# A pump installed on ground takes 15 minutes to fill a water tank of 30m³ volume placed at a height of 50 m. If the efficiency of the pump is 30%, the power consumed by the pump will be ?

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by kunal322

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by kunal322

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Energy needed to pump the water up to the tank is the change in the potential energy of water.

Actual Energy needed = m g h = 30m³ * 1000kg/m³ * 9.8m/sec² * 50m

E = 1.47*10^7 Joules

Due to inefficiency of pump, the actual amount consumed = E / efficiency

= 1.47 * 10^7 J * 100/30 = 4.9 * 10^7 J

**Power consumed by motor = energy utilized / time duration = 4.9*10^7 J/ 15*60 sec**

** = 54.4 kW**

__In case the efficiency is 33 % instead of 30% the Power consumed will be__

__ = 49.45 kW__

Actual Energy needed = m g h = 30m³ * 1000kg/m³ * 9.8m/sec² * 50m

E = 1.47*10^7 Joules

Due to inefficiency of pump, the actual amount consumed = E / efficiency

= 1.47 * 10^7 J * 100/30 = 4.9 * 10^7 J

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