Q 11-06-007NEETNEET 2025Top questionMedium
A sphere of radius $R$ is cut from a larger solid sphere of radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :
Answer: (C) $\dfrac{7}{57}$
Let the small sphere have mass $m$. The large sphere has 8 times its volume, so mass $8m$.
The Y-axis passes through the centre of the large sphere. The small sphere's centre is a distance $R$ from it.
Small sphere: $I_s = \dfrac{2}{5}mR^2 + mR^2 = \dfrac{7}{5}mR^2$
Full large sphere: $I_L = \dfrac{2}{5}(8m)(2R)^2 = \dfrac{64}{5}mR^2$
Remaining part: $I_{\text{rest}} = \dfrac{64}{5}mR^2 - \dfrac{7}{5}mR^2 = \dfrac{57}{5}mR^2$
$$\frac{I_s}{I_{\text{rest}}} = \frac{7}{57}$$
Solution by Sreeraj P, M.Sc Physics