From a disc of radius R a mass M, a circular hole of diameter 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 it

Options

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

Answer

B

Explanation

Moment of inertia of total disc
`I=1//2MR^(2)`
`sigma=M/A`
`sigmaM/(4piR^(2))`
mass `=sigma xx 4pir^(2)`
`m=M/(4piR^(2))xx4pi(R/2)^(2)`
Mass of small part
`m=M/4`
I of removed disc
`I'=1/2(M/4)((R^(2))/4)+(MR^(2))/16`
`=3/32MR^(2)`
`I_("Remaining part")=I-I'`
`=1//2MR^(2)-3//32MR^(2)`
`=13//32MR^(2)`