A disc has mass 9 m. A hole of radius R/3 is cut from it as shown in the figure. The moment of inertia of remaining part about an axis passing through the centre 'O' of the disc and perpendicular to the plane of the disc is :

Options

A. `8 mR^(2)`
B. `4 mR^(2)`
C. `(40)/(9)mR^(2)`
D. `(37)/(9)mR^(2)`

Answer

B

Explanation

`I_(0)=(9mR^(2))/(2)`
`I=I_("com")+M((2R)/(3))^(2)=(M((R )/(3))^(2))/(3)+(4R^(2))/(9)M=(MR^(2))/(2)`
`I^(1)=I_(0)-I=(9)/(2)MR^(2)-(MR^(2))/(2)=4MR^(2)`