Water entering hole at a depth, surface tension The following is the question that very commonly appears in all HS textbooks.
A hollow sphere with a hole is taken to a depth of 40cm when the water starts entering the hole. if the surface tension of water is 70dyne per cm find the radius of the hole.
All the textbooks solve it in a similar manner as given below:-
  When water enters the sphere, radius of the bubble of air formed, will be equal to radius of the hole say r. Then excess pressure inside the bubble equals the external pressure at the depth of 40cm and thus on equating 2T/r to h*d*g, they find the answer.
but i dont find this explanation satisfying at all. why should a bubble form only at that depth and that too of the radius of hole. cant it form at any depth and the radius depend on the depth with maximum being equal to the radius of the hole.?
 A: Assuming I've understood what you're asking (ignore this if I haven't!) the point is that the radius of the hole is equal to the minimum radius of the air-water interface and therefore its maximum pressure.
If you place the sphere just under the surface of the water so the pressure difference is almost zero then the air-water interface forms a portion of a very large sphere with the centre positioned some distance away from the sphere. As you move the sphere farther down in the water the air water interface indents more and the air-water surface radius goes down, with its centre moving closer to the surface of the sphere. When the radius of the bubble equals the radius of the hole, the centre of the air-water interface is in the centre of the hole. Increasing the pressure further causes the radius of the air-water interface to grow again, which reduces the pressure, so the water floods in.
So the last paragraph of your question is quite correct. There is a bubble formed at all depths.
A: Bubble is formed even at less depth ,but the surface tension in the  surface is large enough to hold pressure till a particular depth.
you calculate for the extreme case : $2T/R =\rho g h$.
Here we neglect the the angle of contact of the liquid. 
