Question

The power of a water pump is 2kW. If $g = 10m{s^{ - 2}}$ , the amount of water it can raise in one minute to a height of 10 m is:
(A) 2000 litre
(B) 1000 litre
(C) 100 litre
(D) 1200 litre

Answer 1

Hint: Use the formula for power: $Power = \dfrac{{Work{\text{ }}Done}}{{Time{\text{ for which the work is done}}}}$ to find out the amount of work done to raise water in one minute to a height of 10 m. Further use the formula for work done, $W = mgh$ to find out the amount/ mass of water. Use density of water $ = 1000kg/{m^3}$ to find out the amount of water in terms of volume.

Complete step by step solution
We are given the power of the water pump to be 2kW.
$ \Rightarrow P = 2kW$
We know that $1kW = 1000W$,
$
   \Rightarrow P = 2 \times 1000{\text{ }}W \\
   \Rightarrow P = 2000{\text{ }}W{\text{ or J}}{{\text{s}}^{{\text{ - 1}}}} \\
 $
We also know that power is the work done per unit time or we can say that the work done is equal to the power rating times the time for which the work is done.
$ \Rightarrow W = P \times t$
We are given the time to be one minute, i.e. $t = 1\min = 60\sec $. Substituting the values of power rating, $P = 2000{\text{W or J}}{{\text{s}}^{{\text{ - 1}}}}$ and time,$t = 60\sec $in the above equation yields,
\[
   \Rightarrow W = 2000{\text{ J}}{{\text{s}}^{ - 1}} \times 60{\text{ s}} \\
   \Rightarrow W = 120000{\text{ J}} \\
 \]
We will now use$W = mgh$, where we are given the value of height to which water has to be raised, i.e. 10 m.
$
   \Rightarrow W = mgh \\
   \Rightarrow 120000 = m \times 10 \times 10 \\
   \Rightarrow m = 1200{\text{ }}kg \\
 $
Now we have obtained the amount of water that can be raised to a height of 10 m in one minute with a water pump of power rating 2 kW. We will now convert the amount of water from mass terms to volume terms to match which option is correct.
We know the density of water, \[\rho = 1000{\text{ }}kg{m^{ - 3}}\]and we have mass of water, $m = 1200{\text{ }}kg$. The density is mass per unit volume.
$
   \Rightarrow \rho = \dfrac{m}{V} \\
   \Rightarrow 1000{\text{ k}}g{m^{ - 3}} = \dfrac{{1200{\text{ kg}}}}{V} \\
   \Rightarrow V = \dfrac{{1200}}{{1000}}{\text{ }}{{\text{m}}^{\text{3}}} \\
   \Rightarrow V = 1.2{\text{ }}{m^3} \\
 $
Using the conversion $1{\text{ }}{m^3} = 1000{\text{ L}}$,
$
   \Rightarrow V = 1.2 \times 1000{\text{ }}L \\
   \Rightarrow V = 1200{\text{ L}} \\
 $

Therefore, Option (D) is correct.

Note: We have taken work done, $W = mgh$ because, the work done in moving any object of mass ‘m’ to a height of ‘h’ is equal to the potential energy stored in that object at that particular height. And we know that the potential energy is mgh.