From a circular disc of radius R and mass 9M, a small disc of radius R3is removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

Mass per unit area of disc=9MπR2Mass of removed portion of disc=9MπR2×πR32=MMoment of inertia of removed portion about an axis passing through the centre of disc and perpendicular to the plane of disc, using theorem of parallel axes isI1=M2R32+M2R32=12MR2The moment of inertia of complete disc about the given axis is I2=92MR2So moment of inertia of the disc with removed portion, about the given axis is I=I2−I1=92MR2−12MR2=4MR2