Sum
Water falling from a 50-m high fall is to be used for generating electric energy. If \[1 \cdot 8 \times {10}^5 \text{ kg } \] of water falls per hour and half the gravitational potential energy can be converted into electrical energy, how many 100 W lamps can be lit with the generated energy?
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Solution
Given :
\[\text{ Height, h = 50 m } \]
\[\text{ Mass of water falling per hour, m } = 1 . 8 \times {10}^5 \text{ kg } \]
Power of a lamp,
\[P = 100 \text{ watt } \]
\[\text{ Potential energy of the water }, \]
\[\text{ P . E . = mgh } \]
\[ = 1 . 8 \times {10}^5 \times 9 . 8 \times 50\]
\[ = 882 \times {10}^5 J\]
As only half the potential energy of water is converted into electrical energy,
\[\text{ Electrical energy }= \frac{1}{2}\text{ P . E }. = 441 \times {10}^5 \text{ J/hr }\]
So, power in watt \[\left( J/\sec \right) = \left( \frac{441 \times {10}^5}{60 \times 60} \right)\]
Therefore, the number of 100 W lamps that can be lit using this energy,
\[\text{ n } = \frac{441 \times {10}^5}{3600 \times 100} = 122 . 5 \approx 122\]
Is there an error in this question or solution?
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