Four identical thin rods each of mass M and length l, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is :

\(\frac{2}{3} ML^2\) \(\frac{13}{3} ML^2\) \(\frac{1}{3} ML^2\) \(\frac{4}{3} ML^2\)

\(\frac{4}{3} ML^2\)

\(\frac{mL^2}{12} + \frac{mL^2}{4} = \frac{4mL^2}{12} = \frac{mL^2}{3}\) Total MI = 4 x \(\frac{mL^2}{3}\)