When two soap bubbles of radius $r_1$ and $r_2$ $(r_2 > r_1)$ coalesce,the radius of curvature of the common surface is:

$A$ drop of water of volume $0.05\, cm^3$ is pressed between two glass plates,as a consequence of which it spreads and occupies an area of $40\, cm^2$. If the surface tension of water is $70\, dyne/cm$,then the normal force required to separate out the two glass plates will be in Newton.

The excess pressure in a soap bubble is thrice that in another one. Then the ratio of their volume is

$A$ capillary tube of radius $r$ is dipped in a liquid of density $\rho$ and surface tension $S$. If the angle of contact is $\theta$,what is the pressure difference between the two surfaces in the beaker and the capillary?

There is an air bubble of radius $1.0\,mm$ in a liquid of surface tension $0.075\,N\,m^{-1}$ and density $1000\,kg\,m^{-3}$ at a depth of $10\,cm$ below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is $....Pa \left( g = 10\,m\,s^{-2} \right)$

There is a small hole in a hollow sphere. Water enters the sphere when it is taken to a depth of $40 \, cm$ in water. The diameter of the hole is ....... $mm$ (Surface tension of water $= 0.07 \, N/m$):