System of Particles and Rotational Motion

14 NEET questions on System of Particles and Rotational Motion. A sample is shown below with full explanations; the rest are available free after signing up.

System of Particles and Rotational Motion2025medium

Q1.The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

  • A100 days
  • B105 days
  • C115 days
  • D108 daysAnswer

Explanation

With no external torque the angular momentum Iω is conserved. For a uniform sphere I = (2/5)MR², so doubling the radius makes I four times larger and therefore ω four times smaller, giving a period of 4 x 27 = 108 days.

Source: NEET 2025 (4 May), Code 45, Q17

System of Particles and Rotational Motion2025medium

Q2.A uniform rod of mass 20 kg and length 5 m 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)

  • A100 N
  • B100√3 NAnswer
  • C200 N
  • D200√3 N

Explanation

A smooth wall can only push horizontally, so friction at the floor is the only force that can balance it. Taking torques about the foot of the rod gives N_wall = (Mg/2) cot θ, where θ = 30° is the angle with the horizontal, so f = N_wall = (200/2)√3 = 100√3 N.

Source: NEET 2025 (4 May), Code 45, Q22

System of Particles and Rotational Motion2026hard

Q3.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. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of A is IA and that calculated about a vertical axis passing through the centre of B is IB. The difference IA − IB is:

  • A(M − m)(R + r)²
  • B(m − M)(R + r)²Answer
  • C(m − M)(R − r)²
  • D0

Explanation

The centres are a distance (R + r) apart. About A's centre only sphere B is displaced, so the parallel-axis term is m(R + r)²; about B's centre only sphere A is displaced, giving M(R + r)². The two self terms (2/5)MR² and (2/5)mr² appear in both and cancel in the difference, leaving IA − IB = (m − M)(R + r)², which is negative since m < M.

Source: NEET 2026 re-exam (21 June), Code 50, Q27

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Questions are reformatted from NEET previous year papers with original explanations and topic tagging. NEET Prep is not affiliated with the NEET examining body.