91Ó°ÊÓ

The volume of the water is$\mathrm{V}=1200{\mathrm{m}}^3$

The height of the waterfall is h = 100 m

Mass of of water is 1000 kg

Calculating density from mass and volume, find the Potential Energy of the water at that height usingthecorresponding formula. Using the law of conservation of energy, find the kinetic energy of the water. From this, we can get the energy transferred to the generator. Using this, find the rate of generation of electricity by the generator.

Formula:

$$\begin{gathered} PE=mgh \\ p=\frac{m}{V} \\ P=\frac{Energy}{Time} \end{gathered}$$

The density of the matter with mass m and volume V can be expressed as,

$p=\frac{m}{V}$

Therefore, substituting the given values in the above expression, we get,

$$\begin{gathered} \mathrm{p}=\frac{1000\mathrm{kg}}{1{\mathrm{m}}^3} \\ ={10}^3\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

The potential energy of the water at h = 100 m can be expressed as,

$$\begin{gathered} PE=\mathrm{mgh} \\ =\mathrm{pVgh} \\ ={10}^3\mathrm{kg}/{\mathrm{m}}^3×1200{\mathrm{m}}^3×9.8\mathrm{m}/{\mathrm{s}}^2×100\mathrm{m} \\ =1.17×{10}^9\mathrm{J}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =1.17×{10}^9\mathrm{J} \end{gathered}$$

From the principle of conservation of energy, we can write that the,

PE = KE

And therefore from the above expression, we have,

$\mathrm{KE}=1.17×{10}^9\mathrm{J}$

Only three fourth amount is transferred to electrical energy, therefore, the power generation rate will be,

$$\begin{gathered} \mathrm{P}=\frac{{\mathrm{KE}}_{\mathrm{transferred}}}{\mathrm{time}} \\ =\frac{\left({\displaystyle \frac{3}{4}}\right)\left(\mathrm{KE}\right)}{\mathrm{t}} \\ =\frac{\left({\displaystyle \frac{3}{4}}\right)(1.17×{10}^9\mathrm{J})}{1\mathrm{s}} \\ =8.8×{10}^8\mathrm{J}/\mathrm{s}\left(\frac{1\mathrm{W}}{1\mathrm{J}/\mathrm{s}}\right) \\ =8.8×{10}^8\mathrm{W} \end{gathered}$$