Water flows through the tube as shown in figure. The areas of cross-section of the wide and the narrow portions of the tube are $5{ \mathrm{cm}}^2$ and $2{ \mathrm{cm}}^2$ respectively. The rate of flow of water through the tube is $500{ \mathrm{cm}}^3/8\mathrm{Find}$ the difference of mercury levels in the U-tube.

Applying continuity equation

$$\begin{gathered} {\mathrm{A}}_1{\mathrm{v}}_1={\mathrm{A}}_2{\mathrm{v}}_2=\text{ rate of flow of water } \\ 5{\mathrm{v}}_1=2{\mathrm{v}}_2=500{ \mathrm{cm}}^3/\mathrm{s} \\ {\mathrm{v}}_1=100 \mathrm{cm}/\mathrm{s}=1.0 \mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_2=250 \mathrm{cm}/\mathrm{s}=2.5 \mathrm{m}/\mathrm{s} \\ \mathrm{Now}, \mathrm{applying} \mathrm{Bernoulli}\text{'}\mathrm{s} \mathrm{equation} \mathrm{at} 1 \mathrm{and} 2 \\ {\mathrm{p}}_1+\frac{1}{2}{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{v}}_1^2+{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{gh}}_1={\mathrm{p}}_2+\frac{1}{2}{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{v}}_2^2+{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{gh}}_2 \\ \left({\mathrm{h}}_1={\mathrm{h}}_2\right) \\ \frac{1}{2}{\mathrm{\rho }}_{\mathrm{w}}\left({\mathrm{v}}_2^2-{\mathrm{v}}_1^2\right)={\mathrm{p}}_1-{\mathrm{p}}_2={\mathrm{\rho }}_{\mathrm{Hg}}{\mathrm{gh}}_{\mathrm{Hg}} \\ \frac{1}{2}{\mathrm{\rho }}_{\mathrm{w}}\left({\mathrm{v}}_2^2-{\mathrm{v}}_1^2\right)={\mathrm{\rho }}_{\mathrm{Hg}}{\mathrm{gh}}_{\mathrm{Hg}} \\ {\mathrm{h}}_{\mathrm{Hg}}=\frac{{\mathrm{\rho }}_{\mathrm{w}}\left({\mathrm{v}}_2^2-{\mathrm{v}}_1^2\right)}{2{\mathrm{\rho }}_{\mathrm{Hg}}\mathrm{g}} \\ =\frac{{10}^3(6.25-1)}{2\times 13.6\times {10}^3\times 9.8} \\ =0.0196 \mathrm{m} \\ =1.96 \mathrm{cm} \end{gathered}$$