Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes is $2 \mathrm{cm}$ and the areas of cross-section at A and B are $4{ \mathrm{cm}}^2$ and $2{ \mathrm{cm}}^2$ respectively. Find the rate of flow of water across any section.

Applying Bernoulli's equation at A and B

$$\begin{gathered} {\mathrm{p}}_{\mathrm{A}}+\frac{1}{2}{\mathrm{\rho v}}_{\mathrm{A}}^2+{\mathrm{\rho gh}}_{\mathrm{A}}={\mathrm{p}}_{\mathrm{B}}+\frac{1}{2}{\mathrm{\rho v}}_{\mathrm{B}}^2+{\mathrm{\rho gh}}_{\mathrm{B}} \\ ({\mathrm{h}}_{\mathrm{A}}={\mathrm{h}}_{\mathrm{B}}) \\ \frac{1}{2}{\mathrm{\rho v}}_{\mathrm{B}}^2-\frac{1}{2}{\mathrm{\rho v}}_{\mathrm{A}}^2=\mathrm{\rho g}\left({\mathrm{h}}_{\mathrm{A}}-{\mathrm{h}}_{\mathrm{B}}\right) \\ \text{ or } {\mathrm{v}}_{\mathrm{B}}^2-{\mathrm{v}}_{\mathrm{A}}^2=2\mathrm{g}\left({\mathrm{h}}_{\mathrm{A}}-{\mathrm{h}}_{\mathrm{B}}\right) \\ =2\times 10\times 0.02 \\ \text{ or } {\mathrm{v}}_{\mathrm{B}}^2-{\mathrm{v}}_{\mathrm{A}}^2=0.4{ \mathrm{m}}^2/{\mathrm{s}}^2 \\ {\mathrm{v}}_{\mathrm{A}}{\mathrm{A}}_{\mathrm{A}}={\mathrm{v}}_{\mathrm{B}}{\mathrm{A}}_{\mathrm{B}} \\ 4{\mathrm{v}}_{\mathrm{A}}=2{\mathrm{v}}_{\mathrm{B}} \\ {\mathrm{v}}_{\mathrm{B}}=2{\mathrm{v}}_{\mathrm{A}} \\ \mathrm{Solving} \mathrm{Eqs}. \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get} \\ {\mathrm{v}}_{\mathrm{A}}=0.365 \mathrm{m}/\mathrm{s} \\ \mathrm{Volume} \mathrm{flow} \mathrm{rate}={\mathrm{v}}_{\mathrm{A}}{\mathrm{A}}_{\mathrm{A}} \\ =(0.365)\left(4\times {10}^{-4}\right) \\ =1.46\times {10}^{-4}{ \mathrm{m}}^3/\mathrm{s}=146{ \mathrm{cm}}^3/\mathrm{s} \end{gathered}$$