Question

A water pump of power 2kW is installed in a home. Then, the amount of water (in litres) it can raise in one minute to a height of 10 m is: [Take $g = 10m/{s^2}$ ]
(A) 200 L
(B) 1200 L
(C) 800 L
(D) 350 L

Answer 1

Power can be defined as the work done per unit time. Work done can be given as a change in potential. In this solution we will be using the following formulae;
$P = \dfrac{W}{t}$ where $P$ is the power delivered to do work, $W$ is the work done, and $t$ is the time taken to do the work.
$W = \Delta PE$ where $PE$ is the potential energy of a body, and the delta sign $\Delta$ signifies change in… (in the case, we have a change in potential energy).
$PE = mgh$ where $m$ is the mass of the body, $g$ is the acceleration due to gravity and $h$ is the height of the body.

Complete step by step answer:
To solve the above problem, we define power to be the rate of doing work, and can mathematically be given as
$P = \dfrac{W}{t}$ where $P$ is the power delivered to do work, $W$ is the work done, and $t$ is the time taken to do the work.
But, work can be defined as the change of potential energy when done against gravity.
Hence, we have
$W = \Delta PE$ $W = \Delta PE$ where $PE$ is the potential energy of a body, and the delta sign $\Delta$ signifies change.
But, $PE = mgh$ where $m$ is the mass of the body, $g$ is the acceleration due to gravity and $h$ is the height of the body.
Hence, $m = \rho V$ where $\rho$ is the density of a substance and $V$ is its volume.
Hence, we can write that
$PE = \rho Vgh$
For work done, we have
$W = \Delta PE = \rho Vg\left( {{h_2} - 0} \right)$ (assuming that the height of the pump is at zero position)
Hence,
$W = \Delta PE = 1000V\left( {10} \right)10 = 100000V$
Thus, the power is
$P = \dfrac{W}{t} = \dfrac{{100000V}}{{60}}$ since one minute equal 60 seconds. So,
$2000 = \dfrac{{100000V}}{{60}}$
$\Rightarrow V = \dfrac{{2000 \times 60}}{{100000}} = 1.2{m^3}$ which is 1200 L (because 1000 L make one ${m^3}$ )
Hence, the correct option is B.

Note:
Alternatively, once it is noted that the power can be taken as the rate of change of potential energy with time, we simply write (for exam speed), that
$P = \dfrac{{mg\left( {h - 0} \right)}}{t} = \dfrac{{\rho Vgh}}{t}$
And we proceed to substitute the values as in above and make $V$ subject of the formula.