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An electric pump is used to fill an overhead tank of capacity $9{m^3}$ kept at a height of $10m$ above the ground. If the pump takes $5$ minute to fill the tank by consuming $10kW$ of power, the efficiency of the pump should be (Take $g = 10m{s^{ - 2}}$)
A) $60\% $
B) $40\% $
C) $20\% $
D) $30\% $

Last updated date: 19th Jun 2024
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Hint:This question is based on the concept of efficiency which is defined as output power to the input power for normal practical machines.
It’s always less than $100\% $ for all practical machines.

Formula used:
Potential Energy, $P.E. = mgh$
Where,$m$is the mass in kilograms, $g$ is the acceleration due to gravity in $m{s^{ - 2}}$ and $h$ is the height raised in meters
Power ( $P$ ) = $\dfrac{{energy}}{{time}}$
Efficiency = $\dfrac{{output}}{{input}}$

Complete step by step solution:
Given that the volume of water to be raised $ = 9{m^3}$
Since we know that
Density of water = $1000kg{m^{ - 3}}$
So, mass of water to be raised
Mass = $1000 \times 9$
$ \Rightarrow $ Mass $ = 9000kg$
We also know that the change in potential energy of a body when it is raised a height $h$ above the ground $ = mgh$
Where $m = $ mass of the body in Kg
$g = $ acceleration due to gravity in $m/{s^2}$
$h = $ height raised in $m$
So, the potential energy change for $9000kg$ of water to be raised to a height of $10m$ above the ground is given by;
$PE = 9000 \times 10 \times 10$
$ \Rightarrow PE = 9 \times {10^5}J$
Since power $\dfrac{{energy}}{{time}}$
So output power = $\dfrac{{9 \times {{10}^5}}}{t}$
It is given that time taken to raise the water is 5 min, converting the time in S.I. unit (seconds), we get;
Time = $5 \times 60$
$ \Rightarrow t = 300\sec $
Thus, output power is given by;
$p = \dfrac{{9 \times {{10}^5}}}{{300}} = 3000W$
Given that Input power $ = 10kW = 10 \times {10^3}W$
Hence, efficiency of pump $ = \dfrac{{output}}{{input}}$
$e = \dfrac{{3000}}{{10000}}$
$ \Rightarrow e = 0.3$
But as the options which are given in terms of percentage we have to convert this answer to percentage.
We know that efficiency in percentage Efficiency of pump in percentage
$e = 0.3 \times 100 = 30\% $

So, option (C) is correct.

Additional Information:
The data given in the question is in terms of volume of water to be raised above the ground. This has to be converted to mass in order to calculate the potential energy. This can be done by using density of water whose value is not given in the question, so it should be remembered. Also, the formula of potential energy when a body is raised by height ‘h’ above the ground is ‘mgh’ is to be remembered.

Note: Efficiency of any machine is defined as the ratio between the output powers to the input power.
The potential energy change for anybody of mass ‘m’ to be raised to a height ‘h’ above the ground is equal to ‘mgh’
Where, m = mass of the body
g = acceleration due to gravity
h = height raised above the ground