Three liquids having densities ${\rho }_1, {\rho }_2$ and ${\rho }_3$ are filled in a U-tube. Length of each liquid column is equal to l. ${\rho }_1>{\rho }_2>{\rho }_3$ and liquids remain at rest (relative to the tube) in the position shown in the figure. It is possible that

Since both the vertical limbs of U –tube are open to atmosphere, at surface of liquids, the pressure is equal to the atmospheric pressure. Since heights of the liquids in the two vertical limbs are equal and the liquids have differentdensities, pressure exerted by them will be different from each other. The liquids can remain in static equilibrium relative to the tube only, when the system is accelerating down with acceleration g. In that case, liquids experience weightlessness. Hence, at every point of the tube, the pressure will become equal to atmosphere pressure. But if the tube is not accelerating down under gravity, then the heavier liquid will exert more pressure at the bottom. Hence, at the bottom of the left vertical limb, pressure will be greater than that the bottom of the vertical limb. It means on horizontal limb there will be a resultant horizontal force which will be towards the right as shown in figure. Therefore, the liquid in the horizontal limb will have a rightward acceleration.