From a circular disc of radius R and mass 9 M , a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

Options

A. `4MR^(2)`
B. `(40)/(9)MR^(2)`
C. `10MR^(2)`
D. `(37)/(9)MR^(2)`

Explanation

`Mass prop area`
`m_(1)`: mass of small disc
`(m_(1))/(9M)=(pi(R//e)^(2))/(piR^(2))implies m_(1)=M`
`M.I.` of a circular disc of radius `R` and mass `9M` about the required axis
`I_(1)=(1)/(2)xx9MR^(2)=(9MR^(2))/(2)`
`M.I` of a small disc about the same axis
`I_(2)=[(1)/(2)M((R)/(3))^(2)+M(2R^(2))/(3)]=(1)/(2)MR^(2)`
`M.I.` of the remaining disc is
`I_(1)-I_(2)=4MR^(2)`