A thin wire of length / and uniform linear mass density p is bent into a circular loop with centre O and radius r as shown in figure . The moment of inertia of the loop about the axis XX' is

Options

A. `(3 rho l^(3))/(8 pi^2)`
B. `(rho l^(3))/(16 pi^2)`
C. `(3 rho l^(3))/(8 pi^2 r)`
D. `(rho l^(3))/(8 pi^(2) r)`

Answer

A

Explanation

Since , `2pi r = l` or `r = (l)/(2pi) " " … (i)`
Mass M = `rho l " " … (ii)`
Moment of inertia of the loop about the axis `XX..` = moment of inertia about its diameter + `Mr^(2)` (by theorem of parallel axes)
= `(Mr^(2))/(2) + Mr^(2) = (3)/(2) Mr^(2) = (3)/(2) (rho l) ((l)/(2pi))^(2)` (Using (i) and (ii))
`= (3rho l^(3))/(8 pi^(2))`