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A concentric hole of radius R/2 is cut ...

A concentric hole of radius R/2 is cut from a thin circular plate of mass M and radius R. The moment of inertia of the remaining plate about its axis will be

A

`(13)/( 24) MR^2`

B

`(11)/( 24) MR^2`

C

`(13)/(32) MR^2`

D

`(15)/(32) MR^2`

Text Solution

Verified by Experts

The correct Answer is:
D
Mass of hole `(M.) = ((M)/(pi R^2)) ((pi R^2)/( 4)) =(M )/(4)`

Moment of inertial of remaining plate
`=1/2 MR^2 -1/2 M. ((R )/(2))^2 =(15)/(32) MR^2`
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