8.

GRAVITATION

Single Correct Answer Type

Figure shows a planet in an elliptical orbit around the Sun S. Where is the kinetic energy
of the planet maximum?
1.

a) P b) P c) P d) P

The distance between the centre of the Moon and the earth is D . The mass of the earth is
1 2 3 4

81 times the mass of the Moon. At what distance from the centre of the earth, the
2.

gravitational force will be zero?

a) D b) 2 D c) 4 D d) 9 D
2 3 3 10
3. Suppose, the acceleration due to gravity at the earth''s surface is 10 m s−2 and at the
surface of Mars is 4.0 ms−2. A 60 kg passenger goes from the earth of Mars in a spaceship
moving with a constant velocity. Neglect all other objects in the sky. Which part of the
figure best represents the weight (net gravitational force) of the passenger as a function
of time?

a) A b) B c) C d) D

The charge in the value of g at a height 'h ' above the surface of earth is the same as at a
depth 'd ' below the earth. When both d and h are much smaller than the radius of earth,
4.

then which one of the following is correct?

a) d= h b) d= 3 h c) d=2h d) d=h
2 2
5. The gravitational force between two objects is proportional to 1/ R (and not as 1/ R2)
where R is separation between them, then a particle in circular orbit under such a force
would have its orbital speed v proportional to
a) 1 b) 0 c) 1 d) 1
2 R R
R
A skylab of mass m kg is first launched from the surface of the earth in a circular orbit of
R

radius 2 R (from the centre of the earth) and then it is shifted from this circular orbit to
6.

another circular orbit of radius 3 R. The minimum energy required to place the lab in the
first orbit and to shift the lab from first orbit to the second orbit are
a) 3 m g R , m g R b) 3 m g R , 3 g R c) m g R , m g R d) 2 m g R , m g R
4 6 4 12
7. A tunnel is dug along a diameter of the earth. If M e and Re are the mass and radius,
respectively, of the earth, then the force on a particle of mass m placed in the tunnel at a
distance r from the centre is

Page|1
 a) G M e m r b) G M e m d) G M e m r
3
c) G M e m Re
3 3 2
Re Re r r Re
A comet is in highly elliptical orbit around the Sun. The period of the comet's orbit is 90
days. Some statements are given are given regarding the collision between the comet
8.

and the earth. Mark the correct statement. [Mass of the Sun¿ 2 ×1030 kg, mean distance
between the earth and the Sun¿ 1.5 ×1011 m]
a) Collision is there b) Collision is not possible

c) Collision may or may not be there d) Enough information is not given

9. Two planets have radii r 1 and r 2 and their densities are ρ1and ρ2 respectively. The ratio of
acceleration due to gravity on them will be
a) r ρ :r ρ b) r ρ2 : r ρ2 c) r 2 ρ : r 2 ρ d) r ρ : r ρ

10. A ball of mass m is fired vertically upwards from the surface of the earth with velocity
1 1 2 2 1 1 2 2 1 1 2 2 1 2 2 1

n v e, where v e is the escape velocity and n>1 . To what height will the ball rise? Neglecting
air resistance, take radius of the earth as R
2
a) R R b) c) R n d) 2
2 2 2 Rn
n (1−n ) (1−n )
11. Two bodies of masses M 1 and M 2 are placed at a distance R apart. Then at the position
where the gravitational field due to them is zero, the gravitational potential is
a)
−G
√M1 b)
−G
√M2 c) − M + M 2 G
( √ 1 √ 2) d) −
( √ M 1 +√ M 2 )
2 G
R R R R
12. Two satellite of masses m 1 and m 2 (m1 >m2) are revolving around the earth in a circular
orbit of radii r 1 and r 2 (r 1 >r 2 ), respectively. Which of the following statements is true
regarding their speeds v 1 and v 2?
a) v =v b) v > v c) v < v d) v 1 = v 2
1 2 1 2 1 2
r1 v2
13. A satellite is seen after each 8 h over the equator at a place on the earth when its sense
of rotation is opposite to the earth. The time interval after which it can be seen at the
same place when the sense of rotation of the earth and the satellite is the same will be
a) 8 h b) 12 h c) 24 h d) 6 h

14. A body starts from rest from a point distant r 0 from the centre of the earth. It reaches the
surface of the earth whose radius is R . The velocity acquired by the body is
a) 2 GM
√ 1 1
−
R r0 √
b) 2GM 1 − 1
(R r ) c) GM
√ 1 1
−
R r0
15. A rocket is launched vertically from the surface of earth with an initial velocity v . How
0 √
d) GM 1 − 1
(R R ) 0

far above the surface of earth it will go? Neglect the air resistance

( ) ( ) ( ) ( )
−1 /2 −1 2
a) R 2 g R −1 b) R 2 g R −1 c) R 2 g R −1 d) R 2 g R −1
2 2 2 2
v v v v
16. In problem 86, what is the gravitational field strength at the location of m ?

a) GM b) 4 GM c) 4 GM d) GM
2 2 2 2

17. A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and
l l 3l 3l

radius 10 cm. Find the work done against the gravitational force between them to take
the particle far away from the sphere

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 a) −10 b) −10 c) −9 d) −10
13.34 × 10 J 3.33 ×10 J 6.67 ×10 J 6.67 ×10 J
18. A body is fired with a velocity of magnitude √ g R <V < √ 2 g R at an angle of 30 ° with the
radius vector of the earth. If at the highest point, the speed of the body is V / 4, the
maximum height attained by the body is equal to
a) V
2
b) c) d) None of these
R √2 R
19. Two particles of equal mass go around a circle of radius R under the action of their
8g

mutual gravitational attraction. The speed of each particle is

a)
v=
1
2R (√ GM1 ) b)
v= (√ GM
2R )
c)
v=
1
2 (√ GMR ) d)
v= (√ 4 GM
R )
20. Imagine that you are in a spacecraft orbiting around the earth in a circle of radius 7000
km (from the centre of the earth). If you decrease the magnitude of mechanical energy
of the spacecraft -earth system by 10% by firing the rockets, then what is the greatest
height you can take your spacecraft above the surface of the earth? [ R=6400 km]
a) 6400 km b) 540 km c) 2140 km d) 3000 km

21. A tunnel is dug along the diameter of the earth (radius R and mass M ). There is a
particle of mass 'm ' at the centre of the tunnel. The minimum velocity given to the
particle so that it just reaches to the surface of the earth is
a) GM
√
R
b) GM
√
d) It will reach with the help of a negligible
2R

√
c) 2 GM
velocity
22. Three uniform spheres, each having mass m and radius r , are kept in such a way that
R

each touches the other two. The magnitude of the gravitational force on any sphere due
to the other two is
2 2 2 2
a) Gm b) Gm c) √2
Gm d) √ 3 G m
2 2 2 2

23. Two equal masses each m are hung from a balance whose scale pans differ in vertical
r 4r 4r 4r

height by 'h '. The error in weighing in terms of density of the earth ρ is
a) πGρmh b) 1 πGρmh c) 8 πGρm h d) 4 πGρmh

24. What should be the angular velocity of rotation of the earth about its own axis so that the
3 3 3

weight of a body at the equator reduces to 3/5 or its present value? (Take R as the
radius of the earth)
a)
√ g b)
√ 2g c)
√ 2g d)

25. Consider two solid uniform spherical objects of the same density ρ . One has radius R and
3R 3R 5R √ 2g
7R

te other has radius 2 R. They are in outer space where the gravitational fields from other
objects are negligible. If they are arranged with their surface touching, what is the
contact force between the objects due to their traditional attraction?
a) 2 4 b) 128 G π 2 R4 ρ 2 c) 128 G π 2 d) 128 π 2 R 2 G
Gπ R
26. A satellite of mass m is in an elliptical orbit around the earth. The speed of the satellite
81 81 87

at its nearest position is ( 6 G M e ) /(5 r ) where r is the perigee (nearest point) distance from
the centre of the earth. It is desired to transfer the satellite to the circular orbit of radius
equal to its apogee (farthest point) distance from the centre of the earth. The change in
orbital speed required for this purpose is
Page|3
 d) Zero
a)
0.35
√ G Me b)
0.085
√ G Me c)
√ 2G Me

27. Two satellites of the same mass are launched in the same orbit around the earth so as to
r r r

rotate opposite to each other. If they collide inelastically and stick together as wreckage,
the total energy of the system just after collision is
a) −2GMm b) −GMm c) GMm d) GMm

28. Te radius of the earth is about 6400 km and that of Mars is about 3200 km. The mass of
r r 2r 4r

the earth is about 10 times the mass of Mars. An object weights 200 N on the surface of
the earth. Its weight on the surface of mars would be
a) 6 N b) 20 N c) 40 N d) 80 N

29. Two astronauts have deserted their spaceship in a region of space far from the
gravitational attraction of any other body. Each has a mass of 100 kg and they are 100 m
apart. They are initially at rest relative to one another. How long will it be before the
gravitational attraction brings them 1 cm closer together?
a) 2.52 days b) 1.41 days c) 0.70 days d) 1.41 s

30. The distances from the centre of the earth, where the weight of a body is zero and one-
fourth that of the weight of the body on the surface of the earth are (assume R is the
radius of the earth)
a) 0 , R b) 0 , 3 R c) R , 0 d) 3 R , 0

31. The mass of the earth is 81 times the mass of the Moon and the distance between the
4 4 4 4

earth and the Moon is 60 times the radius of the earth. If R is the radius of the earth,
then the distance between the Moon and the point on the line joining the Moon and the
earth where the gravitational force becomes zero is
a) 30 R b) 15 R c) 6 R d) 5 R

32. A spherical shell is cut into two pieces along a chord as shown in the figure. P is a point
on the plane of the chord. The gravitational field at P due to the upper part is I 1 and due
to the lower part is I 2. What is the relation between them?

a) I > I b) I < I c) I =I d) No definite relation

33. Two bodies with masses M 1, and M 2 are initially at rest and a distance R apart. Then they
1 2 1 2 1 2

move directly towards one another under the influence of their mutual gravitational
attraction. What is the ratio of the distances travelled by M 1to the distance travelled by
M 2?
a) M 1 b) M 2 c) 1 d) 1
M2 M1 2
34. A body of mass m rises to a height h=R/5 from the earth's surface where R is earth's
radius. If g is acceleration due to gravity at the earth's surface, the increase in potential
energy is
a) m g h b) 4 m g h c) 5 m g h d) 6 m g h
5 6 7

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35. In problem 86, what is the gravitational potential energy of the mass m ?

a) −2 GMm (1−2 √ 3) b) −2 GMm (1+2 √ 3) c) −√ 3 GMm (1−2 3) d) −√ 3 GMm (1+2 3)

√ √
√3 l √3 l
36. The earth moves around the Sun in an elliptical orbit as shown in figure. The ratio
2 l 2 l

OA /OB=x . The ratio of the speed of the earth at B to that at A is nearly

a) b) x c) d)
√x x √x x
2

37. A space station is set up in space at a distance equal to the earth's radius from the
surface of the earth. Suppose a satellite can be launched from the space station. Let v 1
and v 2 be the escape velocities of the satellite on the earth's surface and space station,
respectively. Then
a) v =v b) v < v

d) (a), (b) and (c) are valid depending on the

2 1 2 1

mass of satellite
c) v > v
2 1

38. A solid sphere of uniform density and mass M has radius 4 m. Its centre is at the origin
of the coordinate system. Two spheres of radii 1 m are taken out so that their centres are
at P(0 ,−2 , 0) and Q(0 , 2 , 0) respectively. This leaves two spherical cavities. What is the
gravitational field at the origin of the coordinate axes?

a) 31GM b) Gm c) 31 GM d) Zero

39. The value of 'g' at a certain height h above the free surface of the earth is x /4 where x is
1024 1024

the value of 'g' at the surface of the earth. The height h is

a) R b) 2 R c) 3 R d) 4 R

40. The radius of a planet is R . A satellite revolves around it in a circle of radius r with
angular velocity ω 0. The acceleration due to the gravity on planet's surface is
2 3 3 3 2 3 2
a) r ω0 b) r ω0 c) r ω0 d) r ω0
2 2
R R
41. The gravitational potential due to earth at infinite distance from it is zero. Let the
R R

gravitational potential at a point P be −5 J k g−1 . Suppose, we arbitarily assume the

gravitational potential at infinity to be +10 J k g−1, then the gravitational potential at P
will be
a) −1 b) −1 c) −1 d) −1
−5 J k g +5 J k g −15 J k g +15 J k g
42. Two satellite A and B of masses m 1 and m 2 (m 1=2 m 2) are moving in circular orbits of radii
r 1 and r 2 (r 1 =4 r 2), respectively, around the earth. If their periods are T A and T B, then the
ratio T A /T B is

Page|5
 a) 4 b) 16 c) 2 d) 8

43. Four similar particles of mass m are orbiting in a circle of radius r in the same angular
direction because of their mutual gravitational attractive force. Velocity of a particle is
given by

[ ( )] √ √ [ ( )]
1 1
a) Gm 1+2 √ 2 2 b) Gm c) Gm d) 1 Gm 1+2 √ 2 2
(1+2 √ 2)
r 4 r r 2 r 2
44. A spherically symmetric gravitational system of particles has a mass density

ρ=
{ρ0 forforrr>≤RR
0

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

a) b)

c) d)

45. If g is the acceleration due to gravity on the earth's surface, the gain in the potential
energy of an object of mass m raised from the surface of the earth to a height equal to
the radius R of the earth is
a) 1 m g R b) 2 m g R c) m g R d) 1 m g R

46. A satellite of mass m is orbitiing around the earth at a height h above the surface of the
2 4

earth. Mass of the earth is M and its radius is R . The angular momentum of the satellite
is independent of
a) m b) M c) h d) None of these

47. A ring having non-uniform distribution of mass M and radius R is being considered. A
point mass m 0 is taken slowly towards the ring. In doing so, work done by the external
force against the gravitational force exerted by ring is

Page|6
 a) GM m0
√2 R
b) GM m0 1 − 1
R √2 √ 5 [ ]
c) GM m0 1 − 1
R √5 √ 2 [ ]
d) It is not possible to find the required work as the nature of distribution of mass is not
known
48. Four particles, each of mass M , move along a circle of radius R under the action of their
mutual gravitational attraction. The speed of each particle is
a) GM
R
GM
R
b)
√ c) GM
2 √2
R √
( 2 √ 2+1) d) GM 2 √ 2+ 1
R 4
49. Two concentric shells of masses M 1 and M 2 are having radii r 1 and r 2 . Which of the
√ ( )
following is the correct expression for the gravitational filed on a mass m ?

a) F= G( M 1+ M 2 ) , for r< r b) F= G( M 1+ M 2 ) , for r< r

2 1 2 2
r r
c) F= G M 2 , for r <r <r d) F= G M 1 , for r <r <r
2 1 2 2 1 2
r r
Gravitational acceleration on the surface of a planet is
50. √6 g ,where g is the gravitational
11
acceleration on the surface of earth. The average mass density of the planet is times
2
3
that of the earth. If the escape speed on the surface of the earth is taken on be 11kms−1 ,
the escape speed on the surface of the planet in k ms−1will be
a) 5 b) 7 c) 3 d) 11

51. If a man at the equator would weigh (3/5)th of his weight, the angular speed of the earth
is
a)
√ 2 g
5R
b)
√ g
R
c)
√ R
g √
d) 2 R

52. A projectile is fired vertically upwards from the surface of the earth with a velocity K v e ,
5 g

where v e is the escape velocity and k < 1. If R is the radius of the earth, the maximum
height to which it will rise, measured from the centre of the earth, will be (neglect air
resistance)
a) 1−k
2
b) R c) 2 d) R
2 R ( 1−k ) 2

53. A satellite is orbiting around the earth in a circular orbit of radius r . A particle of mass m
R 1−k 1+ k

Page|7
 is projected from the satellite in a forward direction with a velocity v=2/3 times the
orbital velocity (this velocity is given w.r.t. earth). During subsequent motion of the
particle, its minimum distance from the centre of earth is
a) r b) r c) 2r d) 4 r

54. The escape velocity for a body projected vertically upwards from the surface of the earth
2 3 5

is 11.2km s−1. If the body is projected in a direction making an angle 45 ° with the vertical,
the escape velocity will be
a) 11.2 km s−1 b) c) d)
11.2× √ 2 km s
−1 −1 −1
11.2×2 km s 11.2km s
√2
55. If the distance between the earth and the Sun were half its present value, the number of
days in a year would have been
a) 64.5 b) 129 c) 182.5 d) 730

56. If an artificial satellite is moving in a circular orbit around the earth with a speed equal
to half the magnitude of the escape velocity from the earth, the height of the satellite
above the surface of the earth is
a) 2 R b) R c) R d) R

57. A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius
2 4

3 R. The work required to take a unit mass from point P on its axis to infinity is
P
4R

3R 2R

a) 2GM ( 4 √ 2−5) b) −2GM (4 √ 2−5) c) GM d) 2GM ( √ 2−1)

58. The gravitational potential energy of an isolated system of three particles, each of mass
7R 7R 4R 5R

m , at the three corners of an equilateral triangle of side l is

2 2 2 2
a) −G m b) −G m c) −2G m d) −3 Gm

59. If the radius of the earth decreases by 10%, the mass remaining unchanged, what will
l 2l l l

happen to the acceleration due to gravity?

a) Decreases by 19% b) Increases by 19%

c) Decreases by more than 19% d) Increases by more than 19%

60. A body is thrown from the surface of the earth with velocity (g Re )/2, where Re is the
radius of the earth at some angle from the vertical. If the maximum height reached by
the body is Re /4 , then the angle of projection with the vertical is
d) None of these
a)sin−1 √ 5
(4) b) −1 √ 5
cos (4) c) −1 √ 3
sin (2)
61. If g is same at a height h and at a depth d , then

a) R=2 d b) d=2h c) h=d d) None

62. Suppose the gravitational force varies inversely as the nth power of the distance. Then,
the time period of a planet in a circular orbit of radius R around the Sun will be
proportional to

Page|8
 (n+1) (n−1 )
a) n b) c) d) −n
R 2 2 R
63. Two satellites A and B of the same mass are revolving around the earth in the concentric
R R

circular orbits such that the distance of satellite B from the centre of the earth is thrice
as compared to the distance of the satellite A from the centre of the earth. The ratio of
the centripetal force acting on B as compared to that on A is
a) 1 b) 3 c) 1 d) 1
3 9 √3
64. A planet is revolving in an elliptical orbit around the Sun. Its closest distance from the
Sun is r min and the farthest distance is r max . If the velocity of the planet at the distance of

the closest approach is v 1 and that at the farthest distance from the Sun is v 2, then
v1
v2
a) r max b) r min c) r min + r max d) None
r min r max r max −r min
65. A simple pendulum has a time period T 1 when on the earth's surface and T 2 when taken
to a height R above the earth's surface, where R is the radius of the earth. The value of
T 2 /T 1 is
a) 1 b) c) 4 d) 2
2 √
66. A diameter tunnel is dug across the earth. A ball is dropped into the tunnel from one
side. The velocity of the ball when it reaches the centre of the earth is
[Given: gravitational potential at the centre of earth ¿−3/(2 GM /R)]
a) b) c) d)
√R √gR √ 2.5 g R √ 7.1 g R
67. In order to shift a body of mass m from a circular orbit of radius 3 R to a higher orbit of
rdaius 5 R around the earth, the work done is
a) 3GMm b) GMm c) 2 GMm d) GMm
5R 2R 15 R 5R
68. In problem 83, what is the gravitational potential at any point on the circle x 2+ z2 =6?

a) −GM
6
b) −GM
64 √ 10
c) −GM 1 −
32 √10 2 2 [
3 32 √ 10
3
1
] d) −GM 1 +
[ 1
]
69. A cavity of radius R/2 is made inside a solid sphere of radius R . The centre of the cavity
is located at a distance R/2 from the centre of the sphere. The gravitational force on a
particle of mass 'm ' at a distance R/2 from the centre of the sphere on the line joining
both the centres of the sphere and the cavity is (opposite to the centre of the cavity)
[Here g=(GM )/ R2, where M is the mass of the sphere]
a) m g b) 3 m g c) m g d) None of these

70. In problem 80, if x is the distance from the common centre, then what is the
2 8 16

gravitational potential at a point for which r < x < R ?

a) −G M + m
[ x r ] b) −G M − m
[ x r ] c) −G M + m
[ R x ] d) −G M − m
[ R x ]
71. In the solar system, the Sun is in the focus of the system for Sun-earth binding system.
Then the binding energy for the system will be [given that radius of the earth's orbit
round the Sun is 1.5 ×1011m and mass of the earth ¿ 6 ×10 24 kg]
a) 33 b) 33 c) 30 d) 30
2.7 ×10 J 5.4 × 10 J 2.7 ×10 J 5.4 × 10 J
72. Two particles of equal mass go around a circle of radius R under the action of their

Page|9
 mutual gravitational attraction. The speed of each particle is
a)
v=
1
2R (√ Gm1 ) b)
v= (√ Gm
2R)
c)
v=
1
2 (√ GmR ) d)
v= (√ 4 Gm
R )
73. What is the mass of the planet that has a satellite whose time period is T and orbital
radius is r ?
2 3 2 3 2 3 2
a) 4 π r b) 4 π r c) 4 π r d) 4 π T
2 2 3 2
GT GT GT GT
74. The escape velocity corresponding to a planet of mass M and radius R is 50 km s−1. If the
planet's mass and radius were 4 M and R , respectively, then the corresponding escape
velocity would be
a) −1 b) −1 c) −1 d) −1
100 km s 50 km s 200 km s 25 km s
75. A point P is on the axis of a fixed ring of mass M and radius R , at a distance 2 R from the
centre O . A small particle starts from P and reaches O under the gravitational attraction
only. Its speed at O will be
a) Zero
b) 2 GM
√ R
c) 2 GM
√ R
d) 2 GM 1− 1
( √ 5−1)

76. Masses of 1 kg each are placed 1 m, 2 m, 4 m, 8 m,... from a point P . The gravitational
√ R ( √5 )
field intensity at P due to these masses is
a) G b) −G c) 4 G d) 4 G/3

77. 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
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

78. A uniform ring of mass m and radius r is placed directly above a uniform sphere of mass
M and of equal radius. The centre of the ring is directly above the centre of the sphere at
a distance r √ 3 as shown in the figure. The gravitational force exerted by the sphere on
the ring will be

a) GMm b) GMm c) √3
GMm d) GMm
2 2 2 3
√3
79. The minimum energy required to launch a m kg satellite from the earth's surface in a
8r 4r 8r 8r

circular orbit at an altitude 2 R, where R is the radius of earth is

a) 5 m g R b) 4 m g R c) 5 m g R d) 5 m g R
3 3 6 4
80. The masses and radii of the earth and the Moon are M 1 , R1 and M 2 , R2 , respectively. Their
centres are at distance d apart. The minimum speed with which a particle of mass m

P a g e | 10
 should be projected from a point midway the two centres so as to escape to infinity is
a)
√ 2 G(M 1 + M 2)
d
b)
d √
4 G(M 1 + M 2)
√
c) 4 G M 1 M 2
√
d) G( M 1 + M 2 )

81. Two concentric shells have masses M and m and their radii are R and r , respectively,
d d

where R>r . What is the gravitational potential at their common centre?

a) −GM
R
b) −GM
r
c) −G M − m
R r [ ] [
d) −G M + m
R r ]
82. The following figure shows two shells of masses m 1 and m 2. The shells are concentric. At
which point, a particle of mass m shall experience zero force?

a) A b) B c) C d) D

83. Two rings having masses M and 2 M , respectively, having the same radius are placed
coaxially as shown in the figure

If the mass distribution on both the rings is non-uniform, then the gravitational potential
at point P is
a) −GM
R [√1
+
2
2 √5 ] b) −GM 1+ 2
R [ ]
2
c) Zero d) Cannot be determined from the given
information
84. The gravitational force exerted by the Sun on the Moon is about twice as great as the
gravitational force exerted by the earth on the Moon, but still Moon is not escaping from
the gravitational influence of the earth. Mark the option which correctly explains the
above system
a) Escape speed is independent of the direction in which it is projected

b) The rotational effect of the earth plays a role in computation of escape speed, however
small it may be
c) A body thrown in the eastward direction has less escape speed

d) None of the above

85. A satellite of mass m revolves around the earth of radius R at a height x from its surface.
If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the
satellite is

( R+ x )
2 2 1
a) g x b) g R c) g R d) g R 2
R−x R +x
86. In problem 83, what is the gravitational field at the centre of the cavities?

P a g e | 11
 a) 31GM b) Gm c) 31 GM d) GM

87. If the mass of a planet is 10% less than that of the earth and the radius is 20% greater
1024 1024

than that of the earth, the acceleration due to gravity on the planet will be
a) 5/8 times that on the surface of the earth b) 3/4 times that on the surface of the earth

c) 1/2 times that on the surface of the earth d) 9/10 times that on the surface of the
earth
88. A satellite is moving with a constant speed v in a circular orbit about the earth. An object
of mass m is ejected from the satellite such that it just escapes from the gravitational pull
of the earth. At the time of its ejection, the kinetic energy of the object is
a) 1 mv 2 b) 2 c) 3 mv 2 d) 2
mv 2 mv
89. A solid sphere of radius R/2 is cut out of a solid sphere of radius R such that the
2 2

spherical cavity so formed touches the surface on one side and the centre of the sphere
on the other side, as shown. The initial mass of the soild sphere was M . If a particle of
mass m is placed at a distnace 2.5 R from the centre of the cavity, then what is the
gravitational attraction on the mass m ?

R/2
m O
R O

2.5R

a) GMm b) GMm c) GMm d) 23 GMm

2 2 2
100 R2
90. The gravitational potential of two homogenous spherical shells A and B of same surface
R 2R 8R

density at their respective centres are in the ratio 3:4. If the two shells coalesce into a
single one such that surface charge density remains the same, then the ratio of potential
at an internal point of the new shell to shell A is equal to
a) 3:2 b) 4:3 c) 5:3 d) 5:6

91. A satellite moves around the earth in a circular orbit with speed v . If m is the mass of the
satellite, its total energy is
a) −1 m v 2 b) 1 m v 2 c) 3 m v 2 d) 1 m v 2

92. If R is the radius of the earth and g the acceleration due to gravity on the earth's
2 2 2 4

surface, the mean density of the earth is

a) 4 πG b) 3 πR c) 3g d) πR
3g R 4 gG 4 πRG 12G
93. The value of g (acceleration due to gravity) at earth's surface is 10 m s . Its value in m s−2
−2

at the centre of the earth which is assumed to be a sphere of radius R meter and uniform
mass density is
a) 5 b) 10 c) 10 d) Zero

94. A space vehicle approaching a planet has a speed v , when it is very far from the planet.
R 2R

At that moment tangent of its tranjectory would miss the centre of the planet by distance
R . If the planet has mass M and radius r , what is the smallest value of R in order that the
resulting orbit of the space vehicle will just miss the surface of the planet?

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 [ ] [ ] [ ]
1
a) r v 2 + 2 GM 2 b) vr 1+ 2 GM c) r v 2 + 2 GM d) 2GMv
v r r v r r
95. Imagine a light planet revolving around a very massive star in a circular orbit of radius r
with a period of revolution T . If the gravitational force of attraction between the planet
and the star is proportional to r 5 /2 , then the square of the time period will be proportional
to
a) 3 b) 2 c) 2.5 d) 3.5
r r r r
96. A space ship is launched into a circular orbit close to the surface of the earth. The
additional velocity now imparted to the space-ship in the orbit to overcome the
gravitational pull is
a) −1 b) −1 c) −1 d) −1
11.2km s 8 km s 3.2 km s 1.414 × 8 km s
97. A man weighs 80 kg on the surface of earth of radius R . At what height above the
surface of earth his weight will be 40 kg?
a) R b) c) d) ( 2+1)R
√2 R ( √ 2−1 ) R √
98. A planet is revolving around the Sun in an elliptical orbit. Its closest distance from the
2

Sun is r and farthest distance is R . If the orbital velocity of the planet closest to the Sun
is v , then what is the velocity at the farthest point?
a) vr
R
b) vR
r
c)
v
√ r d)

99. A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km.
R
v
√ R
r

Then, the time period of a spy satellite orbiting a few 100 km above the earth's surface (
Rearth =6400 km) will approximately be
a) 1 h b) 1 h c) 2 h d) 4 h
2
100. A projectile is fired vertically upwards from the surface of the earth with a velocity k v e
where v e is the escape velocity and k < 1. If R is the radius of the earth, the maximum
height to which it will rise measured from the centre of earth will be (neglect air
resistance)
a) 1−k
2
b) R c) 2 d) R
2 R(1−k ) 2

101. If the radius of the earth were to shrink by one per cent, its mass remaining the same,
R 1−k 1+ k

acceleration due to gravity on the earth's surface would

a) Decrease b) Remain unchanged c) Increase d) Be zero

102. The orbital velocity of an artificial satellite in a circular orbit just above the earth's
surface is v . For a satellite orbiting at an altitude of half of the earth's radius, the orbital
velocity is

(2)
a) 3 v b)
√( ) 3
2
v c)
√( )2
3
v (3)
d) 2 v

103. A body is released from a point of distance R' from the centre of earth. Its velocity at the
time of striking the earth will be (R > Re )
'

a) 2 g R
√ e
b) R g
√ e c)
√ 2 g (R −R )
'
e

104. If three particles, each of mass M , are placed at the three corners of an equilateral
d)
√ (
2 g R e 1−
Re
R
' )
triangle of side a , the forces exerted by this system on another particles of mass M

P a g e | 13
 placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively,
2 2 2 2
a) 0 , 4 G M b) 4 G M , 0 c) 3G M , G M d) 0 , 0
2 2 2 2

105. An artificial satellite of the earth is launched in circular orbit in the equatorial plane of
3a 3a a a

the earth and the satellite is moving from west to east. With respect to a person on the
equator, the satellite is completing one round trip in 24 h. Mass of the earth is
M =6 × 10 kg. For this situation, the orbital radius of the satellite is
24

a) 4
km b) 6400 km c) 36,000 km d) 29, 600 km
2.66 ×10
106. How many hours would make a day if the earth were rotating at such a high speed that
the weight of a body on the equator were zero
a) 6.2 h b) 1.4 h c) 28 h d) 5.6 h

107. A point mass m is released from rest at a distance of 3 R from the centre of a thin-walled
hollow sphere of rdaius R and mass M as show. The hollow sphere is fixed in position
and the only force on the point mass is the gravitational attraction of the hollow sphere.
There is a very small hole in the hollow sphere through which the point mass falls as
shown. The velocity of a point mass when it passes through point P at a distance R/2
from the centre of the sphere is

d) None of these
a)
√ 2 GM
3R √
b) 5 GM
3R
c)
√ 25 GM
24 R
108. A system of binary stars of masses m A and m B are moving in circular orbits of radii r A and
r B, respectively. If T A and T B are the time periods of masses m A and mB respectively, then

( )
3
a) T A = r A 2 b) T >T (if r >r ) c) T >T (if m >m ) d) T =T
A B A B A B A B A B
TB rB
109. A tunnel has been dug into a solid sphere of non-uniform mass density as shown in the
figure
As one moves from A to B, the magnitude of gravitational field intensity

a) Will continuously decrease

b) Will decrease up to the centre of the sphere and then increase

c) May increase or decrease

d) Will continuously increase

110. The distance of two planets from the Sun are 1013 and 1012m, respectively. The ratio of
time periods of these two planets is
a) 1 b) 100 c) 10 d)
√ 10
√10 √10
P a g e | 14
111. In problem 81, what is the gravitational intensity at a point for which x <r ?

a) Gm b) Gm c) Gm d) Zero
2 2 2

112. The percentage change in the acceleration of the earth towards the Sun from a total
r x R

eclipse of the Sun to the point where the Moon is on a side of earth directly opposite to
the Sun is

( ) ( ) ( )
2 2 2
a) M s r 2 ×100 b) M s r 2 ×100 c) 2 r 1 Mm
× 100 d) r 1 Mm
× 100
Mm r1 M m r1 r2 Ms r2 Ms
113. The two planets with radii R1 , R 2 have densities ρ1 , ρ2, and atmospheric pressure p1 and p2
, respectively. Therefore, the ratio of masses of their atmospheres, neglecting variation
of g and ρ within the limits of atmosphere, is
a) p 1 R2 ρ1 b) p 1 R2 ρ2 c) p 1 R1 ρ1 d) p 1 R1 ρ2
p 2 R1 ρ 2 p 2 R1 ρ 1 p 2 R2 ρ 2 p 2 R2 ρ 1
114. If g is the acceleration due to gravity on the earth's surface, the change in the potential
energy of an object of mass m raised from the surface of the earth to a height equal to
the radius R of the earth is
a) m g R b) 2 m g R c) m g R d) −m g R
2
115. The value of g at a particular point is 10 m s−2. Suppose the earth shrinks uniformly to half
of its present size without losing any mass. The value of g at the same point (assuming
that the distance of the point from the centre of the earth does not change) will now be
a) −2 b) −2 c) −2 d) −2
5ms 10 m s 3ms 20 m s
116. If g is acceleration due to gravity on the earth's surface, the gain in the potential energy
of an object of mass m raised from the surface of earth to a height equal to the radius R
of the earth is
a) 1 m g R b) 2 m g R c) m g R d) 1 m g R
2 4
117. If W 1 , W 2 and W 3 represent the work done in moving a particle from A to B along three
different paths 1, 2 and 3, respectively, (as shown in the figure) in the gravitational field
of a point mass m , find the correct relation between W 1 , W 2 and W 3
B
m
1 2
3

A
a) W >W >W b) W =W =W c) W <W <W d) W >W >W

118. The maximum vertical distance through which a fully dressed astronaut can jump on the
1 2 3 1 2 3 1 2 3 2 1 3

earth is 0.5 m. If mean density of the Moon is two-third that of the earth and radius is
one quarter that of the earth, the maximum vertical distance through which he can jump
on the Moon and the ratio of the time of duration of the jump on the Moon to hold on the
earth are
a) 3 m, 6:1 b) 6 m, 3:1 c) 3m, 1:6 d) 6 m, 1:6

119. A satellite of mass m is revolving around the earth at height R (radius of the earth) from
the earth's surface. Its potential energy will be
a) m g R b) 0.67 m g R c) −m g R d) 0.33 m g R
2
P a g e | 15
120. Three particles, each of mass M , are placed at the three corners of an equilateral
triangle of side l . What is the force due to this system of particles on another particle of
mass m placed at the midpoint of any side?
a) 3GMm b) 4 GMm c) GMm d) 4 GMm
2 2 2 2

121. A solid sphere of uniform density and radius R applies a gravitational force of attraction
4l 3l 4l l

equal to F 1 on a particle placed at a distance 2 R from the centre of the sphere. A

spherical cavity of radius R/2 is now made in the sphere as shown in the figure. The
sphere with the cavity now applies a gravitational force F 2 on the same particle. The
ratio F 1 /F 2 is

a) 1 b) 3 c) 7 d) 9

122. The given figure shows the motion of a planet around the Sun S in an elliptical orbit with
2 4 8 7

the Sun at the focus. The shaded areas A and B are also shown in the figure which can
be assumed to be equal. If t 1 and t 2 represent the time taken for the planet to move from
a to b and c to d , respectively, then

a) t < t
1 2

b) t > t
1 2

c) t =t
1 2

d) From the given information the relation between t and t cannot be determined

123. Imagine a light planet revolving around a very massive star in a circular orbit of radius R
1 2

with a speed of revolution T . If the gravitational force of attraction between the planet
and the star is proportional to R−5 / 2, then
a) 2 is proportional to 2 b) 2 is proportional to 7 /2
T R T R

T is proportional to R T is proportional to R
c) 2 3 /2 d) 2 3.75

124. The atmosphere is held to the earth by

a) Winds b) Gravity c) Clouds d) None of the above

125. Two spherical bodies of masses m and 5 M and radii R and 2 R, respectively, are released
in free space with initial separation between their centres equal to 12 R . If they attract
each other due to gravitational force only, then the distance covered by the smaller body
just before collision is
a) 2.5 R b) 4.5 R c) 7.5 R d) 1.5 R

126. If three uniform spheres, each having mass M and radius R , are kept in such a way that
each touches the other two, the magnitude of the gravitational force on any sphere due
to the other two is

P a g e | 16
 d) √ 3 G M
2 2 2 2
a) G M b) 2G M c) 2G M
2 2 2 2

127. Suppose that the acceleration of a free fall at the surface of a distant planet was found to
4r r 4r 4r

be equal to that at the surface of the earth. If the diameter of the planet were twice the
diameter of the earth, then the ratio of mean density of the planet to that of the earth
would be
a) 4:1 b) 2:1 c) 1:1 d) 1:2

Multiple Correct Answers Type

128. Choose the incorrect statements from the following

a) It is possible to shield a body from the gravitational field of another body by using a
thick shielding material between them
b) The escape velocity of a body is independent of the mass of the body and the angle of
projection
c) The acceleration due to gravity increases due to the rotation of the earth

d) The gravitational force exerted by the earth on a body is greater than that exerted by
the body on the earth
129. If both the mass and radius of the earth decreases by 1%, the value of

a) Acceleration due to gravity would decrease by nearly 1%

b) Acceleration due to gravity would increase by 1%

c) Escape velocity from the earth's surface would decrease by 1%

d) The gravitational potential energy of a body on earth's surface will remain unchanged

130. Suppose a smooth tunnel is dug along a straight line joining two points on the surface of
the earth and a particle is dropped from rest at its one end. Assume that mass of the
earth is uniformly distributed over its volume. Then, which of the following statements
are not correct?
a) The particle will emerge from the other end with velocity (G M e )/(2 Re ), where M e and
Re are earth's mass and radius, respectively
b) The particle will come to rest at the centre of the tunnel because at this position, the
particle is closest to the earth's centre
c) Potential energy of the particle will be equal to zero at centre of the tunnel if it is
along a diameter
d) Acceleration of the particle will be proportional to its distance from the mid-point of
the tunnel
131. The radius and mass of earth are increased by 0.5%. Which of the following statements
are true at the surface of the earth
a) g will increase b) g will decrease

c) Escape velocity will remain unchanged d) Potential energy will remain unchanged

132. For two satellites at distances R and 7 R above the earth's surface, the ratio of their

a) Total energies is 4 and potential and kinetic energies is 2

P a g e | 17
 b) Potential energies is 4

c) Kinetic energies is 4

d) Total energies is 4

133. Figure shows the kinetic energy (E k ) and potential energy (E p ) curves for a two-particle
system. Name the point at which the system is bound system

a) A b) B c) C d) D

134. An orbiting satellite will escape if

a) Its speed is increased by 41% b) Its speed in the orbit is made √ (1.5) times
of its initial value
c) Its KE is doubled d) It stops moving in the orbit

135. Which of the following are correct?

a) An astronaut going from the earth to the Moon will experience weightlessness once

b) When a thin uniform spherical shell gradually shrinks maintaining its shape, the
gravitational potential at its centre decreases
c) In the case of a spherical shell, the plot of V versus r is continuous

d) In the case of a spherical shell, the plot of gravitational field intensity I versus r is
continuous
136. If two satellites of different masses are revolving in the same orbit, they have the same

a) Angular momentum b) Energy c) Time period d) Speed

137. A solid sphere of uniform density and radius 4 units is located with its centre at the
origin O of coordinates. Two spheres of equal radii 1 unit, with their centres at
A(−2, 0 , 0) and B(2 , 0 , 0), respectively, are taken out of the solid leaving behind spherical
cavities as shown in the figure

Then
a) The gravitational force due to this object at the origin zero

b) The gravitational force at point B(2 , 0 , 0) is zero

c) The gravitational potential is the same at all points of circle 2 2

y + z =36

P a g e | 18
 d) The gravitational potential is the same at all points on the circle 2 2
y + z =4
138. An astronaut, inside an earth satellite, experience weightlessness because

a) No external force is acting on him

b) He is falling freely

c) No reaction is exerted by the floor of the satellite

d) He is far away from the earth's surface

139. A body is imparted a velocity v from surface of the earth. If v 0 is orbital velocity and v e be
the escape velocity then for
a) v=v , the body follows a circular track around the earth.
0

b) v> v , but ¿ v , the body follows elliptical path and returns to surface of earth.
0 e

c) v< v , the body follows elliptical path and returns to surface of earth
0

d) v> v , the body follows hyperbolic path and escapes the gravitational pull of the earth

140. Suppose an earth satellite, revolving in a circular orbit experiences a resistance due to
e

cosmic dust. Then

a) Its kinetic energy will increase

b) Its potential energy will decrease

c) It will spiral towards the earth and in the process its angular momentum will remain
conserved
d) It will burn off ultimately

141. The magnitudes of the gravitational field at distance r 1 and r 2 from the centre of a
uniform sphere of radius R and mass M are r 1 and r 2 respectively. Then,
a) F 1 = r 1 for r < R and r =R b) F 1 = r 2 for r > R and r > R
2

1 1 2 1 1
F2 r2 F2 r1
c) F 1 = r 1 for r > R and r > R d) F 1 = r 1 for r < R and r < R
2

1 2 2 1 2
F2 r2 F2 r2
142. Which of the following are correct?

a) Out of electrostatic, electromagnetic, nuclear and gravitational interactions, the

gravitational interaction is the weakest
b) If the earth were to rotate faster than its present speed, the weight of an object would
decrease at the equator but remain unchanged at the poles
c) The mass of the earth in terms of g , R and G is 2
(g R /G)
d) If the earth stops rotating in its orbit around the Sun there will be no variation in the
weight of a body on the surface of earth
143. Which of the following are correct?

a) If R is the radius of a planet, g is the accleration due to gravity, the mean density of
the planet is 3 g/4 πGR
b) Acceleration due to gravity is a universal constant

P a g e | 19
 c) The escape velocity of a body from earth is 11. 2 km s . The escape velocity from a
−1

planet which has double the mass of earth and half its radius is 22.4 km s−1
d) The radio of gravitational mass and inertial mass of a body at the surface of earth is 1

144. An object is taken from a point P to another point Q in a gravitational field

a) Assuming the earth to be spherical, if both P and Q lie on the earth's surface, the work
done is zero
b) If P is on the earth's surface and Q above it, the work done is minimum when it is
taken along the straight line PQ
c) The work done depends only on the position of P and Q and is independent of the path
along which the particle is taken
d) There is no work done if the object is taken from P to Q and then brought back to P
along any path
145. Two satellites S1 and S2 are revolving around the earth in coplanar concentric orbits in
the opposite sense. At t=0 , the positions of satellites are shown in the diagram. The
periods of S1 and S2 are 4 h and 24 h, respectively. The radius of orbit of S1 is 1.28 ×10 4
km. For this situation, mark the correct statement(s)

a) The angular velocity of S as observed by S at t=12 h is −1

0.468 π rad s
b) The two satellite are closest to each other for the first time at t=12 h and then after
2 1

every 24 h they are closest to each other

c) The orbital velocity of S is
0.64 π ×10 km
4

d) The velocity of S1 relative to S2 is continuously changing in magnetic and direction

1

both
146. A particle of mass m lies at a distance r from the centre of earth. The force of attraction
between the particle and earth is (r )
a) F (r )∝ 1 for r < R b) F (r )∝ 1 for r ≥ R
2 2
r r
c) F (r )∝ r for r < R d) F (r )∝ 1 for r < R

147. Which of the following statements are true? For a particle on the surface of the earth
2

a) The linear speed is minimum at the equator

b) The angular speed is maximum at the equator

c) The linear speed is minimum at the poles

d) The angular speed is

7.3 ×10 rad s at the equator
−5 −1

148. A particle of mass m is moved from the surface of the earth to a height h . The work done
by an external agency to do this is
a) m g h for h ≪ R b) m g h for all R c) 1 m g R for h ≫ R d) 1 m g R for h=R

149. A small mass m is moved slowly from the surface of the earth to a height h above the
2 2

surface. The work done (by an external agent) in doing this is

P a g e | 20
 a) m g h , for all values of h b) m g h , for h ≪ R

c) 1/2 m g R , for h=R d) −1/2 m g R , for h=R

150. A planet is revolving round the sun. Its distance from the sun at Apogee is r A and that at
Perigee is r P. The mass of planet and sun is m and M respectively, v A and v P is the
velocity of planet at Apogee and Perigee respectively and T is the time period of
revolution of planet round the sun.
2 2
a) 2 π 3 b) π
2 3 c) v r =v r d) v <v ; r >r
T = ( r A +r P ) T = ( r A +r P ) A A P P A P A P
2Gm 2Gm
151. Two spherical planets P and Q have the same uniform density ρ , masses M P and M Q, and
surface areas A and 4 A , respectively. A spherical planet R also has uniform density ρ
and its mass is (M P + M Q ). The escape velocities from the planets P , Q and R , are V P ,V Q
and V R respectively. Then
a) V >V >V b) V >V >V c) V /V =3 d) V /V = 1
Q R P R Q P R P P Q
2
152. The magnitudes of the gravitational field at distance r 1 and r 2 from the centre of a
uniform sphere of radius R and mass m are F 1 and F 2, respectively. Then
a) F 1 = r 1 if r < R and r < R b) r 2 if r > R andr > R
2

2 2
F 2 r2 1 r2 1
c) F 1 = r 1 if r > R andr > R d) F 1 = r 1 if r < R andr < R
2

1 2 2 1 2
F2 r2 F2 r2
153. Choose the correct statements from the following

a) The magnitude of the gravitational force between two bodies of mass 1 kg each and
separated by a distance of 1 m is 9.8 N
b) The higher the value of the escape velocity for a planet, the higher is the abundance of
the lighter gases in its atmosphere
c) The gravitational force of attraction between two bodies of ordinary mass is not
noticeable because the value of the gravitational constant is extremely small
d) Force of friction arises due to gravitational attraction

154. The gravitation potential on the surface of a planet of radius R mass M is

a) g b) g M c) −GM d) −g R

155. Suppose universal gravitational constant starts to decrease, then

R R

a) Length of the day on the earth will increase

b) Length of the year will increase

c) The earth will follow a spiral path of decreasing radius

d) Kinetic energy of the earth will decrease

156. Mark the correct statements

a) Gravitational potential at the centre of curvature of a thin hemispherical shell of

radius R and mass M is equal to GM /R
b) Gravitational field strength at a point lying on the axis of a thin, uniform circular ring

P a g e | 21
 of radius R and mass M is equal to GMx /[ ( R2 + x 2 ) ], where x is distance of that point
3/ 2

from the centre of the ring

c) Newton's law of gravitation for gravitational force between two bodies is applicable
only when bodies have spherical symmetric distribution of mass
d) None of these

157. A planet of mars m is revolving round the sun (of mass m s ) in an elliptical orbit. If ⃗v is the
velocity of the planet when its position vector from sun r then if the planet rotates in
counter clockwise direction then areal velocity has direction
a) Given by “Right Hand Thumb Rule”

b) Given by “Left Hand Thumb Rule”

c) Normal to the plane of orbit upwards

d) Normal to the plane of orbit downwards

158. A double star consists of two stars having masses M and 2 M . The distance between their
centres is equal to r . They revolve under their mutual gravitational interaction. Then,
which of the following statements are not correct?
a) Heavier star revolves in orbit of radius 2 r /3

b) Both the stars revolve with the same speed, period of which is equal to
3 2
(2 π /r )( 2G M /3)
c) Kinetic energy of the heavier star is twice that of other star

d) None of the above

Consider an attractive central fore of the form F ( r )= , k is constant. For a stable

159. −k
n

circular orbit to exist

r

a) n=2 b) n<3 c) n>3 d) n=−1

160. Choose the correct statements from the following:

a) The gravitational forces between two particles are an action and reaction pair

b) Gravitational constant (G) is scalar but acceleration due to gravity (g) is a vector

c) The value of G and g are to be determined experimentally

d) G and g are constant everywhere

161. Let V and E denote the gravitational potential and gravitational field at a point,
respectively. It is possible to have
a) V =¿and E=0 b) V =0 and E ≠ 0 c) V ≠ and E=0 d) V ≠ 0 and E ≠ 0

162. A double star is a system of two stars of masses m and 2 m, rotating about their centre of
mass only under their mutual gravitational attraction. If r is the separation between
these two starts then their period of rotation about their centre of mass will be
proportional to
a) 32 b) r c) 12 d) −1
2

163. Consider two satellite A and B of equal mass m , moving in the same circular orbit about
r m m

the earth, but in the opposite sense as shown in figure. The orbital radius is r . The
P a g e | 22
 satellites undergo a collision which is perfectly inelastic. For this situations, mark out
the correct statement(s). [Take mass of earth as M ]

a) The total energy of the two satellite plus earth system just before collision is −(GMm)/r

b) The total energy of the two satellite plus earth system just after collision is −(2GMm)/r

c) The total energy of two satellites plus earth system just after collision is −(GMm)/2r

d) The combined mass (two satellites) will fall towards the earth just after collision

164. If the radius of the earth suddenly decreases to 80% of its present value, the mass of the
earth remaining the same, the value of the acceleration due to gravity will
a) Remain unchanged b) Become −2
9.8 /0.64 m s
c) Increase by 36% d) Increase by about 56%

165. Which of the following statements are true about acceleration due to gravity?

a) g decreases in moving away from the b) g decreases in moving away from the
centre if r > R centre if r < R
c) g is zero at the centre of earth d) g decreases if earth stops rotating on its
axis
166. Which of the following are not correct?

a) The escape velocity for the Moon is −1

6 km s
b) The escape velocity from the surface of Moon is v . The orbital velocity for a satellite to
orbit very close to the surface of Moon is v /2
c) When an earth satellite is moved from one stable orbit to a further stable orbit, the
gravitational potential energy increases
d) The orbital velocity of a satellite revolving in a circular path close to the planet is
independent of the density of the planet
167. A satellite is orbiting the earth, if its distance from the earth is increased, its

a) Angular velocity would increase b) Linear velocity would increase

c) Angular velocity would decrease d) Time period would increase

168. A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the
earth's centre. The wall of the tunnel may be assumed to be frictionless. A particle is
released from one end of the tunnel. The pressing force by the particle on the wall, and
the acceleration of the particle vary with x (distance of the particle from the centre)
according to
Pressing
force

a)

x = R/ 2 x = R
x

P a g e | 23
 Pressing
force

b)

x
x = R/ 2 x = R
Acceleration

c)

x
x = R/ 2 x=R
d) Acceleration

x
x = R/ 2 x = R
169. If a body is projected with a speed lesser than escape velocity, then

a) The body can reach a certain height and may fall down following a straight line path

b) The body can reach a certain height and may fall down following a parabolic path

c) The body may orbit the earth in a circular orbit

d) The body may orbit the earth in an elliptical orbit

170. Consider a planet moving in an elliptical orbit around the Sun. The work done on the
planet by the gravitational force of the Sun
a) Is zero in any small part of the orbit b) Is zero in some parts of the orbit

c) Is zero in complete revolution d) Is zero in no part of the motion

171. In case of an orbiting satellite, if the radius of orbit is decreased

a) Its KE decreases b) Its PE decreases c) Its ME decreases d) Its speed decreases

172. 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
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

Assertion - Reasoning Type

This section contain(s) 0 questions numbered 173 to 172. Each question contains
STATEMENT 1(Assertion) and STATEMENT 2(Reason). Each question has the 4 choices (a),
(b), (c) and (d) out of which ONLY ONE is correct.

P a g e | 24
 a) Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for
Statement 1

b) Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for

Statement 1

c) Statement 1 is True, Statement 2 is False

d) Statement 1 is False, Statement 2 is True

173

Statement 1: The speed of satellite always remains constant in an orbit

Statement 2: The speed of a satellite depends on its path

174

Statement 1: If a pendulum is suspended in a lift and lift is falling freely, then its time
period becomes infinite
Statement 2: Free falling body has acceleration equal to acceleration due to gravity

175

Statement 1: The magnitude of the gravitational potential at the surface of solid sphere
is less than that of the centre of sphere
Statement 2: Due to the solid sphere, the gravitational potential is the same within the
sphere
176

Statement 1: Consider a satellite moving in an elliptical orbit around the earth. As the
satellite moves, the work done by the gravitational force of the earth on
the satellite for any small part of the orbit is zero
Statement 2: KE of the satellite in the above described case is not constant as it moves
around the earth
177

Statement 1: The value of escape velocity from the surface of earth at 30 ° and 60 ° is
v 1=2 v e , v2 =2/3 v e
Statement 2: The value of escape velocity is independent of angle of projection

178

Statement 1: Kepler's second law can be understood by conservation of angular

momentum principle
Statement 2: Kepler's second law is related with areal velocity which can further be
proved to be based on conservation of angular momentum as
( dA /dt )=(r 2 ω)/2

P a g e | 25
179

Statement 1: For a mass M kept at the centre of a cube of side ‘a’, the flux of
gravitational field passing through its sides is 4 π GM
Statement 2: If the direction of a field due to a point source is radial and its dependence

on the distance ‘r’ from the source is given as 2 , its flux through a closed
1

surface depends only on the strength of the source enclosed by the surface
r

and not on the size or shape of the surface

180

Statement 1: For a satellite revolving very near to the earth's surface the time period of
revolution is given by 1 h 24 min
Statement 2: The period of revolution of a satellite depends only upon its height above
the earth's surface
181

Statement 1: Two satellites are following one another in the same circular orbit. If one
satellite tries to catch another (leading one) satellite, then it can be done
by increasing its speed without changing the orbit
Statement 2: The energy of earth satellites system in circular orbit is given by

, where r is the radius of the circular orbit

−GMm
E=
2r
182

Statement 1: If earth suddenly stops rotating about its axis then the value of
acceleration due to gravity will becomes same at all the places
Statement 2: The value of acceleration due to gravity is independent of rotation of earth

183

Statement 1: Two satellites are following one another in the same circular orbit. If one
satellite tires to catch another (leading one) satellite, then it can be done
by increasing its speed without changing the orbit
Statement 2: The energy of earth-satellite system in circular orbit is given by
E=−(−Gms ) /(2 a), where r is the radius of the circular orbit
184

Statement 1: The binding energy of a satellite does not depend upon the mass of the
satellite
Statement 2: Binding energy is the negative value of total energy of satellite

185

Statement 1: Two different planets have same escape velocity

Statement 2: Value of escape velocity is a universal constant

186

Statement 1: The earth does not retain hydrogen molecules and helium atoms in its
atmosphere, but does retain much heavier molecules, such as oxygen and
nitrogen

P a g e | 26
 Statement 2: Lighter molecules in the atmosphere have translational speed that is
greater or closer to escape speed of earth
187

Statement 1: The speed of revolution of an artificial satellite revolving very near the
earth is 8 km s−1
Statement 2: Orbital velocity of a satellite, become independent of height of satellite

188

Statement 1: The time period of geostationary satellite is 24 hours

Statement 2: Geostationary satellite must have the same time period as the time taken
by the earth to complete one revolution about its axis
189

Statement 1: There is no effect of rotation of earth on acceleration due to gravity at

poles
Statement 2: Rotation of earth is about polar axis

190

Statement 1: The smaller the orbit of a planet around the Sun, the shorter is the time it
takes to complete
Statement 2: According to Kepler's third law of planetary motion, square of time period
is proportional to cube of mean distance from Sun
191

Statement 1: A force act upon the earth revolving in a circular orbit about the sun.
Hence work should be done on the earth
Statement 2: The necessary centripetal force for circular motion of earth comes from
the gravitational force between earth and sun
192

Statement 1: If time period of a satellite revolving in circular orbit in equatorial plane is

24 h, then it must be a geostationary satellite
Statement 2: Time period of a geostationary satellite is 24 h

193

Statement 1: The difference in the value of acceleration due to gravity at pole and
equator is proportional to square of angular velocity of earth
Statement 2: The value of acceleration due to gravity is minimum at the equator and
maximum at the pole
194

Statement 1: A body becomes weightless at the centre of earth

Statement 2: As the distance from centre of earth decreases, acceleration due to gravity
increases
195

Statement 1: If the earth suddenly stops rotating about its axis, then the acceleration
due to gravity will become the same at all the plates

P a g e | 27
 Statement 2: The value of acceleration due to gravity is independent of rotation of the
earth
196

Statement 1: An astronaut in an orbiting space station above the earth experience

weightlessness
Statement 2: An object moving around the earth under the influence of earth’s
gravitational force is in a state of free fall
197

Statement 1: The force of gravitation between a sphere and a rod of mass M 2 is

¿(G M 1 M 2 )/r
Statement 2: Newton''s law of gravitation holds correct for point masses

198

Statement 1: The principle of superposition is not valid for gravitational force

Statement 2: Gravitational force is a conservative force

199

Statement 1: Even when orbit of a satellite is elliptical, its plane of rotation passes
through the centre of earth
Statement 2: According to law of conservation of angular momentum plane of rotation
of satellite always remain same
200

Statement 1: We can not move even a finger without disturbing all the stars

Statement 2: Every body in this universe attracts every other body with a force which is
inversely proportional to the square of distance between them
201

Statement 1: Gravitational potential of earth at every place on it is negative

Statement 2: Every body on earth is bound by the attraction of earth

202

Statement 1: When distance between two bodies is doubled and also mass of each body
is also doubled. Gravitational force between them remains the same
Statement 2: According to Newton’s law of gravitation, force is directly proportional to
mass of bodies and inversely proportional to distance between them
203

Statement 1: Earth has an atmosphere but the moon does not

Statement 2: Moon is very small in comparison to earth

204

P a g e | 28
 Statement 1: For the planets orbiting around the Sun, angular speed, linear speed and
KE change with time, but angular momentum remains constant
Statement 2: No torque is acting on the rotating planet. So its angular momentum is
constant
205

Statement 1: The time period of revolution of a satellite close to surface of earth is

smaller than that revolving away from surface of earth
Statement 2: The square of time period of revolution of a satellite is directly
proportional to cube of its orbital radius
206

Statement 1: If a particle projected horizontally just above the surface of the earth with
a speed greater than escape speed, then it will escape from gravitational
influence of the earth. Assume that particle has a clear path
Statement 2: Escape velocity is independent of its direction

207

Statement 1: It takes more fuel for a spacecraft to travel from the earth to the Moon
than for the return trip
Statement 2: The point of zero gravitational field intensity due to the earth and the
Moon is lying nearer to the Moon, i.e., in the diagram shown, for r <r 0 , E g is
towards the earth''s centre and for r >r 0 , E g is towards the Moon''s centre,
and at r =r 0 , E g is zero

208

Statement 1: Orbital velocity of a satellite is greater than its escape velocity

Statement 2: Orbit of a satellite is within the gravitational field of earth whereas

escaping is beyouned the gravitational field of earth
209

Statement 1: Gravitational force between two particles is negligible small compared to

the electrical force
Statement 2: The electrical force is experienced by charged particles only

210

Statement 1: In free space a uniform spherical planet of mass M has a smooth narrow
tunnel along its diameter. This planet and another superdense small
particle of mass M start approaching towards each other from rest under
action of their gravitational forces. When the particle passes through the
centre of the planet, sum of kinetic energies of both the bodies is
maximum

P a g e | 29
 Statement 2: When the resultant of all forces acting on a particle or a particle like
object (initially at rest) is constant in direction, the kinetic energy of the
particle keeps on decreasing
211

Statement 1: Space rockets are usually launched in the equatorial line from west to east

Statement 2: The acceleration due to gravity is minimum at the equator

212

Statement 1: Generally the path of a projectile from the earth is parabolic but it is
elliptical for projectiles going to a very great height
Statement 2: Upto ordinary height the projectile moves under a uniform gravitational
force, but for great heights, projectile moves under a variable force
213

Statement 1: The value of acceleration due to gravity does not depend upon the mass of
the body
Statement 2: Acceleration due to gravity is a constant quantity

214

Statement 1: An astronaut in an orbiting space station above the Earth experiences

weightlessness
Statement 2: An object moving around the Earth under the influence of Earth’s
gravitational force is in a state of ‘free-fall’
215

Statement 1: A planet moves faster, when it is closer to the sun in its orbit and vice
versa
Statement 2: Orbital velocity in the orbit of planet is constant

Matrix-Match Type

This section contain(s) 0 question(s). Each question contains Statements given in 2 columns
which have to be matched. Statements (A, B, C, D) in columns I have to be matched with
Statements (p, q, r, s) in columns II.

216. For a planet orbiting about the Sun in a elliptical orbit, some incomplete statements
regarding physical quantities are given in Column I, which can be completed by using
the entries of Column II. Match the entries of Column I with the entries of Column II
Column-I Column- II

(A) Maximum PE of the Sun planet system (p) Is at perihelion

(B) Maximum speed of the planet (q) Is at aphelion

P a g e | 30
 (C) Minimum PE of Sun-planet system (r) Is independent of mass of planet

(D) Minimum kinetic energy of planet (s) Is independent of semi-major axis of

orbit
CODES :

A B C D

a) A,c,d a,d b,c b,d

b) b,d a,c,d a,d b,c

c) a,d b,c b,d a,c,d

d) b,c b,d a,c,d a,d

217. A satellite is revolving around the earth in a circular orbit of radius 'a ' with velocity v 0. A
particle is projected from the satellite in forward direction with relative velocity
V = [ ( √ 5 /4 )−1 ] v 0 , During the subsequent motion of the particle, match the following
Column-I Column- II

(A) Total energy of particle (p) (3 G M e m)/a

(B) Minimum distance of particle from the (q) (5 G M e m)/a

earth
(C) Maximum distance of particle from the (r) 5 a /3
earth
(s) 2 a

(t) 2 a/3

(u) a

CODES :

A B C D

a) a b c

b) c a b

c) c b a

d) a c b

218. Consider the earth to be a homogeneous sphere but keeping in mind its spin, match the
following
Column-I Column- II

(A) Acceleration due to gravity (p) May change from point to point

(B) Orbital angular momentum of the (q) Does not depend on direction of
earth as seen from a distant star projection
(C) Escape velocity from the earth (r) Remains constant

(D) Gravitational potential due to earth at (s) Depend on direction of projection

a particular point
P a g e | 31
 CODES :

A B C D

a) A,b c a c

b) a,c b c d

c) a c,d d a

d) c b d a,b

219.

Column-I Column- II

(A) Gravitational potential (p) On the surface of planets with density

ratio 1:2
(B) Escape velocity (q) Conservation of angular momentum

(C) Ratio of the acceleration due to gravity (r) Varies with the reference point
1:3
(D) Orbiting satellite (s) Does not depend on the angle

(t) Similar to an atom

CODES :

A B C D

a) a b,e c,d d

b) b,e c,d d a

c) c,d d a b,e

d) d a b,e c,d

220. An artificial satellite is in circular orbit around the earth. One of the rockets of the
satellite is momentarily fired, the direction of firing of rocket is mentioned in Column I
and corresponding change (s) are given in Column II. Match the entries of Column I with
the entries of Column II
Column-I Column- II

(A) Towards the earth's centre (p) Orbit changes and becomes elliptical

(B) Away from the earth's centre (q) Orbit plane changes

(C) At right angle to the plane or orbit (r) Semi-major axis of orbit increases

(D) In forward direction (s) Energy of earth-satellite system

increases
CODES :

A B C D

a) a a a,c a,b,c,d

P a g e | 32
 b) a,b,c,d a a a,c

c) a a,c,d a,b,c,d a

d) a,c a,b,c,d a a

221.

Column-I Column- II

(A) The earth moving in an elliptical orbit (p) Conservation of linear momentum
(only earth in system) along any direction

(B) A disc having translation and rotation (q) Conservation of linear momentum
motion both (both slipping) on a rough along specific direction
surface (only disc in system)

(C) A sphere rolling without slipping on a (r) Conservation of angular momentum

curved surface (only the sphere in about any point in the space
system)

(D) Projection of a particle from the (s) Conservation of angular momentum

surface of the earth (only particle in about a specific point in the space
system)

CODES :

A B C D

a) a c d a

b) a b d c

c) b c a d

d) c d no d
match

222.

Column-I Column- II

P a g e | 33
 (A) Geostationary satellite (p) Circular orbit

(B) Total energy of earth-satellite system (q) Elliptical orbit

is constant in
(C) Angular momentum of satellite about (r) Equatorial plane orbit
centre of earth is constant in
(D) Orbital speed of satellite is/may be (s) Non-equatorial plane orbit
constant
CODES :

A B C D

a) A,c b,d a,c b,c

b) a,c a,b,c,d a,b,c a,c,d

c) a,d a,c c,d b,d

d) a,b a,d a,d,c d,c

223. Let V and E denote the gravitational potential and gravitational field, respectively, at a
point due to certain uniform mass distribution described in four different situations of
Column I. Assume the gravitational potential at infinity to be zero. The values of E and V
are given in Column II. Match the statement in column I with the results in Column II
Column-I Column- II

(A) At the centre of thin spherical shell (p) E=0

(B) At the centre of solid sphere (q) E ≠ 0

(C) A solid sphere has a non-concentric (r) V ≠ 0

spherical cavity. At the centre of the
spherical cavity
(D) At the centre of the joining two point (s) V =0
masses of equal magnitude
CODES :

A B C D

a) A,c a,c a,c a,c

b) a,c b,d c,a d,c

c) b,d c,a d,c a,c

d) c,a d,c a,c b,d

Linked Comprehension Type

This section contain(s) 19 paragraph(s) and based upon each paragraph, multiple choice
questions have to be answered. Each question has atleast 4 choices (a), (b), (c) and (d) out of
which ONLY ONE is correct.
Paragraph for Question Nos. 224 to -224

P a g e | 34
A rocket is fired vertically upwards with a speed of v (¿ 5 km s−1) from the surface of earth. It
goes up to a height h before returning to earth. At height h a body is thrown from the rocket
with speed v 0 in such a way so that the body becomes a satellite of earth. Let the mass of the
earth, M =6 × 1024 kg; mean radius of the earth,
R=6.4 ×10 m;
6

−11 2 −2
G=6.67× 10 N m k g ;
−2
g=9.8 m s

224. The value of h is

1.5 ×10 m 3.2 ×10 m 3.2 ×10 m 1.6 ×10 m

a) 5 b) 5 c) 6 d) 6

Paragraph for Question Nos. 225 to - 225

A spaceship is in a circular orbit of radius r 0 around a star of mass M . The spaceship’s rocket
engine can alter its velocity (instantaneously) by an ∆ ⃗v . Amounts direction of firing is
measured by angle θ between the ship’s velocity ⃗v and the vector from the tail to the nose of
the ship. To conserve fuel in a sequence of N firings, it is desirable to minimise

∆ V =∑ |∆ ⃗v i|. ∆ v is known as the specific impulse. We want to use the ship’s engine to cause
N

it to crash into the star (assume the radius of the star to be negligible)
i=1

225. What is the minimum specific impulse required to escape from the star if the engine is
fired in a single rapid burst?

√ √ √ √
2
a) GM b) G M ( 2−1) c) GM d) G M ( 2−1)
√ √2 √
r0 r0 r0 r0

Paragraph for Question Nos. 226 to - 226

The minimum and maximum distances of a satellite from the centre of the earth are 2 R and
4 R, respectively, where R is the radius of the earth and M is the mass of the earth

226. The minimum and maximum speeds are

a)
√ √ GM 2 GM
9R
,
R
b) GM
√ √ 5R
,
3 GM
2R
c)
√ √ GM 2 GM
6R
,
3R
d) GM
√ √ 3R
,
5 GM
2R

Paragraph for Question Nos. 227 to - 227

The orbit of Pluto is much more eccentric than the orbits of the other planets. That is,
instead of being nearly circular, the orbit is noticeably elliptical. The point in the orbit

P a g e | 35
nearest to the Sun is called the perihelion and the point farthest from the Sun is called the
aphelion

227. At perihelion, the gravitational potential energy of Pluto in its orbit has

a) Its maximum value

b) Its minimum value

c) The same value as at every other point in the orbit

d) The value which depends on the sense of rotation

Paragraph for Question Nos. 228 to - 228

A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M
and radius R as shown in the figure. A particle of masses m is placed on the line joining the
two centres at a distance x from the point of contact of the sphere and the shell. Find the
magnitude of the resultant gravitational force on this particle due to the sphere and the shell
if

O
R
x
m O
r

228. r < x <2 r

' ' ' '
a) Gm m ( 2r −x ) b) Gm m ( x−r ) c) Gm m ( x−r ) d) Gm m ( 2 x−r )
2 r3 2 r3 r3 r3

Paragraph for Question Nos. 229 to - 229

The gravitational field in a region is given by ⃗

E =( 5 N k g−1 ) i⃗ +(12 N k g−1) ⃗j

229. Find the magnitude of the gravitational force acting on a particle of mass 2 kg placed at
the origin
a) 26 N b) 30 N c) 20 N d) 35 N

Paragraph for Question Nos. 230 to - 230

P a g e | 36
A uniform metal sphere of radius a and mass M is surrounded by a thin uniform spherical
shell of equal mass and radius 4 a. The centre of the shell falls on the surface of the inner
sphere

230. Find the gravitational field intensity at point P1

a) GM b) GM c) GM d) GM
2 2 2 2
16 a 8a 2a 4a

Paragraph for Question Nos. 231 to - 231

In the graph shown, the PE of earth-satellite system is shown by a solid line as a function of
distance r (the separation between earth's centre and satellite). The total energy of the two
objects which may or may not be bounded to the earth are shown in the figure by dotted
lines

Based on the above information, answer the following questions

231. Mark the correct statements(s)

a) The object having the total energy E is bounded one

1

b) The object having the total energy E is bounded one

c) Both the objects are bounded

2

d) Both the objects are unbounded

Paragraph for Question Nos. 232 to - 232

An unmanned satellite A and a spacecraft B are orbiting around the earth in the same
circular orbit as shown

P a g e | 37
The spacecraft is ahead of the satellite by some time. Let us consider that some technical
problem has arisen in the satellite and the astronaut from B has made it correct. For this to
be done docking of two ( A and B) is required (in layman terms connecting A and B). To
achieve this, the rockets of A have been fired in forward direction and docking takes place
as shown in the figure below :

Take mass of the earth¿ 5.98 ×1024 kg

Radius of the earth¿ 6400 km
Orbital radius¿ 9600 km
Mass of satellite A=320 kg
Mass of spacecraft¿ 3200 kg
Assume that initially spacecraft B leads satellite A by 100 s, i.e., A arrive at any particular
position after 100 s of B's arrival. Based on the above information answer the following
questions

232. To dock A and B in the above-described situation, one can use the rocket system of
either one, i.e., either of A or of B. To accomplish docking in the minimum possible time
which is the best way?
a) To use rocket system of A b) To use rocket system of B

c) Either (a) or (b) d) Information insufficient

Paragraph for Question Nos. 233 to - 233

The satellite when launched from the earth are not given the orbital velocity initially, a
multistage rocket propeller carrier the spacecraft up to its orbit and during each stage
rocket has been fired to increase the velocity to acquire the desired velocity for a particular
orbit. The last stage of the rocket brings the satellite in circular/elliptical (desired) orbit
Consider a satellite of mass 150 kg in a low circular orbit. In this orbit, we cannot neglect
the effect of air drag. This air opposes the motion of satellite and hence the total mechanical
energy of earth satellite system decreases. That means the total energy becomes more
negative and hence the orbital radius decreases which cause the increase in KE When the
satellite comes in the low enough orbit, excessive thermal energy generation due to air
friction may cause the satellite to burn up. Based on the above information, answer the
following questions

233. It has been mentioned in the passage that as r decreases, E decreases but K increases.

P a g e | 38
 The increase in K is [ E=¿ total mechanical energy, r =¿orbital radius, K=¿kinetic
energy]
a) Due to increase in gravitational PE b) Due to decrease in gravitational PE

c) Due to work done by air friction force d) Both (b) and (c)

Integer Answer Type

234. Imagine a new planet having the same density as that of the earth but it is 3 times
bigger than the earth is size. If the acceleration due to gravity on the surface of the earth
is g and that on the new planet is g' , then what is the value of g' / g?
235. The earth (mass¿ 1024 kg) revolves round the Sun with an angular velocity 2 ×107 rad s−1 in
a circular orbit of radius 1.5 ×108 km. Find the force exerted by the Sun on the earth (in
×10 N)
21

236. A particle of mass m is subjected to an attraction central force of magnitude k /r 2 , k being

a constant. If at the instant when the particle is at an extreme position in its closed orbit,
at a distance a from the centre of force, its speed is (k /2ma), if the distance of other
extreme position is 'b ' . Find a /b
237. The density of a newly discovered planet is twice that of the 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 and the radius of the planet is ' R '. Then what is the value of R/ R' ?

P a g e | 39
 8.GRAVITATION

: ANSWER KEY :

1) d 2) d 3) c 4) c a,b,c,d
5) b 6) b 7) a 8) b 13) a,b,d 14) a,b 15) a,b,c,d 16)
9) a 10) c 11) c 12) c a,c,d
13) c 14) b 15) c 16) c 17) a,c,d 18) a,b,c,d 19) b,c 20)
17) d 18) b 19) c 20) c c,d
21) a 22) d 23) c 24) c 21) a,d 22) b,c 23) b,c,d 24)
25) b 26) b 27) a 28) d b,d
29) b 30) a 31) c 32) c 25) a,b 26) b,c 27) c,d 28)
33) b 34) c 35) b 36) b b,d
37) b 38) d 39) a 40) d 29) b,c 30) a, c 31) a,c 32)
41) b 42) d 43) a 44) c a,b,d
45) a 46) d 47) b 48) d 33) a,b,c 34) a,c,d 35) a,d 36)
49) c 50) c 51) a 52) b a,b,d
53) a 54) d 55) b 56) c 37) b,d 38) a,c 39) a,d 40)
57) a 58) d 59) d 60) a c,d
61) b 62) b 63) c 64) a 41) b,c 42) a,b,c,d 43) b,c 44)
65) d 66) b 67) c 68) c b,c
69) b 70) c 71) a 72) c 45) a,c 1) e 2) a 3) c
73) a 74) a 75) d 76) d 4) d
77) a 78) c 79) c 80) b 5) d 6) a 7) a 8) a
81) d 82) d 83) a 84) c 9) d 10) c 11) d 12) d
85) d 86) a 87) a 88) b 13) d 14) a 15) a 16) b
89) d 90) c 91) a 92) c 17) a 18) a 19) e 20) d
93) d 94) a 95) d 96) c 21) b 22) c 23) c 24) a
97) c 98) a 99) c 100) b 25) d 26) e 27) a 28) a
101) c 102) c 103) d 104) b 29) a 30) a 31) b 32) a
105) a 106) b 107) d 108) d 33) a 34) a 35) a 36) d
109) c 110) c 111) d 112) c 37) b 38) a 39) b 40) c
113) d 114) a 115) b 116) a 41) c 42) a 43) c 1) b
117) b 118) a 119) c 120) b 2) d 3) a 4) c
121) d 122) c 123) b 124) b 5) c 6) d 7) b 8) a
125) c 126) d 127) d 1) 1) d 2) d 3) c 4) b
a,c,d 2) b,d 3) a,b,c 4) 5) c 6) a 7) a 8) a
b,c,d 9) c 10) b 1) 3 2) 6
5) b,c,d 6) a,b,c,d 7) a,c 8) 3) 3 4) 2
a,b,c
9) c,d 10) a,c,d 11) b,c 12)

P a g e | 40
 8.GRAVITATION

: HINTS AND SOLUTIONS :

So
Here, angular momentum is conserved.
1 (d) 2
m v GMm 2 0
= ⇒v ∝R
According to it,
R R

Energy of the skylab in the first orbit is

I 1 ω 1=I 2 ω 2 6 (b)

or M R 1 ω 1=M R 1 ω 2 or R1 v 1=R 2 v 2
2 2

At point P4 , the value of R is minimum

−GMm −GMm
=
and hence the velocity is maximum or KE
2(2 R) 4R
Total energy required to place the skylab
is maximum into the orbit of radius 2 R from the
2 (d) surface of earth is
−GMm −GMm 3 GMm
4R
−
R
=
4R ( )
2
3g R m 3
¿ = mgR
Energy of the skylab in the second orbit
81GM GM 4R 4
2
=
x ( D−x )2
¿−(GMm)/6 R
Energy needed to shift the skylab from
9 1
=
the first orbit to the second orbit is
x ( D−x )
9 −GMm GMm GMm 2 m g R
D= x − = × =
10 4R 6R R 24 12
9
x= D
10
Let us first calculate the mass of the
7 (a)

inner solid sphere of radius r

Note that there must be some point
3 (c)
Mass of the inner solid sphere is
where the gravitational field of the earth
is balanced by the gravitational field of
' Me4 2 Me
M= × =π r = 3 3
Mars
4 3 3 Re r
π Re
3

Now, g= or
3
G Mer
At height h above the surface of the
4 (c) G Mer 1
× g=
earth, therefore
3 2 3
Re r Re
Force on the particle of mass m=m g

(
g' =g 1−
2g
R
⇒ ∆ g 1=)2h
R
g ¿
G M e mr

At depth d below the surface of the earth

3
Re
Note :Those layers of the sphere which have radii lar g er th
( d
g' =g 1− ⇒ ∆ g2= g
R ) d
R
the g ravitational force

Using Kepler’s third law, the semi-major

∆ g1 =∆ g2 ⇒ d=2 h 8 (b)

5 (b) axis of the comet’s orbit is given by

(as F G ∝ given)
GMm 1 4π
2
F G= T =
4
×a
3
R R G Ms

P a g e | 41
 [ ] [ G M 1 ( √ M 1+ √ M 2 ) G M 2 ( √ M 1+ √ M 2 )
]
1
T 2 ×G M s 3
a= ¿− +
4π
2
R √ M1 R √ M2

[ ] [ G ( √ M 2 + √ M 1)
]
1
( 90 ×24 × 3600 ) ×6.67 ×10−11 ×2 ×1030 3
¿ ¿− ( √ M 1+ √ M 2 )
4 π2 R
¿ 5.89 ×10 m G ( √ M 1+ √ M 2 )
10 2

So, the major axis of the comet’s orbit

¿−
would be 2 a=1.178× 10 m
R
11

As 2 a<¿ mean distance between the earth 12 (c)

and the Sun, so the collision is not
2 2
m1 v 1 GM m2 v 2 GM
possible
=m1 2 , =m2 2
r1 r1 r2 r2
2
9 (a)
v1 r2 2 2
= ; r <r ⇒ v 1 < v 2
v
2
r1 2 1
Hence, v 1< v 2
3
GM 4 πr ρ 4 2
g= 2
=G 2
=G πrρ
r 3 r 3
g 1 r 1 ρ1
= 13 (c)
g 2 r 2 ρ2 2π 2π
Given
8= =
ω1 +ω2 2 π 2 π
10 (c) +
T1 T2
⇒ T 1=24 h for the earth
1 2 GMm GMm
mv = −
⇒ T 1=12 h (T 2being the time period of the
2 R R+h

satellite, it will remain the same as the

GMmh
¿

distance from the centre of the earth

R ( R+ h )

remains constant)
m g hR
¿
R+h
Since, v=n v e and v e=√ 2 g R , hence ⇒T=
2π
=
2π
=24 h
2
¿(R n )/(1+ n )
2 ω 2+ ω1 2 π 2 π
−
T1 T2
11 (c)
14 (b)
GMm 1 2 GMm
= mv −
G M1 G M2 r0 2 R
=
or
2
x ( R−x )2 1 2 GMm GMm
mv = −
M2 2 2 2
2 R r0
x =R + x −2 Rx
or
2
M1 v GM GM
= −
Let
M2 v R r0
=k
or v =2GM [ ]
M 2 1 1
2 2
−
x ( k−1 ) +2 Rx−R =0 R r0

or v= 2GM
√
−2 R + √ 4 R2 + 4 ( k −1 ) R 2
x=
2( k−1) ( 1 1
−
R r0 )
R √M1
¿
On the surface of the earth,
15 (c)
√ M 1 +√ M 2
Total energy=Kinetic energy + Potential
R √M2
energy
R−x=
√ M 1 +√ M 2
Gravitational potential at point P is 1 2 GmM
¿ mv
− ( G xM + GR−xM )
1 2
At the highest point, v=0,
2 R

P a g e | 42
 Potential energy ¿−(GmM )/(R+ h) orbit of the spacecraft will become
Where h is the maximum height elliptical. Let a be the semi-major axis of
1 2 GmM −GmM this new elliptical orbit
mv − =
2 R R+h −G M m
E final =
+h or
1 2 g Rh R +h 2 g R 2a
v= = 2
, where r =7000 km
2 R h v G Mm
Einitial =
( )
−1
2g R 2r
h=R −1
v
2 G M m −G M m
¿ =
2r 2a
16 (c) r
2 a= =1.11r
F 4 GMm /(3 l ) 4 GM 0.9
r ❑max =2 a [where r max is the distance
l= = =
m m 3 l2
corresponding to aphelion]
17 (d)
Work done = change in GPE ¿ U ∞ −U R
¿ 2.22 r−r =1.22 r
Required greatest height,
h=r max −Re =2140 km
W =0−
GMm GMm
R
=
R ( )
Let the minimum speed imparted to the
6.67 ×10
−11
× 100× 10
−2
−10
21 (a)
¿ =6.67 ×10 J
particle of mass m so that it just reaches
−1
10

the surface of the earth is v . Applying the

Conservation of angular momentum of
18 (b)
principle of conservation of energy,
the body about O yield the following :
V
1
2
m v2 + (
−3 GM
2 R
m=
GM
R
m+ 0 )
Solving, we get v=
30o

A √ GM
R
R
The given system may be regarded as a
B 22 (d)
O n
system of three particles locates at the
V'
three vertices of an equilateral triangle of
'
( mv sin 30 ° ) R=n V ( R +h )
side 2 r
V
2
V
R= ( R+ h ) ∴ V ' =
4
V
4 [ ] Now, F A=F B
Therefore, h=R Gm G M
2 2
¿ =
( 2 r )2 4 r 2
F A and F B are inclined to each other at an
19 (c)

angle of 60 ° . If F is the resultant of F A

2 2 2
GM mv G M
F G= ⇒ =
and F B, then
2 2
4R R 4R

v=
1 GM
2 R √ F=√ 3 ×
GM
4r
2
2

As we decrease the magnitude of

20 (c)
23 (c)
mechanical energy of the spacecraft-
earth system, it means we are increasing
g1 =
g
=g 1−
2h
[ ]
[ ]
2

the energy of the spacecraft-earth system

h R
1+
as the total energy of the bounded system
r
W 2 −W 1=¿error in weighing
is negative
As we change the energy, the circular

P a g e | 43
 ¿2mg [ h1 h2
− =2 m 2
R R ]
GM h
R R
∆ v=v A−v 0=0.085
√ G Me
R

[ GM
∵ g=
R
2
∧h1−h2=h
] Energy of each satellite in the orbit
27 (a)

Therefore, W 2 −W 1=¿error in weighing −GMm

4 h 8π ¿
Total energy of the system before
3
¿ 2 mG π R ρ 3 = Gmρ h 2r
3 R 3
collision,
24 (c)
GMm −GMm
'
m g =m g−mR ω cos ϕ
2 2 Ei =E1=E2 =2 E=−2 × =
As the satellites of equal mass are moving
2r r
Now,
3
in the opposite directions and collide
2
m g=m g−mR ω
5
or mR ω =m g− m g
3 inelastically, the velocity of the wreckage
just after the collision is
2
5
¿ m g orω= 2 g mv−mv=2 mV ,i.e., V =0
3
5 5R √ The energy of the wreckage just after the
collision will be totally potential and will
be
25 (b)
m1=4/3 π R ρ , m2=4 /2 π ( 2 R ) ρ, distance
2 3

between the centres of the two spherical

GM ( 2 m ) −2 GMm
Ef= =
objects, r =3 R
As after collision the wreckage comes to
r r

standstill in the orbit, it will move along

( )( )
G M 1 m2 4 4 1
the radius towards the earth under
3 2
F= =G π R ρ 8 × π R ρ ×
r2 3 3 (3 R )2
128 2 4 2 gravity
¿ Gπ R ρ
81
28 (d)
26 (b) g E=(GM )/ R
2

m 6 G M e G M e m −2 G M e m 1
G M GM /10
E= × − = g M = 12 = 2
Which is the total energy of the earth-
2 5r 2r 3 r
R R /4

satellite system
2 GM
¿
So, semi-major axis of the elliptical orbit
5 R2

is a=
2
5r ¿ gE
5
Speed of the satellite at the apogee
4
2
W M +W E
5
position is v A =
¿ 200 × =80 N
v P ×r
2
2 a−r 5
¿
√
2 6G Me

For orbit to change to a circle of radius

3 5r
a 1=¿acceleration of first
29 (b)

3 r /2=(2 a−r) , the rocket has to be fired Gm1 m2 1 G m2 6.67 ×10−11 × 100
when the satellite is at the apogee
¿ × = 2 =
r2 m1 ( 100 )2
position
r
−13 −1
¿ 6.67 ×10 m s
a 2=¿acceleration of second
New orbital speed is v 0=

Required change in the orbital speed is

√ G Me
3 r /2 √
=
2G M e
3r Gm1 6.67 ×10−11 × 100
¿ 2 =
r ( 100 ) 2
−13
=6.67 ×10 m s
−1

Net acceleration of approach

P a g e | 44
 [ ]
−13 −1
a=a 1+ a2=2 ×6.67 ×10 m s 1 1 1
¿−GMm + +
Now, s= a t
1 2 lsin 60 ° l/2 l 2
2 2 GMm
¿− [1+2 √3 ]
−2 1 −13
1 ×10 = × 6.67 ×10 ×2 ×t
2 √3 l

Solving, we get t=1.41 days

2

Applying conservation of angular

36 (b)

momentum at position A and B

(i)The weight of the body at the centre of
30 (a)
m v A ×OA =m v B ×OB
the earth is equal to zero because
Hence,
v B OA
= =x
gcentre=g 1− ( d
R ) (R
=g 1− =0
R ) v A OB

From the surface of the earth, the escape

37 (b)
( )
g1 d 1 3R
= 1− = ⇒ d=
velocity is √ (2 GM )/ R. From the satellite,
g R 4 4

So from the centre, d '= the escape velocity is calculated as

R

fallows. By conservation of energy,

4

The required distance is

31 (c) 1 2 GMm
m v 2− =0
2 2R

√
d 60 R GM
x× = =6 R ⇒ v 22= ⇒ v 2< v 1

√ m2
m2
+1
√ 81
1
+1 R

Let ρ=¿density of sphere, R=¿radius of

38 (d)

sphere, r =¿radius of the spherical

32 (c)
At point P , we have l 1−l 2=0 (because the
gravitational field inside a shell is zero). cavities
Hence, l 1=l 2 Mass of the complete sphere ¿
4 3
π R ρ=M
3
Mass of the removed sphere ¿ π r ρ=m
Because the gravitational force is the
33 (b) 4 3

mutual force, hence the position of centre

3

of mass remains unaffected Here m= 3 =

3 3
M r M (1 ) M
3
=
R ( 4) 64
∴ M 1 R1=M 2 R2or Now ⃗I R =⃗I + ⃗I P+ ⃗I Q
R1 M 2
=
Here I =0 , also I⃗ P =−⃗I Q ⇒ I R=0
R2 M 1

Increase in gravitational potential energy

34 (c)
39 (a)
is
GM
x= 2

[ ][
R
GMm −GMm GMm GMm ×5
R+
R R
=
R
−
6R ] Again,
x
=
GM
4 ( R+ h ) 2 or x=GM ( )
2
R+ h
2

( )
2
1 2
¿
GMm
R
1− =
6 [ ]
5 GMm 1
R
×
6
∴ 2=
R R +h

or R+h=2 R or h=R
2 2
mgR 1 5 ¿
¿ = m g R= m g h[∵ R=5 h] R +h
6R 6 6
40 (d)
35 (b)
GMm 2
GMm =m ω0 r
U P=−∑ r
2
r

P a g e | 45
 2 3
⇒ GM =ω 0 r
2
⇒ g R =ω0 r
2 3 For r ≤ R :
2 '
2 3 m v Gm m
ω0 r = 2
⇒ g= 2
r r
R
'
(
4 3
Here , m = π r ρ0
3 )
r Substituting in Eq. (i) we get
R v∝r
ie , v – r graph is a straight line passing
through orgine.
For r > R :
According to the problem, as the
41 (b)

potential at ∞ incresees by +10 J k g−1,

2
mv
Gm ( 34 π R ) ρ3
0

hence potential will increase by the same

=
r r
2

amount everywhere (potential gardient 1

will remain constant). Hence, potential at
¿ v∝
√r
point P=10−5=+ 5 J k g−1 The corresponding v – r graph will be as
shown in option (c).
42 (d)

U t =−GMm /R=¿ Initial potential energy

m1 45 (a)
m1=2 m2 ⇒ =2
of the system
m2

U f =−GMm /2 R=¿ Final PE of the system

r1
r 1=4 r 2 ⇒ =4
r2
T A ∝r 1 and T B ∝r 2 (i)
2 2 2 3
∴ ∆ U =U f −U t =−GMm
1
− =
2R R [
1 GMm
2R ]
But g=GM / R2
()
3/ 2
T A r1 3 1 /2
∴ = =( 4 )
T B r2
TA
⇒ =8
TB

43 (a)

dK =−dU +W air friction

∴ GM=g R (ii)
2

From Eqs. (i) and (ii), ∆ U =

2
g R m g Rm
=
2R 2

Centripetal force= net gravitational force

46 (d)
L=mv 0 ( R+h )

√
2 2
2 2Gm 1 Gm GM
⇒ m v 0 =2 F cos 45 ° + F 1= + ¿m ( R +h )=m √ GM ( R+h )
(√ 2) √ 2 4 r
2 2
( R+h )
2
mv G M 2 i.e., L depends on m , M as well as h
[2 √ 2+1]
0
= 2
r 4r
Even through the distribution of the mass
47 (b)

[ ]
1
Gm ( 2 √ 2+1 )
is unknown, we can find the potential due
2
⇒ v 0=
to the ring on any axial point because
4r

44 (c) from any axial point the entire mass is at

P a g e | 46
 the same distance (whatever would be
the nature of distribution)
50 (c)

Potential at A due to the ring is V A =

GM
g=
GM
=
G ( 43 π R ) ρ 3

√2 R R
2
R
2

Potential at B due to the ring is V B=

GM ¿ g∝ρ R
√5 R g
¿ R∝
d U =U f −U f −U i=U B−U A=m0(V B −V A ) ρ
Now escape velocity, v e= √ 2 gR
¿
GM m0 −1 1
R [ +
√5 √ 2 ] ¿ v e ∝ √ gR

√
W gr=−W ext
W g r=dU=−W ext √ g
¿ ve ∝ g × ∝
ρ
g2
ρ
W ext =dU =
GM m0 1
R
−
1
√ 2 √5 [ ] ∴¿
¿ 3 km s
−1

48 (d) 51 (a)
2 2
GM 3/5 m g=m g−mR ω
F 12=

√
2
2R 3 2 g
2 ω 2=g− g ⇒ ω
GM 5 5R
F 14= 2
2R
If a body is projected from the surface of
52 (b)

the earth with a velocity v and reaches a

height h , then according to law of
conservation of energy,
1 2 mgh
mv =
The resultant of these two forces is
2 1+ h/ R
Here v=k v e=k √ 2 g R
( √ 2 G M )/2 R . Now, F 12=(G M )/( 4 R )
2 2 2 2

The combined resultant of all the forces is

m g (r−R)
Therefore, 2
1
m k 2 3 g R=
(r −R)
or
1+
√2 G M 2 + G M 2 G M2 √2 1
+ [ ] = r−R or k 2 r=r−R
R

[ ]
2 2
R2 2 4
Equating this with centripetal force, we
2R 4R r−R
k 2 R 1+
get
R

or r =
R
M v 2 G M 2 2 √ 2+1
[ ]
2
= 2 1−k
R R 4

As the velocity of the particle is less than

53 (a)
or v = or v=
2 GM 2 √ 2+1
R 4 [ ] √
GM 2 √2+1
R 4 ( ) the orbital velocity of the satellite, the
particle goes in the elliptical orbit of the
semi-major axis less than r
Gravitational field inside a shell is zero,
49 (c)

but for points outside it, the shell behaves

as if whole of its mass is concentrated at
its centre. Hence, for a point lying in
between the shells there will be a field
only due to the inner shell (¿ G M 2 /r )
2

because the point will be an external Let r 1 be the minimum distance and v 1 be
point for the inner shell but internal for the velocity of the particle at this
the outer shell

P a g e | 47
 position, then 2 Mr dr
¿
× v 0 r =m0 v 1 r 1 , where m0 is the mass
√
2
2 7R
m0 ×
3 −G . dM
of the particle and v 0 is the orbital speed
dV P=
√ 16 R2 +r 2
equal to √ GM /r 2 GM
4R
r
2 ∫
¿− ∙ dr
v 1 r 1=
√ 2
v r
3 0
From energy conservation,
¿−
7 R 3 R √ 16 R2 +r 2
2 GM
7R
( 4 √2−5 )
2 2 +2 GM
m0 × v 0 2 ∴W= (4 √2−5)
3 Gm0 m0 v 1 GM m0 7R
− = −
2 r 2 r1
Solving the equations, we get r 1=r /2
Gravitational potential energy of a pair of
58 (d)

particles¿−(G m 2)/ l
Change in energy
54 (d)
Since we have three pairs, the total
gravitational potential energy is −3 G m2 /l
2
¿ 1/(2(GMm)/ R)=1/2 M v e
So it is independent of angle as
gravitational field is conservative in 59 (d)
nature 2 (i)
GM
g=
R

According to Kepler’s law,

55 (b) Gm 100 GM
81 R2 (ii)
'
g= =
( )
2
90
2 3 R
T1 R1 R 100
From Eqs. (i) and (ii),
2
= 3
, T 1 =365 days , T 2=? , R1=R , R2=
T2 R 2
2
'

( ) [ ]
R2 3/ 2
R /2
3/ 2
'100 g 100
⇒T 2=T 1 =365 =129 days g= g⇒ =
R1 R 91 g 81
'
g 100
−1= −1
56 (c) g 81

√ √
GM 1 2 GM 19
= ⇒ h=R ∴ ∆ g= g=23 % of g
So increase is more than 19% of g
R+ h 2 R 81

57 (a)

Let the speed at the maximum height be

W =∆ U =U f −U i=U ∞−U P 60 (a)

v 1, then from energy be conservation

¿−U P =−m V P
¿−V P ( as m=1 )
Potential at point P will be obtained by in
integration as given below. Let dM be the
mass of small rings as shown
P

16R2 + r2
4R

r 2 2
dr m v G M m m v 1 GMm
− = −
2 Re 2 5 R e/ 4
From angular momentum conservation,
M 5 Re
dM =
π ¿¿ m v R sin θ=m v 1 ×
e
4

P a g e | 48
 Using v= and solving the equation,
√ g Re g T2
∴ '
=4 ⇒ =2
2 g T1

we get θ=sin ( √45 )

−1

Gravitational potential at a point on the

66 (b)

surface of earth is –
61 (b) GM

If the earth is assumed to be a solid

R
[
g' =g 1−
2h
R ] [ ]
= g 1−
d
R
sphere, then the gravitational potential at
the centre of the earth is −(3/(2 GM )/ R)
2h d
= ⇒ d=2 h
Decreases in gravitational potential is
R R

62 (b) R GM R g
× 2 =
2 πR 2 2
Loss in potential energy is
T= R
v
1 2 GMm R GM
E= m v = n −1 × 2 ×m
2 R 2
Now, gain in kinetic energy = loss in
R

or v=
[ ]
1/ 2

potential energy
2GM

Therefore,
R n−1

∴T=
[√ 2 πR
2 GM /R
(n+1)/2
n−1 ]
=
2π
√ 2 GM
×R
(n+ 1)/ 2
1
2
2 1
m v = m g R∨v =√ g R
2
⇒T ∝ R
67 (c)

[ ][ ]
63 (c) −GMm GMm
andV B=
W= −
V A=
GM
√
rA
Given r B=3 r A
GM
rB √ ¿
5R
GMm GMm
3R
−
5R
3R

Now F A=
GMm 1 1
[ ]
2 GMm
2
m v A M GM GMm ¿ − =
= = 2 R 3 5 15 R
rA rA rA rA
GMm
The circle lies outside the bigger sphere
F B= 2 68 (c)
rB
FB RA 1
2
Hence, V R=V +V P +V Q
Or V =V R +V P +V Q
= =
F A r 2B 9
¿−
Gm −Gm
− (
×2 )
Conserving angular momentum,
64 (a) 6 r
Where r =√ 6 2+ 22=√ 40
m v 1 r min =m v 2 r max or 1 = max
v r
v 2 r min ∴V=
−GM 1
2 [
−
1
3 32 √ 10 ]
65 (d)
We have that T 1=2 π √ l/g andT 2=2 π √ l/g
69 (b)
Mass of sphere M ∝ R3. If mass of
'

(i) complete sphere is M , then, the mass of

√
T2 g
cavity will be M /8
∴ = '
T1 g

Also g= 2
GM
R
' GM GM
∴g= =
( 2 R ) 4 RL
2

P a g e | 49
 In the first case, v e= or50=

In the second case,

√ 2 GM
R √ 2 GM
R

[ ] √
2 1
2 G (4 M ) 2 2 GM
v e= =2
R R
−1
¿ 2 ×50=100 km s
Using the figure, F=¿ force due to whole
sphere−¿force due to cavity
75 (d)

F=
R3 2 ( )
GMm R G(M /8)m
−
( R )2
GMm R G(M /8)m
2

2
− 2
2
Gravitational potential at ‘ P’,
R R

[ ]
GMm 4 1 3 GMm 3
− =
R2 8 8 8 R 2
= mg
8 v P=
−GM
√5 R
70 (c) Gravitational potential at ‘O ’,
V=
−GM −Gm
R
+
x
=−G (
M m
+
R x ) [ ] v o=
−GM

Using work energy theorem,

R

Binding energy¿ GMm /R

71 (a)
1
W =∆ K ⇒ m [ v P−vO ] = m v
2

M =¿ mass of the Sun¿ 1030 kg

2

m=¿ mass of the earth ¿ 6 ×10 24 kg

R=1.5× 10 m
11
m=
[
GM GM
R
−
√5 R 2
1
= m v2
]
Binding energy of the system is
6.67 ×10
−11 30
× 10 × 6 ×10
11
=2.7 ×10 J
24
33
√ 2 GM
R
1−
[
1
√5
=V
]
1.5× 10 76 (d)
G ×1 G ×1 G× 1
72 (c) E= 2
+ 2 + 2 +...+¿
2 2 2 1 2 4
Gm M v Gm
F G= 2
⇒ = 2
4R R 4R

∴v=
2 R√
1 Gm

Suppose that a satellite of mass m

73 (a)

describe a circular orbit around a planet

of mass M [
1 1
E = G 1+ + +...
4 16 ]
GmM 1 4G
F= 2 ¿G =
1 3
This force must be mass times the
r
1−
4
centripetal acceleration

Force on the satellite is always towards

2 2 77 (a)
mv 2 4π
∴ F= =m ω r=m 2 r
the earth, therefore, acceleration of
r T
satellite S is always directed towards the
2 3
4π r
centre of the earth. Net torque of this
∴ M= 2
GT
gravitational force F about the centre of
74 (a)

P a g e | 50
 the earth is zero. Therefore, angular
Potential energy¿−
momentum (both in magnitude and
GMm

direction) of S about the centre of the Minimum energy required

R

earth is constant throughout. Since force

F is conservative in nature, therefore
¿
1 GMm −GMm 5 GMm
− = ( )
mechanical energy of the satellite
6 R R 6 R

remains constant. Speed of S is maximum

5
¿ m gR
when it is nearest to the earth and
6

minimum when it is farthest

Potential energy of mass m when it is
80 (b)

78 (c) midway between masses M 1 and M 2 is

Mdm −G M 1 m G M 2 m
dF=G
4r
2 U= −
d /2 d /2
F=∑ dF cos θ 2 Gm
¿− (M 1+ M 2 )
GMdm
According to law of conservation of
¿∑ cos θ d
2

energy,
4r
¿ 2 × √ ∑ dm
GM 3r
4r 2r 1 2 2Gm
m ve= (M 1 + M 2)
Therefore, escape velocity,
2 d
¿√
3 GMm
8 r2
v e=
√ 4 G( M 1+ M 2 )
d

81 (d)

V=
GM −Gm
R
+
r (
=−G
M m
+
R r ) [ ]
The gravitational field intensity at a point
82 (d)

Alternative solution : inside the spherical shell is zero

The gravitational field due to the ring at a
distance √ 3 r is given by
Note :T h ere is NO g ravitational field inside a sp h erical s

As all the points on the periphery of

Gm ( √ 3 r) 83 (a)
E= ∨E=
√3 Gm
either ring are at the same distance from
3/ 2 2
[ r 2 + ( √ 3 r )2 ] 8r
The required force is EM , i.e., point P , the potential at point P due to
( √ 3 Gm ) M /8 r 2 the whole ring can be calculated as
Note : Follow the central ar g ument ∈this solution:The gVravitational
=−(GM )/(field 2
) where x is the axial
√ R 2+atxany
point on therin g due ¿ a sphere is equal¿ the field due ¿ adistance fromofthe centre of the ring. This
mass M located at the centreof the spher e
sin g≤ particle
expression is independent of the fact
whether the distribution of mass is
uniform or non-uniform
79 (c)

Total energy, E= m v − So, at

1 2 GmM
2 r
¿
GmM GMm
2r
−
r
=−G
mM
2r
P,V=
−GM G× 2 M −GM
√2 R
−
√5 R
=
R [√ 1 2
+
2 √5 ]
r =2 R=R=3 R
A minimum amount of energy equal to
84 (c)
GmM
total energy of the moon-earth system has
E=
6R
to be given to break (unbound) the

P a g e | 51
system. The Sun is exerting force on the
Moon but not providing any energy For the satellite, the gravitational force
85 (d)

provides the necessary centripetal force,

i.e.,

or 2 =g
2
G Me m mv 0 G Me
=
(R +x )
2
(R+ x) R

v 0=
√
g R2
R+ x

At the centre of the cavities, if the smaller

86 (a)

spheres are not removed, then

GM GM GM
I R= × × 2=
R
3
64 32
Also, ⃗I R =⃗I + ⃗I P+ ⃗I Q
I⃗ = I⃗ R + I⃗ P+ I⃗ Q
At P , we have I P =0 and
Gm
I Q=
2
=G
M
64 4 ( )
1
× 2

Hence, I =
31GM
1024

87 (a)

g p=
[
G M−
10
100
M
] =
G× 9 M
×
25 5
= g
[ ]
2 2
20 10 36 R 8
R+ R
100

In circular orbit of a satellite of potential

88 (b)

energy
¿−2× ( kinetic energy )
1 v 2
¿−2× m =−mv
Just to escape from the gravitational pull,
2

its total mechanical energy should be

zero. Therefore , its kinetic energy should
be +mv 2

89 (d)
Let mass of the cavity¿ M '
Density of the sphere ¿ M /(4 /3 π R3)
Mass of the cavity cut out¿ M '
3
4 R M
¿ π ×
3 8 4 3
πR
3

P a g e | 52
 At centre, g =g 1− [ ]
' M R
∴M = ⇒ F net =F Mm−F M m '
'
=0
8 R
GMm GM ' m GMm GMm
¿ = −
From the principle of conserving angular
94 (a)
( )
2 2 2 2
4R 5 4R 50 R
R
momentum, we have
2

mvR =m v r (i)
23 GMm
F net= '

[ v '=¿speed when spaceship is just

100 R 2

90 (c) touching the planet]

From conserving of energy, we have
3 2
M A =σ 4 π R A , M B =σ 4 π R B
Where σ is surface density 1 2 1 '
mv = mv −
GMm
(ii)
2

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

−G M A −G M B 2 2 r
⇒ V A= , V B=
RA RB

[ ]
1/ 2
V A M A R B σ 4 π R A RB R A
2 r 2 2 GM
R= v+
= = = v r
V B M B R A σ 4 π R 2B R A R B

Given
V A MA 3 95 (d)
= =
[ ]
2
V B MB 4 m v GMm 1
= 5 /2 ∵ F ∝ 5 /2
Then R B= R A
4 r r r

For new shell of mass M and radius R , or r ω = 5 /2

3 2 GM

or r 2 = 5 /2
2 2
M =M A + M B=σ 4 π R A +σ 4 π R B 2
4 π GM
2 2 2
σ 4 π R =4 π (R A + R B)
Then
T r
or T ∝ r orT 2 ∝ r 3.5
2 5/ 2

V
= =
2 2
M R A σ 4 π [R A + R B ] R A
=
√ R2A + R 2B 5
=
V e =11.2 km/s
1 2
96 (c)
VA R MA σ 4 π RA RA 3
( R A+ RB)
2 2 2

'2
mv GMm
= 2
Total mechanical energy is given by
91 (a) R R
'2 GM
−GMm GMm v = 2 R
E=K +u= −
v = √ 10 × 6400000=8 km/s
R
2a a '

Therefore, the additional velocity

−GMm
¿
¿ ( 11.2−8 )=3.2 km/s
2a
GM 2 −1 2
=v ⇒ E= mv
a 2 97 (c)
2
92 (c) 40 R
= =√ 2 R=R+ h
Gm 80 ( R+h )2
g=
R
2
h=( √ 2−1 ) R
4 2
M= ω R ρ
As angular momentum is conserved,
3 98 (a)

So ρ= hence
3g
4 πRG
I 1 ω 1=I 2 ω 2
93 (d) Or M R 1 ω 1=M R 1 ω 2 or M R 1 v 1=M R2 v 2
2 2

Inside the earth g =g 1− [ ] h

(∵ ω= Rv )
'
R

P a g e | 53
 Or R1 v 1=R 2 v 2 or VR=vr

∴V=
vr
R
⇒v =
2 2GM
Re ( )
Re
1− ' ⇒ v=
R
2GM R e
2
Re √ Re
1− '
R ( )
99 (c)
T ∝R
2 3
√
∴ v = 2 g Re 1− ( ) Re
R
'

1. Gravitational force on the particle

2 3 2
T =K R ⇒ ( 24 ) =K ( 36000 )
3 104 (b)

placed at the midpoint D of side AB

1
K=
of length a is ¿ ⃗ F3 . But ⃗
9 ×10 √ 10
4

F 1+ ⃗
F 2 +⃗ F1
and ⃗
F 3 are equal in magnitude and
1
'
T= ( 6400 )3 /2 =2h
9× 10 √ 10
4

opposite in direction

Using conservation of energy:

100 (b)

1 2 GMm −GMm
m ( K ve ) − =
2 R R +h
Use v e= √ 2 g R and GM =g R 2
and to solve we get h+ R=R/(1−K 2 )

101 (c)
GM ' GM
[ ]
2 2
g= ,g = 4 GMm 4 G M 2 3a
2 F=F 3= = ∵C D =
R ( 0.99 R )2 3a
2
3a
2
4

( ) 2. Gravitational force on the particle

' 2 2
g R '
∴ = ⇒ g >R
placed at the point of intersection
g 0.99 R

102 (c) of three medians ¿ ⃗

F 1+ ⃗ F3 =0,
F 2 +⃗
since the resultant of ⃗
F 1 and ⃗
F 2 is
(i)
2
m v GMm
equal and opposite to ⃗
= 2
R R F3

and (ii)
2

m v ' GMm
= '
R'
2

Dividing Eq. (i) by Eq. (ii),

v 2 R'
2 =
v' R

⇒
v'
v √ √[
R
= '=
R
2
3
∵ R' =
3R
2 ]
⇒ v'=
√ 2
3
v
Here time period of the satellite w.r.t. an
105 (a)

observer on the equator is 24 h and the

Let the mass of the particle be m
103 (d)
satellite is moving from west to east, so
PE at a distance of R' =(GMm)/ R' angular velocity of the satellite w.r.t.
PE at a distance of Re =−(GMm)/ Re earth’s axis of rotation (considered as
Decrease in PE = Increase in KE fixed) is ω= + , where T s and T e are
2π 2 π
T s Te
time periods of satellite and earth,
GMm GMm 1 2
⇒− + = mv
respectively
R
'
R e 2
2
v =2GM
[ 1 1
Re R
2
− ' ⇒v =
]
2GM
Re
Re
1− '
R [ ] π −1
ω= ( h ) =1.45 ×10 rad s
6
−4 −1

P a g e | 54
 From, v=
√
2 3
GM T ∝a
r 13 3
T 1 ∝ ( 10 )
2

rω=
√
GM
r
2
( T 2 ) ∝ ( 1012 )
3

( )
1 13 3 1
T 10 T
√ ∴ 2 = 12 2 ⇒ 2 =10/ √ 10
3
GM
r 2= T 10 T
r
7 4
r =2.66 ×10 m=2.66 ×10 km
The point lies inside both the shells,
111 (d)

hence gravitational field due to both is

106 (b)

zero
2

R=¿ radius of earth

m g=mR ω

ω=
√ g
R During total eclipse, total attraction due
112 (c)

to the Sun and the Moon,

T =2 π
√
R
g
=2 π √ 64000
2 π ×800
F 1=
G Ms Me G Mm Me
2
r1
+ 2
R2
When the Moon goes on the opposite side
¿ 2 π × 800 s= h=1.36
3600
¿ 0.14 h of the earth, the effective force of
attraction is

Inside the spherical shell, V is constant,

107 (d)
G Ms Me G Mm Me
so from energy conservation,
F 2= 2
− 2
r 1 r2

Change in force, ∆ F=F 1−F 2=

−GMm m v GMm
2 2G MmMe
= − 2
r2
Change in acceleration of the earth,
3R 2 R

[ ] × or v= 4 GM
√
2
v GM 1 GM 2
= 1− =
2 R 3 R 3 3R ∆ F 2 GMm
∆ a= =
Me 2
r2
Average force on the earth,
If the binary stars are rotating about their
108 (d)

common centre of mass, then they have

F 1+ F 2 G M s M e

to be on the same line all the time,

F av = = 2
2 r1
otherwise the centre of mass will be Average acceleration of the earth,
changing. Their angular velocities have to F av G M s
be the same although in the same time,
a av = = 2
Me
the smaller mass will describe a bigger
r1
Percentage change in acceleration is
circle
()
2 2
∆a 2G Mm r1 r1 Mm
×100= 2
× ×100=2 ×100
a av r2 G Ms r2 Ms

113 (d)
GM m
p= ρatm g h= h
But as ω= so, T A=T B
2π 4 3 3 R
2
π [ ( R +h ) −R ]
T 3
4 3
G πR l
As the sphere is having non-uniform mass
109 (c) m 3
⇒ p= × ×h
density, so nothing can be predicted [( ) ]
3 2
4 3 h R
π R 1+ −1
about the variation of gravitational field
3 R

intensity

110 (c)
P a g e | 55
 ( ) Now, h m= or h m ∝
2
m 4 4 u 1
¿ ×G πR lh Here M = π R 2 l
4
3
3 3h
π R 1+ −1
R [ ] 3 3 2g

Hence h m g =h m g
8

Hence, p=
' '
mGρ
or h m=6 hm =6 ×0.5=3 m
3R '
pR
∴ m∝
ii. t= or t ∝
ρ 2u 1
114 (a) g g

∴ t g =t g or
−G M 1 M 2 '
U g= t g 6
R
' '
= =
t g' 1
−GMm GMm
⇒ U f −U i= −
2R R 119 (c)

(at height R )
∆ u=
GMm m g R
2R
=
2
GM
∵ g= 2
R ( ) PE=
−GM
2R
GM R m −1
115 (b) ¿− 2
= mgR
2R 2
2
F=(GMm)/R =m g
As the mass remains the same and radius, 120 (b)
i.e., distance from the centre is also Force on m due to masses at Q and R is
same, therefore g will remain the same, zero. So, the net force is due to the mass
i.e., 10 m s−2 at P
Hence, F=(GMm)/ ( PL )
2

At a distance x from the centre of earth,

116 (a)

the gravitational force

GMm
F= 2
x
GMm
dW =Fdx= dx
Now, PL=l sin 60°=l
x
2
√3
We get W = ∫ [ ]
R +h
GMm 1 1 2
2
dx=GMm − 4 GMm
R x R R+h F= 2
3l
¿ g R2 m
[ 1
−
1
R R+ h ]
Therefore, the gain in potential energy is Mass of the cavity¿ M /8 if mass of the
121 (d)

sphere ¿ M as volume of the cavity is

g R2 m
[ 1
−
R 2R
1
=
mgR
2 ] 1/8th of the sphere

As gravitational force is a conservative

117 (b)

force, work done is independent of path

∴ W 1=W 2=W 3

118 (a)
i. g' =¿ accleration due to gravity on the [ ] [ ]
GMm GMm GMm 1 1 GMm 9−2 G
F 2= = = 2 − = 2 =
Moon ( )
2 2
4R 3 R 4 18 R 36
8 R
2
4 ' ' 4 R 2 g F1 36 9
¿G π R ρ =G π × × ρ= ∴ = =
3 3 4 3 6 F 2 4 ×7 7

P a g e | 56
 Equal time is taken to cover equal area
122 (c) 123 (b)
2
mv −5/ 2 −3/ 4
∝R ,∴v ∝R
R

Now, T = orT ∝ ( )
2
2 πR 2 R
v v

(R ) orT 2 ∝ R 7/ 2
2
2 R
T ∝ −3 / 4

125 (c)
If x 1 and x 2 are the distances covered by
the two bodies, then x 1+ x2 =9 R

Also, M x 1=5 M x 2 ⇒ x 2=
x1
5
x1
x 1+ =9 R ⇒ x 1=7.5 R
5

126 (d)

R
F1
F R
R 30o
30o
R 60o R
R
F2

⃗
F =⃗
F 1+ ⃗
F2
As ¿ ⃗
F 1∨¿∨ ⃗
F 2∨¿
∴|⃗
F|=2 F1 cos 30 °
G M √3 √ 3 G M
2 2
¿2 =
( 2r )2 2 4 r2

127 (d)
GMp
G Me 4 4
g p=g e ; 2
=
; π R p d p = π Re d e
2
R Re 3p
3

or 2 Re d p =Re d e ⇒ =
dp 1
de 2

Statement (a) is incorrect; there is no

128 (a,c,d)

method by which a body can be shielded

from the gravitational field of another
body. Statement (b) is correct. Statement
(c) is incorrect. As the earth rotates
about its axis, a body on the surface of
the earth also rotates with it. Since the
body is in a rotating (non-inertial) frame,
it experiences an outward centrifugal
force against the inward force of gravity.
As a result, the acceleration due to
gravity decreases due to rotation.
P a g e | 57
Statement (d) is incorrect, the forces are
equal in magnitude but opposite in
129 (b,d)

The acceleration due to gravity is g=

GM
direction. Hence, choices (a), (c) and (d)
are wrong
2

The new value of g would be

R

G(0.99 M ) GM
g' = 2
=1.01 2
( 0.99 R ) R
Thus, g would increase by about 1%. The
new escape velocity would be

v e=
√ 2 ×0.99 M × G
√
=
2 MG
=v e
Thus, the escape velocity will remain
0.99 R R

unchanged. The potential energy of a

body of mass m on the earth’s surface
would be
−Gm ( 0.99 M ) −GmM
=
( 0.99 R ) R
Thus, the potential energy will also
remain unchanged. Hence, the correct
choices are (b) and (d)

When the particle is dropped in the

130 (a,b,c)

tunnel at its one end, it starts to execute

simple harmonic motion with mean
position as the centre of the tunnel.
Hence, acceleration of the particle is
directly proportional to its distance from
the centre of the tunnel. It means, option
(d) is correct
The particle has zero velocity at one end
because it is dropped from rest at that
end, therefore, ends of the tunnel are its
extreme positions. Hence, at the other
end, the velocity will become equal to
zero. It means option (a) is incorrect
Velocity of a particle executing S.H.M. is
maximum at its mean position. Hence, at
the centre of the tunnel, its velocity will
be maximum possible. Hence, option (b)
is incorrect
When the particle moves from extreme
position to centre of the tunnel, its
velocity increases. It means KE increases
or PE decreases. But initial potential
energy of the particle is negative. Hence,
potential energy becomes more negative
and it is least at the mid-point of the
tunnel because KE is maximum there. It
means the gravitational PE can never be

P a g e | 58
 equal to zero. Hence, option (c) is also
incorrect If the speed of the orbiting satellite is
134 (a,c)

made √ 2 times, then the satellite will

escape because its velocity becomes
131 (b,c,d)

and U = equal to escape velocity

g=
GM
R 2
, ve=
√ 2 GM
R
−GMm
R This implies that the speed is increased
by 41% and the KE is doubled
and U ∝
M
√
∴ g∝ 2 , v e ∝
M
R
M
R
If both mass and radius are increased by
R
1. For a proper explanation, the
135 (a,b,c)

0.5% then v e and U remains unchanged

earth-Moon system will be
where as g decrease by 0.5%
regarded as an isolated system. At
a particular point in space, the
Distances of the two satellites from the gravitational force of attraction of
132 (b,c,d)

centre of the earth are r 1=2 R and the earth on astronaut will be
balanced by the gravitational force
r 2=−8 R , respectively ( R=¿earth’s)
of attraction of the Moon
radius. Their potential energies are
2. V =−GM / R. As R decreases, GM /R
andV 2=
−GmM −GmM
increases or – GM / R decraeses
V 1=
r1 r2

Their ratio is 3. In case of the spherical shell, plot

V 1 r2 8 R
= = =4
of I versus r is discontinuous
V 2 r1 2 R
The kinetic energy of a satellite can be
obtained from the relation
If two satellites of different masses are
136 (c,d)
2

revolving in the same orbit, then they

m v GmM
= 2
have the same time period and speed
r r
or K= mv =
because
1 2 GmM
2 2r

and K 2=
K 1=
GmM
2 r1
GmM
2 r2
The ratio of their kinetic energies is
v 0=
√ GM
r

and T =2 π
K 1 r2 8 R
= =
K 2 r1 2 R
=4 √ r3
GM
Angular momentum (¿ mvr ) and energy
Their total energies are (¿−(GMm)/2 r ) both depend on the mass
−GmM GmM −GmM of the satellite. So, options (c) and (d) are
correct
E 1= + =
r1 2 r1 2 r1

and E2=
−GmM GmM GmM
+ =
The gravitational field intensity at point O
r2 2r 2 2 r2 137 (a,c,d)

Their ratio is is zero (as the cavities are symmetrical

E1 r 2 8 R
with respect to O )
= = =4
E2 r 1 2 R
Hence, the correct choices are (b), (c)
and (d)

133 (a,b,c,d)
For all the points, E P > E K (numerically).
So, the total energy is negative. Thus, the
system is a bound system is a bound
system corresponding to all the points

P a g e | 59
 (iv) v=v eParabolic path and it escape from
the earth

(v) v> v e Hyperbolic path and escape from

earth

If a satellite is revolving in a circular orbit

140 (a,b,d)

then the magnitude of its KE is equal to

half of the magnitude of gravitational
Now the force acting on a test mass m 0 potential energy. But the potential energy
placed at O is given by is negative. That is why the total energy
F=m0 E=m0 ×0=0 of a revolving satellite is negative. If the
Now, y 2 + z 2=36 represents the equation of satellite experience resistance due to
a circle with centre (0, 0, 0) and radius 6 cosmic dust, then it follows a spiral path
units. The plane of the circle is of decreasing radius. During the process
perpendicular to the x -axis. Since the of motion, its potential energy decreases.
spherical mass distribution behaves as if It means PE becomes more negative or
the whole mass is at its centre (for a the magnitude of PE increase. Hence,
point outside the sphere) and since all the magnitude of KE also increases.
points on the circle are equidistant from Therefore, options (a) and (b) are correct
the centre of the sphere, the circle is a Since cosmic dust exerts a tangential
gravitational equipotential force on the satellite, therefore it
experiences a retarding moment, hence,
its angular momentum does not remain
For an earth satellite moving in a circular
138 (b,c)
conserved. Hence, option (c) is incorrect.
orbit, the centripetal force required for Since KE of the satellite increases,
its circular motion is provided by the therefore speed of the satellite increases.
gravitational force exerted by the earth Power acting against the resisting force is
on it equal to force × speed, therefore it
It means the resultant force on the increases. Hence, thermal power
astronaut is equal to the gravitational generated increases with time. Hence,
force exerted by the earth on him. Hence, ultimately it will turn off. Therefore,
no reaction is exerted by the floor of the option (d) is correct
satellite on him
In other words, his acceleration towards 141 (a,b)
the earth’s centre (centripetal
g= m ρ Gr ∴ g∝ r if r < R
4
acceleration) is exactly equal to 3
acceleration caused by the gravitational GM 1
force
g= 2 ∴ g ∝ 2 if r < R

Hence, options (b) and (c) are correct

r r
if 1
r < R and r 2 < R

then
F 1 g1 r1
Velocity of nature of path of satellite
139 (a,b,c,d) = =
F 2 g2 r2
(i) v=v 0 Circular path around the earth if r 1 > R and r 2 > R

(ii) v< v 0 Elliptical path and body returns then ()

2
F 1 g1 r2
= =
to earth
F 2 g2 r1

(iii) v> v 0 but v eElliptical path around the

1. Factual statement
142 (a,b,c,d)

earth and will not escape

P a g e | 60
 2. or M =
GM gR
2
2 π R 2 2 π × 1.28× 104
g= 2 v 2= =
R G T2 4

3. The weight of a body on the

4

At t=12 h the two satellites are closest to

¿ 0.64 π ×10 km
surface of earth is not determined
each other and after every 24 h they
by the orbital motion of earth
come at the same position relative to
around Sun
each other. It is clear that direction of v 2
143 (a,c,d) w.r.t. v 1 is changing continuously in both
1.
GM G 4 magnitude and direction
Angular velocity of S2 w.r.t. S1 at t=12 h is
3
g= 2 = 2 × π R ρ
R R 3

or ρ=
−1
3g ω=v 1+ v 2 /R 2−R1=0.468 π rad s
4 πGR
146 (b,c)
2. Acceleration due to gravity varies

{
GMm
from plane to place, hence, not
r≥R
correct
2
F= r
4 πG ρ rm
r<R
3.
3
where, ρ is density of earth

√ √
' 2 G(2 M ) 2GM −2 −1
v= =2 =2× 11. 2 km s =22.4 km s (c,d)
Linear speed is v=rω. At the equator, the
e
R R 147

radius of the earth r is maximum.

2

4. Factual statement
Therefore, ω= is minimum. Also,
v
r
Work done is independent of the path
144 (a,c,d)
2π 2 ×3.14
ω= =
chosen and depends only on the initial
T 24 ×60 × 60
and final positions of the object. Also the
−5 −1

Hence, the correct choices are (c) and (d)

¿ 7.3 ×10 rad s
work done on any closed path in a
gravitational field will be zero. Since,
every point on the surface of the earth is
148 (a,d)

at the same potential, no work is done for

−GMm
U i=
points on the surface of the earth. Hence,
R

the correct choices are (a), (c) and (d)

−GMm
U f=
Work done = Change in gravitational
R+h

potential energy
145 (a,b,c,d)
From Kepler’s law, T 2 ∝ R 3

So,
T 1 R1
=
T 2 R2 ( )
3
2 ⇒ W =U f −U i =−GMm ( R+1 h − R1 )
[( ) ]
−1
−GMm h
( )
2 ⇒W = 1+ −1
( )
2
T2 3 24 4 R R
R 2= × R 1= ×1.28 ×10 km
3
T1 6
¿ 3.22 ×10 km
4 ⇒W =
−GMm h
[
1− −1
]
Orbit velocity of S1 is
R R
GMmh
2 π R 1 2 π × 1.28× 104 ⇒W = 2

⇒ W =mg h for h ≪ R
v 1= = R
T1 4

Also, U i= , and
4
¿ 0.64 π ×10 km −GMm
Orbital velocity of S2 is R

P a g e | 61
 √
−GMm −GMm (9 M )
U f ( h=R )= = V R= 2 G
R+ R 2R ( 9 )1 /3 R
GMm

√
⇒ W =U f −U i = 2 GM
2R ¿ 91 /3
R
1
⇒W = m g R ∴ g= 2
2
GM
R { } So V R >V Q >V P and
VP 1
=
VQ 2
149 (b,c)

For r > R , the gravitational field is

152 (a,b)
W =U 2−U 1 =
GMm −GMm
R+ h
−
R
1
=−GMm −
1
R R+h ( ) [ ] F=GM /r
2

¿gR m
h
2

R(R+h) [=
m g Rh
R+ h ] ∴ F1= 2 and F 2= GM
GM
⇒ =
2
F1 r 2

For h≪ R , W =m g h
2
r1 r2 F2 R21
For r > R
For h=R ,W =
mgR
The gravitational field is F=
2 GM
3
×r
R
×r 1and F 2= 3 ×r 2
150 (b,c,d) GM GM
∴ F 1=
( ) {∴ r= r +2 r }
3
4 π r A +r P
2 3
2
T =
A P R R
GM 2 F1 r 1
2 ⇒ =
2 π 3 F2 r 2
⇒T = ( r A +r P )
By law of conservation of angular
2GM

Statement (a) is incorrect; the magnitude

153 (b,c)
momentum
of the force is 6.67 ×10−11 N. Statement (b)
is correct. If the escape velocity for a
m v A r A =m v P r P

planet is high, gases are not able to

⇒ v A r A =v P r P

escape. Statement (c) is also correct.

Statement (d) is incorrect; force of
151 (b,d)

Escape velocity V es =
2 GM
R
Surface area of P= A=4 π R P
2
√ friction arises due to electrical forces.
Hence, correct choices are (b) and (c)

Surface area of Q=4 A=4 π RQ

2
154 (c,d)
⇒ RQ =2 RP GM
V=
If M P =M =ρ π R (Let R be the radius of
4 3
R
GM
P)
3 ⇒V = 2 R
R

So M Q= ρ
4 3 ⇒ V =g R
π ( 2 R ) =8 M
3
So, M R =M P + M Q =9 M
If universal gravitational constant G
155 (b,d)

For planet R , 9 M =ρ π Rr starts to decrease, the gravitational force

4 3

between the Sun and the earth will also

3
So, Rr =( 9 ) . R start to decrease. Therefore, the earth
1 /3

will try to follow a path of larger radius.

Escape velocity from P , V P =
√ 2 GM
R Hence, its period of revolution round the
Sun will increase. Therefore, duration of
V Q=
√ 2 G(8 M )
2R
=2
2 GM
R √ the year will increase. But rotational
motion of the earth about its own axis will
remain unchanged; hence, period of
rotational will remain unchanged or
P a g e | 62
 length of day will remain unchanged. The distance of centre of mass from the
Hence, option (a) is incorrect, (b) is
heavier star is equal to ( Mr+2
correct and (c) is incorrect M +2 M ) 3
M .0 r
=
Since the radius of the circular path of Hence the heavier star revolves in a
the earth will increase or the earth will circle of radius r /3 while the lighter star
follow a spiral path of increasing radius, in a circle of radius 2 r /3. Hence, option
therefore its PE will increase but PE is (a) is incorrect
always negative, so the magnitude of PE To calculate period of revolution of a
will decrease. Hence, KE will also double star system, concept of reduced
decrease. Therefore, option (d) is correct mass can be used
The reduced mass of the system is equal
to
Since every element of the hemispherical
156 (b,c)

shell is at a distance R from the centre of

( M ) (2 M ) 2 M
curvature, therefore gravitational
=
( M +2 M ) 3
potential at its centre¿−GM / R . It is not Hence, the period of revolution will be
equal to GM /R . Hence, option (a) is
3

equal to where r is distance

2π
incorrect
r2

Gravitational field strength at a point,

lying on the axis of a thin uniform circular
√
2
GM

between two stars. Hence, option (b) is

3

ring of radius R , can be calculated by correct

considering two equal elemental lengths Kinetic energy of a star will be equal to
of the ring such that these are lengths 1/2 m v where v is speed of the star which
lying on the ends of a diameter and thus
2

is equal to (radius of its circular orbit)× ω

the expression given in option (b) is Hence, KE of heavier star is
obtained. Hence, option (b) is correct
Newton’s law of gravitation is applicable ( )
2
1( ) r
E = 2 M ω
to only those bodies which have
1
2 3

spherically symmetric distribution of And that of lighter star, E2= M ( )

2
1 2r
mass. For example, if we consider two
ω
2 3
It means, KE of the lighter star is twice of
solid hemispheres of equal radii; one
the heavier star. Hence, option (c) is
made of wood and the other made of iron.
incorrect
These two hemispheres are joined
together to form a shape of complete
sphere, we cannot apply Newton’s
159 (a,b,d)

gravitational law to this sphere. So,

−k
F (r )= n
'

option (c) is correct

r
−k 1
⇒ U ( r ) =−∫ F ( r ) dr =
(n−1) r n−1
According to right Hand Thumb Rule If L is the angular momentum of the
157 (a, c)

“curl the fingers of right hand in the particle of mass m in an orbit of radius r
direction of rotation then thumb gives the then
direction of the areal velocity/angular
Kinetic energy ¿
2 2

momentum”.
L L
= =k (r )
2 I 2 m r2
Since, total energy ¿ E ( r )=U ( r )+ K ( r )

The centre of mass of the double star

158 (a,c) 2
−k 1 L
system remains stationary and both the
⇒ E ( r )= +
(n−1) r n−1 2 m r 2
stars revolve around in circular orbits The criterion that a circular orbit of
which are concentric with the centre of radius r 0 be stable is that E ( r ) is minimum
mass For E(r) to be minimum two conditions
P a g e | 63
 must be fulfilled. −GMm GMm −GMm
E= − =

| | Let the orbital velocity be v , then from

2r 2r r
2
∂E ∂ E
⇒ =0∧ >0
momentum conservation,
2
∂r ∂r
Using both these conditions, we get
r=r0 r=r 0

2 mv−mv=2 m× v 1
( 3−n ) L > 0 v 1=0
This is possible only when n<3 As velocity of combined mass just after
m

We also note that inverse square law collision is zero, the combined mass will
belongs to this category fall towards the earth. At this instant, the
n=−1 also gives stable circular orbits total energy of the system only consists of
the gravitational potential energy given
(Law of direct distance)
by
But n=3 gives circular orbits which are
unstable (Inverse cube law)
GM × 2m
U=
2r

Choices (a), (b) and (c) are correct. When

160 (a,b,c)
164 (b,d)
the particle exerts a force on another Now, g=GM / R2
particle; the second particle exerts an If R reduces to R' =0.8 R , the value of g
equal and opposite force on the first becomes
particle. Choice (d) is incorrect because
−2

although G is constant everywhere, the

' GM GM g 9.8 m s
g= 2 = 2
= =
value of g varies with height and depth
0.64 R 0.64 0.64
Which is choice (b)
R

and also from place to place

Increase in the value of g, ¿
g 0.36 g
=
Therefore, the percentage increase
0.64 0.64
161 (a,c,d)

From relation |E|=

dV
0.36 g
¿ × 100=56.25 %
When V =0 , E=0 ; when V =¿ constant,
dr
Hence, choices (b) and (d) are correct
0.64 g

E=0; when V ≠ 0 , E ≠ 0
So, options (a), (c) and (d) are correct 165 (a,c)

If r > R , g =g
2 2
' R R
162 (a,d) 2
=g 2
( R+ h ) r

If r < R , g =g [ ]
' R−d
R
Where d is the depth below the surface of
earth
If d=R , g=0
Further, due to rotation: g' =g−R ω2 cos2 θ
If ω=0 , then g increases
2 mr 2r
r 2= =
m+2 m 3
2 2 2 3
4 π r2 2 32 π r

1. For Moon, escape velocity is

2
T 2= ⇒ T 2= 166 (a,d)
Gm 27 Gm
and T ∝ m
3 1 −1
2 2 2.4 km s
T 2∝ r 2

2.
Just before collision, the total energy of
163 (a,b,d) v 0=
√ 2 GM
R
=
√
2G 4
× π R3 ρ
M 3
two satellite is
or, v 0 ∝ √ ρ

167 (c,d)

P a g e | 64
 some angle w.r.t. vertical direction,
Linear velocity or orbital velocity is the body will reach up to a certain
v=√ GM /r where r is the distance of the height (where vertical component
satellite from the centre of the earth. of the velocity becomes zero) and
Therefore, v decareses as r is increased then fall done following a parabolic
Also v=rω, where ω is the angular path (this will be a case of
velocity projectile motion)

Therefore, ω= = 3. If v=v 0 =√ GM / R, then the orbit will

v
r r√ √
1 GM
r
=
GM
r3
Thus, ω decreases with increase in r . The
be circular

time period of the satellite is given by But if v 0< v < v e (escape velocity), then
the orbit will be an ellipse
√
3
r
T =2 π
So, all the options are correct
Thus, T incraeses as r is increased.
GM

Hence, the correct choices are (c) and (d) 170 (b,c)
When a planet is on the major axis of the
orbit, gravitational force on the planet is
168 (b,c)

normal to its motion. So, no work is done.

N

As that energy remains the same, no

mg R/ 2
work is done in complete revolution
x

If radius of an orbiting satellite is

171 (b,c)

Net force towards the centre of the earth

decreased, then its PE gets decreased. As
ME = KE + PE, so ME also decreases
2
' mGMx m g R x
¿mg = 3
= 3
=(m g x)/ R

[ ]
R R
Normal force, N=m g' sinθ
−GMm +GMm −GMm
PE= , KE= ∧ME=
r 2r 2r
Thus, pressing force, N=
mgx R

Force on the satellite is always the earth;

R 2x 172 (a,c)

is constant and independent of x

therefore, acceleration of satellite Sis
mg
N=
Hence, option (b) is correct always directed towards the centre of the
2

Tangential force, F=ma=m g' cos θ earth. Net torque of this gravitational
force F about the centre of the earth is
zero. Therefore, angular momentum
√
2
R
−x2
(both in magnitude and direction) of S
gx 4 gx
a=g' cos θ= ⇒ a= √ R 2−4 x 2
about the centre of the earth is constant
R x R
Curve is parabolic and at x= , a=0 throughout. Since force F is conservative
R
in nature, therefore mechanical energy of
Hence, option (c) is correct
2
the satellite remains constant. Speed of S
is maximum when it is nearest to the
earth and minimum when it is farthest
169 (a,b,c,d)
1. If v< v e and the body is projected
vertically upwards, the body will
rise up to that height where its
If the orbital path of a satellite is circular,
173 (e)

velocity becomes zero. After that it

then its speed is constant and if the
will fall freely due to gravity
orbital path of a satellite is elliptical, then
following a straight line path
its speed in its orbit is not constant. In
2. If v< v but the body is projected at that case its areal velocity is constant
e

P a g e | 65
 Gravitational Flux ( ϕ g )=∫ ⃗
If a pendulum is suspended in a lift and
174 (a) E . d ⃗s

lift is moving downward with some For any closed surface ϕ g=4 π GM
acceleration a , then time period of

pendulum is given by, T =2 π and gravitational field E ∝

√
1
l 2
g−a r

In the case of free fall, a=g then T =∞

The time period of the satellite which is
180 (a)

i .e . , the time period of pendulum very near to the earth is given by

becomes infinite

175 (c)
T =2 π
√ R
g
≈ 84 min=1 h 24 min

GM
The value of g at any place is given by the
V ¿=
2 2
[3 R −r ] 182 (c)
3
2R
relation,
At surface, V s =
GM
g =g−ω Re cos λ . When λ is angle of
[at r=R]
R ' 2 2

latitude and ω is the angular velocity of

earth
3
V ¿= V s
2

V ¿> V s If ω=0 , ∴ g ' =g . If there is no rotation

V is not the same everywhere as Assertion is true but reason is false

indicated by V ¿
Here, Statement 1 is incorrect as speed
183 (d)

of one satellite increases, its kinetic

For Statement 1: Gravitational force is
176 (d)

energy and hence total energy increases,

not perpendicular to velocity of the
i.e., total energy becomes less negative
satellite. So for any small part of the orbit
and hence r increases, i.e., orbit changes
work done is not zero, although when
satellite is at perihelion or aphelion
position then work done by gravitational Binding energy is the minimum energy
184 (d)

force for small part would be zero required to free a satellite from the
gravitational attraction. It is the negative
value of total energy of satellite. Let a
The value of escape velocity is derived
177 (d)

satellite of mass m be revolving around

from the method of conservation of total
mechanical energy and energy is earth of mass M e and radius Re total
independent of direction energy of satellite
−G M e m 1 2
178 (a) ¿ PE+ KE= + mv
Re 2
dA 1 2 d θ 1 2
= r ⇒ = r ω
dt 2 dt 2 −G M e m G M e m
¿ +
2
Re 2 Re
mr ω 2
= =constant
2m m GMm
¿−
∴ L=¿constant
2 Re

∴ Binding energy of satellite = −¿(total

energy of satellite)
179 (a)

P a g e | 66
 which depend on mass of the satellite remains unaffected

Assertion is false but reason is true

According to Kepler’s third law of motion,
190 (a)

185 (d) the square of the time period of a planet

As, escape velocity ¿ , so its value about the Sun is proportional to the cube
√
2 GM

depends on mass of planet and radius of

R of the semi-major axis of the ellipse or
mean distance of the planet from the Sun,
the planet. The two different planets have i.e., T 2 ∝ a3. When a is smaller, the time
same escape velocity, when these period is shorter
quantities (mass and radius) are equal

Earth revolves around the sun in circular

191 (e)
186 (a)

From kinetic theory of gas, v= path and required centripetal force is

√ 3 RT
M provided by gravitational force between
earth and sun but the work done by this
So, for lighter gas molecules, v is greater
centripetal force is zero
which is enough to take these molecules
away from the earth’s atmosphere
For a satellite to be geostationary, the
192 (d)

necessary requirements are as follows:

187 (a)

for a satellite revolving very

v 0=Re
√
g
Re + h
near the earth surface Re +h=R e
1. Its orbit must be in equatorial
plane and circular;

v 0= √ R e g 2. Its time period must be 24 h;

3. Its sense of rotation must be the

¿ √ 64 ×10 5 ×10 same as that of the earth about its
3 −1 −1 own axis
¿ 8 ×10 m s =8 km s

Which is independent of height of

Acceleration due to gravity,
193 (b)

satellite

Both Assertion and Reason are true and

' 2 2
g =g−R ω cos λ
reason is the correct explanation of
At equator, λ=0 °i . e . cos 0 °=1 ∴ ge =g−R ω
assertion
2

At poles, λ=90 ° i. e . cos 90 °=0 ∴ g p =g

As the geostationary satellite is
188 (b)

Thus, g p=g e =g−g+ R ω =R ω

established in an orbit in the plane of the
2 2

equator at a particular place, so it move Also, the value of g is maximum at poles

in the same sense as the earth and hence and minimum at equators
its time period of revolution is equal to 24
hours, which is equal to time period of
Variation of g with depth from surface of
194 (c)
revolution of earth about its axis
earth is given by

As a rotation of earth takes place about

189 (a)

polar axis therefore, body places at poles (

g' =g R 1−
d
)
will not feel any centrifugal force and its
R

weight or acceleration due to gravity At the centre of earth, d=R

P a g e | 67
 universe, including stars
(
∴ g' =g 1−
d
R)=0

Because gravitational force is always

201 (a)
∴ Apparent weight of body ¿ m g =0
attractive in nature and every body is
'

Assertion is true but reason is false bound by this gravitational force of

attraction of earth

The value of g at any place is given by

195 (c)

According to Newton’s law of gravitation,

202 (a)
' 2 2
g =g−ω Re cos λ Gm 1 m2
F=
If ω=0 , then g' =g. The value of g will be
2
r
same at all places When m 1 ,m 2 and r all are doubled,

Force acting on astronant is utilised in

196 (a) 2
v GM GM gR
F= = =
providing necessary centripetal force,
2 R (R+ h) R
thus he feels weighlessness, as he in a ie , remains the same.
state of free fall.
Both assertion reason are true and reason
is correct explanation of assertion
We can take the sphere as a point mass
197 (d)

lying at its centre, but the rod will not be

If root mean square velocity of the gas
203 (b)
taken as point mass lying at its centre of
mass molecules is less than escape velocity
from that planet (or satellite) then
atmosphere will remain attached with
The total gravitational force on one
198 (e)
that planet and if v rms >u escape then there
particle due to number of particles is the
will be no atmosphere on the planet. This
resultant force of attraction (or
is the reason for no atmosphere at moon
gravitational force) exerted on the given
particle due to individual particles, 204 (a)
i .e . , ⃗
F =⃗
F 1+ ⃗ F3 + … It means the
F 2 +⃗ The torque on a body is given by τ⃗ =⃗
dL/dt .
principle of superposition is valid In case of a planet orbiting around the
Sun, no torque is acting on it

As no torque is acting on the planet, its

199 (a)
d⃗
angular momentum must remain constant
L
=0 ⇒ ⃗
L =constant
in magnitude as well as direction.
dt

Therefore, plane of rotation must pass 205 (a)

through the centre of earth According to kepler’s law T ∝ r ∝ ( R+h )
2 3 3

i .e . if distance of satellite is more then its

According to Newton’s law of gravitation, time period will be more
200 (a)

every body in this universe attracts every

other body with a force which is inversely
Both the statements are true and
206 (a)
proportional to the square of the distance
Statement 2 correctly explains Statement
between them. When we move our finger,
1
the distance of the objects with respect to
finger changes, hence the force of
attraction changes, disturbing the entire Here, both the statements are correct
207 (a)

P a g e | 68
and Statement 2 correctly explains
Statement 1 The orbital velocity, if a satellite close to
208 (d)

earth is v 0=√ g R e , while the escape

This can be easily proved by writing the
velocity for a body thrown from the
basic equations
earths surface v e= √ 2 g Re

Thus
v 0 √ g Re 1
= = ∨v e= √2 v
v e √ 2 g Re √ 2

Assertion is false but reason is true

If r is the distance between two electrons

209 (b)

then according to Newtons law, the

gravitational force between them is

−31 2
m2 −11 ( 9.1 ×10 ) 5 ×10−71
F G=G 2 =6.67 × 10 × 2
≅ 2
r r r

and according to Coulomb’s law, the

electrical force between electron is

−19 2
1 q×q −9 ( 1.6 ×10 ) 2 ×10−28
F e= =9 ×10 × ≅
4 π ε0 r 2
r
2
r
2

F G 10−71 −43 −43

∴ ≅ ≅ 10 ie , F G =10 F e
F e 10−28

ie , gravitational force between two

particles is negligible compared to the
electrical force.

Both assertion and reason are true but

reason is not the correct explanation of
assertion

Till the particle reaches the centre of the

210 (a)

planet, force on both the bodies are in

direction of their respective velocities,
hence kinetic energies of both keep on
increasing. After the particle crosses the
centre of the planet, forces on both are
retarding in nature. Hence, as the
particle passes through the centre of the
planet, sum of kinetic energies of both
the bodies is maximum. Therefore,
Statement 1 is true, Statement 2 is true;
Statement 2 is a correct explanation for
Statement 1

211 (b)
P a g e | 69
 We know that earth revolves from west to Speed of the planet is maximum at
216 (b)

east about its polar axis. Therefore, all perihelion and minimum at aphelion. So,
the particles on the earth have velocity KE is maximum at perihelion and
from west to east minimum at aphelion
Where KE is maximum, PE is minimum
This velocity is maximum in the
equatorial line, as v=Rω, where R is the
radius of earth and ω is the angular
velocity of revolution of earth about its
polar axis

When a rocket is launched from west to PE and KE of the planet are dependent on
east in equatorial plane, the maximum both mass of planet and on semi-major
linear velocity is added to the launching axis
velocity of the rocket, due to which
launching becomes easier
Angular momentum of the particle
217 (d)

Upto ordinary heights the change in the

212 (c) ¿ m(v 0+ v ) a

distance of a projectile from the centre of

the earth is negligible compared to the
¿
√ 5
4 [ √ ]
m v 0 a v 0=
G Me
a
Total energy of the particle
radius of the earth. Hence, projectile
moves under a nearly uniform gravitation 1 2 G Mem

force and its path is parabolic. But for

¿ m ( v0 + v ) −
2 a
projectile going to great height, the 1 5 G M em

gravitational force decreases in inverse

2
¿ × m v 0−
2 4 a
proportion to the square of the distance G M m G M em −3 G M e m
of the projectile from the centre of the
5 e
¿ − =
earth. Under such a variable force the
8 a a a
At any distance ‘r ’, T.E. ¿ mu −
2 G Mem
path of projectile is elliptical.
1

But angular momentum conservation

2 a
Both Assertion and Reason are true and
gives
reason is the correct explanation of
Assertion. mur=
√ 5G m e
4a

Acceleration due to gravity is given by

√
213 (c)
5 G Mea
⇒ u= a
g=GM /l . Thus, it does not depend on the
4 r2
T.E. at any distance
2

mass of the body on which it is acting.

Also, it is not a constant quality. It
1 5 Gmea G M e m
−
changes with changes in value of both M
‘ r ’= m
2 4 r2 r
But through conservation of total energy,
and R (distance between two bodies)
we have
1 5 G M e a G M e m −3 G M e m
According to Kepler’s law of planetary
215 (c)
m − =
motion, a planet revolves around the sun
2 4 r2 r a
On solving, we get
in such a way that its areal velocity is
constant, i .e . , it move faster, when it is
2 2
3 r −8 ar+ 5 a =0

closer the sun and vice-versa

( r −a ) ( 3 r−5 a )=0

P a g e | 70
 5a So iv → b , c
R=a , r=

Minimum distance ¿ a
3

As rocket has been fired the speed of the

220 (c)

Maximum distance ¿ satellite increases in all these three cases

5a

which leads to the increase in kinetic

3

energy and hence total energy becomes

1. Acceleration due gravity is less negative or we can say the semi-
218 (a)

different at different points and it major axis increases

does not depend upon direction of Due to thrust exerted by firing of the
projection rocket in a direction perpendicular to the
plane of the orbit, the plane of the orbit
2. Angular momentum of the earth changes
remains constant about the Sun iv. Here speed decreases and hence the
3. Escape velocity depends upon g kinetic energy, the total energy and the
and independent of direction of semi-major axis decrease
projection
1. For earth moving in an elliptical
221 (d)
4. Gravitational potential will remain
constant orbit. Centripetal force passes
through centre of mass. So its
torque will be zero at all points of
Gravitational potential is defined as the motion. Here, angular momentum
219 (c)

work done in carrying a unit mass from a can be conserved at all points of
point outside the field (zero potential) to motion. But as external force is
a point inside the field. So when acting, so linear momentum will be
reference point is altered the potential unconserved
will vary and the potential difference will
remain the same. Also the energy is not 2. Angular momentum can be
dependent on the angle by any means and conserved about the point of
depends only on the two points (initial contact as work done about this
and final) in any conservation field like point by force of friction is zero.
gravitational field Linear momentum will not be
So i→c, d conserved as force of friction is an
external force here
Escape velocity is given by √ GM / R and is
dependent on the mass of the planet and 3. External forces acting on the
does not depend on the angle of system are normal reaction,
projection m g sin θ and friction. Torque due to
So ii → d normal reaction and m g sin θ will be
Acceleration due to gravity is given by zero about centre of mass but not
g=GM / R =4 /3 π R ρ G and so its ratio in due to friction. So, angular
2

two planets depends on their densities momentum will not be conserved.

when their mass is not the same Also, in the presence of external
So iii → a forces, the linear momentum will
As the Kepler’s law, the angular not be conserved
momentum will remain conserved for
keeping the area swept in equal to 4. For the projected particle, force of
remain the same. The same concept was gravity which is external in this
proposed by Bohr’s model of atom case will render linear momentum
unconserved. But torque of this

P a g e | 71
 force will be zero as it will pass
through centre of mass. So angular
225 (d)
Let v 0 be the speed of the spaceship n the
momentum is conserved circular orbit of radius r 0 , and v e be the
instantaneously escape velocity for the orbit. Then,

√ √
2 2
m v0 G M m m ve G M m GM 2 GM
1. A geostationary satellite has
222 (b) = 2 , = ∨v 0= , ve=
r0 r0 2 r0 r0 r0
circular orbit over equator As v e=v 0 +|∆ ⃗v|cos θ=v 0 + ∆ v cos θ
The specific impulse required for escape
2. Total energy will remain constant
is the least for θ=0 ie , the initial velocity
in any orbit
of the spaceship and the impulse are in
3. Angular momentum will remain the same direction, and is given by
constant in any orbit

4. In circular orbit, speed remains the

∆ v=v e −v 0 =
√ GM
r0
( √ 2−1)

same, whereas in elliptical orbit

speed changes Applying the principle of conservation of
226 (c)

angular momentum,

1. At centre of thin spherical shell,

223 (a)
m v 1 ( 2 R ) =m v 2 (4 R)
v 1=2 v 2 (i)
From conservation of energy,
V ≠ 0 , E=0

2. At centre of solid sphere,V ≠ 0 , E=0

(ii)
1 2 GMm 1 2 GMm
m v 1− = m v 2−
3. At centre of spherical cavity inside
Solving Eqs. (i) and (ii), we get
2 2R 2 4R
solid sphere, V ≠ 0 , E ≠ 0

4. At centre of two point masses,

V ≠ 0 , E=0
v 2=
√ GM
6R
, v 1=
√2GM

If r is the radius of curvature at point A

3R

According to law of conservation of total

224 (d)

mechanical energy.
Total energy of rocket at the surface of
earth = total energy of rocket at the
highest point
2
m v 1 GMm
= 2
r (2 R )
or ( ) ( )
1 −GMm −GMm
(putting value of v 1 ¿
m v2 + =0+ 4 v1 8 R
2
2 R R+h r= =
or
2 2 2 GM 3
v GM GM gR gR
= − = −
2 R (R+h) R (R+ h) 227 (b)
¿ g R 1−
R
R+h (
=g R
h
R+h ) ( )
or v 2 ( R+h )=2 g R h
or R v 2=2 g R h−v 2 h=( 2 g R−v 2 ) h Gravitational PE at perihelion ¿−GMm /r 1
as r 1 is minimum
or h=
2
Rv
Therefore, PE is minimum
2
(2 g R−v )
2
6.4 ×10 6 ( 5 × 103 )
When 2 r < x <2 R
¿ 228 (c)
2
( 2× 9.8 ×6.4 × 106 ) −( 5 ×103 )
6
¿ 1.6 ×10 m

P a g e | 72
 extent of firing rockets may differ
m'

From work – energy theorem,

233 (b)
M
x
m dK =−d U +W air friction
W air friction is negative, so
Then the force will be due to sphere only
dK =−dU + ¿(a negative quantity)
As K increases, it means U decreases by
'
Gm m
an amount greater than magnitude of
F=
( x−r )2
When x >2 R W air friction
m'
234 (3)

Given R =3 R , M =
' 4 3
πR ρ
M x 3
' ' '2
m g =G M R
Putting M = π R ρ,
4
The sphere and the shell both will
' '3
3
contribute to the g=
' G 4 '3
π R ρ=
27 G 4 3
πR ρ
R 3 9 R 3
' '2 2
GM m
F sphere = GM
( x −r )2 ¿ 3 2 =3 g
GM m
' R
F shell =
( x−R )2 235 (6)
' ' −7 2
GM m GM m F=mr ω =10 ×1.5 × 10 × 10 × ( 2 ×10 )
2 24 8 3
F= +
( x−R )2 ( x−r )2 ¿ 6 ×10 N
21

229 (a)
F=−K /r (negative sign is for attractive
236 (3)
⃗
F =⃗ E m=( 10 i^ +24 ^j ) N 2

¿ √ ( 100 ) +576=26 N force)

Potential energy
230 (a) K −K
U =−∫ F dr =∫
and
GM GM dr=
E P= = r2 r
( 4 a )2 16 a2 Conservation of energy gives (let at other
E P=
GM GM 61GM
+ = extreme position r =b)
( 6 a )2 5 a2 900 a2 K 1 +U 1=K 2+U 2

(i)
1 2 K 1 2 K
For the system to be bounded one, total
231 (a) m v 1− = m v 2−
2 a 2 b
energy of the system must be negative. Where v 1=
So, object having total energy E1 is √ K

Conservation of angular momentum gives

2 ma
bounded and other is unbounded
m v 1 a=mv 2 b

As it is not known that by how much the √

232 (c) a a K
v 2= v 1=
velocity has changed, which will cause Therefore, from Eq. (i)
b b 2 ma

the change in orbital radius and hence

time speed, so we can use either rocket ( ) 2Kma − Kb
2
1 K K 1 a
⇒ m − = m
system to carry out the docking in the
2 2 ma a 2 b

minimum possible time, though the

3
−3 K aK K 2 4a a
= 2 − ⇒b − b+ =0
4a 4b b 3 3

P a g e | 73
 Hence, b=a /3 ⇒ a/b=3 ' ' ' '
g ∝ R ρ ⇒ ρ =2 ρ
Given, ' =1
g
237 (2)
GM 4 g
3
g= =G π R ρ R
R
2
3 '
=2
4 R
g= GπRρ ⇒ g ∝ Rρ
3

P a g e | 74