Question

Sol. Key concept: Useful gravitational power divided by input electrical power.

The pump lifts 240 kg each minute, so the mass flow rate is 240/60 = 4 kg s⁻¹. The useful power equals the rate of increase of gravitational potential energy.

$$P_{out} = \frac{dm}{dt}gh = 4 \times 9.8 \times 10 = 392\ W$$

$$\eta = \frac{392}{500} \times 100\% = 78.4\%$$

The remaining input power is lost rather than converted into useful lifting work. Hence (b).