A thin disc of mass 9M and radius R from which a disc of radius R/3 is cut shown in figure. Then moment of inertia of the remaining disc about O, perpendicular to the plane of disc is -

Options

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

Answer

A

Explanation

Mass density i.e., mass per unit
area of the dic is `(9M)/(piR^(2))`
Mass of removed portion of disc
`=(9M)/(piR^(2))xxpi((R)/(3))^(2)=M`
Moment of inertia of removed portion about an axis passing through centre of disc and perpendicular to the plane of disc, using theorem of parallel axis is
`l_(1)=(M)/(2)((R)/(3))^(2)+M((2R)/(3))^(2)=(1)/(2)MR^(2)`
When portion of disc would not have been removed, then the moment of inertia of complete disc abou the given axis is
`i_(1)=(9)/(2)MR^(2)`
So, moment of inertia of the disc with removed portion, about the given axis is
`l=l_(2)-l_(1)=(9)/(2)MR^(2)-(1)/(2)MR^(2)=4MR^(2)`