A rod of length L and mass M is bent to form a semi-circular ring as shown in figure. The moment of inertia about XY is

Options

A. `(ML^(2))/(2pi^(2))`
B. `(ML^(2))/(pi^(2))`
C. `(ML^(2))/(4pi^(2))`
D. `(2ML^(2))/(pi^(2))`

Answer

A

Explanation

`I= (MR^(2))/(2), pi R= L rArr R= (L)/(pi)`
`rArr I= (ML^(2))/(2pi^(2))`