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

Answer

D

Explanation

mass of the ring =LP
radius of ring `L/(2pi)`

`I_(AB)=(mr^(2))/2`
`I_(x x')=3/2 mr^(2)` (by parallel axis theorem)
By putting value of m and r we get
`I_(x x')=(3 rhoL^(3))/(8pi^(2))`