Variation of atmospheric pressure with depth in U-tube

Question

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 $I$.

Answer 1

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.

Answer 2

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 :

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

Hence,

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

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.