MHTCET 13 Apr 2026 Shift 1 Paper

Concept:For a sliding body, gravitational potential energy converts only into translational kinetic energy. For a rolling ring, it converts into both translational and rotational kinetic energy.Explanation:Let the vertical height of the incline be $h$.For the sliding body, by conservation of mechanical energy:$$mgh = \frac{1}{2}mV^2$$This gives:$$V^2 = 2gh$$Now, for the ring rolling down without slipping, its moment of inertia is $I = mR^2$, and the rolling condition is $v = R\omega$.Using conservation of energy for the ring:$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$Substitute $I = mR^2$ and $\omega = \frac{v}{R}$:$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}mR^2\left(\frac{v}{R}\right)^2$$$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$$Therefore:$$v^2 = gh$$Comparing with $V^2 = 2gh$:$$v^2 = \frac{V^2}{2}$$Hence:$$v = \frac{V}{\sqrt{2}}$$Answer:The linear velocity of the ring at the bottom is $\frac{V}{\sqrt{2}}$.Correct option: A.