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

Options

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

Answer

C

Explanation

The M.I. of a ring (loop) about its diameter, `I_(D)=(1)/(2)MR^(2)`
According to the theorem of parallel axes, M.I. of the loop about XX' is
`I=I_(0)+MR^(2)=(1)/(2)MR^(2)+MR^(2)=(3)/(2)MR^(2)`
But `M=rhoL` where `rho` is the linear density of the wire and `L=2piR therefore R=(L)/(2pi)`
`therefore I=(3)/(2)MR^(2)=(3)/(2)(rhoL).(L^(2))/(4pi^(2))=(3)/(8)(rhoL^(3))/(pi^(2))`