MHTCET 13 May 2023 Shift 2 Solved Paper

Let the length of the rod $A B$ be $l$ and mass of the rod be $m$.Moment of inertia of the uniform rod about an axis passing through one of its end $V$ given as,$\text{MI}=\;\frac{m l^2}{3}$Rotational kinetic energy, $(\text{KE})_{\;\text{rod }\;} = \;\frac{1}{2} l \omega^2$Initially the rod is held, vertical, so potential energy of rod is given by$(\text{PE})_i = m g \;\frac{I}{2} \;\; \left( \because \;\frac{l}{2} = \text{COM}\right)$Initial kinetic energy, $(\text{KE})_i = 0$ The rod is allowed to fall, when it just touches the ground its potential energy becomes zero, i.e.,$(\text{PE})_f = 0$Let, angular velocity of rod at that instant be $\omega$, then kinetic energy at that instant is$(\text{KE})_f = \;\frac{1}{2} / \omega^2$From conservation of energy,$\;(\text{KE})_i + (\text{PE})_i = (\text{KE})_f + (\text{PE})_f$$\Rightarrow \;\; 0 + m g \;\frac{l}{2} = \;\frac{1}{2} l \omega^2 + 0$$\Rightarrow \;\; m g l = l \omega^2$$\Rightarrow \;\; m g l = \;\frac{m l^2}{3} \omega^2$$\Rightarrow \;\; \omega = \sqrt{\;\frac{3 g}{l}}$