Question
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:
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Solution
For larger solid sphere about diameter $Y$-axis,
$$I_{\text{whole}\text{~}} = \frac{2}{5}M(2R)^{2} = \frac{8}{5}MR^{2}$$
Density of sphere is uniform
$$\begin{matrix} & \ \Rightarrow \frac{M}{V_{\text{whole}\text{~}}} = \frac{M_{\text{smaller}\text{~}}}{V_{\text{smaller}\text{~}}} \Rightarrow \frac{M}{\frac{4}{3}\pi(2R)^{3}} = \frac{M^{'}}{\frac{4}{3}\pi R^{3}} \\ & \ \Rightarrow M^{'} = \frac{M}{8} \end{matrix}$$
Using parallel axis theorem for smaller sphere,
$$\begin{matrix} & I^{'} = I_{cm} + M^{'}R^{2} = \frac{2}{5}\frac{MR^{2}}{8} + \frac{MR^{2}}{8} = \frac{7}{40}MR^{2} \\ & \ \therefore\text{~}\text{Ratio}\text{~} = \frac{I_{\text{smaller}\text{~}}}{I_{\text{remaining}\text{~}}} = \frac{I^{'}}{I_{\text{whole}\text{~}} - I^{'}} = \frac{\frac{7}{40}MR^{2}}{\left( \frac{8}{5} - \frac{7}{40} \right)MR^{2}} = \frac{7}{64 - 7} = \frac{7}{57} \end{matrix}$$
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