Glass tube diameter to hold water when turned upside down If I have a glass tube that is closed at one end and has length l, how can I find the range of diameters that the tube must be so that the water does not fall out of the tube when the tube is turned upside down?
I think I need to do a force or energy balance, but I could use some help in approaching this problem. I know at such a small scale, gravity will not be as impactful as say, the surface tension.
the length, l, of the tube should be included in the solution
 A: This is harder than it sounds.
There are three things to consider.


*

*For a very narrow tube, for example $\lt 1~\rm{mm}$ diameter, the surface tension will be able to support a column of liquid - even if the other end of the tube is open. The pressure difference that can be sustained across a curved liquid surface with radius of curvature $R$ is calculated as the force on the circumference ($F = 2\pi R \gamma$) divided by the area $\pi R^2$, giving $\Delta p = \frac{2\gamma}{R}$. As the tube gets narrower, you will be able to sustain longer and longer columns of liquid.

*If you have a slightly larger tube, atmospheric pressure will be the significant term (this requires the other end of the tube to be closed...). In principle, you can support a column of 10 m of water at normal atmospheric pressure; in practice, the saturated vapor pressure of liquid means that the pressure above the liquid will not be 0, and there will be some temperature dependent correction. I discussed this in this earlier answer to a related question.

*There will come a size of tube where atmospheric pressure cannot hold the liquid in, because of instability of the liquid surface. The problem is that if the surface can distort, there will be parts at higher (internal) pressure than others. What this means is that when the tube gets large enough that surface tension no longer defines the shape of the surface, an air bubble will be able to "sneak past" the water surface, and liquid will pour out. This doesn't depend so much on the height of the liquid column, as on the diameter of the orifice.


I have never seen that calculated explicitly, but I believe the following argument should allow us to estimate the approximate diameter. We know the pressure difference across the surface due to surface tension scales with the inverse of the radius of curvature. If we assume, generously, that the surface curves "to the maximum", then when the differential gravitation-induced pressure becomes comparable to the surface-tension induced pressure difference, things become unstable. In other words,
$$\frac{2\gamma}{R}\approx \rho g R\\
R \approx \sqrt{\frac{2\gamma}{\rho g}}$$
Substituting typical values for water, we find that the critical diameter of a tube that won't hold water when turned upside down is 
$$R\approx \sqrt{\frac{2\cdot 0.07}{1000\cdot 10}}\approx 3.7 mm$$
This suggests that a tube with a diameter greater than about 7 mm will no be able to "hold" a column of water when turned upside down, which is certainly on the right order of magnitude. Note that in the above simplified analysis I ignored the effect of "contact angle" which reduces the pressure differential for hydrophilic surfaces (since the radius of curvature would increase to greater than the radius of the orifice). I think that's outside the scope of the question...
A: I am not exactly sure what you wish to show.
You can set up a water barometer in the same way as a mercury barometer is set up.
As it will be about 10 metres high is has to be made of thick wall plastic/rubber tubing with the final metre or so a piece of thick walled glass tubing sealed at the end.  This is so you can see the water meniscus at the top of the column of water.
You need to remove as much dissolved air from the water otherwise the air will come out of solution and prevent the low pressure region at the top being produced.  One way of doing this is to boil the water and then allow it to cool in a completely full sealed bottle/container.
