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.
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1 m
2 m
3 m
4 m
Solution
.
At the centroid
Closer to the heavier mass
At the lightest mass
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).
2 m/s in opposite direction
3 m/s in opposite direction
6 m/s in same direction
12 m/s in opposite direction
Solution
Initial momentum is zero, so final total momentum is zero (conservation):
.
Magnitude 3 m/s, opposite direction.
Solution
Standard result: .
Solution
Standard result for a solid sphere: .
(A hollow sphere has — that is option D.)
Solution
The ring's moment about its central axis is . By the perpendicular axis theorem () and symmetry (), each diameter has .
Solution
Use parallel axis theorem with :
.
Solution
First find the moment about a diameter. Perpendicular axis theorem (, ): .
Now apply parallel axis theorem with (distance from diameter to tangent):
.
2 rad/s
6 rad/s
12 rad/s
18 rad/s
Solution
Conservation of angular momentum: .
.
Linear momentum
Kinetic energy
Angular momentum
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.)
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).
0.5 rad/s
2 rad/s
4 rad/s
8 rad/s
Solution
.
4 J
10 J
20 J
40 J
Solution
.
Solution
Rolling without slipping: .
For a solid sphere, :
.
Solid sphere
Hollow sphere
Disc
Ring
Solution
The body with the smallest wins:
- Solid sphere: (smallest) - Disc: - Hollow sphere: - Ring: (largest)
Solid sphere reaches the bottom first.
Ring
Disc
Solid sphere
Hollow sphere
Solution
.
Largest = smallest → solid sphere ().
0.3 N·m
3 N·m
10 N·m
30 N·m
Solution
.
Solution
.
2 rad/s
4 rad/s
5 rad/s
10 rad/s
Solution
.
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: .
Solution
with , :
.
Solution
The contact point has zero velocity (rolling constraint). The centre moves at . The top point moves at (translation + rotational contributions add).
away from centre, opposite to the hole
away from centre, opposite to the hole
away from centre, opposite to the hole
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.
Solution
Conservation of angular momentum (no external torque on the system):
.
Solution
with :
.
Maximum
Zero
Minimum but non-zero
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 .
Solution
Angular momentum has dimensions .
Planck's constant: from , .
Same dimensions — a frequent NEET trap.
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: .
Solution
.
Disc: , so .
Ring: , so .
Ratio: .
0.5 m
1 m
2 m
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:
.
Solution
About the central axis (perpendicular to plane): .
By parallel axis with :
.
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