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Unit-8: Gravitation
1. Four particles of masses m, 2m, 3m and 4m are kept in 9. The depth at which the effective value of acceleration
sequence at the corners of a square of side a. The g
due to gravity is is (R = radius of the earth)
magnitude of gravitational force acting on a particle of 4
mass m placed at the centre of the square will be 3R R R
(a)R (b) (c) (d)
24 m 2G 6m 2G 4 2 Gm 2 4 2 4
(a) (b) (c) (d) Zero
a2 a2 a2
10. Assuming earth to be a sphere of a uniform density,
2. Acceleration due to gravity on moon is 1/6 of the what is the value of gravitational acceleration in a mine
acceleration due to gravity on earth. If the ratio of 100 km below the earth's surface (Given R = 6400km)
  5 (a) 9.66 m / s 2 (b) 7.64 m / s 2
densities of earth (m ) and moon (e ) is  e  =
 m  3 (c) 5.06 m / s 2 (d) 3.10 m / s 2
then radius of moon Rm in terms of Re will be
11. The angular velocity of the earth with which it has to
5 1 3 1 rotate so that acceleration due to gravity on 60° latitude
(a) Re (b) Re (c) Re (d) Re
18 6 18 2 3 becomes zero is (Radius of earth = 6400 km. At the poles
g = 10 ms–2)
3. A spherical planet far out in space has a mass M 0 and (a) 2.5×10–3rad/sec (b) 5.0×10–1rad/sec
diameter D0 . A particle of mass m falling freely near the (c) 10  10 rad / sec
1
(d) 7.8  10 −2 rad / sec
surface of this planet will experience an acceleration
due to gravity which is equal to 12. If the angular speed of earth is increased so much that
the objects start flying from the equator, then the length
(a) GM 0 / D02 (b) 4 mGM 0 / D02
of the day will be nearly
(c) 4 GM 0 / D02 (d) GmM 0 / D02 (a) 1.5 hours (b) 8 hours
(c) 18 hours (d) 24 hours
4. The moon's radius is 1/4 that of the earth and its mass
is 1/80 times that of the earth. If g represents the
13. The gravitational potential in a region is given by
acceleration due to gravity on the surface of the earth,
V = (3 x + 4 y + 12 z) J /kg. The modulus of the
that on the surface of the moon is
gravitational field at (x = 1, y = 0, z = 3) is
g g g g
(a) (b) (c) (d)
4 5 6 8 (a) 20 N kg −1 (b) 13 N kg −1
(c) 12 N kg −1 (d) 5 N kg −1
5. The acceleration of a body due to the attraction of the
earth (radius R) at a distance 2R from the surface of the
14. Infinite bodies, each of mass 3kg are situated at
earth is (g = acceleration due to gravity at the surface of
distances 1m, 2m, 4m,8m....... respectively on x-axis.
the earth)
The resultant intensity of gravitational field at the origin
g g g will be
(a) (b) (c) (d) g
9 3 4 (a)G (b) 2G (c) 3G (d) 4G

6. The height of the point vertically above the earth's 15. The gravitational field due to a mass distribution is
surface, at which acceleration due to gravity becomes
E = K / x 3 in the x - direction (K is a constant). Taking
1% of its value at the surface is (Radius of the earth = R)
the gravitational potential to be zero at infinity, its value
(a) 8R (b) 9R (c) 10 R (d) 20R
at a distance x is
(a) K / x (b) K / 2 x
7. At surface of earth weight of a person is 72 N then his
weight at height R/2 from surface of earth is (R = radius (c) K / x 2 (d) K / 2 x 2
of earth)
(a) 28N (b) 16N (c) 32N (d) 72N 16. The gravitational potential due to the earth at infinite
distance from it is zero. Let the gravitational potential at
8. Weight of a body of mass m decreases by 1% when it is a point P be −5 J / kg . Suppose, we arbitrarily assume
raised to height h above the earth's surface. If the body the gravitational potential at infinity to be + 10 J / kg ,
is taken to a depth h in a mine, change in its weight is then the gravitational potential at P will be
(a) 2% decrease (b) 0.5% decrease (a) −5 J / kg (b) +5 J / kg
(c) 1% increase (d) 0.5% increase (c) −15 J / kg (d) +15 J / kg

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17. An infinite number of point masses each equal to m are
placed at x =1. x = 2, x = 4, x = 8 ……… What is the total 25. A satellite is launched into a circular orbit of radius ‘R’
gravitational potential at x = 0 around earth while a second satellite is launched into an
(a) − Gm (b) −2Gm orbit of radius 1.02 R. The percentage difference in the
(c) − 4 Gm (d) − 8 Gm time periods of the two satellites is
(a) 0.7 (b) 1.0 (c) 1.5 (d) 3
18. For a satellite escape velocity is 11 km / s . If the satellite
26. Earth is revolving around the sun if the distance of the
is launched at an angle of 60 o with the vertical, then earth from the sun is reduced to 1/4th of the present
escape velocity will be distance then the present length of the day is reduced
(a) 11 km / s (b) 11 3 km / s by
(c) 11 km / s (b) 33 km / s (a)¼ (b)½ (c) 1/8 (d) 1/6
3
27. Find the height above the surface of the earth where
19. If the radius of earth reduces by 4% and density remains weight becomes half.
same then escape velocity will (a) R (b) (2 - 1)R (c) R (d) R
(a) Reduce by 2% (b) Increase by 2% 2 ( 2 + 1) 2
(c) Reduce by 4% (d) Increase by 4%
28. A pendulum clock which keeps correct time at the
20. A rocket of mass M is launched vertically from the surface of the earth is taken into a min then
surface of the earth with an initial speed V. Assuming (a) It keeps correct time (b) It gains time
the radius of the earth to be R and negligible air (c) It loses time (d) None of these
resistance, the maximum height attained by the rocket
above the surface of the earth is 29. Find the velocity of the earth at which it should rotate
R  gR  so that weight of a body becomes zero at the equator.
(a) (b) R − 1
 gR   2V 2
 (a) 1.25 rad s-1 (b) 1.25 × 10-1 rad s-1
 − 1 -2
(c) 1.25 × 10 rad s -1
(d) 1.25 × 10-3 rad s-1
 2V 2 
R  2 gR 
(c) (d) R  2 − 1  30. A satellite of moon revolves around it in a radius n times
 2 gR   V  the radius of moon (R). Due to cosmic dust it
 2 − 1
 V  experiences a resistance F = v2. Find how long it will
stay in the orbit.
21. A body of mass m is situated at a distance 4 R e above (a) m
n (b) m

R
am
( n − 1)
the earth’s surface, where Re is the radius of earth.  GM/R
How much minimum energy be given to the body so that
(c) m ( n − 1) (d)
m vi
it may escape  v  v 2f
(a) mgR e (b) 2mgR e
mgR e 31. Let E1 and E2 denotes gravitational field at distance ‘r1’
(c) mgR e (d)
5 16 and ‘r2’ from axis of infinitely long solid cylinder of radius
‘R’. Which of the following must hold true –
22. The distance of a planet from the sun is 5 times the (a) E1< E2 if r1< r2< R
distance between the earth and the sun. The Time (b) E1> E2 if R < r1< r2
period of the planet is (c) E1> E2 if r1 = R – E, r2 = R + E(E is positive constant < R)
(a) 5 3 / 2 years (b) 5 2 / 3 years (d) All of the above
(c) 5 1 / 3 years (d) 5 1 / 2 years
32. A particle of mass m is projected upwards with velocity
ve
23. A satellite is moving around the earth with speed v in a v= , where ve is the escape velocity then at the
2
circular orbit of radius r. If the orbit radius is decreased
maximum height the potential energy of the particle is
by 1%, its speed will
: (R is radius of earth and M is mass of earth)
(a) Increase by 1% (b) Increase by 0.5%
(c) Decrease by 1% (d) Decrease by 0.5%
−GMm −GMm
(a) (b)
2R 4R
24. If the gravitational force between two objects were −3GMm −2GMm
(c) (d)
proportional to 1/R; where R is separation between 4R 3R
them, then a particle in circular orbit under such a force
would have its orbital speed v proportional to 33. The escape velocity from the earth is about 11 km/s. The
(a) 1 / R 2 (b) R 0 (c) R 1 (b) 1 / R escape velocity from a planet having twice the radius
and the same mean density as the earth is -

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(a) 22 km/s (b) 11 km/s (c) 5.5 km/s (d) 15.5 41. The correct graph representing the variation of total
km/s energy(E), kinetic energy(K) & potential energy (u) of a
satellite with its distance from the centre of earth is -
Energy
34. The distance of Neptune and Saturn from the Sun are Energy
E
nearly 1013 m and 1012 m respectively. Their periodic E
K
times will be in the ratio - (a) U (b)
r r
(a) 10 (b) 100 (c) 10 10 (d) 1000
K U

35. Two spherical stars each of mass M with their centres Energy Energy
distant D apart revolve under mutual gravitational
attraction about the point midway between their K K

centres. The period of revolution will be – (c) r (d) r

3 3 U
D D E
(a)  (b) 2 U
E
GM 2GM
42. A particle of mass m is projected from the surface of
2D 3 GM earth with a speed v0 (v0< escape velocity). The speed of
(c) 2 (d) 
GM D3 particle at height h = R (radius of earth) is : (Here, R =
6400 km and g = 9.8 m/s2)
36. A planet of mass m moves around the sun of mass M in (a) gR (b) v02 − 2gR
an elliptical orbit. The maximum and minimum distance
of the planet from the sun are r1 and r2 respectively. The (c) v02 − gR (d) None of these
time period of the planet is proportional to -
(a) r13 / 2 (b) r23 / 2 (c) (r1 + r2)3/2 (d) (r1 – r2)3/2 43. A satellite revolving around a planet has a time period T
and orbital radius R, the mass of this planet is -
37. Three particles each of mass M are situated at the 4 2 R 3 42 R 3T R 3T 4 2 R 3
vertices of an equilateral triangle of side length a. The (a) (b) (c) (d)
G G G GT 2
only forces acting on the particles are their mutual
gravitational forces. It is desired that each particle 44. Two particles of equal mass m go round a circle of radius
moves in a circle while maintaining the original mutual R under the action of their mutual gravitational attraction.
separation a. What is the net gravitational force on one The speed of each particle is -
of the particles ?
1 1 Gm
GM 2 GM 2 GM 2 2GM 2 (a) V= (b) V =
(a) 2 (b) 2 2 (c) 3 2 (d) 2R Gm 2R
a a a a2
1 Gm 4Gm
(c) V = (d) V =
38. The masses and radii of Earth and Moon are M 1,R1 and 2 R R
M2, R2 respectively. Their centre are at a distance d
apart. The minimum speed with which a particle of mass 45. A body of mass m is placed on the earth's surface. It is
m should be projected from a point mid-way between taken from the earth's surface to a height h = 3R. The
the two centres so as to escape to infinity is : change in gravitational potential energy of the body is -
2G(M1 + M 2 ) G(M1 + M 2 ) 2 3 mgR mgR
(a) (b) (a) mgR (b) mgR (c) (d)
d 2d 3 4 2 4

G(M1 − M 2 ) G(M1 + M 2 ) 46. Two identical solid spheres of radius R placed in contact
(c) 2 (d) 2
2d d with each other, the gravitational attraction between
them is proportional to -
39. A Geostationary satellite is revolving around the earth. (a) R2 (b) R–2 (c) R4 (d) R–4
To make it escape from gravitational field of earth, its
velocity must be increased - 47. The depth d at which the value of acceleration due to
(a) 100 % (b) 41.4% (c) 50% (d) 59.6% 1
gravity becomes times the value at the surface, is [R
n
40. A particle is projected with velocity kve in vertically = Radius of earth] -
upward direction from the ground into the space (v eis R  n – 1 R  n 
escape velocity & k < 1) then the maximum height from (a) (b) R   (c) 2 (d) R  
n  n  n  n +1 
the centre of earth to which it can go, will be -
R R R R 48. A mass m is raised from a distance 2R from surface of
(a) 2 (b) 2 (c) (d)
k +1 k –1 1– k 2
k +1 earth to 3R. Work done to do so against gravity will -

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mgR mgR mgR mgR on point mass ′𝑚′ at 𝑥 = 0, if the mass per
(a) (b) (c) (d) unit length of the rod is 𝐴 + 𝐵𝑥 2 , is given by:
10 11 12 14 1 1
(a) Gm [A ( − ) − BL]
𝑎+𝐿 𝑎
49. There are two bodies of masses 100 kg & 10,000 kg (b) Gm [A ( −
1 1
) − BL]
separated by a distance 1m. At what distance from the 𝑎 𝑎+𝐿
1 1
smaller body, the intensity of gravitational field will be (c) Gm [A ( − ) + BL]
𝑎+𝐿 𝑎
zero ? 1 1
(d) Gm [A ( − ) + BL]
𝑎 𝑎+𝐿
1 1 1 10
(a) m (b) m (c) m (d) m
9 10 11 11 56. Aspaceship orbits around a planetat a height
of 20 km from its surface. Assuming that only
50. If the change in the value of g at a height h above the gravitational field of the planet acts on the
surface of the earth is the same as at a depth x below it spaceship, what will be the number of
when both x and h are much smaller than the radius of complete revolutions made by the spaceship
the earth, then - in 24 hours around the planet? [Given: Mass
h h of Planet = 8 × 1022 kg, Radius of planet =
(a)x = h (b) x = 2h (c) x = (d) x = 2 × 106 m, Gravitational constant G = 6.67 ×
2 3
10−11 Nm2 /kg 2 ]
51. Two bodies of masses m and M are placed a distance d (a) 9 (b) 17
apart. The gravitational potential at the position where (c) 13 (d) 11
the gravitational field due to them is zero is V, then
57. The height ′ℎ′ at which the weight of a body
G Gm
(a) V = − (m + M ) (b) V = − will be the same as that at the same depth
d d ′ℎ′ from the surface of the earth is (Radius
GM of the earth is 𝑅 and effect of the rotation of
(c) V = − (d) V = − G ( m + M )2
d d the earth is neglected)
√5R−R √3R−R
(a) (b)
52. A boy can jump to a height h on ground level. What 2 2
𝑅 √5
should be the radius of a sphere of density d such that (c) (d) R−R
2 2
on jumping on it, he escapes out of the gravitational
field of the sphere 58. The mass density of a planet of radius 𝑅
1/ 2
(a)  4 Gd  (b)  4 gh 
1/ 2
varies with the distance 𝑟 from its centre as
 3 Gd  𝐫2
 3 gh    𝜌(r) = 𝜌0 (1 − 2 ) Then the gravitational field
1/ 2 𝐑
(d)  3 Gd  is maximum at
1/ 2
(c)  3 gh 
 4 Gd  4 gh
    1 3
(a) r = R (b) r = √ R
√3 4
53. The energy required to take a satellite to a 5
(c) r = √ R (d)𝑟 = 𝑅
height ′ℎ′ above Earth surface (radius of 9
Eareth = 6.4 × 103 km ) is E1 and kinetic
energy required for the satellite to be in a 59. The value of the acceleration due to gravity
circular orbit at this height is E2 . The value R
is g1 at a height h = (R = radius of the
of h for which E1 and E2 are equal, is: 2
earth) from the surface of the earth. It is
a) 1.6 × 103 km (b) 3.2 × 103 km
again equal to g1 at a depth 𝑑 below the
(c) 6.4 × 10 km
3
(d) 28 × 104 km 𝑑
surface of the earth. The ratio ( ) equals
𝑅
54. A satellite is revolving in a circular orbit at a (a)
5
(b)
1
9 3
height h from the earth surface, such that 7 4
h << R where R is the radius of the earth. (c) (d)
9 9
Assuming that the effect of earth's
atmosphere can be neglected the minimum 60. The acceleration due to gravity on the
increase in the speed required so that the earth's surface at the poles is 𝑔 and angular
satellite could escape from the gravitational velocity of the earth about the axis passing
field of earth is: through the pole is 𝜔. An object is weighed
(a) √2𝑔𝑅 (b) √𝑔𝑅 at the equator and at a height h above the
poles by using a spring balance. If the
𝑔𝑅
(c) √ (d) √𝑔𝑅(√2 − 1) weights are found to be same, then ℎ is (h <
2
< R where𝑅 is the radius of the earth)
𝑅 2 𝜔2 𝑅 2 𝜔2
55. A straight rod of length 𝐿 extends from 𝑥 = 𝑎 (a) (b)
2𝑔 𝑔
to 𝑥 = 𝐿 + 𝑎. The gravitational force it exerts 𝑅 2 𝜔2 𝑅 2 𝜔2
(c) (d)
8𝑔 4𝑔

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61. Two planets have masses 𝑀 and 16𝑀 and

their radii are 𝑎 and 2a, respectively. The
separation between the centres of the
planets is 10 a. 𝐴 body of mass 𝑚 is fired (d)
from the surface of the larger planet
towards the smaller planet along the line 65. Infinite number of bodies, each of mass 2 kg are
joining their centres. For the body to be able situated on x ‐axis at distances 1 m, 2m, 4m, 8
to reach the surface of smaller planet, the m , respectively, from the origin. The resulting
minimum firing speed needed is
gravitational potential due to this system at the
3 5GM GM2
(a) √ (b) √ origin will be
2 a ma
(a) − 4 G (b) −4G
GM GM
(c) 2√ (d) 4√ 3
(c) − G
a a
(d) − 8 G
3
62. The kinetic energies of a planet in an elliptical
orbit about the Sun, at positions positions A, B 66. Which one of the following plots represents the
and C are K A , K B and KC , respectively. AC variation of gravitational field on a particle with
is the major axis and SB is perpendicular to distance r due to a thin spherical shell of radius
AC at the position of the Sun S as shown in R ?(r is measured from the centre of the spherical
the figure? Then shell)

(a) K A  K B  KC (b) K A  K B  KC (a)

(c) K B  K A  KC (d) K B  K A  KC

63. A black hole is an object whose gravitational

field is so strong that even light cannot escape (b)
from it. To what approximate radius would
earth ( mass = 5.98 10 kg ) have to be
24

compressed to be a black hole?

−9 −6 (c)
(a) 10 m (b) 10 m
−2
(c) 10 m (d) 100 m
64. Dependence of intensity of gravitational field (E)
(d)
of earth with distance (r) from centre of earth is
correctly represented by
67. The figure shows elliptical orbit of a planet m
about the sun S . The shaded area SCD is twice
the shaded area SAB . If
t1 is the time for the
planet to move from C to D and t 2 is the time to
(a)
move from A to B then

(b)

(a) t1 = 4t2 (b) t1 = 2t2

(c) t1 = t2 (d) t1  t2
(c)
68. The largest and the shortest distance of the earth
from the sun are r1 and r2 . Its distance from the

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sun when it is at perpendicular to the major‐axis (c) − Gm 3m2 + 4 2M 
of the orbit drawn from the sun is d  
r1 + r2 r1 + r2 (d) − Gm 6m2 + 4 2M 
(a) (b) d  
4 r1 − r2
2r r r + r2 72. If the radius of earth shrinks by 2% while its
(c) 1 2 (d) 1 mass remains same. The acceleration due to
r1 + r2 3
gravity on the earth's surface will
approximately :
69. Given below are two statements : One is (a) decrease by 2% (b) decrease by 4%
labeled as Assertion A and the other is (c) increase by 2% (d) increase by 4%
labelled as Reason R.
Assertion A : If we move from poles to 73. If the gravitational field in the space is given as
equator, the direction of acceleration due to
 K
gravity of earth always points towards the  − r 2  . Taking the reference point to be at r =
center of earth without any variation in its  
magnitude. 2 cm with gravitational potential V = 10 J/kg.
Reason R : At equator, the direction of Find the gravitational potential at r = 3 cm in
acceleration due to the gravity is towards the SI unit (Given, that K = 6 J cm/kg)
center of earth. (a)9 (b)11
In the light of above statements, choose the (c) 12 (d) 10
correct answer from the options given below
: 74. A simple pendulum doing small oscillations at a place
(a) Both A and R are true and R is the correct 𝐑 height above earth surface has time period of 𝑻𝟏 =
explanation of A. 𝟒 𝐬. 𝑻𝟐 would be it's time period if it is brought to a
(b) Both A and R are true but R is NOT the point which is at a height 𝟐𝐑 from earth surface.
correct explanation of A. Choose the correct relation [𝑹 = radius of earth ] :
(c) A is true but R is false (a) 2 T1 = T2 (b) 2 T1 = 3 T2
(d) A is false but R is true (c) T1 = T2 (d) 3 T1 = 2 T2
70. Given below are two statements : 75. Match List I with List II :
Statement I : The law of gravitation holds
List I List II
good for any pair of bodies in the universe.
(A) Kinetic energy of planet (I) −𝐆𝐌𝐦/𝐚
Statement II : The weight of any person
(B) Gravitation Potential energy (II) 𝐆𝐌𝐦/𝟐𝐚
becomes zero when the person is at the centre
of sun-planet system
of the earth. 𝐆𝐦
In the light of the above statements, choose the (C) Total mechanical energy of planet (III)
𝐫
correct answer from the options given below. (D) Escape energy at the surface (IV) −𝐆𝐌𝐦/𝟐𝐚
(a) Both statement I and Statement II are true of planet for unit mass object
(b) Both statement I and Statement II are false (Where 𝐚 = radius of planet orbit, 𝐫 = radius of
(c) Statement I is true but Statement II are planet, 𝐌 = mass of Sun, 𝐦 = mass of planet) Choose
false the correct answer from the options given below :
(d) Statement I is false but Statement II is true (a) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(b) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
71. Four spheres each of mass m form a square of (c) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
side d (as shown in figure). A fifth sphere of (d) (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
mass M is situated at the centre of square. The
total gravitational potential energy of the 76. An astronaut takes a ball of mass 𝒎 from earth to
system is :
space. He throws the ball into a circular orbit about
earth at an altitude of 𝟑𝟏𝟖. 𝟓 𝐤𝐦. From earth's
surface to the orbit, the change in total mechanical
𝐆𝐌 𝐦
energy of the ball is 𝒙 𝐦 . The value of 𝒙 is (take
𝟐𝟏𝐑𝐞
𝐑𝐞 = 𝟔𝟑𝟕𝟎 𝐤𝐦 ):
(a) 10 (b) 12
(c) 9 (d) 11

d 
( )
(a) − Gm  4 + 2 m + 4 2M 
 77. A satellite of 𝟏𝟎𝟑 𝐤𝐠 mass is revolving in circular orbit
𝟏𝟎𝟒 𝑹
(b) − Gm
( )
 4 + 2 m + 4 2m 
d  
of radius 𝟐𝑹. If
𝟔
𝐉 energy is supplied to the
satellite, it would revolve in a new circular orbit of
radius (use 𝒈 = 𝟏𝟎 𝐦/𝐬 𝟐 , 𝑹 = radius of earth)

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(a) 2.5𝑅 (b) 3𝑅
(c) 4𝑅 (d) 6𝑅

78. A small point of mass 𝒎 is placed at a distance 𝟐𝑹

from the centre ' 𝑶 ' of a big uniform solid sphere of
mass M and radius R. The gravitational force on ' m
' due to M is 𝐅𝟏 . A spherical part of radius 𝐑/𝟑 is
removed from the big sphere as shown in the figure
and the gravitational force on 𝒎 due to remaining
part of 𝑴 is found to be 𝑭𝟐 .

The value of ratio F1: F2 is

(a) 12: 11 (b) 11: 10
(c) 12: 9 (d) 16: 9

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1. (c) g  R   R 
2
1
2

If two particles of mass m are placed x distance apart =  =  =

g  R + h   R + 2R  9
Gmm
then force of attraction = F (Let) g
x2  g = .
Now according to problem particle of mass m is placed 9
at the centre (P) of square. Then it will experience four
forces 6. (b)
D 4m 3m C Acceleration due to gravity at height h is given by
FPD 2
 R 
FPC g = g 
P m  R +h
FPA
2
FPB g  R 
A m 2m B  = g 
100  R + h
Gmm
FPA = force at point P due to particle A = =F R 1
x2  =  h = 9R .
R + h 10
G 2mm G3mm
Similarly FPB = = 2 F , FPC = = 3 F and
x2 x2 2
 R 
G 4 mm 7. (c) Weight of the body at height R, W = W 
FPD = = 4F  R + h
x2 2
Hence the net force on P  
  2
F net = F PA + F PB + F PC + F PD = 2 2 F  =W
R  = W  2  = 4 W = 4  72 = 32 N .
 R  3 9 9
Gmm Gm 2 a R+ 
F net = 2 2 =2 2 [x = =  2 
2 2
x (a / 2 ) 2
half of 8. (b)
the diagonal of the square] Percentage change in g when the body is raised to
4 2 Gm 2 g 2h  100
= . height h ,  100 % = = 1%
a2 g R
Percentage change in g when the body is taken into
2. (a) g d h
depth d,  100 % =  100 % =  100 % [As
4 g R R
Acceleration due to gravity g = GR  g   R
3 d = h]
or  Percentage decrease in weight
gm m Rm gm  1  2h  1
 100  = (1% ) = 0.5% .
1 5
= . [As = and e = (given)] = 
ge e Re ge 6 m 3 2 R  2
Rm  g m   e  1 5 5
 =    =   Rm = Re 9. (b)
Re  ge   m  6 3 18
 d g  d 3R
g = g  1 −   = g 1 −   d =
3. (c)  R  4  R  4
GM GM 4 GM
g= 2
= 2
= 10. (a)
We know R (D / 2) D2
 d
If mass of the planet = M 0 and diameter of the planet Acceleration due to gravity at depth d, g  = g 1 − 
 R
4 GM 0  100   1 
= D0 . Then g = . = g 1 −
D02  = 9 .8 1 − 64 
 6400   
63
4. (b) = 9.8  = 9.66 m / s 2 .
64
GM
Acceleration due to gravity g = 
R2 11. (a)
gmoon M 2
Rearth  1  4 
2 Effective acceleration due to gravity due to rotation of
= moon . =    earth g = g −  2 R cos 2 
 80  1 
2
gearth Mearth Rmoon
 2R
gmoon = gearth 
16 g
= .  0 = g −  2 R cos 2 60 o  =g
80 5 4
4g g 2 rad
= =2 =
5. (a) R R 800 sec
[As g  = 0 and  = 60 o ]

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1 rad  
 = = 2.5  10 −3 .  1 
1 1 1 1 
400 sec = −Gm  + + +  = −Gm   = −2Gm
1 2 4 8  1 − 1 
 
 2
12. (a)
Time period for the given condition m m m m
O
R 1m
T = 2 = 1.40 hr  1.5 hr nearly.
g 2m
4m
13. (b) 8m
 V ˆ V ˆ V  18. (a)
I = − i+ j+ kˆ 
 x y z
(
 = − 3ˆi + 4 ˆj + 12 kˆ [As ) 19. (c)
V = (3 x + 4 y + 12 z ) (given)]
Escape velocity v e  R  and if density remains
It is uniform field Hence its value is same every where
constant v e  R
| I | = 3 2 + 4 2 + 12 2 = 13 Nkg −1 . So if the radius reduces by 4% then escape velocity also
reduces by 4%.
14. (d)
Intensity at the origin I = I1 + I 2 + I3 + I4 + ....... 20. (c)
3kg 3kg 3kg 3kg Kinetic energy given to rocket at the surface of earth =
Change in potential energy of the rocket in reaching
O
1m from ground to highest point
1 mgh v2 g
2m  mv 2 =  = 
2 1 + h/ R 2 1 1
4m +
8m h R
1 1 2g 1 2g 1 1 2 gR − v 2
=
GM GM GM GM
+ 2 + 2 + 2 + .......... + = 2  = 2 −  = 
r12 r2 r3 r4 h R v h v R h v2R
v2R
1 1 1 1  h=
= GM  2 + 2 + 2 + 2 + ....... 2 gR − v 2
1 2 4 8 
R
 1 1 1   h=
= GM 1 + + + + .........  2 gR 
 4 16 64   2 − 1
 v 
 
  21. (c)
1
= GM   Potential energy of the body at a distance 4𝑅𝑒 from the
 1  surface of earth
1 − 
 4  𝑈=−
𝑚𝑔𝑅𝑒
=−
𝑚𝑔𝑅𝑒
=−
𝑚𝑔𝑅𝑒
1+ℎ/𝑅𝑒 1+4 5
a
[As sum of G.P. = ] [As ℎ = 4𝑅𝑒 (given)]
1−r So minimum energy required to escape the body will
4 4 𝑚𝑔𝑅𝑒
= GM  = G  3  = 4G [As M = 3kg be
5
.
3 3
given]
22. (a)
According to Kepler’s law 𝑇 ∝ 𝑅3/2
15. (d)
∴ 𝑇𝑝𝑙𝑎𝑛𝑒𝑡 = (5)3/2 𝑇𝑒𝑎𝑟𝑡ℎ = 5(3/2) × 1𝑦𝑒𝑎𝑟 =
 x
K K
V = − E dx = − dx = . 53/2 𝑦𝑒𝑎𝑟𝑠.
3
2x 2
23. (b)
16. (b)
Gm 1
Potential increases by +10 J / kg every where so it will Orbital velocity v =  v
r r
be +10 − 5 = +5J / kg at P
[If r decreases then v increases]
Percentage change in
17. (b)
1 1
Net potential at origin v = (Percentage change in r) = (1%) = 0.5% 
 Gm Gm Gm  2 2
V = − + + + ......... orbital velocity increases by 0.5%.
 r1 r2 r3 
24. (b)

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1 1
If F  n then v  ; here n = 1
R R n −1
1
v   R0 .
1−1
R E

25. (d)
R–E R R+E
Orbital radius of second satellite is 2% more than first
r
satellite
32. (c)
So from T  (r) 3 / 2 , Percentage increase in time period When only conservative forces are acting, mechanical
3 energy is conserved and at maximum height speed is
= (Percentage increase in orbital radius)
2 zero.
3 − GMm 1  v e 
2
= (2%) = 3%. + m  = U + 0
2 R 2  2 
26. (c) − GMm 1  GM 
+ m =U
T1  r / 4 
3 /2
1 R 2  2R 
T2  r3 =  = −GMm GMm −3GMm
T2  r  8 +
U= R 4R = 4R
27. (b)
1
2 
=
1
2
or h = ( 2 − 1)R 33. (a)
2GM 4
h v= = 2G R 2 d
1 +  R 3
 R
vR d
28. (c) v = 2v0
l
T = 2 as g decreases, T increases,  it loses time. 34. (c)
g
Tn2 R 3n
=
29. (d) Ts2 R 3s
 R2  g 3/ 2
g' = g 1 − Tn 10 
13
= 0 or  =
 g  R =  12 
Ts 10 
=
10
=
1
6400  103 800
= 1.25  10−3 rads −1 = ( 10 ) = 10
3
10

30. (b) 35. (b)

If the velocity is v0 so
dv dv dt
m = v 2dt or 2 = D
dt v m
M C M
GM
and v = mv 02 GMM GM
r = 2
 v0 =
(D / 2) D 2D
GM GM
We know vi = and v f = 2R 2(D / 2) D3
nr R so T = = = 2
vf dV dt v GM / 2D 2GM
 
vi V2
= m

 n − 1
36. (c)
m1 1 m
or t=  − = According to Kepler's law T2  a3
  v i v f   GM/R
r +r 
Here a = semi-major axis =  1 2 
 2 
31. (d)
3
Graph of gravitational field versus distance from axis for r +r 
 T2   1 2 
a solid cylinder  2 
3/ 2
r +r 
T  1 2  (r1 + r2 ) 3 / 2
 2 

37. (c)

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M v
m O m
GM 2 60º GM 2 R
a2 a2 v
the gravitational force provide the necessary centripetal
M M force for circular motion. so
GM 2 Gm2 mv 2 1  Gm 
3 fg = fe  = v=  
F= a2 (2R)2 R 2  R 

38. (d) 45. (b)

Total potential energy at mid point is Change in potential energy
 GM1m GM 2m  mgh
U =
− d / 2 − d / 2  1+ h / R
 
given h = 3R
M1 M2 mg{3R} 3
so U = = mgR
2G 1 + 3R / R 4
or – (M1 + M2)m
d
If v is required escape velocity, the 46. (c)
1 2 2G m m
mv = (M1 + M2)m
2 d
G(M1 + M 2 )
v= 2
d R R
Gmm
F=
39. (b) ( 2R ) 2
ve = 2 v 0 = 1.414 v0 % increase in orbital velocity = 2
4 
G R 3
v0 – v0 Gm 2  3 
100 = 41.4 % F= =
v0 4R 2 4R 2

4
F = G22 R 4
40. (c) 9
Total energy of surface where  is density so F  R4
= Total energy at h height
1 47. (b)
 – GMm   – GMm  Due to depth
 2 m (kve)2 +   =U+  
 R   R+h   d
1 – 
h=
R k2 g R 
1– k2 g = g/n
From the centre g  d d 1  n –1
R  = g 1 –   = 1– = 
r=R+h= n  R R n  n 
1– k2  n – 1
d=R  
41. (c)
 n 
GMm GMm – GMm
U= – ;K= ; E= 48. (c)
r 2r 2r B

42. (c) 3R
A
Total mechanical energy conservation T.Ei = T.Eƒ
2R
GMm 1 GMm 1
– + mv 02 = – + mv2
R 2 2R 2
R
v= v 02 − gR
M
43. (d) GMm
Ui = –
R 3
R 3
4 R 2 3
(2R + R)
T = 2  T 2 = 42 M=
GM GM GT 2 GMm
Uf= –
(R + 3R )
44. (c)
W = U = Uf – Uf

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GMm GM mR GMe m h
= = 2 × ⇒ E1 = ×
12R R 12 (R e + h) R e
= mgR/12 Gravitational attraction
𝑚𝑣 2 𝐺𝑀𝑒 𝑚
𝐹𝐺 = 𝑚𝑎𝑐 = =
49. (c) (𝑅𝑒 + ℎ) (𝑅𝑒 + ℎ)2
100kg 10,000kg GM em
A
O
B mv 2 =
x 1m (R e + h)
mv 2 GMe m
Let at a distance x intensity become zero so E2 = =
2 2(R e + h)
IA = IB
E1 = E2
G(100) G(10,000) h 1
= Clearly, =
x 2
(1 − x) 2 Re 2
Re
1 ⇒h= = 3200 km
x= m 2
11
54. (d) JEE Main 2019
50. (a) For a satellite orbiting close to the earth,
d 2h orbital velocity is given by
× 100% = × 100%
R R v0 = √g(R + h) ≈ √gR
d = 2h Escape velocity (ve ) is
x = 2h ve = √2g(R + h) ≈ √2gR [∵ ℎ << 𝑅]
d=x Δv = ve − v0 = (√2 − 1)√gR
55. (d) JEE Main 2019
51. (d) Given, 𝜆 = (A + Bx 2 )
If P is the point of zero intensity, then Taking small element 𝑑𝑚 of length 𝑑𝑥 at a
M distance 𝑥 from 𝑥 = 0
x= .d
M + m
m
and d−x = d
M+ m
So, dm = 𝜆dx
Now potential at point P, V = V1 + V2
GM GM dm = (A + Bx 2 )dx
=− −
x d−x dF =
Gmdm
Substituting the value of x and d – x we get x2
( )
𝑎+𝐿
G 2 𝐺𝑚
V =− m+ M . ⇒𝐹=∫ 2
(𝐴 + 𝐵𝑥 2 )𝑑𝑥
d 𝑎 𝑥
a+L
A
= Gm [− + Bx]
52. (c) x a
When a boy jumps from a ground level up to height h 1 1
= Gm [𝐴 ( − ) + 𝐵𝐿]
then its velocity of jumping 𝑎 𝑎+𝐿
v = 2 gh …..(i)
and for the given condition this will become equal to 56. (d) JEE Main 2019
escape velocity Time period of revolution of satellite,
2GM 2G  4  2𝜋𝑟
vescape = =   R 3 .d  …..(ii) 𝑇=
R R 3  𝑣
𝐺𝑀
8 𝑣=√
Equating (i) and (ii) 2 gh = R Gd  𝑟
3
1/2
 3 gh  𝑟 𝑟3
R=  . ∴ 𝑇 = 2𝜋𝑟√ = 2𝜋 √
 4 Gd  𝐺𝑀 𝐺𝑀

53. (b) JEE Main 2019 Substituting the values, we get

𝐾. 𝐸. of satellite is zero at earth surface and
(202)3 × 1012
at height ℎ from energy conservation 𝑇 = 2𝜋√
Usurface + E1 = Uh 6.67 × 10−11 × 8 × 1022
GMe m GMe m 𝑇 = 7812.2s
− + E1 = − T ≃ 2.17hr
Re (Re + h)
1 1 ⇒ 11revolutions.
⇒ E1 = GMe m ( − )
Re Re + h

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57. (a)JEE Main 2020
GM GM h
= 2 (1 − )
(h + R)2 R R
R3 = (h + R)2 (R − h)
62. (b)
√5−1 Point A is perihelion and C is aphelion.
Solving h= R
2 So, vA  vB  vc
As kinetic energy k = 1/ 2mv or k  v
2 2
58. (c)JEE Main 2020
𝐫2
𝜌(𝐫) = 𝜌0 (1 − 2 )
So, K A  K B  Kc .
𝐑
𝑟
∴m=∫ 𝜌(𝑟) × 4𝜋𝑟 2 𝑑𝑟 63. (c): Light cannot escape from a black hole,
0 vesc = c
𝑟3 𝑟5
= 4𝜋𝜌0 ( − 2)
3 5𝑅 2GM 2GM
Gm r r3
= c or R = 2
∴ E = 2 = 4𝜋𝜌0 [ − 2 ] R c
r 3 5R
2  6.67 10 Nm2 kg −2  5.98 1024 kg
−11
dE R=
∴
dr
=0 (3 108 ms −1 )2
5 = 8.86 10−3 m = 10−2 m
⇒r=√ R
9
64. (a): For a point inside the earth i.e. r  R
59. (a)JEE Main 2020 GM
𝑅 𝐺𝑀 4𝑔 E=− 3 r
𝑔 (ℎ = ) = = R
2 3𝑅 2 9 where M and R be mass and radius of the earth
( )
2 respectively.
𝑑 4𝑔
𝑔 (1 − ) =
𝑅 9
At the centre, r =0
5R E =0
d=
9 For a point outside the earth i.e. r  R,
On the surface of the earth i.e. r = R,
60. (a)JEE Main 2020
g equator = g − 𝜔2 R GM
E=− 2
2ℎ r
𝑔ℎ = 𝑔 (1 − )
𝑅 The variation of E with distance r from the centre is as
2ℎ 2 shown in the figure.
𝑔( ) = 𝜔 𝑅
𝑅
𝜔 2 R2
h=
2g

61. (a)JEE Main 2020

65. (C): The resulting gravitational potential at the
origin O due to each of mass 2 kg located at
positions as shown in figure is
Velocity should be given so as to reach a
point where field is zero.
16M M G2 G2 G2 G2
x 2
=
(10a − x)2 V= − − − −
1 2 4 8
𝑥 = 8𝑎
By CoE  
𝐺𝑀𝑚 16𝐺𝑀𝑚 1  1 1 1   1 
+ 𝑚𝑣 2 = −2G 1 + + + +. = −2G 
1
− −
8𝑎 2𝑎 2
16𝐺𝑀𝑚 𝐺𝑀𝑚  2 4 8  1 − 
=−
8𝑎
−
2𝑎  2
2
3 5Gm
𝑣= √ = −2G   = −4G
2 𝑎 1

66. (b): Gravitational field due to the thin spherical

shell
Inside the shell, i.e. ( For r  R )

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F =0
On the surface of the shell, i.e. ( For r = R ) Gm2 Gm2 GMm
− 4− 2− 4 2
GM d 2d d
F=
R2 −
Gm2 
d 
( )
4 + 2 m + 4 2M 

Outside the shell, i.e. (For r  R),
GM 72. (d)
F=
r2 GM
g=
R2
as shown in the adjacent figure.
1
M = constant g 
R2
67. (b): Equal areas are swept in equal time. t1 , the
g R
time taken to go from C to D = 2t2 100 = −2 100
g R
where t 2 is the time taken to go from A to B. % change = - 2 (-2)
As it is given that area SCD = 2SAB. % change in g = 4%
increase by 4%
68. (c): Applying the properties of ellipse, we have
73. (b)
V 3
dV k k
− = − 2   dv =  2 dr
dr r 10 2 r

1 1
2 1 1 r1 + r2 v − 10 = k  − 
= + =  2 3
R r1 r2 r1r2 k
V − 10 =  V = 11volts
2r r 6
R= 1 2
r1 + r2
74. (d)
ℓ
69. (d) T1 = 2𝜋 √ (2R)2
GM

ℓ
T2 = 2𝜋√ (3R)2
GM
T1 2
∴ =
T2 3
75. (b)
1 GMm
Effective acceleration due to gravity is the KE = mv 2 =
2 2a
resultant of g & rw2 whose direction & PE = −2KE
magnitude depends upon  . Hence TE = −KE
assertion is false. 76. (d)
When  = 0° (at equator), effective h = 318.5 ≈ ( )
Re
acceleration is radially inward. 20
−GMe m
T ⋅ Ei =
70. (a) Re
Since it is universal law so it hold good for −GMe m −GMe m
any pair of bodies.The value of g at centre is T ⋅ Ef = =
2(R e + h) 2 (R + R e )
zero.So statement I and Statement II are true e 20
71. (a) −10GMe m
⇒ T ⋅ Ef =
21R e
Change in total mechanical energy

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Mahematical tools for Physics DPP 1

= TEf − TEi
GMe m 10 11GMe m
= [1 − ] =
Re 21 21Re

77. (d)

−GMm 104 R
Total energy = if energy = is added then
2(2R) 6
4
−GMm 10 R −GMm
+ =
4R 6 2r
𝐺𝑀
where 𝑟 is new radius of revolving and 𝑔 = 2
𝑅
mgR 104 R mgR2
− + =− (m = 103 kg)
4 6 2r
103 × 10 × R 104 R 103 × 10 × R2
− + =−
4 6 2r
1 1 R
− + =−
4 6 2r
r = 6R
GMm
78. (a) F1 = … … (i)
(2R)2
M
GMm G( )m
F2 = −( 27 )
(2R)2 4R 2
( )
3
11 GMm
F2 = ….(ii)
48 R2
F1 : F2 = 12: 11

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