. 2πr Ans. A, D

µ0 Ia2 . 2(a2 + h2 )3/2

The point X is equidistant from the two wires carrying the equal current I. Hence, magnetic field at X due to two wires is equal in magnitude. We encourage you to show that the resultant magnetic field at X is zero if the currents in wire 1 and 2 are in the directions PQ and SR as shown in the figure.

Q 16. Let [0 ] denotes the dimensional formula of the permittivity of the vacuum and [µ0 ] that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current, then (1998) (A) [0 ] = [M−1 L−3 T2 I] (B) [0 ] = [M−1 L−3 T4 I2 ] (C) [µ0 ] = [MLT−2 I−2 ] (D) [µ0 ] = [ML2 T−1 I]

Q

S

Wire 1

d

d Wire 2

y

a

P

q1 q2 Sol. Coulomb’s law, F = 4π 2 , may be used to get the 0r −1 −3 2 2 dimensions of [0 ] = [M L Q T ] = [M−1 L−3 T4 I2 ]. Speed of light, c = √µ10 0 , gives [µ0 ] = [MLT−2 I−2 ]. Ans. B, C

R X

B2 h

θ

B1

θ d

Paragraph Type Paragraph for Questions 17-18 The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from centre of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in counterclockwise direction if seen from above.

The direction of magnetic field at X due to the wire 1 and 2 are as shown in the figure. The magnitudes of these two fields and their resultant are

Bwires

(2014) Q

S

P

d Wire 2

Wire 1

d a

R

µ I √ 0 , 2π h2 + d2 = B1 cos θ + B2 cos θ µ0 Id µ0 Ia = ≈ . 2 2 π(h + d ) π(h2 + a2 )

B1 = B2 =

where we have used d ≈ a. The field is zero, if Bloop = Bwires , which gives h=

a√ ap 2 π −4= 6 a ≈ 1.2a. 2 2 Ans. C

444

Part V. Electromagnetism

Q 18. Consider d  a, and the loop is rotated about its diameter parallel to the wires by 30◦ from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)

Column I (A) E = 0

Column II (p) Charges are at the corners of a regular hexagon. M is at the centre of the hexagon and PQ is perpendicular to the plane of the hexagon. − Q

+

(A)

µ0 I 2 a2 d

(B)

µ0 I 2 a2 2d

√

(C)

3 µ0 I 2 a2 d

√

(D)

3 µ0 I 2 a2 2d

M

− P +

(B) V 6= 0

Sol. The resultant magnetic field due to the wire 1 ~ =B ~1 + B ~ 2 = µ0 I and 2 at the centre of the loop is B πd (upward normal to the paper). Since a  d, we can as~ to be uniform over the loop. Magnetic moment sume B ~ = πa2 I (upward normal to the loop). of the loop is M The torque on the loop is

+ −

(q) Charges are on a line perpendicular to PQ at equal intervals, M is the mid-point between the two innermost charges. P −+−+−+ M Q

(C) B = 0

~ × B| ~ = M B sin 30◦ |~τ | = |M    µ0 I 1 µo a2 I 2 2 = (πa I) = . πd 2 2d

(r) Charges are placed on two coplanar insulating rings at equal intervals. M is the common centre of the ring. PQ is perpendicular to the plane of the rings. + Q − + M − − P +

(D) µ 6= 0 Ans. B

(s) Charges are placed at the corners of a rectangle of side a and 2a and at the mid points of the longer sides. M is at the centre of the rectangle. PQ is parallel to the longer sides. −

+

−

−

M +

−

P

Matrix or Matching Type

(t) Charges are placed on two coplanar, identical insulating rings at equal intervals. M is the midpoint between the centres of the rings. PQ is perpendicular to the line joining the centres and coplanar to the rings. +

Q 19. Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and µ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current. (2009)

Q

+

P M + − Q

− −

Sol. This problem is a good example of symmetry. The ~ at the point M is zero in the configuraelectric field E ~ due to the positive tions (p), (r) and (s). Note that E and negative charges is separately zero (it is not the cancellation of the field of positive charges by the field of negative charges).

Chapter 32. Magnetic Field due to a Current

445

The electric potential V is zero in the configurations (p), (q) and (t) as potential due to positive charges is canceled by the potential due to symmetrically placed negative charges. When the charges are rotated about the line PQ, the current loops due to positive and negative charges are equal in magnitude but opposite in direction. In the configurations (p), (q) and (t), the positive and negative current loops are equidistant from M giving magnetic field B = 0. In the configurations (r) and (s), positive and negative current loops are at different distances from M giving nonzero B and µ. Ans. A7→(p,r,s), B7→(r,s), C7→(p,q,t), D7→(r,s)

In the case (C), take a small element on the left wire and a symmetrically located element on the right wire. The fields at P due to these elements are equal in magnitude but opposite in directions, making the resultant field zero. In the case (D), the field due outer loop is into the paper and due to the inner loop it is coming out of the paper. The magnitude of these fields are not equal, making the resultant field non-zero. The force on a current carrying wire placed in the magnetic field ~ is not repulsive in any case. of another wire, F~ = i ~l× B, Ans. A7→(q,r), B7→p, C7→(q,r), D7→q

Q 20. Two wires each carrying a steady current i are shown in four configurations in Column I. Some of the resulting effects are described in Column II. Match the statements in Column I with the statements in Column II. (2007)

Q 21. The wire loop PQRSP formed by joining two semicircular wires of radii r1 and r2 carries current i as shown in the figure. The magnitude of the magnetic induction at the centre C is . . . . . . (1988)

Column I (A) Point P is situated midway between the wires. P

(B) Point P is situated at the mid-point of the line joining the centers of the circular wires, which have same radii. P

(C) Point P is situated at the mid-point of the line joining the centers of the circular wires, which have same radii.

P

i

Column II (p) The magnetic field at P due to the currents in the wires are in the same direction. (q) The magnetic field at P due to the currents in the wires are in the opposite directions. (r) There is no magnetic field at P.

r2 S

R

r1 C

Q

P

Sol. The magnetic fields at C due to the straight wire segments PQ and RS are zero because C lies on the line containing these segments. The magnetic field at the centre of an arc of radius r, angle α, and carrying a current i, is given by µ0 iα . (1) 4πr The direction of magnetic field at C due to the arc QR is coming out of the paper and due to the arc SP it is going into the paper. Substitute the values of r and α in equation (1) to get ~ = |B|

~ QR = µ0 i , B 4r1

~ SP = µ0 i ⊗ . B 4r2

Thus, the resultant magnetic field at C is   ~ PQRSP = µ0 i 1 − 1 B . 4 r1 r2

P

(D) Point P is situated at the common center of the wires.

Fill in the Blank Type

(s) The wires repel each other.

Sol. In the case (A), magnetic field at P due to the upper wire is into the paper and due to the lower wire it is coming out of the paper. The fields by two wires are equal in magnitude but opposite in directions making the resultant field zero. In the case (B), magnetic field at P due to the left as well as right circular loop point towards the right.

Ans.

µ0 i 4



1 r1

−

1 r2



Integer Type Q 22. Two parallel wires in the plane of the paper are distance x0 apart. A point charge is moving with speed v between the wires in the same plane at a distance x1 from one of the wires. When the wires carry current of magnitude I in the same direction, the radius of curvature of the path of the point charge is r1 . In contrast, if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is r2 . If xx01 = 3, the value of rr12 is . . . . . . . (2014)

446

Part V. Electromagnetism

Sol. Two cases are shown in the figure.

B1 × × ×

3a 2

a

Case I

Case II

v

v

× × × × × × × × × × a× × 2 P× × × × × × ×

O

× × × × × × ×

B2 x1

× × × × × × × ×

x1 x0

x0

The magnetic fields at an inside point by the two wires of case I are anti-parallel while the magnetic fields at an inside point by the two wires of case II are parallel. ~ 1 and B ~ 2 be the resultant magnetic field at the Let B location of charge in the case I and II. These fields are given by µ0 I (x0 − 2x1 ) µ0 I ~ 1 = µ0 I − = ⊗, B 2πx1 2π(x0 − x1 ) 2π x1 (x0 − x1 ) µ0 I x0 µ0 I ~ 2 = µ0 I + B = ⊗. 2πx1 2π(x0 − x1 ) 2π x1 (x0 − x1 ) The magnetic force on a charge q, moving with speed v, provides the centripetal acceleration for circular motion, qvB = mv 2 /r, which gives r = mv qB . Thus, ~ 2| x0 r1 |B 1 1 = = = = = 3. ~ 1 | x0 − 2x1 1 − 2x1 /x0 1 − 2/3 r2 |B Ans. 3 Q 23. A cylindrical cavity of diameter a exist inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and cavity are infinitely long. A uniform current density J flows along the length. If the magniN tude of magnetic field at point P is given by 12 µ0 aJ, then the value of N is . . . . . . . (2012)

For cylinder 1, consider a circle of radius a with centre on the axis of cylinder 1 as a closed loop and apply Ampere’s law to get I B~1 · d~l = B1 · 2πa = µ0 Ienc = µ0 (πa2 ) J. (1) For cylinder 2, consider a circle of radius 3a/2 centred on the axis of cylinder 2 as a closed loop to get I B~2 · d~l= B2 · 2π(3a/2) = µ0 π(a/2)2 (−J). (2) Use equations (1) and (2) to get B1 = 21 µ0 aJ, B2 = 1 5 1 µ0 aJ, and B = 12 µ0 aJ − 12 µ0 aJ = 12 µ0 aJ. − 12 Ans. 5 Q 24. A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x and QR = 5x. If the magnitude of the mag µ0 I netic field at P due to this loop is k 48πx , find the value of k. (2009) Sol. The magnetic fields at P due to the currents in branch PQ and PR are zero. The magnetic field at P due to the current in branch QR is given by B=

µ0 I (cos θ1 − cos θ2 ). 4πd R

P

O

5x

a

θ2

4x

I

S d Q

2a

Sol. Given configuration is equivalent to an infinitely long solid cylinder of radius a having uniform current density J (cylinder 1) and another infinitely long solid cylinder of radius a/2 having uniform current density −J (cylinder 2). The magnetic field at P is the resultant of the magnetic field at P due to cylinder 1 and 2 ~ =B ~1 + B ~ 2 . By symmetry, both B ~ 1 and B ~ 2 are i.e., B circumferential.

θ1 3x

P

Using triangle QPS and QRP, we get sin θ1 =

d 4x 4 = = . 3x 5x 5

p Solve to get d = 12x/5. Also, cos θ1 = 1 − sin2 θ1 = 3/5. In triangle QRP, cos θ2 = cos(90◦ + θ1 ) = 7µ0 I − sin θ1 = −4/5. Substitute the values to get B = 48πx . Ans. 7

Chapter 32. Magnetic Field due to a Current

447 Ans. (a) 6.54 × 10−5 T (b) Zero, Zero,

Descriptive −6

Q 25. A current of 10 A flows around a closed path in a circuit which is in the horizontal plane as shown in the figure. The circuit consists of eight alternating arcs of radii r1 = 0.08 m and r2 = 0.12 m. Each arc subtends same angle at the centre. (2001) D

8.1 × 10

N

Q 26. A circular loop of radius R is bent along a diameter and given a shape as shown in the figure. One of the semicircles (KNM) lies in the x-z plane and the other one (KLM) in the y-z plane with their centres at origin. Current I is flowing through each of the semicircles as shown in the figure. (2000)

r2 C A

M i

(a) Find the magnetic field produced by this circuit at the centre. (b) An infinitely long straight wire carrying a current of 10 A is passing through the centre of the above circuit vertically with the direction of the current being into the plane of the circuit. What is the force acting on the wire at the centre due to the current in the circuit? What is the force acting on the arc AC and the straight segment CD due to the current at the centre? Sol. Each arc subtends an angle α = π/4 at the centre. The field at the centre by an arc of radius r, subtending 0 iα . The field at the centre an angle α, is given by B = µ4πr by each arc is coming out of the paper and hence adds up to give the total field   µ0 iα µ0 iα µ0 i 1 1 B1 = 4 × +4× = + 4πr1 4πr2 4 r1 r2   −7 1 1 (4π × 10 )(10) + = 4 0.08 0.12 = 6.54 × 10−5 T. The direction of current I in the infinitely long straight wire is anti-parallel to direction of B1 . Thus, ~ 1 , is zero. the force acting on the wire, F~ = I ~l × B The field due to straight wire, at a radial distance 0I r, is B2 = µ2πr in circumferential direction. The force ~ 2 , is zero because the current on arc AC, F~AC = i ~l × B is anti-parallel to the field. To find the force on straight segment CD, consider a small element of length dr at a distance r from the centre. The force on this element is ~ 2 = µ0 iI dr, (into the paper). dF = i d~r × B 2πr Integrate dF from r1 to r2 to get   Z r2 µ0 iI µ0 iI r2 FCD = dr = ln 2π r1 r1 2πr   (4π × 10−7 )(10)(10) 0.12 = ln 2π 0.08 = 8.1 × 10−6 N.

y

L

r1

x

I

z N I

K

(a) A particle of charge q is released at the origin with a velocity ~v = −v0ˆı. Find the instantaneous force F~ on the particle. Assume that space is gravity free. (b) If an external uniform magnetic field B0 ˆ is applied determine the force F~1 and F~2 on the semicircles KLM and KNM due to the field and the net force F~ on the loop. Sol. The magnetic field at the centre of a circular arc of radius R, subtending an angle α, and carrying a current 0 Iα I is given by B = µ4πR . Thus, the magnetic field at ~ KLM = − µ0 I ˆı and the origin by semicircle KLM is B 4R ~ KNM = µ0 I ˆ. The resultant by semicircle KNM it is B 4R ~ = µ0 I (−ˆı + ˆ). The magnetic field at the origin is B 4R magnetic force on a charge q, moving with velocity ~v = ˆ ~ = − µ0 Iqv0 k. −v0 ˆı, is F~ = q~v × B 4R M dF θ R ~ B

θ



N

x

θ

dF K

I z

The force on a current carrying conductor in a mag~ is F~ = I~l × B. ~ Consider the loop KNM. netic field B Take an element of length Rdθ at an angle θ in the upper half (see figure). The force on this element is dF = IB0 Rdθ. The component of dF along x and z directions are dFx = dF cos θ and dFz = −dF sin θ. Consider a symmetrically placed element in the lower half as shown in the figure. The x component of forces on two elements add up but z components cancel out. The toR 90◦ tal force on semicircular loop is F~2 = 0◦ 2dFx cos θ ˆı = ◦ R 90 2IB0 R 0◦ cos θdθ ˆı = 2IB0 R ˆı.

448

Part V. Electromagnetism M

y

L

B0

θ

B

sin 0

(region III) is given by ~ =B ~1 + B ~2 + B ~3 B µ0 i µ0 i ˆ µ0 i =− kˆ − k+ kˆ 2π(d + x) 2πx 2π(d − x)   µ0 i 3x2 − d2 ˆ = k. 2πx d2 − x2

θ

θ

(1)

K

I z

Now, consider the loop KLM. Take an element of length Rdθ at an angle θ (see figure). The force on this element is dF = IB0 R sin θdθ ˆı (coming out of the paper). The total force on semicircular loop is F~1 = R 180 IB0 R sin θdθ ˆı = 2IB0 R ˆı. The net force on the 0◦ loop is F~ = F~1 + F~2 = 4IB0 R ˆı. In general, if a current carrying wire ABC of any shape is placed in a uniform magnetic field then force on ABC is same as the force on a straight wire joining A to C. qv0 I ˆ k Ans. (a) − µ04R (b) F~1 = F~2 = F~ /2 = 2B0 IR ˆı Q 27. Three infinitely long thin wires, each carrying current i in the same direction, are in the x-y plane of a gravity free space. The central wire is along the y-axis while the other two are along x = ±d. (1997)

~ in equation (1) becomes zero when The magnetic field B √ x = d/ 3. Similarly, the magnetic field at P (x,√0), −d < x < 0 (region II) becomes zero when x = −d/ 3. z ⊗2

F

θ

F

θ

⊗

⊗

O

d

1

d

x

3

Let the wire 2 be displaced by a small distance z such that its perpendicular distance from the other √ two wires is r = d2 + z 2 (see figure). The force of attraction per unit length on wire 2 by each wire is 2 0i F = µ2πr . Resolve F along the x and z axes to get the net force per unit length on wire 2 as Fz = −2F cos θ = −

µ0 i2 µ0 i2 z=− z 2 πr π(d2 + z 2 )

(a) Find the locus of the points for which the magnetic field B is zero. (b) If the central wire is displaced along the z direction by a small amount and released, show that it will execute SHM. If the linear density of the wires is λ, find the frequency of oscillation.

µ0 i2 z, πd2 where we have used z  d. The restoring force is proportional to z and hence the wire executes SHM. Apply Newton’s second law on unit length of the wire (mass = λ), to get

Sol. The magnetic field on the y axis (x = 0, z = 0) is zero as the fields due wire 1 and 3 are equal and opposite.

d2 z µ0 i2 =− z = −ω 2 z. 2 dt πλd2 p µ0 ω i The oscillation frequency is f = 2π = 2πd πλ .p µ0 i Ans. (a) (x = 0, ± √d3 , z = 0) (b) 2πd πλ

y

I

II

−d 1

III O 2

IV d

x

3

In region I, the magnetic field by each wire is coming out of the paper and add up to give the non-zero field. In region IV, the field by each wire is going into the paper and there again add up to give the non-zero field. Let the magnetic field be zero at a point P (x, y) lying in the region II or III. It can be seen that if the magnetic field is zero at a point P then it will be zero at all points on the line passing through P and parallel to y axis (since the wires are infinitely long). Thus, it is sufficient to consider a point P (x, 0) lying on the x axis. The net magnetic field at P (x, 0), 0 < x < d

≈−

λ

d2 z µ0 i2 = − 2 z, 2 dt πd

or

Q 28. A straight segment OC (of length L) of a circuit carrying a current I is placed along the x-axis. Two infinitely long straight wires A and B, each extending from z = −∞ to +∞, are fixed at y = −a and y = +a respectively, as shown in the figure. If the wires A and B each carry a current I into the plane of the paper, obtain the expression for force acting on the segment OC. What will be the force on OC if the current in the wire B is reversed? (1992) y B⊗

O

z

A⊗

I

C

x

Chapter 32. Magnetic Field due to a Current Sol. Consider a point P on the segment OC at a distance x from O. The magnetic field due to infinitely long straight conductors is circumferential. The magnetic fields at P due to conductors at A and B are as shown in the figure. y

a θ x θ

P

C

x

θ θ

a A⊗

~B B

P as ~ P = 2B sin θ ˆı = 2 B =

a µ I √ 0 √ ˆı 2π a2 + x2 x2 + a2

µ0 Ia ˆı. π(a2 + x2 )

~ P , is zero. The force on a current element, dF~ = Id~l × B Thus, the net force on segment OC is  zero.  2 2 2 0I ˆ zero ln L a+a k, Ans. F~ = − µ2π 2

B⊗

O

449

~A B

~ A and B ~ B lie in x-y plane and make an Both B angle θ with y axis. Their magnitudes are B = BA = ~ A and B ~ B in the x and BB = 2π√µa02I+x2 . Resolve B y directions. The components along x axis cancel out and components along y axis add up to give the net field at P as

Q 29. Two long parallel wires carrying currents 2.5 A and I (ampere) in the same direction (directed into the plane of paper) are held at P and Q respectively such that they are perpendicular to the plane of paper. The points P and Q are located at a distance of 5 m and 2 m respectively from a collinear point R (see figure). (1990) P

Q

⊗

⊗

2.5 A

IA

R •

x

2m

µ0 I x ~ P = 2B cos θ (−ˆ √ B ) = −2 √ ˆ 2 2 2 2π a + x x + a2 µ0 Ix =− ˆ. π(a2 + x2 ) Now, consider a small current element d~l = dx ˆı at P. The force on this element is ~P = − dF~ = Id~l × B

µ0 I 2 xdx ˆ k. π(a2 + x2 )

Integrate dF from x = 0 to x = l to get the total force on segment OC, Z µ0 I 2 L xdx ˆ F~ = − k 2 2 π 0 a +x  2  a + L2 ˆ µ0 I 2 =− ln k. 2π a2

a

a

Sol. The magnetic field by a current carrying long 0I straight wire is circumferential and is given by B = µ2πr . Thus, the magnetic fields at R due to wire at P, wire at R, and their resultant are given by µ0 IP µ0 × 2.5 µ0 = = , 2πrP 2π × 5 4π µ0 IQ µ0 I BQ = = , 2πrQ 4π µ0 B = BP + BQ = (I + 1) = 10−7 (I + 1). 4π

~B B

B

A⊗

(a) An electron moving with a velocity of 4 × 105 m/s along the positive x direction experiences a force of magnitude 3.2 × 10−20 N at the point R. Find the value of I. (b) Find all the positions at which a third long parallel wire carrying a current of magnitude 2.5 A may be placed, so that the magnetic induction at R is zero.

BP =

y

O

5m

θ C

θ x θ

x

P

Q

⊗

⊗

2.5 A

IA

R •

2m θ

5m

x BP BQ

~A B

Now consider the case when the current direction in the conductor B is reversed. Reversing the current ~ B as shown in direction also reverses the direction of B ~ ~ the figure. Resolve BA and BB along the x and yaxis. The components along y axis cancel out. The components along x axis add up, giving the net field at

The force on a charge q moving with velocity v ~ in a field B is F~ = q~v × B. In the given case, 5 ~ and F = ~v = 4 × 10 m/s ˆı is perpendicular to B −20 3.2 × 10 N. Substitute the values to get 3.2 × 10−20 = (1.6 × 10−19 )(4 × 105 )10−7 (I + 1), i.e., I = 4 A.

450

Part V. Electromagnetism y

BS Q

P ⊗

⊗

S



•

∞

S

⊗

Q i

B ∞

Let the third wire S carrying a current 2.5 A be placed at a distance x from R. The field due this wire, BS = 2.5µ0 µ0 5µ0 2πx is equal and opposite to B = 4π (I + 1) = 4π . This is possible if S is placed to the right of R with current flowing into the paper or S is placed to the left of R with current flowing out of the paper (see figure). Equate magnitude of the magnetic fields, BS = B, to 5µ0 0 get 2.5µ 2πx = 4π , which gives x = 1 m. Ans. (a) 4 A (b) x = ±1 m w.r.t. R. Q 30. A pair of stationary and infinitely long bent wires are placed in the x-y plane as shown in the figure. The wires carry current of i = 10 A each as shown. The segment L and M are along the x-axis. The segment P and Q are parallel to the y-axis such that OS = OR = 0.02 m. Find the magnitude and direction of the magnetic induction at the origin O. (1989) y ∞

L i

R O

∞ x

i M

S

i P

∞

Thus, the resultant magnetic field at O by L, M, P, and Q is −7 ~ =B ~P + B ~ Q = µ0 i kˆ = (4π × 10 ) (10) kˆ B 2πr 2π(0.02) −4 ˆ = 10 k T.

ˆT Ans. 10−4 k Q 31. Two long straight parallel wires are 2 m apart, perpendicular to the plane of the paper. The wire A carries a current of 9.6 A, directed into the plane of paper. The wire B carries a current such that the magnetic field of induction at the point P, at a distance of 10 (1987) 11 m from the wire B, is zero. Find,

Q i ∞

L

R

i

O i i P

∞ x

A⊗

S M

1.6m 2m •

∞

B 10 11

Sol. The magnetic field at a perpendicular distance r from a straight current segment carrying current i, is given by ~ = µ0 i (cos θ1 − cos θ2 ) ⊗, B 4πr

(1)

where θ1 and θ2 are the angles as shown in the figure.

S

1.2m

m •

P

(a) The magnitude and direction of the current in B. (b) The magnitude of the magnetic field of induction at the point S. (c) The force per unit length on the wire B. Sol. Let iA = 9.6 A be the current through the wire at A and iB be the current through the wire at B. The magnetic field at P due to the current iA is

θ2

i

~ B r

⊗P

θ1

~A = − B

µ0 iA ˆı. 2π(rAB + rBP ) A⊗

Apply equation (1) to get the magnetic fields at the point O by the straight segments L and M,

x rAB

~ L | = |B ~ M | = µ0 i (cos 0◦ − cos 0◦ ) = 0. |B 4πr

B

and the straight segments P and Q, ˆ ~P = B ~ Q = µ0 i (cos 90◦ − cos 180◦ )kˆ = µ0 i k. B 4πr 4πr

y

rBP ~A B

•

P

~B B

Chapter 32. Magnetic Field due to a Current The magnetic field at P is zero, if the magnetic field at P due to the current iB is equal and opposite ~ A (see figure). Thus, the direction of current iB to B should be coming out of the paper and µ0 iB µ0 iA = . 2πrBP 2π(rAB + rBP ) Substitute rAB = 2 m, and rBP = 10/11 m to get iB =

(9.6)(10/11) iA rBP = =3A . rAB + rBP 2 + 10/11

Note that the sides of triangle ABS obey Pythago2 2 2 ras theorem, rAB = rAS + rBS , where rAB = 2 m, rAS = 1.6 m and rBS = 1.2 m. Thus, triangle ABS is right-angled triangle with ∠S = 90◦ . The direction of magnetic field at S, due to the current iA , is along SB and due to the current iB it is along SA. A⊗ 6m 1.

~B B

2m 90

2m 1.

◦

S

~A B

B

~ A and B ~ B are perpendicular to each other. Thus, B The magnitudes of these fields are given by µ0 iA (4π × 10−7 )(9.6) = 12 × 10−7 T. = 2πrAS 2π(1.6) −7 ~ B | = µ0 iB = (4π × 10 )(3) = 5 × 10−7 T. |B 2πrBS 2π(1.2)

~ A| = |B

The magnitude of the resultant magnetic field at S is q p ~ A |2 + |B ~ B |2 = 122 + 52 × 10−7 B = |B = 1.3 × 10−6 T. The force per unit length on the wire B is given by f=

µ0 iA iB (4π × 10−7 )(9.6)(3) = 2πrAB 2π(2)

= 2.88 × 10−6 N/m. Ans. (a) 3 A (b) 1.3 × 10−6 T (c) 2.88 × 10

−6

N/m

451

Chapter 33 Permanent Magnets

One Option Correct

B

n ˆ

Q 1. A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector kˆ is coming out of the plane of the paper. The magnetic moment of the current loop is (2012)

(I)

B (IV)

(A) I > III > II > IV (C) I > IV > II > III a

(B) I > II > III > IV (D) III > IV > I > II

x

Sol. The potential energy of a current loop, having ~ magnetic moment µ ~ and placed in magnetic field B, ~ = −µB cos θ. The angles θ is given by U = −~ µ·B for the given orientations are θI = 180◦ , θII = 90◦ , θIII = 45◦ , and θIV = 135◦ . Corresponding potential √ energies are√UI = µB, UII = 0, UIII = −µB/ 2, and UIV = µB/ 2. Ans. C

a

(A) a2 I kˆ
(C) − π2 + 1 a2 I kˆ


(B) π2 + 1 a2 I kˆ (D) (2π + 1) a2 I kˆ

Sol. The magnetic moment of a current loop, having an ~ and current I, is given by µ ~ For the given area A ~ = I A.

ˆ which gives, µ ˆ ~ = a2 1 + π k, loop A ~ = 1 + π2 a2 I k. 2 Ans. B

Q 4. The magnetic field lines due to a bar magnet are correctly shown in (2002) (B) (A) N N

Q 2. Which of the field pattern given below is valid for electric field as well as for magnetic field? (2011) (B) (A)

S

(C)

(C)

B n ˆ (III)

n ˆ

y

I

B n ˆ (II)

(D)

N

S

S

(D)

N

S

Sol. The magnetic lines of force are continuous closed curves coming out of the north pole of a bar magnet. Ans. D Sol. The condition that the magnetic line of forces always form a closed path gives the answer. You may explore the charge distribution that produces circumferential electric field. A time varying magnetic field (Faraday’s law) pointing normal to the paper induces the circumferential electric field. Ans. C

Q 5. A particle of charge q and mass m moves in a circular orbit of radius r with angular speed ω. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on (2000) (A) ω and q (B) ω, q and m (C) q and m (D) ω and m Sol. The angular momentum of the charge particle of mass m, moving in a circular orbit of radius r with angular speed ω, is given by L = mωr2 . The charge q takes time T = 2π/ω to complete one revolution. The average current in the loop and its

Q 3. A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III and IV. Arrange them in the decreasing order of potential energy. (2003) 452

Chapter 33. Permanent Magnets

453

magnetic moment, are given by i=

q qω = , T 2π

µ = iA =

Paragraph Type qω 1 πr2 = qωr2 . 2π 2

Paragraph for Questions 8-9

Thus, ratio of magnetic moment and angular moµ q mentum, L = 2m , depends only on q and m. The ratio is independent of dynamics parameters i.e., ω and r. This ratio for an electron (charge e, mass me ) defines eh h µ Bohr Magneton, µB = 2π L = 4πme , a physical constant. Ans. C Q 6. Two particles, each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2R. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is (1998) (A) q/(2m) (B) q/m (C) 2q/m (D) q/(πm) Sol. Let the rod rotates with an angular speed ω. The time taken to complete one revolution is T = 2π/ω. Each rotating charge is equivalent to a current loop of radius R and current q/T . Total current in the loop and its magnetic moment are given by i=

q qω q + = , T T π

µ = iA = qωR2 .

Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value to zero as their temperature is lowered below a critical temperature Tc (0). An interesting property of superconductors is that their critical temperature become smaller than Tc (0) if they are placed in a magnetic field i.e., the critical temperature Tc (B) is a function of the magnetic field strength B. The dependence of Tc (B) on B is shown in the figure. (2010)

Tc (B) Tc (0)

O

B

Q 8. In the graph below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields B1 (solid line) and B2 (dashed line). If B2 is larger than B1 , which of the following graphs shows the correct variation of R with T in these fields? (A)

R

(B)

R B2

2

The angular momentum of each charge is mωR and total angular momentum is L = 2mωR2 . The ratio, q µ L = 2m , is independent of dynamics parameters ω and R. Ans. A Q 7. A magnetic needle is kept in a non-uniform magnetic field. It experiences (1982) (A) a force and a torque (B) a force but not a torque (C) a torque but not a force (D) neither a force nor a torque Sol. Let N and S be the north and the south poles in the magnetic needle of magnetic strength m. The ~ Let needle is kept in a non-uniform magnetic field B. BN and BS be the strength of magnetic field at the north and south poles, respectively. N θ mBS

mBN ~ B

S

In the non-uniform field, BN 6= BS . Thus, force on the north pole, FN = mBN , is not equal to the force on the south pole, FS = mBS . Hence, the net force on the needle is non-zero. The torque on the needle is also non-zero (except when θ = 0). Ans. A

B2 O

(C)

B1

B1 T

R

(D)

B1 O

O

T

R B1 B2

B2 T

O

T

Sol. From the given figure, Tc (B) is a monotonically decreasing function of B. Thus, B2 > B1 implies Tc (B2 ) < Tc (B1 ). Hence resistance with B2 will become zero at lower temperature in comparison to B1 . Ans. A Q 9. A superconductor has Tc (0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its Tc decreases to 75 K. For this material one can definitely say that when, (A) B = 5 Tesla, Tc (B) = 80 K (B) B = 5 Tesla, 75 K < Tc (B) < 100 K (C) B = 10 Tesla, 75 K < Tc (B) < 100 K (D) B = 10 Tesla, Tc (B) = 70 K Sol. It is given that Tc (0) = 100 K and Tc (7.5) = 75 K. Since Tc (B) is a monotonically decreasing function of B, Tc (5) < Tc (0) and Tc (5) > Tc (7.5). Thus, 75 K < Tc (5) < 100 K. Ans. B

454

Part V. Electromagnetism

Assertion Reasoning Type

Column I

Q 10. Statement 1: The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as core inside the coil. Statement 2: Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized. (2008) (A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

(A) The force exerted by X on Y has a magnitude M g. (B) The gravitational potential energy of X is continuously increasing.

(B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Column II (p) Block Y of mass M left on a fixed inclined plane X, slides on it with a constant velocity. Y X

P

(q) The ring magnets Y and Z, each of mass M , are kept in frictionless vertical plastic stand so that they repel each other. Y rests on the base X and Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings. The whole system is in a lift that is going up with a constant velocity. P

(C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true.

Sol. The sensitivity of a moving-coil galvanometer is defined as the angular deflection θ per unit current i. It is given as θ/i = nAB/k, where n is the number of turns, A is coil area, B is magnetic field, and k is torsional constant of suspension wire. To increase the sensitivity, B is increased by placing a soft iron core inside the coil. The soft iron has a high magnetic permeability and can be easily magnetized or demagnetized. Ans. C

Z Y X

(C) Mechanical energy of the system X + Y is continuously decreasing.

(r) A pulley Y of mass m0 is fixed to a table through a clamp X. A block of mass M hangs from a string that goes over a pulley and is fixed at point P of the table. The whole system is kept in a lift that is going down with a constant velocity. P Y X

(D) The torque of the weight of Y about point P is zero.

(s) A sphere Y of mass M is put in a nonviscous liquid X kept in a container at rest. The sphere is released and it moves down in the liquid. Y X

P

Matrix or Matching Type

(t) A sphere Y of mass M is falling with its terminal velocity in a viscous liquid X kept in a container. Y P

Q 11. Column II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column II. (2009)

X

Sol. Let us consider the five configurations one by one. In (p), the block Y slides with a constant velocity so its acceleration is zero. The forces acting on Y are its weight M g and force F exerted by X (this consists of

Chapter 33. Permanent Magnets frictional force and normal reaction). By Newton’s second law, the resultant of F and M g shall be zero which gives F = M g. The gravitational potential energy of X does not change as X remains fixed. The mechanical energy of the system X + Y decreases because of energy loss due to non-conservative frictional force and decrease in potential energy of Y. The torque of M g about P is M gx, where x is the horizontal distance of Y from P.

455 F ≤ M g (as Y moves down). As Y moves down, the centre of mass of the liquid X moves up (we encourage you to prove it explicitly). Hence gravitational potential energy of X is increasing. Since liquid is non-viscous and there is no non-conservative dissipative force, the total mechanical energy of X+Y is constant.

Y

Y

X P X

P

In (q), the accelerations of X, Y, and Z are zero as lift moves up with a constant velocity. Forces acting on Z are its weight M g downwards and magnetic force FB upwards. Newton’s second law gives FB = M g. The forces on Y are its weight M g downwards, magnetic force FB downwards (by Newton’s third law), and reaction force F exerted by X. Newton’s second law gives F = M g + FB = M g + M g = 2M g. As the lift is going up, the gravitational potential energy of X is increasing. There are no non-conservative forces. The potential energy of X+Y keeps increasing and kinetic energy remains constant. Hence total mechanical energy of X+Y is increasing. The torque of weight of Y about point P is zero because M g passes through P.

In (t), the forces acting on Y are its weight M g and force F exerted by X. Since Y moves with a constant velocity, F = M g. Applying the same argument as in (s), gravitational potential energy of X is increasing. Since liquid is viscous and there is non-conservative dissipative force, the total mechanical energy of X+Y is decreasing.

Y X P

Ans. A7→(p,t), B7→(q,s,t), C7→(p,r,t), D7→q Fill in the Blank Type

P Z Y X

In (r), acceleration is again zero. The forces acting on the block are its weight M g and tension T in the string. Newton’s second law gives T = M g. The forces acting on the pulley Y are tension T towards left, tension T downwards, and force by X. The √ F exerted √ resultant of two tensions is 2T = 2M g, giving us √ F = 2M g. Since the lift is going down, gravitational potential energy of X is decreasing. The kinetic energy of X+Y is constant but their potential energy is decreasing so their mechanical energy decreases. The torque of the weight of Y about P is m0 gx, where x is the horizontal distance of pulley from P.

Q 12. In a hydrogen atom, the electron moves in an orbit of radius 0.5 ˚ A making 1016 revolutions per second. The magnetic moment associated with the orbital motion of the electron is . . . . . . (1988) Sol. The equivalent current, when an electron of charge q = 1.6 × 10−19 C makes 1016 revolutions per second, is i=

q 1.6 × 10−19 = = 1.6 × 10−3 A, T 10−16

where T is the time for one revolution. The magnetic moment of a circular loop of radius r = 0.5 ˚ A= 5 × 10−11 m, carrying a current i, is µ = πr2 i = 3.14(5 × 10−11 )2 (1.6 × 10−3 ) = 1.256 × 10−23 A m2 . Ans. 1.256 × 10−23 A m2

P Y X

In (s), the forces acting on Y are its weight M g and force F exerted by X. Since it is not given whether Y moves with a constant velocity or not, we can only say

Q 13. A wire of length L metre carrying a current i ampere is bent in the form of circle. The magnitude of its magnetic moment is . . . . . . in MKS units. (1987) Sol. The radius of the loop of perimeter L is r = L/(2π) and the area of the loop is A = πr2 = L2 /(4π). Let the loop be placed in x-y plane.

456

Part V. Electromagnetism y

Kirchhoff’s loop law gives ig G − (i − ig )S = 0,

r i

x

O

=⇒

The maximum deflection current of galvanometer sets upper limit on the current measured by this ammeter (imax ). Substitute the values to get

~ and carThe magnetic moment of a loop of area A rying a current i is given by

S=

ig G (0.1) (50) 5 = = = 5.55 Ω. imax − ig 1 − 0.1 0.9 Ans. 5.55

ˆ ~ = iL2 /(4π) k. µ ~ = iA Ans. iL2 /(4π) Integer Type Q 14. A moving coil galvanometer has 50 turns and each turn has an area 2 × 10−4 m2 . The magnetic field produced by the magnet inside the galvanometer is 0.02 T. The torsional constant of the suspension wire is 10−4 N m/rad. When a current flows through the galvanometer, full scale deflection occurs if the coil rotates by 0.2 rad. The resistance of the coil of the galvanometer is 50 Ω. This galvanometer is to be converted into an ammeter capable of measuring current in the range 0 − 1.0 A. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is . . . . . . . (2018) Sol. The magnetic moment of a coil having n turns, ~ The torque area A and current i is given by µ ~ = niA. ~ is given on this coil when placed in a magnetic field B ~ In a moving coil galvanometer, cylinby ~τ = µ ~ × B. ~ and B ~ are perpendicular drical magnets ensure that A to each other. Thus, deflection torque on the coil is τd = niAB. The restoring torque on a coil (at deflection angle θ) due to the wire of torsional constant k is τr = kθ. In equilibrium, deflection torque is balanced by the restoring torque i.e., τd = τr , which gives i=

k θ. nAB

Q 15. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 Ω resistance, it can be converted into a voltmeter of range 2n Ω resistance, it becomes 0 − 30 V. If connected to a 249 an ammeter of range 0 − 1.5 A. The value of n is . . . . . . . (2014)

Sol. Let G be the resistance of the galvanometer. Ig

Ig i − ig

G V

R

G S

Maximum voltage that can be measured by the voltmeter is V = Ig (G + R) which gives 30 = 0.006(G + 4990). Maximum current   that can be measured  G by the  ammeter is I = Ig G + 1 i.e., 1.5 = 0.006 S 2n/249 + 1 . Eliminate G to get n = 5. Ans. 5 Descriptive Q 16. In a moving coil galvanometer, torque on the coil can be expressed as τ = ki, where i is the current through the wire and k is constant. The rectangular coil of the galvanometer having number of turns N , area A and moment of inertia I is placed in magnetic field B. (2005)

The maximum deflection current of the galvanometer (ig ) occurs at full scale deflection θmax i.e., ig =

S = ig G/(i − ig ).

kθmax (10−4 ) (0.2) = = 0.1 A. nAB (50) (2 × 10−4 ) (0.02)

A galvanometer of resistance G is converted to an ammeter by connecting a small shunt resistance S in parallel. i A

ig

G

i − ig S

B i

(a) Find k in terms of given parameters. (b) If for current i0 deflection is π/2, find out torsional constant of spring. (c) If a charge Q is passed suddenly through the galvanometer, find the maximum angle of deflection. Sol. The magnetic moment of a rectangular coil having area A, number of turns N , and current i is |~ µ| = N iA. The magnetic torque on the coil of magnetic moment ~ is τ = µ ~ , placed in a perpendicular magnetic field B, ~ = N ABi = ki, which gives k = N AB. |~ µ × B| In equilibrium, magnetic torque N ABi is equal and opposite to torsional torque Cθ, i.e., N ABi = Cθ. Substitute θ = π/2 and i = i0 to get C = 2N ABi0 /π.

(1)

Chapter 33. Permanent Magnets

457

Let the charge Q flows in a small time interval ∆t. The average current through the galvanometer in this time interval is i = Q/∆t and magnetic torque is N AB(Q/∆t). If I is the moment of inertia of the coil about its axis of rotation and α is its acceleration then by using τ = Iα, we get N AB(Q/∆t) = I(∆ω/∆t),

i.e.,

∆ω = ω = N ABQ/I.

(2)

where we assumed coil to be at rest at t = 0. The rotational kinetic energy just after time ∆t is given by 1 2 2 Iω . At maximum deflection, this energy is converted to potential energy of torsional spring i.e., 2 1 2 Iω

2 = 12 Cθmax .

Substitute C from equation (1) and ω from equation (2) to get 2 1 2 I(N ABQ/I)

2 = 12 (2N ABi0 /π)θmax .

q Aπ Solve to get θmax = Q BN 2Ii0 . Ans. (a) BN A (b)

2Bi0 N A π

q Aπ (c) Q BN 2Ii0

Chapter 34 Electromagnetic Induction

One Option Correct Q 1. The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. I1 and I2 are the currents in the segments ab and cd. Then, (2009) c

(A) e

(B)

e t

t

(C)

e

(D)

e

d a

t

t

b

Sol. By Faraday’s law, the induced emf is given by e = −dφ/dt, where φ is the magnetic flux through the coil. When the bar magnet is far away from the coil the flux φ is constant and hence e is zero. As magnet comes closer to the coil, the strength of magnetic field increases which increases the flux and hence a negative e is induced in the coil. The flux attains a constant maximum value when magnet is inside the coil causing e to become zero. When the magnet comes out from the other end the magnetic field and hence the flux decrease inducing a positive e. When the magnet goes far away from the coil, e becomes zero again.

(A) I1 > I2 (B) I1 < I2 (C) I1 is in the direction ba and I2 is in the direction cd (D) I1 is in the direction ab and I2 is in the direction dc

φ

Sol. The magnetic field is directed into the paper and its magnitude is increasing with time. The situation is similar to as if magnetic north pole is moving towards the loop. By Lenz’s law, the magnetic field due to the induced current, in the region enclosed by the loop, should be coming out of the paper. Thus, the direction of current in the loop is I1 from a to b and I2 from d to c. Ans. D

t

We encourage you to explore this problem in more details. Let M be magnetic moment of the bar magnet moving with a constant velocity v. The separation between the magnet and the coil at time t is given by d = d0 −vt, where d0 is the initial separation. If the coil has N turns and its cross-section area is A, flux φ and induced emf e at time t (for separation much greater than the length of the magnet) are given by

Q 2. An infinitely long cylinder is kept parallel to a uniform magnetic field B directed along positive z-axis. The direction of induced current as seen from the +ve z-axis will be (2005) (A) clockwise (B) anticlockwise (C) zero (D) along the magnetic field

µ0 2M N A , 4π (d0 − vt)3 dφ µ0 6M N Av e=− = . dt 4π (d0 − vt)4

φ=

Sol. The induced current is zero because the magnetic flux does not vary with time. Ans. C

The readers may note the effect of M , N , A and v on induced emf e. The analysis becomes more complicated when the magnet is close to the coil. Ans. B

Q 3. The variation of the induced emf (e) with time (t) in a coil if a short bar magnet is moving along its axis with a constant velocity is best represented as (2004) 458

Chapter 34. Electromagnetic Induction

459

Q 4. A short-circuited coil is placed in a time varying magnetic field. Electric power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled (four times) and the wire radius halved, the electrical power dissipated would be (2002)

(A) halved (C) doubled

(B) the same (D) quadrupled

Sol. Let the coil is made from a wire of radius rw . Let n be the number of turns in the coil and rc be the radius of the coil. Length of the wire used to make the coil 2nrc ρ lw ρ , is lw = 2πrc n and its resistance is R = πr 2 = 2 rw w where ρ is the resistivity of the wire. The magnetic flux through the coil is φ = πrc2 nB, where B is a time varying normal magnetic field. By Faradays’ law, emf is induced in the coil. The induced emf e and the electric power P (Joule’s heating) are given by dB dφ = −πnrc2 , dt dt 2  2  2 3 e2 π 2 n2 rc4 dB π 2 nrw rc dB P = = = . 2 R 2nrc ρ/rw dt 2ρ dt e=−

The power P does not change when the number of turns are quadrupled (4n) and radius of the wire is halved (rw /2). Ans. B Q 5. As shown in the figure, P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current IP flows in P (as seen by E ) and an induced current IQ1 flows in Q. The switch remains closed for a long time. When S is opened, a current IQ2 flows in Q. Then the direction of IQ1 and IQ2 (as seen by E ) are (2002)

current IQ2 will be clockwise to oppose the receding north pole. Ans. D Q 6. A metallic square loop ABCD is moving in its own plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in the figure. Electric field is induced (2001) A

B v

D

(A) (B) (C) (D)

C

in AD, but not in BC in BC, but not in AD neither in AD nor in BC in both AD and BC

Sol. The motional emf and electric field are induced due accumulation of charges. A charge q moving with ~ experiences a magnetic velocity ~v in a magnetic field B ~ This force is same for charges (elecforce F~m = q~v × B. trons) in branch AD and branch BC. Thus, the electric field is induced in both AD and BC. We encourage you to find the magnitude and direction of induced electric field. Ans. D Q 7. Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be (2001)

(i)

(ii)

(iii)

P Q E S

(A) (B) (C) (D)

respectively clockwise and anticlockwise. both clockwise. both anticlockwise. respectively anticlockwise and clockwise.

Sol. When the switch S is closed, an increasing clockwise current is set in the loop P. This situation is analogous to a north pole moving from E towards the loop Q. Using Lenz’s law, the induced current IQ1 will be anticlockwise to oppose an approaching north pole. When switch S is opened, a decreasing clockwise current is set in the loop P. This situation is analogous to a north pole moving away from the loop Q towards E. The induced

(A) (B) (C) (D)

maximum is situation (i). maximum is situation (ii). maximum is situation (iii). the same in all situations.

Sol. The mutual inductance M is given by, φ = M i, where φ is the flux through a coil due to the current i ~ due to a in another coil. The flux through an area A ~ ~ ~ magnetic field B is given by φ = B · A. The directions ~ and A ~ are parallel in configuration (i) but perof B pendicular in configuration (ii) and configuration (iii). Hence, the flux and mutual inductance are maximum in configuration (i). Ans. A Q 8. A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is directed into the plane of the paper as shown. The magnitude of the induced electric field at point P at a distance r from the centre of the circular region, (2000)

460

Part V. Electromagnetism Sol. Let L and R be the self inductance and resistance of the coil connected across a battery of emf E.

B(t)

r

P

L

R

S

I1 (t)

E a

(A) is zero (C) increases as r

(B) decreases as 1/r (D) decreases as 1/r2

Sol. Consider a circular loop of radius r passing through P.

When the switch S is closed, the current in the coil increases with time t as I1 (t) =

i Eh 1 − e−t/τ , R

B(t)

P

a

The flux φ through the loop and the induced emf e in the loop are given by φ = B(t)A = πa2 B(t), dφ dB(t) = −πa2 . dt dt The induced emf is an electric potential caused by ~ By symmetry, E ~ has a conthe induced electric field E. stant magnitude throughout the loop and its direction is circumferential. Thus, I ~ · d~l = −E(2πr), e=− E e=−

i ME h 1 − e−t/τ , R dφ M E −t/τ e=− =− e , dt L M E −t/τ |e| = e , I2 (t) = r rL i kM E 2 −t/τ h I2 (t)B(t) = kI2 (t)I1 (t) = e 1 − e−t/τ . rRL φ = M I1 (t) =

I2 (t)B(t)

r

where τ = L/R is time constant. The magnetic field inside the coil is proportional to the current flowing through it i.e., B(t) = kI1 (t) for some constant k. Let M be the mutual inductance of the coil and the ring, and r be the resistance of the ring. In the ring, flux φ, induced emf e, induced current I2 (t), and the product I2 (t)B(t) are given by

t

2

e a dB(t) = . 2πr 2r dt We encourage you to show that E ∝ r if r < a. Also, draw analogy (E ↔ B, dφ dt ↔ I) between this problem and the problem to find magnetic field by a uniform current carrying cylinder of radius a. Ans. B E=−

Q 9. A coil of wire having finite inductance and resistance has a conducting ring placed co-axially within it. The coil is connected to a battery at time t = 0, so that a time dependent current I1 (t) starts flowing through the coil. If I2 (t) is the current induced in the ring and B(t) is the magnetic field at the axis of the coil due to I1 (t), then as a function of time t > 0, the product I2 (t)B(t), (2000) (A) increases with time. (B) decreases with time. (C) does not vary with time. (D) passes through a maximum.

Differentiate I2 (t)B(t) with respect to t and equate the derivative to zero to get its maximum. The I2 (t)B(t) attains maximum value at t = τ ln 2 = 0.693L/R. We encourage you to plot I1 (t), I2 (t) and I2 (t)B(t) as a function of time t. Ans. D Q 10. A circular loop of radius R, carrying current I, lies in x-y plane with its centre at origin. The total magnetic flux through x-y plane is (1999) (A) directly proportional to I. (B) directly proportional to R. (C) inversely proportional to R. (D) zero. Sol. Let the current in the loop be anticlockwise. The magnetic field inside the loop is coming out of the paper and that outside the loop is going into the paper.

Chapter 34. Electromagnetic Induction y

461 i(A)

x

1.5 1.0 0.5 0

For each magnetic field line coming out of the paper, there is a magnetic field line going into the paper because magnetic field lines form closed loop. Thus, the net magnetic flux through x-y plane is zero. Ans. D Q 11. Two identical circular loops of metal wire are lying on a table without touching each other. Loop A carries a current which increases with time. In response, the loop B (1999) (A) remains stationary. (B) is attracted by the loop A. (C) is repelled by the loop A. (D) rotates about its CM, with CM fixed. Sol. Let an anticlockwise increasing current be setup in loop A. The magnetic field by the loop A is similar to that of an approaching north pole. The approaching north pole introduces time varying flux through loop B thereby inducing a current in loop B. According to Lenz’s law, to oppose the approaching north pole, a north pole will be induced in loop B. Thus, the induced current in loop B is anticlockwise.

A

B⊗

The loop B is placed in a non-uniform magnetic field of loop A. The magnetic field decreases as the distance from loop A increases. Consider a small element on the loop B. This current carrying element is placed in a magnetic field and hence experiences magnetic force. The magnetic force is directed towards the centre of the loop and its magnitude decreases as we go away from A (see figure). The resultant force on the loop B is towards the right i.e., the loop is repelled. Ans. C Q 12. A coil of inductance 8.4 mH and resistance 6 Ω is connected to a 12 V battery. The current in the coil is 1 A at approximately the time (1999) (A) 500 s (B) 20 s (C) 35 ms (D) 1 ms

1

i R V h 1 − e− L t . i= R

t(ms)

Substitute i = 1 A, V = 12 V, R = 6 Ω and L = 8.4 × 10−3 H. Solve to get t = 1.4 × 10−3 ln 2 ≈ 1 ms. Ans. D Q 13. A metal rod moves at a constant velocity in a direction perpendicular to its length. A constant uniform magnetic field exists in space in a direction perpendicular to the rod as well as its velocity. Select the correct statement(s) from the following, (1998) (A) The entire rod is at the same electric potential. (B) There is an electric field in the rod. (C) The electric potential is highest at the centre of the rod and decrease towards its ends. (D) The electric potential is lowest at the centre of the rod and increases towards its ends. ~ and the rod velocity ~v be Sol. Let the magnetic field B as shown in the figure. ~v

+q

~ E

⊗B ~

−q

~ on the positive charges The magnetic force q~v × B is towards the left and on the negative charges it is towards the right. This causes accumulation of charges ~ in the rod. on two ends and an induced electric field E Ans. B Q 14. A small square loop of wire of side l is placed inside a large square loop of wire of side L (L  l). The loops are coplanar and their centres coincide. The mutual inductance of the system is proportional to (1998)

(A) l/L (B) l2 /L (C) L/l (D) L2 /l Sol. Let us find the magnetic field at the centre of the large square loop.

l i

L 45◦

Sol. The growth of current i in a L-R circuit connected to a battery of voltage V is given by

2

L 2

135◦

The field at the centre by each of the four branches is coming out of the paper. Also, by symmetry, the field at the centre by each branch has equal magnitude.

462

Part V. Electromagnetism

Thus, the field at the centre is four times the field by any branch and is given by µ0 i (cos θ1 − cos θ2 ) 4πd √ 4µ0 i 2 2µ0 i (cos 45◦ − cos 135◦ ) = . = 4πL/2 πL

B = 4B1 = 4 ×

Since l  L, we can assume B to be uniform throughout the inner loop. The flux φ through the inner loop and mutual induction M of the two loops are given by √ √ 2 2µ0 i l2 φ 2 2µ0 l2 φ = BA = , M= = . π L i π L

~ on the positive The magnetic force, Fm = q~v × B, charges is towards the right making the end Q at higher potential. We encourage you to generalize this result for wire of any shape. Ans. D Q 16. A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B, constant in time and space, pointing perpendicular to and into the plane of the loop exists everywhere. The current induced in the loop is (1989)

v

Ans. B Q 15. A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic ~ At the position MNQ the speed of the induction B. ring is v and the potential difference developed across the ring is (1996)

BLv/R clockwise BLv/R anticlockwise 2BLv/R anticlockwise zero

Sol. The magnetic flux passing through the square loop of side L placed in a magnetic field normal to its plane ~ ·A ~ = BA = BL2 . The flux through the loop is φ = B does not change with time. By Faraday’s law, induced emf e = −dφ/dt = 0 and hence the induced current i = e/R = 0. Ans. D

~ B N v M

(A) (B) (C) (D)

Q

One or More Option(s) Correct (A) (B) (C) (D)

zero. BvπR2 /2 and M is at higher potential. πBRv and Q is at higher potential. 2BRv and Q is at higher potential.

Sol. Consider a small element of length Rdθ at the angle θ. ~ B ⊗

Rdθ v⊥ R

=

θ

v

M

Q 17. In the figure below, the switches S1 and S2 are closed simultaneously at t = 0 and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf ) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current i in the middle wire reaches its maximum magnitude imax at time t = τ . Which of the following statements is (are) true? (2018)

vk

R

i

V

(1)

where v⊥ is the velocity component perpendicular to the length of the element. Integrate equation (1) from θ = 0 to θ = π to get the motional emf, Z π e= BRv sin θ dθ = 2BRv. 0

R

2L

Q

The motional emf induced across this element is given by de = Bv⊥ (Rdθ) = BRv sin θdθ,

L

S1 V (A) imax = 2R L (C) τ = R ln 2

V

S2 V (B) imax = 4R 2L (D) τ = R ln 2

Sol. Note that the junctions A and B are at the same potential. The circuit on the left of branch AB is independent of the circuit on the right of AB. These are two independent LR circuits. Let i1 and i2 be the currents as shown in the figure.

Chapter 34. Electromagnetic Induction i1

R

L

B

R

i2

The inductors L1 and L2 are connected in parallel. Hence, self induced emf in L1 and L2 are equal i.e.,

2L

i = i1 − i2

V

463

V

E1 = −L1 A

which gives (on integration)

In LR circuit, the current at time t is given by   i(t) = (V /R) 1 − e−(R/L)t . Thus, i h i1 = (V /R) 1 − e−(R/L)t , i h i2 = (V /R) 1 − e−(R/(2L))t , i h i = i1 − i2 = (V /R) e−(R/(2L))t − e−(R/L)t . You can get these results by applying Kirchhoff’s laws. The current i attains its extreme value when   di V R R = − e−(R/(2L))t + e−(R/L)t dt R 2L L i V −(R/(2L))t h =− e 1 − 2e−(R/(2L))t = 0, 2L which gives t = (2L/R) ln 2. You can show that i attains its maxima at t = τ = (2L/R) ln 2. Substitute t in the expression for i to get imax = V /(4R). Ans. (B), (D) Q 18. A source of constant voltage V is connected to a resistance R and two identical inductors L1 and L2 through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t = 0, the switch is closed and current begins to flow. Which of the following option(s) is(are) correct? (2017) S

V

+ −

R

L1

L2

(A) After a long time, the current through L1 will be L2 V R L1 +L2 . (B) After a long time, the current through L2 will be L1 V R L1 +L2 . (C) The ratio of the currents through L1 and L2 is fixed at all times (t > 0). (D) At t = 0, the current through the resistance R is V /R. Sol. Let i1 be the current through L1 , i2 be the current through L2 and i be the current through R. S

V

+ −

di1 di2 = E2 = −L2 , dt dt

R i A i1 L1

i2 L2

L1 i1 = L2 i2 .

(1)

Thus, ratio of the currents through L1 and L2 is fixed. Apply Kirchhoff’s law at the junction A to get i = i1 + i2 .

(2)

When switch is closed at t = 0, the inductors behave like open circuit elements (inductive reactance XL = ωL → ∞ as ω → ∞ at t = 0). Thus, i1 = i2 = 0 and current through the resistor i = i1 + i2 = 0. After a long time, the reactance of the inductors become zero. Apply Kirchhoff’s loop law to get the current through the resistor R as i = V /R. Substitute i = V /R in equation (2) and then solve equations (1) and (2) to get i1 =

L2 V , R L1 + L2

and

i2 =

L1 V . R L1 + L2

We encourage you to find the expressions for i, i1 and i2 as a function of time t and i these h deduce  results from tR(L1 +L2 ) V expressions. Hint: i = R 1 − exp − L1 L2 . Ans. (A), (B), (C) Q 19. A conducting loop in the shape of a right angled isosceles triangle of height 10 cm is kept such that the 90◦ vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in counterclockwise direction and increased at a constant rate of 10 A/s. Which of the following statement(s) is(are) true? (2016) 10 cm

90◦

(A) The magnitude of induced emf in the wire is µ0 /π volt. (B) If the loop is rotated at a constant angular speed about the wire, an additional emf of µ0 /π volt is induced in the wire. (C) The induced current in the wire is in opposite direction to the current along the hypotenuse. (D) There is a repulsive force between the wire and the loop. Sol. We can think of infinite wire as a square loop of infinite dimensions as shown in the figure (note that branches BC, CD, and DA are at infinite distance from the region of interest).

464

Part V. Electromagnetism D

C

D

C ~ 2⊗ B

~1 B

~ dF R

i1

R

A

B r

10cm P

A

φ2 = M21 i1 ,

P

(1)

where M21 is the mutual inductance between the loops. To find the flux φ2 , consider a strip of small width dr at a perpendicular distance r from the infinite wire AB (see figure). The area of the strip is dA = 2rdr. ~ at the strip due to the wire AB The magnetic field B and the magnetic flux through the strip are given by ~ = µ0 i1 ⊗, B 2πr ~ · dA ~ = µ0 i1 (2rdr) = µ0 i1 dr. dφ2 = B 2πr π Integrate dφ2 from r = 0 to r = 0.1 m to get Z 0.1 0.1µ0 i1 µ0 i1 dr = . φ2 = π π 0

(2)

From equations (1) and (2), M21 = 0.1µ0 /π. The magnetic flux φ1 through loop 1 due to the current i2 in loop 2 is given by φ1 = M12 i2 .

B

d~l

dr Q

Let us denote ABCD as loop 1 and PQR as loop 2. Our interest is to find the mutual inductance between the loop 1 and the loop 2. The flux φ2 through the loop 2 due to the current i1 in the loop 1 is given by

i1

i2

Q

Consider a small element d~l on the infinite wire ~ 2 due to the loop PQR at AB. The magnetic field B the location of this element is into the paper. Thus, ~ 2 on this element is as magnetic force dF~ = i1 d~l × B shown in the figure (repulsive). This configuration has rotational symmetry about the wire. Thus, the rotation of the loop PQR at a constant angular speed about the wire will not induce additional emf in the wire. Ans. A, D Q 20. A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis with a constant velocity v0 in the plane of the paper. At t = 0, the right edge of the loop enters a region of length 3L where there is a uniform magnetic field B0 into the plane of the paper, as shown in the figure. For sufficiently large v0 , the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let v(x), i(x) and F (x) represent the velocity of the loop, current in the loop, and force on the loop, respectively as a function of x. Counter-clockwise current is taken as positive. Which of the following schematic plot(s) is(are) correct? [Ignore gravity.] (2016)

(3) L

By the reciprocity theorem of mutual inductance, M12 = M21 . Thus, we can write equation (3) as 0.1µ0 i2 . (4) π By Faraday’s law, the induced emf in loop 1 is given by

R v0

φ1 = M21 i2 =

dφ1 0.1µ0 di2 =− . (using (4)). (5) dt π dt Substitute di2 /dt = 10 A/s in equation (5) to get the magnitude of induced emf in loop 1 (wire AB) as |e1 | = µ0 /π. The current in the loop PQR is counterclockwise ~ 2 due to and increases with time. The magnetic field B this loop in the region of the loop ABCD is into the paper and its magnitude increases with time. By Lenz’s law, the induced current opposes change in magnetic ~ 1 due to the induced curflux. So, the magnetic field B rent in the region of loop ABCD should be out of the paper. This is possible when induced current i1 is from A to B.

x

0

L

(A) vv(x)

2L 3L 4L

(B) i(x)

0

e1 = −

0 L 2L 3L 4L

x

(C) i(x)

0 L 2L 3L 4L

0 L 2L

3L 4L x

(D) F x)

x

0

L 2L 3L 4L

x

Sol. Let the right edge of the loop be at x (< L) i.e., the right edge of the loop is inside the magnetic field B0 and the left edge is outside B0 . Let the velocity of the loop at this instant be v. By Faraday’s law, the induced

Chapter 34. Electromagnetic Induction

465

emf in the loop is e = B0 Lv and induced current in the loop is i = e/R = B0 Lv/R.

(1)

By Lenz’s law, the induced current should oppose the increase in magnetic flux through the loop. Thus, the magnetic field due to the induced current should be coming out of the paper and hence current in the loop should be counterclockwise (see figure). The magnetic ~ ×B ~ 0 = −iLB0 ˆı force acting on the right edge is F~ = iL (towards left). Let m be the mass of the loop. Apply Newton’s second law to get dv B 2 L2 v dv = mv = −iLB0 = − 0 , dt dx R

F =m

(2)

where we have used equation (1) to replace i. Integrate equation (2) from x = 0 to x (< L) to get v(x) = v0 −

B02 L2 x. mR

(3)

Substitute v(x) from equation (3) in equations (1) and (2) to get   B0 L B02 L2 i(x) = v0 − x , (4) R mR   B 2 L2 B 2 L2 (5) F (x) = − 0 v0 − 0 x . R mR Thus, the velocity v(x) and the current i(x) decrease linearly with x but the force F (x) increases linearly with x.

v(x) = v0 −

B02 L3 B 2 L2 − 0 (x − 3L). mR mR

(6)

Substitute v(x) from equation (6) into the expressions for the induced current and the force to get   B 2 L3 B 2 L2 B0 L v0 − 0 − 0 (x − 3L) , R mR mR   2 2 2 3 B L B L B 2 L2 F (x) = − 0 v0 − 0 − 0 (x − 3L) . R mR mR

i(x) =

Thus, the velocity v(x) and the current i(x) decrease linearly with x but the force F (x) increases linearly with x. We encourage you to plot v(x), i(x), and F (x). What is the minimum value of v0 for the loop to come out of the magnetic field? Ans. A, B Q 21. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it. The correct statement(s) is (are) (2012) (A) The emf induced in the loop is zero if the current is constant. (B) The emf induced in the loop is finite if the current is constant. (C) The emf induced in the loop is zero if the current decreases at a steady rate. (D) The emf induced in the loop is finite if the current decreases at a steady rate. Sol. The magnetic field due to an infinitely long conductor is circumferential.

L F

second law and integrate from x = 3L to x (< 4L) to get

i

F

i

R

v 0

x L

2L

3L

4L

Consider the case when loop is completely inside the magnetic field i.e., L < x < 3L. The magnetic flux through the loop is constant. Thus, the induced emf e = 0, induced current i = 0, and magnetic force F = 0 (as i = 0). The velocity of the loop remains constant (as F = 0) at its value at x = L i.e., v = v0 − B02 L3 /(mR). Now, consider the case when the right edge of the loop is at x (> 3L). Let the loop velocity at this instant be v. The induced emf in the loop is e = B0 Lv and induced current is i = B0 Lv/R. By Lenz’s law, the induced current should oppose the decrease in magnetic flux. Thus, the magnetic field due to the induced current should be into the paper and hence induced current is clockwise. The magnetic force acting on the left ~ ×B ~ 0 = −(B 2 L2 /R)v ˆı. Apply Newton’s edge is F~ = iL 0

Since this conductor is placed along the diameter of circular loop (see figure), the total magnetic flux through the circular wire loop is always zero. Ans. A, C Q 22. Two metallic rings A and B, identical in shape and size but having different resistivities ρA and ρB , are kept on top of two identical solenoids as shown in the figure. When current I is switched on in both the solenoids in identical manner, the rings A and B jumps to heights hA and hB , respectively, with hA > hB . The possible relation(s) between their resistivities and their masses mA and mB is (are) (2009)

466

Part V. Electromagnetism A

(A) (B) (C) (D)

ρA ρA ρA ρA

> ρB < ρB > ρB < ρB

and and and and

mA mA mA mA

B

= mB = mB > mB < mB

Sol. When the current is switched on, the emf e is induced in both the rings. Since the rings are identical in shape and size, e is equal in both A and B. The induced current in the ring of perimeter l, cross-section area a and resistivity ρ is i = e/(ρl/a) = ea/(ρl). The magnetic moment of the ring due to induced current is, µ = iA = eaA/(ρl), where A is the area of the ring. By Lenz’s law, the direction of µ is anti-parallel to the magnetic field B of the solenoid. The magnetic potential energy of the ring is ~ = µB = eaAB/(ρl). Um = −~ µ·B The gravitational potential energy of the rings at the maximum height h is Ug = mgh. The conservation of energy, Um = Ug , gives, h = eaAB/(mρlg). Since hA > hB we get mA ρA < mB ρB . Ans. B, D Q 23. A field line is shown in the figure. This field cannot represent, (2006)

Sol. Henry is a unit of inductance (self inductance L and mutual inductance inductance M ). The expression for magnetic flux, φ = M i, gives unit of M as weber/ampere. The expression for emf indi , gives the unit of L as duced in inductor, e = −L dt volt-second/ampere. The expression for energy stored in the inductor, E = 12 Li2 , gives the unit of L as joule/(ampere)2 . The expression for inductive reactance, XL = Lω, gives the unit of L as ohm-second. Ans. A, B, C, D Q 25. Two different coils have self-inductances L1 = 8 mH and L2 = 2 mH. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coils is the same. At that time, the current, the induced voltage and the energy stored in the first coil are i1 , V1 and W1 respectively. Corresponding values for the second coil at the same instant are i2 , V2 and W2 respectively. Then, (1994) W1 V1 1 = (D) = 4 (A) ii12 = 14 (B) ii12 = 4 (C) W 4 V2 2 Sol. Faraday’s law gives induced emf in first coil as V1 = −L1 di1 /dt and in second coil as V2 = −L2 di2 /dt. Use, di1 /dt = di2 /dt, to get L1 8 V1 = = = 4. V2 L2 2 Equate powers given to the two coils, V1 i1 = V2 i2 , to get V2 1 i1 = = . i2 V1 4 The energy stored in the two coils are W1 = 12 L1 i21 and W2 = 12 L2 i22 . Thus, W1 L1 = W2 L2



i1 i2

2 =

8 2

 2 1 1 = . 4 4 Ans. A, C, D

(A) Magnetic field (C) Induced electric field

(B) Electrostatic field (D) Gravitational field

Sol. In electrostatics, the electric field lines cannot form a closed path. Newton’s law of gravitation, F = Gm1 m2 1 q1 q2 , is similar to Coulomb’s law, F = 4π 2 . The r2 0 r properties of gravitational field lines are similar to that of electrostatic field lines. WeR encourage you to apply ~ · d~s = −4πGMenc , to the gravitational Gauss’s law, E find the gravitational field and potential in symmetrical problems like sphere, shell, cylinder etc. Ans. B, D Q 24. The SI unit of the inductance, the Henry can be written as (1998) (A) weber/ampere (B) volt-second/ampere (C) joule/(ampere)2 (D) ohm-second

Q 26. L, C and R represents the physical quantities inductance, capacitance and resistance, respectively. The combination(s) which have the dimensions of frequency is (are) (1984) √ (A) 1/(RC) (B) R/L (C) 1/ LC (D) C/L Sol. The time constant of R-C circuit is τRC = RC and of R-L circuit is τRL = L/R. Thus, 1/(RC) and R/L have the same dimensions as frequency. Thep frequency 1 of LCR circuit at resonance is given by 2π 1/(LC). √ Hence, 1/ LC also has the dimensions of frequency. Ans. A, B, C Paragraph Type Paragraph for Questions 27-28

Chapter 34. Electromagnetic Induction

467

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω. This can be considered as equivalent to a loop carrying a steady current Qω 2π . A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ. (2013)

loop is i = Q/T = Qω/(2π). The magnetic moment of the current loop is

Q 27. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is (A) BR/4 (B) BR/2 (C) BR (D) 2BR

where ω ˜ = vR is the instantaneous angular velocity. Integrate equation (2) from 0 s to 1 s, to get the final QE . The equation (1) gives angular velocity ωf = ω − mR the initial magnetic moment as

Sol. Let the point charge Q is moving in a circle of constant radius R in anticlockwise direction as shown in the figure. y

M = iA =

Qω Q Q πR2 = mωR2 = L = γL, 2π 2m 2m

(1)

where L = mωR2 is the angular momentum of the Q (since, M = γL). The charge charge Q and γ = 2m ~ The tangential Q moves in an induced electric field E. force QE on the charged particle opposes its velocity (v). Apply Newton’s second law in tangential direction, m dv dt = −QE, to get d˜ ω QE =− , dt mR

(2)

Mi = γLi = γmωR2 , and the final magnetic moment as,   QE Mf = γLf = γmωf R2 = γm ω − R2 . mR

v Q

•

R

The change in magnetic moment is, Mf − Mi = 2 . −γQER = −γ BQR 2 Ans. B

E x

Paragraph for Questions 29-31 The magnetic flux through the circular loop at time ~ · (πR2 zˆ) = πB(t)R2 . Faraday’s t is given by φ = B(t) law gives the induced emf e as e=−

dφ dB(t) = −πR2 = −πR2 B. dt dt

(1)

Lenz’s law gives the direction of induced current and ~ as clockwise. The induced emf hence electric field (E) ~ is related to E by I ~ · d~l = −E(2πR). e= E (2) Eliminate e from equations (1) and (2) to get E = BR/2. Ans. B Q 28. The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change is 2 (A) −γBQR2 (B) −γ BQR 2 2 (C) γ BQR (D) γBQR2 2 Sol. The charge Q takes time T = 2π/ω to complete one revolution. Thus, the average current through the

Modern trains are based on Maglev technology in which trains are magnetically levitated. There are coils on both sides of wheels. Due to motion of train, current induces in the coil of track which levitate it. This is in accordance with Lenz’s law. If trains lower down then due to Lenz’s law a repulsive force increases due to which train gets uplifted and if it goes much high then there is a net downward force due to gravity. The advantage of Maglev train is that there is no friction between the train and the track, thereby reducing power consumption and enabling the train to attain very high speeds. Disadvantage of Maglev train is that as it slows down the electromagnetic force decreases and it becomes difficult to keep it leviated and as it moves forward, according to Lenz’s law, there is an electromagnetic drag force. (2006) Q 29. What is the advantage of this system? (A) No friction hence no power consumption. (B) No electric power is used. (C) Gravitation force is zero. (D) Electrostatic force draws the train. Sol. The advantage of Meglev train is the reduced power consumption by eliminating energy dissipation caused by the frictional force. Ans. A

468

Part V. Electromagnetism

Q 30. What is the disadvantage of this system? (A) Train experiences upward force according to Lenz’s law. (B) Friction force creates a drag on the train. (C) Retardation. (D) By Lenz’s law train experiences a drag. Sol. Disadvantage is drag caused by Lenz’s law when the train slows down. Ans. D Q 31. Which force causes the train to elevate up? (A) Electrostatic force. (B) Time varying electric field. (C) Magnetic force. (D) Induced electric field. Sol. The levitation is caused by the magnetic force. Ans. C

Bv Bh dl

~ = Bv vˆ+ Let the magnetic field at this element be B ˆ Bh h, where Bv and Bh are the vertical and horizontal components. Note that Bh is radial due to symmetry. The magnetic force on this element is

ˆ ~ = i d~l × (Bv vˆ + Bh h) F~ = i d~l × B ˆ + iBh dl vˆ. = −iBv dl h

Assertion Reasoning Type Q 32. Statement 1: A vertical iron rod has a coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil. Statement 2: In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction. (2007)

Thus, the induced current in the ring interact with Bh to produce the vertical force that balances the weight of the ring.

I A

B t C

(A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. (B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true. Sol. Let alternating current in the coil be I = I0 sin ωt. This current produce a magnetic field B = B0 sin ωt (since B ∝ I) at the ring. The magnetic flux through the ring is φ = φ0 sin ωt. By Faraday’s law, the induced emf e and induced current i in the ring, are given by dφ = −φ0 ω cos ωt, e=− dt e φ0 ω i= =− cos ωt, r r where r is the resistance of the ring. Thus, ring is a current carrying conductor placed in a magnetic field. Consider a small element d~l of the ring.

D

Exact dynamics of the problem is more complicated and beyond the scope of this book. We encourage you to find the direction of magnetic force at times when the coil current I is at A, B, C, and D as shown in the figure. Ans. A

Matrix or Matching Type

Q 33. Column I gives certain situations in which a straight metallic wire of resistance R is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II. (2007)

Chapter 34. Electromagnetic Induction Column I (A) A charged capacitor is connected to the ends of the wire. (B) The wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion. (C) The wire is placed in a constant electric field that has a direction along the length of the wire. (D) A battery of constant emf is connected to the ends of the wire.

469

Column II (p) A constant current flows through the wire. (q) Thermal energy is generated in the wire.

(r) A constant potential difference develops between the ends of the wire. (s) Charges of constant magnitude appear at the ends of the wire.

Column I (A) Dielectric ring uniformly charged.

(B) Dielectric ring uniformly charged rotating with angular velocity ω. (C) Constant current in ring i0 . (D) i = i0 cos ωt.

Column II (p) Time independent electrostatic field out of system. (q) Magnetic field.

(r) Induced electric field. (s) Magnetic moment.

Sol. A uniformly charged dielectric ring of radius a and charge q produces the time independent electrostatic field given by qx 1 , (on axis). E= 4π0 (a2 + x2 )3/2 When uniformly charged dielectric ring is rotated about its axis with an angular velocity ω, it behaves like a current loop carrying a constant current i = qω 2π . The ~ produced by magnetic moment µ ~ and magnetic field B this loop are given by 1 µ0 qωa2 qωa2 , B= , (on axis). 2 4π x3 The time varying current i = i0 cos ωt produces magnetic moment, magnetic field, and induced electric field (Faraday’s law). Ans. A7→p, B7→(q, s), C7→(q, s), D7→(q, r, s) µ=

Sol. In case (A), the capacitor gets discharged through the wire of resistance R. The energy stored in the capacitor is converted to thermal energy in the wire. In discharging process, the current as well as potential decrease exponentially with time. In case (B), motional emf E = vBl, develops across the two ends of the wire. The charges (free-electrons) in the wire move with an average magnetic force qvB. This movement continues till the magnetic force is balanced by electrostatic force i.e., qE = qE/l = qvB. In case (C), free-electrons in the wire move in the electric field leading to appearance of positive and negative charges of equal magnitude at the two ends. This also establishes a constant potential difference between the two ends. In case (D), Ohm’s law gives a constant current I in the wire. A constant potential difference, IR, is developed between the ends of the wire. The thermal energy is generated in the wire at a rate I 2 R. Ans. A7→q, B7→(r, s), C7→(r, s), D7→(p, q, r)

Q 34. Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II. (2006)

Q 35. Match the physical quantities given in Column I with dimensions expresses in terms of mass (M ), length (L), time (T ), and charge (Q) given in Column II. (1983)

Column I (A) (B) (C) (D)

Angular Momentum Latent Heat Torque Capacitance

(E) Inductance (F) Resistivity

Column II [M L2 T-2 ] [M L2 Q-2 ] [M L2 T-1 ] [M L3 T-1 Q-2 ] (t) [M-1 L-2 T2 Q2 ] (u) [L2 T-2 ]

(p) (q) (r) (s)

~ = Sol. The dimensions of the angular momentum (L 2 −1 m~r × ~v ) are [ML T ] and the torque (~τ = ~r × F~ ) are [ML2 T−2 ]. The latent heat (L = Q/m) has dimensions [L2 T−2 ] . The capacitance (C = Q2 /(2U )) has dimensions [M−1 L−2 T2 Q2 ] and inductance (L = 2U/I 2 ) has dimensions [ML2 Q−2 ]. The dimensions of resistivity (ρ = RA/l) are [ML3 T−1 Q−2 ]. Ans. A 7→ r, B 7→ u, C 7→ p, D 7→ t, E 7→ q, F 7→ s

470

Part V. Electromagnetism

True False Type Q 36. A conducting rod AB moves parallel to the xaxis in a uniform magnetic field pointing in the z direction. The end A of the rod gets positively charged. (1987) y B

A x O

Sol. Let the velocity of the rod moving in a magnetic ~ = B kˆ be ~v = v ˆı. field B y B v A x O

The charges inside the rod move along with the rod. Thus, the force on a charge q inside the rod is ~ = qvB (ˆı × ~k) = −qvb ˆ. Thus, the positive F~ = q~v × B charges move towards the end A of the rod. Ans. T

Q 39. In a straight conducting wire, a constant current is flowing from left to right due to a source of emf. When the source is switched-off, the direction of the induced current in the wire will be . . . . . . (1993) Sol. The current in the circuit decreases when the source is switched off. If the circuit has some inductance then it will oppose the decreasing current by inducing a current from left to right (Lenz’s law). Ans. left to right Q 40. A uniformly wound solenoid coil of selfinductance 1.8 × 10−4 H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The time constant for the current in the circuit is . . . . . . s and the steady state current through the battery is . . . . . . A. (1989) Sol. Let L0 = 1.8 × 10−4 H and R0 = 6 Ω be the inductance and the resistance of the solenoid coil. The self inductance of a solenoid of length l, radius r, and number of turns per unit length n, is given by L0 = µ0 n2 πr2 l. When the solenoid is broken into two equal parts, the length of each half becomes l/2. Thus, self inductance of each half becomes L = L0 /2 = 0.9 × 10−4 H. Also, the total length of the wire in each part is reduced to half. Hence, the resistance of each half becomes R = R0 /2 = 3 Ω. R

i

L

R

i

L

Q 37. A coil of metal wire is kept stationary in a nonuniform magnetic field. An emf is induced in the coil. (1986)

Sol. By Faraday’s law, the emf is induced in the coil if there is a time varying magnetic flux through the coil. When a coil is kept stationary in a non-uniform magnetic field, the flux through the coil is constant. Ans. F Fill in the Blank Type Q 38. The network shown in the figure is part of a complete circuit. If at a certain instant the current (i) is 5 A and is decreasing at a rate of 103 A/s then VB − VA = . . . . . . V. (1997) i

1Ω

15V

5mH

A

B

Sol. The potential at the point B is VB = VA + VR + E + VL = VA − iR + E − Ldi/dt = VA − (5)(1) + 15 − (5 × 10−3 )(−103 ) = VA + 15. Thus, VB − VA = 15 V. Note that di/dt is negative as current is decreasing with time. Ans. 15

2i

V

Two halves of the solenoid are connected in parallel as shown in the figure. If 2i is the current through the battery then, by symmetry, the current through each coil is i. The potential across the resistor is VR = iR and across the inductor is VL = Ldi/dt. Apply Kirchhoff’s law to get VR + VL − V = 0,

i.e.,

di R V + i= . dt L L

(1)

Integrate equation (1) with initial condition i = 0 at t = 0 to get i V h i= 1 − e−tR/L . (2) R The time constant for the current through the circuit is τ = L/R = 0.9 × 10−4 /3 = 3 × 10−5 s. From equation (2), the steady state current (t → ∞) through each solenoid is is = V /R = 15/3 = 5 A. Thus, the steady state current through the battery is 2is = 10 A. Note: In the steady state, inductor becomes a simple conductor. Thus, the effective resistance of the circuit is Re =

Chapter 34. Electromagnetic Induction R k R = R/2 and the steady state current through the battery is V /Re = 2V /R = 2(15)/3 = 10 A. We encourage you to analyse the complications if the solenoid is broken into two unequal halves. Ans. 3 × 10−5 , 10

471 an angle of 45◦ with respect to z-axis. If the mutual µ0 a2 inductance between the loops is given by 2(p/2) , then R the value of p is . . . . . . . (2012) z 45◦

Integer Type Q 41. Two inductors L1 (inductance 1 mH, internal resistance 3 Ω) and L2 (inductance 2 mH, internal resistance 4 Ω), and a resistor R (resistance 12 Ω) are all connected in parallel across a 5 V battery. The circuit is switched on at time t = 0. The ratio of the maximum to the minimum current (Imax /Imin ) drawn from the battery is . . . . . . . (2016)

√

3R

y

R O x

Sol. The circuit is shown in the figure. 1 mH

3Ω

2 mH

4Ω

L1 L2

Sol. The flux through a wire loop having number of ~ and placed in a uniform magnetic field turns n, area S, ~ ~ · S. ~ For a circular loop of radius B is given by φ = n B r, carrying a current i, the magnetic field on an axial point at a distance z is given by

12Ω

~ = |B| 5V

The impedance (effective resistance) of the inductors L1 and L2 is very high immediately after the circuit is switched on (t → 0+ ). Thus, inductors behave as open circuit elements and entire current flows through the resistor R = 12 Ω. Thus, by Kirchhoff’s law, current through the battery is imin = V /R = 5/12 A. In steady state (t → ∞), the impedance of the inductors become zero and they behave as resistors of given values. 3Ω

µ0 ir2 . + z 2 )3/2

2(r2

√ ~ = µ0 i . This Substitute r = R and z = 3R to get |B| 8R field is along the z-axis and may be considered as uniform at the location of square loop (since a  R). Thus, ~ makes an angle of 45◦ with area S ~ (|S| ~ = a2 ) of square B loop. The magnetic flux at the square loop and mutual inductance of the loops are given by 2 ~ |S| ~ cos 45◦ = 2 · µ0 i · a2 · √1 = µ0 ia , φ = n |B| 8R 27/2 R 2 2 µ0 a φ M = = 7/2 . i 2 R

Ans. 7 4Ω

12Ω

imax

5V

The effective resistance of the circuit is Re = (12Ω k 4Ω) k 3Ω = 3Ω k 3Ω = 3/2 Ω. The current through the circuit in steady state is imax = V /Re = 10/3 A. Thus, imax /imin = (10/3)/(5/12) = 8. Ans. 8 Q 42. A circular wire loop of radius R is placed in x-y plane centered at the origin O. A square loop of side a (a  √ R) having two turns is placed with its centre at z = 3R along the axis of the circular wire loop, as shown in the figure. The plane of the loop makes

Q 43. A long circular tube of length 10 m and radius 0.3 m carries a current I along its curved surface as shown. A wire-loop of resistance 0.005 Ω and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of tube. The current varies as I = I0 cos(300t), where I0 is constant. If magnetic moment of the loop is N µ0 I0 sin(300t), then N is . . . . . . . (2011)

I

472

Part V. Electromagnetism

Sol. The current carrying circular tube is equivalent to a solenoid. The magnetic field inside a solenoid is given by B = µ0 ni, where n is the number of turns per unit length, i is the current, and ni is the current per unit length. The circular tube of length L carrying current I has current per unit length I/L. Thus, the magnetic field inside the circular tube is B = µ0 I/L.

By Faraday’s law, the induced emf is given by e = −dφ/dt = −πa2 µ0 ni0 ω cos ωt. The emf e induces the circumferential current in the cylinder. The cylinder’s resistance for the circumferential current is ρl/A where l = 2πR and A = Ld. The induced current is given by i=

The magnetic flux φ, the induced emf e, and the induced current i0 in the circular loop of radius r, placed inside the circular tube, are given by 2

φ = BA = µ0 (I/L)πr , πr2 µ0 dI dφ =− , dt L dt πr2 µ0 dI e i0 = =− . R LR dt e=−

Thus, the magnetic moment of the loop is given by M = i0 (πr2 ) = −

π 2 r4 µ0 dI RL dt

µ0 Ldna2 i0 ω cos ωt e = . ρl/A 2ρR Ans.

µ0 Ldna2 i0 ω cos ωt 2ρR

Q 45. Two infinitely long parallel wires carrying currents I = I0 sin ωt in opposite directions are placed a distance 3a apart. A square loop of side a of negligible resistance with a capacitor of capacitance C is placed in the plane of wires as shown. Find the maximum current in the square loop. Also sketch the graph showing the variation of charge on the upper plate of the capacitor as a function of time for one complete cycle taking anticlockwise direction for the current in the loop as positive. (2003)

π 2 r4 µ0 × 300I0 sin(300t) RL = 5.92µ0 I0 sin(300t) ≈ 6µ0 I0 sin(300t). =

a

Ans. 6 Descriptive Q 44. A long solenoid of radius a and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d (d  R) and length L. A variable current i = i0 sin ωt flows through the solenoid. If the resistivity of the material of cylindrical shell is ρ, find the induced current in the shell. (2005)

a 3a

Sol. The magnetic field at a distance x from an infinitely long wire, carrying a current I, is given by µ0 I B = 2πx . The field inside the square loop by both the wires is coming out of the paper. Consider a small strip of thickness dx placed at a distance x from the left wire. x

dx

R a d

a

a 3a

L

The magnetic field B at the strip and the magnetic flux dφ passing through the strip are given by µ0 I µ0 I + , 2πx 2π(3a − x)   µ0 I 1 1 dφ = + adx. 2π x 3a − x B=

Sol. The magnetic field inside the solenoid is B = µ0 ni (along its axis) and it is zero outside the solenoid. Thus, the flux passing through the cross-section of cylinder is I
~ · dS ~ = (µ0 ni0 sin ωt) πa2 . φ= B

Integrate dφ from x = a to x = 2a to get the flux, φ = π1 µ0 aI ln 2, through the loop. Using Faraday’s law, induced emf in the loop is given by dφ 1 e = − = µ0 aωI0 ln 2 cos ωt. dt π

Chapter 34. Electromagnetic Induction

473

The charge q on the capacitor and the current i through the loop are 1 µ0 aωCI0 ln 2 cos ωt, π dq 1 i= = − µ0 aω 2 CI0 ln 2 sin ωt. dt π

q = Ce =

Maximum current in the loop is given by imax =

1 µ0 aω 2 CI0 ln 2. π qupper q0

T 2 T 4

T

3T 2

(c) The bar is suddenly stopped at time T . The current through resistance R is found i1 /4 at time 2T . Find the value of L/R in terms of the other given quantities. Sol. The magnetic field due to the current carrying 0 I0 pointing into the paper. The rightwire is B = µ2πx ward motion of bar AB increases the flux φ through the loop. Qualitatively, this is analogous to as if south pole of a bar magnet is approaching the loop. By Lenz’s law, a south pole (clockwise current) will be induced to oppose the approaching south pole. The induced emf acts like a battery of voltage dφ dt . A

t R I0

Let us analyse the situation at t = 0. The magnetic field at the loop is coming out of the paper and flux through the loop, φ = π1 µ0 aI0 ln 2 sin ωt, is increasing with time. By Lenz’s law, a field coming out of the paper with increasing flux will induce the clockwise current in the loop. This current is negative as per convention given in the question and it is setup when the upper plate of the capacitor is positively charged at t = 0. Thus, the charge on the upper plate is given by qupper = π1 µ0 aωCI0 ln 2 cos ωt. Ans. imax = π1 µ0 CI0 aω 2 ln 2 Q 46. A metal bar AB can slide on two parallel thick metallic rails separated by a distance l. A resistance R and an inductance L are connected to the rails as shown in the figure. A long straight wire, carrying a constant current I0 is placed in the plane of the rails as shown. The bar AB is held at rest at a distance x0 from the long wire. At t = 0, it is made to slide on the rails away from the wire. Answer the following questions, (2002)

dφ dt

i ⊗ ~ B L

l

B x

Apply Kirchhoff’s loop law in the loop to get dφ di = L + iR, or dt dt dφ = Ldi + R(idt) = Ldi + Rdq, where dq is the charge flown in time dt. Integrate dφ from t = 0 to t = T with conditions, i = 0 at t = 0 and i = i1 at t = T , to get the change in flux ∆φ = Li1 + R∆q.

(1)

The change in flux through the loop from t = 0 to t = T is the flux passing through the region x0 ≤ x ≤ 2x0 and is given by

A

Z

2x0

∆φ =

R

x0

I0

l L B x0

di (a) Find the relation among i, dt and dφ dt , where i is the current in the circuit and φ is the flux of the magnetic field due to the long wire through the circuit. (b) It is observed that at time t = T , the metal bar AB is at a distance of 2x0 from the long wire and the resistance R carries a current i1 . Obtain an expression for the net charge that has flown through resistance R from t = 0 to t = T .

µ0 I0 µ0 I0 l ldx = ln 2. 2πx 2π

(2)

Eliminate ∆φ from equations (1) and (2) to get the charge flowing through the loop   1 µ0 I0 l ∆q = ln 2 − Li1 . R 2π From t = T to t = 2T , the induced emf is zero and circuit is a L-R circuit with the current decaying as t−T

i = i1 e− L/R . i1 L T 4 and t = 2T . Solve toh get R = ln 4 . i di 1 µ0 I0 l (a) dφ dt = iR + L dt (b) R 2π ln 2 − Li1

Substitute i = Ans. (c)

T ln 4

474

Part V. Electromagnetism

Q 47. An inductor of inductance L = 400 mH and resistors of resistance R1 = 2 Ω and R2 = 2 Ω are connected to a battery of emf E = 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at time t = 0. What is the potential drop across L as a function of time? After the steady state is reached, the switch is opened. What is the direction and the magnitude of current through R1 as a function of time? (2001)

Integrate with initial condition i = i0 = 6 A at t = 0 to get i = i0 e−t(R1 +R2 )/L . Substitute R1 = 2 Ω, R2 = 2 Ω and L = 0.4 H to get the current through R1 as a function of time t, i = i0 e−t(R1 +R2 )/L = 6e−t(2+2)/0.4 = 6e−10t . Ans. 12e−5t V, 6e−10t A (clockwise)

E

L R1 R2

S

Sol. Let i be the current through the inductor L and the resistor R2 . i E

L R1 R2

S

di The potential drop across L is L dt and potential drop across R2 is iR2 . Kirchhoff’s loop law gives,

L

di + iR2 − E = 0. dt

Integrate with the initial condition i = 0 at t = 0 to get i E h i= 1 − e−tR2 /L . R2

Q 48. A thermocol vessel contains 0.5 kg of distilled water at 30 ◦ C. A metal coil of area 5 × 10−3 m2 , number of turns 100, mass 0.06 kg and resistance 1.6 Ω is lying horizontally at the bottom of the vessel. A uniform time varying magnetic field is set-up to pass vertically through the coil at time t = 0. The field is first increased from 0 to 0.8 T at a constant rate between 0 s and 0.2 s and then decreased to zero at the same rate between 0.2 s and 0.4 s. The cycle is repeated 12000 times. Make sketches of the current through the coil and the power dissipated in the coil as a function of time for the first two cycles. Clearly indicate the magnitudes of the quantities on the axes. Assume that no heat is lost to the vessel or the surrounding. Determine the final temperature of the water under thermal equilibrium. Specific heat of metal = 500 J/kg K and the specific heat of water = 4200 J/kg K. Neglect the inductance of coil. (2000) Sol. Let τ = 0.4 s be the time period of the cycle. In each cycle, the magnetic field varies as shown in the figure and is given by ( 4t, if 0 < t < τ2 ;
B(t) = τ 0.8 − 4 t − 2 , if τ2 ≤ t < τ .

The potential drop across L is given by VL = L

B(T) 0.8

di = Ee−tR2 /L = 12e−t×2/0.4 = 12e−5t . dt

0.0 0.2 0.4 0.6 0.8 I(A) 1.25

In the steady state (t → ∞), the current through L is i0 = E/R2 = 6 A, VL = 0, and VR2 = 12 V.

0.2 0.4 0.6 0.8

t(s)

t(s)

-1.25

i

P (W) 2.5

L

0.0

R1 R2

Let i be the decaying current in the circuit when the switch S is opened (see figure). The potential across di L is L dt , potential across R1 is iR1 , and across R2 is iR2 . Kirchhoff’s loop law gives L

di + iR1 + iR2 = 0. dt

0.8

t(s)

The flux through the coil of N = 100 turns and area A = 5 × 10−3 m2 is φ = N AB(t). By Faraday’s law, induced emf e and the induced current I in the coil, are given by e=−

dφ dB(t) = −N A , dt dt

I=

e N A dB(t) =− , R R dt

where R = 1.6 Ω is the resistance of the coil. Substitute values of N , A, R, and B(t) in the expression for I to

Chapter 34. Electromagnetic Induction get

475 dφ = B0 ydy. Integrate from y = y0 to y = y0 + a to get the flux through the loop as Z y0 +a B0 a B0 ydy = φ= (2y0 + a). 2 y0

( −1.25 A, if 0 < t < τ2 ; I(t) = 1.25 A, if τ2 ≤ t < τ . The power generated by the coil is independent of current direction and is given by P = I 2 R = 2.5 W. Thus, the heat generated in one cycle is q = P τ = 2.5 × 0.4 = 1 J and in 12000 cycles is Q = 12000q = 12000 J. This energy is used to increase the temperature of the water and the coil. Let the final temperature be T . Energy conservation gives Q = mw Sw (T − T0 ) + mc Sc (T − T0 ), 12000 = 0.5 × 4200(T − 30) + 0.06 × 500(T − 30). Solve to get T = 35.6 ◦ C.

By Faraday’s law, induced emf e and induced current i in the loop are dφ dy0 e B0 av = B0 av, i = = , e = − = B0 a dt dt R R where v = dy0 /dt is the speed of the loop. The flux through the loop increases as it moves downwards. Qualitatively, situation is similar to as if north pole is approaching the loop. By Lenz’s law, a north pole (anticlockwise current) shall be induced to repel the approaching north pole (see figure).

Ans. 35.6 ◦ C Q 49. A magnetic field B = (B0 y/a) kˆ is acting into the paper in the +z direction. B0 and a are positive constants. A square loop EFGH of side a, mass m and resistance R in x-y plane starts falling under the influence of gravity. Note the directions of x and y in the figure. Find, (1999) x

O E

F

H

~g

G

y

(a) the induced current in the loop and indicate its direction. (b) the total Lorentz force acting on the loop and indicate its direction. (c) an expression for the speed of the loop v(t) and its terminal velocity. Sol. Let at time t branch EF is at a distance y0 from the x axis. x

O y0 y E

F ~g ⊗B ~

dy

y

H

G

Consider a small element of thickness dy at a distance y from the x axis. The magnetic field at this element is B = B0 y/a and the flux through it is

Lorentz force on a current carrying conductor in ~ The forces on the branches magnetic field is F~ = i~l× B. parallel to x axis are   B0 y0 ˆ avy0 B02 ~ FEF = i(−aˆı) × k = ˆ, a R   B0 (y0 + a) ˆ av(y0 + a)B02 F~GH = i(aˆı) × k =− ˆ. a R By symmetry, the forces on branch FG and HE are equal and opposite i.e., F~FG = −F~HE . Thus, the net Lorentz force on the loop is F~ = F~EF + F~FG + F~GH + va2 B 2 F~HG = − R 0 ˆ. Other force on the loop is its weight mgˆ . Total force on the loop and its acceleration are given by   va2 B02 ˆ F~ = mg − j, R dv F va2 B02 = =g− = g − κv, dt m mR a2 B 2

where κ = mR0 is a constant. Integrate with initial condition v = 0 at t = 0 to get  g v= 1 − e−κt . κ The velocity v increases with time t and attains its maximum as t → ∞. The terminal velocity of the loop is given by  g g mgR 1 − e−κt = = 2 2 . t→∞ κ κ a B0

vt = lim

(c) v =

mgR B02 a2

B0 av Ans. (a) 2 2R  , anticlockwise (b) − B0 a gmR 1 − e− mR t , vt = B 2 a2 0

B02 a2 v ˆ R 

476

Part V. Electromagnetism

Q 50. A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table. The distance between the rails is L. A conducting massless rod of resistance R can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to the edge of the table. A mass m tied to the other end of the string hangs vertically. A constant magnetic field B exists perpendicular to the table. If the system is released from rest, calculate, (1997)

Sol. The magnetic field due to an infinitesimal bar magnet for end-on position is given by ~ = µ0 2M ˆı, B 4π x3 where M is the magnetic moment of the magnet and x is the distance from the magnet. y M v

~ B L

(a) the terminal velocity achieved by the rod. (b) the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity. Sol. Let v be the velocity of the rod at time t. The rod of length L moving with velocity v in a normal magnetic field B gets a motional emf e = BLv. The current in the rod is i = e/R = BLv/R. The magnetic force on the rod, Fm = iLB = B 2 L2 v/R, opposes the motion of the rod. Let T be the tension in the string and a be the acceleration of the mass m at time t. Since the rod is massless, the tension is equal to the magnetic force i.e., T = B 2 L2 v/R. Apply Newton’s second law on the mass m to get ma = mg − T = mg − B 2 L2 v/R.

(1)

The rod attains the terminal velocity when a = 0. Substitute a = 0 in equation (1) to get the terminal velocity mgR . B 2 L2

Since the distance x is much greater than radius a of the loop, magnetic field can be taken as uniform throughout the loop. For the loop, flux φ, induced emf e, induced current i and the magnetic moment M 0 are given by µ0 a2 M µ0 2M (πa2 ) = , 3 4π x 2x3 dφ 3µ0 a2 M dx 3µ0 a2 M v e=− = = , dt 2x4 dt 2x4 e 3µ0 a2 M v i= = , R 2Rx4 3πµ0 a4 M v M 0 = iS = . 2Rx4 ~ ·S ~= φ=B

According to Lenz’s law, the direction of M 0 is such that it opposes the approaching magnet (see figure). The potential energy of the loop with magnetic moment M 0 ~ is and placed in a magnetic field B ~ 0·B ~ = M 0B U = −M =

3πµ0 a4 M v µ0 2M 3 µ20 a4 M 2 v = , 2Rx4 4π x3 4 Rx7

and the force acting on the loop is

Substitute v = vt /2 in equation (1) to get the required acceleration a=g−

B x

z

R m

vt =

M0

B 2 L2 (vt /2) B 2 L2 mgR g =g− = . mR mR 2B 2 L2 2 Ans. (a)

mgR B 2 L2

(b) g/2

Q 51. An infinitesimally small bar magnet of dipole ~ is pointing and moving with the speed v moment M in the positive x direction. A small closed circular conducting loop of radius a and negligible self-inductance lies in the y-z plane with its centre at x = 0, and its axis coinciding with the x-axis. Find the force opposing the motion of the magnet, if the resistance of the loop is R. Assume that the distance x of the magnet from the centre of the loop is much greater than a. (1997)

F =−

21 µ20 a4 M 2 v dU = . dx 4 Rx8

The positive sign confirms the repulsive nature of the force. µ20 M 2 a4 v , repulsion Ans. F = 21 4 Rx8 Q 52. A solenoid has an inductance of 10 H and a resistance of 2 Ω. It is connected to a 10 V battery. How long will it take for the magnetic energy to reach 1/4 th of its maximum value? (1996) Sol. The L-R circuit with L = 10 H and R = 2 Ω is connected across e = 10 V battery. The current in the circuit varies with time as h i i e h i = i0 1 − e−t/τ = 1 − e−t/τ , (1) R

Chapter 34. Electromagnetic Induction

h i i R e h i = i0 1 − e−t/τ = 1 − e− L t R i Bωr2 h −R t 1−e L . = 2R The steady state current is given by i Bωr2 R Bωr2 h 1 − e− L t = = i0 . t→∞ 2R 2R

lim i = lim

t→∞

Consider an instant t when the rod makes an angle θ with the x axis. y ⊗B ~

i0 C •

r

Q 53. A metal rod OA of mass m and length r is kept rotating with a constant angular speed ω in a vertical plane about horizontal axis at the end O. The free end A is arranged to slide without friction along a fixed conducting circular ring in the same plane as that of rotation. A uniform and constant magnetic induction ~ is applied perpendicular and into the plane of rotation B as shown in the figure. An inductor L and an external resistance R are connected through a switch S between the point O and a point C on the ring to form an electric circuit. Neglect the resistance of the ring and the rod. Initially, the switch is open. (1995)

circuit is given by

2

where i0 = e/R = 5 A is the initial current and τ = L/R = 5 s is the time constant. Maximum current in the circuit is i0 . The magnetic energy, U = 12 Li2 , becomes 1/4th of its maximum value (Umax = 12 Li20 ) when i = i0 /2. Substitute i = i0 /2 in equation (1) and solve to get t = τ ln 2 = 5 × 0.693 = 3.465 s. Ans. 3.465 s

477

θ

•

O

F

x

mg

y A ω S

θ

~ B x

O R

C L

Since the rod was along the positive x axis at t = 0, we get θ = ωt. The forces acting on the rod are its ~ = i0 rB = ωr3 B 2 , weight mg, magnetic force F = |i~l×B| 2R and reaction R at the hinge point O. The torque about O is given by r r + mg cos θ 2 2 ωr4 B 2 mgr = + cos ωt, 4R 2

τ =F (a) What is the induced emf across the terminals of the switch? (b) The switch S is closed at time t = 0, (i) Obtain an expression for the current as a function of time. (ii) In the steady state, obtain the time dependence of the torque required to maintain the constant angular speed. Given that the rod OA was along the positive x-axis at t = 0. Sol. Consider an element of length dx on the rod at a distance x from O. The velocity of this element is v = ωx and motional emf induced across it is de = Bvdx = Bωxdx. Integrate de from x = 0 to x = r to get the induced emf across the two terminals of the rod Z e=

r

Bωxdx = 0

Bωr2 . 2

Note that positive charge in the rod moves from end A to end O. When S is open, emf across the terminals of S is same as emf across the rod. When switch S is closed, given configuration is a series L-R circuit with 2 source of emf e = Bωr 2 . The growth of current in L-R

(clockwise).

The net torque on the rod is zero when it rotates with constant ω. Thus, the external torque required to rotate the rod with constant ω is τext =

ωr4 B 2 mgr + cos ωt, 4R 2 Ans. (a)

(ii)

B 2 ωr 4 4R

+

mgr 2

Bωr 2 2

(b) (i)

(anti-clockwise). Bωr 2 2R

h

R

1 − e− L t

i

cos ωt

Q 54. Two parallel vertical metallic rails AB and CD are separated by 1 m. They are connected at two ends by resistances R1 and R2 as shown in the figure. A horizontal metallic bar of mass 0.2 kg slides without friction vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of 0.6 T perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in R1 and R2 are 0.76 W and 1.2 W, respectively. Find the terminal velocity of the bar and the values of R1 and R2 . (1994)

478

Part V. Electromagnetism R1

A

A

C

R2 B

~ be directed into the paSol. Let the magnetic field B per. The motional emf induced in the rod is e = Blv, where l = 1 m is length of the rod and v is its downward velocity. The equivalent circuit is shown in the figure, where i is the current through the rod. C

×

×

×

×

×

~ B

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

F

C

Sol. The flux through AEFD is φ1 = BA1 and through EBCF is φ2 = BA2 , where A1 = 1 × 1 = 1 m2 and A2 = 0.5 × 1 = 0.5 m2 . E 0.5Ω

1Ω

A

B 1Ω i2

1Ω

e2

i1

e i

⊗ ~ B

D

D R2

The rod attains terminal velocity when upward magnetic force is balanced by its weight i.e., ilB = mg.

(1)

The resistances R1 and R2 are connected in parallel and hence the current i is given by i = e (1/R1 + 1/R2 ) .

(2)

The power dissipated in R1 and R2 are given by P1 = 0.76 W = e2 /R1 , 2

P2 = 1.2 W = e /R2 .

(3) (4)

Solve equations (1)–(4) to get lB(P1 + P2 ) 1 × 0.6(0.76 + 1.2) = = 0.6 V, mg 0.2 × 9.8 R1 = e2 /P1 = 0.62 /0.76 = 0.47 Ω,

e=

R2 = e2 /P2 = 0.62 /1.2 = 0.3 Ω. The terminal velocity is given by v=

×

B

×

1Ω e1

R1

B

×

D

D

A

E ×

1Ω

C

The magnitude of induced emf in these loops are dφ1 = A1 dB = 1 × 1 = 1 V. e1 = dt dt dφ2 = A2 dB = 0.5 × 1 = 0.5 V. e2 = dt dt The magnetic field is into the paper and increasing. According to Lenz’s law, the induced currents in two loops, say i1 and i2 , shall be anticlockwise. The equivalent circuit giving the resistance for each arm, induced emf, and induced current is shown in the figure. Kirchhoff’s law in loops AEFD and EBCF gives 4i1 − i2 = e1 ,

(1)

3i2 − i1 = e2 .

(2)

Solve equations (1) and (2) to get i1 = 7/22 A and i2 = 6/22 A. 7 6 1 Ans. 22 A (E → A), 22 A (B → E), 22 A (F → E) Q 56. A circuit containing a two position switch S is shown in the figure. (1991)

0.6 e = = 1 m/s. lB 1 × 0.6

C

R3 2Ω

Ans. 1 m/s, R1 = 0.47 Ω, R2 = 0.3 Ω Q 55. A rectangular frame ABCD, made of uniform metal wire, has a straight connection between E and F made of the same wire, as shown in the figure. AEFD is a square of side 1 m and EB = FC = 0.5 m. The entire circuit is placed in a steadily increasing uniform magnetic field directed into the plane of the paper and normal to it. The rate of change of the magnetic field is 1 T/s. The resistance per unit length of the wire is 1 Ω/m. Find the magnitudes and directions of the currents in the segments AE, BE and EF. (1993)

F 0.5Ω

R1

2µF E1

1Ω R5

2Ω 1

12V S A E2

2

3V

R2 B 2Ω 3Ω R4

L 10mH

(a) The switch S is in position 1. Find the potential difference VA − VB and the rate of production of joule heat in R1 .

Chapter 34. Electromagnetic Induction

479

(b) If now the switch S is put in position 2 at t = 0. Find, (i) steady current is R4 . (ii) the time when current in R4 is half the steady value. Also calculate the energy stored in the inductor L at that time. Sol. In the steady state, there is no current through the capacitor (infinite resistance). Thus, equivalent circuit when the switch is connected at 1 is as shown in the figure. R3 2Ω

2Ω

A

i2

E1

R1

12V

E2

R2

3V

2Ω

i1 B

Kirchhoff’s law in the upper and lower loop gives E1 − i2 R3 − (i1 + i2 )R1 = 0,

(1)

E1 − i1 R2 − E2 − (i1 + i2 )R1 = 0.

(2)

where time constant τ = L/R = 2 × 10−3 s. The current becomes half of steady state value when i = i0 /2. Substitute i = i0 /2 in equation (6) and solve to get t = τ ln 2 = 2 × 10−3 × 0.693 = 1.386 × 10−3 s. The energy stored in the inductor at this time is given by U = 21 L(i0 /2)2 =

1 2

= 4.5 × 10−4 J. Ans. (a) −5 V, 24.5 W (b) (i) 0.6 A (ii) 1.386 × 10−3 s, 4.5 × 10−4 J Q 57. Two long parallel horizontal rails, a distance d apart and each having a resistance λ per unit length, are joined at one end by a resistance R. A perfectly conducting rod MN of mass m is free to slide along the rails without friction (see figure). There is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper. A variable force F is applied to the rod MN such that, as the rod moves, a constant current i flows through R. Find (a) the velocity of the rod and the applied force F as functions of the distance x of the rod from R, and (b) fraction of the work done per second by F converted to heat. (1988) M

Substitute the values of resistances and battery voltages in equations (1) and (2) to get i1 + 2i2 = 6,

(3)

4i1 + 2i2 = 9.

(4)

Solve equations (3) and (4) to get i1 = 1 A and i2 = 2.5 A. The potentials at point A and B are related by, VA = VB − i1 R2 − E2 = VB − 5 i.e., VA − VB = −5 V. The rate of joule heat production in R1 is given by P = (i1 + i2 )2 R1 = (3.5)2 × 2 = 24.5 W.

R

i

R2

3V

2Ω

i

M

Fm

F

B

v=

3Ω R4

d dx dt

N

L 10mH

x

In the steady state, there is no voltage across the inductor (zero resistance). Kirchhoff’s loop law in the steady state gives E2 − i0 R2 − i0 R4 = 0.

d

Sol. Consider a time t when the rod is at a distance x from the resistance R and it is moving with a speed v = dx/dt towards the right.

R

E2

F

N

The circuit when the switch is connected to 2 is shown in the figure. A

× 10 × 10−3 (3/10)2

(5)

2 Solve equation (5) to get i0 = R2E+R = 35 A. It is an 4 L-R circuit with resistance R = R2 + R4 = 5 Ω. The current in L-R circuit is given by h i i = i0 1 − e−t/τ , (6)

The motional emf induced in the rod is e = dBv = dB dx/dt. The magnetic force on the positive charges inside the rod is towards the end M. Thus, M is at positive potential and current in the circuit is as shown in the figure. The effective resistance of the circuit is Re = R + 2λx. The current through the circuit is i=

e dB dx = , Re R + 2λx dt

480

Part V. Electromagnetism

which gives

Region I

(1) r

The forces on the rod are applied force F~ = F ˆı ~ = −idB ˆı (force on and the magnetic force F~m = id~ × B a current carrying conductor moving in the magnetic field). The resultant force on the rod is F~r = F~ + F~m = (F − idB) ˆı.

C

O

D

(a) Obtain an expression for the magnitude of the induced current in the loop. (b) Show the direction of the current when the loop is entering into the region II. (c) Plot a graph between the induced current and the time of rotation for two periods of rotation.

Apply Newton’s second law on the rod   d dx d2 x , F − idB = m 2 = m dt dt dt to get 

A

ω

dx i(R + 2λx) v= = . dt dB

Region II

Sol. The loop rotates with a constant angular velocity ω. It takes time T = 2π/ω to complete one revolution. Let OP separates the two regions and OA be parallel to OP at t = 0. The angle θ made by OA with OP at time t is given by θ = ωt.



2λi dx dB dt 2λmi2 (R + 2λx) = idB + , d2 B 2

F = idB + m

Region I

where we have used equation (1) to get dx/dt and its derivative. The heat generated per second in the circuit with effective resistance Re is 2

2

dQ/dt = i Re = i (R + 2λx).

Region II

P

ω C

r

θ

A

O

(2) D

The work done per second (power) by the force F is dW =Fv dt    2λmi2 (R + 2λx) i(R + 2λx) = idB + . d2 B 2 dB (3) Divide equation (2) by (3) to get the fraction of the work done per second by F converted to heat as dQ/dt = dW /dt 1+

1 2mλ(R+2λx)i B 3 d3

.

Ans. (a) v = F = idB +

2

2λi m B 2 d2

(R + 2λx) (b)

(R+2λx)i , Bd

1 1+

2mλ(R+2λx)i B 3 d3

Q 58. Space is divided by the line AD into two regions. Region I is field free and the region II has a uniform magnetic field B directed into the plane of the paper. ACD is a semicircular conducting loop of radius r with centre at O, the plane of the loop being in the plane of the paper. The loop is now made to rotate with a constant angular velocity ω about an axis passing through O and perpendicular to the plane of the paper. The effective resistance of the loop is R. (1985)

The area of the loop inside the region II increases with time from t = 0 to t = T /2 and decreases from t = T /2 to t = T i.e., ( 1 θr2 = 1 ωr2 t, if 0 ≤ t < T /2; A = 21 2 12 2 2 πr − 2 ωr t, if T /2 ≤ t < T . The magnetic flux through the loop is a product of magnetic field B and area of the loop inside the magnetic field i.e., φ = BA. Faraday’s law gives the induced emf in the loop as dφ dA e=− = −B dt dt ( 1 2 − Br ω, if 0 ≤ t < T /2; = 12 2 if T /2 ≤ t < T . 2 Br ω, The induced current through the loop of resistance R is given by ( −Br2 ω/(2R), if 0 ≤ t < T /2; e i= = R Br2 ω/(2R), if T /2 ≤ t < T . The magnitude of induced current is Br2 ω/(2R).

Chapter 34. Electromagnetic Induction

481 loop. By Lenz’s law, south pole is induced to oppose the receding north pole. Thus, the current direction is clockwise i.e., OSQPO. Ans. 0.02 m/s, induced current is clockwise

i T 2

T 3T 2

2T

t

As loop enters into the region II, the flux through the loop increases due to increase in portion of the loop inside region II. The magnetic field is pointed into the paper. The situation is similar to as if north pole of a magnet is moving towards the loop. By Lenz’s law, north pole is induced into the loop to repel the approaching north pole. Thus, anticlockwise current is induced into the loop. The situation is opposite when the loop starts coming out of the region II. The current direction and magnitude repeats after each time period T. 2 Ans. (a) 21 BrR ω (b) anticlockwise (c)

Q 60. Write the dimensions of (a) Magnetic flux and (b) Rigidity modulus, in terms of mass (M ), time (T ), length (L) and charge (Q). (1982) Sol. By Faraday’s law, induced emf e = −dφ/dt. The potential energy of a charge q at potential e is U = qe = −qdφ/dt. Thus, the dimensions of the magnetic flux φ are same as the dimensions of U t/q i.e., [ML2 T−2 ][T]/[Q] = ML2 T−1 Q−1 . The modulus of rigidity is defined as the ratio of stress and strain i.e., F/A . As strain is dimensionless, the dimensions of Y = ∆l/l modulus of rigidity are same as the dimensions of the stress i.e., [MLT−2 ]/[L2 ] = [ML−1 T−2 ]. Ans. (a) [ML2 T-1 Q-1 ] (b) [ML-1 T-2 ]

i T 2

T 3T 2

2T

t

Q 59. A square metal wire loop of side 10 cm and resistance 1 Ω is moved with a constant velocity v0 in a uniform magnetic field of induction B = 2 Wb/m2 as shown in the figure. The magnetic field lines are perpendicular to the plane of the loop (directed into the paper). The loop is connected to a network of resistors each of value 3 Ω. What should be the speed of the loop so as to have a steady current of 1 mA in the loop? Give the direction of current in the loop. (1983) Q

v0

P 3Ω

3Ω ~ B

3Ω

A 3Ω S

C

Q 61. Three identical closed coils A, B and C are placed with their planes parallel to one another. Coils A and C carry equal currents as shown in the figure. Coils B and C are fixed in position and coil A is moved towards B with uniform motion. Is there any induced current in B ? If no, give reasons. If yes, mark the direction of the induced current in the diagram. (1982)

A

3Ω

i = e/R = 0.2v0 /4 = 1 mA = 10−3 A, which gives v0 = 0.02 m/s. The magnetic field is directed into the paper. The magnetic flux through the loop, φ = BA, decreases because the area A of the loop within the magnetic field region decreases with time. The situation is similar to as if north pole of a magnet is receding away from the

C

Sol. The magnetic fields at the coil B due to the coil ~ A ) and the coil C (B ~ C ) are from right to left (see A (B figure). ~v

O

Sol. Note that the given network of resistances forms a balanced Wheatstone bridge. The effective resistance of this network is Rn = (3 + 3) k (3 + 3) = 3 Ω. The resistance of the wire loop is Rl = 1 Ω. Thus, the total resistance of the circuit is R = Rn + Rl = 4 Ω. The induced emf in a square loop of side l = 10 cm moving with a constant velocity v0 in a uniform magnetic field B = 2 Wb/m2 is given by e = Blv0 = (2)(0.1)v0 = 0.2v0 . The current through the loop is 1 mA i.e.,

B

~C B ~A B A

B

C

The magnetic flux through the coil B due to the ~ C does not change with time because current field B through the coil C is constant and coils B and C are fixed. When the coil A is moved towards the coil B with ~ A increases with a uniform velocity, the magnetic field B time because of reduction in distance between the coils. Thus, the magnetic flux through the coil B due to the ~ A increases with time. By Faraday’s law, the emf field B is induced in the coil B which sets an induced current in it. This situation is similar to as if south pole of a magnet is approaching the coil B from the left side. By Lenz’s law, south pole is induced in the coil B to oppose the approaching south pole. Thus, the induced current is clockwise when looking from the left (see figure). We encourage you to find the induced emf e in the coil B

482

Part V. Electromagnetism

if coils A and C have the same current i and radius r. Take radius of the coil B as a, separation between the coils A and B as x, and the velocity of the coil A as v. 2 2 0 πivxa r Assume that a  r  x. Hint: e = 3µ . 2(x2 +r 2 )5/2 Ans. Yes, in the direction opposite to A. Q 62. The two rails of a railway track, insulated from each other and the ground, are connected to a millivoltmeter. What is the reading of the millivoltmeter when a train travels at a speed of 180 km/h along the track given that the vertical component of earth’s magnetic field is 0.2 × 10−4 Wb/m2 and the rails are separated by 1 m? Track is south to north. (1981) Sol. The train is moving towards the north with a velocity v = 180 km/h = 50 m/s. N ~v B

E

~v

l

The component of earth’s magnetic field in the vertical direction is Bv = 0.2 × 10−4 Wb/m2 . The separation between the track, l = 1 m, is in the horizontal ~ v , ~v , and plane and is perpendicular to north. Thus, B ~l are perpendicular to each other. The motional emf induced between the tracks is given by e = Bv lv = (0.2 × 10−4 )(1)(50) = 10−3 V = 1 mV. Ans. 1 mV Q 63. A current from A to B is increasing in magnitude. What is the direction of induced current in the loop as shown in the figure? (1979)

A

B

~ due to a straight wire is cirSol. The magnetic field B cumferential. It is coming out of the paper at the location of loop. ~ B A

B

The magnitude of magnetic field increases with increase in current through AB. Thus, the magnetic flux through the loop is increasing with time. The situation

is similar to as if south pole of a magnet is approaching the loop from above the paper. By Lenz’s law, south pole is induced to oppose the approaching south pole. Thus, the induced current is in clockwise direction. Ans. Clockwise

Chapter 35 Alternating Current

One Option Correct

L = 1 µH C = 1µF R = 1 kΩ

Q 1. An AC voltage source of variable angular frequency ω and fixed amplitude V0 is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When ω is increased, (2010) (A) the bulb glows dimmer. (B) the bulb glows brighter. (C) total impedance of the circuit is unchanged. (D) total impedance of the circuit increases.

V0 sin ωt

(A) At ω ≈ 0 the current flowing through the circuit becomes nearly zero. (B) The frequency at which the current will be in phase with the voltage is independent of R. (C) The current will be in phase with the voltage if ω = 104 rad/s. (D) At ω  106 rad/s, the circuit behaves like a capacitor.

Sol. The impedance of the R-C circuit, Z = p R2 + 1/(ω 2 C 2 ), decreases with increase in ω. This increases the current in the circuit as I = V0 /Z. The power of the bulb, I 2 R, increases and hence the bulb glows brighter. Ans. B

Sol. In a series LCR circuit, the current is in phase with the voltage at resonant frequency

Q 2. When an AC source of emf e = e0 sin(100t) is connected across a circuit, the phase difference between the emf e and the current i in the circuit is observed to be π/4, as shown in the diagram. If the circuit consists possibly only R-C or R-L or L-C in series, find the relationship between the two elements. (2003) i

ωr = √

1 1 =p = 106 rad/s. LC (10−6 )(10−6 )

The resonant frequency is independent of R. If the frequency is very low (ω ≈ 0) then inductive reactance XL = ωL → 0 and capacitive reactance XC = 1/(ωC) → ∞. Thus, the impedance of the circuit becomes very large and hence the current flowing through the circuit becomes nearly zero. Thus, at low frequencies, the circuit behaves like a capacitor as its behaviour is largely controlled by XC . If the frequency is very large (ω  ωr ) then inductive reactance XL = ωL → ∞ and capacitive reactance XC = 1/(ωC) → 0. Again, the impedance of the circuit becomes very large and the current flowing through the circuit becomes nearly zero. Thus, at high frequencies, the circuit behaves like an inductor as its behaviour is largely controlled by XL . We encourage you to deduce these results from the expressions of phase φ and current i,

e

t

(A) R = 1 kΩ, C = 10 µF (C) R = 1 kΩ, L = 10 H

˜

(B) R = 1 kΩ, C = 1 µF (D) R = 1 kΩ, L = 1 H

Sol. From the given diagram i is leading e, which is a characteristic of a series R-C circuit. The phase dif1 , gives RC = 1/ω = ference, tan φ = 1 = XRC = ωRC 1/100. Ans. A One or More Option(s) Correct

−1



φ = tan

Q 3. In the circuit shown, L = 1 µH, C = 1 µF and R = 1 kΩ. They are connected in series with an AC source V = V0 sin ωt as shown. Which of the following option(s) is(are) correct? (2017)

i= q

483

1 ωC

− ωL R

V0 R2 +

1 ωC

 ,


2 sin(ωt + φ). − ωL

484

Part V. Electromagnetism π 2

φ O

ωr

ω

− π2 ir i0 O

ω

ωr

The variation of phase difference (between the current and the voltage), φ, with the frequency ω is shown in the first figure. Note that φ → π/2 when ω → 0 (capacitive) and φ → −π/2 when ω → ∞ (inductive). At the resonant frequency ωr , the current is in phase with the voltage i.e., φ = 0 at ω = ωr . The second figure shows the variation of the magnitude of the current (i0 ) with frequency. The magnitude of the current attains its maximum value, ir = V0 /R, at the resonant frequency ωr . Ans. (A), (B) Q 4. A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane ~ points into (of the paper). A uniform magnetic field B the plane of the paper. At t = 0, the loop starts rotating about the common diameter as axis with a constant angular velocity ω in the magnetic field. Which of the following option(s) is(are) correct? (2017)

Area A

Area 2A

ω

(A) The emf induced in the loop is proportional to the sum of the areas of the two loops. (B) The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper. (C) The net emf induced due to both the loops is proportional to cos ωt. (D) The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone.

Sol. Consider the coordinate axes shown in the figure. ˆ Let ~ = B k. The magnetic field is into the paper i.e., B loop 1 be the loop of area A and loop 2 be the loop of area 2A. At t = 0, both loops are in the plane of the paper. They start rotating about the x-axis with a constant angular velocity ω ~ = ωˆı. Let the area vectors ~ 1 and A ~ 2 , respectively. At of loop 1 and loop 2 be A ˆ ˆ ~ ~ t = 0, A1 = Ak and A2 = 2Ak (we chose area into the plane of the paper as positive). In time t, the loops rotate by an angle θ = ωt. Their area vectors make an angle θ = ωt with z-axis (direction of the magnetic field). Thus, the magnetic fluxes passing through the loop 1 and the loop 2 at time t are given by ~ ·A ~ 1 = BA1 cos θ = BA cos ωt, φ1 = B ~ ·A ~ 2 = BA2 cos θ = 2BA cos ωt. φ2 = B y ~2 A

E1

~1 A ωt ~ B E1

z

E2

E2

x z

×

y

ω

By Faraday’s law of electromagnetic induction, induced emfs in the loop 1 and in the loop 2 are given by dφ1 = BAω sin ωt, dt dφ2 E2 = − = 2BAω sin ωt. dt The rate of change of flux is maximum when ωt = π/2 i.e., when the plane of the loops is perpendicular to the plane of the paper. Let T = 2π/ω be the time period to complete one revolution. The fluxes φ1 and φ2 decrease with time if t ∈ (0, T /4). By Lenz’s law, the induced current should oppose reductions in φ1 and φ2 . Thus, induced current in both the loops should be clockwise. Hence, the polarity of ‘equivalent batteries’ are as shown in the figure. These batteries are connected with reverse polarity. Thus, net emf induced due to both the loop is E1 = −

E = E2 − E1 = BAω sin ωt. Note that the net emf E is proportional to the difference in areas of two loops and |E| = |E1 |. We encourage you to relate the given configuration with that of an AC generator (dynamo). Ans. (B), (D)

Chapter 35. Alternating Current

485

Q 5. The instantaneous voltages at three terminals marked X, Y and Z are given by VX = V0 sin ωt, VY = V0 sin(ωt + 2π/3), and VZ = V0 sin(ωt + 4π/3). An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between point X and Y and then between Y and Z. The reading(s) (2017) p of the voltmeter will be rms (A) VYZ = V0 p1/2. rms (B) VXY = V0 3/2. (C) independent of the choice of the two terminals. rms (D) VXY = 0. Sol. The potential difference between point X and Y is given by

Aliter: This problem can be easily solved by phaser diagram (see figure). The potentials VX , VY and VZ form vertices of an equilateral triangle of side V0 . The symmetry makes the subtraction of vectors very easy. Ans. (B), (C) Q 6. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current i(t) = i0 cos(ωt), with i0 = 1 A and ω = 500 rad/s starts flowing in it with initial direction 7π , the key is switched from shown in the figure. At t = 6ω B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 µF, R = 10 Ω and the battery is ideal with emf of 50 V, identify the correct statement (s). (2014) B

VXY = VX − VY
= V0 sin (ωt) − V0 sin ωt + 2π 3

= 2V0 cos ωt + π3 sin − π3 √
√
= − 3V0 cos ωt + π3 = 3V0 sin ωt − π6 . Similarly, the potential difference between point Y and Z is

D A

˜

50V C = 20µF R = 10Ω




(A) Magnitude of the maximum charge on the capaci7π is 1 × 10−3 C. tor before t = 6ω (B) The current in the left part of the circuit just before 7π t = 6ω is clockwise. (C) Immediately after A is connected to D, the current in R is 10 A. (D) Q = 2 × 10−3 C.

and the potential difference between point Z and X is

Sol. The current flowing through the capacitor and its charge are given by

VYZ = VY − VZ
− V0 sin ωt + 4π 3
= 2V0 cos (ωt + π) sin − π3 √ √
= 3V0 cos(ωt) = 3V0 sin ωt + π2 ,

= V0 sin ωt +

2π 3

VZX = VZ − VX
= V0 sin ωt + 4π 3 − V0 sin (ωt) √
√ = 3V0 sin ωt + = 3V0 cos ωt + 2π 3

7π 6




.

The rms value of the√potential V = V0 sin(ωt + φ) is given by V rms = V0 / 2. Thus, rms values p of the porms rms rms tentials are VXY = VYZ = VZX = V0 3/2. Hence, the reading of the voltmeter is independent of the two terminal i.e., reading is same whether it is connected across X-Y or Y -Z or Z-X.

i = i0 cos ωt, Z Z i0 Q = idt = i0 cos ωt dt = sin ωt + k ω i0 = sin ωt, ω

(1)

(2)

where we have used the initial condition, Q = 0 at t = 0, to get the integration constant k = 0. The charge attains the maximum value Qmax at t = π/(2ω), Qmax = i0 /ω = 1/500 = 2 × 10−3 C.

−VZ

VY 2π 3

7π 6

The key is switched from B to D at the time t = 7π/6ω. Substitute t = 7π/6ω in equations (1) and (2) to get current and charge

VYZ

4π 3

i = i0 cos(7π/6) = cos(π + π/6) = − cos(π/6) √ = − 3/2 A,

π 2

π 6

Qi = (i0 /ω) sin(7π/6) = 2 × 10−3 sin(π + π/6)

VX

= −1 × 10−3 C. −VY

VZX VZ −VX

VXY

Note that negative sign in i indicates the counterclockwise direction of current and negative sign in Qi indicates the negative charge on the upper plate.

486

Part V. Electromagnetism

The potential across the capacitor is VC = QCi = 1×10−3 − 20×10 −6 = −50 V (upper plate at lower potential). Immediately after A is connected to D, the potential across the resistor is VR = 100 V and hence the current through it is i = VR /R = 100/10 = 10 A (counterclockwise). The potential across the capacitor when it is fully charged is equal to the battery emf. Thus, the charge on the capacitor in fully charged condition is Qf = V /C = 50/(20 × 10−6 ) = 1 × 10−3 C. The sign of Qf indicates the positive charge on the upper plate. Hence, charge that flows through the battery is

by 1 = 100 Ω, ωC q √ Z1 = XC2 + R12 = 100 2 Ω,

XC =

φ1 = 45◦ . The current I1 =

Q 7. In the given circuit, the AC source has ω = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are) (2012) 100µF

100Ω

0.5H

50Ω

˜

20V

current through the circuit, I is 0.3√ A. current through the circuit, I is 0.3 √ 2 A. voltage across 100 Ω resistor = 10 2 V. voltage across 50 Ω resistor = 10 V.

Sol. Let I1 be the current in branch containing capacitor C = 100 µF and resistor R1 = 100 Ω. We call this branch as capacitive branch. Let I2 be the current in branch containing inductor L = 0.5 H and resistor R2 = 50 Ω (see figure). We call this branch as inductive branch. I1

I

I2

100µF 100Ω 0.5H

˜

20V 1 ωC

√

=

2 10

XL = ωL = 50 Ω, q √ Z2 = XL2 + R22 = 50 2 Ω, φ2 = 45◦ .

Ans. C, D

The The The The

20√ 100 2

A leads voltage by √ 45 . The voltage across R1 is V100Ω = I1 R1 = 10 2 V. The inductive reactance XL , impedance Z2 , and phase difference φ2 for the inductive branch are given by

= 2 × 10−3 C.

(A) (B) (C) (D)

=

◦

Q = Qf − Qi = 1 × 10−3 − (−1 × 10−3 )

I

V Z1

Z1 φ1 R1

50Ω

R2 φ2

ωL Z 2

The current I2 =

=

20 √ 50 2

=

√ 2 2 10

A lags voltage by √ 45 . The voltage across R2 is V50Ω p = I2 R2 = 10√ 2 V. Since I1 leads I2 by 90◦ , I = I12 + I22 = 1/ 10 ≈ 0.3 A. Ans. A, C ◦

Q 8. A series R-C circuit is connected to AC voltage source. Consider two cases (A) when C is without a dielectric medium and (B) when C is filled with dielectric medium of dielectric constant 4. The current IR through the resistor and voltage VC across the capacitor are compared in two cases. Which of the following is (are) true? (2011) B A B A < IR (B) IR > IR (A) IR (C) VCA > VCB (D) VCA < VCB Sol. Introducing the dielectric medium makes the capacitance in Case B four times of the capacitance in Case A. Thus, the circuit impedances in the two cases are given by p Z A = R2 + 1/(ω 2 C 2 ), and p Z B = R2 + 1/(16ω 2 C 2 ). The currents through R (and also through C) in the A B two cases are IR = V /Z A and IR = V /Z B which gives A B A B IR < IR (since Z > Z ). The voltages across C in the two cases are V 1 A A VCA = IR XC = p 2 2 2 R + 1/(ω C ) ωC V =√ , R2 ω 2 C 2 + 1 V 1 B B VCB = IR XC = p 2 2 2 R + 1/(16ω C ) 4ωC =√

The capacitive reactance XC , impedance Z1 , and phase difference φ1 for the capacitive branch are given

V Z2

V 16R2 ω 2 C 2

+1

< VCA . Ans. B, C

Chapter 35. Alternating Current

487

Paragraph Type Paragraph for Questions 9-10 A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers’ usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a stepup transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers’ end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the currents and voltages mentioned are rms values. (2013) Q 9. If the direct transmission method with a cable of resistance 0.4 Ω/km is used, the power dissipation (in %) during transmission is (A) 20 (B) 30 (C) 40 (D) 50 Sol. The current through the cable is given by i = VP = 600×103 = 150 A. The resistance of the cable is R = 4000 0.4 × 20 = 8 Ω. The power loss in the cable is i2 R = 180 kW which amounts to 30% of 600 kW. Ans. B Q 10. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is (A) 200 : 1 (B) 150 : 1 (C) 100 : 1 (D) 50 : 1 Sol. For an ideal transformer with power factor unity, Vs /Vp = ns /np , where V is the voltage and n is the number of turns (subscript s stands for secondary and p for primary). For step-up transformer, Vp = 4000 V and np /ns = 1/10, which gives Vs = 40000 V. For step-down transformer, Vp = 40000 V and Vs = 200 V, which gives np /ns = 200. Ans. A Paragraph for Questions 11-13 The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage source V , by closing switch S1 while keeping switch S2 open. The capacitor can be connected with an inductor L by closing switch S2 and opening S1 . (2006)

V R

C

S1

S2 L

Q 11. Initially, the capacitor was uncharged. Now, switch S1 is closed and S2 is kept open. If time constant of this circuit is τ , then, (A) after time interval τ , charge on capacitor is CV /2. (B) after time interval 2τ , charge on capacitor is
CV 1 − e−2 . (C) the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged. (D) after time interval 2τ , charge on capacitor is
CV 1 − e−1 . Sol. Let i be the current in the circuit at time t. The potential across R is iR and across C is q/C. Kirchhoff’s law gives iR + q/C = V,

or

1 V dq + q= . dt RC R

Integrate with initial condition q = 0 at t = 0 to get t

q = CV (1 − e− RC ) = CV (1 − e−t/τ ), where τ = RC is time constant. Substitute t = 2τ to get q = CV (1 − e−2 ). When capacitor is fully charged (t → ∞), the current i = dq dt = 0 and hence there is no heat dissipation. Ans. B Q 12. After the capacitor gets fully charged, S1 is opened and S2 is closed so that the inductor is connected in series with the capacitor. Then, (A) at t = 0, energy stored in the circuit is purely in the form of magnetic energy. (B) at any time t > 0, current in the circuit is in the same direction. (C) at t > 0, there is no exchange of energy between the inductor and capacitor. (D) at any time t > 0, maximum p instantaneous current in the circuit may be V C/L. Sol. The circuit acts as an oscillator with electrostatic energy 12 CV 2 getting converted into magnetic energy 1 2 2 LI and vice versa alternately. The current becomes maximum when the entire electrostatic energy 12 CV 2 2 is converted to magnetic energy i.e., 21 LImax = 12 CV 2 , p which gives Imax = V C/L. Ans. D Q 13. If the total charge stored in the LC circuit is q0 , then for t ≥ 0, (A) the  charge on =  the capacitor is q π t q0 cos 2 + √LC .

488

Part V. Electromagnetism

(B) the

charge on   the π t q0 cos 2 − √LC .

capacitor

is

q

=

2

(C) the charge on the capacitor is q = −LC ddt2q . 2

1 d q (D) the charge on the capacitor is q = − √LC dt2 .

Sol. In L-C circuit, √ charge oscillates with an angular frequency ω = 1/ LC. Comparing it with SHM, we get q d2 q = −ω 2 q = − . dt2 LC The solution of this equation is q = q0 sin(ωt + φ). The initial condition (when S2 is closed), q = q0 at t = 0, gives q = q0 sin(ωt + π/2). Ans. C

Q 14. You are given many resistances, capacitors and inductors. These are connected to variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current (steady state for DC or rms for AC) flows through the circuit, the corresponding votage V1 and V2 (indicated in circuits) are related as shown in Column I. Match the two, (2010)

Column I

Column II V1

V2

(p) 6mH

3µF V

(B) I 6= 0, V2 > V1

(q)

V1

V2

6mH

2Ω V

(C) V1 = 0, V2 = V

(r)

V1

V2

6mH

2Ω

˜ V

V1

(D) I 6= 0, V2 ∝ I

XL = ωL = 2πνL = 1.88 Ω, q Z = XL2 + R2 = 2.75 Ω.

and

Thus, I = V /Z 6= 0, V1 = XL I = 1.88I, V2 = RI = 2I, and V2 > V1 . In circuit (s), inductive reactance XL , capacitive reactance XC , and impedance Z are XL = 1.88 Ω, XC = 1/(ωC) = 1061 Ω,

and

Z = XC − XL = 1059 Ω.

Matrix or Matching Type

(A) I 6= 0, V1 ∝ I

In circuit (q), inductor will act as zero resistance in steady state giving us I = V /R = V /2, V1 = 0, and V2 = V . In circuit (r ), the inductive reactance XL and impedance Z are

(s)

6mH

V2 3µF

˜ V

V1

(t)

1kΩ

V2

Thus the current in the circuit I = V /Z 6= 0, V1 = XL I = 1.88I, V2 = XC I = 1061I, and V2 > V1 . In circuit (t), XC , R, and Z are XC = 1/(ωC) = 1061 Ω, R = 1000 Ω, and q Z = R2 + XC2 = 1458 Ω. Thus, I = V /Z 6= 0, V1 = RI = 1000I, V2 = XC I = 1061I, and V2 > V1 . Ans. A7→(r,s,t), B7→(q,r,s,t), C7→(p,q), D7→(q,r,s,t) Integer Type Q 15. A series R-C combination is connected to an AC voltage of angular frequency ω = √ 500 rad/s. If the impedance of the R-C circuit is R 1.25, the time constant (in milliseconds) of the circuit is . . . . . . . (2011) Sol. impedance √ of R-C circuit is p The R2 + 1/(ωC)2 = R 1.25. On simplification, we get time constant RC = 2/ω = 4 × 10−3 s. Ans. 4 Descriptive Q 16. In the circuit shown A and B are two cells of same emf E but different internal resistances r1 and r2 (r1 > r2 ) respectively. Find the value of R such that the potential difference across the terminals of cell A is zero a long time after the key K is closed. (2004) R

3µF

˜ V

Sol. In circuit (p), under steady state, capacitor will act like an infinite impedance and inductor will act like a zero impedance. Thus, I = 0, V1 = 0, and V2 = V .

R A

B

r1 r2

L

R R R R

C K

Chapter 35. Alternating Current

489

Sol. A long time after the key K is pressed, the capacitor becomes an open circuit element (XC = ∞) and inductor becomes a closed circuit element (XL = 0). The effective resistance of six resistances, each of value R, is 34 R and total resistance of the circuit is r1 +r2 + 34 R. Kirchhoff’s loop law gives the current through the circuit i = 2E/(r1 + r2 + 34 R). The voltage across cell A is, E − ir1 = 0, which gives R = 43 (r1 − r2 ). Ans. R = 43 (r1 − r2 ) Q 17. In an L-R series circuit, a sinusoidal voltage V = V0 sin ωt is applied. It is given that L = 35 mH, R = 11 Ω, Vrms = 220 V, ω/(2π) = 50 Hz, and π = 22/7. Find the amplitude of current in the steady state and obtain the phase difference between the current and the voltage. Also plot the variation of current for one cycle on the given graph. (2004) V

T 4

T 2

3T 4

T

t

Sol. The amplitude of applied voltage is V0 = √ √ 2Vrms = 220 2 V. The inductive reactance is XL = −3 ωL = 2π × 50 × 35 Ω and impedance √ of the p × 10 ≈ 11 √ circuit is Z = R2 + XL2 = 112 + 112 = 11 2 Ω. The amplitude of the current in the circuit is I0 = V0 /Z = 20 A. V

I

T 4

T 2

3T 4

T

t

The phase difference between the current and voltage is φ = tan−1 (XL /R) = π/4. In L-R circuit, current lags the voltage by φ. The applied voltage is V = V0 sin ωt and current in the circuit is I = I0 sin(ωt − π/4). Ans. 20 A, π/4 Q 18. An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance 5.0 µF and the resulting L-C circuit is set oscillating at its natural frequency. Let Q denotes the instantaneous charge on the capacitor and I the current in the circuit. It is found that the maximum value of Q is 200 µC. (1998) (a) (b) (c) (d)

When Q = 100 µC, what is the value of | dI dt |? When Q = 200 µC, what is the value of I? Find the maximum value of I. When I is equal to one-half of its maximum value, what is the value of |Q|?

Sol. The charge in L-C circuit oscillates with an angular frequency ω given by 1 1 =p −3 LC 2.0 × 10 × 5.0 × 10−6 = 104 s−1 .

ω=√

The charge Q(t) on the capacitor, the current I(t) in the circuit, and time derivative of the current, dI/dt, are given by Q(t) = Q0 sin(ωt + φ), dQ = Q0 ω cos(ωt + φ), I(t) = dt dI(t) = −Q0 ω 2 sin(ωt + φ), dt

(1) (2) (3)

where Q0 = 200 µC is the maximum value of the charge and φ is the phase. Substitute Q(t) = 100 µC in equation (1) to get sin(ωt + φ) = 1/2. Substitute this in equation (3) to get dI(t) −6 8 4 dt = (200 × 10 )(10 )(1/2) = 10 A/s. Substitute Q = 200 µC in equation (1) to get sin(ωt + φ) = 1 and cos(ωt + φ) = 0. Substitute this in equation (2) to get I(t) = 0. Maximum value of I(t) occurs at cos(ωt + φ) = 1. Substitute this in equation (2) to get Imax = Q0 ω = 2 A. The energy of the system is conserved i.e., 1 2 1 1 Q2 LImax = LI 2 + . 2 2 2 C p 2 Simplify to get Q = √ LC(Imax − I 2 ). Substitute I = Imax −4 3 × 10 C. 2 = 1 A to get Q = Ans. (a) 104 A/s (b) zero (c) 2.0 A (d) 1.732 × 10−4 C

Chapter 36 Electromagnetic Waves

One Option Correct

Q 3. The relation between [E] and [B] is (A) [E] = [B][L][T] (B) [E] = [B][L]−1 [T] −1 (C) [E] = [B][L][T] (D) [E] = [B][L]−1 [T]−1

Q 1. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3 × 108 m/s. The final momentum of the object is (2013) (A) 0.3 × 10−17 kg m/s (B) 1.0 × 10−17 kg m/s (C) 3.0 × 10−17 kg m/s (D) 9.0 × 10−17 kg m/s

Sol. In electromagnetic waves, the amplitude of electric field is related to the amplitude of magnetic field by E = cB, where c is speed of light. The dimension of c is [L][T]−1 . Thus, [E] = [B][L][T]−1 . Ans. (C)

Sol. The power is energy per unit time. The pulse of 100 ns duration and 30 mW power will have energy, E = (30 × 10−3 ) × (100 × 10−9 ) = 3 × 10−9 J. For light, the energy and momentum are related by E = −9 pc. Hence, momentum of the pulse is, p = 3×10 3×108 = 10−17 kg m/s. The conservation of momentum gives the final momentum of the object as 10−17 kg m/s. Ans. B

Q 4. The relation between [0 ] and [µ0 ] is (A) [µ0 ] = [0 ][L]2 [T]−2 (B) [µ0 ] = [0 ][L]−2 [T]2 (C) [µ0 ] = [0 ]−1 [L]2 [T]−2 (D) [µ0 ] = [0 ]−1 [L]−2 [T]2 Sol. The speed of an electromagnetic waves in free space is given by √ c = 1/ µ0 0 ,

One or More Option(s) Correct Q 2. In terms of potential difference V , electric current I, permittivity 0 , permeability µ0 and speed of light c, the dimensionally correct equation(s) is (are) (2015) (A) µ0 I 2 = 0 V 2 (B) 0 I = µ0 V (C) I = 0 cV (D) µ0 cI = 0 V

−2 which gives µ0 = −1 . Hence, dimensions are related 0 c −1 −2 by [µ0 ] = [0 ] [L] [T]2 . Ans. (D)

Fill in the Blank Type

Sol. The speed of light in vacuum is given by c = √ 1/ µ0 0 . The impedance (resistance) of free space is p defined as R = µ0 /0 (this quantity is generally used when studying the propagation of plane electromagnetic waves in vacuum). Use these definitions to show that the dimensions of µ0 I 2 are same as dimensions of 0 V 2 . Also, the dimensions of I are same as the dimensions of p 0 cV . We encourage you to show that R = µ0 /0 is dimensionally correct and its value is 376 Ω. Ans. A, C

Q 5. If 0 and µ0 are, respectively, the electric permittivity and magnetic permeability of free space,  and µ the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is . . . . . . (1992) Sol. The speed of light in free space is c = √1 . µ

a medium is v = q is n = vc = µµ . 0 0

Paragraph Type

√1 µ0 0

The refractive index of a medium

Ans.

Paragraph for Questions 3-4 In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In questions below, [E] and [B] stand for dimensions of electric and magnetic fields respectively, while [0 ] and [µ0 ] stand for dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of length and time respectively. All the quantities are given in SI units. (2018) 490

and in

q

µ µ0 0

Part VI

Modern Physics

∞ n=6 n=5 n=4

n=3

n=2

491

Chapter 37 Electric Current through Gases

One Option Correct

The movement of electrons from the cathode to the plate (plate current) is controlled by applying a suitable voltage to the grid. The plate current is maximum when the potential of the grid and the plate is positive. Ans. D

Q 1. A tiny spherical oil drop carrying net charge q is balanced in still air with a vertical uniform electric field 5 of strength 81π 7 × 10 V/m. When the field is switched off, the drop is observed to fall with terminal velocity 2 × 10−3 m/s. Given g = 9.8 m/s2 , viscosity of the air = 1.8 × 10−5 N · s/m2 and the density of oil = 900 kg/m3 , the magnitude of q is (2010) (A) 1.6 × 10−19 C (B) 3.2 × 10−19 C (C) 4.8 × 10−19 C (D) 8.0 × 10−19 C

Q 3. Select the correct statement from the following, (1984)

(A) a diode can be used as a rectifier. (B) a triode cannot be used as a rectifier. (C) the current in a diode is always proportional to the applied voltage. (D) the linear portion of I-V characteristic of a triode is used for amplification without distortion.

Sol. The forces acting on the oil drop are its weight, buoyant force, and electrostatic force. The buoyant force on the oil drop is very small as compared to other two forces. Thus, the weight of the spherical oil drop is balanced by the electrostatic force qE = 43 πr3 ρ g.

Sol. Both diode and triode can be used as a rectifier. The I-V characteristic of diode is not linear. Ans. A

(1)

Q 4. The plate resistance of a triode is 3 × 103 Ω and its mutual conductance is 1.5 × 10−3 A/V. The amplification factor of the triode is (1981) (A) 5 × 10−5 (B) 4.5 (C) 45 (D) 2 × 105

The drop attains terminal velocity in the absence of electric field, when viscous force is equal to its weight i.e., 6πηrv = 43 πr3 ρ g.

(2)

Sol. For the given triode, plate resistance is rp = 3 × 103 Ω and mutual conductance is gm = 1.5 × 10−3 A/V. The amplification factor is given by

Eliminate r from equations (1) and (2) to get the charge  3/2 4πρg 9ηv . (3) q= 3E 2ρg

µ = rp gm = (3 × 103 ) (1.5 × 10−3 ) = 4.5.

Substitute the given parameters in equation (3) to get q = 8.0 × 10−19 C. Note that charge on oil drop is integral multiple of charge on an electron. We encourage you to find the numerical value of buoyant force and compare it with the values of other two forces. Ans. D

Ans. B True False Type Q 5. For a diode the variation of its anode current Ia with the anode voltage Va at two different cathode temperatures T1 and T2 is shown in the figure. The temperature T2 is greater than T1 . (1986)

Q 2. For a given plate voltage, the plate current in a triode valve is maximum when the potential of (1985) (A) the grid is positive and plate is negative. (B) the grid is zero and plate is positive. (C) the grid is negative and plate is positive. (D) the grid is positive and plate is positive.

Ia

Sol. In a triode, the cathode emits electrons by the process of thermionic emission. These electrons are attracted by the plate kept at positive potential.

Va

Grid Cathode Filament

T2 T1

Sol. The number of electrons emitted by the cathode in one second increases with increase in cathode temperature. Thus, for a fixed anode voltage Va , more electrons reach the anode in one second if cathode temperature is

Plate

493

494

Part VI. Modern Physics

high. Thus, the anode current Ia is high at higher cathode temperature. We encourage you to explain why Ia does not vary with cathode temperature for small Va . Ans. T Descriptive Q 6. A triode has plate characteristics in the form of parallel lines in the region of our interest. At a grid voltage of −1 V the anode current I (in mA) is given in terms of plate voltage V by the algebraic relation: I = 0.125V − 7.5. For grid voltage of −3 V, the current at anode voltage of 300 V is 5 mA. Determine the plate resistance (rp ), transconductance (gm ) and the amplification factor (µ) for the triode. (1987) Sol. Let Vp be the plate (anode) voltage, Ip be the plate (anode) current, and Vg be the grid voltage. At Vg = −1 V, the anode current is related to the plate voltage by Ip = (0.125Vp − 7.5) × 10−3 A.

(1)

The plate resistance is the slope of the Ip -Vp plot. The equation (1) gives this slope at Vg = −1 V, 1 dVp = = 8 × 103 Ω. rp = dIp Vg =−1V 0.125 × 10−3 Use equation (1) to get the anode current at the anode voltage Vp = 300 V and the grid voltage Vg1 = −1 V, Ip1 = (0.125(300) − 7.5) × 10−3 = 30 × 10−3 A. Given, Ip2 = 5 × 10−3 A at anode voltage Vp = 300 V and the grid voltage Vg2 = −3 V. Thus, we have two values of anode current at the same anode voltage (Vp = 300 V) and two different grid voltages Vg1 = −1 V and Vg2 = −3 V. Hence, the transconductance of the triode is ∆Ip (30 − 5) × 10−3 gm = = ∆Vg Vp −1 − (−3) Vp =300V = 12.5 × 10−3 A/V. The amplification factor of the triode is given by A = rp gm = (8 × 103 )(12.5 × 10−3 ) = 100. Ans. 8 kΩ, 1.25 × 10−2 A/V, 100

Chapter 38 Photoelectric Effect and Wave-Particle Duality

One Option Correct Q 1. A photoelectric material having work-function φ0 is illuminated with light of wavelength λ (λ < hc/φ0 ). The fastest photoelectron has a de Broglie wavelength λd . A change in wavelength of the incident light by ∆λ results in a change ∆λd in λd . Then the ratio ∆λd /∆λ is proportional to (2017) (A) λ2d /λ2 (B) λd /λ (C) λ3d /λ (D) λ3d /λ2

hc − φ0 . λ

h h =√ , p 2mKmax

(1)

where φ is the work function of the metal. From equation (1), 1/λ versus V0 graph is a straight line with a slope hc/e and intercept −φ/e on V0 axis. V0 2.0

(2)

1.0 0.4 0.0

1 1 1 0.5 0.4 0.3

1 λ

-2.0

(3)

Substitute the values of V0 and λ in equation (1) to get three equations in two unknowns, hc/e and φ/e. Solve any two of the three equations to get h = 6.4 × 10−34 J-s and φ = 2.0 eV. We encourage you to find h and φ if second reading is erroneously measured as (0.4 µm, 1.1 V) instead of (0.4 µm, 1.0 V). Ans. B

Differentiate equation (3) to get (−2)

2.0 1.0 0.4

eV0 = hc/λ − φ, (1)

where m is the mass of the electron and p is its linear momentum. Eliminate Kmax from equations (1) and (2) to get h2 hc = − φ0 . 2mλ2d λ

0.3 0.4 0.5

Sol. In photoelectric effect, the stopping potential (V0 ) is related to the wavelength of incident radiation by

The de Broglie wavelength of the photoelectron having p2 is given by kinetic energy Kmax = 2m λd =

V0 (Volt)

Given that c = 3 × 108 m/s and e = 1.6 × 10−19 C, Planck’s constant (in units of J-s) found from such an experiment is (2016) (A) 6.0 × 10−34 (B) 6.4 × 10−34 (C) 6.6 × 10−34 (D) 6.8 × 10−34

Sol. Here, λ is the wavelength of incident light and λd is the de Broglie wavelength of the fastest photoelectron. The fastest ejected photoelectron has the maximum kinetic energy which is given by Kmax =

λ (µm)

h2 hc ∆λd = (−1) 2 ∆λ, 3 2mλd λ

which gives

Q 3. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2 , respectively. If the ratio u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly (2014) (A) 3.7 eV (B) 3.2 eV (C) 2.8 eV (D) 2.5 eV

∆λd mc λ3d = . ∆λ h λ2 Ans. (D) Q 2. In a historical experiment to determine Planck’s constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength (λ) of incident light and the corresponding stopping potential (V0 ) are given below:

Sol. The maximum kinetic energy of ejected photoelectrons is given by Kmax = 495

hc 1 mu2 = − φ. 2 λ

(1)

496

Part VI. Modern Physics

Substitute the values in equation (1) to get 1240 1 mu21 = − φ, 2 248 1 1240 mu22 = − φ. 2 310

Sol. The de-Broglie wavelength of an electron of mass m and velocity v is given by (2) (3)

Divide equation (2) by (3) and use u1 /u2 = 2 to get φ = 3.7 eV. Ans. A Q 4. Photoelectric effect experiments are performed using three metal plates p, q and r having work functions φp = 2.0 eV, φq = 2.5 eV and φr = 3.0 eV, respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is [Take hc = 1240 eV-nm.] (2009) (A)

(C)

I

I

p q r V r q p

(B)

I

p q r

(D)

V

V I

r q p

V

Sol. The photon energy for wavelength λ is given by E = hc/λ.

(1)

Substitute the values in equation (1) to get the photon energies for three given wavelengths as E1 = 1240/550 = 2.25 eV,

(1)

The fringe width in Young’s double slit experiment is related to wavelength λ and screen distance D by β = λD/d.

(2)

Eliminate λ from equations (1) and (2) to get β = hD/(mvd). Thus, both λ and β decrease when velocity v is increased. Ans. C Q 6. A proton has kinetic energy E = 100 keV which is equal to energy of a photon. Let λ1 be the de-Broglie wavelength of the proton and λ2 be the wavelength of the photon. The ratio λ1 /λ2 is proportional to (2004) (A) E 0 (B) E 1/2 (C) E −1 (D) E −2 Sol. The de-Broglie wavelength of a particle of mass m, momentum p, and kinetic energy E is given by √ (1) λ1 = h/p = h/ 2mE. The wavelength of a photon of energy E is given by λ2 = hc/E.

(2)

Divide equation (1) by (2) to get p λ1 /λ2 = E/(2mc2 ). We encourage you to calculate λ1 and λ2 for the given energy. Ans. B

E2 = 1240/450 = 2.75 eV, E3 = 1240/350 = 3.54 eV. The energies E1 , E2 , and E3 are sufficient to emit the photoelectrons from metal p, E2 and E3 are sufficient to emit the photoelectrons from metal q, and E3 is sufficient to emit the photoelectrons from metal r. Thus, the saturation current, Ip > Iq > Ir . The stopping potential V is related to workfunction φ by V = hc/(eλ) − φ/e.

λ = h/(mv).

Q 7. The figure shows the variation of photocurrent with anode potential for a photosensitive surface for three different radiations. Let Ia , Ib and Ic be the intensities and fa , fb and fc be the frequencies for the curves a, b and c respectively. Then, (2004) I

(2) c

Since φp < φq < φr , equation (2) gives the stopping potential Vp > Vq > Vr . Ans. A Q 5. A beam of electron is used in a Young’s double slit experiment. The slit width is d. When the velocity of electron is increased, (2005) (A) no interference is observed (B) fringe width increases (C) fringe width decreases (D) fringe width remains same

b

a V

(A) fa = fb and Ia 6= Ib (C) fa = fb and Ia = Ib

(B) fa = fc and Ia = Ic (D) fb = fc and Ib = Ic

Sol. In photoelectric effect, the cut off voltage depends on the frequency of incident radiation and saturation current depends on its intensity. The cut-off voltages are equal for a and b but different for c which gives fa = fb 6= fc . The saturation currents are equal for b and c but different for a which gives Ia 6= Ib = Ic . Ans. A

Chapter 38. Photoelectric Effect and Wave-Particle Duality Q 8. A particle of mass M at rest decays into two particles of masses m1 and m2 having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, λ1 /λ2 is (1999) √ √ (A) m1 /m2 (B) m2 /m1 (C) 1 (D) m2 / m1 Sol. The particle of mass M is at rest. Thus, the linear momentum of the system is zero before decay. Before p~i = 0 M

After p ~1 p ~2 m1

497

One or More Option(s) Correct Q 11. Light of wavelength λph falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is φ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electron passing through the anode is λe , which of the following statement(s) is(are) true? (2016)

m2 Light

Let p~1 and p~2 be the linear momenta of the particles after decay. As there is no external force in decay process, the linear momentum of the system is conserved i.e.,

Electrons

V − +

p~1 + p~2 = ~0, which gives |~ p1 | = |~ p2 |. Hence, the de-Broglie wavelengths of two particles of mass m1 and m2 are equal i.e., h/|~ p1 | λ1 = = 1. λ2 h/|~ p2 | Ans. C Q 9. The work function of a substance is 4 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately (1998) (A) 540 nm (B) 400 nm (C) 310 nm (D) 220 nm Sol. In photoelectric effect, the wavelength λ of incident photon, work function φ of the material, and kinetic energy K of the ejected photo-electron are related by hc/λ = φ + K.

(1)

The longest wavelength corresponds to K = 0. Substitute the values in equation (1) to get λmax =

hc (6.63 × 10−34 ) (3 × 108 ) = = 310 nm. φ 4 × 1.6 × 10−19 Ans. C

Q 10. The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV falls on it is 4 eV. The stopping potential is (1997) (A) 2 V (B) 4 V (C) 6 V (D) 10 V Sol. In photoelectric effect, stopping potential is defined as the potential required to stop the photoelectron with maximum kinetic energy. Thus, the stopping potential is 4 V. Note that stopping potential is different from φ, which is 2 eV. Ans. B

(A) λe decreases with increase in φ and λph . (B) λe is approximately halved, if d is doubled. (C) For large potential difference (V  φ/e), λe is approximately halved if V is made four times. (D) λe increases at the same rate as λph for λph < hc/φ. Sol. In photo-electric effect, the maximum kinetic energy of the photo-electron ejected at the cathode, is given by Kmax,c = hc/λph − φ, where λph is the wavelength of the incident light and φ is the work function of the material. The ejected photoelectrons are accelerated from the cathode to the anode by a potential V . Thus, the maximum kinetic energy of the photo-electron at the anode is Kmax,a = Kmax,c + eV = hc/λph − φ + eV. The linear momentum of the electron √ of mass m and kinetic energy K is given by p = 2mK. Thus, the minimum de-Broglie wavelength of the electron at the anode is h h =p p 2mKmax,a h . =p 2m(hc/λph − φ + eV )

λe =

(1)

The wavelength λe increases with increase in φ and λph as denominator of equation (1) decreases with increase in φ and λph . Also, λe is independent of d. The order of magnitudes of hc/λph and φ is almost same. √Thus, if V  φ/e then equation (1) reduces to λe ≈ h/ 2meV . Hence, λe is approximately halved if V is made four times.

498

Part VI. Modern Physics V

λe

metal-1 metal-2

metal-3

λph

The equation (1) is not linear in λph and λe . Thus, the rates of increase of λe and λph will be different. This can be verified by differentiating equation (1). You are encouraged to show that the slope dλe /dλph is different at different values of λph . Ans. C Q 12. For photo-electric effect with incident photon wavelength λ, the stopping potential is V0 . Identify the correct variation(s) of V0 with λ and 1/λ. (2015) (A) V0

(B) V0 λ

(C)

λ

V0

(D)

V0

1/λ

θ 0.001 0.002

eV0 = hc/λ − φ,

where φ is the work function of the metal. From equation (1), 1/λ versus V0 graph is a straight line with a positive slope hc/e. The equation (1) can be written as (eV0 + φ)λ = hc.

(nm−1 )

Sol. The energy of incident photon hc/λ is equal to the sum of minimum energy required to emit the photoelectron (work function φ) and kinetic energy K of the ejected photoelectron. The stopping potential V converts the kinetic energy K of ejected photoelectron into potential energy eV thereby stopping its motion. Thus, hc/λ = φ + K = φ + eV i.e., V =

(1)

1 λ

(A) φ1 : φ2 : φ3 = 1 : 2 : 4. (B) φ1 : φ2 : φ3 = 4 : 2 : 1. (C) tan θ is directly proportional to hc/e, where h is Planck’s constant and c is the speed of light. (D) The violet colour light can eject photoelectrons from metal 2 and 3.

1/λ

Sol. In photoelectric effect, the stopping potential (V0 ) is related to the wavelength of incident radiation by

0.004

φ hc 1 − . e λ e

At V = 0, φ = hc/λ, which gives φ1 = 0.001hc,

φ2 = 0.002hc,

φ3 = 0.004hc.

Hence, φ1 : φ2 : φ3 = 1 : 2 : 4. The slope of V - λ1 graph (which is a straight line) is tan θ = hc/e. The wavelengths corresponding to φ1 , φ2 and φ3 are λ1 = 1/0.001 = 1000 nm, λ2 = 1/0.002 = 500 nm, λ3 = 1/0.004 = 250 nm.

V0 hc e 1 λ

φ hc

− φe V0

hc φ

λ

Thus, λ versus V0 graph is hyperbolic in nature. The details of two graphs are shown in the figures. Ans. A, C Q 13. The graph between 1/λ and stopping potential (V ) of three metals having work functions φ1 , φ2 and φ3 in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is (are) correct? (Here λ is the wavelength of the incident ray.) (2006)

The wavelength of violet colour light is close to 400 nm. Thus, violet light cannot eject the photoelectrons from metal 3. Ans. A, C Q 14. When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength λA . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA − 1.50 eV). If the de-Broglie wavelength of these photoelectrons is λB = 2λA , then, (1994)

(A) (B) (C) (D)

the work function of A is 2.25 eV. the work function of B is 4.20 eV. TA = 2.00 eV. TB = 2.75 eV.

Sol. In photoelectric effect, maximum kinetic energy of the ejected photoelectrons is given by T = hν − φ. Thus, TA = 4.25 − φA ,

(1)

TB = 4.70 − φB .

(2)

Chapter 38. Photoelectric Effect and Wave-Particle Duality Given, TB = TA − 1.50, r √ TB λA h/ 2mTA 1 = √ = = . λB TA 2 h/ 2mTB

(3) (4)

Solve equations (3) and (4) to get TA = 2 eV and TB = 0.5 eV. Substitute TA in equation (1) to get φA = 2.25 eV and substitute TB in equation (2) to get φB = 4.2 eV. Ans. A, B, C Q 15. When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cutoff voltage and the saturation current are 0.6 V and 18.0 mA, respectively. If the same source is placed 0.6 m away from the photoelectric cell, then, (1992) (A) the stopping potential will be 0.2 V. (B) the stopping potential will be 0.6 V. (C) the saturation current will be 6.0 mA. (D) the saturation current will be 2.0 mA. Sol. The intensity of a point source at a distance r varies as I = I0 /r2 , where I0 is a constant. Thus, intensities of the light falling on the photoelectric cell are I1 = I0 /(0.2)2 ,

I2 = I0 /(0.6)2 .

Since saturation current is proportional to the intensity i2 =

(0.2)2 I2 × 18 = 2 mA. i1 = I1 (0.6)2

The cut-off voltage depends on the frequency of incident light and not on its intensity. Thus, the cut-off voltage is 0.6 V in both the cases. Ans. B, D Q 16. Photoelectric effect supports quantum nature of light because (1987) (A) there is a minimum frequency of light below which no photoelectrons are emitted. (B) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on intensity. (C) even when the metal surface is faintly illuminated, the photoelectrons leave the surface immediately. (D) electric charge of the photoelectrons is quantized. Sol. According to Einstein’s quantum theory of the photoelectric effect, light constitutes of photons. The energy of each photon is E = hν, where ν is frequency of light. The photoelectrons are ejected when a photon of energy greater than work function (φ) collides with an electron on metal surface. Maximum kinetic energy of ejected photoelectrons, K = hν − φ, depends on the frequency of incident radiation and not on intensity. The photoelectrons are ejected if photon’s energy is greater than the work function, irrespective of the intensity of light. The numbers of photons striking

499

the metal surface per second and the number of photoelectrons ejected per second depend on the intensity of incident radiation. Ans. A, B, C Q 17. The threshold wavelength for photoelectric emission from a material is 5200 ˚ A. Photoelectrons will be emitted when this material is illuminated with monochromatic radiations from a (1982) (A) 50 W infrared lamp (B) 1 W infrared lamp (C) 50 W ultraviolet lamp (D) 1 W ultraviolet lamp Sol. The emission of the photoelectrons from a metal surface depends on the frequency (wavelength) of incident radiation and not on its intensity. The wavelength of incident radiation should be less than or equal to the threshold wavelength of 5200 ˚ A. The wavelength of ulraviolet radiation is less 5200 ˚ A. Hence, the ultraviolet lamp of 50 W as well as 1 W will emit photoelectrons. Ans. C, D True False Type Q 18. In a photoelectric emission process the maximum energy of the photoelectrons increases with increasing intensity of the incident light. (1986) Sol. The maximum kinetic energy of the photoelectron ejected from the metal of work function φ is given by, Kmax = hν − φ, where ν is the frequency of the incident radiation. The maximum kinetic energy Kmax depends on the frequency of the incident radiation and not on its intensity. It is an important argument in support of quantum nature of light. Ans. F Q 19. The kinetic energy of photoelectrons emitted by a photosensitive surface depends on the intensity of the incident radiation. (1981) Sol. The kinetic energy of photoelectrons emitted by a photosensitive surface depends on the frequency of the incident radiation, Kmax = hν − φ. Ans. F Fill in the Blank Type Q 20. The maximum kinetic energy of electron emitted in the photoelectric effect is linearly dependent on the . . . . . . of the incident radiation. (1984) Sol. The maximum kinetic energy of the ejected photoelectrons is linearly dependent on the frequency of the incident radiation i.e., Kmax = hν − φ. Ans. Frequency

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Integer Type Q 21. In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n × 10−4 N due to impact of the electrons. The value of n is . . . . . . . [Mass of the electron me = 9 × 10−31 kg and 1.0 eV = 1.6 × 10−19 J.] (2018)

Sol. The frequency of incident light is just above the threshold frequency. Hence, the energy of each photon is equal to the work function (Ep = φ = 6.25 eV) and the kinetic energy of emitted photo-electron is zero (Ke0 = 0). The energy incident per second on the cathode is incident power P = 200 W. Thus, number of photons incident per second is Np = P/Ep = P/φ. The photo-electron emission efficiency is 100%. Thus, number of photo-electron emitted per second is equal to the number of photons incident per second i.e., Ne = Np . These photo-electrons are accelerated by a potential difference V = 500 V. Thus, gain in potential energy of each photo-electron is ∆U = eV = 500 eV. By conservation of energy, kinetic energy of the photoelectron when it reaches the anode is Ke = Ke0 + ∆U = 500 eV. The linear momentum of photo-electron of mass me and kinetic energy Ke is given by p pe = 2me Ke . The photo-electron transfer its entire linear momentum to the anode (absorbed by the anode). Thus, gain in linear momentum of the anode by absorbing one photoelectron is ∆pa = pe . The force on the anode is equal to the increase in its linear momentum per second. Since Ne photo-electrons strikes the anode per second, the force acting on the anode is given by Fa = Ne ∆pa = Ne pe = Np pe = (P/φ)pe p = (P/φ) 2me Ke p = (P/φ) 2me eV = 24 × 10

−4

(∵ ∆pa = pe ) (∵ Ne = Np = P/φ) √ (∵ pe = 2me Ke ) (∵ Ke = eV )

N. Ans. 24

Q 22. The work functions of silver and sodium are 4.6 eV and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is . . . . . . . (2013) Sol. In photoelectric effect, the stopping potential (V ) is related to the frequency (ν) of incident radiation and workfunction (φ) of the material by h φ ν− . e e Thus, the slope of V versus ν plot is h/e, a universal constant. Ans. 1 V =

Q 23. A proton is fired from very far away towards a nucleus of charge Q = 120e, where e is electronic charge. It makes a closest approach of 10 fm to the nucleus. The de-Broglie wavelength (in units of fm) of proton at its start is . . . . . . . [Take the proton mass mp = 5/3 × 10−27 kg, h/e = 4.2 × 10−15 J s/C, 9 −15 1 m.] (2012) 4π0 = 9 × 10 m/F, 1 fm = 10 Sol. Let initial and final kinetic energies of the proton be Ki and Kf and corresponding potential energies be Ui and Uf . When proton is far away from the nucleus (r → ∞), its potential energy is Ui = lim

r→∞

1 120e2 = 0. 4π0 r

At closest distance, the proton comes to rest momentarily, giving Kf = 0. The potential energy at closest distance is Uf =

1 120e2 , 4π0 a

where a is the distance of closest approach. Since electrostatic force is conservative, total energy is conserved i.e., Ki + Ui = Kf + Uf . Substitute the values to get Ki = Uf =

1 120e2 . 4π0 a

The de-Broglie wavelength of the proton is given by s h h h 4π0 a λi = = p = pi e 240mp 2mp Ki  1/2 10 × 10−15 −15 = 4.2 × 10 9×109 × 240× (5/3)×10−27 = 7 fm. Ans. 7 Q 24. A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in freespace. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is A×10Z (where 1 < A < 10). The value of Z is . . . . . . . (2011)

Chapter 38. Photoelectric Effect and Wave-Particle Duality Sol. The silver sphere gets positively charged due to emission of photoelectrons. This positively charged sphere attracts (binds) the emitted photoelectrons. The emitted photoelectrons cannot escape if their kinetic energies (hc/λ−φ) are less than or equal to their potential 1 ne2 energies ( 4π ). Thus, in limiting case, 0 r 1 ne2 hc −φ= . λ 4π0 r

(1)

Substitute the values of various parameters in equation (1),

501

Sol. The de-Broglie wavelength of a particle of mass m, momentum p and kinetic energy K is given by √ λ = h/p = h/ 2mK. The total energy is U + K = 2E0 . The kinetic energy and de-Broglie wavelength of the particle in the region 0 ≤ x ≤ 1 are K1 = 2E0 − U = 2E0 − E0 = E0 , p λ1 = h/ 2mE0 ,

(1)

and in the region x > 1 are 1242 n (9 × 109 ) (1.6 × 10−19 ) − 4.7 = , 200 10−2 to get n = 1.04 × 107 . 1242 eV-nm.]

[We have used hc = Ans. 7

Q 25. An α-particle and a proton are accelerated from rest by a potential of 100 V. After this, their deBroglie wavelengths are λα and λp respectively. The ratio λp /λα , to the nearest integer, is . . . . . . . (2010) Sol. The de Broglie wavelength of a particle with momentum p is given by

K2 = 2E0 − U = 2E0 − 0 = 2E0 , p λ2 = h/ 4mE0 . Divide equation (1) by (2) to get λ1 /λ2 =

(2) √

2. √ Ans. 2

Q 27. In a photoelectric experiment set-up, photons of energy 5 eV falls on the cathode having work function 3 eV. If the saturation current iA = 4 µA for intensity IA = 10−5 W/m2 , then plot the variation of photocurrent ip against the anode voltage Va for photon intensity IA = 10−5 W/m2 and IB = 2 × 10−5 W/m2 . (2003) Sol. In photoelectric effect, maximum kinetic energy of the ejected photoelectrons is given by

λ = h/p. The momentum and kinetic energy of a particle of mass m are related by √ p = 2mK.

Kmax = hν − φ = 5 − 3 = 2 eV. ip (µA)

The kinetic energy of a charge q, accelerated through potential V , is given by K = qV . Thus, p √ λ = h/ 2mK = h/ 2mqV ,

-4

-2

8

B

4

A

0

2

4

Va (V)

Thus, anode potential to stop these photoelectrons is 2 V (stopping potential). The saturation current is proportional to the intensity of incident light. Thus,

which gives s r λp 2mα qα V 2 · 4u · 2e · 100 = = λα 2mp qp V 2 · 1u · 1e · 100 √ = 8 = 2.8 ≈ 3.

iB =

(2 × 10−5 ) (4) IB iA = = 8 µA. IA 10−5 Ans. See solution

Ans. 3 Descriptive Q 26. The potential energy of a particle varies as,  E0 for 0 ≤ x ≤ 1, U (x) = 0 for x > 1. For 0 ≤ x ≤ 1, the de-Broglie wavelength is λ1 and for x > 1, the de-Broglie wavelength is λ2 . Total energy of the particle is 2E0 . Find λ1 /λ2 . (2005)

Q 28. Two metallic plates A and B each of area 5 × 10−4 m2 , are placed parallel to each other at separation of 1 cm. Plate B carries a positive charge of 33.7 × 10−12 C. A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0 so that 1016 photons fall on it per square metre per second. Assume that one photoelectron is emitted for every 106 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value of 2 eV. Determine, (2002)

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Part VI. Modern Physics

(a) the number of photoelectrons emitted up to t = 10 s. (b) the magnitude of the electric field between the plates A and B at t = 10 s. (c) The kinetic energy of the most energetic photoelectrons emitted at t = 10 s when it reaches plate B. [Neglect time taken by the photoelectron to reach plate B. Take 0 = 8.85 × 10−12 C2 /N-m2 ] Sol. The number of photons striking the plate A of area A = 5 × 10−4 m2 in a time interval of 10 s is np = (1016 ) (5 × 10−4 ) (10) = 5 × 1013 , and the number photoelectrons emitted by this plate in 10 s is ne = np /106 = 5 × 107 . The positive charge on plate A due to emission of the photoelectrons is qA = ene = (1.6 × 10−19 )(5 × 107 ) = 8.0 × 10−12 C, and, the charge on plate B after collecting ne photoelectrons is qB = 33.7 × 10−12 − 8.0 × 10−12 = 25.7 × 10−12 C. The direction of electric field between the two plates is from plate B to plate A. The magnitude of this electric field is qB − qA qB /A qA /A − = 20 20 2A0 −12 (25.7 − 8) × 10 = 2 × 103 V/m. = 2 (5 × 10−4 ) (8.85 × 10−12 )

E = EB − EA =

In photoelectric effect, the maximum kinetic energy of emitted photoelectrons is given by Kmax = hν − φ = 5 − 2 = 3 eV. These electrons are attracted by plate B. The gain in potential energy of the photoelectron is U = qEd = eEd = e(2 × 103 )(10−2 ) = 20 eV. Thus, kinetic energy of the most energetic photoelectrons when these reach plate B is Kmax + U = 23 eV. Ans. (a) 5 × 107 (b) 2 × 103 N/C (c) 23 eV Q 29. When a beam of 10.6 eV photons of intensity 2.0 W/m2 falls on a platinum surface of area 1.0 × 10−4 m2 and work function 5.6 eV, 0.53% of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies (in eV). [Take 1 eV = 1.6 × 10−19 J.] (2000)

Sol. The energy of each photon is hν = 10.6 eV = 10.6 × 1.6 × 10−19 = 16.96 × 10−19 J.

(1)

The intensity I = 2 W/m2 is the energy falling on 1 m2 area in one second. Thus, the energy of photons falling on area A = 1.0 × 10−4 m2 in one second is E = IA = 2 × 1.0 × 10−4 = 2 × 10−4 J/s.

(2)

The equations (1) and (2) give the number of photons falling in one second as np =

2 × 10−4 E = = 1.18 × 1014 . hν 16.96 × 10−19

The number of photoelectrons emitted in one second is 0.53% of np i.e., ne = np × 0.53/100 = 6.25 × 1011 . Maximum kinetic energy of emitted photoelectrons is related to work function φ by hν = φ + Kmax , Kmax = hν − φ = 10.6 − 5.6 = 5.0 eV. Minimum kinetic energy of photelectrons can be zero i.e., Kmin = 0. Ans. 6.25 × 1011 , zero, 5.0 eV Q 30. In a photoelectric effect set-up, a point source of light of power 3.2 × 10−3 W emits mono-energetic photons of energy 5.0 eV. The source is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work function 3.0 eV and of radius 8.0 × 10−3 m. The efficiency of photoelectrons emission is one for every 106 incident photons. Assume that the sphere is isolated and initially neutral and that photoelectrons are instantly swept away after emission. (1995)

(a) Calculate the number of photoelectrons emitted per second. (b) Find the ratio of the wavelength of incident light to the de-Broglie wavelength of the fastest photoelectrons emitted. (c) It is observed that the photoelectrons emission stops at a certain time t after the light source is switched on. Why? (d) Evaluate the time t. Sol. Let n be the number of photons emitted per second, each with energy Ep = 5 eV. The power of the source, 3.2 × 10−3 W, is equal to the energy of n photons i.e., n (5 × 1.6 × 10−19 ) = 3.2 × 10−3 , which gives n = 4 × 1015 photons/s.

Chapter 38. Photoelectric Effect and Wave-Particle Duality

d r

S

The number of photons per unit area at a distance n d = 0.8 m from the source is nd = 4πd 2 . The photons lying in the area πr2 strike the sphere. Thus, number of photons striking per second on the sphere of radius r = 8 × 10−3 m is ns = nd (πr2 ) =

503

Q 31. A monochromatic point source S radiating wavelength 6000 ˚ A with power 2 W, an aperture A of diameter 0.1 m and a large screen SC are placed as shown in the figure. A photoemissive detector D of surface area 0.5 cm2 is placed at the centre of the screen. The efficiency of the detector for the photoelectron generation per incident photon is 0.9. (1991)

4 × 1015 (8 × 10−3 )2 nr2 = 4d2 4 × (0.8)2

S •

= 1011 photons/s.

(1)

The kinetic energy of the fastest photoelectrons is given by Kmax = hν − φ = 5.0 − 3.0 = 2 eV. The de-Broglie wavelength of the fastest photoelectrons is p (2) λe = h/p = h/ 2mKmax . Divide equation (2) by (1) to get √ λp hc/Ep c 2mKmax = √ = λe Ep h/ 2mKmax p −31 8 3 × 10 2(9.1 × 10 )(2 × 1.6 × 10−19 ) = 5 × 1.6 × 10−19 = 286. (3) The sphere becomes positively charged due to emission of photoelectrons. The emission stops when the potential on the sphere is equal to the stopping potential, which is 2 V. Let q be charge on the sphere when emission stops. Equating potential on the sphere to the stopping potential,

D

6m

(a) Calculate the photon flux at the centre of the screen and photocurrent in the detector. (b) If the concave lens L of focal length 0.6 m is inserted in the aperture as shown, find the new values of photon flux and photocurrent. Assume a uniform average transmission of 80% from the lens. (c) If the work function of the photoemissive surface is 1 eV, calculate the values of the stopping potential in the two cases (without and with the lens in aperture). Sol. The energy of each photon is hc (6.63 × 10−34 ) (3 × 108 ) = λ 6000 × 10−10 −19 = 3.315 × 10 J.

E=

The number of photons emitted per second by the source of power P = 2 W is N1 =

we get q = (4π0 )2r = 1.78 × 10−12 C. Number of photoelectrons emitted till the sphere accumulates charge q is q 1.78 × 10−12 = = 1.11 × 107 . e 1.6 × 10−19 Since the number of photoelectrons emitted per second is 105 /s, required time is t = N/105 = 111 s. Ans. (a) 105 /s (b) 286 (c) Emission of photoelectrons is stopped when its potential is equal to the stopping potential required for the fastest moving electrons. (d) 111 s

2 P = = 6.03 × 1018 . E 3.315 × 10−19 SC

A S •

S0 •

L

D

0.3m

1 q = 2, 4π0 r

N=

L

0.6m

Since one photoelectron is emitted for every 106 incident photons, the number of emitted photoelectrons per second is ns × 10−6 = 105 /s. The wavelength of incident radiations is related to the photons energy by hc/λp = Ep .

SC

A

0.6m

5.4m 6m

The number of photons incident per second on an aperture of diameter d = 0.1 m, placed at a distance r1 = 0.6 m, is N2 =

N1 πd2 6.03 × 1018 × (0.1)2 · = 2 4πr1 4 16 × (0.6)2

= 1.05 × 1016 . The aperture acts as a point source emitting N2 photons per second. The photon flux (number of photons per

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unit area per unit time) at the detector placed at a distance r2 = 6 − 0.6 = 5.4 m from aperture is n=

1.05 × 1016 N2 = = 2.87 × 1013 . 4πr2 2 4 (3.14) (5.4)2

The number of photons incident on the detector of area A per second is N3 = nA = (2.87 × 1013 )(0.5 × 10−4 ) = 1.435 × 109 .

(1)

The number of photoelectron emitted per second with a detector efficiency η = 0.9 is

The photocurrent is charge of emitted photoelectrons per second i.e., i = eNe = 1.6 × 10−19 × 1.2915 × 109 = 2.07 × 10

A.

(2)

Total number of photons incident per second on the lens is same as number of photons incident per second on the aperture i.e., N20 = N2 . The number of photons transmitted by the lens per second is N30 = T N2 = 0.8 × 1.05 × 1016 = 0.84 × 1016 . When a concave lens is introduced, the image of S (say S 0 ) acts as point source of photons. The lens formula, 1/v − 1/u = 1/f, with u = −0.6 m and f = −0.6 m gives v = −0.3 m. The photon flux at the detector placed at a distance r20 = 5.4 + 0.3 = 5.7 m from S 0 is n0 =

N30 4πr20 2

=

0.84 × 1016 = 2.06 × 1013 . 4 (3.14) (5.7)2

Note that the divergence by concave lens reduces the photon flux at the detector. The number of photons incident on the detector is N30 = n0 A = 1.03 × 109 . The number of electrons emitted per second by the detector is Ne0 = ηN30 = 0.927 × 109 , and the photocurrent is i0 = eNe0 = 1.483 × 10−10 A. The maximum kinetic energy of emitted photoelectrons is given by hc 3.315 × 10−19 −φ= − 1 = 1.06 eV. λ 1.6 × 10−19 Thus, the stopping potential is 1.06 V. It is same in both the cases. Ans. (a) 2.87 × 1013 s−1 m−2 , 2.07 × 10−10 A (b) 2.06 × 1013 s−1 m−2 , 1.483 × 10−10 A (c) 1.06 V in both cases. Kmax =

Sol. The energies of the photons for wavelengths λ1 = 4144 ˚ A, λ2 = 4972 ˚ A, and λ3 = 6216 ˚ A are e1 = hc/λ1 = 12420/4144 = 3.0 eV,

Ne = ηN3 = 1.2915 × 109 .

−10

Q 32. A beam of light has three wavelengths 4144 ˚ A, 4972 ˚ A and 6216 ˚ A with a total energy of 3.6 × 10−3 W/m2 equally distributed amongst the three wavelengths. The beam falls normally on an area 1 cm2 of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds. (1989)

e2 = hc/λ2 = 12420/4972 = 2.5 eV, e3 = hc/λ3 = 12420/4216 = 2.0 eV. A photon ejects the photoelectron from a metal of work function φ = 2.3 eV if its energy is greater than φ. Thus, only the photons of wavelengths λ1 and λ2 eject the photelectrons from the given metal. Total intensity, I = 3.6 × 10−3 W/m2 , is equally divided among the three wavelengths. Thus, the intensities of the wavelengths λ1 and λ2 are, I1 = I2 = I/3 = 1.2 × 10−3 W/m2 . The energies falling on an area A = 1 cm2 = 10−4 m2 for wavelengths λ1 and λ2 in time t = 2 s are E1 = I1 At = (1.2 × 10−3 )(10−4 )(2) = 2.4 × 10−7 J, E2 = I2 At = (1.2 × 10−3 )(10−4 )(2) = 2.4 × 10−7 J, and the number of photons corresponding to these energies are 2.4 × 10−7 E1 = = 5 × 1011 , e1 3.0 × 1.602 × 10−19 E2 2.4 × 10−7 n2 = = = 6 × 1011 . e2 2.5 × 1.602 × 10−19

n1 =

Since each photon of wavelength λ1 and λ2 ejects one photoelectron, total number of photoelectrons ejected from the metal are ne = n1 + n2 = 1.1 × 1012 . Ans. 1.1 × 1012

Chapter 39 Bohr’s Model and Physics of the Atom

Divide equation (2) by (3) to get λ = 122 × 9 × 3/4 = 823.5 nm. Ans. B

One Option Correct Q 1. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 ˚ A. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is (2011) (A) 1215 ˚ A (B) 1640 ˚ A (C) 2430 ˚ A (D) 4687 ˚ A

Q 3. A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of few micro-seconds another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector? (2005) (A) Two photons of energy 10.2 eV. (B) Two photons of energy 1.4 eV. (C) One photon of energy 10.2 eV and an electron of energy 1.4 eV. (D) One photon of energy 10.2 eV and another photon of energy 1.4 eV.

Sol. The wavelength of spectral line when an electron in singly ionized atom of atomic number Z makes a transition from mth to nth orbit (m > n) is   1 1 1 = RZ 2 2 − 2 . λ n m In Balmer series, first spectral line corresponds to a transition from m = 3 to n = 2 and second spectral line corresponds to a transition from m = 4 to n = 2. Thus, for hydrogen (Z = 1),   1 1 1 5R =R 2 − 2 = , (1) 6561 2 3 36 and for singly ionized helium (Z = 2),   1 1 1 3R = R × 22 2 − 2 = . λ 2 4 4

Sol. The energy of hydrogen atom when electron is in nth orbit is given by En = −

Thus, the energies of hydrogen atom in its ground state (n = 1) and the first excited state (n = 2) are

(2)

E1 = −13.6/12 = −13.6 eV,

Divide equation (1) by (2) to get λ = 1215 ˚ A. Ans. A

E2 = −13.6/22 = −3.4 eV.

Q 2. The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is (2007) (A) 802 nm (B) 823 nm (C) 1882 nm (D) 1648 nm

The first photon excites the hydrogen atom from its ground state to the first excited state by transferring its energy (E2 − E1 = 10.2 eV). The excited atom returns to the ground state by emitting a photon of energy 10.2 eV. The binding energy of an electron in ground state is Eb = −E1 = 13.6 eV. The second photon of energy E = 15 eV ionizes the hydrogen atom by ejecting the electron with kinetic energy E − Eb = 15 − 13.6 = 1.4 eV. Ans. C

Sol. In hydrogen spectrum, ultraviolet emission corresponds to Lyman series (n = 1, m > 1). The largest wavelength in the ultraviolet region will correspond to minimum energy and hence m = 2. Substitute it in   1 1 1 =R 2 − 2 , (1) λ n m

Q 4. The electric potential between   a proton and an electron is given by V = V0 ln rr0 , where V0 and r0 are constants. Assuming Bohr’s model to be applicable, how does radius rn of the nth Bohr’s orbit varies with the principal quantum number n? (2003) (A) rn ∝ n (B) rn ∝ n1 (C) rn ∝ n2 (D) rn ∝ n12

to get 1/122 = 3R/4.

(2)

The infrared emission corresponds to Paschen series (n = 3, m > 3). The smallest wavelength in the infrared region will correspond to maximum energy and hence m = ∞. Substitute it in equation (1) to get 1/λ = R/9.

13.6 eV. n2

Sol. The potential energy of the electron is

(3)

U = eV = eV0 ln (r/r0 ) . 505

(1)

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Part VI. Modern Physics

Differentiate equation (1) w.r.t. r to get the force on the electron F~ = −dU /dr rˆ = −eV0 /r rˆ.

(2)

This force provides the centripetal acceleration for circular motion i.e., eV0 /r = mv 2 /r.

mvr = nh/(2π).

(2001)

(A) 2 → 1 (B) 3 → 2 (C) 4 → 2 (D) 5 → 4

(3)

Sol. The energy, hν = hc/λ, for infrared is lesser in comparison to ultraviolet. The energy gap for the given transition is   1 7 1 E0 . ∆E4→3 = −13.6Z 2 2 − 2 = 4 3 144

(4)

The energy gap for other transitions are

The quantization of angular momentum gives

Eliminate v from equations (3) and (4) to get r=

Q 7. The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition

∆E2→1 = (3/4)E0 , ∆E3→2 = (5/36)E0 ,

nh √ . 2π meV0

∆E4→2 = (3/16)E0 , ∆E5→4 = (9/400)E0 .

Ans. A Q 5. If the atom 100 Fm257 follows the Bohr’s model and the radius of 100 Fm257 is n times the Bohr radius, then find n. (2003) (A) 100 (B) 200 (C) 4 (D) 1/4

Thus, ∆E5→4 corresponds to infrared radiation. Note that the energy gap between the successive energy levels decreases with increase in n i.e., En+1 − En = E0

Sol. The radius of mth Bohr orbit is given by rm = m2 a0 /Z,

(1)

where a0 = 0.53 ˚ A is Bohr’s radius. For 100 Fm257 , Z = 100 and for its outermost shell m = 5. Substitute the values in equation (1) to get r5 = 25a0 /100 = a0 /4. Ans. D Q 6. A hydrogen atom and a Li2+ ion are both in the second excited state. If lH and lLi are their respective electronic angular momentum, and EH and ELi their respective energies, then, (2002) (A) lH > lLi and |EH | > |ELi | (B) lH = lLi and |EH | < |ELi | (C) lH = lLi and |EH | > |ELi | (D) lH < lLi and |EH | < |ELi | Sol. Second excited state is given by n = 3. The angular momentum is quantized and is given by l = mvr = nh/(2π). Thus, lH = lLi = 3h/(2π). The energy of hydrogen like ion with atomic number Z in nth Bohr’s orbit is given by En = −

13.6Z 2 eV. n2

2n + 1 , n2 (n + 1)2

is a decreasing function of n. Ans. D Q 8. Imagine an atom made up of proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength λ (given in terms of the Rydberg constant R for the hydrogen atom) equal to (2000) (A) 9/(5R) (B) 36/(5R) (C) 18/(5R) (D) 4/R Sol. Coulomb’s attraction provides the centripetal acceleration for circular motion of the electron i.e., Ze2 mv 2 = . 4π0 r2 r

(1)

The angular momentum of the electron is quantized (Bohr’s quantization condition) i.e., mvr = nh/(2π).

(2)

Total energy of the electron in nth orbit is the sum of its kinetic and potential energies i.e., En = K + U =

1 Ze2 mv 2 − . 2 4π0 r

(3)

Eliminate r and v from equation (3) by using equations (1) and (2) to get the total energy 2

13.6×3 eV = 9EH . Thus, EH = − 13.6 32 eV and ELi = − 32 Ans. B

En = −

mZ 2 e4 . 820 h2 n2

(4)

Chapter 39. Bohr’s Model and Physics of the Atom The energy En is proportional to the mass m of the electron. Thus, the energy of hypothetical atom is En0 = 2En , where En is the energy of hydrogen atom. The wavelength of emitted photon for the hypothetical atom is given by E 0 − En0 2(Em − En ) 1 = m = λ hc  hc  1 1 − 2 . = 2R m2 n

(5)

The longest wavelength for transition to first excited level (m = 2) occurs for n = 3. Substitute the values in equation (5) to get   1 5R 1 1 = 2R 2 − 2 = . λ 2 3 18 Ans. C Q 9. The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statement is true? (2000) (A) Its kinetic energy increases and its potential and total energies decrease. (B) Its kinetic energy decreases, potential energy increases and its total energy remains the same. (C) Its kinetic and total energy decrease and its potential energy increases. (D) Its kinetic, potential and total energies decrease. Sol. The potential energy of the electron placed at a distance r from the nucleus of charge Ze is given by U =−

Ze2 . 4π0 r

(1)

Coulomb’s attraction provides the necessary centripetal acceleration for circular motion of the electron i.e., Ze2 mv 2 = , 2 4π0 r r

(2)

The equation (2) gives the kinetic energy of the electron as 1 1 Ze2 K = mv 2 = . 2 2 4π0 r

(3)

The equations (1) and (3) give the total energy as E =U +K =−

507 Sol. The energy in nth orbit of hydrogen like atom is given by En = −

13.6Z 2 . n2

Substitute Z = 3 and n = 1 to get E1 = −122.4 eV. Hence, the energy required to remove an electron from the ground state (binding energy) is 122.4 eV. Ans. D Q 11. If elements with principal quantum number n > 4 were not allowed in nature, the number of possible elements would be (1983) (A) 60 (B) 32 (C) 4 (D) 64 Sol. Allowed values of principal quantum number are n = 1, 2, 3, 4. Maximum number of electron in an orbit with principal quantum number n is Nn = 2n2 . Thus, the number of electrons in an atom with n = 1, 2, 3, 4 is N = N1 + N2 + N3 + N4 = 2(1)2 + 2(2)2 + 2(3)2 + 2(4)2 = 60. Thus, atomic number upto Z = 60 are allowed. Hence, the number of possible elements would be 60. We encourage you to write the azimuthal quantum numbers (l), magnetic quantum numbers (ml ) and spin quantum numbers (ms ) for n = 4. Ans. A Q 12. Consider the spectral line resulting from the transition n = 2 to n = 1 in the atoms and ions given below. The shortest wavelength is produced by (1983) (A) hydrogen atom. (B) deuterium atom. (C) singly ionized helium. (D) double ionized lithium. Sol. The energy gap between n1 = 1 and n2 = 2 levels for hydrogen like atom (atomic number Z) is given by ∆E = −13.6Z

1 Ze2 . 2 4π0 r

2



   1 1 1 2 − 2 = −13.6Z −1 n22 n1 22

= 10.2Z 2 eV.

The orbital radius r decreases when electron makes a transition from excited state to ground state. Thus, K increases but U and E decrease. It is worth noting that K = − 12 U and E = 21 U . Ans. A

The wavelength of the spectral line resulting from this transition is

Q 10. As per Bohr model, the minimum energy required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is (1997) (A) 1.51 eV (B) 13.6 eV (C) 40.8 eV (D) 122.4 eV

For given atoms/ions, doubly ionized lithium has the maximum atomic number (Z = 3). Thus, the wavelength is minimum for doubly ionized lithium. Ans. D

λ=

hc 1242 = nm. ∆E 10.2Z 2

508

Part VI. Modern Physics n=3

One or More Option(s) Correct

λ1

Q 13. Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n  1. Which of the following statement(s) is(are) true? (2016) (A) Relative change in the radii of two consecutive orbitals does not depend on Z. (B) Relative change in the radii of two consecutive orbitals varies as 1/n. (C) Relative change in the energy of two consecutive orbitals varies as 1/n3 . (D) Relative change in the angular momenta of two consecutive orbitals varies as 1/n. Sol. The radius of nth orbital for a hydrogen-like atom of atomic number Z is given by rn = n2 a0 /Z, where a0 = 0.53 ˚ A is the Bohr’s radius. The relative change in the radii of two consecutive orbitals is 2n + 1 (n + 1)2 a0 /Z − n2 a0 /Z rn+1 − rn = = rn n2 a0 /Z n2 2 ≈ (∵ n  1). n The energy of the nth orbital is given by En = −13.6Z2 /n2 eV. The relative change in the energy of two consecutive orbitals is 2

2

2

2

En+1 − En −13.6Z /(n + 1) − (−13.6Z /n ) = En −13.6Z 2 /n2 2 2n + 1 (∵ n  1) ≈− . . =− 2 (n + 1) n The angular momentum of the nth orbital is given by Ln = nh/(2π). The relative change in the angular momentum of two consecutive orbitals is Ln+1 − Ln 1 (n + 1)h/(2π) − nh/(2π) = . = Ln nh/(2π) n Q 14. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5a0 , where a0 is the Bohr ra3h dius. Its orbital angular momentum is 2π . It is given that h is Planck constant and R is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are) (2013) 9 9 4 9 (B) 16R (C) 5R (D) 3R (A) 32R Sol. The quantized orbital angular momentum in nth orbit is given by ln = nh/(2π). Hence, n = 3 for the given orbit. The radius of n orbit is given by

n=1

The possible wavelengths when atom de-excites are λ1 (n1 = 3 to n2 = 2), λ2 (n1 = 3 to n2 = 1), and λ3 (n1 = 2 to n2 = 1) as shown in the figure. The wavelength is given by   1 1 1 (2) = RZ 2 2 − 2 . λ n1 n2 Substitute Z = 2 and appropriate values of n1 and n2 in equation (2) to get λ1 = 9/(5R), λ2 = 9/(32R), and λ3 = 1/(3R). Ans. A, C Q 15. The electron in a hydrogen atom makes a transition n1 → n2 , where n1 and n2 are the principal quantum numbers of two states. Assume the Bohr’s model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of n1 and n2 are (1998) (A) n1 = 4, n2 = 2 (B) n1 = 8, n2 = 2 (C) n1 = 8, n2 = 1 (D) n1 = 6, n2 = 3 Sol. Coulomb attraction provides the centripetal acceleration to the electron for circular motion i.e., mv 2 Ze2 = . 4π0 r2 r

(1)

The angular momentum is quantized i.e., mvr = nh/(2π).

(2)

Solve equations (1) and (2) to get 0 h2 n2 , πmZe2

v=

Ze2 . 20 hn

The time period is the time taken to complete one orbit i.e., T =

42 h3 n3 2πr = 0 2 4. v mZ e

Thus, T1 /T2 = n31 /n32 = 8, which gives n1 /n2 = 2. Ans. A, D Q 16. In the Bohr’s model of the hydrogen atom, (1984)

th

n2 a0 . (1) Z Substitute n = 3 and rn = 4.5a0 in equation (1) to get Z = 2. rn =

n=2 λ3

r=

Ans. A, B, D

λ2

(A) the radius of the nth orbit is proportional to n2 . (B) the total energy of the electron in the nth orbit is inversely proportional to n. (C) the angular momentum of the electron in an orbit is an integral multiple of h/π. (D) the magnitude of the potential energy of the electron in any orbit is greater than its kinetic energy.

Chapter 39. Bohr’s Model and Physics of the Atom Sol. In Bohr’s model of the hydrogen atom, Coulomb’s force provides the centripetal acceleration to the electron of mass m moving with velocity v in an orbit of radius r i.e., mv 2 /r = Ze2 /(4π0 r2 ).

(1)

Also, angular momentum of the electron in n quantized i.e.,

th

orbit is

mvr = nh/(2π).

(2)

Eliminate v from equations (1) and (2) to get the radius of the nth orbit rn = 0 h2 n2 /(πmZe2 ).

(3) th

Total energy of the electron in n orbit is the sum of its kinetic energy K and potential energy U i.e., En = K + U =

Ze2 mZ 2 e4 1 mv 2 − = − 2 2 2, 2 4π0 r 80 h n

where we have used equations (1)–(3) to eliminate r and v. The ratio of kinetic energy to potential energy is given by 1 mv 2 K 1 mv 2 1 mv 2  =− = 2 Ze2 = −  2 U 2 Ze 2 r 2 (mv 2 /r) r − 4π0 r 4π0 r

1 =− . 2

509 Q 18. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to π4 × 1011 Hz. Then the moment of inertia of CO molecule about its centre of mass is close to [Take h = 2π × 10−34 J s.] (A) 2.76 × 10−46 kg m2 (B) 1.87 × 10−46 kg m2 (C) 4.67 × 10−47 kg m2 (D) 1.17 × 10−47 kg m2 Sol. The ground state is represented by n = 1 and first excited state by n = 2. The excitation energy for transition from ground state to first excited state is E2 − E1 = 22 − 12


h2 3h2 = = hν. 2 8π I 8π 2 I

Thus, I = 3h/(8π 2 ν) = 1.87 × 10−46 kg m2 . Ans. B Q 19. In CO molecule, the distance between C (mass = 12 u) and O (mass = 16 u), where 1 u = 53 × 10−27 kg, is close to (A) 2.4 × 10−10 m (B) 1.9 × 10−10 m (C) 1.3 × 10−10 m (D) 4.4 × 10−11 m Sol. Let m1 and m2 be the masses of C and O separated by a distance r.

r1

Ans. A, D

C• m1

r2 •

P

r

•O m2

Paragraph Type Paragraph for Questions 17-19 The key features of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition. (2010) Q 17. A diatomic molecule has moment of inertia I. By Bohr’s quantization condition its rotational energy in the nth level (n = 0 is not allowed) is 2 2 2 2 (A) n12 8πh2 I (B) n1 8πh2 I (C) n 8πh2 I (D) n2 8πh2 I Sol. Bohr’s quantization rule states that angular momentum L is an integral multiple of h/(2π) i.e., L = nh/(2π). The angular momentum of a system rotating with angular velocity ω and having moment of inertia I is L = Iω. and its rotational kinetic energy is  2 1 2 1 L n2 h2 En = Iω = I = . 2 2 I 8π 2 I Ans. D

The molecule rotates about an axis passing through its centre of mass P. The point P lies at a distance 2r 1r r1 = mm from m1 and r2 = mm from m2 . The 1 +m2 1 +m2 moment of inertia of the system about rotation axis is I = m1 r12 + m2 r22 =

m1 m2 2 r , m1 + m2

which gives s I(m1 + m2 ) r= = 1.3 × 10−10 m. m1 m2 Ans. C Paragraph for Questions 20-22 When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimensions, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the deBroglie relation. The energy of the particle of mass m p2 is related to its linear momentum as E = 2m . Thus, the energy of the particle can be denoted by a quantum

510

Part VI. Modern Physics

number ‘n’ taking values 1, 2, 3, . . . (n = 1 called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. [Take h = 6.6 × 10−34 J s and e = 1.6 × 10−19 C.] (2009) Q 20. The allowed energy for the particle for a particular value of n is proportional to (A) a−2 (B) a−3/2 (C) a−1 (D) a2 Sol. The wavelength λ is related to a by a = nλ/2,

(1)

where n = 1, 2, 3 . . . (compare with organ pipe closed at both ends). a

Q 23. The quantum number n of the state finally populated in He+ ion is (A) 2 (B) 3 (C) 4 (D) 5 Sol. The energy of hydrogen like ion with atomic number Z in nth Bohr’s orbit is given by 13.6Z 2 eV. (1) n2 First excited state corresponds to n = 2. Use equation (1) to get ground state energy of H and first excited state energies of H and He+ as En = −

13.6 × 12 = −13.6 eV, 12 2 13.6 × 1 E2,H = − = −3.4 eV, 22 13.6 × 22 E2,He+ = − = −13.6 eV. 22 E1,H = −

E4,H+

E2,H ∆EH E1,H

Use equation (1) to get the linear momentum p = h/λ = hn/(2a),

(2)

and use equation (2) to get the energy 2

2 2

Thus, E ∝ a

∆EH = E2,H − E1,H = −3.4 − (−13.6) = 10.2 eV. En,He+ = E2,He+ + ∆EH

.

= −13.6 + 10.2 = −3.4 eV. Ans. A

Q 21. If the mass of the particle is m = 1.0 × 10−30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to (A) 0.8 meV (B) 8 meV (C) 80 meV (D) 800 meV Sol. The energy in ground state (n = 1) is given by E=

After collision, H comes to ground state by transferring energy ∆EH to He+ , where The energy of He+ after collision becomes

2

E = p /(2m) = h n /(8ma ). −2

E2,H+

h2 . 8ma2

(1)

Substitute the values of various parameters in equation (1) and convert energy into eV to get E ≈ 8 meV. Ans. B Q 22. The speed of the particle, that can take discrete values, is proportional to (A) n−3/2 (B) n−1 (C) n1/2 (D) n Sol. The linear momentum mv = p = h/λ = hn/(2a) gives v ∝ n. Ans. D Paragraph for Questions 23-25 In a mixture of H-He+ gas (He+ is singly ionized He atom), H atoms and He+ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He+ ions (by collisions). Assume that the Bohr model of atom is exactly valid. (2008)

(2)

The equations (1) and (2) give n = 4. Ans. C Q 24. The wavelength of light emitted in the visible region by He+ ion after collisions with H atom is (A) 6.5 × 10−7 m (B) 5.6 × 10−7 m (C) 4.8 × 10−7 m (D) 4.0 × 10−7 m Sol. The energy levels of He+ are given in the figure. n=4

−3.40 eV

n=3

−6.04 eV

n=2

−13.6 eV

n=1

−54.4 eV

The energy corresponding to n = 4 to n = 3 transition is ∆E = −3.40 − (−6.04) = 2.64 eV. The wavelength corresponding to this energy is λ = hc/∆E = 1240 eV-nm/2.64 eV = 469 nm = 4.7 × 10−7 m, which lies in the visible range (4000 ˚ A to 7000 ˚ A). We encourage you to show that the wavelengths corresponding to other transitions do not lie in the visible range. Ans. C

Chapter 39. Bohr’s Model and Physics of the Atom Q 25. The ratio of the kinetic energy of the n = 2 electron for the H atom to that of He+ ion is (A) 1/4 (B) 1/2 (C) 1 (D) 2 Sol. In Bohr’s model, the orbital angular momentum is quantized mvr = nh/(2π),

(1)

and Coulomb attraction provides the centripetal acceleration, 2

2

2

mv /r = kZe /r .

(2)

Eliminate r from equations (1) and (2) to get v = 2πkZe2 /(nh), which gives the kinetic energy 1 2mπ 2 k 2 Z 2 e4 . mv 2 = 2 n2 h2 Thus, ratio of kinetic energy of the n = 2 electron for 2 2 H atom to that of He+ ion is ZH /ZHe = 1/4. Ans. A K=

Fill in the Blank Type Q 26. The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state is . . . . . . m/s. (1997) Sol. The energy in nth orbit of the hydrogen atom is given by 13.6 En = − 2 eV. n The energy of emitted photon when atom makes a transition from n = 5 to n = 1 is   1 1 ∆E = E5 − E1 = −13.6 2 − 2 = 13.056 eV. 5 1 The wavelength of emitted photon is λ = hc/∆E and its momentum is p = h/λ = ∆E/c. By momentum conservation, the momentum of recoiled atom is equal to p and its recoil speed is ∆E 13.056 × 1.6 × 10−19 p = = m mc (1.67 × 10−27 ) (3 × 108 ) = 4.17 m/s.

v=

Ans. 4.17 Q 27. In the Bohr model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in a quantum state n is . . . . . . . (1992) Sol. The kinetic energy K, potential energy U and total energy E are related by K +U = E. It is not difficult to show that K = −E and U = 2E in the Bohr’s model of hydrogen atom. Ans. −1

511 Integer Type Q 28. Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is . . . . . . . (2018) Sol. The energy of hydrogen like ion of atomic number Z in nth Bohr’s orbit is given by 13.6Z 2 E0 Z 2 = − eV, n2 n2 where E0 = 13.6 eV is the ionization energy of the hydrogen atom. The energy of the emitted photon when ion makes a transition from n = 2 to n = 1 is   1 3 1 E2→1 = −13.6Z 2 2 − 2 = 13.6Z 2 × eV. 2 1 4 En = −

The energy of the emitted photon when ion makes a transition from n = 3 to n = 2 is   1 1 5 eV. E3→2 = −13.6Z 2 2 − 2 = 13.6Z 2 × 3 2 36 Given E2→1 is higher than E3→2 by 74.8 eV i.e.,   5 2 3 E2→1 − E3→2 = 13.6Z − = 74.8 eV. 4 36 Solve above equation to get Z = 3. Can you find ionization energy of 73 Li? Ans. 3 Q 29. An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ni to another with quantum number nf . Vi and Vf are respectively the initial and final potential energies of the electron. If VVfi = 6.25, then the smallest possible nf is . . . . . . . (2017) Sol. The energy of a hydrogen atom with an electron in nth orbit is given by En = −

13.6 eV. n2

The potential energy in nth orbit is related to En by Vn = En /2. Thus, potential energies in orbits with quantum numbers ni and nf are given by Vi = −

13.6 13.6 eV, and Vf = − 2 eV, 2n2i 2nf

which gives n2f Vi = 2 = 6.25 (given). Vf ni Take square root to get nf = 2.5ni . Since ni and nf are positive integers, smallest possible integral value of nf is 5 for ni = 2. Ans. 5

512

Part VI. Modern Physics

Q 30. A hydrogen atom in its ground state is irradiated by light of wavelength 970 ˚ A. Taking hc = 1.237 × 10−6 eV m and the ground state energy of hydrogen atom as −13.6 eV, the number of lines present in the emission spectrum is . . . . . . . (2016)

Eliminate rn from equations (1) and (2) to get λ = 2πna0 /Z. For Li2+ , Z = 3. Substitute Z = 3, n = 3, and λ = pπa0 to get p = 2. We encourage you to derive the relation in equation (1). Ans. 2

Sol. The energy of the incident photon of wavelength λ = 970 ˚ A is

Q 32. Consider a hydrogen atom with its electron in the nth orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is . . . . . . . [hc = 1242 eV nm]. (2015)

∆E =

1.237 × 10−6 hc = 12.75 eV. = λ 970 × 10−10

Let the incident photon excites hydrogen atom from the ground state (E1 = −13.6 eV) to a state with principal quantum number n (En = −13.6/n2 eV). Thus, En = E1 + ∆E i.e., − 13.6/n2 = −13.6 + 12.75, which gives n = 4.

Sol. The ionization energy for electron in nth orbital of the hydrogen atom is Eionization = 13.6/n2 eV. The photon energy for radiation of wavelength λ = 90 nm is Ephoton = hc/λ = 1242/90 = 13.8 eV. The photon energy is used to ionize the hydrogen atom and to provide kinetic energy K = 10.4 eV to the ejected electron i.e., Ephoton = Eionization + K.

12.75eV

n=4 n=3

(1)

Substitute the values in equation (1) and solve for n to get n = 2. Ans. 2

n=2 n=1

Descriptive

The hydrogen atom can make C2 = 6 transitions while returning to ground state (see figure). Thus, the emission spectrum will have six lines. Ans. 6

Q 33. In hydrogen-like atom (Z = 11), nth line of Lyman series has wavelength equal to the de-Broglie’s wavelength of electron in the level from which it originated. What is the value of n? (2006)

Q 31. An electron in an excited state of Li2+ ion has angular momentum 3h/2π. The de-Broglie wavelength of the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is . . . . . . . (2015)

Sol. The nth line in Lyman series corresponds to transition (n + 1) → 1. The wavelength of this transition is given by   1 1 2 = RZ 1 − , (1) λ (n + 1)2

n

Sol. The angular momentum of an electron in a state with principal quantum number n is given by

where R = 1.097 × 107 m−1 and Z = 11. The angular momentum in nth orbit is given by

Ln = nh/(2π). Given, Ln = 3h/(2π). Thus, n = 3 for the given excited state.

mvr = nh/(2π).

(2)

The de-Broglie wavelength of electron in (n + 1)th orbit is λ=

O rn

hr 2πr h = = . mv mvr n+1

(3)

The radius of (n + 1)th orbit is r = a0 (n + 1)2 /Z, th

The perimeter of the n Bohr’s orbit is n times the de-Broglie wavelength of the electron in that orbit (see figure) i.e., 2πrn = nλ,

where a0 = 0.529 × 10−10 m is Bohr’s radius. Substitute r from equation (4) into equation (3) and then substitute λ into equation (1) to get

(1) n+1−

where rn is the radius of n rn = n2 a0 /Z.

th

Bohr’s orbit (2)

(4)

1 1 = = 24.93, n+1 2πRZa0

(5)

which gives n = 24. Ans. 24

Chapter 39. Bohr’s Model and Physics of the Atom Q 34. Wavelength belonging to Balmer series, lying in the range of 450 nm to 750 nm, were used to eject photoelectrons from a metal surface whose work function is 2.0 eV. Find (in eV) the maximum kinetic energy of the emitted photoelectrons. [Take hc = 1242 eV nm.] (2004)

Sol. The maximum kinetic energy of emitted photoelectrons is given by Kmax = hc/λmin − φ,

(1)

where φ = 2.0 eV. For the given range of wavelengths, maximum value of the photon energy is hc/λ1 = 1242/450 = 2.76 eV and minimum energy is hc/λ2 = 1242/750 = 1.66 eV. The energy of photon in Balmer series is given by   1 1 (2) ∆Em→2 = 13.6 2 − 2 , 2 m where m ≥ 3. 750nm (1.66eV) 1.88

450nm (2.76eV) E(eV) 2.55 2.85

Substitute the values of m in equation (2) to get ∆E3→2 = 1.88 eV, ∆E4→2 = 2.55 eV and ∆E5→2 = 2.85 eV. Thus, maximum energy of photon lying in the given range is 2.55 eV. Substitute it in equation (1) to get Kmax = 0.55 eV. Ans. 0.55 eV Q 35. A hydrogen-like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between −0.85 eV and −0.544 eV (including both these values). [Take hc = 1240 eV-nm, ground state energy of hydrogen atom = −13.6 eV.] (2002) (a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in these transitions. Sol. Six transitions are possible between four energy levels (k C2 = 6 gives k = 4) as shown in the figure. m+3 (−0.544eV) m+2 m+1 m (−0.85eV)

The energy of nth level is given by En = −

13.6Z 2 eV. n2

513 Thus, 13.6Z 2 = −0.85, m2 13.6Z 2 = −0.544. − (m + 3)2

−

(1) (2)

Solve equations (1) and (2) to get Z = 3 and m = 12. The wavelength is minimum for transitions having maximum energy gap ∆Emax = Em+3 − Em = E15 − E12 = −0.544 − (−0.85) = 0.306 eV. Hence, λmin = hc/∆Emax = 1240/0.306 = 4052 nm. Ans. (a) 3 (b) 4052 nm Q 36. A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during deexcitation. Ground state energy of hydrogen atom is −13.6 eV. (2000) Sol. The energy of hydrogen like atom in state of quantum number n is given by En = −

13.6Z 2 eV, n2

where Z is atomic number of the atom. Thus, the energy in ground state is E1 = −13.6Z 2 ,

(1)

and energy in state 2n is E2n = −

13.6Z 2 . 4n2

(2) 2n ∆Emin 40.8eV

204eV

2n-1 n 1

Maximum energy photon is emitted when the transition is from excited state to ground state (see figure) i.e.,   1 2 E2n − E1 = −13.6Z − 1 = 204 eV. (3) 4n2 Transition from state 2n to state n gives   1 1 E2n − En = −13.6Z 2 − = 40.8 eV. 4n2 n2

(4)

514

Part VI. Modern Physics

Solve equations (3) and (4) to get n = 2 and Z = 4. Substitute these in equation (1) to get E1 = −217.6 eV. The minimum energy photon is emitted when the transition is from state 2n to state 2n−1, ∆Emin = E2n − E2n-1 = E4 − E3 = 10.58 eV. Ans. n = 2, Z = 4, −217.6 eV, 10.58 eV Q 37. Photelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing α-particles. A maximum energy electron combines with an α-particle to form a He+ ion, emitting a single photon in this process. He+ ions thus formed are in their fourth excited state. Find the energies (in eV) of the photons lying in the 2 eV to 4 eV range, that are likely to be emitted during and after the combination. [Take h = 4.14 × 10−15 eV s.] (1999) Sol. The energy of incident photon is 6.64 × 10−34 × 3 × 108 hc = = 3.1 eV. λ 400 × 10−9 × 1.6 × 10−19

(1996)

(a) the kinetic energy. (b) the de-Broglie wavelength of the electron. Sol. The kinetic energy K and potential energy U of an electron in a hydrogen like atom are related by K = −U/2. The total energy is E = U + K = (−2K) + K = −K. Thus,

The de-Broglie wavelength of the electron is given by h h =√ p 2mK 6.63 × 10−34 =p 2(9.1 × 10−31 )(3.4 × 1.6 × 10−19 ) = 6.66 × 10−10 = 6.66 ˚ A.

λ=

Kmax = hc/λ − φ = 3.1 − 1.9 = 1.2 eV. The combination reaction is, He + e → He+ + hν, where hν represents the emitted photon. The ion He+ is hydrogen like with Z = 2. The principal quantum number in fourth excited state of He+ is n = 5. The energy in nth levels of He+ is 13.6Z 2 . n2

Q 38. An electron in a hydrogen like atom is in an excited state. It has a total energy of −3.4 eV. Calculate,

K = −E = −(−3.4) = 3.4 eV.

In photoelectric effect, maximum kinetic energy of the ejected electron is

En = −

The energy of these transitions are ∆E5→4 = E5 − E4 = 1.2 eV, ∆E5→3 = 3.84 eV, ∆E5→2 = 11.4 eV, and ∆E5→1 = 53.2 eV, out of which ∆E5→3 = 3.84 eV lies in the range of 2 eV to 4 eV. There is a possibility that He+ makes multiple transitions during the deexcitation i.e., it can make transition 5 → 4 and then 4 → 3 or 4 → 2 or 4 → 1 and so on. We encourage you to verify that only ∆E4→3 = E4 − E3 = 2.64 eV is less than 4 eV. Ans. During combination 3.4 eV. After combination 3.84 eV, 2.64 eV.

(1)

Ans. (a) 3.4 eV (b) 6.66 ˚ A Q 39. An electron in the ground state of hydrogen atom is revolving in anticlockwise direction in a circular orbit of radius R (see figure). (1996)

Substitute the values in equation (1) to get E1 = −54.4 eV, E2 = −13.6 eV, E3 = −6.04 eV, E4 = −3.4 eV and E5 = −2.2 eV. Using energy conservation in combination reaction

n ˆ

~ B

30◦

EHe + Ee = EHe+ + Ehν , with EHe = 0, Ee = Kmax = 1.2 eV, EHe+ = E5 = −2.2 eV, we get Ehν = 1.2 − (−2.2) = 3.4 eV. After combination, He+ can de-excite to the lower states by emitting photons as shown in the figure. E5 = −2.2eV E4 = −3.4eV E3 = −6.04eV E2 = −13.6eV E1 = −55.4eV

(a) Obtain an expression for the orbital magnetic moment of the electron. (b) The atom is placed in a uniform magnetic induction ~ such that the normal to the plane of electron’s orB bit makes an angle of 30◦ with the magnetic induction. Find the torque experienced by the orbiting electron. Sol. In Bohr’s model, the angular momentum in an orbit is quantized i.e., mvr = nh/(2π).

Chapter 39. Bohr’s Model and Physics of the Atom Substitute n = 1 and r = R to get ground state velocity h . 2πmR The time taken by the electron to complete one revolution is v=

2

2

T = 2πR/v = 4π R m/h. The current through the loop is eh e = , T 4π 2 R2 m and its magnetic moment is i=

~ ×B ~ = M B sin 30◦ rˆ = ehB rˆ, ~τ = M 8πm ~ and where rˆ is a unit vector perpendicular to both M ~ B. eh ehB Ans. (a) 4πm (b) 8πm Q 40. A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV respectively. Alternately, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the values of n and Z. [Ionization energy of H atom = 13.6 eV.] (1994) Sol. The energy of a hydrogen like atom with atomic number Z is given by 13.6Z 2 En = − eV, n2 where n is principal quantum number. The energies in the first excited state (n = 2) and in the second excited state (n = 3) are 2

13.6Z 13.6Z , E3 = − . 22 32 The energy gap between En and E2 is equal to total energy of emitted photons i.e.,   1 1 En − E2 = 13.6Z 2 2 − 2 2 n E2 = −

= 10.2 + 17 = 27.2 eV.

(1)

Similarly, En − E3 = 13.6Z

2



Q 41. A neutron of kinetic energy 65 eV collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90◦ with respect to its original direction. (1993) (a) Find the allowed values of the energy of the neutron and that of the atom after collision. (b) If the atom gets de-excited subsequently by emitting radiation, find the frequencies of the emitted radiation. [Given: mass of He atom = 4 × (mass of neutron), ionization energy of H atom = 13.6 eV.]

eh ~ = iA n M ˆ = i(πR2 ) n ˆ= n ˆ. 4πm The magnetic moment is normal to the plane of orbit. The torque in a magnetic field is given by

2

515

 1 1 − 2 = 4.25 + 5.95 32 n

= 10.2 eV. Solve equations (1) and (2) to get n = 6 and Z = 3. Ans. 6, 3

(2)

Sol. Let K1 = 65 eV be the kinetic energy of the neutron before the collision, m be the mass of the neutron and M = 4m be the mass of the helium atom. The momenta of two particles before the collision are √ p1 = 2mK1 and p2 = 0. Let p01 and p02 be the momenta after the collision. Let p01 be perpendicular to x axis and p02 makes an angle θ with x axis. p01 p1

p2

n

He

y

90◦ x

θ p02

The conservation of linear momentum in x and y directions gives p02 cos θ = p1 ,

(1)

p02

(2)

sin θ =

p01 . 2

2

Square and add equations (1) and (2) to get p02 = p01 + p21 , which gives 4K20 = K10 + K1 .

(3)

Now, let an electron in ground state of singly ionized helium is excited to higher state by absorbing energy ∆E. The conservation of energy gives K1 = K10 + K20 + ∆E.

(4)

Solve equations (3) and (4) to get 3 K1 − 5 2 K20 = K1 − 5 K10 =

4 ∆E = 39 − 5 1 ∆E = 26 − 5

4 ∆E, 5 1 ∆E. 5

The energy of singly ionized helium in nth state is En = −

13.6Z 2 13.6(2)2 54.4 = − =− 2 . 2 2 n n n

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Part VI. Modern Physics

The values of ∆E for  1 ∆E12 = 54.4 2 1  1 ∆E13 = 54.4 2 1  1 ∆E14 = 54.4 2 1

various transitions are  1 − 2 = 40.8 eV, 2  1 − 2 = 48.36 eV, 3  1 − 2 = 51.0 eV 4

and so on. The ∆E12 gives K10 = 6.36 eV and K20 = 17.84 eV, ∆E13 gives K10 = 0.312 eV and K20 = 16.328 eV, and ∆E14 gives K10 = −1.8 eV and K20 = 15.8 eV. Since kinetic energy cannot be negative, only transitions n = 1 → 2 and n = 1 → 3 are possible. n=3

ν32

ν21 n=1

The excited state of electron can be n = 2 or n = 3. The energies of first three levels are E1 = −54.4 eV, E2 = −54.4/(2)2 = −13.6 eV and E3 = −54.4/(3)2 = −6.04 eV. The electron can make three transitions as shown in the figure. The frequencies of these transitions are

ν31 ν21

En = −

13.6 eV. n2

(1)

Substitute the values of n in equation (1) to get E1 = −13.6 eV, E2 = −3.4 eV, E3 = −1.5 eV, E4 = −0.85 eV, etc. The energy gap between n = 4 and n = 2 is ∆E = E4 − E2 = 2.55 eV. Thus, the transition between these levels emits a photon of energy 2.55 eV. The angular momentum of the electron in nth energy level is Ln = nh/(2π).

n=2 ν31

ν32

The photons are emitted when electron makes a transition between the two energy levels. The energy of nth level in hydrogen atom is

E3 − E2 (−6.04 − (−13.6))1.6 × 10−19 = = h 6.63 × 10−34 15 = 1.82 × 10 Hz, E3 − E1 = 11.67 × 1015 Hz, = h E2 − E1 = = 9.84 × 1015 Hz. h

Ans. (a) 6.36 eV, 0.312 eV (of neutron), 17.84 eV, 16.328 eV (of atom) (b) 1.82 × 1015 Hz, 11.67 × 1015 Hz, 9.84 × 1015 Hz Q 42. Light from discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find, [Given, ionization potential of hydrogen is 13.6 eV.] (1992) (a) the energy of the photons causing the photoelectrons emission. (b) the quantum numbers of the two levels involved in the emission of these photons. (c) the change in the angular momentum of the electron in the hydrogen atom in the above transition. (d) the recoil speed of the emitting atom assuming it to be at rest before the transition. Sol. In photoelectric effect, the kinetic energy of the fastest photoelectrons is given by Kmax = E − φ. Thus, the energy of incident photon is E = Kmax + φ = 0.73 + 1.82 = 2.55 eV.

Thus, the change in angular momentum for n = 4 to n = 2 transition is ∆L = L2 − L4 = −h/π. The linear momentum of the atom-photon system is conserved. Initially, linear momentum is zero i.e., pi = 0. Finally, linear momentum of the photon is E/c, linear momentum of the hydrogen atom is mv, and of the total system is pf = E/c + mv. Using, pi = pf , we get 2.55 × 1.6 × 10−19 E =− mc 1.67 × 10−27 × 3 × 108 = −0.814 m/s,

v=−

(recoil).

Ans. (a) 2.55 eV (b) 4 → 2 (c) − πh (d) 0.814 m/s Q 43. Electrons in hydrogen-like atom (Z = 3) make transitions from the fifth to the fourth orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 V. Calculate the work function of the metal, and the stopping potential for the photoelectrons ejected by the longer wavelength. [Rydberg’s constant = 1.094 × 107 m−1 .] (1990)

Sol. The energy of n En = −

th

orbit in hydrogen-like atom is

13.6Z 2 eV. n2

The transition n = 5 → 4 and n = 4 → 3 emit photons of energies   1 1 2 ∆E54 = E5 − E4 = 13.6 × 3 − 2 = 2.754 eV. 42 5   1 1 2 ∆E43 = E4 − E3 = 13.6 × 3 − 2 = 5.95 eV. 32 4 Thus, photon’s energy for shorter wavelength is 5.95 eV. In photoelectric effect, stopping potential corresponds

Chapter 39. Bohr’s Model and Physics of the Atom to maximum kinetic energy photoelectrons. The maximum kinetic energy of ejected photoelectrons is Kmax = ∆E43 − φ,

i.e.,

3.95 = 5.95 − φ,

which gives the work function φ = 2 eV. The maximum kinetic energy of photoelectrons ejected by photons of longer wavelength is Kmax = ∆E54 − φ = 2.754 − 2 = 0.754 eV. Thus, the stopping potential for these photoelectrons is 0.754 eV. Ans. 2 eV, 0.754 eV Q 44. A gas of identical hydrogen-like atoms has some atoms in the lowest (ground) energy level A and some atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level. The atoms of the gas make the transition to a higher energy level by absorbing monochromatic light of photon energy 2.7 eV. Subsequently, the atoms emit radiation of only six different energy photons. Some of the emitted photons have an energy of 2.7 eV, some have more energy and some less than 2.7 eV. (1989) (a) Find the principal quantum number of the initially excited level B. (b) Find the ionization energy of the gas atoms. (c) Find the maximum and the minimum energies of the emitted photons. Sol. Let ni and nf be the principal quantum numbers of the initial and the final (after excitation) states of the atom. An atom in excited state nf can make nf C2 = nf (nf − 1)/2 different transitions while returning to ground state. Since excited atoms emit photons of six different energies, we get nf (nf − 1)/2 = 6.

(1)

2.7eV

Solve the quadratic equation (1) to get nf = 4 and nf = −3 (unacceptable). n=4 n=3 n=2 n=1

Thus, the possible initial states are ni = 1, 2, or 3. Since the atoms reach excited state by absorbing the photons of energy ∆E = 2.7 eV, the energy gap between the states ni and nf is equal to 2.7 eV. If ni = 1 then the energy of all the emitted photons will be less than or equal to 2.7 eV. If ni = 3 then the energy of all the emitted photons will be more than or equal to 2.7 eV. Since some of the emitted photons have energy less than 2.7 eV and some have energy greater than

517 2.7 eV, ni = 1 and ni = 3 are not allowed. Thus, the given conditions can be fully satisfied by ni = 2 only (see figure). The energy of hydrogen like atom in a state of principal quantum number n is given by En = −E0 /n2 , where E0 is a constant. The ground state energy, E1 = −E0 , is the binding energy of the electron. Thus, E0 is equal to the ionization energy of the atom. The energy gap between ni = 2 and nf = 4 is   1 1 E4 − E2 = −E0 2 − 2 = 2.7 eV, 4 2 which gives the ionization energy E0 = 14.4 eV. The energy of the emitted photons is maximum for the transition n = 4 to n = 1 and is given by     1 1 15 Emax = E4 − E1 = −E0 2 − 2 = 14.4 4 1 16 = 13.5 eV. The energy of the emitted photons is minimum for the transition n = 4 to n = 3 and is given by     1 1 7 Emin = E4 − E3 = −E0 2 − 2 = 14.4 4 3 144 = 0.7 eV. Ans. (a) 2 (b) 14.4 eV (c) 13.5 eV, 0.7 eV Q 45. A particle of charge equal to that of an electron, −e, and mass 208 times of the mass of the electron (called a mu-meson) moves in a circular orbit around a nucleus of charge +3e. Assuming that the Bohr model of the atom is applicable to this system, [Take the mass of the nucleus to be infinite and Rydberg’s constant = 1.097 × 107 m−1 .] (1988) (a) derive the expression for the radius of the nth Bohr orbit. (b) find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom. (c) find the wavelength of the radiation emitted when the mu-meson jumps from the third orbit to the first orbit. Sol. Let me be the mass of the electron, m = 208me be the mass of the mu-meson, q = −e be the charge of the mu-meson, and Q = Ze = 3e be the charge of the nucleus. Since mass of the nucleus is infinite, we assume mu-meson to be moving around the nucleus in a circular orbit of radius r. Coulomb’s force on the mumeson provides the necessary centripetal acceleration to it i.e., mv 2 Ze2 = . r 4π0 r2

(1)

518

Part VI. Modern Physics

According to Bohr’s model, the angular momentum of the mu-meson is quantized i.e., mvr = nh/(2π),

(2)

where n is an integer. Eliminate v from equations (1) and (2) to get n2 h2 0 n2 h2 0 = . Zπme2 624πme e2

r=

(3)

The radius of first Bohr orbit for the hydrogen atom is given by h2 0 r0 = . πme e2

(4)

Use equations (3) and (4) to get the value of n for which r = r0 i.e., n=

√

624 ≈ 25.

The energy of the atom is the sum of kinetic and potential energy i.e., E=

1 Ze2 mv 2 − . 2 4π0 r

Eliminate v and r by using equations (1) and (2) to get the energy   mZ 2 e4 (3)3 × 208 m e e4 = − 820 h2 n2 n2 80 h2 25459 = − 2 eV, n

En = −

4

me e where − 8 2 = −13.6 eV is the ground state energy 0h of the hydrogen atom. The wavelength of the radiation emitted when mu-meson jumps from the third orbit to the first orbit, is given by





hc 12420 1 1 = = −25459 2 − 2 , λ λ 3 1 12420 × 9 i.e., λ = = 0.548 ˚ A. 25459 × 8 Ans. (a) rn =

n2 h2 0 624πme e2

(b) n ≈ 25 (c) 0.548 ˚ A

Q 46. A doubly ionised lithium atom is hydrogen-like with atomic number 3. (1985) (a) Find the wavelength of the radiation required to excite the electron in Li2+ from the first to the third Bohr orbit. [Ionization energy of the hydrogen atom equals 13.6 eV.] (b) How many spectral lines are observed in the emission spectrum of the above excited system?

Sol. The energy of hydrogen like atom of atomic number Z = 3 in nth Bohr orbit is given by En = −

13.6Z 2 13.6(3)2 122.4 =− = − 2 eV. 2 n n2 n

A photon can excite the Li2+ ion from the first orbit to the third orbit if its energy is equal to the energy gap between these orbits. The energy gap between the first orbit and the third orbit is   1 1 ∆E = E3 − E1 = −122.4 2 − 2 = 108.8 eV. 3 1 The wavelength of radiation with energy equal to ∆E = 108.8 eV is λ=

˚ hc 12420 eV-A = = 114.15 ˚ A. ∆E 108.8 eV

The electron can make n(n − 1)/2 possible transitions while coming from nth state to the ground state. For n = 3, possible transitions are three i.e., 3 → 2, 3 → 1 and 2 → 1. Thus, the emission spectrum will have three spectral lines. We encourage you to find the wavelengths of these spectral lines. Ans. (a) 114.15 ˚ A (b) 3 Q 47. The ionization energy of a hydrogen like Bohr atom is 4 Rydberg. (1984) (a) What is the wavelength of the radiation emitted when the electron jumps from the first excited state to the ground state? (b) What is the radius of the first orbit for this atom? Sol. The ionization energy is equal to the binding energy of the electron in its first orbit (n = 1). Thus, the binding energy of the electron in its ground state is 4 rydberg i.e., BE = 4 rydberg = 4(hcR∞ ) = 4 (1242 eV-nm) (1.097 × 107 m−1 ) = 4 (1242 × 10−9 ) (1.097 × 107 ) eV = 54.5 eV. Hence, the energy of the electron in its ground state is E1 = −54.5 eV. Bohr’s model gives the energy of the electron in its nth orbit as En = E1 /n2 = −54.5/n2 eV.

(1)

The energy of emitted radiation when electron makes a transition from n1 = 2 to n2 = 1 is equal to the energy gap between these two orbits i.e.,   hc 1 1 = −54.5 2 − 2 = 40.87 eV. (2) λ 2 1 Solve equation (2) to get λ = hc/40.87 = 1242/40.87 = 30.4 nm.

Chapter 39. Bohr’s Model and Physics of the Atom The energy of hydrogen like atom is related to atomic number Z by En = −13.6Z 2 /n2 eV.

(3)

Use equations (1) and (3) to get Z = 2. The radius of hydrogen like atom in nth orbit is given by

519 Sol. The energy of the photon of radiation of wavelength λ = 975 ˚ A is given by ∆E = hc/λ = (12420 eV-˚ A)/(975 ˚ A) = 12.74 eV. The energy of the hydrogen atom in nth state is given by En = −13.6/n2 eV.

2

rn = n a0 /Z,

(4)

where a0 = 0.529 ˚ A is the radius of the hydrogen atom. Substitute n = 1 and Z = 2 in equation (4) to get r1 = 0.529/2 = 0.2645 ˚ A. Ans. (a) 30.4 nm (b) 0.2645 ˚ A Q 48. Ultraviolet light of wavelength 800 ˚ A and 700 ˚ A when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energy 1.8 eV and 4.0 eV respectively. Find the value of Planck’s constant? (1983)

The hydrogen atom is excited from the ground state (n = 1) to the nth state by absorbing a photon of energy ∆E = 12.74 eV if   1 1 (1) En − E1 = −13.6 2 − 2 = 12.74. n 1 Solve equation (1) to get n = 4. The electron in excited state (n = 4) can make n(n − 1)/2 = 4(4 − 1)/2 = 6 different transitions while coming back to the ground state. n=4 n=3

Sol. The photon of the incident ultraviolet light ejects electron from the ground state of hydrogen atom. The energy of the incident photon is equal to the sum of the ionization energy E0 of hydrogen atom and the kinetic energy K of the ejected electron i.e., hc/λ = E0 + K.

(1)

The ionization energy of hydrogen atom is E0 = 13.6 eV = 13.6 × 1.6 × 10−19 = 2.176 × 10−18 J. The kinetic energy of the ejected electron for the wavelength λ1 = 800 ˚ A is K1 = 1.8 eV and for the wavelength λ2 = 700 ˚ A is K2 = 4.0 eV. Substitute it in equation (1) to get hc = λ1 (E0 + K1 ),

(2)

hc = λ2 (E0 + K2 ).

(3)

n=2

n=1

The maximum wavelength in the emission spectrum corresponds to the transition with minimum energy gap i.e., transition from n = 4 to n = 3. The energy gap between these levels is   1 1 ∆E34 = −13.6 2 − 2 = 0.66 eV, 4 3 and corresponding wavelength is λ = hc/∆E3,4 = 12420/0.66 ˚ A = 1.882 × 10−6 m = 1.882 µm. Ans. Six, 1.882 µm

Solve equations (2) and (3) to get λ1 (E0 + K1 ) + λ2 (E0 + K2 ) 2c 800 × 10−10 (13.6 + 1.8)1.6 × 10−19 = 2(3 × 108 ) 700 × 10−10 (13.6 + 4.0)1.6 × 10−19 + 2(3 × 108 )

h=

= 6.6 × 10−34 J s. Ans. 6.6 × 10−34 J s Q 49. Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975 ˚ A. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom as 13.6 eV. (1982)

Q 50. A single electron orbits around a stationary nucleus of charge +Ze, where Z is a constant and e is the magnitude of the electronic charge. It requires 47.2 eV to excite the electron from the second Bohr orbit to the third Bohr orbit. Find, (1981) (a) the value of Z. (b) the energy required to excite the electron from the third to the fourth Bohr orbit. (c) the wavelength of the electromagnetic radiation required to remove the electron from the first Bohr orbit to infinity. (d) the kinetic energy, potential energy and the angular momentum of the electron in the first Bohr orbit. (e) the radius of the first Bohr orbit. [The ionization energy of hydrogen atom = 13.6 eV, velocity of light in vacuum = 3 × 108 m/s, Bohr radius = 5.3 × 10−11 m, and Planck’s constant = 6.6 × 10−34 J s.]

520

Part VI. Modern Physics

Sol. The energy of nth Bohr orbit for hydrogen like atom having nuclear charge +Ze is given by En = −13.6Z 2 /n2 eV. The energy required to excite the electron from the second Bohr orbit to the third Bohr orbit is equal to the energy gap between these orbits i.e.,   1 1 2 − 2 = 47.2. E3 − E2 = −13.6Z 32 2 Solve to get Z = 5. The energy required to excite an electron from the third to the fourth Bohr orbit is equal to the energy gap between these orbits i.e.,   1 1 2 ∆E = E4 − E3 = −13.6(5) − 2 = 16.53 eV. 42 3 The energy required to remove an electron from the first Bohr orbit to infinity (ionization energy) is equal to the binding energy of the electron in first Bohr orbit i.e., BE = −E1 = 13.6(5)2 = 340 eV. The wavelength of radiation of photon energy 304 eV is given by λ=

hc 12420 eV-˚ A = = 36.53 ˚ A. BE 340 eV

The total energy in the first orbit is the sum of the kinetic energy K and the potential energy U i.e., U +K = E1 = −340 eV. Also, K and U in a Bohr orbit are related by K = −U/2. Solve to get K = 340 eV and U = −680 eV. The angular momentum in the nth Bohr orbit is given by Ln = nh/(2π). Substitute n = 1 to get L1 = 1.05 × 10−34 kg m2 s−1 . The radius of the nth Bohr orbit is given by rn = 2 n a0 /Z, where a0 = 5.3 × 10−11 m is the radius of the hydrogen atom. Substitute n = 1 and Z = 5 to get radius of the first orbit as r1 = 1.06 × 10−11 m. Note that the radius of Bohr orbit decreases with increase in the atomic number. Ans. (a) 5 (b) 16.53 eV (c) 36.53 ˚ A (d) 340 eV, −680 eV, −340 eV, 1.05 × 10−34 kg m2 s−1 (e) 1.06 × 10−11 m

Chapter 40 X-rays

One Option Correct

Q 3. Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is (2007) 2 2h 2m2 c2 λ3 (B) (C) (D) λ (A) 2mcλ h mc h2

Q 1. If λCu is the wavelength of Kα X-ray line of copper (atomic number 29) and λMo is the wavelength of the Kα X-ray line of molybdenum (atomic number 42), the the ratio λCu /λMo is close to (2014) (A) 1.99 (B) 2.14 (C) 0.50 (D) 0.48

Sol. The linear momentum of an electron having deBroglie wavelength λ is given by

Sol. The wavelength of Kα X-ray line is related to atomic number Z by   1 1 3 1 2 = R(Z − 1) − 2 = R(Z − 1)2 . 2 λ 1 2 4

p = h/λ. Thus, the kinetic energy of an electron having mass m and de-Broglie wavelength λ is K = p2 /(2m) = h2 /(2mλ2 ).

Substitute the value of Z to get λCu (ZMo − 1)2 (41)2 = = = 2.14. 2 λMo (ZCu − 1) (28)2

The energy of photon corresponding to cut-off wavelength λc is equal to the kinetic energy of the electron i.e.,

Ans. B Q 2. Which of the following statements is wrong in the context of X-rays generated from a X-ray tube? (2008) (A) Wavelength of characteristic X-rays decreases when the atomic number of the target increases. (B) Cut-off wavelength of the continuous X-rays depends on the atomic number of the target. (C) Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube. (D) Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube.

hc/λc = h2 /(2mλ2 ), which gives λc = 2mcλ2 /h. Ans. A Q 4. Kα wavelength emitted by an atom of atomic number Z = 11 is λ. The atomic number of an atom that emits Kα radiation with wavelength 4λ is (2005) (A) 6 (B) 4 (C) 11 (D) 44 Sol. The Kα wavelength is emitted when an electron in L shell makes a transition to the vacant state in K shell. In Mosley’s law, √ ν = a(Z − b), (1)

Sol. The frequency ν of characteristic X-rays is related to atomic number Z by Moseley’s law, √ ν = a(Z − b),

the parameter b ≈ 1 for Kα line because electron from L shell finds nuclear charge Ze shielded by the remaining electron in K shell i.e., effective nuclear charge is (Z − 1)e. With this assumption, substitute the values in equation (1) to get p c/λ = a(11 − 1), (2) p c/(4λ) = a(Z − 1). (3)

which gives c c λ= = 2 . ν a (Z − b)2 Thus, the wavelength of emitted X-rays decreases with increase in Z. The cut-off wavelength of continuous X-rays corresponds to maximum energy of electron in X-ray tube. It is given by hc/λ = eV,

Eliminate λ from equations (2) and (3) to get Z = 6. Ans. A

where V is the accelerating potential. The intensity of X-rays depends on the number of electrons striking the target per second, which, in turn, depends on the electrical power given to the X-ray tube as energy of each electron is eV . Ans. B

Q 5. The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA. Then the number of electrons striking the target per second is (2002)

(A) 2 × 1016 (B) 5 × 106 (C) 1 × 1017 (D) 4 × 1015 521

522

Part VI. Modern Physics

Sol. The current is charge flowing per unit time i.e., i = q/t = ne/t. Substitute the values to get the number of electrons flowing in one second (i.e., number of electrons striking the target per second) n=

3.2 × 10−3 × 1 it = = 2 × 1016 . e 1.6 × 10−19 Ans. A

Q 6. The intensity of X-rays from a Coolidge tube is plotted against wavelength λ as shown in the figure. The minimum wavelength found is λc and the wavelength of the Kα line is λk . As the accelerating voltage is increased, (2001) I

λc

(A) (λk − λc ) increases. (C) λk increases.

λk

λ

Sol. The wavelength λ in continuous X-ray spectrum has a lower limit given by λmin = hc/(eV ). There is no upper limit on λ. Ans. B Q 9. The Kα X-ray emission line of tungsten occurs at λ = 0.021 nm. The energy difference between K and L levels in this atom is about (1997) (A) 0.51 MeV (B) 1.2 MeV (C) 59 keV (D) 13.6 eV Sol. Given λKα = 0.021 nm = 0.21 × 10−10 m. The λKα corresponds to transition of an electron from L shell to K shell. Thus, the energy difference between these levels is EL − EK =

(B) (λk − λc ) decreases. (D) λk decreases.

Sol. The minimum wavelength, λc , is related to accelerating voltage V by hc/λc = eV. The characteristic wavelength, λk , is independent of V . Thus, increase in V decreases λc and hence increases λk − λc . Ans. A Q 7. Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K-shell electrons of tungsten have 72.5 keV energy. X-rays emitted by the tube contains only, (2000) (A) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of ≈ 0.155 ˚ A. (B) a continuous X-ray spectrum with all wavelengths. (C) the characteristic X-ray spectrum of tungsten. (D) a continuous X-ray spectrum with a minimum wavelength of ≈ 0.155 ˚ A and the characteristic Xray spectrum of tungsten. Sol. The energy of incident electron (Ein = 80 keV) is sufficient to knock out K-shell electrons (72.5 keV) thereby emitting characteristic X-rays. The minimum wavelength of continuous spectrum corresponds to Ein and is given by λ=

Q 8. X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from (1998) (A) 0 to ∞. (B) λmin to ∞, where λmin > 0. (C) 0 to λmax , where λmax < ∞. (D) λmin to λmax where 0 < λmin < λmax < ∞.

(6.63 × 10−34 ) (3 × 108 ) hc = = 0.155 ˚ A. Ein (80 × 103 ) (1.602 × 10−19 )

(6.63 × 10−34 )(3 × 108 ) hc = λKα (0.21 × 10−10 )(1.6 × 10−19 )

= 59×103 eV = 59 keV. Ans. C Q 10. The X-ray beam coming from an X-ray tube will be, (1985) (A) monochromatic. (B) having all wavelengths smaller than a certain maximum wavelength. (C) having all wavelengths larger than a certain minimum wavelength. (D) having all wavelengths lying between a minimum and maximum wavelength. Sol. The cutoff wavelength of an X-ray tube operated at a potential V is λmin = hc/(eV ). Ans. C Q 11. The shortest wavelength of X-ray emitted from an X-ray tube depends on (1982) (A) the current in the tube. (B) the voltage applied to the tube. (C) the nature of the gas in tube. (D) the atomic number of the target material. Sol. The shortest wavelength of X-ray emitted by Xray tube depends on the applied voltage V i.e., λmin = hc/(eV ). Ans. B One or More Option(s) Correct Q 12. The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation, (1988)

Ans. D

(A) the intensity increases.

Chapter 40. X-rays

523

(B) the minimum wavelength increases. (C) the intensity remains unchanged. (D) the minimum wavelength decreases. Sol. The minimum wavelength of X-ray is related to the applied potential by λmin =

12420 ˚ hc A. = eV V

Thus, the minimum wavelength decreases when V is increased. The intensity is the rate of energy flow per unit area. The energy of photons increases when V is increased. Thus, intensity also increases with an increase in the applied potential. However, intensity of X-rays is generally controlled by the filament current. An increase in the filament current increases its temperature which in turn increases the rate of electron emissions (thermionic emissions). Thus, more number of electrons strikes the target leading to increase in number of X-ray photons. Ans. A, D Assertion Reasoning Type Q 13. Statement 1: If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change. Statement 2: When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy. (2007) (A) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. (B) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is true, statement 2 is false. (D) Statement 1 is false, statement 2 is true. Sol. The wavelength of characteristic X-rays depends on the energy levels in target material and not on the accelerating potential. Ans. B

Sol. Transition between two atomic energy levels is responsible for hydrogen spectrum as well as characteristic X-rays. Electrons are emitted from a material in photoelectric effect as well as in β-decay. Moseley’s law relates wavelength of characteristic X-rays to atomic number of the target material. Photon energy is converted into kinetic energy of electron in photoelectric effect. Ans. A7→(p,r), B7→(q,s), C7→p, D7→q Fill in the Blank Type Q 15. A potential difference of 20 kV is applied across an X-ray tube. The minimum wavelength of X-rays generated is . . . . . . ˚ A. (1996) Sol. The minimum wavelength of generated X-rays is given by (6.63 × 10−34 ) (3 × 108 ) hc = eV (1.6 × 10−19 ) (20 × 103 ) ˚. = 0.62 × 10−10 m = 0.62 A

λmin =

You may use hc/e = 12431 ˚ A V. Ans. 0.62 Q 16. The wavelength of Kα X-rays produced by an X-ray tube is 0.76 ˚ A. The atomic number of the anode material of the tube is . . . . . . (1996) Sol. The wavelength of Kα X-ray is given by   1 1 1 2 − 2 . = R(Z − 1) λα 12 2

(1)

Simplify equation (1) and substitute the values to get Z − 1= √

2 2 =p 7 3Rλα 3(1.097 × 10 )(0.76 × 10−10 )

≈ 40,

Matrix or Matching Type Q 14. Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II. (2007) Column I

Column II

(A) Transition between two atomic energy levels. (B) Electron emission from a material. (C) Mosley’s law.

(p) Characteristic X-rays. (q) Photoelectric effect. (r) Hydrogen spectrum. (s) β-decay.

(D) Change of photon energy into kinetic energy of electrons.

which gives Z = 41. Ans. 41 Q 17. In an X-ray tube, electrons accelerated through a potential difference of 15, 000 V strike a copper target. The speed of the emitted X-ray inside the tube is . . . . . . m/s. (1992) Sol. The X-rays are electromagnetic waves that travel with the speed of light, c = 3 × 108 m/s, inside the tube (assumed vacuum). Ans. 3 × 108 Q 18. The wavelength of the characteristic X-ray Kα line emitted by a hydrogen like element is 0.32 ˚ A. The wavelength of the Kβ line emitted by the same element will be . . . . . . (1990)

524

Part VI. Modern Physics

Sol. The transition for Kα is n = 2 → 1 and for Kβ is n = 3 → 1. The energy of nth orbit in hydrogen like atom is given by En = −

E1 , n2

where E1 is a constant. Thus, the energies released in n = 2 → 1 and n = 3 → 1 transitions are   1 3 hc 1 = E1 2 − 2 = E1 , (1) λα 1 2 4   hc 1 8 1 = E1 2 − 2 = E1 . (2) λβ 1 3 9 Divide equation (1) by (2) to get λβ =

27 27 λα = × 0.32 = 0.27 ˚ A. 32 32 Ans. 0.27 ˚ A

Q 19. When the number of electrons striking the anode of an X-ray tube is increased the . . . . . . of the emitted X-rays increases while when the speeds of the electrons striking the anode are increased the cut-off wavelength of the emitted X-rays . . . . . . . (1986) Sol. The intensity of X-ray increases when the number of electrons striking the anode increases. The cut-off voltage decreases when the energy (speed) of the electrons striking the anode increases. Ans. intensity, decreases Q 20. To produce characteristic X-rays using a tungsten target in an X-ray generator, the accelerating voltage should be greater than . . . . . . V and the energy of the characteristic radiation is . . . . . . eV. [The binding energy of the innermost electron in tungsten is 40 keV.] (1983)

Sol. To knockout an electron from the innermost shell, energy of the accelerating electron should be more than the binding energy of the innermost electron i.e., 40 keV. Thus, the accelerating voltage should be greater than 40 kV. The binding energy of the innermost electron is equal to the negative of the innermost energy level i.e., E1 = −BE = −40 keV. The Kα characteristic Xray is emitted when atom (with the innermost electron knocked out) makes a transition from L shell (n1 = 2) to K shell (n2 = 1). The energy of Kα characteristic radiation is given by 

1 1 ∆E = E2 − E1 = −40 2 − 2 2 1

Descriptive Q 21. Characteristic X-rays of frequency 4.2 × 1018 Hz are produced when transitions from L-shell to K-shell take place in a certain target material. Use Moseley’s law to determine the atomic number of the target material. [Rydberg’s constant = 1.1 × 107 m−1 .] (2003) Sol. The characteristic X-ray is emitted when an electron in L shell makes a transition to the vacant state in K shell. In Moseley’s law, √

ν = a(Z − b),

the parameter b ≈ 1 for this transition because electron from L shell finds the nuclear charge Ze shielded by the remaining electron in K shell i.e., effective nuclear charge is (Z − 1)e. Thus, by substituting the values,   1 ν 4.2 × 1018 1 1 2 = R(Z − 1) = = − 2 λ c n2 n2 3 × 108   1 1 1 = 1.1 × 107 (Z − 1)2 2 − 2 , 1 2 we get, Z = 42. Ans. 42 Q 22. Assume that the de-Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance d between the atoms of the array is 2 ˚ A. A similar standing wave is again formed if d is increased to 2.5 ˚ A but not for any intermediate value of d. Find the energy of the electron (in eV) and the least value of d for which the standing wave of the type described above can form. (1997) Sol. The standing wave of wavelength λ is formed if the distance d is an integral multiple of λ/2 i.e., d = nλ/2, where, n = 1, 2, · · · . When d is increased from 2 ˚ A to 2.5 ˚ A, n is increased by one. Thus, 2 = nλ/2,

(1)

2.5 = (n + 1)λ/2.

(2)

2˚ A

2.5˚ A

 keV = 30 keV.

Ans. 40 × 103 , 30 × 103

Solve equations (1) and (2) to get n = 4 and λ = 1˚ A. The momentum of an electron with de-Broglie

Chapter 40. X-rays wavelength λ is p = h/λ and its kinetic energy is p2 h2 = 2m 2mλ2 (6.63 × 10−34 )2 = ) 2(9.1 × 10−31 )(10−10 )2 (1.6 × 10−19 = 151 eV.

K=

Minimum value of d occurs for n = 1 i.e., dmin = nλ/2 = 1 × 1/2 = 0.5 ˚ A. Ans. 151 eV (b) 0.5 ˚ A

525

Chapter 41 Semiconductors and Semiconductor Devices

One Option Correct

Sol. In a forward bias, the holes move from p side to n side and electrons from n side to p side. Thus, the charge carriers move from their high density region to low density region, which is a diffusion process. In a reverse bias, movement of charge carriers is opposite of forward bias movement and is caused by the drift mechanism. Ans. B

Q 1. In a p-n junction diode not connected to any circuit, (1998) (A) the potential is same everywhere. (B) the p-type side is at a higher potential than the n-type side. (C) there is an electric field at the junction directed from the n-type side to the p-type side. (D) there is an electric field at the junction directed from the p-type side to the n-type side.

Q 4. The circuit shown in the figure contains two diodes each with a forward resistance of 50 Ω and with infinite backward resistance. If the battery voltage is 6 V, the current through the 100 Ω resistance is (1997)

Sol. In the depletion region, p side has net negative charge and n side has net positive charge.

150Ω 50Ω

p

n 6V ~ E

100Ω

(A) zero (B) 0.02 A (C) 0.03 A (D) 0.036 A Sol. Since the backward resistance of a diode is infinite, the reverse biased diode acts as an open circuit and no current flows through it (see figure).

This charge distribution gives rise to higher potential on n side as compared to p side and an electric field ~ from n side to p side. E Ans. C Q 2. The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap (in eV) for the semiconductor is (1997) (A) 0.9 (B) 0.7 (C) 0.5 (D) 1.1

50Ω

150Ω

6V

100Ω

i

Effective resistance of the circuit is

Sol. The electrical conductivity of a semiconductor increases when the emitted photoelectrons have energy equal to or greater than the band gap. Thus, in the limiting case,

R = 50 + 150 + 100 = 300 Ω, and the current through the circuit is i=

hc 1240 eV-nm ∆E = = = 0.5 eV. λmax 2480 nm

V 6 = = 0.02 A. R 300 Ans. B

Q 5. Which of the following is not true? (1997) (A) The resistance of intrinsic semiconductors decreases with increase of temperature. (B) Doping pure Si with trivalent impurities gives ptype semiconductors. (C) The majority carriers in n-type semiconductors are holes. (D) The p-n junction can act as a semiconductor diode.

Ans. C Q 3. The dominant mechanisms for motion of charge carriers in forward and reverse biased silicon p-n junctions are (1997) (A) drift in forward bias, diffusion in reverse bias. (B) diffusion in forward bias, drift in reverse bias. (C) diffusion in both forward and reverse bias. (D) drift in both forward and reverse bias. 526

Chapter 41. Semiconductors and Semiconductor Devices

527 p

Sol. The majority carriers in n-type semiconductors are electrons. Ans. C

n

p

p

− +

p

n

+ −

Circuit-2 n

p

− +

V

n

+ −

Circuit-1

Q 6. A full wave rectifier circuit along with the output is shown in the figure. The contribution(s) from the diode 1 is (are) (1996)

n

p

− +

Circuit-3

(A) (B) (C) (D)

Output

t

O

1 V

A B C D O

t

(A) C (B) A, C (C) B, D (D) A, B, C, D Sol. In the given circuit, diode 1 contributes to output when it is forward biased. This occurs when input voltage is negative. Thus, contribution from diode 1 are (B, D) or (A, C). V O

A

B

C

D

t

A

B

C

D

t

V O

If (A, C) comes from diode 1 then (B, D) comes from the other diode and vice versa. Ans. B, C Q 7. Read the following statements carefully, Y: The resistivity of semiconductor decreases with increase of temperature. Z: In a conducting solid, the rate of collisions between free electrons and ions increases with increase in temperature. Select the correct statement(s) from the following. (1993)

(A) (B) (C) (D)

n

+ −

Y is true but Z is false. Y is false but Z is true. Both Y and Z are true. Y is true and Z is the correct reason for Y.

Sol. The resistivity of a semiconductor decreases with increase in temperature because the population of charge carriers in conduction band increases. In conductors, the resistivity increases with increase in temperature because the rate of collision between the free electrons and ions increases. Ans. C Q 8. Two identical p-n junctions may be connected in series with a battery in three ways. The potential drops across the two p-n junction are equal in (1989)

circuit-1 circuit-2 circuit-3 circuit-1

and circuit-2 and circuit-3 and circuit-1 only

Sol. The potential across p-n junction reduces in forward bias (Vf ) and increases in reverse bias (Vr ), where Vr > Vf . In circuit 1, one of the p-n junction is forward biased and other is reversed biased. Thus, the potential drops across two p-n junctions are Vf and Vr . In circuit 2, both p-n junctions are forward biased that results in potential drop Vf across the two junctions. In circuit 3, both p-n junctions are reversed biased and hence potential drop is Vr in both the junctions. Ans. B Q 9. A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of (1988) (A) each of them increases. (B) each of them decreases. (C) copper increases and germanium decreases. (D) copper decreases and germanium increases. Sol. The resistance of the copper (metal) decreases and of germanium (semiconductor) increases with decrease in temperature. Ans. D One or More Option(s) Correct Q 10. When a potential difference is applied across, the current passing through (1999) (A) an insulator at 0 K is zero. (B) a semiconductor at 0 K is zero. (C) a metal at 0 K is finite. (D) a p-n diode at 300 K is finite, if it is reverse biased. Sol. At 0 K, a semiconductor becomes a perfect insulator but a metal becomes a superconductor. In reverse bias, a small current flows in p-n diode at 300 K due to movement of minority carriers. Ans. A, B, D Q 11. A transistor is used in common emitter mode as an amplifier, then (1998) (A) the base emitter junction is forward biased. (B) the base emitter junction is reversed biased. (C) the input signal is connected in series with the voltage applied to bias the base-emitter junction. (D) the input signal is connected in series with the voltage applied to bias the base-collector junction.

528

Part VI. Modern Physics

Sol. In common emitter mode as an amplifier, the baseemitter junction is forward biased and the input signal is connected in series with the voltage applied to bias the base-emitter junction. Ans. A, C

Q 16. For the given circuit shown in figure to act as full wave rectifier, the AC input should be connected across . . . . . . and the DC output would appear across ....... (1991) B

A

C

D

Q 12. Holes are charge carriers in (1996) (A) intrinsic semiconductors (B) ionic solids (C) p-type semiconductors (D) metals Sol. The holes are charge carriers in intrinsic semiconductors as well as p-type semiconductors. The number of holes and free electrons in intrinsic semiconductors are equal whereas number of holes is more than that of free electrons in p-type semiconductors. Ans. A, C Q 13. In an n-p-n transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, (1992) (A) the emitter current will be 9 mA. (B) the emitter current will be 11 mA. (C) the base current will be 1 mA. (D) the base current will be −1 mA. Sol. The currents in base, emitter and collector are related by IE = IB + IC .

(1) IE

IC

IB

Given IC = 10 mA and IC = 0.9IE (since 90% of electrons emitted by emitter reach the collector). Substitute these in equation (1) to get IE = IC /0.9 = 10/0.9 = 11 mA and IB = IE − IC = 1 mA. Ans. B, C Q 14. The impurity atoms with which pure silicon should be doped to make a p-type semiconductor are those of (1988) (A) phosphorus (B) boron (C) antimony (D) aluminium Sol. Pure silicon is doped with trivalent atoms to make p-type semiconductor. The atoms of boron and aluminum are trivalent. Ans. B, D Fill in the Blank Type Q 15. In a . . . . . . biased p-n junction, the net flow of holes is from the n region to the p region. (1993) Sol. In reverse bias, the net flow of holes is from the n region to the p region. Ans. reverse

Sol. In full wave rectifier, two diodes are in forward bias in positive cycle and other two diodes are in forward bias in negative cycle. B

D2

D1 C

A D3

D4

D

˜

The full wave rectification can be achieved by connecting AC input between B and D. The DC output appears across A and C. The diodes D2 and D4 conduct in positive cycle (VB > VD ) and D1 and D3 conduct in negative cycle. Ans. B and D, A and C Q 17. . . . . . . biasing of p-n junction offers high resistance to current flow across the junction. The biasing is obtained by connecting the p-side to the . . . . . . terminal of the battery. (1990) Sol. In reverse bias, p side is connected to the negative terminal of the battery and the junction offers very high resistance to current flow. Ans. reverse, negative Q 18. In the forward bias arrangement of p-n junction rectifier, the p end is connected to the . . . . . . terminal of the battery and the direction of the current is from . . . . . . to . . . . . . in the rectifier. (1988) Sol. In forward biasing, the p-end of the rectifier is connected to the positive terminal of the battery. i V

The current in rectifier flows from the p-side to the n-side as shown in the figure. Ans. positive, p-side, n-side

Chapter 42 The Nucleus

Sol. The binding energies of

One Option Correct

15 8O

are given by

= (8 × 1.008665 + 7 × 1.007825 − 15.000109) × 931.5 = 115.49 MeV, BEO = ∆mO c2 = (7mn + 8mp − MO )c2 = (7 × 1.008665 + 8 × 1.007825 − 15.003065) × 931.5

Sol. The activity of a radioactive sample at a time t is given by A = A0 e−λt , where A0 is the initial activity and λ is the decay rate. The decay rate is related to the half life of the sample by λ = 0.693/t1/2 = (ln 2)/t1/2 . Let A1 be the activity measured in the laboratory test at time t1 . Let the laboratory becomes safe at time t2 i.e., activity at time t2 is A2 = A1 /64. Apply the formula for activity to get

= 111.95 MeV. The difference in binding energies of

15 7N

and

15 8O

is

∆BE = BEN − BEO = 115.49 − 111.95 = 3.54 MeV. The electrostatic energies of

(1) 15 7N

and

15 8O

are given by

3 Z(Z − 1)e2 3 7(7 − 1)1.44 = 5 4π0 R 5 R 36.288 = MeV-fm R 48.384 3 8(8 − 1)1.44 = MeV-fm. EO = 5 R R

A2 = A0 e−λt2 .

EN =

Divide A1 by A2 to get A1 /A2 = 64 = eλ(t2 −t1 ) .

and

BEN = ∆mN c2 = (8mn + 7mp − MN )c2

Q 1. An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use? (2016) (A) 64 (B) 90 (C) 108 (D) 120

A1 = A0 e−λt1 ,

15 7N

(1)

Take logarithm on both sides of equation (1) and substitute λ = (ln 2)/t1/2 = (ln 2)/18 to get t2 − t1 = 6(18) = 108 days. Thus, the laboratory can be considered safe after 108 days of the test. Ans. C

The difference in electrostatic energies of is

15 7N

∆E = EO − EN = (12.096/R) MeV-fm.

and

15 8O

(2)

Since ∆E = ∆BE, equations (1) and (2) give R = 12.096/3.54 = 3.42 fm. Ans. C

Q 2. The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by

Q 3. A radioactive sample S1 having an activity of 5 µCi has twice the number of nuclei as another sample S2 which has an activity of 10 µCi. The half lives of S1 and S2 can be (2008) (A) 20 years and 5 years, respectively. (B) 20 years and 10 years, respectively. (C) 10 years each. (D) 5 years each.

3 Z(Z − 1)e2 . E= 5 4π0 R The measured masses of the neutron, 11 H, 157 N, 158 O are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065 u, respectively. Given that the radii of both the 157 N and 15 2 8 O nuclei are same, 1 u = 931.5 MeV/c (c is the speed 2 of light) and e /(4π0 ) = 1.44 MeV-fm. Assuming that the difference between the binding energies of 157 N and 15 8 O is purely due to the electrostatic energy, the radius of either of the nuclei is [ 1 fm = 10−15 m ] (2016) (A) 2.85 fm (B) 3.03 fm (C) 3.42 fm (D) 3.80 fm

Sol. The activity A of radioactive sample having a decay constant λ and number of nuclei N is given by A = |dN/dt| = λN = 0.693N/T, 529

530

Part VI. Modern Physics

where T is half life. Thus, A1 = 0.693N1 /T1 ,

(1)

A2 = 0.693N2 /T2 .

(2)

Divide equation (2) by (1) and substitute the values to get T1 /T2 = (A2 /A1 )×(N1 /N2 ) = (10/5)×(2/1) = 4. Ans. A Q 4. In the options given below, let E denotes the rest mass energy of a nucleus and n a neutron. The correct option is (2007)


137 97 (A) E 236 U > E I + E Y + 2E(n) 92
53
39
137 (B) E 236 E 97 92 U
< E 53 I + 39 Y +
2E(n)
140 94 (C) E 236 92 U
< E 56 Ba
+ E 36 Kr
+ 2E(n) 236 140 94 (D) E 92 U = E 56 Ba + E 36 Kr + 2E(n) Sol. The energies correspond to nuclear fission of 236 92 U. As energy is released in this fission reaction, rest mass energy of 236 92 U is more than the rest mass energy of fission products. Ans. A Q 5. Half-life of a radioactive substance A is 4 days. The probability that a nucleus will decay in two halflives is (2006) (A) 1/4 (B) 3/4 (C) 1/2 (D) 1 Sol. Let t denotes the time and T denotes the half life. The nucleus will decay in two half-lives if ‘it decays in 0 < t ≤ T ’ or ‘it decays in T < t ≤ 2T ’, both of which are mutually exclusive events. The decay in 0 < t ≤ T is an equally likely event with probability of its occurrence P1 = 1/2. The decay will occur in T < t ≤ 2T if it is not occurred in 0 < t ≤ T and occurred in T < t ≤ 2T giving its probability

Hence, the desired probability is

We encourage you to find P1 , P2 and P3 by evaluating the integrals Z T P1 = λe−λt dt, 0 2T

T Z 2T

P3 =

16 8 O.

The

= c2 (4 × 4.0026 − 15.9994)u = c2 0.011 u = c2 0.011 × 931.48 MeV/c2 = 10.24 MeV. Ans. C Q 7. After 280 days, the activity of a radioactive sample is 6000 disintegrations per second. The activity reduces to 3000 disintegration per second after another 140 days. The initial activity of the sample (in disintegration per second) is (2004) (A) 6000 (B) 9000 (C) 3000 (D) 24000 Sol. Let N0 be the initial number of atoms and λ be the decay constant. The activity at time t is given by A = |dN/dt| = λN = λN0 e−λt . Substitute given values to get 6000 = λN0 e−280λ , 3000 = λN0 e

−420λ

.

(1) (2)

Solve equations (1) and (2) to get the initial activity λN0 = 24000 disintegrations per second and T1/2 = 0.693/λ = 140 days. Ans. D Q 8. A nucleus with mass number 220 initially at rest emits an α-particle. If the Q value of reaction is 5.5 MeV, calculate the kinetic energy of the α-particle. (2003)

(B) 5.4 MeV (D) 6.5 MeV

Sol. The mass number of the products are m1 = 216 and m2 = 4. Let K1 and K2 be the kinetic energies of the two particles. The energy conservation gives

P3 = P1 + P2 = 1/2 + 1/4 = 3/4.

P2 =

Sol. The nuclear reaction in star is 4 42 He → Q-value of this nuclear reaction is   Q = 4 × m(42 He) − m(168 O) c2

(A) 4.4 MeV (C) 5.6 MeV

P2 = (1/2) × (1/2) = 1/4.

Z

Q 6. If a star converts all of its He into oxygen, find the amount of energy released per nucleus of oxygen. (Mass of the helium nucleus is 4.0026 u and mass of oxygen nucleus is 15.9994 u). (2005) (A) 7.6 MeV (B) 56.12 MeV (C) 10.24 MeV (D) 23.4 MeV

λe−λt dt,

Q = 5.5 MeV = K1 + K2 . The conservation of linear momentum gives p p 2m1 K1 = 2m2 K2 .

(1)

(2)

Eliminate K1 from equations (1) and (2) and substitute the values of m1 and m2 to get K2 = 5.4 MeV. Ans. B

λe−λt dt,

0

where λ = 0.693/T is the decay constant. Ans. B

Q 9. For uranium nucleus, its mass (m) vary with volume (V ) as (2003) √ (A) m ∝ V (B) m ∝ 1/V (C) m ∝ V (D) m ∝ V 2

Chapter 42. The Nucleus

531

Sol. The nuclear mass m is proportional to mass number A, m ∝ A.

V = 34 πR3 = 43 πR03 A.

(2)

The equations (1) and (2) give m ∝ V . Ans. A 215

Q 10. The half-life of At is 100 µs. The time taken 1 th of for the activity of a sample of 215 At to decay to 16 its initial value is (2002) (A) 400 µs (B) 63 µs (C) 40 µs (D) 300 µs Sol. The number of nuclei after time t is given by

where λ = ln(2)/T1/2 . Substitute the values to get N0 /16 = N0 e

−λt

.

(1)

Take logarithm of equation (1) and solve to get t = 4T1/2 = 400 µs. Note that population reduces to N0 /2n after time nT1/2 . Ans. A Q 11. Which of the following processes represent a γdecay? (2002) (A) A XZ + γ → A XZ-1 + a + b (B) A XZ + 1 n0 → A-3 XZ-2 + c (C) A XZ → A XZ + f (D) A XZ + e−1 → A XA-1 + g Sol. The γ rays are photons (mass-less, charge-less). The mass number A and the atomic number Z of decaying the nuclei do not change in γ-decay. Ans. C Q 12. A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life of one species is τ and that of the other is 5τ . The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figure best represents the form of this plot? (2001) N

N

(A)

(B) t

t

τ

N

N

(C)

(D) t

and

t

NB = N0 e− 5τ .

Thus, the total population after time t is  t  t N (t) = NA + NB = N0 e− τ + e− 5τ . N (t) is a monotonically decreasing function of t, for t > 0. Ans. D Q 13. The electron emitted in beta radiation originates from (2001) (A) inner orbits of atom. (B) free electrons existing in nuclei. (C) decay of a neutron in a nucleus. (D) photon escaping from the nucleus. Sol. In β-decay, a neutron in nucleus decays as, 10 n → 1 0 1 p + −1 e. Ans. C

N = N0 e−λt ,

τ

t

NA = N0 e − τ ,

(1)

The average radius of a nucleus is related to its mass number A by, R = R0 A1/3 , where R0 = 1.1 fm. Thus, volume of the nucleus is given by

τ

Sol. Let N0 be the initial population of both species. The populations of these species after time t become

τ

t

Q 14. Two radioactive materials X1 and X2 have decay constants 10λ and λ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time (2000)

(A) 1/(10λ) (B) 1/(11λ) (C) 11/(10λ) (D) 1/(9λ) Sol. The populations of X1 and X2 after time t are given by N1 = N0 e−10λt , N2 = N0 e

−λt

.

(1) (2)

Divide equation (1) by (2) and substitute the values to get N1 1 e−10λt = = −λt = e−9λt . N2 e e

(3)

Take logarithm of equation (3) and solve to get t = 1/(9λ). Ans. D Q 15. Which of the following is a correct statement? (1999)

(A) (B) (C) (D)

Beta rays are same as cathode rays. Gamma rays are high energy neutrons. Alpha particles are singly ionized helium atoms. Protons and neutrons have exactly the same mass.

Sol. The β particles are electrons emitted by the nucleus whenever a neutron is converted into a proton. The gamma rays are electromagnetic radiations. Alpha particle is a doubly ionized helium atom. The rest mass of a neutron is slightly greater than the rest mass of a proton. Ans. A

532

Part VI. Modern Physics

Q 16. 22 Ne nucleus, after absorbing energy, decays into two α-particles and an unknown nucleus. The unknown nucleus is (1999) (A) nitrogen (B) carbon (C) boron (D) oxygen Sol. Use conservations of atomic number and mass numbers to get Z = 6 and A = 14 for unknown nu14 4 cleus i.e., 22 10 Ne → 6 C + 2 2 He. Ans. B Q 17. Order of magnitude of density of uranium nucleus is [Given R0 ≈ 1.1 × 10−15 m and mp = 1.67 × 10−27 kg.] (1999) (A) 1020 kg/m3 (B) 1017 kg/m3 (C) 1014 kg/m3 (D) 1011 kg/m3

Since Breac > Bprod , energy is not released in this reaction. In reaction W → X + Z, Bprod = 8 × 90 + 5 × 30 = 870 MeV, Breac = 7.5 × 120 = 900 MeV. In reaction W → 2Y , Bprod = 2 × 8.5 × 60 = 1020 MeV, Breac = 7.5 × 120 = 900 MeV. In reaction X → Y + Z, Bprod = 8.5 × 60 + 5 × 30 = 660 MeV, Breac = 7.5 × 120 = 900 MeV.

Sol. Let A be the mass number of uranium. The nuclear mass and the nuclear radius are given by m = Amp , R = R0 A

1/3

−15

≈ 1.1 × 10

A

1/3

mp m = 4 3 = 4 3 3 πR 3 πR0

4 3

m.

1.67 × 10−27 × 3.14 × 1.331 × 10−45

≈ 3 × 1017 kg/m3 . Note that nuclear density is independent of A. Ans. B

Binding energy per nucleon in MeV

Q 18. Binding energy per nucleon versus mass number curve for nuclei is shown in the figure. W , X, Y and Z are four nuclei indicated on the curve. The process that would release energy is (1999)

(A) Y → 2Z (C) W → 2Y

Y

8.5 8.0 7.5 5.0

X W Z 0 30 60 90 120 Mass number of nuclei

(B) W → X + Z (D) X → Y + Z

Sol. The total binding energy of a nucleus, having binding energy per nucleon b and mass number A, is given by B = bA. The energy is released in nuclear reaction when binding energy of the products is more than the binding energy of the reactants. In reaction Y → 2Z, Bprod = 2(5 × 30) = 300 MeV, Breac = 8.5 × 60 = 510 MeV.

Q 19. The half life of 131 I is 8 days. Given a sample of I at time t = 0, we can assert that (1998) (A) no nucleus will decay before t = 4 days. (B) no nucleus will decay before t = 8 days. (C) all nuclei will decay before t = 16 days. (D) a given nucleus may decay at any time after t = 0. 131

The nuclear density density is ρ=

Ans. C

Sol. Nuclear decay is a probabilistic event. We can only give statistical properties of the population and not of an individual nucleus. Ans. D Q 20. Masses of two isobars 29 Cu64 and 30 Zn64 are 63.9298 u and 63.9292 u respectively. It can be concluded from these data that (1997) (A) both the isobars are stable. (B) Zn64 is radioactive, decaying to Cu64 through βdecay. (C) Cu64 is radioactive, decaying to Zn64 through γdecay. (D) Cu64 is radioactive, decaying to Zn64 through βdecay. Sol. The conservation of atomic number gives 64 → 30 Zn64 + −1 e 0 as a feasible decay process. 29 Cu Ans. D Q 21. Fast neutrons can easily be slowed down by (1994)

(A) (B) (C) (D)

the use of lead shielding. passing them through heavy water. elastic collisions with heavy nuclei. applying a strong electric field.

Sol. The neutron slows down by transferring its kinetic energy to colliding particle. Energy transfer is maximized when the masses of colliding particles are comparable. Thus, slow down is most effective when the neutron collides with deuteron present in heavy water (D2 O). Ans. B

Chapter 42. The Nucleus

533

Q 22. Consider α-particles, β-particles and γ-rays each having an energy of 0.5 MeV. In increasing order of penetrating powers, the radiations are (1994) (A) α, β, γ (B) α, γ, β (C) β, γ, α (D) γ, β, α Sol.p The penetration power largely depends on velocity v = 2K/m. Since mα  mβ , we get vα  vβ . The γ rays are electromagnetic waves with vγ ≈ c. Thus, the increasing order of penetration powers is α, β, γ. We encourage you to compare the ionization powers of α, β, and γ. Ans. A Q 23. A star initially has 1040 deuterons. It produces energy via the processes 1H2 + 1H2 → 1H3 + p and 2 3 4 1H + 1H → 2He + n. If the average power radiated by 16 the star is 10 W, the deuteron supply of the star is exhausted in a time of the order of [The nuclei masses are: m(H2 ) = 2.014 u; m(n) = 1.008 u; m(p) = 1.007 u; m(He4 ) = 4.001 u.] (1993) (A) 106 s (B) 108 s (C) 1012 s (D) 1016 s Sol. Given reactions can be written as 31 H2 → 2 He4 + p + n. The Q-value of this reaction is Q = [3m(1 H2 ) − m(2 He4 ) − m(p) − m(n)]c2 = [3(2.014) − 4.001 − 1.008 − 1.007]c2 = 0.026c2 = 0.026 × 931 MeV = 3.87 × 10−12 J. Thus, the energy released by consumption of three deuteron is 3.87 × 10−12 J. Hence, the total energy released by consumption of 1040 deuteron is E=

3.87 × 10−12 × 1040 = 1.29 × 1028 J. 3

At the rate of radiated power P = 1016 W, the star will continue to give energy for a time E 1.29 × 1028 t= = = 1.29 × 1012 s. P 1016 Ans. C Q 24. The decay constant of a radioactive sample is λ. The half-life and mean-life of the sample are respectively given by (1989) (A) 1/λ and (ln 2)/λ (B) (ln 2)/λ and 1/λ (C) λ(ln 2) and 1/λ (D) λ/(ln 2) and 1/λ Sol. The half life and average life of a radioactive sample with decay constant λ are given by, T1/2 = ln(2)/λ = 0.693/λ and τ = 1/λ. Ans. B Q 25. A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is (1988)

(A) 6 h (B) 12 h (C) 24 h (D) 128 h

Sol. Let λ be the decay constant and N0 be the initial population of radioactive nuclei. The activity at a time t is given by A = |dN /dt| = λN = λN0 e−λt .

(1)

From equation (1), initial activity (t = 0) is A0 = λN0 .

(2)

The activity becomes A = A0 /64 after time t and is given by A0 /64 = λN0 e−λt .

(3)

Divide equation (2) by (3) and then take logarithm to get t = ln(64)/λ = 6 ln(2)/λ = 6(0.693/λ) = 6(T1/2 ) = 6(2) = 12 h. Note that the activity decreases to A0 /2n in n half-lives i.e., after time nT1/2 . Ans. B Q 26. During a nuclear fusion reaction, (1987) (A) a heavy nucleus breaks into two fragments by itself. (B) a light nucleus bombarded by thermal neutrons breaks up. (C) a heavy nucleus bombarded by thermal neutrons breaks up. (D) two light nuclei combine to give a heavier nucleus and possibly other products. Sol. The nuclear fusion involves the combination of two light nuclei to form a heavier nucleus. Ans. D Q 27. During negative beta decay, (1987) (A) an atomic electron is ejected. (B) an electron which is already present within the nucleus is ejected. (C) a neutron in the nucleus decays emitting an electron. (D) a part of the binding energy of the nucleus is converted into an electron. Sol. The β decay is a nuclear process given by n → p + e. In this process, a neutron in nucleus decays to a proton by emitting an electron. Ans. C Q 28. The equation 4 11 H → 42 He2+ + 2e− + 26 MeV, represents (1983) (A) β-decay (B) γ-decay (C) fusion (D) fission Sol. The equation 4 11 H → 42 He2+ + 2e− + 26 MeV is an example of nuclear fusion reaction as four hydrogen nuclei are fusing to make a helium nucleus. Ans. C

534

Part VI. Modern Physics

Q 29. Beta rays emitted by a radioactive material are (1983)

(A) (B) (C) (D)

electromagnetic radiation. the electron orbiting around the nucleus. charged particles emitted by the nucleus. neutral particle.

Sol. Beta rays are the electrons emitted by the nucleus 0 when a neutron is converted to a proton 10 n → 11 p+ −1 e. Ans. C Q 30. The half-life of the radioactive Radon is 3.8 days. The time, at the end of which 1/20 th of the Radon sample will remain undecayed, is [Given log10 e = 0.4343.] (1981)

(A) 3.8 days (B) 16.5 days (C) 33 days (D) 76 days Sol. The population of a radioactive sample at time t is given by N = N0 e−λt ,

(1)

where N0 is the initial population and decay constant λ is related to the half-life by λ = 0.693/T1/2 . Substitute N = N0 /20 and λ = 0.693/T1/2 in equation (1) to get N0 /20 = N0 e

−0.693t/T1/2

.

(2)

Take logarithm of equation (2) and simplify to get T1/2 log10 20 T1/2 ln 20 = 0.693 0.693 log10 e T1/2 (1 + log10 2) 3.8 (1 + 0.3010) = = 0.693 log10 e 0.693 0.4343 = 16.43 days.

t=

Substitute Nα = 5 to get Nβ = 2. The nuclear reactions follow few more conservation laws. The β decay reveals many of them. We strongly recommend detailed study of β decay (n → p + e + ν¯). Ans. (A), (C) 140 Q 32. A fission reaction is given by 236 92 U → 54 Xe + 94 Sr + x + y, where x and y are two particles. Con38 sidering 236 U to be at rest, the kinetic energies of the 92 products are denoted by KXe , KSr , Kx (2 MeV) and Ky (2 MeV), respectively. Let the binding energies per 140 94 nucleon of 236 92 U, 54 Xe, and 38 Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s) is (are) (2015) (A) x = n, y = n, KSr = 129 MeV, KXe = 86 MeV (B) x = p, y = e- , KSr = 129 MeV, KXe = 86 MeV (C) x = p, y = n, KSr = 129 MeV, KXe = 86 MeV (D) x = n, y = n, KSr = 86 MeV, KXe = 129 MeV

Sol. Given fission reaction is 236 92 U

A

Ax y 94 → 140 54 Xe + 38 Sr + Zx x + Zy y.

Apply conservations of charge and mass to get 92 = 54 + 38 + Zx + Zy

(1)

236 = 140 + 94 + Ax + Ay .

(2)

The equations (1) and (2) give Zx + Zy = 0 and Ax + Ay = 2. From the given options, these conditions are Ay 1 1 x satisfied if A Zx x = 0 n and Zy y = 0 n. The Q value of the given reaction is Q = BEproducts − BEreactants = (140 × 8.5 + 94 × 8.5) − 236 × 7.5 = 219 MeV.

Ans. B One or More Option(s) Correct Q 31. In a radioactive decay chain, 232 90 Th nucleus decays to 212 82 Pb nucleus. Let Nα and Nβ be the number of α and β − particles, respectively, emitted in the decay process. Which of the following statements is (are) true? (2018) (A) Nα = 5 (B) Nα = 6 (C) Nβ = 2 (D) Nβ = 4 Sol. The α particle (42 He) has +2e charge and four nucleons (two protons and two neutrons). The β − particle (−10 β) has −e charge and zero nucleons. Given radioactive decay chain is represented by 232 90 Th

4 → 212 82 Pb + Nα 2 He + Nβ

0 −1 β.

By the conservation of ‘nucleons number’ 232 = 212 + 4Nα , which gives Nα = 5. By the conservation of charge 90 = 82 + 2Nα − Nβ .

The energy released in the reaction (Q value) is equal to the kinetic energy of the products i.e., KXe + KSr + Kx + Ky = Q, which gives KXe + KSr = 219 − (2 + 2) = 215 MeV.

(3)

(∵ Kx = Ky = 2 MeV). The linear momentum of a particle of mass m and ki√ netic energy K is given by p = 2mK. Since the masses and kinetic energies of x and y are very small in com94 parison to that of 140 54 Xe and 38 Sr, we can neglect the linear momentum of these particles. Initially, the linear momentum of 236 92 U is zero (at rest). Finally, the 94 products 140 54 Xe and 38 Sr will move in opposite direction with equal linear momentum (by conservation of linear momentum). Thus, p p 2MXe KXe = 2MSr KSr , i.e., 140KXe = 94KSr .

(4)

Solve equations (3) and (4) to get KXe = 86 MeV and KSr = 129 MeV. Ans. A

Chapter 42. The Nucleus

535

Q 33. Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice(s) given below. (2008)

dN dt

y x t

t0 B/A 8 6 4 2 0

The decay rates of x and y are dNx /dt = −λx Nx = −λx N0 e−λx t , 100 200

A

dNy /dt = −λy Ny = −λy N0 e

−λy t

.

(2) (3)

Divide equation (2) by (3) to get (A) Fusion of two nuclei with mass number lying in the range of 1 < A < 50 will release energy. (B) Fusion of two nuclei with mass number lying in the range of 51 < A < 100 will release energy. (C) Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments. (D) Fission of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken into two equal fragments. Sol. If B/A of the products is more than the reactants then energy is released in the reaction. Fusion of two nuclei with mass number in 1 < A < 50 produces a nuclei with mass number in 1 < A < 100. The B/A of products and reactants are equal, hence no energy is released. Fusion of two nuclei with mass number in 51 < A < 100 produces a nuclei with mass number in 100 < A < 200. The B/A of products is more than reactants, hence the energy is released. The fission of nucleus with mass number 100 < A < 200 produces the products with mass numbers lying in 1 < A < 200. The B/A of products is less than or equal to reactants, hence no energy is released. The fission of nucleus with mass number 200 < A < 260 produces the products with mass numbers lying in 100 < A < 200. The B/A of products is more than the reactants, hence energy is released. Ans. B, D Q 34. The half-life period of a radioactive element x is same as the mean life time of another radioactive element y. Initially both of them have the same number of atoms. Then, (1999) (A) x and y have the same decay rate initially. (B) x and y decay at the same rate always. (C) y will decay at a faster rate than x. (D) x will decay at a faster rate than y. Sol. The half life is T1/2 = 0.693/λ and mean life is τ = 1/λ. The condition T1/2,x = τy gives λx = 0.693λy .

(1)

dNx /dt λx −(λx −λy )t = e = 0.693e0.307λy t = 0.693, dNy /dt λy where we have used equation (1) and used t = 0 to compare the initial rates. However, the decay rate of x becomes greater than that of y at certain time t0 (see figure). We encourage you to show that t0 = 1.2/λy . Ans. C Q 35. Let mp be the mass of proton, mn the mass of neutron, M1 the mass of 20 10 Ne nucleus and M2 the mass of 40 Ca nucleus. Then, (1998) 20 (A) M2 = 2M1 (B) M2 > 2M1 (C) M2 < 2M1 (D) M1 < 10(mn + mp ) Sol. The mass of a nucleus is less than rest masses of its constituents. This is known as mass defect and is equal to the binding energy of the nucleus. The mass defects in two cases are ∆m1 = 10(mp + mn ) − M1 , ∆m2 = 20(mp + mn ) − M2 . From the binding energy diagram, binding energy per 20 nucleon of 40 20 Ca is more than that of 10 Ne. Thus, ∆m2 /40 > ∆m1 /20. Substitute ∆m1 and ∆m2 to get M2 < 2M1 . Ans. C, D Q 36. The pairs of physical quantities that have the same dimensions is (are) (1995) (A) Reynolds number and coefficient of friction. (B) Curie and frequency of a light wave. (C) Latent heat and gravitational potential. (D) Planck’s constant and torque. Sol. Reynolds number and coefficient of friction are dimensionless. The dimensions of Curie and frequency are [T−1 ]. Latent heat and gravitational potential have same dimensions as energy per unit mass i.e., [L2 T−2 ]. The dimensions of Planck’s constant are same as those of energy-second but dimensions of torque are same as those of energy. Ans. A, B, C

536

Part VI. Modern Physics

Q 37. Which of the following statement(s) is (are) correct? (1994) (A) The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons. (B) The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons. (C) In nuclear fission, energy is released by fusing two nuclei of medium mass (approximately 100 u). (D) In nuclear fission, energy is released by fragmentation of a very heavy nucleus. Sol. For a stable nucleus, binding energy is positive. The rest mass of nucleus is less than the sum of the rest masses of its constituent nucleons. Ans. A, D Q 38. The mass number of a nucleus is (1986) (A) always less than its atomic number. (B) always more that its atomic number. (C) sometimes equal to its atomic number. (D) sometime more than and sometimes equal to its atomic number. Sol. The atomic number is equal to the number of protons in the nucleus. The mass number is the sum of the number of protons and neutrons in the nucleus. If nucleus does not have any neutron (e.g., hydrogen nucleus) then the mass number is equal to the atomic number, otherwise, it is more than the atomic number. Ans. C, D Q 39. From the following equations pick out the possible nuclear fusion reactions. (1984) (A) 6 C13 + 1 H1 → 6 C14 + 4.3 MeV (B) 6 C12 + 1 H1 → 7 N13 + 2 MeV (C) 7 N14 + 1 H1 → 8 O15 + 7.3 MeV (D) 92 U235 + 0 n1 → 54 Xe140 + 36 Sr94 + 2 0 n1 + γ + 200 MeV Sol. The reactions 6 C12 + 1 H1 → 7 N13 + 2 MeV and 14 + 1 H1 → 8 O15 + 7.3 MeV represent the nuclear 7N fusion. Ans. B, C Paragraph Type Paragraph for Questions 40-41 If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z = x/y. If the errors in x, y and z are ∆x, ∆y and ∆z, respectively, then z ± ∆z =

x x ± ∆x = y ± ∆y y

 1±

∆x x

 −1 ∆y 1± . y

 The series expansion for 1 ±

∆y y

−1

, to first power in

∆y/y, is 1 ∓ (∆y/y). The relative errors in independent variables are always added. So the error in z will be   ∆x ∆y ∆z = z + . x y The above derivation makes the assumption that ∆x/x  1, ∆y/y  1. Therefore, the higher powers of these quantities are neglected. (2018) Q 40. Consider the ratio r = 1−a 1+a to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is ∆a (∆a/a  1), then what is the error ∆r in determining r? 2∆a 2∆a 2a∆a ∆a (A) (1+a) 2 (B) (1+a)2 (C) (1−a2 ) (D) (1−a2 ) Sol. Let ∆r be the error in dependent quantity r due to the error ∆a in independent quantity a. Thus, r ± ∆r =

1 − (a ± ∆a) . 1 + (a ± ∆a)

Substitute the expression for r and take the positive sign for ± to get 1 − (a + ∆a) 1 − a − 1 + (a + ∆a) 1 + a (1 − a − ∆a)(1 + a) − (1 − a)(1 + a + ∆a) = (1 + a)(1 + a + ∆a)  −1 −2∆a ∆a = 1 + (1 + a)2 1+a   −2∆a −2∆a ∆a ≈ , 1 − ≈ 2 (1 + a) 1+a (1 + a)2

∆r =

where we have used series expansions and retained the terms up to first power in ∆a. Can you get the results by taking negative sign for ±? Aliter: The method described in the paragraph is essentially differentiation technique we use for error −2∆a analysis. Differentiate r = 1−a 1+a to get ∆r = (1+a)2 . Ans. (B) Q 41. In an experiment the initial number of radioactive nuclei is 3000. It is found that 1000 ± 40 nuclei decayed in first 1.0 s. For |x|  1, ln(1 + x) = x up to first power in x. The error ∆λ, in the determination of the decay constant λ, in s−1 , is (A) 0.04 (B) 0.03 (C) 0.02 (D) 0.01 Sol. The law of radioactive decay gives number of undecayed nuclei N at time t as N = N0 e−λt , where N0 is number of nuclei at t = 0 and λ is the decay constant. The number of nuclei decayed till time t is given by Nd = N0 − N . Thus, the decay constant is given by λ=

1 N0 1 N0 ln = ln . t N t N0 − Nd

Chapter 42. The Nucleus

537

Since N0 = 3000 and t = 1 s are given without any error, we assume that error in λ is due to error in Nd only. Hence, we can write above equation as 1 N0 ln t N0 − (Nd + ∆Nd ) N0 1 = ln t (N0 − Nd )(1 − ∆Nd /(N0 − Nd ))   1 N0 1 ∆Nd = ln − ln 1 − t N0 − Nd t N0 − Nd 1 ∆Nd ≈λ+ , t N0 − Nd

λ + ∆λ =

d which gives ∆λ = t(N∆N . Substitute ∆Nd = 40, 0 −Nd ) t = 1 s, and Nd = 1000 to get ∆λ = 0.02. We encourage you to solve this problem by differentiation also. Ans. (C)

206 4 Sol. In the reaction, 210 84 Po → 82 Pb + 2 He, the Q-value is given by 


 2 206 4 Q = m 210 84 Po − m 82 Pb − m 2 He c

= [209.982876 − 205.974455 − 4.002603] u c2 = 0.005828 × 932 = 5.422 MeV = 5422 keV This energy is in the form of kinetic energy of and 42 He i.e., EHe + EPb = 5422 keV.

206 82 Pb

(1)

Since 210 84 Po was at rest, the conservation of linear momentum gives pHe = pPb .

(2)

The kinetic energy and linear momentum are related by E = p2 /(2m). Use equation (2) to get 2mHe EHe = 2mPb EPb .

(3)

Paragraph for Questions 42-43 The mass of a nucleus A Z X is less than the sum of the masses of (A−Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if (m1 + m2 ) < M . Also two light nuclei of masses m3 and m4 can undergo complete fusion and form a heavy nucleus of mass M 0 only if (m3 +m4 ) > M 0 . The masses of some neutral atoms are given in table below: (2013) 1 1H 3 1H 6 3 Li 70 30 Zn 152 64 Gd 209 83 Bi

: 1.007825 u , : 3.016050 u , : 6.015123 u , : 69.925325 u , : 151.919803 u , : 208.980388 u ,

2 1H 4 2 He 7 3 Li 82 34 Se 206 82 Pb 210 84 Po

: 2.014102 u : 4.002603 u : 7.016004 u : 81.916709 u : 205.974455 u : 209.982876 u

[Given 1 u = 932 MeV/c2 .] Q 42. The correct statement is (A) The nucleus 63 Li can emit an alpha particle. (B) The nucleus 210 84 Po can emit a proton. (C) Deuteron and alpha particle can undergo complete fusion. 82 (D) The nuclei 70 30 Zn and 34 Se can undergo complete fusion. Sol. From the given data and feasibility conditions, reaction 21 H + 42 He → 63 Li is possible but reactions 63 Li → 2 4 210 209 1 70 82 152 1 H+ 2 He, 84 Po → 83 Bi+ 1 H, and 30 Zn+ 34 Se → 64 Gd are not possible. Ans. C Q 43. The kinetic energy (in keV) of the alpha particle, when the nucleus 210 84 Po at rest undergoes alpha decay, is (A) 5319 (B) 5422 (C) 5707 (D) 5818

Substitute mHe = 4 and mPb = 206 and solve equations (1) and (3) to get EHe = 5319 keV. Ans. A Paragraph for Questions 44-45 The β-decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e− ) are observed as the decay product of the neutron. Therefore, considering the decay of a neutron as a two body decay process, it was predicted theoretically that the kinetic energy of the electron should be constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process i.e., n → p + e− + ν¯e , around 1930, Pauli explained the observed electron energy spectrum. Assuming the antineutrino (¯ νe ) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 × 106 eV. The kinetic energy carried by the proton is only the recoil energy. (2012) Q 44. What is the maximum energy of the antineutrino? (A) Zero. (B) Much less than 0.8 × 106 eV. (C) Nearly 0.8 × 106 eV. (D) Much larger than 0.8 × 106 eV. Sol. The energy released in β-decay is given by Q = [mn − (mp + me + mν¯e )] c2 = [mn − (mp + me )] c2

(∵ mν¯e = 0).

This energy is in the form of kinetic energy of the products i.e., Q = Kp + Ke + Kν¯e ≈ Ke + Kν¯e

(∵ Kp ≈ 0).

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Part VI. Modern Physics Before

Q 46. In the core of nuclear fusion reactor, the gas becomes plasma because of (A) strong nuclear force acting between the deuterons. (B) Coulomb force acting between deuterons. (C) Coulomb force acting between deuteron-electron pairs. (D) the high temperature maintained inside the reactor core.

After e−

n •

p • ν¯e

The conservation of linear momentum gives p~p + p~e + p~ν¯e = ~0. The kinetic energy Ke is maximum when Kν¯e = 0 (and hence pν¯e = 0). This is possible in a three body decay without violating the momentum conservation. Thus, Q = Kemax = 0.8 × 106 eV. Similarly, Kν¯e is maximum when Ke = 0 i.e., Kν¯max = Q = 0.8 × 106 eV. e Ans. C Q 45. If the anti-neutrino had a mass of 3 eV/c2 (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron? (A) 0 ≤ K ≤ 0.8 × 106 eV (B) 3.0 eV ≤ K ≤ 0.8 × 106 eV (C) 3.0 eV ≤ K < 0.8 × 106 eV (D) 0 ≤ K < 0.8 × 106 eV Sol. If mν¯e = 3 eV/c2 , then, Q = 0.8 × 106 eV − 3 eV < 0.8 × 106 eV. = Q < 0.8 × 106 eV. The kinetic energy Thus, Kν¯max e min Kν¯e is related to pmin which can be zero in a three ν ¯e body decay making Kν¯min = 0. e Ans. D Paragraph for Questions 46-48

Sol. The gas is ionized at high temperature. Ans. D Q 47. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5kT , when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10−15 m is in the range (A) 1.0 × 109 K < T < 2.0 × 109 K (B) 2.0 × 109 K < T < 3.0 × 109 K (C) 3.0 × 109 K < T < 4.0 × 109 K (D) 4.0 × 109 K < T < 5.0 × 109 K Sol. Initially, kinetic energy of each nuclei is Ki = 1.5kT and Coulomb potential energy is Ui = 0. At the closest distance r, the nuclei comes to rest momentarily, making their kinetic energy Kf = 0 and Coulomb potential energy Uf = e2 /(4π0 r). Using the conservation of energy, Ui + Ki = Uf + Kf , we get 1.5kT + 1.5kT + 0 = 0 + 0 + e2 /(4π0 r).

Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, 21 H, known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D reaction is 2 2 3 1 H + 1 H → 2 He + n + energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of 21 H nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t0 before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 1014 g/cm3 . [Given Boltzmann cone2 stant k = 8.6 × 10−5 eV/K, 4π = 1.44 × 10−9 eV m.] 0 (2009)

Substitute the values to get T = 1.4 × 109 K. Ans. A Q 48. Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion? (A) deuteron density=2.0 × 1012 cm−3 , confinement time=5.0 × 10−3 s (B) deuteron density=8.0 × 1014 cm−3 , confinement time=9.0 × 10−1 s (C) deuteron density=4.0 × 1023 cm−3 , confinement time=1.0 × 10−11 s (D) deuteron density=1.0 × 1024 cm−3 , confinement time=4.0 × 10−12 s Sol. Lawson number nt0 for four different designs are 1.0 × 1010 , 7.2 × 1014 , 4.0 × 1012 and 4.0 × 1012 . Ans. B

Chapter 42. The Nucleus

539

Matrix or Matching Type Q 49. Match the nuclear processes given in Column I with the appropriate option(s) in Column II. (2015) Column I (A) Nuclear fusion (B) Fission in a nuclear reaction (C) β-decay

(D) γ-ray emission

Column II (p) Absorption of thermal neutron by 235 92 U (q) 60 Co nucleus 27 (r) Energy production in stars via hydrogen conversion to helium (s) Heavy water (t) Neutrino emission

Sol. The nuclear fusion is responsible for energy production in stars via fusion of hydrogen nuclei into helium nuclei. In sun, the fusion takes place dominantly by proton-proton cycle, 4 11 H → 42 He + 2e+ + 2ν + 2γ. The neutrino (ν) and γ-rays emissions are parts of this fusion reaction. The uranium based fission reactors involve absorption of thermal neutrons by 235 92 U nuclei to produce the highly fissionable 236 92 U nuclei. This nuclei then fissions 137 97 into two parts e.g., 236 92 U → 63 I + 39 Y + 2n. The fission fragments are unstable and undergo β-decay to reduce their neutron to proton ratio. The fragments are generally formed in excited states and consequently emit γ-rays. The heavy water (D2 O) is used as a moderator to slow down the fast moving neutrons. In β-decay, a neutron is converted into a proton. In this process, an electron and an antineutrino are created and emitted from the nucleus, n → p + e− + ν¯. The 60 60 − β-decay in 60 ¯. 27 Co is given by 27 Co → 28 Ni + e + ν 60 The daughter nuclei 28 Ni is formed in excited state and comes to ground state by γ-ray emission. The γ-rays are high energy electromagnetic rays. These rays are generally emitted when a nuclei in excited state (high energy) makes a transition to a lower state (low energy). Ans. A7→(r, t), B7→(p, s), C7→(q, t), D7→(r) Q 50. Match Column I of the nuclear processes with Column II containing parent nucleus and one of the end products of each process, (2013) Column I (P) (Q) (R) (S)

Alpha decay β + decay Fission Proton Emission

Column II (1) (2) (3) (4)

15 15 8O → 7N + . . . 238 234 U → 92 90 Th + . . . 185 184 Bi → 83 82 Pb + . . . 239 140 Pu → 94 57 La + . . .

Sol. Use the conservations of charge (atomic number) and mass numbers. The β + have positive charge of one

unit. In fission, parent nuclei breaks into two almost equal fragments. Ans. P7→2, Q7→1, R7→4, S7→3 Q 51. Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system. Match the statements in Column I to the appropriate process(es) from Column II. (2009) Column I (A) The energy of the system is increased. (B) Mechanical energy is provided to the system, which is converted into energy of random motion of its parts. (C) Internal energy of the system is converted into its mechanical energy. (D) Mass of the system is decreased.

Column II (p) System: A capacitor, initially uncharged. Process: It is connected to a battery. (q) System: A gas in an adiabatic container fitted with an adiabatic piston. Process: The gas is compressed by pushing the piston.

(r) System: A gas in a rigid container. Process: The gas gets cooled due to colder atmosphere surrounding it (s) System: A heavy nucleus, initially at rest. Process: The nucleus fissions into two fragments of nearly equal masses and some neutrons are emitted. (t) System: A resistive wire loop. Process: The loop is placed in a time varying magnetic field perpendicular to its plane.

Sol. In (p), the capacitor of capacitance C gets charged to charge q increasing its energy by q 2 /(2C). In (q), the work is done on the gas by adiabatic compression (∆Q = 0). The work done by the gas is ∆W < 0. The first law of thermodynamics, ∆Q = ∆U + ∆W , gives ∆U = −∆W > 0 i.e., internal energy of the system is increased. In this process, mechanical energy (work done) is converted into energy of random motion of gas molecules (internal energy). In (r), the energy of the system is decreased through heat loss. There is no mechanical work done.

540

Part VI. Modern Physics

In (s), the internal energy (nuclear energy) of the nucleus is converted into the kinetic energy (mechanical energy) of fission products. In fission, mass is converted to energy leading to decrease of the system mass. In (t), the time varying magnetic field induces emf that produces current in the wire loop. The energy of the wire loop is increased which appears as joule heating of the wire. There is no mechanical work done. Ans. A7→(p,q,t), B7→q, C7→s, D7→s Q 52. Some laws/processes are given in Coulumn I. Match these with the physical phenomenon given in Column II. (2006) Column I

Column II

(A) Nuclear fusion (B) Nuclear fission (C) β-decay (D) Exothermic nuclear reaction

(p) Converts some matter into energy. (q) Generally possible for nuclei with low atomic number. (r) Generally possible for nuclei with higher atomic number. (s) Essentially proceeds by weak nuclear forces.

Sol. The nuclear fusion, fission, and β-decay convert some matter into energy. Nuclear fusion is generally possible for nuclei with low atomic number and nuclear fission for nuclei with high atomic number. The β-decay proceeds by weak forces. Ans. A7→(p,q), B7→(p,r), C7→(p,s), D7→(p,q,r) Q 53. Four physical quantities are listed in Column I. Their values are listed in Column II in a random order. The correct matching of Column I and Column II is given by, (1987) Column I

Column II

(A) Thermal energy of air molecules at room temp. (B) Binding energy of heavy nuclei per nucleon (C) X-ray photon energy (D) Photon energy of visible light

(p) 0.02 eV (q) 2 eV (r) 10 keV (s) 7 MeV

Sol. The thermal energy of air molecule at room temperature is of the order of kT = (1.38 × 10−23 )(298) J = 0.025 eV. The wavelength of X-rays is of the order of 1 ˚ A which corresponds to energy hc/λ = 12420 eV ≈ 10 keV. Binding energy per nucleon of heavy nuclei is of the order of 7 MeV. The energy of photons in the middle of visible spectrum (yellow light with λ = 5500 ˚ A) is hc/λ = 12420/5500 = 2.25 eV.

We encourage you to get familiar with the order of magnitude of these energies. Ans. A7→p, B7→s, C7→r, D7→q True False Type Q 54. The order of magnitude of the density of nuclear matter is 104 kg/m3 . (1989) Sol. Consider a nucleus of mass number A. The masses of the proton and the neutron are approximately equal to m = 1.67 × 10−27 kg. Thus, the mass of the nuclues is M = mA.

(1)

The average radius of the nuclues is given by R = R0 A1/3 ,

where R0 = 1.1 × 10−15 m.

(2)

From equations (1) and (2), the nuclear density is given by ρ= =

M 4 3 3 πR

=

mA

1.67 × 10 4 3 (3.14)

m

= 4 3 4 3 3 πR0 A 3 πR0 −27

1.1 × 10−15

17 3
3 = 3 × 10 kg/m .

Thus, the order of magnitude of the nuclear density is 1017 kg/m3 . Note that the nuclear density is independent of the mass number. Ans. F Fill in the Blank Type Q 55. Consider the reaction: 21 H+ 21 H = 42 He+Q. Mass of the deuterium atom = 2.0141 u and mass of helium atom = 4.0024 u. This is a nuclear . . . . . . reaction in which the energy Q released is . . . . . . MeV. (1996) Sol. Given reaction is a fusion reaction with Q-value   Q = 2m(21 H) − m(42 He) c2 = (2 × 2.014 − 4.0024)931 ≈ 24 MeV. Ans. fusion, 24 Q 56. In the nuclear process, 6 C11 → 5 B11 + β + + X, X stands for . . . . . . . (1992) Sol. The conservations of charge and mass number give X to be chargeless and massless particle. The β-decay contains neutrino as one of the product. Ans. neutrino Q 57.
The binding energies per nucleon for deuteron
2 and helium 2 He4 are 1.1 MeV and 7.0 MeV re1H spectively. The energies released
when two deuterons fuse to form helium nucleus 2 He4 is . . . . . . . (1988)

Chapter 42. The Nucleus

541

Sol. The fusion reaction is 2 1 H2 → 2 He4 . The number of nucleons in 1 H2 and 2 He4 are two and four, respectively. The binding energy of the reactants is BEreac = 2(2×1.1) = 4.4 MeV and that of the products is BEprod = 4 × 7.0 = 28.0 MeV. The energy released in the reaction is ∆E = BEprod − BEreac = 28.0 − 4.4 = 23.6 MeV.

t = 1 s, and t = 3 s in equation (1) to get the decay rates at these times as |dN/dt|t=0 = 103 = λN0 , |dN/dt|t=1 = λN0 e−λ = (103 )(e−0.693 ) = 103 /21 = 500 dps, |dN/dt|t=3 = λN0 e−3λ = (103 )(e−0.693 )3 = 103 /23 = 125 dps.

Ans. 23.6 MeV
Q 58. When boron nucleus 105 B is bombarded by neutrons, α-particles are emitted. The resulting nucleus is of the element . . . . . . and has the mass number . . . . . . . (1986)

Sol. The conservation of atomic number (charge) and mass number give the nuclear reaction as 105 B + 10 n → 7 4 3 Li + 2 He. Ans. Li, 7 Q 59. Atoms having the same . . . . . . but different . . . . . . are called isotopes. (1986) Sol. The isotopes of an element have the same atomic number but different mass number. Ans. Atomic Number, Mass Number Q 60. In the uranium radioactive series the initial nu206 cleus is 238 92 U and the final nucleus is 82 Pb. When the uranium nucleus decays to lead, the number of αparticles emitted is . . . . . . and the number of β-particles emitted is . . . . . . (1985) Sol. The decay process can be written as 238 92 U

→ 206 82 Pb + nα

4 2 He

+ nβ

0 −1 e,

where nα is the number of α-particles and nβ is the number of β-particles decayed in the process. The conservation of mass number and charge gives 238 = 206 + 4nα ,

(1)

92 = 82 + 2nα − nβ .

(2)

Solve equations (1) and (2) to get nα = 8 and nβ = 6. Ans. eight, six Q 61. The radioactive decay rate of a radioactive element is found to be 103 disintegration per second at a certain time. If the half-life of the element is one second, the decay rate after one second is . . . . . . and after three seconds is . . . . . . (1983) Sol. The radioactive decay rate at a time t is given by |dN/dt| = λN = λN0 e−λt ,

(1)

where N0 is the initial population and λ = 0.693/t1/2 = 0.693 s−1 is the decay constant. Substitute t = 0 s,

Note that the decay rate after n half-lives (i.e., after time t = nt1/2 ) is 21n times the initial decay rate. Ans. 500 dps, 125 dps Integer Type Q 62. 131 I is an isotope of Iodine that β decays to an isotope of Xenon with a half-life of 8 days. A small amount of a serum labelled with 131 I is injected into the blood of a person. The activity of the amount of 131 I injected was 2.4 × 105 Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After 11.5 hours, 2.5 ml of blood is drawn from the person’s body, and it gives an activity of 115 Bq. The total volume of blood in the person’s body, in litres, is approximately . . . . . . . [You may use ex ≈ 1 + x for |x|  1 and ln 2 ≈ 0.7]. (2017)

Sol. The decay rate is related to half life by λ = tln1/22 . A sample having N0 radioactive nuclei at time t = 0 will have N = N0 e−λt radioactive nuclei at time t. Thus, activity (A = λN ) of a sample reduces from its initial value A0 to a value A = A0 e−λt at time t. Let V be the total volume of blood and N be the total number of radioactive nuclei at time t. The nuclei are distributed uniformly in blood. A sample of volume v will have n = (v/V )N radioactive nuclei in it. Thus, activity of this sample is v  v v v a = λn = λ N = (λN ) = A = A0 e−λt . V V V V Substitute a = 115 Bq, v = 2.5 ml, A0 = 2.4 × 105 Bq, λ = ln 2/t1/2 ≈ 0.7/(8 × 24) hr−1 and t = 11.5 hr to get (2.5)(2.4 × 105 ) − 0.7×11.5 e 8×24 115   (2.5)(2.4 × 105 ) 0.7 × 11.5 ≈ 1− 115 8 × 24

V =

= 4998 ml ≈ 5 litre. Ans. 5 Q 63. The isotope 125 B having a mass 12.014 u undergoes β-decay to 126 C. 126 C has an excited state of the nucleus (126 C∗ ) at 4.041 MeV above its ground state.

542

Part VI. Modern Physics

If 125 B decays to 126 C∗ , the maximum kinetic energy of the β-particle in units of MeV is . . . . . . . [1 u = 931.5 MeV/c2 , where c is the speed of light in vacuum.] (2016)

Sol. The β-decay is given by the reaction 12 5B

→ 126 C + e−1 + ν¯.

The Q-value of this reaction is given by   Q = m(125 B) − m(126 C) c2 = [12.041 − 12.0] × 931.5 = 13.041 MeV. m(126 C)

Note that = 12 u by definition of atomic mass unit. The energy Q = 13.041 MeV is released in the reaction. Out of this energy, 4.041 MeV is used to excite 12 12 ∗ 6 C to its excited state 6 C . Thus, the kinetic energy available to the β-particle (Kβ ) and the antineutrino (Kν¯ ) is Kβ + Kν¯ = 13.041 − 4.041 = 9 MeV. In βdecay, the kinetic energy of the ν¯ can vary from zero to a maximum value. Hence, the maximum kinetic energy of the β-particle is Kβ,max = 9 MeV (when Kν¯ = 0). Ans. 9 Q 64. For a radioactive material, its activity A and rate of change of its activity R are defined as A = −dN/dt and R = −dA/dt, where N (t) is the number of nuclei at time t. Two radioactive sources P (mean life τ ) and Q (mean life 2τ ) have the same activity at t = 0. Their rates of change of activities at t = 2τ are RP and RQ , respectively. If RP /RQ = n/e, then the value of n is . . . . . . . (2015) Sol. In a radioactive decay, the number of nuclei at time t are given by N (t) = N0 e−λt ,

(1)

where N0 is the number of nuclei at t = 0 and the decay constant λ is related to the mean life τ by λ = 1/τ . Differentiate equation (1) to get the activity A(t) and the rate of change of activity R(t), A(t) = −dN/dt = λN0 e−λt 2

R(t) = −dA/dt = λ N0 e

−λt

(2) .

(3)

Let N0,P and N0,Q be the number of nuclei of P and Q at t = 0, respectively. Given, λP = 1/τ and λQ = 1/(2τ ). The activities of P and Q at t = 0 are equal i.e., λP N0,P = λQ N0,Q ,

(using (2))

(4)

Use equation (3) to get the ratio of rate of change of activities of P and Q at time t = 2τ as   λ2P N0,P e−2λP τ λP λP N0,P e−2λP τ RP = 2 = RQ λQ N0,Q e−2λQ τ λQ λQ N0,Q e−2λQ τ 1/τ e−2(1) = (1) −2(1/2) 1/(2τ ) e 2 = . e

(using (4).)

Ans. 2 Q 65. A nuclear power plant supplying electrical power to a village uses a radioactive material of half life T years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5% of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of nT years, then the value of n is . . . . . . . (2015) Sol. The electrical power produced by a nuclear power plant at time t is proportional to the number of decays per second (activity) at that time i.e.,   dN −0.693t = P0 exp , (1) P = k dt T where P0 is the initial power and T is the half life. The power production decreases with time as shown in the figure. P P0

0.125P0 O

T

2T

3T

t

The power plant will continue to meet power requirement of the village till its power production reduces to the village’s demand i.e., P = 0.125P0 . Substitute P = 0.125P0 in equation (1) and simplify to get t = 3T . Ans. 3 Q 66. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is . . . . . . . (2013) Sol. In a nuclear decay, the number of nuclei at time t is given by N = N0 e−λt ,

(1)

where N0 is the initial number of nuclei and λ is the decay constant. The half life (T1/2 ) is the time when N = N0 /2. Substitute it in equation (1) to get N0 /2 = N0 e−λT1/2 ,

(2)

Take logarithm of equation (2) and substitute the values to get λ=

0.693 0.693 = = 5 × 10−4 s−1 . T1/2 1386

Chapter 42. The Nucleus

543

The fraction of the initial number of nuclei that will decay in first 80 s is −4 N0 − N = 1 − e−λt = 1 − e−5×10 ×80 N0 = 1 − e−0.04 ≈ 1 − (1 − 0.04) (∵ ex ≈ 1 + x for small x).

f=

Thus, the graph between t and |dN/dt| is a straight line with slope −λ. The slope of the given graph is (3 − 4)/(6 − 4) = −0.5. Thus, λ = 0.5 /yr. Substitute λ = 0.5 /yr and t = 4.16 yr in equation (1) to get N = N0 e−λt = N0 e−0.5×4.16 = N0 e−2.08 = 0.125N0 = N0 /8.

= 0.04 = 4%.

Ans. 8

We encourage you to use the additional information (initial activity) to find the numerical values of N0 , N and f . Ans. 4 Q 67. The activity of a freshly prepared radioactive sample is 1010 dis-integration per second, whose mean life is 109 s. The mass of an atom of this radioisotope is 10−25 kg. The mass (in mg) of the radioactive sample is . . . . . . . (2011) Sol. The activity of a radioactive sample of N nuclei, having a decay constant λ, is given by A = |dN/dt| = λN.

(1)

The mean life is given by τ = 1/λ.

(2)

Substitute the values of A and τ in equations (1) and (2) to get N = 1019 . The mass of these atoms is 1019 × 10−25 = 10−6 kg = 1 mg. Ans. 1

dN (t) ln dt

Q 68. To determine the half life of a radioactive ele ment, a student plots a graph of ln dNdt(t) versus t. Here dNdt(t) is the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 years, the value of p is . . . . . . . (2010) 6 5 4 3 2 1

Descriptive Q 69. X-rays are incident on a target metal atom having 30 neutrons. The ratio of nuclear radius of the target atom to that of 42 He nucleus is (14)1/3 . [R = 1.1 × 107 m−1 , c = 3 × 108 m/s.] (2005) (a) Find the mass number of target atom. (b) Find the frequency of Kα line emitted by this metal. Sol. The nuclear radius r of an atom with atomic number A is given by r = r0 A1/3 , where r0 ≈ 1.1 fm. Let A be the atomic number of target nucleus. Using, r0 A1/3 rtarget = = 141/3 , rHe r0 41/3 we get A = 56. The atomic number of target nucleus is Z = A − N = 56 − 30 = 26. In Kα transition, the electron from L shell (n = 2) makes a transition to an electron deficient K shell (n = 1). The shielding from the existing electron in K shell makes effective nuclear charge (Z − 1)e. The frequency of emitted Kα ray is   c 1 1 2 ν = = cR(Z − 1) − 2 . λ 12 2 Substitute the values to get ν = 1.55 × 1018 Hz. Ans. (a) 56 (b) 1.55 × 1018 Hz Q 70. A rock is 1.5 × 109 yr old. The rock contains 238 U which disintegrates to form 206 Pb. Assume that there was no 206 Pb in the rock initially and it is the only stable product formed by the decay. Calculate the ratio of number of nuclei of 238 U to that of 206 Pb in the rock. [Half life of 238 U is 4.5 × 109 yr (Take 21/3 = 1.259).]

2 3 4 5 6 7 8 t (Years)

(2004)

Sol. The activity of a radioactive substance, having a decay constant λ and number of nuclei N at time t, is given by A = |dN/dt| = λN = λN0 e−λt .

(1)

NU = N0 e−λt ,

U. After (1)

where λ = 0.693/T1/2 . The population of Pb at time t is

Take logarithm on both sides of equation (1) to get ln |dN/dt| = ln(λN0 ) − λt.

Sol. Let N0 be the initial population of time t, remaining population of 238 U is

238

(2)

NPb = N0 − NU .

(2)

544

Part VI. Modern Physics

From equations (1) and (2),

Sol. Let Nx , Ny , Nz be the populations of X, Y , Z at time t. Nucleus X decays into Y with a decay rate λx Nx ,

NU 1 1 = = λt . NPb N0 /NU − 1 e −1

dNx /dt = −λx Nx .

(1)

Substitute the values to get e

λt

=e

0.693×1.5×109 4.5×109

= (e

0.693 1/3

)

1/3

=2

Nx

= 1.259,

λxNx

Ny

λyNy Nz

The production rate of Y is same as the decay rate of X. Nucleus Y decays into Z with a decay rate λy Ny . Thus, the rate equation for Y is

and NU /NPb = 1/(1.259 − 1) = 3.861. Ans. 3.861 Q 71. A radioactive element decays by β-emission. A detector records n beta particles in 2 s and in next 2 s it records 0.75n beta particles. Find mean life (in s) correct to nearest whole number. [Given ln 2 = 0.6931, ln 3 = 1.0986]. (2003) Sol. Let N0 be the number of radioactive nuclei at t = 0. The number of nuclei at time t is given by

dNy /dt = λx Nx − λy Ny .

(2)

The production rate of Z is same as the decay rate of Y . Nucleus Z does not decay. Thus, the rate equation for Z is dNz /dt = λy Ny .

(3)

The function,

N = N0 e−λt . The numbers of nuclei at the end of 2 s and at the end of 4 s are N1 = N0 e

−2λ

,

N2 = N0 e

−4λ

.

Thus, the number of nuclei decayed (or the number of β particles detected) in the first 2 s is n = N0 − N1 = N0 (1 − e−2λ ),

(1)

Ny (t) =

N0 λ x [exp(−λy t) − exp(−λx t)] , λx − λy

(4)

attains its maximum when dNy (t)/dt = 0. Differentiate equation (4) and equate derivative of Ny (t) to zero to get     1 0.1 λx 1 t0 = ln ln = λx − λy λy 0.1 − 1/30 1/30 = 15 ln 3 = 16.48 s. The population of X at t0 is Nx = N0 e−λx t0 = 1020 ×e−0.1×16.48 = 1.92×1019 .

and in the next 2 s is 0.75n = N1 − N2 = N0 e−2λ (1 − e−2λ ).

(2)

At t0 , substitute dNy /dt = 0 in equation (2) to get Ny = λx Nx /λy = 0.1 × 1.92 × 1019 /(1/30)

Divide equation (1) by (2) to get

= 5.76 × 1019 .

e−2λ = 0.75 = 3/4.

(3) Total number of nuclei remains constant, N0 = Nx + Ny + Nz . Thus,

Take logarithm of equation (3) to get λ = ln 2 − 0.5 ln 3 = 0.6931 − 0.5×1.0986 = 0.1438.

Nz = N0 − (Nx + Ny ) = 1020 − (1.92 × 1019 + 5.76 × 1019 )

The mean life is τ = 1/λ = 6.95 ≈ 7 s. Ans. 7 s Q 72. A radioactive nucleus X decays to a nucleus Y with a decay constant λx = 0.1 s−1 , Y further decays to a stable nucleus Z with a decay constant λy = 1/30 s−1 . Initially, there are only X nuclei and their number is N0 = 1020 . Set up the rate equations for the populations of X, Y and Z. The population of Y nucleus as a function of time is given by λx [exp(−λy t) − exp(−λx t)]. Find the time Ny (t) = λNx0−λ y at which Ny is maximum and determine the populations of X and Z at that instant. (2001)

= 2.32 × 1019 . Ans. (a) dNx /dt = −λx Nx , dNy /dt = λx Nx − λy Ny , dNz /dt = λy Ny . (b) 16.48 s (c) 1.92 × 1019 , 2.32 × 1019 Q 73. A nucleus at rest undergoes a decay emitting an α-particle of de-Broglie wavelength, λ = 5.76 × 10−15 m. If the mass of the daughter nucleus is 223.610 u and that of the α-particle is 4.002 u, determine the total kinetic energy in the final state. Hence obtain the mass of the parent nucleus in u. [1 u = 931.470 MeV/c2 .] (2001)

Chapter 42. The Nucleus Sol. Let m1 = 4.002 u be the mass of the α-particle and m2 = 223.610 u be the mass of the daughter nucleus. The linear momentum of parent nucleus is zero. The linear momentum of α-particle, having de-Broglie wavelength λ, is p1 = h/λ.

545 The energy Ein is produced by nuclear fission of The energy released by each nucleus is

235

U.

e = 200 MeV = 200 × 106 × 1.6 × 10−19 = 3.2 × 10−11 J. Total number of nuclei required to generate Ein is

By conservation of linear momentum, the linear momentum of daughter nucleus is equal and opposite to that of α-particle i.e.,

n=

3.1536 × 1018 Ein = = 9.855 × 1028 , e 3.2 × 10−11

and their mass is p~2 = −~ p1 .

m = 9.855 × 1028 × 235 × 1.66 × 10−27 = 38444 kg.

The kinetic energies of the α-particle and the daughter nucleus are h2 p21 , = 2m1 2m1 λ2 p2 p2 h2 K2 = 2 = 1 = . 2m2 2m2 2m2 λ2 K1 =

Total kinetic energy in the final state is   1 h2 1 K = K1 + K2 = 2 + . 2λ m1 m2 Substitute h = 6.63 × 10−34 kg m2 /s, λ = −15 5.76 × 10 m, and 1 u = 1.66 × 10−27 kg to get K = 10−12 J = 6.3 MeV. The Q-value of the decay process is equal to the total kinetic energy of the products i.e., Q = [mparent − (m1 + m2 )] c2 = 6.3 MeV. Substitute the values to get mparent = m1 + m2 + 6.3/931.470 = 4.002 + 223.610 + 0.0067 = 227.618 u. Ans. (a) 6.3 MeV (b) 227.618 u Q 74. In a nuclear reactor 235 U undergoes fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 years, find the total mass of uranium required. (2001) Sol. The power generated by the reactor is P = 1000 MW = 109 J/s. The energy generated in time t = 10 years is Eout = P t = 109 × 10 × 365 × 24 × 3600 = 3.1536 × 10

17

J.

The reactor operates at 10% efficiency, η = Eout /Ein = 0.1. Thus, the input energy to the reactor is Ein = Eout /0.1 = 3.1536 × 1018 J.

Ans. 3.84 × 104 kg Q 75. Nuclei of radioactive element A are being produced at a constant rate α. The element has a decay constant λ. At time t = 0, there are N0 nuclei of the element. (1998) (a) Calculate the number N of nuclei of A at time t. (b) If α = 2N0 λ, calculate the number of nuclei of A after one half-life of A and also the limiting value of N as t → ∞. Sol. Let the number of nuclei at time t be N . The number of nuclei produced in time interval t to t + ∆t is ∆Np = α∆t, and the number of nuclei decayed in the same interval is ∆Nd = λN ∆t. Net number of nuclei produced in this interval is ∆N = ∆Np − ∆Nd = (α − λN )∆t.

(1)

If ∆t is very small then equation (1) becomes dN /dt = α − λN.

(2)

Integrate equation (2), Z

N

N0

dN = α − λN

Z

t

dt, 0

to get N=

 1 α − (α − λN0 )e−λt . λ

(3)

Substitute α = 2N0 λ and t = ln(2)/λ in equation (3) to get N = 3N0 /2. The limiting case t → ∞ with α = 2N0 λ gives N∞ = 2N0 .  Ans. (a) λ1 α − (α − λN0 )e−λt (b) (i) 3N0 /2 (ii) 2N0

546

Part VI. Modern Physics

Q 76. In the following, Column I lists some physical quantities and the Column II gives approximate energy values associated with some of them. Choose the appropriate value of energy from Column II for each of the physical quantities in Column I. (1997)

Out of 107 disintegrations per second, 8% are fission and 92% are α decay. Thus, the number of fissions per second and α decays per second are nfission = 0.08 × 107 , nα = 0.92 × 107 .

Column I (A) (B) (C) (D)

Energy of thermal neutrons Energy of X-rays Binding energy per nucleon Photoelectric threshold of a metal

Column II (p) (q) (r) (s)

0.025 eV 0.5 eV 3 eV 20 eV

The energy released per second (power) is given by P = nfission Efission + nα Eα = 0.08 × 107 × 200 + 0.92 × 107 × 5.136 = 2.073 × 108 MeV

(t) 8 MeV (u) 10 keV Sol. The energy of thermal neutron is of the order of kT . Its value is 1.38 × 10−23 × 298 J = 0.025 eV at room temperature T = 298 K. The wavelength of X-rays is of the order of 1 ˚ A which corresponds to energy hc/λ = 12421 eV ≈ 10 keV. Binding energy per nucleon is of the order of 8 MeV. The violet light (λ = 4000 ˚ A) emits the photoelectrons from the metal surface and its energy is hc/λ = 3.1 eV. We encourage you to get familiar with the order of magnitude of these energies. Ans. A7→p, B7→u, C7→t, D7→r Q 77. The element curium 248 96 Cm has a mean life of 1013 s. Its primary decay modes are spontaneous fission and α-decay, the former with a probability of 8% and the latter with a probability of 92%, each fission releases 200 MeV of energy. The masses involved in decay are 248 244 4 96 Cm = 248.072220 u, 94 Pu = 244.064100 u, 2 He = 4.002603 u. Calculate the power output from a sample of 1020 Cm atoms. [1 u=931 MeV/c2 .] (1997) Sol. The α decay of curium is given by 248 96 Cm → 244 4 He. The Q-value of this reaction is Pu + 94 2   2 244 4 Q = m(248 96 Cm) − m( 94 Pu) − m(2 He) c = (248.072220 − 244.064100 − 4.002603) × 931 = 5.136 MeV. Thus, the energies released in each α decay and each fission are Eα = 5.136 MeV, Efission = 200 MeV. The rate of curium dis-integration (fission and α decay) is given by |dN /dt| = λN = N/τmean = 1020 /1013 = 107 .

= (2.073 × 108 )(106 )(1.6 × 10−19 ) = 3.32 × 10−5 W. Ans. 3.32 × 10−5 W Q 78. In an ore containing uranium, the ratio of 238 U to 206 Pb nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of 238 U. Take the half-life of 238 U to be 4.5 × 109 years. (1997) Sol. Let N0 be the number of 238 U nuclei in the ore when it was formed (i.e., at t = 0). Let N be the number of 238 U nuclei at present (at time t). The number of 206 Pb nuclei at present is N/3 (because ratio of 238 U to 206 Pb nuclei is 3). Since all the lead present in the ore is the final stable product of 238 U, we get N0 = N + N/3 i.e., N = 3N0 /4. The law of radioactive decay gives N = N0 e

− 0.693t t 1/2

.

Substitute N = 3N0 /4, t1/2 = 4.5 × 109 years and then solve to get t = 1.86 × 109 years. Ans. 1.86 × 109 years Q 79. At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10 s the number of undecayed nuclei reduces to 12.5%. Calculate, (1996) (a) mean life of nuclei. (b) the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number. Sol. Let λ be the decay constant and N0 be the number of nuclei at t = 0. The number of undecayed nuclei at time t are given by N = N0 e−λt .

(1)

At time t1 , N1 = 0.25N0 and at time t2 = t1 + 10, N2 = 0.125N0 . Substitute in equation (1) to get 0.25N0 = N0 e−λt1 ,

(2)

0.125N0 = N0 e−λt2 = N0 e−λ(t1 +10) = (N0 e−λt1 )e−10λ .

(3)

Chapter 42. The Nucleus

547

Divide equation (2) by (3) and simplify to get λ = ln(2)/10 and mean life τ = 1/λ = 10/ln(2) = 10/0.693 = 14.43 s. Let the number of undecayed nuclei be N3 = 0.0625N2 at time t3 = t2 + t i.e., 0.0625N2 = N0 e−λ(t2 +t) = (N0 e−λt2 )e−λt = N2 e−λt .

(4) 4 ln 2 λ

= 40 s. Solve equation (4) to get t = ln(16)/λ = Note that it takes one half-life to decay from N1 to N2 and four half-lives to decay from N2 to N3 . Ans. (a) 14.43 s (b) 40 s Q 80. A small quantity of solution containing Na24 radio nuclei (half-life = 15 h) of activity 1.0 µCi is injected into the blood of a person. A sample of the blood of volume 1 cm3 taken after 5 h shows an activity of 296 disintegration per minute. Determine the total volume of the blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person. [1 Ci = 3.7 × 1010 disintegration per second.] (1994) Sol. Let N0 be the initial number of radioactive nuclei. Initial activity is given by A0 = |dN /dt| = λN0 , where λ = 0.693/T1/2 . The number of radioactive nuclei at t = 5 h is N = N0 e−λt = (A0 /λ)e−5λ .

Sol. The atomic number of α particle is 2 and its mass number is 4. Conservation of atomic number gives Z = 92 − 2 = 90 and conservation of mass number gives A = 228 + 4 = 232. In magnetic field, the magnetic force on a charged particle provides the centripetal acceleration for circular motion i.e., qvB = mα v 2 /r.

(1)

The equation (1) gives the linear momentum of α particle as pα = mα v = qrB. Thus, kinetic energy of α particle is pα 2 2mα q2 r2 B 2 (2 × 1.6 × 10−19 )2 (0.11)2 (3)2 = = 2mα 2 × 4.003 × 1.67 × 10−27 −13 = 8.34 × 10 J = 5.21 MeV.

Kα =

The conservation of linear momentum, ~0 = p~Y + p~α , gives pY = pα .

(2)

Thus, kinetic energy of Y is pY 2 pα 2 mα = = Kα 2mY 2mY mY 4.003 = 5.21 × = 0.09 MeV. 228.03

KY =

Total energy released in the process is

3

Let V cm be the total volume of the blood. The number of nuclei in 1 cm3 of blood at t = 5 h is n = N/V . Thus, activity of 1 cm3 of blood at t = 5 h is A = |dn/dt| = λn = λN/V = (A0 /V )e

−5λ

,

which gives V =

10−6 × 3.7 × 1010 −(5×0.693/15) A0 −5λ e = e A 296/60

K = Kα + KY = 5.3 MeV. The binding energy of the products is 
BEproducts = 92m 11 H + (232 − 92)m
 −m (Y ) − m 42 He c2

1 0n




= [92 × 1.008 + 140 × 1.009 −228.08 − 4.003] × 931.48 = 1828.5 MeV.

= 5.95 × 103 cm3 = 5.95 litre. Ans. 5.95 litre Q 81. A nucleus X, initially at rest, undergoes alphadecay according to the equation: 92A X → 228 Z Y + α.

The binding energy of the parent is BEparent = BEproducts − K = 1828.5 − 5.3 = 1823.2 MeV.

(1991)

(a) Find the values of A and Z in the above process. (b) The alpha particle produced in the above process is found to move in a circular track of radius 0.11 m in a uniform magnetic field of 3 T. Find the energy (in MeV) released during the process and the binding energy of the parent nucleus X.

[Given m (Y ) =
228.03 u, m 10 n = 1.009 u, m 42 He = 4.003 u, m 11 H = 1.008 u.]

Ans. (a) 232, 90 (b) 5.3 MeV, 1823.2 MeV Q 82. It is proposed to use the nuclear fusion reaction: 2 2 4 1 H + 1 H → 2 He, in a nuclear reactor with an electrical power rating of 200 MW. If the energy from the above reaction is used with a 25 percent efficiency in the reactor, how many grams of deuterium fuel will be needed per day? [The masses of 21 H and 42 He are 2.0141 u and 4.0026 u, respectively.] (1990)

548

Part VI. Modern Physics

Sol. At a power rating of P = 200 MW, total energy output of the reactor in one day is Eout = P t = 200 × 106 × 24 × 3600 = 1.728 × 1013 J. The reactor efficiency is η = Eout /Ein = 0.25. Thus, the total energy produced in a day through nuclear reactions is Ein =

1.728 × 1013 Eout = = 6.912 × 1013 J. η 0.25

The Q-value of given reaction is Q = [2m(21 H) − m(42 He)]c2 = 23.85 × 1.6 × 10−13 = 3.82 × 10−12 J.

Q 85. A uranium nucleus, 238 92 U, emits an alpha particle and the resulting nucleus emits β-particle. What are the atomic number and mass number of final nucleus?

2 × 6.912 × 1013 Ein = 3.62 × 1025 . = Q 3.82 × 10−12

Thus, the required mass of the deuterium is n 3.62 × 1025 M= × 2 = 120 g. N 6.023 × 1023 Ans. 120 g

Q 83. There is a stream of neutrons with kinetic energy of 0.0327 eV. If the half-life of neutrons is 700 s, what fraction of neutrons will decay before they travel a distance of 10 m? (1986) Sol. The speed of the neutron of mass m, having kinetic energy K = 0.0327 eV, is given by s r 2K 2(0.0327)(1.6 × 10−19 ) v= = m 1.67 × 10−27 = 2.5 × 103 m/s. The time taken by the neutron, to travel a distance d = 10 m, is d 10 = = 4 × 10−3 s. v 2.5 × 103 Let N0 be the initial population of the neutrons. The number of neutrons after time t is given by t=

N = N0 exp(−λt), where λ = 0.693 t1/2 is the decay constant and t1/2 = 700 s is the half life of neutrons. Thus, fraction of neutrons that decays in time t is N0 − N = 1 − exp(−λt) N0
= 1 − exp −0.693t/t1/2

f=

= 1 − exp −0.693 × 4 × 10−3 /700

Sol. Given nuclear reactions are 238 92 U

Total number of 21 H nuclei required is

m=

Sol. The atomic number of the given nucleus is Z = 11 and the mass number is A = 24. The number of electrons in a nucleus is zero, number of protons is equal to the atomic number i.e., 11, and number of neutrons is equal to A − Z = 24 − 11 = 13. Ans. zero, 11, 13

(1982)

= [2 × 2.0141 − 4.0026] × 931.8 = 23.85 MeV

n=2×

Q 84. How many electrons, protons and neutrons are there in a nucleus of atomic number 11 and mass number 24? (1982)




= 3.96 × 10−6 . Ans. 3.96 × 10−6

Ax Zx X

4 x →A Zx X + 2 He,

→

Ay Zy Y

+

(1)

0 −1 e.

(2)

Apply conservation of charge and mass number in equation (1) to get 92 = Zx + 2,

238 = Ax + 4,

which gives Zx = 90 and Ax = 234. Now, apply conservation of charge and mass number in equation (2) to get Zx = Zy − 1,

Ax = Ay .

Substitute values of Zx and Ax to get Zy = 91 and Ay = 234. Ans. 91, 234

Appendix A Quick Reference Formulae

Physical Constants c 3 × 108 m/s h 6.63 × 10−34 J s hc 1242 eV-nm G 6.67 × 10−11 m3 kg−1 s−2 k 1.38 × 10−23 J/K R 8.314 J/(mol K) NA 6.023 × 1023 mol−1 e 1.602 × 10−19 C µ0 4π × 10−7 N/A2

Speed of light Planck constant Gravitation constant Boltzmann constant Molar gas constant Avogadro’s number Charge of electron Permeability of vacuum Permitivity of vacuum Coulomb constant Faraday constant Mass of electron Mass of proton Mass of neutron Atomic mass unit Atomic mass unit Stefan-Boltzmann constant Rydberg constant Bohr magneton Bohr radius Standard atmosphere Wien displacement constant

A.2

0

8.85 × 10−12 F/m

1 4π0

F me mp mn u u σ

9 × 109 N m2 /C2 96485 C/mol 9.1 × 10−31 kg 1.6726 × 10−27 kg 1.6749 × 10−27 kg 1.66 × 10−27 kg 931.49 MeV/c2 5.67 × 10−8 W/(m2 K4 )

R∞ µB a0 atm

1.097 × 107 m−1 9.27 × 10−24 J/T 0.529 × 10−10 m 1.01325 × 105 Pa

b

2.9 × 10−3 m K

~a × ~b = (ay bz − az by )ˆı + (az bx − ax bz )ˆ  ˆ + (ax by − ay bx )k |~a × ~b| = ab sin θ

Kinematics Instantaneous velocity: ~vinst = d~r/dt Average velocity: ~vav = ∆~r/∆t Instantaneous acceleration: ~ainst = d~v /dt Average acceleration: ~aav = ∆~v /∆t Motion in a straight line with constant acceleration: v = u + at,

s = ut + 21 at2 ,

Projectile motion: y

u

O

x H θ u cos θ R

Horizontal distance: x = ut cos θ Vertical distance: y = ut sin θ − 21 gt2 Trajectory equation (parabola):

Mechanics y = x tan θ −

Vectors

g x2 2u2 cos2 θ

2u sin θ g

Notation: ~a = ax ˆı + ay ˆ + az kˆ q Magnitude: a = |~a| = a2x + a2y + a2z

Time of flight: T =

Scalar or dot product:

Maximum height: H =

Range: R =

~a · ~b = ax bx + ay by + az bz = ab cos θ

u2 sin 2θ g u2 sin2 θ 2g

Newton’s Laws and Friction Linear momentum: p~ = m~v

Vector or cross product: ~ a × ~b

Newton’s first law defines inertial frame. Newton’s second law: F~ = d~ p/dt, F~ = m~a

ˆ ı ~b θ

v 2 − u2 = 2as

Relative velocity: ~vA/B = ~vA − ~vB

u sin θ

A.1

~ a

ˆ k

ˆ

Newton’s third law: F~AB = −F~BA 549

550

Appendix A. Quick Reference Formulae

Static frictional force:  fapplied , if fapplied ≤ µs N ; fstatic = µs N, otherwise.

Motion of the CM: Define M =

Kinetic frictional force: fkinetic = µk N

Impulse: J~ =

P ~vcm =

Banking of roads :

R

mi~vi , M

Before collision

Centrifugal force is a radially outward pseudo force of magnitude mv 2 /r.

−(v10 − v20 ) v − v2  1 1, completely elastic; = 0, completely in-elastic.

e= l θ θ F

•

Special cases: If v2 = 0 and m1  m2 then v10 = −v1 . If v2 = 0 and m1  m2 then v20 = 2v1 . In elastic collision with m1 = m2 : v10 = v2 and v20 = v1 .

Work, Power and Energy ~ = F S cos θ, W = Work: W = F~ · S Kinetic energy: K = 12 mv 2 =

R

~ F~ · dS.

2

p 2m . 2

Rigid Body Dynamics

Potential energy: Uspring = 21 kx , Ugravitational = mgh. Conservative force and its potential are related by Fconservative = −dU/dx. Work done by conservative forces is path independent i.e., it depends only on the initial and the final points.

Angular velocity: ωav = ∆θ/∆t,

Mechanical energy: E = U + K. It is conserved if forces are conservative in nature. Power: Pav = ∆W /∆t, Pinst = F~ · ~v .

αav = ∆ω/∆t,

P Pxi mi , mi

xcm =

mr 2

CM of few useful configurations: Semicircular Ring

Semicircular Disc

Hemispherical Shell

Hemispherical Solid

r h

C• r

h 3

2r π

C •

C• r •

C• • r

•

4r 3π

r r 2

θ = ωt + 21 αt2 ,

Moment of Inertia: I =

R Rx dm dm

~a = α ~ × ~r

α = dω/dt,

Rotation with constant angular acceleration: ω = ω0 + αt,

Centre of Mass and Collision

~v = ω ~ × ~r

ω = dθ/dt,

Angular acceleration:

Work-energy theorem: W = ∆K.

Equilateral Lamina

• v0 m2 2

Coefficient of restitution:

mg

Centre of mass: xcm =

• v0 m1 1

Energy conservation (in elastic collision):

Time period of a conical pendulum:

p T = 2π l cos θ/g

After collision

• v2 m2

2 2 02 02 1 1 1 1 2 m1 v1 + 2 m2 v2 = 2 m1 v1 + 2 m2 v2

Minimum speed to complete vertical circle: p p vmin, bottom = 5gl, vmin, top = gl

m2 r m1 +m2

F~ext M

m1 v1 + m2 v2 = m1 v10 + m2 v20

Pseudo force in non-inertial frame: F~pseudo = −m~a0 .

• • C m2

~acm =

Momentum conservation:

Centripetal force: Fc = mv 2 /r

• m1

p~cm = M~vcm ,

F~ dt = ∆~ p

• v1 m1

Centripetal acceleration: ac = v 2 /r

C •

mi

Head on collision in one dimension:

v2 µ + tan θ = rg 1 − µ tan θ

Two Point Masses

P

2 1 2 mr

2 2 3 mr

P

2 2 5 mr

i

mi ri 2 ,

2 1 12 ml

ω 2 − ω0 2 = 2αθ

I=

R

r2 dm

2 1 2 2 2 mr m(a +b ) 12

mr 2

a ring

disc

shell

sphere

rod

hollow

solid rectangle

Theorem of parallel axes:

•

Ik

3r 8

The height of CM from the base is h/4 for the solid cone and h/3 for the hollow cone.

Ik = Icm + md2

b

Icm d cm

Appendix A. Quick Reference Formulae

551

Simple Harmonic Motion

Theorem of perpendicular axes: z

Hooke’s law: F = −kx (for small elongation x.)

y

Iz = Ix + Iy

x

Acceleration: a = Time period: T =

p

~τ =

~ dL dt ,

y

√ Velocity: v = Aω cos(ωt + φ) = ±ω A2 − x2

~ = I~ L ω τ = Iα

Potential energy: U

θ

P

~ F

U=

~ r

x −A

Conservation of angular momentum: ~τext = 0 =⇒ ~ = const. L P~ P Equilibrium condition: F = ~0, ~τ = ~0 1 2 Kinetic energy: Krot = 2 Iω Dynamics: F~ext = m~acm ,

p~cm = m~vcm ,

1 2 2 kx

x

O

~τcm = Icm α ~

K = 12 mvcm 2 + 12 Icm ω 2 , ~ = Icm ω L ~ + ~rcm × m~vcm

Gravitation Gravitational force: m1 •

2 F = G mr1 m 2

−ω 2 x

Displacement: x = A sin(ωt + φ)

Radius of gyration: k = I/m ~ = ~r × p~, Angular momentum: L Torque: ~τ = ~r × F~ ,

d2 x k dt2 = − m x = pm 2π ω = 2π k

F

• m2

F

0

A

Kinetic energy: K

K = 12 mv 2

x −A

0

A

Total energy: E = U + K = 12 mω 2 A2 p Simple pendulum: T = 2π l/g q I Physical pendulum: T = 2π mgl , where I is moment of inertia about an axis passing through the point of suspension (O) and l is the distance of the centre of mass from O. Torsional Pendulum:

r

T = 2π

Gravitational potential energy: U = − GMr m

p I/k

Gravitational acceleration: g =

2h R
h −R




Variation of g with depth: ginside ≈ g 1 −

Variation of g with height: goutside ≈ g 1 Effect of non-spherical earth shape on g: gat pole > gat equator (∵ Re − Rp ≈ 21 km) Effect of earth rotation on apparent weight: mgθ0 = mg − mω 2 R cos2 θ

Springs connected in series: 1 keq

=

1 k1

+

k1

1 k2

k2

Springs connected in parallel: k2

keq = k1 + k2

ω ~

k1

Superposition of two SHM’s: •

mg

mω 2 R cos θ

~ A

θ R

δ

~1 A

x1 = A1 sin ωt,

a

•

tan  =

First: Elliptical orbit with sun at one of the focus. ~ Second: areal velocity is constant. (∵ ddtL = 0). Third: T 2 ∝ a3 . In circular orbit T 2 =

x2 = A2 sin(ωt + δ)

x = x1 + x2 = A sin(ωt + ) q A = A1 2 + A2 2 + 2A1 A2 cos δ

vo

•

~2 A



p Orbital velocity of satellite: vo = GM /R p Escape velocity: ve = 2GM /R Kepler’s laws of the planetary motion: •

•

GM R2

2

4π 3 GM a .

A2 sin δ A1 + A2 cos δ

Properties of Matter Modulus of rigidity: Y =

F/A ∆l/l ,

B = −V

∆P ∆V

, η=

F Aθ

552

Appendix A. Quick Reference Formulae

Compressibility: K =

1 B

= − V1

dV dP

Waves on a String

Poisson’s ratio: σ =

lateral strain longitudinal strain

Elastic energy: U =

1 2

=

∆D/D ∆l/l

Speed of waves on a string pwith mass per unit length µ and tension T : v = T /µ

stress × strain × volume

Transmitted power: Pav = 2π 2 µvA2 ν 2

Surface tension: S = F/l

Interference:

Surface energy: U = SA

y1 = A1 sin(kx − ωt),

Excess pressure in a bubble: ∆pair = 2S/R, Capillary rise: h =

∆psoap = 4S/R

2S cos θ rρg

Hydrostatic pressure: p = ρgh Buoyant force is equal to the weight of displaced liquid i.e., FB = ρV g Equation of continuity:

2A cos kx

v1

Bernoulli’s equation: p + 12 ρv 2 + ρgh = constant √ Torricelli’s theorem: vefflux = 2gh

x A

N

A

N

A

λ/4

dv Viscous force: F = −ηA dx

y1 = A sin(kx − ωt),

Stoke’s law:

y2 = A sin(kx + ωt)

y = y1 + y2 = (2A cos kx) sin ωt
 n + 12 λ2 , nodes; n = 0, 1, 2, . . . x= n λ2 , antinodes. n = 0, 1, 2, . . .

F

F = 6πηrv v

Terminal velocity: vt =

A2 sin δ tan  = A1 + A2 cos δ  2nπ, constructive; δ= (2n + 1)π, destructive. Standing Waves:

v2

A1 v 1 = A2 v 2

y2 = A2 sin(kx − ωt + δ)

y = y1 + y2 = A sin(kx − ωt + ) q A = A1 2 + A2 2 + 2A1 A2 cos δ

String fixed at both ends:

2r 2 (ρ−σ)g 9η

L

Poiseuilli’s equation: N

r Volume flow time

A.3

=

πpr 4 8ηl

l

Waves Motion ∂2y ∂x2

=

2 1 ∂ y v 2 ∂t2 .

Notation: Amplitude A, Frequency ν, Wavelength λ, Period T , Angular Frequency ω, Wave Number k, 1 2π T = = , ν ω

v = νλ,

+x;

y = f (t + x/v),

A

N

(i) Boundary conditions: y = 0 at x = 0 and at x = L q n T (ii) Allowed Freq.: L = n λ2 , ν = 2L µ, n = 1, 2, 3, . . .. (iii) Fundamental/1st harmonics: q 1 T ν0 = 2L µ (iv) 1st overtone/2nd harmonics: q 2 T ν1 = 2L µ

2π k= λ

Progressive wave travelling with speed v: y = f (t − x/v),

N

λ/2

Waves

General equation of wave:

A

−x

(v) 2nd overtone/3rd harmonics: q 3 T ν2 = 2L µ (vi) All harmonics are present.

Progressive sine wave:

String fixed at one end:

y

L

A

x λ 2

λ

y = A sin(kx − ωt) = A sin(2π (x/λ − t/T ))

N

A λ/2

N

A

Appendix A. Quick Reference Formulae

553

(i) Boundary conditions: y = 0 at x = 0 (ii) Allowed Freq.: L = (2n + 1) λ4 , q 2n+1 T 4L µ , n = 0, 1, 2, . . ..

ν

=

(iii) Fundamental/1st harmonics: q 1 T ν0 = 4L µ

v 2L

ν0 =

(iv) 1st overtone/2nd harmonics:

(iv) 1st overtone/3rd harmonics: q T 3 ν1 = 4L µ

ν1 = 2ν0 =

2v 2L

(v) 2nd overtone/3rd harmonics:

(v) 2nd overtone/5th harmonics: q 5 T ν2 = 4L µ

ν2 = 3ν0 =

3v 2L

(vi) All harmonics are present. n 2L

T µ

l1 +d

Resonance column: q

l2 +d

(vi) Only odd harmonics are present. √ Sonometer: ν ∝ L1 , ν ∝ T , ν ∝ √1µ . ν =

(i) Boundary condition: y = 0 at x = 0 v (ii) Allowed freq.: L = n λ2 , ν = n 4L , n = 1, 2, . . . st (iii) Fundamental/1 harmonics:

Sound Waves Displacement wave: s = s0 sin ω(t − x/v) Pressure wave: p = p0 cos ω(t − x/v), p0 = (Bω/v)s0 Speed of sound waves: s s s B Y γP vliquid = , vsolid = , vgas = ρ ρ ρ 2

2

0 v Intensity: I = 2πv B s0 2 ν 2 = p2B = Standing longitudinal waves:

p0 2 2ρv

p1 = p0 sin ω(t − x/v), p = p1 + p2 = 2p0 cos kx sin ωt Closed organ pipe: L

(i) Boundary condition: y = 0 at x = 0 (ii) Allowed freq.: L = (2n + 1) λ4 , ν = (2n + v 1) 4L , n = 0, 1, 2, . . . (iii) Fundamental/1st harmonics: v 4L

(iv) 1st overtone/3rd harmonics: ν1 = 3ν0 =

3v 4L

5v 4L

(vi) Only odd harmonics are present. Open organ pipe: A N L

3λ 4 ,

v = 2(l2 − l1 )ν

p1 = p0 sin ω1 (t − x/v),

p2 = p0 sin ω2 (t − x/v)

p = p1 + p2 = 2p0 cos ∆ω(t − x/v) sin ω(t − x/v) ∆ω = ω1 − ω2

ω = (ω1 + ω2 )/2,

A N A

(beats freq.)

Doppler Effect: v + uo ν0 v − us

where, v is the speed of sound in the medium, u0 is the speed of the observer w.r.t. the medium, considered positive when it moves towards the source and negative when it moves away from the source, and us is the speed of the source w.r.t. the medium, considered positive when it moves towards the observer and negative when it moves away from the observer.

Light Waves Plane waves: E = E0 sin ω(t − xv ) I = I0 Spherical waves:

(v) 2nd overtone/5th harmonics: ν2 = 5ν0 =

l2 + d =

Beats: two waves of almost equal frequencies ω1 ≈ ω2

ν=

p2 = p0 sin ω(t + x/v)

ν0 =

l1 + d = λ2 ,

r 0 E = aE r sin ω(t − v ) I = rI02

Young’s double slit experiment P y

S1 d

θ

S2

Path difference: ∆x =

D dy D

554

Appendix A. Quick Reference Formulae 2π λ ∆x

Refraction of Light

Interference conditions: for integer n,  2nπ, constructive; δ= (2n + 1)π, destructive,

p

I = I1 + I2 + 2 I1 I2 cos δ, p p p 2 p 2 I1 + I2 , Imin = I1 − I2 Imax = I1 = I2 : I = 4I0 cos2 2δ , Imax = 4I0 , Imin = 0 λD d

(i) Incident ray, refracted ray and normal lie in the same plane. (ii) µ1 sin i = µ2 sin r

Diffraction from a single slit:

d µ2

r

re

fra

ct

ed

real depth apparent depth

d0 d I O

d d0

=

Critical angle: θc = sin−1

µ

1 µ

θc

Deviation by a prism: y

For Minima: nλ = b sin θ ≈ b(y/D)

θ

b

A y

δ = i + i0 − A

δ

D

Resolution: sin θ =

µ1 i

Apparent depth: µ=

Optical path: ∆x0 = µ∆x Interference of waves transmitted through thin film:  nλ,
constructive; ∆x = 2µd = n + 12 λ, destructive.

te

constructive; destructive

Intensity:

Fringe width: w =

c v

=

Snell’s Law:

nt ide

nλ,
n + 12 λ,

speed of light in vacuum speed of light in medium

inc

 ∆x =

Refractive index: µ =

re fle c

Phase difference: δ =

i

1.22λ b

µ=

Law of Malus:

m sin A+δ 2 A sin 2

i0

r0

r µ

θ

I = I0 cos2 θ

A.4

I0

δ

I 0

i = i for min. deviation δm = (µ − 1)A for small A

δm i

i0

Optics Refraction at spherical surface:

in

Lens maker’s formula: d

d

Q

O u

1 f

= (µ − 1)

h

v

1 R1

−

1 R2

i

Lens formula:

Spherical mirror focal length: f = R/2

f

1 1 1 − = v u f

Mirror formula: 1 1 1 + = v u f

µ2

P

µ1 v m= µ2 u

re

nt

fle

de

i r

µ1

µ2 µ1 µ2 − µ1 − = v u R

ct

normal

ci

Laws of reflection: (i) Incident ray, reflected ray, and normal lie in the same plane (ii) ∠i = ∠r Plane mirror: (i) image and object are equidistant from mirror (ii) virtual image of real object

ed

Reflection of Light

m = v/u

I O

u

v

f

m = −v/u u

v

Power of the lens: P = f1 , P in diopter if f in metre.

Appendix A. Quick Reference Formulae

555

Two thin lenses separated by a distance d: 1 1 1 d = + − F f1 f2 f1 f2

Thermal expansion: L = L0 (1 + α∆T ), A = A0 (1+β∆T ), V = V0 (1+γ∆T ), γ = 2β = 3α ∆l Thermal stress of a material: F A =Y l

d f1

Kinetic Theory of Gases

f2

General: M = mNA , k = R/NA Maxwell distribution of speed:

Optical Instruments

n

Simple microscope: m = D/f in normal adjustment. Compound microscope: Eyepiece

Objective

vp v ¯ vrms

∞

O

u

v

RMS speed: vrms = Average speed: v¯ =

fe D

Resolving power: R =

q

=

v D u fe

2µ sin θ λ

Astronomical telescope: fo

fe

q

3kT m

=

8kT πm

q = 8RT πM q

Most probable speed: vp =

Magnification in normal adjustment: m = 1 ∆d

q

v

3RT M

2kT m

2 Pressure: p = 13 ρvrms Equipartition of energy: K = 12 kT for each degree of freedom. Thus, K = f2 kT for molecule having f degrees of freedoms. Internal energy of n moles of an ideal gas: U = f2 nRT .

Specific Heat In normal adjustment: m = − ffoe , L = fo + fe 1 1 Resolving power: R = ∆θ = 1.22λ

Cauchy’s equation: µ = µ0 +

A λ2 ,

A>0

Dispersion by prism with small A and i: Mean deviation: δy = (µy − 1)A Angular dispersion: θ = (µv − µr )A µv −µr µy −1

≈

θ δy

(if A and i small)

Dispersion without deviation: 0

A

µ

µ

A0

Deviation without dispersion: (µv − µr )A = (µ0v − µ0r )A0

Heat and Thermodynamics

Heat and Temperature Temp. scales: F = 32 + 95 C,

∆Q n∆T



∆Q n∆T

V



p

Relation between Cp and Cv : Cp − Cv = R Ratio of specific heats: γ = Cp /Cv Relation between U and Cv : ∆U = nCv ∆T Specific heat of gas mixture: Cv =

n1 Cv1 + n2 Cv2 , n1 + n2

γ=

n1 Cp1 + n2 Cp2 n1 Cv1 + n2 Cv2

Molar internal energy of an ideal gas: U = f2 RT , f = 3 for monatomic and f = 5 for diatomic gas.

Theromodynamic Processes

(µy − 1)A + (µ0y − 1)A0 = 0

A.5

Specific heat at constant volume: Cv = Specific heat at constant pressure: Cp =

Dispersion

Dispersive power: ω =

Q Specific heat: s = m∆T Latent heat: L = Q/m

K = C + 273.16

Ideal gas equation: pV = nRT ,

n : number of moles
van der Waals equation: p + Va2 (V − b) = nRT

First law of thermodynamics: ∆Q = ∆U + ∆W Work done by the gas: Z V2 ∆W = p∆V, W = pdV V1   V2 Wisothermal = nRT ln V1 Wisobaric = p(V2 − V1 ) p1 V1 − p2 V2 Wadiabatic = γ−1 Wisochoric = 0

556

Appendix A. Quick Reference Formulae

Efficiency of heat engine: ratio of work done by the engine to the heat supplied to it. Q1 − Q2 Q1

η=

Electric field: ~ r) = E(~

T1 Q1

1 q rˆ 4π0 r2

• q

~ E

•

~ r

W

Q2 T2 ηcarnot = 1− = 1− Q1 T1

Q2

1 q1 q2 Electrostatic energy: U = − 4π r 0

T2

Electrostatic potential: V = Coefficient of performance of refrigerator: T1 Q1

Q2 Q2 COP = = W Q1 − Q2

~ · ~r, dV = −E

1 q 4π0 r

W

Electric dipole moment:

T2

Rf

p ~

p~ = q d~

∆Q T

Sf − Si = i Entropy: ∆S = For constant T : ∆S = Q T T For varying T : ∆S = ms ln Tfi

~ · d~r E

∞

Q2

∆Q T ,

~ r

Z V (~r) = −

−q •

• +q d

Potential of a dipole:

Adiabatic process: ∆Q = 0, pV γ = constant

V (r)

1 p cos θ V = 4π0 r2

•

θ r p ~

Heat Transfer Conduction:

∆Q ∆t

x KA

Thermal resistance: R = Req = R1 + R2 = A1

1 Req

=

1 R1

+ R12

Field of a dipole:

= −KA ∆T x



x1 K1

+

x2 K2



K1

K2

x1

x2

1 x

= (K1 A1 +K2 A2 )

A

emissive power absorptive power

=

Ebody abody

1 2p cos θ 4π0 r3

Er θ r Eθ

1 p sin θ Eθ = 4π0 r3

K2

A2

K1

A1

p ~

~ ~τ = p~ × E ~ Torque on a dipole placed in E: ~ U = −~ ~ Potential energy of a dipole placed in E: p·E

x

Kirchhoff’s Law:

Er =

= Eblackbody

Gauss’s Law and its Applications

Wien’s displacement law: Eλ

λm T = b λ

λm

H ~ · dS ~ Electric flux: φ = E H ~ · dS ~ = qin /0 Gauss’s law: E Field of a uniformly charged ring on its axis:

Stefan-Boltzmann law:

∆Q ∆t

Newton’s law of cooling:

A.6

dT dt

= σeAT

4

= −bA(T − T0 )

EP =

Electricity and Magnetism

Coulomb’s law: 1 q1 q2 rˆ 4π0 r2

a q

x

E and V of a uniformly charged sphere: ( E 1 Qr 4π0 R3 , if r < R E= 1 Q if r ≥ R 4π0 r 2 ,

Electrostatics

F~ =

qx 1 2 4π0 (a + x2 )3/2

O

• q1

r

• q2

( V=

1 4π0 1 4π0

Qr 2 R3 , Q r,

P

~ E

R

r

R

r

V

if r < R if r ≥ R

O

Appendix A. Quick Reference Formulae

557

Current electricity

E and V of a uniformly charged spherical shell: E

 E=

0,

if r < R 1 Q , if r ≥ R 4π0 r 2

O

R

r

R

r

V

( V=

1 4π0 1 4π0

Q R, Q r,

if r < R if r ≥ R

Field of a line charge: E =

O λ 2π0 r

Field of an infinite sheet: E =

σ 20

Field in the vicinity of conducting surface: E =

σ 0

Capacitors

Current density: j = i/A = σE i Drift speed: vd = 21 eE m τ = neA Resistance of a wire: R = ρl/A, where ρ = 1/σ Temp. dependence of resistance: R = R0 (1 + α∆T ) Ohm’s law: V = iR Kirchhoff’s Laws: (i) The Junction Law: The algebraic sum of all the currents directed towards a node is zero i.e., Σnode Ii = 0. (ii)The Loop Law: The algebraic sum of all the potential differences along a closed loop in a circuit is zero i.e., Σloop ∆ Vi = 0. Resistors in parallel: A

1 1 1 = + Req R1 R2

Capacitance: C = q/V

R1

R2

B

Parallel plate capacitor: −q

C = 0 A/d

Resistors in series:

+q

A

A d

Req = R1 + R2

R1

A

R2

B

Spherical capacitor: r2

4π0 r1 r2 C= r2 − r1

−q +q

Wheatstone bridge balancing condition: R1

r1

R1 R3 = R2 R4

R3

Cylindrical capacitor:

C=

R4 V

2π0 l ln(r2 /r1 )

r2

Electric Power: P = V 2 /R = I 2 R = IV Galvanometer as an Ammeter:

l

r1 ig G

i

ig G = (i − ig )S

i

i − ig

Capacitors in parallel:

S

Galvanometer as a Voltmeter:

A

Ceq = C1 + C2

R2 ↑ G

C1

C2

B R

VAB = ig (R + G)

G

↑

A ig

Capacitors in series: Charging of capacitors: 1 1 1 = + Ceq C1 C2

C1

C2

A

B

q(t) = CV 1 − e

t − RC

i V

Force between plates of a parallel plate capacitor: F = Q2 2A0

Discharging of capacitors:

Energy stored in capacitor: U = 12 CV 2 =

2

Q 2C

Capacitor with dielectric: C =

0 KA d

C

= 12 QV

Energy density in electric field E: U/V = 12 0 E 2

C

R

h

t

q(t) = q0 e− RC

q(t)

R

B

558

Appendix A. Quick Reference Formulae

Time constant in RC circuit: τ = RC Peltier heat Peltier effect: emf e = ∆H ∆Q = charge transferred . Seeback effect:

Field due to a straight conductor: θ2

B=

e T0

Tn

T

Ti

Thermo-emf: e = aT + 21 bT 2 Thermoelectric power: de/dt = a + bT . Neutral temp.: Tn = −a/b. Inversion temp.: Ti = −2a/b. Thomson heat Thomson emf: e = ∆H ∆Q = charge transferred = σ∆T . Faraday’s law of electrolysis: The mass deposited is m = Zit =

1 F

Eit

µ0 i (cos θ1 − cos θ2 ) 4πd

d

i

⊗

~ B

θ1

Field due to an infinite straight wire: B =

µ0 i 2πd

Force between parallel wires: i1

µ0 i1 i2 dF = dl 2πd

i2

d

Field on the axis of a ring:

where i is current, t is time, Z is electrochemical equivalent, E is chemical equivalent, and F = 96485 C/g is Faraday constant.

BP =

a

µ0 ia2 2 2(a + d2 )3/2

P

i

~ B

d

Magnetism ~ + qE ~ Lorentz force on a moving charge: F~ = q~v × B Charged particle in a uniform magnetic field: mv , r= qB

Field at the centre of an arc: B=

µ0 i 2a

Field inside a solenoid: ~ B ~l ~ F

B = µ0 ni, n = i

~ A

µ ~

~ µ ~ = iA

B= i

~ ~τ = µ ~ Torque on a magnetic dipole placed in B: ~ ×B ~ U = −~ ~ Energy of a magnetic dipole placed in B: µ·B Hall effect: y z

l

x

µ0 N i 2πr

r

Field of a bar magnet:

B1 =

µ0 2M 4π d3

B2 =

µ0 M 4π d3

~ B

l w d

N l

Field inside a toroid:

Magnetic moment of a current loop (dipole):

i

i

θ a

Field at the centre of a ring: B = H ~ · d~l = µ0 Iin Ampere’s law: B

~⊗ r B

~ F~ = i ~l × B

Bi ned

~ B

v q

2πm T = qB

Force on a current carrying wire:

Vw =

a

µ0 iθ 4πa

~2 B d S

~1 B

N d

Angle of dip:

Magnetic Field due to Current Horizontal

Biot-Savart law:

Bv

~ B

~ = dB

µ0 i d~l × ~r 4π r3

δ

Bh = B cos δ

B

⊗

i θ d~l

Bh

~ r

Tangent galvanometer: Bh tan θ =

µ0 ni 2r ,

Moving coil galvanometer: niAB = kθ,

i = K tan θ i=

k nAB θ

Appendix A. Quick Reference Formulae Time period of magnetometer: T = 2π

559

q

Imepedance: Z = e0 /i0

I M Bh

RC circuit:

~ = µH ~ Permeability: B

C

Electromagnetic Induction

p

Z=

R2 + (1/ωC)2

i

˜

~ · dS ~ B

Magnetic flux: φ =

H

Faraday’s law: e =

− dφ dt

e0 sin ωt

tan φ =

Lenz’s Law: Induced current create a B-field that opposes the change in magnetic flux. Motional emf:

1 ωCR

Z

1 ωC

φ R

LR circuit:

+

L

e = Blv

l

⊗B ~

~ v

p Z = R2 + ω 2 L2

˜

e0 sin ωt

di e = −L dt 2

tan φ =

2

Self inductance of a solenoid: L = µ0 n (πr l) h i t Growth of current in LR circuit: i = Re 1 − e− L/R L

ωL R

e i

t

L R

q

i


2 1 ωC − ωL 1 −ωL tan φ = ωC R q 1 1 νresonance = 2π LC Z=

t

R2 +

˜

e0 sin ωt 1 ωC

Z φ

ωL

1 ωC

−ωL

R

i

R

i0

Power factor: P = erms irms cos φ

0.37i0 S

Z

L C R

Decay of current in LR circuit: i = i0 e− L/R L

φ

ωL

LCR Circuit:

e 0.63 R

S

R

i

R

R

i

−

Self inductance: φ = Li,

R

i

Transformer:

t

L R

N1 N2

Time constant of LR circuit: τ = L/R

Mutual inductance: φ = M i,

di e = −M dt

=

e1 i1 = e2 i2

e1

N1

˜

N2

˜

e2

i2

√ Speed of the EM waves in vacuum: c = 1/ µ0 0

B2 2µ0

U V

e1 e2 ,

i1

Energy stored in an inductor: U = 21 Li2 Energy density of B field: u =

=

EMF induced in a rotating coil: e = N ABω sin ωt

A.7

Modern Physics

Alternating current:

Average current in AC: ¯i =

Photo-electric effect

i

i = i0 sin(ωt + φ) T = 2π/ω

t

Photon’s energy: E = hν = hc/λ

T 1 T

RT 0

Photon’s momentum: p = h/λ = E/c

i dt = 0

Max. KE of ejected photo-electron: Kmax = hν − φ

RMS current:

Threshold freq. in photo-electric effect: ν0 = φ/h

i2

" Z #1/2 1 T2 i0 irms = i dt =√ T 0 2

Stopping potential: V0

T

  φ hc 1 − Vo = e λ e

Energy: E = irms 2 RT Capacitive reactance: Xc =

t

1 ωC

Inductive reactance: XL = ωL

de Broglie wavelength: λ = h/p

hc e

−φ e

φ hc

1 λ

560

Appendix A. Quick Reference Formulae

The Atom

Vacuum tubes and Semiconductors Half Wave Rectifier:

Energy in nth Bohr’s orbit:

D

mZ 2 e4 En = − 2 2 2 , 80 h n

13.6Z 2 eV En = − n2

Radius of the nth Bohr’s orbit: 2 2

rn =

0 h n , πmZe2

˜ Full Wave Rectifier:

2

rn =

n a0 , Z

a0 = 0.529 ˚ A

Quantization of the angular momentum: l =

nh 2π

˜

Output

Triode Valve:

Photon energy in state transition: E2 − E1 = hν E2

Grid Cathode Filament

E2 hν

E1

R Output

Emission

Plate

hν E1 Absorption

Plate resistance of a triode: rp =

Wavelength of emitted radiation: for a transition from nth to mth state:   1 1 1 = RZ 2 2 − 2 λ n m

Amplification by a triode: µ = −

X-ray spectrum:

Current in a transistor:

λmin

√

Relation between rp , µ, and gm : µ = rp × gm Ic

α and β parameters of a transistor: α = Ic α Ib , β = 1−α Transconductance: gm =

R0 ≈ 1.1 × 10−15 m

= −λN

Population at time t: N N0 N0 2

O t 1/2

Ic Ie ,

β =

∆Ic ∆Vbe

Logic Gates:

The Nucleus

N = N0 e−λt

∆ip =0

λ

Heisenberg uncertainity principle: ∆p∆x ≥ h/(2π), ∆E∆t ≥ h/(2π)

dN dt

∆Vp =0



∆Vp ∆Vg

Ib λα

X-ray diffraction: 2d sin θ = nλ

Decay rate:



Ie

ν = a(Z − b)

Nuclear radius: R = R0 A1/3 ,

∆Vg =0

∆ip ∆Vg

Ie = Ib + Ic λmin

Moseley’s law:

Transconductance of a triode: gm =

Kα Kβ

I

hc = eV



∆Vp ∆ip

t

Half life: t1/2 = 0.693/λ Average life: tav = 1/λ Population after n half lives: N = N0 /2n . Mass defect: ∆m = [Zmp + (A − Z)mn ] − M Binding energy: B = [Zmp + (A − Z)mn − M ] c2 Q-value: Q = Ui − Uf Energy released in nuclear reaction: ∆E = ∆mc2 where ∆m = mreactants − mproducts .

A 0 0 1 1

B 0 1 0 1

AND AB 0 0 0 1

OR A+B 0 1 1 1

NAND AB 1 1 1 0

NOR A+B 1 0 0 0

XOR ¯ + AB ¯ AB 0 1 1 0

This is your book! Let your contribution count. . . Must have for serious JEE aspirants. Perfect companion for Concepts of Physics. Just like the Russian have separate problems book like the ‘B Bukhhovtsev’ and ‘A A Pinsky’ along with a rigorous theory book like ‘G S Landsberg’. Don’t think and go for it! – Review on Amazon

Report an Error We would be glad to hear your suggestions for the improvement of the book. If you find any conceptual errors or typographical errors, howsoever small and insignificant, please inform us so that these can be corrected in the later editions. We believe, only a collaborative effort from the readers and the authors can make this book absolutely error-free, so please contribute. You may send feedback/corrections to us by, (i) email to [email protected] (ii) website www.concepts-of-physics.com

Amazing and complete collection of Physics problems from IIT JEE. It helped my daughter to consolidate her preparation. If a student religiously attempts each problem and understands their underlying theme, he will definitely reach his goal of cracking the JEE exam. – Gangineni Dhananjhay

We are grateful to Nihal Jain, Akash Goyal, Sai Charan Marriwada, Apoorv Potnis, Nawaz Zafar, Panneer Selvam, Divyanshu Tiwari, Prasad Patankar, Halachandra Kalloli, Prannoy Mehta, Aryamaan Yadav, Kshitiz Chaurasia, and Tribhuvan Narayan Soorya for submitting corrections in previous editions of this book.

Conceptual approach to all problems. I haven’t spotted a single error till now. Much better than similar books from other publishers. The solutions to even the most difficult problems are approached through the basics. Extremely happy with my decision to buy this book. – Karan Singh

Write a Review Dear reader, this book is different from similar books in the market. Major difference is direct relationship between the authors and the readers. To ensure ‘what you get is what we want’, we take care of everything starting from content preparation to the final delivery. The technology helped us to do so. Only hurdle is its promotion! If you like this book then please talk about it.

Excellent Solution Book! Initially, I was bit confused on whether to purchase it as I already had 2 problems and solutions book on IIT. But trust me, this book is much better than the earlier solutions. – Review on Amazon I must thank authors for such a nice and well composed book. This book not only gives the better understanding of the problems but solve it in very simple and artistic way. A recommended book for all IITs aspirants. – Review on Amazon

We sincerely thank reviewers as their reviews helped the book to find more readers. Your reviews on Amazon/Flipkart matter a lot. Here are few reviews of previous editions (slightly paraphrased to correct typos and to remove competitors’ name):

Very good for self study and clearing doubts/concepts. Perfect companion for “Concepts of Physics”. – Shiva

Precise and clear approach. Excellent explanation with necessary diagrams. Definitely a class apart. – Review on Amazon

Much needed problem solver for self-study. Fully answered my doubts on rotational mechanics. – Review on Amazon

Detailed explanations are very helpful to an average student like me. – Review on Amazon

I am a faculty of Physics for JEE. I found the solutions accurate and lucid. I find this book to have great value as it is designed in sync with “Concepts of Physics”. – Nitin Varshney

The language used is simple and very descriptive. Topic wise categorization is the beauty of this book. – Aditya Sen Singh

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562 Worth every rupee you spend on it. This is a wonderful book. The solutions are really good and brings conceptual clarity. This book deserves a really good rating. – Akshansh Bhura It is different from other similar books in the way it gives very detailed and systematic solutions. Really laudable effort. A must for Physics enthusiasts. – Vinod Kumar Sharma Well written and easy to understand. I have two other books which are also JEE solved papers. But they have direct steps without explanation. But in this book explanation is very good. Blindly go for it! – Jani Best in market. Hidden gem! I have other books as well but this one is a gem. Chapters are arranged in order of HCV’s Concepts of Physics. – Ashay Awesome! Must have for JEE aspirants. If you can solve this book then you have a possibility of solving even Irodov or getting into IIT. – Girish At last some new thing to be adored! This book has tremendous concept building capability. – Prasannarout Awesome! Hats off to the quality of material in the book. – Goutam Maji The way of representing the solutions is amazing. Being a physics lover and JEE aspirant I saw Concepts of Physics and this book as complete self study package. – Review on Amazon MERITS: detailed explanation. DEMERITS: black and white (not fancy). CONCLUSION: decide what you want colours or knowledge. – Pratham HC Verma + this book = complete physics teacher. – Review on Amazon The solutions are really good. Far better than other contemporary books of its kind. – Review on Amazon Perhaps the best book there in the market for IIT JEE physics. Supplement this with HC Verma and your concepts will be crystal clear. Perfect tool for self study. Every question is explained in detail. – Anirudh Jayanth

Must for serious IIT JEE aspirant. There are many solution books available in the market for previous year JEE problems. This book stands apart. Solutions are explained in detail. In many questions there are extra points which are highly beneficial. – Akash Goyal Finest written solutions! I m glad to have such a book. – Review on Amazon Excellent book with amazing solutions! It is more than just a solution book, it helps you grasp the key concepts. – Anurag Ghosh Teaching physics for last 15 years and I haven’t seen the book like this. – Review on Amazon Detailed explanations and no errors. Much better than its competitors. – Karthik

Get Connected Many of you are connected with us through email, facebook, and other platforms. This email says a lot: I am Nihal Jain, a JEE aspirant from Mumbai. I had contacted you earlier with regards to an error in your book. Now that JEE Advanced has been conducted, I can with great assurance tell you that your book benefited me a lot. As you know there are several books that have their own take on the JEE syllabus and pattern with relation to the subject. However, IIT JEE Physics is not just a great tool to hone one’s problem solving skills that stands apart from other books but also a concise but complete handbook of concepts. An amazing aspect of the book was it’s appreciation of subtleties of the subject and the scientific method (for example, concept of parallax, vivid usage of vectors throughout the manuscript and so on). I also enjoyed answering questions that “we encourage you” to solve as they were great markers of thrill and challenge. As a result, I recommended your book to all my juniors and friends interested in the subject. I also want to congratulate you on being able to produce a book that is so remarkably typeset that it surpasses any other book in terms of scientific rendering by national standards (graphs, quality of figures, etc). I hope you accept my gratitude and continue to support students like myself. Thanking you, – Nihal Jain

Messages like this inspire us to keep trying for perfection. Best Wishes! – Jitender/ Shraddhesh.