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A rod of lengthL and mass M(0) is bent t...

A rod of length`L` and mass `M_(0)` is bent to form a semicircular ring as shown. The `M.I.` about `X'X` is

A

`(M_(0)L^(2))/(2pi^(2))`

B

`(M_(0)L^(2))/(pi^(2))`

C

`(M_(0)L^(2))/(4pi^(2))`

D

`(2M_(0)L^(2))/(pi^(2))`

Text Solution

Verified by Experts

`L=piRimplies R=(L)/(pi)`
For the complete ring, `M.I.` of the ring about diameter
`=(1)/(2)(2M_(0)R^(2))`. For the semi-circular ring
`M.I.=(1)/(2)[(1)/(2)(2M_(0))R^(2)]`
`=(1)/(2)M_(0)R^(2)`
`=(1)/(2)M_(0)((L)/(pi))^(2)=(M_(0)L^(2))/(2pi^(2))`
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Knowledge Check

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