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A engine pumps up 100 kg of water through a height of 10 m in $5$s . Given that, the efficiency of engine is $60 \%$. If $g=10 \mathrm{~ms}^{-2}$, the power of this engine is
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$3.3$ kW
Efficiency of engine $\eta=60 \%$.
$\begin{aligned} \text { Thus, power } & =\frac{\text { work } / \text { time }}{\eta} \\ & =\frac{100}{60} \times \frac{m g h}{t}
\end{aligned}$
Given, $m=100 \mathrm{~kg}, \quad h=10 \mathrm{~m}, \quad t=5 \mathrm{~s}$ and $g=10 \mathrm{~ms}^{-2}$
$\begin{aligned}\text { Hence, power } & =\frac{100}{60} \times \frac{100 \times 10 \times 10}{5} \\& =3.3 \times 10^3 \mathrm{~W}=3.3 \mathrm{~kW}\end{aligned}$
$\begin{aligned} \text { Thus, power } & =\frac{\text { work } / \text { time }}{\eta} \\ & =\frac{100}{60} \times \frac{m g h}{t}
\end{aligned}$
Given, $m=100 \mathrm{~kg}, \quad h=10 \mathrm{~m}, \quad t=5 \mathrm{~s}$ and $g=10 \mathrm{~ms}^{-2}$
$\begin{aligned}\text { Hence, power } & =\frac{100}{60} \times \frac{100 \times 10 \times 10}{5} \\& =3.3 \times 10^3 \mathrm{~W}=3.3 \mathrm{~kW}\end{aligned}$
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