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. `(rho L^(3))/(8 pi ^(2))`
B. ` ( rho L^(3))/(16pi ^(2))`
C. `(5rho L^(3))/(8 pi ^(2))`
D. `(rho L ^(3))/( 8 pi ^(2))`

Answer

D

Explanation

` I _(XY ) = I_(CM ) +MD ^(2) `
` = (MR ^(2))/( 2) +MR^(2) =(3)/(2) Mr^(2)`
` I_(XY ) = (3)/(2) rho L ((L)/(2pi ))^(2) =(3 rho L^(3))/(8pi ^(2))`