There is a famous trick that goes as follows. Take a cup/glass that is partially filled with water. Place an index card over the top. Turn the entire thing upside-down (while keeping the index card at the cup opening), and then release the card. You should end up with the card "sticking" to the cup, keeping the water inside the cup, even though the cup is upside-down. Various posts about this have been made:
- About an upside down cup of water against atmosphere pressure
- Upside down water cup trick
- Why does the upside down water cup and card trick work?
I have a follow-up question to the questions above. I am trying to understand what role does the surface tension play in this trick (if any at all).
My question is, what would happen if we used a hypothetical liquid that had zero (or near zero) surface tension? There are two sub-questions I want to consider:
- Would the unbalanced pressure argument still hold? As far as I can tell, the answer is yes, so the card would still "stick" to the cup.
- On the other hand, wouldn't the liquid still leak out through the rim of the glass? It seems like it should since there is no surface tension keeping it in. But this seems to contradict the reasoning in #1 above. On top of all this, I have a hard time understanding what would happen to the air trapped at the top of the upside-down glass.
Can anyone help me think through this scenario more clearly. It seems paradoxical.
My tentative thought for now is that if a liquid had zero surface tension, it would be "infinitely porous" so the air molecules would simply penetrate through this substance. Basically, it seems like such a substance would just end up being a gas. I'm not sure if this is correct or if I am completely off, so I hope someone with better knowledge could chime in.
