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))`

Answer

D

Explanation

Here M.I. loop about its axis `OAB=MR^(2)`

M.I. about diameter
`X XY=(MR^(2))/(2)`
M.I. about `X.Y.=MR^(2)+(MR^(2))/(2)=(3)/(2)M^(2)R^(2)`
`=(3)/(2)xxLrho((L)/(2pi))^(2)=(3rhoL^(3))/(8pi^(2))`