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

Answer

A

Explanation

As is clear from Fig.

`I_("disc") = (1)/(2)MR^(2)`
`I_("hole") = (1)/(2) ((M)/(4))(R//2)^(2) + (M)/(4) (R//2)^(2)`
`= (MR^(2))/(32) + (MR^(2))/(16) = (3MR^(2))/(32)`
`:. I_("remaining") = I_("disc") - I_("hole")`
`= (1)/(2)MR^(2) - (3)/(32)MR^(2) = (13)/(32)MR^(2)`