Q 11-05-037NEETJEE MainMedium
A pump lifts $600$ kg of water per minute through a height of $25$ m and delivers it with a speed of $10$ m/s. If the pump is $75\%$ efficient, its input power is ($g = 10\ \text{m/s}^2$)
Answer: (D) $4$ kW
Mass per second $= 10$ kg/s. Useful power:
$$P_{out} = \frac{dm}{dt}\left(gh + \frac{v^2}{2}\right) = 10(250 + 50) = 3000\ \text{W}$$
$$P_{in} = \frac{3000}{0.75} = 4000\ \text{W} = 4\ \text{kW}$$
Solution by Sreeraj P, M.Sc Physics