From a disc of radius R and mass m, a circular hole of diamter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

Options

A. `15 MR^(2)//32`
B. `13MR^(2)//32`
C. `11MR^(2)//32`
D. `9MR^(2)//32`

Answer

B

Explanation

`I_("total disc") = (MR^(2))/2 `
`M_("Removed") =M/4 ("Mass " prop "area ")`
`I_("Removed") ` (above same Perpendicular axis)
`=M/4 ((R//2)^(2))/2 +M/4 (R/2)^(2) =(3MR^(2))/32`
`I_("Remaining disc ") = I_("toatl") -I_("Removed")`
`(MR)^(2))/2 -3/32 MR^(2) = 13/32 MR^(2)`