Question

Water failing from a 50 m high fall is to be used for generating electric energy. If 1.8 $\times$${10}^5$ 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?Take g=10 $m/s^2$

• A. 123
• B. 225
• C. 200
• D. 100

Answer 1

The correct option is A

123

h = 50m, m = 1.8 x ${10}^5$ kg/hr, P = 100 watt,

P.E. = mgh = 1.8 x 10^5 x 9.8 x 50 = 882 x ${10}^5$J/hr

Because, half the potential energy is converted into electricity,

Electrical energy obtained=$\frac{1}{2}$ P. E. = 441 x ${10}^5$ J/hr

So, power in watt (J/sec) is given by = $\frac{441\times {10}^5}{3600}$ watts

$\therefore$Number of 100 w lamps that can be lit = $\frac{441\times {10}^5}{3600\times 100}$=122.5 $\approx$ 123