Determining the height of fluids in a U-shaped tube? A narrow U-shaped glass tube with open ends is filled with 25.0 cm of oil (of specific gravity 0.800) and 25.0 cm of water on opposite sides, with a barrier separating the liquids.

How would one find the final height of the fluid columns on the left and right sides f the tube after the barrier is removed, assuming they don't mix?
I also don't know what the final heights a and b would be if the water and oil were the same density, and if the oils density was significantly less than the waters, using physical reasoning as opposed to calculations.
 A: Removing the barrier altogether seems to make for a really difficult problem because some (unknown quantity) of the oil could rise through the water column and settle at the top.
Let's say we keep the barrier and allow it to move freely. (Is that what is meant with "they don't mix"?)
Then the problem is down to determining a force-free equilibrium at the barrier, which is determined by the hydrostatic pressures in each fluid column. That is, on each side the pressure at the barrier will be the atmospheric pressure at the free surface plus $\rho g h$ with material density $\rho$, height difference $h$ between free surface and barrier, and gravitational acceleration $g$.
If the density of both materials were equal, the equilibrium would be at equal surface heights.
If the density of oil were negligibly small, then the hydrostatic pressure difference through the oil column almost vanishes and the pressure at the barrier has to be identical to atmospheric pressure. I.e. the water would be at the bottom of the U, rising to almost equal heights on both sides, and on one side there would be 25cm of oil on top.
For all cases in between we need some more info on the problem geometry (e.g. the curvature radius of the U) and the do some math.
A: Obviously if densities of liquids are the same, then there will be no difference in heights of fluids in the tube. Otherwise pressures of liquids must come to equilibrium at contact point, as per :
$$ \rho_1 g h_1 = \rho_2 g h_2  $$
From there solve the ratio ${h_1}/{h_2}$ and assuming that lower density liquid height will be $h_1 = h_2 + \Delta h$, you
can extrapolate tube liquid heights difference $\Delta h$.
EDIT
You may need yet another (last) assumption. Assuming liquids are incompressible, and that there must be satisfied mass conservation, then $h_2 = h_0 - \Delta h/2$, where $h_0$ is initial liquid height before barrier removal. This is due to the fact that both liquids must shift from initial position by same amount of displacement, due to mass conservation.
So having these solve for $\Delta h$,- height difference between liquids, then substitute it for finding column height $h_2$ and use that for finding column height $h_1$. You are ready to go now !
