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The moment of inertia of a ring about an...

The moment of inertia of a ring about an axis passing though the centre and perpendicular to its plane is l. It is rotating ring is gently placed `omega` Another identical ring is gently placed on it, so that their centres coincide. If both the rings are rotating about the same axis, then loss in kinetic energy is

A

`(Iomega^(2))/(2)`

B

`(Iomega^(2))/(4)`

C

`(Iomega^(2))/(6)`

D

`(Iomega^(2))/(8)`

Text Solution

Verified by Experts

The correct Answer is:
B

`I_(1)omega _(1) =I_(2) omega_(2)`
`I omega_(1) =2 I omega_(2)`
`omega_(2) =(omega _(1))/(2) =(omega)/(2) `
New `KE =1/2 I omega ^(2)`
`=1/2 2I ((omega)/(2))^(3) =(Iomega^(2))/(4)`
Change in `Ke =1/2 I omega ^(2) -(I omega ^(2))/(4) =(i omega^(2))/(4)`
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