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Suppose you put a drinking glass under water such that it is completely filled with water. Then hold it upside down and slowly raise the drinking glass out of the water. The water in the drinking glass will stay in the glass until the glass is almost completely above the water. This is because in the glass there is only water and no air such that atmospheric pressure causes the water to stay in the glass. However why isn't his the case anymore if the class is completely above the water surface?

If you do this with thin a straw (with is closed on one side) instead of a drinking glass, the water will remain inside. So it seems to depend on the area of the glass opening. Why is that the case?

How can one describe this difference with formulas?

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If someone has an graphical illustration of this experiment, it would be nice if he could add this to the question. – martin Mar 3 '12 at 14:58
Surface tension. – user2963 Mar 3 '12 at 15:27

To take the straw first, this is due to the surface tension of the water. If you pull the straw out of the water then look carefully at the open end of the straw you'll see that the air/water interface has a curved surface. Because there is a surface tension at the air water interface the water surface is acting like a balloon i.e. it's curvature creates a pressure that holds the water in the straw.

The pressure is inversely proportional to the curvature of the air/water interface i.e. the more curved the interface the greater the pressure. If you imagine gradually increasing the width of the straw the curvature of the air/water interface, and hence the pressure it can exert, gets less and eventually it's no longer enough to hold the water in the straw. That's why with a glass, which you can think of as a very wide straw, the water falls straight out.

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Thanks. What's the role of the amospheric pressure? How to put this into formulas? – martin Mar 3 '12 at 15:55

It comes from surface tension. The phenomenon you see in a straw is known as 'capillary action'.

For a thin tube with water, the height of rise is given by $\frac{2T}{r\rho g}$, $r$ is radius of the tube, $T$ is the water-air surface tension constant, and $\rho$ is density of water. This can be easily derived from the fact that pressure difference across a curved surface is $2T/r$ and for water, the radius of curvature of the surface ($r$) just happens to be the same as the radius of tube for small radii.

For a glass, this equation no longer applies, as the meniscus (the curve in the water surface while in a tube) is no longer hemispherical. Neither can we use the alternate derivation normally as different sections of water behave diffently. So, we have nearly no extra pressure decrease due to surface tension. One can look at it as $R$ becoming large in the above equation, which works qualitatively, but not quantitatively. Actually, that formula isn't too accurate for even a straw. It works well when the radius of tube is in millimeters.

As for why the water stays inside a glass till you remove it is because a near-vacuum will be formed otherwise. Vacuum formation is perfectly OK, but the vacuum exerts a greater force on the water than gravity, keeping it up. With a tube of mercury one can actually see a vacuum form due to its large density. With water the vacuum is negligible or nonexistent unless the column is pretty tall. We can look at this from pressures as well. The pressure above the inside water should be 0 (due to the vacuum). But, the pressure at normal water level is atmospheric pressure. So the inside water must rise a height given by $h\rho g=p_0$. If the glass isn't tall enough, then there will be absolutely no vacuum formed (if the water tries to go down, a vacuum will be formed, and it will go up again).

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For a deeper understanding of this topic you should look into Rayleigh–Taylor instability. Rayleigh–Taylor instability can happen when you have an interface between a heavier and lighter fluid in the presence of gravity or some other uniform force field gradient. Flow stability is a very important area of research in general.

As long as the cup is below the water you don't have any change in density. As soon as you take the cup out of the water, despite the fact that statically everything appears to be in balance, gravity and the lower density of air give you an inherently unstable interface.

As the previous answers have noted, surface tension can stabilize the interface. Intuitively you would expect that there will be an upper most diameter for which surface tension can stabilize the interface.

enter image description here

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