A U-tube of base length ‘$l$’ filled with the same volume of two liquids of densities  $\rho$ and  $2\rho$ 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 $2\rho$  should be behind that of  $\rho$.

From the right limb,  $P_A=P_{atm}+\rho gh$
$$\begin{gathered} P_B=P_A+\rho a\frac{1}{2}=P_{atm}+\rho gh+\rho a\frac{1}{2} \\ {P}_C=P_B+\left(2\rho \right)a\frac{1}{2}=P_{atm}+\rho gh+\frac{3}{2}\rho al\mathrm{.......}\left(i\right) \end{gathered}$$

But from the left limb,  $P_C=P_{atm}+\left(2\rho \right)gh\mathrm{.......}\left(ii\right)$
From Eqs.   (i) and (ii),
$$\begin{gathered} P_{atm}+\rho gh+\frac{3}{2}\rho al=P_{atm}+2\rho gh \\ \Rightarrow h=\frac{3al}{2g} \end{gathered}$$