$A$ big drop of radius $R$ is formed by $729$ small drops of water of radius $r$. Then the radius of each small drop will be:

The pressures inside two soap bubbles $A$ and $B$ are $1.02 \text{ atm}$ and $1.04 \text{ atm}$ respectively. The ratio of the volume of bubble $A$ to that of bubble $B$ is (outside pressure = $1 \text{ atm}$)

What should be the diameter of a soap bubble,in order that the excess pressure inside it is $25.6 \ Nm^{-2}$ (in $cm$)? [surface tension of soap solution $= 3.2 \times 10^{-2} \ Nm^{-1}$]

The pressure inside a soap bubble $A$ is $1.01 \text{ atm}$ and that in a soap bubble $B$ is $1.02 \text{ atm}$. The ratio of the volume of bubble $A$ to that of $B$ is (Surrounding pressure $= 1 \text{ atm}$)

$A$ certain number of spherical liquid drops of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid,which one of the following statements is true for the energy $(E)$ in the process?

Two drops of equal radius coalesce to form a bigger drop. What is the ratio of the surface energy of the bigger drop to that of a smaller one?