Question:
Water falling from a $50 \mathrm{~m}$ high fall is to be used for generating electric energy. If $1.8 \times 10^{5} \mathrm{~kg}$ of water falls per hour and half the gravitational potential energy can be converted into electric energy, how many 100 W lamps can be lit.
Solution:
Potential Energy
P.E. $=m g h=1.8 \times 10^{5} \times 9.8 \times 50$
P.E. $=882 \times 10^{5} \mathrm{~J} / \mathrm{hr}$
Electrical energy $=\frac{1}{2}$ P.E. $=441 \times 10^{5} \mathrm{~J} / \mathrm{hr}$
Power $P=\left(441 \times 10^{5}\right) / 3600$
$P=12.25 \times 10^{3} \mathrm{~W}$
No. of $100 \mathrm{~W}$ lamps that can be lit $=\left(12.25 \times 10^{3}\right) / 100$
Number of lamps $=122.5 \approx 123$