Q. 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 (in erg)

Solution:

We have P $\times, 40% =\frac{W}{t}\,\Rightarrow \frac{746}{4} \times \frac{40}{100} =\frac{W}{\left(2 \pi \times \frac{600 \times 2\pi}{60}\right)$$\Rightarrow W=7.46 \,J$

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