From a disc of radius $R$ and mass $M$ a circular hole of diameter$R$ whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

$\text{ Mass per unit area of disc }=\frac{M}{\pi R^2}$

Mass of removed portion of disc,

$M^{\text{'}}=\frac{M}{\pi R^2}\times \pi {\left(\frac{R}{2}\right)}^2=\frac{M}{4}$

Moment of inertia of removed portion about an axis passing through centre of disc O and perpendicular to the plane of disc,

$I_O^{\text{'}}=I_{O^{\text{'}}}+M^{\text{'}}{d}^2$

$=\frac{1}{2}\times \frac{M}{4}\times {\left(\frac{R}{2}\right)}^2+\frac{M}{4}\times {\left(\frac{R}{2}\right)}^2=\frac{M{R}^2}{32}+\frac{M{R}^2}{16}=\frac{3M{R}^2}{32}$

When portion of disc would not have been removed, the moment of inertia of complete disc about centre O is

$I_O=\frac{1}{2}M{R}^2$

 So, moment of inertia of the disc with removed portion is 

$I=I_O-I_O^{\text{'}}=\frac{1}{2}M{R}^2-\frac{3M{R}^2}{32}=\frac{13M{R}^2}{32}$