Q126 / 180 · COMEDK 2014 · Physics
A pump on the ground floor of a building can pump up water to fill the tank of $30 \mathrm{~m}^{3}$ in $15 \mathrm{~min}$. If the tank is $40 \mathrm{~m}$ above the ground, and the efficiency of the pump is $30 \%$, the power consumed by the pump is $\left(g=10 \mathrm{~ms}^{-2}\right)$
- A$4.4 \mathrm{~kW}$
- Correct answerB$44 \mathrm{~kW}$
- C$440 \mathrm{~kW}$
- D$0.44 \mathrm{~kW}$
Solution
Volume of water in tank, $V=30 \mathrm{~m}^{3}$
Time taken to fill the tank,
$t=15 \mathrm{~min}=15 \times 60=900 \mathrm{~s}$
Height of the tank, $h=40 \mathrm{~m}$
$\therefore$ Mass of pumped water, $m=$ Volume $\times$ Density of water
$$
=30 \times 10^{3}=3 \times 10^{4} \mathrm{~kg}
$$
Work done by the pump to fill the tank,
$$
\begin{aligned}
W &=m g h \\
&=3 \times 10^{4} \times 10 \times 40 \\
&=1.2 \times 10^{7} \mathrm{~J}
\end{aligned}
$$
$\therefore$ Output power of the pump,
$$
P_{o}=\frac{W}{t}=\frac{1.2 \times 10^{7}}{900}=\frac{4}{3} \times 10^{4} \mathrm{~W}
$$
$$
\begin{aligned}
&\text { As we know, efficiency }=\frac{\text { Output power }\left(P_{o}\right)}{\operatorname{Input} \text { power }\left(P_{i}\right)} \\
&\Rightarrow \quad 0.3=\frac{4 / 3 \times 10^{4}}{P_{i}} \\
&\Rightarrow \quad P_{i}=\frac{4 \times 10^{4}}{3 \times 0.3}=4.4 \times 10^{4} \mathrm{~W}=44 \mathrm{~kW}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.