A sphere of radius $R$ is cut from a larger solid sphere of radius $2 R$ 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:

• A

  $\frac{7}{8}$
• B

  $\frac{7}{40}$
• Correct answer
  C

  $\frac{7}{57}$
• D

  $\frac{7}{64}$

Solution

$\begin{aligned} & \mathrm{I}_1=\frac{2}{5} \mathrm{M}(2 \mathrm{R})^2 \\ & \mathrm{I}_2=\frac{2}{5} \mathrm{mr}^2+\mathrm{md}^2 \\ & =\frac{2}{5}\left(\frac{\mathrm{M}}{8}\right) \mathrm{R}^2+\frac{\mathrm{M}}{8}(\mathrm{R})^2 \\ & =\frac{7 \mathrm{MR}^2}{40} \\ & \mathrm{I}=\mathrm{I}_1-\mathrm{I}_2=\frac{8 \mathrm{MR}^2}{5}-\frac{7 \mathrm{MR}^2}{40}=\frac{57}{40} \mathrm{MR}^2\end{aligned}$ $\frac{\mathrm{I}_2}{\mathrm{I}_1}=\frac{7}{57}$

Hence, the correct answer is option: 'C'.