How to find out the maximum radius of a hole that can keep water stay in a container by water viscosity? 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?
 A: 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:
$$ \rho g h = \frac{2\gamma}{r} $$
or:
$$ r = \frac{2\gamma}{\rho g h} $$
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.
