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System of Particles and Rotational Motion

System of Particles and Rotational MotionNEET Physics · Class 11 · NCERT Chapter 6

31 NEET previous-year questions on System of Particles and Rotational Motion, each with the correct answer and a step-by-step solution. Filter by topic and expand any question to see how to solve it.

PYQ frequency · topic × year

14
15
16
17
18
19
20
21
22
23
24
Centre of mass
1
1
1
1
1
Torque
2
1
Moment of inertia
3
1
1
Axis theorems
2
1
Rotational dynamics
1
1
1
1
Angular momentum
1
1
1
1
1
Rolling motion
1
1
1
1
1
1

Darker = more questions in our PYQ bank for that topic and year.

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All (31)
Centre of mass (5)
Torque (3)
Moment of inertia (5)
Axis theorems (3)
Rotational dynamics (4)
Angular momentum (5)
Rolling motion (6)

A

1 m

B

2 m

C

3 m

D

4 m

Solution

.

A

At the centroid

B

Closer to the heavier mass

C

At the lightest mass

D

Outside the triangle

Solution

The CoM is the mass-weighted average of the corners. With unequal masses, it is pulled toward the heaviest mass. It is NOT the geometric centroid (that would only happen for equal masses).

A

2 m/s in opposite direction

B

3 m/s in opposite direction

C

6 m/s in same direction

D

12 m/s in opposite direction

Solution

Initial momentum is zero, so final total momentum is zero (conservation):

.

Magnitude 3 m/s, opposite direction.

A

B

C

D

Solution

Standard result: .

A

B

C

D

Solution

Standard result for a solid sphere: .

(A hollow sphere has — that is option D.)

A

B

C

D

Solution

The ring's moment about its central axis is . By the perpendicular axis theorem () and symmetry (), each diameter has .

A

B

C

D

Solution

Use parallel axis theorem with :

.

A

B

C

D

Solution

First find the moment about a diameter. Perpendicular axis theorem (, ): .

Now apply parallel axis theorem with (distance from diameter to tangent):

.

A

2 rad/s

B

6 rad/s

C

12 rad/s

D

18 rad/s

Solution

Conservation of angular momentum: .

.

A

Linear momentum

B

Kinetic energy

C

Angular momentum

D

Speed

Solution

The Sun's gravity is a central force, so the torque about the Sun is zero. By , angular momentum is constant. (This is exactly Kepler's second law.)

A

B

C

D

Solution

, where is the perpendicular distance from the point to the line of motion. So , constant in time (and direction perpendicular to the plane of motion).

A

0.5 rad/s

B

2 rad/s

C

4 rad/s

D

8 rad/s

Solution

.

A

4 J

B

10 J

C

20 J

D

40 J

Solution

.

A

B

C

D

Solution

Rolling without slipping: .

For a solid sphere, :

.

A

Solid sphere

B

Hollow sphere

C

Disc

D

Ring

Solution

The body with the smallest wins:

- Solid sphere: (smallest) - Disc: - Hollow sphere: - Ring: (largest)

Solid sphere reaches the bottom first.

A

Ring

B

Disc

C

Solid sphere

D

Hollow sphere

Solution

.

Largest = smallest → solid sphere ().

A

0.3 N·m

B

3 N·m

C

10 N·m

D

30 N·m

Solution

.

A

B

C

D

Solution

.

A

2 rad/s

B

4 rad/s

C

5 rad/s

D

10 rad/s

Solution

.

A

B

C

D

Solution

Each rod's centre is at distance from the centroid. Moment of inertia of one rod about an axis through its centre, perpendicular to its length and the plane is .

By parallel axis: .

Three rods: .

A

B

C

D

Solution

with , :

.

A

B

C

D

Solution

The contact point has zero velocity (rolling constraint). The centre moves at . The top point moves at (translation + rotational contributions add).

A

away from centre, opposite to the hole

B

away from centre, opposite to the hole

C

away from centre, opposite to the hole

D

At the centre

Solution

Let original mass be . Cut-out mass (area ratio).

Treat as full disc (mass at ) minus cut disc (mass at ):

.

in the direction opposite to the hole.

A

B

C

D

Solution

Conservation of angular momentum (no external torque on the system):

.

A

B

C

D

Solution

with :

.

A

Maximum

B

Zero

C

Minimum but non-zero

D

Equal to F times r

Solution

If the line of action passes through the axis, the perpendicular distance from the axis to the line is zero, so .

A

B

C

D

Solution

Angular momentum has dimensions .

Planck's constant: from , .

Same dimensions — a frequent NEET trap.

A

B

C

D

Solution

Distance from each sphere's centre to the rod's centre is (since each sphere has radius and they touch).

Each sphere about its own diameter: .

By parallel axis (with ): .

Two spheres: .

A

B

C

D

Solution

.

Disc: , so .

Ring: , so .

Ratio: .

A

0.5 m

B

1 m

C

2 m

D

5 m

Solution

No external horizontal force, so CoM stays fixed. Let boat shift by (opposite to man's walk).

Man moves relative to ground. CoM constraint:

.

A

B

C

D

Solution

About the central axis (perpendicular to plane): .

By parallel axis with :

.

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