A disk of certain radius is cut from a disk of mass 9M and radius R as shown in figure. Find its moment of inertia about an axis passing through its centre C and perpendicular to its plane.

Moment of inertia of given disk (before cutting) (w.r.t axis passing through it's centre C and perpendicular to it's plane) or (principle axis) = (9M)R22Mass of given disk = 9M(Assume density and thickness are uniform every where on disk)So mass for area π(R2) = 9MSo mass per unit area = =9MπR2So mass for areaπR32=9MπR2×πR32= M = mass of small diskSo moment of inertia of small disk (w.r.t. its own principle axis) = MR322Now from parallel axis theoremI=Icm+Md2here d is distance between parallel axes,So Ic=MR322+M2R32moment of inertia of small disk wrt principle axis of bigger disk)(here, d=2R3, given in question)Solving this Ic=MR218+4MR29Ic=MR22Now moment of inertia of remaining part w.r.t. given axis=9MR22−MR22=4MR2