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)

An engine pumps up 1 quintal of coal from a mine 100 m deep in 0.5 s.

1.	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)
Answer» 330 kW 100 kW 200 kW 400 kW SOLUTION :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`.