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A pump on the ground floor of a building can pump up water to fill a tank of volume $30 \mathrm{~m}^3$ in 15 minutes. If the tank is $40 \mathrm{~m}$ above the ground and the efficiency of the pump is $30 \%$, how much electric power is consumed by the pump?
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Volume of water $=30 \mathrm{~m}^3, t=15 \mathrm{~min}=15 \times 60=900 \mathrm{~s}$, $h=40 \mathrm{~m}$,
Efficiency $=\eta=\frac{\text { Output power }}{\text { Input power }}=30 \%$
Density of water $\rho_{\mathrm{H}_2 \mathrm{O}}=10^3 \mathrm{~kg} / \mathrm{m}^3$
$\therefore$ Mass of water pumped $=$ Vol $\times$ density $=30 \times 10^3 \mathrm{~kg}$
Output power $=\frac{\mathrm{W}}{t}=\frac{\mathrm{mgh}}{t}$ $=\frac{30 \times 10^3 \times 9.8 \times 40}{900}=13070 \mathrm{~W}$
Input power $=\frac{\text { Output power }}{\text { Efficiency }}=\frac{13070}{31 / 100}$
$=43567 \mathrm{~W}=43.567 \mathrm{~kW}$
Efficiency $=\eta=\frac{\text { Output power }}{\text { Input power }}=30 \%$
Density of water $\rho_{\mathrm{H}_2 \mathrm{O}}=10^3 \mathrm{~kg} / \mathrm{m}^3$
$\therefore$ Mass of water pumped $=$ Vol $\times$ density $=30 \times 10^3 \mathrm{~kg}$
Output power $=\frac{\mathrm{W}}{t}=\frac{\mathrm{mgh}}{t}$ $=\frac{30 \times 10^3 \times 9.8 \times 40}{900}=13070 \mathrm{~W}$
Input power $=\frac{\text { Output power }}{\text { Efficiency }}=\frac{13070}{31 / 100}$
$=43567 \mathrm{~W}=43.567 \mathrm{~kW}$
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