Maximum hole size to stop a fluid passing through a solid In a previous question, the following is answered in a general sense:

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 ?

However, what is not clear to me is whether the material of the vessel matters. The surface tension of water IN AIR is used in the answer $\left(\gamma \approx 7.3{\times}{10}^{-2}\,\frac{\mathrm{N}}{\mathrm{m}}\right)$; however, I would think the surface tension and contact angle between the water and the cup have an effect. 
For example, if I have a one cone cup that is hydrophobic and one cone cup that is hydrophilic, will the hole size necessary to stop the water dripping out be the exact same?
 A: By my understanding overcoming the Laplace pressure $\delta p$ is the only that matters if not disturbed.
$$\delta p = \gamma \cdot \left(\frac{1}{R_x} + \frac{1}{R_y} \right)$$  Where Rx and Ry are radii for a droplet in x,y direction $\gamma$ is surface tension for a liquid at specified circumstances (e.g. temperature). For water at approximately 20 Celsius degrees it is $0.075 \frac{\mathrm{N}}{\mathrm{m}}$.
The pressure build up  $\rho g h$ and $\frac{1}{R} + \frac{1}{R} =$ diameter for round droplets (and hole)
$ p > \delta p$ make the droplet grow bigger than $d$ on outside the hole 
$ \rho g h > \gamma \frac{1}{d} \Rightarrow \rho g h d > \gamma$ make droplet to fall.
Example: Water $\gamma = 0.075 \frac{\mathrm{N}}{\mathrm{m}}$ hole $d = 0.1 \, \mathrm{mm} = 0.0001 \, \mathrm{m}$ in hydrophobic material$$
\Rightarrow h > \frac{\gamma}{\rho g d} = \frac{0.075 \frac{\mathrm{N}}{\mathrm{m}}}{1000 \frac{\mathrm{kg}}{{\mathrm{m}}^3} 9.81 \frac{\mathrm{m}}{{\mathrm{s}}^2} 0.0001 \, \mathrm{m}} = 0.076 \, \mathrm{m}
\,.$$This means that at water level $7.6 \, \mathrm{cm} \, \left(\approx 3"\right)$ and upwards the drops will fall though a $0.1 \, \mathrm{mm}$ hole in hydrophobic material. Hydrophilic materials make droplets to the outer edge of material and the diameter is measured from there. This means that a hole on a hydrophilic surface will leak through very small holes. Tested with an approximately 1 mm hole seems to be a close estimation.
A: The vessel do matters. But you can only loose. If you look the capillarity from Wikipedia, you can notice that the contact angle has an influence in the height of a liquid column. But the equation reveals you, that the influence is $\cos\left(\theta_{\text{contact}}\right)$, and it's a plain multiplier. $\cos\left(0\right)$ is the maximum. 
You will have an effect if you use various materials; you will need even a smaller hole. The other thing which this reveals, is that you need a straight pipe. If you cut the tip of a cone, you actually need a smaller hole than a straight pipe would require. 
A: Could you please give the radius, $R$, of the cone?
Assuming it is given,$\hspace{100px}$,as shown in figure, h could be found from volume of the cone containing liquid. 
$$\frac{1}{3}\pi R^2 h = 200\,\mathrm{mL} ~ \to 2 \times {10} ^{-4} \, \mathrm{m}^{3}$$
So,$$
h=\frac{3 \times 2 \times {10}^{-4}}{\pi R^2} \qquad
\rho = 1000 \frac{\mathrm{kg}}{\mathrm{m}^3} \qquad
g = 9.81 \frac{\mathrm{m}}{\mathrm{s}^{2}}
\,,$$and$$
p = \rho g h ~ = ~1000*9.81*h~=~ \frac{1000 \times 9.81 \times~3~\times~2 \times {10}^{-4}}{\pi R^2}
\,,$$then plug  $p$ and $\gamma$ value in force balance equation,$$
\gamma * 2 \pi r  = ~ \frac{1000\times9.81\times~3~\times~2 \times {10}^{-4}  ~\times~\pi r^2}{\pi R^2}
\,,$$which gives the criteria$$
r  ~ \le ~  0.078 R^2
\,,$$or,$$
\frac{r}{R^2}~ \le ~ 0.078 \, {\mathrm{m}}^{-1}
\,.$$
Assumptions:


*

*droplet is half-sphere;

*volume is neglected;

*applicable only for $200 \, \mathrm{mL}$ of water.
