Assume I have a inverse cone which holds 200ml water. I am going to cut the tip of the cone to create a small hole. How to calculate the maximum radius of the hole that the water will still stay in the container?

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    $\begingroup$ It is going to be surface tension, not viscosity, that may keep your water from flowing out. $\endgroup$ – Jaime Dec 16 '12 at 23:28
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    $\begingroup$ Indeed: viscosity only affects the rate at which it may flow if the surface tension allows it. $\endgroup$ – user10851 Dec 17 '12 at 0:16
  • $\begingroup$ And the amount of water that the surface tension can hold depends on the depth of the water. If the hole in your cone is small enough to hold the water above it, adding more water will cause it to come through the hole. $\endgroup$ – Paul J. Gans Dec 17 '12 at 1:30

If you have a water drop with radius $r$ then the pressure difference between the inside of the drop and the outside is:

$$ \Delta P = \frac{2\gamma}{r} $$

To calculate the hole size you need to work out the pressure at the bottom of the cone and equate this to the pressure calculated using the expression above. The pressure at the bottom of the cone depends on the depth of the water, not the total volume of water in the cone. If the depth of water in the cone is $h$ then the pressure is $\rho g h$, where $\rho$ is the density of the water at the temperature you're working at, and $g$ is the acceleration due to gravity ($\approx$ 9.81 m/sec$^2$). Equating this to the first expression gives:

$$ \rho g h = \frac{2\gamma}{r} $$


$$ r = \frac{2\gamma}{\rho g h} $$

For example at STP $\gamma \approx 7.3 \times 10^{-2}$N/m and $\rho \approx$ 1000kg/m$^3$, so if the depth of the water in your cone is 10cm the maximum radius of the hole is 0.1mm.

Note that this is the maximum radius for which there is no flow at all. For holes a bit bigger than this the flow may be so slow it's difficult to measure.

  • $\begingroup$ I would like to ask, how did you determine radius of curvature of drop?, I mean how to prove mathematically that it's the maxima/minima... and weight of drop was not included? $\endgroup$ – Mrigank Jul 3 '16 at 17:01
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    $\begingroup$ @ELiT: ignoring the weight of the drop is an approximation, but in most circumstances it's a very good approximation. $\endgroup$ – John Rennie Jul 3 '16 at 17:05
  • $\begingroup$ about radius of curvature? if we are ignoring weight, then radius of curvature can literally be anything so, it can tend zero and height may be infinite... $\endgroup$ – Mrigank Jul 3 '16 at 17:21
  • $\begingroup$ @ELiT: I'm not sure what you're asking about the radius of curvature. The highest pressure within the forming droplet is when the droplet is a hemisphere with radius equal to the hole radius. My calculation is to equate this maximum pressure with the pressure at the bottom of the cone. $\endgroup$ – John Rennie Jul 3 '16 at 17:23
  • $\begingroup$ Sorry.... just got it, to fit in the hole, radius of curvature in laplace equation must be more than or equal to radius of hole.. $\endgroup$ – Mrigank Jul 3 '16 at 17:46

protected by AccidentalFourierTransform Aug 19 '18 at 15:29


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