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)`

Options

A. 330 kW
B. 100 kW
C. 200 kW
D. 400 kW

Answer

A

Explanation

Actual power = `W/t = (mgh)/(t) = (100 xx 10 xx 100)/(0.5)`
`= 2 xx 10^5 ` watt
Power required = `("Actual power")/("Efficiency") = (2 xx 10^(5))/((60)/100) = 330 kW`.