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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 the remaining part of disc about a perpendicular axis, passing through the centre?

A
9 MR2/32
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B
15 MR2/32
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C
13 MR2/32
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D
11 MR2/32
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Solution

## The correct option is C 13 MR2/32ITotal disc=MR22 Mass of removed part =M4 (∵ Mass ∝ Area) Moment of inertia of removed part about same perpendicular axis is Iremoved=M4 (R/2)22+M4(R2)2=3MR232 Moment of inertia of remaining part of disc about perpendicular axis passing through centre is Iremaining disc=ITotal−Iremoved =MR22−3MR232=1332MR2

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