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JEE	Mains	Previous	Year	Questions	(2021-2025)	
Gravitation	2025
Page 2


JEE	Mains	Previous	Year	Questions	(2021-2025)	
Gravitation	2025
 
 
Q1: A light planet is revolving around a massive star in a circular orbit of radius ?? with a period of 
revolution ?? . If the force of attraction between planet and star is proportional to ?? -?? /?? then choose 
the correct option : 
A. ?? ?? ? ?? ?? /?? 
B. ?? ?? ? ?? ?? 
C. ?? ?? ? ?? ?? /?? 
D. ?? ?? ? ?? ?? /??            [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
To find the correct option for the relationship between the period of revolution ?? and the radius of the 
orbit R, we will consider the force of attraction and its proportionality to R
-3/2
. 
According to Newton's law of universal gravitation, the force of attraction ?? between two masses ?? 1
 
and ?? 2
 separated by a distance ?? is given by ?? =
?? ?? 1
?? 2
?? 2
, where ?? is the gravitational constant. 
However, in this particular case, the force of attraction is given to be proportional to R
-3/2
, so we can 
write ?? ?
1
?? 3/2
. 
Since the planet is in a circular orbit around the star, the centripetal force required to keep the planet in 
orbit must be provided by this gravitational force. Hence, we can write that 
?? ?? 2
?? ?
1
?? 3/2
, where ?? is the 
mass of the planet and ?? is its orbital speed. 
Simplifying this, we get ?? 2
?
1
?? 1/2
. 
Now, the speed ?? can be related to the period ?? through the circumference of the orbit, which is given 
by 2???? . The orbital speed is the circumference divided by the period: ?? =
2????
?? . 
Substituting this into our proportionality, we get 
(
2????
?? )
2
?
1
?? 1/2
 
which simplifies to 
4?? 2
?? 2
?? 2
?
1
?? 1/2
. 
2024
Page 3


JEE	Mains	Previous	Year	Questions	(2021-2025)	
Gravitation	2025
 
 
Q1: A light planet is revolving around a massive star in a circular orbit of radius ?? with a period of 
revolution ?? . If the force of attraction between planet and star is proportional to ?? -?? /?? then choose 
the correct option : 
A. ?? ?? ? ?? ?? /?? 
B. ?? ?? ? ?? ?? 
C. ?? ?? ? ?? ?? /?? 
D. ?? ?? ? ?? ?? /??            [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
To find the correct option for the relationship between the period of revolution ?? and the radius of the 
orbit R, we will consider the force of attraction and its proportionality to R
-3/2
. 
According to Newton's law of universal gravitation, the force of attraction ?? between two masses ?? 1
 
and ?? 2
 separated by a distance ?? is given by ?? =
?? ?? 1
?? 2
?? 2
, where ?? is the gravitational constant. 
However, in this particular case, the force of attraction is given to be proportional to R
-3/2
, so we can 
write ?? ?
1
?? 3/2
. 
Since the planet is in a circular orbit around the star, the centripetal force required to keep the planet in 
orbit must be provided by this gravitational force. Hence, we can write that 
?? ?? 2
?? ?
1
?? 3/2
, where ?? is the 
mass of the planet and ?? is its orbital speed. 
Simplifying this, we get ?? 2
?
1
?? 1/2
. 
Now, the speed ?? can be related to the period ?? through the circumference of the orbit, which is given 
by 2???? . The orbital speed is the circumference divided by the period: ?? =
2????
?? . 
Substituting this into our proportionality, we get 
(
2????
?? )
2
?
1
?? 1/2
 
which simplifies to 
4?? 2
?? 2
?? 2
?
1
?? 1/2
. 
2024
Solving for ?? 2
, we get 
?? 2
?
?? 2+1/2
4?? 2
,
 so ?? 2
? ?? 5/2
. 
 
Therefore, the correct option is Option C 
?? 2
? ?? 5/2
 
Q2: If ?? is the radius of the earth and the acceleration due to gravity on the surface of earth is 
?? = ?? ?? ?? /?? ?? , then the length of the second's pendulum at a height ?? = ?? ?? from the surface of earth 
will be, : 
A. 
?? ?? ?? 
B. 
?? ?? ?? 
C. 
?? ?? ?? 
D. 
?? ?? ??     [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
To find the length of the second's pendulum at a height h = 2?? from the surface of the Earth, we must 
first understand that the length of a second's pendulum, ?? , is related to the gravitational acceleration, ?? , 
and the period, ?? , by the formula: ?? = 2?? v
?? ?? Since we are talking about a second's pendulum, the 
period, ?? , is 2 seconds (since it takes one second for the pendulum to swing in one direction and 
another second to swing back), thus ?? = 2 seconds. 
Now let's find the gravitational acceleration at height h = 2?? where ?? is the radius of the earth. The 
general formula for gravitational acceleration at a height h above the surface is: ?? h
=
?? (1+
h
?? )
2
 Plugging 
h = 2?? into the formula, we get: 
?? h
=
?? (1 +
2?? ?? )
2
?? h
=
?? ( 1 + 2)
2
?? h
=
?? 3
2
=
?? 9
 
So the gravitational acceleration at height h is one-ninth of the gravitational acceleration at the surface 
of the Earth. Given that ?? = ?? 2
 m/s
2
, we get: ?? h
=
?? 2
9
 m/s
2
 
Now knowing the gravitational acceleration at height h and with the period ?? of 2 seconds, we can 
rearrange the formula for the second's pendulum to solve for the length ?? h
 : 
Page 4


JEE	Mains	Previous	Year	Questions	(2021-2025)	
Gravitation	2025
 
 
Q1: A light planet is revolving around a massive star in a circular orbit of radius ?? with a period of 
revolution ?? . If the force of attraction between planet and star is proportional to ?? -?? /?? then choose 
the correct option : 
A. ?? ?? ? ?? ?? /?? 
B. ?? ?? ? ?? ?? 
C. ?? ?? ? ?? ?? /?? 
D. ?? ?? ? ?? ?? /??            [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
To find the correct option for the relationship between the period of revolution ?? and the radius of the 
orbit R, we will consider the force of attraction and its proportionality to R
-3/2
. 
According to Newton's law of universal gravitation, the force of attraction ?? between two masses ?? 1
 
and ?? 2
 separated by a distance ?? is given by ?? =
?? ?? 1
?? 2
?? 2
, where ?? is the gravitational constant. 
However, in this particular case, the force of attraction is given to be proportional to R
-3/2
, so we can 
write ?? ?
1
?? 3/2
. 
Since the planet is in a circular orbit around the star, the centripetal force required to keep the planet in 
orbit must be provided by this gravitational force. Hence, we can write that 
?? ?? 2
?? ?
1
?? 3/2
, where ?? is the 
mass of the planet and ?? is its orbital speed. 
Simplifying this, we get ?? 2
?
1
?? 1/2
. 
Now, the speed ?? can be related to the period ?? through the circumference of the orbit, which is given 
by 2???? . The orbital speed is the circumference divided by the period: ?? =
2????
?? . 
Substituting this into our proportionality, we get 
(
2????
?? )
2
?
1
?? 1/2
 
which simplifies to 
4?? 2
?? 2
?? 2
?
1
?? 1/2
. 
2024
Solving for ?? 2
, we get 
?? 2
?
?? 2+1/2
4?? 2
,
 so ?? 2
? ?? 5/2
. 
 
Therefore, the correct option is Option C 
?? 2
? ?? 5/2
 
Q2: If ?? is the radius of the earth and the acceleration due to gravity on the surface of earth is 
?? = ?? ?? ?? /?? ?? , then the length of the second's pendulum at a height ?? = ?? ?? from the surface of earth 
will be, : 
A. 
?? ?? ?? 
B. 
?? ?? ?? 
C. 
?? ?? ?? 
D. 
?? ?? ??     [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
To find the length of the second's pendulum at a height h = 2?? from the surface of the Earth, we must 
first understand that the length of a second's pendulum, ?? , is related to the gravitational acceleration, ?? , 
and the period, ?? , by the formula: ?? = 2?? v
?? ?? Since we are talking about a second's pendulum, the 
period, ?? , is 2 seconds (since it takes one second for the pendulum to swing in one direction and 
another second to swing back), thus ?? = 2 seconds. 
Now let's find the gravitational acceleration at height h = 2?? where ?? is the radius of the earth. The 
general formula for gravitational acceleration at a height h above the surface is: ?? h
=
?? (1+
h
?? )
2
 Plugging 
h = 2?? into the formula, we get: 
?? h
=
?? (1 +
2?? ?? )
2
?? h
=
?? ( 1 + 2)
2
?? h
=
?? 3
2
=
?? 9
 
So the gravitational acceleration at height h is one-ninth of the gravitational acceleration at the surface 
of the Earth. Given that ?? = ?? 2
 m/s
2
, we get: ?? h
=
?? 2
9
 m/s
2
 
Now knowing the gravitational acceleration at height h and with the period ?? of 2 seconds, we can 
rearrange the formula for the second's pendulum to solve for the length ?? h
 : 
2 = 2?? v
?? h
?? h
1 = ?? v
?? h
?? h
1
?? = v
?? h
?? h
 
Squaring both sides, we get: 
1
?? 2
=
?? h
?? h
 
Multiplying both sides by ?? h
 gives us the length ?? h
 : 
?? h
=
?? h
?? 2
 
Substituting ?? h
 into the equation yields: 
?? h
=
?? 2
9?? 2
?? h
=
1
9
 m
 
Therefore, the length of the second's pendulum at a height h = 2?? from the surface of the Earth is 
1
9
 
meters. The correct answer is Option A. 
Q3: The mass of the moon is 
?? ??????
 times the mass of a planet and its diameter is 
?? ????
 times the diameter 
of a planet. If the escape velocity on the planet is ?? , the escape velocity on the moon will be: 
A. 
?? ?? 
B. 
?? ?? 
C. 
?? ????
 
D. 
?? ??      [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (d) 
Page 5


JEE	Mains	Previous	Year	Questions	(2021-2025)	
Gravitation	2025
 
 
Q1: A light planet is revolving around a massive star in a circular orbit of radius ?? with a period of 
revolution ?? . If the force of attraction between planet and star is proportional to ?? -?? /?? then choose 
the correct option : 
A. ?? ?? ? ?? ?? /?? 
B. ?? ?? ? ?? ?? 
C. ?? ?? ? ?? ?? /?? 
D. ?? ?? ? ?? ?? /??            [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
To find the correct option for the relationship between the period of revolution ?? and the radius of the 
orbit R, we will consider the force of attraction and its proportionality to R
-3/2
. 
According to Newton's law of universal gravitation, the force of attraction ?? between two masses ?? 1
 
and ?? 2
 separated by a distance ?? is given by ?? =
?? ?? 1
?? 2
?? 2
, where ?? is the gravitational constant. 
However, in this particular case, the force of attraction is given to be proportional to R
-3/2
, so we can 
write ?? ?
1
?? 3/2
. 
Since the planet is in a circular orbit around the star, the centripetal force required to keep the planet in 
orbit must be provided by this gravitational force. Hence, we can write that 
?? ?? 2
?? ?
1
?? 3/2
, where ?? is the 
mass of the planet and ?? is its orbital speed. 
Simplifying this, we get ?? 2
?
1
?? 1/2
. 
Now, the speed ?? can be related to the period ?? through the circumference of the orbit, which is given 
by 2???? . The orbital speed is the circumference divided by the period: ?? =
2????
?? . 
Substituting this into our proportionality, we get 
(
2????
?? )
2
?
1
?? 1/2
 
which simplifies to 
4?? 2
?? 2
?? 2
?
1
?? 1/2
. 
2024
Solving for ?? 2
, we get 
?? 2
?
?? 2+1/2
4?? 2
,
 so ?? 2
? ?? 5/2
. 
 
Therefore, the correct option is Option C 
?? 2
? ?? 5/2
 
Q2: If ?? is the radius of the earth and the acceleration due to gravity on the surface of earth is 
?? = ?? ?? ?? /?? ?? , then the length of the second's pendulum at a height ?? = ?? ?? from the surface of earth 
will be, : 
A. 
?? ?? ?? 
B. 
?? ?? ?? 
C. 
?? ?? ?? 
D. 
?? ?? ??     [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
To find the length of the second's pendulum at a height h = 2?? from the surface of the Earth, we must 
first understand that the length of a second's pendulum, ?? , is related to the gravitational acceleration, ?? , 
and the period, ?? , by the formula: ?? = 2?? v
?? ?? Since we are talking about a second's pendulum, the 
period, ?? , is 2 seconds (since it takes one second for the pendulum to swing in one direction and 
another second to swing back), thus ?? = 2 seconds. 
Now let's find the gravitational acceleration at height h = 2?? where ?? is the radius of the earth. The 
general formula for gravitational acceleration at a height h above the surface is: ?? h
=
?? (1+
h
?? )
2
 Plugging 
h = 2?? into the formula, we get: 
?? h
=
?? (1 +
2?? ?? )
2
?? h
=
?? ( 1 + 2)
2
?? h
=
?? 3
2
=
?? 9
 
So the gravitational acceleration at height h is one-ninth of the gravitational acceleration at the surface 
of the Earth. Given that ?? = ?? 2
 m/s
2
, we get: ?? h
=
?? 2
9
 m/s
2
 
Now knowing the gravitational acceleration at height h and with the period ?? of 2 seconds, we can 
rearrange the formula for the second's pendulum to solve for the length ?? h
 : 
2 = 2?? v
?? h
?? h
1 = ?? v
?? h
?? h
1
?? = v
?? h
?? h
 
Squaring both sides, we get: 
1
?? 2
=
?? h
?? h
 
Multiplying both sides by ?? h
 gives us the length ?? h
 : 
?? h
=
?? h
?? 2
 
Substituting ?? h
 into the equation yields: 
?? h
=
?? 2
9?? 2
?? h
=
1
9
 m
 
Therefore, the length of the second's pendulum at a height h = 2?? from the surface of the Earth is 
1
9
 
meters. The correct answer is Option A. 
Q3: The mass of the moon is 
?? ??????
 times the mass of a planet and its diameter is 
?? ????
 times the diameter 
of a planet. If the escape velocity on the planet is ?? , the escape velocity on the moon will be: 
A. 
?? ?? 
B. 
?? ?? 
C. 
?? ????
 
D. 
?? ??      [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (d) 
V
escape 
 =
v
2GM
R
V
planet 
 =
v
2GM
R
= V
V
Moon 
 =
v
2GM × 16
144R
=
1
3
v
2GM
R
V
Moon 
 =
V
Planet 
3
=
V
3
 
Q4: Four identical particles of mass ?? are kept at the four corners of a square. If the gravitational 
force exerted on one of the masses by the other masses is (
?? v ?? +?? ????
)
????
?? ?? ?? , the length of the sides of the 
square is 
A. ?? ?? 
B. ?? ?? 
C. ?? ?? 
D. 
?? ??      [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (a) 
 
?? net 
= v 2?? + ?? '
?? =
?? ?? 2
?? 2
 and ?? '
=
?? 2
( v 2a)
2
?? net 
= v 2
Gm
2
a
2
+
Gm
2
2a
2
(
2v 2 + 1
32
)
Gm
2
 L
2
=
Gm
2
a
2
(
2v 2 + 1
2
)
a = 4 L

	Physics for JEE Main & Advanced 267 videos\|691 docs\|171 tests

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267 videos|691 docs|171 tests

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FAQs on Gravitation: JEE Main Previous Year Questions (2021-2025) - Physics for JEE Main & Advanced

1. What are the key concepts of gravitation that are important for JEE Mains preparation?

Ans. Key concepts of gravitation that are important for JEE Mains include Newton's law of universal gravitation, gravitational force between two masses, gravitational potential energy, escape velocity, and orbital motion. Understanding these concepts and their applications in problem-solving is crucial for success in the exam.

2. How do I calculate the gravitational force between two objects?

Ans. The gravitational force between two objects can be calculated using Newton's law of universal gravitation, which states that the force \( F \) is equal to the gravitational constant \( G \) multiplied by the product of the masses \( m_1 \) and \( m_2 \), divided by the square of the distance \( r \) between their centers: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \] where \( G \) is approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

3. What is the significance of escape velocity in gravitation?

Ans. Escape velocity is the minimum velocity an object must have to break free from the gravitational attraction of a celestial body without any further propulsion. It is significant because it determines whether a spacecraft can leave a planet's gravitational influence. The escape velocity \( v_e \) can be calculated using the formula: \[ v_e = \sqrt{\frac{2GM}{r}} \] where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( r \) is the radius from the center of the body to the point of escape.

4. How does the gravitational potential energy vary with distance?

Ans. Gravitational potential energy (U) between two masses is defined as: \[ U = -\frac{G \cdot m_1 \cdot m_2}{r} \] where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of the two masses. As the distance \( r \) increases, the gravitational potential energy becomes less negative (increases) and approaches zero, indicating that the objects are less bound by gravity.

5. What are the common problems related to gravitation in JEE Mains, and how can I solve them?

Ans. Common problems related to gravitation in JEE Mains include calculating the gravitational force between two bodies, finding escape velocity, solving problems involving orbits (e.g., circular motion of satellites), and determining gravitational potential energy. To solve them, practice using the relevant formulas, understand the underlying principles, and apply dimensional analysis to check the correctness of your answers. Regular practice with previous years' questions can also help in getting familiar with the types of problems asked.

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