Question

An electric motor drill, rated 350 W has an efficiency of 35%. The torque produced, if it is working at 3000 rpm.

• A. $0.25Nm$
• B. $0.35Nm$
• C. $0.39Nm$
• D. $0.3Nm$

Answer 1

Answer: (C)

Angular velocity of the motor drill,
$\omega =\frac{2\pi \upsilon }{60}=\frac{2\pi \times 3000}{60}$
Let $\tau$ be the torque produced by the motor.
Power produced $=\tau \omega =\tau \times \frac{2\pi \times 3000}{60}$
Now, $\tau \times \frac{2\pi \times 3000}{60}=35\%$ of $350W$
or $\tau \times 2\pi \times 50=\frac{35}{100}\times 350$
or $\tau =\frac{35\times 350}{100\times 2\pi \times 50}=0.39Nm$