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A motor pump lifts 6 tons of water from a well of depth $2\,m$ to the first floor of height $35\,m$ from the ground floor in 20 minutes. The power of the pump (in kW) is ( $g = 10\,m/{s^2}$ )
A. $6kW$
B. $1.85kW$
C. $12kW$
D. $3kW$

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Last updated date: 18th Jun 2024
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Answer
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Hint: The power is the work done per unit time. We know that work done to lift a body through a vertical height is stored as a gravitational potential energy in the body. So, by calculating the gravitational potential energy and dividing it by the given time we can calculate the power.

Complete step by step answer:
It is given that a motor pump lifts 6 tons of water from a well of depth $2\,m$ to a height $35\,m$ .
The time taken is 20 minute.
$ \Rightarrow t = 20 \times 60\,s$
$ \Rightarrow t = 1200\,s$
We need to calculate the power of the pump.
We know that power is the work done per unit time.
In equation form we can write it as
$P = \dfrac{W}{t}$
Where, P is the power, W is the work and t is the time taken.
We know that the work done in lifting a body through a vertical height is stored as the gravitational potential energy.
Gravitational potential energy is given by the equation
$PE = mgh$
Where, m is the mass, g is acceleration due to gravity and h is the height.
This value will give us the value of work.
Let us calculate the value of this work.
$ \Rightarrow W = mgh$
The total height through which it is lifted will be the sum of the depth and the height of the floor from the ground.
$ \Rightarrow h = 2 + 35$
$ \Rightarrow h = 37\,m$
Acceleration due to gravity is given as $g = 10\,m/{s^2}$
Mass, $m = 6\,ton$
$ \Rightarrow m = 6 \times {10^3}kg$
$\because 1\,ton = 1000\,kg$
On substituting the values in equation for work, we get
$ \Rightarrow W = mgh$
$ \Rightarrow W = 6 \times {10^3} \times 10 \times 37\,J$
Now let us substitute the value of work in the equation for power. Then we get,
$ \Rightarrow P = \dfrac{{6 \times {{10}^3} \times 10 \times 37}}{{1200}}$
$ \Rightarrow P = 1.85\, \times {10^3}W$
$ \Rightarrow P = 1.85\,kW$
This is the power of the pump.
So, the correct answer is option B.

Note: Remember that the body is lifted from a depth of $2\,m$ and then to a height of $35\,m$ from the ground. So, we should take the total distance through which it is lifted as the value of h. Don’t substitute just the height from the ground surface. Work is also done to lift the water to the surface of ground from the given depth. so, total work done will be the work done to lift it through a total height of $2\,m + 35\,m = 37\,m$ .