If you have a very thin glass tube and you place it into water, let's say the water in the tube rises to the height of $x$ mm from the surface of the water. What would happen if you poked a hole in the tube at a height $n$ where $n<x$ and coated the sides of the hole with hydrophobic material so water could more easily slip through it. Wouldn't the water just cascade outward? And then the capillary action of the water would pull up more water, causing, in my mind, a perpetual motion machine? Energy conservation is raising its eyebrows at me now but doesn't offer me any help. Please evaluate this scenario and show me why it can't happen.

  • $\begingroup$ The second law is the definition of temperature and you didn't mention temperature anywhere in your question, so obviously it can't apply. What does apply is simple energy conservation. $\endgroup$ – CuriousOne Jun 8 '15 at 15:39
  • $\begingroup$ @CuriousOne :P My bad again, I'll edit that. $\endgroup$ – HyperLuminal Jun 8 '15 at 15:41
  • $\begingroup$ @HyperLuminal : More on capillary-action and perpetual motion. $\endgroup$ – Qmechanic Jun 8 '15 at 16:05
  • $\begingroup$ @Qmechanic The link leads nowhere? $\endgroup$ – HyperLuminal Jun 8 '15 at 16:08
  • $\begingroup$ physics.stackexchange.com/q/89223/8851 "How is the water meniscus at the edge of a capillary tube", physics.stackexchange.com/q/88544/8851 "Capillary tube of insufficient length" may be more relevant. It deals not with the case of a hole on the tube, but with a shorter tube, but that's close enough. $\endgroup$ – b_jonas Jun 8 '15 at 16:56

As soon as you make a hole it will happens one of both: If the diameter of the hole is sufficiently small, then the combination of surface tension (which is caused by cohesion within the liquid) and adhesive forces between the liquid and container act to seal the hole. If the hole is large enough so that water can leak, then the height of the water column will lower to equal the pressures. Remember that the pressure inside the water column in larger that the atmospheric pressure.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.