A motor is used to lift water from a well of depth 10m and to fill a water tank of volume 30m³ in 10 minutes. The tank is at a height of
20m above ground. If 20% of energy is wasted, the power of the motor is (in kW)

Given,

Volume = $30m^3$, the water is raised to by a height of 30m.

We know that,

$$\begin{gathered} \mathrm{Power}=\frac{\mathrm{energy}}{\mathrm{time}} \\ \mathrm{dP}=\frac{\mathrm{dE}}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}}\times \mathrm{g}\times \mathrm{h} \end{gathered}$$

$\frac{\mathrm{dm}}{\mathrm{dt}}=\frac{3000}{10\times 60}\mathrm{kg}/\mathrm{s}=50\mathrm{kg}/\mathrm{s}$

$\mathrm{P}=50\times 10\times 30=15\mathrm{kW}$

$$\begin{gathered} \frac{15}{\mathrm{total} \mathrm{Power}}\times 100=80 \\ \Rightarrow \mathrm{total} \mathrm{power}=\frac{150}{8}\mathrm{kW}=18.75\mathrm{kW} \end{gathered}$$

Hence the correct answer is 18.75kw.