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  4. An engine pumps up 1 quintal of coal from a mine 100 m deep in 0.5 s. If its efficiency is 60 %, power of the engine is (Take .g=10 m / s 2)

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Q. An engine pumps up $1$ quintal of coal from a mine $100\, m$ deep in $0.5\, s$. If its efficiency is $60 \%,$ power of the engine is (Take $\left.g=10\, m / s ^{2}\right)$

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Solution:

Actual power $=\frac{W}{t}=\frac{m g h}{t}$
$=\frac{100 \times 10 \times 100}{0.5}$
$=2 \times 10^{5}$ watt
Power required $=\frac{\text { Actual power }}{\text { Efficiency }}$
$=\frac{2 \times 10^{5}}{\frac{60}{100}}=330\, kW$