ACTIVE SITE TUTORIALS

Date : 09-09-2019 TEST ID: 626

Time : 08:27:00 PHYSICS
Marks : 507
8.GRAVITATION

1. Radius of orbit of satellite of earth is R . Its kinetic energy is proportional to

a) 1 b) 1 c) R d) 1
R √R 3/ 2

A satellite is to revolve round the earth in a circle of radius 8000 km . The speed at which
R

this satellite be projected into an orbit, will be

2.

a) 3 km/ s b) 16 km/ s c) 7.15 km/s d) 8 km/ s

Time speed of revolution of a nearest satellite around a planet of radius R is T . Period of

revolution around another planet, whose radius is 3 R but having same density is
3.

a) T b) 3 T c) 9 T d)
3 √ 3T
The acceleration due to gravity increase by 0.5 % when we go from the equator to the
poles. What will be the time period of the pendulum at the equator which beats seconds
4.

at the poles?
a) 1.950 s b) 1.995 s c) 2.050 s d) 2.005 s

5. Two identical thin rings each of radius R are coaxially placed at a distance R . If the rings
have a uniform mass distribution and each has mass m 1 and m 2 respectively, then the
work done in moving a mass m from centre of one ring to that of the other is

a) Gm m1 ( √ 2+1)
m2 R
b) Gm(m1−m2)( √2+1)
√2 R
c) Gm √ 2(m1 +m2)

Zero
R
d)

Kepler’s second law regarding constancy of aerial velocity of a planet is consequence of

the law of conservation of
6.

a) Energy b) Angular momentumc) Linear momentum d) None of these

7. The periodic time of a communication satellite is

a) 6 hours b) 12 hours c) 18 hours d) 24 hours

8. The curves for potential energy (U ) and kinetic energy (E k ) of a two particle system are
shown in figure. At what points the system will be bound

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 Ek
Energy

O
A B C D
r
U

a) Only at point D b) Only at point A c) At point D and A d) At points A , B and

The acceleration of a body due to the attraction of the earth (radius R ) at a distance 2 R
C

from the surface of the earth is ( g = acceleration due to gravity at the surface of the
9.

earth)
a) g b) g c) g d) g

10. Two equal mass m and m are hung from balance whose scale pans differ in vertical height
9 3 4

by h . Calculate the error in weighing. If any, in terms of density of earth ρ .

a) 2 πρ R 3 Gm b) 8 πρGmh c) 8 πρ R 3 Gm d) 4 πρG m 2 h
3 3 3 3
11. Weight of 1 kg becomes 1/6 on moon. If radius of moon is 1.768 ×10 m , then the mass of
moon will be
6

a) 30 b) 22 c) 24 d) 22
1.99 ×10 kg 7.56 ×10 kg 5.98 ×10 kg
7.65 ×10 kg
12. A satellite of mass m is placed at a distance r from the centre of earth (mass M ¿. The
mechanical energy of the satellite is
a) −GMm b) GMm c) GMm d) −GMm

13. A satellite S is moving in an elliptical orbit around earth. The mass of the satellite is very
r r 2r 2r

small compared to the mass of the earth?

a) The acceleration of S is always directed towards the centre of the earth
The angular momentum of S about the centre of the earth changes in
direction but its magnitude remains constant
b)

c) The total mechanical energy of S varies periodically with time

d) The linear momentum of S remains constant in magnitude
14. The escape velocity of a body on the surface of the earth is 11.2km/ s. If the earth’s mass
increases to twice its present value and the radius of the earth becomes half, the escape
velocity would become
a) 5.6 km/ s b) 11.2km/ s (remain unchanged)

c) 22.4 km/s d) 44.8 km/s

15. A geostationary satellite is orbiting the earth at a height of 5 R above the surface of the
earth, R being the radius of the earth. The time period of another satellite in hours at a
height of 2 R from the surface of the earth is
a) 5 b) 10 c) d) 6
6 √2
√2
16. The escape velocity of a sphere of mass m from earth having mass M and radius R is

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 given by
a)
√ 2 GM b)
2
√ GM 2 GMm
c)
√ d) GM
√
17. Gravitational potential on the surface of earth is ( M =¿ mass of the earth, R=¿ radius of
R R R R

earth)
a) −GM /2 R b) −g R c) g R d) GM /R

18. If orbital velocity of planet is given by v=G a M b R c, then

a) a=1/3 , b=1 /3 , c=−1/3 b) a=1/2 , b=1/2 , c=−1/2

c) a=1/2 , b=−1 /2 , c=1/2 d) a=1/2 , b=−1 /2 , c=−1 /2

19. If both the masses and radius o the earth, each decreases by 50%, the acceleration due
to gravity would
a) Remain same b) Decrease by 50% c) Decrease by 100% d) Increase by 100%

20. A spherical planet for out in space has a mass M 0 and diameter D 0 . A particle of mass m
falling freely new the surface of this planet will experience an acceleration due to gravity
which is equal to
a) G M /D 2 b) 4 mG M / D2 c) 4 G M / D 2 d) G mM /D 2

21. Kepler discovered

0 0 0 0 0 0 0 0

a) Laws of motion b) Laws of rotational motion

c) Laws of planetary motion d) Laws of curvilinear motion

22. If r represents the radius of the orbit of a satellite of mass m moving around a planet of
mass M , the velocity of the satellite is given by
a) v 2=g M b) v 2= GMm c) v= GM d) v 2= GM
r r r r
The ratio , where gand gh are the accelerations due to gravity at the surface of the
23. g
gh
earth and at a height h above the earth’s surface respectively, is

( ) ( ) ( ) ( )
2 2 2 2
a) 1+ h b) 1+ R c) R d) h
R h h R
24. To an astronaut in a spaceship, the sky appears

a) Black b) White c) Green d) Blue

25. In planetary motion the areal velocity of position vector of a planet depends on angular
velocity (ω) and the distance of the planet from sun (r ). If so the correct relation for areal
velocity is
a) dA ∝ωr b) dA ∝ω 2 r c) dA ∝ω r 2 d) dA ∝ √ ωr

26. Earth binds the atmosphere because of

dt dt dt dt

a) Gravity b) oxygen between earth and atmosphere

c) Both (a) and (b) d) None of the above

27. A point mass is placed inside a thin spherical shell of radius R and mass M at a distance
R/2 from the centre of the shell. The gravitational force exerted by the shell on the point

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 mass is
a) GM b) −GM c) Zero d) GM
2 2 2

28. A man weighs 80 kg on earth surface. The height above ground where he will weigh
2R 2R 4R

40kg, is (radius of earth is 6400 km)

a) 0.31 times r b) 0.41 times r c) 0.51 times r d) 0.61 times r

29. A particle falls towards earth from infinity. It’s velocity on reaching the earth would be

a) Infinity b) c) d) Zero
√ 2 gR 2 √ gR
30. What is the height the weight of body will be the same as at the same depth from the
surface of the earth? Radius of earth is R
a) R b) c) √5 R−R d) √3 R−R
2 √ 5 R−R
31. The time period of geostationary satellite at a height 36000 km is 24 h. A spy satellite
2 2

orbits earth at a height 6400km. What will be the time period of sky satellite?
(Radius of earth = 6400 km )
a) 5 h b) 4 h c) 3 h d) 12 h

32. The acceleration to gravity at a height 1/20th of the radius of the earth above the earth
surface is 9 m s−2. Its value at a point at an equal distance below the surface of the earth
in m s−2 is about below the surface of the earth in m s−2 is about
a) 8.5 b) 9.5 c) 9.8 d) 11.5

33. Two satellite A and B go round a planet orbits having radii 4 R and R , respectivly. If the
speed of satellite A is 3 v , then speed of satellite B is
a) 3 v b) 4 v c) 6 v d) 12 v
2 2
34. An asteroid of mass m is approaching earth, initially at a distance of 10 Re with speed v i. It
hits the earth with a speed v f ( Re ∧M e are radius and mass of earth), then

i (
a) v 2=v 2 + 2Gm 1− 1
f
Me R 10 ) f i
Re (
b) v 2=v 2 + 2G M e 1+ 1
10 )
f i
Re (
c) v 2=v 2 + 2G M e 1− 1
10 ) i
Re (
d) v 2=v 2 + 2Gm 1− 1
f
10 )
35. The distance of a geo-stationary satellite from the centre the earth (Radius R=6400 km¿ is
nearest to
a) 5 R b) 7 R c) 10 R d) 18 R

36. A satellite moves in elliptical orbit about a planet. The maximum and minimum velocities
of satellites are 3 ×10 4 ms−1 and 1 ×103 ms−1 respectively. What is the minimum distance of
satellite from planet, if maximum distance is 4 ×104 km?
a) 3 b) 3 c) 3 d) 3
4 ×10 km 3 ×10 km 4 /3 ×10 km 1 ×10 km
37. Assuming earth to be a sphere of radius R, if g30 ° is value of acceleration due to gravity at
latitude of 30 ° and g at the equator, the value of g−g 30° is
a) 1 ω 2 R b) 3 ω 2 R c) 2 d) 1 ω 2 R
ω R
4 4 2
38. Two satellites of earth, S1∧S 2 , are moving in the same orbit. The mass of S1 is four times
the mass of S2 . Which one of the following statements is true?

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 a) The time period of S1 is four times that of S2
b) The potential energies of earth and satellite in the two cases are equal

c) S1 and S2 are moving with the same speed

d) The kinetic energies of the two satellites are equal

39. The ratio of acceleration due to gravity at a height h above the surface of the earth and
at a depth h below the surface of the earth for h< ¿ radius of earth
a) Is constant

b) Increases linearly with h

c) Decreases linearly with h
d) Decreases parabolically with h
40. The weight of an object in the coal mine, sea level, at the top of the mountain are W 1 , W 2
and W 3 respectively, then
a) W <W >W b) W =W =W c) W <W <W d) W >W >W

41. As we go from the equator to the poles, the value of g

1 2 3 1 2 3 1 2 3 1 2 3

a) Remains the same b) Decreases

c) Increases d) Decreases upto a latitude of 45 °

42. Spot the wrong statement:

The acceleration due to gravity ' g ' decreases if
a) We go down from the surface of the earth towards its centre

b) We go up from the surface of the earth

c) We go from the equator towards the poles on the surface of the earth

d) The rotational velocity of the earth is increased

43. The earth revolves about the sun in an elliptical orbit with mean radius 9.3 ×10 7 m in a
period of 1 year. Assuming that there are no outside influences
a) The earth’s kinetic energy remains b) The earth’s angular momentum remains
constant constant
c) The earth’s potential energy remains d) All are correct
constant
44. The escape velocity of a body from earth’s surface is v e. The escape velocity of the same
body from a height equal to 7 R from earth’s surface will be
a) v e b) v e c) ve d) v e
√2 2 2 √2 4
45. The escape velocity from the earth is 11km s . The escape velocity from a planet having
−1

twice the radius and the same mean density as the earth would be
a) −1 b) −1 c) −1 d) −1
5.5 km s 11km s 15.5 km s 22 km s
46. If density of earth increased 4 times and its radius become half of what it is, our weight
will

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 a) Be four times its present value b) Be doubled

c) Remain same d) Be halved

47. The time period of a geostationary satellite is

a) 12 hours b) 24 hours c) 6 hours d) 48 hours

48. The weight of a body on surface of earth is 12.6 N. When it is raised to a height half the
radius of earth its weight will be
a) 2.8 N b) 5.6 N c) 12.5 N d) 25. 2N

49. Imagine a light plant revolving around a very massive star in 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 correct relation is
a) 2 5/ 2 b) 2 7/ 2 c) 5 /2 d) 2 7/ 2
T ∝r T ∝r T ∝r T ∝r
50. If V , R and g denote respectively the escape velocity from the surface of the earth radius
of the earth, and acceleration due to gravity, then the correct equation is
a)
V = √ gR b)
V=
√ 4
g R3 c)
V =R √ g
51. In an elliptical orbit under gravitational force, in general
3
d)
V = √ 2 gR

a) Tangential velocity is constant b) Angular velocity is constant

c) Radial velocity is constant d) Areal velocity is constant

52. A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius
3 R. The work required to take a unit mass from point P on its axis to infinity is

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

53. If then radius of earth R, then the height h at which the value of g becomes one-fourth,
7R 7R 4R 5R

will be
a) R b) 3 R c) 3 R d) R

54. Two small and heavy spheres, each of mass M , are placed a distance r apart on a
8 8 4 2

horizontal surface. The gravitational potential at the mid-point on the line joining the
centre of the spheres is
a) Zero b) −GM c) −2GM d) −4 GM
r r r
If the mass of moon is of earth^' s mass, its radius is of earth^' sradius and if g is
55. 1 1

acceleration due to gravity on earth, then the acceleration due to gravity on moon is..
90 3

a) g b) g c) g d) g

56. Which of the following graphs represents the motion of a planet moving about the sun
3 90 10 9

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 a) b) c) d)
T2 T2 T2 T2

57. The escape velocity for a body projected vertically upwards from the surface of earth is
R3 R3 R3 R3

11kms . If the body is projected at an angle of 45 ° with the vertical, the escape velocity
−1

will be
a) b) c) d)
11/ √ 2 ms
−1
11 √ 2 kms
−1 −1 −1
22 kms 11kms
58. A rocket is launched with velocity 10 km/s . If radius of earth is R , then maximum height
attained by it will be
a) 2 R b) 3 R c) 4 R d) 5 R

59. The mass of the earth is 6.00 ×10 22 kg. The constant of gravitation g=6.67 ×10−11 N m 2 k g−2.
The potential energy of the system is −7.73 ×1028 J. The mean distance between earth and
moon is
a) 8 b) 6 c) 4 d) 2
3.80 ×10 m 3.37 ×10 m 7.60 ×10 m 1.90 ×10 m
60. Gravitational mass is proportional to gravitational

a) Field b) Force c) Intensity d) All of these

61. The acceleration due to gravity is g at a point distant r from the centre of earth of radius
R . If r < R , then
a) g ∝r b) 2 c) −1 d) −2
g ∝r g ∝r g ∝r
62. In the following four periods
(i) Time of revolution of a satellite just above the earth’s surface (T st )
(ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth
(T ma)
(iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform
field of 9.8 N /kg (T sp )
(iv) Period of an infinite length simple pendulum in the earth’s real gravitational filed
(T is )
a) T >T b) T > T
st ma ma st

c) T >T d) T =T =T =T

63. The height from the earth surface at which the value of acceleration due to gravity
sp is st ma sp is

reduces to 1/4 th of its value at earth’s surface (assume earth to be sphere of radius
6400 km )
a) 6400 km b) 2649 km c) 2946 km d) 1600 km

64. A space ship moves from earth to moon and back. The greatest energy required for the
space ship is to overcome the difficulty in
a) Entering the earth’s gravitational field

b) Take off from earths field

c) Take off from lunar surface

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 d) Entering the moon’s lunar surface

65. If the mass of earth is 80 times of that of a planet and diameter is double that of planet
and ' g ' on earth is 9.8 m/ s2, then the value of ' g ' on that planet is
a) 2 b) 2 c) 2 d) 2
4.9 m/ s 0.98 m/s 0.49 m/s 49 m/ s
66. An iron ball and a wooden ball of the same radius are released from a height ' h ' in
vacuum. The time taken by both of them to reach the ground is
a) Unequal b) Exactly equal c) Roughly equal d) Zero

67. If gravitational force on a body of mass 1.5 kg at point is 45N, then the intensity of the
gravitational field at that point is
a) −1 b) −1 c) −1 d) −1
67.5 N k g 45 N k g 30 N k g 15 N k g
68. A man inside an artificial satellite feels weightlessness because the force of attraction
due to earth is
a) Zero at that place

b) Is balanced by the force of attraction due to moon

c) Equal to the centripetal force

d) Non-effective due to particular design of the satellite

69. A geostationary satellite is orbiting the earth at a height of 6 R above the surface of the
earth; R being the radius of the earth. What will be the time period of another satellite
at a height 2.5 R from the surface of the earth?
a) b) c) d) 12 h
6 √2 h 6 √ 2.5 h 6 √3 h
70. If the radius of the earth shrinks by 1%, its mass remaining same, the acceleration due to
gravity on the surface of earth will
a) Decrease by 2% b) Decrease by 0.5% c) Increase by 2% d) Increase by 0.5%

71. If a body describes a circular motion under inverse square field, the time taken to
complete one revolution T is related to the radius of the of the circular orbit is
a) T ∝ r b) 2 c) 2 3 d) 4
T ∝r T ∝r T ∝r
72. Which of the following graphs between the square of the time period and cube of the
distance of the planet from the sun is correct?

a) b) c) d)

73. Two identical solid copper spheres of radius R are placed in contact with each other. The
gravitational attraction between them is proportional to
a) 2 b) −2 c) 4 d) −4
R R R R
74. If the density of a small planet is the same as that of earth, while the radius of the planet
is 0.2 times that of the earth, the gravitational acceleration of the surface of that planet
is
a) 0.2 g b) 0.4 g c) 2 g d) 4 g

75. Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If

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 the ratio of densities of earth (ρ¿¿ e)¿ and moon (ρ¿¿ m)¿ is = then radius of moon
( )
ρe 5
ρm 3
(Rm ) in terms of Re will be
a) 5 R b) 1 R c) 3 R d) 1
Re
2 √3
e
18 6 e 18e

76. Gravitational field is

a) Conservative b) Non-conservative c) Electromagnetic d) Magnetic

77. The mass of a planet is six times that of the earth. The radius of the planet is twice that
of the earth. If the escape velocity from the earth is v , then the escape velocity from the
planet is
a) b) c) v d)
√3 v √2 v √5 v
78. Acceleration due to gravity is maximum at ( R is the radius of earth)

a) A height R from the earth’s surface b) The centre of the earth

c) The surface of the earth

2
d) A depth R from the earth’s surface

79. An artificial satellite is revolving round the earth in a circular orbit. Its velocity is half the
2

escape velocity. Its height from earth’s surface is

a) 6400 km b) 12800 km c) 3200 km d) 1600 km

80. A satellite is revolving round the earth in an orbit of radius r with time period T . If the
satellite is revolving round the earth in an orbit of radius r + Δr (Δ r ≪ r ) with time period
T + Δ T (Δ T ≪ T ) then
a) Δ T = 3 Δr b) Δ T = 2 Δr c) Δ T = Δr d) Δ T =− Δr

81. Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the
T 2 r T 3 r T r T r

acceleration due to gravity on the surface of the earth. If Re is the maximum range of a
projectile on the earth’s surface, what is the maximum range on the surface of the moon
for the same velocity of projection
a) 0.2 R b) 2 R c) 0.5 R d) 5 R

82. The escape velocity of a body from the earth is v e. If the radius of earth contracts to
e e e e

1/4th of its value, keeping the mass of the earth constant, the escape velocity will be
a) Doubled b) Halved c) Tripled d) Unaltered

83. The ratio of the distances of two planets from the sun is 1.38. The ratio of their period of
revolution around the sun is
a) 1.38 b) 3 /2 c) 1 /2 d) 3
1.38 1.38 1.38
84. A planet in a distant solar system is 10 times more massive than the earth and its radius
is 10 times smaller. Given that the escape velocity from the earth is 11kms−1, the escape
velocity from the surface of the planet would be
a) −1 b) −1 c) −1 d) −1
1.1 kms 11kms 110 kms 0.11 kms
85. A satellite moves in a circle around the earth. The radius of this circle is equal to one-half
of the radius of the moon’s orbit. The satellite completes one revolution in
a) 1 lunar month b) 2 lunar month c) −3 /2 d) 3 /2
2 lunar month 2 lunar month
2 3
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86. A body is released from a point distance r from the centre of earth. If R is the earth and
r > R , then the velocity of the body at the time of striking the earth will be
a)
√ gR b)
√ 2 gR c)
√ 2 gR
r −R
d)
√ 2 gR(r −R)

87. Two equal masses m and m are hung from a balance whose scale pans differ in height by
r

h. If ρ is the mean density of earth, then the error in weighing is

a) Zero b) 4 πGρ mh/3 c) 8 πGρ mh/3 d) 2 πGρ mh/3

88. A 20 cm long capillary tube is dipped in water. The water rises upto 8 cm. if entire
arrangement is put in a freely falling elevator lengths of water column in the capillary
tube will be
a) 4 cm b) 8 cm c) 10 cm d) 20 cm

89. A satellite revolves around the earth in an elliptical orbit. Its speed

a) Is the same at all points in the orbit

b) Is greatest when it is closest to the earth

c) Is greatest when it is farthest from the earth

Goes on increasing or decreasing continuously depending upon the

mass of the satellite
d)

90. A geostationary satellite

a) Revolves about the polar axis b) Has a time period less than that of the
near earth satellite
c) Moves faster than a near earth satellite d) Is stationary in the space

91. A shell of mass M and radius R has a point mass m placed at a distance r from its centre.

a) b) c) d)

92. The angular velocity of the earth with which it has to rotate so that acceleration due to
gravity on 60 ° latitude becomes zero is (Radius of earth ¿ 6400 km . At the poles g=10 ms−2)
a) −3 b) −1 c) 1 d) −2
2.5 ×10 rad /s 5.0 ×10 rad / s 1 ×10 rad /s 7.8 ×10 rad /s
93. In a satellite, if the time of revolution is T , then KE is proportional to

a) 1 b) 1 c) 1 d) −2/ 3
2 3 T
T
94. Three identical bodies of mass M are located at the vertices of an equilateral triangle of
T T

side L. They revolve under the effect of mutual gravitational force in a circular orbit,
circumscribing the triangle while preserving the equilateral triangle. Their orbital
velocity is
a)
√ GM b)
√ 3 GM c)
√ 3 GM d)
√ 2 GM

95. Imagine a new planet having the same density as that of earth but it is 3 times bigger
L 2L L 3L

than the earth in size. If the acceleration due to gravity on the surface of earth is g and
that on the surface of the new planet is g ' , then

P a g e | 10
 a) ' b) ' c) ' d) '
g =2 g g =3 g g =4 g g =5 g
96. An artificial satellite of the earth moves at an altitude to h=670 km along a circular orbit.
The velocity of the satellite is
a) −1 b) −1 c) −1 d) −1
7.5 km s 8.5 km s 11.2km s 4.5 km s
97. Escape velocity of a body of 1 kg mass on a planet is 100 m/sec . Gravitational Potential
energy of the body at the Planet is
a) −5000 J b) −1000 J c) −2400 J d) 5000 J

98. A particle of mass m is placed at the centre of a uniform spherical shell of mass 3 m and
radius R . The gravitational potential on the surface of the shell is
a) −Gm b) −3 Gm c) −4 Gm d) −2Gm

99. The work that must be done in lifting a body of weight P from the surface of the earth to
R R R R

a height h is
a) PRh b) R +h c) PRh d) R−h
R +h
100.Two spherical planets A and B have same mass but densities in the ratio 8:1. For these
R−h PRh PRh

planets, the ratio of acceleration due to gravity at the surface of A to its value at the
surface of B is
a) 1 : 4 b) 1 : 2 c) 4 : 1 d) 8 : 1

101.A satellite which is geostationary in a particular orbit is taken to another orbit. Its
distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time
period in the second orbit is
a) 4.8 hours b) c) 24 hours d)
48 √ 2 hours 24 √ 2 hours
102.An object weighs 10N at the north-pole of the earth. In a geostationary satellite distance
7 R from the centre of earth (of radius R ) what will be its true weight?
a) 3 N b) 5 N c) 2 N d) 0.2 N

103.The value of g decreases inside the surface of earth because

a) A force of upward attraction is applied by the shell of earth above

b) The shell of earth above exerts no net force

c) The distance from the centre of the earth decreases

d) The density of the material at the centre of the earth is very small

104.What is the escape velocity for a body on the surface of a planet on which the
acceleration due to gravity is ( 3.1 ) m s and whose radius is 8100 km
2 −2

a) b) c) 27.9 km . s−1 d)
27.9 √ 5 km. s
−1 −1 −1
2790 km . s 27.9 km . s
√5
105.If satellite is shifted towards the earth. Then time period of satellite will be

a) Increase b) Decrease c) Unchanged

d) Nothing can be
said
106.A spherically symmetric gravitational system of particles has a mass density

ρ=
{ρ 0 for r ≤ R
0 for r > R

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 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)

107. If the angular speed of the earth is doubled, the value of acceleration due to gravity (g)
at the north pole
a) Doubles b) Becomes half c) Remains same d) Becomes zero

108.One can easily “weight the earth” by calculating the mass of earth using the formula (in
usual notation)
a) G R2 b) g R2 c) g R d) G R3
E E E E

109.Pick out the most correct statement with reference to earth satellites
g G G g

a) Geostationary satellites are used for remote sensing

b) Polar satellites are used for telecommunications

c) INSAT group of satellites belong to geostationary satellites

d) Polar satellites are at a height of about 36,000 km

110.The radius of orbit of a planet is two times that of the earth. The time period of planet is

a) 4.2 years b) 2.8 years c) 5.6 years d) 8.4 yeasrs

111. The earth (mass = 6 ×10 24kg) revolves around the sun with angular velocity 2 ×10−7 rad s−1
in a circular orbit of radius 1.5 ×108 km. The force exerted by the sun on the earth in
newton is
a) Zero b) 25 c) 39 d) 21
18 ×10 27 ×10 36 ×10
112.If the Earth losses its gravity, then for a body

a) Weight becomes zero, but not the mass b) Mass becomes zero, but not the weight

c) Both mass and weight become zero d) Neither mass nor weight become zero

113.The mass of the moon is about 1.2% of the mass of the earth. Compared to the
gravitational force the earth exerts on the moon, the gravitational force the moon exerts
P a g e | 12
 on earth
a) Is the same b) Is smaller c) Is greater d) Varies with its
phase
114. Select the proper graph between the gravitational potential ( v g) due to hollow sphere
and distance (r ) from its centre

a) b) c) d)

115.A person sitting on a chair in a satellite feels weightless because

a) The earth dose not attract the object in a satellite

The normal force by the chair on the person balances the earth’s
attraction
b)

c) The normal force is zero

d) The person in satellite is not accelerated

116.A missile is launched with a velocity less than the escape velocity. The sum of its kinetic
and potential energy is
a) Positive b) Negative

c) Zero d) May be positive or negative depending

upon its initial velocity
117. The ratio of the radii of the planets P1 and P2 is a . The ratio of their acceleration due to
gravity is b . The ratio of the escape velocities from them will be
a) ab b) c) d)
√ ab √ a /b √ b /a
118.Weight of a body is maximum at

a) Moon b) Poles of earth c) Equator of earth d) Centre of earth

119.Three weights w ,2 w∧3 w are connected to identical spring suspended from a rigid
horizontal rod. The assembly of the rod and weights fall freely. The positions of the
weight from the rod are such that
a) 3 w will be farthest b) w will be farthest

c) All will be at the same distance d) 2 w will be farthest

120. Potential energy of a satellite having mass ' m ' and rotating at a height of 6.4 × 106 m from
the earth surface is
a) −0.5 mg R b) −mg R c) −2 mg R d) 4 mg R

121.The kinetic energy needed to project a body of mass m from the earth surface (radius R )
e e e e

to infinity is
a) mgR /2 b) 2 mgR c) mgR d) mgR / 4

The ratio of radii of earth to another planet is and the ratio of their mean densities is
122. 2
3
. If an astronaut can jump to a maximum height of 1.5 m on the earth, with the same
4
5

P a g e | 13
 effort, the maximum height he can jump on the planet is
a) 1 m b) 0.8 m c) 0.5 m d) 1.25 m

123.There are two planets. The ratio of radius of the two planets is K but ratio of acceleration
due to gravity of both planets is g. What will be the ratio of their escape velocity
a) 1 /2 b) −1 /2 c) 2 d) −2
( Kg ) ( Kg ) ( Kg ) ( Kg )
124.The binding energy of a satellite of mass m in a orbit of radius r is
(R=radius of earth , g=acceleration due ¿ gravity)
2 2 2 2
a) mgR b) mgR c) −mgR d) −mgR

125.Two satellites of mass m and 9 m are orbiting a planet in orbit of radius R . Their periods
r 2r r 2r

of revolution will be in the ratio of

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

126.At what depth below the surface of the earth, acceleration due to gravity g will be half its
value 1600 km above the surface of the earth
a) 6 b) 6 c) 6 d) None of these
4.2 ×10 m 3.19 ×10 m 1.59 ×10 m
127.If ρ is the density of the planet, the time period of nearby satellite is given by

a)
√ 4π
3G ρ
4π
b)
Gρ √ c) 3 π
Gρ √ d) π
√
128.Two spheres of mass m and M are situated in air and the gravitational force between
Gρ

them is F . The space around the masses is now filled with a liquid of specific gravity 3.
The gravitational force will now be
a) F b) F c) F d) 3 F

129.For a body to escape from earth, angle at which it should be fired is?
3 9

a) 45 ° b) ¿ 45 ° c) ¿ 45 ° d) any angle

If g ∝ 3 (instead of 2 ), then the relation between time period of a satellite near earth’s
130. 1 1

surface and radius R will be

R R

a) 2 3 b) 2 c) 2 d) T ∝ T
T ∝R T∝R T ∝R
131. If the force inside the earth surface varies as x n, where r is the distance of body from the
centre of earth, then the value of n will be
a) −1 b) −2 c) 1 d) 2

132.The time period of a satellite of earth is 5h. If the separation between the earth and the
satellite is increased to 4 times the previous value, the new time period will become
a) 10 h b) 18 h c) 40 h d) 20 h

133. The escape velocity from the earth is 11kms−1. The escape velocity from a planet having
twice the radius and same mean density as that of earth is
a) −1 b) −1 c) −1 d) None of these
5.5 kms 11kms 22 kms
134. If the value of g acceleration due to gravity at earth surface is 10 m s−2. Its value in m s−2 at
the centre of the earth, which is assumed to be a sphere of radius R metre and uniform
mass density is

P a g e | 14
 a) 5 b) 10/ R c) 10/2 R d) Zero

135.If the earth rotates faster than its present speed, the weight of an object will

a) Increase at the equator but remain unchanged at the poles

b) Decrease at the equator but remain unchanged at the poles

c) Remain unchanged at the equator but decrease at the poles

d) Remain unchanged at the equator but increase at the poles

136. The distance of neptune and saturn from sun are nearly 1013 and 1012 m respectively.
Assuming that they move in circular orbits, their periodic times will be in the ratio
a) b) 100 c) d)
√ 10 10 √ 10 1/ √ 10
137.At a distance 320 km above the surface of earth, the value of acceleration due to gravity
will be lower than its value on the surface of the earth by nearly (radius of earth = 6400
km)
a) 2% b) 6% c) 10% d) 14%

138.If an object of mass m is taken from the surface of earth (radius R ) to a height 2R , then
the work done is
a) 2 mgR b) mgR c) 2 mgR d) 3 mgR

139.A thief stole a box full of valuable articles of weight w and while carrying it on his head
3 2

jumped down from a wall of height h from the ground. Before he reaches the ground, he
experienced a load
a) Zero b) w /2 c) w d) 2 w

140. The gravitational potential energy of a body of mass ' m ' at the earth’s surface is −mg Re .
Its gravitational potential energy at a height Re from the earth’s surface will be (Here Re
is the radius of the earth)
a) −2 mg R b) 2 mg R c) 1 mg R d) −1 mg R
e e e e

141.An astronaut orbiting the earth in a circular orbit 120 km above the surface of earth,
2 2

gently drops a spoon out of space-ship. The spoon will

a) Fall vertically down to the earth b) Move towards the moon

c) Will move along with space-ship d) Will move in an irregular way then fall
down to earth
142. Two bodies of masses m 1 and m 2 are initially at rest at infinite distance apart. They are
then allowed to move towards each other under mutual gravitational attraction. Their
relative velocity of approach at a separation distance r between them is

[ ] [ ] [ ] [ ]
1/ 2 1 /2 1/ 2 1 /2
a) ( m1−m2 ) b) 2 G m +m c) r d) 2 G m m
2G ( 1 2)
r 2 G ( m1 m2 ) r1 2
r
143.In the solar system, which is conserved

a) Total Energy b) K.E. c) Angular Velocity d) Linear Momentum

144.The total energy of a circularly orbiting satellite is

a) Twice the kinetic energy of the satellite b) Half the kinetic energy of the satellite

P a g e | 15
 c) Twice the potential energy of the satellite d) Half the potential energy of the satellite

145.The time period of a simple pendulum on a freely moving artificial satellite is

a) Zero b) 2 sec c) 3 sec d) Infinite

146.A satellite is revolving around the planet. The gravitational force between them varies
with R−5 / 2 , where R is the radius of the satellite. The square of the time period T will be
directly proportional to
a) 3 b) 7 /2 c) 3 /2 d) 5 /7
R R R R
147.A body is orbiting very close to the earth’s surface with kinetic energy KE. The energy
required to completely escape from it is
a) KE b) 2 KE c) KE d) 3 KE

148.A man is standing on an international space station, which is orbiting earth at an altitude
2 2

520 km with a constant speed 7.6 km/ s. If the man’s weight is 50 kg , his acceleration is
a) 2 b) 2 c) 2 d) 2
7.6 km/ s 7.6 m/s 8.4 m/s 10 m/s
149.Three or two planets. The ratio of radius of the two planets is K but ratio of acceleration
due to gravity of both planets is g. What will be the ratio of their escape velocity?
a) 1/ 2 b) −1/ 2 c) 2 d) −2
( Kg) ( Kg) ( Kg) ( Kg)
150.A planet revolves around the sun in an elliptical orbit. The linear speed of the planet will
be maximum at

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

151.If a man weighs 90 kg on the surface of earth, the height above the surface of the earth
of radius R , where the weight is 30 kg is
a) 0.73 R b) c) R/3 d)
R/ √ 3 √3 R
152.Correct form of gravitational law is

a) F= −Gm 1 m 2 b) ⃗
F=
−Gm1 m2 c) ⃗
F=
−Gm1 m2
r^ d) ⃗
F=
−Gm1 m2 r⃗
2 2 3 3

153.At what height above the earth’s surface does the force of gravity decrease by 10%? The
r r r r

radius of the earth is 6400 km?

a) 345.60 km b) 687.20 km c) 1031.8 km d) 12836.80 km

154.The satellite of mass m revolving in a circular orbit of radius r around the earth has
kinetic energy E . Then its angular momentum will be

√
a) E b) E c) d)
mr2 2mr
2 √ 2 Em r 2 √ 2 Emr
155.A satellite of mass m is circulation around the earth with constant angular velocity. If
radius of the orbit is R0 and mass of the earth M , the angular momentum about the
centre of the earth is
P a g e | 16
 a) m GM R
√ 0
b) M GM R
√ 0
√
c) m GM
R0
d) M GM
√ R0
156.Two satellite A and B, ratio of masses 3 : 1 are in circular orbits of radii r and 4 r . Then
ratio of total mechanical energy of A to B is
a) 1 : 3 b) 3 : 1 c) 3 : 4 d) 12 : 1

157.The required kinetic energy of an object of mass m , so that it may escape, will be

a) 1 mgR b) 1 mgR c) mgR d) 2 mgR

4 2
158. The acceleration due to gravity on a planet is 1.96 m s−2. If it is safe to jump from a height
of 3 m on the earth, the corresponding height on the planet will be
a) 3 m b) 6 m c) 9 m d) 15 m

159. If ge , g h and gd be the accelerations due to gravity at earth’s surface, a height h and at
depth d respectively. Then
a) g > g > g b) g > g < g c) g < g < g d) g < g > g

160.The radius of a planet is 1/4 of earth’s radius and its acceleration due to gravity is double
e h d e h d e h d e h d

that of earth’s acceleration due to gravity. How many times will the escape velocity at
the planet’s surface be as compared to its value on earth’s surface
a) 1 b) c) d) 2
√2 2 √2
√2
161.The mass of diameter of a planet are twice those of earth. The period of oscillation of
pendulum on this planet will be (if it is a second’s pendulum on earth)
a) 1 s b) c) 2 s d) 1 s
2 √2 s
√2 2
162.At what temperature, the hydrogen molecule will escape from earth’s surface?

a) 1 b) 2 c) 3 d) 4
10 K 10 K 10 K 10 K
163.If the radius of a planet is R and its density is ρ , the escape velocity from its surface will
be
a) v ∝ ρR b) v ∝ ρR c) v ∝ √ ρ d) v ∝ 1
e √ e
e e
√ρR
164.A body is orbiting around earth at a mean radius which is two times as greater as
R

parking orbit of a satellite, the period of body is

a) 4 days b) 16 days d) 64 days
c)
2 √ 2 days
165.If M is the mass of the earth and R its radius, the ratio of the gravitational acceleration
and the gravitational constant is
b) M
2
a) R c) 2 d) M
2 MR
R
166.What would be the velocity of earth due to rotation about its own axis so that the weight
M R

at equator become 3/5 of initial value. Radius of earth on equator is 6400 km

a) −4 b) −4 c) −4 d) −4
7.4 × 10 rad / sec 6.7 ×10 rad / sec 7.8 ×10 rad /sec 8.7 × 10 rad / sec
167.The rotation period of an earth satellite close to the surface of the earth is 83 min. the
satellite in a orbit at a distance of three times earth radii from its surface will be
a) 83 min c) 664 min d) 249 min
b)
83 × 8 min√
168.The escape velocity on earth is

P a g e | 17
 a) −1 b) −1 c) −1 d) −1
1.12 kms 11.2ms 11.2kmh 11.2kms
169.In a gravitational field, at a point where the gravitational potential is zero

a) The gravitational field is necessarily zero b) The gravitational field is not necessarily
zero
c) Nothing can be said definitely about the d) None of these
gravitational field
170.Orbital velocity of earth’s satellite near the surface is 7 km/ s. When the radius of the orbit
is 4 times than that of earth’s radius, then orbital velocity in that orbit is
a) 3.5 km/ s b) 7 km/ s c) 72 km/s d) 14 km/s

171. The time period T of the moon of planet Mars (mass M m) is related to its orbital radius R ¿
= Gravitational constant) as
2 3 2 3 3
a) T 2= 4 π R b) T 2= 4 π GR c) T 2= 2 π R G d) T 2=4 π M G R3
m
G Mm Mm Mm
172.A body is projected vertically upwards from the surface of a planet of radius R with a
velocity equal to half the escape velocity for that planet. The maximum height attained
by the body is
a) R/3 b) R/2 c) R/ 4 d) R/5

173.If mass of a body is M on the earth surface, then the mass of the same body on the moon
surface is
a) b) Zero c) d) None of these
M /6 M
174.A synchronous satellite goes around the earth once in every 24 h. What is the radius of
orbit of the synchronous satellite in terms of the earth’s radius (Given mass of the earth,
me =5.98 ×10 kg . radius of earth, r e =6.37 × 10 m, Universal constant of gravitation,
24 6

−11 2 2
G=6.67× 10 N m /k g ¿
a) 2.4 r b) 3.6 r c) 4.8 r d) 6.6 r

175.Four particles each of mass M , are located at the vertices of a square with side L. The
e e e e

gravitational potential due to this at the centre of the square is

a) GM b) GM c) Zero d) GM
−√ 32 −√ 64 2 √ 32
L L
176.An astronaut on a strange planet finds that acceleration due to gravity is twice as that on
L

the surface of earth. Which of the following could explain this

a) Both the mass and radius of the planet are half as that of earth

Radius of the planet is half as that of earth, but the mass is the same as
that of earth
b)

c) Both the mass and radius of the planet are twice as that of earth

Mass of the planet is half as that of earth, but radius is same as that of
earth
d)

177.If g is the acceleration due to gravity on earth’s surface, the gain of 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) 2 mgR b) mgR c) 1 mgR d) 1 mgR

178.An object weighs 72 N on earth. Its weight at a height of R/2 from earth is
2 4

P a g e | 18
 a) 32 N b) 56 N c) 72 N d) Zero

179.A point mass m is placed inside a spherical shell of radius R and mass M . at a distance
R/2 from the centre of the shell. The gravitational force exerted by the shell on the point
mass is
a) GMm b) −GMm c) Zero d) GMm
4

180.Two particles of equal mass m go around a circle of radius R under the action of their
2 2 2
R R R

mutual gravitational attraction. The speed of each particle with respect to their center of
mass is
a)
√ Gm
R
b)
√ Gm
4R
Gm
3R
c)
√ d) Gm

181.A straight rod of length L extends from x=a to x=L+a . Find the gravitational force it,
2R √
exerts on a point mass m at x=0 if the linear density of rod μ= A +B x 2
a) Gm A + BL
a [ ] b) Gm A 1 − 1 + BL
a a+ L [( ) ]
c) Gm BL+ A
[ a+ L ] [
d) Gm BL− A
a ]
182. The escape velocity of an object from the earth depends upon the mass of the earth (M ),
its mean density, (ρ), its radius (R) and the gravitational constant (G). Thus the formula
for escape velocity is
a)
v=R
√ 8π
3
Gρ b)
v=M
√ 8π
3
GR c)
v=√ 2GMR
183.Two spheres of radius r and 2 r are touching each other. The fore of attraction between
√
d) v= 2 GM
2
R

them is proportional to
a) 6 b) 4 c) 2 d) −2
r r r r
184.If the radius of the earth were to shrink by two percent, its mass remaining the same, the
acceleration due to gravity on the earth’s surface would
a) Decrease by 2% b) Increase by 2% c) Increase by 4% d) Decrease by 4%

185. A projectile is projected with velocity k v e in vertically upward direction from the ground
into the space. ( v e is escape velocity and k < 1). If resistance is considered to be negligible
then the maximum height from the centre of earth to which it can go, will be :¿ radius of
earth)
a) R b) R c) R d) R
k +1
2
k −1
2
1−k
2
k +1
186. Mass of moon is 7.34 × 1022 kg. If the acceleration due to gravity on the moon is 1.4 m s−2,
the radius of the moon is (G=6.667 ×1011 N m 2 k g−2)

0.56 ×10 m 1.87 ×10 m 1.92 ×10 m 1.01 ×10 m

a) 4 b) 6 c) 6 d) 8

187.Choose the correct statement from the following :

Weightlessness of an astronaut moving in a satellite is a situation of
a) Zero g b) No gravity c) Zero mass d) Free fall

188.The distance between centre of the earth and moon is 384000 km . If the mass of the earth
is 6 ×10 24 kg and G=6.66× 10−11 N m2 /k g 2. The speed of the moon is nearly
a) 1 km/sec b) 4 km/sec c) 8 km/ sec d) 11.2 km/ sec

189.If Gravitational constant is decreasing with time, what will remain unchanged in case of
P a g e | 19
 a satellite orbiting around earth
a) Time period b) Orbiting radius c) Tangential velocity d) Angular velocity

190.Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in
free space with initial separation between their centres equal to 12R. 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

191.A body weighs w newton at the surface of the earth. Its weight at a height equals to half
the radius of the earth, will be
a) w b) 2 w c) 4 w d) 8 w

192.Suppose the law of gravitational attraction suddenly changes and becomes an inverse
2 3 9 27

cube law i.e. F ∝ 1/r 3, but still remaining a central force. Then
a) Keplers law of areas still holds

b) Keplers law of period still holds

c) Keplers law of areas and period still hold

d) Neither the law of areas, nor the law of period still holds

193.A planet moves around the sun. At a given point P, it is closest from the sun at a distance
d 1 and has a speed v 1. At another point Q , when it is farthest from the sun at a distance d 2
, its speed will be
2 2
a) d 1 v 1 b) d 2 v 1 c) d 1 v 1 d) d 2 v 1
2 2
d2 d1 d2 d1
194.A body of mass m rises to a height h=R/5 from the surface of earth, where R is the radius
of earth. If g is the acceleration due to gravity at the surface of earth, the increase in
potential energy is
a) (4/5)m g h b) (5/6)m g h c) (6 /7 )m g h d) m g h

195.Orbital velocity of an artificial does not depend upon

a) Mass of the earth b) Mass of the satellite

c) Radius of the earth d) Acceleration due to gravity

196.A body has weight 90 kg on the earth’s surface, the mass of the moon is 1/9 that of the
earth’s mass and its radius is 1/2 that of the earth’s radius. On the moon the weight of
the body is
a) 45 kg b) 202.5 kg c) 90 kg d) 40 kg

197.A research satellite of mass 200 kg circles the earth in an orbit of average radius 3 R/2
where R is the radius of the earth. Assuming the gravitational pull on a mass of 1 kg on
the earth’s surface to be 10 N , the pull on the satellite will be
a) 880 N b) 889 N c) 890 N d) 892 N

198.A pendulum clock is set to give correct time at the sea level. This clock is moved to hill
station at an altitude of 2500 m above the sea level. In order to keep correct time of the
hill station, the length of the pendulum

P a g e | 20
 a) Has to be reduced b) Has to be increased

c) Needs no adjustment d) Needs no adjustment but its mass has to

be increased
199.The mean radius of the earth is R , its angular speed on its own axis is ω and the
acceleration due to gravity at earth’s surface is g. The cube of the radius of the orbit of a
geostationary satellite will be
a) 2 b) 2 2 c) 2 d) 2 2
R g /ω R ω /g Rg/ω R g /ω
200.The Earth is assumed to be a sphere of radius R . A platform is arranged at a height R
from the surface of the Earth. The escape velocity of a body from this platform is fv ,
where v is its escape velocity from the surface of the Earth. The value of f is
a) 1 b) 1 c) d) 1
√2
201.A satellite is revolving around the earth with a kinetic energy E . The minimum addition
3 2 √2

of kinetic energy needed to make it escape from its orbit is

a) 2 E b) c) E /2 d) E
√E
202.How high a man be able to jump on surface of a planet of radius 320 km, but having
density same as that of the earth if he jumps 5 m on the surface of the earth (Radius of
earth = 6400 km)
a) 60 m b) 80 m c) 100 m d) 120 m

203.The change in the value of g at a height h above the surface of the earth is the same as at
a depth d below the surface of earth. When both d and h are much smaller than the
radius of earth, then which one of the following is correct?
a) d= h b) d= 3 h c) d=2h d) d=h
2 2
204. A particle is projected vertically upwards from the surface of earth (radius Re ) with a
kinetic energy equal to half of the minimum value needed for it to escape. The height to
which it rises above the surface of earth is
a) R b) 2 R c) 3 R d) 4 R

205. The ratio of the radii of planets A and B is k 1 and ratio of acceleration due to gravity on
e e e e

them is k 2. The ratio of escape velocities from them will be

a) k k
1 2
b) k k
√ 1 2 c)
√ k1
k2
d)
√ k2
k1
206. P is point at a distance r from the centre of a solid sphere of radius r . The variation of
gravitational potential at P(ie , V ) and distance r from the centre of sphere is represented
by the curve.

a) b) c) d)

207.If the escape velocity of a planet is 3 times that of the earth and its radius is 4 times that
of the earth, then the mass of the planet is (Mass of the earth ¿ 6 ×10 24 kg )
a) 22 b) 22 c) 26 d) 22
1.62 ×10 kg 0.72 ×10 kg 2.16 ×10 kg 1.22 ×10 kg
208.Two identical satellite A and B are circulating round the earth at the height of R and 2 R
respectively. (where R is radius of the earth). The ratio of kinetic energy of A to that of B

P a g e | 21
 is
a) 1 b) 2 c) 2 d) 3

209.What is the binding energy of earth-sun system neglecting the effect of other planets and
2 3 2

satellites? (Mass of earth M e =6 ×10 kg, mass of the sun M x =2× 10 k g )

24 30 −2

a) 10 b) 3 c) 33 d) 33
8.8 ×10 J 8.8 ×10 J 5.2 ×10 J 2.6 ×10 J
210.Which one of the following statements regarding artificial satellite of the earth is
incorrect
a) The orbital velocity depends on the mass of the satellite

A minimum velocity of 8 km/ sec is required by a satellite to orbit quite

close to the earth
b)

c) The period of revolution is large if the radius of its orbit is large

d) The height of a geostationary satellite is about 36000 km from earth

211.There are two bodies of masses 100,000 kg and 1000 kg separated by a distance of 1 m.
At what distance (in metre) from the smaller body, the intensity of gravitational field will
be zero?
a) 1/9 b) 1/10 c) 1/11 d) 10/11

212.The centripetal force acting on a satellite orbiting round the earth and the gravitational
force of earth acting on the satellite both equal F . The net force on the satellite is
a) Zero b) F c) d) 2 F
F 2 √
213.The angular velocity of rotation of star (of mass M and radius R ) at which the matter
start to escape from its equator will be
a)
√ 2G M2
R
2 GM
g
b)
√ c) 2 GM
R 3
√ d) 2 GR
M √
214.Two bodies of masses m and 4 m are placed at a distance r . The gravitational potential at
a point on the line joining them where the gravitational field is zero is
a) −4 Gm b) −6 Gm c) −9 Gm d) zero

215.A body is acted upon by a force towards a point. The magnitude of the force is inversely
r r r

proportional to the square of the distance. The path of body will be

a) Ellipse b) Hyperbola c) Circle d) Parabola

216.Three equal masses of 1 kg each are placed at the vertices of an equilateral triangle PQR
and a mass of 2 kg is placed at the centroid O of the triangle which is at a distance of √ 2 m
from each of the vertices of the triangle. The force, in newton, acting on the, mass of 2 kg
is
a) 2 b) c) 1 d) Zero
√2
217.The change in potential energy when a body of mass m is raised to a height nR from the
centre of earth ( R = radius of earth)
2
a) m g R (n−1) b) nm g R c) m g R n d) m g R n
n 2
n+1
218.If three particles each of mass M are placed at the three corners of an equilateral
n +1

triangle of side a , the forces exerted by this system on another particle of mass M placed
(i) at the mid point of a side and (ii) at the centre of the triangle are respectively

P a g e | 22
 a) 0, 0 b) 4 G M , 0
2
c) 0 , 4 G M
2
d) 3G M , G M
2 2

2 2 2 2

219.How much energy will be necessary for making a body of 500 kg escape from the earth
3a 3a a a

[ g = 9.8 m s−2 , radius of earth = 6.4 × 106m]

a) About
9.8 ×10 J
b) About 6.4
×10 J
c) About 3
.1 ×10 J
d) About
27.4 × 10 J
6 8 10 12

220. Two satellites S1 and S2 revolve around a planet in coplanar circular orbits in the same
sense. Their periods of revolution are 1 h and 8 h respectively. The radius of orbit of S1 is
10 km. When S2 is closest to S1, the speed of S2 relative to S1 is
4

a) 4 −1 b) 4 −1 c) 4 −1 d) 4 −1
π ×10 km h 2 π × 10 km h 3 π × 10 km h 4 π × 10 km h
221.The tidal waves in the sea are primarily due to

a) The gravitational effect of the moon on the earth

b) The gravitational effect of the sun on the earth

c) The gravitational effect of venus on the earth

d) The atmospheric effect of the earth itself

222.If mass of earth is M , radius is R and gravitational constant is G , then work done to take
1 kg mass from earth surface to infinity will be
a)
√ GM b) GM
R
c)
√ 2 GM

223.Which of the following graphs represents correctly the variation of the intensity of
2R R
d) GM
2R

gravitational field (I ) with the distance (r ) from the centre of a spherical shell of mass M
and radius a ?

a) b) c) d)

224.The orbital angular momentum of a satellite revolving at a distance r from the centre is L
. If the distance is increased to 16 r , then the new angular momentum will be
a) 16 L b) 64 L c) L d) 4 L

225.A body of mass m is moved to a height h equal to the radius of the earth. The increase in
4

potential energy is
a) 2 m g R b) m g R c) m g R /2 d) m g R /4

226.The ratio of acceleration due to gravity at a height 3 R above earth’s surface to the
acceleration due to gravity on the surface of the earth is (R = radius of earth)
a) 1 b) 1 c) 1 d) 1
9 4 16 3
227. The mean radius of the earth’s orbit round the sun is 1.5 ×10 . The mean radius of the
11

orbit of mercury round the sun is 6 ×10 10 m . The mercury will rotate around the sun in
a) A year b) Nearly 4 years c) Nearly 1 year d) 2.5 years

228.The mass of the moon is 1/8 of the earth but the gravitational pull is 1/6 of the earth. It
4

is due to the fact that

P a g e | 23
 a) Moon is the satellite of the earth b) The radius of the earth is 8.6 the moon

c) The radius of the earth is √ 8/6 of the d) The radius of the moon is 6/8 of the earth
moon
229.Where will it be profitable to purchase 1 kilogram sugar

a) At poles b) At equator
At 45 ° latitude At 40 ° latitude
c) d)

230.An artificial satellite is moving in a circular orbit around the earth with a speed equal to
half the magnitude of escape velocity from the earth. The height of the satellite above
the earth’s surface will be
a) 6000 km b) 5800 km c) 7500 km d) 6400 km

231.Three particles each of mass m rotate in a circle of radius r with uniform angular speed ω
under their mutual gravitational attraction. If at any instant the points are on the vertex
of an equilateral of side L, then angular velocity ω is

a)
√ 2 Gm
L3
b)
√ 3 Gm
L3
c)
√ 5 Gm
L3
Gm
d)
L3 √
A solid sphere of mass M and radius R has a spherical gravity of radius such that the
232. R

centre of cavity is at distance R/ 2 from the centre of the sphere. A point mass m is
2

placed inside the cavity at a distance R/4 from the centre of sphere. The gravitational
pull between the sphere and the point mass m is
a) 11GMm b) 14 GMm c) GMm d) GMm
2 2 2 2

233.The changes in potential energy when a body of mass m is raised to a height nR from
R R 2R R

earth’s surface is ( R=¿ radius of the earth)

2
a) mgR n b) mgR c) mgR n d) mgR n
(n−1) (n+1) (n¿¿ 2+1)¿
234.A man starts walking from a point on the surface of earth (assumed smooth) and reaches
diagonally opposite point. What is the work done by him?
a) Zero b) Positive c) Negative d) Nothing can be
said
235.The radius of the earth is R . The height of a point vertically above the earth’s surface at
which acceleration due to gravity becomes 1% of its value at the surface is
a) 8 R b) 9 R c) 10 R d) 20 R

236.The gravitational potential difference between the surface of a planet and a point 20 m
above it is 14 Jk g−1. The work done in moving a 2.0 kg mass by 8.0 m on a slop of 60 °
from the horizontal is equal to
a) 7 J b) 9.6 J c) 16 J d) 32 J

237.The radius of the earth is about 6400 km and that of the mars is 3200 km. The mass of
the earth is about 10 times the mass of the mars. An object weighs 200 N on the surface
of earth, its weight on the surface of mars will be

P a g e | 24
 a) 8 N b) 20 N c) 40 N d) 80 N

238.At what distance from the centre of the earth, the value of acceleration due to gravity g
will be half that on the surface ( R = radius of earth)
a) 2 R b) R c) 1.414 R d) 0.414 R

239.Which of the following astronomer first proposed that sun is static and earth rounds sun

a) Copernicus b) Kepler c) Galileo d) None

240. The gravitational field in a region is given by ⃗I =( 4 i^ + ^j ) Nk g−1. Work done by this field is
zero when a particle is moved along the line
a) x + y=6 b) x +4 y=6 c) y + 4 x=6 d) x− y =6

241. A sphere of mass M and radius R2 has a concentric cavity of radius R1 as shown in figure.
The force F exerted by the sphere on a particle of mass m located at a distance r from
the centre of sphere varies as (0 ≤ r ≤ ∞)

a)

b)

c)

d)

242. R is the radius of the earth and ω is its angular velocity and g p is the value of g at the
poles. The effective value of g at the latitude λ=60 ° will be equal to
a) g − 1 R ω 2 b) g − 3 R ω 2 c) 2 d) g + 1 R ω2
g p−R ω
243.The escape velocity of a particle of mass m varias as
p p p
4 4 4

P a g e | 25
 a) 2 b) m c) 0 d) −1
m m m
244.The value of g on the surface of earth is smallest at the equator because

a) The centripetal force is maximum at equator

b) The centripetal force is least at equator

c) The angular speed of earth is maximum at equator

d) The angular speed of earth is least at equator

245.The gravitational potential energy of a body of mass m at a distance r from the centre of
the earth is U . What is the weight of the body at this distance?
a) U b) Ur c) U d) U

246.If g is the acceleration due to gravity at the earth’s surface and r is the radius of the
r 2r

earth, the escape velocity for the body to escape out of earth’s gravitational field is
a) gr b) c) g/r d) r / g
√ 2 gr
247.At what height in km over the earth’s pole the free fall acceleration decreases by one
percent? (Assume the radius of the earth to be 6400 km)
a) 32 b) 64 c) 80 d) 1.253

248. According to Kepler’s law T 2 is proportional to

a) 3 b) 2 c) R d) −1
R R R
249. The height at which the weight of a body becomes 1/16th, its weight on the surface of
earth (radius R ), is
a) 5 R b) 15 R c) 3 R d) 4 R

250.Let g be the acceleration due to gravity at earth’s surface and K be the rotational kinetic
energy of the earth. Suppose the earth’s radius decreases by 2% keeping all other
quantities same, then
a) g decreases by 2% and K decreases by
g decreases by 4% and K increases by 2%
4%
b)

c)
g increases by 4% and K increases by 4% d) g decreases by 4% and K increases by 4%
251.A geostationary satellite is revolving around the earth. To make it escape from
gravitational field of earth, its velocity must be increased
a) 100% b) 41.4% c) 50% d) 59.6%

252. The correct graph representing the variation of total energy (E) kinetic energy (K ) and
potential energy (U ) of a satellite with its distance from the centre of earth is
a) b) c) d)
E E
Energy

Energy

Energy

Energy

U K K
K
O O O O
r r E r r
K U U U

253.Two identical satellites are at R and 7 R away from earth surface, the wrong statement is
E

( R=¿Radius of earth)
a) Ratio of total energy will be y

P a g e | 26
 b) Ratio of kinetic energies will be y
c) Ratio of potential energies will be y
Ratio of total energy will be y but ratio of potential and kinetic energy
will be z
d)

254.If the earth suddenly shrinks (without changing mass) to half of its present radius, the
acceleration due to gravity will be
a) g/2 b) 4 g c) g/ 4 d) 2 g

255. The gravitational force F g between two objects does not depends on

a) Sum of the masses b) Product of the masses

c) Gravitational constant d) Distance between the masses

256. A point P(R √ 3 , 0 ,0) lies on the axis of a ring of mass M and radius R . The ring is located
in y−z plane with its centre at origin O . A small particle of mass m starts from P and
reaches O under gravitational attraction only. Its speed at O will be
a)
√ GM
R
b)
√ Gm
R
c)
√ GM
2R
d)
√ Gm
2R
257. In the previous question, the angular speed of S2 as actually observed by an astronaut is
S1
a) π rad h−1 b) −1 c) 2 π rad h−1 d) π rad h−1
π rad h
2 3 3
258. The mass of the moon is 7.34 × 10 kg and the radius is 1.74 × 10 m. the value of
22 6

gravitational field intensity will be

a) −1 b) −1 c) −1 d) −1
1.45 Nk g 1.55 Nk g 1.7 Nk g 1.62 Nk g
259.A body of mass m kg. starts falling from a point 2 R above the Earth’s surface. Its kinetic
energy when it has fallen to a point ' R ' above the Earth’s surface [ R -Radius of Earth, M -
Mass of Earth, G -Gravitational Constant]
a) 1 GMm b) 1 GMm c) 2 GMm d) 1 GMm

260.If acceleration due to gravity on the surface of a planet is two times that on surface of
2 R 6 R 3 R 3 R

earth and its radius is double that of earth. Then escape velocity from the surface of that
planet in comparison to earth will be
a) 2 v b) 3 v c) 4 v d) None of these

261.Planetary system in the solar system describes

e e e

a) Conservation of energy b) Conservation of linear momentum

c) Conservation of angular momentum d) None of these

262.The ratio of the radius of the earth to that of the moon is 10. The ratio of acceleration
due to gravity on the earth and on the moon is 6. The ratio of the escape velocity from
the earth’s surface to that from the moon is
a) 10 b) 6 c) Nearly 8 d) 1.66

263.The depth d at which the value of acceleration due to gravity becomes 1/n times the
value of the surface, is [ R = radius of the earth]

P a g e | 27
 a) R
n (n)
b) R n−1 c) R
2
d) R
( n+1n )
264. If v e and v o represent escape velocity and orbital velocity of a satellite corresponding to a
n

circular orbit of radius R , then

a) v =v b) 2 v =v
e o √ o e
c) v =v / 2 d) v and v are not related
o √

265.If the distance between two masses is doubled, the gravitational attraction between them
e e o

a) Is doubled b) Becomes four times

c) Is reduced to half d) Is reduced to a quarter

266. 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 a gravitational field of point mass m ,
then

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

267.A satellite whose mass is M , is revolving in circular orbit of radius r around the earth.
1 2 3 1 2 3 1 2 3 1 3 2

Time of revolution of satellite is

√ √ √
5 3
a)
T∝
r b) r3 c) T ∝ r d)
T∝
r
T∝ 2
GM GM G M /3 1
G M /4
268. Escape velocity on a planet is v e. If radius of the planet remains same and mass becomes
4 times, the escape velocity becomes
a) 4 v b) 2 v c) v d) 1 v
e e e e

269.If g is the acceleration due to gravity on the surface of the earth, the gain in potential
2

energy of an object of mass m raised from the earth’s surface to a height equal to the
radius R of the earth is
a) mgR b) mgR c) mgR d) 2 mgR

270.A geostationary satellite is orbiting the earth at the height of 6 R above the surface of
4 2

earth, R being radius of earth. The time period of another satellite at a height of 2.5 R
from the surface of earth, is
a) 10 h b) 6 c) 6 h d)
h 6 √2 h
√2
271.A satellite moves round the earth in a circular orbit of radius R making 1 rev/day. A
second satellite moving in a circular orbit, moves round the earth ones in 8 days. The
radius of the orbit of the second satellite is
a) 8 R b) 4 R c) 2 R d) R

272.A satellite in launched in a circular orbit of radius R around the earth. A second satellite
is launch in to an orbit of radius 1.01R . The period of second satellite is longer than the
first one (approximately) by
a) 1.5 % b) 0.5% c) 3% d) 1%

P a g e | 28
 Gravitational acceleration on the surface of a planet is
273. √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

274.A solid sphere is of density ρ and radius R . The gravitational field at a distance r from the
centre of the sphere, where r < R , is
3 2 3
a) ρπ G R b) 4 π G ρr c) 4 π G ρ R d) 4 π G ρr
2
r 3 3
275.A body revolved around the sun 27 times faster than the earth. What is the ratio of their
3r

radii
a) 1/3 b) 1/9 c) 1/27 d) 1/4

276.Satellite A and B are revolving around the orbit of earth. The mass of A is 10 times of

mass of B .The ratio of time period

( )
TA
TB
is

a) 10 b) 1 c) 1 d) 1
5 10
277. A mass M is split into two parts m and (M −m), which are then separated by a certain
distance. The ratio m/ M which maximizes the gravitational force between the parts is
a) 1 : 4 b) 1 : 2 c) 4 : 1 d) 2 : 1

278. The mass and radius of the sun are 1.99 ×1030 kg and R=6.96 ×108 m. The escape velocity of
a rocket from the Sun is
a) 11.2km/ s b) 2.38 km/ s c) 59/5 km/ s d) 618 km/ s

The height at which the acceleration due to gravity becomes (where g=¿the
279. g

acceleration due to gravity on the surface of the earth) in terms of R, the radius of the
9

earth, is
a) 2 R b) R c) R d)
√2 R
√3 2
280. The escape velocity for the earth is v e. The escape velocity for a planet whose radius is
four times and density is nine times that of the earth, is
a) 36 v b) 12 v c) 6 v d) 20 v

281.The ratio of the radius of a planet ' A ' to that of planet ' B ' is ' r ' . The ratio of acceleration
e e e e

due to gravity on the planets is ' x ' . The ratio of the escape velocities from the two
planets is
a) xr b)
√ r c)
√ rx d)

282.A satellite of the earth is revolving in a circular orbit with a uniform speed v . If the
x √ x
r

gravitational force suddenly disappears, the satellite will

a) Continue to move with velocity v along the original orbit
b) Move with a velocity v , tangentially to the original orbit

P a g e | 29
 c) Fall down with increasing velocity

d) Ultimately come to rest somewhere on the original orbit

283.If the earth stops rotating, the value of g at the equator

a) increases b) decreases c) no effect d) None of these

284.If a planet of given density were made larger (keeping its density unchanged) its force of
attraction for an object on its surface would increase because of increased mass of the
planet but would decrease because of larger separation between the centre of the planet
and its surface. Which effect would dominate?
a) Increase in mass b) Increase in radius

c) Both affect the attraction equally d) None of the above

285.If different planets have the same density but different radii, then the acceleration due to
gravity on the surface of the planet is related to the radius (R) of the planet as
a) 2 b) c) g ∝ 1 d) g ∝ 1
g∝R g∝R 2
R R
286. A mass of 6 ×10 kg is to be compressed in a sphere in such a way that the escape
24

velocity from the sphere is 3 ×10 8 m/s . What should be the radius of the sphere?
(G=6.67× 10−11 N – m2 /kg 2 ¿
a) 9 km b) 9 m c) 9 mc d) 9 mm

287. The gravitational force between two point masses m 1 and m 2 at separation r is given by
m1 m2
F=k
The constant k
2
r

a) Depends on system of units only b) Depends on medium between masses only

c) Depends on both (a) and (b) d) Is independent of both (a) and (b)

288.For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential
energy is
a) 2 b) 1 c) 1 d)
√2
2 √2
289.A satellite A of mass m is at a distance of r from the centre of the earth. Another satellite
B of mass 2 m is at a distance of 2 r from the earth’s centre. Their time periods are in the
ratio of
a) 1 : 2 b) 1 : 16 c) 1 : 32 d)
1 :2 2 √
290. The largest and the shortest distance of the earth from the sun are r 1 and r 2, its distance
from the sun when it is at the perpendicular to the major axis of the orbit drawn from the
sun
a) r 1+ r 2 b) r 1 r 2 c) 2r 1 r 2 d) r 1+ r 2
4 r 1+ r 2 r 1 +r 2 3
291.The total energy of satellite moving with an orbital velocity v around the earth is

a) 1 mv 2 b) −1 mv 2 c) 2 d) 3 mv 2
mv
292.The potential energy of gravitational interaction of a point mass m and a thin uniform rod
2 2 2

of mass M and length l , if they are located along a straight line at distance a from each
P a g e | 30
 other is

(a)
a) U = GMm log a+l
e
a
b) U =GMm 1 − 1
(a a+ l )
l (a)
c) U =−GMm log a+ l
e
d) U =−GMm
a
293.Three identical bodies of mass M are located at the vertices of an equilateral triangle of
side L. They revolve under the effect of mutual gravitational force in a circular orbit,
circumscribing the triangle while preserving the equilateral triangle. Their orbital
velocity is
a)
√ GM b)
√ 3 GM c)
√ 3 GM

294.If the diameter of mars is 6760 km and mass one-tenth that of earth. The diameter of
L 2L L
d)
√ 2 GM
3L

earth is 12742 km. If acceleration due to gravity on earth is 9.8 m s−2, the acceleration due
to gravity on mass is
a) −2 b) −2 c) −2 d) −2
34.8 m s 2.84 m s 3.48 m s 28.4 m s
295.A clock S is based on oscillation of a spring and clock P is based on pendulum motion.
Both clock run at the same rate on earth. On a planet having the same density as earth
but twice the radius,
a) S will run faster than P
b) P will run faster than S
c) Both will run at the same rate as on the earth

Both will run at the same rate which will be different from that on the
earth
d)

296.The correct option is

a) The time taken in travelling DAB is less than that for BCD

b) The time taken in travelling DAB is greater than that for BCD

c) The time taken in travelling CDAD is less than that for ABC

d) The time taken in travelling CDA is greater than that for ABC

297.In the above Question find apparent weight of the object?

a) 3 N b) Zero c) 2 N d) 0.2 N

298.The earth revolves round the sun in one year. If the distance between them becomes
double, the new period of revolution will be
a) 1/2 year b) c) 4 years d) 8 years
2 √ 2 years
299. At what height h above earth, the value of g becomes g/2 ?(R=radius of earth)

a) 3 R b) c) ( √ 2−1 ) R d) 1 R
√2 R
√2
300.When of the following graphs correctly represents the variation of g on earth?

P a g e | 31
 a) b) c) d)

301.Rockets are launched in Eastward direction to take advantage of

a) The clear sky on Eastesn side b) The thinner atmosphere on this side

c) Earth’s rotation d) Earth’s tilt

302.If the diameter of mass is 6760 km and mass one-tenth that of earth. The diameter of
earth is 12742 km. If acceleration due to gravity on earth is 9.8 ms−2 , the acceleration due
to gravity on mass is
a) −2 b) −2 c) −2 d) −2
34.8 ms 2.48 ms 3.48 ms 28.4 ms
303.An artificial satellite is moving in a circular orbit around the Earth. The height of the
satellite above the surface of Earth is R. Suppose the satellite is stopped suddenly in its
orbit and allowed to fall freely. On reaching Earth, its speed will be
a) b) c) d) 5
√ gR 2 √ gR 3 √ gR √ gR
304. A planet moving along an elliptical orbit is closest to the sun at a distance r 1 and farthest
away at a distance of r 2. If v 1 and v 2 are the linear velocities at these points respectively.

Then the ratio is

v1
v2

() ()
2 2
a) r 1 b) r 1 c) r 2 d) r 2
r2 r2 r1 r1
305.A spherical hollow is made in a lead sphere of radius R such that its surface touches the
outside surface of the lead sphere and passes through the centre. The mass of the lead
sphere before hollowing was M . The force of attraction that this sphere would exert on a
particle of mass m which lies at a distance d ( ¿ R ) from the centre of the lead sphere on
the straight line joining the centres of the sphere and the hollow is
a) GM m b) GM m
2 2
d 8d

[ ] [ ]
GM m 1 GM m 1
c) 1+ d) d 2 1−
( ) ( )
2
d2 R R
8 1+ 8 1−
2d 2d

A planet has twice the radius but the mean density is thas comparsed to earth.What
306. 1

is the ratio of escape velocity from earth to that from the planet?
4

a) 3:1 b) 1:2 c) 1:1 d) 2:1

307.A clock S is based on oscillation of a spring and a clock P is based on pendulum motion.
Both clocks run at the same rate on earth. On a planet having the same density as earth
but twice the radius
a) S will run faster than P b) P will run faster than S

c) They will both run at the same rate as on d) None of these

the earth
308.Select the correct statement from the following

P a g e | 32
 a) The orbital velocity of a satellite increases with the radius of the orbit

Escape velocity of a particle from the surface of the earth depends on

the speed with which it is fired
b)

c) The time period of satellite does not depend on the radius of the orbit

The orbital velocity is inversely proportional to the square root of the

radius of the orbit
d)

309. A body of mass 500 g is thrown upward with a velocity 20 ms−1 and reaches back to the
surface of a planet after 20 s. Then the weight of the body on that planet is
a) 2 N b) 4 N c) 5 N d) 1 N

310.A planet of mass m moves around the sun of mass M in an elliptical orbit. The maximum
and minimum distance of the planet from the sun are r 1 and r 2, respectively. The time
period of the planet is proportional to
a) (r +r ) b) (r +r )1/ 2 c) ¿ d) ¿ ¿

311.Halley’s comet has a period of 76, had distance of closest approach to the sun equal to
1 2 1 2

8.9 ×10 m. the comet’s farthest distance from the sun if the mass of sun is 2 ×10 kg and
10 30

G=6.67× 10 in MKS units is

11

a) 12 b) 13 c) 12 d) 13
2 ×10 m 2.7 ×10 m 5.3 ×10 m 5.3 ×10 m
312.If the earth were to spin faster, acceleration due to gravity at the poles

a) increase b) decreases

c) remain the same d) depends on how fast it spins

313.Which force in nature exits every where

a) Nuclear force b) Electromagnetic force

c) Weak force d) Gravitation

314.Kepler’s second law states that the straight line joining the planet to the sun sweeps out
equal times. This statement is equivalent to saying that
a) Total acceleration is zero b) Tangential acceleration is zero

c) Longitudinal acceleration is zero d) Radial acceleration is zero

315.Which of the following statement about the gravitational constant is true?

a) It is a force

b) It has no unit

c) It has same value in all system of units

It does not depend on the nature of the medium in which the bodies are
kept
d)

316.A satellite is orbiting around the earth with orbital radius R and time period T . The
quantity which remain constant is
a) T /R b) 2 c) 2 2 d) 2 3
T /R T /R T /R
317.Assuming earth to be a sphere of a uniform density, what is the value of gravitational
acceleration in a min 100 km below the earth’ surface (Given R=6400 km)
P a g e | 33
 a) 2 b) 2 c) 2 d) 2
9.66 m/ s 7.64 m/ s 5.06 m/s 3.10 m/s
318.The atmosphere is held to the earth by

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

319.The diagram showing the variation of gravitational potential of earth with distance from
the centre of earth is
a) V b) V c) V d) V

R R R R
O O O O
r r r r

320.The escape velocity of an object on a planet whose g value is 9 times on earth and whose
radius is 4 times that of earth in km/ s is
a) 67.2 b) 33.6 c) 16.8 d) 25.2

321.Infinite number of masses, each 1 kg, are placed along the x -axis at
x=± 1 m, ± 2m , ± 4 m ,± 8 m ,± 16 m ….. The magnitude of the resultant gravitational potential
in terms of gravitational constant G at the origin (x=0) is
a) G/2 b) G c) 2 G d) 4 G

322.There is a mine of depth about 2.0 km. In this mine the conditions as compared to those
at the surface are
a) Lower air pressure, higher acceleration due to gravity

b) Higher air pressure, lower acceleration due to gravity

c) Higher air pressure, higher acceleration due to gravity

d) Lower air pressure, lower acceleration due to gravity

323.A satellite of mass m is orbiting close to the surface of the earth (Radius R ¿ 6400km) has
a kinetic energy k . The corresponding kinetic energy of the satellite to escape from the
earth’s gravitational field is
a) K b) 2 K c) mgR d) mK

324.What is the intensity of gravitational field at the centre of a spherical shell

a) 2 b) g c) Zero d) None of these

Gm/r
325.Two particles each of mass m are moving around a circle of radius R due to their mutual
gravitational force of attraction, velocity of each particle is
d) None of these
a)
v=
Gm
√ b)

326.Average density of the earth

2R
Gm
v=
√ R
c) Gm
v=
√ 4R

a) does not depend on g b) is a complex function of g

c) is directly proportional to g d) is inversely proportional g

327.Venus looks brighter than other planets because

a) It is heavier than other planets b) It has higher density than other planets

P a g e | 34
 c) It is closer to the earth than other planets d) It has no atmosphere

328. A spherical planet has a mass M P and diameter D P. A particle of mass m falling freely
near the surface of this planet will experience an acceleration due to gravity, equal to
a) 4 G M / D 2 b) G M m/ D2 c) G M / D 2 d) 4 G M m/ D 2

329.If the radius of earth decreases by 1% and its mass remains same, then the acceleration
P P P P P P P P

due to gravity
a) increases by 1% b) decreases by 1% c) increases by 2% d) decreases by 2%

330.Suppose the gravitational force varies inversely as the n th power of distance. Then the
time period of a planet in circular orbit of radius R around the sun will be proportional to
a) ( n+1
2 )
b) ( n−1
2 )
c) n d) ( n−2
2 )
R
331.The additional kinetic energy to be provided to a satellite of mass m revolving around a
R R R

planet of mass M , to transfer it from a circular orbit of radius R1 to another of radius

R2 (R2 > R 1) is
a) GmM 1 − 1
2
(2
R1 R2 ) b) GmM 1 − 1
(R R )
c) 2 GmM 1 − 1
(R R )
d) 1 GmM 1 − 1
2 (R R2 )
332.If the radius of earth’s orbit is made 1/4th ,then duration of an year will become
1 2 1 2 1

a) 8 times b) 4 times c) 1/8 times d) 1/4 times

333.A particle of mass M is situated at the centre of a spherical shell of same mass and
radius a . The magnitude of the gravitational potential at a point situated at a /2 distance
from the centre, will be
a) 4 GM b) GM c) 2GM d) 3GM

334.Read the following statements

a a a a

S1 : An object shall weigh more at pole than at equator when weighed by using a physical
balance
S2 : It shall weigh the same at pole and equator when weighed by using a physical
balance
S3 : It shall weigh the same at pole and equator when weighed by using a spring balance
S4 : It shall weigh more at the pole than at equator when weighed using a spring balance
Which of the above statements is/are correct
a) S and S b) S and S c) S and S d) S and S

335. The distance between the earth and the moon is 3.85 ×10 8m. At what distance from the
1 2 1 4 2 3 3 4

earth’s centre, the intensity of gravitational field will be zero? The masses of earth and
moon are 5.98 ×1024 kg∧7.35 ×1022 kg respectively
a) 8 b) 8 c) 8 d) None of these
3.47 ×10 m 0.39 ×10 m 1.82 ×10 m
336.A satellite of mass m revolves around the earth of radius R at a height x from its surface.
If gis the acceleration due to gravity on the surface of the earth, the orbital speed the
satellite is

( )
2 2 1 /2
a) gx b) gR c) gR d) gR
R−x R +x R+ x

The gravitational field due to a mass distribution is 1= 2 in x direction. Hence C is

337. C
x

P a g e | 35
 constant. Taking the gravitational potential to be zero at infinity, potential at x is
a) 2C b) C c) 2C d) C
2 2
x x
338.Reason of weightlessness in a satellite is
x 2x

a) Zero gravity b) Centre of mass

c) Zero reaction force by satellite surface d) None

339.The relay satellite transmits the T.V. programme continuously from one part of the world
to another because its
a) Period is greater than the period of rotation of the earth

b) Period is less than the period of rotation of the earth about its axis

c) Period has no relation with the period of the earth about its axis

d) Period is equal to the period of rotation of the earth about its axis

340.A comet of mass m moves in a highly elliptical orbit around the sun of mass M . The
maximum and minimum distances of the comet from the centre of the sun are r 1 and r 2
respectively. The magnitude of angular momentum of the comet with respect to the
centre of sun is

[ ] [ ] ( ) ( )
1 /2 1/ 2 2 1 /2 2 1 /2
a) GM r 1 b) GM mr 1 c) 2G m r 1 r 2 d) 2G Mm r 1 r 2
( r 1+ r 2 ) ( r 1+ r 2 ) r 1 +r 2 r 1 +r 2
341.Sun is about 330 times heavier and 100 times bigger in radius than earth. The ratio of
mean density of the sun to that of earth is
a) −6 b) −4 c) −2 d) 1.3
3.3 ×10 3.3 ×10 3.3 ×10
342.Astronaut is in a stable orbit around the earth when he weighs a body of mass 5 kg. What
is reading of spring balance?
a) Spring will not be extended

b) Spring will be extended according to Hook’s law

c) Less than 5 kg-wt

d) More than 5 kg-wt

343.If the change in the value of ' g ' at a height h above the surface of the earth is the same
as at a depth x below it, then (both x and h being much smaller than the radius of the
earth)
a) x=h b) x=2 h c) x= h d) 2
x=h
344.A satellite is launched into a circular orbit of radius R around the earth. A second
2

satellite is launched into an orbit of radius 4 R. The ratio of their respective periods is
a) 4:1 b) 1:8 c) 8:1 d) 1:4

345.When earth moves around the sun, the quantity which remains constant is

a) Angular velocity b) Kinetic energy c) Potential energy d) Areal velocity

346.If distance between earth and sun become four times, then time period becomes

P a g e | 36
 a) 4 times b) 8 times c) 1/4 times d) 1/8 times

347.The escape velocity for a body projected vertically upwards from the surface of the earth
is 11.2km s−1. If the body is projected in a direction making an angle of 45 ° with the
vertical, the escape velocity will be
a) b) c) d)
11.2 × √ 2 km s 11.2/ √ 2 km s
−1 −1 −1 −1
11.2km s 11.2×2 km s
348.The period of revolution of planet A around the sun is 8 times that B. The distance of a
from the sun is how many times greater than that of B from the sun?
a) 2 b) 3 c) 4 d) 5

349.What does not change in the field of central force

a) Potential energy b) Kinetic energy c) Linear momentum d) Angular momentum

350.If a planet of given density were made larger its force of attraction for an object on its
surface would increase because of planet’s greater mass but would decease because of
the greater distance from the object to the centre of the planet. Which effect
predominate?
a) Increases in mass b) Increase in radius

c) Both affect attraction equally d) None of the above

351.The potential energy of 4-particalse each of mass 1 kg placed at the four vertices of a
square of side length 1 m is
a) + 4.0G b) −7.5 G c) −5.4 G d) +6.3 G

352.In the above problem, the ratio of the time duration of his jump on the moon to that of his
jump on the earth is
a) 1 : 6 b) 6 : 1 c) d)
√ 6 :1 1 :√6
353.A mass m is placed at a point B in the gravitational field of mass M . When the mass m is
brought from B to near point A , its gravitational potential energy will
a) Remain unchanged b) Increase c) Decrease d) Become zero

354.An artificial of mass ' m ' revolves around the earth near to its surface then its binding
energy is [ Re , g are radius and acceleration due to gravity respectively of the earth]
a) 1 mg R b) −1 mg R c) mg R d) −mg R
e e e e

355.Energy required to move a body of mass m from an orbit of radius 2 Rto 3 R is

2 2

a) 2 b) 2 c) GMm/8 R d) GMm /6 R
GMm /12 R GMm /3 R
356.The distance of a planet from the sun is 5 times, the distance between the earth and the
sun. the time period of the planet is
a) 3 / 2 b) 3 /2 c) 3 /1 d) 1 /2
6 T yr 5 T yr 5 T yr 5 T yr
357.The orbital velocity of the planet will be maximum at

P a g e | 37
 a) A b) B c) C d) D

358.Acceleration due to gravity g for a body of mass m on earth’s surface is proportional to

(Radius of earth = R, mass of earth = M)
a) 2 b) 0 c) mM d) 3 /2
M/R m 1/ R

The mass of the moon is of the earth but the gravitational pull is of the earth. It is
359. 1 1

due to the fact that

81 6

a) The radius of the moon is 81 of the earth b) The radius of the earth is 9 of the moon
6 √6
c) Moon is the satellite of the earth d) None of the above

360. Two planets have radii r 1 and r 2 and densities d 1 and d 2 respectively. Then the ratio of
acceleration due to gravity on them will be
a) r d :r d b) r d :r d c) r 2 d :r 2 d d) r :r

361. The escape velocity from earth is v es. A body is projected with velocity 2 v es with what
1 1 2 2 1 2 2 1 1 1 2 2 1 2

constant velocity will it move in the inter planetary space

a) v b) 3 v c) 3 v d) 5 v
√ es √ es
362.The moon’s radius is 1/4 that of the earth and its mass is 1/80 times that of the earth. If
es es

g represents the acceleration due to gravity on the surface of the earth, that on the
surface of the moon is
a) g/ 4 b) g/5 c) g/6 d) g/8

363.When a satellite going round the earth in a circular orbit of radius r and speed v loses
some of its energy, then r and v change as
a) r and v both will increase b) r and v both will decrease

c) r will decrease and v will increase d) r will decrease and v will decrease

364.A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and
radius 10 cm. Find the work to be done against the gravitational force between them, to
take the particle far away from the sphere.
(You may take G=6.67× 1 0−11 N m2 kg 2 ¿
a) −10 b) −10 c) −9 d) −10
13.34 × 10 J 3.33 ×10 J 6.67 ×10 J 6.67 ×10 J
365.Two equal masses m and m are hung from a balance whose scale pan differs in vertical
height by h /2. The error in weighing in terms of density of the earth ρ is
a) 1 π G ρ mh b) π G ρ mh c) 4 π G ρ mh d) 8 G ρ mh

366.In a certain region of space, the gravitational field is given by – k /r , where r is the
3 3 3

distance and k is a constant. If the gravitational potential at r =r 0 be V 0 , then what is the

expression for the gravitational potential V?

P a g e | 38
 a) k log(r /r ) b) k log(r ¿ r) c) V +k log(r /r ) d) V +k log(r /r )

367. The density of earth in terms of acceleration due to gravity ( g ) , radius of earth (R) and
0 0 0 0 0 0

universal gravitational constant ( G ) is

a) 4 πRG b) 3 πRG c) 4g d) 3g

368.The velocity with which is projectile must be fired so that it escapes earth’s gravitation
3g 4g 3 πRG 4 πRG

does not depend on

a) Mass of the earth b) Mass of the projectile

c) Radius of the projectile’s orbit d) Gravitational constant

369. Two planets have the same average density but their radii are R1 and R2. If acceleration
due to gravity on these planets be g1 and g2 respectively, then
2 3
a) g 1 = R1 b) g 1 = R2 c) g 1 = R1 d) g 1 = R1
2 3
g2 R2 g2 R1 g2 R2 g2 R2
370.A solid sphere of uniform density and radius r applies a gravitational force of attraction
equal to F 1on a particle placed at P, distance 2 R from the centre O of the sphere. A
spherical cavity of radius R/2 is now made in the sphere as shown in figure. The sphere
with cavity now applied an gravitational force F 2 on same particle placed at P. The ratio
F 2 /F 1 will be

a) 1/2 b) 7/ 9 c) 3 d) 7

371.The height at which the acceleration due to gravity decreases by 36% of its value on the
surface of the earth. (The radius of the earth is R )
a) R b) R c) R d) 2 R

372.If a planet was suddenly stopped in its orbit, k suppose to be circular, find how much
6 4 2 3

time will it take in falling onto the sun?

a)
√ 2/8 times the period of the planet’s revolution
b)
4 √2 times the period of the planet’s revolution
c)
3 √ 2 times the period of the planet’s revolution
d) 9 times the period of the planet’s revolution

373. A body is projected upwards with a velocity of 4 ×11.2kms−1 from the surface of earth.
What will be the velocity of the body when it escapes from the gravitational pull of earth?
a) b) c) d)
11.2kms
−1 −1
2 ×11.2 kms
−1
3 ×11.2 kms √ 15 ×11.2 km s
−1

374.A body falls freely under gravity. Its speed is v when it has lost an amount U of the
gravitational energy. Then its mass is
a) Ug c) 2U
2
b) U d) 2
2 2 2 Ug v
375.If the density of the earth is doubled keeping radius constant, find the new acceleration
v g v

due to gravity? (g=9.8 m/s ¿¿ 2)¿

P a g e | 39
 a) 2 b) 2 c) 2 d) 2
9.8 m/ s 19.6 m/s 4.9 m/ s 39.2 m/s
376.Three particles each of mass m are kept at verities of an equilateral triangle of side L .
The gravitational field at centre due to these particle is
a) Zero b) 3GM c) 9 GM d) 12 GM
√ 3 L2
377.The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity
2 2
L L

on planet B. A man jumps to height of 2 m on the surface of A . What is the height of

jump by the same person on the planet B ?
a) 6 m b) 3 c) 2/9 m d) 18 m
m
378.A satellite is moving with a constant speed v in a circular orbit about the earth. An object
2

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
379.Two planets of radii in the ratio 2:3 are made from the material of density in the ratio
2 2

3:2. Then, the ratio of acceleration due to gravity at the surface of the two planets will
g1
g2
be
a) 1 b) 2.25 c) 4/9 d) 0.12

380.A particle of mass m is located at a distance r from the centre of a shell of mass M and
radius R . The force between the shell and mass is F (r ). The plot of F ( r ) versus r is

a) b) c) d)

381.The condition for a uniform spherical mass m of radius r to be a black hole is [G =

gravitational constant and g = acceleration due to gravity]
a) 1 /2 b) 1 /2 c) 1 /2 d) 1/ 2
( 2 Gm/ r ) ≤c ( 2 Gm/ r ) =c ( 2 Gm/ r ) ≥c ( gm/ r ) ≥ c
382.If mass of a satellite is doubled and time period remain constant the ratio of orbit in the
two cases will be
a) 1 : 2 b) 1 : 1 c) 1 : 3 d) None of these

383.Periodic time of a satellite revolving above Earth’s surface at a height equal to R , radius
of Earth, is
[ g is acceleration due to gravity at Earth’s surface]
a)
2π
√ 2R b)
4 √2 π
√ R
g
c)
2π
√ R
g
d)
8π
√
384.An earth satellite of mass m revolves in a circular orbit of a height h from the surface of
g
R
g

the earth. R is the radius of the earth and g is acceleration due to gravity at the surface
of the earth. The velocity of the satellite in the orbit is given by

√
2
a) g R b) gR c) gR d) g R2
R +h R +h
385.At what depth below the surface of the earth, the value of g is the same as that at a
R+ h

height of 5 km?
a) 1.25 km b) 2.5 km c) 5 km d) 10 km

P a g e | 40
386.Two bodies of masses 2kg and 8kg are separated by a distance of 9 m. the point where
the resultant gravitational field intensity is zero is at a distance of
a) 4.5 m from each b) 6 m from 2 kg c) 6 m from 8 kg d) 2.5 m from 2 kg
mass
387.A satellite is launched into a circular orbit of radius ' R ' around earth while a second
satellite is launched into an orbit of radius 1.02 R . The percentage difference in the time
periods of the two satellites is
a) 0.7 b) 1.0 c) 1.5 d) 3

388.Which of the following statements is correct in respect of a geostationary satellite

a) It moves in a plane containing the Greenwich meridian

b) It moves in a plane perpendicular to the celestial equatorial plane

Its height above the earth’s surface is about the same as the radius of
the earth
c)

Its height above the earth’s surface is about six times the radius of the
earth
d)

389.If g is the acceleration due to gravity on the surface of earth, its value at a height equal
to double the radius of earth is
a) g b) g c) g d) g

390.A spaceship is launched into a circular orbit close to earth’s surface. The additional
2 3 9

velocity that should be imparted to the spaceship in the orbit to overcome the
gravitational pull is (Radius of earth = 6400 km and g=9.8 m s−2)
a) −1 b) −1 c) −1 d) −1
11.2km s 8 km s 3.2 km s 1.5 km s
391.If r denotes the distance between the sun and the earth, then the angular momentum of
the earth around the sun is proportional to
a) 3 /2 b) r c) d) 2
r √r r
392.Escape velocity on the surface of earth is 11.2km/ s. Escape velocity from a planet whose
mass is the same as that of earth and radius 1/4 that of earth is
a) 2.8 km/ s b) 15.6 km/ s c) 22.4 km/s d) 44.8 km/s

393.A satellite is orbiting around the earth. By what percentage should we increase its
velocity, so as to enable it escape away from the earth?
a) 41.4% b) 50% c) 82.8% d) 100%

394. The escape velocity from the earth is 11.2kms−1. The escape velocity from a planet having
twice the radius and the same mean density is (¿ kms¿ ¿−1)¿
a) 11.2 b) 5.6 c) 15 d) 22.4

395. In a certain region of space gravitational field is given by I (Kr). Taking the reference
point to be at r =V 0, find the potential.
a) K log r +V b) r0 c) K log r −V d) log r −V r
0 K log +V 0 0 0
r0 r r0 r0
396. The escape velocity of projectile on the earth’s surface is 11.2kms . A body is projected
−1

out with thrice this speed. The speed of the body for away from the earth will be
a) −1 b) −1 c) −1 d) None of these
22.4 kms 31.7 kms 33.6 kms

P a g e | 41
397.The change in the gravitational potential energy when a body mass m is raised to a
height nR above the surface of the earth is (here R is the radius of the earth)
a)
( n+1n ) mgR b)
( n−1
n
) mgR c) nmgR d) mgR
n
398. Mass M is divided into two parts xM and ( 1−x ) M . For a given separation, the value of x
for which the gravitational attraction between the two pieces becomes maximum is
a) 1 b) 3 c) 1 d) 2

399.A satellite is moving around the earth with speed v in a circular orbit of radius r . If the
2 5

orbit radius is decreased by 1%, its speed will

a) Increase by 1% b) Increase by 0.5% c) Decrease by 1% d) Decrease by 0.5%

400.If the height of a satellite from the earth is negligible in comparison to the radius of the
earth R , the orbital velocity of the satellite is
a) gR b) gR/2 c) d)
√ g/ R √ gR
401. The escape velocity for a body of mass 1 kg from the earth’s surface is 11.2kms−1. The
escape velocity for a body of mass 100 kg would be
a) 2 −1 b) −1 c) −1 d) −2 −1
11.2×10 kms 112 kms 11.2kms 11.2×10 kms
402.Weight of a body of mass m decreases by 1% when it is raised to height h above the
earth’s surface. If the body is taken on a depth h in a mine, change in its weight is
a) 0.5% decrease b) 2% decrease c) 0.5% increase d) 1% increase

403.An earth satellite is moved from one stable circular orbit to farther stable circular orbit.
Which one of the following quantities increase?
a) Linear orbit speed b) Gravitational force

c) Centripetal acceleration d) Gravitational potential energy

404. ge and g p denote the acceleration due to gravity on the surface of the earth and another
planet whose mass and radius are twice to that of the earth, then
a) ge b) g =g c) g =2 g d) g = ge
g p= p e p e p
√2
405.Who among the following gave first the experimental value of G
2

a) Cavendish b) Copernicus c) Brook Teylor d) None of these

406.Assuming the earth to have a constant density, point out which of the following curves
show the variation of acceleration due to gravity from the centre of earth to the points
far away from the surface of earth

a)

b)

P a g e | 42
 c)

d) None of these

407.The masses of two planets are in the ratio 1:2. Their radii are in the ratio 1:2. The
acceleration due to gravity on the planets are in the ratio
a) 1:2 b) 2:1 c) 3:5 d) 5:3

408.According to Kelper’s law of planetary motion if T represent time period and r is orbital
radius, then for two planets these are related as

( ) () ( ) ( ) ( ) ( ) ()
3 3 3 4 3 2 3
a) T 1 = r 1 b) T 1 2
=
r1 c) T 1 = r 1 d) T 1 = r 1
T2 r2 T2 r2 T2 r2 T2 r2
409. The maximum and minimum distances of a comet from the sun are 8 ×10 12 m and
1.6 ×10 m. If its velocity when nearest to the sun is 60 m/s , what will be its velocity in m/s
12

when it is farthest
a) 12 b) 60 c) 112 d) 6

410.A satellite is placed in a circular orbit around earth at such a height that it always
remains stationary with respect to earth surface. In such case, its height from the earth
surface is
a) 32000 km b) 36000 km c) 6400 km d) 4800 km

411.Out of the following, the only correct statement about satellites is

A satellite cannot move in a stable orbit in a plane passing through the

earth’s centre
a)

b) Geostationary satellites are launched in the equatorial plane

We can use just one geostationary satellite for global communication

around the globe
c)

d) The speed of satellite increases with an increase in the radius of its orbit

412.Acceleration due to gravity is g on the surface of the earth. Then the value of the
acceleration due to gravity at a height of 32 km above earth’s surface is (Assume radius
of earth to be 6400 km)
a) 0.99 g b) 0.8 g c) 1.01 g d) 0.9 g

413.In the above problem, if the shell is replaced by a sphere of same mass and radius then
the graph of F ( r ) versus r will be

a) b) c) d)

414.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
a) 3 V
2
b)
√ 3
2
V c)
√ 2
3
V d) 2 V
3
415.A body is projected with velocity of 2 ×11.2km/ s from the form the surface of earth. The

P a g e | 43
 velocity of the body when it escapes the gravitational pull of earth is

a) b) 11.2km/ s c) d) 0.5 ×11.2 km/s

√ 3 ×11.2km/ s √ 2× 11.2km/s
416. The escape velocity of a body on the earth’s surface is v e. A body is thrown up with a
speed √ 5 v e . Assuming that the sun and planets do not influence the motion of the body,
velocity of the body at infinite distance is
a) Zero b) v c) d) 2 v
√ 2 ve
417.The period of a satellite in a circular orbit of radius R is T , the period of another satellite
e e

in a circular orbit of radius 4 R is

a) 4 T b) T /4 c) 8 T d) T /8

418.Two point masses A and B having masses in the ratio 4:3 are separated by a distance of
1 m. When another point mass C of mass M is placed in between A and B, the force

between A and C is rd of the force between B and C . Then the distance of C from A is
1
3
a) 2 m b) 1 m c) 1 m d) 2 m

419.A rocket is sent vertically up with a velocity v less than the escape velocity from the
3 3 4 7

earth. Taking M and r as the mass and radius of earth, the maximum height h attained by
the rocket is given by the following expression
a) 2 2 b) 2 2 2
v R /(2GR−Mv) v R /(2GR + v R)
c) 2 2 2 d) 2 2
v R /(2GR−v R) v R /( 2GRv + RM )
420.A particle of mass m is thrown upwards from the surface of the earth, with a velocity u.
The mass and the radius of the earth are, respectively, M and R . G is gravitational
constant and g is acceleration due to gravity on the surface of the earth. The minimum
value of u so that the particle does not return back to earth, is
a)
√ 2 g R2 b)
√ 2 GM
R2
c)
√ 2 GM
R
d)
√ 2 gM

421.At the surface of a certain planet, acceleration due to gravity is one-quarter of that on
R2

earth. If a brass ball is transported to this planet, then which one of the following
statements is not correct
The mass of the brass ball on this planet is a quarter of its mass as
measured on earth
a)

The weight of the brass ball on this planet is a quarter of the weight as
measured on earth
b)

c) The brass ball has the same mass on the other planet as on earth

d) The brass ball has the same volume on the other planet as on earth

422.Distance between the centres of two stars is 10 a . The masses of these stars are M and
16 M and their radii a and 2 a respectively. A body of mass m is fired straight from the
surface of the larger star towards the smaller star. The minimum initial speed for the
body to reach the surface of smaller star is

P a g e | 44
 a) 2 GM
3 √ a
b) 3 5 GM
2 √ a
c) 2 5 GM
3 √ a
d) 3 GM
2 √ a
423. At a given place where, acceleration due to gravity is g m s , a sphere of lead of density
−2

d kg m is gently released in a column of liquid of density ρ kg m . If d > ρ , the sphere will

−2 −3

a) Fall vertically with an acceleration of b) Fall vertically with no acceleration

−2

Fall vertically with an acceleration g

gms

c) d− ρ d) Fall vertically with an acceleration ρ/d

( d )
424.The orbital speed of Jupiter is

a) Greater than the orbital speed of earth b) Less than the orbital speed of earth

c) Equal to the orbital speed of earth d) Zero

425.A body is at rest on the surface of the earth. Which of the following statement is correct?

a) No force is acting on the body

b) Only weight of the body acts on it

c) Net downward force is equal to the net upward force

d) None of the above statement is correct

426.A body is taken to a height of nR from the surface of the earth. The ratio of the
acceleration due to gravity on the surface to that at the altitude is
a) 2 b) ¿ c) −1 d) (n+1)
(n+1) (n+1)
427.If suppose moon is suddenly stopped and then released (given radius of moon is one-
fourth the radius of earth)and the acceleration of moon with respect to earth is 0.0027
m s ), then the acceleration of the moon just before striking the earth’s surface is (Take
−2

g = 10 m s−2)
a) −2 b) −2 c) −2 d) −2
0.0027 m s 5.0 m s 6.4 ms 10 m s
428.The effect of rotation of the earth on the value of acceleration due to gravity is

a) g is maximum at the equator and maximum at the poles

b) g is minimum at the equator and maximum at the poles
c) g is maximum at the both poles
d) g is minimum at the both poles
429.The period of a planet around sun is 27 times that of earth. The ratio of radius of planet’s
orbit to the radius of earth’s orbit is
a) 4 b) 9 c) 64 d) 27

430. The value of g on the earth’s surface is 980 cm s−2. Its value at a height of 64 km from the
earth’s surface is
a) −2 b) −2 c) −2 d) −2
960.40 cm s 984.90 cm s 982.45 cm s 977.55 cm s
431. What should be the angular speed of earth in ra d −1 so that a body 5kg weighs zero at the

P a g e | 45
 equator? (Take g = 10 m s−2 and radius of earth = 6400 km)
a) 1/1600 b) 1/800 c) 1/400 d) 1/80

432.The acceleration due to gravity about the earth’s surface would be half of its value on the
surface of the earth at an altitude of ( R=4000 mile )
a) 1200 mile b) 2000 mile c) 1600 mile d) 4000 mile

433.An artificial satellite moving in circle orbit around the earth has a total (kinetic +
potential) energy E0 . Its potential energy and kinetic energy respectively are
a) 2 E and −2 E b) −2 E and −3 E c) 2 E and −E d) −2 E and −E

434.Two identical trains P and Q move with equal speeds on parallel tracks along the
0 0 0 0 0 0 0 0

equator. P moves from east to west and Q from west to east

a) Data is sufficient to arrive at a conclusion

b) Both exert equal force on track

c) Train Q exerts force on track

d) Train P exerts greater force on track
435.A person will get more quantity of matter in kg-wt at

a) Poles b) at latitude of 60 ° c) Equator d) Satellite

436. A satellite with kinetic energy E k is revolving round the earth in a circular orbit. How
much more kinetic energy should be given to it so that it may just escape into outer
space
a) E b) 2 E c) 1 E d) 3 E
k k k k

437.Force of gravity is least of

2

a) The equator b) The poles

c) A point in between equator and any pole d) None of these

438.The speed of earth’s rotation about its axis is ω . Its speed is increased to x times to make
the effective acceleration due to gravity equal to zero at the equator, then x is around
−2
(g=10 m s , R=6400 km)
a) 1 b) 8.5 c) 17 d) 34

439.A body weighs 700 g wt on the surface of the earth. How much will it weigh on the surface

of a planet whose mass is and radius is half that of the earth

1
7
a) 200 g wt b) 400 g wt c) 50 g wt d) 300 g wt

440.The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is

a) b) 3 V c) d) 2 V
√3 V e √ 2V e
441.The bodies situated on the surface of earth at its equator, becomes weightless, when the
e e

earth has KE about it axis

a) m g R b) 2 m g R/5 c) M g R /5 d) 5 M g R/2

442.The depth from the surface of the earth of radius R at which the acceleration due to

P a g e | 46
 gravity will be 75% of the value on the surface of the earth is

a) R/ 4 b) R/2 c) 3 R/4 d) R/8

443.Two balls, each of radius R , equal mass and density are placed in contact, then the force
of gravitation between them is proportional to
a) F ∝ 1 b) F ∝ R c) 4 d) F ∝ 1
2 F∝R
R
444.If a planet consists of a satellite whose mass and radius were both half that of the earth,
R

the acceleration due to gravity at its surface would be ( g on earth ¿ 9.8 m/s ec 2)
a) 2 b) 2 c) 2 d) 2
4.9 m/ s ec 8.9 m/s ec 19.6 m/s ec 29.4 m/ s ec
445.At what height from the ground will the value of ' g ' be the same as that in 10 km deep
mine below the surface of earth
a) 20 km b) 10 km c) 15 km d) 5 km

446.The mass of a spaceship is 1000 kg . It is to be launched from the earth’s surface out into
free space. The value of ' g ' and ' R ' (radius of earth) are 10 m/s 2 and 6400 km respectively
a) 11 b) 8 c) 9 d) 10
6.4 × 10 Joules 6.4 × 10 Joules 6.4 × 10 Joules 6.4 × 10 Joules
447.The acceleration due to gravity on a planet is same as that on earth and its radius is four
times that of earth. What will be the value of escape velocity on that planet if it is v e on
earth
a) v b) 2 v c) 4 v d) v e
e e e
2
448. The gravitational field due to a mass distribution is I =k /x in the x -direction (k is a
3

constant). Taking the gravitational potential to be zero at infinity, its value at a distance
x / √ 2 is
a) k /x b) k /2 x c) 2 d) 2
k /x k /2 x
449.A uniform ring of mass M and radius r is placed directly above a uniform sphere of mass
8 M and of same radius R . The centre of the ring is at a distance of d= √ 3 R from the
centre of the sphere. The gravitational attraction beween the sphere and the ring is
a) G M
2
b) 3G M
2
c) 2G M
2
d) √3 G M 2
2 2
√2 R2 2

450.A small satellite is revolving near earth’s surface. Its orbital velocity will be nearly
R 2R R

a) 8 km/ sec b) 11.2 km/ sec c) 4 km/sec d) 6 km/ sec

451.A satellite in a circular orbit of radius R has a period of 4 h. Another satellite with orbital
radius 3 R around the same planet will have a period (in hour)
a) 16 b) 4 c) d)
4 27 √ 4 8 √
452.What will be the acceleration due to gravity at height h if h> ¿ R . Where R is radius of
earth and g is acceleration due to gravity on the surface of earth
g g
a)
( )
1+
h
R
2
(
b) g 1− 2 h
R)
c)
( 1−
h
R )
2
(
d) g 1− h
R )
453.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−3 / 2, then T 2 is proportional to

P a g e | 47
 a) 3 b) 5 /2 c) 3 /2 d) 7 /2
R R R R
454.The period of moon’s rotation around the earth is nearly 29 days . If moon’s mass were 2
fold, its present value and all other things remained unchanged, the period of moon’s
rotation would be nearly
a) b) c) 29 ×2 days d) 29 days
29 √ 2 days 29 √ 2 days
455.Which is constant for a satellite in orbit

a) Velocity b) Angular momentumc) Potential energy d) Acceleration

456.Two bodies of masses 100 kg and 1000 kg are separated by distance of 1 m. What is the
intensity of gravitational field at the mid point of the line joining them?
a) −11 2 −2 b) −8 −1 c) −7 −1 d) −6 −1
6.6 ×10 Nm kg 2.4 × 10 Mk g 2.4 × 10 Nk g 2.4 × 10 Nk g
457.What will be the effect on the weight of a body placed on the surface of earth, if earth
suddenly starts rotating with half of its angular velocity of rotation?
a) No effect

b) Weight will increase

c) Weight will decrease

d) Weight will become zero

458.One goes from the centre of the earth to a distance two third the radius of the earth,
where will the acceleration due to gravity be the greatest?
a) At the centre of the earth

b) At a height half the radius of the earth

c) At a height one-third the radius of the earth

d) At a height two-third the radius of the earth

459.The gravitational attraction between the two bodies increases when their masses are

a) Reduced and distance is reduced b) Increased and distance is reduced

c) Reduced and distance is increased d) Increased and distance is increased

460.A particle of mass m is placed inside a spherical shell, away from its centre. The mass of
the shell is M
The particle will move towards the centre if m< M , and away from the
centre if m> M
a)

b) The particle will move towards the centre

c) The particle will oscillate about the centre of shell

d) The particle will remain stationary

461. Two planets of mean distance d 1 and d 2 from the sun and their frequencies are n1 and n2
respectively then
a) n2 d 2=n d 2 b) n2 d 3=n 2 d 3 c) n d 2=n d 2 d) n2 d =n 2 d

462.If satellite is revolving around a planet of mass M in an elliptical orbit of semi-major axis
1 1 2 2 2 2 1 1 1 1 2 2 1 1 2 2

a , find the orbital speed of the satellite when it is at a distance r from the focus
P a g e | 48
 a) v 2=GM 2 − 1
[ ]
r a
b) v 2=GM 2 − 1
2 [ ]
r a
c) v 2=GM 2 − 1
2 2 [ r a ] d) v 2=G 2 − 1
[ ]
r a
463. Two stars of mass m 1 and m 2 are parts of a binary system. The radii of their orbits are r 1
and r 2 respectively, measured from the C.M. of the system. The magnitude of
gravitational force m 1 exerts on m 2 is
a)
m1 m2 G b)
m1 G c)
m2 G d) (m ¿ ¿ 1+m2) ¿
2 2 2 2
( r 1 +r 2 ) ( r 1 +r 2 ) ( r 1 +r 2 ) ( r 1 +r 2 )
If the earth were ¿ suddenly contract ¿ thof its present radius without any change
464. 1

in its mass, the duration of the new day will be nearly

n

a) 24 h b) 24 n h c) 24 h d) 2
2 24 n h
n
465.A man can jump to a height of 1.5 m on a planet A . What is the height he may be able to
n

jump on another planet whose density and radius are, respectively, one-quarter and one-
third that of planet A
a) 1.5 m b) 15 m c) 18 m d) 28 m

466. The value of escape velocity on a certain planet is 2 km s−1. Then, the value of orbital
speed for a satellite orbiting close to its surface is
a) b) c) d)
2 √ 2kms
−1
−1
112 kms
−1
1 kms √ 2 Kms −1

467.The maximum vertical distance through which a full dressed astronaut can jump on the
earth is 0.5 m. Estimate the maximum vertical distance through which he can jump on
the moon, which has a mean density 2/3rd that of earth and radius one quarter that of
the earth
a) 1.5 m b) 3 m c) 6 m d) 7.5 m

468. Assuming that the earth is a sphere of radius R E with uniform density, the distance from

its centre at which the acceleration due to gravity is equal to ( g is the acceleration due
g

to gravity on the surface of earth) is

3

a) R E b) 2 R E c) R E d) R E

469.Radius of earth is around 6000 km . The weight of body of height of 6000 km from earth
3 3 2 4

surface becomes
a) Half b) One-fourth c) One third d) No change

470.The mass of the moon is 1/81 of earth’s mass and its radius 1/4th that of the earth. If the
escape velocity from the earth’s surface is 11.2km s−1, its value for the moon will be
a) −1 b) −1 c) −1 d) −1
0.15 km s 5 km s 2.5 km s 0.5 km s
471.Geostationary satellite

a) Falls with g towards the earth b) Has period of 24 hrs

c) Has equatorial orbit d) Above all correct

472.If a new planet is discovered rotating around the sum with the orbital radius double that
of earth, then what will be its time period (in earth’s days)?
a) 1032 b) 1023 c) 1024 d) 1043

P a g e | 49
473.The mass of a planet that has a moon whose time period and orbital radius are T and R
respectively can be written as
a) 2 3 −1 −2 b) 2 3 −1 −2 c) 2 3 −1 −2 d) 2 3 −1 −2
4π R G T 8π R G T 12 π R G T 16 π R G T
474.Choose the correct statement from the following. The radius of the orbit of a
geostationary satellite depends upon
a) Mass of the satellite, its time period and the gravitational constant

b) Mass of the satellite, mass of the earth and the gravitational constant

Mass of the earth, mass of the satellite, time period of the satellite and
the gravitational constant
c)

Mass of the earth, time period of the satellite and the gravitational
constant
d)

475.The diameters of two planets are in the ratio 4:1 and their mean densities in the ratio
1:2. The acceleration due to gravity on the planets will be in ratio
a) 1 : 2 b) 2 : 3 c) 2 : 1 d) 4 : 1

476.Given radius of Earth ' R ' and length of a day ' T ' the height of a geostationary satellite is
[G-Gravitational Constant, M-Mass of Earth]

( ) ( ) ( ) ( )
2 1 /3 1/ 3 2 1/3 2 1/3
a) 4 π GM b) 4 πGM −R c) GM T −R d) GM T +R
2 2 2 2
T R 4π 4π
477. According to Kepler, the period of revolution of a planet (T ) and its mean distance from
the sun (r ) are related by the equation
T r =¿ constant
b) 2 −3= constant
T r =¿ constant T r =¿ constant
a) 3 3 c) 3 d) 2
T r
478.Gas escapes from the surface of a planet because it acquires an escape velocity. The
escape velocity will depend on which of the following factors :
I. Mass of the planet
II. Mass of the particle escaping
III. Temperature of the planet
IV. Radius of the planet
Select the correct answer from the codes given below :
a) I and II b) II and IV c) I and IV d) I, III and IV

479.The escape velocity of a projectile from the earth is approximately

a) 11.2 m/sec b) 112 km/ sec c) 11.2km/ sec d) 11200 km/sec

480.Escape velocity on the earth

a) Is less than that on the moon b) Depends upon the mass of the body

c) Depends upon the direction of projection d) Depends upon the height from which it is
projected
481.The weight of an astronaut, in an artificial satellite revolving around the earth, is

a) Zero b) Equal to that on the earth

c) More than that on the earth d) Less than that on the earth

482.
The acceleration due ¿ gravity becomes ( g2 )
¿ acceleration due to gravity on the surface of the earth) at a height equal to
P a g e | 50
 a) 4 R b) R c) 2 R d) R

483.The orbit of geostationary satellite is circular, the time period of satellite depends on
4 2

(i) mass of the satellite

(ii) mass of the earth
(iii) radius of the orbit
(iv) height of the satellite from the surface of earth
Which of the following correct?
a) (i) only b) (i) and (ii) c) (i), (ii) and (iii) d) (ii), (iii) and (iv)

484.If suddenly the gravitational force of attraction between earth and a satellite revolving
around it becomes zero, then the satellite will
a) Continue to move in its orbit with same velocity

b) Move tangentially to the original orbit with the same velocity

c) Become stationary in its orbit

d) Move towards the earth

485. The value of ' g ' at a particular point is 9.8 m/ s2. Suppose the earth suddenly shrinks
uniformly to half its present size without losing any mass. The value of ' g ' at the same
point (assuming that the distance of the point from the centre of earth does not shrink)
will now be
a) 2 b) 2 c) 2 d) 2
4.9 m/ s ec 3.1 m/s ec 9.8 m/ s ec 19.6 m/s ec
486.A spring balance is graduated on sea level. If a body is weighed with this balance at
consecutively increasing heights from earth’s surface, the weight indicated by the
balance
a) Will go on increasing continuously b) Will go on decreasing continuously

c) Will remain same d) Will first increase and then decrease

487.The acceleration due to gravity near the surface of a planet of radius R and density d is
proportional to
a) d b) 2 c) dR d) d
2 dR
R
488.320 km above the surface of earth, the value of acceleration due to gravity is nearly 90%
R

of its value on the surface of the earth. Its value will be 95% of the value on the earth’s
surface
a) Nearly 160 km below the earth’s surface

b) Nearly 80 km below the earth’s surface

c) Nearly 640 km below the earth’s surface

d) Nearly 320 km below the earth’s surface

489.In some region, the gravitational field is zero. The gravitational potential in this region

a) Must be variable b) Must be constant c) Cannot be zero d) Must be zero

490.A body of weight 500 N on the surface of the earth. How much would it weigh half-way
below the surface of the earth?

P a g e | 51
 a) 125 N b) 250 N c) 500 N d) 1000 N

491.LANDSAT series of satellites move in near polar orbits at an altitude of

a) 3600 km b) 3000 km c) 918 km d) 512 km

492.For the moon to cease to remain the earth’s satellite, its orbital velocity has to increase
by a factor of
a) 2 b) c) d)
√2 1/ √ 2 √3
493.The force of gravitation is

a) Repulsive b) Electrostatic c) Conservative d) Non-conservative

494.Two astronauts have deserted their space ships 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) 0.41 days

495. Distance of geostationary satellite from the surface of earth radius(R e =6400 km) in terms
of Re is
a) 13.76 R b) 10.76 R c) 6.56 R d) 2.56 R

496.The gravitational force between a point like mass M and an infinitely long, thing rod of
e e e e

linear mass density perpendicular to distance L from M is

a) MGλ b) 1 MGλ c) 2 MGλ d) Infinite
2
L 2 L
497.If the radius of the earth were to shrink by 1% its mass remaining same, the acceleration
L

due to gravity on the earth’s surface would

a) Decrease by 2% b) Remain unchanged c) Increase by 2% d) Become zero

498.For a body lying on the equator to appear weightless, what should be the angular speed
of the earth?(Take g=10 ms−2 ; radius of earth = 6400 km)
a) −1 b) −1 c) −3 −1 d) −2 −1
0.125 rad s 1.25 rad s 1.25 ×10 rad s 1.25 ×10 rad s
499.Two metallic spheres each of mass M are suspended by two strings each of length L. The
distance between the upper ends of strings is L . The angle which the strings will make
with the vertical due to mutual attraction of the spheres is

[ ]
a) tan−1 GM
gL [ ]
b) tan−1 GM
2 gL
c) tan−1 GM
2 [ ]
gL [ ]
d) tan−1 2 GM
2
gL
500.Hubble’s law states that the velocity with which milky ways is moving away from the
earth is proportional to
a) Square of the distance of the milky way from the earth

b) Distance of milky way from the earth

c) Mass of the milky way

d) Product of the mass of the milky way and its distance from the earth

501.If the moon is to escape from the gravitational field of the earth forever, it will require a
velocity

P a g e | 52
 a) 11.2 −1 b) Less than −1
km s 11.2km s
c) Slightly more than −1 d) −1
111.2 km s 22.4 km s
502.The time period of an earth satellite in circular orbit is independent of

a) The mass of the satellite

b) Radius of its orbit

c) Both the mass and radius of the orbit

d) Neither the mass of the satellite nor the radius of its orbit

503. The orbital speed of an artificial satellite very close to the surface of the earth is V o . Then
the orbital speed of another artificial satellite at a height equal to three times the radius
of the earth is
a) 1 V b) 2 V c) 0.5 V d) 4 V

504.A particle is fired vertically upwards from the surface of earth and reaches a height 6400
o o o o

km. The initial velocity of the particle is ( R=6400 km, g=10 m s−2 )
a) −1 b) −1 c) −1 d) None of these
11.2m s 8 km s 3.2 km s
505. The masses and radii of the earth and moon are M 1 , R1 and M 2 , R2 respectively. Their
centres are distance d apart. The minimum velocity with which a particle of mass m
shoule be projected from a point midway between their centres so that it escapes to
infinity is
a)
2
√ G
d
(M 1+ M 2 ) b)
2
√ 2G

506.Where can a geostationary satellite be installed

d
(M 1 + M 2) c)
2
√ Gm
d
(M 1+ M 2) d)
2
√ Gm(M 1 + M 2)
d (R1 + R2 )

a) Over any city on the equator b) Over the north or south pole

c) At height R above earth d) At the surface of earth

507.At some point the gravitational potential and also the gravitational field due to earth is
zero. The speed is
a) On earth’s surface b) Below earth’s surface

c) At a height Re from earth’s surface ( Re =¿ d) At infinity

radius of the earth)

P a g e | 53
 ACTIVE SITE TUTORIALS
Date : 09-09-2019 TEST ID: 626
Time : 08:27:00 PHYSICS
Marks : 507
8.GRAVITATION

: ANSWER KEY :

1) a 2) c 3) a 4) d 153) a 154) c 155) a 156) d

5) b 6) b 7) d 8) d 157) c 158) d 159) b 160) a
9) a 10) b 11) d 12) d 161) b 162) d 163) b 164) c
13) a 14) c 15) c 16) a 165) b 166) c 167) c 168) d
17) b 18) b 19) d 20) c 169) a 170) a 171) a 172) a
21) c 22) d 23) a 24) a 173) c 174) d 175) a 176) a
25) c 26) a 27) c 28) b 177) c 178) a 179) c 180) b
29) b 30) c 31) b 32) b 181) b 182) a 183) d 184) c
33) c 34) c 35) b 36) c 185) c 186) b 187) d 188) a
37) b 38) c 39) c 40) a 189) c 190) c 191) c 192) d
41) c 42) c 43) b 44) c 193) c 194) b 195) b 196) d
45) d 46) b 47) b 48) b 197) b 198) a 199) d 200) d
49) b 50) d 51) d 52) a 201) d 202) c 203) c 204) a
53) b 54) d 55) c 56) c 205) b 206) c 207) c 208) d
57) c 58) c 59) a 60) d 209) d 210) a 211) c 212) b
61) a 62) c 63) a 64) b 213) c 214) c 215) a 216) d
65) c 66) b 67) c 68) c 217) a 218) b 219) c 220) a
69) a 70) c 71) c 72) c 221) a 222) b 223) b 224) d
73) c 74) a 75) a 76) a 225) c 226) c 227) c 228) c
77) a 78) c 79) a 80) a 229) b 230) d 231) b 232) b
81) d 82) a 83) b 84) c 233) c 234) a 235) b 236) b
85) c 86) d 87) c 88) d 237) d 238) c 239) a 240) c
89) b 90) a 91) b 92) a 241) b 242) a 243) c 244) c
93) d 94) a 95) b 96) a 245) c 246) b 247) a 248) a
97) a 98) c 99) c 100) c 249) c 250) c 251) b 252) c
101) b 102) d 103) b 104) b 253) d 254) b 255) a 256) a
105) b 106) c 107) c 108) b 257) d 258) d 259) b 260) a
109) c 110) b 111) d 112) a 261) c 262) c 263) b 264) b
113) a 114) c 115) c 116) b 265) d 266) a 267) b 268) b
117) b 118) b 119) c 120) a 269) b 270) d 271) b 272) a
121) c 122) c 123) a 124) b 273) c 274) d 275) b 276) b
125) b 126) a 127) c 128) a 277) b 278) d 279) a 280) b
129) d 130) b 131) c 132) c 281) c 282) b 283) a 284) b
133) c 134) d 135) b 136) c 285) b 286) d 287) a 288) b
137) c 138) c 139) a 140) d 289) d 290) c 291) b 292) c
141) c 142) b 143) a 144) d 293) a 294) c 295) b 296) b
145) d 146) b 147) a 148) c 297) b 298) b 299) c 300) a
149) a 150) c 151) a 152) d 301) c 302) c 303) a 304) c

P a g e | 54
305) d 306) c 307) b 308) d 409) a 410) b 411) b 412) a
309) d 310) d 311) c 312) c 413) b 414) c 415) a 416) d
313) d 314) b 315) d 316) d 417) c 418) a 419) c 420) c
317) a 318) b 319) c 320) a 421) a 422) b 423) c 424) b
321) c 322) b 323) b 324) c 425) c 426) a 427) c 428) a
325) c 326) b 327) c 328) a 429) b 430) a 431) b 432) c
329) c 330) a 331) d 332) c 433) c 434) d 435) c 436) a
333) d 334) d 335) a 336) d 437) a 438) c 439) b 440) a
337) b 338) c 339) d 340) d 441) c 442) a 443) c 444) c
341) b 342) a 343) b 344) b 445) d 446) d 447) b 448) c
345) d 346) b 347) a 348) b 449) d 450) a 451) c 452) a
349) d 350) a 351) c 352) b 453) b 454) d 455) b 456) c
353) c 354) a 355) d 356) b 457) b 458) d 459) b 460) d
357) c 358) a 359) b 360) a 461) b 462) a 463) a 464) c
361) c 362) b 363) c 364) d 465) c 466) c 467) b 468) a
365) c 366) c 367) d 368) b 469) b 470) c 471) d 472) a
369) a 370) b 371) b 372) a 473) a 474) d 475) c 476) c
373) d 374) c 375) b 376) a 477) b 478) c 479) c 480) d
377) d 378) b 379) a 380) d 481) a 482) b 483) d 484) c
381) c 382) b 383) b 384) d 485) c 486) b 487) c 488) d
385) d 386) c 387) d 388) d 489) b 490) b 491) c 492) b
389) d 390) c 391) c 392) c 493) c 494) b 495) c 496) c
393) a 394) d 395) a 396) b 497) c 498) c 499) c 500) b
397) a 398) a 399) b 400) d 501) a 502) a 503) c 504) b
401) c 402) a 403) d 404) a 505) a 506) a 507) d
405) a 406) c 407) b 408) d

P a g e | 55
 ACTIVE SITE TUTORIALS
Date : 09-09-2019 TEST ID: 626
Time : 08:27:00 PHYSICS
Marks : 507
8.GRAVITATION

: HINTS AND SOLUTIONS :

1 (a) ¿ 2+0.005=2.005 s
GMm
K . E .=
V A =¿( Potential at A
2R 5 (b)

2 (c) due to A ) + (Potential

at A due to B)
√ √ √
5 2
GM gR
2
10 × ( 64 ×10 )
v 0= = = 3 −G m1 G m2
r r 8000 ×10 ⇒V A = −
R √2 R
Similarly,
2
¿ 71.5 ×10 m/s=7.15 km/ s

V B=¿(Potential at B
Time period of satellite
3 (a)
due to A ) + (Potential
which is very near to at B due to B)
planet −G m2 G m1

√
⇒ V B= −

√ √
R √2 R
3 3

Since,
R R 1
T =2 π =2 π ∴T ∝
GM 4 ρ
G π R3 ρ
W A → B=m(V B−V A ) ⇒W A → B
i .e . time period of
3

nearest satellite does

Gm ( m1 −m2) ( √ 2−1)
¿
not depends upon the
√2 R
radius of planet, it
only depends upon the
6 (b)
=¿ constant
density of the planet.
dA L
=
In the problem,
dt 2 m

density is same so
The system will be
8 (d)
time period will be
bound at points where
same
total energy is
negative. In the given
curve at point A, B and
4 (d)

T =2 π
∆T ∆g
l
g √ C the P.E. is more than
K.E.
=
or
T 2g
9 (a)

( ) ( )
2 2
R R g
( )
'
−∆ g −1 −0.5 g =g =g =
∆T= ×T = × ×2=+ 0.005 s R+ h R+2 R 9
2g 2 100
∴ Time period at
equator As with height g varies
10 (b)

P a g e | 56
 as locations.
The acceleration of S(¿
g =
'' g
[
=g 1−
2h
] centripetal
acceleration) is always
2
[ 1+h /R ] R

directed towards the

centre of earth

14 (c)

and in according with

figure h1 >h 2 , so W 1 will
v e=
√ 2 GM
∴ve∝
If M becomes double
R
M
R √
be lesser than W 2 and and R becomes half
then escape velocity
W 2 −W 1=m g 2−m g 1=2 m g [ h1 h 2
−
R R ] becomes two times

or W 2 −W 1=2 m
GM h 15 (c)
R R
2
2 3 3
T1 R1
(6 R )
[ ]
GM 2
= 3= 3
=8
as g= 2
∧( h1−h 2 )=h T 2 R 2 (3 R )
R
or
24 ×24
2
=8
T2
2 mhG 4
W 2 −W 1=
R
3
3
π R3 ρ ( ) 2
T 2=
24 ×24
8
8
3
4
3[
¿ π ρGmh as M = π R3 ρ
] 2
2
T 2=72
T 2=36 ×2
11 (d) T 2=6 √ 2
and
G Mm
gm =
Escape velocity does
Rm
2 16 (a)

not depend on the

ge 9.8 2 2
gm = = m/ s =1.63 m/ s
Substituting mass of the projectiles
6 6
6 2
Rm =1.768× 10 m, gm =1.6317
and G=6.67× 10 N -
m/s
Gravitational potential
(b)

m /k g We get
−11

at a point on the
2 2

surface of earth
22
M m=7.65 ×10 kg

When a satellite is
13 (a) 2
−GM −g R
V= = =−g R
moving in on elliptical
R R

orbit, it’s angular

momentum ( )
18 (b)

√
¿ ⃗ ⃗
about the centre of
r × p GM
v= =G1/ 2 M 1/ 2 R−1/ 2
earth dos not change
R

its direction. The

Here, g=GM / R and
19 (d)
linear momentum (
¿ m ⃗v ) does not remain G( M /2) 2 GM
constant as velocity of
g' = = 2 =2 g
( R/2 )2
satellite is not
R
∴ % increase in
constant. The total
mechanical energy of ( )
g' −g
g= × 100
S is constant at all
g

P a g e | 57
 ¿ ( 2 g−g
)× 100=100 %
28
Weight of body at
(b)

height above the

g

20 (c) earth’s surface is

GM G M 0 4 G M 0 ' w
g= = = w=
R 2 ( D0 /2 ) 2 D 20
( )
2
h
1+
r
The value of
23 (a)
80
acceleration due to
⟹ 40=

gravity g at height h ( )
2
h
1+
above the surface of
r

earth is
⟹ h=0.41r

This should be equal

g 29 (b)
gh =
( ) to escape velocity
2
h
1+
R
Where R is radius of i .e . , √ 2 gR
earth. 30 (c)

( )
2
g h
( )
2
∴ = 1+ gR h
gh R 2
=g 1−
( R +h ) R
25 (c) or

( )( )
2
dA L dA 2 h h 2h
= ⇒ ∝ vr ∝ω r 1− 1+ 2 + =1
dt 2 m dt R R R

or
3 2
h h h
Earth is surrounded by
26 (a) + 2 − =0
3
R R R
an atmosphere of
or ( )
2

gases (air). The reason

h h h
+ −1 =0
is that in earth’s
R R2 R
or
atmosphere the
average thermal
h −1 ± √ 1+ 4 √ 5−1
= =
velocity of even the
R 2 2

highest molecules at or h=
√5 R−R
the maximum possible
2

temperature is small
compared to escape From Kepler’s third
31 (b)

velocity which in turn law of planetary

depends upon gravity ( motion
v e= √ g Re ¿. Therefore, T ∝R
2 3

the molecules of gases 2 3

cannot escape from

T 2 R2
∴ 2= 3
the earth. Hence, an
T 1 R1

atmosphere exists
2

( )
3
T2 6400+ 6400
around the earth.
¿ 2
=
( 24 ) 36000+ 6400

( )
3
2 2 16
¿ T 2=(24 ) ×
Gravitational field
27 (c) 53

inside hollow sphere

⟹ T 2 =4 h

will be zero
Given,
32 (b)

P a g e | 58
 gR
2
20 ×20 1 2 −G M e m
gh=9= 2
= g K f = mv f ∧U f =
( R+ R/20 ) 21 ×21 2 Re
Substituting these
or g=
values in Eq. (i), we
9 × 21× 21

get
20 ×20

Now, gd =g 1− ( )
d
R 1 2 G Mem 1 2 G Mem
mv − = mv −

[ ]
9 ×21 ×21 R /20 2 i 10 R e 2 f Re
¿ 1− =9.5 m s−2
20× 20 R 1 2 1 2 G Mem G Me
⟹ mv f = mvf + −
2 2 Re 10 R

According to Kepler’s
33 (c)
2 2 2G Me 2G M e
⟹ v f =v i + −
third law, we have
Re 10 R e

T ∝R
2 3

2
2 2 G Mem
∴ v f =v i +
Re
1−
1
10 ( )
( )
3
TA
4R 64
Hence , 2 = =
TB R 1
6 R from the surface of
35 (b)

earth and 7 R from the

TA 8
¿ =
centre
TB 1
2 π ωB 8
¿ =
2 π ωA 1
If no external torque
36 (c)
vB × 4 R 8
acts on a system, then
¿ =
angular momentum of
R×vA 1

the system does not

vB
¿ =2
change.
3v
¿ v B=6 v
ie , If τ=0

Applying law of
34 (c) dL
⟹ =0
conservation of energy
dt

for asteroid at a
∴ L=constant
Hence , mv max r min =mvmin r max
distance 10 Re and at
earth’s surface.
v min × r max
⟹ r min =
v max
3 4
1× 10 × 4 × 10 4 3
¿ 4
= ×10 km
3 × 10 3

The value of
37 (b)

acceleration due to
gravity at latitude λ is
given by
2 2
g λ =g−R ω cos λ
2 2
∴ g−g λ =R ω cos λ
At λ=30° ,
K i +U i=K f +U f … .(i) 2 2
g−g 30° =R ω cos 30 °
1 2 −G M e m
(2)
2
Now , K f = mvi ∧U i =
2 10 R e ¿ Rω
2 √3

P a g e | 59
 3 2 to equator
¿ Rω
Kinetic and potential
4 43 (b)

energies varies with

When two satellite of
38 (c)
position of earth w .r . t .
earth are moving in sun. Angular
same orbit, then time momentum remains
period of both are constant every where
equal. From Kepler’s
third law 44 (c)

where r is a
2 3 1
Time period is
T ∝r ve ∝
√r
independent of mass, position of body from
hence their time the surface
periods will be equal.
The potential energy
and kinetic energy are
v1
v2
r
= 1=
r2√ √
R +7 R
R
v
⇒ v 2= 1
2 √2

mass dependent,
hence the PE and KE Escape velocity
45 (d)

of satellites are not

equal.
But, if they are
v e=
√ 2 GM
R

orbiting in a same
√
4
2G π R3× d
orbit, then they have
3
v e=
equal orbital speed.
R

39
Acceleration due to
(c) √ 4
2Gπ R 3 × d=R
where d=¿ mean
3
8
3
πGd
√
gravity at height h , density of earth
∵ ve ∝ R √ d
g1=g 1− ( 2h
)
√
R
Acceleration due to
v e Re d e
∴ =
gravity at depth h ,
vp Rp dp

g1=g 1− (
h
R ) ¿
Re de
2 Re de √
¿ v p =2 v e
( )( ) =(1− hR )
g1 1−2 h/ R 2h h
−1
∴ = = 1− 1− ¿ 2 ×11=22 km s
−1
g 2 1−h /R R R

∴ decreases linearly 46
g1 (b)
g2 g ∝ ρR
with h

Weight of body on the

48 (b)

Because value of g surface of earth

40 (a)

decreases when we
move either in coal At height h ,the value
mg=12.6 N

mine or at the top of of g ' is given by

mountain ' R
2
g =g
Value of g decreases
42 (c) (R+h)
2

when we go from poles

P a g e | 60
 R dA 1 2 d θ 1 2
Now , h= = r = r ω=constant

where ω is angular
2 dt 2 dt 2

( ) speed of planet and r

2
' R 4
∴ g =g =g
its radius.
R+(R /2) 9
4
Weight at height h=mg
9 52 (a)
4 W =∆ U =U f −U i=U ∞−U P
¿ 12.6 × =5.6 N
9
¿−U P =−m V P

Since, gravitation
49 (b) ¿−V P ( as m=1 )
Potential at point P
provides centripetal will be obtained by in
force integration as given
below. Let dM be the
2

mass of small rings as

mv k 2 k
= 5 /2 ie , v = 3/ 2
r
So that shown
r mr

T=
2 πr
v
=
2
k
∴ T ∝r
√
mr 3 /2

7/2
2
ie , T =
4 π 2 m 7/ 2
k
r

From Kepler’s second

51 (d)

law of planetary
motion, a line joining
M
any planet to the sun
dM =
π ¿¿
sweeps out equal
2 Mr dr
areas in equal times,
¿ 2
7R
that is, the aerial −G . dM
velocities of the planet
dV P=
√ 16 R2 +r 2
remains constant dA = 4R

area of the curved

2 GM r
2 ∫
¿− ∙ dr
triangle SAB
7 R 3 R √ 16 R2 +r 2
2 GM
¿− ( 4 √2−5 )
7R
+2 GM
∴W= (4 √ 2−5)
7R

The value of
53 (b)

acceleration due to
1
gravity at height h
≈ ( AB × SA)
2
above the surface of
1
the earth is given by
≈ (rd θ ×r )
2
1 2
≈ r dθ
Thus, the areal
2

(instantaneous)
velocity of the planet
is

P a g e | 61
 ' g R
g= h=
( )
2 2
h v e
1+ −1
R v
2

g =g ( 1+ ) =g (1− )
−2
' h 2h R
⇒ h= =4 R(¿ .)
( )
R R 11.2
2
−1
g
' 10
Given , g =
4
59 (a)
g
4 (
=g 1−
2h
R ) U=
−GMm
r
or
1 2h
⟹ =1− −GMm
4 R r=
U
2h 3
⟹ = −11 24
−6.67 × 10 ×6 × 10 ×7.4 ×1
R 4 r= 38
3R −7.79 ×10
⟹ h= 8
8 ¿ 3.8 ×10 m

Inside the earth

54 (d) 61 (a)

4
Gravitational potential
'
g = πρGr ∴ g ' ∝ r
3
of A at

(i)
62 (c)
−GM −2 GM
O= =
r /2
For B, potential at
r

O=
−GM −2 GM
=
T st =2 π
GM √
( R+h )3
=2 π

[As h≪ R and GM =g R 2
R
g √
]
r /2
∴ Total potential
r

(ii) T ma=2 π
¿−
4 GM
r
(iii)
√ R
g
55 (c)

√ √
G × M /90 1 1 R
Accelerationdue ¿ gravity on moon gm= = Tgsp=2 π =2 π
10
( )
2
(R/3) 1 1 2g
g +
l R
[As l=R ]
56 (c)
Kepler’s law T 2 ∝ R 3
(iv) T is =2 π [As
57
The escape velocity is
(c)
l=∞]
√ R
g

independent of angle
of projection, hence, it
63 (a)

will remain same

' g
g=
( )
2
−1 h
ie .11 kms . 1+
R

If the body is projected

58 (c) g g
=
( )
4 2

with velocity v (v < v e )

h
1+
R
then height up to
where it rises,
h h
1+ =2⇒ =1
R R

P a g e | 62
 ⇒ h=R GM
g=
∴ h=6400 km r
2

∴ log g=log G+ log M −2 log r

Differentiating both
When the spaceship is
64 (b)

sides w.r.t. t
to take off,
gravitational pull of (
1 dg 1 dr dr
earth requires more
= =0−2 × ×100=−
g dt r dt dt
energy to be spent to
overcome it
⇒
1 dg
g dt(× 100 =−2 × )
1 dr
r dt
×100 (
dg
65 (c) ⇒ ×100=−2 × (−1 )=2
∴ g increasing by 2%
dt

( )( ) ( )
2
Mp Re 1
g p=g e =9.8 ( 2 )2
Me Rp 80

According to kepler’s
2
72 (c)
¿ 9.8 /20=0.49 m/s
third law T 2 ∝ r 3 ; At
r =0 ,T =0. It shows
66 (b)

Time of decent t= .

In vacuum no other
√ 2h
g that the graph
between T 2 and r 2 is a
force works except straight line passing
gravity so time period through origin
will be exactly equal
73 (c)
67 (c)
( ) =4
2
4 3
F=mI G× πR ρ
G× m× m 3
F 45 −1 F= =
∴ I= = =30 N k g ( 2 R )2 4R
2
3
m 1.5
4
∴F∝R

According to Kepler’s
69 (a)
Let R be the radius of
74 (a)

law of periods earth and ρ its density,

T ∝ a ¿-major axis] then since shape of
2 3

Here, in case I a is 7 R earth is assumed

as satellite is 6R spherical
above the earth and we have
for a geostationary
satellite T = 24 h
Mass of earth=volume × density
4 3
2 3 M = π R × ρ … .(i )
∴ ( 24 ) ∝ ( 7 R ) … . ( i )
Similarly for case II The acceleration due
3

to gravity which arises

in the body due to
2 3
T ∝ ( 3.5 R ) … . ( ii )
Dividing Eq. (i) by Eq.
(ii), we get gravitational force of
( 24 )2 ( 7 R )3 attraction is given by
= GM
T2 ( 3.5 R )
3
g= 2
… . (ii )

Putting the value of M

R
( 24 )2
from Eq.(i), we get
⟹ T2=
8
¿ T =6 √ 2 h

70 (c)

P a g e | 63
 4
G
3
πR ρ
3
4
gd =g 1− ( R
R)=0

Thus, the acceleration

g= =G πR ρ … . ( iii )
R
2
3
Given , ρ p= ρ , R p =0.2 R e due to gravity is
maximum on the
earth’s surface
4 4
∴ g p=G π R p ρ p =G× π × 0.2 R ρ=0.2 g
3 3

75 (a) 79 (a)

√ √
4 g e ρ e Re GM 1 2 GM
g= πGρR ⇒ g ∝ ρR⇒ = × v= =
3 gm ρ m R m R+ h 2 R
6 5 Re 5 ⇒ 4 R=2 ( R+ h ) ⇒ h=R=6400 km
⇒ = × ⇒ Rm = R e
1 3 Rm 18
80 (a)
77 (a) Since, T 2=k r 3

√
2G M e 2∆T 3∆r ∆T 3 ∆r
⇒ = ⇒ =
Escape velocity , v e = T r T 2 r
Re
Given , M p=6 M e , R p =2 R e 81
Range of projectile
(d)

∴ v p=
√ 2 G∙ 6 M e
(2 R¿¿ e )
= √3 v e ¿
R=
2
u sin 2θ

If and θ are constant

g

Acceleration due to
78 (c) u
then R ∝
gravity at a height h
1

above the earth’s

g
R m g e Rm 1 Re
g = ⇒ = ⇒ Rm = ⇒R
surface is
gh = Re g m Re 0.2 0.2

( )
2
h
1+ 82 (a)
R
Where g is the
acceleration due to
gravity on the earth’s
Escape velocity v e=
R
√ 2 GM
R

surface
'
If R =
4
At
R
h= , g h=
g
=
4g
2GM
v ' e =2
R √
Since, G and M are
( )
2 2
9
constant hence,
R
1+
2R
g g v ' e =2 v e
At
h=R , g h= =
( )
R
2
4
1+
Mass of planet,
84 (c)
R
Acceleration due to
M p=10 M e , where M e is
gravity at a depth d
mass of earth. Radius
below the earth’s
of planet,
surface is gd =g 1− ( )
d
Re
R Rp= , where R e is radius of earth
At
Escape speed is given
10

by,
R
d= , g d=g 1−
2
2
=
2R 2
g
( )
At the centre of earth, v=
√ 2 GM
R

P a g e | 64
 √ 2 G× M p
√
100× 2 GM e

{
So , for planet v p= =
Rp Re −GMm
,r ≥R
¿ 10 × v e r
U (r)=
−1 −1 −GMm
¿ 10 ×11kms =110 kms ,r<R
R

The period of
85 (c)
92 (a)
revolution of a satellite
' 2 2 2
g =g−ω R cos λ ⇒ 0=g−ω R co
at a height h from the
surface of earth is
given by
0=g−
ω2 R
4
⇒ ω=2
g
=
1 rad
R 400 sec √
93 (d)

√
3
( Re + h )
T =2 π
g Re
Given , T m=1 lunar
2 Velocity of satellite v=
1
√ GM
r

month,
2
KE ∝ v ∝
r

√
2 3
¿T ∝r
( )
2
h
R+ KE ∝T
−2/ 3
2
∴ T sat =2 π 2
gR 94 (a)
1 GMM M V
2
⟹ T sat = 3/ 2 =
2
lunar month
2
L L
−3/ 2

√
T moon=2 GM
⇒V =
L
Using law of
86 (d)

conservation of energy
−GMm 1 2 GMm
= mv −
r 2 R
2
v GM GM
= −
The acceleration due
95 (b)
2 R r

( ) ( ) to gravity on the new

r −R r−R
¿ GM =gR
planet can be using
rR r
v=√ 2 gR(r−R)/r the relation
GM
… . (i )
Error in weighing
87 (c) g= 2
R
' 4 3
¿ m g−m g =m g−m g(1−2 h/ R) but M = π R ρ , ρbeing density .
Thus, Eq. (i) becomes
3
m2 h g
¿ m g 2 h/ R=
R
4 3
4 2 G× πR ρ
G πR ρ 3
m2 h 3 8 πG ρ mh ∴ g=
¿ × = R
2
R R
2
3
4
¿G× πR ρ
3
Water fills the tube
88 (d)
⟹ g∝ R
entirely in gravity less
' '

condition.
g R
∴ =
g R

91 (b)

P a g e | 65
 To move it by a small
distance dx ,
'
g 3R
⟹ = =3
Work done
'
g R
'
⟹ g =3 g
GMm
¿ F dx = dx
For the satellite to
96 (a) 2

Total work done

x

move along closed

orbit (a circle with a [ ]
R+h R +h
dx −GMm
¿ GMm ∫ =
radius R+h ) it should
2
R x x R

be acted upon by a
force directed towards
¿ GMm −
1
[
1
R R+h ]
the centre. In this
case, this is the force
¿
[
( R+h )−R
R(R +h)
=
GMmh
R(R+h) ]
of earth’s attraction.
According to Newton’s
GM mhR g mhR PRh
× = =
Second law
R
3
R+ h R+ h R +h

Mass of two planets is

mv
2
GMm 100 (c)
=
same, so
R +h ( R+h )2
At the earth’s surface,
4 3 4 3
GMm π R 1 ρ 1= π R 2 ρ 2
=m g
or
2 3 3

Therefore,
R

( ) ()
1 /3 1 /3
R 1 ρ2 1 1

√
gR 2 = = =
v= =7.5 km s−1 R 2 ρ1 8 2
R+ h

( )
2 2
g 1 GM /R 1 R2 2
97 (a)
= = =( 2 ) =4
g 2 GM /R 22 R1
v e=
√ 2 GM

Potential energy
R
=100 ⇒
GM
R
=5000
101
T ∝ r . If r becomes
(b)
3 /2

double then time

period will become
−GMm
U= =−5000 J
R
( 2 )3 /2 times
So new time period
Gravitational potential
98 (c)

will be
on the surface of the
shell is
24 × 2 √ 2 hr i. e . T =48 √ 2
V =¿ Gravitational
potential due to The true weight of a
102 (d)

particle (V 1 ) body is given by m g

+ Gravitational and with height g
potential due to shell decrease
particle ¿)
So,
W s m g' 1
= =
¿−
R
+
R
=(
Gm −G 3 m −4 Gm
R ) But here,
W E mg ¿¿

h=7 R−R=6 R , ie, h /R=6

So,
Force on the body
99 (c)

WE 10
GMm W S= = =0. 2 N
¿ 2 ( 1+6 ) 2
49
x

P a g e | 66
 Consider that the
103 (b) 106 (c)

earth is sphere of
For r ≤ R :

radius R and mass M.

2 '
m v Gm m
= 2
Then, value of
r r

acceleration due ( )
4
Here , m' = π r 3 ρ0
3
Substituting in Eq. (i)
we get
v∝r
ie , v – r graph is a
straight line passing
through orgine.
For r > R :
to gravity at the point
A on the surface of
earth is given by mv
=
2
Gm ( 34 π R ) ρ 3
0

GM r r
2
g= 2

If ρ is density of the
R 1
¿ v∝
material of earth, then
√r
The corresponding
4 v – r graph will be as
shown in option (c).
3
M= π R ρ
3
4 3
G× π R ρ
Acceleration due to
3 107 (c)
∴ g=
gravity at poles is
2
R
independent of the
4
angular speed of earth
¿ g= πGR ρ

Let g ' be acceleration 108

3

due to gravity at the

(b)

point B at a depth x ; where M E

G MEm
mg=
below the surface of
2
RE
and R E is the mass and
earth. A body at point
radius of the earth
B will experience force
only due to the portion respectively. M E =
g 2
of the earth of radius
R
G E
OB (¿ R−x ). The outer
110 (b)

spherical shell, whose

( )
3 /2
R2
thickness is x , will not
3/ 2
T 2=T 1 =1× ( 2 ) =2.8 year
R1
exert any force on
body at point B.
111 (d)
2
F=mR ω
104 (b) ¿ 6 ×10 × ( 1.5× 10 ) ( 2 ×10 )
24 11 −7 2

v=
√
2 GM
R
=√ 2 gR=√ 2 × ( 3.1 ) × 8100 × ( 10 )
2 3
¿ 36 ×10 N
21

Force between earth

¿ 27.9 km/s ec
−1 113 (a)

and moon F=
105 (b) Gmm me
2 3
T ∝r r
2

P a g e | 67
 This amount of force, that attractive force is
both earth and moon working on the missile
will exert on each
other i .e . they exert
117 (b)
same force on each
other
= =
√
( v e ) p 1 √ 2 g 1 R1 g 1 R 1
× =√ ab
( v e ) p 2 √ 2 g 1 R2 g 2 R 2

Gravitational potential
114 (c)
For w 2 w ,3 w apparent
119 (c)

at a pint outside the weight will be zero

sphere V g=
−GM
. But because the system is
r falling freely. So the
V s is same at a point distances of the
inside the hollow weights from the rod
sphere as on the will be same.
surface of sphere.
Hence, graph (c) is
Potential energy
120 (a)
correct.
−GMm GMm −GMm
¿ = =
A person feels
115 (c) r R e +h 2 Re

weightlessness in
2
g Re m −1
¿− = mg R e=−0.5 mg R
satellite orbit because
2 Re 2

he is in free fall along

with the satellite and
121 (c)

experiences no force
1 2 1
m v e = m 2 gR=mgR
of support from the
2 2

satellite. The 122 (c)

perception of weight Re 2
comes from the
Given =
Rp 3
support force exerted
on one by the floor, a
de 4
=
chair etc. If that
dp 5

support is removed
2
As MG=g Re

and one is in free fall,

4 3
¿ M =d e × π R e
we feel no experience
3
of weight.
4 3 2
d e × π R e ×G=g e Re
3
4
If missile is launched
116 (b) ¿ d e × π R e ×G=g e … . ( i )
Similarly for planet
3
with escape velocity,
then it will escape 4
from the gravitational
dp× π R p G=g p … . ( ii )
Dividing Eq. (i) by Eq.
3
field and at infinity its
total energy becomes (ii), we get
zero g e Re d e
But if the velocity of
= ×
projection is less than
gp Rp dp

escape velocity then

ge 2 4 8
= × = =0.5
sum of energies will be
g p 3 5 15

negative. This shows 123 (a)

P a g e | 68
 √
g R 127
Time period of nearby
v1 (c)
v=√ 2 gR ∴ = 1 × 1 =√ g × K= ( Kg )1 /2
satellite
v2 g2 R 2

The energy required to √

124 (b) r3
T =2 n
remove the satellite
GM

from its orbit around

the earth to infinity is
¿2π
R3
GM √
called binding energy
¿2π ¿¿¿
of the satellite. It is
equal to negative of Gravitational force
128 (a)

total mechanical dsesnot depend on the

energy of satellite in medium.
its orbit.

The body can be fired

129
GMm (d)
Thus , binding energy=−E=
at any angle because
2r

the energy is sufficient

GM
but , g=
to take the body out of
2
R

the gravitational field

2
⟹ GM=gR
of earth
2
gm R
∴ BE=
2r

Gravitational force
130 (b)

Time period is
125 (b)
provides the required
independent of mass.
centripetal force
Therefore their
periods of revolution
2
2 GMm 4 π GM
will be same.
m ω R= 3
⇒ 2 = 4 ⇒T ∝
R T R

Force acting on a body

131 (c)
Radius of earth
126 (a)

of mass M at a point at
depth d . Inside the
R
R=6400 km∴ h=
Acceleration due to earth is
4

gravity at a height h F=m g ' =m g 1− ( d

R )
( ) ( )
2 2
R R 16
gh=g =g = g
R+ h
R+
R 25 ¿
mGM R−d GM m
R
2
R
= (R
3 )
r (∵ R−d=

So, F ∝ r ; Given F ∝ r n
4
At depth ' d ' value of
acceleration due to n=1
gravity

(According According to Kepler’s

132 (c)
1
gd = gh
law
to problem)
2
2 3
T ∝r
⇒ g d=
1 16
2 25( ) d
g ⇒ g 1− =
1 16
R 2 25
g( ) ( ) 2 3
¿ 5 ∝r …(i)
By solving we get
' 2
¿ ( T ) ∝ ( 4 r ) … ( ii )
3

From Eqs.(i) and (ii),

we have
6
d=4.3 ×10 m

P a g e | 69
 ( ) ( )
3 3/ 2 3 /2
25
=
r T 1 R1 1013
2 3 = = =( 1000 )1/ 2=
(T ) '
64 r T 2 R2 1012
T =√ 1600=40 h
137 (c)
g' 2h 2× 320 1
Escape velocity,
133 (c)
=1− =1− =1− =
∴ % decrease in
g R 6400 10

v ascape=
√ 2 GM
R g= (
g−g '
)
× 100

√
8 g
¿R πG ρ 1
3 ¿ ×100=10 %
∴ v e ∝ R if ρ=constant . 100
Since the planet is
having double radius
Work done
138 (c)

in comparision to
earth, therefore
mgh
W =∆ U =
escape velocity
−h
1+
becomes twice ie ,
R

22 kms .
h
−1 Substituting R= we get
L
mg× 2 R
Let the gravitational
134 (d) ∆U=
1+ 2
force on a body mass
2mgR
m at O due to moon of
∆U=
3
mass M and earth of
mass 8 /M be zero, When the thief with
139 (a)

where EO=x and box on his head

MO=( r−x ) . Then, jumped down from a
G81 M × m GM m wall, he along with box
is falling down with
2
=
( r −x )2
acceleration due to
x

or gravity, so the
81 1
=
apparent weight of box
2 2
x ( r −x )

or = becomes zero,
9 1

(because,
x ( r−x )
On solving; x=9 r /10
R=m g−m g=0), so he
experiences no load
till he reaches the
135 (b)

ground
' 2 2

Rotation of the earth

g =g−ω R cos λ

results in the
decreased weight
140 (d)

apparently. This
mgh mg Re m
∆ U =U 2−U 1= = =
decrease in weight is
h Re
1+
not felt at the poles as
1+
Re Re
the angle of latitude is ⇒ U 2−(−mg R4 ) =
mg R e
⇒ U 2=
−
90 ° 2 2
136 (c)

The velocity of the

141 (c)

spoon will be equal to

P a g e | 70
 the orbital velocity Divide (iii) by (ii), we
when dropped out of
get = or E=
the space-ship
E 1 U
U 2 2

Let velocities of these Time period of simple

142 (b) 145 (d)

masses at r distance
pendulum T =2 π
from each other be v 1
and v 2 respectively In artificial satellite
√ 1
g'

By conservation of g =0 ∴ T =¿ infinite
'

momentum
Gravitational force
146 (b)
m1 v 1−m2 v 2=0
⇒ m1 v 1=m2 v 2 …(i) provides the required
By conservation of centripetal force ie ,
energy GMm
Change in P . E .=¿
2
m ω R= 5

change in K . E .
R2
2
m4 π GMm
Gm1 m2 1 2 1 2 ⟹ 2
= 7
= m1 v 1 + m2 v 2 T
r 2 2 R2
2 2 2 2 2 7/2
m1 v 1 m2 v 2 2 Gm1 m2 ⟹T ∝R
⇒ + =
m1 m2 r
…(ii)
The energy given to
147 (a)

On solving equation (i)

the body so as to
and (ii)
completely escape
from its orbit is equal
v 1=
√ 2 G m22
r (m1 +m2 )
and
to its kinetic energy
KE.
v 2=
√ 2 G m21
r (m1 +m2 ) 148
Acceleration due to
(c)

∴ v app =|v 1|+|v 2|=

√ 2G
r
(m1+ m2) gravity at an altitude h
is

Kinetic energy of the 2; where Re is

144 (d) g Re
2
gh =
( R e +h )
satellite is K= the radius of the earth
GMm

…(i)
2r
2
9.8 m/s × ( 6400× 10 m )
2 3

Potential energy of the

gh = 2
=
( 6400 ×10 3 m+520 ×103 m )
satellite is U =
−GMm
Escape velocity
149 (a)

…(ii)
r

Total energy of the

v=√ 2 gR

satellite is E=
−GMm
v g R
∴ 1= 1 × 1
v2 √
g2 R2
…(iii)
2r
¿ √ g× K=(Kg)
1 /2

Divide (iii) by (i), we

get =−1 or E=−K From Kepler’s second

150 (c)
E

law of planetary
K

P a g e | 71
 motion, the linear ¿ mv=√ 2 Em … . ( i )
speed of a planet is If r is the radius of the
maximum, when its orbit of the satellite,
distance from the sun then its angular
is least, ie , at point A . momentum

Using Eq. (i),

L=mvr

L=( √ 2 Em ) r =√ 2 Emr 2

Angular momentum ¿
155 (a)

Mass × Orbital
151 (a) velocity × Radius

Given, or (√ )
m g ' 30 GM
= ¿ m× × R 0=m √ GM R 0
m g 90 R0
g' 1
=
Total mechanical
g 3 156 (d)

Now, or energy of satellite

2
' R
g =g
( R +h )2
2 −GMm E A m A r B 3
g' R 1 E= ⇒ = × ⇒ ×
= = 2r E B mB r A 1
g ( R+ h )2 3

or or
R 1 157 (c)
=
R +h √ 3 1 2
kinetic energy= mve
( R+h )=√ 3 R 2
or h=( √ 3−1 ) R=0.73 R 1
¿ m× 2 gR
2
153 (a) ¿ mgR
' 10 g 90
g =g− − g
A person is safe, if his
100 100 158 (d)

velocity while reaching

2 2
' R 9 R
g =g ∨ =
the surface of moon
( R +h )2 10 ( R+ h )2
or from a height h ' is
3 R
equal to its velocity
=
√10 R+h
or h=( √ 10−3 ) R /3 while falling from
( √ 10−3)×6400 height h on earth. So
=345. 60 km
3 √ 2 g ' h ' =√ 2 g h
or
If m is the mass and v
154 (c) ' '
h =g h /g =9.8 ×3 /1.96=15 m
is the orbital velocity
of the satellite, then When going above at a
159 (b)

kinetic energy. height h or at a depth

1 2 d below earth’s
surface, in any case
E= mv
2
acceleration due to
1 2 2
gravity decrease.
¿ Em= m v
2
Therefore,
2 2
¿ m v =2 Em

P a g e | 72
 ge > gh∧g e > g d 4

Moreover gh < gd ,if h=d .

¿ 10 K

163 (b)
160 (a)
v e=R
√ 8
Gπρ ∴ v e ∝ R √ ρ

√ √
vp g R 1 1 3
v=√ 2 gR ⇒ = p × e = 2× =
ve g e R p 164 4 √ 2
From Kepler ‘s third
(c)
v
law of planetary
∴ v p= e
√2
motion,
161 (b) 2 3

Gravity, g= 2
T ∝a
GM
Given , T 1 =1 day ( geostationary )
R
2 a 1=a , a2=2 a
gearth M e R p
∴ = × 2 T1
2
a1
3
g planet M p R e ∴ =
2 3
ge 2 T2 a2
⇒ = (2 a ) a2
3 3
gp 1 2
⟹ T = 3 T = 3 ×1=8
2
2 1

Also, T ∝
√
T g a1 a
1
⇒ e= p ⟹ T 2 =2 √ 2 days
√ g T p ge
⇒
2
Tp
1
√
= ⇒T p=2 √ 2 s
2
165
Gravitational
(b)

acceleration is given
by
As we know gas
162 (d)

molecules cannot
GM
g=
escape from earth’s
where G =
2
R
atmosphere because
gravitational constant
their root mean square
velocity is less than
g M
escape velocity at
∴ =
G R2
earth’s surface. If we
fill this requirement,
then gas molecules Weight of the body at
166 (c)

can escape from

earth’s atmosphere. equator ¿ of initial
3

weight
5
ie , v rms =v es
∴ g = g (because
√
' 3
3 RT
¿ = √2 g R e
mass remains
M
5

constant)
2 Mg R e
¿T= … .(i)
3R ' 2 2 3 2
−3 −2 g =g−ω R cos λ ⇒ g=g−ω R
Given , M =2 ×10 kg , g=9.8 ms 5

6
Re =6.4 ×10 m , R=8.31 J mol – K
−1 −1
⇒ ω 2=
2g
5R
⇒ω=
2g
5R
=
2 ×1
5 ×6400 √ √
Substituting in Eq. (i), −4 rad

we have
¿ 7.8 ×10
sec
−3 167
6 (c)
2× 2× 10 ×9.8 × 6.4 ×10
( )
T= 2 /3

( )
3 /2
3 ×8.31 r R+3 R
T 2=T 1 2 =83
r1 R

P a g e | 73
 ¿ 83 ×8=664 min synchronous satellite
around the earth is
169 (a) given by
−dV
( )
1 /3
I= 2 3
4π r T G me
2

If V =0 then
dx T =
2
⇒r=
G me 4 π2
gravitational field is Substituting the given
necessarily zero values, we get

[
170 (a)
( 24 ×60 ×60 )2+ 6.67 ×10−11 × 5
r=
. If orbital radius
1 22 22
v∝ 4× ×
7 7
√r
becomes 4 times then
6
r =42.08 ×10 m
orbital velocity will
6

becomes half,
r 42.08 ×10 m
∴ = =6.6 ⇒ r=6.6
r e 6.37 ×10 6 m
7
i .e . =3.5 km/ s
Potential at the centre
2 175 (a)

due to single mass

Time period,
171 (a)
−GM
3 /2 ¿
2 πR 2π R L / √2
Potential at the centre
T= =

√ G M m √G M m

Where the symbols

R due to all four masses
GM GM
have their meaning as
¿−4 =−4 √2
L/ √ 2 L
given in the question GM
Squaring both sides,
¿−√ 32×
L
we get m m
2 3
2 4π R
T = L
G Mm
m m

If body is projected
172 (a)

with velocity v (v < v e ) Acceleration due to

176 (a)

then gravity on the surface

Height up to which it of the earth is
R
will rise,
h= G Me
v
2 ge = 2
e Re
Where M e and Re are
2
−1
v

v= (Given) the mass and the

ve
2 radius of the earth
R R R respectively
Acceleration due to
∴ h= = =

( )
(ve )
2
4−1 3
gravity on the surface
−1
of the planet is
v e /2

Mass does not vary

173 (c) GMp
g p=
from place to place
2
RP
Where M p and R p be
Time period (T ) of a the mass and the
174 (d)

P a g e | 74
 radius of the planet shell, i .e . ,r < R
respectively Einside =0
If both mass and ∴ The gravitational
radius of the planet force acting on a point
are half as that of the mass m at a distance
earth, then R/2 is
G(M e /2) G Me F=m E inside=0
g p= 2
=2 2
=2 g e
( R e /2 ) R
Gravitational force
e 180 (b)

provides necessary
The potential energy
177 (c)
centripetal force
of an object at the
surface of the earth
−GMm
U 1= … . (i)
The potential energy
R

of the subject at a
height h=R from the
2 2
Gm mv
surface of the earth
2
=
(2 R ) R

√
−GMm −GMm Gm
U 2= = … ( ii ) ⟹v=
Hence, the gain in
R+h R+ R 4R

potential energy of the 181 (b)

object Gm(μ dx )
∴ dF=
∆ U =U 2−U 1 x2
−GMm GMm
∆U= +
R+ R R
−GMm GMm
∆U= + a+ L
2R R dx
F=Gm ∫ ( A +Bx ) 2
1 GMm a x
∆U=
But we know that [( ]
2 R

2
F=Gm A
1
−
1
a a+ L
+ BL )
GM =gR
2
1 gR m 183 (d)
Hence , ∆ U = Gm1 m2 G m1 m2
2 R F= =
−2
, ie, F ∝ r
2 2
1 ( r +2 r ) 9r
Note that F ∝ r 4 by
¿ ∆ U = mgR
2

taking m= π r ρ and
4 4
178 (a)

then
3
( ) ( )
2 2
' R R 4
g =g =g = g
R+ h R 9
R+ r r
3 3
4
2 F∝ 2
,ie , F ∝r

is not correct because

r
' 4 4
∴ W = ×W = ×72=32 N
the gravitational law
9 9
obeys inverse square
Gravitational field due law and is not related
179 (c)

to a spherical shell with densities

At a point inside the
184 (c)

P a g e | 75
 Let at O there will be a
2 ; If R decreases
collision. If smaller
GM
g=

then g increases. sphere moves x

R

Taking logarithm of distance to reach at O,

both the sides; then bigger sphere
log g=log G+ log M =−2 log R will move a distance of
Differentiating it we ( 9 R−x )
get;
dg 2 dR
=0+0−
g R
¿−2 ( )
−2
=
100 100
4

∴ % increase in
dg 4
g= × 100= ×100=4 % F G ×5 M
g 100 a small= =
M ( 12 R−x )2

Kinetic energy =
185 (c) F GM
a big= =
Potential energy
5 M ( 12 R−x )2
1 2
1 mgh 1 mgh Rk
2 x= a small t
2
m ( k ve ) =
2
⇒ mk 2 gR= ⇒ h= 2
2
2 h 2 h 1−k 1 G ×5 M
1+ 1+ ¿ … (i )
Height of Projectile
R R 2 (12 R−x )2

from the earth’s

1
( 9 R−x )= abig t 2
surface ¿ h
2

Height from the centre

1 GM 2
¿ t …(ii)
2 (12 R−x )2
Thus, dividing Eq. (i)
2
Rk
r =R +h=R +
by Eq. (ii), we get
2
1−k
By solving r =
R x
1−k
2 ∴ =5
9 R−x
186 (b)
⟹ x=45 R−5 x
⟹ 6 x=45 R
g=
GM
R 2
∨R=
√
GM
g
⟹ x=7.5 R

¿ √ 6.67 ×10
The value of
−11
191
× 7.34 ×10 22/1.4=1.87 ×10 6 m (c)

188 (a) acceleration due to

gravity at a height h
v=
√ √
GM
r
=
6.67 ×10−11 ×6 ×10 24
384000 ×10
3
=1 km/ s
above the earth’s
surface is given by
189 (c)

r . If G is
2
2 4π 3
T =
variable then time
GM

period, angular
velocity and orbital
radius also changes
accordingly

190 (c)
P a g e | 76
 ( Here , M = M9 , R = R2 )
' g
g= e e

( )
2
h m m
1+
R
where R is radius of (
2
ge M e Rm 9 M e R
Hence , = × 2= ×
earth.
gm M m Re M e 2
ge 9
R ¿ =
Whenh= gm 4
2
' g 4g gm 4
g= = So , =
ge 9
( ) h
2
9
1+
R ∵ Weight of body on moon
' ' 4 4 ¿ weight of body on earth× gm / ge
Hence , weight w =m g = mg= w .
9 9 4 4
¿ 90 × =90 × =40 kg
9 9

Angular momentum
193 (c)

remains constant
197 (b)

( ) ( )
2 2
' R R 4
g =g =g ¿ = g[
v 1 d1 R+ h 3 R/2 ¿ 9
m v 1 d 1=m v 2 d2 ⇒ v 2=
d2 ' 4 4 × 200× 10
∴ W = × mg= =889
9 9

Gravitational force on
194 (b)

a body at a distance x
198 (a)
l . At the hill g
from the centre of
T =2 π
√
will decrease so to
g
earth F=
GMm
keep the time period
2

Work done, same the length of

x

pendulum has to be
reduced
R +h R +h
GM m
W = ∫ F dx= ∫ 2
dx
R R x

[ ] [ ]
R +h
−1 1 1991
Orbital velocity
¿ GMm =m g R2 − (d)
x R R+h
This work done
R

g R2 and
appears as increase in
potential energy
v 0=
GM

v 0=rω
r√ √
=
r

∆ E p=m g R 2
[ 1
−
1
R R +h ] This gives r =
3
2
R g
ω
2

¿ m g ( 5 h )2
[ 1
−
1 5
= m gh
]
Escape velocity of the
5h 6 h 6 200 (d)

195 (b) body from the surface

of earth is v=√ 2 gR
v=
√ GM
r Escape velocity of the
body from the
platform
Acceleration due to
196 (d)

Potential energy +
gravity on earth is
Kinetic energy = 0
given by g=
GM
R
2
⇒−
GMm 1
+ m v 2p=0 ⇒ v p=
2R 2
GM
R2 √
P a g e | 77
 1 1 1 From Eq. (i) and (ii),
¿ √2 gR= ; ∴ f =
√2 √2 √2 g 1−(2h
R
=g 1−
d
g ) ( )
Kinetic energy of
201 (d)
⟹ 2 h=d
satellite in its orbit 204 (a)
1 2
(√ )
2
E= mvo 1 2 GM GMm
2 ( KE )escape = m =
2 Re Re
1
¿ E= m
2
GM
r
=
GMm
2r ( ) ( KE ) =
1 GMm
kinetic energy at
body
2 Re
By law of conservation
initially

escape velocity
of energy
' 1 2
E = mv e

( )( )
2 Total Total final
mechanical = mechanical
1
¿ m
2 (
2GM
r
=
GMm
r ) energy energy
( KE+ PE )surface =( KE+ PE )at height h
¿2 E
Therefore, additional
1 GMM GMm GMm
kinetic energy
⇒ − =0−
2 Re Re R e+ h
required (∴ velocity at
maximum height is
zero)
¿ 2E−E=E

For earth,
202 (c) ⇒ v=Re

GM 4 205 (b)
g= = πR ρ G
3
√
2
vA g R
For the planet,
R
v=√ 2 gR ⇒ = A × A =√ k 1 ×
vB gB RB
G M1 4
g1 = 2
= π R1 ρ G 206 (c)
R 1
3
−GM 2 2
g R 6400 V P= (3 R −r )
= = =20 3

inside the sphere and

2R
g 1 R1 320
Let h and h1 be the
distance upto which
−GM
V P=
the man can jump on outside the sphere
r

surface of the earth

and planet, then
Escape velocity of the
207 (c)
m g h=m g 1 h1

planet is v p=
g
∴ h1= h=20 ×5=100 m
g1
2GM P
RP
Where M p and R p be
√
the mass and radius of
Acceleration due to
203 (c)

the planet respectively

gravity at height h
Escape velocity of the
(
2h
) earth is v e=
√
gh=g 1− … . (i ) 2G Me
R
and depth d Re
Where M e and Re be
(
d
gd =g 1− …. ( ii )
R ) the mass and radius of

P a g e | 78
 the earth respectively
According to given
problem, v p=3 v e and
v 1 2 GM
ω= =
R R R
=
√
2GM
R3 √
Let gravitation field is
R p =4 R e 214 (c)

zero at P as shown in
∴
√ 2GM p
4 Re
=3
Re √
2G Me M p 9Me
⇒
4 Re
⇒ M p=36 M e =36 × 6 ×10 kg
=
Re
24
figure.

24 26
¿ 216 ×10 kg=2.16 × 10 kg

208 (d)
K A rB
= =
R+h B
K B r A R+ h A
=
(
R+2 R 3
R+ R
=
2 )( ) ∴
Gm G ( 4 m)
x
2
=
( r −x )2
2 2
⟹ 4 x = ( r−x )
Binding energy of the
209 (d)
⟹ 2 x =r−x
r
system ¿
G Me M s ⟹ x=
3
2r
Gm G ( 4 m )
−11 24
6.6 ×10 × 6 ×10 ×2 ×10
30 ∴ V p= −
¿ 11
x r −x
2× 1.5× 10 3 Gm 6 Gm −9 Gm
33 ¿− − =
¿ 2.6 ×10 J r r r

Let x be the distance

211 (c)
When a body is acted
215 (a)

of point from the on by the force

smaller body where towards a point and
gravitational intensity the magnitude of force
is zero. is inversely
G m1 G m2 proportional to the
square of distance. It
∴ 2
= 2

means it obeys inverse

( 1−x ) x
or
square law and
x
1−x
m
= 2=
m1
or 10 x=−x
√ √
1000
=
1
100,000 10
represents ellipse, for
example path of the
planet around the sun
or x=(1/11)m
and the force acts
between sun and
Actually gravitational planet proportional to
212 (b)

force provides the 1

centripetal force
2
r

Escape velocity
213 (c)
216 (d)

v=
√ 2 GM

If star rotates with

R

angular velocity ω
Then

P a g e | 79
 ∆ E=GMm ( r1 − r1 )
In this problem; r 1=R
1 2

and r 2=nR

∆ E=GMm ( R1 − nR1 )
R ( n )
GMm n−1
¿

Here, OP=OQ=¿=√ 2m ( R )
¿ m g R( )
n−1 Gm
∵ g=
The gravitational force
n 2

on mass 2 kg at O due 218

to mass (i)Gravitational force
(b)

1 kg at P is on the particle placed

G× 2× 1
=G along at mid point D of side
BC of length a is
F OP= 2
( √2 )

The gravitational force

OP

on mass 2 kg at O due
to mass
1 kg at Q is
=G along
G× 2× 1
F OQ= 2
( √2 )
⃗
F =⃗
F 1+ ⃗
F 2 +⃗
F3
The gravitational force
OQ

on mass 2 kg at O due Here, ⃗

F 2=−⃗
F3
to mass
1 kg at R is
∴⃗
F =⃗
F1 +0=⃗
F1

=G along ¿ or
G ×2 ×1
F ¿= 2
( √2 )
Resolve forces F OQ and
2 2
GMM GM 4G M
F=F 1= = =
F ¿ into two
2 2 2
[ AD ] (3 a /4) 3a

rectangular (ii)gravitational force

components on the particle placed
F OQ cos 30 ° and at the point O , ie the
F ¿ cos 30 ° are equal in intersection of three
magnitude of equal medians is
and opposite direction
⃗
F =⃗
F 1+ ⃗ F3 =⃗0 or
F 2 +⃗
¿ F OP−¿
F=0
(1
2
1
¿ G− G × +G × =G−G=Zero N
2 )
The change in
217 (a)

potential energy in
gravitational field is
given by

P a g e | 80
 Since, the resultant of
surface ¿−
GM
F 2 and ⃗
⃗ F 3 is equal and
Its potential energy at
R
opposite to ⃗
infinite = 0
F1

∴ Work done = change

Potential energy of a in potential energy
219 (c)

body at the surface of GM

the earth
¿
R
2
−GMm −9 R M
Intensity of
PE= = =−m
223g R (b)
R R
gravitational field at a
6
¿ 500 ×9.8 × 6.4 ×10
point inside the
10

So, if we give this

¿−3.6 ×10 J
spherical shell is zero
amount of energy in and outside the shell is
the form of kinetic
energy then body
2
I ∝ 1/r

escape from the earth 224 (d)

220 (a)
2
m v RMm 2 GM
L=mvr=m (√ GMr )r =m √GMr ∴ L
= 2 ⇒v =
R R R
Increase in potential
225 (c)
2 2

energy,
2 πR 2 4π R GM
v= ⇒v = 2
=
T T R
GMm
(
−GMm
)
2 3
2 4π R ∆U= −
∴T = R
GM ( R+ R)
If T 1 and T 2 are the
time periods for
¿
1 G Mm 1 GM
2 R
=
2 R( )
2
1
mR = m g R
2
satellite S1 and S2
respectively
The value of
226 (c)

acceleration due to
( ) ( ) ( )
2 /3
T 1 2 R1 3 T
⇒ R2= 2
gravity changes with
= R1
T2 R2 T1
T 1=1 h ,T 2=8 h=10 km height (ie , altitude). If
4

g ' is the acceleration

()
3 /2

due to gravity at a
8 4 4
R 2= ×10 km=4 × 10 km
point, at height h
1

above the surface of

2 π R 1 2 π × 104 4 −1
v 1= = =2 π × 10 km h
earth, then
T1 1
2 π R 2 2 π × 4 × 104 4 −1
v 2= = =π ×10 kmh ' GM
T2 8 g=
Relative velocity of S2
( R+h )2

with respect to S1 is
GM
but , g= 2
R
v=v 2−v 1 ( π ×10 4−2 π ×104 ) kmh−1 ' 2 2
g GM R R
|v|=π ×10 4 kmh−1 ∴ = × =
g ( R+h )2 GM ( R+ h )2
GM GM
Potential energy of the
'
222 (b) Here , g = =
( R+h ) ( R+3 R )2
2

1 kg mass which is
placed at the earth

P a g e | 81
 GM GM g e G m2 2 mω 2 L
¿ = = a cos 30 °=m ω r= ∴r
( 4 R )2 16 R2 16 L2 √3

To calculate the force

227 (c) 232 (b)

( ) ( ) of attraction on the
3 /2 10 3/2
T mercury r mercury 6 × 10 1
= = =
point mass m we
11
T earth r earth 1.5× 10 4
(approx.)
should calculate the
force due to the solid
1
∴ T mercury = year
sphere and subtract
4
from this the force
which the mass of the
228 (c)

hollow sphere would

2
g m G( M /8) Re
= = ; …(i)
have exerted on m ie,
ge GM / R2e 8 R2m

Given,
m gm 1
=
m ge 6

or = …(ii)
gm 1
ge 6
From Eqs. (i) and (ii);
2
Re 1
2
=
8R 6 GmM Gm M
'

or Re = √ 8 /6 R m
m
F= 2
− 2
x y
[ x=R/ 4 , x+ y=R/2 ]

Weight is least at the

229 (b)

equator
M=
4
3() π R3 ρ

( )
3
230 (d) ' 4 R M
¿M = π ρ=
escape velocity 3 2 8
Given that , the orbital velocity of satellite=
2 GMm Gm(M /8) 14 GmM
F= − =
ve ( R /4 )2 ( R/4 )2 R2
⟹ v o = … . ( i)
But we know that,
2

Change in potential
233 (c)

energy
v o=
√ gR 2

On putting these
R+ h
∧v e= √ 2 gR
∆ U =U 2−U 1

values in Eq. (i)

−GMm GMm
∴∆U = +
( R+ nR ) R

√ gR 2 √2 gR
=
On squaring both
R+ h 2 ¿∆U=
−GMm GMm
R ( 1+ n )
+
R

sides, we obtain
gR
2
2 gR
¿∆U=
GMm −1
R 1+n [
+1
]
[
= ( R2 g ) m n GM
R +h 4 ¿∆U= × ∴ g= 2
2
¿ 2 gR =gR ( R+h ) R (1+n) R
¿ 2 R=R+ h∨R=h
¿ h=R=6400 km
¿ ∆ U =mgR ( n+1n )
The earth behaves for
231 (b) 234 (a)

all external points as if

P a g e | 82
 its mass M were
concentrated at its Weight on mars
centre. When man of
massm walks from a
mG (m/10)
¿ m g'=
point on earth’s
( R /2 )2

surface and reaches

4 4
¿ m× m g= ×200=80 N
diagonally opposite
10 10

point, then 238 (c)

gravitational potential
( )
2

energy given by
' R 1 R
g =g ⇒ =
R+ h √ 2 R +h
−GMm ⇒ R+ h=√ 2 R ⇒ h=( √ 2−1 ) R=0.4
Hence, distance from
U=
Will remain same.
R
centre
Hence, no work is
done by the man
¿ R+0.414 R=1.414 R

against gravity.
Work done by the
240 (c)

gravitational field is
Acceleration due to
235 (b)
zero, when
gravity at a height h displacement is
from earth’s surface perpendicular to
GM gravitational field.
Here, gravitational
'
g=
( R+h )2
g field, ⃗I =4 i^ + ^j . if θ1 is
the angle which makes
'
Since , g =
100
g GM with positive x -axis,
then
¿ =
100 ( R+ h )2

or
( R +h )2 GM tanθ 1=
1
¿ =
100 g 4

[ ]
−1 1
()
2
( R +h ) GM θ1=tan =14 ° 6 '
¿ =R2 ∴ g= 2 4
If θ2 is the angle
100 R

which the line y + 4 x=6

¿ R+h=10 R

makes with positive x -

⟹ h=9 R

axis, then
Gravitational intensity,
236 (b)
−1
θ2=tan (−4)=75° 56 '
so θ1 +θ2 =90 °
dV 14 −1
I= = =0.7 Nk g
ie , the line y + 4 x=6 is
Acceleration due to
dx 20
perpendicular to I
gravity,
−1

Work done under this

g=I =0.7 Nk g 241 (b)
F=0 when 0 ≤ r ≤ R1
field in displacing a Because intensity is
body on a slope of 60 ° zero inside the cavity
through a distance s F increase when
¿ m(g sin 60 °)s R1 ≤r ≤ R2
F ∝ 2 when r > R 2
¿ 2 × ( 0.7 × √ 3 /2 ) × 8=9.6 J 1
r
237 (d)

P a g e | 83
242 (a) km
2 2 2 21 2
g=g p−R ω cos λ=g p−ω R cos 60 °=g p− R ω
249 (c)
4
g
The escape velocity of
243 (c) '
g=
( )
2
h
a particle
1+
R
v e= √ 2 gR g g
Hence, the escape
=
( )
16 h
2

velocity is independent
1+
R
of mass of the particle.
(1+ hR ) =16
2

Due to rotation of
244 (c)
h
earth the effective
1+ =4
R
acceleration due to
h
gravity
=3
R
' 2 2 h=3 R
For a given point on
g =g−R ω cos λ .

the surface of earth g

250 (c)

decreases as ω and
GM L
2
g= K=
increases. The angular If mass of the earth
2
R 2I

speed of earth is and its angular

maximum at equator momentum remains
hence, the value of g constant then g ∝
1
on the surface of the
2
R
earth is smallest. and K ∝
1

i .e . , if radius of earth
2
R

decreases by 2% then
245 (c)

g and K both increases

GMm
Gravitational potential energy ,U =
by 4%
r
GMm
¿ U = 2 ×r
The period of
251 (b)
r
revolution of
¿ U =g × mr
geostationary satellite
¿ U =( mg ) r
U is the same as that of
the earth.
¿ mg=
r
Orbital velocity v o=√ g R e
At a height h , (Taking
247 (a)
Escape velocity v e= √ 2 g Re
h< ¿ R ) from the where Re is radius of
surface of earth earth.

( or
)
2h v e −v o
gh=g 1− % increase= ×100
R vo
gh 2h 90
=1− = % increase=
√2 g Re −√ g R e × 100
g R 100
√ g Re
or
2h 99 1
=1− = ¿ ( √ 2−1 ) ×100
R 100 100
or g=
R 6400 ¿ ( 1.141−1 ) ×100=41.4 %
= =32
100 200
252 (c)

P a g e | 84
 −GMm GMm
U= , K=
r 2r |v| π ×10 π
4

and E=
−GMm ω= = = r
R 2−R1 4 × 10 −1×10
4 4
3

For a satellite U , K
2r

Gravitational field
258 (d)
and E varies with r
and also U and E intensity
remains negative
−11 22

whereas K remain
GM 6.6 ×10 × 7.34 ×10
I= 2
= 2

always positive
R ( 1.74 ×10 6 )
−1
¿ 1.62 Nk g

Orbital radius of
253 (d)
Potential energy
259 (b)

satellites r 1=R+ R=2 R −GMm −GMm

r 2=R+ 7 R=8 R U= =
r R+ h
and and
−GMm −GMm
U 1= U initial =
r1 3R
−GMr −GMm
U 2= U final=
r2
Loss in PE=¿ gain in
2R

and
GMm
K 1=
2 r1 GMm GMm GMm
KE= − =
GMm 2R 3R 6R
K 2=
2 r2
v=√ 2 gR . If
260 (a)
and
GMm
acceleration due to
E 1=
2r 1
GMm gravity and radius of
the planet, both are
E 2=
2r 2
U 1 K1 E1 double that of earth
then escape velocity
∴ = = =4
will be two times,
U 2 K2 E2

254 (b) i .e . v p=2 v e

g= 2 . If radius
GM
262 (c)
shrinks to half of its
R

present value then g

√
ve g e Re
= g R =√ 6 ×10= √ 60=8 ¿
will becomes four
vm m m

times (nearly)
¿

256 (a)
Total final263 (b)

( )( )
Total
mechanical = mechanical
energy P energy O (
g' =g 1−
d
R) g d
(
⇒ =g 1− ⇒ d=
n R )
1 2642 GMm (b)
v e= √ 2 gR and
1 2 GMM
⇒ m ( 0) − = mv −
2 √
( √ 3 R ) R2 2
2 R
v 0=√ gR ∴ √ 2 v 0=v e
GMm 1 2 GMm
⇒− − mv −
2R 2 R 265 (d)
⇒ v=
√ GM
R
F∝
1
2 . If r becomes

double then F reduces

r
257 (d)
P a g e | 85
 to
F Given , T 1 =1, T 2=8 , R1=R
4 T1
2 3
R1
∴ =
Since the gravitational
266 (a) 2 3
T2 R2
field is conservative
2
T2
field, hence, the work
3 3
R =R2 1 2
T1
done in taking a
particle from one point
3 3 2
R2=R 1 × ( 8 )

to another in a
3 2
R2=R × ( 2 )
3 3

gravitational field is ⇒ R2 =R× 4=4 R

path independent

From Kepler’s law,

272 (a)
268 (b)

⇒ v e ∝ √ M if
√
2 3
2 GM T ∝R
v e=
( ) ( )(
2 3

)
3
T2 R2
R=¿ constant
R 1.01 R
¿ = = =¿
If the mass of the
T1 R1 R

planet becomes four

T2
¿ =¿
times then escape
T1

velocity will become 2

T 2−T 1 1.5
times
¿ = =1.5 %
T1 100

269 (b) 273 (c)

∆U=
mgh mgh mgR
1+
h
=
1+
R
=
2 g=
GM
=
G ( 43 π R ) ρ3

2 2
R R R R
¿ g∝ρ R
270 (d)
By Kepler’s law T 2 ∝ R 3
g
¿ R∝
Now escape velocity,
ρ
T 2 2 R2
( ) ( )
3
Hence , =
T1 R1 v e= √ 2 gR

( ) ¿ v e ∝ √ gR
3
2.5 R+ R
¿

√ √
6 R+ R g g2
¿ ve ∝ g × ∝
¿( )
3
1 ρ ρ
2 ∴¿
T1 ¿ 3 km s
−1
T 2=
( 2 )3/ 2
For a geostationary
When r < R ,
274 (d)

satellite
Gravitational field
intensity,
T 1=24 h
24
=6 √ 2 h
So ,T 2=
2 √2 I=
GM
R
3
Gr 4
r= 3
R 3
π R3 ρ =(4 πG ρ r
3 )
From Kepler’s third
271 (b)
275 (b)
law of planetary
ω body=27 ωearth
motion 2 3 2 1 1 1
T ∝ r ⇒ ω ∝ 3 ⇒ ω ∝ 3/ 2 ∴ r ∝ 2
T ∝R
2 3 r r ω

P a g e | 86
 ( ) ( ) √
2 /3 2 /3
r body ω earth 1 1 vp Rp ρ
⇒ = = = v ∝ R √ρ ∴ = × p =4 × √ 9=
r earth ωbody 27 9 v e Re ρe
⇒ v p=12 v e

Time period of satellite 281

276 (b)
(c)

√
T =2 π √ ¿ ¿ ¿ vA g R
v e= √ 2 gR ⇒ = A × A =√ x ×
vB g B RB
Where , Re =Radius of earth ,

Due to inertia of
h=Height ¿ earth surface . 282 (b)
Time period not
depend on mass. So, direction
time period of both
The earth possesses
283 (a)
satellite will be equal.
rotational motion
277 (b) about an axis through
Gm( M −m)
; For its poles. The value of
acceleration due to
F=
x2
maxima, gravity at a place (at
dF G given latitude) is
affected due to its
= ( M −2 m )=0
dm x 2
rotational motion. If
or
earth ceases to rotate,
m 1
=
the weight of body at
M 2

278 (d) equator will increase.

However, there will be
v e=
√ 2 GM
R no effect on the weight
at poles. The effect of
√
−11

rotation of the earth

30
2 ×6.67 × 10 ×1.99 × 10
¿
on acceleration due to
8
6.96 ×10

gravity is to decrease
¿ 618 km/ sec

279 (a) its value. Therefore, if

GM the earth stops
rotating, the value of g
'
g= ,
( R+h )2
acceleration due to will increase.
gravity at heighth

The relation between

2 284 (b)
g GM R
mass and density of
⟹ = 2 ∙
9 R ( R+h )2
earth is given form
( )
2
R
Newton’s law of
¿g
R+h
gravitation, according
⟹ =(
9 R+h )
2

to which
1 R

2
R 1 gR e
⟹ = M e=
R+ h 3 G
⟹ 3 R=R+ h where M e is mass of
⟹ 2 R=h earth, G the
gravitational constant,
Re the radius of earth
280 (b)

and g the acceleration

P a g e | 87
 due to gravity. the system of units

For a moving satellite

Also , mass=volume ×density
288 (b)
G × volume × density
kinetic energy ¿
g= 2 GMm
Assuming spherical
R

Potential energy
2r
shape of earth volume
4 3 −GMm Kinetic energy 1
¿ πR ¿ ⇒∴ =
3 r Potential energy 2
3
4 πR
Mass of the satellite
g=G × ρ 289 (d)
3 R2
4 does not affect the
time period
⟹ g=G ∙ πR ρ
Hence, increases in
3

radius would () ( ) ()
3 /2 3/ 2 1/ 2
T A r1 r 1 1
= = = =
dominate.
T B r2 2r 8 2√ 2

The earth moves

285 (b) 290 (c)

around the sun is

GM G 4 3 4
g= = 2 π R ρ= πG ρ R ie , g=R
elliptical path, so by
R 3 3
2
R
using the properties of
Let escape velocity be ellipse
286 (d)

v e, then kinetic energy r 1= (1+ e ) a and

is r 2= (1−e ) a

and
1 2 r 1 +r 2
¿ mv e … . ( i ) ⇒ a=
2 2
+GM e m
… ( ii ) r 1 r 2=( 1−e2 ) a 2
Where a=¿ semi major
¿ escape energy=
Re
Equating Eqs. (i) and axis
(ii), we get b=¿ semi minor axis
1 2 GM e m e=¿ eccentricity
Now required distance
m=
2 e Re
= semi latusrectum
⟹ ve=
√ 2GM e
Re
2 GM e
¿
b
a
2

⟹ R= 2 2
(r 1 r 2) 2r r
2
ve a (1−e )
Given,
¿ = = 1 2
a (r 1 +r 2 )/2 r 1 +r 2
−11 2
G=6.67× 10 N – m /kg , 291
Let a satellite is
(b)
24 8 2
M e =6 ×10 kg , v e =3 ×10 m/s
−11 24
revolving around earth
with orbital velocity v .
2 ×6.67 × 10 ×6 × 10
R=
¿¿
The gravitational
−3
R=8.89 ×10
potential energy of
−3

satellite is
R ≈ 9 ×10 m=9 mm

k represents
287 (a) −G M e m
U= … (i)
gravitational constant
Re
The kinetic energy of
which depends only on

P a g e | 88
 satellite is
1 G Mem
K= …(ii)
2 Re
∴Total energy of
satellite
E=U + K
G Mem 1 G Mem
¿− +
Re 2 Re
1 G Mem Resultant force=√ 3 F
¿− ∴ Force on mass at A due ¿ mass at B
2 Re
…(iii)
( )
2

But we know that

GM
¿√3 2

necessary centripetal
L
Centripetal force for
force to the satellite is circumscribing the
provided by the triangle in a circular
gravitational force. ie , orbit is provided by
mutual gravitational
m v G Mem interaction.
2
=

( )
2
Re Re Mv
2
GM
2
ie , =√ 3
2 G Me m ( L/ √ 3 ) L
2
¿mv = …(iv)

√
Re
Hence, from Eqs. (iii)
GM
¿ v=
and (iv), we get
L

−1 2
E= mv 294 (c)
2
GM
292 (c)
g= 2

So,
R

gM M M
gE
=
ME ( )( )
R 2 1
× E = ×
RM 10
12742
6760 (
gM

⇒ dU=
Gm ( M
l
dx ) ∴
gE
=0.35 ⇒ g M =9.8 × 0.35=3.48

x 295 (b)
a+l
GmM dx 4
⇒ U ∫ dU = ∫ x 3
πR ρ G×
l a GM 3 4
g= 2 = = πG ρ R i
( )
2
3
For pendulum clock, g
−GmM a+l R R
⇒U = log e
will increase on the
l a

planet, so time period

Given,
293 (a)
will decrease. But for
F 1=F2=F∧θ=60 ° spring clock, it will not
change. Hence, P will
run faster than S

From Kepler’s third

296 (b)

law of planetary

P a g e | 89
 motion also known as
law of periods Below the surface of
the earth g ∝r and
above the surface of
2 3

Where T is time period

T =k r

and r the mean earth g ∝1/r 2.

distance from the sun. Therefore, the graph
Hence, greater is the (a) is correct
distance of planet
from sun, greater is its
Launching the rocket
301 (c)
period of revolution.
in the direction of
earth’s rotation allows
If g is the acceleration it to exploit the earth’s
297 (b)

due to gravity of earth rotational velocity ie ,

at the position of launching it from West
satellite, the apparent to East. (It gains speed
weight of a body in the from velocity addition
satellite will be with the earth’s
W app=m ( g' −a ) rotational velocity.)
But as satellite is a
freely falling body,
302 (c)
GM
'
ie , g =a Accelerationdue ¿ gravity , g=
So, W app=0
2
R

( )( )
2
gM M M R
= × E
298 (b) gE ME RM

( ) ( )
3/ 2 2
T2 r 2 1 12742
=( 2 ) =2 √ 2⇒ T 2=2 √ 2 years
3 /2
= ¿ ×
T 1 r1 10 6760
gM
=0.35
The value of
299 (c) gE

acceleration due to
−2
g M =9.8 × 0.35=3.48 ms
gravity at height h
(when h is not Using conservation of
303 (a)

negligible as energy.
compared to R ) 1 2 GMm GMm
2 mv − =0−
' R 2 R 2R
g =g 2
(R+h) 1 2 GMm GMm
¿ mv = −
Here, g =
' g 2 R 2R

g
∴ =g
R
2
2
¿ v2 =2
GM
R [ ]
1−
1
2
2 ( R+h )2 2
¿v =
GM
1 R
2 R
¿ = 2
2 ( R+ h )2 But gR =GM

√
¿ =
1
2 R+h
R ∴v=
√gR2

¿ v=√ gR
R
¿ R+h=√ 2 R
∴ h= ( √ 2−1 ) R 304 (c)
300 (a)
P a g e | 90
 [∵ angular
√
v 1 r 1=v 2 r 2
momentum is
2G M e
v es=
constant]
Re

√
4
2G∙ π R 3e d e
305
Gravitational force
(d)
¿
3
Re
4
(
∴ M = π R 3e d e
3 )
between sphere of ¿ v es ∝ √ d e × R e … . ( i )
mass M and the similarly, for a planet
particle of mass m at B
is
v es ∝ √ d p × R p … ( ii )
'

( )
1 /2
GM m v es de Re
F 1= So , '
= ×
2 v dp Rp
d
If M 1 is the mass of the
es
1
removed part of
Given , d p= d e , R p=2 Re
4
sphere, then

()
1
d e 21

( )
4 1 4 M v es d e R
M 1= π ( R /2 )3 ρ= π R2 ρ = = × e
3 8 3 8 v es 4 2 Re
Gravitational force
between the removed
1 /2 1
¿ (4 ) ×
part and the particle
2

of mass m at B is
1
¿ 2 × =1
2
G M 1m G ( M /8)m GM m v es
F 2= 2
= 2
= 2 So , =1:1
( d−R/2 ) ( d−R/2 ) 8 ( d−R/2 ) v ' es

307 (b)
g= πρGR . If density is
4

same then g ∝ R
3

∴ Required force, According to problem

GM m GMm R p =2 R e ∴ g p=2 g e
For clock P (based on
F=F 1−F2= 2
− 2

pendulum motion)
d 8 [ d−( R/2 ) ]

[ ]
GM m 1

√
¿ 1− l
d2
( )
2
R T =2 π
8 1−
Time period decreases
2d g

on planet so it will run

The maximum velocity faster because g p > ge
306 (c)

with which a body For clock S (based on

must be projected in oscillation of spring)
the atmosphere, so as
to enable it to just
overcome the
T =2 π
√ m

So it does not change

k

gravitational pull, is
known as escape 308 (d)
velocity.
Escape velocity from
earth’s surface is
v 0=
√ GM
r

Here,
309 (d)

P a g e | 91
 If earth were to spin
faster, that is angular
−1

Using Newton’s
u=20 ms , m=500 g=0.5 kg ,t=20 s

equation of motion velocity increases,

then except at poles,
the weight of bodies
1 2
s=ut+ ¿
will decrease at all
2

places.
1 2
0=20 ×20+ (−g ) ( 20 )
2
−2
¿ g=2m s
Earth and all other
314 (b)
∴ Weight of body on planet=mg
planets move around
¿ 0.5 ×2=1 N
the sun under the
According to Kepler’s effect of gravitational
310 (d)

third law, T 2 is force. This force

always acts along the
proportional to cube of
line joining the centre
semi-major axis of the
of the planet and the
elliptical orbit.
sun and is directed
towards the sun. In
r 1+ r 2

other words, a planet

Semi – major axis=
2
moves around the sun
[ ]
3
r 1 +r 2
under the effect of a
2
∴T ∝
2
¿T ∝¿ purely radial force.
Therefore, areal
velocity of the planet
It is self-evident that
311 (c)
must always remain
the orbit of the comet
constant.
is elliptic with sun
begin at one of the
∆A L
∴ = =a constant vector
focus. Now, as for
Therefore, Kepler’s
∆ t 2m

elliptic orbits,
2nd law is the
according to kepler’s
consequence of the
third law,
principle of
conservation of
( )
2 3 2 1 /3
4π a T GM
angular momentum
2
T = ⇒ a= 2
GM 4π

[ ]
( 76 ×3.14 × 107 ) ×6.67 × 10−11
1 /3
(L)
τ =0
× 2× 1010
a= Now , τ=I α
4 π2
But in case of ellipse,
∴ I α =0∨α =0
¿ α T =r α=0
ie , tangential
2 a=r min + r max
acceleration is zero.
12 10
∴ r max =2 a−r min =2× 2.7 ×10 −8.9 ×10
12
≅ 5.3 ×10 m

From Kepler’s third

316 (d)

The variation of g with law of planetary

312 (c)

angular velocity (ω) is motion,

given by T ∝R
2 3

' 2
g =g−Rω

P a g e | 92
 escape ( K )= m×2
T
2
' 1 GM
⟹ =constant
R
3 2 R
'
K =2 K
317 (a)

Here the force of

100 325 (c)
(
g' =g 1−
d
) (
=9.8 1− )
=9.66 m/s 2
attraction between
R 6400

319 (c) them provides the

necessary centripetal
[ ( )] force
2
−GM r −GM −GM
V ¿= 3− ,V surface= , V out =
2R R R r

320 (a)
vp
ve
g
√R
= p × p =√ 9 ×4=6 ∴ v p=6 × v e =67.2 km/s
ge R e
2 2

Gravitational potential
321 (c) mv Gm
∴ =
R ( 4 R )2
V =GM ( 1 1 1
+ + +…
r1 r2 r3 ) ∴v=
√ Gm
4R
¿ G ×1( 11 + 12 + 14 + 18 + 161 +…326.) (b)
GM
¿ G(
1−1/2 )
1 Accelerationdue ¿ gravity g= 2
R
3
4G π R
(∴ ∑ of GP= 1−ra ) ∴ g=
3 R2
ρ

¿2G ⟹ g=
3(
4 GπR
)
ρ ( ρ=average dens

Below the sea level the

322 (b) ⟹ g∝ ρ∨ ρ∝ g

pressure is increasing 328

Gravitational
(a)
with depth in mine due
to presence of attraction force on
atmospheric air there. particle B
The acceleration due G MP m
to gravity below the
F g= 2
( D P /2 )
surface of the earth Acceleration of
decreases with the particle due to gravity
distance from the
surface of the earth
Fg 4GM P
a= = 2
m DP
as ( )
d
g' =g 1−
R
If M be mass of earth
329 (c)

323 (b) and R its radius, the

GMm acceleration due to
gravity is given by
K=
Escape velocity
2R
GM
g= … .(i)
√
2GM 2

Where, G is
V e= R

Kinetic energy to
R

P a g e | 93
 gravitational constant.
Given , R=0.99 R
' GM
∴g= … ( ii )
( 0.99 R )2

¿ 1.02
( )
GM
R
2 V p=V sphere +V partical

From Eq. (i), we get

GM GM 3 GM
¿ + =
'
a a/2 a

Hence, acceleration
g =1.02 g

due to gravity S2 is correct because

334 (d)

increases by whatever be the g, the

same force is acting on
'

Hence, percentage both the pans. Using a

g −g=1.02−1=0.02 g

increases =2%. spring balance, the

value of g is greater at
the pole. Therefore mg
The necessary
330 (a)
at the pole is greater.
centripetal force
S4 is correct. S2 and S4
required for a planet
are correct
to move round the sun
= gravitational force
335 (a)

exerted on it
G Me G Mm
2
=
x ( r −x )2
or
m v G Mem
2
ie , = n
R R

v=
( )
G Me
R
n−1
1 /2 r−x
x
=
Mm
Me √ √
=
7.35 ×1022

or r =0.11 x+ x=1.11 x
5.98× 10
24

( )
n−1 1 /2
2 πR R 8
Now ,T = =2 πR × x=r /1.11=3.85 ×10 /1.11
v G Me 8
¿ 3.47 ×10 m

( )
2 n−1 1 /2
R ×R
⟹=2 π
The gravitational force
G Me 336 (d)

exerted on satellite at
( )
(n+1 )/ 2
R
a height x is
¿2π 1/ 2
(G M e )
⟹T ∝R
(n+1 )/ 2 G Me m
F G=
( R+ x )2
where
331 (d)
−GMm −GMm
+ KE= M e =mass of earth .
Since, gravitational
2 R1 2 R2

force provides the

KE=
GMm 1
2
−
1
R 1 R2 [ ] necessary centripetal
force, so,
332 (c) 2
G Me m mv o
( )
3/2

( 14 )
3 /2
r 1 =
T 2=T 1 2 =T 1 = ׿ ( R + x ) (R+ x)
2

where v o is orbital
r1 8

333 (d) speed of satellite.

P a g e | 94
 [ ]
G Mem 2 2 GM r 2
1 /2
⟹ =m v o v 1=
( R+ x ) r 1 ( r 1+r 2)

( )
2
gR m GMe

[ ]
2 1 /2
⟹ =m v o ∴ g= 2 2 GM r 1 r 2
( R+ x ) R ∴ L=m v 1 r 1=m
( r 1+ r 2 )

√[ ][ ]
2 2 1 /2
gR gR
⟹ vo = =
(R+ x ) (R+ x )341 (b)

As mass, M =
4 2
πR ρ
337 (b) 3
or
x x
C C 3M
V =−∫ I dx=−∫ dx= 3
ρ=
∞ ∞ x
2
x 4πR
3

( )
3
ρ s M s Re 1
∴ = × 3 =330 × =3.3
Telecommunication
339 (d)
ρs M e R s 100

satellites are
geostationary satellite As there is no gravity
342 (a)

in space so spring will

not be extened.
The gravitational force
340 (d)

of sun on comet is
radial, hence angular The value of g at the
343 (b)

momentum is constant height h from the

over the entire orbit. surface of earth
Using law of
conservation of ( )
2h
g' =g 1−
angular momentum, at
R
The value of g at depth
locations A and B
x below the surface of
earth

g '=g 1− ( Rx )
These two are given
L=mv 1 r 1=m v 2 r 2 or equal, hence

…(i) (1− 2Rh )=(1− Rx )

v1 r1
v 2=
On solving, we get
r2
Using the principle of
conservation of total x=2 h
energy at A and B
We have
344 (b)
1 2 GMm 1 2 GMm
m v 1− = m v 2− 2 3
2 r1 2 r2 T ∝R
or R1=r

( r1 − r1 ) …
¿ R2=4 r
v 22−v 21=2 GM 2 3
T1 (r )
(ii)
2 1
∴ =
Putting the values
2 3
T2 (4 r )

from Eq. (i) in Eq. (ii)

T1 1
¿ =
and solving, we get
T2 8

When earth moves

345 (d)

P a g e | 95
 round the sun then
according to Kepler’s
348 (b)
As T 2 ∝ r 3,
second law, the radius
vector drawn from the so,
2 3
TA rA
=
sun to earth, sweeps
2 3
TB rB

out equal areas in or ( )

2/ 3
rA T A
equal time, ie , its areal
= =( 8 )2 /3=4
rB TB
velocity (or the area or r A =4 r B ;
swept out by it per so r A −r B=4 r B −r B =3 r B
unit time) is constant.
While in such motion, 349
For central force,
(d)
angular velocity,
kinetic energy and torque is zero
potential energy dL
change.
∵τ= =0⇒ L=¿
constant
dt

i .e . Angular
According to Kelper’s
346 (b)
momentum is constant
third law (law of
periods),
Let R be the original
350 (a)

where T is time taken radius of a planet.

2 3
we have T ∝ R

by the planet to go Then attraction on a

once around the sun body of mass m placed
and R is semi-major on its surface will be
axis (distance) of the GM m
elliptical orbit.
F= 2

If size of the planet is

R
∴ T =k R …(i)
made double
3 3

Where k is constant of
proportionality. ie , R =2 R , then mass
'

When R becomes 4 of the planet becomes

times let time period 4 3
be T ' .
' 3 2
M = π ( 2 R ) ρ=8 × π R ρ=8 M
New force
3 4
'2
∴ T =k ¿
'
T
2
1 ' −G M m −G8 M ×m
∴ '2 = F= '2
= =2 F
R (2 R )2
ie , force of attraction
T 64

increases due to the

T 1
¿ '=
increase in mass of the
T 8

planet
'

So, time period

¿ T =8 T

becomes 8 times of
previous value.
351 (c)
Here , AB=BC =CD=DA=1 m
BD= AC
Escape velocity of a
347 (a)
¿ √ 12+ 12
body from the surface ¿√2 m
of earth is 11.2 km s−1 Total potential energy
which is independent
of the angle of project
U=
[ −G× 1× 1 −G ×1 ×1
AB
+
BC
+
−
][ ][
P a g e | 96
 +
[ −G ×1 ×1
DA
+
][ BD
+
][
−G× 1× 1 −G ×1 ×1
AC ] E=
G Mem
r

[ ] [ ]
−G ×1 ×1 −G ×1 ×1 At r=2 R ,
¿4 × +2 =−5.4 G
1 √2 −G M e m
E1=
(2 R)
At r=3 R
−G M e m
E2=
(3 R)
Energy required to
move a body of mass m
from on orbit of radius
2 R ¿3 R is
352 (b)
ge R e ρ e 2 4 ge
∆ E=
G Me m 1 1 G Mem
R
− =
2 3 6R [ ]
= = × =6∨g m=
gm Rm ρ m 3 1 6
For motion on earth, From Kepler’s third
356 (b)

using the relation, law of planetary

1 2
s=ut+ a t motion

We have,
2 T ∝R
2 3

Given , R1=R , R2=5 R

=0+ ×9.8 r or
1 1 2 2 3
2 2 T1 R
∴ 2
= 3
t=1/ √ 9.8 s T2 (5 R )
For motion on moon, T1 1
⟹ = 3/ 2
1 2 T 2 (5 )
3=0+ ( 9.8/6 ) t 1
2 T 2=5
3 /2
T1
or t 1=6 √ 9.8 s ∴ =6 or
t1 3

t ∴ T 2 =5 T [∴ T 1=T ]
2

t 1=6 t

Let m be mass of
357 (c)

Gravitational potential planet and M that of

353 (c)

energy of a body in the sun, r the radius

gravitational field, between the two. Let
the planet be moving
. When r
−GM m
with velocity v o , then
E=
decreases negative
r
Gravitational force =
value of E increase ie , centripetal force
E decreases

Binding energy ¿∨E∨¿

354 (a)

1 GMm 1
¿ = gm Re
2 Re 2
2

Gravitational potential
355 (d) GMm m v o
=
energy of body will be
2
r r

P a g e | 97
 √
GM 9
⟹ vo = ∴ Re = Rm
r √6
1
⟹ vo ∝
The relation between
360 (a)
√r
Hence, larger the
density ( d ) and
distance, smaller the
acceleration due to
orbital velocity. At
gravity (g) is
point C distance from
sun is maximum,
3g
hence orbital velocity
d=
4 π Re G
is lowest. At point A d 1 g1 r 2
distance from sun is
∴ = ×
d 2 r 1 g2
minimum, hence
orbital velocity is
g1 d 1 r 1
⟹ =
maximum.
g2 d2 r2

Velocity of body in
361 (c)

According of law of
358 (a)
inter planetary space
gravitation, the force
of attraction acting on
v '=√ v 2−v 2es
Where v es= escape
the body due to earth
is given by velocity and
v=¿ velocity of
projection
Mm
F=G 2
… .(i )

The acceleration due

R
'
√
∴ v = ( 2 v es ) −v es=√ 3 v es ⇒ v =√ 3
2 2 2 '

to gravity g in the
body arises due to the
362 (b)
force F from Newton’s Using g= 2 we get
GM
second law of motion, R
we have gm =g /5
F=mg … ( ii ) 363 (c)
From Eqs. (i) and (ii), −GMm
. If B . E .
we get
B . E .=
decreases then r also
r

decreases and v
Mm
mg=G 2
R
increases as v ∝
GM 1
⟹ g= 2 √r
R
364 (d)
Gravitational pull
359 (b)
−GMm
depends upon the
U i=
r
acceleration due to
−11 −2
6.67 × 10 ×100 ×10
gravity on that planet
U i=
0.1
−11
1 1 −6.67 × 10
M m= M e , g m = ge U i=
81 6 0.1
−10

( ) ( We know
1/ 2 ¿−6.67 ×10 J
)
1/ 2
GM Re M e gm 1
g= 2 ⇒ = × = 81 ×
R Rm M m ge 6
∴ W =∆ U
¿ U f −U i ( ∴ U f =0 )

P a g e | 98
 mass of the solid
sphere and particle
−10
∴ W =U i=6.67 ×10 J

respectively and R is
the radius of the
sphere. The
gravitational force on
particle due to sphere
with cavity = force due
to solid sphere
creating cavity,
Error in weight =
365 (c)

assumed to be present
difference in weight at
above at that position
two different heights
GM m G(M /8) m 7 G
[
¿ m g 1−
2 h1
R ] [
− m g 1−
2 h2
R ] ie , F 2=

So,
4 R2
−
( 3 R /2 )2
=
36

2m g 2 m GM h
¿ ( h 2−h1 )= × 2 ×
( )
F 2 7 GM m GM m 7
R R R 2 = / =
[where, h2 −h1=h 2]
2 2
F1 36 R 4R 9
2m 4 h 4 371
The value of
2
¿ 3 × G× π R ρ × = πGm ρh (b)
R 3 2 3
acceleration due to
gravity at a height h
366 (c)

Here, I = reduces to
dV
=−k /r
dr
or dV =k
64
dr ¿ 100−36=64 %= g
100
Integrating it, we get
r 2
64 gR
∴ g=
v r 100 ( R+h )2
dr
or
∫ dV =∫ k r 8
=
R
∨h=
R

or V =V 0+ k log r /r 0
V0 r0
10 R+ h 4

If the mass of sun is M

372 (a)

Acceleration due to
367 (d)
and radius of the
gravity planet’s orbit is r ,
then as v 0=√ GM /r
GM G 4 3
g= = 2× πR ρ
R 3
2
R
∴ ρ=
3g
4 πGR
T=

…(i)
2 πr
v0
=2 πr
√r
GM
, ie ,T 2=
4 π2 r
GM

Now, if the planet

(When stopped in the
369 (a)
g= πρGR . If ρ =
4
3 orbit) has velocity v
when it is at a distance
constant then =
g 1 R1
x from the sun, by
conservation of
g 2 R2

Gravitational force due mechanical energy,

370 (b)

to solid sphere,
GM m
1
2
m v2 +
x(
−GMm
=0−
GMm
r )
or
F 1= ,
( 2 R )2
where M and m are
P a g e | 99
 ( ) [ ] 4
2
−dx 2 GM r −x Accelerationdue ¿ gravity g= π ρ
= ,
dt r x 3

√ √
¿ g∝ρ
dx 2 GM (r−x)
ie ,− = g1 ρ 1
or
dt r x ∴ =
g 2 ρ2

√ [ ]
t 0
r x
∫ dt =− 2 GM ×∫ dx g1 ρ
r (r −x) = [ ∵ ρ2=2 ρ ]
Substituting x r sin 2 θ
0
g2 2 ρ

and solving the RHS,

g2=g1 ×2=9.8 ×2
2
g2=19.6 m/ s
T=
r
√
2 GM
×
πr
2
In the light of Eq. (i)
( )
The net force acting
376 (a)

reduces to on a unit mass placed

at O due to three equal
t=
1
√4 √2
T ,ie , t=
8 ( )
√2 T
masses M at verities
A , B∧C is the
gravitational field
The minimum velocity
373 (d)
intensity at point O .
of projection to
The gravitational force
achieve escape
on the particle placed
velocity can be
at the point of
calculated as
intersection of three
medians.
1 2
Intial KE= mv
2
1 2 1 2
¿ ×m ( 4 × 11.2 ) =16 × mve
2 2
As mve energy is used
1 2

up in coming out from

2

the gravitational pull

of the earth, so
final KE should be
1 2 Since, the resultant of
F 1∧F2 is equal and
15 × mv e
2
opposite to F 3.
1 2 1 2
Hence , mve =15 × mve
2 2
'2 2

It is given that,
∴ v =15 v e 377 (d)

acceleration due to
¿ v = √ 15 v e
'

gravity on planet A is
¿ √ 15 ×11.2kms
−1

9 times the
U =¿ Loss in acceleration due to
374 (c)

gravitational energy = gravity on planet Bie ,

gain in K.E. g A =9 gB … ( i )
From third equation of
So, U = m v ⇒ m=
1 2U
motion
2
2 v
2

2
v =2 gh
375 (b)

P a g e | 100
 v
2
3
At planet A , h A = … ( ii ) ρ 1 : ρ 2=
2 gA 2
v
2 g1 ρ 1 R 1 3 2
At planet B , hB = … ( iii ) ∴ = = × =1
2 gB g 2 ρ2 R 2 2 3
Dividing Eq. (ii) by Eq.
(iii), we have
380 (d)

hA gB
=
hB gA
F=
{ }GMm
r2
,r≥ R

From Eq. (i), g A =9 gB

Escape velocity for
381 (c)
hA gB 1

that body v e=
∴ = =
hB 9 gB 9
¿ h B=9 h A=9× 2=18 m(∴ h A =2 m)
v e should be more than
√ 2 Gm
r

or equal to speed of
In circular orbit of a
378 (b)
light
satellite of potential
energy
¿−2× ( kinetic energy )
i .e .
√ 2Gm
r
≥c

Mass of satellite does

1 v 2 382 (b)
¿−2× m =−mv
Just to escape from the not affect its orbital
2

gravitational pull, its radius

total mechanical
energy should be zero.
383 (b)
Therefore , its kinetic
√ √
3 3
( R +h ) (2 R )
energy should be +mv 2
T =2 π 2
=2 π =4 √ 2
gR g R2

384 (d)
The acceleration due
379 (a)

to gravity (g) is given

by
v 0=
√ √ GM
r
=
g R2
R+h

Acceleration due to
GM 385 (d)
g=
gravity at depth d
2

where M is mass, G
R

the gravitational below the surface

constant and R the earth
radius.
Since, planets have a
gd =g 1− ( d
R )
spherical shape Acceleration due to
gravity at height h
from the surface of the
4 3
V= π r
earth
3
Also , mass ( M )=volume ( V ) × density ( ρ )

g=
4
G πR ρ
3
3
gh=g g−
2h
R ( )
R
2 Given gh=g d
4 Gπ ρ R 2h d
⟹ g= ∴ =
3 R R
Given , R1 : R 2=2 :3 d=2h

P a g e | 101
 √
d=10 km GMm 1 2 GM
= m v 2t , v e = =√ 2 g R
R 2 R
If x is the distance of
386 (c)
v e= √ 2 × v 0=1.41 ×8 km s =11.2 k
−1

point on the line ∴ the additional

joining the two masses velocity to be imparted
from mass m 2 ' where to the orbiting satellite
gravitational field for escaping is
intensity is zero, then 11.2 −8=3.2 km s−1

= 2 or
Gm G m2
Angular momentum of
391 (c)
2
( r−x ) x
2 8 the earth around the
sun is
2
= 2
( 9−x ) x
or
1 2 L=M E v 0 r
=
On solving, x=6 m
√ ( √ )
9−x x G Ms G Ms
⇒ L=M E r ∵ v 0=
r r
1/ 2
387 (d) ⇒ L= [ M 2E G M s r ]
% change in T = (% Where, M E =¿ Mass of
3
the earth
change in
2
M s=¿ Mass of the sun
r =¿ Distance between
3
R ¿= × ( 2 ) %=3 %
the sun and the earth
2

Acceleration due to
389 (d) ∴ L∝ √ r

gravity at a height h 392 (c)

form the surface of 1
. If R becomes
the earth
ve ∝
√R
then v e will be 2
' 1 1
g =g
times
4
( )
2
h
1+
R
Given , h=2 R
The velocity with
393 (a)
1
which satellite is
'
∴ g =g
( 1+2 )2
g orbiting around the
earth is the orbital
'
¿g=
9
velocity (v o) and that
required to escape out
For orbiting the earth
390 (c)

of gravitational pull of
close to its surface
earth is the escape
velocity (v e ).
¿
m v 2 GMm
R
= 2 , ie, v 0=
R
GM
R√=√ g R
We know that
∴ v 0 =√ (9.8 ×6.4 ×10 )=8 km s −1
6

For escaping from

v e= √ 2 gR∧v o= √ gR

close to the surface of

∴ Increase∈ velocity required

earth,
v e −v o √ 2 gR−√ gR
¿ =
vo √ gR
¿ √ 2−1=0.414
Percent increase in
P a g e | 102
 velocity required r r
V −V 0=k log so , V =k log +V
¿ 0.414 × 100=41.4 % r0 r0

Escape velocity from

394 (d)
By the law of
396 (b)

the earth conservation of energy

−1
(v¿ ¿ e)=11.2 kms ¿ (U + K )surface =(U + K )∞
Let the mass, radius
and density of earth be
GMm 1 2 1 2
⟹− + m(3 v e ) =0+ m v
M , R∧ρ respectively
R 2 2

and for given planet

2
GM 9 v e 1 2
⟹− + = v
mass, radius and
R 2 2

density are M ' , R' ∧ρ ' ,

2 2GM
Since , v e =
respectively.
R

∴ Escape velocity from

2 2
ve 9 ve 1 2 2 2
∴− + = v ⟹ v =8 v e
the earth
2 2 2
v=2 √ 2 v e

√ ( )
4 ¿ 2 √ 2 ×11.2=31.7 kms
−1
2 G× π R3 ρ
3
v e=
Gravitational potential
R 397 (a)

v e=
√ 8 Gπ R2 ρ

Similarly, escape
3
…. ( i ) energy of mass m at
any point at a distance
r from the centre of
velocity from the given
earth is
planet
−GMm

√
8 Gπ R ' 2 ρ U=
At the surface of earth
v ' e= … . ( ii ) r

Dividing Eq. (i) by Eq.

3

(ii) we get
r =R
−GMm GM
(
√ √ √
ve 8 Gπ R ρ 3 2
R 2 ∴ Us= =−mgR ∵ g= 2
R R
At the height h=nR
= × = '2
'
ve 3 '2
8 Gπ R ρ R
11.2 R from the surface of
earth
¿ ' = '
ve R
11.2 R r =R +h=R +nR=R (1+n)
¿ ' =
v e 2R −GMm −mgR
∴ U h= =
R ( 1+n ) (1+ n)
Change in
−1
∴ v ' e =22.4 kms

gravitational potential
We know that intensity energy is
395 (a)

is negative gradient of −mgR

potential,
∆ U =U h−U s= −(−mgR )
(1+ n )
ie , m I =−(dV /dr ) and
as here I =−(k /r ), so
¿−
mgR
1+n
+mgR=mgR 1− (
1
1+n
=m )
v r
dV k dr
= , ie ,∫ dV =k ∫ 398 (a)
dr r r r

or
0 2
F ∝ xm× ( 1−x ) m=x m (1−x )
For maximum force
0

P a g e | 103
 ( )
dF 99 2h
=0 w=w 1−
dx 100 Re
dF 2 2
ie , h=0.005 R e
⇒ =m −2 x m =0
At point below the
dx
surface of earth at
⇒ x=1 /2
depth h . The weight of
399 (b)

body given by
1
v∝
√r
% increase in speed
¿ 1/2 (% decrease in
'
w =w 1−
( 2h
Re )
radius)
'
w
=0.995
¿ 1/2(1 %)=0.5 % w
i .e . speed will ( 1−0.995 ) w
increase by 0.5%
% ∆ w= ×100
w

At a certain velocity of
401 (c)
% ∆ w=0.5 % (decreases)
projection of the body
will go out of the
Gravitational potential
403 (d)

gravitational field of
energy is given as
earth and never to
return to earth. The
−GMm
U=
initial velocity is called
r
escape velocity
−GMm −GMm
U 1= , U 2=
r1 r2
v e= √ 2 gR
Where g is
As r 2 >r 1 ,hence ,

acceleration due to
gravity and R the
U 1−U 2=GMm
[ ]
r 2−r 1
r1 r2
is positie

radius. As is clear ie ,U 1> U 2

from above formula,
that escape velocity
¿ U 2 <U 1
ie , gravitational
dose not depends upon potential energy
mass of body hence, it increases.
will be same for a body
of 100kg as for 1kg
Acceleration due to
404 (a)
body.
gravity is given by

At an altitude h the
402 (a) GM
g=
acceleration due to
2

where G is
R

gravity is
gravitational constant.
g' =g 1− ( 2h
Re ) For earth: g e=
G Me
2
Re
'
¿ m g =mg 1−
( 2h
Re ) For planet : g p=
G Mp
Rp
2

ie , w' =w 1− ( 2h
Re ) Therefore ,
ge G M e /R e
=
g p G M p /R 2p
2

P a g e | 104
 2 12
ge M e Rp 60 ×1.6 ×10 60
¿ = × … (i) ∴ v min = 12
= =12 m/
g p M p R2e 8 ×10 5
Given , M p=2 M e , R p=2 Re
Putting the values in
410 (b)

the Eq. (i), we obtain ( )

2 2 1 /3
T R
h= 2
−R
2 4π
g e M e ( 2 R e) 1 4

[
= × 2
= × =2 ( 24 × 60× 60 )2 × ( 6.4 × 106 ) × 9.8
2
gp 2 Me Re 2 1 ¿ 2
4 × ( 22/7 )
¿ 3.6 ×10 m = 36000
ge
∴ g p= 7

km
2

406 (c)

g ∝r (if r < R ) and g ∝

Orbital speed,
1 411 (b)
2

(if r > R )
r
GM ; so speed of
v 0=
√
satellite decreases
r

Acceleration due to
407 (b)
with the increase in
gravity on earth is
the radius of its orbit.
given by
We need more than
one satellite for global
GM e
communication. For
g= 2
Re
Me stable orbit it must
pass through the
¿ g∝
centre of earth. So,
2
Re

only (b) is correct

( )
2
gp M p Rp
Hence , = × 1 1 1
…( i )
gp M p Rp 2 2 2

Acceleration due to
1 412 (a)
M p 2 ∧R p 1
gravity at height h is
1

Given , = = 1

Mp Rp 2
Substituting the given
g g
2 2
'
g= = =0.99 g
value in Eq. (i), we get
( ) ( )
2 2
h 32
1+ 1+
gp Re 6400
1 2 2 2
= ×1

gp 2 1
=
1 ()
F ( r )=¿where ρ is
2 413 (b)
∴ g p : g p =2 :1
density of sphere)
1 2

According to Kepler’s
408 (d)

law T 2 ∝ r 3
414 (c)

⟹ ( ) ( )
T1 2
r
= 1
3 v=
GM
R √=V ,

T2 r2
v '=
√
GM
(R+ R /2)

By conservation of
√ √
409 (a)
2 GM 2
¿ = V
angular momentum
3 R 3

mvr=¿ constant 415 (a)

v min ×r max =v max ×r min 1 2 1
KE= mv − m ¿
2 2

P a g e | 105
 1 4 x
2
x
2
¿ m¿ ⇒ = ∨4=
2 3 3 ( 1−x )2
( 1− x )2
Or 2= or 2−2 x=x
1 2 1 2
x
mv =3 × m× ( 11.2 )
2 2 1−x
3 x=2 or x= m
v=√ 3× 11.2 2
3
Since, velocity of
416 (d)

If m is the mass of
419 (c)

projection (v)is
racket, M that of earth
greater than the
and R is the radius of
escape velocity ( v e ) ,
earth, then
therefore at infinite gravitational potential
distance the body energy of racket near
moves with a velocity the surface of earth
v '= √ v 2−v 2e GMm
U 1=
Gravitational potential
'
∴ v = √¿ ¿ R

417 (c) energy of racket at a

heighth from earth’s
( ) ( )
3/ 2 3 /2
T 1 R1
surface
R
= = ⇒ T 2=8 T 1
T 2 R2 4R
−GMm
U 2=
Let a point mass C is
418 (a) ( R+h )
Increase in
placed at a distance of gravitational potential
x m from the point energy of racket
mass A as shown in
the figure
−GMm GMm
∆ U =U 2−U 1= +
R+ h R
GMmh
¿∆U=
( R+ h ) R
If v is the escape
velocity of racket, then
Here, = , Force
MA 4
MB 3 1 2
between A and C is
∆ U = mv
2

…(i)
1 2 GMmh
GM M A ⟹ mv =
F AC = 2
2 ( R+h ) R
Force between B and
x 2 2 2
⟹ mv R +mv Rh=2 GMm h
C is ⟹ v 2 R2=( 2 GM −v 2 R ) h

…(ii)
2 2
GM M B v R
F BC = ∴ h= 2
(1−x )2 2 GM−v R
According to given 420 (c)
problem F AC = F BC
1 2
GM =g R
3

∴
G MA M 1 GMBM
x2
=
3 ( 1−x )2 ( )
u=√ 2 gR= 2
√ GM
R 2
R=
√
2 GM
R

[Using (i) and (ii)]

Mass of the ball
421 (a)

always remain
MA MB MA x
2

constant. It does not

2
= 2
∨ =
x 3 ( 1− x ) M B 3 ( 1−x )2

P a g e | 106
 depend upon the
acceleration due to
gravity
1
⇒ m v 2min =
2
GMm
8a
( 45 ) ⇒ v min= 3
2 √
First we have to find a
422 (b)
When a sphere of mass
423 (c)

point where the m is released in a

resultant field due to liquid, it falls vertically
both is zero. Let the down with
point P be at a
distance x from centre acceleration ¿
m g−F B

of bigger star.
m
4 3 4 3
G(16 M ) GM π r d g− π r ρg
⇒ = ⇒ x=8 a 3 3 (d−ρ)g
x
2
( 10 a−x )2 =
(from O 1)
4 2 d
πr d
3

Orbital radius of
424 (b)

Jupiter > Orbital

radius of Earth
ie , once the body
= . As r J > r e
vJ re
reaches P , the ve rJ
gravitational pull of therefore v J < v e
attraction due to M
takes the lead to make 426
Acceleration due to
(a)
m move towards it
automatically as the gravity at a height
gravitational pull of above the earth
attraction due to 16 M surface
vanishes ie , a
minimum KE or
velocity has to be
imparted to m from
surface of 16 M such
that it is just able to
overcome the
gravitational pull of ( )
2
' R
g =g
16 M . By law of
R+ h

conservation of =(
R )
2
g R +h
energy.
'
g
(Total mechanical =(
R )
2
g R +nR
energy at A ) = (Total
'
g
mechanical energy at g
P)
2
'
=( 1+ n )
g
1
⇒ m v 2min +
2 [
G (16 M ) m GMm
2a
− 427
8a ] Just before striking,
(c)

the distance between

¿ 0+ [ GMm G(16 M )m
2a
−
8a ] the centre of earth and
moon is,

P a g e | 107
 Re 5 R e Te
2
r =R e + =
So, acceleration of
4 4 ¿¿
Rp
moon at this moment
1 /2
=( 27 )
is
Re
Rp 2
G Me =3
16 Re
a= 2
= × 10=6.4 m s−2
( 5 R e /4 ) 25 R p =9 R e

428 (a) 430 (a)

' 2 2 2

The latitude at point

As g =g−ω R cos λ ' gR
g=
on the surface of the
( R+h )2

earth is defined as the ( )

2
6400 −2
¿ 980 × =960 cm s
angle, which the line
6400+64

joining that point to

the centre of earth
431 (b)
Hence, g' =g−R ω2=0;
makes with equatorial
plane. It is denoted by
ω=√ g/ R= √ 10 /(6400 ×103 )=1/8

λ. For the poles λ=90 ° 432

and for equator λ=0 ° .
(c)

(i) Substituting λ=90 ° ( ) ( ) (

' 2 2
g R 1 R 1
= ⇒ = ⇒ =
in the above
g R +h 2 R+h 2 4
By solving we get
expression, we get h=1656.85 mile≈ 1600 mile
2 2
g pole=g−ω R cos 90 °
∴ g pole =g 433 (c)
ie , there is no effect of =− E0, and
GMm
KE=
rotational motion of
2r
the earth on the value
−GMm
PE= =2 E 0
of g at the poles.
r
(ii) Substituting λ=0 °
−GMm
⇒ TE=KE + PE= =E 0
in the above
2r
expression, we get
Since, earth from west
434 (d)
2 2

to east, so train Q has

gequator =g−ω R cos 0 °

effectively more
2
∴ gequator =g−ω R
ie ,the effect of rotation
angular velocity in
of the earth on the
comparison to train P
value of g at the
and hence,
equator is maximum.
experiences a greater
centrifugal force
From kepler’s third directed radially
429 (b)

law of planetary outwards. So, train Q

motion: will exert a lesser
force on track Q in
comparison to train P .
2 3
T ∝R

Hence, P exerts
Given , T p =27 T e '

greater force on track

2 3
T e R e
2
= 3
T p R p

P a g e | 108
 The earth is not a solid
435 (c) GM /7 4
g p= = g
sphere but is
R /4 7
2

Hence weight on the

somewhat flattened at planet
the poles and bulged
at equator, its
4
¿ 700 × =400 gm wt
equatorial radius is 21 440
7

km larger than its

(a)

polar radius, since,

vp
ve
=
√
M p Re
Me R p
1
√
× = 6 × = √ 3 ∴ v p=
2

When there is a
441 (c)

weightlessness in the
body at the equator,
then g' =r −R ω2=0
GM
g=
or ω=√ g/ R and linear
2

Hence, value of g is
R
velocity
least at equator and
maximum at poles.
¿ ω R=( √ g / R ) R=√ g R
Also, W =mg , therefore ∴ KE of rotation of
a person will get more earth ¿ I ω
1
quantity of matter in
2
2
kg-wt at equator. 1 2
¿ × M R ×ω
2 2
2 5

Binding energy = −¿
436 (a) 2 2 1
¿ M (ω R) = M g R
kinetic energy
5 5

And if this amount of

At depth d from the
442 (a)
energy (E k ) given to
surface of the earth.
satellite then it will
escape into outer
space
(
g' =g 1−
d
R )
Given, g =
'75 3
g= g
438 (c)
At equator, g' =g−R ω2 .
100 4

Then, ( )
When angular velocity
3g d
=g 1−
be
4 R
On solving, d=R /4
ω ( ¿ x ω ) , then,
'

0=g−R ω or
'2 443 (c)

ω =√ g /R=x ω
'

or x=( √ g /R)/ω
or
√10 /(6.4 × 106 ) ×24 × 60 ×60=17
Let masses of two
x=
2π
balls are m 1=m2=m
(given) and the density
439 (b)

We know that g= be ρ .
GM

Distance between
2

On the planet
R

P a g e | 109
 their centres¿ AB=2 R
Thus, the magnitude of
the gravitational force
v=√ 2 gR ⇒
vp
ve
g R
√
= p × p =√ 1× 4
ge Re

F that two balls

∴ v p =2 v e

separated by a
distance 2R exert on
448 (c)

each other is
dV =−Edx
or
(m) (m) x /√ 2 x /√ 2
F=G V =− ∫ Edx=− ∫ k x−3 dx=k / x
( 2 R )2
∞ ∞

( )
2
4 3
πR ρ
From the figure the
m
2
3 449 (d)
¿G =G
gravitational intensity
2 2
4R 4R

due to the ring at a

4
∴F∝R
distance d= √ 3 R on its
axis is
444 (c)
GM M
g= 2
∴g∝ 2

According to problem
R R

and R p =
Me Re
M p=
2 2

( )( ) ( )
2
g M p Re 1
∴ p= = × ( 2 )2=2
ge Me R p 2 GM GM × √ 3 R √3G
I= = =
2 3 /2 2 3/ 2
2 (d2+ R ) ( 3 R 2+ R ) 8R
Force on sphere
⇒ g p =2 g e =2× 9.8=19.6 m/s

445 (d)
¿(8 M )I = ( 8 M ) × √
3GM
d
(
g 1− =g' 1−
'
R
2h
R ) ( ) 8 R2

d = depth of mine ¿
√3 G M 2
h = height from
2
R

surface
According to Kepler’s
451 (c)

∴ g' 1−
d
R (
=g ' 1−
2h
R ) ( ) third law
2 3
⇒ d=2 h T ∝R

( )
3 /2
⇒ 10=2 h T 2 R2
⇒ h=5 km ⟹ =
T 1 R1

( )
446 (d) T 2 3 R 3 /2
∴ =
T1 R
W =0−
R
=(
−GMm GMm
R ) T2
⟹ =√ 27
2 m T1
¿ g R × =mgR
R ∴ T 2 =√ 27 T 1=√ 27 × 4=4 √ 27 h
3
¿ 1000 ×10 ×6400 × 10
9 452 (a)
¿ 64 × 10 J
( )
2
10 ' R g
¿ 6.4 × 10 J g =g =
( )
2
R+ h h
1+
447 (b) R

Gravitational force
453 (b)

P a g e | 110
 where, R−d=r =¿
( provides the
) distance of a place
GM m
¿
from the centre of
3 /2
R
necessary
centripetal force earth, therefore, g' ∝ r
( ie , m R ω 2 )
So,
459 (b)
F=G m1 m2 /r , thus on
2

increasing masses and

( )
2 2
GM m 2π 4 π mR
reducing distance r ,
2
3 /2
=mR ω =mR = 2
T
or
R T
force of gravitational
attraction F will
2 5/ 4
4π R
increase
2 2 5/2
T = ie , T ∝ R
GM

Time period does not

454 (d)
The gravitational
460 (d)

depends upon the intensity at a point

mass of satellite inside the spherical
shell is zero

Angular momentum is
455 (b)

conserved in central
461 (b)

field
2 2
T T 1
3
= 3 = 2 3 =¿

constant
R d n d

Resultant gravitational ∴ n1 d 1=n2 d 2 [where

456 (c)
2 3 2 3

intensity at a mid- n=¿ frequency]

point on the line
joining the two bodies
As in case of elliptic
462 (a)
is
orbit of a satellite
mechanical energy
G m2 G m1 4G
I= 2
– 2
= 2
(m2−m1 )
( r /2 ) ( r /2 ) r
E=−(GMm /2 a)
remains constant, at
−11
4 × 6.6 ×10
any position of
¿ 2
(1000−100)
1
satellite in the orbit,
−7 −1
¿ 2.4 × 10 Mk g
…
−GMm
KE + PE=
The value of g at
457 (b)
(i)
2a

latitude λ is ;
Now, if at position r , v
g =g−R ω cos λ . If is the orbital speed of
' '2 2

earth stops rotating, satellite

ω=0 ; g =g. It means
KE= m v and
'

the weight of body will

1 2

increase
2
…(ii)
−GMm
PE=
So, from Eqs. (i) and
r
The acceleration due
458 (d)

(ii), we have
to gravity at a depth d
inside the earth is
1 GMm −GMm
m v 2− = ,ie , v 2 =G
2 r 2a
g' =g 1− ( d
R ) (
=g
R−d
R
=g
r
R
463
) (a)

P a g e | 111
 v o √ g Re
Both the stars with
1
Thus , = =
same angular velocity
v e √ 2 g Re √ 2

ω around the centre of

∴ v e = √2 v o

mass (CM ) in their

ve 2
¿ v o= = =√ 2kms
−1

respective orbits as √2 √ 2
shown in figure
The magnitude of Here to point 7 of
467 (b)

gravitational force m 1 problem Solving skills

exerts on m 2 is h 1 g2 h1 g1 0.5 × g
[ = ∨h2= = =3.0]
G m1 m2 h 2 g1 g2 g /6
Energy spent =
|F|= 2
( r 1+ r 2 )
m g h e =m gm h m
or h m=ge h e /g m …(i)

¿ G
( 4
3
3 2 4
(2
π R e ρ/R e ¿ h e ¿¿ G π Rm ρm
3
Re ρ e 3 4
¿ × ×he = × ×0.5=3 m
R m ρm 2 1
464 (c)
469 (b)

⇒ when
I 1 ω 1=I 2 ω 2
( )
2
' R
g =g
( ) ( )
2
2 2 2π 2 R 2π R+ h
MR = M∙ 2
h=R then g =
5 T1 5 n T2 ' g

So the weight of the

T1 24 4
T 2= =
body at this height will
2 2
n n

465 (c) become one-fourth

470 (c)
1 H B gA
2
u
( )
H= ⇒H ∝ ⇒ = gm M m Re 1 16
2g g H A gB = ×
2
= ×( 4) =
Now g B= as g ∝ ρR
gA ge M e Rm 81 81
12 16
gm = g
H B gA 81 e
∴ = =12 ∴ v e = √ 2 ge R e= √ 2 ×9.8 × ( 6400 ×1
H A gB
−1
⇒ H B=12 × H A =12× 1.5=18 m ≈ 11.2 km s

466
The orbital velocity of
(c) v m=√ 2 g m R m= 2×
√ 16
81
1
ge × R e
4

satellite close to the

2 2 −1
¿ √ 2 g e Re = × 11≈ 2.5 km s
earth is
9 9

From Kepler’s third

v 0= √ g R e … . ( i ) 472 (a)
where Re is radius of
law of planetary
the earth. The escape
motion
velocity for a body
thrown from the
2 3
T ∝R
earth’s surface is Given , R p=2 Re
2 3
v e= √ g Re … . ( ii ) Te Re
∴ 2
= 3
Tp Rp

P a g e | 112
 T 2e R3e GM
⟹ = g= 2

At a height h above the

T 2p ( 2 R e )3 R

() earth’s surface, the

Te 1 3/2
⟹ =
acceleration due to
Tp 2
⟹ T p =2 √ 2 T e gravity is
Since ,T e =365 days=1 year , we have ' GM
g=
T p=2 √ 2× 365 days (R+ h)
2

T p=1032.37
( ) ( )
−2 2
g h h
∴ ' = 1+ = 1+
T p=1032 days . g R R

( ) ( )
' −2
g h 2h
473 (a) = 1+ = 1−
g R R
( )
2 2 3
2 GMm 2π GM 4π R
m ω R= ⇒ R= ⇒ M= ' g
R
2
T R
2
GT
2 but g = ( given)
2
474 (d) g /2 2h
∴ =1−
g R

√ [ ]
1 /3
r3 GM T 2 GM T 2
T =2 π ⇒ r 3= ⇒ r= 2h 1
=
GM 4 π2 4 π2 R 2
R
475 (c) h=
4
4 g1 ρ1 R1 1 4 2
g= GπRρ ⇒ = = × =
3 g2 ρ2 R 2 2 4831 1
Time period of satellite
(d)
476 (c)

T =2 π
√ r3
GM
⇒ T 2=
4 π2
GM
( R+ h )3 T∝
1
, where M is mass of earth

∝ ¿ where R is radius
M
1 /2

[ ] [ ] of the orbit, h is the

1 /3 1
GM T 2 GM T 2 3
⇒ R+ h= ⇒h= −R
height of satellite from
4π2 4 π2

477 (b) the earth’s surface.

=¿ constant
2
T
When gravitational
484 (c)
3
r
⇒ T r =¿ constant force becomes zero,
2 −3

then centripetal force

on satellite becomes
478 (c)

i . e . escape zero and therefore, the

v e=
√ 2 GM

velocity depends upon

R satellite will become
stationary in its orbit.
the mass and radius of
the planet 485 (c)

2 . Since M and r
GM
480 (d) g=

are constant, so
r
v e=
√ 2GM
(R+ h) g=9.8 m/s
2

Because value of g
486 (b)
The acceleration due
482 (b)

to gravity decreases with

increasing height
487 (c)

P a g e | 113
 completing 14 orbits a
day.
4
g= πρGR ⇒ g ∝ dR ¿
given in the problem)
3

492 (b)
v e= √ 2 v o , i . e . if the
At height
488 (d)
orbital velocity of
g' 2 h 90 moon is increased by
factor of √ 2 then it will
'
h , =1− =
or
g R 100
escape out from the
2h 90 10 1 gravitational field of
earth
=1− = =
or
R 100 100 10

R=20 h=20 ×320=6400 km494 (b)

At dept Here, m 1=m2=100 kg ;
r =100m
Acceleration of first
g' d 95
d, =1− =
or
g R 100
astronaut,
d 95 5 1 G m1 m2 1 G m1
=1− = = a 1= × = 2
or
R 100 100 20 2
m1
Acceleration of second
r r

R 6400
d= = =320 km astronaut,
20 20 G m1 m2 1 Gm2
a 2= 2
× = 2
m2
Net acceleration of
489 (b) r r

. If I =0 then approach
−dV
I=
V =¿ constant
dr
Gm2 G m1 2 G m1
a=a 1+ a2= 2
+ 2 = 2
r r r

Weight on surface of
490 (b) −11
2×(6.67 × 10 )× 100
earth, mg=500 N and
¿
( 100 )2
weight below the
−13 −2
¿ 2 ×6.67 ×10 ms
surface of earth at
As s= a t
1 2
R 2
d=

[ ]
1 /2
2
( )
1 /2
2s 2 × ( 1 /100 )
∴ t= =
( ) d −13
a 2 ×6.67 × 10
second
m g ' =mg 1−
R

¿ mg (1− )
1 On solving we get
2 t=1.41 days
mg
¿ =250 N 496 (c)
2

Landsats 1 through 3
491 (c)

operated in a near
polar orbit at an
altitude of 920 km
with an 18 day repeat
coverage cycle. These
satellites circled the Let the mass M be
earth every 103 min
P a g e | 114
 placed symmetrically GM
∞ ∞
∴ tan θ= 2
GM ( λdx) L gL
⇒ F net =∫ dF sin θ=¿ ∫ ¿
X + L √ X 2 + L2 −1 GM
( )
2 2
−∞ −∞
¿ θ=tan 2
∞
dx gL
⇒ F net =GMλL ∫ 3/ 2
−∞ ( X 2 + L2 )
GMλL
The escape velocity at
⇒ F net = (2) 501 (a)
2

the surface of earth is

L
2 GMλ
11.2 km s−1
⇒ F net = 2
L

Time period of satellite

497 (c) 502 (a)
2
GM ' GM GM (100 )
g= ;g = '2 =
R2
√
2 3
R ( 99 ) R 2 ( R +h )
% increase in
T =2 π
G Me
where
'
(g −g)×100
g=
g R+h=orbital radius of satellite,

( ) [( ) ]
M e =mass of earth .
Thus, time period does
2
g' 100
¿ −1 ×100= −1 × 100
not depend on mass of
g 99

([ 1+ 991 ) −1]× 100 ≈ 2 % satellite.

2

503 (c)

At equator, , If r =R then
498 (c) 1
v∝
g ’=g−R ω =0 or
2
√r
v=V 0
ω=√ g/ R If r =R +h=R +3 R=4 R
or
then v= =0.5 V 0
V0
ω=√ 10/(6.4 ×106 )=1.25 ×10−3 rad s−1 2

The metallic spheres

499 (c)
If a body is projected
504 (b)

will be at positions as from the surface of

shown in the figure. earth with a velocity v
and reaches a height h
, applying
conservation of energy
(relative to surface of
earth)
1 2 mgh
mv =
2 [1+(h/ R)]
h=R=6400 km,
−2

So,
GM × M g=10 ms
T sin θ=F= 2
L
GM
2 v 2=g hie , v =√ 10 × 6400 ×103=8 k
¿ 2
L
Gravitational potential
505 (a)
T cos θ=Mg
at mid point

P a g e | 115
 −G M 1 −G M 2 KE=¿ PE∨¿
V= +
Now, √
d /2 d /2 1 2 2 Gm
⇒ mv = ( M 1 + M 2 ) ⇒ v=2
2 d
−2 Gm
PE=m×V = (M 1+ M 2)
507 (d)
¿ mass of particle)
d
and I = 2
−GM GM
So, for projecting
V=
r
V =0 and I =0 at r =∞
r
particle from mid point
to infinity

P a g e | 116