Question

A pump on the ground floor of a building can pump up water to fill a tank of volume $30m^3$ 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 $g=10m{s}^{-2})$.

• A. $36.5kW$
• B. $44.4kW$
• C. $52.5kW$
• D. $60.5kW$

Answer 1

Answer: (B)

$44.4kW$

Here, Volume of water $=30m^3$
$t=15min=15\times 60s=900s;Height,h=40m$
Efficiency, $\eta =30$%
Density of water $={10}^{3}kg{m}^{-3}$
$\therefore$ Mass of water pumped $=Volume\times Density$
$=\left(30m^3\right)\left({10}^{3}kg{m}^{-3}\right)=3\times {10}^{4}kg$
$P_{output}=\frac{W}{t}=\frac{mgh}{t}=\frac{(3\times {10}^{4}kg)\left(10m{s}^{-2}\right)\left(40m\right)}{900s}$
$=\frac{4}{3}\times {10}^{4}W$
Efficiency, $\eta =\frac{P_{output}}{P_{input}}$
$P_{input}=\frac{P_{output}}{\eta }=\frac{4\times {10}^4}{3\times \frac{30}{100}}=\frac{4}{9}\times {10}^5$
$=44.4\times {10}^{3}W=44.4kW$.