In a hydroelectric power plant, waterfalls at a rate of 1000 kg/s from a height of 100 m. Assuming that 50% of the energy of falling water is converted into electrical energy, calculate the power generated per second. $(t a k e g = 9.8 m s^{- 2})$

200000 W

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195000 W

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490000 W

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380000 W

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Solution

The correct option is C
490000 W

Mass of water falling per second = 1000 kg
Height = 100 m
Energy possessed by water per second = Potential energy = m $\times$ g $\times$ h = 1000 $\times$ 9.8 $\times$ 100 = 980000 J
It is mentioned that only 50% of the energy is converted into electrical energy.
So, the amount of Electrical energy generated per second= Total Energy $\times$ $\frac{50}{100}$ = 490000 J
we know that the power generated = energy generated per second = 490000 W

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In a hydroelectric power plant, waterfalls at a rate of 1000 kg/s from a height of 100 m. Assuming that 50% of the energy of falling water is converted into electrical energy, calculate the power generated per second.(take g=9.8 ms−2)

In a hydroelectric power plant, waterfalls at a rate of 1000 kg/s from a height of 100 m. Assuming that 50% of the energy of falling water is converted into electrical energy, calculate the power generated per second. $(t a k e g = 9.8 m s^{- 2})$