Question

A pump on the ground floor of a building can pump up water to fill the tank of $30\, m^3$ in $15\,min$. If the tank is $40\,m$ above the ground, and the efficiency of the pump is $30\,%$, the power consumed by the pump is $(g = 10 \, m \, s^{-2})$

• A. 4.4 kW
• B. 44 kW
• C. 440 kW
• D. 0.44 kW

Answer 1

Answer: (B)

Here, Capacity of the tank = $30\, m^3$
Time taken by the pump to fill the tank,
$t = 15\, min= 15 \times 60\, s = 900 \, s$
Height through which water is lifted, $h = 40 \, m$
Efficiency of the pump, $ \eta = 30\% = \frac{30}{100} = 0.3$
As density of water $ = 10^3 \, kg \, m^{-3}$
$ \therefore $ Mass of water pumped
$m$ = volwne $\times$ density
$m = 30\, m^3 \times 10^3\, kg\, m^{-3} = 3 \times 10^4\, kg$
Work done by the pump,
$W = mgh = (3 \times 10^4\, kg) (10\, m\, s^{-2}) (40 \,m)$
$ = 12 \times 10^6 \, J$
$ \therefore $ Output power of the pump
$= \frac{W}{t} = \frac{12 \times10^{6} J}{900 s } =\frac{4}{3} \times10^{4} W = 1.33 \times10^{4} W $
As , Input power = $ \frac{Output}{\nu}$
$= \frac{1.33 \times10^{4}W}{0.3} = 4.4 \times10^{4} W = 44 kW$