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. `(219MR^(2))/(256)`
B. `(237MR^(2))/(512)`
C. `(19MR^(2))/(512)`
D. `(197MR^(2))/(256)`

Answer

B

Explanation

Moment of inertia along the axis passing through centre of mass and perpendicular to the plane of full disc having radius R about .O. point

`I_("total")=(MR^(2))/(2)`
Radius of removed disc =R/4
Mass of removed dics `=(MR^(2))/(piR^(2))xxpi((R)/(4))^(2)=M//16`
Moment of inertia of removed disc about its own axis O. is
`I_(cm)=(1)/(2)(M)/(16)((R)/(4))^(2)=(MR^(2))/(512)`
Moment of inertia of removed disc about O point
`I_("removed disc")=I_(cm)+(M)/(16)x^(2)`
[using parallel axis theorem]
`=(MR^(2))/(512)+(M)/(16)((3R)/(4))^(2)=(19MR^(2))/(512)`
Moment of inertia of remaining disc
`I_("remaining disc")=I_("total")-I_("removed disc")=(MR^(2))/(2)-(19MR^(2))/(512)`
`=(237)/(512)MR^(2)`