IIT JEE Advanced Comprehensive Physics N K Bajaj MHE Mc Graw Hill Education

Table of contents :
Cover......Page 1
Title page......Page 4
Copyright page......Page 5
A Word to the Reader......Page 6
Syllabus......Page 8
Contents......Page 10
MODEL TEST PAPER—I......Page 1063
MODEL TEST PAPER—II......Page 1075
IIT-JEE 2012: PAPER-I......Page 1087
IIT-JEE 2012: PAPER-II......Page 1095
JEE ADVANCED 2013: PAPER-I......Page 1105
JEE ADVANCED 2013: PAPER-II......Page 1113
Physics Jee Advanced—2014 Paper-I—Model Solutions......Page 1122
Physics Jee Advanced—2014 Paper-II—Model Solutions......Page 1133
JEE ADVANCED 2015: PAPER–I (MODEL SOLUTIONS)......Page 1143
JEE ADVANCED 2015: PAPER–II (MODEL SOLUTIONS)......Page 1154
JEE ADVANCED 2016: PAPER-I (MODEL SOLUTIONS)......Page 1166
JEE ADVANCED 2016: PAPER-II (MODEL SOLUTIONS)......Page 1181
JEE Advanced 2017: Paper - I (Physics)......Page 1200
JEE Advanced 2017: Paper - II (Physics)......Page 1209
Back Cover......Page 1220

Citation preview

JEE ADVANCED Comprehensive

Physics

2019

JEE ADVANCED

2019

Comprehensive

Physics

McGraw Hill Education (India) Private Limited CHENNAI

McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San  Juan Santiago  Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur,Chennai - 600 116, Tamil Nadu, India Comprehensive Physics—JEE Advanced Copyright © 2018 by McGraw Hill Education (India) Private Limited No Part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers. McGraw Hill Education (India) Private Limited 1 2 3 4 5 6 7 8 9 D102542 22 21 20 19 18

ISBN (13) : 978-93-87572-57-7 ISBN (10) : 93-87572-57-9 Information contained in this work has been obtained McGraw Hill Education (India), from sources believed to be reliable. However, neither, McGraw Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.

at Printers, 312 EPIP, HSIDC, Kundli, Sonepat, Haryana Cover Design: Neeraj Dayal Visit us at: www.mheducation.co.in

A Word to the Reader Comprehensive Physics—JEE Advanced is highly useful for aspirants appearing for JEE Advanced. It is as per the syllabus of JEE Advanced. Divided into 29 core chapters, it covers the entire gamut of the subject through a combination of conceptual theory supported by solved problems and practice exercises. Each chapter opens with a review of the basic concepts, formulae, laws and definition that is linked to the concepts and problems of that chapter. There are plenty of fully solved questions of all types as per the latest pattern and syllabus. Questions in each chapter have been classified as follows: Section Section Section Section Section

I: II : III : IV : V:

Multiple Choice Questions with Only One correct choice. Multiple Choice Questions with One or More correct choices. Multiple Choice Questions based on passage. Matching of Column type questions. Assertion-Reason type questions.

Key Features

New Feature

THE PUBLISHERS

Syllabus General: Units and dimensions, dimensional analysis; least count, significant figures; Methods of measurement and error analysis for physical quantities pertaining to the following experiments: Experiments based on using Vernier

specific resistance of the material of a wire using meter bridge and post office box. Mechanics: tion; Relative velocity.

-

Law of gravitation; Gravitational potential and field; Acceleration due to gravity; Motion of planets and satellites in circular orbits; Escape velocity. Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of uniform bodies with simple geometrical shapes; Angular momentum; Torque; Conservation of angular momentum; Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres; Equilibrium of rigid bodies; Collision of point masses with rigid bodies. Linear and angular simple harmonic motions. theorem and its applications.

(in sound). Thermal physics: Thermal expansion of solids, liquids and gases; Calorimetry, latent heat; Heat conduction in one heats (Cv and Cp for monoatomic and diatomic gases); Isothermal and adiabatic processes, bulk modulus of gases;

Electricity and magnetism: of point charges and of electrical dipoles in a uniform electrostatic field; Electric field lines; Flux of electric field; charged infinite plane sheet and uniformly charged thin spherical shell. in a capacitor. applications; Heating effect of current.

viii Syllabus

circular coil and inside a long straight solenoid; Force on a moving charge and on a current-carrying wire in a uniform magnetic field. Magnetic moment of a current loop; Effect of a uniform magnetic field on a current loop; Moving coil galvanometer, voltmeter, ammeter and their conversions. d.c. and a.c. sources. Optics: Magnification. Modern physics: Atomic nucleus; Alpha, beta and gamma radiations; Law of radioactive decay; Decay constant; these processes.

Contents A Word to the Reader Syllabus

v vii

1. Units and Dimensions

1.1–1.24

2. Motion in One Dimension

2.1–2.36

3. Vectors

3.1–3.13

4. Motion in Two Dimensions

4.1–4.34

5. Laws of Motion and Friction

5.1–5.61

6. Work, Energy and Power

6.1–6.39

7. Conservation of Linear Momentum and Collisions

7.1–7.38

8. Rigid Body Rotation

8.1–8.68

9. Gravitation

9.1–9.36

10. Elasticity

10.1–10.24

11. Hydrostatics (Fluid Pressure and Buoyancy)

11.1–11.32

12. Hydrodynamics (Bernoulli’s Theorem and Viscosity)

12.1–12.25

13. Simple Harmonic Motion

13.1–13.47

14. Waves and Doppler’s Effect

14.1–14.48

15. Thermal Expansion

15.1–15.14

16. Measurement of Heat

16.1–16.11

17. Thermodynamics (Isothermal and Adiabatic Processes)

17.1–17.41

18. Kinetic Theory of Gases

18.1–18.16

19. Transmission of Heat

19.1–19.23

20. Electrostatic Field and Potential

20.1–20.50

21. Capacitance and Capacitors

21.1–21.35

22. Electric Current and D.C. Circuits

22.1–22.50

23. Heating Effect of Current

23.1–23.23

24. Magnetic Effect of Current and Magnetism

24.1–24.52

25. Electromagnetic Induction and A.C. Circuits

25.1–25.54

26. Ray Optics and Optical Instruments

26.1–26.52

27. Wave Optics

27.1–27.34

28. Atomic Physics

28.1–28.43

29. Nuclear Physics

29.1–29.27

x Contents

Model Test Paper I Model Test Paper II IIT-JEE 2012 Paper–I IIT-JEE 2012 Paper–II JEE JEE JEE JEE JEE JEE

Advanced Advanced Advanced Advanced Advanced Advanced

2013 Paper–I 2013 Paper–II 2014: Paper–I 2014: Paper–II 2015: Paper–I (Model Solutions) 2015: Paper–II (Model Solutions)

   JEE Advanced 2016: Paper-I (Model Solutions)    JEE Advanced 2016: Paper-II (Model Solutions)    JEE Advanced 2017: Paper-I    JEE Advanced 2017: Paper-II

MTPI.1–MTPI.12 MTPII.1–MTPII.12 IJI.1–IJI.8 IJII.1–IJII.10 JAI.1–JAI.8 JAII.1–JAII.9 JAI.1–JAI.11 JAII.1–JAII.10 JAI.1–JAI.11 JAII.1–JAII.12 P-I.1–P-I.15 P-II.1–P-II.19 P-I.1–P-I.9 P-II.1–P-II.11

1

Units and Dimensions

Chapter

REVIEW OF BASIC CONCEPTS 1.1

THE SI SYSTEM OF UNITS

The internationally accepted standard units of the fundamental physical quantities are given in Table 1.1. Table 1.1

Fundamental SI Units

Physical Quantity

Name of the Unit

Symbol

Mass

kilogram

kg

Time

second

s

Electric current

ampere

A

Temperature

kelvin

K

Luminous intensity

candela

cd

Amount of substance

mole

mol

Physical Quantity

Name of the Unit

Symbol

Angle in a plane

radian

rad

Length

metre

m

Solid angle

steradian

sr

Table 1.2 Physical Quantity Area

Dimensional Formulae of some Physical Quantities

Dimensional Formula

Physical Quantity

Dimensional Formula

0 2 0

Heat energy

ML2T–2

0 3 0

Entropy

ML2T–2K–1

ML T

Volume

ML T

Density

ML–3T 0

Velocity

M0LT –1 0

M0L2T –2K–1 Latent heat

–2

Acceleration

M LT

Momentum

MLT–1 2 –1

M0L2T–2 ML2T–2K–1 mol–1

Thermal conductivity

MLT–3K–1

Angular momentum

ML T

Wien’s constant

M0LT0K

Force

MLT–2

Stefan’s constant

ML0T–3K–4

Energy, work

ML T

Boltzmann’s constant

ML2T–2K–1

Power

ML2T–3

Torque, couple Impulse Frequency

2 –2

Molar gas constant

ML2T–2K–1 mol–1

2 –2

Electric charge

TA

–1

Electric current

A

Electric potential

ML2T–3 A–1

ML T MLT

0 0 –1

MLT

0 0 –1

Angular frequency

MLT

Angular acceleration

M0L0T–2

MLT–3 A–1 Capacitance

M –1L–2T4A2

Pressure

–1 –2

ML T

Inductance

ML2T–2A–2

Elastic modulii

ML–1T–2

Resistance

ML2T–3A–2

Stress

–1 –2

ML T

ML2 T–2A–1 (Contd.)

1.2 Comprehensive Physics—JEE Advanced

Physical Quantity

Dimensional Formula

Physical Quantity

Dimensional Formula

2 0

ML0T–2A–1

Moment of inertia

ML T

Surface tension

ML0T–2

Permeability

MLT –2A–2

Viscosity

ML–1T –1

Permittivity

M–1L–3T4A2

Planck’s constant

ML2T –1

Gravitational constant

1.2

–1 2 –2

M LT

PRINCIPLE OF HOMOGENEITY OF DIMENSIONS

Consider a simple equation, A + B = C. If this is an equation of physics, i.e. if A, B and C are physical quantities, then this equation says that one physical quantity A, when added to another physical quantity B, gives a third physical quantity C. This equation will have no meaning in physics if the nature (i.e. the dimensions) of the quantities on the left-hand side of the equation is not the same as the nature of the quantity on the right-hand A is a length, B must also be a length and the result of addition of A and B In other words, the dimensions of both sides of a physical equation must be identical. This is called the principle of homogeneity of dimensions.

1.3

Dimensions of at = dimensions of x [ x] [L] or [a] = [LT 1 ] [M 0 LT 1 ] [t ] [T] Dimensions of bt2 = dimensions of x L x or b= 2 = [LT–2] = [M0 LT–2] T2 t 1.2 The pressure P, volume V and temperature T of a gas are related as a P V b = cT V2 where a, b, and c are constants. Find the dimensions a of . b

USES OF DIMENSIONAL ANALYSIS

Dimensional equations provide a very simple method of deriving relations between physical quantities involved in any physical phenomenon. The analysis of any phenomenon carried out by using the method of dimensions is called dimensional analysis. This analysis is based on the There are four important uses of dimensional equations: 1. Checking the correctness of an equation. 2. Derivation of the relationship between the physical quantities involved in any phenomenon. 3. Finding the dimensions of constants or variables in an equation. 4. Conversion of units from one system to another. 1.1 The distance x travelled by a body varies with time t as x = at + bt2, where a and b are constants. Find the dimensions of a and b. SOLUTION The dimensions of each term on the right hand of the given equation must be the same as those of the left hand side. Hence

SOLUTION Dimensions of

a

= dimensions of P. V2 Dimensions of a = dimensions of PV2 Also dimensions of b = dimensions of V Dimensions of

a [ PV 2 ] = b [V ] = [PV] = [ML–1 T–2] = [ML2 T – 2]

[L3]

NOTE 1. Trigonometric function (sin, cos, tan, cot etc) are dimensionless. The arguments of these functions are also dimensionless nents are also dimensionless

1.3 When a plane wave travels in a medium, the displacement y of a particle located at x at time t is given by y = a sin(bt + cx) where a, b and c are constants. Find the dimensions b of . c

Units and Dimensions 1.3

this equation must be the same. Equating the powers of M, L and T, we have 1 and a = 0, b + c = 0 and – 2c = 1, which give b = 2 1 c= . 2 Hence t = k m0 l1/2 g(– 1/2)

SOLUTION Terms bt and cx must be dimensionless. Hence 1 [T 1] [b] = [t ] and

[c] =

1 [ x]

1 [L 1 ] [ L]

Thus t is independent of the mass of the bob and is directly proportional to proportional to

1.4 2

a e ax b P is pressure, x is a distance and a and b are constants. Find the dimensional formula for b. P=

SOLUTION a2 b

= [P]

Also ax is dimensionless. Hence [a] = [L – 1]. [b] =

[a 2 ] [L 2 ] = [ P] [ML 1 T 2 ]

= [M–1 L–1 T2] The principle of homogeneity of dimensions can also on other physical quantities.

SOLUTION Let

t

m al bg c

or t = k m al bg c, where k is a dimensionless constant. Writing the dimensions of each quantity, we have 2 c

[T] = [M ]a [L]b [LT or

[M 0 L0 T] = [M a Lb

c

T

2c

]

]

According to the principle of homogeneity of dimensions, the dimensions of all the terms on either side of

l and inversely

g.

The dimensional method can also be used to convert a physical quantity from one system to another. The method is based on the fact that the magnitude of a physical quantity X remains the same in every system of its measurement, i.e. (1) X = n1 u1 = n2 u2 where u1 and u2 are the two units of measurement of quantity X and n1 and n2 are their respective numerical values. Suppose M1, L1 and T1 are the fundamental units of mass, length and time in one system of measurement and M2, L2 and T2 in the second system of measurement. Let a, b and c be the dimensions of mass, length and time of quantity X, the units of measurement u1 and u2 will be u1 = M1a Lb1 T1c and

u2 = M a2 Lb2 T2c

Using these in Eq. (1), we have n1 M1a Lb1 T1c

1.5 The time period (t) of a simple pendulum may depend upon m the mass of the bob, l the length of the string and g the acceleration due to gravity. Find the dependence of t on m, l and g.

l g

t= k

b = [LT –1] = M0 LT –1] c Note that the dimensions of a are the same as those of y.

= n2 M a2 Lb2 T2c

n2 = n1

M1 M2

a

L1 L2

b

T1 T2

c

(2)

Knowing (M1, L1, T1), (M2, L2, T2), (a, b and c) and the value of n1 in system 1, we can calculate the value of n2 in system 2 from Eq. (2). 1.6 Dyne is the unit of force in the c.g.s. system and newton is the unit of force in the SI system. Convert 1 dyne into newton. SOLUTION The dimensional formula of force is [F] = [M1 L1 T–2] Therefore, a = 1, b = 1 and c = –2

1.4 Comprehensive Physics—JEE Advanced

System 1 (Given System)

System 2 (Required System)

n1 = 1 dyne

n2 = ? (number of newtons)

M1 = 1g

M2 = 1 kg

L1 = 1 cm

L2 = 1 m

T1 = 1 s

T2 = 1 s

n2 = n1 =1 =1

a

M1 M2

1

1g 1 kg

1 cm 1m 1

1g

=1

1

T1 T2

2

(2)

1

1

102 cm

1.4

1.7 Convert 72 kmh–1 into ms–1 by using the method of dimensions.

quantity on other quantities of a given system, it has its own limitations, some of which are listed as follows: 1. In more complicated situations, it is often not easy

2.

SOLUTION Given System

Required System

n1 = 72 units

n2 = ?

L1 = 1 km

L2 = 1 m

T1 = 1 h

T2 = 1 s

= 72 = –1

M1 M2

1

L1 L2

1000 m 1m

3.

1

60

60 s 1s

1

72 1000 = 20 60 60

Hence 72 km h = 20 ms

LIMITATIONS OF DIMENSIONAL ANALYSIS

Though the dimensional method is a simple and a very

10–5 newtons in 1 dyne, i.e. 1

n2 = n1

(3)

Using L = 10 m in Eq. (3), we get T = 2 s. Using T = 2 s and L = 10 m in either Eq. (1) or Eq. (2), we get M = 4 kg.

10–5

Hence there are 1 dyne = 10–5 N.

[ML2 T–2] = 100 J

Dividing Eq. (2) by Eq. (1), we get 100 J 100 Nm L= = = 10 m 10 N 10 N

c

1s 1s

1 cm

103 g

(1)

[LT–1] = 5 ms–1

b

L1 L2

SOLUTION [MLT–2] = 10 N

–1

4.

5.

NOTE

will depend. In such cases, one has to make a guess which may or may not work. This method gives no information about the dimensionless constant which has to be determined either vation. This method is used only if a physical quantity varies as the product of other physical quantities. It fails if a physical quantity depends on the sum or difference of two quantities. Try, for instance, to 1 2 at using the method obtain the relation S = ut + 2 of dimensions. This method will not work if a quantity depends on another quantity as sin or cos of an angle, i.e. if the dependence is by a trigonometric function. The method works only if the dependence is by power functions only. This method does not give a complete information in cases where a physical quantity depends on more than three quantities, because by equating the powers of M, L and T, we can obtain only three equa-

Sometimes it is more convenient to use units rather than 72 km h–1 =

72 km 72 1000 m = = 20 ms–1 1h 60 60 s

1.8 If the units of force, energy and velocity are 10 N, 100 J and 5 ms–1 time.

1.5

SIGNIFICANT FIGURES

indicates the degree of precision of that measurement. The degree of precision is determined by the least count of the measuring instrument. Suppose a length measured by a metre scale (of least count = 0.1 cm) is 1.5 cm, then

Units and Dimensions 1.5

with a vernier callipers (of least count = 0.01 cm) the 0.001 cm) the same length may be 1.536 cm which has

1.9

It must be clearly understood that we cannot increase the accuracy of a measurement of changing the unit. For 39.4 kg. It is understood that the measuring instrument has g or 39400000 mg, we cannot change the accuracy of measurement. Hence 39400 g or 39400000 mg still have

SOLUTION M

only the magnitude of measurement. M 2. We use the following rule to determine the number and division of various physical quantities. Do not worry about the number of digits after the decimal place. Round off the result so that it has the

The least accurate quantity determines the accuracy of the sum or product. The result must be rounded off to the appropriate digit.

accurate quantity.

(preceding) digit to be retained is left unchanged. 4 which is less than 5. 2. If the digit to be dropped is more than 5, the preced3. If the digit to be dropped happens to be 5, then (a) the preceding digit to be retained is increased by 1 if it odd, or (b) the preceding digit is retained unchanged if it is even. result of the indicated rounding-off is therefore, 5.3. off to 3.4. 1. For addition and subtraction, we use the following rule. Find the sum or difference of the given measured that it has the same number of digits after the decimal place as in the least accurate quantity (i.e., the

1.10 A man runs 100.2 m in 10.3 s. Find his average speed

SOLUTION Average speed (v) =

100.2 m 10.3s

–1

the time 10.3 s has only three. Hence the value of the The correct result is v = 9.73 ms–1

1.6

LEAST COUNTS OF SOME MEASURING INSTRUMENTS

1. Least count of metre scale = 1 mm = 0.1 cm 2. Vernier constant (or least count) of vernier callipers = value of 1 main scale division – value of 1 vernier scale division = 1 M.S.D. – 1 V.S.D Let the value of 1 M.S.D = a unit If n vernier scale divisions coincide with m main scale divisions, then value of m of 1 M.S.D 1.V.S.D = n ma = unit n

1.6 Comprehensive Physics—JEE Advanced

ma m a unit 1 n n 3. Least count of a micrometer screw is found by the formula Least count = Pitch of screw Total number of divisions on circular scale Least count = a –

where pitch = lateral distance moved in one complete rotation of the screw.

1.7

ORDER OF ACCURACY: PROPORTIONATE ERROR

The order of accuracy of the result of measurements is determined by the least counts of the measuring instruments used to make those measurements. Suppose a length x is measured with a metre scale, then the error in x is x, where x = least count of metre scale = 0.1 cm. If the same length is measured with vernier callipers of least count 0.01 cm, then x = 0.01 cm. x . x x 100. x 1. Suppose a quantity is given by a=x+y Then a = x + y a x = a (x 2.

mum error is

y y)

If a = x – y

-

a= x+ y We take the worst case in which errors add up. a x y = a ( x y) 3. Suppose we determine the value of a physical quantity u by measuring three quantities x, y and z whose true values are related to u by the equation u = x y z– quantities x, y and z be respectively ± x, ± y and ± z so that the error in u by using these observed quantities is ± u. The numerical values of x, y and z are given by the least count of the instruments used to measure them. Taking logarithm of both sides we have log u = log x + log y – log z Partial differentiation of the above equation gives

x y z u = x y z u The proportional or relative error in u is u/u. The values of x, y and z may be positive or negative and in some uses the terms on the right hand side may counteract each other. This effect cannot be relied upon and it is necessary to consider the worst case which is the case when all errors add up giving an error u given by the equation: x y z u = x y z u ma u, multiply the proportional errors in each factor (x, y and z) by the numerical value of the power to which each factor is raised and then add all the terms so obtained. proportional error in the result of u. When the proportional error of a quantity is multiplied by 100, we get the percentage error of that quantity. 1.11 ) of a rectangular metal block, a student makes the following measurements. Mass of block (m) = 39.3 g Length of block (x) = 5.12 cm Breadth of block (y) = 2.56 cm Thickness of block (z) = 0.37 cm The uncertainty in the measurement of m is ± 0.1 g and in the measurement of x, y and z is ± 0.01 cm. Find the value of (in g cm – 3) up to appropriate of . SOLUTION =

m = xyz 5.12

39.3 2.56 0.37 –3

y m x z + + + y m x z ma 0.1 0.01 0.01 0.01 = 39.3 5.12 2.56 0.37 = 0.0353 = 0.0353 = 0.0353 Round off error = 0.3 gcm – 3. =

–3

Units and Dimensions 1.7 –3 Hence is not accurate to the fourth decimal place. In fact, it is accurate only up to the much be

is written as

SOLUTION –3

1.12 The time period of a simple pendulum is given by T = 2 L/ g . The measured value of L is 20.0 cm using a scale of least count 1 mm and time t for 100 oscillations is found to be 90 s using a watch of least count 1 s. Find the value of g (in m s–2) up to appro-

A B 1 C D X +3 = 2 + + A B 3 C D X =2

1% + 2% +

= 17%

1 3

3% + 3

4%

NOTE 2. The quantity which is raised to the highest power conured to a high degree of accuracy.

the value of g. SOLUTION If t is the time for n oscillations, then T = T=2

of quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Find the percentage error in the measurement of X.

L . Squaring, we get g 4 2 Ln 2 g= t2

t . Given n

(1)

Putting L = 0.200 m, n = 100 and t = 90 s, g=

4

(3.14) 2

0.2

(100) 2

(90) 2

= 9.74 m s–2 From Eq. (1), the relative error in g is g L 2 t = g L t Note that there is no error in counting the number (n) of oscillations. Thus g 0.1 cm 2 1s = g 20.0 cm 90s = 0.005 + 0.022 = 0.027 g = 0.027 9.74 = 0.26 m s–2 Rounding off g g = 0.3 m s–2. Hence the value of g must be rounded off as g = 9.7 m s–2. Hence g = (9.7 ± 0.3) m s–2 1.13 In the measurement of a physical quantity X = A2 B , the percentage errors in the measurements C1/3 D3

1. For a simple pendulum T = 2 T 1 = T 2 2. For a sphere of radius r,

A 2 r = A r

Surface area A = 4 r2 Volume V =

g

4 3 r 3

3 r r

V V

3. Acceleration due to gravity g =

GM R

g 2 R M = g R M 4. For resistances connected in series Rs Rs

Rs = R1 + R2

R1 R1

R2 R2

5. For resistances connected in parallel 1 1 = R1 Rp

1 R2 Rp RP2

–

=

R1

R12

Rp R 2p

R1

R12

R2

R22

R2

R22

6. Kinetic energy K and linear momentum p are related as p2 K 2 p K = 2m K p

1.8 Comprehensive Physics—JEE Advanced

I Multiple Choice Questions with Only One Choice Correct 1. The pressure P is related to distance x, Boltzmann constant k and temperature as a –ax/k e P= b The dimensional formula of b is (a) [M –1 L–1T–1] (b) [MLT 2] (c) [M 0 L2T 0] (d) [M0 L0 T 0] 2. The magnitude of induced emf e in a conductor of length L B is given by 1 ( BL2) a The dimensional formula of a is (a) [M 0 L0T] (b) [ML2T –2] (c) [M 2 LT –1] (d) M 0 L0 T 0] 3. Two resistors R1 = [3.0 ± 0.1] and R2 = (6.0 ± 0.3) are connected in parallel. The resistance of the combination is (a) (2.0 ± 0.4) (c) (2.0 ± 0.2) (d) (2.00 ± 0.04) 4. If the resistances in Q. 3 above were connected in tance of the combination will be (a) 1.1% (b) 2.2% (c) 3.3% (d) 4.4% 5. Which of the following pairs of physical quantities do not have the same dimensions? (a) Pressure and Young’s modulus (b) Emf and electric potential (c) Heat and work e=

6. Which of the following pairs of physical quantities have different dimensions? (a) Impulse and linear momentum (b) Planck’s constant and angular momentum (c) Moment of inertia and moment of force (d) Torque and energy 7. A = A0 e–a/kT, k is Boltzmann constant and T is the absolute temperature. The dimensions of a are the same as those of (a) energy (b) time (c) acceleration (d) velocity 8. A cube has a side of 1.2 cm. The volume of the cube 3

(b) 1.73 cm3

9.

10.

11.

12.

13.

14.

(c) 1.7 cm3 (d) 17.3 cm3 The quantities L/R and RC (where L, C and R stand for inductance, capacitance and resistance respectively) have the same dimensions as those of (a) velocity (b) acceleration (c) time (d) force Which one of the following has the dimensions of ML–1T –2? (a) torque (b) surface tension (c) viscosity (d) stress The dimensions of angular momentum are (a) MLT –1 (b) ML2T–1 (c) ML –1T (d) ML0T –2 According to the quantum theory, the energy E of a photon of frequency is given by E=h where h is Planck’s constant. What is the dimensional formula for h? (b) M L2 T –1 (a) M L2 T –2 2 (c) M L T (d) M L2 T2 The dimensions of Planck’s constant are the same as those of (a) energy (b) power (c) angular frequency (d) angular momentum The volume V of water passing any point of a uniform tube during t seconds is related to the cross-sectional area A of the tube and velocity u of water by the relation V A u t which one of the following will be true? (a) = = (b) =

(c) = (d) 15. The frequency n of vibrations of uniform string of length l and stretched with a force F is given by p F n= 2l m where p is the number of segments of the vibrating string and m is a constant of the string. What are the dimensions of m? (b) M L–3 T0 (a) M L–1 T–1 –2 0 (c) M L T (d) M L–1 T0

Units and Dimensions 1.9

16. If velocity (V), acceleration (A) and force (F) are taken as fundamental quantities instead of mass (M), length (L) and time (T), the dimensions of Young’s modulus would be (a) FA2V –2

(b) FA2V–3

(c) FA2V–4 (d) FA2V –5 17. The dimensions of permittivity ( 0) of vacuum are (a) M–1 L–3 T4 A2 (b) ML–3 T2 A2 (c) M–1 L3 T4 A2 (d) ML3 T2 A2 18. What are the dimensions of permeability ( 0) of vacuum? (b) MLT–2 A–2 (a) MLT–2 A2 (c) ML–1 T–2 A2 (d) ML–1 T–2 A–2 19. The dimensions of 1/ 0 0 are the same as those of (a) velocity (b) acceleration (c) force (d) energy 20. (b) ML2 T–2 K–1 (a) MLT –2 K–1 (c) M0L2T–2 K–1 (d) M0LT–2 K–1 21. What are the dimensions of latent heat? (a) ML2 T –2 (b) ML–2 T–2 0 –2 (c) M LT (d) M 0 L2T –2 22. What are the dimensions of Boltzmann’s constant? (b) ML2T–2 K–1 (a) MLT –2 K–1 0 –2 –1 (c) M LT K (d) M0L2T–2 K–1 23. The dimensions of potential difference are (a) ML2T – 3 A – 1 (b) MLT–2 A–1 (c) ML2T – 2 A (d) MLT–2 A 24. What are the dimensions of electrical resistance? (b) ML2 T–3 A–2 (a) ML2T –2 A2 (c) ML2 T –3 A2 (d) ML2 T–2 A–2 25. (a) MLT –3 A–1 (b) MLT –2 A–1 –1 –1 (c) MLT A (d) MLT0 A–1 26. (b) M0L T –1 A–1 (a) ML0 T –1 A–1 –2 –1 (c) MLT A (d) ML 0 T–2 A–1 27. (b) ML2 T–2 A–2 (a) ML2 T –2 A–1 –2 –2 –1 (c) ML T A (d) ML–2 T–2 A–2 28. The dimensions of self inductance are (b) ML2 T–2 A–2 (a) ML2 T–2 A–1 (c) ML–2 T–2 A–1 (d) ML–2 T–2 A–2

29. The dimensions of capacitance are (a) M–1 L–2 TA2 (b) M–1 L–2 T2 A2 –1 –2 3 2 (c) M L T A (d) M–1 L–2 T4 A2 30. If velocity (V), force (F) and energy (E) are taken as fundamental units, then dimensional formula for mass will be (a) V –2F 0 E (b) V 0 FE 2 (c) VF –2E 0 (d) V–2 F 0E 31. Frequency (n) of a tuning fork depends upon length (l) of its prongs, density ( ) and Young’s modulus (Y) of its material. Then frequency and Young’s modulus will be related as (a) n Y (b) n Y 1 1 (d) n (c) n Y Y 1 2 32. The dimensions of E ( = permittivity of free 0 0 2 space and E (a) MLT –1 (b) ML2T –2 –1 –2 (c) ML T (d) ML2T–1 IIT, 2000 33. Of the following quantities, which one has dimensions different from the remaining three (a) Energy per unit volume (b) Force per unit area (c) Product of voltage and charge per unit volume (d) Angular momentum 34. If the time period t of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula t = d a r b s c and if a = 1, c = – 1, then b is (a) 1 (b) 2 (c) 3 (d) 4 35. In the measurement of a physical quantity X = A2 B . The percentage errors introduced in the C 1/ 3 D 3 measurements of the quantities A, B, C and D are 2%, 2%, 4% and 5% respectively. Then the minimum amount of percentage of error in the measurement of X is contributed by: (a) A (b) B (c) C (d) D 36. Which of the following has the dimensions ML–1 T–1? (c) Bulk modulus (c) Angular momentum 37. Pressure gradient dp/dx is the rate of change of pressure with distance. What are the dimensions of dp/dx?

1.10 Comprehensive Physics—JEE Advanced

(a) ML–1 T–1 (b) ML–2 T–2 –1 –2 (c) ML T (d) ML–2 T–1 38. If E, M, J and G respectively denote energy, mass, angular momentum and gravitational constant, then EJ 2 has the dimensions of M 5G 2 (a) length (b) angle (c) mass (d) time IIT, 1990 39. If e, 0, h and c respectively represent electronic charge, permittivity of free space, Planck’s cone2 has the dimenstant and speed of light, then 0 hc sions of (a) current (b) pressure (c) angular momentum (d) angle 40. If L, R, C and V respectively represent inductance, resistance, capacitance and potential difference, L are the same as those then the dimensions of RCV of 1 (a) current (b) current (c) charge

(d)

1 charge

41. If E and B respe E has the B

dimensions of (a) displacement (b) velocity (c) acceleration (d) angle 42. If C and V respectively represent the capacitance of a capacitor and the potential difference between its plates, then the dimensions of CV2 are (a) ML2T–2 (b) ML3T–2 A–1

(c) ML2T–1A–1 (d) M0L0T0 43. If h and e respectively represent Planck’s constant h and electronic charge, then the dimensions of e are the same as those of

44. If energy E, velocity V and time T are chosen as the fundamental units, the dimensional formula for surface tension will be

45.

(a) E V2T–2

(b) E V–1T–2

(c) E V–2T–2

(d) E2V–1T–2

oscillates with a period proportional to Pa db Ec where P is the static pressure, d is the density of

and c respectively are 5 1 1 (a) , , 6 2 3 (c)

1 1 , , 3 2

5 6

a, b (b)

1 , 2

5 1 , 6 3

(d) 1, 1, 1

46. In a system of units in which the unit of mass is a kg, unit of length is b metre and the unit of time is c second, the magnitude of a calorie is 4.2c 4.2c 2 (b) (a) ab 2 ab 2 abc 4.2 (c) (d) 4.2 abc 47. The error in the measurement of the radius of a sphere is 1%. The error in the measurement of the volume is (a) 1% (b) 3% 48. If the error in the measurement of the volume of a sphere is 6%, then the error in the measurement of its surface area will be (a) 2% (b) 3% (c) 4% (d) 7.5% 49. The moment of inertia of a body rotating about a 2 in the SI system. What is the value of the moment of inertia in a system of units in which the unit of length is 5 cm and the unit of mass is 10 g? (a) 2.4 × 103 (b) 2.4 × 105 (c) 6.0 × 103 (d) 6.0 × 105 V 50. A quantity X is given by 0L where 0 is the t 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 (a) resistance (b) charge (c) voltage (d) current IIT, 2001 51. ) of a liquid by the the formula

R4 P l Q

where R = radius of the capillary tube, l = length of the tube, P = pressure difference between its ends, and

Units and Dimensions 1.11

Q Which quantity must be measured most accurately? (a) R (b) l (c) P (d) Q 52. The mass m of the heaviest stone that can be

dent also measures the diameter of the wire to be 0.4 mm with a uncertainty of ± 0.01 mm. Take g 2

from the reading is (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

v, the speed of water, density (d) of water and the acceleration due to gravity (g). Then m is proportional to (b) v4 (a) v2 6 (c) v (d) v 53. The speed (v) of ripples depends upon their wavelength ( ), density ( ) and surface tension ( ) of water. Then v is proportional to (b)

(a) (c)

IIT, 2007 59. In a vernier callipers, one main scale division is x cm and n divisions of the vernier scale coincide with (n – 1) divisions of the main scale. The least count (in cm) of the callipers is n 1 nx x (b) (a) n (n 1) x x (c) (d) n (n 1)

1

(d)

1

54. The period of revolution (T) of a planet moving round the sun in a circular orbit depends upon the radius (r) of the orbit, mass (M) of the sun and the gravitation constant (G). Then T is proportional to (a) r1/2 (c) r3/2

(b) r (d) r2

55. If the velocity of light (c), gravitational constant (G) and planck’s constant (h) are chosen as fundamental units, the dimensions of time in the new system will be (a) c –5/2G2h–1/2 (b) c–3/2G–2h2 2

–2 1/2

–5/2

1/2 1/2

(c) c G h (d) c G h 56. The amplitude of a damped oscillator of mass m varies with time t as A = A0 e( at m ) The dimensions of a are (b) M0LT–1 (a) ML0T–1 –1 (c) MLT (d) ML–1T 57. A student measures the value of g with the help of a simple pendulum using the formula g=

58.

60.

4

2

L

2

T The errors in the measurements of L and T are L and T respectively. In which of the following cases is the error in the value of g the minimum? (a) L = 0.5 cm, T = 0.5 s (b) L = 0.2 cm, T = 0.2 s (c) L = 0.1 cm, T = 1.0 s (d) L = 0.1 cm, T = 0.1 s long, by Searle’s method. In a particular reading,

IIT, 2007 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. Least count for length = 0.1 cm Least count for time = 0.1 s

Student Length of the Number of Total time for pendulum oscillations (n) (cm) (n) oscillations (s)

Time period (s)

I

64.0

16.0

II

64.0

4

64.0

16.0

III

20.0

4

36.0

9.0

If EI, EII and EIII are the percentage errors in g. i.e. g 100 for student I, II and III, respectively, g (a) E1 = 0 (c) EI = EII

(b) EI is minimum (d) EII is minimum

61. The density of a solid ball is to be determined in an 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 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is (a) 0.9% (b) 2.4% (c) 3.1% (d) 4.2% IIT, 2011

1.12 Comprehensive Physics—JEE Advanced

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61.

(c) (a) (d) (a) (a) (a) (b) (c) (b) (d) (c)

(a) (c) (b) (c) (d) (c) (b) (c) (d) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(b) (c) (d) (d) (a) (d) (d) (a) (a) (d)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

1. k x

1

JK

=

K

m

=Jm = [M L2T–2] = M L T –2 [a] Also [P] = [b] [a] = [ P]

3.

ML1 T 1

ML T

2 2

= [M 0 L2T 0]

BL2 [M 0 L0 T 0 ] [ML0 T 2 A 1 ] = e [ML2 T 3 A 1 ] = [M0 L0T1] 1 R2

1 1 = R1 R

R= R R

2

= =

R1 R2 3.0 6.0 = = 2.0 R1 R2 9.0 R1

R12

R2

R22

0.1

0.3

2

( 6) 2

(3)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(c) (b) (b) (b) (d) (b) (a) (c) (c) (b)

L = 1.2 cm,

[L–1]

So the correct choice is (c). 2. a =

(d) (b) (a) (a) (d) (c) (b) (b) (d) (c)

3

–1

[b] =

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

5. The correct choice is (d). 6. The correct choice is (c). 7. Since kT has dimensions of energy, the correct choice is (a). 8. V = L3 = 1.2 cm 1.2 cm 1.2 cm

SOLUTIONS

[a] =

(d) (d) (c) (b) (b) (c) (b) (b) (c) (b)

= 0.019

R = 0.019 R2 = 0.019 (2)2 = 0.076 Hence the correct choice is (b). 4. R = R1 + R2 = 3.0 + 6.0 = 9.0 R = R1 + R2 = 0.1 + 0.3 = 0.4 R 100 R 0.4 = 100 9.0 = 4.4% So the correct choice is (d).

[L2 ]

Hence the correct choice is (c). 9. L/ R is the time constant of an L-R circuit and CR is the time constant of a C-R circuit. The dimension of the time constant is the same as that of time. Hence the correct choice is (c). 10. ML–1T –2 are the dimensions of force per unit area. Out of the four choices, stress is the only quantity that is force per unit area. Hence the correct choice is (d). 11. The angular momentum L of a particle with respect to point whose position vector is r is given by L=r p where p is the linear momentum of the moving particle. Dimensions of L = dimension of r dimensions of p = L MLT –1 = ML2T–1 Thus the correct choice is (b). dimension of E ML2 T 2 = 12. Dimensions of h = dimension of T 1 = ML2 T –1 Thus the correct choice is (b). 13. The correct choice is (d). 14. The dimensions of the two sides of proportionality are L 3 = L 2 (LT –1 ) T = L2 + T – Equating the powers of dimensions on both sides, we have 2 + =3 – =0

Units and Dimensions 1.13

1 (3 – ), i.e. = . 2 Thus the correct choice is (b). 15. Squaring both sides of the given relation, we get p2 F p2 F or m = n2 = 2 4l m 4 l 2 n2 dimensions of m dimensions of F = dimensions of l 2 dimensions of n 2 ( p is a dimensionless number) which give

=

= and

=

2

MLT L2

T

–1

= ML T

1 2

0

Hence the correct choice is (d). 16. Dimensions of Young’s modulus Y are ML–1 T–2. The dimension of V, A and F in terms of M, L and T are (V) = LT –1, (A) = LT –2 and (F) = MLT –2 Let (Y ) = (V a A b F c ) Putting dimensions of Y, V, A and F. We have (ML–1 T –2 ) = (LT –1) a ( LT – 2 ) b (MLT – 2) c or M1 L – 1 T – 2 = M c La + b + c T – a – 2b – 2c Equating powers of M, L and T we have c = 1, a + b + c = – 1 and – a – 2b – 2c = – 2 which give a = – 4, b = 2 and c = 1. 2 –4 Thus (Y) = (FA V ) Thus the correct choice is (c). 17. According to Coulomb’s law of electrostatics, force F between two charges q1 and q2 a distance r apart in vacuum, is given by q1q2 1 F= 4 0 r2 or

0

=

1 4

Dimensions of

q1q2 F 0

r2

Q2

=

MLT 2 L2 –1 –3 2 2 =M L T Q –1

–3

4

2

Q T

=M L T A A The correct choice is (a). 18. The force per unit length between two long wires carrying currents I1 and I2 a distance r apart in vacuum, is given by II 4 rf f = 0 1 2 or 0 = r 4 I1 I 2

Dimensions of

0

=

L MLT

2

L1

A2 = MLT A – 2 Therefore, the correct choice is (b). 19. Dimensions of 1 1 –2

0

0

MLT =

2

A2

1 1 2 2 2 T

M 1L 3 T 4 A

1 2 2

= LT – 1

L which are the dimensions of velocity. Hence the correct choice is (a). 20. The heat energy content H of a body of mass m at temperature is given by H = ms where s H s= m Dimensions of s dimensions of heat energy dimension of mass dimension of temperature ML2 T 2 = M 0 L 2T – 2 K – 1 M K Thus the correct choice is (c). 21. Latent heat L is the amount of heat energy H required to change the state of a unit mass without producing any change in temperature. Thus H L= m =

ML2 T 2 M 2 –2 = L T = M 0 L 2T – 2 Thus the correct choice is (d). 22. According to the law of equipartition of energy, the energy per degree of freedom of a gas atom or molecule at a temperature kelvin is given by 1 2E k or k = E= 2 where k is the Boltzmann’s constant. dimensions of E Dimensions of k = dimension of Dimensions of L =

ML2 T 2 ML2 T 2 K 1 K 23. The potential difference V between two points is the amount of work done in moving a unit charge from one point to the other. Thus,

V=

work done charge moved

W q

1.14 Comprehensive Physics—JEE Advanced

ML2 T 2 = ML2 T –2 Q –1 Q = ML2 T –3 A –1 ( Q = AT) Hence the correct choice is (a). 24. From Ohm’s law, resistance R is given by potential difference R= current Dimensions of V =

ML2 T 3 A 1 A 2 – 3 –2 = ML T A Thus the correct choice is (b). 25. Force F q in an electric E is given by F F = qE or E = q dimensions of F MLT 2 Dimensions of E = dimensions of Q AT Dimensions of R =

= MLT–3A–1.

26. The force F q moving with speed v perpendicular to the direction of a B is given by F F = qvB or B = qv MLT 2 Dimensions of B = = ML0 T –1Q–1 Q LT 1 = ML0T –2A–1 ( Q = AT) Hence the correct choice is (d) 27. linked with a circuit of area A B is given by = BA cos where vectors. Dimensions of = dimensions of BA ( cos is dimensionless) = ML0 T –2 A–1 L2 = ML2 T–2 A–1 Thus the correct choice is (a). 28. The self inductance L of a coil in which the current dI varies at a rate is given by dt dI e=–L dt where e is the e.m.f. induced in the coil. Now, the dimensions of e.m.f. are the same as those of potential difference, namely, ML2 T –3 A–1. e Now, L=– dI dt Dimensions of L dimensions of e = dimensions of I / dimensions of t

ML2 T 3 A 1 = ML2 T –2 A–2 A/T Thus the correct choice is (b). 29. When a capacitor of capacitance C is charged to a potential difference V, the charge Q on the capacitor plates is given by Q Q = CV or C = V dimensions of Q Dimensions of C = dimensions of V AT = 2 ML T 3 A 1 = M –1 L –2 T 4 A2 Hence the correct choice is (d). 30. Let (M) = V a F b E c Putting the dimensions of V, F and E, we have (M) = (LT –1 )a (MLT –2 )b (ML2T –2 )c =

M 1 = M b +c La+b+2c T –a–2b–2c

or

Equating the powers of dimensions, we have b+c=1 a + b + 2c = 0 – a – 2b – 2c = 0 which give a = – 2, b = 0 and c = 1. Therefore (M) = (V –2 F 0 E). Thus the correct choice is (d). 31. Let n l a b Yc Putting dimensions of all the quantities, we have (T –1) La (ML–3 ) b (ML–1 T –2)c Equating powers of M, L and T on both sides, we get b + c = 0, a – 3b – c = 0 and – 2c = – 1 1 1 and c = . Thus which give a = – 1, b = – 2 2 n l –1 1/2 Y 1/2 Hence the correct choice is (a). 32. We know that q1 q2

F= 0

Hence

1 2

0

=

E2 =

4

and E =

F q

q1 q2

4 F r2 q1 q2

F2

F r2

q2

q1 q2

F

= Dimensions of

r

0

2

1 2

q 0

2

r2

E 2 = dimensions of

F r2

=

Units and Dimensions 1.15

M LT

2

41. The force a velocity

= ML–1T –2

L2 Hence the correct choice is (c).

= q (E +

33. Energy per unit volume, force per unit area and product of voltage and charge density all have dimensions of ML2T–2, but the dimensions of angular momentum are ML2T–1. Hence the correct choice is (d). 34. Given t = da/2 rb/2 sc/2. Substituting dimensions, we have (T) = (ML–3)a/2 (L)b/2 (MT–2)c/2 = M(a + c)/2 · L(–3a/2 + b/2) T–c 3 a Equating powers of L, we have, 2 Given a = 1.

on a particle of charge q moving with in E and B

b = 0. 2

3 2

b = 0 or b = 3, which is choice(c). 2 A2 B 35. Given X = 1/ 3 3 C D Taking logarithm of both sides, we have 1 log X = 2 log A + log B – log C – 3 log D 3 Partially differentiating, we have A B 1 C D x = 2 3 A B 3 C D x A Percentage error in A = 2 = 2 2% A = 4% B Percentage error in B = = 2% B 1 C 1 = 4% Percentage error in C = 3 C 3 4 = % 3 D Percentage error in D = 3 = 3 5% D = 15% contributed by C. Hence the correct choice is (c). 36. The correct choice is (b) 37. The correct choice is (b) 38. Dimensions of J and G are ML2T–1 and M–1L3 T–2 respectively. 39. Dimensions of o and h are M–1L–3 T4 A2 and ML2T–1 respectively. 40. RC has the dimensions of time (T). V has the dimensions of emf which has the dimensions of dI . L dt

B)

Hence the dimensions of E are the same as those of vB. 42. Energy stored in a capacitor of capacitance C having a potential difference V between its plates is given by 1 U = CV2 2 Hence, the dimensions of CV 2 = dimensions of energy. Hence the correct choice is (a). (ML2 T 1 ) = ML2 T–2 A–1 AT

h e

43. Dimensions of

Dimensions of B = MT–2 A–1 B area 44. Let surface tension = Ea Vb Tc. Using the dimensions of , E, V and T and equating powers of M, a, b and c. The correct choice is (c). 45. The correct choice is (a). 46. Let n1 be the magnitude (i.e. numerical value) of a physical quantity when the fundamental units are (M1, L1, T1) and n2 the magnitude of the same physical quantity when the fundamental units are (M2, L2, T2), then, it is obvious that n1 (M1x L1y T1z ) = n2 (M 2x Ly2 T2z )

(i)

where x, y and z are the dimensions of the physical quantity in mass, length and time respectively. Now, we know that 1 calorie = 4.2 joule = 4.2 kg m2s–2 n1 = 4.2, x = 1, y = 2 and z = – 2. Hence, in the second system of units in which M2 = a kg, L2 = b m and T2 = c s, we have from (i) M1 M2

x

n2 = n1

L1 L2

1 kg a kg

x

= n1

1m bm

= 4.2

1 a

= 47. V = have

1

1 b

2

y

T1 T2 y

1 c

1s cs

z

z

2

4.2c 2 ab 2

4 3 r . Taking logarithm of both sides, we 3 log V = log 4 + log

+ 3 log r – log 3

1.16 Comprehensive Physics—JEE Advanced

Differentiating, we get V r =3 = 3 × 1% = 3% V r V r r r 48. =3 or 6% = 3 or = 2%. V r r r Now surface area s = 4 r2 or log s = log 4 + 2 log r s r =2 = 2 × 2% = 4%. s r 49. The dimensions of moment of inertia are (ML2). We have n1(u1) = n2(u2) or

n1(M1 L21 ) = n2 (M 2 L22 )

2

n1 (M1L22 )

M1 L1 n2 = = n1 2 M2 L2 (M 2 L 2 ) Given n1 = 6.0, M1 = 1 kg, L1 = 1 m, M2 = 10 g and L2 = 5 cm. Therefore, 2 1 kg 1m n2 = 6.0 × × 10 g 5 cm 2

1000 g 100 cm = 6.0 × × 10 g 5 cm = 6.0 × 100 × (20)2 = 2.4 × 105

50. The capacitance of a parallel plate capacitor is given by C = 0A/d. Hence the dimensions of 0L are the same as those of capacitance. V Dimension of 0L t dimension of C dimensions of V time dimension of Q = ( Q = CV) time Hence the correct choice is (d). =

51. The correct choice is (a). relation

is given by the

R l P Q R l P Q It is clear that the error in the measurement of R is = 4

of R4 in the formula. Hence the radius (R) of the capillary tube must be measured most accurately. Thus the quantity which is raised to the highest power needs the most accurate measurement. 52. Take m 53. Take v 54. Take T

v a d b g c and show that a = 6.

1 . 2 3 r a Mb G c and show that a = – . 2 a

b

c

and show that a = –

55. The correct choice is (d). 56. at/m is dimensionless. Therefore, dimension of m M = Dimension of a = dimension of t T = M L0 T–1 57. The proportionate error in the measurement of g is T g L = + 2 T g L Hence g will be minimum if L and T are minimum. Thus the correct choice is (d). FL 4MgL = (1) 58. Y= Al d 2l where –2 M g L l 10–3 m l = ± 0.05 mm, d = 0.4 mm = 0.4 10–3 m d = ± 0.01 mm Substituting the values of M, g, L, d and l in Eq. (1) we get Y = 2.0 1011 Nm–2 From Eq. (1) the proportionate uncertainty in Y is given by Y M = Y M

g g

L L

2 d d

Since the values of M, g and L g = 0 and L = 0. Hence Y 2 d = Y d

l l

=

2 0.01 mm 0.4 mm

l l M = 0, 0.05 mm . mm

= 0.05 + 0.0625 = 0.1125 Y = 0.1125 Y = 0.1125 2.0 1011 = 0.225 1011 Nm–2 Since the value of Y decimal place, the value of Y must be rounded Y = 0.2 1011 –2 Nm Y + Y = (2.0 0.2) 1011 Nm–2 Hence the correct choice is (b). 59. Vernier constant = value of 1 M.S.D – value of 1 V.S.D. Now n V.S.D = (n – 1) M.S.D = (n – 1) x cm n 1 x cm 1 V.S.D = n n 1 x V.C. = x cm – x cm = cm n n Hence the correct choice is (c).

Units and Dimensions 1.17

60. T = 2

l g

g=

4

2

T2

l

Thus the percentage error is minimum for student I. 61. Least count of screw gauge

Therefore,

=

2 T T g For student I, EI = 100 g g l = g l

pitch number of divisions on circular scale

0.5 mm = 0.01 mm 50 diameter of ball = 2.5 mm + 20 = 2.7 mm M Density = 4 3 r 3 is =

0.1 2 0.1 100 64.0 .0 5 = % 16 0.1 2 0.1 For student II, EII = 100 64 64.0 15 = % 32 0.1 2 0.1 For student III, EIII = 100 20.0 36.0 19 = % =

=

M M

0.01 mm

3 r r

0.01 100 2.7 = 2% + 1.1% = 3.1%

= 2% + 3

II Multiple Choice Questions with One or More Choices Correct 1. Which of the following are not a unit of time? (a) parsec (b) light year (c) micron (d) second 2. Choose the pair of physical quantities which have identical dimensions. (a) Impulse and linear momentum (b) Planck’s constant and angular momentum (c) Moment of inertia and moment of force (d) Young’s modulus and pressure 3. The dimensions of energy per unit volume are the same as those of (a) work (b) stress (c) pressure (d) modulus of elasticity 4. When a wave traverses a medium, the displacement of a particle located at x at time t is given by y = a sin (bt – cx) where a, b and c are constants of the wave. Which of the following are dimensionless quantities? (a) y/a (b) bt b (c) cx (d) c

5. In Q.4, the dimensions of b are the same as those of (a) wave velocity (b) wave frequency (c) particle frequency (d) wavelength b 6. In Q.4, the dimensions of are the same as those c of (a) wave velocity (b) angular frequency (c) particle velocity (d) wave frequency 7. The Van der Waal equation for n moles of a real gas is a P (V – b) = nRT V2 where P is the pressure, V is the volume, T is the absolute temperature, R is the molar gas constant and a, b are Van der Waal constants. The dimensions of (a) a are the same as those of PV2 (b) b are the same as those of V a (c) are the same as those of RT V (d) bP are the same as those of RT.

1.18 Comprehensive Physics—JEE Advanced

8. In Q.7, which of the following have the same dimensions as those of PV? a (a) nRT (b) V ab (c) Pb (d) V2 9. In Q.7, the dimensions of nRT are the same as those of (a) pressure (b) energy (c) work (d) force 10. Which of the following are dimensionless? (a) Boltzmann constant (b) Planck’s constant (c) Poisson’s ratio (d) relative density 11. For a body in uniformly accelerated motion, the distance x of the body from a reference point at time t is given by x = at + bt2 + c where a, b and c are constants of motion. (a) The dimensions of c are the same as those of x, at and bt2. (b) The dimensional formula of b is [M0 LT–2]. a (c) is dimensionless. b (d) The acceleration of the body is 2b. 12. The side of a cube is L = (1.2 ± 0.1) cm. The volume of the cube is 3 (b) (1.73 ± 0.02) cm3

0.3) 0.2) (c) The minimum resistance obtainable is (2.0 ± 0.3) (d) The minimum resistance obtainable is (2.0 ± 0.2) 14. A physical quantity P is given by P=

a 3b 2

d c The percentage errors in the measurements of a, b, c, and d are 1%, 3%, 4%, and 3% respectvely. P is 14% P is 10% measurement of b. (d) The minimum error is contributed by the measurement of c. 15. When a plane wave travels in a meduim, the diplacement y of a particle located at x at time t is given by y = a sin (bt – cx)

and R2 = (6.0 ±

where a, b, and c are constants. (a) The unit a is the same as that of y. (b) The SI unit of b is Hz. (c) The dimensional formule of c is [M0L–1T0] b are the same those of (d) The dimensions of c velocity.

1. Choices (a), (b) and (c) are units of length 2. The dimensions of moment of inertia are ML2T 0 and of moment of force are ML2T–2. All other pairs in (a), (b) and (d) have identical dimensions. 3. Dimensions of energy per unit volume are = dimensions of energy / dimensions of volume = ML2T –2 / L3 = ML–1T –2. Stress, pressure and modulus of elasticity all have the dimensions of ML–1T–2. The dimensions of work are ML2T–2. Hence the correct choices are (b), (c) and (d). 4. Since the sine function is dimensionless, sin (bt – cx) is dimensionless. Therefore, y and a must have the same dimensions, i.e. y/a is dimensionless. Since the argument of a sine function (or any trigonometric function) must be dimensionless, bt

and cx are also dimensionless. Hence the correct choice are (a), (b) and (c). 5. Since bt is dimensionless, the dimensions of b = dimensions of 1/t = T–1, which are the dimensions of angular frequency as well as wave frequency. Hence the correct choices are (b) and (d). 6. Dimensions of bt = dimensions of cx. Therefore b x = dimensions of = LT –1. Dimensions of c t Hence the correct choices are (a) and (c). 7. a ab = n RT PV – Pb + V V2 From the principle of homogeneity, it follows that all the four choices are correct.

(c) (1.7 ± 0.4) cm3

(d) (1.7 ± 0.3) cm3

13. Two resistances R1 = (3.0 ± 0.1) 0.2) are to be joined together. ANSWERS AND SOLUTIONS

Units and Dimensions 1.19

8. The correct choices are (a), (b), (c) and (d). 9. The dimensions of nRT = dimensions of PV = ML–1 T –2 L3 = ML2 T–2 which are dimensions of energy as well as work. Hence the correct choices are (b) and (c). 10. The correct choices are (c) and (d). 11. From the principle of homogeneity of dimensions, the dimensions of c must be the same as those of x at and bt2. Therefore, choice (a) is correct. Also dimension of bt2 = dimension of x. Hence [b]= [LT–2]. Hence choice (b) is also correct. Velocity of the body is dx d = [at + bt2 + c] = a + 2bt v= dt dt and acceleration is dv d = (a + 2 bt) = 2b, which is choice (d) dt dt a has dimension of choice (c) is wrong since b time [T] 12. Volume of cube (V = L3) = 1.2 cm 1.2 cm 3 . Now L = (1.2 ± 0.1) cm has V must

Rs ±

Rs = (9.0 ± 0.3)

Thus choice (a) is correct and choice (b) is wrong. The minimum value is obtained when the resistances are joined in parallel. 1 1 1 = Rp R1 R2 R1 R2 3.0 6.0 = = 2.0 R1 R2 3.0 6.0 RR Rp = 1 2 ( R 1 + R 2 = R s) Rs Rp Rs R1 R2 = R1 R2 Rs Rp Rp =

Now

0.1 0.2 0.3 3.0 6.0 9.0 = 0.033 + 0.033 + 0.033 = 0.099 0.1 Rp = 0.1 Rp = 0.1 2 = 0.2 =

Minimum value is (Rp ± Rp) = (2.0 ± 0.2)

.

Hence choice (c) is wrong and choice (d) is is correct. 1 14. log P = 3 log a + 2 log b – log d – log c 2 P 3 a 2 b d 1 c = P ma a b d 2 c 1 = 3 1% + 2 3% + 3% + % 2 = 3% + 6% + 3% + 2% = 14% Hence the correct choices are (a), (c) and (d). 15. The value of any trignomatric function is a dimensionless number. Hence choice (a) is correct. The argument of a trignometric function is also dimensionless. Hence (bt – cx) is dimensionless. Hence b has dimension [T–1] the same as that of frequency and c has dimension of [L–1]. Thus choices (b), (c) and (d) are all correct.

value of V = 1.7 cm3. Now V = L3 V 3 L 3 0.1 = = = 0.25 V L 1.2 V = 0.25 V = 0.25 1.7 cm3 = 0.425 cm3 The error in V is in the value of V should be rounded off as V = 0.4 cm3. Thus the correct result is V ± V = (1.7 ± 0.4) cm3, which is choice (c). 13. value is Rs= R1+ R2= 3.0 + 6.0 = 9.0 Error in Rs= Rs= R1+ R2= 0.2 + 0.1 = 0.3

III Multiple Choice Questions Based on Passage Passage-I The dimensional method is a very convenient way of

physical quantities of a given system. This method has its own limitations. In a complicated situation, it is often not easy to guess the factors on which a physical quantity will depend. Secondly, this method gives no information about the dimensionless proportionality constant. Thirdly, this

1.20 Comprehensive Physics—JEE Advanced

method is used only if a physical quantity depends on the product of other physical quantities. Fourthly, this method will not work if a physical quantity depends only on Finally, this method does not give complete information in cases where a physical quantity depends on more than three quantities in problems in mechanics. 1. The dimensional method cannot be used to obtain denpendence of (a) the height to which a liquid rises in a capillary tube on the angle of contact (b) speed of sound in an elastic medium on the modulus of elactricity. (c) height to which a body, projected upwards with a certain velocity, will rise on time t. (d) the decrease in energy of a damped oscillator on time t. 2. In dimensional method, the dimensionless proportionality constant is to be determined (b) by a detailed mathematical derivation (c) by using the principle of dimensional homogeneity. (d) by equating the powers of M, L and T.

least count of the measuring instrument used. The number degree of precision of that measurement. The importance calculation cannot increase the precision of a physical in the sum or product of a group of numbers cannot be greater than the number that has the least number of weakest link. 3. 0.0123 kg. What is the total mass supported by the ures? (c) 0.0124 kg

(d) 0.012 kg

4. The radius of a uniform wire is r = 0.021 cm. The value is given to be 3.142. What is the area of (a) 0.0014 cm2

2

(b) 0.00139 cm2

2

5. A man runs 100.5 m in 10.3 s. Find his average Passage-II In the study of physics, we often have to measure the physical quantities. The numerical value of a measured

(a) 9.71 ms–1

–1 –1

–1

ANSWERS AND SOLUTIONS

1. The correct choices are (a), (c) and (d). The height of a liquid in a capillary tube depends on cos , where is the angle of contact. The height 1 S to which a body rises is given by S = ut + at2, 2 1 which is a sum of two terms ut and at2. The 2 tially with time. 2. The correct choices are (a) and (b). 3. curate only upto the fourth decimal place. Hence

the sum must be rounded off to the fourth decimal place. Therefore the correct choice is (c). 2 . Now, 4. A = r2 = 3.142 (0.021)2 A = 0.0014 cm2, which is choice (a). 100.5 m –1 5. Average speed = 10.3 s the time has only three. Hence the result must 9.71 ms–1. Thus the correct choice is (a).

Units and Dimensions 1.21

IV Matrix Match Type 1. Match the physical quantities in column I with their SI units in column II Column I Column II (a) (b) (c) (d)

Stefan’s constant Universal gas constant Electrical permittivity Magnetic permeability

(p) (q) (r) (s)

JK–1 mol–1 Fm–1 Hm–1 Wm–2 K–4

ANSWER 1. (a) (c)

(s) (q)

(b) (d)

(p) (r)

2. Column I (b) 0.034 (c) 0.002504 (d) 1.25 1.07

Column II (q) 4 (r) 2 (s) 5

ANSWER 2. (a) (c)

(s) (q)

(b) (d)

(r) (p)

3. Match the quantities in column I with their order of magnitude given in column II Column I Column II (a) 2.6 (b) 3.9 (d) 4.2

104 104 10–24 10–24

(p) (q) (r) (s)

105 10–23 10–24 104

ANSWER 3.

x = 4.3 103, take its logarithm to the base 10. Log x = 3.633 and round if off as log x = 4 . So the order of magnitnde of x is 104. (a) (c)

4.

(s) (r)

(Q) given in column II. Column I (a) (b) (c) (d)

Angular momentum Latent heat Torque Capacitance

(b) (d)

(p) (q)

Column II (p) (q) (r) (s)

M L–2 T–2 M L2 Q–2 M L2 T–1 M L3 T–1 Q–2

1.22 Comprehensive Physics—JEE Advanced

(t) M–1 L–2 T2 Q2 (u) M0 L2 T–1

(e) Inductance (f) Resistivity

ANSWER 4. (a) (r) (c) (p) (e) (q)

(b) (d) (f)

(u) (t) (s)

SOLUTION (a) Angular momentum L = r p = r (mv) [L] = [L M L T–1] = [M L2 T–1] Q heat energy [M L2 T 2 ] = = [M0 L2 T–2] (b) Latent heat = m mass [M] (c) Torque

= r

[ ]= [L (d) Q = CV [C] = (e) |V | =

=

M L T–2] = [M L2 T–2] C= Q2

RA L

Q V

[M L2 T 2 ]

LdI dt

[L] = (f)

F

W Q

Q W /Q

Q2 W

= [M–1 L–2 T2 Q2] LQ T2

WT 2

[M L2 T 2 ] [T 2 ]

Q2

Q2

VA IL

WA QIL

= [M L2 Q–2]

[M L2 T 2 L2 ] = M L3 T–1 Q–2 Q Q L T

5. Column I gives three physical quantities. Select the appropriate units given in Column II. Column I Column II (a) Capacitance

(p) Ohm second

(b) Inductance

(q) (coulomb)2 (joule)–1 (r) coulomb (volt)–1 (s) newton (ampere metre)–1

(c) Magnetic induction or

–1

IIT, 1990

SOLUTION 5. (a) From Q = CV, unit of C = From U =

unit of Q = coulomb (volt)–1 unit of V

Q2 , unit of C = (coulomb)2 (volt)–1 2C

Units and Dimensions 1.23

(b) From |e| = L

dI , unit of L = volt second (ampere)–1 = ohm second dt

(c) From F = BIL, unit of B = newton (ampere metre)–1 Hence (a)

(q), (r)

(b)

(p), (t)

(c) (s) 6. Some physical quantities are given in Column I and some possible SI units in which these quantities may be indicate your answer by darkening appropriate bubbles in the 4 Column I Column II (p) (volt) (coulomb) (metre)

(a) GMeMs G – universal gravitational constant, Me – mass of the earth, Ms – mass of the Sun (b)

(c)

(d)

3RT M R – universal gas constant, T – absolute temperature, M – molar mass

(q) (kilogram) (metre)3 (second)–2

F2

(r) (metre)2 (second)–2

q2 B2 F – force, q – charge, B GM e Re

(s) (farad) (volt)2 (kg)–1

G – universal gravitational constant Me – mass of the earth, Re – radius of the earth IIT, 2007

SOLUTION 6. (a) F =

Gm1m2 . Therefore the SI unit of G is Nm2 kg–2. r SI unit of GMeMs = (Nm2kg–2) kg2 = Nm2 = kg ms–2

(b) SI unit of

3RT SI unit of PV SI unit of work = = = Nm kg–1 = kg ms–2 M SI unit of M kg

(c) From F = q v B (d) SI unit of

m2 = kg m3 s–2.

GM e Nm 2 kg 2 = Re m

kg

=

kgms

2

m

kg–1 = m2 s–2

F F2 = SI unit of v. Hence SI unit of 2 2 = (ms–1)2 = m2 s–2 qB q B m 2 kg 2 kg = m2 s–2 m

(p) Since volt coulomb = work, SI unit of (volt) (coulomb) (metre) = SI unit of work = Nm2 = kg ms–2 m2 = kg m3s–2

metre = Nm

coulomb and coulomb volt = work, the SI unit of (farad)(volt)2 (kg–1) = (coulomb) volt kg–1 = SI unit of work kg–1 = kg ms–2 m kg–1 = m2s–2

(s) Since farad =

m

(volt)

1.24 Comprehensive Physics—JEE Advanced

Hence the correct choices are as follows (a) (p), (q) (b) (r), (s)

(c)

(r), (s)

(d)

(r), (s)

V Assertion-Reason Type Questions In the following questions, Statement-1(Assertion) is followed by Statement-2 (Reason). Each question has the following four options out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and State(b) Statement-1 is true, Statement-2 is true but Statement-2 is not ment-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 The order of accuracy of measurement depends on the least count of the measuring instrument.

Statement-2 The smaller the least count the greater is the num2. Statement-1 The dimensional method cannot be used to obtain the dependence of the work done by a force on the angle between force and displacement x. Statement-2 All trignometric functions are dimensionless. 3. Statement-1 The mass of an object is 13.2 kg. In this measureStatement-2

SOLUTIONS 1. The correct choice is (b). 2. Work done is W = F x cos . since cos is dimensionless, the dependence of W on cannot be determined by the dimensional method. Hence, the correct choice is (a)

3. The correct choice is (c). The degree of accuracy measurement cannot be increased by changing the unit.

2

Motion in One Dimension

Chapter

REVIEW OF BASIC CONCEPTS 2.1

SCALAR AND VECTOR QUANTITIES

A scalar quantity has only magnitude but no direction, such as distance, speed, mass, area, volume, time, work, energy, A vector quantity has both magnitude and direction, such as displacement, velocity, acceleration, force,

2.2

2.1 A particle moves from A to B along a circle of radius R placement in terms of R and

POSITION VECTOR AND DISPLACEMENT VECTOR

The position vector of a particle describes its instantaneous position with respect to the origin of the chosen frame of and is denoted by vector r For one-dimensional motion (say along x-axis), r = x i , where x is the distance of the particle from origin O For two-dimensional motion (say in the x–y plane), r = xi

Fig. 2.2

SOLUTION Path length = arc AB = R

y j , where (x, y) are the x and y coordinates of

For three-dimensional motion, r = x i

yj

zk

Displacement vector r1 is the position vector of a particle at time t1, and r2 at time t2, then the displacement vector is given by s = r2 – r1 Vector s is the resultant of vectors r2 and – r1 The displacement vector particle after a given interval

Fig. 2.3

Fig. 2.1

s = AB = AC + CB = 2 AC s = 2R sin ( /2)

2.2 Comprehensive Physics—JEE Advanced

2.3

The rate of change of displacement with time at a given instant is called instantaneous velocity and is given by dx v= dt vav =

total displacement time interval

2.2 The position of a particle moving along x-axis is given by x = 2t t2 + t , where x is in metre and t (a) Find the velocity of the particle at t (b) Find the average speed of the particle in the time interval from t = 2 s to t SOLUTION (a) x = 2t

t2 + t

dx = 2 – 6t dt v at (t = 2 s) = 2 – 6

t2

v=

(b) Position at t = 2 s is x1 = 2

(2)2 = 2 ms–1

(2)2 + (2) = 0

Position at t x2 = 2

Average velocity =

2

12 ms

1 2 at 2

and

1 2 at 2 v2 – u2 = 2 a(x – x0)

or

v2 – u2 = 2a s

or

s = ut +

in a straight horizontal direction, we will consider only the magnitudes of u, v, a and s and take care of their direction by assigning positive or negative sign a will mean acceleration – a Quantities directed vertically upwards are taken to be positive and those directed vertically downwards to gravity is directed downward for a body moving vertically up or falling vertically down, we take a –2 =–g nth second is given by sn = displacement in n seconds – displacement in (n – 1) seconds 1 1 a (n 1) 2 = un + a (n) 2 u (n 1) 2 2 sn = u +

a (2n 1) 2

Applications 1

INSTANTANEOUS ACCELERATION

The rate of change of velocity with time at a given instant is called instantaneous acceleration and is given by dv a= dt

2.5

x = x0 + ut +

2

x2 – x1

2.4

v = u + at

INSTANTANEOUS VELOCITY AND AVERAGE VELOCITY

EQUATIONS OF ONE DIMENSIONAL MOTION WITH CONSTANT ACCELERATION

Let x0 be the position of a particle at time t = 0 and let u be its velocity at t a for time t x and acquires a velocity v The particle suffers a displacement s = x – x0 in time t The equations of motion of the particle are

from A with initial velocity u and reaches B with a velocity v, then the velocity midway between A and B is u 2 v2 v = 2 (ii) A body starting from rest has an acceleration a for a time t1 and comes to rest under a retardation b for s1 and s2 are the distances travelled in a time t2 t1 and t2, (a)

s1 s2

b a

t1 t2

(b) Total distance travelled (s1 + s2) = where T = t1 + t2

1 ab T2, 2 a b

(c) Maximum velocity attained is vmax =

ab a

b

T

2.3

(d) Average velocity over the whole trip is v vav = max 2 (iii) At time t = 0 a body is thrown vertically upwards with a velocity u At time t = T, another body is thrown vertically upwards with the same velocity u T u t= 2 g (iv) A body is dropped from rest and at the same time another body is thrown downward with a velocity u from the same point, then (a) the acceleration of each body is g, (b) their relative velocity is always u, x (c) their separation will be x after a time t = u (v) From the top of a building, body A is thrown upwards with a certain speed, body B is thrown downwards with the same speed and body C is dropped t1, t2 and t are their respective times of reach the ground, then t =

t1 t2

(vi) A body of mass m is dropped from a height h on a x (a) the average retardation in sand is given by gh a= x because loss in PE (mgh) = work done against the resistive force of sand (max (b) total average force exerted by sand is F = mg + ma = m(g + a)

(f) the speed with which the body hits the ground is 2 g

v=

a h

u

g g

a a

NOTE Note that t1 is less than t2

Some useful tips …… are in the ratio 12 : 22

2

acceleration, the distance covered by it in the lst, air resistance is nelected, the speed with which it u, the maximum height attained is proportional to u2 and the time of ascent is proportional to u (v) For a freely falling body, (a) velocity time (b) distance fallen (time)2 (c) velocity distance fallen

2.6

GRAPHICAL REPRESENTATION Displacement – time (x – t) graphs (Fig. 2.4)

(vii) A body is thrown vertically upward with a velocity u a constant acceleration (or retardation) a (< g) (a) the net acceleration during upward motion = g + a, (b) the net acceleration during downward motion = g – a, (c) the maximum height attained is

ating

a > 0) a < 0) -

u2 2 g a (d) the time taken to reach the maximum height is h=

t1 =

2h g

u a

g

Fig. 2.4

a

NOTE

(e) the time of descent is t2 =

2h g

The slope of x – t graph gives velocity for uniform mo-

u a

g2

a2

1/ 2

2.4 Comprehensive Physics—JEE Advanced

v – t) graphs for uniformly acceler-

2.3 The displacement x (in metres) of a body varies with time t (in seconds) as 2 2 t 16t 2 x = How long does the body take to come to rest?

O

t a

O

o

u

o a

t o

O

t u

o a

o

SOLUTION v=

O

t

–u u

o a

O

o

t u

o a

O

O

u

t

–u u

o a

o

Fig. 2.5

0=

t

o

o a

o

Acceleration = slope of (v – t) graph v – t) graph

t 16

t 16

t = 12 s

2.4 vertically upwards with a velocity of 10 ms –1 (a) After how long will the ball hit the ground? (b) After how long will the ball pass through the point c will it hit the ground? Take g = 10 ms–2 SOLUTION u = + 10 ms –1, a = – 10 ms–2

(a) s

NOTE

dx = dt

Now

s = ut + t+

a–t

1 2 at 2 1 2

(– 10)t2

t2 – 2t (t + 2) (t t

Fig. 2.6

a – t), (v – t), (x – t

t is not

(b) s = 0, u = + 10 ms –1 and a = – 10 ms –2 0 = 10t t2 t=2s –1 (c) v = u + at = 10 – 10 The negative sign indicates the velocity v is di2.5

Fig. 2.7

elled by the body in 20 s, (b) the displacement of the body in 20 s and (c) the average velocity in the time interval t = 0 to t

2.5

S = u

2.7

29

29a 2

2 2

m

RELATIVE VELOCITY IN ONE DIMENSION

A and B are moving in a straight line with velocities vA and vB respectively, the relative velocity of A with respect to B vAB = vA – vB The relative velocity of B with respect to A will be vBA = vB – vA

Fig. 2.8

SOLUTION (a

2.7 A police van moving on a highway with a speed of 10 ms –1 –1

–1

Speed (ms–1) 6

, with what speed will

SOLUTION 5

10

15

20

( )

–1

Fig. 2.9

= area of OAB +

area of BCD

car is vC –1 respect to van is vBV vBV = vB – vV –1

vB be the velocvB = vBV + vV =

a speed =

1 2

(b

6 10

1 2

6 10

OAB – area of

(c) Average velocity =

displacement time

0

2.6 A body moving in a straight line covers a distance of

2.8 From the top of a tower 60 m tall, a body is thrown vertically down with a velocity of 10 ms –1 same time, another body is thrown vertically upward from the ground with a velocity of 20 ms –1 how long will the two bodies meet? (b) At what height above the ground do they meet? Take g = 10 ms –2

SOLUTION Sn = u S = u S = u

a 2n 1 2 9a 2 a 2 a = 2 ms –2

–1

vBC = vB – vC

60 m

BCD

–1

vV = 10 ms–1

Fig. 2.10

SOLUTION (a (1) (2) and

u=

C and let t be the

For body 1: s = – h1 , u1 = – 10 ms –1, a = – 10 ms –2 1 – h1 = – 10t + 10 t 2 2 h1 t (t + 2) (1)

2.6 Comprehensive Physics—JEE Advanced

For body 2: s = + h2, u2 = + 20 ms –1, a = – 10 ms –2 h2 = 20t t2 t (– t t or 60 = Adding (1) and (2), h1 + h2 t t (b) Using t h2 = 20 m Alternative method –1

u12 = u1 – u2

v

and obtain the expression for v(t x, use v = dx = v(t) dt

1 2 at 2

t

t h2 = 20t

t 2 = 20

(2)2 = 20 m

dx which gives dt

v and integrate both

sides t

x

dx =

a12 = a1 – a2 = – g – (– g s = ut

a dt 0

u

v dt 0

x0

s12 = – h1 – (h2) = – (h1 + h2) = – 60 m Using

t

dv =

2.10 A particle starts from rest at x –2 which varies with time as shown time t ticle and (b) its displacement in time interval from t = 0 to t –2

2.9 –1 The driver of train A sights another train B moving on the same track at a speed of 10 ms –1 ately applies brakes and achieves a uniform retardation of 2 ms –2 minimum distance between trains A and B when the driver of A sights B?

5ms

6s

Fig. 2.11

SOLUTION (a)

SOLUTION B is,

uAB

A = uA – uB

aAB

A B is = aA – aB = – 2 – 0 = – 2 ms –2

–1

m =

–2 c y = mx + c, the acceleration a varies with time as

a= dv = dt

A B vAB of A B is sAB which is found by using the relation v2 – u2 = 2 as which gives sAB = 100 m 0 – (20)2 = 2 (– 2) sAB

2.8

SOLVING PROBLEMS INVOLVING NONUNIFORM ACCELERATION

(a) Finding velocity and displacement if the dependence of acceleration on time is given. dv Use a = which gives dv = a dt dt given expression for a in terms of t and integrate

2

ms

6

v

6

t

6

t

t

dv =

0

6

0

v= zero at t Using t

12

t

dt

t2

(1)

t

v will be maximum at t –1 vmax

(b) From v =

dx , we have dx = v dt dt

x

t

dx = 0

v dt = 0

2 0

12

t2

t dt

2.7

t

x=

t 2

12

2

a=

0

2 2 = m 12 2 9 (c) Finding velocity and displacement if the dependence of acceleration on displacement is given dv dx d v dv Use a(x) = = =v dt dt dx dx dx v dt or v dv = a(x) dx

v

=

a(x) in terms

of x v

v dv =

t

dt 0

t

x

dx =

v dt 0

x0

Hence we get an expression for x 2.12

a x dx

dx = v (x)

ity u ates it at a rate a = neous velocity and k

t = 0) moving with a velock v where v is the instanta-

SOLUTION

t

dt (a) a =

0

where v 2.11 A particle is moving along the x-axis with an acceleration a = 2x where a is in ms –2 and x the particle starts from rest at x when it reaches the position x

dv dt

k v = v

dv v

u 1/ 2

2 v

1/ 2

u

dv dx v 0

2x = v

dv dx

v dv = 2x dx

v d v = 2 x dx

dv dt

2 1

–1 which gives v (c) Finding velocity and displacement if the dependence of acceleration on velocity is given

v

= – k dt

= – k dt 0

= – kt

Putting v = 0, we get t =

2 u k

dv dv = v dx dt dv k v = v dx = dx a=

1

v2 x2 = 2 2 2

dv

t

SOLUTION a =v

dt

Hence we obtain the expression for v(t dx v= dx = vdt dt

x0

Hence we get an expression for v(x) in terms of x dx v(x) = dt dx dt and integrate v x

x0

dv a

x

v

x

dv = a u

dv dt

x

dx = 0

x=

0

1 k 2u

v dv u /

k

v dv k

2.8 Comprehensive Physics—JEE Advanced

I Multiple Choice Questions with Only One Choice Correct 1.

last second of its fall, the ball covers a distance g = 10 ms–2, the height of the tower is

2. A ball is thrown vertically upwards from the foot of

(c) 100 m (d) 120 m 3. Two cars travelling on a straight road cross a kilometer stone A at the same time with velocities 20 ms–1 and 10 ms–1 with constant accelerations of 1 ms–2 and 2 ms–2 kilometer stone B at the same instant, the distance between A and B is (c) 1000 m (d) 1200 m 4. The acceleration a of a body moving with initial velocity u changes with distance x as a = k2 x , where k elled by the body when its velocity becomes 2u is u 2k

(c)

u 2k

/

(b)

u 2k

(d)

u 2k

/

2px ( p q)

(b)

k2 (d) 2 k2 2 7. The velocity of a particle at time t (in second) is related to its displacement x (in metre) as v= x (a) 1 ms–1 (b) 2 ms–1 –1

8. A car, starting from rest, has a constant accelera–2 for some time and then has a constant retardation of 2 ms–2 for some time and The maximum velocity of car during its motion is –1 (a) 12 ms–1 –1 (d) 21 ms–1 9. A freely falling body, falling from a tower of height h covers a distance h/2 in the last second of its moThe height of the tower is (take g = 10 ms–2) nearly 10.

/

2q x ( p q)

(d) – q x

6. The velocity of a particle moving along the x-axis is given by v = k x where k The acceleration of the particle is

k 2

A ms–1 towards a bigger ball B collision with ball B, ball A retraces the path and –1

instant is

(c)

(b)

(c)

/

5. A particle is moving along the x neous velocity when it is at a distance x from the origin is v = p q x 2 , where p and q are positive

(a) zero

k 2

–1

Take g = 10 ms–2

(a)

(a)

time interval 0 to 6 s? (a) zero –1

11.

A during the (b) 2 ms–1 –1 –1

A monkey is running northwards on the roof of the –1

of the monkey as observed by a person standing on the ground? –1 in the southward direction –1 in the northward direction –1 in the southward direction –1 in the northword direction 12. A police van moving on a highway with a speed –1 ing away in the same direction with a speed of –1

2.9 –1

, with what speed will the bullet hit the

(a) 120 ms–1

19.

(x-t)

–1

–1

–1

the velocity-time (v-t) graph of the motion of the

–1

13. Car A

on a B and C, each moving with –1 in opposite directions on the other lane are approaching car A stant when the distance AB = distance AC = 1 km, the driver of car B decides to overtake A before C car B so as to avoid an accident? (b) 2 ms–2 (a) 1 ms–2 –2

–2

14. The driver of a train A ms–1 sights another train B moving on the same track towards his train at a speed of 10 ms–1 immediately applies brakes and achieves a uniform

Fig. 2.12

–2

what must be the minimum distance between the trains? (a) 100 m (b) 200 m 15.

g = 10 ms–2, the

16. A body, starting from rest, moves in a straight line with a constant acceleration a for a time interval t during which it travels a distance s1 move with the same acceleration for the next time interval t during which it travels a distance s2 relation between s1 and s2 is (a) s2 = s1 (b) s2 = 2s1 (c) s2 s1 (d) s2 s1 17. v1 is the velocity of the body at the end v2 that at the end of the second time interval, the relation between v1 and v2 is (a) v2 = v1 (b) v2 = 2v1 (c) v2 v1 (d) v2 v1 18. A body dropped from the top of a tower hits the the tower? (a) 1 s (c) 2 2 s

Fig. 2.13

20.

g = 10 ms–2, what is the ratio of the distances travmotion? (a) 1 : 9

21.

–1

g = 10 ms–2, what is the ratio of the distances travelled by the bullet during the (a) 9 : 1

(b) 2 s (d)

s

(b) 2 : 9

(b) 9 : 2

22. A body moving in a straight line with constant acceleration of 10 ms–2

2.10 Comprehensive Physics—JEE Advanced th

in the 6 second? 23. A body, moving in a straight line with an initial –1 and a constant acceleration, covrd

distance will it cover in the next 2 seconds? (c) 90 m (d) 100 m 24. A body, moving in a straight line, with an initial velocity u and a constant acceleration a, covers a th second and a distance u and a of 60 m in the 6th respectively are –2 (b) 10 ms–1, 10 ms–2 (a) 10 ms–1 –1 –2 –1 , 10 ms–2 25. A car, starting from rest, has a constant acceleration a1 for a time interval t1 during which it covt2, the ers a distance s1 car has a constant retardation a2 and comes to rest after covering a distance s2 in time t2 following relations is correct? (a)

a1 a2

s1 s2

t1 t2

(b)

a1 a2

s2 s1

t1 t2

(c)

a1 a2

s1 s2

t2 t1

(d)

a1 a2

s2 s1

t2 t1

(a)

1 1 a1t1 = a2 t2 2 2

(b)

1 (a1 t1 2

27.

1

a1

a2 t1

t2

(a)

(c)

s a1a2 2 a1 a2

1 2

1 2

g = 10 ms–2

a2 t2 )

(d)

aa 2s 1 2 a1 a2 s a1a2 2 a1 a2

35.

(b)

v 2

–1

h1 during

s, 1 2

1 2

28. A car, starting from rest, is accelerated at a constant rate until it attains a speed v a constant rate (a) zero

(b) na

na

(c) n a (d) n a 34. Two balls are dropped from the same point after an

(d) zero

(b)

(b) s2 = 2s1 (a) s2 = s1 (c) s2 s1 (d) s2 s1 33. A car moving at a speed v is stopped in a certain distance when the brakes produce a deceleration a nv, what must be the deceleration of the car to stop it in the same distance and in the same time? 2

the maximum speed attained by it will be aa 2s 1 2 a1 a2

(d) v 2 2 29. The displacement of a body from a reference point, is given by x = 2t where x is in metres and t that the body is (a) at rest (b) accelerated (c) decelerated (d) in uniform motion 30. –1 (a) 2 ms–1 –1 (c) 6 ms (d) 12 ms–1 31. –2 (a) 2 ms–2 –2 –2 (c) 6 ms 32. A car, starting from rest, at a constant acceleration covers a distance s1 in a time interval t tance of s2 in the next time interval t at the same ac-

(a)

26.

(c)

v

(c)

th

h2 and h2 will be related as h2 (a) h1 (c) h1 = h2

g = 10 ms–2, h1

(b) h1 = 2h2 h2 (d) h1 =

36. A ball is thrown vertically downward with a velocity u u reach the ground is given by u 2u (b) (a) g g (c)

u g

(d)

u g

2.11

37. (a)

u2 g

(c)

u2 g

(b)

2u 2 g

(d)

u2 g -

38. –1 –1

train? 39.

and leaves it at a

–2

–2

–2

–2

platform?

(c) 10 s (d) 12 s 40. The motion of a body is given by the equation dV t dt

41.

V(t)

where V(t) is the speed (in ms–1) at time t (in sect = 0, the magnitude of the initial acceleration is –2 (b) 6 ms–2 –2 (c) 9 ms (d) zero (a) V(t) = (1 – e t ) (b) V(t) = 2 (1 – e t) (c) V(t) =

2

t

1 e

2

2t

(d) V(t) = 42.

2

1 e

tion is half the initial value is (b) 2 ms–1 (a) 1 ms–1 –1

-

–1

43. Two stones are thrown up simultaneously with initial speeds of u1 and u2 (u2 > u1 variation of

x = (x2 – x1), the relative position

t = 10 s? Assume that the stones do not rebound

Fig. 2.14

44. A body starts from rest at time t = 0 and undergoes

2.12 Comprehensive Physics—JEE Advanced

45. Acc (ms 2)

+3 0

1

2

3

3

4 Time (s)

46.

Fig. 2.15

the velocity-time (v–t) graph of the motion of the body from t = 0 s to t

47.

time t

–1

–1

–1

–1

from t = 0 to t (a) 6 m

(b) 9 m

represents the displacement-time (x – t) graph of the motion of the body from t = 0 s to t

Fig. 2.17

48. A body, moving in a straight line, covers half the distance with a speed V, the remaining part of the distance was covered with a speed V for half the time and with a speed V for the other half of the (a)

2V V V 2V V V

(b)

V V V 2V V V

(c)

2V V V V V

(d)

V V V V V

49. A particle moving in a straight line covers half the the distance is covered in two equal time interFig. 2.16

motion is

2.13

motion and air resistance, its velocity v varies with the height h v – t) graphs

50. (a) (b) (c) (d)

(i), (ii) and (iv) only (i), (ii) and (iii) only (ii) and (iv) only all

Fig. 2.19

54. A car, starting from rest, accelerates at a constant –2

constant rate of 10 ms–2 Fig. 2.18

51. A stone dropped from a building of height h reaches the ground after t ing if two stones are thrown (one upwards and the other downwards) with the same velocity u and they reach the ground after t1 and t2 seconds respectively, then the time interval t is

52.

(a) t = t1 – t2

(b) t =

(c) t =

(d) t =

t1 t2

t1

2 t12

t2 t22

x) of a particle is related to time (t) as x = at + bt2 – ct where a, b and c velocity of the particle when its acceleration is zero is given by 2 (a) a + b c

(b) a +

b2 2c

b2 c

(d) a +

b2 c

(c) a +

53. A ball is dropped vertically from a height h above vertically to a height h

attained by the car? –1

55.

(c) 20 ms

–1

(b) 10 ms–1 –1

car is 6 s?

(c) 100 m (d) 120 m 56. The distance x covered by a body moving in a straight line in time t is given by the relation 2x2 x=t v is the velocity of the body at a certain instant of time, its acceleration will be (a) – v (b) – 2v v v 57. The distance x covered by a body moving in a straight line in time t is given by x2 = t2 + 2t The acceleration of the body will vary as 1 1 (a) (b) 2 x x (c)

1 x

(d)

1 x

58. A body is thrown vertically up with a velocity u passes three points A, B and C

2.14 Comprehensive Physics—JEE Advanced

u u u with velocities , and 2 AB is BC 20 (a) (b) 2 (c)

10

average velocity of the particle is –1

–1 –1

(d) zero

(d) 1

59. A body is thrown vertically up with a velocity u passes a point at a height h above the ground at time t1 while going up and at time t2 Then the relation between u, t1 and t2 is 2u 2u (a) t1 + t2 = (b) t2 – t1 = g g (c) t1 + t2 =

u g

(d) t2 – t1 =

(a) t1t2 =

2h g

(c) (t1 + t2)2 =

(b) t1t2 = 2h g

h g

(d) (t1 + t2)2 =

t= x where x is in metres and t ment of the particle when its velocity is zero is 67.

the maximum time to reach the ground if H H (b) h = (a) h = 2 2 H H (d) h = 2 2 62. A body of density enters a tank of water of density after falling through a height h depth to which it sinks in water is (a) (c)

(

)

h

(b) (d)

h

(a) T =

2 u k

(b) T =

2u k

(c) T =

2u / k

(d) T =

2u 2 k

68.

–2

t

(

)

h

63. A body, falling freely under gravity, covers half the

will be

which varies with time as shown

ms–1 ms–1

ms–1 ms–1

for n seconds, then the value of n is (c) 2 – 64.

2

(d) 2 +

-

t = 0) moving with a velocity u result of which it decelerates at a rate a=–k v where v is the instantaneous velocity and k is a T taken by the particle to come to rest is given by

(c) h =

h

-

66. The displacement x of a particle moving in one dimension is related to time t by the equation

h g

61. A body dropped from a height H above the ground strikes an inclined plane at a height h above the

Fig. 2.20

zontal and with speed u travelled along the incline is (a) 2h (b) h h (d) h (c) 2

u g

t1, t2 and h is

60.

65. A body of mass m1 cally upwards with an initial velocity u reaches a maximum height h Another body of mass m2

2 A to point B, Fig. 2.21

2.15

69.

a body with position (x) from the origin O

v) of

a – t)

71.

v – t) of the motion of the body? Assume that x = 0 and v = 0 at t

variation of the acceleration (a) with position (x)? v v

Fig. 2.22

Fig. 2.25

v (ms–1) 0

v (ms–1)

2

4

0

( )

2

(a)

4

( )

(b)

Fig. 2.23

70. The velocity (v) of a body moving along the postive x-direction varies with displacement (x) from the origin as v = k x , where k the graphas show the displacement-time (x – t) graph of the motion?

v (ms–1) 0

v (ms–1)

2

4

0

( )

2

(c)

4

( )

(d)

Fig. 2.26

72.

-

val t = 2s to t (a) 2 m

(b) – 2 m

73. A particle of mass m moving with initial velocity u enters a medium at time t a resistive force F = kv wher k is a constant of the medium and v velocity of the particle varies with time t as kt kt (b) v = u – (a) v = u + m m (d) v = u ekt/m 74. x of the particle varies with time t as (assume that x = 0 at t = 0) (c) v = u e

Fig. 2.24

kt/m

2.16 Comprehensive Physics—JEE Advanced

mut k mu (b) x = t k (a) x =

75. 1 2 t 2

(c) x =

mu 1 e k

kt/m

(d) x =

mu 1 e 2k

kt/m

u at 2

time t given by m (a) t = ln(2) k (c) t =

(b) t =

m k

(d)

m ln(2) 2k

m 2k

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73.

(c) (b) (a) (d) (d) (d) (d) (a) (d) (a) (c) (a) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74.

(a) (c) (b) (a) (a) (c) (c) (d) (c) (d) (a) (b) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75.

(a) (a) (c) (a) (a) (c) (c) (c) (c) (c) (d) (d) (a)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70.

(b) (b) (c) (b) (d) (c) (b) (c) (c) (a) (b) (c)

(d) (c) (b) (c) (b) (c) (b) (b) (a) (a) (a) (c)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71.

6. 12. 18. 24 30. 36. 42. 48. 54. 60. 66. 72.

(c) (a) (c) (d) (b) (a) (a) (c) (a) (a) (d)

SOLUTIONS 1. Let h be the height of the tower and t be the time h= and

1 2 gt 2

h = ut +

(1)

2+ = 60 m

1 h = g (t – 1)2 2

Now h = h – 16 h

9h

=

=

1 2 gt 2

1 (– 10) (2)2 2

16 h

1 g(t – 1)2 2

(2) t

1 h= 2

Fig. 2.27

10

2

2. Let h = AB be the height of the tower and P be the by the ball to go from B to P time taken to go from A to P fore, time taken by the ball to go from A to B is t u –1 0 = u – 10 u

3. Let s be the distance between A and B 1 1 1 t2 = 20t + t2 2 2 For the second car, 1 (2) t2 = 10t + t2 s = 10t + 2 Equating (1) and (2), we get t value of t in either (1) or (2) gives s s = 20t +

(1)

(2)

2.17

a=

2u

2

1 1 1 h 1 gn2 = g g 2 2 h 2 2h 1 1 h2 1 = g = g g2 g

h=

v d v = a dx = k2 x dx v d v = k2

x dx

u

x= 5. a =

Hence h is given by

dv dv d v dx v = = dx dt dx dt

u 2k

2

1 (p – qx2)–1/2 2

10. The time taken by ball A to reach ball B =

12 m = 2 ms–1 which is 6s

11. –1

-

Therefore, the velocity of the monkey as observed

d (k x ) dx k k2 = 2 x 2

–1

The negative sign indicates that the direction of this

The car attains the maximum velocity at the end of t1 –1 vmax t1 1 1 h g (2n – 1) so that n = 2 2 g

–1

12.

The time during which the car decelerates is t2 = (t – t1) where t is the total time taken for the car t2, the initial velocity is v t1 t1 – 2t2 t1 – 2 (t – t1) t1 – 2t 2 2 t1 = t =

h 2

×2s

–1

7. Given v2 x v2 = u2 + 2ax, 2 u = 2 ms–1 we have u 8. Let t1 The velocity at the end of t1 is t1 t1 v = u + at1

9.

Average velocity =

v= k x

= k x

–1

(– 2 q x)

qx v = (p – qx2)1/2 v Hence a = – qx 6. The acceleration is dv dv d v dx = = v a= dx dt dx dt

a= k x

20

A covers a

distance of 20 m upto ball B

=

Given

g

g = 10 ms–2, we get h – 60h + 100 = 0 The positive root of this quadratic gives h

/

dv dv d v dx = = v dx dt dx dt dv d = p – qx2)1/2 dx dx =

2h

1

= 10 ms–1

moving, the net speed of the bullet = speed of the –1 –1

–1 –1 –1

and the

the bullet will hit the car with a speed which is the relative speed of the bullet with respect to the –1

13. Let us suppose that cars A and B are moving in the positive x C is moving in the –1 negative x vA –1 –1 –1 = + 10 ms , vB and vC –1 –1 B –1 with respect to A is vBA = vB – vA The relative velocity of C with respect to A is vCA –1 = vC – vA t = 0, the distance between A and B = distance between A and C C will cover a distance

2.18 Comprehensive Physics—JEE Advanced

A at a time t given

AC by t=

AC 1000 m = vCA ms 1

s2 = ut +

Car B will overtake car A C does and avoid an accident, if it acquires a minimum acceleration a such that it covers a distance s = AB = 1000 m in time t –1 speed u = vBA tion 1 2 s = ut + at 2 1 2 ×a 2 which gives a = 1 ms–2 14. The relative speed of train A with respect to train B –1

is given by

15.

for the next time interval t during which the body

s2

=2

18. Let h 1 g 2

which gives s

tance of s1 =

1 2 1 gt = 2 2

10

(2)2

h= 2

g

h The time t taken to fall through 2 h 1 2 1 2 = gt g= gt or t2 2 2 2

s

(u) of the bullet is ( –1 u = gt = 10 maximum height (h) attained by the bullet is h 1 2 1 2 gt = 10 = 2 2 time taken by the bullet to return to the point of

20.

t is

-

1 2 1 2 at = at 2 2 The velocity of the body at the end of this time interval is v = 0 + at = at s1 = 0 +

g is given by t= 2 2

total time taken by the stone to hit the ground is given by h=

1 2 gt or t = 2

2h g

2 10 h1

h1 =

1 g g(1)2 = 2 2

tance h given by

-

1 g 2 g 2 h2 through which the stone falls in the

h=

which gives t 16.

1 2 gt = 2

ment x increases linearly with time t velocity v is a positive constant between t = 0 and t t t v) is zero between t t t t = 20 s, displacement (x) decreases linearly with time (t v) is constant but negative between t t

given by

t seconds, where t is the time taken by the bullet to rise to a height h of t is given by 1 1 gt 2 t– 10 t2 h = ut – 2 2 or t2 – 10 t + 21 = 0 or (t t

u = at)

17. Here v1 = 0 + at = at and v2 = v1 + at = at + at = 2 at v2 = 2v1

19. 2

1 2 1 2 at = at2 + at = at2 ( 2 2 2 s1

Now h1/h2 21. The maximum height h attained by the bullet is given by u2 ( v = 0) v2 – u2 = – 2gh or h = 2g

2.19

or h =

2 10 en by the stone to attain this height is given u g 10 (t = 1 s), the stone covers a distance h1 given by

by t =

h1 = ut –

1 2 gt 2

1–

ers a height h given by 1 2

h

1 2 t

10

(1)2

s), the stone cov2

10

= 120 m

22. The distance covered in the nth second is given by 1 sn = u + a n 2 1 s =u+a =u+ a 2 2 u+

2

6

s6

23. Now, sn = u + a n a s and s

s2 =

2 2 1 a12 2 1 a2 t2 t1 = 2 a2 2 a2

or

s2 =

1 a 2 t22 2

Thus we have

26. Average speed =

1 2

which gives

1 which gives a = 10 ms–2 2 1 2 1 2

s =s +s a 11a and 60 = u + 24. u+ 2 2 –1 These equations give u and a = 10 ms–2

a 1t 1 = a 1t 1 time interval t2 have, from v = u + at, 0 = a1 t1 – a2 t2

a 1t 1 = a 2t 2)

a2 t22

a2 a12 t12 a1 a22 t22

=

a2 ( a1

a 1t 1 = a 2t 2)

t2 t1

total distance total time

s1 t1

s2 t2

1 1 a1 t12 + a 2 t22 2 2

1 a1t1 (t1 + t2) 2

(

a 1t 1 = a 2t 2)

1 a t t t 2 1 1 1 2 Average speed = t1 t2 1 1 = a 1t 1 = a 2t 2 2 2 27. The maximum speed v attained by the car = speed it attains at the end of time interval t1 during which v = a 1t 1 = a 2t 2 1 v2 Now s1 = a1 t 12 = 2a1 2 and s2 =

v2 1 a 2 t22 = 2a2 2

s = s1 + s2 = or

u = a 1 t 1)

a1 a2

=

t1 is v = 0 +

(

a1 t12

s1 s2

As shown above, s1 + s2 =

1 2

25.

s2 s1

(

(ii)

=

–1

10 or u

s6

or

From (i) and (ii) we get

h2

h1/h2

t1 is given by 2a1s1 = v2 – u2 = a12 t12 ( v = a 1t1 and u = 0) 1 a 1t 12 (i) or s1 = 2 The distance covered in the next time interval t 2 is given by – 2a2 s2 = 0 – a21 t21 ( v = 0 and u = a1t1 now)

v = 2s

v2 1 2 a1

a1 a2 a1 a2

1/ 2

(

v = a 1t 1)

(

v = a 2t 2) 1 a2

2.20 Comprehensive Physics—JEE Advanced

28. The distance s1 covered by the car during the time it is accelerated is given by 2 s1 = v2, which gives s2 covered during the time s1 = v2/2 the car is decelerated is, similarly given by s2 = v2/2 v2 1 1 s = s1 + s2 = (i) 2 t1 is the time of acceleration and t2 that of deceleration, then v = t1 = t2 or t1 = v/ and t2 = v/ Therefore, the total time taken is 1 1 (ii) t = t1 + t2 = v From (i) and (ii), the average speed of the car is given by total distance s v total time t 2

a* = (nv)2/2s = n2v2 34.

t1 1 1 falls a distance h1 = gt 12 = 10 2 2 The second ball falls for t2 1 1 it falls a distance h2 = gt 22 = 2 2 h1 – h2

2

2

10

h1 = ut –

1 2 gt 2

1–

1 2

10

(1)2 –1

2 seconds is v = u – gt x = 12t t2 + 9 x changes with time t, the body by

dx t dt v changes with time t, the body is not in uniform motion; it is accelerated because v increases with t 30. v t v = u + at u = 12 ms–1 v=

31. Now v given by

t a a=

or a

dv dt

d 12 dt

t

10

2

h1 = h2 36. From v = u + gt u = u + gt or 2u = gt or t = 2u/g u)2 – u2 u2 37. From 2gh = v2 – u2, we have 2gh 2 or h u /g 38. The total distance covered in order to cross the platform is s = length of train + length of platform = speed is u ms–1 2 = 2a 39.

–1

v

v2 – u2 = 2as a

2

– which is

–2

a

the relation, v = u + at

–2

t or t

–2

t is v = at distance covered during this time interval is s1 = 1 2 at v = at is the initial velocity for the 2 next time interval t elled in the next time interval t is s2 = a t2 + Thus s2

1 2

seconds is h2

v = u + at, we a is

–2

=

35. The total time taken by the bullet to reach the highest point (where its velocity becomes zero) is given by 0 = u – gt or t = u/g

29.

32.

a * = n 2a

1 2 at = at2 2 2

s1

33. The distance over which the car can be stopped is given by 2 as = v2 or a = v2/2s v becomes nv, the value a* of a to stop the car in the same distance is

40. Given t = 0 is

dV t V(t) (i) dt t = 0, V(0) = 0 (given), the acceleration at a(0) =

41.

dV t dt

–2 t 0

V(t) with respect to t given in each pression for dV(t)/dt dV t d 21 e = dt dt

t

= 2

d (1 – e dt

t

) = 6e

t

2.21

= 6 – 6 (1 – e

time interval = 6 ms–1 time interval from t = 2 s to t v = u + at t Therefore, velocity at t s after t = 2 s) – and velocity at t 2 s after t velocity time graph from t = 2 s to t -

V (t)

t

V(t) = 2 (1 – e t)

(ii)

42.

–1

at time t * given

V(t *) or V(t*) = 1 or

2(1 – e

t*

) = 1 or e

t*

t * ln e–1

The graph is again linear but its slope is negative BC in

t*

*

or t Putting this value of t V at t

)

e

–1

= 2(1 – e 43.

stones are given by 1 2 gt x 1 = u 1t – 2 and x2 = u2 t –

–1

46. The distance covered = area under the velocity time graph from t = 0 to t ABC in (

g is negative)

area of triangle ABD 1 BD AD = BD AD = 6 ms–1 2

2 s = 12 m

47.

x = x2 – x1 = (u2 – u1)t x2 – x1) varies linearly with t, the graph is a straight line upto t At t x1 = 0; so we have for time between 6 1 2 x = x 2 – x 1 = u 2t – gt 2 Thus (x2 – x1) versus t graph is not linear; it is a Now at t = 10 s, x = 0, therefore 1 g (10)2 0 = 10u2 – 2 –1 or u2 g 2 Hence x t t t = 0, u t = 0 to t = 2 s, acceleration a

=2 =2

1 2 gt 2

stones:

44.

45. As shown above, the velocity of the body in time interval t = 2 s to t v t At t v

x–t) t = 0 s and t t = 0 s to t = 2 s, the slope of the graph must increase with t AB t = 2 s to t BC of graph (c), the slope is increasing

48. Let the total distance be 2 s and let t1 and t2 be t1 = s/V –2

t = 0 to t = 2 s), the velocity of the body is v = 0 + at or v t –1 and at t = 2 s, v = 6 ms–1 at t = 1s, v from t = 0 to t = 2 s, the velocity–time graph is AB of For the next time interval from t = 2 s to t –2 acceleration a initial velocity u

t2 distance s1 covered in time with speed V is s1 2 t2 t2 =V and the distance s2 covered in time with 2 2 t2 , so that speed V is s 2 = V 2 s = s1 + s2 = V

t2 2

V

= (V + V )

t2 2 t2 , which gives 2

2.22 Comprehensive Physics—JEE Advanced

2s V V s Total time taken = t + t2 = V

t2 =

2s V V

s 2V V V V (V V ) Hence, average speed= =

2s V (V s 2V V

V ) V

total distance total time

2V V V 2V V V

49. Let the total distance be S S is 2 S /2 S t1 = = 6 Let t2 be the time taken to cover a distance S1 with t that to cover distance S 2 with Then Now

S1 S1 + S2 =

Therefore

t1 t12 t2 = t 2 which gives t = 52. Velocity is

t2 and S 2

t2 or

S 6

total distance = total time

S 2

S S S/

t22

v =

dx = a + 2bt dt

ct2

(i)

d2 x

= 2b – 6ct d t2 Acceleration is zero at time t given by 0 = 2b – 6ct b t or t = c b b2 b2 c a v = a + 2b c c 9c 2 Acceleration is

S = S1 + S 2 2 S t2 =

t2

t1 t2

t

S and t2 = t 2

Total time taken t = t1 + t2 =

1 2 gt1 (ii) 2 For the stone thrown downwards with velocity u, we have 1 2 gt (iii) h = ut 2 + 2 Notice that the displacement of the stone in the three case is the same, equal to h and (iii) we have 1 2 1 2 1 2 1 2 gt = – ut1 + g t 1 ut1 = g t 1 – gt 2 2 2 2 1 2 1 2 1 2 1 2 and gt = ut2 + g t2 ut2 = gt – g t2 2 2 2 2 h = – ut1 +

a =

50. Motion corresponding to graph (i) cannot be realised because if we draw a line parallel to the v-axis, the body will have two different velocities at a given value of t Graph (iii) is also not possible because at some values of t, the graph is parallel to the v-axis which responding to graphs (ii) and (iv) are possible in

53. The velocity at a height h is given by v 2 = u2 + 2gh For downward motion, u = 0 and the value of g is negative and h becomes more and more negav 2 increases with h vector is directed downwards, v becomes more and v2 h, the graph of v versus h is g is For upward motion, v 2 = u2 + 2gh directed downwards and h ly, v2 decreases with h velocity vector is positive, v becomes less and less v with h is parav becomes less and less positive, graph

51. For the stone dropped with zero initial velocity, we have 1 gt 2 (i) h= 2 For the stone thrown upwards with velocity u, we have

54. Let t1 be the time during which the car has an accel–2 and t2 be the time during which eration a1 the car has a deceleration a2 = – 10 ms–2 car starts from rest, the maximum speed attained t2 by the car v = 0 + a1t1 = a1t1 this v

Average speed =

2.23

Therefore, 0 = v – a2t2 or v = a2 t2 a1 t1 = t1 = 2t2 a2 t2 or t1/t2 = a2/a1 t1 + t2 = 6, we get t1 t2 55. The distance moved in t1 1 s1 = 0 × t1 + a1 t21 2 The distance moved in t2 = 2s is 1 a t2 s2 = vt2 – 2 2 2 56. x2 x = t with respect to t we have dx dx x =1 (i) dt dt dx =v xv v x v Now dt t, we have dx dt

2

x v2

d2x

d2x

dt 2 xa

dt 2 a=0

or

a=

where a =

=0 v2

(ii)

x

2h g 61. The time t1 taken by the body to strike the inclined plane is given by 2( H h) t1 = g The time t2 taken by the body to reach the ground after striking the plane is 2h t2 = g 2h 2( H h) Total time t = t1 + t2 = + g g 60. Product of roots is t1 t2 =

v

57. Given, x2 = t2 + 2t

dx dt or

2

or 62.

d x dt

2

x a = x2

(ii)

(iii)

x2 = t2 + 2t a = 2/x u

– 2g(BC) =

u

2

2

u u

H

h

=

d dh

2( H h ) g

1 (H 2

h)

1 h

1/ 2

2h g 1 h 2

or H – h = h

2

= 2

=

u2

u2

=–

u2

(i)

u2 u2 u2 – =– (ii) 16 9

= 0

1/ 2

= 0

or h = u=

Effective retardation in water is geff =

t2 + 2t + 1 + x a = x2

58. – 2g(AB) =

1

dt dh

2

d2x a= 2 we have dt (t + 1)2 + x a = x2 or

2 g

or

(i)

2

x

dx dt

x2

dx x =t+1 dt

1 2 gt 2

The roots of this quadratic equation are t1 and t2 2u The sum of the roots is t1 + t2 = g

v

x

dt 2

or

2h 2u t+ =0 g g

Time t will be maximum if

d x

dx 2x = 2t + 2 dt

t2 –

or

2

a

h = ut –

59.

g(

H 2

2gh )

v Find the maximum depth x from the relation v2– u2 = 2geff x 63. h is the total distance travelled in n seconds, then 1 gn2 h= 2 n – 1)th second, the h 1 = gn2 distance fallen is h = 2 1 1 Hence g(n – 1)2 = h = gn2 2 or n2 n+2=0 The positive root of this equation gives the required value of n

2.24 Comprehensive Physics—JEE Advanced

AB not displacement = t time 65. For the body of mass m1, we have 64. Average velocity =

becomes zero at t particle will be maximum at t

u2 2g For the body of mass m2, if S is the maximum distance travelled along the incline then v2 – u2 = 2aS Now, when S is maximum, v a = – g g sin = – g 2

vmax = –

h=

t=

66

x x = t2 – 6t + 9

velocity v = Find t

(i)

67. Given a = – k v v–1/2 dv = – k dt

or

–2

6

ms–2 per second

and its intercept is c y = mx + c, the acceleration a (in ms–2) as a function of time t is given by

or

x + v0

dv =– dt

v0 x0

dx dt

a=–

v0 x0

v

a=–

v0 x0

having a positive slope =

70. Given v = k x have

v=

or

v=–

t

6

0

12

t

t2

t+k

where k particle starts from rest, v = 0 at t (1) we get k v=–

12

t

2v

dt

2

t

(1)

(2)

v0 x x0

v0

2

dv =– t dt 6

or

-

v0 v02 or a= x x– x0 0 Thus the graph of a versus x is a straight line

t

6

(1)

where v0 and x0 tiating with respect to time t, we have

v02 intercept = – x0

a=–

v0 x0 v varies with x as

v0 x0

v=–

v1/2 = – kt + c given initial condition (v = u at t = 0), we get c= 2 u

68. The slope of the line is m = –

6 –1

and intercept is c = + v0

or

2 v1 2 u1 2 = – kt Now, use t = T and v

(6)2

12

69. The slope of the given v versus x graph is m = –

d 2 dx = (t – 6t + 9) = 2 t – 6 (ii) dt dt v t dv =–k v dt

t=6s

dv k2 = 2 dt

this in (2)

v=

2

and negative

v 2 = k 2x

dv dx = k2 = k 2v dt dt

dv =

v0 x0

k2 dt 2 k 2t 2

v

dx dt

2.25

dx k 2t = 2 dt dx = x= Thus x

k2 tdt 2

dv = dt

k v m

k2

dv = v

k dt m

t

2

t2

v

a – t graph is m =

71. line is

2

–2

c a=

2

2

ms–2

v

t

dt

kt m

v u

kt m

dv =

2

0

t

v=

t

v = ue

kt/m

dx = ue dt dx = u e

kt/m

x

2

t

t

t

(t

0

)

x=

t t2

t dt

x

dx =

t dt 2

t dt 2

2

t 2

or

2

1 2

e

dt

2

2

kt/m

dt

kt/m t

k /m

0

mu (1 e k

kt/m

)

u 2 u = ue 2

2

t

x= =

75. Putting v =

2

0

(1)

0

x= u

2

dx =

kt/m

dx = u e v–t

dx t = dt

=

dt

Thus v = 0 at t

72.

74.

t

0

k dt m0

|ln|vu = ln

t

2

t

dv = v u

dv = t 2 dt dv =

kv m

F m

73. Acceleration a =

kt/m

1 = e kt/m 2 kt ln(2) = m

2 = ekt/m t=

m ln(2) , k

2.26 Comprehensive Physics—JEE Advanced

II Multiple Choice Questions with One or More Choices Correct 1. At time t

–1

t

is neglected, which of the following statements will be true? (a) The two bullets will be at the same height above the ground at t (b) The two bullets will reach back their starting (c) The two bullets will have the same speed at t (d) The two bullets will attain the same maxi2. The ratios of the distances covered by a freely nth seconds of its motion (a) form an arithmetic progression (b) form the series corresponding to the squares n natural numbers (d) form a series corresponding to the differences of the squares of the successive natural numbers 3. The displacement x of a particle varies with time a bt according to the relation x = 1 e b (a) At t = 1/b, the displacement of the particle a/b) (b) The velocity and acceleration of the particle at t = 0 are a and – ab respectively (c) The particle cannot reach a point at a distance x from its starting position if x > a/b (d) The particle will come back to its starting point as t 4. of the following statements are correct? Take g = 10 ms–2 (a) The net displacement of the bullet in 10 s is zero (b) The total distance travelled by the bullet in (c) The rate of change of velocity with time is constant throughout the motion of the bullet

–1

5. Two bodies of masses m1 and m2 are dropped from heights h1 and h2 ground after time t1 and t2 and strike the ground with speeds v1 and v2 correct relation from the following: (a)

t1 t2

h1 h2

(b)

t1 t2

m2 h1 m1h2

(c)

v1 v2

h1 h2

v1 (d) v 2

m2 h1 m1h2

v-t) graphs shown in

6. motion of a particle?

Fig. 2.28

7. Two balls of different masses are dropped from the is neglected and the value of g remains constant, which of the following statements are true? (a) The heavier ball reaches the ground before time (c) The heavier ball hits the ground with a higher speed speed

2.27

8. A body is dropped from the top of a tower of height h h g = 10 ms–2, how long does it take to reach the ground? )s

(a) (2 +

(b) (2 –

(a) the stone passes the point from where it was (b) the speed with which it passes the point of –1

)s

6 )s 9. t = 0) located at x = 0 moves along the positive x-direction under the 6

with x as

(d) the stone hits the water with a speed of –1

13. A balloon is rising vertically upwards at a velocity of 10 ms–1

v=k x

opens his parachute and decelerates at a constant –2 g = 10 ms–2

where k (a) the displacement x varies with time t as x = k 2t 2 k 2t (b) the velocity v varies with time t as v = 2 k2 (c) the acceleration of the particle is 2 (d) the distance s travelled by the particle in time k 2T 2 T is s = 10. The motion of a body is given by dv v dt where v is the velocity (in ms–1) at time t (in sect (a) the velocity of the body when its acceleration is zero is 2 ms–1 (b) the initial acceleration of the body is 6 ms–2 (c) the velocity of body when the acceleration is half the initial value is 1 ms–1

11. A stone falls freely from rest and the total distance covered by it in the last second of its motion is

–1

(c) He hits the ground with a speed of 10 ms–1

14.

a constant reterdation a which varies with instantaneous velocity v as a = – kv where k of the particle is u at time t = 0, then (a) the velocity at time t is given by v = u – at (b) the velocity decreases exponentially with

1 u in time k 2 (d) the total distance covered by the particle before coming to rest is u/k 15. A body moves from point A to point B with a velocity v1 and returns to point A with a velocity v2 1 (a) the average speed is (v1 v2 ) 2 (c) the velocity will decrease to

g = 10 ms–2,

(b) the average speed is

2 v1v2 (v1 v2 )

(c) the stone hits the ground with a speed of –1

(d) the acceleration of the stone during the last -

12.

–1

g = 10 ms–2,

(d) the average velocity is 16.

1 (v1 2

v2 )

of a tower of height h with a speed u reaches the ground after time t1 cally downwards from the top of the tower with the same speed, it reaches the ground after time t2

2.28 Comprehensive Physics—JEE Advanced

Then

1 2 (a) h = g t1 2 (c) u =

17.

t22

1 g(t – t ) 2 1 2

where k of the particle is u at time t = 0, then (a) the velocity at time t is given by v = u – at (b) the velocity decreases exponentially with

1 (b) h = g t1t2 2 (d) u =

1 g(t + t2) 2 1

u 1 in time 2 k (d) the total distance covered by the particle before coming to rest is u/k (c) the velocity will decrease to

a constant ratardation a which varies with instantaneous velocity v as a = – kv

ANSWERS AND SOLUTIONS

1. The two bullets will attain the same height at time 1 t=n n– n2 n 2 1 2 n which gives n 2 t or t is 2 10 = 20 s while that of the second bullet is

placement x is maximum when t a a 1 e b b 4.

ms–1

–1

-

dx d a 1 e bt ae dt dt b Acceleration of the particle is given by

5.

a 1 e b

1

a 1 1 b

2a b

e

1

1

t = 0, the values v and respectively are v = ae–0 = a and = – abe–0 = – ab -

h1 = 0

t1 +

1 2 1 2 gt = gt 2 1 2 1

and h2 = 0 Therefore

t2 +

1 1 gt 22 = gt 2 2 2 2

t1 t2

bt

h1 h2

Also, we have v1 = 0 + gt1 = gt1 and v2 = gt 2

dv d ae bt abe bt = dt dt At t = 1/b, the displacement of the particle is x=

–1

For heights h cos or > /4. Hence the correct choice is (b). 32. Given, h = 2 m, R = 1.25 m and horizontal distance s = 10 m. When the string breaks, the stone is projected in the horizontal direction, which means that there is no initial vertical velocity. From s = ut + 1 gt 2, we have ( u = 0), 2 1 2 gt (i) 2 The horizontal distance travelled in time t is s = vt (ii) where v is the velocity of the stone in the horizontal direction which is the same as its velocity in circular motion. Eliminating t from (i) and (ii) we get h =

g s2 2h Now, centripetal acceleration is v2 =

ac =

v2 R

2

10 100 = 200 ms –2 2 1.25

Thus, the correct choice is (b). 33. Refer to Fig. 4.17. AB and CD represent the two velocity vectors. The change in velocity v = v2 – v1. which can written as v = v2 + (– v1). Thus, to v, we reverse the direction of vector AB as v2 and – v1 by triangle or parallelogram law. This is shown in Fig. (c). The magnitude of vector v is given by v = [(v1)2 + v22 + 2(v1)(v2)cos(180° – )]1/2 = [2v2(1 – cos )]1/2 = 2 v sin

is the horizontal range

g s2 2hR

2 Hence the correct choice is (d).

(

v1 = v2 = v)

Motion in Two Dimensions 4.17

Fig. 4.19

37. If r is the radius of the track, then distance moved in 5s = r = 20 cm. Therefore, speed along the 20 circle (v) = = 4 cms –1. Now, angular speed 5 v 4 rad s –1. Hence the correct choice = r 20 5 is (d). 2

–

1

Fig. 4.17

34. Since the force is always perpendicular to the velocity (i.e. the direction of motion) of the particle, no work is done by the force on the particle. Hence the kinetic energy of the particle remains constant. The particle will move in a circle in a plane. Thus the correct choice is (c). 35. As shown in Fig. 4.18, at diametrically opposite points A and B, the magnitude of the velocity is same (= V) but the directions of the velocity are opposite. Hence the change in momentum is MV – (– MV) = 2 MV. Thus the correct choice is (c).

38. Referring to Fig. 4.20, the cyclist is moving on a straight road from A to B with a velocity v = 6 ms–1. As he approaches the circular turn, he decelerates at rate at, represented by vector BD. The magnitude of deceleration is at = 0.4 ms –2. At point B, two accelerations t and c, the centripetal acceleration directed towards the centre C act on the cyclist. v 2 (6) 2 = 0.3 ms –2. Using the law of R 120 parallelogram of vector addition, vector BE gives the resultant acceleration a whose magnitude is Now ac =

(

DE =

a =

(a t2

c)

+ a c2)1/2 = {(0.4)2 + (0.3)2}1/2 = 0.5 ms –2

Hence the correct choice is (a).

Fig. 4.20 Fig. 4.18

36. When the train is at rest or moving with a uniform velocity, the plumb line hangs vertically along OB (Fig. 4.19). If the train moves with an acceleration a, the plumb line gets inclined along OC, the direction of the resultant of accelerations a and g. It is = a/g. Hence the correct choice is (a).

39. In Fig. 4.21 vc represents the velocity of the car and vP that of the parcel. M is the position of the man. From parallelogram law, the direction of the resultant velocity vr must be along the direction along which the man is standing. It follows from is given by vc 10 1 or = 45º sin = vp 10 2 2

4.18 Comprehensive Physics—JEE Advanced

Fig. 4.23 Fig. 4.21

Hence the correct choice is (a). 40. Velocity of rain (vr) = 4 ms –1 vertically downwards. Velocity of wind (vw ) = 3 ms–1 from north to south direction. A rain drop is acted upon by two velocities vr and vw as shown in Fig. 4.22. From the triangle law, the resultant velocity of the rain drop is v = OW. In order to protect himself from rain, he must hold his umbrella at an angle with the vertical (towards north) given by tan

=

RW OR

vw vr

43. Refer to Fig. 4.24. Let PR be the direction along which he should row his boat. The boat is acted upon by two velocities – boat velocity (vb) and water velocity (vw). The angle should be such that that the resultant velocity (v) is along PQ, i.e. sin

vw vb

=

2 = 0.5 4

= 30° (upstream).

3 4

Thus the correct choice is (b).

Fig. 4.24

Hence the correct choice is (a). 44. Time taken to cross the river by the shortest path PQ is

Fig. 4.22

41. The magnitude v of the resultant velocity gives the speed with which the rain strikes the umbrella, which is given by v=

[v2r

+

v2w]1/2

= [16 + 9]

1/2

= 5 ms

–1

Hence the correct choice is (c). 42. In order to cross the river in the shortest time, the resultant velocity v of the swimmer must be perpendicular to the velocity vw of water, as vs2 = v2 + v 2w or v 2w = v 2s – v2 = 25 – 9 = 16 –1 or vw = 4 ms which is choice (b).

t=

PQ v

PQ vb2

36 vw2

(4)

2

(2)2

6 3s

Hence the correct choice is (c). 45. Since the direction of the velocity changes from point to point on the circle, choices (a), (b) and (d) are incorrect. 46. From energy conservation, [see Fig. 4.25] 1 2 1 mu = mv 2 + mgL 2 2 v=

u2

2 gL

v=

v2

( u )2

2(u 2

gL)

Motion in Two Dimensions 4.19

47.

v O

,R

u2. So the correct graph is (d).

B v

L

A

v

–u

u

Fig. 4.25

II Multiple Choice Questions with One or More Choices Correct 1.

2.

3.

4.

5.

different angles of projection, say, and , that give it the same range. Then, and are such that (a) cosec = sec (b) tan ( + ) (c) sin2 – cos2 = sin2 – cos2 (d) cot = cos sec A ball is projected upwards at a certain angle with the horizontal. Which of the following statements are correct? At the highest point (a) the velocity of the projectile is zero (b) the acceleration of the projectile is zero (c) the velocity of the projectile is along the horizontal direction. (d) the acceleration of the projectile is vertically downwards. Choose the correct statements from the following. The range of a projectile depends upon (a) the angle of projection (b) the acceleration due to gravity (c) the magnitude of the velocity of projection (d) the mass of the projectile A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that: (a) its velocity is constant (b) its acceleration is constant (c) its kinetic energy is constant (d) it moves in a circular path A simple pendulum of length r and bob mass m swings in a vertical circle with angular frequency . When the string makes an angle with the

vertical, the speed of the bob is v. The radial acceleration of the bob at this instant is given by (a)

v2 r

(b)

v2 r

v2 (d) r 2 r 6. A body is moving in a circle of radius r with a uniform speed v, angular frequency , time period T and frequency . The centripetal acceleration is given by (c)

(a)

v2 r

(c)

v

(b) 4 (d)

4

r 2

2

r

2

T 7. Which of the following statements are true about a body moving in a circle with a uniform speed? (a) The speed of the body is constant but its velocity is changing (b) The acceleration is directed towards the centre (c) The velocity and acceleration vector are perpendicular to each other. (d) Elastic, frictional, gravitation and magnetic forces can cause a uniform circular motion. 8. from a place in the enemy country at a distance x half-way point. Then (a) the velocity with which the missile was projected is

gx

4.20 Comprehensive Physics—JEE Advanced

x 2g (c) the speed of the missile when it was detected is gx 2 (d) the maximum height attained by the missible x is . 4 (b) you have a warning time of

9. v. An armyman with an anti-aircraft gun on the ground sights the enemy plane when it is directly u. Then (a) the angle with the vertical at which the gun v u (b) the angle with the vertical at which the gun

(d) the initial speed of the projectile is (a2 + c2)1/2. 12. A projectle thrown at an angle of 30º with the horizontal has a range R1 and attains a maximum height h1. Another projectile, thrown with the same speed, at an angle of 30º with the vertical has a range R2 and attains a maximum height h2. Then (a) R2 = 2R1 (b) R2 = R1 (c) h2 = 2h1 (d) h2 = 3h1 13. The maximum height attained by a projectile is increased by 1% by increasing its speed of projection without changing the angle of projection. Then the percentage increase in the (a) horizontal range will be 2% (b) horizontal range will be 1%

cos–1

v . u (c) the maximum height at which the enemy sin–1

u2

v2

. 2g (d) the maximum height at which the enemy (u v) 2 . 2g 10. From the top of a tower of height 40 m, a ball is projected upwards with a speed of 20 ms–1 at an angle of elevation of 30º. The total time taken by the ball to hit the ground is T and the time taken to come back to the same elevation) is t. The horizontal distance covered by the ball is x. If g = 10 ms–2, then T T =2 (b) = 2 (a) t t (c) x = 40 2 m (d) x = 40 3 m 11. The horizontal distance x and the vertical height y of a projectile at time t are given by x = at

and

14. The speed of projection of a projectile is increased by 1% without changing the angle of projection. Then, the percentage increase in the (a) horizontal range will be 1%. (b) maximum height attained will be 2%

y = bt2 + ct

where, a, b and c are constants. Then (a) the speed of the projectile 1 second after it is a2 + b2 + c2)1/2 (b) the angle with the horizontal at which the –1 c a (c) the acceleration due to gravity is –2b.

15. A body is projected at time t = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in meters) respectively vary with time t (in seconds) as x = 10 3 t y = 10 t – t2 (a) The acceleration due to gravity on the surface of the planet is 10 ms–2. (b) The maximum height attained by the body is 25 m. (d) The horizontal range is 100 m. 16. For a particle moving in a circle with a constant speed, (a) the velocity vector is always along the tangent to the circle. (b) the acceleration vector points towards the centre of the circle. (c) the velocity and acceleration vectors are perpendicular to each other. (d) the velocity and acceleration vectors are parallel to each other.

Motion in Two Dimensions 4.21

17. A particle is acted upon by a constant force which is always perpendicular to its velocity. The motion of the particle takes place in a plane. It follows that (a) its speed is constant (b) its acceleration is constant (c) its kinetic energy is constant (d) its momentum is constant. 18. A stone of mass 250 g is tied to the end of a string of length 1.0 m. It is whirled in a horizontal circle with a frequency of 30 rev./min. (a) The tension in the string changes as the stone moves in the circle. (b) The tension in the string is constant equal to

linear speeds of points A, B and C respectively at that instant, then (a) vA = vB = vC (b) vA > vB > vC (c) vA = 0, vB =

Fig. 4.26

(a) vA = vB = vC (b) vA = vB > vC (c) vA = vC > vB (d) vA < vB < vC 20. A uniform disc of radius R is rolling (without slipping) on a horizontal surface with an angular speed as shown in Fig. 4.27. O is the centre of the disc, points A and C are located on its rim and R from O. During rolling, point B is at a distance 2 the points A, B and C lie on the vertical diameter at a certain instant of time. If vA, vB and vC are the

(d)

vB vC

3 4

Fig. 4.27

2

newton. 4 (c) The speed of the stone is ms–1. (d) The maximum speed with which the stone can be whirled is 20 ms–1. 19. A uniform disc of radius R is rotating about its axis with angular speed . It is gently placed on a horizontal surface which is perfectly frictionless (Fig. 4.26). If vA, vB and vC are the linear speeds of points A, B and C respectively, then

3R 2

IIT, 2004 21. The trajectory of a projectile in a vertical plane is y = ax – bx2, where a and b are constants and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. a . (a) The horizontal range of the projectile is 2b (b) The maximum height attained by the projeca2 tile is . 4b 2a , where g is the bg acceleration due to gravity. (d) The angle of projection from the horizontal is = tan–1 (a – 2bx). IIT, 1990 22. The coordinates of a particle moving in a plane are given by x = a cos (pt) and

y = b sin (pt)

where a, b and p are positive constants and b < a. Then (a) the path of the particle is an ellipse (b) the velocity and acceleration of the particle are perpendicular 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 interval t = 0 to t = /2p is a. IIT, 1999

4.22 Comprehensive Physics—JEE Advanced ANSWERS AND SOLUTIONS

1. For the same range + = 90º or = 90º – . Choices (a), (b) and (d) satisfy this relation between and but choice (c) does not. 2. The correct choices are (c) and (d). u 2 sin 2 , is independent of the mass g of the projectile. Hence choices (a), (b) and (c) are correct.

9. Let G be the position of the gun and E that of the v, when u in a direction with the horizontal (Fig. 4.28). The horizontal component of u is

3. Range R =

vx

u cos

v

4. The magnitudes of velocity and acceleration remain constant but their directions are changing continuously. In uniform circular motion the force is radial (centripetal) and is always perpendicular to the velocity which is tangential. Thus, choice (c) and (d) are correct.

v

5. The radial component of acceleration is ar = centripetal acceleration v

v2 = =r 2 ( v=r ) r Hence, the correct choices are (c) and (d). 6. Since v = r = 2 correct.

r=

Fig. 4.28

Let the shell hit the plane at point P and let t be the time taken for the shell to hit the plane. It is clear that the shell will hit the plane, if the horizontal distance EP travelled by the plane in time t = the distance travelled by the shell in the horizontal direction in the same time, i.e. v t = vx v t or v = v x = u cos or cos = . u To avoid being hit, the plane should have a minimum altitude = maximum height attained by the shell which is

2 r , all the four choices are T

7. All the four choices are correct. 8. For maximum range = 45° for which Rmax = v20/g gx which is choice (a). Hence v0 = g Rmax The warni warning time is t=

tf 2

v0 sin g

gx sin 45 g

hmax =

x 2g

u2 1

Hence choice (b) is also correct. At half-way point, the missile is at its maximum height. Therefore, the vertical component of velocity is zero at this point. Hence the velocity is given only by the horizontal component which is vx gx = v0 cos = gx cos 45° = , which is choice 2 (c). The maximum height attained is hmax =

v02 sin 2 2g

gx sin 2 45 2g

Thus, all the four choices are correct.

u 2 sin 2 2g

x 4

=

2g

u 2 (1 cos 2 ) 2g v2 u2

(u 2

v2 ) 2g

Hence the correct choices are (a) and (c). 10.

t =

2u sin g

2 20 sin 30 =2s 10

Initial downward velocity = u sin = 20 sin 30° = 10 ms–1. The time taken to fall through a height of 40 m is given by 1 40 = 10t1 + 10 t21 which gives t1 = 2 s. Hence, 2 the total time taken to hit the ground is T = 2 + 2 = 4 s. Therefore T/t = 2. Also, the horizontal distance

Motion in Two Dimensions 4.23

travelled in 4 s = (u cos ) T = 20 cos 30° = 40 3 m. Hence the correct choices are (a) and (c). 11. The horizontal component of velocity is dx d at = a dt dt The vertical component of velocity is vx =

4

T= (i)

dy d (ii) b t 2 ct = 2 bt + c dt dt The value of vy at t = 1 s is (2b + c). Therefore, the magnitude of velocity at t = 1 s is vy =

v = (v 2x + v 2y) 1/2 = [a2 + (2b + c)2] 1/2 If a projectile is projected with an initial velocity v0 at an angle with the horizontal, the horizontal and vertical components of its velocity at time t are given by (iii) vx = v0 cos and vy = v0 sin – gt (iv) Comparing (iii) and (iv) with (i) and (ii) above we have v0 cos = a and v0 sin = c. Dividing, we get, tan = c/a. Compairing (iv) with (ii) we have g = – 2b We have seen above that v0 cos = a and v0 sin = c. Squaring and adding we get: v 20 = a2 + c2 or v0 = (a2 + c2) 1/2. Hence the correct choice are (b), (c) and (d). 12. The range is the same for and (90° – ). Hence R1 = R2 for = 30° or 60 °. v 2 sin 2 Since hmax = 0 2g h1 h2

and v0 is the same, we have 2

sin 30

1 3

sin 2 60

or

h h1 = 2 . 3

Thus the correct choice are (b) and (d). v02 sin 2 2g

13. We know that h =

. The increase

h in h

when v0 changes by v0 can be obtained by partially differenting this expression. Thus h=

2 v0 v0 sin 2 2g

v0 1 h = v0 2 h Now range R =

1 2

v02 sin 2 g

,

h h

0.01 = 0.005

2 v0 v0

2 v0 R = =2 v0 R

0.005 = 0.01 = 1 %

2v0 sin g

v0 T = = 0.005 = 0.5 % v0 T Hence the correct choices are (b) and (c). 14. The correct choices are (a) and (d). Use

R = R

v0 2 v0 h 2 v0 T , and . v0 v0 v0 h T 15. The horizontal and vertical displacements are given by x = (v0 cos )t (1) 1 2 (2) and y = (v0 sin )t – gt 2 Given x = 10 3 t (3) and y = 10 t – t2 (4) Comparing (3) with (1) and (4) with (2), we have (5) v0 cos = 10 3 v0 sin = 10 (6) 1 and g =1 g = 2 m s–2 2 Equations (5) and (6) give v0 = 20 ms–1 and = 30°. hmax =

v02 sin 2 2g

(20)2

sin 2 30 = 25 m 2 2

tf =

2v0 sin g

2 20 sin 30 = 10 s 2

R=

v02 sin 2 g

(20)2

sin 60 2

100 3 m

Hence the correct choices are (b) and (c). 16. The correct choices are (a), (b) and (c). 17. Since the force is always perpendicular to the velocity (i.e. the direction of motion) of the particle, no work is done by the force on a particle. The particle will move in a circle in a plane. Its speed and hence its kinetic energy will remain constant. Since the direction of the velocity is along the tangent to the circle, it will keep changing with time. Hence, the momentum will not remain constant. Since velocity is changing with time, the acceleration (which is perpendicular to velocity) will also keep changing. Hence the correct choices are (a) and (c). 18. Since the stone is whirled in a horizontal plane, the weight mg of the stone (which acts vertically down-

4.24 Comprehensive Physics—JEE Advanced

wards) is perpendicular to the plane of the circular motion and, therefore, has no component along this plane. Hence the tension in the string is constant. Given m = 250 g = 0.25 kg, R = 1.0 m, frequency ( ) 30 = 30 rev./min = = 0.5 Hz. Angular frequency 60 ( ) = 2 = 2 0.5 = rad s –1. The necessary cen-

dx dx dy 2bx = a dt dt dt vy = avx – 2bx vx = (a – 2bx)vx

At the maximum height vy = 0. Using this in Eq. (2), we get (a – 2bx) = 0 or x = a/2b. Putting this value of x in Eq. (1), we have (since y = hmax at this value of x)

tripetal force is provided by the tension in the string. Therefore, m v2 R

m

= 0.25

2

T=

R R

hmax = a

2

=m

2

R

4 Speed of the stone is v = R = 1.0 The maximum speed is given by

N =

ms–1.

100 R m = 20 ms –1

vmax =

100 1.0 0.25

vB vc

3R 2

1 2R

tf = 2t = a

2

a 2b

a2 4b

3 . 4

Hence the correct choices are (c) and (d). 21. Given y = ax – bx2 (1) (a) The value of y is zero at x = 0 and x = R (horizontal range). Putting y = 0 and x = R in Eq. (1), we get R = a/b. (b) Differentiating Eq. (1) with respect to time t, we have

2a 2 4bg

a 2bg

2 bg

dy d (ax – bx2) = a – 2bx dx dx Hence the correct choices are (b) and (d). 22. Given x = a cos (pt) (1) y = b sin (pt) (2) (d) tan

Hence the correct choices are (b), (c) and (d) 19. Since the surface is perfectly frictionless, the disc will not roll on the surface; it will simply keep on rotating at point A where it is placed. Now, linear speed = distance from centre angular speed. Therefore, R and vC = R vA = R, vB = 2 Hence the only correct choice is (c). 20. The disc is rolling about the point O. Thus the axis of rotation passes through the point A and is perpendicular to the plane of the disc. From the relation v = r where r is the distance of the point on the rim about the axis of rotation, we have 3R vA = 0, vB = (AB) = 2 and vC = (AC) = 2R Hence

b

2hmax g

t=

2 m vmax = 100 R

or

a 2b

(c) The time t to reach the maximum height is given by 1 hmax = gt 2 2

2

1.0 =

(2)

=

From (1) and (2)

x2

y2

2

2

= cos2 (pt) + sin2 (pt) = 1

a b Hence the path of the particle is an ellipse. Let the position vector of the particle at time t be r = xi Velocity v =

yj

dr dt

dx i dt

dy j dt

v = – a p sin (pt) i + bp cos (pt) j At

t = /2p, v = – ap sin

2

i + bp cos

2

= – ap i 2

j (3)

2

Acceleration a = – ap cos (pt) i – bp sin (pt) j At

t = /2p, a = – bp2 j At

t = /2p, v a = abp3 ( i j ) = 0.

Hence v a . It is easy to see that choices (c) and (d) are incorrect.

Motion in Two Dimensions 4.25

III Multiple Choice Questions Based on Passage

Two objects are projected from the same point with the same speed u at angles of projection and with the horizontal respectively. They strike the ground at the same point at a distance R from the point of projection. The respective maximum heights attained by the objects are h1 and h2 and t1 and t2 1. R, h1 and h2 are related as (a) R =

h1h2

(b) R =

(c) R = 2 2h1h2

(a) sin (b) (c) tan (d) 3. The ration h1/h2 is equal to (b) (a) sin2 (c) tan2 (d) 4. The sum (h1 + h2) is equal to

2h1h2

cos2 cot2

(a)

u2 sin2 g

(b)

u2 cos2 g

(c)

u2 g

(d)

u2 2g

(d) R = 4 2h1h2

t 2. The ratio 1 is equal to t2

cos cot

SOLUTIONS Since the horizontal range in the same Therefore = 90° – and we have

+

= 90°.

1. From Eqs. (1), (2) and (3), we have 2 gh1 2 gh2 u2 2 4 h1h2 g u2 u2 Hence the correct choice is (d) 2. From Eqs. (4) and (5) it follows that the correct choice is (c). 3. From Eqs. (1), (2), (4) and (5), we have R=

u2 sin2 h1 = 2g

(1)

h2 =

u2 u2 sin2 = cos2 2g 2g

(2)

R=

u2 2g

(3)

t1 =

2u sin g

t2 =

2u sin g

2 sin

cos

(4) 2u cos g

(5)

The position vector r with respect to the origin of a particle varies with time t as r = (at) i + (bt – ct2) j where a, b and c are constants. 5. The trajectory of the particle is a (a) straight line (b) circle (c) parabola (d) none of these

h1 = tan2 h2 Hence the correct choice is (c). 4. h1 + h2 =

u2 , which is choice (d). 2g

6. The magnitude of the initial velocity of the particle is (a)

a2

c2

(b)

b2

c2

(c) a 2 b 2 (d) (a + b – 2c) 7. The angle with the horizontal along which the particle is projected is given by

4.26 Comprehensive Physics—JEE Advanced

(a) sin (c) tan

= =

b c

(b) cos

b a

(d) tan

10. The maximum height to which the particle rises is

a c

b c

(b)

(a)

2c

= a

8. (a)

=

2

b2

ab c

2b 2 c

b2 2c

(b)

b2 4b 2 (d) 4c c 11. The horizontal range of the particle is (c)

c b (d) a ac 9. The acceleration due to gravity at that place is (a) 2 a (b) 2 b (c) 2 c (d) none of these

(a)

ab c

(b)

(c)

bc a

(d) abc

(c)

ac b

SOLUTIONS 5. Comparing Eq. r = (at) i + (bt – ct2) j with Eq. r = x i + y j , we get x = at (1) (2) and y = bt – ct2 b x– Eliminating t from (1) and (2) we get y = a c x2 which is the equation of a parabola. Hence 2 a the correct choice is (c). 6. From Eqs. (1) and (2), we have dx =a (3) dt dy = b – 2ct (4) vy = dt Putting t = 0 in Eqs. (3) and (4), the initial values of vx and vy are a and b respectively. The initial speed of the particle is vx =

2

1 2 gt (6) 2 Comparing Eqs. (5) and (6) with Eqs. (3) and (4) we have u cos = a and u sin = b which give tan = b/a, which is choice (c). 8. When t = tf, y = 0. Putting y = 0 and t = tf in Eq. (2) we get 0 = tf (b – ctf ) gives tf = 0 and tf = b/c. But tf = 0 is not possible. Hence the correct choice is (a). 9. Comparing Eq. (6) with Eq. (2), we get g = 2c, which is choice is (c). 1 b . Putting y = hmax 10. Now y = hmax when t = tf = 2 2c and t = b/2c in Eq. (2) we get y = (u sin )t –

and

11.

2

b2 4c

R = (u cos )tf a2

Now u =

2

u = a b , which is choice (c). 7. If the particle is projected with an initial velocity u at an angle with the horizontal, then the horizontal displacement x and vertical displacement y at time t are x = (u cos )t (5)

b b c 2c 2c

hmax = b

b 2 , tan

a cos

=

a

2

b

2

and tf =

(7) =

b which gives a

b . Putting these values c

ab . Hence the correct choice in Eq. (7), we get R = c is (a).

An object of mass m is whirled with a constant speed v in a vertical circle with centre O and radius R. T1, T2, T3 and T4 are the tensions in the string when the object is at A (top of the circle), B, C (the lowermost point of the circle) and D respectively (Fig. 4.29) Fig. 4.29

Motion in Two Dimensions 4.27

12. Tensions T1, T2 and T3 are related as (a) T1 = T2 = T3 (b) T1 < T2 < T3 (c) T1 > T2 > T3 (d) T1 = T3 < T2

(a)

Rg 2

(b)

13. Tension T4 is given by

(c)

2Rg

(d) 2 Rg

(a) T4 =

mv 2 + mg cos R

(b) T4 =

mv 2 – mg cos R

(c) T4 =

mv R

Rg

Here g is the acceleration due to gravity. 15. The minimum speed the object must have at the lowest point C to complete the circle is (a)

Rg

(b)

2Rg

(d)

5 Rg

2

+ mg sin

mv 2 – mg sin R 14. The minimum speed the object must have at the highest point A to complete the circle is (d) T4 =

SOLUTIONS 12. Refer to Fig. 4.30

(c) 2 2Rg

16. At the minimum speed of Q.15, the tension in the string is (a) 4 mg (b) 5 mg (c) 6 mg (d) mg

mv 2 + mg R Hence the correct choice is (b). 13. At point D, the force towards the centre is F4 = T4 – mg cos T3 =

mv 2 = T4 – mg cos R mv 2 + mg cos , which is choice (a). R 14. In order to keep a body of mass m in a circular path, the centripetal force, at the highest point A, must at least be equal to the weight of the body. Thus T4 =

Fig. 4.30

When the object is at A, since the weight mg acts vertically downwards, the force towards centre O is. F1 = T1 + mg mv 2 = T1 + mg R mv 2 – mg R At point B, weight mg has no component along BO. Hence the force towards O is T1 =

mv 2 = T2 R At point C, the weight mg acts in opposite direction to tension T3. Thus at C, the force towards centre O is F3 = T3 – mg

mv 2A = mg or vA = Rg R gives the minimum speed the body must have at the highest point so that it can complete the circle. Hence the correct choice is (b). 15. The minimum speed vC of the body must have at the lowest point C is given by v2C = v2A + 2 2 Rg where we have used v2 = u2 + 2gh, with h = 2 R. Thus

F2 = T2

mv 2 = T3 – mg R

v2C = Rg + 4 Rg = 5 Rg or vC =

5 Rg , which is choice (d).

16. The tension at this point is given by vC2 g = m(5 g + g) = 6 mg R Hence the correct choice is (c). T3 = m

4.28 Comprehensive Physics—JEE Advanced

The kinetic energy of a particle moving along a circle of radius R depends on distance (s) as K = as2 where a is a constant. 17. The centripetal force is given by (a)

as 2 2R

(b)

(a) mas (b) 2mas (b) as (d) 2as 20. The net force acting on the particle is

as 2 R

2as 2 4as 2 (d) R R 18. The speed of the particle around the circle is (c)

a (a) 2s m

1/ 2

a (b) s m

2a 1 / 2 a 1/ 2 (d) s m 2m 19. The tangential force acting on the particle is (c) s

s (a) 2as 1 R

(b) as 1

(c) 2as 1

1/ 2

R2

1/ 2

s2

1/ 2

s2

(d) zero

R2

SOLUTIONS 17. Given KE = force is

1 mv2 = as2. Therefore, the centripetal 2 1 2 mv 2 R

2 mv 2 R which is choice (c). fc =

2as 2 , R

or

2a v=s m

2a ds =v = s m dt

f = (f 2c + f 2t )1/2

dv dt

2a d s dt m

2a m

1/ 2

ds dt

A conical pendulum consists of a string of length L at one end carrying a body of mass m at the other end. The mass is revolved in a circle in the horizontal plane about-a

= 2 as 1

The angular frequency of revolution of the body is . The string makes an angle with the vertical axis. 21. The tension in the string is m 2 L (c) m 2L

L 2 m (d) m L2 (b)

1/ 2

2

(2 a s ) s2

2

1/ 2

R2 Thus the correct choice is (c).

22. The angle of inclination of the string with the vertical is given by (a) cos

(a)

2 a s2 R

=

Hence the correct choice is (c). 19. The tangential acceleration is at =

. Therefore,

2 a 1/ 2 2 a 1/ 2 2 a s s m m m Tangential force is ft = mat 2a s =m = 2as, which is choice (d). m 20. Net force acting on the particle is

1/ 2

1/ 2

1/ 2

at =

18. The speed v of the particle around the circle is given by 1 mv2 = as2 2

But

=

g 2 2

(c) cos

=

g

L L

(b) sin (d) sin

23. The linear speed of the body is (a) L (b) (c)

L cos

(d)

= =

L sin L tan

g 2

L

2

L

g

Motion in Two Dimensions 4.29

SOLUTIONS 21. Let T be tension in the string. Figure 4.31 shows the forces acting on the system. Tension T can be resolved into two mutually perpendicular components. The horizontal component T sin provides the centripetal force for circular motion and the vertical component T cos balances the weight mg.

Thus T cos T sin

and But T sin

=m

= mg

(1)

m v2 = m 2r r r = L sin . Therefore, =

2

L sin

or T = m

2

L

(2)

Hence the correct choice is (c). mg (3) 22. From (1), we have cos = T Using (2) in (3), we get mg g cos = , which is choice (a). 2 2 m L L 23. Linear velocity is v = r = L sin , which is choice (b).

Fig. 4.31

IV Matching 1. For a body projected at time t = 0 horizontally with velocity u from a height h. Column I Column II 2h g

(a) Horizontal displacement

(p)

(b) Vertical displacement

(q) u

(c) Time taken to hit the ground (d) Horizontal range

2h g (r) is proportional to t (d) is proportional to t2

ANSWER (a) (r) (c) (p)

(b) (d)

(s) (q)

2. A body is projected from the ground with velocity u such that its range is maximum. Column I Column II (a) Maximum height attained

(p)

(b) Horizontal range

(q)

2u g u g 2 u2 g

(d) Time to reach maximum height

(s)

u2 4g

4.30 Comprehensive Physics—JEE Advanced

ANSWER For maximum range (a) (c)

= 45°.

(s) (p)

(b) (d)

3. A body is projected with velocity u at angle Column I

(r) (q)

= 30° with the horizontal. Column II

(a) Velocity at maximum height

(p)

13u 4

(b) Velocity at half the maximum height

(q)

2 7 u 3

(c) Average velocity between the point of projection and highest point 2 (d) Velocity at t = 3

(r)

3u 2 7 u 8

ANSWER

Maximum height h =

u 2 sin 2 30 2g

u2 8g

(a) At maximum height, vx = ucos

and vy = 0. Therefore, v = ucos h . 2 vy2 = (u sin )2 – 2 gh = u2 sin2 2

= u2 sin2 30° – g vx = u cos 30° = v= Average velocity =

OA = Now

Now

v 2x

v 2y

2

u 8g

u 4

– gh u2 8

u2 8

3u 2 3u 2 4

u2 8

7 u 8

net displcemrnt OA = (see Fig. 4.32) 1 time taken tf 2 R2 4

h2

u2 R= g OA =

3u . 2

h =

(b) For half the maximum height,

(c)

= ucos 30° =

sin 60° = u4

3u 4

64 g 2

16 g 2

1 u sin tf = 2 g

u 2g

3u 2 2g 1/ 2

13u 2 8g Fig. 4.32

Motion in Two Dimensions 4.31

Average velocity =

At t =

(d)

OA 1 t 2 f

13u 2 2 g 8g u

2 t f, 3

vy = u sin

–g

2 tf = u sin 30º – g 3 u 2

2u 3

= u cos 60° =

3u 2

= vx = u cos v=

13u 4

v 2x

Thus the answer is as follows: (a) (r) (c) (p)

v 2y

3u 2 4 (b) (d)

u2 36

1/ 2

2 3

u g

u 6

2 7 u 3

(s) (q)

4. Match objects in circular motion listed in column I with the sources that provide the necessary centripetal force listed in column II. Column I Column II (a) A boy whirling a stone tied to a string in a circle (p) Frictional force (b) The moon revolving around the earth (q) Muscular force (c) The electron revolving around the proton (r) Gravitational force in a hydrogen atom (d) A car negotiating a curved road (s) Electrostatic force

ANSWER (a) (c)

(q) (s)

(b) (d)

(r) (p)

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each questions has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1.

(c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. A body is projected horizontally with a velocity u from the top of a building of height h. It hits the ground after a time t = 2h / g . The vertical and horizontal motions can be treated independently.

4.32 Comprehensive Physics—JEE Advanced

A body is projected from the ground with kinetic energy K at an angle of 60° with the horizontal. If air resistance is neglected, its kinetic energy when it is at the highest point of its trajectory will be K/4.

The velocity vector at a point is always along the tangent to the trajectory at that point.

At the highest point of the trajectory, the directions of the velocity and acceleration of the body are perpendular to each other.

Then the force does no work on the body and its kinetic energy remains constant.

One end of a string of length R is tied to stone of mass m and the other end to a small pivot on a frictionless vertical board. The stone is whirled in a vertical circle with the pivot as the centre. The minimum speed the stone must have, when it is at the topmost point on the circle, so that the string does not slack is gR . At the topmost point on the circle, the centripetal force is provided partly by tension in the string and partly by the weight of the stone.

In a uniform circular motion, the centripetal force is always perpendicular to the velocity vector.

In a non-uniform circular motion, the particle has two acceleration-one along the tangent to the circle and the other towards the centre of the circle. In a non-uniform circular motion, the magnitude and the direction of the velocity vector both change with time. In a non-uniform circular motion, the acceleration of the particle is equal to sum of the tangential acceleration and the centripetal acceleration. The two accelerations are perpendicular to each other.

The maximum range on an inclined plane when a body is projected upwards from the base of the plane is less than that when it is projected downwards from the top of the same plane with the same speed. The maximum range along an inclined plane is independent of the angle of inclination of the plane. In projectile motion, the velocity of the body at a point on it trajectory is equal to the slope at that point.

In a uniform circular motion, the kinetic energy of the body remains constant. The momentum of the body does not change with time. In a uniform circular motion, the acceleration is always directed towards the centre of the circle. Otherwise the speed of the body moving along the circle will change with time.

SOLUTIONS 1. The correct choice is (a). The time taken by the body to hit the ground is the same as if it was dropped from that height and fell freely under gravity. 2. The correct choice is (b). If m is the mass of the body and u its velocity of projection, the initial kinetic energy is 1 K = mu2 2 At the highest point, the horizontal velocity is (u cos 60°) and vertical velocity is zero. Hence the kinetic energy at the highest point is K =

1 1 m (u cos 60°)2 = 2 4

1 K mu2 = 2 4

At the highest point of the trajectory, the velocity of the body is horizontal (parallel to the ground) but its acceleration is g directed vertically downwards. 3. The correct choice is (a). When the stone is at the topmost point A on the circle, the centripetal force is provided by (mg + T) as shown in Fig. 4.33. mv 2 = mg + T R When the stone is at A, the string will not slack if Thus

tension T = 0, which gives

mv 2 = mg R

v=

Rg

Motion in Two Dimensions 4.33

R max =

5.

. Fig. 4.33

4. The range along the inclined plane when a body is projected with velocity u at an angle with the horizontal is given by R =

=

2u 2 sin(

) cos

g cos u

6. 7. 8.

g cos

2

ac2

a=

2

g cos 2

u2

Thus R max > Rmax. Since the range R depends on angle , Statement is false. Hence the correct choice is (c). The correct choice is (d). At the highest point on the trajectory, the slope is zero but velocity is u cos . The correct choice is (a). The correct choice is (a). The correct choice is (d). The acceleration of the particle is given by

2

[1 – sin ] =

at2

where ac = centripetal acceleration and at = tangential acceleration.

[sin (2 – ) – sin ]

9. The correct choice is (c). The speed of the body remains constant but the momentum changes with time because the direction of the velocity vector changes with time.

where is the angle of inclination of the plane. Range R will be maximum of sin(2 – ) = 1 or 2 – = 90° in which case Rmax =

u2 g (1 sin )

u2 g (1 sin )

10. The correct choices is (a). If the acceleration vector is directed towards the centre of the circle, it will have a component along the tangent, as a result the speed of the body will change and the motion no longer remains uniform.

If the body is projected downwards from the top of the same inclined plane, the maximum range will be

VI Integer Answer Type 1. 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, will the body take the maximum time to reach the ground? IIT, 1986 2. On a frictionless horizontal surface, assumed to be the x-y plane, a small trolley A is moving along a straight line parallel to the y-axis with a constant velocity vT = ( 3 1) ms–1 as shown Fig. 4.34. At a particular instant when OA makes an angle of 45° with the x-axis, a ball is thrown from origin O. Its velocity makes an angle of 60° with the x-axis and its velocity is such that it hits the trolley. Find the magnitude of the veolcity of the ball with respect to the horizontal surface. IIT, 2002

v

Fig. 4.34

3. A train is moving along a 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 of 60° to the 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 IIT, 2011

4.34 Comprehensive Physics—JEE Advanced

SOLUTIONS 1. A body falling freely from point P at a height H hits the inclined plane at point Q at a height h. As a result, the velocity becomes horizontal and the the ground at point R. The motion of the body from P to R via Q is shown in Fig. 4.35. P

H d 2t It is easy to check that when = 2, is negah dh 2 tive. 2. Refer to Fig. 4.36. It follows that v B is the resultant of v and vT . In triangle OAB vB vT = sin135 sin15 vB =

H–h Q H

=

Q h

=

vT sin (90 45 ) sin (60 45 ) ( 3 1) cos 45 sin 60 cos 45 cos 60 sin 45 ( 3 1) 1 / 2 3 / 2 2 1/ 2 2

= 2 ms–1

R

Fig. 4.35

v

If t is the time taken by the body to fall from P to Q through a height (H – h), then we have (H – h) =

1 gt 2

2

2( H h) g

or t =

1/2

v v

The time t taken by the body to fall from Q to R is given by ( the initial velocity at O is zero) h=

1 gt 2

2

1/2

2h g

or t =

The total time taken by the body to reach the ground is 1/2

2( H h ) g

t =t + t =

2h g

(1) t=

dt d t = 0 and is dh dh 2 negative. Therefore, differenting Eq. (1) w.r.t. h and dt = 0, we have setting dh

Time t will be maximum if

= = which gives

2( H h) g

2 1 (H g 2 1 g

1 (H 1

(H

1/2

h)

1/2

h)

1/2

( 1)

2h g 1 h 2

1/2

1/2

1 1/2

h) =

1 1/2

h

3. u = 10 ms ,

= 60°

1/2

2

dt d =0 = dh dh

Fig. 4.36 –1

h1/2 H =2 h

2u sin g

2 10 sin 60 10

3s

Let v be the velocity of the train. The horizontal velocity of ball at the instant it is thrown = (v + ux) = (v + u cos ). Therefore, the horizontal range of the ball with respect to the ground is R = (v + ucos )t, where t =

3s

It is clear that Distance travelled by ball in time t + 1.15 = R 1 i.e. vt + at 2 + 1.15 = (v + u cos )t 2 1 2 at + 1.15 = (u cos )t 2 1 a ( 3 )2 + 1.15 = (10 cos 60°) 3 2 a = 5 ms–2

5

Laws of Motion and Friction

Chapter

REVIEW OF BASIC CONCEPTS 5.1

SOLUTION

NEWTON’S FIRST LAW OF MOTION -

every body continues in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by an external unbalanced force.”

5.2

NEWTON’S SECOND LAW OF MOTION

5 2

v=

–1

–2 Using u v –2 we get a Force F = ma = 1.5

the rate of change of linear momentum of a body is directly proportional to the applied force and the change takes place in the direction in which the force acts.

t = 0.5 s in v = u + at 5 = 7.5 N.

5.2

Linear Momentum p = mv dv dp = m dt dt

SOLUTION F1 = F2 (

F=

m

= ma a=

=

dv dt

= 120 F12

F22

2 F1 F2 cos

102 102

2 10 10

1 2

= 10 N

NOTE

a= p–t

(F – t

F 10 = m 5

–2

5.3 5.1 v1

(2 i

3j

k)

1

to v 2

( 3i

2j

3k)

1

5.2 Comprehensive Physics—JEE Advanced

SOLUTION

2 10

v=

= m[ ( 3 i

2j

3 k ) (2 i

3j

v= 2

k

p = 2 ( 5)2 = 2 p t

F

=

( 1)2

(4)2

42 12 96

s

3

1 x

= 1

–1

v

12.96 = 4.32 ~ 4.3 N 3

5.3

1/ 2

1 2

m = 10–3

1/ 2

1

v= 2

x

= m ( 5 i j 4k)

1 x

k = 10–3

[

p = mv2 – mv1

or

10

3

1 1 0.5 –1

2 2

1/ 2

1 x

1/ 2

x=

1

NEWTON’S THIRD LAW OF MOTION

whenever one body exerts a force on a second body, the second body

EXAMPLE 5.4 x

action there is an equal and opposite reaction k

F= k = 10–3 v

x2 2

5.4

x x –1

law of conservation of linear momentum when no net external force acts on a system consisting of several particles, the total linear momentum of the system is conserved, the total linear momentum being the vector sum of the linear momentum of each particle in the system’.

.

SOLUTION F = ma = m

dv dv d v dx =m = mv dx dt dx dt

k

F=

Recoil of a Gun

x2 k x

2

LAW OF CONSERVATION OF LINEAR MOMENTUM

= mv

dv dx

k x m v2 k = c 2 mx

v dv =

v dv

2

k mx 2

dx

vb

dx

mb vb + mg vg

(i) v=0 k c= . m

c x v2 k k = 2 mx m 2k 1 v= 1 m x

mb vb + mg vg = 0 or

1/ 2

mb vg

vg = –

mb v b mg

mg

5.3

vg =

5.5

mb vb mg

IMPULSE A

=

v2

v2

2v 2 cos

=

2v 2 (1 cos )

2v cos

2

B

t

5.6

CONTACT FORCES

(1) Normal Reaction Impulse of a force is the product of the average force and the time for which the force acts and it is equal to the change in momentum of the body during that time. –1 or N s. I = Fav t = p

normal reaction (R). m R = mg

5.5

[Fig.

v

m

R = mg cos

SOLUTION v1 v2

= mv2 – mv1 = m(v2 – v1)

Fig. 5.2

= m[v2 + (– v1 m v v = v2 + (– v1

v2

v1

(2) Tension T

Fig. 5.1

v is [

v1

v2 = v v=

v12

v22

2v1v2 cos

5.7

FRICTION

5.4 Comprehensive Physics—JEE Advanced

or

–1

=

5.8

( )

BANKING OF ROUND TRACKS

limiting friction limiting static friction (fs s s

=

fs R

R

F

F= kinetic or sliding friction (fk ( k k

=

m R

fk R

mv 2 R v

F s.

k

F

Angle of Friction

= mg v v

N; N

= ( Rg)1/2

BANKING OF CURVES

f R

-

=

f R

Fig. 5.3 –1

( )

Angle of Repose

m Fig. 5.5

R N1

Car on a banked curved road

N2 N = N1 + N2 N cos

mg N sin

Fig. 5.4

f ) = mg sin R) = mg cos =

f = R

mv 2 – F cos R N cos = mg + F sin F= N N sin

=

5.5

F

5.10 =

v2

Rg

Rg

v2

v2 = Rg

1

MOTION IN A VERTICAL CIRCLE m

n n

v R

O A T1. Since

v R

mg centre O

R

. R v = (Rg

(

=

5.9

)1/2

v2 Rg

A CYCLIST NEGOTIATING A CURVED LEVEL ROAD

Fig. 5.7

F = T1 + mg = T1 =

or N

B

N = mg m

string is T2

mv 2 R

mv 2 – mg R

(i)

OB OB

mg

B

F = N = mg F = T2 =

mv 2 R

(ii)

At C

mg T3

C F = T3 – mg = T3 =

or

mv 2 + mg R

C

Fig. 5.6 2

mv /R

mg

T3

mv 2 R

F mv 2 > mg R R

A T 1. m

v=

A gR

mv 2 R

5.6 Comprehensive Physics—JEE Advanced

m v 2A = mg R

SOLUTION

v 2A = Rg

or

vC C vC2 = v 2A + 2

2 Rg

v2 = u2 + 2gh vC2

Fig. 5.8

= Rg + 4 Rg = 5 Rg

vC =

or

h = 2R

dx is m =

5 Rg

M dx L

L

m

2

x

o

vC2 R

T1 = m

g

= m (5g + g) = 6 mg

=

5.6

= m L

M L

2 L

1 ML 2

xdx o 2

5.8

k=

–1

l m r

SOLUTION 30 = 60 F = kx

=2

m(L + x)

2

mr = kx

x= =

2

m

0.02

.

SOLUTION

x

2

0.5

3.14 3.14

0.05

2

2

Fig. 5.9

0.1 N

5.7

OQ = r M

AB

–2

2

2 0.02

r= L + x F = kx = 2

Tension

g

.

r = L + x. =F

mL k

–1

=2

L

A i.e. B.

PQ = l O = T sin

mv 2 = T sin r

(i)

5.7

mg = T cos

(ii)

SOLUTION OP OC 1 = = OP 2

v2 rg v=

rg n

=

0.2 10

t

–2

g

mg 0.2 10 = =4N T= cos cos 60

.

OC = 60°. r = CP = OP sin

60 =

–1

5

= 5 sin 60°

3 2

5.9 m

SOLUTION Fig. 5.11

mg

2

m 2

r

N

r

g

C mr = N sin mg = N cos

N sin

2

2

2

r

g 30 60 r 4

2 2

v r g

–2

4

3.14

mg 0.5 10 = 10 N cos cos 60 R sin = N sin

N=

g =

m

.

2

0.5 9.8

2

0.3

= 0.3

2

T= R OC O

Fig. 5.10

N mR

=

5.10 m

(i) (ii)

5.11

2

=

10 0.5 5

2 ds

1

second = 3.14 s

SOLVING PROBLEMS IN MECHANICS BY FREE BODY DIAGRAM METHOD

5.8 Comprehensive Physics—JEE Advanced

R F – R = m 1a

(i)

R = m 2a

(ii)

free body diagram (1) Two masses tied to a string going over a frictionless m2 (m1 pulley m1 m2

F = (m1 + m2)a a=

F (m1

m2 )

m2 is F2 = m2 a = (3) Three masses in contact m1 m2

m2 F (m1 m2 ) m3 F

m1 a

a

m2

m3

Fig. 5.12

m1 is m1g – T. m1 m 1g – T = m 1a

(i)

T – m 2g = m 2a

m2 is (T – m2g). m2 is (ii) Fig. 5.14

a=

T=

m1 m1

m2 g m2

R m2

R

2m1m2 g m1 m2

m3 m2

m3 F – R = m 1a R – R = m 2a R = m 3a

(2) Two masses in contact m2 m1 F

m2 m1

m1

(i) (ii) (iii)

a. To a

m2 a=

(m1

F m2

m3 )

m2 is F2 = R F 2 = R = R + m 2a Fig. 5.13

F 2 = m 3a + m 2a

5.9

= (m2 + m3)a F2 =

F

m1 a

(m2 m3 ) F (m1 m2 m3 )

T m2

m1

m3 is F 3 = R = m 3a =

(m1

a

m3

m2 m2

m3 F m2 m3 )

T

m 3.

(4) Two masses connected with a string m2 m1 F

m2 a

a

m1 T

Fig. 5.17

Fig. 5.15

T = m 1a

(i)

F – T = m 2a

(ii)

a=

(i)

T – T = m 2a

(ii)

F – T = m 1a

(iii)

a=

F m1

T = m 3a

(m1

F m2

m3 )

m2

T = m 1a =

m1

m1 F m1 m2

T = (m2 + m3)a =

(m2 m3 ) F (m1 m2 m3 )

m2 T = m3 a Fig. 5.16

F

m1 a=

T=

m3 is T

m3 F m2 m3 )

(6) Two masses connected by a string and suspended m2 from a support m1 m2

F m1

(m1

m2 is T

m2

m2 F m1 m2

(5) Three masses connected by strings m3 m1 m2

F tension T

T

m1 m2 m1

T = T + m 1g

(i)

T = F + m 2g

(ii)

T = F + (m1 + m2)g

5.10 Comprehensive Physics—JEE Advanced

T = m 1a =

m1m2 g (m1 m2 )

m 2 < m 1. m1 f = R = m 1g

Fig. 5.20

m1 T – f = m 1a

Fig. 5.18

T – m 1g = m 1a

(7) Two blocks connected by a string passing over a

(8) Two blocks connected by a string passing over a

table

plane

m1

m2

T

(a) Mass m1 moving up along the incline with acceleration a

a. 1

2

(a) F.B.D. of

F.B.D. of

1

2

2

1

(b)

2

Fig. 5.19 Fig. 5.21

m1

m2

m1 T = m 1a m 2g – T = m 2a

a=

m2 g (m1 m2 )

(i) (ii)

m2 [see Fig.

= m 1a

(i)

m 2g – T = m 2a

(ii)

T – mg sin

a=

(m2 m1 sin ) g (m1 m2 )

5.11

T = m2 (g – a) m1 f = R = m1g cos m1 T – m1g sin T – m1g sin

– f = m 1a

– m1g cos

= m 1a

(b) Mass m1 moves down the incline with acceleration a T – m 2g m1g sin – T = m1a = m 2a a=

m1 sin m1

m2 g m2

T = m2 (g + a) (9) Two blocks connected by a string passing over a plane

m1 Fig. 5.23

1

m2

f

T

2

m1

m2 T – m1g sin m2g sin

2

1

m2

– T = m 2a

m1

f

m1 f = m 1a

m2 sin

2

m1

m1 sin m2

T = m1(a + g sin

1)

1

= m2 (g sin

R

(i)

2

m2

– a)

R + m 2g = R

F

F F Fig. 5.22

(10) One block placed on top of the another m2 m1 . Case 1: The maximum force that can be applied on the lower block so that the upper block does not slip [Fig. F

g=a

m2

F

m2

m 1g

a

g

F

F

m2

= R = m 1g

= m 1a

T a=

m1

m1 does

(ii)

– f = m 2a – R = m 2a

(iii)

– m 1g = m 2a = (m1 + m2) g

F

m1

F < F

m1

m2 m2

Case 2: The maximum force that can be applied on the upper block so that it does not slip on the lower block. [Fig. 5.24(a)] F F m1 m2 m1 m2 f = R = m 1g m1 on m2

5.12 Comprehensive Physics—JEE Advanced

–2

g

.

SOLUTION

Fig. 5.24

R

m2

a m1 F or

F

= R = m 1a

R = m 1g

– m 1g = m 1a m2 f = m 2a a=

F

=

(i) 1m 1g

= m 2a

m1 g m2 (m1

(ii)

m2 )m1 g m2

F < F F

Fig. 5.26

F 5.11

Fcos m

F = 10 N

= ma a=

F cos m

10 cos 60 1

5

2

Fcos – f = ma Fcos – mg = ma a= = Fig. 5.25

F cos

mg m

10 cos 60 –2

0.2 1 10 1

5.13

Fsin

F sin < mg

= mg – R. Since F sin

+ R = mg or

SOLUTION

5.12 m1

m2

AB CD Fig. 5.27. String AB CD Fig. 5.29

P

AB Q

CD

R cos R sin

CQ –2

g

Fig. 5.27

.

SOLUTION

= mg = ma a= g g

=g

CD is m = 0.5 Since string AB

AB

3

5.14 m

P = m1 + m + m2 P = 5.25

a

10 = 52.5 N

Q

QD

+ m2 = 0.5 Q = 3.4

10 = 34 N

Fig. 5.30

a. 5.13

SOLUTION

m a

Fig. 5.31

Fig. 5.28

T sin direction is

5.14 Comprehensive Physics—JEE Advanced

T sin

= ma

(i)

T cos

= mg

(ii)

a g a= g =

= 9.8

= 9.8

3 17

2

Fig. 5.33

5.15 m1

m2

T2 = 2T1

(iii)

-

m2

m1 .

–2

g

a1 = 2a2

m2g – 2T1 =

m2 a1 2

T1 a1 =

2m2 g 2 2 10 = m2 4m1 2 4 1.5

a2 =

a1 5 = 2 2

T1 = m1a1 = 1.5 T2 = 2T1 = 2

–2

–2

5 = 7.5 N 7.5 = 15 N

5.16 Fig. 5.32

m1 m1

SOLUTION a1 a2

m1 x1

m1

m2 m1 m2

x2 t

m2

x1 or x1 = 2x2 2

x2 =

d 2 x1 dt 2 m2 T2

= 2

d 2 x2 dt 2

(i) m1 m1

T1

m 2.

m1

m2

a1 = 2a2

m2 is F

m1 x1

m2

m2

Fig. 5.34

m 2. m1

SOLUTION T 1 = m 1a 1

m2 m 2g – T 2 = m 2a 2

(i) (ii)

m1 T m1

m2

m2 is f = R1 = m1g.

5.15

R = mg = 50 R 1 = m 1g

(i)

F – f – T= 0

T – f= 0

(ii)

T=f

(iii)

R 2 = R 1 + m 2g Using (iii) in (ii) F – f – f= 0 F = 2f = 2 m1g m2 F m1 F F = 2 m 1g =2

0.5

0.1

R1 – mg = ma R1 = m(g + a) = 50 (c) mg – R2 = ma R2 = m(g – a) = 50

(9.8 + 2.2) = 600 N (9.8 – 2.2) = 380 N

NOTE a

5.18 AB

10 = 1 N

L

A

M B

F1 = 4 N

F2 P

x

A.

Fig. 5.37

SOLUTION Since F1 F2 a

Fig. 5.35

T AP

5.17 m –1

a

–2

AP is m1 =

Mx L

PB is m2 =

M (L L

. AP

SOLUTION

P we con-

PB

x)

PB.

R -

Fig. 5.38

AP F1 – T = m1a

(i)

PB T – F2 = m2a

(ii)

F1 T m = 1 T F2 m2 Fig. 5.36

Mx L M L L

x x

L

x

5.16 Comprehensive Physics—JEE Advanced

F1 L

T= =

x L

F2 x

5.20 m

4 1 0.2

3 0.2 1 F

= 3.8 N

5.19 m

F

–2

a

–2

g M

Fig. 5.41

.

SOLUTION R

-

f –2

g

.

Fig. 5.39

SOLUTION A T A A – a)

Fig. 5.42

f = mg F= R f Fmin =

Fig. 5.40

mg – T = m(A – a) T – Mg = MA

R or mg

(i) (ii)

mg

=

F or F

mg

2 10 = 80 N 0.25

5.21 m M

mg – Mg = m (A – a) + MA

F

M

m g a Mg A= M m =

30 10 5 15 10 15 30 –2

T 250 N

A–a = M(g + A) = 15

–2

(10 + 6.7)

Fig. 5.43

5.17

F –2

g

m

M

v

.

SOLUTION Fig. 5.45

a SOLUTION P P

Q is f = mg Q.

P is aP = Q is aQ =

f m

mg m

f M

mg M P is

Q aQP = aQ – aP = Fig. 5.44

m

=

F – R = ma

(i)

f = mg

(ii)

R = Ma

(iii)

M

g

mg M

g m M

g 1

= 0.3 10 1

1 6

–2

Mg + f = R 5.23

a R=

MF M m

f

R

or

mg

or

F F

m = 500 g

a

MF M m mg M m M =

mg M m M

=

2 10 10 2 0.4 10

SOLUTION g

= 60 N

mg

P M

m

= mg

sin

cos

5.24

Q

m M

P –1

-

P Q

=g+a

0.58

5.22

v

–2

Q is P.

-

5.18 Comprehensive Physics—JEE Advanced

1 cos 45

=

sin 45

0.2

2

2 1 cos 45

5.25

Fig. 5.46

1

a

2

2

SOLUTION -

Fx Fy

SOLUTION A

Fx = (mg cos ) sin Fy = (mg cos ) cos

Mg

a is (A cos

Fig. 5.47

–a

m (A cos – a) = Ma F = mg cos sin A= N = Mg + mg cos2 F = N

=

m cos sin M

m cos 2

M m a m cos 2 1 1/ 2 1 cos 45 –2

I Multiple Choice Questions with Only One Choice Correct 1.

m

(c) g m

3 mg 2 (c)

2u cos

(d) u cos

5 mg 3

7 mg 6

(d)

9 mg 7

2.

P Q u

B

C

M u cos

u cos

Fig. 5.48

5.19

3.

P Q

M

7.

m

m F

-

P v P

Q

Q

P is g (c)

g 1

mg (c) F

mg M m M

(d)

g 1

mg

8.

m M

m = 10–2 x F= – k = 10–2

k 2 x2 2

v

x x

Fig. 5.49

4.

–1

–1

–1

–1

a 9.

= (c)

= 1

5.

a g

(d)

a g

= 1

a g

m

(c)

m -

(c) 0.3

(d)

=

2 n

10.

(d) A

=

1 n

B

1 2

A (c) A (d) A

B B B

11.

Fig. 5.50

6. a

2 a

12. N

m v

2 2 a nNmv

(c)

3 2 a

(d)

4 2 a

–1

n N mv n

5.20 Comprehensive Physics—JEE Advanced

(c) 13.

nN m v

(d)

nN v m

r

(c) 18.

( (c) 10 N 14. A

-

r v r2 v

(d)

r v2 r2 v2

m v

(d) 20 N

mv cos (

B

(c) 2 mv cos

mv sin ( ) (d) 2 mv sin

2

19.

2

x–t

A B.

B A. 15. h

2h g 1 (c) sin

Fig. 5.52

2h g ·

16.

2h g M

1 (d) · cos

(c) 8 s

2h g

(d) 16 s

20. (c) 0.8 Ns

(d) 1.6 Ns

21.

v

M s

F M g sin

Mg cos

(c) 2 Mg cos

(d) 2 Mg sin

M

(c) M 22.

v2 2s

g

2v 2 s

2 g

M

(d) M

v2 2s

g

2v 2 s

2 g

m u

Fig. 5.51

17.

v

5.21

u2 2 g sin (c)

u2 2 g cos

u2 4 g sin

(d)

u2 4 g cos

23. –2 –2

g

Fig. 5.55

27.

n

Fig. 5.53

(c) 60 N

(d) 600 N

24. (c)

k

= 1/(1 – n2

k

=

1 / 1 n2

(d)

k

= 1 – 1/n2

k

=

1 1 / n2

28. –2

–2

–2

–2

x k

25.

(x) = kx

x x0

–2

g

–2

(c)

.

k

k

. At

cot k

(d) k cot

29. = 0.5 x

x x x

(c) x = 26.

m1

(d) x =

g l2 1

m2

g

(c) –2

.

2 cot

in l

m2 m1

g

2 cot

30.

Fig. 5.54

2

x=

l

2

g (l2 – 1) (d)

1

g l

2

1

5.22 Comprehensive Physics—JEE Advanced

31. (c)

R h

32.

sin sin

1

(d)

2

R 10 R (d) 30

–2

g

)

(c) 40 N

m

(d) 60 N

37.

M m

33.

M

m

a. M

M Ma M m Ma (d) m m M

ma M m ma (c) M

t2 =1 t1

36.

10 = 0.95) R ) 5 R (c) 20

t2 t1

F –2

g

M

.

M F

m is a M Fig. 5.57

(c) 24 N 38.

Fig. 5.56

F

ma M

F

F M

(c)

(d)

(d) 96 N m2

m1

-

ma M

m1

am M

m2

m1 m

34.

m1 + m

m1 g

g 3 (c) g

2g 3

35. 1

t1 Fig. 5.58

2

t 2.

two t2 is

t1 t2 t1

sin sin

1/ 2 1 2

t2 t1

sin 2 sin

1

2 2

39. -

5.23

tion

a g

(c) t 1 is

to it.

1/ 2

(d) t 1

a g

1/ 2

45. –2

g –2

–2

–2

–2

) F

40. x 1. x2

(c)

x1/x2 is 2 (c)

3

2F 3 5F (d) 6

F

-

3F 5

46.

F

m a

(d) 2 F1 = F2 = F3 = F

41. a a g 2

g g

(c)

g

(d)

2

42.

3

m F1 -

Fig. 5.59

F2

2 1 a F1

1/ 2 3 F2 = F1

F2 is

F2 = 2F1

(c) F2 = 3F1

(c)

2 1 a

2 a

(d) a l1 l2

47.

(d) F2 = 4F1

43.

l1 + l2 1 (l1 + l2) 2 (c) 3l2 – 2l1

44. h

t

(d) 3l1 – 2l2 48.

-

a

A

h t

a t g

Fig B

B A Fig. 5.60

5.24 Comprehensive Physics—JEE Advanced

10 2 49.

2 10

53.

M strings A B

C R tension in string B is

W

-

(c) 0.75

(d) 0.85

54. 50

–1

–1

.

(c) 300

(d) 480

55.

M

Fig. 5.61 –2

100 g (c) 100 2 g newton

100

(d)

2

(c) 60°

g newton

)

(d) 90°

56.

50.

F = 600 – 2 F

(c) 0.9 Ns

i

105 t t

j –5

(d) 1.8 Ns

(2 i

3j)

10–5

(c) (3 j

2i )

105

(2 i (d) (2 i

3j) 3j)

105 10–5

57.

51.

–1

L A A 58.

(c)

52.

1

–2

F = 6 i – 8 j + 10 k –2

(c) 6 59.

(d) 8

g

–2

)

5.25

(d) cot

(c) sec

60.

=3

(d) cosec

=3 =3

64.

(c) 45°

(d) 60°

descending

descending 61.

m1

m2

1

2

t m1

1

1

t0 m2

2)

(m1

1

m2

2

2

is

Fig. 5.64

65.

m1 + m2)gt0 (c) 2(m1 + m2)gt0

(d)

m

1 (m1 + m2)gt0 2

62.

M

2 Mg

F

2 mg g

–2

)

(See Fig. 5.62) (c) 12 N

(c)

(M

m) 2

m2 g

(d)

(M

m) 2

M2 g

(d) 15 N

Fig. 5.62

63.

Fig. 5.65

66.

Fig. 5.63

m L

M

5.26 Comprehensive Physics—JEE Advanced

mL M mL (c) M m 67.

ML m mL (d) M m u

m

72.

m f

m

u 2 sin 2 g (c)

(c) g +

3u 2 sin 2 2g

2u 2 sin 2 g

(d)

f m

g f m

(d) g –

73.

3u 2 sin 2 g

68.

f m

–2

g a

-

(c) 500 N

(d) 600 N

74. m2

m1

v

av2 (c) 2 av2

av2 (d) av2

cos sin

cos sin

v1 x1

69. t seconds to

t

m1 m2

(c) 0.5 (c) 0.5

0.75 (d) 0.75

x2 x1/x2 is

m2 m1

m1 m2

-

v2

m2 m1

(d)

1

70. v

Fig. 5.66

v/n

-

75.

n = 1 = 1 (c)

= 1

(d)

= 1

71.

M tion a Ma M

1 n2 1 n

2

1 n

cot

(c)

-

m mg m

ma Mg M m

Ma M (d)

ma Mg M m

76.

1/ 2

-

2

1 n2

1/ 2

cot n –1

n g

–2

)

mg m

(c) 3

(d) 4

5.27

77.

m F

(c)

= sin

F = cot = cos

(d) F

78. mg 1 2

(c)

(d) N

mg 1

mg

mg

(d)

2

1

is N = F

2

1

79. T Fig. 5.68

82.

80.

g 4 3g (c) 4

g 2

–1

R

–2

g

R is

(d) g P

m Q

83. A

P k

–1

g

–2

.

A P

Q P

mg

Q is kA 2

84. –1

(c) kA

–2

g

–1 –1

(c)

–1

(0.35)

–1

(d)

(0.25) (0.45)

85. R

gR 2 (c)

gR (d) 2 gR

2gR

86.

Fig. 5.67

81.

-

–2

g

m F –1

f = mg

(c) 10 3

–1

2 –1

–1

5.28 Comprehensive Physics—JEE Advanced

87.

A

B

m

90.

m

3 -

m m g g g 2

g (c)

g g 2

(d)

g 2

g 2

Fig. 5.69

88.

x–y

91.

p (t) = A[ i cos (kt) – j sin (kt A k

(c) 45°

.

-

y = kx2 (y

(d) 90°

89.

m on

x

a

m y

a a gk

a’ P (c) F

F a 2 2m a x2 (c)

F x 2m a

2a gk

92.

x

(d)

a 4 gk

m L

F x 2 2m a x2 (d)

a 2 gk

-

F a2 x2 2m x

Fig. 5.71

(c) 27 Fig. 5.70

(d) 36

5.29 ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73. 79. 85. 91.

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74. 80. 86. 92.

(c) (c)

(c) (d)

(c) (c) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75. 81. 87.

(c) (c) (c) (c) (c) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76. 82. 88.

(d) (c)

(d) (d) (d) (d) (c)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71. 77. 83. 89.

(d) (d) (c)

(c)

(c) (c) (d)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72. 78. 84. 90.

(d) (c) (c) (d) (c) (c) (d)

(c) (c) (c) (c) (c) (c) (d) (c) (d) (d) (d)

(d)

SOLUTIONS 1.

T

a

Q is aQ =

3g 2

f = M

Q

a .

mg M P = aP + aQ =

T – mg = ma T–

(1)

mg m = 2 2

3g 2

a

(2) m (g – a) sin

= =

F = mg cos

OA = y Now F = =

dr dy = 0 + 2y dt dt

dy dt

cos sin 2

cos 2

6.

A tion a is (A cos – a

. P

Q is f = mg Q.

P is aP =

F mg cos sin = N 2 mg mg cos 2

For

u cos P

N

=

dr = u y = r cos dt M

3.

sin

N = 2 mg + mg cos2

dy r dr = dt y dt Now

m(g – a) cos

5.

t 2r

m M

4. g = (g – a

T= 7 mg 6 2. AB = AC = r OB = OC = x AOB r2 = x2 + y2

g 1

f = m

mg = g m

N N sin

= (2 m) a = m (A cos 3ma A= m cos

– a)

5.30 Comprehensive Physics—JEE Advanced

3a = cos 45

2aA sA = v2A

= 3 2 a

7. f = mg

f)

sA + sB

= F

sA

mB mA

70 50

7 . Since 5

sB

11.

12.

k x–2 dx 2m v2 k = +C 2 2 mx

v dv = –

(1) n

C v x we get C = – k/2m v2 k 1 = 2 2m x

sA sB

mA aA = mB aB

= N

dv d v dx dv 8. F = ma = m =m = mv dt dx dt dx k F=– 2 x2 k dv k – = mv v dv = – x–2 dx 2 dx 2m 2x

a B s B = v 2B

Nm

n

Nmv

m AB = L

AC = h

n -

nNmv . v=

1

k = 10–2

2

m = 10–2 v

13.

1/ 2

k 1 m x

1 x=

–1

.

9. g sin .

14.

-

g sin

g cos

– g cos g sin

mg cos R mg sin .

g sin = – (g sin cos = 2 sin

or

10.

mB

mA aA

– g cos )

is a = g sin

A

B

aB vA v = mB B t t 2 2 mA vA = mBvB or mA vA = m2B v2B

mA aA = mB aB or mA or

vA

(i)

vB

Fig. 5.72

t sA

A

sB

B v2 – u2 = 2 as

5.31

or

v2 – 0 = 2aL

or

v=

2aL

2 g sin L

v=

2 gh

2 0 2 0 s

–1

At t h = L sin

–1

v = u + at or v = 0 + at. v t= a

2 gh g sin

15. 16.

= mv – (– mv) = 2 mv –1 =2 –1 = 0.8 Ns

2h g

1 sin

Mg Mg sin

Mg sin . To -

21. v

s R = Mg a v u s v2 – u2 = 2as v2 = 2 as or a = v2/2s

nent Mg sin Mg sin

= 2 Mg sin

17.

A) = v

F1

r2 -

=

Av = r2v r 2v

M v2 2s F2 = Mg F1 + F2 = M

= r2 v2

v2 2s

22.

mg sin F = mg sin

r 2 v 2.

mg

a = – g sin – g sin = – 2g sin

r 2 v 2. 18. 19.

t

t = 2s is

x = 0 to x t = 0 to 2s. Between t = 2s

t

t x

Fig. 5.73

t

s t 0 – u2 = 2

20. Between t is v t

t x–t

(– 2g sin ) s=

t

g

v2 – u2 = 2as s or

u2 4 g sin

-

5.32 Comprehensive Physics—JEE Advanced

23.

t –2

a

a=

–2

s

mg

u = 0. Now t s = ut +

a s F = ma = 60

1.0 = 60 N

t=

u

1 2 at 2 a=a

–2

10 s. t =

24.

s

Since

=

f mg

n n

nor

60

s

1 2 at 2 1 =0+ 3 2 m2 = 6

10 = 120 N 26. m2

mac f 120 60

–2

25.

–2 a is F = ma = 10 3 = 30 N mg R

F

T = 60 N

nent F1 = m1g sin m1g. Now F1 = m1g sin = 5 10 sin 30° = 25 N. Since F1 T T F m1 f m1 m1

ac

f mac = f or ac = m

u = 0) s = ut +

mg =

s

f= 0.2

10 s is (

F = 30 N

f f f R mg f = mg = 0.2 = 20 N

Fig. 5.74

=

or

10

or

10

T – m1g sin 30° – f = 0 f = T – m1g sin 30° = 60 – 5

10

sin 30°

= 60 – 25 = 35 N F f

27.

-

F =F–f= 30 – 20 = 10 N a =

F m

10 10

–2

a1 = g sin 45° = a2 = (g sin 45° – g = (1– k) 2

g 2 k g cos 45°)

5.33

t22

= n2

t12

a1 a2

1

or

1

k

=1–

k

1 n2

31.

A

. h. Now OA = OC = R mg sin

28.

F = mg x is

cos mg sin = mg cos OAB =

R2

mg cos mg sin = OB OB = x AB AB

2 x2 .

Fig. 5.75

f (x) = mg sin – mg cos = mg (sin – cos ) = mg (sin – kx cos ) x = x0 sin – kx0 cos = 0

f (x

x0 =

n k Fig. 5.77

29. g sin – g cos 0.5x cos x x

= g (sin

–

cos ) = g (sin

–

R x

or

30.

AB = 1 AC = l BC =

l

2

x

=

a a cos g sin

g

x2

x R2

(

x2

=

3R

= 0.95 R 10 R – x = R – 0.95 R = 0.05 R x =

= AB/BC = 1/ l 2 1 . A

1

3=

2

h =

32. m is F

a cos or

=

m

g

F on

l2 1

M.

F = ma a = 33. M

Fig. 5.76

ma. m.

M

= g sin a=g

1 ) 3

M = F – f = F – ma F ma M = M

F M

on m2 = M is

ma M

m is f = ma m f = ma.

5.34 Comprehensive Physics—JEE Advanced

34.

a

m

a=

mg – T = ma

ss m

or T = mg – ma T=

2 mg 3

m Using (ii) in (i) we get

2 mg = mg – ma 3 a=g

F=

(M + m)g +

1

+

=(

h s1 = s2 =

h sin

h sin

2)

t22 t12

a2 = g sin

1

10

38 .

1

-

2

a1 s2 a2 s1

=

g sin = g sin sin 2

=

sin

m tension T

1 2

h sin

2

sin h

m 2g (m1 + m)g (m1 + m)g = m2g or m =

m

m1

m 6 – m1 = –4 0.4

k

F=

s

mg = 0.7

5

10 = 35 N.

1

2

f

2

= F – mk mg = 35 – 0.5

mg l

a=

(l – x) or

3 1 = 40 N

37.

40.

m (M + m)g M

2

=

F M

f 10 m 5

–2

v = 30° is a1 = g sin 1 1 v2 – u2 2 = 60° is a2 = g sin 2 u2 = 2a1x1 = 2a2x2 x1 a = 2 x2 a1

= 2ax

1

2

10 = 35 – 25

u

m

M

5

= 10 N

x

6 10 3

m2

s

1

l

T=

m1

(m1 + m

39.

l–x

1

(5 + 3)

= 96 N

v12 = 2a1 s1 v22 = 2a2 s2 v 1 = a 1t 1 v2 = a2t2 or a12 t11 2 2 a2 t2 = 2a2 s2

mg

(M + m)g

m

a1 = g sin

M=

2

(ii)

(M + m) g

2

= 2a1 s1

1g

=

1

= (0.5 + 0.7)

35.

36.

1mg

=

g sin g sin

2 1

sin 60 sin 30

3

41. a

a M + m)a F = (M + m)a +

2

(M + m)g m is

1mg

(i) -

g sin g sin Now

a cos g a cos

=

or a = g a = g

g 3

.

5.35

F

a=

M

T= M

m F 2 M m

=

F1 F1 = mg sin F2 F2 = mg sin

mg cos

–

+

m=

(i)

mg cos

(ii)

n

t

30

1/ 2 3

M 2

46.

F2 + F1 = 2 mg sin F2 – F1 = 2 mg cos F2 F1 = F2 F1 or F2 = 3 F1 43. m l L L–l

m a 2

M

Fig. 5.78

42.

m

T=

a= F2

5F 6

F m

F1

F

F = (F12 + F22)1/2 =

2F(

F1 = F2 = F)

F

F 3.

=2 =

F – F3 2 F–F

= ( 2 – 1) F 2 – 1) l = ml

F = ( 2 – 1)a. m

L– l) mlg

= m (L – l m(L – l) mlg = m (L – l) g L 1

l=

or

44.

h=

g+a

0.5 120 0.5 1 Fig. 5.79

1 2 gt 2 a

t

g = h=

1 gt2 2

45.

t = t 1 M

a g

F x is F = kx

1 1 2 gt2= gt or (g + a)t 2 = gt2 2 2 or

47.

1/ 2

m

k

l0 F = k(l – l0)

l F 2 = k(l1 – l0) k(l2 – l0)

(i) (ii)

5.36 Comprehensive Physics—JEE Advanced

3 l2 l0 2 l1 l0 Which gives l0 = 3l1 – 2l2. Using this value of l0 in 1 . either (i) or (ii) we get k = l2 l1 When a stretching force of 5 N is applied, let l3 be the length of the spring. Then 5 = k(l3 – l0) Substituting the values of l0 and k, and solving we get l3 = 3l2 – 2l1 Hence the correct choice is (c). 48. If the springs A and B are massless or their mass is negligible compared to the mass with which they are loaded, the tension is the same everywhere on the spring. Hence each spring balance will read 4 kg. Thus the correct choice is (c). 49. Let T be the tension in string C and T in string B. The y-component T cos balance with the weight Mg and the x-component T sin balances with tension T . Thus (see Fig. 5.80) T = T sin and Mg = T cos Dividing the two we get T = Mg tan = 100 g tan 45° = 100 g newton Hence the correct choice is (a).

= (600 t

105 t 2

t 3 10

3

s

0 5

= 600 3 10–3 – 10 (3 10–3)2 = 1.8 – 0.9 = 0.9 Ns Hence the correct choice is (c). 51. The radius of the circular motion of the bead is r = L. The linear acceleration of the bead is a = r = L. If m is the mass of the bead, then Force acting on the bead = ma = m L Reaction force acting on the bead is R = m L The bead starts slipping when frictional force between the bead and the rod becomes equal to centrifugal force acting on the bead, i.e. m v2 r L = mr 2 = mL 2 = 2 = ( t)2 R=

m

or or

( (

v=r ) = t)

= ( t)2 or t =

or

52. The magnitude of the force is F = F F = [(6 i – 8 j + 10 k ) (6 i – 8 j + 10 k )]1/2 = {(6)2 + (8)2 + (10)2]1/2 = (200)1/2 = 10 2 N F 10 2 N = = 10 2 kg. Hence the a 1 ms 2 correct choice is (b). 53. Refer to Fig. 5.81. Since the block moves with a constant velocity, no net force acts on it. Therefore, the horizontal component F cos of force F must balance with the frictional force, i.e. fr = F cos . Mass =

Fig. 5.80

50. Now F = 600 – 2 105 t. F will be zero at time t given by 600 – 2 105 t = 0 or t = 3 10–3 s

Fig. 5.81

t

F dt

Therefore, impulse =

Also

0 t

(600 – 2

= 0

105 t)dt or

fr = (mg – F sin ) = (f – F sin ) (f – F sin ) = F cos (200 – 100 sin 30°) = 100 cos 30°

Laws of Motion and Friction 5.37

or

200 100

1 2

= 100

0.866

= 86.6 86.6 = = 0.58, 150

or

which is choice (b). 54. Change of momentum of one bullet = m (v – u) = 0.03 {50 – (–30)} = 2.4 kg ms–1 Average force = rate of change of momentum of 200 bullets = 200 2.4 = 480 N, which is choice (d). 55. Let the body leave the surface at point B as shown in Fig. 5.82. When the body is between points A and B, we have M v2 Mg cos – N = r

Fig. 5.82

When the body leaves the surface at point B, the normal reaction N becomes zero. Thus M v2 Mg cos = r or

cos

=

v2 (5)2 = rg 5 10

1 or = 60° 2 Hence the correct choice is (c). 56. Mass of each piece (m) = 1 kg. Initial momentum = 0. Final momentum = p1 + p2 + p3. From the principle of conservation of momentum, we have p1 + p2 + p3 = 0 or p3 = – (p1 + p2) = – (mv1 + mv2) = – m (v1 + v2) =

(2 i

= –1 kg = (2 i Force

(2 i + 3 j) kgms 10

5

1

s

= (2i + 3 j) 105 newton Hence the correct choice is (b). 57. Magnitude of recoil momentum of the gun = forward momentum of the bullet = mbvb = (40 10–3 kg) 1200 ms–1 = 48 kg ms–1 n bullets per gun is F

=n =n

change in momentum time

48 kg ms 1 1s = 48 n newton Given F = 144 N. Thus 144 = 48 n which gives n = 3. Hence the correct choice is (d). 58. Force F = ( s – k) mg = (0.8 – 0.6) 4 10 = 8 N =

n

F 8N Acceleration = = 2ms–2, which is m 4 kg choice (a). 59. No net force acts on the block as it moves at a constant velocity. Therefore, downward force = upward force or mg sin = mg cos or = tan , which is choice (c). 60. When a cylinder rolls up or down an inclined plane, its angular acceleration is always directed down the plane. Hence the frictional force acts up the inclined plane when the cylinder rolls up or down the plane. Thus, the correct choice is (b). 61. Linear momentum of the system at time t = 0 is p1 (m1 v1 m2 v2 ) and at time t = 2t0 it is p2 (m1 v1 m2 v2 ) . Change in linear momentum in time 2t0 = p2 p1 . The rate of change of linear momentum is p2 p1 /2t0. From Newton’s second law of motion, the rate of change of momentum equals the force acting on the two particles, which is (m1 g + m2 g ). Hence ( p2 p1 ) = (m1 + m2) g 2t 0

3 j) ms–1

3 j) kg ms–1

p F= 3 t

=

(m1 v1 + m2 v2 ) – (m1 v1 + m2 v2 ) = (m1 + m2) g (2t0) Hence the correct choice is (c). or

5.38 Comprehensive Physics—JEE Advanced

62. The horizontal component of F parallel to the surface is F sin F is given by F sin = mg 3 10 or F sin 60° = 0.5 3 = 0.5 3 10 2 which gives F = 10 N. Hence the correct choice is (b). 63. As shown in Fig. 5.83, the insect will crawl without slipping if the value of is not greater than that given by the condition: force of friction f = mg sin . Now f = N, where N is the normal reaction. Thus N = mg sin or mg cos = mg sin or 1 = 3, which is choice (a). cot = F

or

64. Let T be the tension in the string. When the system is in equilibrium, then for the two equal masses m, we have T = mg (1) and for the mass 2 m, we have =

2 mg

(2) 1

or = 45°, Dividing (2) by (1), we get cos = 2 which is choice (c). 65. The force F on the pulley by the clamp is given by the resultant of two forces: tension T = Mg acting horizontally and a force (m + M)g acting vertically downwards. Thus F=

( Mg )2

{(m

mL M m Hence the correct choice is (c). 67. At the highest point of trajectory, the projectile has only a horizontal velocity which is u cos . After L =

or

horizontal velocity. If u is the horizontal velocity of the other fragment, the law of conservation of momentum gives (2m) u cos = m 0 + mu which gives u = 2u cos Now, the time taken to reach the highest point (as well as the time taken to fall down from this point) u sin is . Therefore, the horizontal distance travg elled by the other fragment is u sin u sin u cos + 2 u cos g g

Fig. 5.83

2T cos

mv – (M + m)V = 0 V m = or v M m Since the distance moved is proportional to speed, the displacement L of the plank is given by L V m = M m L v

M ) g}2

= [M2 + (m + M)2]1/2 g which is choice (d). 66. Before the boy starts walking on the plank, both the boy and the plank are at rest. Therefore, the total momentum of the boy–plank system is zero. If the boy walks with a speed v on the plank and as a result if the speed of the plank in the opposite direction is V, then the total momentum of the system is mv – (M + m)V. From the principle of conservation of momentum, we have

3u 2 sin 2 u 2 sin 2 u 2 sin 2 2g 2g g Hence the correct choice is (b). 68. The mass of water stream striking against the wall in 1 second = av . Hence, the change in its momentum per second is (av )v – (– av )v = 2a v2. The normal component of the rate of change of momentum and, therefore, force is 2a v2 cos . Hence the correct choice is (a). 69. The acceleration of the block sliding down the smooth inclined plane is a1 = g sin and down the rough inclined plane is a2 = g sin – g cos . Given t1 = t and t2 = 2t. If the length of the inclined plane is s, we have 1 1 s = a1t12 = a2t 22 2 2 =

a1t 21 = a2t 22

or or

t2 = (g sin

g sin

or

sin

= 4 (sin

3 3 tan = 4 4 Hence the correct choice is (d).

which gives

=

– g cos ) –

(2t)2

cos ) (

= 45°)

Laws of Motion and Friction 5.39

70. We use the relation v2 – u2 = 2as. Since u = 0, we have v2 = 2as. s ( v1= v) Now v 21 = 2a1s or v2 = 2g sin 2 v and v22 = 2 a2s or 2 = 2(g sin – g cos ) s n Dividing, we get or n2 (sin – cos ) = sin 1 which gives = 1 tan , which is choice (a). n2 71. The force of friction between the block and the belt is f = mg, where m is the mass of the object. This force produces an acceleration of the block which is given by force mg = g a= mass m The block will slide on the belt without slipping until its speed (v) becomes equal to the speed of the belt. Since u = 0, we have v2 = 2 as 52 v2 v2 or s = = 2.5 m 2a 2 g 2 0.5 10 Hence the correct choice is (b). 72. Normal reaction R = f. Therefore, force of friction = R = f. The net downward force F = mg F mg f = – f. Hence, the acceleration a = m m f g – . Hence the correct choice is (d). m 73. As the boy is climbing the pole at a constant speed (no acceleration), the force of friction must be just mg balanced by his weight, i.e. R = mg or R = = 40 10 = 500 N. Hence the correct choice is (c). 0.8 74. From the principle of conservation of momentum, we have m2 v m1v1 = m2v2 or 1 (i) v2 m1

75. The forces acting on the balloon are its weight acting downwards and upthrust F acting upwards. Thus F – Mg = Ma (i) When mass m is removed, we have F – (M – m) g = (M – m)a

(ii)

where a is the new acceleration. Eliminating F from (i) and (ii) and simplifying we get Ma mg a = M m which is choice (a). 76. Let m be the mass per unit length of the rope. Let x t. The weight of this part is F1 = mgx Now, if a small part dx dt, F2 = rate of change of momentum mdx v = mv2 = dt dx Now = v, where v is the velocity of that part dt of the rope at that instant. But v2 = 2gx. Hence F2 = mv2 = m (2gx) = 2mgx. Total force F = F1 + F2 = mgx + 2mgx = 3mgx = 3F1 Hence the correct choice is (c). 77. Refer to Fig. 5.84. Vertical component of F is F sin and the horizontal component is F cos .

Fig. 5.84

opposite force F on each block. Let a1 and a2 be the accelerations of blocks m1 and m2 respectively. Then a m1 F = m1a1 = m2a2 or 2 (ii) a1 m2 Also v21 = 2a1x1 and v22 = 2a2x2, which give x1 = x2

v12 a2 . v22 a1

m2 m1

2

[Use Eqs. (i) and (ii)] Hence the correct choice is (b).

m1 m2

=

m2 m1

Thus R + F sin

= mg R = mg – F sin

or Frictional force R =

(mg – F sin ). Also

(mg – F sin ) = F cos or F mum, i.e. if

F=

mg sin cos

(i)

5.40 Comprehensive Physics—JEE Advanced

d ( sin + cos ) = 0 d cos – sin = 0

or or

= tan , which is choice (a).

78. Now tan

= . Therefore, cos

1

= 1

sin

=

2

and

2

1

Using these in Eq. (i) above and simplifying, we get mg F= 2 1 Hence the correct choice is (d). 79. Force required to accelerate the body of mass m is F = ( s – k) mg = (0.75 – 0.5) mg = 0.25 mg F 0.25 mg Acceleration = = = 0.25 g, which is m m choice (a). 80. Let a be the acceleration at a time t of the blocks

v2 10 10 = = 12.5 m g 0.8 10 This is the minimum radius the curve must have for the car to negotiate it without sliding at a speed of 10 ms–1. Hence the correct choices is (b). 83. Speed of train (v) = 36 km h–1 = 10 ms–1 Radius of the curve (R) = 200 m Distance between rails (x) = 1.5 m Let the outer rails be raised by a height h with respect to the inner rails so that the angle of banking is (Fig. 5.85). or

Rmin =

Then

tan

=

h=

or

h v2 = Rg x xv 2 1.5 (10)2 = Rg 200 10

= 0.075 m = 7.5 cm Thus, the correct choice is (a).

acceleration is F = (m + m) a = 2 ma F =2ma

displacement, i.e. F = kA

(1)

Fig. 5.85

(2)

Equating (1) and (2), we get a

=

kA 2m force of friction = ma

kA kA 2m 2 Hence the correct choice is (b). 81. Since the block is held stationary, it is in translational as well as rotational equilibrium. Hence no net force and no net torque acts on the block. No net force will act on the block if f = mg and N = F. No net torque will act on the block, if torque by frictional force f about centre O = counter torque by normal reaction N about centre O. Hence choice (d) is false. 82. Speed of car (v) = 36 km h–1 = 10 ms–1 mum centripetal force that friction can provide is =m

f

=

mg =

mv 2 R

84. Now v = 54 km h–1 = 15 ms–1, R = 50 m. The required angle of banking is given by v2 15 15 = = 0.45 Rg 50 10 Thus, the correct choice is (d). 85. The horizontal velocity v must be such that the centripetal force equals the weigth of the body, i.e. mv 2 = mg or v = gR , which is choice (b). R 86. The motor cyclist can leave the ground only at the highest point on the bridge. At this point, the centripetal force is mv2/R. He will not leave the ground if the centripetal force equals the weight mg. Thus mv 2 = mg or v = gR = 10 10 = 10 ms–1. R Hence, the correct choice is (a). 87. When the system is in equilibrium, the spring force = 3 mg. When the string is cut, the net force on block A = 3 mg – 2 mg = mg. Hence the acceleration of this block at this instant is tan

=

Laws of Motion and Friction 5.41

force on block A mg g = = mass of block A 2m 2 When the string is cut, the block B falls freely with an acceleration equal to g. Hence the correct choice is (c). dp F = 88. Force dt d = [A{ i cos (kt) – j sin (kt)] dt a=

Fig. 5.87

The block will topple if the torque due to normal reaction N about O mg sin about O, i.e. N OA = mg sin OB

= Ak [– i sin (kt) – j cos (kt)] Now F p = Ak[– i sin (kt) – j cos kt] A[ i cos (kt) – j sin (kt)]

mg cos

= A2k[– sin (kt) cos (kt) + cos (kt) sin kt] =0(

F = T sin

+ T sin

T cos

15 cm 2

=

for sliding, the

91.

= 2 T sin

F 2sin = mf

T= Also

2 34°. 3 Since for toppling is less than correct choice is (b). tan

i i = j j = 1 and i j = 0)

Hence the angle between F and p is 90° which is choice (d). 89. Refer to Fig. 5.86. Let f be the force producing the acceleration of each mass. It follows from the

5 cm = mg sin

(1) (2)

Using (1) and (2), we get f=

F cos 2m sin

=

F 2m tan

=

Fx

2m a 2 Hence the correct choice is (b).

x2 Fig. 5.88

For the bead to stay at rest, (see Fig. 5.88) N cos = mg N sin

= ma a which give tan = . Now g dy d tan = slope of the curve = = (kx2) = 2 kx dx dx 2kx = Fig. 5.86

90. The block will just begin to slide if the downward force mg sin just overcomes the frictional force, i.e. if mg sin = N = mg cos tan = = 3 = 60° (see Fig. 5.87)

a g

x=

a 2 gk

92. Radius of the circular path is BC = r = L sin , where L = AB is the length of the string. The vertical component T cos of tension T balances with the weight mg and the horizontal component T sin provides the necessary centripetal force for circular motion. Hence

5.42 Comprehensive Physics—JEE Advanced

T sin = mr

2

T = mL

2

T

= mL

324 = 0.5

= m(L sin )

2

2

0.5

2

= 36 rad s–1

Fig. 5.89

II Multiple Choice Questions with One or More Choices Correct 1. Which of the following statements are true? No net force acts on (a) a drop of rain falling vertically with a constant speed

(c) can go up to a speed of 100 ms–1 in 10 s. (d) after acquiring a speed of 50 ms–1, can come to rest, with the engine shut off and brakes not applied, in a time of 10 s. 4.

(c) a car moving with a constant velocity on a rough road (d) a body moving in a circular path at constant speed. 2. In which of the following situations would a force of 9.8 N act on a stone of mass 1 kg? Neglect air (a) Just after it is dropped from the window of stationary train (b) Just after it is dropped from the window of a train running at a constant speed of 36 km h–1. (c) Just after it is dropped from the window of a train accelerating at 1 ms–2. which is accelerating at 1 ms–2. 3. a car and the ground is 0.5. The car starts from rest and moves along a perfectly horizontal road. If g = 10 ms–2, the car 5 ms–2 without slipping. (b) can attain a speed of 20 ms–1 in a minimum distance of 40 m.

speed of 15 ms–1 gushes out of a tube of cross sectional area 1 cm2, strikes against a kachha ver-

N on it. The impact of the water stream on the wall will ( to the water stream is only about 1500 N. (b) will break it. 105 pascal on the wall. 5. A block is released from the top of a smooth inclined plane. Another block of the same mass is allowed to fall freely from the top of the inclined plane. Both blocks reach the bottom of the plane (a) in equal time (b) with equal speed (c) with equal momentum (d) with equal kinetic energy 6. A body of mass 200 g is moving with a velocity of 5 ms–1 along the posotive x- direction. At time t = 0 when the body is at x = 0, a constant forse of 0.4 N

Laws of Motion and Friction 5.43

7.

8.

9.

10.

directed along the negative x-direction is applied to the body for 10s. (a) At time t = 2.5 s, the body will be at x =1.25 m (b) At time t = 2.5 s, the speed of the body will be zero. (c) At time t = 30 s, the body will return to x = 0. (d) At time t = 30 s, the speed of the body will be 15 ms–1 A train starts from rest with a constant acceleration a = 2 ms–2. After 5 seconds, a stone is dropped from the window of the train. If g = 10 ms–2. (a) the magnitude of the valocity of the stone 0.2 second after it is dropped is 2 ms–1. (b) the angle between the resultant velocity vector of the stone and the horizontal 0.2 second after it is dropped is = tan–1 (0.2). (c) the acceleration of the stone after it is dropped is a = 2 ms–2. (d) the acceleration of the stone after it is dropped is g = 10 ms–2. A man of mass m His weight when the lift is (a) stationary is mg (b) moving up with a uniform speed of 2 ms–1 is 5 mg. (c) moving up with a uniform acceleraiton a (< g) is m (g + a). (d) moving down with a uniform acceleration a (< g) is m (g – a). Two identical blocks, each of mass m, connected by a light string, are placed on a rough horizontal surface. When a force F is applied on a block in the horizontal direction, each block moves with an acceleration a. Assuming that the frictional forces on the two blocks are equal, (a) the tension in the string will be F. (b) the tension in the string will be F/2. (c) the frictional force on each block will be (F – ma). (d) the frictional force on each block will be F ma . 2 A body of weight W is suspended from a rigid support P by means of a massless string as shown in Fig. 5.90. A horizontal force F is applied at point O of the rope. The system is in equilibrium when the string makes an angle with the vertical. If the tension in the string is T, (a) F = T sin

(b) W = T sin

W2

(c) T =

F2

(d) F = W tan

Fig. 5.90

11. Two blocks of masses m1 and m2 are connected by a string of negligible mass which passes over a plane as shown in Fig. 5.91. m1 and the plane is . (a) If m1 = m2, the mass m1 up the inclined plane when the angle of inclination is , then = tan . (b) If m1 = m2, the mass m1 up the inclined plane when the angle of inclination is , then = sec – tan . (c) If m1 = 2 m2, the mass m1 down the plane if = 2 tan . (d) If m1 = 2 m2, the mass m1 1 down the plane if = tan – sec . 2

m1 m2 q

Fig. 5.91

12. Two blocks A and B are connected to each other by a string and a spring of force constant k, as shown in Fig. 5.92. The string passes over a frictionless pulley as shown. The block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C of friction between the surfaces of the blocks is . If the mass of block A is m,

5.44 Comprehensive Physics—JEE Advanced

(a) (b) (c) (d)

the the the the

mass of block B is m. mass of block B is m/ . energy stored in the spring is m2g2/2k. energy stored in the spring is m2g2/k. B

T T k

C A

Fig. 5.92

13. Two blocks A and B of equal masses m each are connected to each other by a string passing over a frictionless pulley as shown in Fig. 5.93. The A and the surface below is 0.5. When the system is released, (a) the acceleration of the blocks is 3g/4. (b) the acceleration of the blocks is g/4. (c) the tension in the string is 3mg/4. (d) the tension in the string is mg/4.

(b) the tension in the rope at t = 6 s will be 10,000 N. (c) the tension in the rope at t = 11 s will be 8,000 N. (d) the height upto which the lift takes the passengers is 40 m. 15. Two blocks of masses m1 = 2 m and m2 = m are connected by a light string passing over a friction less pulley as shown in Fig. 5.95. When they are released, g (a) the acceleration of m1 is 3 vertically downwards. (b) the acceleration of m2 is 2g vertically upwards. 3 (c) the tension in the string is 4g Fig. 5.95 . 3 (d) the distance through which m2 rises in t secgt 2 onds after the blocks are released is . 6 16. Two blocks of masses m1 and m2 connected by a frictionless surface as shown in Fig. 5.96. A force F is applied to m2 in the horizontal directon as shown. F . (a) the acceleration of each block is (m1 m2 )

Fig. 5.93

14. A lift is going up. The variation of the speed of the lift with time is shown in Fig. 5.94. The total mass of the lift and passengers is 1000 kg. If g = 10 ms–2,

m1 F . (m1 m2 ) (c) the tension in the string is F.

(b) the tension in the string is

(d) the force on mass m1 is

m2 F . (m1 m2 )

Fig. 5.96

Fig. 5.94

(a) the tension in the rope of the lift at t = 1 s will be 12,000 N.

17. Ten coins, each of mass m, are placed on top of each other on a horizontal table. (a) The force on the 7th coin (counted from the bottom) due to all the coins above it is 3mg vertically downwards. (b) The force on the 7th coin by the 8th coin (both counted from the bottom) is 3mg vertically downwards.

Laws of Motion and Friction 5.45

(c) The force on the 7th coin by the 8th coin (both counted from the bottom) is 2mg vertically upwards. (d) The reaction force of the 6th coin from the bottom on the 7th coin from the bottom is 4mg vertically upwards. 18. Two blocks of masses m1 and m2 are placed in contact on a horizontal frictionless surface as shown in Fig. 5.97. A force F is applied to mass m1 as shown. (a) The acceleration of mass m2 is F/m2 m2 F m2 is . (m1 m2 ) m2 is F. F (d) The acceleration of mass m1 is . (m1 m2 )

Fig. 5.97

19. Two blocks of masses m1 and m2 (m2 < m1) are placed on an inclined plane of inclination and joined by a string as shown in Fig. 5.98.

l = 20% L l = 25% (d) If = 0.25, L 21. A block starts sliding from the top of an inclined plane of inclination between the block and the plane varies as = kx where x is the distance moved down the plane and k is a positive constant. (a) The block has a uniform acceleration along the plane. (b) The acceleration of the block increases for tan . x< k (c) If

= 0.25,

tan . k tan (d) The block starts decelerating for x > . k 22. A block of mass m is lying at x = 0 on a smooth horizontal surface. A variable force F = kx is applied to it as shown in Fig. 5.99 where k is a constant. Then (a) the block will move on the surface with a uniform acceleration. (b) the block will move on the surface with a variable acceleration. (c) the block will lose contact with the surface mg . after travelling a distance x0 = k sin (d) the block will always remain in contact with the surface. x=

Fig. 5.98

and the plane is and the blocks are released, (a) the acceleration of the blocks is g(sin – cos ). (b) the acceleration of the blocks is zero. (c) the tension in the string is zero. (d) the tension in the string is (m1 + m2)g (sin – cos ). 20. A uniform chain of length L is placed on a rough tween the chain and the table is length of the chain that can hang from the edge of the table is l. Then L (a) l = (1 ) L (b) l = (1 )

Fig. 5.99

23. A pendulum of length L and bob mass M is oscillating in a plane about a vertical line between angular limits – and + . For an angular displacement | | < , the tension in the string is T and the velocity of the bob is V. Then (a) T cos = Mg MV 2 (b) T – Mg cos = L (c) The magnitude of the tangential acceleration of the bob is |at| = g sin (d) T = Mg cos IIT, 1986

5.46 Comprehensive Physics—JEE Advanced

24. When a bicycle is in motion, the force of friction that it acts (a) in the backward direction on the front wheel and in the forward direction on the rear wheel. (b) in the forward direction on the front wheel and in the backward direction on the rear wheel.

(c) in the backward direction on both the front and the rear wheels. (d) in the forward direction on both the front and the rear wheels. IIT, 2007

ANSWERS AND SOLUTIONS 1. Choice (a) is true. Since the direction of motion as well as the speed of the drop of rain are constant, its velocity is constant and hence no net force acts on the drop. The downward gravitational force (i.e. weight) of the drop is completely cancelled by the upward buoyant force on it due to air and the viscous force due to its motion. Choice (b) is also true. The weight of the cork is balanced by the upward buoyant force due to water. Choice (c) is also true. The force due to the car’s acceleration is balanced by the backward force of friction between its tyres and the rough road. Choice (d) is false. Although the speed of the body is constant, its velocity is changing all the time because the direction of motion keeps changing. Hence, the motion of the body is being accelerated. A force (called centripetal force) must act on the body to produce the acceleration. Thus, choices (a), (b) and (c) are correct. 2. In choice (a), a force F = mg = 1 9.8 = 9.8 N acts on the stone vertically downwards. In choice (b), the velocity of the train is constant. Hence, there is no acceleration (and therefore, force) along the direction of motion. When the stone is dropped, the only force acting on it is F = mg = 9.8 N vertically downwards. In choice (c), before the stone is dropped, a force = ma = 1 kg 1 ms–2 = 1 N acts on it in the horizontal direction. But, after it is dropped, this force ceases to act because the stone is no longer located in the accelerating system (i.e. train). Hence, in this case also, the net force on the stone is 9.8 N vertically downwards. In choice (d), the weight of the stone is balanced by the normal accelerated in the forward direction along with the train. Hence, the force acting on the stone = ma = 1 kg 1 ms–2 = 1 N in the direction of motion of the train. Thus, the correct choices are (a), (b) and (c). 3. ping is given by ma = mg or a = g = 0.5 10 = 5 ms–2 tance s covered to acquire a speed v = 20 ms–1 is

given by v2 – u2 = 2as or (20)2 – 0 = 2 5 s or s = 40 m. Further, since the frictional retardation is 5 ms–2, the time needed to come to rest from a speed of 50 ms–1 is t = v/a = 50/5 = 10 s. Hence, choice (a), (b) and (d) are correct. 4. The rate of change of momentum of water stream = av2 – 0 = av2 . This is the force F water stream on the wall. F = av2 = (1 10–2) (15)2 1000 = 2250 N The pressure on the wall is F 2250 = 2.25 105 pascal a 1 10 2 Hence, the correct choices are (b), (c) and (d). 5. Since the acceleration along the inclined plane (g sin ) is less than g, the blocks take different times to reach the bottom. The speed of each block on reaching the bottom is v = 2g h , where h is the height of the inclined plane. Thus choice (b) is correct. The directions of the velocity are different, hence their momenta are not the same. Their 1 kinetic energy mv2 is the same. Hence the cor2 rect choices are (b) and (d). 6. Given u = + 5 ms–1 along positive x-direction F = – 0.4 N along negative x-direction P=

m = 200 g = 0.2 kg F 0.4 = = –2 ms–2. The The acceleration a = m 0.2 negative sign shows that the motion is retarded. The position of the body at time t is given by 1 x = x0 + ut + at2 2 At t = 0, the body is at x = 0. Therefore, x0 = 0. Hence 1 x = ut + at2 2 Since the force acts during the time interval from t = 0 to t = 10 s, the motion is accelerated only between t = 0 and t = 10 s. The position of the body t = 2.5 s is given by

Laws of Motion and Friction 5.47

1 (– 2) (2.5)2 = 1.25 m 2 The velocity of the body at t = 2.5 s is v = u + at = 5 + (– 2) (2.5) = 5 – 5 = 0 t = 0 to t = 10 s) the motion is accelerated. During this time a = – 2 ms–2. Putting u = 5 ms–1, a = – 2 ms–2 1 and t = 10 s in equation x = ut + at2. We have 2 1 x1 = 5 10 + (– 2) (10)2 = – 50 m (i) 2 The velocity of the body at t = 10 s is v = u + at = 5 + (– 2) 10 = – 15 ms–1 During the remaining 20 seconds, i.e. from t = 10 s to t = 30 s, the acceleration a = 0, because the force ceases to act after t = 10 s. The velocity of the body remains constant at – 15 ms–1 during the last 20 seconds. The distance covered by the body during the last 20 seconds is x2 = – 15 20 = – 300 m Position of the body at t = 30 s is x = x1 + x2 = – 50 – 300 = – 350 m The magnitude of the velocity (i.e. speed) of the body at t = 30 s is 15 ms–1. Hence the correct choices are (a), (b) and (c). 7. Given u = 0, a = 2 ms–2. Since the stone is located in the train, the acceleration of the stone is a = 2 ms–2. At time t = 5 s, the velocity of the stone is v = u + at = 0 + 2 5 = 10 ms–1. Before the stone is dropped, its motion is accelerated with the train. But, the moment it is dropped, its acceleration due to the motion of the train ceases. Therefore, after the stone is dropped, it has the following two motions: (a) a uniform motion with velocity 10 ms–1 parallel to the ground, i.e. vx = 10 ms–1 (the horizontal velocity) (b) an accelerated motion vertically downwards due to gravity. In time t = 0.2 s, the vertical velocity of the stone is vy = 0 + gt = 10 0.2 = 2 ms–2. The resultant velocity of stone at t = 0.2 s is x=5

v=

v 2x

2.5 +

v 2y =

(10) 2

( 2) 2 =

104

= 2 26 ms–1 The angle, which the resultant velocity vector, makes with the horizontal is given by vy 2 = = 0.2 tan = vx 10

After the stone is dropped, the horizontal velocity vx remains unchanged because the acceleration is zero along the horizontal direction. The only acceleration of the stone is the acceleration due to gravity. Hence the correct choices are (b) and (d). 8. mg) on the on him an upward reaction for (R) which it measures. When the lift is stationary, R = mg. When the lift is moving up or down with a uniform speed, it has no acceleration of its own. Hence, the reading of the machine will still be mg. When the machine is moving downwards with an acceleration a, a force F = ma acts downwards. But the reaction R = mg acts upwards. Hence, the effecting reading will be Reff = F – R = mg – ma = m(g – a) In case (c), R and mg both act in the same direction (upwards). Hence, the effective reading will be Reff = mg + ma = m(g + a) Hence the correct choices are (a), (c) and (d). 9. Refer to Sec. 5.11 and Fig. 5.15 on page 5.9 Equations of motion of m1 and m2 are m 1a = T – f (i) and m 2a = F – T – f (ii) Subtracting the two equations, we have (m1 – m2)a = 2T – F Since m1 = m2, we get, we get 0 = 2T – F F 2 Putting T = 10 N in Eq. (i) above, we have f = T – m 1a F – ma = 2 Hence the correct choices are (b) and (d). 10. The mass is in equilibrium at point O under the action of the concurrent forces F, T and W = mg. Therefore, as shown in Fig. 5.100, the horizontal component T sin of tension T must balance with force F and the vertical component T cos must balance with weight W = mg. or

T=

Thus

F = T sin

(i)

and

W = T cos

(ii)

Squaring Eqs. (i) and (ii) and adding we get T 2 = W 2 + F 2. Dividing (i) by (ii) we get F = W tan . Hence the correct choices are (a), (c) and (d).

5.48 Comprehensive Physics—JEE Advanced

and T – mg = ma (ii) Adding (i) and (ii) we get 1 1 0 .5 g g= g= a= 2 2 4 Using a = g/4 in Eq. (i) gives T = 3mg/4 Hence the correct choices are (b) and (c). 14. Between t = 0 and t = 2 s, the acceleration of the lift is a=

Fig. 5.100

11. The block m1 will just begin to move up the plane if the downward force m2g due to mass m2 trying to pull the mass m1 up the plane just equals the force (m1g sin + m1g cos ) trying to push the mass m1 down the plane, i.e. when m2g = m1g(sin + cos ) Now, it is given that m1 = m2 = m. Therefore, we have 1 = sin + cos = sec – tan The block m1 will just begin to move down the plane if the downward force (m1g sin – m1g cos ) on m1 just equals the upwards force m2g acting on m1 due to m2, i.e. if m2g = m1g(sin

–

cos )

m1 1 = m2 sin cos m1 = 2m2, then we have

or If

2 =

PE =

–

1 1 mg kx2 = k 2 2 k

1

= 2 ms–2

BE 4 ms 1 = 2 ms–2 EC 2s Tension = m (g – a) = 1000 (10 – 2) = 8,000 N The height to which the lift rises = area of OABC = 40 m. Hence all the four choices are correct. 15. The correct choices are (a), (c) and (d). The acceleration of each mass is the same and is given by a=

(m1 m2 ) g (m1 m2 )

2m1 m2 g (m1 m2 ) Since the masses start from rest, the distance moved by m1 in time t is 1 2 1 2 1 2 s = ut + at = 0 + at = at 2 2 2 16. The correct choices are (a), (b) and (d). The acceleration of each mass is given by The tension in the string is T =

cos

1 sec 2 Hence the correct choices are (b) and (d). 12. Since the blocks slide at the same uniform speed, no net force acts on them. If M is the mass of block B, then the tension in the string is T = Mg. Also T = mg. Equating the two, we get M = m m or M = F mg x= = . Therefore, k k potential energy stored in the spring is = tan

4 ms 2s

Since the lift is accelerating upward, the tension in the rope at t = 1 s (between t = 0 and t = 2 s) is T = m (g + a) = 1000 (10 + 2) = 12,000 N Between t = 2 s and t = 10 s, the speed of the lift is constant. Hence a = 0 and T = mg = 1000 10 = 10,000 N Between t = 10 s and t = 12 s, the lift is decelerating. Its deceleration is given by

a=

1 sin

AD OD

2

2

=

m g 2k

2

Hence the correct choices are (b) and (c). 13. If the acceleration of the blocks is a, then we have mg – T = ma (i)

a=

F (m1

m2 )

The tension in the string is T = m1 a. The force on mass m1 is m2 F (m1 m2 ) th 17. The 7 coin from the bottom has 3 coins above it. Hence, the force on the 7th coin = weight of 3 coins = 3 mg, vertically downwards. Since the 8th coin has 2 coins above, it supports the weight of two coins. Hence the force on the 7th F1 =

Laws of Motion and Friction 5.49

coin by the 8th coin = weight of 8th coin + weight of two coins above it = weight of three coins = 3 mg vertically downwards. From Newton’s third law, the reaction force th coin on the 7th coin is equal and th coin th th on the 6 th th coin on the 6 coin = weight of 7 coin + weight of 3 coins above it = weight of 4 coins = 4 mg vertically downwards. Hence, the reaction of the 6th coin on the 7th coin = 4 mg vertically upwards. Hence the correct choices are (a), (b) and (d). 18. The correct choices are (b) and (d). The acceleration of both the masses is the same and is given by net force F = a= total mass (m1 m2 ) Force on mass m2 = m2 a. 19. Figures 5.101 (a) and 5.101 (b) show the free body diagram of the two blocks.

Solving Eqs. (i) and (ii), we get T = 0 and a = g(sin

–

cos )

Hence the correct choices are (a) and (c). 20. Let M be the mass of the chain and L its length. If a length l hangs over the edge of the table, the Ml force pulling the chain down is g. The force L of friction between the rest of the chain of length M (L l) (L – l) and the table is g. L For equilibrium, the two forces must be equal, i.e. Ml M (L l) g= g L L or

l=

or

l=

(L – l) L 1

l 0.25 1 = = = or 20%. L 1 1 0.25 5 Hence the correct choices are (a) and (c). 21. The acceleration of the block is given by a = g(sin

–

cos

)

= g(sin – kx cos ) Acceleration a is not uniform; it varies with x. For sin > kx cos , a is positive, i.e. a is positive for x < tan /k. For x > tan /k, a is negative. For x = tan /k, a after which it begins to decrease as the block is decelerated. Hence the correct choices are (b), (c) and (d). 22. The horizontal and vertical components of F are F cos and F sin (see Fig. 5.102).

Fig. 5.101

T is the tension in the string and f1 and f2 are the frictional forces. It follows from the diagrams that N1 = m1g cos and f1 = m1g cos N2 = m2g cos and f2 = m2g cos If a is the acceleration of the blocks down the plane, the equations of motion are m1a = m1g sin – T – f1 = m1g sin – T – m1g cos (i) + T – f2 and m2a = m2g sin = m2g sin + T – m2g cos (ii)

Fig. 5.102

N + F sin

= mg N = mg – F sin

= mg – kx sin where N is the normal reaction.

5.50 Comprehensive Physics—JEE Advanced

The block will lose contact with surface at x = x0 for which N = 0. Putting N = 0 and x = x0, we mg have 0 = mg – kx0 sin x0 = k sin The acceleration a of block along the horizontal surface is given by ma = F cos

It follows from Fig. 5.103, that MV 2 = T – Mg cos L Tangential force ft = Mg sin Tangential acceleration at = g sin . So the correct choices are (b) and (c).

= kx cos

24.

k x cos a= m which depends upon x. Hence the correct choices are (b) and (c).

the force of friction acts in the opposite direction. But when a body itself applies a force in order to move, the force of friction acts in the direction of plied to the rear wheel and as a result the front wheel moves by itself. So, while pedalling, the choice (a) is correct. When pedalling is stopped, the choice (c) is correct as long as bicycle remains in motion.

23.

Fig. 5.103

III Multiple Choice Questions Based On Passage Question 1 to 3 are based on the following passage of reference; it holds only for inertial frames of reference. Passage I An inertial reference frame is a frame which moves with Reference Frames a constant velocity. 1. A reference frame attached to the earth reference. Rest and motion are relative; there is nothing like absolute rest or absolute motion. The position or state (b) cannot be an inertial frame because the earth of motion of a body may appear different from different is revolving round the sun (c) is an inertial frame because Newton’s laws everything else in a moving train are at rest in a reference of motion are applicable in this frame (d) cannot be an inertial frame because the earth frame situated in the train but they are in motion in a reference frame situated on the platform. Similarly, a stone dropped by a passenger from the window of a railway 2. Which of the following observers are inertial? (a) A child revolving in a merry-go-round carriage in uniform motion appears to him to fall vertically (b) A driver in a car moving with a constant downwards but to a person outside the carriage, it appears velocity to follow a parabolic path.

Laws of Motion and Friction 5.51

(c) A pilot in an aircraft which is taking off (d) A passenger in a train which is slowing down to a stop. 3. Choose the correct statements from the following. (a) An inertial frame is non-accelerating. (b) An inertial frame is non-rotating.

(c) A reference frame moving at a constant velocity with respect to an inertial frame is also an inertial frame. (d) Newton’s laws of motion hold for both inertial and non-inertia frames.

SOLUTION 1. The velocity of the earth changes with time (due to a change in its direction) as it revolves round the sun. Therefore, a frame attached to the earth is accelerated. Accelerated frames and rotating frames of reference are not inertial frames. Hence the correct choices are (b) and (d). Question 4 to 6 are based on the following passage Passage II Two bodies A and B of masses m and 2 m respectively are moving with equal linear momenta. They are subjected to the same retarding force. 4. If x1 and x2 are the respective distances moved by them before stopping, then x1/x2 is 1 (a) (b) 1 2 (d) 2 (c) 2

2. The observers in (a), (c) amd (d) are all accelerating. Hence, they are non-inertial. Only the driver in (b) is inertial since his motion is not accelerated. 3. The correct choices are (a), (b) and (c).

5. If t1 and t2 are the respective times taken by them to stop, then t1/t2 is (a) 1 (b) 2 1 1 (c) (d) 2 4 6. If a1 and a2 are their respective decelerations, then a1/a2 is (a) 1 (c)

(b) 2 2

(d)

1 2

SOLUTION (mu ) 2 p 2 = m m where F = ma is the retarding force p = mu is linear momentum. Thus

4. 2 ax = u2

2 max = mu2

x=

2 Fx =

p2 2 Fm

1 Since p and F are constants, x . Hence the m correct choice is (d). u mu p 5. 0 = u + at t=– =– =– . Hence a ma F t1 = t2. Thus the correct choice is (a). 6. a1 = F/m and a2 = F/2m. Hence the correct choice is (b).

Questions 7 to 9 are based on the following passage Passage III Two bodies of masses m and 2 m respectively are moving with equal kinetic energies. They are subjected to the same retarding force. 7. If x1 and x2 are the respective distances moved by them before stopping, then x1/x2 is (a) 2 (b) 2 1 (c) (d) 1 2

8. If t1 and t2 are the respective times taken by them to stop, then t1/t2 is 1 1 (a) (b) 2 2 (c) 2 (d) 2 9. If a1 and a2 are their respective decelerations, then a1/a2 is (a) 4 (b) 2 1 (c) (d) 1 2

SOLUTION 1 mu2. If K 2 is the kinetic energy, then Fx = K x = K/F.

7. 2ax = u2

2 max = mu2

Fx =

Since K and F are constants, the correct choice is (d).

5.52 Comprehensive Physics—JEE Advanced

8. 0 = u + at

t=–

u mu p =– =– . Now a ma F

p2 = m2u2 = (2m)

p = mu

1 2 mu 2

= 2mK.

Passage IV A block of masses m is initially at rest on a frictionless horizontal surface. A time-dependent force F = at – bt2 acts on the body, where a and b are positive constants. 10. t1 given by a 2a (a) (b) b b a a (c) (d) 2b 2b 11. F is given by

(c)

(d)

I

(a)

(c)

a3

(b)

3b 2 a3 9b 2

13.

d 2F

dt negative.

2

=

a3

a3

(b)

4mb 2 a3

(d)

12mb 2

a3 8mb 2 a3 16mb 2

12. t1

I

a –b 2b Hence the correct choice is (b).

a 2b

bt 2 )dt

0

=

2

=

a2 . 4b

a a 2 b a 3 a3 – = 2 2b 3 2b 12b 2 Hence the correct choice is (d). 13. Now impulse = change in momentum = mv – 0 = mv I a3 , which is choice (c). v = ma = m 12mb 2

(c) 2

Passage V A body of mass m is initially at rest. A periodic force F = a cos(bt + c) is applied to it, where a, b and c are constants.

at12 bt 3 – 1 2 3

=

Questions 14 to 17 are based on the following passage

14. The time period T of the force is 1 2 (b) (a) b b

(at

=

d (a – 2 bt) = – 2 b, which is dt

= at1 – bt21 = a

Fdt

= 0 t1

Hence the correct choice is (c). 11. F

6b 2

12b 2 attained by the block is

v (a)

a3

(d)

dF d Now = (at – bt2) = a – 2bt dt dt dF = 0 and t = t1, we get Putting dt a 0 = a – 2 bt1 t1 = 2b Also

imparted to the block

is given by

4a 2 b dF d 2F = 0 and dt dt 2

< 0.

1 Hence the correct choice is (b). m

(c)

SOLUTION 10.

F , i.e. a m

12.

a2 (b) 4b

2a 2 b

m . Hence the

9. a =

Questions 10 to 13 are based on the following passage

a2 (a) 2b

2mK . Thus t F correct choice is (a).

Hence t = –

a b

(d) 2

b a

15. a mb c (c) ma (a)

b mc b c (d) ma

(b)

Laws of Motion and Friction 5.53

16. The smallest value of t after t = 0 when the velocity of the body becomes zero is given by a (b) t1 = (a) t1 = a c c

(c) t1 =

b

(d) t1 =

b

17. The distance travelled by the body from time t = 0 to t = t1 is given by (a)

(c)

a mb

cos c

2

(b)

a2 cos c mb

(d)

2a mb 2

sin c

2a 2 sin c mb

SOLUTION 14. F will repeat itself at values of t given by cos(bt + c) = + 1, i.e. bt + c = 0, 2 , 4 , t=–

c 2 c 4 c , , , b b b

2 . Hence the The smallest time interval is T = b correct choice is (b). dp 15. From Newton’s second law of motion, F = dt dv =m dt

16. From Eq (i) it follows that v = 0 at values of t given by sin(bt + c) = 0 or (bt + c) = 0, , 2 , ... c c 2 c or t = – , , . Therefore, b b b –

b

c b

=

t1

t

vdt =

x= 0

a cos(bt + c)dt m

a 1 sin (bt + c)dt mb 0 =–

a a cos (bt + c)dt = sin(bt + c) (i) m0 mb

=–

bt + c) = 1, a v = . mb Hence the correct choice is (a).

=–

a mb 2 a mb 2 a mb

2

cos(bt1 + c) cos b

b

cos( + c) =

c a cos c mb 2

Hence the correct choice is (a).

Questions 18 to 20 are based on the following passage Passage VI

(c) a1 > a2 = a3

(d) a1 = a2 = a3

19. The tension in the string between masses m2 and m3 is

Three masses m1 = m, m2 = 2 m and m3 = 3 m are hung on a string passing over a frictionless pulley as shown in Fig. 5.104. The mass of the string is negligible. The system is then released. 18. If a1, a2 and a3 are the accelerations of masses m1, m2 and m3 respectively, then

(a) mg

Fig. 5.104

(b) 3 mg

5mg 3 20. The tension in the string between masses m1 and m2 is 2mg (a) 4 mg (b) 3 5mg (c) (d) 2 mg 3 (c) 4 mg

(a) a1 < a2 < a3 (b) a1 > a2 > a3

, which is choice (d).

dx dx = v dt. Therefore, the distance dt moved between t = 0 and t = t1 is

t

v=

b

17. Now v =

dv Thus m = a cos(bt + c) dt dv =

c

t1 =

(d)

SOLUTION 18. When the masses are released, mass m1 moves upward and masses m2 and m3 move downward with a common acceleration given by

(m2 m3 m1 ) g (2m 3m m) g 2g = = (m1 m2 m3 ) (m 2m 3m) 3 The correct choice is (d).

a=

5.54 Comprehensive Physics—JEE Advanced

19. The let T be the tension in the string between m1 and m2 and T be the tension in the string between m2 and m3 [see Fig. 5.105 (a)]. Figure 5.105 (b) shows the free-body diagram of mass m3.

m 3 g – T = m 3a T = m3(g – a) = 3 m

g

2g 3

= mg

Hence the correct choice is (a). 20. Figure 5.105(c) shows the free-body diagram of mass m1. T – m 1g = m 1a 2g 3 Hence the correct choice is (c). T = m1(g + a) = m

g

=

5mg 3

Fig. 5.105

Questions 21 to 23 are based on the following passage Passage VII Three blocks of masses m1 = m, m2 = 2m and m3 = 3m connected by two strings are placed on a horizontal frictionless surface as shown in Fig. 5.106. A horizontal force F is applied to mass m1 as shown.

F F (d) 5m 6m 22. The force on mass m2 is 5F (a) (b) 2 F 6 (c)

2F (d) F 3 23. The force on mass m3 is F (a) F (b) 2 F F (c) (d) 3 6 (c)

Fig. 5.106

21. The common acceleration of the blocks is F F (b) (a) m 3m

SOLUTION 21. If F2 and F3 are the forces on masses m2 and m3 respectively, then the free-body diagrams of m1, m2 and m3 are as shown in Fig. 5.107 where a is the common acceleration of the system.

Adding Eqs. (1), (2) and (3) we get a=

(m1

F m2

m3 )

=

F F = (m 2m 3m) 6m

Hence the correct choice is (d). 22. Adding Eqs. (2) and (3) we have F2 = (m2 + m3)a = (2m + 3m) Fig. 5.107

F – F 2 = m 1a

(1)

F2 – F 3 = m 2a

(2)

F 3 = m 3a

(3)

which is choice (a). 23. From Eq. (3), we have F F = F3 = 3ma = 3m 6m 2 Hence the correct choice is (b).

F 5F = , 6m 6

Laws of Motion and Friction 5.55

Questions 24 to 26 are based on the following passage Passage VIII Two blocks of masses m1 = m and m2 = 2 m are connected by a light string passing over a frictionless pulley. The mass m1 is placed on a smooth inclined plane of inclination = 30° and mass m2 hangs vertically as shown in Fig. 5.108.

(c)

g 2

(d) g

25. If the system is released the tension in the string is 3mg (a) mg (b) 2 (c) 2mg

(d)

2mg 3

26. If the inclined plane was rough, it was found that when the system was released, block m1 remained at rest. The frictional force between block m1 and the inclined plane is (a)

3mg 2

(b) 3 mg

(c)

4mg 3

(d)

Fig. 5.108

24. If the system is released, the blocks move with an acceleration equal to g g (a) (b) 4 3

2mg 2

SOLUTION 24. Since the inclined plane is smooth and m2 > m1, block m1 will up the plane and block m2 will move vertically with a common acceleration a. If T is the tension in the string, the free-body diagrams of masses m1 and m2 are as shown in Fig. 5.109

a=

(m2

m1 sin ) g (2m m sin 30 ) g g = = m1 m2 m 2m 2

Hence the correct choice is (c). 25. From Eqs. (1) and (2), we get g = mg 2 Hence the correct choice is (a). 26. Since the blocks remain at rest, the equations of motions of blocks m1 and m2 are (here f is the frictional force on m1) T = m2(g – a) = 2m

T – m1g sin and

f = m2g – m1g sin

The equations of motion of the blocks are and

–f=0 T = m 2g

These equations give

Fig. 5.109

T – m1g sin

g

= m 1a

(1)

m 2g – T = m 2a

(2)

Equations (1) and (2) give Questions 27 to 29 are based on the following passage Passage IX Two blocks of masses m1 = 3 m and m2 = 2 m are suspended

=2m

g – m g sin 30° mg 3mg = 2 mg – = , 2 2 which is choice (b). A and B. Wire A has negligible mass and wire B has a mass m3 = m, as shown in Fig. 5.110. The whole system of blocks, wires and the support have an upward acceleration a.

5.56 Comprehensive Physics—JEE Advanced

27. The tension 1 (a) m(g 2 3 m(g (c) 2 28. The tension

at the mid-point C of wire B is 3 + a) (b) m(g – a) 2 5 + a) (d) m(g + a) 2 at point O of wire B is

(a) 3m(g + a)

(b) 3m(g – a)

(c) 2m(g + a)

(d) 2m(g – a)

29. The tension at the mid-point D of wire A is

Fig. 5.110

(a) 2m(g + a)

(b) 4m(g – a)

(c) 6m(g + a)

(d) 8m(g – a)

SOLUTION 28. Let T1 be the tension in wire A. Since this wire has negligible mass, the tension is the same (= T1) at every point on this wire. Let T2 be the tension at point O of wire B. Then, we have for wire A.

27. Refer to Fig. 5.111. Let T be the tension at the midpoint C of wire B. Then T– m2

m3 g 2

= m2

m3 a 2

T = m2

m3 (g + a) 2

= 2m

m (g + a) 2

=

5 m(g + a), 2

T 1 – T 2 – m 1g = m 1a where T2 is given by T2 – (m2 + m3)g = (m2 + m3)a T2 = (m2 + m3) (g + a) = (2m + m) (g + a) = 3m(g + a) Hence the correct choice is (a).

Fig. 5.111

29. Putting T2 = 3m(g + a) in Eq. (1), we get T1 = 6 m(g + a).

which is choice (d).

Hence the correct choice is (c).

IV Matching 1. Three blocks of masses m1 = 3 m, m2 = 2 m and m3 = m are placed in contact on a horizontal frictionless surface as shown in Fig. 5.112. A horizontal force F is applied to m1 as shown. Match items in column I with those in column II Column I Column II (a) (b) (c) (d)

Force acting on m2 if F = 12 N Force acting on m2 if F = 6 N Force acting on m3 if F = 12 N Forcer acting on m3 if F = 6 N

(p) (q) (r) (s)

1 3 2 6

N N N N

Fig. 5.112

(1)

Laws of Motion and Friction 5.57

SOLUTION The contact forces acting on m2 and m3 respectively are (m2 m3 ) F F2 = (m1 m2 m3 ) F3 =

and

(m1

m3 F m2 m3 )

Hence the correct matching is as follows (a)

(s)

(b)

(q)

(c)

(r)

(d)

(p)

2. Two blocks of masses M = 5 kg and m = 3 kg are placed on a horizontal surface as shown in Fig. 5.113. 1 = 0.5 and that between the blocks M and the horizontal surface is 2 = 0.7. Taking g = 10 ms–2, match items in column I with those in column II Column I Column II (a) Frictional force between the blocks (p) 5 ms–2 (b) Acceleration of the upper block (q) 96 N F applied to M so that the two blocks move together without slipping. (r) 4 ms–2 (d) The common acceleration of the Fig. 5.113 blocks if F = 32 N. (s) 15 N

SOLUTION 1mg

(a) Force of frictional between blocks =

= 0.5

(b) Acceleration of the upper block of mass m is a =

3

10 = 15 N.

1mg/m

=

1g

= 0.5

10 = 5 ms–2.

(c) The force of friction between block M (with block m placed on top of it) and the horizontal surface = (M + m) 2g = (5 + 3) 0.7 10 = 56 N a = 5 ms–2. The force due to this acceleration = (M + m)a = (5 + 3) F

5 = 40 N

= 56 + 40 = 96 N

(d) If F = 32 N, the common acceleration of the blocks is a =

F (M

m)

=

32 = 4 ms–2 (5 3)

Hence the correct matching is as follows: (a)

(s)

(b)

(p)

(c)

(q)

(d)

(r)

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has four choices out of which only one choice is correct

(a) Statement-1 is true, Statement-2 is true and ment-1.

5.58 Comprehensive Physics—JEE Advanced

(b) Statement-1 is true, Statement-2 is true but Statement-1. (c) Statement-1 is true; Statement-2 is false. (d) Statement-1 is false; Statement-2 is true. 1. Statement-1 A block is pulled along a horizontal frictionless surface by a thick rope. The tension in the rope will not always be the same at all points on it. Statement-2 The tension in the rope depends on the acceleration of the block-rope system and the mass of the rope. 2. Statement-1 A truck moving on a horizontal surface with a uniform speed u is carrying sand. If a mass m of the sand ‘leaks’ from the truck in a time t, the force needed to keep the truck moving at its uniform speed is u m/ t. Statement-2 Force = rate of change of momentum. 3. Statement-1 Two blocks of masses m and M are placed on a horizontal surface as shown in Fig. 5.114. The 1

and that between the block M and the horizontal surface is 2 applied to block M so that the two blocks move without slipping is F = ( 1 + 2) (M + m)g.

Fig. 5.114

Statement-2 acceleration. 4. Statement-1 A shell of mass m three fragments having masses in the ratio 2:2:1. mutually perpendicular directions with speed v.

The speed of the third (lighter) fragment will be 2 2 v. Statement-2 The momentum of a system of particles is conserved 5. Statement-1 F such that the block shown in Fig. 5.115 does not move is mg/cos , where block and the horizontal surface.

Fig. 5.115

Statement-2 normal reaction. 6. Statement-1 A ball of mass m is moving towards a batsman at a speed v it by an angle without changing its speed. The impulse imparted to the ball is zero. Statement-2 Impulse = change in momentum 7. 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. IIT, 2007 8. Statement-1 When a ball dropped from a certain height hits the momentum. Statement-2 reaction forces, being equal and opposite, cancel each other.

ANSWERS/SOLUTIONS 1. The correct choice is (a). 2. the leaking sand on the truck = rate of change of momentum = u m/ t. The sand falling vertically

the vertically upward direction. This perpendicular force can do no work on the truck. Since the truck moves with a uniform velocity, the force

Laws of Motion and Friction 5.59

3. The correct choice is (c). The force of friction between block m and block M = 1mg, where

From the principle of conservation of momentum, the momentum of the third (lighter) fragment of m must be 2 p but opposite in direction. mass 5 Thus, if V is the speed of the lighter fragment, we have mV 2mv = 2p = 2 5 5

1

blocks. Now, the force of friction between block M (with block m on top of it) and the horizontal surface = 2(M + m)g, where 2 of friction between block M and surface. The F applied to block M must be enough to overcome this force of friction and the force due to acceleration of the system. If the acceleration of the system is a then this force = (M + m)a. Thus (i) F = (M + m)a + m2(M + m)g Now, since the force on block m is 1mg, its acceleration a is mg force on mass m = 1 = 1g (ii) a= m mass m Using (ii) in (i) we get F = 1(M + m)g + 2(M + m)g = ( 1 + 2) (M + m)g

or V = 2 2v 5. The correct choice is (a). The component of F parallel to the horizontal surface is F cos . F F cos just overcomes the frictional force f = mg. Thus mg cos 6. The correct choice is (d). Refer to the solution of Q.27 of section I. F

= mg

F

=

7. motion also called the law of inertia. The dishes are not dislodged even when the cloth is suddenly pulled because the dishes have the inertia of rest. Statement-2 is Newton’s third law of motion, it

4. The correct choice is (a). The mass of two frag2m ments of equal masses = each. The mass 5 m . The momenta of of the lighter fragment = 5 2mv . The resultant of heavier fragments are p = 5 momenta p and p is p = (p2 + p2)1/2 =

cos

choice is (b). 8. The assertion is true but the reason is not correct because action and reaction forces do not act on the same body and hence do not cancel each other. Hence the correct choice is (c).

2p

VI Integer Answer Type 1. The magnitude of force f (in newton) acting on a body varies with time t (in millisecond) as shown in Fig. 5.116. Find the magnitude of the total impulse (in Ns) of the force on the body from t = 4 ms to t = 16 ms. IIT, 1994

2. A block of mass 0.2 kg is held against a wall by applying a horizontal force of 5 N on the block. is 0.5. Find the magnitude (in newton) of the frictional force acting on the block. Take g = 10 ms–2. IIT, 1995 3. Block A of mass m and block B of mass 2m are massless string and a frictionless pulley as shown

Fig. 5.116

block A and the wedge is 2/3 and that between block B and the wedge is 1/3. If the blocks are A (in ms–2). IIT, 1997

5.60 Comprehensive Physics—JEE Advanced

Fig. 5.117

4. A piece of uniform string hangs vertically so that its free end just touches the horizontal surface of a

table. The upper end of the string is now released. At any time during the falling of the string, the total force on the surface of the table is n times the weight of the part of the string lying on the surface. Find the value of n. IIT, 1989 5. A uniform rope of mass M and length L is pulled by a constant force of 10 N. Find the tension (in newton) in the rope at a point at a distance L/5 from the end where the force is applied. IIT, 1978

SOLUTIONS 1. Impulse from t = 4 ms to t = 16 ms = area under the F – t graph = area of EBCD = area of trapezium EBCF + area of CDF =

1 2

(200 + 800) N +

1 2

(2

800 N

10–3 s) (10

10–3 s)

= 1 + 4 = 5 Ns 2. Normal reaction R = 5 N. At equilibrium, the force of friction = weight of the block (see Fig. 5.118)

Fig. 5.119

The equations of motion of blocks A and B are T – mg sin 45° – A mg cos 45° = ma, where A = 2/3 and 2 g sin 45°– B 2 mg cos 45° – T = 2 ma, where B = 1/3. Adding these equations and solving we get a= –

Fig. 5.118

= mg = 0.2 10 = 2 N 3. Case (a): Let us assume that block A moves up the plane and block B moves down the plane. The free body diagrams of the blocks are as follows (See Fig. 5.119)

g 9 2

Case (b): If we assume that block A moves down 7g . and block B moves up, we would get a = – 9 2 Thus in both cases, the acceleration has a negative value which implies that the blocks will decelerate. This is not possible because the blocks start from rest. Hence when the blocks are released, they move with zero acceleration. Thus acceleration of block A = 0. 4. Let x be the length of the string lying on the surface of the table at an instant of time t. If an additional length dx of the string falls on the surface in time dt, the velocity v of this element when it strikes the surface is given by ( u = 0) v2 = u2 + 2gx = 0 + 2gx or

v2 = 2gx

(1)

Laws of Motion and Friction 5.61

The total force on the surface is F = rate of change of momentum of element of length dx + weight of a length x of the string lying on the table. If m is the mass per unit length of the string, then d dx (mdxv) + mxg = mv F= dt dt + mxg = mv2 + mxg (2) dx v dt Using (1) in (2) we get F = 2 mgx + mgx = 3 mgx But mx = M, the mass of the string lying on the table. Hence F = 3 Mg Thus n = 3. M . Let us 5. Mass per unit length of the rope is m = L P at a distance x from the end x = 0. Let T be the tension in the rope at point P. A

P

B

x=

x=L x

L–x

Fig. 5.120

x

For part AP the tension is towards the positive x-direction and for part BP the tension is towards the negative x-direction. If a is the acceleration produced in the rope by the constant force F, then for part AP, T = (mass of AP) a or T = mxa (1) For part BP, we have F – T = (mass of BP) a = m(L – x)a T From (1), we have a = mx Using this in (2), we get T ( L x)T F – T = m(L – x) mx x (L

or

F= T

At

x= L – T=

F L

x) x

L 5 4L 5

1

T

L or T x

(2)

Fx L

4L , 5 4F 5

4 10 5

8N

6

Work, Energy and Power

Chapter

REVIEW OF BASIC CONCEPTS 6.1 1.

WORK

force F in moving the body from a position r1 to a position r2 is given by r2

Work done by a Force

(a) Work done by a constant force When a constant force F acting on a body produces a displacement S, then the work done by the force is given by W = F S = FS cos where is the angle between the force vector F and the displacement vector S [see Fig. 6.1]. F and S are the magnitudes of F and S respectively.

F dr = area under the (F – r) graph

W= r1

6.1 makes an angle of 60° with the horizontal. Find the work done if a force of 150 N is applied to drag the box through a distance of 10 m. SOLUTION W = FS cos = 150

10

cos 60° = 750 J

6.2 A horizontal force F pulls a 20 kg box at a constant Fig. 6.1

(i) If is acute, cos is positive. Hence work is positive for acute . In this case the force increases the speed of the body. (ii) If = 90°, W = 0, i.e. if the force is perpendicular to displacement work done by the force is zero. (iii) If is obtuse, W is negative. In this case the force decreases the speed of the body. (iv) If = 0, i.e. force F is in the same direction as displacement S, then W = FS (v) If =180°, force F is opposite to S (example frictional force), W = – FS. Work done by frictional and viscous force is always negative. (b) Work done by a variable force Suppose a force F is not constant but depends on the position vector r of the body, then the work done by the

the work done by force F in moving the box through a distance of 2 m. SOLUTION Since the box is moved at a constant velocity, the applied force F just overcomes the frictional force f, i.e. F = f = mg Work done W = FS cos = 0.25

= mgS cos 0° 20

9.8

2 = 98 J

6.3 A block of mass m = 5 kg slides down from the top of an inclined plane of inclination = 30° with the

6.2 Comprehensive Physics—JEE Advanced

the block and the plane is 0.25. The length of the plane is 2 m. Find the work done by the (a) gravitational force, (b) frictional force and (c) normal reaction if the block slides to the the bottom of the plane. SOLUTION Refer to Fig. 6.2.

Fig. 6.2

Displacement S = AB = 2 m from A to B. (a) Angle between F and S is (90° – ) = 90° – 30° = 60° Work done by gravitational force is W1 = FS cos 60° = mgS cos 60° = 5 9.8

2

1 = 49 J 2

(b) Work done by frictional force is W2 = f S cos 180° = – f S = – RS = – mg (cos )S = – 0.25 5 9.8 cos 30° = – 21.2 J

2

(c) Since the normal reaction R is perpendicular to displacement S, work done by normal reaction is W3 = RS cos 90° = 0 6.4 A block of mass m = 2 kg is raised vertically upwards by means of a massless string through a distance of S = 4 m with a constant acceleration a = 2.2 ms–2. Find the work done by (a) tension and (b) gravity.

= 2 (2.2 + 9.8) = 24 N Work done by tension is ( T and S are in the same direction) W1 = TS cos 0° = 24 4 1 = 96 J (b) Since the gravitational force mg and displacement S are in opposite directions, work done by gravity is W2 = mgS cos 180° Fig. 6.3 = – 2 9.8 4 = – 78.4 J (c) Net work done W = W1 + W2 = 96 – 78.4 = 17.6 J 6.5 A block of mass m = 2 kg in suspended by a light string from the ceiling of a lift. The lift starts moving down with an acceleration a = 1.8 ms –2. Find the 5 seconds. SOLUTION Tension T = m(g – a) = 2 (9.8 – 1.8) = 16 N Distance moved in t = 5 s is 1 1 S = at 2 = 1.8 5 2 2 2 = 22.5 m Since the tension and displacement are in opposite directions, the work done by tension is W = TS cos180° = – TS = – 16 6.6 A constant force F = (2 i

T = m(a + g)

3 j) newton displaces

a body from position r1 = (4 i r2 = ( i

5 j) metre to

3 j) metre. Find the work done by the force.

SOLUTION Displacement S = r2 – r1 = (i

SOLUTION (a) From the free body diagram (Fig. 6.3) T – mg = ma

22.5 = – 360 J

3 j)

W = F S = (2 i

(4 i

5 j)

3 j) ( 3 i

3i

8j

8 j)

= – 6 + 24 = 18 J [ i i = j j 1 and i j = 0]

Work, Energy and Power 6.3

6.7 A body of mass m = 0.5kg travels in a straight line with a velocity v = 5x3/2 where v is in ms–1 and x is in metre. Find the work done in displacing the body from x = 0 to x = 2 m. SOLUTION Acceleration

dv d = dt dt 3 1/ 2 = 5 x 2 15 1/ 2 = x 2 75 2 = x 2

5 x3 / 2

a=

dx dt 5 x3 / 2

x 0

= 0.5

v

2

Fdx =

= 0.5

madx 0

75 2 75 2

2

x 2 dx 0

x3 3

2

0

0.5 75 8 0 2 3 = 50 J =

6.8 A block of mass 5 kg slides down from the top of an inclined plane of angle of inclination 30°. The the plane is 0.3. The length of the plane is 2 m. Find work done by (a) by gravity, (b) frictional force and (c) normal reaction. SOLUTION

= – 14.7 3 = – 25.5 J (c) Work done by normal reaction = RS cos 90° = 0 ( R S)

6.2

x 2

Work done W =

dx dt

Displacement S = 2 m down the inclined plane. (a) Work done by gravity = mg sin AB = (mg sin ) S = 5 9.8 sin 30° 2 = 49 J (b) Work done by friction = f S cos 180° =–fS = – RS = – mg cos S = – 0.3 5 9.8 cos 30° 2

ENERGY

So, energy is measured in the same units as work, namely, joule. Like work, energy is a scalar quantity. Energy can exist in various forms, such as heat energy, electrical energy, sound energy, light energy, chemical energy, nuclear energy, mechanical energy, etc. Mechanical energy is of two types, and Kinetic Energy: Energy due to Motion A moving object can do work on another object when it strikes it. In other words, an object in motion has the ability to do work

An initially motionless body can move and acquire a velocity only if a force acts on it. The work done by the force in causing the body to move measures the kinetic energy (written as KE) of the moving body, i.e. KE = W The kinetic energy of a body of mass m, moving with a velocity v is given by 1 KE = mv2 2 This relation holds even if the force is variable, i.e. if the force varies both in magnitude and direction. Work-Energy Principle Suppose a body of mass m moves with an initial velocity . A force F acts on it, as v. The work done by the force is given by Fdx

W=

v

=m

Fig. 6.4

=

ma dx

dv 1 dx = m(v2 – dt 2

1 1 mv2 – 2 2

2

2

)

6.4 Comprehensive Physics—JEE Advanced

Thus, This is the work-energy principle. Thus, when a force does work on a body, its kinetic energy increases; the increase in kinetic energy being equal to the amount of work done. The converse of this is also true. When the kinetic energy of a body is decreased by a retarding force, the decrease is equal to the work done by the body against the retarding force. Thus kinetic energy and work are equivalent quantities and are, therefore, measured in the same units, namely, joule. Potential Energy: Energy due to Position or An object can have energy not only by virtue of its motion, but also because of its position or

the reaction of the spring and is called the which is proportional to the displacement x and acts in a direction opposite to the displacement, i.e. F – x or F = – where is the force constant of the spring. The negative sign indicates that the force acts in a direction opposite to displacement. To stretch a spring by a displacement x, we must exert a force F on it, equal but opposite to the force F exerted by the spring on us. Therefore, the applied force is F =–F= Notice that F is a variable force as it depends on x. Therefore, the work done by the applied force in stretching the spring through a distance x is given by x

x

F dx =

W= An object held at a position above the surface of the earth has potential energy by virtue of its position. Consider a body of mass m. It is lifted vertically to a height above the earth by applying a force F vertically upward. The force F must be just enough to overcome the gravitational attraction, i.e. F = mg where g is the acceleration due to gravity at that place. For bodies not too far above the surface of the earth, the value of g is practically constant. Hence the work done by a constant force F in displacing a body by a height can be calculated by the product F = . Thus gravitational potential energy of a body of mass m at a height above the surface of the earth is . Gravitational PE = Consider a perfectly

a frictionless horizontal surface as shown in Fig. 6.5. We assume that the mass of the spring is negligible compared to the mass of the block.

Fig. 6.5

If we stretch the spring by a distance x, the spring will exert a force on us during stretching. This force is due to

0

( )dx 0

x

x dx =

= 0

x2 2

x

0

1 2

2

It is evident that the work done in compressing the 1 2 spring by an amount x is also given by W = . 2 Law of Conservation of Energy The total energy of an isolated system remains constant, the energy can only change from one form to another. 6.9 A block of mass 0.5 kg is taken from the bottom of an inclined plane to its top and then allowed to slide down to the bottom. The length of the inclined plane friction between the block and the plane is 0.2. Find (a) work done by the gravitational force over the round trip, (b) work done by the applied force over the upward journey, (c) work done by the frictional force over the round trip and (d) the kinetic energy of the block when it reaches the bottom of the plane. What conclusion will you draw from your answers to (b), (c) and (d)? Take g = 10 ms–2. SOLUTION Refer to Fig. 6.6. m = 0.5 kg, = 0.2, (= AC) = 2.5 m and (= AB) = 1.5 m

Work, Energy and Power 6.5

W4 = –

Fig. 6.6

sin

=

cos

=

AB 1.5 = = 0.6 AC 2.5 1 sin 2

= 0.8

(a) Work done by gravitational force to take the block from C to A is W1 = (– mg sin ) The negative sign indicates that the gravitational force mg sin is opposite to the displacement which is from C to A. Work done by the gravitational force to move the block from A to C is (because now the gravitational force is in the direction of the displacement) W2 = (+ mg sin Total work done by the gravitational force over the round trip is WG = W1 + W2 = 0 This shows that gravitational force is conservative. (b) Frictional force f = R = mg cos . The applied force to move the block from C to A is F = mg sin + f = mg sin + mg cos = mg (sin +

cos )

Work done by applied force to move the block from C to A is Wa = F AC = mg (sin + cos ) = 0.5 10 (0.6 + 0.2 0.8) 2.5 = 9.5 J (c) Friction always opposes motion. When the block is moved from C to A, frictional force f is opposite to direction. Hence work done by frictional force in the upward journey is W3 = – When the block slides from A to C, f acts upwards along the plane and is opposite to the displacement. Hence work done by frictional force in the downward journey is

Total work done by frictional force over the round trip is Wf = W3 + W4 = – 2 = – 2 mg cos = – 2 0.2 0.5 10 0.8 2.5 =– 4 J (d) When the block is at A, its initial velocity = 0. Let v be the velocity when it reaches C. Since f acts upwards, the net force on the block when it slides down is F = mg sin – f = mg (sin – cos ) F = g (sin – cos ) m = 10 (0.6 – 0.2 0.8) = 4.4 ms –2 = 2 , we have

Acceleration a =

From v2 –

2

v2 – 0 = 2 2

4.4

2.5

2 –2

v = 22 m s

1 mv2 2 1 = 0.5 2 = 5.5 J

Kinetic energy at C =

22

Conclusion Initial kinetic energy at A = 0. Therefore, change in K.E.= 5.5 – 0 = 5.5 J. Now total work done is W = WG + Wa + Wf = 0 + 9.5 – 4 = 5.5 J Thus, work done = change in kinetic energy. This is the work-energy principle. 6.10 A block of mass m = 500 g is placed at the top of an inclined plane of inclination = 60°. The length of the plane is 2 m. The block is released from rest. Find its speed when it reaches the bottom of the plane if (a) the inclined plane is smooth the plane is 0.4. Take g = 10 ms–2. SOLUTION Refer to Fig. 6.6 of Example 6.9 above. Height of the inclined plane is = AB = AC sin = 2 sin 60° = 3 m = 1.73 m

6.6 Comprehensive Physics—JEE Advanced

(a) As the block slides down the plane, it loses potential energy and gains kinetic energy. From the principle of conservation of energy, Gain in K.E = loss in P.E. 1 mv2 = or 2 = 2 10 1.73 5.9 ms–1 v= 2 (b) As the block slides down, loss of P.E. = gain in K.E. + work done against friction, i.e., 1 mg cos AC = m v2 2 1 1.73 10 = v 2 0.4 10 cos 60 2 2 v 5.2 ms–1 6.11 An elastic spring of negligible mass has a force constant 4 Nm–1 the wall and the other end touches a block of mass m = 250 g placed on a horizontal surface. The spring is compressed by an amount x = 5 cm as shown in and the horizontal surface is

= 0.2. If the system is

the spring.

the same plank before coming to rest if it were moving with a speed of 200 ms–1 ? SOLUTION Since the bullet and the plank are the same, the resistive force F exerted on the bullet is the same in the two cases. The kinetic energy is spent in doing work against friction. Hence 1 mv12 = Fx1 2 1 mv22 = Fx2 2 These equations give and

x2 = x1

v22 v12

200 100

= 2 cm

2

= 8 cm 6.13 A uniform chain of length L and mass M lies on a frictionless horizontal table with a very small part hanging from the edge of the table. The chain begins to fall under the weight of the hanging part. Obtain the expression for the velocity of the chain at the instant when the length of the hanging part becomes L/n where n > 1. SOLUTION M Mass per unit length of chain = . The mass of L L of the chain is length n

Fig. 6.7

SOLUTION Loss in P.E. of spring + work done against friction = gain in K.E. 1 2 1 mgx = mv 2 or 2 2 1 4 0.05 2 0.2 0.250 10 0.05 2 1 = 0.250 v 2 2 which gives v = 0.5 ms–1 6.12 A bullet moving with a speed of 100 ms–1 travels a distance of 2 cm in a plank of wood before coming to rest. How much distance will the same bullet travel in

m=

M L

L n

M n

As the chain slips down from the table, gravitational L (hanging part) depotential energy of the length n creases while the part of the chain left on the table does not lose any gravitational potential energy. The loss of P.E. of the hanging part gets converted into K.E. of the entire chain, i.e. Gain of K.E. of the complete chain = loss of P.E. of the hanging part. The mass m of the hanging part can be assumed to be concentrated at its centre of mass L below the edge of the which is at a height = 2n table. If v is the velocity of the slipping chain, then 1 M v2 = 2

=

M n

g

L 2n

MgL 2n 2

Work, Energy and Power 6.7

v2 =

gL n2 gL n

v=

6.3

CONSERVATIVE AND NON-CONSERVATIVE FORCES

(a) Conservative force A force is conservative if (i) the work done by it on a body in moving it from one position to another depends only on the initial

NOTE F is negative if r is opposite to F and positive if r is in the same direction as F.

6.14 The force between two point charges q1 and q2 1q2 separated by a distance r is F = where is a r2 constant. Find the potential energy of the system of charges. SOLUTION F=

followed by it between the two positions. or (ii) the net work done by the force on a body that moves through any closed path is zero. The above two conditions are equivalent. Examples of conservative forces are gravitational force, electrostatic force and spring force. (b) Non-conservative force A force is non-conservative if (i) The work done by it on a body in moving it from one position to another depends on the path followed by the body between the two positions. or (ii) The work done by the force on a body that moves through a closed path is non zero. Examples of non-conservative forces are frictional and viscous forces. Conservative Force and Potential Energy For a conservative force F that depends upon position r, there is a potential energy function U which also depends on r. When a conservative force does positive work, the potential of the system decreases, i.e. Work done = decrease in potential energy or or

Fdr = – dU dU F= dr

Hence

The change in potential energy when the body is displaced from r = a to r = b is b

Ub – Ua =

Fdr a

dU dr

dU = – Fdr. Integrating

r

r

Fdr =

U= 0

U=

1

r 2 dr

1 q2 0

q2

r

6.15 The potential energy U with position r as a b U= 2 r r where a and b are positive constants. Find the position r0 where the particle will be in stable equilibrium. SOLUTION If no force acts on a particle, it will be in stable equilibrium, i.e. F = 0 or dU =0 F= dr d a b =0 d r r2 r 2a

b

3

2

r = r0 =

=0

2a b

r r Stable equilibrium corresponds to minimum potendU d 2U 2a tial energy, i.e. = 0 and > 0. If r = r0 = , 2 dr b dr dU = 0, then U can be minimum or maximum. If dr d 2U d r2 Now

> 0 at r = r0, U will be minimum. dU = dr

2a

b

3

r2

r

6.8 Comprehensive Physics—JEE Advanced

d 2U d r2 At

r4

r0

3 b4 8 a3

=

4

– 2b

b4 4a

3

b 2a

3

b4

, which is posi8a tive.

NOTE For stable equilibrium U is minimum and for unstable equilibrium, U is maximum.

dU d 2U = 0 and >0 dr d r2

For unstable equilibrium;

6.4

m v2A – mg (i) r The body will revolve in the circle if the string does not sag, i.e. TA 0. From Eq. (i) it follows that m v2A – mg r v2A

3

2a corresponds to stable equilibHence r = r0 = b rium.

For stable equilibrium;

TA = tension in the string when the body is at A. TA =

b 2a

= 6a at r

and

2b

r3 2a r = r0 = , b

d 2U d r2

6a

=

dU d 2U = 0 and < 0. dr d r2

MOTION IN A VERTICAL CIRCLE

Consider a body of mass tied to a string of length r revolved in a vertical circle with centre O at the other end of the string as shown in Fig. 6.8. At all positions of the body, there are two forces acting on it: its own weight and the tension in the string. v

0 rg

Therefore, the minimum speed at the top that the body must have so that it can complete the circle is given by (vA)min =

rg

(ii)

Case (b): When the body is at bottom B of the circle When the body is at B, the net force on the body towards the centre is (TB – mg). Hence m vB2 r where vB = speed of the body at bottom B and TB = tension in the string when the body is at B. TB – mg =

m vB2 + mg (iii) r The minimum speed at B that the body must have so that it can complete the circle is obtained from the conservation of energy. As the body goes up from B to A, it K.E. decreases and P.E. increases. Loss in K.E. = gain in P.E. TB =

1 2 1 mvB – mv2A = mg 2 2 vB = (vB)min =

v2A

AB = mg

2r

4gr

( vA )2min

4 gr

(iv)

Using (ii) in (iv) we get (vB)min = v

Case (a): When the body is at top A of the circle When the body is at top A of the circle, the net force towards centre O is TA + mg. Hence [see Fig. 6.8] m vA2 r where vA = speed of the body at top A TA + mg =

(v)

The net force towards the centre acting on the body is obtained from (iii) by using (v). TB =

Fig. 6.8

5gr

m r

5gr + mg = 6 mg

NOTE 1. When a body moves in a vertical circle, the speed decreases as it goes up and increases as it goes down. Hence the body has a non-uniform circular motion. 2. The tension in the string is different positions of the body on the circle. The tension is minimum when the body is at the top of the circle and maximum when it is at the bottom of the circle.

Work, Energy and Power 6.9

6.16 A small block of mass m, starts from rest at A and slides on a frictionless track which ends in a circular loop of radius r. If = 6r when it reaches C as shown in Fig. 6.9. What is the force exerted on the block by the track when it is at C so that the block is able to complete the circle.

min

6.5

=

1 2 mvmin 2

min

= 2.5r.

POWER i.e.

Work Time The faster a given amount of work is done, the greater is the power of the agent that does the work. In the SI system, the unit of power is the watt (symbol W). i.e. Power =

1 W = 1 Js–1

Fig. 6.9

SOLUTION Let v be the speed of the block when it reaches C. From conservation of energy, gain in K.E. = loss in P.E., i.e. 1 OB = – mgr mv2 0 = 2 1 2 v = g 6r – gr = 5gr 2 v = 10gr When the block is at C, the track exerts a normal reaction N on the block. Since the block is moving in a circular path, the necessary centripetal force for circular motion is provided by the normal reaction (Fig. 6.10).

Fig. 6.10

m v2 m 10 gr = = 10 mg r r Thus the track exerts a force on the block equal to 10 times the weight of the block. To complete the circle, the minimum speed at D must be vmin = 5gr . Hence N=

Since the watt is a small unit for the measurement of power, larger units, namely kilowatt (kW) and megawatt (MW) are often used. 1 kW = 1000 W = 103 W 1 MW = 1,000,000 W = 106 W The power of an agent can also be expressed in terms of the force applied and the velocity of the object on which the force is applied. Now, power is given by W F S = = Fv t t S ( = rate of change of displacement = v) t Power is a scalar quantity as it is the ratio of two scalars W and t, or a scalar product of two vectors F and v. 6.17 An engine pulls a car of mass 1000 kg on a level road at a constant velocity of 5 ms–1 . If the frictional force is 500 N, what power does the engine generate? What extra power must the engine supply to maintain the same speed up an inclined plane having a gradient of 1 in 10? SOLUTION Since the car moves at a constant velocity, its acceleration is zero. Hence the engine has to do work only to overcome the frictional force f . Power = f v = 500 5 = 2500 W = 2.5 kW For an inclined plane having a gradient of 1 in 10, sin 1 . To maintain the same speed up the inclined plane, = 10 the engine has to do extra work against the force mg sin .

6.10 Comprehensive Physics—JEE Advanced

Therefore, v

Extra power = mg sin 9.8

= 100

1 3 v 3

1 10

5 = 4900 W

3

=

=

with water. If the tank is 40 m above the ground and Take g = 10 ms–2 . SOLUTION Volume of tank V = 2000 litre = 2000 10–3 m3 = 2 m3 Mass of water m = V = 1000 2 = 2 103 kg Work done to lift this mass to a height = 40 m is W= = 2 103 10 40 = 8 105 J W 8 105 4 Power needed = = = 103 W t 10 60 3 is the total power consumed, the useful power 0.4 . Hence 4 103 3 = 3.33 103 W = 3.33 kW

3000

5

3

2

3

SOLUTION Since F is constant, acceleration a =

= Fv = mav dv a= dt

dv d x dx dt

dv = dx mv 2

v dv =

m

v

a= vd v dx

At time t, the velocity of the car is v=

+ at = 0 + at = at =

Kinetic energy E at time t = =

F2 2 t . m

Since F is constant, E Power Thus

m

Ft 1 1 mv2 = m m 2 2

Ft = m

at time t = Fv = F

F2 t m

mv

displacement in time t is proportional to (b) t (a) t1/2 3/2 (d) t2 (c) t SOLUTION = Fv = mav = m

dx

2

t. So the correct choices are (b) and (c).

dx

0

Ft . m

t2 .

x

v2 d v =

F is constant. m

6.21 A body, initially at rest, moves in a straight line

SOLUTION

v

3x

6.20 A car of mass m starts from rest at time t = 0 and is driven on a straight horizontal road by the engine which exerts a constant force If friction is negligible, the car acquires kinetic energy E at time t and develops a power . Which of the following is/are correct ? (a) E t (b) E t2 (c) t (d) t2

=

6.19 A constant power is supplied to a car of mass m = 3000 kg. The velocity of the car increases from = 2 ms–1 to v = 5 ms–1 when the car travels a distance x = 117 m. Find the value of . Neglect friction.

Now

3

3 117 = 1000 W = 1 kW

6.18

0.4

v3

=

= 4.9 kW

If

m

vdv =

m

dt

dv dt

v

Work, Energy and Power 6.11

v

t

vd v =

Integrating 0

m0

dt

(

dx =

= constant) x

v2 = t 2 m

2 1/2 t dt m t

2 dx = t1/2 dt m 0 0

v=

2 1/2 t m

dx = dt

2 1/2 t m

x=

4 3/2 t 3m

So the correct choice is (c).

I Multiple Choice Questions with Only One Choice Correct 1. A particle of mass m is moving in a circular path of a constant radius r such that its centripetal acceleration ac is varying with time t as ac = 2 r t2 where is a constant. The power delivered to the particle by the force acting on it is 2 2 2 2 rt (b) rt (a) 2 4 2 5

(c)

r t 3

(d) zero

IIT, 1994 2. A stone is tied to a string of length L and whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at the lowest position and has a speed . The magnitude of the change in its velocity as it reaches a position where the string is horizontal is (a)

2

2 L

(b)

2 gL

(c)

2

L

(d)

2(

2

L)

IIT, 1998 3. The power supplied to a body initially at rest varies with time t as = 2 where is a constant. The velocity of the body at an instant of time t will be proportional to (a) t (b) t3/2 (c) t2 (d) t3 4. A force F = (3 i 4 j) newton acts on a particle moving along a line 4y + = 3. The work done by the force is zero if the value of is (a) 1 (b) 2 (c) 3 (d) 4

5. Force F acting on a body moving in a straight line varies with the velocity v of the body as F = /v where is a constant. The work done by the force in time t is proportional to (a) t (b) t3/2 –1/2 (c) t (d) t–3/2 6. The force F acting on a body varies with its displacement x as F = –2/3. The power delivered by the force will be proportional to (b) x–1/2 (a) x–3/2 (c) x1/2 (d) x3/2 7. of 200 ms–1. After passing through the plank, its speed reduces to 180 ms–1. Another bullet, of the same mass and size but moving with a speed of 100 ms–1 speed of this bullet after passing through the plank? Assume that the resistance offered by the plank is the same for both the bullets? (a) 48 ms–1 (b) 49 ms–1 (c) 50 ms–1 (d) 51 ms–1 8. If the mass of either bullet in Q. 10 is 7 g and the thickness of the wooden plank is 1 m, what is the average resistance offered by the plank? (a) 36 N (b) 38 N (c) 40 N (d) 42 N 9. An engine pulls a car of mass 1500 kg on a level road at a constant speed of 5 ms–1. If the frictional force is 1500 N, what power does the engine generate?

6.12 Comprehensive Physics—JEE Advanced

(a) 5.0 kW (b) 7.5 kW (c) 10 kW (d) 12.5 kW 10. In Q. 9, what extra power must the engine develop to maintain the same speed up an inclined plane having a gradient of 1 in 10? Take g = 10 ms–2. (a) 2.5 kW (b) 5.0 kW (c) 7.5 kW (d) 10 kW 11. Two identical cylindrical vessels, with their bases at the same level, each contain a liquid of density . The height of the liquid in one vessel is 1 and that in the other is 2. The area of either base is A. What is the work done by gravity in equalizing the levels when the vessels are interconnected? (b) A g ( 1 + 2)2 (a) A g ( 1 – 2)2 2

2

(c) A g

1

2

2

(d) A g

1

2

2

17. ing a thickness of 3.5 cm. The total thickness penetrated by the bullet is (a) 8 cm (b) 10 cm (c) 12 cm (d) 14 cm 18. etrating a thickness x of the plank. What is the total thickness penetrated by the bullet? (a) 2x (b) 4x (c) 6x (d) 8x 19. A body of mass m, having momentum , is moving on a rough horizontal surface. If it is stopped in a distance x body and the surface is given by 2

12. 3

with water. If the tank is 60 m above the ground the tank? Take g = 10 ms–2. (a) 100 kW (b) 150 kW (c) 200 kW (d) 250 kW 13. The distance x moved by a body of mass 0.5 kg by a force varies with time t as x = 3t2 + 4t + 5 where x is expressed in metre and t in second. What (a) 25 J (b) 50 J (c) 75 J (d) 100 J 14. In a hydroelectric power station, the height of the dam is 10 m. How many kg of water must fall per second on the blades of a turbine in order to generate 1 MW of electrical power? Take g = 10 ms–2. (a) 103 kgs–1 (b) 104 kgs–1 (c) 105 kgs–1 (d) 106 kgs–1 15. A uniform steel rod of mass m and length is pivoted at one end. If it is inclined with the horizontal at an angle its potential energy will be 1 1 cos (b) sin (a) 2 2 (c) cos (d) sin 16. A bullet, incident normally on a wooden plank, loses one-tenth of its speed in passing through the plank. The least number of such planks required to stop the bullet is (a) 5 (b) 6 (c) 7 (d) 8

(a)

=

(c)

=

2

2 gm 2 x

(b)

=

(d)

=

2mgx

2 gm 2 x 20. A uniform chain of mass M and length L is held on 1 a horizontal frictionless table with th of its length n hanging over the edge of the table. The work done is pulling the chain up on the table is Mg Mg (b) (a) 2n n Mg Mg (c) (d) 2 n 2 n2 21. A body of mass m = 1 kg is dropped from a height 2mgx

base, as shown in Fig. 6.11. As a result the spring is compressed by an amount x = 10 cm. What is the force constant of the spring. Take g = 10 ms–2. (a) 600 Nm–1 (b) 800 Nm–1 –1 (c) 1000 Nm (d) 1200 Nm–1

Fig. 6.11

Work, Energy and Power 6.13

22. A force acts on a particle of mass 3 g in such a way that the position of the particle as a function of time is given by x = 3t – 4t2 + t3 where x is in metres and t is in seconds. The work (a) 570 mJ (b) 450 mJ (c) 490 mJ (d) 530 mJ 23. A sphere of mass m is tied to one end of a string of length and rotated through the other end along a horizontal circular path with speed v. The work done in one full horizontal circle is (a) zero (b) mg 2 (c)

mv 2

2

(d)

mv 2

24. The kinetic energy acquired by a mass m in travelling a certain distance d, starting from rest, under the action of a constant force is (a) directly proportional to m (b) independent of m 1 (c) directly proportional to m (d) directly proportional to m 25. A position dependent force F = 7 – 2x + 3x 2 newton acts on a body of mass 2 kg and displaces it from x = 0 to x = 5 m. The work done in joules is (a) 70 (b) 270 (c) 35 (d) 135 26. A particle is moved from a position r1 = (3 i + 2 j – 6 k ) metre to a position r2 = (14 i + 13 j + 9 k ) metre under the action of a force F = (4 i + j + 3 k ). What is the work done? (a) 10 J (b) 100 J (c) 0.01 J (d) 1 J

(a) zero

(b) 20 2 m/s

(c) 20 3 m/s

(d) 40 m/s

29. A force F = – K ( y i + x j ) where K is a positive constant, acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particle is (b) 2 Ka2 (a) – 2 Ka2 2 (c) – Ka (d) Ka2 IIT, 1998 30. A wind-powered generator converts wind energy into electrical energy. Assume that the generator tercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to (a) v (b) v2 3 (c) v (d) v4 IIT, 2000 31. A body of mass 6 kg is acted upon by a force which t2 metre causes a displacement in it given by x = 4 where t is the time in second. The work done by the force is 2 seconds is (a) 12 J (b) 9 J (c) 6 J (d) 3 J 32. A ladder 2.5 m long and of weight 150 N has its centre of gravity 1 m from its bottom. A weight of 40 N is attached to the top end. The work required to raise the ladder from the horizontal position to the vertical position is (a) 190 J (b) 250 J (c) 285 J (d) 475 J 33. A body of mass 5 kg rests on a rough horizontal

27. energy increases by 28. A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in Fig. 6.12. If it starts its journey from rest at x = 0, its velocity at x = 12 m is

Fig. 6.12

pulled through a distance of 10 m by a horizontal force of 25 N. The kinetic energy acquired by it is (take g = 10 ms–2) (a) 200 J (b) 150 J (c) 100 J (d) 50 J 34. A body is moving up an inclined plane of angle with an initial kinetic energy E friction between the plane and the body is . The work done against friction before the body comes to rest is: E cos (b) E cos (a) cos sin (c)

E cos cos sin

(d)

E cos cos sin

6.14 Comprehensive Physics—JEE Advanced

35. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to (a) x2 (b) ex (c) x (d) loge x 36. A body of mass m accelerates uniformly from rest to velocity v1 in time t1. The instantaneous power delivered to the body as a function of time t is mv1t (a) t1 (c)

(b)

mv12t 2

(d)

t12

37. A force F

5 i 3 j 2k

mv12t t12 mv1t 2 t1

newton is applied to

a particle which displaces it from its origin to the point r

(2 i

j) metre. The work done on the

particle (in joule) is (a) – 7 (b) + 7 (c) + 10 (d) + 13 38. Two masses M and m (with M > m) are connected by means of a pulley as shown in Fig. 6.13. The system is released. At the instant when mass M has fallen through a distance , the velocity of mass m will be (a) (b) (c)

(d)

2 M

2 m

M

2 M

m m

M

2

Fig. 6.14

40. An escalator is moving downwards with a uniform speed . A man of mass m is running upwards on it at a uniform speed v. If the height of the escalator is , the work done by the man in going up the escalator is (a) zero (b) mg mg v (c) (d) v v 41. The potential energy (in joule) of a body of mass 2 kg moving in the x – y plane is given by U = 6x + 8y where the position coordinates x and y are measured in metre. If the body is at rest at point (6 m, 4 m) at time t = 0, it will cross the y-axis at time t equal to (a) 1 s (b) 2 s (c) 3 s (d) 4 s 42. In Q. 41 above, the speed of the body when it crosses the y-axis is (a) zero (b) 5 ms–1 (c) 10 ms–1 (d) 20 ms–1 43. 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 (as shown in Fig. 6.15) in the gravitational m between W1, W2 and W3.

M

Fig. 6.13

m m

39. A mass m, lying on a horizontal frictionless sur-face is connected to mass M as shown in Fig. 6.14. The system is now released. The velocity of mass m when mass M as descended a distance is (a)

(c)

2 Mg m 2 Mg M m

(b)

(d)

2 mg M 2

Fig. 6.15

(a) W1 > W3 > W2

(b) W1 = W2 = W3

(c) W1 < W3 < W2

(d) W1 < W2 < W3 IIT, 2003

Work, Energy and Power 6.15

44. A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = – + ax3. Here and a are positive constants. For x 0, the functional form of the potential energy U(x) of the particle is (see Fig. 6.16)

Fig. 6.17

(a) 3 ms

–1

(b) 4 ms–1

(c) 5 ms–1 (d) 6 ms–1 48. A body of mass m is dropped from a height above the ground. The velocity v of the body when it has lost half its initial potential energy is given by (b) v =

(a) v =

2

(c) v =

Fig. 6.16

IIT, 2002 45. A raindrop of radius r falls from a certain height above the ground. The work done by the gravitational force is proportional to (a) r (b) r2 (c) r3 (d) r4 46. A smooth steel ball is moving to and fro about the lowest position O of a frictionless hemispherical bowl. The ball attains a maximum height of 20 cm on either side of O. If g = 10 ms–2, the speed of the ball when it passes through O will be (a)

(d) v = 2 2 49. A body of mass m is thrown vertically upwards with a velocity v. The height at which the kinetic energy of the body is half its initial value is given by v2 v2 (b) = (a) = g 2g (c)

=

v2 3g

(d)

=

v2 4g

50. A car of mass m moving at a speed v is stopped in a distance x by the friction between the tyres and the road. If the kinetic energy of the car is doubled, its stopping distance will be (a) 8x (b) 4x (c) 2x (d) x 51. A body is allowed to fall freely under gravity from

2 ms–1

(b) 2 ms–1 (c) 0.2 ms–1 (d) 0.02 ms–1 47. The bob of a pendulum is released from a horizontal position A as shown in Fig. 6.17. The length of of the bob is dissipated as heat due to the friction of air, what would be the speed of the bob when it reaches the lowermost point B? Take g = 10 ms–2.

impact with the ground, to what height will it rise after one impact? (a) 2.5 m (b) 5.0 m (c) 7.5 m (d) none of these 52. In Q. 51, to what height will the body rise after two such impacts with the ground? (a) 2.5 m (b) 5.0 m (c) 7.5 m (d) none of these 53. A body of mass m thrown vertically upwards attains a maximum height . At what height will its kinetic

6.16 Comprehensive Physics—JEE Advanced

displaced towards wall 1 by a small distance x (a)

(b)

6

5

a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equix is librium position of the block B. The ratio y (a) 4 (b) 2 1 1 (c) (d) 2 4

(c)

(d) 4 3 54. A body, having kinetic energy , moving on a rough horizontal surface, is stopped in a distance x. The force of friction exerted on the body is (a)

(b)

x

(c)

x

(d)

x

55. force F = , where is a positive constant. If the potential energy U is zero at x = 0, the variation of potential energy with the coordinate x is represented by [see Fig. 6.18]

Fig. 6.19

IIT, 2008 57. A bob of mass m is suspended by a massless string of length L. The horizontal velocity v at position A B. The angle at which the speed of the bob is half of that at A

Fig. 6.18 v

IIT, 2004 56. A block (B) is attached to two unstretched springs S1 and S2 with spring constants and 4 , respectively (see Figure 6.19). The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The blook B is

Fig. 6.20

(a) (c)

=

(b)

4 3 4

2

(d)

4 3 4

IIT, 2008

ANSWERS

1. 7. 13. 19. 25. 31. 37.

(b) (b) (c) (a) (d) (d) (b)

2. 8. 14. 20. 26. 32. 38.

(d) (b) (b) (d) (b) (b) (c)

3. 9. 15. 21. 27. 33. 39.

(b) (b) (b) (c) (a) (b) (c)

2

4. 10. 16. 22. 28. 34. 40.

(c) (c) (b) (d) (d) (d) (d)

5. 11. 17. 23. 29. 35. 41.

(a) (c) (a) (a) (c) (a) (b)

6. 12. 18. 24. 30. 36. 42.

(b) (a) (b) (b) (c) (b) (c)

Work, Energy and Power 6.17

43. (b) 49. (d) 55. (a)

44. (d) 50. (c) 56. (b)

45. (c) 51. (c) 57. (d)

46. (b) 52. (d)

47. (d) 53. (c)

48. (a) 54. (a)

SOLUTIONS v2 = 2 r t2 v= r dv Tangential acceleration at = = dt Tangential force F = m at =

1. ac =

2

Work done is zero if the force is perpendicular to the displacement, i.e. if lines (1) and (2) are perpendicular each other. Thus the product of their slopes = – 1, i.e. 4 =–1 =3 3 4

r t2

Power = F v = 2. Refer to Fig. 6.21.

2 2

rt

=

5. Power = F v = v in time t is

v = . Therefore, work done

t

t =

W=

(

= constant)

0

Hence the correct choice is (a). 6. Given F

x–2/3. Therefore, acceleration x–2/3, i.e. dv v = K x–2/3 (K = constant) dx vdv = K

Fig. 6.21

v2

From conservation of energy, 1 1 2 = mv2 + mgL 2 2 v=

2

2 L

Change in velocity v = v = v ( ) . Thus v is the resultant of v and – . It follows from Fig. 6.21 (b) that v =

v2

(

=

2

2

)2 2

=

2(

2

L)

3

t =

t 2 dt =

2 v= t3 3m Hence the correct choice is (b). 4. The force F is parallel to the line 4 y= x + C 3 The particle moves along the line 3 y= – 4 4

3

(1)

(2)

or

2/3

dx

v

x1/6

= F v x–2/3 x1/6 x–1/2 7. Let m be the mass of each bullet. Since the resistance offered by the plank is the same for the two bullets, the amount of work done by the plank is the same for the two bullets. From work-energy principle, the decrease in the kinetic energy is the same for the two bullets. 1 1 2 mv 12 1– 2 2 1 1 = m(200)2 – m (180)2 (i) 2 2 If v2 is the speed of the second bullet after passing through the plank, then Decrease in KE of second bullet 1 1 mv 22 = m 22 – 2 2 1 1 = m(100)2 – mv 22 (ii) 2 2 Equations (i) and (ii) we have, 1 1 m(200)2 – m (180)2 2 2 1 1 = m(100)2 – mv 22 2 2 =

3. From work-energy principle, change in K.E. = work done. 1 mv2 = 2

x1/3

x

6.18 Comprehensive Physics—JEE Advanced

which gives v 22 = 2400 or v

49 ms

–1

Hence

Hence the correct choice is (b).

PE after connection =

8. Thickness of plank (S ) = 1.0 m Mass of bullet (m) = 10 g = 10–2 kg If F is the average resistive force exerted by the plank on the bullet, the work done by the plank on the bullet is W = FS Work done = decrease in KE of the bullet. Hence FS =

= Change in PE =

Putting S = 1 m and m = 10–2 kg, we get F = 38 N. Hence the correct choice is (b). 9. Since the car moves at a constant velocity, its acceleration is zero. Hence the engine has to do work only to overcome the frictional force (f). Since the distance moved in 1 second is v metres, the work done per second or the power of the engine is = f v = 1500 5 = 7500 W = 7.5 kW Hence the correct choice is (b). 10. When the car is being pulled along an inclined plane 1 , the engine has of gradient 1 in 10 i.e. sin = 10 to do extra work against the component Mg sin of the weight Mg of the car. The extra work per second or extra power the engine has to develop to maintain the same speed v is = Mg sin v 1 5 = 1500 10 10 = 7500 W = 7.5 kW Hence the correct choice is (c). 11. The work done by gravity equals the change in the potential energy of the system after the vessels are interconnected. We may regard the liquid in each vessel as equivalent to a point mass kept at their respective centres of gravity. Remembering that the mass of the liquid is given by ( ) and that the PE of a mass at a height in earth’s gravity is , we have Total PE at start = (A

1

)g

1

2

+ (A

2

)g

2

2

A g ( 21 + 22) 2 After the vessels are connected, the height of liquid in each vessel is ( 1 + 2)/2. =

A g ( 4

2

g

1

– 2(

2 1

2 +

1

A g {( 4

=–

1 1 m(200)2 – m(180)2 = 3800 m 2 2

1

A

+

1

A g ( 4

2)

1

2

2)

–

2

2

2

2)

+

2 2)}

2

2 1

=–A g

2

2 This must be equal to the work done ‘by’ gravity on the liquid. Thus the work done ‘by’ gravity is 2

A g

1

2

2 Hence the correct choice is (c). 12. Volume (V) = 30 m3, density of water ( ) = 1000 kg m–3. Therefore, mass of water to be lifted is m=

V = 1000

30 = 3

104 kg

Work done to lift this mass of water to a height = 60 m is W= = 3 104 10 60 = 1.8

107 J

work done by the pump is W 100 1.8 107 100 = W = 30 30 = 6

107 J

Time taken t = 10 min = 600 s. Therefore, power consumed is W 6 107 = = t 600 = 100,000 W = 100 kW Hence the correct choice is (a). dx d = (3t2 + 4t + 5) = 6t + 4. dt dt dv d Acceleration is a = = (6t + 4) = 6 ms–2. dt dt Therefore, applied force is F = ma = 0.5 6 = 3 N. Now t = 2s, the distance moved is

13. Velocity (v) =

x =3

(2)2 + 4

Work done W = Fx = 3 correct choice is (c).

2 + 5 = 25 m 25 = 75 J. Hence the

Work, Energy and Power 6.19

14. Let M kg of water fall per second. The power is = rate at which work is done = mass per second g = But = 1 MW = 106 W, = 10 m. Therefore M=

=

106 = 104 kg s–1, 10 10

15. The weight of the rod acts at the centre of gravity which is at a distance of /2 from the pivoted end. When the rod makes an angle with the horizontal, the vertical height of the centre of gravity is sin 2 i.e. the centre of gravity rises by an amount . Therefore =

1 2

=

sin

Hence the correct choice is (b). 16. Let v 9v plank. Its speed after it passes the plank = . If x 10 is the thickness of the plank, the deceleration a due to the resistance of the plank is given by 9v 10

2ax = v2 –

2

19 v 100

2

=

(i)

Suppose the bullet is stopped after passing through n such planks. Then the distance covered by the bullet is = nx. Thus, we have v2 – 0 = 2 n=

or

= 2anx

v2 v 2 100 100 = = = 5.26 2 2a x 19 19 v

– v2 = 2ax or

2

–

3 4

2

= 2a

3.5

2

cms–2. The bullet will come to 16 rest when its velocity v = 0. If x is the thickness

which gives a =

2a

2

. But a = 2

16

2

2

16

cms–2.

= 8 cm

Hence the correct choice is (a). 18. Since the wood offers a constant deceleration and hence a constant retardation force, the bullet after penetrating a further distance of 3x. Therefore, the total distance penetrated by the bullet before it comes to rest = x + 3x = 4x. Hence the correct choice is (b). 19. Force of friction = mg. Therefore, retardation a = mg/m = g. Also 2ax = v2 or 2am2x = m2v2. But = mv. Therefore, 2 am 2x = 2 But a = g. Therefore, 2

g m 2x =

2

2

or

=

2 g m2 x

. Hence the correct choice is (a).

M . L The mass of the hanging portion of the chain is mL m = . This mass can be assumed to be conn centrated at the centre of the hanging portion of the

20. The mass per unit length of the chain m =

L from the edge 2n of the table. Therefore, the work done in pulling the hanging portion of the chain on to the table top is mL L W = m gx = g n 2n chain which is a distance of x =

Thus the minimum number of planks required is 6. Hence the correct choice is (b). 17. Let cms–1 be the speed of the bullet. Since the mass of the bullet remains unchanged, its speeds 3 cms–1 after it penetrates a distance becomes v = 4 x = 3.5 cm. The retardation a due to the resistance of the wooden plank is given by 2

2

x =

or

Therefore x =

which is choice (b).

PE =

penetrated by the bullet, then 2 – v 2 = 2ax

=

mg L2 2n

2

=

M gL 2n2

Hence the correct choice is (d). 21. Since the platform is depressed by an amount x, the total work done on the spring is mg ( + x). This work is stored in the spring in the form of 1 2 . Equating the two, we potential energy 2 have 1 2m g 2 = mg ( + x) or = 2 x2

6.20 Comprehensive Physics—JEE Advanced

Given, = 0.4 m, x = 0.1 m, m = 1 kg and g = 10 ms–2. Substituting these values, we get = 1000 Nm–1. Hence the correct choice is (c). 22. The instantaneous velocity of the particle is dx d = (3t – 4t2 + t3) = 3 – 8t + 3t2 v= dt dt The instantaneous acceleration of the particle is dv d a= = (3 – 8t + 3t2) = – 8 + 6t dt dt

26. The displacement of the particle is r = r2 – r1 = (14 i + 13 j + 9 k ) – (3 i + 2 j – 6 k ) = (11 i + 11 j + 15 k ) Work done = F r = (11 i + 11 j + 15k ) (4 i + j + 3k ) = 4 11 + 11 + 3 15 = 100 J Hence the correct choice is (b).

4

4

Fd x =

W=

ma

0

0

dx dt dt

4

(– 8 + 6t) (3 – 8t + 3t 2) dt

=m 0 4

(– 24 + 82t – 72t 2 + 18t3) dt

=m 0

9 44 t 2 0 = m (– 96 + 656 – 1536 + 1152) =m

24t

41t 2

10–3 ( m = 3 10–3 kg) = 528 mJ ( 1 mJ = 10–3 J)

= 176 m = 176 10–3

= 528

24t 3

3

The closest choice is (d). 23. The centripetal force, being directed towards the centre is always perpendicular to the direction of displacement which is the direction of the velocity. Thus the dot product F S = 0, i.e. work done is zero, which is choice (a). 1 24. Kinetic energy K = mv 2. If a is the accelera2 tion, then v2 = 2ad. But a = force/mass = F/m. 2F d 1 2F d . Hence K = m Therefore, v2 = m 2 m = Fd, which is independent of m. Thus the correct choice is (b). 25. Work done is

i.e K increases by 0.44 K. The percentage increase 0.44 K in K is K 28. Work done = area under the (F – x) graph =

1 2

0

= 7x

7 2 x 3x d x x3

5 0

= 7 5 – (5)2 + (5)3 = 135 J Hence the correct choice is (d).

4+

1 2 4 = 80 J

Now, work done = increase in kinetic energy. If v is the velocity at x = 2 m, then increase in K.E. = 1 mv 2. Therefore 2 1 80 2 80 2 mv 2 = 80 or v2 = = 2 m 0.1 = 1600 or v = 40 ms–1, which is choice (d). 29. In going from (0, 0) to (a, 0), the x-coordinate varies from 0 to a while the y-coordinate remains zero. Work done by force F along this path is ( y = 0) a

F dx =

W1 = 0

0

x2

4 + 10

a 2

Fdx =

10

10

5

5

W=

2 1 = 27. Momentum = mv or 2 = m2v2 or 2 m 1 mv2. Thus the kinetic energy is 2 2 K = 2m If = 1.2 . Therefore the new kinetic energy will be 2 1.2 2 = 1.44 = 1.44 K K = 2m 2m

( K x j) dx i = 0 0

(

j i = 0)

In going from (a, 0) to (a, a), the x-coordinate remains constant at x = a while the y-coordinate changes from 0 to a.

Work, Energy and Power 6.21

Work done by force F along this path is ( a

F dy =

W2 =

x = a)

=

a

0

a

–K

yi

aj

dy j

1 m 2

2

=

(ii)

0

dy = – Ka2

= – Ka 0

(

i j = 0, j j = 1)

Since work is a scalar quantity, the total work done is W = W1 + W2 = 0 – Ka2 = – Ka2 Hence the correct choice is (c). 30. The power output of the generator is directly proportional to (i) the velocity v of the air molecules in wind and (ii) the average kinetic energy KE = 1 mv2 of the striking molecules (which is propor2 tional v2). Hence Power output v v2 v3 which is choice (c). 31. The velocity of the body at time t is given by dx d t2 t v= = = dt dt 4 2 At t = 0, v = = 0 and t = 2 s, v = 1 ms–1, Now, work done = increase in KE 1 1 1 2 mv2 – = mv2 – 0 = 2 2 2 1 1 = mv2 = 6 (1)2 2 2 = 3 J. Hence the correct choice is (d). 32. Work done = increase in potential energy in (i) raising the weight 150 N of the ladder through a height 1 m and (ii) raising a weight 40 N through 2.5 m = 150 N 1 m + 40 N 2.5 m = 250 Nm = 250 J Hence the correct choice is (b). 33. Friction force = mg = 0.2 5 10 = 10 N. Effective force F = applied force – frictional force = 25 – 10 = 15 N. Kinetic energy = work done by force F in pulling the body through a distance S (= 10 m) = 15 10 = 150 J, which is choice (b). 34. The retardation is given by [see Fig. 6.22] a = g ( cos + sin ) (i) Let be the initial velocity of the body. If it is stopped after moving a distance up the plane, then 2 =2 1 2 Kinetic energy = E = 2

Fig. 6.22

Now, work done is W = gain in PE = It is clear from fore, W=

= sin

sin . There(iii)

From (i), we have g= Also

a ( cos

= tan

or

sin

=

=

sin )

(iv)

sin cos

cos

(v)

Using (iv) and (v) in (iii), we have ma ( cos ) ( cos sin ) Using (ii) in (vi), we get W=

W=

(vi)

E cos , which is choice (d). ( cos sin )

35. Retardation (– a) is proportional to displacement (x), i.e. – a x or a – x. Hence the motion of the particle is simple harmonic. When the displace1 ment is x, the kinetic energy = m 2 (A2 – x2), 2 where m, and A are the mass, angular frequency and amplitude respectively. When displacement 1 x = 0, the kinetic energy = m 2 A2. Therefore, 2 the loss of kinetic energy for a displacement x is 1 1 1 m 2A2 – m 2 (A2 – x2 ) = m 2x2 2 2 2 which is proportional to x2 (since m and are constants of the motion. Hence the correct choice is (a). 36. Power delivered in time t1 is 1 = F v1 = m a v1 v1 Now, acceleration vector is a . Therefore t1 1

=

m v1 v1 t1

mv12 t1

(

v1 v1 = v12)

6.22 Comprehensive Physics—JEE Advanced

Power delivered per unit time = Power delivered at time t =

1

t=

Therefore, the x and y components of acceleration are F 6 = 3 ms–2 ax = x m 2

1

t1 mv12t

t12 t1 Hence the correct choice is (b). Notice that choices (a), (c) and (d) do not have the dimensions of power. 37.

W = F r = 5i

3j

2k

2i

and ay =

Fy

8 = 4 ms–2 2

m

Resultant acceleration a =

j

= = 10 i i

3j j

i j

k i

k j

0

= 10 – 3 = 7 Hence the correct choice is (b). 38. If mass m falls through a distance , mass m rises up through the same distance . Let v be the common velocity of the masses when this happens. Now, loss in PE = gain in KE, i.e. 1 (M + m) v2 – = 2 2 M m which gives v = , which is choice M m (c). 39. When M has descended a distance , loss of PE = . If v is the common velocity of the masses, 1 (M + m) v2. Hence gain in KE = 2

Work done = mg

v v

v

, which is choice (d)

41. Given U = 6x + 8y joule and mass m = 2 kg. Force along x-axis is dU d (6x + 8y) = 6 newton |Fx| = dx dx Force along y-axis is dU d (6x + 8y) = 8 newton |Fy| = dy dy

a 2y

2

4

3

2

= 5 ms–2

The x and y coordinates of the body at time t are 1 1 x = x0 – 3 t2 a x t2 = 6 – 2 2 =

3 2 t 2

6

metre

1 ay t 2 2

y = y0

and

2t 2

= 4

4

1 2

4 t2

metre

The body will cross the y-axis when x = 0, i.e. at 3 2 t = 0 or t = 2 s. Hence the time t given by 6 2 correct choice is (b). 42. vx = ax t = 3 ms–2 2 s = 6 ms–1 vy = ay t = 4 ms–2 2 s = 8 ms–1

1 (M + m) v2 = 2 2Mg , which is choice (c). or v= M m 40. Relative speed of man with respect to escalator = (v – ). Actual displacement of man per second = (v – ). Hence, the actual displacement of man in going up v . Therefore, the escalator of hight is (v )

ax2

v 2x

v=

v 2y

6

2

8

2

= 10 ms–1

Hence the correct choice is (c). 43. Gravitational force is conservative. The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle. Hence the correct choice is (b). 44. The potential energy of the particle is given by U=–

Fdx = –

(–

+ ax3) dx

x2 x4 x2 = (2 – ax2) (1) a 2 4 4 From Eq. (1) it follows that U = 0 at two values of x which are x = 0 and x = 2 / . Hence graphs (b) and (c) are not possible. Also U is maximum or dU minimum at a value of x given by = 0, i.e. dx 2 d a x4 0= = – ax3 4 dx 2

or

U=

= x( – ax2) or

x=

/ . At this value of x,

Work, Energy and Power 6.23

U is maximum if Now

d 2U dx

=

2

x=

At

d 2U dx

d 2U d x2

d ( dx

< 0,

– ax3) =

– 3ax2.

or

/ ,

=

– 3a

=

– 3 = – 2 , which is negative.

2

a

/ . Hence U is maximum at x = Hence graph (a) is also not possible. Also U is negative for x >

2 / . Therefore, the correct graph

is (d). 45. Mass of the drop m = volume density of water = 4 3 r , where is the density of water. Work done 3 by gravitational force is 4 3 r 3 Thus W r3. Hence the correct choice is (c). 46. At the highest point, the energy of the ball is entirely potential = and at the lowest point, the 1 mv2. Since friction energy is entirely kinetic = 2 is absent, the principle of conservation of energy requires 1 mv2 = 2 W=

1 49. Initial KE = mv2. Now, gain in PE = loss in KE. 2 Thus 1 = mv2 4

=

or v = 2 2 10 correct choice is (b).

0.2 = 2 ms–1. Hence the

=

Hence the correct choice is (d). 50. If a is the deceleration due to the force of friction f, then 2ax = v2 1 or mv 2 = max 2 or KE = f x ( f = ma) Thus if KE is doubled, x is also doubled. Hence the correct choice is (c). 51. Height = 10 m. PE at this height = . On reaching the ground, KE = . Since the body after one impact = 0.75 . If v1 is the initial upward velocity after the impact, we have 1 3 mv 21 = 0.75 = 2 4 v 21 = 1.5

or The height 1

90 = 0.9 100

v=

Hence the correct choice is (a).

2g

1

to which the body will rise is =

1.5 2g

= 0.75

1 mv 22 = 2 . Therefore,

1 mv2 = 0.9 2 = 1.8 10 2 = 36 or v2 = 1.8 which gives v = 6 ms–1. Hence the correct choice is (d). 48. Initial PE = . Now, gain in KE = loss in PE. Thus 1 1 mv2 = 2 2 or

=

v12

= 0.75 10 = 7.5 m ( = 10 m) Hence the correct choice is (c). 52. After the second impact, the initial KE of body = 3 3 2 = , i.e. 4 4

47. PE at A = at point B =

v2 4g

3 4

2

9 8 The height 2 to which the body will rise after the second impact is v 22 =

or

2

=

v22 9 9 = = 2g 8 2g 16

=

9 10 45 = m 16 8

Hence the correct choice is (d). 53. As the body rises, the initial kinetic energy is converted into potential energy. At the maximum height , the energy is entirely potential = , which is equal to the initial kinetic energy. Let its initial value. At this height, the potential energy

6.24 Comprehensive Physics—JEE Advanced

, PE = 0.25

Thus

= 0.25

or

1 2

=

. Hence the correct choice is (c). 4 54. Let f be the force of friction and m be the mass of the body. The retardation a = f/m. If v is the initial speed of the body, then 2ax = v2 1 or max = mv2 = 2 But ma = f. Therefore fx = or f = /x. Hence the correct choice is (a). 55. Potential energy function is x

U(x) =

x

Fd 0

xd 0

2

which givens

=2

2

y 1 = , which is choice (c) x 2

57. Refer to the Fig. 6.23. Here OA = OB = OC = L and OD = OC cos = L cos Therefore = OA – OD = L – L cos

v

1 2 x 2

v

The value of U(x) is always negative for both positive and negative values of x. Thus the variation of potential energy with x is an inverted parabola as shown in choice (a). 56. Potential energy stored in spring S1 when the block 1 2 B is moved through a distance x is U1 = 1x 2 1 2 = . When the block is released, it moves to the 2 left, compressing the spring S2 through a distance y. The potential energy stored in spring S2 when its 1 1 2 (4 ) y2 = 2 2. compression is y is U2 = 2y = 2 2 Since y is the maximum compression of spring S2, from conservation of energy, we have U1 = U2, i.e.

Fig. 6.23

or = L (1 – cos ). From conservation of energy, total energy at A = total energy at C, i.e. 1 1 v mv2 = m 2 2 2

2

+ mgL (1 – cos

)

8 gL (1 – cos ) (1) 3 The minimum velocity the bob must have at A so v2 =

as to reach B is v = 5gL . Putting this in Eq. (1), 7 3 we get cos = – . Therefore lies between 8 4 and

II Multiple Choice Questions with One or More Choices Correct 1. In which of the following is no work done by the force? (a) A man carrying a bucket of water, walking on a level road with a uniform velocity. (b) A drop of rain falling vertically with a constant velocity. (c) A man whirling a stone tied to a string in a circle with a constant speed (d) A man walking up on a staircase. 2. The work done by a force on a body does not depend upon (a) the mass of the body

(b) the displacement of the body (c) the initial velocity of the body (d) the angle between the force vector and the displacement vector 3. A simple pendulum of length L and having a bob of mass M is oscillating in a plane about a vertical line between angular limits – and + . At a time when the angular displacement is (< ), the tension in the string is T and the velocity of the bob is v. Which of the following relations will hold? (a) T cos = Mg (b) T – Mg cos = Mv2/L

Work, Energy and Power 6.25

(c) The magnitude of the tangential acceleration of the bob is |aT| = g sin (d) T = Mg cos 4. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that (a) the velocity of the particle is constant (b) the acceleration of the particle is constant (c) the kinetic energy of the particle is constant (d) the particle moves in a circular path. IIT, 1987 5. Two inclined frictionless tracks of different inclinations meet at A from where two blocks and Q of different masses are allowed to slide down from rest at the same time, one on each track, as shown in Fig. 6.24.

Fig. 6.24

(a) Both blocks will reach the bottom at the same time (b) Block Q will reach the bottom earlier than block (c) Both blocks reach the bottom with the same speed (d) Block Q will reach the bottom with a higher speed than block 6. Choose the correct statements from the following: (a) When a conservative force does positive work on a body, its potential energy increases. (b) When a body does work against friction, its kinetic energy decreases. (c) The rate of change of total momentum of a many–particle system is proportional to the net external force acting on the system. (d) The rate of change of total momentum of many–particle system is proportional to the net internal force acting on the system. 7. Which of the following forces are conservative? (a) Coulomb force between charged particles at rest (b) Force of a compressed elastic spring (c) Gravitational force between two masses (d) Frictional force. 8. A body of m is moving in a straight line at a constant speed v. Its kinetic energy is and the magni-

tude of its momentum is . Which of the following relations is/are correct? (a)

=

2

(c) 2 = v

(b)

=

(d) v =

2 m 2

9. A block of mass m is taken from the bottom of an inclined plane to its top and then allowed to slide down to the bottom again. The length of the inclined plane is L block and the plane is . The inclination of the plane is . (a) The work done by the gravitational force over the round trip is zero. (b) The work done by the applied force over the upward journey is mgL (sin + cos ). (c) The work done by the frictional force over the round trip is zero. (d) The kinetic energy of the block when it reaches the bottom is mgL (sin – cos ). 10. A uniform rod has a mass m and a length . The potential energy of the rod when (a) it stands vertically is zero. (b) it stands vertically is /2 (c) it is inclined at an angle with vertical is 1 cos . 2 (d) it is inclined at an angle with the vertical is 1 sin . 2 11. A body is subjected to a constant force F in newton given by F = – i + 2 j + 3k where i , j and k are unit vectors along x, y and z axes respectively. The work done by this force in moving the body through a distance of (a) 4 m along the z-axis is 12 J. (b) 3 m along the y-axis is 6 J. (c) 4 m along the z-axis and then 3 m along the y-axis is 18 J. (d) 4 m along the z-axis and then 3 m along the y-axis is (12)2 (6) 2 J. 12. Figure 6.25 shows the force F (in newton) acting on a body as a function of x. The work done in moving the body (a) from x = 0 to x = 1 m is 2.5 J. (b) from x = 1 m to x = 3 m is 10 J. (c) from x = 0 to x = 4 m is 15 J. (d) from x = 0 to x = 4 m is 12.5 J.

6.26 Comprehensive Physics—JEE Advanced

Fig. 6.25

13. A block of mass 2 kg, initially at rest on a horizontal g = 10 ms–2 (a) the work done by the applied force in 4 s is 240 J. (b) the work done by the frictional force in 4 s is 96 J. (c) the work done by the net force in 4 s is 336 J. (d) the change in kinetic energy of the block in 4 s is 144 J. 14. In which of the following cases is no work done by the force? (a) A satellite revolving around the earth in a circular orbit. (b) The electron revolving around the proton in a hydrogen atom. (c) A charged particle moving in a circle in a (d) A man carrying a load on his head and walking on a level road at a uniform velocity. 15. A raindrop falls from a certain height above the ground. Due to the resistance of air, its acceleration gradually decreases until it becomes zero when the drop is at half its original height. If W1 and W2 are the amounts of work done by the of the journey, then (b) W1 < W2 (a) W1 > W2 (c) W1 = W2 0 (d) W1 = W2 = 0 16. A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force F = /r2 where is a constant. (a) The kinetic energy of the particle is /2r. (b) The potential energy of the particle is – /2r. (c) The total energy of the particle is – /2r. (d) The total energy of the particle is zero. 17. The displacement x (in metres) of a particle of mass 100 g moving in a straight line under the action of

a constant force is related to time t (in seconds) as x =t–2 (a) The acceleration of the particle is 1 ms–2. (b) The acceleration of the particle is 2 ms–2. (c) The velocity of the particle at t = 3 s is 2 ms–1. (d) The work done by the force in 5 s is 1.8 J. 18. Two springs 1 and 2 have spring constants 1 and 2 ( 1 > 2). W1 and W2 are the amounts of work done to increase the lengths of springs 1 and 2 by the same amount and W3 and W4 are the amounts of work done when springs 1 and 2 are stretched with the same force. Then (b) W1 < W2 (a) W1 > W2 (c) W3 > W4 (d) W3 < W4 19. A box of mass m is dragged along a horizontal surface with a uniform speed with a force F directed at an angle with the horizontal as shown between the box and the surface is . (a) The normal reaction on the box is F cos . R= (b) The normal reaction on the box is R = (mg – F sin ) (c) The work done on the box in dragging it through a distance x is mgx . W= (sin cos ) (d) The work done on the box in dragging it through a distance x is mgx . W= (cos sin )

Fig. 6.26

20. A force F = (3 i + 4 j ) newton acts on a particle located at the origin O. (a) The work done in taking the particle from O along the x-axis to a point A (2 m, 0) is 6 J. (b) The work done in taking the particle from A parallel to the y-axis to a point B (3 m, 2 m) is 8 J. (c) The work done in taking the particle from O to A and then to B is 14 J.

Work, Energy and Power 6.27

(d) The work done in taking particle from O directly to B is 10 J. 21. The potential energy of a system varies with distance x as U = ax2 – bx where a and b are positive constants. Then (a) the potential energy is minimum at x = b/2a

(c) T1 = (d) T1 =

mg , T2 = 2

3mg 2

3mg mg , T1 = 2 2

b2 . 4a (b) the potential energy is maximum at x = b/2a and Umin = –

b2 . 4a (c) The force acting on the system decreases linearly with x. (d) The force acting on the system is proportional to x2. IIT, 2005 22. A car is of mass m moving along a circular track of radius r with a speed which increases linearly with time t as v = , where is a constant. Then (a) the instantaneous power delivered by the 3 3 t /r. centripetal force is (b) the power delivered by the centripetal force is zero. (c) the instantaneous power delivered by the 2 t. tangential force is (d) the power delivered by the tangential force is zero. 23. A body of weight mg is suspended from a rigid support by two light strings AB and AC as shown in Fig. 6.27. The tension in string AB is T1 and in string AC the tension is T2. Then and Umax =

(a) T1 =

3 T2

(b) T2 =

3 T1

Fig. 6.27

24. The potential energy function of a particle executing linear simple harmonic motion is given by 1 2 U (x) = 2 where x is the displacement of the particle from the equilibrium position x = 0 and is the force constant of the oscillator. Figure 6.28 shows the graph of U(x) against x for = 0.5 Nm–1. If the total energy of the particle is 1 J, it will turn back when it reaches the position (a) x = – 1 m (b) x = – 2 m (c) x = + 2 m (d) x = + 1 m

Fig. 6.28

ANSWERS AND SOLUTIONS 1. In choices (a) and (c), no work is done because the force of gravity in choice (a), and the centripetal force in choice (c) are perpendicular to the direction of the displacement. In choice (b), no net force acts on the raindrop since it is falling with a uniform velocity, hence no work is done. In choice (d), the man has to do work against gravity. Hence the correct choices are (a), (b) and (c). 2. The correct choices are (a) and (c). 3. Referring to Fig. 6.29, the net force along the string is F = T – Mg cos and this force provides the centripetal force Mv2/L necessary for circular motion of the bob. Hence choice (b) is correct.

Relation (a) i.e. T cos = Mg cannot hold for all values of because if = 0, then T = Mg and the net force along the string is F = T – Mg cos 0 = T – Mg = 0 Since F = 0, there would be no centripetal force and the bob would not oscillate. Hence choice (a) is incorrect. The tangential acceleration aT of the bob is caused by the tangential component Mg sin . Therefore, aT = Mg sin /M = g sin . Hence choice (c) is correct. The relation (d) cannot hold because if T = Mg cos , the net force along the string will be zero. Therefore, there will be no centripetal force.

6.28 Comprehensive Physics—JEE Advanced

Hence the correct choices are (b) and (c).

Fig. 6.30 Fig. 6.29

4. The particle moves in a circular path with a uniform speed, because the force has a constant magnitude and is perpendicular to the velocity of the particle. The velocity and acceleration of the particle are not constant because their directions are changing as the particle moves in a circular path. The magnitude of the velocity (i.e. speed) is 1 mv2 is constant. Therefore, the kinetic energy 2 constant. Hence choices (c) and (d) are correct and choices (a) and (b) are incorrect. 5. Refer to Fig. 6.30. The accelerations of blocks and Q are m g sin a1 = 1 m1

1

= g sin

1

m2 g sin 2 = g sin 2 m2 Since 2 > 1; a2 > a1. Now PE of block P at A = 1 m1 . Its KE on reaching the bottom = m1v 21. 2 Equating the two we get 1 m1v 21 = m1 2

and

or

a2 =

v1 =

6. The work done by a conservative force is equal to the negative of the potential energy. When the work done is positive, the potential energy decreases. Thus choice (a) is incorrect. Friction always opposes motion. Hence, when a body does work against friction, its kinetic energy decreases. Thus choice (b) is correct. The rate of change of total momentum of a many–particle system is proportional to the net force external to the system; the internal forces between particles cannot change the momentum of the system. Hence the correct choices are (b) and (c). 7. The correct choices are (a), (b) and (c). 2

8. Now

= mv or

2

= m2v2 or

2m

=

1 mv2 = 2

or

1 mv2. 2 Dividing the two we get 2 = v. Hence the correct relations are (a) and (c). 9. The gravitational force is conservative. Therefore, the work done by the gravitational force over the round trip is zero. Refer to Fig. 6.31. =

2

. Also

= mv and =

2

Similarly, for block Q, v2 = 2 . Since v1 = v2, both blocks will reach the bottom with the same = 0) and v2 = a2t2. But speed. Now, v1 = a1t1 ( v1 = v2. Therefore a 1t 1 = a 2t 2 t1 a = 2 t2 a1 Since a2 > a1; t1 > t2; i.e. block takes a longer time to reach the bottom. Hence the correct choices are (b) and (c).

Fig. 6.31

When the block is moved from C to A, the force of friction f acts along the plane in the downward direction which is in the direction of the component mg sin of the gravitational force. Hence the applied force F is F = mg sin + f

Work, Energy and Power 6.29

frictional force f f = = normal reaction R mg cos f = mg cos Therefore F = mg sin + mg cos = mg(sin + cos ) Work done by the applied force over the upward journey is Wa = F L = mg(sin + m cos ) L The frictional force is non-conservative. Hence the work done by the frictional force over the round trip is not zero. When the block is at a point A, it is at rest and its initial velocity = 0. It is allowed to slide down the plane. Let v be the velocity when it reaches the bottom C of the plane. Since the frictional force now acts upwards, the net force acting on the block when it slides down is Fn = mg sin – f = mg sin – mg cos Now

=

= mg(sin

–

cos )

fn m = g(sin – cos ). The velocity v is given by v2 – 2 = 2aL (where L = AC) or v2 = 2aL ( = 0) Kinetic energy, 1 KE = mv2 2 1 = m 2AL = maL 2 or KE = mg(sin – cos )L Acceleration of the block, a =

Hence the correct choices are (a), (b) and (d). 10. The potential energy in the vertical position = work done in raising it from horizontal position to vertical position. In doing so, the mid-point of the rod is raised through a height = /2. Since the entire mass of the rod can be assumed to be concentrated at the mid-point (centre of gravity), the work done = = /2. Refer to Fig. 6.32. AD = AB = . In the inclined position, let the centre of gravity C of the rod be at a height above the ground, so that AC = /2. In triangle ACE, Fig. 6.32 we have = AC sin(90° – ) =

2

cos

PE =

=

1 2

cos . The correct choices are

(b) and (c). 11. The correct choices are (a), (b) and (c). Displacement along the z-axis is S = 4 k metres. Therefore, work done is W1 = F S = (– i + 2 j + 3 k ) (4 k ) = – 4 i k + 8 j k + 12 k k Now i k = 0 and j k = 0 because j and j are perpendicular to k . But k k = 1. Therefore, W1 = 12 J. Displacement along the y-axis is S = 3 j metres. Therefore, the work done is W2 = F S = (– i + 2 j + 3 k ) (3 j ) = 6 j j = 6 J Since work is a scalar, the total work done is just the algebraic sum of W1 and W2, i.e. W = W1 + W2 = 12 + 6 = 18 J. 12. The correct choices are (a), (b) and (c). Work done = area under the F–x graph. Work done by the force in moving the body from x = 0 to x = 1 m is (see Fig. 6.33) W1 = area of triangle OAD 1 Fig. 6.33 = AD OD 2 1 = 5 N 1m = 2.5 J 2 Hence the correct choice is (a). Work done in moving the body from x = 1 m to x = 3 m is W2 = area of rectangle ABDE = AD DE = 5 N 2 m = 10 J, which is choice (c) Work done in moving the body from x = 0 to x = 4 m is W3 = area of (triangle OAD + rectangle ABDE + triangle BEC) = 2.5 + 10 + 2.5 = 15 J, which is choice (d). 13. The correct choices are (a), (b) and (d). Force of friction (f) = mg = 0.2 2 10 = 4N. Applied force (F) = 10 N. Since friction opposes motion, the net force acting on the body when it is moving is

6.30 Comprehensive Physics—JEE Advanced

dv d = (2t – 4) = 2 ms–2 dt dt Now F = ma = 0.1 kg 2 ms–2 = 0.2 N Now distance moved in t = 5 s = (5)2 – 4 5 + 4 = 9 m. Work done W = 0.2 N 9 m

F = F – f = 10 – 4 = 6 N 6N F Acceleration a = = = 3 ms–2 M 2 kg The distance travelled by block in 4s is 1 2 1 at = 0 + 3 (4)2 = 24 m S= + 2 2 Work done by applied force in 4 s is

= 1.8 Nm or J

W = applied force distance moved in 4 s = 10 N 24 m = 240 J Work done by the force of friction in 4 s is W = f S = 4 N 24 m = 96 J. Work done by the net force in 4 s is W = F = 6 N 24 m = 144 J Velocity acquired by the block in 4 s is v = + at = 0 + 3 4 = 12 ms–1 Kinetic energy of the block at t = 4 s is 1 1 mv2 = 2 (12)2 = 144 J KE = 2 2 Since the initial KE = 0, the change in KE = 144 J. 14. All the four choices are correct. In choices (a), (b) and (c), the centripetal force (being radial) is perpendicular to the velocity (and hence displacement) vector. In choice (d), the gravitational force (being vertically downwards) is perpendicular to the displacement. 15. The correct choice is (c). Work done is each half of the journey = where = H/2; H being the original height from which the drop fell. 16. The correct choices are (a) and (c). mv 2 = 2 or mv2 = . Therefore, F= r r r KE =

1 2 18. If the increase in length is x, W1 = 1x and 2 W1 1 2 W2 = = 1 . Since 1 > 2; 2x . Therefore, W2 2 2 W1 is greater than W2. Let a force F x1 and the x second by x2. Then F = 1x1 = 2 x2 or 1 = 2 . x2 1 1 1 2 2 Now W3 = x and W = x . Therefore, 1 1 4 2 2 2 2 2

PE =

=–

r

Fdr =

r

r

2

dr =

1r =– r r

dr Fig. 6.34

r2

Work done W = Fx Thus the correct choices are (a), (b) and (d). 20.

Total energy = KE + PE =

2

W3 x1 2 = 1 = 1 = 2 < 1. x2 W4 2 2 1 1 Thus the correct choices are (a) and (d). 19. The different forces acting on the block are shown in Fig. 6.34. It follows that R = F cos and R + F sin = mg Eliminating R we get mg F= cos sin

1 mv2 = 2 2r r

Now

a=

Acceleration

–

=–

2r r 2r 17. The correct choices are (b), (c) and (d). x = (t – 2)2 = t2 – 4t + 4 dx d 2 v= = (t – 4t + 4) = 2 t – 4 dt dt

WO

= (3 i + 4 j ) (2 i ) = 6 J

WO

A

WA

B

A

B

= (3 i + 4 j ) (2 j ) = 8 J = 6 + 8 = 14 J

WO

B

= (3 i + 4 j ) (2 i + 2 j ) = 6 + 8 = 14 J

Hence the correct choices are (a), (b) and (c) dU d 2U = 0 and > 0. Now 21. U is minimum if dx dx 2

Work, Energy and Power 6.31

dU d dU = (ax2 – bx) = 2ax – b. So = 0 at x dx dx dx d 2U d = b/2a. Also = (2ax – b) = 2a, which is 2 dx dx

Equations (1) and (2) give T1 = =

mg and T2 = 2

3 T1

3mg . Thus the correct choices are (b) and (c). 2

positive. Hence choice (a) is correct and choice (b) is wrong. For x = b/2a, b 2 b b2 –b =– . 4a 2a 2a Force acting on the system is dU F= – = b – 2ax dx Hence choice (c) is correct and choice (d) is wrong. Umin = a

Fig. 6.35

24. At any instant during its motion, the energy of the particle is partly kinetic and partly potential, but the total energy remains constant. If v is the velocity of the particle and x its displacement from the mean position at an instant of time, then at that instant the total energy is 1 1 2 E = K.E. + P.E. = mv2 + 2 2 where m is the mass of the particle. The particle will turn back when v = 0 momentarily. At

2 2 mv 2 t = . Since Fc is r r = Fc v = 0. Tangential force

22. Centripetal force Fc =

perpendicular to v, md v d Ft = = m ( ) = . Since Ft is parallel to dt dt v, = Ft v = = 2t. Hence the correct choices are (b) and (c). 23. and

T1 sin 60° = T2 sin 30°

(1)

T1 cos 60° + T2 cos 30° = mg

(2)

1 2E 2 1 2 x= = 2 0.5 = ± 2 m. Hence the correct choices are (b) and (c). that moment E =

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I Invariance Newton’s laws of motion are applicable in all inertial reference frames. Some physical quantities, when measured by observers in different reference frames, have exactly the same value. Such physical quantities are called ant. In Newtonian mechanics mass, time, and force are invariant quantities. On the other hand, some physical quantities, when measured by observers in different reference frames, do not have the same value. Such physical quantities are called not invariant. In Newtonian mechanics displacement, velocity and work (which is the dot product of force and displacement) are not invariant. Also kinetic 1 2 mv is not invariant. energy 2

Physicists believe that all laws of physics are invariant in all inertial reference frames. For example, the workenergy principle states that the change in the kinetic energy of a particle is equal to the work done on it by the force. Although, work and kinetic energy are not invariant in all reference frames, the work-energy principle remains invariant. Thus even though different observers measuring

energy principle holds in their respective frames. 1. Choose the invariant quantities from the following (a) mass (b) time (c) velocity (d) displacement 2. Which of the following quantities is/are not invariant (a) Work (b) Kinetic energy (c) Torque (d) Displacement

6.32 Comprehensive Physics—JEE Advanced

3. Choose the correct statements form the following. (a) Kinetic energy is not invariant. (b) Potential energy is not invariant.

(c) Laws of conservation of energy and momentum are invariant. (d) All laws of physics are invariant.

ANSWERS 3. The correct choices are (c) and (d).

1. The correct choices are (a) and (b). 2. All choices are correct. Questions 4 to 8 are based on the following passage Passage II Mechanical energy exists in two forms: kinetic energy and potential energy. Kinetic energy is the energy possessed by a body by virtue of motion. Potential energy is the energy These two forms of energy are inter-convertible. If no other form of energy is involved in a process, the sum of kinetic energy and potential energy always remains constant. 4. Two particles of masses m1 and m2 have equal linear momenta. The ratio of their kinetic energies is (a) 1

(b)

m2 m1 2

m1 m2

SOLUTION

(d)

2 1 1 mv2 = (mv)2 = , 2m 2 2m = mv is the linear momentum. Thus 2

K1 =

2m1

2

and K2 =

(b) 1 : 2

(c) 1 : 3 (d) 1 : 4 7. In Q.6, what is the original speed of the heavier particle? (a) 2(1 +

2 ) ms–1

(b) 2(1 –

2 ) ms–1

(d) (2 2 – 1) ms–1

8. A uniform rod of mass m and length is made to stand vertically on one end. The potential energy of the rod in this position is mg mg (b) (a) 4 3 (c)

mg 2

(d)

2

m1 m2

4. Kinetic energy K = where

(a) 1 : 1

(c) (2 2 + 1) ms–1

m2 m2 (c) (d) m1 m1 5. Two particles of masses m1 and m2 have equal kinetic energies. The ratio of their linear momenta is m1 (a) 1 (b) m2 (c)

6. A particle of mass m has half the kinetic energy of another particle of mass m/2. If the speed of the heavier particle is increased by 2 ms–1, its new kinetic energy equals the original kinetic energy of the lighter particle. The ratio of the original speeds of the lighter and heavier particles is

2m2

K1 m = 2 , which is choice (c). K2 m1 = 2mK . Hence the correct choice 5. K = 2/2m is (b). 6. Let v and v be the original speeds of the heavier and the lighter particles respectively. We then have 1 1 1 m mv2 = v2 2 2 2 2 1 v2 = v or v = 2v 4

Hence the correct choice is (b). 7. When the heavier particle is speeded up by 1 2.0 ms–1, its kinetic energy becomes m(v + 2)2. 2 Since this equals the original kinetic energy of the lighter particle, we have 1 1 m(v + 2)2 = (m/2)(4v2) 2 2 2 v + 4 + 4v = 2v2 or v2 – 4v – 4 = 0 v= =

4

16 16 2

4 2 8 = 2 2 2 2

The positive root is v = 2+ 2 2 = 2(1 + Hence the correct choice is (a).

2 ).

Work, Energy and Power 6.33

8. The potential energy in the vertical position = work done in raising it from horizontal position to vertical position. In doing so, the mid-point of the rod is raised through a height = /2. Since the entire Questions 9 to 11 are based on the following passage Passage III A light rod of length L having a body of mass M attached to its end hangs vertically. It is turned through 90° so that it is horizontal and then released. 9. The centripetal acceleration when the rod makes an angle with the vertical is (a) g cos

(b) 2g cos

(c) g sin

(d) 2g sin

mass of the rod can be assumed to be concentrated at the mid-point (centre of gravity), the work done = = /2. Hence the correct choice is (c).

10. The tension in the rod when it makes an angle with the vertical is (a) Mg cos (b) 2 Mg cos (c) 3 Mg cos (d) zero 11. The value of when the tension in the rod equals the weight of the body is given by 1 1 (b) = cos–1 (a) = cos–1 2 3 1 2

= sin–1

(c)

(d)

= sin–1

1 3

SOLUTION 9. The rod is released from the horizontal position OA. Let OB be the position of the rod when tension in the rod is T (Fig. 6.36).

v 2 2gL cos = L L = 2g cos ,

centripetal acceleration =

which is choice (b). 10. The centripetal force when the body is at B is M v2 Fc = L Thus, we have M v2 (2) T – Mg cos = L Using (1) in (2), we get Fig. 6.36

Let be the angle with the vertical at this position. The loss of PE when the body falls from A to B = Mg OC = MgL cos . If v is the velocity of the body at B, then 1 Mv2 = MgL cos 2

or v2 = 2gL cos

(1)

Questions 12 to 14 are based on the following passage Passage IV A small roller coaster starts at point A with a speed on a curved track as shown in Fig. 6.37. The friction between the roller coaster and the track is negligible and it always remains in contact with the track.

M 2gL cos = 2 Mg cos L or T = 3 Mg cos Thus the correct choice is (c). 11. T = Mg. Therefore, 1 Mg = 3 Mg cos or cos = , which is choice (b). 3 T – Mg cos

=

12. The speed of the roller coaster at point B on the track will be 1/ 2 2 2 (b) (a) ( 2 + )1/2 3 (c) (

2

)1/2

+2

2

(d)

2 13. The speed of the roller coaster at point C on the track will be 1/ 2

(a)

Fig. 6.37

1/ 2

3

(c)

2

3 2

2

(b)

1/ 2

2 3

1/ 2

4 3

(d) (

2

+2

)1/2

6.34 Comprehensive Physics—JEE Advanced

14. The speed of the roller coaster at point D on the track will be

SOLUTION 1 2 + . If vb 2 1 is the speed at point B, the total energy at B = mv2b 2 + mg(2 /3). From the principle of conservation of energy, we have 1 1 2mg 2 + = mv2b + 2 2 3

12. Total energy at A = KE + PE =

which gives vb =

2

1/ 2

2 3

(a) (

2

+

(c) (

2

+3

Questions 15 to 17 are based on the following passage Passage V The displacement x of a particle moving in one dimension, under the action of a constant force is related to time t by the equation t = x +3 where x is in metre and t is in second. 15. The displacement of the particle when its velocity is zero is

SOLUTION 15. Given t =

x + 3 or x = t – 3 or (1) x = (t – 3)2 Differentiating (1) with respect to t, we get dx = 2 (t – 3) dt or v = 2 (t – 3) (2) From (2) it follows that v = 0 at t = 3 s. Using t = 3 s in (1), we get x = 0. Thus, the displacement of the particle is zero when its velocity is zero. Thus the correct choice is (a). 16. From Eq. (2), we have dv d = [2(t – 3)] = 2 ms–2. a = dt dt Hence the correct choice is (d). Questions 18 to 20 are based on the following passage Passage VI The work done by a constant force acting on a body is given by W=F r where F is the force vector and r is displacement vector. The displacement vector r = r2 – r1 where r1 is the initial

)1/2

(b) (

2

+2

(d)

(

2

+4

)1/2 )1/2

13. Similarly, the speed at point C is given by 1 1 mg 2 + = mv2c + which gives 2 2 3 vc =

4

2

1/ 2

, which is choice (c). 3 14. At point D, the energy is entirely kinetic. If the speed of the roller coaster at point D is vd, then we have

,

which is choice (b).

)1/2

or

1 mv2d = 2 vd = (

(a) zero

+ 2

1 2

2

)1/2, which is choice (b).

+2

(b) 1 m

(c) 2 m

(d) 3 m

16. The acceleration of the particle (a) increases with time (b) decreases with time (c) increases with time up to t = 3 s and then decreases with time. (d) remains constant at 2 ms–2. 17. (a) 1 J

(b) 3 J

(c) 6 J

(d) zero

17. From Eq. (2), the initial velocity, i.e., velocity at t = 0 is v0 = 2(0 – 3) = – 6 ms–1 Final velocity, i.e., velocity at t = 6 s is v = 2(6 – 3) = 6 ms–1

=

1 1 1 mv2 – mv20 = m(v2 – v02 ) 2 2 2

=

1 m[(6)2 – (– 6)2] = 0, which is choice (d). 2

position vector and r2 If the force is variable, the work done in moving a body from a position r1 to a position r2 is given by r2

F dr

W = r1

where dr

Work, Energy and Power 6.35

20. A body of mass m is projected from the ground with a velocity at an angle above the horizontal. The work done by the gravitational force in time sin t= is g

18. A particle is moved from a position r1 = (3 i + 2 j – 4 k ) metre to a position r2 = (5 i + 6 j + 9 k ) metre under the action of a force F = ( i + 3 j + k ) newton. The work done is (a) zero

(b) 13 J

(c) 27 J

(d) 35 J

(1 + cos )

(c)

(b)

sin2

(b)

2

sin2

2

sin 2 (d) zero 2 21. If = 45°, the work by the gravitational force in 2 sin is Q.20 above in time t = g (c)

19. A body of mass m is projected from a tower of height at an angle above the horizontal. The work done by the gravitational force during the time it takes to hit the ground is (a)

2

(a) 2

2

(1 + sin )

(a)

(d) zero

2

(b)

4

2 (d) zero

2

(c)

SOLUTION 18. W = F r = F (r2 – r1)

Work done W = sin 20. t = g

= ( i + 3 j + k ) [(5 i + 6 j + 9 k ) – (3 i + 2 j – 4 k )]

, which is choice (c).

the body attains maximum height

= ( i + 3 j + k ) (2 i + 4 j + 13 k )

2

max

=

sin 2 2g

2 sin 2 = which is Work done = max 2 choice (c). 21. Net displacement = 0. Hence W = 0. Thus the correct choice is (d).

= 2 + 12 + 13 = 27 J, which is choice (c) 19. Net displacement from the time the body is projected to the time it hits the ground is = vertically downwards.

IV Matching 1.

Column I (a) (b) (c) (d)

Force Impulse Energy stored in a spring Force constant of a spring

ANSWER (a) (s) (c) (r)

.

Column II (p) (q) (r) (s)

Slope of force-extension graph Area under force-time graph Area under force-extension graph Slope of linear momentum-time graph

(b) (d)

(q) (p)

6.36 Comprehensive Physics—JEE Advanced

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each questions has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 A simple pendulum of length is displayed from its mean position O to position A so that the string makes an angle 1 with the vertical and then released. If air resistances is neglected, the speed of the bob when the string makes an angle 2 with the vertical is v =

2 (cos

2

cos 1 ) .

Statement-2 The total momentum of a system is conserved if no external force acts on it. 2. Statement-1 A uniform rod of mass m and length is held at an angle with the vertical. The potential energy of 1 cos . the rod in this position is 2 Statement-2 The entire mass of the rod can be assumed to be concentrated at its centre of mass. 3. Statement-1 A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30° with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. Statement-2 the surface decreases with the increase in the angle of inclination. IIT, 2007

4. Statement-1 A man carrying a bucket of water and walking on a rough level road with a uniform velocity does no work while carrying the bucket. Statement-2 The work done on a body by a force F in giving it a displacement S W = F S = FS cos where is the angle between vectors F and S. 5. Statement-1 A crane lifts a car up to a certain height in 1 min. Another crane Q lifts the same car up to the same height in 2 min. Then crane consumes two times more fuel than crane Q. Statement-2 Crane supplies two times more power than crane Q. 6. Statement-1 Two inclined frictionless tracks of different inclinations 1 and 2 meet at A from where two blocks and Q of different masses m1 and m2 are allowed to slide down from rest, one on each track as shown in Fig. 6.38. Then blocks and Q will reach the bottom with the same speed.

Fig. 6.38

Statement-2 Blocks and Q have equal accelerations down their respective tracks. 7. Statement-1 In Q.6 above, block will take a longer time to reach the bottom than block Q. Statement-2 Block Q has a greater acceleration down the track than block . 8. Statement-1 Comets move around the sun in highly elliptical orbits. The work done by the gravitational force of the sun on a comet over a complete orbit is zero.

Work, Energy and Power 6.37

Statement-2 The gravitational force is conservative. 9. Statement-1 The total energy of a system is always conserved irrespective of whether external forces act on the system. Statement-2 If external forces act on a system, the total momentum and energy will increase. 10. Statement-1 The rate of change of the total linear momentum of a system consisting of many particles is proportional to the vector sum of all the internal forces due to inter-particle interactions.

Statement-2 The internal forces can change the kinetic energy of the system of particles but not the linear momentum of the system. 11. Statement-1 An elastic spring of force constant is stretched by a small length The work done in extending the spring by a further length x is 2 2. Statement-2 The work done in extending an elastic spring by a length x is proportional to x2.

SOLUTIONS 1. The correct choice is (b). It is clear from Fig. 6.39 = cos 2. Therefore, 1 = that = cos 1 and – cos 1 = (1 – cos 1) and 2 = (1 – cos 2). Let m be the mass of the bob and v be its speed when it reaches position B. Then, from the principle of conservation of energy, K.E. at B = loss of P.E. as the bob moves from A to B. Hence 1 mv2 = 1– 2 2 = mg[ (1 – cos 1) – (1 – cos 2)] = (cos 2 – cos 1) v=

2 (cos

2

cos 1 )

Fig. 6.39

2. The correct choice is (a). Let C be the centre of mass of the rod AB so that AC = /2. Let be the height of C above the ground. In triangle ACD, we have CD = AC sin (90° – ) (see Fig. 6.40). Or

=

cos . Since the en2 tire mass of the rod can be assumed to be concentrated at the centre of mass, therefore, potential energy = work done

Fig. 6.40

to raise the rod from horizontal position on the ground to the position show = 1 cos . 2 3. Statement-1 is true. The decrease in mechanical energy is smaller when the block made to go up on the inclined surface because some part of the kinetic energy is converted into gravitational potential friction does not depend on the angle of inclination of the plane. Hence the correct choice is (c). 4. The correct choice is (a). Since the velocity is uniform, the man exerts no net force on the bucket in the direction of motion. The only force he exerts on the bucket is against gravity (to overcome) the weight mg of the bucket) and this force is perpendicular to the displacement (i.e. = 90°). Hence W = FS cos 90° = 0. 5. The two cranes do the same amount of work = . Hence they consume the same amount of fuel. Crane does the same amount of work in half the time. Hence crane supplies two times more power than crane Q. Thus the correct choice is (d). 6. The acceleration of blocks and Q respectively are m g sin 1 a1 = 1 = g sin 1 m1 and

a2 =

m2 g sin m2

2

= g sin

2

Since 2 > 1; a2 > a1. The potential energy of block at A = m1 . When it reaches the bottom B, its 1 kinetic energy is m1v21 where v1is its speed when 2 it reaches B. Now P.E. at A = K.E. at B. Hence 1 v1 = 2 . m1 = m1v21 2

6.38 Comprehensive Physics—JEE Advanced

9. Statement-1 is false; the total energy of an isolated system is conserved. Statement-2 is true. Hence the correct choice is (d). 10. Statement-1 is false and Statement-2 is true. The rate of change of momentum is proportional to the net external force acting on the system. Hence the correct choice is (d). 11. The correct choice is (d). Potential energy stored in 1 2 the spring when it is extended by x is U1 = 2 Potential energy stored in the spring when it is further extended by x is 1 U2 = (x + x)2 = 2 2 2 Work done = gain in potential energy = U2 – U1 1 2 3 2 =2 2– = 2 2

1 m2v22 v2 = 2 = v 1. 2 Hence the correct choice is (c). 7. The correct choice is (a). If t1 and t2 are the times taken by and Q to reach the bottom, then Similarly

m2

=

v1 =

1

+ a 1t 1 = a 1t 1

(

1

= 0)

and

v2 =

2

+ a 2t 2 = a 2t 2

(

2

= 0)

Now

v1 = v2. Hence a1t1 = a2t2. Thus

t1 a = 2 t2 a1 Since a2 > a1; t1 > t2. 8. The correct choice is (a). For a conservation force, the work done in moving a body from one point to another does not depend on the nature of the path and the work done over a closed path is zero, irrespective of the nature of the path.

VI Integer Answer Type 1. A particle of mass 1g executes an oscillatory motion on the concave surface of a spherical dish of radius 2 m placed on a horizontal plane. If the motion of the particle begins from a point on the dish at a height of 2 cm from the horizontal plane and the in metre covered by the particle before it comes to rest. IIT, 1990 2. A light inextensible string that goes in Fig. 6.41 connects two blocks of masses 0.36 kg and 0.72 kg. Taking in g = 10 m/s2 joules) by the string on the block

system is released from rest. IIT, 2009 3. A block of mass 0.18 kg is attached to a spring of

block is at rest and the spring is unstretched. An impulse is given to the block as shown in Fig. 6.42. The block slides a distance of 0.06 m and comes to rest for Fig. 6.42

initial velocity of the block in m/s is V = N/10. Then N is

IIT, 2011

Fig. 6.41

SOLUTION

1. Refer to Fig. 6.43 in which is the angle which the normal to the surface at the location of the particle makes with the verti-

cal. The particle will keep on oscillating about O till its initial potential energy is completely used up in doing work against the frictional force f and the particle comes to rest. It follows from the f= Since Fig. 6.43

mg cos

u2) colliding with each other. Let v1 and v2 be their respective velocities after the collision. If velocities u1, u2, v1 and v2 are all along the same straight line, the collision is known as one-dimensional or head-on collision (Fig. 7.1)

v

v

Fig. 7.1

From the law of conservation of momentum m 1 u 1 + m 2 u 2 = m 1v 1 + m 2 v 2

Coefficient of Restitution

If momentum along positive x-axis is taken to be positive, the momentum along the negative x-axis is taken to be negative. Two-dimensional or Oblique Collision If the velocities of the colliding bodies are not along the same straight line, the collision is known as two-dimensional or oblique collision (Fig. 7.2) v

Newton proved experimentally that, when two bodies collide, the ratio of the relative velocity after collision to the relative velocity before collision is constant for the two bodies. This constant is known as restitution and is denoted by letter e. velocity of separation after collision e =– velocity of approach before collision

(i) (ii) (iii)

(iv) v

v 2 v1 u 2 u1 For a perfectly elastice collision, e = 1. For a perfectly in inelastic collision, e = 0, because the two bodies stick together and hence v2 = v1. Perfectly elastic or perfectly inelastic collisions do not occur in nature. Hence, for any collision, e lies between 0 and 1. For a head-on collision (Fig. 7.1) velocity of separation e =– velocity of approach or

e =–

v2 v1 u2 u1 (v) For an oblique collision (Fig. 7.2) Velocity of approach = u1 cos 1 – u2 cos 2 Velocity of separation = v2 cos 2 – v1 cos 1 v cos 2 v1 cos 1 e=– 2 u1 cos 1 u2 cos 2 =

Fig. 7.2

In this case, we apply the law of conservation of momentum separately for x and y components of momenta. The components of momentum along the positive x-axis and positive y-axis are taken to be positive and components of momentum along negative x-axis and negative y-axis are taken to be negative. Momentum conservation of x-components gives m1u1 cos 1 + m2u2 cos 2 = m1v1 cos 1 + m2v2 cos 2 Momentum conservation of y-components gives – m1u1 sin 1 + m2u2 sin 2 = m1v1 sin 1 – m2v2 sin 2

Velocities after Head-on Elastic Collision Refer to Fig. 7.1 again. From the law of conservation of momentum, we have m 1u 1 + m 2u 2 = m 1v 1 + m 2v 2 (1)

Conservation of Linear Momentum and Collisions 7.3

v1 = – u1 v v2 e= 1 u2 u1 v1 – v2= e (u2 – u1)

and

m1 em2 u1 m1 m2 Using (3) in (2), we get

m2 1 e u2 m1 m2

(3)

m1 1 e u1 m1 m2

m2 em1 u2 m1 m2

(4)

v2 =

height from where it was dropped while the much more massive ground remains at rest. Finally, if the body at rest is much lighter than the colliding body, i.e. if m2 > m1, such that m1 is negligibly small, then

v12

K =1 Ki

v1 m = 1 m1 u1 m1 m1

m2 m2

m2 . Therefore, m2 2

=

4m1m2 m1

m2

2

NOTE

The fraction of kinetic energy lost by mass m1 is maximum . if m1 = m2 and minimum if m2 (iv) Change in kinetic energy of a system in a perfectly inelastic head-on collision. In a perfectly inelastic collision, the two bodies stick together after the collision. Hence v1 = v2 and e = 0. Putting e = 0 in Eqs. (3) and (4), we get v1 =

m1 u1 m1 m2

m2 u2 m1 m2

7.4 Comprehensive Physics—JEE Advanced

m2 m1 u2 u1 m1 m2 m1 m2 If mass m2 is stationary, u2 = 0. Then

e=

m1 u1 m1 m2

e=

and

v2 =

v1 =

(7) vcos

m1 u1 and v2 = m1 m2 Notice that v1 = v2 = v (say) Total K.E. of the system before collision is 1 Ki = m1u12 2 and after the collision is 1 Kf = m1 m2 v 2 2 Loss in K.E. of the system is 1 1 Ki – Kf = m1u12 m1 m2 v 2 2 2 From Eq. (7)

v = u1

j

v cos

j

u cos

=

v cos u cos

= eu cos

(10)

Since the impulsive force acts along the normal, the momentum along the normal is not conservved. Since the component of the impulsive force along the horizontal is zero, the momentum along the horizontal is conserved. Hence u sin = v sin (11) From Eqs. (10) and (11), we get v = (e2 cos2

(8)

m1 . Using this in m1 m2

Eq. (8) we get

+ sin2 )1/2 u

(12)

tan (13) e For a perfectly elastic collision, e =1 and Eqs. (12) and (13) give v =u and tan

=

and

=

i.e. for a perfectly elastic collision, the body re-

m1m2u12 Ki – Kf = 2 m1 m2 In general, if u2 0, we have Ki – Kf =

velocity of separation velocity of approach

m1m2 2 m1 m2

and at the same angle on the other side of the normal. 2

u1 u2

(9)

Oblique Impact on a Fixed Horizontal Plane Consider a body of mass m moving with a velocity u making an angle with the normal ON

Direct Impact on a Fixed Plane after impact, then, in this case (10) we get [Fig. 7.4] v = eu

=

= 0. Using this in Eq.

v

v making an angle with the normal. Since the horizontal place along the normal. The normal component of u is u cos along – y direction and the normal component of v is v cos along the +y directon. Now Fig. 7.4 v

7.1 A body is dropped from rest from a height h = 5.0 m.

is 0.8? Fig. 7.3

Conservation of Linear Momentum and Collisions 7.5

SOLUTION Refer to Fig. 7.5. = 2gh S e 2 gh . This is also the speed just before the second impact. Therefore,

v

speed just after second impact = e2 2 gh . This is the initial speed for the upward motion of the body

v

after the second impact, i.e. u = e2 2 gh . Therefore, height attained after two impacts is h2 =

u2 1 2 e 2 gh = 2g 2g

= (0.8)4

2

= e 4h

From conservation of momentum, mu = mv1 + mv2 u = v1 + v2 (i) Taking the scalar product of u with itself, we have u u = (v1 + v2) (v1 + v2)

5 = 2.05 m

NOTES (1) Height attained after n impacts is hn = e2nh (2) Speed of rebound after nth impact is vn = e n 2 gh (3) Total distance travelled before the body comes to rest = h

1 e

2

1 e2

.

7.2 A steel ball of mass m moving with velocity u1 undergoes a perfectly elastic head-on collision with another identical steel ball moving with velocity u2. Show that, after the collision, they merely exchange their velocities.

Refer to Fig. 7.1 again. From conservation of momentum, mu1 + mu2 = mv1 + mv2 u1 + u2 = v1 + v2

(i)

v2 – v1 = e (u1 – u2) For a perfectly elastic collision, e = 1. Hence v2 – v1 = u1 – u2 From (i) and (ii) we get

Fig. 7.5

(ii)

v1 = u2 and v2 = u1 7.3 A steel ball of mass m moving with a velocity u undergoes a perfectly elastic oblique collision with another indentical steel ball initially at rest. Show that, after the collision, they move at right angles to each other.

u2 = v12

2 v1 v 2

v22

(ii)

Since kinetic energy is also conserved in an elastic collision, we have 1 2 1 1 2 mu = mv12 mv2 2 2 2 u2 = v12 + v22

(iii)

Using (iii) in (ii), we get 2 v1 v2 = 0 v1 v2 = 0 v1 v2 cos

=0

cos

=0

= 90°

7.4 Two steel balls of the same mass m moving in opposite directions with the same speed u collide head-on. If the collision is perfectly elastic, predict the result of the collision.

m1 = m2 = m, u1 = u and u2 = – u. Let v1 and v2 be their velocities after collision. Total momentum before collision = m1u1 + m2u2 = m (u – u) = 0 Total momentum after collision = mv1 + mv2 = m (v1 + v2) From conservation of momentum, 0 = m(v1 + v2) v2 = – v1

7.6 Comprehensive Physics—JEE Advanced

Since e = 1, we have v2 – v1 = u1 – u2 = u – (– u) = 2u Putting v1 = – v2, we get v2 = u. Also v1 = – u. Thus, after the collision, the two balls move in opposite directions with equal speeds, each equal to u but their directions are reversed. 7.5 A ball of mass 2 kg moving with a velocity of 8 ms–1 collides head-on with another ball of mass of 4 kg moving with a velocity of 2 ms–1 moving in the same restitution is e = 0.5. (a) Find the velocities of the balls after the collision. (b) Calculate the loss of kinetic energy due to collision.

7.6 In Example 7.5, what is the loss of kinetic energy if the ball of mass 4 kg is moving towards the mass of mass 2 kg, their speeds being the same?

In this case u2 = – 2 ms–1. Equations (i) and (ii) become (iii) 4 = v1 + 2v2 (iv) and v2 – v1 = 5 Equations (iii) and (iv) give v1 = – 2 ms–1 and v2 = 3 ms–1 1 1 Ki = 2 (8)2 + 4 (–2)2 = 72 J 2 2 1 1 2 (–2)2 2 2 Loss of K.E = 72 – 22 = 50 J Kf =

Refer to Fig 7.1 again. (a) Given m1 = 2 kg, m2 = 4 kg, u1 = 8 ms–1, u2 = 2 ms–1 and e = 0.5 From conservation of momentum m 1u 1 + m 2u 2 = m 1v 1 + m 2v 2 2

8+4

2 = 2v1 + 4v2

12 = v1 + 2v2

(i)

Since e = 0.5, we have v2 – v1 = e (u1 – u2) = 0.5 (8 – 2) = 3 (ii) Eliminating v2 from (i) and (ii) we get v1 = 2 ms–1. Using this in (i) or (ii), we get v2 = 5 ms–1 (b) Kinetic energy before collision is Ki =

1 2 2 = 72 J

=

between blocks B and C.

1 2

(8)2 +

Fig. 7.6

4

(2)2

Kinetic energy after collision is Kf =

7.7 Two blocks B and C of masses 1 kg and 2 kg respectively are connected by a massless elastic spring of spring constant 150 Nm–1 and placed on a horizontal frictionless surface as shown in Fig. 7.6. A third block A of mass 1 kg moves with a velocity of 3 ms–1 along the line joining B and C and collides with B. If the collision is perfectly elastic and the natural length

1 1 m 1u 12 + m 2u 22 2 2

=

1 1 m 1v 12 + m 2v 22 2 2 1 2

2

(2)2 +

1 2

4

(3)2 = 22J

(5)2 = 54 J

Loss of K.E. = Ki – Kf = 72 – 54 = 18 J

Given mA = mB = 1 kg, mC = 2 kg and u = 3 ms–1. Block A will collide with block B. Since they have equal masses and the collision is perfectly elastic, A will come to rest and B will move to the right with a velocity u. Block B will compress the spring. Hence block C will accelarate and block B will retard until both B and C move with the same velocity. Let this common velocity be v. Since no external force acts, the momentum of B and C is conserved, i.e. mB u = (mB + mC) v 1

3 = (1 + 2) v

v = 1 ms–1

Conservation of Linear Momentum and Collisions 7.7

If x is the maximum compression, then from the principle of conservation of energy, 1 1 1 mAu2 = (mB + mC) v2 + kx2 2 2 2 1 1 1 (3)2 = (1 + 2) (1)2 2 2 1 + 150 x2 2 which gives x = 0.2 m = 20 cm Minimum separation between B and C = 80 cm – 20 cm = 60 cm 7.8 A block of m1= m is moving on a frictionless horizontal surface with velocity u1 = 2u towards another block of mass m2 = 3m moving on the same surface with velocity u2 = u in the same direction. A massless spring of force constant k is attached to m2 as shown in Fig. 7.7. When block m1 collides with the spring, show that the maximum compression of the spring is given u 3m by x = . 2 k

Fig. 7.7

SOLUTION When block m1 collides with spring, it begins to get compressed. As a result m2 gains speed. The compression of the spring is maximum at the instant when the relative velocity of m1 with respect to m2 is zero, i.e. when both m1 and m2 have equal velocities. Let v be the common velocity of the blocks. From conservation of momentum, m1u1 + m2 u2 = (m1 + m2)v

1 2

m

1 m2u22 2 (2u)2 +

– 1 2

1 m1 2 3m

5u 4

u 3m 2 k

7. Useful Formulae and Tips 1. A body of mass m is dropped from a height h. Due to the friction of air, it will hit the ground with a speed less than 2gh . If v is the speed with which it hits the ground, the work done by friction is 1 1 mv2 – mgh = m(v2–2gh) Wf = 2 2 If friction is absent, Wf = 0, then v = 2gh . 2. Two block A and B of masses m1 and m2 are released from the same height at the same time. Block A slides along an inclined plane of inclination and block B falls vertically downwards (Fig. 7.8) If the inclined plane is frictionless, gain in KE = loss in PE, i.e. 1 m1v21 = m1gh v1 = 2gh 2 If the air friction is neglected. 1 m2v22 = m2gh v2 = 2gh 2 Thus both block will hit the ground with the same speed independent of the mass. But the times taken to reach the ground will be different.

v

v

2h For block A, t1 =

g sin 2

2h g 3. If a block of mass m in contact with a spring compressed by a distance x is released, the block will leave the spring with a velocity v determined from For block B, t2 =

m2 v 2 = u2 –

x=

Fig. 7.8

2mu + 3mu = (m + 3m)v 5u v= 4 From the law of conservation of energy Loss in K.E. = gain in P.E of spring If x is the maximum compression, then 1 m1u12 2

3 mu2 = kx2 4

1 2 kx 2

1 (m + 3m) 2 2

=

1 kx2 2

1 1 kx2 = mv2 2 2 which gives v = constant.

k x, where k is the spring m

7.8 Comprehensive Physics—JEE Advanced

4. If a block of mass m moving with speed u comes in contact with a relaxed spring of spring constant k, its velocity v when the spring is compressed by an amount x is obtained from. 1 1 1 mu2 = mv2 + kx2 2 2 2 k x2 m

which gives v =

7. A chain has a length L and mass M. A part L/n is hanging at the edge of the table. The length of the chain lying on the table is (L – L/n). Then work done against gravity to pull the hanging part on the MgL table = 2n 2 8. If a body of mass m moving with velocity v is stopped in a distance x by a retarding force F, then 1 mv2 = Fx 2 (a) If two bodies of masses m1 and m2 moving with the same velocity are subjected to the same retarding force, the ratio of the stopping distance is x1 m1 = x2 m2 (b) If the two bodies are moving with equal kinetic energy and are stopped by the same retarding force, then x1 = x2

1/ 2

u

2

5. If two springs of spring constants k1 and k2 are stretched by the same force F, then F = k1x1 = k2 x2. Potential energy stored is 1 1 U1 = k 1x 21 = Fx1 2 2 1 1 and U2 = k 2x 22 = Fx2 2 2 which give

U1 x = 1 U2 x2

k2 k1

6. If the two springs are stretched by the same amount x, then F1 = k1x and F2 = k2x. U1 =

(c) If the two bodies are moving with equal liner momentum and are stopped by the same force, then. p2 = Fx 2m

1 1 k1x2 and U2 = k 2x 2. 2 2

U1 k = 1 U2 k2

F1 F2

and

m2 x1 = m1 x2

I Multiple Choice Questions with only One Choice Correct 1. A ball P moving with a velocity u suffers a onedimensional collision with another ball Q of the same mass but at rest. After the collision the velocity of Q is found to be three times that of P. The (a)

1 2

(b)

1 3

(c)

1 4

(d)

2 3

2.

= 45°.

the collision is 1 (a) 2

(b)

1 2 2

3 5 (d) 8 8 3. Two particles of the same mass m moving in different directions with the same speed v collide and stick together. After the collision, the speed of the composite particle is v/2. The angle between the velocities of the two particles before collision is (a) 60° (b) 90° (c) 120° (d) 150° 4. Two blocks of masses m1 = m and m2 = 3 m are connected by a spring of force constant k and placed on a horizontal frictionless surface as shown in Fig. 7.9. The spring is stretched by an amount x and released. The system executes simple harmonic motion. The relative velocity of the blocks when the spring is at its natural length is (c)

Conservation of Linear Momentum and Collisions 7.9

(a) x (c)

x 2

3k 2m

(b) 2 x

k m

k 3m

(d) 2 x

k 3m

Fig. 7.9

5. A block P moves with an initial velocity of 4 ms–1 towards a block Q of the same mass at rest at a distance of 2 m on a rough horizontal surface. The cois 0.2. An elastic one-dimensional collision occurs between the two blocks. If g = 10 ms–2, the separation between the blocks when both have come to rest is (a) 1 m (b) 2 m (c) 3 m (d) 4 m 6. A perfectly elastic oblique collision occurs between a ball A moving along the x-axis and a ball B at rest and of the same mass as ball A. After the collision, ball A moves at an angle of 30° with the x-direction and ball B at an angle with the x-axis. The value of is (a) 15° (b) 30° (c) 45° (d) 60° 7. A particle A moving with momentum p suffers a one-dimensional collision with a particle B of the same mass but at rest. During the collision B imparts an impulse I to A tion between A and B is 2I 2I (b) –1 (a) p p 2I I +1 (d) 2 1 p p 8. A man of mass m stands on one end of a wooden plank of length L and mass M kept initially at rest on a horizontal frictional surface. If the man walks from one end of the plank to the other end at a constant speed, the resulting displacement of the plank is mL ML (b) (a) M m (c)

(c)

mL ( M m)

(d)

mL ( M m)

9. A body P of mass m moving with kinetic energy k and momentum p undergoes a one-dimensional

elastic collision with a body of mass 2 m at rest. After the collision, the ratio of the kinetic energy of P to that of Q is 1 1 (b) (a) 8 4 1 8 (c) (d) 9 9 10. A shell of mass m initially at rest explodes into three fragments of masses in the ratio 2 : 2 : 1. mutually perpendicular directions with speed v. The speed of the third (lighter) fragment will be 2v

(a) v

(b)

(c) 2 2 v

(d) 3 2 v

11. A shell is fired with a speed of 100 ms –1 at an angle of 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms–1. The speed of the heavier fragment immediately after the explosion is (b) 150 ms–1 (a) 125 ms–1 (c) 175 ms–1 (d) 200 ms–1 12. A body of mass m moving with a certain speed suffers an inelastic collision with a body of mass M system to the initial kinetic energy is M m (b) m M (a) m M (c)

m

M

(d)

m

M

m M 13. A neutron moving at a speed v undergoes a head-on elastic collision with a nucleus of mass number A at rest. The ratio of the kinetic energies of the neutron after and before collision is (a) (c)

A 1 A 1

2

A

2

(b) (d)

A 1 A 1

2

A

2

A 1 A 1 14. A radioactive nucleus of mass number A, initially at rest, emits an -particle with a speed v. What will be the recoil speed of the daughter nucleus? 2v 2v (a) (b) A 4 A 4 4v 4v (c) (d) A 4 A 4

7.10 Comprehensive Physics—JEE Advanced

15. A ball is dropped from a height of 10 m. It is embedded 1 m in sand. In this process (a) only momentum is conserved (b) only kinetic energy is conserved (c) both momentum and kinetic energy are conserved (d) neither momentum nor kinetic energy is conserved. 16. n small balls, each of mass m, impinge elastically each second on a surface with velocity u. The force experienced by the surface will be (a) mnu (b) 2 mnu 1 mnu (c) 4 mnu (d) 2 17. A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of 16 2 (b) (a) 25 5 3 9 (d) 5 2 18. An isolated particle of mass m is moving in a horizontal plane (x – y), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragment is at y = + 15 cm. The larger fragment at this instant is at (a) y = – 5 cm (b) y = + 20 cm (c) y = + 5 cm (d) y = – 20 cm IIT, 1997 19. A body P strikes another body Q of mass that is p times that of body P and moving with a velocity 1 that is of the velocity of body P. If body P comes q (c)

p (a) p (c)

q q

p q p q 1

(c)

h(1 e2 ) 2(1 e2 )

(d)

h(1 e2 ) 2(1 e2 )

23. A bullet of mass m velocity v on a wooden block of mass M suspended from a support and gets embedded in it. The kinetic energy of the bullet + block system is (a)

1 mv2 2

(b)

1 (M + m)v2 2

(c)

Mmv 2 2 M m

(d)

m2 v2 2 M m

24. A body of mass m moving with a speed v suffers an inelastic collision and sticks with another body of mass M = 2m at rest. The speed of the composite body will be v (a) 3v (b) 3 (c)

2v 3

(d)

3v 2

of the system to the initial kinetic energy is 1 2 (a) (b) 3 3

p q p q 1

20. Two equal spheres A and B lie on a smoot horizontal circular groove at opposite ends of a diameter. Sphere A is projected along the groove and at the end of time T impinges on sphere B. If e is the occur after a time equal to (a) T (b) eT 2T (d) 2 eT (c) e

K 1 M KM (d) M M 4 22. A particle falls from a height h plate and rebounds. If e tution, the total distance travelled by the particle before it stops rebounding is h(1 e2 ) h(1 e2 ) (b) (a) (1 e2 ) 1 e2 (c)

25.

p q (b) q p 1 (d)

21. A nucleus of mass M amu emits an -particles with a energy K MeV. The total energy of disintegration (in MeV) is KM (a) K (b) M 4

3 2 26. A body of mass 5 kg is moving along the x-axis with a velocity 2ms–1. Another body of mass 10 kg is moving along the y-axis with a velocity 3 ms–1. They collide at the origin and stick together. The (c) 3

(a) IIT, 1997

(c)

(d)

3 ms–1 4 ms–1 3

(b)

3 1 ms–1

(d) none

Conservation of Linear Momentum and Collisions 7.11

27. A block of wood of mass M is suspended by means of a thread. A bullet of mass m into the block with a velocity v. As a result of the impact, the bullet is embedded in the block. The block will rise to vertical height given by mv 1 (a) 2g M m

2

mv 1 (b) 2g M m

2

mv 2 mv 2 1 1 (d) 2g M m 2g M m 28. A moving particle of mass m makes a head-on collision with a particle of mass 2m initially at rest. If the collision is perfectly elastic, the percentage loss of energy of the colliding particle is (a) 50% (b) 66.7% (c) 88.9% (d) 100% (c)

29. A body of mass m moving with a velocity v in the x-direction collides with a body of mass M moving with a velocity V in the y-direction. They stick together during collision. Then (a) the magnitude of the momentum of the composite body is

mv

2

MV

2

(b) the composite body moves in a direction MV making a angle = tan–1 with the mv x-axis. (c) the loss of kinetic energy as a result of col1 Mm (V2 + v2) lision is 2 M m (d) all the above choices are correct. 30. A body falls from a height h on a horizontal surface and rebounds. Then it falls again and again rebounds 1 , the 3 total distance covered by the body before it comes to rest is h 5h (b) (a) 4 4 (c) 2h (d) 3h 31. In Q. 30 above, the total time taken by the body to come to rest is (a) (c) 3

2h g 2h g

(b) 2

2h g

(d) 4

2h g

32. A body of mass m moving with a velocity v in the x-direction collides and sticks with another body of

mass M moving with a velocity V in the y-direction. The magnitude of the momentum of the composite body is (a) (mv MV )

(b) (m + M) (v + V)

(c) [(mv)2 + (MV)2]1/2

(d) (Mv + mV)

33. In Q. 32 above, the angle subtended by the velocity vector of the composite body with the x-axis is given by mv MV (b) = tan–1 MV (a) = tan–1 mv mV Mv (c) = tan–1 (d) = tan–1 Mv mV 34. A body P of mass m1 moving with a certain velocity collides head-on with a stationary body Q of mass m2. It the collision is elastic, the fraction of kinetic energy transferred from body P to body Q is (a) (c)

2(m1m2 ) (m1

m2 )

2

2m12 (m1

m2 )2

(b) (d)

4(m1m2 ) (m1

m2 )2

2m22 (m1

m2 )2

35. Two particles, each of mass m, moving along different directions with a velocity u making the same angle with the x-axis collide and stick together. The composite particle moves along the x-axis with a velocity u/2. The angle between their directions of motion before collision is (a) 60° (b) 90° (c) 120° (d) 150° 36. A bullet of mass m moving with a horizontal velocity u strikes a stationary wooden block of mass M suspended by a string of length L = 50 cm. The bullet emerges out of the block with speed u/4. If M = 6 m, the minimum value of u so that the block can complete the vertical circle is (take g = 10 ms–2) (b) 20 ms–1 (a) 10 ms–1 –1 (c) 30 ms (d) 40 ms–1 37. A compound pendulum consists of a uniform rod of length L of negligible mass. A body of mass m1 = m m2 = 2 m middle of the rod as shown in Fig. 7.10 The horizontal velocity v that must be given to mass m1 to rotate the pendnlum to the horizontal position OC is

Fig. 7.10

7.12 Comprehensive Physics—JEE Advanced

2 gL 3

(a) 2 (c)

(a) 4 (c) 2

(b) 2 gL

2gL

(d)

IIT, 2009 40. A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The velocity V of the bullet is (see Fig.7.13) (a) 250 m/s (b) 250 m/s (c) 400 m/s (d) 500 m/s

gL

38. A ball is thrown will a velocity v1 towards a vertical wall at an angle with the wall. It rebounds with a velocity v2 making an angle with the wall as between the ball and the wall is e, then v2 is given by (a) v2 = v1(cos + e sin ) (b) v2 = v1(sin + e cos ) (c) v2 = v1 sin 2

e2 cos 2

(d) v2 = v1 cos 2

e2 sin 2

(b) 3 (d) 1

Fig. 7.13

IIT, 2011 41. A ball is thrown from a point O with a velocity u = 20 ms–1 at an angle = 30° with the horizontal. It hits a vertical wall which is at a distance x from O as shown in Fig. 7.14. After rebounding from the wall, the ball returns to O without retracing its path. e = 0.5, If g = 10 ms–2 the value of x is (a) 9.6 m (b) 10.3 m (c) 11.5 m (d) 12.8 m

Fig. 7.11

39. Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v, respectively, as shown in Fig. 7.12. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, will these two particles again reach the point A? v

v

Fig. 7.12

Fig 7.14

ANSWERS 1. 7. 13. 19. 25. 31. 37.

(a) (b) (a) (d) (a) (b) (a)

2. 8. 14. 20. 26. 32. 38.

(c) (c) (c) (c) (c) (c) (d)

3. 9. 15. 21. 27. 33. 39.

(c) (a) (a) (b) (a) (a) (c)

4. 10. 16. 22. 28. 34. 40.

(d) (c) (b) (a) (c) (b) (d)

5. 11. 17. 23. 29. 35. 41.

(b) (a) (b) (d) (c) (c) (c)

6. 12. 18. 24. 30. 36.

(d) (a) (a) (b) (b) (d)

Conservation of Linear Momentum and Collisions 7.13

SOLUTIONS 1. Let v1 and v2 be the velocities of P and Q after the collision. v2 v1 = v2 v1 (1) u 0 u From conservation of momentum, we have mu + 0 = mv1 + mv2 (2) u = v1 + v2 From Eqs. (1) and (2), we get u u (1 – e) and v2 = (1 + e) 2 2 v2 1 e = v1 1 e 1 e 1 Given v2 = 3v1. Hence 3 = e= 1 e 2 2. Let u be the speed of the ball before the collision. After the collision, its speed will be v1 =

u

v = =

2

2 u2 2

2

eu

5 u 8

e

1 2

1 1 mu 2 mv 2 2 2 Fraction of K.E. lost = 1 mu 2 2 v2 5 3 =1– 2 =1– = 8 8 u 3. Let be the angle between the velocities of the two particles before collision. If p1 and p2 are the momenta of the particles before collision and p is the momentum of the composite particle, then the conservation of momentum gives p2 = p 21 + p 22 + 2 p1 p2 cos 2m

v 2

2

= (mv)2 + (mv)2 + 2 (mv) (mv) cos

1 or = 120° 2 4. If v is the relative velocity of the two blocks when the spring is at its natural length, then from the conservation of energy, we have 1 1 v2 = kx2 (1) 2 2 where is the reduced mass of the system and is given by cos

=–

m1 m2 m = (m1 m2 ) (m

3m 3m = 3 m) 4

Substituting in Eq. (1) we get 3 k mv2 v = 2x 4 3m 5. Frictional force f = mg. The retardation due to friction is f = g = 0.2 10 = 2 ms–2 a= m Since the blocks have the same mass and the collision is elastic, after the collision, block P will come to rest at the position previously occupied by block Q and Q will begin to move with the velocity at which P strikes Q which is given by kx2 =

v2 = 2as = 2

2

v=

2=8

8 ms–1

Moving with velocity of 8 ms–1, the block Q will come to rest after travelling a distance x given by v2 8 = =2m 2a 2 2 Hence the correct choice is (b). 6. In an oblique elastic collision between two body of the same mass, they move at right angles to each other after the collision. Hence the correct choice is (d). 7. Let p1 and p2 be the linear momenta of A and B after the collision. Now, impulse = change in momentum. For particle B : I = p1 For particle A : I = p – p2 p2 = p – I x=

2 u2 = 8

=

e=

v1

v2 u

=

=

m v1 p1

m v2 mu p2

p where m is the mass of each particle and u is the velocity A before collision and v1 and v2 are the velocities of A and B after the collision. Hence p p I I p I 2I e= 1 = = –1 p p p So the correct choice is (b). 8. Total initial momentum of the man-plank system is zero. If he walks with a speed v on the plank, as a result, the plank moves with a speed, say, v in the system = mv – (M + m) v . From conservation of momentum,

7.14 Comprehensive Physics—JEE Advanced

0 = mv – (M + m) v v = (M v

V = 2 2 v.

m m)

Since the distance moved in proportional to speed (since there is no acceleration), the displacement L of the plank is given by choice (c). 9. Total momentum before collision is p = mu + 0 1 = mu and kinetic energy is K = mu2. If v1 and 2 v2 are the velocities of P and Q after the collision, then, from momentum conservation, mu = mv1 + (2 m) v2 p = p1 + p2

(1)

From conservation of kinetic energy, K = K1 + K2 where K1 = Thus

Fig. 7.15

m p1 = v1, where v1 = 200 ms–1. If v2 is the ve3 locity of the heavier fragment, its momentum is 2 mv2 /3. Conservation of momentum along x and y-directions gives p = p2 cos

1 1 mv 12 and K2 = (2 m) v 22 2 2 1 1 mu2 = mv12 + mv 22 2 2

p2 p2 = 1 2m 2m

mv = (3)

From Eq. (1), p2 = p – p1. Using this in Eq. (3) and solving we get p 4p . Hence p2 = p1 = – 3 3

K2 =

and

p2 1 p2 mv 12 = 1 = 2m 18 m 2 p22

2

1 4p mv 22 = = 4m 2 9m

K1 1 = , which is choice (a). K2 8 10. Mass of each heavier fragment = 2 m/5 and of lighter fragment = m/5. Momentum of each heavier fragment is p = 2 mv perpendicular directions, their resultant momentum =

p2

p2 =

2 m v2 cos 3 3v = 2v2 cos

p22 4m

2p2 = 2p 21 + p 22

Now K1 =

11. Velocity of the shell at the highest point is v = 100 sin 30° = 50 ms–1 parallel to x-direction. Its momentum is p = mv (Fig. 7.15).

2 p . If V is the speed of the lighter

fragment, from the conservation of momentum, we have 2 mv mV = = 2 2 p 5 5

and

(1)

p1 = p2 sin 2 m v2 m v1 = sin 3 3 v1 = 2v2 sin

From (1) and (2), we get v2 = =

(2) 1 (v12 + 9v2)1/2. 2

1 [(200)2 + 9(50)2]1/2 2

= 125 ms–1 12. Initial momentum of the system = mv, since body of mass M is at rest. After the inelastic collision, the bodies stick together and the mass of the composite body is (m + M). If V is the speed of the composite body, its momentum will be (m + M)V. From the principle of conservation of momentum, we have mv = (m + M)V or V = Initial KE =

m v m M 1 1 mv 2. Final KE = (m + M)V2. 2 2

Conservation of Linear Momentum and Collisions 7.15

Therefore, 2

Final KE m+M V m = = 2 Initial KE m m M v Hence the correct choice is (a). 13. Mass of neutron (m1) = 1 unit. Mass of nucleus (m2) = A units. Refer to page 7.2. Here u1 = u and u2 = 0. Therefore the velocity of the neutron after the collision is v1 =

m1 m1

m2 u= m2

1 A u 1 A

KE of neutron after collision = =

1 A 1 A

1 2

KE of neutron before collision = =

1 2

u2 =

1

1 m1v 12 2 2

u2 1 m u2 2

1 2 u. 2

1 A 2 , which is choice (a). 1 A 14. The total number of nucleus (i.e. protons + neutrons) in a nucleus is called its mass number. An -particle is a helium nucleus having 2 protons and 2 neutrons. So the mass number of an -particle = 4. When a nucleus of mass number A emits an -particle, the mass number of the daughter nucleus reduces to (A – 4). If V is the recoil speed of the daughter nucleus, we have, from the law of conservation of momentum, (A – 4)V – 4v = 0 Their ratio is

4v A 4 Hence the correct choice is (c). V=

or

15. The collision is inelastic because the two bodies stick to each other after collision. In an inelastic collision, only the momentum is conserved; there being a loss in kinetic energy. Hence the correct choice is (a). 16. Since the collision (impact) is elastic, the ball rebounds with the same speed. Therefore, the change in momentum of each ball = 2 mu. The change in momentum per second due to n balls = 2 mnu. But the change in momentum per second is the force. Hence the correct choice is (b). 17. A ball dropped from a height h1 on reaching the planet’s surface will have a velocity given by v1 =

2 gh1

Let v2 be the velocity with which the ball bounces. It will attain a height h2 given by v22 = v2 = v1

2 gh2 h2 h1

18 . = 0.6 5

or

1–

v2 = 1 – 0.6 v1

or

v1

v2 v1

= 0.4 =

2 5

Hence the correct choice is (b). 18. Let m and M be the masses of the lighter and heaviers fragments respectively. Since the particle is moving along the x-axis, the y-component of momentum will be zero immediately after and before explosion, i.e. mvy + MVy = 0 where vy and Vy are the velocities of the lighter and heavier fragments respectively immediately after explosion. Thus m m/ 4 vy = vy Vy = M 3m / 4 = 1 vy 3 Since y = + 15 cm, the direction of vy is along the positive y-axis and that of Vy will be along the negative y-axis. An instant later (say, at time t), it is given that y = 15 cm = vy t 1 1 Y = Vy t = vy t = y 3 3 1 15 cm = – 5 cm = 3 19. Given mQ = p mP and vQ = vP /q. From the principle of conservation of momentum, we have (since body P comes to rest after collision) mP vP + mQ vQ = mQ v where v is the velocity of body Q after collision. Thus v mP vP + p mP P = p mPv. q v p q which gives = (i) vP pq e=

v vp

v vQ

vp

vp q

7.16 Comprehensive Physics—JEE Advanced

which gives

v e = (q – 1) vP q

(ii)

22. The total distance travelled is S = h + e2h + 2e4h + 2e6h + = h + 2h(e2 + e4 + e6 +

p q which is Equating (i) and (ii), we get e = p q 1 choice (d). 20. Refer to Fig. 7.16. If sphere A is projected with velocity v, the time taken by it to strike B is equal r = T or r = Tv. Now, to v titution is given by v vA e= B v

-

= h + 2h

= 1

e 1 e2

2e 2

=

h(1 e2 )

1 e2 1 e2 23. Initial momentum (p) = momentum of bullet + momentum of block = mv + 0 = mv. From the momentum of bullet + block system of mass (M + m) = Initial momentum p. Now

r A

)

KE =

B

2

p2 M m

m2 v2 2 M m

Hence the correct choice is (d). 24. The correct choice is (b). In an inelastic collision, Fig. 7.16

where vA and vB are the velocities of A and B after the collision. Thus, vB – vA = ev. The spheres travel with this relative velocity. It is clear that one will overtake the other after travelling a distance = 2 r. 2 r 2 r 2T v 2T Time taken = vB v A ev ev e (since r = Tv). Hence the correct choice is (c). 21. Let V be the velocity of the nucleus and v that of the -particle after disintegration, then from the principle of conservation of momentum, we have (since the mass of an -particle is 4 amu) 4v (i) (M – 4) V = 4v or V = M 4 Total KE = Now

1 2

4

Total KE =

1 1 (M – 4) V 2 + 2 2

4

v2 (ii)

v2 = K. Using (i) in (ii), we have 1 (M – 4) 2

16v 2 M

4

2

=

8v 2 + 2v2 M 4

=

4K MK +K= ( M 4 M 4

Hence the correct choice is (b).

+ 2v2

composite body, use the principle of conservation of linear momentum. 25. The correct choice is (a). Find kinetic energies before and after the collision. 26. Momentum of 5 kg mass (p1) = 5 2 = 10 kg ms–1 along the x-axis. Momentum of 10 kg mass (p2) = 10 3 kg ms–1 along the y-axis. These two momenta are perpendicular to each other. Therefore, the resultant initial momentum is p=

p12

p22

10

2

10 3

2

= 20 kg ms–1

If v ms–1 is the velocity of the combined mass, v = 15 v kg ms–1. Now, from the principle of conservation 4 of momentum, we have 15 v = 20 or v = ms–1, 3 which is choice (c). 27. Let V be the velocity of the block with the bullet embedded in it at the time of impact. Then from the principle of conservation of momentum, we have mv = (M + m) V mv (i) V= M m If the block, with the bullet embedded in it, rises to a vertical height h, then from the principle of conservation of energy, we have 1 (M + m) V 2 = (M + m) gh 2 or

2v2 = K)

or

V=

2gh

(ii)

Conservation of Linear Momentum and Collisions 7.17

Using (ii) in (i), we get mv 2gh = M m

31. Total time =

2h g

2e

2h g

=

2h g

2

2h e g

=

2h g

2

2h e g 1 e

=

2h g

h is given correctly by choice (a). 4mM

28. Percentage loss of energy =

=

4m

M

2m

m

100 =

2m m 2 Hence the correct choice is (c).

2

100

800 = 88.9% 9

29. Refer to Fig. 7.17. Here p = mv and P = MV. The resultant of p and P is pr =

p2

P2

mv

2

MV

1 3 1 1 3

2e 2

1

2

2h g

e2 2h 1 e g 1 e

2h , g

which is choice (b). 32. Refer to Fig. 7.18.

2

y

pr

P

P

q

x

p

Fig. 7.17

which is choice (a). The angle which the resultant momentum pr subtends with the x-axis is given by P MV tan = , which is choice (b). p mv Loss of KE =

1 mv 2 2

1 m 2 v 2 M 2V 2 M m 2

1 MV 2 2

1 Mm = (V2 + v2), which is choice (c). 2 M m 30. Total distance = h + 2 e2 h + 2 e4 h + ... = h + 2 e2 h (1 + e2 + ...) 2e2 h

=h +

h

1 e2 1

=h 1 which is choice (b).

1 3 1 3

1 e2 1 e2

2

2

5h , 4

Fig. 7.18

Let v be the velocity of the composite body at an angle with the x-axis (Fig. 7.11). Equating the x and y-axes, we have mv= (m + M) v cos = p cos (i) and

MV = (m + M) v sin

= p sin

(ii)

where p = (m + M) v is the momentum of the composite body. Find p by squaring and adding (i) and (ii). The correct choice is (c). 33. Divide (ii) by (i). The correct choice is (a). 34. Let u be the velocity of P before collision and v1 and v2 the velocities of P and Q after collision. From conservation of momentum, we have m1u + 0 = m1v1 + m2v2 which gives (1) m1(u – v1) = m2v2 From the conservation of kinetic energy we have 1 1 1 m1u12 = m1v12 + m2v22 2 2 2 2 2 m1(u – v1) = m2v22 m1(u – v1)(u + v1) = m2v22

(2)

7.18 Comprehensive Physics—JEE Advanced

Dividing Eq. (2) by Eq. (1) we get (3) u + v1 = v2 From Eqs. (1) and (3) we get m1 m2 u (4) v1 = m1 m2 1 Initial K.E. of P(K) = m1u2, 2 1 m 1v 12 P= 2 1 1 Decrease in K.E. of P( K) = m1u2 – m1v12. 2 2 Fractional decreases in K.E. of P is 1 1 m1u 2 m1v12 K 2 2 = 1 K m u2 2 1 =

u2

v12 u2 m1 m1

=1 – =

=1– m2 m2

2

u2 [use Eq. (4)]

= 2mv

u = mu ( 2

=2 m 2 cos which gives cos

2

2 which is choice (c)

2 =

3mu = 4M

5gL

4M 5 gL 3m 4 = 6 5 10 0.5 = 40 ms–1. 3

u=

2

37. Since mass m2 is at a distance L/2 from the axis of rotation, it speed will be v/2 (half that of mass m1). From the principle of conservation of energy we have 1 m 1v 2 + 2 1 mv2 + 2

1 v 2 m2 = m1gOB + m2gOA 2 2 1 v2 L 2m = mgL + 2mg 4 2 2

2 gL , which is choice (a). 3 38. The component of velocity parallel to the wall remains unchanged and the component of velocity perpendicular to the wall reduces by e times its value before collision. Thus we have v2 cos = v1 cos which gives v = 2

v = u/2)

=1 1 2

5gL

Hence the correct choice is (d).

2

4m1m2

+ mucos

V=

v12

(m1 m2 )2 Thus the correct choice is (b). 35. Refer to Fig. 7.19. From conservation of x-component of momentum, we have mucos

36. Let V be the speed of the block after the bullet emerges out of it. From conservation of momentum we have u mu = MV + m 4 3mu which gives V = 4M Now refer to Q. 27 of Section III of Chapter 4. The minimum speed the block must have to complete the vertical circle is

60°

= 120°,

and v2 sin

= ev1 sin

Squaring and adding we get v2 = v1 (cos2

e2 sin 2 )

Hence the correct choice is (d). 39. Refer to Fig. 7.20. First collision will occur when angle r 2 = v 2v which gives Fig. 7.19

r

=120°.

B, m2 will move back with a speed v and will make a second collision with m1

Conservation of Linear Momentum and Collisions 7.19

Horizontal velocity after rebounding from the wall is

at C. After the second collision at C, m1 will move back with a speed v and meet m2 at A. If the third collision at A is neglected, the particles will make two collisions before they reach A. Hence the correct choice is (c). v

u x = e ux = e u cos

Horizontal displacement from O to A or from A to O = x. Time taken to go from O to A is x (i) u cos Time taken to return from A to O is x t2 = (ii) eu cos Since the horizontal and vertical motions are independent of each other, the net vertical displacement Sy = 0 since the ball returns to O. If t is the total time taken by the ball to go from O to A and return to O, then from 1 2 gt we have S y = u yt – 2 1 2 gt 0 = (u sin ) t – 2 2u sin t= (iii) g Now t = t1 + t2. Using (i) and (ii) in (iii), we have t1 =

v

v

v

v

v

Fig. 7.20

40.

2h 2 5 = = 1s g 10 Horizontal range (R) = horizontal velocity t f) =

time

Horizontal velocities of the bullet and of the ball after the collision respectively are 100 = 100 ms–1 (v)bullet = 1 20 (v)ball = = 20 ms–1 1 From conservation of momentum,

2u sin g

= x=

(m)bullet

V = (m)bullet

0.01 V = 0.01

.

(v)bullet + (m)ball (v)ball 100 + 0.2

20

=

V = 500 ms–1 41. Refer to Fig 7.14 on page 7.12. Horizontal velocity before hitting the wall is ux = u cos

=

x u cos

x eu cos

u 2 sin(2 ) 1 1 e 20 2 sin 60 1 1 0.5 20 3

11.5 m, which is choice (c)

II Multiple Choice Questions with one or More Choices Correct 1. In an inelastic collision of two bodies, which of the following do not change after the collision? (a) total kinetic energy (b) total linear momentum (c) total energy (d) total angular momentum 2. Which of the following statements are true?

(a) In a elastic collision of two bodies, the momentum and energy of each body is conserved. (b) The total energy of a system is always conserved irrespective of whether external forces act on the system. (c) The work done by a force in nature on a body, over a closed loop, is not always zero.

7.20 Comprehensive Physics—JEE Advanced

(d) In an inelastic collision of two bodies, the kinetic energy of the system. 3. A molecule in a gas container hits the wall with speed v at an angle with the normal and rebounds with the same speed as shown in Fig. 7.21. Which of the following statements are true? (a) The momentum of the system is conserved in the collision. (b) The momentum of the molecule before collision with the wall is equal to the momentum of the molecule after collision. (c) The collision is elastic. (d) The collision is inelastic. Wall

v

v

Fig. 7.21

4. A particle A suffers an oblique elastic collision with a particle B that is at rest initially. If their masses are the same, then, after the collision (a) they will move in the opposite directions (b) A continues move in the original direction while B remains at rest (c) they will move in the mutually perpendicular directions (d) A comes to rest and B starts moving in the direction of the original motion of A 5. Choose the correct statements from the following: (a) The general form of Newton’s second law of motion is Fext = ma. (b) A body can have energy and yet no momentum. (c) A body having momentum must necessarily have kinetic energy. (d) The relative velocity of two bodies in a head– on collision remains unchanged in magnitude and direction 6. A ball of mass m moving horizontally at a speed v collides with the bob of a simple pendulum at rest. The mass of the bob is also m. (a) If the balls stick together, the height to which the two balls rise after the collision is

v2 . 8g

(b) If the balls stick together, the kinetic energy of the system immediately after the collision becomes half of that before collision. (c) If the collision is perfectly elastic, the bob of v2 . the pendulum will rise to a height of 2g (d) If the collision is perfectly elastic, the kinetic energy of the system immediately after the collision is equal to that before collision. 7. A body of mass 1 kg, initially at rest explodes into three fragments of masses in the ratio of 1 : 1 : 3. lar to each other with a speed of 30 ms–1, one along the + x direction and the other along the + y direction. Then (a) the speed of the heavier fragment will be 10 2 ms–1. (b) the speed of the heavier fragment will be 15 2 ms–1. (c) the direction of motion of the heavier fragment will be at angle of 135° with the + y direction (d) the direction of motion of the heavier fragment will be at an angle of 45° with the + x direction. 8. Two bodies A and B of masses m and 2m respecnected by a spring of spring constant k. A third body C of mass m moves with a velocity v0 along the line joining A and B and collides elastically with A as shown in Fig. 7.22. At a certain instant of time t0 after the collision, it is found that A and B have the same velocity v and at this instant, the compression of the spring is x0. Then (a) v =

v0 2

(b) v =

v0 3

(c) k =

2 mv02 3 x02

(d) k =

3 mv02 2 x02

v0

Fig. 7.22

9. A block of mass M attached to a light spring of force constant k rests on a horizontal frictionless surface as shown in Fig. 7.23. A bullet of mass m moving with a horizontal velocity v strikes the block and gets embedded in it. The velocity of the block with the bullet in it just after impact is V. If the impact compresses the spring by an amount x, then

Conservation of Linear Momentum and Collisions 7.21

(a) v = [k(M + m)]1/2 (b) v =

x m

1/ 2

2k M m

x

(c) V = [2k (M + m)]1/2 (d) V =

1/ 2

k M

x m

Fig. 7.25

x

m

Fig. 7.23

10. A body P of mass 1 kg moving with a velocity of 3 ms–1 along the + x direction collides head-on with a body Q of mass 2 kg at rest. The collision is elastic. After the collision (a) P moves along the +x direction with a velocity of 1 ms–1. (b) P moves along the –x direction with a velocity of 1 ms–1. (c) Q moves along the +x direction with a velocity of 2 ms–1. (d) Q moves along the –x direction with a velocity of 2 ms–1. 11. A block of mass M with a massless spring of force constant k is resting on a horizontal frictionless surface (Fig. 7.24). A block of mass m projected horizontally with a speed u collides and sticks to the spring at the point of maximum compression of the spring.

Fig. 7.24

If v is the velocity of the system after mass m sticks to the spring and n is the fraction of the initial kinetic energy of mass m that is stored in the spring, then (a)

v u

(c) n =

M (M

m) M

(M

m)

(b)

v u

(d) n =

The track is frictionless and the collision is elastic. If vA and vB are the velocities of A and B after the collision, then (a) vA = 0, vB = 1.4 ms–1 (b) vA = 1 ms–1, vB = 1 ms–1 (c) vA = – 1.4 ms–1, vB = 1.4 ms–1 (d) vA = – 1 ms–1, vB = 1 ms–1 13. Two blocks, each of mass m, moving in opposite directions with the same speed u, on a horizontal frictionless surface, collide with each other, stick together and come to rest. Then (a) work done by external force on the system is zero. (b) work done by the external force on the system is mu2. (c) work done by the internal force on the system is zero. (d) work done by the internal force on the system is – mu2. 14. A ball P of mass m1 moving with velocity u collides head-on with a stationary ball Q of mass m2. The collision is perfectly elastic. After the collision (a) if m1 = 2 m2, balls P and Q move in the same direction with speeds in the ratio of 1 : 4. (b) if m1 = 3 m2, balls P and Q move in the same direction with speeds in the ratio of 1 : 3. (c) if m2 = 2 m1, balls P and Q move in opposite directions with speeds in the ratio of 1 : 2. (d) if m2 = 3 m1, balls P and Q move in opposite directions with equal speeds. 15. A ball P of mass m1 moving with a velocity u collides obliquely with a stationary ball Q of mass m2. The collision is perfectly elastic. After the coloriginal direction of ball P as shown in Fig. 7.26.

m (M

m) m

(M

m)

12. A small ball A slides down the quadrant of a circle as shown in Fig. 7.25 and hits the ball B of equal mass which is initially at rest.

Fig. 7.26

7.22 Comprehensive Physics—JEE Advanced

(a) if m1 = m2, balls P and Q angles to each other with the same speed. (b) if m1 = m2, balls P and Q 60° with each other with the same speed. (c) if m1 = 2 m2, balls P and Q angles to each other with speeds in the ratio of 1 : 2. (d) if m1 = 2 m2, balls P and Q angle of 60° with each other with speeds in the ratio of 1 : 2. 16. Which of the following statements is/are incorrect in the case of an elastic collision between two bodies? (a) If two balls of the same mass moving with the same speed in opposite directions, collide head-on, then after the collision they move in opposite direction with the speed each ball had before collision. (b) If a body suffers a head-on collision with another body of the same mass but at rest, then dead and the second body moves with the e = 1 for a perfectly elastic collision. (d) If a body P collides head-on with a body Q of the same mass but at rest, then the percentage fraction of kinetic energy transferred from P to Q is 50%. 17. Which of the following statements is/are true in the case of an inelastic collision between two bodies? (a) The vector sum of the linear momenta of the two bodies before collision is equal to the vector sum of the linear momenta after the collision in the case of both one-dimensional and two-dimensional collisions. (b) The total energy of the system is conserved. (c) The two bodies stick together after inelastic collision. (d) If a body collides with another body of the same mass but at rest and two bodies stick together, the ratio of the total kinetic energy before and after collision is 2 : 1. 18. A body P of mass 1 kg moving a velocity of 15 ms–1 collides head-on with a stationary body Q 1/3, then (a) velocity of P after collision will be 5 ms–1. (b) velocity of Q after collision will be 10 ms–1. (c) the loss of kinetic energy of the system is 50 J. (d) the percentage fractional decrease in the kinetic energy of body P is 50%.

19. A body P of mass m1 moving with a velocity u1 = (a i + b j ) collides with a stationary body Q of m2. After the collision body P is found to move with velocity v1 = (c i + d j ) where a, b, c and d are constants. Then (a) Impulse received by P is m1[(a – c) i + (b – d) j ] (b) Impulse received by P is m1[(c – a) i + (d – b) j ] (c) Impulse imparted to Q is m1[(a – c) i + (b – d) j ] (d) Impulse imparted to Q is m2[(a – c) i + (b – d) j ] 20. A billiards ball C of mass m moving with velocity u collides two identical balls A and B in contact and at rest. After the collision, ball C is stopped dead and balls A and B move along directions shown in Fig. 7.27 with the same speed v. Then

Fig. 7.27

(a) v = (b) v =

u 3 u

2 1 (c) Loss of kinetic energy = mu2 3 1 (d) Loss of kinetic energy = mu2 6 21. A U-238 nucleus emits an alpla particle and changes into Th-234. In this process (a) the momentum of Th-234 is equal and opposite to that of the alpha particle. (b) the magnitude of the momentum of Th-234 is greater than that of the alpha particle. (c) the kinetic energy of the alpha particle is equal to that of Th-234. (d) the kinetic energy of the alpha particle is greater than that of Th-234.

Conservation of Linear Momentum and Collisions 7.23

22. collision. In this case (a) the momentum of the ball just after the collision is the same as that just before collision, (b) the mechanicial energy of the ball remains the same in the collision. (c) the total momentum of the ball and the earth is conserved. (d) the total energy of the ball and the earth is conserved. IIT, 1986 23. Two balls having linear momenta p1

p i and

is no external force acting on the balls. If p 1 and p 2 are option (s) is/are not allowed for any non-zero value of p, a1, a2, b1, b2, c1 and c2 (a) p 1 = a1 i

b1 j

(b) p 1 = c1 k ; p 1

c1 k ; p 2

a2 i

b2 j

c2 k ;

(c) p 1 = a1 i

b1 j

c1 k ; p 1

(d) p 1 = a1 i

b1 j ; p 2

a2 i

a2 i

b2 i

c1 k

b1 i IIT, 2008

p2 = – p i undergo a collision in free space. There SOLUTIONS

1. The correct choices are (b) and (c). 2. Choice (a) is false. In an elastic collision of two bodies, the speeds of the bodies change due to collision. Therefore, the momentum and energy of each body will change but the total momentum and total energy of the system of two bodies are conserved. Choice (b) is also false. The total energy of an isolated system is conserved. If external forces act on the system, the total momentum and energy will change. Choice (c) is true. For a non-conservative force such as friction, the work done over a closed loop is not zero. Choice (d) is also true. In an inelastic collision, the two bodies stick together after colliding. This results in heat or sound energy which is dissipated at the expense of kinetic energy. Hence choices (c) and (d) are correct. 3. The system consists of the molecule and the wall. Let us assume that initially the wall is stationary so that its momentum and kinetic energy are both zero before the collision. Therefore, the total momentum of the wall + molecule system before the collision is P = 0 + mv, where m is the mass of the molecule. After the collision the wall acquires a recoil velocity, say, V and a recoil momentum MV where M is the mass of the wall. After the collision, the recoil momentum of the wall + momentum of the outgoing molecule = momentum of the incoming molecule so that the total momentum of the system is conserved. Notice that the momenta of outgoing and incoming molecules are not the same, their directions are different. mentum produces a negligible velocity so that the kinetic energy of the wall is negligible after the collision. Since the speed v of the molecule is the same before and after collision, its kinetic energy

remains unchanged. Hence the total kinetic energy is also conserved. Therefore, the collision is elastic. Remember, in an inelastic collision, although the total momentum is conserved, the total kinetic energy is not conserved, it decreases. Hence the correct choices are (a) and (c). 4. From the principle of conservation of momentum, mentum. Momentum conservation is possible in cases (c) and (d). In case (c), the two masses should move in mutually perpendicular directions with velocity v/ 2 each inclined at 45° with the original direction of motion of particle A. In case (d), particle B must move with velocity v in the original direction of motion of A. Hence the correct choices are (c) and (d). 5. The general form of Newton’s second law is Fext =

dp d dv dm = (mv) = m +v dt dt dt dt

The form Fext = ma is valid only if

dm = 0, i.e. dt

if mass does not change with time. Hence choice (a) is incorrect. Choice (b) is correct because a body at rest may have potential energy and yet no momentum. Choice (c) is also correct. A body has momentum if it has mass and velocity and a body having a mass and velocity must have kinetic energy. Choice (d) is incorrect because the relative velocity remains unchanged in magnitude and gets reversed in direction; (v2 – v1) = – (u2 – u1). Hence the correct choices are (b) and (c).

7.24 Comprehensive Physics—JEE Advanced

6. In the inelastic collision, two bodies stick together. After the collision, the speed of the ball and the bob (sticking together) is v = v/2. The height to which they will rise is given by v = or

h =

2gh v2 2g

v2 8g

Mass of the ball and the bob sticking together is 1 1 2m m = 2 m. KE after collision = m v 2 = 2 2 v 2 1 1 = mv2. KE before collision = mv2. 2 4 2 Therefore, their ratio is 1: 2. In an elastic collision between two bodies of the same mass with one of them initially at rest, the moving body is brought to rest and the other moves in the same direction with the same speed. Thus the ball will come to rest and the bob of the pendulum acquires a speed v. At this speed, it will rise to height h given by h = v2/2 g. Thus all four choices are correct. 7. Let m1, m2 and m3 be the masses of the three fragments. As the total mass is 1 kg and m1 : m2 : m3 = 1 : 1 : 3, we have m1 = m2 = 0.2 kg and m3 = 0.6 kg. The linear momentum of m1 is p1 = m1 v1 = 0.2 30 = 6 kg ms–1 and let it be directed along the x-axis (Fig. 7.28).

Fig. 7.28

The linear momentum of m2 is p2 = m2 v2 = 0.2 30 = 6 kg ms–1 and let it be directed along the y-axis. The magnitude of the resultant momentum is p = ( p 12 + p 22)1/2 = [(6)2 + (6)2]1/2

or = 45° with the x or y axes. From the principle of conservation of linear momentum, the magnitude of the momentum of the third fragment is (here v3 is the speed of the heavier fragment) p3 = m3 v3 = 6 2 But m3 = 0.6 kg. Therefore, 6 2 10 2 = 14.1 ms–1 0.6 The direction of the velocity of the heavier fragment is inclined with x or y axes at an angle of 135° (see Fig. 7.28). The correct choices are (a) and (c). 8. Initially (i.e. before collision) bodies A and B are at rest and C is moving to the right (towards A) with a velocity v0. At a certain instant, say t = 0, C collides with A. Since the collision is elastic and A and C have equal masses, the entire momentum (mv0 ) and 1 kinetic energy m v02 of C are transferred to A 2 and hence C comes to rest. Thus at t = 0, A moves to the right with a velocity v0 and at this instant the spring is uncompressed and B is at rest. Hence the momentum of the system at t = 0 is (mv0). When A moves to the right, it compresses the spring and as a result body B begins to move to the right. It is given that at time t = t0, the compression of the spring is x0. Let v be the common velocity of A and B at this instant. From the principle of conservation of linear momentum, we have momentum of C before collision = momentum of A after collision + momentum of B after collision v (1) or mv0 = mv + (2m)v or v = 0 3 From the principle of conservation of energy, we have KE of C before collision =(KE of A + KE of B) after collision + PE in stored spring v 3=

1 1 1 1 mv20 = mv2 + (2m)v2 + kx 20 2 2 2 2 where k is the spring constant. Thus mv 20 = 3mv 2 + kx 02 Using (1) in (2), we get or

= 6 2 kg ms–1 The direction of the resultant momentum is given by p tan = 2 = 1 p1

mv 20 = 3m

v0 3

2

+ kx 02

2 m v02 2 m v02 = kx 02 or k = 3 3 x02 Thus the correct choices are (b) and (c). or

(2)

Conservation of Linear Momentum and Collisions 7.25

9. PE stored in the spring =

1 kx2. 2

or

mv = (M + m)V or V =

mv ( M + m)

(1)

After collision, KE of block + bullet in it = PE of the spring. Thus 1 1 (M + m)V 2 = kx2 2 2 k which gives V = x ( M m) Using Eq. (2) in (1), we have v= =

(M

m) V m

(M

m) m

1/ 2

k M

(2)

m

x m Thus the correct choices are (a) and (d). 10. From conservation of momentum, we have m 1u 1 + m 2u = m 1v 1 + m 2v 2 u1 + 0 = 1

v1 + 2

v2

1 (u1 – v1) (1) 2 From conservation of kinetic energy, we have v2 =

1 1 1 1 m1u12 + m2u22 = m1v12 + m2v22 2 2 2 2 2 2 2 u1 = v1 + 2v2 v12 = u21 – 2v22

(2)

Solving Eqs. (1) and (2) we get v1 = – 1 ms–1 and v2 = + 2 ms–1. Hence the correct choices are (b) and (c). 11. For conservation of momentum and conservation of total energy, we have mu = (M + m) v (1) 1 1 1 mu2 = (M + m)v2 + kx2 2 2 2 1 mu2, we get Dividing Eq. (2) bv 2 Also

1=

(M

m) v 2 mu 2

1 2 kx 2 1 mu 2 2

(3)

v m u ( M m) Using this in Eq. (3), we get

From Eq. (1), we have

1 2 kx M n= 2 = 1 ( M m) mu 2 2 Thus the correct choices are (b) and (c). 12. If uA is the velocity with which A strikes B, then 1 muA2 = mgh 2 uA = 2gh = 2 9.8 0.1 = 1.4 ms–1. Since the masses of the balls are equal and the collision is elastic, they exchange their velocities after collision. Hence the only correct choice is (a). 13. Total initial momentum before collision = mu + (– mu 0, as the blocks come to rest. Since the change in momentum is zero, no external force acts on the system. Hence no work is done by the external force on the system. From work-energy principle,

x

= [k (M + m)]1/2

1

1 2 kx ( M m) v 2 2 =1– 1 mu 2 mu 2 2

(2)

1 2 1 2 mu mu = – mu2. Hence 2 2 the correct choices are (a) and (d). 14. Let v1 and v2 be the velocities of balls P and Q respectively after the collision. Then we have (1) m 1u = m 1v 1 + m 2v 2 initial K.E = 0 –

and

1 1 1 m1u12 = m1v21 + m2v22 2 2 2 2 2 m 1u 1 = m 1v 1 + m 2v 22

(2)

(a) Putting m1 = 2m2 in Eqs. (1) and (2) and u 4u and v2 = . solving them, we get v1 = 3 3 Thus v1 = v2/4. u 3u (b) For m1 = 3m2, we get v1 = and v2 = . 2 2 Thus v1 = v2 /3. u 2u (c) For m2 = 2 m1, we get v1 = – and v2 = 2 3 giving v1 = – v2 /3 (d) For m2 = 3m1, we get v1 = –

u u and v2 = 2 2

giving v1 = – v2 Hence all the four choices are correct.

7.26 Comprehensive Physics—JEE Advanced

15. From conservation of x and y components of momentum we have m1u = m1v1 cos + m2v2 cos 0 = m1v1 sin

and

(1)

m 1v 1 = m 2v 2

(2)

Percentage fractional decrease in K.E. of P is 1 2 1 2 mu v K 2 1 = 2 1 2 K mu 2

If m1 = m2, then from Eqs. (1) and (2), we get v1 = v2 and u = 2v1 cos

= =

1 2

=

Hence choice (a) is correct and choice (b) is wrong.

v2 3 and cos = = 30°. 2 2 Hence choice (d) is correct and choice (c) is wrong. 16. The only incorrect statement is (d). Refer to the solution of Q.34 of Section I. v1 =

4m1m2 K 4m m = = = 1 or 100 % 2 K ( m m) 2 (m1 m2 ) 17. The only incorrect statement is (c). Statements (a), (b) and (d) are true. 18. Given m = 1 kg, u = 15 ms–1 and e = 1/3. Let v1 and v2 be the velocities of P and Q after the collision. mu + 0 = m v1 + m v2 u = v1 + v2 (1) v1

(2)

u

From Eqs. (1) and (2), we get v1 =

u 15 (1 – e) = 2 2

u 15 and v2 = (1 + e) = 2 2 Total K.E. before collision =

1

1 = 5 ms–1 3

1 = 10 ms–1 1 3 1 mu2 2

v12

100

u2 (15) 2

(5)2

19.

Similarly putting m1 = 2 m2 in Eqs. (1), (2) and (3), we get

v2

u2

100

100 89% (15) 2 Hence the correct choices are (a), (b) and (c).

= 45°.

e=

1 1 m v12 + m v22 2 2

1 1 (52+ 102) = 62.5 J 2 Loss in K.E. = 112.5 – 62.5 = 50 J.

(3)

Using these in Eq. (3), we get cos2

(15)2 = 112.5 J

1

=

From conservation of kinetic energy, we have 1 1 1 m1u2 = m1 v12 + m2 v22 2 2 2 m1 u2 = m1 v12 + m2 v22

1 2

Total K.E. after collision =

– m2v2 sin

which give m1u = (m1v1 + m2v2) cos and

=

v2 is the velocity of ball Q after the collision, then from the conservation of momentum, we have m 1u + 0 = m 1 v 1 + m 2 v 2 m2 v2 = m1(u – v1) = m1[(a i + b j ) – (c i + d j )] = m1[(a – c) i + (b – d) j ] Impulse received by P is IP = m1 v1 – m1 u = m1(v1 – u) = m1[(c – a) i + (d – b) j ] Impulse imparted to Q is IQ = m2 v2 – 0 = m1[(a – c) i + (b – d) j ] Hence the correct choices are (b) and (c). 20. Conservation of x-component of momentum gives mu = mv cos 30° + mv cos 30° = 2 mv v=

u 3

3 = 2

3 mv

. Hence choice (c) is wrong.

K.E. before collision =

1 mu2 2

Conservation of Linear Momentum and Collisions 7.27

K.E. after collision = =

1 1 mv2 + mv2 = mv2 2 2

=

mu 2 3

1 1 1 mu2 – mu2 = mu2. 2 3 6 Hence the correct choices are (a) and (d). 21. Initially U-238 is at rest and has zero momentum. When it emits an -particle, the sum of the momenta of -particle and Th-234 nucleus must be zero, i.e. v m m v + mTh vTh = 0 = – Th m vTh

Hence

K K TH

mTH m

2

=

mTH m

Since mTH > m ; K > KTh. Hence the correct choices are (a) and (d). 22. In a collision, the momentum of indivdual bodies is not conserved. In an inelastic collision, there is a loss of kinetic energy. Hence choices (a) and (b) are incorrect. The correct choices are (c) and (d). 23.

Loss of K.E. =

Now KTh =

m mTH

momentum is pi

p1

p2

pi

p i = 0. There-

c1 = 0 and in choice (b) and (c) for non-zero values of the

1 1 2 mTh vTh and K = m v2 2 2 m v 2 = mTH vTH

i and j . Hence the choices (a) and (d) are not allowed.

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I Collisions: In physics we come across many examples of collisions. The molecules of a gas collide with one another and with the walls of the container. The collision of a neutron with an atom is well known. In a nuclear reactor fast neutrons down. They are, therefore, made to collide with hydrogen atoms. The term collision does not necessarily mean that a particle or a body must actually strike another. In fact, two particles may not even touch each other and yet they the other. When two bodies collide, each body exerts an equal and opposite force on the other. The fundamental conservation laws of physics are used to determine the velocities of the bodies after the collision. Collision may be elastic or inelastic. Thus a as an event in which two or more bodies exert relatively strong forces on each other for a relatively short time. The forces that the bodies exert on each other are internal to the system. Almost all the knowledge about the sub-atomic particles such as electrons, protons, neutrons, muons, quarks, etc. is

obtained from the experiments involving collisions. There are certain collisions called nuclear reactions in which new particles are formed. For example, when a slow neutron collides with a uranium-235 nucleus, new nuclei baruim-141 and krypton-92 are formed. This collideuterium and trituim collide (or fuse) to form a helium nucleus with the emission of a neutron. 1. Which one of the following collisions is NOT elastic? (a) A hard steel ball dropped on a hard concrete (b) Two balls moving in the same direction collide and stick to each other. (c) Collisions between molecules of an ideal gas (d) Collisions of fast neutrons with hydrogen 2. Which one of the following statements is true about inelastic collisions? (a) The total kinetic energy of the particles after collision is equal to that before collision. (b) The total kinetic energy of the particles after collision is less than that before collision. (c) The total momentum of the particles after collision is less than that before collision.

7.28 Comprehensive Physics—JEE Advanced

(d) Kinetic energy and momentum are both conserved in the collision. 3. In elastic collisions (a) only energy is conserved

(b) only momentum is conserved (c) neither energy nor momentum is conserved (d) both energy and momentum are conserved.

ANSWERS

1. The correct choice is (b) 2. The correct choice is (b) Questions 4 and 5 are based on the following passage Passage II Two balls marked 1 and 2 of the same mass m and a third ball marked 3 of mass M are arranged over a smooth horizontal surface as shown in Fig. 7.29. Ball 1 moves with a velocity v1 towards balls 2 and 3. All collisions are assumed to be elastic.

3. The correct choice is (d) 4. If M < m, the number of collision between the balls will be (a) one (b) two (c) three (d) four 5. If M > m, the number of collisions between the balls will be (a) one (b) two (c) three (d) four

Fig. 7.29

SOLUTION 4. both have the same mass, after the collision ball 1 with come to rest and ball 2 will move with speed v1. The ball will collide with the stationary ball 3. After this second collision, let v2 and v3 be the speeds of balls 2 and 3 respectively. Since the collision are elastic, v2 and v3 are given by (see page 7.2) m M v2 = v1 (i) m M 2m v1 (ii) m M If M < m, it follows from (i) and (ii) that v2 < v3 and both have the same direction. Therefore, ball and

v3 =

Questions 6 to 8 are based on the following passage Passage-III A body of mass m1 = m moving with a velocity v1 = v in the x-direction collides with another body of the same mass m2 = m moving in the y-direction with the same speed v2 = v. They coalesce into one body during the collision. 6. The magnitude of the momentum of the composite body is (a) mv (b) 2 mv (c) 2mv

(d) 2 2 mv

2 cannot collide with ball 3 again. Hence there are only two collisions. Thus the correct choice is (b). 5. If M > m, we have from Eq. (i) M m v1 M m The negative sign indicates that, after the second collision, ball 2 will move in opposite direction lision. Therefore, ball 2 will make another collision with ball 1. Hence, in this case, there are three collisions in all between the balls. Thus the correct choice is (c). v2 = –

7. The angle which the direction of the momentum vector of the composite body makes with the x-axis is (a) 30° (b) 45° (c) less than 30° (d) greater than 45° 8. The fraction of initial kinetic energy transformed into heat during the collision is 1 1 (b) (a) 2 4 (c)

2 3

(d)

1 3

Conservation of Linear Momentum and Collisions 7.29

SOLUTION 6. Refer to Fig. 7.30. Let v be the velocity of the composite body and let be the angle which the velocity vector v makes with the x-axis.

7. Diving Eq. (2) by Eq. (1), we have tan 45°, which is choice is (b). 8. Initial kinetic energy is 1 1 m1v21 + m 2v 22 Ki = 2 2 1 1 mv2 + mv2 = 2 2 From Eq. (3), we have v = v/ energy is 1 Kf = (m1 + m2)(v )2 2 1 v2 = 2m = 2 2 Loss in K.E. = Ki – Kf =

Fig. 7.30

Conservation of x and y components of momentum gives m1v1 = (m1 + m2)v cos mv = 2mv cos (1) m2v2 = (m1 + m2)v sin

mv = 2mv sin

(2)

Squaring and adding Eqs. (1) and (2), we get 2(mv)2 = (2mv )2

2mv =

2 mv

(3)

Momentum of composite body is 2 mv . Hence the correct choice is (b). Questions 9 to 11 are based on the following passage Passage IV A ball P moving with a velocity u strikes an identical stationary ball Q such that after the collision, the direction of motion of balls P and Q make an angle of 30° with the original direction of motion of ball P, as shown in Fig. 7.31.

= mv2 –

=

mv2 2 . Final kinetic

1 mv2 2

1 mv2 2

1 mv2 2 Fraction of initial K.E. transformed into heat is 1 2 mv K 1 = 2 2 = . Ki 2 mv Hence the correct choice is (a). =

9. The speed v1 of ball P after the collision is u u (b) (a) 2 3 (c)

u

(d)

u

2 3 10. The speed v2 of ball Q after the collision is u 3 3 2u 2u (c) (d) 3 3 11. The ratio of the total kinetic energy of the balls after collision to that before collision is 1 1 (b) (a) 3 3 2 1 (c) (d) 3 2 (a)

Fig. 7.31

=1

u

(b)

SOLUTION From conservation of x and y components of momentum, mu = mv1 cos 30° + mv2 cos 30°

3 2 0 = mv1 sin 30° – mv2 sin 30°

u = (v1 + v2) and

(1)

7.30 Comprehensive Physics—JEE Advanced

v1 = v2 9. Eqs. (1) and (2) give v1 = 10. v2 = v1 =

(2) u 3

, which is choice (d).

u

and after collision 1 1 Kf = mv21 + mv22 2 2

. Hence the correct choice is (a). 3 11. Total kinetic energy before collision is 1 Ki = mu2 2

Kf Ki

Questions 12 to 14 are based on the following passage Passage V A body P of mass m moving with a velocity u along the + x-direction makes a head-on elastic collision a body Q of mass 2 m at rest. 12. After the collision, body P moves along the (a) positive x-direction with speed u/3. (b) negative x-direction with speed u/3. (c) positive x-direction with speed 2u/3. (d) negative x-direction with speed 2u/3.

SOLUTION Let V be the velocity of body Q after the collisions. From the principle of conservation of linear momentum, we have mu = mv + (2m)V or

u – v= 2 V

(1)

The conservation of kinetic energy gives

or

u2 – v2 = 2V2

or (u – v) (u + v) = 2 V 2

(2)

Using Eq (1) in Eq (2), we have 2V(u + v) = 2V2 or u + v = V or

2(u + v) = 2V

=

2 , which is choice (c). 3

Passage VI A body A of mass m1 moving with a velocity u makes a headon elastic collision with a body B of mass m2 initially at rest.

=

mu 2 3

14. What fraction of its kinetic energy does body P lose after the collision? (a)

8 9

(b)

7 8

(c)

6 7

(d)

5 6

1 mu2 2 1 Final kinetic energy, Kf = mv2 2 Loss in kinetic energy is 1 1 K = Ki – Kf = mu2 – mv2 2 2 Ki =

K Ki

1 2 1 2 mu mv 2 = 2 1 2 mu 2

(3)

Questions 15 to 17 are based on the following passage

+

13. After the collision, body Q moves along the (a) positive x-direction with speed u/3. (b) negative x-direction with speed u/3. (c) positive x-direction with speed 2u/3. (d) negative x-direction with speed 2u/3.

=

12. From Eqs. (1) and (3), we get v = – u/3. Hence the correct choice is (b). 13. Using v = – u/3 in Eq. (1), we get V = 2u/3. Hence the correct choice is (c). 14. Initial skinetic energy of the colliding mass is

1 u2 m 3 2

1 m 2

Fractional loss =

1 1 1 mu2 = mv2 + (2m)V2 2 2 2

u2 3

=

u2

v2 u2

=1– =

1 3

v u

=1–

2

2

(

v = – u/3)

8 , which is choice (c). 9

15. After the collision, body B will move with the greatest speed if (b) m1 > m2 (c) m1 = m2 (d) m1 = 2 m2

Conservation of Linear Momentum and Collisions 7.31

16. After the collision, body B will move with the greatest momentum if (b) m1 > m2 (c) m1 = m2 (d) m1 = 2 m1

17. After the collision, body B will move with the greatest kinetic energy if (b) m1 > m2 (c) m1 = m2 (d) m2 = 2 m1

SOLUTION

Since the collision is elastic, both linear momentum and kinetic energy are conserved. Thus (1) m1 u1 = m1 v1 + m2 v2 1 1 1 m1 u21 = m1v21 + m2v22 2 2 2 Solving for v1 and v2, we get and

and

v1 =

m1 m1

m2 m2

v2 =

2m1 u1 m1 m2

m1

2

(4)

2

– 2 m1m2 + 2 m1m2

Questions 18 to 20 are based on the following passage Passage VII A body P of mass m moving along the positive x-direction with velocity u collides with a body Q of mass M initially at rest. After the collision body P moves along the positive y-direction and body Q moves along a direction making an angle below the x-axis. The collision is assumed to be elastic. 18. The angle is give by M m (b) = tan–1 (a) = tan–1 m M (c)

= tan–1

M M

m m

(d)

2

+ 2 m1m2

2m1m2u m1

m2

2

2 m1m2 2

(3)

m2

m2

Momentum p2 will be maximum if the denominator

u1

+

m1

p2 =

(2)

15. Equation (4) gives the recoil speed of body B which can rewritten as 2u1 v2 = m2 1 m1 Now v2 will be maximum if the denominator is the m minimum, i.e. if 2 0.0969 n > 3.1 or (0.8)n
r (d) 0 x r 5. Two particles of equal mass have velocities v1 = a i and v2 = a j

6.

j) where a and b

tre of mass of the two particles moves along a (a) straight line (b) circle (c) ellipse (d) parabola

-

an inclined plane of inclination 1 sliding and reaches the bottom with speed v1 and its time of descent is t1. The same sphere is then released from rest from the top of another inclined plane of inclination 2 without sliding and reaches the bottom with speed v2 and its time of descent is t2 2> 1 (a) v2 > v1 ; t2 < t1 (b) v2 = v1 ; t2 < t1 (c) v2 < v1 ; t2 > t1 (d) v2 = v1 ; t2 = t1 7. The moment of inertia of a uniform rod of mass M and length L about an axis passing through its centre and inclined to it at an angle = 60° is (a)

ML2 3

ML2 (c) 12

(c)

R

(d) (2 1) 2(2 1) 3. A carpet of mass M is rolled along its length in the form of a cylinder of radius R and kept on a rough

a1 = b ( i

8. A solid cylinder of mass M and radius R is rolling without slipping on a horizontal plane with a speed v to a maximum height given by v2 v2 (b) sin (a) 2g 2g

(b)

ML2 4

ML2 (d) 16

3v 2 4g

(d)

3v 2 sin 4g

9. A uniform rod AB of mass m and length L is suspended by two strings C and D of negligible mass as shown in Fig. 8.39. When string D tension in string C will be

mg 4 (c) 2 mg

(b) mg

(a)

(d) 4 mg

Fig. 8.39

10. A body of mass m is projected with a velocity u at an angle of 60° with the horizontal. The magnitude of the angular momentum about the point of projection when the body is at the highest point of its trajectory is

mu 3 2mu 3 (b) (a) 2g 5g 3mu 3 mu 3 (c) (d) 4g 16 g 11. A cubical block of side L and mass m rests on a rough horizontal surface. A horizontal force F is applied normal to one of its faces at a point that is directly above the centre of the face at a height 2L/3 above the base. The minimum force F required to topple the block before sliding is (a)

2mg 3

(b)

3mg 2

(c)

3mg 4

(d)

4mg 3

12. duration of the new day will be (a) 6 hours (b) 12 hours (c) 18 hours (d) 30 hours

Rigid Body Rotation 8.23

of friction between the coin and the record is the minimum angular frequency of the record for

13. A circular ring of mass M and radius R is rotating about its axis at an angular frequency . Two m the opposite ends of a diameter of the ring. The angular frequency of the ring becomes . The ratio / is 2M M (a) (b) M 2m M 2m

2m M (d) M 2m 14. A solid sphere rolls down from the top of an in(c)

the bottom is v . The ratio v /v is

(c)

g r

(d) 2

(a)

g r

(b)

(b)

g r

(c) 2 20.

(d) 2

g 2r g r

1 (ai + bj) 3 2 (c) (ai + bj) 3

g 2 (d) g 3 3 16. A block of mass M is released from the top of an (c)

2 g r 2 g r

m corners of a right angled triangle as shown in Fig. OA = a and OB = b the centre of mass is (here i and j are unit vectors along x and y axes respectively).

1 (ai – bj) 3 2 (d) (ai – bj) 3

(a)

acceleration of the disc down the inclined plane is g (a) g (b) 2

of the plane is v. A circular disc of the same mass M velocity on reaching the bottom is v . The ratio v /v will be 1 2 (b) (a) 3 3 2 2 (c) 1 (d) 3 17. Two circular loops A and B of radii R and 2R respectively are made of the same wire. Their moments of inertia about the axis passing through the centre and perpendicular to their plane are IA and IB respectively. The ratio IA/IB is 1 (a) 1 (b) 2 1 1 (c) (d) 4 8 18. A small coin is placed at a distance r from the centre of a gramophone record. The rotational speed of

2 g r

19.

the plane is v. When the same sphere slides down

3 (b) 1 (a) 5 3 7 (d) (c) 5 5 15. A circular disc is rolling down an inclined plane

(a)

(b)

Fig. 8.40

21. A sphere of mass M and Radius R is released from the top of an inclined plane of inclination . The and the sphere so that it rolls down the plane without sliding is given by 2 tan (a) = tan (b) = 3 (c) 22.

=

2 tan 5

(d)

=

2 tan 7 M and length

L moment of inertia of the composite structure about an axis passing through the centre of mass of the structure and perpendicular to its plane is

8.24 Comprehensive Physics—JEE Advanced

(a)

ML2 2

(b)

ML2 8

(d)

ML2 4

26. I1 I2 I3 and I4 are respectively the moments of inertia of a thin square plate ABCD of uniform thickness

ML2 12 23. M and length L are welded to form a square ABCD as shown in Fig. 8.41. What is the moment of inertia of the composite structure about a line which bisects rods AB and CD (c)

ML2 6 ML2 (c) 2

the plate (Fig. 8.43). The moment of inertia of the plate about an axis passing through thecentre O and perpendicular to the plane of the plate is (b) 2(I3 + I4) (a) 2(I1 + I2) (d) I1 + I2 + I3 + I4 (c) I1 + I3

ML2 3 2 ML2 (d) 3 (b)

(a)

Fig. 8.43 Fig. 8.41

24 . A thin uniform metallic triangular sheet of mass M has sides AB = BC = L. What is its moment of inertia about axis AC (Fig. 8.42) (a) M L2

27. The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height h from rest without sliding is

12 (b)

25.

M L2 6

28.

2

(c)

ML 3

(d)

2M L2 3

Fig. 8.42

M and radius R rolling and partly sliding. During this kind of motion of the sphere (a) total kinetic energy is conserved (b) the angular momentum of the sphere about the point of contact with the plane is conserved (c) only the rotational kinetic energy about the centre of mass is conserved (d) the angular momentum about the centre of mass is conserved.

(a)

10 gh 7

(b)

gh

(c)

6 gh 5

(d)

4 gh 3

energy to the total kinetic energy is given by (a) 7 : 10 (b) 2 : 5 (c) 10 : 7 (d) 2 : 7 29. A cart of mass M is tied at one end of a massless rope of length 10 m. The other end of the rope is in the hands of a man of mass M. The entire system is on a smooth horizontal surface. The man is at x = 0 and the cart at x a point (a) x = 0 (b) x = 5 m (c) x = 10 m (d) they will never meet. 30. A mass m is moving with a constant velocity along a line parallel to the x angular momentum with respect to the origin

Rigid Body Rotation 8.25

(a) is zero (c) goes on increasing

(b) remains constant (d) goes on decreasing.

31. Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle with AB. The moment of inertia of the plate about the axis CD is then equal to (a) I (b) I sin2 (c) I cos2

(d) I cos2

2

32. A smooth sphere A is moving on a frictionless horizontal surface with angular speed and centre of mass velocity v on with an identical sphere B at rest. Neglect friction are (a) (c)

A

and A < A =

B B

respectively. Then (b) A = (d) B =

Fig. 8.44

1 MR2 2 3 (c) MR2 2

(b) MR2 (d) 2 MR2

34. A cubical block of side a is moving with a velocity v on a horizontal smooth plane as shown in O. The angular speed of the block after it hits the ridge at O is 3v 3v (b) (a) 4a 2a (c)

3v 2a

(d) zero

35. A thin wire of length L and uniform linear mass density is bent into a circular loop with centre at O as shown in Fig. 8.46. The moment of inertia of the loop about the axis XX is (a) (c)

L3 8

(b)

2

5 L3 16 2

(d)

L3 16

2

3 L3 8 2

B

33. A disc of mass M and radius R is rolling with angular speed on a horizontal plane as shown in Fig. 8.44. The magnitude of angular momentum of the disc about the origin O is

(a)

Fig. 8.45

Fig. 8.46

36. A tube of length L incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity . The force exerted by the liquid at the other end is (a)

M

(c)

M

2

2

2

L

(b) M

L

(d)

4

M

2

2

L

L2

2

37. An equilateral triangle ABC formed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and Fig. 8.47

one along AB and the other along AC as shown in Fig. 8.47. Neglecting

8.26 Comprehensive Physics—JEE Advanced

(a) angular velocity and total energy (kinetic and potential). (b) total angular momentum and total energy. (c) angular velocity and moment of inertia about the axis of rotation. (d) total angular momentum and moment of inertia about the axis of rotation. 38. A cubical block of side L rests on a rough horizontal . A horizontal force F is applied on the block as shown in Fig. minimum force required to topple the block is mg/4 (c) mg/2 (d) mg(1 — )

(a) M

(c) A M (d) AM2 43. One end of a thin uniform rod of length L and mass M1 is rivetted to the centre of a uniform circular disc of radius ‘r’ and mass M2 so that both are coplanar. The centre of mass of the combination from the centre of the disc is: (Assume that the point of attachment is at the origin)

L ( M1 M 2 ) 2 M1 2 ( M1 M 2 ) (c) LM1 (a)

39. The angular velocity of a body changes from 1 to 2 without applying a torque but by changing the moment of inertia about its axis of rotation. The ratio of the corresponding radii of gyration is (b) (a) 1 : 2 1: 2 (c) (d) : : 2 1 2 1 40. A thin uniform rod AB of mass m and length L is hinged at one end A stands vertically and is allowed to fall freely to the

(a)

mg L

(b)

1/ 2

3 g 1/ 2 g (d) L L 41. Moment of inertia of uniform horizontal solid cylinder of mass M about an axis passing through its edge and perpendicular to the axis of the cylinder when its length is 6 times its radius R is: (c)

(a)

39 MR 2 4

(b)

39 MR 4

49 MR 49 MR 2 (d) 4 4 42. A is the areal velocity of a planet of mass M angular momentum is (c)

(b) 4 (d) 8

45. A body of mass ‘m’ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one elongation in the spring is 5 cm. The original length of the spring is: (a) 16 cm (c) 14 cm

(b) 15 cm (d) 13 cm

46. A particle performs uniform circular motion with an angular momentum L

the rod when its end B

mg 3L

LM1 2( M1 M 2 ) 2 LM1 (d) ( M1 M 2 ) (b)

44. Two circular loops A and B of radii rA and rB respectively are made from a uniform wire. The ratio of their moments of inertia about axes passing through their centres and perpendicular to their planes is IB rB is equal to IA rA (a) 2 (c) 6

Fig. 8.48

(b) 2MA

2

ular momentum becomes: (a) 4L (b) 2L L (d) L (c) 2 4 47. A uniform rod of length 1 metre is bent at its midpoint to make 90° angle. The distance of the centre of mass from the centre of the rod is (a) 36.1 cm

(b) 25.2 cm

(c) 17.7 cm

(d) zero

48. A mass is whirled in a circular path with constant angular velocity and its angular momentum is L the string is now halved keeping the angular veloc(a) L 4 (c) L

(b) L 2 (d) 2L

Rigid Body Rotation 8.27

-

49.

(a) (b) (c) (d)

Moment of inertia Angular momentum Angular velocity Rotational kinetic energy

50. A solid sphere is rotating about its diameter. Due to

two beads are at the centre of the rod and the system is rotating with angular velocity 0 about its axis perpendicular to the rod and passing through its mid point (see Fig. 8.50). There are no external the angular velocity of the system is M 0 M 0 (b) (a) M 3m M 6m (c)

M

6m M

0

(d)

0

speed of the sphere will

1 % 3 1 (b) decrease by nearly % 3 1 (c) increase by nearly % 2 2 (d) decrease by nearly % 3 51. The height of a solid cylinder is four times its radit = 0 on a belt which is moving in the horizontal direction with a velocity v = 2.45 t2 where v is in ms–1 and t (a) increase by nearly

t equal to (a) 1 s (b) 2 s (c) 3 s (d) 4 s 52. A circular portion of diameter R is cut out from a uniform circular disc of mass M and radius R as shown in Fig. 8.49. The moment of inertia of the remaining (shaded) portion of the disc about an axis passing through the centre O of the disc and perpendicular to its plane is (a)

15 MR2 32

(b)

7 MR2 16

(c)

13 MR 2 32

(d)

3 MR2 8

Fig. 8.49

53. A smooth uniform rod of length L and mass M has m

Fig. 8.50

54. Two particles A and B each other under a mutual force of attraction. At the instant when the speed of A is V and that of B is 2V the speed of the centre of mass of the system is (a) 0 (b) V (c) 1.5V

(d) 3V

55. One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M made to rotate about a line perpendicular to its plane and passing through the center of the rotation is (Fig. 8.51) 1 MR2 (a) 2 1 (b) MR2 4 1 (c) MR 2 8 (d) 2 MR 2

Fig. 8.51

56. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the center of mass is (a) 30 m/s (b) 20 m/s (c) 10 m/s (d) 5 m/s

8.28 Comprehensive Physics—JEE Advanced

57. throughout these motions). The directions of the frictional force acting on the cylinder are: (a) up the incline while ascending and down the incline while descending (b) up the incline while ascending as well as descending (c) down the incline while ascending and up the incline while descending (d) down the incline while ascending as well as descending. 58. The angular momentum of a particle moving in a circular orbit with a constant speed remains conserved about (a) any point on the circumference of the circle (b) any point inside the circle (c) any point outside the circle (d) the centre of the circle

A

(a)

B

(b)

B

(c)

A

A

(d)

B

B A

63. with speed vr

out rolling and reaches the bottom with speed vs. Then (a) vr > vs (c) vr = vs 64.

(b) vr < vs (d) vr = vs = 0

ABCD shown in a = AB = BC/2. The moment of inertia of the lamina is the minimum along the axis passing through (a) BC (b) AB (c) HF (d) EG

59. A particle moves in a circular orbit with uniform orbit is itself rotating at a constant angular speed. We may then say (a) the angular velocity as well as the angular acceleration of the particle are both constant (b) neither the angular velocity nor the angular acceleration of the particle are constant (c) the angular velocity of the particle varies but its angular acceleration is a constant (d) the angular velocity of the particle remains constant but its angular acceleration varies 60. A cylinder of mass m and radius r is rotating about its axis with a constant speed v (b) mv2 (a) 2 mv2 1 1 (c) mv2 (d) mv2 2 4 61. A circular disc of mass m and radius r is rolling on a smooth horizontal surface with a constant speed v

1 1 mv2 (b) mv2 4 2 3 (c) mv2 (d) mv2 4 62. Two solid spheres A and B R made of materials of densities A and B respectively. Their moments of inertia about a diameter are IA and IB respectively. The ratio IA /IB is (a)

Fig. 8.52

65. A uniform rod of length L is suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length. What minimum speed must be imparted to the lower end so that the rod completes one full (a)

2gL

(b) 2 gL

(c) 6gL (d) 2 2gL 66. A circular disc of radius R is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through when it reaches the equilibrium position will be (a)

g 3R

(b)

2g 3R

2g 2g (d) 2 R R 67. A massless and inextensible cord is wound round the circumference of a circular ring of mass M and radius R. The ring is free to rotate about an axis passing through its centre and perpendicular to its (c)

Rigid Body Rotation 8.29

plane. A mass m is attached at the free end of the cord and is at rest. The angular speed of the ring when mass m has fallen through at height h is

2 gh R2

(a)

(b)

2mgh MR 2

2mgh 2mgh (d) 2 M m R M 2m R 2 68. The moment of inertia of a hollow sphere of mass M and internal and external radii R and 2R about an axis passing through its centre and perpendicular to its plane is 3 13 MR 2 (b) MR 2 (a) 2 32 31 62 (c) MR2 (d) MR2 35 35 69. angular frequency with his arms outstretched. He (c)

71. A particle is moving in the x – y plane with a constant velocity along a line parallel to the x away from the origin. The magnitude of its angular momentum about the origin. (a) is zero (b) remains constant (c) goes on increasing (d) goes on decreasing 72. A thin uniform disc has mass M and radius R. A circular hole of radius R/3 is made in the disc as shown in Fig. 8.54. The moment of inertia of the remaining portion of disc about an axis passing through O and perpendicular to the plane of the disc is 1 2 MR2 (b) MR2 (a) 9 9 1 4 (c) MR2 (d) MR2 3 9

with folded arms is 75% of that with outstretched (a) (b) (c) (d)

increase by 33.3% decrease by 33.3% increase by 25% decrease by 25%

70. A uniform disc of radius R is rolling (without slipping) on a horizontal surface with an angular speed as shown in Fig. 8.53. O is the centre of A and C are located on its rim and R from O point B is at a distance 2 the points A B and C lie on the vertical diameter vA vB and vC are the linear speeds of points A B and C respectively at (a) vA = vB = vC (c) vA

(b) vA > vB > vC 4 vC = vB (d) vA vC = 2 vB 3

Fig. 8.53

Fig. 8.54

73. A solid metallic sphere of radius R having moment of inertia equal to I about its diameter is melted and recast into a solid disc of radius r of a uniform thickness. The moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is also equal to I. The ratio r/R is (a)

2 15

(b)

2 10

(c)

2 5

(d)

1 2

74. A small object of uniform density rolls up a curved surface with an initial velocity v 3v 2 a maximum length h = 4g initial position. The object is (see Fig. 8.55) (a) ring (b) solid sphere (c) hollow sphere (d) disc

8.30 Comprehensive Physics—JEE Advanced

Fig. 8.55

75. The mass per unit length of a non-uniform rod OP of length L varies as m=k x L where k is a constant and x is the distance of any point on the rod from end O. The distance of the centre of mass of the rod from end O is (b) 2 L (a) L 3 3 2L L (c) (d) 3 2 76. A tube of length L compressible liquid of mass M and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity . The force exerted by the liquid at the other end is. (b) M 2L (a) 2 M 2L 1 3 (c) M 2L (d) M 2L 2 2 77. A uniform thin rod of mass M and length L is hinged by a frictionless pivot at its end O 8.56. A bullet of mass m moving horizontally with a velocity v strikes the free end of the rod and gets embedded in it. The angular velocity of the system about O just after the collision is mv 2m v (b) (a) L ( M m) L ( M 2m) (c)

3mv L( M 3m)

(d)

Fig. 8.56

78. A gramophone record of mass M and radius R is rotating at an angular velocity . A coin of mass m is gently placed on the record at a distance r = R/2 from its centre. The new angular velocity of the system is 2 M 2 M (b) (a) ( 2 M m) ( M 2m) (c)

M M 10 cm and height 15 cm (d)

79. A bolck of base 10 cm

friction between them is 3 . The inclination of this inclined plane from the horizontal plane is gradually increased from 0°. Then (a) at the plane (b) the bock will remain at rest on the plane up to certain and then it will topple (c) at the plane and continue to do so at higher angles (d) at the plane and on further increasing topple at certain .

mv LM

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

(d) (d) (a) (a) (b) (a) (b) (b) (b) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

(d) (c) (c) (a) (c) (c) (c) (a) (b) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(d) (a) (c) (d) (a) (c) (c) (b) (a) (b)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

(d) (c) (b) (a) (d) (a) (d) (d) (c) (d)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

(a) (c) (d) (d) (b) (d) (d) (c) (b) (c)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(b) (a) (c) (a) (b) (a) (b) (a) (a) (d)

Rigid Body Rotation 8.31

61. 67. 73. 79.

(c) (c) (a) (b)

62. (c) 68. (d) 74. (d)

63. (b) 69. (a) 75. (b)

64. (c) 70. (c) 76. (c)

65. (c) 71. (b) 77. (c)

66. (b) 72. (d) 78. (a)

SOLUTIONS 1. Since there is no external force acting on the gun-

4. Since the mass is not distributed uniformly over the

remains at rest. 2. Let = mass per unit area of the disc. Mass of the cut-out portion m1 = R2 /2 and mass of the remaining portion is (see Fig. 8.57)

lie anywhere between its centre and the circumference. Hence the correct choice is (d). m v m2 v 2 5. vCM = 1 1 m1 m2

m2 =

R2 2

R2

Let O1 be the centre of mass of the remaining portion. The centre of mass of the square is at O2. Taking moments of m1 g and m2 g about O m1 g x1 = m2 g x2 2

R 2

x1

R2

g = x2 =

x2

R 2(2

(

1)

1 (v1 + v2) 2

=

a i 2

aCM =

2

R 2

=

g

j

m1a1 m2 a2 a = 1 m1 m2 2

is conserved. Since both the inclined planes are of v2 = v1. The acceleration of the sphere rolling down the plane is

Since

2

>

R

M 4

M and its P.E. = mg (R/2) = 4

g

MgR Decrease in P.E. = MgR – 8

R 2

MgR 8

7 MgR 8

1;

g sin I 1 MR 2 a2 > a1. Hence

t1 sin = t2 sin

Fig. 8.57

2

a2 = 0)

line.

a=

M ( R / 2) 2

(

b i j 2 Since vectors vCM and aCM are parallel to each 6.

m=

m 1 = m 2)

=

x1 = R /2)

3. The entire mass M of the carpet can be assumed to be concentrated at its centre of mass which is originally at a height R P.E. = MgR. When the carpet is unrolled to a radius R R/2 above

(

2 1

Thus t1 > t2. Hence the correct choice is (b). M dx. 7. Mass of element of length dx is dm = L Perpendicular distance of the element from the axis of rotation = x sin . Therefore moment of inertia about the axis of rotation AB is (Fig. 8.58) I=

L/2

dm( x sin ) 2

L/2

=

M sin2 L

L/2 L/2

x 2 dx

8.32 Comprehensive Physics—JEE Advanced

=

ML2 sin2 12

ML2 ML2 sin2 (60°) = 12 16

=

L 2

Also a =

(3) a=

3mg 4 So the correct choice is (a). mg – T =

10.

Fig. 8.58

8. K.E. =

1 1 Mv 2 + I 2 2

=

1 1 Mv 2 + 2 2

1 MR 2 2

=

1 1 3 Mv 2 + Mv2 = Mv2 2 4 4

v = u cos 60° = u/2 along the horizontal direction. Hence the perpendicular distance of its linear momentum from the point of projection = maximum height attained. r1 = hmax =

2

v R

2

L . 2

= F

1

u 2

2L 3

Since the weight mg torque of the weight about A is = mg

2

L 2

The minimum force required to topple the block is given by 2L L = mg ( 1)min = 2 or Fmin 3 2 3mg 4

Fig. 8.60

Fig. 8.59

Moment of inertia of the rod about A is I = ML2/3. A is I

3u 2 8g

11. Torque of F about A is [see Fig. 8.60]

Fmin =

g 2L

=

3 mu 3 16 g

=

(1)

Torque of force mg about A is = Mg

3u 2 8g

=m

mg – T = ma

=

u 2 sin 2 60 2g

Magnitude of angular momentum = m r1 v

Loss in K.E. = gain in P.E. 3 or Mv 2 = Mgh 4 3v 2 h= 4g 9. When string D A. Let a be the linear acceleration of the centre of mass and T the tension in string C

=

3g . Then from Eq. (1) 4 mg T= 4

(2)

12. Let M be the mass and R the initial radius of the is the angular velocity of the rotation of T of the day is T=

2

Rigid Body Rotation 8.33

Let R be the radius of the earth after contraction and its angular velocity. From the conservation a=g

I =I 2 2 M R 2 and I M R 2 are the where I 5 5 moments of inertia of the earth before and after 2 2 MR 2 = MR 2 5 5

or

R2

=

2

R (

=4

R = R/2)

The duration T of the new day will be T = T =

2

I=

=

2 4

=

T 4

24 hours = 6 hours 4

16. The acceleration of the block sliding down the plane is a = g sin where l is the length reaching the bottom is given by v2 = 2 al = 2g sin l or v = 2gl sin The acceleration of the disc rolling down the plane is (as shown above) 2 a = g sin 3 bottom is given by v 2 = 2a l =

13. From the law of conservation of angular momen-

I =I Here I = MR2 and I = (M + 2m) R2. Therefore I M = I M 2m

Hence the correct choice is (a). 1 1 14. For rolling : Mgh = Mv2 + I 2 2 1 1 = Mv2 + 2 2 =

2 M R2 5

For sliding : Mgh =

R2

1 7 Mv 2 = Mv2 2 10 v 7 or = 5 v 15. The acceleration down the plane is given by g sin a= I 1 MR 2

IB = MBR 2B = 4 R = 16

v2

1 Mv 2. Therefore 2

v = 2

and

2

1 1 7 Mv2 + Mv2 = Mv2 2 5 10 2 I = MR 2 and 5

4 gl sin 3

gl sin v 2 = 3 v 3 Hence the correct choice is (b). 17. Let be the mass per unit length of the wire. The mass of loop A is MA = 2 R and mass of loop B is MB = 4 R Their moments of inertia respectively are IA = MA R A2 = 2 R R2 = 2 R3 or

=

1 MR2. Using this and 2

=

v R

(2R)2

R3

IA 1 = IB 8 Hence the correct choice is (d). 18. Let m the centripetal force mr 2 just exceeds the force of friction mg. The minimum is given by mr

2

= mg or

=

g r

Hence the correct choice is (c). 19. The minimum angular frequency is independent of g/r the mass. Hence the correct answer is still which is choice (a). 20. The (x y) co-ordinates of the masses at O A and B respectively are (refer to Fig. 8.40 on page 8.23) y1 x2 = a y2 = 0) and (x3 y3 = b) (x1 The (x y) co-ordinates of the centre of mass are xCM =

m1 x1 m2 x2 m3 x3 m1 m2 m3

8.34 Comprehensive Physics—JEE Advanced

= yCM = =

m

0 m a m m m m

0

=

a 3

m1 y1 m2 y2 m3 y3 m1 m2 m3 m

0 m 0 m m m m

b

=

b 3

The position vector of the centre of mass is xCM i + yCM j =

a b 1 i+ j = (ai + bj) 3 3 3

Fig. 8.61

-

21.

g sin I 1 MR 2 where K is the radius of gyration of the sphere

PS = PQ sin 60° = L sin 60° =

a=

the sphere about its diameter is 2 I= MR2 5 a=

g sin 2 1 5

=

5 g sin 7

(i)

f provides the necessary torque which is given by = force moment arm = fR But = I is the angular acceleration of I = fR a= R f =

I Ia 2 = 2 = Ma R R 5

= Mg cos . Thus Mg cos 5 g cos 2 Equating (i) and (ii) we have or

a=

2 MR 2 5 normal reaction 2 = Ma 5 I

(ii)

5 5 2 g sin = g cos or = tan 7 2 7 Hence the correct choice is (d). 22. Given PQ = QR = RP = L. The centre of mass is located at centroid C which cuts lines PS QT and UR in the ratio 2 : 1. Let h = CS = CT = UC PQS

h=

PS 1 = 3 3

3 L. 2

L 3 L= 2 3 2

its moment of inertia about an axis passing through its centre of mass C and perpendicular to its plane Ic = 3 (I + Mh2) where I is the moment of inertia of each rod about the axis passing through its centre and perpendicuI=

M L2 L Also Mh2 = M 12 2 3

2

=

M L2 12

2 2 M L2 M L2 =3 ML = ML 12 12 6 2 Hence the correct choice is (a). 23. Refer to Fig. 8.41 on page 8.24. Moment of inertia is a scalar quantity. So the moment of inertia of the structure is the sum of the moments of inertia of the

Ic = 3

I = I1 + I2 + I3 + I4 where I1= moment of inertia of rod 1 about an axis passing through its centre E and perpendicular to its M L2 12 I2= moment of inertia of rod 2 about an axis passing through its centre F and perpendicular to its plane plane =

M L2 12 I3 = moment of inertia of rod 3 about a parallel axis L 2 L M L2 at a distance from it = M = 2 4 2 =

Rigid Body Rotation 8.35

I4 = moment of inertia of rod 4 about a parallel axis L M L2 at a distance from it = . 4 2 I= = 24.

2 ML2 3

M L2 12

M L2 12

M L2 4

linear velocity v and an angular velocity kinetic energy at any time t = translational KE + rotational KE 1 1 or KE = mv2 + I 2 2 2

M L2 4

Here

1 2 moment of inertia of a square sheet ABCD about 1 Is its diagonal AC or It = 2 sheet = M + M = 2M L2 M L2 = Is = (2M) 12 6

the moment of inertia of triangular sheet ABC =

It =

Is M L2 = 2 12

I=

2 MR2 and 5

KE =

1 1 Mv2 + 2 2

=

v R

2 MR2 5

v2 R2

1 1 7 Mv2 + Mv2 = Mv2 2 5 10

7 Mv2 which gives v = 10 Hence the correct choice is (a). or

28.

=

Mgh =

10 g h 7

the total (translational + rotational) kinetic energy respectively are 1 1 1 I 2 and Kt = mv2 + I 2 2 2 2 The moment of inertia of a sphere of mass m and 2 radius r about its centre is I = mr2. Also = 5 2 v2 2 v/r. Therefore I 2 = mr2 = mv2. Thus 2 5 5 r 1 1 1 7 Kr = mv2 and Kt = mv2 + mv2 = mv2. 5 2 5 10 Hence 1 m v2 Kr 2 5 = 7 Kt 7 m v2 10 Hence the correct choice is (d). Kr =

Fig. 8.62

Hence the correct choice is (a). 25. The correct choice is (b) because the angular momentum of the sphere about the point of contact with plane surface also includes the angular momentum about the centre of mass. 26. I 1 = I 2. I3 = I4. From perpendicular axes theoplate about an axis passing through the centre and perpendicular to the plane of the plate is I = I1 + I2 = I3 + I4 = 2I1 = 2I3 ( I 1 = I 2 I 3 = I 4) or I 1 = I 3. Thus I = I 1 + I 2 = I 3 + I 4 = I 1 + I 3. Hence the correct choice is (c). 27. Let the mass of the sphere be M and R its radius. energy is entirely potential given by PE = Mgh

29.

man will meet at their centre of mass. Since their at x = 5 m. Hence the correct choice is (b).

30. Refer to Fig. 8.63. The angular momentum of the mass at point P(x y) about origin O v = m (x i

L = mr

y j)

(v i )

= myv ( k ) (

i

i = 0 and j

i = –k )

8.36 Comprehensive Physics—JEE Advanced

32. Since there is no friction between the sphere and the horizontal surface and also between the spheres momentum from sphere A to sphere B due to the collision. Since the collision is elastic and the A only transfers its linear velocity v to sphere B. Sphere A will continue to rotate with the same angular speed Fig. 8.63

Now m and v are constants. Also y remains constant as the mass moves parallel to the x-axis. Hence L remains constant. Thus the correct choice is (b). 31. Refer to Fig. 8.64. According to the perpendicular about the z-axis is Iz = Ix + Iy

is (c). 33. Refer to Fig. 8.65. Let OC = RC and let vc be the velocity of the centre of mass of the disc. The linear momentum of the centre of mass is pc = Mvc Lc is the angular momentum of the disc about C then the angular momentum about origin O is L0 = Lc + Rc pc

with Ix = Iy. The square plate lies in the x-y plane. Since the directions of the x and y the only restriction being that the angle between will not change if the axes are rotated through any angle in the plane of the plate. This can be proved as follows. Fig. 8.65

Magnitude of L0 = Ic =

+ Rc

1 MR2 2

Mvc sin

+ M Rcvc sin 1 M R2 2

Ic =

Fig. 8.64

Consider a particle of the plate of mass mn located at a point P(xn yn) in the x-y plane. Then IAB = mn y2n The moment of inertia about line CD will be ICD = mn y 2n The coordinates (xn yn) and (xn yn) are related as xn = xn cos – yn sin yn = yn sin – yn cos Now ICD = mn y 2n = mn(xn sin – yn cos )2 = (mn x2n) sin2 + (mn y 2n ) cos2 – 2 (mn xn yn) sin cos =

mn y2n

and mn xn yn = 0. Hence

IAB =

( =

1 MR2 2 Rc sin

R

= R and vc = R )

3 MR2 2

Hence the correct choice is (c). 34. When the block hits the ridge at point O start rotating about an axis passing through O and perpendicular to the plane of the paper (Fig. 8.66).

mn x2n

ICD = IAB (sin2 + cos2 ) + 0 = IAB = I Thus the correct choice is (a).

+ MR

Fig. 8.66

Rigid Body Rotation 8.37

an-gular momentum is conserved. Angular momentum of the block before it hits the ridge is a 1 Li = Mv AC = Mv = Mva (1) 2 2 Angular momentum of the block after it hits the ridge is (2) Lf = IO where IO is the moment of inertia of the block about an axis passing through O and perpendicular to the plane of the block and is the angular speed of rotation of the block. From the parallel axes IO = Ic + Mr

a2 . Hence 2 1 1 2 IO = Ma2 + Ma2 = Ma2 6 2 3 2 Ma2 Lf = 3 a 2

2

+

a 2

2

1 Ma2 6

=

(3)

Li = Lf

3v 2 2 Mva = Ma2 or = 4a 3 3 Hence the correct choice is (a). 35. Let m be the mass of the loop and r its radius. The moment of inertia of the loop about an axis passing through the centre O is (Fig. 8.67)

Fig. 8.67

IO =

1 mr2 2

ertia bout XX is I = IO + mr2 = Hence I=

3 2

-

1 3 mr2 + mr2 = mr2 2 2 m = L and radius r = L/2 . L

M v 2 M (r )2 ML = Mr 2 = 2 r r Hence the correct choice is (a).

Fc = 37.

2

gular momentum L is conserved. As the beads slide

2

C is Ic = and r2 =

36. The entire mass of the liquid can be regarded as being concentrated at the centre of mass of the tube L from the axis of which is at a distance of r = 2 revolution. The force exerted by the liquid at the other end of the tube is the centripetal force of a L mass M revolving in a circle of radius r = . Thus 2

L 2

2

Thus the correct choice is (d).

3 L3 8

2

change. From L = I

will also

correct choice is (b). 38. Torque due to F about A is 1 = FL Since the weight mg acts through the centre of mass of the block (which is at a distance of L/2 from the slide of the block) the torque due to weight mg about A is L 2 = mg 2 The minimum force required to topple the block is obtained when 1 is slightly greater than 2 limit L ( 1)min = 2 or Fmin L = mg 2 mg or Fmin = 2 Hence the correct choice is (c). 39. The magnitude of angular momentum of a rotating body is given by L = I I = conK1 and K2 are the corstant. Hence I1 1 = I2 2 I1 = MK12 and I2 = MK22. Hence MK21 1 = MK22 2 or

K1 = K2

2 1

40. Since the centre of gravity of the rod is at its cenL 1 1 = mg L. Gain in KE = I 2 where 2 2 2 I is the moment of inertia of the rod about an axis

mg

passing through its end and perpendicular to its 1 length which is given by I = mL2. Now gain in 3

8.38 Comprehensive Physics—JEE Advanced

1 2

1 mL2 3

which gives

2

=

mB r = B mA rA

1 = mgL 2

r Using (ii) in (i) and putting B rA have

3g L

41. Given: l = 6R moment of inertia about the given axis is given by I= M

R2 4

l2 3

=M

R2 4

(6 R ) 2 3

R2 =M 4

36 R 2 3

2 = 49 MR 4

area swept by radius vector time taken Assuming that the orbit of the planet is a circle of radius R

42. Areal velocity A =

A= T= A= or

=

Angular momentum

2

2

. Hence

R2 2

= (MR ) = 2 MA

or

rB rA

45. Let L cm be the original length of the spring and k be the spring constant. Then m(L + x1) 21 = kx1

L L

2 2

m(L + x2) x1 x2

1

2

=

2

Given x1 choice (b). 46. L = I

= kx2 x1 x2

(i)

x2 = 5 cm and L

1 I = 2 2

2

=

K 2

2

=2

K=

1.

1 I 2

2

Using

is the

OP = OQ = l/2

2A R2

A

(0, l/4)

Hence the correct choice is (b). 44. IA = mA rA2 and IB = mB rB2 . Hence IB mB rB = (i) IA mA rA Let k be the mass per unit length of the wire. Then the masses of loops A and B are mA = (2 rA)k and mB = (2 rB)k

C (l/8, l/8)

D O

M L/2 0 LM1 = 1 = M1 M 2 2 ( M1 M 2 )

2

3

P

Hence the correct choice is (b). 43. The mass of the rod can be considered to be concentrated at its centre (x = L/2) where x = 0 is the origin. Hence RCM

rB rA

is doubled and K value of L becomes one-fourth. Hence the correct choice is (d). 47. Rod POQ of length l = 100 cm is bent at its midpoint O so that POQ = 90° (see Fig. 8.68).

R2 R2 = 2 / 2 2A

L =I

8=

and

Hence the correct choice is (d).

R T

(ii)

E

(l/4, 0)

Q

Fig. 8.68

The mass of part PO of length l/2 can be taken to be concentrated at its mid-point A whose coordinates l/4) and of part OQ of length l/2 at its midpoint B whose coordinates are (l of mass of these two equal masses is at mid-point C between A and B. The coordinates of C are (l l/8). OC =

(OE ) 2

(CE )2 =

l 8

2

l 8

2

Rigid Body Rotation 8.39

=

l

=

Now v = 2.45 t2 dv d a= 2 . 45t 2 = 4.9t dt dt

100 cm

32 32 2 48. L = mr . For given m and L r2 r the angular momentum L becomes one-fourth. Hence the correct choice is (a). 49. momentum L = (I I is the moment of inertia and is the angular velocity I and hence 1 2 I will also change. Hence 2 the correct choice is (b). 4 4 50. V = r3 or log V = log + 3 log r. 3 3 Differ We have V r r 1 V 1 1 3 or 0 . 5% % V r r 3 V 3 6 2 I = constant or 2 2 5 mr = constant or r = constant (c) kinetic energy

or 2 log r + log

= log c 2 r r

=0

r 1 1 2 % % r 6 3 The negative sign indicates that decreases. Hence the correct choice is (b). 51. The cylinder will topple when the torque mgr equals h (see Fig. 8.69) the torque ma 2 or

2

t=

Ic = I0 + mr2 = =

or 54.

2 gr h

g 2

(

h = 4r)

(i)

1 MR 2 + 32

M 4

R 2

2

3 MR 2 32

Moment of inertia of the shaded portion about 1 3 13 MR 2 – MR 2 = MR2 O is Is = I – Ic = 2 32 32 which is choice (c). 53. From the principle of conservation of angular I0 and 0 are the I0 0 = I moment of inertia and angular velocity when the beads are at the centre of the rod and I and those when the beads are at the ends of the rod. I0 =

ML2 12

I=

ML2 12

ML2 12

or a =

g 9 .8 = = 1 s. 2 4 .9 9 .8

Hence the correct choice is (a). 52. Moment of inertia of complete disc about O is M 1 MR 2. Mass of the cut-out part is m = . I= 4 2 The moment of inertia of the cut-out portion about 1 R 2 = its own centre I0 = mr2 = 1 M 2 2 4 2 1 2 MR because r = R/2. From the parallel axes 32 tion about O is

and

Fig. 8.69

(ii)

mL2 4 0

= =

M

mL2 4

L2 (M + 6m) 12

6m 12

L2

M 0 M 6m

Hence the correct choice is (b).

centre of mass will remain at rest. Hence the correct choice is (a). 55. Area of complete disc = R 2. Area of one quarter 1 R 2. Mass of this sector = M. sector OAB = 4 Mass per unit area of the sector is

8.40 Comprehensive Physics—JEE Advanced

M

4M

2

2

(1)

energy =

Divide the sector into a large number of cylindrical shells. Consider an element of mass dm at a distance r from the centre O and having thickness dr (see Fig. 8.70). Then 2 rdr Fig. 8.70 mo dm = 4 The moment of inertia of the sector about the axis of rotation is

KE =

mo =

R

I=

r 2 dm

0

R /4

R

R

2 m0 4

m0 R 4 8

r 3 dr

0

I=

(2)

1 I 2

2

1 1 mv 2 + I 2 2

or vCM = 57.

MV M m

10 kg 14 ms 10 4 kg

v r (

I=

IA MA = = IB MB 63.

A B

Hence the correct choice is (c).

= 10 ms–1

-

2 g sin 3 -

Hence the correct choice is (c).

celeration is

its angular acceleration is always directed down the plane. Hence the frictional force acts up the inclined plate when the cylinder rolls up or down

where is the inclination of the plane. h on reaching the bottom are given by

-

64. IBC =

m AB 3

.

IAB =

1 2 mr and 2

m BC 3

I =

IHF =

m AB 12

v r

1 2

IEG =

m BC 12

1 mr2 2

1 I 2

v r

2

2

1 mv2 4 61. The kinetic energy of a rolling disc consists of two 1 parts: translational energy = mv2 and rotational 2

2ar h and vs =

Since as > ar choice (b).

-

60. The kinetic energy (which is rotational) is

=

as = g sin

vr =

stant. Hence the correct choice is (c).

1 mr 2 ) 2

1 1 3 mv2 + mv2 = mv2 2 4 4 Hence the correct choice is (c). 4 62. Mass of sphere A MA = R3 A 3 4 2 2 B MB = R3 B IA = MAR2 and IB = 3 5 5 2 MB R

ar =

58. The correct choice is (d). 59.

2

=

MV = (M + m) vCM 1

2

1 2 mr 2

1 1 = mv2 + 2 2

1 MR2 2

(a). 56. The velocity vCM of the centre of mass can be obtained by using the principle of conservation of

.

=

65.

where M

2as h vs > vr

2

ma 2 3

2

4 ma 2 3

2

ma 2 12

2

ma 2 3

HF is the miniMgL

Rigid Body Rotation 8.41

MgL =

1 I 2

1 ML2 2 3

2

2

6g 6g . Now v = L = L 6 gL L L Hence the correct choice is (c). 66. PE at = 60° is Mgh (1 – cos ) where h is the distance between the axis of rotation and the centre of mass of the disc. Thus h = R. Gain in KE when 1 I 2 the disc reaches the equilibrium position = 2 1 3 2 2 2 where I = IC.M. + Mx = MR MR MR2. 2 2 Here x is the distance between the centre of mass x = R. Now PE = KE gives 3 MgR (1 – cos 60°) = MR 2 2 4 2g which gives = 3R 67. Loss in PE = gain in rotational KE. Thus 1 1 mgh = (I + mR 2) 2 = (MR2 + mR2) 2 2 2 1 2 = R (M + m) 2 2 2mgh or = . Hence the correct choice is (c). M m R2 or

=

68. We obtain the given hollow sphere as if a solid sphere of radius R has been removed from a solid sphere of radius 2R. The mass of the given hollow sphere is (here is the density of the material of the sphere) M = M1 – M2 4 4 (2R)3 and M2 = R3 are the 3 3 masses of spheres of radii 2R and R respectively.

where M1 =

M = 28 R3 (i) 3 The moment of inertia of the given hollow sphere is I= =

2 2 M1 (2R)2 – M2 R2 5 5 2 5

4 3

2R

3

2 4 = (32 – 1) 5 3 choice (d).

2R

2

2 5

4 3

R3

R5 62 MR 2 35

Given I2 = or

3 I1. Hence I1 4 =

2

4 3

K1 = K2 = =

R2

1

3 I1 4

=

1 I1 2

1 I2 2

2

2 1

1 2

=

2

4 1 I1 3 2

3I1 4

100 =

4 3

1

2

4 K1 3

2 1

Percentage increase in KE = 4 K1 K1 = 3 K1

2

1.

K2

K1 K1

100

100 = 33.3% 3

Hence the correct choice is (a). 70. The disc is rolling about the point O. Thus the axis of rotation passes through the point A and is perpendicular to the plane of the disc. From the relation v = r where r is the distance of the point vB = (AB) vA vC = (AC) = 2R

and

v B 3R vc 2 choice is (c).

1 2R

Hence

=

=

3R 2

3 . Thus the correct 4

71. Refer to Fig. 8.71. The angular momentum of the particle at point P(x y) about origin O is given by L=mr v =m

xi

yj Fig. 8.71

= – myv k ( (ii)

I =

69. Let I1 and 1 be the moment of inertia and angular frequency when his arms are outstretched and I2 and 2 those when his arms are folded. Then I1 1 = I2 2

i

i = 0 and j

i =– k)

m and velocity v are constant. Also y remain constant as the particle moves parallel to the x-axis. Hence L remains constant. Thus the correct choice is (b).

8.42 Comprehensive Physics—JEE Advanced

M

72. Mass per unit area of the disc =

2

. Therefore,

R mass of the removed portion (hole of radius R/3) is

R 2 M = R 3 9 The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is 1 I= MR2 2 Using the parallel axes theorem, the moment of inertia of the removed portion of the disc about the axis passing through O and perpendicular to the plane of the disc is I = MI of mass m about O + m OO M

m=

2

R 1 m = 3 2 1 M = 2 9

2

2R 3

+m

R2 M + 9 9

2

mR 2 . Hence the object is a 2 disc, which is choice (d). 75. Consider a rod OP of length L lying along the x-axis with O as the origin (Fig. 8.72). Consider a small element AB of length dx at a distance x from O. On solving, we get I =

Fig. 8.72

k (xdx) L The distance of the centre of mass from O is given by L

xCM =

4R2 1 = MR2 9 18

Therefore, the moment of inertia of the remaining 1 MR2 – 2 1 4 MR2 = MR2. Hence the correct choice is (d). 18 9 73. Let M be the mass of the sphere. The mass of the disc will also be M. The moment of inertia of the sphere about its diameter is 2 Is = MR2 5 The moment of inertia of the disc about its edge and perpendicular to its plane is (using parallel axes theorem) Id = Icm + Mh2 = 1 Mr2 + Mr2 = 3 Mr2 2 2 Given Is = Id. Hence, we have portion of the disc about O = I – I =

AB (= dm) = mdx =

Mass of element

=

k x2 d x L0

( d m) x

L

( d m) x3 3

L

0 2 L

x 2

k xd x L0 L3 / 3 2

L /2

0

The correct choice is (b). 76. When the tube AB is rotated about its end A in a horizontal plane with a uniform angular velocity, all points of the tube rotate with same angular velocity. Consider a small element of the liquid of length dr at a distance r from the axis of rotation. The mass of this element is (see Fig. 8.73). M dr m= L

2 3 MR2 = Mr2 5 2 2 r = , which is choice (a). 15 R 74. From the principle of conservation of mechanical energy, we have 1 1 mv2 + I 2 = mgh 2 2 which gives

1 1 mv2 + I 2 2

v R

2

= mg

3v 2 4g

2L 3

Fig. 8.73

Rigid Body Rotation 8.43

Force exerted by the element is

Since no external torque acts on the system, the angular momentum is conserved, i.e. L = L or (I + mr2) = I

M 2 r dr L Total force exerted by the liquid at end B is L M M 2 2 L F= r dr r dr 0 0 L L dF = m r

=

M L

2

r2 2

L

0

2

=

1 M 2

2

or

L ,which is choice (c).

or

=

1 M R2 2

I m r2

I

=

m r2

2 m r2 M R2

1

77. Let be the angular velocity acquired by the system (rod + bullet) immediately after the collision. Since no external torque acts, the angular momentum of the system is conserved. Thus mvL = I (1)

1 M R2 2

Putting r = R 79. The block will just begin to slide if the downward force mg sin just overcomes the frictional force, i.e. if mg sin = μN = μ mg cos tan = μ = 3 = 60° [see Fig. 8.74] The block will topple if the torque due to normal reaction N about O just exceeds the torque due to mg sin about 0, i.e.

where I is the moment of inertia of the system about an axis passing through O and perpendicular to the rod. Thus I = M.I. of rod about O + M.I. of bullet stuck at its lower end about O 1 1 ML2 + mL2 = (M + 3m)L2 (2) = 3 3 Using Eq. (1) in Eq. (2), we have 1 m vL = (M + 3m) L2 3 3m v or = L ( M 3m ) Hence the correct choice is (c).

N mg cos

OA = mg sin 5 cm = mg sin

2 3 Since for toppling is less than correct choice is (b). tan

78. The initial angular momentum of the rotating record is L =I 1 where I = MR2. 2 Let be the angular velocity of the record when the coin of mass m is placed on it at a distance r from its centre. The angular momentum of the system becomes L = (I + mr2)

=

OB 15 cm 2 34° for sliding, the

Fig. 8.74

II Multiple Choice Questions with One or More Choice Correct 1. In the HCl molecule, the separation between the nuclei of hydrogen and chlorine atoms is 1.27 Å. If the mass of a chlorine atom is 35.5 times that of a hydrogen atom, the centre of mass of the HCl molecule is at a distance of

35.5 1.27 Å from the hydrogen atom 36.5 35.5 1.27 Å from the chlorine atom (b) 36.5 (a)

8.44 Comprehensive Physics—JEE Advanced

1.27 Å from the hydrogen atom 36.5 1.27 (d) Å from the chlorine atom 36.5 2. Choose the correct statements from the following: (a) The position of the centre of mass of a system of particles does not depend upon the internal forces between particles. (b) The centre of mass of a solid may lie outside the body of the solid. (c) A body tied to a string is whirled in a circle with a uniform speed. If the string is suddenly cut, the angular momentum of the body will not change from its initial value. (d) The angular momentum of a comet revolving around a massive star, remains constant over the entire orbit. 3. Which of the following statements are correct? (a) When a body rolls on a surface, the force of friction acts in the same direction as the direction of motion of the centre of mass of the body. (b) During rolling, the instantaneous speed of the point of contact is zero. (c) During rolling, the instantaneous acceleration of the point of contact is zero. (d) A wheel moving down a perfectly frictionless inclined plane will slip and not roll on the plane. 4. In which of the following is the angular momentum conserved? (a) The planet Neptune moves in an elliptical orbit round the sun with the sun at one of the foci of the ellipse (c)

elliptical orbit round the nucleus (c) An -particle, approaching a nucleus, is scattered by the force of electrostatic repulsion between the two (d) A boy whirls a stone, tied to a string, in a horizontal circle 5. Four tiny masses are connected by a rod of negligible mass as shown in Fig. 8.75. (a) The moment of inertia of the system about axis AB is 50 ma2 (b) The radius of gyration od the system about axis AB is 5 a. (c) The moment of inertia of the system about axis CD is 10 ma2 (d) The radius of gyration of the system about axis CD is a.

Fig. 8.75

6. The moment of inertia of a uniform circular disc of mass M and radius R about its centre and normal 1 to its plane is MR2. Then 2 (a) its radius of gyration about the centre is R. (b) the moment of inertia of the disc about its di1 ameter is MR2. 4 (c) the moment of inertia of the disc about an axis passing through a point on its edge and normal 3 to the disc is MR2. 2 (d) the moment of inertia of the disc about a tangent 5 in the plane of the disc is MR2. 4 7. A molecule consists of two atoms, each of mass m, separated by a distance a. The rotational kinetic energy of the molecule is K and its angular frequency is . I is the moment of inertia of the molecule about its centre of mass. Then 1 (b) I = ma2 (a) I = ma2 2 1 K 2 K (d) = a m a m 8. A circular ring of mass m and radius r rolls down an inclined plane of height h. When it reaches the bottom of the plane its angular velocity is and its rotational kinetic energy is K. Then 1 1 (b) = gh gh (a) = r 2r 1 (c) K = mgh (d) K = mgh 2 9. A rope is wound round a solid cylinder of mass M and radius R. If the rope is pulled with a force F, the cylinder acquires an anglular acceleration and the rope acquires a linear acceleration a. Then F 2F (a) = (b) = MR MR 2F F (c) a = (d) a = M M (c)

=

Rigid Body Rotation 8.45

10. A solid sphere rotating about its diameter at an angular frequency has rotational kinetic energy K. 1 When it is cooled so that its radius reduces to of n its original value, the new values of and K become and K respectively. Then (a)

=n

(b)

= n2

K K =n (d) = n2 K K 11. Three forces act on a wheel of radius 20 cm as shown in Fig. 8.76. (c)

(c) Frictional force = mg cos , where m is the mass of the cylinder and friction between the cylinder and the plane. (d) Frictional force helps rotational motion of the cylinder but opposes its translational motion. 13. A spherical ball is released from rest from point A on a hemispherical surface and it rises up to a point C as shown in Fig. 8.78. Part AB of the surface is rough and the ball rolls from A to B without slipping. Part BC of the surface is frictionless. KA, KB and KC are kinetic energies of the ball at points A, B and C respectively. Which of the following is/are correct? (b) KB > KA, hA < hC (a) KA = KC, hA = hC (c) KB > KC, hA > hC (d) KC > KA, hA > hC

Fig. 8.78

Fig. 8.76

(a) The torque produced by the force of 8N is 0.8 Nm clockwise. (b) The torque produced by the force of 4 N is 0.8 Nm anticlockwise. (c) The torque produced by the force of 9 N is zero. (d) The net torque produced by the forces is 1.8 Nm clockwise. 12. A solid cylinder rolls down a rough inclined plane without slipping as shown in Fig. 8.77. Choose the correct statement (s) from the following.

Fig. 8.77

(a) If is decreased, the force of friction will decrease. (b) Frictional force is dissipative.

14. The moment of inertia of a uniform disc about its diameter is I. Then the moment of inertia about an axis (a) passing through its centre and perpendicular to its plane is 2I. (b) tangential to its plane is 5I. (c) tangential and perpendicular to its plane is 6I. (d) all the above choices are correct. 15. A wire of mass M, length L and density is bent to form a circular ring of radius R. Then, the moment of inertia of the ring about its diameter is 1 MR2 (b) R4L /2 (a) 2 ML2

R 2 L3 8 2 8 16. A uniform solid sphere and a solid cylinder of the same mass and the same diameter are released from rest on the top of an inclined plane of inclination . If they roll down the plane without slipping, then (a) the acceleration of each down the plane is g sin . (b) the ratio of the accelerations of the sphere and the cylinder is 15:14. (c) the ratio of the times taken by the sphere and the cylinder to reach the bottom of the plane is 14 : 15 . (d) the sphere and the cylinder will reach the bottom with the same speed. (c)

(d)

8.46 Comprehensive Physics—JEE Advanced

17. A rope is wound around a hollow cylinder of mass 3 kg and radius 20 cm. The rope is pulled with a constant force of 30 N. If is the angular acceleration of the cylinder and a the linear acceleration of the rope, then (b) = 40 rad s–2 (a) = 50 rad s–2 –2 (c) a = 30 ms (d) a = 10 ms–2 18. A rope of negligible mass is wound around the circumference of a bicycle wheel (without tyre) of diameter 1 m. A mass of 2 kg is attached to the end of the rope and is allowed to fall from rest. The mass falls 2 m is 4 s. The axle of the wheel is horizontal and the wheel rotates in the vertical plane. Take g = 10 ms–2 and neglect the friction due to air. (a) the linear acceleration of the wheel is 0.25 m s–2. (b) the angular acceleration of the wheel is 0.5 rad s–2. (c) the magnitude of the torque acting on the wheel is 10 Nm. (d) the moment of inertia of the wheel about the horizontal axle is 20 kg m2. 19. A smooth sphere A is moving on a horizontal frictionless surface with angular speed and centre of mass velocity v. It collides head-on with an identical sphere B at rest. After the collision their angular speeds are A and B respectively. If the collision is elastic and the friction is neglected, then (b) B = 0 (a) A = (d) A = B (c) A < B IIT, 1999 20. The torque acting on a body about a given point is given by

= A

L where A is a constant vector

and L is the angular momentum of the body about that point. It follows that dL is perpendicular to L at all instants of (a) dt time. (b) the component of L in the direction of A does not change with time. (c) the magnitude of L does not change with time. (d) All the above choices are correct. IIT, 1998 21. A uniform bar of length 6a and mass 8 m lies on a horizontal frictionless table. Two point masses m and 2m moving in opposite directions but in the same horizontal plane with speeds 2v and v respectively strike the bar at distance a and 2a from one end and stick to the bar after the collision. Then after the collision

(a) the velocity of centre of mass is zero. (b) the angular speed of the bar with the masses v . stuck to it is 5a (c) the moment of inertia of the bar with masses stuck to it about the axis passing through the end of the bar and perpendicular to its plane is 30 ma2. 3 (d) the total energy of the bar mv2. 5 IIT, 1991 22. A disc of mass M and radius R is rolling with angular speed on a horizontal surface as shown in Fig. 8.79. The magnitude of angular momentum of the disc about the origin O is (here v is the linear velocity of the disc)

Fig. 8.79

(a)

3 MR2 2

(c) MRv

(b) MR2 (d)

3 MRv 2

IIT, 1999 23. Which of the following statements is/are correct about a particle moving in a circle with a constant speed? (a) The linear velocity and acceleration vectors are perpendicular to each other. (b) The linear velocity vector is always perpendicular to the angular velocity vector. (c) The force acting on the particle is radial. (d) The force does no work on the particle. 24. The position vector of a particle with respect toorigin O is r . If the torque acting on the particle is zero, then (a) the linear momentum of the particle remains constant. (b) the angular momentum of the particle about O remains constant. (c) the force applied to the particle is perpendicular to r . (d) the force applied to the particle is parallel to r.

Rigid Body Rotation 8.47

25. A block of mass m is connected to a spring of spring constant k r as shown in Fig. 8.80. The mass of the pulley is 2m. The block is pulled down by x0 from the equilibrium position and released. The spring has negligible mass. Then 1 2 (a) the total energy of the system is kx0 . 2 (b) the velocity of the block when it is at a distance x from the equilibrium position is 1/ 2 k . x02 x 2 2m (c) the velocity of the block is maximum when x = x 0. (d) the velocity of the block is zero when x = x0.

Fig. 8.80

26. A rod of mass M = 0.9 kg and length L = 1 m is suspended at O as shown in Fig. 8.81. A bullet of mass m = 100 g moving with velocity v = 80 ms–1 in the horizontal direction strikes the end P of the rod and gets embedded in it. If I is the moment of inertia of the system and is the angular velocity immediately after the collision, then (a) I = 0.4 kgm2 (b) I = 0.3 kgm2 (c) = 20 rad s–1 (d) = 26.7 rad s–1

Fig. 8.81

27. A wheel is initially at rest. A constant torque acts on it for a time t. As a result, the wheel acquires an angular acceleration and angular velocity . If the angular displacement produced is , then (a) t2 (b) = constant (c) t (d) power 28. A string of length L and of negligible mass hangs from a support O. The other end of the string carries

a mass m which is moved in a horizontal circle of radius R to form a conical pendulum. If the string makes an angle = 60° with the vertical, then (a) the speed of the body along the circle is v = Rg . (b) the tension in the string is 2mg. (c) the horizontal component of the angular mom3 Lg mL . entum of the body point O is 2 2 (d) the magnitude of the torque acting on the 3 mgL. 2 29. A solid cylindrical roller of mass M and radius R is rolled on a rough horizontal surface by applying a horizontal force F. If aCM is the linear acceleration of the centre of mass and f is the frictional force between the roller and the surface, then F 2F (b) aCM = (a) aCM = M 3M F (c) f = (d) f = zero 3 30. If the resultant of all the external forces acting on a system of particles is zero, then for an inertial frame, one can surely say that (a) linear momentum of the system does not change in time (b) kinetic energy of the system does not change in time (c) angular momentum of the system does not change in time (d) potential energy of the system does not change in time IIT, 2009 31. horizontal plane surface. In Fig. 8.82, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then, body about point O is

(a) VC

VA = 2 V B

(b) VC

VB = VB

VC VA

(c) VC

VA

= 2 VB

(d) VC

VA

= VB

VC

Fig. 8.82

IIT, 2009

8.48 Comprehensive Physics—JEE Advanced

32. A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision [See Fig. 8.83] (a) the ring has pure rotation about its stationary CM. (b) the ring comes to a complete stop. (c) friction between the ring and the ground is to the left.

(d) there is no friction between the ring and the ground.

Fig. 8.83

IIT, 2011

ANSWERS AND SOLUTIONS

1. Since, most of the mass of an atom is concentrated in its nucleus and the size of a nucleus (which is of the order of 10–15 m) is very small compared to the separation between the hydrogen and chlorine atom (which is 1.27 10–10 m), the atoms can be treated as point masses. The centre of mass of HCl molecule will be on the line joining the two atoms. Let us say that the H atom is located at, say, x = 0 and the Cl atom at x = x. Let xCM be the position of the centre of mass between x = 0 and x = x. If m1 and m2 are the masses of H and Cl of centre of mass, we have m 0 m2 x m2 x xCM = 1 = m1 m2 m1 m2 Now x = separation between H and Cl atoms = 1.27 Å and m2 = 35.5 m1. Hence 35.5 m1 1.27 Å xCM = = 35.5 1.27Å m1 35.5 m1 36.5 This gives the distance of the centre of mass from the hydrogen atom. The distance of the centre of mass from the chlorine atom is 1.27 Å – 35.5 1.27 Å = 1.27Å 36.5 36.5 Hence the correct choices are (a) and (d). 2. The only incorrect statement is (c). Since no external torque acts on the body even after the string is cut, the angular momentum will remain unchanged. 3. Statement (a) is correct. The direction of motion of the centre of mass is the direction along which the body rolls. Since the force of friction is opposite to the direction of the velocity of the point of contact, the force of friction acts in the direction of motion of the centre of mass. Statement (b) is also correct. At each instant of time, the point of contact is momentarily at rest. Statement (c) is incorrect. Since the body is rotating while it is rolling, the direction of the velocity is changing with time. Hence the

acceleration of the point of contact is not zero. Statement (d) is correct. Rolling cannot take place in the absence of friction because it is the frictional force that provides the necessary torque which makes the body roll on a surface. Hence the correct choices are (a), (b) and (d). 4. The angular momentum is conserved in the four cases. The object, in each case, is moving under the action of a central (radial) force. The torque dL ; the due to a radial force is zero. Since = dt angular momentum L does not change with time. 5. All four choices are correct. The moment of inertia about AB is I = m1r 21 + m2r 22 + m3r23 + m4r24 = m 0 + 2m (a)2 + 3m (2a)2 + 4m (3a)2 = 0 + 2 ma2 + 12ma2 + 36ma2 = 50ma2 The radius of gyration K is given by I = MK2 = (m1 + m2 + m3 + m4)K2 = (m + 2m + 3m + 4m)K2 = 10mK2 I = 50ma 2 = 5a 10m 10m The moment of inertia about CD is I = m (2a)2 + 2 m (a)2 + 3m 0 + 4 m = 4 ma2 + 2ma2 + 0 + 4ma2 = 10ma2 The radius of gyration is or

K=

(a)2

I 10ma 2 = =a 10ma 10m 6. The correct choices are (b), (c) and (d). Let us consider two perpendicular diameters, one along the x-axis and the other along the y-axis. Then 1 MR2 Ix = Iy = 4 According to the perpendicular axes theorem, the moment of inertia of the disc about an axis passing through the centre is =

Rigid Body Rotation 8.49

1 1 MR2 + MR2 4 4 1 = MR2 2

Ic = Ix + Iy =

which is choice (b). Since the disc is uniform, its centre of mass coincides with its centre. Therefore, the moment of inertia of the disc about an axis passing through its centre of mass and normal to its plane is 1 MR 2 ICM = IC = 2 According to the theorem of parallel axes, the moment of inertia of the disc about an axis passing through a point on its edge and normal to its plane is given by 1 MR 2 + MR2 Ie = ICM + M h2 = 2 ( h = R) 3 2 = MR . 2 From the parallel axes theorem, the moment of in1 ertia of the disc about a tangent is MR2 + MR2 = 4 5 2 MR . 4 7. The correct choices are (b) and (d). Since the two atoms have the same mass, the centre of mass is at a distance of a/2 from each atom. Therefore, the moment of inertia of the molecule about its centre of mass is

Kinetic energy is k = = 8.

mgh =

=

2

a 2

I=m

1 I 2

2

2

+

ma 2

2

=

ma 2 2

, which gives 2k

2k = I

1 1 mv2 + I 2 2 1 2 mr 2

a 2

=m

=

1 2 mr 2

2

2 2 = 1 a

k m

1 1 mv2 + (mr2) 2 2 2 ( I = mr2) 2

(

= mr2 2 1 gh . Also = r 1 2 1 gh 1 I = (mr2) mgh 2 = 2 r 2 2 Hence the correct choices are (a) and (d). 1 9. I = MR2 and torque = FR = I . Hence 2 K=

FR FR 2F = = 1 2 I MR MR 2 2F Also a= R= M Thus the correct choices are (b) and (c). 10. The correct choices are (b) and (d). r 2 I 2 2 mr2, I = m = 2. I =I ,I= n n 5 5 I Hence I = 2 n or = n2 I 1 1 1 K = I 2, K = I 2 = (n2 )2 2 n 2 2 2 1 2 2 = I n = n2K. 2 11. The torques produced by forces 8 N, 4 N and 9 N respectively are 1 = 8 N 0.2 m sin 30° = 0.8 Nm (clockwise) 2 = 4 N 0.2 m sin 90° = 0.8 Nm (anticlockwise) 3 = 9 N 0.2 m sin 90° = 1.8 Nm (clockwise) Net torque = 0.8 – 0.8 + 1.8 = 1.8 Nm clockwise Hence the correct choices are (a), (b) and (d). 12. Refer to the Fig. 8.84. Here f is the frictional force. The linear acceleration of the centre of mass of the rolling cylinder is given by g sin (1) acm = I cm 1 mR 2 where R is the radius of the cylinder and Icm is the moment of inertia of the cylinder about the centre of mass which is given by 1 Icm = mR2 2 =

v=r ) Fig. 8.84

Using this in Eq. (1), we have g sin 2 g sin acm = mR 2 = 1 3 2mR 2 Now, for linear motion, we have mgsin – f = macm

(2)

(3)

8.50 Comprehensive Physics—JEE Advanced

Using Eq. (2) in Eq. (3), we get 2mg sin mgsin – f = 3 mg sin which gives f = (4) 3 It follows from Eq. (4) that if is decreased, f will also decrease. Hence choice (a) is correct. As the cylinder is rolling down, the point of application of the frictional force is at rest at any given instant. Hence no work is done by the frictional force, i.e. the frictional force is not dissipative. Therefore, statement (b) is wrong. Statement (c) is also wrong as f is given by Eq. (4). Statement (d) is correct because the frictional force provides torque = f R to help rotational motion but it will oppose translational motion. Hence the correct choice are (a) and (d). 13. The ball is at rest at point A. Hence its kinetic energy KA (rotational + translational) is zero, it has only gravitational potential energy mghA. As it is released at A, it begins to roll (due to friction in part AB) thus acquiring kinetic energy at the expense of gravitational potential energy. When it reaches point B, its kinetic energy consists of both rotational and translational energy. At point B its potential energy 1 is zero. At point B, the rotational kinetic energy is 2 I 2, where is the angular velocity at point B and I is the moment of inertia of the ball about its centre. Since part BC is frictionless, the torque on the ball is zero. Hence its angular momentum L = I remains constant in part BC. Hence, the angular velocity of the ball remains constant in part BC. The translational kinetic energy is converted into gravitational potential energy. At point C, the translational kinetic energy of the ball is zero; it has rotational kinetic 1 energy I 2 and gravitational potential energy 2 mg hC. Thus KA = 0 (1) 1 1 KB = mv2 + I 2 (2) 2 2 where v is the linear velocity of the centre of mass of the ball and 1 KC = I 2 (3) 2 It follows from (1), (2) and (3) that KB > KC > KA. Now total energy at A is EA = 0 + mghA = mghA and 1 I 2 + mg hC. From the total energy at C is EC = 2

law of conservation of energy EA = EC, i.e. mg hA 1 1 = I 2 + mg hC or mg (hA – hC) = I 2. Since the 2 2 right hand side of this equation is positive, hA > hC. Hence the correct choices are (c) and (d). 14. From perpendicular axes theorem, Ix + Iy = Ic. Hence Ic = I + I = 2 I. [see Fig. 8.85(a)] 1 we knows that Ic = MR2 MR2 = 2Ic = 4I. 2

Fig. 8.85

From parallel axes theorem [see Fig. 8.84 (b)], we have IAB = Ix + MR2 = I + 4 I = 5 I. Using parallel axes theorem [see Fig. 8.84 (c)], we have IPQ = Ic + MR2 = 2I + 4I = 6 I. Hence the correct choice is (d). M M = R2L (1) 15. = R2 L L L =2 R R= (2) 2 The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane = MR2. From the parallel axes theorem, the moment of inertia of the ring about its diameter = 1 MR2 2 four choices are correct. g sin 16. Acceleration a = I 1 MR 2

Rigid Body Rotation 8.51

If s is the distance along the inclined plane, the velocity when the body reaches the bottom of the plane is given by v2 – 0 = 2as v = 2as and the time taken to reach the bottom is 1/ 2 I 2s 1 MR 2 t = 2s = g sin a For cylinder: For sphere: Hence giving

1 MR2 2 2 Is = MR2 5 2 g sin ac = 3 as 15 = ac 14 Ic =

dL dL ; L = 0. Hence dt dt

dL = 0 L = constant. Hence choice dt dL (c) is correct. Since A , choice (b) is also cordt rect. Thus the correct choice is (d). 21. Since no external force is applied, the linear momentum is conserved. Hence L

and as =

5 g sin 7

vs as t a = and s = c vc ac tc as Thus the correct choices are (b) and (c). 17. I = MR2 and = FR. FR F 30 = = = 50 rad s–2 = 2 = MR I MR 3 0.2 a = R = 0.2 50 = 10 ms–2 The correct choices are (a) and (d). 1 1 2=0+ a (4)2 18. s = ut + at2 2 2 a = 0.25 ms–2. a 0.25 = = 0.5 rad s–2 = R 0.5 = mg moment arm = 2 10 0.5 = 10 Nm 10 I= = = 20 kg m2 0.5 Hence all the four choices are correct. 19. Since the collision is elastic and head-on and the spheres have the same mass, they will exchange their velocity after the collision, i.e. A comes to rest and B moves with velocity v. Since there is no friction, the torque on each sphere is zero. Hence their angular speeds remain unchanged on collision. Thus the correct choices are (a) and (b). (1) 20. Given = A L dL dL . Hence = A L . This = dt dt

dL is perpendicular to both L and A . dt Hence choice (a) is correct. Now L L = L2, where L is the magnitude of L . means that

dL dL dL L =2L + dt dt dt dL dL 2L =2L dt dt

L

Since L

Also

We know that

Differentiating, we have d d 2 (L L) = (L ) dt dt

dL =0 dt

(8 m + m + 2 m)vCM = 2m(– v) + m(2v) + 8m

0

where vCM is the velocity of the centre of mass. This gives vCM = 0. The moment of inertia of the system is 1 8m (6a)2 I = 2ma2 + m(2a)2 + 12 = 2ma2 + 4ma2 + 24ma2 = 30 ma2 Since no external torque is applied, the angular momentum of the system is conserved. If is the angular speed, then 2 mv a + m 2v 2a = I 2 v 6mva = 30ma2 2 = 5a The system has no translational K.E. Hence Total K.E. = K.E. of rotation =

1 I 2

2

1 v 2 = 3 mv2 30 ma2 2 5 5a Hence all the four choices are correct. 22. The angular momentum about O is =

LO = LCM + M R v Its magnitude is

R v and LCM = I

LO = I + MRv 1 MR 2 + MR = 2 =

3 MR2 2

R

(

v=R )

8.52 Comprehensive Physics—JEE Advanced

v 3 3 MR2 = MRv R 2 2 Hence the correct choices are (a) and (d). 23. All the four choices are correct. The linear velocity v , angular velocity and radius vector r are related as r v = =

Hence v is perpendicular to both and r . Hence choice (b) is correct. The acceleration (and hence force) is centripetal (towards the centre) and v is r . Hence choice (a) and (c) are correct. Since F is r , the work done is zero. Hence choice (d) is also correct. dL 24. = r F and = . Hence the correct choices dt are (b) and (d). 1 25. Total energy before the block is released = kx20. 2 After the block is released, the total energy when it is at a distance x from the equilibrium position = K.E. of block + rotational K.E. of pulley + P.E. 1 1 1 stored in the spring = mv2 + Ip 2 + kx2. 2 2 2 1 1 m pr 2 = (2m)r2 = mr2. 2 2 From the principle of conservation of energy, we have ( v = r ) v 2 1 2 1 2 1 1 kx0 = mv2 + (mr2) + kx r 2 2 2 2 where Ip =

1/ 2

k . x02 x 2 2m Thus v is maximum when x = 0. Hence the correct choices are (a), (b) and (d). which gives v =

1 1 26. I = mL + ML2 = 0.1 (1)2 + 0.9 (1)2 3 3 = 0.4 kgm2 From conservation of angular momentum, we have mvL = I mv L 0.1 80 1 = = = 20 rad s–1. I 0.4 The correct choices are (a) and (c). = 0 + t = t. Since is constant 27. = 0+ t

28. Refer to Fig. 8.86. OA = OB = OC = L R = l sin , OD = L cos

Fig. 8.86

T sin

= mg

(1)

mv 2 R Dividing (1) by (2), we get T cos

=

(2)

v = Rg tan = Rg tan 60 = 1.73Rg Hence choice (a) is incorrect. From (1) and (2) we get T 2 = (mg)2 + = (mg)2 +

m2

(v2)2

R2 m2

(Rg tan 60°)2 R2 = (mg)2 + 3(mg)2 = 4(mg)2 T = 2 mg Horizontal component of angular momentum about O = mv OD = mv L cos Rg tan

=m

L cos

= mL cos

Lg tan sin

= mL sin

Lg cos

2

and = I , it follows that is constant. Hence t. 1 2 t . Hence t2. Also power P = . Also = 2 Hence P . Thus all the four choices are correct.

= mL sin 60°

Lg cos 60

3 Lg mL , which is choice (c). 2 2 Torque about O = mgR = mgL sin =

3 mgL 2 Hence the correct choices are (b), (c) and (d). 29. Since the frictional force acts opposite to the direction of motion, the equation of translational motion is = mgL sin 60° =

F – f = MaCM

(1)

Rigid Body Rotation 8.53

For rotational motion, we have =I Since the torque is due to frictional force, = f R. a CM 1 MR 2 Hence fR = 2 R aCM 1 I MR 2 , 2 R f=

or

MaCM 2

VB

V A = Vi

Hence the correct choice are (b) and (c) 32. Let M be the mass of the ring and m that of the ball and let V and v be their velocity before collision. The initial momentum of the system (ring and ball) in the horizontal direction is

(2)

2F F and f = . 3M 3 So the correct choices are (b) and (c).

Equations (1) and (2) give aCM =

pi = MV

mv

=2

1 + 0.1

(–20)

=2 – 2 = 0 tum of the system pf = 0 in the horizontal direction. Hence Vcm = 0 for the ring, i.e. the ring has pure rotation about its centre of mass. So choice (a) is correct. The total initial angular momentum of the system about the point of collision is Li = mvr – I

30. The only correct choice is (a) 31. Refer to Fig. 8.87.

V R = mvr – MRV

= mvr – MR2 Fig. 8.87

= 0.1

20

0.75 – 2 2

VA = 0, Let V B = V i , then V C = 2V i

= 1.5 – 1 = 0.5 kg m s

0.5

1

–1

V C – V A = 2V i – 0 = 2V i

From the conservation of angular momentum, the

V C – V B = 2V i V i = V i

the friction between the ring and the ground is to the left. So the correct choices are (a) and (c).

VB

VC = Vi

III Multiple Choice Questions Based on Passage Questions 1 to 2 are based on the following passage Passage I Quantization In physics some measurable quantities are quantized. A physical quantity is said to be quantized if it can have only discrete (not continuous) values. Some examples of quantities which are quantized are mass, charge and energy. For sub-atomic particles (called fundamental particles) the angular momentum is quantized. Fundamental particles such as electrons and protons have a certain intrinsic angular momentum of their own. This angular momentum is called spin angular momentum. The spin

angular momentum of a fundamental particle is quantized and its value is given by h S=n 2 where h is the Planck’s constant = 6.63 10–34 Js and n is a number called the spin quantum number. The value of n for electrons, protons, positrons and antiprotons can 1 1 be + and – . Pions have n = 0. 2 2 1. Which of the following is not quantized ? (a) Mass (b) Energy (c) Linear momentum (d) Charge

8.54 Comprehensive Physics—JEE Advanced

2. Choose the correct statements from the following. (a) Electric charge on a charged body can only be an integral multiple of the smallest possible charge. (b) Energy can have only discrete values. (c) The spin angular momentum of an electron

can be +

h h or – . 4 4

(d) The spin angular momentum of a pion is +

h h or – . 2 2

ANSWERS 1. The correct choice is (c). 2. The correct choices are (a), (b) and (c). The spin angular momentum of a pion is zero. Questions 3 to 5 are based on the following passage Passage II A hollow sphere of mass M and radius R is initially at rest on a horizontal rough surface. It moves under the action of a constant horizontal force F as shown in Fig. 8.88.

Fig. 8.88

3. The frictional force between the sphere and the surface

(a) retards the motion of the sphere (b) makes the sphere move faster (c) has no effect on the motion of the sphere (d) is independent of the velocity of the sphere. 4. The linear acceleration of the sphere is 10 F 7F (b) a = (a) a = 7M 5M 6F F (c) a = (d) a = 5M M 5. The frictional force between the sphere and the surface is F F (b) (a) 2 3 F F (c) (d) 4 5

SOLUTIONS 3. If the horizontal force F is applied at the centre of mass of the sphere, then the frictional force opposes the translational motion of the sphere. If force F is applied above the centre of mass, the torque due to frictional force tends to rotate the sphere faster. Hence, in this case, frictional force f acts in the direction of motion, as shown in Fig. 8.89. Thus the correct choice is (b).

Fig. 8.89

4. Let a and be the linear and angular accelerations of the sphere respectively. For translational motion,

F + f = Ma (1) The magnitude of the net torque acting on the sphere = FR – f R. Hence, for rotational motion the equation is Ia ( a = R) FR – f R = I = R 2 For a hollow sphere, I = MR2. Hence 3 2 a 2 FR – f R = MR2 = MRa 3 R 3 2 F – f = Ma (2) 3 6F , which is choice Equations (1) and (2) give a = 5M (c) Ma F 5. From Eqs. (1) and (2) we get f = = . Hence 6 5 the correct choice is (d).

Rigid Body Rotation 8.55

Questions 6 to 11 are based on the following passage Passage III Four solid spheres each of mass m and radius r are located with their centres on four corners of a square ABCD of side a as shown in Fig. 8.90.

8. The moment of inertia of sphere B about side AD is 2 2 m mr (b) (5a2 + 2r2) (a) 5 5 m m (c) (2r2 + a2) (d) (3a2 + 5r2) 5 5 9. The moment of inertia of sphere D about side AD is 2 2 2 mr (b) m(r2 + a2) 5 3 m m (c) (3a2 + 4r2) (d) (5a2 + 2r2) 5 5 10. The moment of inertia of the system of four spheres about diagonal AB is m m (8r2 + 5a2) (b) (7r2 + 4a2) (a) 5 5 m m (5r2 + 8a2) (d) (3r2 + 5a2) (c) 5 5 11. The moment of inertia of the system of four spheres about side AD is 2m m (2r2 + 5a2) (b) (7r2 + 5a2) (a) 5 5 (a)

Fig. 8.90

6. The moment of inertia of sphere A about diagonal AB is 2 2 2 2 mr (b) mr (a) 3 5 a2 a2 (d) m r2 4 4 7. The moment of inertia of sphere C about diagonal AB is 2 2 2 (a) mr (b) (2r2 + 3a2) 5 5 (c) m

(c)

r2

m (5r2 + 3a2) 5

(d)

m (4r2 + 5a2) 10

(c)

2m (4r2 + 5a2) 5

(d)

m (3r2 + 5a2) 5

SOLUTION Refer to Fig. 8.91.

2 mr2 each. The distance 5 of this axis (shown by broken lines) from the diagonal AB = a/ 2 . From the parallel axes theorem, the moment of inertia of spheres C and D about diagonal AB is a 2 2 2 2 m a2 mr2 + m(CO)2 = mr2 + m = mr2 + 2 2 5 5 5 6. The correct choice is (b). 7. The correct choice is (d). 8. The correct choice is (b). 9. The correct choice is (a). 10. The moment of inertia of the system of four spheres about diagonal AB is IAB = MI of A about AB + MI of B about AB + MI of C about AB + MI of D about AB 2 2 2 1 m r2 m r2 m a2 = m r2 5 5 5 2 2 1 + mr 2 ma 2 5 2

their centre and parallel to AB =

Fig. 8.91

The moment of inertia of spheres A and B about their 2 mr2 each. Also the moment of common diameter AB = 5 inertia of spheres C and D about an axis passing through

8.56 Comprehensive Physics—JEE Advanced

IAD = MI of A about AD + MI of D about AD + MI of B about AD + MI of C about AD

2

8r 8 a2 = mr2 + ma2 = m 5 5 The correct choice is (a). 11. Moment of inertia of sphere A about side AD = moment of inertia of sphere D about side AD = 2 mr2. Using the parallel axes theorem, moment of 5 inertia of sphere C about AD = moment of inertia 2 of sphere B about AD = mr2 + ma2. Hence the 5 moment of inertia of the system of four spheres about side AD is Questions 12 to 15 are based on the following passage Passage IV A small sphere rolls down without slipping from the top of a track in a vertical plane. The track has a elevated section and a horizontal part. The horizontal part is 1.0 m above the ground and the top of the track is 2.4 m above the ground. (See Fig. 8.92)

Fig. 8.92

12. If g = 10 ms , the horizontal velocity when the sphere reaches point A is (b) 2 5 ms–1 (a) 5 ms–1 (c) 7 ms–1 (d) 2 7 ms–1 –2 13. If g = 10 ms , the time taken by the sphere to fall through h = 1.0 m is –2

=

2 m r2 5

2 m r2 5

2 m r 2 m a2 5 2 m r 2 m a2 5

=

8 mr2 + 2 ma2 = m 5

8r2 5

2 a2

The correct choice is (c). 1

s

(b)

2

s 5 5 (c) 0.1 s (d) 0.2 s –2 14. If g = 10 ms , the distance on the ground with respect to point B (which is vertically below the end A of the track) is (a) 1.0 m (b) 1.4 m (c) 2.0 m (d) 2.8 m 15. Choose the correct statement/statements from the following. (a) During its motion as a projectile after the sphere leaves the track at A, it will stop rotating. (b) During its motion as a projectile after point A, the sphere will continue to rotate about its centre of mass. (c) Due to rotation, the horizontal range of the sphere will be less than that found in Q.14 above. (d) The rotation of the sphere has no effect on the horizontal range found in Q.14. (a)

SOLUTION 12. The loss in potential energy when the sphere moves from the top of the track to point A = gain in total kinetic energy (translational and rotation), i.e. 1 1 Mv2 + I 2 2 2 2 v where I = MR2 and = . Thus 5 R 1 1 2 Mg(H – h) = Mv2 + MR2 2 2 5

= or

Mg(H – h) =

v = 10( H

2

h) g

1/ 2

7

10 (2.4 1.0) 10 1 / 2 =2 5 ms–1 7 The correct choice is (b). 13. Since the vertical component of velocity is zero, the =

v R

1 1 7 Mv2 + Mv2 = Mv2 2 5 10

Rigid Body Rotation 8.57

2h = g

1 2 1.0 = s, which is choice (a). 5 10 1 14. Horizontal range = vt = 2 5 = 2.0 m, which 5 is choice (c). 15. A t=

-

and (c).

Questions 16 to 18 are based on the following passage Passage V

in case (b) is

M and radius R s and

(a)

2

(b)

2 3

(c)

5 7

(d)

7 3

inclination 16.

18.

(a) to that in case (b) is

(c)

the bottom in case (a) to that in case (b) as

2 3 7 (d) 9

(a) 1

(b)

5 7

(a) 1 (c)

17.

(b) (d)

2

3 2 2 3

SOLUTION 16. g sin 5 I = g sin a1 = 1 2 7 MR In case (b), the acceleration is a2 = g sin Hence the correct choice is (c).

I

2 MR 2 5

18. From s = is (d).

Questions 19 to 22 are based on the following passage Passage VI m1 = 3 m and m2 = m are attached to M = 2 m and radius R) as shown in Fig. 8.93. The masses are then released. 19. the system is (a) 2 g 2g (b) 3 2g (c) 5 3g (d) 7

17. Using v2 = 2as v1 = v2 1 2 at 2

20. Tension T1 is mg (a) 5

7 mg 5 21. Tension T2 is 9 mg (a) 5 (c)

(c) mg 22. (a) 2 mgR (c) Fig. 8.93

5 , which is choice (c). 7

2 mgR 5

3 mg 5 9 mg (d) 5 (b)

7 mg 5 3 mg (d) 5 (b)

(b)

2 mgR 3

(d) 3 mgR

8.58 Comprehensive Physics—JEE Advanced

SOLUTION = (T1 – T2)R Also,

= I

where I =

a R 1 = MR2 2

(3)

1 MR2 and 2

is the angular

=

a 1 = MRa = m Ra R 2 ( M = 2m)

T1 – T2 = ma

(5)

19.

2g . Hence the correct choices is (c). 5 20. T1 = 3m(g – a) = 2g 9 mg = ; which is choice (d). 3m g 5 5 2g 21. T2 = m(g + a) = m g 5 7 mg = so choice (b) is correct. 5 get a =

Fig. 8.94

tensions T1 and T2

and

a is the acceleration

m 1g – T 1 = m 1a T 2 – m 2 g = m 2a

3mg – T1 = 3ma T2 – mg = ma

(1) (2)

22.

The resultant tension (T1 – T2

= mRa = mR

2 g 2mgR = , which 5 5

is choice (c). Questions 23 to 26 are based on the following passage Passage VII m

As R at a height H

P

(a)

2gR

(b)

3gR

(c)

5gR

(d)

7gR

24. Q is

R

(a) mg

(b) 2 mg

(c) 3 mg

(d) 5 mg

25. Q is (a)

. mg

(c)

26 mg

(b)

5 mg

(d) 3 3 mg

26.

H so that the A

Fig. 8.95

23.

Q on the track is

R

(b) 2.5 R

(c) 2.6 R

(d) 2.7 R

Rigid Body Rotation 8.59

SOLUTION 23.

Q = Gain in K.E. 1 1 mg(H – R) = mv2 + I 2 2 mg

R – R) = = v=

1 2 1 mv + 2 2

Fh =

2

25.

2 mR 2 5

v R

mv2 m = R R

2

1 2 1 2 7 mv + mv = mv2 2 5 10

Fh2

Fv2

F= = Thus the correct choice is (c). 26. A A

5gR . Hence the correct choice

m v 2A = mg R

Q

vA =

26 mg

Rg

H

7 7 mRg m v 2A = 10 10 H = 2 R + 0.7R = 2.7R, which is choice (d).

Q the centre O

mg(Hmin – 2R) =

Questions 27 to 30 are based on the following passage Passage VIII

(c) Mg sin 29. (a) MgR

M and radius R with the horizontal. 27.

2 g sin 3 2 g cos (d) 3

(a) g sin (c)

mg

Fv

is (c). 24.

(5 gR) = 5 mg, which is choice (d).

(c) 30.

(a)

28.

(c)

Mg sin (a) 3

2 MgR sin 3

MgR sin 3

(d)

M and the same radius R

(b)

g sin 3

(b) MgR sin

2 Mg sin (b) 3

3 2

(b) 2 (d) 1

2

SOLUTION Mg sin f

f a is the linear acceleration

– f = Ma

(1)

is the angular acceleration and I acting on it is =I

is (see Fig. 8.96) Now

I=

1 MR2, 2

=

a R

fR= f=

or 27. Fig. 8.96

and

= f R. Hence

1 M R2 2

a 1 = M Ra R 2

1 Ma 2

(2) a=

is choice (b).

2 g sin , which 3

8.60 Comprehensive Physics—JEE Advanced

28. f =

MgR sin , which is choice (d). 3 30. For a ring, I = M R2. The correct choice is (a).

1 Mg sin Ma = , which is choice (a). 2 3

29.

Questions 31 to 33 are based on the following passage Passage IX M and radius R is mounted on

=fR=

mg ( M m)

(c)

2 Mmg ( M 2m)

(c)

m is hung

(b)

Mmg ( M m)

(d)

Mmg ( M 2m)

33.

Fig. 8.97

to hn, where h is the height through which the mass n is (a) zero (b) 1 1 (d) 2 (c) 2

rest. 31. (b)

2mg ( M 2m)

32. The tension in the string is (a) mg

(a) g

(d)

mg M

SOLUTION a

32.

ma = mg – T where T is the tension in the string.

(1)

T= is choice (d).

33.

= TR = I I T= R =

1 MR 2 2

mgh = =

a R2

=

1 = Ma 2

1 2 1 mv + I 2 2 1 m R2 2 1

2

+

2

1 2

(2m + M) R2

1 MR 2 2

2

2

(2) Fig. 8.98

31. a=

mM g , which ( M 2m)

2 mg , which is choice (d). ( M 2m)

=

mgh ( M 2m) R 2

1/ 2

h1/2. Hence the correct choice is (c).

Thus

Questions 34 to 36 are based on the following passage

x2. Both the discs rotate in the clockwise direction.

Passage X

IIT, 2007

Two discs A and B I and 2I A

34. The ratio x1/x2 is (a) 2

x1. Disc B

(c)

1 2 1 (d) 2 (b)

2

Rigid Body Rotation 8.61

35. When disc B is brought in contact with disc A, they t. The

36. is (a)

2I 3t 9I (c) t

9I 2t 3I (d) 2t (b)

(a)

(c)

2

I

(b)

2 I

2

(d)

2

I 3

2

I 6

SOLUTION 34.

A

A is

1 1 kx12 = I (2 )2 2 2 For disc B, the energy is

(1)

2I 3

(2)

From (1) and (2) we get

x12 x22

=2

=

x1 = x2

= I (2 ) + (2I)

=

2.

35. When disc B is brought in contact with disc A, let I the

changein angular momentum 2I = Time 3t

36. E1 =

I

=

1 1 I(2 )2 + (2I) 2 2

2

1 1 I ( )2 = (3I) 2 2 3 E 1– E 2 8 1 = 3I 2 – I 2= I 3 3

= 3I

2

E2 =

I I = I + 2I = 3I

3I

3

= A is

1 2 1 kx 2 = (2I)2 2 2

I

= I (2 ) – I

3

=

2

8 I 3

2

2

Questions 37 to 39 are based on the following passage Passage XI M and radius R constant k

Fig. 8.99

L

IIT, 2008

L

V0

V0 i .

38.

μ. 37.

to x (a) –kx 2kx (c) 3

(b) – 2kx kx (d) 3

(a)

k M

(b)

2k M

(c)

2k 3M

(d)

k 3M

8.62 Comprehensive Physics—JEE Advanced

39.

V0 (c) μg (a) μg

M k

(b) μg

3M k

(d) μg

5M 2k

M 2k

SOLUTION 37. Let translational motion (see Fig. 8.100) Ma = f – 2kx

38. Acceleration a = (1)

Let

= 39. When V0

kx =– 3M

2

x. Hence the motion

k . 3M x

x

) and f

= μMg. Ma = μMg – 2kx and

Fig. 8.100

– fR =

1 MRa 2

Eliminating a

=I or

– μMgR =

1 MR 2 2

a ( R

a=

R)

kx

(2)

=

3 Mg 2

Eliminating f a =–

(3)

kx 3M F = Ma =

kx . 3

V

= μg

(5)

3

MV2

= kx2

3M k

IV Assertion-Reason Type Questions Mg h/3. Statement-2 correct. -

throughout its motion. 2. Statement-1 Two bodies A and B and m

m A is v

Statement-2 1. Statement-1 M and radius R rolls down h. The rotational kinetic

B is 2v, v/3.

Rigid Body Rotation 8.63

3. Statement-1 B -

is

3g / L .

Statement-2

circle. Statement-2

8. Statement-1 4. Statement-1 M and m (with M > m) V

v

9. Statement-1

MV ( M m)

=

anticlockwise along the circle. Statement-2

Statement-2 Statement-2 5. Statement-1 L

M and closed at both the ends. The tube is then rotated in a horizontal

other end is ML Statement-2

2

.

angular momentum. 10. Statement-1 mass M and the same radius R

Statement-2 11. Statement-1

6. Statement-1

-

L and mass m is bent into a r as shown in Fig. 8.101. X X is 3mL2/8 2.

Statement-2 12. Statement-1

Fig. 8.101

same time. Statement-2

Statement-2

-

XX 2

about YY + mr . 7. Statement-1 AB hinged at one end A

13. Statement-1 M and length L is -

8.64 Comprehensive Physics—JEE Advanced

Statement-2

14. Statement-1 suddenly .

Fig. 8.102

Statement-2

Statement-2 17. Statement-1

15. Statement-1 -

zero. Statement-2

Statement-2

18. Statement-1 -

16. Statement-1

tact is non zero. Statement-2

SOLUTIONS 1. The correct choice is (a).

4.

-

Potential energy = Translational kinetic energy + Rotational kinetic energy or or

1 1 Mv2 + I 2 2 2 1 1 1 M R2 Mgh = MR2 2 + 2 2 2 3 = MR2 2

MV = (M + m)vCM MV vCM = ( M m)

Mgh =

5. 2

r=

gh 3R 2 1 Now the rotational kinetic energy = I 2. 2 2 and I gh 1 1 MR 2 Rotational kinetic energy = 2 2 3R 2

or

2

=

= 2.

M gh 3

L 2 M

is the circle or radius r = Thus Fc =

M v2 M (r )2 = = Mr r r

2

=

ML 2

2

6. O is IO =

initially at rest) will remain at rest. 3.

L . 2

1 mr2 2 -

ertia about X X is

1 2 3 mr + mr2 = mr2 2 2 L = 2 r or r = L/2 I = IO + mr2 =

Now

Rigid Body Rotation 8.65

I=

3m 2

L 2

2

=

3mL2 8 2

a=

7. The correct choice is (c). Loss in P.E. = gain in through a distance L/2, the loss in P.E. =

1 I 2

Gain in K.E. =

2

=

MgL ML2 2 = 6 2 8. The correct choice is (d). 9.

1 2

2

ML 3

MgL . 2

s

>

5 g sin 7

a = g(sin =

14.

2 MR 2 5

cos )

and . I

I

>

–

g,

3g L

s

I

13.

2

L = I is constant, .

15. The correct choice is (a). 16. A

-

at A 17. -

R

c

=

radius will reach at the same time.

angular momentum will remain unchanged. 1 MR2 10. The correct choice is (a). For cylinder, Ic = 2 2 2 Is = MR . Also = I or = . 5 I , I. Hence Ic s = Is c Ic > Is ; both, as > ac

g sin I 1 MR 2

Hence no work is done.

c.

18.

11.

ity is changing with time. Hence the instantaneous

12. The correct choice is (a). The linear acceleration is

V Integer Answer Type 1.

2.

-

L and mass 300 g rests on F is L

F g = 10 ms–2 8.103. Find the distance 3. Fig. 8.103

IIT, 1980

8.66 Comprehensive Physics—JEE Advanced

v

Fig. 8.106

IIT, 1997

Fig. 8.104

7.

g = 10 ms–2

0.5 m with a stick as shown in Fig. 8.107. The stick B

2

A IIT, 1987 4.

.

ring is large enough that rolling always occurs and the

m

P. Take g = 10 ms–2.

(P

the circle is r, the tension in the string is T = Ar–n, where A n. IIT, 1993 5. A block X m = 0.5 kg is held by a string on = 30°. drum Y

M = 2 kg and radius R = 0.2 m as

Fig. 8.107

IIT, 2011

X –2

8.

g = 10ms newton) in the string during the motion.

5 cm and -

N

10 kg m2, then N is (see Fig. 8.108)

Fig. 8.105

6.

M and radius R u = 6 ms–1 on a rough sliding motion at t

t0 v

–1

)

t 0. Fig. 8.108

IIT, 2011

SOLUTION 1. O O1

O2 W1

Rigid Body Rotation 8.67

W2

3L

Fmin

L 2

= mg

Fmin =

2mg 2 0.3 10 = = 2N. 3 3

Fig. 8.110

3.

Fig. 8.109

p

(21)2 cm2

A = gain in total kinetic energy (translational and rotation), i.e. 1 1 Mg(H – h) = Mv2 + I 2 2 2 2 v Where I = MR2 and = . Thus 5 R

cm2

Mg(H – h) =

(28)2 cm2 [(28)2 – (21)2 cm2 area. Hence

=

or

W1 mg m = 1 = 1 = W2 m2 g m2 Taking moments about O W1 OO1 = W2 OO2 OO2 =

W1 W2

v=

= 9 7

mg

( 1)min =

10 H

h g

2

12

7 .

= 2 5 ms

–1

t=

1.0

10

12

7

L 2

2

2h = g

1 2 1.0 = s 5 10

Horizontal range = vt = 2 5 = 2.0 m. 4. Angular momentum is mvr = constant (say k). k . v= mr mv 2 mk 2 = 2 3 = m–1k2r–3 = Ar–3 r m r –1 2 where A = m k T = Ar–n, we get n = 3 5. X is as shown in Fig. X, then ma = T – mg sin (1) Also

A is = mg

v R

OO1

9 7 cm = 9 cm 7 F about A is (Fig. 8.110) 3L 1 = F

2

2 MR2 5

1 1 7 Mv2 + Mv2 = Mv2 2 5 10

10

=

=

2.

1 1 Mv2 + 2 2

T=

8.68 Comprehensive Physics—JEE Advanced

f = μmg a =R

P is

= RT = I 1 MR 2 I 2 T= = R R 1 T = Ma 2 Using (2) in (1)

F =

1 a MR 2 R

1 ma = Ma – mg sin 30° 2 mg a= M 2m M mg T= 2 M 2m =

2 0.5 10 =5N 2 2 2 0.5

(2)

Where

I = mR2 + mR2 = 2mR2

and

a =R

Also

f = μmg

F

(3)

R – fR = I

a R

R – μmgR = (2 mR2)

= 2maR

F – μmg = 2 ma 2 – μ 2 10 = 2 2 0.3 0.8 . = . Hence P μ= 2 10 10 8. Let M R its radius. diagonal AB I = I1 + I2 + I3 + I where I1, I2, I3 and I AB. I1 = I2 =

2 MR2 5

I3 = I =

2 MR2 + Mr2 5

Fig. 8.111

6.

=

1 I 2 2 1 v = MvR + (MR2) 2 R v u =v + 2 2u 2 6 –1 v= = 3 3

I =2

MuR = MvR +

7.

Fig. 8.112

2

a

2 MR2 + M 5

2

2 MR 2 5

2

2 MR 2 5

Ma 2 2

= 8 MR 2 + Ma2 5 8 = 5

0.5

5 10 2

2 2

+ 0.5 10–2)2

=9

10

kgm2. Hence N = 9.

9

Gravitation

Chapter

REVIEW OF BASIC CONCEPTS 9.1

NEWTON’S LAW OF GRAVITATION

Newton’s law of universal gravitation states as follows: ‘Any two particles of matter anywhere in the universe attract each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, the direction of the force being along the line joining the particles, i.e. (Fig. 9.1)

to all other masses is obtained from the principle of superposition which states that ‘the gravitational force experienced by one mass is equal to the vector sum of the gravitational forces exerted on it by all other masses taken one at a time.’

Fig. 9.1

F

m1 m2 r2

where F is the magnitude of the force of attraction between two particles of masses m1 and m2 separated by a distance r. In the form of an equation the law is written as F=

Gm1 m2 r

2

where G is a constant called the universal gravitation constant. The value of this constant is to be determined experimentally and is found to be G = 6.67 10–11 N m2 kg–2.

9.2

GRAVITATIONAL FORCE DUE TO MULTIPLE MASSES

If a system consists of more than two masses, the gravitational force experienced by a given mass due

Fig. 9.2

The gravitational force on mass m due to masses m1, m 2, m3, ... mn is given by (Fig. 9.2) + Fn F = F1 + F2 + F3 + NOTE (1) Gravitational force is always attractive. (2) Gravitational force between two masses does not depend the medium between them. (3) Gravitational force acts along the straight line joining the centres of the two bodies.

9.1 Two bodies A and B of masses m 1 = 1 kg and m 2 = 16 kg respectively are placed 1.0 m apart. A third body C of mass m = 3 kg is placed on the line joining A and B. At what distance from A should C be placed so that it experiences no gravitational force?

9.2 Comprehensive Physics—JEE Advanced

SOLUTION Let x metre be the distance between A and C (Fig. 9.3)

The magnitude of the resultant force is Fr =

F12

F22

2 F1 F2 cos

=

F2

F2

2 F 2 cos 60

=

3F =

3

Gm 2 a2

Fig. 9.3

Force exerted by A on C is G m 1m F1 = directed towards A x2 Force exerted by B on C is G m2 m F2 = directed towards B (1 x) 2 C will experience no force if F1 = F2, i.e. G m1m x

2

=

G m2 m

(1 x) 2

m2 (1 x) 2 = m1 x2 16 = 4=

(1 x) 2 x2 1 x x

Gm 2 a2

= F (say)

F = mg =

GmM

R2 where M is the mass of the earth and R its radius (nearly constant for a body in the vicinity of the earth) GM g= R2 All bodies near the surface of the earth fall with the same acceleration which is directed towards the centre of the earth.

VARIATION OF G

1. Variation with altitude The acceleration due to gravity of a body at a height h above the surface of the earth is given by gh = g

2

R R

h

where g is the acceleration due to gravity on the surface of the earth. If h is very small compared to R, we can use binomial expansion and retain terms of order h/R. We then get 2h gh = g 1 R Fig. 9.4

SOLUTION The forces exerted on the body at A by bodies at B and C are shown in Fig. 9.5. F1 = F2 =

ACCELERATION DUE TO GRAVITY

Considering the earth as an isolated mass, a force is experienced by a body near it. This force is directed towards the centre of the earth and has a magnitude mg, where g is the acceleration due to gravity.

9.4

which gives x = 0.2 m. Note that if body C is placed to the left of body A or to the right of body B, it will 9.2 Three bodies, each of mass m, are placed at the vertices of an equilateral triangle of side a as shown in Fig. 9.4. Find the magnitude and direction of the force experienced by the body at vertex A.

9.3

, directed vertically downwards.

Fig. 9.5

Thus, the acceleration due to gravity decreases as the altitude (h) is increased. 2. Variation with depth The acceleration due to gravity at a depth d below the surface of the earth is given by d gd = g 1 R This equation shows that the acceleration due to gravity decreases with depth. At the centre of the earth where d = R, gd = 0. Thus the acceleration due to gravity is maximum at the surface of the earth, decreases with increase in depth and becomes zero at the centre of the earth.

Gravitation 9.3

3. Variation with Latitude Due to the rotation of earth about its axis, the value of g varies with latitude, i.e. from one place to another on the earth’s surface. At poles, the effect of rotation on g is negligible. At the equator, the effects of rotation on g is the maximum. In general, the value of acceleration due to gravity at a place decreases with the decrease in the latitude of the place. The acceleration due to gravity at a place on earth where the latitude is is given by g = g – 2 R cos2 ; = angular velocity of rotation of earth. At equator, =0 g e = g – R 2 (minimum) ° gp = g (maximum) At poles, = 90 g = g – 0.0337 cos 2 Thus the value of g varies slightly from place to place on earth. Variation of g with altitude and depth is shown in Fig. 9.6.

SOLUTION g = g 1 = 2 1 2 which gives

=

2

R R

h 2

R R

h

R R

h=

h 2 1 R.

9.5 A body weighs 63 N on the surface of the earth. How much will it weigh at a height equal to half the radius of the earth? SOLUTION W = mg = 63 N W = mg = mg

R

h

h R R R/2

= 63 Fig. 9.6

R 2

R

W =W

2

R

2

= 28 N

Variation of g (gravitational acceleration)

9.3 The acceleration due to gravity on the surface of the moon is 1.67 ms–2. The mass of the earth is about 80 times that of the moon. Estimate the ratio of the radius of the earth to that of the moon.

9.6 Assuming the earth to be a sphere of uniform mass of a body when it is taken to the end of a tunnel 32 km below the surface of the earth. Radius of earth = 6400 km. SOLUTION

SOLUTION ge =

G Me

Re = Rm

Me Mm

=

Re2

80

and gm = gm gm 1.67 9.8

G Mm Rm2

1/ 2

d 32 =g 1 R 6400 Decrease in weight = mg – mg g =g 1

= mg 1 1/ 2

3.7

9.4 Assuming that the earth is a sphere of radius R, at what altitude will the value of acceleration due to gravity be half its value at the surface of the earth?

Percentage decrease =

199 200

=

=

199 g 200 mg 200

mg / 200 100 = 0.5% mg

9.7 At what depth below the surface of the earth will the value of acceleration due to gravity become 90% of its value at the surface? R = 6.4 106 m.

9.4 Comprehensive Physics—JEE Advanced

SOLUTION g = 0.9 g.

g =g 1

d R d = 0.1 R = 6.4

intensity is given by

d R

I=

Gravitational Field due to some continuous Mass Distributions

0.9 = 1 –

9.5

105 m

M and radius R at a point at a distance r from the centre and on the axis of the ring is given by GM r I= 2 (R r 2 )3 / 2

GRAVITATIONAL FIELD INTENSITY

of mass M and radius R at a point P at a distance r > R from the centre of shell, GM I = 2 (outside the shell) r Inside the shell, I = 0 M and radius R is GM I = 2 for r > R r GM I= for r = R R2 GMr I= for r < R R3

experienced by a unit mass placed at that point. M. a point P at a distance r from M, we place a small mass m at P. The gravitational force exerted on m by M is (Fig. 9.7) GM m F= r2

Fig. 9.7

M at

P is given by

F GM = 2 m r I is a vector quantity. In vector form GM I= – r r2 where r is a unit vector directed from M to P, i.e radially away from M. The negative sign indicates that I directed radially inwards towards M. The SI unit of I is N kg–1. In three dimensions, if mass M is located at the origin, P (x, y, z) is given by I=

I= –

GM r2

Fig. 9.8

of a solid sphere.

9.6

r

where r = x i + y j z k represents the position of point P with respect to mass M at the origin. For a many body system, the principle of superposition gravitational forces, i.e. I = I1 + I2 + I3 +

dI

GRAVITATIONAL POTENTIAL ENERGY

Gravitational potential energy of a system of two masses M and m held a distance r respective locations along any path and without any acceleration. (see Fig. 9.9)

+ In

where I1, I2, ... In point due to bodies of masses M1, M2, ... Mn. For continuous mass distributions (i.e rigid bodies), Fig. 9.9

Gravitation 9.5

Work done to bring mass M Work done to bring mass m from r =

A is W1 = 0. to r = r is

=–

r

W2 =

F dr

=

GMm – ( R h)

GmM h R ( R h)

r

=

F dr cos

If h < < R ;

Since mass M will attract mass m, angle and dr is zero. Hence r

W2 =

GM m r2 r

= GMm

GMm r

GMm r Gravitational potential energy of the system is

Total work

GmM h R2

W = W1 + W2 = –

GMm r The zero of potential energy is assumed to be at r = . The negative sign indicates the potential energy is negative U= W = –

Expression for Increase in Gravitational Potential Energy If the body m is moved away from M, the potential energy of the system increases. [see Fig. 9.10]

= mgh GM R

9.7

r 2dr = –

U=

between F

dr

GMm R

2

Fig. 9.11

g

GRAVITATIONAL POTENTIAL

Gravitational potential at a point P in M Fig. 9.12

that point along any path and without any acceleration, i.e., (Fig. 9.12) V=

W GM m =– m r m

GM r Potential V is a scalar quantity. Hence the gravitational potential at a point P due to a number of masses m 1, m2, ... mn at distances r 1, r 2, ... rn respectively from P is given by V = V1 + V2 + + Vn V= –

=–G

m1 r1

m2 r2

mn rn

The SI unit of V is J kg–1. Relation between Gravitational Field Intensity (I) and Gravitational Potential (V)

Fig. 9.10

P.E. at B = –

GMm r1

a point are related as dV dr Gravitational Potential due to a Spherical Shell I=

GMm P.E. at C = – r2 Increase in P.E. = –

GM m r1

GM m – r2

= GMm

1 r1

1 r2

If the body of mass M is the earth, then the increase in gravitational P.E. when a body of mass m is taken from the surface of the earth to a height h above the surface is given by (see Fig. 9.11), R = radius of the earth. U = P.E. at Q – P.E at P

GM (for r r > R) where M is the mass and R is the radius of the shell. GM (ii) At a point on the surface of the shell, V = R GM (iii) At a point inside the shell, V = (for r < R) R Figure 9.13 shows the variation V with r for a spherical shell. (i) At a point outside the shell, V =

9.6 Comprehensive Physics—JEE Advanced

P will be zero if Gm1 Gm2 I1 = I2 r 12 r 22 or

Fig. 9.13

9=

Gravitational Potential due to a Solid sphere of mass M and radius R (i) For points outside the sphere (r > R), GM r (ii) For points inside the sphere (r < R), V=

V=

3G M R 2 2 R3

m2 = m1

r2 6

(ii) At the centre of the sphere (r = 0) 3G M V= 2R (iv) On the surface of the sphere (r = R) GM V= R 9.8 Two masses m1 = 100 kg and m2 = 8100 kg are held 1 m apart. (a) At what point on the line joining them is the tational potential at that point. (b) Find the gravitational potential energy of the system. Given G = 6.67 10–11 Nm2 kg–2.

r1 r2

2

81 =

r1

r1 = 0.1 m

1 r1

Gravitational potential at P is V = V1 + V2 = – G =

6.67 10

= – 6.67

=

Gm2 r 22

8100 0.9

11

100 8100 1

= – 5.4

10–5 J

9.9 Three equal masses, each equal to m, are kept at the vertices of an equilateral triangle of side a. Find the centroid of the triangle. SOLUTION Refer to Fig. 9.15.

directed towards m1

directed towards m2

100 0.1

11

m2 r2

10–7 J kg–1

6.67 10

P due to m2 is I2 =

m1 r1

(b) Gravitational potential energy of the system is G m1 m2 G.P.E. = r

Fig. 9.14

r 12

1 r1

r2 = r – r1 and r = 1 m)

P due to m1 is [Fig. 9.14]

Gm1

2

(

SOLUTION

I1 =

r1

Fig. 9.15

Gravitation 9.7

The centroid O divides the lines AD, BE and CF in the ratio 2:1. Also AO = BO = CO = r (say). Now AO 2 = AD and 3 a2 = 4

a2

AD =

3a 2

2 3

3a a a = , i.e. r = 2 3 3 O due to masses m at A, B Gm and C are IA = IB = IC = 2 = I. Their directions are r shown in Fig. 9.15. The angle between any two of them is = 120°. The resultant of IB and IC is AO =

I2

I =

I2

9.11 Two particles of masses m1 and m2 are initially at rest wards each other due to gravitational attraction. Find the ratio of their accelerations and speeds when the separation between them becomes r. SOLUTION Since no external force acts on the system, the acceleration of their centre of mass is zero, i.e.

2 I 2 cos 120

Gm r

=

3Gm = r

Gm r 3 3

Gm a

a

r

3

9.10 The gravitational potential at a height h = 1600 km above the surface of the earth is – 4.0 107 J kg–1. Assuming the earth to be a sphere of radius R = 6400

0=

m1a1 m1

m2 a2 m2

m2 m1

The negative sign indicates that they move in opposite directions. Let v1 and v2 be the speeds of two masses when they are at a distance r. Due to gravitational attraction, they gain speed as they approach each other. Hence their kinetic energy increases and gravitational potential energy (G.P.E) decreases. From the conservation of energy, Loss in G.P.E. = gain in K.E. G.P.E.)i – (G.P.E.)f = (K.E.)f – (K.E.)i 0

Gm1m2 r

=

1 m1 v 12 2

G m1 m2 1 = m1v12 r 2

at that height.

V= g =

GM r GM r

2

g GM / r 2 1 1 = = = V GM / r r R h g =

V R

h

1 m2 v 22 2 1 m2 v 22 2

0 (i)

Since no external force acts, the total momentum of the system is conserved, i.e., pi = pf or (ii) 0 = m 1v 1 – m 2v 2 From Eqs. (i) and (ii), we get

SOLUTION Let r = R + h. Then

and

m2 a2 m2

a1 = a2

I will cancel with IA at O is zero. The gravitational potential at O is V = V1 + V2 + V3 Gm r

m1 a1 m1

m 1a 1 = – m 2 a 2

= I directed vertically down.

=

aCM =

=

7

and

4.0 10 J kg 6.4 1.6

1

106 m

= 5.0 J kg–1 m–1 = 5.0 ms–2

v1 =

2G m22 r m1 m2

v2 =

2G m 12 r m1 m2

v1 m = 2 v2 m1

1/ 2

1/ 2

9.8 Comprehensive Physics—JEE Advanced

9.8

ESCAPE VELOCITY

The escape velocity is the minimum velocity with which a body must be projected in order that it may escape the earth’s gravitational pull. The magnitude of the escape velocity is given by 2MG R where M is the mass of the earth and R its radius. Substituting the known values of G, M and R, we get ve = 11.2 kms–1. The expression for the escape velocity can be written in terms of g as ve =

ve =

2gR

The escape velocity is independent of the mass of the body and the direction of projection.

9.9

v=

Orbital Velocity Let us assume that a satellite of mass m goes around the earth in a circular orbit of radius r with a uniform speed v. If the height of the satellite above the earth’s surface is h, then r = (R + h), where R is the mean mv2 necessary radius of the earth. The centripetal force r to keep the satellite in its circular orbit is provided by GmM between the earth and the the gravitational force r2 satellite. This means that mv2 mM =G 2 r r where M is the mass of the earth. Thus v=

GM r

T=

GM = gR2

If h ve, the satellite will escape the gravitational pull of the earth following a hyperbolic path.

9.10

Fig. 9.16

According to Kepler’s second law, if the time interval between P1 and P2 equals the time interval between P3 and P4, then area A1 must be equal to area A2. Also the planet has the greater speed in its path from P1 to P2 than in its path from P3 to P4. 3. Law of periods The squares of the periods of the planets are proportional to the cubes of their mean distances from the sun. If T1 represents the period of a planet about the sun, and r1 its mean distance, then T12 r13 If T2 represents the period of a second planet about the sun, and r2 its mean distance, then for this planet T 22 r32 These two relations can be combined since the factor of proportionality is the same for both. Thus T12

KEPLER’S LAWS OF PLANETARY MOTION

Johannes Kepler formulated three laws which describe planetary motion. They are as follows: 1. Law of orbits Each planet revolves about the sun in an elliptical orbit with the sun at one of the focii of the ellipse. The orbit of a planet is shown in Fig. 9.16(a) in which the two focii F1 and F2, are far apart. For the planet earth, F1 and F2 are very close together. In fact, the orbit of the earth is practically circular. 2. Law of areas A line drawn from the sun to the planet (termed the radius) sweeps out equal areas in equal intervals of time. In Fig. 9.16(b) P1, P2, P3 and P4 represent positions of a planet at different times in its orbit and S, the position of the sun.

T22

=

r13 r23

9.12 The mass of Jupiter is 318 times that of the earth and its radius is 11.2 times that of the earth. Calculate the escape velocity from Jupiter’s surface. Given the escape velocity from earth’surface = 11.2 km s–1. SOLUTION For Jupiter :

vJ =

2M J G RJ

For Earth :

vJ =

2M E G RE

9.10 Comprehensive Physics—JEE Advanced

vJ = vE

MJ ME

At the highest point, v = 0. If h is the maximum height attained, the energy of the rocket at height h is

RE RJ

Ef =

1 5.33 = 318 11.2 vJ = 5.33 vE = 5.33 11.2 = 59.7 km s–1

From conservation of energy, Ei = Ef , i.e.,

9.13 Calculate the escape velocity of a body at a height 1600 km above the surface of the earth. Radius of earth = 6400 km. SOLUTION Work required to move a body of mass m from r = R + h to r = is W=

dr r

2

=

Gm M R h

If ve is the escape velocity, then

h R

ve v . Hence e = ve 2 2

= ve

h R

R

h

h R

h

h=

R 3

9.15

2 GM R h

Gm M (i) R h Total energy when it hits the surface of the earth is

2 gR 2 R h GM 2

R 6 10 m and

106 m, and h = 1.6 ve =

Given v =

2GM R

Ei =

g

Given R = 6.4 g = 9.8 ms–2

v=

GmM ( R h)

SOLUTION Total energy of the body at height h is

1 Gm M mve2 = 2 R h ve =

GmM = R

R , where 2 R is the radius of the earth. Show that it will hit the surface of the earth with a speed v = ve / 3 , where ve is the escape velocity from the surface of the earth.

R h

R h

1 2 mv 2

A body is dropped from a height h equal to

Fdr

= Gm M

GmM ( R h)

1 Gm M m v2 2 R From conservation of energy, Ei = Ef , i.e. Ef =

Gm M 1 = m v2 R h 2

6 2

2 9.8 (6.4 10 ) (6.4 1.6) 106

= 10 10 ms = 10 km s–1

v=

–1

9.14 A rocket is launched vertically from the surface of the earth with an initial velocity equal to half the escape velocity. Find the maximum height attained by it in terms of R where R is the radius of the earth. Ignore atmospheric resistance. SOLUTION On the surface of the earth, the total energy of the rocket is 1 2 GmM Ei = K.E. + P.E. = mv R 2

For h =

GmM R

2GM R

= ve

(ii)

h R

h

R R

h

v R ,v= e . 2 3

9.16 Show that the minimum energy required to launch a satellite of mass m from the surface of the earth in a circular orbit at an altitude h = R, where R is the 3mgR where M is the mass of radius of the earth is 4 the earth.

Gravitation 9.11

SOLUTION Total energy of the satellite orbiting the earth is Gm M E1 = 2r Gm M 2( R h)

=

GmM 4R

(

h = R)

Total energy when the satellite was at rest on the surface of the earth is E2 = K.E. + P.E. Gm M R

Gm M = R Minimum energy required is Emin = E1 – E2 = 0

Gm M 4R

=

NOTE Since Fg = Fc, the satellite is a freely falling body and is, therefore, weightless.

9.18 A body projected vertically upwards from the surface of the earth with a certain velocity rises to a height of 10 m. How high will it rise if it is projected with the same velocity vertically upwards from a planet whose density is one-third that of the earth and radius half that of earth? Ignore atmospheric resistance. SOLUTION Since the kinetic energy of the body is the same in both the cases, loss in K.E. = gain in P.E. will be equal, i.e., mgphp = mgehe

GmR R

3Gm M 3 = = mg R 4R 4 9.17 A satellite of mass m = 100 kg is in a circular orbit at a height h = R above the surface of the earth where R is the radius of the earth. Find (a) the acceleration due to gravity at any point on the path of the satellite, (b) the gravitational force on the satellite and (c) the centripetal force on the satellite. SOLUTION (a) g = g

2

R R

h

2

R

= 9.8

R

R

=

9.8 4

(b) Gravitational force on satellite is Fg = mg = 100 2.45 = 245 N m v2 Gm M = r r2 g= or

Fc =

R

= mg

2

R2

r2

GM R

2

h

=

ge R = e gp Rp

G R

e

2

=2

4 R3 3

=

4 GR 3

3=6

p

hp = 6he = 6

10 = 60 m

9.19 A satellite of mass 2000 kg is orbiting the earth at an altitude R/2, where R is the radius of the earth. What extra energy must be given to the satellite to increase its altitude to R? Given R = 6.4 106 m.

r1 = R

= mg = 245 N.

R 3R = and in the second 2 2

case, r2 = R + R = 2R. Required energy = E2 – E1 =

Gm M 2 r2

=

Gm M 4R

=

GmM 12 R

=

mgR 12

2

R R

g=

GM

R2

Gm M

Now

SOLUTION

= 2.45 ms–2

(c) Centripetal force Fc =

ge gp

hp = he

Gm M 2 r1 GmM 3R

g=

GM R2

9.12 Comprehensive Physics—JEE Advanced

6.4 106

2000 9.8

=

SOLUTION From Kepler’s law of periods, T2

12 10 J 8

= 1.04

Tp2

9.20 Two bodies of masses m1 and m2 are held at a distance r apart. Show that at the point where the gravitational is given by G m1 r

V=

m2

2 m1m2

SOLUTION P (Fig.

9.17). Then G m1

=

G m2

r1 = r2

m1 m2

=

m1

r 12

r1 r

r1

r1 = Also

Fig. 9.17

= (10)3 = 1000 31.6 = 31.6 years

9.22 A satellite is revolving in a circular orbit close to the surface of the earth with a speed v. What minimum additional speed must be imparted to it so that it escapes the gravitational pull of the earth? Radius of earth = 6.4 106 m. SOLUTION g R and ve = 2 1

= 0.414

2g R

gR 9.8 6.4 106

= 3.28 103 ms–1 = 3.28 km s–1

m2 r m1 m1

(i)

m2

r m1 m1

m2

r m2 m1

(ii)

m2

Gm1 r1

Gm2 r2

(iii)

Using (i) and (ii) in (iii) and simplifying, we get V=

r 3e

1000 = 1 year

ve – v =

Gravitational potential at P is V=

rp3

Additional speed required is

r2 = r – r1

=

Tp = Te

v=

r 22

= r

Te2

=

r3. Therefore

G m1 r

m2

2 m1m2

9.21 The distance of a planet from the sun is 10 times that of the earth. Find the period of revolution of the planet around the sun.

9.23 A body of mass m is placed at the centre of a spherical shell of radius R and mass M. Find the gravitational potential on the surface of the shell. SOLUTION Gravitational potential on the surface of the shell due to body of mass m is Gm Vb = R Gravitational potential on the surface of the shell due to shell itself is GM Vs = R G V = Vb + Vs = m M R 9.24 A tunnel is drilled from the surface of the earth to its centre. A body of mass m is dropped into the tunnel. Find the speed with which the body hits the bottom of the tunnel. The mass of earth is M and its radius is R.

Gravitation 9.13

Hence

SOLUTION Let v be the required speed. Gain in K.E. = loss in P.E. = P.E. at the surface – P.E. at the centre. The poten3 GmM tial energy at the centre of the sphere = . 2 R

1 mv2 = 2 v=

Gm M R GM = R

3 GmM 2 R gR

g=

GM R2

I Multiple Choice Questions with Only One Choice Correct 1. A body of mass m is placed at the centre of a spherical shell of radius R and mass M. The gravitational potential on the surface of the shell is G G (M + m) (b) – (M – m) (a) – R R (c) –

G mM R M m

(d) –

G mM R M m

2. The magnitude of angular momentum of the earth revolving round the sun is proportional to R n where R is the distance between the earth and the sun. The value of n is (assume the orbit to be circular). (a) 1 2

(b) 1

(c) 3 (d) 2 2 3. A tunnel is drilled from the surface of the earth to its centre. The radius of the earth is R and g is the acceleration due to gravity on its surface. A body of mass m is dropped into the tunnel. If M is the mass of the earth, the speed with which the body hits the bottom of the tunnel is (a) (c)

m M

gR

2 gR

(b) M m (d)

gR

gR

4. A satellite is orbiting at a height R above the surface of the earth where R is the radius of the earth. By what percentage must the energy of the satellite be increased so that it orbits at a height 2 R above the surface of the earth? (a) 25% (b) 33.3% (c) 50% (d) 66.7% 5. If E1 is the energy required to raise a satellite to a height h = R (radius of the earth) above the surface

of the earth and E2 is the energy required to put it into a circular orbit at that height, then the ratio E1/ E2 is 1 (a) 1 (b) 2 1 2 (c) (d) 3 3 6. If g is the acceleration due to gravity on the surface of the earth, the gain in potential energy of a satellite of mass m raised from the earth’s surface to a height equal to the radius R of the earth is (a) mgR/4 (b) mgR/2 (c) mgR (d) 2 mgR 7. The escape velocity of a body projected vertically upwards from the surface of the earth is v. If the body is projected at an angle of 30° with the horizontal, the escape velocity would be (a) v/2 (b) 3 v/2 (c) 2 v (d) v 8. The escape velocity of a body at a height h above the surface of the earth is (g = acceleration due to gravity on the surface of the earth and R = radius of the earth) is (b) 2 g R h (a) 2 gR 2 gR 2 gR 2 (d) 2 R h R h 9. A small planet is revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force between the planet and the star were proportional to r –5/2, then T would be proportional to (b) r 5/3 (a) r 3/2 7/4 (c) r (d) r 3 10. If both the mass and the radius of the earth decrease by 1%, the acceleration due to gravity on the surface of the earth will (c)

9.14 Comprehensive Physics—JEE Advanced

(a) decrease by 1% (b) increase by 1% (c) increase by 2% (d) remain unchanged. 11. Two masses M1 and M2 are separated by a distance r. A particle of mass m is placed exactly mid-way between them. The minimum speed with which the particle should be projected so as to escape to G M1 M 2 r

(a) v = 2 (b) v =

2 G M1 M 2 r G M1 M 2 mr

(c) v = 2 (d) v =

2 G M1 M 2 mr

1/ 2

1/ 2

2 1/ 2

2 1/ 2

12. The areal velocity of a planet of mass m moving along an elliptical orbit around the sun is (a) proportional to m (b) proportional to m 1 (c) proportional to (d) independent of m m 13. A planet of mass m revolves around the sun in an elliptical orbit of semimajor axis a. If M is the mass of the sun, the speed of the planet when it is at a distance x from the sun is (a) (c)

GM GM

1 x

1 2a

(b)

2 x

1 a

(d)

GM

1 2x

1 a

GM a 2 x2

14. A satellite revolves around a planet with a speed v in a circular orbit of radius r. If R is the radius of the planet, the acceleration due to gravity on its surface is v2 r v2 R (b) g = 2 (a) g = 2 R r v2 v2 (c) g = (d) g = R r 15. Three spheres, each of mass M and radius R, are kept such that each touches the other two. The magnitude of the gravitational force on any one sphere due to the other two is 3 GM 2 GM 2 (b) (a) 2 R2 2R2 (c)

3 GM 2 2 R2

(d)

3 GM 2 4 R2

16. An extremely small and dense neutron star of mass M and radius R is rotating at an angular frequency . If an object is placed at its equator, it will remain stuck to it due to gravity if R R2 2 (a) M > (b) M > G G 3 2 2 R R 3 (c) M > (d) M > G G 17. What is the minimum energy required to launch a satellite of mass m from the surface of the earth of radius R in a circular orbit at an altitude of 2R? 5GmM 2GmM (b) (a) 6R 3R GmM GmM (c) (d) 2R 3R 18. Two stars, each of mass m and radius R are approaching each other for a head-on collision. They start approaching each other when their separation is r >> R. If their speeds at this separation are negligible, the speed with which they collide would be (a) v =

Gm

1 R

1 r

(b) v =

Gm

1 2R

(c) v =

Gm

1 R

(d) v =

Gm

1 2R

1 r 1 r 1 r

19. Consider a particle of mass m suspended vertically by a string at the equator. Let R and M denote the radius and the mass of the earth respectively. If is the angular velocity of earth’s rotation about its own axis, the tension in the string is equal to mM mM (b) G (a) G 2R2 R2 (c) G 20.

mM R

2

–m

2

R

(d) G

mM R

2

+m

2

R

m, are placed along a straight line at distances of r, 2r, 4r, 8r, etc. from a reference point O. The gravitational O will be (a) (c)

5 Gm 4r

2

3 Gm 2r

2

(b) (d)

4 Gm 3r 2 2 Gm r2

Gravitation 9.15

21. In Q. 20, the magnitude of the gravitational potential at point O will be Gm Gm (b) (a) 2r r 3 Gm 2 Gm (c) (d) 2r r 22. A satellite in force-free space sweeps stationary interplanetary dust at a rate dM/dt = v, where M is the mass and v is the velocity of the satellite and is a constant. The acceleration of the satellite is 2 v v2 (b) (a) M M 2

v (d) – v2 M 23. The time of revolution of a satellite is T. Its kinetic energy is proportional to 1 1 (b) 2 (a) T T 1 (c) 3 (d) T –2/3 T 24. A solid sphere of uniform density and radius 4 units is located with its centre at 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 leaving behind spherical cavities as shown in Fig. 9.18. Choose the incorrect statement from the following. (c)

Fig. 9.18

(a) the gravitational force due to this object at the origin is zero. (b) the gravitational force at point B (2, 0, 0) is zero. (c) the gravitational potential is the same at all points of the circle y2 + z2 = 36. (d) the gravitational potential is the same at all points of the circle y2 + z2 = 4. IIT, 1993 25. Two bodies of masses m1 and m2 are initially at rest move toward each other under mutual gravitational

attraction. Their relative velocity of approach at a separation distance r between them is (a)

1/ 2

2G m1 m2 r

2G m1 m2 r 2

(b) (c)

r 2G m1 m2

(d)

2G m1 m2 r

1/ 2

1/ 2

1/ 2

26. If the distance between the earth and the sun were half its present value, the number of days in a year would have been (a) 64.5 (b) 129 (c) 182.5 (d) 730 IIT, 1996 27. around the earth has a total (kinetic + potential) energy E0. Its potential energy is (a) – E0 (b) 1.5 E0 (c) 2 E0 (d) E0 IIT, 1997 28. 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. Which of the following statements is correct? (a) The acceleration of S is always directed towards the 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 remains constant. (d) The linear momentum of S remains constant in magnitude. IIT, 1998 29. The distance between the sun and the earth is r and the earth takes time T to make one complete revolution around the sun. Assuming the orbit of the earth around the sun to be circular, the mass of the sun will be proportional to (a) (c)

r2 T r3

(b) (d)

r2 T2 r3

T2 T3 30. A meteor of mass M breaks up into two parts. The mass of one part is m. For a given separation r the

9.16 Comprehensive Physics—JEE Advanced

mutual gravitational force between the two parts will be the maximum if M M (a) m = (b) m = 2 3 M M (c) m = (d) m = 2 2 2 31. A body of mass m is raised to a height h above the surface of the earth of mass M and radius R until its gravitational potential energy increases by 1 mgR. The value of h is 3 R R (b) (a) 3 2 mR mR (c) (d) M m M 32. Two balls A and B are thrown vertically upwards from the same location on the surface of the earth gR 2 gR and respectively, with velocities 2 3 3 where R is the radius of the earth and g is the acceleration due to gravity on the surface of the earth. The ratio of the maximum height attained by A to that attained by B is (a) 2 (b) 4 (c) 8 (d) 4 2 33. A uniform sphere of mass M and radius R exerts a force F on a small mass m situated at a distance of 2R from the centre O of the sphere. A spherical portion of diameter R is cut from the sphere as shown in Fig. 9.19. The force of attraction between the remaining part of the sphere and the mass m will be 7F 2F (a) (b) 9 3 4F F (c) (d) 9 3

Fig. 9.19

34. The centres of a ring of mass m and a sphere of mass M of equal radius R, are at a distance 8 R apart as shown in Fig. 9.20. The force of attraction between the ring and the sphere is

(a) (c)

2 2 GmM 27 R 2

(b)

GmM 9R

(d)

2

GmM 8R 2 2 GmM 9 9R2

Fig. 9.20

35. Two objects of masses m and 4m are at rest at under mutual gravitational attraction. Then, at a separation r, which of the following is true? (a) The total energy of the system is not zero. (b) The force between them is not zero. (c) The centre of mass of the system is at rest. (d) All the above are true. IIT, 1994 36. A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01 R. The period of the approximately (a) 0.5% (c) 1.5%

(b) 1.0% (d) 3.0%

IIT, 1995 37. 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 (a) 1 (b) 2 (c) 4 (d) 2 IIT, 2001 38. 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 (a) 4 Mg/k (b) 2 Mg/k (c) Mg/k (d) Mg/2k IIT,2002 39. A geo-stationary 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’s surface (REarth = 6400 km) will approximately be

Gravitation 9.17

(a) (1/2) h (c) 2 h

(b) 1 h (d) 4 h

IIT, 2002 40. A satellite of mass m is moving in a circular orbit of radius R above the surface of a planet of mass M and radius R. The amount of work done to shift the satellite to a higher orbit of radius 2R is (here g is the acceleration due to gravity on planet’s surface) mgR (a) mgR (b) 6 mMgR mMgR (c) (d) M m 6 M m 41. The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is (here R is the radius of the earth) n n mgR (b) mgR (a) n 1 n 1 mgR (c) nmgR (d) n 42. A body of mass m is dropped from a height nR above the surface of the earth (here R is the radius of the earth). The speed at which the body hits the surface of the earth is (a)

2 gR n 1

(b)

gR

(a) –

(b) –

10 3gR

(c) –

2 gR 10 4 gR

(d) – 10 10 46. The radius of the earth is R and g is the acceleration due to gravity on its surface. What should be the angular speed of the earth so that bodies lying on the equator may appear weightless? (a)

g R

(b)

2g R

g g (d) 2 R 2R 47. 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 (as shown in Fig. 9.21) in m correct relation between W1, W2 and W3. (c)

(a) W1 > W3 > W2

(b) W1 = W2 = W3

(c) W1 < W3 < W2

(d) W1 < W2 < W3 IIT, 2003

2 gR n 1

2 gRn 2 gRn (d) n 1 n 1 43. Two solid spheres of radii r and 2r, made of the same material, are kept in contact. The mutual gravitational force of attraction between them is proportional to 1 1 (b) 2 (a) 4 r r (c) r2 (d) r4 44. A comet is moving in a highly elliptical orbit round the sun. When it is closest to the sun, its distance from the sun is r and its speed is v. When it is farthest from the sun, its distance from the sun is R and its speed will be (c)

(a) v

r R

1/ 2

(b) v 3/ 2

r R

48. A binary star system consists of two stars of masses M1 and M2 revolving in circular orbits of radii R1 and R2 respectively. If their respective time periods are T1 and T2, then (a) T1 > T2 if R1 > R2 (b) T1 > T2 if M1 > M2 (c) T1 = T2 (d)

2

r r (d) v R R 45. The value of the acceleration due to gravity at the surface of the earth of radius R is g. It decreases by 10% at a height h above the surface of the earth. The gravitational potential at this height is (c) v

Fig. 9.21

T1 T2

R1 R2

3/ 2

IIT, 2006 49. A spherically symmetric gravitational system of R 0 for r where particles has a mass density = 0 for r R 0 is a constant. A test mass can undergo circular of particles. Its speed v as a function of distance

9.18 Comprehensive Physics—JEE Advanced

50. A satellite is moving with a constant speed ‘V ’ in a circular orbit about the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is

r(0 < r < ) from the centre of the system is represented by (see Fig. 9.22) IIT, 2008

1 mV 2 2 3 (c) mV 2 2

(b) mV2

(a)

(d) 2mV2 IIT, 2011

Fig. 9.22

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49.

(a) (d) (c) (c) (a) (b) (d) (d) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50.

(a) (c) (a) (b) (b) (c) (b) (b) (b)

3. 9. 15. 21. 27. 33. 39. 45.

(d) (c) (d) (d) (c) (a) (c) (c)

4. 10. 16. 22. 28. 34. 40. 46.

(b) (b) (c) (b) (a) (a) (b) (a)

5. 11. 17. 23. 29. 35. 41. 47.

(d) (a) (a) (d) (c) (d) (a) (b)

6. 12. 18. 24. 30. 36. 42. 48.

(b) (d) (b) (b) (a) (c) (d) (c)

SOLUTIONS 1. Gravitational potential on the surface of the shell due to the body of mass m is Gm Vb = – R Gravitational potential on the surface of the shell due to shell itself is GM Vs = – R G V = Vb + Vs = – (M + m), which is choice (a). R GmM m v2 GM 2. = v= R2 R R Angular momentum L = mvR = m = m(GMR)1/2. i.e. L R1/2. So the correct choice is (a).

GM R

R

3. Let v be the required speed. Gain in K.E. = loss in P.E. = P.E. at the surface – P.E. at the centre of the earth 1 GmM mv2 = – – 2 R

3 GmM 2 R

GM GM g = gR R2 R 4. The total energy of a satellite in orbital radius r is 1 GMm E = K.E. + P.E. = mv2 – 2 r v=

GMm GMm 2r r GMm = 2r =

v

GM r

Gravitation 9.19

r = R + R = 2R. Hence GMm E= 4R

k r 3/ 2 m 2 r T= =2 r v v=

In the second case, r = R + 2R = 3R. Hence GMm 6R

E =

=

GMm 12 R

E Percentage increase = |E|

GMm 4R

100

P.E. at a height h (= R) = GMm 2R

GMm R

GMm 2R GMm R

GM GMm 1 g = mgR 2R R2 2 7. The correct choice is (d) because the escape velocity is independent of the direction along which the body is projected. 8. Escape velocity at height h = 2g R h , where =

2

R

. Hence the correct choice is (c). R h 9. Since the gravitational force provides the necessary centripetal force, g =g

mv2 r

r

R2

G m M1 G m M2 r/2 r/2 2Gm = (M1 + M2) r If v is the required velocity of projection, the total initial energy is Total P.E. =

1 2Gm mv 2 – (M1 + M 2) 2 r Ef = 0. Putting Ei = Ef , we get

Ei =

v= 2

G M1

M2

k = constant

1/ 2

r

dA L = , where L is the magnidt 2m tude of angular momentum of the planet about the sun. L = mvr sin . Hence,

12. A real velocity

dA mvr sin v r sin = = dt 2m 2 So the correct choice is (d). 13. Total energy of the planet in an elliptical orbit of semimajor axis a is GmM 2a Total energy of the planet when it is at a distance x from the sun (here v = speed of the plane at that instant) is E2 = K.E. + P.E. E1 = –

1 GmM mv2 – 2 x From conservation of energy, E1 = E2, i.e. =

–5/2

mv2 = kr–5/2, r

GM g m 2 R = – g M R = – 1% – 2 (– 1%) = + 1%

GMm 3 GMm = R 4R

E1 2 = E2 3 6. P.E. on the surface of the earth =

Gain in P.E. =

10. g =

3/ 2

Hence g will increase by 1%. 11. Distance of m from M1 or M2 = r/2. Therefore,

GMm / 12 R = 100 = 33.3% GMm / 4 R 5. E1 = P.E. at h = R – P.E. at h = 0 = P.E. at r = 2R – P.E. at r = R GMm GMm GMm = 2R R 2R E2 =

kr 7/4 r , which is choice (c)

i.e. T

Increase in energy E = E – E GMm GMm = – 6R 4R

m

–

GmM 1 GmM = mv2 – 2a 2 x

9.20 Comprehensive Physics—JEE Advanced

which gives v =

2 x

GM

1 , which is choice (c). a

mv 2 GmM 14. = r r2 GM = v2 r g=

GM

=

R2

v2 r R2

So the correct choice is (a). 15. Force between any two spheres is F=

GM 2 2R

2

=

GM 2 4R2

This is the force exerted on any sphere (say A) by the other two spheres B and C (Fig. 9.23). Thus the resultant force on sphere A is Fr = =

F2 3F =

F2

2 F 2 cos 60 3 GM 2 4R2

, which is choice (d).

Fg > Fc or

GmM

2

> mR

mv 2 GmM = r r2

v2 =

or

off if

2

GM r

1 GmM mv2 = 2 2r

or

GmM 2r Thus the KE of a satellite in a circular orbit is numerically half its PE but opposite in sign. The total energy of the satellite in orbit is

i.e.

KE =

GmM GmM – 2r r

=–

Fig. 9.23

R3 G

R2 Hence the correct choice is (c). 17. Consider a satellite of mass m moving with a speed v at an altitude r (measured from the centre of the earth). Then 1 mv 2 Kinetic energy (KE) = 2 GmM Gravitational potential energy (PE) = – r where M is the mass of the earth. For a satellite in circular orbit, we have

E = KE + PE =

16. An object of mass m, placed at the equator of the star, will experience two forces: (i) an attractive force due to gravity towards the centre of the star and (ii) an outward centrifugal force due to the rotation of the star. The centrifugal force arises because the object is in a rotating (non-inertial) frame; this force is equal to the inward centripetal force but opposite in direction. Force on object due to gravity is GmM Fg = R2 Centrifugal force on the object is Fc = mR 2

or M >

GmM 2r

It is given that r = 2R + R = 3R, where R is the radius of the earth. GmM E= – 6R GmM Now PE on the surface of the earth = – R Minimum energy required (E min) GmM GmM =– – R 6R 5GmM 6R Hence the correct choice is (a). 18. The speeds of stars at separation r are negligible. Therefore, their energy is entirely potential at this separation (since KE = 0) Gm1 m2 E1 = (PE at r) = – r =

Gm 2 r As the stars approach each other under gravitational attraction, they begin to acquire speed and hence =–

Gravitation 9.21

kinetic energy at the expense of potential energy. When they eventually collide, the separation between their centres is r = R + R = 2R At

r = 2R, the total energy is E2 = PE at (r = 2R) + KE at (r = 2R) =–

2

Gm 2R

Gm 2 + mv2 2R From the principle of conservation of energy, E1 = E2, i.e. Gm 2 Gm 2 – =– + mv2 2R r or

E2 = –

1 1 which gives v = Gm 2R r Hence the correct choice is (b). 19. The acceleration due to gravity at a place on the surface of the earth is given by g = g – 2R cos where is the latitude of the place. At the equator, = 0. Therefore g = g – 2R Therefore, the tension in the string will be mg = mg – m 2R GmM = – m 2R R2 Hence the correct choice is (c). 20. O will be I = Gm = =

Gm r

2

1

1

r2

2r

1

2

4r

1

1

1

2

2

2

2

Gm 1 r

2

2

1

0

4

2

1

1

1

2

4

26

2

2

8r

2

4Gm Gm 1 = 2 = 1 r 1 3r 2 r 1 1 2 2 4 Hence the correct choice is (b). 21. The gravitational potential, in magnitude, at point O is 1 1 1 1 V = Gm r 2r 4r 8r =

Gm 2

1

8

1

Gm 1 1 r 2

=

Gm 1 r 20

22.

1 4

1 8

1

1

1

1

2

23

2

2

1

2Gm = 1 r 1 2 Hence the correct choice is (d). From Newton’s second law of motion, force is the rate of change of momentum, i.e. =

1 1 mv2 + mv2 2 2

+

=

Gm r

d dM (Mv) = v= dt dt

F=

Retardation =

F M

v2 M

v2

dM v dt or acceleration =

v2 . Hence the correct choice is (b). M 23. The orbital speed of a satellite at a height r from the GM centre of the earth is given by v = r where M is the mass of the earth. If m is the mass of the satellite, its kinetic energy is K=

1 1 GM mv2 = m 2 2 r

1 . Now, the time period r r3 of the satellite is given by T = 2 g R2

Thus, K is proportional to

where R is the radius of the earth. Thus T r3/2. But 1 K . Hence K T–2/3, which is choice (d). r 24. The distance of each cavity from the centre O is the same. Since the two cavities are symmetrical with respect to the centre O and the mass of the sphere can be regarded as being concentrated at the centre O, the gravitational force due to the sphere is zero at the centre. Hence choice (a) is correct. For the same reason, the gravitational potential is the same at all points of the circle y2 + z2 = 36 whose radius is 6 units and at all points of the circle y2 + z2 = 4. Hence choices (c) and (d) are also correct. But the gravitational force at point B cannot be zero. 25. tance from each other, their gravitational potential energy is zero. When they are at a distance r from each other the gravitational P.E. is

9.22 Comprehensive Physics—JEE Advanced

G m1 m2 r The minus sign indicates that there is a decrease in P.E. This gives rise to an increase in kinetic energy. If v1 and v2 are their respective velocities when they are a distance r apart, then, from the law of conservation of energy, we have PE = –

G m1 m2 1 m1v21 = r 2 or

2 G m2 r

v1 =

G m1 m2 1 m2v22 = r 2

and

2 G m1 r Therefore, their relative velocity of approach is or

v2 =

2 G m2 r

v1 + v2 =

2G (m2 r Hence the correct choice is (a). 26. According to Kepler’s law of periods, =

T1 = T2 = T2 =

R1 R2 2

3/ 2

3/ 2

T1

2 2 27. For a satellite, we have

R1 R1 / 2

2 G m1 r

2 2

M=

Also v =

4 2 r3 2 r . Using this, we get M = or T 2G T

r3

. T2 Hence the correct choice is (c). 30. Mass of the second part = M m. Gravitational force between the two parts is M

F=

3/ 2

= 129 days.

GmM 2r GmM Potential energy = – r Total energy E0 = KE + PE Gm M Gm M Gm M = 2r r 2r Kinetic energy =

PE 2

or PE = 2E0. Hence the correct choice is (c). 28. For elliptical orbit, the earth is at one focus of the ellipse. For spherical bodies, the gravitational force is central (or radial). Hence statement (a) is correct. The gravitational force exerts no torque on the satellite. Hence the angular momentum of S remains constant in magnitude as well as direction. Hence choice (b) is incorrect. For elliptical orbit, the distance of the satellite from the earth varies periodically. Hence potential energy, kinetic energy and

v2 r G

or

m1 )

2 2 365 days

linear momentum vary periodically. Hence choices (c) and (d) are also incorrect. 29. Let m and M be the masses of the earth and the sun respectively and v the speed of the earth in circular orbit. To keep the earth in circular orbit, the graviGmM must balance the centripetal tational force r2 2 mv , i.e. force r GmM m v2 = r2 r

G(M

m) m r

2

=

G r

2

(Mm

m 2)

dF d 2F F will be maximum if = 0 and is negadm d m2 tive. dF dF G = 2 (M 2m). Setting = 0, we Now, dm dm r M . Now get M 2m = 0 or m = 2 d 2F 2G = – 2 , which is negative. d m2 r Hence the correct choice is (a). GM m 31. PE on the surface of earth = – R PE at a height h above the surface of earth = GM m – ( R h) GM m GM m Increase in PE = – – ( R h) R = GMm =

by

1 R

1 R h

g Rmh ( R h)

PE will increase by

= GMm

h R ( R h) g

GM R2

1 mgR at a value of h given 3

g Rmh 1 = mgR ( R h) 3

Gravitation 9.23

1 R or h = , which is choice (b). R h 3 2 32. If h is the maximum height attained, then we have GM m 1 GM m mv2 – =– ( R h) 2 R h

or

v2 =

which gives

2g h R ( R h)

GM

g

R2

For ball A, we have

2g hA R 4g R = R hA ) ( 3

hA = 4R

For ball B, we have

2g hB R 2g R = ( R hB ) 3

hB =

R 2

hA = 8, which is choice (c). hB 33. The force of attraction between the complete sphere and mass m is GmM GmM (i) F= 2 = (2 R) 4R2 4 3 Mass of complete sphere is M = R . Mass 3 4 R 3 of the cut out portion is m0 = . Thus, 3 2 M . The distance between the centre of the m0 = 8 R 3R cut out portion and mass m = 2R = . 2 2 Hence the force of attraction between the cut out portion and mass m is f=

G m0 m (3R / 2)

2

=

G ( M/8) m 2

=

GmM 2

2 9

9 R /4 4R 2F Using (i), we get f = . Therefore, the force of 9 attraction between the remaining part of the sphere 2F 7F and mass m = F f = F = which is 9 9 choice (a). 34. Refer to Fig. 9.24. Let be the mass per unit length of the ring. L = 2 R is the length of the ring. Consider a small element of length dx of the ring located at C. Then G M dx . Therefore, force Force along BC is f = (3R) 2 G M dx 8R along BA is dF = f cos = = 2 9R 3R 8 GM dx 27

because dx = × L = m, the mass of the ring. Hence the correct choice is (a).

=

R2

Total force =

8 GM 27 R 2

dx =

8 GMm 27 R 2

Fig. 9.24

35.

the system of two masses and the force between

rest initially and there is no external force, the centre of mass cannot move. Hence the correct choice is (d). 36. According to Kepler’s law of period, T2 = kR3 where k is a constant. Taking logarithm of both sides, we have 2 log T = log k + 3 log R Differentiating, we get 2 or

T R =0+3 T R T 3 = T 2

3 1.01 R R = × 2 R R

R

100

= 1.5% Hence the correct choice is (c). 37. The acceleration due to gravity at a height h above the surface of the earth is given by g2 = g1

R

2

R h where g1 is the value at the surface of the earth. Now T2 = 2 T2 = T1

l and T1 = 2 g2

l g1

g1 R h R R = = =2 g2 R R

(

h = R)

Hence the correct choice is (d). 38. Let x be the extension in the spring when it is loaded with mass M. The change in gravitational potential energy = Mgx. This must be the energy stored in the 1 2 kx . Thus spring which is given by 2 2Mg 1 2 kx = Mg x or x = , which is choice (b). k 2

9.24 Comprehensive Physics—JEE Advanced

39. For a satellite of mass m moving with a velocity v in a circular orbit of radius r around the earth of mass M, we have mv 2 GmM = or v = r r2 Now

v= T2 = T1

GM r

2 r 2 r . Thus = T T r2 r1

GM or T r

r3/2.

3/ 2

(1)

Given r2 = 6400 km and r1 = 36000 km. For a geostationary satellite T1 = 24 h. Using these values 64 3 / 2 = 1.8 h. in (1), we have get T2 = 24 × Hence the closest choice is (c). 360

40. PE at a distance r from the centre of the planet GM m =– r GM m GM m Initial PE = – =– R R 2R GM m GM m =– R 2R 3R Now, work done = increase in PE Final PE = –

GM m 1 1 GM m 1 = = mgR 2 3 R 6R 6 Hence the correct choice is (b). =

g

GM R2

GM m n GM m – = mgR ( n 1 ) R 1 n R Hence the correct choice is (a). 42. From the principle of conservation of energy, we have GM m 1 GM m mv2 =– ( R n R) 2 R 41. Change in PE =

which gives v2 =

2 n R GM 2n R g = 2 (n 1) R (n 1) g

GM R2

Hence the correct choice is (d). 43. If is the density of the material of each sphere, then the mass of the sphere of radius r is 4 3 and the mass of the sphere of radius M1 = r 3 4 (2r)3 . 2r is M2 = 3

Distance between their centres is d = r + 2r = 3r. 4 4 G r3 ( 2 r )3 G M1 M 2 3 3 = Now F = d2 9r 2 which gives F r4, which is choice (d). 44. The angular momentum of the comet is constant over the entire orbit. Hence vr = VR or r V=v , which is choice (b). R 45. gh =

GM ( R h) 2

Also g =

GM

. Thus

R2

Given gh = R2 ( R h) 2

gh R2 = . g ( R h) 2

90 g 100 =

9 or (R + h) = 10

10 R 3

GM GMR 2 gR 2 =– 2 =– ( R h) ( R h) R ( R h) 3gR =– 10 Hence the correct choice is (c). 46. At the equator, the value of g is g =g–R 2 Potential = –

where is the angular speed of the earth. For bodies to appear weightless at the equator, g = 0, i.e. g – R 2= 0 which gives

=

g . Hence the correct choice is (a). R

47. Gravitational force is conservative. The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle. Hence the correct choice is (b). 48. In a binary star system, the two stars move under their mutual gravitational force. Therefore, their angular velocities and hence their time periods are equal. Thus the correct choice is (c). 49. If M is the total mass of the system of particles, the orbital speed of the test mass is v=

For r

R, v =

GM r G

4 3 r 3 r

0

which gives v

r,

Gravitation 9.25

50. Let M be the mass of the earth and r be the orbital radius of the satellite. The energy needed so that the object of mass m escapes from x = r to x = is

i.e. v increases linearly with r up to r = R. Hence choices (b) and (d) are wrong. For r > R, the whole mass of the system is 4 R3 0, which is constant. Hence for M = 3 r > R, v= i.e, v

1 r

dx

E = GMm r

x

2

GMm r

(i)

The orbital speed V is given by

GM r

mV 2 GMm = r r2 Using (ii) in (i), E = mV2.

. Hence the correct choice is (c)

mV2 =

GMm r

(ii)

II Multiple Choice Questions with one or More Choices Correct 1. A satellite is moving around the earth in a stable circular orbit. Choose the correct statements from the following. (a) It is moving at a constant speed. (b) Its angular momentum remains constant. (c) It is acted upon by a force directed away from the centre of the earth which counter balances the attraction by the earth. (d) It behaves as if it were as freely falling body. 2. The escape velocity from a planet depends upon (a) the mass of the body (b) the mass of the planet (c) the average radius of the planet (d) the average density of the planet 3. The orbital velocity of a body in a stable orbit around a planet depends upon (a) the average radius of the planet (b) the height of the body above the planet (c) the acceleration due to gravity (d) the mass of the orbiting body 4. Choose the correct statements from the following: (a) The equivalence of inertial and gravitational mass has provided a clue to the deeper understanding of gravitation . (b) At poles, the effect of rotation of earth on the value of g is the minimum. (c) Very massive rockets and extremely tiny particles, such as the molecules of a gas, require the same initial velocity to escape from the earth. (d) A geostationary satellite, if imparted the necessary velocity, can be put in orbit at any height above the earth.

5. Choose the correct statements from the following: (a) The magnitude of the gravitational force between two bodies of mass 1 kg each and separated by a distance of 1 m is 9.8 N. (b) Higher the value of the escape velocity for a planet, the higher is the abundance of lighter gases in its atmosphere. (c) The gravitational force of attraction between two bodies of ordinary mass is not noticeable because the value of the gravitation constant is extremely small. (d) Force of friction arises due to gravitational attraction. 6. Choose the wrong statements from the following. (a) It is possible to shield a body from the gravishielding material between them. (b) The escape velocity of a body is independent of the mass of the body and the angle of projection. (c) The acceleration due to gravity increases due to the rotation of the earth. (d) The gravitational force exerted by the earth on a body is greater than that exerted by the body on the earth. 7. A comet is revolving around the sun in a highly elliptical orbit. Which of the following will remain constant throughout its orbit? (a) Kinetic energy (b) Potential energy (c) Total energy (d) Angular momentum

9.26 Comprehensive Physics—JEE Advanced

8. A satellite is orbiting the earth. If its distance from the earth is increased, its (a) angular velocity would increase (d) linear velocity would increase (c) angular velocity would decrease (d) time period would increase. 9. For two satellites at distance R and 7R above the earth’s surface, the ratio of their (a) total energies is 4 and potential and kinetic energies is 2 (b) potential energies is 4 (c) kinetic energies is 4 (d) total energies is 4. 10. A satellite is orbiting the earth in a circular orbit of radius r. Its (a) kinetic energy varies as 1/r (b) angular momentum varies as 1/ r (c) linear momentum varies as 1/ r (d) frequency of revolution varies as 1/r3/2. 11. If both the mass and radius of the earth decrease by 1%, the value of (a) acceleration due to gravity would decrease by nearly 1% (b) acceleration due to gravity would increase by 1% (c) escape velocity from the earth’s surface would decrease by 1% (d) the gravitational potential energy of a body on earth’s surface remains unchanged. 12. An object is taken from a point P to another point Q (a) assuming the earth to be spherical, if both P and Q lie on earth’s surface the work done is zero. (b) If P is on earth’s surface and Q above it, the work done is minimum when it is taken along the straight line PQ. (c) The work done depends only on the positions of P and Q and is independent of the path along which the particle is taken. (d) there is no net work done if the object is taken from P to Q and then brought back to P, along any path. 13. A satellite is moving in a circular orbit around the earth with a speed equal to half the escape velocity from the earth. The radius of the earth is R and h is the height of the satellite above the surface of the earth. If the satellite is suddenly stopped in its orbit and allowed to fall freely, it will hit the surface of the earth with a speed v. Then R (b) h = R (a) h = 2 (c) v =

2gR

(d) v =

gR

14. Two stars of masses m and 2 m are co-rotating about their centre of mass. Their centres are at a distance r apart. If r is much larger than the size of the stars, then their (a) common period of revolution is proportional to r3/2. (b) orbital velocities are in the ratio 2 : 1. (c) kinetic energies are in the ratio 1 : 2 . (d) angular momenta are in the ratio 1 : 4. 15. A space-ship is orbiting close to the surface of the earth at a speed v. The radius of the earth is R and g is the acceleration due to gravity close to the surface of the earth. An additional speed of v0 is to be imparted to the space-ship so that it overcomes the gravitational pull of the earth. Then (b) v = 2Rg (a) v = Rg (c) v0 =

Rg

2 1

(d) v0 =

Rg

2 1

16. A satellite of mass m is moving in a circular orbit of radius r around a planet of mass M. (a) The magnitude of angular momentum with respect to the centre of the orbit is m GMr , where G is the gravitation constant. (b) The magnitude of the angular momentum is mR 2gr where g is the acceleration due to gravity on the surface of the planet. (c) The direction of angular momentum is parallel to the plane of the orbit. (d) The direction of angular momentum is perpendicular to the plane of the orbit. IIT, 1997 17. Two bodies of masses m1 = m and m2 = 4 m are placed at a distance r is zero at a point P at a distance x from mass m1. The gravitational potential at point P is V. Then (U is the gravitational P.E. of the system) (a) x =

r 3

(c) U = –

(b) x = 7G r

2r 3

(d) U = –

9G r

18. A small planet is revolving with speed v around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force between the planet and the star were inversely proportional to r, then (a) v is independent of r. (b) v decreases if r is increased. (c) T increases if r is increased. (d) T is independent of r. 19. The total energy of a satellite of mass m moving with speed v around the earth of mass M in a

Gravitation 9.27

20.

21.

22.

23.

24.

circular orbit of radius r is directly proportional to (a) m (b) M (c) v (d) r Assuming the earth to be a sphere of radius R and uniform density, if the acceleration due to gravity at a point at a distance r from the centre of the earth is g, then 1 for r < R (a) g r for r < R (b) g r 1 (c) g r2 for r > R (d) g for r > R r2 A satellite is revolving around a planet in a circular orbit. During its motion, the satellite begins to experience a resistive force (possibly due to cosmic dust). As a result (a) its kinetic energy will decrease (b) its angular speed will decrease (c) its time period of revolution will increase (d) its angular momentum will decrease. If a satellite revolving around the earth is moved from one stable orbit to another stable orbit of a higher orbital radius, then its (a) time period will increase (b) centripetal force will decrease. (c) gravitational potential energy will increase (d) angular momentum will remain unchanged A comet is moving around the sun in a highly elliptical orbit. Its (a) kinetic energy and gravitational potential energy both change over the orbit (b) total energy remains constant throughout the orbit (c) linear momentum changes in magnitude as well as direction over the orbit (d) angular momentum remains constant over the orbit. IIT, 1998 A body of mass m is released from rest from above at a distance r from the centre of the earth of mass M and radius R. If the air resistance is neglected, it will strike the surface of the earth with a velocity v. If g is the acceleration due to gravity on earth s surface, (a) v =

2Rg if r >> R

(b) v =

Rg if r = 2R

(c) v =

Rg if r = 4R 2

(d) v =

1 Rg if r = 8R 2

25. A satellite of mass m is launched from the surface of the earth in a circular orbit at a height h = R above the surface of the earth; R being the radius of the earth. If g is the acceleration due to gravity on the surface of the earth and the air resistance is neglected, (a) the increase in gravitational potential energy is mgR (b) the velocity with which it was projected from the earth is gR (c) the total energy spent in launching the satel3 mgR lite in the circular orbit is 4 (d) the acceleration due to gravity at the site of the satellite is g/4. 26. Two satellites of the same mass are put in the same orbit round the earth but they revolve in opposite directions. They undergo an inelastic collision and stick together. (a) The ratio of the potential energy of the system before and just after the collision is 2 : 1 (b) The ratio of the total energy of the system before and just after the collision is 1 : 2 (c) After the collision, the combined mass will continue moving in the same orbit (d) After the collision, the combined mass will fall freely under the gravity of the earth. 27. A solid sphere of uniform density and radius 4 units is located with its centre at 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 leaving behind spherical cavities as shown in Fig. 9.25.

Fig. 9.25

(a) the gravitational force due to this object at the origin is zero. (b) the gravitational force at point B (2,0,0) is zero. (c) the gravitational potential is the same at all points of the circle y 2 + z2 = 36. (d) the gravitational potential is the same at all points of the circle y2 + z2 = 4. IIT, 1993

9.28 Comprehensive Physics—JEE Advanced

28.

tances r1 and r2 from the centre of a uniform sphere of radius R and mass M are F1 and F2 respectively. Then F (a) 1 F2 (b)

F1 F2

r1 if r1 < R and r2 < R r2 r22 r12

(c)

F1 F2

r12

(d)

F1 F2

r2 if r1 > R and r2 > R r1

r22

if r1 < R and r2 < R

IIT, 1994

if r1 > R and r2 > R

ANSWERS AND SOLUTIONS The correct choices are (a), (b) and (d). The correct choices are (b), (c) and (d). The correct choices are (a), (b) and (c). The only incorrect statement is (d). A geostationary satellite can be put in orbit only at a height of 35,870 km above the earth. 5. Statement (a) is incorrect; the magnitude of the force is 6.67 10–11 N. Statement (b) is correct. If the escape velocity for a planet is high, the thermal speeds of lighter gases are less than the escape velocity. Therefore, lighter gases are not able to escape. Statement (c) is also correct. Statement (d) is incorrect; force of friction arises due to electrical forces. Hence correct choices are (b) and (c). 6. Statement (a) is incorrect; there is no method by which a body can be shielded from the gravi-

1. 2. 3. 4.

correct. Statement (c) is wrong. As the earth rotates about its axis, a body on the surface of the earth also rotates with it. Since the body is in a rotating (non–inertial) frame, it experiences an outward centrifugal force against the inward force of gravity. As a result, the acceleration due to gravity decreases due to rotation. Statement (d) is incorrect, the forces are equal in magnitude but opposite in direction. Hence choices (a), (c) and (d) are wrong. 7. The correct choices are (c) and (d). The kinetic energy of the comet changes because its speed in the orbit keeps changing. The potential energy changes because the distance of the comet from the sun keeps changing for an elliptical orbit. GM 8. Linear velocity or orbital velocity is v = , r where r is the distance of the satellite from the centre of the earth. Therefore v decreases as r is increased. Also v = r , where is the angular velocity. Therefore =

v 1 GM = = r r r

GM r3

Thus decreases with increase in r. The time period of the satellite is given by r3 GM Thus T increases as r is increased. Hence the correct choices are (c) and (d). 9. Distances of the two satellites from the centre of the earth are r1 = 2R and r2 = 8R respectively. R = earth’s radius. Their potential energies are T =2

and

V1 = –

GmM r1

V2 = –

GmM r2

V1 r2 8R = = = 4. V2 r1 2R The kinetic energy of a satellite can be obtained from relation mv 2 GmM = r r2 1 GmM or K= mv2 = 2 2r GmM GmM Thus K1 = and K2 = 2r1 2r2 The ratio of their kinetic energies is K1 r2 8R = = = 4. K2 r1 2R Their total energies are GmM GmM GmM E1 = – =– 2r1 r1 2r1 Their ratio is

and

E2 = –

GmM r2

GmM GmM =– 2r2 2r2

E1 r2 8R = = = 4. E2 r1 2R Hence the correct choices are (b), (c) and (d).

Their ratio is

Gravitation 9.29

1 GmM 1 mv2 = or KE . Thus 2 2r r choice (a) is correct. Angular momentum = mvr = GM r = m GM r , which is proportional m r to r . Hence choice (b) is wrong. Linear momenGM , which is proportional to tum mv = m r 1 . Hence choice (c) is correct. The frequency r of revolution is

10. Kinetic energy

1 GM 1 1 = i.e. 3 3/ 2 2 r r T Hence the correct choices are (a), (c) and (d). 11. The acceleration due to gravity is GM g= R2 The new value of g would be G 0.99 M GM g = 1.01 1.01 g 2 0.99 R R2 =

Thus g would increase by about 1%. The new escape velocity would be 2 0.99 M 0.99 R

2M G = ve R Thus the escape velocity will remain unchanged. The potential energy of a body of mass m on earth’s surface would be ve =

GM 0.99 M 0.99 R

G

=

GmM R Thus the potential energy will also remain unchanged. Hence the correct choices are (b) and(d). 12. Work done is independent of the path chosen and –

=–

object. Also the work done on any closed path in a the surface of the earth is at the same potential, no work is done for points on the surface of the earth. Hence the correct choices are (a), (c) and (d). 13. If M is the mass of the earth, the escape velocity is 2GM ve = R For a satellite of mass m and orbital radius r (= its distance from the centre of the earth), the orbital speed v is given by GmM mv 2 GM = or v = 2 r r r

But

1 ve = 2

v=

1 2

2GM R

GM 2R

=

GM GM = r 2R or r = 2R. Height above earth = 2R – R = R. Potential energy of the satellite in its orbit is GmM GmM =– ( r = 2R) E1 = – r 2R The kinetic energy is zero because the satellite is stopped. Potential energy of the satellite on the surface of the earth is GmM E2 = – R Loss of PE = E1 – E2 GmM GmM – 2R R GmM = 2R This is converted into kinetic energy. If v is the speed with which the satellite hits the surface of the earth, then from the law of conservation of energy, we have 1 GmM mv2 = 2 2R GM GM or v2 = = gR g R R2 The correct choices are (b) and (d). 14. The distance x of the star of mass m from the centre of mass is given by m 2m = x (r x) =–

r . The orbital speed v1 of the star 3 of mass m1 = m is given by (here m2 = 2 m)

which gives x =

Gm1 m2 r2

=

m1v12 m1v12 = r /3 x

Gm2 2Gm = 3r 3r 2 x 2 r Time period (T) of m = = v1 3

which gives v1 =

=

(i) 3r 2GM 2 (r)3/2 3GM

9.30 Comprehensive Physics—JEE Advanced

or T r3/2. The orbital speed v2 of the star of mass m2 = 2m is given by Gm1m2 m2 v22 2m v22 Gm(2m) = or = (r x) 2r / 3 r2 r2 or

v2 =

v

2Gm = v1 3r

[see Eq. (i)]

1 1 1 m1 v12 and K2 = m2 v22 = (2m) v22 2 2 2 2 K1 v1 1 = = , since v1 = v2. K2 2v22 2 mv1r Angular momentum of m1 is L1 = m1v1 x = 3 Angular momentum of m2 is L2 = m2v2(r – x) Now K1 =

= L1 L2

2 mv 2 2 r 3

1 4

(since v1 = v2)

The correct choices are (a), (c) and (d). 15. The orbital velocity of the space-ship of mass M and orbital radius r is given by GM r where r = R + h; R being the radius of the earth and h the height of the space-ship above the surface of the earth. For a space-ship close to the earth s surface r = R. Therefore v=

GM GM = Rg g R2 R The space-ship will overcome the gravitational pull of the earth if it is given an additional v0 such that v + v0 = ve where ve is the escape velocity which is given by v=

2GM = 2Rg R The required addition velocity is ve =

v0 = ve – v = =

2Rg –

The magnitude of angular momentum is L = mrv sin where is the angle between vectors r and v. For a circular orbit, = 90°. Therefore L = mrv (1) The gravitational force of attraction on the satellite is GMm/r2 which provides the necessary centripetal force mv2/r, i.e. GMm r2

=

m v2 r

or

(mrv) = m GMr From Eqs. (1) and (2), we have

Rg

(2)

L = m GMr This gives the magnitude of angular momentum. The direction of angular momentum is perpendicular to the plane of the orbit. The correct choices are (a) and (d). 17. Consider a point P at a distance x from mass m1; its distance from mass m2 is (r – x). The net gravitaP will be zero if Gm1 x

2

=

Gm2

x)2

(r

or m2 m1 = (r x) x which gives

2 1 The correct choices are (a) and (d). 16. At a certain instant of time let r be the radius vector of the satellite from the centre of its circular orbit. If the velocity of the satellite is v as shown in Fig. 9.26, its angular momentum is given by L = r (mv) Rg

Fig. 9.26

r m1

x=

m1

(1)

m2

r m2

(2) m1 m2 The gravitational potential at point P is given by Gm1 Gm2 (3) U=– x r x Using (1) and (2) in (3), we have (r – x) =

U = – G m1

m1 r m1

m2

m2

m1 r m2

m2

Gravitation 9.31

G m1 m1 m2 m2 m1 m2 r G =– (4) m1 m2 2 m1m2 r Putting m1 = m and m2 = 4m in Eqs. (1) and (4) we =–

18. F =

k , where k is a constant. r

mv 2 k = r r

v=

k , which is independent of r. m

2 r m = 2 r T r. v k Hence the correct choices are (a) and (c). T=

GmM . Hence the correct choic19. Total energy E = – es are (a) and (b). 2r 20. The correct choices are (a) and (d). 21. Due to friction, the satellite will slow down. Hence its kinetic energy will decrease. Now v = R . Therefore, a decrease in v results in a decrease in . Since T = 2 / , time period T will increase. The angular momentum will remain unchanged because the torque due to a radial (central) force is zero. Hence the correct choices are (a), (b) and (c). GM . If r increases, v will 22. Orbital speed v = r r3 2 r =2 . Hence decrease. Time period T = GM v T will increase. Centripetal force = gravitational GmM force = . Hence if r increases, centripetal r2 GM force will decrease. Gravitational P.E. = – . r If r increase, P.E. becomes less negative, i.e. it increases. Since no external torque acts, the angular momentum is conserved. Hence all the four choices are correct. 23. The speed of the comet is greater when it is closer to the sun than when it is farther from it. Hence both v and r change for an elliptical orbit. Thus choice (a) is correct. The total energy is conserved. The linear momentum changes because the velocity changes in magnitude as well as direction. Since the force is central, the angular momentum is conserved. Hence all the four choices are correct. 24. Since the initial velocity of the body is zero, its total energy is Mm Ei = – G r

When the body reaches the earth s surface, its velocity is v and its distance from the centre of the earth is the earth s radius R. Therefore 1 Mm Ef = mv2 – G 2 R From energy conservation, Ei = Ef, i.e. Mm 1 Mm mv2 – G =–G R 2 r which gives v = R 2g

1 R

choices are (a) and (b). 25. (a) Increase in P.E. = – =

1 r

1/ 2

GmM ( R h)

GmM R

GMm 2R

(

h = R)

GM 1 mgR g R2 2 (b) If v is the velocity of projection, then from energy conservation, we have 1 GMm GMm mv2 – =– 2 R R h =

Using h = R, this equation gives v = gR . (c) Total energy of the satellite orbiting at h = R is GMm GMm GMm – =– E1 = K.E. + P.E. = 4R 4R 2R Total energy when the satellite is at rest on the surface of the earth is GMm E2 = – R Energy required is E = E1 – E2 =–

GMm – 4R

3 3 GMm GMm = = mgR 4 4 R R 2

g . 4 R h Hence the correct choices are (b), (c) and (d).

(d) Value of g at h (= R) = g

R

=

GMm . Therer fore, P.E. of two satellites before collision is GMm 2GMm (P.E.)i = 2 =– r r

26. (a) P.E. of a satellite in orbit = –

After the collision, they stick together. So the combined mass is 2m. Therefore, P.E. just after collision is

9.32 Comprehensive Physics—JEE Advanced

the combined mass will not move in an orbit; it will fall freely under the gravity of the earth. Hence the correct choices are (b) and (c). 27. The distance of each cavity from the centre O is the same. Since the two cavities are symmetrical with respect to the centre O and the mass of the sphere can be regarded as being concentrated at the centre O, the gravitational force due to the sphere is zero at the centre. Hence choice (a) is correct. For the same reason, the gravitational potential is the same at all points of the circle y 2 + z 2 = 36 whose radius is 6 units and at all points of the circle y2 + z2 = 4. Hence choices (c) and (d) are also correct. But the gravitational force at point B cannot be zero.

GM 2m 2GMm =– r r Hence the ratio of P.E. before and after collision is 1 : 1. (b) Total energy of a satellite in orbit is (P.E.)f = –

GMm GMm GMm + =– 2r 2r r Therefore, the total energy of the two satellites before collision is GMm GMm Ei = 2 =– 2r r Let V be the velocity of the combined mass. Final momentum = (2m)V. Since the satellites have opposite velocities, initial momentum = mv – mv = 0.

K.E. + P.E. =

28. (i) For r > R, F =

2 mV = 0 or V = 0. Hence the combined mass has no kinetic energy; it has only potential energy. Thus, total energy of the system just after the collision is Ef = –

GM r2

. Therefore,

F1 r22 = F2 r12 r22 r12 So choice (b) is correct and choice (d) is wrong. GMr . Therefore, (ii) For r < R, F = R3 F1 =

GM (2m) 2GMm =– r r Ei 1 = Ef 2

GM

F1 =

and F2 =

GMr1 3

, F2 =

GM

. Hence

GMr2 3

. Hence

F1 r = 1 F2 r2

R R So choice (a) is correct and choice (c) is wrong.

Since the combined mass comes to rest, the centripetal force disappears ( velocity V = 0). Therefore,

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I A satellite of mass m is revolving in a circular orbit of radius r around the earth of mass M. The speed of the satellite in its orbit is one-fourth the escape velocity from the surface of the earth. 1. The height of the satellite above the surface of the earth is (R = radius of earth) (a) 2R (b) 3R (c) 5R (d) 7R

2. The magnitude of angular momentum of the satellite is m (b) (a) m GMR GMR 2 m (c) (d) 2m GMR GMR 2 2 3. If the total energy of the satellite is E, its potential energy is (a) – E (b) E (c) 2E (d) – 2E

SOLUTION 1. v =

GM = r

GM ( R h)

(1)

v=

ve 1 2GM = 4 4 R

(2)

Gravitation 9.33

Equations (1) and (2) give h = 7R, which is choice (d). 2. L = mvR = m

GM 8R

R=

m 2 2

The correct choice is (c). 3. The correct choice is (c).

GMR (

h = 7 R)

Questions 4 to 8 are based on the following passage Passage II Considering the earth as an isolated mass, a force is experienced by a body at any distance from it. This force is directed towards the centre of the earth and has a magnitude mg, where m is the mass of the body and g is the acceleration due to gravity. The value of the acceleration due to gravity decreases with increase in the height above the surface of the earth and with increase in the depth below the surface of the earth. Even on the surface of the earth, the value of g varies from place to place and decreases with decrease in the latitude of the place. 4. Assuming the earth to be a sphere of uniform mass density, which of the graphs shown in Fig. 9.27 represents the variation g with distance r from the centre of the earth (R = radius of earth)

5. A body weighs 63 N on the surface of the earth. How much will it weigh at a height equal to half the radius of the earth? (a) 63 N (c) 28 N

(b) 32.5 N (d) none of these

6. Assuming the earth to be a sphere of uniform mass density, the weight of a body when it is taken to the end of a tunnel 32 km below the surface will (radius of earth = 6400 km) (a) decrease by 0.5% (b) decrease by 1% (c) increase by 0.5% (d) increase by 1% 7. If gp is the acceleration due to gravity at the poles and ge that at the equator, then (a) gp < ge

(b) gp > ge

(c) gp = qe

(d) ge = 0

8. If a tunnel is dug along a diameter of the earth and a body is dropped from one end of the tunnel, (a) it will fall and come to rest at the centre of the earth where its weight becomes zero. (b) it will emerge from the other end of the tunnel. (c) it will execute simple harmonic motion about the centre of the earth. (d) it will accelerate till it reaches the centre and decelerate after that eventually coming to rest at the other end of the tunnel.

Fig. 9.27

SOLUTION 4. At a height h above the surface of the earth, gh =

g0 R 2

( R h) 2

At a depth d below the surface of the earth gd = g0 1

d R

where g0 = acceleration due to gravity at the surface of the earth. Hence the correct choices is (d).

5. W = W0

2

R R h

2

R

= 63 R

R 2

= 28 N

199 g0 32 d = g0 1 = 200 6400 R Decrease in weight = mg0 – mg

6. g = g0 1

199 mg 0 = 200 200 Hence the correct choice is (a). = mg0 1

9.34 Comprehensive Physics—JEE Advanced

7. Due to the rotation of the earth about its axis, g p > g e. Questions 9 to 12 are based on the following passage Passage III The escape velocity on a planet or a satellite is the minimum velocity with which a body must be projected from that planet so that it escapes the gravitational pull of the planet goes into outer space. We obtain the expression for the escape velocity by equating the work required to with the initial kinetic energy given to the body. The escape velocity from a planet of mass M and radius R is given by ve =

2MG = R

2gR

where g is the acceleration due to gravity on the surface of the planet and G is the gravitation constant. 9. Choose the only incorrect statement from the following. The escape velocity from a planet (a) is independent of the mass of the body which is projected. (b) is independent of the direction in which the body is projected. (c) depends on the mass and radius of the planet. (d) will be less than the value given by the ex2MG if the planet has an pression ve = R atmosphere.

SOLUTION 9. The only incorrect statement is (d). Due to atmospheric friction, a part of the initial kinetic energy is lost as heat. Hence the actual value of the escape velocity is greater than the value obtained from the given expression. vJ = 10. vE

MJ ME

1 = 29 319 11 Hence the correct choice is (a). 11. The escape velocity at a height h is given by RE = RJ

ve =

2g R h

8. The correct choice is (c). 10. The mass of Jupiter is about 319 times that of the earth and its radius is about 11 times that of the earth. The ratio of the escape velocity on Jupiter to that on earth is (a) (c)

29 1

(b) 29 (d)

1

29 29 11. If R is the radius of the earth and g the acceleration due to gravity on its surface, the escape velocity of a body projected from a satellite orbiting the earth at a height h = R from the surface of the earth will be (a)

gR

(b)

2gR

(c)

3gR

(d) 2 gR

12. A body is dropped from a height equal to half the radius of the earth. If ve is the escape velocity on earth and air resistance is neglected, it will strike the surface of the earth with a speed ve v (b) e (a) 2 2 (c)

ve 3

(d)

ve 3

where g is the acceleration due to gravity at height h, R 2 g =g R h For h = R, we get ve = gR , which is choice (a). 12. The correct choice is (c). Use conservation of energy, i.e. Total energy at h (= R/2) = Total energy when the body strikes the earth –

GmM 1 GmM = mv2 – ( R h) 2 R

Gravitation 9.35

IV Assertion Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 A body is projected up with a velocity equal to half the escape velocity from the surface of the earth. If R is the radius of the earth and atmospheric resistance is neglected, it will attain a height h = R/3. Statement-2 The gravitational potential is – GM/R on the surface of the earth and it increases with height; M being the mass of the earth. 2. Statement-1 The total energy (kinetic + potential) of a satellite moving in a circular orbit around the earth is half its potential energy. Statement-2 The gravitational force obeys the inverse square law of distance. 3. Statement-1 Two bodies of masses m1 = m and m2 = 3m are iniallowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach when they are at a separation r is v =

2Gm r

Statement-2 The gain in the kinetic energy of each body equals the loss in its gravitational potential energy. 4. Statement-1 An astronaut inside a massive space-ship orbiting gravitational force. Statement-2 The centripetal force necessary to keep the spaceship in orbit around the earth is provided by the gravitational force between the earth and the spaceship. 5. Statement-1 The escape velocity varies with altitude and latitude of the place from where it is projected. Statement-2 The escape velocity depends on the gravitational potential at the point of projection. 6. Statement-1 A comet orbits the sun in a highly elliptical orbit. Its potential energy and kinetic energy both change over the orbit but the total energy remains constant throughout the orbit. Statement-2 For a comet, only the angular momentum remains constant throughout the orbit. 7. Statement-1 The acceleration due to gravity decreases due to rotation of the earth. Statement-2 A body on the surface of the earth also rotates with it in a circular path. A body in a rotating (non-inertial) frame experiences an outward centrifugal force against the inward force of gravity.

SOLUTIONS 2Gm and R total energy at r (= R + h) = total initial energy, i.e. GmM 1 GmM – = mv2 – r 2 R 2. The correct choice is (a). The centripetal force needed for circular motion is provided by the 1. The correct choice is (b). Use ve =

gravitational force. Since the gravitational force obeys the inverse square law of distance, the orbital velocity of the satellite is given by v=

GM r

where M = mass of earth and r = orbital radius.

9.36 Comprehensive Physics—JEE Advanced

Therefore

1 m1v12 2

Gm1m2 or v1 r

2Gm2 r

Gm1m2 or v2 r

2Gm1 r

1 GmM Kinetic energy = mv2 = 2 2r where m = mass of the satellite. From the inverse

and

the satellite is given by

Therefore, their relative velocity of approach is

GmM r Total energy E = K.E. + P.E. GmM GmM = + 2r r

Potential energy = –

GmM P.E. =– = 2r 2 3. The correct choice is (d). Initially when the two masses are at an distance from each other, their gravitational potential energy is zero. When they are at a distance r from each other the gravitational P.E. is Gm1m2 P.E = – r The minus sign indicates that there is a decrease in P.E. This gives rise to an increase in kinetic energy. If v1 and v2 are their respective velocities when they are at a distance r apart, then, from the law of conservation of energy, we have

1 m1v22 2

v = v1 + v2 = =

2Gm2 + r 2G (m2 r

2Gm1 r m1 )

Putting m1 = m and m2 = 3m, we get v = 2

2Gm r

4. The correct choice is (b). Because the centripetal force equals the gravitational force exerted by the earth on the space-ship, the astronaut does not experience any gravitational force of the earth. The only force of gravity that an astronaut in an orbiting space-ship experiences is that which is due to the gravitational force exerted by the space-ship. Since but very small. 5. The correct choice is (a). The gravitational potential at a point varies with the altitude and latitude of the place. 6. The correct choice is (c). 7. The correct choice is (a).

10

Elasticity

Chapter

REVIEW OF BASIC CONCEPTS 10.1

ELASTICITY

The ability of a body to regain its original shape and size when the deforming force is withdrawn, is known as elasticity.

10.2

STRESS

When a deforming force is applied to a body, it reacts to the applied force by developing a reaction (or restoring) force which, from Newton’s third law, is equal in magnitude and opposite in direction to the applied force. The reaction force per unit area of the body which is called into play due to the action of the applied force is called stress. Stress is measured in units of force per unit area, i.e. Nm–2. Thus F Stress = A where F is the applied force and A is the area over which it acts.

10.3

STRAIN

When a deforming force is applied to a body, it may ratio of the change in size or shape to the original size or shape of the body. Strain is a number; it has no units or dimensions. The ratio of the change in length to the original length is called longitudinal strain. The ratio of the change in volume to the original volume is called volume strain. The strain resulting from a change in shape is called shearing strain.

10.4 HOOKES’ LAW Hookes’ law states that, within the elastic limit, the stress developed in a body is proportional to the strain produced

in it. Thus the ratio of stress to strain is a constant. This constant is called the modulus of elasticity. Thus stress Modulus of elasticity = strain Since strain has no unit, the unit of the modulus of elasticity is the same as that of stress, namely, Nm–2.

10.5

YOUNG’S MODULUS

Suppose that a rod of length L and a uniform crosssectional area A is subjected to a longitudinal pull. In other words, two equal and opposite forces are applied at its ends. F Stress = A The stress in the present case is called linear stress, tensile stress, or extensional stress. If the direction of the force is reversed so that L is negative, we speak of compressional strain and compressional stress. If the elastic limit is not exceeded, then from Hooke’s law Stress strain or Stress = Y strain stress F L (1) strain A L where Y, the constant of proportionality, is called the Young’s modulus of the material of the rod and may be ratio of the linear stress to linear strain, provided the elastic limit is not exceeded. Since strain has no unit, the unit of Y is Nm–2.

or

10.6

Y =

BULK MODULUS

Solids, liquids and gases can be deformed by subjecting them to a uniform normal pressure P in all directions. Stress and pressure have the same dimension (force per unit area), but pressure is not the same thing as

10.2 Comprehensive Physics—JEE Advanced

stress. Pressure is the force per unit area acting on the surface of a system, the force being everywhere perpendicular to the surface so that, for a uniform pressure, the force per unit area is the same. Pressure is a particular kind of stress which changes only the volume of the substance and not its shape. The substance may be a solid, liquid or gas. A small increase in pressure P applied to a substance decreases its volume from, say, V to V – V so that V is the small decrease in volume. The volume strain is given by V Volume strain = – V

Suppose the lower face PQRS F is applied parallel to the upper face MNUV. As a result of this, the lines joining the two faces turn through an angle . We say that the face MNRS is sheared through an angle (measured in radians). The angle is called the shear strain or the angle of shear and is a measure of the degree of deformation. If A is the area of the face MNUV, the ratio F/A is the shearing stress. It is found that for small deformation, the shearing stress is proportional to the shear strain, i.e. F F F , = ,n= A A A

pressure and the corresponding volume strain, i.e. P PV B= (2) V V V If P is positive, V will be negative and vice versa. The B ensures that B is always positive. The SI unit of B is Nm–2. The reciprocal of B is known as compressibility. The bulk modulus of a gas depends on the pressure. Under isothermal conditions (i.e. when the temperature is kept constant), the bulk modulus of a gas is equal to its pressure P. Under adiabatic conditions (i.e. when heat is not allowed to leave or enter the system), the bulk modulus is equal to P, where (= Cp/Cv) is the ratio of the molar heat capacities of the gas at constant pressure and constant volume. Thus Isothermal bulk modulus = P Adiabatic bulk modulus = P.

The quantity n is called the shear modulus or the modulus of rigidity. Referring to Fig. 10.1, if is small,

10.7

SHEAR MODULUS OR MODULUS OF RIGIDITY

Shear is a particular kind of stress which only solids can withstand. The solid is deformed by changing its shape without changing its size. The body does not move or rotate as a whole: there is a relative displacement of its contiguous layers. Consider a solid in the form of a rectangular cube as in Fig. 10.1.

tan

L L

F L A L This equation looks similar to Eq. (1) for Young’s modulus with the difference that F/A here is the tangential stress and not longitudinal stress. so that

10.8

=

POISSON’S RATIO

When a wire is stretched with a force, apart from an increase in its length, there is a slight decrease in its diameter, i.e. both shape and volume change under longitudinal stress. The ratio of the decrease D in diameter to the original diameter D is called lateral strain, i.e. strain at right angles to the deforming force. Thus change in diameter D Lateral strain = original diameter D change in length L Also Longitudinal strain = original length L The ratio of the two is called Poisson’s ratio and is denoted by . Hence, =

Lateral strain Longitudinal strain

D/ D L/ L

D L D L Since it is a ratio between two types of strain, is dimensionless. Theoretically, one can show that it must be less than 0.5. For most solids it lies between 1/4 and 1/3, and for rubber it is very nearly 0.5. =

10.9 Fig. 10.1

=

ENERGY STORED IN A STRAINED WIRE: STRAIN ENERGY

If a wire is stretched, the energy stored per unit volume is given by

Elasticity 10.3

U = =

1 stress 2 1 S 2

where

S = stress,

Since

Y = U =

10.10

10.1 A steel wire (original length = 2 m, diameter = 1mm) and a copper wire (original length = 1m, diameter = 2 mm) are loaded as shown in Fig. 10.2. Find the ratio of extension of steel wire to that of copper wire. Young’s modulus of steel = 2 1011 Nm–2 and that of copper = 1 1011 Nm–2.

strain

= strain

S 1 S 2

=

1 Y 2

2

=

1 S2 2 Y

THERMAL STRESSES

then due to expansion or contraction, tensile or compressive stress is set up in the rod. These stresses are called thermal stresses. If a rod of length L is free to expand or contract and its temperature is changed by T, the change in its length is given by L =

L

F =

FL A L FL 4F L L= = YA Y d2 Y=

For steel wire Fs = (6 + 4) 9.8 = 98 N For copper wire Fe = 6 9.8 = 58.8 N

FL AY

( L)s =

AY T

Thus, the thermal stress in the rod is F = A

SOLUTION

T

where Now from Eq. (1), we have L =

Fig. 10.2

4 Fs Ls Ys

d Ls

Y T

Lc

its volume cannot change, then a change in temperature results in a change in pressure. The thermal stress is then given by P = B T where B

10.11

is its

TORQUE

If a rod or wire of length l and radius r and the other end is twisted through an angle (in radian) by applying a torque, the restoring torque is given by =

r4 2l

Where is the shear modulus of the material of the rod or wire.

and ( L)c =

2 s

F = s Fc

Ls Lc

=

98 58.8

=

20 3

2 1

Yc Ys 1 2

4 Fc Lc

d c2

Yc dc ds 2

2

2

10.2 When the pressure on all sides of a metal cube is increased by 107 Pa, its volume decreases by 0.015% . Find the bulk modulus of the metal. SOLUTION P = 107 Pa,

0.015 100

V = V B=

P = V /V

= 6.67

107 0.015 /100

1010 Pa or Nm–2

10.4 Comprehensive Physics—JEE Advanced

10.3 A rubber cube is subjected to a pressure of 106 Pa on all sides. As a result, each side of the cube decreases by 1%. Calculate the bulk modulus of rubber. SOLUTION Volume of cube is V = L3. Therefore V 3 L = =3 V L

– 1% = – 3% = 106 = 3.33 3/100

P = V/V

B=

3 100 107 Nm–2

10.4 The base of a rubber cube of side 3.0 cm is clamped. A horizontal force F of 2.7 N is applied on the top face. Calculate (a) the angle of shear and (b) the horizontal displacement of the top face of the eraser. Shear modulus of rubber = 1.5 105 Nm–2. SOLUTION Refer to Fig. 10.1 and also to Fig. 10.3.

Fig. 10.3

Area of top face (A) = 3 cm 3 cm = 9 cm2 = 9 10–4 m2 F 2.7 Shearing Stress = = 3 103 Nm–2 A 9 10 4 (a) Angle of Shear

= =

(b) tan

x = L

10.5 In this example, Statement-I is followed by Statement-II. Choose (a) if statement-II is true and statement-II is true and statement-II is the correct explanation for statement-I. (b) if statement-I is true and statement-II is true but statement-II is not the correct explanation for statement-I. (c) if statement-I is true and statement-II is false. (d) if statement-I is false and statement-II is true. (i) Statement-I: A wire of length L extends by an amount L when a block of weight W is hung from it as shown in Fig. 10.4(a). If the same wire goes over a frictionless pulley and two blocks each of weight W are hung at its ends as shown in Fig 10.4(b), the wire will extend by 2 L, Fig. 10.4 if the elastic limit is not exceeded. Statement-II: For a given load, the extension is proportional to the length of the wire. (ii) Statement-I: Steel is more elastic than rubber. Statement-II: For a given deforming force, a steel wire is extended by an amount less than rubber band of the same radius and the same length. (iii) Statement-I: Figure 10.5 shows the stress-strain curves for two materials 1 and 2. It follows from the graphs that material 1 has higher Young’s modulus then material 2. Statement-II: Within elastic limit, the slope of the stress-strain graph is greater for material 1 than for material 2.

shearing stress shear modulus 3 103 5

1.5 10

= 2.0

10–2 rad

x = L tan = L ( is small) = 3 cm 2.0 10–2 = 6 10–2 cm = 0.6 mm

Fig. 10.5

Elasticity 10.5

(iv) Statement-I: It follows from Fig. 10.5 than material 1 is more brittle than material 2. Statement-II: The plastic range is more for material 1 than for material 2. (v) Statement-I: A hollow shaft is stronger than a solid shaft both made of the same material and the same external dimensions. Statement-II: To produce a given twist, a greater torque is needed for the hollow shaft than for the solid shaft. SOLUTIONS (i) Statement-II is true. Since the length of the wire and tension in wire in cases (a) and (b) shown in Fig. 10.4 are the same, the extension in the wire will be the same equal to L in both case. So Statement-I is false. The correct choice is (d) (ii) Ys > Yr, i.e.

FL FL < or ( L)s < ( L)r AYs AYr

So Statement-II is true. This means than steel develops less strain that rubber and hence steel, can withstand greater stress than rubber. So Statement-I is also true. The correct choice is (a). (iii) Young’s modulus = slope of the straight portion of the stress – strain graph. Hence Young’s modulus of material 1 is greater than that of material 2. So the correct choice is (a). (iv) The plastic range AB of material 2 is less than that of material 1. So the breaking strain of material 2 is less than that of material 1. Hence material 2 is more brittle than material 1. So, Statement-I is false and Statement-I is true. (v) The torque per unit angular twist for a solid shaft is r4 (1) 2L If r1 and r2 are the external and internal radii of the hollow shaft, the torque per unit angular twist is s

h

=

=

2L

(r12

r22 )(r12

i.e.

h

=

r2 2 (r1 2L

=

s

r22 )

r22

1

r2

(

r1 = r)

> s. So the correct choice is (a).

10.6 A heavy and thick rubber rope is hung frame support. It is stretched by its own weight. The strain in the rope is (a) the maximum near the support. (b) the maximum near the free end of the rope (c) the maximum at the centre of the rope (d) the same everywhere. SOLUTION If M is mass of the rope and L its length, then the tension at a point P at a distance y from the free end is (see Fig. 10.6) Tp =

Mg L

y

Thus the tension is maximum at A and zero at B. Now

Fig. 10.6

L T L AY Hence the strain is maximum near A and minimum near B. So the correct choice is (a). Strain =

10.7 A metal rod of length 1.0 m and cross-sectional area 1.0 mm2 supports. The tension in the rod is zero when the temperature is 40°C. Find the tension in the rod when the temperature falls to 20°C. Young’s modulus of metal = 2.0 1011 Nm–2 pansion = 1.0 10–5 K–1. SOLUTION

(r14 r24 ) 2L

=

h

r22 )

Since the volumes are the same, = r2L. Using this eq. (2), we get

(2) (r12

r22 ) L

Thermal stress = Y T T = Y T A T =A Y T = (1.0 10–5) (1.0 10–4) (2.0 1011) (40 – 20) = 40 N

10.6 Comprehensive Physics—JEE Advanced

10.8

Now

A bob of mass 1 kg is suspended by a rubber cord 2.0 m long and of cross-sectional area 1.0 cm2 . It is revolved in a horizontal circle of radius 1.0 m at a speed of 4 ms–1. If Young’s modulus of rubber is 5 108 Nm–2, calculate the extension of the cord. Take g = 10 ms–2.

L=

TL AY

= 7.5

18.87 (1.0 10

4

)

2.0 (5 108 )

10–4 m = 0.75 mm

10.9 A wire of length L is suspended from a support. Its length becomes L1 when its is loaded with a block of mass m1 and L2 when it is loaded with a block of mass m2. Then 1 L1 L2 (b) L ( L1 L2 ) (a) L 2

SOLUTION Refer to Fig. 10.7. Let T be the tension in the cord.

(c) L

L1m2 (m1

L2 m1 m2 )

(d) L

L1m2 (m2

L2 m1 m1 )

v

SOLUTION L=

mgL AY

L1 =

m1 gL AY

L1 – L =

m1 gL AY

L2 =

m2 gL AY

L2 – L =

m2 gL AY

Fig. 10.7

Figure 10.7(b) shows the free body diagram of the bob. From the diagram, it follows that mv 2 (centripetal force) T sin = r and T cos = mg Squaring and adding these equations, we get T=

=

mv 2 r

2

1.0 42 1.0

(mg ) 2

and

2

(i)

(ii)

Dividing (i) by (ii) L1 L2

(1.0 10) 2

m1 L = m2 L

L

L1m2 m2

L2 m1 m1

So the correct choice is (d).

= 18.87 N

I Multiple Choice Questions with Only One Choice Correct 1. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied? (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 IIT, 1981

Elasticity 10.7

2. When a wire of length L is stretched with a tension F, it extends by l. If the elastic limit is not exceeded, the amount of energy stored in the wire is 1 Fl (a) F l (b) 2 (c)

Fl 2 L

(d)

Fl 2 2L

IIT, 1990 3. A wire, suspended from one end, is stretched by attaching a mass of 2 kg to the lower end. The wire stretches by 1 mm. The ratio of the energy gained by the wire to the gravitational energy lost by the mass in dropping a distance of 1 mm is (Take g = 10 ms–2) (a) 1 (b) 1/2 (c) 1/3 (d) 1/4 4. When the pressure on a metal cube is increased by 107 Pa, its volume decreases by 0.015%. The bulk modulus of the metal (in Nm–2) is (a) 1.5 1010 (b) 3.33 1010 10 (c) 6.67 10 (d) 7.5 1010 5. A rubber cube is subjected to a pressure of 106 Pa on all sides. As a result, each side of the cube decreases by 1%. The bulk modulus of rubber is (in Nm–2) (a) 1 108 (b) 3 107 7 (c) 3.33 10 (d) 6.67 107 6. Two wires A and B made of the same material have their lengths in the ratio 1 : 2 and their diameters in the ratio 2 : 1. If they are stretched with the same force, the ratio of the increase in the length of A to that of B is 1 (a) 1 (b) 2 1 1 (c) (d) 4 8 7. A copper wire of length 1.0 m and a steel wire of length 0.5 m having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by 1 mm. If the Young’s modulii of copper and steel are 1.0 1011 Nm–2 and 2.0 1011 Nm–2, the total extension of the composite wire is (a) 1.25 mm

(b) 1.50 mm

(c) 1.75 mm

(d) 2.0 mm

8. Two springs of equal lengths and equal crosssectional areas are made of materials whose Young’s modulii are in the ratio of 3 : 2. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of

(a) 3 : 2 (c) 3 3 : 2 2

(b) 3 : 2 (d) 9 : 4

9. A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. The work one in stretching the wire by an amount x is given by YA x 2 L YA L2 (c) x (a)

YA x 2 2L YA L2 (d) 2x (b)

IIT, 1987 10. A solid sphere of radius R and made of a material of bulk modulus K is completely immersed in a liquid in a cylindrical container. A massless piston of area A M is placed on the piston to compress the liquid, the fractional change in the radius of the sphere, R/R is given by Mg Mg (a) (b) KA 2K A Mg Mg (c) (d) 3K A 4K A 11. A steel wire of cross-sectional area 3 10–6 m2 can withstand a maximum strain of 10–3. Young’s modulus of steel is 2 1011 Nm–2. The maximum mass the wire can hold is (Take g = 10 ms–2) (a) 40 kg (b) 60 kg (c) 80 kg (d) 100 kg 12. A rubber eraser 3 cm 1 cm 8 cm is clamped at one end with the 8 cm edge vertical. A horizontal force of 2.4 N is applied at the free end (the top face). If the shear modulus of rubber is 1.6 105 Nm–2, the horizontal displacement of the top face will be (a) 1 mm (b) 2 mm (c) 3 mm (d) 4 mm 13. The density of water at the surface of the ocean is . If the bulk modulus of water is B, what is the density of ocean water at a depth where the pressure is nP0, where P0 is the atmospheric pressure? (a)

B

B n 1 P0

(b)

B

B n 1 P0

B B (d) B nP0 B nP0 14. One end of a uniform rod of mass M and crosssectional area A is suspended from rigid support and an equal mass M is suspended from the other end. The stress at the mid-point of the rod will be (c)

10.8 Comprehensive Physics—JEE Advanced

2M g A Mg (c) A (a)

(b)

3M g 2A

(d) zero

15.

of linear expansion 1 and 2 and Young’s modulii Y1 and Y2 rods are heated to the same temperature. There is no bending of rods. If 1 : 2 = 2 : 3, the thermal stresses developed in the two rods will be equal provided Y1 : Y2 is equal to (a) 2 : 3 (b) 1 : 1 (c) 3 : 2 (d) 4 : 9 16. One end of a uniform wire of length L and of weight W is attached rigidly to a point in the ceiling and a weight W1 is suspended from its lower end. If S is the area of cross-section of the wire, the stress in the wire at a height 3L/4 from its lower end is W (a) W1/S (b) W1 S 4 3W (c) W1 S (d) (W1 + W)/S 4 17. A uniform wire (Young’s modulus 2 1011 Nm–2) is subjected to a longitudinal tensile stress of 5 107 Nm–2. If the overall volume change in the wire is 0.02%, the fractional decrease in the radius of the wire is (b) 1.0 10–4 (a) 1.5 10–4 –4 (c) 0.5 10 (d) 0.25 10–4 IIT, 1993 18. An elastic spring of unstretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is (a)

1 ky2 2

(b)

1 k(x + y)2 2

(d)

(c) 2b –

a 2

(a) 2 (c) 3

(d) 4a – 3b IIT, 1987

1011 N/m2 1012 N/m2

(b) 2 (d) 3

10–11 N/m2 10–12 N/m2 IIT, 2003

22. A light rod of length L is suspended from a support horizontally by means of two vertical wires A and B of equal length as shown in Fig. 10.9. The crosssectional area of A is half that of B and the Young’s modulus of A is twice that of B. A weight W is hung as shown. The value of x so that W produces equal stress in wires A and B is (a)

L 3

(b)

L 2

(c)

2L 3

(d)

3L 4

1 k (x2 + y2) 2

1 ky(2x + y) 2 IIT, 1994 19. The length of an elastic string is a metre when the longitudinal tension is 4 N and b metre when the tension is 5 N. The length of the string (in metre) when the longitudinal tension is 9 N is (a) a – b (b) 5b – 4a (c)

20. The Poisson’s ratio of a material is 0.4. If a force is applied to a wire of this material, there is a decrease of cross-sectional area by 2%. The percentage increase in its length is : (a) 3% (b) 2.5% (c) 1% (d) 0.5% 21. Fig. 10.8 shows a graph of 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 crosssectional area of the wire is 10–6 m2, the Young’s modulus of the material of the Fig. 10.8 wire is

A

B

TA

TB L

x

C C W

Fig. 10.9

23. In Q. 22 above, the value of x at which W produces equal strain in wires A and B is (a)

L 4

(b)

L 2

Elasticity 10.9

2L 4L (d) 3 5 24. A wire breaks when subjected to a stress S. If is the density of the material of the wire and g, the acceleration due to gravity, then the length of the wire so that it breaks by its own weight is (c)

(a)

gS

(b)

(c)

gS

(d)

(c)

mg cos AY

(b)

g S S g

1 ML 2 1 (c) ML 8

(a)

1 1 p/ B

mg AY cos

(b) 1

2

2

1 ML 4 3 (d) ML 8

(b)

2

2

28. A rubber cord of mass M, length L and cross-sectional area A is hung vertically from a ceiling. The Young’s modulus of rubber is Y. If the change in the diameter of the cord due to its own weight is neglected, the increase in its length due to its own weight is

mg sin mg (c) (d) AY AY sin 26. The density of a metel at a normal pressure is . Its density when it is subjected to an excess pressure p is . If B is the bulk modulus of the metal, the ratio / is (a)

(d) 1 + B/p

27. A thin uniform metallic rod of mass M and length L is rotated with a angular velocity in a horizontal plane about a vertical axis passing through one of its ends. The tension in the middle of the rod is

25. A stone of mass m is attached to one end of a wire of cross-sectional area A and Young’s modulus Y. The stone is revolved in a horizontal circle at a speed such that the wire makes an angle with the vertical. The strain produced in the wire will be. (a)

1 1 p/ B

(a)

MgL AY

(b)

MgL 2 AY

(c)

2MgL AY

(d)

MgL 4 AY

29. The volume of a wire remains unchanged when the wire is subjected to a certain tension. The Poisson’s ratio of the material of the wire is (a) 0.25 (b) 0.3 (c) 0.4 (d) 0.5

p B

ANSWERS 1. 7. 13. 19. 25.

(a) (a) (a) (b) (b)

2. 8. 14. 20. 26.

(b) (a) (b) (b) (a)

3. 9. 15. 21. 27.

(b) (b) (c) (a) (d)

4. 10. 16. 22. 28.

(c) (c) (c) (c) (b)

5. 11. 17. 23. 29.

(b) (b) (d) (b) (d)

6. 12. 18. 24.

(d) (d) (d) (d)

SOLUTIONS FL 4 FL = AY d 2Y Since the wires are made of the same material, Y is the same for all wires. For a given tension F, L L d2 L The value of 2 is the largest for L = 50 cm and d = 0.5 mm. d 1 2. Energy stored per unit volume = (stress strain). 2 If A is the cross-sectional area of the wire, the total 1. Extension

L=

energy store in the wire is

1 (stress strain) 2 1 F l = AL 2 A L

U=

volume of wire

1 Fl 2 3. Energy gained by the wire is 1 1 mg l U1 = Fl = 2 2 1 = 2 10 (1 2 =

10–3) = 0.01 J

10.10 Comprehensive Physics—JEE Advanced

Gravitational P.E. lost by the mass is U2 = mgh = 2 10 (1 10–3) = 0.02 J Hence the correct choice is (b). 0.015 V = = – 1.5 10–4 4. P = 107 Pa, 100 V Bulk modulus B = –

= Mg is applied. Now, the time period of vertical oscillations is given by T= 2

P V /V

107

= 6.67 1010 Nm–2 1.5 10 4 5. V = L3. Therefore V 3 L = = 3 (– 1%) = – 3% V L =–

P 106 =– = 3.33 V /V 3%

B=– 6.

L=

107 Nm–2

FL 4 FL = AY d 2Y

L

A

L

B

=

LA LB

dB dA

2

1 2

=

1 2

2

=

1 8

ls =

F Ls AYs

ls = lc = 1 mm

0.5 m 1.0 m

1 2

Y

x L

2

A

L

Y Ax 2 2L Hence the correct choice is (b). 10. Pressure exerted by the piston on the liquid when a mass M is placed on the piston, P = Mg/A. This pressure is exerted by the liquid equally in all directions. Therefore, the surface of the sphere experiences a force P per unit area. The stress on the sphere is P = Mg/A. Now, the volume of the sphere is =

7. When a wire of length L, cross-sectional area A and Young’s modulus Y is stretched with a force F, the extension l in the wire is given by FL l = AY Since F and A are the same for the two wires, we have F Lc For copper wire lc = AYc For steel wire,

ML YA

T1 Y2 3 = = T2 Y1 2 Hence the correct choice is (a). 9. Work done per unit volume of the wire 1 = (stress strain). 2 stress Y = strain stress = Y strain Work done per unit volume 1 1 x 2 Y (strain)2 = Y = 2 2 L But, volume of the wire = A L Work done =

Since F and Y are the same for wires A and B,

M =2 k

4 R3 3 Due to stress, the change in the volume of the sphere is V =

4 R3 4 = . 3R2 R 3 3 = 4 R2 R V 3 R Volume strain = V R V =

Ls Yc Lc Ys 1.0 1011 Nm

2

2.0 1011 Nm

2

= 0.25 mm Total extension = 1 + 0.25 = 1.25 mm. Hence the correct choice is (a). F L 8. Young’s modulus Y = A l F YA Force constant k = l L where l is the extension in the spring of original length L and cross-sectional area A when a force F

stress M g /A = strain 3 R/R R Mg or = R 3K A Hence the correct choice is (c). 11. Maximum stress = Young’s modulus strain K=

=2

1011

10–3 = 2

maximum 108 Nm–2

Elasticity 10.11

Maximum force (F) = maximum stress 8

=2

area

–6

10 3 10 = 600 N F 600 Maximum mass = = = 60 kg, which is g 10 choice (b). 12. Figure 10.10 shows a rubber eraser ABCD clamped at end AB and a horizontal F applied at the free face DC resulting in a displacement DD = CC . The shear angle is given by DD tan = AD

V P B Volume of the same mass M of water at the given depth is V P V =V– V=V– B V =

P B

=V 1

=

V (B – B

P)

Density of water at that depth is =

=

M V V = = V V V B P B B B

P

=

B B

n 1 P0

Hence the correct choice is (a). 14. Since the rod is uniform, half its weight can be taken to act at its mid-point. Therefore, stress at mid-point is weight of suspended mass + weight of half the rod cross-secttional area Fig. 10.10

DD AD Area of top face = 3 cm 1 cm = 3 cm2 = 3 10–4 m2

Since

is small, tan =

Shearing stress =

=

F 2.4 = A 3 10

=8 Shear angle

= =

1 Mg 3M g 2 = = , which is choice (b). 2A A stress l 15. Young’s modulus Y = = , where = strain L Now, if the temperature of a rod is increased by , the increase in its length due to thermal expansion is l = L l = . Now stress is Strain = L =Y =Y For the two rods, the stress is and 2 = Y2 2 1 = Y1 1 But 1 = 2 (given). Hence Y1 1 = Y2 2 Y1 3 or = 2 = . Hence the correct choice is (c). Y2 2 1 16. Since the wire is uniform, i.e. its mass per unit length is constant over its entire length L, the total donward weight = the weight of the sus3L of the wire or pended mass + weight of length 4 3W F = W1 + . 4 3 W1 W force F 4 = = Stress = area S S Hence the correct choice is (c) Mg +

4

103 Nm–2

shearing stress shear modulus 8 103 1.6 105

= 0.05 rad

DD = 0.05 or DD = 0.05 AD AD = 0.05 8 cm = 0.4 cm = 4 mm Hence the correct choice is (d).

or

13. Pressure at the surface of the ocean = P0, the atmospheric pressure. Pressure at a depth = nP0 (given). Increase in pressure ( P) = nP0 – P0 = (n – 1)P0 Let V be the volume of a certain mass M of water at the surface, so that M = V. The decrease in volume under pressure P is

10.12 Comprehensive Physics—JEE Advanced

17.

L L

stress Y

5 107

= 2.5

2 1011

10 –4. Given

V 0.02 = 2 10–4 V 100 Now, V = r2L. Therefore, ( r 2 L)

V = V

r2 L

=

2 r

= 2 r r or

2 r V = r V

rL

r2L L

r2 L L L

L = 2 10–4 – 2.5 L = – 0.5 10–4

10–4

1 kx2 U1 = 2 Potential energy stored in the spring when it is further extended by y is 1 U2 = k(x + y)2 2 1 1 Work done = U2 – U1 = k(x + y)2 – kx2 2 2 1 ky(2x + y) 2 19. If L is the initial length, then the increase in length by a tension F is given by FL l= r 2Y 4L Hence a= L + l = L = L + 4c (1) r 2Y =

b= L +

r 2Y

= L + 5c

(2)

L

. Solving (1) and (2) for L and r 2Y c, we get L = 5a – 4b and c = b – a. For F = 9N, we have where c =

x= L +

9L

log A = log

4 Differentiating, we have

r = – 0.25 10–4 r Fractional decrease in r = 0.25 10–4 which is choice (d). 18. Potential energy stored in the spring when it is extended by x is

and

d /d l d /d or (i) l /l l where d is the diameter and l is the length of the wire. The area of cross-section is d2 A = r2 = 4 =

or

or

9L

20.

= L + 9c r 2Y = (5a – 4b) + 9(b – a) = 5b – 4a Hence the correct choice is (b).

A d =2 A d d 1 A = d 2 A

or

+ 2 log d

1 × 2% = 1% 2 A = 2%, given) A

( Using this in (i) we have 1% = 2.5% l 0.4 Hence the correct choice is (b). l

=

force area lus is given by

(

= 0.4)

W l . Strain = . Young’s moduA l

21. Stress =

W/A W l stress = (i) l /l l A strain l 4 10 4 = m/N. Using Now, slope of graph is W 80 this in (i), we get (given l = 1 m and A = 10–6 m2) Y=

80

1

= 2 1011 N/m2 4 10 10 6 22. Let TA and TB be the tensions in wires A and B respectively. If aA and aB are the respective crosssectional areas, then Y=

4

Stress in wire A =

TA aA

TB aB The stress in wires A and B will be equal if Stress in wire B =

TA T T a = B or A = A = 1 (given). aA aB TB aB 2 Since the system is in equilibruim, the moments of forces TA and TB about C will be equal (see Fig. 10.3), i.e. TA x = TB (L – x)

Elasticity 10.13

TA 1 L x = L x or = TB 2 x x 2L which gives x = . Hence the correct choice 3 is (c). T Stress 23. Strain = = , aY Young s modulus Strain in wires A and B will be equal if TA T = B a AYA aBYB or

TA Y = A TB YB

or

aA =2 aB

1 =1 2

which gives TA = TB. Equating moments about C, we have TA x = TB (L – x) L ( TA = TB). Hence Which gives x = L – x or x = 2 the correct choice is (b).

24. Let L be the required length of wire. If A is its area of cross-section and is the density of its material, the weight of the wire is W = mg = Al g

26. Let V be the volume of the metal at normal pressure. p , the decrease in the volume when From B = – V/V the metal is subjected to an excess pressure p is given by pV | V| = B p pV New volume of metal is V = V – =V 1 B B Now mass of the metal is m = V. Therefore, its new density is m V = = = p p V V 1 1 B B 1 or = , which is choice (a). 1 p/ B 27. Let m be the mass per unit length of the rod. Then M = mL. Consider a small element of length dx located at C at a distance x from A. (Fig. 10.12)

weight AL g = =L g area A S = L g L = S/ g

Stress = or

25. Refer to Fig. 10.11. For vertical equilibruim T cos

= mg mg or T= cos If L is the original length of the wire, the increase in length is TL l= AY Strain =

l T mg = = L AY AY cos

Fig. 10.12

The mass of element of length dx = mdx. The centripetal force at C is (1) dF = (mdx) x 2 The force (or tension) at point C is due to the outer portion CB of the rod, i.e. portion from x = x to x = L. Therefore tension at point C is x L

(mdx) x

F=

2

x x

= m

2

x L

xdx = x x

1 m 2

M Now, m = . Therefore, L 1M 2 2 (L – x2) F= 2 L Fig. 10.11

=

1 ML 2

2

1

x2 L2

2

(L2 – x2)

10.14 Comprehensive Physics—JEE Advanced

If r is the radius of the rod and its density, the mass of the rod M = r2L . Therefore, x2 1 r2 L2 2 1 2 (2) 2 L Thus the tension in the rod varies with x, it is zero at x = L, i.e. at point B and maximum at x = 0, i.e. at A where the rod is pivoted. L of the rod is Tension in the middle x 2 1 1 r2 L2 2 1 F = 4 2 3 = r2 L2 2 8 Thus the correct choice is (d).

To obtain the total elongation l of the cord, we integrate from y = 0 to y = L. Thus

28. Let M be the mass of the rubber cord, L its length, A its cross-sectional area (assumed constant). Let tion dl of an element AB of length dy at a distance y (Fig. 10.13). The force due to the weight of the cord is maximum at P(y = 0) and zero at Q(y = L). Therefore, the force Fig. 10.13 acting at the lower end B of the element is Mg (L – y) (1) F= L This force is responsible for the elongation of element AB. Now stress F / A Fdy Y = strain dl / dy Adl Fdy or dl = (2) AY Using Eq. (1) in Eq. (2), we get dl =

Mg (L – y)dy LAY

Mg LAY

L

Mg = Ly LAY

y2 2

L

Mg 2 L LAY

L2 2

l =

F =

dl =

=

(L – y) dy 0

0

MgL 2 AY

The correct choice is (b). 29. Poisson’s ratio =

lateralstrain = longitudinalstrain

r /r l /l

where r is the radius of the wire and l its length and r is the change in r and l the change in l when the wire is subjected to tension. Volume of wire before elongation is V 1 = r 2l Volume of wire after elongation is V2 = (r – r)2 (l + l) Given

V1 = V2. Thus r 2l = (r – r)2 (l + l) = [r2 – 2r( r) + ( r)2] (l + l) = r 2(l + l) – 2 r r (l + l) + ( r)2 (l + l)

Since r and l are very small, terms of order r l and ( r)2 and higher can be ignored. Then, we have r2l = r 2l + r2 l – 2 rl r or

r l=2l r =

which is choice (d).

or

l l

=2

r /r 1 = = 0.5, l /l 2

r r

Elasticity 10.15

II Multiple Choice Questions with One or More Choices Correct 1. Figure 10.14 shows the stress–strain graphs for materials A and B.

Fig. 10.14

From the graph it follows that (a) material A has a higher Young’s modulus (b) material B is more ductile (c) material A is more brittle (d) material A can withstand a greater stress. 2. Two wires A and B have the same cross-section and are made of the same material, but the length of wire A is twice that of B. Then, for a given load (a) the extension of A will be twice that of B (b) the extensions of A and B will be equal (c) the strain in A will be half that in B (d) the strains in A and B will be equal. 3. Two wires A and B have equal lengths and are made of the same material, but the diameter of wire A is twice that of wire B. Then, for a given load (a) the extension of B will be four times that of A (b) the extensions of A and B will be equal (c) the strain in B is four times that in A (d) the strains in A and B will be equal. 4. In Q. 3, above (a) Young’s modulus of A is twice that of B (b) Young’s modulus is the same for both A and B

(c) A can withstand a greater load before breaking than B (d) A and B will withstand the same load before breaking 5. Choose the correct statements from the following: (a) Steel is more elastic than rubber. (b) The stretching of a coil spring is determined by the Young’s modulus of the wire of the spring. (c) The frequency of a tuning fork is determined by the shear modulus of the material of the fork. (d) When a material is subjected to a tensile (stretching) stress, the restoring forces are caused by interatomic attraction. 6. A uniform wire of Young’s modulus Y is stretched by a force within the elastic limit. If S is the stress produced in the wire and is the strain in it, the potential energy stored per unit volume is given by 1 1 2 S (b) Y (a) 2 2 S2 1 (d) Y S 2Y 2 7. A wire AB of length 2l and cross-sectional area a A and B (Fig. 10.15). The wire is pulled at the centre into shape ACB such that d R0, the interatomic forces are attractive. 6. The correct choices are (a), (b) and (c). 7. AC + CB = 2 AC = 2(l2 + d2)1/2. Increase in length is L = AC + CB – AB = 2(l2 + d2)1/2 – 2l T Stress = . a Strain =

L 2(l 2 = L

If d R0, the interatomic force is attractive and this force provides the restoring force under which the material regains its original shape and size when the stress is removed. 10. The correct choice is (c). When the material is subjected to a compressional stress, R becomes less than R0 and in this case the interatomic force is repulsive which causes the restoring force.

FL . Since the two wires are made of the AY same material, the Young’s modulus Y is the same. Since F and A also the same, L L. Hence the correct choice is (a). 6. The correct choice is (d).

5.

L = L

L=

L F = L AY Since F, A and Y are the same for the two wires, the strains in them are equal. 7. Area of cross-section A = d 2/4, where d is the diameter of the wire. Thus 4F L 1 or L L = 2 d2 d Y Strain =

V Integer Answer Type 1. A light rod AB of length 2m is suspended from the ceiling horizontally by means of two vertical wires as shown in Fig. 10.22. One of the wires is made of steel of cross-section 0.1 cm2 and the other of brass of cross-section 0.2 cm2. The Young’s modulus of brass is 1.0 1011 Nm–2 and of steel is 2.0 1011 Nm–2. A weight W is hung at point C at a distance x from end A. It is found that the stress in the two n wires is the same when x = metre. Find the value 3 of n. IIT, 1980

Fig. 10.22

2.

the strains in the two wires are the same.

IIT, 1980 3. A metal wire of negligible mass, length 1 m and cross-sectional area 10–6 m2 is kept on a smooth ball of mass 2 kg is attached to the other end of the wire. When the wire and the ball are rotated with angular velocity of 20 rad s–1, it is found that the wire is elongated by 10–3 m. If the Young’s modulus n. of the metal is n 1011 Nm–2 IIT, 1992 4. In Q.10 above, if the angular velocity is gradually increased to 100 rad s–1, the wire breaks. If the x. breaking stress is x 1010 Nm–2 IIT, 1992 5. A body of mass 3.14 kg is suspended from one end of a wire of length 10.0 m. The radius of the wire is changing uniformly from 9.8 10–4 m at one end to 5.0 10–4 m at the other end. Find the change in the length of the wire in mm. Young’s modulus of the material of wire is 2 1011 Nm–2. IIT, 1994 6. Steel wire of length ‘L’ at 40°C is suspended from the ceiling and then a mass ‘m’ is hung from its free

Elasticity 10.23

end. The wire is cooled down from 40°C to 30°C to regain its original length ‘L linear thermal expansion of the steel is 10–5/°C, Young’s modulus of steel is 1011 N/m2 and radius

of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of ‘m’ in kg is nearly. IIT, 2011

SOLUTIONS 1. If Ts and Tb are the tensions in steel and brass wires, then the stress in them will be the same if Tb Ts = Ab As

Ts Tb

As Ab

0.1 cm 2 0.2 cm

2

1 2

Since the system is in equilibrium, the moments of forces Ts and Tb about C will be equal, i.e. (see Fig. 10.23) Ts x = Tb (2 – x) x = 2(2 – x)

x=

Thus the value of n = 4

4 metre 3 ( Tb = 2Ts)

Given m = 2 kg, R = 1 m, = 20 rad s–1, A = 10–6m2 and L = 10–3 m. Substituting these values in (1) and solving, we get Y = 8 1011 Nm–2 Thus n = 8. 4. If the wire breaks at breaking stress is

max

mR 2 max 2 1 (100)2 Fmax 2 1010 Nm 2 . = A 10 6 A Thus x = 2. 5. Figure 10.24 shows a wire PQ of length L having radius r1 at end P increasing uniformly to a radius r2 at end Q. Consider an element AB of the wire of length dy at a distance y from end P. Let r be the radius of the wire at y and (r + dr) at (y + dy). As the radius increases uniformly from P to Q. dr = constant, say C dy where

C=

r2

dy =

dr C

r1 L

. Thus

Fig. 10.23

stress T Y AY The strain in the two wires will be equal if

2. Strain =

Ts Tb = AsYs AbYb Ts A = s Tb Ab

Ys Yb

0.1 0.2

2.0 1011 1.0 1011

1

Equating moments about C, we have Ts x = Tb (2 – x) which gives x = 1 metre ( Ts = Tb) 3. Let R = L = length of the wire = radius of the circle. The centrifugal force F = mR 2 produces the stress as at result of which the wire elongates. Thus

Y=

FL A L

mR 2 L A L

mR 2 2 A L

(1)

= 100 rad s–1, then the

Fig. 10.24

10.24 Comprehensive Physics—JEE Advanced

dl produced in the element of length dy due to a force F = Mg. Stress in the element = F/ r2 and strain in it = dl/dy. Hence stress Y= strain

F / r2 dl / dy

r 2 dl

dy =

l=

dl

=– =

F CY

r2 r1

F 1 = CY r r1 r2

F (r2 r1 ) CYr1r2

dr r2 F 1 CY r1

r2

r1 L

. Hence

MgL Yr1r2 Substituting the given values of M, L, Y, r1 and r2 and g = 9.8 ms–2, we have l=

Fdy

dr . Therefore, C Fdr Fdr Y= or dl = 2 r Cdl CYr 2 Integrating from r = r1 to r = r2, we obtain the total extension l produced in the wire by a force F which is given by But

Now F = Mg and C =

1 r2

l=

3.14 9.8 10.0 3.14

2 1011

9.8 10

4

5.0 10

4

= 10–3 m = 1 mm. 6. Change in length L = L T mgL mgL Also Y = L= A L YA Equation (i) and (ii) we get ( A = r2) m= =

TY g (10 5 )

= 3.2 kg

(i) (ii)

r2 (10) 3 kg

(1011 ) 3.14 9.8

(1 10 3 ) 2

Hydrostatics (Fluid Pressure and Buoyancy) 11.1

11

Hydrostatics (Fluid Pressure and Buoyancy)

Chapter

REVIEW OF BASIC CONCEPTS 11.1

FLUID PRESSURE 105 Pa

P0 P= SI

F A

11.5

GAUGE PRESSURE

–2 –2

11.2

P=

g g

PASCAL’S LAW

pressure

11.6 11.3

HYDR0STATIC PRESSURE

DENSITY AND RELATIVE DENSITY

P0 g

–3 –3

P = P0 +

–3

g

11.1 density density

e

g

11.4

ATMOSPHERIC PRESSURE

–2

SOLUTION 103

–3

g

11.2

Comprehensive Physics—JEE Advanced

4

10 Pa

103)

10

SOLUTION (

2

2

=5

104) 104 10 1 2

cal side =

10–4

1g

p1 = P0 +

sure)

104

A

2

p1 = 4

–3

1

–3

P2 = P0 +

104 P

1g

(P0

2g

P2 – P1 = ( 2 – P 2 – P 1) A = ( 2 – 1)

2

= (1500 – 1000)

104) 104

1)

5

10–4

11.4

11.2

OP A P

A

r

L O

O is P0

P L

SOLUTION

SOLUTION F = (P0 + P=

g) A +

F = P0 + A

g+

at

a distance r =

O

A

11.3

Fig. 11.2

2

=

2

=A

r = L/2 to r = L F=A

2

L

r =A L/2

=

Pressure at P = Fig. 11.1

2

r2 2

F 3 = A

L

= L/2 2 2

L

3

A

2 2

L

Hydrostatics (Fluid Pressure and Buoyancy) 11.3

L

11.5

A

= L A

acceleration a P1 A – P2

= ( L A) a

P1 and P2

–3 –3

A=

–3

P1 – P2 = La SOLUTION

(

1

–

2)

–

1

(i)

g = La 2

La g

=

(ii)

P1 > P2 tan

1

=

2

L

=

a g

11.6 A

°

Fig. 11.3

Pressure at A = P0 +

0 0g

Pressure at B = P0 +

g

0

0

+

0

0

+(

+

g

(ii)

= –

0)

=

a g

tan 30° =

= 1300

(

g

a g

a

tan 30° =

=

11.8

0

11.7

SOLUTION tan

0)

0

(i)

PRESSURE DIFFERENCE IN AN ACCELERATED LIQUID

3

–2

=

ARCHIMEDES' PRINCIPLE

‘

a Law of Floatation

11.9

APPLICATIONS V g

Fig. 11.4

V

11.4

Comprehensive Physics—JEE Advanced

Case

Vi

< V0 outside liquid and Vi inside liquid V g =V g Vi = V

and

V0 = V

(

V = V0 + Vi )

= V0 >

Case Case T a liquid ( > Case

V

and V1 in liquid 1 and V2 in liquid 2 < 2) 1< 1

T= V

g–V g

= V(

– )g

T =V (

– )g

2

Case Fig. 11.5

a g = g + a and T = V ( a (< g) g = g – a and T = V(




1

2 1)

2

Hydrostatics (Fluid Pressure and Buoyancy) 11.11

(a) 12.

(c)

r 2 = 2r

=

17.

=r (d)

density

= 4r

>

r

2

2/ 1

2 r

4 (c) r

+

2 r

–

+

4 (d) r

–

(d) A and B

18.

(c)

is

2 A

4 A

(d)

B 2 3

19.

A

and (a)

14.

( > g

(c)

1

1 ( 2

20.

and

(d) g

15.

1

(a)

T

(c)

+

2

1 2 1

g

g

A and B is

2

(c) (a)

1/ 2

(c)

2

2

1 2

1

2

1 2

1 ( 2

1

+

2

1

1

2

r

T

r

1/ 2

1/ 2

T

(d)

T

and density is tied to a

22.

r

> (a) (c)

(d)

23.

(a) (c)

(d) 2

2)

1 2

(d)

1 2

+

1/ 2

T

16.

1 2

(d)

21. (a)

1 4

A

and area A

A

1

(c)

13.

(a)

is

1

(a) (a)

1

2

11.12

Comprehensive Physics—JEE Advanced

30.

24. 2

E=

31.

25. 32.

–3

26.

–3

–3

rises

33.

is

(c)

27.

2

(d)

34. (a)

28. 0

e–

(a) (c)

k is a con

2

2k

0g

k

(d)

2

2

and density

(a)

k (d)

g

r

or equal to

0g

0g

2

g

35.

0=

2

2

g

(c) 2 =

29.

2

2

(a) 2

0g

(c)

2k

2

2 g

g

(d)

2 g

36.

10–2

–1

)

Hydrostatics (Fluid Pressure and Buoyancy) 11.13 –5

–5

–7

–7

44.

r1 and r2 (r2 > r1)

37.

(a)

r1 r2 r2 r1

(c)

1 (r1 + r2) 2

38.

r1 r2 r2 r1

2

(c)

2

–1

–2

1 2

1

(d)

2

radius r R

(c) 40.

(a) g

D/ is ( (

1)

1) 1/ 3

1

(d)

1

1 s r

1 R

1 s r W

1 R

ion) 3 1 s r

1/ 3

(d)

3

1 s r

s its 1 R 1 R

47.

R

V is

V 2 W (d) (4 ) W R

(a) 2 W (c) (21/3) W

(a)

1/3

1

1

R

(c)

(a) R (c) 4 R

R R

R

(d)

48.

42.

1)

46.

(a)

41.

(2

D

(c) 39.

2

45.

(d) (r2 – r1)

–2

(a)

r 5r

r

43.

(a) R (c) 4 R

R R

49. a r1 and r2

a (a)

(a) 2 (c) 4

(r1 + r2) (r1 + r2)

(r1 – r2) (d) 4 (r1 – r2)

4 (c) 2 50. W

2 (d) 4 V

11.14

Comprehensive Physics—JEE Advanced

V W (d) 16 W

(a) 2 W W 51.

Fig. 11.21 2 4

6

54. R

52.

R

A

V M

densities

L L

H 2

and 2

H 2

A 5

Fig. 11.22

(a) Mg Mg – V g (c) Mg + R2 (d) g (V + R2 )

L 4

liq 3 2 5 (c) 4

4 3

(a)

55.

and

(d) 3d

Fig. 11.23

(a) Fig. 11.20

53.

decreases and increases increases and decreases and increase and decrease

56. R T P0 ABCD

Hydrostatics (Fluid Pressure and Buoyancy) 11.15

(a) | 2P0 P0 (c) | P0 R2 (d) | P0 R2

+ R2 – 2 RT| 2 +R – 2 RT| 2 +R – 2 RT| 2 +R + 2 RT|

r

Fig. 11.25

Fig. 11.24

57.

r

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

(d) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

(d) (a) (a) (d)

(c) (c) (a) (c) (a) (c) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(c) (c) (a) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52.

(c) (c)

(a) (d)

(c)

5. 11. 17. 23. 29. 35. 41. 47. 53.

(d) (c) (c)

6. 12. 18. 24. 30. 36. 42. 48. 54.

(a) (c) (d) (c) (a) (c)

(c) (a) (c) (d) (a) (d) (c) (c)

SOLUTIONS 1.

V – and

=V

g

(1)

– z =V

g

(2)

v=

2.

U W=V

V g

U

W

=

V

( g

a=

=

v=

a=

2

=

2 g 1 =

–1

2g V

g

V) V V

1 0 75 0 75

g

g=

2g 0 – 2g = 2

H 3. Let

1

=

/

g

/

g 3 H v2 – u2 = 2aS

g 3

H V

V=

–3

11.16

Comprehensive Physics—JEE Advanced

8.

is a=

ss V

=

v=

g V V 2

3

g

=

1000 g 250 g = =3g 250

2 3g 1 =

= 200 or

9.

V

6g

g=

k

H

k=

H H=

=

)

/k

a 1 k

1

= or

/k –

1 1 k

=

g= (

/

v2 6g = 2g 2g

4. Let

Vg =

a =g 1

F=

1 k

10. is F = A 5.

1 ( 2

A

2

–

1

1 ( 2

g

6.

1 ( 2

W=

= 7. Let p1 and p2 A and B p 1 – p 2= A

W= g

(

F = (p1 – p2) A = F= F=

= (1)

gA

2

–

1 2

2

( 2 – 1) A g

1 4

Ag(

2

–

1)

1 ( 2

V=

2

–

1)

–

1)

A

1 ( 2 – 1) 2

2

11.

r

= AL L ALa

p1 = p2 = 0

(2) =

La g

p=

p1

2

p2

1)

Hydrostatics (Fluid Pressure and Buoyancy) 11.17

or

p=

/2 area

=

2

2 r

=

2

=

or

g =

g

2

p1 r2 = r

or

r2 or

=

r2

T =2

=2

g

=r

1/ 2

=

12. 13.

r

p=

r

=

r=

/2

g

T

2

T

g

16.

2 r

/ /

g=

=2 =

F = 14. Let

=

=

W=

17.

V

= V W=

–

T=

2 A

=

U=

Vg

Vg

Vg

F =U – W = ( –

V

) Vg

= V

U

uid is

U =V g

a=

F

Vg

=

V

F = W – U = ( – ) Vg

=

g

is a= =

F

v=

Vg

=

2

(i)

1

V 2

v=

g

2

(ii)

2

a 15.

or

2 1

F = ( – ) Vg =

18. Let V1 (

=

V)

a = = g

=

1

=

2

1 A and

1

11.18

Comprehensive Physics—JEE Advanced

1

1 4

V1 g = 1

V1

r2

g

0

0

3 1 = 4

or

= r2 g = r2 g =( r

B 2

V2 g = 1

=

2

2V2 3 3 4

g or

2

2 3

=

=

3 = 2

=

1

=

2

3

2

3

n

3

n

2

1

2

1

=

2

= P0

1 2

1 2

=

P1V1 = P2V2 or (P0 +

2V1 =

2 1

=

2

V2

P2 = P0 V2 = 2V1

n

2

1

g

P2

1

2

2

1 2

2

P0 0 75 = g

1

2

V1 V2

1V1

=

V1 = V2 = V

g

=

r

22.

+

is

=

2

r = 2r

r r2

=2 r

1 ( 2

=

2

=2

1

+

2

25.

r2

=

=

(2r)2

r

26.

2 =

2

=2

V g=

or

=

=

r2 is

or

V 2 13 6

g+ 0 2

27. and

=

R

2 V2

V1 and V2

=

40 3

g p = 2 /R

V1 V2

21.

g) V1

24.

1

20. =

/2

P0

1 1

/2 =

P1 P1 = P0 +

total ss total lu 1

2 )

23. Let V1

19. =

2

2

28.

2 cos r g

V 2

g –3

Hydrostatics (Fluid Pressure and Buoyancy) 11.19

=(

) g

r r2 = r 21 + r 22 or r =

as

=

e–

=

0

=

0g

e–

(1)

33.

P

0g

=

0g

0g

k

k

2

=

PE =

1

4

2

(

= r2

g

g

or

=

2

2

2

2

g

gr = 2

r=

2

=

=

or

= 2r)

is

k

g

=

2

2

0g

(e– – e–0) =

g

4

g

34.

0

2

=

g

W =2

is 2

2

e

29. Let

W=

r22 W) =

= r2

=

0

=–

(4 0

=

= 0 to P=

3 0) 2

r12

2

PE =

1000 2

35.

30.

=

2g

2

=

(

g )2 2g

2

(2)

2

2 g

liquid is P1 =

P2 =

r =

(2 )

2

g

2

=

r g 2

P1 = P2 31.

r or

32. 4 4 and P2 = P1 = r1 r2

air r1

36. F = 2 =2

r g 2

=

2

r=

10

g W = FS = 2 10 7

2

37. Let r

r2

P1 V1 + P2 V2 = PV or

4 r1

4 3 r1 3

4 r2

4 3 4 r2 = 3 r

4 3 r 3

(1)

4 r1

–

4 r2

=

4 r

(1)

11.20

Comprehensive Physics—JEE Advanced

r1 r2 r2 r1

r= 38.

2 r 2

39.

1 2

and g 10 2 2

W =4

R (n

or

=

3

r

1/3

1) = 4

R

= 103

4 T= 3

Q=

R

3

s T

(2)

40. W

r2 W

3V 4

42. Tension = 43. Force = 2 r1

(R3

r 3)

or

(R3

r 3)

or

3

(R

r2 2

R=4

r ) = R3 1/ 3

1 A

R =

47.

=

0

or v

g 2 g = g

4 3

or

r3 = (10)

( + 10) = 10

=g 1

R g R

2

2

=g 0

2

4 3 125 64

a = r2

49. v2 =g 2

g 1

P1 V1 = P2 V2

=g 1 v

2

48.

or ( + 10)

v

g

=1–

=g 1 or

= R3

0

0

=g –

R3

0

3

= g

4 3

g=

V2/3

W

r=2 r + 2 r2 = 2 (r1 + r2)

2

r

3

46. Let

2/3

and

=

2

r3

44.

v

(R

4 3

R = r

L=2 R L=2

41.

0

4 3 R3

T 4 V= 3

1 2

4 3

(1)

R = n1/3 r)

(

1

3

R 1 r

2

2

45. Let

2 = r g

g or

v=

r=

a/

a

5r 4

3

0

g

Hydrostatics (Fluid Pressure and Buoyancy) 11.21

2 cos r g

=

2

=

cos

FB

a g a is

a r is

50. W =4

r

=

2

4 3

V=

=

r3

L 4

1 10

2 3 20

3L 4

A 5

g+ =

g

1 4

W = FB 2/3

3V 4

W=4

A 5

or

1 1 ALDG = 5 4

or D =

5 4

V

is V

W =4

53.

2/3

V

4 2/3

3V 4

=4

2/3

51. Let r

and

=4W

R

V = 106

6

a F T F

)

4 r3 4 R3 = 3 3

R3 = 106 r 3 or R = 100 r

r is

S = 4 r2 Fig. 11.26

E1 = 4 r 2

54.

106

E2 = 4 R 2 E2 = E1 52.

R r

2

density A = 5

1 106

L

=

100 r r

2

1 106

=

1 D= ALD 5 W) =

1 ALDg 5

1

55.

= Mg +

g = Mg + R2

g

102

56.

in ABCD

F

1

ABCD 2

due

11.22

Comprehensive Physics—JEE Advanced

57.

ABCD

1

= P0

g

2

p

2 r AB

2

=T

2R = 2RT

F = P0

g

2

+R

2

2

= [2P0

– 2RT – 2RT

II

1.

–1

Multiple Choice Questions with One or More Choices Correct 6.

T–2

2. 7.

3.

4.

8. (a) rises

5.

A B

9. (a) rises

A B A A B

10.

1 r

Hydrostatics (Fluid Pressure and Buoyancy) 11.23

(a) PA = PB (c)

1

>

PC = PD (d)

2

15.

1

2

2

1

L M

11. Mg n is Mg 1

12. 16.

1 n

loaded is L1 = L – (d) L1 = L – /n

1 2

103 (a) 13.

–3 1

and

2

103

are

1

1

2

2

–3

17. P is F

W1

(a) P P and F

reads W2

14.

(a) W1 (c) W2 densities 1 and PC and PD

W1 W2 2

F

18.

a

PA PB A B C and D F U

Fig. 11.28 Fig. 11.27

A until

11.24

19.

Comprehensive Physics—JEE Advanced

(a) F = 2a (c) U = 2a

F=a (d) U = a and

2

aera A 1

liquid are

1 and

and

=

2

2

and

g 2sin

F

11.28

Comprehensive Physics—JEE Advanced

III Multiple Choice Questions Based on Passage 3.

Questions 1 to 4 are based on the following passage

P

Passage I

5

as

P P = P0 e

1.

P

/

0

0)

P0

and

0 0 0 0

0

LT

0 is

L T0

Fig. 11.35

0 0

T

2.

/

n

1 2 (d) 2

(c) 1

n is

4. (a) (c)

= =

0

e(2)

0

(d)

SOLUTION 3. P

1. 0

2.

4. P = P0

=–

P0

e

/

0

=–

0

0

k k P= = V ( = /V)

PV = k

=–

P0 = P0 e 2

P

k 0

Questions 5 to 10 are based on the following passage Passage II

1 = e 2 2= e e(2)

/

/ /

0

0

= 0

=

0

e

0

=2 0 = 0e–1/2

Hydrostatics (Fluid Pressure and Buoyancy) 11.29

9.

5.

10. 1

and

2

1 and

2 is

6.

7.

Fig. 11.36 –3 3

3

3

3

8.

(a)

=

1 ( 2

=2

R r

1

1

+

+

(c)

=

1 (2 3

(d)

=

1 ( 3

2 1

1

2)

+

2)

+ 2 2)

SOLUTION 5.

or

=

5

=

30 30 5

7. 6. 3

1

is

g = ( – 10)

or

g = ( – 5) = ( – 5)

264

g

– 15 or 1

=

3 3

8. Let let g

its density and

=

4 (R3 – r3) 3

11.30

Comprehensive Physics—JEE Advanced

and

= =

4 3 r 3 R3

1 V 3

r3

(i)

r3 V=

V g=

4 R3 3

1 V 3

or

=

3

Let (1 –

V or

+

g = V 0g or

0

4 ( R3 3

4 3 r 3

r3 )

(R3 – r3) + r3

or

= /

=

=

( (

) 1)

1=

( (

) –1 1)

(1 (

) 1)

r3 R3 r R

or

3

3

r3 r

3

0

=

=V =

0

4 R3 3

= R3

and

R3

or

+

/

or

=

1 3

0

or

= 1 3

3 + (1 – ) = (1 – )

1 3

(iii)

)

10.

1 0 3) 2 4 1) V and V g

9.

+ (1 – ) =

(ii)

) 1)

g

0

= (1 = (

g + (1 – ) V

V g=

24 03

1

4

(3 )3 g = (3 )2 =

its den

1 (2 3

2

2 1

+

1g

+ (3 )2

2)

Questions 11 to 13 are based on the following passage Passage III H 4r liquid is

r B H

1

h1

h2

+

4r B H 2r

Fig. 11.37

h

2g

Hydrostatics (Fluid Pressure and Buoyancy) 11.31

11.

4

(a)

1

2 3 5 (c) 3

2 3

(c)

3 4

(a)

(d)

3 5

13.

2

(d) 2

12.

2

=

2

=

2

=

2 2

SOLUTION

11.

=4 r =A = 4A

(4r)2 = 16

W

r2

2

3A =

4A

3

Liquid level

A – A = 3A h2

Fig. 11.38

1

+ ) 3A –

W= U =W g(

1

3

4 + )3A – 4 g 1A = 3 5 1 = 3

1

W

U

U = F1 – F2 = g(

4 5 U equals

12.

(2r)2

2

3

Fig. 11.39

4A

4A

2

13.

=

4 2

g

IV Assertion-Reason Type Questions I 1. Statement-1

Statement-2

11.32

Comprehensive Physics—JEE Advanced

5. Statement-1

2. Statement-2 n

Statement-2 6. Statement-1

n Statement-2

Statement-2 3. Statement-3 7. Statement-1

R is

2 /R

R is 4 /

Statement-2 Statement-2 8. Statement-1 4. Statement-1

us R

n

P0

4 R

and P0 Statement-2

n Statement-2

SOLUTION 1.

5.

2. g 6.

g 3.

4.

7. 8.

.

g n

= P0 +

2 + R

g

12

Hydrodynamics (Bernoulli’s Theorem and Viscosity)

Chapter

REVIEW OF BASIC CONCEPTS 12.1

vt

VISCOSITY

r

r2 g

2

vt =

9 vt

F= –

A

r2

dv dx dv dx

A

–2

Fig. 12.1

[ML

–1

12.2

–1

NOTE

STOKES’ LAW

F= 6

12.3

-

v

r

rv

12.4

POISEUILLI’S FORMULA r

TERMINAL VELOCITY

p Q= F

W–U

W

U

l

pr4 8 l

12.2 Comprehensive Physics—JEE Advanced

Capillaries Connected in Series l1

l2

r1

r2 p

R F= R = R1 + R2 =

8 l1

8 l2

r14

r24

2

v 15 = x

–1 3

v x

A

–4

12.2 3

Q = Q1 = Q 2 = p1

p R1

–3

–5

R2

p2 P = p1 + p 2

SOLUTION

Capillaries Connected in Parallel p

vt =

1 1 = R1 R

1 R2

8 l1

R1 =

r 14

R=

R2 =

R1 R 2 R1

R2

=

2 r2 g 9 2

12 1 9

8 l2

–2

r1 p R1

p R2

–1

r vt –5 –3

8

4 r3 3

R

=

4 R3 3

R = 2r 2

SOLUTION 2

A

–1

–1

2

12.4 v x

–2

-

SOLUTION r

–1

–1

–1

–1

-

–2

18 1

-

12.1

–4

98

5

12.3

r2 Q2 =

3 2

r 42 F=6 =6

Q1 =

3

–1 –3

= 15

12.3

h1

–2

g

h2

SOLUTION –3 –3

V

m W mg = V g U V g F=W–U a=

F F = = m V

–

V

Vg

Vg V Fig. 12.3

g

=

A B

1 2 s = ut + at 2 1 a t2 2

h

2h = a

t=

h1 – h2

g

a 2h

h1 – h2 a g

AB F = a p d m = adx

1/ 2

x1

x1

1/ 2

2 1

=

p1 = h 1 g p2 = h 2 g p = p1 – p2

Fe =

2

dm x

2

=

= 12.5

a

xd x x2

x2

1 2

2

a x12

x 22

= –3

2

h1 – h2 = x1 g

–2

x2

2g

x12

x 22

2

2

2

2 1

]

–1

12.6 –3

3

–1

–3

–2

Fig. 12.2

SOLUTION AB dx

x

SOLUTION l

r

–2

–2

12.4 Comprehensive Physics—JEE Advanced

Q=

=

p=

3

3 14 1 3

12 3 14 1 12

12.8

1 3

the total energy of an incompressible and non-viscous liquid in a streamline

6

3 –1

energy being the sum of pressure energy, potential energy and kinetic energy of the liquid.

8 lQ

1 mv 2 2

PV + m g h +

r4 2

p

12.5

BERNOULLI’S THEOREM

P+

1 2

gh +

v2

=

m V

STREAMLINE OR LAMINAR FLOW 12.9

VELOCITY OF EFFLUX H h

12.6

CRITICAL VELOCITY AND REYNOLD’S NUMBER Fig. 12.4

v vc =

V

k r

A PA +

r k

B

1 V2 + 2

gH = PB +

1 2 v + g H–h 2 AV = a v

PA = PB = P A a k

P +

k k

1 2

a 2v 2 A

2

+ gH = P +

1 2 v + g H–h 2 1/ 2

12.7 a1

EQUATION OF CONTINUITY OF FLOW

2g h

v =

a2 A2

1

a2 v1

v2 A>>a v =

a 1 v1 = a2 v 2

2g h

av

vice versa

t=

H h g

R = vt = 2 h H

h

12.5 NOTE

SOLUTION H 2

h = R

=H

h

v

h

=

k r –1

=

12.7

1 1

2

–1

12.10 SOLUTION

–2

g 3

31 4 1

Q

SOLUTION

3

v =

–2 2

Q = A

3 14 1 3 14 1

25

–1

3 –1

r2

A=

2 1

4

3 14 1

=

2gh =

2

12.11

4 2 2

h = PQ

–1

A

a

12.8 –1

SOLUTION a 1= r 21

r 22

a2 =

Fig. 12.5

a 1 v 1 = a 2 v2 v2 =

=

= 12.9

a1 v1 = a2 r1 r2 1

r 12 r 22

SOLUTION V

v1

2

v1

1 1 V 2 = PQ + 2 2 PP = PQ = P

PP + 2

–1

–1

gh +

P +

gh +

AV = a v v =

g

1 1 V2 = P + 2 2

V 2 = v 2 – 2g h –2

v

P AV = a v

AV a

Q

1 2 v2

v2

12.6 Comprehensive Physics—JEE Advanced 1/ 2

2 g ha 2

V=

A2

AV = aA

av

Q=

a2 1/ 2

2g h A

2

a

= av = a 2gh

F = Qv = Q 2gh

2

12.12

12.13

A

-

a

a = 2

h

Q v Q A >> a av

2

L vQ

agh

Q 2gh

O Fig. 12.6

SOLUTION

SOLUTION

v= v

A >> a

2gh

Q a Q2 a

F = av2 = av av

3

36

Q v = av2

=6 O=F

–5

3

L 2

av2

F

=

Q L a 3

2

F = av = a

=

2gh = 2a gh v=

5 2

6 1 2 1

14

4

–3

2gh

I Multiple Choice Questions with Only One Choice Correct 1.

T g

T 2 T

–2

2 2 2.

3.

T 2 T 4 -

12.7

v v v3

4. 10.

v

v2 v4 r v

v 2 v

r

v

v 4

v

5.

v 2 v

h

v

11.

V vt V

R h g h g

vt 2 vt

h g h g

6.

vt vt

12.

A

L L

a

L

h

-

ghA gha

W1

ghA gha

7. r P

W2

t1 < t2 ; W1 > W2

t1 > t2 ; W1 < W2

t1 = t2 ; W1 = W2

t1 > t2 ; W1 = W2

13.

Q

r

P

Q Q 4

t1

t2

14.

Q Q 8

8.

A l

B

r

l P

-

A

Q 15.

R

R

l

-

P

Q

2Q 3

Q 3

3Q 4

16.

V P d

l d

9.

P a

12.8 Comprehensive Physics—JEE Advanced

h1

V 4 V 16

V V 8

h2 h1 h2

17. –1 –4

2

23.

d

–2

g –4

2

–5

2

–5

2

–5

2

h

R d

18. L

2

y R

y

R h

4 R2 h

R2 4h

h2 R

24.

R L

A

L

2

A

19.

-

A

L 2

L

B

25.

H

m

x y

h R h

R= 2 h H 20.

mg x y

R h H

R=

R= h

h H

R= 2 h H

mg y x

h

y x

mg 1

h

mg 1

26.

R H h= 4 H h= 2

r

H h= 3

r p

h=H

l

21.

1/4

r

r r

r 27. g

22.

–2

h

2

H

B B

x y l

12.9

2 1

3 2

R= V

2 3

1 2

R=

28.

2h g

aV A

g 2h

R= V

2h g

R=

AV a

2h g

A a m t ga A

gA a

2ga A

gA 2a

Fig. 12.7

32.

29.

2

3mg 4A

2mg A

mg A

mg 2A

r

r2 r4

30.

r3 r5

A 33.

r

a

h

a A g

h 2

–1

–1

–1

r3 r5

r r4

–2

–1

34.

31.

h A

–2

g

a V

R

2

–1

4

–1

5 6

–1

5 8

–1

5

5

ANSWERS

1. 7. 13. 19. 25. 31.

2. 8. 14. 20. 26. 32.

3. 9 15. 21. 27. 33.

4. 10 16. 22. 28. 34.

5. 11. 17. 23. 29.

6. 12. 18. 24. 30.

12.10 Comprehensive Physics—JEE Advanced

SOLUTIONS 1. W1 = mgh

V

gh =

P1 – P2 =

ghV

1 2

v 22 – v 12

1 2 3 = 2

v

P = W2 =

P

V = h gV

W = W1 + W2 = 2 h gV P= t=

=

2.

W t

= 6.

W 2 h gV = P P 2

3

3

3

h1

– v 2]

v2

3 2

2gh = 3 hg

h2

3

v1 =

av2

av1 2 gh 2 m1 = av1

3

2

2 gh 1

v2 =

m2 = av2

h v =

p1 = m1 v1

2gh

p2 = m2 v2 v 2

va =

gh 2

1 2 gh 2

F m 1v 1 – m 2v 2

V V T = Av a

2V

=

av 12 –

=

a

av22

2 gh1 – a

2gh2

= 2 ag h1 – h2

A gh

A

=2 a h V V = 2

T

=

2V / 2 A gh / 2

h h = 2

V A gh

Pr 4 ; 8 l l

7. Q =

l r

T 2

Q

Pr4

P

Q 8. A

2

3.

4.

–1

=

ML 1

1

B

]

Pa = 2P

Pb = P l

Q = Qa + Qb

r r

r v

5. v

v =

2gh

2

Pr 4 = 8 l 1 1 = l 3l

2P / 3 r 4 8 2l 1 3l

P /3 r 4 8 l

-

12.11

Q =

mgh = mgL

3l 2

l =

Pr 4 3l / 2

8

Pr 4 8 l

2 3

13.

2Q 3

l x 2

lx

9.

l

V

l

l 2 x g d x = l2 g

W= o

W t

P=

W V

V t

= l2

=

e 1 2 v 2

=

1 = 2 P 10.

g

l2 2

l4 g 14 = 2

=

xd x o

98 2

av 14.

av

-

3

v3 a 1v 1 = a 2v 2

r 21v1 =

v2 r2 = 12 v1 r2

r2 r 2

2

r 22v2 15.

=4

m

v2 = 4v1 = 4v 11. gr 2

2 vt = 9 k=

2 9

V=

4 3

r = vt = k

= mg = kr2 6

rvt = mg –

vt W–U F = 6 rvt

W

r3

g

4 r3 3

g

= mg 1

r3

3 4

-

-

g

3V 4

2

4 r3 U = 3

r

4 r3 3 m

2/3

2/3

m

vt

rv t

V2/3

1 r

V Vt = k

3 4

2/3

V [

12.

v 2

2/3

16.

= 4vt 2/3

= 4] -

v 2 v

t1 > t2

V= V

d4 l d V 2

P r4 8 l l

Pd4 128 l

r= l

d 2

12.12 Comprehensive Physics—JEE Advanced

V V

V

17. –1

v1

–4

a1

v 1a 1 = v 2 a 2

2

v2 v2

h v22 = v21 + 2 gh –1

v2 a2 =

v1 a1 v2

4

–5

2

18.

Fig. 12.8

V1 v a = 1 1 V2 v2 a2 v1 =

v1 L v2

2 gy 8 gy

R

R

2

R = vt

v2 =

2gy

v1 = v2

2

1 2

R=

1 L2 V1 = 2 R2 V2

19.

t 2 H h 2 g

2g h

20. R

h H

h dR dh

h d2 R

1 L2 2 R2

V1 = V2 R =

v

2g 4 y

d h2 hH – h2

R

1/2

L

h

2

dR =2 dh

h

=

v

1 hH – h2 2 H 2h hH

v=

2g h

t 1 2 gt 2

1 2 H–h= gt 2 t=

2 H h g

H – 2h

1/ 2

dR dh H – 2h H h = 2 d 2R

S=H–h

S=

h2

–1/2

h

d h2 h

h d 2R dh

2

=–

2h H

h

1/ 2

1

1 H 2h 4h H h

12.13

h= H

P1r14 P2 r24 = 8 l1 8 l2

2

d R d h2

=–

2H

P1 l = 1 P2 l2

h H /2

R h=H

R

r2 r1

4

1 2

=

24 = 8

25.

21. 1 2

-

v2 = p = gh

r 4 F= 3

v 2g 2

h= v=r =r

2

h=

26.

2 2 2

r

g r

r x

y g=

y x

y g = mg 1

p R4 p r4 = 8 l 8 l R4 = r4 + 16r4

22.

m x x

R

2

g h

2

3

1/4

R

p 2r 8 l

4

r

27. h R=2 h H

=

h H 2

h1

h1

1

H 2

h2

2

h1

H h1 = 2 2 h1 = h2

23.

v t=

H 2

h2

H 2

2

=

v2 v1

A1 A2

= 2v2 A1v1 = A2v2 A1 1 2 A1 2

h

h2 A

2 gd

m = Ah

2h g

Fig. 12.9

vt

R =

2 gd

2h = g

v=

4dh

2gh av

2

R2 = 4 dh

d=

P1r14 8 l1

Q2 =

24. Q1 =

1v1

h2

28. H 2

=

= v H 2

H 2

e

R 4h

av av v = av2

P2 r24 8 l2 Q1 = Q2

t

12.14 Comprehensive Physics—JEE Advanced

e

=

AV = av AV v = a

av2 a v2 = Ah Ah

a 2 gh Ah 2ga = A

v

=

t

2 gh

t R = vt 1 h = gt2 2

29. h =h v = h= v =

2g

h 4

gh 2 R = v

m A gm 2A

30.

2h g

t=

2gh

R=

v V

AV a

2h g

2h g

32. a v = AV av V= A

vt =

r2g

2 9

P = f vt

r

=6

Fig. 12.10

P P +

1 2

1 v2 + gh 2 v2 = V 2 + 2g H – h

V 2 + gH = P +

rvt

r vt × vt

P= 6

=

f=6

r2g

2 9

8 g2

2

2

r5

r5

33. mg

v = 2

2g H a A

1 v 31.

h

2 98

2

2 2

= 25

rv = mg

6 rv

–1

v

v m=

4 3

r3

6

rv =

4 3

r3 g

12.15

v2 –

h

2 r2 g 9

v= gh

ag + hk

k

ag

h

-

v=

2gh

h

-

=

k

h

r4

k

h

h=

v2 h = 2g h

34.

ag + hk kag

ag = h

hk a

2 1

8

64

–1

5

h

II Multiple Choice Questions with Two or More Choices Correct 1. 6.

2. 3.

7.

8.

4. 9.

OP a

L O P

5.

-

O L P

p v

12.16 Comprehensive Physics—JEE Advanced

3 p= 4

3 p= 8

2 2

L

3

p2 – p1

2 2

L

4 4

v=

3 L 2

3 L 2

v=

14. h

10.

–1

–2

g

–1

R –1

5 2 11.

PA vA vA

A

vB

PB

PA vB

vA

A B A B PA = PB

-

B

PB vB

Fig. 12.12

15.

-

Fig. 12.11

r

12. v1

p1 v1 > v2 p1 > p2 13.

–1

–3

v 2=

ANSWERS AND SOLUTIONS 1. 2. 3.

pc

Fig. 12.13

ps –1 2

3

h

p2

v1 < v2 p1 < p2 v1

p2 – p1

,

v2

pc > ps h=

r2 2 2g

pc < ps h=

r2 g

2

12.17 2

dF = mr

-

2

=a

rdr

F l r

L–l

r=

L v

av

L

F= a

a

2

-

L

= -

1 a 2

l

2

[L2

L

L

L – l 2] = a

l=L

4.

r2 2

2

rdr = a

F=

2

l

l

L

l 2

2 2

3a

L

8

-

5. A

B A

a1

v1

Fig. 12.14

p1

a2 v2 B

p2

P

1 2 1 2 v1 = p2 + v2 2 2 1 p1 – p2 v22 v12 2

v=

v1 > v2 10. 11.

6. p r4 8 l

Q=

F 3 = a 8

1 2 3 v =p= 2 8

p1 +

a2 < a1 a1v1 = a2v2 p2 < p1

p=

2 2

L

v

2 2

L

3 L 2

vA < vB 1 1 2 PA + v2A = PB + vB 2 2

-

vB > vA pB < pA

7.

12. rv

F= 6 8.

vt =

r2g

2 9

=

13. p = p2 – p1 =

9. dr m = a dr

r

O =

1 2 p

1 2

v21 – v22 2

2

3

12.18 Comprehensive Physics—JEE Advanced 3

4

14.

D2 ; D 4

A=

vs = r r

a=

d2 ;d 4 AV = av

v= t=

AV a

2h g

D

ps +

2

V

d2

2

–1

ps

vc vc

2 1 25

1 2 1 2 v = r 2 2 ps h

pc – ps =

R = vt = 5 t

vx = v v = gt

–1

pc

2

–1

v2x +vy2 1/2

2

= 5 2

+

2

]

pc – ps =

–1

15. 1 p+ 2

1 2 1 2 v s = pc + vc 2 2 pc

gh =

v2

1 2 r 2

gh

2

h=

r2 2 2g

III Multiple Choice Questions Based on Passage Questions 1 to 4 are based on the following passage Passage I

F=6 1.

F=6

rv rv

–1

–1

–1

–2

–2

–1

–2

–2

2.

-

terminal velocity r

v F

-

12.19

3.

4. -

ANSWERS 1. 2. 3.

mg

vt = -

r

6

4.

Questions 5 to 7 are based on the following passage Passage II

f

-

f F F 6.

-

dv F=– A dx dv dx

A

–1

–2

–2

–1 –1

] –2

]

7.

–1 –3

5.

–1

–2 –4

f

–2 –4

F

–2

–4

–2

–4

–2

SOLUTION 5. -

2

ML

=

1

2

L

= [ML–1

–1

]

L

7. 6. =

F A d v dx

F = A

Questions 8 to 11 are based on the following passage Passage IV

dv = dx a

–3

2 =4 5

–4

–2

A m

A.

h

12.20 Comprehensive Physics—JEE Advanced

8.

10. agh

Ag 2a

Agh

2ag A

agh

Agh

a mg A

A mg a

gh 2

gh

2a mg A

2A mg a

3gh 4

3gh 2

11.

9.

SOLUTIONS 8.

10.

P=h g F

P

a

h ga =

v=

2gh

2a mg A

e

av = av

2a g A

av v = av2 av2 = a

11. h =h

2gh = 2 agh

9.

v

agh = 2a mg A

m = Ah

=

Questions 12 to 14 are based on the following passage Passage V

gh 2

h

2 gH

13.

A

R

H

a

12.

h 4

g H2 + h2

2gh 2g H

h.

2g

2gh

hH

2hH

2 hH

h2

A h a 2g

A 2h a g

A H a 2g

a 2H A g

14.

Fig. 12.15

H2

1/2 1/2

]

=

12.21

SOLUTION 12.

vx =

A dx/dt

2gh -

vy =

x

u2 = v2 + 2 as

2gH

2g x = – A

a v=

v 2x

v 2y =

2g H

dt = –

h

13. 2H g R = vx

dx dt Ad x

a 2 gx

=– t=

A

A a 2g

x–1/2 dx

t x=h t=

2H g

2gh

x=h

2 hH

t= –

14.

-

2g x

= –

a =

= avx = a 2g x Questions 15 to 17 are based on the following passage Passage VI

2A a 2g

h 4

h/4

h 1/ 2

h1/ 2

A h a 2g

1 gh 2 3 gh 2

a1 = 3a2

15.

h

16.

h. a2

x–1/2 dx

x1 / 2 =– a 2 g 1/ 2

t

a1

a 2g

h/4

A

x

vx =

A

2 gh 2 2 gh

17. 2gh

gh

gh 2

1 gh 2

2h g 6

2h g

4

h g

8

2h g

SOLUTIONS 15.

v1

a 1v 1 = a 2v 2 v1 =

v2

a2 v v2 = 2 a1 3 A

P + gh + Fig. 12.16

1 2

1 2 v2 2 v12 = v22 – 2gh

v12 = P +

B,

12.22 Comprehensive Physics—JEE Advanced

v1 =

1 gh 2

h=h t=–

3 16. v2 = 3v1 = gh 2 17. h

=–

dh dt dh 2 dt = – = – v1 g

v1 = –

g

h

1/ 2

dh

h

2

2 h

g

4

h g

18. 2 2 gH

2 gH

2 gH

gH

d

H/2 a

2

h–1/2 dh

Questions 18 to 20 are based on the following passage Passage VII A d

h

19.

A

x

h=H

H

H

2

H 2 20.

H 2 2 h x

Fig. 12.17

2H 3

3H 2

3H 4

H 2

SOLUTION 18.

v1

v2 =

h = H v2 Av1 = av2

v1 =

a v2 A

a 0. Therefore, the maximum downward acceleration cannot exceed g, i.e. amax = g a =g –

A2 – x2 = x2 x=

max

(a) the maximum downward acceleration of the oscillating body and (b) the maximum amplitude for which the string remains taut.

=

(A2 – x2) =

9.8 = 3.13 rad s–1 1.0

g A

13.9 A body of mass m is attached by a string to a suspended spring of spring constant k. Both the string and the spring have negligible mass. The body is pulled down a distance A and released. Assuming that the string

SOLUTION The energy will be half kinetic and half potential at a value of displacement x when K.E. = P.E., i.e.

13.5

A 2

13.8 A horizontal platform is executing SHM in the vertical direction with an amplitude of 1.0 m. A block of mass 5 kg is placed on the platform. What is the maximum frequency of platform’s SHM so that the block is not detatched from the platform?

2

Amax = g Amax =

g 2

=

mg k

2

k m

13.6 Comprehensive Physics—JEE Advanced

13.10 At time t = 0, the displacement of a simple harmonic oscillator from the mean position is x0 and the velocity is v0. Obtain the expressions for the amplitude and phase constant of the oscillator in terms of x0, v0 and where is the angular frequency of the oscillator. SOLUTION For a simple harmonic oscillator x = A sin ( t + )

(i)

v=

From (iii) and (iv), we get

and

A=

x02

tan =

x0 v0

v02

a=

1/ 2

k x m

F = m

(i)

Since a (–x), the motion of the block is simple harmonic. Comparing Eq. (i) with a = – 2x, we get 2

dx = A cos ( t + ) (ii) dt Putting t = 0 and x = x0 and v = v0 in (i) and (ii), we have (iii) x0 = A sin (iv) and v0 = A cos and

The block is pulled to the right by a small distance x from the equilibrium position and released. The restoring force on the block is F = – kx. The acceleration of the block is

=

k m

Time period T =

k m

= 2

=2

m k

(2) Vertical Oscillations of a Mass-spring System Consider a massless spring suspended from a support. [Fig. 13.4 (a)]. A block of mass m is attached at the lower end, as a result, the string extends by an amount d given by [Fig. 13.4 (b)]. F = kd mg = kd (i) This is the equilibrium state of the system.

2

= tan

1

x0 v0

NOTE If at t = 0, x = x0 and v0 = 0, then A = x0 and

. On 2 v the other hand, if at t = 0, x = 0 and v = v0 then A = 0 and = 0

13.5

=

EXPRESSIONS FOR TIME PERIOD OF MASS-SPRING SYSTEM

(1) Horizontal Oscillations of a Mass-Spring System Consider a block of mass m placed one horizontal frictionless surface and attached to a spring of negligible mass and spring constant k as shown in Fig. 13.3.

Fig. 13.4

When the body is pulled through a distance y from this position and released [Fig. 10.4 (c)], the restoring force is F = – ky and the acceleration of the block is a=

F k =– y m m

Hence the motion is simple harmonic whose time period is T= 2

Fig. 13.3

m k

(ii)

Equation (i) determines k. Time period given by Eq. (ii) is the same as for horizontal oscillation. It depends only on m and k and is independent of gravity.

Simple Harmonic Motion

(3) Parallel Combination of Springs Figure 13.5 shows the equilibrium state of a block connected to two springs which are joined in parallel. If the block is pulled down through a distance x, the extension produced in each spring will be x. The restoring forces in the springs are F1 = – k1 x and Fig. 13.5 F2 = – k2 x. Total restoring force F = F1 + F2 = – (k1 + k2)x = – (kp) x where kp = k1 + k2 is the effective force constant of the parallel combination. The time period is given by m =2 kp

T= 2

m (k1

k2 )

(4) Series Combination of Springs Figure 13.6 shows the equilibrium state of a block connected to two springs which are joined in series. The block is pulled down by a distance x. Let x1 and x2 be the extensions produced in the springs. The restoring force in each spring will be the same equal to F = – k 1x 1 = – k 2x 2 so that Total extension

or

x1 = –

Fig. 13.6

= –F

1 k1

= –F

k1 k2 k1k2

F= –

k1k2 k1 k2

1 k2

x = – ks x

k1k2 is the effective force constant of the k1 k2 series combination. The time period is

where ks =

T= 2

m =2 ks

m(k1 k2 ) k1k2

(5) A Block Connected between Two Springs Figure 13.7 shows the equilibrium state of a block connected between two springs

Fig. 13.7

If the block is displaced through a distance x, say to the right, the spring k1 is extended by x and spring k2 is compressed by x so that the restoring force exerted by each spring on the block is in the same direction (along the left). If F1 and F2 are the restoring forces, F1 = – k1x and F2 = – k2 x, the total restoring force is F = – (k1 + k2)x = – kx where k = (k1 + k2) is the effective force constant of the system. The time period is T= 2

m =2 k

m k1

k2

13.11 If a spring of force constant k is cut into two equal halves, what is the force constant of each half. SOLUTION

F F and x2 = – k1 k2

x = x1 + x2

13.7

If a force F produces an extension x in the spring, then F = kx (i) Since the extension produced by a force is proportional to the length of the spring, if a spring is cut into two equal halves, the same force F will produce an extension x = x/2 in half the spring. If k is the force constant of half the spring, x (ii) F= k x =k 2 From Eqs. (i) and (ii), we get k = 2k. 13.12 Two blocks, each of mass m, are connected by a spring of force constant k and placed on a horizontal frictionless surface as shown in Fig. 13.8. Equal force F is applied to each block as shown. Find time period of the system when the force is removed. SOLUTION

Fig. 13.8

When a force F is applied at each end of a spring, every coil of the spring is not elongated. The coil at point O in the middle of the spring is not elongated. This situation is the same as two springs each of length l/2 (where l is the length of the complete spring) joined to each other at point O. If k is the force constant of the complete spring, the force constant of each half = 2k. Hence

13.8 Comprehensive Physics—JEE Advanced

T= 2

m 2k

13.13 A spring has a natural length of 50 cm and a force constant of 2.0 103 Nm–1. A body of mass 10 kg is suspended from the spring. (i) What is the stretched length of the spring? (ii) If the body is pulled down further stretching the spring to a length of 58 cm, and then released, what is the frequency and amplitude of oscillation. Neglect the mass of the spring.

The radius of the circle along which the trolley moves is r = 40 cm = 0.4 m When the table is rotated, the tension in the spring is equal to the centripetal force, i.e. mv 2 = mr 2 r = 2.0 0.4 (10 )2 = 790 N

T=

Now, extension produced in the spring by this force = 40 – 35 = 5 cm = 0.05 m

SOLUTION Natural length of the spring = 50 cm = 0.50 m k = 2.0 103 N m–1, m = 10 kg (i) The extension produced in the spring is mg 10 9.8 y= = k 2.0 103 = 0.049 m = 4.9 cm Stretched length of the spring = 0.50 + 0.049 = 0.549 m = 54.9 cm (ii) The frequency of oscillation is given by =

1 2

k m

1 2.0 103 2 10 = 2.25 Hz

Force constant of spring =

force extension

790 N 0.05 m

= 1.58 104 Nm–1 1.6 104 Nm–1 13.15 A tray of mass M = 12 kg is supported on two identical springs as shown in Fig. 13.9. When the tray is depressed a little and released, it executes an SHM of period 1.5s. (a) Find the force constant of each spring. (b) When a block of mass m is placed on the tray, the period of the SHM becomes 3.0s. Find the value of m.

The length of the spring in the equilibrium position = 54.9 cm. If it is pulled to a length of 58 cm, the maximum displacement from the equilibrium position = 58 – 54.9 = 3.1 cm, which is the amplitude of oscillation. 13.14 A small trolley of mass 2.0 kg resting on a horizontal frictionless turntable is connected by a light spring to the centre of the table. The relaxed length of the spring is 35 cm. When the turntable is rotated at a speed of 300 rev/min, the length of the spring becomes 40 cm. Find the force constant of the spring.

Fig. 13.9

SOLUTION (a) The time period of oscillation of M is given by M 2k

T =2

SOLUTION Mass of trolley (m) = 2.0 kg –1

300 = = 60

Frequency of rotation ( ) = 300 rev. min 5 rev. s–1 Angular frequency ( ) = 2 = 2 5 = 10 rad s–1

where k is the force constant of each spring. Therefore k=

2

2

M

T2

=

2 (3.142) 2 12

(1.5) 2 = 105.3 N m–1

Simple Harmonic Motion

(b) The new period T1 is now given by M m 2k

T1 = 2 so that m + M = or

m + 12 =

2

2

(3)2 105.3

2 (3.142) 2 m = 36 kg

or

13.6

T12 k

= 48

EXPRESSIONS FOR TIME PERIOD OF SOME OTHER SYSTEMS

(1) A Ball Oscillating in a Concave Mirror A small spherical steel ball is placed a little away from the centre of a concave mirror whose radius of curvature is R. When the ball is released, it begins to oscillate about the centre. Place a small steel ball at A, a little away from the Fig. 13.10 centre O of a concave mirror of radius of curvature R (= OC = AC) as shown in Fig. 13.10. Let ACO = . If m is the mass of the ball, its weight mg acts vertically downwards at A. This force is resolved into two rectangular components: mg cos (which is balanced by the reaction of the mirror) and mg sin (which provides the restoring force F). Thus F = – mg sin = – mg (since =

mg x R

(

is small, R being very large)

x = R ; x being the arc OA)

= – Kx where force constant K = mg/R. Thus the motion is harmonic and the angular frequency is given by = T= 2

k = m

g R

R g

(2) Oscillation of a Liquid in a U-tube The column of the liquid is displaced through y by gently blowing into the tube (Fig. 13.11). The columns exhibit

13.9

vertical oscillations. Let L, A and be respectively, the length of the liquid column, area of cross-section of the tube and density of the liquid. We shall neglect viscous effects. Since the right-hand side column is higher by 2y, with respect to the column on the left-hand side, the mass of this column of liquid is m = 2A y. The restoring force (which is a gravitational force) is given by F = – mg = – 2A gy = – Ky where the force constant K = 2A g. The angular frequency of the harmonic oscillation is K M where M = AL is the total mass of the liquid in oscillation. Thus =

=

Fig. 13.11

2A g AL

2g L

Oscillations of a liquid column

The time period of oscillation is T =2

L 2g

It is interesting to note that the period of oscillation does not depend on the density of the liquid or the area of crosssection of the tube. (3) Oscillation of Floating Vertical Cylindrical Body A cylindrical piece of cork of height h and density c a liquid of density l. The cork is depressed slightly and released. Let A be the cross-sectional area of the cork and M its mass. Fig. 13.12 (a) shows the static equilibrium, the weight of the cork being balanced by the Fig. 13.12 weight of the liquid it displaces. If the cork is depressed through a distance x, as shown in Fig. 13.12 (b), the buoyant force on it increases by l Agx, because l Ax is the mass of the liquid displaced by dipping, g being the acceleration due to gravity. If viscous effects are neglected, the restoring force on the cork is given by

13.10 Comprehensive Physics—JEE Advanced

F= –

l

Agx = – Kx

where K = l Ag. Since F – x, the motion of the cork is simple harmonic. The time period of the motion is M K where M is the mass of the cork = Ah c. Hence

If I is the moment of inertia of the system about the axis passing through O and perpendicular to the plane of the rod, the angular acceleration is =

T= 2

Ah c l Ag

T= 2

h g

2

Time period T = 2

T is independent of the mass of the bob.

restoring torque = – mgL sin = – mgL

=

mgL I

Thus

T= 2

where

I=

Thus

T= 2

If M B. 480 – A = 10 Case (i) A < B ; B – A = b A = 470 Hz. On loading with a little wax, the frequency of a fork decreases slightly, i.e. A becomes slightly less than 470 Hz. Hence the number of beats per second must increase. But b decreases to 5. Hence A cannot be less than B. Case (ii) A > B. In this case A = B + b = 480 + 10 = 490 Hz. On loading A, A decreases. Hence b = A – B will decrease. Since b is observed to decrease to 5, A must be greater than B. (a) Hence before loading, A = 490 Hz (b) After loading A = 490 – 5 = 485 Hz 14.18 A metal wire of diameter 1.5 mm is held on two knife edges separated by a distance of 50 cm. The tension in the wire is 100 N. The wire vibrating with its fundamental frequency and a vibrating tuning fork together produce 5 beats per second. The tension in the wire is then reduced to 81 N. When the two are excited, beats are heard at the same rate. Calculate (a) the frequency of the fork, and (b) the density of the material of the wire. SOLUTION

2

Therefore frequency of maxima is b

Hence the frequency of beats is b = 1 – 2

b.

(a) Let N be the frequency of the tuning fork. Then, the frequency of the wire, when the tension is 100 N will be (N + 5) and when the tension is 81 N, it is (N – 5); since in each case 5 beats are heard per second. Hence

14.12 Comprehensive Physics—JEE Advanced

N+5=

and N – 5 =

1 T1 1 100 10 = = 2L m 2 0.5 m m (i) 1 T2 1 81 = = 2L m 2 0.5 m

9 m (ii)

Subtracting (ii) from (i) we have 1

10 =

m

or

m = 0.01 kg m–1

Using this values of m in (i) or (ii) gives N = 95 Hz. (b) Now m = r2 = Putting d = 1.5 we get = 5.7

14.11

d2 4 10–3 m and m = 0.01 kg m–1, 103 kg m–3

DOPPLER EFFECT IN SOUND

The apparent change in frequency of sound heard by an observer due to a relative motion between the observer and the source of sound is called the Doppler effect. The expressions for the apparent frequencies are as follows: 1. Source approaching a stationary observer

where

v

, v us = real frequency v = velocity of sound us = velocity of source 1

=

2. Source receding from a stationary observer v v + us 3. Observer approaching a stationary source of sound v + u0 , 3 = v where u0 = velocity of observer 2

=

4. Observer receding from a stationary source of sound v u0 4 = v 5. Both approaching each other 5

=

v v

u0 us

6. Both receding from each other v u0 v us 7. Source approaching a receding observer 6

=

v u0 v us 8. Observer approaching a receding source 7

=

v u0 v us When the source of sound goes pass stationary observer, the apparent change in the frequency of sound is given by 2 vus = v 2 us2 8

=

2us v When the observer goes past a stationary source of sound, the apparent change in frequency of sound is given by If us 2. If is the frequency of the L tuning fork, then + 4 and 2 = – 4 1 = 4 1 = 4 2 1

But

=

2

L2 50 = L1 48

25 24

25 4 = 24 4 which gives = 196 Hz. Hence the correct choice is (a). 29. Let T1 be the tension in each string when they are in unison. Let T2 be the tension in each string when T ; 2> 1 they are not in unison; then since such that T 2 = 2 T1 1 Now T2 = 1.01 T1. Therefore Thus

2

=

1.01 = 1

1

1 100

1/ 2

Also, or

1 100

2

1/ 2

1 201 = . Thus 200 200 201 2 = 200 1 – 1= 4 2 = 1 + 4. Therefore, we have 4

1 1

which gives choice is (b).

1

1+

201 = 200 = 800 Hz. Hence the correct

30. Let v1 be the speed of sound at 27°C and v2 at 31°C. Then v2 v1

273 31 273 27 1

1/ 2

1 2

304 1 / 2 1 300 4 151 300 150

4 300

Thus the correct choice is (b). 31. If a source emitting light of wavelength goes away from the earth, the apparent wavelength of the light reaching the earth is given by v =1+ c where v is the speed of the source of light and c the speed of light. The increase in wavelength = – is given by v = c 5 Here = 5% = and c = 3 108 ms–1. 100 Therefore, 5 = 1.5 107 ms–1 100 Hence the correct choice is (d). 32. If the aircraft is approaching the radar station with a speed u, the apparent frequency of radiowaves rev=3

108

is given by

2u c Apparent increase in frequency is 2u = – = c Given, = 600 MHz = 600 106 Hz and = 6 kHz = 6 103 Hz. Thus c 3 108 6 103 u= = 2 2 600 106 =

Expanding binomially, we have 1

Now, frequency speed of sound. Hence v 151 2 = 2 = v1 150 1 151 300 151 = = 302 Hz. 2 = 1 150 150 Hence beat frequency = 302 – 300 = 2

1/ 2

= 1.5

1

103 ms–1 = 1.5 kms–1

Hence, the correct choice is (a). u0 1 1 . Given u0/v = . Therefore 33. = v 10 11 = 10 11 or – = – = . 10 10 The percentage increase in is 100 = 10% which is choice (c).

Waves and Doppler’s Effect 14.27

34. To form a stationary wave, waves y and y must travel in opposite directions. Wave y = a cos (kx – t) travels along the positive x-direction. Waves y = – a cos (kx – t) and y = – a sin (kx – t) in choices (b) and (d) travel along positive x-direction. Hence choices (b) and (d) are not possible. Choice (a) is also incorrect because at x = 0 y = a sin t and y = a cos (– t) = a cos t Therefore, the resultant displacement at x = 0 which is y + y = a sin t + a cos t is not zero, i.e. these waves do not produce a node at x = 0. Choice (c) is correct because at x = 0, y + y = 0. 35. The frequency of a string of length L, mass m per unit length, stretched with a tension T and vibrating in p segments, is given by p T 2L m If the radius of the wire is r and mass = length

r2 L L

its density, then =

r2 .

p T 2Lr Since , T, L and are the same for both wires, p = constant or p r. Hence the number of loops r formed in the thicker wire will be two times that in the thinner wire. Hence the correct choice is (b). =

36.

the rope is different at different points on the rope. sion = weight of the rope + the weight attached to the free end of the rope = 6 kg + 2 kg = 8 kg wt. Tension at the free end of the rope = 2 kg wt. Since T , if the tension becomes 4 times, the frev 1 quency is doubled. Since = ; . Hence

the wavelength is halved. Thus the correct choice is (b). v 37. For a closed tube = . For an open tube = 4 L v . Hence = 2 = 2 512 = 1024 Hz. Thus the 2L correct choice is (a). 38. When two waves of amplitudes a1 and a2 superpose to produce beats, the resultant amplitude of the maxima of intensity is A = a1 + a2 2

=

. k 40. The amplitude of the three waves are a1 = 10 m, a2 = 4 m and a3 = 7 wave be zero. Then the wave of the second wave = /2 and of third wave =

2

. Therefore,

2

third wave is . Hence their resultant is

=

m=

i.e. a1 = a2 = a. Thus A = 2a. Therefore, A2 = 4a2 or Imax = 4I. Hence the correct choice is (b). 39. Distance between adjacent nodes (or antinodes) = /2. Also 2 x in the argument of the sine function = k 2 . or = k Hence, the distance between adjacent nodes

Now, intensity (amplitude) . Since the two waves have the same intensity, their amplitudes are equal,

a = a1 – a3 = 10 – 7 = 3 m The phase difference between this resultant (of /2. Therefore, the resultant of a and a3 is A = (a2 + a 22)1/2 = (32 + 42)1/2 = 5 m. Hence the correct choice is (c). 41. Wave velocity = v. Particle velocity is dy V= dt = y0 Vmax = y0

2 v

cos

2

vt

x

2 v 2 v

Now, Vmax = 2 v, if y0

= 2v, which gives

= y0. Hence the correct choice is (d). 42. If the train is going away from the observer, the apparent frequency is 1

=

v

v u

1

It is observed that = 1.2 apparent frequency is 2

or

= 2

v u v

=

1 1

u v

(i)

u v

1. In the second case, the

=

1

u v (ii)

14.28 Comprehensive Physics—JEE Advanced

u = 1 + u or 1.2 = 1 + , v v 1 u = 0.2. Using this in (ii), we get or u = 0.2 v or v 5 = 1.25. Hence the correct choice is (b). 4 2 v 43. For pipe P1 : 1 = 4 L1 Now, from (i) we have

For pipe P2 :

3

3v 2L2

=

It is given that 1 = 3. Therefore, L1/L2 = 1/6 which is choice (b). 44. The speed of the wave in the string is given by T m According to Hookes’law, tension (T) extension (x). x . Therefore Hence v

47. For a gas, adiabatic elasticity E = P where = Cp/Cv and P is the pressure. The speed of sound in the gas is given by RT M where M is the molecular mass and R is the gas constant. Thus vN M He 4 1 = vHe MN 28 7 48. Frequency of the fundamental mode is given by

46. Standing waves are formed on the string. Particle displacements are given by 2 x

y = a sin

cos (2

1 T ; 2L m T = tension and m = mass per unit length

= (– 2

fa) sin

2 x

or Now

fa =

f=

v 20 a

sin (2

ft)

=

v . Therefore, 10

10 3 m

20 v f

1

10 ms

1

3

10 /2 Hz

Hence the correct choice is (a).

103 Hz 2 =2

T / m1

1/ 4L

T / m2

Now

m1 =

M1 = L

and

m2 =

M2 2L

1

Hence

r2

= 2

49. Beat frequency

0

=

1

b

v=

(

2

2r 2 L L r2

4 r2

2

2L 2L

m2 m1 =4 =

r2 r2

=1

30 = 10 per second. Now 3 v v – 2=

=

1

2

5 6 10 (6 5) 1) = 300 ms–1

1 2 b 2

50. Let the frequency of fork C be n. Then nA = n + 0.03 n = 1.03n and nB = n – 0.02 n = 0.98 n. The beat frequency is nb = nA – nB or

v 10 10 ms

1/ 2L

Hence the correct choice is (a).

(V)max = 2 fa

2

=

2

or

dy V= dt

Given(V)max =

1

f t)

Particle velocity

P

=

v=

v 15 . x = 15 . = 1.22 v x Hence the correct choice is (a). dy d 45. Particle velocity V = [A sin (kx – t)] dt dt = – A cos (kx – t) Hence Vmax = A

E

v=

10–2 m

5 = 1.03 n – 0.98 n = 0.05 n

which gives n = 100 Hz. Hence nA = 1.03 100 = 103 Hz which is choice (c). 51. Given t1 = 10ºC, n1= 1.001 kHz, t2 = 30ºC, n2 = 1 kHz. Let l1 and l2 be the vibrating lengths of the wire at 10ºC and 30ºC respectively. Since tension T is kept constant. (i) n 1l 1 = n 2 l 2 l2 n = 1 l1 n2

1.001 1.001 1

Waves and Doppler’s Effect 14.29

If l2 = l1 (1 +

55. For train A: uA = 0.1 v

t)

l2 = 1 + a t = 1 + a (30–10) l1 = 1+ 20a From (i) and (ii) we have 1 + 20 = 1.001 or

(ii)

–4

which gives = 0.5 × 10 per ºC, which is choice (d). 52. Let L1 and L2 be the lengths of the two parts of sonometer wire. Given or Thus

1

1

=

L1 = L1 =

1 2 L1

T and m

L2 =

2

k

=

1 2 L2

T m

1 T = constant, say k. 2 m

and L2 =

1

2

k 2

The fundamental frequency of the complete sonometer wire is T 1 k = = 2( L1 L2 ) m ( L1 L2 ) or or

1

=

L2 1 = k 1

L1 k

1 2

= 1

1 2

, which is choice (d). 2

53. Let W1 be the weight of the bob in air and W2 when it is immersed in water. Given 1

=

W1 = W2

1 W1 and 2L m

2

=

1 W2 2L m

2 1 2 2

Relative density = =

weight in air loss of weight in water

W1 = W1 W2

Hence the correct choice is (a).

2 1

2 1

2 2

54. After 2 seconds, both the pulses will be at the same location on the string and will superpose on each other. Since their amplitudes are equal and opposite, they cancel each other and the string becomes straight. Hence the string has no potential energy, i.e. the total energy is purely kinetic. Thus the correct choice is (b).

For train B:

A

=

B

=

uB = v/5 = 0.2 v

v

uA v

or 5.5 = 5

v

uB v

or 6.0 = 5

v

uA v

v

uB v

uB 0.2 v = = 2, which is choice (b). uA 0.1v 56. Let L be the length of the wire between the bridges and let m be the mass per unit length of the wire. 5 Five antinodes on a length L implies that L = 1 2 2L or 1 = . Thus in this case we have 5 1

5 2L

v1

=

1

T1 , where T1 = M1 g m

Three antinodes on a length L implies that 3 2L . In this case, we have L= 2 or 2 = 2 3 2

=

3 T2 , where T2 = M2 g 2L m 2 = 2, M1 = 9 kg and M2 = M. Therefore,

v2

Given 1 we have

5 9g 3 Mg 2L m 2L m which gives M = 25 kg. Hence the correct choice is (a). 57. Given v = 330 ms–1 and up = 22 ms–1. The apparent frequency of the police man’s horn of frequency 176 Hz as heard by the motorcyclist is given by 330 um 176 = (330 – um) (i) 330 22 308 The apparent frequency of the stationary siren of frequency 165 Hz as heard by the motorcyclist is given by 330 um (ii) 2 = 165 330 Since the motorcyclist does not observe any beats, 1 = 2. Equating (i) and (ii) and solving for um we get um = 22 ms–1. Hence the correct choice is (b). 58. Let be the frequency of the tuning fork and e the end correction. Given L1 = 0.1 m and L2 = 0.35 m 1

= 176

= =

=

1

3

=

1 4 ( L1

P e)

3 4 ( L2 e)

P

(i) (ii)

14.30 Comprehensive Physics—JEE Advanced

Equating (i) and (ii), we get 1 3 = L1 e L2 e 1 3 = 01 . e 0.35 e which gives e = 0.025 m. Hence the correct choice is (b). 59. The density of a wire of mass M, length L and diameter d is given by 4M 4m = = 2 d L d2 or

Now

vA = vA = vB

T mA

and vB =

T mB

490 =

(n 1) 2L

1 T 1 1.44T 1.2 = = 2L m 2L m 2L Comparing this with Eq. (1), we have =

= 1.2 –

= 6 Hz. Hence

+ 6 = 1.2

1 2L

T 1 = m 1.2 2 L

T = m 1.2

30 = 25 Hz. Thus, 1.2 the frequency decreases by 5 Hz. Hence the correct choice is (a). nv ; n = 1, 3, 5, etc 64. For a closed pipe, n = 4L nv For an open pipe, n = ; n = 1, 2, 3, etc 2L Now = 30 Hz. Therefore,

where v =

P

=

.

(n = 1), we have

Dividing, we have 490 (n 1) = which gives n = 6. 420 n 61. The fundamental frequency of the vibration of a wire of length L, mass m per unit length and under tension T is given by 1 T (1) = 2L m If the tension is increased by 44%, the new tension is 44 T = T + 0.44 T = 1.44 T T =T+ 100 Since L is kept constant, the new fundamental frequency is

Given

1 T 2L m Since T and m are the same, the new frequency will be =

T m

=

=

mB d = B mA dA

but vA = n A and vB = n B, n being the frequency of the source. v d 0.5 10 3 A Hence = A = B = = 0.5, vB dA 10 3 B which is choice (a). 60. The frequency of the nth hormonic is 420 Hz. The frequency of the (n +1)th harmonic is 490 Hz. Therefore, n T 420 = 2L m and

6 = 30 Hz, which is choice (b). 0.2 62. A travelling wave is characterized by wave functions of the type y = f (vt + x) or y = f (vt – x) where f stands for sine or cosine function. Hence the correct choice is (d). Choices (a), (b) and (c) represent a stationary or standing wave. 63. If the length is increased by 20%, the new length is 20 L L =L+ = 1.2 L 100 The original frequency is which gives

T m

1 P (1) 4 L1 1 For the open pipe vibrating in the third harmonic (n = 3), we have 1

3 P 2 L2 2 . Equating (1) and (2), we get 3 3

Given

1

=

=

L1 1 = L2 6

2

(2)

.

1

Hence the correct choice is (d). 65. The frequency of sound is a characteristic of its source. Hence frequency of sound is the same in air as is water. Therefore, the observer in air will receive a sound of frequency 600 Hz. If a and w are the wavelengths in air and water resptively, then v vw n= a a

w

Waves and Doppler’s Effect 14.31

300 = 0.5 m 600 Hence the correct choice is (a). The wavelength is water is vw 1500 = = 2.5 m w= 600 66. Refer to Fig. 14.12. 3 L1 = and L2 = 4 4 v N= 1 = (1) 4 ( L1 e) which gives

a=

va

3v (2) 4 ( L2 e) where e is the end correction. Eliminating e from Eqs (1) and (2), we get (3) v = 2N (L2 – L1) and

N=

3

=

Lengths L1 and L2 are measured with a metre scale whose least count is 0.1 cm. Thus L1 = (17.7 0.1) cm and L2 = (53.1 ± 0.1) cm. The maximum error in (L2 – L1) is ± 0.2 (for maximum error, the errors in individual measurements add up). Thus L2 – L1 = (53.1 – 17.7) = 35.4 cm. Hence L2 – L1 = (35.4 ± 0.2) cm. Using this in Eq. (3), we have v = 2 cms–1

480

(35.4 ± 0.2) = (33984 ± 192)

Hence maximum error = 192 cm s–1, which is choice (a).

For rotational equilibrium of the rod about O, we have T1 AO = T2 BO or T1 x = T2 (L – x), T T ( L x) . But 1 = 4. Hence which gives 1 = T2 T2 x ( L x) 4= x L which gives x = . Hence the correct choice 5 is (a). 68. When the source moves towards from the stationary observer, its apparent frequency is v

(1) v u When the source moves away from the stationary observer, its apparent frequency is =

v

(2) v u When both the forks are moving relative to stationary observer, the number of beats heard by him per second = – Since us T2. They produce 6 beats per second when vibrated. If the tension in one of them is changed slightly, it is observed that the beat frequency remains unchanged. Which of the following is/are possible? (b) T1 was decreased (a) T1 was increased (d) T2 was decreased (c) T2 was increased IIT, 1991 16. A plane progressive wave of frequency 25 Hz, amplitude 2.5 10–5 m and initial phase zero propagates in a non-absorbing medium along the negative x-direction with a velocity of 300 ms–1. Then (a) Wavelength of the wave is 12 m (b) The phase difference between the oscillations at two points 6 m apart is (c) The corresponding amplitude difference is 2.5 10–5 m (d) The corresponding amplitude difference is zero. IIT, 1997 17. The (x, y) co-ordinates 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 a certain instant of time, the possible expression(s) for u(x, y) is/are (a = positive constant) x y cos (a) u(x, y) = a cos 2L 2L x y (b) u(x, y) = a sin sin L L x 2 y (c) u(x, y) = a sin sin L L 2 x y (d) u(x, y) = a cos sin L 2L IIT, 1998

Waves and Doppler’s Effect 14.35

18. A transverse sinusoidal wave of amplitude a, wavelength 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 ms–1, then (b) = 10–3 m (a) =2 10–2 m 103 Hz (d) f = 104 Hz (c) f = 2 IIT, 1998 19. As a wave propagates in a non-absorbing medium, (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) the total intensity of a spherical wave over a spherical surface centred at the source remains constant at all times. IIT, 1999 20. A moving pulse is represented by the expression 0.8 y(x, t) = [(4 x 5t ) 2 5] where x and y are in metre and t in second. Then (a) the pulse is moving in the + x direction (b) in 2s it will travel a distance of 2.5 m (c) its maximum displacement is 0.16 m (d) it is a symmetric pulse. IIT, 1999

21. In a wave motion y = a sin (k x – t), y can represent (c) displacement

(d) pressure IIT, 1999

22. Standing waves can be produced (a) on a string clamped at both the ends (b) on a string clamped at one end and free at the other wall (d) when two incident waves with a phase difference of are moving in the same direction. IIT, 1999 23. 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 and that with the longer air-column is the second resonance. Then, 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 nance was somewhat shorter than 1/4th of the wavelength of the sound in air IIT, 2009

ANSWERS AND SOLUTIONS 1. When a wave travelling in a medium falls on the into the second medium. Therefore, the intensity refracted waves will be less than that of the incident wave. The velocity of a wave depends upon the medium in wave will be the same as that of the incident wave. But the velocity of the refracted wave will be different from that of the incident wave. The frequency same as that of the incident wave. From v = have.

, we

=

v

same as that of the incident wave. But the wavelength of the refracted wave will be different from that of the incident wave. Furthermore, when a wave travelling in a rarer medium, it undergoes a phase change of

or 180°.

a rarer medium, it does not undergo any phase change. The refracted wave, in both cases, does not undergo any phase change. Thus the correct choice are (a) and (d). 2. The correct choices are (a) and (b).

14.36 Comprehensive Physics—JEE Advanced

3. Statement (a) is correct. Let us write y (x, t) = f (vt + x) = f (z) Differentiating with respect to t, we have

u v Expression (i) may be written as

y f z f = =v t z t z Differentiating again w.r.t time t we have

u 1 v Expanding binomially and retaining terms upto order u2/v2, we have

2

y 2

= v2

2

and

f 2

t z Similarly differentiating twice with respect to x we have 2 2 y f = 2 2 x x Hence, 2

y 2

= v2

2

y 2

t x which is the standard equation (in differential form) of a travelling wave. Statement (b) is also correct. Because the wave is remains unchanged. The wavelength cannot change Statement (c) is incorrect. The ultrasonic wave bends away from the normal because the speed of the wave (being a sound wave) is greater in water than in air. Statement (d) is correct. The reason is that solids have a much higher modulus of elasticity than gases at STP. 4. Statement (a) is incorrect. A change in pressure has no effect on the speed of sound. The decrease in the speed of sound at high altitudes is due to a fall in temperature. Statement (b) is correct. Standing waves are produced due to superposi-

dent wave. Consequently, the resultant amplitude at nodes is not exactly zero. Thus the nodes are not Statement (c) is also correct. To observe beats the difference between the two interfering frequencies must be less than about 10–16 Hz. Since visible light waves have very high frequencies, beats are not observed due to persistence of vision. Statement (d) is correct. We know that 1

=

1

u v

2

(i)

=

1

1 =

1

=

1

1

u v

(ii)

u2

(iii)

v2

1 > 2. 5. A stationary wave is characterized by a function of the type y = f (t) g(x). Hence choices (a) and (b) represent a stationary wave. Choice (d) is a superposition of two oppositely travelling waves of the same amplitude and the same frequency which gives rise to a stationary wave. Hence choice (d) also represents a stationary wave. 6. Comparing y = a sin 2 (bt – c x) with 2 vt x y = a sin

we have 2 v 2 2 b= and 2 c = which give v = b and Particle velocity is

=

1 . Thus v = b/c. c

dy d = [a sin 2 (bt – cx)] dt dt = 2 ab cos 2 (bt – c x) Vmax = 2 ab. Now Vmax = 2v, if 2b 2 ab = c 1 which gives c = . a V=

b/c =b 1/ c Thus the correct choices are (b), (c) and (d). 7. Person A hears the sound of his own source whose frequency is . He also hears the sound of the source carried by person B, towards whom he is moving with a speed u. The apparent frequency of this sound is given by Frequency

=

v

=

= Beat frequency

u v

1 b

=

or –

= 9 Hz

–

=

u v

u 90v = v 10v ( u = v /10).

=

Waves and Doppler’s Effect 14.37

Person B hears the sound of his own source of frequency in . He also hears the sound of the source carried by person A, who is approaching with a speed u. The apparent frequency of this sound is given by v =

or

–

=

u v

1

v

=

Beat frequency

=

b

–

P 2L

T m

490 =

(2)

p T 600 = 2L m L

2v v = (3) 2 L0 L0 In a closed pipe, only odd harmonics are present. Hence, the frequencies of the overtones are 3, 5, 7, ... etc. times the fundamental frequency. Hence the is

=2

0

=

(4)

330 Lc = 0.75 m. 4Lc = ±10. There are the following two 2

= 10. Thus

3v = 10 4 Lc

33 m 32 Hence the correct choices are (a) and (d). 10. Including end correction, we have for the fundamental mode, L =

10 which gives L = m. Hence the correct choices are (a) and (d). 7 9. The fundamental frequencies of the open and closed pipes respectively are v (1) 0= 2 L0 v and (2) c= 4Lc where v is the speed of sound. In an open pipe, all harmonics are present. Hence the frequencies of the overtones are 2, 3, 4, ..... etc. times the fundamental frequency. Hence the fre-

1

3v 4 Lc

33 m 34 Case (b) : 1 – 2 = – 10. In this case, we get

490 p 1 = giving p = 6. 220 p Substituting this value of p in Eq. (1) we get 420 =

=

Lo =

(1)

p 1 T 2L m Dividing (1) and (2) we have

and

c

Putting v = 330 ms–1 and Lc = 0.75 m, we get

u = v u

90 v /10 = 10 Hz v v /10 Hence the correct choices are (a) and (d). 420 =

=3

110 =

v Lo

u v u

=

8.

Now Given 1 – 2 possibilities. Case a : 1 –

v v±u 1 =

v u

2

= L + 0.3 D or = 4 (L + 0.3 D) 4 where D is the diameter of the tube. Now v = v = or

4 (L + 0.3 D) =

v

(i)

Given L = 16 cm = 0.16 m, v = 336 ms–1 and = 480 Hz. Using these values in (i) and solving we get D = 5 10–2 m = 5 cm. The correct choice is (b). 11. The correct choice is (a). dx , dx = vdt = k T dt. As the temdt perature increases linearly, the rate of change of temperature with distance is given by

12. Since v =

T T dT = 2 1 L dx where L is the distance between locations A and B. Thus dx =

LdT T2 T1

But dx = k T dt. Therefore k T dt =

LdT T2 T1

14.38 Comprehensive Physics—JEE Advanced

or

dt =

If T1 is kept constant and T2 is increased such that the new value 2 becomes greater than 1 by 6, again 6 beats will be heard per second. Hence choice (c) is correct. If T2 is kept constant and T1 is decreased such that the new value of 1 becomes less than 2 by 6, again 6 beats will be heard per second. Hence choice (b) is also correct. v 300 16. = = = 12 m 25 2 path difference Phase difference =

LdT k (T2

T1 ) T

Integrating from T = T1 to T = T2, we have t=

t=

=

T2

L k T2

T

T1

T1 1/ 2

L

T T1 ) 1 / 2

k (T2

–1/2

2L k (T2 T1 )

T2

T1

T2

T1

2L

=

k T2 T1 Thus the correct choices are (a) and (c). 13. Linear velocity of source (whistle) is us = r = 1.5 20 = 30 ms–1. The observer will hear a range of frequencies lying between a minimum value vmin and a maximum value max, which are given by min

v

=v

v us

and vmax = v

v v us

; source receding ; source approaching

Substituting v = 440 Hz, v = 330 ms–1 and us = 30 ms–1 and solving we get vmin = 403.3 Hz and vmax = 484 Hz. The correct choice is (d). 14. The standard equation of a wave travelling in the negative x-direction is y = A sin ( t + kx + 0) 2 v 2 where = ,k= and 0 is the phase at x = 0 and t = 0. Comparing the given equation with this equation, we have 2 = 10 = 0.2 m k = 10 and

2 v

=15

= 15

15 15 0.2 = = 1.5 ms–1 2 2 Thus the correct choices are (b), (c) and (d). v=

15.

=

1 T . Since the wires are identical, 2L T . Since T1 > T2;

1

>

2. Also

2 6= 12 In a non-absorbing medium, the amplitude of the wave remains constant as the wave propagates. Thus the correct choices are (a), (b) and (d). 17. The expression for u(x, y) must satisfy the following boundary conditions: (i) u = 0 at x = 0 and at y = 0 (ii) u = 0 at x = L and at y = L. The choices (b) and (c) satisfy these conditions. 18. y = a sin ( t – kx) Speed at a point on the string is dy = a cos( t – kx) V= dt v Vmax = a = 2 af. Given Vmax = . 10 Hence 10 v v = 2 af f = 10 3 10 20 a 20 =

1

–

2

= 6.

= =

v 10 = f 103 2

=2

103 Hz 2 10–2 m

So the correct choices are (a) and (c). 19. The amplitude (and hence intensity) of a plane wave remains constant as the wave propagates in a non-absorbing medium. For a spherical wave, the same energy crosses a spherical surface of area 4 r2 where r is the distance from the source. The unit area per second. Hence, for a spherical wave, the intensity decreases as 1/r2. But the total intensity spread over the spherical surface is the same at all times. Hence the correct choices are (a), (c) and (d). 20. The displacement y(x, t) is maximum when the denominator is minimum, i.e.

Waves and Doppler’s Effect 14.39

[(4x + 5t)2 + 5] is minimum. Its minimum value is 5 when (4x + 5t)2 = 0 or 4x + 5t = 0 x = t

or

ymax =

5 4

22. Standing waves are produced due to a superposi-

5 ms–1 4

v=

0.8 = 0.16 m 5

Distance travelled by pulse in t = 2 s is

23.

5 2 = – 2.5 m 4 The negative sign shows that the pulse is travelling in negative x-direction. Since y is not a symmetric function of x, the form of the pulses changes as it travels. Hence the correct choices are (b) and (c). 21. In an electromagnetic wave, y represents electric pendicular to each other as well as perpendicular to the direction of propagation of the wave. In a mechanical wave, y represents displacement. In a sound wave, y represents pressure. Thus all the four choices are correct. vt =

clamped end of the string. In case (c) the incident ing in the same direction cannot produce standing waves; they give rise to interference. Thus the correct choices are (a), (b) and (c). tuning fork. At second resonance, the frequency of the third harmonic equals the frequency of the tuning fork. As the amplitude of oscillation of the fundamental mode is the highest, the intensity of Hence choice (a) is correct. Choice (b) is wrong, the prongs are kept in a vertical plane. Choice (c) is also incorrect as the amplitude of vibration of the ends of the prongs is typically around 1 mm. Choice (d) is correct. Due to end-correction (= e), (L1 + e) = choices are (a) and (d).

III Multiple Choice Questions Based On Passage Questions 1 to 3 are based on the following passage Passage I When two sound waves travel in the same direction in a medium, the displacements of a particle located at x at time t is given by y1 = 0.05 cos (0.50 x – 100 t) and

y2 = 0.05 cos (0.46 x – 92 t)

where y1, y2 and x are in metre and t in second IIT, 2006

1. What is the speed of sound in the medium? (b) 100 ms–1 (a) 332 ms–1 –1 (c) 92 ms (d) 200 ms–1 2. How many times per second does an observer hear the sound of maximum intensity? (a) 4 (b) 8 (c) 12 (d) 16 3. At x = 0, how many times between t = 0 and t = 1 s does the resultant displacement become zero? (a) 46 (b) 50 (c) 92 (d) 100

SOLUTION 1. The two displacements can be written as y1 = A cos (k1x –

1t)

and y2 = A cos (k2x –

2t)

where A = 0.05 m, k1 = 1=2

1 = 100

2

(1) (2) –1

= 0.50 m , 1

rad s1, k2 =

2

= 0.46 ms–1 and 2

is

2

=2

2

= 92

v1 = =

rad s–1. The speed of either wave

1 1

100 0.50

=2

1

2

1

= 200 ms–1

=

1

k1

14.40 Comprehensive Physics—JEE Advanced

92 k2 0.46 Hence the correct choice is (d). or

v2 =

2 2

2. Beat frequency = 1=

2

=

1

2

1

– =

=

2.

= A cos (k1x –

= 200 ms–1.

1t)

+ A cos (k2x –

2t)

For x = 0, we have y = A cos

Now

= 2 A cos

100 = 50 Hz 2

92 2 = = 46 Hz and 2= 2 2 Beat frequency = 50 – 46 = 4 Hz. Hence the correct choice is (a). 3. The resultant displacement is given by y = y1 + y2

Questions 4 to 6 are based on the following passage Passage II A string 25 cm long and having a mass of 2.5 g is under of tension. A pipe closed at one end is 40 cm long. When the pipe in its fundamental frequency, 8 beats per second heard. It is observed that decreasing the tension in the string decreases the beat frequency. The speed of sound in air is 320 ms–1. IIT, 1982

1t

+ A cos 1 ( 2

1

2t 2 )t

1 ( 1 2 )t 2 y = 0.10 cos (96 t) cos (4 t) cos

Between t = 0 and t = 1 s, cos (96 t) becomes zero 96 times and cos (4 t) becomes zero 4 times. Hence the resultant displacement y at x = 0 becomes zero 100 times between t = 0 and t = 1 s. The correct choice (d).

4. The frequency of the fundamental mode of the closed pipe is (a) 100 Hz (b) 200 Hz (c) 300 Hz (d) 400 Hz 5. overtone is (a) 92 Hz (b) 108 Hz (c) 192 Hz (d) 208 Hz 6. The tension in the string is very nearly equal to (a) 25 N (b) 27 N (c) 28 N (d) 30 N

SOLUTION 4. Frequency of fundamental mode of the closed pipe is v 320 np = = = 200 Hz 4L 4 0.40 The correct choice is (b). 5. Since the beat frequency is 8, the frequency of the ns = np ± 8 = 200 ± 8 = 192 Hz or 208 Hz 1 T ns = l m

(1)

Questions 7 to 11 are based on the following passage Passage III ends are represented by the equation y = 4 sin

x 15

cos (96 t)

where x and y are in cm and t in second.

IIT, 1985

It is given that the beat frequency decreases if the tension in the string is decreased. As the frequency decreases with decrease of tension, it is obvious that ns > np. Hence ns = 208 Hz and not 192 Hz. Thus the correct choice is (d). 6. Substituting the values of l, m and ns in Eq. (1), we get T = 27.04 N. Hence the correct choice is (b).

7. The frequency of vibrations of the string is (a) 48 Hz (b) 50 Hz (c) 96 Hz (d) 100 Hz 8. The maximum displacement of a point at x = 10 cm is (a) 2 cm (b) 4 cm (c) 2 3 cm

(d) 4 3 cm

Waves and Doppler’s Effect 14.41

9. How many nodes are formed on the string? (a) 2 (b) 3 (c) 4 (d) 5 10. In which harmonic mode is the string vibrating? (a) Fundamental (b) third

11. The velocity of the particle at x = 7.5 cm at t = 0.25 s is (a) zero (b) 320 ms–1 –1 (c) 60 ms (d) 96 ms–1

SOLUTION 7. 2 = 96 = 48 Hz. Thus the correct choice is (a) 8. Displacement is maximum when cos (96 t) = 1. 10 2 = 4 sin At x = 10 cm, ymax = 4 sin 15 3

10. The correct choice is (c) because 5 nodes are formed on the string. 11. The velocity of the string at a point x at time t is obtained by differentiating y = 4 sin

= 2 3 cm

The correct choice is (c). 9. At nodes the displacement is always zero. Hence nodes are located at values of x given by x x = 0 or =p sin 15 15 where p = 0, 1, 2, 3, ... etc. Thus x = 15 p = 0, 15, 30, 45 and 60 cm. Thus the correct choice is (d). Questions 12 to 15 are based on the following passage Passage IV The displacement of the medium in a sound wave is given by y1 = A cos (ax + bt) where A, a and b are positive constants. The wave is x = 0. The intensity of IIT, 1991 12. The wavelength and frequency of the incident wave respectively are (a)

2 b , a 2

2 b (c) , a 2

(b)

a 2 , 2 b

x 15

cos (96 t)

which respect to t. Velocity

dy = – (4 dt

96 )

sin

x sin (96 15

t)

At x = 7.5 cm and t = 0.25 s, the velocity is zero because at t = 0.25 s, sin (96 t) = sin (24 ) = 0. Hence the correct choice is (a).

13. (a) y2 = 0.8 A cos (– ax + bt) (b) y2 = – A cos (– ax + bt) (c) y2 = – 0.64 A cos (– ax + bt) (d) y2 = – 0.8 A cos (– ax + bt) 14. In the standing wave formed due to the superposimum values of the particle speed in the medium is (a) Ab (b) 1.64 Ab (c) 1.8 Ab (d) 2 Ab 15. The minimum value of the particle speed in the medium is

1 (d) a, b

(a) zero

(b) 0.2 Ab

(c) 0.64 Ab

(d) 0.8 Ab

SOLUTION 12. The incident wave is given by y1 = A cos (ax + bt) (1) The standard wave equation is y = A cos (kx + t) (2) where k is the wave number and , the angular frequency. Comparing (1) and (2) we get k = a and =

13.

2 2 b. Hence wavelength = = and frequency k a b = = . 2 2 The correct choice is (a). times that of the incident wave, the amplitude Ar of

14.42 Comprehensive Physics—JEE Advanced

0.64 = 0.8 times that

Differentiating Eq. (3) with respect to t, the particle

of the incident wave, i.e. Ar = 0.8 A

V2 =

suffers a reversal of amplitude (which implies a phase change of radian), i.e. Ar = – 0.8 A. Since the incident wave is travelling in the negative x dix wave can be obtained from Eq. (1) by replacing A by Ar = – 0.8 A and x by – x wave is given by (3) y2 = – 0.8 A cos (– ax + bt) Thus the correct choice is (d). 14. Differentiating Eq. (1) will respect to time t, we get the expression for the particle speed in the medium due to the incident wave, which is d y1 d {A cos (a x + bt)} V1 = dt dt = – Ab (a x + bt) (4) Maximum particle speed due to the incident wave is (V1)max = Ab Questions 16 to 19 are based on the following passage Passage V The air column in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 400 Hz. The speed of sound in air is 320 ms–1. The end correction may be neglected. Let P0 denote the mean pressure at any point in the pipe and P0 the maximum amplitude of pressure variation. IIT, 1998 16. The length L of the air column is (a) 20 cm (b) 60 cm (c) 1.0 m (d) 1.4 m 17. The amplitude of pressure variation at the middle of the air column is

d y2 dt d [– 0.8 A cos (bt – ax)] dt

= 0.8 Ab sin (bt – ax) (V 2)max = 0.8 Ab From the superposition principle, the maximum particle speed in the medium due to both the incisum of the individual maximum particle speeds, i.e. Vmax = (V1)max + (V2)max = Ab + 0.8 Ab = 1.8 Ab So the correct choice is (c). 15. Since A and b are positive constants, the minimum particle speed is Vmin = (V1)max – (V2)max = Ab – 0.8 Ab = 0.2 Ab The correct choice is (b). (b) 2 P0 P0 (c) 2 P0 (d) 2 18. The maximum and minimum pressures at the open end of the pipe respectively are (a) P0 + P0, P0 – P0 (b) P0 + P0, P0 (c) P0, P0 – P0 (d) P0, P0 19. The maximum and minimum pressures at the closed end of the pipe respectively are (a) P0 + P0, P0 – P0 (b) P0 + P0, P0 (c) P0, P0 – P0 (d) P0, P0 (a)

P0

SOLUTION 16. Figure 14.13 shows the longitudinal displacement y as a function of x for x lying between x = 0 and x = L, where L is the length of the pipe.

Fig. 14.13

The fundamental frequency of a closed pipe is given by v = 4L In a closed pipe, only odd harmonics are present, i.e. the frequency v1 the fundamental frequency, that of the second overtone v2 is 5 times the fundamental frequency and so on. Thus

Waves and Doppler’s Effect 14.43

5v 4L 5v 5 320 L= = = 1.0 m 4 2 4 400 The correct choice is (c). 17. Since the distance between two consecutive 2

nodes is

2

antinode is

=5 =

and that between a node and the next 4

, it follows from Fig. 14.7 that

5 2 2 4 4 We know that the pressure variation (excess pressure) is maximum at a node and minimum (equal to zero) at an antinode. Therefore, the pressure variation at a distance x from a node is given by 2 x P = P0 cos (1) L=

L 5 The centre C of the tube is at x = or x = 2 8 5 since L . The second node N2 is at x = 4 Questions 20 to 22 are based on the following passage Passage VI A wire of mass 9.8

10

–3

kg per metre passes over a

plane which makes an angle of 30° with the horizontal. Two masses M1 and M2 are tied at the two ends of the wire. Mass M1 rests on the inclined plane and mass M2 hangs freely vertically downwards. The whole system is in equilibrium. Now a transverse wave propagates along the wire with a speed of 100 ms–1.

. Therefore, the distance of C from N2 is x = 2 5 . Using this value of x in Eq. (1) we 8 2 8 have P at C =

P0 cos

=

P0 cos

2 8 4

=

P0

2 Thus the correct choice is (d). 18. At an antinode, the pressure variation is zero, i.e. P0 = 0. Hence, at an antinode Pmax = Pmin = P0 So the correct choice is (d). 19. At a node, the pressure variation is maximum equal to P0. Hence, at a node Pmax = P0 + P0 and Pmin = P0 – P0 Thus the correct choice is (a).

20. The tension T in the wire is (a) 0.98 N (b) (c) 98 N (d) 21. The value of mass M1 is (a) 2 kg (b) (c) 10 kg (d) 22. The value of mass M2 is (a) 5 kg (b) (c) 15 kg (d)

9.8 N 980 N 5 kg 20 kg 10 kg 20 kg

SOLUTION 20. Refer to Fig. 14.14. Let T be the tension in the string when the system is in equilibrium. position, the component M1g cos

Fig. 14.14

of weight M1 g

balances with the normal reaction N and the other component M1 g sin will balance with tension T in the string. Also weight M2 g of mass M2 will balance with tension T. Thus (1) M1 g sin = T and M2 g = T (2) Now, the speed of a transverse wave in a wire of mass m per unit length and stretched with a tension T is given by T v= m or T = v2 m (3)

14.44 Comprehensive Physics—JEE Advanced

Given m = 9.8 10–3 kg m–1 and v = 100 ms–1. Using these values in Eq. (3), we have T = (100)2 9.8 10–3 = 98 N 21. From Eq. (1), we get T 98 M1 = = g sin 9.8 sin 30

Thus the correct choice is (d). 22. From Eq. (2), we have M2 =

= 20 kg

Questions 23 to 25 are based on the following passage Passage VII A source of sound of frequency 90 Hz is moving towards a wall with a speed u = v/10, where v is the speed of sound in air. IIT, 1981 23. The beat frequency of the sound heard by an observer between the wall and the source is (a) 20 Hz (b) 10 Hz (c) 5 Hz (d) zero

T 98 = = 10 kg, which is choice (b). g 9.8

24. The beat frequency of the sound heard by an observer behind the source is 200 (a) Hz (b) 20 Hz 9 200 Hz (d) zero 11 25. The beat frequency of the sound heard by the observer moving with the source is (a) 11 Hz (b) 9.9 Hz (c) 10 Hz (d) 20 Hz (c)

SOLUTION 23. The observer hears two sounds–one coming directly from the approaching source and the other considered as coming from the mirror image of the source). The apparent frequency of the approaching source is v = 100 Hz v us v v/10 When the observer is between the wall and the =

v

v v v /10

= 90 Beat frequency =

–

= 100 – =

= 90

from the wall is also . Therefore, frequency of beats = – = 0. The observer will not hear any beats. So the correct choice is (d). 24. When the observer is behind the source, i.e. when the source is between the wall and the observer, the apparent frequency of the sound coming directly from the receding source is =

v v us

=

900 Hz 11

900 11

200 Hz. 11

Hence the correct choice is (c). 25. If the observer is moving with the source, the frequency of the direct sound is v = 100 Hz. The ap=

= 90

v us v us v + v /10 v v /10

= 110 Hz

Beat frequency = – = 110 – 90 = 20 Hz So the correct choice is (d).

Waves and Doppler’s Effect 14.45

IV Matrix Match Type 43. Column I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as f. Match each system with statements given in Column II describing the nature and wavelength of the standing waves. Column I Column II (a) Pipe closed at one end (p) Longitudinal waves

(b)

Pipe open at both ends

(q)

Transverse waves

(c)

Stretched wire clamped at both ends

(r)

f

=L

(d)

Stretched wire clampeed at both ends and at mid-point

(s)

f

= 2L

(t)

f

= 4L

IIT, 2011

SOLUTION (a) The closed end of a pipe is a node and the open end is an antinode. The distance between a node and the next antinode is /4. Hence f = 4L. In pipes, the standing waves are due to superposition of oppositely travelling sound (longitudinal) waves. (a) (p, t) (b) Here L = (b) /2,

f

2 (p, s)

f

(c)

f

= 2L (

distance between consecutive antinodes = /2)

= 2L. Also standing waves on a string are due to superposition of transverse waves. (q, s)

L = f (d) Here the mid point is a node. Hence 2 2 (d) (q, r)

f

=L

ANSWERS (a)

(p, t)

(b)

(p, s)

(c)

(q, s)

(d)

(q, r)

14.46 Comprehensive Physics—JEE Advanced

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation of Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 Only longitudinal mechanical waves can propagate in gases. Statement-2 Gases have only bulk modulus. 2. Statement-1 Two sound waves of equal intensity I produced beats. The maximum intensity of sound produced in beats is 4I. Statement-2 If two waves of amplitudes a1 and a2 superpose, the maximum amplitude of the resultant wave = a1 + a2. 3. Statement-1 A medium must possess elasticity in order to support wave motion. Statement-2 Restoring force does not exist in a medium which does not have elasticity. 4. Statement-1 Solids can support both longitudinal and transverse mechanical waves but only longitudinal mechanical waves can propagate in gases. Statement-2 Gases do not have shear modulus. 5. Statement-1 In standing sound waves, a displacement node is a pressure antinode and vice versa. Statement-2 In a standing wave, the restoring force is the maximum at a node and minimum at an antinode.

6. Statement-1 Our ears cannot distinguish two notes, one produced by a violin and other by a sitar, if they have exactly the same intensity and the same frequency. Statement-2 When a musical instrument is played, it produces a fundamental note which is accompanied by a number of overtones called harmonics. 7. Statement-1 Doppler’s effect does not occur in case of a supersonic source. Statement-2 A supersonic source produceds a shock wave. 8. Statement-1 If a source of sound moves always from a stationary observer, the apparent frequency of sound as heard by the observer is greater than the actual frequency. Statement-2 The cause of the apparent change in frequency is the change in the wavelength brought about by the motion of the source. 9. Statement-1 If an observer moves towards a stationary source of sound, the frequency of the sound as heard by him is greater than the actual frequency. Statement-2 The apparent increase in frequency is due to the fact that the observer intercepts more waves per second when the moves towards the source. 10. Statement-1 If a source of sound is in motion and the observer is stationary, the speed of sound relative to him remains unchanged. Statement-2 The apparent change in frequency is due to the change in the wavelength brought about by the motion of the source. 11. Statement-1 If the observer is in motion and the source of sound is stationary, the speed of sound relative to him is changed.

Waves and Doppler’s Effect 14.47

tionary observer with a certain velocity and (ii) observer approaching a stationary source of sound with the same velocity. Statement-2 The cause of the apparent change in the frequency is different in the two cases.

Statement-2 The wavelength of sound received by the observer does not change due to his motion. 12. Statement-1 The apparent frequency is not the same in the following two cases– (i) source approaching a sta-

SOLUTION 1. The correct choice is (a). Gases cannot withstand a shearing stress or longitudinal stress. Hence they do not have shear modulus and Young’s modulus; they have only bulk modulus. 2. The correct choice is (a). When two waves of amplitudes a1 and a2 superpose to produce beats, the resultant amplitude of the maximum of intensity is A = a1 + a2 Now, intensity (amplitude)2. Since the two waves have the same intensity, their amplitudes are equal, i.e. a1 = a2 = a. Thus A = 2a. Therefore, A2 = 4a2 or Imax = 4I. 3. The correct choice is (a). 4. The correct choice is (a). Gases cannot withstand a shearing stress. Hence gases do not have any shear modulus; they have only bulk modulus. Solids have Young’s modulus, bulk modulus and shear modulus. Therefore, solids can support both transverse and longitudinal waves. 5. The correct choice is (c). 6. The correct choice is (d). When a musical instrument is played, it produced a fundamental note

7.

8. 9. 10. 11. 12.

which is accompanied by a number of overtones called harmonics. The number of harmonics is not the same for all instruments. It is the number of harmonics which distinguishes the note produced by a sitar and that produced by a violin. The correct choice is (a). If the source of sound is moving at a speed greater than the speed of sound, then in a given time the source advances more than the wave. The resultant wave motion is a conical wave called a shock wave which produces a sudden and violent sound. The correct choice is (d). The correct choice is (a). The correct choice is (a). The correct choice is (a). The correct choice is (a). In case (i) the speed of sound relative to the observer remains unchanged; the change in frequency is due to a change in wavelength brought about by the motion of the source. In case (ii) the wavelength of sound remains unchanged; the change in frequency is due to a change in the speed of sound relative to the observer.

VI Integer Answer Type 1. An ambulance sounding a horn of frequency 256 Hz is moving towards a vertical wall with a velocity of 5 ms–1. If the speed of sound is 330 ms–1, how many beats per second will be heard by an observer standing a few metres behind the ambulance? IIT, 1981 2. A steel wire of length 1 m, mass 0.1 kg and uniform cross-sectional area 10–7 m2 ends. The temperature of the wire is decreased by 80/3°C. If transverse waves are set up in the wire,

3.

vibration in Hz. The Young’s modulus of steel = 2 1011 Nm–2 of steel = 1.2 10–5 per °C. IIT, 1981 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. IIT, 2009

14.48 Comprehensive Physics—JEE Advanced

SOLUTIONS 1. The observer will hear a sound of the source moving away from him and another sound after of these sounds are v 256 330 = = 252 Hz 1 = v u (330 5) v 256 330 = = 260 Hz 2 v u (330 5) No. of beats per second (beat frequency) = 260 – 252 = 8 2. Contraction L = L TL ; T = tension Young’s modulus Y = A L T=

1011

= 6.4 N

10–7

1 T ; 2L

1.2

10–5

80 3

= mass per unit length of wire

1 6.4 = 4 Hz 2 1 0.1

3. Mass per unit length of the string is 1.0 10

3

= 5 10–3 kg m–1 20 10 2 Speed of waves in the string is m=

T = m

v= Now

YA L = YA L

=2

= =

=

and

Frequency of fundamental mode is

v=

0.5 3

= 10 m s –1

5 10 v 10 = = = 0.1 m 100 = 10 cm

Separation between successive nodes =

2

= 5 cm

Thermal Expansion 15.1

15

Thermal Expansion

Chapter

REVIEW OF BASIC CONCEPTS 15.1

THERMAL EXPANSION

15.2 (i)

mass or = volume Since mass remains constant, Thus = 0 0 Density =

If the temperature of a body is increased, its length, surface area and volume all increase. (i) If the body is in the form of a rod, the increase in its length when the temperature is increased by T is proportional to ( ) the original length and (b) increase in temperature T, i.e. T = T Thus

=

(ii)

=

where =

0

(1 +

Strain Now

=

(iv)

0

(1 +

: i.e.

:

0 0

=

=

Young’s modulus (Y) = Thermal stress =

T)

and is (°C)–1 or K–1. , =1 : 2 : 3 =2 = 3 .

The SI unit of ,

1

(1

0

T)

If a rod is held between two rigid supports and its temperature is increased or decreased, the rigid supports prevent the rod from expanding or contracting. As a result, a stress (called thermal stress) is developed in the rod. The change in length of the rod is = T

T)

(iii)

=

=

= constant.

Thus density of a substance decreases with increase in temperature.

(ii) Similarly

0 0

0

=

where material of the body. Thus T) = 0 (1 +

APPLICATIONS

(iii)

T stress strain =Y

T

A linear metallic scale expands when heated and contracts when cooled. A reading of 1 unit of a heated scale is equivalent to a an actual length of 1 unit (1 + T) where T is the

15.2 Comprehensive Physics—JEE Advanced

rise in temperature. If the reading of the heated scale is units, the actual length = (1 + T) units. If the temperature is decreased by T, the actual reading = (1 – T) units. (iv) Consider two metal rods of lengths 1 and 2. Let 1 and 2 be their lengths when their temperature is increased by T, then and

1

=

1(1

+

1

T))

2

=

2(1

+

2

T)

where 1 and 2 linear expansion. The difference of their lengths is 2

–

1

=

2(1

+

T) –

2

1(1

+

1

15.1 At 20°C, a brass rod has a length 50.0 cm. It is joined to a steel rod of the same length and the same diameter at the same temperature. Find the change in the length of the composite rod when it is heated to 220°C. For brass = 2 10–5 K–1 and for steel = 1 10–5 K–1. SOLUTION For brass: (

T)

Their difference in lengths will remain constant if 2

i.e. if

–

1

=

2

–

1 1

=

2 2

1

)b =

b

(

(v)

= (2

10 )

=2

–3

10

) =

The time period of a simple pendulum is given by

=(

(1

=

= 1+

1 2

T(

T

stress =

1 2

T

T

1011) (2

(5

10–5)

10–6) [30 – (– 20)]

= 500 N 15.3 (60

60 24)

1 86400 s T 2 If the room temperature falls by T, then 1 86400 s T Time gained in one day = 2 =

) = 3 mm

T

= (1

=

Time lost in one day =

(220 – 20)

F

F=

T is small)

1 T T 2 T 1 = T T 2 This gives the time lost per second. T

10–5)

15.2 A metal wire of cross-sectional area 5 10–6 m2 is held taut at 30°C between two rigid supports with negligible tension in it. Find the tension developed in the wire if it is cooled to –20°C. Given of metal = 2 10–5 K–1 and Y = 1 1011 N m–2.

Also

T)1/2

= (1 +

0.5

m

)b + (

SOLUTION Thermal stress = Y

)

(220 – 20)

= 1 mm

T= 2

T = T

0.5

T

= (1

T = 2

–5

= 2 mm

For steel:

If the room temperature rises by T, the length of the pendulum increases. Hence the time period increases which implies that a metallic pendulum clock slows down. If is the length of heated clock, then its time period becomes

T

b

5

10–4 K–1. If its temperature is increased by 30°C,

SOLUTION = (1 +

T) =

(1 0

0

T) –

0

=–

0

T

Thermal Expansion 15.3

0

or 0

=–

T 10–4

=–5

30 = – 1.5

10–2.

The negative sign indicates that the density decreases with increase in temperature. Percentage change in density = (1.5 10–2) 100 = 1.5% 15.4 A block of mass 248 g and volume 205 cm3 liquid contained in a vessel. The density of the liquid at 0°C is 1.248 g cm–3. It is found that the block just sinks in the liquid if the temperature is raised to 50°C. expansion of the block is negligible compared to that of the liquid. SOLUTION The block will just sink in the liquid if its density becomes equal to the density of the liquid at 50°C. Density of block = 0

248 g 205 cm

= (1 +

= 1.210 g cm–3

3

T)

1.248 = 1.210 (1 + = 6.28

10

–4

50) (°C)–1 or K–1

The quantity T Hence the value of must be rounded off to two sig-

15.5 same material. At 25°C, the diameter of the shaft is 8.00 cm and the diameter of the hole is 7.99 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip into the shaft. Give for steel = 1 10–5 K–1. SOLUTION Decrease in temperature =

0.01cm 1 10

T2 – T1 = –125

SOLUTION T = 40 – 15 = 25°C Actual distance = measured distance (1 + T) –5 25) = 3152 (1 + 1.2 10 = 3152.9 cm

3153 cm

15.7 A metal pendulum clock gives correct time at a temperature of 20°C. How much time does it lose in a day if the room temperature is 35°C. Given of metal = 1.25 10–5 K–1. SOLUTION 1 T 2 1 (24 60 60)s Time lost in a day = T 2 1 = 1.2 10 5 15 2 (24 60 60) = 8.1 s

Time lost per second =

15.8

10–4 K–1

= 6.3

T=

15.6 A steel tape gives correct readings at a temperature of 15°C. On a day when the temperature is 40°C, this tape measures the distance between two points as 3152 cm. What is the actual distance between the two points? Given for steel = 1.2 10–5 K–1.

5

8.00 cm

= – 125°C

T2 = –125 + T1 = –125 + 25 = –100°C

bottle and water are heated to 80°C if (a) the expansion of bottle is neglected (b) the expansion of bottle is not neglected of glass of water = 6.0 10–4(°C)–1, –5 –1 = 0.6 10 (°C) . SOLUTION Volume of bottle = volume of water = 1 litre = 1000 cm3

(a) If the expansion of bottle is neglected, the = 1000 (1 + T) – 1000 = 1000 (1 + 6.0 10–4 60) – 1000 = 1036 – 1000 = 36 cm3

15.4 Comprehensive Physics—JEE Advanced

(b)

= 3 = 3 0.6 10–5 = 1.8 10–5(°C)–1 If the expansion of bottle is not neglected, the = 1000

(1 +

T) – 1000 (1 +

= 1000 = 1000

( (6.0

= 34.9 cm

T)

–

)

T –4

10

10–5)

– 1.8

60

3

I Multiple Choice Questions with Only One Choice Correct 1. When a solid metallic sphere is heated, the largest percentage increase occurs in its (a) diameter (b) surface area (c) volume (d) density 2. A steel scale measures the length of a copper rod as cm when both are at 20°C, the calibration linear expansion for steel and copper are and respectively, what would be the scale reading (in cm) when both are at 21°C? 1 (b) (a) 1 (c)

(d)

3. Two uniform brass rods and B of lengths and 2 and radii 2 and respectively are heated to the same temperature. The ratio of the increase in the length of to that of B is (a) 1 : 1 (b) 1 : 2 (c) 1 : 4 (d) 2 : 1 4. of thermal expansion 1 and 2 and Young’s modulii Y1 and Y2 walls. The rods are heated to the same temperature. If there is no bending of the rods, the thermal stresses developed in them are equal provided Y (a) 1 = Y2 (c)

Y1 = Y2

1 2 1 2

Y (b) 1 = Y2 (d)

Y1 = Y2

2 1 2 1

IIT, 1989 5.

of volume expansion of the metal and mercury are 1 and 2 respectively. If their temperature is increased by T, the fraction of the volume of metal submerged in mercury changes by a factor

(a) (c)

1 1 1 1

T T

(b)

T T

(d)

2 1 2 1

1 1

2 1

T T

2 1

IIT, 1991 6. A thin copper wire of length increases its length by 1% when heated from temperature T1 to T2. What is the percentage change in area when a thin copper plate having dimensions 2 is heated from T1 to T2? (a) 1% (b) 2% (c) 3% (d) 4% 7. direction ( -axis) is 2.0 10–6 K–1 and that in the other two perpendicular ( -and z-axes) directions is 1.6 10–6 K–1 expansion of the crystal? (b) 1.8 10–6 K–1 (a) 1.6 10–6 K–1 (c) 2.0 10–6 K–1 (d) 5.2 10–6 K–1 8. A metal ball immersed in alcohol weighs W1 at 0°C and W2 expansion of metal is less than that of alcohol. If the density of the metal is large compared to that of alcohol, then (a) W1 > W2 (b) W1 = W2 (c) W1 < W2

(d) W2 =

W1 2

IIT, 1980 9. A rod of length 20 cm made of a metal expands by 0.075 cm when its temperature is raised from 0°C to 100°C. Another rod of a different metal B having the same length expands by 0.045 cm for the same change in temperature. A third rod of the same length is composed of two parts, one of metal and the other of metal B. This rod expands by 0.060 cm for the same change in temperature. The portion made of metal has length

Thermal Expansion 15.5

(a) 20 cm (b) 10 cm (c) 15 cm (d) 18 cm 10. A metallic circular disc having a circular hole at its centre rotates about an axis passing through its centre and perpendicular to its plane. When the disc is heated, its (a) moment of inertia increases but angular speed decreases (b) moment of inertia decreases but angular speed increases (c) moment of inertia and angular speed both increase (d) moment of inertia and angular speed both decrease. 11. A uniform metal rod of length and mass M is rotating with angular speed about an axis passing through one of the ends and perpendicular to the rod. If the temperature increases by ºC, then the change in its angular speed is proportional to 2

(d)

16.

(c)

1

2

1

2

2

1 2

(d)

1

1 2 2

1

+ 1

2)

(b)

1 ( 2

(d) (

2

1

+

2)

2)

1

17. Three rods of the same length are arranged to form an equilateral triangle. Two rods are made of the 1

and the third rod which forms the base of the trian2. The altitude of the triangle will remain the same at all temperatures if the 1/ 2 is nearly 1 (a) 1 (b) 2 1 (c) (d) 4 4

12. A steel metre scale is to be ruled so that the millimetre intervals are accurate to about 5 × 10–5 m at a certain temperature. The maximum temperature

(c) (d) 2 14. Two spheres made of the same material have the same diameter. One sphere is hollow and the other is solid. If they are heated through the same range of temperature, (a) the hollow sphere will expand more than the solid sphere (b) the solid sphere will expand more than the hollow sphere (c) both spheres will expand equally (d) the hollow sphere will not expand at all. 15. Two rods of lengths 1 and 2 are welded together to make a composite rod of length ( 1 + 2). If the

1

-

neous rod changes linearly from 1 to 2 from one end to the other end of the rod. The effective coef(a) (

1

of linear expansion of steel = 10 × 10–6 K–1) (a) 2 ºC (b) 5 ºC (c) 7 ºC (d) 10 ºC 13. When the temperature of a rod increases from to + , the moment of inertia of the rod increases from I to I + I I sion of the rod is , the ratio is I 2 (b) (a)

1

(c)

(b)

(a) (c)

and 2 ear expansion of the composite rod will be 1 (a) ( 1 + 2) (b) 1 2 2

18.

19.

80% of its volume immersed in a liquid at 0ºC. When the temperature of the liquid is raised to 62.5ºC, the cal expansion of liquid is (b) 2 × 10 3 K 1 (a) 1 × 10 3 K 1 (c) 3 × 10 3 K 1 (d) 4 × 10 3 K 1 fraction 1 of its volume is submerged, while at the temperature 60°C, a fraction 2 is seen to be subiron is Fe and that of mercury is 1/ 2 can be expressed as (a) (c)

1 60 1 60 1 60 1 60

Fe

(b)

Hg Fe Hg

(d)

Hg,

then the ratio

1 60 1 60

Hg

1 60

Hg

1 60

Fe

Fe

IIT, 2001 20. Two rods, one made of aluminium and the other made of steel, having initial lengths 1 and 2 respectively are connected together to form a single rod of length ( 1 + 2 expansion for aluminium and steel are 1 and 2

15.6 Comprehensive Physics—JEE Advanced

respectively. If the length of each rod increases by the same amount when their temperature is raised by °C, then the ratio 1/( 1 + 2) (b) 2/ 1 (a) 1/ 2 (c) 2/( 1 + 2) (d) 1/( 1 + 2) IIT, 2003 21.

22.

18 10–5/°C. A thermometer has a bulb of volume 10–6 m3 and the cross-sectional area of the stem is 0.002 cm2 mercury when the temperature is 0°C. When the temperature rises to 100°C, the length of the mercury column in the stem will be (a) 9 cm (b) 18 cm (c) 9 mm (d) 18 mm when determined using two different vessels B are 1 and 2 linear expansion of vessel is linear expansion of vessel B is 1

(a) 1

(b)

1

2 2

2

3

(d)

2

(c) increase by 2 (d) decrease by 2 26. A vertical glass tube, closed at the bottom, contains a mercury column of length 0 at 0°C. If is the the of the mercury column when the temperature rises to °C is (assuming that not more than 100°C)

(b)

(a)

=

0[1

+ ( – 3 )]

(b)

=

0[1

+(

(c)

=

0[1

+ ( + 2 )]

is 20 10–6 per °C. The barometer reads 75 cm at 40°C. The atmospheric pressure at 40°C is (a) 75 cm of Hg (b) 74.94 cm of Hg (c) 75.06 cm of Hg (d) none of these 28.

. When the temperature is raised by T, the depth upto which the cube is submerged in the liquid remains unchanged. If the expansion of the beaker is ignored, the relation between and is (a)

(c)

1

(d)

1

24. A steel rod and a copper rod have lengths and respectively at a certain temperature. It is found that the difference between their lengths remains constants at all temperatures. If and are their / (a) (c)

is given by 1

(b)

1

(c)

(b)

=

3 =3

(d)

=

2 =2

IIT, 2004 29. A copper wire of length and cross-sectional area is held at the ends by two rigid supports. At temperature T the wire is just taut with negligible tension. If the temperature reduces to (T – T), the speed of transverse waves in the wire is (here Y is Young’s modulus, density and linear expansion of copper) Y T T (b) (a)

(d)

25. A uniform metallic circular disc, mounted on frictionless bearings, is rotating at an angular frequency about an axis passing through its centre and

+ 3 )]

(d) = 0[1 + ( – 2 ) ] 27. The brass scale of a barometer gives correct reading

1

and mercury respectively, the volume of mercury (a)

(b) increase by

and

2 + 3 23. contains some mercury. It is found that at different temperatures, the volume of and

(c)

1

2

expansion of the metal is . If the temperature of the disc is increased by , the angular frequency of rotation of the disc will (a) remain unchanged

(c)

T

(d)

T IIT, 1979

Thermal Expansion 15.7

30. A metal ball immersed in alcohol weighs W1 at 0°C and W2 expansion of metal is less than that of alcohol. If the density of the metal is large compared to that of alcohol, then

(a) W1 > W2

(b) W1 = W2

(c) W1 < W2

(d) W2 =

W1 2 IIT, 1980

ANSWERS

1. 7. 13. 19. 25.

(c) (d) (d) (a) (d)

2. 8. 14. 20. 26.

(c) (b) (c) (c) (d)

3. 9. 15. 21. 27.

(b) (b) (c) (a) (c)

4. 10. 16. 22. 28.

(d) (a) (b) (d) (b)

5. 11. 17. 23. 29.

(a) (b) (c) (a) (a)

Y1

1

6. 12. 18. 24. 30.

(b) (b) (d) (d) (b)

SOLUTIONS 1. On heating, the diameter, surface area and volume of the sphere will all increase. Since the mass remains unchanged, the density decreases. The percentage increase in the volume is the largest beexpansion. Hence the correct choice is (c). 2. The length of 1 cm division of the steel scale at 21°C is (1 cm) [1 + (21 – 20)] = (1 + ) cm Length of copper rod at 21°C will be ( cm) [1 + (21 – 20)] = (1 + ) cm 1 1

Scale reading at 21°C =

=

=

=2

.

(

=2)

1 2

and

stress = Y

Y1 = Y2

2 1

1

1 100

2

=2

=

2

1

1

2 100

=

102 100

Thus the area increases by 2%, which is choice (b).

Hence the correct choice is (b). 4. If a rod of length expansion is heated to a temperature , the increase in the length is given by = Strain =

or

1 2 100 Now 2 2 = area of the plate at temperature T2 and 2 2 = area of the plate at temperature T1. Therefore, 1 2 = 1 100 2

3. The increase in length due to heating is independent of the radius of the rod. The increase in the length of rod is = =

2

Hence the correct choice is (d). 5. The correct choice is (a). It follows from the fact that the volumes of metal and mercury increase to 0 (1 + 1 T) and 0 (1 + 2 T) respectively; 0 being the initial volume. 6. Length of wire at temperature T2 is

Hence the correct choice is (a).

and of rod B is

= Y2

= strain = Y

Since the value of is the same for the two rods, the stresses in them will be equal if

7. =

+

+ –6

z

=

+2 (

z)

= –6

= 2.0 10 + 2 1.6 10 = 5.2 10–6 K–1 Hence the correct choice is (d). 8. Let 0 be the volume of the metal at 0°C and its volume at °C. At temperature the upthrust is U = where is the density of alcohol at temperature . Now

15.8 Comprehensive Physics—JEE Advanced

= 0 (1 + ) where alcohol and 0 is the volume of alcohol displaced at temperature = 0°C. Now the density of alcohol at temperature is 0

=

where Hence

0

1 is the density of alcohol at U =

0

(1 +

)

=

0

0

= U0

1

= 0°C.

0

where is the length of metal and ( – ) that of metal B in rod C. Now ( )1 = 0.075 cm, ( )2 = 0.045 cm and ( )3 = 0.060 cm. Notice that 1 [( )1 + ( )2]. This is possible only if ( )3 = 2 20 cm = = 10 cm. Hence the correct choice 2 2 is (b). 10. Due to thermal expansion, the diameter of the disc as well as that of the hole will increase. Therefore, the moment of inertia will increase resulting in a decrease in the angular speed. Hence the correct choice is (a). 11. At ºC, the length of the rod becomes = (1 + ), where From the law of conservation of angular momentum, we have I =I

or

1 3

2

=

1 3 2

1

(1 )2 Now, for a given value of , (1 + constant, say . = or

=

–1

–

= ( – 1)

)

. Hence the corect choice is (b).

12. The maximum temperature variation T allowed is given by = T, which gives 5

5 10

= 5ºC 1 10 10 6 Hence the correct choice is (b). 13. The moment of inertia of a rod of mass M and length is given by 2 (1) I= 1 if the axis of rotation is through the 12 1 centre and = if the axis of rotation is through 3 the end of the rod. Partially differentiating (1), we have I= 2

where

=

Now

=

. Therefore,

I= 2

× 2

=2(

)×

= 2I

I =2 , which is choice (d). I 14. The correct choice is (c). 15. Total length of the composite rod at 0°C is 0 = 1 + 2. When the composite rod is heated to °C, its length at °C will be = 1 (1 + 1 ) + 2 (1 + 2 ) = ( 1 + 2) + 1 1 + 2 2 , or or

=

0

+(

1 1

+

2 2)

Effective coefficient of expansion of the composite rod is 0

=

1 1

2

1

2

0

2

,

which is choice (c). 16. Consider a small element of length at a distance from one end of the rod. Let be the length

2

=

i.e. (

–

T=

where U0 is the upthrust at 0°C. Since the upthrust is independent of temperature, W1 = W2. Hence the correct choice is (b). 9. Here ( )1 = 1 ( )2 = 2 ( )3 = +( – ) 2 1

or

or

)–2 is

of linear expansion by unit length of the rod is at the ( 2 – 1)/ . Therefore, the value of element located at is =

1

+

2

1

Increase in the length of the element = , where T is the rise in temperature. Therefore, the increase in the length of rod is

Thermal Expansion 15.9

=

18. Weight of cylinder = weight of the liquid displaced, i.e. (0.8 ) b = 0 5 b b or (1) 0 = 0 .8 4 Here 0 is the density of the liquid at 0°C. For the block to sink in the liquid at °C, the density of the liquid must change from 0 to at °C. Now

T 0

=

T

2

1

1

0

=

T

=

T

where

eff

2

2

1 ( 2

=

2

1

2

1 1

=

2

1

2

2

+

2).

0 (2) 1 For the block to sink = b. Using (1) and (2), we have 5 b /4 b = 1

=

1

T= 1

0

T

eff

Hence the correct

or

choice is (b). 17. Let be the length of each rod at 0°C and altitude at 0°C. Then (Fig. 15.1),

0

be the

1+

=

5 4

Given = 62.5°C. Using this value, we get = 4 10–3 (°C)–1 or K–1, which is choice (d). 19. Let 0 be the total volume of the block of iron at 0°C and its total volume at 60°C. Then =

(1 + 60

0

Fe)

(1)

Let v0 be the volume of the block submerged in mercury at 0°C and v the volume submerged at 60°C. Then (2) v = v0 (1 + 60 Hg) Fig. 15.1

0

2

=

Dividing (1) by (2), we have

2 1/ 2

4 When the rods are heated to a temperature , the altitude becomes 2

=

1

=

2

2

1

Since 1 and can write (1 + 1 + 2 2 . Thus

2

1

4

2

2

1/ 2

1

1 2

2 1

4

1 2

2 1/ 2

4

2

2

(2)

2

1

1 2

1

=

1

or

4

1/ 2

1 2

2

1 , which is choice (c). 4

and

v

=

(3)

Hg

2.

Using these in (3), we

2

or

1 2

+ 1=

1 2

1 2

=

Fe

20. The length of each rod increases by the same amount if 1 1 = 2 2 or

1/ 2

Equating (1) and (2), we get 2

v0

Given

0

are much less than unity, we )2 1 + 2 1 and (1 + 2 )2

2

1 60 1 60

/v / 0 v0

(1)

1

1

2

=

1

1

1

2 2

1

or

+1

2

2

= 2

1

, which is choice (c). 2

15.10 Comprehensive Physics—JEE Advanced

21. Increase in volume of mercury when the temperature increases by 100°C is v= T = (18 10–5) (10–6) 100 10–9 m3 v or =

= 18 Now v =

9

(18 10

=

m3 ) 4

= 0.09 m = 9 cm

2

i.e. For vessel ; given)

v

=

:

+3 =

For vessel B:

1

=

2

+3 +3

v 1

=

+3

=

which gives

2

+3 2

=

1

(

v

+3

(

= 3 v) 1

23. Let

3

2

=

is the same.

+ , which is choice (d).

J=I

(3)

Since no external torque acts, J remains constant. Partially differentiating (3), we have +

I

I=0

, which is choice (a).

= = The difference between their lengths will remain constant at all temperatures if = , i.e. if = =

Hence the correct choice is (d). 25. The moment of inertia of the disc about the given axis of rotation is 1 MR2 (1) I= 2 where M is the mass of the disc and R its radius.

I

=–

(4)

I Using (1) and (2) in (4) we get = –2

The negative sign indicates that the angular frequency decreases due to increase in temperature. Hence the correct choice is (d). 26. If 0 is the volume of mercury at 0°C and 0 the cross-sectional area of the tube at 0°C, then the length of the mercury column at 0°C is 0

24. When the rods are heated by °C, the increase in length of steel rod and copper rod are

or

(2)

Now, the angular momentum of the disc is given by

= or

M = constant)

. Therefore,

or

2 1

R=R

=

2

Since the liquid is the same, Hence 1

But

(

I = MR2 -

22.

+

or

1 M 2R R 2 I = MR R I=

(0.002 10 m ) Hence the correct choice is (a).

=

If the disc is heated, it expands. Hence R increases. The resulting increase in I is obtained by partially differentiating (1).

=

0

(1)

0

The cross-sectional area at °C is given by =

0

(1 +

)

where of glass. Since = 2 , we have = 0 (1 + 2 ) If is the volume of mercury at °C, the length of the mercury column at °C is 0 1 = = 0 1 2 Using (1) we have = 0 (1 + ) (1 + 2 )–1 Since is not too high (~ 100°C) and and are of the order of 10–5, and will be very small compared to unity. Hence we can expand (1 + 2 )–1 binomially and retain terms upto order . Thus (1 + 2 or

)–1 = (1 – 2 ) = 0 (1 + ) (1 – 2 =

0

[1 + ( –2 ) ]

)

Thermal Expansion 15.11 2 where we have neglected term of which is negligibly small. Thus the correct choice is (d). 27. Due to rise in temperature, the brass scale expands. It will give lower readings because the graduations on the scale will be farther apart. If H is the barometric height at 0°C, the error in the reading of the scale at 40°C is H= H = 75 20 10–6 40 = 0.06 cm Atmospheric pressure at 27°C = H + H = 75 + 0.06 = 75.06 cm of Hg. Thus the correct choice is (c). 28. Let the initial temperature be T and let M be the mass of the cube. Let 0, 0 and 0 respectively be the base area of the cube, the density of the material of the cube and the depth upto which it is submerged in the liquid, the upthrust = 0 0 0 . From

or

= M=

0 0 0

Using (2) and (3) we have 0 = 0 0 (1 T) which gives 2 = . Thus the correct choice is (b). 29. = T. Since the wire is free to contract, the tension F must increase to produce the same change in length. Now / Y= /

0

Also

=

(1

0

or

where Now

. From

= M=

or

=

0 0

(

=

0;

=

F

;

= mass per unit length = / = T

=

=

Y

is the density of alcohol at temperature . =

0

(1 +

)

0

U =

0

(1 +

)

=

0

0

= U0

1

0

where U0 is the upthrust at 0°C. Since the upthrust is independent of temperature, W1 = W2.

given)

II Multiple Choice Questions with One or More Choices Correct 1. Choose the wrong statements from the following. A metallic circular disc having a circular hole at its centre rotates about an axis passing through its centre and perpendicular to its plane. When the disc is heated.

T

1 where 0 is the density of alcohol at = 0°C. Hence

From (1) and (4), we get 0 0 0

F

=

(4)

=

v=

where alcohol and 0 is the volume of alcohol displaced at temperature = 0°C. Now the density of alcohol at temperature is

(3)

The upthrust at temperature (T + DT) =

T

30. Let 0 be the volume of the metal at 0°C and its volume at °C. At temperature the upthrust is U =

(2)

T)

=

Speed of transverse waves in the wire is

When the temperature is raised to (T + T), let , and be the base area, density and depth at this (area) expansion is = 2 . Hence T) = 0(1 + T) = 0 (1 + 2

T)

F=

(1)

0 0 0

(1 + 2

(a) (b) (c) (d)

its its its its

speed will decrease diameter will decrease moment of inertia will increase speed will increase

15.12 Comprehensive Physics—JEE Advanced

2. Choose the wrong statements from the following. Two spheres made of the same material have the same diameter. One sphere is hollow and the other is solid. If they are heated through the same range of temperature, (a) the hollow sphere will expand more than the solid sphere (b) the solid sphere will expand more than the hollow sphere (c) both spheres will expand equally (d) the hollow sphere will not expand at all. 3. Two rods and B of different metals have lengths and B at a certain temperature. It is observed that rod is 5 cm longer than rod B at all temperatures. If = 1.0 10–5 per °C and B = 1.5 10–5 per °C, then (a) = 10 cm (b) = 15 cm (c) B = 10 cm (d) B = 5 cm 4. A uniform metallic circular disc of mass M and radius R, mounted on frictionless bearings, is rotating an angular frequency about an axis passing through its centre and perpendicular to its plane. The temperature of the disc is then increased by . If . (a) the moment of inertia increases by MR2 (b) the moment of inertia remains unchanged.

(c) the angular frequency increases by 2 (d) the angular frequency decreases by 2

. .

5. A clock with a metallic pendulum gains 6 seconds each day when the temperature is 20°C and loses 12 seconds when the temperature is 40°C. Then (a) the clock will keep correct time at tempera80 °C ture 3 (b) the clock will keep correct time at tempera100 °C. ture 3 is 1.2

10–5 per °C.

is 2.1 10–5 per °C. 6. A bimetallic strip is formed out of two identical strips one of copper and the other of brass. The are C and B. When the temperature of the strip is increased by T, it bends to form an arc of radius of curvature R. Then R is (a) proportional to T (b) inversely proportional to T (c) proportional to ( C – B) (d) inversely proportional to ( C – B). IIT, 1999

ANSWERS AND SOLUTIONS 1. Due to thermal expansion, the diameter of the disc as well as that of the hole will increase. Therefore, the moment of inertia will increase resulting in an increase in the angular speed. Hence the correct choices are (a) and (c). 2. Statements (a), (b) and (d) are wrong. 3. When the rods are heated by °C, the increase in the length of the rods is = and ( – =

= B B B) will remain the same for all , if B, i.e. = B B B

=

=

B

= 1.5

B. Also

B

(

–

1.5 10 1.0 10 B)

5

= 1.5

= 5 cm. Thus

1.5 B – B = 5 cm B = 10 cm, and The correct choices are (b) and (c).

= 15 cm.

4. The moment of inertia of the disc about the given axis of rotation is 1 MR2 (1) I = 2 If the disc is heated, it expands. Hence R increases. The resulting increase in I is obtained by partially differentiating (1). 1 M 2R R ( M = constant) I= 2 or I = MR R But

R=R I = MR2

. Therefore, , which is choice (a)

(2)

Now, the angular momentum of the disc is given by =I (3) Since no external torque acts, remains constant. Partially differentiating (3), we have I

+

I= 0

Thermal Expansion 15.13

or

I

=–

At 20°C, the gain in time is

(4)

1 ( – 20) 2 At 40°C, the loss in time is

I Using (1) and (2) in (4) we get = –2 The negative sign indicates that the angular frequency decreases due to increase in temperature. Thus the correct choices are (a) and (d).

6 =

1 (40 – ) 2 Dividing (4) by (3), we have 12 =

5. Time taken for one oscillation of the pendulum is

or

2

T =4

which gives =

(1)

we have

Partially differentiating, we get 2

2T T = 4

(3)

86400

(4)

12 40 = 6 20

T=2 2

86400

1 2 which gives

80 °C. Using this value in Eq. (3), 3 80 3

6=

(2)

Dividing (2) by (1), we get T 1 = = = ( = ) T 2 2 2 where is the change in temperature. Now, one day = 24 hours = 86400 s. Therefore, gain or loss of time in one day is 1 86400 seconds T= 2 Let be the temperature at which the clock keeps correct time.

= 2.1

20

86400

10–5 per °C.

Thus the correct choices are (a) and (d). 6. C = 0 (1 + C T) and B = 0 (1 + B T) C – B = 0 ( C – B) T. Initially the bimetallic strip is straight, i.e. R is R decreases. The amount of bending is proportional to ( C – B) and T. Greater the bending, the smaller is the value of R. Hence the correct choices are (b) and (d).

III Multiple Choice Questions Based On Passage Question 1 to 3 are based on the following passage Passage I 20,000 J of heat energy is supplied to a metal block of mass 500 g at atmospheric pressure. The initial tempera400 J kg–1 °C–1 of volume expansion of metal = 8 spheric pressure = 105 Pa.

10–5 °C–1 and atmoIIT, 2005

1.

(a) 120°C (b) 130°C (c) 140°C (d) 150°C 2. Work done by the block on the surroundings is (a) 0.05 J (c) 1.0 J

(b) 0.1 J (d) 10 J

3. The change in internal energy is (a) zero (b) equal to 20,000 J (c) slightly greater than zero (d) slightly less than 20,000 J.

SOLUTION 1.

Q= T=

Final temperature = 100 + 30 = 130°C, which is choice (b).

T. Therefore, Q

=

20, 000 J (0.5 kg ) (400 J kg

1

C 1)

= 100°C

2. Density of metal ( ) = 8000 kg m–3. Volume of the block is

15.14 Comprehensive Physics—JEE Advanced

=

=

0.5 kg 8000 kg m

3

=

Increase in volume = = (8

1 10 16

10–5 )

1 16 =

3

=

10–3 m3

1 2

10–6 m3

T

1 10 2

= (105)

Work done W = P

6

= 0.05 J, which is choice (a). 3. Change in internal energy U = Q – W = 20,000 – 0.05 = 19999.95 J Thus the correct choice is (d).

100

(a) 2 : 1 (c) 3 : 1

Questions 4 to 6 are based on the following passage Passage II A rod of metal X of length 50.0 cm elongates by 0.10 cm when it is heated from 0°C to 100°C. Another rod of metal Y of length 80.0 cm elongates by 0.08 cm for the same rise in temperature. A third rod of length 50 cm, made by welding pieces of rods X and Y placed end to end, elongates by 0.03 cm when its temperature is raised from 0°C to 50°C. 4. X and of metal Y are in the ratio of

(b) 3 : 2 (d) 4 : 3

5. The length of the rod of metal X in the composite piece is (a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm 6. The length of the rod of metal Y in the composite piece is (a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm

SOLUTION = (1 + Therefore,

0.10 50.0 100 = 2.0 10– 6 per °C. For rod Y, we get = 1.0 10–6 per °C. So the correct choice is (a). 5. In the composite rod, + = 50.0 cm. When this rod is heated by 50°C, let the new lengths be and . Given + = 50.0 + 0.03 = 50.03 cm. Here

4. For rod X,

=

=

50) and

50.03 =

+

= 50 + (

=

+(

(1 + +

+

50). ) 50

) 50

Substituting the values of and and noting that + = 50 cm, we get = 10 cm, which is choice (a). 6. = 50 – = 50 – 10 = 40 cm, which is choice (d).

IV Integer Answer Type 1. A composite rod is made by joining a copper rod, end to end, with a second rod of a 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. = 1.7 10–5 per °C and that of the second rod is = 10–5 per °C. Find the value of n. IIT, 1979

SOLUTION 1. Length of the second rod at 25°C = 70 cm. Length of copper rod at 125°C = 30 (1 + 1.7 10–5 100) = 30.051 cm Increase in the length of copper rod = 0.051 cm

Increase in the length of second rod = 70 100 = 7000 cm Total increase in length = 0.051 cm + 7000 cm = 0.191 cm (given) which gives = 2 10–5 per °C. Thus the value of = 2.

16

Measurement of Heat

Chapter

REVIEW OF BASIC CONCEPTS 16.1

HEAT ENERGY

Heat is a form of energy. It is, therefore, measured in energy units. The SI unit of heat is joule (J). Another unit commonly used is the calorie. A calorie is the amount of heat required to raise the temperature of 1 g of water through 1°C. Experiments have shown that 4.18 J of mechanical work produce one calorie of heat. Thus 1 calorie = 4.18 joules or The ratio

1 cal = 4.18 J J=

work done = 4.18 J per cal heat produced

Specific Heat Capacity If m = 1 unit and T = 1 unit, then s = Q heat of a substance is the amount of heat required to raise the temperature of a unit mass of the substance through a unit degree. A commonly used unit of s is cal g–1 °C–1. In the SI system s is expressed in J kg–1 K–1. The two units are related as 1 cal g–1 °C–1 = 4.18 Jg–1 °C–1 ( 1 cal = 4.18 J) 1cal g–1 °C–1 = 4180 J kg–1 °C–1

or

Since the size of a degree on the celsius scale is equal to that on the kelvin scale, a temperature difference of, say, 1 °C is equal to a temperature difference of 1 K. Thus 1 cal g–1 °C–1 = 4180 J kg–1 K–1

is called the mechanical equivalent of heat. capacity of water = 1 cal g–1 °C–1 or 4180 J kg–1 K–1.

16.2

CALORIMETRY

If two substances having different temperatures are

energy will continue till the temperatures are equalized. The common temperature at thermal equilibrium is called the equilibrium temperature. If no heat energy is allowed to escape to the surroundings, the amount of heat energy gained by the initially colder body is equal to the amount of heat energy lost by the initially hotter body, i.e. Heat gained by one body = heat lost by the other body. This is the basic principle of calorimetry. The heat energy Q needed to raise the temperature through T of a mass m capacity s is given by Q = ms T

Molar Specific Heat C of a substance is the amount of heat energy required to raise the temperature of 1 mole of the substance through 1 K. It is expressed in J mol–1 K–1. s C) are related as C s = m where m is the number of kilograms per mole in the substance. ) is v

p

16.2 Comprehensive Physics—JEE Advanced

SOLUTION

Cp – Cv = R where R is the universal gas constant and its value is R = 8.315 J mol–1 K–1

16.3

LATENT HEAT

The heat energy supplied to a substance to change from solid to liquid state or from liquid to gaseous state is not registered by a thermometer as the heat energy is used up in bringing about a change of state. Hence it is called latent (or hidden) heat. A substance has two latent heats. of a substance is the heat energy required to convert a unit mass of a substance from the solid to the liquid state, without change of temperature. The latent heat of fusion of ice is 3.36 105 J kg–1 or 80 cal g–1. of a substance is the heat energy required to convert a unit mass of the substance from the liquid to the gaseous state, without change of temperature. The latent heat of vaporisation of steam is 2.26 106 J kg–1 or 540 cal g–1. 16.1 A 12 kW drilling machine is used to drill a hole in a metal block of mass 10 kg. Assuming that 25% power is lost in the machine, calculate the rise in tempera-

The amount of heat energy in copper block at 500°C is Q = ms T = (3.35 390 500) J Now L = 335 J g–1 = 335 103 J kg–1. The maximum mass of ice that can melt is Q 3.35 390 500 m= = L 335 103 = 1.95 kg 16.3 16 g of oxygen is heated at constant volume from 25°C to 35°C. Find the amount of heat energy required. Given Cv = 20 J mol–1 K–1. SOLUTION Mass of 1 mole of oxygen = 32 g Number of moles in 16 g of oxygen is 16 1 = n= 32 2 Heat energy required is Q = n Cv T =

of metal = 0.4 J g–1 K–1. SOLUTION

20

(35 – 25) = 100 J

16.4

3 Useful power available = 75% of 12 kW = 4 = 9 kW = 9000 W

12 kW

Heat energy consumed in 2 minutes (= 120 s) is Q = (9000 120) J Now

1 2

s = 0.4 J g–1 K–1 = 0.4

103 J kg–1 K–1

Q Rise in temperature is T = ms =

9000 120 10

3

(0.4 10 )

= 270°C

16.2 A copper block of mass 3.35 kg is heated to 500°C and then placed on a large block of ice. What is the maximum mass of ice that can melt? –1 K–1 and latent heat of fusion of water = 335 J g–1.

290 J of heat energy is required to raise the temperaheat of nitrogen at constant pressure. SOLUTION 7 1 Number of moles in 7 g of oxygen is n = = 28 4 Q = n Cp T Cp =

Q n T

=

290 1 4

= 29 J mol–1 K–1

40

16.5 200 g of water at 25°C is added to 75 g of ice at 0°C of the mixture?

Measurement of Heat 16.3 T2

SOLUTION

Q=

Heat energy of 200 g of water at 25°C is Q1 = ms T = 200 1 25 = 5000 cal Heat energy required to melt 75 g of ice at 0°C is Q2 = mL = 75 80 = 6000 cal

If 1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C, the equilibrium temperature is 100°C, 0.665 kg of steam will be left and 1.335 kg of water will be formed. Water of mass m at °C is mixed with ice of mass mi at 0°C. L f mi (a) If m = temperature = 0°C. L f mi (b) If m < temperature = 0°C and mass of ice melted is mi=

. Amount of ice left = mi – m i.

Lf

(c) If m >

L f mi

temperature =

L mi m

m

> 0°C

16.6 very low temperatures (close to absolute zero), the s of a solid varies with absolute temperature T as s= 3 where is a constant whose value depends upon the material of the solid. Find the heat energy required to raise the temperature of 200 g of the solid from 1 K to 4 K. SOLUTION The amount of heat energy required to raise the temperature of mass m of the solid by dT kelvin is given by dQ = msdT Heat energy required to raise the temperature of the solid from T1 = 1 K to T2 = 4 K is

T 3 dT

T1

=

4

T1

T24

0.2 4 = 12.75 =

Since Q1 < Q2, the whole of ice will not melt. Hence, NOTES

T2

msdT = T14 ( 4) 4

(1)4

joule

16.7 Three liquids 1, 2 and 3 of masses m1 = m, m2 = 2m and m3 = 3m are at temperatures of 10°C, 18°C and 30°C. When liquids 1 and 2 are mixed, the equilibrium temperature is 16°C. When liquids 2 and 3 are mixed, the equilibrium temperature is 22°C. Find the equilibrium temperature when liquids 1 and 3 are mixed. Assume that there is no loss of heat to the surroundings. SOLUTION Let s1, s2 and s3 uids 1, 2 and 3 respectively. When liquids 1 and 2 are mixed, heat gained by 1 = heat lost by 2 m1s1(16 – 10) = m2s2(18 – 16) ms1(6) = 2ms2(2) 2 s 3 2 When liquids 2 and 3 are mixed heat gained by 2 = heat lost by (3) 2ms2(22 – 18) = 3ms3(30 – 22) s s3 = 2 3 s1 =

-

(1)

(2)

Let T be the equilibrium temperature. When liquids 1 and 3 are mixed. Heat gained by 1 = heat lost by 3 (3) ms1(T – 10) = 3ms3(30 – T) Using (1) and (2) in (3), we have 2 3

s2(T – 10) = 3

s2 (30 – T) 3

T = 22°C

16.4 Comprehensive Physics—JEE Advanced

I Multiple Choice Questions with Only One Choice Correct 1. 300 g of water at 25°C is added to 100 g of ice at (a) –

5 °C 3

(c) – 5 °C

(b) –

5 °C 2

(d) 0 °C

IIT, 1989 2. 100 g of ice at 0°C is mixed with 100 g of water (a) 0°C (b) 20°C (c) 40°C (d) 60°C 3. 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 rises to 80°C. The mass of steam condensed in kilogram is (a) 0.13 (b) 0.065 (c) 0.260 (d) 0.135 IIT, 1986 4. minute, from 27°C to 77°C. If the heat of combustion of LPG is 4.0 104 J g–1, how much fuel in grams is consumed per minute? (a) 15.25 (b) 15.5 (c) 15.75 (d) 16 5. A copper block of mass 2 kg is heated to a temperature of 500°C and then placed in a large block of ice at 0°C. What is the maximum amount of ice that –1

°C–1 and latent heat of fusion of water is 3.5 105 J kg–1. IIT, 2005 4 6 (a) kg (b) kg 3 5 8 10 kg (d) kg (c) 7 9 6. How much heat energy is joules must be supplied to 14 grams of nitrogen at room temperature to raise its temperature by 40°C at constant pressure. Molar mass of nitrogen = 28 and R J K–1 mol–1 is the gas constant. (a) 50 R (b) 60 R (c) 70 R (d) 80 R

7. The temperature of a liquid does not increase during boiling. The heat energy supplied during this process, (a) increases the kinetic energy of the molecules of the liquid (b) increases the potential energy of the molecules (c) increases both the kinetic and potential energy of the molecules (d) is merely wasted since no increase occurs in the total energy of the molecules. 8. 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?

Fig. 16.1

IIT, 2000 9. 2 kg of ice at –20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible

and ice are 1 kcal/kg/°C and 0.5 kcal/kg/°C re-spectively and the latent heat of fusion of ice is 80 kcal/kg. (a) 7 kg (b) 6 kg (c) 4 kg (d) 2 kg IIT, 2003 10. A metal sphere of radius r S is rotated about an axis passing through its centre at a speed of n rotations per second. It is suddenly stopped and 50% of its energy is used in increasing

Measurement of Heat 16.5

its temperature. Then the rise in temperature of the sphere is: (a)

2

2

2 2 2

n r 5S

(b)

n2

10 r 2 S

7 5 ( rn)2 r2 n2 S (d) 8 14 S 11. If there are no heat losses, the heat released by the condensation of x grams of steam at 100°C into water at 100°C converts grams of ice at 0°C into water at 100°C. The ratio /x is (a) 1 (b) 2 (c) 3 (d) 4 12. 5 g of water at 30°C and 5 g of ice at –20°C are

15. A 1 kW electric kettle contains 2 litre water at 27°C. The kettle is operated for 10 minutes. If heat is lost to the surroundings at a constant rate of 160 J/sec, the temperature attained by water in 10 minutes (a) 57°C (c) 77°C

(c)

ice = 0.5 cal g–1 (°C)–1 and latent heat of fusion of ice = 80 cal g–1. (a) –5°C (b) 0°C (c) + 5°C (d) + 10°C 13. A metal block of mass 10 kg is dragged on a horizontal rough road with a constant speed of 5 ms–1. the road is 0.5, the rate at which heat is generated in (J s–1) is (a) 100 (b) 245 (c) 9.8 (d) 10 14. 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 represents the veriation of temperature with time? (see Fig. 16.2). IIT, 2004

(b) 67°C (d) 87°C

IIT, 2005 16. In an industrial process 10 kg of water per hour is to be heated from 20°C to 80°C. To do this, steam at 150°C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90°C. How many kg of steam are required per hour? heat of steam = 1 kilo cal kg–1 °C–1. Latent heat of steam = 540 kilo cal kg–1. (a) 1 kg (b) 2 kg (c) 3 kg (d) 4 kg 17. With what velocity should a lead bullet at an initial temperature of 30°C strike a target so that it just melts? Assume that 84% of the heat produced is cal kg–1 (°C)–1, latent heat of lead = 6 kilo cal kg–1 and melting point of lead = 330°C. (1 kilo calorie = 4.2 103 joule)

18.

(a) 100 5 ms–1

(b) 100 2 ms–1

(c) 100 10 ms–1

(d) 100 15 ms–1

ids at extremely low temperature (close to absos of a solid varies with absolute temperature T as s = cT 3 where c is a constant depending on the material of the solid. The heat energy required to raise the temperature of 0.1 kg of the solid from 0 K to 4 K is. (a) 4.2 c joule (b) 6.4 c joule (c) 8.4 c joule

(d) 12.6 c joule

19. 400 g of ice at 253 K is mixed with 0.05 kg of steam at 100°C. Latent heat of vaporisation of steam = 540 cal/g. Latent heat of fusion of ice = 80 cal/g. temperature of the mixture is (a) 273 K (b) 300 K (c) 330 K (d) 373 K IIT, 2007 Fig. 16.2

16.6 Comprehensive Physics—JEE Advanced

20. Two litres of water (density = 1 g/ml) in an openlid insulated kettle is heated by an electric heater of power 1 kW. The heat is lost from the lid at the rate of 160 J/s. The time taken for heating water

–1

to 75°C is (a) 340 s (c) 620 s

K–1) from 20°C

(b) 550 s (d) 760 s IIT, 2005

ANSWERS

1. 7. 13. 19.

(d) (b) (b) (a)

2. 8. 14. 20.

(a) (a) (c) (b)

3. (a) 9. (b) 15. (d)

4. (c) 10. (a) 16. (a)

5. (c) 11. (c) 17. (d)

6. (c) 12. (b) 18. (b)

SOLUTION 1. Let the temperature of the mixture be C. Heat lost by water in calories = 300 1 (25 – ) = 7500 – 300 Heat in calories required to melt 100 g of ice = 100 80 = 8000 Now Heat lost = heat gained 5 °C or 7500 – 300 = 8000 or = – 3 Since is negative, the water at 25°C cools to 0°C and melts a part of ice at 0°C. Heat lost = 300 1 (25 – 0) = 7500 cal. Hence only a part of the ice melts and resulting temperature is 0°C. Hence the correct choice is (d). 2. The amount of heat required to convert 100 g of ice at 0°C into water at 0°C = 100 80 = 8000 calories. This is precisely the amount of heat lost by 100 g of water at 80°C to bring its temperature down to 0°C. Therefore, the temperature of the mixture remains 0°C. Hence the correct choice is (a). 3. Let the mass of steam condensed be m kg. The latent of vaporisation of water = 556 k cal/kg. Therefore, heat lost by steam = 556 m + m (100 – 80) = 576 m kcal. Heat gained by calorimeter and water = (1.1 + 0.02) (80 – 15) = 72.8 kcal. Now, heat lost = heat gained, i.e. 576 m = 72.8 or m = 0.13 kg 4. min–1 = 3000 cm3 min–1 Density of water = 1 g cm–3.

Heat energy supplied to the geyser per minute = 3.0 4200 50 = 63

104 J min–1

Now, heat of combustion of LPG = 4.0 104 J g–1 = 4.0 107 J kg–1 Amount of fuel consumed per minute =

63 104 J min

1

4.0 107 J kg

1

= 15.75 10–3 kg min–1 = 15.75 g min–1 5. Heat energy in copper block = 2 400 500 = 4 105 J. The amount of ice that melts will be maximum if the entire heat energy of the copper block is used up in melting ice. Now, 3.5 105 J of heat energy is needed to melt 1 kg of ice into water. Therefore, the amount of ice melted by 4 105 J of heat energy is 4 105 J 8 = kg 5 1 7 3.5 10 J kg Hence the correct choice is (c). 6. Number of moles in 14 grams of nitrogen (n) = 1 14/28 = . Since nitrogen is diatomic Cp = 7R/2. 2 Therefore, amount of heat energy supplied 1 7R = n Cp = 40 = 70R joules 2 2 Hence the correct choice (c) 7. The correct choice is (b).

= 3000 g min–1 = 3.0 kg min–1

8.

Rise in temperature = 77 – 27 = 50°C –1

°C

–1

ice increases from – 10°C to 0°C. At 0°C, ice starts melting at a constant temperature. When the whole

Measurement of Heat 16.7

of ice has melted into water, the temperature of water will increase from 0°C to 100°C. At 100°C, again the temperature becomes constant due to the conversion of liquid water into water vapour (steam) at 100°C. Hence the correct graph is (a). 9. Let m kg be the mass of ice melted into water. Heat lost by 5 kg of water = 5 kg 1 kcal/kg/°C 20°C = 100 kcal. Heat gained = m kg 80 kcal/kg + 2 kg 0.5 kcal/kg/°C 20°C = 80 m kcal + 20 kcal. Now, heat gained = heat lost. Therefore, 80 m + 20 = 100 or m + 1 kg = 6 kg. Hence the correct choice is (b). 10. Moment of inertia of the sphere I =

2 Mr2. Given 5

= n rotations per second = 2 n rad s–1. The kinetic energy is 1 2 Mr2 (2 n)2 2 5 4 = M 2r 2n 2 5 Since half of KE is converted into heat energy, we have 1 2 dQ = KE = M 2 r2 n2 2 5 KE =

1 I 2

2

=

Heat required for step (ii) = 5 80 = 400 cal When water at 30°C is allowed to cool to 0°C, the heat given out = 5 1 (30 – 0) = 150 cal Thus it is clear that all the ice cannot melt and the system will remain at 0°C. Since only (150 – 50) = 100 cal are available for melting ice, the mass of ice melted = 100/80 = 1.25 g. 6.25 g of water at 0°C and (5 – 1.25) = 3.75 g of ice Hence the correct choice is (b). 13. Heat energy generated per second = work done per second against friction = force of friction distance moved in 1s = = 0.5 10 9.8 5 = 245 J s–1 Hence the correct choice is (b). 14. Liquid oxygen will undergo a change of phase when heated from 50 K to 300 K. During the phase change the temperature will not change. When all the liquid oxygen has changed to gaseous state, the temperature will increase because rate of heating is constant (see Fig.16.3). Hence the correct graph is (c).

Now dQ = MSdT which gives 2 M 2 r 2 n2 2 2 r 2 n2 dQ 5 dT = = = 5S MS MS Hence the correct choice is (a). 11. The latent heat of vaporisation of water is very nearly 540 calories per gram. Therefore heat released in the condensation of x gram of steam = 540 x calories. The latent heat of fusion of ice is very nearly 80 calories. Therefore, heat required to convert gram of ice at 0°C to water at 100°C = 80 + 100 = 180 calories. Thus 180 = 540 x

15. Power (P) = 1 kW = 1000 W, mass of water (m) = s) = 4200 J/kg °C. Time = 10 min = 600 s. Initial temperature T1 = 27°C. Heat energy supplied by kettle in 600 s = = 1000 600 = 600,000 J. Heat energy lost in 600 s = 160 600 = 96,000 J. Therefore, heat energy used up in heating the water is H = 600,000 – 96,000 = 504,000 J

=3 x Hence the correct choice is (c).

or

Now H = ms T = ms (T2 – T1) = 2 27), where T2

12. temperature of ice from – 20°C to 0°C and (ii) melt the ice at 0°C into water at 0°C. Heat required for step (i) = 5 50 cal

Fig. 16.3

0.5

{0 – (–20)} =

2

4200

4200

(T2 –

(T2 – 27) = 504,000

which gives T2 = 87°C, which is choice (d). 16. Let the mass of steam required per hour be m kg. Heat gained by water in boiler per hour is

16.8 Comprehensive Physics—JEE Advanced

=10 kg

18.

1 kilo cal kg–1 °C–1

(80 – 20)°C = 600 kilo cal (1) Heat lost by steam per hour is = heat needed to cool m kg of steam from 150°C to 100°C + heat needed to convert m kg of steam at 100°C into water at 100°C + heat needed to cool m kg of water from 100°C to 90°C = m 1 (150 – 100) + m 540 + m 1 (100 – 90)

constant. The amount of heat energy required to raise the temperature of the solid of mass m through dT kelvin is dQ = msdT Heat energy needed to raise the temperature of the solid from 0 K to 4 K is 4

0

= 50 m + 540 m + 10 m = 600 m kilo cal

17. Let the mass of the bullet be m kg and its velocity be v ms–1. Before striking the target, the kinetic en1 ergy of the bullet is mv2 joule which is converted 2 into heat energy when the bullet strikes the target. Thus heat energy produced is 1 mv2 joule Q = 2

8.4 103

kilo cal

Heat energy absorbed by the bullet is Q = 84% of Q = 0.84 Q =

0.84

mv

2

8 4 103

kilo cal

In order that heat energy Q melts the bullet, it bullet from 30°C upto its melting point (330°C) and then to supply the latent for melting. Hence 0.84

m v2

8. 4 103

=m

0.03

(330 – 30)

= 9 m + 6 m = 15 m

0

T 3dT = mc

= mc 0

44 4

= mc

T4 4 c

= 0.1

4

0

(4)3

= 6.4 c joule. So the correct choice is (b). 19. Heat required to melt the whole of ice is Q1 = ms T + mL = 400

0.5

20 + 400 (

80 253 K = –20°C)

= 4000 + 32000 = 36000 cal The maximum heat released by steam when the whole of it (= 50 g) is converted into water at 0°C is Q2 = ms T + mL = 50 1 100 + 540 50 = 5000 + 27000 = 32000 cal Since Q2 is less than Q1, the whole of ice will not

1 m v 2 joule 2 = 4.2 103 joule/kilo cal m v2

cT 3dT

4

(2)

Heat lost = heat gained. Equating (1) and (2) we have 600 m = 600 or m = 1 kg, which is choice (a).

=

4

msdT = m

Q =

m

6

will be 0°C or 273 K. 20. Mass of 2 litres of water = 2 kg. Heat energy needed to raise the temperature of 2 kg of water from 20°C to 75°C is Q = 2 (4.2 103) 55 = 4.62 105 J If is the time taken, heat energy supplied by the heater in time is Q1 = (power time) = 1000 joule Heat energy lost in time is Q2 = 160 joule

which gives v = 100 15 ms–1

Heat energy available for heating water is Q = Q1 – Q2 = 840 J

Thus the correct choice is (d).

Equating Q = Q , we get

550 s.

Measurement of Heat 16.9

II Multiple Choice Questions with One or More Choices Correct 1. A source of heat supplies heat at a constant rate to a solid cube. The variation of the temperature of the cube with heat supplied is shown in Fig. 16.4. (a) Portion BC of the graph represents conversion of solid into liquid. (b) Portion BC of the graph represents conversion of solid into vapour. (c) Portion DE of the graph represents conversion of vapour into liquid. (d) Portion DE of the graph represents conversion of liquid into vapour

(c) thermal capacity of liquid (d) thermal capacity of vapour. 4. In Fig. 16.4, it is observed that DE = 3 BC. This means that (a) the thermal capacity of the vapour is 3 times that of the liquid of the liquid (c) the latent heat of vaporisation of the liquid is 3 times the latent heat of fusion of the solid. (d) the latent heat of fusion of the solid is 3 times the latent heat of vaporisation of the liquid. 5. 800 J kg–1 K–1 is initially at a temperature 20°C. It –1

Fig. 16.4

2. In Q.1 above, the slope of portion CD of the graph shown in Fig. 16.4 gives (a) latent heat of fusion (b) latent heat of vaporisation (c) thermal capacity of liquid (d) thermal capacity of vapour 3. In Q.1 above, the slope of the portion EF of graph shown in Fig. 16.4 gives (a)

and on returning to the starting point strikes a lump of ice at 0°C and gets embedded in it. Assume that all the energy of the bullet is used up in melting. Neglect the friction of air. Latent heat of fusion of ice = 3.36 105 Jkg–1. (a) Energy of bullet used in melting is 1000 J. (b) The mass of ice method = 5 g (c) The mass of ice melted is slightly greater than 5 g. (d) The mass of ice melted is less than 5 g. 6. ity s falls from a height of 10 cm and bounces to a height of 4 m. If all dissipated energy is absorbed by the ball as heat, its temperature rises by 0.075 K. Take = 10 ms–2 . (a) Q = 60 J (b) Q = 100 J (c) s = 800 J kg–1 K–1 (d) s = 1333 J kg–1 K–1

ANSWERS AND SOLUTIONS 1. The correct choices are (a) and (d). The heat supplied is the latent heat. 2. The slope of portion CD of the graph gives the amount of heat supplied by unit rise in temperature of of the liquid. Hence the correct choice is (c). 3. The correct choice is (d). 4. The portion BC of the graph represents the conversion of solid into liquid, temperature remaining the

same. The portion DE represents the conversion of liquid into vapour at the same temperature. The heat supplied in the two cases is latent (hidden). Hence the correct choice is (c). 5. If the friction offered by air is neglected, the speed of the bullet on returning to the starting point will be equal to its initial speed v = 200 ms–1. The kinetic energy of the bullet is

16.10 Comprehensive Physics—JEE Advanced

K.E. =

x = 5.3 10–3 kg = 5.3 g. Hence the correct choices are (a) and (c).

1 mv2 2

6. Q =

1 = (50 2

–3

2

10 )

(200) = 1000 J

Also

Heat lost by bullet for its temperature to fall from 20°C to 0°C = (50 10–3) 800 20 = 800 J. If x kg is the mass of ice melted, then x

(3.36

(h – h ) = 1

s=

10

(10 – 4) = 60 J.

Q = ms T. Hence

Q 60 = m T 1 0.075

= 800 J kg–1 K–1. Thus the correct choices are (a) and (c).

105) = 1000 + 800

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I The basic principle of calorimetry is heat gained by one body = heat lost by the other body. This follows from the principle of conservation of energy according to which the total heat energy of the two substances must remain constant. Hence heat lost by one body must be gained by the other, provided no part of heat energy is allowed to escape. An aluminium container of mass 100 g contains 200 g of ice at – 20°C. Heat is added to the system at ice = 2100 J kg–1 K–1 aluminium = 840 J kg–1 K–1 and latent heat of fusion of ice = 3.36 105 J kg–1.

1. The time taken to raise the temperature of the container and ice from – 20°C to 0°C is (a) 12 s (b) 24 s (c) 36 s

(d) 48 s

2. The time taken to melt ice at 0°C into water at 0°C is (a) 40 s (b) 80 s (c) 120 s

(d) 160 s

3. The temperature of the system after 4 minutes is (a) 15.45°C (b) 20.45°C (c) 25.45°C

(d) 30.45°C

SOLUTION 1. Heat energy needed to raise the temperature of the container and ice from – 20°C to 0°C is Q1 = (100

10–3)

+ (200

840 –3

10 )

20

2100

20

= 1680 + 8400 = 10080 J 10080 Time needed is 1 = = 24 s 420 So the correct choice is (b). 2. Heat energy required to melt ice at 0°C into water at 0°C is Q2 = (200 10–3) (3.36 105) = 6.72

4

10 J

Time needed is

= 2

6.72 104 = 160 s, which is 420

choice (d). 3. Total time 1 + 2 = 24 + 160 = 184 s is less than 4 minutes (= 240 s). Hence heat energy supplied during (240 – 184) = 5 s will be used up in raising the temperature of the system (container + 200 g plied is 56 s = 420 56 = 23520 J. If temperature, then 23520 = (100 10–3) 840 + (200

10–3)

4200

which gives = 25.45°C. So the correct choice is (c).

Measurement of Heat 16.11

4. The rate at which heat is produced is (a) 35.8 Js–1 (b) 36.8 Js–1 –1 (c) 37.8 Js (d) 38.8 Js–1 5. The power of the drill is (a) 40 W (b) 42 W (c) 48 W (d) 56 W 6. The torque required to drive the drill is (a) 1.2 Nm (b) 2.2 Nm (c) 3.2 Nm (d) 4.2 Nm

Questions 4 to 6 are based on the following passage Passage II A steel drill making 180 revolutions per minute is used in during a hole in a block of steel. The mass of the steel block is 180 g. 90% of the entire mechanical energy is used up in producing heat and the rate of rise of temperature of of steel = 420 J kg–1 K–1.

SOLUTION 4.

Q

= mass

sp. heat

second = (180 10–3) 420 So the correct choice is (c). 5. Power of drill =

180 60 = 3 r.p.s. Angular frequency of rotation ( ) = 2 3 = 6 rad s–1 Now power = torque angular frequency or P = P 42 = = = 2.2 Nm 6 3.14

6. Number of revolutions per second =

rise in temperature per 0.5 = 37.8 Js–1.

37.8 = 42 W, which is choice 0.9

(b).

Thus the correct choice is (b).

IV Integer Answer Type 1. 2 kg of ice at – 20°C is mixed with 5 kg of water at 20°C in an insulating vessel having negligible heat

and ice are kcal/kg/°C and 0.5 kcal/kg/°C and latent heat of fusion of ice is 80 kcal/kg. IIT, 2003

SOLUTION 1. mi si [0 – (– 20)] + m L = m s (20 – 0) 2 0.5 20 + m 80 = 5 1 20

m = 1 kg. Mass of water in container = 5 + 1 = 6 kg

17

Chapter

Thermodynamics (Isothermal and Adiabatic Processes)

REVIEW OF BASIC CONCEPTS 17.1

EQUATION OF STATE

In the case of ideal gases, the equation of state is PV = RT, for one mole and PV = nRT, for n moles where P, V and T are respectively the pressure, volume and temperature of the gas and R is the universal gas constant.

17.2

MOLAR SPECIFIC HEAT

C of a substance is the amount of heat energy required to raise the temperature of 1 mole of the substance through 1 K. It is expressed in J mol–1 K–1. s C) are related as C s= m where m is the number of kilograms per mole in the substance. v) is

p

17.3

The Zeroth law of thermodynamics which states that

Thus if

17.4

when R is the universal gas constant and its value is R = 8.315 J mol–1 K–1

T = TC and TC = T , then T = T .

INTERNAL ENERGY

amount of energy. This energy is called the and is usually denoted by the symbol U. The internal energy of solid, liquid or gas consists of two rotational and vibrational) of the molecules, and of the molecules. If the intermolecular forces are extremely weak or absent, then the change in internal energy is given by U = mcv T where m is the mass of the gas, cv

T

heat Cv Mcv where M is the molecular mass), we have, for n moles of an ideal gas m Cv T U = nCv T = M

17.5 Cp – Cv = R

ZEROTH LAW OF THERMODYNAMICS

FIRST LAW OF THERMODYNAMICS

When heat energy is supplied to a system, a part of this energy is used up in raising the temperature of the system the rest is used up in doing external work against the

17.2 Comprehensive Physics—JEE Advanced

surroundings. Thus, if Q is the heat energy supplied to a gas and if W is the work done by it, then from the law of conservation of energy, the increase U in the internal Q – W) or Q= U+ W

law of thermodynamics which may be stated in words as ‘

’

where

=

Cp Cv

.

For an adiabatic process, Q thermodynamics, Q = U + W, it follows that U + W = 0 or W = – U W is positive. hence U is negative, i.e. the temperature of the gas falls. W is negative. Hence U is positive, i.e. the temperature of the gas increases. 3. Isochoric Process

Sign convention for

Q,

W and

U

1.

Q is positive if heat is supplied to the system and negative if heat is taken out of the system. 2. W is positive if work is done by the system and negative if work is done on the system. 3. U is positive if the temperature of the system increases and negative if the temperature of the system decreases.

17.6

THERMODYNAMIC PROCESSES

1. Isothermal Process

between P and V is PV = constant

. From PV = nRT, it follows n = constant), the relation

an isochoric process. The relation between P and T is P1 P2 P = constant P T T1 T2 T The work done in a process is W = P V. For an isochoric process, V = 0. Hence W law of thermodynamics it follows that Q= U Q is positive. Hence U is positive, i.e. the temperautre of the gas rises. Hence pressure also increases. Q is negative. Hence U is negative, i.e. the temperature of the gas falls. Henc pressure also falls. 4. Isoboric Process

P 1V 1 = P 2V 2 T = constant)

If T is kept constant T = 0. Hence U = 0, i.e. for an isothermal process, the internal energy of the gas remains that Q= W i.e. all the heat energy supplied to the gas is used up in doing work. gas. Hence both W and Q are positive. on the gas. Hence both W and Q are negative. 2. Adiabatic Process . For an adiabatic process, the realtion between P and V is PV = constant P 1V 1 = P 2V 2

V

. The relation between V and T is V1 V2 V = constant T T1 T2 T

from V T, that the temperature rises. Hence W and U thermodynamics, that Q is also positive. Hence Q, U and W are all positive in isobaric expansion. Q, U and W are all negative.

17.7

WORK DONE IN A PROCESS

from volume V1 to volume V2 is given by W =

V2 V1

1. Work Done in an Isothermal Process When an

17.3

at constant temperature) work is done by it. For an isothermal process, the equation of state for n moles of ideal gas is PV = nRT where T is the constant absolute temperature and R is the universal gas constant. The value of R for all gases is R = 8.31 J K–1 mol–1 P=

The work done W = nRT

V2 V1

or

V

n RT V = nRT ln

W = 2.303 nRT log

V2 V1

e)

4. Work Done in an Isochoric Process In an isochoric process, volume V is constant, i.e. = 0. Hence, work done W = 0.

17.8

INDICATOR DIAGRAM

P – V diagram) is a graph in which P) is plotted on the V) on the -axis. Figure 17.1 shows indicator diagrams for expansion, compression and for a closed cyclic process. The initial P1, V1) is represented by point P2, V2) by point . The intermediate states are represented by points between and on the curve .

V2 V1

where V1 is the initial volume and V2 P1 and P2 P 2V 2)

P 1V 1 =

P1 P2 2. Work Done in Adiabatic Process When a gas undergoes an adiabatic change, the pressurevolume changes obey the relation PV = constant = C where = Cp/Cv the gas at constant pressure to that at constant volume. W = 2.303 nRT log

C

, where C is a constant V The work done is given by 1 W= P 2V 2 – P 1V 1) 1 P=

= =

1 1 nR 1

P 1V 1 – P 2V 2) T 1 – T 2)

since P1V1 = nRT1 and P2V2 = nRT2, T1 and T2 being the absolute temperatures before and after the adiabatic change. 3. Work Done in an Isobaric Process For an isobaric process, pressure P is constant. Therefore, the work done is given by W =P

V2 V1

= P V 2 – V 1)

Fig. 17.1

The work done in a process is given by W = area enclosed by P – V curve and volume axis. Thus, the work done is given by the area of the shaded portion in Fig. 17.1.

17.9 given by

EFFICIENCY OF AN IDEAL HEAT ENGINE ) of an ideal reversible heat engine is =1 –

where

T2 T1

T1 = absolute temperature of the source which supplies heat and T2 = absolute temperature of the sink which takes in the part of heat not converted into useful work,

17.4 Comprehensive Physics—JEE Advanced

T1 is always greater than T2. If T1 = T2 = 0 implying that an engine working under isothermal conditions can produce no useful work. Complete conversion of heat into = 1) is possible only if T2 = 0, i.e. the sink is at absolute zero, which is unattainable.

17.10

Cv, Cp AND GAS

= Cp /Cv FOR AN IDEAL

1. For a monoatomic gas, 3R 5R , Cp = and Cv = 2 2 2. For a diatomic gas, 5R 7R , Cp = and Cv = 2 2 3. For a triatomic or polyatomic gas

5 = 1.67 3

=

7 = 1.4 5

= 1.4) at 127°C are expanded adiabatically to twice the original volume. done in the process. Given R = 8.3 J K–1 mol–1 and 0.4 = 0.76. SOLUTION T1 V1

and

W=

=

=

– 1)

V1 V2 nR 1

1

0.4

= 400

0.76 = 304 K

T 1 – T 2)

2 8.3 1.4 1

17.3

CP = Cv Cp = Cv Cv =

= T2 V2

T2 = T1

Cp – Cv = R

or

– 1)

4 = 1.33 3 RELATION BETWEEN Cp, Cv AND Cv = 3R, Cp = 4R and

17.11

=

17.2

takes up 746 J of heat from the high temperature res-

R 1

and Cp =

R 1

17.1 5 moles of an ideal gas are compressed to half the initial volume at a constant temperature of 27.0°C. Calculate the work done in the process. Given R = 8.3 J K–1 mol–1. Write your result up to appropriate

SOLUTION T1 = 100°C = 373 K, T2 = 0°C = 273 K, Q1 = 746 J T W =1– 2 T1 Q1

W = Q1

T1

T2 T1

= 746 = 200 J Q2 = Q1 – W = 746 – 200 = 546 J

SOLUTION Given T = 27.0°C = 300 K. Since T is constant, the process is isothermal W = nRT loge

V2 V1

=5

8.3

300

=5

8.3

300

373 273 373

=

W 200 = = 0.268 = 26.8% Q1 746

NOTE

loge

1 2

= – 8628 J Since the value of R s. .), the value of W must be rounded off to 2 s. . as W = – 8.6 103 J The negative sign indicates that work is done on the gas.

=1–

T2 273 =1– = 0.268 T1 373

17.4 Figure 17.2 shows the P-V diagram of a cyclic process . Calculate the work done in process to to C C to

17.5

17.6

8

= 1.4) are heated at constant volume. If 280 J of heat energy is supplied

6 (Nm–2)

4

gas and the work done.

2

2

4

6

8

SOLUTION Since V = 0, W modynamics, =

10 (litre)

= 280 J

Fig. 17.2

SOLUTION W + area of rectangle 1 2 1 = 2

=

SOLUTION

+ 4

6

10

17.7 The pressure P of an ideal gas varies with volume V as P = where is a constant. The volume of n moles of the gas is increased from V to mV. Find the work done and the change in internal energy.

–3

+4

6

10

–3

mV

mV

W=

1 litre = 10–3 m3)

V = V

V

10–3 J

= 36

W C = –24 10–3 J Negative sign shows that the work is done on the gas. volume is constant) WC W = 36 10–3 – 24 10–3 = 12 10–3 J which is equal to area enclosed by the closed loop . 17.5 = 1.4) are heated at constant pressure. If 280 J of heat energy is sup-

=

nCp n C

Q=

U+W

T=

U+W

T=

U+W

U=

U+W W

U=

1

V2 2

mV

= V

2

2

m2 – 1)

Cp C U=nC =

2

2

m2

T)

1 1

17.8 Two moles of a monoatomic gas undergo a cyclic process as shown in Fig. 17.3. 2 x 105

SOLUTION U=nC Given

(Nm–2)

C T=n Cp Q = n Cp U= W=

Cp T =

nCp T

T = 280 J. Hence

2.80 = 200 J 1.4 Q–

1 x 105

U = 280 – 200 = 80 J

0

1

2

3

4

5 (

3)

Fig. 17.3

Process is isobaric, process C is adiabatic and process C is isothermal. Find

17.6 Comprehensive Physics—JEE Advanced

Hence the sope of P-V graph for an adiabatic process is greater than for an isothermal process. Figure 17.4 shows two P-V curves, one for adiabatic expansion and the other for isothermal expansion.

C . SOLUTION P = P = 2 105 Nm–2 V = 2 m3, V = 5 m3 and C P V = PC V C For isothermal process C P V = PC VC P =P ) 1 1

V V

VC =

Substituting the values of V , V and , we get 3 VC PV PC = = 0.2 105 N m–2 VC W=W =P

+W

+ WC 1 V –V)+ 1 C

P V – P CV C)

+ P V loge Substituting the values, we get W = 6 105 N m–2

V VC

17.9 Show that the slope of P-V curve for an adiabatic process is greater than that for an isothermal process. SOLUTION For an isothermal process, PV = constant. Differentiating w.r.t. volume V we have P + =0 = V iso For an adiabatic process, PV = constant, Cp . where = Cv Differentiating w.r.t. volume V, we have +V =0 PV – 1 adia

Since

> 1,

adia

P V

=

iso

.

Fig. 17.4

17.10 Two moles of a diatomic ideal gas, initially at pressure 5.0 104 Pa and temperature 300 K are expanded isothermally until the volume of the gas is doubled and then adiabatically expanded until the volume is again doubled. Find end of the complete proces

process. P-V graph for the complete process. Given R = 8.3 JK–1 mol–1. SOLUTION Let P1, V1, T1 be the initial pressure, volume and temperature of the gas For isothermal process P1V1 = P2V2 where V2 = 2V1 P2 = P1

V1 V2

= 5.0

104

= 2.5

104 Pa

1 2

In an isothermal process, the temperature remains constant. Hence T2 = T1 = 300 K. Therefore change U)1 = 0. Work done is W1 = nRT1ln

V2 V1

=2

8.3

300

17.7

= 3.45 103 J For adiabatic process P 2V 2 = P 3V 3 P3 = P2

5 8.3 2 5R for diatomic gas) Cv = 2 = – 3.03 103 J =2

V2 V3

= 2.5

1 2

104

1.4

P3 103 Pa, T3 = 227 K W = W1 + W2 = 3.45 103 + 3.03 103 = 6.48 103 J U)2 = 0 – 3.03 103 = U U)1 – 3.03 103 J. The negative sign shows that there is a decrease in internal energy. P-V graph for the complete process is shown in Fig. 17.5.

3

10 Pa T 2V 2

– 1)

– 1)

= T 3V 3

1

V2 V3

T3 = T2

1 2

= 300

0.4

= 227 K

Work done is

Isothermal expansions

W2 = =

nR 1

T2

T3

2 8.3 1 4 1)

Adiabatic expansions

= 3.03 103 J Change in internal energy is )2 = nCv T = n

5R 2

1

T

2

3

Fig. 17.5

I Multiple Choice Questions with Only One Choice Correct 1. If 2 moles of an ideal monoatomic gas at temperature T are mixed with 3 moles of another monoatomic gas at temperature 2T, the temperature of the mixture will be 8T 6T 5 5 4T 3T 3 2 2. If heat energy Q is supplied to an ideal diatomic gas, the increase in internal energy is U and the work done by the gas is W. The ratio Q : U : W is 3. Figure 17.6 shows a cyclic process. When a given mass of a gas is expanded from state to state , it absorbs 30J of heat energy. When the gas is adiabatically compressed from state to state ,

the work done on the gas is 50 J. The change in internal energy of the gas in the process is

Fig. 17.6

4. Figure 17.7 shows a cyclic process for an ideal diatomic gas. The ratio of the heat energy absorbed in the process to the work done on the gas in the process C is

17.8 Comprehensive Physics—JEE Advanced

10. For a thermodynamic process, the P-V graph for a monoatomic gas is a straight line passing through the origin and having a positive slope. The molar heat capacity of the gas in this process is R R

11.

7 2 ln 2

5 ln 2 4

5 ln 2 2

5. The internal energy of 3 moles of hydrogen at temperature T is equal to the internal energy of n moles of helium at temperature T/2. The value of n ideal gases) 3 2 6. The temperature of n moles of an ideal gas is increased from T to 3T in a process in which the temperature changes with volume as T = 2 where is a constant. The work done by the gas in this process is n RT n RT 3 n RT n RT 2 7. In a certain process, pressure P, volume V and temperature T of a gas are related as PV = n where and n are constants. The work done by the gas when the pressure is kept constant, is proportional to T)n T )1/n T n –1) T n + 1) 8. If the pressure of an ideal gas in a closed container is increased by 2%, the temperature of the gas increases by 5°C. The initial temperature of the gas is 9.

pressure P in the balloon is related to the volume V as PV2/3 = , where is a constant. If T is the temperature of the mixture, volume V is proportional to T T2 3 T4 T

R

600 K has a work output of 800 J per cycle. How much heat energy is supplied to the engine from the source in each cycle?

Fig. 17.7

7 4 ln 2

3 R 2

12. should the temperature of the source be increased

13. The pressures and volumes corresponding to P = 3 104 Pa, –3 3 4 V = 2 10 m , P = 8 10 Pa, V = 5 10–3 m3. In process , 600 J of heat and in process , 200 J of heat is added to the system. The change in the internal energy in process would be

Fig. 17.8

14. One mole of an ideal gas requires 207 J heat to raise its temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same R, the gas constant = 8.3 JK–1 mol–1)

17.9

15. during the cycle is PV PV

PV

Fig. 17.10

21. In rising from the bottom of a lake to the top, the temperature of an air bubble remains unchanged, but its diameter is doubled. If h is the barometric density ) at the surface of the lake, the depth of the

Fig. 17.9

16. The equation of state corresponding to 8 g of O2 is 2 to be an ideal gas) RT PV = 8 RT PV = 4 RT PV = RT PV = 2 17. The equation of state of a gas is P

aT V

2

Vc

h h h h 22. Entropy of a thermodynamic system does not change when the system is used for cold reservoir

RT + )

where a, , c and R are constants. The isotherms can be represented by m n – P= where and depend only on temperature and m = – c, n m = c, n = 1 m = – c, n m = c, n = – 1

-

chorically

23. Heat energy absorbed by a system in going through a cyclic process shown in Fig.17.11 is 7

4

2

–3

18. When an ideal monoatomic gas is heated at constant pressure, the fraction of heat energy supplied which increases the internal energy of the gas is 2 3 5 5

19.

3 7

3 4

heats is 4150 J kg–1 K–1 and the ratio of the two gas at constant volume in units of J kg–1 K–1?

Fig. 17.11

24. will be value of

20.

to by three different paths 1, 2 and 3 as shown in Fig. 17.10. If W1, W2 and W3 respectively be the work done by the gas along the three paths, then W1 < W2 < W3 W1 > W2 > W3 W1 = W2 = W3

W1 < W2 W1 < W3

for the mixture?

= 5/3) is mixed = 7/5). What

25. If the ratio Cp/Cv = , the change in internal energy of the mass of a gas, when the volume changes from V to 2V at constant pressure P is R PV 1

17.10 Comprehensive Physics—JEE Advanced

PV PV 1 1 26. Two cylinders and equal amounts of an ideal diatomic gas at 300 K. The piston of is free to move, while that of is gas in each cylinder. If the rise in temperature of the gas in is 30 K, then the rise in temperature of the gas in is

Fig. 17.12

T 1,

30.

piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the pison suddenly. If L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1/T2 is given by

27. Two identical containers and tionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in is m and that in is m . The gas in each cylinder is now allowed to expand V. The changes in pressure in and are found to be P and 1.5 P respectively. Then m

m =3m

m =2m

m =4m

m

28. Two monoatomic 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 gas 2 is M1 M2 M1 M2

L1 L2

2/3

L2 L1

L1 L2 L2 L1

2/3

IIT, 2000 31. 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

M2 M1

W2 > W1 > W3

W2 > W3 > W1

W1 > W2 > W 3

W1 > W3 > W2 IIT, 2000

M2 M1

32.

IIT, 2000 29. T and volume V. Its volume is increased by V due to an increase in temperature T, pressure remaining constant. The quantity = V V T) varies with IIT, 2000

expansion. If its temperature falls by 2K, its internal energy will

7 initially at S.T.P. 5 are compressed adiabatically so that its temperature becomes 400ºC. The increase in the internal energy

33. 5 moles of Hydrogen

R = 8.30 J mol–1 K–1)

34. During an adiabatic process, the pressure of a gas is proportional to the cube of its absolute temperature. The value of Cp/Cv for that gas is:

17.11

35.

3 5 5 3

IIT, 2002

4 3 3 2 2

32) at a temperature T. The pressure of the gas is P T has a pressure of P P 8 P P Fig. 17.15

36. In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas

ill increase 37. P-V plots for two gases during adiabatic processes are shown in Fig. 17.13. Plots 1 and 2 should correspond respectively to

IIT, 2001

40. Figure 17.16 shows the P-V mass of an ideal gas undergoing cyclic process. represents isothermal process and represents adiabatic process. Which of the graphs shown in Fig. 17.17 represents the P-T diagram of the cyclic procee?

Fig. 17.16

IIT, 2003

2 2

and He

2

and N2

Fig. 17.13

IIT, 2001 38.

C , as shown in Fig. 17.14. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C is

IIT, 2002

Fig. 17.17

Fig. 17.14

39. Which of the graphs shown in Fig. 17.15 correctly represents the variation of V/dp)/V with p for an ideal gas at constant temperature?

41. The pressure and density of a diatomic gas p1, 1) to p p2, 2). If 2 = 32, then 2 is p1 1 1 128

17.12 Comprehensive Physics—JEE Advanced

42.

43.

= 1.4) is adiabatically compressed so that its temperature rises from 27°C to 35°C. The change in the internal energy of the R = 8.3 J/mole/K)

Va V = Vc V

Va T = 2 T1 V

Va V = V Vc

Va T = 1 T2 V

V, pressure P and temperature T. The mass of each molecule of the gas is m is the P P T

44.

45.

mP

a monoatomic ideal gas which expands at constant pressure. What fraction of the heat energy is converted into work? 2 3 2 5 5 7

Fig. 17.18

49.

in the V – T diagram. Which of the diagrams shown in Fig. 17.20 shows the same process on a P – V diagram.

gas at 0°C and also contains an insulated piston of negligible weight and negligible thickness at the middle point. The gas on one side is heated to 100°C. If the piston moves through 5 cm, the length of the hollow cylinder is Fig. 17.19

46.

volume V1 to volume V2 and then compressed adiabatically to original volume V1. The initial pressure is P1 P3. If the net work done is W, then P 3 < P 1, W < 0 P 3 > P 1, W P 3 > P 1, W

47.

P 3 = P 1, W = 0 IIT, 2004

resistance 50 for 10 minutes by connecting it to a d.c. source of 10 V. The change in the internal energy is Fig. 17.20

48. Two different adiabatic paths for the same gas intersect two isothermals at T1 and T2 as shown in the P–V

50.

= 1.4) expands from 5 10–3 m3 to 25 10 m at a constant pressure of 1 105 Pa. The heat energy supplied to the gas in this process is –3

3

17.13

internal energy of the gas is

51. Three moles of an ideal gas are taken through a cyclic process as shown on T–V diagram in Fig. 17.21. The gas loses 2510 J of heat in the complete cycle. If T = 100 K and T = 200 K, The work done by the gas during the process is

. The change in the

IIT, 2005 54. The P–V diagram for n moles of an ideal gas The maximum temperature of the gas during the process is 3P0V0 P0V0 nR 4nR 2 P0V0 nR

3P0V0 2nR

Fig. 17.21

52. Liquid oxygen at 50 K is heated at 300 K at constant pressure of 1 atmosphere. The rate of heating is constant. Which of the graphs shown in Fig. 17.22 represent the variation of temperature T )? IIT, 2004

Fig. 17.23

55.

lid insulated kettle is heated by an electric heater of power 1 kW. The heat is lost from the lid at the rate of 160 J/s. The time taken for heating water –1 –1 K ) from 20°C to 75°C is

IIT, 2005 2

PT = constant.

56.

Fig. 17.22

53. One mole of a monoatomic ideal gas is contained in a insulated and rigid container. It is heated by

1 T

2 T

3 T

4 T IIT, 2008

ANSWERS

1. 7. 13. 19. 25.

2. 8. 14. 20. 26.

3 9. 15. 21. 27.

4. 10. 16. 22. 28.

5. 11. 17. 23. 29.

6. 12. 18. 24. 30.

17.14 Comprehensive Physics—JEE Advanced

31. 37. 43. 49. 55.

32. 38. 44. 50. 56.

33. 39. 45. 51.

34. 40. 46. 52.

35. 41. 47. 53.

36. 42. 48. 54.

SOLUTIONS 1. Let T0 be the temperature of the mixture. Since the total internal energy remains unchanged, we have U of mixture = U1 + U2 n 1 + n 2) C v T 0 = n 1 C v T 1 + n 2 C v T 2 n 1 + n 2) T 0 = n 1 T 1 + n 2 T 2 T0 = 2T + 3 T ) = 8T 8T Which gives T0 = 5 7R 5R 2. For a diatomic gas Cp = and Cv = 2 2 7 Q = n Cp T = nR T 2 5 U = n Cv T = nR T 2 7 5 nR T – nR T 2 2 = nR T Q: U: W=7:5:2 3. In the process , work is done on the gas. = –50 J. Since this process is W) Q) thermodynamics, the change in internal energy in this process is Q) W) U) W= Q– U=

Since the process is cyclic, there is no net change in internal energy. Hence U) = –50 J U) 4. In the process , V is proportional to T. Hence pressure P remains constant. Therefore, heat C p = 7 R/2 for a diatomic gas) 7R = n Cp T = n T 0 – T 0) Q) 2 7 n RT0 = 2 Process C is isothermal in which the gas is compressed. Hence work done on the gas in this process is V0 W ) C = – n R T0) ln 2V0

1 2

= – 2 n RT0 ln = 2 nRT0 Q

=

W

7 4 ln 2

5. The internal energy of n moles of an ideal gas at temperature T is given by U=

n RT 2 where = number of degrees of freedom. For hydrogen, = 5. Therefore 5 15 3 RT = RT U1 = 2 2 For helium, = 3. Therefore 3 3 n R T/2) = n RT U2 = 2 4 Given U1 = U2, i.e. 15 3 RT = nRT 2 4 which gives n = 10. 2 . Therefore =2 or 6. Given T = =

2 PV = n RT

nR T V

P=

Work done W =

3T

V T

= =

3T T 3T

nRT V nR 2

T

=

nR 2

2 T 2

3T T

= n RT n

7. Given V = P

. Since P = constant,

2

= T)

17.15

=

n – 1)

T

T2 =

P V=

Work done W =

n 1

T2 – T2

T

n = +c where c = constant of integration. Hence the

8. Since the volume of the gas is constant, P1 T = 1 P2 T2

13. Process is isochoric, i.e. the volume remains constant. Thus V = 0. Hence work done P V = 0. Process is isobaric, i.e. the pressure remains constant and external work has to be done. The 10–3 work done = P V – V ) = 8 104 – 2 10–3) = 240 J. Therefore, change in internal energy is

Now P2 = P1 + 0.02 P1 = 1.02 P1 and T2 = T1 + 5. P1 T1 = T1 5 1.02 P1

=

Heat energy required to raise the temperature n moles of a gas by T at constant volume is Qv C Qv = n Cv T, = v Qp Cp

n RT P= . Using V

V 2/3 =

T V –1/3 =

nR

or

= constant

Hence V T 3 10. Since the P – V graph is a straight line with a positive slope, P V or PV –1 = constant For a process in which PV n = constant, the molar heat capacity is given by R R C= 1 1 n Putting n = – 1 and = we have C=

11.

= 800 – 240 = 560 J

Qp = n Cp

Equation of state is PV = nRT

or

–

14. Heat energy required to raise the temperature of n moles of a gas by T at constant pressure is

T1 = 250 K

9. PV 2/3 =

n RT V

300 = 750 K 0.4

R 5 1 3

5 3 R 3R = 1 1 2

R = 2R 2

W T 300 =1– 2 =1– = 0.5. Therefore Q = 2W Q T1 600 =2

12. T2 = 300 K. Now = 1 – T2/T1. When 0.4, the value of T1 is given by T2 = 1 – 0.4 = 0.6 T1 T 300 = 500 K. When or T1 = 2 = 0.6 0.6 = 0.6, the value of T2 should be

= 40% =

= 60%

Qv = =

Cv Cp 3 5

Qp =

3R / 2 5R/2

207 = 124.2

Qp 124 J

15. Work done = area enclosed by the indicator diagram 1 = 2 1 = P – P) V – V) 2 = 3 PV 16.

1 . Now 4 the equation of state for n moles of an ideal gas is 1 RT RT = PV = n R T = 4 4 n) =

17. Expanding the equation of state we have PV c + aT2 V c–1 = RT + or or where given that

–c P = – aT 2 V –1 + RT V–c + –c –1 P= – = RT + and = aT2. We are

P=

m

–

n

17.16 Comprehensive Physics—JEE Advanced

Comparing the powers of V = – c and n

m

18. Now Qp = n Cp T and Qv = n Cv T Qv gives the heat energy which increases the internal energy of the gas. Thus the required fraction is Qv C 1 3 1 = v = = = Qp Cp 5 5/3 For monoatomic gas

=

5 3

19. Given Cp – Cv = 4150 and Cp/Cv = 1.4 or Cp = 1.4 Cv. Therefore, 1.4 Cv – Cv = 4150 or 20.

Cv =

4150 = 10375 J kg–1 K–1 0.4

m

=7h

23. Heat energy absorbed = work done = area of the loop r2 =

2

/4 =

4

U = n Cv T We know that Cp – Cv = R Cp

or

Cv

2

= 102

24. For a monoatomic gas, Cv = 3R/2 and for a diatomic gas, Cv = 5R/2. Since one mole of each gas is mixed together, the Cv of the mixture will 1 3R 5R Cv = = 2R 2 2 2 Now Cp – Cv = R. Therefore, for the mixture, Cp = R + Cv = R + 2R = 3R heats of the mixture is

Cp

R Cv

=1+ =1+

Cv

=

R , which gives Cv

R 1

PV = n RT volume changes by V, the change in temperature T is given by P V = nR T or

T=

P V PV = nR nR V = 2V – V = V

U=n

R

m

22. When work is converted into heat at a constant temperature, the entropy of the system remains con-

=

Cv

3R = 3 = 1.5 2R 2

=

Cv =

3 . Since the diameter of the 21. Volume bubble is doubled in rising from the bottom to the top of the lake, its volume becomes 8 times. Now PV = constant. Therefore, the pressure at the bottom of lake = 8 times that at the top. Let H be the depth of the lake. H h – h) m

H = 7h

Cp

25. Let T be the increase in temperature when the volume of the gas is changed by V at constant pressure. The change in internal energy of n moles of a gas is given by

Therefore,

P V P-V) curve, which is the largest for curve 3 and the smallest for

or

=

1

PV = nR

PV 1

26. Heat is given to the gas in cylinder at constant pressure while the same amount of heat is given to the gas in cylinder at constant volume. Heat given to gas in is Q = n Cp T Heat given to gas in Since nCp or

is Q = nCv T

Q = Q , we have T = nCv T Cp 7 T = 30 K = 42 K T = Cv 5 for a diatomic gas, Cp/Cv = 7/5)

27. The equation of state for an ideal gas of mass m and molecular mass M is m RT PV = M For an isothermal process, T = constant. DifferenT, we get

17.17

P V+V P=0

1 V 1 1 = or = V T T T Thus, the value of decreases as T is increased. or

V P= – P V mRT P= MV

or

m RT

P =–

MV

Hence

m P = m P

Given

P = 1.5

or

– 1)

30. For adiabatic process, T1 V 1

mRT P= – MV

V = 2V – V = V) and

P =–

P . Therefore,

T1 V2 = T2 V1 For a monoatomic gas, Hence

m RT MV

T1 = T2

m 1 = m 15

3m =2m .

28. The speed of sound in a gas of bulk modulus density is given by

and

L2 L1

5 3

= T2 V 2

– 1)

. Thus

1

V2/V1 = L2/L1. 1

=

L2 L1

2/3

31. Since the slope of the P–V graph for adiabatic expansion is times that for isothermal expansion, curves and represent isothermal and adiabatic expansions of V 2. the gas from initial volume V1

v= V P V Now, for a perfect gas, PV = nRT. Differentiating at constant T, we get V P P V + V P = 0 or = –P V is given by

Hence

=–

P

v=

If m is the mass of the gas and M its molecular mass, then m mRT RT or PM = = RT PV = M V or

P

=

RT M

v=

RT M

Hence v1 =

RT M1

v1 = v2

M2 M1

or

give

or v2 =

Fig. 17.24

between volumes V1 beand V2 is greater than the area under curve tween V1 and V2, it follows that W1 > W3. P–V graph for isobaric V2 ume V1 and pressure P1 sure remaining unchanged at P1 between volumes V1 and V2 is greater than the area under curves and . Hence W2 is greater than W1 and W3

RT M

and v2 =

RT which M2

29. For an ideal gas, PV = nRT. Since pressure P is kept constant, P V = nR T nRT V nR nRV V or = = = P V T P nRT T

32.

= Hence

+ . In an adiabatic process, = 0. = – = – 4.5 J. Hence the correct

33. Given T1 = 0ºC = 273 K, T2 = 400ºC = 673 K Work done W =

nR T2 1

T1

5 8.3 400 7 1 5

17.18 Comprehensive Physics—JEE Advanced

= 41500 J = 41.5 kJ

for a diatomic gas, atomic gas for which = 5/3. Therefore, the slope of the P-V curve is less for a diatomic gas than for a monoatomic gas. Hence curve 1 corresponds to diatomic gas and curve 2 to monoatomic gas. Thus

to be negative, i.e. W law of thermodynamics = + adiabatic process, = 0. Hence

. For an = – = implies that the internal energy increases. Hence the correct

38. Process of Q. 38)

34. For an adiabatic process TPn = 1

where n =

W Cp

,

Cv

Therefore,

and

W =W

gives

3 2

=

39. For an ideal gas, pV = nRT. Differentiating, we have since T is kept constant

1 = 3 or n

1

= 3, which

p

P)O2 =

1 mole RT V

P)He =

1 mole R 2T V

He

P

O2

1 p

–

p1

Given = 0 and < 0. Hence the change in internal energy < 0. Now, for an ideal gas, the internal energy can decrease only by decrease in

Since at any instant PV = constant,

P V

or

the slope of P-V curve is proportional to . Now,

p2 2

p2 = p1

2

7/5

7

or

Cv =

= 128

1

U = Cv T. Now Cp – Cv = R or

42. , i.e.

=

1

37. For an adiabatic process, PV = constant, Differentiating, we have V = 0 or

1

41. pV = constant. For a given mass of gas, V Hence p = constant

P)O2 =

+

1 V

=

of versus p. 40. Since is an isothermal process, the temperature of the gas remains constant between . Hence the P-T diagram must be perpendicular to the T-axis between and . Hence the correct choice

36.

–1

V p

=2

P)He

PV

+ V = 0 or

Hence

35. For a gas, PV = nRT. Hence

P

+W

or W

T –1/n

T 3. Hence

+W

or 5 = 10 + 0 + W

T

Given P

= P V2 – V1) = 10

Process C, occurs at constant volume. Hence W = 0. Given Q = 5 J, i.e. total work done is W = 5 J. Therefore, we have

is a constant.

Since n = constant for a given gas, P

=

1/ n

P=

occurs at constant pressure.

R 1

, where

=

Cp Cv

Cp Cv

–1=

. Hence

R Cv

.

17.19

U=

43. PV =

R T 8.3 8 = = 166 J 1 1 4 1)

m RT. Therefore, the density of the gas is M m PM PmN mP = V RT RT

to V1, P3 are as shown in Fig. 17.25. Let W1 and W2 be the work done in isothermal expansion and adiabatic compression respectively. Therefore, net work done is W 2) = W 1 – W 2 W = W1

44. Heat energy supplied = Cp . Change in internal energy = Cv . Therefore, work done = – Cp – Cv ) . =

Cp

Cv p

=1–

T

=1–

1

2 1 = 5 5/ 3 = 5/3 for a monoatomic gas)

45. Let L and r its radius. Since the mass of the gas remains unchanged and the pressures of the gas in both sides are equal, we have, from Charles’ law, V1 V = 2 T1 T2 L L r 2, V 2 = r 2, 5 5 2 2 T1 = 0°C = 273 K and T2 = 100°C = 373 K. Using

Given V1 =

L L 5 5 2 = 2 273 373 which gives L = 64.6 cm. 46. For an isothermal process: PV = constant and for an adiabatic process: PV = constant, where is the Cp /Cv) of the gas. When a gas is compressed from a volume V to a V – V), the increase in pressure is P)adia =

VP for an adiabatic compression V

VP for an isothermal compression. V Hence P3 will be greater than P1. Therefore, the P–V diagrams of isothermal expansion from V1, P1 to V2, P2 and adiabatic compression for V2, P2

Fig. 17.25

Now, the area under the adiabatic curve is more than that under the isothermal curve. Hence W2 > W1. Therefore, W < 0. Hence the correct . 47. Heat produced is given by 2

10

2

10 60 = 1200 J R 50 Since the container is rigid, the change in volume = 0. Hence work done = = 0. From the =

energy is

=

–

=

= 1200 J. Hence the

48. The two adiabatic paths and for the gas intersect the two isothermals and at temperatures a and lie T1 and T2 on the same adiabatic path, we have T1Va – 1) =T2V – 1) 1

Va V

or Since points path,

=

T2 T1

and c also lie on the same adiabatic T1 V )

– 1)

= T2 Vc

1

V Vc

or

P)iso =

Va V

= 1

=

T2 T1 V Vc

1

– 1)

17.20 Comprehensive Physics—JEE Advanced

= P V = nR T = nR T – T ) = 3 8.3

W

Va V = V Vc

or

49. Since the temperature T remains constant along the path , P will be inversely proportional to V along this path. Hence, as P increases, V must decrease in a nonlinear fashion. This is represented by the curve in Fig. 17.26. , the volume V is constant. Hence the graph of P against V is a straight line perpendicular to the V-axis. On a P – V diagram, the corresponding path is as shown in Fig. 17.26. For the path , V is directly proportional to T pressure remaining constant. The corresponding path is, therefore a straight line parallel to the V-axis. Thus the cyclic process on a P–V diagram

PV = nRT)

Process work done in this process W = 0. Since the whole process is cyclic, the change in internal energy in the complete cycle is zero, i.e. U = 0. Q = – 2510 J) Q= U+W=

U+W +W W +0

+W

or W The negative sign shows that the work is done 52.

oxygen is greater than 50 K. Therefore, between 50 K and 300 K, liquid oxygen undergoes a change of

53. The heat energy supplied is Q = I2

2

100 60) 3 10 J = 240 kJ

= 240

Since V W of thermodynamics, U = Q = 240 kJ. Thus the

Fig. 17.26

50. Work done on the gas is W = P V = P V – Vi) = 1 105 = 2000 J

10 –3

The internal energy is given by U =

PV 1

PV PVi ,U = . 1 1 Therefore, change in internal energy is P U = U – Ui = V – Vi) 1 Ui =

=

1 105

25 5 1.4 1

10

3

51. In process , the volume V increases linearly with temperature T. Hence process is isobaric

P0 and c = 3 P0. V0

These equations give m = – Now PV = nRT we have

P= T=

= 5000 J

energy supplied to the gas is Q = W + U = 2000 + 5000 = 7000 J

process is

54. The equation of straight line is P = mV + c where m is the slope and c is the intercept. For points and , we have 2 P0 = mV0 + c V 0) + c and P0 = m

nRT V

1 mV2 + cV nR

T will be maximum if

= 0 and

2 2

< 0.

V and putting = 0, we get

2 mV + c = 0 c which gives V=– 2m c2 1 Tmax = – =– 4nRm 4nR

3P0 2 = P0V0 P0 /V0 4nR

17.21

55. Mass of 2 litres of water = 2 kg. Heat energy needed to raise the temperature of 2 kg of water from 20°C to 75°C is Q=2 103) 55 = 4.62 105 J

P=

Differentiating we have

If is the time taken, heat energy supplied by the heater in time is Q1 time) = 1000

3nRT 2 T =

Heat energy lost in time is Q2 = 160 Heat energy available for heating water is Q = Q1 – Q2 = 840 J Equating Q = Q , we get 550 s. Thus the 56. Given PT 2 =

nRT . Hence V nRT 3 = V T

V

V V T

=

3nRT 2

3nRT 2

nRT and PT 2 = P 3 = . T

Using V = we get

PV = nRT, we have

II Multiple Choice Questions with One or More Choices Correct 1. Figure 17.27 is the P-V diagram for a Carnot cycle. In this diagram, represents isothermal process and adiabatic process Fig. 17.27 represents adiabatic process and isothermal process represents isothermal process and adiabatic process represents adiabatic process and isothermal process. 2. Figure 17.28 shows the P-V diagram of a cyclic process. If is the heat energy supplied to the system, is the internal energy of the system and is the work done by the system, then which of the following relations is/are correct = =0 Fig. 17.28

= =– 3. If Q represents the heat energy supplied to a system, U the increase in internal energy and W the work done by the system, then which of the following are correct? Q = W for an isothermal process U=– U=

W for an adiabatic process Q for an isochoric process

U = – Q for an isobaric process. 4. The initial state of n moles of an ideal gas is repP 2, V 2, resented by P1, V1, T1 T2. Wi is the work done by the gas in an isothermal T1 = T2 = T) and Wa in an adiabatic process, then V2 Wi = nRT loge V1 Wi = nRT loge Wa = Wa =

1 1 nR 1

P1 P2 P 1V 1 – P 2V 2) T 1 – T 2)

17.22 Comprehensive Physics—JEE Advanced

5.

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. If is the number of degrees of freedom of gas molecules and W is the work done by the gas during the expansion, then =5 W=

6.

5PV 4

W=

3PV 2

namic process involving four steps. The amounts of heat involved in these steps are Q1 Q4 = 3645 J respecQ2 = – 5585 J, Q3 tively. The corresponding amounts of work done are W1 = 2200 J, W2 = – 825 J and W3 = – 1100 J and W4 . Then

Fig. 17.29

is 0.036 J. C is – 0.024 J. C is zero. is 0.06 J. 11. Figure 17.30 shows the P–V diagram for an ideal gas. From the graph, we conclude that

W4 = 275 J

W4

16% 7.

P, volume V and temperature T. The ratio CP/Cv = and U is the internal energy. If R is the gas constant, then R Cv = U = nCv T 1 U=

PV 1

zero.

U = nCpT

8. Fig. 17.30

of the source should be increased by 28.6°C. of the sink should be decreased by 28.6 °C.

9.

12. Figure 17.31 shows the P – V diagram for an ideal gas. From the graph we conclude that

800 J per cycle, the work output per cycle is 200 J. minutes. The room temperature is 27°C. The latent heat of fusion of ice is 80 cal/g. 105 J. to 10. Fig. 17.31

104 J. 10.

process

is adiabatic. P–V diagram of a cyclic

process.

17.23

to

, the temperature T of

decreases. in the process. 13. n moles of an ideal monoatomic gas is kept in a closed vessel of volume 0.0083 m3 at a temperature of 300 K and a pressure of 1.6 106 Pa. Heat 104 J is supplied to the gas. Given –1 –1 R = 8.3 J mol K . n=5 Cp = 20.75 J mol–1 K–1 106 Pa

R

v=

14. For an ideal gas

sure when the temperature of n moles of the gas changes by T is n Cv T.

adiabatic process is equal in magnitude to the work done by the gas. isothermal process. process. 15.

16. One mole of oxygen at 27 °C is enclosed in a vessel which is thermally insulated. The vessel is moved with a constant speed v and is then suddenly stopped. The process results in a rise of temperature of the gas by 1 °C. Then, if M = molecular mass of oxygen. 5 7 = Cp/Cv) = 3 5

P, volume V) to state P/2, Volume 2V) along a straight line in the P-V diagram as shown in Fig. 17.32. Then

M

1

v=

2R M

1

17. Cv and Cp of a gas at constant volume and constant pressure, respectively. Then Cp – Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas Cp + Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas Cp/Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas Cp Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas 18. Figure 17.33 shows the P-V plot of an ideal gas taken through a cycle . The part is a semi-circle and is half of an ellipse. Then,

Fig. 17.33 Fig. 17.32

to exceeds the work that would be done by it if the system were taken from to along the isotherm. T-V diagram, the path becomes a part of a parabola. P-T diagram, the path becomes a part of a hyperbola.

is isother-

mal C zero

C is

17.24 Comprehensive Physics—JEE Advanced

ANSWERS AND SOLUTIONS 1. For adiabatic process, PV = constant. Differentiating w.r.t V we get V +P V

–1

or

=0 =

P V

For isothermal process, PV = constant. Hence P = V Now, / P-V) graph. Thus, P-V) graph for an adiabatic process is times that for an isothermal process. Hence curves and both represent adiabatic process and curves and both represent isothermal process. Thus the correct choices are 2. In a cyclic process, the system returns to its initial state. Hence the change in internal energy = 0. of thermodynamics, = + 3.

=

= 0)

W = P V and U = nCv T and = Cp/Cv Q = U + W. For an isothermal process, T = 0. Therefore U = 0. Hence Q = W adiabatic process Q = 0. So U = – W, which is W = 0, so Q = U Q= U+ W

4. 5. For an adiabatic change the relation between T and V is Cp = TV – 1) Cv 1 V T = V T T Given V = 5.66 V and T = . Therefore, 2 – 1) =2 Taking logarithm of both sides, we have

TV

or

– 1)

– 1)

=T V

=1+

l l

or

2) 0.3010 =1+ 5.66) 0.7528 = 1 + 0.4 = 1.4

Since = 1.4, the gas is diatomic. For a diatomic gas, the number of degrees of freedom of the molecules = 5. We know that the work done by the gas during adiabatic expansion is given by 1 W= PV – P V 1 where pressure P after expansion is obtained from the relation PV PV = T T V T or P =P V T T /2 P = T 11.32 P Putting = 1.4, V = 5.66 V and P = in 11.32 We have W= =

1 1 4 1) 1 0.4

PV

=P

V 5.66V

PV

P 11.32

1 PV 2

5.66V

= 1.25 PV

6. Since W2 and W3 are negative, it means that the work is done on the gas. Hence Q2 and Q3 are negative which implies that heat is evolved in processes 2 and 3. Since Q1 and Q4 are positive, heat is Q2 + Q3), the gas absorbs Q1 + Q4 a net amount of heat energy in a complete cycle, which is given by Q = Q1 + Q2 + Q3 + Q4

The net work done by the gas is W = W1 + W2 + W3 + W4 = 2200 – 825 – 1100 + W4 W4 Since the process is cyclic, the change in internal energy U namics, we have W= Q – U = Q or 275 + W4 = 1040 or W4 = 1040 – 275 = 765 J

17.25

Q = 800 net work done by the gas h= total heat absorbed by the gas

=

80 = 80,000 J = 80,000

275 765 3645

W = = Q1 Q4 1040

9.

=

= 0.1083 = 10.83%

7. The internal energy of n moles of an ideal gas at absolute temperature T is given by U = nCvT where Cv ume. We know that Cp Cv

–1=

R or – 1 = or Cv = Cv

R Cv

8.

PV RT

R 1

=

1

T=

T2 T1

=1–

0.25 = 1 –

300 which gives T1

= 0.25 + 0.05 = 0.30. The new temperature of the source should beT 1 so that T 300 0.30 = 1 – 2 = 1 – T1 T1

1 2

1 2

+

10–3 m3) 4 Nm–2 + 4 Nm–2 6 10–3m3 = 0.012 + 0.024 = 0.036 J

11. For point , PV = 5 10–3) = 10 10–3 J For point =2 10–3) = 10 10–3 J Since PV = constant, the process is isothermal. For an isothermal process, U = 0. Since the gas undergoes expansion W is positive and Q = W. 12. The P–V graph for an adiabatic process is not a P V = n R T and P V = n R T . Therefore PV T T 6 1 3 = = , PV T T 2 4 4 i.e. T > T . Hence the internal energy increases. Work done = area under upto the volume axis. Heat energy is absorbed in the process. Hence the

which gives T = 428.6 K. So, increase in temperature of the source = T1 – T1 = 428.6 – 400

0.30 = 1 –

104 J, which

W = – 0.024 J. The negative C sign shows that the work is done on the gas.

PV 1

T1

such that

3.36

WC = P V = 0 because V = 0. Work done in complete cycle = 0.036 – 0.024 + 0 = 0.012 J = 0.25.

T2

273 300 273

=

10. W

R

Q2 = mL = 103 4.2 = 3.36 105 J. So

Q1 = Q2 + W = 3.36 105 + 3.36 104 37 104 J Q 37 104 J Power = 1 = 20 kW. So choice 180 s

Now, the ideal gas equation for n nodes is PV PV = nRT or n = RT U=

=

T1 T2 is correct. Q 3.36 105 W= 2 = 10

11%

Cp – Cv = R or

T2

0.25 = 200 J. So

T2 of the sink should be T2 T =1– 2 T1 400

which gives T2 = 280 K = 7°C. Decrease in temperature of the sink is 27°C – 7°C = 20°C,

13.

PV = nRT =

n=

PV RT

1 6 106 ) 0.0083 16 = 8.3 300 3

17.26 Comprehensive Physics—JEE Advanced

Cp = = 20.75 J mol–1 K–1 Q = n Cv

5R 5 = 2 2

P0 and V0 be the intercepts on the P and V axes. The equation of straight line is P0 P= V – V 0) V0 P V P0 V0

8.3

3R 2

T Cv =

Q 2 Q 2 2. 104 = = 16 nCv 3nR 8.3 3 3 = 375 K Final temperature = 300 + 375 = 675 K = 402°C. T=

P2 P = 1 T2 T1 = 3.6 14.

15.

P2 =

P1

6

T2 T1

=

Since P =

P V P0 V 2 RT V =1 T= 0 R RV0 V P0 V0 which represents a parabola on the T-V graph.

1.6 106 675 300

V=

10 Pa

U = n Cv T Q = U + W. For an adiabatic process, Q = 0. Hence 0 = U + W or | U| = | W|. T = 0, Hence U = 0. Q = 0. to W1 = area of trapezium P 3 PV = P V= 2 2 =

3 RT 2

RT V

P P0

RT P

RT =1 PV0

which does not represent a hyperbola. So is correct. 16.

Therefore, Cv = 5 R/2 and Cp = 7 R/2. So = Cp/Cv = 7/5. The kinetic energy of oxygen mol1 M v2, where M = ecules with a velocity v0 = 2 molecular weight of oxygen. = Cv 1 = Cv Now heat energy = Cv Cp R Cp – Cv = R or –1= Cv Cv – 1) =

R or Cv = Cv

Therefore,

1 M v2 = 2

or

v=

If the process to work done would be W2 = RT loge Thus W1 > W2

V2 V1

were isothermal, the = RT loge

RT

R 1

R 1 2R

M

1

17. Cp + Cv) for diatomic gas = Fig. 17.34

V0 P 2 R P0

T = V0 P –

7R 2

Cp + Cv) for monoatomic gas = Cp Cv for diatomic gas =

5R =6R 2 5R 2

35 R 2 4

Cp Cv for monoatomic gas =

15 R 2 4

3R =4R 2

17.27

Cp

= =1+

2

U=

is smaller for diatomic gas than

Cv for monoatomic gas. Cp – Cv = R. 18.

PV

V 1

is negative because V < V .

From Q = U + W

Q is negative, C .

C the cyclic process is clockwise. Hence the correct

process increases with V rect. For the process C , volume is decreasW

III Multiple Choice Questions Based on Passage Questions 1 to 4 are based on the following passage Passage I

2. Which of the graphs shown in Fig. 17.36 represents the P – T diagram for the complete process?

7R at pressure P 2 and temperature T are isothermally expanded to twice the original volume. The gas is then compressed at constant pressure to its original volume. Finally the gas is heated at constant volume to its original pressure P . Three moles of an ideal gas

Cp

1. Which of the graphs shown in Fig. 17.35 represents the P – V diagram for the complete process?

Fig. 17.36

3. The net work done

W by the gas during the R is the gas constant) RT RT RT RT 4. The net heat energy supplied to the gas in the complete process is W W W

Fig. 17.35

SOLUTION 1 . During V C), P = constant and during ), V = constant. Hence the

1. During isothermal process

correct P – V

C

,P

2. During isothermal pro

), T = constant. C), P = constant and C ), P T. Hence the correct P – T diagram for the complete process

17.28 Comprehensive Physics—JEE Advanced

3. For process , P V = P V which gives P P = because V = 2V 2 V V = C . Since V = 2 V and For process C, T TC T PC = P VC = V and T = T , we get TC = 2 P . = 2 P P P . Since PC = For process C , C = T TC 2 T = 2TC. Work done is isothermal process W

= nRT loge

V V

is

= 3 RT loge

Questions 5 to 9 are based on the following passage Passage II Two moles of an ideal gas at volume V, pressure 2 P and temperature T undergo a cyclic process as shown in Fig. 17.37.

Work done in isobaric process C is P W = P VC – V ) = V – 2V ) 2 PV 3 =– =– RT . 2 2 Work done is isochoric process C is W Total work done 3 RT 2

W = 3 RT loge = 3 RT

RT

Thus th 4. Since the process is cyclic, U = 0. From Q = U + W, we get Q = W VC) of the gas in state C is

6. 8V 3 4V 3 7.

V 2V 3 V ) of the gas in state

V V

V W) in the complete cycle is RT loge

RT 4 RT 3

Fig. 17.37

9. V 3 V

V ) of the gas in state is 2V 3 4V 3

is

V

8.

5.

2 RT loge 3 Q) in the heat energy in the com-

plete process is

W W

W.

SOLUTION 5. For isobaric process V T

=

V T

V =

V T T

6. For isothermal process P V = PC VC

7. For isobaric process C

, =

V

4T / 3 4V = T 3

C,

PV 2 P 4V / 3 8V VC = = = , PC P 3

=0

VC V = T TC 8. W W

V =

V T TC

=P V –V )=2P = nRT loge

VC V

, =

8V / 3 T = 2V 4T / 3 4V 3

V

=

2 PV 3

17.29

=2

R

4T log e 3

W

= P C V – V C) = P

W

= nRT loge =2

R

Total work done is

8V / 3 8 = RT loge 4V / 3 3 8V 3

2V

=–

W=W

2 PV 3

+W

+W

2 PV 8 + RT loge 3 3

=

2 PV – 2 RT loge 3

2 RT loge 3 9. For a cyclic process, U = 0. Hence Q = W. So

V V

T loge

+W

=

V 2V

1 = – 2 RT loge 2

= 2 RT loge

Questions 10 to 12 are based on the following passage Passage III V 5 consists of one gram mole of a gas with Cp/Cv) = 3 7 at a certain temperature T. and another gas with = 5 The gram molecular weights of gases and are 4 and 32 respectively. The gases do not react with each other and are assumed to be ideal. The gaseous mixture follows the = constant in an adiabatic process. relation PV – 10. The number of moles of gas mixture is

11. The adiabatic compressibility of the mixture at pressure P is 3 5 5P 7P 13 7 P P 12. If the temperature T of the mixture is raised from 300 K to 301 K, the percentage change in the speed of sound in the gaseous mixture is 1 3

in the gaseous

1 % 6

1 2

SOLUTION 10. Let n and n be the number of moles of gases Cv) be their and C v) Since the gases do not react with each other and are assumed to be ideal, the internal molecular kinetic energy before and after the gases are mixed must be the same. Hence n +n Cv )m T n C v ) T + n C v) T Cv ) m ture at constant volume. Now R Cv ) = Cv ) = 1 R

C v) m = Therefore, n + 1

1

m

n 1

=

n m

n 1

5 7 , = and m is given by 3 5 the adiabatic relation for the mixture,

Given n = 1,

PV Given PV

5 1 3

R

m

= constant

= constant. Hence 1

-

1

=

+

m

=

13

. Using

n 1 n = 7 1 1 5 13

5n 13 1 n 3 + = 2 6 2 n = 13 + 13 n or 2 n = 4 or n = 2 Thus the mixture contains 2 moles of gas . or

17.30 Comprehensive Physics—JEE Advanced

of the mixture is

11. 1 . V We are given that PV where

m

=

13

m

= constant

. Partially differentiating, we have

PV

m

+

or

–

1 V

V m

V = P

V=0

m 1

PV

m

V =– P

or

1 m

V P

=–

=

m

PV

m 1

=–

12. We know that the speed of sound in a gas is proportional to the square root of absolute temperature, i.e. v = 1/2 = constant 1 Taking logarithm, log v = log + log T 2 v 1 T =0+ Differentiating, v 2 T 1 1K 1 = 2 300 K 600 Percentage change v 100 1 100 = = % v 600 6

V m

=

P

1 m

P

Questions 13 to 15 are based on the following passage Passage IV Two moles of a monoatomic ideal gas occupy a volume V at 27°C. The gas is expanded adiabatically to volume 2 2 V. Gas constant R = 8.3 JK–1 mol–1.

P

13 , P

=

2 14. The change in the internal energy of the gas in this process is

15. The work done by the gas during the process is

13. 150 2

SOLUTION 13. T1 = 300 K, V1 = V, V2 = 2 2 V. Let T2 obtained from the adiabatic relation T1 V1 or

– 1)

V1 V2

T2 = T1

For a monoatomic gas T2 = 300

= T2 V2

1 2 2

2/3

=

14. T2 is

– 1) 1

5 . Therefore, 3

U = nCv T2 – T1) For a monoatomic gas Cv =

3R . Therefore, 2

3 8.3 2 The negative sign implies that the internal energy U=2

15. For a adiabatic process, Q = 0. Hence W = – U

= 150 K

Questions 16 to 21 are based on the following passage Passage V is taken through the process and another sample of 2 kg of the same gas is taken through the process as

shown in Fig. 17.38. The molecular mass = 4 and R = 8.3 J K–1 mol–1. 16. The temperature of state

is

17.31

18. The temperature of state C is 19. The temperature of state

20. The work done in process 104 103 21. The heat supplied in process to 106 106

Fig. 17.38

17. The temperature state

is

is

is

104 J 103 J is nearly equal 106 J 106 J

SOLUTION 16. Number of moles of helium is 2000 mass in gram n= = = 500 4 molecular mass From equation state at , PV P V =nRT T = nR =

4.15 104 10 = 100 K 500 8.3

17. For isochoric process T T

=

P P

=

8.3 10

4

4.15 104

= 2.

19. For isobaric process T P 20 = = =2 T P 10 which gives T = 2 T = 200 K. So the correct 20. Work done in process W= W + W = 8.3

104

103 J

21. From U = n Cv T, heat energy in process for a monoatomic gas Cv = 3 R/2) Q

Thus T = 2 T 18. For isobaric process

is = 0 + P VC – V )

U)

W)

= n Cv TC – T ) + P V – V ) C,

= 500

TC V 20 = C = =2 T V 10 Thus TC = 2 T Questions 22 to 25 are based on the following passage Passage VI One mole of an ideal monoatomic gas is taken round the cyclic process

= 1.87

400 100 104

106

22. The work done by the gas is P 0V 0 P 0V 0

23.

P 0V 0 P0V0

is P0V0 3P0V0 2 2 5P0V0 P 0V 0 2 24. The heat energy absorbed by the gas in the process is 3P0V0 P0 V0 2 C

Fig. 17.39

3 83 2 + 4.15

17.32 Comprehensive Physics—JEE Advanced

5P0V0 2 25.

P0V0 2 5P0V0 2

P0 V0

in the process

C is

P0 V0 P0 V0

SOLUTION 22. Work done by the gas in the cyclic process W = area enclosed in the P–V diagram = area of triangle 1 = P 0 – P 0) 2

Using the ideal gas equation for points have P V = nRT or P0 V0 = nRT

is

1 2 V0 – V0) = P0V0, =

23. In the isobaric process by the gas is given by Q = nCp T = nCp T – TC Using ideal gas equation PV = nRT for points C, we have P V = nRT or P0 V0 = nRT and PC VC = nRTC or P0 V0) = nRTC

and

P V = nRT

and

T – TC = –

Q

=–

= 3 P0 V0

Q 25.

3R . Therefore 2 -

ergy absorbed by the gas is given by

where W = W = work done by the gas = P0V0. Since the process is cyclic, there is no change in the internal energy of the gas, i.e. U = 0. Hence Q = W = P 0V 0

C p P0 V0

If Q

R

is the heat absorbed in the process Q =Q

5R . Hence 2

5 P0 V0 2 24. The heat energy absorbed in the isochoric process is given by Q = nCv T = nCv T – T Q

= 2 Cv P0 V0 R

Q= U+ W

P0 V0 nR

Now, for a monoatomic gas Cp =

2 P0 V0 nR

Now, for a monoatomic gas, Cv =

P0V0 = nR TC – T ) or

P0)V0 = nRT

which give T – T =

Q

and , we

=–

Questions 26 to 30 are based on the following passage Passage VII Two moles of an ideal monoatomic gas, initially at pressure P1 = P and volume V1 = 2 2 V, undergo an adiabatic compression until its volume is V2 = V and the pressure is P2. Then the gas is given heat energy Q at constant volume V2.

+Q

+Q

P0V0 = 3P0V0 + Q or

Q

=

, then

–

5P0V0 2

P0V0 2

26. Which of the graphs shown in Fig. 17.40 represents the P–V diagram of the complete process? 27. Pressure P2 is 2P

2P

2 2P 28. The total work done by the gas is PV 2 PV

2 PV PV

17.33

29. The change in internal energy due to adiabatic compression is 2 PV PV 2 PV PV 30. The temperature T2 of the gas after it is adiabatiR is the gas constant). PV 2PV R R 2PV R

2 2PV R

Fig. 17.40

SOLUTION 26. 27. For an adiabatic change, P1V1 = P2 V2 . Therefore, P2 = P1 For a monoatomic gas, P2 = P

2 2V V

V1 V2 = 5/3. Hence

5/3

= P 2 2

5/3

. So the correct

28. In an adiabatic process, heat Q = 0. Hence from the W1 = – U1 = – n Cv T = – 2 Cv T2 – T1

n = 2)

Now for 2 moles P1 V1 = 2 RT1 and P2V2 = 2 R T2. Thus PV PV T1 = 1 1 and T2 = 2 2 2R 2R Using these, we have C W = – v P 2 V 2 – P 1 V 1) R C PV 1 1 =– v V2 PV 1 1 R V2 C =– v R

P 1V 1

V1 V2

1

1

=–

3 P1 V1 2

=–

3P 2 2V 2

V1 V2

2/3

1

2 2

2/3

1

=–3 2 The work done at contant volume = 0. Hence the 29. In adiabatic compression, Q = 0. Hence U = – W, 30. Since U = nCv T = 2 Cv T2 – T1) U we have T2 = T1 + 2Cv Using T2 = =

T1 =

P1 V1 3R and Cv = , we get 2R 2

P1 V1 3 P1 V1 + 2 2 3R / 2 2R P1 V1 2R =

V1 V2

V1 V2

2/3

P 2 2V 2 2PV = R R

2/3

1

17.34 Comprehensive Physics—JEE Advanced

Questions 31 to 35 are based on the following passage Passage VIII Two moles of an ideal mono-atomic gas is taken through a cycle as shown in the P–T During the process , pressure and temperature of the gas vary such that PT = , where is a constant. IIT, 2000

Fig. 17.41

31. Constant is given by PT 1 1 P 1T 1 2 P 1T 1 2 P 1T 1 32. The work done in process R is the gas constant) RT1 RT1 RT1 RT1 33. The heat energy released in process is RT1 RT1 RT1 RT1 34. The heat energy absorbed in process C is RT1 RT1 RT1 RT1 35. The heat energy absorbed in process C is RT1 RT1 e e RT1 RT1 e e

SOLUTION 31. Using the ideal gas equation PV = nRT, the volumes of the gas in states , and C are n RT n R 2T1 2 nRT1 V = = = P P1 P1 V =

n RT P

n R T1 2 P1

=

=

PV = nR

, we have

P=

V

1/ 2

nR V

V

=

nR

V

V

= 2 nR

W

V

= 2 2 nRT1

V

1 nRT1 2 P1

= 2 nR 2 PT 1 1 1 2

2nRT1 P1 = – 2 nRT1

2

= – 4 RT1. The negative sign indicates that heat is absorbed in 33. For a monoatomic gas Cv = 3R/2. The internal energy of the gas in the process is U) = n Cv T = n Cv T – T ) =2

or P V =

Q is given by

3R 2

U)

T 1 – 2 T 1) T1) = –3RT1

= 3R

1/ 2

nR V The work done in process or

=

2

P

V

= V

1 n RT1 2 P1

nRTC n R 2T1 nRT1 and VC = = = PC P1 2 P1 It is given that in the process , the pressure and temperature of the gas vary such that PT = where is a constant. Thus for point , we have = P T = P1 T 1) = 2 P 1T 1 32. For the process PV = nRT and PT = Eliminating T

V

W

+W

= – 3 RT1 – 4 RT1 = – 7 RT1 which

17.35

34. The process C takes place at a constant pressure P = 2 P1. Therefore, the work done in this process is W =P V P1 VC – V ) nRT1 P1

= 2P1

W

35. The process C takes place at a constant temperature T = 2 T1. Therefore, the work done in this P = nRT/V) V

W

V

= VC

1 nRT1 2 P1

VC

= nRT1 = 2 RT1

= nCv T1 = 2 Q

+W

U)

3R 2

V = 2. Therefore, VC

C is W

=2

2T1

R

= 4RT1loge

T1 = 3RT1

Questions 36 to 38 are based on the following passage Passage IX

log e T = 2 T 1)

= 0 as the temperature is constant.

U) Therefore Q

= 3 RT1 + 2 RT1 = 5 RT1

cyclic process starting from

V VC

= nRT log e

The change in the internal energy is process U) = nCv T = nCv TC – T ) T 1 – T 1) = nCv

nRT V

=

+W

U)

V V = 2 and V V R is the gas constant.

Given

= 4. The temperature T = 27°C.

as shown in Fig. 17.42.

IIT, 2001 36. The temperature of the gas at point

is

37. The total heat absorbed in the complete cycle is 38. The total work done by the gas in the complete cycle is

Fig. 17.42

SOLUTION 36. For process Hence V T

=

37. Process Hence Process

W

V T

V T = V

Q

U = 0. Hence Q = W.

, the plot of V versus T is linear.

T =2

300 K = 600 K

occurs at constant pressure =2

C

V

T.

5R T –T 2 C occurs at constant temperature. Q = U + W.

= nRT loge

VC V

= nRT loge

VD V

= nRT loge

VD V

=2 Q

C

R

V D = V C) V V

600 loge 4

= 1200 loge = 1200 loge

R R

1 2

17.36 Comprehensive Physics—JEE Advanced

Process C occurs at constant volume. Hence QC D = nCv T – T ) 3R =2 2 R Process Hence

= nRT loge

300

R

loge

= – 1200 loge

occurs at constant temperature. = WD

QD

=2

V VD

Questions 39 to 41 are based on the following paragraph Passage X 5 3 is trapped inside a liquid of density l

1 4

R

Total heat absorbed in the complete cycle is Q = 1500R + 1200 loge R R – 1200 loge R = 600R 38. In a cyclic process U = 0. Hence W = Q = 600R

40. When the gas bubble is at height its temperature is T0

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 IIT, 2008

h

0

from the bottom,

2/ 5

0 2/ 5

0

T0

H

0

H

0

T0

3/ 5

0 3/ 5

0

T0

H

0

41. The buoyancy force acting on the gas bubble is R is the universal gas constant) Fig. 17.43

39.

ancy force the following forces are acting on it

l nR

-

0

the pressure of the liquid

0

pressure of the liquid and force due to viscosity of the liquid

7/5

0

0

2 /5

H

0

SOLUTION

41.

39. 1

40. For an adiabatic process TP For T0

= 5/3, 0

1 H

Which gives T

=– 2/5

T0

= constant.

2 . Therefore , 5

=T

2/5

0

8/ 5

0

0

H

3 /5

0

2/5

P0

F = weight of liquid displaced , V = volume of the bubble. nRT From PV = we have V = Therefore, P nR F= P =

nR

=

T0

0

2/5

0

3 /5

H

0

H

0

3/ 5

P0

l

to viscosity of the liquid

2/5

H

0

2/5

0

h

0

nR

= 0

H

2/5

P0

)]3 / 5

17.37

IV Matching 1.

P – V) graph of a cyclic process of an ideal gas is shown in Fig. 17.44. Match the process listed in column I with the consequences listed in Column II Column I

Column II

P

>0 0 0 and = 0. In process , volume decreases and pressure increases. Hence < 0 and < 0. 2. 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. Column I Column II a valve. Chamber I contains an ideal gas and the

Fig. 17.45

its original volume such that its pressure P where V is the volume of the gas. its original volume such that its pressure P where V is its volume

1 V2

.

1 V 4/3

or remains constant

.

pressure P and volume V follow the behaviour shown in the graph in Fig. 17.46.

Fig. 17.46

IIT, 2008

17.38 Comprehensive Physics—JEE Advanced

SOLUTION = 0. Since the system is completely insulated, no heat can = – = 0, i.e.there is no

enter or leave the sysytem, i.e.

nRT Given PV2 = constant. PV = n RT or P = n RT/V. Therefore, PV2 = V is increased, T will decrease, i.e. is negative. nR. . Hence For PV , the work done = 1 +

= nCv = n Cv

nCp

Cp = Cv + For monoatomic gas Cv =

+

V 2 = nRTV. Hence T

1 . Thus if V V

nR 1 nR 1

R 1

3R and for = 2 2 3R R 3R R + = –R= 2 1 2 2 2 is negative and Cp is positive, is negative, i.e. the gas loses heat. Hence the cor-

Cp = Now

= nCp

. Since

PV4/3 = constant, Then T Putting Cp = Cv +

1 V

1/ 3

. Thus if V is increased, T will decrease, i.e.

R 3R 3R = – 3R = – , which is negative. Now 1 4/3 2 2

ture of the gas increases. Therefore, = + . Since both and are positive,

= nCp

is negative.

. Since Cp and

are both negative,

is positive. Now will be positive, i.e. the gas gains heat. Hence the correct

3. One mole of a monatomic gas is taken through a cycle as shown in the P-V II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I. IIT, 2011

Fig. 17.47

Column I C

Column II

17.39

C

SOLUTION is isobaric. Hence V T. Therefore T > T . U = nCv T = nCv T – T ). Since T < T , U is negative, i.e. internal energy decreases. Q = nCp T is also negative. Hence heat is lost. W = 3P V – V ) = – 6PV. Which is negative. Hence work is done on the gas. C is isochoric. Hence P internal energy decreases. W=P V V = 0)

T. Therefore T > TC. U = nCv T = nCv TC – T ) is negative, i.e. Q=

U + W),

Q=

U. Since

U is negative, heat is lost.

C is isobaric, i.e. V T. Hence TD > TC U = nCv TD – TC) is positive. Hence internal energy increases. Q = nCp TD – TC) is positive. Hence heat is gained by the gas. W = P V = P V – V) = 8 PV which is positive. So work is done by the gas.

gas is compressed, work is done on the gas, i.e. by the gas.

U = 0. Therefore Q = W. Since the W is negative. Hence Q is negative. Hence heat is lost

ANSWER

V Assertion-Reason Type Questions

following four choices out of which only one choice is correct. ment-2 is the correct explanation for Statement-1. ment-2 is the correct explanation for Statement-1.

Fig. 17.48

Statement-2 1. Statement-1 PV versus P graph for a cerFigure 17.48 shows T tain mass of oxygen gas at two temperatures T1 and T2. It follows from the graph that T1 > T2.

an ideal gas. 2. Statement-1 If two bodies of equal mass and made of the same material at different temperature T1 and T2 are

17.40 Comprehensive Physics—JEE Advanced

3.

4.

5.

6.

brought in thermal contact, the temperature of each T1 + T2)/2 when thermal equilibrium is attained. Statement-2 They have the same thermal capacity. Statement-1 Two vessels and of equal capacity are contains a gas at 0°C and 1 atmosphere pressure and vessel is evacuated. If the stopcock is suddenly and will be 0.5 atmosphere. Statement-2 If the temperature is kept constant, the pressure of a gas is inversely proportional to its volume. Statement-1 Two vessels and are connected to each other by a stopcock. Vessel contains a gas at 0°C and 1 atmosphere pressure and vessel is evaluated. The two vessels are thermally insulated from the surroundings. If the stopcock is suddenly opened, there will be no change in the internal energy of the gas. Statement-2 No transfer of heat energy takes place between the system and the surroundings. Statement-1 Two vessels and are connected to each other by a stopcock. Vessel contains a gas at 300 K and 1 atmosphere pressure and vessel is evacuated. The two vessels are thermally insulated from the surroundings. If the stopcock is suddenly opened, the expanding gas does no work. Statement-2 Since Q = 0 and U law of thermodynamics that W = 0. Statement-1 Heating system based on circulation of steam are

Statement-2 The latent heat of steam is high. 7. Statement-1 V-T graphs of a certain mass of an ideal gas at two pressures P1 and P2. It follows from the graphs that P1 is greater than P2.

Fig. 17.49

Statement-2 The slope of V-T graph for an ideal gas is directly proportional to pressure. 8. Statement-1 The curves and in Fig. 17.50 show P-V graphs for an isothermal and an adiabatic process for an ideal gas. The isothermal process is represented by curve .

Fig. 17.50

Statement-2 The slope of the P-V graph is less for an isothermal process than for an adiabatic process.

or circulation of hot water.

SOLUTION 1.

2.

is parallel to the P-axis. This means that PV/T is a constant, independent of pressure. Hence line corresponds to an ideal gas for which PV/T temperatures, a real gas behaves more like an ideal gas. Hence T1 is greater than T2. if the two bodies have the same thermal capacity which is equal to mass of the body heat capacity. Since the two bodies have the same

3.

4. 5.

mass and are made of the same material, they have the same thermal capacity. of equal capacity, the volume occupied by the gas is doubled when the stopclock is opened. Hence, pressure becomes half. mally insulated from the surroundings, no heat Q = 0. Since U W = 0.

17.41

6. 7. From PV = RT, we have PV R) T

an adiabatic process PV = constant. Differenting, we get + V =0 P V –1 P which gives =– = Cp/Cv V Since > 1, the slope of the P-V curve for an adiabatic process is greater than that for an isothermal process. Thus both the statements are true and statement-2 is the correct explanation for statement-1.

V R V 1 = , i.e. T P T P Hence Statement-1 is true but Statement-2 is false. 8. For an isothermal process PV = constant. DifferenP = – . For tiating we get + =0 V

VI Integer Answer Type 1. Two soap bubbles and are kept in a closed chamber where the air is maintained at pressure N/m2. The radii of bubbles and are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio

n /n , where n and n are the number of moles of air in bubbles and , respectively. [Neglect the effect of gravity.]

SOLUTION Using P V = n RT and P V = n RT, we have

1. If P0 is the external pressure, then P = P0 +

4 r

=8+

P = P0 +

4 r

=8+

4 0.04 2 10

2

4 0.04 4 10

2

= 16 N m–2 = 12 N m–2

n n

=

P P

V V

P P =

r r 12 16

3

V= 4 2

3

=6

4 r 3) 3

18

Kinetic Theory of Gases

Chapter

REVIEW OF BASIC CONCEPTS 18.1

18.3

PRESSURE EXERTED BY AN IDEAL GAS

The pressure of a gas in a container is a result of the continuous bombardment of the gas molecules against the walls of the container and is given by 2

2

1 m n vr ms 1 M vr ms 1 2 2U vr ms P= V 3 3 V 3 3V where m = mass of each molecule, n = number of molecules in the container, vrms = root mean square speed of molecules, V = volume of container, M = mass of gas in the container, = density of the gas and U = internal energy of the gas.

18.2

vrms =

v22

v32

vn2 )

1/ 2

where v1, v2, v3, vn are the speeds of the molecules 1, 2, 3, n respectively. In terms of P and , vrms is given by vrms =

3P

3k T m

The mean translational kinetic energy of a molecule of a gas is given by 1 mv2rms E= 2 In terms of E, the pressure of the gas is given by 2 nE 2 U 1 2 P= v rms = 3 V 3V 3 where U = nE is the total translational kinetic energy of all the n molecules of the gas. It is also called the internal energy of the gas. R Boltzman’s constant k = N0 where N0 is Avagdro number.

ROOT MEAN SQUARE SPEED 1 2 ( v1 n

MEAN TRANSLATIONAL KINETIC ENERGY

3 RT M

where m is the mass of each molecule, T the absolute temperature of the gas and k a constant called Boltzman’s constant. Its value is k = 1.38 10–23 J K–1 per molecule R is the universal gas constant and its value is R = 8.315 J K–1 mol–1

18.4

EQUATION OF STATE OF AN IDEAL GAS

The relationship between pressure P, volume V and absolute temperature T of an ideal gas is called the equation of state. For n moles of a gas, this relation is PV = nRT where R is the molar gas constant. From Avogadro’s law, it follows that one mole of all gases, at the same temperature and pressure, occupies gas occupies 22.4 litres at STP. Consequently, for on mole PV is constant for all gases. This constant is the the ratio T molar gas constant and can be evaluated as follows: At STP, V = 22.4 litre = 22.4 10–3 m3

18.2 Comprehensive Physics—JEE Advanced

P = 0.76 m Hg = 1.013 T = 273 K R=

18.5

PV T

105 Nm–2

1.013 105

22.4 10 273 = 8.315 J mol–1 K–1

According to Van der Waal the true pressure exerted by a gas is greater than P by an amount a/V 2 (where a is a constant) due to attractive forces between molecules and the true volume of the gas is less than V by an amount b (where b is another constant) because molecules themselves occupy a a

(V – b) = RT

V2 At high pressures, when the molecules are too many and too close together, the correction factors a and b both become important. But at low pressures, when they are not too many and not too close together, a gas behaves like an ideal gas and obeys the equation PV = RT.

18.6

DEGREES OF FREEDOM AND EQUIPARTITION OF ENERGY

The total number of coordinates or independent quantities tion of a dynamical system is called the degrees of freedom of the system. The molecules of a monoatomic gas consist of single atoms. Therefore, the molecules of a monoatomic gas have three degrees of freedom corresponding to translational motion. The molecules of a diatomic gas have degrees of freedom-three corresponding to translational motion and two for rotational motion. A polyatomic molecule has six degrees of freedom including one of vibrational motion. The law of equipartition of energy is stated as follows. In any dynamical system with a uniform absolute temperature T, the total energy is distributed equally among all the degrees of freedom and the average energy 1 kT, where per degree of freedom per molecule equals –23 –1 2 k = 1.38 10 J K .

If the molecules of a gas have f degrees of freedom, 1 kT. Therefore, then kinetic energy per molecule = f 2 kinetic energy per mole is 1 f U = Nf kT = RT 2 2 U f Now Cv = = R T 2

=

f 2

Cp

1 R

1

R

2 f

1

f R 2

Cv

f 2

f R+R= 2

Cp = Cv + R =

3

VAN DER WAAL’S EQUATION OF STATE

P

and

2 2 5 = for a monoatomic gas = 1 + 3 5 3 2 7 4 = for a diatomic gas = 1 + = for a triatomic 6 5 3 or polyatomic gas Thus

=1 +

Relation between Cp, Cv and R For a monoatomic gas;

Cv =

3R 5R and Cp = 2 2

For a diatomic gas;

Cv =

5R 7R and Cp = 2 2

For a polyatomic gas;

Cv = 3R and Cp = 4R

Relation between Cp, Cv and Cp – Cv = R =

Cp

Cp =

Cv

Cv – Cv = R

R

Cv =

Cp = Cv + R =

Cv 1

R 1

+R=

R 1

18.1 Calculate the root mean square speed of the molecules of hydrogen gas at S.T.P. Density of hydrogen at S.T.P. is 9 10–2 kg m–3. SOLUTION At S.T.P., pressure P = 1.01 = 9 10–2 kg m–3. vrms = =

105 Pa and density

3P 3 1.01 105 9 10

2

= 1840 ms–1

18.2 Calculate the temperature (in kelvin) at which the root mean square speed of a gas molecule is half its value at 0°C.

Kinetic Theory of Gases

SOLUTION

SOLUTION (a) Average translational K.E. of a molecule of an ideal gas is 3 E = kT, where T = temperature 2 in kelvin

3kT m

vrms = v rms = vrms

T T

1 = 2

T 273

T 273

T = 68.25 K

18.3 Find the mean translational kinetic energy of an oxygen molecule at 0°C. Given Avogadro number N = 6.03 1023 per mole and R = 8.3 JK–1 mol–1. SOLUTION 1 3 3 RT E = mv2rms = kT = 2 2 2 N E=

3 8.3 273 2

18.3

6.03 1023

= 5.64

10–21 J

18.4 Calculate the mean translational kinetic energy of 1 mole of hydrogen at S.T.P. Density of hydrogen at S.T.P. is 0.09 kg m–3. SOLUTION

At T = 0°C = 273 K, 3 E= (1.38 10–23) 2 = 5.65 10–21 J

273

At T = 100°C = 373 K, 3 E= (1.38 10–23) 2 = 7.72 10–21 J

373

(b) Number of molecules in 1 mole of a gas is N = 6.02 1023 K.E. of 1 mole at 273 K = (5.65 10–21) (6.02 1023) = 3.40 J K.E. of 1 mole at 373 K = (7.72 10–21) (6.02 1023) = 4.65 J 18.6 The speed of sound in a gas at S.T.P. is 330 ms–1 and the density of the gas is 1.3 kg m–3. Find the number of degrees of freedom of a molecule of the gas. SOLUTION

vrms =

3P

= 1.84 Mass of 1 mole is m = 22.4 = 2.016

3 1.01 105 0.09

=

103 ms–1 = 10

–3

m

0.09 kg m

–3

f = 10–3)

(1.84

v2 P

(330) 2

1.3

1.01 105

= 1.4

If f is the number of degrees of freedom, then

10–3 kg

1 K.E. = mv2rms 2 1 = (2.016 2 = 3.41

3

P

v =

103)2

103 J

18.5 Calculate (a) the average translational kinetic energy of the molecules of an ideal gas at 0°C and at 100°C and (b) the energy per mole of the gas at 0°C and 100°C. Given Avogadro’s number N = 6.02 1023 and Boltzmann’s constant k = 1.38 10–23 JK–1.

2 1

2 =5 (1.4 1)

18.7 The volume of 2 moles of a diatomic ideal gas at 300 K is doubled keeping its pressure constant. Find the change in the internal energy of the gas. Given R = 8.3 JK–1 mol–1. SOLUTION According to kinetic theory, there are no internal forces of interaction between the molecules of an ideal gas. This implies that the potential energy is

18.4 Comprehensive Physics—JEE Advanced

zero. Hence, for an ideal gas, the internal energy is only due to kinetic energy of the molecules. For n moles of an ideal gas at absolute temperature T, the internal energy is f nRT U= 2 where f is the number of degrees of freedom. For a diatomic gas f = 5. Therefore 5 U = nRT 2 Since pressure is kept constant, V T (Charles’ law), i.e. V1 V = 2 T1 T2 T2 =

V2 T1 = 2T1 = 600 K ( V1

U = U2 – U1 =

V2 = V1)

5 nR(T2 – T1) 2

5 2 8.3 2 = 1.245 104 J

=

(600 – 300)

Alternative Method U = nCv T 5R . Therefore 2 5R U= 2 (600 – 300) 2 = 1.245 104 J

For diatomic gas, Cv =

18.8 Hydrogen gas is contained in a vessel of volume 10 litres at 27°C. The gas pressure is 106 Pa. Find (a) total translational kinetic energy of hydrogen molecules, (b) mean (average) kinetic energy of the molecules and (c) total kinetic energy of the molecules. SOLUTION Volume of gas = 10 litres = 10

10–3 m3 = 10–2 m3) 3 3 106 (a) Total K.E. of molecules = PV = 2 2 10–2 = 1.5 104 J (b) Average K.E. of molecules 5 ( Hydrogen is diatomic) = kT 2

5 1.38 10–23 (273 + 27) 2 = 1.035 10–20 J 5 5 106 10–2 = 2.5 104 J (c) Total K.E. = PV = 2 2 =

18.9 n1 moles of a monoatomic gas are contained in a vessel A of volume V1 at pressure P1 and temperature T1. n2 moles of the same gas are contained in a vessel B of volume V2 at pressure P2 and temperature T2. The two vessels are now connected by a tube. Obtain the expression for (a) common temperature T and (b) common pressure P in the Vessels. SOLUTION If f = number of degrees of freedom, f f f n1RT1 + n2RT2 = (n1 + n2)RT 2 2 2 Here f = 3 (monoatomic gas) Equation (1) gives n T n2T2 T= 11 n1 n2 Final pressure is (n n2 ) RT PV P2V2 1 1 P= 1 V1 V2 V1 V2

18.7

IDEAL GAS LAWS

Boyle's Law At any given temperature, the volume of a given mass of a gas is inversely proportional to pressure, i.e. 1 V (T = constant and n = constant) P PV = constant P 1V 1 = P 2V 2 Charle's Law If the pressure of a gas is kept constant, the volume of a given mass of the gas is directly proportional to its absolute temperature, i.e. V T (P = constant and n = constant) V = constant T V1 V = 2 T1 T2

Kinetic Theory of Gases

If the volume of a gas is kept constant, the pressure of a given mass of the gas is directly proportional to its absolute temperature, i.e. P = constant (V = constant and n = constant) T P1 P = 2 T1 T2 Avogadro Law Equal volumes of all gasses at the same temperature and pressure contain an equal number of molecules. The number of molecules in one mole of any gas is N0 = 6.02 1023. N0 is called the Avogadro number. If volume V1 of one gas contains N1 molecules and volume V2 of another gas contains N2 molecules at the same temperature and pressure, then V1 V = 2 N1 N2 Equation of State in terms of Boltzmann constant (k) universal gas constant R Avogadro number N0 R PV = nRT = nN0 T = nN0kT N0 PV = NkT

k=

or where

N = nN0 = number of molecules

18.10 An electric bulb of volume 250 cm3 was sealed during manufacture at a pressure of 10–3 mm of Hg at 27°C. Find the number of air molecules in the bulb. SOLUTION Let N be the number of air molecules in the bulb. It is given that P = 10–3 mm of Hg = 10–4 cm of Hg, V = 250 cm3 and T = 273 + 27 = 300 K. Now PV = NkT (1) At STP, one mole of a gas occupies a volume V0 = 22400 cm3 and contains N0 = 6.02 1023 molecules, P0 = 76 cm of Hg and T0 = 273 K. (2) P0V0 = N0kT0 Dividing (1) by (2), we get N = N0 = (6.02 = 8.04

T0 T

P P0 1023)

V V0 273 300

1015 molecules

10 4 76

250 22400

18.5

18.11 A cylinder of volume 30 litres contains oxygen at a gauge pressure of 15 atm at 27°C. When some oxygen is ejected out from the cylinder, the gauge pressure falls to 11 atm and temperature falls to 17°C. Find the mass of oxygen ejected. R = 8.3 J K–1 mol–1 and molecular mass of oxygen = 32. SOLUTION Let N1 be the number of molecules, P1 be the pressure and T1 be the temperature of oxygen before some oxygen is ejected and N2, P2 and T2 be their values after the gas is ejected. Since the volume V remains unchanged (= volume of cylinder), (1) P1V = N1kT1 P2V = N2kT2

(2)

8.3

R N0

= 1.38 10–23 JK–1 per 6.02 1023 molecules is Boltzmann constant. Given, V = 30 litres = 30 10–3 m3, P1 = 15 atm = 15 1.01 105 Pa, T1 = 273 + 27 = 300 k, P2 = 11 atm = 11 1.01 105 Pa and T2 = 273 + 17 = 290 K. Substituting these values in Eqs. (1) and (2), we get N1 = 10.87 1024 and N2 = 8.25 1024 Number of molecules of oxygen ejected = N1 – N2 = 2.62 1024. The molecular mass of oxygen = 32, i.e. one mole of oxygen (which contains 6.02 1023 molecules) has a mass of 32 g = 0.032 kg. Hence 2.62 1024 molecules will have a mass of where k =

0.032

2.62 1024

6.02 1023

= 0.14 kg

18.12 A narrow glass tube of length 100 cm is closed at both ends. It lies horizontally with 20 cm of mercury column in the middle dividing the tube into two compartments I and II of equal length. The air in each compartment is at standard temperature and pressure. The tube is then turned to a vertical position. By what distance will the mercury column be displaced? SOLUTION Let A cm2 be the cross-sectional area of the tube. Then P0 = 76 cm of Hg and V0 = 40 A cm3 [Fig. 18.1(a)]. When the tube is turned to vertical position [Fig. 18.1(b)], Let x cm be the displacement of mercury column and let P1, V1 be the air pressure in compartment I and P2, V2 those in compartment II.

18.6 Comprehensive Physics—JEE Advanced

and

P1 =

P0V0 P 40 A = 0 V1 (40 x) A

40 P0 (40 x)

P2 =

P0V0 P 40 A = 0 V2 (40 x) A

40 P0 (40 x)

The mercury column will be in equilibrium if P2 (in cm of Hg) + 10 cm of Hg = P1(in cm of Hg), i.e. if 40 P0 40 P0 + 10 = 40 x (40 x) where P0 = 76 cm of Hg. Thus 40 76 40 76 + 10 = (40 x) (40 x)

Fig. 18.1

Using Boyle’s Law, we have

x2 + 608x – 1600 = 0 Solving for positive root, we get x = 2.6 cm

P0V0 = P1V1 and P0V0 = P2V2

I Multiple Choice Questions with Only One Choice Correct 1. In an adiabatic process, the root mean square speed of the molecules of a monoatomic gas becomes twice its initial value. The ratio of the initial volume (a) 2 (b) 23/2 (c) 4 (d) 8 2. The root mean square speed of hydrogen molecules at a certain temperature is v. If the temperature is doubled and the hydrogen gas dissociates into atomic hydrogen, the rms speed will become (a)

v 4

(b)

v 2

(c) 2v (d) 4v 3. A gas in a closed container has temperature T and pressure P. If the molecules of the gas undergo inelastic collisions with the walls of the container, then (a) both P and T will decrease (b) P decreases and T increases (c) P increases and T decreases (d) both P and T remain the same 4. Two moles of hydrogen are mixed with n moles of helium. The root mean square speed of the gas molecules in the mixture is 2 times the speed of sound in the mixture. The value of n is (a) 1 (b) 2 (c) 3 (d) 4

5. A vessel contains 4 moles of oxygen and 2 moles of argon at absolute temperature T. The total internal energy of the gas mixture is (a) 6 RT (b) 9 RT (c) 11 RT (d) 13 RT 6. The average translational kinetic energy of a molecule of a gas at absolute temperature T is proportional to 1 (b) T (a) T (c) T (d) T 2 7. The root mean square speed of the molecules of an enclosed gas is v. What will be the root mean square speed if the pressure is doubled, the temperature remaining the same? (a) v/2 (b) v (c) 2 v (d) 4 v 8. The mass of an oxygen molecule is about 16 times that of a hydrogen molecule. At room temperature the rms speed of oxygen molecules is v. The rms speed of the hydrogen molecule at the same temperature will be (a) v/16 (b) v/4 (c) 4 v (d) 16 v 9. The average kinetic energy of hydrogen molecules at 300 K is E. At the same temperature, the average kinetic energy of oxygen molecules will be

Kinetic Theory of Gases

(a) E/16 (b) E/4 (c) E (d) 4E 10. At room temperature (27°C) the rms speed of the molecules of a certain diatomic gas is found to be 1920 ms–1. The gas is (a) H2 (b) F2 (c) O2 (d) Cl2 IIT, 1984 11. The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K, the room mean square speed of the gas molecules is v, then at 480 K it will be (a) 4 v (b) 2 v v v (d) (c) 2 4 IIT, 1996 12. Three closed vessels A, B and C are at the same temperature. Vessel A contains only O2, B only N2 and C a mixture of equal quantities of O2 and N2. If the average speed of O2 molecules in vessel A is v1, that of N2 molecules in vessel B is v2, the average speed of O2 molecules is vessel C is 1 (v1 + v2) (b) v1 (a) 2 3k T (c) v1 v2 (d) M IIT, 1992 13. The average translational energy and the rms speed of molecules of a sample of oxygen gas at 300 K are 6.21 10–21 J and 484 ms–1 respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour) (a) 12.42 10–21 J, 968 ms–1 (b) 8.78 10–21 J, 684 ms–1 (c) 6.21 10–21 J, 968 ms–1 (d) 12.42 10–21 J, 684 ms–1 IIT, 1997 14. A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at a temperature 2T has a pressure of P (b) P (a) 8 (c) 2P (d) 8P IIT, 1997 15. A vessel contains a mixture of 1 mole of oxygen and two moles of nitrogen at 300 K. The ratio of

18.7

the rotational kinetic energy per O2 molecule to that per N2 molecule is (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) depends on the moment of inertia of the two molecules. IIT, 1998 16.

with air at temperature (T1, T2), volume (V1, V2) and pressure (P1, P2) respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be (a) T1 + T2 (b) (T1 + T2)/2 (c) T1T2 (P1V1 + P2V2)/(P1V1T2 + P2V2 T1)

(d) T1T2 (P1V1 + P2V2)/(P1V1T1 + P2V2T2) 17. If the temperature of a gas is increased from 27°C to 927°C, the root mean square speed of its molecules (a) becomes half (b) is doubled (c) becomes 4 times (d) remains unchanged 18. At what temperature will oxygen molecules have the same root mean square speed as hydrogen molecules at 200 K? (a) 527°C (b) 1327°C (c) 2127°C (d) 2927°C 19. An enclosure of volume V contains a mixture of 8 g of oxygen, 14 g of nitrogen and 22 g of carbon dioxide at absolute temperature T. The pressure of the mixture of gases is (R is universal gas constant) 3RT RT (b) (a) 2V V 5RT 7 RT (c) (d) 4V 5V 20. At what absolute temperature T is the root mean square speed of a hydrogen molecule equal to its escape velocity from the surface of the moon? The radius of moon is R, g is the acceleration due to gravity on moon’s surface, m is the mass of a hydrogen molecule and k is the Boltzmann constant mgR 2mgR (b) (a) 2k k 3mgR 2mgR (c) (d) 2k 3k

18.8 Comprehensive Physics—JEE Advanced

23. 0.014 kg of nitrogen is enclosed in a vessel at a temperature of 27 °C. The amount of heat energy to be supplied to the gas to double the rms speed of its molecules is approximately equal to (a) 6350 J (b) 7350 J (c) 8350 J (d) 9350 J 24. A vessel of volume V contains an ideal gas at absolute temperature T and pressure P. The gas is allowed to leak till its pressure falls to P . Assuming that the temperature remains constant during leakage, the number of moles of the gas that have leaked is V V (P + P ) (b) (P + P ) (a) RT 2 RT

21. Two perfect gases at absolute temperatures T1 and T2 are mixed. There is no loss of energy in this process. If n1 and n2 are the respective number of molecules of the gases, the temperature of the mixture will be (a)

n1T1 n2T2 n1 n2

(c) T1 +

(b)

n2 T n1 2

n2T1 n1T2 n1 n2 n1 T n2 1

(d) T2 +

22. An insulated box containing a diatomic gas of molar mass M is moving with a velocity v. The box is suddenly stopped. The resulting change in temperature is (R is the gas constant) (a)

M v2 2R

(b)

M v2 3R

(c)

M v2 5R

(d)

2M v 2 5R

(c)

V (P – P ) RT

V (P – P ) 2 RT

(d)

IIT, 2003 ANSWERS

1. 7. 13. 19.

(d) (b) (d) (c)

2. 8. 14. 20.

(c) (c) (c) (d)

3. 9. 15. 21.

(d) (c) (a) (a)

4. 10. 16. 22.

(b) (a) (c) (c)

5. 11. 17. 23.

(d) (b) (b) (d)

6. 12. 18. 24.

(c) (b) (d) (c)

SOLUTIONS 1. Let T1 be the initial temperature. Since vrms T, T2 = 4 T1. For an adiabatic process, T1 V1( – 1) = T2 V2( – 1) V1 V2 V1 V2

1

5 1 3

T2 = =4 T1 =4

V1 = (4)3/2 = 8 V2 3 RT ; M = molecular mass. Now T is M doubled and M is halved. Hence the rms speed will become twice, which is choice (c). 3. Since the temperature of the molecules is the same as that of the walls, there is no exchange of

energy in a collision. Hence the molecules rebound with the same average speed whether the collision is elastic or inelastic. Therefore, both P and T will not change. 3 RT ; M = molecular mass of mixture M The speed of sound in the mixture is

4. vrms =

RT Cp ; = of the mixture M Cv Given vrms = 2 v. Therefore

v=

3 RT M

2. vrms =

= =

2

RT M

3 2

For the mixture, Cv =

n1 Cv

1

n1

n2 Cv n2

2

Kinetic Theory of Gases

and

Cp = Cp Cv

=

n1 C p

n2 C p

1

n1 n1 C p

n2 C p

2

(1)

n2 (C v ) 2

For hydrogen, n1 = 2, (Cv)1 = For helium, n2 = n, (Cv)2 =

n2

1

n1 (C v )1

2

3R 5R and (Cp)1 = 2 2

5R 7R and (Cp)2 = 2 2

12. For a given M, vav T only. Therefore, the correct choice is v1 as the temperatures of vessels A and C are the same. 13. E T and vrms T . Hence at 600 K, 600 E = 6.21 10–21 300 = 12.42 10 –21 J and

8. v = vh = Thus or

3 kT / m0 and

3kT / m . Therefore, v0 = 3kT / mh . v0 mh = = vh m0 v h = 4 v 0.

1 16

1 4

Hence the correct choice is (c). 9. The correct choice is (c) as explained above. 10. We know that vrms =

= 2 10 –3 kg = 2 g Since M = 2, the gas is hydrogen. 3RT or vrms M choice is (b).

11. vrms =

T . Hence the correct

684 ms–1

1 mole RT and V 1 mole R 2T (P)He = V P He = 2 or (P)He = 2 (P)O2 PO (P)O2 =

2

15. Since both the gases are diatomic, each has two degrees of freedom associated with rotational motion. According to the law of equipartition of energy, the rotational kinetic energy per degree of freedom is (1/2)kT. Since the temperatures of the two gases are equal, their rotational kinetic energies will be equal. Hence the correct choice is (a). 16. According to the kinetic theory, the average kinetic 3 kT. Let n1 energy (KE) per molecule of a gas = 2 and n2 be the number of moles of air in vessels 1 and 2 respectively. Before mixing, the total KE of molecules in the two vessels is 3 3 E1 = n1kT1 + n2 kT2 2 2 3 = k(n1T1 + n2T2) 2 After mixing, the total KE of molecules is

3 RT which gives M

3 RT 3 8.3 300 = M = v2 1920 2 r ms

600 300

14. For a gas, PV = nRT. Hence

U =

3 6. E = kT. So the correct choice is (c). 2 7. The rms speed = 3kT / m which is independent of pressure. Hence the correct choice is (b).

vrms = 484

which is choice (d).

Using these values and = 3/2 in Eq. (1) and solving, we get n = 2. 5. The internal energy of n moles of a gas temperature T is given by f (nRT) 2 where f = number of degrees of freedom. For oxygen f = 5 and for argon f = 3. Hence U = U1 + U2 5 3 = (4RT) + (2RT) 2 2 = 13RT

18.9

E2 =

3 (n1 + n2) kT 2

where T is the temperature when equilibrium is established. Since there is no loss of energy (because the vessels are insulated), E2 = E1 or 3 3 (n1 + n2)kT = k (n1T1 + n2T2) 2 2 or

T=

n1T1 n2T2 (n1 n2 )

(1)

18.10 Comprehensive Physics—JEE Advanced

Now P1V1 = n1RT1 and P2V2 = n2RT2 which give PV P2V2 1 1 and n2 = RT1 RT2 Using these in Eq. (1) and simplifying, we get

20. The root mean square speed is given by

n1 =

T1T2 ( PV P2V2 ) 1 1 T= ( PV P2V2T1 ) 1 1T2 17. The root mean square speed is given by 3kT m i.e. vrms T. Initial temperature T1 = 27 + 273 = 300 K. Final temperature T2 = 927 + 273 = 1200 K. Since temperature is increased by 4 times, the speed is doubled. Hence the correct choice is (b). vrms =

18. For oxygen : vrms = For hydrogen : vrms =

3 kT0 m0 3kTh mh

m0 Th = 16 200 = 3200 K = 2927°C mh Hence the correct choice is (d).

or T0 =

19. The pressure exerted by a gas is given by nRT P= V mass RT = molecular weight V 8 RT 1 RT = 32 V 4 V

14 RT 1 RT = 28 V 2 V Pressure exerted by carbon dioxide Pressure exerted by oxygen P2 =

P3 =

22 RT 1 RT = 44 V 2 V

From Dalton’s law of partial pressures, the total pressure exerted by the mixture is given by P = P1 + P2 + P3 1 RT 1 RT 1 RT + + 4 V 2 V 2 V 5RT = , which is choice (c). V =

3kT m

The escape velocity is given by ve =

2gR

For vrms = ve, we require 3kT m

2 gR or T =

2mg R , which is choice (d). 3k

21. Average kinetic energy per molecule of a perfect 3 gas = kT. 2 3 n1 kT1 2 Average KE of molecules of the second gas 3 = n2 kT2 2 =

Total KE of the molecules of the two gases before they are mixed is

The value of vrms will be the same if T0 Th = m0 mh

Pressure exerted by oxygen P1 =

vr ms =

3 3 n kT + n kT 2 1 1 2 2 2 3 = (n1 T1 + n2 T2)k (1) 2 If T is the temperature of the mixture, the kinetic energy of the molecules (n1 + n2) in the mixture is 3 K = (n1 + n2)kT (2) 2 Since there is no loss of energy K = K . Equating (1) and (2) we get n1T1 n2T2 n1 T1 + n2 T2 = (n1 + n2) T or T = n1 n2 Hence the correct choice is (a). 22. Let n be the number of moles of the gas in the box. 1 The kinetic energy of the gas = n M v 2 . When 2 the box is suddenly stopped, this energy is used up in changing the internal energy, as a result of which the temperature of the gas rises. The change in internal energy is given by 5 U = nCv T = n R T 2 5 For a diatomic gas Cv = R. 2 1 5R T=n Mv2 Hence n 2 2 K=

or

T=

M v2 , which is choice (c) 5R

Kinetic Theory of Gases

23. The root mean square speed is related to absolute temperature T as crms =

5R 2

U = 0.5

3kT m

18.11

(1200 – 300)

0.5 5 8.314 900 = 9353 J. 2 So the correct choice is (d) =

For a given gas, m crms T. Hence in order to double the root mean square speed, the absolute temperature must be increased to four times the initial value. Initial temperature T1 T2 = 4 T1 = 1200 K. V = 0. Hence the heat energy supplied to the gas does no work on the gas; it only increases the internal energy of the molecules. The increase in internal energy is U = n Cv T 5R Since nitrogen is diatomic Cv = . The number 2 of moles of nitrogen in the vessel is

PV . Let RT n be the number of moles of the gas that leaked till the pressure falls to P . Since volume V of the vessel cannot change and temperature T remains constant during leakage, we have

24. Number of moles present initially is n =

n =

PV RT

Number of moles that leaked is n=n–n =

V PV PV – = (P – P ) RT RT RT

So the correct choice is (c).

mass in kg 0.014 103 = = 0.5 n= molecular mass 28

II Multiple Choice Questions with One or More Choices Correct 1. The root mean square speed of the molecules of a gas depends upon (a) the pressure of the gas (b) the density of the gas (c) the temperature of the gas (d) the mass of a molecule of the gas 2. The average translational kinetic energy of a molecule of a gas at absolute temperature T is E and the root mean square speed of the molecules is v. Then (a) E T (b) E T (c) v T (d) v T 3. The root mean square speeds of the molecules of hydrogen, oxygen and carbon dioxide at the same temperature are vh, vo and vc respectively. Then (b) vh > vo (a) vh = vo = vc (c) vo > vc (d) vh < vo < vc 4. One mole of oxygen at 27 °C is enclosed in a vessel which is thermally insulated. The vessel is moved with a constant speed v and is then suddenly stopped. The process results in a rise of temperature

of the gas by 1 °C. Then, if M = molecular mass of oxygen. (a)

(= Cp /Cv) = R

(c) v =

M(

1)

5 3

(b)

=

(d) v =

7 5 2R M ( 1)

IIT, 1983, 1996 5. In a vessel a gas at temperature T has a pressure P. The density of the gas is and vrms is the average root-mean-square speed of a molecule. If N is the number of molecules per unit volume and m the mass of a molecule, then (k = Boltzmann constant) (a) P =

1 2 vrms 3

(b) P =

2 2 vrms 3

3kT (d) P = N kT m 6. N molecules of a gas, each of mass m, strike per A at an angle to the vertical and rebound with a speed v. (c) vrms =

18.12 Comprehensive Physics—JEE Advanced

The collisions are assumed to be elastic. (a) The magnitude of change in momentum is | p| = 2mv sin (b) | p| = 2mv cos | p| (c) The pressure exerted on the wall is P = NA N | p| (d) P = A 7. The pressure P of n moles of a gas varies with volume V as P = a – bV2

of the piston from the base of the cylinder is h1 = 20 cm as shown in Fig. 18.2. When the temperature is raised to T2 = 177 °C, the new height of the piston above the base becomes h2. The system is then insulated from the surroundings and the piston is brought back to its original height. The new equilibrium temperature becomes T3. Given (1.5)0.4 = 1.18. IIT, 2004

where a and b are positive constants. The highest absolute temperature to which the gas can be heated is Tmax. (a) Tmax = (b) Tmax =

8.

a3 / 2

;

nR 3b

R = gas constant

2a 3 / 2 3nR 3b 1/ 2

(c) At Tmax, V =

a b

(d) At Tmax, V =

a 3b

1/ 2

Fig. 18.2

gas at temperature T1 = 27 °C. Initially the height

(a) h2 = 30 cm (c) T3 = 81°C

(b) h2 = 24.5 cm (d) T3 = 258 °C

ANSWERS AND SOLUTIONS 1. vr ms =

3P

3kT . Hence all the four choices m

are correct. 3 3kT 2. E = kT and vr ms = . Hence the correct 2 m choices are (a) and (d). 3. The rms speed is the maximum for the gas which is the lightest. Hence the correct choices are (b) and (c). 4. Oxygen is diatomic; it has 5 degrees of freedom. Therefore, Cv = 5 R/2 and Cp = 7 R/2. So = Cp /Cv = 7/5. The kinetic energy of oxygen molecules with a 1 velocity v0 = M v2, where M = molecular weight of oxygen. 2 Now heat energy = Cv dT = Cv 1 = Cv Cp R But Cp – Cv = R or –1= Cv Cv or

R ( – 1) = or Cv = Cv

R 1

Therefore, or

1 M v2 = 2 v=

R 1 2R

M

1

So the correct choices are (b) and (d). 5. The average pressure of a gas is given by 1 v 2rms P= 3 where density = mN; here m is the mass of a molecule and N is the number of molecules per unit volume of the gas. Thus 1 2 mN vrms (1) P= 3 The average kinetic energy of a molecule is given by 3 1 mv 2r ms = kT (2) 2 2 Using (2) in (1), we get P = NkT Thus the correct choices are (a), (c) and (d).

Kinetic Theory of Gases

6. Refer to Fig. 18.3. Since the collision is elastic, the speed v of the molecule before collision = speed after rebound. Therefore, the magnitude of the change in momentum normal to the wall is | p| = mv cos – (– mv cos ) = 2 mv cos

or

a 3b

1/ 2

a a 3b

or or

b

1/ 2

a

a 3b

= nR Tmax

a 3

= nR Tmax

2a 3 / 2

Tmax =

18.13

(4)

3 3b nR d 2T

It is easy to check that

is indeed negative at dV 2 a = 3 bV . Hence Eq. (4) gives the maximum value of T. So the correct choices are (b) and (d). 8. Let A be the cross-sectional area of the base of the cylinder. Given T1 = 273 + 27 = 300 K and T2 = 273 + 177 = 450 K. At constant pressure, we have from Charles’ law, V1 V2 = T1 T2 2

Fig. 18.3

Number of collisions per second = N. Therefore, time interval between successive collisions is 1 N Force exerted on the wall by the molecules is = rate of change of momentum t=

F =

| p| t

F Pressure P = . Hence the correct choices are (b) A and (c). 7. For n moles of an ideal gas, PV = nRT Using the given relation between P and V, we have (a – bV2) V = nRT or aV – bV3 = nRT (1) Differentiating with respect to V, we get dT dV

a – 3 bV2 = nR

dT 1 = (a – 3 bV2) dV nR

or

a 3b

1/ 2

b

a 3b

3/ 2

= nR Tmax

or

Ah1 Ah2 = T1 T2 h2 =

T2 T1

h1

(2)

450 K × 20 cm 300 K = 30 cm

Since the system is completely insulated from the surroundings, it cannot take heat from or give heat to the surroundings. Hence the second process is adiabatic for which T 2V 2(

d 2T dT = 0 and is negative. T will be maximum if dV 2 dV dT = 0 in Eq. (2), we get Setting dV 1/ 2 a 2 (3) a = 3 bV or V = 3b Using (3) in (1) we get a

or

–1)

= T 3V 3(

–1)

where V2 = Ah2 and V3 = Ah1. Also, for a diatomic gas = 1.4. Therefore, ( 1) V2 T3 = T2 × V3 = 450 ×

30 20

(1.4 1)

= 450 × (1.5)0.4. = 450

1.18 = 531 K = 258 °C

So the correct choices are (a) and (d).

18.14 Comprehensive Physics—JEE Advanced

III Multiple Choice Questions Based on Passage Questions 1 to 5 are based on the following passage Passage I Kinetic Theory of Gases The molecules of a gas move in all directions with various speeds. The speeds of the molecules of a gas increase with rise in temperature. During its random motion, a fast molecule often strikes against the walls of the container of the gas. The collisions are assumed to be perfectly elastic, i.e. the molecule bounces back with the same speed with which it strikes the wall. Since the number of molecules is very large, billions of molecules strike against the walls of the container every second. These molecules exert a sizeable force on the wall. The force exerted per unit area is the pressure exerted by the gas on the walls. According to the kinetic theory, the pressure of a gas of density at absolute temperature T is given by 1 2 vr ms 3 where vr ms is the root mean square speed of the gas molecule and is given by P=

vr ms =

3kT m

where m is the mass of a molecule and k is Boltzmann constant. 1 2 1. From the relation P = vr ms , 3 dimensions of pressure are (a) ML–1T–1 (b) ML–1T–2 –2 –1 (c) ML T (d) ML–2T–2.

3kT , it follows that the m constant k should be expressed in units of (a) newton per metre per kelvin (b) newton per kelvin (c) joule per kelvin (d) joule per kilogram per kelvin 3. Choose the only correct statement from the following. (a) The pressure of a gas is equal to the total kinetic energy of its molecules per unit volume of the gas. (b) The product of pressure and volume of a gas is always constant. (c) The average kinetic energy of the molecules of a gas is proportional to its absolute temperature. (d) The root mean square speed of a molecule is proportional to the absolute temperature of the gas. 4. If the temperature of a gas is increased from 27°C to 927°C, the root mean square speed of its molecules (a) becomes half (b) is doubled (c) becomes four times (d) remains unchanged 5. The root mean square speed of oxygen gas molecule at T = 320 K is very nearly equal to (the molar mass of oxygen is M = 0.0320 kg per mole and gas constant R = 8.31 J mol–1 K–1) (a) 300 ms–1 (b) 500 ms–1 (c) 700 ms–1 (d) 900 ms–1 2. From the relation vr ms =

ANSWERS AND SOLUTIONS 1. Dimensions of P = dimensions of density sions of (velocity)2 = ML–3

dimen-

(LT–1)2 = ML–1T–2

Hence the correct choice is (b). 2. Squaring we have k=

2 mvrms unit of energy = 3T unit of temperature = joule per kelvin

3. The correct choice is (c) 4. The correct choice is (b) since vr ms is proportional to the square root of absolute temperature 5. vrms =

3kT = m

3RT = M

3 8.31 320 0.032

= 499.3 ms–1 Hence the correct choice is (b)

Kinetic Theory of Gases

Questions 6 to 12 are based on the following passage Passage II Van der Waals Equation of State The equation of state PV = nRT holds for an ideal gas. The behaviour of real gases shows departures from an ideal gas behaviour especially at high pressures. The model of an ideal gas is based on a number of assumptions. Van der PV = nRT by taking into account two of those assumptions which may not be valid. He argued that (i) the volume of the molecules may not be negligible compared to the volume V occupied by the gas and (ii) the attractive forces between the molecules may not be negligible. He said that pressure P in equation PV = nRT is less than the true pressure by an amount p because of attractive forces between the molecules. According to him, the pressure ‘defect’ p is inversely proportional to the square of volume, i.e. 1 p V2 a or p= 2 V where a is constant depending on the nature of the gas. a Thus the true pressure of the gas is P = P + p = P + 2 . V He further argued that V is not the true volume of the volume. According to him, the true volume of the gas is V = (V – b) where b is a factor depending on the actual volume of the molecules themselves. Thus Van der Waals’ equation for real gases is P V = nRT, i.e. a P V b = nRT V2

18.15

At high pressures, when the molecules are too many and too close together, the correction factors a and b become important. 6. In Van der Waals’ equation of state for real gases, the product PV has the same dimensions as those of a (a) bP (b) V ab (c) (d) nRT V2 7. The dimensions of a are the same as those of (a) PV (b) PV2 (c) P2V (d) P/V 8. The dimensions of b are the same as those of (a) P (b) V (c) PV (d) nRT a 9. The dimensions of are the same as those of b (a) work (b) force (c) pressure (d) power 10. The dimensional formula for ab is (a) ML2T–2 (b) ML4T–2 6 –2 (c) ML T (d) ML8T–2 11. The correction factors a and b depend upon (a) the pressure of the gas (b) the volume of the gas (c) the temperature of the gas (d) the nature of the gas 12. The equation of state PV = nRT holds if the gas has (a) low pressure and low density (b) low pressure and high density (c) high pressure and low density (d) high pressure and high density.

ANSWERS AND SOLUTIONS 6. From Van der Waals’ equation, we have a ab = nRT V V2 From the principle of homogeneity of dimensions, PV – bP +

a

= dimensions of P V2 Dimensions of a = dimensions of PV2. Hence the correct choice is (b).

7. Dimensions of

8. Dimensions of b = dimensions of V, which is choice (b)

dimensions of PV 2 a = dimensions of V b = dimensions of PV = dimensions of work Hence the correct choice is (a). 10. Dimensions of ab = dimensions of PV2 dimensions of V = dimensions of (PV3) 9. Dimensions of

= ML–1T–2 (L3)3 = ML8T–2 Hence the correct choice is (d). 11. The correct choice is (d). 12. The correct choice is (a).

18.16 Comprehensive Physics—JEE Advanced

Questions 13 and 14 are based on the following passage Passage III Mean Free Path During their random motion, the molecules of a gas often come close to each other. When the distance between two molecules is comparable with the diameter of a molecule, the forces between them become very strong. As a result, their individual momenta before and after the encounter are different. When this happens a ‘collision’ is said to have occurred. The average distance a molecule travels before it suffers a collision with another molecule is called the mean free path (lc), which can be estimated as follows. Suppose the average speed of a molecule of diameter d is v . In one second, this molecule sweeps out a volume d2 v this volume, it will suffer collisions with them. If n is the number density (i.e. number per unit volume) of the molecules, then the number of molecules in this volume = d2 v n. The number of collisions per second = c = d2 v n. Therefore, the average time between two collisions (called collision period Tc) is

Tc =

1

=

1

d 2 vn Hence the mean free path (i.e. the average distance the molecule travels between two successive collisions) is c

lc = v Tc =

1

d 2n For air at S.T.P. the value of lc 3 10–7 m. 13. The mean free path of a molecule of a gas depends (a) only on its diameter (d) (b) only on the number density (n) of the molecules (c) on both d and n (d) neither on d nor on n. 14. The average collision period in a gas (a) increases if the pressure is increased (b) decreases if the pressure is increased (c) increases if the temperature of the gas is increased. (d) decreases if the temperature of the gas is increased.

ANSWERS AND SOLUTIONS 13. The mean free path is given by lc =

1

d 2n Hence the correct choice is (c). 14. The collision period is given by 1 Tc = d 2 vn

If the pressure is increased, the volume of the gas decreases. Hence number density n increases. Therefore, Tc will decrease. If the temperature of the gas is increased, the average speed v of the molecules increases. Hence Tc will decrease. Thus the correct choices are (b) and (d).

19

Transmission of Heat

Chapter

REVIEW OF BASIC CONCEPTS 19.1

THERMAL CONDUCTIVITY

If a steady temperature difference (T1 – T2) is to be maintained between the ends of a rod, heat must be supplied at a steady rate at one end and the same must be taken out at the other end. Suppose an amount of heat Q t so that Q/t is the rate of (i) Q/t will be proportional to the area A of the crosssection of the rod. (ii) Q/t will be proportional to (T1 – T2). (iii) Q/t will be inversely proportional to l, the distance between ends of the rod.

H=

Q t

A T1 T2 l

or

k A T1 T2 Q = l t

where k is a constant of proportionality called the thermal conductivity of the substance. It is a measure of how stance. In the SI system, k is expressed in Js–1 m–1 °C–1 or J s–1 m–1 K–1 or Wm–1 K–1. –3 –1 K ] [k

19.2

CONDUCTION THROUGH A COMPOSITE SLAB

Fig. 19.1

temperatures T1 and T2 (T1 > T2 Q1 t Q2

and

t

= =

T2 )

(1)

L k 2 A 2 (T1

T2 )

(2)

L

(A1 + A2

L. If keq is the equivalent

the composite slab is k eq ( A1 Q = t Since

A2 ) (T1 L

T2 )

(3)

Q1 Q 2 Q = + t t t keq (A1 + A2) = k1 A1 + k2 A2

Case 1. Two slabs placed one on top of the other Suppose we have a composite slab made up of two different slabs of materials of thermal conductivities k1 and k2, and L cross-sectional areas A1 and A2

k 1 A1 (T1

keq = If A1 = A2, then keq =

k 1 A1 A1

k 2 A2 A2

1 (k1 + k2). 2

(4)

19.2 Comprehensive Physics—JEE Advanced

Case 2. Two slabs placed in contact one after the other L1 and L2 but of the same cross-sectional area A placed in contact as

T0 =

19.3

( k 1 T1

k 2 T2 )

(k 1

k 2)

THERMAL RESISTANCE

slab becomes much easier if we use the concept of thermal

R=

V I

R=

temperature difference rate of flow of heat

R=

(T1 T2 ) L k A(T1 T2 )

Fig. 19.2

temperatures T1 and T2 (T1 > T2). Let T0 be the temperature Q1 Q2 of the junction. In the steady state, = , i.e. t t k 1 A(T1

T0 )

L1 k1 L1

=

(T1 – T0) =

T0 =

k 2 A(T0

T2 )

potential difference , e

Q/t

posite slab is Req = R1 + R2

(T0 – T2)

L2

T2

L kA

L2 k2

T1

k 1 T1

k 2 T2

L1 k1

L2 k2

L1

L2

L1 L2 = L1 keq A k1 A (5) keq =

L2 k2 A

k 1k 2 ( L1 ( L1k2

L2 ) L2 k1 )

(b) If the two slabs are joined in parallel as shown in Q1 Q = t t

or

Q2

1 1 = R1 R

t

k 1 A(T1 T0 ) Q = L1 t

(6)

k 1k 2 A(T1 T2 ) Q = k 1 L2 k 2 L1 t

(7)

keq = If L1 = L2, keq =

k 1k 2 ( L1 k 1 L2

2 k 1k 2 (k 1

k 2)

and

L2 ) k 2 L1

A2 )

L

L1 + L2) but its cross-sectional area is A. If keq is the equivalent thermal conductivity of the composite slab, Q k A (T1 T2 ) = eq t ( L1 L2 )

keq ( A1

1 R2

(8)

keq =

( k 1 A1 ( A1

=

k1 A1

k2 A 2

L

L

k 2 A2 ) A2 )

19.1 is maintained at 100°C and the other end is immersed minutes. Calculate the thermal conductivity of metal. –1 . SOLUTION Given L = 0.3 m, A = r2 = (10–2)2 = T1 = 100°C and T2 = 0°C. Now

10–4 m2,

19.3 –1

Q = mLf = 31.4 = 31.4 Also

80 cal 80 4.2 J (

1 cal = 4.2 J)

k A (T1 T 2 ) Q = L t k= =

Fig. 19.3

SOLUTION

QL t A(T1 T2 ) (31.4 80 (5 60)

= 336 W m

0.3

10 4 )

( –1

4.2)

K

In the steady state,

(100

0)

Q1

k1 A1 (T1

–1

L1

19.2 A cylindrical metal boiler of radius 10 cm and on an electric heater. If the water boils at the rate –1 102 W m–1 K–1 and latent heat of vaporisation = 2.26 103 –1.

50

0.02

t

=

T0 )

Q2 t =

, i.e. k2 A 2 (T 0

T2 )

L2

(300 T0 ) 400 0.01 (T0 = 0.1 0.2 T0 = 100°C

0)

19.4 A steel rod (L1 = 10 cm, A1 = 0.02 m2, k1 = 50 W m–1 K–1) and a brass (L2 = 10 cm, A2 = 0.02 m2, k2 = 110 W m–1 K–1 ends of the composite rod are maintained at 403 K

SOLUTION Q = mLv Q t

–1

= 50

)

(2.26

2.26

103

–1

)

103 Js–1

Base area of boiler A = r2 = (0.1)2 = 10–2 m2 L = 3.14 cm = 3.14 10–2 m k A (T f T w ) Q = L t 3 50 2.26 10 = (1.13 102 )

(

10 2 )

3.14 10

(T f

Tw)

2

Tf – Tw = 1000°C Tf = 1000 + Tw = 1000 + 100 = 1100°C 19.3 A steel rod (L1 = 10 cm, A1 = 0.02 m2 and k1 = 50 J s–1 m–1 K–1) is welded to a silver rod (L2 = 20 cm, A2 = 0.01 m2, k2 = 400 J s–1 m–1 K–1 junction in the steady state.

Fig. 19.4

SOLUTION 1 1 (k1 + k2) = (50 + 110) 2 2 = 80 W m–1 K–1

keq =

k eq ( A1 A 2 ) (T1 T2 ) Q = 0.1 t 80 0.04 (403 273) = 0.1 = 4.16 103 Js–1 19.5 A metal cylinder of radius r and thermal conductivity k1 = 2k is surrounded by a cylindrical metallic shell of inner radius r and outer radius 2r conductivity k2 = k posite system are maintained at constant temperatures

19.4 Comprehensive Physics—JEE Advanced

T1 and T2 (T1 > T2). Find the equivalent thermal conductivity of the system.

Fig. 19.6 Fig. 19.5

SOLUTION L and cross-sectional area A1 = r2 of conductivity k1 and a rod of the same L and cross-sectional area A2 = [(2r)2 – r2] 2 = 3 r and conductivity k2 placed one on top of the other (in parallel). In the steady state Q1 Q2 Q = + t t t =

k1 A1 (T1

T2 )

k2 A 2 (T1

L

k1 ( r 2 ) (T1 Q = L t

Consider a small element of thickness dr at a distance r from the axis of the shell. Let dT be the temperature between the inner and outer surfaces of the element. If H 2 rL dr dr 2 kL dT = r H H=

r2

T2 )

r1

L T2 )

k2 (3 r 2 ) (T1

T2 )

L

ln

(i)

T2 )

(ii)

4 keq = k1 + 3 k2 keq =

k1

3k2 2k 3k 5k = = 4 4 4

SOLUTION

1

=

4 cm 2 cm

=

2 kL H

(T2 – T1)

2 3.14

.3 0.5 H

H = 3.14

19.4

(100 – 0)

.3 0.5 100 H 4 10 Js–1

CONVECTION

In convection heat is transferred by the physical movement of matter by molecules via collisions in their local regions.

19.5 19.6 A cylindrical metallic shell has inner radius r1 = 2 cm L = 50 and outer radius r2 T1 = 0°C and T2 –1 –1 K from the outer to the inner surface.

T

2 3.14

Cross-sectional area of the composite system is A = (2r)2 = 4 r2. If keq is the equivalent conductivity, then keq (4 r 2 ) (T1 Q = L t

dT

dr 2 kL 2 = dT r H T r2 r1

ln

k

RADIATION

All bodies emit heat from their surfaces at all temperaradiant heat or thermal radiation. (1) Black Body A perfect black body is one which

emitter of radiations. Consequently, a black body, when

19.5

black body radiation. (2) Emissive Power of a (3) Absorptive Power

e) of a body is area SI unit of e is Js–1 m–2 or Wm–2.

lo

e

T

T T0 ) T2 1

emissive power to the absorptive power is the same for all substances, i.e. e = constant a (5) Stefan’s Law by a unit area of a black body is proportional to the fourth power of its absolute temperature, i.e. E T 4 or E = T 4 where is a constant known as Stefan’s constant. Its value is = 5.735 10–8 Wm–2 K–4 When a black body at absolute temperature T is surround by another black body at a lower absolute temperature, T0,

=

t =K

T1 T2

e

T0 T0

ms k

K=

where by a completely absorbs all radiations, a = 1 for a black body. (4) Kirchhoff’s Law

T1 and T2, we have k t, ms

T1 to T2 when placed in a medium of temperature T0. An approximate formula is T1

T2

1 T1 T2 T0 t K 2 (7) Wien’s Displacement Law As the temperature of a black body increases, the maximum intensity of emission =

other words, mT

= b = constant

where m takes place at absolute temperature T constant b is b 10–3 mK (metre kelvin) 19.7 2

E=

4

(T –

T40)

a perfect black body, then E= (T 4 – T 40) where emissivity is always less than unity. power to that of the black body. (6) Newton’s Law of Cooling by a body is directly proportional to the excess of its temperature difference is small, i.e. dQ (T – T0) dt where T is the temperature of the body and T0 that of the m is the mass of the body, s heat and dT dt, then dQ = msdT dQ dT = ms = – k (T – T0) dt dt

If his skin temperature is 37°C, calculate the rate at 0.8 and Stefan’s constant

= 5.7

SOLUTION T = 273 + 37 = 310 K and T0 = 273 + 27 = 300 K (T4 – T04)

Rate of loss of heat = A = 0.8

2

10–8) [(310)4 – (300)4]

(5.7 102 W

= 1.03 19.8

constant b

10–3 mK.

SOLUTION

where – k

m

dT (T T0 ) =

k dt ms

10–8 W m–2 K–1.

T=

b m

=

T = b. 2.8

0

475.3 10

3

= 6080 K

19.6 Comprehensive Physics—JEE Advanced

If the temperature at the end of next 6 minutes is T, then

19.9 A body cools from 80°C to 50°C in 6 minutes in a room where the temperature is 20°C. What is the temperature of the body at the end of next 6 minutes?

e

50 20 T 20

=6K

30 T 20

e

(ii)

From (i) and (ii) SOLUTION e

80 50

20 20 e

e

=6K

(2)

30 = 2(T – 20)

(2) = 6 K

(i)

T = 35°C

I Multiple Choice Questions with Only One Choice Correct 1. their ends kept at the same temperatures 1 and 2 with 2 > 1. If A1 and A2 are their cross-sectional areas and k1 and k2 their thermal conductivities, same if A1 k1 = (a) A2 k2 (c)

A1 k1 = A2 k2

1

(b)

A1 k2 = A2 k1

(d)

A1 k2 = A2 k1

2

(a)

k1 = k2

1 s1

k1

3k2 4

(d)

k1 k2 k1 k2 3k1

2 s1

k1 1 = (d) k1 = k2 k2 2 4. A slab of stone of area 0.34 m2 and thickness 10 cm is exposed on the lower face to steam at 100°C. A block of ice at 0°C rests on the upmelted. Assume that the heat loss from the sides –1 . What is the thermal conductivity 3.4 105 of the stone in units of Js–1m–1°C–1?

(a) 1.0

(b) 1.5

(c) 2.0

(d) 2.5

5. face area A and

of the system is

(c)

1 s2

1

heat across the cylindrical surface and the system is

(b)

k1 = k2

(c)

2

2. A cylinder of radius R made of a material of thermal conductivity k, is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity k2 two ends of the combined system are maintained

(a) k1 + k2

(b)

2 s2

k2 4

3. areas of cross-section have their ends kept at the same temperatures 1 and 2. If k1 and k2 are their thermal conductivities, 1 and 2 their densities and s1 and s2 of heat in the two rods will be the same if

(a) T = (c) T =

P

2

(b) T =

A P

P. If the emissivity is Stefan’s constant, the

1/ 2

P A P

1/ 4

(d) T = A A 6. What are the dimensions of Stefan’s constant? (a) ML–2 –2K –4 (b) ML–1 –2K–4 –3 –4 K (d) ML0 –3K–4 7. upon (a) the nature of its surface (b) the area of its surface

19.7

(c) the temperature of its surface (d) all the above factors

14.

8.

are joined end to end. But when they are joined depends upon (a) the nature of its surface (b) the area of its surface (c) the temperature of its surface (d) all the above factors

9. A composite slab consists of two slabs A and B of different materials but of the same thickness placed of A and B are k1 and k2 respectively. A steady temperature difference of 12°C is maintained across the composite slab. If k1 = k2/2, the temperature difference across slab A will be (a) 4°C (b) 8°C (c) 12°C (d) 16°C 10. l1 and l , radii r1 and r2 have thermal conductivities k1 and k2 at the same temperature difference. If l1 = 2l2 and r1 = r2 same if k1/k2 is (a) 1 (b) 2 (c) 4 (d) 8 11. A solid sphere and a hollow sphere of the same ature and allowed to cool in the same surround-

heat, in the same conditions, in (a) 24 s (b) 3 s

(b) 16 (d) 2

13. A body cools from 60°C to 50°C in 10 minutes. If of the body at the end of next 10 minutes will be (a) 38.5°C (b) 40°C (c) 42.85°C

(d) 45°C

(d) 48 s

(a) 2.5 s

(b) 10 s

(c) 20 s

(d) 5 s

15.

16. made from the same metal form the sides of an ABC B points A and B are maintained at temperatures T and 2 T respectively. In the steady state, the temperature of point C is TC heat conduction takes place, the ratio TC /T is 1 3 (a) (b) 2( 2 1) 2 1 1

(c)

3( 2

17.

T, then (a) the hollow sphere will cool at a faster rate for all values of T (b) the solid sphere will cool at a faster rate for all values of T (c) both spheres will cool at the same rate for all values of T (d) both spheres will cool at the same rate only for small values of T. 12. If the temperature of a black body increases from increases by a factor of 287 4 (a) 7 (c) 4

(c) 1.5 s

1)

(d)

1 2

1

S1 and S2 are made of the same S1 is three times that of S2. Both the spheres are heated

S1 to that of S2 is 1 (a) 3 (c)

3 1

(b) (d)

1 3 1 3

1/ 3

18. A spherical black body of radius 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be (a) 225 (b) 450

19.

19.8 Comprehensive Physics—JEE Advanced

U1, between

(a) 1 (c) 2/3

U2 and 1500 nm is U3 106 (a) U1 = 0 (c) U1 > U2

(b) 1/2 (d) 1/3

b = 2.88 (b) U3 = 0 (d) U2 > U1

20. black bodies at temperatures T1, T2 and T3 respecare such that Fig. 19.8

24. the same cross-section have been joined as shown Fig. 19.9 Fig. 19.7

(a) T1 > T2 > T3 (c) T2 > T3 > T1

the junction of the three rods will be (a) 45°C (b) 60°C (c) 30°C (d) 20°C

(b) T1 > T3 > T2 (d) T3 > T2 > T1

21. When the temperature of a black body increases, m to 0.13 body at the respective temperature is: 16 (a) 1 (c)

1 4

(c) 32

(b) it is the darkest body at all times

4 (b) 1 (d)

1 16

22. If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio was previously will be (a) 4

25. An ideal black-body at room temperature is thrown into a furnace. It is observed that (a) initially it is the darkest body and at later

(b) 16 (d) 64

23. K and 2K and thickness x and 4x, respectively, are T2 and T1 (T2 > T1 A T2 T1 K slab, in a steady state is f, with f x

(d) initially it is the darkest body and at later

26. fall of temperature (T ) of two bodies x and y same surface area, with time (t) due to emission of radiation. Find the correct relation between Fig. 19.10 emissive power (E) and absorptive power (a) of the two bodies. (a) Ex > Ey; ax < ay

(b) Ex < Ey; ax > ay

(c) Ex > Ey; ax > ay

(d) Ex < Ey; ax < ay

19.9

27.

temperature of the sun. Given Wien’s constant b = 2.88 10–3 mK. (a) 5000 K (b) 6000 K (c) 8000 K (b) 106 K

conductivity 420 W/m/K has one of its ends in area of cross-section is 10 cm2, the amount of ice that melts in 1 minute is

28. A body cools from 75°C to 65°C in 5 minutes in a ture of the body at the end of next 5 minutes will be (a) 55°C (b) 56°C (c) 57°C (b) 58°C 29. A liquid takes 6 minutes to cool from 80°C to 50°C.

35. the rate of 1400 Wm–2 the sun from the surface of the earth is 1.5 1011 m and the radius of the sun is 7.0 108 sun as a black body, it follows from the above data that the surface temperature of the sun is about (c) 6000 K

(d) 6100 K

36. rods are joined one after the other and this combi-

(a) 6 min. (b) 8 min. (c) 10 min. (d) 12 min. 30. A body initially at 80°C cools to 64°C in 5 minutes

rods are placed one on top of the other and connected to the same vessels. If q1 and q2 q1 is q2 2 (b) 1 1 (d) 8

the two cases, then the ratio (a) 15°C (b) 16°C (c) 20°C (d) 25°C 31. In Q. 30 above, the temperature of the body at the end of 15 minutes will be (a) 41°C (b) 43°C (c) 45°C (b) 47°C 32. of thermal expansion 1, 2 lii Y1, Y2

1 2 1 (c) 4

(a)

1 : 2 = 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to: (a) 2 : 3 (b) 1 : 1

33. A sphere, a cube and a thin circular plate have the same mass and are made of the same material. All

Fig. 19.11

37. (a) the for (b) the for (c) the the (d) the 34.

maximum for the sphere and minimum the plate. maximum for the sphere and minimum the cube. maximum for the plate and minimum for sphere. same for all the three.

A and B respectively are coated with carbon black on their tensity of emission of radiation are 300 nm and 500 them are in the ratio of (a)

5 3

(c)

5 3

(b) 2

(d)

5 3 5 3

4

19.10 Comprehensive Physics—JEE Advanced

38. transfer promarily due to radiation? 42. A layer of ice at 0 °C of thickness x1 pond of water. L, and k respectively are the latent heat of fusion of water, density of ice and thermal conductivity of ice. If the atmospheric temperature is – T °C, the time taken for the thickness of the layer of ice to increase from x1 to x2 L L (a) (x1 + x2)2 (b) (x2 – x1)2 2kT kT L 2 L 2 2 (c) (d) x2 x1 x2 x12 2kT kT 43. AB and CD L, cross-sectional area A and thermal conductivity k A, C and D are maintained at temperatures T1 = 20°C, T2 = 30°C and T3 temperature at B is (a) 32 °C (b) 33 °C (c) 34 °C (d) 35 °C

39. A spherical body of emissivity , placed inside a perfectly black body (emissivity = 1), is maintained at absolute temperature T a unit area of the body per second will be ( is Stefan’s constant) (b) T4 (a) T4 4 (c) (1 – ) T (d) (1 + ) T 4 40.

A and B made of the same

diameter of A is twice that of B, the ratio of rates of A and B will be (a) 1 : 1 (b) 2 : 1 (c) 4 : 1 (d) 1 : 4 41. A cubical ice box is made of thermocole of thick-

thermocole is 1.0 10–2 Wm–1 K–1 and latent heat –1 , the mass of of fusion of water is 3.35 105 ice left unmelted in 3 hours is very nearly equal to

Fig. 19.12

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43.

(b) (d) (c) (d) (a) (b) (c) (a)

2. 8. 14. 20. 26. 32. 38.

(c) (c) (d) (b) (c) (c) (d)

3. 9. 15. 21. 27. 33. 39.

(d) (b) (b) (d) (c) (c) (b)

4. 10. 16. 22. 28. 34. 40.

(a) (d) (b) (d) (c) (b) (c)

5. 11. 17. 23. 29. 35. 41.

(d) (c) (d) (c) (d) (a) (a)

6. 12. 18. 24. 30. 36. 42.

(d) (b) (d) (b) (b) (d) (c)

SOLUTION 1.

kA 2 Q 1 = l t Since ( 2 – 1) and l are the same for the two rods, Q/t will be the same if the product kA is the same for the two rods, i.e. if

k1A1 = k2A2 or 2.

A1 k2 = A2 k1

l and let its ends be maintained at temperatures 1 and 2. Area of

19.11

the cross-section of the inner cylinder = R2. Area of cross-section of outer cylinder = (2R)2 – R2 = 3 R 2. 2 Q1 = k1 R

1

2

k (4 R 2 ) l Q = Q1 + Q2

Q= Now

(i)

1

2

(iii) (iv) k = k1 + 3k2

k=

or

k1

where is the Stefan’s constant and T is the absolute temperature. E = 4 T Dimensions of =

l

k (3 R 2 ) 1 2 (ii) Q2 = 2 l Let the effective thermal conductivity of the compound cylinder be k across the compound cylinder is

3k2 4

dime

per unit area per second dimension of T 4

=

ML2

2

K

)

1

l is (d).

or Now

4.

=

=

mLd tA

2

or

–3

K–4

(

1

(

1

=

k2 A (

2)

l

– ) = k 2( – k2 = 2k1

2)

– ) = 2( – 2) 1 – 2 = 12°C or 2 = – ) = 2{ – (

3( 1 – ) = 24 or ce the correct choice is (b).

1 1

(i) 1 – 12 (ii)

– 12)} –

= 8°C

A and B are and

Q2 k r2 = 2 2 t l2

Q1 = Q2, if k1r12 k r2 k l = 2 2 or 1 = 1 l1 l2 k2 l2

perature T

r22 r12

=2

(2)2 = 8

11. upon its material, its surface area and its tempera-

A is P= A

1

Q1 k r2 = 1 1 t l1

T4

= ML0

4

10.

3.6 3.4 105 0.1 = 1 Js–1 m–1 °C–1 3600 0.34 100 0

=

k 1(

Also,

1

5.

1

9. Let 1 and 2 be the temperatures at the two faces of the composite slab and let be the temperature at the common face of the slab. If l of each slab and A the area of their face, then, in A= rate B, i.e. k1 A (

= mL, where m is the mass of ice melted and L its latent heat. Now time (t) = 1 hour = 3600 s and thickness d heat is mL Q= t Qd k= A 2 1

2

7. 8.

3.

6.

T4

E=

or T =

P A

1/ 4

at the same temperature, will cool at the same rate. 12. From Stefan’s law, E = T 4 E1 = (273 + 7)4

19.12 Comprehensive Physics—JEE Advanced

E2 =

E2 560 4 = E1 280 = (2)4 = 16

(273 + 287)4

13. Q = k(T – T0) t where k is a constant, T in time interval t and T0 is the temperature of the m is the mass of the body and s its is the temperature at the end of the next 10 minutes) ms 60 50 50 60 =k 25 and 10 min 2 ms 50 10 min

=k

Fig. 19.13

or

kA(TB TC ) l

2

25

14. Let Q be the heat transferred. If k is the thermal conductivity of each rod, their equivalent conductivity, when they are joined in series (end to end) is 2k. If t1 is time of transfer of heat, then 2k A l

3 2

tween the temperature of the body and that of the difference = 50 – 30 = 20° and in the second case the temperature difference = 40 – 30 = 10°. Since the temperature difference in the second case is half

TB > TA B to A and from B to C. In the steady state, rate B to C Q2 Q3 from C to A, i.e. t t

2 BC )

(

TB = T 2 )

d dt

r2 . Since ms

1

r3 , r

r 2)

T4

m1/3 m2 / 3 m 1

d dt

m1 / 3 d for S1 dt = d for S2 dt

1/ 3

m of S2 m of S1

1 3

=

1/ 3

18. P = (4 r2) T4 or P P2 = P1 or

r2 r1

2

P2 = 4P1 = 4

T2 T1

4

r2 T 4 1 2

2

2 1

4

=4

450 = 1800 W

19. From Wien’s displacement law

15.

16.

4 3

d = (4 dt

t1

equivalent conductivity is k/2. k A t2 2 Q2 = l Now Q1 = Q2 2k A t1 kA t2 = or t2 = 4t1 = 4 12 l 2l = 48 s

AC =

(

2l

17.

m=

Q1 =

T)

TC T

ms

50

kA(TC

m

T = b, the maxi-

b 2.88 106 nmK = 1000 nm T 2880 K U2 is the maximum. Since a black body U1 0 and

m =

U3 20. For black body radiations,

T = constant. It is > 2 3 > 1 follows that T1 > T3 > T2, which is choice (b). 21. T = constant, i.e. T2 0.26 1 =2 1 T1 = 2 T2 or T1 0.13 2 m

19.13

T2 = 2T1.

or

E1 = E1 E2

enters O

E = T4 T 41 and E2 = T 42 T14 T24

=

T1 T2

4

O, i.e. QA + QB = QC

4

1 2 (

or

1 = 16

kA

t

kA

l

l t) = t

t

T2 = 2T1)

or

t

kA t 0 l

3t = 180 or t = 60°C.

22. E=

T4

where R at absolute temperature T Q = 4 R2 T 4 If R and T are both doubled, we have Q = 4 (2R) 2 (2T ) 4 = 64 4 R2 T4 = 64 Q

Fig. 19.14

25.

23. Let A be the area of each slab. In the steady state, 26. T2

Q t

l1 K1 A

A (T2 T1 ) T1 = l1 l2 l2 K1 K 2 K2 A

(1)

Given l1 = x, l2 = 4x, K1 = K and K2 = 2K Q t

A (T2 T1 ) = x 4x K 2K

A (T2

T1 ) K x

1 3

1 3 24. Let A and l be the area of cross-section and the k conductivity and t°C the temperature of the junction O O from rods A and B kA t QA = l kA t and QB = l C is kA t 0 QC = l f=

of body x falls more rapidly with time than that of body y Ex > Ey ax > ay correct choice is (c). 27. Given k = 420 W/m/K, A = 10 cm2 = 10 10–4 m2 = 10–3 m2, 1 = 100°C, 2 = 0°C, t = 1 minute = 60 s and l k A( 1 2)t Q= l 2520 Q = 2520 J = = 600 cal. 4.2 m=

Q 600 cal = L 80 cal

28. We have e

or

e

T1 T0 T2 T0 75 25 65 25 5 e 4

= Kt =K = 5K

5 or

e

50 40

= 5K (1)

19.14 Comprehensive Physics—JEE Advanced

If the temperature at the end of next 5 minutes is T , we have 65 25 = 5K e T 25 or

e

e

or

T

40 25

= 5K

40 T 25

=

40 T 25

=

(2)

e

Stresses are equal for the two rods if Y1 1 t = Y2 2 t Y1 3 or = 2 = Y2 2 1

5 4

5 4

T = 57°C. 29. Given e

80 20 50 20

32. Let l heated so that the increase in temperature is t°C. lt where rod Strain = l t/l = t Stress = Y strain = Y t

= 6K

e

(2) = 6K

(1)

If the body takes t minutes to cool from 60°C to 30°C, then 60 20 = tK (2) e e (4) = tK 30 20

33. Since the material is the same the density is the same. Since the mass is the same and density is the same, all three have the same volume. For the same

heat by radiation is proportional to the surface area. 34.

m

T= lo t = lo 6 or

e e

4) 2)

lo 10 4) 0.602 lo 10 2) 0.301 t = 12 minutes.

b

2.88 10

T=b 3

= 6000K

480 10

m

2 35. Let R be the radius of the sun and let r be the radius of the earth’s orbit round the sun. If the surface temperature of the sun is T (in kelvin), the ener-

30. If T0 we have e

80 T0 64 T0

= 5K

(1)

4 R2 T 4, where is Stefan’s constant whose value is 5.67 10–8 Wm–2 K–4. Now the area of the spherical surface of radius r is 4 r2

e

64 T0 52 T0

= 5K

(2)

earth’s surface is 4 R2 T 4 4 r2

80 T0 64 T0 = 64 T0 52 T0

=

T4 =

R2 T 4 r2

= 1400

1400 r 2 R2

T0 = 16°C, which is choice (b). 31. If T is the temperature after 15 minutes, then 52 T0 = 5K e T T0 e

52 16 T 16

= 5K

=

1.5 1011

1400 5.67 10

= 1.1338 (3)

80 16 52 16 = T 16 64 16 T = 43°C, which is choice (b).

or

8

2

7 108

2

1015

T = 5803 K

36. If a steady temperature difference ( L and cross-sectional area A

1

–

2)

is

19.15

q=

k A(

1

2)

L

b4 A

=

4 m 2

m

where k

For a sphere of radius r, A = 4 r

the two rods are connected in series. If two rods L1 and L2 and conductivities k1 and k2 are joined in series, the equivalent conductivity ks L1

4

b

Q= A

Q= where k = 4

L2

L L2 = 1 (1) ks k1 k2 For two identical rods, L1 = L2 = L and k1 = k2 = k, in ks = k. Further, when two

and

Q1 = Q2

1 2 (2) 2L In the second case, the two rods are connected in

=

= k

r2

(3)

4 m

b4 r12 4 m 1

r22

Q2 = k

kA

k1 and k2 are joined in parallel, the equivalent condu-

4 m

Q1 = k

composite rod is (2L) but its cross-sectional area is A q1 =

b4 4 r 2

4 m 2

r1 r2

4 m 2 4 m 1

2

3 cm 5cm

2

500 nm 300 nm

4

=

5 3

2

38.

kp = k1 + k2 For two identical rods, k1 = k2 = k kp = (2k). Furthermore, the cross-sectional area of the composite rod is (2A q2 =

2k 2 A L

1

2

due to conduction or convection. 39. body of emissivity

(3)

q1 1 = . Now, the rate q2 8 37. black body at absolute temperature T (1) Q = AT 4 where A is the surface area of the body and is ment law, m T = b

b m

T 4, irrespective of the

40. proportional to the surface area. 41. 0.30 m. Since a cube has 6 faces, the total surface area of the cube exposed to air is A = 6 (0.3)2 = 0.54 m2. of thermacole (d) = 5.4 cm = 5.4 10–2 m of exposure (t) = 3 hours = 3 60 60 = 1.08 104 s –2 conductivity (k) = 1.0 10 J s–1 m–1°C–1

where m maximum emission of radiation and b is Wien’s T=

is

Ta) = 40°C Ti)= 0°C t is

(2) Q=

kA(Ta Ti )t d

19.16 Comprehensive Physics—JEE Advanced

=

1.0 10

2

0.54

5.4 10

= 4.32

0) 1.08 104

(40

kATdt L = A Ldx or dt = · xdx x kT

2

104 J

Now, heat of fusion of water, L = 3.35 –1 = 3.35 105

–1 102 6 10 J of

x

t

L 2 dt = xdx kT x

0

Q m= L

4.32 104

t=

or

3.35 105

L x2 kT 2

x2

L (x22 – x21 ), 2kT

= x1

which is choice (c). 43. Let TB be the temperature at B of heat from C towards B is Q1 k A (T2 TB ) = t L/2

42. When the temperature of the air is less than 0°C, the cold air near the surface of the pond takes heat

x be the thickness of the ice layer at a certain time. If the thickness increases by dx in time dt, then the dt

k A (T3 Q2 = t L/2

D towards B is TB ) B towards A is

Q3 k A (TB T1 ) = L t In the steady state, the rate at which heat enters B = rate at which heat leaves B, i.e. Q Q1 Q2 = 3 t t t

( T )]dt kATdt = (1) x x where A is the area of the layer of ice and – T °C is dm is the Q = dm L. But dm = A dx, where Q=

1

kA[0

or

2 k A (T2 L

2 k A (T3 L

TB )

TB )

k A (TB T1 ) L or 2(T2 – TB) + 2(T3 – TB) = TB – T1 =

TB = =

Fig. 19.15

Q = A Ldx

(2)

T1

2 T2 5

2 T3

20 2 30 2 40 5

= 32 °C, which is choice (a).

II Multiple Choice Questions with One or More Choices Correct 1. (a) All bodies emit thermal radiations at all temperatures

mirror a velocity of 3

108 ms–1

19.17

2. depends upon (a) the nature of its surface (b) the area of its surface (c) the temperature of its surface

body increases.

7. A composite block is made of slabs A, B, C, D and E

3. does not depend upon (a) the density of the body (b) the nature of its surface (c) the area of its surface (d) the temperature of its surface 4. mum emission are T1 and T2 respectively. If Wien’s constant is b 10–3 mK, then (b) T1 (a) T1 = 3718 K (c) T2 = 2318 K (d) T2 = 4677 K 5. A and B of equal surface area are placed one on top of the other to form a composite plate of A and B are

a constant K

L)

Q A and E slabs are same. E is maximum. (c) temperature difference across slab E is smallest. C B D.

the exposed surface of plate A is – 10 °C and that of the exposed surface of plate B the temperature of the contact surface is T1 if the plates A and B are made of the same material and T2 if their thermal conductivities are in the ratio 2 : 3 then (b) T1 = – 2°C (a) T1 = – 4°C (d) T2 = 0°C (c) T2 = – 3°C 6. Initially a black body at absolute temperature T is kept inside a closed chamber at absolute temperature T0 allow sun rays to enter. It is observed that temperatures T and T0 remains constant. Which of the

Fig. 19.16

8. Initially a black body at absolute temperature T is kept inside a closed chamber at absolute temperature T0 allow sun rays to enter. It is observed that temperatures T and T0 remains constant. Which of the

body remains the same. body increases. body increases.

body remains the same. body increases.

ANSWERS AND SOLUTIONS 1. 2. All the four choices are correct.

3. From Wien’s law, and m

m

T = b, where b is a constant m

temperature of the body.

depends only on T, the

19.18 Comprehensive Physics—JEE Advanced

4. From Wien’s law,

max

the thermal resistances are connected as shown

T=b 3

T1 =

7.8 10 .

T2 =

10

7

= 3718 K

3

= 4677 K 6.2 10 7 So the correct choices are (a) and (d). 5. Let T be the temperature of the contact surface. k A( 10 T ) k A(T 10) Q = A = B 4.0 6.0 t or

k A ( 10 T ) k (T 10) = B 4 6

(a) If kA = kB (b) If

(1)

T1 = – 2°C.

kA 2 = kB 3

Fig. 19.17

Reff 1 1 = Reff RB

T2 = 0°C.

6.

=

4

from a black body is proportional to (T – T0) . Since T and T0 remain constant, the rate of emission of

Reff =

rect choices are (a) and (d). 7. Let W L, area A (= LW) and thermal conductivity K L L = R= KA K ( LW ) A, B, C, D and E respectively are RA = where

L R = (2 K )(4 LW ) 8

3 4R

1 RC 2 R

1 RD 5 4 = 4R R

R 4 QA = (Q)eff = Q .

T , where T is the temperature R difference between the ends of the slab. Also the temperature difference between the ends of slabs B, C and D is the same = ( T)eff Q R ( T)A = QARA = A 8 Q R ( T)B = ( T)C = ( T)D = QAReff = A 4 Q R and ( T) = A ( Q = Q A) 24 E is the minimum. Q=

R=

L 1 = K ( LW ) KW

RB =

4L 4R = 3K ( LW ) 3

RC =

4L R = 4 K (2 LW ) 2

Now QB =

RD =

4L 4R = 5K ( LW ) 5

QC =

( T )C Q R/4 1 = A = QA R/2 RC 2

R =

L R = 6 K (4 LW ) 24

QD =

( T )D Q R/4 5 = A = QA RD 4R /5 16

Since slabs B, C and D are in parallel and slabs A and E are in series with this parallel combination,

( T )B Q R/4 3 = A = QA RB 4R /3 16

And Q = QA

B is

19.19

Now QB + QD =

T and T0 remain constant, the rate of emission of

3 5 1 QA + QA = QA which 16 16 2

-

is equal to QC the correct choices are (a), (c) and (d).

rect choices are (a) and (d).

8. 4

from a black body is proportional to (T – T0) . Since

III Multiple Choice Questions Based on Passage Questions 1 to 6 are based on the following passage Passage I Thermal Radiations

is transferred to us by a process which, unlike conduction or convection, does not require the help of a medium in the

1.

-

fan’s constant are (b) ML–1 –2K–4 (a) ML–2 –2K–4 –3 –4 K (d) ML0 –3K–4 2. What is the SI unit of Stefan’s constant? (b) W m–1K–4 (a) J s–1K–4 –2 –4 (c) W m K (d) J m–2K–4 3. thermal radiations lie? 4. thermal radiations? (a) Constant volume air thermometer (b) Platinum resistance thermometer

in terms of that emitted from a reference body (called the black body) at the same temperature. A black body E emitted by a unit area of a black body E = T4 where T is the absolute temperature of the body and is a constant known as Stefan’s constant. If the body is not a perfect black body, then E= T4 where is the emissivity of the body.

5. When a body A T1 is surrounded by another body B at a lower temperature T2, then the rate of loss of heat from body A will be proportional to (b) (T1 – T2)4 (a) T 41 (c) (T1 – T2) (d) (T 41 – T 42) 6. pends upon (a) the surface area of the body (b) the temperature of the body (c) the nature of the surface of the body (d) the emissivity of the surface of the body

ANSWERS AND SOLUTIONS 1.

E=

T4 where E

E = dimensions of 2

area

time

Dimensions of choice (d).

=

ML

2

–3

2

=

ML 3 K4

–3

K–4, which is

2. 3. 4. 5. 6. All the four choices are correct.

19.20 Comprehensive Physics—JEE Advanced

Questions 7 to 13 are based on the following passage

9. revealed that the atmospheres of stars contain

Passage II Stellar Spectra Like the solar spectrum, the spectra of stars show a continuous spectrum on which dark aborption lines are photosphere) of

10.

(c) uranium

(d) helium

(a) surface temperature (b) mass

the outer, relatively cooler, layer of the star, the radiations

(d) all the above factors 11. Wien’s displacement law tells us that an extremely hot star should look

in the outer layer of the star. T of a star can be estimated by m at which the intensity of

12. In Wien’s displacement law the SI unit of Wien’s constant b is (a) metre per kelvin (b) metre per kelvin squared (c) metre kelvin (d) metre kelvin squared 13. E of black body radiations where

displacement law which states that m T=b where b is a constant called Wien’s constant and the above relation is called Wien’s Displacement Law which states that as the temperature increases, the maximum intensity of emission shifts (or is displaced) towards the shorter b has been found 10–3 mK. 7. (a) continuous emission spectrum (b) emission line spectrum (c) emission band spectrum (d) absorption line spectrum 8.

the outer layers of the sun

(d) destructive interference between waves of Fig. 19.18

SOLUTION 7. 8. 9. 10. 11.

-

m

T = constant, if T

m

12. From m T = b, the SI unit of b = SI unit of m unit of T = metre kelvin, which is choice (c).

SI

13.

Questions 14 to 16 are based on the following passage Passage III A, B and C equal diameters are joined in series as shown in the folk, k and 0.5 k

Fig. 19.19

In the steady state, the free ends of rods A and C are at curved surfaces of rods.

19.21

14.

A and B is (a) 55.7°C (c) 75.7°C

16.

(b) 65.7°C (d) 85.7°C

15.

B and C is (a) 57.1°C (c) 37.1°C

nation is 7k (a) 3 (c)

(b) 47.1°C (d) 27.1°C

5k 3

(b)

2k 7

(d)

3k 5

SOLUTION 14.

T1 = 85.7°C. So the corsame for all rods. If T1 and T2 are the temperatures at the junction points between A and B and between and C respectively, then k A(100 T1 ) Q = A d t = Given

rect choice is (d). 15. 16.

T1 T2 = 57.1°C, which is choice (a). diameters, the equivalent thermal conductivity of

k A T2 0 k B A T1 T2 = C d d kA = 2k, kB = k and kC = 0.5k

1 1 = ke kA

1 kB

2(100 – T1)= (T1 – T2) = 0.5(T2 – 0) T1 = T1 – T2 and

ke =

(1)

T1 – T2 = 0.5 T2

1 1 = kC 2k

1 k

1 0.5k

2k . So the correct choice is (b). 7

(2)

Questions 17 to 19 are based on the following passage Passage IV

17. (a) 26.5°C (c) 24.5°C

(b) 25.5°C (d) 23.5°C

(a) 2.5°C (c) 1.5°C

(b) 2.0°C (d) 0.5°C

18. area 1 m2 and thickness 0.01 m separated by 0.05 m thick 19. –1

–1

K .

is nearly equal to (a) 1000 J s–1 (c) 3000 J s–1

(b) 2000 J s–1 (d) 4000 J s–1

SOLUTION 17.

T2 and T3 be the temperature k g A(T1 T2 ) dQ1 0.8 1 (300 T2 ) = = dt 0.01 dg

(1)

k A(T2 T3 ) 0.08 1 (T2 dQ2 = a = dt da 0.05

(2)

T3 )

k g A(T3 T4 ) dQ3 0.8 1(T3 273) = = dt 0.01 dg

(3)

In the steady, dQ3 dQ1 dQ2 = = dt dt dt Fig. 19.20

50

(300 – T2) = T2 – T3

(4)

19.22 Comprehensive Physics—JEE Advanced

19. (5) (300 – T2) = T3 – 273 26.5°C. T3 So the correct choice is (a). 18. T3 T2 = 273.5 K = 0.5°C, which is choice (d).

T2 = 273.5 K and T3 dQ1 dt

2000 J s–1, which is choice (b).

IV Assertion-Reason Type Questions Statement 2

correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1is false, Statement-2 is true. 1. Statement 1 tion. Statement 2 e) and absorptive power (a temperature and for radiation of the same wave-

convection. 4. Statement 1 uid. Statement 2

5. Statement 1 netic waves. Statement 2 6. Statement 1 and 4 m and termperatures 4000 K and 1000 K

2. Statement 1

two spheres will be the same. Statement 2

same (room) temperature. Statement 2 tivity than plastic. 3. Statement 1 If the earth did not have an atmosphere, it would become extremely cold.

is directly proportional to its absolute temperature and inversely proportional to its surface area.

SOLUTION Conversely, if a body is a poor emitter of a radia-

1. law,

e

e = constant a a

2. -

temperature is lower than our body temperature. Since metal is a better conductor of heat than plastic, when we touch the metal cap and the plastic

19.23

4. 5. 6.

the metal cap much more quickly than to the plastic body. 3.

E = T 4A = T 4

4 R2; R = radius of sphere

is a poor conductor of heat, the atmosphere acts as a blanket for the earth and keeps the earth warn

V Integer Answer Type 1.

A is 3 times that of B. In the steady state, the temperature difference across the wall is 36°C. Find the temperature difference (in °C) across the layer A.

rod, end to end, with a second rod of a different material but of the same cross-section. At 25°C, the

10–5 per °C and that of the second rod is 10–5 per °C. Find the value of n.

= 1.7 =n

2. A wall has two layers A and B each made of different material. Both layers have the same thickness.

3.

A and B have thermal emissivities of of the two bodies are the same. If they emit total raratio TA/TB.

SOLUTIONS Given

1. = 30 (1 + 1.7 = 30.051 cm

10–5

T1 – T2 = 36

T2 = T1 – 36 T1 – T0

100)

100 = 7000

cm =2

cm = 10–5 per °C.

n = 2.

Fig. 19.21

2. layers A and B is the same, i.e. (a = area of crosskA a (T1 T0 ) k a(T0 T2 ) = B x x kA = 3 kB 3(T1 – T0) = T0 – T2

3. E1 = eA A E2 = eB Given E1 = E2 TA4,

TA e = B TB TA (1)

A TB4

1/4

=

0.81 0.01

1/4

=3

(2)

Electrostatic Field and Potential 20.1

20

Electrostatic Field and Potential

Chapter

REVIEW OF BASIC CONCEPTS 20.1

COULOMB’S LAW

On the basis of his measurements, Coulomb arrived at a law, known after his name as Coulomb’s law, which states that the magnitude of the electric force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them, i.e. F or

q1 q2

F12 =

k=

4

0

0

r2

q1 q2 n

F21 = –

4

0

r2

9

109 Nm2 C–2

The force F is attractive for unlike charges (q1 q2 < 0) and repulsive for like charges (q1 q2 > 0). Coulomb’s Law in vector form

F21 =

20.2

4

0

r2

q1 q2 n 4

0

r2

RELATIVE PERMITTIVITY (OR DIELECTRIC CONSTANT)

the permittivity of the medium to permittivity of vacuum, i.e. r r

= 0

is also called the dielectric constant (K) of the medium.

Thus K =

or 0

Fig. 20.1

q1 q2 n

F12 = –

q1 q2

1

4

where n is a unit vector directed from q1 to q2. Force exerted by q1 on q2 is

r2 In the SI system, k is written as 1/4 0 where 0 is called the permittivity of vacuum and its value is 10–12 C 2 N–1 m–2 0 = 8.854 Then

q1 q2 n

Case (b): Like charges (q1 q2 > 0) [Fig. 20.1(b)]

r2

F= k

Case (a): Unlike charges (q1 q2 < 0) [Fig. 20.1(a)] Force exerted on q2 by q1 is

=K

0

K for air = 1. If

charges q1 and q2 are situated in a medium other than air or vacuum, the magnitude of force between them is given by q1 q2 q1 q2 F= 2 4 r 4 0 Kr 2

20.2 Comprehensive Physics—JEE Advanced

20.3

PRINCIPLE OF SUPERPOSITION

If many charges are present, the total force on a given charge is equal to the vector sum of the individual forces exerted on it by all other charges taken one at a time. 20.1 Two point charges q1 = +9 C and q2 = –1 C are held 10 cm apart. Where should at third charge +Q be placed from q2 on the line joining them so that charge Q does not experience any net force? SOLUTION Charge Q will not experience any net force if the forces exerted on it by charges q1 and q2 are equal and in opposite directions.

Fig. 20.2

It follows from Fig. 20.2 that charge Q will not experience forces in opposite direction if it lies at any point between AB. Let x be the distance of Q from q2. Then forces exerted on Q by q1 and q2 respectively are F1 =

and

q1 Q i 4

F2 = –

0

(0.1

q2 Q i 4

0

x

x)

2

=

4

0 6

1 10 2

4

6

9 10

0

9 10 4 9=

0

(0.1

(0.1

Qi x)

2

Fig. 20.3

Equilibrium of charge Q at O [See Fig. 20.3] Since Q is at the same distance from equal charges q1 and q2, it will be equilibrium for any positive or negative value, because it will experience equal and opposite forces. Equilibrium of charge q1 at A If Q is negative, it will repel q1. Also q2 is repel q1. Hence q1 cannot be in equilibrium if Q is negative. So Q must be positive. Force exerted on q1 by Q is F=

(0.1

x)

2

Qi x2

F =–

1 10 4

0

Qi x2

=0

x) 2

x2

0.1 x 3= x x = 0.05 m = 5 cm 20.2 Two charges, each equal to – 4 C, are held a certain distance apart. A charge Q is placed exactly mid-way between them. Find the magnitude and sign of Q so that the system of three charges is in equilibrium.

Qi

r 2

2

4 10

6

4 10

6

i

2

4 0r Net force on q1 will be zero if F + F = 0, i.e. if

4 –

0

6

Force exerted on q1 by q2 is

4 10

6

4 10 4

Qi

Net force on Q = F1 + F2 Net force on Q = 0 if F1 + F2 = 0 6

SOLUTION A system of charges is in equilibrium if no charge of the system experiences any net force.

0

6

Qi

r 2

2

–

Q= 1

4 10

6

4

4 10 0

r

2

6

i

=0

10–6 C = 1 C

It is easy to check that charge q2 will also be in equilibrium. Hence the system of three charges will be in equilibrium if Q = +1 C. 20.3 Four point charges, each equal to q = 4 C, are held at the corners of a square ABCD of side a = 10 cm. Find the magnitude and sign of a charge Q placed at the centre of the square so that the system of charges is in equilibrium. SOLUTION AC (= r) = 2a . Let us consider the equilibrium of charge q at A (Fig. 20.4)

Electrostatic Field and Potential 20.3

kq

i.e. if

a

q

q

2

2 2

2Q

=0

q Q = – (1 2 2 ) 4 =–

20.4

Force exerted on charge at A by charge at B is FAB =

kq a

i,

2

force F

1

where k =

4

FAD = FAC = = and

FOA =

=

kq

o

kq 2 (a 2 )2 kq 2 a2 2 2 kqQ a 2 2

(cos 45 i (i

sin 45 j)

(cos 45 i 2

sin 45 j)

k qQ

Fx = where

kq 2 a kq a2

=

kq

Fy =

j

2

kq 2

2 k qQ

2

a2

a 2 2 q

q 2 2

(1)

Electric r from a source charge q is given by E=

j)

( i j) a2 Net force on charge q at A in the x-direction is Fx =

F q0

If a charge q due to other charge or charges is E, then the charge q will experience a force F given by F = qE

j

a2

E is then given by E=

Similarly 2

ELECTRIC FIELD

positive point charge q0 at the point in space where the

Fig. 20.4

2

4 C (1 2 2 ) = – (1 2 2 ) C 4

2Q i

i i

1 4

q 0

r2

For a positive charge (+q), vector E is directed radially outwards from it and for a negative charge (–q), E is E is individual charges. (2) A pair of equal and opposite point charges separated by a certain distance is called an electric dipole. Let 2a be the separation between point charges –q and +q (Fig. 20.5).

q

2Q 2 2 a Similarly net force on charge q at A in the y-direction is 2

q

Fig. 20.5

Resultant force on charge q at A is F=

Fx2

Fy2

2

P due to +q and –q respectively are 2

Charge q will be in equilibrium if F = 0 i.e. if =0

2 E+ =

qi 4

0 (r

a)2

20.4 Comprehensive Physics—JEE Advanced

NOTE

qi

E– = –

0 (r

4

a)2

line of a dipole is along the dipole moment.

P is Ea = E+ + E–

torial line of a dipole is antiparallel to the dipole moment.

2q (2a)r

=

4

0 (r

2

a 2 )2

2p r

=

4

0 (r

2

a 2 )2

where p = q(2a) is the dipole moment and 2a is the vector distance between charges –q and +q. Dipole moment p is a vector quantity directed from –q to +q. For a very short dipole (a 0) is positive, it is negative in the neighbourhood of an isolated negative charge (q < 0). V=

to each charge, assuming that all other charges are absent, and then simply add these individual contributions. Since, addition here is the ordinary sum, not a vector sum. The potential at any point due to two point charges q1 and q2 is, therefore, simply given by 1

q1 q2 4 0 r1 r2 where r1 and r2 are the distances of the point in the question from charges q1 and q2 respectively. V=

Electrostatic Field and Potential 20.7

The potential at any point due to a system of N point charges in given by V = V1 + V2 +

20.10

+ VN =

1 4

N 0 n 1

qn rn

charge on an electron is 1.6 10–19 C, 1 eV = 1.6 10–19 J

20.13

POTENTIAL ENERGY OF AN ELECTRIC DIPOLE IN AN EXTERNAL ELECTRIC FIELD

RELATION BETWEEN E AND V

means that the potential decreases along the direction of E =–

20.11

dV dr

ELECTRIC POTENTIAL ENERGY

The electric potential energy of a system of point charges Fig. 20.12

We assume that the charges were at rest when kinetic energy. The electric potential energy of two point charges q1 and q2 separated by a distance r12 as shown in Fig. 20.11 (a) is given by 1 q1 q2 U12 = 4 0 r12

E, as shown in Fig. 20.12, it experiences a torque given by = p E sin where

is the angle between the line joining the two =p

The torque tends to rotate the dipole to a position where = 0, i.e, p is parallel to E. The electric potential energy of a dipole is U =– p E

20.14

Fig. 20.11

The electric potential energy of a system of three point charges as shown in Fig. 20.11 (b) is given by U = U12 + U23 + U31 =

1 4

0

q1 q2 r12

q2 q3 r23

E

ADDITIONAL USEFUL FORMULAE

(i) Charge q at each vertex of an equilateral triangle of side a (Fig. 20.13).

q1 q3 r13

This expression can be generalized for any number of charges.

20.12

THE ELECTRON-VOLT

The SI unit of potential energy is the joule. In atomic physics a more convenient unit called the electron-volt (written as eV) is used. An electron-volt is the potential energy gained or lost by an electron in moving through Since the magnitude of

Fig. 20.13

At centroid O, E0 = 0 and V0 = where r =

a 3

.

3q 4

0r

20.8 Comprehensive Physics—JEE Advanced

(ii) Charge q at each vertex of a square of side a (Fig. 20.14).

q

=

4

0r

1 4

1

2

1

q

=

4

1 16

0r

2

q

1

1 4

1 2

1 4

3

0r

2

Potential at O is q

V0 = Fig. 20.14

At center C, Ec = 0 and Vc =

4q 4

0r

; r

4 4

In the above two cases, if one of the charges is O and C is E = q/4 or2, directed towards the empty vertex.

1 0r

2

NOTE

1

q

=

a

0r

1

q 1 2

2

0r

(vi) A short electric dipole of dipole moment p (Fig. 20.17)

(iii) For Fig. 20.15,

Fig. 20.17 Fig. 20.15

Vc = 0

and

Ec =

2 2q 4

0r

2 p cos

At point A, Er =

4

2

0r

3

= Also tan = (

+ ).

p 4

3 0r

q, placed on the x-axis at x = r, x = 2r, x = 4r… and so O is q 1 1 1 E0 = 2 2 2 4 0r 1 2 42

2p 4

= 90°)

Electric potential at A is VA = At point B; VB =

p 4

At point C; VC = 0

0r

2

0r

3

0x

3

and EA is (

At point C on equatorial line, EC = (

4

E2

1 tan . Angle between 2

Fig. 20.16

p sin

1 1 /2

3 cos 2

At point B on axial line, EB =

Ec = 0

Er2

A is EA =

(iv) For Fig. 20.16,

Vc = 0

and E =

p cos 4

0r

2

= 0°) p 4

0y

3

Electrostatic Field and Potential 20.9

=

1

Q

At point P, EP =

Q ) L (i) Charge Q distributed uniformly on a rod of length L (Fig. 20.18) (Linear charge density

4

VP =

r (r

0

Q 4

0L

L) r

log e

L r

(ii) Charge Q distributed uniformly on a conducting sphere or shell of radius R (Fig. 20.19) (Surface Q charge density = ) 4 R2 Fig. 20.18

Fig. 20.19

At point C outside the sphere or shell (r > R), EC =

Q 4

0r

At centre O, V0 =

2

Q 4

0

R

4

0

At point B just outside the surface (r = R), EB =

Q 4

0

R2

At point A inside the sphere or shell (r < R), EA = 0 Potential at center O of sphere or shell, Q V0 = 4 0R Q At points inside (r < R), VA = = VB (at 4 0R surface) Q 4 0R (iii) Charge Q distributed uniformly on a semicircular wire of radius R (Fig. 20.20) (Linear charge Q density = ) R

Fig. 20.20

(iv) Charge Q distributed uniformly on a ring of radius R P on the axis is

At point C outside (r > R), VC =

At center O, E0 = 4 tion =

0R

along negative y-direc-

Q 4

2

0

R2

Fig. 20.21

EP =

Q 4

r 0

R2

r2

3 /2

20.10 Comprehensive Physics—JEE Advanced

R

E is maximum at r = Emax =

1 4

2

(ii) Charge Q kept at each vertex of a square of side a (Fig. 20.23)

and

2Q 0

3 3R 2

U=

Electric potential at point P is Q 1 VP = 4 0 R2 r 2

=

1 /2

(i) Charge Q kept at each vertex of an equilateral of side a. Potential energy is (Fig. 20.22). 3Q 2 4

0

a

Q2 4

0a

2Q 2 4 (4

0 (a

2)

2)

Fig. 20.23

(iii) An electric dipole of dipole moment p placed in E with angle between p and and potential energy U = – p.E. E. Torque = p The zero of potential energy is taken at = 90°. (a) When = 0°, = 0, U = minimum = –pE (stable equilibrium) (b) When = 90°, = maximum = pE, U = 0 (c) When = 180°, = 0, U = maximum = pE (unstable equilibrium) (d) Work done in turning a dipole from angle 1 to angle 2 is W = pE (cos 1 – cos 2)

V is maximum at r = 0 (i.e. at centre O) and Q Vmax = 4 0R

U=

4 Q2 4 0a

Fig. 20.22

If

= 0° and

1

2

= 180°, W = 2pE

I Multiple Choice Questions with Only One Choice Correct 1. Three point charges Q, – 2Q and – 2Q are placed at the vertices of an equilateral triangle of side r. The work done to increase their separation to 2 r is (a) zero (c)

(b)

2 Q2 4

0

r

(d)

Q2 4

0

(c) r

2 Q2 4 0r

2. A point charge Q is placed at point P at a distance R from the centre O of a metallic spherical shell of inner radius 2R and outer radius 2.5 R. The electric potential at the centre of the shell will be (a)

Q 4

(c) zero

0

R

(b) (d)

1 4

0

5Q 6R

0

9Q 10 R

1 4

(a)

3. Two concentric metallic shells of radii R and 2R are given charges Q and 2Q respectively. If the two shells are connected by a metallic wire, the change in electric potential on the outer shell is

Q 4 4

0

R

3Q 0 (2 R)

(b)

4

Q 0 (2 R)

(d) zero

4. A metal sphere of radius R carries a charge Q. The E and the electric potential is V. If R is doubled keeping Q the same, the new values of E and V will be (a)

E V and 4 2

(b)

E V and 2 4

(c) 4E and 2V (d) 2E and 4V 5. A metal sphere of radius R has surface charge density E and the electric potential is V. If R is halved, keeping the same, the new values of E and V will be (a) 4E and 2V (b) 2E and 4V (c) E and

V 2

(d)

E and V 2

Electrostatic Field and Potential 20.11

6.

at a point P on its axial line at a distance r from it is EP and at a point Q on its equatorial line at a distance 2 r from it is EQ . Then (a) EP = – 16 EQ

(b) EP = –8 EQ

(c) EP = 8 EQ

(d) EP = – EQ

7. A hollow metal sphere is charged such that the potential at its centre is V. The potential on the surface of the sphere is (a) zero (b) V (c) more than V (d) less than V 8. The work done in carrying a charge q once round a circle with a charge Q at the centre is W1. The work done is W2 if charge q is moved from one end of a diameter to the other. Then (a) W1 > W2 (b) W1 < W2 (c) W1 = W2 0 (d) W1 = W2 = 0 9. Three point charges 4q, Q and q are placed in a straight line of length l at points distant 0, l/2 and l respectively. If the net force on charge q is zero, the magnitude of the force on charge 4q is (a)

q2 0

(c)

l2

3 q2

(b)

2 q2 0

l2

4 q2

(d) 2 l2 0 l 10. 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 Q Q (b) – (a) – 2 4 Q Q (c) + (d) + 2 4 IIT, 1987 11. One thousand spherical water droplets, each of radius r and each carrying a charge q, coalesce to form a single spherical drop. If v is the electrical potential of each droplet and V that of the bigger drop, then V 1 V 1 = (b) = (a) v 1000 v 100 V V (c) = 100 (d) = 1000 v v 12. Two metallic identical spheres A and B carrying equal positive charge + q are a certain distance apart. The force of repulsion between them is F. A third uncharged sphere of the same size is brought in contact with sphere A and removed. It is then 0

brought in contact with sphere B and removed. What is the new force of repulsion between A and B? 3F (a) F (b) 8 F F (c) (d) 2 4 13. Two small identical balls P and Q, each of mass 3 /10 gram, carry identical charges and are suspended by threads of equal lengths. At equilibrium, they position themselves as shown in Fig. 20.24. 1 What is the charge on each ball. Given = 4 0 9 2 –2 –2 9 10 Nm C and take g = 10 ms . (b) 10–5 C (a) 10 –3 C –7 (d) 10–9 C (c) 10 C

Fig. 20.24

14. Two point charges q1 = 2 C and q2 = 1 C are placed at distances b = 1 cm and a = 2 cm from the origin on the y and x axes as shown in Fig. 20.25. P (a, b) will subtend an angle with the x-axis given by (a) tan = 1 (b) tan = 2 (c) tan = 3 (d) tan = 4

Fig. 20.25

15. An electric dipole placed with its axis in the direc(a) a force but no torque (b) a torque but no force (c) a force as well as a torque (d) neither a force nor a torque. 16. An electric dipole placed with its axis inclined at experiences

20.12 Comprehensive Physics—JEE Advanced

(a) a force but no torque (b) a torque but no force (c) a force as well as a torque (d) neither a force nor a torque 17. An electric dipole placed in a non-uniform electric (a) a force but no torque (b) a torque but no force (c) a force as well as a torque (d) neither a force nor a torque. 18. A cube of side b has a charge q at each of its vertices. What is the electric potential at the centre of the cube? 4q 3q (b) (a) 3 0b 0b 2q

(c)

4q 3

0b

2q

(c)

0b

3q

(b)

2

0b

2

(d) zero

2

(a)

20. Two point charges – q and + q are located at points (0, 0, – a) and (0, 0, a) respectively. What is the electric potential at point (0, 0, z)? qa q (b) (a) 2 4 4 0z 0a (c)

2qa 4

z2

0

a2

(d)

2qa 4

0

z2

a2

21. In Q. 20, how much work is done in moving a small test charge q0 from point (5, 0, 0) to a point (– 7, 0, 0) along the x-axis? (a) (c)

5 7 2 12

4

q0 q

4

(b) W1 < W2

0

a

q0 q 0

a

(b)

7 5

Fig. 20.26

(d) W1 = W2 = 0 25. Two concentric spheres of radii r1 and r2 carry charges q1 and q2 respectively. If the surface charge density ( ) is the same for both spheres, the electric potential at the common centre will be

19. (a)

24. A positive charge (+ q) is located at the centre of a circle as shown in Fig. 20.26. W1 is the work done in taking a unit positive charge from A to B and W2 is the work done in taking the same charge from A to C. Then (a) W1 > W2 (c) W1 = W2

(d) zero

0b

23. In a hydrogen atom, the electron and the proton are bound together at a separation of about 0.53 Å. nite separation of the electron from the proton, the potential energy of the electron-proton system is (a) – 54.4 eV (b) – 27.2 eV (c) – 13.6 eV (d) zero

4

q0 q 0

a

(d) zero

22. A neutral hydrogen molecule has two protons and two electrons. If one of the electrons is removed we get a hydrogen molecular ion (H +2). In the ground state of H +2 the two protons are separated by roughly 1.5 Å and the electron is roughly 1 Å from each proton. What is the potential energy of the system? (a) – 38.4 eV (b) – 19.2 eV (c) – 9.6 eV (d) zero

0

(c) 0

26.

r1 r2

(b)

(r1 – r2)

(d)

r2 r1

0

(r1 + r2)

0

a sphere of radius r having a uniform surface charge density is (b)

(a) 0

(c)

0r

(d)

2

0

2 0r

27. Two equal negative charges – q (0, a) and (0, – a). A positive charge + Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will (a) execute SHM about the origin (b) move to the origin and remain at rest there. (d) execute oscillations but not SHM. IIT, 1985 28. Four charges q, 2q, 3q and 4q are placed at corners A, B, C and D of a square as shown in Fig. 20.27. P of the square has the direction along

Electrostatic Field and Potential 20.13

(a) zero q

(b)

4

2

(d) (a) AB (b) CB (c) AC (d) DB 29. Particle A has a charge + q and particle B has a charge + 4q, each having the same mass m. When allowed to fall from rest through the same potential difference, the ratio of their speeds vA/vB will be (a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1 30. Four equal charges Q are placed at the four corners of a square of side a. The work done in removing a charge – Q is 2 Q2 (a) zero (b) 4 0a (c)

2 Q2 0a

(d)

Q2 2

a

31. A point charge + q is placed at the mid point of a cube of side L cube is q (b) zero (a)

(Q1

Q2 ) ( 2

1)

1 q Q1 24

0

Q2 R

IIT, 1992 34. An electron of mass me, initially at rest, moves in time t1. A proton of mass mp, also initially at rest, takes time t2 to move through an equal distance in gravity, the ratio t2/t1 is nearly equal to (a) 1

(b) (mp/me)1/2

(c) (me/mp)1/2

(d) 1836

IIT, 1997 35. A nonconducting ring of radius 0.5 m carries a total charge of 1.11 10–10 C distributed non-uniformly E everywhere in space. The value of the line integral l l

0

2R

q 2 (Q1 + Q2) 4 0R

(c)

Fig. 20.27

0

0

– E dl (l = 0 being centre of the ring) in

volts is (a) + 2 (c) – 2

(b) – 1 (d) zero

IIT, 1997 36. A metallic solid sphere is placed in a uniform

0

(c)

6 q L2 0

(d)

q 2

6L

0

32. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surface of the solid sphere and the outer surface of the hollow shell is V. If the shell is now given a charge of – 3Q, the new potential difference between the same two surfaces is (a) V (b) 2 V (c) 4 V (d) – 2 V IIT, 1989 33. 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 the centre of the other is

lines of force follow? (a) 1 (c) 3

(b) 2 (d) 4 IIT, 1996

Fig. 20.28

37. A charge + q x = 3 x0, x = 5x0

x = x 0, q is x = 2x0, x = 4x0, x = 6x0 x0 is a positive constant. The potential at the origin to this system of charges is (a) zero

(b)

4

0

q x0 ln 2

20.14 Comprehensive Physics—JEE Advanced

q ln 2 4 0 x0 IIT, 1998 38. Three charges Q, + q and + q are placed at the vertices of a right-angled isosceles triangle as shown in Fig. 20.29. The net electrostatic energy Q is equal to (a)

q 1

(b)

2

(c) – 2q

2q 2

2

(d) + q IIT, 2000

43. Two identical charges are placed at the two corners of an equilateral triangle. The potential energy of the system is U. The work done in bringing an (a) U (b) 2 U (c) 3 U (d) 4 U 44. The magnitude of electric intensity at a distance x from a charge q is E. An identical charge is placed at a distance 2x from it. Then the magnitude of the force it experiences is (a) qE (b) 2qE (c)

qE 2

(d)

qE 4

45. A particle carrying a charge q is shot with a speed v Q. It approaches Q up to a certain distance r and then returns as shown in Fig. 20.30.

Fig. 20.29

Fig. 20.30

39. Eight dipoles of charges of magnitude e are placed the cube will be 8e (a) 0

(c)

e

(b)

16 e

(c)

0

(d) zero

0

40. A point Q lies on the perpendicular bisector of an electrical dipole of dipole moment p. If the distance of Q from the dipole is r (much larger than the size Q is proportional to (b) p and r–2 (a) p–1 and r–1 2 –3 (c) p and r (d) p and r–3 41. A particle of mass m and charge q is released E. The kinetic energy attained by the particle after moving a distance x is (a) qEx2 (b) qE 2x (c) qEx (d) q2Ex 42.

3

Vm–1 along the y-axis. A body of mass 1 g and charge 10–6 C –1

positive x-axis with a velocity of 10 ms . Its speed (in ms–1) after 10 second will be (neglect gravitation) (a) 10 (c) 10 2

(b) 5 2 (d) 20

If charge q were moving with a speed 2v, the distance of the closest approach would be (a) r (b) 2r r 2

(d)

r 4

46. Four charges, each equal to – Q, are placed at the corners of a square and a charge + q is placed at its centre. If the system is in equilibrium, the value of q is (a)

Q 1 2 2 4

(b)

Q 1 2 2 4

(c)

Q 1 2 2 2

(d)

Q 1 2 2 2

47. A charge having magnitude Q is divided into two parts q and (Q – q) which are held a certain distance r apart. The force of repulsion between the two parts will be maximum if the ratio q/Q is (a)

1 2

(b)

1 3

1 1 (d) 4 5 48. A charge Q is given to a hollow metallic sphere of radius R. The electric potential at the surface of the sphere is (c)

Electrostatic Field and Potential 20.15

(a) zero

(b)

1

(c)

1 4

Q R

0

Q

(d) 4 0 Q/R 4 0 R2 49. In Q. 48, the potential at a distance r from the centre of the sphere where r < R is (a) zero (c)

(b)

1 4

Q 0

R

r

(d)

1 4 4 R

55. 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 (see Fig. 20.31).

Q R

0

r

0Q

r

50. The electric potential V at any point (x, y, z) in space is given by V = 4x2 volt where x, y and z are 2 m) in Vm–1 is (a) 8 along negative x-axis (b) 8 along positive x-axis (c) 16 along negative x-axis (d) 16 along positive x-axis

Fig. 20.31

IIT, 2001 IIT, 1992

51. Two spheres of radii r and R carry charges q and Q respectively. When they are connected by a wire, there will be no loss of energy of the system if (a) qr = QR (b) qR = Qr (c) qr2 = QR 2 (d) qR 2 = Qr 2 52. Two equal point charges of 1 C each are located at points ( i j k ) m and (2 i 3 j k ) m. What is the magnitude of electrostatic force between them? (a) 10–3 N (b) 10–6 N (c) 10–9 N (d) 10–12 N 53. Three equal point charges q are placed at the corners of an equilateral triangle. Another charge Q is placed at the centroid of the triangle. The system of charges will be in equilibrium if Q equals (a)

q 3

(b) –

q 3

q q (d) – 3 3 54. A metallic sphere A of radius a carries a charge Q. It is brought in contact with an uncharged sphere B of radius b. The charge on sphere A now will be (c)

aQ (a) b (c)

bQ a b

bQ (b) a (d)

aQ a b

V where 0 is the t 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 (a) resistance (b) charge (c) voltage (d) current IIT, 2001 57. 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: (a) VA < VB (b) VA > VB 56. A quantity X is given by

(c) VA < VC

0L

(d) VA > VC

IIT, 2001 58. Two equal poi 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 (a) x (b) x 2 (c) x3

(d) 1/x IIT, 2002

59. A metallic shell has a point charge q kept inside its circular cavity. The charge is not exactly at the centre of the cavity. Which of the diagrams in Fig. 20.32 correctly represents the electric lines of force?

20.16 Comprehensive Physics—JEE Advanced

q

(a) + q 3 alone

(b) + q1 and + q3

(c) + q1, + q3 and – q2

(d) + q1 and – q2 IIT, 2004

62.

charge densities 1 = – , 2 = + 2 and 3 = + 3 are placed parallel to the x–z plane at y = a, y = 3a and y = 4a at point P is

Fig. 20.32

IIT, 2003 60. Three negative point charges – q each, and three positive point charges + q, + q and + Q are placed at the vertices of a regular hexagon as shown in Fig. 20.33. For what value of Q O A, B, D, E and F be O due to charge Q at C alone? q (a) q (b) 2 (c) 2q

(d) 5q

Fig. 20.35

(a) zero

2

(b)

j 0

3

(c)

j

(d)

0

IIT, 2004

3

j 0

IIT, 2005 63. A metallic sphere of radius R is charged to a potential V distance r (> R) from the center of the sphere is V r

(b)

(c)

VR

(d) zero

r2 64. Two point charges q1 = 1 C and q2 = 2 C are placed at points A and B 6 cm apart as shown in Fig. 20.36. A third charge Q = 5 C is moved from C to D along the arc of a circle of radius 8 cm as shown. The change in the potential energy of the system is

Fig. 20.33

61. Figure 20.34 shows a spherical Gaussian surface and a charge distribution. When the Gaussian surface, be due to

(a)

Fig. 20.34

Vr R2

Fig. 20.36

(a) 3.0 J

(b) 3.6 J

(c) 5.0 J

(d) 7.2 J

Electrostatic Field and Potential 20.17

65. A partical of mass m and charge + q is midway

when another positive point charge is moved from (– a, 0, 0) to (0, a, 0) is (a) positive (b) negative (c) zero (d) depends on the path connecting the initial and

a charge + q and at a distance 2L apart. The middle charge is displaced slightly along the line joining oscillation is proportional to. (a) L1/2

(b) L

(c) L3/2

(d) L2

70.

66. Five point charges, each equal to + q, are placed L. The magnitude of the force on a point charge – q placed at the centre of the haxagon is (a) (c)

5q2 4

2 0L

q2 4

2 0L

(b)

3q 2 4

2 0L

(d) zero

IIT, 1992 67. Two isolated metal spheres of radii R and 2R are charged such that both have the same surface charge density . The spheres are located far away from each other. When they are connected by a thin conducting wire, the new surface charge density on the bigger sphere will be 2 3 (b) (a) 3 5 5 (c) (d) 6 2 IIT, 1996 68. A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in Fig. 20.37. the emptied space is (a) zero everywhere (b) non-zero and uniform (c) non-uniform (d) zero only at its centre

71.

72.

Fig. 20.37

IIT, 2007 69. Positive and negative point charges of equal a a magnitude are kept at 0, 0, and 0, 0, 2 2

73.

IIT, 2007 A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. (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 IIT, 2007 Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then, (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 IIT, 2008 Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, 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 (a) 1 : 2 : 3 (b) 1 : 3 : 5 (c) 1 : 4 : 9 (d) 1 : 8 : 18 IIT, 2009 A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (– a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x = 5a/4. Two point charges – 7C and 3C are placed at (a/4, – a/4, 0) and (– 3a/4,3a/4, 0), respectively. Consider a cubical surface formed by

20.18 Comprehensive Physics—JEE Advanced

six surfaces x = ± a/2, y = ± a/2, z = ± a/2. The elec-

E = E0 x , where E0 is area (as

74.

20.38] 2C

(a)

(b)

0

0

10 C

(c)

2C

(d)

0

12 C 0

Fig. 20.39

Fig. 20.38

(a) 2E0a2

(b)

(c) E0a2

(d)

2 E 0a 2 E0 a 2 2

IIT, 2009 ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73.

(a) (b) (c) (d) (d) (a) (d) (b) (a) (c) (c) (c) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74.

(d) (d) (b) (c) (a) (a) (b) (d) (a) (d) (c) (b) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69.

(d) (c) (d) (d) (d) (b) (d) (d) (b) (b) (c) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70.

(a) (d) (b) (b) (b) (b) (d) (a) (a) (b) (b) (a)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71.

(c) (c) (c) (b) (b) (d) (c) (a) (b) (c) (c) (d)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72.

(a) (b) (a) (d) (c) (d) (c) (b) (d) (b) (c) (b)

SOLUTIONS 1. or

W = Uf – Ui

Ui =

1 4

0

r

+

Q –Q

[(q) (–2q) + q(–2q) + (–2q) (–2q)]

2.5

=0 Uf =

2

1 4

0

(2 r )

[(q) (–2q) + q(–2q) + (–2q) (–2q)]

=0 W = 0. So the correct choice is (a). 2. Charge Q at point P will induce a charge – Q on the inner surface and a charge + Q on the outer surface of the shell (Fig. 20.40). The electric potential at O is

Fig. 20.40

V=

1 4

0

Q R

Q 2R

Q 2.5 R

Electrostatic Field and Potential 20.19

=

1 4

4q2 + 4qQ = 0

9Q 10 R

0

Hence Q = – q. The force on charge 4q is

3. Before connection, charge Q on the inner shell induces a charge – Q on the inner surface of the outer shell and a charge + Q on its outer surface. Therefore, the total charge on the outer surface of the outer shell = Q + 2Q = 3Q. When the two shells are connected by a conducting wire, the entire charge Q on the inner shell is conducted to the outer shell. Therefore, the charge on the outer shell now is Q + 2Q = 3Q, the same as before. Hence there will be no change in its electric potential. Q

4. E =

2

4 0R 4 0R Hence if R is doubled, E becomes E/4 and V becomes V/2. So the correct choice is (a). Q . In terms of , 5. = 4 R2 E=

and V = 0

R 0

If R is halved, E remains the same but V becomes V . So the correct choice is (c). 2

4q 2

4 qQ

F= 4

2

l 2

0

4

0

l2

Putting Q = – q, we get F =– | F |=

4q 2

16 q 2 4

0

3q 2 0

Q

and V =

4q (q + Q) = 0

l2

l2

4

0

l2

=–

3q 2 0

l2

, which is choice (c).

10. Refer to Fig. 20.42. The three charges will be in equilibrium if no net force acts on each charge. Charge q is in equilibrium because the forces acting on it by charge Q at A and charge Q at B are equal and opposite. Charge Q at A will be in equilibrium if Q2

qQ 4

0

l

2

4

0

(2 l )

2

=0

q=–

Q . 4

Similarly charge Q at B will be in equilibrium if Q . Hence the correct choice is (d). q=– 4

6. If p is the electric dipole moment, then EP =

2p 4

EQ = –

0

r3

Fig. 20.42

p 4

0

11. If R is the radius of the big drop, we have

( 2 r )3

EP = – 16 EQ, which is choice (a). 7. The potential inside a spherical conductor is constant including that on its surface. Hence the correct choice is (b). 8. The potential at every point on the circle due to charge Q is the same. Work done = q (potential difference). Hence the correct choice is (d). 9. Refer to Fig. 20.41.

Fig. 20.41

The net force on q will be zero if q (4 q) 4

0

l

2

qQ 4

0

l 2

2

=0

4 R3 4 r3 = 1000 3 3 which gives R = 10 r. The electrical potential of each droplet is q v = 4 0r and that of the big drop is V =

1000 q 4 0R

V 1000 r = = 100 ( R = 10r) v R Hence the correct choice is (c). 12. When two identical metallic spheres are brought in contact, the charges on them are equalized due to sphere C is brought in contact with sphere A having a charge + q, and then removed, the total charge q

20.20 Comprehensive Physics—JEE Advanced

is equally shared between the two so that the charge left on A is +q/2 and that developed on C is +q /2. The sphere C carrying a charge +q/2 is now brought in contact with sphere B which is already carrying a charge +q. The total charge is q / 2 + q = +3q/2 which must distribute equally on B and C. Thus when C is removed, B will have a charge of +3q/4 and C also has a charge of +3q/4. Hence when C is removed from both A and B, q New charge on A = + 2 3q New charge on B = + 4 Since force is proportional to the product of the charges, it follows that the new force of repulsion between A and B is 3/8 of the earlier force (F). Hence, the new force of repulsion between A and B is 3F/8. 13. Refer to Fig. 20.43. Let us consider forces on a ball, say, Q. Three forces act on it: (i) tension T in the thread, (ii) force mg due to gravity and (iii) force F due to Coulomb repulsion along + x-direction. For equilibrium, the sum of the x and y components of these forces must be zero, i.e. T cos 60° – F = 0 T sin 60° – mg = 0

and

1

E1 =

q1

4 0 a2 and is directed along + x E2 at (a, b) due to q2 has a magnitude 1 q2 E2 = 4 0 b2 and is directed along + y-axis. The angel E with the x-axis is given by

Fig. 20.44

tan

=

E2 q a2 = 2 2 E1 q1 b

=

1 2

2 1

2

=2

Hence the correct choice is (b). 15.

E exerts a force q E on charge + q and a force – qE on charge – q of the dipole. Since these forces are equal and opposite, they add upto zero. 16. The correct choice is (b). A torque acts on the 17. The correct choice is (c). In a non-uniform electric

Fig. 20.43

These equations give F = mg cot 60° = 10

1 3

= 10–3 N. Now F =

Putting 1 and 4 14.

F = 10 =9

0

–3

10 –3

V =

q2

1 4

3 10

translational motion and a torque which gives it a rotational motion. 18. The distance of a vertex from the the centre of the cube of side b is r = 3 b / 2. Now the potential due to charge q at the centre is q/4 0r. Hence the potential due to the arrangement of eight charges (each of magnitude q) at the centre is

r2

19.

N, r = 0.3 m

10 9, we get q = 10–7 coulomb.

0

due to q1 has a magnitude

E1 at (a, b)

8q 4

0r

=

4q 3

0b

torially. From the symmetry of the eight charges with respect to the centre of the cube, it is evident charges cancel in pairs (being equal and opposite). will be zero.

Electrostatic Field and Potential 20.21

20. Refer to Fig. 20.45. The distance of point P1 from charge + q is r1 = z – a and from charge – q is r2 = z + a.

Here q1 = q2 = q = + 1.6 10–19 C (proton), q3 = – q = – 1.6 10–19 C (electron), r12 = 1.5 Å = 1.5 10–10 m, r13 = r23 = 1 Å = 1 10–10 m and 1/4

=9

109 Nm2C–2. Thus

U= –

4 q 2 1010 joule 3 4 0

0

=– =–

=

1 4

0

q 4

0

q r1

q r2

=

4

0

q1 q2 r12

q1 q3 r13

1010

9 109

e2 joule 4 0 r 1 e =– eV 4 0 r

r2 r1 r1 r2

U= –

, 4 0 z 2 a2 which is choice (c). 21. Refer to Fig. 20.45 again. Any point on the perpendicular bisector passing through the centre of the dipole is at the same distance from the two charges. Hence the potentials at point P2(5, 0, 0) and at point P3(–7, 0, 0) are zero. Since P2 and P3 are at the same potential (zero), the potential difference between them is zero. Hence no work will be done in moving a charge from P2 to P3. The answer will not change if the path of the charge is changed because the work done is independent of the path taken. 22. Refer to Fig. 20.46. The total potential energy of the arrangement of charges is the sum of the energies of each pair of charges. The potential energy of the system comprising the three charges q1, q2 and q3 is U = W1 + W2 + W3 1

19

potential energy of the electron-proton system is

2qa

=

4 1.6 10

3 = – 19.2 eV 23. Charge on electron (– e) = – 1.6 10 –19 C, charge on proton (e) = 1.6 10–19 C, separation r = –10 0.53 Å = 0.53 10 m. If the zero of potential

Fig. 20.45

Potential at P1 =

4 q 1010 eV 3 4 0

=–

1

9 109

19

1.6 10

0.53 10 10 = – 27.2 eV Hence the correct choice is (b). 24. Points A, B and C are at the same distance from charge + q; hence electrical potential is the same at these points, i.e. there is no potential difference between A, B and C. Hence W1 = W2 = 0. 25. The electric potential at the common centre is q1 q2 V = + 4 0 r1 4 0 r2 Now

= V =

q1 4

r12

1

q2

4 r22

q1 r1 4

0

q2 q3 r23

=

=

r12

q2 r2

4 r22

(r1 + r2)

0

Hence the correct choice is (d). 26. If q surface is q E = 4 0 r2 Fig. 20.46

But

=

q 4 r2

. Therefore q = 4 r 2 .

20.22 Comprehensive Physics—JEE Advanced

Hence

E =

4 r2 4

0

r2

= 0

Thus the correct choice is (a). 27. Refer to Fig. 20.47. Forces exerted by charges – q at A and B on charge Q are F1 and F2 which are equal and have a magnitude F=

qQ 4

0

r2

Fig. 20.48

P due to charge 4 q at D is 1 4q E2 = along PB 4 0 r2 PB is E = E2 – E1 = q and 3q at A and C will be E =

1

; Fr is also proportional to (1/r2). r2 Hence charge Q will move towards the origin and because of its inertia it will overshoot the origin O. Thus, charge Q will oscillate about O but its motion is not simple harmonic because the force Fr is not proportional to its instantaneous distance from the origin as Fr 1/r2. Hence the correct choice is (d). 28. r from a point charge Q is given by Since F

E=

1 4

Q 0

r2

If Q from Q. Refer to Fig. 20.48. Let PA = PB = PC = PD = r P due to charge 2q at B is 1 2q along PD E1 = 4 0 r2

4

2q 0

r2

2q

4 0 r2 P due to charges

directed along PA.

Thus E = E , but they are mutually perpendicular to each other. therefore, their resultant will be along PQ (see Figure) which is parallel to CB. Hence the correct choice is (b)

Fig. 20.47

The resultant of these equal forces equally inclined with the x-axis is along the negative x-direction towards the origin. The magnitude of the resultant is given by F 2r = F 2 + F2 + 2 F2 cos = 2 F2 (1 + cos )

1

1

29.

E. The force on charge q is

FA = qE

qE m Similarly, the acceleration of charge 4q is 4q E aB = = 4 aA m Let s be the distance travelled by A and B to acquire speeds vA and vB. Then (since u = 0) v2A = 2 aA s and vB2 = 2 aB s

Therefore, its acceleration is aA =

v 2A vB2

=

aA aB

1 4

or vA/vB = 1/2. Hence the correct choice is (b). 30. Refer to Fig. 20.49. a AO = BO = CO = DO = r = 2 Since electric potential is a scalar and since the charges at corners are equal, the magnitude of the electric potential at point O due to the four charges = 4 times that due to a single charge. Thus 1 Q V=4 4 0 r

Electrostatic Field and Potential 20.23

1

=

Q a/ 2

0

=

34. Force F = qE. Therefore, acceleration a = qE/m. Hence the distance travelled by the particle in time t is 1 2 1 qE 2 at = t s= 2 2 m

2Q 0a

For electron,

1 2

se =

For proton,

qE me

1 2

sp =

t 21

qE mp

t 22

Given se = sp. Therefore t12 t2 = 2 me mp

Fig. 20.49

Work done = Q V = 31.

choice is (c). q

2Q 2 . Hence the correct 0a E

=

36.

1

V1 =

4 Potential at C2 is V2 =

0

1 4

0

Q1 R Q2 R

Q2

1

37. V =

4

+

0

1 4

1 4

q x0

0

q 3 x0

q 2 x0

0

=

q 5 x0

q 4 x0

1 2

q 1 x0

Q1 2R

gives the potential at the

to the surface and directed towards the centre of the sphere. Hence the correct choice is (d).

=

2R

me

surface of a conductor. On the surface of a metal-

0

32. When any additional negative charge is given to a hollow spherical shell, the potential on its surface falls, but the potential at each point within the shell also falls by the same amount. Hence the potential difference between the given surfaces remains unchanged. Thus, the correct choice is (a). 33. Refer to Fig. 20.50. Potential at C1 is

–E

1/ 2

centre of the ring, which is zero.

s

. Hence the correct choice is (a).

0

35. The integral

mp

t2 t1

or

1 3

q 6 x0

1 4

q 4

upto infinity upto infinity

1 5

1 6

Qq a

q log e 2 4 0 x0

loge (1 + 1) =

0 x0

38. The net electrostatic energy is ( of triangle = 2 a) U=

upto infinity

Qq

hypotenuse side

qq a

2a

For U = 0, we require Qq a

Fig. 20.50

Work done W = q (V1 – V2) = =

q 4

0

Q1 R

q 4

0

2R

Qq 2a

q2 = 0 or Q + Q + q = 0 a 2

which gives Q = – q Q2 2R

Q2 R

Q1 2R

(Q1 – Q2) ( 2 – 1)

2 2

1

=

2q 2

2

39. A dipole consists of two equal and opposite charges separated by a certain distance. Hence the total charge enclosed in the cube is zero. Therefore, the

20.24 Comprehensive Physics—JEE Advanced

40.

dicular bisector of a dipole is given by (for r >> a), here p is the dipole moment p

E=

4 0 r3 Hence the correct choice is (d). 41. Initial kinetic energy of the particle is zero. The gain in kinetic energy in distance x = decrease in to move the particle through a distance x = force distance = q Ex. Hence the correct choice is (c). 42. Given vx = 10 ms–1 directed along the y-axis, the acceleration of the body along the y-axis is ay =

qE m

10

6

10

103

45. Charge q will momentarily come to rest at a distance r from charge Q when all its KE is converted to PE, i.e. 1 qQ 1 mv2 = r 4 2 0 Therefore, the distance of closest approach is given by qQ 2 r= 4 0 mv 2 1 Thus r . Hence if v is doubled, r becomes v2 one-fourth. Thus the correct choice is (d). a 46. Let the side of the square be a. OA = OC = r = 2 (see Fig. 20.51).

= 1 ms–2

3

Therefore, the velocity of the body along the y-axis at time t = 10 s is vy = ayt = 1 × 10 = 10 ms–1 Resultant velocity v = (10) 2

=

vx2

(10)2

v 2y 10 2 ms

1

Hence the correct choice is (c). 43. Let q be the magnitude of each charge and a the length of each side of the triangle ABC. The potential energy of the system of two equal charges placed at vertices A and B is U (given). This means that U is the work done in bringing a charge q from B with the charge q at vertex A. Hence the work done in bringing an identical charge q C = work done to overcome the force of repulsion of q placed at A placed at a distance a from it + work done to overcome the force of repulsion of q placed at B at the same distance a from it = U + U = 2U, which is choice (b). q . Hence the magnitude of the 44. Given E = 4 0 x2 electric intensity at a distance 2x from charge q is E =

q 4

0 (2 x)

q 2

4

0x

2

1 4

E 4

Therefore, the force experienced by a similar charge q at a distance 2x is qE F = qE = 4 Hence the correct choice is (d).

Fig. 20.51

1. Stability of charge + q at the centre Charges – Q at corners A and C will attract charge + q with equal and opposite forces. Similarly charges – Q at corners B and D will attract charge + q with equal and opposite force. Hence no net force acts on charge – q. 2. Stability of charge –Q at any corner Q at corner A. This charge will experience four forces: (i) Force of repulsion F1 due to charge – Q at B (ii) Force of repulsion F2 due to charge – Q at D (iii) Force of repulsion F3 due to charge – Q at C (iv) Force of attraction F due to charge + q at O. Now

F1 = F2 =

and

F3 =

Q2 4

0a

2

Q2 4

0 ( 2r )

Q2 2

4

0 ( 2a

2

)

Electrostatic Field and Potential 20.25

and

q ( Q)

1

F =

2qQ

2

4 0 r 4 The resultant of F1 and F2 is given by F12

F=

F22

0a

2Q

2 F1 =

2

4 0a2 The forces F and F3 act along AP. Hence the net force acting on charge – Q at A due to charges – Q at B, C and D is F = F + F3 Q2

2 Q2

=

2qQ 4

0a

2

=

4

0 ( 2a

2

)

4

0 ( 2a

or

q r

dF For F to be maximum, = 0, i.e. dq q Q 0

r

q 2

=0

Since r d [q(Q – q)] = 0 dq q 1 which gives Q 2 Hence the correct choice is (a). 48. For points on the surface of the sphere or outside the sphere, a charged sphere behaves as if the charge is concentrated at its centre. Therefore, the potential at the surface of the sphere is given by 1

Q , which is choice (b). 4 0 R 49. At points inside a charged metallic sphere, i.e. for r < R, the potential is zero. Hence the correct choice is (a). dV 50. E = – i where i is a unit vector along the dx positive x-axis. Hence E at a point whose x-coordinate is x = 1 m is

3j

r=

q=

V=

4

0r

0R

Q . Hence the correct choice is (b). R

52. r = (2 i

)

d 1 dq 4

4

Q

k)

(i

k)

j

(i

2j

2 k ) m.

The magnitude of r is

Q 1 2 2 4 Hence the correct choice is (a). 47. The force of repulsion between the two parts is given by 1 q Q q F= 4 0 r2 or

q

V=

Q 2 (1 2 2 ) 2

8i Vm 1 .

8 xi

The negative sign shows that E is along the negative x-axis. Hence the correct choice is (a). 51. There will be no loss of energy if the potential of the spheres is the same i.e. if

Q 2 (1 2 2 )

4 0 a 2 4 0 ( 2a 2 ) For equilibrium, F = F , i.e.

d 4 x2 i dx

E=–

2

F= =

12

22

1

q1 q2

4

22

1 4

4 =3m

r2

0

9 109

6

10

10

6

= 10–3 N

2

3 Hence the correct choice is (a).

53. The system will be in equilibrium if the net force on charge q at one vertex due to charges q at the other two vertices is equal and opposite to the force due to charge Q at the centroid, i.e. (here a is the side of the triangle) 3 q2 4

0a

Qq

2

4

which gives Q = – is (b). 54.

q 3

0

a 3

2

. Hence the correct choice

A to B until their potentials become equal. If charge q A to B, then Q 4

q 0a

q 4

0b

a bQ q which gives q = . Hence b a b bQ aQ charge left on A = Q – q = Q – . a b a b Hence the correct choice is (d).

or Q – q =

55. Since all the three charges are of the same polarity, around them cannot be closed. Hence choices (a),

20.26 Comprehensive Physics—JEE Advanced

(b), the line in the middle is a closed circular loop. Thus, the only correct choice is (c). 56. The capacitance of a parallel plate capacitor is given by C = 0 A/d. Hence the dimensions of 0 L are the same as those of capacitance. V Dimensions of 0L t dimension of C dimensions of V = time dimension of Q = ( Q = CV) time charge = = current time Hence the correct choice is (d). 57.

59. Because the charge is not located at the centre of the cavity, inside the cavity the lines of force are skewed. Hence choice (a) and (b) are incorrect. Outside the shell, the lines of force are the same as if the charge were located at the centre of the cavity. Also there can be no line of force in the metallic body of the shell. Hence choice (d) also incorrect. Thus the correct pattern is shown in (c). 60. O due to charges – q at A and D are equal and opposite. Hence they cancel each other. Similarly charges + q at B and E do not contribute to electric at O. Due to charge – q at F O will be E1 =

Thus V decreases as dx increases in the direction VA VB , which is choice (b). 58. Potential energy of the system when charge Q is at O is

q 4

qQ a x

=

2qQ a

1

2qQ a

1

U = U – U0 = =

2qQ

Hence U

a

3

x2

0r

2

=

2Q 4

0r

a2

61. E.ds E is due to all the charges, both inside and outside the Gaussian surface. Hence the correct choice is (c). plane sheet carrying a uniform charge density given by E=

x2

1

a2 x2

(

a2 2qQ 1 a

x a

2 2

(x 2)

x 2 which is choice (b).

2

q . Hence the correct choice is (b). 2

qQ 2a

qQ a x

x

a)

O

directed from O to F 4 0 r2 E1 = 2E2, we require

For

62. Fig. 20.52

directed from O to F

2

Q

which gives Q =

When charge Q is shifted to position O , the potential energy will be (see Fig. 20.52).

U=

0r

E2 =

2qQ a

qQ a

4

O charges at A, B, D, E and F due to charge + Q at C is given by

dV E=– dx

qQ U0 = a

q

is

2 0 It is independent of the distance of point P from the sheet and is, therefore, uniform. The direction of dicular to it if is positive and is towards the sheet and perpendicular to it if is negative. Hence E1 =

2qQ a and

j along –ve y–direction

2

0

E2 =

2 2

0

E3 =

3 2

0

j along –ve y–direction j along – ve y–direction

Electrostatic Field and Potential 20.27

From the superposition principle, the net electric P is E = E1 + E2 + E3 2 = j 2 0 2 0 =–

3

3 2

j

F=

q 4

o

q (OD)2

=

q2 4

o

L2

j 0

j , which is choice (c). 0

63. Let the charge on the sphere be Q. Then Q V= 4 0R which gives Q = 4 0RV r is Q

E=

=

4

0 RV 2 0r

=

Fig. 20.53

RV

4 0r 2 r2 4 Thus the correct choice is (c). 64. If charge Q is moved from C to D along the arc, the potential energy between pairs (q1, Q) and (q1, q2) will not change as the distance between them remains unchanged ( AC = AD). The potential energy of the pair of chages q2 and Q will change. Now, distance BC = (8)2 (6) 2 = 10 cm and BD = 8 – 6 = 2 cm. Therefore, change in P.E. is

10–6)

= (2

U=

q2Q 1 4 0 BD

(5

10–6)

(9

1 BC 1 0.02

10–9)

1 0.1

65. If the middle charge is displaced by a distance x, the net force acting it, when it is released, is

=

q2

1 4

0

(L

q2

1

x)2

4

0

(L

x)2

4q 2 Lx 4

2 0 (L

For x R, the entire charge Q may be assumed to be concentrated at the centre of the sphere. (c) the electric potential is maximum at centre of the sphere. (d) the electric potential is zero at centre of the sphere. 11. A sphere of raduis R is made of a conducting material and carries a charge Q. Choose the correct statements from the following. (a) Charge Q resides on the surface of the sphere.

0r

Fig. 20.60

for r > r0

(d) No work is needed to move a charge from one point to another on the surface of the sphere. 12. A proton and an electron are placed in a uniform enced by proton and electron are Fp and Fe and ap and ae are the respective magnitudes of their accelerations. Then (a) Fe > Fp (b) Fe = Fp (c) ae > ap (d) ae < ap 13. A metal ring of radius R carries a charge Q distributed uniformly on it. A point P lies on the axis of the ring at a distance r from its center. Then (a) The potential at the centre of the ring is zero. zero (c) For r >> R, the potential varies as 1/r. (d) For r >> R, r 2.

Which of the following statements are true? discontinuous at r = r0. (b) The net charge enclosed in a sphere of radius r = 2r0 is (c) No charge exists at any point in a spherical region of radius r < r0. (d) Electrostatic energy inside the sphere of radius r = r0 is zero. IIT, 2006 10. A sphere of radius R is made of a nonconducting material and carries a positive charge Q. Figure 20.61 shows the variation of E with distance r from the Fig. 20.61 centre of the sphere.

14.

ged. A point P lies at a perpendicular distance r from it Then r. r 2. istant straight lines prependicular to the wire.

15.

P lises at a perpendicular distance r from it. Then r. r 2. distant straight lines prependicular to the plane sheet sheet.

Electrostatic Field and Potential 20.31

16. In Gauss’s theorem

E ds s

17.

18.

19.

20.

q

. The surface

0

integral is evaluated by choosing a closed surface called the Gaussian surface. Here (a) the closed surface can be of any shape or size. (b) q is the net charge enclosed inside the Gaussian surface; charges outside the surface are not considered. (c) E both inside and outside the surface. (d) The exact location of the charges inside the surface does not affect the value of the integral. Two equal point charges q each are held at x = + a and x = – a. A third charge Q is placed at x = 0. The potential of the system will (a) decrease if Q is displaced by a small distance along the x-axis (b) increase if Q is displaced by a small distance along the x-axis (c) decrease if Q is displaced by a small distance along the y-axis (d) increase if Q is displaced by a small distance along the y-axis The electric potential at a point P at a distance x from a point charge is given by k V= r where k is a constant. (a) k is dimensionsless (b) the dimensions of k are [ML–3T–3A–1] k P= 2. r P = kr2. Two point charges q1= 4 C and q2 = –1 C are placed at x = 0 and x = 15 cm on the x-axis. (a) Electric potential is zero at x = 3 cm. (b) Electric potential is zero at x = 12 cm. x = 0 and x = 15 cm. x = ± 30cm. Choose the correct statement(s) from the following. (a) The electric potential due to a point charge at a distance r from it varies as 1/r (b) Electric potential at a distance r from the centre of a charged sphere varies as 1/r provided r is less than the radius of the sphere r from a point charge varies as 1/r2 IIT, 1980

21. Two small balls, each having a charge + Q are suspended by two insulating strings each of length L taken in a satellite orbiting the earth. In the satellite the angle between the strings is and the tension in each string is T. Then (a) = zero (b) = 180° (c) T =

Q2 16

2 0L

(d) T =

Q2 4

2 0L

IIT, 1986 22. A positively charged thin metal ring of radius R is x-y plane with its centre at origin O. A negatively charged particle P is released from rest at the point (0, 0, z) where z > 0. Then the motion of P is (a) periodic for all values of z satisfying 0 VQ. Hence statement (a) is correct. charge – q is given by V=–

1 4

0

q r

Referring to Fig. 20.35(b), the potentials at points A and B are q q 1 1 and VB = – VA = – 4 0 OA 4 0 OB Since OA < OB, the potential at A is more negative than at B, i.e. VB > VA. Hence statement (b) is in-correct. We know that the potential energy of a charge q2 q1 at a distance r from it is given by 1 q1 q2 U= 4 0 r Referring to Fig. 20.21(a) the potential energy of a small negative charge (– q0 tive charge (+ q) at points P and Q is (setting q1 = q and q2 = – q0) U p= –

1 4

0

q q0 1 and UQ = – OP 4

0

q q0 OQ

Since OP < OQ, it is clear that UQ > UP. Hence statement (c) is correct. Referring to Fig. 20.21(b), the potential energy of a small negative charge (– q0 tive charge (– q) at points A and B is (setting q1 = – q and q2 = – q0) UA =

1 4

0

q q0 1 and UB = OA 4

0

q q0 OB

Since OA < OB, UA > UB. Hence statement (d) is correct.

3. The small positive charge (+ q0) will tend to move from P to Q due to the force of repulsion exerted on it by the charge (+ q charge (+ q) does positive work on charge q0 to move it from P to Q. Hence the work done by the q) in moving the charge (+ q0) from Q to P will be negative. Hence statement (a) is incorrect. The work done by the external agency to move a negative charge (– q0) from B to A is positive, since the external agency has to overcome the force of repulsion exerted by the negative charge (– q) on the small negative charge (– q0). Hence statement (b) is correct. In going from Q to P, a small negative charge is speeded up due to the force of attraction exerted on it by the positive charge (+ q). Hence statement (c) is correct. In going from B to A, the small negative charge (– q0) is slowed down due to the force of repulsion exerted on it by the negative charge (– q). Hence the kinetic energy of the small negative charge decreases. Hence statement (d) is correct. 4. E in moving a charge (– q) around a closed path of any shape (circular or elliptical) is given by W = q E d1 Now, we know that the line integral of an electroE d1 = 0 Hence the work done W = 0 irrespective of whether the path is circular or elliptical. Hence statement (a) is correct. We know that the potential at points equidistant from a point charge is the same. Hence, the equicharge are concentric spheres with the point charge as the common centre. Hence statement (b) is correct. The derivation of Gauss’s law assumes 1/r2 dependence of distance between charges in Coulomb’s law. Gauss’s law will not hold if Coulomb’s law involved 1/r3 or any other power of the distance r. Hence statement (c) is incorrect. A single conductor can have capacitance. It is a caa single spherical conductor of radius r has a capacitance C = 4 0r. Hence statement (d) is incorrect.

20.34 Comprehensive Physics—JEE Advanced

5. Refer to Fig. 20.64. Since the bob is in equilibrium For equilibrium, we have mg = T cos and qE = T sin mg qE Thus T = = . cos sin

Now, AF = BF = 5 a/2 and CF = DF = a/2. Putting these values, we get VF =

1

q

1 5 Work done in carrying a charge –e from O to F is W = – e (VF – V0) = – eVF 1 qe = 1 a 5 0

Hence the correct choices are (c) and (b).

0

a

Hence the correct choice is (a) and (c). 7. The distance of the point of intersection of diagonals = side of the hexagon = a. The potential 1 q . Thereat this point due to each charge = 4 0 a 1 6q fore, total potential = 4 0 a point due to equal charges at opposite corners will cancel each other in pairs. So the correct choices are (b) and (c). 8. Refer to Fig. 20.66. Since the charge on sheet 1 is

Fig. 20.64

6. Refer to Fig. 20.65. Potential at O is V0 =

1 4

q r

0

q r

q r

q r

=0

E = /2 0 which points away from it; to the left in region I and to the right in regions II and III. Sheet E = /2 0. Since the charge of sheet 2 is negative, the direction E is towards it; i.e. to the right in regions I and II and to the left in region III.

Fig. 20.65

Refer to Fig. 20.65 again. (AE)2 = a2 + =

5 a2 , giving AE = 4

and BE =

5a . Similarly DE = 2

a 2

2

5a 2

a a , CE = . Potential at E is 2 2

VE =

1 4

0

q AE

q BE

q DE

q CE

Fig. 20.66

=0

Work done in carrying a charge – e from O to E is W = – e (VE – VO) = – e (0 – 0) = 0 Potential at F is q q q q 1 VF = 4 0 AF BF DF C F

Region I Region II

: E1 = E – E = – : E2 = E + E =

2

2 0

2

0

2

=0 0

= 0

directed to the right. Region III : E3 = E – E = 0. Thus the correct choices are (a); (b) and (c).

0

Electrostatic Field and Potential 20.35

r > r0 is given by dV d Q E= – =– dr dr 4 0 r

9.

= For r

Q 4

r 0, E = –

0r

2

d Q dr 4 0 r0

=0 (

0

E

4 r2 =

Q 0

or

Q 4

0r

2

4 r2 =

Q 0

or Q = Q, which is independent of r as long as r is greater than r0. Hence statement (b) is also true. All the four choices (a), (b), (c) and (d) are correct 10. For r R, E =

Qr 4

0R

3

, i.e. E

and

r0 = constant) r = r 0.

Therefore, statement (a) is true. For r < r0, E = 0. Hence the charge resides only on the spherical surface of radius r = r0. No charge exists in the region for which r < r0. Therefore, statement (c) is also true. Electric energy density is given by 1 2 u= 0E 2 Since for r < r0, E = 0; u = 0 for r < r0. Hence statement (d) is true. Let Q be the net charge enclosed inside the spherical surface of radius r = 2r0. Then from Gauss’s theorem, we have Q E = or

12. Force on proton is Fp = eE, in the direction of E and force of electron is Fe = eE, opposite to the direction of E. Also ap = Fp /mp and ae = Fe /me. Since mp > me; ap < ae. So the correct choices are (b) and (c). 13. P are Q 1 V= 2 4 0 ( R r 2 )1 / 2 E=

1 4

Qr 0

(R

2

r 2 )3 / 2

,

which is zero for r = 0. Q Q and E = . So the For r >> R, V = 4 0r 4 0r 2 correct choices are (b), (c) and (d). P is given by

14. E=

n 2

0r

Where is the linear charge density and n is a unit vector perpendicular to the wire. So the correct choices are (a) and (d). 15. P is n

E=

2 0 where n is a unit vector perpendicular to the sheet. Since E is independent of r form. Hence the correct choices are (c) and (d). 16. All the four choices are correct. 17. Initial P.E. of the system [Fig. 20.67(a)] is U= =

1 4

0

q 4

0a

qQ a

qQ a

2Q

q 2

q2 2a (1)

r

For r > R, the entire charge Q may be assumed to be concentrated at the centre of the sphere. Hence for r > R, Q , i.e. E 1/r2 E = 4 0r 2 Since the sphere is made of a non-coducting meterial, charge Q is uniformly distributed over the entire volume of the sphere. For a unifrom distribution of charge, the potential is maximum at the dicular to the surface of a conductor. Hence the correct choices are (b) and (c). 11. For a conductor, all the four choices are correct.

Fig. 20.67

If charge Q is given a displacemet x along the x-axis the P.E. of the system becomes [Fig. 20.67 (b)]

20.36 Comprehensive Physics—JEE Advanced

U = =

1 4

qQ (a x)

0

q 4

2Q

0a

2

2

a (a

2

2

q 2

2

x )

state of weightlessness. The distance between the balls = 2 L. Hence

q2 2a

qQ (a x)

(2)

2

Since a > (a – x ); it follows from Eqs. (1) and (2) that U > If charge Q is given a displacement y along the y – axis, the P.E of the system will be [Fig. 20.67(c)] U = =

1 4

0

q 4

qQ r

2

qQ r

2aQ r 0a

q 2a

q 2

(3)

Since r > a, it follows from Eqs. (1) and (3) that U < U. So the correct choices are (b) and (c). work [ML2 T 2 ] . Therefore [V] = charge =[ML2T–3A–1] [k] = [V] [r]

18. V =

= [ML2T–3A–1] 3 –3

T=

Q2 (2 L) 2

4

So the correct choices are (b) and (c). 22. If Q1 is the charge on the ring and – Q2 is the charge on particle P, the force due to the ring on particle P when it is at (0, 0, z) is F =–

1

Q1 Q2 z

(R2

4

z 2 )3 / 2

When z > 0, F is in the – z direction. When z < 0, F is in the + z direction. So the motion of P will be periodic for 0 < z < . 1 Q1 Q2 z , i.e. F – z. Hence When z QB. So choice (b) is correct. cal shell is zero. So choice (a) is correct.

2

(

are wrong because, for elliptical orbit, the speed of charge – q is the highest when it is closest to charge Q (as in planetary motion). Hence the only correct choice is (a). 28. Let QA and QB are the charges on metal shell A and metal sphere B after they are connected by a wire. Since their electric potentials will be equal, VA = VB

Q

Now

A

=

QA 4 A

and

RA2

=

B

QA QB

B

RB RA

QB

=

4 RB2 2

RA RB =

RB RA

2

RB RA

Hence choice (c) is also correct. are A

EA = And

0 B

EB =

0

EA = EB

A B

< 1, i.e. EA < EB.

So choice (d) is also correct. All the four choices are correct. 29. Gauss’s law is valid only if Coulomb's law holds, i.e. if E r–2. Hence choice (a) is wrong. Gauss’s

0

or Q = Q, which is independent of r as long as r is greater than r0. Hence statement (b) is also true. All the four choices (a), (b), (c) and (d) are correct 27. The torque of Coulomb force (which is radial) on dL charge – q is zero. Hence = 0 L = constant. dt Hence the angular momentum of charge – q is constant. So choice (a) is correct. All other choices

distribution around an electric dipole. So choice (b) is also wrong. Choice (c) is correct becausse the directions of similar charges. Work done WA

= q(VB – VA) = (VB – VA) ( q = +1 C) Hence choice (d) is correct. So the correct choices are (c) and (d). B

20.38 Comprehensive Physics—JEE Advanced

III Multiple Choice Questions Based on Passage Questions 1 to 4 are based on the following passage Passage I q, are placed along the x-axis at x = 1, x = 2, x = 4, x = 8 .... and so on. 1. The electric potential at the point x = 0 due to this set of charges is q q (b) (a) 2 0 0 q q (c) (d) 3 0 4 0

3. If the consecutive charges have opposite sign, the electric potential at x = 0 would be q

(a)

3

(a) (c)

q 3

(b) 0

q 5

(d) 0

q 4

0

0

q q (d) 5 0 6 0 4. If the consecutive charges have opposite sign, the x = 0 would be q

(a)

3

=

=

0

q 4

q 1 1

0

1

q 4

0

q 2 1 2

q 4 1 4

q 8 1 8

6

5

0

q

(d)

6

0

0

0

to to

1 2 1 1 2

(1 0) q = = , 1 4 0 2 0 2 which is choice (b). 2. Since the charges are placed along the same straight x = 0 will be directed along the x-axis and its magnitude is given by q

4

0

E=

1 4

q

q

q

2

2

2

82

1

0

q 4

q

=

=

2 1 4

1 0

Fig. 20.69

1

q

(b)

q

(c)

=

4

0

q

SOLUTION 1. The potential at x number of charges placed on the x-axis as shown in Fig. 20.69, is

V=

4

(c)

x = 0 is

2.

q

(b)

1

q 4

0

1 16

0

to

1 64

to

1 4 1 1 4

(1 0) q = 3 3 4

q 4

4

, which is choice (a). 0

3. If the consecutive charges have opposite sign, the potential at x = 0 is given by V= = –

1 4

0

q 4 1 2

q 1

q 2

q 4

1

1 4

1 16

0

1 8

1 32

to

q 8

q 16 to

q 32

to

Electrostatic Field and Potential 20.39

=

=

1

q 4

4 3

q 0

1 2

1

1 1 4

=

4 q = 3 6

–

1 1 4

0

4

1 2

0

Hence the correct choice is (d). 4. E =

1 4

q

q

2

2

2

1

0

q (16)

q

2 q

2

(32)

2

4

=

q 4

0

1 4

1 64

=

to

Question 5 to 9 are based on the following passage Passage II A rigid insulated wire frame in the form of a right-angled triangle ABC, is set in a vertical plane as shown in Fig. 20.70. Two beads of equal masses m each and Fig. 20.70 carrying charges q1 and q2 are conn-ected by a cord of length l and can slide without friction on the wires. The beads are stationary. IIT, 1978 5. The value of angle is (a) 30° (b) 45° (c) 60° (d) 75° 6. The tension in the chord is q1q2 (a) mg (b) 4 0l 2

1 256

1 1024

to

1

q 4

0

q (8)2

1 16

1

q 4

0

1 4

1 1 16 16 15

1 16 4 15

to

1

1 16

1

q 5

0

Hence the correct choice is (c).

(c) mg +

q1q2

(d) zero 4 0l 2 7. The normal reaction on bead P is (a)

2 mg

(b) 2 mg

(d) 3 mg (c) 3 mg 8. The normal reaction on bead Q is (a) mg

(b)

2 mg

(c) 2 mg

(d)

3 mg

9. If the cord is cut, the magnitude of the product q1 q2 of the charges for which the beads continue to remain stationary is (a)

mgl 2 4 0

(c)

3 (4

(b) )mgl2

(d) (4

3mgl 2 4 0 )mgl2

SOLUTION F =

1 4

q1 q2 0

l2

(iv) Normal reaction N1.

Fig. 20.71

Let us consider forces acting on bead P as shown in Fig. 20.71. These forces are: (i) Weight mg vertically downwards (ii) Tension T in the string (iii) Electric force between P and Q given by

The net force along the string is (T – F). Bead P will be in equilibrium, if the net force acting on it is zero. Resolving forces mg and (T – F) parallel and perpendicular to plane AB, we get, when the bead P is in equilibrium, mg cos 60° = (T – F) cos (1) and

N1 = mg cos 30° + (T – F) sin

(2)

For the bead at Q, we have mg sin 60° = (T – F) sin and N2 = mg cos 60° + (T – F) cos

(3) (4)

20.40 Comprehensive Physics—JEE Advanced

5. Dividing Eq. (3) by Eq. (1), we get tan = tan 60° or = 60°, which is choice (c). 6. Using = 60° in (3), we have mg sin 60° = (T – F) sin 60° 1

q1 q2

+ mg 4 0 l2 So the correct choice (c). 7. From Eq. (2) we have (since T – F = mg) N1 = mg cos 30° + mg sin 60° or T = F + mg =

= 2 mg cos 30° =

(5)

3 mg;

which is choice (c). 8. From Eq. (4) we have

N2 = mg cos 60° + mg cos 60° = mg Thus is the correct choice is (a). 9. When the string is cut, T = 0. Putting T = 0 in Eq. (5), we get 1 q1 q2 mg = – 4 0 l2 The right hand side of this equation should be positive which is possible if q1 and q2 have opposite signs. Thus, for equilibrium the beads must have unlike charges. The magnitude of the product of the charges is |q1 q2| = (4 0) mgl2, which is choice (d).

Questions 10 to 13 are based on the following passage (a) T = 2

Passage III Two charges, each equal to q, are kept at x = – a and x = a on the x-axis. A particle of mass m and charge q0 = q/2 is placed at the origin. 10. The charge q0 is given a small displaceplacement x (R), E=

Fig. 20.76

=

4 3

U=

Q2 8

0

R

So the correct choice is (c). 24. For points outside the sphere, it behaves if the entire charge Q is concentrated at its centre. Hence the correct choice is (a).

Electrostatic Field and Potential 20.45

25. Take the electronic charge as ‘e’ and the permittivity as ‘ 0’. Use dimensional analysis to determine the correct expression for p.

Questions 25 to 26 are based on the following passage Passage VII A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing treated as neutral plasma. Let ‘N’ be the number density of free electrons, each of mass ‘m’. When the electrons are becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency ‘ p’ which is called the plasma frequency. To sustain the oscillations,

26.

an angular frequency

, where a part of the energy is approaches all the free electrons are set to resonance together and p

(a)

Ne m 0

(b)

m 0 Ne

(c)

Ne2 m 0

(d)

m

0

Ne2

tion will occur for a metal having the density of electrons N 4 1027 m–3. Taking 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

IIT, 2011

SOLUTION 25. From F =

q1q2 4

= [ML3T–2]

0r

2

, dimensions of

26.

e2

= [Fr2]

p=

0

=

[N] = [L–3]. Hence Ne2 m 0 2 1/ 2

Ne m 0

= [L–3] –1

Ne2 m 0

4 1027 10

= 3.2

[ML3T–2]

[M–1] = [T–2]

= [T ] = dimension of

p

= =

p.

1/ 2

p

2 c

10

(1.6 10 30

15

10

)

11

rad s–1

. Since c = =

19 2 1/ 2

,

3 108

2 15

3.2 10

p

6

10–7 m = 600 nm

IV Assertion–Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1 (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1.

(c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 Figure 20.77 shows the tracks of two charged partices A and B two charged plates. The charge to mass ratio of B is greater than that of Neglect the effect of gravity.

20.46 Comprehensive Physics—JEE Advanced

7. Statement-1 In Q. 6 above, the work done to move an electron from P to Q and then back to P is zero. Statement-2 8. Statement-1 Fig. 20.77

Statement-2 The vertical acceleration of particle B is greater than that of particle 2. Statement-1 A positive charge + q is located at the centre of a circle as shown in Fig. 20.78. W1 is the work done in taking a small positive charge + q0 from A to B and W2 is the work done Fig. 20.78 in taking the same charge from A to C. Then W2 > W1 Statement-2 Work done = Charge potential difference.

9. Statement-1 Electrons move from a region of higher potential to a region of lower potential. Statement-2 An electron has less potential energy at a point where the potential is higher and vice versa.

3. Statement-1 A small test charge is initially at rest at a point in released, it will move along the line of force passing though that point. Statement-2 The tangent at a point on a line of force gives the 4. Statement-1

-

tial must also be zero at that point. Statement-2 potential.

5. Statement-1 If electric potential is constant in a certain region of Statement-2 potential. 6. Statement-1 If an electron is moved from P to Q, its potential energy increases (see Fig. 20.79) Statement-2 Potential at Q is less than that at P.

Fig. 20.79

moving an electron around it in a complete orbit is greater if the orbit is elliptical than if the orbit is circular. Statement-2

10. Statement-1 The equipotential surfaces corresponding to a x-direction are equidistant planes parallel to the y-z plane. Statement-2 Electric is normal to every point on an equipotential surface 11. Statement-1 is uniform. Statement-2 point charge is a sphere with the charge at its center. 12. Statement-1 If a metallic sphere A of radius r carrying a charge Q is brought in contact with an uncharged metallic sphere B of radius 2r, the charge on sphere A reduces to Q/3. Statement-2 A to B until their potentials are equalised. 13. Statement-1 A metal ring of radius R carries a charge + Q distributed uniformly. A point charge – q is placed on the axis of the ring at a distance x = 2R from the centre of the ring. If the charge is released from rest, it will execute a simple harmonic motion along the axis of the ring. Statement-2 In simple harmonic motion, the restoring force acting on the oscillator is proportional to (– x), where x is the displacement from the mean position. IIT, 1988

Electrostatic Field and Potential 20.47

charge Q uniformly distributed on the surface is Q . given by 4 0R

14. Statement-1 For practical purposes, the earth is used as a reference at zero potential in electrical circuits. Statement-2 The electrical potential of a sphere of radius R with

IIT, 2008

SOLUTION 1. The correct choice is (a). Let E be the electric

it from Q back to P an equal positive work is done

downwords. If q is the charge of the particle, it will experience a force F = qE. Hence its acceleration (in the vertical direction) is

So the correct choice is (a). 8. The correct choice is (d). The work done by the

F qE m m where m is the mass of the particle. If t is time spent by the particle between the plates (i.e in the region

path of any shape (circular or elliptical) is zero. 9. The correct choicce is (d). Since the electron has a negative charge, it has less energy at a point where the potenial is higher and vice versa. Hence in an

a=

lower potential to a region of higher potential. y=

1 2 qEt 2 at = 2m 2

q y. Since particle B m tion, it has a higher charge to mass ratio. The correct choice is (d). Points A, B and C are at the same distance from charge + q. Hence electric potential difference between points A, B and C is zero. Hence W1 = W2 = 0. The correct choice is (d). The test charge will move along the line of force if the line of force is straight (as in the case of a single charge). If the lines of force is curved, the charge will not move along the line of force. The reason is that the line of force does not give the direction of velocity, it gives the direction of the force which is along the tangent to the curve at that point. The correct choice is (d). Since E = – dV/dr, if E = 0, V = constant not necessarily equal to zero. The correct choice is (a). The correct choice is (c). Since charge of an electron is negative, P.E at P and Q is eq UP = – 4 0 (OP)

10.

Thus 2.

3.

4. 5. 6.

UQ = –

4

eq 0 (OQ )

Since OQ > OP, UQ is less negative than UP, i.e.UQ > UP. For the same reason, VQ > VP. 7. Since charge + q will attract the electron, work is done to move the electron from P to Q is negative

11.

tential surface. Therefore, for a constant electric x-direction, the equipotential surfaces are planes parallel to the y-z constant, the equipotential surfaces are equidistant from each other. The correct choice is (a). region around a point charge varies with distance E=

12.

4

0r

2

A to B until thier potentials become equal. If charge q A to B, then Q q q = 4 0r 4 0 ( 2r ) which gives Q – q = left on A = Q –

13.

Q

q 2

q=

2Q . Hence charge 3

2Q Q = . 3 3

x from its centre is given by E=

1

Qx

4

(R

2

x 2 )3 / 2

Force on charge – q is F=–qE=–

1 4

qQ x (R

2

x 2 )3 / 2

The motion will be simple harmonic if F – x which is true only if x C

-

6

4 3 6

Capacitance and Capacitors 21.13

29. capacitor A has a charge q on it whereas B is B

Fig. 21.29

Fig. 21.27

34.

q q

exists in the region

q -

30. plates is

C V and the other to V of the capacitors are connected together. When the positive ends are also

4 4

C V –V

4

C V –V

4

31.

35.

-

36. A capacitor of capacitance C

-

C V +V C V +V

-

and this capacitor is connected to a second uncharged capacitor of capacitance C 32. C = C = C3 = C4 = 4

C

of the total energy stored in the capacitors after

-

connection is

3 37. plate is A is d

Fig. 21.28

33.

A and B in

A d 3 A d

A d 4 A d

21.14 Comprehensive Physics—JEE Advanced

43.

Fig. 21.30

44.

38. plate is A d A d

3 A d

3 A d

3

4 A d

45.

A and B and C = 3

C

3 Fig. 21.31

39. A capacitor of capacitance C V pacitor of capacitance C V

Fig. 21.32

46. R ,R

,C

C =4

40. K and conductivity K K K K Fig. 21.33

41. plates of a parallel plate capacitor increases its d is the separation of the two plates of the capacitor, the thickness of the d 3 7d 9

d 9 d

47. A capacitior of capacitance C = C is charged to

capacitances C = C and C3 = C

42. CV 3 CV

CV 3

Capacitance and Capacitors 21.15

48.

R is increased

d initially. 3

constant K radius R

R /R is

3

speed V

t is

4 3 49. A perallel plate rapacitor of plate area A and plate separation d V. d is AV d

AV d

Fig. 21.35

6 R d Vt

AV

3AV d d 50. One plate of a parallel plate capacitor of plate area A and plate separation d is connected to the positive V

d d

dVt

R 9V t

6 R d Vt d

plates is

d AV d

9Vt dVt

R 9V t

52.

AV d

Vd A

9Vt

the switch S

Fig. 21.34

51. A parallel plate capacitor C with plates of unit area and separation d

Fig. 21.36

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49.

2. 8. 14. 20. 26. 32. 38. 44. 50.

3. 9. 15. 21. 27. 33. 39. 45. 51.

4. 10. 16. 22. 28. 34. 40. 46. 52.

5. 11. 17. 23. 29. 35. 41. 47.

6. 12. 18. 24. 30. 36. 42. 48.

21.16 Comprehensive Physics—JEE Advanced

SOLUTIONS P.D. across C C

rr

1. C = 4

r

r

C =4

r

r

C C

=

E = r r = = r r

Q = CV

Equivalent capacitance is CC C C

=

C is given to the positive plate, C. Let Q tive plate loses a charge Q and the negative plate gains a charge Q such that the total positive charge on the positive plate = total negative charge on the negative plate, i.e. Q Q

C

3. C CC C = C C C

K is open, the total C C = C 3

Q

Charge on capacitors is Q =

C C and

3 9 9 = 3 9 4

9 Q = CV = 4

capacitance =

V =V

=

5.

Q

C =

C V

E C = E C

2.

C

C V

E =

r

CV 3

V =

Q = C

C

6. C =4

r

C =4

r

r

Before connection, the total energy is q q = 8 C r After connection, the entire charge q of the inner ergy after connection is q q = U = C r U =

Fig. 21.37

When key K is closed, capacitor C is shortC. C is Q = CV Q –Q = CV –

CV CV = 3 3

=

4. . Current in the circuit is I =

U –U q

q

6 r r 7. Charge on the inner sphere is q = CV = 4 rV

6 V

=V –V

=

q r

Capacitance and Capacitors 21.17

= =

q r

4 q 4

.

r

=

=n

n =n Q = CV = C V

12. capacitors, each of capacitance C connected in par13.

KC C = KC

=

C V

E C V = E C V C Q = 3CV. When -

C = 3C

Q C

CV , E =

V

8. C=C capacitor C

11. E =

r

C K C. Q, the potential -

3CV 3V = K C K

-

9. method of virtual displacement. W d in the plate separation d to the resulting change U in the stored energy, i.e. W= U F plates, then the work W done to increase the plate d W= F d Now we know that the energy U of a parallel-plate capacitor of plate area A and capacitance C is U=

F =

Q

-

15. Energy stored in the capacitor is CV =

4

C

C

C

–4

16. Q on the plates does C Q = CV, the potential difference V

Q A r its radius, 4 r3 m= 3

n 4 R3 . 3 r. Now, the capacitance of a sphere C =n C

QV, a decrease in V results in a

stored energy =

17.

nm =

R=n

4

C

mc

d in d is,

d A

10.

R

their total capacitance is 6 capacitors in series each of capacitance 6

Q Q d = C A

where Q increase U in U due to an increase U=

14.

C

=

A

d When a dielectric of dielectric constant K is intro-

21.18 Comprehensive Physics—JEE Advanced

K

C =

A d

21. Q = CV and Q C V CV capacitors are connected in parallel such that the plates of opposite polarity are connected together,

C =K C

,

Q = C V and Q = CV

Now

C V = = C V K But V/V = 18.

V =

Q 4CV = C C

CV =V C

Equivalent capacitance C = C

K = 8.

8

Q C

t and dielectric constant K a parallel plate capacitor, the potential difference

U =

C V

=

C = 3C 3

V =

3C

CV

22. region of a charged cylindrical capacitor is given

d

V =E

t

K

E=

4 r where is the charge per unit length and r is the dis-

V, the separad

E d

=t

or K =

K

=

t t

23.

d

.

R = RC is zero.

.6

19.

r

24. 25.

Q and Q are E =

Q

and E =

A

Fig. 21.38

Q A

Q >Q

3 A and B 20.

C = K d = C A

K

E= E – E =

A 8.8

=

A

Q –Q Q

Q –Q

C

–3 6

.

Q C A d

26. When switch S3 is closed, the potential difference across C and C of V and V -

V E= d d

d

V = Ed =

d

A

=

V E

= x

V = dielectric strength

= 6.67

6

27. We have C =

A d

–6

C =

K

A

K d

= =

AK d AK d

,

Capacitance and Capacitors 21.19

AK AK = d d C and C are in parallel and their equivalent capacitance is A K +K C =C +C = d C3 capacitance is d d = + = + C3 A K K AK C C

U

C3 =

and

=

U Given

U

Q

Q Q

U U

get Q

Q

Q –Q

Given Q – Q

Q

Q. Q =

32.

d

=

K

A AK d

or C =

K

where

28. Given C + C =

K3

K

=

C C C C

CC

or

6C + 6C

CC

or

6C + 6C

CC

Let

C =xC 6C

+ 6x C

6x

x

which gives x = 29.

K

+

K

K3 Fig. 21.39

6

C +C

or

U

C C

CC

C , it is thus ineffective. Capacitors C and C4 are in series, hence equivalent capacitance of capacitors C and C3

xC

points A and B 3

or

C >C ,x=

3

3

is

33.

A to B Ui = U + U =

CV +

voltage is V=V –V As the capacitors C and C are in series, the

B

30.

C4 C3

CV

When they are connected, the potential across each is V =

V +V

Uf =

CV +

C CV = CV = C

V

V

or

C=

C

C

3

6

6

Decrease in energy = Ui – Uf C V +V

= =

31. U =

4

4

Q = CV C Potential difference across A and B = potential difference across capacitor C Q C C

C V +V

C V –V

Q Q and U = C C

34. V=E

d

21.20 Comprehensive Physics—JEE Advanced

A d

C=

is

A

A

or

.

C Uf =

K and thickness

Q C

Q C

t A

C = d

. C

t

K Now Q = CV = C V CV CV V = C C/

.

.

But

C

Ui = Uf

4

Ui

Q

Q C

C

Q C

Q C

=

37.

V

parallel, each of capacitance A C= d

35. A d Q is the charge on the plates, the potential difference is Q Qd V= C A Let d t and dielectric constant K is introduced, the new capacitance is is C =

capacitance is 3 A C = 3C = d

A

C = d

t

K

Q difference is V =

Fig. 21.40

Q C

Q d

t

38.

K A

Given V = V d=d –t

K

or d – d = t

Given d = d

K

capacitance C = A/d the equivalent capacitance is A C= C= d

t K

which gives K 36.

Q is the initial charge on capacitor C , the initial Ui = Q C When the two capacitors are connected together, on each capacitor is Q

Fig. 21.41

C =C

C = Q/V Q =Q =Q

C =

39. Q =C V

–6

–3

C

Capacitance and Capacitors 21.21

.

hold is Q =C V

–6

–3

C 44. Given

–3

pacitor cannot hold a charge of 8

C; it can

C

C

and

C3

C

C

3

C3 45.

–3 –3 6 capacitor is 3

V =

C

6

stand = V + V = 6 kilovolts + 3 kilovolts = Fig. 21.42

40. Let A

d the plate separation. RA d

= where R conductivity

d RA

C=

K

or

C=C +C

R=

d A

C=

A d = RC =

d A

K

CV , U =

U

=

C C

C C

4 8 = 3 9

3

R R

R = R =

=

3

4 3

s R R

=

=

3

C=C +C = RC = C

C

R R

For Circuit 3 R =

A d

thickness t A . Given C C = d t

s

K

d

C=

42. U =

A

= RC = 3 For Circuit 2 R =

capacitor is = RC =

-

For Circuit 1 R = R + R

is

=

41.

46. Let R and C tance of the circuit.

V =

3

6= 4 s

47. Charge on C is Q = C V = CV C and C3 is

CV

U

C =

U

CC C = C C

C = C3 = C C and C is

43. U = dU U

CV

U = CV V CV V CV

V V

V =

Q C

C

=

C

CV CV = C C C

C is Q =C V =

CV 3

= V 3

21.22 Comprehensive Physics—JEE Advanced

d 3

x

and the thickness of water col-

d =

d 3

x

is d =

wires is Q =Q =Q

= CV –

48.

CV CV = , which is 3 3

C =4 C =

4

capacitance is

R

Ceq =

RR R R

Given C = 3C 4

which gives

RR R R

=3

4 e R

R 3 = R

Now C =

W= 50.

Q C

x ,K

get Ceq =

Q 49. Energy stored initially is Ui = . if d C C C Q Uf = C Work needed is Q W = Uf – Ui = C

d 3

d =

Q C

CV

A d d k = RCeq Putting d = A

Vt

we

6 R d Vt

=

52. When switch S d is

V CV

U =

V =V

J

When switch S

Q = CV

A d

potential is V =

A V d

CV C C

q C

C

V

=

V

volt

drawn as shown in U = on the capacitor plates is A V Q = CV = d

C

C V V

= Percentage loss of energy is Fig. 21.43

51.

x ,

x = Vt

6 d

d 3

t

x

U

x = Vt.

U U

V

V =

V

II Multiple Choice Questions with One or More Choices Correct 1. A parallel plate capacitor is charged and the of insulating handles

V

J

Capacitance and Capacitors 21.23

QE

QE

6. When a capacitor of capacitance C is charge to a potential V , the energy stored in it is U . When this charged capacitor is connected to an uncharged capacitor of capacitance C is V U.

increases 2.

V V

point

U U

point 3. A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and

C C

C

V V

C

U U

C C

C C C C

7. V

constant K

-

the potential difference Q, E and

W

the charge on the plates is increased increased the energy stored in the capacitor increases. 8. A parallel plate capacitor is charged to a voltage V AV d

Q= E=

V Kd

Q= W=

V KA d AV d

K decreases

4. A parallel plate ai

-

and energy associated with this capacitor are givQ , V , E and U respectively. A dielectric

creases the energy stored in the capacitor decreases. 9. A parallel plate capacitor is charged to a voltage V.

Q, V, E and U are related to the Q>Q

V>V

E>E

U>U

5. A parallel plate capacitor of plate area A has a charge Q E Q

Q A

A

the charge on the plates is halved unchanged the energy stored in the capacitor is dou10.

V halved,

21.24 Comprehensive Physics—JEE Advanced

C C is 8

halved unchanged the energy stored in the capacitor is dou-

14.

C. C. A and B is C, are con-

V

11 S

-

Fig. 21.46

is C and the energy stored in the capacitors is U, then C C C C = Fig. 21.44

A

CV. B

U= 15.

V/3. CV . 3 CV . exists in the region

12.

4

CV

U=

CV

A and B of capacitances C and C 3 respectively are connected in series. When this supply, the potential difference across capacitor A A A and B are V and V C C 3

constant 4 is inserted parallel to the plates, the V thickness x

C C

V. V x

V x

13.

3

V

V

V

V

16.

Fig. 21.45

Fig. 21.47

Capacitance and Capacitors 21.25

A and B x and of the other x total capacitance –4

J.

17. attains a steady value, then

20.

A are arranged d apart. V Q is the charge on plate

fc is zero. capacitor is

3

Q4 on plate 4 then

rgy stored in the capacitor is nearly –8 J.

Fig. 21.50 Fig. 21.48

Q =

18. attains a steady value, then

Q4 = – C. –4

J.

AV d AV d

21. Capacitor C

AV d

Q = Q4 =

4

AV d C

separately allowed to discharge through equal t circuits is zero at t at t t Fig. 21.49

19.

C sooner then capacitor C initial charge.

ANSWERS AND SOLUTIONS 1.

voltage V across capacitor plates increases due to its capacitance C decreases. Now Q = CV charge Q

21.26 Comprehensive Physics—JEE Advanced

V increases and Q

5. on each plate of the capacitor is

QV also

F=

2.

A V E= d Q Qd Now V = = C A Q E= A

tween the plates it is E=

Q

Q 8.8 = A 8.8

3.

F= AV d

Q = CV =

V C = V C C

V Kd

A V

d A

d

U =

Q C

=

=

KA d

where C =

energy stored U = U>U are correct.

U =

CV

U U

K

=

+

CV

C C

C

V, the voltage of Q = CV, the charge on the plated C increases and V V/d V and d re

4.

Q

CV

7. Due to the introduction of dielectric the capacitance C

AV d

Work done W = U – U =

U = AV d

AV Kd

V=V increases capacitance C Q>Q unchanged, and V E = V/d

Q =C V . they are connected. Q = Q + Q . Using Q = CV, we have C V = C V + C V, which gives

K

Q U = = C

A d

QE.

6. V/d

E=

C

in the capacitor is U =

CV

C increases and V

d E=E CV increases

CV . U

C

8. Due to the introduction of dielectric, the capacitance C disconnected, the charge Q on the capacitor plates

Capacitance and Capacitors 21.27

V = Q/C C increases, V E = V/d

Q =

V 3

3C

energy

QV

stored in the capacitor is U =

V +

3C U –U=

Q

3

CV – CV =

3

=

3

CV

CV .

V decreases, U 12.

all the four choices are correct. 9. Charge Q capacitance C = A/d V = Q/C the capacitor is U = Q Q

V=E

d

d is C. U

C=

C

A = d

A .

A

or

K and thickness t A

C = d

10.

Q = CV E = V/d, E d is halved. Energy

CV

stored U =

K . C

=

unchanged equal to V, the capacitance C that Q V

t

.

.

CV C/

=

=

C

4

Now Q = CV = CV V =

C

and V

CV = C

V

=

11. When switch S closed, the potential differnce across capacitors A and B V the charges on capacitors A and B are Q = Q = CV. When the dielectric is introduced, the capacitance

x x

CV CV V = = volt C 3C 3

introduced, is U=

CV +

CV = CV

CV +

CV

-

pacitance is C =

C = KC = 3C. After the switch is opened, the potential difference across capacitor A V volt. Let V potential difference across capaciteo B. When the S charge on capacitor B Q. Q = CV = C V or V =

x . C

A

=

x

x

Now Q = C V = C V = Given V = V C =C . C C or = = or = x x which givens x 13. effective voltage is V=V –V As the capacitors C and C are in series, the effec-

introduced, is U =

C

=

C

=

C

=

3

6

21.28 Comprehensive Physics—JEE Advanced

or C =

17.

6

tors

cf

is Q = CV

C

Potential difference across A and B = potential difference across capacitor C Q C = = C Fig. 21.51

14. pacitance is C = C circuit is U=

CV =

4

15. Given V

CV

I=

V

V R

V = R

=

3

3

A

V is the potential difference across the capacitor, then applying Kirchhoff’s loop rule to loop abcfa Q, then we have

Q=C V =C V

– V – IR + V + V

C V = C V 3 When the dielectric of capacitor A

3

+V

V=

C =

volt 3 Charge on capacitor = CV

C

–6 –6

B C =C C C = C C 3 3 V and V are the new potential differences across A and B, then V C = V C 3 Also V +V V

CV

Energy stored is U = =

3 C.

–6

–8

3

J

18. BC and AD of the current is A

B

D

C

A through the

current in the circuit is 16.

I= capacitors each of capacitance 6

Now, p.d across A and C AB

is, therefore C 6

BC C.

Energy stored in the circuit = =

–6

CV

AD. Charge on each capacitor is Q = CV = C. Energy stored =

–4

J

CV + –4

J

CV = CV

–6

Capacitance and Capacitors 21.29

19.

A A ,C = d d

C =

Cb =

Ca = C + C =

A d

C x, then

C C =

A = Cb d

20.

x and of

A A and C = d x d x Ca = C

+ C

=

Q = CV =

A d d

AV d

x

C

A

=

x d Ca > Ca wrong. d

Cb =

Q4 = – C V = –

C C =

And

C +C =

AV d -

CC C C

A d

=

V ; V = voltage R

21. At t

Cb = C C C C Now

C.

e–t/ where = RC,

of

d

A x d

d

Ad x d x

x

III Multiple Choice Questions Based on Passage Questions 1 to 4 are based on the following passage Passage I

C=

A parallel plate capacitor C= each of area A, separated d. A dielectric

AK d t A t d K A

C=

area A and thickness t and dielectric constant K is introduced with its faces parallel to the plates as

d

t

A

C= d t

1. Fig. 21.52

K

K

21.30 Comprehensive Physics—JEE Advanced

2.

K = 3, for what value of t/d will the capacitance 3

3.

3

3 4

3 4 K = 3 and t/d

4

4.

capacitors is

SOLUTION 1. of two capacitor one of thickness t dielectric of dielectric constant K and the other of d–t tances respectively are KA C = t A and C = d t C CC = + or C = C C C C C

C = Ca

Given K = 3 and C/Ca t d 3 which gives

C= A d

t

3 A . d

Now Ca =

K

Ua =

2. Ca =

A d

t 3 = d 4

3. Putting K = 3 and t/d =

we get C=

t d K

A d

Ca C

Q Q and U = Ca C

4.

Questions 5 to 7 are based on the following passage Passage II C. 5.

A and B is

6.

A and B is Fig. 21.53

Q U = , which is Ua 3

Capacitance and Capacitors 21.31

7.

A and C is 8 3

3

SOLUTION 5.

As the 3 parallel, the potential difference across each is the

C of the parallel C C3

-

–6

=6

C3

–6

=8

=

C

C

4

or C = 3

C +6 C+8

3 C of the

C=8 C C and C =

C.

Potential difference across C =

C3 and C4

C 4 AB

=

C3 C or C A and B C and C = C + C

C4

6

3 7.

6.

C4

C = –6

=6 C

–6

Charge on 4

Now C and C capacitance C C

–6

Charge on 3

–6

C.

C3 and C4 is C C3 and

–6

A and C is =

C 3

6

=

6

Questions 8 to 12 are based on the following passage Passage III

8. through the circuit is C C

A and B with capacitances 3 9. capacitor A

capacitor C throught a switch S

C C 10. capacitor B C

C C

11. capacitor C C C Fig. 21.54

=

3

21.32 Comprehensive Physics—JEE Advanced

SOLUTION 8.

Q through the circuit. When the switch is pressed, the voltages developed on capacitors A, B and C are Q Q volt, VB = volt and VA = 6 6 VC =

Q

Q 6

6

A

volt.

3 Qf

A

Qi

6

–6

which gives Q 9.

Q

Q

6

Qi

Vi Ci

A

A

A is –Q

C,

10.

B is Qf

B

Qi

B

–Q

C C

11. Fig. 21.55

Applying Kirchhoff’s law to the loop, we have VA + VB – VC

E R =6 r

E C

A

Qf

C

Qf

C

C

Qf

A

C,

Questions 12 to 15 are based on the following passage Passage IV r

Qf

,R

12.

R is

13.

R is

C =4 14. 3 4 3 15.

C is C C

Fig. 21.56

SOLUTION 12. Current through R is I =

E

C C

14.

R r Potential difference across R is V = I R 13. Current throught R is E = =3A I = R r . . Potential difference across R is V = I R =3

V=V +V Effective capacitance C =

=

6

=

=

A

C

CC C C

4 3

15. Charge Q = CV =

4 3

–6

–6

C

C

Capacitance and Capacitors 21.33

Questions 16 to 18 are based on the following passage Passage V ra and rb + Q and the inner sphere is earthed. 16. Out of Q, a part Q which will appear on the outer surface of the outer sphere is Q rb ra Q rb ra rb ra Qrb rb ra 17.

Qrb ra

Qra rb

Q rb ra rb

Q rb ra ra

18. C=

Qra rb ra

rb

4 ra

ra ra

4 rb

C=

ra

4

C=

rb rb ra

4 rb

C=

Q which appears on the inner surface of the outer sphere is

SOLUTION 16. Out of charge Q, a part Q surface and another part Q surface of the outer shere such that Q+Q =Q

Q ra

Q rb

Q

A is earthed, VA Q or Q rb rb Q =Q–Q ra rb ra rb

Q–Q

Electric potential at shere B is

which given Q =

Q rb

Q rb ra rb

18. V = VB – VA = VB

Q Q Q rb rb rb Q = rb 4 Electric potential at the inner sphere A is Q Q Q VA = ra rb rb 4 VB =

4

V=

4

Q rb

Q rb ra rb

17. Q = Q – Q = Q –

Fig. 21.57

ra

Q = rb 4

C=

4 Q = rb V

=

Qra rb

A and B is VB Q rb

ra

rb rb ra

IV Assertion–Reason Type Questions following four choices out of which only one choice is correct.

-

21.34 Comprehensive Physics—JEE Advanced

Statement-2 1. Statement-1 4. Statement-1 source supplying a constant voltage V are kept connected to the source and the space charge on the plates will increase. Statement-2

Statement-2

on the shape and size of the conductor.

plates. 2. Statement-1

5. Statement-1

voltage V plates of a parallel plate capacitor charged to a

the energy stored in the capacitor will decrease. Statement-2

3. Statement-1

Statement-2 -

Fig. 21.58

in the capacitor will decrease.

SOLUTION 3.

1.

Q

constant voltage V capacitance C the plates is increased. Now, energy stored U = Q C is decreased, Q C U will increase.

V C increases due Q = CV, the charge Q on the capacitor plates will increase. 2.

4. 5.

Q on the

E charge q is F = qE

C increases due to the introduction of the dielectric. Now, energy Q stored U = Q C C increases, U will decrease.

V Integer Answer Type 1. -

E is constant, force F is the

Capacitance and Capacitors 21.35

2. R C

R

,

C =4

SOLUTION 1.

Given V = V

Q is the charge on the plates, the potential difference is Q V = C

d =d –t

Qd A

d –d =t

Let d t and dielectric constant K is introduced, the new capacitance is

d

or

K t

Given d – d

A

C =

K

K Which gives K = 3.

t

K

2.

Q difference is

R =

R R

R R

3

C =C +C V =

Q C

d

t

K A

= RC =

3

6=4 s

Electric Current and D.C. Circuits 22.1

22

Electric Current and D.C. Circuits

Chapter

REVIEW OF BASIC CONCEPTS 22.1

positive charge

ELECTRIC CURRENT

Conductor

negative charge

(+)

(–)

Fig. 22.2

current is said to be steady or constant

22.2

q I= t

DRIFT SPEED OF ELECTRONS IN A CONDUCTOR

q t.

vd E Fig. 22.1

vd =

I=

eE m

dq dt

Convention regarding direction of current -

Relations between drift speed and current I in a conductor as vd =

I ne A

n e A = cross-sectional area of conductor

22.2 Comprehensive Physics—JEE Advanced NOTE

=

R

R

R

R3

.

22.3

+…

OHM’S LAW

Ohm’s Law states that

22.6

EMF, TERMINAL VOLTAGE AND INTERNAL RESISTANCE OF A CELL

. I or V = R I

V

E

R

22.4

ELECTRICAL RESISTIVITY

V r

r

A R or

=

or A RA

R=

A

E connected to an external resistance R.

is

called the

Unit of

r

=

unit of R unit of A unit of

tre

Hence

Fig. 22.3

Total resistance of the circuit = R + r. The current in the circuit is I=

22.5

RESISTORS IN SERIES AND PARALLEL -

tance R

E R

r

Potential difference across r is v = Ir Potential difference across R is V = IR. V is called the v internal resistance. Thus E= V + v = IR + r

R = R + R + R3 + …

V = E – Ir

22.7

GROUPING OF CELLS

resistances r and r

E and E and internal

Electric Current and D.C. Circuits 22.3

nE R nr

I=

n

Fig. 22.4

E nE nR r

I= E

=E + E

r

=r + r

and

n m

For n cells in series E = E + E + … + En and

mnE nR m r

I=

rn = r + r + … + rn

22.1

SOLUTION N e Fig. 22.5

E r r

= =

E r r

E r

N=

It e

Ne . Thus t electrons

.

r

For n

22.2 E r

and

q t

q = Ne and I =

r

= =

E r

r

resistance is nr each cell.

E r

r

En rn

... ...

rn

E n nE and the total internal r is the internal resistance of

–7

free electrons

3

SOLUTION vd =

.

=

E internal resistance. 3. Cells should be connected in series if the external resistance R r. R < r. R r. n E

I en A 7

–3

t=

22.3

L vd

. 3

22.4 Comprehensive Physics—JEE Advanced

-

tance of the tube.

t

SOLUTION

SOLUTION I=

dq dt

dq = Idt

t q=

t

=

V R= I

t + 3t dt

to t

N + Ne e

I

t

.

t

dt 3 3 t 3

t

3

=

22.4

3

=

3

22.6

7

SOLUTION L RA

= L R= A I=

SOLUTION L increases and A

L V and I = A R

V A L L

But I = enAvd enAvd = vd =

V A L V Lne

R =

A

L A

7

=

R L = R L

–3

22.5 The current I t I t + 3t

AL = constant. Thus AL =AL AL L . Therefore A = L L R = A

R

A A

R

NOTE R is stretched to n n R.

Electric Current and D.C. Circuits 22.5

22.7

-

R L A = R L A AL = constant A A

L L

R = R

L L

A A

L L

SOLUTION R =

3

= 3 3=

nR

22.9

=

R

resistance and

R 22.8

SOLUTION AL = constant

A L =A L .

L =L +

A =A R =

L =L L L

A Fig. 22.6

L L and R = . Therefore A A R R

=

L L

A A

SOLUTION of R R3 and R is R R

R

=

R

R R=R –R

R R R

Simple Method

Fig. 22.7

Current in the circuit is

L L used.

I= L R= A R

R

E R

r

Potential difference across R is V = IR L

A

22.6 Comprehensive Physics—JEE Advanced

Potential difference across R is V = IR R R3 and R since these resisR3 is V I3 = R3

resistor. There-

v = Ir Fig. 22.9

V =E – v 22.10 Calculate the steady state current in the 3

Current I = Potential difference across R is V = IR

resistor in

resistor. Hence the . resistor is 3 22.11 and a resistance of the battery. SOLUTION Fig. 22.8

E =IR + r E

SOLUTION 3

and

and

is R =

r 22.12

SOLUTION Let R .

Fig. 22.10

r

E

r

r

r

r E

Electric Current and D.C. Circuits 22.7

PQ

be R

22.14 A and B S S is closed.

Fig. 22.11

RPQ =

R Fig. 22.14

R A and B is

R

RAB

R R

S

=R

R R

SOLUTION

R

R R

Current I =

Fig. 22.15

E R

r

RAB =

22.13

S

A and B has a resistance of 3

.

Fig. 22.16

RAB

Fig. 22.12

SOLUTION

22.15 I

Fig. 22.13

RAB

=

3

3

3

RAB

Fig. 22.17

22.8 Comprehensive Physics—JEE Advanced

SOLUTION

IR A and B is R =

.

Sign Convention for emfs and Voltage drops

=3

bcdeb and

C and D is R= Current I =

3 3 3 3 V R

. Fig. 22.20

Fig. 22.18

22.8

KIRCHHOFF'S LAWS that cell.

First Law or Junction Rule

IR junction A. junctions at b and e b or e I + I = I3 and bcdeb

Fig. 22.19

E – I 3R 3 – I R and

– E + I R + I 3R 3

I + I – I3 – I – I

E E R R and R3 I I and I3

I + I = I3 + I + I NOTE

Second Law or Loop Rule I and I

I3

Electric Current and D.C. Circuits 22.9

I3

I

BCDEB –I R + E – I3R3 – 3I I3

-

I +I

I 7I

22.16

and – 3I

I

I +I

I.

I

I I =

I =

and I3 = I + I = B and E B

E.

VB – V E = I 3R 3 = Fig. 22.21

22.18

SOLUTION P PQ Q

QR

as

A and B A and C

R I

-

E E nal resistances r and r are connected to an external resistor R

I R. 22.17 I I and I3 E

E

B and E R = 3 and

R

R3

Fig. 22.23

SOLUTION We choose any direction for the current I in the cirI I Fig. 22.22

SOLUTION B I + I = I3 ABEFA I 3R 3 – E + I R

Fig. 22.24

22.10 Comprehensive Physics—JEE Advanced

B and D are

ABCDEA – E + Ir + E + Ir + IR I

I

I

I I is

I I I

Fig. 22.25

Fig. 22.27

VA – VB = – E – Ir

Condition for balanced Wheatstone’s Bridge

A B. VB –VC = E – Ir B VR = IR

ABDA and BCDB C. IP+IG–IR I –I Q

22.19

G I

I

and

I +I S–IG I

IP–IR

I I

R P

IQ–IS

I I

S Q

P R = Q S

or

P Q = R S

Fig. 22.26

SOLUTION VA – VD

I – 3I

I

I But VA = VD I

22.9

A and D are earthed. I

WHEATSTONE’S BRIDGE P Q R and S A and C B and D

P Q R and S

Fig. 22.28

the resistor R =

AD = DC S

AC

Electric Current and D.C. Circuits 22.11

AD and DC and . For a balanced

22.21 A and B

P R = Q S 22.20 A and B R.

Fig. 22.31

Fig. 22.29

SOLUTION

SOLUTION

C and D because

C and D

rent I

A and B V across A and B

V

V = IR.

A and B R

R.

Fig. 22.32 Fig. 22.30

RAB =

R R

R =R R

and DEBFD V I–I

ABFGA ACDFA I–I +I

22.12 Comprehensive Physics—JEE Advanced

I +I I –I

I–I

I–I +I

3I – 3I

R R

I

I =V

I – 3I

and

I I =

I I =

I

7I

I

= R =

.

I.

22.10

=

3

3

3

R R =

3

THE POTENTIOMETER AB

I and I V 7 I

V

R R

3

I – 3I – I

R R

E is a

E RAB

22.22 resistances R and R

AC resistance is

A R

D such that

R and R . Fig. 22.34

AB and L

R = L/A

A V = IR or V = I L/A = KL V

R is the resistance

K = I /A

L

V = K or L

Fig. 22.33

difference v

SOLUTION

v

cell E or E = v =

R = R

3 V

R

R R

R R

=

R

=

Applications of potentiometer

R R

R =

3 3

E

and E and E E = E = E

and E =

v = . The E of the

Electric Current and D.C. Circuits 22.13

SOLUTION

E S

R=3

E=

AB = L AC =

and V

r

Fig. 22.36

AB I=

Fig. 22.35

S V of E V=

R is connected across the cell E is E

V

r

AB = I

22.24

v

=

resistance is 3

AB

.

v = V Ir = IR r= R R

and R

r is

22.23 Fig. 22.37

resistance of 3

E

SOLUTION AB

E.

r

E=

E = V + v = IR + Ir I E and r is its internal resistance. Thus v

R

AB is K =

E = V

V

V

I =

. 3

22.14 Comprehensive Physics—JEE Advanced

Potential difference across AB due to E is

Potential difference across AC cell is r VAC = IRAC

Potential difference across due to E . Vl = resistance due to E is .

I =

E is

tial difference across AC

VAC

resistance . due to E is

r

r

R connected across E

V V =V

I =

.

R

VR = I R = 22.25

R R

VR across AC VR = resistance of AC

current I

= R

R

R VR =

SOLUTION AC tance of AB.

AB AC

I

r be the resis-

22.11

R R

AMMETER

Fig. 22.38

AB =

r

AC is RAC = r

. Therer

Fig. 22.39

Electric Current and D.C. Circuits 22.15

I G I

V I R

S = shunt resistance G

S

I = I + Is

I G = I sS I–I S

IG S=

I I

G

I

22.26 Fig. 22.40

V = VG + VR =IG+IR V =G+R I SOLUTION I S=

I I I

I

G

G

.

R=

V –R I

22.27

. .

. .

RA =

GS G S

.

SOLUTION

resistance of the circuit and hence the current current.

22.12

G

I R=

V –G= I

VOLTMETER

V . .

RV = R + G tor of resistance R > R

R is con-

R

-

tion is

R.

R=

22.28 are connected resistance. sure the current in the circuit. Find the error in -

Fig. 22.42

SOLUTION

Total resistance of the circuit is R Current in the circuit is I =

. -

Fig. 22.41

V =

.

I=

. The current I =

.

V =I

.

I Multiple Choice Questions with Only One Choice Correct 1.

and +

are connected in series. +

2.

. The

Electric Current and D.C. Circuits 22.17

3.

resistor is Fig. 22.45

6. 3

Fig. 22.46 Fig. 22.43

4.

7. P and

3R

Q 3 3

A and B is

P to Q

R

Q to P

7R

P to Q

3

R Fig. 22.47

Q to P

3

8.

A and B is 3R R R R

5.

E is

is used

9.

Fig. 22.44

Fig. 22.48

I

across it is

22.18 Comprehensive Physics—JEE Advanced

10.

.

15.

16. 11.

L E E

-

L

L

L 3

L

L 3

R is

n n R R n

nR R n is bent into a circle. The

17. circle is

18.

12. tion. What is the resistance X

current I in the circuit is

Fig. 22.49

each are connected as D -

13. A and D

Fig. 22.51

is bent into the

19. r

3

3 3

20. Fig. 22.50

14.

A and B is

Electric Current and D.C. Circuits 22.19

22. R -

Fig. 22.52

21. t

I in the resistor AB t

I I I

K is -

23.

E and different internal resistances r and r are connected in series to an external resistance R R

t t

I

Fig. 22.54

Fig. 22.53

Fig. 22.55

24.

26. 25. the conductor is/are

r R= r

R=r +r

R=r –r

R=r =r

22.20 Comprehensive Physics—JEE Advanced

Fig. 22.56

27.

3 28.

resistances is S the total resistance is P n is

S = nP

Fig. 22.57

32. -

29.

in the ratio of

3

and

other. One cell has an and the other cell has

3

internal resistance difference across the X and Y is

3

30.

-

tance X X rB. 58.

R= R=

59.

rA rB rA + rB R

A and B in

A 62.

RC circuit consists of a resistance R C F connected in series

Electric Current and D.C. Circuits 22.25

e

AB and BC

63.

L AB

BC is r. The current I

r

Fig. 22.78

-

65. X

BC

across AB.

BC AB. AB and BC AB and BC

X

64. S

Y to X is C

Fig. 22.79

C

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61.

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62.

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63.

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64.

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65.

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

SOLUTIONS 1.

A = cross-sec-

L R= R +R =

L A

d L L = A A

+

R=

L A

d

22.26 Comprehensive Physics—JEE Advanced

=

4.

+

R=

2. E = V + Ir = I R + r r is the internal resistance of the cell and the current in the circuit is I=

E R

r

I=

R V = IR and E= E–V

E E

V E

IR I R r R R

r

=

r R

r

r R

Fig. 22.81

r

= 3.

E. Points A B and C at A B and C

I +I =I APQDA

V

D D and E is V E and B is V = V V. Therefore V I = 3 V I = 3 V I3 =

I

I

I

V E and C =

I I =

I =

I =

3

I =

3

3

P I = I3 + I I3 = I – I =

3

3

=

P to Q 5. I = I +E I

I

I

6. Fig. 22.80

I = I + I3 V 3 Hence I3 =

=

V /3

V

=

V=

3 Fig. 22.82

E

3 I3

3

Electric Current and D.C. Circuits 22.27

A and B

.

V–V =

Current I =

I 3 = 3 3

10.

–3

-

7. 3

I=

–3

R –3

R

R 11.

E /L. Hence

Fig. 22.83

RAB =

=E

8.

L

E E = L

L

E=

R

x E=

xE L x=

L

.

12. =

Fig. 22.84

RAB =

X

R

7 X . Hence the correct choice is

13.

9. Current I =

=

A and D resistor is R

V=

or

=

+

R=3

R=

I =

= Fig. 22.85

V =

14.

I

=

3

22.28 Comprehensive Physics—JEE Advanced

I+

I=

I

R =

R

R

R I+R I

R

R =R

. Hence the correct

R

15. Let the three resistances be X X R = +

R

3

R

X and Y. Then X

or

=

R

+ R

R =

R

+

R

=

R

=

20.

X. 3 + X Y

Y

X X = 3. Hence Y

R circuit is

. Hence the

=

=

R V R

the

16.

R

=

R

+

R

+

n R

+n

resistor in

or R = R/n

A and B

17.

21. AB is

. Hence the 18.

t

I=

E. Hence no current

B

t as I= I e I=I

A and D is the

R or

=

+

R=3

=

–t/

t >>

t

I =

3 I

resistor

22.

19.

r R

r

R r r R =

R r

r=

R

r

r resistors connected in

Electric Current and D.C. Circuits 22.29

23.

A and B of the

26. E

AB I=

E r r

R

resistance r is E – I r I=

E=Ir

Ir r r

R

27.

R = r – r . Hence the correct choice

A and B Hence current in AC CD CD

.

sectional area A ity

-

-

24.

R=

-

A

V

V=

R

V

R Resistance R

R. The -

AB

resistances R bined resistance is =

. Current I

or R =

R

tional to

R

. Thus if

R

. a

R R

AB R R

R

R

R R of the R R R

Re =

R R

R R

3

R

R or

The roots are R . R

Fig. 22.87

28. Let x and y Fig. 22.86

25. of A

xy x y nxy x+y= x y

S = x + y and P = A

-

S = nP or

and

22.30 Comprehensive Physics—JEE Advanced

x y x=

xy + y

n

n

[n

y

or x =

x are

n

y

+

y

y

n

x

n

n n 29. = or

R = R

R A

R A

=

r

A = A =

3

R = R 3

and

=

r r r

3

Fig. 22.88

r r

=

E

R=

3

=

.

E

=

3

I R =I R or

and

A and B

. Hence

E Therefore

I I

3

R = R

3

E

R=

=

=

=7

3 32.

X 30. Y

= I= X = Y X = Y

X and Y

31.

is

33.

=

V = IR =

= or

E=

E ACDA CBDC and ADBGFA –

3

3

+ +

3

3

3

E

34. B

V

A =

I . = A

.

R

Electric Current and D.C. Circuits 22.31

CC C C

Ceff =

B.

3

F

F

3

Q is the

D and

R E =

R or R =

3

R

Q = VAB

3

Ceff

3 C AC is

VAC =

A and C

.

Q C

C and D is

VCD = VAD – VAC

CD is

37. R

R R by Fig. 22.89

R or

=

+

3

=

I= current due to E IR = E I R R

=

E R

=

R =

R

P Q

R S

R

R

R

38.

/

-

R

R = IR

and

. Thus R R

R R

36.

V

I=

7 35. Current due to E is

=

R

7

R=

R R + R and R + R R

R

. Hence the correct choice is I=

Potential difference across AB is VAB AD is VAD

. Hence the correct choice

39. For series connection x = nR R tion y = . Therefore xy = nR n 40.

R

R

Rn

Rn

R

R

Rn

y

n

R = R . Hence n

x

22.32 Comprehensive Physics—JEE Advanced

Subtracting (2) from (1), we have 1 1 = Rn x which gives Rn =

1 y

4 A, which is choice (b). 5

xy

which is choice (b). y x 41. Let m cells be connected in series and n such groups are connected in parallel. If the emf of each cell is E and internal resistance r, then the total emf of m cells in series in mE and the total internal resistance is mr. When n such groups are in parallel, the effective internal resistance is mr/n. Then the current through an external resistance R is I=

mE mr R n

mnE nR mr

(1)

mnE

= nR

mr

2

2 mnRr

Now, I will be maximum if the denominator is the minimum, i.e. if nR = mr. Given R = 3 and r = 1 . Using these values, we have 3n = m. But mn m m = 48 (given). Therefore = 48, which gives 3 m = 12. Thus n = 4. Hence the correct choice is (b). 42. The given circuit is a Wheatstone’s bridge. The current through the galvanometer will be zero if the bridge is balanced, i.e. if P R where P = 2 + 3 = 5 , Q = 10 Q S R = 4 . The value of S is given by

Fig. 22.90 where R1 =

R1

R2

I1 I

I2 Fig. 22.90

70 70 and R2 = 3 4

10 70 / 4

44. The emfs of cells connected in reverse polarity cancel each other. Hence cells marked 2, 3 and 4 together cancel the effect of cells marked 5, 6 and 7 and the circuit reduces to that shown in Fig. 22.91. Now cells Fig. 22.91 1 and 8 are in reverse polarity. Hence the voltmeter reading = 5 – 5 = 0 V. Hence the correct choice is (d). 45. containing the capacitor. Thus, the current, say I, R and 2R. Applying Kirchhoff’s second rule to the loop abcdefa, we have (see Fig. 22.92) V 2V – I(2R) – IR – V = 0 or I = 3R Potential drop across capacitor = 2V – V – I(2R) = V – =V–

2V 3

V 3R

2R

V , which is choice (c). 3

and

5 4 10 S or S = 8 . Thus the effective resistance of the parallel combination of 12 and x ohm must be 8 . Therefore 1 1 1 12 x 8 which gives x = 24 . Hence the correct choice is (d). 43. Since the seven resistances are in parallel, the effective resistance is R = 70/7 = 10 . Therefore, the current in the circuit is I = 14/10 = 7/5A. The given circuit can be redrawn as shown in

R 7 = R2 5

current I2 is given byI2= I

. The

Fig. 22.92

46.

R 6, resistances R1, R2, R3 and R4 constitute the four arms of a balanced Wheatestone’s bridge. Hence R1 R3 or R1 R4 = R2 R3 R2 R4 Thus the correct choice is (c). 47. Refer to Fig. 22.93. The branches ABPQ and PQCD are a balanced Wheatstone’s bridge. Therefore, resistances (each equal to 2R) between E and F and between F and G do not contribute and the circuit simplies to the Re between P and Q is given by

Electric Current and D.C. Circuits 22.33

Fig. 22.93

1 Re

1 4R

1 2r

1 4R

2Rr which gives Re = . Hence the correct choice R r is (a)

53. The correct choice is (b). 54. Let the value of each resistance be r. The network can be redrawn as shown in the Fig. 22.95.

48. is R R1 = AC R2 RCB where RAC and RCB are the resistances of the bridge wire of length AC and CB respectively. If the radius of the wire AB is doubled, the ratio RAC/RCB will remain unchanged. Hence the balance length will remain the same. Thus, the correct choice is (b). 49. The voltmeter must be connected in parallel with the resistor and the ammeter must be connected in series with the resistor. Hence the correct circuit is (a). 50. The potential at Q with respect to R is 15 V and R is at 25 V higher potential than S. Thus Q is 40 V higher than S. When Q is grounded, its potential becomes zero. thus, Vs = – 40 V. Hence the correct choice is (d). 51. Since the cells are in opposition, the effective emf = the positive to the negative terminal of the battery, a current of 4 = 0.5 A 5 3 B to C. Hence the correct choice is (b). 52. Resistance of each side of the square = 10 0.1 = 1 . As shown in Fig. 22.94, the square forms a W h e a t s t o n e ’s bridge which satisfies the balancing Fig. 22.94 condition. Thus, no the diagonal BD. Hence the correct choice is (a).

Fig. 22.95

(i) Net resistance R1 between points A and B r The series combination of resistances and 3 r which has an equivalent resistance r1 = 3 5 r = , is in parallel with resistance r. Hence 6 5r r r r1 6 = 5r R1 = 5r r r1 11 r 6 (ii) Net resistance R2 between points B and C The series combination of resistances r and

r , 2 r 2

r , 3

r = which has an equivalent resistance r2 = r + 3 4r r , is in parallel with resistance . Hence 3 2 r 4r r r2 2 3 = 4r R2 = 2 r r 4r 11 r 2 2 2 3 (iii) Net resistance R3 between points A and C r The series combination of resistances r and , 2

22.34 Comprehensive Physics—JEE Advanced

r = 2

which has an equivalent resistance r3 = r + 3r r , is in parallel with resistance . Hence 2 3 r 3 R3 = r 3

r3 r3

r 3 r 3

3r 2 = 3r 3r 11 2

5r 4r 3r Hence R1 : R2 : R3 = = 5 : 4 : 3. Thus : : 11 11 11 the correct choice is (c). 55. At instant of time t, the charge on the capacitor is given by q = q0 (1 – e–t /RC) and the potential drop across the capacitor is given by ( V = q/C) VC = V0 (1 – e–t/RC) where V0 is the voltage of the battery. The potential drop across the resistor is –t/RC

VR = V0 – VC = V0 – V0 (1 – e

–t/RC

) = V0 e

The resistance of wire of resistivity

is

2

l dl = kl2 (1) A m where k = d/m is a constant of the wire. Taking logarithm of both sides of (1) we have log R = log k + 2 log l Differentiating R=

R 2 l =0+ R l

R = 2 0.1% l R = 0.2%. Thus, the resistance of the increases by 0.2%. which is choice (b). 59. The resistance between points A and E is given by Given

l

2 l l

= 0.1%. Therefore,

1 1 = RAE 6

1 6

by giving RAE = 3 . The network reduces to that shown in Fig. 22.96 (a). Similarly the resistance

t / RC

VC 1 e = VR e t / RC Given

= e t /RC – 1

VC = 8. Therefore, VR 8 = et/RC – 1

et/RC = 9 = (3)2 t or = 2 loge (3) RC or t = RC 2 loge (3) = (5 106) (1 10–6) 2 1.1 = 11 s Hence the correct choice is (b). 56. The two sub circuits are closed loops. They cannot send any current through the 3 resistor. Hence the potential difference across the 3 resistor is zero, which is choice (a). 57. Given I = 1 mA = 10–3 A, G = 20 and R = 4980 . V Now I= R G or

or

V = I(R + G) = 10–3 = 5.0 V

(4980 + 20)

Hence the correct choice is (c). 58. The mass of a wire of length l, cross sectional area A and density d is given by m m = Ald or A = ld

Fig. 22.96

between points A and D in Fig. 22.96 (a) is RAD = 3 . The network reduces to that shown in Fig. 22.96 between points A and C is RAC = 3

. The network

effective resistance between points A and B is 2 So the correct choice is (b). 60. Let the internal resistance of each battery be r. Let R be the unknown resistance and G be the resistance of the galvanometer. Let E be the emf of each battery. When the batteries are connected in series, the total emf = 2E = 2 1.5 = 3 V and total internal resistance is 2r. The current in the circuit will be I=

R

3 G

2r

Given I = 1 A. Therefore

Electric Current and D.C. Circuits 22.35

1=

3 G

R

2r

or R + G = (3 – 2r) (1)

When the batteries are connected in parallel, the total emf = E = 1.5 V and the total internal resistance is r/2. Hence the current in the circuit will be r 2 Given I = 0.6 A. Therefore, R

15 . R

G

r 2

G

or R + G =

2.5

r 2

(2)

VC = 8. Therefore, VR 8 = et/RC – 1 et/RC = t = RC t= = =

or or or

r 3 – 2r = 2.5 – 2 1 which gives r = ohm, which is choice (d) 3 61. When the key K is inserted, the current starts growing and after some time it acquires a steady value. At

through the inductor (because an ideal inductor offers zero resistance to a steady current). Now, the network of resistors is a balanced Wheatstone’s bridge. Hence, no current flows through the resistance 2 R. Therefore, this resistance can be ignored. The net resistance between points X and Y = resistance of the parallel combination of 2R 2R = R. (2 R 2 R)

Hence the current in the circuit is I=

VC 1 e t / RC = = e t /RC – 1 VR e t / RC Given

From Eqs. (1) and (2), we have

2 R and 2 R =

VR = V0 – VC = V0 – V0 (1 – e–t/RC) = V0 e–t/RC

15 .

I =

0.6 =

VC = V0 (1 – e–t/RC) where V0 is the voltage of the battery. The potential drop across the resistor is

E

E

R rA

rB

=

2E R rA rB

Now, the terminal voltage of cell A is 2ErA VA = E – IrA = E – R rA rB

63. R =

9 = (3)2 2 loge (3) RC 2 loge (3) (5 106) (1 10–6) 11 s

1.1

l

. Since the two wires are made of the same r2 material, resistivity is the same for wires AB and BC. Since the wires have equal lengths, it follows that R 1/r2 . Hence

RAB 1 = , i.e RBC = 4RAB RBC 4 Since the current, is the same in the two wires, it follows from Ohm’s law (V = IR) that VBC = 4 VAB. Hence choice (a) is wrong. Now power dissipated is P = I2 R. Since I is the same, P R. Hence PBC PAB

=

RAB =4 RBC

Hence chioce (b) is correct. Choice (c) is wrong because current density (i.e. current per unit area) is different in wires AB and BC because their crosswire is E = V/l. Since the two wires have the same length (l), E is proportional to potential difference (V). Since VBC = 4 VAB, EBC = 4EAB. Hence choice (d) is also incorrect. 64. Refer to Fig. 22.97.

2ErA =0 R rA rB which gives 2 rA = R + rA + rB or R = rA – rB VA = 0, if E –

So the correct choice is (a). 62. At instant of time t, the charge on the capacitor is given by q = q0 (1 – e–t /RC) and the potential drop across the capacitor is given by ( V = q/C)

2

Fig. 22.97

22.36 Comprehensive Physics—JEE Advanced

Charge on capacitor C1 in Q1 = C1 V1 = 3 F 3V =9 C Charge on capacitor C2 is Q2 = C2 V2 = 6 F 6V = 36 C Now, when the switch is open, the initial charge Y to X = 18 – 18 = 0 C because the right plate of C1 has a charge – 18 C and left plate of C2 has a charge + 18 C. When the switch S is Y to X = – 9 + 36 = + 27 from Y to X when the switch is closed = 27 C – 0 = 27 C. 65. Corrected length L1 (= AJ) = 52 + 1 = 53 cm Corrected length L2 (BJ) = (100 – 52) + 2 = 50 cm For a balanced Wheatstone bridge, L X 53 = 1 L2 50 10

When the switch S is open, capacitors C1 and C2 are in series and their combined capacitance is 3 6 C1 C2 = =2 F 3 6 C1 C2 Charge on each capacitor = 2 F 9V = 18 C When the switch S is closed, in the steady state, C=

resistors R1 and R2 will be in series and their combined resistance is R = R1 + R2 = 3 + 6 = 9 . Therefore, the current in each resistor is 9V I= =1A 9 Potential difference across R1 is V1 = IR1 = 1 3 =3V Potential difference across R2 is V2 = IR2 = 1 6 =6V Since capacitor C1 is connected across R1, the potential difference across C1 is V1. Similarly, the potential difference across C2 is V2.

X = 10.6

II Multiple Choice Questions with One or More Choices Correct 1. A galvanometer has a resistance of 100 and fullscale range of 50 A. It can be used as a voltmeter or an ammeter, provided a resistance is connected to it. Pick the correct range and resistance combination (s): (a) 50 V range with 10 k resistance in series (b) 10 V range with 200 k resistance in series (c) 5 mA range with 1 resistance in parallel (d) 10 mA range with 2 resistance in parallel IIT, 1991 2. Choose the correct statements from the following. (a) A low voltage supply of, say, 6 V must have a very low internal resistance. (b) A high voltage supply of, say, 6000 V must have a very high internal resistance. (c) A wire carrying current stays electrically neutral. (d) A high resistance voltmeter is used to measure the emf of a cell. 3. The terminal voltage of a battery is (a) always equal to its emf (b) always less than its emf

(c) greater or less than its emf depending on the direction of the current through the battery (d) greater or less than its emf depending on the magnitude of its internal resistance 4. The internal resistance of the cell shown in Fig. 22.98 is negligible. On closing the key K, the ammeter reading changes from 0.25 A to 5/12 A, then (a) R1 = 10 (b) R1 = 15 (c) the power drawn from the cell increases (d) the current through R decreases by 40%

Fig. 22.98

5. A current I flows in the circuit shown in Fig. 22.99. Then

Electric Current and D.C. Circuits 22.37

(a) If a resistance R 2 = R is connected in parallel with R1 = R, the current through R1 will remain equal to I. (b) If a resistance R2 = 2R is connected in parallel with R1 = R, the current through R1 will remain equal to I. (c) If a resistance R2 = 2R is connected in parallel with R1 = R, the current through R1 will become I /3. (d) If a resistance R2 = 2R is connected in parallel with R1 = R, the current through R2 will be I/2.

(a) The potential at point P is 6 V. (b) The potential at point Q is – 0.5 V (c) If a voltmeter is connected across the 6 V battery, it will read 7 V. (d) If a voltmeter is connected across the 6 V battery, it will read 5 V. 8. In the circuit shown in Fig. 22.102, (a) the current through NP is 0.5 A (b) the value of R1 = 40 (c) the value of R = 14 (d) the potential difference across R = 49 V

Fig. 22.99

6. Two equal resistances R1 = R2 = R are connected with a 30 resistor and a battery of terminal voltage E. The currents in the two branches are 2.25 A and 1.5 A as shown in Fig. 22.100. Then (b) R2 = 60 (a) R2 = 15 (c) E = 36 V (d) E = 180 V

Fig. 22.100

7. Which of the following statements are correct about the circuit shown in Fig. 22.101 where 1 and 0.5 are the internal resistances of the 6 V and 12 V batteries respectively?

Fig. 22.101

Fig. 22.102

9. Choose the correct statements from the following. (a) If n identical cells, each of emf E and internal resistance r are connected in series, the emf of the combination is nE and the internal resistance of the combination is nr. (b) If n identical cells, each of emf E and internal resistance r are connected in parallel, the emf of the combination is E/n and the internal resistance of the combination is r/n. (c) Cells should be connected in series if the external resistance R is greater than internal resistance r. (d) Cells should be connected in parallel of R is smaller than r. 10. The resistance network shown in Fig. 22.103 is connected to a battery of emf 30 V and internal resistance of 1 . Then

Fig. 22.103

22.38 Comprehensive Physics—JEE Advanced

(a) (b) (c) (d)

the voltage drop across the 2 resistor is 12 V. the voltage drop across the 12 resitor is 12 V. the terminal voltage of the battery is 24 V. the voltage drop across the internal resistance of the battery is 6 V.

11. nonuniform cross-section. Which of the following quantities remain constant along the length of the conductor? (a) Current (b) drift speed

resistances r1 = 1 and r2 = 2 respectively, with polarities as shown in Fig. 22.105. IIT, 1997 5 V 6 (c) r = 1.5

12. A cell of emf E and internal resistance r supplies a current of 0.9 A through a 2 resistor and a current of 0.3 A through a 7 resistor. Then (a) r = 1.0 (b) r = 0.5 (c) E = 2.0 V (d) E = 2.25 V 13. A voltmeter graded as 6000 per volt reads 3 R is connected in series with it the reading of the

(b) E = 1.2 V

(a) E =

(d) r =

2 3

Fig. 22.105

16. In the circuit shown in Fig. 22.106, cells E1 and E2 have emfs 4 V and 8 V and internal resistances 0.5 and 1 respectively. IIT, 1978

resistance of the new voltmeter is R . Then (a) R = 5.4 104 (b) R = 3.6 104 V (c) R = 5.4 104 (d) R = 7.2 104 V 14. with 1 and 2 resistances as shown in Fig. 22.104. A 6 V battery of negligible internal resistance is connected between A and B. IIT, 1987 A 6V

1

1

2

1

2

1

2

2

To infinity

Fig. 22.106

(a) The potential difference across E1 is 4.25 V (b) The potential difference across E1 is 3.75 V (c) The potential difference across E2 is 8.5 V (d) The potential difference across E2 is 7.5 V. 17. In the circuit shown in Fig. 22.107, if the galvanometer resistance is 6 , then in the steady state B

B

Fig. 22.104

(a) The effective resistance between A and B is 3 . (b) The effective resistance between A and B is 2 . (c) The current in the 1 resistance closest to the battery is 3.0 A. (d) The current in the 2 resistance closest to the battery is 1.5 A. 15. A single battery of emf E and internal resistance r is equivalent to a parallel combination of two batteries of emfs E1 = 2 V and E2 = 1.5 V and internal

C1 = 10 m F

C2 = 20 m F

A

G=6 R1 = 8

R1 = 16 D

12 V

Fig. 22.107

C

Electric Current and D.C. Circuits 22.39

(b) the current through R2 is 4 A. (c) the charge on C1 is 80 C. (d) the charge on C2 is 80 C. 18. In the cirucit shown Fig. 22.108, the current through (a) the 3 resistor is 1.0 A 9 A (b) the 3 resistor is 15 (c) the 4 resistor is 0.50 A (d) the 4 resistor is 0.25 A

Fig. 22.108

IIT, 1998

ANSWERS AND SOLUTIONS 1. For voltmeter, the resistance R to be connected in series with the galvanometer is given by V –G R= Ig For 50 V range, R =

50 50 10

6

– 100

= 10 –6 – 100 106 1000 k . Hence choice (a) is incorrect. For 10 V range, R =

10

– 100 50 100 6 = 2 10 5 – 100 2

in the external circuit V < E. However, if the cur105

200 k . Hence choice (b) is correct. For ammeter, the shunt resistance is given by Ig G S= Is For 5 mA range, S =

50 10

5 10 Hence choice (c) is correct. For 10 mA range, S =

50 10

across the battery would show V > E. Hence the correct choice is (c). 4. Before closing the key, E E 0.25 = E = 2.5 V I1 = R 10 After closing the key,

6 3

ter one end of the wire as leave it from the other end. Statement (d) is incorrect. A voltmeter does not measure the emf; it measures only the potential difference because it draws some current from the cell. A potentiometer is used for emf measurement because at balance point, no current is drawn from the cell. The potential difference measured with a voltameter is always less than the emf. Thus the correct choices are (a), (b) and (c). 3. Terminal voltage V = E – I r, where E is the emf and r the internal resistance of the battery. If the current

100 = 1

Effective resistance is R2 =

6

100 = 0.5 10 10 3 Hence choice (d) is incorrect. Thus the correct choices are (b) and (c). 2. Statement (a) is correct. The current I drawn from a supply of voltage E and internal resistance r is given by I = E/r. So r must be very small so that a high current can be drawn. Statement (b) is also correct. At such a high voltage, the current drawn from the supply will become dangerously large if its internal resistance is small. Hence a high voltage supply must have a very high internal resistance so that the current drawn from it does not exceed the safe limit. Statement (c) is correct. A wire carrying current is not charged. It stays neutral because as many electrons en-

I2 =

RR1 10 R1 = . (10 R1 ) ( R R1 )

E R2

2.5 (10 R1 ) 5 = 10 R1 12 which gives R1 = 15 . Before closing the key, or

Power P1 = I12 R2 = (0.25)2 After closing the key,

10 = 0.625 W

5 2 10 15 = 1.04 W (10 15) 12 Hence power drawn from the cell increases. Before closing the key, the current through R is I1 = 0.25 A. After closing the key, the current through R is

Power P2 = I 22 R2 =

22.40 Comprehensive Physics—JEE Advanced

5 3 1 = = 0.25 A,which is equal to I1. 12 5 4 Hence the correct choices are (b) and (c). 5. When R2 = R is connected in parallel with R1 = R, the resistance of the combination is R/2. Therefore, the current in the circuit is V/R/2 = 2V/R = 2I, where I = V/R was the current in the circuit when R2 was not connected. Current 2 I divides equally among two equal parallel resistors. Hence the current through R1 will still be I. Thus choice (a) is correct. When R 2 = 2R is connected in parallel with R1 = R, the total resistance in the circuit is 2 R/3 and the current in the circuit is V 3V 3I = = 2R/3 2R 2 3I 2 Current through R1 = = I. Hence choice 2 3 (b) is also correct. The remaining current I through R2. Hence the correct choices are (a), (b), and (d). 6. Using Kirchhoff’s Ist law current through R2 is 2.25 – 1.5 = 0.75 A. Also since R2 is in parallel with the 30 resistance, R2 must be 60 since only half I1 =

through 30 resistor. Total resistance in the circuit becomes 60 + 20 = 80 Potential drop across the battery E = I R = 2.25 80 = 180 V. Hence the correct choices are (b) and (d). 7. Total resistance = 4 + 1 + 0.5 + 0.5 = 6 . Net volt6 age in the circuit is 6 V. Current I = = 1 A in the 6 anticlockwise direction VPR = 1

4=4V

Since R is connected to earth, VR = 0. Hence VP = 4 V VS Q = 0.5

69 yields R = 14 R 40 / 7 Hence the correct choices are (b) and (c). 9. The correct choices are (a), (c) and (d). In choice (b) the emf of the combination E and not E/n. 10. Total resistance of parallel combination is given by 1 1 1 1 = R1 12 6 4 or R1 = 2 . Total resistance in circuit is R = R1 + 2 = 4 . Therefore, current in the circuit is 3.5 =

=

upon the drift speed of free electrons. Resistivity depends only on the material of the conductor. Hence the correct choices are (a) and (c). 12. If E is the emf of the cell and r its internal resistance, then E E = 0.9 and = 0.3 2 r 7 r Dividing the two equations, we get 7 2

1 = 0.5 V. S is at a

higher potential than Q

E

30 =6A R r 4 1 Potential drop across 2 resistor = 2 6 = 12 V. The voltage drop across the 2 resistor is 12 V. Since the resistance of the parallel combination is also 2 , the voltage drop across this combination is also 12 V. Therefore, the total voltage drop across the network = 12 + 12 = 24 V. Thus the terminal voltage of the battery is 24 V. Voltage drop across the battery = 30 – 24 = 6 V. All the four choices are correct. 11. Current does not depend on the cross-section of the conductor. Drift speed is inversely proportional I =

or

r 9 = r 3 r = 0.5 .

VQ = – 0.5 V Current is being forced into the 6 V battery in the opposite direction. Hence V6 = E + I r = 6 + 1 1 = 7 V. Hence the correct choices are (b) and (c).

13. A voltmeter is graded according to its resistance

8. Potential difference across MP = p.d. across N O = p.d. across NP (see Fig. 22.104)

a voltmeter has a resistance R ohms and it reads V

Current across NP, INP

10 = 20

Across MP, 0.5 R1 = 20

or

1 or INP = 2 A R1 = 40

Total current = 2 + 0.5 + 1.0 = 3.5 A

Using r = 0.5

E

= 0.9, we get E = 2.25 V. (2 r ) Hence the correct choices are (b) and (d). in

as R/V ohm per volt. It is given that V = 3.0 V and the voltmeter is graded as 6000 /V. Hence the resistance of the voltmeter is R = 6000 /V 3.0 V = 1.8 10 4

Electric Current and D.C. Circuits 22.41

or (R + 1) (R – 2) = 0 voltmeter is V 3.0 = = 1.67 I= R 1.8 104

which gives R = – 1 or 2 . Since negative value of R is not admissible, R = 2 .

10 –4 A

In order to convert this instrument into a voltmeter tance R that must be connected in series with it is given by (here V = 12 V) R =

V 12 –R= I 1.67 10

= 5.4

4

– 1.8

10 4

10 4

The resistance of the new voltmeter = R + R = 1.8 104 + 5.6 104 = 7.2 104 . Hence the correct choice are (a) and (d). 14. Let circuit be broken as shown in Fig. 22.109(a). tance remains unaffected by removing one mesh nite network be R. The effective resistance of the remaining part of the circuit beyond CD is also R. The circuit can be recombined as shown in Fig. 22.109 (b). The resistance R and 2 are in parallel. Their combined resistance is 2R R = R 2 R is the series with remaining 1 total combined resistance is

Fig. 22.110

Applying Kirchhoff’s loop rule to the two meshes in Fig. 22.110, we have ( R = 2 ) 1 I + 2I = 6 (1) and 2(I – I ) – 2I = 0 which gives I = 2I . Therefore, using Eq. (1), we have 6 = 1.5 A. 2I + 2I = 6 or I = 4 So the correct choices are (b), (c) and (d). 15. Refer to Fig. 22.111. Applying Kirchhoff’s loop rule to loop abcda, we have – Ir2 + E2 + E1 – Ir 1 = 0 which gives I =

E1 r1

E2 2 1.5 7 = = A r2 1 2 6

resistance. The

2R +1 R 2 which must be equal to the total resistance of the Fig. 22.111

The emf across A and B is

Fig. 22.109

2R 3R 2 +1= R 2 R 2 2 or R + 2 R = 3 R + 2 or R2 – R – 2 = 0 or R2 – 2 R + R – 2 = 0 or R (R – 2) + (R – 2) = 0 R=

E = – E2 + Ir2 7 5 = – 1.5 + 2= V 6 6 This is the emf of the single battery. The internal resistance of the single battery is the resistance r of the parallel combination of internal resistances r1 and r2 which is r1 r2 1 2 2 r= = r1 r2 1 2 3 So the correct choices are (a) and (d). 16. Equivalent resistance of the parallel combination of 3 and 6 is R = 3 6/(3 + 6) = 2 . As the cells are in opposition, net emf E = E2 – E1 = 8 – 4 = 4 V.

22.42 Comprehensive Physics—JEE Advanced

= 2. Also V1 + V2 = 12. Therefore, V1 = 8 V and

Therefore, current is I=

R

E 4.5 r1

r2

=

4 2

4.5

0.5 1

V2 = 4V. Thus Q = C1V1 = 10 F 8 V = 80 C = C2. Thus the correct choices are (a), (c) and (d). 18. A and B of the AB is 4 . Therefore, total resistance = 3 + 4 + 2 = 9 . Current I = 9 V/9 = 1 A. The current is equally divided in the 8 A and B and the remainder 8 . Hence current in AC = 0.5 A. This current is equally CD CD. Therefore, current in the 4 resistor = 0.25 A. Hence the corrects choice are (a) and (d).

= 0.5 A

Potential difference across E1 is V1 = E1 + I r1 = 4 + 0.5 0.5 = 4.25 V Potential difference across E2 is V2 = E2 – Ir2 = 8 – 0.5 1 = 7.5 V Hence the correct choices are (a) and (d). 17. branch ABC ADC. So choice (a) is R2 R1 =I A and C = V C2 12 V. Therefore Q = V1C1 = V2C2. Thus 1 V2 C1

III Multiple Choice Questions Based on Passage Questions 1 to 4 are based on the following passage

R3 = 2R2 = 4 ohm C= 5 F

Passage I

IIT, 1988 1. The current in resistance R1 is (a) 0.5 A (b) (c) 1.5 A (d) 2. (a) 1.5 A (b) (c) 0.9 A (d) 3. (a) 0.3 A (b) (c) 0.2 A (d) 4. (b) (a) 4.8 10–6 J (d) (c) 1.44 10–5 J

Fig. 22.112

E1 = 3E2 = 2E3 = 6 volt R1 = 2R4 = 6 ohm

1.0 A zero R3 is 1.2 A 0.6 A R4 is 0.25 A zero 9.6 10–6 J 1.92 10–5 J

SOLUTION 1.

I2 R3 = E1 or I2 = R1 is zero,

2.

EBCDE

The correct choice is (a). 3. FDHGF

E1 = 1.5 A R3

Electric Current and D.C. Circuits 22.43

I 3 R 2 – I 2 R 3 + I 3 R 4= – E 2 – E 3

1 5 10–6 (2.4)2 = 14.4 2 Hence the correct choice is (c). =

or I3 (R2 + R4) = I2 R3 – E2 – E3 or

I3 (2 + 3) = 1.5

4–2–3

or 5I3 = 1 or I3 = 0.2 A So the correct choice is (c). 4. F and E is V= E2 + I3 R2 = 2 + 0.2 and A F and A 1 CV2 2

R1 = 6

A

E1 = 6V

B

C

I1

C=5 F

E

10 –6 J

F

I3

R2 = 2 E2 = 2V

I1 R3 = 4

E

I2 I3

G E3 = 3V

D

H

R4 = 3

Fig. 22.113

Questions 5 to 8 are based on the following passage Passage II E1 = 3 V, E2 = 2 V, E3 = 1 V and r1 = r2= r3= R = 1 .

Fig. 22.114

IIT, 1981

5. Current I1 (a) zero (c) 1.0 A 6. Current I2 (a) zero (c) 2.0 A 7. Current I3 (a) 1.0 A (c) 2.0 A 8. If r2 B (a) 1.0 A (c) 3.0 A

r1 is (b) 0.5 A (d) 1.5 A r2 is (b) 1.0 A (d) 3.0 A r3 is (b) 1.5 A (d) 2.5 A A is connected to R (b) 2.0 A (d) zero

SOLUTION 5.

abcda and

6

I2 choice (a).

abcdefa 7.

– I 1r 1 + E 1 – E 2 – I 2 r 2= 0 and – I1r1 + E1 – E3 – I3 r3= 0 E1 – I1r1=E2 + I2r2 = E3 + I3r3

(i) a

have I1 = I2 + I3

(ii)

E1 – (I2 + I3) r1= E3 + I3r3 or 2I3 + I2 = 2 (iii) Also E2 + I2r2= E3 + I3r3 or I3 – I2 = 1 (iv) I1 = 1 A. So the correct choice is (c). Questions 9 to 12 are based on the following passage Passage III

I3 = I1 – I2 = 1 – 0 = 1 A. So the correct choice is (a). 8. Since I2 a and d = emf E2 = 2 V and remains equal to 2 V r2 tial difference across a and d E1 and E2 do not the currents I1 and I3 A is conI1 = I3 B R

The direction of one of the currents in the branches is IIT, 1991 9. In branch ab, the current i1 is a to b

22.44 Comprehensive Physics—JEE Advanced b

b to a a to b b to a 10. In branch ad, the current i2 is a to d d to a a to d d to a 11. In branch bd, the current I3 is b to d d to b b to d d to b. 12. (a) 8.0 (b) 8.5 (c) 17 (d) 17.5

I4

I3

10 W

5W

I1

c 5

a

I5

I2 5W

I

10 W

I

d

f

10 W

e

8.5 V

Fig. 22.115

SOLUTION a, b and d I = I1 + I2, I4 = I1 – I3 and I5 = I2 + I3. abda, bcdb and adcefa 10 I1 + 5 I3 – 5 I2 = 0 or

I2 = 2 I1 + I3

(i)

5(I1 – I3) – 10(I2 + I3) – 5 I3 = 0 or and or

I1 – 2 I2 = 4 I3

I1 = 0.2 A, I2 = 0.3 A and I3 = – 0.1 A. Since I1 and I2 I3 indicates that the direction I3 should be from d to b and not from b to d. 9. The correct choice is (c). 10. The correct choice is (c). 11. The correct choice is (d). 12. Total current is I = I1 + I2 = 0.2 + 0.3 = 0.5 A. Hence

(ii)

5 I2 + 10(I2 + I3) – 8.5 + 10(I1 + I2) 2 I1 + 5 I2 + 2 I3 = 1.7

R= (iii)

So the correct choice is (c).

Questions 13 to 16 are based on the following passage Passage IV E, F, G and H of emfs 2 V, 1 V, 3 V and 1 V and internal resistances 2 , 1 , 3 and 1

13.

(+) E (–)

14.

A

V 8.5 = = 17 I 0.5

B

resistor is 1 A 7 1 (c) A 11 (a)

B and D is 2 V 7 2 (c) V 11 (a)

(+) F (–)

2W

(–) H (+)

15. D

(–)

G

(+)

cell G is (a) equal to 3 V (b) more than 3 V

C

Fig. 22.116

IIT, 1984

1 A 9 1 (d) A 13

(b)

2 V 9 2 (d) V 13 (b)

Electric Current and D.C. Circuits 22.45

16. cell H is (a) equal to 1 V (b) more than 2 V

SOLUTION 13. Let I1 and I2 be the currents in branches BAD and DCB I3 DB resistor. E

2V

I1

A

I1

I1

2W

5 6 A and I2 = A 13 13 1 A I3 = I1 – I2 = – 13 The current 2 BD assumed. So the correct choice is (d). 14. B and D = 2

H 1W

1V I3

D

I2

I2 3V

C

I2

G 3W

G = 3 – (6/13) choice is (d).

D have BADB and DCBD Questions 17 to 19 are based on the following passage Passage V are connected to a 60

resistor reads 30 V. IIT, 1980 Voltmeter

(1/13)

= 1 + 6/13 = 19/13 is (d).

1.6 V. So the correct

H 1.46. Thus the correct choice

17. The resistance of the voltmeter is (a) 600

(b) 800

(c) 1000

(d) 1200

18. The current in the circuit is 3 3 (a) A (b) A 32 16 3 3 A (d) A 8 4 19. When the same voltmeter is connected across the 300 (c)

400

300

3 = 21/13

16.

I1 = I2 + I3 or I3 = I1 – I2

connected across the 400

the DB as

15.

Fig. 22.117

and 400

(ii)

I1 =

1V

1W

I1

(i)

B I2

I3

2W

F

2I1 + I1 + 2(I1 – I2) = 2 – 1 = 1 or 5I1 – 2I2 = 1 and 3I2 + I2 – 2(I1 – I2) = 3 – 1 = 2 or 6I2 – 2I1 = 2

60 V

Fig. 22.118

(a) 40 V

(b) 22.5 V

(c) 37.5 V

(d) 25 V

SOLUTION 17. Potential difference across the 400

resistance

resistance, their combined resistance is

the 300 resistance = 60 – 30 V = 30 V. Let R be the resistance of the voltmeter. As the voltmeter is

R=

400 R 400 R

22.46 Comprehensive Physics—JEE Advanced

and 400 R should be equal to 300 . Thus

Total resistance in the circuit = 400 + 240 =640 Current in the circuit is

resistances,

So the correct choice is (a).

R = 1200 18. When the voltmeter is connected across the 300 resistance, their combined resistance is 300 R R = 300 R

300 1200 300 1200

19. 240

resistance

3 240 = 22.5 V 32 Thus the correct choice is (b). =

= 240

Questions 20 to 23 are based on the following passage Passage VI

23. 10– 4 J 10–4 J

(a) 2 (c) 6

state. The currents, the values of resistances and emfs of

10–4 J 10–4 J

(b) 4 (d) 8 1A

C = 4 F.

3

IIT, 1986 20. The value of current i1 is (a) 1 A (c) 3 A 21. The value of current i2 is (a) 1 A (c) 3 A 22. The value of current i3 is (a) 1 A (c) 3 A

60 V 3 = A 640 32

I=

400 R 300 = 400 R

4V

3

5

A

i1

E

2A

(b) 2 A (d) 4 A

4 F 3V

1

2A

(b) 2 A (d) 4 A

1

B

2

i2 D

i3

4

3 1A

(b) 2 A (d) 4 A

Fig. 22.119

SOLUTION 20.

A,

V = 5 i1 + 1 i1 + 2 i2 = 15 + 3 + 2 = 20 V

i1 21.

B, i2 + 1 = 2 or i2

C = 20 V

choice (a). 22.

D, i1 = i2 + i3

1 4 10 – 6 (20)2 = 8 2 Thus the correct choice is (d).

= 2 A. So the correct choice is (b). 23. AEDB

=

Questions 24 to 26 are based on the following passage Passage VII

A voltmeter of resistance 400 the 400 resistance.

1 CV2 2

i3 = i1 – i2 = 3 – 1

is connected across

IIT, 1996

10– 4 J

24. The value of current i1 is 1 1 (a) A (b) A 10 20 1 1 A (d) A (c) 30 40 25. The value of current i2 is 1 1 (a) A (b) A 30 15

Electric Current and D.C. Circuits 22.47

(c)

1 A 10

(d)

2 15

(a)

10 V 3

(b) 5 V

(c)

20 V 3

(d) 4 V

26.

SOLUTION

Fig. 22.120

24.

– 200 I3 + 200(I1 + I2 – I3) + 100(I2 – I3) = 0 Resistance R = 200 of the 400 resistor and the resistance of 400 the voltmeter.

or 2 I1 + 3 I2 – 5 I3 = 0 (3) Simultaneous solution of Eqs. (1), (2) and (3) yields 1 A I1 = I2 = I3 = 30 So the correct choice is (c). 25. The correct choice is (a). 26. R(= 200 ) 1 20 200 = V. = I3 R = 30 3 Hence the correct choice is (c).

of

– 100 I2 – 100(I2 – I3) + 100 I1 = 0 or

I1 – 2 I2 + I3 = 0

(1)

– 100 I1 – 200(I1 + I2 + I3) + 10 = 0 or

3 I1 + 2 I2 – 2 I3 – 0.1 = 0

(2)

Fig. 22.121

IV Assertion-Reason Type Questions correct.

22.48 Comprehensive Physics—JEE Advanced

(a) Statement-1 is True, Statement-2 is True; ment-1. (b) Statement-1 is True, Statement-2 is True; Statement-1. (c) Statement-1 is True, Statement-2 is False. (d) Statement-1 is False, Statement-2 is True. 1. Statement-1

5. Statement-1 E1 and E2 are the emfs of cells C1 and C2 E1 > E2. Cell C1 R is x x

-

AB is

R C1

-

Statement-2

D

x

A

C2

B

G

Fig. 22.123

2. Statement-1 Statement-2 –1

electron is also very small (= 1.6

10

– 19

C), yet

Statement-2 7

3. Statement-1

AD due to cell C1 = E2, the emf of cell C2. 6. Statement-1 AB tor R decrease. Statement-2

x

Statement-2 A and D due to cell C1 = emf E2 of cell C2. 7. Statement-1 Electrons in a metallic conductor have no motion if direction. 4. Statement-1

Statement-2 AC x. If the radius AB

IIT, 1982 8. Statement-1

becomes 4 x.

Statement-2 R1

R2 G

IIT, 1993 C

A

B

9. Statement-1 -

x

Fig. 22.122

Statement-2 the square of its radius.

of the standard resistance.

Electric Current and D.C. Circuits 22.49

Statement-2

IIT, 2008

SOLUTIONS 1. The correct choice is (c). The electrons suffer a

A and B due to cell C1 x. 6. The correct choice is (a). If the value of R is A and B due to cell C1 x. 7. Electrons in a conductor have random thermal motion. Statement-1 is false and Statement-2 is true. 8. Both the statements are true and Statement-2 is the

decelerated by collision. The net acceleration

2. The correct choice is (b). The current in a metal de-

1029 3.

3

9.

.

R is the standard resistance and X

rent stays neutral because as many electrons enter

4. The correct choice is (d). The condition for no R1 R2

RAC RCB

Fig. 22.124

RAC and RCB AC and CB AB is doubled, the ratio RAC/RCB 5.

X (100 l ) = R l The value of X

l the same, the value of R must be increased. So Statement-1 is false, Statement-2 is true.

AB

V Integer Answer Type Questions 1. emf of 5 V and an internal resistance of 0.2

2.

. Find

IIT, 1997 L internal resistances are connected in series. Due to

Fig. 22.125

22.50 Comprehensive Physics—JEE Advanced

T in a time t. A number N L

T in

the same time t. Find the value of N.

AJ = 60 cm. Find the value of X (in ohm). IIT, 2002 4.

IIT, 2001

AB in volts is

Fig. 22.126

3.

Fig. 22.127

AB resistance X and a resistance of 12

IIT, 2011 are connected

SOLUTIONS 1. All the cells are in series. Therefore, total emf = 40 V and total internal resistance = 1.6 . Therefore, current in the circuit is I=

40 = 25 A 1.6

3. In the balanced condition X BJ 40 = = 12 AJ 60

X=8

4. E – Ir = 5 – 25

0.2

= 5 – 5 = 0 volt. 2.

t is Q = V 2t/R, V Here Q = ms T

m R its resistance.

3V 2 t ms T = R m R second case, (2m) s T =

NV 2 t 2R

s Fig. 22.128

(1)

CDFEC 6–1

I–2

I – 3= 0 I= 1A

(2) VA – 6 + 1

I – VB = 0 VA – VB = 6 – 1

2

2=

N /2 or N 2 = 36 or N = 6, 9

1=5V

23

Current

Chapter

REVIEW OF BASIC CONCEPTS 23.1

23.2

HEATING EFFECT OF CURRENT

If a current I R V

t

POWER-VOLTAGE RATING OF ELECTRICAL APPLIANCES

(P – V P = VI

H I=

2

V t R J

H = VIt = I 2 Rt = H Electrical Power r R

V V V2 = = I P /V P 1. Power of Electrical Appliances Connected in Parallel R=

If E

Let R1 R2 R3 = VI = I2 R =

V

2

2

P

P V

E R V = R R r 2 [

Electrical Energy

P1 P2 P3

E = I(R + r A

V2 R1 = P1 P1 =

(60

V2 R1

V2 R2 = P2 P2 =

V2 R2

V2 R3 = P3 P3 =

V2 R3

R 1 1 1 1 = + + + ... R1 R2 R3 R

P=

= 1000 = 3.6

106

V2 = V2 R

1 R1

1 R2

1 R3

V2 V2 V2 + + + ... R1 R2 R3 P = P1 + P2 + P3 + ... =

...

23.2 Comprehensive Physics—JEE Advanced

2. Power of Electrical Appliances Connected in Series R1

R R = R1 + R2 + R3 + ...

R

R1 V

I=

P = I 2 R = I2 (R1 + R2 + R3

R

=

R1

300 1000 R1

P = P1 + P2 + P3 + ... 23.1

Fig. 23.1

R1 = IR1 =

SOLUTION P

V 2

V 200 200 = = 40 R= P 1000 R V 2 160 160 V = P = R 40 P P 100 P 1000 640 = 100 = 36% 1000

300 R1 = 100 1000 R1 23.3 A

300 R1 1000 R1

R1 = 500

B

23.2 SOLUTION A

RA =

V2 PA

= 200

200 = 1000 40

SOLUTION P

2

R=

P1 = P1 >> P

B

V

RB =

V2 200 200 = = 1000 R 40

V 12 300 300 = R 1000

=

V PB

200

200 60

= 666.7

I. A B

PA = I2 RA PB = I2 RB

PA I2 R R = 2 A = A PB RB I RB RA > RB; PA > PB

A B.

23.3

P2 = I 22 R = (5

V. A

PA =

V2 RA

B

PB =

V2 RB

R PA = B RA PB

105 2 10 = 2.5 P1 W2 = W3 W1 > W2 > W3 W1 < W2 = W3 W1 < W2 < W3

29.

30. R R

1 5

Fig. 23.8

34.

Fig. 23.7

31.

32.

L

T

B2

B3

t

N L T t

N

Fig. 23.9

23.8 Comprehensive Physics—JEE Advanced

35.

A B

B A

42. 36.

43.

37.

AB

BC

L AB

10–1

–1

BC

K–1

r

I

BC AB. BC AB. AB

38. AB

Fig. 23.10

44. 2 39.

A

B

A

B A

B 1 4 40.

1 2 A

B Fig. 23.11

45. B A

1

41.

Fig. 23.12

BC BC

r

23.9

46. R1 R2

P1 > P2 > P3 P2 > P1 > P3

R3 R1 R2

P2

R3

P3 > P2 > P1 P1 = P2 = P3

P1

P3

Fig. 23.13

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43.

2. 8. 14. 20. 26. 32. 38. 44.

3. 9. 15. 21. 27. 33. 39. 45.

4. 10. 16. 22. 28. 34. 40. 46.

5. 11. 17. 23. 29. 35. 41.

6. 12. 18. 24. 30. 36. 42.

SOLUTIONS 1. If E

r R1 E

I

R1

r R1

Q1 = I R1 =

Q2 = Q1

R1

r

R1 Fig. 23.14

2

E R2

2

E

2

4 5

R2

r

Q2 r =

I = R1 R2 . 4 5

V 2 4 5

2. 1

6 =2 5

6 5

.

V1 =

= 2V 5

V 2

4 5

2V 5

23.10 Comprehensive Physics—JEE Advanced

V4 =

2V 5

V2 = V3 =

P1 =

V12 R1

3V 5

=

3V . 5

Y =

= 4V 25

2 V2 9V 2 P2 = 2 = = 4.5 V R2 25 2 25

V2Y kR

S = 5.

2 V2 9V 2 P3 = 3 = = 3V R3 25 3 25

S = stress

S Y

=

2

L L S

G

S G

S=V t

2 V2 4V 2 P4 = 4 = = V R4 25 4 25

H =

P2

V2t R

H

1 R

HG S = HS G

3. I1 =

4I 5

I2 =

3I 2I I3 = 5 5

I4 =

2 S = 3 G

I 5

P 1 = I 12 R 1 =

16 I 25

P 2 = I 22 R 2 =

9I2 25 4I 25

P 4 = I 42 R 4 =

I2 25

2 G 3

6. W = V × I

R = 12/2 = 6 2

2

1 =

16 I 25

2 =

18 I 2 25

2

P 3 = I 32 R 3 =

S=

3 = 4 =

12 I 25

I = W/V = 24/12

. If n

n 1.0 =2 6

n = 12

7. I =

2

4I2 25

500 =5A 100 100 R= = 20 5 R

P1 8 = P2 9

200 =5 R 20

4.

R = 20

8. V2 = k (T – T0 R L =

L T

k L V2 k = = L R

k T T=

L L

V I = 100

R I 2R = 900

23.11

R =

900 = 4.0 15 15

9.

V = IR = V–V =

10. P

10 6

3V 2 R P = 9 P. P = 16.

R 17. Let R

I1 =

100 106 = 5000 A 20 000

P1 = I 21 R =

2

R

t 2

Q =

V t 2R

R = 2.5 107 R R =R t

I2 =

100 106 =5 200

P2 =

I 22 R

105 A

Q =

V 2t 2V 2t = R R

Q = 4Q = 2.5

10

11

18.

R

2.5 1011 P2 = = 10 4 2.5 107 P1

I1 =

E r1

r

11.

2

P=I R

I

r2

2

P=

V R P 2 V = P V

V V

P=

r1

2V V R

r2 t

r

r1 r1 (r2 + r

r2

=

2

r

2

r2

r1 r2

R2

13.

R1 + R2 Q

14. 15. Let R

V

P =

V2 3R

2

r1 r2 (r1 – r2

r = 19. Let R1

2

r2 (r21 + 2r r1 + r2

r 2 (r1 – r2 2=

r

= r2 (r1 + r

r1 (r 22 + 2r r2 + r 2

P =–2 P

r 1t

r

2

E

Q2 = P P

I

12.

r1

P = 2I IR

P 2 I = P I

2

E

Q1 = I 2 r1 t =

Q= t=

V 2t R1 R2 Q R1

V = t1 + t2

R2 2

=

V

V 2 t1 V 2 t2 = R1 R2 =

Q R1 V

2

+

Q R2 V2

23.12 Comprehensive Physics—JEE Advanced

25.

20.

I 21 R1t = I 22 R2 t E I1 = R1 r

1 1 1 = + R1 R2 R 1 V2 = t1 Q R1 2 1 = V Q t

1 V2 = t2 Q R2 1 R1

1 R2

=

E2

1 1 + t1 t2

V2t R

t1 H

R2

R2

R1

R=

. r R2

2

r

R2 = (R2 + r

H=

t2 t1

R2 /R1

2

E

E2

R1 =

r=

26. H =

R2 = R0(1 + t2

R1 + r

R1

t1 t2 t = t1 t2 21. R1 = R0 (1 + t1 t2 R2 1 = R1 1

r

2

I2 =

2

R1

R2 l = A

l r2

V 2 t r2 l r l

27. Let I

t1

I 5

t2 22. 2

Q = V t R

2205 80 4.2 23. Let R

210

210

P1 = I2

–1

20

P2 = 1 P1 5

I 2

P2 =

2

P2 = P1 = 10 5 5

4 = I2 –1

28.

1

R1 = R + R + R = 3R R2 = R + R + 4R = 6R R1 R2 3R 6R R = = = 2R R1 R2 3R 6R

R R = R 4 V 2 = 4V 2 R R

4

= R = 2R R=2

–1

=4

29. 30.

24.

R 10 I=

R = 0.05

V2 P

100 10 R 10 10 R

I P = I 2R

2

100 100 = 1000

I = I

R 10

R

23.13

R

100 10 R 10 10 R

=

10 R R 10 10 R

= 1

R

10

5

R

5R R

W3 =

250 2 1042

34. Let R

10

R

25R 2 R

156.3 2 1042

W1 < W2 < W3

Q = I2 5

W2 =

R

R

R

2

R = 5 35.

V = 22

P

31. 32.

t V Q = ms T

Q = V 2t/R R=

m

P = VI V I= P

V R= I

s

220 2 V2 = = 484 100 P

R

V 2

ms T = 3V t R R

m

(2m s T =

2

2 = N /2 9

N2

P =

V R

2

=

110 2 484

36.

NV 2 t 2R

R = 20 P = V2/R V

N

P V2 = RP R

=

Pma

=

V=

RP

20 1000

= 100 2 V

33.

B1 B2

B3

are

R2 =

37.

250 2 = 625 100

V2 R1 = W1 2

250 60

= 1042

B3 B1

Q = KA 2 d t

= R3

V3

250 625 625 1042 V2

W1 =

V12 R1

A A=6

VR1 V1 = R1 R2

B2

1

93.7 2 625

=6

2

(0.5 m

1 1.5 Q = 1.68 10 1 10 3 t = 2.52 104 –1

100

23.14 Comprehensive Physics—JEE Advanced

H=

R=

V2 = 2.52 R 420 420 2.52 104

IB =

104 IA

IA

IB

IB

=7

V2 A V2 = l R

38.

V = 400 = 1 A RB 400

42.

200 2 = 80 500

R=

.

l V

A L

39. RA =

2L

RB =

r2 RA RB

2r

P=

2

RA = 2RB

Q 40.

l

43. R =

1 R

V2 50 2 = R 80

r2 AB

A V2 200 = RA = PA 40 RB =

BC

B are

1/r2

R

2

= 1000

RAB = 1 RBC 4

V2 200 2 = = 400 PB 100

V = IR P = I2 R

IA =

PA 40 = = 0.2 A V 200

IB =

PB 100 = = 0.5 A V 200

RBC = 4RAB

I

P

VBC = 4 VAB. R

PBC R = AB = 4 RBC PAB

AB

BC

R = RA + RB = 1000 + 400 = E = V/l l E VBC = 4 VAB EBC = 4EAB

1400 I=

V 400 = = 0.286 A R 1400

I

IA

(V IB

A 44.

B

2 ven V 2

41. V A IA =

B are

V 400 = = 0.4 A RA 1000

45. cur

V 2 = 30 3

V2

V

23.15 2

V2 R P 2 > P 1 > P 3.

V

P=

rr 46.

1 R

P R1

R1 = 1

R2 R3 = 2 .

R2 = 0.5

II Multiple Choice Questions with One or More Choices Correct 1. m 2m

I

5.

l

A

V 2.

V

A l

6. tance 0.5 R=8 3.

7. R

4. R = 0.5 E

8.

.

R

r P

P P

R = r. R = 2r. E2/2r. E2/4r.

23.16 Comprehensive Physics—JEE Advanced

9.

E&

A

B

r

A

P

B

I E I= r

A

2

P

=

B

E I= 2r

14. 2

E r

P

=

E 4r

10.

11. –1

. Fig. 23.16

BD

15. Fig. 23.15

–1

. A

12.

E

A

r sup

E r=1

D D

E r=2

13.

A B

A

B

Fig. 23.17

ANSWERS AND SOLUTIONS 1.

l A

d m = Ald A

d

I t H = I 2Rt

H H

R

23.17

R ence V

1/P

H 2 H = V t R

H

1 R

R1 R2 R

V 2 = 200 200 W 40 = 1000

2.

200

1/P R1

R2. If R V

I1 =

200

V R

R1

60 = 666.7

I2 =

200 200 100 = 400

V R

R2

R1 > R2 I2 > I1

= 1000 + 666.7 + 400 = 2066.7 I =

200 = 0.097 A 2066.7

V 2/R

3. R = If l

l = A r

2 V2 = V A l R

5.

l r2 R

6. (r

H = V 2/R

ce E = V l

0.5 = 3

E

E I=

4. Rb =

2 V2 = 200 = 400 Pb 100

E = 88 = 8 A R r 8 3 I2 (R + r

2 R h = 200 = 40 1000

7. V

I

r = 0.1

2

23.18 Comprehensive Physics—JEE Advanced

P = V I = 1.5 2

d 2P dR 2

2

Pc = I r

at R r

E2 8r 3

P

R

R=r

R=r

PR = P – Pc P

R 1.5 R R 0.10

2 = E 4r

9. R

R

=–

E

I

R

=

1.5 R 2.0 R 0.10

P = (E – E I E = Ir. P = EI – I2r

PR 1.5 R 2.0 = 2.6 R 0.10 R = 0.65 .

I dP = E – 2 Ir dI dP dI

P 8.

I= E

I =

E 2r d 2P

R r

dI 2

=–2r P

I = E/2r. I = E/2r

2

P = I 2R =

E R R r

E dP/dR

2

r

P

P d2P/dR2 R

E

10.

r dP = dR

E2 R r

2

2R R r

1

11.

dP/dR

d P dR

2

=

2E

R r

I2

P Q = I22 5 I1 4

3

3R R r

I2 C

I1

d2P/dR2 at R = r

2

I1

B

P R=r

R=r P 2

I

A

2R 1– =0 R r R=r

2 = E 4r

2

I 12

D

5

I1 = I 2 . 2 2 20 = I2 5

I2

2

–1

I1

4=1

4=4

23.19

14. A

B

E

P = I 2R =

R r

E 2 R1

P1 = P1 = P2

R1

r

R1 R2

R2 R1

E2

6

6 12

13. P1 = ce

100 R1

r r

R r

D

C

BD 2

2

3

E 2 R2 R2

r

2

15.

2

C CB B 6 + 2 = 8 A.

2

4 r 6 r

1.5 = 6 1.96 4

E2R

P2 =

2

D

I= 6 =6A 1

Fig. 23.18

12. I =

B

r 1

C

B

E

2

P2 =

200 R2

2

P 1 = P 2.

R1 = 1 R2 4

Fig. 23.19 2

current I

2

A V1 IR R = 1 = 1 = 1 V2 IR2 R2 4

D = (2 A

3

1

5

I 2 R1 P1 R = 2 = 1 = 1 I R2 P2 R2 4

III Multiple Choice Questions Based on Passage Questions 1 to 5 are based on the following passage passage I

B

A B

Joule’s Law

I q V

A

A

B

t

23.20 Comprehensive Physics—JEE Advanced

I= q t A

B V W = qV

3.

V R causes a current I

2 H = qV = I tV = I 2Rt = V t R P

1.

V = IR

(

2 P = H = IV = I2R = V t R A B

V R R

R V I R

V

4. 5. A B

P

B. A.

2.

P 4

P 2

P

P

ANSWERS AND EXPLANATIONS 1.

I2 (2R

I2 R = 2P

2V 2 V2 =2 = 2P 2R R 2.

4. 5. Let R

V P = V2/R

3.

2V R

2

=4

V2 = 4P. R

P V2 = 2 2R

R

V2 = P = 4P. R/4 2

23.21

Questions 6 to 9 are based on the following passage

7.

Passage II 8. 1/8 9. 6.

SOLUTION 6.

P

P1 + P2 = 16 + 80 = 96 7. 2

P1 2

8

1 8

8.

9. 2

P2 2

Questions 10 to 12 are based on the following passage

10.

I

Passage III 11.

12.

Fig. 23.20

SOLUTION 10.

3I 4

I2 = 15 I = 3I 20 4 I1 = 5 I = I 20 4 I 22

2

11.

5 = 45

I = 4 A.

I2

2 =

2

12.

I1 I = 4

4 6= 4

6

23.22 Comprehensive Physics—JEE Advanced

Questions 13 to 16 are based on the following passage

13.

R1

14.

R2

15.

R3

16.

R4

Passage IV R1 = 4

R2 = 6

R4 = 10

R3 = 12

S1

C = 500 S1 S1

S2

S2

Fig. 23.21

SOLUTION P = I 2R

S1

P1 P2 P3 = I 12 R1 I 22 R2 I 32 R3 U = 1 CV2 = 1 500 2 2 S1 R4 P4 = 0. Let I1 t R2 I3 I3 I 2R 2 = I 3R 3 I1 = I2 + I3 I1 = 2I3 + I3

I3 =

I1 I2 I3 = I1

2 I1 3

I1 3 I1 3

10–6

2

2

2

2

S2 R4

U R1 R2

I2

13. P1 = 3.6 3 6

R3

I1 = I2 + R2 R3 I2 = 12 I3 I2 = 2I3 I2 = 2 3

14. P2 = 3.6 2 6 3 . 6 1 15. P3 = 6

2 I1 3

16.

1 3

IV Assertion–Reason Type Questions

one

R3

R4 P4

12

23.23

Statement-2 1. Statement 1 3. Statement 1 (I – V T2 T 1.

T1 T2

Statement-2

4. Statement 1

Fig. 23.22

Statement-2

2. Statement 1 Statement-2

SOLUTIONS 1.

I1 =

I–V

2.

T2

R

R1

I2 =

V R

R2

R1 > R2 I2 > I1

T2 ture T1

V

T2 T 1. 4. R

1/P

R

P V

3.

R

R1 R2 R

V

V2/R

24

and Magnetism

Chapter

REVIEW OF BASIC CONCEPTS 24.1

BIOT-SAVART LAW

dB at a point whose position vector with respect to a current element dl is r is given by I (dI r ) (1) dB = 0 4 r3 where

0

24.2

10–7 Hm–1

=4

AMPERE'S CIRCUITAL LAW

Fig. 24.1

a closed curve is proportional to the current threading or passing through the closed circuit i.e. B·dI = where

24.3

0

B=

0I

directed into the page if I is clockwise 2r and outside the page if I is anticlockwise.

0I

is the permeability of free space.

MAGNETIC FIELD DUE TO CURRENT CARRYING CONDUCTORS

For a coil of N turns. NI B= 0 2r (iii) (Fig. 24.3). B=

(i) I (Fig. 24.1) B=

0I

directed into the page (away from the

2 r reader) if the I is upwards and towards the reader if I is downwards. At points Q or S, B = 0 (ii) radius r (Fig. 24.2)

Fig. 24.2

0I

2r

2

directed into the page. For a semi-circular element ( = ) I B= 0 4r

Fig. 24.3

24.2 Comprehensive Physics—JEE Advanced

straight conductor are parallel. Therefore, dl = 0. Hence B = 0 at point .

(iv) (Fig. 24.4) B=

0I

(sin

4 r directed into the page Special Cases (a) If the conductor XY

and point lies near the centre of the conductor (as in Fig. 24.1), = = 90° so B= =

0I

4 r

+ sin )

r

(v) (Fig. 24.7) O due to straight portions and ST is zero and due to semicircular part is I B = 0 directed into the page. 4r

Fig. 24.4

(sin 90° + sin 90°)

Fig. 24.7

0I

If the current in anticlockwise, B is directed towards the reader.

2 r (vi)

(Fig. 24.8)

(b) If the conductor XY is of

a 2 b2 directed into the page ab For a square coil (b = a) B=

lies near the end X or Y as shown in Fig. 24.5, then = 90° and = 0°, Fig. 24.5 then I B = 0 (sin 90° + sin 0°) 4 r =

2

0I

B=

0I

2 2 a

0I

4 r

Note: near one of its ends. (c) If the conductor XY L and point lies on the right bisector of the conductor, as shown in Fig. 24.6, then = and L/2 sin = sin = x L/2 = L 2 r2 2 L = 2 4r L2

so B =

0I

2 r

Fig. 24.8

llow metal pipe of (Fig. 24.9)

(vii) At point

,B=

0I 2 r

L 4r

2

L2

Fig. 24.6

(d) If the point lies on the straight conductor or on its axis, then dl and r for each element of the

Fig. 24.9

Magnetic Effect of Current and Magnetism 24.3

(Fig. 24.9) Inside the pipe at a point at a distance r from the axis, Ir B= 0 2 2 Outside the pipe at a point at a distance r from the axis I B= 0 2 r (viii) (Fig. 24.10) At point

4

, B =

0nI,

where n =

N N ; N = total no. of turns. Outside the L 2 solenoid, B = 0. 24.1 bent wires and STU lying in the x- plane and each carrying a current I as shown. Find the magniO. Given OQ = OT = a

2M

0

,B=

For a toroid of radius

(

2

2 3/2

) where M = IA = I is the magnetic moment. If current I is anticlockwise B is directed from O to . For clockwise current B is form to O. 2

Fig. 24.10

(ix) revolving in a circular orbit of radius r with speed r with speed v (Fig. 24.11) e ev = en = Current along orbit I = T 2 r The direction of I is opposite to the direction of motion of the electron. O is I B= 0 4r ev and 2 r is directed into the page.

where I =

SOLUTION As point O is along the line segments and ST, the O due to and ST is zero. The O due to wires and TU respectively are I (k ) I (k ) and B2 = 0 B1 = 0 4 (OQ) 4 (OT ) both directed along the positive z-axis. The resultant t O is ( OQ = OT = a)

Fig. 24.11

e vr r2 = 2

Magnetic moment M = IA = I (x) solenoid In the middle region B = per unit length.

Fig. 24.12

0nI;n

At the ends of solenoid, B =

= no. of turns

0 nI

2

B = B1 + B2 = 2

0I

4 a

(k )

0I

2 a

k

24.2 I in the opposite direction are placed perpendicular to the x- plane. One wire is located at point (0, a, 0) and the other wire at Q (0, –a, 0). Find the magnitude and direction of at point A(x, 0, 0).

24.4 Comprehensive Physics—JEE Advanced

SOLUTION Refer to Fig. 24.13. Wire 1 carries a current I along the positive z-direction and wire 2 carries a current I along the negative z-direction. = OQ = a, OA = x,

= QA = r.

SOLUTION (a) (i) Radius r of wire A when it is bent into a circle is given by 2 r=L

L 2

r=

loop is B1 =

0I 2r

0I

(1)

L

O is (OT = a)

due to wire B

= =

Fig. 24.13

=

A due to wire 1 is

M

0I

B1 =

2 (

0I (sin 45° + sin 45°) 4 a 0I

2 2 a 4

0I

a

2 L

L 8

0I 2

)

2 a x2 According to Biot-Savart law, B1 is perpendicular to both and wire 1 and therefore in the x- plane. A due to wire 2 is 0I 2

B2 =

2 a x2 The -components of B1 and B2 cancel each other but the x-components add up. These components are B1 cos and B2 cos both along the positive x-direcA is B = B1 + B2 = (B1 cos + B2 cos ) i = (a

0I 2

a 2

x )

a cos

=

(a

0I 2

a 2

x )

2

x2 a r

Since centre O of the square is at the same distance from each side of the square and each arm carries the same current, the mag-

(i ) a a

2

Fig. 24.14

x

2

(i)

24.3 Two wires A and B have the same length L and carry equal currents I. Wire A is bent into a circle and wire B is bent into a square. (a) Obtain expression for the i) the centre of the circular loop and (ii) the centre of the square. Which

of the same and in the same direction. Hence O is B2 = 4B

=

16

0I

2 L

0I

8 2 L

(2)

(b) Dividing (2) by (1) we get B2 B1

8 2 2

8 1.41 (3.14) 2

= 1.16

Hence B2 > B1 centre due to the square loop will be greater than that due to the circular loop.

Magnetic Effect of Current and Magnetism 24.5

24.4 Figure 24.15 shows a wire loop ABCDEA carrying a current I as shown. Given AE = ED = a and AB = CD = a/2. Find the magnitude and direction of the magnetic F where BF = CF = a/2.

24.5 A wire ABCDE is bent as shown in Fig. 24.16. The wire carries a current I and the radius of the bent coil BCD is r. Find the magnitude and direction of the O. SOLUTION The straight line segments AB and DE are collinear with O AB and DE at O is zero. Angle subtended at O by arc BCD = 2 – . BCD at O is B =

0I 2 2r 2

Fig. 24.15

SOLUTION F is B = BAB + BBC + BCD + BDE + BEA Since point F lies in line with current elements AB and CD, BAB = BCD = 0 Also BDE = BEA =

0I

0I

(sin 0° + sin 45°) = 4 a 4 2 a directed out of the page and towards the reader. BBC =

0I

(sin 45° + sin 45°)

4 BC/2

directed into the page and away from the reader. Now BF 2

BC =

0I

BBC =

1

a

4

2

a 2

FC 2

2

a 2

2

a 2

0I

1

a

2

0I

0I

a

24.6 A wire ABCEF is bent as shown in Fig. 24.17 and caries a current I. The radius of the smaller arc ABC is r1 = r and that of the bigger arc is r2 = 2r. Find the magnitude of the Fig. 24.17

SOLUTION ABC at O is

directed out of the page.

2 2 a Since BBC > BDE + BEA the page and has a magnitude B =

The current through BCD is clockwise. Therefore, the O is into the page and away from the reader.

O.

2 2 directed into the page. Now BDE + BEA =

Fig. 24.16

0I

2 2 a

B is directed into 0I

a

1

B1 =

1 2 2

B2 =

0I 2 2r1 2 DEF at O is 0I

2r2 2

24.6 Comprehensive Physics—JEE Advanced

since B1 and B2 are both directed into the page, the O is B = B1 + B2 =

0I

2r1

B

2 r=

0I

2r

1

0 Ir 2

=

2

0I

2 2

2r2 2

B=

Putting r1 = r and r2 = 2r, we get B =

0i

2

2

r

NOTE

4

24.7 A long straight cylinder of radius carries a current I which is uniformly distributed across its cross-secr from the axis of the cylinder in cases (a) r > (b) r
; B

shows the variation of B with r.

and

SOLUTION Figure 24.18 shows the cross-sectional view of the cylinder.

Fig. 24.19

walls. Therefore, in the case r < [Fig. 24.18 (b)], no current threads the Amperian loop. Hence B = 0 for points inside a hollow cylinder.

Fig. 24.18

Case (a) r > . For this case, the Amperian loop is a circle of radius r concentric with the cross-section [Fig. 24.18 (a)]. For this loop, L = 2 r and the current threading the loop is i = I. From Ampere’s circuital law. I B 2 r = 0I B= 0 BL = 0i 2 r Case (b) r < . For this case, L = 2 r [Fig. 24.18 (b)] and the current threading the loop is i = current per unit cross-sectional area of the cylinder cross-sectional area of the Amperian loop =

I 2

r2

Ir

2 2

From Ampere’s law,

24.8 The current density J (current per unit area) in a solid cylinder of radius varies with distance r from its axis as J = kr where k is a constant. Find the magnetic where (a) r > and (b) r < . SOLUTION Current I =

JdA

kr

(2 rdr )

Case (a) We take the Amperian loop of radius r > . Since the loop is outside the cylinder, the current through the loop is I=

kr

(2 rdr )

2 k r 2 dr

0

B

2 r =

3

0

0I

=

2

3

2

3

3

B

0

r 3 Case (b) For r < , the current through the Amperian loop is

Magnetic Effect of Current and Magnetism 24.7

r

I =

kr (2 rdr ) 0

B

24.4

2 r =

0I

=

2 kr 3 3 0

(vi) Frequency of revolution is v =

2 kr 3 3

B

0 kr

2

3

FORCE ON A MOVING CHARGE IN A MAGNETIC FIELD

The force on a charge q moving with a velocity v in a B is given by F = q(v B) The direction of F is perpendicular to both v and B. The magnitude F of vector F is given by F = qvB sin where is the angle between vectors v and B. (1) F = 0 if v = 0, i.e. a charge at rest does not experience any magnetic force. (2) F = 0 if = 0 or 180°, i.e. the magnetic force vanishes if v is either parallel or antiparallel to the direction of B. = 90°, i.e. if v is (3) F is maximum = Fmax if perpendicular to B, the magnetic force has a maximum value given by Fmax = qvB The direction of the force when v B is given by Fleming’s left hand rule. (4) If v is perpendicular to both E and B and E is E . perpendicular to B, then F = 0 if v = B

24.5

MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELD

Case (a): If v is perpendicular to B, the particle describes a circle in the region of the magn F v. (i) The speed along the circular path is constant. (ii) The kinetic energy is constant. (iii) Velocity and momentum continually change. mv 2 (iv) The radius r of the circular path is given by r qvB r=

mv qB

qB which is indepen2 m

dent of both v and r.

Case (b): If v is inclined to B at an angle , the particle mv sin moves in a helical path. The radius of helix is r = , qB time period T =

2 m and pitch of the helix = v cos qB

Applications (i)

-

region of magnetic

describes a semi-circle of radius. mv r= qB

Fig. 24.20

T m 2 qB (ii) If the particle enters the t=

as shown in Fig. 24.21, then

and

r=

mv qB

t=

2 m qB

Fig. 24.21

where is in radian. (iii) In Fig. 24.22, the particle wall not be able to hit the wall if d > r, i.e. d>

mv qB

B

mv qd

2mK qB

where m = mass of particle and K = kinetic energy. If the particle is accelerated through a potential difference V, then K = qV. (v) Time period of revolution is T =

2 m qB

T

Fig. 24.22

Fig. 24.23

24.8 Comprehensive Physics—JEE Advanced

(iv)

SOLUTION Refer to Fig. 24.24. d qBd r mv Linear defection x = r(1 – cos ) sin

=

24.9 through a potential difference of 18 kV and enters a certain initial velocity. What is the trajectory of the Fig. 24.24

initial velocity and (b) makes an angle of 30° with the 10–31 kg.

A is

SOLUTION V = 18 103 V 1 mv2 = eV 2 =

2

B= 2eV m

v=

(1.6 10

19

1/2

8 107 ms

31

9 10

1

(a) Since v is to B, = 90°, the trajectory of the electron is circular having a radius r =

mv eB

(9 10

31

)

(1.6 10 = 4.5

(8 107 ) 19

)

0.1

–3

10 m = 4.5 mm

(b) The trajectory of the electron is helical. The radius of heix is r=

mv eB

mv eB

sin

= 4.5 mm

2 r

4

10 2

7

10 0.05

4 10

5

T

directed inwards along the negative z-direction (a) = 90°. Therefore, force on proton is F = qvB sin

1/2

(18 103 )

)

0I

sin 30° = 2.25 mm

EXAM 24.10 A long straight wire lying along the -axis carries a current of 10 A along the positive -direction. A proton moving with a velocity of 107 ms–1 is at a distance 5 cm from the wire at a certain instant. Find the magnitude and direction of the force acting on the proton at that instant if its velocity is directed (a) along the negative x-direction (b) along the positive -direction and (c) along the positive z-direction

= (1.6 10–19) 107 (4 10–5) sin 90° = 6.4 10–17 N According to Fleming's L.H. rule, the direction of the force is parallel to the wire and opposite to the direction of current I, i.e. F is along the negative -direction (b) = 90°, F = qvB = 6.4 10–17 N. The force is directed towards the wire, i.e. along negative x-direction (c) = 180°. F = qvB sin 180º = 0 24.11 A proton and an -particle move perpendicular to a -particle is four times that of a proton and its charge is twice that of a proton. Find the ratio of radii of the circular path followed by them if both (a) have equal velocities, (b) have equal linear momenta, (c) have equal kinetic energies and (d) are accelerated through the same potential difference. SOLUTION Given

m =4 mp

and

q =2 qp

Magnetic Effect of Current and Magnetism 24.9

mv qB

(a) r =

mpv

rp

rp

mp

r

m

qp B q qp

m v q B

and r

1 4

1 2

2

mv qB

(b) r =

Kinetic energy K = r=

m qB

2K m

q qp

mp

rp r mv qB

(c) r =

rp r

m

1 2 mv 2

2K m

v

Fig. 24.27

1 2 mK qB 1 4

2

(iv) Force per unit length between two long straight 1 and I2 and

1

p qB q qp

f =

2 r attractive if I1 and I2 are in the same direction and repulsive if I1 and I2 are in opposite directions. Force on a segment of length l of either wire is F = f l. (v) 1 placed at a

2

(d) K = qV. Therefore 1 1 2mV r= 2mqV qB B q

24.6

rp

mp

r

m

q qp

1 4

0 I1 I 2

2

current I2 as shown in Fig. 24.28. II L F = 0 1 2 loge 1 vertically upwards. 2 r x

1 2

FORCE ON A CURRENT CARRYING CONDUCTOR IN A MAGNETIC FIELD

(i) Force on a straight conductor placed perpendicuF = BIL upwards if current I is from left to right and downwards if I is from right to left (given by Fleming’s left hand rule) Fig. 24.28

(vi)

1

2

Fig. 24.25

as shown in Fig. 24.29.

Fig. 24.26

(ii) (Fig. 24.26) F = BI(2 ) = 2 vertically upward for clockwise current and downward for anticlockwise current (iii) Net force F = F1 – F2 = 0

Fig. 24.29

24.10 Comprehensive Physics—JEE Advanced

Force F1 and F2 being equal and opposite cancel and F3 and F4 are given by expression above. Net force on coil is F = F3 – F4 =

0 I1 I 2 a

0 I1 I 2 a

2 x

2 (x

L)

0 I1 I 2 aL 2 x( x L) directed towards the wire (attractive)

F=

5

20 5 10

2

2 5 10 2 = 2 10–5 N (repulsive since I1 and I2 are in opposite directions) Force exerted by on CD is F2 =

24.12 The battery of a car is connected to the motor by 50 cm long wires which are 1.0 cm apart. If the current

10 7 )

= (4

=

0 I1 I 2

2 r2 (4

CD

10 7 ) 2

5

20 5 10

2 10

2

2

= 5 10–5 N (attractive since I1 and I2 are in the same direction)

Is the force attractive or repulsive. SOLUTION Force per unit length is f=

0 I1 I 2

10 7 )

(4

2 r

2

200

200 2

(1.0 10 )

0.8 Nm

1

F = f l = 0.8 0.5 = 0.4 N Since the currents in the wires are in opposite direction, the force is repulsive. 24.13 A small rectangular loop ABCD of sides 5 cm and 3 cm carries a current of 5 A. It is placed with its longer side parallel to a long straight conductor of length 5 m at a distance of 2 cm from it as shown in Fig. 24.30. If the current in force on the loop. Is the loop attracted towards or

to current in is directed outwards (towards the reader) and perpendicular to the plane of the coil. Therefore, forces F3 and F4 on BC and AD are equal and opposite and hence cancel each other. Therefore, the net force on coil ABCD is F = F2 – F1 = 5 10–5 – 2 10–5 = 3 10–5 N (attractive). Hence coil is attracted towards . 24.14 A particle of charge q and mass m moving in region I with a velocity v enters normally a region II of width d where a uniform B (directed inwards) exists as shown in Fig. 24.31. There is no and III. (a) What is the maximum speed (vmax) of the particle so that it returns back Fig. 24.31

(b) What will happen if v = 2 vmax SOLUTION

Fig. 24.30

SOLUTION Force exerted by F1 =

0 I1 I 2 2 r1

(a) Refer to Fig. 24.32(a). The particle describes a circular path of radius r = mv/qB in region II. It will return to region I if it describes a semicircle in region II. This happens if r vmax, the particle is cross over to region III after describing a circular trajectory in rejoin II with O as the centre. In region III, te particle is move along the tangent at Q. The particle will suffer a deviation . In triangle sin

=

sin

=

If v =

OQ

d r

qBd mv

vmax v

2 vmax, then sin

24.7

TORQUE ON A CURRENT CARRYING COIL IN A MAGNETIC FIELD

The torque on a coil of N turns, area A carrying a current I B is given by =M B Magnitude of torque is = MB sin = NIAB sin where M = NIA is the magnetic moment and is the angle The magnitude of torque on a coil in radial magnetic =k where k is the restoring couple per unit twist and

=

1 2

= 45°

Then k or I NAB Current sensitivity of the galvanometer is NAB Cs = I k NIAB = k

24.15 A straight horizontal conducting rod of length 50 cm and mass 60 g is suspended by two vertical wires at its ends. A current of 5 A set up in the rod. to the conductor in order that the tension in (b) What will be the tension in the wires if the direction of the current is reversed, keeping Ignore the mass of the wires and take g = 10 m/s–2. SOLUTION Refer to Fig. 24.33.

Fig. 24.33

is the = 90°.

24.8

I=

TORQUE ON A BAR MAGNET IN A MAGNETIC FIELD

The magnetic dipole moment of a bar magnet of pole strength q and length (2a M = q(2a) It is a vector pointing from the south to the north pole of a magnet. Force on north pole N of magnet = qB (in the direction of B) Force on south pole S of the magnet = –qB (opposite to B) opposite forces on the magnet. The two forces, therefore, constitute a coupe which tends to rotate the magnet in the clockwise direction. The arm of the couple is 2 a. The torque is given by = arm of the coupe force = 2a qB = q(2a) B or =M B

24.12 Comprehensive Physics—JEE Advanced

where M = q(2a) is called the magnetic moment of the bar magnet. The direction of is perpendicular to both M and B. If M and B are both in the plane of the paper then the torque will be perpendicular to the plane of the paper and directed into it away from the reader. The magnitude of the torque is = MB sin where is the angle between M and B. The SI unit of M is Nm T–1 or JT–1 (joule per tesla).

24.9

POTENTIAL ENERGY OF A MAGNETIC DIPOLE

Fig. 24.34

The magnetic potential energy of a magnetic dipole in any orientation B is the dipole from its zero energy position ( = 90°) to the given position. . U = –MB cos In vector notation, U = – (M·B) For stable equilibrium U is minimum. Hence = 0 and = 0. For unstable equilibrium, U is maximum i.e. = 180°. Hence = 0

24.10

SOME USEFUL TIPS

1. Magnetic dipole moment of a bar magnet is M = m l, where m is pole strength and l is the length of the magnet. The value of M depends on the volume of the magnet. (a) If a magnet is cut into two equal parts by cutting it by a plane along its length, its volume is halved, Hence the magnetic dipole moment of a piece is halved = M/2. The pole strength M is also halved as length l remains m = l the same. (b) If a magnet is cut into two equal parts by cutting it by a plane transverse to its length, the volume and length are both halved. Hence the magnetic moment becomes M/2 but pole strength m remains the same. (c) If a wire of magnetic dipole moment M and length l is bent as shown in Fig. 24.34, the 1 and distance between the pole becomes 2 magnetic moment becomes l M M =m 2 2 If the wire is bent as shown in Fig. 24.35, the magnetic moment becomes

M =m

Fig. 24.35

l 2

M 2

NOTE Magnetic dipole moment M is a vector quantity directed from south pole to north pole.

experiences no net force but experiences a torque = M B. The magnitude of is = MB sin where is the angle between M and B. (a) when = 90°, max = MB (b) when = 0°, min = 0 (stable equilibrium) (c) when = 180°, min = 0 (unstable equilibrium) Potential energy is U = –M.B = –MB cos . When = 0°, P.E is minimum Umin = –MB. U is max = Umax = MB when = 180° (d) Work done in rotating the magnet from 1 to 2 is W = MB (cos 1 – cos 2) experiences a force as well as a torque. 3. The time period of a bar magnet oscillating in a I , where I T=2 MB ml 2 , 12 m = mass of magnet and l = length of magnet. (a) If a bar magnet is cut into two equal parts by cutting along its length, then each part has M = M/2 and I = I/2. Hence T = T. (b) If a bar magnet is cut into two equal parts by cutting perpendicular to its length, then each part has M = M/2 and I = I/8. Hence T = T/2. (c) If two bar magnets of magnetic moments M1 and M2 are placed one on top of the other as shown n Fig. 24.36, then time period is given by (since I = I1 + I2 and M = M1 + M2)

is the moment of inertia of the bar magnet =

Magnetic Effect of Current and Magnetism 24.13

24.16

( I1 I 2 ) ( M1 M 2 ) B

T1 = 2

A closely wound solenoid of 1000 turns and area of cross-section 5 cm2 carries a current of 3 A. It is suspended through its centre (a) what is the magnetic 10–2 T is set up at an angle of 30° with the axis of the

Fig. 24.36

If the magnets are placed as shown in Fig. 24.37, then I = I1 + I2 but M = M1 – M2 and ( I1 I 2 ) T2 = 2 ( M1 M 2 ) B M T1 and T2 are related as 1 M2

(T22

T12 )

(T22

T12 )

SOLUTION M = e r2 = 3.14 (1.6 10–10)2 = 1.26 10–23 Am2

e to a bar magnet (a) At a point on axial line (Fig. 24.38) 2Mr

0

(r 2

4

l 2 )2

parallel to M = m

For a very short magnet (l rp (a) r = rp < rd (c) r = rd > rp (d) rp = rd = r IIT, 1997 36. Two particles each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2l. The rod is rotated at a 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 q q (b) (a) 2m m 2q q (c) (d) m m IIT, 1998 37. Two very long straight parallel wires carry steady currents I and – I. 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 at this instant is (a)

I qv 2 d 0

(b)

0

I qv d

Magnetic Effect of Current and Magnetism 24.19

(c)

2

0

I qv d

42. An electron revolves in a circle of radius 0.4 Å with a speed of 106 m/s in Hydrogen atom. The magnetic

(d) zero

IIT, 1998 38. A charged particle is released from rest in a region which are parallel to each other. The particle will move in a (a) straight line (b) circle (c) helix (d) cycloid IIT, 1999 39. An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric x along the + z direction, then y direction and negative ions towards – y direction y direction y direction y direction and negative ions towards + y direction IIT, 2000 40. Two long parallel wires are at a distance 2d apart. plane of the paper, as shown in Fig. 24.51. The B along the line XX is given by IIT, 2000

10–7 motion of the electron in tesla is [ 0 = 4 –19 H/m; Charge on the electron = 1.6 10 C] (a) 0.1 (b) 1.0 (c) 10 (d) 100 43. A proton of velocity (3 i 2 j) ms–1

of

magnetic induction (2 j 3k) tesla. The acceleration produced in the proton is (charge to mass ratio of proton = 0.96 108 C kg–1) (a) 2.8 (b) 2.88 (c) 2.8 (d) 2.88

108 (2 i 3 j) 108 (2 i 3 j 2k) 108 (2 i 3k) 108 ( i 3 j 2k)

44. A long wire carries a steady current. It is bent into at the centre of the loop is B. The wire is then bent into a circular coil of n turns and the same current of coil will be (a) nB (c) 2nB

(b) n2 B (d) 2n2 B

45. circular loop of radius 3 cm at a point on its axis at a distance of 4 cm from the centre is 54 T. The magT) at the centre of the loop will be (a) 250 (b) 150 (c) 125 (d) 72 46. A wire ABCDEF (with each side of length L) bent as shown in Fig. 24.52 and carrying a current I is B parallel to the positive y–direction. What is the magnitude and direction of the force experienced by the wire?

Fig. 24.51

41. An electron moves with a speed of 2 along the positive x

105 ms–1 B

= (i 4 j 3k ) tesla. The magnitude of the force (in newton) experienced by the electron is (the charge on electron = 1.6 10–19 C) (a) 1.18 10–13 (b) 1.28 10–13 (c) 1.6 10–13 (d) 1.72 10–13

Fig. 24.52

(a) BIL along positive z-direction

24.20 Comprehensive Physics—JEE Advanced

(b) BI 2/L along positive z-direction (c) BIL along negative z-direction (d) BL/I along negative z-direction 47. long bent wires are placed in the x-y plane as shown in Fig. 24.53. The wires carry a current I = 10 A each as shown. The segments RL and SM are along the x-axis. The segments PR and QS are along the y-axis, such that OS = OR = 0.02 m. What is the magnitude and direction of the magnetic induction at the origin O? (a) 100 Wb m–2 vertically upward (b) 10–4 Wb m–2 vertically downward (c) 10–4 Wb m–2 vertically upward (d) 10–2 Wb m–2 vertically downward IIT, 1989

(a) 1 MeV (b) 2 MeV (c) 4 MeV (d) 0.5 MeV 51. A loosely wound helix made of stiff wire is mounted vertically with the lower end just touching a dish of mercury. When a current from the battery is started in the coil through the mercury (a) the wire oscillates (b) the wire continues making contact (c) the wire breaks contact just when the current is passed (d) the mercury will expand by heating due to passage of current. IIT, 1981 52. A non-planar loop of conducting wire carrying a current is placed as shown in Fig. 24.54. Each of the straight sections of the loop is of length 2a. The P(a, 0, a) points in the direction (a) (b) (c)

Fig. 24.53

(d)

48. A particle of charge q and mass m moves in a circular orbit of radius r with angular speed . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on (a) and q (b) , q and m (c) q and m (d) and m IIT, 2000 49. Two long parallel wires P and Q are held perpendicular to the plane of the paper at a separation of 5 m. If P and Q carry currents of 2.5 A and 5 A respectively in the same direction, then the magnetic P and Q is (a)

0

(b)

3

0

3 0 (c) (d) 2 2 50. A proton of mass m and charge +e is moving in 0

MeV. What should be the energy of an -particle (mass 4 m and charge + 2e) so that it revolves in a circular orbit of the same radius in the same mag-

1 2 1 3 1 3

i

1 2

j

k

j

k j

i

i k

k

Fig. 24.54

IIT, 2001 53. Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. A uniform magnetic this plane. The speeds of the particles are vA and vB respectively and the trajectories are as shown in Fig. 24.55. Then (a) vA < mB vB (b) mA vA > mB vB (c) mA < mB and vA < vB (d) mA = mB and vA = vB

Fig. 23.55

IIT, 2003 54. 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

(a)

0 NI

b

(b)

2

0 NI

a

Magnetic Effect of Current and Magnetism 24.21

(c)

0 NI

2 b

a

ln

b a

(d)

0 NI

2 b

a

ln

a b

(b) E = a i ; B = c k (c) E = 0; B = c j

IIT, 2001 55. A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It 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 (a) qbB/m (b) q(b – a)B/m (c) qaB/m (d) q(b + a)B/2m IIT, 2002 56. Which pattern shown in Fig. 24.56 correctly repre-

(d) E = a i ; B = c k

bi bk bj

IIT, 2003 58. A magnetized wire of magnetic moment M is bent into an arc of a circle that subtends an angle of 60° at the centre. The equivalent magnetic moment is (a) (c)

M

(b)

3M

(d)

2M 4M

59. are held 0.1 m apart and carry a current of 5 A each in the same direction. The magnitude of the magis (a) 10–5 T

(b)

2

10–5 T

10–5 T (d) 2 10–5 T (c) 3 60. A proton moving with a speed u along the positive x-axis enters at y = 0 a region of uniform magnetic B = B0 k which exists to the right of y-axis as shown in Fig. 24.58. The proton leaves the region after some time with a speed v at co-ordinate y. Then (a) v > u, y < 0 (b) v = u, y > 0 (c) v > u, y > 0 (d) v = u, y < 0 Fig. 24.56

57. An electron is moving in the x-y plane along the positive x-axis. There is a sudden change in its path due to the presence of electric and/or magP as shown in Fig. 24.57. The curved path lies in the x-y plane and is found to be non-circular. Which of the following combinations is possible?

Fig. 24.58

IIT, 2004 61. A particle of charge q moves with a velocity v = ai B = b j + c k where a, b and c are constants. The magnitude of the force experienced by the particle is (a) zero (b) qa(b + c) 2 2 1/2 (c) qa(b – c ) (d) qa(b2 + c2)1/2 62. A current I lateral triangle of side a. The magnitude of the mag-

Fig. 24.57

(a) E = 0; B = b i

ck

(a)

3 0I 2 a

(b)

9 0I 2 a

24.22 Comprehensive Physics—JEE Advanced

(c)

3 3 0I 2 a

(d) zero

63. A particle of mass m and charge q, accelerated by a potential difference V enters a region of a uniform B. If d is the thickness of the region of B, the angle through which the particle deviates from the initial direction on leaving the region is given by 1/2

(a) sin

q = Bd 2mV = Bd

q 2mV

1/2

(b) cos

q 2mV

1/2

(c) tan

= Bd

67.

q 1/2 2mV 64. A metal wire of mass m slides without friction on two rails spaced at a distance d apart. The track B. A constant current I and back down the other rail. If the wire is initially at rest, the time taken by it to move through a distance x along the track is (d) cot

Fig. 24.59

(a)

(b)

(c)

(d)

= Bd

(a) t =

BId 2 xm

(b) t =

2xm BId

(c) t =

BIdm 2x

(d) t =

2dm BIx

65. A particle of charge q and mass m is released from

Fig. 24.60

IIT, 2011 68. 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 the XY plane and a steady current ‘I Z the spiral is [see Fig. 24.61] (a)

the origin with a velocity v = a i in a uniform magB = b k . The particle will cross the y-axis at a point whose y-coordinate is ma 2ma (b) y = (a) y = qb qb (c) y = –

ma qb

(d) y = –

(c)

0 NI

2 b a 0 NI

2b

ln

ln

b a

b a

(d)

0 NI

2 b a 0 NI

2b

ln

ln

b a b a

b a b a IIT, 2011

2ma qb

66. A thin wire loop carrying a current I is placed in a B pointing out of the plane of the coil as shown in Fig. 24.59. The loop will tend to (a) move towards positive x-direction (b) move towards negative y-direction (c) contract (d) expand

(b)

Fig. 24.61

Magnetic Effect of Current and Magnetism 24.23

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67.

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68.

(d) (c) (b) (a) (d) (d) (d) (b) (c) (b) (d) (c)

(a) (a) (d) (c) (a) (c) (a) (b) (a) (d) (b) (a)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63.

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64.

(c) (d) (c) (c) (b) (c) (c) (a) (a) (b) (a)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65.

(c) (c) (a) (d) (c) (c) (b) (a) (d) (c) (b)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66.

(c) (c) (d) (d) (d) (a) (c) (c) (b) (c) (d)

(d) (c) (c) (c) (d) (a) (c) (c) (c) (d) (c)

SOLUTIONS 1.

B2 = magnetic energy density. Therefore 2 0 B2

energy [ML2 T 2 ] = = [ML–1 T–2] 3 volume [L ] 0 RC = time constant = [T] B 2 RC

=

= [ML–1 T–2] [T] = [ML–1 T–1], which

0

Fig. 24.62

are the dimensions of viscosity. 2.

In Fig. (b), r = L,

of the wire and is given by I B= 0 (1) 2 R R = 1 mm = 1 10–3 m, 0 = 4 10–7 Hm–1 and –3 B = 2 10 T. Using these values in Eq. (1), we get I = 10 A. 2 mv cos 30 2 mv 3 = (1) 3. Pitch x = qB qB 2 Radius r =

mv sin 60° mv = qB 2qB

x From Eqs. (1) and (2), we get r = 2 3 choice (c).

= 0° and

6. Figure 24.63 shows the cross-sectional view of the rod. From Fleming left-hand rule, the magnetic force F acting on the rod is directed to the right and is given by F = B I L. Since the rod moves with a constant velocity, no net force acts on it. Hence the components of Mg and F parallel to the inclined plane must balance, i.e.

(2) , which is

4.

CD is nonuniform, wire AB will experience as force as well as a torque. Hence it will have both translational as well as rotational motions. 5. Refer to the Fig. 24.62. R in Fig. (a) is given by I B = 0 (sin + sin ) (1) 4 r

= 45°. Using these

Fig. 24.63

or or

F cos

= Mg sin

B I L cos

= Mg sin B=

Mg tan IL

24.24 Comprehensive Physics—JEE Advanced

Mg

Putting = 30°, we get B =

3 IL

, which is choice

=

I (cos 45° – cos 90°) 4 r

=

I 0I (cos 45° – 0) = 4 r 4 2 r

(d). 7. centre is B=

0

0

0I

2r

45

ADB at centre O is B1 =

90

0I

2r

2

L

directed out of the page.

E

r

O

ACB at centre O is B2 =

0I 2r

2

directed into the page

2 O is

Fig. 24.64

B = B2 – B1 = Putting

0I

( – ) directed into the page.

2 r

= 60° =

3

, we get B =

O due to current element DE is the same as that due to AE. Hence, O due to one side AD is

0I . 3r

8. The magnitude of magnetic induction due to a circular loop of radius R carrying a current I is I given by B = 0 2R

BAD =

2

0

I

4 2 r

2 0I 4 r

=

Since the centre of the square is equidistant from the ends A, B, C and D of each side of the square and each side produces at the centre O the same that due to one side. Hence (because r = L/2)

B=

I 4R

B = 4BAD =

0

The direction of B is normal to the plane of the loop. Since the current in the bigger loop is clockwise and that in the smaller loop is anticlockwise centre C are in opposite directions. Therefore, the magnetic induction at the centre is given by 0

B = B1 – B2 =

I

1 R1

4

1 R2

The wires PQ and S R the centre C. Hence the correct choice is (a). L 9. Refer to Fig. 24.64. Here r = OE = . Referring to 2 O due to the current element AE is given by BAE = –

I 4 r

45

0

sin 90

d

=

I 45 | cos |90 4 r 0

2

0

r

I

=

2 2

0

L

I

.

Hence the correct choice is (d). 10. Let the masses of X and Y be m1 and m2 and let their velocities after being accelerated be v 1 and v2 respectively. Since the particles have equal charges and have been accelerated through the same potential difference, their kinetic energies are equal, i.e. 1 1 m1v 21 = m2v 22 2 2 B, the radii of the circular paths are given by m1v12 mv = q B v 1 or q B = 1 1 R1 R1 and

m2 v22 mv = q B v 2 or q B = 2 2 R2 R2

Therefore

m1v1 mv m2v2 = 2 2 or 1 2 1 R1 R2 R1

m22 v22 R22

Magnetic Effect of Current and Magnetism 24.25

But

m1v 21 = m2v 22. Therefore, we have m1 = m2

R1 R2

2

helix whose radius is r=

Hence the correct choice is (c). 11.

tre due to each coil is B. Since the planes of the coils are at right angles to each other, the directions Br =

B

2

B

2

= 2 B Hence the correct choice is (c). pendicular to the direction of motion of the particle, the speed of the particle cannot change but its velocity changes. Hence the correct choice is (c). 13. The radius of the circular orbit is given by 2m K r= qB The charge of an –particle is twice that of a proton and its mass is four times the mass of a proton. Therefore m /q is the same for both. Hence r will the same for both particles. Thus the correct choice is (b). 14. The cyclotron frequency is given by qB = 2 m It is independent of the speed of the particle and the radius of its circular path. Now q /m. The charge of a proton is half that of an -particle and the mass of a proton is one-fourth. Therefore, will be doubled. Hence the correct choice is (d). 15. V = 2.88 10 3 V. The velocity of the electron is given by 1 m v2 = e V 2 v=

=

=

2eV m

3.2 107

1.6 10 =9

10

–4

9 10 2 1.6 10

19

9 10

sin 30

0.1

Hence the correct choice is (c). 16. The correct choice is (a). 17. The correct choice is (d). 18. The velocity when the potential difference is V is 2 eV m and force F = e v B When the potential difference is doubled, i.e. V = 2V, the velocity is v =

2 eV m

=

2e

2V m

2.88 103 2.88 103

I 1/2

31

10 7 ms–1

with the velocity v, then v = v sin and v11 = v cos . The electron has two motions: a linear motion parallel to magnetic

2 v

21. Referring to Fig. 24.65, the forces acting on arms BC and AD are equal and opposite. The force on arm AB is given by Ii F1 = 0 2 a which is directed towards the wire. The force on arm CD is given by 0I i F2 = 2 a b which is directed away from the wire. Since F1 > F2, the loop will move towards the wire. Hence the correct choice is (c).

1/2

31

=

Force F = e v B = 2 evB = 2 F. Hence the correct choice is (c). 19. The correct choice is (a). 20. produced by the current in the loop and the external

i 19

19

m = 0.9 mm

1/2

2 1.6 10

= 3.2

31

9 10

v=

12.

or

m v sin eB

Fig. 24.65

24.26 Comprehensive Physics—JEE Advanced

22. Magnetic moment m = AI = r 2I, where r is the radius of the circular loop. Now, the circumference of the circle = length of the wire, i.e. l2

2

2 r = l or r =

l2 I

r 2I =

Therefore, m =

2

4

4

l2 I , 4

=

2

26. Electric intensity E =

V d

where V is the potential difference between the plates and d, the separation between them. d = 3 mm = 3 10 –3 m E=

V 600 = d 3 10

=2

3

10 5 V m–1

which is choice (d). 23. Magnetic moment m = IA = I r 2. m = I1 r 21 = I2 r 22 I1 r2 = 22 = I2 r1

6 3

2

B e v = e E or B =

=4

Hence the correct choice is (d). 24. An electron moving in a circular orbit is equivalent to a current carrying loop. As explained above, the current is e I= e = T where T is the time period of the motion of the electron around the nucleus. If v is the speed of the electron, 2 r T= v I=

B=

ev 2 r

e 2

v =r )

(

I e = 0 2r 4 r 0

or

=

4 rB 0e

Hence the correct choice is (c). 25.

2I 10 7 2 30 = =2 r 0.03

0

4

10–4 T

which is directed into the page. G at C is 10

B = BD – BG = 2 = 1.6

10 –4 T

and is directed into the page. The force on 25 cm of wire C is F = B I l sin 90° = 1.6 10–4 10 =4

10 –4 T

10 –4 – 0.4

10 –4 N

0.25

B dl =

0I

Since no current exists in the medium (air) inside the pipe I = 0. Hence B = 0. Hence the correct choice is (b). 28. P due to current I1 in conductor AOB is I B1 = 0 1 2 a P due to current I2 in conductor COD is I B2 = 0 2 2 a Since the two conductors are perpendicular to each B1 and B2 will be perpendicular to each P is

C is 10–4

(I21 + I22)1/2

0

2 a

Hence the correct choice is (c). 29. The work done is given by M H sin

W=

7

2 20 = 0.4 0.1 which is directed out of the page. BG =

27. From Ampere’s law, we have

B = (B21 + B22)1/2 =

D at wire C is BD =

E 2 105 = = 0.1 T v 2 106

d

= MH |– cos | 0

0

= MH (1 – cos ) Hence the correct choice is (d). 30. The kinetic energy of proton is K = 2 MeV = 2 106 eV =2

106

1.6

10–19 J = 3.2

1 mv 2 = 3.2 10–13 2 Now, mass of proton is m = 1.67 fore,

10–13 J

10–27 kg. There-

Magnetic Effect of Current and Magnetism 24.27

v2 =

2 3.2 10 1.67 10

v = 1.96

or

13

27

= 3.83

m

1014

It is given that the charge q is moving in a circular path of radius 2l. Therefore, the time period = 2 (2l)/v. Hence qv (2l)2 = qvl m = 2 (2l )

107 ms–1.

Now force on proton is F = evB = 1.6 10–19 1.96 = 7.84 10–12 N

107

mv 2 = evB or r = r

The angular momentum L = mv(2l). Therefore, qvl q m = L mv(2 l ) 2m

2.5

Hence the closest choice is (d). 31. The point x = + a lies along the line of the straight section PQ at point x = a is zero. 32. The radius of the circular path of a particle of mass m, charge e moving with a speed v perpendicular to B is given by m v e B

e , the m charge to mass ratio. Hence the correct choice is (c). 33. The total Lorentz force on the electron is F = – e (E + v B) Thus, r is inversely proportional to

37. point equidistant from the two wires will be equal is zero. Hence the force on a charge at this point is zero. 38. E, the force on a particle of charge q is F = qE in the direction of the electric E is parallel to B, the velocity v of the particle is parallel to B. Hence B will not affect the motion of the particle since v B = 0. Thus the correct choice is (a). 39. E on a particle of charge q is to impart to it a velocity v which is proportional to qE, i.e.

v B. If E is along + z-direction, the force – e E will be along – z-direction. If B is along + x direction, force – e (v B) will be along + z direction. When eE = evB, the electron moves along + y-direction

v

r=

mv 2 = qB

2m K 1 , where, K = mv 2. qB 2

m since K and B are the same for q the three particles. If mp is the mass of a proton and qp its charge, then md = 2mp and qd = qp and m = 4 mp and q = 2qp. From these it follows that r = rp < rd. 36. According to Ampere’s Law, the magnetic moment of a current I cross-section A is given by Thus r

qE i

qE

B on a charge moving with a velocity v is to exert on it a force F = q(v

for an electron moving along + y direction, the elecz direction and magnetic x direction, then the electron will be 34. Refer to the solution of Q. 15. The correct choice is (c). 35. The radius of the circular path is given by

= IA

B)

q (qE i )

Bk

Thus

F

q2EB ( j )

Since F be

j k j) q2, both positive and negative ions will y direction.

k

q2EB i

40. and 2 at a point between them act in opposite directions. But at a point to the left of wire 1 and to the B along the line XX is as shown in choice (b). 41. Given v = (2 given by

105 i ) ms–1. The force vector is

F = q(v

B)

= q {2 =2

105

105 i

(i 4 j

q ( 4k

3k ) }

3 j)

Therefore, the y and z components of the force are

24.28 Comprehensive Physics—JEE Advanced

Fy = 6 and

105

Fz = –8

10

q 5

q Fy2

Magnitude of force = =q

2R coil of n turns and radius r is

Fz2

B =

105

10

= 1.6

10–19

105

= 1.6

10–13 N, which is choice (c).

10

2 r 2 = v

10 7 )

(4

2

2 0.4 10

2

=

10

j + 9i

= q (6i

= 10 tesla

(2j + 3k)

k + 4j

j + 6j

k)

= 3q (2i – 3j + 2k) newton

=3

F 3q = (2 i 3 j 2 k ) m m

(0.96

= 2.88

(1)

2(r 2 x 2 )3 / 2

(2)

Hence the correct choice is (a).

= q (6k – 9j + 0 + 6i )

Acceleration =

r2

Substituting the values of r and x, we get B0 125 or B 27 125 125 B0 = B = 54 T = 250 T 27 27

43. Given v = (3i + 2j) ms–1 and B = (2j + 3 k ) tesla. Force experienced by the proton is B) = q (3i + 2j)

oI

oI 2r Dividing (1) by (2), we get

10–3 A

Hence the correct choice is (c).

F = q(v

2

B0 r3 = B (r 2 x 2 )3/2

3

10

0 In

B0 =

10–17 s

=8

19 17

B=

10

106

e 1.6 10 = t 8 10

Current I =

I B= 0 = 2r

0.4 10

0 nI

is given by

42. Given r = 0.4 Å = 0.4 10–10 m, v = 106 ms–1 Speed of electron in the orbit is 2 r v= ; here t is time taken by the t electron to complete one revolution. Thus t=

0 nI

= n 2B 2r 2R / n 2R Hence the correct choice is (b). 45. Given x = 4 cm = 0.04 m and r = 3 cm = 0.03 m.

(6 105 ) 2 ( 8 105 ) 2

= q

0I

Now B =

46. Wires AB and EF experience no forces since curforces on BC and DE are equal in magnitude but are directed in opposite directions. Hence their resultant is zero. Only force acting is on CD. Hence the correct choice is (a). 47. Since point O lies along the segments LR and SM, at point O. As point O is close to R and S, the net O due to segments PR and QS is B = BP + BQ =

108) (2i – 3j + 2k)

108 (2i – 3j + 2k) ms–2

Hence the correct choice is (b). 44. Let L be the length of the wire and let R be the radius of the circle when the wire is bent into one circular turn and r be the radius of the coil of n turns. Then R L = 2 R = 2 nr or R = nr or r = n

=

0

4

2I d

0I

4 RO

= 10–7

0I

4 SO

0I

4

1 d

1 d

2 10 = 10–4 Wb m–2 0.02

outside the plane of the paper. Hence the correct choice is (c). 48. Magnetic moment M = current area charge area = time period

Magnetic Effect of Current and Magnetism 24.29

=

r2

q T

Angular momentum L = m r2 1 q r2 M 2 q . Hence the correct choice 2 L m 2 m r is (c). 49.

B=

0

4

·

2 I1 r1

0

4

·

2I 2 r2

Given I1 = 2.5 A, I2 = 5 A and r1 = r2 = 2.5 m. Using 0 . The magnitude of B 2 is 0/2 . Hence the correct choice is (c). mv 50. For proton: r = eB m v 4mv 2mv For -particle r = eB 2eB eB v Given r = r . Hence v = . 2 1 2 Energy of proton E = mv . Energy of -particle 2 is

these values, we get B = –

E =

1 mv 2

2

=

1 2

4m

v 2

mAv A mB vB

2 T

1 q r2 2

2

=

1 mv 2 = E 2

Hence E = 1 MeV which is choice (a). 51. When a current is passed through the helix, the neighbouring coils of the helix attract each other due to which it contracts. As a result the contact is broken and the coils will recover their original state is made again and the process continues. Thus the wire oscillates. Hence the correct choice is (a). 52.

in the y-z plane, which is along the x-direction and (ii) the other due to the planar loop in the x-y plane, which is along the z-direction. The direction of the

( i k ) . Hence the correct choice is (d). 53. The radius r of the circular path of a particle of mass m and charge q moving with velocity v B is given by 2 mv = qvB r or mv = qrB. Hence mA v B = qrA B and mB v B = q rB B.

rA rB

rA > rB. Hence mA v A > mB v B. Thus the correct choice is (b). 54. The correct choice is (c). For derivation of the expression, refer to a Textbook of Physics. 55. The radius r of the circular path is given by (see Fig. 24.66) mv 2 = qvB r qB or v= (r) m qB vmin= (rmin) m qB (b – a), m which is choice (b). =

Fig. 24.66

56. The correct choice is (d) because the lines of force are continuous inside the magnet. 57. Since the path of the particle beyond P is non-circular, both E and B P. Hence choices (a) and (c) are incorrect. Since the curved path lies in the x-y must be in the x-z plane. Hence choice (d) is also incorrect. Thus, the correct choice is (b). 58. Let r be the radius of the circle. The length of the 60 r arc = (2 r) = . Therefore, the length 2l 360 3 of the magnet is r 6l 2l = or r = 3 If m is the pole strength of each pole of the magnet, 6l the magnetic moment of the arc = m r = m =

3 (2 ml )

=

3M

Hence the correct choice is (c). 59. Wires A and B carry current I = 5 A each coming out of the plane of the page as shown in Fig. 24.66. The P due wire A is equal to that due to wire B, i.e. 2I BA = BB = 0 4 a =

10

7

2 5 01 .

24.30 Comprehensive Physics—JEE Advanced

= 10–5 T BA is perpendicular to PA and BB is perpendicular to PB. Therefore, = 60°. The P is given by BR2 = B 2A + B 2B + 2BA BB cos which gives BR = 2BA cos

The correct choice is (d). 62. Refer to Fig. 24.68. Let AB = BC = AC = a. Let OD = r.

2

3 = 2 Hence the correct choice is (c). 10–5

=2

Magnitude of F = [(q ab)2 + (q ac)2]1/2 = qa(b2 + c2)1/2

3

10–5 T

Fig. 24.68

O due to current I AB of the triangle is given by BAB = It is clear that OD = r =

BAB= Fig. 24.67

60. When the proton enters the region of the magnetic F given by F = q (u B) where q is the charge of the proton. The force F is perpendicular to both u and B. Since the force is perpendicular to the velocity of the particle, it does not do any work. Hence the magnitude of the velocity of the particle will remain unchanged; only the direction of the velocity changes. Hence v = u. Since u is perpendicular to B, the proton moves in a circular path. Since the charge of proton is positive, u is along posotive x-axis and B is directed out of the page, the proton will move in a circle in the x-y plane in the clockwise direction. Hence its y coordinate will be negative, when it leaves the region. Thus the correct choice is (d). 61.

F = q(v

B) = q{a i

= q(ab i

j + ac i

(bj + c k)} k)

= q(ab k – acj) = qa(b k – cj)

=

0I

4 r =

(sin

+ sin )

= 60° and

AD a/ 2 = tan tan 60

0I 4 r

2 3 a

=

a/2 3

=

a 2 3

(sin 60° + sin 60°)

3 0I 2 a

sides BC and AC = that due to side AB. Hence, the O due to the current in the three sides of triangle ABC is B = BAB + BBC + BCA = 3BAB Hence the correct choice is (b). 63. Refer to Fig. 24.69. Let v be the velocity of the particle. Its kinetic energy is 1 2 2qV 1 / 2 mv = qV or v = (1) 2 m The particle follows a circular path from A to B of radius r which is given by mv 2 mv = qvB or r = r qB

(2)

Using (1) and (2), we have r=

m 2qV qB m

1/2

1 2mV B q

1/2

Magnetic Effect of Current and Magnetism 24.31

B = b in the positive z-direction, and the charge of the particle is positive, the path of the particle is a circle as shown in the circular path is r=

mv ma = qB qb

Thus y = – 2 Fig. 24.69

In triangle BCD, sin

=

BD d = . Therefore, BC r

q 1/ 2 , which is choice (a). 2mV 64. Refer to Fig. 24.70. Wire PQ of length d, the spacing between rails carries a current I vertically sin

= Bd

the reader and perpendicular to the length PQ of the wire. Thus angle between I and B is 90°. The force exerted on the wire of length d by the F = BId sin 90° = BId Using Fleming’s left hand rule, the direction of the force is to the left. The acceleration of the wire is force F BId = = a= mass m m 1 2 at 2 choice is (b).

Now x =

t =

2x . Hence the correct a

r=–

Fig. 24.71

2ma qb

So the correct choice is (d). 66. A circular metal loop carries a current. Hence charge, say, q moves along the circle with a velocity, say v which is tangential to the circle at every point (Fig. 24.59 on page 24.22). The force experienced by the charge is F = q(v B). Since v is along the tangent and B is directed out of the x – y plane, the direction of the force is towards the centre O of the loop. Hence the force tends to contract the loop. Further, since F is perpendicular to v, no work is done on the loop. Hence it cannot have any translational motion. Thus the correct choice is (c). 67.

Hence the correct choice is (c). 68. 0I . The direc2r z-direction if the current is

radius r and carrying a current I =

anticlockwise. Consider a small element of width dr. The current through the element is total current in spiral dI = width of element total width of spiral =

Idr b a b

B= a

Fig. 24.70

65. Refer to Fig. 24.71. Since the velocity of the particle is v = a along the positive x-axis and the direction of

=

b

0 NdI

=

2r 0 NI

b

0 NI

dr 2 b a r a

dr b 0 NI = ln 2 b a a r 2 b a a

24.32 Comprehensive Physics—JEE Advanced

II Multiple Choice Questions with One or More Choice Correct 1. A straight wire carrying current is parallel to the y–axis as shown in Fig. 24.72. The P is parallel to the x– axis z–axis (c) magnetic lines are concentric circles with the wire passing through their common centre wire are oppositely directed.

(a) E = 0, B = 0 (c) E 0, B = 0

(b) E = 0, B (d) E 0, B

0 0 IIT, 1985 6. The force F experienced by a particle of charge q moving with a velocity v B is given by F = q (v B). Which pairs of vectors are at right angles to each other? (a) F and v (b) F and B (c) B and v (d) F and (v B) 7. Choose the correct statements from the following. in magnitude from point to point but has a constant direction, is set up in a region of space. A charged particle enters that region

Fig. 24.72

2.

dal solenoid (a) is independent of the radius of the solenoid (b) depends on the number of turns and the current in the solenoid (c) is constant in magnitude inside the solenoid (d) is always radial inside the solenoid.

3. (a) is independent of the radius of the solenoid (b) depends on the number of turns and current in the solenoid (c) is uniform throughout the solenoid (d) is perpendicular to the axis of the solenoid. 4. A charged particle of mass m and charge q enters a B with a velocity v at an angle with B. (a) The kinetic energy of the particle will not change if 0. (b) The momentum of the particle will not change if = 0. (c) The particle moves in a circle of radius mv/qB if = 90°. (d) The frequency of circular motion is independent of the speed of the particle if = 90°. 5. 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 may have

a constant speed. The initial velocity of the particle is either along the direction of the

from point to point in magnitude and direction is set up in a certain region of space. A charged particle enters the region with a certain initial velocity. The direction of the the initial velocity. (c) A electron travelling in the positive x–direction enters a region of space having an y– direction. positive z–direction, the electron will travel (d) A charged particle moves in a uniform magof the particle will change during this time. 8. Which of the following statements are correct? (a) A current carrying coil, free to rotate, when itself such that its plane becomes perpendicu(b) The trajectory of a charged particle moving -

between two long parallel wires carrying equal currents in the same direction is zero.

Magnetic Effect of Current and Magnetism 24.33

(d) An electron is moving in the anticlockwise sense in a horizontal circular orbit. The

(a) M =

of the orbit will be vertically upward.

(c) B =

9.

us r carries a current I. It is placed in a uniform B. The tension in the loop will be doubled if (a) I is doubled (b) B is doubled (c) r is doubled (d) Both B and I are doubled. 10. A particle having a mass of 0.5 g carries a charge of 2.5 10 –8 C. The particle is given an initial horizontal velocity of 6 10 4 ms –1. To keep the particle moving in a horizontal direction the direction of the velocity tion of the velocity of 3.27 T 11. A wire is bent into a circular loop of radius R and carries a current I of the loop is B. The same wire is bent into a double loop. If both loops carry the same current in the the double loop is B1. If they carry the same current I centre is B2. Then (b) B1 = 4B (a) B1 = 0 B (c) B2 = 0 (d) B2 = 4 12. A straight horizontal conducting rod of mass m and lenght l is suspended by two vertical wires (of negligible mass) at its ends. A current I is set up in the B normal to the conductor is required to keep the tension in the wires equal to zero. If the direction of the current is reversed B, a tension T is developed in the wires. Then mg 2mg (b) B = (a) B = Il Il (c) T = BIl + mg (d) T = BIl – mg 13. In the hydrogen atom the electron (charge e) moves around the proton with a speed v in a circular orbit of radius r. The magnetic dipole moment of the circulating electron is M the site of the proton (i.e. at the centre of the orbit) is B. Then

e vr 4 0 ev 2

4 r

e vr 2

(b) M = (d) B =

0 ev 2

2 r

14. a circular coil of radius r, having n turns and carrying a current I can be doubled by (a) changing I to 2I, keeping n unchanged (b) changing n to 2n, keeping I unchanged (c) changing I to 2I and n to 2n (d) changing n to n , keeping I unchaged. 2 15. The force experienced by a charged particle moving B with a velocity v is zero if (a) v = 0 (b) v is parallel to B (c) v is perpendicular to B (d) v is antiparallel to B 16. pends upon (a) the number of turns in the solenoid (b) the current in the solenoid (c) the radius of the toroid (d) the permeability of the core of the solenoid. 17. The plates of a parallel plate capacitor are in the y – z plane. The separation between the plates is 3 mm and a potential difference of 600 V is applied across the plates. An electron is projected between the plates with a velocity of 2 106 ms– 1 along the positive y-direction. The electron moves unde105 V m–1. region between the plates is 0.1 T. the positive z-direction. the negative z IIT, 1981 18. An alpha particle and a deuteron, after being accelerated through the same potential difference, enter dicular to their velocities. If r and rd are the radii of the circular paths of the alpha particle and the deuteron respectively and and d their respective frequency of revolution, then (b) r = rd (a) r = 2 rd (c)

=

d

2

(d)

=

d

IIT, 2004

24.34 Comprehensive Physics—JEE Advanced

19.

rent I in the same direction are in the x-y plane in a gravity free space (see Fig. 24.73). The central wire is at x = 0 while the other two wires are at x = d. IIT, 1997

(b) the magnitude of the angular momentum about the centre is doubled. (c) the radius r becomes half (d) the frequency remains unchanged. 22. A long, thin and hollow cylindrical metal pipe of radius R carries a current I along its length. For such a pipe, the pipe and increases as we go towards the wall. 0I . (c) On the surface of the pipe, B = 2 R (d) at a point outside the pipe at distance r from r 2. 23. A particle of charge q moving with a velocity B = B0 ( i

v = v0 i

j k) .

(a) The particle describes a circle in the magnetic

Fig. 24.73

(a) cannot be zero beyond B (i.e. x > d) (b) cannot be zero beyond A (i.e. – x > – d) (c) will be zero between x = 0 and x = d. (d) will be zero between x = 0 and x = – d. 20. Two charged particles 1 and 2 of masses m1 and m2, charges q1 and q2 enter a uniform magnetic v1 and v2 shown in Fig. 24.74.

(b) The magnitude of the force acting on the particle is qv0B0 (c) The magnitude of the force on the particle is 2 qv0B0. (d) The force vector lies in the y-z plane. 24. A thin rod AB of length l carries a current I1. It is PQ carrying a current I2 as shown in Fig. 24.75.

Fig. 24.75

(a) The force experienced by the rod is F = Fig. 24.74

(a) If q1 = q2, m1v1 > m2v2 (b) If q1 = q2, m1v1 < m2v2 (c) If v1 = v2,

m1 m2 > q1 q2

(d) If m1 = m2, v1q1 < v2q2. 21. A charged particle is accelerated through a potential difference V. It then enters a region of uniform r and its frequency of revolution is . If V is doubled. (a) the kinetic energy is doubled

0 I1 I 2

4

loge 1

l 2a

(b) The force experienced by the rod is F = l 0 I1 I 2 loge 1 4 a (c) The rod experiences no torque (d) The rod experiences a force as well as a torque. 25. Choose the correct statements from the following. is [ML0 T–2 A–1] (b) 0 is dimensionless.

Magnetic Effect of Current and Magnetism 24.35

(c) The dimensions of 0 0 are the same as those of speed. E are the same as those (d) The dimensions of B of speed. 26. An annular wire loop ABCD carries a current I1 as shown in Fig. 24.76. O is the common centre of the curved parts AB and CD of the loop. A straight wire passing through O and perpendicular to the plane of the loop carries a current I2 directed towards the reader. Then

(d) The proton follows track D and electron follows track A. IIT, 1984 28. A particle of charge + q and mass m moving under E i and B k follows a trajectory from P to Q as shown in Fig. 24.78. The velocity at P is v i and at Q is – 2v j . Which of the following statements is/are correct ?

Fig. 24.76

(a) the net force on the loop is zero. (b) the net torque on the loop is zero. (c) As seen from O the loop will rotate in clockwise sense about axis OP. (d) As seen from O the loop will rotate in anticlockwise sense about axis OP. IIT, 2006 27. A neutron, a proton, an electron and an alpha par-

inward normal to the plane of the paper. The tracks of the particles are labeled as shown in Fig. 24.77. (a) The electron follows track D and alpha particle follows track B (b) The proton follows track A and alpha particle follows track B

Fig. 24.78 2

(a) E =

3mv 4qa

P 3

is

3mv . 4a P

is zero. Q is zero. IIT, 1991 29. A particle of mass m and charge q moving with velocity v enters Region II normal to the boundary as shown in Fig. 24.79. Region II has a uniform B perpendicular to the plane of the paper. The length of the Region II is l. Choose the correct choice(s)

Fig. 24.79

(a) The particle entres Region III only if its qlB velocity v > m Fig. 24.77

(c) The electron follows track A and neutron follows tracks C

(b) The particle enters Region III only if its qlB velocity v < m

24.36 Comprehensive Physics—JEE Advanced

(c) Path length of the particle in Region II is qlB maximum when velocity v = m (d) Time spent is Region II is the same for any velocity v as long as the particle returns to Region I IIT, 2008 30. H+, He+, and O++ all having the same kinetic energy pass through a region in which there is a uniform masses of H+, He+ and O++ are lu, 4u and 16u respectively. Then (a) H+

(b) O++ (c) He+ and O++ IIT, 1994 31. A proton and an electron moving with the same velocity v B which is perpendicular to their velocity. If rp and re are the radii of their circular trajectories and tp and te the time after which each particle comes out (a) rq < re (c) tp < te

(b) rp > re (d) tp > te IIT, 2011

ANSWERS AND SOLUTIONS 1. to the plane containing the point P and the current element I, which is the z-direction. Hence choice (a) the left and to the right of the current element are oppositely directed. Hence choice (d) is correct. concentric circles with their common centre at the wire. Hence choice (c) is also correct. 2. The correct choices are (a), (b) and (c). The mag3. uniform at the edges of the solenoid and is parallel to the axis. 4. perpendicular to the velocity of the particle; the speed of the particle cannot change; only the direction of motion, i.e. velocity (and hence momentum) will change. The correct choices are (a), (c) and (d). 5. The force on a charge q moving with a velocity v is given by F = q (E + v B) There will be no change in velocity if F = 0. This can happen in three cases. (i) Both E and B are zero which is choice (a). (ii) E = 0 and v and B are parallel so that v B = 0 in which case B 0 which is choice (b). (iii) The electric force qE is equal and opposite to magnetic force q (v B) in which case the net force is zero which is choice (d). Hence the correct choices are (a), (b) and (d). 6. The pairs F and v and F and B are always at right angles to each other, because F is always perpendicular to the plane containing B and v. Vectors B and v may have any angle between them.

7. Statement (a) is correct. Since the direction of the velocity of the particle remains unchanged, no magnetic force acts on the particle. The force experienced by a particle of charge q moving with a velocity v B is F = q (v B) Since F = 0, (v B) = 0, i.e. v and B are parallel to each other. Thus, the initial velocity of the particle is either Statement (b) is also correct. Since the magnetic force is always perpendicular to particle velocity, it cannot change the magnitude of the velocity (i.e. speed); it can only change the direction of is equal to its initial speed. The direction of the initial velocity. y-direction, the electron y-direction. It will y-direction. Since the magnetic force is perpendicular to the magnetic

the negative z-direction (use Fleming’s left-hand rule). Statement (d) is also incorrect. No work is done Hence its kinetic energy remains constant. 8. Statement (a) is correct. The loop will rotate until equilibrium state is attained. It will then come to rest because the torque acting on it becomes zero. We know that the torque is given by = BIA sin

Magnetic Effect of Current and Magnetism 24.37

where

is the angle between the direction of magB and the normal to the plane of the loop. It is clear that = 0 when = 0. Thus, in the equilibrium state, the orientation of the loop is such

of the loop. Statement (b) is incorrect. If v is parallel to B, the particle does not experience any force. Hence its trajectory will be a straight line. Statement (c) is correct. Since currents are equal and in the same point midway between the wires will be equal and opposite and hence they will cancel each other. Statement (d) is incorrect. The direction of the hand rule, will be vertically downward. Note that the charge of an electron is negative. 9. The force acting on the loop is given by F = m B sin where m = r2I. Force will be doubled if I or B are doubled. Hence the correct choices are (a) and (b). 10. experience gravitational force mg. As a result the particle will not continue moving in the horizontal direction but will describe a parabolic path. So must be perpendicular to the direction of the velocity. The magnetic force experienced by the particle is given by F = q (v B) The magnitude of the force is F = q v B sin . If the particles is to move in the horizontal direction, this force must balance the force of gravity, i.e. mg = q v b sin The minimum value of B corresponds to sin = 1 or = 90°. Thus mg = v B mg 0.5 10 3 9.8 = 3.27 T or B = qv 2.5 10 8 6 10 4 Hence the correct choices are (a) and (c). 11. The radius of the double loop r = R/2. Now I B= 0 2R r at the centre of the loop is I I ( r = 1/2) B1 = 0 = 0 = 2 B 2r 2R Similarly for the second loop of the double loop, B2 = 2 B Since the currents in the two loops are in the same B1 + B2 = 4B.

Since the currents in the two loops are in opposite B1 and B2 are equal and opposite. the double loop = B1 – B2 = 0. Hence the correct choices are (b) and (c). 12. In order that the tension in the supporting wires is zero the downward gravitational force mg on the rod must be balanced by an upward force BIl due BIl = mg mg or B= Il If the current is reversed, the direction of the force due to B becomes downwards, in the direction of the gravitational force. Hence the tension in the string is T = BIl + mg The correct choices are (a) and (c). 2 r e ev . Current I = = . 13. Time period T = v T 2 r Magnetic moment M = current area of orbit ev e vr =I r2 = r2 2 r 2

B=

0I

=

0 ev 2

2r 4 r So the correct choices are (b) and (c). 14. the coil is given by 0 nI 2r Hence the correct choices are (a) and (b). 15. F = q (v B). The magnitude of the force is F = q v B sin Thus F = 0 if v = 0 or = 0° or 180°. Hence the correct choices are (a), (b) and (d). 16. The correct statements are (a), (b) and (d). The

B=

B = μnI where is the permeability of the core of the solenoid. B is independent of the radius of the toroid. 17. plates is E =

V 600 = d 3 10

3

Fig. 24.80

=2

105 V m–1.

24.38 Comprehensive Physics—JEE Advanced

The electron will experience a force qE in the negative x ed in the region between the plates, the magnetic force experienced by it must be along the positive x-direction and its magnitude must be equal to qE. Since magnetic force = q(v B), it follows that the v (as well as E) and directed along the negative z-direction. The magnitude of the magnetic force =qvB sin 90° = qvB. Thus 2 10 5 E = = 0.1 T qvB = qE or B = v 2 10 6 So the correct choices are (a), (b) and (d). 18. If a particle of mass m and charge q is accelerated through a potential difference V, it acquires kinetic energy = qV. Hence 1 mv2 = qV 2 2qV or v= m where v is the velocity acquired by the particle. The radius of the circular path is given by r=

mv m 2qV = = qB qB m

q qd

q m

q . Hence m

For given V and B, r r = rd

2V B

md m

2q p

2m p

qp

4m p

=1

The frequency of revolution is given by qB q = 2 m m Hence

III are along the positive z-axis. Hence the net magB cannot be zero beyond A, (i.e. – x > – d). For Points between O and B: The directions of the along the negative z-axis but the direction of the the positive z B will be zero for some value of x lying between x = 0 and x = d. For Points between O and A: The directions of the along the positive z-axis but that due to current in wire III is along the negative z-axis. Hence the net B will be zero for some value of x lying between x = 0 and x = – d. So all the four choices are correct. 20. The radius r of the circular path of a particle of mass m and charge q moving with a velocity v perpenB is given by 2 mv = qvB mv = qrB (1) r mv r If q1 = q2, then 1 1 = 1 m2 v2 r2 It follows from Fig. 24.66 that r1 > r2. Hence m1v1 > m 2v 2. If v1 = v2, then from Eq. (1), we have m1 qr = 11 m2 q2 r2

(1)

Now, alpha particle has 2 protons and 2 neutrons and a deuteron has 1 proton and 1 neutron. Therefore, m = 4 mp, md = 2 mp, q = 2 qd and qd = qp. Using these in (1), we get r = rd

For Points on the x-axis beyond A: The directions of

=1 d

Thus the correct choices are (b) and (d). 19. By using the right hand rule, the following conclusions can be drawn: For Points on the x-axis beyond B: The directions III are along the negative z-axis. Hence the net magB cannot be zero beyond B (i.e. x > d).

r1 m m > 1, m1q2 > m2q1 or 1 > 2 . If m1 = m2, r2 q1 q2 v1q2 > v2q1. So the correct choices are (a) and (c). 21. Kinetic energy = qV. Hence if V is doubled, K.E. is also doubled. The magnitude of the angular momentum about the centre is L = mvr. Now r = mv/qB. Therefore, m2 v 2 2m 1 2 2m = (K.E.) L= mv = qB qB 2 qB Thus if V is doubled, K.E. is doubled and L is also doubled. The radius of the circular path is given by m2 v 2 2m 1 2 = 2m qV r2 = 2 2 = 2 2 mv q B q B q2 B 2 2 Since

=

2mV qB 2

Magnetic Effect of Current and Magnetism 24.39

Thus r

V . The frequency of revolution is

qB 2 m Hence the correct choices are (a), (b) and (d). 22. Consider a point P at a distance r from the axis of the pipe. For all points inside the pipe, r < R P, we choose an Amperean loop to be a cylindrical loop of radius R as shown in Fig. 24.81. Since the current enclosed in the Amperean surface is zero, it follows from Ampere’s =

24. Consider a small element of length dx of the rod at a distance x from wire PQ as shown in Fig. 24.82. PQ at the element is 0 I2 B= 2 x Therefore, force exerted on the element is

points inside the pipe. For points outside the pipe, r > R, we have from Ampere’s law,

Fig. 24.82 0 I1 I 2 dx

dF = BI1 dx =

2 x Total force experienced by the rod is F=

Fig. 24.81

B .dl =

0I

B

dl =

0I

B

2 r=

0I 0I

B=

0 I1 I 2

= (B is || dl)

2

, i.e. B

0I

2 R Hence the correct choices are (a) and (c). 23. Since v is not perpendicular to B, the particle will not describe a circle. The force acting on the particle is F = q(v B) = q[v0 i

= qv0B0 k j Thus the magnitude of the force is F = qv0B0 = qv0B0(12 + 12)1/2 = 2 qv0B0 and it lies in the y-z plane. Hence the correct choices are (c) and (d).

dx x

a l 2 a B is not uniform; it decreases as we go from and A to end B of the rod. Hence the rod will also experience a torque. Thus the correct choices are (b) and (d). 25. From F = qvB, it is easy to see that the dimensional formula of B is [ML0 T–2 A–1], which is choice (a). Choice (b) is wrong. From = 0/(2 ), the dimensions of 0 are B r [ 0] = I =

B0 i j k ]

= qv0B0 i i i j i k

x (a l )

x a

0 I1 I 2

=

1 2 r r At the surface of the pipe (r = R) B=

dF

ML0 T 2 A A

From F = [ 0] = [

0

r2

0

q1q2 2

L

q1q2

1 4

1

=

log e

= [MLT–2 A–2] , the dimensions of

A2T2 2

2

0

are

= M–1 L–3 T4 A2

MLT L F r –2 –2 –1 –3 4 2 [M L T A ] = [L–2 T2] 0] = [MLT A ]

0 0

= [L–1 T], which is reciprocal to speed.

Hence choice (c) is incorrect.

24.40 Comprehensive Physics—JEE Advanced

E = v. Hence choice B

From qE = qvB (d) is correct. 26.

I2 is tangential to the curved parts AB and CD of the loop. Hence every current element dl of parts AB and CD is parallel or antiparallel to B. The magnetic force on AB or CD is zero since = 0° or 180° in the expression dF = Bldl sin . The magnetic force on straight parts AD and BC is not zero. The magnetic force on AD is directed towards the reader which is equal and opposite to the force on BC which is directed away from the reader. These equal and opposite forces cancel each other. Therefore, the net force on the loop ABCD is zero. Since these equal and opposite forces do not act at the same point, they will exert a net torque on the loop which will rotate it in the clockwise sense when viewed from O. Hence the correct choices are (a) and (c). 27. The radius of the circular track is given by mv r= qB For a neutron q = 0. For an electron q = –e and for a proton q = +e. For an alpha particle q = 2e and m = mass of four protons. Hence the correct choices are (a) and (b). 28. In going from P to Q, the change in the kinetic 1 1 3 energy of the particle = m(2v)2 – mv2 = 2 2 2 mv2 ing the particle from P to Q = F·d where d is the displacement in the direction of E. So work done W = qE i 2a i 2qaE i i = change in K.E. Hence 2qaE =

2qaE . Now work done

3 2 mv 2

3mv 2 4qa

E

Thus choice (a) is correct. P is F vp

qE i v i

qE v

q

3mv 2 4qa

r=

mv qB

Therefore, the particle can enter region III if r > l, qBl . i.e. if v > m In region II, the maximum path length is r = l, qBl which gives v = m The time period of the circular motion is 2 r 2 mv 2 m T= v v qB qB The particle will return to region I if the time spent T m by it in region II is , which is independent 2 qB of the velocity. Hence the correct choices are (a), (c) and (d). 1 1 1 30. mHv2H = mHev2He = mOv2O. Given mH = 1u, 2 2 2 1 mHe = 4u and mO = 16u. Hence vHe = vH and 2 1 v O = v H. 4 The radius of the circular path is given by mv ; q = charge R= qB Now charge of H+ = e, charge of He+ = e and charge of O++ = 2e. It is easy to check that rHe = rO and 1 rH = rHe 2 Smaller the value of r Thus the correct choices are (a) and (c). 31. the electron and proton in the region of magnetic Since

r =

mv , rp > re qB

v

3mv3 4a So choice (b) is also correct. Q is zero 0 . Hence the correct

because i j 0 and i k choices are (a), (b) and (d). 29. In region II, the particle follows a circular path of radius

Fig. 24.83

Since v B, the charged particle describes a semicircular trajectory. Hence the time after which period, i.e.

Magnetic Effect of Current and Magnetism 24.41

t=

T 2

Since mp > me and q and B are the same, tp > te. Hence the correct choices are (b) and (d).

m qB

III Multiple Choice Questions based on Passage Questions 1 to 3 are based on the following passage Passage I Two long parallel wires carrying currents 2.5 amperes and I ampere in the same direction (directed into the plane of the paper) are held at P and Q respectively such that they are perpendicular to the plane of the paper. The points P and Q are located at a distance of 5 m and 2 m, respectively, from a collinear point R (see Fig. 24.84).

Fig. 24.84

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. IIT, 1990 1. R is (b) 5.0 10–7 T (a) 2.5 10–7 T (d) 2.5 10–6 T (c) 5.0 10–6 T 2. R due to current I = 2.5 A in wire P is (b) 2 10–7 T (a) 1 10–7 T –7 (d) 4 10–7 T (b) 3 10 T 3. The current I in wire Q is (a) 1 A (b) 2 A (c) 3 A (d) 4 A

SOLUTION 1. The magnitude of the force experienced by a particle of charge q moving with a velocity v in a F = qvB sin where is the angle between v and B. Given F = 3.2 10–20 N, v = 4 105 ms–1 and = 90°. For electron q = 1.6 10–19 C. Using these value we get B = 5 10–7 T, which is choice (b) 2. R due to currect I in wire P is 7 I 2.5 = 1 10–7 T B1 = 0 = 4 10 2 r1 2 5 The correct choice is (a).

3.

R due to currect I in wire Q is B2 =

0I

2 r2

=

4

10 2

7

2

I

=I

10–7 T

B1 and B2 will be in the downwar direction, parallel and colliner. Hence the resultant R is B = B1 + B2 = (1 + I) 10–7 T Now B = 5 10–7 T. Therefore (1 + I) 10–7 = 5 10–7 or 1 + I = 5 or I = 4 A. So the correct choice is (d).

Questions 4 to 7 are based on the following passage

4. The force experienced by the charged particle in the

Passage II

(a) along the positive y-direction (b) along the negative y-direction (c) in the x-y plane (d) in the y-z plane. 5. If the particle emerges from the region of magnetic

The region between x = 0 and x = L B0 k . A particle of mass m, positive charge and velocity v0 i travels along x-axis and enters IIT, 1999

initial velocity, the value of L is

24.42 Comprehensive Physics—JEE Advanced

(a)

2mv0 qB0

(b)

(c)

mv0 2qB0

(d)

mv0 qB0

(c) v0 j (d) – v0 j 7. In Q. 6, the time spent by the particle in the mag-

3mv0 2qB0

6.

(a) t =

2 m qB0

(b) t =

(c) t =

3 m 2qB0

(d) t =

x = 2.1 L,

(a) v0 i

(b) – v0 i

2 m qB0 m qB0

SOLUTION 4. The force experienced by the charged particle is given by F = q(v

B) = (v0 i )

= qv0B0( i

mv0 (2) qB0 Since the particle emerges from the region of the r=

(B0 k )

k)

= qv0B0(– j )

angle of 30° with the initial vector, it follows from triangle ABC that AB = AC sin 30° or L = r sin 30° m v0 m v0 sin 30 = (3) = 2qB0 qB0

(1)

The force is along the negative y-direction, which is choice (b). 5. Refer to Fig. 24.85.

Thus the correct choice is (c). 6. will continue to move in a circular path till it completes half the circular path and emerges out of the v0 i moving along the negative x-axis as shown in Fig. 24.39. 7. Distance travelled by the particle in the magnetic . Therefore, time t=

Fig. 24.85

The particle describes a circle of radius Questions 8 to 11 are based on the following passage

9.

(c)

2 3 r

(d)

4 r

m qB0

[Use Eq. (2)]

B2 at O due to the circular arc ACB is (a) 5 0 I 12r

A wire loop consists of a straight segment AB and a circular arc ACB of radius r. The segment AB subtends an angle of 60° at the centre O of the circular arc. The wire loop carries a current I in the clockwise direction (Fig. 24.86). 8. B1 at O due to the straight segment AB is 0I 0I (b) (a) 2 r 2 2 r 0I

v0

=

So the correct choice is (d).

Passage III

0I

r

0I

(b)

2r (c) 3

0I

8r 7 0I (d) 18r 10.

Fig. 24.86

B at O due to the whole wire loop is

Magnetic Effect of Current and Magnetism 24.43

(a) B = B1 + B2 B12

(b) B = B2 – B1

(d) B = B22 11. B is (a) parallel to the plane of the coil (c) B =

B22

(b) perpendicular to the plane of the coil and directed out of the page (c) perpendicular to the plane of the coil and directed into the page. (d) inclined at an angle of 60° with the plane of the coil.

B12

SOLUTION 8.

O due to current I in AB is given by (use Biot-Savart law)

9. (n = 1 turn) circular loop of radius r and carrying a current I is 0 nI

B=

Here loop ACB is a fraction of a circle i.e. n < 1. Since ACB subtends an angle (360° – 60°) = 300° at O, hence the fraction n is

Fig. 24.87

BAB =

0I

4 x

(sin

+ sin

=

0I

4 0I

2 3 r

3r/ 2

(sin 30

(0.5 + 0.5) =

sin 30 ) 0I

2 3 r

n=

)

3r Here = = 30°. Also x = r cos = r cos 30° = . 2 Therefore, B1 =

2r

,

which is choice (c). plane of the paper directed into the page. Questions 12 to 15 are based on the following passage Passage IV A moving coil galvanometer consists of a coil of N turns and area suspended by a thin phosphor bronze strip in B. The moment of inertia of the coil about the axis of rotation is I and C is the torsional constant of the phosphor bronze strip. When a current is (in radian). IIT, 2004 12. Choose the correct statement from the following. The magnitude of the torque experienced by the coil is independent of (a) N (b) B (c) i (d) I 13. The current sensitivity of the galvanometer is increased if (a) N, A and B are increased and C is decreased.

300 360

=

5 6 ACB is

5 I 5 0I 6 = B2 = 12r 2r As the current in ACB is clockwise, the direction of 0

the paper and directed into the page. The correct choice is (a). 10. Since B1 and B2 are in the same direction, the net B = (B1 + B2), which is choice (a). 11. The correct choice is (c). (b) N and A are increased and B and C are decreased (c) N, B and C are increased and A is decreased (d) N, A, B and C are all increased. 14. When a charge is passed almost instantly through the coil, the angular speed acquired by the coil is NAB BAQ (b) (a) QI NI NABQ NAQI (c) (d) I B 15. radian) of the coil is I 1 (b) max = (a) max = I C C (c)

max

= I

C

(d)

max

=

IC

24.44 Comprehensive Physics—JEE Advanced

SOLUTION 12. The magnitude of torque experienced by the coil is given by = iNAB sin where is the angle which the normal to the plane of the coil makes with the direction of the magnetic coil is always parallel to the direction of the magnet= 90°. Hence = iNAB = Ki = where K = NAB So the correct choice is (d). 13. Let current i is passed through the coil. Then, restoring torque = C . When the coil is in equilibrium, iNAB = NAB i C Hence the correct choice is (a). Current sensitivity is

=

Questions 16 to 18 are based on the following passage Passage V 10–19 C E along the direction shown in Fig. 24.88. The speed of the particle is 107 ms–1 the inward normal to the plane of the paper. The particle F. IIT, 1984 16. The value of angle is (a) 30° (b) 45° (c) 60° (d) 75° 17. The radius of the circular path of the particle in the A particle of mass 1.6

(a) 0.1 m (c) 0.3 m

14. If is the angular speed acquired by the coil when a charge is passed through it for very short time t, then angular momentum = = I timeinterval t Q or = t = = KQ i t NABQ or I = NABQ or = , which is choice I (c). 15. From the principle of conservation of energy, we have 1 2

=

1 2

which gives

2 max

max

=

I , which is choice (a) C

18. The distance EF is (a)

2m

(b)

(c)

2 m 10

(d)

10–27 kg and charge 1.6

(b) 0.2 m (d) 0.4 m

2

Fig. 24.88

SOLUTION 16. Let O be the centre of the circular path. It is obF such that BF is tangent to the circle. OE and OF are normals, which meet at O (see Fig. 24.89). Therefore, angle OFE = angle OEF. Hence = 45°. mv . Substituting the given values and solving 17. r = qB we get r = 0.1 m 2 m 18. EF = 2 r cos 45° = 2 r = 2 0.1 10

Fig. 24.89

2 m 5 1 m 2

Magnetic Effect of Current and Magnetism 24.45

IV Matrix Matching Type 1. In Column I are listed some charged bodies and current carrying conductors. Match them with the effects they produce listed in column II Column I Column II

with a constant angular velocity (c) A coil carrying a current I = I0 sin

t

(r) Magnetic moment IIT, 2006

ANSWER (a) (c)

(p) (p), (r), (s)

(b) (d)

(p), (q), (r) (q)

2. Two wires each carrying a steady current I Column I. Some of the resulting effects are described in Column II. Match the statemens in Column I with those in Column II. Column I Column II (a) Point P B) at P due to the currents in the wires are in the same direction. (b) Point P joining the centers of the circular wires which have same radii.

B) at P due to the current in the wires are in opposite directions.

(c) Point P joining the centers of the circular wires, which have same radii

(d) Point P is situated at the common centre of the wires.

P.

(s) The wires repel each other.

IIT, 2007

24.46 Comprehensive Physics—JEE Advanced

SOLUTION (a) At point P the currents are equal but in opposite directions. Hence the correct choices are (q) and (r). (b) The currents in the two coils are in the same sense (anticlockwise). Hence, at point, P the current are the same direction, which is choice (p). P due to the currents are in the opposite directions. Since point P is the same distance from the centres of the coils, and their radii are point P is zero. Hence the correct choices are (q) and (r). P of the coils are not equal (because their radii are different). Further, direction. If the currents are in opposite directions in coils (or wires), they repel each other. Hence the correct choices are (q) and (s). (a) (q), (r) (b) (p) (c) (q), (r) (d) (q), (s) 3. 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 V be the electric potential at M Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B the magnetic M and μ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current. IIT, 2007 Column I Column II

(a) E = 0 (p)

(p) Charges are at the corners of a regular hexagon. M is at the centre of the hexagon. PQ is perpendicular to the plane of the hexagon.

(b) V

0 (q)

(q) Charges are on a line perpendicular to PQ at equal intervals. M is the mid-point between the two innermost charges.

(c) B = 0 (r)

(r) Charges are placed on two coplanar insulating rings at equal intervals. M is the common centre of the rings. PQ is perpendicular to the plane of the rings.

(d) μ

(s) Charges are placed at the corners of a rectangle of sides a and 2a and at the mid point of the longer sides. M is at the centre of the rectangle. PQ is parralled to the longer sides.

(t)

0 (s)

(t) Charges are placed on two coplanar, identical insulating rings at equal intervals. M is the mid-point between the centres of the rings. PQ is perpendicular to the line joining the centres coplanar to

Magnetic Effect of Current and Magnetism 24.47

SOLUTION M due to charge at A and B =

(p)

2q 4

0r

2

where r = AM = BM directed from M to B. The electric

M due to other charges are net V = 0. Since the total charge of the system is zero, on rotation net current = 0. Hence B= 0 and μ = 0

(q) The net charge on the right of M is 2q – q = q and on the left of M is –q M and directed to the left of M. But net V = 0. Net charge = 0. On rotation, current = 0. Hence B = 0 and μ = 0.

not zero. Hence B

M 0 and μ

V

0. On rotating, net current is

0.

E = 0. But net V 0. On rotating, net current 0. Hence B 0 and μ 0. E 0 and is directed to the right but net V = 0. On rotation, net current = 0. Hence B = 0 and μ = 0.

ANSWERS

(a) (c)

(p), (r), (s) (p), (q), (t)

(b) (d)

(r), (s) (r), (s)

24.48 Comprehensive Physics—JEE Advanced

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by statement-2 (Reason). Each question has the following four options out of which only one choice is correct. (a) Statement-1 is True, Statement-2 is True and State-ment-2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; but 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. 1. Statement 1 nitude from point to point but has a constant direction, is set up in a region of space. If a charged particle enters the region in the direction of the uniform rate in the region. Statement 2 The force F experienced by a particle of charge B is moving with a velocity v given by F = ( v B ). 2. Statement 1 A charged particle moves in a uniform magnetic energy of the particle cannot change but its momentum can change. Statement 2 The magnetic force is always perpendicular to the velocity of the particle. IIT, 1993 3. Statement 2 A current carrying loop is free to rotate. It is placed Statement 2 The torque on the coil is zero when its plane is 4. Statement 1 An electron moving in the positive -direction enters a region where uniform electric and magnetic

Statement 2 If a charged particle moves in a direction perpenacting on it is given by Fleming’s left-hand rule. 5. Statement 1 If a charged particle is released from rest in a region each other, it will move in a straight line. Statement 2 6. Statement 1 A proton and an alpha particle having the same kinetic energy are moving in circular paths in a paths will be equal. Statement 2 Any two charged particles having equal kinetic energies and entering a region of uniform magnetB in a direction perpendicular to B , will ic describe circular trajectories of equal radii. 7. Statement 1 Two particles having equal charges and masses m1 and m2, after being accelerated by the same potential difference (V), enter a region of uniform r1 and r2 respectively. Then m1 = m2

r1 r2

Statement 2 Gain in kinetic energy = work done to accelerate the charged particle through potential difference . 8. Statement-1 No net force acts on a rectangular coil carrying a steady current when a suspended freely in a uniStatement-2 The magnitude of force experienced by a straight conductor of length L carrying a current I and placed B is BIL. IIT, 1981

Magnetic Effect of Current and Magnetism 24.49

9. Statement-I There is no change in the energy of a charged partiforce is acting on it. Statement-2 The magnetic force acting on a moving charged particle is always perpendicular to its velocity. IIT, 1983 10. Statement-I A charged particle enters a region of uniform magpath of the particle is a circle. Statement-2 allel and equidistant. IIT, 1983 11. Statement-I An electron and a proton are moving with the same kinetic energy along the same direction. When to their direction of motion, they describe circular path of the same radius. Statement-2 The radius of the circular pathof a particle of charge q, mass m and moving with velocity v perpendicuB is given by

r=

mv qB IIT, 1985

12. Statement-1 tude from point to point but has a constant direction, is set up in a region of space. If a charged particle enters the region in the direction of the magnetic the region Statement-2 The force F experienced by a particle of charged B q moving with a velocity v is given by F = q( v B ) IIT, 1989 13. Statement-I The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. Statement-2 Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized. IIT, 2008

SOLUTIONS 1. The correct choice is (d). If v is parallel to B , F = 0. Hence the particle does not experience any force and is, therefore, not accelerated in the region. It

plane of the coil is perpendicular to the magnetic 4. The correct choice is (a). Because electron has a

2. The correct choice is (a). Since the magnetic force is always perpendicular to the velocity, no work particle. Hence magnetic force cannot change the magnitude of velocity (i.e. speed); it can only change the direction of velocity. Hence kinetic en1 mv 2 remains unchanged but momentum ergy 2 p = v will change. 3. The correct choice is (a). The loop will rotate and come to rest when the torque acting on it becomes zero. The magnitude to torque acting on a loop of area A and carrying a current I B is given by = BIA sin where

the magnetic force is perpendicular to the magnetic ing’s left-hand rule) should be along the negative -direction. 5. force is F = E in the direction of E . Since E is parallel to B , the particle velocity v (acquired due to force F ) is parallel to B . Hence B will not exert any force since v B = 0 and the motion of the particle is not affected by B . 6. The correct choice is (c). The radius of the circular path is given by

is the angle between the direction of the =

coil. It is clear that

= 0 when

= 0, i.e. when the

mv = qB

2mK 1 ; where = v2 qB 2

24.50 Comprehensive Physics—JEE Advanced

These forces do not exert any net force on the coil. The force on arms AD and BC are F2 = B I b sin . These forces merely compress the coil and are resisted by its rigidity. These forces also exert no net force on the coil. Thus, both the statements are true but Statement-2 is not the correct explanation for statement-1. 9. F = q(v B). Since F is perpendicular to v, power p = F v = 0. Hence both the statements are true and statement-2 is the correct explanation for Statement-1. 10. The path of the Particle is a helix; it is a circle if

m . q Now, the charge of an alpha particle is twice that of a proton and its mass is four times the mass of a proton, m / q will be the same for both particles. Hence will be the same for both particles. 7. The correct choice is (d). Kinetic energy K = qV. Since and are the same for the two particles, r

Therefore r1 = r1 Hence = r2

2m1qV and r2 = qB m1 m2

m1 = m2

2m2 qV qB r1 r2

2

.

8. A coil ABCD B. Its planes makes an angle with B and it carries a current I as shown in the Fig. 24.90. The forces F1 = B Il on AB and CD are equal and opposite but constitute a couple which tends to rotate the coil.

is true. mv 11. r = qB

2m K , where K qB

1 mv 2 2

Since K and q are the same, r m . Hence the electron will describe a circle of a circle of a smaller radius. So. Statement-1 is false but Statement-2 is true. 12. If v is parallel to B, F = 0. Hence the particle does not experience any force and is, therefore, not with a constant speed. Statement-1 is false and Statement-2 is true. 13. Placing a soft iron core inside the coil makes the

in the coil. The sensitivity increases because, for Fig. 24.90

when a core is inserted the coil. Statement-1 is true and Statement-2 is false.

VI Integer Answer Type 1. A potential difference of 600 V is applied across the plates of a parallel plate capacitor. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates with a velocity of 2 105 ms-1 in the region between the capacitor plates. Neglect edge effects. IIT, 1981

2. Two long straight parallel wires are 2 m apart, perpendicular to the plane of the paper (Fig. 24.91) Wire A carries a current of 9.6 A directed into the plane of the paper. Wire B carries a current P at a distance of 10/11 m from the wire is zero. Find the magnitude of the current (in ampere) in wire B. 3. A steady current current I goes through a wire loop PQR having shape of a right angle triangle with

Magnetic Effect of Current and Magnetism 24.51

PQ = 3x. PR = 4x and QR = 5x. If the magnitude of 0I , P due to this loop is k 48 x k. IIT, 1990

4. 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 ohm and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as I = I 0cos(300 t) where I0 is constant. If the magnetic moment of the loop is Nμ0I0sin (300 t), then ‘N’ is IIT, 2011

Fig. 24.92

Fig. 24.91

SOLUTION V 1. E = d

600 3 10

3

=2

5

10 Vm

–1

directed to the right in the plane of the paper. Its magnitude is given by (see Fig. 24.94 on page 24.52) B1 =

24.93] e v B = eE

E B= v

2 105 2 105

= 1T

or Given

Fig. 24.93

2 R1

P B2 due to I2 is equal and opposite to B1. Therefore, the current in wire B should be normal to it and directed out the plane of the paper. Therefore B1 = B2 or

2. Let the currents in A and B be I1 and l2 respectively. Let AP= R1, BP = R2. According to Ampere’s P due to I1 is

0 I1

Thus

0 I1

2 R1

=

I2 =

0 I2

2 R2 I1 R2 2 R1

10 m and 11 10 32 R1 = AB + BP = 2 + m. 11 11 I1 = 9.6 A, R2 = BP =

I2 =

9.6 10/11 32/11

9.6 10 = 3.0 A 32

24.52 Comprehensive Physics—JEE Advanced

25 r 2

1 =

sin

144 x

r=

2

3 and cos 5

=

0I

3 4 r 5 k =7

B(at P) =

=

12 x 5

4 5

7 0I 20 r

4 5

7 0I 48 x

4. inside the tube is B= =

P is (Fig. 24.95) B (at P) = =

0I

4 r 0I

4 r

[sin

L

; L = length of tube

0 I0

cos (300 t ) L

If r through it is

Fig. 24.94

3.

0I

0 I0

= BA =

+ sin 90° – ]

cos (300 t ) r 2 L

Magnitude of induced emf is (sin

+ cos )

|e| =

r2

0 I 0 (300)

d dt

sin (300 t )

L

The induced current through the loop is i=

0 I 0 (300)

e R

r2

sin (300 t )

LR

Magnetic moment of the loop is M=i

Fig. 24.95

Now and

sin cos

r 4x r = 3x

=

These equations give sin2

+ cos2

=

r2

r2

16 x 2

9 x2

r2 =

300

2 4

r RL

0 I0

sin (300t)

Given M = Nμ0I0 sin (300t)

(i)

Comparing (i) and (ii) we get

(ii)

N =

300 2 r 4 RL

300 = 5.92

(3.14) 2 (0.1)4 0.005 10 6

Electromagnetic Induction and A.C. Circuits

25

Electromagnetic Induction and A.C. Circuits

Chapter

REVIEW OF BASIC CONCEPTS 25.1

e

B is

A =B

A = B A cos

where

d dt

Lenz’s Law

MAGNETIC FLUX

Thus d dt

e= – k where k

N turns, the

k = 1 and one

can write e= –

= N B A cos

d dt e

=

25.2

25.1

B dA

N turns

FARADAY’S LAWS OF ELECTROMAGNETIC INDUCTION

The

d dt

e

N

d dt

i=

1 d induced em = total resistance circuit R dt

The direct law.

Faraday’s Law of Electromagnetic Induction

Flow of Induced Charge

-

circuit. q = idt =

q 1 d 1 dt = d R dt R

=

change in lu Resistance

25.2 Comprehensive Physics—JEE Advanced

Heat Dissipation d idt = i d dt = induced current

H = eidt =

Fig. 25.3

Fleming’s Right Hand Rule

I1 and I2

25.3

I1 and I2 will decrease. I1 and I2

EXPRESSION FOR INDUCED EMF

(B). B increases with time, the induced current i is

Applications of Lenz’s Law

B

towards the magnet, the induced current i is

Fig. 25.4

Fig. 25.1

d d BA dt dt B decreases with time, I B dR increases at a rate , then dt

i

R2

e

e Fig. 25.2

B

d dt

R2

B

dB dt

2 R

PQRS

I

or I increases with time. or I decreases with time. I1 and I2

v

Fig. 25.5

dR . dt

v, the

B

-

The current I

-

Electromagnetic Induction and A.C. Circuits

e Blv where l = PS = QR Induced current i R is the resistance i= Force F

e = Blv v

Bl v R

e R

v is F = Bil =

25.3

v

v

v

B 2l 2 v R

Power needed is P = F v =

B 2l 2 v2 R

Fig. 25.7

e = Blv sin R is B as

v NOTE -

ends P and Q 2BvR

e = Bv

R

v

XY sliding on metallic rails PQRS to the right as shown in Fig. Fig. 25.8

PQ

l

B

v

e=

1 B l2 2

Fig. 25.6

Change in

) N turns is rotated , the

A B e = e0 sin where e0 = NBA

25.4

= e0 sin

t

Fig. 25.9

MOTIONAL EMF v

l B as shown in Fig. P and Q Fig. 25.10

P is

25.4 Comprehensive Physics—JEE Advanced

R centre O

25.5

A solen

P

e=

25.2

B as

1 B R2 2

-

ELECTRIC MOTOR SOLUTION I in a solenoid

V

n e

i=

V

B=

R is

V

0nI

e

R Vi and heat loss = i2R = Vi – i2R = ei. ei e = Vi V Some important points about a d.c. motor e and hence current i V V and e= 2 and initial current = V/R

ing the coil is = NAB cos

= NAB cos ° = NAB = N r2 0nI r

= 0.1 m. e=0

I1 = 0 to

I2 = 2 A is = N r2

0n

I2 – I1 0.12

= 100

i

is minimum.

= 0°. N the -

A

2

10 e=

25.1

10

t

, 100 turns and side

=

2

10

0 3

2

EXAMPLE 25.3 -

1 2

t

SOLUTION R=2

T

, N = 100, A = 0.1

0.1 = 10–2 m2,

B = 0.1 T and t

NBA 2 – cos 1 = 100 0.1 10 –2 1 = 0.1 0 2 change in lu 0.0 In e= = 0.0 time e 1 Induced Current i = R 2

t t2 Find the induced current in the coil at t = 10 s. SOLUTION e| =

d d = dt dt

t2

t

10– 3

At t = 10 s, e Induced current at t = 10 s is I=

e = R

t

10

3

= 10–2A

Electromagnetic Induction and A.C. Circuits

25.5

25.4

EX

10 T.

Fig. 25.12

SOLUTION 100

=

3

l = 0. =2 1 2 Bl 2 1 = 2

e=

3

-

2

10

10

3

10

its diameter. Assume that the normal to the = 0 with t = 0. SOLUTION coil changes with time.

Fig. 25.11

the coil does not change with time, hence no

10

coil does not change with time, hence the magnetic

25.5 –1

in the eastward direction at a 10 T. Find the

Fig. 25.13

a diameter, as shown in

SOLUTION –1

v e = Blv

= 300 ms–1 10 A =

25.6 r

N turns is B

r2

Fig. 25.14

then at time t, = t this time is = NBA cos = NBA cos

t

25.6 Comprehensive Physics—JEE Advanced

Q acquires a negad d =– dt dt

e= – =

NBA sin

=

NB

NBA cos

sin

dF = qvB = qv T

t t

25.7 A metal rod PQ

dE = v

CD at distances a and b

Force on the element is

t

r2 sin

= Nr2B

t

P and Q

Now

2 x

qv 0 I Iv dF = = 0 q 2 x 2 xq

dE = –

I as

0I

dV dx

dV = – dE ere dV

dx = –

0Iv dx 2 x

PQ is 0 Iv

V

2

EXAMPLE 25.8 A metal rod PQ

b

dx = x a

l AB and CD

straight wire XY

0 Iv

2

ln

b a

v I as shown in Fig.

R PQ

Fig. 25.15

SOLUTION dx. Consider one such element CD

at distance x

Fig. 25.17

resistance R. on rod PQ

v.

SOLUTION -

Fig. 25.16

I in

wire CD is B=

ACPQ with time.

0I

2 x

The directi v

PQ

P to Q P to Q. Hence end P

Fig. 25.18

Electromagnetic Induction and A.C. Circuits

dx at a distance x

where r is th

SOLUTION

PQ is 0 Ir

d = BdA = Brdx =

2 x

a

= Induecd

e= =

a l

0 Ir

2

a

dx Ir a l = 0 ln x 2 a

0Iv

ln

a l a

v=

dr dt

e=

d dt

v=

i R 2kx Bl

dx = Blv dt R ACPQ = R + 2kx induced in the circuit is e Bl v = i= R 2kx R 2kx

I a l dr d = 0 ln dt 2 a dt

2 ent induced in R is i=

t. 0 Ir dx 2 x

= d =

ACPQ when the rod is at a R is = BA = Blx

distance x

dx

PQ ACPQ is a l

25.7

= Bl

25.10 A metal rod PQ

AB and CD

m l

Iv a l e = 0 ln R 2 R a

R -

width dx

ACPQ dF = Bidx =

0I

2 x PQ is

F F= = = 25.9 A metal rod PQ AB and CD B

dF =

idx

0 Ii

2

a

0I v

0I

2

a l

2 R

ln

v 0I a l ln R 2 a

u

distance x dx x a l

a l a

a

dx x

2

Fig. 25.20

SOLUTION

l k R is connected

F PQ so that it is accelerated

Induced currnet i =

Force F = Bil = – ma = – m

dv dt

The ne m

v PQ when it is at a distance x R.

Bvl R

dv =–B dt =–

Fig. 25.19

B 2 vl 2 R

Bvl R

l

25.8 Comprehensive Physics—JEE Advanced

dv B 2l 2 =– dt = –kdt v mR k=

k dt 0

v u

ln

dqR = d d dq = R

t

dv v u

d dt

dq d R= dt dt

B 2l 2 mR

v

Intergrating

iR =

– kt

Q=

v = e–kt u

R

=

2 NBA 2 NB = R R

25.12 Two circular coils A and B b>a

v = ue–kt dx = ue –kt dt The rod comes to rest when t = Integrating

r2

a and b

-

3b

x= sient current I

B

A is R

A

x

dx = u e 0

kt

dt

0

x=– x=

u e k

kt

0

u k

=–

u k

umR B 2l 2

25.11 r has N turns and a resis-

tance R

Fig. 25.21

B Q

SOLUTION A due to current

I in coil B is SOLUTION A

r

BAB =

2

NBA cos 0°

= NBA

2b

2

0 Ib

2

x

2 3/2

=

Sin

= NBA

NBA

NBA

= BAB Induc

d e| = dt e Induced current is i = R

d dt

1 R

b

x = 3b

A, hence A is

NBA

– NBA

0I

area o e

A d dt

d dt IRdt = d IR =

cos 0° =

A, = 0°. a2

0I

b

Electromagnetic Induction and A.C. Circuits

or

where V

1 d = R R

Idt = Q=

R

a2

0I

=

d d =– BA dt dt d =– kt x2 dt

V= –

bR

25.13 R and mass M

Q

E= B = kt

k is a

kx 2

Force acting on the element is dF = dq

t

E= =

Q R2 kQ 2

R Torque acting on the disc is

t.

R

SOLUTION

=

R2

xdF = 0

Q dx at a distance xdx

Q R2

2 xdx

kQR 2

kx 2

x2dx

x3 dx =

kQR 2

0

is the angu-

lar acceleration I=

dq =

R

kQ R2

2 xdx

= I , where I

R2 x

kx2

2 x = – kx2

–E

constant and t

25.9

1 MR2 and 2

=

1 MR2 2

d =

kQ dt 2M

=

kQ dt 2M 0

=

kQt 2M

=

d . Hence dt

d dt

t

d 0

25.6

i = Mi where M i is changing with time, the

Fig. 25.22

E. Since E= –

dV dl

dV = – V= – E

MUTUAL INDUCTANCE

di dt Expressions of M in some situations l N2 turns. e= – M

Edl = – E 2 x

2 x

N1 wound

25.10 Comprehensive Physics—JEE Advanced

M=

0

R >> r

N1 N 2 A ; A = common cross-sectional area l R and r r2 2R

0

M=

25.7

SELF INDUCTANCE = Li,

i where L di e= L . dt

N turns, cross-

sectional area A and length l 0N

L=

2

A

l

Fig. 25.23

r a and b with a, b >> r Fig. 25.26

M=

2

0

r

2

a ab

2

b

2

U=

1 2 Li . 2

L = L 1 + L2 M

Fig. 25.27

Fig. 25.24

a and b

tance x

L = L1 + L2 + 2M

-

L = L1 + L2 – 2M M=

0

2

a

x

log e

b x

1 L

1 L

1 L2

M

M = k L1 L2 ; k

25.8

Fig. 25.25

GROWTH AND DECAY OF CURRENT IN A D.C L—R CIRCUIT (FIG. 25.28) S1 is closed at t = 0, with switch S2

NOTE =

0

r,

-

0

where

r

t, i = io

e–t/

L = is the time constant. R

Electromagnetic Induction and A.C. Circuits

25.11

SOLUTION = LI

L=

I

=

0 1 10

3

–3

25.15 Fig. 25.28

io = E/R At

t =

t = , i = io 1

1 e

SOLUTION

i 0.

dI = = 20 As–1 dt 0.1

1

e

L

–3

=L

dI dt H

L

25.16 Fig. 25.29

Decay of current: At time t = 0, let i0 = E/R S1 S2 i = i0 e–t/ At

t= , i =

i0 e

SOLUTION

i 0.

Mutual inductance M = 0

=

Fig. 25.30

25.9

M

dI = dt

2

H –

0.01

–2

25.17

ENERGY STORED IN AN INDUCTOR

are connected in series to a

L is increased I U=

l

0. 1.0 –

e

0 AN1 N 2

through a switch. The switch is closed at time t = 0

1 LI2 2

current.

25.14 -

tor at t

e–2

-

25.12 Comprehensive Physics—JEE Advanced

25.19

SOLUTION i = i0 – e t/

i0

i

, i = i0 =

t

L R

Time constant

100

e–t/

i0 = i0 1 2

E = R 100

SOLUTION L = 100 mH = 0.1 H, R 2 E i0 = = R 0.1 L = = = 0.002 s R i = e–t/ i0

– e–t/

e–t/ = 1–

1 1 = 2 2 t

e t/ = 2

1 = e–t/ e–t/ 2 t = ln2 = 0.002

t VL = – L

di d =–L [i0 dt dt 1

= – Li0 Li0

e–t/

25.10

e–t/

,E

ln2 =

t 10–3 s

TRANSFORMER

e–t/ .

e–2

e

primary secondary

25.18

EX

-

SOLUTION i0 =

12 E 3 = = R

1 2 1 Li 0 = 2 2 3 P = i 0E = U=

or

P = i 02 R =

10

3

2 s p

=

Ns or Np

s

=

Ns Np

p

where Ns 3

Np

2

ep

d p/dt

es

d s/dt

Electromagnetic Induction and A.C. Circuits

Thus d Ns d t Np

es = or

SOLUTION Pi = epIp =

Ns d p Np d t

p

Ip =

Ns ep Np

es =

Ns > Np, then es > ep N s < N p, step-down

step-up

then es < ep

=

and es

, ep =100

Pi = 100 ep Po Pi

Po =

Is =

Po 3000 = es 200

=

25.13

Pi =

25.22

I2R and that

SOLUTION Np Ns = 2000, Po es Is = ep Ip

or

Ip

es = ep Is

Ns Np

ep

Np

es

Ns

=

25.20

cienc

ep =

Po Pi

=

es =

Ns

1000

2000

Po ep I p 12000 200 = A . 200 3

Ip =

Po = ep

Is =

Po 12000 = = 12 A es 1000 2

Pow Ip Pi ep Ip = 220 Po es Is = 10

Np

= 200

where Ip and Is

SOLUTION ep es

es

Rp =

Ip

Is

2

Is

Po = 1100 Pi

Rs

200 3

2

2

25.23

25.21 -

25.14 Comprehensive Physics—JEE Advanced

SOLUTION Ns =

es ep

Is =

Po Pi = = es es

Np =

1000 = 20,000

220 .

= 1.2 A

Simila I = I0

t

= where

2 T

=

Vo2 1 T 02

=

V02 1 dt T 20

=

V02 T T 2

1 2

2 t 2

=

V02 T T 2

0

V02 2

t dt T

1 20

2 t dt T 0

Roo

=2 –1

T

V0

V2 =

Vrms =

T I = I0 sin

t dt

T

as

reads

1 V02 T T

t

sinusoidally varying current whose magnitude changes continuously with time and whose direction reverses periodically is called an alternating current T

=0

I =0

V2 =

V = V0 sin t resistance R, the current I in the circuit is V V I= = 0 sin t = I0 sin R R where I0 = V0 /R, is the maximum or peak value I

0

t

V = V0

ALTERNATING CURRENT

T

V0 2

=

Is is called th 25.11

V0 2 t cos T T

=–

Simila ing current I = I0

2 t T

Irms =

Root Mean Square Voltage and Current

2

I0

I2 =

2

1 2

=

V -

t 1 2

=

I

X t T

EXA

25.24 V

T

X t dt X =

0

T

=

T

dt

t

1 X t dt T0

V

t

0

V = V0

V0 sin

t

SOLUTION

2 t T

V = V0

T

1 V = V0 T0 V0 sin T 0

V0

t dt

Vrms =

T

=

t dt = –

V0 co T

t

t

T 0

and or

2

V0 2

=

100 2

Electromagnetic Induction and A.C. Circuits

25.25 A 100

mum.

Z=R

25.15

-

Special Cases (a) A.C. circuit containing only a pure resistor (Fig. 25.32) SOLUTION

Vrms V0 =

2 Vrms =

Irms = I0 =

25.12

2

200 Vrms = =2A 100 R Fig. 25.32

V0 . = 100 R

SERIES LCR CIRCUIT V VR

C

VC

VL

L

R

VR = V0 sin

t

I = I0 sin

t

V0 R e across R current in the circuit. (b) A.C. circuit containing only an ideal inductor (Fig. 25.33) where

I0 =

Fig. 25.31

V = VR2

Z=

R

VL VC

2

XL

2

XC

2

=

R

2

L

The current in the circuit is t– I = I0 tan

=

L

1 , i.e. C

L


1 C

where 1 , then tan LC

>

< =

1 LC 1 , then LC

I0 =

t t

2

V0 V = 0 XL L

XL = L is call across the inductance leads the current in the circuit /2. (c) A.C. circuit containing only an ideal capacitor (Fig. 25. 34) VC = V0 sin

=0

I = I0 sin

t t

2

25.16 Comprehensive Physics—JEE Advanced

where I0 =

V0 = XC

is cal

edance. I = I0

CV0

t+

1

where = tan–1

R C

Fig. 25.34

XC = age a

C

i

Fig. 25.36

/2. (d) A.C. circuit containing an ideal inductor and a pure resistor (Fig. 25.35) V0 = I0 Z V0 =

VR2 VL2

25.13

POWER IN LCR CIRCUIT

In a series LCR V = V0 sin t, the current in the circuit is t± I = I0 XL < XC or XL > XC.

VR = IR, VL = IXL V and Z = 0 = I0 nce.

R

2

X L2

=

R

2

L

2

I0 =

is called and

Vo ; Z = R2 Z

= tan

Insta source is

2

R

VI = V0 sin

Pt

1 C

1 C

L

–1

L

= V0 I0 sin

t

I0

t t+

t

T

Fig. 25.35

I = I0

t–

L is t where = tan R nd current in the circuit. (e) A.C. circuit containing an ideal capacitor and a pure resistor ( Fig. 25.36) V0 = I0 Z –1

where

V0 = VR2 VC2 VR = IR, VC = IXC

V and Z = 0 = I0

R2

1 P t dt T0

P

X C2 =

R2

1 C

2

= =

1 T

T

sin t

V0 I0

V0 I 0 cos T

t cos

cos

t sin

dt

0 T

T

sin 2

t dt

sin f sin

0

t

0

t dt VI = 0 0 cos T

T 2

0

V0 V0 I 0 cos = 2 2 or P = Vrms Irms cos =

I0 2

cos

Electromagnetic Induction and A.C. Circuits

Power Factor of an A.C. Circuit on Vrms and Irms

is

tan and

cos

=

V0 R

I0

I0 = R2 1/2

1 tan 2

I0

R

R

ma

2 1/2

1 C

L

1 Figure 2

.

I0

1 1 C

L

1

2

R R

= R2

L

1 C

2 1/2

R Z

Fig. 25.37

resis tance im e

P

and i.e. when 1

Special Cases For an A.C. circuit containing only a resistor, R Z = R and cos = = 1 = 0 and R P =V I

when I0 =

2

I0 R2

ma

R 1 C

L

rms rms

For an A.C. circuit containing only an inductor or a capacitor, P =0 LCR circuit. Hence P =

R2 +

2

R – L

2=

Bandwidth and Quality Factor of LCR Circuit For an LCR V = V0 sin t

2 r

2

–

1 R = LC L

= 0, where

Case 2: L –

R + 1 2L

r

=

1 LC

V0 2

L

1 C

2 1/2

2 2 1/2 rL 2

R

1 = –R C 2

R

2

= R2

1 =+R C –

R

Th

P = Vrms Irms

V0 Z

2

ma

1 =±R C

L–

Wattless Current

I0

=

2 1/2

1 C

L

Case 1: L –

I0 =

V0

, I0

I0

R

=

=

In ter

1 C

L

=

=

I0

25.17

+

R – L

R + 1 1 = – 2L

2 r

=0 2 2 1/2 rL 2

R

I0

ma

2

,

25.18 Comprehensive Physics—JEE Advanced

= Qualit

2

–

1

R L

=

Q Q=

qu width

1

=

t1 =

LCR

resonant

=

r

=

t1 =

2

2 t–

I

1 L L = R R C

LC

t=+1

V

t2 =

2

2

t2 –

– -

Q i t = t1 – t2 =

Q.

P

Now tan

–

=

100 L = 100 R . °=

=

t=

rad

.

=

= 3.2

100

10–3 s

25.27

Fig. 25.38

Q. The resonance Q, the

25.26 -

tance 100

SOLUTION Vrms L

tan

V0 2

R L = R

2 Vrms =

2 2 1/2

,

L

100

100

2 1/2

, tan

=

1 CR

1/2 2

120 10

t=+1

V

t1 =

2

2 t+

t2 + .

, Vrms =

–1

2

I is

2 2

1/2

1

2

t1 =

2

F, R

2 110

I0 =

t–

1 C2

R2

rad s–1,

=2 =2 R = 100 . V = V0 sin t

I0 =

I0 =

V0

I0 =

I = I0

V0 =

SOLUTION C = 100 F = 100 10 =2 =2 V = V0 sin t t+ I = I0

=

2

t2 =

2

– -

Electromagnetic Induction and A.C. Circuits

t = t1 – t2 = Now tan

VR

1 = CR 120

=

1

33.

=

VL,C

=

33.

10–3 s

120

rms

230

R=

= Irms

1 rC

rL

= Vrms

Irms

= Vrms

Irms

=

25.28 A series LCR circuit with L R a.c. source.

= Irms

=

rad P

t=

rms

25.19

F and

C

cos

2 230 230 Vrms = R

25.29 A A and B. L, C

and R at resonance.

A and B. SOLUTION

L and C at resonance.

A is a resistor and B Irms Vrms

-

=

r

=

1

r

2

1

=

LC

Z= R

=

L

VL

rms

Im is 1 C

VC

rms

R2

Irms =

1 L= C

25.14

X C2 =

2

2

Vrms 220 = Z .3

LC OSCILLATIONS L and

XL = Irms

C

rL

230

= Irms =

Z=

2 1/2

2 230

= Irms =

220 Vrms = I rms . A and B are connected in series

Reactance o B is XC =

2

2

V0 = Z

–1

0 10

=R I0 =

220 Vrms = I rms .

A is R =

SOLUTION r

B, again the -

230

XC = Irms 1 10

1 LC

1 rC

and tim =

q = q0

T= 2 t+

LC I = I0

t+

25.20 Comprehensive Physics—JEE Advanced

I Multiple Choice Questions with Only One Choice Correct 1. to

v v1/2 v2

v v3/2

t = t1 t1 and t2.

g less than g g

B

t = t2

PQR

a is at B at t = 0 as

v and its edge R t = t0

-

4.

induced current i with time t

Fig. 25.39

2.

Fig. 25.41 (a)

the other coil the current will decrease. 3. t = t2

t = t1

Fig. 25.41 (b)

5. Fig. 25.40

g g

t t

PQ and RS CD CD

v as

Electromagnetic Induction and A.C. Circuits

25.21

9. v

Fig. 25.42

6.

-

I. The centre O x >> a

v

I = 3 sin

2

2

2

10. A series LCR tance L tance R

A A L, C and R, the

C

L C R R or L or C.

v

1 x 1 x

t

2

x x2

t

3

a tance x

t as

11.

2

Fig. 25.43

7. as shown. The ratio I1 / I2 L1 L2

t is L2 L1

L1 L2

12. In a series LCR

L2 L1

-

2 13. Fig. 25.44

V = 200 2

I2 = 3A in the

8.

where V the ammeter is

resistor is

14.

Fig. 25.45

t+ t

25.22 Comprehensive Physics—JEE Advanced

F F -

15. line is 1

16. at t = 0

t

I in the circuit shown K

Fig. 25.48

18.

19.

f is connected

to an LCR

I in f

Fig. 25.46

Fig. 25.49

20. Fig. 25.47

17. V across the inductor L with time t K t = 0, in the circuit shown in 21.

-

LC 2

22. L, C and R

= 10.

Electromagnetic Induction and A.C. Circuits

28. RC L/R

25.23

inductance and a resistance connected in series

LC C/L

23. V = 200 2 is connected to 1

t

29.

V = V0sin t a circuit. As a result a current I = I0

t– V 0I 0

24.

V 0I 0

30.

V 0I 0 L and resistance v B,

R

A and B is

Fig. 25.50

25.

V and

current I V and

t

I = 100 sin 100 t

3

mA Fig. 25.51

BLv/R BLv/R BLv/R

26.

31.

27.

A

B. A

connected across the ends such that the total resisR BR A

AB R B2 A

ABR

R2

32. B

R is MNQ the

25.24 Comprehensive Physics—JEE Advanced

across the ring is

P at a distance r region

v

1 Bv R2 and M 2 RBv and Q RBv and Q

1 r 1

. r2 IIT, 2000

r

Fig. 25.53 Fig. 25.52

37.

33.

l

B r

L L >> l

-

R

34.

2

l L

l L

L l

L2 l

in the x-y

r2 B 2R

I, lies

R

38.

r B 2R

r2 B R 2

r

2

B

2

2

R –1

x-y is 0.2 × 10

I R R

35.

is

36.

Bt a and is directed

39.

F is charged to a

40.

r B.

Electromagnetic Induction and A.C. Circuits

25.25

t Br 2 1 10 t Br 2 1 t

Br 2 1 t Br 2 1 t l, mass m and resistance

41. R

Fig. 25.55

IIT, 2000

height h 45. m 2 gR 2

mgR 2 Bl mgR

2B l 2

m 2 gR 2 is induced AD BC

2B l

42.

r1 and r2 r22

0

2r1

0

v

2 2

B 3l 3

0

ABCD

r1 > r2

BC

r12

2r2

r1 r2 2r1

2

0

r1 r2 2r2

2

v in B = 2T as shown in Fig.

43. tance 1

BC AD AD nor in

Fig. 25.56

AD and BC IIT, 2001 P and Q

46.

-

S Ip IQ1 are:

P Q S

E IQ1 and IQ2

IQ2

Q. E

equal to 3 –1

–1

–1

–1

Fig. 25.57

Fig. 25.54

44. Two IIT, 2002

25.26 Comprehensive Physics—JEE Advanced

47.

51. An LCR series circuit with R = 100

2

E = E0

48.

IIT, 2002 t

is connected to

3 A 2

3 52. v

E and the current I in the circuit

current I its ends at distances r1 and r2

R-C or R-L or L-C

r2 > r1 0 Iv

R R R R

, , , ,

2

C = 10 F C=1 F L = 10 H L=1H

0 Iv

2 53. IIT, 2003

loge

r1 r2

0 Iv

r2 r1

log e 1

increasing according to the equation I

r1 r2 t2

54. s. The Fig. 25.58

49. An air ms–1 is 2

a and b

55. B = kx where k

10 1 10

2 10

3 10

1

coil as shown in Fig.

kab2 1 kab2 2

L and resistance R is con-

50. 1 L loge R

L loge R

L loge R

L loge R

-

56.

2 kab2

Fig. 25.59

x as

Electromagnetic Induction and A.C. Circuits

57.

58. -

resented as

along the x

25.27

x x x x direction

59.

I1 and I2 are the currents in the segments AB and CD. Then, I1 > I2 I1 < I2 I1 is in the direction BA and I2 is in the direction CD I1 is in the direction AB and I2 is in the direction DC

Fig 25.61 Fig. 25.60

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

2. 8. 14. 20. 26. 32. 38. 44. 50. 6.

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

5. 11. 17. 3. 29. 35. 41. 47. 53. 59.

6. 12. 18. 24. 30. 36. 42. 48. 54.

SOLUTIONS 1.

l e

Bl v

R

i=

Bl v R

25.28 Comprehensive Physics—JEE Advanced

F = Bil

B2 l 2 v = R Power needed is P = F v =

i=

B 2 l 2 v2 . So the R

Bva R

i against t

Bva R

-

2.

B v2 a t Rb

B v2 a . Rb

5. As the rod CD in each coil will decrease.

C to D C to D and, as a result, end C

3.

CD D

t1 and t2 no current is ing g. 4.

t

right a distance CD = v t

charge. 6. = BA = B a I Now B = 0 2 x

RD = b.

=

I a2 2 x

=

Fig. 25.62

7.

Now

across L2

RC CT = RD QD

I a 2 dx d = 0 2 dt dt 2 x

vt

b b

=

Bv

2

x

2

1 x2 L1 t. Hence.

Integrating, we get L1 I1 = L2 I2

CT =

e

I a2 v

dI1 dI 2 = L2 dt dt L1 dI1 = L2 dI2

CT a/2

a b vt 2b a ST = 2CT = b

0

L1

RD CD CT = RD QD

8.

I1 L = 2. I2 L1

b – vt

ST = Bv a

a vt b

R i=

e Bv = a R R

a vt b

A = a2

0

e=

Hence e

2

Fig. 25.63

v

dx dt

Electromagnetic Induction and A.C. Circuits

I2 tance = 3

I1 =

resis-

12

VL = VC

= 2 A. There-

VR2

V=

I = I1 + I2

VL

2

VC

= VR

Current in the circuit is I=

9. 2

I0 = Irms =

25.29

I0 2

2R

2

Hence =

2

VL = IXL =

A 1 > C

XC > XL or

10. L R = Z

R

Irms = Vrms

2

=

L

C

100

2

10 -

tor, i.e. 1 1 LI2 CV 2 = LI 2 or C = 2 2 V2

1 1

200 2

14.

2

1 C

= 100 rad s–1.

V0 = 200 2

1/ 2

=

XL = R

R

13.

2

1 C

R2

2R 2

1/ 2

=

V = Z

L

=

1

2

2

10

2

R

F

15. Current in the transmission line is I=

R, C and L are 11.

nance, i.e. when Z = R, where R Vrms R=

I= 2= 12.

E R 2

r

LR or

R = XL = XC XC = 0 and the X L2 =

–R t /L

I = I0

e

I = I0

e –t /

where = L/R is the time constant. At t = 0, I = 0. At t >> t, I = I0.

r = 1.0

R2

2

16.

;r

VR = VL = VC

= I2R

= 100 A

10000

Vrms 12 = =2 I rms

r

Z=

Power loss

-

=

ltage

R2

R2 = 2 R R = XL

17. V=–L

dI d = – LI0 dt dt = – LI0

R –Rt /L e L

e –Rt/ L

25.30 Comprehensive Physics—JEE Advanced

V0 e –t / , where V0 = I0R is the initial V = V0 at t t >> . Hence

V

22.

L R 100 tan = 2 1000 =

18. R tan

L=

10

ltage current

RC 23.

2

=2 cL

V0 = 200

L

I=

. Thus I increases with increase in = c 1

=

cC

or

3

LC > c, I decreases

25.

V0

I P=

1 = or e t/ = 2 or t = 2

=

=

2

=

LC circuit is 1

21. 2

LC

2

100

1

10

10 –3 A = 20 mA

10 –3

=

-

3

V0 I 0 cos 2 100 100 10 2

3

27.

L 10

For

10

H.

due to which it contracts. As a result the contact is

3

1 103 10

2

current is induced in the coil. 2

where L

2

200 2

1 C

26.

1

C=

C=

I0 = 100 mA = 100 I and V is

I0 / 2 is

Zc

24.

I = I0 e –t/

or

C=

= 20

10 –3 s

=2

Current at time t

e –t/

2

C

inductance L 1 1 1 1 3 3 = + + = = =1 L L L L L 3 or L

= 2 millisecond.

where I0

V0

=

20. L 100 10 = R

= 100 rad s –1

2

Vrms = Vrms Zc

1

=

c

t t,

2 1/ 2

1 C

when I with increase in =

V = 200 2 equation with V = V0 sin

V R

where

charge charge = = time ltage charge/time

–3

19. The current in an LCR I=

charge /

RC

2

10 28.

–12

C

wire oscillates. the circuit is

Electromagnetic Induction and A.C. Circuits

V R2 + V 2L

V

1/2

l l

1 2R

r B

3

t

t = CV0

r B

2

L A=

sin

Q0

2

1 V02 2 R

=

I=

I

circular coil. Hence

2