Question

A fan is making 600 rpm. If it makes 1200 rpm, what is the increase in its angular velocity?
$A.10\pi {\text{ rad/sec}} \\\ B{\text{.20}}\pi {\text{ rad/sec}} \\\ C.60\pi {\text{ rad/sec}} \\\ D.40\pi {\text{ rad/sec}} \\\$

Answer 1

In the question, we need to determine the increase in the angular velocity (change in the angular velocity) of the fan. For this, we will use the relation between the angular velocity and the speed which is given as $\omega = \dfrac{{2\pi n}}{{60}}$.

Complete step by step answer:

The ratio of the product of speed in rpm and the central angle to 60 results in the angular velocity of the revolving body (here fan). Mathematically, $\omega = \dfrac{{2\pi n}}{{60}}$ were, ‘n’ is in revolution per minute (rpm).

Here, the speed is varying from 600 rpm to 1200 rpm. So, ${n_1} = 600{\text{ rpm and }}{n_2} = 1200{\text{ rpm}}$

Substitute ${n_1} = 600{\text{ rpm}}$ in the formula $\omega = \dfrac{{2\pi n}}{{60}}$ to determine the angular velocity of the fan at 600 rpm.

${\omega _1} = \dfrac{{2\pi n}}{{60}} \\\ = \dfrac{{2\pi (600)}}{{60}} \\\ = 20\pi {\text{ rad/sec}} - - - - (i) \\\$

Again, substitute ${n_2} = 1200{\text{ rpm}}$ in the formula $\omega = \dfrac{{2\pi n}}{{60}}$ to determine the angular velocity of the fan at 1200 rpm.

${\omega _2} = \dfrac{{2\pi n}}{{60}} \\\ = \dfrac{{2\pi (1200)}}{{60}} \\\ = 40\pi {\text{ rad/sec}} - - - - (ii) \\\$

Now, according to the question, we need to determine the increase in the angular velocity of the fan, which is given as $\vartriangle \omega = {\omega _2} - {\omega _1}$

So, substitute ${\omega _1} = 20\pi {\text{ rad/secrad/sec and }}{\omega _2} = 40\pi {\text{ rad/sec}}$ in the equation $\vartriangle \omega = {\omega _1} - {\omega _2}$ to determine the increase in the angular velocity of the fan.
$\vartriangle \omega = {\omega _2} - {\omega _1} \\\ = 40\pi - 20\pi \\\ = 20\pi {\text{ rad/sec}} \\\$

Hence, the increase in the angular velocity of the fan such that its speed increases from 600 rpm to 1200 rpm is $20\pi {\text{ rad/sec}}$.

Option B is correct.

Note: It is worth noting down here that while using the formula $\omega = \dfrac{{2\pi n}}{{60}}$, the unit of the speed of the revolution should be in revolution per minute and nothing else. If any other unit has been given then, we convert the quantity accordingly.