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

Options

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

Answer

A

Explanation

Considering the information given in the question, let us draw the figure

If the above figure is considered, then moment of inertia o disc will be given as `I=I_("remain")+I_((R//2))rArr I_("remain")=I-I_((R//2))` Putting the values, we get
`=(MR^(2))/(2)-[((M)/(4)((R)/(2))^(2))/(2)+(M)/(2)((R)/(2))^(2)]`
`=(MR^(2))/(2)-[(MR^(2))/(32)+(MR^(2))/(16)]=(MR^(2))/(2)-[(MR^(2)+2MR^(2))/(32)]`
`=(MR^(2))/(2)-(3MR^(2))/(32)=(16 MR^(2)-3MR^(2))/(32)`
`I_("remain")=(13MR^(2))/(32)`.