Question

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

• A. 7.46 j
• B. 7400 j
• C. 74.6 j
• D. 746 j

Answer 1

Answer: (A)

7.46 j

Given,

$P=\frac{1}{4}hp=\frac{746}{4}=186.5W$

efficiency$=40$%

$\omega =\frac{2\pi \times 600}{60}=20\pi rad/sec$

The power used for doing work with efficiency 40% is,

$P^{\prime }=186.5\times \frac{40}{100}=74.6W$

Torque, $\tau =\frac{P^{\prime }}{\omega }=\frac{74.6}{20\pi }$

The work done in one rotation,

$W=\tau \times 2\pi =\frac{74.6\times 2\pi }{20\pi }=7.46J$

The correct option is A.