The excess pressure due to surface tension in a spherical liquid drop of radius $r$ is directly proportional to

Two small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is

Two small drops of mercury,each of radius $R$,coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

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 $....Pa \left( g = 10\,m\,s^{-2} \right)$

When two soap bubbles of radii $4 \, cm$ and $5 \, cm$ coalesce,what is the radius of the common interface surface in $cm$?

An air bubble of radius $0.1\, cm$ is in a liquid having surface tension $0.06\, N/m$ and density $10^3\, kg/m^3$. The pressure inside the bubble is $1100\, N/m^2$ greater than the atmospheric pressure. At what depth $h$ (in $m$) is the bubble below the surface of the liquid? $(g = 9.8\, m/s^2)$