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A pump on the ground floor of a building can pump up water to fill a tank of volume $36 \mathrm{~m}^3$ in $30 \mathrm{~min}$. If the tank is $50 \mathrm{~m}$ above the ground, and the electric power consumed by the pump is $40 \mathrm{~kW}$, the efficiency of the pump is (Use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ and density of water $=1000 \mathrm{~kg} / \mathrm{m}^3$ )
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1984 Upvotes
Verified Answer
The correct answer is:
$25 \%$
We have
$$
\begin{aligned}
& \mathrm{P}_{\text {input }}=40,000 \text { watt } \\
& \mathrm{P}_{\text {output }}=\frac{\mathrm{mgh}}{\mathrm{t}}=\frac{(\mathrm{V} \rho) \mathrm{gh}}{\mathrm{t}} \\
& =\frac{36 \times 1000 \times 10 \times 50}{30 \times 60}=10,000 \mathrm{watt}
\end{aligned}
$$
$$
\text { So, } \eta=\frac{P_{\text {output }}}{P_{\text {input }}}=\frac{10,000}{40,000}=0.25
$$
$$
\begin{aligned}
& \mathrm{P}_{\text {input }}=40,000 \text { watt } \\
& \mathrm{P}_{\text {output }}=\frac{\mathrm{mgh}}{\mathrm{t}}=\frac{(\mathrm{V} \rho) \mathrm{gh}}{\mathrm{t}} \\
& =\frac{36 \times 1000 \times 10 \times 50}{30 \times 60}=10,000 \mathrm{watt}
\end{aligned}
$$
$$
\text { So, } \eta=\frac{P_{\text {output }}}{P_{\text {input }}}=\frac{10,000}{40,000}=0.25
$$
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