A thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them is $90^{\circ}$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

• A

  $\frac{M L^2}{24}$
• Correct answer
  B

  $\frac{M L^2}{12}$
• C

  $\frac{M L^2}{6}$
• D

  $\frac{\sqrt{2} M L^2}{24}$

Solution

Since rod is bent at the middle, so each part of it will have same length $\left(\frac{L}{2}\right)$ and mass $\left(\frac{M}{2}\right)$ as shown. Moment of inertia of each part through its one end $=\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2$ Hence, net moment of inertia through its middle point $O$ is $\begin{aligned} I & =\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2+\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2 \\ & =\frac{1}{3}\left[\frac{M L^2}{8}+\frac{M L^2}{8}\right]=\frac{M L^2}{12} \end{aligned}$

Hence, the correct answer is option: 'B'.