Question

If the power of the motor of a water pump is $3kW$, then the volume of water that can be lifted to the height of $10m$ in $1$ minute by the pump is (take $g=10m/{{s}^{2}}$)
$A.1800l$.
$B.180l$.
$C.18000l$.
$D.18l$.

Answer 1

Hint: We should start solving this problem by using the mathematical concept of power. Power is defined as the rate of doing work. During the process we have to apply the relationship among volume, mass and density of water. Mass of an object per unit volume is called its density. We have to convert the units accordingly also.
Formula used:
In this particular problem we are using two formulae, they are as follows:-
\[P=\dfrac{W}{t}\]and $m=\rho V$

Complete answer:
From the question above, we get the following parameters:-
Power, $P=3kW=3000W$.
Height, $h=10m$
Time, t= $1$ minute = $60s$
Acceleration due to gravity, $g=10m/{{s}^{2}}$
We have to find Volume of water lifted, $V$.
Now, applying the basic formula of power, we get,
$P=\dfrac{W}{t}$
It can also be written as,
$P=\dfrac{PE}{t}$, where PE is the potential energy in lifting water. Work done is equal to the change in potential energy in this case. Therefore, we have replaced work done with potential energy.
$P=\dfrac{mgh}{t}$…………. (i)
But we know that, $m=\rho V$, where $\rho =$density and m=mass of water.
Putting $m=\rho V$in equation (i), we get,
$P=\dfrac{\rho Vgh}{t}$………………….. (ii)
Now, putting values of the parameters in the equation (ii), we get,
$3000=\dfrac{{{10}^{3}}\times V\times 10\times 10}{60}$
$3000\times 60=1000\times 10\times 10\times V$
$180000=100000V$
$V=\dfrac{180000}{100000}$
$V=1.8$
Therefore, Volume of water lifted is $1.8{{m}^{3}}$.
So, the volume of water in litres is $1800$.

So, the correct answer is “Option A”.

Additional Information:
Power can be defined as the rate of doing work. It can also be defined as the rate of energy consumed. SI unit of power is Watt. For higher values, Horse Power (HP) is also used as the unit of power.

Note:
Always, take care of units of the given parameters to get the result in these types of problems. To convert ${{m}^{3}}$ into litres we have to multiply it with$1000$. Start solving with a basic equation and then put values accordingly. Formulae should be used correctly with proper calculations. When the value of g is not given, we can also use $g=9.8m/{{s}^{2}}$.