A U-tube of base length 'l' filled with the same volume of two liquids of densities ρ and 2ρ is moving with an acceleration 'a' on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by
 

For the given situation, liquid of density 2ρ should be behind that of ρ.
 From the right limb, ${\mathrm{P}}_{\mathrm{A}}={\mathrm{P}}_{\mathrm{atm}}+\mathrm{\rho gh}$
$$\begin{gathered} {\mathrm{P}}_{\mathrm{B}}={\mathrm{P}}_{\mathrm{A}}+\mathrm{\rho a}\frac{1}{2}={\mathrm{P}}_{\min }+\mathrm{\rho gh}+\mathrm{\rho a}\frac{1}{2} \\ {\mathrm{P}}_{\mathrm{C}}={\mathrm{P}}_{\mathrm{B}}+\left(2\mathrm{\rho }\right)\mathrm{a}\frac{1}{2}={\mathrm{P}}_{\min }+\mathrm{\rho gh}+\frac{3}{2}\mathrm{\rho al} \left(\mathrm{i}\right) \\ \mathrm{But} \mathrm{from} \mathrm{the} \mathrm{left} \mathrm{limb},{\mathrm{P}}_{\mathrm{C}}={\mathrm{P}}_{\mathrm{atm}}+\left(2\mathrm{\rho }\right)\mathrm{gh} \left(\mathrm{ii}\right) \\ \mathrm{From} \mathrm{Eqs}. \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \\ {\mathrm{P}}_{\mathrm{atm}}+\mathrm{\rho gh}+\frac{3}{2}\mathrm{\rho al}={\mathrm{P}}_{\mathrm{atm}}+2\mathrm{\rho gh} \Rightarrow \mathrm{h}=\frac{3\mathrm{al}}{2\mathrm{g}} \end{gathered}$$