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Water is pumped out of a well 10m deep by means of a pump rated at 10 kW. Find the efficiency of the motor if 4200 kg of water is pumped out every minute. Take `g = 10m//s^2.`

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Work done by the pump per minute, `W =mgh = (4200) (10 (10)) = 4.2 xx 10^5 J`
Output power of the pump `= (W)/(t) = (4.2xx 10^5J)/(60s) = 7 xx 10^3 W = 7kW`
Efficiency of the pump `= ("output power")/("input power" (i.e. "rated power")) = (7kW)/(10kW) = 0.7 = 70%`

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