Water falls down a 500.0 m shaft to reach a turbine which generates electricity. How much water must fall per second in order to generate $1.00 \times 10^{9}$ watts of power?(Assume 50% efficiency of conversion and $g = 10 m / s^{2}$ )

250 m^{3}

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400 m^{3}

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500 m^{3}

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200 m^{3}

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Solution

The correct option is B $400 m^{3}$
Let m be the mass of water required per second.
Power generated by water just before striking the turbine $P = m g h = 5000 m$ $W a t t s$
Power transferred to the turbine $P_{t} = 0.5 \times P = 2500 m$ $W a t t s$
Now,
$\Rightarrow P_{t} =$ Power to be generated
$\Rightarrow 2500 m = 10^{9}$
$\Rightarrow m = 4 \times 10^{5} k g$
$\Rightarrow v o l l u m e = m a s s / d e n s i t y = 400 m^{3}$ . $[∵ D e n s i t y o f w a t e r = 10^{3} k g / m^{3}]$

Hence correct answer is option $B$

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Water falls down a 500.0 m shaft to reach a turbine which generates electricity. How much water must fall per second in order to generate 1.00×109 watts of power?(Assume 50% efficiency of conversion and g=10m/s2)

Water falls down a 500.0 m shaft to reach a turbine which generates electricity. How much water must fall per second in order to generate $1.00 \times 10^{9}$ watts of power?(Assume 50% efficiency of conversion and $g = 10 m / s^{2}$ )