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(C) The moment of inertia of a point mass is given by $I = m r^2$,where $r$ is the perpendicular distance from the axis of rotation.Let the axis pass

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$3 m l^2$

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(C) The moment of inertia of a point mass is given by $I = m r^2$,where $r$ is the perpendicular distance from the axis of rotation.Let the axis pass through $A$ and be parallel to the diagonal $DB$.The perpendicular distances of the four masses from this axis are:$1$. For mass at $A$: $r_A = 0$ (since the axis passes through $A$).$2$. For mass at $B$: $r_B = l \sin(45^\circ) = l / \sqrt{2}$.$3$. For mass at $D$: $r_D = l \sin(45^\circ) = l / \sqrt{2}$.$4$. For mass at $C$: $r_C = l \sin(45^\circ) + l \sin(45^\circ) = l / \sqrt{2} + l / \sqrt{2} = 2l / \sqrt{2} = l \sqrt{2}$.The total moment of inertia is $I = m(r_A^2 + r_B^2 + r_D^2 + r_C^2)$.$I = m [0^2 + (l / \sqrt{2})^2 + (l / \sqrt{2})^2 + (l \sqrt{2})^2]$$I = m [0 + l^2/2 + l^2/2 + 2l^2]$$I = m [l^2 + 2l^2] = 3 m l^2$.

Four equal masses,$m$ each,are placed at the corners of a square of side length $l$ as shown in the figure. The moment of inertia of the system about an axis passing through $A$ and parallel to $DB$ would be:

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