When two soap bubbles of radii $a$ and $b$ $(b > a)$ coalesce,the radius of curvature of the common surface is:

$A$ glass capillary tube of inner diameter $0.28 \ mm$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that the water level in the tube is the same as that in the vessel (in $N/m^2$) is:
Surface tension of water $= 0.07 \ N/m$
Atmospheric pressure $= 10^5 \ N/m^2$

The pressure inside a small air bubble of radius $0.1 \, mm$ situated just below the surface of water will be equal to. [Take surface tension of water $T = 70 \times 10^{-3} \, N/m$ and atmospheric pressure $P_0 = 1.013 \times 10^5 \, N/m^2$]

The radii of two mercury drops are $R_1$ and $R_2$. Under isothermal conditions,a single drop of radius $R$ is formed from them. The relation between $R, R_1$ and $R_2$ 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?

Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to: