Without cavity I1=2MR2
$\begin{aligned}
\text { Mass of removed disc }= & \frac{\mathrm{M}}{\pi \mathrm{R}^2} \times\left(\frac{\mathrm{R}}{3}\right)^2 \pi \
& =\left(\frac{\mathrm{M}}{9}\right)
\end{aligned}$
M.I. of removed disc I2=29M(3R)2+9M×(32R)2
$\begin{aligned}
& =\frac{\mathrm{MR}^2}{18} \
& \mathrm{I}=\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{18}=\frac{4 \mathrm{MR}^2}{9} \
& (\mathrm{n}=9)
\end{aligned}$
