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

Options

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))`

Explanation

`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))`