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Question
An electric motor pump is 60% efficient and is rated 2.5 H.P. Calculate the maximum load it can lift through a height of 10m in 8 sec. (1 H.P. = 750 W).
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Solution
Useful power supplied by the motor =`(2.5xx750 xx60)/100` watt
= 1125 watt
If maximum load lifted is W kg, then :
Work done in lifting up the load = W × 10 × 10 J
And Work done by the pump = 1125 × 8
Or Load lifted; W =`(1125xx8)/(10xx10)` = 90 kg.
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