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?

Given, the volume of the tank is 30  $m^3$, time of pumping is 15 min, height of the tank from the ground is 40 m and the efficiency of the pump is 30% .

The density of the water is $\rho ={10}^{3}kg/m^3$.

Let m be the mass of water and V be the volume of water, then

$m=\rho V$

Substitute the given values in the above expression.

$m={10}^3\times 30=30\times {10}^{3}kg$

Let W be the work done and h be the height of the tank.

$W=mgh$

The above expression is the work against the potential energy by the pump.

Substitute the given values in the above expression.

$W=30000\times 9.8\times 40=11760kJ$

Let $P_0$ be the output power and t be the time of pumping.

$P_0=mgh\times t$

Substitute the given values in the above expression.

$P_0=11760\times 1000\times 900=13066.67W=13.066kW$

Let $P_i$ be the input power and η be the efficiency of the pump, then

$\eta =\frac{P_0}{P_i}$

Substitute the given values in the above expression.

$\frac{30}{100}=\frac{13.066}{P_i}$

$P_i\cong 43.6kW$

Hence, the power consumed by the pump is 43.6  kW.