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(D) Volume of the tank, $V = 30 \; m^{3}$.Time of operation, $t = 15 \; min = 15 	imes 60 = 900 \; s$.Height of the tank, $h = 40 \; m$.Efficiency of

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$43.6$

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(D) Volume of the tank, $V = 30 \; m^{3}$.Time of operation, $t = 15 \; min = 15 	imes 60 = 900 \; s$.Height of the tank, $h = 40 \; m$.Efficiency of the pump, $\eta = 30 \% = 0.3$.Density of water, $ho = 10^{3} \; kg/m^{3}$.Mass of water, $m = ho 	imes V = 10^{3} 	imes 30 = 30,000 \; kg$.Output power $(P_{out})$ is the rate of doing work against gravity:$P_{out} = \frac{mgh}{t} = \frac{30,000 	imes 9.8 	imes 40}{900} = \frac{11,760,000}{900} = 13,066.67 \; W \approx 13.067 \; kW$.Input power $(P_{in})$ is related to output power by efficiency: $\eta = \frac{P_{out}}{P_{in}}$.$P_{in} = \frac{P_{out}}{\eta} = \frac{13.067 \; kW}{0.3} = 43.556 \; kW$.Rounding to one decimal place, the power consumed is $43.6 \; kW$.

$A$ pump on the ground floor of a building can pump up water to fill a tank of volume $30 \; m^{3}$ in $15 \; min$. If the tank is $40 \; m$ above the ground, and the efficiency of the pump is $30 \%$, how much electric power (in $kW$) is consumed by the pump?

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