Q 11-06-167JEE MainJEE Main 2021 (27 Aug, Shift 1)Easy
Moment of inertia of a square plate of side $l$ about the axis passing through one of the corner and perpendicular to the plane of square plate is given by:
Answer: (B) $\frac{2}{3}Ml^2$
About the perpendicular axis through the centre: $I_c = \dfrac{M(l^2 + l^2)}{12} = \dfrac{Ml^2}{6}$.
The corner is $\dfrac{l}{\sqrt{2}}$ from the centre:
$$I = \frac{Ml^2}{6} + M\frac{l^2}{2} = \frac{2}{3}Ml^2$$
Solution by Sreeraj P, M.Sc Physics