Question

A pump is used to lift 500 kg of water from a depth of 80 m in 10s (take g=10m/s^2).Calculate:
(i) work done by the pump (ii) the power at which pump works
(iii) the power rating of the pump if it’s efficiency is 80%.

Answer 1

Explanation:

work done =mgh =400000J

power= mgh/t= 400000/10=40000

Answer 2

a) W=FS

W=mg×S

500×10×80

4,00,000J

b) P=W/t

=400000/10=40000W=40kW

c) efficiency=useful power/power input

40%=40/Power input

Power input = 40 × 100/40

Power input = 40kW