Question

An automobile engine develops $100KW$ when rotating at a speed of $1800rev/min$. The torque it delviers (in N-m)

• A. 350
• B. 440
• C. 531
• D. 628

Answer 1

The correct option is C 531

Let the torque delivered by the engine be $\tau$.

$1800$ $\frac{rev}{min}$ = $30$ $\frac{rev}{sec}$

= $60\pi$ $\frac{rad}{sec}$

Now , $P=\tau \omega$

Given, $P$ = $100$ $KW$ = ${10}^5$ $W$

Therefore ,

${10}^5$ = $\tau$ * $60\pi$

$\tau$ = $\frac{{10}^5}{60\pi }$ $Nm$

= 531 $Nm$