From a disc of radius Rand mass M a circular hole of diameterR 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?

Mass per unit area of disc =MπR2Mass of removed portion of disc,M'=MπR2×πR22=M4Moment of inertia of removed portion about an axis passing through centre of disc O and perpendicular to the plane of disc,IO'=IO'+M'd2=12×M4×R22+M4×R22=MR232+MR216=3MR232When portion of disc would not have been removed, the moment of inertia of complete disc about centre O isIO=12MR2 So, moment of inertia of the disc with removed portion is I=IO-IO'=12MR2-3MR232=13MR232