In a hydroelectric power station water is allowed to fall at the rate of $2000$ kg per second on a turbine kept $100$ m below the water level. Calculate the potential energy of water falling every one second at the highest point.

2 \times 10^{6} J^{2}

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2 \times 10^{8} J

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2 \times 10^{7} J

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2 \times 10^{6} J

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Solution

The correct option is C $2 \times 10^{6} J$
Potential energy, $P E = m g h = 2000 \times 10 \times 100 = 2 \times 10^{6} J$ .

From law of conservation of energy, total energy at the highest point.

$E_{1} = p o t e n t i a l e n e r g y + k i n e t i c e n e r g y E_{1} = m g h + 0 (K E a t t h e h i g h e s t p o i n t = 0) \Rightarrow E_{1} = 2 \times 10^{6} J$

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In a hydroelectric power station water is allowed to fall at the rate of 2000 kg per second on a turbine kept 100 m below the water level. Calculate the potential energy of water falling every one second at the highest point.

The correct option is C 2×106JPotential energy, PE=mgh=2000×10×100=2×106J.From law of conservation of energy, total energy at the highest point.E1=potentialenergy+kineticenergyE1=mgh+0(KEatthehighestpoint =0)⇒E1=2×106J

In a hydroelectric power station water is allowed to fall at the rate of $2000$ kg per second on a turbine kept $100$ m below the water level. Calculate the potential energy of water falling every one second at the highest point.