From a uniform circular disc of radius R and mass 9 M, a small disc of radius $\frac{R}{3}$  is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing, through centre of disc is

Let  $\sigma$  be the mass per unit area.

The total mass of the disc 

$=\sigma \times \pi R^2=9M$   The mass of the circular disc cut   $=\sigma \times \pi {\left(\frac{R}{3}\right)}^2=\sigma \times \frac{\pi R^2}{9}=M$

Let us consider the above system as a complete disc of mass 9M and a negative mass M super imposed on it.  Moment of inertia  $\left(I_1\right)$  of the complete disc $=\frac{1}{2}9M{R}^2$   about an axis passing through $O$   and perpendicular to the plane of the disc.

$M.I.$  of the cut out portion about an axis passing through   and perpendicular to the pale of disc 

$=[\frac{1}{2}\times M\times {\left(\frac{R}{3}\right)}^2+M\times {\left(\frac{2R}{3}\right)}^2]$    (Using perpendicular axis theorem)  

The total  $M.I.$  of the system about an axis passing through  $O$  and perpendicular to the plane of the disc is 

$I=I_1+I_2=\frac{1}{2}9M{R}^2-[\frac{1}{2}\times M\times {\left(\frac{R}{3}\right)}^2+M\times {\left(\frac{2R}{3}\right)}^2]$

$=\frac{9M{R}^2}{2}-\frac{9M{R}^2}{18}=\frac{(9-1)M{R}^2}{2}=4M{R}^2$