Solve with detailed explanation

Solution:
Let the input power be P.
So 80 % of this power is used to lift the water,
so, used power, P' =0.80 P
Density of water is 1 Kg/L
So mass of 800 L water, m  = 800 Kg
Power to lift the water,
$$\begin{gathered} P\text{'}=\frac{mgh}{t}=\frac{800\times 10\times 20}{20}=8000 \mathrm{W} \\ \mathrm{But}, P\text{'}=0.80 P \\ \Rightarrow P=\frac{P\text{'}}{0.80}=\frac{8000}{0.80}=10000 W \\ =10\times {10}^3 \mathrm{W}=10 \mathrm{kW} \left(1\mathrm{kW}={10}^3\mathrm{W}\right) \end{gathered}$$
So, input power = 10 kW
Hence, the correct answer is option (2).