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2015-11-08T10:45:40+05:30

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Power = 0.5 HP = 0.5×745.7 J/s = 372.85 J/s
efficiency = 90%
Work done in 1 second = 372.85×(90/100) = 335.5 J

Mass of 1l water = 1Kg
Work done to raise 1l water by 10m = mgh = 1×9.8×10 = 98J
in 1s, the pump can raise = 335.5/98 liter
to raise 500l, time = 500 ÷ (335.5/98) = 500×98/335.5 = 146 second

time required is 146 second.
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but the answer provided is 1.1 * 10^5 sec
Then either you have written the question wrong or the answer is wrong.
are the dimensions of the tank given ?
2015-11-08T12:43:04+05:30

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Change in potential energy of 500 litres from ground level to tank:
   = m g h
   = 500 kg * 10 m/sec² * 10 m = 50, 000 J

Effective mechanical power of the pump that is available for pumping water
       = 0.5 HP * 90 % = 0.5 HP * 745.7 W/HP  * 0.9  = 335.45 W

Time duration to fill the tank = energy / power
   = 50, 000 / 335.45   sec. =  149 sec  = 2 min and 29 sec.

If  the value of g is taken as 9.81 m/s/s  then the answer will be a little different.
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