Question

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 ?
A. `3 MR^2//32`
B. `15 MR^2//32`
C. `13 MR^2//32`
D. `11 MR^2//32`

Answer 1

Correct Answer - C
( c) `I_(Total disc) = (MR^2)/(2)`
`M_(Removed) =(M)/(d) (Mass prop area)`
`I_(Removed)` (about the same perpendicular axis)
=`(M)/(4)((R//2)^2)/(2) + (M)/(4) ((R)/(2))^2 = (3 MR^2)/(32)`
`I_(Removed disc) = I_(Total) - I_(Removed)`
=`(MR^2)/(2) -(3)/(32) MR^2 = (13)/(32) MR^2`.
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