PHYSICS BOOK-7

HOME ASSIGNMENT SOLUTIONS

GRAVITATION

SOLUTIONS

1. (D)
GM1M 2 G
2. (C) F= = 100 100
R R
G F
= ………(i)
R 10000
G G
F  = 125  75 =  9375
R R
F
F =  9375 [Put from Equ. (i)]
1000
15
F  = F.
16
4 g R
3. (A) g =  GR. If  = constant then 1 = 1 .
3 g2 R 2
4. (A) Gravitational force does not depend on the medium.
GM g M
5. (B) Acceleration due to gravity g =  = .
R2 G R2
g M  R2 1 4 2
6. (B) =  =  =
g M R 2 2 1 1
GM L2
7. (C) g = 2 and K =
R 2I
1 1
If mass of the earth and its angular momentum remains constant then g  2
and K  2
R R
i.e. if radius of earth decreases by 2% then g and K both increases by 4%.
u2 1 H g
8. (C) H= H  B = A
2g g H A gB
g
Now g B = A as g   R
12
H g
 B = A = 12  H B = 12  H A = 12 1.5 = 18m.
H A gB
9. (A) Gravitational attraction force on particle B
2
GM p m .Acceleration of particle due to gravity a = F /m = 4GM / Dp
Fg = 2
g p/
 Dp 
 
 2 
GM e
10. (A) Acceleration due to gravity on the surface of the earth is ge = . Where Me and Re are the mass
Re2
and the radius of the earth respectively.

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 GM p
Acceleration due to gravity on the surface of the planet is gp = . Where Mp and Rp be the mass
R p2
and the radius of the planet respectively.
If both mass and radius of the planet are half as that of the earth, then
M 
G e 
11. (D)
g p =  2  = 2 2 e = 2 ge
2 GM
 Re  Re
 2 
 

Force will be zero at the point of zero intensity

m1 81M 9
x= D= D = D.
m1 + m2 81M + M 10
12. (A) k represents gravitational constant which depends only on the system of units.
13. (A)
4 g  R
14. (A) g= G R  g  R  e = e  e
3 gm  m Rm
6 5 Re 5
 =   Rm = Re
1 3 Rm 18
15. (A) g
16. (C) Areal velocity of the planet remains constant. If the areas A and B are equal then t 1 = t 2 .
17. (C) Angular momentum remains constant
v1 d 1
mv 1 d 1 = mv 2 d 2  v 2 =
d2
18. (C) The earth moves around the sun is elliptical path. so by using the properties of ellipse
r1 = (1 + e ) a and r2 = (1 − e ) a
r1 + r2
a = and r1 r2 = (1 − e 2 ) a 2
2
where a = semi major axis
b = semi minor axis
e = eccentricity
b2
Now required distance = semi latus rectum =
a
a 2 (1 − e 2 ) (r1 r2 ) 2r r
= = = 1 2
a (r1 + r2 ) / 2 r1 + r2
T2
19. (B) = constant  T 2 r −3 = constant
r3
Gm (M − m )
20. (B) F=
r2
dF
For maximum force =0
dm
 GmM Gm 2 
d
  
 r2 − r2  = 0
dm
 
m 1
 M − 2m = 0  =
M 2
21. (a) No. the shell does not shield other bodies lying outside it from exerting gravitational forces on a
particle lying inside it.
(b) yes. If the size of spaceship orbiting around the earth is large enough, the astronaut inside the
spaceship can detect the variation in g.
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 (c) the tidal effect depends inversely upon the cube of the distance whereas the gravitational force varies
inversely as the square of the distance. As the moon is closer to earth than the sun, so its tidal effect is
greater than that of the sun.
The ratio of these two effects is
3 3
Tm  d s   1.5 1011 
=  = 8 
= 61.5 106.
Ts  d m   3.8 10 
GMm
22. Weight of body, w = mg =
R2
When the diameter or radius of the earth becomes double its present value, the weight of the body will be
GMm 1
w' = = W
( 2R )
2
4
i.e., weight will become one-fourth of the present value.
Gm1m2
23. for inverse square law, F =
r2
F Gm2
 Acceleration of body A = = 2 =a
m1 r
Gm1m2
For inverse fourth power law, F ' =
r4
F ' Gm2 a
Acceleration of body A = = 4 = 2
m1 r r
24. When we move our finger, the distance between the objects and our finger changes. Hence, the force of
attraction changes, disturbing the entire universe, including the stars.
1
25. The value of ‘g’ on the surface of the moon is about th of its value on the surface of earth.
6
For the same gain of P.E. in both cases, we have
ge he gh
mgm hm = mge he or hm = = e e = 6he
gm ge / 6
26.
27. (a) Gravitational force does not depend upon the
medium between the two particles.
(b) Kepler’s law will be applicable whenever
inverse square law is involved.
(c) Tides arise in the oceans due to the force of
attraction between the moon and sea water.
(d) No, while electrical forces depend upon the
medium, the gravitational forces do not depends upon medium. To sum up, the ‘gravity screens’ are not possible.
(e) The tidal effect depends inversely on the cube
of the distance unlike force which depend inversely on the square of the distance.
(f)

(g) At perihelion because the earth has to cover

greater linear distance to keep the areal velocity constant.

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 1
28. (a) F  .So, if r is increased by a factor of 3, F will be reduced by a factor of 9. Thus, the new force will be
r
x
.
9
 1 1 
(b) F = G
m1m2
r 2 (
= 6.67 10−8  2  dyne
 1 
)
6.67 10−8
−8
= 6.67 10 dyne =
980
= 7 10−11 g-wt
−13 2
(c) Given, k = 10 s / m
3

1
As, 1s = day and 1 m
24  60  60
1
= km
1000
1 1
 k = 10−13  ( day ) km−3
2

( 24  60  60) (1/1000 )
2 3

= 1.33 10−14 (day)2 km–3

For the moon, r = 3.84 105 km

( )
3
T
2
= kr 3 = 1.33 10−14  3.84 105
= 753.087
T = 27.3 day
(d) If FL and FR are the forces exerted on mass m lying at O  (after displacing it from O to O  ) by masses M, m
lying on left and right respectively then from figure.
Mm Mm
FL = G and FR = G
( x + x ) ( x − x )
2 2

Since ER> FL , the net force on m is towards right. Hence, the equilibrium is unstable.
(e) (i) According to law of conservation of linear momentum L = mvr = constant, therefore the comet moves
faster when it is close to the sun and moves slower when it is farther away from the sun. Therefore, the speed
of the comet does not remain constant.
(ii) As the linear speed varies, the angular speed also varies. Therefore, angular sped of the comet does not
remain constant.
(iii) As no external torque is acting on the comet, therefore, according to law of conservation of angular
momentum, the angular momentum of the comet remain constant.
1
(iv) Kinetic energy of the comet = mv 2 As the linear speed of the comet changes its kinetic energy also
2
changes. Therefore, its KE does not remain constant.
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 (v) Potential energy of the comet changes as its kinetic energy changes.
(vi) only angular momentum and total energy of a comet remain constant throughout its orbit.
GM G 4 3 4
29. Here, g= 2
= 2 .  R  =  GR
R R 3 3
g  R
g1 R
 = = 1:1
g 2 2 R. 
2
30. (a) Law of conservation of angular momentum.
(b) According to Kepler’s third law of periods, the square of the period of revolution of a planet around the
sun is proportional to the cube of the semimajor axis of its elliptical orbit.

GM GM 1 GM g
31. (D) g= , g = =  g = .
( R + 2R ) 9 R
2 2 2
R 9
32. (B) Weight is least at the equator.
33. (B) Intensity inside a shell is zero.
34. (B) Acceleration due to gravity at latitude  is given by g  = g − R 2 cos2 
3
At 30o , g30o = g − R 2 cos 2 30o = g − R 2
4
3
 g − g30 =  2 R.
4
35. (B) g = g =  R cos2 
 2

Rotation of the earth results in the decreased weight apparently. This decrease in weight is not felt at the
poles as the angle of latitude is 90o .
GM
36. (C) g= . Since M and r are constant, so g = 9.8 m/s2
r2
37. (D) S2 is correct because whatever be the g, the
same force is acting on both the pans. Using a spring balance, the value of g is greater at the pole.
Therefore mg at the pole is greater. S4 is correct. S2 and S4 are correct.
38. (A) Due to three particles net intensity at the centre

I = I A + I B + IC = 0
Because out of these three intensities one equal in magnitude and the angle between each other is 120 0
39. (D) Potential at origin
 1 1 1 
V = 2  GM 1 + + + ..........
 2 4 8  GP
 
 1 
= 2GM  = 4GM = 4G
1
1 − 
 2
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40. (C) AB = BC = CD = DA = 1 m
AC =BD = 12 + 12 = 2 m
Total potential energy of the system of four particles each of mass 1 kg placed at the vertices A, B, C and
D of a square is

 −G 11   −G 11   −G 11  +  −G 11   −G 11   −G 11 

U = + +   + + 
 AB   AC   AD   BC   BD   CD 
−G 11   −G 11   −G 11   −G 11   −G 11   −G 11 
=  + + +  + + 
 1   2   1   1   2   1 

= −4G −
2G  1 
= −2G  2 +  = −5.4 G
2  2
41. (D)

Gm G ( 4m )
=
x 2 ( r − x )2
1 2
=  r − x = 2x
x r−x
r
 3x = r  x =
3
Gm G ( 4m )
 The gravitational potential = − −
r / 3 2r / 3
3Gm 6Gm 9Gm
= − − =−
r r r
42. (C) If the body is projected with velocity v(v < ve) then height upto where its rises,
R R
h= 2  h= 2
= 4 R ( approx.)
v
−1  11.2 
 −1
e
2 
v  10 
GM 2GM −GMm
43. (B, C, D) g = , ve = and U=
2 R
R R
M M M
 g ,ve  and U
R2 R R
If both mass and radius are increased by 0.5% then v e and U remains unchanged where as g decrease
by 0.5%.
44. (B) g  = g −  2 R cos 2 
For weightlessness at equator  = 0 and g' = 0
g 1 rad
 0 = g −2R   = =
R 800 s
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 mgh 1
45. (B) U = = mgR ( h = R)
h 2
1+
R
46. (B) Let velocities of these masses at r distance from each other be v 1 and v 2 respectively.
By conservation of momentum
m 1v1 − m 2 v 2 = 0
 m 1v1 = m 2 v 2 … (i)
By conservation of energy change in P.E.=change in K.E.
Gm 1 m 2 1 1
= m 1 v 12 + m 2 v 22
r 2 2
m 12 v 12 m 22 v 22 2 Gm 1 m 2
 + = …(ii)
m1 m2 r
On solving equation (i) and (ii)
2 Gm 22 2 Gm 12
v1 = and v2 =
r(m 1 + m 2 ) r(m 1 + m 2 )

2G
 v app =| v1 | + | v 2 | = (m 1 + m 2 )
r
2 GM
47. (D) ve =
(R + h)

vp gp Re 1 1
48. (A) v = 2 gR  =  = 2 =
ve ge Rp 4 2
ve
 vp =
2

49. (C) Escape velocity v = 2GM If star rotates with angular velocity 
R

v 1 2GM 2GM
then  = = =
R R R R3

 2h 
50. (i) At altitude h1 g h = g 1 − 
 R
Thus acceleration due to gravity decreases with increasing altitude.
 d
(ii) At depth d, g d = g 1 − 
 R
Thus acceleration due to gravity decreases with increasing depth.
Rw2 cos 2   
(iii) At latitude  , g  = g 1 − 
 g 
For maximum variation (decrease) of g , cos  = 1 or  = 0o
Thus the effect of rotation on the effective value g is greatest at the equator (for which  = 00 .)
GM
(iv) As g = , where M is the mass of the earth. Thus acceleration due to gravity is independent of mass
R2
m of the body.
51. (a) at poles because distance of poles from the centre of earth issmaller than the distance of any other point on
earth’s surface from its centre.
(b) More over At poles, no centrifugal force acts on the body.
52. The weight of a person can become zero under the following conditions:
(i) when the person is at the centre of the earth (as g=0 at the centre of the earth).
(ii) When the person is at the null points in space (At these points, the gravitational forces due to different
masses cancel out).
(iii) When the person is standing in a freely falling lift.
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 (iv) When the person is inside a spacecraft which is orbiting around the earth.
53. (a) The weight of an object below surface of the earth is zero.
(b) Since, gravitational potential difference is zero therefore the work done is zero.
(c) Acceleration due to gravity is independent of the mass of the body.
(d) At the mountain, g decreases and time period of the pendulum clock increases at T = 2 l /g . On the
other hand, a spring in the wrist watch remains unaffected by the variation of g.
(e) mg p = mg e since, g p  g e ,  m  m. so, we shall have greater mass of sugar at the equator.
GMm
(f) By making use of the relation g m =
R 2m
54. (a) Let A and B be the two given spheres and O be the mid-point on the line joining their centres.

If I1 and I2 are the gravitational fields at O due to A and B respectively,

G 100 G 100
I1 = , I2 =
( 0.5) ( 0.5)
2 2

Since, I1 and I2 are equal and opposite, then resultant gravitational field at O is zero.
If V is the gravitational potential at O due to A and B
G 100 G 100 2G 100
V=− − =−
0.5 0.5 0.5
−2  6.67 10−11 100
or V = = −2.7 10−8 J/kg
0.5
The object placed at O is in equilibrium as there is no net force acting on the object. But it will be in an
unstable equilibrium as once displaced from O, it will not come back to O.
GMm  GMm 
(b) Gain in PE ΔU = U f − U i = − −− 
2R  R 
GMm  1  GMm
=  − + 1 =
R  2  2R

=
( gR ) m = 1 mgR  as g = GM 
2

 
2R 2  R2 
(c) Mass of the earth, M = 6.0 1024 kg,m = 67kg
G = 6.67 1011 N-m2Kg−2
GMm
Gravitational potential energy, U = −
R
6.67 10−11  6 1024  67J = −4.1105 J
=−
6.6 1010
GMm V −GM
As we know V = − = =
R m R

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 4.1105
=− J kg −1 = −6.1103 J kg −1
67
55. (a) Consider a small element of the ring of mass dM gravitational force between dM and m, distance x apart in
figure i.e.,

dF can be resolved into two rectangular components

(i) dF cos  along PO and
(ii) dF sin  perpendicular to PO (given figure)
The total force (F) between the ring and mass (m) can be obtained by integrating the effects of all the elements
forming the ring. Whereas all the components perpendicular to PO cancel out, i.e.
G ( dM ) m  b 
 dF sin  = 0, the component along PO and add together to give F i.e. F =  dF sin   x2
 
x
GMmb
=
(r )
3/2
2
+ b2
 dm = M and x = r 2 + b 2
  ( ) 
1/2



 r  GMm
Since, b = r , F = GMm  3 
=
 2 2r  2 2 r
2

GMm ( 2r ) 2GMm
When b becomes 2r , F  = =
(r )
3/2
2
+ 4r 2 5 5r 2

F   2GMm   2 2r 2  4 2
=  =
F  5 5r 2   GMm  5 5
Thus,

G  5 1 U1 = −GJ
(b) U1 = − J or
5

−G  5 1 5G
U2 = J=− J
6 6
 5G  5G G
Work done =  −  − ( −G ) = G − = .
 6  6 6
56. (a) (i) Acceleration due to gravity at altitude b from the earth’s surface is given by
g
g'= 2
 b 
1 + 
 Re 
Where Re is the radius of t he earth.

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 Therefore, acceleration due to gravity decrease with increasing altitude.
(ii) Acceleration due to gravity at depth b from the earth’s surface is given by
(iii) Acceleration due to gravity is independent of the mass of the body.
 1 1
(iv) The formula – GMm  2 −  is more accurate than the formula mg (r 2 – r1 ) for the difference of potential
r r1 
energy between two points r2 and r1 distance away from the centre of the earth.
According to principle of the energy conservation Eqs. (i) and
(b) Let us consider that the rockey sphere has mass
M and radius R. The escape speed for such a sphere is given by
 4  3
2G  R  8 G 
e =
2GM
=  3 
= R
R R 3
3
or R = e
8 G 
Here,  = 3.0 g / cm3 = 3.0 103 kg / m3 ,e = 40m / s

3
Thus, R = 40 m
8  3.14  6.67 10−11  3 103
or R = 40  772.6m = 30904 m = 30.904 km
 GMm  GMm = gR 2  m = mgR
57. (D) W =0 −− =
 R  R R
= 1000 10  6400 103 = 64 109 J = 6.4 1010 J
1 2 1
58. (C) mve = m 2 gR = mgR
2 2
2GM
59. (B) ve =  ve  M if R = constant
R
60. (B)
GMm
61. (A) K .E =
2R
2GM GM
62. (A) ve = = 100  = 5000
R R
GMm
Potential energy U = = −5000 J
R
GM 1 2GM
63. (A) v= =
R+h 2 R
 4R = 2 ( R + h )  h = R = 6400 km.
64. (B) ve = 2gR and v0 = gR  2v0 = ve
GM gR 2
65. (D) v0 = =
r R+h
Using T = 2 R / g , where R = 6.4 10 m
6
66. (D)
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 g = 9.8 m / s 2 , we get T  1.5 hrs
67. (C) Angular momentum
L = mvr = 2mE.r = 2mEr 2
ve 2
68. C) v0 = = = 2 km / s
2 2

( R + h) ( 2R )
3 3

T = 2 = 2 R
gR 2 = 4 2
69. (B)
gR 2 g
T12 R13 ( 6 R )
3

70. (C) = = =8
T22 R23 ( 3R )3
24  24
=8
T22
24  24
T22 =
8
T22 = 72
T22 = 36  2

T2 = 6 2
71. (B) ve = 2v0 = 1.414v0
v  ve − v0
Fractional increase in orbital velocity  = v = 0.414
 v  0

 Percentage increase = 41.4%

72. (B) Due to inertia of direction.

73. (C)
74. (C) Kinetic energy = Potential energy
1 mgh 1 mgh
m (kv e ) 2 =  mk 2 2 gR =
2 h 2 h
1+ 1+
R R
Rk 2
h =
1−k2
Height of Projectile from the earth's surface = h
Rk 2
Height from the centre r = R+h = R+
1−k2
R
By solving r=
1−k2
75. (C) In the problem orbital radius is increased by 1%.
Time period of satellite T  r 3 / 2
Percentage change in time period
3 3
= (% change in orbital radius) = (1 %) = 1 . 5 %
2 2
mgh 1
76. (B) U = = mgR ( h = R)
h 2
1+
R
77. (a) No, we cannot determine the mass of satellite by measuring its time period.
(b) The value of escape speed on the moon is small as compared to the value of the earth.

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 (c) No energy is required. This is because the work done by centripetal force is zero.
(d) Parking orbit is that orbit in which the period of revolution of a satellite is equal to the period of rotation
the earth about its axis.
(e) The weight of a body is zero in a geostationary satellite.
(f) It is not possible. This is because New Delhi is not in the equatorial plane.
(g) The bomb would merely act as another satellite. It would never hit the earth.
78. (a) The food packet will not fall on the earth. As the satellite as well as astronaut were in a state of
weightlessness, hence, the food packet, when dropped by mistake, will also start moving with the same
velocity as that of satellite and will continue to move along with the satellite in the same orbit.
(b) If suddenly the gravitational force of attraction between the earth and a satellite revolving around it
becomes zero, satellite will not be able to revolve around the earth. Instead, the satellite will start moving
along a straight line tangentially at that point on its orbit, where it is at the time of gravitational force
becoming zero.
(c) The change in gravitational potential energy of a body between two given points depends only upon the
position of the given points and is independent of the path followed. It is due to the fact that the
gravitational force is a conservative force and work done by a conservative force depends only on the
position of initial and final points and is independent of path followed.
(d) (i) No, escape velocity is independent of the mass of the body.
(ii) Yes, escape velocity depends (through slightly on the location from where the body is projected
because with location g changes and so should ve = 2gR ( ) change.
(iii) No, escape velocity is independent of the direction of projection.
(iv) Yes, escape velocity depends (through slightly) on the height of location from where the body is
projected as g depends on height.
79. (a) Step I: Height h = 1000 km, R= 6.4  106 m g = 9.8 ms-2
Step II : h = 10001000 = 106 m
Step III: Escape velocity at a height h above the earth’s surface ve = 2 gh ( R + h )

gR 2 2  gR 2 2 gR 2
2 (
Step IV: gh =  ve = R + h) =
( R + h) ( R + h) R+h
2

( )
2
2  9.8  6.4 106
Step IV: ve = ms −1 = 10.42kms −1
( 6.4 + 1)106
(b) (i) Given, T = 7 h 39 min = 459×60 s
R = 9.4103 km = 9.4106 m, Mm= ?

4 2 R3
 Mass of mass Mm = .
G T2
4  ( 3.14 )  ( 9.4 106 )
2 3

=
6.67 10−11  ( 459  60 )
2

Tm2 RMS
2
(ii) From Kepler’s third law, = Where RMS is the mass -sun distance and RES is the earth -sun distance
T 2 RES
2

 Tm = (1.52 )  365 = 684 days

3/2

80. (i) As, according to Kepler’s third law, we get

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 1/3
4 2 r 3  GMT 2 
T = 2
;r =  2  (1)
GM  4 
As we know r = R + b
1/3
 GMT 2 
b 2  –R
 4 
b = 4.23 107 m − 6.4 106 m
b = 3.59 107 m
OA R
(ii) In OES , cos  = =
OS R + b
1 1
= = = 0.1513
 b (1 + 5.609 )
1 + 
 R
 n 3.59 107 m 
 as , = = 5.609 
 R 6.4 10 m
6


Where, 810 or 2 = 1620

Number of satellites required to cover entire the earth.
3600
= = 2.2  3
1620
81. When r increases:
 GM 
(i) Linear velocity  v =  decreases.
 r
 
 v GM 
(ii) Angular velocity   = =  decreases.
 r r 3 

 1 2 GMm 
(iii) Kinetic energy  K = mv =  increases.
 2 r 
 GMm 
(iv) Potential energy  U = −  increases.
 2r 
 GMr 
(v) Total energy  E = −  Increases. (or becomes less negative)
 2r 

(
(vi) Angular momentum L = mvr = GMm2 r increases. )
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 PHYSICS BOOK-7
OBJECTIVE QUESTIONS
SOLUTIONS
GRAVITATION

SOLUTIONS
2
4 
G   R 3  
MM  3 6
  F  R  R4
1. (C) F=G =
R2 (2 R) 2
R2
2. (A) In the absence of gravity weight of the bodies will become zero but mass will not change.
−GMm
3. (B) U= . If r increases then U also increases.
r
4. (D) According to Kepler's law T 2  R 3
If N is the frequences then N 2  (R)−3
−3 / 2 2/3
N 2  R2  R1  N 2 
or =   = 
N1  R1  R2  N1 
5. (D)
6. (A)
Gm m m e
7. (A) Force between earth and moon F=
r2
This amount of force, both earth and moon will exert on each other i.e. they exert same force on each
other.
4
8. (B) g=  GR . If density is same then g  R
3
According to problem R p = 2 Re  g p = 2 g e
l
For clock P (based on pendulum motion) T = 2
g
Time period decreases on planet so it will run faster because g p  g e
m
For clock S (based on oscillation of spring) T = 2
k
So it does not change.
 d
9. (D) For scientist A which goes down in a mine g' = g 1 − 
 R 
For scientist B, which goes up in a air  2h 
g ' = g 1 − 
 R
So it is clear that value of g measured by each will decreases at different rates.
10. (B) Gravitational pull depends upon the acceleration due to gravity on that planet.
1 1
Mm = M e , gm = ge
81 6
1/2 1/2
M g 
=  81 
GM Re 1
g=  = e  m  
R2 R m  M m g e   6
9
 Re = Rm
6
11. (D) Reading of spring balance R = m(g − a)
If the lift falls freely then a = g  R = 0
GM M
12. (C) g= 2
g
R R2
Me Re
According to problem Mp = and Rp =
2 2
2
 Mp   Re 
 =  1   (2) 2 = 2
gp
 =  
R  2
ge  Me  p 

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  g p = 2 g e = 2  9 .8 = 19 .6 m /s 2
13. (C) Apparent weight = actual weight – upthrust force
Vdg ' = Vdg − Vg

 g' =  d −   g
 d 
M ge
14. (B) g 2
. If mass and radius of the planet are twice then g  will be half that of g e i.e. g p =
R 2
1
15. (A) The value of g on the surface of the earth g
R2
1
At height h from the surface of the earth g 
(R + h) 2
R2 9 .8  (6400 )2
 g = g = = 8 .4 m /s 2
(R + h) 2
(6400 + 480 )2
2 2 2
g'  R  1  R  1  4000 
16. (C) =   =   = 
g R +h 2 R +h 2  4000 + h 
By solving we get h = 1656 .85 mile  1600 mile
l
17. (A) T = 2 . At the hill g will decrease so to keep the time period same the length of pendulum has to
g
be reduced.
−GM GM
18. (D) V = and I=
r r2
V = 0 and I = 0 at r = 
19. (C) U = Loss in gravitational energy = gain in K.E.
1 2U
So, U= mv 2  m = 2
2 v
ve g e Re
20. (C) = = 6  10 = 60  8 (nearly)
vm g m Rm
21. (B)
vA gA RA vA
22. (C) v e = 2 gR  =  = x r  = rx
vB g B RB vB
23. (A) Time period of satellite which is very near to planet
R3 R3 1
T = 2 = 2 T 
GM 4
G R 3 

3
i.e. time period of nearest satellite does not depends upon the radius of planet, it only depends upon
the density of the planet.
In the problem, density is same so time period will be same.
24. (B) Otherwise centrifugal force exceeds the force of attraction or we can say that gravitational force won't
be able to keep the satellite in circular motion.
25. (C)
K A rB  R + h B   R + 2R  3
26. (D) = = =
  R + R  = 2
K B rA  R + h A 
3/2 3/2
Tmercury  rmercury   6  10 10  1
27. (C) =  
 =  
 = (approx.)
 1.5  10
11
Tearth  rearth   4
1
 Tmercury = year
4
x x
A A
28. (C) V ( x ) =  Edx = −  2
dx =
 
x x
29. (A)   G ( m )( m ) 
  −G ( m )( m ) 
W = U f − U i = 3 −  − 
 
 rf 
  ri 

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 N − m2
Here G = 6.67  10−11
kg 2
m = 0.1kg, rf = 0.4 m and ri = 0.2 m substituting the values we get W = 0.5  10–12 .
GM
30. (B) Potential due to particle V1 = −
R
2GM
and potential due to shell itself V2 = −
R
3GM
 Total potential is V = V1 + V2 = −
R
31. (B) Let R be the radius of earth and g the acceleration due to gravity on earth’s surface. Then the desired
ratio (say x) is:
 h
g 1 −  2
R   h  h 
x=  = 1 − 1 + 
g  R  R 
2
 h
1 + 
 R
 h  2h 
= 1 − 1 +  ( h  R )
 R  R
h
= 1+
R
From this expression we see that x increases linearly with h.
32. (D) Gravitation field due to the ring at a distance d = 3R on its axis is:

3
GMd 2
E= 2
(R + d2 )
Substituting the values we get;
−10 = 2 (VB − VA ) + 4
or VB − VA = −7 J / kg
GMm
33. (C) Kinetic energy of satellite is K = while magnitude of potential energy
2r
GMm
U=
r
U
K=
2
GMm
34. (B) On earth’s surface potential energy of particle is U = −
r
GMm
And kinetic energy at a velocity equal to its escape velocity is K =
R
i.e. K+U=0
or total mechanical energy is zero.
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35. (D) Work done will be zero when displacement is perpendicular to the field. The field makes an angle
  1
1 = tan −1   with positive x-axis while the line y + 4x = 2 makes an angle
  4
2 = tan ( −4) with positive x-axis
−1

1 +  2 = 90 i.e. the line y + 4x = 2 is perpendicular to E .

3GM
=
8R 2
3GM 2
Force on sphere = 8ME =
R2
GM
36. (B) g=
R2
4
M =  R3 .
3
1
 3M  3
R= 
 4 
Substituting in (1) we have:
2
 4  3
g = GM  
 3M 
1 2
or g  M 3  3
1 2
g planet = g (8) 3 (8) 3 = 8g
1 2
g planet = 9 (8) 3 (8) 3 = 8 g
37. (C) Work done = increase in gravitational potential energy
 
mgR mgR  mgh 
 W1 = =  U =
R 2 h
1+  1+ 
R  R
mgh
and W2 =
h
1+
R
Given that W1 = 2W2
mgR 2mgh
=
2 h
1+
R
R
h=
3
38. (A) WA _ B = (U B − U A ) + ( KB − K A )
WA→B = m (VB − VA ) + ( KB − K A )
G ( M )( M ) GM 2
39. (A) Force between any two spheres will be : F = =
( 2R )
2
4R2
The two forces of equal magnitudes F are acting at angle 60 on any of the sphere.

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  Fnet = F 2 + F 2 + 2 ( F )( F ) cos 60 = 3F
3 GM 2
or Fnet =
4 R2
Gm
40. (B) When − mass of the shell is m, the potential is
R
Gm
To add an elementary mass dm, the work done is − dm
R
M
Gmdm
 Net work done = self potential energy = − 
0
R
GM 2
=−
2R
41. (D) Net gravitational field inside a shell is zero.
Hence, net force on the particle will be zero i.e. the particle stays at rest in its original position.
GM
42. (A) E (r ) = .r for r  R
R3
Gr  4  4Gr
= 3   R3 .  =
R 3  3
1 1
From conservation of energy mv 2 = m ( 20 ) − Ve2 
2
43. (B)
2 2  
Here Ve = escape velocity  8 2km / hr


2
( )
y 2 = ( 20) − 8 2
2

or V 16.5km / hr
44. (A) Potential energy at height h:

U = - [work done by gravity + work done by electrostatic force]

= mgh + qEh
= (mg + qE)h
or U = kh
where k = mg + qE i.e. U-h graph is a straight line passing through origin.
GMm
45. (C) Kinetic energy K .E =
2r
GMm
Potential energy P.E = −
r
GMm
and the total energy E = −
2r
1
Kinetic energy is always positive and K .E 
r
1
Potential energy is negative and P.E 
r
1
Similarly total energy is also negative and E 
r
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 Also E  P.E
 A is kinetic energy, B is potential energy and C is total energy of the satellite.
46. (A) From conservation of mechanical energy:
Decrease in gravitational potential energy = Increase in kinetic energy.
Gm1m2 1
or − Vr2
r 2
mm
Here  = 1 2 = reduced mass
m1 + m2
and Vr = relative velocity of approach
2G ( m1 + m2 )
vr =
r
4 
G   R3  
g= 2 =  2  = 4  G R
GM 3
47. (C)
R R 3
i.e. g   R
R is increased by a factor of 2 i.e. to keep the value of g to be constant the value of  has to b e
1
changed by a factor of .
2
 2h 
48. (B) g h = g 1 −  ( h  R, radius of earth)
 R
dW d ( mg ) d (g)
= = =m
dh dh dh
2mg
= = constant
R
49. (A) Time period of a satellite above the earth’s surface is:
4 R3 3 3
T2 = = =
GM   G
 M 
G 
4
  R3 
3 
3
T 2 = = a universal constant.
G
v 1
50. (C) v = v0 = e (orbital speed v0 of a satellite near the earth’s surface is equal to times the
2 2
escape velocity of a particle on earth’s surface)
Now from conservation of mechanical energy:
Decrease in kinetic energy = increase in potential energy
1 ve2 mgh
or m =
2 2  h
1 + 
 R
1  2 gR  mgh
or m  = or h = R
2  2  1+ h
R
51. (D) Let r be the radius of the satellite. Then

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 GM
v02 = ….(1)
r

R = radius of earth
Applying conservation of mechanical energy between points A and B, we have:
1 2 GMm  GMm 
mv = − −− 
2 r  R 
GM 2GM
or v 2 = 2 − = ve2 − 2v02
R r
or v = ve2 − 2v02
52. (C) Energy required to raise a satellite upto a height h:
mgh
E1 =  U =
h
1+
R
1
E2 = mv0 2
2
1  GM 
= m 
2  r 
1  GM 
m 
2  R+h
1  GM  R
= m 2 
2  R  1+ h
R
mgR
or E2 = ….(2)
 h
2 1 + 
 R
From equations (1) and (2)
E1 2h
=
E2 R
53. (A) A lighter body inside a denser medium behaves like a negative mass so far as gravitational force is
concerned. Two air bubbles i.e. two negative masses will attract each other.
54. (C) By the principle of superposition of fields
E = E1 + E2
Here E = net field at the centre of hole due to entire mass
E1 = field due to remaining mass
 GM  R
E2 = field due to mass in hole = 0 E1 = E =  3 r where r =
 R  2
GM
E=
2R2
GMm GMm
55. (B) Binding energy of satellite in the first case is and in the second case is
2r 3r
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 GMm  1 1  GMm
Energy increased E =  − =
R  2 3 6r
GMm
 % increase = 6r 100 = 33.33%
GMm
2r
56. (B) The whole space can be divided into three regions
(i) 0  r  R1 F (r ) = 0
4  R3 
(ii) R1  r  R2 F ( r ) =  G  m  r − 12 
3  r 
4  R3 − R3 
(iii) R2  r  F ( r ) =  Gm  2 2 1 
3  r 
Here  is the density of material of the sphere.
57. (A) As we know that areaL velocity of planet is constant. Therefore, the desired time is:

 Area SAB 
t AB =  T
 Area of ellipse 
  ab 1 
=T  − ( b )( ea )  /  ab
 4 2 
1 e 
=T  − 
 4 2 
58. (B) W = increase in potential energy of system
=Uf – Ui
= m (Vf – Vi )
(V = gravitational potential)
  GM  
= m 0 −  − 
  R 
GMm
=
R
GM
Note: Even if mass is nonuniformly distributed potential at centre would be − .
R
5
−
59. (B) Centripetal force  R 2
5
−
mR  R 2 2

5
−
 R
2 2

7
−
2  R 2

7
T 2  R2
dA L
60. (C) Areal velocity = = constant
dt 2m
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 dA vr sin 
Here L = mvr = sin  (angular momentum)
dt 2
dA
i.e. m
dt
61. (A) L=mvr …(1)
2
mv GMm
= 2 ….(2)
r r
From (1) and (2), we can see that;
1
L = m ( GMr ) 2
1
L  r2
62. (D) The fastest possible rate of rotation of a planet is that for which the gravitational force on material at
equator just barely provides the centripetal force needed for the rotation. Let M be the mass of thepalnet
R its radius and m the mass of a particle on its surface. Then
GMm
mR 2 max =
R2
4 
G   R3 . 
max =
GM
= 3 
3 3
R R
 G  2
=2 =
3 Tmin
3
 Tmin =
G
63. (A) Total energy of a planet in on elliptical orbit is:
GMm
E=− (m = mass of planet)
2a
From conservation of mechanical energy K.E. + P. E. = E
1 2 GMm GMm
or mv − =−
2 r 2a
2 1
or v = GM  − 
r a
GMm
64. (A) Energy of the satellite on the surface of earth is E1 = −
R
and energy of satellite at a distance ( 2R + R ) = 3r from centre of earth is
GMm GMm
Ef = − =−
2 ( 3r ) 6R
5GMm
 Energy required to launch the satellite is E = E f − Ei =
6R
65. (A) Magnitude of potential energy per unit mass = E
 magnitude of total potential energy of the body = mE
Which is also equal to its binding energy
1 2
Hence mve = mE
2
or ve = 2E

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66. (B) Angular momentum of planet about the sun is constant.
i.e. mvr sin = constant
At positions P and Q,  = 90 and m = mass of planet = constant
vr = constant
v1 r2
or =
v2 r1
67. (C) Let M be the mass of the planet and m the mass of satellite. Then
GMm
mr 2 =
r2
or Gm = r 3 2
GM r 3 2
g= 2 = 2
R R
3
68. (C) Distance of second satellite from the centre of earth has become ( 4 ) 2 or 8 items

Hence, T’ = 8T = (8)(90) = 720 min

69. (D)

From conservation of mechanical energy

1   Gm 2 Gm 2 
3  mv 2  = 3  − 
2   2R d 
1 2
v 2 = Gm  − 
R d 
1 2
v = Gm  − 
R d 
GMm
70. (B) Potential energy at a height of 2R from surface of earth = −
3r
GMm
and total energy of a satellite at a height h = −
2 ( R + h)
GMm GMm
Given that − =−
3R 2 ( R + h)
R
 h=
2
2
1 gm Re M m Re2 Rm3 Rm
71. (A) = = = =
6 ge Re2 M m Rm2 Re3 Re
72. (D) Applying Newton’s second law to the circular orbit, we have
mv2 4 2 rm GMm
= = 2
r T2 r
Where m is the mass of satellite, u its orbital speed and T is its period.
3
2 r 2
T=
GM

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SCO 101, Phase-1, Urban Estate, Dugri, Near LIC Building, Ludhiana
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 For rR
4 3
and M =  R  (  = density of planet )
3
3
We see that T =
G
i.e. T is independent of R.
73. (C) T 2  r3
3
or T  r 2
3
−
or   r 2
(T = 2 /  )
−2

 is halved. Therefore, distance will become   or ( 2 ) 3

2
1 3
2
74. (AD) At two positions, when the planet is closet to the sun (perigee) and when it is farthest from the sum
(apogee), velocity vector is perpendicular to force vector i.e. work done is zero. In one complete
revolution work done is zero.
75. (ACD) The gravitational field is zero at the centre of a solid sphere. The small spheres can be considered as
negative mass m located at A and B. The gravitational field due to these masses at O is equal and
opposite, Hence the resultant field at O is zero, Therefore, option (a) is correct. Options (c) and (d) are
correct because plane of these circles is y-z i.e. potential at any point on these two circles will be
equal due to the positive mass M and negative mass-m and –m.
GM
76. (AB) For r < R; F = .r or F r
R3
F1 r1
 = for r1  r and r2  R
F2 r2
GM 1
and for r  R F = 2 or F  2
r r
2
F1 r2
= for r1  r and r2  R
F2 r12
1  2 4 2 3 
77. (AC) Time period T   T = .a 
G GM 2 
 GMm 
and kinetic energy K  G  K = 
 2a 
 Time period will increase while kinetic energy will decrease.
78. (AD) Time period in both the cases comes out to be
R3
T1 = T2 = 2  84.2
GM
But v1  v2 . Because the difference in potential energy between the extreme position and mean
position will be more in the first case.
79. (ABCD) Initially potential and kinetic both energies are zero. Thus form conservation of mechanical energy
total energy of the two objects at separation r will be zero. Therefore, option (A) is correct.
Further decrease in gravitational potential energy = increase in kinetic energy
G ( m )( 4m ) 1 2
or = vr ….(1)
r 2
( m )( 4m ) = 4m
Here  = reduced mass =
m)4m 5
Substituting in equation (1), we get
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SCO 101, Phase-1, Urban Estate, Dugri, Near LIC Building, Ludhiana
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 vr =relative velocity of approach
10Gm
=
r
Option (B) is also correct
G ( m )( 4m ) 4Gm2
From equation (1) total kinetic energy = =
r r
 option (C) is also correct. Net torque of two equal and opposite forces acting on two objects is zero.
Therefore, angular momentum will remain conserved. Initially both the objects were stationary i.e.
angular momentum about any point was zero. Hence angular momentum of both the particles about
any point will be zero at all instants. Therefore, option (D) is also correct.
80. (ABD) E-r and V-r graphs for a spherical shell and a solid sphere are as follows:

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SCO 101, Phase-1, Urban Estate, Dugri, Near LIC Building, Ludhiana
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