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, their line of centres horizontal. The moment of inertia of the system about a vertical axis through the centre of A is I_A, and about a vertical axis 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.(m-M)(R-r)²
- D.0
Correct Answer
(B) (m-M)(R+r)²
Solution & Explanation
Centre-to-centre distance of the two attached spheres is d = R + r (radii touch externally). Moment of inertia of a solid sphere about a diameter (axis through its own centre) = (2/5)MR². About a vertical axis through the centre of A: A contributes (2/5)MR² (axis through its own centre); B is shifted by (R + r), so by the parallel-axis theorem it contributes (2/5)mr² + m(R + r)². I_A = (2/5)MR² + (2/5)mr² + m(R + r)². About a vertical axis through the centre of B: B contributes (2/5)mr²; A is shifted by (R + r), contributing (2/5)MR² + M(R + r)². I_B = (2/5)MR² + (2/5)mr² + M(R + r)². Subtract: I_A − I_B = m(R + r)² − M(R + r)² = (m − M)(R + r)². This matches option B, (m − M)(R + r)². Since m < M, the difference is negative — I_B > I_A, as expected because the heavier sphere M sits at the far distance for the B-axis.
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