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A portion of a ring of radius R has been...

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 `I_(A) and I_(O)` be the moment of inertia passing through points A and O are perpendicular to the plane of the ring. Then,

A

`I_(O)=mR^(2)`

B

`I_(O)=I_(A)`

C

`I_(O) gt I_(A)`

D

`I_(A) gt I_(O)`

Text Solution

Verified by Experts

The correct Answer is:
D
Whole mass has equal distance from the centre O. Hence, `I_(0) mR^(2)`. 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, `I_(A) gt I_(0)`
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