From a circular disc of radius R and 9M , a small disc of mass M and radius `(R )/(3)` is removed concentrically .The moment of inertia of the remaining disc about and axis perpendicular to the plane of the disc and passing through its centre is

Options

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

Answer

A

Explanation

`I=(9M(R)^(2))/(2)-(M((R)/(3))^(2))/(2)`
`=(9MR^(2))/(2)-(MR^(2))/(18)=(80MR^(2))/(18)`
= `(40MR^(2))/(9)`