A dam is situated at a height of $550 m$ above sea level and supplies water to a power house which is at a height of $50 m$ above sea level. $2000 k g$ of water passes through the turbines per second. What would be the maximum electrical power output of the power house if the whole system were $80 %$ efficient.

8 M W

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10 M W

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12.5 M W

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16 M W

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Solution

The correct option is A $8 M W$
Maximum electrical power output,

$P = \frac{m g Δ h}{t}$

$\Rightarrow P = \frac{m}{t} (g Δ h)$

$= 2000 \times 9.80 \times (550 - 50)$

$= 9.80 \times 10^{6}$

At 80% efficiency, $0.80 \times 9.80 \times 10^{6} = 7.84 \times 10^{6} M \approx 8 M W$

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The correct option is A 8 MWMaximum electrical power output,P=mgΔht⇒P=mt(gΔh)=2000×9.80×(550−50)=9.80×106At 80% efficiency, 0.80×9.80×106=7.84×106M≈8MW

The correct option is A $8 M W$
Maximum electrical power output,

$P = \frac{m g Δ h}{t}$

$\Rightarrow P = \frac{m}{t} (g Δ h)$

$= 2000 \times 9.80 \times (550 - 50)$

$= 9.80 \times 10^{6}$

At 80% efficiency, $0.80 \times 9.80 \times 10^{6} = 7.84 \times 10^{6} M \approx 8 M W$