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

Answer

D

Explanation

The M.l of a ring (loop) about its diameter, `l_(D)=(1)/(2)MR^(2)` According to the theorem of parallel axes, M.l. of the loop about XX' is
`l=l_(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" "l=(3)/(2)MR^(2)`
`=(3)/(2)(rhoL)(L^(2))/(4pi^(2))=(3)/(8)(rhoL^(3))/(pi^(2))`