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
Concept. Useful power $= \dfrac{dm}{dt}\left(gh + \dfrac{v^2}{2}\right)$; input $=$ useful/efficiency.
Step 1. Useful. $10(250 + 50) = 3000$ W.
Step 2. Input. $\dfrac{3000}{0.75} = 4$ kW.
Answer. $4$ kW (Option D).
Solution by Sreeraj P, M.Sc Physics