Question

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

• A. $m l^{2}$
• B. zero
• C. $m l^{2} / \pi^{2}$
• D. $m l^{2} / 2 \pi^{2}$

Answer 1

Answer: (D)

As $\pi r=l \therefore r=\frac{l}{\pi}$
Moment of inertia of a ring about its diameter $=\frac{1}{2} m r^{2}$
$\therefore $ Moment of inertia of semicircle $=\frac{1}{2}\left[m\left(\frac{l}{\pi}\right)^{2}\right]=\frac{m l^{2}}{2 \pi^{2}}$