A thin wire of length `L` and uniform linear mass density `rho` is bent into a circular loop with centre at `O` as shown. The moment of inertia of the loop about the axis `XX'` is :
.

Options

A. `(rhoL^(3))/(8pi^(2))`
B. `(rhoL^(3))/(16pi^(2))`
C. `(5rhoL^(3))/(16pi^(2))`
D. `(3rhoL^(3))/(8pi^(2))`

Explanation

Mass of ring `m=rho L`
`L= 2piimplies r=(L)/(2pi)`
`M.I.` of the ring about an axis passing through `O` and parallel to `XX'`
`I_(0)=(1)/(2)mr^(2)`
`I_(XX')=I_(0)+mr^(2)=(3)/(2)mr^(2)`
`=(3)/(2)rho L ((L)/(2pi))^(2)=(3rho L^(3))/(8pi^(2))`