An engine pumps up $500$ kg of water through height of $10$ m in $50$ s. If efficiency of engine is $75 %$ , then find power of engine.

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Solution

Output power $= \frac{m g h}{t}$
$= \frac{500 k g \times 10 \frac{m}{s ²} m / \times 10 m}{5 s}$
$= 10 \times 10 ³ W a t t$
$= 10 k W$

$E f f e c i e n c y = [\frac{(O u t p u t p o w e r)}{(I n p u t p o w e r)}] \times 100$
$75 = [\frac{10 k W}{(I n p u t p o w e r)}] \times 100$
$I n p u t p o w e r = 1.33 k W$

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An engine pumps up 500 kg of water through height of 10m in 50s. If efficiency of engine is 75%, then find power of engine.

Output power =mght=500kg×10ms²m/×10m5s =10×10³Watt =10kWEffeciency=[(Outputpower)(Inputpower)]×100 75=[10kW(Inputpower)]×100 Inputpower=1.33kW

An engine pumps up $500$ kg of water through height of $10$ m in $50$ s. If efficiency of engine is $75 %$ , then find power of engine.