91Ó°ÊÓ

The length of the dam is L = 1270 m.

The height of the dam is h = 170.

The power output is P = 2000 MW.

The relationship between the power and work is given by,

$\mathit{P}\mathbf{=}\frac{\mathbf{W}\mathbf{ }}{\mathbf{t}\mathbf{ }}\mathbf{ }\mathbf{}\mathbf{}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, P, is power, W is work done, and t is time.

The work done in the form of potential energy is given by,

W = mgh ........(2)

Convert the power output in watt.

$$\begin{gathered} \mathrm{P}=\left(2000\mathrm{MW}\right)\left(\frac{{10}^6\mathrm{W}}{1\mathrm{MW}}\right) \\ =2×{10}^9\mathrm{W} \end{gathered}$$

It is given that 92% of the work done on water by gravity is converted to electrical energy.

The work done on the water is calculated as follows.

$$\begin{gathered} W=Pt \\ 0.92W=\left(2×{10}^{9}W\right)\left(1s\right) \\ W=\frac{\left(2×{10}^{9}W\right)}{0.92} \\ =2.174×{10}^{9}J \end{gathered}$$

The mass of water flowing from the top of the dam per second is calculated as follows.

$$\begin{gathered} \mathrm{W}=\mathrm{mgh} \\ 2.174×{10}^9\mathrm{J}=\mathrm{m}\left(9,8\mathrm{m}/{\mathrm{s}}^2\right)\left(170\mathrm{cm}\right) \\ \mathrm{m}=\frac{2.174×{10}^9\mathrm{J}}{\left(9,8\mathrm{m}/{\mathrm{s}}^2\right)\left(170\mathrm{cm}\right)} \\ =1.3×{10}^6\mathrm{kg} \end{gathered}$$

Find the volume of the water as follows.

$$\begin{gathered} V=\frac{m}{P_{water}} \\ =\frac{1.3×{10}^{6}kg}{1000kg/m^3} \\ =1.3×{10}^3{m}^3 \end{gathered}$$

Therefore, the required volume of water is $1.3×{10}^3{m}^3$.