A thin rod of mass `M` and length `L` is bent into a circular ring. The expression for moment of inertia of ring about an axis passing through its diameter is

Options

A. `(ML^(2))/(2pi^(2))`
B. `(ML^(2))/(4pi^(2))`
C. `(ML^(2))/(8pi^(2))`
D. `(ML^(2))/(pi^(2))`

Answer

C

Explanation

`I = (MR^(2))/2 but 2piR = L rArr R = L/(2pi)`
` :. I = (ML^(2))/(8pi^(2))`