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The moment of inerta of a ring of mass 1...

The moment of inerta of a ring of mass `1kg` about an axis passing through its centre perpendicular to its surface is `4kgm^(2)`. Calculate the radius of the ring.

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Verified by Experts

`l = MR^(2)`
`R = sqrt((l)/(M))=sqrt((4)/(1))=2m`
Hint : `l = MR^(2)`
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