Physics - Fluid Question with Solution | TestHub

The pressure at the interface must be same, calculated via either tube. Since, both tube are open to the atmosphere, we must have

$h_1{\mathrm{\rho }}_{1}g=h_2{\mathrm{\rho }}_{2}g$

or$h_1{\mathrm{\rho }}_1=h_2{\mathrm{\rho }}_2$

or$h\mathrm{\rho }=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$

or$h\propto \frac{1}{\mathrm{\rho }}$