Question

A pump delivers water at the rate of $V$ cubic metre per second. By what factor should its power be raised so that it delivers water at the rate of $nV$ cubic metre per second?
(1) $\sqrt n$
(2) $n$
(3) ${n^3}$
(4) ${n^2}$

Answer 1

We will first obtain the relation between the new velocity of flow and the initial velocity of flow of water. We will then obtain the relation between power and the velocity, which will eventually lead to the solution.

Complete step by step answer:
First, we express the volume flow rate of water as
$V = Av$
Let the new volume flow rate be $V'$. Then we can write
$V' = Av'$
where $v'$ be the new velocity of flow of water.
It is given that the new volume flow rate is increased by $n$ times. Hence, we can write
$V' = nV$
$\begin{array}{c} Av' = nAv\\\ v' = nv \end{array}$
Now, we express the kinetic energy of the water flowing out of the pump as
$KE = \dfrac{1}{2}m{v^2}$
where $m$ is the mass of water flowing out of the pump and $v$ is the velocity of flow of water.
Now, the mass of water flowing out in $t$ time can be written as
$\begin{array}{c} m = \rho Vt\\\ = \rho Avt \end{array}$
where $\rho$ is the density of water and $A$ is the area of cross section of the pump.
Using the relation for mass $m$ in the kinetic energy relation, we get
$\begin{array}{c} KE = \dfrac{1}{2}\rho Avt \times {v^2}\\\ = \dfrac{1}{2}\rho At{v^3} \end{array}$
We express the relation for the power of the pump as
$P = \dfrac{{KE}}{t}$
We now insert the equation for $KE$ in this equation. Hence,
$\begin{array}{c} P = \dfrac{{\dfrac{1}{2}\rho At{v^3}}}{t}\\\ = \dfrac{1}{2}\rho A{v^3} \end{array}$
Since the density $\rho$ and the area of cross section $A$ of the pump is constant, we can say that the power is proportional to the cube of the velocity of flow of water.
$P\;\propto \;{v^3}$
If $P'$ is the power when water flows at the rate of $nV$ cubic metre per second, we can express the ratio of $P$ to $P'$ as
$\dfrac{P}{{P'}} = \dfrac{{{v^3}}}{{{{\left( {v'} \right)}^3}}}$
Since $v' = nv$, we can write
$\dfrac{P}{{P'}} = \dfrac{{{v^3}}}{{{{\left( {nv} \right)}^3}}}\\\ = \dfrac{1}{{{n^3}}}\\\ \implies P' = {n^3}P$
Hence, the power of $n$ has to be raised by a factor of $3$ so that the volume flow rate is $nV$ cubic metre per second.

Therefore, the option (3) is the correct answer.

Note:
The equation $V = Av$ comes from the equation of continuity.
According to the equation of continuity, the volume flow rate of a liquid remains the same at any point inside the pipe if the liquid is incompressible.