AIIMS 2019 Physics Torque and Angular Momentum MCQ Question

To find the torque required to stop the wheel's rotation, we first calculate its angular velocity in radians per second: $\omega = 60 \, \text{rpm} \times \frac{2\pi \, \text{rad}}{1 \, \text{min}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = 2\pi \, \text{rad/s}$. The angular deceleration needed to stop the wheel in one minute (60 seconds) is $\alpha = \frac{-\omega}{t} = \frac{-2\pi}{60} \, \text{rad/s}^2$. Using the relation $\tau = I \alpha$, where $I = 2 \, \text{kg-m}^2$, we find $\tau = 2 \times \frac{-2\pi}{60} = -\frac{2\pi}{30} = -\frac{\pi}{15} \, \text{Nm}$, indicating that a torque of $\frac{\pi}{15} \, \text{Nm}$ is required to stop the wheel, thus confirming option D.

Other options are incorrect as they either miscalculate the angular velocity, the time for deceleration, or the resulting torque based on the moment of inertia.