AIIMS Specialist-Physics(26 Years' Chapterwise Solved Papers) [2 ed.] 9789313197126

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26 YEARS'

Arihant Prakashan (Series), Meerut

Arihant Prakashan (Series) Meerut All Rights Reserved

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FIRST & FOREMOST Whenever a student decides to prepare for any examination his/her first and foremost curiosity is to know about the type of questions that are expected in the exam. This becomes more important in the context of competitive examinations where they have to face neck-to-neck competition. We feel great pleasure in presenting before you this book containing Error Free Solutions of previous years’ (1994-2019) AIIMS (Medical) questions specifically designed according to latest trend of questions. Going through this book, you will get an exact idea of the questions generally asked in AIIMS. We have made maximum efforts to provide correct solutions to the best of our knowledge and opinion. Detailed explanatory discussions follow the answers. Discussions are not just sketchy–rather, have been drafted in a manner that the students will surely be able to solve some other related problems too. In this way, the students shall be able to judge the extent upto which they have been able to comprehend the methods. We hope, these sets of papers along with error free solutions would be highly beneficial to the students. We would be grateful if any discrepancy or mistake in the questions or answers is brought to our notice so that these could be rectified in subsequent editions.

PUBLISHER

CONTENTS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

Physical World and Measurement Kinematics Laws of Motion Work, Energy and Power Rigid Body and Rotational Motion Gravitation Mechanical Properties of Solids Mechanical Properties of Fluids Thermal Properties of Matter Thermodynamics Kinetic Theory of Gases Oscillations Waves Electric Charges and Fields Electrostatic Potential and Capacitance Current Electricity Moving Charges and Magnetic Effect of Current Magnetism and Matter Electromagnetic Induction Alternating Current Electromagnetic Waves Ray Optics Wave Optics Dual Nature of Matter and Radiation Atoms Nuclei Semiconductor and Electronic Devices Communication System

!

AIIMS Solved Paper 2019

1-7 8-19 20-31 32-42 43-53 54-64 65-68 69-77 78-88 89-96 97-103 104-114 115-125 126-135 136-145 146-159 160-170 171-178 179-185 186-194 195-199 200-218 219-229 230-239 240-248 249-262 263-275 276-280 1-15

CHAPTER

1 Physical World and Measurement 1. In an experiment to measure the height of a bridge by dropping stone into water underneath. If the error in measurement of time is 0.2 s at the end of 4 s, then the error in estimation of height of bridge will be (neglect the water resistance, i.e., thrust) [AIIMS 2018] h

(a) ± 19.68 m (c) ± 7.84 m

(b) ± 17.22 m (d) ± 12.22 m

2. A physical quantity X is given by X=

2k 3 l 2 m n

.

The percentage errors in the measurements of k , l, m and n are 1%, 2%, 3% and 4% respectively. The value of X is uncertain by [AIIMS 2017] (a) 8% (c) 12%

(b) 10% (d) None of these

3. In terms of basic units of mass ( M ), length ( L), time (T ) and charge (C ), the dimensions of magnetic permeability of vacuum (m 0) would be [AIIMS 2015] (a) [MLC -2 ] (c) [ML2 T-1C -2 ]

(b) [LT-1C -1 ] (d) [LTC -1 ]

4. The dimensional formula for electric flux is 3

[AIIMS 2015] -1 -3

(a) [ML A T ] (c) [ML3 A1T-3 ]

2 2

-1 -2

(b) [M L A T ] (d) [ML-3 A -1T-3 ]

5. The pressure on a square plate is measured by measuring the force on the plate and the length of the side of the plate by using F the formula p = 2 . If the maximum errors l in the measurement of force and length are 4% and 2% respectively, then the maximum error in the measurement of pressure is [AIIMS 2014] (a) 1% (c) 8%

(b) 2% (d) 10%

6. The force F is given by expression F = A cos( Bx) + C sin( Dt ), where x is the displacement and t is the time. Then, D dimension of is same as that of B (a) velocity [LT-1 ] (b) angular velocity [T-1 ] (c) angular momentum [ML2 T-1 ] (d) velocity gradient [T-1 ]

[AIIMS 2013]

7. The unit of coefficient of thermal conductivity is (a) Wm-1K -1 (b) JK -1

[AIIMS 2012] (b) WmK

(d) JK

2

AIIMS Chapterwise Solutions ~ Physics 8. If C and L denote the capacitance and inductance respectively, the dimension of [AIIMS 2011] LC is (b) [M0L2 T-2 ] (d) [M0L0 T0 ]

(a) [M0L0 T2 ] (c) [MLT-2 ]

9. The dimension of universal gas constant is (a) [ML2 T-2 q-1 ] 3 -2

-1

(c) [ML T q ]

[AIIMS 2010] (b) [M2LT-2 q] (d) None of these

10. The ratio of the dimensions of Planck’s constant and that of the moment of inertia is the dimension of [AIIMS 2010]

11. Percentage error in the measurement of mass and speed are 2% and 3%, respectively. The error in the estimation of kinetic energy obtained by measuring mass and speed will be [AIIMS 2009] (b) 10% (d) 8%

constant (G) and Planck’s constant ( h) are taken as fundamental units in a system. The dimension of time in this new system should be [AIIMS 2008] (b) G -1/ 2 h1/ 2c1/ 2 (d) G1/ 2 h1/ 2c1/ 2

13. Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L , of time T and of current A, would be [AIIMS 2007] (a) [ML2 T-3 A -1 ] (c) [ML2 T-1A -1 ]

(b) [ML2 T-2 ] (d) [ML2 T-3 A -2 ]

14. If the time period (T ) of vibration of a liquid drop depends on surface tension ( s), radius ( r) of the drop and density (r) of the liquid, then the expression of T is [AIIMS 2007] (a) T = k rr 3 / s 3

1/ 2

(c) T = k rr / s

(b) m = – n (d) m = – n2

16. Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimensional formula of permittivity is (a) [ML-2 T-2 A]

[AIIMS 2004] (b) [M-1L-3 T4 A 2 ]

(c) [MLT-2 A]

(d) [ML2 T-1A 2 ]

17. The difference in the lengths of a mean solar day and a sidereal day is about (a) 1 min (c) 15 min

(b) 4 min (d) 56 min

18. The dimensional formula for Reynold’s number is 0

0

[AIIMS 2002]

0

(a) [M L T ]

(b) [MLT]

(c) [ML-1T]

(d) [MLT-1 ]

19. Of the following quantities, which one has

12. The speed of light (c), gravitational

(a) G1/ 2 h1/ 2c -5 / 2 (c) G1/ 2 h1/ 2c -3 / 2

(a) m = n (c) m2 = n

[AIIMS 2003]

(a) frequency (b) velocity (c) angular momentum (d) time

(a) 12% (c) 2%

liquid. Which of the following relation is correct? [AIIMS 2005]

(b) T = k r1/ 2 r 3 / s (d) None of these

15. The mass of the liquid flowing per second per unit area of cross-section of the tube is proportional to (pressure difference across the ends) n and (average velocity) m of the

dimensions different from the remaining three [AIIMS 2002] (a) energy per unit volume (b) force per unit area (c) product of voltage and charge per unit volume (d) angular momentum per unit mass

20. Which of the following pairs does not have similar dimensions?

[AIIMS 2001]

(a) Tension and surface tension (b) Stress and pressure (c) Planck’s constant and angular momentum (d) Angle and strain

21. The length and breadth of a metal sheet are 3.124 m and 3.002 m respectively, the area of this sheet upto four correct significant figures is [AIIMS 2001] (a) 9.378 m2

(b) 9.37 m2 2

(c) 9.378248 m

(d) 9.3782 m2

22. What is the dimensional formula of gravitational constant? (a) [ML2 T-2 ] (b) [ML-1T-1 ] (c) [M-1L3 T-2 ] (d) None of the above

[AIIMS 2000]

3

Physical World and Measurement 23. The dimensional formula of the constant a in van der Waal’s gas equation a ö æ ç p + 2 ÷(V - b) = RT is V ø è 4 -1

[AIIMS 1999]

(b) [ML T ]

(c) [ML5 T-3 ]

(d) [ML5 T-2 ]

one metre?

[AIIMS 1998]

(a) [ML0 T-2 A -1 ]

(b) [ML2 T - 2 A -1 ]

(c) [ML2 T -1A -3 ]

(d) [M0L-2 T -2 A -2 ]

(c) h / e 2

(d) h2e 2

26. Which of the following physical quantity unit is not a fundamental unit? [AIIMS 1996] (a) Length (c) Magnetic field

(b) Mass (d) Current

æ 1 ö y , where p is the ÷÷ = b p k è ø BT pressure, y is distance, k B is Boltzmann constant and T is the temperature. Dimension of b are [AIIMS 1995]

27. In the equation çç

–1

1 2

(a) [M L T ] 1

–1

(c) [M L

–2

T ]

Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

31. Assertion When we change the unit of measurement of a quantity, its numerical value changes.

0

Reason Smaller the unit of measurement, smaller is its numerical value. [AIIMS 2015]

(d) [M0 L0 T0 ]

32. Assertion The error in the measurement of

0

2

(b) [M L T ]

radius of the sphere is 0.3%. The permissible error in its surface area is 0.6%.

28. The dimension of the modulus of rigidity, is

[AIIMS 1994]

(a) [ML-2 T-2 ] (c) [ML-1T-1 ]

[AIIMS 1994] (b) 1553164.13 (d) 1650763.73

Direction (Q. Nos. 31 and 32) Read the

(h-Planck’s constant, e-charge) [AIIMS 1997] (b) h2 /e

(a) 2348123.73 (c) 652189.63

Assertion & Reason

25. Dimensions of ohm are same as (a) h/e

(b) x = y2 z (c) y2 = xz (d) z = x2 y

30. How many wavelengths of Kr86 are there in

24. The dimensional formula of magnetic flux is

have units gcm 2s –5, gs –1 and cms –2 , respectively. The relation between x, y and [AIIMS 1994] z is (a) x = yz2

2 -2

(a) [ML T ]

29. The three physical quantities x, y and z

(b) [MLT-2 ] (d) [ML-1T-2 ]

Reason The permissible error is calculated D A 4D r . by the formula = A r [AIIMS 2008]

Answers 1. 11. 21. 31.

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

2. 12. 22. 32.

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

3. (a) 13. (d) 23. (d)

4. (a) 14. (a) 24. (b)

5. (c) 15. (b) 25. (c)

6. (a) 16. (b) 26. (c)

7. (a) 17. (b) 27. (b)

8. (a) 18. (a) 28. (d)

9. (a) 19. (d) 29. (a)

10. (a) 20. (a) 30. (d)

AIIMS Chapterwise Solutions ~ Physics

Explanations 1. (c) From second equation of motion, we have

or

1 s = ut + at 2 2 1 h = (0)t + ´ 9.8 ´ (4)2 2

æQ a = g ö çç ÷÷ èQ u = 0 ø

= 78.4 m Given, Dt = 02 . s and t = 4 s Now, error in the estimation of height, Dh . ö æ Dt ö æ 02 = ± 2 ç ÷ = ± 2 ç ÷ = ± 01 . h è t ø è 4 ø or Dh = ± 01 . ´h = ± 01 . ´ 78.4 = ± 7.84 m

2. (c) Given, X =

2k3 l 2 m n

Percentage error in X, DX Dm 1 Dn ù é 3Dk 2Dl ´ 100 = ê + + + × ´ 100 2 n úû X l m ë k 1 = 3 ´ 1% + 2 ´ 2% + 3% + ´ 4% 2 = 3% + 4% + 3% + 2% = 12% Hence, the value of X is uncertain by 12%.

3. (a) The force per unit length experienced due to two wires in which current is flowing in the same direction is given by dF m 0 2I1 I2 = × 4p d dl where I1 and I2 are the currents flowing in the two wires. The dimension of above given quantities are as follows F = [MLT -2 ] current, I1 I2 = [A2 ] length, dl = [L] distance between two wires, d = [L] Now, after substituting these dimensions in above equation, we get -2

2

Þ

[MLT ] [A ] = [m 0 ] [L] [L]

Þ

é C2 ù [MLT -2 ] =m0ê 2 ú [L] ëT Lû -2

Þ Dimension of m 0 = [MLC ]

é 1C ù êQ 1A = ú Tû 1 ë

4. (a) Electric flux is the measure of flow of the electric field through a given area. It is given by F fE = Eds = ds q where, E is electric field, F is force, q is charge and ds is area of surface projected in a plane perpendicular to the field. é MLT -2 ù 2 3 -1 -3 \Dimension of fE = ê ú [L ] = [ML A T ] ë AT û

5. (c) Pressure is given by p =

F

l2 Hence, percentage of maximum error in the measurement of pressure is Dp é DF 2Dl ù ´ 100% = ê + ´ 100% p l úû ë F ù é DF êQ F ´ 100% = 4%,ú = 4% + 2 ´ 2% ê ú Dl ´ 100% = 2% ú ê û ë l

\Maximum percentage error in p = 8%.

6. (a) Force is given by F = A cos(Bx) + C sin(Dt ). In trigonometric functions like sin q, cos q, etc, q is dimensionless. So, [Bx] = [M 0 L0 T 0 ] Þ Dimension of [B] = [L-1 ] (Q Dimension of x = [L] ) Also, [Dt ] = [M 0 L0 T 0 ] Þ Dimension of [D] = [T -1 ] (Q Dimension of t = [T]) éDù Now, dimension of ê ú = [LT -1 ] ëBû which is same as the dimension of velocity = [LT -1 ]

7. (a) The coefficient of thermal conductivity is the amount of heat energy any material allows to get transferred per metre length for per unit change in temperature. DQ æ Dx ö 1 \ K = ç ÷´ Dt è DT ø A where, DQ is the amount of heat energy, Dx is change in length of conductor, DT is change in temperature and

5

Physical World and Measurement Dt is change in time. J -m Unit of K = 2 = Wm-1 K-1 m sK

1 Jù é êëQ 1 W = 1s úû

8. (a) Resonating frequency is given by 1 2p LC where, L and C are inductance and capacitance respectively. 1 LC = 2 2 4p f 1 \Dimension of LC = Dimension of 2 2 4p f f =

0 0 2

= [M L T ]

9. (a) Ideal gas equation is given by pV = nRT where p, V and T are pressure, volume and temperature respectively and n is number of moles. \Dimension of universal gas constant, pV (ML-1 T -2 ][L3 ] = R= [q] nT (Q n is a dimensionless quantity) = [ML2 T -2 q-1 ]

10. (a) We know that, energy, E = hn E [where, n is frequency] n 2 -2 [E] [ML T ] [h] = = = [ML2 T -1 ] [n] [T -1 ]

Planck’s constant, h = \

Dm Dv ´ 100% = 2% and ´ 100% = 3% m v \Percentage error in kinetic energy = 2% + 2 ´ 3% = 8% Given,

Also, I = moment of inertia, which is quantitative measure of the rotational inertia of a body and equal to the product of mass of each particle in the body with the square of its distance from the axis of rotation. I = MR2 Þ Dimension of [l ] = [M][L2 ] =[ML2 ] Hence, ratio of dimension of [h] [ML2 T -1 ] = = [T -1 ] [l ] [ML2 ] 1 = = dimension of frequency [T] 1 11. (d) Kinetic energy, E K = mv 2 2 Percentage error in kinetic energy, DE k Dm 2Dv ´ 100% = ´ 100% + ´ 100% Ek m v

12. (a) According to the method of dimensional analysis, the dimension of each term on both sides of an equation must be same. Time µ c x G y h z T = kc x G y h z

Þ

Putting the dimensions in above relation Þ [M 0 L0 T1 ] =[LT -1 ]x [M -1 L3 T -2 ] y [ML2 T -1 ]z Þ [M 0 L0 T1 ] = [M - y + z Lx

+ 3 y + 2z

T-x

-2y - z

]

Comparing the powers of M, L and T, …(i) -y + z = 0 …(ii) x + 3y + 2 z = 0 …(iii) - x - 2y - z = 1 On solving Eqs. (i), (ii) and (iii), we have 5 1 x=- ,y=z= 2 2 Hence, dimension of time is [G1 /2 h1 /2c -5/2 ].

13. (d) From Ohm’s law, V = IR Potential difference V W = = \Resistance, R = I qI Current (Q potential difference is equal to work done per unit charge) [ML2 T -2 ] So, dimension of R = = [ML2 T -3 A -2 ] [AT][A]

14. (a) Let T µ sx r yr z Þ T = ksx r yr z Substituting the dimension of symbols and equating [T] = [MT –2 ]x [L] y [ML–3 ]z x+ z = 0 y – 3z = 0 –2 x = 1 Solving Eqs. (i), (ii) and (iii), we get x = –1/2, y = 3/2 and z = 1/2 rr 3 s where, k is proportionality constant.

\

T = ks-1 /2 r 3 /2r1 /2 = k

15. (b) According to the question, m µ pn × v m t×A m or = kp nv m t×A [where, k is proportionality constant] We have,

…(i) …(ii) …(iii)

6

AIIMS Chapterwise Solutions ~ Physics Using principle of homogeneity, we get [ML–2 T –1 ] = [ML–1 T –2 ]n [LT –1 ]m [ML–2 T –1 ] = [M]n[L]n+ m [T]-2 n- m Equating both sides, we get n = 1 , - 2n - m = - 1 and – n + m = –2 –1 + m = –2 \ n = – m or m = – n

16. (b) From Coulomb’s law, two stationary point charges q1 and q2 attract / repel each other with a force F which is directly proportional to the product of charges and inversely proportional to the square of distance r between them. 1 q1q2 1 q1q2 i.e. F = × × Þ e0 = 4p F r 2 4pe0 r 2 \Dimension of permittivity, [A2 T2 ] [ e0 ] = = [M -1 L-3 T 4 A2 ] [MLT -2 ][L2 ]

17. (b) A solar day is the time it takes for the earth to rotate about its axis, so that the sun appears in the same position in the sky. The sidereal day is the time it takes for the earth to complete one rotation about its axis w.r.t the ‘fixed’ stars. The sidereal day is ~ 4 minutes shorter than the solar day.

18. (a) Reynold’s number describes the ratio of inertial force per unit area to viscous force per unit area for a flowing fluid. Thus, Reynold's number is the ratio of two physical quantity of same dimension which cancel out each other. Hence, Reynold’s number is dimensionless [M 0 L0 T 0 ] quantity.

19. (d) Energy per unit volume =

[ML2 T –2 ] [L3 ] –2

Force per unit area =

[MLT ] 2

[L ]

= [ML–1 T –2 ]

= [ML–1 T –2 ]

Product of voltage and charge per unit volume =

V ´Q VIt Power ´ Time = = = [ML-1 T -2 ] Volume Volume Volume

Angular momentum per unit mass [ML2 T –1 ] = = [L2 T –1 ] [M] Hence, angular momentum per unit mass has different dimension from other three.

20. (a) Dimension of surface tension is =

[MLT -2 ] force = [MT -2 ] = [L] length

Dimension of tension = force = [MLT -2 ] Hence, dimensions of surface tension and tension are not equal. While dimensions of stress and pressure are MLT -2 F =[ML-1 T -2 ] = A L2 F Stress = = [ML-1 T -2 ] A \Dimensions of stress and pressure are equal. E Planck’s constant = h = n [ML2 T -2 ] Dimension of [h] = = [ML2 T -1 ] [T -1 ] p=

Angular momentum = Moment of inertia ´ Angular velocity Dimension of [L] = [M1 L2 T 0 ][M 0 L0 T -1 ] = [ML2 T -1 ] \Dimensions of angular momentum and Planck’s constant are same. Also, angle and strain are dimensionless quantities.

21. (a) Significant figure is a method of expressing error in measurement. Significant figures follow certain set of rules. i.e., all non-zero digits are significant. Any zeros that are between non-zeros are also considered significant. Hence, length (3.124m) and breadth (3.002m) have four significant figures. \ Area = Length ´ Breadth= 3124 . ´ 3.002 = 9 . 378248 will also have four significant figures. Hence, area = 9378 . m2 .

22. (c) From Newton’s law of gravitation, ‘‘Every object in the universe attracts every other object with a force F which is directly proportional to the product of masses (m1 and m2 ) and inversely proportional to the square of distance between them (r ).’’ Gm1 m2 i.e. F = r2 (where, G is gravitational constant) Fr 2 G= \ m1 m2

7

Physical World and Measurement \ Dimensions of gravitational constant =

[MLT -2 ][L]2 [M]2

= [M -1 L3 T -2 ]

23. (d) Quantities having same dimensions can be added and subtracted. In the given equation, a ö æ ç P + 2 ÷ (V - b ) = RT è V ø P , V , T are pressure, volume and temperature respectively. a As, unit of 2 = unit of P V Þ unit of a = unit of P ´ unit of V 2 newton = ´ metre 6 = newton-metre 4 metre2 Dimensional formula of a = [MLT -2 ][L4 ] = [ML5T -2 ]

24. (b) Magnetic flux is a measurement of the total magnetic field which passes through a given area. It is given by f = BA where, B is magnetic field and A is area. Also, F = ILB, (where, I is current) FA \ f= IL [MLT -2 ][L2 ] Hence, dimension of (f) = [A][L] 2

-2

-1

= [ML T A ]

25. (c) Dimension of ohm =

volt = [ML2 T –3 A –2 ] ampere

E ML2 T –2 = = [ML2 T –1 ] n T –1 And dimension of e = [AT] [ML2 T –1 ] = [ML2 T –3 A –2 ] Þ Dimension of h /e 2 = [A2 T2 ] Hence, ohm has same dimension as h /e 2 have. Dimension of h =

26. (c) Standards which are used to measure the fundamental quantities are called fundamental units which are length, mass, time, temperature, current, luminous intensity, quantity of matter. Hence, magnetic field is not a fundamental quantity. y 1 27. (b) Given equation, = pb k BT Dimension of [b] =

[ML2 T –3 ][T] [ML–1 T –2 ][L]

= [M 0 L2 T 0 ]

28. (d) The modulus of rigidity, h=

t xy shearing stress F/A Fl = = = shearing strain g xy Dx / l ADx

F = shear stress, F is force and A A is area on which force acts. Dx = tan q = shear strain. g xy = l where, Dx is the transverse displacement and l is the initial length. \Dimensional formula of where, t xy =

h=

[MLT -2 ][L] [L]2 [L]

=[ML-1 T -2 ]

29. (a) Given, x = gcm2s –5 = [ML2 T –5] y = gs –1 = [ML0 T –1 ] and

z = cms –2 = [M 0 L T –2 ]

Now, z2 = [M 0 L2 T –4 ] and yz2 = [ML0 T –1 ] [M 0 L2 T –4 ] = [ML2 T –5] = x i.e.

x = yz2

30. (d) One metre = 165076373 . wavelengths of the orange-red emission line in the electromagnetic spectrum of krypton-86.

31. (c) Assertion is true but Reason is false. We know that, Q = n1u1 = n2u2 , where u1 and u2 are the two units of measurement of the quantity Q and n1 , n2 are their respective numerical values. \When we change the unit of measurement of a quantity, its numerical value changes. Also, nu = constant 1 nµ Þ u i.e., smaller the unit of measurement, greater is its numerical value.

32. (c) Assertion is true but Reason is false. Area of sphere, A = 4pr 2 (error will not be involved in constant 4p) where, r is radius of sphere. DA 2Dr Fractional error, = A r DA ´ 100 = 2 ´ 03 . % = 0.6% A DA 4Dr But is false. = A r

AIIMS Chapterwise Solutions ~ Physics CHAPTER

2 Kinematics 1. What will be the a versus x graph of the given plot from the following graphs? [AIIMS 2018] v(m/s) v0

3. A bus starts from rest and moves with

constant acceleration 8 ms -2 . At the same time, a car travelling with a constant velocity 16 m/s overtakes and passes the bus. After how much time and at what distance, the bus overtakes the car?

(a) t = 4 s, s = 64 m (c) t = 8 s, s = 58 m x0

a

x(m)

4. Two vectors A and B have equal magnitudes. If magnitude of A + B is equal to n times the magnitude of A - B, then the angle between A and B is [AIIMS 2016]

a

(a)

(b) x

æ n - 1ö (a) cos -1 ç ÷ è n + 1ø

x

æ n - 1ö (c) sin-1 ç ÷ è n + 1ø

a

a (c)

(d)

x

[AIIMS 2016] (b) t = 5 s, s = 72 m (d) None of these

x

æ n2 - 1 ö ÷ (b) cos -1 çç 2 ÷ è n + 1ø 2 æ n - 1ö ÷ (d) sin-1 çç 2 ÷ è n + 1ø

5. A boat is sent across a river with a velocity 2. A block is dragged on a smooth plane with the help of a rope which moves with a velocity v as shown in the figure. The horizontal velocity of the block is [AIIMS 2017] v

of 8 kmh -1. If the resultant velocity of the boat is 10 kmh -1, the river is flowing with a velocity of [AIIMS 2016]

(a) 12.8 km h-1 (c) 8 km h-1

(b) 6 km h-1 (d) 10 km h-1

6. Two balls are thrown horizontally from the q

v sinq v (c) cos q

(a)

top of a tower with velocities v 1 and v2 in opposite directions at the same time. After how much time, the angle between velocities of balls becomes 90º? [AIIMS 2016] (b) vsinq (d) vcos q

(a) (c)

2 v1 v 2 g g v1 v 2

(b) (d)

v1 v 2 g v1v 2 2g

9

Kinematics 7. If an object is thrown at an angle of 60º with horizontal, find elevation angle of the object at its highest point as seen from the point of projection. [AIIMS 2016] (a) tan-1 (c) tan-1

3 2 1 2

(b) tan-1

1 2

above a river. After the ball has been falling for 2s, a second ball is thrown straight down after it. What must the initial velocity of the second ball be so that both hit the water at the same time? [AIIMS 2015] (b) 55.5 m/s (d) 9.6 m/s

9. A particle is projected with an angle of projection q to the horizontal line passing through the points ( P , Q) and (Q, P ) referred to horizontal and vertical axes (can be treated X-axis and Y -axis, respectively). The angle of projection can be given by [AIIMS 2015] 2 ù é P 2 + PQ + Q 2 ù é 2 -1 P + Q - PQ (a) tan-1 ê ú (b) tan ê ú PQ PQ û û ë ë 2 2ù 2 2 ù é é P +Q P + Q + PQ (d) sin-1 ê (c) tan-1 ê ú ú 2 PQ û ë 2 PQ û ë

10. What will be ratio of speed in first 2s to [AIIMS 2014]

s

2

3

surface such that it creates an acceleration of 19.6 m/s 2 . If after 5 s, its engine is switched off, the maximum height of the rocket from earth’s surface would be [AIIMS 2011] (a) 245 m (c) 980 m

(b) 490 m (d) 735 m

14. If a vector 2$i + 3$j + 8 k$ is perpendicular to the vector 4$j - 4i$ + ak$ , then the value of [AIIMS 2010] a is (a) -1 (c) -

(b)

1 2

1 2

(d) 1

15. A balloon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to height h. The time taken by the stone to reach the ground is [AIIMS 2010] h (a) 4 g 2h (c) g

h (b) 2 g g (d) h

16. There are N -coplanar vectors each of

s0

1

(b) b1 (d) 2 b2

13. A rocket is fired upward from the earth

8. A ball is dropped from a bridge 122.5 m

the speed in next 4s ?

position x of particle with respect to time t from origin given by x = b0 + b1 t + b2t 2 . The acceleration of particle is [AIIMS 2012] (a) b0 (c) b2

(d) None of these

(a) 40 m/s (c) 26.1 m/s

12. A particle moves along with X-axis. The

4

5

6

t

magnitude V . Each vector is inclined to the preceding vector at angle 2p / N . What is the magnitude of their resultant? [AIIMS 2009]

(a) 2 : 1 (c) 2 : 1

(b) 3 : 1 (d) 1 : 2

(a) V / N

11. A body starts from rest and moves with a uniform acceleration. What is the ratio of the distance covered in the nth second to the distance covered in n second? [AIIMS 2013] 2 1 n n2 2 1 (c) 2 - 2 n n

(a)

(b)

1 2

-

1 n 1

n 2 (d) + 2 n n

(b) V

(c) Zero

(d) N / V

17. A car starting from rest, accelerates at the rate f through a distance s, then continue at constant speed for time t and then decelerates at the rate f /2 to come to rest. If the total distance travelled is 15 s, then [AIIMS 2008] (a) s = ft (c) s =

1 2 ft 72

1 (b) s = ft 2 6 1 (d) s = ft 2 4

10

AIIMS Chapterwise Solutions ~ Physics

18. A projectile can have the same range R for two angles of projection. If t 1 and t2 be the times of flights in the two cases, then the product of the two times of flights is proportional to [AIIMS 2008] (a) R 2

(b)

1

(c)

R2

1 R

(d) R

23. A ball is thrown vertically upwards. Which of the following plots represents the v-t graph of the ball during its flight, if the air resistance is not ignored? [AIIMS 2003]

(a) v

19. A car travels 6 km towards north at an angle of 45º to the east and then travels distance of 4 km towards north at an angle 135º to east. How far is the point from the starting point? What angle does the straight line joining its initial and final position makes with the east? [AIIMS 2008] (a) 50 km and tan-1 (5)

t

(c)

20. A particle starting from the origin (0,0)

(d)

v

v

t

t

24. A body starting from rest moves along a

(a)

moves in a straight line in the xy-plane. Its coordinates at a later time are ( 3, 3). The path of the particle makes with the X-axis an angle of [AIIMS 2007] (b) 45º

t

straight line with a constant acceleration. The variation of speed ( v ) with distance ( s) is represented by the graph [AIIMS 2003]

(b) 10 km and tan-1 ( 5 ) (c) 52 km and tan-1 (5) (d) 52 km and tan-1 ( 5 )

(a) 30º

(b) v

(c) 60º

(d) 0º

(b)

v

v

s

(c)

s

(d) v

v

21. When a ball is thrown up vertically with velocity v 0, it reaches a maximum height of h. If one wishes to triple the maximum height, then the ball should be thrown [AIIMS 2005] with velocity (a) 3 v 0 (c) 9 v 0

(b) 3 v 0 (d) 3 / 2 v 0

22. Which of the following v-t graphs shows a realistic situation for a body in motion? [AIIMS 2004]

s

25. At the uppermost point of a projectile, its velocity and acceleration are at an angle of (a) 180º (c) 60º

v t (a)

(b)

Q = ai$ - 2$j - k$ are perpendicular to each other, then the positive value of a is

t (c)

(b) 1 (d) 3

[AIIMS 2002]

200 m/s to 100 m/s, while travelling through a wooden block of thickness 10 cm. Assuming it to be uniform, the retardation will be [AIIMS 2001] (a) - 15 ´ 104 m / s 2

v t

[AIIMS 2002]

27. The velocity of a bullet is reduced from t

v

(b) 90º (d) 45º

26. If the vectors P = a$i + a$j + 3k$ and

(a) zero (c) 2 v

s

(d)

(b) 15 ´ 104 m / s 2 (c) 12 ´ 104 m / s 2 (d) 14.5 m / s 2

11

Kinematics 28. Two projectiles are projected with the same velocity. If one is projected at an angle of 30º and the other at 60º to the horizontal. The ratio of maximum heights reached, is [AIIMS 2001] (a) 1 : 3 (c) 3 : 1

(b) 2 : 1 (d) 1 : 4

29. A body A starts from rest with an acceleration a1. After 2 s, another body B starts from rest with an acceleration a2 . If they travel equal distances in 5 s, after the start of A, the ratio a1 : a2 will be equal to (a) 9 : 5 (c) 5 : 9

(b) 5 : 7 (d) 7 : 9

[AIIMS 2001]

30. A body is released from the top of the tower H metre high. It takes t second to reach the ground. Where is the body after [AIIMS 2000] t /2 second of release? (a) At 3H / 4 metre from the ground (b) At H / 2 metre from the ground (c) At H / 6 metre from the ground (d) At H / 4 metre from the ground

speed of 144 km/h in 20s, then it covers a distance [AIIMS 1997] (a) 400 m

(b) 1440 m (c) 2880 m (d) 25 m

36. Rain is falling vertically downwards with a velocity of 3 km/h. A man walks in the rain with a velocity of 4 km/h. The rain drop will fall on the man with a velocity of (a) 5 km/h (c) 1 km/h

(b) 4 km/h [AIIMS 1997] (d) 3 km/h

37. Which of the following is constant in a [AIIMS 1996]

projectile motion?

(a) Velocity of projection (b) Horizontal component of the velocity (c) Vertical component of the velocity (d) All of the above

38. Angle between two vectors of magnitudes 12 and 18 units, when their resultant is 24 units is [AIIMS 1996] (a) 82 º 31¢

(b) 63º 51¢

(c) 89º16¢

(d) 75º 52 ¢

39. A body A is dropped vertically from the

31. The angle between P + Q and P - Q will be

35. If a car at rest accelerates uniformly to a

[AIIMS 1999]

(a) 90º only (b) between 0º and 180º (c) 180º only (d) None of these

32. A particle is thrown vertically upwards.

top of a tower. If another identical body B is projected thrown from the same point at the same instant, then [AIIMS 1994] (a) Both A and B will reach the ground simultaneously (b) A will reach the ground earlier than B (c) B will reach the ground earlier than A (d) Either A or B

The velocity at half of the height is 10 m/s, then the maximum height attained by it will be (Take, g = 10 m/s2 ) [AIIMS 1999]

Assertion & Reason

(a) 10 m (c) 15 m

and Reason carefully to mark the correct option from those given below

(b) 20 m (d) 25 m

33. A body is projected at such angle that the horizontal range is three times the greatest height. The angle of projection is [AIIMS 1998] (a) 42 º 8¢ (c) 33 º7 ¢

(b) 53 º7 ¢ (d) 25º 8¢

34. If the water falls from a dam into a turbine wheel 19.6 m below, then the velocity of water at the turbine is (Take, g = 9.8 m/s2 ) (a) 19.6 m/s (c) 98.8 m/s

[AIIMS 1998] (b) 39.0 m/s (d) 9.8 m/s

Direction (Q. Nos 40-48) Read the Assertion

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

40. Assertion When q = 45° or 135°, the value of R remains the same, only the sign changes. Reason R =

u2 sin 2 q g

[AIIMS 2017]

12

AIIMS Chapterwise Solutions ~ Physics

41. Assertion Displacement of a body may be zero, when distance travelled by it is not zero. Reason The displacement is the longer distance between initial and final position. [AIIMS 2014]

42. Assertion A body can have acceleration even if its velocity is zero at that instant of time. Reason The body will be momentarily at rest when it reverses its direction of motion. [AIIMS 2013]

43. Assertion When a body is dropped or

45. Assertion The average velocity of the object over an interval of time is either smaller than or equal to the average speed of the object over the same interval. Reason Velocity is a vector quantity and speed is a scalar quantity. [AIIMS 2008]

46. Assertion The dimensional formula for relative velocity is same as that of the rate of change is velocity. Reason Relative velocity of P w.r.t. Q is the ratio of velocity of P and that of Q. [AIIMS 2002]

47. Assertion The average speed of an object

thrown horizontally from the same height, it would reach the ground at the same time. Reason Horizontal velocity has no effect on the vertical direction. [AIIMS 2012]

44. Assertion The projectile has only vertical component of velocity at the highest point of its trajectory. Reason At the highest point, both components of velocity present.

may be equal to arithmetic mean of individual speeds. Reason Average speeds is equal to total distance travelled per total time taken. [AIIMS 2000]

48. Assertion A body with constant acceleration may not speed up. Reason A body with constant acceleration always moves along a straight line. [AIIMS 1998]

[AIIMS 2011]

Answers 1. (c)

2. (a)

3. (a)

4. (b)

5. (b)

6. (b)

7. (a)

8. (c)

9. (a)

10. (d)

11. (a)

12. (d)

13. (a)

14. (c)

15. (b)

16. (c)

17. (c)

18. (d)

19. (c)

20 (c)

21. (a)

22. (b)

23. (d)

24 (c)

25. (b)

26. (d)

27. (a)

28. (a)

29. (c)

30. (a)

31. (b)

32. (a)

33. (b)

34. (a)

35. (a)

36. (a)

37. (b)

38. (d)

39. (a)

40. (a)

41. (c)

42. (a)

43. (a)

44. (d)

45. (b)

46. (d)

47. (a)

48 (c)

Explanations 1. (c) Equation of given line when velocity is plotted against position with intercept = v 0 and slope v 0 / x0 is v x …(i) v = mx + v 0 Þ v = - 0 + v 0 x0 é v0 ù êQ m = tan q = – ú x0 û ë (negative sign is because the slope is decreasing) On differentiating w.r.t. t, we get v dx v dv …(ii) =- 0× + 0 Þa= - 0 v dt x0 dt x0 Putting the value of v from Eq. (i) in Eq. (ii), we get ù v2 v é v v2 a = - 0 ê - 0 x + v 0 ú = 20 x - 0 x0 ë x0 x0 û x0 which represents a straight line with positive æ – v2 ö slope (v 20 / x20 ) and negative intercept ç 0 ÷. ç x ÷ è 0 ø Only graph (c) has positive slope and negative intercept.

2. (a) Let at any instant of time, the length AB be l. Here, angle q and length l vary with time, then using Pythagoras theorem in DABC, AB2 = AC2 + BC2 2 x + y2 = l 2 C y m

q

x

B

3. (a) Let the bus overtakes the car after time t and at distance s. As the bus starts from rest, initial velocity, u = 0 Acceleration, a = 8 ms -2 According to second equation of motion, 1 1 s = ut + at 2 = ´ 8 ´ t 2 2 2 Uniform motion of the car, v = 16 ms -1 , a = 0ms –2

A

On differentiating both sides w.r.t.t, we get dy dx dl = 2l 2x + 2y dt dt dt As there is no vertical motion of the block, so dy dx dl = 0, = v x and =v dt dt dt \ 2 x v x = 2lv l or vx = v x v v or = vx = æ x ö sin q ç ÷ èl ø

…(ii)

Similarly, s = vt = 16t Solving Eqs. (i) and (ii), we get 1 Þ 16t = ´ 8 ´ t 2 2 Þ 16t = 4t 2 t = 4 s, s = 64 m

4. (b) Let q be the angle between A and B. \ Þ

|A + B|= n|A - B| 2

2

A + B + 2 AB cos q = n A2 + B2 + 2 AB cos (180°- q)

Þ

| A | = | B |= A = B = x 2

2 x (1 + cos q) = n2 × 2 x2 (1 - cos q)

Þ

1 + cos q = n2 - n2 cos q (1 + n2 )cos q = n2 - 1

v

I

…(i)

cos q =

n2 - 1

n2 + 1 æ n2 - 1 ö ÷ Þ q = cos -1 çç 2 ÷ è n + 1ø 5. (b) From Fig, we have vb = velocity of boat = 8 kmh–1 X q

vrb

vb

Y

vr

Z

v rb = resultant velocity of boat = 10 kmh–1 v r = velocity of river Þ v 2rb = vb2 + v 2r Þ vr =

v 2rb – vb2 = 100 – 64 = 6 kmh–1

14

AIIMS Chapterwise Solutions ~ Physics For second ball,

6. (b) Let t be the time after which the angle between two velocities become 90°. Then, velocities of v A and v B at time t are [From, v = u$i + at$j] v A = v1 $i+gt$j v = – v $i + gt$j B

2

v2

O

v1

B A

1 ´ 9.8 ´ (5 - 2)2 2 [ t = (5 - 2) because second ball has fallen after 2 s] 122.5 = 3u + 441 . Þ 3u = 78. 4 Þ u = 261 . m/s 122.5 = u (5 - 2) +

9. (a) The equation of trajectory, y

B

v2

b

x2

vB gt

x1 d

Q P

A a v1 v gt A

q Q

Since, v A ^vB , v A × vB = 0 Þ (- v1 i$ + g t$j )× (- v2 $i + g t$j ) = 0 - v1 v2 + g 2t 2 = 0 Þ t =

7. (a)

y

xù é y = x tan a ê1 - ú gives Rû ë

v1 v2 g and

a

é Qù P = Q tan q ê1 - ú Rû ë

…(ii)

Q2 P2 x

R 2

R=

3 H max 3u2 / 8g = R 2 3u2 / 4g 2 æ 3ö ÷ b = tan-1 çç ÷ è 2 ø æ 3ö ÷. Thus, the angle of elevator is tan-1 çç ÷ è 2 ø 8. (c) Let the ball hit water in t second. For first ball, 1 s = ut + at 2 2 1 122.5 = 0 + ´ 9.8 ´ t 2 = 4.9 t 2 2 122. 5 Þ t = = 25 = 5 s 4.9

[1 - P / R] [1 - Q / R]

P 3 - Q3 P 2 - Q2

=

P 2 + PQ + Q 2 P+Q

Put the value of R in Eq. (i), we get ù é Q Q2 = tan q ê 2 2ú P P + PQ + Q û ë

3 u2 R /2 = 4 g tanb =

=

1 3 [P - Q 3 ] = P 2 - Q 2 R

u2 sin2 60° 3u2 = 2g 8g

3 u2 u2 sin(2 ´ 60° ) And range, R = = × 2 g g

\

…(i)

Hmax

Maximum height, H max =

\

Pù é Q = P tan q ê1 - ú Rû ë

On dividing Eq. (i) by Eq. (ii), we get

60° b O

x

P

Þ

é P 2 + PQ + Q 2 ù q = tan-1 ê ú PQ û ë

10. (d) Let v1 and v2 be the speed in first 2s and in next 4s, respectively. As we know, speed = area under the s -t graph 1 v1 2 ´ 2 ´ s0 Þ = v2 1 ´ 4 ´ s 0 2 v1 1 = Þ v2 2 or

v1 : v2 = 1: 2

15

Kinematics 11. (a) Distance travelled in nth second, a a sn = u + (2n - 1) = 0 + (2n - 1) 2 2 Also, distance covered in n seconds, 1 1 s = ut + at 2 = 0 + an2 2 2 On dividing Eq. (i) by Eq. (ii), we get a (2n - 1) sn 2 2 1 = = - 2 n n s an2 /2

As acceleration because …(i)

13. (a) Speed of rocket after 5 s, v = u - gt 0 = u - 9.8 ´ 5Þ u = 49 m/s 1 Also, h1 = ut - gt 2 2 1 245 m = 0 - ´ 9.8 ´ (5)2 = 2 2 When engine is turned off, v 2 = u2 - 2gh2 0 = u2 - 2gh2 u2 49 ´ 49 245 m = = 2g 2 ´ 9.8 2

Maximum height from earth surface 245 245 = h1 + h2 = + = 245m 2 2 $ $ $ 14. (c) Let a = 2i + 3 j + 8k b = 4$j - 4$i + a k$ = -4$i + 4$j + a k$ According to question, a ^ b Þ a × b = 0 Þ (2 i$ + 3$j + 8k$ ) × (-4 i$ + 4$j + ak$ ) = 0 Þ -8 + 12 + 8 a = 0 Þ 8 a = -4 4 1 \ a=- =8 2

æg ö v = 2ç ÷ h= è 8ø

…(ii)

æ dx ö \ Velocity, v = ç ÷ = b1 + 2b2t è dt ø dv d 2 x = 2 = 2 b2 \ Acceleration, a = dt dt

h2 =

æg ö v2 = 2 ç ÷ h è 9ø

so,

12. (d) Given, distance, x = b 0 + b1t + b2t 2

gh 2

When the stone released from this balloon. It gh will go upward with velocity, v = . 2 In this condition, time taken by stone to reach the ground, 2gh ù v é t = ê1 + 1 + 2 ú g ë v û gh é 2gh ù h 2 t = ê1 + 1 + ú=2 g êë gh / 4 úû g

16. (c) Since, each of N- coplanar vector is inclined at 2p / N to the preceding hence, they will form a closed polygon. Therefore, their resultant must be zero.

17. (c) Let car starts from point A from rest and moves up to point B with acceleration f . A

s

B

x

C

y

D

[As, v 2 = u2 + 2as]

Velocity at B, v = 2 fs

Car moves with this constant velocity to C, then distance, x = 2 fs × t [As, s = ut ] …(i) So, velocity at C is 2fs and car stops after y distance at D. Then é As v 2 = u2 + 2asù ( 2 fs )2 Distance, y = ú ê = 2s 2( f /2) êëÞ 0 = u2 – 2as úû Total distance, AD = AB + BC + CD = 15s (given) Þ s + x + 2 s = 15s Þ x = 12s Put value of x in Eq (i), we get 1 2 12s = 2 fs × t Þ 144s2 = 2 fst 2 Þ s = ft 72

18. (d) A projectile can have same range, if angles of projection are complementary i.e., q and (90º -q).

15. (b) Since, balloon rises from rest, therefore, u

u = 0. The velocity of balloon at height h, using equation, v 2 - u2 = 2gh

u

v 2 - (0)2 = 2gh v 2 = 2gh

g , 8

O

90°– q q

16

AIIMS Chapterwise Solutions ~ Physics In both cases, 2 u sin q t1 = g 2u sin (90º - q) 2ucos q t2 = = g g

20. (c) Draw the situation as shown. OA represents …(i) …(ii)

the path of the particle starting from origin O (0,0). Draw a perpendicular from point A to X-axis. Let path of the particle makes an angle q with the X-axis, then y

From Eqs. (i) and (ii), we get 4u2 sin qcos q t1 t2 = g2

A

(Ö3,3)

2

t1 t2 = =

2u sin2q

[Q sin2q = 2sin qcos q]

g2 2 g

æ u2 sin2q ö 2R æç u2 sin2q ö÷ ÷= ç R \ = ÷ ç ÷ g g g çè ø è ø

q O (0, 0)

Hence, t1t2 µ R

19. (c) The given condition has been shown in the figure given below, Þ

B 4 135° A q

6 a

B

x

tan q = slope of line OA AB 3 tan q = = = 3 OB 3 {Q tan 60°= 3} tan q = tan 60° q = 60º

21. (a) At maximum height, velocity is zero. From equation of motion, we have v 2 = u2 - 2gh where, v is final velocity and u is initial velocity. v = 0, u = v 0

45°

O

0 = v 20 - 2gh Þ From calculations, Applying pythagoras theorem in DOAB, we get OB2 = OA2 + AB2 or

OB =

52 km

Hence, car is at 52 km from the starting point. In DOAB, AB 4 2 æ2ö tan q = = = or q = tan-1 ç ÷ OA 6 3 è3ø tan 45° + tan q \tan a = tan (45° + q) = 1 - (tan 45° ) tan q 1 + tan q 1 + 2/3 5/3 = = = 1 - tan q 1 - 2/3 1/3 or

v 0¢ = 2g ´ 3h = 3 v 0 Note: If ball has to be thrown to a greater height, its initial velocity should be more than the original one.

62 + 42

= 36 + 16 =

v 0 = 2gh

When h¢= 3h, then

=5 a = tan-1 (5)

Hence, the angle straight line joining its initial and final position makes with the east is tan-1 (5.)

22. (b) Except graph (b), the other three graphs shows the motion of the body with more than one velocity at a particular time which is not possible for realistic situation.

23. (d) Speed of body at highest point is zero. From equation of motion, we have v 2 = u2 - 2gh When u = 0, then v 2 = -2gh As ball is thrown upwards, velocity decreases as height increases. At the highest point, velocity is zero. After that, ball comes downwards and its speed increases. If air resistance is considered, then it is represented by graph (d).

17

Kinematics 24. (c) From third equation of motion, we have v 2 = u2 + 2as Since, body starts from rest u = 0 \ v 2 = 2as Which is general equation of parabola symmetrical along X-axis.

28. (a) If body is projected with initial velocity u at an angle q with the horizontal, then the height H reached by projectile is given by u

u

P

ux

g q

H1

60° 30°

25. (b) When a body is projected at an angle q, then it follows a projectile motion. At the highest point P of its path, the vertical component of velocity becomes zero, while horizontal component prevails, also acceleration due to gravity g always acts vertically downwards, hence angle between velocity and acceleration is 90º.

H2

u

A

B 2

H=

2

u sin q 2g

where, g is acceleration due to gravity. Given, q1 = 30º and q2 = 60º 2 æ1ö ç ÷ 2 2 H1 sin q1 sin (30º ) 1 2 \ = = = è ø = H2 sin2 q2 sin2 (60º ) æ 3 ö2 3 ÷ ç ç 2 ÷ è ø

29. (c) The distance covered by the body in the nth a (2n - 1) 2 where, u is initial velocity and a is acceleration. Distance covered by the body A in fifth second after its start with acceleration a1 , is a 9a (s5 )A = 0 + 1 (2 ´ 5 - 1) = 1 2 2 Time taken by second body = 5 - 2 = 3 s a 5a (s3 )B = 0 + 2 (2 ´ 3 - 1) = 2 2 2 Given, (s5 )A = (s3 )B a1 5 \ = a2 9 second of motion is sn = u +

26. (d) The scalar product of two vectors is given by P × Q = PQ cos q where, q is angle between them. Given, P = ai$ + a$j + 3k$ Q = ai$ - 2$j - k$ q = 90º \

P × Q = PQ cos 90º = 0 $ × (ai$ - 2$j - k) $ =0 (a$i + a$j + 3k)

a2 - 2a - 3 = 0 2 ± 4 + 12 a= = 3, - 1 Þ 2 Hence, positive value of a is 3.

27. (a) From equation of motion, …(i) v 2 = u2 + 2as where, v is final velocity, u is initial velocity, a is retardation and s is distance travel. Given, v = 100 m/s, u = 200 m/s, s = 10 cm = 10 ´ 10-2 m Putting the numerical values in Eq. (i), we have \ (100)2 = (200)2 + 2 ´ a ´ 10 ´ 10-2 Þ Þ

- 30000 = 20a ´ 10-2 30000 a= 20 ´ 10-2 = -15 ´ 104 m/s2

Negative sign shows negative acceleration, i .e . retardation.

A

30. (a) Since, the body is released from rest, initial velocity, u = 0. From equation of motion, we have 1 H = ut + gt 2 2 where, g is acceleration due to gravity. We get, u = 0 1 …(i) H = gt 2 \ 2 t Height of body at time is 2 h=

1 æ t ö2 1 t 2 H g ç ÷ = g = 2 4 4 2 è2ø

h H

C

B

[from Eq. (i)]

Hence, height of body from the ground H 3H metre =H= 4 4

18

AIIMS Chapterwise Solutions ~ Physics

31. (b) For a parallelogram with sides P and Q, the diagonals are P + Q and P - Q. Angle between diagonals can be any value between 0º and 180º.

32. (a) At the highest point of its path, final velocity is zero. From third equation of motion, we have v 2 = u2 - 2gH For v = 0, at highest point u2 = 2gH Þ H =

u2 2g

Also, velocity at half the height is 10 m/s From v 2 = u2 - 2gh, we have 2 1æu ö (10)2 = u2 - 2 ´ g ´ ç ÷ 2 çè 2g ÷ø

Hö æ çQ h = ÷ 2ø è

u2 Þ u = 200 m/s 2 Maximum height obtained is u2 200 = = 10 m H max = 2g 2 ´ 10

Þ

100 =

Since u = 0, we have v = 2gh v = 2 ´ 9.8 ´ 19.6 = 38416 . = 19.6 m/s

35. (a) Since car starts from rest, so initial velocity is zero. From first equation of motion, we have v = u + at where, v is final velocity, u is initial velocity, a is acceleration and t is time. 5 m/s = 40 m/s Given, v = 144 km/h = 144 ´ 18 u = 0 and t = 20 s \ 40 = 0 + 20 ´ a Þ a = 2 m / s2 From second equation of motion, 1 s = ut + at 2 2 1 s = 0 + ´ 2 (20)2 = 400 m \ 2 36. (a) The situation is shown in figure below. 4 km/hr

vm

33. (b) Horizontal range is R=

u2 sin2 q g

3 km/hr

…(i)

vrm vr

and height H is H=

u2 sin2 q 2g

…(ii)

where, u be the initial velocity with which the body is projected at an angle q. Given, R = 3H u2 sin2q u2 sin2 q = 3´ \ 2g g

\

(also, sin2q = 2 sin qcos q) u2 2 sin q cos q u2 sin2 q = 3´ 2g g

Þ

2cos q = 1.5 sin q 2 tan q = = 133 . Þ 1.5 Þ q = 53º7¢ Hence, angle of projection is 53º7¢.

Relative velocity of rain w.r.t man, v rm = =

v 2r + v 2m = (3)2 + (4)2 = 5 km/h

37. (b) If a body is in the projectile motion, then its velocity can be resolved into horizontal and vertical components. And the body is subjected to acceleration due to gravity (g), which is vertically downwards. Therefore, horizontal component remains constant, while vertical component first decreases, becomes zero and then increases in the magnitude.

38. (d) Given, magnitude of first vector (A ) = 12 Magnitude of second vector (B) = 18 and resultant of the given vectors (R ) = 24. We know that resultant vector

34. (a) From equation of motion, we have

| R |= 24 =

v 2 = u2 + 2gh where, v is final velocity, u is initial velocity, g is gravity and h is height.

v 2r + v 2m + 2v r v m cos 90º {Qcos 90°= 0}

A2 + B2 + 2 AB cos q

= (12)2 + (18)2 + 2 ´ 12 ´ 18 cos q or

2

(24) = 144 + 324 + 432cos q

19

Kinematics or or or or or

576 = 468 + 432 cos q 576 - 468 = 432 cos q 432 cos q = 108 108 cos q = = 025 . 432 q = cos -1 0.25 = 75º 52¢

39. (a) As the body A is dropped from rest, \

h=

2h 1 2 gt Þ t A = 2 A g B

h

the earth because vertical downward component of velocity for both the bodies will 2h be zero and time of decent, t = . g Horizontal velocity has no effect on the vertical direction.

44. (d) Both Assertion and Reason are false. At the . highest point, the projectile possesses velocity only along horizontal direction.

45. (b) As displacement is either smaller or equal to distance but never be greater than distance, so average velocity of object is smaller than or equal to its average speed over the same interval.

A

O

43. (a) Both the bodies will take same time to reach

Y

As B is projected from top of tower, its initial velocity is zero. \ v 2B = 0 + 2gh Þ v B = 2gh Also, v B = 0 + gt B 2gh 2h Þ tB = = =tA g g u2 sin 2q 40. (a) Horizontal range (R ) = g When q = 45°, maximum horizontal range, u2 u2 sin 90° = Rmax = g g When q = 135°, maximum horizontal range, -u2 u2 sin 270° = R= g g Negative sign implies opposite direction.

41. (c) Here, Assertion is true but Reason is false. The displacement is the shortest distance between initial and final positions. When final position of a body coincides with, its initial position displacement is zero, but the distance travelled is non-zero.

42. (a) When a body while going vertically upward reaches at the highest point, then it will be momentarily at rest and then it reverses its direction. At highest point, velocity is zero but its acceleration is equal to acceleration due to gravity.

46. (d) Both Assertion and Reason are false. Relative velocity of P with respect to Q = Velocity of P - Velocity of Q Hence, relative velocity will have dimension of velocity [LT -1 ]. Also, rate of change of velocity is acceleration. Hence, it has dimension of [LT -2 ].

47. (a) When a particle moves with speeds v1 , v2 , v3 ..... during time intervals t1 , t2 , t3 ... respectively, then total distance travelled v1t1 + v2t2 + v3t3 +... and total time taken by particle = t1 + t2 + t3 + ... Hence, average speed is given by v t + v2t2 + v3t3 + ... vav = 1 1 t1 + t2 + t3 + ... (Let, t1 = t2 = t3 ... = t ) (v + v2 + v3 + ... ) t Then, vav = 1 nt Therefore, average speed is the arithmetic mean of individual speeds.

48. (c) Here, Assertion is true but Reason is false. When a body is projected up with an initial velocity making an angle q with the horizontal, then it follows a projectile motion path. In the uniform motion, acceleration is constant, also its speed first decreases, vertical component becomes zero at highest point and then increases.

AIIMS Chapterwise Solutions ~ Physics CHAPTER

3 Laws of Motion 1. A lift is moving in upward direction. The total mass of the lift and the passengers is 1600 kg. The variation of the velocity of the lift is as shown in the figure. The tension in the rope at t = 8 s will be [AIIMS 2018] v (ms–1) 12 0

3

(a) 11200 N (c) 48000 N

6 10 t(s)

(b) 16000 N (d) 12000 N

2. In the given system, calculate the acceleration of the light and smooth pulley. [AIIMS 2017] C

B F A M Smooth

(a) F / M (c) F / 4M

(b) F / 2 M (d) F / 8M

3. A box is placed on an inclined plane and has to be pushed down. The angle of inclination is [AIIMS 2017] (a) equal to the angle of friction (b) more than the angle of friction (c) equal to the angle of repose (d) less than the angle of repose

4. A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth’s surface. A block B placed at the top of the wedge takes time T to slide down the length of the wedge. We place the block at the top of the wedge and the cable supporting the chamber is broken at the same instant, the block will [AIIMS 2016] (a) take longer time than T to slide down the wedge (b) take shorter time than T to slide down the wedge (c) remain at the top of the wedge (d) jump off the wedge

5. A boy of mass M is applying a horizontal force to slide a box of mass M¢ on a rough horizontal surface. The coefficient of friction between the shoes of the boy and the floor is m and that between the box and the floor is m¢. In which of the following cases is certainly not possible to slide the box? [AIIMS 2016] (a) m < m ¢ , M < M¢ (c) m < m ¢, M > M ¢

(b) m >m ¢ , M < M ¢ (d) m > m ¢ , M > M ¢

6. A body of mass 40 kg resting on rough horizontal surface is subjected to a force P which is just enough to start the motion of the body. If m s = 0.5, m k = 0.4, g = 10 m/s2 and the force P is continuously applied on the body, then the acceleration of the body is [AIIMS 2015] (a) zero (c) 2 m/s 2

(b) 1 m/s 2 (d) 2.4 m/s 2

21

Laws of Motion 7. A hemispherical bowl of radius r is set rotating about its axis of symmetry in vertical plane. A small block kept in the bowl rotates with the bowl without slipping on its surface. If the surface of the bowl is smooth and the angle made by the radius through the block with the vertical is q, then find the angular speed at which the bowl is rotating. (a) w =

rg sin q

[AIIMS 2015] g (b) w = r cos q

(c) w =

gr cos q

(d) w =

gr tan q

12. A monkey of mass M kg slides down a light rope attached to a fixed spring balance, with an acceleration a. The reading of the balance is [where, g = acceleration due to gravity] [AIIMS 2009] æ aö ç (b) M = w ç 1 + ÷÷ gø è (c) the force of friction exerted by the rope on the monkey is M (g - a) N (d) the tension in the rope is wg N wg (a) M = g -a

13. The graph shows the variation of velocity of a rocket with time. The time of burning of fuel from the graph is [AIIMS 2009] v(m/s) 1000

8. A weight w is suspended from the mid-point of a rope, whose ends are at the same level. In order to make the rope perfectly horizontal, the force applied to each of its ends must be [AIIMS 2014] (b) equal to w (d) infinitely large

(a) less than w (c) equal to 2w

9. Pulling force making an angle q to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is a, then the magnitude of force required to move the body is equal to [AIIMS 2014] W sin a g tan(q - a) W sin a (c) cos(q - a)

W cos a cos(q - a) W tan a (d) sin(q - a)

(a)

(b)

10. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is [AIIMS 2012] (a) 0.01

(b) 0.02

(c) 0.03

(d) 0.06

11. Consider the situation as shown. The maximum value of force F such that the block does not move [AIIMS 2012] F 60° Ö3 kg

(a) 5 N

(b) 10 N

(c) 15 N

(d) 20 N

O

A 110 120 B C

10

(a) 10 s (c) 120 s

t(s)

(b) 110 s (d) Data insufficient

14. A uniform rope of length L meters is lying over a table. If the coefficient of friction be m, then the maximum length L1 of this rope which can overhang the edge without sliding is [AIIMS 2009] (a)

mL (m + 1)

(b)

(m + 1) mL

(c)

mL (m - 1)

(d)

(m -1) L m

15. A smooth block is released from rest on a 45° incline and then slides a distance d. The time taken to slide is n times as much to slide on rough incline than on a smooth incline. The coefficient of friction is [AIIMS 2008] (a) m k = 1 (c) m s = 1 -

1 n2 1 n2

(b) m k = 1 (d) m s = 1 -

1 n2 1

n2

16. A person is standing in an elevator. In which situation he finds himself weight less? [AIIMS 2005] (a) When the elevator moves upward with constant acceleration equals to g. (b) When the elevator moves downward with constant acceleration equals to g. (c) When the elevator moves upward with uniform velocity (d) When the elevator moves downward with uniform velocity

22

AIIMS Chapterwise Solutions ~ Physics

17. In the figure given the position-time graph of a particle of mass 0.1 kg is shown. The impulse at t =2s is [AIIMS 2005] 6 4

x(m)

2 O

2

t (s)

4

6

(a) 0.2 kg m s - 1 (b) - 0.2 kg m s - 1 (c) 0.1 kg m s - 1 (d) - 0.4 kg m s - 1

plane of inclination q and friction coefficient m. If the time of ascent is half of the time of descent, the friction coefficient [AIIMS 2005] m is equal to 3 tanq 4 4 (c) tanq 5

(a)

3 tan q 5 1 (d) tan q 2

(b)

7N

The coefficient of friction between the blocks is 0.5. The acceleration of the lower block is [AIIMS 2005] (b) 1.5 m/s 2 (d) 2.5 m/s 2

20. A stone tied to a string is rotated with a uniform speed in a vertical plane. If mass of the stone is m, the length of the string is r and the linear speed of the stone is v, when the stone is at its lowest point, then the tension in the string at its lowest point, will be (g = acceleration due to gravity) [AIIMS 2001] (b)

on levelled road can take a sharp circular turn of radius 4 m, then the coefficient of friction between cycle tyre and road will be [AIIMS 1999] (b) 0.41

(c) 0.71

(d) 0.61

some velocity. A constant air resistance acts. If the time of ascent is t 1 and that of descent is t2 , then [AIIMS 1999] (b) t 1 > t 2 (d) None of these

24. A 1 kg particle strikes a wall with velocity

Smooth

mv 2 + mg r mv (c) r

22. If a cyclist moving with a speed of 4.9 m/s

(a) t 1 < t 2 (c) t 1 = t 2

5 kg

(a)

(b)

23. A ball is thrown vertically upward with

diagram.

(a) 1 m/s 2 (c) 2 m/s 2

m3

g(1 - 2m) 3 g (1 - gm) (d) 9

2 gm 3 g(1 - 2m) (c) 2

(a) 0.81

19. Consider the situation as shown in the 2 kg

m2

three masses m1, m2 and m3 connected by a string passing over a pulley P. The mass m1 hangs freely and m2 m1 and m3 are on a rough horizontal table (the coefficient of friction = m). The pulley is frictionless and of negligible mass. The downward acceleration of mass m1 is [AIIMS 2001] (Assume, m1 = m2 = m3 = m) (a)

18. A particle is projected up a rough inclined

P

21. A system consists of

mv 2 - mg r

(d) mg

1 m/s at an angle 30° with horizontal direction and reflects at the same wall in 0.1s, then the force will be [AIIMS 1999] (a) 30 3 N (b) zero

(c) 40 3 N (d) 10 3 N

25. The rocket engine lift a rocket from the earth, because hot gases

[AIIMS 1998]

(a) push it against the air with very high velocity (b) push it against the earth with very high velocity (c) heat up the air which lifts the rocket with very high velocity (d) react against rocket and push it up with very high velocity

26. If the force on a rocket, moving with a velocity 500 m/s is 400 N, then the rate of combustion of the fuel will be [AIIMS 1997] (a) 0.8 kg/s (c) 8 kg/s

(b) 10.8 kg/s (d) 1.6 kg/s

23

Laws of Motion 27. If the radii of circular paths of two particles of same masses are in the ratio 1 : 2, then to have a constant centripetal force, their velocities should be in a ratio of [AIIMS 1996] (a) 4:1 (c) 1 : 4

(b) 1 : 2 (d) 2 : 1

only on the rate of decrease of mass. Reason Thrust also depends upon exhaust speed of the gases. [AIIMS 2012]

33. Assertion The centripetal forces and

28. A boy of mass 40 kg is standing in a lift, which is moving downwards with an acceleration 9.8 m/s2 . The apparent weight of the boy is (take g = 9.8 m/s2 ) [AIIMS 1996] (a) 40 ´ 9.8 N 40 (c) N 9.8

32. Assertion Thrust on a rocket depends not

centrifugal forces never cancel out. Reason They do not act at the same time. [AIIMS 2010]

34. Assertion A safe turn by a cyclist should neither be fast nor sharp. Reason The bending angle from the vertical would decrease with increase in velocity. [AIIMS 2010]

(b) 0 N (d) 40 N

Assertion & Reason

35. Assertion Angle of repose is equal to the

Direction (Q. Nos. 29-39) Read the Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion (c) Assertion is true, but Reason is false (d) Both Assertion and Reason are false

29. Assertion Linear momentum of a body changes even when it is moving uniformly in a circle. Reason In uniform circular motion, velocity remains constant. [AIIMS 2015]

30. Assertion If large number of concurrent forces acting on the same point, then the point will be in equilibrium, if sum of all the forces is equal to zero. Reason A body subjected to three concurrent forces cannot be in equilibrium. [AIIMS 2012]

31. Assertion Two similar trains are moving along the equatorial line with the same speed but in opposite direction. They will exert equal pressure on the rails. Reason In uniform circular motion the magnitude of acceleration remains constant but the direction does not change. [AIIMS 2012]

angle of limiting friction. Reason When the body is just at the point of motion, the force of friction in this stage is called limiting friction. [AIIMS 2008]

36. Assertion The acceleration of a body down a rough inclined plane is greater than the acceleration due to gravity. Reason The body is able to slide on a inclined plane only when its acceleration is greater than acceleration due to gravity. [AIIMS 2007]

37. Assertion The driver in a vehicle moving with a constant speed on a straight road is in a non-inertial frame of reference. Reason A reference frame in which Newton’s laws of motion are applicable is non-inertial. [AIIMS 2004]

38. Assertion A reference frame attached to earth is an inertial frame of reference. Reason The reference frame which has zero acceleration is called a non inertial frame of reference. [AIIMS 2004]

39. Assertion On a rainy day, it is difficult to drive a car or bus at high speed. Reason The value of coefficient of friction is lowered due to wetting of the surface. [AIIMS 2004]

AIIMS Chapterwise Solutions ~ Physics

Answers 1. 11. 21. 31.

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

2. 12. 22. 32.

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

3. 13. 23. 33.

4. 14. 24. 34.

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

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

5. 15. 25. 35.

(a) (a) (b) (b)

6. 16. 26. 36.

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

7. 17. 27. 37.

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

8. 18. 28. 38.

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

9. 19. 29. 39.

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

10. (d) 20. (a) 30. (c)

Explanations FBD of the block of mass M,

1. (a) In given v-t graph,

N

a T

M v 12 A (m/s) 0

Mg

B D 6

3

M

C 10 t(s)

8

The acceleration of the lift and the passengers at t = 8 s, (In DBDC) a = slope of v - t graph at t = 8 s = - BD / DC 12 == - 3 m/s2 4 Negative sign implies that the acceleration is in vertically downward direction.

T

a = 3 m/s2

Free body diagram of the lift and the passengers at t = 8 s is shown below,

Mg

Applying Newton’s 2nd law Mg - T = Ma \

Smooth

Applying Newton’s 2nd law, N = Mg and T = Ma

...(i)

FBD of the pulley, a1

T

F T

F - 2 T = 0 ´ a1 Þ

(Q Pulley is light)

F = 2T = 2Ma

...(ii) [using, Eq. (i)]

Þ a = F /2M We know that, if acceleration of a pulley is a, then the acceleration of it’s one free arm will be 2a while its second arm is fixed. Thus, the acceleration of the pulley, a a1 = (acceleration of the block) 2 F F [from Eq. (ii)] = = 2 ´ 2M 4M

3. (d) Free body diagram of the box,

T = M(g - a)

(given, M = 1600 kg)

= 1600(10 - 3) = 1600 ´ 7 = 11200 N

2. (c) Let the acceleration of the block be a and acceleration of the pulley be a1 in the given system. T a M

N

(putting values)

T T Smooth

a1 F

q sin Mg q

fr q Mg cos q Mg

Q The box has to be pushed down, this means f r ³ Mg sin q (Q f r = mN ) Þ mN ³ Mg sin q (Q N = Mg cos q) Þ m Mg cos q ³ Mg sin q Þ m ³ tan q Since we know, m = tanl where, l angle of repose.

25

Laws of Motion Þ tan l ³ tan q \ l³q So, angle of repose ³ angle of inclination

In 2nd case, when the body is moving. FBD of the body, N a

4. (c) According to the question, when the

P

supporting cable is broken, the whole system (the chamber, the wedge and the block) is in the state of free fall.

g

mg

Let the acceleration of the body be a. Applying Newton’s 2nd law,

B

A

fk

g (free fall)

Þ

g (free fall)

As it is known that free fall is the motion of an object, where the gravity is only force acting upon it. So, the block being under the state of free fall does not move from its original position. Thus, the block remains at the top of the wedge.

5. (a) Given, M M¢ mass of the boy = M mass of the block = M ¢ f1 f2 The coefficient of friction between the boy and the surface is m and that between the block and the surface is m ¢. As, given situation where, f1 = frictional force between the boy and the surface. f2 = frictional force between the block and the surface. For the box pushed by the boy, f r1 > f r2 [Q f r = mN = m Mg] Þ m Mg > m ¢M ¢g Þ mM > m ¢M ¢ If m < m ¢ and M < M ¢, the block will not move certainly.

6. (b) Given, mass of the body, m = 40 kg, m s = 0.5 and m k = 0.4 N In Ist case, when motion of the body has just been started by the force P. P FBD of the body, fs ...(i) \ N = mg mg and P = fs =m s N [Q from Eq. (i)] = m s mg (Putting values) = 0.5´ 40 ´ 10 ...(ii) = 200 N

P - f k = ma 200 - 160 (putting values) a= 40 40 = = 1 m/s2 Þ a= 1 m/s2 40

7. (b) Let the angular speed of the bowl be w. The situation according to the question, On applying Newton’s 2nd law for the given FBD, P - Fk …(i) Ma = P - f k Þ a= M P - m k Mg …(ii) and Ma = P - m k M g Þ a = M from Eqs. (i) and (ii), we get P – fk P – m k Mg (Q f r = m k Mg ) = \ a= M M N cos q O N

q A

q

N sin q

B

Mg

where, Mg = weight of the block N, Normal reaction force between the block and the hemisphere. AO, the radius of hemisphere = r AB AB In DAOB, sin q = = \ AB = r sin q ...(i) AO r This is the radius of the circle at which the block is rotating. Resolving the normal reaction force N, along AB and perpendicular to AB, we have, N sin q = The centripetal force N sin q = M( AB )w2 = Mr sin q w2 N = Mrw2 and

N cos q = Mg

...(ii) ...(iii)

26

AIIMS Chapterwise Solutions ~ Physics 10. (d) Given, m = 2 kg, v = 6 m/s and t = 10 s

From Eqs. (ii) and (iii), we get Mr w2 N = N cos q Mg Þ \

g 1 w2 r = Þ w2 = cos q g r cos q w=

g r cos q

8. (d) Let the force T is applied on the both ends of the rope. According to the question, A

B 2T cos q T

q

q

T

T sin q

T sin q W

Thus, we have, Fnet in horizontal direction is zero. 2T cos q= Wg Wg T= \ 2 cos q For the rope to the horizontal, q will be equal to 90°. Wg Wg [Qcos 90° = 0] We have, T = = 2 cos 90° 2 ´ 0 = Wg / 0 Thus, T will be infinitely large.

Let the coefficient of friction be m and the acceleration of the block due to friction be a. N Due to the motion of v=6 m/s the block, the frictional force produces an acceleration ‘a’ opposite to the velocity ‘v’ of the block. fr FBD of the block mg according to the question, This acceleration would stop the block. Applying Newton’s 2nd law, f r = ma and N = mg Þ mN = ma Þ m mg = ma ...(i) \ a = mg Using v = u - at , Q v = 0, 0 = 6 - mg ´ 10 6 (Q g = 10 m /s2 ) Þ m= 10 ´ 10 = 0.06

11. (d) FBD of the block, F sin 60° N

F 60°

Ö3 kg f

9. (c) According the question,

Ö3 g

FBD of the block, N + F sin q = W Þ N = W - F sin q The block will move if F cos q ³ f max

3F and 2 3Fö ÷ 2 ÷ø

N = 3g + F sin 60° = 3g +

F N

Þ F cos q ³ m(W - F sin q) Since we know, sin a m = tan a = f cos a sin a W (W - F sin q) \F cos q ³ cos a F(cos qcos a + sin q sin a) ³ W sin a W sin a F ³ cos(q - a) W sin a Fmin = cos(q - a)

q

æ ç 3g + ç è 1æ Fö F = ç 10 + ÷ = 5 + 2è 2ø 4

f max = mN =

F sin q

F cos q

1 2 3

The block will not move if F cos 60° £ f max . F F £ 5+ 2 4 F F - £5 Þ 2 4 F Þ £5 4 F £ 20 \ Fmax = 20 N

27

Laws of Motion 12. (c) Given, mass of monkey = M kg Acceleration of the monkey = a ms - 2

15. (a) As given in the question, Let the mass of the block is m, For smooth inclined plane, N fr

Spring balance T M T Monkey

m

a

Mg

Here, the reading of the spring balance will be equal to the tension in the rope. Also, the tension in the upper part of the rope is due to frictional force between the monkey and the rope. Applying Newton’s 2nd law, we get Mg - T = Ma \ T = Mg - Ma = M(g - a) N

g

q mg cos q q n mg i q s

N = mg cos q and mg sin q= ma \a = g sin q Q The block starts from rest 1 Using, s = ut + at 2 2 Given, s = d 1 d = 0 + g sin q × t12 2 1 d = g sin qt12 2 For rough inclined plane, N

13. (a) Until the fuel is burning the rocket is accelerating in upward direction. After the fuel is exhausted, the rocket starts moving under the gravity. From the given velocity vs time graph of the rocket, the rocket is accelerating in upward direction from t = 0 to t = 10 sec. Thus, the time of burning of fuel is 10 sec.

14. (a) According to the question, N f m (L–L )g 1 L

L1 m Lg L 1

m L g £ f max L 1 m m Q f max = mN = m × (L - L1 ) g [Q N = (L - L1 ) g ] L L m m L g £ m (L - L1 ) g \ L L 1 Þ L1 £ m (L - L1 ) Þ L1 (m 1 + 1) £ mL mL L1 £ (m + 1) mL L1(max) = \ (m + 1) For no sliding,

...(i)

q sin mg q

fr q mg mg cos q

N = mg cos q and mg sin q - f r = ma1 mg sin q - m mg cos q = ma1 [Q f r = mN = m mg cos q] \ a1 = g sin q - mg cos q 1 Using, s = ut + at 2 , we have 2 1 d = 0 + g (sin q - m cos q) t22 2 1 ...(ii) d = g (sin q - m cos q) t22 2 From Eqs. (i) and (ii), 1 1 g sin q× t12 = (sin q - m cos q) gt22 2 2 Q t2 = nt1 \

sin qt12 = (sin q - m cos q) n2t12

Þ

sin 45° = (sin 45° - m cos 45° )n2 (Q q = 45°) 1 1 = (1 - m)n2 2 2 1 =1-m n2 1 m k =1 - 2 n

Þ Þ Þ\

28

AIIMS Chapterwise Solutions ~ Physics

16. (b) Let the elevator moving with acceleration a in downward direction.

a

\Retardation, a1 = g sin q + mg cos q and time, t1 = t For downward motion, accelerating force on the block,

...(i)

mg sin q - f r2 = ma2 Q f r2 = mN and N = mg cos q Free body diagram of the man in that elevator, Applying Newton’s 2nd law, mg - N = ma \ N = m(g - a) N a mg

N will be zero, only when a = g , at that condition, the man finds himself in weightlessness.

17. (b) If a constant force F is applied on a body for a short interval of time Dt, then impulse of this force is F ´ Dt . From Newton’s second law, Dv F = ma = m Dt Þ F Dt = mDv = Dp Þ I = Change in linear momentum, Dp (Q mass = constant) \ I = mDv = m(v2 - v1 ) Dx ö é Dx ù æ = mç 0 ÷ Q v = 0 and v1 = Dt ø êë 2 Dt úû è æ 4ö = 0.1 ´ ç - ÷ = - 0.2 kg-m/s è 2ø

18. (b) For upward motion, g vin mo

m

g

u

d ar pw

q fr sin q

fr

N

m

Þ Þ\

g

s in

q

q mg

mgcos q

mg sin q - mN = ma2 mg sin q - m mg cos q = ma2 a2 = g sin q - mg cos q

and time t2 = 2t 1 1 s = a1t12 and s = a2t22 Q 2 2 \ a1t12 = a2t22 Þ

a1t 2 = a2 (2t )2 Þ a1 = 4a2

a1 = 4a2 From Eqs. (i), (ii) and (iii) we get g sin q + mg cos q = 4(g sin q - mg cos q) Þ 5 mg cos q = 3g sin q 3 tan q m= Þ\ 5

mg cos q

mg

The retarding force on the block, mg sin q+ f r = ma1 Q f r = mN = m mg cos q (Q N = mg cos q)

…(iii)

19. (a) According to given situation, N 2 kg

7N

f 20

f max = mN = 0.5 ´ 20 = 10 N Since 7 N < 10 N, the upper block will not move with respect to the lower block. 2

q

...(ii)

7N

5 Smooth

Both the blocks will move together with same acceleration, 7 = (2 + 5)a a = 1 m/s2

29

Laws of Motion 20. (a) According to the question, for a stone of mass m and length of string r at the point B, weight mg acts vertically downwards, while tension T B acts vertically upwards. Their resultant provides the necessary centripetal force, that is A

vA

mg r

mv 2 = fr r From Eqs. (i) and (ii), we get, mv 2 = m mg r v2 = mg Þ r v 2 4.9 ´ 4.9 = 0.61 m= = \ 4 ´ 9.8 rg Thus,

23. (a) Upward journey, R : air resistance

TB B

motion Ý mg

mg vB

Applying Newton’s 2nd law, at point B mv 2 mv 2 + mg T B - mg = Þ TB = r r

21. (b) According to the question, a mmg

a

mg + R R =g + m m Let h be the height attained by the ball 1 ...(i) h = a1t12 2 R motion Downward journey, ß mg - R R Acceleration, a2 = =g mg m m 1 ...(ii) h = a2t22 2 From Eqs. (i) and (ii), we get R g + t2 a1 m = = R t1 a2 g m Thus, t2 > t1 Retardation, a1 =

a mmg

m mg

the type of connection between blocks and pulley is shown above Applying Newton’s 2nd law, Fnet (in the downward direction) = mtotal a Fnet = mg - mmg - mmg and m total = m + m + m mg - mmg - mmg = (m + m + m)a Þ mg (1 - 2m ) = 3ma g (1 - 2m) g (1 - 2m) = 3a Þ a = 3 22. (d) For uniform circular motion, centripetal force on the body mv 2 ...(i) = r where v is velocity, r is radius and m is mass. Also, the frictional force ...(ii) F =m mg where, m is coefficient of friction, m is mass and g is acceleration due to gravity. In this case, the required centripetal force is generated from friction force.

24. (d) According to the question, Wall 1 m/s 30° 30° 1 m/s

Q Velocity of the particle in vertical direction remain unchanged (during just before collision and just after collision). \Impulse in vertical direction is zero.

30

AIIMS Chapterwise Solutions ~ Physics The change in linear momentum in horizontal direction, Dp = mDv = m(- v cos 30° - vcos 30° ) = - 2mv cos 30° or | Dp | = 2mv cos 30° As, we know that, Impulse = Force ´ Time interval I = FDt and I = Dp Thus, we have Dp 2 mv cos 30° F = = Dt Dt 3 2 ´ 1 ´ 1´ ì 3ü 2 = íQ cos30°= ý 2 0.1 î þ 3 = 20 ´ = 10 3 N 2 Moving upward

25. (b) In a rocket, highly compressed heat gases are ejected from the tail of the rocket in the form of jet with a very high velocity. These ejecting gases push the rocket in upward direction against the gravity.

Þ

æ v1 ç çv è 2

2

ö r ÷ = 1 ÷ r2 ø v1 = v2

r1 r2

=

r 2r

[Q r2 = 2r1 and r1 = r ]

v1 : v2 = 1: 2

28. (b) According to the question, Lift

a = 9.8 m/s2

Free body diagram of the body in that lift. N

dm dt

dm : rate of combustion of fuel dt Given, F = 400 N, v = 500 m/s dm F 400 = = \rate of combustion, dt v 500 = 0.8 kg/s where,

27. (b) Given, radius of first particle (r1 ) = r and radius of second particle (r2 ) = 2r . We know that when a particle is moving in a circular path, then the centripetal force, mv 2 F = r Q Both particles have same centripetal force, then we have F1 = F2 mv12 mv22 Þ = r1 r2 (Q masses of the both particles are equal)

a = 9.8 m/s2

mg

Hot gases

26. (a) In a rocket propulsion, the thrust force F =v×

Þ\

Applying Newton’s 2nd law, mg - N = ma \ N = mg - ma = m (g - a) = m(9.8 - 9.8) (putting values) Þ N= 0 Thus, in a free fall, the apparent weight of the body is zero.

29. (c) In uniform circular motion, the direction of motion changes, continuously therefore velocity changes. As, p = mv therefore, momentum of a body also changes in uniform circular motion.

30. (c) A body subjected to three concurrent forces is found to in equilibrium if sum of these forces is equal to zero. i.e.,

F1 + F2 + F3 + K = 0

31. (d) Due to earth axial rotation, the speed of the trains relative to earth will be different and hence, the centripetal forces on them will be mv 2 different. Thus, their effective weights mg r mv 2 and mg + will be different. So, they exert r different pressure on the rails.

31

Laws of Motion 32. (a) In a motion of a rocket in a gravity free space. dm The thrust on the rocket, Fth = v rel dt where, v rel is the exhaust speed of gases relative the rocket. dm is the rate of mass at which gases are dt ejecting from rocket.

33. (c) A centripetal force is a real force due to which a particle or a body moves a circular path. While a centrifugal force is a pseudo force which acts on a particle or a body moving in a circular path. When, we describe its motion with respect to an uniformly rotating frame.

m mg cos a = mg sin a or tan a = m or a = tan- 1 (m)

36. (d) Let a body which has mass m is placed on an inclined plane whose inclination is greater than angle of repose. The force on the body down the inclined plane is given by F = mg sin q - mR ma = mg sin q - m mg cos q or a = g sin q - mg cos q ...(i) = g(sin q - m cos q) R

34. (c) For safe turn, tan q= v 2 / rg It is clear that for the safe turn, v should be small and r should be large. Also, bending angle from the vertical would increase with increase in velocity.

q

slipping is about to occur, the two forces acting on each object are normal reaction N and frictional force m N. Thus, limiting angle N F

l

tan l =

...(i)

Now, at the point of sliding down the angle of inclination of the plane with the horizontal is called the angle of repose (a ). From figure R

mg

sin

a

a

m

mg

o gc

sa

m

co

sq

37. (c) If we take a body resting with respect to two frames with origin O and O ¢, then the body will be at rest. But the frame (vehicle) is moving with constant speed, so, this frame can not be an inertial frame of reference or it is a non inertial frame of reference. But the frame in which Newton’s law of motion are applicable, is an inertial frame.

38. (d) An inertial frame of reference is one which

mN

mN =m N l = tan- 1 (m)

mg

g

It is obvious from Eq. (i) that, a < g .

35. (b) At the point of a rough contact where

or

...(ii)

From Eqs. (i) and (ii), l =a

has zero acceleration and in which law of inertia hold good i.e., Newton’s law of motion are applicable equally. Since, earth is revolving around the sun and earth is rotating about its own axis also, so the forces are acting on the earth and hence, there will be acceleration of earth due to these factors. That is why, earth cannot be taken as inertial frame of reference.

39. (a) On a rainy day due to wetness of the surface, the coefficient of friction decreases which might easily leads to skidding of a bus or car. That’s why, it is difficult to run a bus or a car with high speed on a rainy day.

AIIMS Chapterwise Solutions ~ Physics

CHAPTER

4 Work, Energy and Power 1. The particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac = k 2 rt 2 , where k is a constant. The power delivered to particle by the forces acting on it is [AIIMS 2017] (a) 2 p mk 2 r 2t

3. A body moves from rest with a constant acceleration. Which one of the following graphs represents the variation of its kinetic energy K with the distance travelled x? [AIIMS 2016] K

K

(a)

(b) mk 2 r 2t 1 (c) mk 4 r 2t 5 3 (d) zero

(b) O

x

O

K

2. A person of mass 70 kg wants to loose 7 kg

x

K

(c)

(d)

by moving up and down on the stairs of 12 m height. Assume he burns twice as much fat while going up than going down. If 1 kg of fat is burnt on expending 9000 kcal. How many times must he go up and down to reduce his 7 kg weight?

4. If the linear momentum is increased by

(Take, g = 10 m /s2 )

5. Consider the situation shown in figure.

(a) 18 ´ 103 times (b) 24 ´ 103 times (c) 30 ´ 103 times (d) 21 ´ 103 times

[AIIMS 2016]

O

x

O

x

50%, then kinetic energy will increase by (a) 50%

(b) 100%

(c) 125%

[AIIMS 2015] (d) 25%

One end of a spring of spring constant 400 N/m is attached 0 to a fixed rigidly support, horizontally. A 40 g mass is released from rest while situated at a height 5 m on a smooth curved track.

33

Work, Energy and Power The maximum deformation in the spring is nearly equal to (Take, g = 10 m/s2 ) [AIIMS 2015]

12. A 12 HP motor has to be operated 8 h/day. How much will it cost at the rate of 50 paise/kWh in 10 days? [AIIMS 2010] (a) ` 350 (c) ` 375

5m

(b) ` 358 (d) ` 397

13. Two springs A and B are identical but A is (a) 9.8 m

(b) 9.8 cm

(c) 0.98 m

(d) 0.009 km

6. The force on a particle as the function of displacement x (in x-direction) is given by F = 10 + 0.5 x The work done corresponding to displacement of particle from x = 0 to x = 2 unit is [AIIMS 2014] (a) 25 J

(b) 29 J

(c) 21 J

(d) 18 J

7. A body is moved by a machine delivering power at constant rate in a straight line. The distance travelled by it in time duration t is proportional to [AIIMS 2014] (a) t 3 / 2

(b) t 1/ 2

(c) t 2

(d) t

8. A bullet of mass m moving with velocity v strikes a suspended wooden black of mass M. If the block rises to a height h, the initial velocity of the block will be

(a) WA > WB and W ¢A = W ¢B (b) WA > WB and W ¢A < W ¢B (c) WA > WB and W ¢A > W ¢B (d) WA < WB and W ¢A < W ¢B

14. A force F acting on an object varies with distance x as shown in the figure. The force is in N and x in m. The work done by the force in moving the object from x = 0 to x = 6 m is F(N) 3 2 1

[AIIMS 2013] M+m (b) 2 gh m M+m (d) 2 gh M

(a) 2gh (c)

harder than B( k A > k B ). Let W A and WB represent the work done when the springs are stretched through the same distance and W ¢ A and W ¢ B are the work done when these are stretched by equal forces, then which of the following is true [AIIMS 2010]

m 2 gh M+m

O

9. The particle of mass 50 kg is at rest. The work done to accelerate it by 20 m/s in 10 s is [AIIMS 2012] (a) 103 J

(b) 104 J

(b) 2 ´ 103 J (d) 4 ´ 104 J

10. The potential energy of a certain spring when stretched through a distance ‘s’ is 10 joule. The amount of work (in joule) that must be done on this spring to stretch it through an additional distance ‘s’ will be (a) 30 J

(b) 40 J

(c) 10 J

(a) 13.5 J (c) 15 J

d 4

(b) 3Mg

d 4

(c) -3Mg

d (d) Mgd 4

(b) 10 J (d) 20 J

spring constant k versus length l of a spring correctly? [AIIMS 2008] k

k (a)

(b)

O

11. A chord is used to vertically lower a block

(a) Mg

[AIIMS 2010]

15. Which of the following graphs depicts

[AIIMS 2011] (d) 20 J

of mass M by a distance d with constant g downward acceleration . Work done by 4 the chord on the block is [AIIMS 2011]

1 2 3 4 5 6 x (m)

O

l

l

k

k (c)

(d) O

l

O

l

AIIMS Chapterwise Solutions ~ Physics

34 16. A body of mass 5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface collides with a nearly weightless spring of force constant k = 5 N/m. The maximum compression of the spring would be 1.5 m/s

[AIIMS 2008] (a) 0.25 m

(b) 0.75 m (c) 0.5 m

(d) 1.5 m

17. If a spring extends by x on loading, then energy stored by the spring is (if T is the tension in the spring and k is the spring constant) [AIIMS 2007] T2 (a) 2x

T2 (b) 2k

(c)

2k T2

2T 2 (d) k

18. A vertical spring with force constant k is fixed on a table. A ball of mass m at a height h above the free upper end of the spring falls vertically on the spring, so that the spring is compressed by a distance d. The net work done in the process is 1 (a) mg (h + d ) + kd 2 2 1 (c) mg (h - d ) - kd 2 2

[AIIMS 2007] 1 (b) mg (h + d ) - kd 2 2 1 (d) mg (h - d ) + kd 2 2

19. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is [AIIMS 2006] (a) 40 m/s (c) 10 m/s

(b) 20 m/s (d) 10 30 m/s

20. A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/s. It is subjected to a retarding force F = - 0.1 x J/m during its travel from x = 20 m to x = 30 m. Its final kinetic energy will be [AIIMS 2005] (a) 475 J

(b) 450 J

(c) 275 J

(d) 250 J

21. A ball is released from the top of a tower.

22. The kinetic energy of a body becomes four times, its initial value. The new linear momentum will be [AIIMS 2004] (a) eight times that of initial value (b) four times that of initial value (c) twice of the initial value (d) remain as the initial value

23. 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) x3 / 2

(c) x

[AIIMS 2004] (d) x3

24. A force (3i$ + 4$j) N acts on a body and displaced it by (3$i + 4$j) m. The work done by the force is [AIIMS 2003] (a) 5 J

(b) 25 J

(c) 10 J

(d) 30 J

25. A body of mass, 5 kg has momentum of 10 kg-m/s. When a force of 0.2 N is applied on it for 10 s, the change in its kinetic energy is [AIIMS 2003] (a) 4.4 J

(b) 3.3 J

(c) 5.5 J

(d) 1.1 J

26. A bullet of mass, 10 g leaves a rifle at an initial velocity of 1000 m/s and strikes the earth at the same level with a velocity of 500 m/s. The work in overcoming the resistance of air will be [AIIMS 2002] (a) 500 J

(b) 5000 J

(c) 3750 J

(d) 475 J

27. Two bodies of masses m and 4m are moving with equal kinetic energy. Then, the ratio of their linear momentum will be (a) 1 : 1

(b) 2 : 1

(c) 4 : 1

[AIIMS 2002] (d) 1 : 2

28. A bullet is fired from a rifle. If the rifle recoils freely, then the kinetic energy of the rifle will be [AIIMS 2001] (a) same as that of bullet (b) more than that of bullet (c) less than that of bullet (d) None of the above

29. A spring 40 mm long is stretched by

The ratio of work done by force of gravity in first, second and third second of the motion of the ball is [AIIMS 2005]

applying a force. If 10 N force is required to stretch the spring through 1 mm, then work done in stretching the spring through 40 mm is [AIIMS 2001]

(a) 1 : 2 : 3 (c) 1 : 3 : 5

(a) 24 J (c) 56 J

(b) 1 : 4 : 9 (d) 1 : 5 : 3

(b) 8 J (d) 64 J

35

Work, Energy and Power 30. Consider the situation shown in figure. Mass of block A is m and that of block B is 2m. The force constant of the spring is k . Friction is absent everywhere. System is released from rest with the spring unstretched. The maximum extension of [AIIMS 1999] the spring xm is

A

Assertion & Reason Direction (Q. Nos. 35-43) Read the Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

35. Assertion For looping a vertical loop of B

(a)

2mg k

(b)

4mg k

(c)

mg k

(d)

3 mg × 2 k

31. A

frictionless track A D ABCDE ends in a circular loop of radius R. h 2R C E A body slides down the track from point A B which is at a height, h = 5 cm. Maximum value of R for the body to successfully complete the loop is [AIIMS 1999] (a) 5cm

(b)

15 cm 4

(c)

10 cm 3

(d) 2 cm

32. A body of mass 5 kg is raised vertically to a height of 10 m by a force of 170 N. The velocity of the body at this height will be [AIIMS 1998] (a) 15 m/s (c) 9.8 m/s

(b) 37 m/s (d) 22 m/s

kp ö æ ÷ are stretched by and k Q çç where k Q = 2 ÷ø è applying forces of equal magnitude. If the energy stored in Q is E, then the energy stored in P is [AIIMS 1997] (b) 2E

(c) E/4

(d) E/2

34. The potential energy of a body in a force

b , where a and b are x x positive constants and x is the distance of the body from the centre of the field. For stable equilibrium, the value of x will be.

field is U =

(a) b /2 a

a

2

–

(b) 2a /b

(c) a /b

Reason In this event, the velocity at the highest point will be zero. [AIIMS 2017]

36. Assertion The rate of change of total momentum of a many particle system is proportional to the sum of the external forces of the system. Reason Internal forces can change the kinetic energy but not the momentum of the system. [AIIMS 2016]

37. Assertion A light body and heavy body have same momentum. Then, they also have same kinetic energy. Reason Kinetic energy does not depend on mass of the body. [AIIMS 2015]

38. Assertion A spring has potential energy,

33. Two springs P and Q of force constants k p

(a) E

radius, r the minimum velocity at lowest point should be 5gr.

[AIIMS 1996] (d) b / a

both when it is compressed or stretched. Reason In compressing or stretching, work is done on the spring against the restoring force. [AIIMS 2013]

39. Assertion Work done in moving a body over a closed loop is zero for every force in nature. Reason Work done does not depend on nature of force. [AIIMS 2013]

40. Assertion Potential energy of a stretched or compressed spring, proportional to square of extension or compression. Reason Graph between potential energy of a spring versus the extension or compression of the spring is a straight line. [AIIMS 2012]

AIIMS Chapterwise Solutions ~ Physics

36 41. Assertion According to law of conservation of mechanical energy change in potential energy is equal and opposite to the change in kinetic energy.

Reason Change in kinetic energy of particle is equal to the work done only in case of a system of one particle. [AIIMS 2010]

43. Assertion Water at the foot of the water fall

Reason Mechanical energy is not a conserved quantity. [AIIMS 2011]

is always at different temperature from that at the top. Reason The potential energy of water at the top is converted into heat energy during falling. [AIIMS 2008]

42. Assertion The change in kinetic energy of a particle is equal to the work done on it by the net force.

Answers 1. 11. 21. 31. 41.

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

2. 12. 22. 32. 42.

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

3. 13. 23. 33. 43.

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

4. 14. 24. 34.

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

5. 15. 25. 35.

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

6. 16. 26. 36.

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

7. 17. 27. 37.

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

8. 18. 28. 38.

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

9. 19. 29. 39.

10. 20. 30. 40.

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

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

Explanations = 126 . ´ 104 J

1. (b) If v is instantaneous speed, centripetal acceleration is given by v2 ac = r v2 = k2 r t 2 [Q ac = k2 r t 2 is given] r Þ v = krt In circular motion, work done by centripetal force (Fc ) is always zero because Fc ^ v and work is done only by tangential force. Q Tangential acceleration, dv d at = = (kr t ) = k r dt dt \Tangential force, Ft = mat = mkr Thus, power delivered,

\

P = Ft v = (mkr )(krt ) = mk2 r 2t

2. (d) Given, m = 70 kg, g = 10 m/s2 , h = 12 m In going up and down once. mgh Calories burnt = mgh + 2 (as given in the question) 3 = ´ 70 ´ 10 ´ 12 2

(Q 4.2 J = 1 cal) = 3 k-cal Since, given 1 kg of fat is burnt on expending 9000 k-cal \ 7kg of fat will be burnt on expending = 7 ´ 9000 kcal = 63000 k-cal. Thus, number of times the person has to go up 63000 and down on the stairs = = 21000 3 = 21 ´ 103 times

3. (c) Q Kinetic energy of a body having mass m and moving with velocity v, KE = (1/2)mv 2

…(i)

In given question, body moves from rest \u = 0 and acceleration is a. \Velocity of body after moving distance x, v 2 = u2 + 2as Þ

v 2 = 0 + 2ax v 2 = 2ax

From Eqs. (i) and (ii), we get 1 KE = m × 2ax = max Þ KE µ x 2 Hence, option (c) is correct.

...(ii)

(distance)

37

Work, Energy and Power 4. (c) Since we know, the kinetic energy 2

1p 2m where symbols have their usual meaning 3 Q p2 = p1 + 50% of p1 = p1 2 p1 = initial momentum

1 /3

æ 3P ö v=ç x÷ èm ø

\

KE =

Thus, D KE = KE2 – KE1 ù 1 2 1 é æ 3 ö2 1 5p22 2 . [ p2 – p12 ] = = ê ç p1 ÷ – p1 ú = 2m 2m êë è 2 ø úû 2m 4 5 \ D KE% = ´ 100 = 125% 4

5. (b) Applying the conservation of energy to the system, Loss of potential energy of mass = Gain in potential energy of the spring. 1 mgh = kx2 2 Given, m = 0.04 kg, h = 5m, k = 400 N/m and x : deformation (compression) in the spring Þ

2mgh 2 ´ 0.04 ´ 10 ´ 5 = 400 k

x= =

1 m = 10 cm » 9.8 cm 10

6. (c) Since we know, work done W =

x2

x2

x1

x1

ò F × dx = ò F × dxcos q

Q Force and displacement in same direction \ q = 0° and cos 0°= 1 x1 = 0 and x2 = 2 2

W = ò Fdx = 0

2

ò0 (10 + 0.5x)dx 2

é x2 ù = ê10 x + 0.5 ú = 21 J 2 û0 ë

7. (a) Q Power, P = F × v Þ Þ Þ

P = ma × v vdv ×v P = m× dx P dx = ò v 2dv mò v3 v2 + 1 P = ×x= 3 m 2+ 1

[Q F = ma ] vdv ù é êëQ a = dx úû

dx æ 3P ö1 /3 =ç x÷ dt è m ø

Þ

Þ

ò

dx = k x1 /3 , dt dx = k ò dt x1 /3

dx ö æ çQ v = ÷ dt ø è

æ 3P where k = ç èm

1 /3

ö x÷ ø

1 - +1

Þ

x 3 x2 /3 = kt Þ = kt -1 2/3 +1 3 3 /2

Þ

æ2 ö x = ç kt ÷ è3 ø

Þ

x µ t 3 /2

8. (a) Initial KE of the block, when bullet strikes to it 1 KE = (m + M )v 2 2 when m and M are masses of bullet and block. Similarly, its potential energy (PE) = (m + M ) gh. By law of conservation of energy, KE = PE 1 (m + M )v 2 = (m + M )gh Þ 2 Þ v = 2gh

9. (b) Given, m = 50 kg, v1 = 0, v2 = 20 m/s, t2 = 10 s Since, we know from work-energy theorem, Wnet = Change in KE 1 1 = mv22 – mv12 2 2 1 1 = m(v22 – v12 ) = ´ 50(20 ´ 20 – 0) 2 2 = 104 J

10. (a) Q Work done in stretching a spring from distance x1 to distance x2 , 1 1 W = k x22 - k x12 2 2 1 2 Case (i) W = k s - 0 2 [Q x1 = 0 (unstretched), x2 = s] 1 2 …(i) (given) ks = 10 J 2 Case (ii) The spring is stretched an additional distance s.

AIIMS Chapterwise Solutions ~ Physics

38 \Net extension of the spring, x2 = s + s = 2 s and x1 = s \Work done in this case, 1 1 W = k(2s)2 - ks2 2 2 3 1 2 [using Eq. (i)] = ks [4 - 1] = ks2 = 30 J 2 2

11. (c) When the block moves vertically downward with acceleration g / 4, then tension in the chord Applying Newton’s second law, Mg - T = Mg / 4 gö 3 æ T = M ç g - ÷ = Mg \ 4ø 4 è

\

W A k A x2 k A = = WB kB k B x2

[Q k A > k B (given)] Þ W A > WB Case (ii) Stretching both springs by same force. Q Force, F = k × x 2

æ F ö ÷ k A çç 2 ÷ W ¢A k A x A è k A ø = kB Thus, = = 2 2 kA W ¢B k B xB æ F ö ÷ k B çç ÷ è kB ø Thus, W ¢B > W ¢A [Q k A > k B ]

14. (a) Q Area under F versus x graph gives work done by the force.

T

F(N) M

3 2 1

g/4

Mg

Work done by the chord, W = F × s = Fscos q = Td cos180° Since, the torque on the block and displacement of the block are in opposite direction. \ q = 180° d æ 3 ö = ç - Mg ÷ ´ d = - 3Mg 4 è 4 ø

12. (a) Since, a motor of 12 HP works for 10 days at the rate of 8 h/day, then energy consumption = power ´ time = 12 ´ 746 ´ (8 ´ 60 ´ 60) ´ 10 = 2.5 ´ 109 J Rate of energy = 50

Paise kWh

Q 3.6 ´ 106 J energy cost = ` 0.5 \ 2.5 ´ 109 J energy cost =

2.5 ´ 109 3.6 ´ 106

A

B O

C 3

6 x (m)

Area under F versus x graph = Area of ODAB + Area of DABC 1 W = 3 ´ 3 + (3 ´ 3) 2 = 9 + 4.5 = 13.5 J

15. (d) For a given material, the different size of spring k1 l1 = k2 l2 = constant where, k is spring constant and l is length of the spring. 1 kµ \ l So, option (d) is correct.

16. (d) Applying conservation of energy, ´ 0.5

(Q 1kWh = 3. 6 ´ 106 J) = ` 347. 2 » ` 350 (Approx)

13. (b) Q Work done in stretching a spring to a distance x, 1 2 kx 2 where, k is force constant of the spring. Case (i) Stretching springs through same distance. W =

D

kinetic energy of the body = potential energy of the spring 1 1 mv 2 = kx2 Þ 2 2 where, x is maximum compression of the spring. 1 1 (putting values) ´ 5 ´ (1.5)2 = ´ 5 ´ x2 Þ 2 2 Þ

x2 = (1.5)2

Þ

x = 1.5 m

17. (b) According to the question,

39

Work, Energy and Power O

1 mv 2 + mgh2 2 1 or mg (H - h2 ) = mv 2 or v = 2g (100 - 20) 2 mgH =

O

= 2 ´ 10 ´ 80 = 40 m/s

A x

20. (a) Given, F = –01 . x J / m, x1 = 20m and x2 = 30m

T=kx

B

u = 10 m/s Applying work-energy theorm, Wnet = Change in kinetic energy

mg

]

The energy stored in the spring, U =

1 2 kx 2

x2

…(i)

ò F × dx = K f – K i

x1

Q Tension in the spring, T = mg = kx T \ x= k From Eqs. (i) and (ii),

Þ …(ii)

2

30

Þ

1 æT ö T2 k×ç ÷ = 2 èkø 2k

–1 é x2 ù 1 ´ ê ú = K f – ´ 10 ´ (10)2 10 ë 2 û20 2 [Q putting u = 10 m/s]

–1 [(302 ) – (20)2 ] = K f – 500 10 ´ 2 (900 – 400) Þ K f = 500 – = 500 – 25 20 = 475J Þ

18. (b) According to the question, when mass m falls vertically on spring, then spring is compressed by distance d. m

21. (c) Let the mass of the ball be m.

h m d

Thus, Wnet = Work done by the gravity + Work done by the spring 1 = mg (h + d ) – kd 2 2 1 Work done by the spring = – kd 2 because 2 spring force and change in the spring length are in opposite direction.

19. (a) According to conservation of energy, 30

h 1= m

H=100 m

1

ò –01. xdx = K f – 2 mu

20

2

U =

30

h2=20 m

Potential energy of ball at height H = Total kinetic energy of ball + Potential energy of ball at height h

\Force due to gravity, F = mg For 1st second, the ball is released, u = 0 \Distance moved by the ball in first second 1 s1 = u + g (2 t – 1) 2 1 = 0 + ´ 10 (2 ´ 1 – 1) = 5 m 2 …(i) \ Work done, W1 = F × s = mg × 5 = 5mg For 2nd second, distance moved by the ball in 2nd second, 1 s2 = u + g (2 t - 1) 2 1 ...(ii) s2 = 0 + 10 ´ (2 ´ 2 – 1) = 15 m 2 \Work done, W2 = F × s2 = mg × 15 = 15mg For 3rd second, distance moved by the ball in 3rd second, 1 s3 = u + g (2t - 1) 2 1 s3 = 0 + 10 ´ (2 ´ 3 – 1) = 25 m 2

AIIMS Chapterwise Solutions ~ Physics

40 …(iii) \Work done, W3 = F × s3 = 25mg From Eqs. (i), (ii) and (iii), we get W1 : W2 : W3 = 5 mg : 15 mg : 25 mg = 1 : 3 : 5

22. (c) Since, we know 1 p2 × 2 m where, p = linear momentum and m = mass of the object. kinetic energy, K =

Thus,

K2 p2 = 22 Þ K1 p1 p22 =

p22 =

K2 2 p K1 1

4K 1 2 p K1 1

[Given, K 2 = 4K 1 ]

Thus, p2 = 2 p1 Means the new linear momentum will be twice of the initial value.

23. (a) Let the retardation of the particle be a and its displacement be x. According to the condition, aµ x a= kx where, k is constant. Q Kinetic energy of a moving body, KE =

…(i) 1 mv 2 2

1 …(ii) mu2 2 Velocity of the particle moving with retardation a after its displacement x,

Initial kinetic energy, KE1 =

v 2 = u2 - 2ax \

KE2 =

1 1 mv 2 = m(u2 - 2ax ) 2 2

…(iii)

\ Loss of KE = KE1 - KE2 1 1 mu2 - m(u2 - 2ax ) 2 2 1 DKE = m × 2ax = max 2 [using Eq. (i)] = m × kx × x =

= mkx2 \ Loss of KE µ x2

24. (b) Work done is given by scalar product of

25. (a) Momentum ( p ) is defined as product of mass (m) and velocity (u). i.e. p = mu Given, m = 5kg, p = 10 kg -m/s p 10 = = 2 m /s u= 5 m From Newton’s second law, F = ma where, a is acceleration. F 02 . a= = = 0.04 m/s Þ m 5 From equation of motion, we have v = u + at v = 2 + 0.04 ´ 10 = 2.4 m/s Change in kinetic energy is 1 1 DK = mv 2 - mu2 2 2 1 1 = ´ 5 ´ (2.4)2 - ´ 5 ´ (2)2 2 2 1 . = ´ 5 ´ 176 2 = 4 .4 J

26. (c) Work done in overcoming the resistance of air is equal to change in kinetic energy. Kinetic energy acquired by the body of mass m due to velocity v is 1 K = mv 2 2 Change in kinetic energy 1 = m(v12 - v22 ) 2 1 = ´ 10 ´ 10-3 [(1000)2 - (500)2 ] 2 1 = ´ 10 ´ 10-3 ´ 750000 2 = 3750 J 1 27. (d) Kinetic energy (K ) = mv 2 …(i) 2 where, m is mass and v is velocity. Also, momentum ( p ) = mv …(ii) From Eqs. (i) and (ii), we have p = 2mK

force and displacement. \ Given, \

W = F×d F = 3i$ + 4$j, d = 3$i + 4$j W = F × d = (3i$ + 4$j )× (3i$ + 4$j ) = 9 + 16 = 25J

Given,

m1 = m, m2 = 4m p1 = p2

1 m = 4m 2

41

Work, Energy and Power 28. (c) From the principle of conservation of momentum, if no external force acts upon a system of two (or more) bodies, then the total momentum of the system remains constant. p2 = 2 mK where p is momentum, K is kinetic energy and m is mass. Þ p1 = p2 Þ 2 m1 K 1 = 2 m2 K 2 1 Þ m1 K 1 = m2 K 2 = constant or K µ m Since, mass of rifle is very much greater than that of bullet, so kinetic energy of rifle will be less than kinetic energy of bullet.

From Eqs. (i) and (ii),

Þ

Also,

or Þ\

32. (d) Given, m = 5kg, h = 10 m and Fext = 170 N, u = 0 m Fex = 170 N

30. (b) Let the block B moves x (below) and due to this the extension in the spring be x. Applying law of conservation of energy, increase in potential energy of the spring = decrease in potential energy of the block B 1 2 kx = 2mgx Þ 2 4mg x= \ k

31. (d) Condition for a body to complete a vertical circular loop. The speed of the body at the lowest point of the vertical circle, …(i) v L ³ 5 gR where, R is radius of the loop. \

v Lmini =

5 gR

According to the problem, applying conservation of energy from A to B, 1 mgh = mv 2L 2 …(ii) v L = 2gh Þ

Fin = mg

10 m

29. (b) Q Spring constant, k = Force required to change the length of the spring by a unit length 10 = -3 N/m = 104 N/m 10 Work done in stretching the spring, 1 1 W = k x22 - kx12 2 2 where symbols have their usual meaning. 1 1 = ´ 104 ´ (40 ´ 10-3 )2 - 104 ´ 0 2 2 1 4 -6 = ´ 10 ´ 1600 ´ 10 = 8 J 2

2gh ³ 5gR 2h ³R 5 2h R£ 5 2h 2 ´ 5 Rmax = = = 2 cm 5 5

Ground

Applying work-energy theorem, Wnet = DKE or

Wex + Win = DKE

Þ

Fex × h - mgh = DKE

1 1 mv 2 - mu2 2 2 and displacement in opposite direction. Q mg 1 2 170 ´ 10 - 5 ´ 10 ´ 10 = 5v - 0 Þ 2 1 2 1700 - 500 = 5v Þ 2 1200 ´ 2 2 = 480 v = Þ 5 \ v = 21.9089 m/s = 22 m/s =

33. (d) Q Since we know, spring force F µ –x Þ F = – kx where, k = spring constant. kp Given, kQ = 2 Fp k p xP Thus, = FQ kQ xQ \

(displacement)

xQ kP = [Q FP = FQ ] kQ xP

Energy stored in a spring, U =

1 2 kx 2

...(i)

AIIMS Chapterwise Solutions ~ Physics

42 UP k x2 = P 2P UQ kQ xQ

\

kQ kp

=

kP kQ

=

1 2

Up =

\

[using Eq. (i)]

æ kQ × çç è kP

KE =

2

ö ÷ ÷ ø

2

=

E 2

(Q UQ = E )

34. (b) Given, the potential energy of the body U =

a x2

–

b at x x

Since, we know –dU –2a b Fin = = 3 + 2 dx x x For stable equilibrium, –dU =0 Fin = dx –2a

\

x3

+

b x2

=0Þ

x=

\

1 é 2a ù b– =0 x úû x2 êë

2a b

35. (c) At the lowest point of a vertical circle, the minimum velocity at the bottom umin = 5 gr Velocity at highest point, v = gr

36. (b) Rate of change of momentum is proportional to external forces acting on the system. The total momentum of whole system remain constant when no external force is acted upon it. dp ext = F net dt If

ext Fnet

p2 2m

1 , when p = momentum = constant m Thus, when two bodies have same momentum, then lighter body possess more kinetic energy. Hence, Assertion and Reason both are false.

\ KE µ [Q k p = 2kQ ]

UQ

37. (d) The kinetic energy in terms of momentum is

= 0, then p = constant

Since, the derivative of any constant term is always zero. It can also be said that, if the derivative of any term is zero, then that particular term is constant in nature. So, here net external force is zero, therefore linear momentum of the system becomes constant. A system has potential energy due to internal forces. This potential energy can convert into kinetic energy.

38. (a) The work done on the spring against the restoring force is stored as potential energy in both conditions when it is compressed or stretched. Thus, the spring will have potential energy in both the conditions.

39. (d) Work done due to moving a body over a closed loop is zero only when the body is moving under the action of conservation forces (like gravitational or electrostatic forces). i.e., work done depends upon the nature of force. 1 40. (c) Potential energy, U = kx2 i.e., U µ x2 2 This is a equation of parabola, so graph between U and x is a parabola, not a straight line.

41. (c) For conservation forces, the sum of kinetic and potential energies at any point remains constant throughout the motion. This is known as law of conservation of mechanical energy. According to this law, E1 = E2 Q E = U + KE \ U1 + KE1 = U2 + KE2 Þ U2 - U1 = KE1 - KE2 Þ DU = - DKE

42. (c) ò Fnet × ds = DKE Change in kinetic energy = Work done by net force The relationship is valid for a particle as well as for system of particles.

43. (c) When the water is at top of the fall it has potential energy mgh (where, m is the mass of the water and h is the height of the fall). On falling, this potential energy is converted into kinetic energy and further a very small part of its kinetic energy is converted into heat energy and so temperature of water increases.

CHAPTER

5 Rigid Body and Rotational Motion 1. A boy is pushing a ring of mass 3 kg and

3. A tube of length L is filled completely with

radius 0.6 m with a stick as shown in figure. The stick applies a force of 3N on the ring and rolls it without slipping with an accelertation of 0.4 m/s2 .

an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity w. The force exerted by the liquid at the other end is

Stick

[AIIMS 2016] 1 (a) Mw2 L2 2 1 (c) Mw2 L 4

a O

2

(b) Mw L (d)

1 Mw2 L 2

4. A uniform metallic rod rotates about its The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of F friction between the stick and the ring is . 10 The value of F is [AIIMS 2017] (a) 2 N

(b) 4 N

(c) 6 N

(d) 3 N

2. A solid sphere rolls down two different inclined planes of same height, but of different inclinations. In both cases [AIIMS 2017] (a) speed and time of descent will be same (b) speed will be same, but time of descent will be different (c) speed will be different, but time of descent will be same (d) speed and time of descent both are different

perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly, then [AIIMS 2016] (a) its speed of rotation increases (b) its speed of rotation decreases (c) its speed of rotation remains same (d) its speed increases because its moment of inertia increases

5. If A is the top most point of a wheel of radius 2m rolls on ground without slipping. Then the value of displacement of point A, when wheel completes half of rotation. [AIIMS 2015] (a) 7.45 m (b) 9 m (c) 6.3 m (d) 8.6 m

AIIMS Chapterwise Solutions ~ Physics

44 6. A massless rod S having length 2l has equal

10. A thin hollow sphere of mass m is

point masses attached to its two ends as shown in figure. The rod is rotating about an axis passing through its centre and making angle q with the axis. The magnitude of change of momentum of i.e., dL / dt equals

completely filled with a liquid of mass m. When the sphere rolls with a velocity v, kinetic energy of the system is (neglect friction) [AIIMS 2012]

[AIIMS 2015] w

(b) mv 2 (d)

4 mv 2 5

11. A solid sphere rolls without slipping on the

l

roof. The ratio of its rotational kinetic energy and its total kinetic energy is [AIIMS 2012]

m q

(a) 2/5 (c) 2/7

rod l

(b) 4/5 (d) 3/7

12. A disc is rolling (without slipping) on a horizontal surface. C is centre Q and P are two points equidistant from C. Let v P , vQ and vC be the magnitude of velocities of points P, Q and C respectively, then

m axis 3

1 mv 2 2 4 (c) mv 2 3

(a)

2

(a) 2 m I w sin q . cos q (b) ml 2 w2 sin 2 q (c) mI 2 sin 2 q (d) m1/ 2 I1/ 2 w sin q . cos q

[AIIMS 2012]

P

C

Q

7. A uniform sphere of mass 500 g rolls without slipping on a plane surface so that its centre moves at a speed of 0.02 m/s. The total kinetic energy of rolling sphere [AIIMS 2015] would be (in J) (a) 14 . ´ 10-4 J (c) 575 . ´ 10-3 J

(b) 075 . ´ 10-3 J (d) 4.9 ´ 10-5 J

8. A rod of length L is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod, when it is in vertical position is 2g (a) L

(b)

3g L

g (b) 2L

[AIIMS 2014] g (d) L

9. The moment of inertia of a circular loop of radius R, at a distance of R / 2 around a rotating axis parallel to horizontal diameter of loop is [AIIMS 2012] (a) MR 2 (b) 2 MR 2

1 MR 2 2 3 (d) MR 2 4

(b)

(a) vQ > vC > v P (c) vQ = v P , vC =

1 v 2 P

(b) vQ < vC < v P (d) vQ = vC = v P

13. The radius of gyration of a body about an axis at a distance of 6 cm from its centre of mass is 10 cm. Then, its radius of gyration about a parallel axis through its centre of mass will be [AIIMS 2011] (a) 800 cm (c) 0.8 cm

(b) 8 cm (d) 80 m

14. A uniform cylinder has a radius R and length L. If the moment of inertia of this cylinder about an axis passing through its centre and normal to its circular face is equal to the moment of inertia of the same cylinder about an axis passing through its centre and perpendicular to its length, then (a) L = R R (c) L = 3

[AIIMS 2010] 3R 3 (d) L = R 2

(b) L =

45

Rigid Body and Rotational Motion 15. A wire of mass m and length l is bent in the

19. A particle of mass m moves in the XY plane

form of a circular ring. The moment of inertia of the ring about its axis is

with a velocity v along the straight line AB. If the angular momentum of the particle with respect to origin O is L A, when it is at A and [AIIMS 2007] LB , when it is at B then,

[AIIMS 2010] æ 1 ö (a) ç 2 ÷ml 2 è 8p ø æ 1 ö 2 (c) ç 2 ÷ml è 4p ø

æ 1 ö (b) ç 2 ÷ml 2 è 2p ø

Y

(d) ml 2

B A

16. Four point masses, each of value m, are placed at the corners of square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is [AIIMS 2008] 2

(a) 2 ml (c) 3 ml 2

(b) 3 ml (d) ml 2

2

17. For the given uniform square lamina ABCD whose centre is O, the correct relation in between the moments of inertia about the position AC, AD and EF is [AIIMS 2008] F

D

C

O A

B

E

O

(a) LA > LB (b) LA = LB (c) the relationship between LA and LB depends upon the slope of the line AB (d) LA < LB

20. The moment of inertia of a rod about an axis through its centre and perpendicular to it is 1 ML2 (where M is the mass and L, the 12 length of the rod). The rod is bent from the middle so that the two halves make an angle of 60°. The moment of inertia of the bent rod about the same axis would be. [AIIMS 2007] 1 ML2 48 1 (c) ML2 24

1 ML2 12 ML2 (d) 8 3

(a)

(a) 2 I AC = I EF

(b) I AD = 4 I EF

(c) I AC = I EF

(d) I AC = 2 I EF

18. In the diagram shown all three rods are of equal length L and equal mass M. The system is rotated such that the rod B is the axis, what is the moment of inertia of the system?

X

(b)

21. A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity v m/s. If it is to climb the inclined surface, then v should be [AIIMS 2005]

[AIIMS 2007] h

A

v

B

10 gh 7 (b) ³ 2 gh

(a) ³ C

4 2 ML 3 ML2 (c) 6

(a)

ML2 3 2 2 (d) ML 3

(b)

(c) 2 gh (d)

10 gh 7

AIIMS Chapterwise Solutions ~ Physics

46 22. A ladder is leaned against a smooth wall and

27. A constant torque of 31.4 N-m is exerted on a

it is allowed to slip on a frictionless floor. Which figure represents the track of its centre of mass? [AIIMS 2005]

pivoted wheel. If the angular acceleration of the wheel is 4p rad/s2 , then the moment of inertia, will be [AIIMS 2001] (a) 5.8 kg-m 2 (b) 4.5 kg-m 2 (c) 5.6 kg-m 2 (d) 2.5 kg-m 2

Time (a)

28. The moment of inertia of a regular circular

Time (b)

disc of mass 0.4 kg and radius 100 cm about an axis perpendicular to plane of the disc and passing through its centre is [AIIMS 1999] (a) 0.2 kg-m 2 (c) 0.002 kg-m 2

(b) 0.02 kg-m 2 (d) 2 kg-m 2

29. If there is change of angular momentum Time (c)

Time (d)

form J to 5JJ in 5s, then the torque is [AIIMS 1997]

23. A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass, m is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period [AIIMS 2005] (a) decreases continuously (b) decreases initially and increases again (c) remains unaltered (d) increases continuously

24. The direction of the angular velocity vector is along

[AIIMS 2004]

(a) the tangent to the circular path (b) the inward radius (c) the outward radius (d) the axis of rotation

4J (b) 5 (d) None of these

30. When the axis of rotation passes through its centre of gravity, then the moment of inertia of a rigid body is [AIIMS 1996] (a) reduced to its minimum value (b) zero (c) increased to its maximum value (d) infinity

31. Radius of gyration of a body depends upon (a) shape of the body (c) area of the body

[AIIMS 1995] (b) axis of rotation (d) translation motion

32. The moment of inertia of a disc of mass M

25. In an orbital motion, the angular momentum vector is

3J (a) 5 5J (c) 4

[AIIMS 2004]

(a) along the radius vector (b) parallel to the linear momentum (c) in the orbital plane (d) perpendicular to the orbital plane

26. A body is projected from the ground from the horizontal making any angle. What happens to the angular momentum about the initial position in this motion? [AIIMS 2002] (a) Decreases (b) Increases (c) Remains constant (d) First increases and then decreases

and radius R about an axis which is tangential to the circumference of the disc and parallel to its diameter, is [AIIMS 1995] 5 MR 2 4 4 (c) MR 2 5

(a)

3 MR 2 2 2 (d) MR 2 3 (b)

33. If the earth is treated as a sphere of radius R and mass M its angular momentum about the axis of its rotation with period T , is [AIIMS 1994] MR 2T (a) 2p p MR 3 (c) T

4pMR 2 (b) 5T 2 p MR 2 (d) T

47

Rigid Body and Rotational Motion

Assertion & Reason

39. Assertion The velocity of a body at the bottom of an inclined plane of given height is more when it slides down the plane, compared to, when it rolls down the same place. Reason In rolling down a body acquires both, kinetic energy of translation and rotation. [AIIMS 2010]

Direction (Q. Nos. 34-44) Read the Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

40. Assertion The speed of whirlwind in a tornado is alarmingly high. Reason If no external torque acts on a body, its angular velocity remains conserved.

34. Assertion The total kinetic energy of a rolling solid sphere is the sum of translational and rotational kinetic energies. Reason For all solid bodies, total kinetic energy is always twice of translational kinetic energy. [AIIMS 2017]

[AIIMS 2008]

41. Assertion A judo fighter in order to throw his opponent on to the mat tries to initially bend his opponent and then rotate him around his hip. Reason As the mass of the opponent is brought closer to the fighters hip, the force required to throw the opponent is reduced.

35. Assertion A ladder is more apt to slip, when you are high up on it, then when you just begin to climb. Reason At the high up on a ladder, the torque is large and on climbing up the torque is small. [AIIMS 2016, 2008]

[AIIMS 2006]

42. Assertion For system of particles under central force field, the total angular momentum is conserved. Reason The torque acting on such a system is zero. [AIIMS 2005]

36. Assertion If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant. Reason The linear momentum of an isolated system does not remain constant. [AIIMS 2015]

43. Assertion There are very small sporadic changes in the speed of rotation of the earth. Reason Shifting of large air masses in the earth’s atmosphere produce a change in the moment of inertia of the earth causing its speed of rotation to change. [AIIMS 2004]

37. Assertion If the ice on the polar caps of the earth melts, then length of day will increase. Reason Moment of inertia of the earth increases, as ice on polar caps melts. [AIIMS 2014]

38. Assertion The centre of mass of an electron

44. Assertion The inertia of rotational motion is analogous to the inertia of an object in translational motion. Reason Moment of inertia is the rotational inertia of the body. [AIIMS 2002]

and proton, when released moves faster towards proton. Reason Proton is heavier than electron. [AIIMS 2012]

Answers 1. 11. 21. 31. 41.

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

2. 12. 22. 32. 42.

(b) (a) (a) (a) (a)

3. 13. 23. 33. 43.

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

4. 14. 24. 34. 44.

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

5. 15. 25. 35.

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

6. 16. 26. 36.

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

7. 17. 27. 37.

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

8. 18. 28. 38.

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

9. 19. 29. 39.

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

10. 20. 30. 40.

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

AIIMS Chapterwise Solutions ~ Physics

Explanations 1. (a) Here, mass of ring, M = 3 kg.

2. (b) In rotational motion, mechanical energy

radius of ring, R = 0.6 m

remains conserved.

force applied, f = 3N

Therefore, when height of inclinations are equal, then speed of spheres will be same. But acceleration, a µ sin q, where q = angle of inclination. Hence, at different inclination, acceleration of spheres will be different and hence, time of descent will be also different.

acceleration, a = 0.4 m/s2 coefficient of friction between F stick and ring = 10 Now, according figure, f = Ma

A

a

a

D

3 - fs = f = Ma or

fs = 3 - Ma = 3 - 3 ´ 0.4 = 1.8 N

h

Q Torque, t = Moment of inertia (I ) ´ Linear acceleration (a) Þ

t = Ia

…(i)

Also, torque about point O is t = fs R - f k R

…(ii)

h q1

q1

C

B F

E

3. (d) Taking an element of mass dm of width dx at distance of x from the rotational axis, w

from Eqs. (i) and (ii), We have, fs R - f k R = I . a

x

Stick

Then, mass of this element, M dm = dx L

a=0.4m/s2 O fk

Instantaneous force on small element dm is dF = dmw2 x

fs

= MR2 ´

a R a and R I (inertia ) = MR2 )

(Q a = angular acceleration =

or 1.8 ´ 0.6 - f k ´ 0.6 = 3 ´ 0.6 ´ 0.4 f k = 1.8 - 12 . = 0.6 N If m is the coefficient of kinetic friction, then fk = m ´ f Þ

fk = m ´ 3 f 0.6 m= k = 3 3 F 2 . = = 02 = 10 10

Hence, on compairing, we get F =2N

dx L

Total force exerted by the liquid about the length L, M 2 L w ò x dx F = 0 L 1 2 = Mw L 2

4. (b) When a metallic rod is heated, it expands. Due to this its moment of inertia (I) about a perpendicular bisector increases as, or

I µ r2

According to law of conservation of angular momentum, Þ

J = I w (w = angular speed) 1 [Q J is constant] wµ I

Thus, as the inertia (I) increases, the angular speed decreases.

49

Rigid Body and Rotational Motion 5. (a) When the wheel completes half of rotation without slipping, then point A reaches at A¢ . A d B

[\ R = 2 m]

42 + 4p2 = 7.45 m

6. (b) The radius of the circle followed by the masses w I

9. (d) According to theorem of parallel axis, moment of inertia of a circular loop of radius R, at a distance of R / 2 is 2

æRö I = ICM + M ç ÷ è2ø

m q

1 MR2 MR2 + 2 4 3 2 I = MR 4

ro

d

I=

l

10. (c) As the hollow sphere of mass m is filled with

m

liquid of mass m, total mass of sphere becomes 2m. axis

As, angular momentum, L = r ´ p = r ´ mv Þ

1 2 Iw 2

According to law of conservation of energy, 3g L 1 æ ML2 ö÷ 2 w or w = Mg = çç L 2 2 è 3 ÷ø

= (2R )2 + (pR )2

is r = l sin q.

Gain in rotational kinetic energy (K.E) = where, I = ML2 / 3 and w = angular speed

Displacement, d = ( AB )2 + (BA ¢)2

=

distance through which centre of gravity of the rod falls = L / 2. L \ Loss of potential energy = Mg , 2 where, M = mass of rod.

A¢

AB = 2R = 4 m

8. (b) When upper end of the rod hits the ground, the

| L | = lsin q (mw lsin q)

On differentiating, we get d |L| dq = mwl2 2 sin q. cos q dt dt dL 2 2 = 2ml w sin q . cos q Þ dt

[Q v = rw]

Total energy = KE + rotational KE 1 1 Total energy = × (2m)v 2 + Iw2 2 2 = mv 2 +

dq ù é êëQ w = dt úû

= ml2 w2 sin2q [Q sin 2q = 2 sin q cos q]

7. (a) The total kinetic energy = Translational kinetic energy + Rotational kinetic energy 1 1 1 2 1 = Iw2 + Mv 2 = . Mr 2 w2 + Mv 2 2 2 2 5 2 2 é 2ù êQ for sphere, I = 5 Mr ú ú ê û ë(about diameter ) 1 1 7 2 2 2 Mv [Q v = rw] = Mv + Mv = 5 2 10 7 1 = ´ ´ (0.02)2 = 1.4 ´ 10-4 J 10 2

1 æ 2 2 ö v2 4 2 ç mr ÷ × 2 = mv 2è3 3 ø r

11. (c) Kinetic energy of sphere, Kr =

1 2 Iw 2

\ Moment of inertia of sphere, I =

2 MR2 5

\ Rotational kinetic energy of sphere, 1 K r = MR2 w2 5 Total energy of sphere, 1 1 K t = Iw2 + Mv 2 2 2 1 2 1 = ´ MR2 w2 + MR2 w2 [Q v = Rw] 2 5 2 7 2 2 MR w = 10

AIIMS Chapterwise Solutions ~ Physics

50 Total energy of sphere, K t = Kr

\

=

Kt

7 MR2 w2 10

1 / 5 MR2 w2 2 2

7 / 10 MR w

=

(ii) Moment of inertia about its centre and perpendicular to its length æ L2 R2 ö÷ = M çç + 4 ÷ø è 12 According to question,

2 7

12. (a) Q C

C

vC w

rC

P

vP rP

v C rQ

vQ

15. (c) Length of the wire = l

rC

If r be radius of circular ring, then l 2pr = l Þ r = 2p

O

Fig (i)

ML2 MR2 MR2 or L = 3 R + = 12 4 2

w

Fig. (ii)

\ Moment of inertia, I = mr 2 2 æ l ö æ 1 ö = m × ç ÷ = ç 2 ÷ ml2 è 2p ø è 4p ø

From Fig. (i), OC = rC

let \

Velocity, v = rC × w

16. (c) The situation is shown in figure.

Since, angular velocity (w) is constant. \

X

v µ rC

Q

From Fig. (ii), it is clear that rQ > rC > rP \

A

vQ > vC > v P

According to theorem of parallel axis,

X¢

I AB = ICM + Mh2

C

D h

2 MK 2AB = MK CM + Mh2

I xx ¢ = m1 r12 + m2 r22 + m3 r32 + m4 r42

CM

m = mass of body

where,

P

A

13. (b)

B

K AB , K CM = radius of gyration about AB and about centre of mass.

I xx ¢ = m ´ DP 2 + m ´ BQ 2 + m ´ CA2 + O

B

h = distance from centre.

[Q Point mass at A has zero distance from XX ¢] As,

2 K 2AB = K CM + h2

Given, K AB = 10 cm, h = 6 cm Putting then, we get, K CM = 8 cm

2 1 l BD = 2 2

and

DP = BQ =

Þ

æ 2l ö ÷ + m ´ ( 2l)2 I xx ¢ = m ´ 2 ´ çç ÷ è 2 ø

2

14. (b) (i) Moment of inertia of a cylinder about its centre and parallel to its length =

AB = l BD = AC = 2 l

MR2 . 2

Q Point mass at A has zero distance from xx¢. = 3 ml 2

R

17. (b) Let the each side of square D

F

C

lamina be d. L

So, I EF = IGH (due to symmetry) (ii)

(i)

and I AC = I BD (due to symmetry)

G

H O

A

E d

B

51

Rigid Body and Rotational Motion Now, according to theorem of perpendicular axis, Q Moment of inertia about axis AC + Moment of inertia about axis BD = MI about O Þ

I AC + I BD = Io

{Q I AC = I BD }

2I AC = Io

…(i)

Similarly, MI about EF + MI about GH = MI about O Þ

I EF + IGH = Io

Þ

2I EF = Io

…(ii) {Q I EF = IGH }

According to theorem of parallel axis,

So,

I AD =

md 2 4

length = 0. Now, we have to find the moment of inertia of rods A and C about a perpendicular axis through their centre. Therefore, the moment of inertia of the system, ML2 ML2 ML2 + = 12 12 6 Y

angular momentum, L= r ´p = rmv sin f (- k$ ) Therefore, the magnitude of L is, L = mvr sin f = mvd

A

f

B

P r

d O

22. (a) Centre of mass remains at the centre of ladder.

N (x, 4) N q

M/ 2

20. (b) Since, rod is bent at

Moment of inertia of each part through

2 æ2 2ö v [Q M = m] ç MR ÷ 2 = mgh è5 øR 10 Þ v= gh 7 Hence, to climb the inclined surface velocity 10 should be greater than or equal to gh. 7

X

where, d = r sin f is the distance of closest approach of the particle to the origin. As d is same for both the particles, hence L A = L B . the middle, so each part of it will have æLö same length ç ÷ and è2ø æMö mass ç ÷ as shown. O è2ø

energy)

Let ( x, 4) be the co-ordinate of centre of mass and l be the length of rod. l l \ x = cos q and y = sin q 2 2 l2 x2 + y 2 = (sin2 q + cos2 q) \ 4 l2 2 2 x + y = Þ 4 Hence, track of centre of mass of ladder is circle.

I = MI about A + MI about C

19. (b) From the definition of

(Rotational energy)

1 1 Þ mv 2 + 2 2

md 2 = 4 I EF 3

I = I A + IC =

Also, moment of inertia of solid sphere is 2 I = MR2 5 1 1 2 mv 2 + Iw = mgh \ 2 2 (Potential (Translational kinetic energy)

18. (c) The moment of inertia of rod B about its own

Þ

Hence, net moment of inertia through its middle point O is 2 2 ML2 1 æMö æLö 1 æMö æLö I= ç ÷ç ÷ + ç ÷ç ÷ = 3 è 2 ø è2ø 3 è 2 ø è2ø 12 rotation and translation is converted to potential energy.

I AC = I EF I AD = I EF +

1 æMö æLö ç ÷ç ÷ 3 è 2 ø è2ø

21. (a) In the given case, the sum of kinetic energy of

From Eqs. (i) and (ii), we get

\

2

its one end =

mg

(Centre of mass)

So, option (a) is correct.

23. (b) From law of conservation of angular

L/2

momentum J = Iw = constant 60° L/2

M/2

Hence, if I decreases, w increases and vice- versa. When liquid is dropped, mass increases hence, I increases (I = mr 2 ). So, w decreases, but as soon as the liquid starts falling, w increases again.

AIIMS Chapterwise Solutions ~ Physics

52 24. (d) The right hand rule tells about the direction of angular velocity vector which states that the direction of angular velocity vector is always along the axis of rotation. v

25. (d) Angular momentum,

J = Iw = mrv Since, direction of velocity is perpendicular to orbital O plane and J µ v therefore, in an orbital motion, the angular momentum vector v is perpendicular to the orbital plane.

26. (b) About the initial position, the angular momentum of projected body is in clockwise direction. y Mv

Mg

x

Also, the torque on the body due to its weight (Mg) is in clockwise direction. Hence, the angular momentum increases.

27. (d) We know that, t = Ia where, t = torque, I = moment of inertia and a = angular acceleration. Given,

a = 4p rad / s2 , t = 31.4 N-m

Þ

I=

t 31.4 = = 2. 5 kg -m2 . a 4 ´ 314

28. (a) Moment of inertia of a regular circular disc about an axis perpendicular to plane is 1 I = MR2 2 where, M is mass and R is radius. Given, m = 0.4 kg, R = 100 cm =1 m 1 I = ´ 0.4 ´ (1)2 = 0.2 kg -m2 2

\

29. (b) We know that, \

dJ =t dt

Given,

J1 = J, J2 = 5J

\

DJ = J2 - J1 = 5J - J = 4 J

and

dt = 5s 4 t= J 5

So,

30. (a) The moment of inertia of a particle about an axis is given as, I = mr 2 , Þ I µ r 2 At centre of gravity, the distance would be minimum. Thus, the moment of inertia reduced to its minimum value.

31. (b)

I = Mk2 k=

v

I = M

Mr 2 M

k µr Thus, radius of gyration of a body depends upon axis of rotation.

32. (a) Given, mass of the disc = M and radius = R. We know that, moment of inertia of a disc about 1 diameter = MR2 4 And from the theorem of parallel axes, the required moment of inertia 1 5 = MR2 + MR2 = MR2 4 4

33. (b) The radius of earth = R and mass of earth = M The angular momentum about the axis of rotation with period T is 2 2p é 2 ù Iw = MR2 ´ Q for solid sphere, I = MR2 ú 5 T ëê 5 û 4p MR2 = 5T

34. (c) The kinetic energy of a rolling solid sphere = K trans + K rot =

1 1 mv 2 + Iw2 2 2

1 1 2 mv 2 + ´ mR2 w2 2 2 5 2 é 2ù êë As, v = Rw and I = 5 MR úû = 7 / 10 mv 2

=

Different solid bodies have different kinetic energy as for different solid bodies, I is different. Thus, the Assertion is true while Reason is false.

35. (a) When a person is high up on the ladder, then a large torque is produced due to his weight about the point of contact between the ladder and the floor. Whereas, when he starts climbing up, the torque is small. Due to this reason, the ladder is more apt to slip, when one is high up on it. Hence, Assertion and Reason both are true and Reason explanation Assertion.

53

Rigid Body and Rotational Motion 36. (d) If a force is applied at the centre of mass of a rigid body, its torque about centre of mass will be zero but acceleration will be non-zero.

C F

Hence, velocity will change. Hence, Assertion and Reason both are false.

37. (a) When the ice on the polar caps melts, the mass concentrated near the axis of rotation is spread out. The moment of inertia of system increases. As, I = mr 2 Thus,

I µ r2

Since, no external torque acts on the system. Iw = constant. Therefore, if I increases, the angular speed will decrease. So, the time period of rotation of the 2p ö æ earth ç T = ÷ increases i.e., the duration of the wø è day will increase. Hence, Assertion and Reason both ane true and Reason given explanation of Assertion.

38. (d) The position of centre of mass of electron and proton remains at rest. As their motion is due to internal force of electrostatic attraction, which are independent of the mass of particle. Hence, Assertion and Reason both are false.

39. (a) In sliding down, the entire potential energy of body is converted only into translational energy. While in rolling motion, some part of potential energy is converted into kinetic energy of rotation and rest into kinetic energy of translation. Therefore, in sliding motion, the velocity acquired by the body is more. Hence, Assertion and Reason both are true and Reasons explaion of Assertion.

40. (b) In a whirlwind in a tornado, the air from nearby regions gets concentrated in a small space there by decreasing the value of its moment of inertia considerably. Since, Iw = constant, its angular speed increases to high value.

If no external torque acts, then dL = 0 or L = constant t = 0 or dt or Iw = constant Hence, if I changes, then w will also change. Hence, Assertion and Reason both are true but Reason does not explain of Assertion.

41. (a) The judo fighter in order to throw his opponent to the mat tries to initially bend his opponent and then rotate him around his hip because by doing this, the weight of opponent is made to pass through the hip of judo fighter to make its torque zero.

42. (a) Torque is rate of change of angular momentum of a body. dJ =t dt If external torque is zero (t = 0), then Þ

J = constant

Hence, Assertion and Reason both are true and Reason explains Assertion.

43. (a) When large air masses in the earth’s atmosphere shift due to rotation of particle of atmosphere around the axis of rotation of earth. They cause a change in moment of inertia and since angular momentum is to be maintained constant, the angular velocity or speed of rotation changes.

44. (a) The moment of inertia of a rigid body, I = Smr 2 Also, the moment of inertia of a body rotating about an axis with unit angular velocity (w) equals twice the kinetic energy (K) of rotation of that axis. i.e., I=

2K w2

Hence, moment of inertia quantifies the rotational inertia of an object that is, its inertia with respect to rotational motion in a manner some what analogous to how mass quantifies the inertia of an object with respect to translational motion.

AIIMS Chapterwise Solutions ~ Physics CHAPTER

6 Gravitation 1. A spaceship is launched into a circular orbit close to earth’s surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull? (Take, radius of earth = 6400 km, g = 9.8 m/s2 ) [AIIMS 2018] (a) 3.28 km/s (c) 10 km/s

(b) 12 km/s (d) 40 km/s

body projected with a velocity equal to one-third of the escape velocity from the surface of the earth? (Take, radius of the earth = R) [AIIMS 2017] (b) R/3

(c) R/5

(d) R/8

3. Two satellites S 1 and S2 are revolving round a planet in coplanar circular orbits of radii r1 and r2 in the same direction, respectively. Their respective periods of revolution are 1h and 8h. The radius of orbit of satellite S 1 is equal to 10 4 km. What will be their relative speed (in km/h) when they are closest? [AIIMS 2017]

p (b) p ´ 104 ´ 104 2 (c) 2 p ´ 104 (d) 4p ´ 104 (a)

4. Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing w on the earth will weight on the planet is (a) w (c) w/2

surface with kinetic energy KE. The energy required to completely escape from it is [AIIMS 2016] (a) 2 KE (c) 3 KE

(b) 2 KE (d) None of these

6. The reading of a spring balance

2. What is the maximum height attained by a

(a) R/2

5. A body is orbiting very close to the earth

(b) 2w [AIIMS 2016] (d) 21/ 3 w at the planet

corresponds to 100 N while situated at the north pole and a body is kept on it. The weight record on the same scale if it is shifted to the equator, is (Take, g = 10 m/s2 and radius of the earth, R = 6.4 ´ 10 6 m) [AIIMS 2015] (a) 99.66 N (c) 97.66 N

(b) 110 N (d) 106 N

7. The maximum vertical distance through which a full dressed astronaut can jump on the earth is 0.5 m. Estimate the maximum vertical distance through which he can jump on the moon, which has a mean density 2/3rd that of the earth and radius one quarter that of the earth. [AIIMS 2014] (a) 1.5 m (c) 6 m

(b) 3 m (d) 7.5 m

8. The value of g at a particular point is 9.8 m/s2 . Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass. The value of g at the same point (distance of the point from the centre of earth does not change) will now be [AIIMS 2013] (a) 9.8 m/s 2 (c) 19.6 m/s 2

(b) 4.9 m/s 2 (d) 39.2 m/s 2

55

Gravitation 9. If the earth stops moving around its polar axis, then what will be the effect on the value of g for body placed at south pole? [AIIMS 2012]

(a) Remain same (b) Increase (c) Decrease but not zero (d) Decrease zero

year. The work which would have to be done on the earth to bring it to rest relative to the sun is, (ignore the rotation of earth about its own axis) given that the mass of the earth = 6 ´ 1024 kg and distance between sun and earth is 15 . ´ 10 8 km [AIIMS 2011] (b) 2.7 ´ 1031 J (d) +2.7 ´ 1033 J

11. If the value of g at the surface of the earth is 9.8 m/s2 , then the value of g at a place 480 km above the surface of the earth will be (radius of the earth is 6400 km) [AIIMS 2010] 2

(a) 8.4 m/s (c) 7.2 m/s 2

2

(b) 9.8 m/s (d) 4.2 m/s 2

12. At what height above the surface of earth the value of acceleration due to gravity would be half of its value on the surface of earth? (Take, radius of the earth is 6400 km) [AIIMS 2009] (a) 2561 km (c) 3200 km

(b) 2650 km (d) 9800 km

2Gm æ 1ö ç1 ÷ Me R è 10 ø 2GMe æ 1ö (b) v f2 = v i2 + ç1 + ÷ 10 ø Re è 2GMe æ 1ö (c) v f2 = v i2 + ç1 ÷ 10 ø Re è 2Gm æ 1ö (d) v f2 = v i2 + ç1 ÷ Re è 10 ø

[AIIMS 2007]

16. Two spheres of same size, one of mass 2 kg and another of mass 4 kg, are dropped simultaneously from the top of Qutab Minar (height = 72 m). When they are 1 m above the ground, the two spheres have the same [AIIMS 2006] (a) momentum (c) potential energy

(b) kinetic energy (d) acceleration

17. The condition for a uniform spherical mass m of radius r to be a black hole is (Take, G = gravitational constant and g = acceleration due to gravity) [AIIMS 2005] 1/ 2

æ 2Gm ö (a) ç ÷ £c è r ø 1/ 2 æ 2Gm ö (c) ç ÷ ³c è r ø

1/ 2

æ 2 gm ö (b) ç ÷ =c è r ø 1/ 2 æ gm ö (d) ç ÷ ³c è r ø

18. The velocity with which a projectile must

13. The period of revolution of an earth’s satellite close to surface of earth is 90 min. The time period of another satellite in an orbit at a distance of four times the radius of earth from its surface will be [AIIMS 2009] (a) 90 3 min (c) 720 min

initially at a distance of 10Re with speed v i. It hits the earth with a speed v f (Re and Me are radius and mass of earth), then (a) v f2 = v i2 +

10. The earth circles around the sun once a

(a) 2.7 ´ 1030 J (c) -2.7 ´ 1033 J

15. An asteroid of mass m is approaching earth,

be fired so that it escapes earth’s gravitation does not depend on [AIIMS 2003] (a) mass of the earth (b) mass of the projectile (c) radius of the projectile’s orbit (d) gravitational constant

19. The motion of planets in the solar system is

(b) 270 min (d) 360 min

an example of the conservation of

14. A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them, to take the particle far away from the sphere (Take, G = 6.67 ´ 10 -11 Nm2 /kg2 ) [AIIMS 2008] (a) 13.34 ´ 10-10 J

(b) 3.33 ´ 10-10 J

(c) 6.67 ´ 10-9 J

(d) 6.67 ´ 10-10 J

(a) mass (b) linear momentum (c) angular momentum (d) energy

[AIIMS 2003]

20. If vo be orbital velocity of a satellite in a circular orbital close to the earth’s surface and ve is escape velocity from earth, then relation between the two is [AIIMS 2002] (a) ve =2 vo (c) ve = vo 2

(b) ve = 3 vo (d) vo = ve

AIIMS Chapterwise Solutions ~ Physics

56 21. The value of acceleration due to gravity at earth’s surface is g. Its value at the centre of the earth, which we assume as a sphere of radius R and of uniform mass density, will be [AIIMS 2002] (a) 10Rm/s 2 (c) 5 Rm/ s 2

28. The earth rotates about the sun in an elliptical orbit as shown in figure. At which point its velocity will be maximum? B

(b) zero (d) 20Rm/s 2

A

C

S

22. The escape velocity from the earth is about 11 km/s. The escape velocity from a planet having twice the radius and the same mean density as the earth is [AIIMS 2001] (a) 22 km/s (c) 5.5 km/s

[AIIMS 2000] (b) 8R (d) 6R

24. Potential energy of a satellite having mass m and rotating at a height of 6.4 ´ 10 6 m from the earth centre is [AIIMS 2000] (a) - 0.2 mg Re (c) - 0. 5 mgRe

(b) -2 mgRe (d) - mgRe

25. Escape velocity of a body when projected from the earth’s surface is 11.2 km/s. If it is projected at an angle of 50° from the horizontal, then escape velocity is [AIIMS 1999] (a) 12.8 km/s (c) 11.2 km/s

(b) 16.2 km/s (d) 11.8 km/s

(a) intensity (c) force

[AIIMS 1998] (b) field (d) None of these

27. If mass of a body is M on the earth’s surface, the mass of the same body on moon’s surface will be [AIIMS 1997] (a) M (c) zero

(b) at A

(c) at D

(d) at B

planet sweeps out equal areas in equal time interval. This is the statement of [AIIMS 1996] (a) Kepler’s third law (b) Kepler’s first law (c) Newton’s third law (d) Kepler’s second law

30. A body weighed 250 N on the surface assuming the earth to be a sphere of uniform mass density, how much would it weigh half way down to the centre of the earth? [AIIMS 1995] (a) 195 N

M (b) 6 (d) None of these

(b) 240 N

(c) 125 N

(d) 210 N

31. Two satellites of mass m1 and m2 ( m1 > m2 ) are going around the earth in orbits of radius r1 and r2 ( r1 > r2 ). Which statement about their velocities is correct? [AIIMS 1994] (a) v1 < v 2 (c) v1 / r1 = v 2 / r2

(b) v1 > v 2 (d) v1 = v 2

32. In what manner does the escape velocity of a particle depend upon its mass? (a) m0

26. Gravitational mass is proportional to the gravitational

(a) at C

29. The radius vector, drawn from the sun to a

(b) 11 km/s (d) 15.5 km/s

M 23. If the mass of moon is , where M is the 81 mass of earth, find the distance of the point from the moon, where gravitation field due to earth and moon cancel each other. Given that distance between earth and moon is 60R, where R is the radius of earth. (a) 4R (c) 2 R

[AIIMS 1997]

D

(b) m2

(c) m–1

[AIIMS 1994] (d) m

33. If the radius of the earth shrinks by one percent and its mass remaining the same, then acceleration due to gravity on the earth’s surface will [AIIMS 1994] (a) remains constant (c) increase

(b) decrease (d) Either (b) or (c)

34. A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energies, is (a) positive (c) negative

(b) zero [AIIMS 1994] (d) first (b), then (c)

57

Gravitation 35. All the known planets move in [AIIMS 1994] (a) elliptical path (c) hyperbolic path

(b) straight path (d) circular path

36. There is no atmosphere on the moon, because [AIIMS 1994] (a) escape velocity of gas molecules is less than their root mean square velocity (b) it is closer to the earth and also it has the inactive inert gases in it (c) escape velocity of gas molecules is greater than their root mean square velocity (d) it is too far from the sun and has very low pressure in its outer surface

37. The orbital speed of jupiter, is

[AIIMS 1994]

(a) equal to the orbital speed of earth (b) greater than the orbital speed of earth (c) proportional to the distance from the earth (d) less than the orbital speed of earth

Assertion & Reason Direction (Q. Nos. 38-50) Read the Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

38. Assertion An astronaut in an orbiting space station above the earth experience weightlessness. Reason An object moving around the earth under the influence of earth’s gravitational force is in a state of ‘free fall’. [AIIMS 2015]

39. Assertion The ratio of inertial mass to gravitational mass is equal to one. Reason The inertial mass and gravitational mass of a body are equivalent. [AIIMS 2014]

40. Assertion A satellite moving in a circular orbit around the earth has a total energy E 0, then its potential energy is - E 0. Reason Potential energy of the body at a point in a gravitational field of orbit is [AIIMS 2013] GMm / R .

41. Assertion At the centre of earth a body has centre of mass, but no centre of gravity. Reason This is because g = 0 at the centre of earth. [AIIMS 2013]

42. Assertion The length of day is slowly increasing. Reason The dominant effect causing a slowdown in the rotation of the earth is the gravitational pull of other planets in the solar system. [AIIMS 2011] 43. Assertion There is no effect of rotation of earth on acceleration due to gravity at poles. Reason Rotation of earth is about polar axis. [AIIMS 2010]

44. Assertion A tennis ball bounces higher on hills than in plains. Reason Acceleration due to gravity on the hill is greater than that on the surface of earth. [AIIMS 2009]

45. Assertion Generally the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very large height. Reason The path of a projectile is independent of the gravitational force of earth. [AIIMS 2008]

46. Assertion A man in a closed cabin falling freely does not experience gravity. Reason Inertial and gravitational mass have equivalence. [AIIMS 2006]

47. Assertion The earth is slowing down and as a result the moon is coming nearer to it. Reason The angular momentum of the earth moon system is not conserved. [AIIMS 2003]

48. Assertion A balloon filled with hydrogen will fall with acceleration g /6 on the moon. Reason Moon has no atmosphere. [AIIMS 2001]

49. Assertion The square of the period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse. Reason Sun’s gravitational field is inversely proportional to the square of its distance from the planet. [AIIMS 1999]

50. Assertion The comets do not obey Kepler’s laws of planetary motion. Reason The comets do not have elliptical orbits. [AIIMS 1997]

AIIMS Chapterwise Solutions ~ Physics

58

Answers 1. 11. 21. 31. 41.

2. 12. 22. 32. 42.

(a) (a) (b) (a) (a)

3. 13. 23. 33. 43.

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

4. 14. 24. 34. 44.

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

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

5. 15. 25. 35. 45.

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

6. 16. 26. 36. 46.

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

7. 17. 27. 37. 47.

8. 18. 28. 38. 48.

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

9. 19. 29. 39. 49.

(a) (b) (b) (a) (a)

10. 20. 30. 40. 50.

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

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

Explanations 1. (a) The orbital velocity of spaceship in circular

3. (b) From Kepler’s third law,

orbit,

T12

vo = GM / r If spaceship is very close to earth surface,

T22

r = radius of earth = R \

vo =

GM = R

R g = R

GM ö æ Rg ç since, g = 2 ÷ è R ø

vo =

2 /3

The escape velocity of spaceship, ve = (2Rg ) = 2v 0 = 2 ´ 7.92 = 1120 . km/s \ Additional velocity required = 11.20 - 7.92 = 3.28 km/s

é 4 ´ 104 104 ù = 2p ê ú km/h 1 úû êë 8 1 = 2p ´ ´ 104 = p ´ 104 km/h 2 4. (d) The mass of an object is

2. (d) From conservation of energy, total energy at surface = total energy at height H. GMe m 1 1 GMe m mv12 = mv22 R 2 R+ H 2

4 mass = average density ´ volume = r ´ pR3 3 Þ

where, Me , m = mass of earth, body and v1 , v2 = velocity at initial and final position Here v2 = 0, at maximum height,

v12 R Here, \

gp -1

æ 2 GM çQ ve = ç R -1 è

R ve2 v12

1 ve ve Þ =3 3 v1 R R H= = 9-1 8

v1 =

3

æ Rp ö ÷ = çç ÷ Me è Re ø

Mp

…(i)

where, R p and Re are the radius of planet and earth, respectively. The ratio of gravity of two is GM p

GMe m GMe m 1 =mv12 R R+ H 2

=

´ r1

æ 8h ö 4 4 r2 = ç ÷ ´ 10 = 4 ´ 10 km è 1h ø The speed of satellite S2 relative to S1 , ér r ù 2pr2 2pr1 |v2 - v1| = = 2p ê 2 - 1 ú T2 T1 ë T2 T1 û

6.4 ´ 106 ´ 9.8

Solving this, we get R H= 2GMe

2 /3

ö ÷ ÷ ø

Þ

= 7.92 ´ 103 m/s = 7.92 km/s

\

r23

Here, T1 = 1 h, T2 = 8 h and r1 = 104 km

Here, R = 6400 km = 6.4 ´ 106 m, g = 9.8 m / s2 \

r13

æT r2 = çç 2 è T1

Þ

2

=

ö ÷ ÷ ø

ge

2

=

=

R2p M p æ Re ö ÷ = ´ç GMe Me çè R p ÷ø Re2 2 /3

æM ö ´ç e ÷ Me çè M p ÷ø

Mp

[from Eq. (i)]

1 /3

æ Mp ö ÷ = çç ÷ è Me ø

= (2)1 /3

(\M p = 2Me )

59

Gravitation Therefore, weight is w p mg p = = (2)1 /3 or w p = 21 /3 w e we mge

5. (d) For body close to earth, GM GM ö æ = gR v= çQ g = 2 ÷ R è R ø mgR 1 2 Kinetic Energy, K = mv = 2 2 KE required to remove body to infinity K ¢ = mgR mgR =K \Additional KE required = K ¢ - K = 2

6. (a) At the pole, the weight is same as the true one. Thus, 100 N = m(10 m/s2 ) Þ m = 10 kg At the equator, the apparent weight is given by

But

W¢= W mgh¢ = mg ´ 0.5 Þ h¢ = 3 m 6

8. (a) There is no change in the mass of earth and distance of the point from the centre of earth does not change, so value of g will remain same, i.e. 9.8 m/s2 .

9. (a) Variation in g due to rotation of earth at latitude l is g ¢ = g - w2 R cos2 l At poles, l = 90°, we get g pole = g - w2 R cos2 90° \ g pole = g i.e., there is no effect of rotational motion of the earth on the value of g at poles.

10. (c) Angular frequency (w) i.e, given by, w=

mg ¢ = mg - mw2 R Also, the angular speed of an equatorial point on the earth’s surface is 2p 2p w= = T 24 ´ 60 ´ 60 Þ Now,

w = 727 . ´ 10-5 rad/s mg ¢ = 100 - 10(727 . ´ 10-5 )2 ´ 6.4 ´ 103 = 99.66 N

7. (b) For a spherical body, value of g is g =

GM R2

4 3 pR ´ r, 3 R = radius of sphere and r = density. G 4 4 g = 2 ´ pR3r = pGR r Þ 3 3 R For moon, R 2 given that R = e and r = r e 3 4 and g = gm R 4 2 g m = pG ´ e ´ r e Þ 3 4 3 1æ4 ö 1 = ç pG Re r e ÷ = g 6è3 ø 6

= 1.99 ´ 10-7 rad/s Work done, W = Final KE - Initial KE W = K f - Ki 1 1 (Q v = wr ) = 0 - mv 2 = m (rw)2 2 2 1 = - ´ 6 ´ 1024 ´ (1.5 ´ 1011 ´ 1.99 ´ 10-7 )2 2 = - 27 . ´ 1033 J

11. (a) The value of g on the surface of the earth is, g =

where, M = mass of body =

Work done on jumping 0.5 m on earth, W = m ´ g ´ 0.5 Work done on jumping h¢ m on moon, æg ö W ¢ = mç ÷ × h ¢ è 6ø

2p 2p = T 365 ´ 24 ´ 3600

GM R2 1

So,

g µ

\

g¢ = g

R2 At height h from the surface of the earth, 1 g¢ µ (R + h)2 R2 2

(R + h )

=

9.8 ´ (6400)2 (6400 + 480)2

= 8.4 m/s2

12. (b) As The ratio, or or or

g =

GM R2

i.e.,

g µ

1 R2

g¢ 1 R2 = = g (R + h)2 2 R + h = 2R h = ( 2 - 1)R h = (0.414) ´ 6400 = 2650 km

AIIMS Chapterwise Solutions ~ Physics

60 13. (c) From Kepler’s law, T 2 µ R3

Substituting these values in Eq. (i), we get

3 /2

GMe m 1 1 GMe m mv 2i = mv 2f 2 10Re Re 2

T µR

or

3 /2

T ¢ æ R¢ ö =ç ÷ T èRø

Þ

v 2f = v 2i +

2GMe 2GMe Re 10Re

\

v 2f = v 2i +

2 GMe æ 1 ö ç1 ÷ 10 ø Re è

3 /2

or

T ¢ æ 4R ö =ç ÷ T è R ø

\

T ¢ = 8T = 8 ´ 90 = 720 min

= (4)3 /2 = 8

14. (d) The potential energy of particle (initially) is Ui = –

GMm r 6.67 ´ 10-11 ´ 100 ´ 10-2 01 .

Ui = –

R

=

1

m

M = 100 kg

Ui = –

6.67 ´ 10-11 = - 6.67 ´ 10-10 J 01 .

We know that, \ \

…(i)

1 …(ii) mv 2 2 The potential energy is GMm …(iii) U = R The acceleration due to gravity is GM …(iv) g = 2 R From Eqs. (i), (ii), (iii) and (iv), it is clear that momentum, kinetic energy and potential energy are dependent on the mass of body, while acceleration due to gravity is independent of mass of body and hence remains same.

17. (c) A black hole is an object which is, so massive that even light cannot escape from it.

W = DU = U f - U i (Q as, r

• ¥ than U f

= 0)

W = - U i = 6.67 ´ 10-10 J

15. (c) Applying law of conservation of energy for asteroid at a distance 10Re and at earth’s surface, Ki + Ui = K f + U f

…(i)

vi 10 Re

Re Me Earth

Ki =

1 mv 2i 2

Kf =

1 GMe m mv 2f and U f = 2 Re

and U i = -

2Gm r Thus, for a spherical body of mass (m) to be a black hole, if the escape velocity must be greater than and equal to the velocity of light. i.e., ve ³ c 2Gm or ³c r

The escape velocity is ve =

18. (b) To leave the gravitational field of the earth,

vf

Now,

p = mv The kinetic energy is given by, KE =

m =10×10 -13 kg 0.

16. (d) The momentum of a body of mass (m) is

GMe m 10Re

kinetic energy is equal to potential energy. i.e., potential energy = kinetic energy GMe m 1 + = mve2 2 R where, m is mass of projectile, Me is mass of earth, G is gravitational constant and R is radius. 2GMe \ ve = Re The above formula shows that escape velocity is independent of the mass of the projectile.

61

Gravitation 19. (c) From Kepler’s second law, a line joining any planet to the sun sweeps out equal areas in equal intervals of time i.e., the areal speed of the planet remains constant. dA J = = constant dt 2m From Kepler’s law, areal speed is constant therefore, angular momentum J is constant. Hence, Kepler’s second law is equivalent to conservation of angular momentum.

Since, gravitation field cancel each other hence, the force of attraction is equal and opposite.

…(ii)

F2 Earth 60R - x

x

\

earth is given by

ve = 2gRe

M F1

Moon

20. (c) The orbital velocity of a satellite close to …(i) vo = gRe The escape velocity of a body thrown from earth’s surface is

P

M 81

F1 = F2 M æ ö Gç ÷M è 81 ø = GM ´ M x2 (60R - x )2

Taking square root of the above expression, we have 1 1 = 9 x 60R - x

Thus,

gRe 1 vo = = 2gRe 2 ve

Þ Þ

\

ve = 2vo

Hence, distance of that point from moon is 6R.

21. (b) The value of acceleration due to gravity decreases on going below the surface of the earth. Let value of g below the surface of earth be g ¢, then æ h ö g ¢= g çç 1 - ÷÷ è Re ø At the centre of earth, h = Re æ R ö g ¢= g çç 1 - e ÷÷ = 0 \ è Re ø

22. (a) The escape velocity (ve ) is ve

Þ

v vp ve

Þ

4 Gr × pR3 GM 3 = = R R 4 2 prGR = 3 µR 2R = =2 R

v p = 2ve = 2 ´ 11 = 22 km/s

23. (d) From Newton’s law of gravitational, F=

Gm1 m2

r2 where m1 , m2 are masses and r is the distance between the two masses.

9 x = 60R - x x = 6R

24. (c) Gravitational potential energy of a body at a height h from earth's surface is given by GMe m U =Re + h Height of satellite from earth is h = 6.4 ´106 m = 6400 km = Re Also, GMe = gRe2 \

(Qg = GM / R2 )

gmRe2 mgRe GMe m ==2Re 2Re 2 = - 0.5mgRe

U =-

25. (c) Escape velocity is given by ve = 2GM / r where, G is gravitational constant, M is mass and r is radius of orbit. From the equation, it is clear that escape velocity is independent of angle of projection. Hence, it remains same when projected at an angle of 50°.

26. (c) From Newton’s law of gravitation, the force of attraction acting between the gravitational mass (m) and earth's mass (M ) is directly proportional to the product of masses and inversely proportional to the square of distance (r ) between both the masses. Hence,

F=G

Mm

r2 where, G is gravitational constant.

AIIMS Chapterwise Solutions ~ Physics

62 Fr 2 GM

Þ

m=

Þ

m µF

31. (a) According to Newton’s law of gravitation, F =

Hence, gravitational mass is proportional to gravitational force.

27. (a) Mass is a property of a physical object that quantifies the amount of matter it contains. Unlike weight, the mass of something stays the same regardless of location. Hence, mass of body on moon is M.

28. (b) From law of conservation of angular,

32. (a) Escape velocity (ve ) = 2gRe .

where, I is moment of inertia and w is angular velocity. \ J = mr 2 w = constant 1 wµ 2 Þ r

Therefore, it is independent of the mass of the particle or it will depend on m0 .

33. (c) Acceleration due to gravity on the earth’s

Hence, w (angular velocity) is maximum when r is minimum that is, earth is nearest to the sun i.e., at A.

29. (d) According to the Kepler’s second law, the line joining the sun and a planet sweeps out equal area in equal interval of time. A planet takes the same time to travel from A to B as from C to D as shown in figure. B A q

Sun D

The shaded area are equal.

30. (c) Given, weight of the body on the

R/2

surface of the earth.

Now, Re shrinks by 1%, so the new value for radius of the earth is 0.09 Re . g GMe = >g g¢ = \ 2 2 0.9801 (0.99) Re Thus, acceleration due to gravity will increase.

34. (c) When missile is launched with a velocity

ö ÷ ÷ ø

But if the velocity is less than escape velocity, then the potential energy becomes more than the kinetic energy and hence, total energy of system becomes negative.

35. (a) According to Kepler’s law of orbit, each P

4 æ R ö3 M ¢ = p ç ÷ ´ r [mg = 250 N] 3 è2ø 1 M 4 = pR3r ´ = 8 3 8 where, M = mass of earth on the surface. GM ¢ GM ¢ GM g Now, g ¢ = = 4 2 = 4´ 2 = 2 2 8R R R æ ö ç ÷ è2ø mg 250 mg ¢ = = = 125 N Þ 2 2

surface is given by g = GMe / Re2

equal to escape velocity, then sum of kinetic energy and potential energy is, 1 æ - GMm ö E = mve2 + ç ÷ 2 è R ø æ 2GM 1 2GM GMm E= m = 0 çç as, ve = R 2 R R è

C

Then, the effective mass of earth at a depth R / 2 below is

r2 In equilibrium, the gravitational force provide the centripetal force to the satellite, GM mv 2 GMm So = Þ v= r r r2 1 vµ Þ r For two masses, v1 / v2 = r2 / r1 Since r1 > r2 , therefore v1 < v2 .

J = Iw = constant

dq

GMm

planet moves around the sun in an elliptical orbit with the sun at of the foci shown in figure. Minor axis r A Major axis

S F1

a ea

O b

F2

B

63

Gravitation 36. (a) Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on surface of moon is small. Therefore, the value of escape speed on the surface of moon is small. The molecules of gases on the surface of moon have thermal speed greater than the escape velocity. That is why all the molecules of gases have escaped and hence, there is no atmosphere on the moon.

37. (d) Orbital velocity (vo ) =

GM r

1ö æ ç vo µ ÷ where, r is the distance from sun. rø è Since, the distance of jupiter from the sun is greater than the distance between the sun and the earth, therefore orbital speed of jupiter is less than the orbital speed of earth.

38. (a) When a person is sitting in a satellite, the gravitational force or weight is count balanced by the centrifugal force acting on the person and hence, the net force is zero and the person feels weightlessness. Hence, option (a) is correct.

39. (a) The ratio of gravitation force on a body due to two gravitational masses m1 and m2 is, F1 GMm1 / R2 m1 = = F2 GMm2 / R2 m2

… (i)

For the inertial masses m1 and m2 , the ratio of force of inertia is F1 m1 g m1 … (ii) = = F2 m2 g m2 From Eqs. (i) and (ii), it is clear that both are identical. Thus, the ratio of inertial mass to gravitation mass is equal to one. Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.

40. (d) The kinetic and potential energy for satellite of mass m are given by, KE =

- GMm GMm and PE = R 2R

Total energy, Eo = KE + PE GMm GMm = 2R R =

- GMm PE = 2R 2

PE = 2Eo Hence, both Assertion and Reason are false.

41. (a) At distance (h), the acceleration due to gravity is,

hö æ g ¢ = g ç1 - ÷ Rø è At centre, h = R, then g ¢ = g (1 - 1) = 0 \A body has no weight at the centre of earth, and hence, no centre of gravity. But centre of mass of a body is not related to gravity. So, centre of mass exist. Hence, both Assertion and Reason are true and Reason is the correct explanation of Assertion.

42. (d) Length of the day depends upon the angular speed of earth about it’s own axis, the earth is not slowing down due to conservation of angular momentum. Thus, the rotational speed remains constant. But the revolutional speed increases when it is closer to sun and decreases when it is away from sun. Thus, the total momentum remains same. Hence, option (d) is correct.

43. (a) The variation in g due to rotation of earth, g ¢ = g - w2 R cos2 l where, l = lattitude of point on earth. At poles, l = 90° \ g ¢ = g - w2 R cos 90° g¢ = g Thus, no effect of rotation on g.

44. (c) Suppose that the tennis ball bounces with a velocity u. It will go up, till its velocity becomes zero. If h is the height up to which it rises on the hill, then (0)2 - u2 = 2(- g ¢)h where, g ¢is acceleration due to gravity on the hill. u2 h= \ 2g ¢ Since, the acceleration due to gravity on the hill (g ¢) is less than that on earth (effect of height), it follows that tennis ball will bounce higher on hills than in plains.

45. (c) Upto ordinary heights, the change in the distance of a projectile from the centre of earth is negligible compared to the radius of earth. Hence, the projectile moves under a nearly uniform gravitational force and the path is parabolic. But for the projectiles moving to a large height, the gravitational force decreases with decreasing (variable) force and the path of the projectile becomes elliptical.

AIIMS Chapterwise Solutions ~ Physics

64 46. (a) In a closed cabin falling freely, the forces are acting on the man as shown mg - R = mg or R = mg - mg = 0 Since, reaction on the man is zero hence, man does not experience gravity. R

Man

g

mg

Inertial mass of a body is the ratio of net force acting on the body to the acceleration produces in the body, i.e. F m= a Gravitational mass of a body is, the ratio of gravitational force acting on the body to the acceleration due to gravity, w i.e. m= g Inertial mass and gravitational mass are equivalent. Hence, option (a) is true.

47. (d) Earth is not slowing down. Angular momentum of earth-moon system is conserved, had it not been so, then Kepler’s second law of planetary motion would not have been valid. Kepler’s second law states that dA J = = constant dt 2m dA is areal speed, J is angular momentum dt and m is mass. Hence, angular momentum is constant.

where

48. (a) Earth is surrounded by an atmosphere of gases, while there is no such atmosphere around the moon. Reason is that escape velocity at moon is only 2.38 km s -1 . At the temperature of moon the average velocity of the gas molecules is greater than this, therefore gas molecules cannot stay at moon. Also, gravity on moon is one-sixth of gravity on earth. Hence, there will be no resistance faced by a body falling on moon, the balloon will fall g with acceleration . 6

49. (b) Let’s assume that the semi-major axis of the ellipse be equal to the average distance of the sun from the planet. By applying Newton’s law, GMm a2 where,

\ Þ Þ

m a M

= m (w2 a)

w = angular velocity of the planet. 2p (T = time period of the planet) = T (2p )2 GMm =m 2 a 2 a T 2 ö æ p 4 ÷ a3 T 2 = çç ÷ è GM ø T 2 µ a3

50. (b) The gravitational force on a comet due to sun is not normal to the comet’s velocity, but the work done by the gravitational force over every complete orbit of the comet is zero. Because the gravitational force is a conservative force and the work done by a conservative force over a closed path is always zero. Hence, comets do not have elliptical orbits and do not obey Kepler’s law.

CHAPTER

7 Mechanical Properties of Solids 1. The length of a metal wire is l1, when the tension in it is T1 and length is l2 when the tension is T2 . The natural length of the wire is [AIIMS 2015]

(c)

(b) l1 l2

l1 T2 - l2T1 T2 - T1

(d)

(a) 133 . ´ 1011 N/m2

(b) 133 . ´ 1012 N/m2

(c) 7.5 ´ 10-13 N/m2

(d) 3 ´ 1010 N/m2

6. The given graph shows, the extension ( Dl)

l1 T2 + l2 T1 T1 + T2

2. The Young’s modulus of a wire of length L and radius r is Y N / m 2 . If the length is L r reduced to and radius . Its Young’s 2 2 modulus will be [AIIMS 2013] Y 2 (c) 2Y

at the breaking point are 2 ´ 10 11 N/ m2 and 0.15 respectively. The stress at the breaking point for steel is therefore [AIIMS 2010]

of a wire of length 1 m suspended from the top of a roof at one end with a load W connected to the other end. If the cross-sectional area of the wire is 10 -6 m2 , calculate the Young’s modulus of the material of the wire. l (H 10 –4 )m

l +l (a) 1 2 2

5. In steel, the Young’s modulus and the strain

(b) Y

(a)

(d) 4Y

4 3 2 1

3. In designing a beam, for its use to support a load. The depression at centre is proportional to (where, Y is Young’s modulus) [AIIMS 2012] (a) Y 2

(b) Y

(c)

1 Y

(d)

1 Y2

4. A force F is applied on the wire of radius r and length L and change in the length of the wire is l. If the same force F is applied on the wire of the same material and radius 2r and length 2L, then the change in length of the other wire is [AIIMS 2010] (a) l

(b) 2l

(c)

l 2

(d) 4l

0 11

(c) 3 ´ 10

[AIIMS 2008] -11

2

(a) 2 ´ 10 N/ m -12

W(N)

20 40 60 80

(b) 2 ´ 10 2

N/ m

N/m2

-13

(d) 2 ´ 10

N/m2

7. For a constant hydraulic stress on an object the fractional change in the object’s volume ( DV / V ) and its bulk modulus ( B) are related as [AIIMS 2005] DV µB V DV (c) µ B2 V

(a)

DV 1 µ V B DV 1 (d) µ 2 V B

(b)

AIIMS Chapterwise Solutions ~ Physics

66 8. The bulk modulus of a metal is 10 10 N/ m2 and Poisson’s ratio 0.2. If average distance between the molecules is 3 Å, then the interatomic force constant is [AIIMS 2002] (a) 5.4 N/m (c) 7.5 N/m

(b) 75 N/m (d) 30 N/m

strain X is produced, then the potential energy stored in its unit volume will be [AIIMS 2001] (b) 0.5 Y 2 X (d) Y X 2

10. According to Hooke’s law of elasticity, if stress is increased, then the ratio of stress and strain is [AIIMS 2001]

(b)

S 2Y

(c)

2Y S

2

(d)

S2 2Y

Direction (Q.Nos. 14-16) Read the Assertion and Reason carefully to mark the correct option from those given below. (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

represents incompressibility of the material. Dp Reason B = , where symbols have DV / V their usual meaning. [AIIMS 2011]

11. Which one of the following affects the [AIIMS 1999]

15. Assertion Stress is the internal force per

(a) Change in temperature (b) Hammering and annealing (c) Impurity in substance (d) All of the above

unit area of a body. Reason Rubber is more elastic than steel. [AIIMS 2002]

12. Longitudinal strain is possible in (a) liquids (b) gases (c) solids (d) All of the above

(a) 2S 2 Y

14. Assertion Bulk modulus of elasticity B

(a) becomes zero (b) remains constant (c) decreases (d) increases

elasticity of a substance?

material of wire, then energy stored in the wire per unit volume is [AIIMS 1997]

Assertion & Reason

9. If a wire of Young’s modulus Y , longitudinal

(a) 0.5 Y X 2 (c) 2 Y X 2

13. If S is stress and Y is Young’s modulus of a

16. Assertion The bridges are declared unsafe

[AIIMS 1998]

after a long time. Reason Elastic strength of the bridges decreases with time. [AIIMS 2001]

Answers 1. (c) 11. (d)

2. (b) 12. (c)

3. (c) 13. (d)

4. (c) 14. (a)

5. (d) 15. (c)

6. (a) 16. (a)

7. (b)

8. (a)

9. (a)

10. (b)

67

Mechanical Properties of Solids

Explanations 1. (c) As we know, Young’s modulus of elasticity, Y=

Stress Force / Area = Strain Change in length Original length

l1 =

Þ

(Q given, l1 = l )

F / A (stress ) Dl (strain) L Dl \Breaking stress, F/A = Y × = 2 ´ 1011 ´ 015 . L

5. (d) Young’s modulus, Y =

In first case, Þ

l2 ×4 2 l l l2 = 1 = 2 2

Þ

Y=

T1 ´ L A ´ (l 1 - L )

…(i)

where, L is the original length of the wire and A is cross -sectional area of the wire.

6. (a) From the graph, l = 10-4 m, F = 20 N,

In second case, T2 ´ L Y= A ´ (l 2 - L )

Q Young’s modulus, Y = \

2. (b) As we know that, Young’s modulus is a So, it is independent from length, width, etc. 3. (c) As we know that, the depression in a rectangular beam, is given by W depression, d = .l3 4Y bd 3 where, W is the load at centre, l is the length, b is the width and d is the depth of beam. 1 Thus, for a given shape of a beam, d µ Y

4. (c) As we know that, the Young’s modulus,

From Eqs. (i) and (ii), we get Yl1 Y l2 A = A L 1 2L 2 l l1 A1 = 2 A2 Þ 2 l Þ l1 pr 2 = 2 . p (2r )2 2

F ×L 20 ´ 1 = 2 ´ 1011 N / m2 = A × l 10–6 ´ 10–4

stress to the volume strain is called the bulk modulus of the material. Normal stress Dp B= =Volume strain DV / V Q Given, stress is constant, Dp = constant 1 1 DV \ Þ Bµ µ DV /V V B

8. (a) The relation between bulk modulus (B ), Poisson’s ratio (s) and Young’s modulus (Y ) is Y = 3B (1 - 2s) Given,

Stress F / A Y= = Strain Dl / L

Similarly for second case, Y l2 F = A 2L 2

Y=

Stress (F / A ) Strain (l / L )

7. (b) When strain is small, the ratio of the normal

property of a material.

F / A1 Y × l1 A1 ÞF = l1 / L L

A = 10-6 m2 , L = 1 m

…(ii)

From Eqs. (i) and (ii), we get T l - T1 l2 L= 2 1 T2 - T1

In first case, Y =

= 3 ´ 1010 N / m2

\ …(i)

B = 1010 N / m2 , s = 02 .

Y = 3 ´ 1010(1 - 2 ´ 02 . ) = 1.8 ´ 1010 N / m2

Interatomic force constant is k = Yr = 1.8 ´ 1010 ´ 3 ´ 10-10 = 5.4 N / m

9. (a) When a wire is stretched, work is done …(ii)

against the interatomic forces. This work is stored in the wire in the form of elastic potential energy. Elastic potential energy per unit volume is 1 1 u = ´ stress ´ strain = ´ Y ´ (strain)2 2 2 stress ù é êëQ Y = strain úû 1 [Q strain = X] = Y X 2 = 0.5 Y X 2 2

AIIMS Chapterwise Solutions ~ Physics

68

u=

U 1 = ´ stress ´ strain V 2

\

u=

1 (stress )2 × 2 Y

\

u=

1 S2 2 Y

10. (b) Hooke’s showed experimentally that, if strain is small, then the stress is proportional to strain. The ratio of stress and strain is constant for material of the given body and is called modulus of elasticity E. Thus, stress E= = constant strain Hence, if stress is increased, strain also increases. Thus, the ratio of stress and strain remains constant.

11. (d) Elasticity is a property of an object by virtue of which it undergoes elastic deformation in response to stress. Hammering increases elasticity while anealing decreases it. When temperature increases, elasticity also increases. Also, impurity in a substance increases, the elasticity of the substance.

12. (c) When the size or shape of a body is changed under an external force, the body is said to be strained. The change occured in the unit size of the body is called strain. When the length of a wire increases under an applied load, the ratio of increase in length to original length is called longitudinal strain. This is possible only in solids.

13. (d) When a wire is stretched, work is done against the interatomic forces. This work is stored in the wire in the form of elastic potential energy.

stress ù é êëQ Y = strain úû [Q given, stress = S]

14. (a) Bulk modulus of elasticity measures, how good the body is to regain its original volume on being compressed. Therefore, it represents incompressibility of the material.

15. (c) The internal restoring force acting per unit area of cross-section of the deformed body is called stress. Also, elasticity is the property of the material of a body by virtue of which it opposes any change in its size or shape. Greater the force required, more elastic the body is. Hence, steel is more elastic than rubber.

16. (a) A bridge during its use undergoes alternating strains for a large number of times each day, depending upon the movement of vehicles on it. When a bridge is used for long time, it losses its elastic strength. Due to which, the amount of strain in the bridge for a given stress will become large and ultimately, the bridge may collapse. This may not happen, if the bridges are declared unsafe after long use.

CHAPTER

8 Mechanical Properties of Fluids 1. A ring of radius R is kept on water surface. Surface tension of water is T and mass is m. To lift the ring from water surface, force will be [AIIMS 2018]

3. Determine the height above the dashed line XX¢ attained by the water stream coming out through the hole situated at point B in the diagram given below. Given that h = 10 m, L = 2 m and a = 30º. [AIIMS 2016]

(b) mg + pTR (d) mg + 4pTR

(a) 0 (c) mg + 2pTR

2. Water is flowing in a non-viscous tube as shown in the diagram. The area of cross-section at point A and B are A1 and A2 , respectively. The pressure difference between point A and B is Dp. The rate of flow is A B A1

A2

[AIIMS 2018] (a) A1 A2 (b) A1 A2 (c) A1 A2

h L X¢

a=30°

B

(a) 10 m (c) 5 m

(b) 7.1 m (d) 3.2 m

4. A spherical ball is dropped in a long column of viscous liquid. Which of the following graphs represent the variation of (i) gravitational force with time (ii) viscous force with time (iii) net force acting on the ball with time? [AIIMS 2015]

2Dp r( A12

-

X

A22 )

F

Dp r( A12 - A22 )

Q

P

2Dp A12 - A22

(d) None of the above

R t

(a) Q, R, P

(b) R,Q, P

(c) P,Q, R

(d) R, P,Q

AIIMS Chapterwise Solutions ~ Physics

70 5. A boy of mass m stands on one end of a wooden plank of length L and mass M. The plank is floating on water. If the boy walks from one end of the plank to the other end at a constant speed, the resulting displacement of the plank is given by [AIIMS 2013] mL (a) M mL (c) (M + m)

ML (b) m mL (d) (M - m)

density 9 has a concentric spherical cavity and just sinks in water. If the radius of sphere be R, then the radius of cavity ( r) will be related to R as [AIIMS 2013] 8 (a) r = R 3 9 8 3 (c) r 3 = R 3

(density = 19.5 kg/m3) is 0.2 m/s in viscous liquid (density = 1.5 kg / m3), find the terminal speed of a sphere of silver (density =10.5 kg / m3) of the same size in the same liquid. [AIIMS 2009] (a) 0.4 m /s (b) 0.133 m/s (c) 0.1 m/s (d) 0.2 m/s

12. A small ball of densityr is dropped from a

6. A sphere of solid material of relative

3

11. If the terminal speed of a sphere of gold

2 (b) r = R 3 3 2 3 (d) r 3 = R 3 3

7. The ratio of radius of two bubbles is 2 : 1.

height ‘h’ into a liquid of density s(s > r). Neglecting damping forces, the maximum depth to which the body sinks is [AIIMS 2008] (a)

(b) 1 : 4

(c) 2 : 1

(d) 4 : 1

8. The coefficient of viscosity for hot air is [AIIMS 2012] (a) greater than the coefficient of viscosity for cold air (b) smaller than the coefficient of viscosity for cold air (c) same as the coefficient of viscosity for cold air (d) increase or decrease depending on the external pressure

9. A rectangular vessel when full of water, takes 10 min to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water? [AIIMS 2011] (a) 9 min (c) 5 min

(b) 7 min (d) 3 min

(b)

hs r-s

(c)

(s - r ) h(s - r ) (d) h s r

13. When a body falls in air, the resistance of air depends to a great extent on the shape of the body. Three different shapes are given. Identify the combination of air resistances which truly represents the physical situation? (the cross-sectional areas are the same.) [AIIMS 2007] R

What is the ratio of excess pressure inside them? [AIIMS 2012] (a) 1 : 2

hr r-s

R

(1)

R

(2) w

(3) w

w

(a) 1 < 2 < 3 (c) 3 < 2 < 1

(b) 2 < 3 < 1 (d) 3 < 1 < 2

14. By sucking through a straw, a student can reduce the pressure in his lungs to 750 mm of Hg (density = 13.6 g / cm3). Using the straw, he can drink water from a glass upto a maximum depth of (Take, pressure of atmosphere = 760 mm of Hg) [AIIMS 2006] (a) 10 cm

(b) 75 cm

(c) 13.6 cm (d) 1.36 cm

15. A candle of diameter d is floating on a liquid in a cylindrical container of diameter D ( D > > d ) as shown in figure.

following will not be there for a fluid?

If it is burning at the rate of 2 cm/h, then the top of the candle will

(a) Viscosity (b) Surface tension (c) Pressure (d) Archimedes’ upward thrust

(a) remain at the same height (b) fall at the rate of 1 cm/h (c) fall at the rate of 2 cm/h (d) go up at the rate of 1 cm/h

10. If there were no gravity, which of the [AIIMS 2011]

L L d

D

[AIIMS 2005]

71

Mechanical Properties of Fluids 16. A given shaped glass tube having uniform cross-section is filled with water and is mounted on a rotable shaft as shown in figure. If the tube is rotated with a constant angular velocity w, then [AIIMS 2005] A

B

20. A soap bubble in vacuum has a radius 3 cm and another soap bubble in vacuum has radius 4 cm. If two bubbles coalesce under isothermal condition, then the radius of the new bubble will be [AIIMS 2002] (a) 7 cm

(b) 5 cm

(c) 4.5 cm

(d) 2.3 cm

21. If work done in increasing the size of a

L

soap film from 10 cm × 6 cm to 10 cm × 11 cm is 2 ´ 10 -4 J, what is the surface tension? [AIIMS 2000]

2L

(a) 2 ´ 10-8 Nm-1 (c) 2 ´ 10-4 Nm-1

(a) water levels in both sections A and B go up (b) water level in section A goes up and that in B comes down (c) water level in section A comes down and that in B it goes up (d) water level remains same in both sections

17. In old age arteries carrying blood in the human body become narrow resulting in an increase in the blood pressure. This follows from [AIIMS 2004] (a) Pascal’s law (b) Stokes’ law (c) Bernoulli’s principle (d) Archimedes’ principle

18. Bernoulli’s equation is a consequence of conservation of (a) energy (c) angular momentum

[AIIMS 2003] (b) linear momentum (d) mass

19. A lead shot of a 1 mm diameter falls through a long column of glycerine. The variation of its velocity v with distance covered is represented by [AIIMS 2003]

(b) v

(a) v Distance covered

(c)

(d) v

22. A hole is made at the bottom of the tank filled with water (density 1000 kg/m3 ). If the total pressure at the bottom of the tank is 3 atmosphere (Take, 1 atmosphere = 10 5 N /m 2 ), then the velocity of efflux is

v

[AIIMS 2000] (b) 400 m/s (d) 800 m/s

(a) 200 m/s (c) 500 m/s

23. The work done in splitting a drop of water of 1 mm radius into 10 6 droplets is (Take, surface tension of water = 72 ´ 10 -3 N / m) [AIIMS 1998] (b) 10.98 ´ 10-5 J (d) 8.95 ´ 10-5 J

(a) 5.98 ´ 10-5 J (c) 16.95 ´ 10-5 J

24. The excess of pressure inside the first soap bubble is three times that inside the second bubble, then the ratio of volume of the first to the second bubble will be [AIIMS 1998] (a) 1 : 27

(b) 3 : 1

(c) 1 : 3

(d) 1 : 9

25. The relative humidity on a day, when partial pressure of water vapour is 0.012 ´ 10 5 Pa at 12º C is (Take, vapour pressure of water at this temperature is 0.016 ´ 10 5 Pa) [AIIMS 1996] (a) 70%

Distance covered

(b) 2 ´ 10-2 Nm-1 (d) None of these

(b) 40%

(c) 75%

(d) 25%

26. A cylindrical vessel is filled with water as shown in figure. A hole should be bored so that the water comes out upto maximum distance from the surface

H

[AIIMS 1996] Distance covered

Distance covered

3H (a) 4

H (b) 4

(c) H

(d)

H 2

AIIMS Chapterwise Solutions ~ Physics

72 27. The surface tension of a liquid decreases with a rise in

[AIIMS 1994]

(a) diameter of container (b) temperature of the liquid (c) thickness of container (d)viscosity of the liquid

Assertion & Reason Direction (Q.Nos. 28-43) Read the Assertion and Reason carefully to mark the correct option from those given below.

33. Assertion A small drop of mercury is spherical but bigger drops are oval in shape. Reason Surface tension of liquid decreases with increase in temperature. [AIIMS 2011]

34. Assertion To float, a body must displace liquid whose weight is greater than actual weight of the body. Reason During floating the body will experience no net downward force in that case. [AIIMS 2011]

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

35. Assertion A ship floats higher in the water

28. Assertion Viscous force is measurement of

36. Assertion A bubble comes from the bottom

resistance of liquid. Reason It converts kinetic energy into heat energy of liquid. [AIIMS 2018]

29. Assertion The stream of water flowing at high speed from a garden hose pipe, tends to spread like a fountain when held vertically up but tends to narrow down when held vertically down. Reason In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant. [AIIMS 2016]

30. Assertion When height of a tube is less than liquid rise in the capillary tube, the liquid does not overflow. Reason Product of radius of meniscus and height of liquid in capillary tube always remains constant. [AIIMS 2014]

31. Assertion If a liquid in a vessel is stirred and left to itself, the motion disappear after few minutes. Reason The moving liquid exerts equal and opposite force. [AIIMS 2013]

32. Assertion A large force is required to be drawn apart normally two glass plates enclosing a thin water film. Reason Water works as glue and sticks two glass plates. [AIIMS 2012]

on a high pressure day than on a low pressure day. Reason Floating of ship in the water is not possible because of buoyancy force which is present due to pressure difference. [AIIMS 2009] of a lake to the top. Reason Its radius increases.

[AIIMS 2008]

37. Assertion In taking into account, the fact that any object which floats must have an average density less than that of water, during world war I, a number of cargo vessels are made of concrete. Reason Concrete cargo vessels were filled with air. [AIIMS 2007]

38. Assertion A thin stainless steel needle can lay floating on a still water surface. Reason Any object floats when the buoyancy force balances the weight of the object. [AIIMS 2006]

39. Assertion For Reynold’s number Re > 2000, the flow of fluid is turbulent. Reason Inertial forces are dominant compared to the viscous forces at such hight Reynold’s numbers. [AIIMS 2005]

40. Assertion In a pressure cooker, the water is brought to boil. The cooker is then removed from the stove. Now on removing the lid of the pressure cooker, the water starts boiling again. Reason The impurities in water bring down its boiling point. [AIIMS 2004]

73

Mechanical Properties of Fluids

Reason The material of the balloon can be easily stretched. [AIIMS 2000]

41. Assertion Smaller drops of liquid resist deforming forces better than the larger drops.

43. Assertion A needle placed carefully on the

Reason Excess pressure inside a drop is directly proportional to its surface area.

surface of water may float, whereas a ball of the same material will always sink.

[AIIMS 2002]

Reason The buoyancy of an object depends both on the material and shape of the object. [AIIMS 1998]

42. Assertion The size of a hydrogen balloon increases as it rises in air.

Answers 1. 11. 21. 31. 41.

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

2. 12. 22. 32. 42.

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

3. 13. 23. 33. 43.

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

4. 14. 24. 34.

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

5. 15. 25. 35.

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

6. 16. 26. 36.

7. 17. 27. 37.

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

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

8. 18. 28. 38.

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

9. 19. 29. 39.

(b) (a) (a) (a)

10. 20. 30. 40.

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

Explanations 1. (d) The required force to lift the ring from water surface F = weight of ring + force due to surface tension

3. (d) Let the velocity at point B be v B . From conservation of total mechanical energy, potential energy = kinetic energy

F

L Sin a

h

Mg

X

= mg + 2T (2pR ) = mg + 4pTR

2. (a) By Bernoulli’s theorem, 1 1 (for horizontal tube) p1 + rv12 = p2 + rv22 2 2 1 ...(i) p1 - p2 = r(v22 - v12 ) 2 But, by the principle of continuity, Q = A1 v1 = A2 v2 From Eqs. (i) and (ii), 1 æ Q2 Q2 ö p1 - p2 = r ç 2 - 2 ÷ 2 çè A2 A1 ÷ø Þ

Q = A1 A2 = A1 A2

2( p1 - p2 ) r ( A12 - A22 ) 2Dp r( A12 - A22 )

...(ii)

a = 50° A

B

L Cos a

1 …(i) mg (h - L sin a ) = mv 2B 2 where, Lsin a is the vertical component of L at point B. Þ v 2B = 2g (h - L sin a ) 1ö æ = 2g ç 10 - 2 ´ ÷ 2ø è 2 Þ v B = 18g Now, let maximum height attained by water stream be h . From Eq. (i), 1 mg (H - L sin a ) = mv 2B sin2 a 2 where, v B sin a is the vertical component of velocity at B. 1 2 v sin2 a Þ H - L sin a = 2g B

AIIMS Chapterwise Solutions ~ Physics

74 \Þ

Þ

v 2B sin2 a 2g 2 æ1ö 18g ç ÷ 1 è2ø H = 2´ + 2g 2 18 . m = 1+ = 325 8

H = L sin a +

8. (a) Viscosity of gases is due to diffusion of gas molecules from one moving layer to another. Since, rate of diffusion of gas molecule is directly proportional to the square root of absolute temperature, hence coefficient of viscosity (h) also proportional to the square root of temperature. Hence, it increases with temperature.

4. (c) Gravitational force remains constant for the falling spherical ball. It is represented by straight line P. Now, the viscous force (F = 6phrv ) increases as the velocity increases with time. Hence, it is represented by curve Q. Net force = Gravitational force – Viscous force As viscous force increases, net force decreases and finally becomes zero. Then, the body falls with a constant terminal velocity. It is thus represented by curve R.

5. (c) Before the boy starts walking on the plank, both the boy and the plank are at rest. So, total momentum of (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 system is æV m ö çç = ÷ mv - (M + m)V = 0 or v M + m) ÷ø ( è Since, distance moved is proportional to speed, the displacement L¢ of the plank is given by, L¢ V m mL or L ¢= = = L v ( M + m) ( M + m)

6. (a) As, the sphere is just sink in water, then weight of the sphere = weight of water displaced Also, weight of whole sphere = weight of solid part – weight of cavity Given, the density of solid = 9 and we know that, density of water = 1 4 4 æ4 ö \ ç pR3 - pr 3 ÷ 9 ´ g = pR3 ´ 1 ´ g 3 3 è3 ø Þ

3

3

3

3

3

9R - R = 9r Þ 9r = 8R æ 8ö r 3 = ç ÷ R3 è 9ø

7. (a) The excess pressure inside the bubble in air is 1 r where, T = surface tension and r = radius of bubble, p1 r2 1 = = p2 r1 2 pµ

9. (b) If A0 is the area of orifice at the bottom below the free surface and A be the area of rectangular vessel, time t taken to the emptied the tank, A 2H t = A0 g \

t1 = t2

H1 H2

Þ

t1 = t2

H1 t Þ 1 = 2 t2 H1 / 2

\

t2 =

t1 10 = = 5 2 » 7 min 2 2

10. (d) The magnitude of upward buoyant force (B) is given by Archimedes’ principle as, …(i) B = Vrg Thus if g = 0, then B = 0 or the fluid does not experience Archimedes’ upward thrust, if there were no gravity.

11. (c) Terminal speed of spherical body in a viscous liquid is given by 2r 2 (r - s)g vT = 9h where, r = density of substance of body and s = density of liquid. From given data, r Ag - sl v T (Ag ) = v T (Gold) rGold – sl 10.5 - 1.5 9 v T (Ag ) = ´ 02 . = ´ 02 . = 01 . m/s Þ 19.5 - 1.5 18

12. (a) Velocity of small ball when reaches at surface, ...(i) v = 2gh If g ¢ be the apparent gravitational acceleration on the ball inside the liquid, then Vrg - Vsg æ r - s ö ÷÷ g ...(ii) = çç g ¢= Vr è r ø Now,

Þ

v 2 - u2 = -2g ¢h¢ (r - s) 0 - 2gh = -2 gh¢ r [from Eqs. (i) and (ii)] hr h¢= r-s

75

Mechanical Properties of Fluids 13. (c) In this case, the air resistance is dependent on the area of body in contact with air. The resultant resistance, RµA Means larger the area of body, the more area available for air to exert upward force on it and hence, the resistance is more, so 3 < 2 < 1 will follow.

14. (c) Pressure difference between lungs of student and atmosphere = (760–750) mm of Hg i .e , hdg = 10 mm of Hg = 1 cm of Hg or h ´ 1 = 1 ´ 13.6 \ h = 13.6 cm Thus, he can drink water from a glass upto a maximum depth of 13.6 cm.

15. (b) Weight of candle is equal to weight of liquid displaced. Weight of candle = Weight of liquid displaced Vrg = V ¢r ¢g æ d2 ö æ d2 ö Þ çç p ´ L ÷÷ r ¢ ´ 2L ÷÷ r = çç p 4 4 è ø è ø r 1 = Þ r¢ 2 Since, candle is burning at the rate of 2cm/h, then after an hour, candle length is 2L - 2. \ (2L - 2)r = (L - x )r ¢ L-x r \ = r¢ 2(L - 1) L-x 1 = Þ x = 1 cm Þ 2 2 (L - 1) Hence, in one hour it melts 1 cm and so it falls at the rate of 1 cm/h.

16. (a) According to Bernoulli’s principle, p + rgh + 1 / 2rv 2 = constant The B section is at further distance, hence the velocity is higher and pressure is low. But section A is close to the centre and hence rotates with lower velocity and have high pressure than B. Thus to compensate for pressure, the level of water rises in both section, but more in section B than in section A.

17. (c) From Bernoulli’s principle for a flowing liquid, where the velocity of flow is less, the pressure is large and vice-versa. Hence, arteries carrying blood become narrow resulting in increase in the blood pressure.

18. (a) According to Bernoulli’s theorem, the total energy per unit volume for a moving fluid is constant. 1 i.e., p + rv 2 + rgh = constant 2 Thus, Bernoulli’s equation is a consequence of conservation of energy.

19. (a) Initially the velocity of lead shot will increase due to effect of gravity, but after travelling certain distance, the net force on lead shot is zero due to viscosity of glycerine and it starts moving with constant terminal velocity.

20. (b) Since, process is isothermal, the total pressure of air inside the bubble is same as excess of pressure given by, 4T p= R where, T is surface tension and R is radius. Also, an isothermal process obey’s Boyle’s law hence, pV = constant Let R be the radius of coalesce system, then p1V1 + p2V2 = pV æ 4T ö æ 4 3 ö æ 4T ö æ 4 3 ö æ 4T ö æ 4 3 ö ÷ ç ÷ ç ç R ÷ çè 3 pR1 ÷ø + ç R ÷ çè 3 pR2 ÷ø = çè R ÷ø çè 3 pR ÷ø è 2ø è 1ø R12 + R22 = R

Þ

Given, R1 = 3 cm and R2 = 4 cm R = 32 + 42 Þ R = 5 cm

\

21. (b) Initial area of film, A1 = 10 ´ 6 ´ 10-4 = 60 ´ 10-4 m2 Final area of film, A2 = 10 ´ 11 ´ 10–4 = 110 ´ 10–4 m2 Increase in surface area is DA = A2 – A1 = 50 ´ 10–4 m2 Since, soap film has two surfaces, hence total increase in surface area. = 2 ´ 50 ´ 10-4 m2 = 100 ´ 10-4 m2 2 ´ 10-4 W T = = DA 100 ´ 10-4

\

= 2 ´ 10-2 Nm-1

22. (b) Velocity of efflux v is given by v=

2( p - pa ) 2(3 pa - pa ) = = r r

v=

4 ´ 105 = 1000

400 m/s

4 pa r

AIIMS Chapterwise Solutions ~ Physics

76 23. (d) We know that, the surface tension (T ) of a liquid is equal to the work (W ) required to increase the surface area of the liquid film by unity at constant temperature. W \ T = Þ W = TDA DA = 4p ´ (10-3 )2 ´ 72 ´ 10-3 (n1 /3 - 1) = 4p ´ (10-3 )2 ´ 72 ´ 10-3 [(106 )1 /3 - 1] = 8.95 ´ 10-5 J

24. (a) Given, p1 = 3 p2 Þ

continuity, Av = constant Thus, when the pipe held vertically upward, the speed decreases due to potential energy and hence the pressure of stream increases and spread like a fountain, while in case of vertically down position, the velocity increase and hence the pressure of stream is low and narrow stream flows.

30. (a) The height to which a liquid rises is given by

R 3 4T 4T = 3× Þ 2 = R1 1 R1 R2

V1 R13 R13 1 = = 3 = V2 R2 (3R1 )3 27 m …(i) 25. (c) Relative humidity (RH) = M Now, as the vapour pressure of a gas at constant volume and temperature is proportional to its mass, partial pressure of water vapour RH = vapour pressure of water Also,

0.012 ´ 105

´ 100 = 75% 0.016 ´ 105 26. (d) The horizontal range, x = horizontal velocity × time = 2gh ´ 2 (H - h) / g = 2 h (H - h) =

29. (a) As we know that, according to equation of

h H

h=

2T cos q rRg

where, T = surface tension, q = angle of contact, r = density of liquid, R = radius of capillary and g = acceleration due to gravity. Hence, when the tube is of insufficient length, curvature radius of the liquid meniscus increases, so as to maintain the product hR a finite constant. i .e . as h decreases, R increases and the liquid meniscus becomes more and more flat, but the liquid does not overflow. Hence, both Assertion and Reason are true and Reason explains the Assertion.

31. (c) This is due to the viscosity of liquid. When liquid is stirred, the liquid comes in motion. The layer of liquid close to the vessel wall exert opposite force to inner layer and due to this, the liquid comes to rest after some time and the motion disappears. Hence, Assertion is true while Reason is false.

32. (c) The two glass plates stick together due to surface tension. x

dx For range x to be maximum, we have, =0 dh d H [h (H - h)] = 0 Þ h = dh 2

27. (b) For small range of temperatures, the variation of surface tension is given by the formula S t = S 0 (1 - at ) where, S 0 and S t are the surface tensions of liquid at 0ºC and tºC respectively and a is the coefficient of surface tension with temperature. The surface tension of a liquid decreases with a rise in temperature of the liquid.

28. (a) Viscous force acts in opposite direction of motion of liquid tangentially on the layers which tends to destroy its motion. Since motion of fluid destroys due to viscous force, hence its kinetic energy converts into heat energy of liquid.

Surface tension =

Force Length

Thus, the force required is directly proportional to the surface tension. Hence, Assertion is true while Reason is false.

33. (b) In the case of small drop of mercury, the force due to surface tension is large as compared to its weight, so it has spherical shape. In the case of bigger drops, the force of gravity (weight) exceeds the force of surface tension, hence it is oval shape.

34. (a) During floating, total upward thrust on the body is equal or greater than the weight of the body, hence during floating, net downward force is zero. Hence, both Assertion and Reason are true and Reason explains Assertion.

77

Mechanical Properties of Fluids 35. (d) The level of floating of a ship in the water is unaffected by the atmospheric pressure. It depends on buoyancy force which results from the pressure difference in the fluid. If the atmospheric pressure changes, the pressure at all points in the water changes by same amount keeping pressure difference in the water same. Hence, both Assertion and Reason are false.

36. (b) Since, the fluid move from higher pressure to lower pressure and in a fluid, the pressure increases with increase of depth. Hence, the pressure p0 will be lesser at the top than that at the bottom ( p0 + hrg ). So, the air bubble moves from the bottom to the top and does not move sideways, since the pressure is same at the same level. Further in coming from bottom to top the pressure decreases. According to Boyle’s law pV = constant. Therefore, if pressure decreases, the volume increases, it means radius increases. Hence, both Assertion and Reason are true, but Reason does not explain the Assertion.

37. (a) The density of concrete of course, is more than that of water and a block of concrete will sink like a stone, if dropped into water. Concrete cargo were filled with air and as such, average density of cargo vessels =

mass of concrete + mass of air volume of concrete + volume of air

It follows that the average density of cargo vessels must be less than that of water. As a result, the concrete cargo vessels did not sink. Hence, Assertion and Reason both are true and Reason explains the Assertion.

38. (c) When a thin steel needle is laid on surface of water, it sinks as upward force due to buoyancy does not support its weight. But when needle is laid carefully on a still water surface, then due to surface tension of water molecules, it diplaces as much amount of water as its weight. Thus, due to buoyancy force and surface tension, it floats on water surface.

39. (a) Reynold’s number is the ratio of inertial forces (vs r ) to viscous force (m / L ) and is used for determining whether a flow will be laminar or turbulent. It is established from experiments that flow is steady of laminar when Reynold’s number is less than about 2000, where in the region between 2000 to 3000, the flow is unstable i .e .

may change from laminar to turbulent or vice-versa. This case arises only when internal forces are dominant compared to viscous force. Hence, Assertion and Reason both are true and Reason given explanation of Assertion.

40. (c) We know that the boiling point of liquid increases with increase of vapour pressure. In a cooker, water with the food to be cooked, is heated, such that confined water vapours raise the superincumbent pressure. As a result, water boils at the temperature higher than boiling point 100°C. Now, when the cooker is removed from the stove and its lid is also removed, pressure decreases due to which its boiling point comes down but it is still at 100ºC so, water again begins to boil. When we add small impurities in water, the boiling point is not affected and remains same. Thus, Assertion is true while Reason is false.

41. (c) Since, excess pressure is inversely proportional to radius, hence excess pressure inside a smaller drop is large as compared to larger drop due to which smaller drop resists deforming forces better than a larger drop. 2T W ö 2W æ Also, = p= çQ T = ÷ R R × DA DA ø è 1 pµ Þ DA Hence, excess pressure is inversely proportional to surface area. Thus, Assertion is true while Reason is false.

42. (b) The balloon itself produces the lift and is usually made of a highly flexible latex material. The ascent rate can be controlled by the amount of gas the balloon is filled with weather balloon’s reach sufficient heights due to diminishing pressures causing the balloon to expand to such a degree that it disintegrates.

43. (c) The force of buoyancy depends upon the area of contact of the body with the fluid. The force of buoyancy is more in case of the needle than in the case of the ball. Now, it might so happen that the force of buoyancy (B ) is greater than the weight (mg ) of the needle, in which case the needle might float. But in the case of the ball, B < mg and so the ball sinks. Hence, Assertion is true while Reason is false.

AIIMS Chapterwise Solutions ~ Physics CHAPTER

9 Thermal Properties of Matter 1. The coefficient of cubical expansion of mercury is 0.00018/°C and that of brass 0.00006/°C. If a barometer having a brass scale were to read 74.5 cm at 30°C, find the true barometric height at 0°C. The scale is supposed to be correct at 15°C. [AIIMS 2017] (a) 74.122 cm (c) 42.161 cm

(b) 79.152 cm (d) 142.39 cm

2. A composite slab is prepared by pasting two plates of lengths L1 and L2 and thermal conductivities K 1 and K2 . The slabs have equal cross-sectional area. Find the equivalent conductivity of the slab. L1 + L2 L1 L2 + K1 K 2 L + L2 = 1 K1 K 2 L1 L2 = + K1 + K 2 K1 + K 2 L L = 1 2 L1 + L2

[AIIMS 2016]

(a) Keq =

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

(c) 9 : 1

(a) e = 0.18 (b) e = 0.02 (c) e = 0.2

[AIIMS 2014] (d) e = 0.15

5. A black body emits heat at the rate of 20 W, when its temperature is 727°C. Another black body emits heat at the rate of 15W, when its temperature is 227°C. Compare the area of the surface of the two bodies, if the surrounding is at NTP. (a) 16 : 1

(b) 1 : 4

(c) 12 : 1

[AIIMS 2014] (d) 1 : 12

its one end A is maintained at 100°C and other end B at 10°C, the temperature at a distance of 60 cm from point B is (a) 64°C

(b) 36°C

(c) 46°C

[AIIMS 2012] (d) 72°C

7. A non-conducting body floats in a liquid

2 of its volume immersed in 3 the liquid. When liquid temperature is 3 increased to 100°C, of body’s volume is 4 immersed in the liquid. Then, the coefficient of real expansion of the liquid is (neglecting the expansion of container of the liquid) [AIIMS 2012]

at 20°C with

such that its radiant intensity, (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object O2 has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source O1 to that of source O2 is [AIIMS 2015] (b) 1 : 9

surface area 5 cm2 , radiates 300 J of energy each minute. The emissivity is (given, Boltzmann constant = 5.67 ´ 10 -8 Wm 2K 4)

6. A rod AB is 1m long. The temperature of

3. The black body spectrum of an object O1 is

(a) 1 : 81

4. A body at a temperature of 727°C and has

(d) 81 : 1

(a) 15.6 ´ 10-4 °C -1 -4

(c) 156 . ´ 10 °C

-1

(b) 156 ´ 10-4 °C -1 (d) 0156 . ´ 10-4 °C -1

79

Thermal Properties of Matter

but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of A and B are K 1 and K2 respectively. A steady temperature difference of 12°C is maintained across the K composite slab. If K 1 = 2 , then 2 temperature difference across slab A is [AIIMS 2012] (a) 4°C (c) 8°C

(b) 6°C (d) 10°C

9. A black body at a temperature of 2600 K has the wavelength corresponding to maximum emission 1200 Å. Assuming, the moon to be a perfectly black body. The temperature of the moon, if the wavelength corresponding to maximum emission is 5000 Å is [AIIMS 2011] (a) 780 K (c) 524 K

(b) 624 K (d) 364 K

10. The rate of dissipation of heat by a black body at temperature T is Q. What will be the rate of dissipation of heat by another body at temperature 2T and emissivity 0.25? [AIIMS 2011] (a) 16Q (c) 8Q

(b) 4Q (d) 4.5Q

11. A metal plate 4 mm thick has a temperature difference of 32°C between its faces. It transmits 200 k-cal/h through an area of 5 cm2 . Thermal conductivity of the material is [AIIMS 2010] (a) 58.33 W/m-°C (b) 33.58 W/m-°C (c) 5 ´ 10-4 W/m-°C

of two liquids A B A and B contained in vessels of negligible heat capacity are supplied heat at t O Time the same rate. The temperature-time graph for the two liquids are shown. If S represents specific heat and L represents latent heat of liquid, then [AIIMS 2009] (a) S A > S B , LA < LB (c) S A < S B , LA < LB

(b) S A > S B , LA > LB (d) S A < S B , LA > LB

14. Black holes in orbit around a normal star are detected from the earth due to the frictional heating of infalling gas into the black hole, which can reach temperatures greater than 10 6K. Assuming that the infalling gas can be modelled as a black body radiator, then the wavelength of maximum power lies [AIIMS 2009] (a) (b) (c) (d)

in the visible region in the X-ray region in the microwave region in the g-ray region of electromagnetic spectrum

15. Three objects coloured black, gray and white can withstand hostile conditions upto 2800°C. These objects are thrown into furnace, where each of them attains a temperature of 2000°C. Which object will glow brightest? [AIIMS 2007] (a) White object (b) Black object (c) All glow with equal brightness (d) Gray object

16. Flash light equipped with a new set of

(d) None of the above

12. A clock with a metal pendulum beating seconds keeps correct time at 0°C. If it loses 12.5 s a day at 25°C, the coefficient of linear expansion of metal pendulum is 1 (a) /°C 86400 1 (c) /°C 14400

T

13. Equal masses Temperature

8. Two slabs A and B of different materials

[AIIMS 2010] 1 (b) /°C 43200 1 (d) /°C 28800

batteries, produces bright white light. As, the batteries wear out [AIIMS 2006] (a) the light intensity gets reduced with no change in its colour (b) light colour changes first to yellow and then red with no change in intensity (c) it stops working suddenly, while giving white light (d) colour changes to red and also intensity gets reduced

80

AIIMS Chapterwise Solutions ~ Physics

17. A bimetallic strip consists of metals X and Y. It is mounted rigidly at the base as shown in figure. The metal X has a higher coefficient of expansion compared to that for metal Y. When bimetallic strip is placed in a cold bath [AIIMS 2006]

X

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

it will bend towards the right it will bend towards the left it will not bend but shrink it will neither bend nor shrink

(b) 625 (d) 16

19. A black body at a temperature of 227°C,

radiates heat at a rate of 20 cal m-2 s-1. When its temperature is raised to 727°C, the heat radiated by it in cal m-2 s-1 will be [AIIMS 2003] (c) 320

(d) 640

20. Figure shown below are the black body radiation curves at temperatures T1 and T2 (T2 > T1). Which of the following plots is correct? [AIIMS 2003] I

T2 T1

(a)

O

(b)

T2 T1

O

l

(c)

l

I T1

O

T2 l

(d)

O

(d) 102

22. The temperature of the cold junction of a

(b) 460°C

(c) 600°C

[AIIMS 2001] (d) 400°C

T1 T2 l

[AIIMS 2000]

(a) line absorption spectrum (b) line emission spectrum (c) emission of band spectrum (d) None of the above

24. The real coefficient of volume expansion of glycerine is 0.000597 per°C and linear coefficient of expansion of glass is 0.000009 per°C. Then, the apparent volume coefficient of expansion of glycerine is [AIIMS 2000] (a) 0.000558 per°C (c) 0.00027 per°C

(b) 0.00057 per°C (d) 0.00066 per°C

25. When a solid is converted into a gas directly by heating, then this process is known as [AIIMS 1999] (a) sublimation (c) condensation

(b) vaporisation (d) boiling

26. A centigrade and fahrenheit thermometers are dipped in boiling water. The water temperature is lowered until the fahrenheit thermometer registers a temperature of 140° C. The fall of temperature as registered by the centigrade thermometer is [AIIMS 1998] (a) 40°

I

(c) 10-3

example of

[AIIMS 2004]

I

(b) 10-2

23. Fraunhoffer line of the solar system is an

becomes 100 times its present radius and its surface temperature becomes half of its present value. The total energy emitted by it, then will increase by a factor of

(b) 160

(a) 10-4

(a) 500°C

18. Suppose the sun expands, so that its radius

(a) 40

and at 100°C, its density is 9.7 g/cc. The coefficient of linear expansion of the substance is [AIIMS 2002]

thermocouple is 0°C and the temperature of the hot junction is T ° C. The relation for 1 the thermo emf is given by, E = AT - BT 2 2 (when A = 16 and B = 0.08). The temperature of inversion will be

Y

(a) 104 (c) 256

21. The density of a substance at 0°C is 10 g/cc

(b) 80°

(c) 50°

(d) 90°

27. Heat travels through vacuum by [AIIMS 1998] (a) convection (c) conduction

(b) radiation (d) All of these

81

Thermal Properties of Matter 28. A quantity of heat required to change the unit mass of a solid substance to its liquid state, while the temperature remains constant, is known as [AIIMS 1998] (a) latent heat (c) heat

(b) latent heat of fusion (d) specific heat

29. On a cold morning, a metal surface on touching is felt colder than a wooden surface, because the metal has [AIIMS 1997] (a) (b) (c) (d)

low thermal conductivity high thermal conductivity high specific heat low specific heat

Direction (Q. Nos. 34-52) Read the Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false.

34. Assertion It is hotter over the top of a fire than at the same distance on the sides.

30. The instrument used to measure the temperature of the source from its thermal radiation is [AIIMS 1997] (a) hydrometer (c) thermopile

Assertion & Reason

(b) barometer (d) pyrometer

31. Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C, respectively. The temperature of the junction of the three rods will be [AIIMS 1996]

Reason In the upward direction, the heat propagate through convection. [AIIMS 2017]

35. Assertion A brass tumbler feels much colder than a wooden tray on a chilly day. Reason The thermal conductivity of brass is less than that of wood. [AIIMS 2016]

36. Assertion A hollow metallic closed container maintained at a uniform temperature can act as a source of a black body radiation. Reason All metals act as a black body.

90°C B

37. Assertion For higher temperature, the peak emission wavelength of a black body shifts to lower wavelength.

0°C A C 90°C

(a) 45° C (c) 30° C

(b) 60° C (d) 20° C

earth by convection.

substance gradually, its colour becomes (b) red

(c) white

[AIIMS 1996] (d) green

33. The bulb of one thermometer is spherical, while that of other is cylindrical. If both of them have equal amounts of mercury, which one will respond quickly to the temperature? [AIIMS 1994] (a) Elliptical (c) Cylindrical

Reason Peak emission wavelength of a black body is proportional to the fourth power of temperature. [AIIMS 2014]

38. Assertion Heat from the sun reaches the

32. On increasing the temperature of a (a) yellow

[AIIMS 2014]

(b) Spherical (d) Both (b) and (c)

Reason Air can be heated by conduction and convection. [AIIMS 2013]

39. Assertion Soft steel can be made red hot by continued hammering on it, but hard steel cannot. Reason Energy transfer in case of soft is large as in hard steel. [AIIMS 2012]

40. Assertion It is hotter over the top of a fire than at the same distance of the side.

82

AIIMS Chapterwise Solutions ~ Physics Reason Air surrounding the fire conducts, more heat upward. [AIIMS 2011, 2003]

47. Assertion Woollen clothes keep the body warm in winter. Reason Air is a bad conductor of heat.

41. Assertion Like light radiation, thermal

[AIIMS 2002]

radiations are also electromagnetic radiation. Reason The thermal radiations required no medium for propagation. [AIIMS 2010]

48. Assertion A spaceship while entering the earth's atmospheres is likely to catch fire. Reason The temperature of upper atmosphere is very high. [AIIMS 2000]

42. Assertion The knowledge of Albedo helps

49. Assertion Fahrenheit is the smallest unit

us to estimate the atmosphere of a planet. Reason The clouds are not good reflectors of light. [AIIMS 2009]

measuring temperature. Reason Fahrenheit was the first temperature scale used for measuring temperature. [AIIMS 1999]

43. Assertion If earth did not have atmosphere, its average surface temperature would be lower than what is now. Reason Green house effect of the atmosphere would be absent, if earth did not have atmosphere. [AIIMS 2007]

50. Assertion Bodies radiate heat at all temperatures. If temperature of the body T > 0. Reason Rate of radiation of heat is proportional to the fourth power of absolute temperature. [AIIMS 1999]

44. Assertion While measuring the thermal

51. Assertion The equivalent thermal

conductivity of liquid experimentally, the upper layer is kept hot and the lower layer is kept cold. Reason This avoids heating of liquid by convection. [AIIMS 2007]

conductivity of two plates of same thickness in contact is less than the smaller value of thermal conductivity. Reason For two plates of equal thickness in contact, the equivalent thermal conductivity is given by 1 1 1 = + K K 1 K2 [AIIMS 1997]

45. Assertion Perspiration from human body helps in cooling the body. Reason A thin layer of water on the skin enhances its emissivity. [AIIMS 2005]

52. Assertion A hollow metallic closed container maintained at a uniform temperature can act as a source of black body radiation. Reason All metals act as black bodies.

46. Assertion Temperatures near the sea coast are moderate. Reason Water has a high thermal conductivity. [AIIMS 2003]

[AIIMS 1996]

Answers 1. 11. 21. 31. 41. 51.

(a) (a) (a) (b) (b) (a)

2. 12. 22. 32. 42. 52.

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

3. 13. 23. 33. 43.

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

4. 14. 24. 34. 44.

(a) (b) (b) (b) (a)

5. 15. 25. 35. 45.

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

6. 16. 26. 36. 46.

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

7. 17. 27. 37. 47.

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

8. 18. 28. 38. 48.

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

9. 19. 29. 39. 49.

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

10. 20. 30. 40. 50.

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

83

Thermal Properties of Matter

Explanations 1. (a) Given, g brass = 0.00006/ °C and

3. (d) Given, lm1 = 200 nm and lm2 = 600 nm

g Hg = 0.00018/ °C g 0.00006 a brass = brass = 3 3

As,

= 0.00002 = 2 ´ 10-5 / °C The brass scale is true at 15°C, therefore at 30°C its graduations will increase in length and so observed reading will be less than actual reading at 30°C. \The change in reading, Dl = la brass (DT ) = 74.5 ´ 2 ´ 10- 5(30 - 15)

According to Wien’s displacement law, lmT = constant where, lm is wavelength corresponding to the highest radiant energy and T is absolute temperature of the emitter. lm1 T …(i) Thus, = 2 lm2 T1 and according to Stefan-Boltzmann's law. The power of radiation emitted by an emitter per unit area,

= 0 × 02235cm \Actual reading at 30°C

E µT 4 E1 æ T1 =ç E2 çè T2

l30 = lobserved + Dl = 74.5 + 0.02235 = 74.522 cm Assuming area of cross-section to be constant, we have V0r 0 = V30r30 or ah0r 0 = ah30r30 Therefore, true height at 0°C h30 r h0 = h30 30 = r0 (1 + g Hg DT ) 74.522 = 1 + 0.00018 ´ 30 74.522 = = 74.122 cm 1.0054

2. (a) According to the given situation, M

K1

K2

L1

L2

N

Since, both plates are in series combination. Thus, the heat current will be same in both plates. The equivalent thermal resistance, R = R1 + R2 L But, R= KA L1 + L2 L L \ = 1 + 2 K eq A K1 A K2 A K eq =

L1 + L2 L1 L + 2 K1 K2

4 æl ö ÷ = ç m2 ÷ ç lm ø è 1

ö ÷ ÷ ø

4

From Eq. (i), putting the values E1 æ 600 ö 4 =ç ÷ Þ E2 è 200 ø

E1 81 = E2 1

4. (a) Given, E = 300 J/min Þ

T = 727°C T = 273°C + 727°C = 1000 K A = 5cm2 = 5 ´ 10-4 m2

As we know that, the energy emitted by a body per unit time , E = esAT 4 where, e = emissivity of the body, s = stefan constant, A = surface area of the body and T = Temperature of the body. Putting the values, 300 e= 5.67 ´ 10-8 ´ 5 ´ 10-4 ´ (103 )4 ´ 60 300 . = = 018 300 ´ 5.67

5. (d) Given, P1 = 20 W \

TC1 = 727° C T K 1 = 273 + 727°C = 1000 K P2 = 15 W TC2 = 227°C T K 2 = 273 + 227 = 500 K

84

AIIMS Chapterwise Solutions ~ Physics Since, energy emitted by a black body in one second, P = sAT 4 P1 AT4 = 1 14 P2 A2T2

Thus,

Putting the values,

The temperature difference across slab, A = (12 - x ) = (12 - 4) = 8°C

Þ

20 A1 (1000)4 = 15 A2 (500)4

Þ

A1 4 æ 1 ö 4 1 4 1 = ç ÷ = ´ = A2 3 è 2 ø 3 16 12

6. (a) Given, L = 1 m, T A = 100°C and T B = 10°C. Let the temperature at distance of 60 cm from point B be q. A

B 10°C

1m 100°C 40 cm

60 cm

According to the thermal conduction, KA(100 - q) KA(q - 10) = 40 60 where, A = area of rod and K = coefficient of thermal conductivity of the material. 100 - q q - 10 or = 2 3 or q = 64°C Let the value of the body = V, then 2 3 V1 = V and V2 = V 3 4 As we know that, V2 = V1 (1 + gDt ) V - V1 V2 - V1 Thus, = g= 2 V1 Dt V1 (t2 - t1 ) æ3 2ö ç - ÷V 1 / 12 4 3ø Putting the values, è = 2 2 3 ´ 80 / V(100 - 20) 3 = 15.625 ´ 10-4 / °C K2 2 The given situation can be shown as,

8. (c) Given, T1 = 12°C, T2 = 0 °C, K 1 =

A K1

B x

K2

9. (b) Given, T1 = 2600 K, T2 = ? lm1 = 1200Å, lm2 = 5000Å According to Wein’s displacement law, lm T = constant or

lm1 T1 = lm2 T2

Putting the values, 1200 ´ 2600 = 5000 ´T2 12 ´ 260 T2 = = 624 K Þ 5

10. (b) Given, T1 = T , T2 = 2T 1 4 As we know that, the energy emitted by a body per unit time. P1 = Q , P2 = ?, e2 =

P = esAT 4 , where symbols has their usual meaning. \

P1 e1T14 = P2 e2T24

Putting th values,

7. (a) Given, t1 = 20°C and t2 = 100°C

12°C

Rate of flow of heat will be equal in both the slabs, because they are in series combination. \ (12 - x )K 1 = K 2 ( x - 0) (12 - x )K 2 = 2K 2 ( x - 0) K ö æ 12 - x = 2 x Þ x = 4°C ç\ K 1 = 2 ÷ 2 ø è

0°C

Q 1 æT ç = P2 1 / 4 çè 2T P2 =

ö ÷ ÷ ø

4

16 Q Þ P2 = 4Q 4

11. (a) Given, DT = 32°C, Dx = 4mm = 4 ´ 10-3 m A = 5 cm2 = 5 ´ 10-4 m2 H = KA

DT = 200 k-cal/h Dx

Putting the values, . 200 ´ 103 ´ 42 5 ´ 10-4 ´ 32 K = 60 ´ 60 4 ´ 10-3 K = 58.33W / m- °C

12. (a) Given, t1 = 0°C, t2 = 25°C , \ t2 - t1 = Dt = 25 - 0 = 25°C As we know that, change in time period of a pendulum with change in its length, DT 1 Dl ...(i) = × T 2 l

85

Thermal Properties of Matter Since, l ¢ = l (1 + aDt ) \ l ¢ - l = laDt Dl ...(ii) = a. Dt Þ l From Eqs. (i) and (ii), we get 1 DT = a × Dt × T 2 2DT a= Þ DtT 2 ´ 12.5 1 / °C = = 25 ´ 24 ´ 60 ´ 60 86400

13. (d) From the graph, (i) the temperature of liquid A rises faster than that of liquid B, while both liquids are given same amount of heat and their masses are also equal.

17. (b) Coefficient of thermal expansion a, a X > a Y (given) On placing bimetallic strip in a cold bath, metal X will shrink more than Y, because a X >a Y . Hence, the strip will bend towards left.

18. (b) From Stefan's law, if the emissive power of a body at absolute temperature T be e, then the energy emitted by its unit area per second is sT 4 ´ e , also, if A is the surface area of the body, then E = sT 4eA When R ¢ = 100 R and T ¢ = emitted is æT ö E ¢ µ 4p (100R )2 ç ÷ è2ø

Thus, S A < S B

4

2

(ii) the horizontal portion of the graphs represent change in phase, ( from the graph) Q A >QB mL A > mL B (Q Q = mL; for change in phase) \

T , then energy 2

L A > LB 6

14. (b) Given, T = 10 K

æ 100 ö 2 4 E ¢µ ç ÷ ´ 4pR T è 4 ø 2

\ \

19. (c) Given, T1 = 227°C, P1 = 20cal / m2 - s

Applying Wien's displacement law, lm × T = b (constant) where, b = 2.9 ´ 10-3 m-K

T2 = 727°C, P2 = ? According to Stefan’s law, P = sT 4 ( for black body)

Thus, we have lm =

æ 100 ö E¢ = ç ÷ ´E è 4 ø E¢ = 625 E E ¢ = 625E

2.9 ´ 10-3

\

106

= 2.9 ´ 10-9m = 2.9 nm This wavelength lies in the X-ray region of the electromagnetic spectrum.

15. (b) According to the Kirchhoff's law, ‘A good absorber is a good emitter’. Thus, the black object will be a good emitter. Therefore, among the given options, the black object will glow with highest brightness.

16. (d) As we know that, when batteries wear out, the temperature of filament of the flash light decreases, therefore the intensity of the radiation reduces. Thus, the dominating wavelength in radiation spectrum increases, (according to Wien's displacement law).

Þ

P1 T14 = P2 T24 20 æ 273 + 227 ö ÷ =ç P2 çè 273 + 727 ÷ø

4

4

Þ

æ 1000 ö 4 P2 = 20 ´ ç ÷ = 20 ´ 2 è 500 ø = 320 cal/m2 -s

20. (a) According to the Wien’s displacement law, (lm × T = constant) where, lm = wavelength corresponding to the highest radiant energy and T = absolute temperature of the body. 1 lm µ Þ T Given, T1 < T2 Thus, lm1 > lm2

86

AIIMS Chapterwise Solutions ~ Physics

21. (a) Given, T1 = 0°C

26. (a) Given, T F = 140°C , TC = ?

e1 = 10g / cc, d =? T2 = 100° C, e2 = 97 . g/cc As we know that, coefficient of volume expansion, e1 - e2 g= e1 (T2 - T1 ) Putting the values, then 10 - 97 . g= 10 (100 - 0) =

. 03 = 3 ´ 10-4 /°C 1000

g 3 ´ 10-4 Since, a = = / °C = 10-4 / °C 3 3

22. (d) According to seebeck effect, the thermo emf, 1 E = AT - BT 2 2 Given, A = 16, B = 0.08 1 E = 16T - ´ 0.08 ´ T 2 \ 2 Since, emf is zero at temperature of inversion, we have 0 = 16T - 0.04T 2 T =

16 = 400°C 0.04

23. (a) Fraunhoffer lines, in astronomical spectroscopy, any of the dark (absorption) lines in the spectrum of the sun or other star, caused by selective absorption of the sun's or stars radiation at specific wavelengths by the various elements existing as gases in its atmosphere. Thus, Fraunhoffer lines are absorption spectrum.

24. (b) As we know that, coefficient volume

Since, we know TC T F - 32 = 5 9 140 - 32 (putting values) TC = 5 ´ Þ 9 = 60°C Q Boiling point of water is 100°C. \Fall in temperature (in °C) DT = 100°C - 60°C = 40°C

27. (b) Radiation does not require any medium for heat transfer. In radiation, heat is transmitted in the form of electromagnetic waves from the hot body.

28. (b) Latent heat of a material is the amount of heat which converts its unit mass into liquid from solid state. This amount of heat is also known as latent heat of fusion.

29. (b) When we touch metal, the metal rapidly conducts heat from our hand and gives a cold feeling. Wood on the other hand is a bad conductor of heat, it conducts heat slowly from our hand and appears less cold.

30. (c) A thermopile is a group of thermocouples connected in series. Thermocouples measure the temperature difference between two points are not absolute temperature.

31. (b) Let the temperature of junction be q, since rods B and C are parallel to each other (because both having the same temperature difference). Hence, given figure can be redrawn as follows [R be the resistance of each rod].

expansion of the glass = 3 ´ linear coefficient of expansion = 3 ´ 0.000009 = 0.000027 per° C Given, volume expansion of glycerine is = 0.000597 per°C Since, we know \ g apparent = g real - g vessel = 0.000597 - 0.000027 = 0.00057 / °C

25. (a) Sublimation is the change from solid to gas or gas to solid without passing through the liquid state.

90°

B RP=

R 2

R

0° A R C

R 90°

Q

Q (q1 - q2 ) = t R

87

Thermal Properties of Matter and Þ Þ Þ

æQ ö æQ ö ç ÷ =ç ÷ è t ø BC è t ø AB (90 - q) (q - 0) = R /2 R 180 - 2q = q q = 60°C

32. (a) According to the Wien's displacement law 1 T On increasing temperature, wavelength decreases. Since, yellow colour has minimum wavelength, the substance gradually becomes yellow. lm µ

33. (c) Since, the heat exchange between two bodies by conduction depends on the surface area in contact each other. As we know, a cylinder has maximum surface area for a given mass.

34. (b) Over the top of a fire, heat is transferred by both convection and radiation, while on the sides, heat is transferred by radiation only.

35. (c) The thermal conductivity of brass is high, i.e. brass is a good conductor of heat, so when a brass tumbler is touched, heat quickly flows from human body to tumbler. Consequently, the tumbler appears colder, on the other hand, wood is a bad conductor. So, heat does not flow from the human body to the wooden tray in case. Thus, it appears comparatively hotter.

36. (c) A hollow metallic closed container maintained at a uniform temperature can act as source of black body radiation. It is also well know that all metals cannot act as black body, because if we take a highly metallic polished surface, it will not behave as a perfect black body.

37. (c) According to Wien's law, 1 (for a black body) lm µ T where, lm = wavelength corresponding to highest radiant energy and T = absolute temperature of the emitter. When temperature is increased, the peak emission wavelength of a black body shifts to lower wavelength.

38. (d) Heat from sun reaches to earth by radiation, so Assertion is wrong and air is heated only by convection, so Reason is wrong statement.

39. (a) As we know, the soft steel is easily deformed on hammering, it means that it absorbs energy easily, when work is done on it during hammering.

40. (a) It is hotter over the top of a fire than at the same distance on the sides, because at the top of the fire heat is transferred by both convection (by air) and radiation, while on the sides heat is transferred by radiation only. Therefore, at top of the fire is hotter as compare to the sides.

41. (b) Light radiations and thermal radiations both belongs to electromagnetic spectrum. Light radiation belongs to visible, while thermal radiation belongs to infrared region of EM spectrum. Also, EM radiations required no medium for propagation.

42. (c) The term Albedo explains, the reflecting power of planet to reflect the light which is incident on it. As the clouds are good reflectors of light, thus the presence of clouds on a planet increases the reflecting power of planet. Hence, with the knowledge of Albedo we can confirm the presence of atmosphere clouds at the planet.

43. (a) Earth is heated by sun’s infrared radiation, the earth also emits radiation most in infrared region. These radiations are reflected back due to heavy gases, like CO2 by atmosphere. These back radiation keep the earth’s surface warm at night. This phenomenon is called green house effect. When the atmosphere were absent then temperature of earth falls.

44. (a) We know that to measure thermal conductivity of liquids experimentally, they must be heated from the top, i.e. upper layer is kept hot and lower layer is kept cold, so as to prevent convection in liquids.

45. (c) When water leaves the body through perspiration energy content of molecules remained in body decreases, therefore temperature also decreases. Thus, perspiration from human body helps in cooling the body.

88

AIIMS Chapterwise Solutions ~ Physics

46. (b) Temperatures near sea coast are moderate because thermal conduction never takes place at sea coast causing moderate temperature. Also water has the highest thermal conductivity of any liquid. As temperature of water is lowered, the rate at which the energy is transferred is reduced to an ever increasing extent. Instead of energy being transferred it is stored.

47. (b) The fibres of woollen clothes have large interspaces and air is filled in these spaces. Since, this air is immovable and convection currents are not set up in it so heat transmission by convection does not take place. Also, since air is a bad conductor of heat, there is no heat transmission by conduction also.

51. (a) When two conductors of same cross-section area A are joined in series with the same thickness d, then equivalent thermal conductivity K is given by d + d2 K = 1 d1 d + 2 K1 K2 Given, d1 = d2 = d 2K 1 K 2 d +d 2 K = = = K1 + K2 d d K1 + K2 + K1 K2 K1K2 K =

2K 1 K 2 K1 + K2

Thus, we have K < K 1 and K < K 2 .

48. (c) A spaceship is a vehicle that travels through

52. (c) All metals have got different physical

space. When spaceship enters earth's atmosphere, it experiences air friction due to which it will catch fire, if proper precaution will not be taken. Also, temperature of upper atmosphere is low.

characteristics, due to which they will be having different emissive and abosorptive powers, so all metals cannot act as black bodies. But a hollow metallic closed container can act as a black body.

49. (c) Fahrenheit is the smallest unit measuring temperature. The relation between fahrenheit (F) and celsius (C) scale is C F - 32 = 5 9 Also, celsius was the first temperature scale. Also, used for measuring temperature.

50. (a) From Stefan's law the energy (E ) radiated per second by a black body is proportional the fourth power of absolute temperature (T ) E = sT 4 Body does not radiate heat at 0 K.

If a small hole is made on the surface of the hollow container, whose inner surface is painted black and the shape is as shown in the figure. Then any radiation that enters the hole has a good chance of being absorbed after multiple reflections taking place on the inner surface of the container and thus it can act as a black body.

CHAPTER

10 Thermodynamics 1. If coefficient of performance of a refrigerator is b and heat given to surrounding is Q2 , then what is heat absorbed? [AIIMS 2018] 1 (b) Q2 b Q (b – 1) (d) 2 b

(a) bQ2 (c) (1 + b ) Q2

2. One mole of an ideal diatomic gas undergoes transition from A to B along a path AB as shown below. The change in internal energy of the gas during the transition is [AIIMS 2017]

4. In an isobaric process of an ideal gas, the ratio of heat supplied and work done by æQ ö the system i. e, ç ÷ is èW ø [AIIMS 2013] (a)

g -1 g

(b) g

(c)

g g -1

(d) 1

5. The freezer in a refrigerator is located at the top section, so that

[AIIMS 2011]

(a) the entire chamber of the refrigerator is cooled quickly due to convection (b) the motor is not heated (c) the heat gained from the environment is high (d) the heat gained from the environment is low

p(kPa)

6. When an ideal monoatomic gas is heated 5

of constant pressure, fraction of heat energy supplied which increases the internal energy of gas, is [AIIMS 2010]

A

B

2 0

(a) 20 kJ (c) - 20 kJ

4 V(in m3)

6

(b) - 12 kJ (d) 20 J

3. When 20 J of work was done on gas, 40 J of heat energy was released. If the initial internal energy of the gas was 70 J, then what is the final internal energy? (a) 50 J (c) 90 J

(b) - 150 J [AIIMS 2015] (d) 110 J

(a) 2/5 (c) 3/7

(b) 3/5 (d) 3/4

7. A perfect gas goes from state A to state B by absorbing 8 ´ 10 5 J of heat and doing 6.5 ´ 10 5J of external work. It is now transferred between the same two states in another process in which it absorbs 10 5J of heat. In the second process, [AIIMS 2009] (a) work done on gas is 105 J (b) work done on gas is 0.5 ´ 105 J (c) work done by gas is 105 J (d) work done by gas is 0.5 ´ 105 J

90

AIIMS Chapterwise Solutions ~ Physics

8. The temperature-entropy diagram of a reversible engine cycle is shown in the figure. Its efficiency is [AIIMS 2008] T

S0

(b) 1/4

(c) T

(d) p

(a) the pressure decreases [AIIMS 2000] (b) the pressure increases (c) the pressure remains the same (d) the pressure may increase or decrease depending upon the nature of the gas

S 2S0

15. During the adiabatic expansion of two

(c) 1/3

(d) 2/3

9. When you make ice cubes, the entropy of water

[AIIMS 2006]

(a) does not change (b) increases (c) decreases (d) may either increase or decrease depending on the process used

10. n moles of a monoatomic gas is carried round, the reversible rectangular cycle ABCDA as shown in the diagram. The temperature at A is T0. The thermodynamic efficiency of the cycle is [AIIMS 2004] B

2p0

A

(a) - 2 J

(b) 3 J

C

(c) 1J

[AIIMS 2000] (d) 2 J

16. In an adiabatic process, the quantity which remains constant is

[AIIMS 1999]

(a) total heat of system (b) temperature (c) volume (d) pressure

17. A sample of gas expands from volume V 1 to V2 . The amount of work done by the gas is greatest, when the expansion is [AIIMS 1998] (b) adiabatic (d) All of these

18. The volume of a gas expands by 0.25 m 3 at a constant pressure of 103 N/m2 . The work done is equal to [AIIMS 1996]

D

T0

V0

(b) 50%

V

2V0

(c) 20%

(d) 25%

11. The latent heat of vaporisation of water is 2240 J. If the work done in the process of vaporisation of 1 g is 168 J, then increase in internal energy will be [AIIMS 2002] (b) 2072 J

(c) 2240 J

(d) 2408 J

12. 1 mm 3 of a gas is compressed at 1 atmospheric pressure and temperature 27 °C to 627 °C. What is the final pressure under adiabatic condition? (g for gas = 1.5) (a) 27 ´ 105 Nm–2 (c) 36 ´ 105 Nm–2

moles of a gas, the internal energy of a gas is found to decrease by 2 J.The work done during the process on gas will be equal to

(a) isothermal (c) isobaric

p

(a) 1904 J

(b) R

by a piston, which is then maintained at the same position with the passage of time

T0

(a) 15%

thermodynamical coordinate? [AIIMS 2001] (a) V

14. Air in a cylinder is suddenly compressed

2T0

(a) 1/2

13. Which one of the following is not a

[AIIMS 2002] (b) 80 ´ 105 Nm–2 (d) 56 ´ 105 Nm–2

(a) 250 W

(b) 2.5 erg (c) 250 N

(d) 250 J

19. Heat is supplied to a diatomic gas at constant pressure. The ratio of [AIIMS 1996] DQ : DU : DW is (a) 5 : 3 : 2 (b) 5 : 2 : 3 (c) 7 : 5 : 2 (d) 7 : 2 : 5

20. A perfect gas is contained in a cylinder kept in vacuum. If the cylinder suddenly bursts, then the temperature of the gas [AIIMS 1996] (a) is increased (b) becomes zero K (c) remains unchanged (d) is decreased

21. A certain mass of gas at 273 K is expanded to 81 times its volume under adiabatic conditions. If g = 125 . for the gas, then its final temperature is [AIIMS 1995] (a) - 182 °C (b) 0°C

(c) - 235°C (d) - 91°C

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Thermodynamics

Assertion & Reason Direction (Q. Nos. 22-40) Read the Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

22. Assertion In adiabatic expansion, the product of p and V always decreases. Reason In adiabatic expansion process, work is done by the gas at the cost of internal energy of gas. [AIIMS 2017]

23. Assertion Molar heat capacity cannot be defined for isothermal process. Reason In isothermal process, p-V versus [AIIMS 2017] T graph is a dot.

24. Assertion In an adiabatic process change in internal energy of a gas is equal to work done on or by the gas in the process. Reason Temperature of gas remains constant in an adiabatic process. [AIIMS 2016]

25. Assertion Thermodynamic process in nature is irreversible. Reason Dissipative effects cannot be eliminated. [AIIMS 2014]

26. Assertion It is impossible for a ship to use the internal energy of sea water to operate its engine. Reason A heat engine is different from a refrigerator. [AIIMS 2011]

27. Assertion The isothermal curves intersect each other at a certain point. Reason The isothermal changes takes place rapidly, so the isothermal curves have very little slope. [AIIMS 2008]

28. Assertion The temperature of the surface of the sun is approximately 6000 K. If we take a big lens and focus the sun rays, then we can produce a temperature of 8000 K. Reason This highest temperature can be produced according to second law of thermodynamics. [AIIMS 2007]

29. Assertion Air pressure in a car tyre increases during driving. Reason Absolute zero temperature is not zero energy temperature. [AIIMS 2007]

30. Assertion In an isolated system, the entropy increases. Reason The processes in an isolated system are adiabatic. [AIIMS 2006]

31. Assertion The Carnot cycle is useful in understanding the performance of heat engines. Reason The Carnot cycle provides a way of determining the maximum possible efficiency achievable with reservoirs of given temperatures. [AIIMS 2006]

32. Assertion Reversible systems are difficult to find in real world. Reason Most processes are dissipative in nature. [AIIMS 2005]

33. Assertion Air quickly leaking out of a balloon becomes cooler. Reason The leaking air undergoes adiabatic expansion. [AIIMS 2005]

34. Assertion In a pressure cooker, the water is brought to boil. The cooker is then removed from the stove. Now, on removing the lid of pressure cooker, the water starts boiling again. Reason The impurities in water bring down its boiling point. [AIIMS 2004]

35. Assertion When a bottle of cold carbonated drink is opened, a slight fog forms around the opening. Reason Adiabatic expansion of the gas causes lowering of temperature and condensation of water vapours. [AIIMS 2003]

36. Assertion First law of thermodynamics does not forbid flow of heat from lower temperature to higher temperature. Reason Heat supplied to a system always equal to the increase in its internal energy. [AIIMS 2002]

37. Assertion In adiabatic compression, the internal energy and temperature of the system get decreased.

92

AIIMS Chapterwise Solutions ~ Physics Reason In isothermal process, whole of the heat energy supplied to the body is converted into internal energy. [AIIMS 1997]

Reason The adiabatic compression is a slow process. [AIIMS 2001]

38. Assertion Melting of solid causes change

40. Assertion it is not possible for a system,

in its internal energy.

unaided by an external agency to transfer heat from a body at a lower temperature to another at a higher temperature.

Reason Latent heat is the heat required to melt a unit mass of solid. [AIIMS 1998]

39. Assertion According to the first law of

Reason It is not possible to violate the second law of thermodynamics. [AIIMS 1994]

thermodynamics, DQ = DU + pDV .

Answers 1. 11. 21. 31.

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

2. 12. 22. 32.

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

3. 13. 23. 33.

(a) (b) (b) (a)

4. 14. 24. 34.

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

5. 15. 25. 35.

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

6. 16. 26. 36.

(b) (a) (b) (a)

7. 17. 27. 37.

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

8. 18. 28. 38.

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

9. 19. 29. 39.

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

10. 20. 30. 40.

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

Explanations 1. (d) The performance of a refrigerator is expressed by means of ‘‘coefficient of performance’’ b which is the ratio of the heat extracted from the cold body to the work needed to transfer it to the hot body. Q2 Heat extracted Q2 i.e. b = …(i) = = W Q2 – Q1 Work done where, Q2 = heat given to surrounding (given) and Q1 = heat absorbed from system. \

Q2 = bQ2 – bQ1

Þ

Q1 =

Q2 (b – 1) b

2. (c) From graph, p1 = 5 kPa, p2 = 2 kPa , 3

3

V1 = 4m and V2 = 6m

f = degree of freedom = 5(for diatomic) p(kPa)

5

A B

2

0

4

6

V(m3)

The change in internal energy, f 5 DU = ( p2V2 – p1V1 ) = (12 – 20) = –20kJ 2 2 3. (a) From the first law of thermodynamics, dQ = dU + dW where, dQ = heat supplied to the system (40J), dU = change in internal energy and dW = work done by system (–20J ). or dU = dQ - dW = 40 - 20 = 20 J Since, heat energy is released by the gas, its internal energy decreases from 70 J to (70 - 20 = 50 J) 50 J.

4. (c) From ideal gas equation, …(i) pV = nRT where, symbols have their usual meaning. For isobaric process (in which pressure remains constant), amount of heat, Q = nC p DT …(ii) Work done, …(iii) W = pDV From Eqs. (i), (ii) and (iii), we get nC p DT C p Cp Q = = = [Q R = C p - C V ] W nRDT R C p - CV C ö C p / CV æ g ç\g = p÷ = = ç Cp C V ÷ø g -1 è -1 CV

93

Thermodynamics 5. (a) The freezer in a refrigerator is located at the top section so that the entire chamber of the refrigerator is cooled quickly due to convection. when the freezer is placed on top, the cold air produced from it is denser than the warmer air in the bottom. So, cold air being dense sinks down and the warm air is forced to rise up. So, when the warm air rises up, it gets cooled in the freezer.

6. (b) For ideal monoatomic gas, specific heat ratio r =

Cp

=

CV

We know that, DQ = nC p DT and DU = nC V DT

5/2R 5 = 3/2R 3

[at constant pressure] [at constant volume]

Therefore, fraction of heat energy required is DU C V 3 = = DQ C p 5

9. (c) The entropy function gives us a numerical measure of the irreversibility of a given process, i.e., it is a measure of disorder of a system. During formation of ice cubes orderedness increases, i.e, disorderness decreases, hence entropy decreases.

10. (b) Maximum efficiency of reversible process is given by

7. (b) In first process,

T T¢ For n moles of monoatomic gas, At A, p0V0 = nRT 0 At B, 2 p0V0 = nRT ¢ From Eqs. (i) and (ii), we get T ¢= 2T 0 T 1 \ h = 1 - 0 = = 50% 2T 0 2 h=1-

dQ = 8 ´ 105 dW = 6.5 ´ 105 J From first law of thermodynamics, dU = dQ - dW dU = 8 ´ 105 - 6.5 ´ 105 = 1.5 ´ 105 J In the second process, change in internal energy dU remains the same. \ dW = dQ - dU = 1 ´ 105 - 1.5 ´ 105 dW = - 0.5 ´ 105 J Negative sign shows that work is done on the gas.

8. (c) According to the figure,

11. (b) From first law of thermodynamics, we have DU = Q - W Amount of heat = Mass ´ Latent heat \ We have, Q = mL = 1 ´ 2240 W = 168 J (given) \ DU = 2240 - 168 = 2072 J T2 = 627 + 273 = 900 K, g = 1.5 Tg For adiabatic change, g–1 = constant, p

2T0 Q3

3

1 Q1

Þ 2 Q2

S0

…(i) …(ii)

12. (a) Given, p1 = 105 Nm–2 , T1 = 27 + 273 = 300 K,

T

T0

we know that, amount of heat = entropy ´ temperature This step is perfectly fine need not to be explained. 1 3 \ Q1 = T 0S 0 + T 0S 0 = T 0S 0 2 2 Q2 = T 0 (2 S 0 - S 0 ) = T 0S 0 Q3 = 0 W Q1 - Q2 Now, efficiency, h = = Q1 Q1 Q2 2 1 =1=1- = 3 3 Q1

S 2S0

Þ Þ

æ p2 ç çp è 1

( 1.5-1 )

ö ÷ ÷ ø

æT = çç 2 è T1

1.5

ö ÷ ÷ ø

1 /2 3 /2 æ p2 ö æ 900 ö ç 5÷ = ç ÷ è 300 ø è 10 ø

p2 = 27 ´ 105 Nm–2

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AIIMS Chapterwise Solutions ~ Physics

13. (b) The process of change of state of a system involves change of thermodynamic variables such as pressure ( p ), volume (V ) and temperature (T ) of the system. The process is known as thermodynamic process. Hence, R is not a thermodynamical coordinate.

14. (a) Due to compression, the temperature of the system increases to a very high value. This causes the flow of heat from system to the surroundings, thus decreasing the temperature. This decrease in temperature results in decrease in pressure (as, p µ T ).

5 DU = mC V DT = mRDT 2 5 ù é Q CV = Rú 2 û ëê and \

DW = DQ – DU = mRDT DQ : DU : DW = 7 : 5 : 2

20. (c) During the free expansion of a perfect gas,no work is done. Since, no heat is supplied from outside and there is no change in internal energy.Therefore, there is no change in temperature.

15. (a) In adiabatic expansion, there is no transfer of

21. (a) Given, initial temperature (T1 ) = 273 K,

heat into or out of the system Q=0 According to first law of thermodynamics, DU = Q - W \ DU = - W (Q Q = 0) Hence, work done on the gas is - 2 J.

initial volume (V1 ) = V , final temperature = T2 , final volume (V2 ) = 81 V, g = 1.25 We know that, under adiabatic condition, T1V1g- 1 = T2V2g- 1

16. (a) When a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between system and surroundings, the process is known as adiabatic process. Thus, in this process p,V and T changes but DQ = 0.

17. (c) The amount of work done is equal to the area under the curve in a pV graph. Here, curve 1 is for isobaric, curve 2 for isothermal and curve 3 for adiabatic process.

or

æV T2 = T1 çç 1 è V2

ö ÷ ÷ ø

g -1

1.25- 1

æ V ö = 273 ç ÷ è 81V ø

T2 = 273 (1 / 81)1 / 4 273 = = 91K 3 = 91 - 273 = - 182°C

22. (b) From first law of thermodynamics, 1

p

2 3

V1

V2

V

As, the area under curve 1 is maximum, hence work done is maximum in isobaric expansion.

18. (d) Given, change in volume of gas (DV ) = 025 . m3 , constant pressure ( p ) = 103 N/m2 We know that, work done (W ) = pDV = 103 ´ 025 . = 250 J 7 2

19. (c) DQ = mC p DT = mRDT [QC p =

7 R (for diatomic) ] 2

DQ = DU + DW or DQ = DU + p DV For an adiabatic expansion, DQ = 0, DW = positive \DU is negative. Also, DU µ pDV \Internal energy DU or product of pV will decrease in adiabatic expansion. Also, in adiabatic expansion neither heat enters into nor goes out. Therefore, work is done at the cost of internal energy of gas. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion. Q 23. (b) Molar specific heat is given by C = mDT In isothermal process, if heat supplied is compensated by p -V work done, then the temperature of the gas will not increase so, DT = 0. Therefore, C is not defined.

95

Thermodynamics Also, in isothermal process, ( p -V ) and T both are constants. Therefore, ( p - V ) versus T graph is a dot. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

24. (c) In an adiabatic process, no exchange of heat is permissible, i.e.,DQ = 0 As, from first law of thermodynamics, DQ = DU + DW = 0 Þ DU = - DW \In an adiabatic process, change in internal energy of a gas is equal to work done on or by the gas in the process. Also, for an adiabatic process, temperature is related to DU by DU = nC V DT So for all infinitely possible values of DU, there will be infinitely possible temperature. So, temperature of gas changes in adiabatic process. Assertion is true but Reason is false.

25. (c) The thermodynamic process is irreversible, as there always occurs a loss of energy due to energy spent in working against. The dissipative force is not recovered back.

26. (b) For using the internal energy of sea water to operate the engine of a ship, the internal energy of the sea water has to be converted into mechanical energy. Since, whole of the internal energy cannot be converted into mechanical energy, a part has to be rejected to a colder body (sink). Since, no such body is available, the internal energy of the sea water cannot be therefore, used to operate the engine of the ship. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

27. (d)

p

Isothermal curve

T1 T2 V

Isothermal curves are drawn for constant temperatures, hence do not intersect each other.

Also, slope of isothermal curve is minimum as isothermal change takes place slowly. Both Assertion and Reason are false.

28. (d) According to second law of thermodynamics, there is a net transfer of heat from a body at lower temperature to a body at higher temperature without the aid of an external agent. Since, the given information produces a contradiction in second law of thermodynamics, therefore it is not possible to produce temperature of 8000 K by collecting the sun rays with a lens. Both Assertion and Reason are false.

29. (b) When a person is driving a car, then the temperature of air inside the tyre is increased because of frictional force. From the Gay Lussac’s law, pressure is directly proportional to temperature. Hence, when temperature increases, the pressure also increases. The absolute zero is the temperature at which molecular motion ceases i.e., translational kinetic energy of molecules becomes zero, but the molecules possess potential energy according to kinetic theory of gases. Thus, the absolute zero temperature is not the zero energy temperature. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

30. (b) According to second law of thermodynamic in a closed (or isolated) system, heat can only flow from hotter to colder region. Since, entropy is a measure of this energy dispersion, entropy must always increase. The processes in an isolated system are adiabatic. Since, an isolated system is a physical system where neither matter nor energy can be exchanged between surrounding and system but it can only move around inside the system. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion

31. (a) An ideal engine called a Carnot engine serves as ideal heat engines. This ideal heat engine turns put to be the best at using energy as heat to do useful work. Carnot engine comprises four cycles. During each cycle of engine, the working substance

96

AIIMS Chapterwise Solutions ~ Physics absorbs energy |Q ¢H | as heat from a thermal reservoir at constant temperature T H and discharges energy |Q t| as heat to a second thermal reservoir at a constant lower temperature T L . Thus, Carnot cycle provides a way of determining the maximum possible efficiency achievable with reservoirs of given temperature. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

32. (a) Heat and work can cause temperature rise or change of state of a body. This heat given to body is used in increasing the temperature of body and rest against frictional forces. It is impossible by any means to recover the energy lost in doing work against dissipative forces. Usually all processes are dissipative in nature. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

33. (a) The leaking air of balloon undergoes adiabatic expansion. In this expansion, due to work done against external pressure, the internal energy of air reduces. Thus, it becomes cooler. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

34. (c) The pressure cooker works on the principle that boiling point of a liquid increases with increase of vapour pressure above the liquid. When the cooker is removed from the stove and the lid is removed, pressure again decreases due to which its boiling point decreases and water starts boiling again. Assertion is true but Reason is false.

35. (a) Cold drinks are produced by injecting carbon dioxide into the drink at several atmospheres. Carbon dioxide dissolves readily even at normal atmospheric pressure. When the pressure is released (i.e., bottle is opened) carbon dioxide comes out of solution forming numerous bubbles and releasing the carbon dioxide back into the atmosphere. Also, in this

process, no heat is lost or gained, hence it is an adiabatic expansion, which causes lowering of temperature and condensation of water vapours.

36. (a) First law of thermodynamics tells only about the conversion of mechanical energy into the heat and vice-versa. It does not put any condition as to why heat cannot flow from lower temperature to higher temperature. First law of thermodynamics given dQ = dU + dW If heat is supplied as such, its volume does not change i.e., dV = 0, then whole of the heat energy supplied to the system will increase in its internal energy only.

37. (d) In adiabatic process, there is no exchange of heat between the system and the surrounding. This can be possible only if the gas under adiabatic process is allowed to expand or compressed very quickly. Thus, it is a quick process. When the gas is compressed adiabatically, the heat produced cannot escape to the surrounding through the insulating walls. As a result, the temperature of the gas and hence, the internal energy increases. Both Assertion and Reason are false.

38. (b) Melting of solid is associated with change in its internal energy, infact internal energy increases when a solid melts (in accordance with first law of thermodynamics), while latent heat is defined as the heat required to melt a unit mass of solid. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

39. (c) If an amount of heat Q is given to a system a part of it is used in increasing the internal energy (DU ) of the system and rest in doing work (W ). \ Q = DU + W = DU + pDV Hence, first law of thermodynamics is a form of law of conservation of energy. Assertion is true but Reason is false.

40. (a) The statement given in the Assertion is itself a form of the second law of thermodynamics which cannot be violated. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

CHAPTER

11 Kinetic Theory of Gases 1. The molar specific heat of a gas as given

6. 1 mole of gas occupies a volume of 200 mL

5 from the kinetic theory is R. If it is not 2 specified whether it is C p or C V , one could conclude that the molecules of the gas

at 100 mm pressure. What is the volume occupied by two moles of gas at 400 mm pressure and at same temperature?

(a) are definitely monoatomic [AIIMS 2015] (b) are definitely rigid diatomic (c) are definitely non-rigid diatomic (d) can be monoatomic or rigid diatomic

(a) 50 mL

2. The root mean square velocity of hydrogen molecule at 27°C is v H and that of oxygen at 402°C is vO , then [AIIMS 2014] (a) vO > vH (c) 2 vO = 3 vH

(b) 4vO = 9 vH (d) 9 vO = 134 vH

3. A balloon is filled at 27°C and 1 atm pressure by 500 m 3 He. At -3°C and 0.5 atm pressure,the volume of He will be (a) 700 m3 (c) 1000 m3

(b) 900 m3 [AIIMS 2013] (d) 500 m3

4. Root mean square (rms) speed of the molecules of ideal gas is v. If pressure is increased two times at constant temperature, then the rms speed will become [AIIMS 2013] v (a) 2

(b) v

(c) 2v

(d) 4v

5. A gas mixture contains one mole O2 gas and one mole He gas, find the ratio of specific heat at constant pressure to that at constant volume of the gaseous mixture. (a) 2

(b) 1.5

(c) 2.5

[AIIMS 2013] (d) 4

[AIIMS 2012] (b) 100 mL (c) 200 mL (d) 400 mL

7. The heat required to increase the temperature of 4 moles of a monoatomic ideal gas from 273 K to 473 K at constant volume is [AIIMS 2012] (a) 200 R

(b) 400 R

(c) 800 R

(d) 1200 R

8. The velocity of sound in air at NTP is 330 m/s. What will be its value when temperature is doubled and pressure is halved ? [AIIMS 2011] (a) 330 m/s (c) 330/ 2 m/s

(b) 165 m/s (d) 333/ 2 m/s

9. If pressure of CO2 (real gas) in a container RT a , then mass of 2V - b 4V 2 the gas in container is [AIIMS 2010]

is given by p =

(a) 11 g

(b) 22 g

(c) 33 g

(d) 44 g

10. By what percentage should the pressure of a given mass of a gas be increased, so as to decrease its volume by 10% at a constant temperature? [AIIMS 2009] (a) 5%

(b) 7. 2%

(c) 12.5%

(d) 11.1%

11. Two rigid boxes containing different ideal gases are placed on table. Box A contains one mole of nitrogen at temperature T0,while box B contains one mole of

98

AIIMS Chapterwise Solutions ~ Physics helium at temperature (7/ 3)T0.The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases T f , in terms of T0 is [AIIMS 2008] 3 T 7 0 3 (c) Tf = T0 2

(a) Tf =

7 T 3 0 5 (d) Tf = T0 2 (b) Tf =

pV nT versus p for oxygen gas at two different temperatures.

pV (J-mol-1K-1) nT

average and most probable speeds of molecules of a gas obeying Maxwellian velocity distribution. Which of the following statements is correct? [AIIMS 2004] (a) vrms < vav < vmp (c) vmp < vrms < vav

(b) vrms > vav > vmp (d) vmp > vrms > vav

15. At constant pressure thermometer gave a

12. The figure below shows the plot of

T2

14. v rms , vav and v mp are root mean square,

T1

reading of 47.5 units of volume when immersed in ice-cold water and 67 units in a boiling liquid. The boiling point of the liquid, is [AIIMS 2001] (a) 125°C

(b) 100°C

(c) 135°C

16. The average kinetic energy of a gas

molecule at 27°C is 6.21 ´ 10 -21 J, then its average kinetic energy at 227°C is (a) 10.35 ´ 10-21 J

[AIIMS 1999] (b) 1135 . ´ 10-21 J

(c) 52. 2 ´ 10-21 J

(d) 5 . 22 ´ 10-21 J

17. Rate of diffusion is p

Read the following statements concerning the above curves

(d) 112° C

[AIIMS 1998]

(a) faster in liquids than in solids and gases (b) faster in gases than in liquids and solids (c) faster in solids than in liquids and gases (d) None of the above

(i) The dotted line corresponds to the ‘ideal’ gas behaviour. (ii) T1 > T2 pV at the point, where the (iii) The value of nT curves meet on the Y-axis is the same for all gases.

18. A given sample of an ideal gas occupies a

Which of the above statements is true?

19. If at same temperature and pressure, the

[AIIMS 2007] (a) (i) only (b) (i) and (ii) only (c) All of these (d) None of these

13. Two balloons are filled, one with pure He gas and the other by air, respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is [AIIMS 2006] (a) more in the He filled balloon (b) same in both balloons (c) more in air filled balloon (d) in the ratio of 1 : 4

volume V at a pressure p and absolute temperature T . The mass of gas is m. Which of the following gives the density of the gas ? [AIIMS 1998] (a) p/(kT )

(b)

pm kT

(c)

p kTV

(d) mkT

densities for two diatomic gases are d 1 and d2 respectively, then the ratio of velocities of sound in these gases will be [AIIMS 1996] (a)

d2 2d1

(b)

d2 d1

(c)

2d1 d2

(d)

d1 d2

20. The temperature of a gas is held constant, while its volume is decreased. The pressure exerted by the gas on the walls of the container increases, because its molecules [AIIMS 1996] (a) strike the walls more frequently (b) strike the walls with higher velocities (c) are in contact with the walls for a shorter time (d) strike the walls with larger force

99

Kinetic Theory of Gases 21. When we heat a gas-sample from 27°C to 327°C, then the initial average kinetic energy, of the molecules was E. What will be the average kinetic energy after heating? [AIIMS 1995] (a) 2E (c) 2E

(b) 327 E (d) 300 E

22. An ideal gas is heated from 27° C to 627°C at constant pressure. If initial volume was 4 m 3 , then the final volume of the gas will be [AIIMS 1995] (a) 6 m3 (c) 12 m3

(b) 2 m3 (d) 4 m3

Direction (Q. Nos 27-32) Read the Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

27. Assertion The molecules of a monoatomic gas has three degrees of freedom.

23. For Boyle’s law to hold good, the gas should be

Assertion & Reason

[AIIMS 1995]

(a) perfect and at constant temperature but variable mass (b) perfect and of constant mass and temperature (c) real and at constant temperature but variable mass (d) real and of constant mass and temperature

Reason The molecules of diatomic gas has five degrees of freedom. [AIIMS 2017]

28. Assertion The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume. Reason The molecules of gas collide with each other and the velocities of the molecules change due to the collision. [AIIMS 2015]

24. If C p and C V are the specific heats for a gas at constant pressure and at constant volume respectively, then the relation [AIIMS 1994] C p - C V = R is exact for (a) ideal gas and nearly true for real gases at high pressure (b) ideal and real gases at all pressures (c) ideal gas and nearly true for real gases at moderate pressure (d) ideal gas at all pressure and real gas at moderate pressure.

25. In a vessel, the gas is at a pressure p. If the mass of all the molecules is halved and their speed is doubled, then the resultant pressure will be [AIIMS 1994] (a) p (c) p / 2

(b) 4p (d) 2 p

26. A bulb contains one mole of hydrogen mixed with one mole of oxygen at temperature T . The ratio of rms values of velocity of hydrogen molecules to that of oxygen molecules, is [AIIMS 1994] (a) 4 : 1 (c) 1 : 4

(b) 1 : 16 (d) 16 : 1

29. Assertion C p can be less than C V . Reason C p - C V = R is valid only for ideal gases.

[AIIMS 2011]

30. Assertion If a gas container in motion is suddenly stopped, the temperature of the gas rises. Reason The kinetic energy of ordered mechanical motion is converted into the kinetic energy of random motion of gas molecules. [AIIMS 2010]

31. Assertion The root mean square and most probable speeds of the molecules in a gas are the same. Reason The Maxwell distribution for the speed of molecules in a gas is symmetrical. [AIIMS 2006]

32. Assertion For gas atom, the number of degrees of freedom is 3. Cp Reason =g CV

[AIIMS 2000]

100

AIIMS Chapterwise Solutions ~ Physics

Answers 1. 11. 21. 31.

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

2. 12. 22. 32.

3. (b) 13. (b) 23. (b)

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

4. (b) 14. (b) 24. (c)

5. (b) 15. (d) 25. (d)

6. (b) 16. (a) 26. (a)

7. (d) 17. (b) 27. (b)

8. (c) 18. (b) 28. (b)

9. (b) 19. (b) 29. (b)

10. (d) 20. (a) 30. (a)

Explanations 1. (d) A monoatomic gas has 3 degrees of freedom and diatomic gas has 5 degrees of freedom. For monoatomic gas, 3 specific heat at constant volume, C V = R 2 (R is universal gas constant) and specific heat at constant pressure, 3 5 C p = CV + R = R + R = R 2 2 For a diatomic gas, 5 7R C V = R and C p = C V + R = 2 2 Hence, we can conclude that the molecules of the gas can be monoatomic or rigid diatomic.

2. (c) We know that, 3RT M v rms µ T vH TH = = vO TO

v rms = \

or

27 + 273 300 = = 402 + 273 675

4 9

vH 2 or 3v H = 2vO = vO 3

\

p1V1 pV = 2 2 T1 T2

3 5 R+ R 2 2 = 2R CV = 1+ 1

\

C p = 2R + R = 3R C p 3R = = 1.5 C V 2R

5 ù é êQ For O2 , C V = 2 R ú ú ê êand for He, C = 3 R ú V êë 2 úû [QC p = C V + R]

6. (b) Given, p1 = 100 mm, V1 = 200 mL and p2 = 400 mm From Boyle’s law, (at constant temperature) pV = constant or

p1V1 = p2V2 pV 100 ´ 200 V2 = 1 1 = 400 p2

V2 = 50 mL (volume of 1 mole gas) Volume of 2 mole gas = 2 ´ 50 = 100 mL

3. (b) According to gas equation, pV = nRT \

5. (b) For mixture of gases,

or V2 =

p1V1T2 p2T1

V2 =

1 ´ 500 ´ (273 - 3) 0.5 ´ (273 + 27)

V2 =

1 ´ 500 ´ 270 = 900 m3 0.5 ´ 300

4. (b) Root mean square (rms) speed of gas molecules does not depend on the pressure of gas (if temperature remains constant) because pressure (p) µ density (r ) (Boyle’s law). If pressure is increased n times, density will also increase by n times, but v rms remains constant.

7. (d) Specific heat for an ideal monoatomic gas at constant volume is given by 3 CV = R 2 Also, at constant volume, 1 æ DQ ö CV = ç ÷ n è DT ø V (where, n is number of moles) \Heat, DQ = nC V DT 3 (Q n = 4) DQ = 4 ´ ´ R (473 - 273) 2 3 = 4 ´ ´ R ´ 200 = 1200 R 2

8. (c) As, there is no effect of change in pressure on the velocity of sound in air, but velocity depends on temperature as, vµ T Therefore,when temperature is doubled, velocity becomes 330 2 m/s.

101

Kinetic Theory of Gases 9. (b) van der Waals’ gas equation for n mole of real gas,

2 ö æ ç p + n a ÷ (V - nb ) = nRT ç V 2 ÷ø è

(where a and b are van der Waals’ constants) æ nRT ö n2 a ÷or …(i) p = çç 2 ÷ è V - nb ø V Given equation, ù éæ 1 ö RT ÷ æ RT 1 ö2 a ú a ö êç 2 æ ÷ = êç ÷- ç ÷ × ú p = çç 2 ÷ 2 è 2V - b 4V ø ê çç V - b / 2 ÷÷ è 2 ø V ú úû êë è ø …(ii) On comparing Eqs. (i) and (ii), we get 1 n= 2 m We know that, n = M [where, m = mass of gas and M = molecular mass of gas] 1 or m = nM = ´ 44 = 22 g 2 [MCO2 = 12 + 2 ´ 16 = 44 g]

10. (d) From Boyle’s law, 1ö æ when T is constant, pV = constant ç or p µ ÷ . Vø è When volume is decreased by 10% 90 , the pressure must i.e. Volume becomes 100 become 100/90. (100 - 90) ´ 100 \% increase in pressure = 90 = 111 . %

7T ö 5R 3R æ (T f - T 0 ) + çT f - 0 ÷ = 0 2 2 è 3 ø 12 3 Tf = T0 = T0 \ 8 2

12. (c) (i) The dotted line in the diagram shows that there is no deviation in the value of

pV for nT

different temperatures T1 and T2 for increasing pressure. So, this gas behaves ideally. Hence, dotted line corresponds to ‘ideal’ gas behaviour. (ii) At high temperature, the deviation of the gas should be less and at low temperature, the deviation of gas should be more. In the graph, deviation for T2 is greater than for T1 . Thus, T1 > T2 . (iii) Since, the two curves intersect at dotted line. pV So,the value of at that point on the Y-axis nT is same for all gases. So, all of above statements are true.

13. (b) Ideal gas equation can be written as …(i) pV = nRT where, symbols have their usual meaning. From Eq. (i), we have p n = = constant V RT So, at constant pressure and temperature, all gases will contain equal number of molecules per unit volume.

14. (b) Maxwell gave the distribution of molecules at different speed by the curve given below. dN dv

At a particular temperature

11. (c) Internal energy of a system remains constant or change in internal energy of the system is zero, We know that, DU = nC V DT 5R For N2 , DU A = 1 ´ (T f - T 0 ) 2 5R ö æ for diatomic gas ÷ çQ C V = 2 ø è 3R æ 7 ö For He, DU B = 1 ´ çT - T ÷ 2 è f 3 0ø 3R æ ö for monoatomic gas ÷ çQ C V = 2 è ø Now, total internal energy of the system is zero. \ DU A + DU B = 0

vmp vav vrms

From graph between

v(m/s)

dN and v, it can be seen dv

dN is maximum at most probable speed. dv The graph also represent that v rms > vav > v mp. that

As, Þ

3RT 8RT 2RT > > M pM M RT RT RT > 1.6 > 1.41 177 . M M M

102

AIIMS Chapterwise Solutions ~ Physics

15. (d) Given, initial volume (V1 ) = 47.5 units, Temperature of ice-cold water (T1 ) = 0°C = 273 K and final volume (V2 ) = 67 units. We know from the Charle’s law, at constant pressure, V1 V2 = T1 T2 V or T2 = 2 ´ T1 (T2 is boiling point of liquid) V1 67 = ´ 273 = 385K or 112°C 47.5

16. (a) The average kinetic energy of a gas molecule at temperature T is given by 3 E = kT 2 Þ E µT where, k is Boltzmann’s constant. Given, E1 = 621 . ´ 10-21 J, T1 = 273 + 27 = 300 K, \ Þ

T2 = 273 + 227 = 500 K, E2 = ? E1 T1 = E2 T2 T 500 E2 = 2 × E1 = ´ 6. 21 ´ 10-21 T1 300 E2 = 1035 . ´ 10-21 J

17. (b) Diffusion refers to the process by which the molecules intermingle as a result of their kinetic energy of random motion. Gases have large intermolecular space and they possess more kinetic energy than liquids and gases. \Rate of diffusion is faster in gases than in liquids and solids. T Alternatively, diffusion rate = k m where, k is constant, T is temperature and m is mass. 1 nm 2 18. (b) Pressure, p = × v rms 3 V where, nm = mass of the gas, V = volume of the gas mn and = density of the gas. V 1 1 3RT rRT Thus, p = rv 2rms = r = 3 3 M0 M0 é 3RT ù êQ v rms = ú M0 û ë

pM 0 pmN A [Q R = N A k and M 0 = mN A ] = RT kN AT pm Þr = kT r=

19. (b) The velocity of sound in a gas is given by v=

Cp gp where, g = CV d

\ For two gases at constant pressure, v1 d g = 2 1 v2 d1 g 2 For a diatomic gas, g 1 = g 2 = \

7 5

v1 d2 = v2 d1

20. (a) The pressure exerted by gas on the walls of container increases because the molecules of enclosed gas strike the walls of container more frequently due to lesser effective distance between the walls as there is decrease in volume of gas.

21. (a) Given, initial temperature (T1 ) = 27° C = 273 + 27 = 300 K, Final temperature (T2 ) = 327° C = 327 + 273 = 600 K and initial average kinetic energy (E1 ) = E. We know that, -3 average KE = kT 2 \ E µT E1 T1 300 1 = = = \ E2 T2 600 2 E2 = 2E1 = 2E

22. (c) Given, initial temperature (T1 ) = 27° C = 300 K, final temperature (T2 ) = 627°C = 900 K and initial volume (V1 ) = 4 m3 . We know from the Charle’s law, (at constant pressure), V V V µ T or 1 = 2 T1 T2 T 900 or V2 = V1 ´ 2 = 4 ´ = 12 m3 T1 300

23. (b) According to Boyle’s law, the volume (V ) of a given mass of a perfect gas at constant temperature is inversely proportional to its pressure ( p ). Mathematically, V µ (1 / p ) or pV = K (constant). Therefore, for Boyle’s law to hold good, the gas should be perfect and of constant mass and temperature.

103

Kinetic Theory of Gases 24. (c) C p - C V = R is exact for ideal gas and nearly

28. (b) The total translational kinetic energy of gas

true for real gases at moderate pressure because real gases behave ideally at low pressure and high temperature.

molecules is given by 3 3 [Q pV = nRT ] E = nRT = pV = 1.5 pV 2 2 Also, the velocity of molecules changes on collision.

25. (d) Given, initial pressure of the gas ( p1 ) = p;

Initial mass of the gas molecules (m1 ) = m, final mass of gas molecules (m2 ) = 0.5 m, initial speed of molecules (v1 ) = v and final speed of the molecules (v2 ) = 2 v . Let resultant pressure be ‘p2 ’.

We know that pressure of the gas ( p ) 1 mn = ´ ´ v2 3 V \ p µ mv 2 2

or

p1 m1 æ v1 ö = ´ç ÷ p2 m2 çè v2 ÷ø 2

=

1 æ1ö 1 ´ç ÷ = 0.5 è 2 ø 2

p2 = 2 p1 = 2 p

26. (a) Given, one mole of hydrogen mixed with one mole of oxygen. We know that, 3RT M 1 v rms µ M (where M is molecular mass) (v rms ) H MO 32 = = 2 (v rms )O MH

rms velocity, v rms = \

or

or

[Q M (mass of O2 ) = MO = 2 ´ 16 = 32 and M (mass of H2 ) = 1 ´ 2 = 2] (v rms ) H = 16 = 4 (v rms )O

\ (v rms ) H : (v rms )O = 4 : 1

27. (b) A monoatomic gas molecule like He, consists of single atom. It can have translational motion in any direction in space. Thus, it has three (translational) degree of freedom, f = 3. It can also rotate but due to its small moment of inertia, rotational kinetic energy is neglected. The molecules of a diatomic gas (like O2 ,CO2 , H2 ) cannot only move, but also rotate about any one of the three coordinates. Hence, it can have two rotational degree of freedom. Thus, a diatomic molecule has 5 degrees of freedom i.e., 3 translational and 2 rotational.

29. (b) C p can be less than C V for those substances which contract on heating like for water from 0 to 4°C. Assertion and Reason both are true but Reason is not a correct explanation of Assertion.

30. (a) The motion of the container is known as the ordered motion of the gas and zig-zag motion of gas molecules within the container is called disordered motion (random motion). When the container suddenly stops, ordered kinetic energy gets converted into disordered kinetic energy which increases the temperature of the gas.

31. (d) Maxwell distribution Refer to Q. 14. 3RT 2RT and v mp = M M It is obvious that v rms >v mp Correct Reason Also, Maxwell distribution for the speed of molecules in a gas is asymmetrical. v rms =

n

O vprobable vrms V

Both Assertion and Reason are false.

32. (b) The molecule of a monoatomic gas (such as He, Ar, etc) consists of a single atom. Its translational motion can take place in any direction in space. Thus, it can be resolved along three coordinate axes and can have three independent motions. Hence, it has three degrees of freedom (all translational). By definition of specific heat specific heat at constant pressure C p (g ) = = specific heat at constant volume C V Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

AIIMS Chapterwise Solutions ~ Physics CHAPTER

12 Oscillations 1. In forced vibration, m = 10 gm, f = 100 Hz and driver force, F = 100 cos (20pt ), then what is the amplitude of particle? (Take p2 = 10 ) [AIIMS 2018] (a) 10 m

(b) 0.5 m

(c) 0.025 m (d) 0.25 m

2. A load of mass m falls from a height h on the scale pan hung from a spring as shown. If the spring constant is k and mass of the scale pan is zero and the m mass m does not bounce h relative to the pan, then the amplitude of vibration is [AIIMS 2017] 2 hk mg (b) 1+ k mg

(a) mg (c)

mg mg 1 + 2 hk + k k mg

(d) None of these

represented by the equations pö æ y 1 = 0.1 sinç 100 pt + ÷ and y2 = 0.1 cos pt . 3ø è The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is at t = 0 [AIIMS 2016] -p 3

(b)

p 6

(c)

-p 6

(d)

in the forward direction in the backward direction in the backward direction in the forward direction

5. The velocity vector v and displacement vector x of a particle executing SHM are related as vdv = - w2 x dx with the initial condition v = v 0 at x = 0. The velocity v, when displacement is x, is [AIIMS 2015] (a) v = (c) v =

3. Two simple harmonic motions are

(a)

a g a (b) tan-1 g g (c) tan-1 a g (d) tan-1 a (a) tan-1

3

v 02 + w2 x2

(b) v =

v 03 + w3 x3

(d) v = v 0 - (w3 x3e x )1/ 3

motion oscillator having identical four springs has time period [AIIMS 2014] k1 k2 k3

k4 m

4. A simple pendulum is setup in a trolly

[AIIMS 2016]

3

6. As shown in figure a simple harmonic

p 3

which moves to the right with an acceleration a on a horizontal plane. Then, the thread of the pendulum in the mean position makes an angle with the vertical

v 02 - w2 x2

m 4k m (c) T = 2p k (a) T = 2 p

m 2k m (d) T = 2 p 8k

(b) T = 2 p

105

Oscillations 7. The velocity of a particle moving in the xy-plane is given by dx dy = 8 p sin 2pt and = 5p cos 2pt dt dt where t = 0, x = 8 and y = 0, the path of the particle is [AIIMS 2014] (a) a straight line (c) a circle

(b) an ellipse (d) a parabola

8. Two simple harmonic motions are

pö æ represented by y 1 = 4 sin ç 4p t + ÷ and 2ø è y2 = 3 cos( 4pt ). The resultant amplitude is

(a) 7

(b) 1

(c) 5

[AIIMS 2013] (d) 2 + 3

9. A particle is executing linear simple harmonic motion of amplitude A. What fraction of the total energy is kinetic when the displacement is half the amplitude? (a)

1 4

(b)

1 2 2

(c)

1 2

[AIIMS 2013] 3 (d) 4

10. Average value of kinetic energy and potential energy over entire time period in a SHM is [AIIMS 2013] (a) 0, (c)

1 mw2 A 2 2

1 1 mw2 A 2 , mw2 A 2 2 2

1 mw2 A 2 ,0 2 1 1 (d) mw2 A 2 , mw2 A 2 4 4

(b)

11. A simple harmonic oscillator consists of a particle of mass m and an ideal spring with spring constant k . The particle oscillates with a time period T . The spring is cut into two equal parts. If one part oscillates with the same particle, the time period will be (a) 2T

(b) 2T

(c)

T 2

[AIIMS 2012] T (d) 2

12. The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (if it is a second’s pendulum on earth) [AIIMS 2010] 1 (a) s 2

(b) 2 2 s

(c) 2s

1 (d) s 2

13. Two simple pendulums first of bob mass M 1 and length L1, second of bob mass M2 and length L2 × M 1 = M2 and L1 = 2L2 . If the vibrational energies of both are same. Then, which is correct? [AIIMS 2010] (a) Amplitude of B greater than A (b) Amplitude of B smaller than A (c) Amplitude will be same (d) None of the above

14. A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency w. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time [AIIMS 2008] (a) at the mean position of the platform (b) for an amplitude of g /w2 (c) for an amplitude of g 2 /w2 (d) at the highest position of the platform

15. The functionsin2 ( wt ) represents [AIIMS 2008] (a) a periodic, but not simple harmonic motion with a period 2p / w (b) a periodic, but not simple harmonic motion p with a period w 2p (c) a simple harmonic motion with a period w p (d) a simple harmonic motion with a period w

16. A particle of mass m is executing oscillations about the origin on the X-axis. Its potential energy is U ( x) = k ( x)3 , where k is a positive constant. If the amplitude of oscillation is A, then its time period T is [AIIMS 2008] 1 (a) proportional to A (c) proportional to A

(b) independent of A (d) proportional to A 3 / 2

17. A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T . The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is T (a) 4

T (b) 8

T (c) 12

[AIIMS 2007] T (d) 2

AIIMS Chapterwise Solutions ~ Physics

106 18. The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is [AIIMS 2007] (a) 0.5 p

(b) p

(c) 0.707 p (d) zero

23. The frequency of oscillator of the springs as shown in figure will be [AIIMS 2001] (a)

1 (k1 + k2 )m k1k2 2p

(b)

k1k2 1 2 p (k1 + k2 )m

19. Which of the following functions represents a simple harmonic oscillation? [AIIMS 2005] (a) sin wt - cos wt

(b) sin2 wt

(c) sin wt + cos 2 wt

(d) sin wt - sin 2 wt

20. Two springs are connected to a block of mass M placed on a frictionless surface as shown below. If both the springs have a spring constant k , the frequency of oscillation of block is [AIIMS 2004]

k

k M

(a)

1 k 2p M

(b)

1 k 2p 2M

(c)

1 2k 2p M

(d)

1 M 2p k

k2

1 k 2p m k (d) 2p m

(c)

24. A horizontal platform with an object placed on it, is executing simple harmonic motion in the vertical direction. The amplitude of oscillation is 392 . ´ 10 -3 m. What should be the least period of these oscillations, so that the object is not detached from the platform? [AIIMS 1999] (a) 0.145 s (c) 0.1256 s

(b) 0.1556 s (d) 0.13565 s

25. Which one of the following statement is not correct for a particle executing SHM? [AIIMS 1999]

21. Two springs of force constants k and 2k are connected to a mass as shown below. The frequency of oscillation of the mass is [AIIMS 2003] k

2k m

(a)

1 k 2p m

(b)

1 2k 2p m

(c)

1 3k 2p m

(d)

1 m 2p k

motion with an angular velocity of 3.5 rad/s and maximum acceleration 7.5 m /s2 . The amplitude of oscillation will be [AIIMS 1999]

harmonic motion with time period T . If the length of the pendulum is increased by 21%, then the increase in the time period of the pendulum of the increased length is (b) 13%

(c) 50%

(a) Acceleration of the particle is maximum at the mean position (b) Restoring force is always directed towards a fixed point (c) Total energy of the particle always remains the same (d) Restoring force is maximum at the extreme position

26. A particle executes simple harmonic

22. Simple pendulum is executing simple

(a) 22%

k1

[AIIMS 2001] (d) 10%

(a) 0.53 cm (c) 0.61 cm

(b) 0.28 cm (d) 0.36 cm

27. If the time period of oscillation of mass m suspended from a spring is 2 s, the time period of mass 4m will be [AIIMS 1998] (a) 2 s (c) 4 s

(b) 3 s (d) 5 s

107

Oscillations 28. If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2 s, then its maximum velocity is [AIIMS 1998] (a) 0.6 m / s (c) 0.4 m / s

(b) 0.16 m / s (d) 0.32 m / s

29. Two identical springs of spring constant k are connected in series and parallel as shown in figure. A mass m is suspended from them. The ratio of their frequencies of vertical oscillation will be [AIIMS 1997]

34. The composition of two simple harmonic motions of equal periods at right angles to each other and with a phase difference of p, results in the displacement of the particle along a [AIIMS 1994] (a) straight line (c) hexagon

(b) circle (d) ellipse

Assertion & Reason Direction (Q. Nos. 35 to 44) Read the Assertion and Reason carefully to mark the correct option from those given below

k

k

k

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

k

m

m

(a) 1 : 2 (c) 4 : 1

(b) 2 : 1 (d) 1 : 4

35. Assertion A spring of force constant k is

30. If a spring extends by x on loading, then energy stored by the spring is (where, T is the tension in spring and k is spring constant) [AIIMS 1997] (a)

2T 2 k T2 (d) k

2k

Reason The spring connected in series are represented by k = k 1 + k 2 . [AIIMS 2017]

(b)

2

T T2 (c) 2k

36. Assertion Soldiers are asked to break steps

31. A lightly damped oscillator with a frequency ( w) is set in motion by harmonic driving force of frequency ( n). When n < w, then response of the oscillator is controlled by [AIIMS 1996] (a) oscillator frequency (b) spring constant (c) damping coefficient (d) inertia of the mass

32. The tension in piano wire is 10 N. What should be the tension in the wire to produce a note of double the frequency? (a) 40 N

(b) 5 N

(c) 80 N

[AIIMS 1995] (d) 20 N

33. The periodic time of a body executing SHM is 4 s. After how much interval from time t = 0, its displacement will be half of its amplitude? [AIIMS 1995] 1 (a) s 4

1 (b) s 2

1 (c) s 6

cut into two pieces having lengths in the ratio 1 : 2. The force constant of series 3k combination of the two parts is . 2

1 (d) s 3

while crossing the bridge. Reason The frequency of marching may be equal to the natural frequency of bridge and may lead to resonance which can break the bridge. [AIIMS 2016]

37. Assertion The percentage change in time period is 1.5%, if the length of simple pendulum increases by 3%. Reason Time period is directly proportional to length of pendulum. [AIIMS 2012]

38. Assertion The periodic time of a hard spring is less as compared to that of a soft spring. Reason The periodic time depends upon the spring constant and spring constant is large for hard spring. [AIIMS 2009]

AIIMS Chapterwise Solutions ~ Physics

108 39. Assertion Water in a U-tube executes

42. Assertion The time period of pendulum on

SHM, the time period for mercury filled up to the same height in the U-tube be greater than that in case of water. Reason The amplitude of an oscillating pendulum goes on increasing. [AIIMS 2007]

satellite orbiting the earth is infinity. Reason Time period of a pendulum is inversely proportional to g . [AIIMS 2002]

43. Assertion Speed of wave =

40. Assertion The amplitude of an oscillating

Reason Wavelength is the distance between two nearest particles in phase.

pendulum decreases gradually with time. Reason The frequency of the pendulum decreases with time. [AIIMS 2003]

41. Assertion In SHM, the motion is to and fro

[AIIMS 2002]

44. Assertion In simple harmonic motion, the velocity is maximum when the acceleration is minimum.

and periodic. Reason Velocity of the particle,

Reason Displacement and velocity of SHM p differ in phase by . [AIIMS 1999] 2

v = w a2 - x2 where, x is displacement.

Wavelength Time period

[AIIMS 2002]

Answers 1. 11. 21. 31. 41.

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

2. 12. 22. 32. 42.

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

3. 13. 23. 33. 43.

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

4. 14. 24. 34. 44.

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

5. 15. 25. 35.

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

6. 16. 26. 36.

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

7. 17. 27. 37.

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

8. 18. 28. 38.

9. 19. 29. 39.

(a) (a) (b) (a)

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

10. 20. 30. 40.

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

Explanations 1. (c) In forced vibration or oscillation, the driving force is given by F = F0 cos wd t Compare with F = 100cos(20pt ) Þ F0 = 100 N, wd = 20p (given) f0 The amplitude, A = m(w20 - w2d ) where, Þ

w0 = 2pf = 200p, m = 10 gm 100 A= 0.01[(200p )2 - (20p )2 ] 10000 = = 0.025 m 10 ´ 39600

2. (b) Given, spring constant = k Potential energy of spring at height, h = mgh 1 Kinetic energy of spring at x0 = kx2 2 Potential energy of spring at x0 = mgx0

From law of conservation of energy, Potential energy of spring at height h = Kinetic energy of spring at x0 - Potential energy of spring at x0 1 mgh = kx20 - mg x0 2 where, x0 is maximum elongation in spring. 1 Þ kx20 - m gx0 - mgh = 0 2 2mg 2mg h=0 x0 Þ x20 k k Since, the above equation is a quadratic equation, therefore, for calculating the value of x0 , we will use x0 =

x0 =

2mg ± k

-b ± b2 - 4ac 2a 2

2mg æ 2mg ö h ç ÷ + 4´ k è k ø 2

109

Oscillations So, amplitude = Elongation in spring for lowest extreme position - Elongation in spring for equilibrium position mg 2hk = x0 - x1 = 1+ k mg

3. (c) Equation of SHM is given by pö æ y1 = 01 . sinç 100pt + ÷ 3ø è dy1 pö æ = 01 . ´ 100p cos ç 100pt + ÷ v1 = dt 3 è ø pö æ = 10p cos ç 100pt + ÷ 3ø è pö æp = 10p sinç + 100pt + ÷ 3ø è2 Also, y2 = 01 . cos pt dy2 v2 = = - 01 . p sin pt dt

6. (c) In the given figure, spring with spring constants k1 and k2 are in parallel and that are with k3 and k 4 are also in parallel. The two combinations are in series with each other. Q k1 = k2 = k Also, k3 = k 4 = k Þ k1 + k2 = k + k = 2k and k3 + k 4 = k + k = 2k 1 1 1 = + keq 2k 2k keq = k m k where, m is mass and k is spring constant.

Time period is given by, T = 2p

7. (b) yx-graph gives the shape of path of particle, dx = 8p sin2pt dt Integrating within the limits,

= 01 . p sin(p + pt )

x

8p [cos 2pt ]0t = 4 (1 - cos 2pt ) 2p x - 12 …(i) cos2pt = 4 dy = 5p cos 2pt dt Integrating within the limits, x- 8= -

4. (b) In accelerated frame of reference, a fictitious force (pseudo force) ma acts on the bob of pendulum as shown in figure. A pseudo force is an apparent force which acts on all masses whose motion is described using non-inertial frame of reference.

y

t

ò0 dy = 5 pt ò0cos 2pt 5 sin2pt 2 y sin2pt = (5 / 2) y=

q

a ma

t

ò8dx = ò0 8p sin2ptdt

At t = 0, phase difference of v1 w.r.t. v2 p æp pö =ç + ÷-p=6 è2 3ø

q mg

Þ

…(ii)

We know that, sin2 x + cos2 x = 1. The reaction force on the pendulum produces a backward horizontal acceleration a in the pendulum. ma a Hence, tan q = = mg g æ aö q = tan-1 çç ÷÷ in the backward direction. èg ø dv 5. (b) Given, v = - w2 x dx v x On integrating within the limit, ò vdv = ò -w2 xdx v

Þ Þ

x

v0

0

é 2ù é v2 ù 2 2 2 2 2 x ê ú = - w ê ú Þ v - v0 = - w x ë 2 û0 ë 2 ûv0 v=

v 20 - w2 x2

Put the value of Eqs. (i) and (ii), y2 ( x - 12)2 + 2 =1 2 (4) (5) (2)2 This is the equation of an ellipse y2 x2 i.e., + 2 =1 2 a b \The path of the particle is an ellipse.

8. (a) The given two SHM’s are ...(i) y1 = 4 sin(4pt + p / 2) = 4 cos 4pt and y2 = 3 cos(4pt ) ...(ii) It can be seen from Eqs. (i) and (ii) that the phase difference between the two vibration is zero. i.e., f = 0°

AIIMS Chapterwise Solutions ~ Physics

110 The resultant amplitude is given by A2 = a2 + b2 + 2ab cos f Here,

a = 4,b = 3 A2 = 42 + 32 + 2 ´ 4 ´ 3cos 0° A2 = 49 Þ

A =7

9. (d) Kinetic energy of particle at a distance x from mean position executing linear simple harmonic motion is given by 1 KE = mw2 ( A2 - x2 ) 2 Potential energy of particle executing SHM is given by 1 PE = mw2 x2 2 1 Total energy (E ) = KE + PE = mw2 A2 2 A When, x = (given) 2 æ 1 A2 ö÷ 3 KE = mw2 çç A2 = mw2 A2 2 4 ÷ø 8 è \

KE 3 = E 4

\

T = 2p l / g 1 T Þ e = T µ g Tp

1 2

Minimum KE = 0 æ 0 + 1/2 mw2 A2 ö 1 ÷ = mw2 A2 Average KE = çç ÷ 4 2 ø è æ 0 + 1/2 mw2 A2 ö ÷ Similarly, average PE = çç ÷ 2 ø è =

1 mw2 A2 4

11. (c) Given, mass of the particle = m and spring constant = k m k 1 As spring is cut into two equal parts and k µ l (where, l is the length of spring) \ k ¢ = 2k m 1 \ = T T ¢ = 2p 2k 2 The time period of oscillator, T = 2p

12. (b) Given, mass of earth = me , mass of planet = 2me , radius of earth = R p , radius of planet = 2 R p GM (where, G is constant) R2 ge R2p 2 M = e ´ 2 = \ g p M p Re 1

gp ge

2 1 = ÞTp = 2 2s Þ Tp 2 1 g 1 13. (b) Frequency, n = or n µ 2p l l n1 l2 L2 1 \ = = = (Q L1 = 2L2 ) 2L2 2 n2 l1 n2 = 2n1 n2 > n1 1 Energy, E = mw2 A2 2 E = 2p2 mn2 A2 1 and A2 µ mn2 A12 M2 n22 n22 = = \ A22 M1 n12 n12 Þ

Þ

10. (d) In a SHM, maximum KE = mw2 A2

Gravity, g =

Also,

(where, A is amplitude) (\w = 2pn) (Q E is same) [Q M1 = M2 ]

A1 > A2

[Q n2 > n1 ]

14. (b) As the amplitude is increased, the maximum acceleration of the platform (along with coin as long as they do not get separated) increases. If we draw the FBD for coin at one of the extreme positions as shown Then, from Newton’s law mg - N = mw2 A

w2A mg N

For loosing contact with the platform, N = 0 g So, A= 2 w

15. (b) Given, y = sin2 (wt ) It can be shown as, y O

p w

2p 3p 4p w w w

t

dy = 2wsin wt cos wt = wsin2wt dt 2 d y = 2w2 cos 2wt dt 2 d2 y For SHM, µ-y dt 2 Hence, the given function does not satisfy the above relation.

111

Oscillations \This function is not SHM, but periodic and from the y - t graph, time period is p t = w 3

16. (a) Potential energy, U = k ( x) and dU dx F = - 3k( x )2 F =-

force,

Also, for SHM x = A sin wt and

d2 x dt 2

…(i) + w2 x = 0

d2 x

= - w2 x dt 2 d2 x …(ii) Also, F = ma = m 2 = - mw2 x dt From Eqs. (i) and (ii), we get 3kx w= m 2p m m = 2p T = = 2p Þ w 3kx 3k( A sin wt ) 1 T µ Þ A Þ Acceleration, a =

17. (c) Let displacement equation of particle executing SHM is y = a sin wt As particle travels half of the amplitude from a the equilibrium position, so y = 2 1 p a Therefore, = a sin wt or sin wt = = sin 2 6 2 p p or or t = wt = 6w 6 T p 2p ö æ or t = = ç as, w = ÷ æ 2p ö 12 T ø è 6ç ÷ èT ø

18. (a) The displacement equation of particle executing SHM is x = acos(wt + f)

…(i)

Velocity

+wa 0 –wa

t (i)

Acceleration

+w2a 0 – w2 a

(ii)

t

dx = - awsin(wt + f) dt

Velocity, v =

…(ii)

Acceleration, dv …(iii) = - aw2 cos(wt + f) dt From the curve, it is noted that the acceleration 1 curve of A is shifted (to the left) by T relative 4 to the velocity curve of v. This implies that the acceleration is 90° (0.5p ) out of phase with the velocity. A=

19. (a) One of the conditions for SHM is that restoring force (F ) and hence acceleration (a) should be proportional to displacement ( y ). i.e., aµ y Let y = sin wt - cos wt dy = wcos wt + wsin wt dt 2 d y = - w2 sin wt + w2 cos wt dt 2 or a = - w2 (sin wt - cos wt ) a = - w2 y Þ a µ - y This satisfies the condition of SHM. Other equations do not satisfy this condition.

20. (b) Let when the oscillating mass is at a distance x towards right from its equilibrium position, the instantaneous extensions in the springs of force constants k, is x = x1 + x2 k

k m

Since, the springs are in series the restoring force exerted by each spring on mass m is same. Then, F = - k x1 = - k x2 æ 1 1 ö -2F and x = x1 + x2 = - F ç + ÷ = k èk kø k F =- x \ 2 k Þ Effective force constant is . 2 Hence, frequency of oscillation is 1 k n= 2p 2M

21. (c) Let F1 and F2 be the restoring forces produced when the mass m is displaced by a distance x towards right.

AIIMS Chapterwise Solutions ~ Physics

112

For a body executing simple harmonic motion, maximum acceleration is given by amax = - Aw2

Then, F1 = - kx and F2 = - 2kx

where, A is amplitude and w is angular velocity. In order that body is not detached, this acceleration should be equal to acceleration due to gravity. Therefore, - w2 A = g

Total restoring force is F = F1 + F2 = - kx - 2kx = - (3k )x Effective force constant is 3k. 1 3k Hence, frequency, n = 2p m

22. (d) The time period of a simple pendulum of a length l is given by T = 2p

l g

…(i)

When length of pendulum increases by 21%. We have, total increased length of pendulum. l ¢ = 121 . l 121 . l ¢ …(ii) \ T = 2p g Dividing Eq. (i) by Eq. (ii) , we get 1 T = 121 . T¢ 121 T¢= T = 11 . T 100 Percent increase in time period 11 . T -T .T 01 = ´ 100 = ´ 100 = 10 % T T Alternative Time period of simple pendulum l Þ T µ l T = 2p g DT 1 Dl = × \ 2 l T Dl Since, = 21% l Increase in the time period of pendulum is DT 1 = ´ 21 % » 10 % 2 T

23. (b) As springs are connected in series, effective force constant k1 k2 1 1 1 , k= = + k k1 k2 k1 + k2 Hence, frequency of oscillation is k1 k2 1 k 1 = Þ n= 2p m 2p (k1 + k2 )m

24. (c)

mg

k1

k2

g A [on neglecting (- ) sign] A Also, time-period, T = 2p g

Þ

w=

= 2p

3.92 ´ 10-3 s = 01256 . 9.8

25. (a) In case of simple harmonic motion, restoring force is always directed towards a fixed point, total energy of the particle remains constant and restoring force is maximum at extreme positions. Also, acceleration of particle in SHM is proportional to displacement. a = - w2 y and is zero (minimum) at mean position.

26. (c) The maximum acceleration (a ) of a particle in simple harmonic motion with amplitude (a) and angular velocity w is a = w2 a Given,

a = 7.5 m / s2

w = 3.5 rad / s a 7.5 = 0.61 cm Þ Amplitude, a = 2 = (3.5)2 w

27. (c) The time period of a spring of mass m and spring constant k is given by m ÞT µ m T = 2p k T1 = 2s, we have 2 m 1 = = Þ T2 = 4 s T2 4m 2

28. (b) Maximum velocity of simple pendulum is the product of its amplitude and angular velocity. Amplitude of oscillation, a = 50 mm = 0.05 m Time period, T = 2s The angular velocity of simple pendulum is 2p 2p w= = = p rad /s T 2

113

Oscillations Therefore, maximum velocity, v max = aw = 0.05 ´ p = 0157 . m/s » 016 . m/s

We know that frequency (n) =

effective spring constant is 1 1 1 2 = + = k¢ k k k k k¢ = Þ 2 Hence, frequency, 1 k¢ 1 k …(i) = n¢ = 2p m 2p 2m When springs are connected in parallel, the effective force constant is k ¢¢ = k + k = 2k Therefore, frequency is 1 2k n ¢¢ = 2p m

k

…(ii)

k m

Dividing Eq. (i) by Eq. (ii), we get n¢ k / 2m 1 = = 2k / m 2 n ¢¢

30. (c) Energy stored by spring of force constant k, on giving displacement x is 1 …(i) E = kx2 2 Tension in spring is T = kx T …(ii) x= Þ k Putting the value of x from Eq. (ii) in Eq. (i), we get 1 æ T ö2 T 2 energy, E = k ç ÷ = 2 èkø 2k

31. (b) Given, frequency of damped oscillator = w, frequency of harmonic driving force = n and n < w. We know that, if n < w, then vibrations are nearly in phase with the impressed force or the response of the oscillator is controlled by spring constant.

32. (a) Given, initial tension in the piano wire (T1 ) = 10N, initial frequency of note (n1 ) = n Final frequency of the note (n2 ) = 2n

nµ T n1 T1 = n2 T2

\

29. (a) When springs are connected in series the

1 T 2p m

\

2

10 æ nö ç ÷ = T2 è 2n ø

or \

Final tension, T2 = 40 N.

33. (d) Given, time period (T ) = 4 s and displacement y 1 a or = . a 2 2 We know that displacement equation 2p y = a sin t T y 2p 1 2p or = sin t or = sin t a T 2 T p 2p -1 1 or t = sin = T 2 6 (where, t is the interval of time) T 4 1 t = = = s 12 12 3

(y) =

34. (a) Let us consider the equations of SHM, x = asin wt and y = b sin(wt + p ) = - b sin wt where, a and b are amplitudes. y - b sin wt b or = =x a sin wt a b æ ö or y = ç- ÷x è aø This is an equation of a straight line.

35. (d) Spring constant (k) is given by k= Þ \

f 1 Þ kµ l l k1 l2 2 = = k2 l1 1 k1 = 2k, k2 = k

Spring connected in series are represented by 1 1 1 1 1 3 = + = + = k ¢ k1 k2 2k k 2k 2k k¢= \ 3 Both Assertion and Reason are false.

36. (a) If the soldiers while crossing a suspended bridge march in steps, the frequency of marching steps of soldiers may match with the natural frequency of oscillations of the

AIIMS Chapterwise Solutions ~ Physics

114 suspended bridge. In that situation, resonance will take place, the amplitude of oscillation of the suspended bridge will increase enormously, which may cause the collapsing of the bridge. To avoid such situations, the soldiers are advised to break steps on suspended bridge. Both Assertion and Reason are true and Reason is correct explanation of Assertion.

37. (c) Time period of simple pendulum of length l is T = 2p

l g

T µ l \Time period is directly proportional to square root of length of pendulum. DT 1 Dl Percentage change in time period, = × T 2 l DT 1 = ´ 3 = 1.5 % 2 T k Assertion is true but Reason is false .

38. (a) The time period of a oscillating

41. (b) One of the conditions for SHM to take place are that the motion of the particle should be in a straight line to and fro about a fixed point. Also, velocity of particle is, v = w a2 - x2 where, w is angular velocity, a is amplitude and x is displacement. Both Assertion and Reason are true but Reason is not correct explanation of Assertion.

42. (a) The time period of a pendulum is T = 2p

l 1 Þ T µ g g

Since, g = 0 in satellite l \ T = 2p = ¥ (infinite) 0 Both Assertion and Reason are correct and Reason is correct explanation of Assertion.

43. (a) The distance moved by the wave in time of

spring is given by m Þ T = 2p k

Both Assertion and Reason are true but Reason is not correct explanation of Assertion.

k

m 1 k Since, the spring constant is large for hard spring, therefore hard spring has a less periodic time as compared to soft spring. Both Assertion and Reason are correct and Reason is correct explanation of Assertion.

T µ

39. (d) The period of the liquid executing SHM in a U-tube does not depend upon the density of the liquid. Therefore, time period will be the same, when mercury is filled up to the same height as the water in the U-tube. Now, as the pendulum oscillates, it drags air along with it. Therefore, its kinetic energy is dissipated in overcoming viscous drag due to air and hence, its amplitude goes on decreasing. Both Assertion and Reason are false.

40. (b) Oscillation’s damp down after sometime because of frictional forces acting on it. This is known as damping. Damping is the term used to describe the loss of energy with each cycle of a oscillation.

one complete oscillation of a particle of the medium, is called wavelength. Alternatively, the distance between two nearest medium particles oscillating in the same phase is called the wavelength. Also, wave speed is defined as the ratio of distance travelled by wave (wavelength) to time period. Both Assertion and Reason are correct and Reason is correct explanation of Assertion.

44. (b) In SHM, if a is amplitude, y is displacement, w is angular velocity, then velocity is given by v = w a2 - y 2 Also, angular acceleration, a = - w2 y When displacement is zero ( y = 0) that is when the particle passes through its equilibrium position, then the velocity is maximum and acceleration is minimum (zero). Also, in simple harmonic motion, displacement and velocity p differ in phase by . 2 Both Assertion and Reason are correct but Reason is not correct explanation of Assertion.

CHAPTER

13 Waves 1. Frequency of the wave is 50 Hz. Length is 1 m and mass of string is 10 gm. What is the tension of the string? [AIIMS 2018]

(a) 5N

(b) 10N

(c) 100N

(d) 50N

2. The driver of a car travelling with speed

30 ms - 1 towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 ms - 1, the frequency of reflected sound as heard by driver is [AIIMS 2017] (b) 555.5 Hz (d) 500 Hz

(a) 550 Hz (c) 720 Hz

3. If the intensity is increased by a factor of 20, then how many decibels is the sound level increased? [AIIMS 2015] (a) 18

(b) 13

(c) 9

(d) 7

4. On the same path, the source and observer are moving in such a way that the distance between these two increases with the time. The speeds of source and observer are same and equal to 10 ms - 1 with respect to the ground while no wind is blowing. The apparent frequency received by observer is 1950 Hz, then the original frequency must be (Take, the speed of sound in present medium is 340 m/s) [AIIMS 2015] (a) 2068 Hz (c) 1903 Hz

(b) 2100 Hz (d) 602 Hz

5. A source and an observer are moving towards each other with a speed equal to v , where v is the speed of sound. The 2 source is emitting sound of frequency n. The frequency heard by the observer will be [AIIMS 2013] (a) zero n (c) 3

(b) n (d) 3n

6. Velocity of sound waves in air is 330 m/s. For a particular sound in air, a path difference of 40 cm is equivalent to phase difference of 1.6p. The frequency of the wave is [AIIMS 2013] (a) 165 Hz (c) 660 Hz

(b) 150 Hz (d) 330 Hz

7. What will be the wave velocity, if the radar gives 54 waves/min and wavelength of the given wave is 10 m? [AIIMS 2012] (a) 4 m/s (c) 9 m/s

(b) 6 m/s (d) 5 m/s

8. A longitudinal wave is represented by xö æ x = xo sin 2p ç nt - ÷ lø è The maximum particle velocity will be four times the wave velocity, if [AIIMS 2011]

p xo 4 px (c) l = o 2 (a) l =

(b) l = 2 p xo (d) l = 4p xo

116

AIIMS Chapterwise Solutions ~ Physics

9. A fork A has frequency 2% more than the standard fork and B has a frequency 3% less than the frequency of same standard fork. The forks A and B when sounded together produced 6 beats/s. The frequency of fork A is [AIIMS 2010] (a) 116.4 Hz (c) 122.4 Hz

(b) 120 Hz (d) 238.8 Hz

10. A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (The velocity of the sound in the medium is 350 m/s.) [AIIMS 2010] (a) 750 Hz (b) 857 Hz (c) 1143 Hz (d) 1333 Hz

11. The equation y = 4 + 2 sin(6t - 3 x) represents a wave motion. Then, wave speed and amplitude, respectively are [AIIMS 2009] (a) wave speed 1 unit, amplitude 6 unit (b) wave speed 2 unit, amplitude 2 unit (c) wave speed 4 unit, amplitude 1/2 unit (d) wave speed 1/2 unit, amplitude 5 unit

12. A closed organ pipe of length 1.2 m vibrates in its first overtone mode. The pressure variation is maximum at [AIIMS 2009] (a) 0.4 m from the open end (b) 0.4 m from the closed end (c) Both (a) and (b) (d) 0.8 m from the open end

13. A resonance in an air column of length 40 cm resonates with a tuning fork of frequency 450 Hz. Ignoring end correction, the velocity of sound in air is [AIIMS 2009] (a) 1020 ms - 1 (c) 620 ms - 1

(b) 720 ms - 1 (d) 820 ms - 1

14. A vehicle with a horn of frequency n, is

moving with a velocity of 30 ms - 1 in a direction perpendicular to the straight line joining the observer and the vehicle. The observer receives the sound to have a frequency n + n1. Then (if the sound velocity in air is 300 ms - 1), [AIIMS 2009]

(a) n1 = 10 n (c) n1 = 0 .1 n

(b) n1 = 0 (d) n1 = - 0 .1 n

15. A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is [AIIMS 2008] (a) 105 Hz (b) 1.05 Hz (c) 1050 Hz (d) 10.5 Hz

16. A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of [AIIMS 2007] (a) 1000

(b) 10000

(c) 10

(d) 100

17. A boat at anchor is rocked by waves whose crests are 100 m apart and velocity is 25 m/s. The boat bounces up once in every (a) 2500 s

(b) 75 s

(c) 4 s

[AIIMS 2006] (d) 0.25 s

18. Two tuning forks P and Q when set vibrating, give 4 beats/s. If a prong of the fork P is filed, the beats are reduced to 2/s. What is frequency of P, if that of Q is 250 Hz? [AIIMS 2006] (a) 246 Hz (b) 250 Hz (c) 254 Hz (d) 252 Hz

19. An organ pipe closed at one end has fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe which a normal person can hear is? [AIIMS 2004] (a) 4

(b) 13

(c) 6

(d) 9

20. The waves produced by a motor boat sailing in water are

[AIIMS 2004]

(a) transverse (b) longitudinal (c) longitudinal and transverse (d) stationary

21. An earthquake generates both transverse (S ) and longitudinal ( P ) sound waves in the earth. The speed of S waves is about 4.5 km/s and that of P waves is about 8.0 km/s. A seismograph records P and S waves from an earthquake. The first P wave arrives 4.0 min before the first S wave. The epicentre of the earthquake is located at a distance of about [AIIMS 2003] (a) 25 km

(b) 250 km (c) 2500 km (d) 5000 km

117

Waves 22. If equation of a sound wave is y = 0.0015 sin(62.8 x + 314 t ), then its wavelength will be [AIIMS 2003] (a) 2 unit (c) 0.1 unit

(b) 0.3 unit (d) 0.2 unit

504 Hz, the frequency of horn of car A will be [AIIMS 2000] 15 m/s

23. A siren emitting sound of frequency 800 Hz is going away from a static listener with a speed of 30 m/s. Frequency of sound to be heard by the listener is (Take, velocity of sound = 330 m/s) [AIIMS 2002] (a) 286.5 Hz (c) 733.3 Hz

24. A string in a musical instrument is 50 cm long and its fundamental frequency is 800 Hz. If the frequency of 1000 Hz is to be produced, then required length of string is [AIIMS 2002] (b) 40 cm (d) 62.5 cm

25. The tension in a piano wire is 10 N. The tension in a piano wire to produce a node of double frequency is [AIIMS 2001] (a) 20 N

(b) 40 N

(c) 10 N

(d) 120 N

26. A sings with a frequency n and B sings with a frequency 1/8 that of A. If the energy remains the same and the amplitude of A is ‘A’, then amplitude of B will be [AIIMS 2001] (a) 2 A (c) 4 A

(b) 8 A (d) A

B

(b) 295.2 Hz (d) None of these

30. If the vibrations of a string are to be

(a) twice (c) eight times

(b) 63 m/s (d) 161 m/s

(b) four times (d) half

31. Standing waves are produced in 10 m long stretched string. If the string vibrates in 5 segments and wave velocity is 20 m/s, then its frequency will be [AIIMS 1998] (a) 5 Hz (c) 10 Hz

(b) 2 Hz (d) 4 Hz

32. The equation of a travelling wave is y = 60 cos(1800t - 6 x) where, y is in microns, t in second and x in metre. The ratio of maximum particle velocity to the velocity of wave propagation is [AIIMS 1998] (a) 3.6 ´ 10- 4 (c) 3.6 ´10

with the equation y = 10 sin p (0.02 x - 200 . t) where, x is in metre and t in second. The maximum velocity of the particle in wave motion is [AIIMS 2000]

(b) 3.6 ´10- 6 (d) None of the above

33. The equation of displacement of two waves are given as pö æ y 1 = 10 sin ç 3pt + ÷ and 3ø è y2 = 5(sin 3pt + 3 cos 3pt )

28. A wave is represented by the equation y = a sin(0.01 x - 2t ) where, a and x are in cm and t in second. Velocity of propagation of the wave is [AIIMS 2000] (a) 200 cm/s (c) 25 cm/s

(a) 529.2 Hz (c) 440.5 Hz

-8

27. A transverse wave passes through a string

(a) 100 m/s (c) 120 m/s

A

increase by a factor of two, then tension in the string should be made [AIIMS 1999]

(b) 481.2 Hz (d) 644.8 Hz

(a) 37.5 cm (c) 50 cm

30 m/s

(b) 10 cm/s (d) 50 cm/s

29. Two cars A and B approach a stationary observer from opposite sides as shown in figure. Observer hears no beats. If the frequency of the horn of the car B is

Then, the ratio of their amplitudes will be (a) 2 :1 (c) 1 : 2

[AIIMS 1997] (b) 1 : 1 (d) None of these

34. The air column in a pipe which is closed at one end will be in resonance with a vibrating tuning fork at a frequency 260 Hz. The length of the air column is (a) 12.5 cm (c) 31.92 cm

[AIIMS 1997] (b) 35.75 cm (d) 62.5 cm

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AIIMS Chapterwise Solutions ~ Physics

35. The equation of a wave is given by ö æ 2pt y = 10 sinç + a ÷. If the displacement is ø è 30 5 cm at t = 0, then the total phase at t = 7. 5 s will be (a)

2p rad 3

(b)

p rad 3

(c)

p rad 2

[AIIMS 1996] 2p (d) rad 5

36. An observer standing by the side of a road hears the siren of an ambulance, which is moving away from him. If the actual frequency of the siren is 2,000 Hz, then the frequency heard by the observer will be [AIIMS 1996] (a) 2000 Hz (c) 4000 Hz

(b) 1990 Hz (d) 2100 Hz

37. A closed organ pipe and an open organ pipe of the same length produce four beats, when sounded together. If the length of the closed organ pipe is increased, then the number of beats will [AIIMS 1996] (a) remains the same (c) decrease

(b) increase (d) first (d), then (a)

38. A tube closed at one end containing air produces fundamental node of frequency 512 Hz. If the tube is open at both the ends, the fundamental frequency will be [AIIMS 1995] (a) 1024 Hz (c) 1280 Hz

a free end changes its direction of motion, but phase remains constant after reflection. Reason When string wave reaches the free end, there is no medium present in front of it. [AIIMS 2018]

41. Assertion Our ears can distinguish two notes, one produced by a violin and other by a guitar, if they have exactly same intensity and same frequency. Reason When a musical instrument is played, it produces a fundamental note which is accompanied by a number of overtones called harmonics. [AIIMS 2014]

42. Assertion In stationary wave, there is no transfer of energy. Reason The ratio of kinetic energy to potential energy is independent of the position. [AIIMS 2013]

43. Assertion It is not possible to have interference between the waves produced by two violins. Reason For interference of two waves, the phase difference between the waves must remain constant. [AIIMS 2012]

44. Assertion In a stationary wave, there is no

(b) 256 Hz (d) 768 Hz

transfer of energy.

39. The frequency of a tuning fork is 256 Hz. It will not resonate with a fork of frequency [AIIMS 1994] (a) 738 (c) 768

40. Assertion A string wave travelling towards

(b) 256 (d) 512

Assertion & Reason Direction (Q. Nos. 40-49) Read the Assertion and Reason carefully to mark the correct option from these given below. (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

Reason There is no outward motion of the disturbance from one particle to adjoining particle in a stationary wave. [AIIMS 2011]

45. Assertion The fundamental frequency of an open organ pipe increases as the temperature is increased. Reason As the temperature increases, the velocity of sound increases more rapidly than the length of the pipe. [AIIMS 2010]

46. Assertion To hear distinct beats, difference in frequencies of two sources should be less than 10. Reason More the number of beats per second, more difficult to hear them. [AIIMS 2009]

119

Waves 47. Assertion Ocean waves hitting a beach

Reason The same tuning fork will not be in resonance with open pipe of same length due to end correction of pipe. [AIIMS 2000]

are always found to be nearly normal to the shore. Reason Ocean waves hitting a beach are assumed as plane waves. [AIIMS 2007]

49. Assertion When two vibrating tuning forks having frequencies 256 Hz and 512 Hz are held near each other, beats cannot be heared.

48. Assertion A tuning fork is in resonance with a closed pipe. But the same tuning fork cannot be in resonance with an open pipe of the same length.

Reason The principle of superposition is valid only if the frequencies of the oscillators are nearly equal. [AIIMS 1994]

Answers 1. 11. 21. 31. 41.

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

2. 12. 22. 32. 42.

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

3. 13. 23. 33. 43.

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

4. 14. 24. 34. 44.

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

5. 15. 25. 35. 45.

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

6. 16. 26. 36. 46.

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

7. 17. 27. 37. 47.

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

8. 18. 28. 38. 48.

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

9. 19. 29. 39. 49.

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

10. 20. 30. 40.

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

Explanations 1. (c) The frequency of string is n =

1 T 2l ml

where, T is the tension, ml is the mass per unit length and l is the length of string. 10 ´ 10–3 So, ml = = 10–2 kg -m–1 1 1 T 2500 ´ 4 Þ T = = 100 N 50 = Þ 2 10–2 100

2. (c) When the sound is reflected from the hill, it approaches the driver of the car. Therefore, the driver acts as an observer and the distance between source and observer is decreasing (Q driver is moving towards the hill). In this case, the apparent frequency heard by æ v + vo ö ÷ the driver (observer), f ¢= f çç ÷ è v - vo ø Given, f = 600 Hz, Velocity of sound in air, v = 330 ms - 1 , Velocity of car, vo = 30 ms - 1 æ 330 + 30 ö ÷ = 720 Hz f ¢= 600çç ÷ è 330 - 30 ø

3. (b) If the intensity of sound in watts per square metre is I, then intensity level in decibel is æ I ö given by b = 10 log çç ÷÷ (where, the base of log è Io ø is 10). æ I ö 20 I Thus, b1 = 10 log çç ÷÷ and b2 = 10 log I Io è o ø Þ Change in sound level, Db = b2 - b1 éI ù é 20 I ù = 10 log ê ú - 10 log ê ú ë Io û ë Io û Db = 10 log (20) = 13 dB

4. (a) Let the original frequency be f , then apparent frequency æ v - u0 ö ÷´f f ¢= çç ÷ è v + us ø where, u0 = speed of observer, us = speed of source and v = speed of sound wave. 340 - 10 f Þ 1950 = 340 + 10 35 Þ f = ´ 1950 = 2068 Hz 33

120

AIIMS Chapterwise Solutions ~ Physics

5. (d) When the source and observer are

9. (c) Let the frequency of standard fork be n.

approaching towards each other, then apparent frequency heard by the observer is æ v + vo ö ÷ n (where, vs is speed of sound and n¢= çç ÷ è v - vs ø vo is the speed of source or observer) vö æ çv + ÷ æ vö 2ø = è n çç Q Given, vs = vo = ÷ vö æ 2ø è çv - ÷ 2ø è æ3 2ö n¢= ç ´ ÷ n = 3n è2 1ø

6. (c) We know that, phase difference =

2p ´ path l

difference (where, l is wavelength) Here, phase difference = 1.6p and path difference = 40 cm 2p 1.6p = ´ 40 l 2 l= ´ 40 = 50 cm = 0.5m 1.6 Also, v = nl (where, v = velocity of sound and n= frequency) v 330 n= = = 660 Hz l 0.5 54 Hz, 7. (c) Given, frequency of wave, n = 60 wavelength of wave = 10 m We know that, 54 wave velocity, v = nl = ´ 10 60 v = 9 m/s xö æ 8. (c) Given, x = xo sin 2p ç nt - ÷ lø è Comparing the given equation with y = asin(wt – kx ), we get 2p a = xo , w = 2pn and k = l Hence, maximum velocity of particle (v max )particle = aw = xo ´ 2pn w 2pn and wave velocity, v wave = = = nl k 2p / l We are given that (v max )particle = 4v wave Þ 2pnxo = 4nl px l= o Þ 2

2 n 100 3 and the frequency of B, nB = n n 100 According to question, n A - nB = 6 2 3 æ ö æ ö \ çn + n÷ - ç n n÷ = 6 100 ø è 100 ø è Þ n = 600/ 5 = 120 Hz The frequency of A, 2 n A = 120 + ´ 120 = 122 . 4 Hz 100 The frequency of A , n A = n +

10. (a) When the source is coming to the stationary observer apparent frequency, æ v ö æ 350 ö ÷ n or 1000 = ç ÷ n¢= çç ÷ ç 350 - 50 ÷ n ø è v - vs ø è or n = (1000 ´ 300 / 350) Hz When the source is moving away from the stationary observer the apparent frequency, æ v ö ÷n n¢¢ = çç ÷ è v + vs ø æ 350 ö æ 1000 ´ 300 ö ÷ç = çç ÷ =750 Hz ÷ 350 ø è 350 + 50 ø è

11. (b) Given, y = 4 + 2sin(6t – 3 x) from the equation motion of wave, we see that amplitude of wave is 2 unit. coefficient of t 6 = = 2 units \Wave speed, v = coefficient of x 3

12. (a) Let l be the length of air column. A

A l/4 N l/4

l = 1.2 m

A N

In first overtone mode, 3l (where, l is wavelength) l = 4 l l 1.2 = 0.4 m = = \ 3 4 3 Pressure variation will be maximum at displacement nodes, i.e., at 0.4 m from the open end.

121

Waves

17. (c) The distance between the two consecutive

13. (b) Length of the air column, -2

l = 40 cm = 40 ´10

m

Frequency of tuning fork, n = 450 Hz In resonance, the frequency of tuning fork will be equal to frequency of air column. v (where, v is the velocity of light) n= 4l or v = n ´4l = 450 ´ 4 ´ 0.4 ms - 1 = 720 ms - 1

14. (b) When velocity of source is perpendicular to the line joining observer and source, then there is no Doppler’s effect. Hence, the frequency of sound heard to the observer is same. So, n1 = 0.

15. (a) It is given that 315 Hz and 420 Hz are the two consecutive resonant frequencies, let these are nth and (n + 1) th harmonics. nv ...(i) 315 = 2L (n + 1)v ...(ii) 420 = 2L Dividing Eq. (i) by Eq. (ii), we get 315 n = Þn=3 420 n + 1 v Lowest resonant frequency, fo = …(iii) 2L From Eqs. (i) and (iii), we get 315 315ü ì v fo = = ý íQ n n þ î 2L 315 Þ fo = = 105Hz 3

16. (d) Let the intensity of sound be I and I ¢. Loudness of sound initially, æ I ö b1 = 10 log çç ÷÷ è I0 ø Later, Given,

\

æ I ö 20 = 10 log çç ÷÷ è I0 ø æ I¢ö 2 = log ç ÷ èIø I¢ 2 [Q log = 10 I

18. (a) There are four beats between P and Q, therefore, the possible frequencies of P are 246 or 254, (i.e., 250 ± 4) Hz. When the prong of P is filed, its frequency becomes greater than the original frequency. The beats between P and Q will be more than 4. But it is given that the beats are reduced to 2, therefore, 254 is not possible. Therefore, the required frequency must be 246 Hz. (This is true, because on filing the frequency may increase to 248, giving 2 beats with Q of frequency 250 Hz).

19. (c) A closed pipe produces

Anti-node only odd harmonics. The frequency of note emitted from the pipe for v being velocity of sound in l/4 air, is f ¢= n( v / 4l ) and l is length of pipe. f ¢= n ´ fundamental Node frequency æ v ö ççQ fundamental frequency = ÷ 4 lø è

\ We know that human ear can hear frequencies upto 20,000 Hz, hence

æ I¢ ö b2 = 10 log çç ÷÷ è I0 ø b2 - b1 = 20

crests in transverse wave motion is called wavelength. Wavelength = Distance between two consecutive crests i.e., l = 100 m Velocity of wave = 25m/s Hence, time in one bounce of boat l T = v 100 = = 4s 25

æ I¢ ö - 10 log çç ÷÷ è I0 ø

a æ aö c ç ÷ = c Þ = 10 ] b èb ø

So, intensity decreases by a factor of 100.

20,000 = n ´1500 20,000 Þ n= » 13 1500 Maximum possible harmonics obtained are 1, 3, 5, 7, 9, 11, 13. Hence, man can hear upto 13th harmonic =7-1= 6 Note First overtone of the closed pipe is called the third harmonic.

122

AIIMS Chapterwise Solutions ~ Physics

20. (c) When the motor boat sails in water, the motor boat disturbs the surface of water forming bow waves on the surface of water. Actually, these are transverse waves which are produced on the surface of water. But inside the water, the longitudinal waves produced due to vibration of the rudder. So, waves produced by a motor boat sailing in water are both longitudinal and transverse.

21. (c) Distance travelled by both the waves is same. Let the time taken by the S and P waves to reach the seismograph be t1 and t2 , the t1 - t2 = 4 min \ t1 - t2 = 60 ´ 4 = 240 s Let distance of epicentre be x, then x = v1t1 = v2t2 Þ 4.5 ´ t1 = 8 t2 4.5 t2 = t Þ 8 1 4.5 ö æ t ÷ = 240 ç t1 8 1ø è 240 ´ 8 t1 = = 548.58 s Þ 3.5 \ x = v1t1 = 4.5 ´ 548.5 = 2468.25 » 2500 km

22. (c) The standard equation of sound waves is t ö æx y = A sin 2p ç + ÷ èl T ø where, A is amplitude, l is wavelength and T is time period. Given, equation of wave is ...(i) y = 0.0015 sin(62.8 x + 314t ) Comparing given Eq. (i) with standard wave equation, we get 2p = 62.8 l 2 ´3.14 2p l= = = 0.1 unit Þ 62.8 62.8

23. (c) The perceived frequency (n¢) when listener is static and source is moving away is given by æ v ö ÷ n¢ = nçç ÷ è v + vs ø where, n is frequency of source, v is velocity of sound and vs is velocity of source. æ 330 ö ÷ = 733.3 Hz n¢= 800 ´ çç ÷ è 330 + 30 ø

Note In the limit, when speed of source and observer are much lesser than that of sound v, the change in frequency becomes independent of the fact whether the source is moved or the detector.

24. (b) The frequency produced in a string of length l, mass per unit length m and tension T is 1 T 1 Þnµ n= 2l m l Given, l1 = 50 cm, n1 = 800 Hz and n2 = 1000 Hz n1 l2 = \ n2 l1 nl 800 ´ 50 Þ l2 = 1 1 = = 40 cm n2 1000

25. (b) For mass m per unit length of wire and tension T , the frequency of node emitted by the wire is 1 T n= 2l m where, l is length of wire. Given, T1 = 10 N, n1 = n and n2 = 2 n n1 T1 n 10 = Þ = n2 T2 2n T2 Þ

T2 = 10 ´ 4 = 40 N

26. (b) The energy of a wave travelling with velocity v is given by 1 mv 2 2 Also, v = amplitude ´ angular velocity = Aw 1 1 E = m( Aw)2 = mA2 w2 \ 2 2 Also, w = 2pn where, n is frequency. 1 \ E = mA2 (2pn)2 Þ E µ A2 n2 2 It is given that energy remains the same. Hence, E A = EB E=

\

æ AA ç çA è B

2

2

ö æ nB ö ÷ ÷ =ç ÷ çn ÷ ø è Aø

Given,

n A = n, nB =

\

A A n/ 8 1 = = 8 AB n

n 8

AB = 8 A A = 8 A

123

Waves 27. (b) Standard equation of a wave travelling with amplitude a and angular frequency w is given by y = a sin(wt - kx ) Given, equation of wave is y = 10 sin p(0.02 x - 2t ) Comparing this equation with standard wave form, we get a = 10 m, w = 2p rad/s Maximum velocity, v max = aw \ v max = 10 ´ 2p = 20p = 63 m/s

28. (a) The standard equation to wave is …(i) y = a sin(kx - wt ) where, w is angular velocity and t is time. Given, equation is ...(ii) y = a sin(0.01 x - 2 t ) Comparing the given equation with standard waveform, we get kx = 0.01 x and wt =2t \ k = 0.01 and w = 2 w 2 Thus, velocity = = = 200 cm/s k 0.01 \Velocity of propagation of the wave is 200 cm/s.

29. (a) The apparent frequency of the two cars should be equal for no beats. From Doppler’s effect, the apparent change in frequency of a wave that is perceived by stationary observer is æ v ö ÷ n¢= no çç ÷ è v - vs ø where, v is speed of sound. Since observer is not hearing any beats, therefore, the apparent frequency of the two cars should be equal. \ n¢ = n ¢¢ æ v ö æ v ö ÷ ÷ = nç 504 çç ç v - 15 ÷ ÷ v 30 ø è ø è (330 - 15) ´ 504 \ n= (330 - 30) = 529.2 Hz

30. (b) The frequency of vibration of a string is given by n=

1 2l

T m

where, l is length, T is tension and m is the mass per unit length. Given, n1 = n, n2 = 2n n1 T n \ = = 1 n2 2n T2 Þ

1 T1 = 4 T2

Þ T2 = 4T1 Hence, when frequency is 2n, then tension must increase four times.

31. (a) If string vibrates in p loops, then pl 5l (Q p = 5) = 2 2 l ´2 l= \ 5 10 (Q l = 10 m) = ´2 = 4m 5 Speed 20 \Frequency = = = 5Hz Wavelength 4 l =

32. (a) The standard equation of a wave of amplitude a, angular velocity w is y = a cos(wt - kx ) Given, equation is y = 60 cos(1800t - 6 x ) Comparing Eq. (i) with Eq. (ii), we get a = 60m = 60 ´ 10- 6 m

...(i) ...(ii)

w = 1800 rad/s and k=6 Also, maximum velocity is umax = aw = 60 ´ 10- 6 ´ 1800 w 1800 Wave velocity, v = = = 300 m/s k 6 Hence, ratio of maximum particle velocity to velocity of wave propagation. umax 60 ´ 10- 6 ´ 1800 = 300 v = 3.6 ´10- 4

33. (b) The standard equation of wave is y = a sin(wt - f) where, a is amplitude, w is angular velocity, t is time and f is phase difference. Given, equations are pö æ ...(i) y1 = 10sinç 3pt + ÷ 3ø è and y2 = 5 (sin3pt +

3 cos 3pt )

...(ii)

124

AIIMS Chapterwise Solutions ~ Physics

Multiplying and dividing Eq. (ii) by 2, we have ö æ1 3 cos 3pt ÷÷ y2 = 5´ 2 çç sin 3pt + 2 ø è2 Using sin( A + B ) = sin A cos B + cos A sin B p 1 p 3 Also, cos = and sin = 3 2 3 2 p p æ ö \ y2 = 5 ´ 2 ç cos sin3pt + sin cos 3pt ÷ 3 3 è ø pö æ y2 = 10 sin ç 3pt + ÷ 3ø è On comparing with standard equation, we have a1 = 10, a2 = 10 \ a1 : a2 = 1 : 1

34. (c) In a closed organ pipe, there is an anti-node ( A ) at the open end and a node (N ) at the closed l end. Distance between anti-node and node is . 4 If l be the length of pipe, then l l = 4 v v and frequency, n = = l l/4 l 4l Given, n = 260 Hz, v = 332 m/s v 332 l = = \ 4n 4 ´ 260 = 0.3192 m = 31.92 cm

37. (b) The fundamental frequency of the open and closed organ pipe is given by v v nopen = , nclosed = 2L 4L The beat frequency is nopen - nclose = 4 Now, if we increase the length of the closed pipe, then the frequency nclose is going to decrease and the beat frequency nopen - nclose is going to increase. \The number of beat will increase.

38. (a) Given, frequency of fundamental node, when tube is closed (n1 ) = 512 Hz. We know that, frequency of fundamental node for closed organ pipe (n2 ) = v / 4l and for open pipe (n1 )= v /2l \ n2 = 2n1 Þ n2 = 2 ´ 512 = 1024 Hz

39. (a) Given, frequency of tuning fork ( f ) = 256Hz We know that, fork of the frequency 256 will resonate only with those forks, whose frequency is an integral multiple of 256 (i.e., 512 and 738). Since, the frequency 738 is not an integral multiple of 256, therefore it will not resonate with the fork having frequency 256.

40. (a) Consider the free end is attached to a light N

35. (a) Displacement ( y ) = 5cm, initial time (t1 ) = 0, final time (t2 ) = 7.5s æ 2pt ö Equation of wave, y = 10 sin ç + a÷ è 30 ø Displacement of wave at t = 0, y = a sin a = 10 sin a 5 = 10 sin a 5 1 sin a = = 10 2 p a= Þ 6 Therefore, total phase at t = 7.5s 2p ´ 7.5 p 2p 2pt rad +a= + = f= 30 30 6 3

36. (b) When the source is moving away from the observer, then the frequency of sound heard by him, must be less than the actual frequency. \Frequency heard by the observer will be 1990 Hz.

ring. When a crest produced at A, reaches B end, the ring rises above its equilibrium position. As the ring moves up, it streches the string and produces a reflected crest, which travel towards A. There is no phase difference as crest is reflected as a crest. A

B

A

B

Rod Light ring

This happens as, there is no medium to oppose the motion of pulse. Hence, option (a) is correct.

41. (a) When a musical instrument is played, it produces a fundamental node which is accompanied by a number of overtones called harmonics. The number of harmonics is not the same for all instruments. It’s the number of harmonics which distinguishes the note produced by a guitar and that produced by violin. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

125

Waves 42. (c) A stationary wave consists of two travelling waves of equal amplitude propagating in opposite directions. Therefore, total energy transfer is zero. But the form of energy keeps on changing from KE changes. So, the one form to another i.e., the PE ratio of KE to PE is dependent on the position. Assertion is true but Reason is false.

43. (a) The initial phase difference between the two waves coming from different violins changes, therefore the waves produced by two different violins does not interfere because two waves interfere only when the phase difference between them remain constant throughout. Both Assertion and Reason are true and Reason is correct explanation of Assertion.

44. (b) In stationary wave, total energy associated with it is twice the energy of each of incident and reflected wave. Large amount of energy are stored equally in standing waves and trapped within the waves. Hence, there is no transmission of energy through the waves. Both Assertion and Reason are true but Reason is not correct explanation of Assertion.

45. (a) The fundamental frequency of an open organ v . 2l As temperature increases, both v and l increase but v increases more rapidly than l. Hence, the fundamental frequency increases as the temperature increases. Both Assertion and Reason are true and Reason is correct explanation of Assertion.

pipe is n =

46. (b) According to the property of persistence of hearing, the impression of a sound heard persists in our mind for 1/10 s. Therefore, number of beats per second should be less than 10. Hence, difference in frequencies of two sources must be less than 10.

47. (a) Waves produced on the surface of water are transverse in nature. When such waves are produced in water they spread out. Till the ocean waves reach the beach-shore, they acquire such a large radius of curvature that they may be assumed as plane waves. Hence, ocean waves hit the beach normally to the shore.

48. (c) If a closed pipe of length L is in resonance æ v ö with a tuning fork of frequency n ç = ÷. è 4L ø An open pipe of same length L produces v vibration of frequency . Obviously, it 2L cannot be in resonance with the given tuning æ v ö fork of frequency n ç = ÷. è 4L ø Assertion is true but Reason is false.

49. (c) The principle of superposition does not state that the frequencies of the oscillation should be nearly equal. And for beats to be heard, the condition is that the difference in frequencies of the two oscillations should not be more than 10 times per sec for a normal human ear to recognise it. Hence, we cannot hear beats in the case of two tunning forks vibrating at frequencies 256 Hz and 512 Hz, respectively. Assertion is true but Reason is false.

AIIMS Chapterwise Solutions ~ Physics

CHAPTER

14 Electric Charges and Fields 1. A uniformly charged non-conducting disc 2 cm with surface charge 2 density 10nC /m + + + + + ++ ++ + + + having radius + + + + + ++ + R = 3cm. Then, the value of electric field intensity at a point on the perpendicular bisector at a distance of x = 2 cm is [AIIMS 2018] (a) 200 N/C (c) 251.2 N/C

(b) 280.8 N/C (d) 215.2 N/C

equal to the weight of the particle. Which statement is true? [AIIMS 2016] (a) In Fig (a), - q is in stable equilibrium. (b) In Fig (a), - q is in neutral equilibrium. (c) In Fig (b), - q is in stable equilibrium. (d) Neither in Fig (a) nor in Fig (b) - q is in stable equilibrium.

4. A metallic solid sphere is placed in a uniform electric field. Which path, the lines of force follow as shown in figure?

2. An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge upto a radius R. The atom as a whole is neutral.The electric field at a distance r from the nucleus is( r < R) [AIIMS 2017] Ze é 1 r ù (a) 4pe0 êë r 2 R 3 úû (c)

Ze é r 1ù 4pe0 ëê R 3 r 2 ûú

Ze é 1 r ù (b) 4pe0 êë r 3 R 2 úû (d)

Ze é r 1ù + 2ú 4pe0 ëê R 3 r û

3. In the following figures, a particle –q +Q with small charge –Q -q -q is free to move up or down, but Fig (a) Fig (b) not sideways near a larger fixed charge Q. The small charge is in equilibrium in the positions shown, the electrical upward force is

1

1

2

2

3

3 4

(a) 1

(b) 2

4

(c) 3

[AIIMS 2015] (d) 4

5. In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is however not constant, but increases uniformly along the positive z-direction at the rate of 10 5 NC -1 m -1. The force experienced by the system having a total dipole moment equal to 10 -7 Cm in the negative z-direction is [AIIMS 2015] (a) - 10-2 N (b) 10-2 N

(c) 10-4 N

(d) -10-4 N

6. Three charged particles are collinear and are in equilibrium, then

[AIIMS 2014] (a) all the charged particles have the same polarity (b) the equilibrium is unstable (c) all the charged particles cannot have the same polarity (d) Both (b) and (c) are correct

127

Electric Charges and Fields 7. In air, the value of the total electric flux emitted from unit positive charge is [AIIMS 2012] (c) (4pe0 )-1 (d) 4pe0

(b) (e0 )-1

(a) e0

positioned exactly at a point midway between two equal and opposite point charges. If the sphere is slightly displaced towards the positive charge and released, then [AIIMS 2011] (a) it will oscillate about its original position (b) it will move further towards the positive charge (c) its electric potential energy will decrease and kinetic energy will increase (d) its total energy remains constant but is non-zero

9. A charge Q is divided into two parts of q and Q - q. If the coulomb repulsion between them when they are separated is Q to be maximum, the ratio of should be q [AIIMS 2011] (b) 1/2

(c) 4

(d) 1/4

10. Charge q is uniformly distributed over a thin half ring of radius R.The electric field at the centre of the ring is [AIIMS 2009] (a)

q 2 p 2 e0 R 2

(b)

q 4p 2 e0 R 2

(c)

q 4pe0 R 2

(d)

q 2 pe0 R 2

11. A hollow cylinder has a charge q within it. If f is the electric flux associated with the curved surface B, the flux linked with the plane surface A will be [AIIMS 2008]

1 (a) 2 f (c) 3

ö æq ç - f÷ ÷ çe ø è 0

charged particle 1 through a rectangular region of uniform electric field as shown in the figure. What is the direction of electric field and the direction of deflection of particles 2,3 and 4? [AIIMS 2007] Top 2s

r3

1r

s4 Down

(a) down,top, down (c) top, top, down

(b) down, down, top (d) top, down, down

14. Two parallel large thin metal sheets have equal surface charge densities ( s = 26.4 ´ 10 -12 C / m 2 ) of opposite signs.The electric field between these sheets, is [AIIMS 2006] (a) 1.5 N/C

(b) 15 . ´ 10-10 N/C

(c) 3 N/C

(d) 3 ´ 10-10 N/C

15. The spatial distribution of the electric field due to charges ( A, B) is shown in figure. Which one of the following statements is correct? [AIIMS 2006] A

B C

(d) 2qa along + x direction

13. The figure shows the path of a positively

8. A small uncharged metallic sphere is

(a) 2

(c) qa along the line joining points ( x = 0, y = 0, z = 0) and ( x = a, y = a, z = 0)

B

A

q (b) 2 e0 q (d) -f e0

12. Three point charges +q, -2q and +q are placed at points ( x = 0, y = a, z = 0), ( x = 0, y = 0, z = 0) and ( x = a, y = 0, z = 0), respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [AIIMS 2008] (a) 2qa along + y direction (b) 2qa along the line joining points ( x = 0, y = 0, z = 0) and ( x = a, y = a, z = 0)

(a) A is positive and B is negative and| A| >|B| (b) A is negative and B is positive and| A| = |B| (c) Both are positive but A > B (d) Both are negative but A > B

16. Two infinitely long parallel conducting plates having surface charge densities + s and -s respectively, are separated by a small distance.The medium between the plates is vacuum. If e 0 is the dielectric permittivity of vacuum, then the electric field in the region between the plates is (a) zero (c) s / e0 V/m

[AIIMS 2005] (b) s / 2 e0 V/m (d) 2 s / e0 V/m

AIIMS Chapterwise Solutions ~ Physics

128 17. In the basic CsCl crystal structure, Cs+ and

Cl - ions are arranged in a bcc configuration as shown in the figure.The net electrostatic force exerted by the eight Cs+ ions on the [AIIMS 2004] Cl - ion is

20. Three charges are placed at the vertices of an equilateral triangle of side a as shown in the following figure. The force experienced by the charge placed at the vertex A in a direction normal to BC is [AIIMS 2003]

Cs+

Cs+

Cs+ Cl

–

a

Cs+

Cs+ a

Cs+

–Q

Cs+

a

(a)

1 4e 2 × 4pe0 3a2

(b)

(c)

1 32e 2 × 4pe0 3 a2

(d) zero

B

O

r

(a)

21. An electric dipole placed in a non-uniform electric field experiences

[AIIMS 2003]

(a) Both, a torque and a net force (b) only a force but no torque (c) only a torque but no net force (d) no torque and no net force

22. A conducting sphere of radius 10 cm is

E R

C

(c) zero (d) Q 2 / (2 pe0 a2 )

charged non-conducting sphere of radius R as a function of the distance from its centre if represented graphically by [AIIMS 2004]

E

+Q

a

(a) Q 2 (4pe0 a2 ) (b) - Q 2 (4pe0 a2 )

1 16e 2 × 4pe0 3 a2

18. The electric field due to a uniformly

O

+Q

A

R

r

(b)

charged with 10mC. Another uncharged sphere of radius 20 cm is allowed to touch it for sometime. After that if the spheres are separated, then surface density of charges on the spheres will be in the ratio of [AIIMS 2002]

E O

(a) 1 : 1 (c) 1 : 3

E R

O

r

(c)

R

r

(d)

19. The figure shown below is a distribution of charges. The flux of electric field due to these charges through the surfaces S is [AIIMS 2003] +q

–q

23. The point charges Q and -2Q are placed at some distance apart. If the electric field at the location of Q is E, the electric field at the location of - 2Q will be [AIIMS 2001] 3E 2 E (c) 2

(a) -

(b) - E (d) - 2E

24. Let E a be the electric field due to a dipole in its axial plane distance l and let Eq be the field in the equatorial plane distant l, then the relation between E a and[AIIMS be Eq will2000]

S +q

(a) 3 q /e0 (c) q /e0

(b) 2 : 1 (d) 1 : 4

(b) 2q /e0 (d) zero

(a) Ea = 4Eq (c) Ea = 2 Eq

(b) Eq = 2 Ea (d) Eq = 3Ea

129

Electric Charges and Fields 25. A ring shaped conductor with radius a carries a net positive charge q uniformly distributed on it as shown in figure. A point P is situated at a distance x from its centre. Which of the following graph shows the correct variation of electric field [AIIMS 1997] ( E ) with distance ( x)? E

E x

(a)

(b)

29. Assertion When charges are shared between any two bodies, no charge is really lost, some loss of energy does occurs. Reason some energy disappears in the form of heat, sparking, etc. [AIIMS 2014]

30. Assertion A metallic shield in the form of a x

E (c)

Reason Gauss’s law is independent of size of the Gaussian surface. [AIIMS 2015]

hollow shell may be build to block an electric field. Reason In a hollow spherical shield, the electric field inside it is zero at every point. [AIIMS 2012]

x

(d) None of these

Assertion & Reason Direction (Q. Nos. 26-36) Read the Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

26. Assertion The electric field due to a dipole on its axial line at a distance r is E. Then, electric field due to the same dipole on the equatorial line and at the same distance will be E/2. Reason Electric field due to dipole varies inversely as the square of the distance. [AIIMS 2017]

27. Assertion Acceleration of charged particle in non-uniform electric field does not depend on velocity of charged particle. Reason Charge in an invariant quantity, i.e. amount of charge on particle does not depend on frame of reference. [AIIMS 2017]

28. Assertion A charge q is placed on a height h/ 4 above the centre of a square of side b. The flux associated with the square is independent of side length.

31. Assertion If the bob of a simple pendulum is kept in a horizontal electric field, its period of oscillation will remain same. Reason If bob is charged and kept in horizontal electric field, then the time period will be decreased. [AIIMS 2011]

32. Assertion The specific charge for positive rays is a characteristic constant. Reason The specific charge depends on charge and mass of positive ions present in positive rays. [AIIMS 2010]

33. Assertion A charged particle free to move in an electric field always moves along an electric field line. Reason The electric field lines diverge from a negative charge and converge at a positive charge. [AIIMS 2009]

34. Assertion In a cavity within a conductor, the electric field is zero. Reason Charges in a conductor reside only at its surface. [AIIMS 2007]

35. Assertion The Coulomb force is the dominating force in the universe. Reason The Coulomb force is weaker than the gravitational force. [AIIMS 2003]

36. Assertion Electric lines of force never cross each other. Reason Electric field at a point superimpose to give one resultant electric field. [AIIMS 2002]

AIIMS Chapterwise Solutions ~ Physics

Answers 1. 11. 21. 31.

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

2. 12. 22. 32.

3. 13. 23. 33.

(a) (b) (b) (b)

4. 14. 24. 34.

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

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

5. 15. 25. 35.

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

6. 16. 26. 36.

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

7. (b) 17. (d) 27. (a)

8. (b) 18. (b) 28. (a)

9. (a) 19. (d) 29. (c)

10. (a) 20. (c) 30. (a)

Explanations 1. (c) Given, R = 3 cm = 3 ´ 10-2 m

3. (c) In first diagram, if we displace -q in

-2

x = 2 cm = 2 ´ 10 m æ x 1 . s × 2p çç 1 2 4pe0 R + x2 è é 2 ´ 10-2 ù . ê1 = 9 ´ 109 ´ 10 ´ 10-9 ´ 628 ú 13 ´ 10-2 úû êë

Electric field, E =

ö ÷ ÷ ø

= 2512 . N /C

2. (a) Given, charge on nucleus = +Ze, Total negative charge = - Ze (atom is electrically neutral) Charge Negative volume charge density, r = Volume -Ze r= 4 p (R )3 3 -3 Ze or …(i) r= × 4 pR3 Consider a gaussian surface with radius r. By Gauss theorem, q ò E × dA = e 0 E ´ 4pr 2 =

q e0

downward direction, electric force on it decreases, because in this case, the downward force (mg ) gets dominated over the electric force (Fe ) and it will move down (unstable equilibrium). In second diagram, if we displace - q in downward direction, electric force on it increases, because in this case, the electric force (Fe ) gets dominated over the downward force (mg ) and net force restores - q to its original position (stable equilibrium).

4. (d) Since, electric field inside the metallic conductor is zero. Therefore, the correct field line is represented by line 4, i.e. no electric line of force enters in metallic sphere.

5. (a) Consider an electric dipole with -q charge at A and +q charge at B, placed along Z-axis such that its dipole moment is in negative z-direction. i.e. pz = - 10-7 cm. The electric field is along positive direction of Z-axis such that, dE = 105 NC -1 m-1 dz From F = qdE, z

…(ii)

E

Charge enclosed by Gaussian surface, 4 q = Ze + pr 3r 3 Zer 3 [Using Eq. (i)] = Ze - 3 R From Eq. (ii), Zer 3 Ze - 3 q R = E= 4pe0 r 2 4pe0 r 2

A

or

=

Ze é 1 r ù - 3ú ê 2 4pe0 ë r R û

(– q)

O B y

x p (+q) z¢

Multiply and divide by dz. dE dE = q ´ dz ´ = p dz dz (Q dipole, p = moment charge ´ distance) dE F = p \ dz

131

Electric Charges and Fields

electrostatic force, then that particle never be in stable equilibrium. Hence, option (b) is correct. Since, all charges are in equilibrium so, net force any one of charges are in equilibrium should be zero for this, two equal and opposite forces will act on any one of the charges. Therefore, all the charged particles cannot have the same polarity.

7. (b) According to Gauss’s law, electric flux

1 times, e0 the net charge q enclosed by the surface. q Electric flux, f = e0 1 f= e0

q ö æ çl = ÷ pR ø è where, l is linear charge density and R is radius. Electric field at centre due to dl is kq ö k × lR dq æ dE = çQ E = 2 ÷ è r ø R2 kldq dE = R

q(charge on dl ) = lR dq

+ + + + + + + + + + + + +

6. (d) If a charged particle is in equilibrium under

10. (a) From figure,

dl¢ + + + + + + +

Force experienced by the system in negative z-direction æ - dE ö F = - p´ç ÷ è dz ø -7 5 = 10 ´ (- 10 ) = - 10-2 N

q dq R q

through any closed surface is equal to

[Qq = 1 (unit positive charge)] f = (e0 )-1

8. (b) Initially, the force on the sphere is equal due to both negative and positive charge. \ Net force = 0 On displacing the sphere towards the positive charge, force on sphere due to positive charge will be more than due to the negative charge, because positive charge is nearer. So, sphere will move further to the positive charge.

9. (a) Let distance between two charges is x The force between them, 1 q(Q - q ) F = × 4pe0 x2 where, x is constant, so for maximum force q(Q - q ) should be maximum. d {q(Q - q )} = 0 \ dq Q - 2q = 0 Þ

Q =2 q

dl + + + + + + +

dE

dE cos q

Now, electric field will have two components dE cos q and dE sin q. But we need to consider only the component dE cos q, as the component dE sin q will cancel out because of the field due to the symmetrical element dl. \Total field at centre =2ò

p/2

0

dE cos q

2kl p / 2 cos q dq R ò0 2kl 2kl = [sin q]p0 / 2 = [1 - 0] R R

=

= =

2q 2kl 2kq = = R pR2 4pe0 . pR2 q

æ çQ k = 1 ç 4pe0 è

2p2 e0 R2

11. (a) Apply Gauss’s law to calculate the charge associated with plane surface A . Gauss’s law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by e0 . q i.e. ftotal = e0 where, q is net charge and e0 is electric permittivity.

ö ÷ ÷ ø

AIIMS Chapterwise Solutions ~ Physics

132 Let electric flux linked with surfaces, A, B and C be f A , fB and fC , respectively. i.e. ftotal = f A + fB + fC Since, fC = f A q \ 2f A + fB = ftotal = e0 1 2

ö æq ÷ ç ç e - fB ÷ ø è 0

or

fA =

But

fB = f ö 1æq f A = çç - f ÷÷ 2 è e0 ø

Hence,

Particle (3) has positive charge, so deflection is in upward direction and particle (4) is negatively charged, so will move in downward direction.

14. (c) The situation is shown in the figure. Plate 1

has surface charge density s and plate 2 has surface charge density-s. The electric fields at point P due to two charged plates add up giving s

(given)

1 ++++++++++++++

E

12. (b) O is the origin at which -2q charge is placed. The system is equivalent to two dipoles along x and y-directions respectively. The dipole moments of two dipoles are shown in figure.

P

–––––––––––––– –s 2

E=

s = 26.4 ´ 10-12 C / m2 ,

Given,

y

e0 = 8.85 ´ 10-12 C2 / N-m2

(0,a,0) q

P(a,a,0)

p2 + p2 2

» 3 N/C

16. (c) Resultant electric field between the plates E ¢ = E + E = 2E s s = 2× = 2e0 e0

= (qa) + (qa) = 2qa

13. (a) The figure shows the path of a positive charged particle (1) through a rectangular region of uniform electric field. Top – – – – – – – – – – –

1r

8.85 ´ 10-12

diverge out) from positive charge and end (i.e. converge) on negative charge or extends to infinity. Thus, A is positive charge and B is negative charge. Also, density of lines at A is more than that at B, i.e. |A| >|B|.

2

E

26.4 ´ 10-12

15. (a) Electric lines of force usually start (i.e. x

The resultant dipole moment will be directed along OP, where P = (a, a, 0). The magnitude of resultant dipole moment is p¢ =

E=

Hence,

ap 2q (a,0,0) (0,0,0) O p z a

2s

s s s + = 2e0 2e0 e0

r3 s4

+++++++++++ Down

Since, positive charged particle moves in parabolic path in electric field along its direction. So, the electric field is in upward direction. And it is always from positive potential to negative potential. The particle (2) is negatively charged, so it is deflected in downward direction.

E

+s

–s

+ + + + + + + + +

– – – – – – – – – – –

E

E

E

17. (d) The electrostatic force is given by, F =

1 q1q2 × 4pe0 r 2

The force due to one Cs + ion is balanced by diagonally opposite Cs + ion. Hence, net electrostatic force on Cl - ion due to eight Cs + ions is zero.

133

Electric Charges and Fields 18. (b) Outside the spherical charge, the intensity of electric field at a point P situated at a distance r from the centre of the charge is E=

q 1 × 4pe0 r 2

(if r > R)

21. (a) An electric dipole is a system of two equal and opposite point charges separated by a small distance. When placed in a non-uniform electric field, the two poles experience unequal forces EQ and EQ ¢. EQ¢

On the surface of spherical charge, the electric field is given by q 1 (if r = R) E= × 4pe0 R2

q 2l

and inside the spherical charge, the electric field is 1 qr (if r < R) × E= 4pe0 R3 Hence, the variation of E - r is a follows +

+

+

+

R

+

+

E Eµr O

Eµ 12 r r

R

Torque, t = EQ × 2l sin q = pE sin q Thus, in addition to torque, there is also an external unbalanced force acting on the dipole. Hence, there would be translatory as well as rotatory motion.

22. (b) r1 = 10 cm, r2 = 20 cm, q1 = 10mC

+

+

EQ

Note At the centre of sphere, r = 0 so, E = 0.

19. (d) Since, total charge enclosed by Gaussian surface is zero. \Total electric flux, f = 0 Alternative The total flux is zero because number of lines entering the surface are same as those leaving it.

When both spheres touches, then charge flows from one charge sphere to another uncharged sphere till, when potential is same. i .e . V1 = V2 1 q1 1 q2 = × 4 pe0 r1 4 pe0 r2 r2 q2 …(i) or = r1 q1 q1 2 s1 4pr12 q1 æ r2 ö ç ÷ Q = = q2 s2 q2 çè r1 ÷ø 4pr22

=

Since, F1 sin 60° and F2 sin 60° cancel to each other. Hence, net force normal to BC is zero. F2 sin 60°

F2

3 0° F1 sin 60°

a

Also,

60°

–Q B

a

[from Eq. (i)]

s1 : s2 = 2 : 1

charges -2Q and Q at distance r is given by,

+Q

60° F1 a

\

r1 r2

23. (c) The intensity of electric field at a point due to

A G

2

æ r2 ö ç ÷ çr ÷ è 1ø r2 20 2 = = = r1 10 1

q2 1 20. (c) |F1| = |F2| = × 2 4 p e0 a

+Q C

E=

1 (-2Q ) × 2 4pe0 r

E¢ =

1 (Q ) × 4pe0 r 2

Dividing Eq. (i) by Eq. (ii), we get 2Q E E =Þ E¢ = E¢ Q 2

…(i) …(ii)

AIIMS Chapterwise Solutions ~ Physics

134 24. (c) Intensity of electric field at a point on the axial plane of dipole is given by 1 2p × Ea = 4pe0 r 3

...(i)

Intensity of electric field at a point on the equatorial line of dipole is given by p 1 ...(ii) Eq = × 4pe0 r 3 Dividing Eq. (i) by Eq. (ii), we get Ea 2 = Eq 1 Þ

E a = 2Eq

25. (c) The net electric field at point P is given by E=

qx 4 pe0 ( x + a2 )3 /2 2

i$

E Emax –a Ö2 O a Ö2 Emax

x

E=

29. (c) During the sharing of charges between any two bodies no net loss of charge takes place as charge is always conserved. During this process, some energy is disappeared in the form of heat. But some amount of energy is scattered in sharing between these charges. Assertion is true but Reason is false.

26. (c) We know that, for an electric dipole

and

Eequatorial

Hence, from Eqs. (i) and (ii), we get Eaxial = Eequatorial 2

non-uniform electric field is given by, qE a= m (where q is charge, E is electric field and m is mass.) As E change, a also changes, \Acceleration does not depend on velocity. Charge is an invariant quantity. We know that energy and mass vary with motion, but charge does not, when an electron is moving close to speed of light, its mass varies (according to equation of relativity), but there is no such equation for the charge of an electron. The charge remains the same even when electron is moving with the speed of light. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Thus, it can be seen from the equation that the flux is independent of side of square or in general of Gaussian surface. Hence, both Assertion and Reason are ture and Reason is the correct explanation of Assertion.

and E will be maximum, where, dE =0 dx q Þ x=± 2 2q 1 and E max = × 4pe0 3 3a2 1 æ 2p ö ç ÷ 4pe0 è r 3 ø 1 æ pö = ç ÷ 4pe0 è r 3 ø

27. (a) Acceleration of charged particle in

centre of square. Thus, according to Gauss’s law, the flux through the square is q ò EdA = e 0

q 1 × 4 pe0 x2

Eaxial =

Eequatorial =

28. (a) We can assume the charge is placed at the

\At centre of ring, x = 0 Þ E = E0 If x >> a,

E 2 Reason is false as electric field due to dipole varies inversely as cube of distance, 1 i.e., Eµ 3 r Assertion is true but Reason is false. Hence,

...(i) ...(ii)

30. (a) In a hollow spherical shield, the charge is present only on its surface but charge is zero at every point inside the hollow sphere. Hence, the metallic shield in the form of hollow shell may be build to block an electric field. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

135

Electric Charges and Fields 31. (a) When the bob is placed in an electric field, the time period of simple pendulum will remain same as the bob is not charged. If simple pendulum having charged bob is placed in a horizontal electric field, then the time period will be decreased because there will be an increase in restoring force, as due to the charge present in the bob, it will move along the electric field and the field tries to keep it to its normal position along the field and hence, the time period decreases, as two forces act on its electric force and gravitational force. Hence, both Assertion and Reason are true and Reason is the correct explanation of Assertion.

32. (b) Specific charge of a positive ion corresponding to one gas is fixed but it is different for different gases. Hence, both the Assertion and Reason are true but Reason is not the correct explanation of Assertion.

33. (d) Only a charged particle initially at rest moves along the electric field lines. But when the charged particle is in motion, then it makes an angle with the electric field line and thus follow a parabolic path. The electric lines of force converge at a negative charge and diverge from a positive charge. Hence, both Assertion and Reason are false.

34. (a) It is because the charges only reside at the surface of a conductor, the charge enclosed in the Gaussian surface in the cavity is zero.

In accordance with Gauss’s law, q i.e., ò E ×dA = e 0 If q = 0, then electric field E is zero. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

35. (d) The magnitude of the electrostatic force

æ 1 q1q2 ö ÷ ç ç 4pe × r 2 ÷ acting between the two charged ø è 0 particles is very large as compared to the æ Gm1 m2 ö gravitational force ç ÷ acting between them. è r2 ø The electrostatic force between the two protons is nearly 1036 times the gravitational force between them. The electrostatic force between the two electrons is nearly 1043 times the gravitational force between them. The electrostatic force between a proton and an electron is nearly 1039 times the gravitational force between them. But gravitational force is dominating in the universe. Both Assertion and Reason are false.

36. (a) Electric field lines of force can never intersect each other, because if they do so, then at the point of intersection two tangents can be drawn which would mean two directions of force at that point which is impossible. If there are n point charges q1 , q2 , ..., q n, resultant intensity E = E1 + E2 + ... + En. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

AIIMS Chapterwise Solutions ~ Physics CHAPTER

15 Electrostatic Potential and Capacitance 1. Electric field inside a parallel plate capacitor is 100 V/m and dielectric constant inside it is 5.5. Then, the value of polarisation is [AIIMS 2018] (a) 450 (c) 550

4. Charges + q and -q are placed at points A and B respectively which are a distance 2L apart, C is the mid-point between A and B. R

(b) 350 (d) 250

C are connected to a 2 battery of V volts, as shown below

2. Two capacitors C and

V

C 2

C

A

C

B

D

The work done in moving a charge +Q along the semicircle CRD is [AIIMS 2017] (a)

qQ 4pe0 L

(b)

qQ 2 pe0 L

(c)

qQ 6pe0 L

(d)

- qQ 6pe0 L

5. A parallel plate capacitor of capacitance C The work done in charging both the capacitors fully is [AIIMS 2017] (a) 2CV 2 1 (c) CV 2 2

3 (b) CV 2 4 1 (d) CV 2 4

3. A parallel plate capacitor has an electric field of 10 5 Vm -1 between the plates. If the charge on the capacitor plates is 1mC, the force on each capacitor plate is [AIIMS 2017] (a) 0.5 N (c) 0.005 N

(b) 0.05 N (d) None of these

is connected to a battery and is charged to a potential difference V . Another capacitor of capacitance 2C is connected to another battery and is charged to potential difference 2V . The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is [AIIMS 2016] (a) zero (c)

3 CV 2 2

25CV 2 6 9 CV 2 (d) 2

(b)

137

Electrostatic Potential and Capacitance 6. A charged spherical conductor of radius a and charge q is surrounded by another charged concentric sphere of radius b (b > a). The potential difference between conductors is V. When, the spherical conductor of radius b is discharged completely, then the potential difference between conductor will be [AIIMS 2014] (a) V (c)

(b)

q1 q2 4pe0 a 4pe0 b

10. Two spherical conductors A and B of radii a and b (b>a) are placed concentrically in air. The two are connected by a copper wire as shown in figure. The equivalent capacitance of the system is [AIIMS 2012] b A

Va b

B

A

(d) None of these

7. Three plates of common surface area A are connected as shown in figure. The effective capacitance will be d A

B

[AIIMS 2013] 3 (c) 2 e0 A / d (d) e0 A / d 2

8. In which of the states shown in figure, is potential energy of a electric dipole maximum? [AIIMS 2013]

–q

+q

(b)

E –q

E

(c)

(d)

+q

9. There are four point charges +q, -q, +q and -q are placed at the corners A, B, C , and D respectively of a square of side a.

2)

1 (b) m F 4 (d) 8mF

–q

A (+q)

a

B (– q)

[AIIMS 2011] (a) 1.06 pF (c) 6.36 pF

D (– q) a C (+q) The potential energy of 1 the system is times. 4pe 0 [AIIMS 2013]

q2 ( -4 + a 4q 2 (c) a

[AIIMS 2012] 1 (a) m F 2 1 (c) m F 8

E

–q

(a)

broken into eight drops of equal radius. Then, the capacitance of each small drop is

a number of plates as shown in figure. The distance between the consecutive plates is 0.885 cm and the overlapping area of the plates is 5 cm 2 . The capacity of the unit is

E

+q

(d) 4pe0 a

12. A gang capacitor is formed by interlocking

+q

(a)

(b) 4pe0 (a + b )

11. A spherical drop of capacitance 1 mF is

d

(a) 3e0 A / d (b) e0 A / d

4pe0 ab b-a (c) 4pe0 b

(a)

q2 ( -4 + 2a -4 2q 2 (d) a

(b)

2)

(b) 4 pF (d) 12.72 pF

13. Two capacitors of 10 mF and 20 mF are connected in series with a 30 V battery. The charge on the capacitors will be respectively. [AIIMS 2011] (a) 100 mC, 100 mC (b) 200 mC, 100 mC (c) 200 mC, 200 mC (d) 100 mC, 200 mC

AIIMS Chapterwise Solutions ~ Physics

138 14. A capacitor of capacitance 1 mF is charged to a potential of 1 V. It is connected in parallel to an inductor of inductance 10 -3 H. The maximum current that will flow in the circuit has the value [AIIMS 2010] (a) 1000 mA (c) 1 mA

from a point A where potential is 600 V to the point B where potential is zero. Velocity of the ball at the point B is 20 cm/s. The velocity of the ball at the point A will be [AIIMS 2009]

(b) 1 A (d) 1000 mA

15. A charged oil drop of mass 25 . ´ 10 -7 kg is in space between the two plates, each of area 2 ´ 10 -2 m 2 of a parallel plate capacitor. When the upper plate has a charge of 5 ´ 10 -7 C and the lower plate has an equal negative charge, the oil remains stationary. The charge of the oil drop is (Take, g = 10 m /s2 ) [AIIMS 2010] -1

18. A ball of mass 1 g and charge 10 -8 C moves

(a) 22.8 m/s (c) 16.8 m/s

(b) 22.8 cm/s (d) 16.7 cm/s

19. In the figure, a proton moves a distance d in a uniform electric field E as shown in the figure. Does the electric field do a positive or negative work on the proton? Does the electric potential energy of the proton increase or decrease? [AIIMS 2007] E

-6

(a) 9 ´ 10 C (c) 8.85 ´ 10-13 C

(b) 9 ´ 10 C (d) 18 . ´ 10-14 C

P d

16. In the circuit shown in the figure, the potential difference across the 4.5 mF capacitor is [AIIMS 2010] 3 mF 4.5 m F

stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C, then the resultant capacitance is [AIIMS 2007]

12 V

8 V 3

(a) Negative, increase (b) Positive, decrease (c) Negative, decrease (d) Positive, increase

20. A parallel plate capacitor is made by 6 mF

(a)

(b) 4 V

(a) (n - 1) C (c) C

(c) 6 V

(d) 8 V

17. A network of four capacitors of capacities equal to C 1 = C , C2 = 2C , C3 = 3C and C 4 = 4C are connected to a battery as shown in the figure. [AIIMS 2009]

(b) (n + 1) C (d) nC

21. Two concentric conducting thin spherical shells A and B having radii rA and rB ( rB > rA) are charged to Q A and -QB (|QB| >|Q A|). The electric field along a line (passing through the centre) is [AIIMS 2005] E

C2

E

(a)

(b)

C1

C3

0

C4

rA

rB

x

0

E

V

22 3

(b)

3 22

(c)

7 4

(d)

4 7

rA

rB

x

rA

rB

x

E

(c)

The ratio of the charges on C 2 and C 4 is (a)

r

(d) 0

rA

rB

x

0

139

Electrostatic Potential and Capacitance 22. Equipotential surfaces associated with an

27. An insulated charged sphere of radius 5 cm

electric field which is increasing in magnitude along the x- direction are

has a potential of 10 V at the surface. The potential at the centre will be [AIIMS 1998]

(a) planes parallel to yz-plane [AIIMS 2004] (b) planes parallel to xy-plane (c) planes parallel to xz-plane (d) coaxial cylinders of increasing radii around the X-axis

(a) same as that at 5 cm from the surface (b) same as that at 25 cm from the surface (c) 10 V (d) zero

23. A 40 mFcapacitor in a defibrillator is charged to 3000 V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2 ms. The power delivered to the patient is [AIIMS 2004] (a) 45 kW (c) 180 kW

(b) 90 kW (d) 360 kW

24. A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 kV, its kinetic energy will be [AIIMS 2003] (a) 1840 keV (c) 1 keV

(b) 1/1840 keV (d) 920 keV

25. Two material having the dielectric constants K 1 and K2 are filled between two parallel plates of a capacitor. Where area of each plate is A and the distance between the plates is d.

K1

K2

The capacity of the capacitor is [AIIMS 2001] (a)

Ae0 (K1 ´ K 2 ) 2d (K1 + K 2 )

Ae0 (K1 - K 2 ) 2d Ae0 (K1K 2 ) (c) 2 (K1 + K 2 ) Ae ( K + K 2 ) (d) 0 1 2d

(b)

capacitors are used to make a combination of 16mF and 1000 V are [AIIMS 1996] (a) 4

(b) 32

(c) 8

(d) 3

29. It is possible to have a positively charged body at (a) positive potential (c) negative potential

[AIIMS 1995] (b) zero potential (d) All of these

30. If an electron is brought toward another electron the electric potential energy of the system [AIIMS 1995] (a) becomes zero (c) remains the same

(b) increases (d) decreases

Assertion & Reason Direction (Q.Nos. 31-43) Read the Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) Assertion is true but Reason is false. (d) Both Assertion and Reason are false.

31. Assertion Net electric field inside a conductor is 0. Reason Charge is present on the surface of conductor. [AIIMS 2018]

32. Assertion If a point charge q is placed in

26. A particle of mass 2g and charge 1 mC is held at a distance of 1m from a fixed charge 1 mC. If the particle is released, it will be repelled. The speed of particle when it is at a distance of 10 m from the fixed charge is (a) 90 m/s (c) 45 m/s

28. Minimum number of 8mF and 250 V

(b) 100 m/s [AIIMS 2000] (d) 55 m/s

front of an infinite grounded conducting plane surface, the point charge will experience a force. Reason This force is due to the induced charge on the conducting surface which is at zero potenital. [AIIMS 2016]

33. Assertion Dielectric polarisation means formation of positive and negative charges inside the dielectric. Reason Free electron are formed in this process. [AIIMS 2014]

AIIMS Chapterwise Solutions ~ Physics

140

39. Assertion If three capacitors of capacitances

34. Assertion Electrons move from a region of

C 1 < C2 < C3 are connected in parallel then their equivalent capacitance C p > C3 1 1 1 1 Reason = + + C p C 1 C2 C3 [AIIMS 2003]

higher potential to a region of lower potential. Reason An electron has less potential energy at a point where potential is higher and vice-versa. [AIIMS 2013]

40. Assertion A metallic shield in form of a

35. Assertion Conductors having equal

hollow shell may be built to block an electric field. Reason In a hollow spherical shield, the electric field inside it is zero at every point.

positive charge and volume, must also have same potential. Reason Potential depends only on charge and volume of conductor. [AIIMS 2011]

[AIIMS 2001]

41. Assertion If the distance between parallel

36. Assertion A and B are two conducting

plates of a capacitor is halved and dielectric constant is three times, then the capacitance becomes 6 times. Reason Capacity of the capacitor does not depend upon the nature of dielectric material. [AIIMS 1999]

spheres of same radius. A being solid and B hollow. Both are charged to the same potential. Then, charge on A = charge on B. Reason Potentials on both are same. [AIIMS 2009]

37. Assertion A parallel plate capacitor is

42. Assertion The surface charge densities of

connected across battery through a key. A dielectric slab of dielectric constant K is introduced between the plates.The energy which is stored becomes K times. Reason The surface density of charge on the plate remains constant or unchanged.

two spherical conductors of different radii are equal. Then the electric field intensities near their surface are also equal. Reason Surface charge density is equal to charge per unit area. [AIIMS 1997]

43. Assertion In the absence of an externally

[AIIMS 2007]

applied electric field, the displacement per unit volume of a polar dielectric material is always zero. Reason In polar dielectrics, each molecule has a permanent dipole moment but these are randomly oriented in the absence of an externally applied electric field. [AIIMS 1995]

38. Assertion The lighting conductor at the top of high building has sharp pointed ends. Reason The surface density of charge at sharp points is very high resulting in setting up of electric wind. [AIIMS 2005]

Answers 1. 11. 21. 31. 41.

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

2. 12. 22. 32. 42.

(b) (b) (a) (a) (b)

3. 13. 23. 33. 43.

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

4. 14. 24. 34.

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

5. 15. 25. 35.

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

6. 16. 26. 36.

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

7. 17. 27. 37.

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

8. 18. 28. 38.

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

9. 19. 29. 39.

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

10. 20. 30. 40.

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

Explanations 1. (a) Electric constant, K = 5.5

Therefore, work done, W =

Electric Susceptibility, x = K - 1 = 5 .5 - 1 = 4.5 \Polarisation, P = xE = 4.5 ´ 100 = 450

5. (c) According to the question,

2. (b) The two capacitors are in parallel.

+

Their equivalent capacitance C 3C C¢ = C + = 2 2 Work done in charging the equivalent capacitor is stored as potential energy, 3C ù 1 é W = U = C ¢V 2 QC¢ = \ 2 ûú 2 ëê 3 2 CV 4 3. (b) Net charge on the parallel plate capacitor is 1 mC. Electric field near one of the plates due to the 105 other plate is E = Vm-1 . 2 Now, force is given by, F = qE Substituting, q = 1 m C = 1 ´ 10-6 C \

=

We get,

F = 0.05 N

4. (d) In Ist case, when charge + Q is situated at C +q

+Q

A

C

–q B

2L

L

Electric potential energy of system, 1 q(-q ) 1 (-q )Q 1 qQ × + × + × U1 = 4pe0 2L 4pe0 L 4pe0 L 1 q1q2 ) (QU = 4pe0 r In IInd case, when charge +Q is moved from C to D. +q

–q

A

+Q

B 2L

D L

Electric potential energy of system in this case 1 q(-q ) 1 (q )Q 1 (-q )Q U2 = × + × + × 4pe0 2L 4pe0 3L 4pe0 L Work done, W = DU æ 1 qQ 1 qQ ö÷ qQ é 1 ù U2 - U1 = çç × × ÷ 4pe L êë 3 - 1úû 4 3 L 4 L pe pe ø è 0 0 0 2qQ qQ qQ 1 æ -2 ö = × ç ÷= =12pe0 L 6pe0 L 4pe0 L è 3 ø

- qQ 6pe0 L

C– V

2C –

+

2V

\ Common potential of capacitors Net charge V1 = Net capacitance 2 C ´ 2V - C ´ V V1 = 2C + C (Q negative terminal connected with positive terminal) 3CV = =V 3C Now, the energy stored in the system, 1 1 3 U = C net V12 = (2C + C )× V 2 = CV 2 2 2 2

6. (a) Since, potential difference between spheres depends upon electric field present in the space between spheres, therefore the electric field in the space between two spheres due to outer sphere is zero. So, potential difference between spheres does not depend upon the charge on outer sphere.

7. (c) The given circuit is equivalent to parallel combination of two identical capacitors. As the air is present between the capacitor plates, so each one will have e A capacitance, C = 0 d 2e A Thus, Ceq = 2C = 0 d d C

A

A

B

C

A d

AIIMS Chapterwise Solutions ~ Physics

142 8. (a) Since, we know potential energy of a electric dipole p in an external electric field E U = - p × E = - pE cos q For U max q, will be equal to p. i.e., angle between direction of dipolen moment (-q ® + q ) of electric field should be 180° a p.

9. (a) According to the question, there will be six pairs. The potential energy of the system, 1 æ q (-q ) (-q )q (-q )q U = + + ç 4pe0 è a a a +

(-q )q qq qq ö + + ÷ a a 2 a 2ø

Hence,the capacity of 8 capacitors = 8C = 8 ´ 0.5 = 4 pF

13. (c) According to the question, the equivalent capacitance for capacitors in series is 1 1 1 1 1 = + = + Cs C1 C2 10 20 20 Þ Cs = mF 3 20 Also, q = CV = ´ 30 = 200 mC 3 This charge on the two capacitors in series is same.

14. (a) Charge on the capacitor, q 0 = CV = 1 ´ 10-6 ´ 1 = 10-6 C

2

=

q qq æ 2 ö ç -1 - 1 - 1 - 1 + ÷ = ( 2 - 4) 2ø a 4pe0 a è

10. (c) Since, the capacitance of a spherical capacitor, C = 4pe0 r where, r is radius of the sphere. Q Potential of a capacitor, q q V = = C 4pe0 r

…(i)

Hence, q = q 0 sin wt or I 0 = wq 0 = maximum current 1 1 Now, w= = = (109 )1 /2 LC 10-9 \

I 0 = (109 )1 /2 ´ (1 ´ 10-6 ) = 1000 mA

[from Eq. (i)]

where, q = charge on capacitor. For given conductors A and B, V A > VB Therefore, charge will flow from A to B. So, the net capacitance will be C = 4pe0b

11. (a) Let R and r be the radii of bigger and each smaller drop respectively. 4 3 4 pR = 8 ´ pr 3 \ 3 3 (Q net volume is constant) …(i) Þ R = 2r The capacitance of a smaller spherical drop is …(ii) C = 4pe0 r The capacitance of bigger drop is C ¢ = 4pe0 R (Q R = 2r ) = 2 ´ 4pe0 r [From Eq. (ii)] = 2C C¢ 1 (Q C ¢ = 1 mF) C= = mF 2 2

12. (b) The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor, e A 8.854 ´ 10-12 ´ 5 ´ 10-4 = 0.5 pF C= 0 = d 0.885 ´ 10-2

15. (c) As given in question, q = 5 ´ 10-7 C

Fe –q1

q = -5 ´ 10-7 C

+

E Since, the drop mg – remains stationary Thus, Fe = mg Þ q1 E = mg (where, q1 is charge on the drop) ö æ q ç\E = s = q ÷ = mg Þ q1 ç Ae0 e0 Ae0 ÷ø è mg Ae0 Þ q1 = q

=

2.5 ´ 10-7 ´ 10 ´ 2 ´ 10-2 ´ 8.85 ´ 10-12 -13

= 8.85 ´ 10

5 ´ 10-7 C

16. (d) According to the question, C net of 3 mF and 6 mF, C1 = 3 mF + 6 mF = 9 mF The total capacitance of the circuit, 4.5 ´ 9 C= = 3mF 4.5 + 9 Charge on 4.5 mF and C1 , Q = CV = 3 ´ 12 mC = 36 mC \Potential difference across 4.5 mF, Q V = = 8V 4.5 mF

143

Electrostatic Potential and Capacitance 17. (b) According to the question, the charge flowing through C 4 is q 4 = C 4 ´ V = 4 CV

21. (a) According to the …(i)

The series combination of C1 ,C2 and C3 gives 1 1 1 1 = + + C ¢ C 2C 3C 6C C¢ = Þ 11 Since C1 , C2 and C3 are in series. \C1 , C2 and C3 will have equal charge q. 6C Therefore, V q = C ¢V = 11 6CV …(ii) Charge on C2 , q2 = 11 Thus,

6CV q2 3 = 11 = 4CV 22 q4

18. (d) Given, m =1g =10–3 kg q = 10-8 C, v1 = ? v2 = 20 cm/s, V1 = 600 V and V2 = 0 Applying work-energy theorem, DKE = Net work done 1 Þ m(v22 - v12 ) = - q (V2 - V1 ) 2 1 ´ 10-3 [(20 ´ 10-2 )2 - v12 ] Þ 2

B question, inside the x¢ shell A, electric field, A E in = 0 x rB rA At the surface of shell A, kQ E A = 2 A ¾® (a fixed rA EA positive value) Between the shell A and B, at a distance x from the common EB centre, kQ A E = 2 ¾® (as x increases E decreases) x At the surface of shell B, k (Q A - Q B ) EB = ¾® (a fixed negative rB2 value because |Q A| (Eg )Si > (Eg )Si > (Eg )Ge = (Eg )Ge = (Eg )Si

17. Application of a forward bias to a 11. The Boolean expression P + PQ, where P and Q are the inputs of the logic circuit, represents [AIIMS 2015] (a) AND gate (c) NOT gate

(b) NAND gate (d) OR gate

p-n junction

[AIIMS 2011] (a) increases the number of donors on the n-side (b) increases the electric field in the depletion zone (c) increases the potential difference across the depletion zone (d) widens the depletion zone

265

Semiconductor and Electronic Devices 18. The curve between charge density (r) and distances near p-n junction will be r

r p

n

p

(a)

r

n

(b)

r

22. A common-emitter amplifier has a voltage gain of 50, an input impedance of 100 W and an output impedance of 200 W. The power gain of the amplifier is [AIIMS 2008] (a) 500

(b) 1000

(c) 1250

(d) 100

23. The circuit given below represents which of r n

p

(c)

p r

r

the logic operations? n

Inputs

(d)

r

[AIIMS 2010]

19. If in a p-n junction diode, a square input signal of 10V is applied as shown in the figure 5V RL

–5V

Then, the output signal across RL will be [AIIMS 2010] 10V (a)

–10V +5V

(c) OR

(d) NOR

The voltage gain in dB is (a) 30 dB (c) 3 dB

[AIIMS 2007]

(b) 60 dB (d) 20 dB

25. A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA. When it operates with a 6 V battery through a limiting resistor R. The value of R is [AIIMS 2006] (a) 40 k W (b) 4 k W

(c) 200 W

(d) 400 W

26. The minimum potential difference between

(c) 5 V

[AIIMS 2006] (d) 4.2 V

of electrons and holes are m e and m h , respectively. Which of the following is true ?

20. The input resistance of a common-emitter transistor amplifier, if the output resistance is 500kW, the current gain a = 0.98 and power gain is 6.0625 ´ 10 6, is [AIIMS 2009] (c) 100 W

(b) 3 V

27. In a semiconducting material, the mobilities

–5V

(d) 400 W

21. In the energy band diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is a/an Ec

[AIIMS 2005] (b) m e < m h (d) m e < 0 ; m h > 0

(a) m e > m h (c) m e = m h

28. Consider an n-p-n transistor amplifier in common-emitter configuration. The current gain of the transistor is 100. If the collector current changes by 1 mA, what will be the change in emitter current? [AIIMS 2005] (a) 1.1 mA (c) 0.01 mA

(b) 1.01 mA (d) 10 mA

29. Which of the following logic gates is an universal gate ?

Eg Ev

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

(b) NOT

24. An amplifier has a voltage gain AV = 1000.

(a) 1 V

(d)

(a) 198 W (b) 300 W

(a) AND

Output

the base and emitter required to switch a silicon transistor ‘ON’ is approximately

(b)

(c)

[AIIMS 2007]

p-type semiconductor insulator metal n-type semiconductor

(a) OR [AIIMS 2009]

(b) NOT

(c) AND

[AIIMS 2005] (d) NAND

30. A Ge specimen is doped with Al. The concentration of acceptor atoms is ~ 1021 atoms/m3 . Given that the intrinsic concentration of electron-hole pairs is

AIIMS Chapterwise Solutions ~ Physics

266 ~ 10 19 / m3 , the concentration of electrons in the specimen is [AIIMS 2004] (a)

1017 3

m

(b)

1015 3

m

(c)

104 3

m

(d)

102 3

m

31. Which logic gate is represented by the following combination of logic gates? A Y B

[AIIMS 2004]

(a) OR (c) AND

37. Which one of the following is true about the p-type and n-type semiconductor? [AIIMS 2000] (a) n-type semiconductor have holes in majority (b) The concentration of electrons and holes are equal in both n-type semiconductors (c) n-type semiconductors have free electrons in majority (d) n-type semiconductor has excess negative charge

38. The truth table given for which of the following gates is correct ?

(b) NAND (d) NOR

32. To a germanium sample, traces of gallium are added as an impurity. The resultant sample would behave like [AIIMS 2003] (a) (b) (c) (d)

a conductor a p -type semiconductor an n-type semiconductor an insulator

[AIIMS 2000]

A

B

Q

0 0 1 1

0 1 0 1

0 1 1 1

(a) NAND gate (c) AND gate

(b) OR gate (d) NOT gate

39. In a full wave rectifier circuit operating

33. In the following common-emitter configuration an n- p- n transistor with current gain b = 100 is used. The output voltage of the amplifier will be

from 50 Hz mains frequency, then the fundamental frequency in the ripple will be [AIIMS 2000] (a) 50 Hz (c) 70 Hz

(b) 100 Hz (d) 25 Hz

40. When n-p-n transistor is used as an 1kW

10kW Vout

amplifier, then

10 m V

[AIIMS 2003] (a) 10 mV (b) 0.1 V

(c) 1.0 V

(d) 10 V

34. The value of current gain a of a transistor is 0.98. The value of b will be (a) 490 (c) 59

[AIIMS 2002]

(b) 49 (d) 69

35. When added an impurity into the silicon which one of the following produces n-type of semiconductors? [AIIMS 2001] (a) Iron (c) Aluminium

(b) Magnesium (d) Phosphorus

36. The current gain for a transistor working a common-base amplifier is 0.96. If the emitter current is 7.2 mA, the base current will be [AIIMS 2001] (a) 0.42 mA (c) 0.29 mA

(b) 0.49 mA (d) 0.35 mA

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

[AIIMS 1999] electrons move from base to collector holes move from emitter to base electrons move from collector to base holes move from base to emitter

41. The transfer ratio b of a transistor is 50. The input resistance of the transistor when used in the common-emitter configuration is 1 kW. The peak value of the collector AC current for an AC input voltage of 0.01 V, is (a) 500 mA (c) 0.01 mA

(b) 0.25 mA [AIIMS 1998] (d) 100 mA

42. In a diode, when there is a saturation current, the plate resistance will be

[AIIMS 1997] (a) data insufficient (b) zero (c) some finite quantity (d) infinite quantity

43. When two semiconductors p and n-type are brought into contact they form p-n junction, which acts like a/an (a) rectifier (c) conductor

a

(b) amplifier [AIIMS 1997] (d) oscillator

267

Semiconductor and Electronic Devices 44. Which of the following, when added as impurity into the silicon, produces n-type semiconductor? [AIIMS 1995] (a) B

(b) P

(c) Mg

(d) Al

45. The logic circuit given in the figure performs the logic operation

[AIIMS 1995]

A B

Y

Y

C

(a) ABC

(b) ABC

(c) ABC

(d) ABC

46. If a p-n diode is reverse biased, then the resistance measured by an ohm-metre, will be [AIIMS 1995] (a) high

(b) zero

(c) infinite

(d) low

47. A certain logic circuit has A and B as the two inputs and Y as the output. What is the logic gate in the circuit, if the truth table of the circuit is as shown [AIIMS 1994]

(a) XOR

A

B

Y

0

0

0

0

1

1

1

0

1

1

1

0

(b) OR

(c) NOR

Assertion & Reason Direction (Q.Nos. 51-70) Read the Assertion and Reason carefully to mark the correct option from those given below. (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

51. Assertion A transistor amplifier in common emitter configuration has a low input impedance. Reason The base to emitter region is forward biased. [AIIMS 2014]

52. Assertion In a transistor amplifier, the output voltage is always out of phase with the input voltage. Reason The emitter base junction is reverse biased and the base collector junction is forward biased. [AIIMS 2013]

53. Assertion When base region has larger width, the collector current increases. Reason Electron hole combination in base result in decrease of base current. (d) NAND

[AIIMS 2011]

48. A pure semiconductor has a/an [AIIMS 1994]

54. Assertion NAND or NOR gates are called

(a) finite resistance which decreases with temperature (b) infinite resistance at 0°C (c) finite resistance which increases with temperature (d) finite resistance which does not depend upon temperature

digital building blocks. Reason The repeated use of NAND (or NOR) gates can produce all the basic or complicated gates. [AIIMS 2010]

49. An intrinsic semiconductor, at the absolution zero temperature, behaves like a/an [AIIMS 1993] (a) n-type semiconductor (b) insulator (c) p-type semiconductor (d) superconductor

50. If the feedback voltage is increased in a negative feedback amplifier, then [AIIMS 1992] (a) both gain and distortion increases (b) both the gain and distortion decreases (c) the distortion increase (d) the gain increase

55. Assertion At a fixed temperature, silicon will have a minimum conductivity when it has a smaller acceptor doping. Reason The conductivity of an intrinsic semiconductor is slightly higher than that of a lightly doped p-type. [AIIMS 2009]

56. Assertion The value of current through p-n junction in the given figure will be 10 mA. +5V 300W

+2V z

Reason In the above figure, p-side is at higher potential than n-side. [AIIMS 2008]

AIIMS Chapterwise Solutions ~ Physics

268 57. Assertion Light emitting diode (LED) emits spontaneous radiation. Reason LED are forward biased p-n junctions. [AIIMS 2007]

58. Assertion The energy gap between the

64. Assertion A transistor amplifier in common-emitter configuration has a low input impedance. Reason The base to emitter region is forward biased. [AIIMS 2004]

valence band and conduction band is greater in silicon than in germanium. Reason Thermal energy produces fewer minority carriers in silicon than in germanium. [AIIMS 2007]

65. Assertion In a transistor, the base is made

59. Assertion In common base configuration, the current gain of the transistor is less than unity. Reason The collector terminal is reverse biased for amplification. [AIIMS 2006]

increases with temperature. Reason The atoms of a semiconductor vibrate with larger amplitude at higher temperatures thereby increasing its resistivity. [AIIMS 2003]

60. Assertion A p-n junction with reverse bias

67. Assertion An n-type semiconductor has a

can be used as a photo-diode to measure light intensity. Reason In a reverse bias condition, the current is small but it is more sensitive to change in incident light intensity.

large number of electrons but still if is electrically neutral. Reason An n-type semiconductor is obtained by doping an intrinsic semiconductor with a pentavalent impurity. [AIIMS 2001]

[AIIMS 2006]

61. Assertion The number of electrons in a p-type silicon semiconductor is less than the number of electrons in a pure silicon semiconductor at room temperature. Reason It is due to law of mass action. [AIIMS 2005]

62. Assertion In a common-emitter transistor amplifier, the input current is much less than the output current. Reason The common-emitter transistor amplifier has very high input impedance. [AIIMS 2005]

63. Assertion The logic gate NOT can be built using diode. Reason The output voltage and the input voltage of the diode have 180° phase difference. [AIIMS 2005]

thin. Reason A thin base makes the transistor stable. [AIIMS 2004]

66. Assertion The resistivity of a semiconductor

68. Assertion Zener diode works on a principle of breakdown voltage. Reason Current increases suddenly after breakdown voltage. [AIIMS 2000]

69. Assertion If the temperature of a semiconductor is increased, then its resistance decreases. Reason The energy gap between conduction band and valence band is very small. [AIIMS 1998]

70. Assertion The temperature coefficient of resistance is positive for metals and negative p-type semiconductors. Reason The effective charge carriers in metals are negatively charged whereas in p-type semiconductors they are positively charged. [AIIMS 1996]

Answers 1. 11. 21. 31. 41. 51. 61.

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

2. 12. 22. 32. 42. 52. 62.

3. 13. 23. 33. 43. 53. 63.

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

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

4. 14. 24. 34. 44. 54. 64.

(a) (a) (a) (b) (b) (a) (b)

5. 15. 25. 35. 45. 55. 65.

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

6. 16. 26. 36. 46. 56. 66.

7. 17. 27. 37. 47. 57. 67.

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

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

8. 18. 28. 38. 48. 58. 68.

(a) (a) (b) (b) (a) (b) (b)

9. 19. 29. 39. 49. 59. 69.

10. 20. 30. 40. 50. 60. 70.

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

(a) (a) (a) (a) (b) (a) (b)

Explanations 1. (b) Given that, VCE = 12 V Current gain, b = 100,

4. (a) Step-I Draw the give expression using basic

IC = 100 IB

gates -3

IC = 100 ´ 0.04 ´ 10 Again,

A B

= 4 mA

Y

VCC = VCE + IC RC

C D

20 = 12 + 4 ´ 10-3 RC Þ

Step-II Make the bubble on the output of AND gate and input of OR gate with NOT gate.

RC = 2kW

2. (c) For first transistor, Voltage gain = b

Rout Rin

or, G = b

Transconductance, g m = Þ

Rin = G=

b 25 = g m 0.03

b Rin

A B

Rout Rin

Rout = 0.03 Rout (25 / 0.03)

C D

Step-III Double inversion eliminated and bubbled OR gate converted into NAND gate. …(i)

Similarly, for second transistor with b¢= 20, and

g m = 0.02 Rout = 0.02Rout 20 0.02 G [from Eq. (i)] = 0.02 0.03 2 G ¢= G 3

Voltage gain, G ¢= 20.

3. (d) Breakdown voltage, VZ = 6 V \Current in 1kW resistor = 6 / (1 ´ 103 ) = 6 mA \Current in diode = 5 ´ 6 = 30 mA \Total current in R = 30 + 6 = 36 mA Potential difference across R = 30 - 6 = 24 V 24 2 R= = kW 36 ´ 10-3 3

A B

Y=AB+CD

C D

Hence, number of NAND gate = 3

5. (a) Given, current gain in common emitter mode b = 49 DIC = bDI B = 49 ´ 5 = 245mA DI E = DI B + DIC = 250 mA

6. (c) Given, for one indium atom to be doped in 5 ´ 107 silicon atoms. Number density of silicon = 5 ´ 1022 atom/m3 Number of acceptor atom/cm3 =

5 ´ 1022 5 ´ 107

= 1 ´ 1015 cm- 3

AIIMS Chapterwise Solutions ~ Physics

270

On differentiating,

7. (a) Output Y = A × B × C If

A = 0, B = 1, C = 1

\

Y = 0 × 1 × 1 = 1× 0× 0 = 0 = 1

8. (a) Given, current gain factor, a = 0.98. We know

ds =e dn p mp =

Þ

that current gain in common emitter mode, a 0.96 Current gain in b = = = 24 1 - a 1 - 0.96 0.8 = 10-3 A 800

DIC =

13. (c) Input current =

DIC 10 1 A= mA = 24 b 24

9. (c) Conductivity, s = nem ee + nhm h e

= (8 ´ 1013 ´ 25000 + 5 ´ 1012 ´ 100) ´ 1.6 ´ 10-19

But current density, J = s E \

J = ni e (m e + m h )E

15. (a) In question, the given truth table is for NAND gate. X = AB

ne > nh

16. (c) Here, carbon has the highest resistivity and germanium has the lowest resistivity.

\Hence, semiconductor is n-type.

10. (a) Y = ( A + B ) = AB = AB. which is AND gate.

11. (d) The logic circuit can be figured as P

P+PQ PQ

P

Q

P ×Q

P + PQ

0

0

0

0

0

1

1

1

1

0

0

1

1

1

0

1

Clearly, the gate is an OR gate as from table.

12. (a) The overall conductivity of a semiconductor is

\Carbon has lowest conductivity and germanium has highest conductivity. So, carbon has highest band gap energy and germanium has lowest band gap energy.

17. (a) In forward bias, more number of electrons enter in n-side from battery thereby increasing the number of donors on n-side.

18. (a) In the p-region, the concentration of holes is more than the number of electrons, while in n-region the electron’s concentration is more than holes. The charge density (r) is zero at the junction. It is positive to the right (n-side) and negative to the left (p-side) of the junction as shown in the curve. r

s = ne e m e + n p e m p s = e [m e ne + n p m p ] ne = n2i / n p Now,

s=e

êë n p

Space charge region n

Also, ne n p = n2i (for an intrinsic semiconductor)

é n2i ê

Vi 0.01 = 3 = 10-5A Ri 10 = 50 ´ 10-5 = 500 m A

mho/cm

Q

mn mp

s = nie (m e + m h )

m e = mobility of electron = 25000 cm2 V -1 s -1 .

= 320 ´ 10

m n,

Output collector current = bI B

3

= 8 ´ 10 per cm and

-3

n2p

14. (a) The overall conductivity of semiconductor is

where, ne = concentration of electron 13

n2i

n p = ni

-3

DI B =

ù é n2 ê- i m n + m p ú = 0 ú ê n2p û ë

Distance from junction

ù m n + m pn p ú úû

p

271

Semiconductor and Electronic Devices 19. (d) For Vi < 0 , the diode is reverse biased and hence offer infinite resistance, So Vo = 0 For Vi > 0, the diode is forward biased Vo = Vi . Hence, the option (d) is correct. 0.98 a = = 49 1 - a 1 - 0.98 æR ö Power gain = b2 ç 2 ÷ çR ÷ è 1ø é 500 ´ 103 ù 6 ú ´ 49 Þ 6.0625 ´ 10 = 49 ´ ê R1 êë úû \ R1 = 198 W

20. (a) Current gain, b =

21. (a) The given figure represents p-type semiconductor because majority charge carriers holes are present in valence band. p-type semiconductor are formed due to adding trivalent impurities, hence large number of holes are produced in valence band of semiconductor.

22. (c) Also, AV = b AC ´ resistance gain ( = Ro / Ri ) Given, AV = 50, Ro = 200 W, Ri = 100 W 200 Hence, 50 = b AC ´ 100 \

b AC = 25

25. (d) Current in the circuit = 10 m A = 10 ´ 10–3 A and voltage in the circuit = 6 - 2 = 4 V From Ohm’s law, V = IR 4 V \ = 400 W R= = I 10 ´ 10-3

26. (a) The cut-off voltage for silicon is ~1 V, so to switch on a silicon transistor a potential difference of 1 V approximately is required between the base and emitter.

27. (a) Since, electrons are lighter in mass compared to positively charged particles, they move easily in the semiconducting material. Hence, the mobility of holes in p-type semiconductor is less than mobility of electrons in n- type semiconductor. i.e., m e > m h .

28. (b) Here, b = 100, DIC = 1 mA DI B =

\

DI E = DIC + DI B = 1 + 0.01 = 1.01 mA

29. (d) The NAND gate and the NOR gate can be said to be universal gates, since combinations of them can be used to accomplish any of the basic operations and can thus, produce an inverter, an OR gate or an AND gate.

30. (a) From law of mass-action, n2i = ne ´ nh

Now, AC power gain = AV ´ b AC = 50 ´ 25 = 1250

23. (a) Let two inputs are A and B. The circuit can be

where, ni is concentration of electron-hole pair and nh is concentration of acceptor or holes. Given, ni = 1019 per m3 , nh = 1021 per m3

shown as A B

DIC 1 = = 0.01 mA b 100

\

1

Y1

2

The two gates shown are NAND gates. The output of gate –1, Y1 = A × B

1038 1021

= 1017 / m3

==

24. (a) Voltage gain, AV = 1000

==

Hence,

== ==

Y = A ×B Y = A× B

In dB, voltage gain = 30 dB

ne =

Use Demorgan's theorem, A + B = A × B

Y = A×B× A×B = A×B

A = 10 log 10 1000 dB

Þ

31. (c) The output, Y = A + B

The output of gate –2, Y = Y1 × Y1 or

(1019 )2 = ne ´ 1021

Output

This is the Boolean expression of AND gate. [\ A = 10log AV ] (Q log 10 10 = 1)

32. (b) When trivalent impurities (like gallium) are added into germanium sample, then p-type semiconductor is formed.

AIIMS Chapterwise Solutions ~ Physics

272 33. (c) Current gain, b =

iC Vo = i B Vi

where, Vo is output voltage and Vi is input voltage. b = 100, Vi = 10-2 V

Given, \

Vo = bVi = 100 ´ 10-2 = 1 V

34. (b) The relation between a and b is b = a / (1 - a ) Given a = 0.98, hence we have 0.98 b= = 49 1 - 0.98

35. (d) For n-type semiconductors impurity atom

hence it attracts to electrons from collector region. Hence, electrons move from base to collector.

41. (a) From the relation, V = iR We have, \ Also, b= Þ

36. (c) Current gain of transistor is a=

D iC D iE

where, D iC is change in collector current and D i E is change in emitter current. Given, a = 0.96, Di E = 7.2 mA DiC = 0.96 ´ 7.2 = 6.91 mA Also,

D i E = D iC + D i B

\

D i B = D i E - D iC = 7.2 - 6.91 = 029 . mA

37. (c) n-type semiconductor has free electrons in majority.

38. (b) In question, the given truth table is for OR gate. Q = A+ B

D iC Di B

DiC = bDi B = 50 ´ 10-5 = 500 mA

should have 5 valence electrons. Silicon has 4 valence electrons, when phosphorous with 5 valence electrons in its outermost orbit is added, then four of the five valence electrons of the impurity atom form covalent bonds with each valence electron of four silicon atoms and fifth valence electron becomes free to move in the crystal.

V = 0.01 V, R = 1000 W V 0.01 Di B = = = 10-5 A R 1000

42. (d) We know that plate resistance is given by r p = dV / dI where, dV is change in voltage and dI is change in current. Also, at saturation change in current is zero. dV rp = \ =¥ 0 Hence, plate resistance will be infinite.

43. (a) A p-n junction can be used as a rectifier because it permits current in one direction only.

44. (b) When silicon is doped with pentavalent imprurities like phosphorous, semiconductor is formed.

then

n-type

45. (c) The output of the given logic circuit is Y= A+ B+C Þ

Y = A+ B+C

So, applying de-Morgan’s theorem, ==

Y = A × (B + C ) Y = A × B ×C

Þ Y= ABC

39. (b) As in full wave rectifier, the current flows for

46. (a) When a p-n junction diode is reverse biased,

both the cycle through the load. Output voltage is obtained is not constant but pulsating in nature.

then minimum current (due to thermally generated minority carriers) is flowing, hence resistance is high.

Hence, fundamental frequency in the ripple is twice of operating frequency that is 100 Hz.

40. (a) When n-p-n transistor is used as an amplifier, then emitter-base junction is in forward bias and collector-base junction is in reverse bias. Due to reverse biasing of collector base, collector is connected with positive terminal of battery,

47. (a) In question, the given truth table is for XOR gate. XOR gate gives the output for dissimilar values of inputs. The Boolean expression for XOR gate is Y = A × B + A × B. It is read as ‘‘Y equals A and B negated OR A negated AND B’’.

273

Semiconductor and Electronic Devices 48. (a) This is because as the temperature rises, more

54. (a) These gates are called digital building blocks

and more of the covalent bonds in the crystal lattice break, thus creating more and more charge carriers. This results in increasing conductivity or decreasing resistivity.

because using these gates only (either NAND or NOR) we can compile all other gates also (like OR, AND, NOT, XOR).

49. (b) At temperature close to zero, all shared electrons (4 valence electrons) are tightly bound and so, no free electrons are available to conduct electricity through the crystal. Thus, they act like a insulator.

50. (b) If the feedback voltage is increased in negative feedback amplifier, then the gain decrease as, Vo A= Vi + mVo As the feedback is provided, then the distortion decreases.

51. (b) Input impedance of common-emitter configuration is given by æ DV = ç BE ç DI B è

ö ÷ ÷ ø VCE

This is why they are also called universal gates. Hence, Assertion and Reason both are true and Reason explains Assertion.

55. (d) The trivalent impurity atoms are called acceptor atoms because they create holes which accept electrons. When the doping is small the number of charge carriers are small. nh > ne i.e., the number of holes is greater than the number of electrons. The conductivity (s = nhm he ) is dependent on the number of carrier and mobility (m h ). As holes have less mobility than electrons, so the conductivity is less. Although, the conductivity of an intrinsic semiconductor is less than of a lightly doped p-type. Hence, option (d) is correct.

= constant

Here, DVBE = voltage across base and emitter DI B = base current of the order of few microampere. Hence, impedance of common-emitter is low. In a transistor, emitter-base region should be forward biased for good amplification. Hence, Assertion and Reason both are true, but Reason does not explain Assertion.

52. (d) The output voltage of common-emitter amplifier is 180° out of phase with the input voltage. In the all configuration of transistor, the emitter-base junction is always forward biased whereas the collector base-junction is reverse biased. Hence, Assertion and Reason both are false.

53. (d) When base region has larger width electron hole combination increase the base current. The output collector current decreases by the relation. I E = I B + IC Hence, the Assertion and Reason both are false.

56. (b) The p-side of p-n junction is taken at higher potential than n-side so, p-n junction is forward biased. Taking its resistance to be zero and applying Ohm’s law. V 5- 2 I= = = 10-2 A = 10 mA R 300 Hence, Assertion and Reason both ane true but Reason is not the correct explanation of Assertion.

57. (a) When a junction diode is forward biased as shown in figure, energy is released at the junction due to recombination of electrons and holes. In the junction diode made of gallium arsenide or indium phosphide, the energy is released in visible region. Hence, Assertion and Reason both are true and Reason explains Assertion.

58. (b) The energy gap between valence band and conduction band in germanium is 0.76 eV and the energy gap between valence band and conduction band in silicon is 1.1 eV. Also, it is true that thermal energy produces fewer minority carriers in silicon than in germanium.

AIIMS Chapterwise Solutions ~ Physics

274 This is due to the fact that the atomic number of silicon is less than germanium and hence, the binding energy or force of attraction of nucleus to electron is more in silicon than germanium. Hence, Assertion and Reason both are true but Reason does not explain Assertion.

59. (b) For amplification, emitter terminal is forward biased and collector terminal is reverse biased. In this configuration, the current gain is ratio of change in collector current to the change in emitter current at constant collector-base voltage (VCB ). æ DI a=ç C ç DI è E

\

ö ÷ ÷ ø VCB

silicon semiconductor that is for p-type semiconductor, ne < nh . Hence, Assertion and Reason both are true and Reason explains Assertion.

62. (c) The common-emitter transistor amplifier has input resistance equal to 1 kW (approx) and output resistance equal to 10 k W (approx). Thus,

æDI b AC = ç C çDI è B

ö ÷ ÷ ø VCE

= Constant

The value of b AC is from 15 to 50, which means the output current is larger as compared to the input current. Hence, option (c) is correct.

Current gain (a ) is less than unity because collector current (IC ) is always less than emitter current (I E ).

63. (d) NOT gate inverts the signal applied to it. But in

Hence, option (b) is true.

64. (b) The input (base-emitter) circuit is forward

60. (a) Photo-diode is a reverse biased p-n junction diode as shown in figure. At the p-n junction, there exists a junction field. When such a p-n diode is illuminated with light photons having energy hn > E g and intensities I1 , I2 , I3 , etc., the electron and hole pairs generated in the depletion layer (or near the junction) will be separated by the junction field and made to flow across the junction. There would be a change in the reverse saturation current as shown in figure. Hence, a measurement of the change in the reverse saturation current on illumination can give the values of the light intensity. mA

diode, the input and output are is same phase. Thus, NOT gate cannot be built by a diode. biased by a low voltage battery VBE , so that the resistance of the input circuit is small. The output (collector-emitter) circuit is reverse biased by means of a high voltage battery so that resistance of output is high. The weak input AC signal is applied across the base-emitter circuit and the amplified output signal is obtained across the collector-emitter circuit. Input impedance of common-emitter æ DV ö configuration = ç BE ÷ ç DI ÷ B ø V = constant è CE Here, DVBE = voltage across base and emitter, DI B = base current of the order of few microampere. So, the input impedance is low in common-emitter configuration.

Reverse bias Volts

I1 I2 I3 I4

Hence, both Assertion and Reason are true but Reason does not explain Assertion.

65. (c) In a transistor, the base is lightly doped and

mA

Hence, option (a) is true.

61. (a) According to law of mass-action, n2i = ne nh For p-type semiconductor, number of electrons is less than the number of electrons in a pure

very thin. This feature is main concept of transistor action on account of which only few holes (less than 5%) combine with the electrons in base region. Thickness of base region is so small that most of electrons diffusing through base region cross over through collector region. By reducing, the base width the power of transistor increase but it does not mean that transistor becomes stable. Hence, Assertion is true white Reason is false.

Semiconductor and Electronic Devices 66. (d) The temperature coefficient of resistance of semiconductors is negative that is their electrical resistance decreases (or conductivity increases) with rise in temperature. Hence, Assertion and Reason both are false.

67. (b) An n-type semiconductor is formed by doping pure germanium or silicon crystal with suitable impurity atoms of valency fire. As the impurity atoms take the position of Ge atom in germanium crystal, its four electrons form covalent bonds by sharing electrons with the neighbouring four atoms of germanium whereas the fifth electron is left free. Since, the atoms on the whole is electrically neutral, the n-type semiconductor is also neutral. Hence, Assertion and Reason both are true but Reason does not explain Assertion.

68. (b) When diode is reverse biased, small current flows due to minority carriers. The small reverse current remains almost constant over a sufficiently long range of reverse bias. When the applied reverse voltage reaches the breakdown voltage of the Zener diode, there is a large change in the current. Hence, Assertion and Reason both are true but Reason does not explain Assertion.

69. (a) When temperature increases, then due to breaking of more no. of covalent bonding, charge carrier increases, hence resistance decreases.

275 A semiconductor has a band gap which is small enough, such that it’s conduction band is appreciably thermally populated with electrons at room temperature. Hence, Assertion and Reason both are true and Reason explains Assertion.

70. (b) In case of metals when we increase the temperature, the relative movement of the atoms as well the free electrons increase. Because of which there is more collisions and more loss of energy. Due to which the resistance increases and it gives positive temperature coefficient of resistance for metals. But in the case of a semiconductor (p-type or n-type) the charge carriers are electrons and holes. So, the conductivity will depend upon the number of such electron-hole pairs. At any time, this number of electron-hole pairs is proportional to T 3 /2 e - DE / KT , where DE is the band gap. Now, with increase of temperature, the concentration of the electron-hole pairs increases which leads to an increase of conductivity or conversely a decrease of resistivity. So, a p-type semiconductor has a negative temperature coefficient of resistance. Hence, Assertion and Reason both are true but Reason does not explain Assertion.

AIIMS Chapterwise Solutions ~ Physics CHAPTER

28 Communication System 1. The modulation in which pulse duration varies in accordance with the modulating signal is called [AIIMS 2014] (a) PAM

(b) PCM

(c) PWM

(d) PPM

2. In an amplitude modulated wave, the maximum amplitude is 10 V and the modulation index is 2/3 m, then the minimum amplitude is (in volt) [AIIMS 2014] (a) 7

(b) 9

(c) 6

(d) 2

3. To double the covering range of a TV transmitter tower, its height should be made [AIIMS 2013] (a) 2 times (c) 2 times

(b) 4 times (d) 8 times

4. What fraction of the surface area of earth can be covered to establish communication by one geostationary satellite? [AIIMS 2012] 1 2 1 (c) 4 (a)

1 3 1 (d) 8

(b)

[AIIMS 2011]

[AIIMS 2010] (b) Radio (d) Walky-talky

8. The skip zone in radio wave transmission is that range where

[AIIMS 2009]

(a) there is no reception of either ground wave or sky wave (b) the reception of ground wave is maximum, but that of sky wave is minimum (c) the reception of ground wave is minimum, but that of sky wave is maximum (d) the reception of both ground and sky wave is minimum.

9. A 1000 kHz carrier wave is modulated by an audio signal of frequency range 100 - 5000 Hz. Then, the width of the channel in kHz is [AIIMS 2008] (b) 20

(c) 30

(d) 40

works on a wavelength of 1.3 mm. The number of subscribers it can feed if a channel requires 20 kHz are [AIIMS 2008] (a) 2.3 ´1018

(b) 1.15 ´ 1010

(c) 1 ´105

(d) 2 ´ 105

11. If the highest modulating frequency of the

6. The area covered by a transmitting antenna (a) 320 p km 2 (c) 120 p km 2

(a) TV (c) Telephone

10. An optical fibre communication system

(a) ground wave propagation (b) sky wave propagation (c) space wave propagation (d) optical communication

of height 50 m is

transmission system?

(a) 10

5. The frequency range of 2 MHz to 30 MHz comes under

7. Which one of the following is a full duplex

[AIIMS 2011] (b) 640 p km 2 (d) 1440 km 2

wave is 5 kHz, then the number of stations that can be accommodated in a 150 kHz bandwidth [AIIMS 2007] (a) 15

(b) 10

(c) 5

(d) 30

277

Communication System 12. Given below is a circuit diagram of an AM demodulator.

regions can be established by space wave and tropospheric wave.

AM Signal

C

R Output

For good demodulation of AM signal of carrier frequency f , the value of RC should be 1 1 (b) RC < (a) RC = f f

1 (c) RC ³ f

[AIIMS 2006] 1 (d) RC >> f

13. For sky wave propagation of 10 MHz signal, what should be the minimum electron density in ionosphere? [AIIMS 2005] (a) ~1.2 ´ 1012 m - 3

(b) ~106 m - 3

(c) ~1014 m - 3

(d) ~1022 m- 3

14. Audio signal cannot be transmitted because (a) the signal has more noise [AIIMS 2004] (b) the signal cannot be amplified for distance communication (c) the transmitting antenna length is very small to design (d) the signal is not a radio signal

15. In short wave communication, waves of which of the following frequencies will be reflected back by the ionospheric layer having electron density 10 11 per m 3 ? [AIIMS 2003] (a) 2 MHz

17. Assertion Communication in UHF/VHF

(b) 10 MHz (c) 12 MHz (d) 18 MHz

Assertion & Reason Direction (Q. Nos. 16-23) Read the Assertion and Reason carefully to mark the correct option from those given below (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion (c) Assertion is true but Reason is false (d) Both Assertion and Reason are false

16. Assertion Transducer in communication system converts electrical signal into a physical quantity. Reason For information signal to be transmitted directly to long distances, modulation is not a necessary process. [AIIMS 2010]

Reason Communication in UHF/VHF regions is limited to line of sight distance. [AIIMS 2010]

18. Assertion In a communication system based on amplitude modulation, the modulation index is kept less than one. Reason It ensures minimum distortion of signal. [AIIMS 2009]

19. Assertion Optical fibre communication has immunity of cross-talk. Reason Optical interference between fibres is zero. [AIIMS 2007]

20. Assertion A portable AM radio set must be kept horizontal to receive the signals properly. Reason Radio waves are polarised electromagnetic waves. [AIIMS 2007]

21. Assertion Electromagnetic waves with frequencies more than the critical frequency of ionosphere cannot be used for communication using sky wave propagation. Reason The refractive index of the ionosphere becomes very high for frequencies higher than the critical frequency. [AIIMS 2006]

22. Assertion Television signals are received through sky wave propagation. Reason The ionosphere reflects electromagnetic waves of frequencies greater than a certain critical frequency. [AIIMS 2005]

23. Assertion Microwave communication is preferred over optical communication. Reason Microwaves provide large number of channels and bandwidth compared to optical signals. [2003]

278

AIIMS Chapterwise Solutions ~ Physics

Answers 1. (c) 11. (a) 21. (a)

2. (d) 12. (d) 22. (d)

3. (b) 13. (a) 23. (a)

4. (b) 14. (d)

5. (b) 15. (a)

6. (b) 16. (d)

7. (c) 17. (b)

8. (a) 18. (a)

9. (a) 19. (a)

10. (b) 20. (b)

Explanations 1. (c) PAM ® Pulse amplitude modulation PCM ® Pulse code modulation PWM ® Pulse width modulation PPM ® Pulse position modulation PWM is used as the samples of the message signal are used to vary the width or duration of the individual pulse. 2 2. (d) Given, modulation index (ma ) = and 3 E max = 10 V The modulation index is given by E - E min ma = max E max + E min 2 10 - E min = Þ 3 10 + E min Þ

E min = 2 V

3. (b) The maximum distance upto which a TV transmit signal is given by d = 2hR where, d = covering range, h = height of tower and R = radius of earth. So, d µ h To double the covering range its height should be made 4 times. 4. (b) To cover the entire earth three geostationary satellites are required to be placed at the vertices of an equilateral triangle covering 120° area by each. 120° 1 Thus, each satellite will cover = = 360° 3

5. (b) The sky waves are the radio waves of frequency between 2 MHz to 30 MHz. These waves can propagate through atmosphere but reflected by the ionosphere of the earth’s atmosphere.

6. (b) The area covered by a transmitting antenna is A = pd 2 where, d = distance upto which the signal can be transmitted and d = 2hR . Þ Here,

A = p ´ 2hR h = 50 m, R = 6.4 ´ 106 m (radius of earth)

Area covered = p ´ 2 ´ 50 ´ 6.4 ´ 106 = 640p ´106 m2 = 640p km2

7. (c) A full duplex transmission system is one in which the transmission and receiving of signal takes place in both directions simultaneously. In TV and radio only the received signals are observed. It is called simplex transmission. But in walky-talky two or more person can talk on a frequency, only one at a time. In telephone, we can transmit and receive signal simultaneously. Thus, it is based on full duplex transmission system.

8. (a) A skip zone is an annular region between the farthest point at which the ground wave can be received and the nearest point at which the refracted wave from sky can be received. With in this region, no signal can be received as there are no radio waves to receive.

9. (a) The width of the channel is equal to twice the maximum frequency of the band i.e. width of channel = 2 ´ f m (max) = 2 ´ 5000 = 10000 Hz =10 kHz

279

Communication System

16. (d) In any communication system, information

10. (b) The frequency of optical source is 3 ´ 108 c = 23 . ´ 1014 Hz f = = l 13 . ´ 10-6 Therefore, the number of subscribers are, . ´ 1014 23 . ´ 1010 = 115 n= 20 ´ 103

11. (a) The number of stations in a bandwidth, N=

Bandwidth Bandwidth / Stations

17. (b) UHF/VHF are very high frequency signal and do not reflected by ionosphere and travel in straight line.

The bandwidth per station is twice the maximum modulating frequency due to two (lower and upper) band. Thus, number of stations in 150 kHz bandwidth are 150 N= = 15 2´5

The space wave following the line of sight propagation get blocked at some point by the curvature of the earth.

12. (d) For good demodulation of AM signal, the value of RC (which is a time constant) is chosen 1 such that RC >> f where, f is the frequency of the carrier signal.

13. (a) The critical frequency of sky wave undergoing reflection from a layer of atmosphere is fc = 9 N max where, N is electron density per m3 . f 2 (10 ´106 )2 \ N max = c = 81 81 ~ - 12 . ´ 1012 m - 3 20 Hz to 20 kHz which is less than the radio signal frequency (104 to 108 Hz). So, they cannot be transmitted in this form. They have to be converted to other form and then amplified to get desired signal which can be transmitted in radio signal range.

15. (a) The perceived frequency of sky wave for reflection from an ionospheric layer is

where, n is the number density of electrons/m . Given, n = 1011 /m3 1

nc = 9 ´ (1011 )2 = 2.8 MHz » 2 MHz

18. (a) The modulation index is the ratio of amplitude of modulating signal to the amplitude of carrier signal i.e. m = Am / Ac The value of m is kept to be less than 1 as the amplitude of carrier is more than the modulating signal. This is done due to for m