Question

An automobile engine develops 100kW when rotating at a speed of $1800rev/min$. The torque it delivers is

& \text{A}\text{. }3.33Nm \\\ & \text{B}\text{. 200}Nm \\\ & \text{C}\text{. 530}\text{.5}Nm \\\ & \text{D}\text{. 2487}Nm \\\ \end{aligned}$$

Answer 1

Torque is produced due to the force applied on the object which results in the rotational motion. Here we have given the power produced by an engine and power can be defined in terms of torque. By using the formula of power we can find the torque which is produced by the automobile engine.

Formula used:
$\tau =\dfrac{P}{\omega }$

Complete answer:
Power is the amount of work done per unit time and in terms of torque it can be given as a product of torque and angular velocity. The mathematical expression for it, is given as
$P=\tau \omega$
Here we have given the power of the engine and the angular velocity and we have to calculate the torque. Above equation can be rewritten as
$\tau =\dfrac{P}{\omega }$
Before substituting the values we have to change the units into SI units. The SI unit of angular velocity is rad/s. We can change it as shown below

& \omega =1800rev/min=\dfrac{1800\times 2\pi rad}{60s} \\\ & \omega =\dfrac{3600\times 3.14}{60}rad/s \\\ & \omega =188.4rad/s \\\ \end{aligned}$$ And the SI unit of power is watt so we can write $$\begin{aligned} & P=100kW \\\ & P=100\times {{10}^{3}}W \\\ & P={{10}^{5}}W \\\ \end{aligned}$$ Now substituting these values in the formula for torque we get, $$\begin{aligned} & \tau =\dfrac{{{10}^{5}}}{188.4}Nm \\\ & \tau =530.5Nm \\\ \end{aligned}$$ **So, the correct answer is “Option C”.** **Note:** The answer may vary considering the decimal points. Also it is essential to change the unit to avoid error. 1 revolution is equal to 2π in radians and 360 in degrees. The formula for power in terms of torque can be verified as the power is given by the product of force and the velocity, condition is the angle of rotation should be perpendicular to the applied force.