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Question

If I1 is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and I2 is the moment of inertia of the ring (formed by bending the rod) about an axis perpendicular to the plane of ring and passing through its centre, then the ratio I1I2 is

A
3π2
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B
2π2
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C
π22
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D
π23
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Solution

The correct option is D π23
let L be the length of rod,
I1=mL212
Mass of the ring formed by bending the rod will be m and for the radius of ring,
2πr=L
r=L2π
The moment of inertia of ring about an axis to its plane and passing through its centre is,
I2=mr2
I2=mL24π2
I1I2=π23

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