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(D) Let the three rods be $R_1$,$R_2$,and $R_3$ as shown in the figure. We want to calculate the moment of inertia of the system about rod $R_1$.$1$.

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$\frac{4 M L^2}{3}$

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(D) Let the three rods be $R_1$,$R_2$,and $R_3$ as shown in the figure. We want to calculate the moment of inertia of the system about rod $R_1$.$1$. For rod $R_1$: Since the axis of rotation lies along the rod itself,the moment of inertia $I_1 = 0$.$2$. For rod $R_2$: This rod is perpendicular to the axis of rotation $R_1$ and is attached at one of its ends. The moment of inertia of a rod about an axis passing through its end and perpendicular to its length is $I_2 = \frac{M L^2}{3}$.$3$. For rod $R_3$: This rod is parallel to the axis of rotation $R_1$ at a distance $L$. Using the parallel axis theorem,$I_3 = I_{CM} + M d^2$. Since the axis passes through the center of mass of $R_3$ (which is at distance $L$ from $R_1$),$I_{CM}$ about the axis parallel to the rod is $0$. Thus,$I_3 = 0 + M L^2 = M L^2$.Total moment of inertia $I = I_1 + I_2 + I_3 = 0 + \frac{M L^2}{3} + M L^2 = \frac{4 M L^2}{3}$.

Three identical rods each of mass $M$ and length $L$ are joined to form a symbol $H$. The moment of inertia of the system about one of the sides of $H$ is

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