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The M.I. of a disc of mass M and radius ...

The M.I. of a disc of mass M and radius R, about an axis passing through the centre O and perpendicular to the plane of the disc is `(MR^(2))/(2)` . If one quarter of the disc is removed, the new moment of inertia of the disc will be

A

`(MR^(2))/(3)`

B

`(MR^(2))/(4)`

C

`(3)/(8)MR^(2)`

D

`(3)/(2)MR^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
M.I. of the disc of mass `M=I=(1)/(2)MR^(2)`
When a quarter of the portion is removed, the new mass
`=(3)/(4)M=M'`
`therefore" New M.I. "=(1)/(2)M'R^(2)=(1)/(2)((3)/(4)M)R^(2)=(3)/(8)MR^(2)`
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