1.

Calculate the power of an engine required to lift `10^5`kg of coal per hour from a mine 360 m deep.

Answer» Correct Answer - A
Power of the engine required, `P = (W)/(t) = (mgh)/(t) = ((10^5kg)(10m//s^2)(360m))/(60xx60s) = 10^5W =100kW`


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