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 `40m` above the ground and the efficiency of the pump is `30 %` , how much electric power is consumed by the pump? `("Take" g = 10 ms^(2))`

Options

A. 36.5 kW
B. 50 kW
C. 52.5 kW
D. 60.5 kW

Answer

B

Explanation

Here, Volume of water = `30 m^3`
t=15 min = 15 x 60 s = 900 s, Height, h = 60 m
Efficiency, `eta`=40 %
Density of water = `10^3 kg m^(-3)`
`therefore` Mass of water pumped = Volume x Density
`= (30 m^3) (10^3 kg m^(-3)) = 3 xx 10^4 kg`
`P_("output")=W/t=(mgh)/t=((3xx10^4 kg)(10 m s^(-2))(60 m))/(900 s)`
`=2xx10^4 W`
Efficiency, `eta=(P_("output"))/(P_("input"))`
`P_("input")=(P_("output"))/eta=(2xx10^4)/(40/100)=5xx10^4`
`=50xx10^3 W =50 kW`