Q. An automobile engine develops $100\ kW$ when rotating at a speed of $1800\ rev/min$. What torque does it deliver
A) $350\ N$-$m$
B) $440\ N$-$m$
C) $531\ N$-$m$
D) $628\ N$-$m$
Explanation
(c) $\omega = 2\pi n = \frac{2\pi \times 1800}{60} = 60\pi\ rad/s$
$P = \tau \times \omega \Rightarrow \tau = \frac{P}{\omega} = \frac{100 \times 10^{3}}{60\pi} = 531\ N - m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.