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Question
A pump is used to lift 600 kg of water from a depth of 75 m in 10s. Calculate:
- The work done by the pump,
- the power at which the pump works, and
- The power rating of the pump if its efficiency is 40%. (Take g = 10m s-2).
`["Hint" : "Efficiency" = "useful power"/"power input"]`
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Solution
Mass of water = 600 kg
Height to which the water has to be raised = 75 m
Time = 10 s
(a) Work done by the pump = mgh
W = mgh
W = 600 × 10 × 75
W = 450000
W = 4.5 × 105 J
(b) Power at which pump works `= "work done"/"Time taken"`
`= "mgh"/"t"`
`= (600 xx 10 xx 75 "J")/"10 s"`
= 45000 W
= 45 kW ...[1 kW = 1000 W]
(c) `"Efficiency" = "useful power"/"power input"`
Efficiency = 40 % = 0.4
`=> 0.4 = "45 kW"/"Power input"`
Power input = `"45 kW"/0.4`
Power input = 112.5 kW
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