NEET System of Particles and Rotational Motion PYQs & Practice Questions

10 free NEET Physics questions on System of Particles and Rotational Motion, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Rigid Body Equilibrium / Torque · Moment of Inertia of Carved Body · Centre of Mass · Rotational Kinematics · Centre of Mass - Continuous Body · Pure Rolling - Velocity of Lowest Point · Moment of Inertia of a Rod - Finding Length · Angular Acceleration Direction · Radius of Gyration - Solid vs Hollow Sphere · Moment of Inertia - Parallel Axis Theorem · Angular Momentum About Different Reference Points · Angular Momentum vs Rotational KE

1. Q1NEET 2026Moment of Inertia - Parallel Axis Theorem

   A solid sphere A of radius R and mass M is attached at a point to a smaller solid sphere B of radius r < R and mass m < M, with the line joining their centres lying along the horizontal. The moment of inertia of the system about a vertical axis passing through the centre of A is I_A and that about a vertical axis passing through the centre of B is I_B. The difference I_A − I_B is:

   • A. (m − M)(R + r)²
   • B. (m − M)(R − r)²
   • C. 0
   • D. (M − m)(R + r)²
   Show answer

   Answer: —

   I_A = (2/5)MR² [A about its own centre] + (2/5)mr² + m(R+r)² [B shifted by parallel axis theorem, since centre-to-centre distance is R+r]. I_B = (2/5)mr² [B about its own centre] + (2/5)MR² + M(R+r)² [A shifted by the same distance R+r]. The common terms (2/5)MR² and (2/5)mr² cancel in the difference, leaving I_A − I_B = m(R+r)² − M(R+r)² = (m−M)(R+r)².

2. Q2NEET 2026Angular Momentum About Different Reference Points

   A thin horizontal disc is rotating about a vertical axis passing through its fixed centre O. Its angular momentum is L_A and L_B computed about points A and B, respectively, which lie along a line through O, with OB = 2 × OA. The value of L_A/L_B is:

   • A. 1/2
   • B. 1
   • C. 2
   • D. 1/4
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   Answer: A — 1/2

   Since the disc spins about a fixed axis through its centre O (centre of mass at rest, not translating), the total linear momentum of the system is zero. The angular momentum about any point P is given by L_P = L_cm + r × P_total; since P_total = 0, the angular momentum is the same about every point in space, including A, B, and O. Hence L_A = L_B, and L_A/L_B = 1.

3. Q3NEET 2026Angular speed and revolutions

   The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s. The number of revolutions completed by the flywheel during this time is:

   • A. 600
   • B. 300
   • C. 900
   • D. 150
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   Answer: C — 900

   Average ω=(600+1200)/2=900 rpm = 15 rev/s; in 10s → 150 revolutions.

4. Q4NEET 2026Moment of inertia of a ring

   A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in the x-y plane) with centre 'C'. The moment of inertia of the ring about an axis yy' will be:

   • A. 3mL³/8π
   • B. 3mL²/8π²
   • C. 3mL³/8π²
   • D. 3mL²/8π
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   Answer: B — 3mL²/8π²

   Radius r=L/2π, total mass M=mL. I about diameter = ½MR² = 3mL³/8π² (per official key).

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5. Q5NEET 2025Rigid Body Equilibrium / Torque

   A uniform rod of mass 20kg and length 5m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is: (take g = 10 m/s²)

   • A. 100N
   • B. 100√3N
   • C. 200N
   • D. 200√3N
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   Answer: A — 100N

   N1 = Mg (vertical, from floor), N2 = f (horizontal, from wall equals friction by equilibrium). Taking torque about the floor contact point A: Mg(L/2)sinθ = N2·L·cosθ, so N2 = (Mg/2)tanθ = (200/2)×tan60° = 100√3 N.

6. Q6NEET 2025Moment of Inertia of Carved Body

   A sphere of radius R is cut from a larger solid sphere of radius 2R, tangent internally (the smaller sphere's surface passes through the centre of the larger sphere, both centres on the x-axis, 2R marks the diameter of the small cavity). The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :

   • A. 7/8
   • B. 7/40
   • C. 7/57
   • D. 7/64
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   Answer: B — 7/40

   Mass ratio M1/M = R³/(2R)³ = 1/8, so M1 = M/8. Using parallel axis theorem, total MOI about Y-axis of full sphere (2/5)M(2R)² = MOI of cavity sphere about Y-axis [M/8×(2/5)R² + M/8×R²] + MOI of remaining part. Remaining IT = (8/5)MR² − (M/8)R²(1+2/5) = (64−7)/(40)MR² = (57/40)MR². Cavity I = (M/8)R²(7/5) = (7/40)MR². Ratio I(small)/I(rest) = (7/40)/(57/40) = 7/57.

7. Q7NEET 2024Pure Rolling - Velocity of Lowest Point

   A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel respectively)? [Figure: wheel rolling on ground, P at top, Q at bottom, moving with velocity v]

   • A. Both the points P and Q move with equal speed.
   • B. Point P has zero speed.
   • C. Point P moves slower than point Q.
   • D. Point P moves faster than point Q.
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   Answer: C — Point P moves slower than point Q.

   In pure rolling, the point of contact with the ground (Q, the lowest point) always has zero instantaneous speed, while the topmost point (P) moves with speed 2v (twice the speed of the center). So P moves faster than Q.

8. Q8NEET 2024Moment of Inertia of a Rod - Finding Length

   The moment of inertia of a thin rod about an axis passing through its mid-point and perpendicular to the rod is 2400 g cm². The length of the 400 g rod is nearly:

   • A. 20.7 cm
   • B. 72.0 cm
   • C. 8.5 cm
   • D. 17.5 cm
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   Answer: B — 72.0 cm

   I = ML²/12, so L² = 12I/M = (12×2400)/400 = 72. L = √72 ≈ 8.49 cm ≈ 8.5 cm.

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9. Q9NEET 2023Angular Acceleration Direction

   The angular acceleration of a body, moving along the circumference of a circle, is:

   • A. along the radius towards the centre
   • B. along the tangent to its position
   • C. along the axis of rotation
   • D. along the radius, away from centre
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   Answer: B — along the tangent to its position

   Angular acceleration is a vector quantity whose direction is along the axis of rotation, as determined by the right-hand rule, similar to angular velocity.

10. Q10NEET 2023Radius of Gyration - Solid vs Hollow Sphere

    The ratio of radius of gyration of a solid sphere of mass M and radius R about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is:

    • A. 5 : 3
    • B. 2 : 5
    • C. 5 : 2
    • D. 3 : 5
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    Answer: C — 5 : 2

    For solid sphere: (2/5)MR² = Mk1², so k1² = (2/5)R². For hollow sphere: (2/3)MR² = Mk2², so k2² = (2/3)R². The ratio k1²:k2² = (2/5):(2/3) = 3:5. Note: this question was officially marked as Bonus (full marks awarded to all candidates) due to ambiguity between the ratio of radii of gyration versus their squares.