Water falls from a 40 m high dam at the rate of 9 $\times$ 104 kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of 100 W lamps, that can be lit, is :
(Take g = 10 ms$-$2)
Solution
<p>Total gravitational PE of water per second $= {{mgh} \over T}$</p>
<p>$= {{9 \times {{10}^4} \times 10 \times 40} \over {3600}} = {10^4}$ J/sec</p>
<p>50% of this energy can be converted into electrical energy so total electrical energy $= {{{{10}^4}} \over 2} = 5000$ W</p>
<p>So total bulbs lit can be $= {{5000\,W} \over {100\,W}} = 50$ bulbs</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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