A thin wire of length (l) and mass (m) is bent in a form of... | Physics | SolveeIt | Solveeit

Today, August 18

Question

A thin wire of length $l$ and mass $m$ is bent in a form of semicircle. Its moment of inertia about an axis joining its free ends will be

A

$\frac {2ml^2}{\pi^2}$

B

$ml^2$

C

$\frac {ml^2} {2}$

D

$\frac {ml^2}{2\pi^2}$

Answer

$\frac {ml^2}{2\pi^2}$

Explanation

Solution

Here $\pi r = l \, i.e., \, r = \frac{l}{\pi}$ M.O.I. = $\frac{Mr^2}{2} = \frac{Ml^2}{2 \pi^2}$