JEE Main Previous Year Questions (2025): Circular Motion | Physics for JEE Main & Advanced PDF Download

 Page 1


JEE Main Previous Year Qs (2025): 
Circular Motion 
Q1: A tube of length 1 m is filled completely with an ideal liquid of mass 2 M, 
and closed at both ends. The tube is rotated uniformly in horizontal plane 
about one of its ends. If the force exerted by the liquid at the other end is ?? 
then angular velocity of the tube is v
?? ?? ?? in SI unit. The value of ?? is _ _ _ _ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Answer: 1 
 
Solution: 
 
 
 
F = 2M ?? 2
l
2
= Mw
2
l 
?? =
v
F
M l
 
 
Q2:  
 
 
A string of length ?? is fixed at one end and carries a mass of ?? at the other 
end. The mass makes (
?? ?? ) rotations per second about the vertical axis 
Page 2


JEE Main Previous Year Qs (2025): 
Circular Motion 
Q1: A tube of length 1 m is filled completely with an ideal liquid of mass 2 M, 
and closed at both ends. The tube is rotated uniformly in horizontal plane 
about one of its ends. If the force exerted by the liquid at the other end is ?? 
then angular velocity of the tube is v
?? ?? ?? in SI unit. The value of ?? is _ _ _ _ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Answer: 1 
 
Solution: 
 
 
 
F = 2M ?? 2
l
2
= Mw
2
l 
?? =
v
F
M l
 
 
Q2:  
 
 
A string of length ?? is fixed at one end and carries a mass of ?? at the other 
end. The mass makes (
?? ?? ) rotations per second about the vertical axis 
passing through end of the string as shown. The tension in the string is _ _ _ _ 
ML. 
JEE Main 2025 (Online) 24th January Evening Shift 
Answer: 36 
Solution: 
 
 
Tcos ? ?? = mg 
Tsin ? ?? = M ?? 2
R 
Using equation (2) 
Tsin ? ?? = M ?? 2
( L s in ? ?? ) 
T = M ?? 2
 L = M (
3
?? × 2 ?? )
2
 L 
T = 36M L 
 
Q3: A particle of charge ?? . ?? ?? ?? and mass ???? ?? ?? is present in a strong 
magnetic field of 6.28 ?? . The particle is then fired perpendicular to 
magnetic field. The time required for the particle to return to original 
location for the first time is _ _ _ _ s. ( ?? = ?? . ???? ) 
JEE Main 2025 (Online) 4th April Evening Shift 
Answer: 0 
Solution: 
Page 3


JEE Main Previous Year Qs (2025): 
Circular Motion 
Q1: A tube of length 1 m is filled completely with an ideal liquid of mass 2 M, 
and closed at both ends. The tube is rotated uniformly in horizontal plane 
about one of its ends. If the force exerted by the liquid at the other end is ?? 
then angular velocity of the tube is v
?? ?? ?? in SI unit. The value of ?? is _ _ _ _ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Answer: 1 
 
Solution: 
 
 
 
F = 2M ?? 2
l
2
= Mw
2
l 
?? =
v
F
M l
 
 
Q2:  
 
 
A string of length ?? is fixed at one end and carries a mass of ?? at the other 
end. The mass makes (
?? ?? ) rotations per second about the vertical axis 
passing through end of the string as shown. The tension in the string is _ _ _ _ 
ML. 
JEE Main 2025 (Online) 24th January Evening Shift 
Answer: 36 
Solution: 
 
 
Tcos ? ?? = mg 
Tsin ? ?? = M ?? 2
R 
Using equation (2) 
Tsin ? ?? = M ?? 2
( L s in ? ?? ) 
T = M ?? 2
 L = M (
3
?? × 2 ?? )
2
 L 
T = 36M L 
 
Q3: A particle of charge ?? . ?? ?? ?? and mass ???? ?? ?? is present in a strong 
magnetic field of 6.28 ?? . The particle is then fired perpendicular to 
magnetic field. The time required for the particle to return to original 
location for the first time is _ _ _ _ s. ( ?? = ?? . ???? ) 
JEE Main 2025 (Online) 4th April Evening Shift 
Answer: 0 
Solution: 
To solve this problem, note that when a charged particle with mass ?? and charge ?? is fired perpendicular 
to a magnetic field of strength ?? , it undergoes uniform circular motion. The period (time to complete one 
full circle) is given by: 
?? =
2 ????
????
 
Here's how to calculate it step by step: 
Convert the given quantities to SI units: 
Charge: ?? = 1 . 6 ?? C = 1 . 6 × 10
- 6
C 
Mass: ?? = 16 ?? g = 16 × 10
- 9
 kg 
Magnetic field: ?? = 6 . 28 T 
Write down the period formula: 
?? =
2 ????
????
 
Substitute the values: 
?? =
2 ?? × ( 16 × 10
- 9
 kg )
( 1 . 6 × 10
- 6
C ) ( 6 . 28 T )
 
Notice that 2 ?? ˜ 6 . 28, which cancels with the given magnetic field value, simplifying the expression: 
?? =
6 . 28 × 16 × 10
- 9
1 . 6 × 10
- 6
× 6 . 28
=
16 × 10
- 9
1 . 6 × 10
- 6
 
Simplify the fraction: 
?? =
16
1 . 6
×
10
- 9
10
- 6
= 10 × 10
- 3
= 0 . 01 s 
Thus, the time required for the particle to return to its original location is: 
 
Q4: A bob of mass ?? is suspended at a point ?? by a light string of length ?? 
and left to perform vertical motion (circular) as shown in figure. Initially, 
by applying horizontal velocity ?? ?? at the point ' ?? ', the string becomes slack 
when, the bob reaches at the point ' ?? '. The ratio of the kinetic energy of the 
bob at the points ?? and ?? is _ _ _ _ . 
 
 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 4 
B. 1 
C. 2 
D. 3 
Page 4


JEE Main Previous Year Qs (2025): 
Circular Motion 
Q1: A tube of length 1 m is filled completely with an ideal liquid of mass 2 M, 
and closed at both ends. The tube is rotated uniformly in horizontal plane 
about one of its ends. If the force exerted by the liquid at the other end is ?? 
then angular velocity of the tube is v
?? ?? ?? in SI unit. The value of ?? is _ _ _ _ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Answer: 1 
 
Solution: 
 
 
 
F = 2M ?? 2
l
2
= Mw
2
l 
?? =
v
F
M l
 
 
Q2:  
 
 
A string of length ?? is fixed at one end and carries a mass of ?? at the other 
end. The mass makes (
?? ?? ) rotations per second about the vertical axis 
passing through end of the string as shown. The tension in the string is _ _ _ _ 
ML. 
JEE Main 2025 (Online) 24th January Evening Shift 
Answer: 36 
Solution: 
 
 
Tcos ? ?? = mg 
Tsin ? ?? = M ?? 2
R 
Using equation (2) 
Tsin ? ?? = M ?? 2
( L s in ? ?? ) 
T = M ?? 2
 L = M (
3
?? × 2 ?? )
2
 L 
T = 36M L 
 
Q3: A particle of charge ?? . ?? ?? ?? and mass ???? ?? ?? is present in a strong 
magnetic field of 6.28 ?? . The particle is then fired perpendicular to 
magnetic field. The time required for the particle to return to original 
location for the first time is _ _ _ _ s. ( ?? = ?? . ???? ) 
JEE Main 2025 (Online) 4th April Evening Shift 
Answer: 0 
Solution: 
To solve this problem, note that when a charged particle with mass ?? and charge ?? is fired perpendicular 
to a magnetic field of strength ?? , it undergoes uniform circular motion. The period (time to complete one 
full circle) is given by: 
?? =
2 ????
????
 
Here's how to calculate it step by step: 
Convert the given quantities to SI units: 
Charge: ?? = 1 . 6 ?? C = 1 . 6 × 10
- 6
C 
Mass: ?? = 16 ?? g = 16 × 10
- 9
 kg 
Magnetic field: ?? = 6 . 28 T 
Write down the period formula: 
?? =
2 ????
????
 
Substitute the values: 
?? =
2 ?? × ( 16 × 10
- 9
 kg )
( 1 . 6 × 10
- 6
C ) ( 6 . 28 T )
 
Notice that 2 ?? ˜ 6 . 28, which cancels with the given magnetic field value, simplifying the expression: 
?? =
6 . 28 × 16 × 10
- 9
1 . 6 × 10
- 6
× 6 . 28
=
16 × 10
- 9
1 . 6 × 10
- 6
 
Simplify the fraction: 
?? =
16
1 . 6
×
10
- 9
10
- 6
= 10 × 10
- 3
= 0 . 01 s 
Thus, the time required for the particle to return to its original location is: 
 
Q4: A bob of mass ?? is suspended at a point ?? by a light string of length ?? 
and left to perform vertical motion (circular) as shown in figure. Initially, 
by applying horizontal velocity ?? ?? at the point ' ?? ', the string becomes slack 
when, the bob reaches at the point ' ?? '. The ratio of the kinetic energy of the 
bob at the points ?? and ?? is _ _ _ _ . 
 
 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 4 
B. 1 
C. 2 
D. 3 
Answer: C 
Solution: 
 
 
As the string becomes slack when the bob reaches at the point D so tension is zero at this point. And the 
bob just completes the circle. 
We know, to complete a vertical loop of radius R , the minimum velocity required at the bottom of loop is 
v 5???? . 
So, ?? 0
= v 5???? = ?? ?? ( as R = ?? ) 
So, using energy conservation, 
1
2
?? ?? ?? 2
=
1
2
?? ?? ?? 2
+ ???? h 
?
1
2
?? ( 5???? ) =
1
2
?? ?? ?? 2
+ ???? (
?? 2
) 
? ?? ?? ?? =
?? ???? 2
( 5 - 1 ) = 2 ?? ???? 
For point C, 
1
2
?? ?? ?? 2
=
1
2
?? ?? ?? 2
+ ???? ( ?? +
?? 2
) 
?
5
2
?? ???? = ?? ?? ?? +
3 ?? ???? 2
 
?? ?? ?? =
?? ???? 2
( 5 - 3 ) = ?? ???? 
So, 
?? ?? ?? ?? ?? ?? =
2 ??????
?? ?? ?? = 2 
Q5: A body of mass ???????? is moving in circular path of radius ?? ?? on vertical 
plane as shown in figure. The velocity of the body at point ?? is ???? ?? /?? . The 
ratio of its kinetic energies at point ?? and ?? is : 
 
Page 5


JEE Main Previous Year Qs (2025): 
Circular Motion 
Q1: A tube of length 1 m is filled completely with an ideal liquid of mass 2 M, 
and closed at both ends. The tube is rotated uniformly in horizontal plane 
about one of its ends. If the force exerted by the liquid at the other end is ?? 
then angular velocity of the tube is v
?? ?? ?? in SI unit. The value of ?? is _ _ _ _ . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Answer: 1 
 
Solution: 
 
 
 
F = 2M ?? 2
l
2
= Mw
2
l 
?? =
v
F
M l
 
 
Q2:  
 
 
A string of length ?? is fixed at one end and carries a mass of ?? at the other 
end. The mass makes (
?? ?? ) rotations per second about the vertical axis 
passing through end of the string as shown. The tension in the string is _ _ _ _ 
ML. 
JEE Main 2025 (Online) 24th January Evening Shift 
Answer: 36 
Solution: 
 
 
Tcos ? ?? = mg 
Tsin ? ?? = M ?? 2
R 
Using equation (2) 
Tsin ? ?? = M ?? 2
( L s in ? ?? ) 
T = M ?? 2
 L = M (
3
?? × 2 ?? )
2
 L 
T = 36M L 
 
Q3: A particle of charge ?? . ?? ?? ?? and mass ???? ?? ?? is present in a strong 
magnetic field of 6.28 ?? . The particle is then fired perpendicular to 
magnetic field. The time required for the particle to return to original 
location for the first time is _ _ _ _ s. ( ?? = ?? . ???? ) 
JEE Main 2025 (Online) 4th April Evening Shift 
Answer: 0 
Solution: 
To solve this problem, note that when a charged particle with mass ?? and charge ?? is fired perpendicular 
to a magnetic field of strength ?? , it undergoes uniform circular motion. The period (time to complete one 
full circle) is given by: 
?? =
2 ????
????
 
Here's how to calculate it step by step: 
Convert the given quantities to SI units: 
Charge: ?? = 1 . 6 ?? C = 1 . 6 × 10
- 6
C 
Mass: ?? = 16 ?? g = 16 × 10
- 9
 kg 
Magnetic field: ?? = 6 . 28 T 
Write down the period formula: 
?? =
2 ????
????
 
Substitute the values: 
?? =
2 ?? × ( 16 × 10
- 9
 kg )
( 1 . 6 × 10
- 6
C ) ( 6 . 28 T )
 
Notice that 2 ?? ˜ 6 . 28, which cancels with the given magnetic field value, simplifying the expression: 
?? =
6 . 28 × 16 × 10
- 9
1 . 6 × 10
- 6
× 6 . 28
=
16 × 10
- 9
1 . 6 × 10
- 6
 
Simplify the fraction: 
?? =
16
1 . 6
×
10
- 9
10
- 6
= 10 × 10
- 3
= 0 . 01 s 
Thus, the time required for the particle to return to its original location is: 
 
Q4: A bob of mass ?? is suspended at a point ?? by a light string of length ?? 
and left to perform vertical motion (circular) as shown in figure. Initially, 
by applying horizontal velocity ?? ?? at the point ' ?? ', the string becomes slack 
when, the bob reaches at the point ' ?? '. The ratio of the kinetic energy of the 
bob at the points ?? and ?? is _ _ _ _ . 
 
 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 4 
B. 1 
C. 2 
D. 3 
Answer: C 
Solution: 
 
 
As the string becomes slack when the bob reaches at the point D so tension is zero at this point. And the 
bob just completes the circle. 
We know, to complete a vertical loop of radius R , the minimum velocity required at the bottom of loop is 
v 5???? . 
So, ?? 0
= v 5???? = ?? ?? ( as R = ?? ) 
So, using energy conservation, 
1
2
?? ?? ?? 2
=
1
2
?? ?? ?? 2
+ ???? h 
?
1
2
?? ( 5???? ) =
1
2
?? ?? ?? 2
+ ???? (
?? 2
) 
? ?? ?? ?? =
?? ???? 2
( 5 - 1 ) = 2 ?? ???? 
For point C, 
1
2
?? ?? ?? 2
=
1
2
?? ?? ?? 2
+ ???? ( ?? +
?? 2
) 
?
5
2
?? ???? = ?? ?? ?? +
3 ?? ???? 2
 
?? ?? ?? =
?? ???? 2
( 5 - 3 ) = ?? ???? 
So, 
?? ?? ?? ?? ?? ?? =
2 ??????
?? ?? ?? = 2 
Q5: A body of mass ???????? is moving in circular path of radius ?? ?? on vertical 
plane as shown in figure. The velocity of the body at point ?? is ???? ?? /?? . The 
ratio of its kinetic energies at point ?? and ?? is : 
 
 
 
(Take acceleration due to gravity as ???? ?? / ?? ?? ) 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 
3 - v 2
2
 
B. 
2 + v 3
3
 
C. 
2 + v 2
3
 
D. 
3 + v 3
2
 
Answer: D 
Solution: