The motor of an engine is rotating about its axis with an angular velocity of $100\dfrac{{rev}}{m}$. It comes to rest in $15s$, after being switched off. Assuming constant angular deceleration, what are the numbers of revolution made by it before coming to rest?

Hint: In the given question, we are given with the angular velocity of the motor engine and the time for which it works. We can also see that there will be angular declaration, which is given as constant in the problem. Now, to find the number of revolutions we will use the relation between all these properties.

Formula used: We will use formula for angular displacement $\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}$ and angular acceleration $\omega = {\omega _0} + \alpha t$

Complete step by step answer:
In the above question, we are given that
Initial angular velocity is $100\dfrac{{rev}}{m}$
Now, converting angular velocity to rad/sec,
$rad/s = \dfrac{{rev/m}}{{60\sec /m}} \times 2\pi rad/rev$
Now, substituting the value,
$
   \Rightarrow \dfrac{{100}}{{60}} \times 2\pi rad/\sec \\
   \Rightarrow \dfrac{{10}}{3}\pi rad/\sec \\
 $
Hence, the initial angular velocity in rad/sec is $\dfrac{{10}}{3}\pi rad/\sec $
Total time interval is $15s$ .
Now, we will use formula for angular acceleration,
That is $\omega = {\omega _0} + \alpha t$, where $\omega $ is the final velocity, ${\omega _0}$ is the initial velocity, $\alpha $ is the angular acceleration and $t$ is the time interval.
Now, substituting the values given in the problem,
\[
  \omega = {\omega _0} + \alpha t \\
   \Rightarrow 0 = \dfrac{{10}}{3}\pi + \alpha 15 \\
   \Rightarrow \alpha = - \dfrac{2}{9}\pi \\
 \]
Now, the angular acceleration is $ - \dfrac{2}{9}\pi rad/{s^2}$
Now, using the formula for angular displacement $\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}$
$
   \Rightarrow \theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2} \\
   \Rightarrow \theta = \dfrac{{10}}{3}\pi \left( {15} \right) - \dfrac{1}{2} \cdot \dfrac{2}{9}\pi {\left( {15} \right)^2} \\
   \Rightarrow \theta = \pi \left( {15} \right)\left( {\dfrac{{30 - 15}}{9}} \right) \\
   \Rightarrow \theta = 25\pi rad = 12.5rev \\
 $

Hence, the answer for the above problem is $12.5$ revolutions.

Note: In the given question, we know that when the engine is switched off the final velocity will be zero, as the motor goes to rest. We also know that the angular acceleration will be also negative as the body is decelerating. Now, we will use the certain formulas to find the number of revolutions made by the motor before coming to rest.