Four identical rods (each of length I) are joined end to end to form a square as shown in fig. The mass of each rod is M. The moment of inertia of the square about the median line is

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$\frac{2 {ML}^{2}}{3}$

$\frac{{ML}^{2}}{4}$

$\frac{{ML}^{2}}{6}$

$\frac{{ML}^{2}}{5}$

answer is A.

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Detailed Solution

$\begin{array}{l} I_{median line} = I_{A} + I_{B} + I_{C} + I_{D} \\ = \frac{2 {ML}^{2}}{12} + 2 M {(\frac{L}{2})}^{2} \\ = {ML}^{2} [\frac{1}{6} + \frac{1}{2}] = \frac{2}{3} {ML}^{2} \end{array}$

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