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

Options

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

Answer

A

Explanation

Mass of the disc = 9 M
Mass of removed portion of disc = M
The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is
`I_(1) = (9)/(2) MR^(2)`
Now, the moment of inertia of the disc with removed portion
`I_(2) = (1)/(2) M ((R)/(3))^(2) = (1)/(18) MR^(2)`
Therefore , moment of inertia of the remaining portion of disc about O is `I = I_(1) - I_(2) = (9 MR^(2))/(2) = (MR^(2))/(18) = (40 MR^(2))/(9)`