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Question

# From a uniform circular disc of radius R and mass 9M, a small disc of radius R3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the disc is?

A
10MR2
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B
379MR2
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C
4MR2
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D
409MR2
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Solution

## The correct option is D 4MR2Given:Mass of dics 9M Mass of removed part m=9MπR2×π(R/3)3=MMoment of inertia of Disc without hole about axis perpendicular to the plane passing through center isI1=9MR22Moment if inertia of hole about axis passing through center of disc I2=M(R3)22+M(2R3)2The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of discI=I1−I2I=9MR22−⎡⎢ ⎢ ⎢⎣M(R3)22+M(2R3)2⎤⎥ ⎥ ⎥⎦=4MR2

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