MHTCET 15 May 2024 Shift 1 Paper

Concept:By the law of conservation of energy, the maximum kinetic energy of rotation is converted entirely into gravitational potential energy at the highest point.Explanation:The rod swings about a horizontal axis through its end, so its moment of inertia is taken about that end.For a uniform rod about its centre, $I_{\text{cm}} = \frac{ML^2}{12}$.Using the parallel axis theorem, $I = I_{\text{cm}} + M\left(\frac{L}{2}\right)^2$.Thus, $I = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2}{3}$.At maximum angular velocity $\omega$, the kinetic energy is $\frac{1}{2}I\omega^2$.At maximum height $h$, the potential energy gained is $Mgh$.Equating them: $\frac{1}{2}I\omega^2 = Mgh$.Substituting $I = \frac{ML^2}{3}$, we get $\frac{1}{2}\cdot \frac{ML^2}{3}\omega^2 = Mgh$.Simplifying: $h = \frac{L^2 \omega^2}{6g}$.Answer:$h = \frac{L^2 \omega^2}{6g}$, so the correct option is C.