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RE-NEET2026Physics-Rotational Motion

RE-NEET 2026 Physics Moment of Inertia MCQ Question

Type: MCQ-numerical-Hard-Class 11

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 Iᵦ. The difference Iₐ - Iᵦ is:

Question diagram
A

(m − M)(R + r)²

B

(m − M)(R − r)²

C

0

D

(M − m)(R + r)²

Correct Answer

Option A

Detailed Explanation

To find the difference in moments of inertia IA−IBI_A - I_B, we use the parallel axis theorem. The moment of inertia of sphere A about its center is IA=25MR2I_A = \frac{2}{5}MR^2. For sphere B, using the parallel axis theorem, we have:

IB=25mr2+m(R+r)2I_B = \frac{2}{5}mr^2 + m(R + r)^2

Calculating the difference:

IA−IB=25MR2−(25mr2+m(R+r)2)I_A - I_B = \frac{2}{5}MR^2 - \left( \frac{2}{5}mr^2 + m(R + r)^2 \right)

This simplifies to:

IA−IB=(m−M)(R+r)2I_A - I_B = (m - M)(R + r)^2

Thus, the correct answer is option A. The other options fail because they either miscalculate the contributions from the masses or the distances involved.

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