JEE

Manzil JEE Comeback (2025)

Physics
DPP: 9
Gravitation

Q1 If the distance between the centres of earth Q5 Following curve shows the variation of
and moon is D and mass of earth is 81 times →
that of moon. At what distance from the intensity of gravitational field (I ) with distance
centre of earth gravitational field will be zero: from the centre of solid sphere(r):
(A) D (A)
2

(B) 2D

(C) 4D

(D) 9D (B)
10

Q2 Two identical satellites are moving around the

earth with constant velocity in circular orbits at
height R and 3R from surface of earth
respectively, R being the radius of the earth. (C)
The ratio of their kinetic energies is:
(A) 2 : 1 (B) 4 : 3
(C) 8 : 1 (D) 16 : 1

Q3 A body weighs 144 at the surface of earth.

N
(D)
When it is taken to a height of h = 3R , where
R is radius of earth, it would weigh:
(A) 48 N

(B) 36 N

(C) 16 N

(D) 9 N
Q6 The ratio of the K.E. required to be given to the
satellite to escape earth's gravitational field to
Q4 If the angular speed of the earth is doubled, the K.E. required to be given so that the
the value of acceleration due to gravity g at the satellite moves in a circular orbit just above
north pole:- the earth's surface is:
(A) Doubles (B) Becomes half (A) 1 : 1 (B) 2 : 1
(C) Remains same (D) Becomes zero (C) 1 : 2 (D) ∞

Q7 Given below are three statements. Based on

their truthness/falseness, choose correct

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option. (A) 1 (B) 1

m m
Statement I: Angular momentum of planets 9 10

revolving around sun is conserved. (C) 1 (D) 10

m m
Statement II: The total energy of a satellite is
11 11

negative. Q11 Two stars of masses m and 2m at a distance d

Statement III: Linear momentum of planets
rotate about their common centre of mass in
revolving around sun is also conserved.
free space. The period of revolution is:
(A) Statement I, Statement II are incorrect,
(A)
Statement III is correct. 2π
√
3Gm

(B) Statement I, Statement II are correct,

3
d

(B)
Statement III is incorrect. 1
√
3Gm

(C) Statement I, Statement III are incorrect, 2π d

3

Statement II is correct. (C) 3

d
√
2π
(D) Statement II, Statement III are correct, 3Gm

Statement I is incorrect. (D) 3

1 d
√

Q8 Three equal masses of 1

2π 3Gm
kg each are placed at
the vertices of an equilateral triangle PQR and Q12 A satellite of mass M is in a circular orbit of
a mass of 2 kg is placed at the centroid O of radius R about the centre of the earth. A
meteorite of the same mass, falling towards
the triangle which is at a distance of √2 m
the earth, collides with the satellite completely
from each of the vertices of the triangle. The
inelastically. The speeds of the satellite and the
gravitational force, in newton, acting on the
meteorite are the same, just before the
mass of 2 kg is:-
collision. The subsequent motion of the
(A) 2 (B) √ combined body will be
2

(C) 1 (D) Zero (A) In the same circular orbit of radius R

(B) Such that it escapes to infinity
Q9 The value of acceleration due to gravity at a
(C) In an elliptical orbit
height of R/2 above the Earth's surface will be
(D) In a circular orbit of a different radius
(Take acceleration due to gravity on the
surface of earth to be g) Q13 Two satellites of masses m and 3m revolve
(A) g
around the earth in circular orbits of radii r &
3r respectively. The ratio of orbital speeds of
2

(B)
the satellites respectively is:
3g

4
(A) 1 : 1
(C) 4g

(B) 3 : 1
9

(D) g (C)
√3 : 1

(D) 9
3
: 1

Q10 There are two bodies of masses 100 kg and

Q14 1
10000 kg separated by a distance 1 m. At what The mass of a planet is th that of the earth
10

distance from the smaller body, the intensity

and its diameter is half that of the earth. The
of gravitational field will be zero.

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acceleration due to gravity on the surface of

that planet is:
(A) −2
19. 6 m s

(B) 9. 8 m s
−2

(C) −2
4. 9 m s
A star in its prime age is said to be under
(D) 3. 92
equilibrium due to gravitational pull and
−2
m s

Q15 Two particles of equal mass m go around a outward radiation pressure (p). Consider the

circle of radius R under the action of their shell of thickness dr as in the figure. If the
mutual gravitational attraction. The speed v of pressure on this shell is dp then the correct
each particle is
equation is ( G is universal gravitational
(A)
1
√
Gm constant)
2 R
(A) dp −GM r ρ r

(B) dr
=
r
2
4Gm
√
R (B) dp −4GM r ρ r

=
(C) dr r
GM
√
2R
(C) dp −3GM r ρ r

=
(D) dr
r

Gm
√
(D) dp −GM r ρ r
R
=
r
dr

Q16 A geostationary satellite is orbiting around an

Q18 If ν e and ν 0 represent the escape velocity and
arbitrary planet P at a height of 11R above
orbital velocity of a satellite corresponding to a
the surface of P , R being the radius of P .
circular orbit of radius R, then:
The time period of another satellite in hours at
(A) ν e = ν0 (B) √
a height of 2R from the surface of P is 2ν 0 = ν e

________. (C) v0 (D) ν e = 2v 0

νe =
√2
P has the time period of 24 hours.
(A) Q19 The mass of a body on the surface of the earth
6√ 2

(B) 3 is 10 kg. The mass of the same body on the

surface on the moon is (gm = 1/6 ge, where gm,
(C) 5
(D) 6
ge is acceleration due to gravity on the surface
√2 of the moon and the earth, respectively)
(A) 10 kg (B) 20 kg
Q17 A star can be considered as a spherical ball of (C) 3.33 kg (D) 15 kg
hot gas of radius R . Inside the star, the density
Q20 The figure shows elliptical orbit of a planet m
of the gas is ρ r at a radius r and mass of the
about the sun S. The shaded area SCD is
gas within this region is M r .
twice the shaded area SAB. If t 1 is the time
for the planet to move from C to D and t 2 is
the time to move from A to B then :-

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Q23 The work done in slowly lifting a body from

earth's surface to a height R (radius of earth) is
equal to two times the work done in lifting the
same body from earth's surface to a height h,
then
(A) R (B) R
h = h =
6 4

(C) R (D) R
h = h =
3 2

(A) t 1 = t2
Q24 Imagine a light planet revolving around a very
massive star in a circular orbit of radius R with
(B) t 1 = 3t 2
time period of revolution T. If the gravitational
(C) t 1 = 4t 2
force of attraction between the planet and the
(D) t 1 = 2t 2
star is proportional to R–3/2, then
(A) T is proportional to R5/4
Q21 Two bodies of masses m1 and m2 are initially
(B) T is proportional to R7/4
at rest at infinite distance apart. They are then
allowed to move towards each other under (C) T is proportional to R3/2
mutual gravitational attraction. Their relative (D) T is proportional to R
velocity of approach at a separation distance r
Q25 A large spherical mass M is fixed at one
between them is
(A) 1/2
position and two identical point masses m are
(m 1 −m 2 )

[2G
r
]
kept on a line passing through the centre of M
(B) 1/2
(see figure). The point masses are connected
2G
[ (m 1 + m 2 )]
r by a rigid massless rod of length l and this
(C) r assembly is free to move along the line
[ ]
2G(m 1 m 2 )
connecting them. All three masses interact
(D) 2G 1/2 only through their mutual gravitational
[ m1 m2 ]
r
interaction. When the point mass nearer to M
Q22 Four particles of masses m, 2m, 3m and 4m are is at a distance r = 3 l from M , the tension in
kept in sequence at the corners of a square of M
the rod is zero for m = k( . The value of k
)
side a. The magnitude of gravitational force 288

acting on a particle of mass m placed at the is

centre of the square will be:
(A) 24m G
2

2
a

(B) 2
6m G
Q26 Gravitational acceleration on the surface of a
2
a

(C)
√ 6g

4√ 2Gm
2
planet is where g is the gravitational
11
2
a
acceleration on the surface of the earth. The
(D) Zero
average mass density of the planet is 2/3
times that of the earth. If the escape speed on

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the surface of the earth is taken to be Q30 Imagine a new planet having the same density
as that of the earth but its radius is 3 times
11 kms
−1
, the escape speed on the surface of
bigger than the earth’s radius. If the
the planet in kms will be
−1

acceleration due to gravity on the surface of

Q27 The gravitational potential energy of a satellite the earth is g and that on the new planet is g′,
revolving around the earth in circular orbit is then what is the value of g′/g?
4MJ . Find the additional energy ( in MJ) that
should be given to the satellite so that it
escapes from the gravitational force of the
earth (consider only gravitational force on the
satellite and no atmospheric resistance ).

Q28 The initial velocity v i required to project a

body vertically upward from the surface of the
earth to reach a height of 10R, where R is the
radius of the earth, may be described in terms
of escape v e velocity such that
x
vi = √ × ve . The value of x will be_______.
11

Q29 A particle is projected from point A, that is at a

distance 4R from the centre of the earth, with
speed v 1 in a direction making 30 ∘ with the

line joining the centre of the earth and point A

, as shown. Find v 1 if the particle passes
grazing the surface of the earth. Consider
gravitational interaction only between these
two. Express your answer in the form
1000X
m/s and fill value of X. (use
√2

GM
7 2 2
= 6.4 × 10 m /s )
R

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Answer Key
Q1 D Q16 B

Q2 A Q17 A
Q3 D Q18 B
Q4 C Q19 A

Q5 A Q20 D
Q6 B Q21 B
Q7 B Q22 C

Q8 D Q23 C
Q9 C Q24 A
Q10 C Q25 7

Q11 C Q26 3
Q12 C Q27 2
Q13 C Q28 10
Q14 D Q29 8

Q15 A Q30 3

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:
(4)

Given, m E = 81m m

let ' p ' be the point at which gravitational field Q2 Text Solution:
will be zero (1)
∴ OP = x & O′P = D − x

now, for gravitational field to be zero at point '

p '

Gm e
∴ Ee = at P
2
Q3 Text Solution:
x
(4)
Gm m
′
Em = at P g 2
R 1
(D − x)
2
= =
g 2
16
(4R)

Gm E Gm m
Video Solution:
∴ =
2 2
x (D − x)

81m m mm
=
2 2
x (D − x)

81 1
=
2 2
x (D − x)
Q4 Text Solution:

taking square root Acceleration due to gravity at poles is

independent of the angular speed of earth.
9 1 Video Solution:
=
x (D − x)

⇒ 9D − 9x = x ⇒ 9D = 10x

⇒ x = 9D/10

Video Solution:
Q5 Text Solution:
The variation of gravitational field intensity
with distance from the centre of the solid
sphere shows

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 2

2
∣
→

mv

orbit

1
mv
Outside the sphere, intensity decreases

with the square of the distance

Inside the sphere, the intensity increases

linearly with distance from the centre.

Q6 Text Solution:

0
∝ r

Video Solution:

K.E. required for satellite to escape from

earth's gravitational field

1
2

=
1

2
m(

m(
√

√
2GM

GM

R
R
)

The ratio between these two energies = 2

Q7 Text Solution:
2

Statement I is correct because the force of

gravitation on the planet by sun is central.
Statement II is correct because total
mechanical energy of an object is the sum of

(relative to infinity is always negative) and

since |KE|
negative.

at every point.

Video Solution:
< |P E|,
=

K.E. required for satellite to move in circular

its K.E. (which is always positive) and the P.E.

GM m

GM m

2R

hence total energy is

Statement III is incorrect because p = mv and

velocity of planets around sun is not constant

Android App
→

I ∝
1

R
2

Q8 Text Solution:
The correct option is D zero
Given:

OP = OQ = OR = √ 2 m

m = 1 kg; M = 2 kg

The gravitational force on mass 2 kg due to

mass 1 kg at P.

F OP = G

Similarly,

F OQ = G

F OR = G

F OQ cos30

each other.
2×1

(√ 2)

2×1

(√ 2)

2×1

(√ 2)

∘
2

2
= G

= G
along OP

along OQ

= G along OR

Net force at O in direction parallel to the base

of triangle is due to force F OQ and F OR .

and F OR cos30 ∘ are equal and

acting in opposite directions, thus balance

Therefore net force at O will be in vertical

direction.
The resultant force on the mass 2 kg at O is
given by
F net

⇒ F net
= F OP − (F OQ sin30

= G − (

Video Solution:

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G

2
+
G

2
∘ ∘
+ F OR sin30 )

) = 0
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2
⇒ r 1 = 2 (d − r 1 ) = d
3

Since, gravitational force between masses =

Centripetal force

2d G × m × 2m
2
Q9 Text Solution: ∴ mω ⋅ =
2
⇒ ω =
3 d
Acceleration due to gravity at a height from
the earth's surface is given by, √
3Gm

3
g d
g =
h 2
(1+ ) 3
R
2π d
√
T = = 2π
R ω 3Gm
At height, h =
2

g =
g

=
4g
Video Solution:
1
9
(1+ )
2

Video Solution:

Q12 Text Solution:

Q10 Text Solution:

By momentum conservation,
G×100 G×10000 1 10 1

2
=
2
⇒
x
= ⇒ x = m
x 1–x 11
(1–x)
v
vx =
Video Solution:
2

v
vy =
2

2 2
v v
√
⇒ v N et = +
4 4

v
=
√2

∵ v t < v N et < v 0
Q11 Text Solution:
⇒ Path is elliptical

Video Solution:

From the center of mass formula,

mr 1 = 2mr 2

Q13 Text Solution:

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GM
Orbital speed, v =
√
r
(M : Mass of Earth

⇒ Orbital speed is independent of mass of the

satellite.
1
v ∝
√r
2
mv Gmm 1 Gm
√
= ⇒ V =
v1 r2 3r R 2 2 R
√ (2R)
⇒ = √ =
v2 r1 r

Video Solution:
Video Solution:

Q16 Text Solution:

Q14 Text Solution:
T 3 3/2
M = ( ) ⇒ T = 3 hours
mass of planet (M = mass of earth) 24 12
10

R
Video Solution:
Radius of planet = (R = radius of earth)
2

Acceleration due to gravity on planet is

GM /10 2 GM
g = = = 0.4 × g = 0.4 × 9.8
2 5 2
(r/2) R

2
= 3. 92 m/s

Video Solution: Q17 Text Solution:

dp −GM r ρ r

=
2
dr r

Video Solution:

Q15 Text Solution:

Both the particles moves diametrically
opposite position along the circular path of
radius R and the gravitational force provides Q18 Text Solution:
required centripetal force.
ν e = √ 2gR and ν 0 = √ gR

∴ √ 2ν = νe
0

Q19 Text Solution:

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Mass of the body is independent on the Video Solution:

gravitational acceleration, i.e., it remains
unchanged.

Video Solution:

Q22 Text Solution:

Q20 Text Solution:

Planets sweep an equal area in equal intervals
of time Area under SCD is twice the area under From ΔABC
SAB so it must take double the time to sweep o a
sin 45 =
the area SCD than SAB. 2r

So t 1 = 2t 2 .
1
=
a

√2 2r

Video Solution:
√ 2a

⇒ r =
2

now, force on mass 'm' at O, due to 'A'

G(m)(m) 2
Gm
FA = =
2 2
r r

Force due to 'B' on 'O'

Q21 Text Solution: G(m)(2m) 2Gm
2

FB = = = 2F A
Let velocities of these masses at r distance r
2
r
2

from each other be v1 and v2 respectively. Force due to 'C ' on 'O'
By conservation of momentum G(m)(3m) 3Gm
2

FC = = = 3F A
m 1v 1 – m 2v 2 = 0 r
2
r
2

Þ m 1v 1 = m 2v 2 ….(i) Force due to 'D' on 'O'

By conservation of energy G(m)(4m) 4Gm
2

FD = = = 4F A
Change in P.E. = change in K.E. r
2
r
2

Gm 1 m 2 1
2
1
2
now the directions of forces are
= m1 v + m2 v
r 1 2
2 2

2 2 2 2

m v m v 2Gm 1 m 2
….. (ii)
1 1 2 2

⇒ + =
m1 m2 r

On solving equation (i) and (ii)

FA & FC are in opposite direction, resultant will
be 2FA.
2 2

√ 2Gm
2
√ 2Gm
1

v1 = and v 2 =
r(m 1 +m 2 ) r(m 1 +m 2 )

2G
√
∴ v app = |v 1 | + |v 2 | = (m 1 + m 2 )
r

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2 2
F resultant = √ (2F A ) + (2F A )

2 2
√
= 4F + 4F
A A

2 2
√ Gm
= 8F = 2√ 2F A = 2√ 2 ⋅
A
Q25 Text Solution:
2
r

Due to gravitational interaction connected

2 2
Gm Gm
⇒ F R = 2√ 2 ⋅ ⇒ F R = 4√ 2
2 2

masses have some acceleration.

a /2 a

Video Solution: Let both small masses are moving with

acceleration 'a' towards larger mass M

Force eq. for mass nearer to larger mass

GM Gm
− = ma . . . (i)
Q23 Text Solution:
2 2
(3l) l

Work
Force eq. for mass far to larger mass
( ) =
GM Gm
done + = ma . . . (ii)
2 2
(4l) l
Increase in Gravitational
( ) from equation (i) & (ii)
Potential Energy
GM Gm GM Gm
− = +
mgh 9l
2
l
2
16l
2
l
2

Since, ΔU =
h M M
1+ ⇒ − = m + m
R
9 16

mgR mgR 7M
= 2m
⇒ W1 = = 144
R
2
1+
R
7M M
⇒ m = = k( )
288 288
mgh

and similarly, W 2 =
h
⇒ K = 7
1+
R

Since, W 1 = 2W 2
Q26 Text Solution:
mgR 2mgh 4
3
⇒ = (G)ρ( πR )
h GM 3
2
1+ g = = ; g ∝ ρR
R 2 2
R R

R ′ ′ ′ ′
g ρ R 2 R √6
⇒ h =
3 = ( )( ) = ( )( ) =
g ρ R 3 R 11

Q24 Text Solution: R

′
3√ 6

n+1
Given =
1 R 22
T ∝ R 2 , F ∝
n
R
4
3
√ (G)((ρ)( πR ))
3 2GM 3
√
2
+1
Ve = =
R R
T ∝ R 2

′
5 Ve ′ R′ ρ
√
⇒ V e ∝ R√ ρ ⇒ = ×
T ∝ R 4 ρ
Ve R

Video Solution:

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2 2
3√ 6 GMm 1 GMm 1
2
√ − + mv = − + mv
∴ V e ′= 11 × × 1 2
4R 2 R 2
22 3

2 2
1 1 3 GM
V e ′= 3 km / sec v − v =
2 1
2 2 4 R

2
1 GM
v =
1
Q27 Text Solution:
2 R

1
v1 = √ 64 × 10 6
√2
P E = −4M J
8000
v1 = m/s
T E = −2M J √2

So, X = 8

The additional energy required to make the Video Solution:

statellite escape = +2MJ

Q28 Text Solution:

Using conservation of energy:
2
1 GmM GmM
⇒ mv − = 0 −
i
2 R 11R

2
20 GM Q30 Text Solution:
⇒ v =
i
11 R
4
Given R′= 3R, M = πR ρ
3

3
2GM
√
ve =
R GM ′
g′=
2
R′
10
√
∴ vi = ve 4
11 Putting M ′= πR′ ρ
3

Video Solution: G 4 27 G 4
3 3
g′= πR′ ρ = ( πR ρ)
2 2
R′ 3 9 R 3

GM
= 3
2
=3g
R

Video Solution:

Q29 Text Solution:

Conserving angular momentum
m v 1 cos 60° × 4R = mv 2 R

Conserving energy of the system

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