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 is consumed by the pump? (Take $g=10 m s^{-2}$)

Here, Volume of water $= 30 m^3$

$t= 15 min = 15 x 60 s = 900 s ;$ Height ,$h=40 m$

Efficiency,$\eta =30\%$

Density of water $= {10}^{3}kg m^{-3}$

$\therefore$Mass of water pumped = Volume x Density

 $$\begin{gathered} =\left(30m^3\right)({10}^{3}kg m^{-3})=3\times {10}^4 kg \\ {P}_{output}=\frac{W}{t}=\frac{mgh}{t}=\frac{\left(3\times {10}^{4}kg\right)\left(10 m s^{-2}\right)\left(40 m\right)}{900 s} \\ =\frac{4}{3}\times {10}^{4 }W \end{gathered}$$

Efficiency,$$\begin{gathered} \eta =\frac{P_{output}}{P_{input}} \\ {P}_{input}=\frac{P_{output}}{\eta }=\frac{4\times {10}^4}{3\times {\displaystyle \frac{30}{100}}}=\frac{4}{9}\times {10}^5=44.4\times {10}^3 W=44.4 kW \end{gathered}$$