Q 11-06-164JEE MainJEE Main 2021 (25 Jul, Shift 2)Easy
A solid disc of radius $20$ cm and mass $10$ kg is rotating with an angular velocity of $600$ rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in $10$ s is ______ $\pi\times10^{-1}$ N m
Numerical value type. Enter your answer.
Answer: 4
$I = \frac{1}{2}MR^2 = \frac{1}{2}\times10\times0.04 = 0.2$ kg m$^2$.
$\omega_0 = 600$ rpm $= 20\pi$ rad s$^{-1}$, so $\alpha = \dfrac{20\pi}{10} = 2\pi$ rad s$^{-2}$.
$$\tau = I\alpha = 0.4\pi = 4\pi\times10^{-1}\ \text{N m}$$
Solution by Sreeraj P, M.Sc Physics