In my physics book "Fundamentals of Physics" by Jearl Walker there is a figure that shows a U-tube with uniform cross-sectional area. The U-tube contains two liquids in static equilibrium: water in the right arm and oil in the left. $l$ is 135mm and $d$ is 12.3mm (see image). My book says that liquids are in static equilibrium so pressure at point $I$ would be same as pressure at point $J$. My question is how can pressure at $I$ and $J$ be same? $J$ is at more depth than $I$ so the atmospheric pressure at $J$ should be higher than atmospheric pressure at $I$. enter image description here

  • $\begingroup$ Are you thinking that the atmospheric pressure at J should be the same as the pressure in the oil column at the same height? $\endgroup$ Commented Nov 20, 2020 at 3:27

2 Answers 2


The pressure at I and J obviously is not the same, but can be reasonably assumed it is. The pressure difference is

$$\Delta p = g \cdot \rho_\mathrm{air} \cdot \Delta h$$

With air density ( about $1.2\ \mathrm{kg/m^3}$ ) roughly 1000 times lower than water density, $100\ \mathrm{mm}$ difference in the level heights makes about $0.1\ \mathrm{mm}$ of the water column. This is very well comparable with the error of the liquid level height determination and of water and oil densities determination.


This experiment is designed for showing liquid pressure differences based on their density differences. So atmospheric pressure can be safely ignored.

At the interface point liquids reaches static pressure balance :

$$ \rho_{oil}\cdot~g\cdot~h_{oil} = \rho_{water}\cdot~g\cdot~h_{water} \tag 1$$


$$ \frac {\rho_{oil}}{\rho_{water}} = \frac {h_{water}}{h_{oil}} = \frac {\ell}{\ell+d} \tag 2$$

This ratio is $$ \frac {135~mm}{135~mm+12.3~mm} \approx 0.92 \tag 3$$

So material in left side of tube has density $$ 0.92 \times 1000 ~kg/m^3 = 920 ~kg/m^3 \tag 4$$

Which by looking at the standard liquid densities we see that it applies for Cotton seed oil, Heating oil, Menhaden oil, Rape seed oil, Soybean oil.

Thus, problem is well-defined. No need to account for tiny atmosphere pressure differences here.


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