MHTCET 11 Apr 2026 Shift 2 Paper

Concept:When the second ring is placed gently, no external torque acts, so angular momentum is conserved, but kinetic energy decreases.Explanation:The second ring is identical and initially at rest, so its angular momentum is zero.Initial angular momentum of the system is $L_i = I\omega$.When both rings rotate together, the total moment of inertia becomes $I + I = 2I$.Let the common final angular velocity be $\omega'$.Using conservation of angular momentum:$$I\omega = 2I\omega'$$Thus,$$\omega' = \frac{\omega}{2}$$Initial kinetic energy is only due to the first ring:$$K_i = \frac{1}{2}I\omega^2$$Final kinetic energy of both rings is:$$K_f = \frac{1}{2}(2I)\left(\frac{\omega}{2}\right)^2 = \frac{I\omega^2}{4}$$Loss in kinetic energy:$$K_i - K_f = \frac{1}{2}I\omega^2 - \frac{1}{4}I\omega^2 = \frac{I\omega^2}{4}$$Answer:The loss in kinetic energy is $\frac{I\omega^2}{4}$, which matches option C.