From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centreis cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre? [NEET 2016]

Considering the information given in the question, let us draw the figure.If the above figure is considered, then moment of inertia of disc will be given asI=Iremaining +I(R/2)⇒ Iremaining =I−I(R/2)Putting the values, we get=MR22−M4R222+M4R22=MR22−MR232+MR216=MR22−MR2+2MR232=MR22−3MR232=16MR2−3MR232Iremaining =13MR232