From a circular disc of mass `M` and radius `R`, a part of `60^(@)` is removed. The `M.I.` of the remaining portion of disc about an axis passing through the center and perpendicular to plane of disc is

Options

A. `(5)/(6)MR^(2)`
B. `(5)/(3)MR^(2)`
C. `(5)/(12)MR^(2)`
D. `(5)/(24)MR^(2)`

Explanation

Mass of the removed portion `=Mxx(60^(@))/(360^(@))=(M)/(6)`
Its `M.I. I_(1)=(1)/(2)(M)/(6)R^(2)`
`M.I.` of the complete disc `I_(2)=(1)/(2)MR^(2)`
`M.I.` of the remaining portion `=I_(2)-I_(1)=(5)/(12)MR^(2)`