Question

A pump is used to lift 500 kg of water from a depth
of 80 m in 10 s. Calculate :
(a) the work done by the pump,
(b) the power at which the pump works,
(c) the power rating of the pump if its efficiency is
40%. (Take g = 10 m s2)
useful power
[Hint: Efficiency=-
power input​

Answer 1

Step-by-step explanation:

Given,

mass of water = 500 kg

depth = 80 m

g = 10 m/s²

t = 10 sec

Hence work done in lifting water = weight x distance

=> W = mgh

=> W = 500 x 10 x 80 = 400000

Power = work/time

=> P = w/t = 400000/10 = 40000 watt = 40 KW

Since efficiency of the pump = 40% = 0.4

=> power output/power input = 0.4

=> 40/Power input = 0.4

=> Power input = 40/0.4 = 100 KW

Hope it helps you