# A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Given,

Volume of the tank, V = 30 m3

Operation time of pump, t = 15 min = 15×60 sec = 900 sec

Height of the tank from ground, h = 40 m

Efficiency of the pump, η = 30%

We know, density of water, ρ = 103 kg m-3

So, Mass of water, m = ρV

m = 103 m3 × 30 kg m-3

m = 3 × 104 kg

Output power, Po = Work done/Time of operation

Po = mgh/t

Po = (3×104 kg × 9.8 m s-2 × 40 m)/(900 sec)

Po = 1.307×104 W = 13.07 kW

We know,

Efficiency (η) = Output power (Po)/Input power (Pi)

Input power (Pi)= Output power (Po)/Efficiency (η)

Pi = (1.307×104 W)/(0.3)

Pi = 4.355×104 W = 43.55 kW

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Inelastic collisions in 1 D44 mins  Check Your progress Part 2| Interactive Quiz: Work, Energy & Power44 mins  Work, Energy & Power (Lecture 6)41 mins  Work, Energy & Power (Lecture 4)50 mins  Complete Revision of Work , Power and Energy | Mastering important chapters46 mins  Work, Energy & Power (Lecture 5)54 mins  Conservation of Linear Momentum, Visualize the concept (Theory + Interactive Quiz)41 mins  Interactive Quiz of Work Power energy | Check your progress53 mins  Work, Energy & Power (Lecture 2)44 mins  Work, Energy & Power (Lecture 3)47 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 