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An installation consisting of an electric motor driving a water pump left $75 \mathrm{~L}$ of water per second to a height of $4.7 \mathrm{~m}$.If the motor consumes a power of $5 \mathrm{~kW}$, then the efficiency of the installation is
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The correct answer is:
$69 \%$
$\begin{array}{l}
\begin{array}{l}
\text { Power consumed by motor }=5 \mathrm{~kW} \\
=5 \times 10^{3} \mathrm{~W}=5000 \mathrm{~W}
\end{array} \\
\text { Power used in lifting water }=\frac{\mathrm{mgh}}{\mathrm{t}} \\
=7.5 \times 9.8 \times 4.7=3454.5 \mathrm{~kW} \\
\text { Efficiency }=\frac{\text { Power used }}{\text { Power consumed }} \times 100 \% \\
=\frac{3454.5}{5000} \times 100=69 \%
\end{array}$
\begin{array}{l}
\text { Power consumed by motor }=5 \mathrm{~kW} \\
=5 \times 10^{3} \mathrm{~W}=5000 \mathrm{~W}
\end{array} \\
\text { Power used in lifting water }=\frac{\mathrm{mgh}}{\mathrm{t}} \\
=7.5 \times 9.8 \times 4.7=3454.5 \mathrm{~kW} \\
\text { Efficiency }=\frac{\text { Power used }}{\text { Power consumed }} \times 100 \% \\
=\frac{3454.5}{5000} \times 100=69 \%
\end{array}$
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