91Ó°ÊÓ

The given data can be listed below as,

• The cross-section area at C is one half the area D ${\mathrm{A}}_{\mathrm{D}}=\frac{{\mathrm{A}}_{\mathrm{c}}}{2}$.
• The distance of D is h₁ below the level of the liquid in A.

The value of total energy between points in a laminar uniform fluid flow will be same. Here, the total energy refers to the sum of pressure, kinetic, and potential energy.

At point D, the speed of efflux is express as,

${\mathrm{v}}_{\mathrm{D}}={\sqrt{2\mathrm{gh}}}_1$

Applying the equation of continuity at the point C and D,

$$\begin{gathered} {\mathrm{A}}_{\mathrm{c}}{\mathrm{V}}_{\mathrm{c}}={\mathrm{A}}_{\mathrm{D}}{\mathrm{V}}_{\mathrm{D}} \\ =\frac{{\mathrm{A}}_{\mathrm{D}}{\mathrm{V}}_{\mathrm{D}}}{{\mathrm{A}}_{\mathrm{C}}} \end{gathered}$$

Here, A_C and A_D are the cross-sectional area at C and D, and v_C and v_D are the velocity at point C and D.

Substitute all the value in the above equation.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{c}}=\frac{{\mathrm{A}}_{\mathrm{D}}{\mathrm{V}}_{\mathrm{D}}}{{\displaystyle \frac{{\mathrm{A}}_{\mathrm{D}}}{2}}} \\ =2\sqrt{2{\mathrm{gh}}_1} \\ =\sqrt{8{\mathrm{gh}}_1} \end{gathered}$$

Applying the Bernoulli’s equation at point A and C,

${\mathrm{P}}_{\mathrm{A}}+\frac{1}{2}{\mathrm{pV}}_{\mathrm{A}}^2+{\mathrm{pgh}}_{\mathrm{A}}=\mathrm{pc}+\frac{1}{2}{\mathrm{pV}}_{\mathrm{c}}^2{\mathrm{pgh}}_{\mathrm{c}}$

Here, pA is the atmospheric pressure at A, vA is the velocity at A that can be considered as zero (since the tank is very large), $p$is the density of liquid, g is the gravitational acceleration, p_C is the pressure at C.

Substitute all the value in the above equation.

$$\begin{gathered} {\mathrm{p}}_{\mathrm{A}}+\frac{1}{2}{\mathrm{pV}}_{\mathrm{A}}^2+{\mathrm{pgh}}_{\mathrm{A}}={\mathrm{p}}_{\mathrm{c}}+\frac{1}{2}{\mathrm{pV}}_{\mathrm{c}}^2{\mathrm{pgh}}_{\mathrm{c}} \\ \left({\mathrm{p}}_{\mathrm{A}}-{\mathrm{p}}_{\mathrm{c}}\right)+\frac{1}{2}\mathrm{p}\left(0\right)+{\mathrm{pgh}}_1=\frac{1}{2}{\mathrm{pV}}_{\mathrm{c}}^2+\mathrm{pg}\left(0\right) \\ {\mathrm{p}}_{\mathrm{g}}+{\mathrm{pgh}}_1=\frac{1}{2}\mathrm{p}{\left(\sqrt{8{\mathrm{gh}}_1}\right)}^2 \\ {\mathrm{p}}_{\mathrm{g}}+{\mathrm{pgh}}_1=4{\mathrm{pgh}}_1 \\ {\mathrm{p}}_{\mathrm{g}}=3{\mathrm{pgh}}_1 \end{gathered}$$

This is the gauge pressure at point C and the surface of the fluid at E.

The pressure at height point E can also be written as,

${\mathrm{P}}_{\mathrm{E}}={\mathrm{pgh}}_2$

On equation both the above equations as,

$$\begin{gathered} {\mathrm{p}}_{\mathrm{g}}={\mathrm{p}}_{\mathrm{E}} \\ 3{\mathrm{pgh}}_1={\mathrm{pgh}}_2 \\ {\mathrm{h}}_{2=3{\mathrm{h}}_1} \end{gathered}$$

Hence, the height of the fluid in the column is$3{\mathrm{h}}_1$.