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# Water Falling from a 50-m High Fall is to Be Used for Generating Electric Energy. If 1 ⋅ 8 × 10 5 Kg of Water Falls per Hour and Half the Gravitational Potential Energy - Physics

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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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 8 Work and Energy
Q 19 | Page 133
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