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)

Question

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) (A) 330 kW (B) 100 kW (C) 200 kW (D) 400 kW

Tardigrade Admin · Accepted Answer

Actual power =(W/t)=(m g h/t) =(100 × 10 × 100/0.5) =2 × 105 watt Power required =( text Actual power / text Efficiency ) =(2 × 105/(60)100)=330 kW

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 $g = 10 m / s^{2})$

330 kW

100 kW

200 kW

400 kW

Actual power $= \frac{W}{t} = \frac{m g h}{t}$
$= \frac{100 \times 10 \times 100}{0.5}$
$= 2 \times 1 0^{5}$ watt
Power required $= \frac{Actual power}{Efficiency}$
$= \frac{2 \times 1 0 ^{5}}{\frac{60}{100}} = 330 kW$

Q. An engine pumps up 1 quintal of coal from a mine 100m deep in 0.5s. If its efficiency is 60%, power of the engine is (Take g=10m/s2)