Gravitation: JEE Mains Previous Year Questions (2021-2024) | Physics for JEE Main & Advanced PDF Download

 Page 1


JEE Mains Previous Year Questions 
(2021-2024): Gravitation 
2024 
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
. 
Page 2


JEE Mains Previous Year Questions 
(2021-2024): Gravitation 
2024 
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
. 
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 3


JEE Mains Previous Year Questions 
(2021-2024): Gravitation 
2024 
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
. 
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 4


JEE Mains Previous Year Questions 
(2021-2024): Gravitation 
2024 
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
. 
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
 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Gravitation 
2024 
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
. 
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
 
Q5: Escape velocity of a body from earth is ???? . ?? ???? /?? . If the radius of a planet be one third the 
radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is : 
A. ?? . ?? ???? /?? 
B. ?? . ?? ???? /?? 
C. ?? . ?? ???? /?? 
D. ???? . ?? ???? /??      [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (a) 
R
P
=
R
E
3
, M
P
=
M
E
6
 V
c
= v
2GM
e
R
e
… . (i) 
V
P
= v
2GM
P
R
P
… (ii) 
V
e
V
P
= v 2
 V
P
=
V
e
v 2
=
11.2
v 2
= 7.9 km/sec
 
Q6: The gravitational potential at a point above the surface of earth is -?? . ???? × ????
?? ?? /???? and the 
acceleration due to gravity at that point is ?? . ?? ?? /?? ?? . Assume that the mean radius of earth to be 
???????? ???? . The height of this point above the earth's surface is : 
A. ???????? ???? 
B. ???????? ???? 
C. ?????? ???? 
D .???????? ????    [JEE Main 2024 (Online) 30th January Morning Shift] 
Ans: (a) 
-
?? ?? ?? ?? ?? + h
 = -5.12 × 10
-7
… . ( i)
?? ?? ?? ( ?? ?? + h)
2
 = 6.4 … . ( ii)
 
By (i) and (ii) 
? h = 16 × 10
5
 m = 1600 km 
Q7: A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet 
from Sun is reduced to one fourth of the original distance, how many days will it take to complete one 
revolution: 
A. 20 
B. 50 
C. 100 
D. 25     [JEE Main 2024 (Online) 29th January Evening Shift]