Explanation:
If the motor pumps water (density \(=\rho\) ) continuously through a pipe of area of cross-section A with velocity \(v\), then mass flowing out per second.
\(m = Av\rho {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Rate of increase of kinetic energy
\( = \frac{1}{2}\frac{{dm}}{{dt}}{v^2} = \frac{1}{2}(Av\rho ){v^2} = \frac{1}{2}A\rho {v^3}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (2)\)
\(\dfrac{P^{\prime}}{P}=\dfrac{\dfrac{1}{2} A \rho v^{13}}{\dfrac{1}{2} A \rho v^{3}} \Rightarrow \dfrac{P^{\prime}}{P}=\left(\dfrac{v^{\prime}}{v}\right)^{3}\)
Now, \(\dfrac{m^{\prime}}{m}=\dfrac{A \rho v^{\prime}}{A \rho v}=\dfrac{v^{\prime}}{v}\)
As, \(\quad m^{\prime}=n m \Rightarrow v^{\prime}=n v\)
\(\therefore \dfrac{P^{\prime}}{P}=n^{3} \Rightarrow P^{\prime}=n^{3} P\)