Awater pump pumps 3 kg of water every second from the bottom of the well to the top of the well 50 m above if the pump uses 2000w of electrical power to accomplish the task what is the efficiency of the pump

m = mass of water pumped = 3 kg

t = time taken to pump mass "m" of water = 1 sec

h = depth from which the water is pumped = 50 m

P = electrical power used by the pump = Input power = 2000 Watt

P' = output power

U = potential energy gained by water

η = efficiency of the pump

potential energy gained by water is given as

U = mgh

output power is given as

Output power = Potential energy gained by water / time

P' = mgh/t

inserting the values

P' = (3) (9.8) (50)/1

P' = 1470 Watt

efficiency of the pump is given as

efficiency = output power/Input power

η = P'/P

η = 1470/2000

η = 0.735

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explanation: