A small sphere of radius $R$ is cut out and completely removed from a larger uniform solid sphere of

Question

A small sphere of radius $R$ is cut out and completely removed from a larger uniform solid sphere of radius $2R$ and mass $M$, as shown in the figure. The ratio of the moment of inertia of the removed smaller sphere (about the $Y$-axis) to that of the original uncut larger sphere (about the $Y$-axis) is

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(d) : The volume of a sphere is proportional to $R^3$. Since the radius of the small sphere is half of the large one, its mass is $m = M(R/2R)^3 = M/8$. The moment of inertia of the original large sphere about the Y-axis is $I_{large} = \frac{2}{5}M(2R)^2 = \frac{8}{5}MR^2 = \frac{64}{40}MR^2$. Using the parallel axis theorem, the MOI of the small removed sphere about the Y-axis is $I_{small} = I_{cm} + md^2 = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$. Substituting $m = M/8$, we get $I_{small} = \frac{7}{5}\left(\frac{M}{8}\right)R^2 = \frac{7}{40}MR^2$. The required ratio is $\frac{I_{small}}{I_{large}} = \frac{7/40}{64/40} = \frac{7}{64}$.