A thin wire of length / and mass m is bent in m the form of a semicircle as shown in the figure. Its moment of inertia about an axis joining its free ends will be

Options

A. `ml^2`
B. zero
C. `ml^2 // p^2`
D. `ml^2 // 2pi^2`

Answer

D

Explanation

As `pi r = l therefore r = (l)/(pi)`
Moment of inertia of a ring about its diameter = `(1)/(2) mr^(2)`
`therefore` Moment of inertia of semicircle = `(1)/(2) [ m ((l)/(pi))^(2) ] = (ml^2)/(2 pi^(2))`