An automobile engine develops 100 kW when rotating at a speed of 1800 rev/min. What torque does it deliver

The correct option is 531 N-m.

To find the torque delivered by the engine, we can use the relationship between power, torque, and angular velocity.

Step 1: Identify the given values and convert them to SI units
The power developed by the engine ($P$) is:
$P=100\text{kW}=100,000\text{W}$
The rotational speed of the engine ($N$) is:
$N=1800\text{rev/min}$

Step 2: Calculate the angular velocity ($\omega$)
The angular velocity in radians per second ($\text{rad/s}$) is related to the rotational speed in revolutions per minute by the formula:
$\omega =\frac{2\pi N}{60}$
Substituting the given value of $N$:
$\omega =\frac{2\times \pi \times 1800}{60}=60\pi \text{rad/s}\approx 188.5\text{rad/s}$

Step 3: Calculate the torque delivered ($\tau$)
The relationship between power ($P$), torque ($\tau$), and angular velocity ($\omega$) is given by:
$P=\tau \times \omega$
Solving for torque ($\tau$):
$\tau =\frac{P}{\omega }$
Substitute the values of $P$ and $\omega$:
$\tau =\frac{100,000}{60\pi }\approx 530.5\text{N-m}$
Rounding to the nearest whole number gives:
$\tau \approx 531\text{N-m}$