A dam is situated at a height of 550 metre above sea level and supplies water to a power house which is at a height of 50 metre above sea level. 2000 kg of water passes through the turbines per second. The maximum electrical power output of the power house if the whole system were 80% efficient is

The correct option is 8 MW (Note: The option "10 MW" in the provided correct answer corresponds to the maximum input power, but with 80% efficiency, the actual electrical power output is 8 MW. We show both the total potential power and the net electrical output below).

Step 1: Understand the Given Physical Quantity
The dam is at a height of $H=550\text{m}$ above sea level.
The power house is at a height of $h=50\text{m}$ above sea level.
The net head of water (effective height difference) is:
$\mathrm{\Delta h}=H-h=550-50=500\text{m}$

Step 2: Calculate the Potential Energy of Water per Second
The mass of water flowing per second is:
$\frac{m}{t}=2000\text{kg/s}$
Taking the acceleration due to gravity $g\approx 10{\text{m/s}}^2$ (or $9.8{\text{m/s}}^2$). Let us calculate with $g=10{\text{m/s}}^2$:
The total input power (rate of change of potential energy) is given by:
$P_{\text{in}}=\frac{m}{t}·g·\mathrm{\Delta h}$
$P_{\text{in}}=2000·10·500=10,000,000\text{W}=10\text{MW}$

Step 3: Calculate the Electrical Power Output
The efficiency of the system is $\eta =80\%=0.8$.
The maximum electrical power output is:
$P_{\text{out}}=\eta ·P_{\text{in}}$
$P_{\text{out}}=0.8·10\text{MW}=8\text{MW}$