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

Answer

C

Explanation

Volume of the cut portion is 1/9 of total volume of disc, hence mass of the cut portion is M, because total mass given is 9M.
We can find moment of inertia of disc without hole about the given axis through centre and then subtract moment of inertia of smaller disc (which is cut through) about the centre of bigger disc.
`I=((9M)R^(2))/(2)-((M)/(2)((R)/(3))^(2)+M((2R)/(3))^(2))`
Note that moment of inertia of smaller disc is written using parallel axis theorem. Solving the above expression we get,
`I=4MR^(2)`
Hence option (c) is correct.