Four identical rods are joined end to end to form a square. The mass of each rod is M. The moment of inertia of the system about an axis passing through the point of intersection of diagonals and perpendicular to the plane of the square is

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

$\frac{13 M l^{2}}{3}$

$\frac{13 M l^{2}}{6}$

$\frac{M l^{2}}{6}$

answer is A.

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Detailed Solution

$\begin{matrix} I_{median line} = I_{1} = I_{A} + I_{B} + I_{C} + I_{D} \\ = 2 \times \frac{M l^{2}}{12} + 2 M {(\frac{l}{2})}^{2} \\ = \frac{M l^{2}}{6} + \frac{M l^{2}}{2} = \frac{2}{3} M l^{2} \end{matrix}$

Using perpendicular axis theorem,

$I = I_{1} + I_{1} = 2 \times \frac{2}{3} M l^{2} = \frac{4}{3} M l^{2}$

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