Moment of inertia of a ring about a tangent perpendicular to the plane of the ring is I. Then moment of inertia of the same ring about a tangent lying in the plane of the ring is
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a
3I2
b
I4
c
2I3
d
3I4
answer is D.
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Detailed Solution
By parallel axis theorem, I=mR2+mR2=2mR2Now moment of inertia about a diameter =12mR2Moment of inertia about tangent ABIAB=12mR2+mR2=32mR2=32 x I2=3I4