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An engine pumps up 200 kg of water through a height of 12 m in 4 s. The efficiency of the engine is 60%.

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 so first we have to find power that is

P=W/T

W= mgh

m=200kg

h=12m

t=4s

P  =  mgh/t

    =200x10x12/4

    =6000J

we know

if the pump is 100% effecient this is the rate at which the pump is working to raise the water. in practice some power is lost and pump works at less than 100% efficiency ,here

efficiency=power output/power inputx100%

if the pump is only 60% efficient the

  =6000 x 100/60  = 10000W

                               = 10kw

since some power is lost 10 kw is the minium power rating of the pump

HOPE THIS HELPS YOU .

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