Download NEET UG Physics Gravitational MCQs

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Unit - 6
Gravitational
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SUMMARY
1.
Gravitational force between two point masses is
F
2.
Gm1 m2
r2
Acceleration due to Gravity
(I)
(II)
GM
 9.81ms 2
R2
At a height h from surface of earth
on the surface of earth =
g
g1 
h

1  R 


(III)
2
 g (1 
2h
) if h << R
R
At a depth d form the surfce of earth
d
)
R
g1 = g if d = R i. e. on the surface of earth
(IV) Effect of rotation of earth at latitude 
g1 = g – R2 cos2 
- at the equator  = 0
g1 = g – R2 = minimum value
- At the pole  = 900
g1 = g – R2 = maximum value
- At the equator effect of rotation of earth is maximum and value of g is minimum.
- At the pole effect of rotation of earth is zero and value of g is maximum.
Field Strength
 Gravitational field strength at a point in gravirtational field is defined as,
g 1  g /(1 
3.

F
= gravitational force per unit mass
m
 Due to point mass

E
GM
1
(towards the mass) E  2
2
r
r
 Due to solid sphere
E
GM
r
R3
At r = 0, E = 0 at the center
inside points Ei 
At r = R, E 
GM
i.e. on the surface
R2
out side points Eo =
GM
1
or E o  2
2
r
r
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At r   , E  0
 Due to a sphericell shell inside points E=0
GM
r2
outside points E0 =
GM
R2
on the surface E – r graph is discontinuous
 on the axis of a ring
just outside surface E 
Er 
GMr
(R2  r 2 )
3
2
At r = 0, E = 0 i.e. at the center
If r >> R, E 
4.
GM
, i.e. ring behaves as a points mass
r2
As r   E  0
Gravitational potential :(i) Gravitational potential at a point in a gravitational field is defined as the negative of work
done in moving a unit mass from infinity to that point per unit mass, thus
Vp  
w  wp
m
(ii) Due to point mass
V 
Gm
rm
v   as r  0 and v  0 as r  
(iii) Due to solid sphere
 inside points Vi  –
GM
(1.5 R2  0.5r 2 )
R
GM
i.e. on the surface
R
V - r graph is parabola for inside points
as r = R V  
GM
r
Due to sphercal shell
 out side points v  
(iv)
inside points Vi  
GM
R
outside points Vi  
GM
R
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(v)
on the axis of a ring
Vr  
GM
R2  r 2
GM
i.e. at center
R
Gravitational potential Energy
(i) This is the work done by gravitational forces in arranging the system from infinite sepration
in the present position
(ii) Gravitational potential energy of two point massess is
at r = 0, V  
5.
Gm1m2
r
(iii)
To find the gravitational points energy of more than two points masses we have to make
pairs of masses. Neighter of the pair should be repeated. For example in case of four point
masses.
U 
 m4 m3 m4 m2 m4 m1 m3 m2 m3 m1 m2 m1 
U G 






r23
r41
r32
r31
r21
 r43

for n point masses total number of pairs will be
(iv)
n( n  1)
2
If a point mass m is plaled on the surface of earth the potential energy here is Uo
Uo  
GMm
R
and potential energy at height h is
Uh  
GMm
(R  h)
the difference in potential energy would be
mgh
U = Uh – Uo =
1 + h/r
If h << R, U = mgh
6.

Relation between field strength E
and potential V
(i) if V is a function of only one variable (Say r) then
dV
E
  slope of U  r graph
dr
(ii)
If V is funtion at three coordinates variable; e x, y, and z then

 v v v 
E    iˆ  ˆj  kˆ 
 x y z 
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7.
Escape velocity
(i)
From the surface of earth
Ve  2 gR 
8.
2GM
GM
as g  2
R
R
=11.2 km / sec
(ii) Escape velocity does not depend upon the angle which particle is projected form the surface
and the mass of body
Motion of satellites
GM
r
(i) orbital speed Vo 
(ii) time period T 
2
r 3/ 2 T 2  r 3
GM
(iii) Kinetic energy K 
GMm
2r
(iv) Potential energy U   
GMm
...
r
(v) Total Mechanical energy. E  
9.
GMm
r
Kepler’s laws
- First law :
Each planet moves in an elliptical orbit with the sun at one focus of ellipse
- Second law :
The radius vectors drawn form the sun to a planet, sweeps out equal area in equal time interval
i.e. areal Velocity is constant.
this law is derived from the law of conservation of angular momentum
dA
L

dt
2m = constant here L is the angular momentum and m is mass of planet
- Third law
-
-
T 2  r3
where r is semi-major axis of elliptical path
The gravitational force acting between two bodies is always attractive. It is independent of medium
between bodies. It holds good over a wide range of distance. It is an action and reaction pair.
It is conservative force. It is a central force and obey inverse square law as F  1/ r 2
The value of G is never zero any where but the value of g is zero at the center of earth.
the acceleration due to gravity is independent of mass, shape, size etc of falling body.
the rate of decrease of the acceteration due to gravity with height is twice as compared to that
with depth.
It the rate of rotation of earth increases the value of acceleration due to gravity decreases at
all points on the surface of earth except at poles.
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-
If the radius of planet decreasees by n% keeping its mas unchanged, the accelreotion due to
gravity on its surface increases by 2n%.
If the mass of a planet increases by n% keeping its radius unchanged the acceleration due to
gravity on its surface increases by n%.
The value of g at a location gives the value of intensity of gravitational field at the location.
The orbital velocity of a satellite is independent of mass of the satellite but depends upon the
mass and radius of planet around which the rotation is taking place
The value of orbital velocity for a satellite near the surface of earth is 7.92 kms–1.
The direction of orbital velocity of satellite at an instant is along the tangent to the orbital path
at that instant.
The work done by a satellite in a complete orbit is zero.
For a satellite orbiting close to the surface of earth (h << R), the time period of revolution
2
 g / R  0.001237rad / sec
T
A geostationary satellite revolves around the earth from west to east. Its period of revoluton
is one day i.e. 24 hours. The orbital velocity of geostationary satellite is 3.08 kms–1. Its height
above the surface of earth is about 36000 km. The relative angular velocity of geostationary
satellite w.r.t earth is zero.
When a satelilte is orbiting in its orbit, no energy is required to keep it in its orbit.
When the total energy of a satellite is negative, it will be moving in either a circular or an elliptical
orbit.
When the total energy of a satellite is zero, it will escape away from its orbit and its path becomes
parabolic.
When the height of satellites is increased its potential energy will increase and K.E. will decrease.
When the velocity of satellite is increased, its total energy will increase and it will start orbiting
in a circular path of larger radius.
For a satellite orbiting in a circular orbit, the value of potenial energy is always greater than
its K.E.
If the velocity of a satellite orbiting the earth is increased by 41.4% or its K.E. is doubled,
then it will escape away from the gravitational field of earth
is 2 R / g  84.6 minutes and angular velocity.  
-
-
2  orbital velocity
If the gravitational force is inversely proportional to the nth power of distance r, then the orbital
velocity of a satellite
Escape velocity =.
V0  r
-
n/ 2
and time period is T  r
( n  2)
2
When a body is projected horizontally with velocity v, from any height from the surface of earth,
then the following possibilities are there.
(i) If v < v0, the body fails to revolve around the earth and finally falls to the surface of earth.
(ii) If v = v0, the body will revolve around the earth in circular orbit.
(iii) If v < ve the body will revolve around the earth in elliptical orbit.
(iv) If v = ve, the body will escape from the gravitational field of earth.
(v) If v > ve the body will escape, following a hyperbolic path.
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-
1
th of its present size without any change in mass, the duration
n
of the day will be nearly 24/n2 hours.
Force function F(r) is related with potential energy function U(r) by a relaion
-
dU
dr
A given planet will have atmosphere if the root mean square velocity of molecules in its atmosphere
-
If earth suddently contracts to
F 
(i.e. Vrms =
-
In the weightlessness state, the bodies donot have weight but they do possess inertia on account
of their mass. the bodies floating inside the space craft may collide with each other and crash.
If a body is released from a height aqual to n times the radius of earth, then its striking velocity
on the surface of earth is

-
3RT
) is smaller than escape velocity for that planet.
M
2ngR
n 1
If polar ice caps melt then moment of inertia, Angular velocity will decrease and period of rotaiton
of earth increase.
The, line joining the places on earth having same value of g are called isogams.
Gravity meter and Etvos gravity balance are used to measure changes in accelaration due to
gravity.
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Newton’s Law of Gravitation
For the answer of the following questions choose the correct alternative from among the
given ones.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Two identical solid copper spheres of radius R are placed in contact with each other. The
gravitational force between them is proportional to
(A) R2
(B) R-2
(C) R-4
(D) R4
The gravitational force Fg between two objects does not depend on
(A) sum of the masses
(B) product of masses
(C) Gravitational constant
(D) Distance between the masses
The atmosphere is held to the earth by
(A) clouds
(B) Gravity
(C) Winds
(D) None of the above
Two sphere of mass m1 and m2 are situated in air and the gravitational force between them
is F. The space around the masses is now filled with liquid of specific gravity 3. The gravitational
force will now be
(A) F
(B) 3F
(C) F/3
(D) F/9.
A satellite of the earth is revolveing in a circular orbit with a uniform speed v. If the gravitational
force suddennly disappears, the satellite will
(A) Contineue to move with velocity v along the original orbit.
(B) Move with a Velocity v, tangentially to the original orbit.
(C) Fall down with increasing velocity.
(D) Ultimately come to rest somewhere on the original orbit.
Correct form of gravitational law is



Gm1m 2
Gm1m 2
Gm1m 2
Gm1m 2 
ˆ
F
=
–
F
=
–
r
F
=–
r
(A) F = –
(B)
(C)
(D)
2
2
3
r
r
r
r3
Mass M is divided into two parts xM and (1 - x) M. For a given separation, the value of
x for which the gravitational force between the two pieces becomes maximum is
(A) 1
(B) 2
(D) 1/2
(D) 4/5
24
The earth (mass = 6 x 10 kg) revolves around the sun with angular velocity 2  10–7 rad/
sec in a circular orbit of radius 1.5 x 108 km. The force exerted by the sun on the earth is
= ................... N
(A) 18  1025
(b) zero
(C) 27  1039 (D) 36  1021
Two particle of equal mass go round a circle of radius r. Under the action of their mutual
gravitational force.
The speed of each particle is =..................
(A) 
10.
11.
1
1
2r Gm
(B)  
Gm
2r
(C)  
1 Gm
2 r
(D) 
4Gm
r
The distance of the moon and earth is D the mass of earth is 81 times the mass of moon.
At what distance from the center of the earth, the gravitational force will be zero
(A) D/2
(B) 12D/3
(C) 4D/3
(D) 9D/10
One can easily “ Weight the earth ” by calculating the mass of earth using the formula
(in usual notation)
(A)
g
Re
G
(B) g/G Re2
(C)
G 2
Re
g
(D)
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G 3
Re
g
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12.
Three equal masses of m kg each are plced the vertices of an equilateral triangle PQR and
a mass of 2m kg is placed at the centroid 0 of the triangle which is at a distance of 2 m
from each of vertices of triangle. The force in newton. acting on the mass 2m is = ............
13.
14.
15.
(A) 2
(B) 1
(C) 2
(D) zero
Which of the follwing 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 unit
(D) It depends on the value of the masses.
Two point masses A and B having masses in the ratio 4 : 3 are seprated by a distance of
lm. When another point mass c of mass M is placed in between A and B the forces A and
C is 1/3rd of the force between band C, Then the distance C form A is = ............... m
(A) 23
(B) 1/3
(C) 1/4
(D) 2/7
The gravitational force between two point masses m1 ans m2 at separation r is given by
mm
F  G 1 2 2 The constant k ...............
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)
Acceleration Due to Gravity
16.
As we go from the equator to the poles, the value of g ...............
(A ) Remains constant (B) Decreases (C) I ncreases (D) Decreases upro latitude of 45
17.
If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface,
the mean density of the earth is = ...............
(A) 4 G / 3gR
18.
19.
20.
22.
(B) 3 R / 4 gG
(C) 3g / 4 RG
(D)  RG / 12 g
g=10ms-2.
The radius of the earth is 6400 km and
In order that a body of 5 kg weights zero
at the equator, the angular speed of the earth is = ............... rad/sec
(A) 1/80
(B) 1/400
(C) 1/800
(D) 1/600
The time period of a simple pendulum on a freely moving artificial satellite is ............... sec
(A) 0
(B) 2
(C) 3
(D) Infinite
A spherical planet far out in space has mas Mo and diameter Do. A particle of m falling near
the surface of this planet will experience an acceleration due to gravity which is equal to
(A) GM 0 / Do 2
21.
o
(B) 4mGMo / Do 2
(C) 4GMo / Do 2
(D) GmMo / Do 2
A body weights 700 g wt on the surface of earth How much it weight on the surface of planet
whose mass is 1/7 and radius is half that of the earth
(A) 200g wt
(B) 1400g wt
(C) 50 g wt
(D) 300g wt.
2
The value of g on he earth surface is 980 cm/sec . Its value at a height of 64 km from the
earth surface is ............... cm5–2
(a) 960.40
(B) 984.90
(C) 982.45
(D) 977.55
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23.
24.
25.
If earth rotates faster than its present speed the weight of an object will.
(A) increases at the equator but remain unchanged of the poles.
(B) Decreases at the equator but remain unchanged at poles.
(C) Remain unchanged at the equator but decreases at poles.
(D) Remain unchanged at the equator but increases at the poles.
The moon’s radius is 1/4 that of earth and its mass is 1/80 times that of the earth. If g represents
the acceleration due to gravity on the surface of earth, that on the surface of the moon is
..............
(A) g/4
(B) g/5
(c) g/6
(D) g/8
The depth of at which the value of acceleration due to gravity becomes 1/n the time the value
of at the surface is (R = radius of earth)
26.
27.
28.
29.
30.
31.
32.
33.
34.
( n  1)
n
 n 
(D) R 

 n 1
If the density of smalll planet is that of the same as that of the earth while the radius of the
planet is 0.2 times that of lthe earth, the gravitational acceleration on the surface of the planet
is ...............
(A) 0.2g
(B) 0.4g
(c) 2g
(D) 4g
If mass of a body is M on the earth surface, than the mass of the same body on the moon
surfae is
(A) M/6
(B) 56
(C) M
(D) None of these
An object weights 72 N on the earth. Its weight at a height R/2 from earth is = .............. N
(A) 32
(B) 56
(C) 72
(D) zero
If the radius of earth is R then height ‘h’ at which value of ‘g’ becomes one - fourth is
(A) R/4
(B) 3R/4
(C) R
(D) R/8
If the mass of earth is 80 times of that of a planet and diameter is double that of planet and
‘g’ on the earth is 9.8 ms-2 , then the value of ‘g’ on that planet is = ............... ms-2
(A) 4.9
(B) 0.98
(c) 0.49
(D) 49
Assuming earth to be a sphere of a uniform density, what is value of gravitational acceleration
in mine 100 km below the earth surface = ............... ms-2
(A) 9.66
(B) 7.64
(C) 5.00
(D) 3.1
Let g be the acceleration due to gravity at earth’s surface and k be the rotational K.E. of earth
suppose the earth’s radius decreases by 2% keeping alt other quantities same then
(A) g decreases by 2% and K decreases by 4 %
(B) g decreases by 4% and K increases by 2%
(C) g increases by 4% and K increases by 4%
(D) g decreases by 4% and K increases by 4%
A body weight 500 N on the surface of the earth. How much would it weight half way below
the surface of earth
(A) 125N
(B) 1250N
(C) 500N
(D) 1000N
The radii of two planets are respectively R1 and R2 and their densities are respectively 1 and
2 the ratio of the accelerations due to gravity at their surface is ...............
(A) R/n
(B) R
(A) g1 : g 2 
1  2
. 2
2
R1 R2
(C) g1 : g2  R1 2 : R21
(C) R/n2
(B) g1 : g2  R1 R2 : 12
(D) g1 : g2  R1 1 : R22
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35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
At what height over the earth’s pole, the free fall acceleration decreases by one percent
= .................. km (Re = 6400 km).
(A) 32
(B) 80
(C) 1.253
(D) 64
Weight of a body is maximum at
(A) moon
(B) poles of earth
(C) Equator of earth (D) Center of earth
At what distance from the center of earth, the value of aceeleration due to gravity g will be
half that of the surfaces (R = Radius of earth)
(A) 2R
(B) R
(C) 1.414 R
(D) o.414 R
The acceleration due to gravity near the surface of a planet of radius R and density d is proportional
to
(A) d/R2
(B) dR2
(C) dR
(D) d/R
The acceleration due to gravity is g at a point distance r from the center of earth R. if r < R
then
(A) g  r
(B) g  r2
(c) g  r-2
(D) g  r-1
The density of a newly discoverd planet is twice that of earth. The acceleration due to gravity
at the surface of the planet is equal to that at the surface of earth. If the radius of the earth
is R, the radius of planet would be .....................
(A) 2R
(B) 4R
(C) 1/4 R
(D) .......
Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will
be .................. ms-2(g = 9.8 ms2)
(A) 19.6
(B) 9.8
(C) 4.9
(D) 2.45
Weight of body of mass m decreases by 1% when it is raised to height h above the earth’s
surface. If the body is taken to a depth h in a mine. change in its weight is
(A) 2% decreases
(B) 0.5% decreases
(C) 1% increases
(D) 0.5% increases
If density of earth increased 4 times and its radius becomes half of then out weight will be...
(A) Four times it present calue
(B) doubled
(C) Remain same
(D) halved
A man can jump to a height of 1.5 m on a planet A what is the height ne may be able to
jump on another planet whose density and radius are respectively one- quater and one- third
that of planet A
(A) 1.5 m
(B) 15 m
(C) 18 m
(D) 28 m
If the value of ‘g’ acceleration due to gravity, at earth surface fis 10ms–2. its value in ms–2
at the center of earth, which is assumed to be a sphere of Radius ‘R’ meter and uniform density is
(A) 5
(B) 10/R
(C) 10/2R
(D) zero
A research satellite of mass 200 kg. circles the earth in an orbit of avrage radius 3R/2 where
R is radius of earth. Assuming the gravitational pull 10 N, the pull on the satellite will be =..........N
(A) 880
(B) 889
(C) 890
(D) 892
Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the
e
ratio of densities of earth e and moon m is   5 3 then radius of moon Rm in terms
m
of Re will be ...............
(A)
5
Re
18
(B)
1
Re
6
(C)
3
Re
16
(D)
1
2 3
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Re
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48.
49.
50.
The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from
the surface of the earth is = ...............
(g = acceleration due to gravity at the surface of earth)
(A) g/9
(B) g/3
(C) g/4
(D) 9
The height at which the weight of a body becomes 1/16 th its weight on the surface of
(radius R) is
(A) 3R
(B) 4R
(C) 5R
(D) 15R
A spherical planet has a mass Mp and diameter Dp A particle of mass m falling freely near
the surface of this planet will experience an acceleration due to gravity, equal to
(A)
51.
4 GMp
Dp 2
(B)
GMpm
Dp 2
(C)
GMp
Dp 2
(D)
4GMpm
Dp 2
Assuming the earth to have a constant density, point out which of following curves show the
variation acceleration due to gravity from center of earth to points far away from the surface
of earth ...............
(D) None of these
Gravitational potential, Energy and Escape Velocity
52.
53.
54.
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 definetely, about the gravitational field
(D) None of these
The mass of the earth is 6.00  1024 kg and that of the moon is 7.40  1022 kg. The constant
of gravitation G = 6.67  10-11 Nm2 kg–2. The potential energy of the system is -7.79  1028
Joules. the mean distance between the earth and moon is = ............. meter.
(A) 3.80  108
(B) 3.37  108
(C) 7.60  108
(D) 1.90  102
The masses and radii of earth and moon are M1, R1 and M2, R2 respectively. Their centres
are d distance of apart. The minimum velocity with which a particle of mass m should be projected
from a point midway between their centres so that it esacapes to infinity is...............
(A) 2
G
(M1  M 2 )
d
(B) 2
(C) 2
Gm
( M1  M 2 )
d
(D) 2
2G
(M1  M 2 )
d
Gm (M 1  M 2 )
d ( R1  R2 )
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55.
A rocket is launched with velocity 10 kms-1. If radius of earth is R then maximum height attained
56.
by it will be = ..............
(A) 2R
(B) 3R
(C) 4R
(D) 5R
What is the intensity of gravitational field at the center of spherical shell
Gm
(B) g
(C) zero
(D) None of these
r2
Escape velocity of a body of 1 kg. on a planet is 100 ms-1. Gravitational potential energy of
the body at the planet is =............... J
(A) -5000
(B) -1000
(C) -2400
(D) 5000
A body of mass m kg starts falling from a point 2R above the earth’s surface. Its K.E. 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)
57.
58.
1 GMm
1 GMm
2 GMm
1 GMm
(B)
(C)
(D)
2 R
6 R
3 R
3 R
The Gravitational P.E. of a body of mas m at the earth’s surface is -mgRe. Its gravitational
potential energy at a height Re from the earth’s surface will be = ............ here
(Re is the radius of the earth)
(A)
59.
61.
1
1
mg Re
(D)  mg Re
2
2
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
Energy required to move a body of mass m from an orbit of radius 2R to 3R is ................
62.
GMm
G m
G m
G m
(B)
(C)
(D)
2
2
12R
3R
8R
6R
Radius of orbit of satellite of earth is R. Its K.E. is proportional to
(A)
60.
–2 mgRe
(B) 2 mgRe
(C)
(A)
1
1
1
(B)
(C) R
(D)
R
R
R 3 /2
A particle falls towards earth from infinity. It’s velocity reaching the earth would be.............
(A)
63.
(A) infinity
64.
66.
(C) 2 gR
2gR
(D) zero
The escape velocity of a sphere of mass m from earth having mass M and Radius R is given by
(A)
65.
(B)
2GM
R
(B) 2
GM
R
(C)
2GMm
R
(D)
GM
R
The escape velocity for a rocket from earth is 11.2 kms–1 value on a planet where acceleration
due to gravity is double that on earth and diameter of the planet is twice that of earth will be
= ........... kms–1
(A) 11.2
(B) 22.4
(C) 5.6
(D) 53.6
–1
The escape velocity from the earth is about 11 kms . The escape velocity from a planet having
twice the radius and the same mean density as the earth is =............ kms–1.
(A) 22
(B) 11
(C) 5.5
(C)15.5
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67.
If g is the acceleration due to gravity at the earth’s surface and r is the radius of the earth,
the escape velocity for the body to escape out of earth’s gravitational field is.................
(A) gr
68.
69.
(B)
(D)  
(B) 
GM
r
(C) 
2GM
r
(D) 
Ve
2
(B)
Ve
2
(C)
Ve
2 2
(D)
4GM
r
Ve
4
(B) 12800
(C) 3200
(D) 1600
The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is..........
3 Ve
(B) 3 Ve
(C)
2 Ve
(D) 2Ve
There are two planets, the ratio of radius of two planets is k but the acceleration due to gravity
of both planets are g what will be the ratio of their escape velocity.
1/ 2
 kg 
(B)  kg 
1/ 2
(C)
 kg 
2
(D)
 kg 
2
The escape velocity of a body on the surface of the earth is 11.2 km/sec. If the mass of the
earth is increases to twice its present value and the radius of the earth becomes half, the escape
velocity becomes = ............... kms–1
(A) 5.6
77.
2GM
R2
An artificial satellite is revolving round the earth in a circular orbit. its velocity is half the escape
velocity. Its height from the earth surface is = ............... km
(A)
76.
(C)   2GMR
The escape velocity of a body from earth’s surface is Ve. The escape velocity of the same
body from a height equal to 7 R from earth’s surface will be
(A)
75.
8
8
GR
G  (B)   M
3
3
Two small and heavy sphere, each of mass M, are placed distance r apart on a horizontal surface
the gravitational potential at a mid point on the line joining the center of spheres is
(A) 6400
74.
(D) m-1
The escape velocity of an object from the earth depends upon the mass of earth (M), its mean
density (), its radius (R) and gravitational constant (G), thus the formula for escape veloctiy is
(A)
73.
(C) m0
(B) m
(A) zero
72.
(D) 1120 kms–1
The escape velocity of a particle of mass m varies as...................
(A)   R
71.
(D) r/g
The escape velocity of a projectile from the earth is approximately
(A) 11.2 kms–1
(B) 112 kms–1
(C) 11.2 ms–1
(A) m2
70.
(C) g/r
2gr
(B) 11.2
(C) 22.4
(D) 494.8
1
1
Given mass of the moon is
of the mass of the earth and corresponding radius is
of
81
4
–1
the earth, If escape velocity on the earth surface is 11.2 kms the value of same on the surface
of moon is = ....... kms–1.
(A) 0.14
(B) 0.5
(C) 12.5
(D) 5
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78.
3 particle each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational
field at center due to these particies is ................
(A) zero
79.
(B)
(C)
9Gm
L2
12 Gm
(D) 3 L2
Escape velocity on the surface of earth is 11.2 kms–1 Escape velocity from a planet whose masses
the same as that of earth and radius 1/4 that of earth is = ....... kms–1
(A) 2.8
80.
3GM
L2
(B) 15.6
(C) 22.4
(D) 44.8
The velocity with which a projectile must be fired so that it escapes earth’s gravitational does
not depend on ..................
(A) mass of earth
(B) Mass of the projectile
(C) Radius of the projectiles’s orbit
(D) Gravitational constant
81.
The escape velocity for a body projected vertically upwards from the surface of earth is 11
kms-1. If the body is projected at an angle of 450 with the vertical, the escape velocity will
be ............kms-1.
(A)
82.
11
2
(B) 11 2
84.
(B) 2 Ve
(D) Ve/2
(A) 6.67  10-9 J
(B) 6.67  10-10 J
(C) 13.34  10-10 J (D) 3.33  10-10 J
A particle of mass M is situated at the center of a spherical shell of same mass and radius
a the magnitude of gravitational potential at a point situated at a / 2 distance from the center
will be
4GM
a
(B)
GM
a
(C)
2GM
a
(D)
3GM
a
The mass and radius of the sun are 1.99 x 1030 kg and R = 6.96 x 108 m. The escape
velocity of rocket from the sun is =.......... km/sec
(A) 11.2
86.
(C) 4 Ve
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radiius
10 cm. Find the work to be done aginst the gravitational force between them to take the particle
is away from the sphere (G = 6.67  10-11 SI unit)
(A)
85.
(D) 11
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 Ve on the earth
(A) Ve
83.
(C) 22
(B) 12.38
(C) 59.5
(D) 618
The mass of a space ship is 1000 kg. It is to be lauched from earth’s surface out into free
space the value of g and R (radius of earth) are 10ms-2 and 6400 km respectively the required
energy for this work will be = ................. J
(A) 6.4  1011
(B) 6.4  108
(C) 6.4  109
(D) 6.4  1010
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87.
88.
The diagram showing th variation of gravitational potential of earth with distance from the center
of earth is
(A)
(B)
(C)
(D)
A shpere of mass M and Radius R2 has a concentric cavity of Radius R1as shown in figure.
The force F exerted by the shpere on a particle of mass m located at a distance r from the
center of shhere varies as (O  r   )
(A)
(B)
(C)
(D)
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89.
90.
91.
Which one of the following graphs represents correctly the variation of the gravitational field
with the distance (r) from the center of spherical shell of mass M and radius a
(A)
(B)
(C)
(D)
Which of the following graphs represents the motion of a planet moving about the sun.
(A)
(B)
(C)
(D)
The curves for P.E. (U) K.E. (Ek) of two particle system ae shown in figure . At what points
systemwill be bound.
(A) Only at point D
Ek
(B) Only at point A
(C) At point D and A
(D) At points A, B and C
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Energy
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...........
Energy
92.
Energy
(B)
Energy
(A)
(C)
93.
94.
(D)
A shell of mass M and radius R has a point mass m placed at a distance r from its center.
The gravitational potential energy U(r) –v will be
(A)
(B)
(C)
(D)
Motion of satellite
If Ve and Vo are represent the escape velocity and orbital velocity of satelllite correspoinding
to a circular orbit of radius r, then
(A) Ve = Vo
(B) 2 Vo  Ve
(C) Ve  Vo / 2 (D)ve and vo are not related
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95.
If r represents the radius of the orbit of a saltellite of mass m moving around a planet of mas
M, the velocity of the satellite is given by
(A)  2 
96.
gM
r
99.
100.
101
102
103.
(C)  
GM
r
(D)  2 
GM
r
(B) V1 < V2
V1 V2
(D) r  r
1
2
(C) V1 > V2
A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from
the centere of earth in new orbit is two times of the earlier orbit. The time period in second
orbit is .........hours.
(a) 4.8
98.
GMm
r
Two satellites of mass m1 and m2 (m1 > m2 ) are revolving round the earth in cirular orbits
of r1 and r2 ( r1 > r2) respectively. Which of the following statement is true regarding their
speeds V1 and V2
(A) V1 = V2
97.
(B)  2 
(C) 24
(B) 48 2
(D) 24 2
As astronaut orbiting the earth in a circular orbit 120 km above the surface of earth,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 sapace - ship
(D) Will move in an irregualr way then fall down to earth
The period of a satellite in cirular orbit around a planet is independent of
(A) the mass of the planet
(B) the radius of the planet
(C) mass of the satellite
(D) all the three parameters (A), (B) and (C)
Two satellites A and B go round a planet p in circular orbits having radii 4 R and R respectively
if the speed of the satellite A is 3V, the speed if satellite B will be
(A) 12V
(B) 6V
(C) 4/3 V
(D) 3/2 V
A small satellite is revolveing near earth’s surface. Its orbital velocity will be nearly
= .......... kms–1.
(A) 8
(B) 4
(C) 6
(D) 11.2
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 to the earth
(D) goes on increasing or decresing continuously
depending upon the mass of the satellite
If the height of a satellite from the earth is negligible in comparison of the radius of the earth
R, the orbital velocity of the stellite is ............
(A) gR
(B) gR/2
(C)
g/R
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(D)
gR
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104
105.
106.
107.
108.
109.
110.
111.
A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit
radius is decreasd by 1% its speed will
(A) increase by 1%
(B) increase by 0.5%
(C) decrrease by 1%
(C) Decrrease by 0.5%
orbital velocity of an artifiicial satellite does not depend upon
(A) mass of earth
(B) mass of satellite
(C) radius of earth
(D) acceleration due to gravity
orbital velocity of eath’s satellite near the surface is 7 kms–1. when the radius of orbit is 4 times
that of earth’s radius, then orbital velocity in that orbit is =........kms-1
(A) 3.5
(B) 17
(C) 14
(D) 35
Two identical satellites are at R and 7R away from each surface, the wrong statement is
(R - Radius of earth)
(A) ratio of total energy will be 4
(B) ratio of kinetic energes will be 4
(C) ratio of potential energies will be 4
(D) ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2
Which one of follwoing sttements regarding artificial satellite of earth is incorrect
(A) The orbital velocity depends on the mass of the satellite
(B) A minimum velocity of 8kms-1 is required by a satellite to orbit quite close to the earth.
(C) The period of revolution is large if the radius of its orbit is large
(D) The height of geostationary satellite is about 36000 km from earth
The weight of an astronuat, in an artificial satellite revolving around the earth is
(A) zero
(B) Equal to that on the earth
(C) more than that on earth
(D) less than that on the earth
The distance of a geo-stationary satellite from the center of the earth (Radius R= 6400km)
is nearest to
(A) 5R
(B) 7R
(C) 10R
(D) 18R
A geo-stationary satellite is orbiting the earth of a height of 6R above the surface of earth, R
being the radius of earth. The time period of another satellite at a height of 2.5 R from the
surface fo earth is = ...............hr
(A) 6
112.
(B) 6 2
(C) 10
(D) 6 / 2
1
1
(and not as 2 )where
R
R
R is separation between them, then a particle in circular orbit under such a force would have
its orbital speed v proportional to
If the gravitational force between two objects were proportional to
1
1
(B) R0
(C) R1
(D)
2
R
R
A satellite moves around the earth in a circular orbit of radius r with speed v, If mass of the
satellite is M , its total energy is
(A)
113.
(A) 
1
MV 2
2
(B)
1
MV 2
2
(C)
3
MV 2
2
(D) MV2
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114.
A satellite with K.E. Ek is revolving round the earth in a circular orbit. How much more K.E.
should be given to it so that it may just escape into outerspace ?
(A) Ek
115.
116.
117.
118.
119.
120.
121.
(B) 2 Ek
(D) 3 Ek
Potential energy of a satellite having mass m and rotating at a height of 6.4 x 106 m from
the surface of earth
(A) -0.5 mg Re
(B) -mg Re
(C) -2mgRe
(D) 4 mgRe
When a satellite going round the earth in a circular obrit of radius r and speed v loses some
of its energy, then r and v changes as
(A) r and v bothe will increase
(B) r and v both will decease
(C) r will decrease and v will increase
(D) r will increase and v will decrease
The time period of a satellite of earth is 5 hours If the sepration between the earth and the
satellite is increased to four times the previous value, the new time period will become ...... hours
(A) 10
(B) 120
(C) 40
(D) 80
A person sitting in a chair in a satellite feels weightless because
(A) the earth does not attract the objects in a satellite
(B) the normal force by the chair on the person balances the earth’s attraction
(C) the normal force is zero
(D) the person in satellite is not accelerated
Two satellites A and B go round a planet in cirular orbits having radii 4R and R respectively
If the speed of satellite A is 3v, then speed of satellite B is
(A) 3v/2
(B) 4v/2
(C) 6v
(D) 12v
A satellite moves in a circle around the earth, the radius of this circlr is equal to one half of
the radius of the moon’s orbit the satellite completes one revolution in .........lunar month
(A) 1/2
(B) 2/3
(C) 2–3/2
(D) 123/2
The additional K.E. to be provided to a satellite of mass m revolving around a planet of mass
M, to transfer it from a circular orbit of radius R1 to another radius R2 (R2 > R1) is
 1
1 

2
2 
R2 
 R1
(A) G M m 
 1
1 


R2 
 R1
(C) 2 G M m 
122.
1
Ek
2
(C)
 1
1 


R2 
 R1
(B) G M m 
(D)
 1
1
1 
GMm 


2
R2 
 R1
Rockets are launched in eastward direction to take advantage of
(A) the clear sky on eastem side
(B) the thiner atmosphere on this side
(C) earth’s rotation
(D) earth’s tilt
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123.
124.
A satellite of mass m is orbiting close to the surface of the earth (Radius R = 6400 km) has
a K.E. K. The corresponding K.E. of satellite to escape from the earth’s gravitational field is
(A) k
(B) 2 k
(C) mg R
(D) m k
A planet moving along an elliptical orbit is closest to the sun at a distance r 1 and farthest away
at a distance of r2. If v1 and v2 are the liner velocities at these points respectively, then the
v1
ratio v is ............
2
r1
(A) r
2
125.
 r1 
(B)  
 r2 
2
r2
(C) r
1
 r1 
(D)  
 r2 
2
The time period T of the moon of planet Mars(Mm) is related to its orbital radius R as
(G = Gravitational constant)
4π 2GR 2
(B) T 
Mm
4 R 2
(A) T 
GMm
2
2
2 R 2G
(d) T2 = 4 Mm GR2
Mm
A geostationary satellite is orbiting the earth at a height of 5 R above that of surface of the
earth. R being the radius of the earth. The time period of another satellite in hours at a height
of 2R from the surface of earth is ............ hr
(C) T 2 
126.
127.
128.
129.
130.
(A) 5
(B) 10
(C) 6 2
(D) 6 / 2
The figure shows ellipticall orbit of a planet m about the sun s. the shaded area SCD is twice
the saded area SAB. If t1 is the time for the planet to move from C and D and t2 is the time
to move from A to B then
(A) t1 > t2
(B) t1 > 4t2
(C) t1 = 2t2
(D) t1 = t2
The period of a satellite in a circular orbit of radius R is T. the period of another satellite
in a circular orbit of radius 4R is
(A) 4T
(B) T/4
(C) 8T
(D) T/8
If the earth is at one- fourth of its present distance from the sun the duration of year will be
(A) half the pesent Year
(B) one-eight the present year
(C) one-fourth the present year
(D) one-sixth the present year
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
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131.
132.
133.
134.
Kepler’s second law regarding constancy of aerial velocity of a palnet is consequence of the
law of conservation of
(A) energy
(B) angular Momentum
(C) linear momentum
(D) None of these
The largest and shortest distance of the earth from the sun are r1 and r2 its distance from
the sun when it is at the perpendicular to the major axis of the orbit drawn from the sun
r1 r2
2r1 r2
r1  r2
r1  r2
(B) r  r
(A)
(C) r  r
(D)
4
3
1
2
1
2
According to keplar, the period of revolution of a planet (T) and its mean distance from the
sun (r) are related by the equation
(A) T 3 r 3  cons tan t
(B) T 2 r 3  cons tan t
(C) Tr 3  cons tan t
(D) T 2 r  cons tan t
A satellite of mass m is circulating around the earth with constant angular velocity. If radius of
the orbit is Ro and mass of earth M , the angular momentum about the center of earth is
(A) m GMRo
(C) m
135.
136.
137.
138.
139
(B) M GMRo
GM
Ro
(D) M
GM
Ro
The earth E moves in an elliptical orbit with the sun s at one of the foci as shown in figure.
Its speed of motion will be maximum at a point ..........
(A) C
(B) A
(C) B
(D) D
The period of revolution of planet A around the sun is 8 times that of B. The distance of A
from the sun is how many times greater than that of B from the sun.
(A) 2
(B) 3
(C) 4
(D) 5
The earth revolves round the sun in one year. If distace between then becomes double the new
period will be .......... years.
(A) 0.5
(B) 2 2
(C) 4
(D) 8
The maximum and minimum distance of a comet from the sun are 8 x 1012m and 1.6 x 1012
m. If its velocity when nearest to the sun is 60 ms-1, What will be its velocity in ms-1 when
it is farthest ?
(A) 6
(B) 12
(C) 60
(D) 112
The period of moon’s rotation around th earth is nearly 29 days. If moon’s mass were 2 fold
its present value and all other things remained unchanged the period of moons’s rotation would
be nearly .....days
(A) 29 2
(B) 29 / 2
(C) 29  2
(D) 29
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140.
141.
142.
143.
144.
145.
If the velocity of planet is given by   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
The radius of orbit of a planet is two times that of earth. The time period of planet is.........years.
(A) 4.2
(B) 2.8
(C) 5.6
(D) 8.4
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) r3/2
(B) r
(C) r1/1
(D) r2
What does not change in the field of central force
(A) potential energy
(B) Kinetic energy
(C) linear momentum
(D) Angular momentum
A thin uniform annular disc (See figure) of mass M has outer
radius 4R and inner radius 3R. The work required to take a
unit mass from point P on its axix to infinity is..............
(A)
2GM
(4 2  5)
7R
(B) 
2GM
(4 2  5)
7R
(C)
GM
4R
(D)
2GM
( 2  1)
5R
Suppose the gravitational force varies inversely as the nth power of distance the time period
of planet in circular orbit of radius R around the sun will be proportional to
 n  1


 2 
146.
147.
Rn
n 2


2 
(D) R 
(A) R
(B) R
(C)
If the radius of the earth were to shrink by 1% its mass remaing the same, the accelration due
to gravity on the earth’s surface would
(A) decrease by 2%
(B) remain Unchanged
(C) increase by 2%
(D) increases by 1%
A body of mass m is taken from earth surface to the height h equal to radius of earth, the
increase in potential energy will be
(A) mg R
148.
 n 1


 2 
1
(B) 2 mgR
(C) 2 mgR
1
(D) 4 mgR
An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential lenergy)
Eo, its potential energy is
(A) -Eo
(B) 1.5 Eo
(C) 2 Eo
(D) Eo
170
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149.
Two bodies of masses m1 and m2 are initially at rest at infinite distance apart. They are then
allowed to move towards each other under mutual agravitational attraction Their relative velocity
of apporach at sepration distance r between them is
 2 G ( m1  m 2 ) 
(A) 

r

1

1
2
1

2
r
(C) 

 2 G m1 m 2 
150.
1
 2 G ( m1  m 2 )  2
(B) 

r

(D)
 2 G m1 m 2  2
r


A geostationary satellite orbits around the earth in a circular orbit of radus 3600 km the time
period of a satellite orbiting a few hundred kilometers above the earth’s surface (Rearth = 6400
km) will approximately be = ...... h
(A) 1/2
(B) 1
(C) 2
(D) 4
KEY NOTE
1.C
2.A
3. B
4. A
5. B
6. C
7. A
8. D
9. C
10. D
11. B
12. D
13. A
14. A
15. A
16. C
17. C
18. C
19. D
20. C
21. B
22. A
23. B
24. B
1.C
25. B
2 . A 25. B
26. A
3. B
27. C
26. A
4. A
28. A
5. B 27. C
29. C
6. C 28. A
30. C
7. A 29. C
31. A
8. D
32. C
9. C 30. C
33. B
10. D 31. A
34. D
11. B 32. C
35. A
12. D
26. B
13. A 33. B
27. C
14. A 34. D
38. C
15. A 35. A
39. A
16. C
40. D
17. C 26. B
41. A
18. C 27. C
42. B
19. D 38. C
43. B
20. C
44. C
21. B 39. A
45. D
40.
D
22. A
46. B
23. B 41. A
47. A
24. B
48. A
42. B
43. B
44. C
45. D
46. B
47. A
48. A
49. C
50.C
A
49.
51. C
50.
A
52. A
51.
53.C
A
54.A
A
52.
55.
53. AC
56. C
54.
57.A
A
58.CB
55.
59.CD
56.
60. A
57.
61.A
D
58.
62.B
A
63.
59. DB
64. A
60.
A
65.B
61.
66. D
A
67. A
B
62.
68. C
63.
69. BC
64.
70. A
A
71.
65.BD
72. C
66. A
67. B
68. C
69. C
70. A
71. D
72. C
73. A
74.73.
A A
75. A
74. A
76. C
77.75.
C A
78.76.
A C
79.77.
C C
80. B
81.78.
D A
82.79.
B C
83.80.
B B
84. D
85.81.
D D
86.82.
D B
87.83.
C B
88. B
89.84.
D D
90.85.
C D
91.86.
D D
92. C
93.87.
C C
94.88.
B B
95.89.
D D
96. B
90. C
91. D
92. C
93. C
94. B
95. D
96. B
97. B
121 D
98. C97. B122. C
99. C
98. C123 B
100. B
124. C
99.
C
101. A
125. A
102. B100.126.
B C
103. D
127.
101. A C
104. B
128. C
B B
105. B102.129.
106. A103.130.
D B
107. D
104.131.
B B
108. A
132. C
B B
109. A105.133.
110. B106.134.
A A
111. D107.135.
D B
112. B
136. C
A B
113. A108.137.
A A
114. A109.138.
115. A110.139.
B D
116. C
140. B
D C
117. D111.141.
B C
118. C112.142.
119. C113.143.
A D
120. C
144.A
114. A
115. A
116. C
117. D
118. C
119. C
120. C
145. A
146.DC
121
147. B
122.
C
148. C
123
149.BB
150.CC
124.
125. A
126. C
127. C
128. C
129. B
130. B
131. B
132. C
133. B
134. A
135. B
136. C
137. B
138. A
139. D
140. B
141. C
142. C
143. D
144.A
171
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145. A
146. C
147. B
148. C
149. B
150. C
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HINT
4
G (  R 3.  )2
4
3
  2  2 R4
2
4R
3
(1)
)
(c) F  G (m)(m

2
(2 R)
(4)
(5)
(6)
(A) Gravitational force does not depend upon the medium
(B) Due to inertia to direction
(D) the force exerted by sun on the earth
F = m2R
 (6  1024 ) (2  107 ) 2 (1.5  1011 )
 36  1021 N
(7)
F  xm ( x  1) m  m 2 x ( x  1)
for maximum force 
dF
0
dx
dF
 m2  2 m2 x  0  x  1/ 2
dx
(C) cenripetal force provided by the gravitational force of attraction between two particites

(9)

m 2 G ( m )( m )

r
(2 r ) 2
 
Gm
r
1
2
(D - X)
(10)
(D)
For will be zero at the point of zero intensity
x
11.
12.
m1
m1 
m2

81m
9
D
D
10
81m  m
B mg 
GM e m
where Me and Re is the mass and radius of the earth respectivel
Re 2
 Me 
g
Re 2
G
(D) Here FOA = FOB = FOC =
   
F  FOA  FOB  FOC
G (m) (2m)
r2
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=
14.
= 0
(A) let a point mass C is placed at a distance of x m from the point mass A as show in figure
mA
here
x
m
C
(1–x)
mB
ma 4

3
mb
G (m)(mA)
.........(i )
x2
G (m)(mB)
FBC 
.........(ii)
(1  x )2
FAC 
According to given problem FAC 
1
FBC
4
with the help of equ (i) and (ii)

G (m)(mA) 1 G (m)(mB )

x2
3 (1  x) 2

mA
x2
4
x2
x2




4

mB 3(1  x) 2
3 3(1  x) 2
(1  x) 2
2
x
 2  2x  x
1 x
 x  23 m
 3x  2
GM
4
and M   R 3 
2
R
3
17.
(c) g 
18.
G 4
3g
.  R3    
2
R 3
4 RG
(c) for condition of weight lessness at equator
g 
 g

19.
R
10
1

rad sec
3
6400  10
800
(D) Time peripd of simple pendylym T  2
l g'
In artificial satellite g’ = 0  T  
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20.
GM
GMo
4GMo
(C) g  R 2  ( Do 2) 2  Do 2
21.
(B) we know that g 
on the planet gp 
GM
R2
GM 7
4 7g
R2 4
Hence the weight on the planet = 700 
4
= 400 gm t
7
2
2
22.
g '  R   6400 
2

  g '  960.400 ms
 
g  ( R  h)   6400  64 
23.
g 1  g   2 R Cos 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 90’
24.
GM
(B) using g  R 2
25.
1
(B) g  g (1 
26.
4
(A) g  πGR
3

we get gm = g/5
d
(n 1)
)d 
R
R
n
and g'  4/3πGR'
g' R'

 0.2  g'  0.2g
g
R
27.
(C) mass does not vary from place to place
28.
 R  4
 R 
(a) g '  g 
  g
  g
 Rh
 R  R2  9
2
2
4
W'  4/9 w  (72)  32 N
9
2
29.
 R 
  9 / 4 by solving h  R
(c) g '  g 
 Rh
30.
 Mp   Re 
 1  2 9.8
 0.49ms 2
(c) g p  g e 

  9.8   (2) 
Me
Rp
80
20


 

31.
d
100 


2
(A) g'  g  1    9.8  1 
  9.66 ms
 R
 6400 
2
174
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32.
(C) g 
GM
2
and k = L 2I
2
R
1
1
k 2
2 and
R
R
i.e. if radius of earth decreases by 2% then g and k both increase by 4%
(B) Weight on surface of earth, mg = 500N and weight below the surface of earth at
If mass of the earth and its angular momentum ramains constant then g 
33.
d =
R
mg
 d
 250N
, mg1 = mg 1    mg 1  1/2  
2
2
 R
34.
4
(D) g  π GR
3
35.
(A) g α
g1
g2

R1 1
R 2 2
GM g α 1 or r α 1
r2
g
r2
If g decreases by one percent then r should be increase by 1/2 % i.e. R 
1
 6400  32 km
2  100
2
37.
1
R
 R 

 R  h  2 R  h 
 
(C) g '  g 
2 Rh
Rh


2  1 R  0.414 R
Hence distance form center = R + 0.414R
= 1.414R
4
π GR  g α d R (  d given in the problem)
3
38.
(C) g 
39.
(A) insided the earth g' 
40.
(D) g 
4
π Gr  g ' r
3
Rp  g p
4

π  GR 
Re  g e
3
 Rp 
41.
(A) g  
42.
(B) For height
Re R

2
2
Δg
24
100% 
100 %
g
4
For depth
43.
  e 
1
  (1)  
 

2
  p 
Δg
d h
100%    1/2  0.5
g
R R
(B) g  R
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44.
(C) H 
V2
1
H
g
 Hα
 B  A
2g
g
HA gB
Now g B 

gA
as g  R
12
HB g A

 12  HB  12 HA  12 1.5  18 m
HA gB
46.
2
 R 
 R 
2


(B) g'  g  R  h   g  3 R   4/9.g(g  10 ms )



2
47.
4
g e e Re
(A) g  3 πGR  g α R  g   . Rm
m
m

6 5 Re
 .
1 3 Rm
 Rm 
5
Re
18
48.
 R 
 R 
(A) g '  g 
  g
  g /9
 R  2R 
 R  2R 
49.
(C) g ' 
g
(1  h / R) 2
g
g

16 (1  h / R ) 2
2
 h
1    16
 R
1
h
4
R
h
3
R
h  3R
50.
(A) Gravitational attraction fore on particle B
Fg 
GMpm
(Dp / 2) 2
Acceleration of paritcle due to particle due to gravity a 
Fg 4GMp

m
Dp 2
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51.
(C) g α r it(r  r) and g α
52.
(A) I  
53.
(A) U  
1
(it r  R)
r2
dv
dx
GMm
r
6.67 10 11  7 10 22  6 10 24
r
7.79 10 28 
r  3.8 108 m
54.
(A) V  
GM 1
GM 2

d /2
d /2
Now P.E  mv  
2GM
(m1  m2 ) (m  mass of particie)
d
K.E = P.E

1 2 2GM
m 
(m1  m2 )
2
d
  2
55.
(C) If the body is projected with velocity υ (υ  Ve) then height up to where it rises
h
57.
G(M1  M 2 )
d
R
2
Ve
1
V2

 11.2 


 10 
2
(A) Ve  2GM
R
100 

58.
R
 4 R(approx)
P.E. U  
GMm
 5000 J
R
GM
R
GM
 5000
R
(B) P.E. U  
Uinitial = –
GMm GMm

r
Rh
GMm
GMm
and Ufinal = –
3R
2R
loss of P.E. = gain in K.E =
GMm
GMm
GMm
–
=
2R
3R
6R
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59.
(D) ΔU  U 2  U1 
mgh
mgRe mgRe


1  h/Re 1  Re
2
Re
 U 2  (  mg Re) 
60.
mg Re
2
(A) If body is projected with velocity υ (υ  Ve) then height up to which it will rise
R
but u  ve/2  h 
2

 Ve 

 1
 Ve/2 
61.
1
U 2   mgRe
2
R
 R/3
4 1
(D) change in potential energy in displacing a body from r 1and r2 is given by
1 1 
1  GMm
 1
 U  GMm     GMm 


6R
 2 R 3 R 
 r1 r2 
GMm
1
 K.E. α
2R
R
62.
(A) K.E =
63.
(B) this should be equal to escape velocity i.e. = 2gR
64.
(A) A Escape velocity does not depend on the mass of the projectile
65
(B) Ve 
Vp
g p Rp
.
 22  2
g e Re
 Ve  2Ve  2 11.2  22.4Kms 1
2GM
R
R
 G
66.
(A) Ve 
69.
since the planet is having double radius incomparision to earth therefore the escape velocity
becomes twice i.e. 22 kms-1
(C) Because it does not depend on the mass of projetile
71.
(D)
8
3
 Ve α R if  constant
Gravitational potential of A at 0  
GM
2GM

r /2
r
of B at 0  
GM
2GM

r /2
r
Total potential at 0  
4GM
r
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72.
V1

V2
73.
1
where r is the position of body from the surface
r
(C) Ve α
r2

r1
R  7R
V
 2 2  V2  1
R
2 2
GM
1 2GM

R h 2
R
(A) V 
4R  2(R  h)
 h  R  6400 km
Vp

Ve
1
Mp Re

 6   3 Ve  3 Ve
Me Rp
2
74.
(A)
75.
(A)
76.
(C) Ve 
77.
(C)
78.
2GM 4 2 2GM 2 11.2
. 

 2.5 kms1
R R 9
R
9
(A) Due to three particles net intensity at the center
υ  2gR 
V1

V2
g1 R1
.
 gk  (kg)1/2
g2 R2
2GM
M
 Ve α
R
R
If M becomes double and R becomes half then escape velocity; becomes two times
on earth Ve =
2GM
 11.2 kms 1
R
on moon Vm 
 


I  I A  IB  I C  0
because out of these three intensities ARE equal in magnitude
and between each other is 120’
80.
1
1
if R becomes
then Ve will be 2 times
4
R
(D) Escape velocity does not depends upon the angle of projection
82.
(B)  
83.
Ve = 2Ve
(B) potential energy of system of two mass
79.
(C)
Ve α
2gR 
g Rp
Vp
 p.
 1 4  2
Ve
g e Re
GMm  6.67 10 11 100 10 10 3
U

 6.67 10 10 J
2
R
10 10
so, the amount of work done to take the particle up to infinte will be 6.67  10-10 J
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84.
Vp = Vsphere + Vpartical 
(D)
85.
(D) Ve =
86.
(D)
2GM
=
R
GM GM 3GM


a
a/2
a
2  6.67 1011  1.99  1030
 618 kms1
6.98  108
m
 GMm  GMm
W  0-   gR 2  mgR

R
R
 R 
 1000  10  6400  103
 64 109
 6.4  1010 J
GM
R
87.
(C)
Vin  
88.
(B)
F  0 When 0  r  R1
Vsurface 
GM
R
Vout  
GM
r
beccause intensity is zero inside Caviy
Fincrease When R1  r  R 2
Fα
89.
(D)
1
when r  R 2
R
Intensity will zero inside the spherical shell
I = upto r = a and I
1
when r > a
r2
90.
91.
(C) kepler’s law T2 α r3
(D) System will be bound at points where total energy is negative. In the given curve at point
A, B, and C the P.E. is more than K.E.
92.
(C)
93.
U-
GMm
GMm
GMm
K
and E  
r
2r
2r
For satellite U, K, and E varies with r and also U and E remains negativw where K reamin
always positive
(C) Gravitational P.E. = m x gravitational potential U = mv so the graph of U will be same
as that of V for a spherical shell
94.
(B) Ve  2gR
96.
(B) υ 
and V0  gR
 Ve 
2 V0
GM
it r1  r2 then V1  V2
r
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97.
(B) T α r
3
2
if r becomes double then time period will become (2)3/2 times so new time period
will be 24 x 2 2 hr i.e. = 48
2
98.
(C)
The velocity of the spoon will be equal to the orbital velocity when dropped out of the
space ship
100.
(B)
VA 3V

VB VA

104.
(B)
GM
V
 A 
R
VB
υ
υα
RB

RA
R
1

4R 2
 VB  6V
1
r
% increase in speed =
=
1
% decrease in radius
2
1
(1%) = 0.5 %
2
105.
(B)
υ
GM
r
106.
(A)
υα
1
it orbital radius becomes 4 times then orbital velocity will becomes halt
r
107.
(D)
orbital rudius of satellites
U1  
K1  

110.
111.
(B)
(D)
r1 = R+R = 2R
r2 = R+7R = 8R
GMm
GMm
and U 2  
r1
r2
GMm
2r1
and K2  
GMm
2r2
U1 K 1 E 1


4
U2 K 2 E2
6R from the surface of earth and 7R from the center
Distance of satellite from the center are 7R and 3.5R respectively
3/2
3/2
T2  r2 
 3.5R 
    T2  24 

T1  r1 
 7R 
112.
(B
 6 24 hr
Gravitational force provides the required centripetal force for orbiting the satellite
mυ 2 K

R
R
becuse F α
1
R
 υ α Ro
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Total energy = -K.E. = 
1
2
2 m
113.
(A)
114.
(A)
B.E. = -K.E.
And it this amount of energy(Ek) given to satellite it will escape into outer space
1
GMm
g Re 2 m
GMm
GMm
= –
= –
= –
= – mg Re = 0.5 mg Re
P.E. =
2
Re  h
2 Re
2R e
r
115.
(A)
116.
(C) B. E. =
117.
R 
(D) T2  T1  2 
 R1 
118.
B
(C) V 
A
120.
time period of revolution of moon around the earth = 1 lunar month
GMm
1
if B.E. decreases the r also decreases and V increases as  
r
r
3/2
V
 r 
Te
 e 
Tm  rm 
121
123.
124.
125.
3 /2
1
 
2
3 /2
 Te  2  2 /3 lunar month
GMm  1
1 
  
2  R1 R2 
Because Earth rotation from west to east direction
GMm
2R
(C) v1 r1 = v2r2 (angular momentum is constant)
(B) k 
(A) Time period T 
T 2 
2R
2R

GMm
GM
R
3/ 2
4 2 R 3
GMm
2
126.
4R
2
R
GMm
GMm
(D)  2R  K.E.   2R
1
2
K .E 
122.
rA

rB
 T1 (4)3/2  8T1  40 hr
3
T1
R1
(6R ) 3


8
(C)
2
3
(3R ) 3
T2
R2
2
 T2 
24  24
 72
8
 T2  6 2
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127.
(C)
ΔA
L

ΔAαΔt
Δt 2m
ΔASCD t1 2A
 
 t1  2t 2
ΔASAS t 2 A

3
 R 
T1
  1 
T2
 R2 
2
 R 
 

 4R 
3
2
128.
(C)
129.
 T1   1 
1
1
2
3 
(B) since T α r
 T    4   T  8 T
   
130.
VJ re
(B) orbital radius of Jupiter > orbital radius of Earth V  r As rj < re there fore Vj < Vee
e
j
131.
dA L
(B) dt  2m = constant
132.
(C) The earth moves around the sun in elliptical path, so by using the properties of ellipse
2
 T2  8T1
3
r1  (1  e) a and r2 = (1-e) r2, a =
r1  r2
2
 r1r2  (1  e 2 ) a 2
where a= semi major axis
b= semi minor axis
e= eccentircity
Now required distance = sem latysrectum = b2/9
 a2
133.
134.
135.
(1  e 2 )
r1r2
2r r

 12
a
(r1  r2 )/2 r1  r2
(B) T2/ r3 = constant T2 r-3 = constant
(A) Angular momentum = Mass  orbital velocity x Radius
 m
GM
 R0
R0
 m
GMR 0
(B) speed at the earth will be maximum when its distance from the sun is minimum because
m r = constant
3
136
(C)
TA  rA 
 
TB  rB 
(B)
T2  r2 
 
T1  r1 
2
3
137.
3
4 
 8   A
 4B 
2
 (2)
3
2
2
 rA  4B  8
3
2
 4rB
 2 2  T2  2 2 years
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138.
(A) By conservation of angular momentum (mr) = constant
Vmin X rmax = Vmax X rmin
60 1.6 1012 60

 12 ms1
12
8 10
5
(D) Time period does not depends upon the mass of satellite
Vmin 
139.
1
1
1
GM
 G 2 M 2R 2
R
140.
(B) υ 
141.
 R2 

(B) T2  T1 
 R1 
3
142.
2
3
 1 2  2  2.8 yar
(C) Angular momentum of earth around the sun
L = MEVo r
GMS
r
 ME
2
.r  M E .GMS .r
ME = mass of earth
MS = mass of Sun
r = Distance between sun and the earth
143.
L α r
(D)
For central force toraqe is zero
  
144.
dL
 0  L  Constant
dt
(A) Wext U  U F
 Gdm 
 0 
(1)
 x

 G
M
π  7R 2
2π r dr
16R 2  r 2

2 GM
7R2

GM
7R2

2 GM
7R2
 r2


2 GM
4 2R  5R
7R


2 GM
4 2 5
7R

rdr
16 R2  r2
GM
 π ad2  7R
(2)
2
 16R
2


4R
3R

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145.
(A)
m 2 Rα
T α R
(C) g 
147.
(B) ΔU 
α
1
 T 2 α R n 1
Rn
n+1
2
GM
R2
146.
148.
 4π 2 
1

m
 2  R
Rn
 T 
is mass ramains constant then  
1
R2
mgh
1
 mgR
2
1 h
R
(C) P.E. = 2. total energy = 2E0
GMM
GMM
and Eo = –
r
2r
(B) let velocities of these masses at r distance from each other be v1 and v2 respectively
By conservation of momentum
Because we know U = –
149.
 m 1 V1  m 2 V2 ____________ 1 
m 1 V1  m 2 V2  0
By conservation of energy
change in P.E. = change in K.E.
Gm1m 2 1
1
2
2
 m1V1  m 2 V2
r
2
2
2

2
m1V1 m 2 V2
2 Gm1m 2


_______ ii 
m1
m2
r
on solving (i) and (ii)
1 
2Gm 2
2
r m1  m2 
and
V ap p  V1  V 2 
3
150.
T2  r2 
 
(C)
T1  r1 
2
υ2 
2Gm 1
2
r m1  m 2 
2G m 1  m 2 
r
3
 6400  2
 T2  24 
  2 hours
 3600 
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Assertion - Reason Type Questions
1.
2.
Direction (Read the following uestions and choose)
(A) If both Assertion and Reason are true and the Reason is correct explanation of assertion
(B) If both Assertion and Reason are true, but reason is not correcte explanation of the Assertion
(C) If Assertion is true, but the Reason is false
(D) If Assertion is faluse, but the Reason is true
Assertion : The value of acc. due to gravity (g) does not depend upon mass of the body
GM
Reason : This follows from g  2 , where M is mass of planet (earth) and R is radius of
R
planet (earth)
(a) A
(b) B
(c) C
(d) D
-2
Assertion : Unit of gravitational field intensity is N/kg or ms
Reason : Gravitational field intensity
Force N kg.m/sec 2


 ms  2
mass kg
kg
3.
4.
5.
(a) A
(b) B
(c) C
(d) D
Assertion : The time period of a geostationary satellite is 24 hours
Reason : Such a satellite must have the same time period as the time taken by the earth to
complete one revolution about its axis
(a) A
(b) B
(c) C
(d) D
Assertion : Even when orbit of a satellite is elliptical, its plane of rotaiion passes through the
center of earth
Reason : This is in accordance with the principle of conservation of angular momentum
(a) A
(b) B
(c) C
(d) D
Assertion : The time Period of pendulum, on a satlellite orbiting the earth is infinity
Reason :
Time period of a pendulum is inversely proportional to
(a) A
6.
(b) B
(c) C
g
(d) D
Assertion : The escape velocity on the surface of a planet of the same mass but
1
times the
4
radius of earth is 5.6 kms-1
7.
8.
9.
Reason :
(a) A
Assertion :
Reason :
(a) A
Assertion :
The escape velocity Ve = 2gR
(b) B
(c) C
(d) D
The comet does not obey kepler’s law of planetaty motion
The comet does not have ellitical orbit
(b) B
(c) C
(d) D
The square of the period of revolution of a planet is proportional to the cube of
its distance from the sun.
Reason : Sun’s gravitational field is inversely proportional to the square of its distance from
the planet
(a) A
(b) B
(c) C
(d) D
Assertion : Space ship while entering the earth’s atmoshere is likely to catch fire
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10
Reason : Temperature of upper atomosphere is very high
(a) A
(b) B
(c) C
(d) D
Assertion : The earth is slowing down and as a result the moon is coming nearer to it
Reason : The angular momentum of earth - moon system is not conserved
(a) A
(b) B
(c) C
(d) D
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Assertion - Reason Type question :
1. A
2. A
3. A
Both the asseration and reason are true and the latter is correct expalnation of the former.
Both the asseration and reason are true and the latter is correct expalnation of the former.
As the satellite is to be stationary over a particular place, its time period of revolution = 24
hours= time peroid of revolution of earth about its axis.
4. A
T  2π  on a satellite, there is a weightless -ness, so g= 0 hence T  
g
5. A
Thus both the asseration and reason are correct
As no torque is acting on the planet, its angular momentum must stay constant in a magnitude
as well as direction there for planet of rotation must pass throught the center of earth
6. D
Ve 
2 GM
R
Ve' = Ve
R
i.e.
R
Ve α
1
R
 2 Ve  2  11.2  22.4 kms1
4
thus assertion is wrong but Reason is correct
7. A Both the asseration and reason are correct and the Reason is correct explanation of Asseration
8. A A Both the asseration and reason are correct and the Reason is correct explanation of Asseration
9. C Here Asseration is correct but Reason is wrong because the space ship while entering the earth’s
atmosphere may catch fire due to atmoshperic air friction
10. D Here Both the asseration and reason are wrong because the angular momentum of earth
moon system is conserved in the absence of extermal touque.
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Comperehensions Type Questions
(1)
If a smooth tunnel is dug across a diameter of earth and a particle is related from the surface
of earth, the particle oscillates simple harmonically along it.
(1) Time period of the particle is not equal to
3
2π
R 2
GM
(C) 84.0 min
(D) Nome of these
(2) Maximum speed of the these
(A) 2π R g
(B)
2 GM
R
(B)
(A)
2.
3.
GM
R
(C)
3 GM
2R
(D)
GM
2R
When a paricle is projected from the surface of earth, it mechanincal energy and angular momentum
about center of earth at all time is constant
(i) A particle of mass m is projected from the surface of earth with velocity vo at angle 
with horizontal suppose h be the maximum height of particle from surface of earth and v
its speed at that point them v is
(A) v0coso
(B) >v0coso
(C) <v0coso
(D) zero
(ii) Maximum height h of the paritcle is
(A) 
Vo 2 sin 2θ
2g
(C) 
Vo 2 sin 2θ
2g
(B) 
Vo 2 sin 2θ
2g
Vo 2 sin 2θ
(D) can be greater than or less than
2g
A solid sphere of mass M and radius R is surrounding by a spherical
shell of same mass M and raius 2R as shown. A small particle
of mass m is relased from rest from a height (h <<R) aobove the
shell there is a hole in the shell.
(i) in what time will it enter the hole at A
hR 2
GM
(A)
2 hR 2
GM
(C)
(B)
2hR 2
GM
(D) None of these
(ii) what time will it take to move from A to B ?
(A) 
(B) 
R2
(C) 
GMh
R2
R2
GMh
(D) None of these
GMh
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(iii) with what apporximate speed will it colide at B ?
4.
5.
(A)
2 GM
R
(C)
GM
2R
(B)
3 GM
2R
(D)
GM
R
A planet is revolving round the sun in elliptical orbit. Velocity at perigee position (nearest) is
v1| and at apogee position (farthest) is v2 Both these velocities are perpendicular to the joinging
center of sun and planet r is the minimum distance and r2 the maximum distance.
(i) when the planet is at perigee position, it wants to revolve in a circular orbit by itself. For
this value of G
(A) Should increase
(B) Should decrease
(C) data is in sufficeint
(D) will not depend on the value of G
(ii) At apogee position suppose speed of planer is slightly decreased from v2, then what will
happen to minimum distance r1 in the subsequent motion
(A) r1 and r2 bothe will dicreases
(B) r1 and r2 bothe will increases
(C) r2 will remain as it is while r1 will increase
(D) r2 will remain as it is while r1 will decrease
Garvitational potential at any point inside a spherical shall is uniform
GM
where M is the mass of shell and R its
and is given by 
R
3GM
radius. At the center solid sphere, potential is 
2R
(i)
There is a concentric hole of radius R in a solid sphere of radius 2R Mass of the remaining
portion is M What is the gravitation potential at center?
(A)
 3GM
7R
(B)
 5GM
7R
(C)
 7GM
14
(D)
 9GM
14 R
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Solution
(1)
(i) T  2π
R
g
putting g 
GM
we have
R2
T
3
2π
R2
Gm
(ii)maximum speed is at centre from conservation mechanicall energy(from cerface to center)
increase in K.E. = decrease in P.E.
1
mn 2  U i  U f  m(Vi - Vf )
2
υ  2(vi  vf )
 GM 3 GM 
 2 


12 
 R
(2)
(i)
GM
R
From conservation of angular momentum at A and B
mVocosθ  m  R  h 
 R 
  
 V0 cosθ
 R h 
   Vo cosθ
(ii)
From consercation of mechanical energy, Decrease in K.E. = increase in P.E.

1
2
m V0  V 2
2

2
 V0  V 2 


2 gh
1 h
R
2
mgh
1 h
R
but here V  V0cos θ
2
2
 V0  V 2  V0 sin 2θ so let V0  V 2  x
here x  V0
x
h 
2
sin 2θ
2gh
1 h
R
x
2g  x
R
2
x
 O sin 2 
i.e. h >
i.e. h >
2g
2g
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(3)
(i)
Acceleration due to gravity near the surface of shell can be assumed to be uiform
g
G(2M) GM

(2R) 2
2R 2
From h 
t =
(ii)
2h
hR 2
2
g
GM
GMh
 GM 
VA  2 gh  2  2  h 
R2
 2R 
From A to B field due to shell is zero, but field due to sphere is not-zero
hence tAB
(iii)
1 2
gt
2
R
R2


VA
GMh
KA=0 polential between A and B due to shell is
From energy conservation
KA + UA = KB + UB
KB = (UA = UB) = m(VA – VB)
1
mVB2 = m(VA – VB)
2
 VB 
(4)
2(VA  VB ) =
GM
R
At perigee position v1 > v0 where v0 is the orbital velocity for circular motion
V0 
GM
 G
r
so value of G should incease , so that v0 will increase for this position and which
will become equal to v1
(5)
(ii)
path will becomemore elliptical , keeping r2 constant and r1 to decrease
Density of given material

M
3M

3
3
4
28πR 3
3 π[(2R)  R ]
Vwhole = Vhole + Vremaing
Vremaing = Vwhole – Vhole
3  GM1 GM2 
 

2  2R
R 
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4
8
3
here M1  π (2R)  M
3
7
4
M
M 2  ( ) R 3 
3
7
Vremaining  
9GM
14R
Match the Column
(1)
(2)
on the surface of earth acceleration due to gravity is g and gravitational potential is V match
the following
Table - 1
Table -2
(A)
At height h = R value of g
(P)
decreases by a factor
1
4
(B)
At height h = R/2 value of g
(Q)
decreases by a factor
1
2
(C)
(D)
At height h = R value of v
At depth h = R/2 value of V
(R)
increases by a factor 11/8
(S)
increases by a factor 2
(T) None
Density of planet is two times the density of earth Radius of this planet is half (As compared
to earth)
Match the following
Table-1
Table-2
(A)
Acceleration due to gravity on
(P)
Half
this planet’s surface
(B)
Gravitational potential
(Q) same
on the surface
(C)
(3)
Gravitational potential
(R)
Two times
at centre
(D)
Gravitational field strength
(S)
four times
at centre
let V and E denote the gravitational potential and gravitational field at a point. Then the match
the follwing
Table-1
Table-2
(A)
E = 0, V = 0
(P)
At center of Spherical shell
(Q)
At centre of solid sphere
(B)
E  0, V= o
(C)
V  0, E -0
(R)
At centre of circular ring
(D)
V  0, E  0
(S)
At centre of two point masses of equal magnitude
(T)
None
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(4)
(5)
Match the following
Table-1
(A)
time period of an earth satellite in circular orbit
(B)
Orbital velocity of satellite
(C)
Mechnical energy of stellite
Table -2
(P) Independent of mas of satellite
(Q) independent of radius of orbit
(R) independent of mass of earth
(S) none
Match the following (for a satellite in circular orbit)
Table-1
Table-2
(A)
kinetic energy
(P)
GMm
2r
(B)
potential energy
(Q)
GM
r
(C)
Total energy
(R)

(D)
orbital velocity
(S)
GMm
2r
GMm
r
solution :
(1)
A-p, B-Q, C-s D-t
(2)
A-Q, B-p, C-p D-o
(3)
A-t, B-t, C-p,q,r,s, D-t
(4)
A-q, B-t, C-r, D-s
(5)
A-s, s-B, c-p, D-q
• • •
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