Q. No. 1 The gravitational force between two objects does not depend on

Option 1 Sum of the masses

Option 2 Product of the masses

Option 3 Gravitational constant

Option 4 Distance between the masses

Correct Answer 1

Explanation Gm mF =

r

1 2

2

Q. No. 2 The mass of the moon is about 1.2% of the mass of the earth. Compared to the

gravitational force the earth exerts on the moon, the gravitational force the moon

exerts on earth

Option 1 Is the same

Option 2 Is smaller

Option 3 Is greater

Option 4 Varies with its phase

Correct Answer 1

Explanation Newton’s IInd

law : Every action has equal and opposite reaction.

Q. No. 3 Three identical point masses, each of mass 1 kg lie in the x-y plane at points (0, 0), (0,

0.2m). The net gravitational force on the mass at the origin is

Option 1 1.67 10 i + j N

∧ ∧∧ ∧∧ ∧∧ ∧ ××××

-9

Option 2 3.34 10 i + j N

∧ ∧∧ ∧∧ ∧∧ ∧ ××××

-10

Option 3 1.67 10 i - j N

∧ ∧∧ ∧∧ ∧∧ ∧ ××××

-9

Option 4 3.34 10 i - j N

∧ ∧∧ ∧∧ ∧∧ ∧ ××××

-10

Correct Answer 1

Explanation

F1 = F2 = F

F = F i + F j∧ ∧∧ ∧∧ ∧∧ ∧

net 1 2

→

= F i + j∧ ∧∧ ∧∧ ∧∧ ∧

(((( )))) (((( ))))

(((( ))))

6.67 10 1 1Gm mF = =

R 0.2

× ×× ×× ×× ×-111 2

2 2

= 1.67 10××××-9

(((( )))) F = 1.67 10 i + j N→ ∧ ∧→ ∧ ∧→ ∧ ∧→ ∧ ∧

∴ ×∴ ×∴ ×∴ ×

-9net

Q. No. 4 Four particles of masses m, 2m, 3m, and 4m are kept in sequence at the corners of a

square of side a. The magnitude of gravitational force acting on a particle of mass m

placed at the centre of the square will be

Option 1 24m G

a

2

2

Option 2 6m G

a

2

2

Option 3 4 2Gm

a

2

2

Option 4 Zero

Correct Answer 3

Explanation

(((( ))))G 2m m 4GmF = =

aa

2

2

2 2

4 2GmF = F 2 =

a

2

net 2

Q. No. 5 If the distance two masses is doubled, the gravitational attraction between them

Option 1 Is doubled

Option 2 Becomes four times

Option 3 Is reduced to half

Option 4 Is reduced to a quarter

Correct Answer 4

Explanation 1F

r∝∝∝∝

2

Q. No. 6 The gravitational force between two stones of mass 1 kg each separated by a distance

of 1 metre in vacuum is

Option 1 Zero

Option 2 6.675 10 newton××××-5

Option 3 6.675 10 newton××××-11

Option 4 6.675 10 newton××××-8

Correct Answer 3

Explanation (((( )))) (((( )))) (((( ))))

(((( ))))

6.67 10 1 1Gm mF = = = 6.67 10 N

r 1

××××××××

-11

-111 2

2 2

Q. No. 7 The distance of the centres of moon and earth is D. The mass of earth is 81 times the

mass of the moon. At what distance from the centre of the earth, the gravitational

force will be zero

Option 1 D

2

Option 2 2D

3

Option 3 4D

3

Option 4 9D

10

Correct Answer 4

Explanation

(((( ))))

(((( ))))

G 81M GM =

x D - x2 2

(((( ))))

81 1 =

x D - x⇒⇒⇒⇒

2 2

9 1 = x D - x

⇒⇒⇒⇒

⇒⇒⇒⇒ 9D - 9x = x

9D x =

10⇒⇒⇒⇒

Q. No. 8 Four identical point masses each equal to m are placed at the corners of a square of

side a. The force on a point mass m’ placed at the point of intersection of the two

diagonals is :

Option 1 (((( ))))4Gmm'

a2

Option 2 (((( ))))2Gmm'

a2

Option 3 (((( ))))Gmm'

a2

Option 4 zero

Correct Answer 4

Explanation

Net force = 0

Q. No. 9 Two planets have the same average density but their radii are R1 and R2. If acceleration

due to gravity on these planets be g1 and g2 respectively, then

Option 1 g R=

g R

1 1

2 2

Option 2 g R=

g R

1 2

2 1

Option 3 1

2

g R=

g R

21

22

Option 4 g R=

g R

31 1

32 2

Correct Answer 1

Explanation 4g = GR

3π ρπ ρπ ρπ ρ

as G, ρρρρ are constant here

g R∝∝∝∝

g R =

g R⇒⇒⇒⇒ 1 1

2 2

Q. No. 10 An iron ball and a wooden ball of the same radius are released form a height ‘h’ in

vacuum. The time taken by both of them to reach the ground is

Option 1 Unequal

Option 2 Exactly equal

Option 3 Roughly equal

Option 4 Zero

Correct Answer 2

Explanation Exactly equal

Q. No. 11 The correct answer to above question is based on

Option 1 Acceleration due to gravity in vacuum is same irrespective of size and mass of the body

Option 2 Acceleration due to gravity in vacuum depends on the mass of the body

Option 3 There is no acceleration due to gravity in vacuum

Option 4 In vacuum there is resistance offered to the motion of the body and this resistance

depends on the mass of the body

Correct Answer 1

Explanation Acceleration due to gravity in vacuum is same irrespective of size and mass of the body

Q. No. 12 A body weighs 700 gm wt on the surface of the earth. How much will it weigh on the

surface of a planet whose mass is 1

7 and radius is half that of the earth

Option 1 200 gm wt

Option 2 400 gm wt

Option 3 50 gm wt

Option 4 300 gm wt

Correct Answer 2

Explanation GM Mg = g

R R⇒⇒⇒⇒ ∝∝∝∝

2 2

(((( ))))g M R 1 4

= = 2 = g M R 7 7

⇒⇒⇒⇒ × ×× ×× ×× ×

2

22 2 1

1 1 2

(((( )))) (((( ))))4g 4

g = weight of body = weight of body7 7

⇒ ⇒⇒ ⇒⇒ ⇒⇒ ⇒12 2 1

4 weight of body = 700 gmwt

7∴ ×∴ ×∴ ×∴ ×

= 400 gm wt

Q. No. 13 Assume that the acceleration due to gravity on the surface of the moon is 0.2 times

the acceleration due to gravity on the surface of the earth. If Re is the maximum range

of a projectile on the earth’s surface, what is the maximum range on the surface of the

moon for the same velocity of projection

Option 1 0.2 Re

Option 2 2 Re

Option 3 0.5 Re

Option 4 5 Re

Correct Answer 4

Explanation (((( ))))

2u sinRange of projectile = u cos

g

θθθθ θθθθ

1R

g∝∝∝∝

gR 1 = = = 5

R g 0.2⇒⇒⇒⇒ eM

e M

⇒⇒⇒⇒ RM = 5 Re

Q. No. 14 The escape velocity for a body projected vertically upwards from the surface of the

earth is 11.2 km s-1

. if the body is projected in a direction making an angle 450 with the

vertical, the escape velocity will be :

Option 1 11.2km s

2

-1

Option 2 11.2 2 km s××××-1

Option 3 11.2 2km s××××-1

Option 4 11.2 km s-1

Correct Answer 4

Explanation Escape velocity does not depend on angle of projection.

Q. No. 15 A spring balance is graduated on sea level. If a body is weighed with this balance at

consecutively increasing heights from earth’s surface, the weight indicated by the

balance

Option 1 Will go on increasing continuously

Option 2 Will go on decreasing continuously

Option 3 Will remain same

Option 4 Will first increase and then decrease

Correct Answer 2

Explanation Acceleration due to gravity decreases with altitude, so weight will also decrease with

altitude.

Q. No. 16 An object weights 72 N on earth. Its weight at a height of

R

2 from earth is

Option 1 32 N

Option 2 56 N

Option 3 72 N

Option 4 Zero

Correct Answer 1

Explanation

(((( ))))W

GMwWeight of body =

R2

1 w

R∴ ∝∴ ∝∴ ∝∴ ∝

2

w r R 4 = = =

3Rw r 9

2

⇒⇒⇒⇒

2

2

2 1

1 2

4 4 w = w = 72 N

9 9⇒⇒⇒⇒ ××××2 1

= 32 N

Q. No. 17 The depth d at which the value of acceleration due to gravity becomes

1

n times the

value at the surface, is [R = radius of the earth]

Option 1 R

n

Option 2 n - 1R

n

Option 3 R

n2

Option 4 nR

n + 1

Correct Answer 2

Explanation dg' = g 1 -

R

g' d = 1 -

g R

⇒⇒⇒⇒

1 d = 1 - n R

⇒⇒⇒⇒

1 n - 1 d = 1 - R = R

n n

⇒⇒⇒⇒

Q. No. 18 At what distance from the centre of the earth, the value of acceleration due to gravity

g will be half that on the surface (R = radius of earth)

Option 1 2 R

Option 2 R

Option 3 1.414 R

Option 4 0.414 R

Correct Answer 4

Explanation

g x∝∝∝∝

GMxg =

R3

x = distance from centre.

g x =

g R⇒⇒⇒⇒ 1

2

1 x = 2 R

⇒⇒⇒⇒

R x =

2⇒⇒⇒⇒

Q. No. 19 Suppose a vertical tunnel is dug along the diameter of earth assumed to be a sphere of

uniform mass having density ρρρρ If a body of mass m is thrown in this tunnel, its

acceleration at a distance y from the centre is given by

Option 1 4

G ym3

ππππρρρρ

Option 2 3G y

4π ρπ ρπ ρπ ρ

Option 3 4y

3πρπρπρπρ

Option 4 4G y

3π ρπ ρπ ρπ ρ

Correct Answer 4

Explanation 4G R y

GMy 3g = =

R R

π ρπ ρπ ρπ ρ

3

3 3

4

= G y3

π ρπ ρπ ρπ ρ

Q. No. 20 At what height above the earth’s surface does the acceleration due to gravity fall to1%

of its value at the earth’s surface?

Option 1 9 R

Option 2 10 R

Option 3 99 R

Option 4 100 R

Correct Answer 1

Explanation 1g

r∝∝∝∝

2

g r g r

⇒ ⇒⇒ ⇒⇒ ⇒⇒ ⇒

2

1 2

2 1

R + h 100 =

R

⇒⇒⇒⇒

2

On solving : h = 9 R

Q. No. 21 At what height above the earth’s surface is the acceleration due to gravity 1% less

than its value at the surface? [R = 6400km]

Option 1 16 km

Option 2 32 km

Option 3 64 km

Option 4 32 2km

Correct Answer 2

Explanation 2hg' = g 1 -

R

g - g ' 200h 100 =

g R

⇒⇒⇒⇒ ××××

200h 1 =

R⇒⇒⇒⇒

R 6400km h = = = 32 km

200 200⇒⇒⇒⇒

Q. No. 22 At what depth below the surface of the earth acceleration due to gravity will be half its

value 1600 km above the surface of the earth?

Option 1 4.3 10 m××××6

Option 2 2.4 10 m××××6

Option 3 3.2 10 m××××6

Option 4 1.6 10 m××××6

Correct Answer 1

Explanation Acceleration due to gravity at 1600 km height.

(((( )))) (((( ))))

GM gR Rg '= = = g

R + hR + h R + h

22

2 2

6400

= g6400 + 1600

2

= 0.64 g

Acceleration due to gravity is at depth d

dg ''= g 1 -

R

g g''=

2∴∴∴∴

'

d 0.64g g 1 - =

R 2

⇒⇒⇒⇒

d = 0.68R

⇒⇒⇒⇒

⇒⇒⇒⇒ d = (0.68) (6400) (1000) m

= 4.3 10 m××××6

Q. No. 23 The rotation of the earth about its axis speeds up such that a man on the equator

becomes weightless. In such a situation, what would be the duration of one day?

Option 1 R2

gππππ

Option 2 1 R

2ππππ g

Option 3 2 Rgππππ

Option 4 1Rg

2ππππ

Correct Answer 1

Explanation For weightlessness at equator due to rotation.

g - w2R = 0

g w =

2⇒⇒⇒⇒

2 R T = = 2

w g

ππππ⇒⇒⇒⇒ ππππ

Q. No. 24 There are two bodies of masses 103 kg and 10

5 kg separated by a distance 0f 1 km. At

what distance from the smaller body, the intensity of gravitational field will be zero

Option 1 1km

9

Option 2 1km

10

Option 3 1km

11

Option 4 10km

11

Correct Answer 3

Explanation

Acceleration due to gravity should be equal and opposite.

(((( )))) (((( ))))(((( ))))

G 10 G 10 =

x 1 - x

1003 5

2 2

(((( ))))

1 100 = x 1 - x

⇒⇒⇒⇒2 2

1 10 = x 1 - x

⇒⇒⇒⇒

⇒⇒⇒⇒ 1 - x = 10x

1 x = km

11⇒⇒⇒⇒

Q. No. 25 In some region, the gravitational field is zero. The gravitational potential in this region

Option 1 Must be variable

Option 2 Must be constant

Option 3 Cannot be zero

Option 4 Must be zero

Correct Answer 2

Explanation -dvE =

dr

If V = constant, E = 0.

Q. No. 26 The magnitudes of the gravitational field at distance r1 and r2 from the centre of a

uniform sphere of radius R and mass M F1 and F2 respectively. Then

Option 1 if r R and r

F r=

FR

r

1 12 2

2 2

< < ☒

Option 2 if r R and

F r=

Fr

rR

21 2

1 222 1

< > ☒

Option 3 if r R and r

F r=

FR

r

1 11 2

2 2

> > ☐

Option 4 if r R and

F r=

Fr

rR

21 1

1 222 2

< < ☐

Explanation Magnitude of gravitational field

Distance from centre of uniform sphere.

Q. No. 27 Two concentric shells of mass M1 and M2 are having radii r1 and r2. Which of the

following is the correct expression for the gravitational field on a mass m?

Option 1 G(M + M )

I=r

for r r1 2

2 1<

Option 2 G(M + M )I=

rfor r r1 2

2 2<

Option 3 for r

MI= r rG

r 1

2

2 2< <

Option 4 for r

GMI = r r

r 1

1

2 2< <

Correct Answer 4

Explanation Gravitational field due to outer shell = 0

GM I =

r∴∴∴∴ 1

2

Q. No. 28 A spherical shell is cut into two pieces along a chord as shown in the figure. P is a point

on the plane of the chord. The gravitational field at P due to the upper part is I1 and

that due to the lower part is I2. What is the relation between them

Option 1 I I1 2>

Option 2 I I1 2<

Option 3 I1 = I2

Option 4 No definite relation

Correct Answer 3

Explanation Gravitation field at inner point of shell = 0

Q. No. 29 A particle of mass m is placed inside a spherical shell, away from its centre. The mass

of the shell is M.

Option 1 The particle will move towards the centre.

Option 2 The particle will move away from the centre, towards the nearest wall.

Option 3 The particle will move towards the centre it m < M, and away from the centre if m >

M.

Option 4 The particle will remain stationary.

Correct Answer 4

Explanation Gravitational field at inner point of shell = 0.

Q. No. 30 A uniform ring of mass m and radius r is placed directly above a uniform sphere of

mass M and of equal radius. The centre of the ring is at a distance 3 r from the

centre of the sphere. The gravitational force exerted by the sphere on the ring will be

Option 1 GMm

8r2

Option 2 GMm

4r2

Option 3 3 GMm

8r2

Option 4 GMm

16r2

Correct Answer 3

Explanation

dFsin θθθθ will be cancelled

F = dFcos∴ θ∴ θ∴ θ∴ θ∫∫∫∫

GMdm 3

= 24r

××××∫∫∫∫ 2

3 GM

= dm8 r ∫∫∫∫2

3GMm

= 8r

2

Q. No. 31 A solid sphere of radius

R

2is cut out of a solid sphere of radius R such that the spherical

cavity so formed touches the surface on one side and the centre of the sphere on the

other side, as shown. The initial mass of the solid sphere was M. If a particle of mass m

is placed at a distance 2.5 R from the centre of the cavity, then what is the

gravitational attraction on the mass m?

Option 1 GMm

R2

Option 2 GMm

2R2

Option 3 GMm

8R2

Option 4 23 GMm

100 R2

Correct Answer 4

Explanation

Mass of extracted part from gravity

4 RM

M3 2m' = =

4 8R

3

ππππ

ππππ

3

3

Using principle of superposition.

(((( )))) (((( ))))

MG m

GMm 8Net force = -

2R 2.5 R

2 2

23 GMm

= 100 R

2

Q. No. 32 A Solid sphere of uniform density and radius R applies a gravitational force of

attraction equal to F1 on a particle placed at a distance 2R from the centre of the

sphere. A spherical cavity of radiusR

2 is now made in the sphere as shown in the

figure. The sphere with the cavity now applies a gravitational force F2 on the same

particle. The ratio F

F

1

2

is

Option 1 1

2

Option 2 3

4

Option 3 7

8

Option 4 9

7

Correct Answer 4

Explanation

(((( ))))

GMm GMmf = =

4R2R1 2 2

(((( ))))

MG m

GMm 7 GMm8f = - =

36 R32RR

2

2 2 2 2

F 9 = F 7

∴∴∴∴ 1

2

Q. No. 33 The following figure shows two shells of masses m1 and m2. The shells are concentric.

At which point, a particle of mass m shall experience zero force?

Option 1 A

Option 2 B

Option 3 C

Option 4 D

Correct Answer 4

Explanation Gravitational field inside shells is zero. So at point D, gravitational field will be zero,

due to both shells. That’s why the force experienced by any particle at D will be zero.

Q. No. 34 A solid sphere of uniform density and mass M has radius 4 m. Its centre is at the origin

of the coordinate system. Two spheres of radii 1 m are taken out so, that their centres

are at P (0, -2, 0) and Q (0, 2, 0), respectively. This leaves two spherical cavities. What

is the gravitational field at the origin of the coordinate axes?

Option 1 31 GM

1024

Option 2 GM

1024

Option 3 31 GM

Option 4 Zero

Correct Answer 4

Explanation Zero

Q. No. 35 Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and

length l is :

Option 1 GM

along + x axis2����

Option 2 GM along + y axis

ππππ2����

Option 3 2 GM along + x axis

ππππ2����

Option 4 2 GM along + y axis

ππππ2����

Correct Answer 4

Explanation

m

mass of element dm = dθθθθππππ

Gravitational field due to element, Gdm Gmd

dg = = R R

θθθθ

ππππ2 2

dg cos θθθθ of each element will be cancelled out.

sindg = dgsin = sin

R

ππππ

θθθθ θ θθ θθ θθ θ

ππππ ∫ ∫∫ ∫∫ ∫∫ ∫ 2

0

Gm

= sin dR

ππππ

θ θθ θθ θθ θππππ ∫∫∫∫2

0

2Gm

= as r = or R = R

ππππ

ππππππππ 2

����� � � �

2 Gm

= along + y axisππππ

2����

Q. No. 36 Two identical thin uniform rods of mass m and length L are placed as shown in figure.

The gravitational interaction force between the two rods is

Option 1 Gm

4L

2

2

Option 2 Gm

2L

2

2

Option 3 Gm

9L

2

2

Option 4 Gm 4n

3L

2

2����

Correct Answer 4

Explanation

GM 1 1

g due to rod (1) = - L x x + L

x

Mmass of element dm = dx

L

Force on element due to rod (1)

dF = gx dm

GM 1 1

= - dxx x + LL

2

2

GM 1 1

f = df = - dxx x + L L

∴∴∴∴

∫ ∫∫ ∫∫ ∫∫ ∫2L

2

2

L

(((( ))))GM

= Ln - Ln x + 1L

22L

2

(((( )))) (((( ))))GM

= Ln2L - Ln 3L - LnL - Ln 2LL

2

2

GM 4

= n3L

2

2����

Q. No. 37 Two rings having masses M and 2M, respectively, having same radius are placed

coaxially as shown in figure.

If the mass distribution on both the rings is non-uniform, the gravitational potential at

point P is

Option 1 GM 1 2- +

R 2 5

Option 2 GM 2- 1 +

R 2

Option 3 Zero

Option 4 Cannot be determined from given information

Correct Answer 1

Explanation Potential at axial points due to ring doesn’t depend on distribution of mass.

For given

GM GM

v = - = -2RR + R

p1 2 2

(((( ))))

(((( ))))

G 2M 2GMv = - = -

5RR + 2Rp

2 22

v = v + v∴∴∴∴ net p p1 2

GM 1 2= +

R 2 5

Q. No. 38 The magnitude of gravitational potential energy of a body at a distance r from the

centre of earth is u. Its weight at a distance 2r from the centre of earth is

Option 1 u

r

Option 2 u

4r

Option 3 u

2r

Option 4 4r

u

Correct Answer 2

Explanation GMmu = -

r

(((( ))))

GMm GMmF = =

4r2r2 2

-u F =

4r∴∴∴∴

Q. No. 39 Three particles each of mass 100 gm are brought from a very large distance to the

vertices of an equilateral triangle whose side is 20 cm in length. The work done will be

Option 1 0.33 10 Joule××××-11

Option 2 -0.33 10 Joule××××-11

Option 3 1.00 10 Joule××××-11

Option 4 -1.00 10 Joule××××-11

Correct Answer 4

Explanation (((( ))))w = U = U - U = U as u = 0∆∆∆∆net f i f i

(((( )))) (((( ))))-G 0.1 0.2 -6.67 10 0.1 0.1U = 3 = 3

0.2 0.2

× × ×× × ×× × ×× × ×× ×× ×× ×× ×

-11

f

= -1.0005 10 J××××-11

Q. No. 40 The change in potential energy, when a body of mass m is raised to a height nR from

the earth’s surface is (R = Radius of earth)

Option 1 nmgR

n-1

Option 2 nmgR

Option 3 nmgR

n +1

2

2

Option 4 nmgR

n+1

Correct Answer 4

Explanation GmmU = - = -mgR

Ri

(((( )))) (((( ))))

Gmm -GMm -mgRU = - = =

nR + R R n + 1 n + 1f

(((( ))))(((( ))))

-mgR U = U - U = - -mgR

nH∴ ∆∴ ∆∴ ∆∴ ∆ f i

(((( ))))

1= mgR 1-

n + 1

n

= mgR n + 1

Q. No. 41 What impulse need to be given to a mass m, released from the surface of earth along a

straight tunnel passing through centre of earth, at the centre of the earth, to bring it to

rest. (Mass of earth M, radius or earth R)

Option 1 GMm

R

Option 2 GMmm

R

Option 3 GMm

2R

Option 4 Zero

Correct Answer 1

Explanation Conservation of mechanical energy.

Ks + Ps = Kc + Pc

-GMm P -3GMm = +

R 2m 2R

⇒⇒⇒⇒

2

P -GMm GMm = P = 2m 2R R

⇒ ⇒⇒ ⇒⇒ ⇒⇒ ⇒2 2

2

(((( ))))GM

P = mR

⇒⇒⇒⇒

Q. No. 42 Two bodies of masses m and M are placed a distance d apart. The gravitational

potential at the position where the gravitational field due to them is zero is V, then

Option 1 GV = - (m+M)

d

Option 2 GmV = -

d

Option 3 GMV = -

d

Option 4 (((( ))))

GV = - m + M

d

2

Correct Answer 4

Explanation

(((( ))))

Gm GM =

x d - x2 2

m M =

x d - x⇒⇒⇒⇒

(((( )))) (((( )))) d - x m = M x⇒⇒⇒⇒

(((( )))) (((( )))) x M + m = m d⇒⇒⇒⇒

m x = d

M+ m

⇒⇒⇒⇒

m d - x = d

M+ m

∴∴∴∴

-GM -GM

V = + x d - x

p

(((( ))))

(((( ))))(((( ))))

M M + mm M + m= -G +

m d M d

(((( ))))

(((( ))))G M + m

= - m + Md

(((( ))))-G

= M + md

2

Q. No. 43 P is a point at a distance r from the centre of a spherical shell of mass M and radius a,

where r < a. The gravitational potential at P is

Option 1 GM-

r

Option 2 GM-

a

Option 3 r-GM

a2

Option 4 a - r-GM

a

2

Correct Answer 2

Explanation GM-

a

Q. No. 44 P is a point at a distance from the centre of a solid sphere of radius a. The gravitational

at P is V. If V is plotted as a function of r, which is the correct curves?

Option 1

Option 2

Option 3

Option 4

Correct Answer 3

Explanation

Q. No. 45 The gravitational potential due to the earth at infinite distance from it is zero. Let the

gravitational potential at a point P be -5 J/kg. Suppose, we arbitrarily assume the

gravitational potential at infinity to be + 10 J/kg, then the gravitational potential at P

will be

Option 1 -5 J/kg

Option 2 +5 J/kg

Option 3 -15 J/kg

Option 4 +15 J/kg

Correct Answer 2

Explanation Work done by ext force on unit mass

= v = v - v = -5 - 0 = -5∞∞∞∞∆∆∆∆ p

w = v - v∞∞∞∞ext p

⇒⇒⇒⇒ -5 = vp - 10 ⇒⇒⇒⇒ vp = +5 J/kg

Q. No. 46 A person brings a mass of 1 kg from infinity to a point A. Initially the mass was at rest

but it moves with a speed of 2 m/s as it reaches A. The work done by the person on

the mass is -3 J. The potential of A is

Option 1 -3 J/kg

Option 2 -2 J/kg

Option 3 -5 J/kg

Option 4 -7 J/kg

Correct Answer 3

Explanation Work energy theorem:

Wext + wc = ∆∆∆∆ KE

(((( )))) (((( )))) (((( )))) (((( ))))- v1

-3 + m v = 1 2 = 22

∞∞∞∞⇒⇒⇒⇒2

f

(((( ))))- v m v = 5∞∞∞∞⇒⇒⇒⇒ f

⇒⇒⇒⇒ - vf = 5 ⇒⇒⇒⇒ vf = -5 J/kg

Q. No. 47 The escape velocity for the earth is 11.2 km/sec. The mass of another planet is 100

times of the earth and its radius is 4 times that of the earth. The escape velocity for

this planet will be

Option 1 112.0 km/s

Option 2 5.6 km/s

Option 3 280.0 km/s

Option 4 56.0 km/s

Correct Answer 4

Explanation 2GMV =

Res

MV

R∝∝∝∝es

V M R = V M M

∴ ×∴ ×∴ ×∴ ×es es Earht

Earth Earth es

1

= 100 = 54

××××

V = 5 V = 5 11.2 km/s∴ × ×∴ × ×∴ × ×∴ × ×es Earth

= 56 km/s

Q. No. 48 The ratio of the radii of planets A and B is k1 and ratio of acceleration due to gravity on

them is k2. The ratio of escape velocities from them will be

Option 1 k1k2

Option 2 k k1 2

Option 3 k

k

1

2

Option 4 k

k

2

1

Correct Answer 2

Explanation V = 2gResc

(((( ))))

(((( ))))

V g R = = K K = K K

V g R× ×× ×× ×× ×

esc 1 111 2 1 2

esc 2 22

Q. No. 49 For a satellite escape velocity is 11 km/s. If the satellite is launched at an angle of 600

with the vertical, then escape velocity will be

Option 1 11 km/s

Option 2 11 3 km/s

Option 3 11km/s

3

Option 4 33 km/s

Correct Answer 1

Explanation Escape velocity does not depend on angle of projection.

Q. No. 50 A person sitting in a chair in a satellite feels weightless because

Option 1 The earth does not attract the objects in a satellite

Option 2 The normal force by the chair on the person balances the earth’s attraction

Option 3 The normal force is zero

Option 4 The person is satellite is not accelerated

Correct Answer 3

Explanation Feeling of weight comes from normal force.

Q. No. 51 Planetary system in the solar system describes

Option 1 Conservation of energy

Option 2 Conservation of linear momentum

Option 3 Conservation of angular momentum

Option 4 None of these

Correct Answer 3

Explanation In case of central forces, angular momentum is always an served along the axis passing

there anybody or line joining them.

Q. No. 52 A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is

very small compared to the mass of earth

Option 1 The acceleration of S is always directed towards the centre of the earth

Option 2 The angular momentum of S about the centre of the earth changes in direction but its

magnitude remains constant

Option 3 The total mechanical energy of S varies periodically with time

Option 4 The linear momentum of S remains constant in magnitude

Correct Answer 1

Explanation The acceleration of S is always directed towards the centre of the earth

Q. No. 53 Inside a satellite orbiting very close to the earth’s surface, water does not fall out of a

glass when it is inverted. Which of the following is the best explanation for this?

Option 1 The earth does not exert any force on the water.

Option 2 The earth’s force of attraction on the water is exactly balanced by the force created by

satellite’s motion.

Option 3 The water and the glass have the same acceleration, equal to g, towards the centre of

the earth, and hence there is no relative motion between them

Option 4 The gravitation attraction between the glass and the water balances the earth’s

attraction on the water.

Correct Answer 3

Explanation The water and the glass have the same acceleration, equal to g, towards the centre of

the earth, and hence there is no relative motion between them

Q. No. 54 If a small part separates from an orbiting satellite, the part will

Option 1 Fall to the earth directly

Option 2 Move in spiral and reach the earth after a new rotations

Option 3 Continue to move in the same orbit as the satellite

Option 4 Move further away from the earth gradually

Correct Answer 3

Explanation Continue to move in the same orbit as the satellite

Q. No. 55 If ve and v0 represent the escape velocity and orbital velocity of a satellite

corresponding to a circular orbit of radius R, then

Option 1 ve = v0

Option 2 2v = v0 e

Option 3 vv =

2

0e

Option 4 ve and v0 are not related

Correct Answer 2

Explanation 2GM GMv = , v =

r re 0

Q. No. 56 Two satellite of masses m1 and m2 (m1 > m2) are revolving round the earth in circular

orbits of radius r1 and r2 (r1 >r2) respectively. Which of the following statements is true

regarding their speeds v1 and v2?

Option 1 v1 = v2

Option 2 v v1 2<

Option 3 v v1 2>

Option 4 v v=

r r

1 2

1 2

Correct Answer 2

Explanation GMv =

r0

M = mass of planet = constant

1 v

r∴ ∝∴ ∝∴ ∝∴ ∝0

v v∴∴∴∴ 1 2<

Q. No. 57 A satellite is around the earth with speed v in a circular orbit of radius r. If the orbit

radius is decreased by 1% , its speed will

Option 1 Increase by 1%

Option 2 Increase by 0.5%

Option 3 Decrease by 1%

Option 4 Decrease by 0.5%

Correct Answer 2

Explanation GM

v = v rr

⇒⇒⇒⇒ ∝∝∝∝

-1

20 0

v r-1

= v 2 r

∆∆∆∆ ∆∆∆∆⇒⇒⇒⇒ 0

0

(((( ))))-1 r

change in v = change in r2 r

∆∆∆∆⇒⇒⇒⇒ 0% %

(((( ))))-1

= -12

%

0.5 Incre= . ase%

Q. No. 58 Two satellite A and B go round a planet P in circular orbits having radii 4R and R

respectively. If the speed of the satellite A is 3v,the speed of the satellite B will be

Option 1 12v

Option 2 6v

Option 3 4v

3

Option 4 3v

2

Correct Answer 2

Explanation 1 v r 4Rv = =

v r Rr∝∝∝∝ ⇒⇒⇒⇒ B A

0

A B

v = 2v = 2 3v = 6v⇒⇒⇒⇒ ××××B A

Q. No. 59 If orbital velocity of planet is given by v = Ga M

b R

c, then

Option 1 1 1 1a = , b = , c = -

3 3 3

Option 2 1 1 1a = , b = , c = -

2 2 2

Option 3 1 1 1a = , b = - , c =

2 2 2

Option 4 1 1 1a = , b = - , c = -

2 2 2

Correct Answer 2

Explanation GM

v = = G M rr

1 1 -1

2 2 20

Q. No. 60 Two satellite of same mass are launched in the same orbit of radius r around the earth

so as to rotate opposite to each other. If they collide in elastically and stick together as

Wreckage, the total energy of the system just after collision is

Option 1 2GMm-

r

Option 2 GMm-

r

Option 3 GMm

2r

Option 4 Zero

Correct Answer 1

Explanation Due to inelastic collision of two satellite of same mass and same orbit (orbits pear will

be same)

So, net momentum = 0

∴∴∴∴ (KE)after collision = 0

(((( ))))-GM 2mPE =

r

Q. No. 61 Two Identical satellites are at R and 7R away from earth surface, The wrong statement

is (R = Radius of earth)

Option 1 Ratio of total energy will be 4

Option 2 Ratio of kinetic energies will be 4

Option 3 Ratio of potential energies will be 4

Option 4 Ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2

Correct Answer 4

Explanation r1 = 2R, r2 = 8R (from centre of planet)

GMm -GMm -GMmKE = , PE = , TE =

2r r 2r

So in all cones ratio will be 4.

Q. No. 62 Energy required in moving a body of mass m from a distance 2R to 3R from centre of

earth of mass M is

Option 1 GMm

12R2

Option 2 GMm

3R2

Option 3 GMm

8R

Option 4 GMm

6R

Correct Answer 4

Explanation -GMm -GMmEnergy required = -

3r 2r

GMm 1 1

= - R 2 3

GMm

= 6R

Q. No. 63 A satellite whose mass is M, is revolving in circular orbit of radius r around the earth.

Time of revolution of satellite is

Option 1 rT

GM∝∝∝∝

5

Option 2 r

TGM

∝∝∝∝3

Option 3 rT

GM

3

∝∝∝∝2

Option 4 r

TGM

4

∝∝∝∝3

1

Correct Answer 2

Explanation 2 r

T = GM

ππππ

3

2

Q. No. 64 The distances of two satellites from the surface of the earth are R and 7R. Their time

periods of rotation are in the ratio :

Option 1 1 : 7

Option 2 1 : 8

Option 3 1 : 49

Option 4

1 : 7

3

2

Correct Answer 2

Explanation T r r = 2R, r = 8R∝∝∝∝

3

21 2

T r 1 1 = = =

T r 4 8

∴∴∴∴

3 3

2 21 1

2 2

Q. No. 65 The figure shows the motion of a planet around the sun in an elliptical orbit with sun

at the focus. The shaded areas A and B are also shown in the figure which can be

assumed …….t1 and t2 represent the time for the planet to move from a to b and d to c

respectively, then

Option 1 t t1 2<

Option 2 t t1 2>

Option 3 t1 = t2

Option 4 t t≤≤≤≤1 2

Correct Answer 3

Explanation Kepler’s second law ⇒⇒⇒⇒ Line joining the sun and planet sweeps out equal areas in

equal time interval.

Q. No. 66 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 the point

Option 1 C

Option 2 A

Option 3 B

Option 4 D

Correct Answer 2

Explanation According to conservation of angular momentum :

mvr = constant.

Speed will be maximum at A.

Q. No. 67 Four particles, each of mass m move along a circle of radius R under the action of their

mutual gravitational attractions. The speed of each particle is

Option 1 GM

R

Option 2 GM2 2

R

Option 3 (((( ))))

GM2 2 + 1

R

Option 4 GM 2 2 +1

R 4

Correct Answer 4

Explanation

(((( ))))GM GM

F = = 2RR 2

2 2

1 2 2

(((( ))))

GM GMF = =

4R2R

2 2

2 2 2

Net force on (1) towards centre

= 2F1 cos450 + F2

GM 1 GM

= 2 + 22R 4R

2 2

2 2

GM Gm GM 1 1

= + = + 422R 4R R

2 2 2

2 2 2

GM 2 2 + 1

= 4R

2

2

And,

GM 2 2 + 1 mv

= 4 RR

2 2

2

Gm 2 2 + 1

v = R 4

⇒⇒⇒⇒

Q. No. 68 Which of the following statements is/are true about the gravitational constant G?

Option 1 G is a dimensionless number ☐

Option 2 The value of G is the same anywhere in the universe ☒

Option 3 G has the same value in all systems of units ☐

Option 4 The value of G does not depends on the nature of the medium between the

two bodies

☒

Explanation The value of G is the same anywhere in the universe, The value of G does not depends

on the nature of the medium between the two bodies

Q. No. 69 The value of the acceleration due to gravity g on earth depends upon

Option 1 The mass of the earth ☒

Option 2 The average radius of the earth ☒

Option 3 The average density of the earth ☒

Option 4 None of the above quantities ☐

Explanation

(((( ))))

(((( ))))GM R - hGMg = , g =

RR + h

e eeout in2 3

ee

4M = R

3

π ρπ ρπ ρπ ρ

3e e

Q. No. 70 Choose the correct statement (s) from the following

Option 1 The gravitational forces between two particles are an action and reaction

pair

☒

Option 2 Gravitation constant (G) is scalar but acceleration due to gravity (g) is a

vector

☒

Option 3 The values of G and g are to be determined experimentally ☒

Option 4 G and g are constant everywhere ☐

Explanation ‘g’ varies.

Q. No. 71 Choose the correct statements (s) from the following

Option 1 The magnitude of the gravitational force between two bodies of mass 1 kg

each and separated by a distance of 1 m is 9.8.N.

☐

Option 2 Higher the value of the escape velocity for a planet, the higher is the

abundance of lighter gases in its atmosphere.

☒

Option 3 The gravitational force of attraction between two bodies of ordinary mass

is not noticeable because the value of the gravitation constant is extremely

small.

☒

Option 4 Force of friction arises due to gravitational attraction. ☐

Explanation Higher the value of the escape velocity for a planet, the higher is the abundance of

lighter gases in its atmosphere., The gravitational force of attraction between two

bodies of ordinary mass is not noticeable because the value of the gravitation constant

is extremely small.

Q. No. 72 Choose the wrong statements (s) from the following

Option 1 It is possible to shield a body from the gravitational field of another body by ☒

using a thick shielding material between them.

Option 2 The escape velocity of a body is independent of the mass of the body and

the angle of projection.

☐

Option 3 The acceleration due to gravity increases due to the rotation of the earth. ☒

Option 4 The gravitational force exerted by the earth on a body is greater than that

exerted by the body on the earth.

☒

Explanation It is possible to shield a body from the gravitational field of another body by using a

thick shielding material between them, The acceleration due to gravity increases due

to the rotation of the earth, The gravitational force exerted by the earth on a body is

greater than that exerted by the body on the earth

Q. No. 73 A comet is revolving around the sun in a highly elliptical orbit. Which of the following

will remain constant throughout its orbit?

Option 1 Kinetic energy ☐

Option 2 Potential energy ☐

Option 3 Total energy ☒

Option 4 Angular momentum ☒

Explanation Total energy and Angular momentum

Q. No. 74 The weight of an object will be

Option 1 Zero at the centre of earth ☒

Option 2 One-fourth of its value at sea level at a height equal to the radius of the

earth above its surface.

☒

Option 3 Same in all satellites ☒

Option 4 Same at all points on the surface of the earth ☐

Explanation GM x(a) g =

R

ein 3

e

(x = distance from centre of earth)

∴∴∴∴ g at centre = 0

GM(b) g surface =

R∞∞∞∞ 2

(((( ))))

GM GMg heights R = =

4R2R∞∞∞∞ 2 2

1 g at height R = g surface

4∞∞∞∞∴ ×∴ ×∴ ×∴ ×

(c) Weightlessness condition.

Q. No. 75 For two satellites at distance R and 7R above the earth’s surface, the ratio of their

Option 1 Total energies is 4 and potential and kinetic energies is 2 ☐

Option 2 Potential energies is 4 ☒

Option 3 Kinetic energies is 4 ☒

Option 4 Total energies is 4 ☒

Explanation Potential energies is 4, Kinetic energies is 4, Total energies is 4

Q. No. 76 A satellite is orbiting the earth in circular orbit of radius r. Its

Option 1 1Kinetic energy varies as

r

☒

Option 2 Angular momentum varies as

1

r

☐

Option 3 Linear momentum varies as

1

r

☒

Option 4 1Frequency of revolution varies as

r

3

2

☒

Explanation GMm 1KE = KE

2r r∴ ∝∴ ∝∴ ∝∴ ∝

Angular momentum = mvr

(((( ))))GM

= m rr

angular momentum r∴ ∝∴ ∝∴ ∝∴ ∝

Linear momentum = mv

GM

= mr

1 Linear momentum

r∴ ∝∴ ∝∴ ∝∴ ∝

GM 1freaquency = f

2 r r

∴ ∝∴ ∝∴ ∝∴ ∝

ππππ

3 3

2 2

Q. No. 77 An object is taken from a point P to another point Q in a gravitational field

Option 1 Assuming the earth to be spherical, if both P and Q lie on earth’s surface

the work done is zero

☒

Option 2 If P is on earth’s surface and Q above it, the work done is minimum when it

is taken along the straight line PQ

☐

Option 3 The work done depends only on the position of P and Q and is independent

of the path along which the particle is taken

☒

Option 4 There is no net work done if the objects is taken from P to Q and then

brought back to P, along any path

☒

Explanation w + w = KE∆∆∆∆ext c

w = U⇒⇒⇒⇒ ∆∆∆∆ext

Q. No. 78 Consider a planet moving in an elliptical orbit around the sun. The work done on the

planet by the gravitational force of the sun

Option 1 Is zero in any small part of the orbit ☐

Option 2 Is zero in some parts of the orbit ☒

Option 3 Is zero in complete revolution ☒

Option 4 Is zero in no part of the motion ☐

Explanation In elliptical orbit, only at some part of motion

F V→ →→ →→ →→ →

⊥⊥⊥⊥

In complete revolution, s = 0→→→→

and

gravitational force = conservative force

wc = 0

Q. No. 79 Let V and E denote the gravitational potential and gravitational field at a point. It is

possible to have

Option 1 V = 0 and E= 0 ☒

Option 2 V = 0 and E 0≠≠≠≠ ☐

Option 3 V 0 and E = 0≠≠≠≠ ☒

Option 4 V 0 and E 0≠ ≠≠ ≠≠ ≠≠ ≠ ☒

Explanation V = 0 and E = 0, V 0 and E = 0,≠≠≠≠

V 0 and E 0≠ ≠≠ ≠≠ ≠≠ ≠

Q. No. 80 An orbiting satellite will escape if

Option 1 Its speed is increased by 41% ☒

Option 2 (((( ))))Its speed in the orbit is made time of its initial 1.5 value ☐

Option 3 Its KE is doubled ☒

Option 4 It stops moving in the orbit ☐

Explanation Escape vel of satellite = 2 orbital vel. of satellite.××××

Q. No. 81 An astronaut, inside an earth satellite, experiences weightless-ness because

Option 1 No external force is acting on him ☐

Option 2 He is falling freely ☒

Option 3 No reaction is exerted by floor of the satellite ☒

Option 4 He is far away from earth’s surface ☐

Explanation He is falling freely, No reaction is exerted by floor of the satellite

Q. No. 82 If the radius of the earth suddenly decreases to 80 % of its present value, the mass of

the earth remaining the same, the value of the acceleration due to gravity will

Option 1 Remain unchanged ☐

Option 2 9.8Become ms

0.64

-2

☒

Option 3 Increase by 36% ☐

Option 4 Increase by about 56% ☒

Explanation Acceleration due to gravity at surface

GM

g = R

2

1g R = R, R = 0.8 R

R∝∝∝∝ 1 22

g R 1 1 = = =

g R 0.8 0.64

2 2

2 1

1 2

g 9.8 g = = m/s

0.64 0.64∴∴∴∴

212

g increase in g = - 1 100

g

∴ ×∴ ×∴ ×∴ ×

2

1

% %

1

= - 1 1000.64

××××

%

0.36

= 1000.64

××××

%

= 5 6 .2 5 %

Q. No. 83 Which of the following are correct?

Option 1 Escape velocity tells us how fast an object needs to go to escape the

planet’s gravity.

☒

Option 2 The time period of a geostationary satellite is 24 h. ☒

Option 3 If a satellite revolves close to the earth’s surface, then velocity of the

satellite is gR.

☒

Option 4 When a satellite is revolving around the earth in a fixed orbit, its linear

momentum changes continuously.

☒

Explanation Escape velocity tells us how fast an object needs to go to escape the planet’s gravity,

The time period of a geostationary satellite is 24 h, If a satellite revolves close to the

earth’s surface, then velocity of the satellite is gR.,

When a satellite is revolving

around the earth in a fixed orbit, its linear momentum changes continuously.

Q. No. 84 Assertion: If earth suddenly stops rotating about its axis, then the value of acceleration

due to gravity will become same at all the places.

Reason: The value of acceleration due to gravity is independent of rotation of earth.

Option 1 If both ASSERTION and REASON are true and reason is the correct explanation of the

assertion.

Option 2 If both ASSERTION and REASON are true but reason is not the correct explanation of

the assertion.

Option 3 IF ASSERTION is true but REASON is false.

Option 4 IF both ASSERTION and REASON are false.

Option 5 IF ASSERTION is false but REASON is true.

Correct Answer 3

Explanation IF ASSERTION is true but REASON is false

Q. No. 85 Assertion: Orbital velocity of a satellite is greater than its escape velocity.

Reason: Orbit of a satellite is within the gravitational field of earth whereas escaping is

beyond the gravitational field of earth.

Option 1 If both ASSERTION and REASON are true and reason is the correct explanation of the

assertion.

Option 2 If both ASSERTION and REASON are true but reason is not the correct explanation of

the assertion

Option 3 IF ASSERTION is true but REASON is false

Option 4 IF both ASSERTION and REASON are false

Option 5 IF ASSERTION is false but REASON is true

Correct Answer 4

Explanation IF both ASSERTION and REASON are false

Q. No. 86 Assertion: The time period of revolution of a satellite close to surface of earth is

smaller than that revolving away from surface of earth.

Reason: The square of time period of revolution of a satellite is directly proportional to

cube of its orbital radius.

Option 1 If both ASSERTION and REASON are true and reason is the correct explanation of the

assertion.

Option 2 If both ASSERTION and REASON are true but reason is not the correct explanation of

the assertion

Option 3 IF ASSERTION is true but REASON is false

Option 4 IF both ASSERTION and REASON are false

Option 5 IF ASSERTION is false but REASON is true

Correct Answer 1

Explanation 2 r

T = GM

ππππ

3

2

Q. No. 87 Assertion: Generally the path of a projectile from the earth is parabolic but it is

elliptical for projectiles going to a very larger height

Reason: The path of a projectile is independent of the gravitational force of earth.

Option 1 If both ASSERTION and REASON are true and reason is the correct explanation of the

assertion.

Option 2 If both ASSERTION and REASON are true but reason is not the correct explanation of

the assertion

Option 3 IF ASSERTION is true but REASON is false

Option 4 IF both ASSERTION and REASON are false

Option 5 IF ASSERTION is false but REASON is true

Correct Answer 3

Explanation IF ASSERTION is true but REASON is false

Q. No. 88 Assertion: We cannot move even a finger without disturbing all the stars.

Reason: Everybody in this universe attracts every other body with a force which is

inversely proportional to the square of distance between them.

Option 1 If both ASSERTION and REASON are true and reason is the correct explanation of the

assertion.

Option 2 If both ASSERTION and REASON are true but reason is not the correct explanation of

the assertion

Option 3 IF ASSERTION is true but REASON is false

Option 4 IF both ASSERTION and REASON are false

Option 5 IF ASSERTION is false but REASON is true

Correct Answer 1

Explanation If both ASSERTION and REASON are true and reason is the correct explanation of the

assertion

Q. No. 89 Statement-1: Escape velocity is independent of the angle of projection

Statement-2: Escape velocity from the surface of earth is gR Where R is radius of

earth.

Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for

Statement-1.

Option 2 Statement-1 True, Statement-2 is true; Statement-2 is NOT correct explanation for

Statement-1

Option 3 Statement-1 is true, Statement-2 is false.

Option 4 Statement-1 is False, Statement-2 is True.

Correct Answer 3

Explanation V from earth surface = 2gResc

Q. No. 90 Statement-1: Gravitational potential is zero inside a shell.

Statement-1: Gravitational potential is equal to the work done in bringing a unit mass

from infinity to a point inside gravitational field.

Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for

Statement-1.

Option 2 Statement-1 True, Statement-2 is true; Statement-2 is NOT correct explanation for

Statement-1

Option 3 Statement-1 is true, Statement-2 is false.

Option 4 Statement-1 is False, Statement-2 is True.

Correct Answer 4

Explanation (((( ))))

-GMV = R = radius of shell

Ri

Q. No. 91 Statement-1: A spherically symmetric shell produces no gravitational field anywhere.

Statement-1: The field due to various mass elements cancels out, everywhere inside

the shell.

Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for

Statement-1.

Option 2 Statement-1 True, Statement-2 is true; Statement-2 is NOT correct explanation for

Statement-1

Option 3 Statement-1 is true, Statement-2 is false.

Option 4 Statement-1 is False, Statement-2 is True.

Correct Answer 4

Explanation GMg = 0, g = for shell.

rin out 2

Q. No. 92 Statement-1: Rate of change of weight near the earth’s surface with height h is

porportional to h0.

Statement-2: Since gravitational potential is given by v = -GM/r.

Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for

Statement-1.

Option 2 Statement-1 True, Statement-2 is true; Statement-2 is NOT correct explanation for

Statement-1

Option 3 Statement-1 is true, statement-2 is false.

Option 4 Statement-1 is False, statement-2 is True.

Correct Answer 2

Explanation (((( ))))Near earth surface h Re<<

2hg = g 1-

R

h se

dgh -2gs =

dh R∴∴∴∴

e

Q. No. 93 Statement-1: Two particles are to be projected from the surface of earth so that

particles just leave the gravitational field of earth. One particle is projected vertically

upward and another is at an angle of 450 with vertical. Speed given to both particles is

same.

Statement-2: Escape speed does not depend upon angle of projection.

Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for

Statement-1.

Option 2 Statement-1 True, Statement-2 is true; Statement-2 is NOT correct explanation for

Statement-1

Option 3 Statement-1 is true, Statement-2 is false.

Option 4 Statement-1 is False, Statement-2 is True.

Correct Answer 1

Explanation Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for

Statement-1

Q. No. 94 Statement-1: For a satellite revolving very near to earth’s surface the time period of

revolution is given by 1 h 24 min.

Statement-2: The period of revolution of satellite depends only upon its height above

the earth’s surface.

Option 1 Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for

Statement-1.

Option 2 Statement-1 True, Statement-2 is true; Statement-2 is NOT correct explanation for

Statement-1

Option 3 Statement-1 is true, Statement-2 is false.

Option 4 Statement-1 is False, Statement-2 is True.

Correct Answer 1

Explanation 2 r

T = GM

ππππ

3

2

For satellite near earth surface

h 0, v = R→→→→

GMg =

R2

(((( )))) (((( ))))2 R 2 R T = =

GM g

π ππ ππ ππ π∴∴∴∴

2 232

R 6.4 10 R T = 2 = 2

g 10

××××∴ π π∴ π π∴ π π∴ π π

6

2 800se cπ ×π ×π ×π ×����

= 84.6 min

= 1.4 hr.

Passage Text In the graph shown, the PE of earth-satellite system is shown by solid line as a function

of distance r (the separation between earth’s centre and satellite). The total energy of

two objects which may or not be bounded to earth are shown in figure dotted lines.

Q. No. 95 Mark the correct statement(s)

Option 1 The object having total energy E1 is bounded one.

Option 2 The object having total energy E2 is bounded one.

Option 3 Both the objects are bounded one.

Option 4 Both the objects are unbounded one.

Correct Answer 1

Explanation For bounded objects, total energy = (-) ve.

Q. No. 96 If object having total energy E1 is having same PE curve as shown in figure, then

Option 1 r0 is the maximum distance of object from earth’s centre

Option 2 This object and earth system is bounded one

Option 3 The KE of the object is zero when r = r0

Option 4 All the above

Correct Answer 4

Explanation (a) At r0, Total energy = potential energy

So KE = 0, speed = 0, r0 = maximum distance

(b) As total energy = (-) ve : bounded system.

Q. No. 97 If both the object have same PE curve as shown in figure, then

Option 1 For object having total energy E2 all values of r are possible

Option 2 For object having total energy E2 values of r < r0 are only possible

Option 3 For object having total energy E1 all values of r are possible

Option 4 None of these

Correct Answer 1

Explanation As TE = (+) ve an PE = (-)ve so KE = (+)ve for all ‘r’.

Passage Text The satellites when launched from earth are not given the orbital velocity initially, in

practice, a multi-stage rocket propeller carries the space-craft upto its orbit and during

each stage rocket has been fired to increase the velocity to acquire the desired velocity

for a particular orbit. The last stage of the rocket brings the satellite in

circular/elliptical (desired) orbit.

Consider a satellite of mass 150 kg in low circular orbit, in this orbit, we can’t neglect

the effect of air drag. This air drag opposes the motion of satellite and hence total

mechanical energy of earth-satellite system decreases means total energy becomes

more negative and hence orbital radius decreases which cause the increase in KE.

When the satellite comes in enough low orbit, the excessive thermal energy

generation due to air friction may cause the satellite to burn up.

Q. No. 98 What is the reason that during launching of satellite, while crossing the atmosphere it

don’t get burnt, but while falling down towards earth or orbiting in lower orbit, it gets

burnt up?

Option 1 While going up air friction force doesn’t come in to existence

Option 2 While going up satellite is with launching vehicle whose speed is controllable

Option 3 While going up space-craft is protecting the satellite from air friction by itself getting

burned

Option 4 None of these

Correct Answer 2

Explanation Air friction depends on speed of object.

Q. No. 99 What would be the motion of satellite if air drag has to be considered?

Option 1 Moves with uniform speed in the launching orbit

Option 2 Orbital radius decreases continuously as a result moves with non-uniform velocity in

elliptical orbit

Option 3 Orbital radius decreases continuously and hence collapses with earth after some time

in random manner and there is equal chance of burning up the satellite due to air

friction also.

Option 4 Moves with non-uniform speed in the launching orbit

Correct Answer 3

Explanation Orbital radius decreases continuously and hence collapses with earth after some time

in random manner and there is equal chance of burning up the satellite due to air

friction also

Q. No. 100 It has been mentioned in passage that as r decreases, E decreases but K increases. The

increases in K is [E = Total mechanical energy, r = orbit radius, K = kinetic energy] is

Option 1 Due to increases in gravitational PE

Option 2 Due to decreases in gravitational PE

Option 3 Due to work done by air friction force

Option 4 Both (b) and (c)

Correct Answer 2

Explanation Work done by air friction = (-)ve

So, will not increase KE.

Q. No. 101 If due to air drag, the orbital radius of earth decreases from R to R - R, R R,∆ ∆∆ ∆∆ ∆∆ ∆ << then

the expression for increase in orbital velocity v∆∆∆∆ is

Option 1 R GM

2 R

∆∆∆∆3

Option 2 R GM

2 R-

∆∆∆∆3

Option 3 GMR

R∆∆∆∆

3

Option 4 GMR

R-∆∆∆∆

3

Correct Answer 1

Explanation v RGM 1v = = -

R v 2 R

∆ ∆∆ ∆∆ ∆∆ ∆⇒⇒⇒⇒

(((( )))) v = GM R⇒⇒⇒⇒1

2

(((( ))))veR = decrease in R

++++∆∆∆∆

- R GM V =

2R R

∆∆∆∆ ⇒⇒⇒⇒ ∆∆∆∆

R GM

= 2 R

∆∆∆∆3

Q. No. 102 For information given in above question, The change in KE, ∆∆∆∆ K is

Option 1 GMm R

R× ∆× ∆× ∆× ∆-

2

Option 2 GMm R

R× ∆× ∆× ∆× ∆

2

Option 3 GMmR

2R∆∆∆∆

2

Option 4 GMm- R

2R× ∆× ∆× ∆× ∆

2

Correct Answer 3

Explanation

1K = mv

2

2

k v R R1 = 2 = 2 =

k v 2 R R

∆ ∆ ∆ ∆∆ ∆ ∆ ∆∆ ∆ ∆ ∆∆ ∆ ∆ ∆ ⇒⇒⇒⇒

R R GMm GMm k = k = = R

R R 2R 2R

∆ ∆∆ ∆∆ ∆∆ ∆ ⇒⇒⇒⇒ ∆ ∆∆ ∆∆ ∆∆ ∆

2

Q. No. 103 For information given in question no. 101, the change in PE ∆∆∆∆ U is

Option 1 GMmR

R∆∆∆∆-

2

Option 2 GMmR

R∆∆∆∆

2

Option 3 GMmR

2R∆∆∆∆

2

Option 4 GMmR

2R∆∆∆∆

2-

Correct Answer 1

Explanation P R GMmP = = -

R P R

∆ ∆∆ ∆∆ ∆∆ ∆⇒⇒⇒⇒

- R -GMm P =

R R

∆∆∆∆ ⇒⇒⇒⇒ ∆∆∆∆

GMm= - R

R

∆∆∆∆

2

Q. No. 104 For information given in question no. 101, the work done by air friction force W, is

(Where M is the mass of earth and m is mass of satellite.)

Option 1 GMmR

R∆∆∆∆

2-

Option 2 GMmR

R∆∆∆∆

2

Option 3 GMmR

2R∆∆∆∆

2

Option 4 GMmR

2R∆∆∆∆

2-

Correct Answer 4

Explanation Work energy theorem :

w + w = KE∆∆∆∆c NC

w = KE + PE⇒⇒⇒⇒ ∆ ∆∆ ∆∆ ∆∆ ∆NC

GMm -GMm= R + R

2R R

∆ ∆∆ ∆∆ ∆∆ ∆

2 2

GMm= - R

2R

∆∆∆∆

2

Passage Text The minimum and maximum distance of satellite from the centre of the earth are 2R

and 4R, respectively, where R is the radius of earth and M is the mass of the earth.

Q. No. 105 The minimum and maximum speeds

Option 1 GM 2GM,

9R R

Option 2 GM 3GM,

5R 2R

Option 3 GM 2GM,

6R 3R

Option 4 GM 5GM,

3R 2R

Correct Answer 3

Explanation Conservation of angular momentum :

mV r = m V rmax min min max

⇒⇒⇒⇒ Vmax = 2 Vmin ….(1)

Conservation of mechanical energy :

1 GMm 1 GMm m V - = m V -

2 V 2 r

2 2max min

min max

By putting value Vmin = 2R, Vmax = 4r

(((( ))))GM

V - V = --- 22R

∴∴∴∴2 2max min

From (1) and (2)

GM 2GMV = V =

6R 3Rmin max

Q. No. 106 Radius of curvature at the point of minimum distance is

Option 1 8R

3

Option 2 5R

3

Option 3 4R

3

Option 4 7R

3

Correct Answer 1

Explanation mv GMm =

R r

2max

2c min

(((( )))) (((( ))))v r R =

GM⇒⇒⇒⇒

2 2

max minc

(((( ))))2R2GM

= 3R GM

××××

2

8R

= 3

Q. No. 107 For a planet orbiting about sun in elliptical orbit, some incomplete statements

regarding physical quantities are given in column - I, which can be completed by using

entries of column - II. Match the entries of column - I with the entries of column - II.

No. Column A Column B Column C Id of Additional

Answer

1 (a) Maximum PE of

sun planet system

(Q) is at aphelion

(S) is dependent on

semi-major axis

orbit

2 (b) Maximum

speed of planet

(P) Is at perihelion (R) is independent

of mass of planet,

(S) is dependent on

semi-major axis

orbit

3 (c) Manimum PE of

sun planet system

(P) Is at perihelion

(S) is dependent on

semi-major axis

orbit

4 (d) Minimum

kinetic energy of

planet

(Q) is at aphelion

(S) is dependent on

semi-major axis

orbit

Explanation

vmin = perigee, vmax = Apogee.

-GMmPE =

r

1KE = mv

2

2

m vmax rmin = m vmin rmax

Q. No. 108 For identical satellites are orbiting in four elliptical orbits having same semi-major axis

but different eccentricities. In Column - I some quantities associated with four orbits

are given and in column - II the words which can give the information about physical

quantities mentioned in column - I. Match the entries of column - I with the entries of

column - II.

No. Column A Column B Column C Id of Additional

Answer

1 (a) Total energy of

all four orbits

(P) Same (R) Constant

2 (b) Speed of

satellite in all four

orbits

(Q) Different

(S) Varying

3 (c) Velocity of

satellite in all four

orbits

(Q) Different

(S) Varying

4 (d) Angular

momentum of

satellite about

centre of earth in

all four orbits

(Q) Different

(R) Constant

Explanation Total energy is conserved, as there is no non-conservative force.

For given orbit, angular momentum is conserved for different orbit.

Q. No. 109 Considering earth to be a homogeneous sphere but keeping in mind its spin, match the

following :

No. Column A Column B Column C Id of Additional

Answer

1 (a) Acceleration

due to gravity

(P) May change

from point to point

(Q) Does not

depend on

Direction of

projection

2 (b) Orbital angular

momentum of the

earth as seen from

a distant star

(R) Remain

constant

3 (c) Escape velocity

from the earth

(P) May change

from point to point

4 (d) Gravitational

potential due to

earth at a particular

point

(R) Remain

constant

Explanation a-P, b-R, c-P, d-R

Q. No. 110 A sky lab of mass 2 10 kg××××3 is first launched from the surface of earth in circular orbit

of radius 2R and then it is shifted from this circular orbit to another circular orbit of

radius 3R. Calculate the minimum energy (in 108 J) required to shift the lab from first

orbit to the second orbit. R = 6400 km, g = 10m/s2.

Correct Answer 0106

Is Integer Type ☒

Explanation (((( )))) (((( ))))

-GMm -GMmTE = TE =

4R 6Ri f

∴∴∴∴ Energy required = (TE)f - (TE)i

-GMm -GMm

= = 6R 4R

GMm mgR

= = 12R 12

(((( )))) (((( )))) (((( ))))2 10 10 6400 1000

= 12

× ×× ×× ×× ×3

1280

= 10 J12

××××8

0106 10 J××××8

����

Q. No. 111 Distance between the centres of two stars is 10a. The masses of these stars are M and

16 M and their radii a and 2a respectively. A body of mass m is fired straight from the

surface of the larger star towards the smaller star. What should be its minimum initial

speed (in km/s) to reach the surface of the smaller star?

(((( ))))M = 6 10 kg, a = 1200 km××××22

Correct Answer 3100

Is Integer Type ☒

Explanation

At P, gnet = 0

(((( ))))

(((( ))))

G 16MGM =

x 10a - x2 2

1 4 = 10a - x = 4xx 10a - x

⇒ ⇒⇒ ⇒⇒ ⇒⇒ ⇒

x = 2 a⇒⇒⇒⇒

Conservation of mechanical energy :

(((( )))) (((( )))) (((( )))) (((( ))))G 16M m G M m G 16M m G M mmv - - = 0 - -

2 2a 8a 8a 2a

2

mv 8GMm GMm 2GMm GMm - - = - =

2 a 8a a 2a⇒⇒⇒⇒

2

mv 8GMm GMm 2GMm GMm

= + - - 2 a 8a a 2a

2

mv 64GMm + GMm - 16GMm - 4GMm =

2 8a⇒⇒⇒⇒

2

mv 35GMm =

2 8a⇒⇒⇒⇒

2

35GM v =

4a⇒⇒⇒⇒

Q. No. 112 A man can jump vertically to a height of 1.5 m on the earth. Calculate the radius of a

planet of the same mean density as that of the earth from whose gravitational field he

could escape by jumping. Radius of earth is 6.41 10 m.××××6

Correct Answer 3100

Is Integer Type ☒

Explanation Conservation of mechanical energy at earth :

(((( ))))mv GMm GMm

- = -2 R R + h

2

(((( ))))v GM GM GM 1

= - = 1 - h2 R R + h R

1 + R

⇒⇒⇒⇒

2

2GM h v = 1 - 1 +

R R

⇒⇒⇒⇒

-12

2GM h

v = R R

××××2

2GMh 4 v = M = R

3R

⇒⇒⇒⇒ π ρπ ρπ ρπ ρ

3

2∵∵∵∵

8 v = G Rh

3⇒⇒⇒⇒ π ρπ ρπ ρπ ρ

v = vese

8 8 G Rh = G R

3 3⇒⇒⇒⇒ π ρ π ρπ ρ π ρπ ρ π ρπ ρ π ρ 2

p

R = Rh⇒⇒⇒⇒ p

Q. No. 113 Two satellites A and B of equal mass moves in the equatorial plane of the earth, close

to the earth’s surface. Satellite A move s in the same direction as that of the rotation

of the earth while satellite B moves in the opposite direction. Calculate the ratio of the

kinetic energy of B to that of A in the reference frame fixed to the earth.

(((( ))))g = 9.8 m/s , R = 6.37 10 m××××2 6

Correct Answer 0001

Is Integer Type ☒

Explanation Speed of satellite close to earth surface.

GMv = = gR = 9.8 6.37 10

R× ×× ×× ×× × 6

sat

= 62.426 10 m / s××××3

= 62.426 km/s

Speed of particle of earth surface VE = wR

(((( ))))2

= RT

ππππ

2

= 6.37 10 m/s24 3600

ππππ × ×× ×× ×× ×

××××

6

= 0.463 km/s

Vel of sat (1) w.r.t. Earth

v2 = vset - ve = 62.426 - 0.463 = 61.963 km/s

KE v Ratio of KE = = 1.030

KE v

∴∴∴∴

2

1 1

2 2

����

Q. No. 114 Three particles of mass m each are placed at the three corners of an equilateral

triangle of side a. Find the work (in 10-10

J) which should be done on this system to

increase the side of the triangle to 2a. (m = 5 kg, a = 10 cm)

Correct Answer 0250

Is Integer Type ☒

Explanation

-3GmPE =

x

2

systm

(((( ))))-3Gm

PE = a

2

i

(((( ))))-3Gm

PE = 2a

2

f

(((( )))) (((( ))))(((( )))) (((( ))))

(((( ))))

6.67 10 53 Gm 3w = PE - PE = =

2 a 2 0.1

××××2-11

2

ext f i

0.250 10××××-10

�

= 0250

Q. No. 115 Earth is a sphere of uniform mass density. How much would a body weight (in×××× 10N)

half ways down the centre of earth if it weight 100N on the surface?

Correct Answer 0005

Is Integer Type ☒

Explanation GMmweight on surface = = 100 N

R2

Weight at a point at R

2 from centre

RGMm

GMm2= =

2RR

2

= 50 N

Q. No. 116 An infinite collection of equal masses of 2 kg are kept on a horizontal line (x-axis) at

positions x = 1, 2, 4, 8,….Find the gravitational potential at x = 0 (in - Gj units)

Correct Answer 0004

Is Integer Type ☒

Explanation 2kg 2kg 2kg 2kg 2kg

x = 0 x = 1m 2m 4m 8m 16m

• • • • • •• • • • • •• • • • • •• • • • • •

(((( ))))

(((( ))))

(((( )))) (((( )))) (((( ))))-G 2 G 2 G 2 G 2Pot at x = 0 = - - -

1 2 4 8

1 1 1= -2G 1 + + + +....

2 4 8

= -4G

Q. No. 117 Two satellite of mass ratio 1 : 2 are revolving around the earth in circular orbits such

that the distance of the second satellite is four times as compared to the distance of

the first satellite. Find the ratio of their centripetal force.

Correct Answer 0008

Is Integer Type ☒

Explanation GMm mF = f

r r∴ ∝∴ ∝∴ ∝∴ ∝

2 2

f m r 1 4 = = = 8

f m r 2 1

⋅⋅⋅⋅

2 2

1 1 2

2 2 1

Q. No. 118 Distance between the centres of two stars is 10a. The masses of the stars are M and 16

M their radii are a and 2a, respectively. A body of mass m is fired straight from the

surface of the larger star towards the smaller star. What should be its minimum speed

to reach the surface of the smaller star (round off to the nearest integer in the unit of

GM

a)

Correct Answer 0003

Is Integer Type ☒

Explanation

At P, gnet = 0

(((( ))))

(((( ))))

G 16MGM =

x 10a - x2 2

1 4 = 10a - x = 4xx 10a - x

⇒ ⇒⇒ ⇒⇒ ⇒⇒ ⇒

x = 2 a⇒⇒⇒⇒

Conservation of mechanical energy :

(((( )))) (((( )))) (((( )))) (((( ))))G 16M m G M m G 16M m G M mmv - - = 0 - -

2 2a 8a 8a 2a

2

mv 8GMm GMm 2GMm GMm - - = - =

2 a 8a a 2a⇒⇒⇒⇒

2

mv 8GMm GMm 2GMm GMm

= + - - 2 a 8a a 2a

2

mv 64GMm + GMm - 16GMm - 4GMm =

2 8a⇒⇒⇒⇒

2

mv 35GMm =

2 8a⇒⇒⇒⇒

2

35GM v =

4a⇒⇒⇒⇒