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Q.

From a complete ring of mass M and radius R, a 300 sector is removed. The moment of inertia of the incomplete ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is

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a

912MR2

b

11.312MR2

c

MR2

d

1112MR2

answer is B.

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Detailed Solution

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Mass of incomplete ring =m-M2π×π6=M-M12=11M12 Moment of inertia of incomplete ring =11M12R2=1112MR2

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