MHTCET 25 Apr 2025 Shift 1 Paper

Concept:When two rotating discs are coupled along the same axis, angular momentum is conserved, but rotational kinetic energy is not conserved.Explanation:Let the final common angular velocity be $\omega$.Before coupling, total angular momentum is:$$L = \mathrm{I}_1 \omega_1 + \mathrm{I}_2 \omega_2$$After coupling, the total moment of inertia becomes:$$\mathrm{I} = \mathrm{I}_1 + \mathrm{I}_2$$Using conservation of angular momentum:$$(\mathrm{I}_1 + \mathrm{I}_2)\omega = \mathrm{I}_1 \omega_1 + \mathrm{I}_2 \omega_2$$So the final angular velocity is:$$\omega = \frac{\mathrm{I}_1 \omega_1 + \mathrm{I}_2 \omega_2}{\mathrm{I}_1 + \mathrm{I}_2}$$The rotational kinetic energy of the system after coupling is:$$K = \frac{1}{2}(\mathrm{I}_1 + \mathrm{I}_2)\omega^2$$Substituting the value of $\omega$:$$K = \frac{1}{2}(\mathrm{I}_1 + \mathrm{I}_2)\left(\frac{\mathrm{I}_1 \omega_1 + \mathrm{I}_2 \omega_2}{\mathrm{I}_1 + \mathrm{I}_2}\right)^2$$$$K = \frac{(\mathrm{I}_1 \omega_1 + \mathrm{I}_2 \omega_2)^2}{2(\mathrm{I}_1 + \mathrm{I}_2)}$$Answer:The final rotational kinetic energy of the system is:$$\frac{(\mathrm{I}_1 \omega_1 + \mathrm{I}_2 \omega_2)^2}{2(\mathrm{I}_1 + \mathrm{I}_2)}$$This corresponds to Option D.