If the water falls from a dam into a turbine wheel 19.6 m below, then the velocity of water at the turbine is:

9.8 m / s

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19.6 m / s

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39.2 m / s

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98.0 m / s

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Solution

The correct option is B $19.6 m / s$
Here PE is converted into KE
$∴ m g h = \frac{1}{2} m v^{2}$
$∴ 2 g h = v^{2} ⟹ v = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 19.6} = 19.6 m / s$
$∴ v = 19.6 m / s$

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If the water falls from a dam into a turbine wheel 19.6 m below, then the velocity of water at the turbine is:

The correct option is B 19.6m/sHere PE is converted into KE∴mgh=12mv2∴2gh=v2⟹v=√2gh=√2×9.8×19.6=19.6m/s∴v=19.6m/s

The correct option is B $19.6 m / s$
Here PE is converted into KE
$∴ m g h = \frac{1}{2} m v^{2}$
$∴ 2 g h = v^{2} ⟹ v = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 19.6} = 19.6 m / s$
$∴ v = 19.6 m / s$