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:

1. The rod of mass $M$ and length $L$ is divided into two halves.2. Each half has mass $m=\frac{M}{2}$ and length $l=\frac{L}{2}$.3. The axis passes through the end of each half-rod. The moment of inertia of a rod about its end is given by $I=\frac{1}{3}m{l}^2$.4. Total moment of inertia $I_{\text{total}}$ is the sum of the moments of inertia of the two halves:
$$I_{\text{total}}=I_1+I_2=\frac{1}{3}\left(\frac{M}{2}\right){\left(\frac{L}{2}\right)}^2+\frac{1}{3}\left(\frac{M}{2}\right){\left(\frac{L}{2}\right)}^2$$
$$I_{\text{total}}=2\times \left(\frac{1}{3}\cdot \frac{M}{2}\cdot \frac{L^2}{4}\right)=\frac{M{L}^2}{12}$$
Note: The distribution of mass relative to the axis remains the same regardless of the $90^{\circ }$ angle.