point of \(AB\) perpendicular to the
plane of the paper is \(\frac{{M{l^2}}}{{12}}.\)
Moment of inertia about the axis through the center of the square and parallel to this axis,
\(I = {I_0} + M{d^2} = M\left( {\frac{{{l^2}}}{{12}} + \frac{{{l^2}}}{4}} \right) = \frac{{m{l^2}}}{3}\).
For all the four rods, \(I = \frac{4}{3}\,m{l^2}.\)
$(g=$ ગુરુત્વીય પ્રવેગ $\theta=$ આકૃતિમાં દર્શાવ્યા મુજબ કોણ)