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.

If two glass plates have water between them and are separated by a very small distance (see figure),it is very difficult to pull them apart. This is because the water in between forms a cylindrical surface on the side that gives rise to a lower pressure in the water in comparison to the atmosphere. If the radius of the cylindrical surface is $R$ and the surface tension of water is $T$,then the pressure in the water between the plates is lower by:

The excess of pressure inside a soap bubble compared to the outer pressure is:

$A$ spherical soap bubble has internal pressure $P_0$ and radius $r_0$ and is in equilibrium in an enclosure with pressure $P_1 = \frac{8P_0}{9}$. The enclosure is gradually evacuated. Assuming temperature and surface tension of the soap bubble to be fixed, find the value of $\frac{\text{final radius}}{\text{initial radius}}$ of the soap bubble.

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:

When a big drop of water is formed from $n$ small drops of water,the energy loss is $3E$,where $E$ is the energy of the bigger drop. The radius of the bigger drop is $R$ and that of the smaller drop is $r$. Then the value of $n$ is: