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 calculated about a vertical axis passing through the centre of $B$ is $I_B$. The difference $I_A - I_B$ is :
Answer: (B) $(m - M)(R + r)^2$
The spheres touch, so their centres are $d = R + r$ apart.
About the axis through the centre of A (parallel axis theorem for B):
$$I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m(R + r)^2$$
About the axis through the centre of B:
$$I_B = \frac{2}{5}MR^2 + M(R + r)^2 + \frac{2}{5}mr^2$$
$$I_A - I_B = (m - M)(R + r)^2$$
Solution by Sreeraj P, M.Sc Physics