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Printed at: Repro Knowledgecast Ltd., Mumbai

© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical

including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

P.O. No. 176624 TEID: 13590

A compilation of 32 years of AIPMT/NEET questions (2019-1988) Includes solved questions from NEET 2019 and Odisha NEET 2019 Includes ‘1603’ AIPMT/NEET MCQs Topic - wise and Subtopic - wise segregation of questions Year-wise flow of content beginning with the latest questions Relevant solutions provided Graphical analysis of questions – Topic - wise and Subtopic - wise

Salient Features

• TOPIC - WISE AND SUBTOPIC - WISE •

NEET PHYSICS

PREVIOUS SOLVED PAPERS PSP

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Target’s ‘NEET: Physics PSP (Previous Solved Papers)’ is a compilation of questions asked in the past 32 years (1988-2019) in the National Eligibility cum Entrance Test (NEET), formerly known as the All India Pre-Medical Test (AIPMT). The book is crafted in accordance with the Std. XI and Std. XII NCERT textbook. The book consists of topic - wise categorization of questions. Each chapter is further segregated into subtopics and thereafter all the questions pertaining to a subtopic are arranged year-wise starting with the latest year. To aid students, we have also provided hints for questions wherever deemed necessary. A graphical (% wise) analysis of the subtopics for the past 32 years as well as 7 years (2013 onwards) has been provided at the onset of every topic. Both the graphs will help the students to understand and analyse each subtopic’s distribution for AIPMT (32 years) and NEET-UG (7 Years). We are confident that this book will comprehensively cater to needs of students and effectively assist them to achieve their goal.

We welcome readers’ comments and suggestions which will enable us to refine and enrich this book further. All the best to all Aspirants!

Yours faithfully, Authors

Edition: First

Why this book?

• This book acts as a go-to tool to find all the AIPMT/NEET questions since the past 32 years at one place.

• The subtopic wise arrangement of questions provides the break-down of a chapter into its important components which will enable students to design an effective learning plan.

• The graphical analysis guides students in ascertaining their own preparation of a particular topic.

Why the need for two graphs?

Admission for undergraduate and post graduate medical courses underwent a critical change with the introduction of NEET in 2013. Although it received a huge backlash and was criticised for the following two years, NEET went on to replace AIPMT in 2016. The introduction of NEET brought in a few structural differences in terms of how the exam was conducted. Although the syllabus has majorly remained the same, the chances of asking a question from a particular subtopic is seen to vary slightly with the inception of NEET. The two graphs will fundamentally help the students to understand that the (weightage) distribution of a particular topic can vary i.e., a particular subtopic having the most weightage for AIPMT may not necessarily be the subtopic with the most weightage for NEET.

How are the two graphs beneficial to the students?

• The two graphs provide a subtopic’s weightage distribution over the past 32 years (for AIPMT) and over the past 7 years (for NEET-UG).

• The students can use these graphs as a self-evaluation tool by analyzing and comparing a particular subtopic’s weightage with their preparation of the subtopic. This exercise would help the students to get a clear picture about their strength and weakness based on the subtopics.

• Students can also use the graphs as a source to know the most important as well as least important subtopics as per weightage of a particular topic which will further help them in planning the study structure of a particular chapter.

(Note: The percentage-wise weightage analysis of subtopics is solely for the knowledge of students and does not guarantee questions from subtopics having the most weightage, in the future exams.

Question classification of a subtopic is done as per the authors’ discretion and may vary with respect to another individual.)

Disclaimer Utmost care has been taken in compiling and checking the information to ensure that the content is useful and accurate. However, the publisher and the authors shall not be responsible for any loss or damages caused to any person on account of errors or omission which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors.

Frequently Asked Questions

PREFACE

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Index

02 Kinematics

09 Kinetic Theory of gases

04 Work, energy and power

05 System of Particles and Rotational Motion

06 Gravitation

07 Properties of Matter

08 Thermodynamics

10 Oscillations

11 Wave Mechanics

12 Electrostatics

13 Capacitors

14 Current Electricity

15 Magnetic Effect of Electric Current

17 Electromagnetic Induction and Alternating Current

19 Ray Optics

20 Interference and Diffraction

21 Dual Nature of Matter and Radiation

22 Atoms and Nuclei

100

01 Physical World and measurement

03 Laws of Motion

16 Magnetism

18 Electromagnetic Waves

No. Topic Name Page No.

109

115

127

140

154

160

183

200

206

222

226

242

249

263

282

1

9

27

45

57

75

86

23 Electronic Devices

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Topic-wise Weightage Analysis of past 7 Years (2013 Onwards)

Number of Questions 0 5 10 15 20 25 30 35 40 45

8

21

30

18

27

18

18

15

12

18

16

8

25

21

8

23

10

24

21

19

21

24

Physical World and measurement

Kinematics

Laws of Motion

Work, energy and power

System of Particles and Rotational Motion

Gravitation

Properties of Matter Thermodynamics

Kinetic Theory of gases

Oscillations

Wave Mechanics

Electrostatics

Capacitors

Current Electricity

Magnetic Effect of Electric Current Magnetism

Electromagnetic Induction and Alternating Current

Electromagnetic Waves Ray Optics

Interference and Diffraction

Dual Nature of Matter and Radiation

Atoms and Nuclei

Electronic Devices

40

Total No. of Questions: 445

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75

2199

1. The time period of a geostationary satellite is

24 h, at a height 6RE (RE is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 RE from surface will be, [Odisha 2019]

(A) 122.5 h (B) 6√2 h

(C) 12√2 h (D) 242.5 h

2. The kinetic energies of a planet in an elliptical

orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then [2018]

(A) KA < KB < KC (B) KA > KB > KC (C) KB < KA < KC (D) KB > KA > KC 3. A geostationary satellite is orbiting the earth at

a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is

[2012]

(A) 6√2 (B) 6

√2

(C) 5 (D) 10 4. A planet moving along an elliptical orbit is

closest to the sun at a distance r1 and farthest away at a distance of r2. If v1 and v2 are the linear velocities at these points respectively, then the ratio

v1 v2

is: [2011]

(A) (r1/r2)2 (B) r2/r1 (C) (r2/r1)2 (D) r1/r2

S A

B

C

6.1 Kepler’s laws of planetary motion

Gravitation 6

32 Years NEET/AIPMT Analysis (Percentage-wise weightage of sub-topics)

7 Years NEET Analysis (2013 Onwards) (Percentage-wise weightage of sub-topics)

6.1 Kepler’s laws of planetary motion

6.2 Universal law of gravitation, universal gravitational constant (G)

6.3 Acceleration due to gravity, acceleration due to gravity below and above the Earth’s surface

6.4 Gravitational potential and gravitational potential energy

6.5 Escape velocity

6.6 Earth satellites and orbital velocity of satellite

6.7 Energy of orbiting satellite

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NEET: Physics PSP

5. The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t1 is the time for the planet to move from C to D and t2 is the time to move from A to B then [2009]

(A) t1 = 4t2 (B) t1 = 2t2 (C) t1 = t2 (D) t1 > t2 6. The period of revolution of planet A around

the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun? [1997]

(A) 4 (B) 5 (C) 2 (D) 3 7. The distance of two planets from the sun are

1013 m and 1012 m respectively. The ratio of time periods of the planets is [1994, 1988]

(A) √10 (B) 10√10 (C) 10 (D) 1√10 8. The largest and the shortest distance of the

earth from the sun are r1 and r2. Its distance from the sun when it is at perpendicular to the major – axis of the orbit drawn from the sun is

[1988]

(A) r1+ r2

4 (B)

r1+ r2r1– r2

(C) 2r1r2r1+ r2

(D) r1+ r2

3 9. Two astronauts are floating in gravitational

free space after having lost contact with their spaceship. The two will: [2017]

(A) keep floating at the same distance between them.

(B) move towards each other. (C) move away from each other. (D) will become stationary. 10. Kepler’s third law states that square of period

of revolution (T) of a planet around the sun, is proportional to third power of average

distance r between sun and planet i.e., T2 = Kr3 here K is constant. If the masses of sun and planet are M and m respectively then as per Newton’s law of gravitation force

of attraction between them is F = GMm

r2 , here G is gravitational constant. The relation between G and K is described as [2015]

(A) GK = 4π2 (B) GMK = 4π2

(C) K = G (D) K = 1G

11. Which one of the following plots represents

the variation of gravitational field on a particle with distance r due to a thin spherical shell of radius R? (r is measured from the centre of the spherical shell) [2012]

(A) (B) (C) (D) 12. Two spheres of masses m and M are situated

in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be [2003]

(A) 3 F (B) F

(C) F3 (D)

F9

13. A body of weight 72 N moves from the surface of earth at a height half of the radius of earth, then gravitational force exerted on it will be [2000]

(A) 36 N (B) 32 N (C) 144 N (D) 50 N 14. Gravitational force is required for [2000] (A) stirring of liquid (B) convection (C) conduction (D) radiation 15. The earth (mass = 6 × 1024

kg) revolves round the sun with angular velocity 2 × 10−7 rad/s in a circular orbit of radius 1.5 × 108 km. The force exerted by the sun on the earth in newtons, is [1995]

(A) 18 × 1029 (B) 36 × 1028 (C) 27 × 1028 (D) 36 × 1021

F

r R O

F

r R O

F

r R O

F

r R O

dA

B m

D

C

S A

v

6.2 Universal law of gravitation, universal gravitational constant (G)

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Chapter 6: Gravitation

16. Two particles of equal mass go around a circle of radius R under the action of their mutual gravitational force of attraction. The speed v of each particle is (M = mass of the particle)

[1995]

(A) �GMR

(B) 12 �GM

R

(C) 13 �GM

R (D)

14 �GM

R

17. A body weighs 200 N on the surface of the

earth. How much will it weigh half way down to the centre of the earth? [2019]

(A) 250 N (B) 100 N (C) 150 N (D) 200 N 18. If the mass of the Sun were ten times smaller

and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct? [2018]

(A) Raindrops will fall faster. (B) Walking on the ground would become

more difficult. (C) Time period of a simple pendulum on

the Earth would decrease. (D) ‘g’ on the Earth will not change. 19. The acceleration due to gravity at a height 1 km

above the earth is the same as at a depth d below the surface of earth. Then: [2017]

(A) d = 12 km (B) d = 1 km

(C) d = 32 km (D) d = 2 km

20. Starting from the centre of the earth having

radius R, the variation of g (acceleration due to gravity) is shown by [Phase-II 2016]

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

21. A spherical planet has a mass Mp and diameter Dp. A particle of mass m falling freely near the surface of this planet will experience acceleration due to gravity, equal to [2012]

(A) 4 GMP

DP2 (B)

GMPmDp

2

(C) GMPDp

2 (D) 4GMpm

Dp2

22. The height at which the weight of a body

becomes 1th

16, its weight on the surface of earth

(radius R), is [2012] (A) 3R (B) 4R (C) 5R (D) 15R 23. The dependence of acceleration due to gravity

g on the distance r from the centre of the earth, assumed to be a sphere of radius R of uniform density is as shown in figure below [2010]

(1) (2) (3) (4) The correct figure is (A) (4) (B) (1) (C) (2) (D) (3) 24. Imagine a new planet having the same density

as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g', then [2005]

(A) g' = g/9 (B) g'= 27g (C) g' = 9g (D) g' = 3g 25. The density of a newly discovered planet is

twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be [2004]

(A) 2R (B) 4R

(C) 14 R (D)

12 R

6.3 Acceleration due to gravity, acceleration due to gravity below and above the Earth’s surface

R r

g

O

R r

g

O

R r

g

O

R r

g

O

R r

g

R r

g

R r

g

R r

g

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NEET: Physics PSP

26. The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2 m on the surface of A. What is the height of jump by the same person on the planet B? [2003]

(A) �29�m (B) 18 m

(C) 6 m (D) �23�m

27. What will be the formula of mass of the earth

in terms of g, R and G? [1996]

(A) G Rg

(B) g R2

G

(C) g2 RG

(D) G gR

28. The acceleration due to gravity g and mean

density of the earth ρ are related by which of the following relations? (where G is the gravitational constant and R is the radius of the earth.) [1995]

(A) ρ = 3g

4π GR (B) ρ =

3g4π GR3

(C) ρ = 4π gR2

3G (D) ρ =

4π gR3 3G

29. The radius of earth is about 6400 km and that

of mars is 3200 km. The mass of the earth is about 10 times mass of mars. An object weighs 200 N on the surface of earth. Its weight on the surface of mars will be [1994]

(A) 20 N (B) 8 N (C) 80 N (D) 40 N. 30. Assuming that the gravitational potential

energy of an object at infinity is zero, the change in potential energy (final – initial) of an object of mass m, when taken to a height h from the surface of earth (of radius R), is given by, [Odisha 2019]

(A) GMmR + h

(B) − GMmR + h

(C) GMmh

R(R +h) (D) mgh 31. The work done to raise a mass m from the

surface of the earth to a height h, which is equal to the radius of the earth, is : [2019]

(A) 12 mgR (B)

32 mgR

(C) mgR (D) 2mgR

32. At what height from the surface of earth the gravitation potential and the value of g are −5.4 × 107 J kg−2 and 6.0 ms−2 respectively? Take the radius of earth as 6400 km

[Phase-I 2016] (A) 1400 km (B) 2000 km (C) 2600 km (D) 1600 km 33. Infinite number of bodies, each of mass 2 kg, are

situated on X-axis at distance 1 m, 2 m, 4 m, 8 m,..... respectively, from the origin. The resulting gravitational potential due to this system at the origin will be [2013]

(A) − G (B) – 83 G

(C) – 43 G (D) − 4G

34. A body of mass 'm' is taken from the earth's

surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be [2013]

(A) mg2R (B) 23 mgR

(C) 3mgR (D) 13 mgR

35. A particle of mass M is situated at the centre

of a spherical shell of same mass and radius a. The gravitational potential at a point situated

at a2 distance from the centre, will be

[2011, 2010]

(A) − 3GMa (B) − 2 GM

a

(C) − GMa (D) − 4GM

a 36. A body of mass m is placed on the earth’s

surface. It is taken from the earth’s surface to a height h = 3R. The change in gravitational potential energy of the body is [2003]

(A) 34 mgR (B)

23

mgR

(C) mgR

2 (D)

mgR4

37. The ratio of escape velocity at earth (ve) to the

escape velocity at a planet (vp) whose radius and mean density are twice as that of earth is

[Phase-I 2016] (A) 1 : 4 (B) 1 : 2 (C) 1 : 2 (D) 1 : 2 2

6.4 Gravitational potential and gravitational potential energy

6.5 Escape velocity

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38. A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = 5.98 × 1024 kg) have to be compressed to be a black hole? [2014]

(A) 10−9 m (B) 10−6 m (C) 10−2 m (D) 100 m 39. If ve is escape velocity and v0 is orbital

velocity of a satellite for orbit close to the earth’s surface, then these are related by

[2012] (A) v0 = √2 ve (B) v0 = ve (C) ve = �2v0 (D) ve = √2 v0

40. A particle of mass m is thrown upwards from

the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is [2011]

(A) �2GMR2 (B) �2GM

R

(C) �2gMR2 (D) �2gR2

41. The earth is assumed to be a sphere of radius

R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is [2006]

(A) 1/2 (B) √2 (C) 1/√2 (D) 1/3 42. With what velocity should a particle be

projected so that its height becomes equal to radius of earth? [2001]

(A) �GMR

�1/2

(B) �8GMR

�1/2

(C) �2GMR

�1/2

(D) �4GMR

�1/2

43. For a planet having mass equal to mass of the

earth but radius is one fourth of radius of the earth. Then escape velocity for this planet will be [2000]

(A) 11.2 km/s (B) 22.4 km/s (C) 5.6 km/s (D) 44.8 km/s 44. The escape velocity of a sphere of mass m is

given by (G = Universal gravitational constant; Me = Mass of the earth and Re = Radius of the earth ) [1999]

(A) �2GM emRe

(B) �2GM e

Re

(C) �GM e

Re (D) �2GM e + Re

Re

45. The escape velocity of a body on the surface

of the earth is 11.2 km/s. If the earth’s mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes [1997]

(A) 22.4 km/s (B) 44.8 km/s (C) 5.6 km/s (D) 11.2 km/s 46. The escape velocity from earth is 11.2 km/s. If

a body is to be projected in a direction making an angle 45° to the vertical, then the escape velocity is [1993]

(A) 11.2 × 2 km/s (B) 11.2 km/s (C) 11.2 /√2 km/s (D) 11.2√2 km/s 47. For a satellite escape velocity is 11 km/s. If

the satellite is launched at an angle of 60° with the vertical, then escape velocity will be

[1989] (A) 11 km/s (B) 11 √3 km/s

(C) 11√3

km/s (D) 33 km/s 48. A remote – sensing satellite of earth revolves

in a circular orbit at a height of 0.25 × 106 m above the surface of earth. If earth’s radius is 6.38 × 106 m and g = 9.8 ms–2, then the orbital speed of the satellite is [Re-Test 2015]

(A) 6.67 km s–1 (B) 7.76 km s–1

(C) 8.56 km s–1 (D) 9.13 km s–1 49. A satellite S is moving in an elliptical orbit

around the earth. The mass of the satellite is very small compared to the mass of the earth. Then, [Re-Test 2015]

(A) the acceleration of S is always directed towards the centre of the earth.

(B) the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant.

(C) the total mechanical energy of S varies periodically with time.

(D) the linear momentum of S remains constant in magnitude.

50. The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A is 3V, then the speed of satellite B will be [2010]

(A) 3V4

(B) 6V (C) 12V (D) 3V2

6.6 Earth satellites and orbital velocity of satellite

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NEET: Physics PSP

51. If the gravitational force between two objects

were proportional to 1R

(and not ∝ 1

R2 ), where R is the separation between them, then a particle in circular orbit under such a force would have its orbital speed ‘v’ proportional to

[1994, 1989]

(A) 1

R2 (B) R0 (C) R1 (D) 1R

52. A satellite A of mass m is at a distance of r

from the surface of the earth. Another satellite B of mass 2m is at a distance of 2r from the earth’s centre. Their time periods are in the ratio of [1993]

(A) 1 : 2 (B) 1 : 16 (C) 1 : 32 (D) 1 : 2√2 53. The mean radius of earth is R, its angular

speed on its own axis is ω and the acceleration due to gravity at earth’s surface is g. What will be the radius of the orbit of a geostationary satellite? [1992]

(A) (R2g/ ω2) 1/3 (B) (Rg/ ω2) 1/3 (C) (R2ω2 / g) 1/3 (D) (R2g/ω) 1/3 54. A satellite of mass m is orbiting the earth (of

radius R) at a height h from its surface. the total energy of the satellite in terms of g0, the value of acceleration due to gravity at the earth’s surface, is [Phase-II 2016]

(A) – 2mg0R2

R + h (B)

mg0R2

2(R + h)

(C) − mg0R2

2(R + h) (D)

2mg0R2

R + h

55. The additional kinetic energy to be provided to

a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R1 to another of radius R2 (R2 > R1) is

[2010]

(A) GmM � 1R1

2 – 1R2

2�

(B) GmM � 1R1

– 1R2

�

(C) 2GmM � 1R1

– 1R2

�

(D) 12 GmM � 1

R1 – 1

R2�

56. Two satellites of earth, S1 and S2 are moving

in the same orbit. The mass of S1 is four times the mass of S2. Which one of the following statements is true? [2007]

(A) The potential energies of earth and satellite in the two cases are equal.

(B) S1 and S2 are moving with the same speed.

(C) The kinetic energies of the two satellites are equal.

(D) The time period of S1 is four times that of S2.

57. For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is [2005]

(A) 12 (B)

1√2

(C) 2 (D) √2 58. The satellite of mass m is orbiting around the

earth in a circular orbit with a velocity v. What will be its total energy? [1991]

(A) (3/4) mv2 (B) (1/2) mv2 (C) mv2 (D) –(1/2) mv2 59. A planet is moving in an elliptical orbit. If T,

V, E and L stand for kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of gravity respectively, then which of the following statements is CORRECT?

[1990] (A) T is conserved. (B) V is always positive. (C) E is always negative. (D) L is conserved but direction of vector L

changes continuously.

1. (B) 2. (B) 3. (A) 4. (B) 5. (B) 6. (A) 7. (B) 8. (C) 9. (B) 10. (B) 11. (B) 12. (B) 13. (B) 14. (B) 15. (D) 16. (B) 17. (B) 18. (D) 19. (D) 20. (C) 21. (A) 22. (A) 23. (A) 24. (D) 25. (D) 26. (B) 27. (B) 28. (A) 29. (C) 30. (C) 31. (A) 32. (C) 33. (D) 34. (B) 35. (A) 36. (A) 37. (D) 38. (C) 39. (D) 40. (B) 41. (C) 42. (A) 43. (B) 44. (B) 45. (A) 46. (B) 47. (A) 48. (B) 49. (A) 50. (B) 51. (B) 52. (D) 53. (A) 54. (C) 55. (D) 56. (B) 57. (A) 58. (D) 59. (C)

6.7 Energy of orbiting satellite

Answers to MCQ's

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Chapter 6: Gravitation

1. By Kepler’s third law, T2 ∝ r3

∴ T2 ∝ (RE + h)3

∴ T2T1

= �RE+ h2RE+ h1

�32

∴ T2T1

= �RE+ 2.5RERE+ 6RE

�32

∴ T224

= 1

2√2

∴ T2 = 6√2 h 2. Point A indicates perihelion position while

point C represents aphelion position. This means point A is closest to the sun

followed by point B and C. Hence, vA > vB > vC ∴ KA > KB > KC 3. From Kepler's third law, T2 ∝ r3

∴ T2T1

= �r2r1

�3/2

Here, r1 = R + h1 = R + 5R = 6R and r2 = R + h2 = R + 2R = 3R

∴ T2 = T1 �3R6R

�3/2

= 24 × �12�

3/2

= 24√8

= 6√2 hrs. 4. According to principle of angular momentum

conservation, L1 = L2 mr1v1 = mr2v2 ∴ r1v1 = r2v2

∴ v1 v2

= r2r1

5. According to Kepler’s law of area of planetary

areal velocity is constant.

∴ dAdt

= constant

∴ A1t1

= A2t2

Now, A1 = 2A2

∴ 2A2t1

= A2t2

∴ t1 = 2t2

6. From Kepler’s 3rd law of planetary motion,

T2 ∝ R3. (TA) = 8TB.

∴ �rArB

�3= �TA

TB�

2 = �8TB

TB�

2= 64

∴ rArB

= 4

∴ rA = 4rB. 7. From Kepler’s 3rd law of planetary motion,

T2 ∝ r3

∴ T ∝ r3/2.

∴ T1T2

= �r1r2

�3/2

= �1013

1012�3/2

= 103/2 = 10√10 . 8. Applying the properties of ellipse, we have

2R

= 1r1

+ 1r2

= r1 + r2

r1r2

R = 2r1r2r1+ r2

9. Under mutual gravitational force, astronauts

move towards each other with very small acceleration.

10. FCP = FG

mv2

r = GMm

r2

v = �GMr

T = 2πrv

T2 = 4π2r3

GM

T2 = Kr3

K = 4π2

GM GMK = 4π2

Hints to MCQ's

Sun

Instant position of satellite

R major axis

r2 r1

SAMPLE C

ONTENT

82

82

NEET: Physics PSP

11. For a thin spherical shell: For r > R

F = GMr2

For r = R

F = GMR2

For r < R F = 0 12. Gravitational force does not depend on the

medium between the masses. 13. Fsurface = G Mm

Re2

At height h =

Re2

,

Fh = Re/2 = G Mm

(Re + Re / 2 )2 = G Mm

�32Re�

2

= 49

× Fsurface = 49

× 72 = 32 N 15. The force exerted by the sun on the earth, F = Mω2R = 6 × 1024 × (2 × 10−7)2 × 1.5 × 1011

∴ F = 36 × 1021 N

16. GM2

(2R)2 = Mv2

R or

GM4R

= v2

∴ v = 12 �GM

R

17. Acceleration due to gravity at depth d,

gd = g �1 – dR� = g �1– 1

2� �∵ d = R2�

∴ gd = g2

Weight of the body at depth d = R/2,

Wd = mgd = m × g/2 = 12 × 200

∴ Wd = 100 N 18. Gravitational acceleration of earth, g =

GMR2

Where, M is mass of the earth. As g is independent of mass of the Sun,

increase in G will increase value of g. Hence, statement (D) is incorrect.

Also terminal velocity of raindrop depends on g therefore increase in g will cause raindrops to fall faster.

Hence, statement (A) is correct. Increased value of g will make walking on

ground more difficult. Hence, statement (B) is correct.

Time period of simple pendulum will decrease

as T ∝ 1

�g . Hence, statement (C) is correct.

19. Given: gd = gh ….(i) But, gd = g�1– d

R� and

gh = g�1– 2hR �

∴ g�1– d

R� = g �1–2hR �

….[From (i)]

∴ d = 2h ∴ d = 2 × 1 ….( h = 1 km)

∴ d = 2 km 20. Acceleration due to gravity is given by

g = 43 πρGr ; r ≤ R

g = 43

πρR3Gr2 ; r > R

21. F =

GMP m

�Dp / 2�2

∴ g = Fm

= GMP

�Dp / 2�2 =

4GMPDp

2 22.

WhW

= mghmg

∴ gh = g16

As gh =gR2

(R + h)2 ,

Comparing, � RR + h�

2

= 116

⇒ R

R + h =

14

∴ R + h = 4R ∴ h = 3R 23. Acceleration due to gravity at a depth d below

surface of earth is

g′ = GMR2 �1– d

R� = g �1– dR�

at the depth d, distance of point from centre of

the earth is (R − d) i.e., r = R − d In this case, gd ∝ R − d ∴ gd ∝ r At height h distance from centre of the earth is

(R + h) i.e., r = R + h In this case,

gh = g� RR + h�

2 =

gR2

r2

⇒ gh ∝ 1r2

24. g =

GMr2 =

Gr2 �4

3× πr3ρ� =

43 × πρGr

∴ g′

g =

3rr

⇒ g′ = 3g.

F

r r = R O

R r

g

O

r = R X

g →

gd ∝ r

gh ∝ 1r2

r →

Y

SAMPLE C

ONTENT

83

Chapter 6: Gravitation

25. ρp = 2ρe, gp = ge

g = 43 πρGR

∴ Rp

Re = �

gp

ge� �ρe

ρp� = (1) × �1

2�

∴ R p=

Re2

= R2

26. given, ga = 9 × gb

∴ gb = ga9

on planet A: v2 = u2 – 2gaha

∴ ha = u1

2

2ga = 2 m

on planet B:

hb = u1

2

2ga =

u12

2�ga9 �

= 9 × u1

2

2ga = 9 × ha

∴ hb = 9 × 2 = 18 m 27. g =

GMeR2

∴ Me = gR2

G

28. g = G ×

MR2

∴ g = G (4/3) πR3 × ρ

R2 = G × 43 πR × ρ

∴ ρ = 3g

4π GR

29. Radius of earth (Re) = 6400 km; Radius of

mars(Rm) = 3200 km; Mass of earth (Me) = 10 Mm and weight of the object on earth (We) = 200 N.

WmWe

= mgmmge

= MmMe

× � ReRm

� = 110

× (2)2 = 25

or Wm = We × 25 = 200 × 0.4 = 80 N.

30. Potential energy of object of mass m on the

surface of earth, P.E. = –GMmR

Potential energy of object of mass m at a height h from the surface of the earth,

P.E.′ = –GMmR + h

∴ Change in potential energy = P.E.′ − P.E.

= –GMmR + h + GMm

R

= GMm hR(R+h)

31. Initial potential energy on earth’s surface,

Ui = –GMm

R

Final potential energy at height h = R

Uf = –GMm

2R

Work done,

W = Uf – Ui = –GMm

2R – �–GMm

R �

= GMm

R –

GMm2R

= GMm

2R =

�gR2�m2R

∴ W = 12 mgR

32. V = − GM

(R + h) and g =

GM(R + h)2

Taking ratio of both,

⇒ |V|g

= R + h

∴ 5.4 ×107

6.0 = R + h

∴ 9 × 106 = R + h ∴ h = (9 – 6.4) × 106 = 2.6 × 106 = 2600 km

33. Gravitational potential is given as, V = –GM

R ∴ V = – GM �1

1+ 1

2+ 1

4+ 1

8+…+ ∞�

For M = 2 kg,

V = − G × 2�1+ 12

+ 122 + 1

23 +…+ ∞�

= − 2G 1

1 – 12

(∵ For an infinite GP, sum = 1th term1 – Common Ratio

)

∴ V = − 4G 34. The change in potential energy is given as,

∆U = Uf − Ui = –GMmR + 2R

– –GMm

R

= GMm

R �1 – 1

3� = 23 GMm

R

= 23

GMm × RR2 =

23 �GM

R2 � mR

∴ ∆U = 23 mgR

35. Point B is at a distance a2 from the centre of the shell

as shown in figure. Gravitational potential at point B due to particle at O is

V1 = − GM(a / 2)

a

B a/2

O

SAMPLE C

ONTENT

84

84

NEET: Physics PSP

Gravitational potential at point B due to spherical shell is

V2 = − GMa

Hence, total gravitational potential at the point B,

V = − GM(a / 2)

+ �− GMa � = − 2GM

a− GM

a

= − 3GM

a 36. ∆U =

mgh

�1 + hR�

= mg ×3R

�1+ 3RR �

= 34 mgR

37. Escape velocity, ve = R�8

3 πGρ

⇒ vevp

= R�ρ

Rp�ρP

Given: Rp = 2R and ρp = 2ρ

∴ vevp

= 1

2√2

38. ve = �2GMR

= c

⇒ R = 2GM

c2 = 2 × 6.67 × 10–11 × 5.98 × 1024

�3 × 108�2

= 2 × 6.67 × 5.98

9 ×10–3 m

= 8.86 × 10−3 m ≈ 10−2 m

39. Escape velocity , ve = �2 GMR

….(i)

Where M and R be the mass and radius of the earth respectively.

The orbital velocity of a satellite close to the earth’s surface is

v0 = � GMR

....(ii)

From (i) and (ii), we get ve = √2v0 40. For the particle to not return back to earth, u = escape velocity = �2gR

∴ u = �2GM

R2 × R [∵ g = GM R2 ]

∴ u = �2GMR

41. At a platform at a height h, escape velocity = binding energy of sphere

∴ 12 mve′

2 = GMmR +h

∴ ve' = �2GMR + h

= �2GM2R

(∵ h = R)

At surface of earth, ve = �2GMR

given, ve' = f ve

∴ �2GM2R

= f �2GMR

∴ 1

2R = f2

R

∴ f = 1

√2

42. v2 =

2gh

1+ hR

given h = R

∴ v = �2gR

1+ �RR�

= �2× (GM)

R2 ×R

2 = �GM

R

43. ve = �2gR =�2GMR

given, Rp = 14 Re

∴ vescape (p) = 2 vescape (e) = 2 × 11.2 = 22.4 km/s 45. ve = �2GM e

Re ∴ ve ∝ �Me

Re

∴ vev′e

=�Me

Re × 0.5 Re

2Me = �1

4 =

12

∴ v′e = 2ve = 22.4 km/s.

46. ve = �2GM

R = �2gR

∴ Escape velocity does not depend on the angle of projection, so it will not change.

47. ve = �2GM

R = �2gRe

∴

Escape velocity is independent of angle of projection, so it will not change.

48. GMm

r2 = mv2

r also, r = R + h

∴ v = �GMR

= �GMR2

R2r = �g

r R

∴ v = ��

9.80.25 × 106+ 6.38 ×106� × 6.38 × 106

= �1.47106 × 6.38 × 106

= 7.76 × 103 m/s = 7.76 km/s 49. Force on satellite is only gravitational force,

which will always be towards the centre of earth.

SAMPLE C

ONTENT

85

Chapter 6: Gravitation

50. Orbital speed of the satellite around the earth

is v = �GMr

⇒ v ∝ �1r

∴ vBvA

= �rArB

∴ vB = vA�rArB

= 3V�4RR

⇒ vB = 6V

51. F ∝1R

⇒ F =kR

∴ FCP = F mv2

R=

kR

∴ v2 = km

⇒ v = �km

This shows v is independent of R. i.e., Speed is proportional to R0

52. T2 ∝ r3

TATB

= r3/2

23/2r3/2 = 1 : 2 √2

Time period is independent of mass.

53. Centripetal force = gravitational forcemv2

r =

GMmr2

∴ m(rω)2

r =

GMmr2 (∵ v = rω)

∴ r3 = GMω2

∴ r3 = gRM

2×M

ω2 (∵ G =gRM

2).

∴ r = �gRω2

2�

1/3

54. Total energy of a satellite is,

T.E. = –GMm

2(R + h) ….(i)

∴ Multiplying and dividing the equation (i) by R2.

T.E. = – GMmR2

2 (R + h)R2

∴ T.E. = –g0mR2

2(R + h)....( g0 =

GMR2 )

55. ∆K.E. = (K.E.)2 − (K.E.)1

∴ ∆K.E. = �– GMm2R2

� – �– GMm2R1

� = GMm

2� 1

R1– 1

R2�

56. P.E. of satellite u = – GMm

r.

∴ u ∝ mgiven, Ms1 = 4 Ms2

∴ US1US2

= 4MS1MS2

∴ US1= 4US2 ∴ option (a) is wrong.

Orbital speed of satellite, v = �GMr

∴ v is independent of mass of satellite ∴ option (b) is correct.

K.E. of satellite = GM m

2r

∴ K.E. ∝ m , given s1 = 4s2

∴ K.E.S1K.E.S2

= 4MS1MS2

∴ K.E.S1 = 4 K.E.S2. ∴ option (c) is wrong,

Time period of satellite = �4π2r3

GMe

∴ T is independent of mass of satellite. ∴ option (d) is wrong. 57. K. E =

12 mv0

2

= 12 m

GMr

= GMm

2r(∵ v =�GM

r)

P.E = –GMm

r

∴K.EP.E

=GMm

2r ×

rGMm

[ignoring (–) sign]

∴ K.EP.E

= 12

58. Centripetal force = gravitational force.mv2

r =

GMmr2

∴ v2 = GM

r

Now K.E. = 12 mv2 =

12 m

GMr

P.E. = – GMm

r = – mv2

∴ Total energy = K.E + P.E = 12 mv2 – mv2

= –12

mv2

59. In elliptical orbits, linear speed changes fromposition to position. Hence T is not conserved.V is always negative (attraction).Angular momentum is conserved, both inmagnitude and direction.Total energy is always negative. Hence,correct statement is C.

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