Question

A quarter horse power motor runs at a speed of $ 600\, rpm $ . Assuming $ 40\% $ efficiency, the work done by the motor in the one rotation will be

• A. $ 7.46\,J $
• B. $ 74.6\,J $
• C. $ 7400\,J $
• D. $ 7.46\,erg $

Answer 1

The Correct Option is A

Given, power is $\frac{1}{4} nP = \frac{764}{4}W = 186.5\,W$
Since, efficiency of motor is $40\%$.
The power used in doing work is $40\%$ of $186.5W$, so
$P = 186.5 \times \frac{40}{100} = 74.6\,W$
Angular velocity of the motor, $\omega = 600\, rpm = 10 \,rps$
$\Rightarrow \omega = (600)(2\pi)60\,rad/s$
$\Rightarrow \omega = 20\,\pi \,rad /s$
Let the torque be $T$.
So, $P = T\omega$
$\Rightarrow T = \frac{P}{\omega} = \frac{74.6}{20\,\pi}$
Now, work done in $1$ rotation $= T(2n)$
$W = \frac{74.6}{20\,\pi} \times 2\pi$
$W = 7.46\,J$