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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.