Ncert Exercises & Solutions

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 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Volume of the tank, V = 30 $m^{3}$
Time of operation, t = 15 min = 15 × 60 = 900 s
Height of the tank, h = 40 m
Efficiency of the pump, $η$ = 30%
Density of water, ρ = $10^{3}$ kg/m3
Mass of water, m = ρV = 30 × $10^{3}$ kg
Output power can be obtained as:

$P_{o} = \frac{W o r k d o n e}{T i m e} = \frac{m g h}{t}$
$= \frac{30 \times 10^{3} \times 9.8 \times 40}{900} = 13.067 \times 10^{3} W$

For input power $P_{i}$ , efficiency $η$ is given by the relation:

$η = \frac{P_{0}}{P_{i}} = 30 %$
$P_{i} = \frac{13.067}{30} \times 100 \times 10^{3}$
$= 0.436 \times 10^{5} W$
$= 43.6 kW$

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