Question: A water pump lifts \[18 \times {10^3}\] liters of water to a height of 30 m in one hour. If the effi...

Question

A water pump lifts $18 \times {10^3}$ liters of water to a height of 30 m in one hour. If the efficiency of the pump is 75% power of the pump is ? [g= $10\,m{s^{ - 2}}$ ]
A. 2 KW
B. 3 KW
C. 6 KW
D. 4 KW

Answer 1

Solution

Potential energy is the energy that an item has as a result of its location in relation to other items. Potential energy is frequently related to restoring forces like springs and gravity. An external force that acts against the force field of the potential performs the operation of stretching a spring or lifting a load. The force field, which is considered to be stored as potential energy, stores this work. We use the following formula here.

Formula used:
Work done to lift the pump = $mgh$
Here, $m$ = mass, $g$ = acceleration due to gravity and $h$ = height.

Complete step by step answer:
Forces and potential energy are inextricably related. If the work done by a force on a body moving from point A to point B is independent of the path between these points (if the work is done by a conservative force), the work done by this force measured from point A assigns a scalar value to every other point in space and defines a scalar potential field. In this scenario, the force may be described as the inverse of the potential field's vector gradient.

The gravitational potential energy of an item, which is dependent on its mass and distance from another object's centre of mass, is a common form of potential energy.
Now mass of water = $18 \times {10^3}$ litre
Since 1l = 1000 c
It is now mass of water = $18 \times {10^6}$ litre
Water lifted to the height of 30 m
Time = 1 hour = 3600 s
Efficiency of pump being 75%
Now, using the formula,
Power of the pump = $\dfrac{{mgh}}{t} \times \dfrac{{75}}{{100}}$
$P = \dfrac{{18 \times {{10}^6} \times 10 \times 30}}{{3600}} \times \dfrac{{75}}{{100}} \\\ \Rightarrow P = 2000 watt = 2 KW \\\ \therefore P = 2KW$

Hence option A is correct.

Note: Because the work of potential forces acting on a body moving from a start to an end position is defined only by these two positions and not by the body's trajectory, there is a function known as potential that can be assessed at these two points to determine this work.