A solid sphere A of radius R and mass M is attached at a point to a… | NEET 2026 (Re-neet)

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Moment of Inertia & Radius of Gyration

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 $I_A$ and that about a vertical axis passing through the centre of $B$ is $I_B$ . The difference $I_A - I_B$ is:

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Answer: (c)

Distance between centres $= R + r$ .
$I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m(R+r)^2$
$I_B = \frac{2}{5}MR^2 + M(R+r)^2 + \frac{2}{5}mr^2$
$I_A - I_B = (m - M)(R+r)^2$ , so the answer is (c).

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