A pump on the ground floor of a building can pump up water to fill a tank of volume `30 m^3` in 15 min. If the tank is 60 m above the ground, and the efficiency of the pump is 40%, how much electric power is consumed by the pump?
(Take `g = 10 m s^(-2)`)

Explanation

Mass of water lifted,
m = volume `xx` density `= 30 m^3 xx 1000 kg//m^3 = 3 xx 104 kg`
Height through which water is lifted, `h = 40 m`
Work done by the pump, i.e., output `= mgh = (3 xx 104) xx 9.8 xx 40 J = 1.176 xx 10^7 J`
We know that efficiency, `eta = ("output")/("input")` Here, `eta = 30% = 30/100 = 0.3`
Input (i.e., work done on the pump) = `("output")/(eta) = (1.176 xx 10^7 J)/(0.3) = 3.92 xx 10^7 J`
Power consumed by the pump = `("work done on the pump")/("time") =(3.92 xx 10^7 J)/(15 xx 60 s) = 43.6 xx 103 W = 43.6 kW`.