In the formula for the mass of a drop of water on the end of a tube, is r internal or external radius? In the formula for the mass $m$ of a drop of water forming on the end of a vertical capillary tube,
$$m = \frac{2\pi r \gamma}{g},$$
does $r$ refer to the internal or external radius of the capillary tube?

In the picture, the black part is meant to be the capillary tube, filled with water with a drop forming at the bottom.
 A: r is the internal radius. It is clear if you look at the wikipedia page.
Think of it like this: if the tube was 5 meters thick, would that affect the drop of water at all? The answer is no: the water droplet is affected only by the diameter of the tube that is in contact with -> i.e the internal radius
A: r would be the internal radius. After all, is the thickness of the tube ever provided?
A: $mg=2πrγ$
That is the formula for the capillary force. You get this value by multiplying the surface tension $\gamma$ (force per unit length) by the length under tension. For a meniscus inside a capillary, this length is the circumference of that meniscus (which is at the top of the tube not shown in the diagram). Clearly the diameter of the meniscus is the inner diameter of the tube. This force keeps the exposed water droplet hanging, and so is equivalent to the weight of that drop. The mass of that drop should be straightforward to get if you know the density of water.
A: Strictly speaking $r$ doesn't refer to the tube at all. It is the radius of the 'base' of the droplet (the base being the circular area where the droplet is attached to the capillary tube).
For a well designed tube the rim of the tube is hydrophobic, it repels water, causing the droplet to detach from the rim while hanging, thus having a base equal to the internal tube. In that case $r$ is indeed equal to the internal tube radius.
However, for a rim of the tube that is wetted by water (like the one you've drawn in your question) the $r$ can basically be anything between the internal radius of the tube and the external radius of the tube, as determined by the contact angle of the water with the rim. Indeed, as mentioned by @PhotonicBoom, for a tube with a very thick wall it will never be the external radius, but for a thin tube it is certainly possible.
