MHT-CET Physics and Chemistry Full Test-17 PHY-QUE-12 Solutions

Question-12 A disc of mass 25\ kg and radius 0.2\ m is rotating at 240\ r.p.m. A retarding torque brings it to rest in 20\ second. If the torque is due to a force applied tangentially on the rim of the disc then the magnitude of the force is

(A) \frac{\pi}{2}\ N

(B) 2\pi\ N

(C) \pi\ N

(D) 4\pi\ N

Answer : (C)

Explanation:

Frequency f=240\ rpm=\frac{240}{60}\ rps=4\ rps

∴ \omega=2\pi f=2\pi\times4=8\pi\ rad/s

Angular acceleration

\alpha=\frac{\omega_2-\omega_1}{t}=\frac{0-8\pi}{20}=-\frac{2\pi}{5}\ rad/s^2

Moment of inertia

I=\frac{MR^2}{2}=\frac{25\times(0.2)^2}{2}=0.5\ kg\ m^2

Torque \tau=I\alpha=0.5\times\left(-\frac{2\pi}{5}\right)=-0.2\pi\ Nm

\tau=Fr

∴ |F|=\frac{|\tau|}{r}=\frac{0.2\pi}{0.2}=\pi\ N