The amount of energy consumed by a 10hp water pump in 30 minutes to lift the water to overhead tank is :

26.856 \times 10^{6} J

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0.8949 \times 10^{6} J

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53.712 \times 10^{6} J

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13.428 \times 10^{6} J

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Solution

The correct option is D $13.428 \times 10^{6} J$
We know that, $E n e r g y = P o w e r \times t i m e$
and Power, $P = \frac{W o r k d o n e / E n e r g y}{T i m e}$
Therefore energy consumed by 10hp water pump in 30 minutes $= p o w e r \times t i m e = 10 \times 746 \times 30 \times 60 J$
$1 h p = 746 W, 1 m i n = 60 s e c$
$= 13.428 \times 10^{6} J$

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The amount of energy consumed by a 10hp water pump in 30 minutes to lift the water to overhead tank is :

The correct option is D 13.428×106JWe know that, Energy=Power×timeand Power, P=Work done/EnergyTimeTherefore energy consumed by 10hp water pump in 30 minutes =power×time=10×746×30×60J1hp=746W,1min=60sec=13.428×106J