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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The given situation can be shown as

Let us take the moment of inertia of the system about rod $R_{1}$ then total moment of inertia is
For rod $R_{1} l_{1} = 0$
For rod $R_{2}$ , using perpendicular axis theorem,
$l_{2} = \frac{M L^{2}}{3}$
For rod $R_{3}$ , using parallel axis theorem,
$l_{3} = l_{C M} + l_{(a t L)} = 0 + M L^{2} = M L^{2}$
Now, putting the values of $l_{1}, l_{2}$ and $l_{3}$ in Eq. (i), we get
$l_{τ} = 0 + \frac{M L^{2}}{3} + M L^{2} \Rightarrow l_{τ} = \frac{4 M L^{2}}{3}$

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