Two soap bubbles with radii $r_1$ and $r_2$ $(r_1 > r_2)$ come in contact. Their common surface has a radius of curvature $r$. Find $r$.

The surface tension and vapour pressure of a liquid at $25^{\circ} C$ are $8 \times 10^{-2} \text{ Nm}^{-1}$ and $2.5 \times 10^3 \text{ Pa}$ respectively. The radius of the smallest spherical water droplet which can form without evaporating at $25^{\circ} C$ is (in $\mu m$)

Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension,the surface energy released in this process is

$A$ spherical drop of water has a radius of $1\, mm$. If the surface tension of water is $70 \times 10^{-3}\, N/m$,the difference of pressure between the inside and outside of the spherical drop is ........ $N/m^2$.

What should be the radius of a water drop so that the excess pressure inside it is $72 \ Nm^{-2}$ (in $mm$)? (The surface tension of water is $7.2 \times 10^{-2} \ Nm^{-1}$)

If the internal pressures of two soap bubbles are $1.01 \text{ atm}$ and $1.02 \text{ atm}$ respectively, find the ratio of their volumes.