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A portion of a ring of radius R has been removed as shown in figure. Mass of the remaining portion is m. Centre of the ring is at origin O. Let IA and IO be the moments of inertia passing through points A and O are perpendicular to the plane of the ring. Then,

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I0=mR2

IO>IA

Both 1 and 4 are correct

answer is D.

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Detailed Solution

Whole mass has equal distance from the centre O. Hence, IO = mR2 . Further, centre of mass of the remaining portion will be to the left of point O. More the distance of axis from centre of mass, more is the moment of inertia. Hence, IA > IO.

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