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

Answer

D

Explanation

The moment of inertia of the loop about `XX'` axis is `I_(XX) = (mR^(2))/2 + mR^(2) = 3/2 mR^(2)`
where, m = mass of the loop and R = radius of the loop
Here, `m=L rho` and `R = L/(2pi)`, therefore, `I_(XX') = 3/2(Lrho)(L/(2pi))^(2) =(3L^(2)rho)/(8pi^(2))`