Due to surface tension, the excess pressure inside a smaller drop is $9$ units. If $27$ smaller drops combine, then the excess pressure inside the bigger drop is:

$A$ glass tube of uniform internal radius has a valve separating two identical ends. Initially,the valve is in a tightly closed position. End $1$ has a hemispherical soap bubble of radius $r$. End $2$ has a soap bubble with a radius of curvature $R$ $(R > r)$ as shown in the figure. Just after opening the valve,

$A$ cylinder with a movable piston contains air under a pressure $p_1$ and a soap bubble of radius $r$. The pressure $p_2$ to which the air should be compressed by slowly pushing the piston into the cylinder for the soap bubble to reduce its size by half will be: (The surface tension is $\sigma$, and the temperature $T$ is maintained constant)

What is the ratio of the number of moles of air in two soap bubbles of radii $R_1$ and $R_2$?

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

$64$ small drops,each of radius $r$ and charge $q$,coalesce to form a single large drop. If the charge is conserved,what is the ratio of the surface charge density of a small drop to that of the large drop?