A circular hole of radius R/4 is made in a thin uniform disc having mass M and radius R . The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc

I0=mr^2/2

1/2*m/16*r^2/16+m/16*9r^2/16(m+md)

=mr^2+18mr^2/(512)

=19mr^2/512

I remaining=mr^/2-19/512*mr^2

=237/512 mr^2