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A thin wire of length L and uniform line...

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
`

A

`(rhoL^3)/(8pi^2)`

B

`rhoL^3/(16pi^2)`

C

`(5rhoL^3)/(16pi^2)`

D

`(3rhoL^3)/(8pi^2)`

Text Solution

Verified by Experts

The correct Answer is:
D
(d) Moment of inertia about the diameter of the circular
loop (ring) `= (1)/)2) MR^2`
Using parallel axis theorem
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 = Lrho and R = (L)/(2pi),`
`:.I_(XX) = (3)/(2)(Lrho)((L)/(2pi))^2 = (3L^3rho)/(8pi^2)`
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