The angular velocity of a particle moving on a circular path of radius $50\text{cm}$ is increased in $5\text{min}$ from $100\text{rpm}$ to $400\text{rpm}$. Find the magnitude of tangential acceleration of the particle.

The correct option is D $\frac{\pi }{60}{\text{ms}}^{-2}$
Angular velocity is increased from $100\text{rpm}$ to $400\text{rpm}$ in $\Delta t=5\text{min}$
Radius of circular path is $r=50\text{cm}=0.50\text{m}$

Then, angular acceleration $\alpha =\frac{\Delta \omega }{\Delta t}$
$\Rightarrow \alpha =\frac{{\omega }_2-{\omega }_1}{\Delta t}=\frac{400-100}{5}$
$\alpha =60\text{rpm/min}$
$\therefore \alpha =\frac{60\times 2\pi \text{rad}}{(60{)}^2{\text{s}}^2}=\frac{2\pi }{60}{\text{rad/s}}^2$
Hence, tangential acceleration is given by:
$a_t=\alpha \times r$
$\therefore a_t=\frac{2\pi }{60}\times 0.50=\frac{\pi }{60}{\text{rad/s}}^2$