91Ó°ÊÓ

Rank of KE:$K{E}_A>K{E}_C>K{E}_B$

This problem is based on the continuity equation. Kinetic energy can be written in the form of the area by using a continuity equation. Thus, using this relation, we can get the required rank value.

Formulae:

The rate of flow of the fluid in terms of area and speed,$R_v=Av$ (i)

The kinetic energy of a body in motion,$KE=\left(\frac{1}{2}\right)m{v}^2$ (ii)

The mass of a body in terms of density,$m=\mathrm{ÒÏ}V$ (iii)

Substituting the value of mass from equation (iii) and velocity from equation (i) in equation (ii), we can get the kinetic energy equation as follows:

$$\begin{gathered} KE=\left(\frac{1}{2}\right)pV×{\left(\frac{R_v}{A}\right)}^2 \\ =\left(\frac{\mathrm{ÒÏ}V{R}_v^2}{2{\mathrm{Ï€}}^2}\right)×\left(\frac{1}{r^4}\right)...............\left(b\right)\left(âˆ\mu A={\mathrm{Ï€°ù}}^2\right) \end{gathered}$$

This equation shows that KE is inversely proportional to$4^{th}$ power of the radius of the pipe. Thus, KE is minimum when radius will be maximum.

Therefore, rank of sections according to radius, greatest first is$B>C>A$ .