The angular velocity of a particle moving in a circle of radius $50cm$ is increased in $5$ min from $100$ revolutions per minute to $400$ revolutions per minute. Find the tangential acceleration of the particle.

The correct option is D $\frac{\pi }{60}m/s^2$
Let angular acceleration be $\alpha$
and initial and final angular velocities be ${\omega }_1$ and ${\omega }_2$.
Then, $\alpha =\frac{({\omega }_2-{\omega }_1)}{t}=\frac{(400-100)}{5}$
$=60rev/mi{n}^2$
$=\frac{60\times 2\pi }{(60{)}^2}rad/s^2=\frac{2\pi }{60}rad/s^2$
Therefore, tangential acceleration,
$a_t=\alpha r=\frac{2\pi }{60}\times \frac{50}{100}=\frac{\pi }{60}m/s^2$