A disc of mass .m. and radius R has a concentric hole of radius .r.. Its M.I. about an axis through centre and normal to its plane is

Explanation

Consider a ring of radius .x. and of thickness dx of the disc.
The M.I. of the ring about a normal axis through its centre is `dI=(m)/(pi(R^(2)-r^(2)))2pi x.dx.x^(2)`
The total M.I. of the disc is obtained by integrating the above expresion.
`int dI= int_(r )^(R )=(m)/(pi(R^(2)-r^(2)))2pi x.dx.x^(2)`
on solving `I=(m)/(2)(R^(2)+r^(2))`