From a disc of mass ' \(M\) ' and radius ' \(R\) ', a circular hole of diameter ' \(R\) ' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

Moment of inertia of disc is given by
\(\mathrm{I}_{\text {disc }}=\mathrm{I}_{\mathrm{f}}+\mathrm{I}_{\text {hole }}\)
.... \(\left\{\mathrm{I}_{\mathrm{f}}=\right.\) M.I. of remaining part \(\}\)
\(\begin{aligned}
& \therefore \quad \mathrm{I}_{\mathrm{r}}=\mathrm{I}_{\text {dise }}-\mathrm{I}_{\text {hole }} \\
& \mathrm{I}_{\text {disc }}=\frac{\mathrm{MR}^2}{2}
\end{aligned}\)
By parallel axes theorem we get,
\(\mathrm{I}_{\text {hole }}=\left[\frac{\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}\right)^2}{2}+\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}\right)^2\right]\)
\(\ldots\left\{\begin{array}{l}\because \mathrm{M}_{\text {bole }}=\frac{\mathrm{M}_{\text {dive }}}{4} \\ \because \text { the surface density is same }\end{array}\right\}\)
\(\therefore \quad \mathrm{I}_{\text {hole }}=\left[\frac{\mathrm{MR}^2}{32}+\frac{\mathrm{MR}^2}{16}\right]\)
Substituting eq (iii) and eq (ii) in eq (i) we get,
\(\begin{aligned}
\mathrm{I}_{\mathrm{r}} & =\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{32}-\frac{\mathrm{MR}^2}{16} \\
& =\mathrm{MR}^2\left[\frac{1}{2}-\frac{1}{32}-\frac{1}{16}\right] \\
& =\frac{13}{32} \mathrm{MR}^2
\end{aligned}\)