MHTCET 19 Apr 2026 Shift 1 Paper

Concept:As the rod falls, its gravitational potential energy is converted into rotational kinetic energy about the hinge.Explanation:For a uniform rod of length $l$, the centre of mass is at the midpoint, i.e. at distance $\frac{l}{2}$ from hinge $A$.When the rod falls from vertical to horizontal, the centre of mass descends by height $\frac{l}{2}$.Loss in gravitational potential energy is$$m g \cdot \frac{l}{2}$$This energy becomes rotational kinetic energy of the rod.Moment of inertia of a uniform rod about one end is$$I = \frac{1}{3} m l^2$$Using energy conservation,$$m g \frac{l}{2} = \frac{1}{2} I \omega^2$$Substitute $I = \frac{1}{3} m l^2$:$$m g \frac{l}{2} = \frac{1}{2} \left(\frac{1}{3} m l^2\right) \omega^2$$$$m g \frac{l}{2} = \frac{1}{6} m l^2 \omega^2$$Cancelling $m$ from both sides:$$g \frac{l}{2} = \frac{1}{6} l^2 \omega^2$$$$3 g l = l^2 \omega^2$$$$\omega^2 = \frac{3 g}{l}$$Therefore,$$\omega = \sqrt{\frac{3 g}{l}}$$Answer:The angular velocity of the rod when end $B$ strikes the ground is $\sqrt{\frac{3 g}{l}}$, which is Option B.