The rate of flow of water through the tube shown is 1000 cm³/s. The area of cross-section at point A and B are 10 cm² and 4 cm². The difference in mercury level (h) in the tube is ______cm. $\left(\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2, {\mathrm{\rho }}_{\mathrm{Hg}}=13600\mathrm{kg}/{\mathrm{m}}^3, {\mathrm{\rho }}_{{\mathrm{H}}_2\mathrm{O}}={10}^3\mathrm{kg}/{\mathrm{m}}^3\right)$

Applying equation of continuity ${\mathrm{A}}_1{\mathrm{v}}_1={\mathrm{A}}_2{\mathrm{v}}_2=\mathrm{Q}$

As $\mathrm{Q}=1000{\mathrm{cm}}^3/\mathrm{s},{\mathrm{A}}_1=10{\mathrm{cm}}^2,{\mathrm{A}}_2=4{\mathrm{cm}}^2$

${\mathrm{v}}_{\mathrm{A}}=100\mathrm{cm}/\mathrm{s},{\mathrm{v}}_{\mathrm{B}}=250\mathrm{cm}/\mathrm{s}$

Applying Bernoulli's theorem at A and B.

${\mathrm{P}}_{\mathrm{A}}+\frac{1}{2}{\mathrm{\rho v}}_{\mathrm{A}}^2={\mathrm{P}}_{\mathrm{B}}+\frac{1}{2}{\mathrm{\rho v}}_{\mathrm{B}}^2$ (both points are at the same level)

$\therefore {\mathrm{P}}_{\mathrm{A}}-{\mathrm{P}}_{\mathrm{B}}=\frac{1}{2}\mathrm{\rho }\left({\mathrm{v}}_{\mathrm{B}}^2-{\mathrm{v}}_{\mathrm{A}}^2\right)$

${\mathrm{P}}_{\mathrm{A}}-{\mathrm{P}}_{\mathrm{B}}$ can also be approximately written as ${\mathrm{h\rho }}_{\mathrm{Hg}}\mathrm{g}$

$\therefore {\mathrm{h\rho }}_{\mathrm{Hg}}\mathrm{g}=\frac{1}{2}{\mathrm{\rho }}_{{\mathrm{H}}_2\mathrm{O}}\left({\mathrm{v}}_{\mathrm{B}}^2-{\mathrm{v}}_{\mathrm{A}}^2\right)$

$\Rightarrow \mathrm{h}\left(13600\right)10=\frac{1}{2}1000\left((2.50{)}^2-(1.00{)}^2\right)\Rightarrow \mathrm{h}\simeq 1.93\mathrm{cm}$