A small hole is made in a disc of mass `M` and radius `R` at a distance `R//4` from centre. The disc is supported on a horizontal peg through this hole. The moment of inertia of the disc about horizontal peg is

Options

A. `(MR^(2))/9`
B. `5/16MR^(2)`
C. `9/16MR^(2)`
D. `5/4MR^(2)`

Answer

C

Explanation

Applying theorem of parallel axes.
`I=(MR^(2))/2+M(R/4)^(2),I=(MR^(2))/2+(MR^(2))/16, I+(9MR^(2))/16`