The moment of inertia of a rod about an axis passing through its centre and perpendicular to its length is…
The moment of inertia of a rod about an axis passing through its centre and perpendicular to its length is $\frac{1}{12} M L^2$, where $M$ is the mass and $L$ is the length of the rod. The rod is bent in the middle so that the two halves make an angle of $60^{\circ}$. The moment of inertia of the bent rod about the same axis would be
$\frac{1}{48} M L^2$
$\frac{1}{12} M L^2$
$\frac{1}{24} M L^2$
$\frac{1}{8 \sqrt{3}} M L^2$
Solution
The moment of inertia of bend rod is
$\begin{aligned}
& \mathrm{I}=\frac{1}{3}\left(\frac{\mathrm{M}}{2}\right)\left(\frac{\mathrm{L}}{2}\right)^2+\frac{1}{3}\left(\frac{\mathrm{M}}{2}\right)\left(\frac{\mathrm{L}}{2}\right)^2 \\
& =\frac{1}{24} \mathrm{ML}^2+\frac{1}{24} \mathrm{ML}^2=\frac{1}{12} \mathrm{ML}^2
\end{aligned}$