A circular plate of radius R/2  is cut from one edge of a thin circular plate of radius R . The moment of inertia of the remaining portion about an axis through O perpendicular to plane of the plate is (M =mass of remaining circular plate.)

Let us assume the mass of the plate without the portion being removed be m₀

Let m' be the total mass of plate of removed plate ${\mathrm{m}}^{\text{'}}=\frac{{\mathrm{m}}_0}{4}$
Mass of remaining plate is $\mathrm{m}=\frac{3{\mathrm{m}}_0}{4}$
${\mathrm{I}}_{\mathrm{total} }={\mathrm{I}}_{\mathrm{removed} }+{\mathrm{I}}_{\mathrm{remaining}}$

$\Rightarrow {\mathrm{I}}_{\mathrm{remaining} }={\mathrm{I}}_{\mathrm{total} }-{\mathrm{I}}_{\mathrm{removed} }$

${\mathrm{I}}_{\mathrm{remaining} }=$${\mathrm{m}}_0\frac{{\mathrm{R}}^2}{2}-\left(\frac{\frac{{\mathrm{m}}_0}{4}{\left(\frac{\mathrm{R}}{2}\right)}^2}{2}+\frac{{\mathrm{m}}_0}{4}\times {\left(\frac{\mathrm{R}}{\mathrm{L}}\right)}^2\right)=\frac{13}{32}{\mathrm{m}}_0{\mathrm{R}}^2=\frac{13}{24}{\mathrm{mR}}^2$