Explanation

(b) The mass of the water flowing out per second
$m = Avp$ …(i)
Where p=density of water
A=area of cross-section of pipe.
V= velocity of water
Rate of increase of kinetic energy
$= \frac{1}{2}mv^{2} = \frac{1}{2}(Avp)v^{2}$ …(ii)
Mass m, flowing out per second, can be increasing to m’ by increasing v to v’, then power increase from P to P’.
$\ \ \frac{P^{'}}{P} = \frac{\frac{1}{2}A^{'}pV^{'3}}{\frac{1}{2}Apv^{3}}\ or\frac{P^{'}}{P} = \left( \frac{v^{'}}{v} \right)^{3}$
Now, $\ \ \ \ \ \ \ \frac{m^{'}}{m} = \frac{Apv^{'}}{Apv} = \frac{v^{'}}{v}$
As, $\ \ \ \ \ \ \ \ \ \ m^{'} = nm,v^{'} = nv$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{P^{'}}{P} = n^{3}\ \Rightarrow P^{'} = n^{3}P$