Pressure inside two soap bubbles are $1.01 \, atm$ and $1.02 \, atm$. The ratio between their volumes is:

Small drops of liquid of the same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is:

An air bubble of radius $1.0 \ mm$ is observed at a depth of $20 \ cm$ below the free surface of a liquid having surface tension $0.095 \ J/m^2$ and density $10^3 \ kg/m^3$. The difference between pressure inside the bubble and atmospheric pressure is . . . . . . $N/m^2$. (Take $g = 10 \ m/s^2$)

The surface tension of a soap bubble is $0.03 \text{ N/m}$. The work done in increasing the diameter of the bubble from $2 \text{ cm}$ to $6 \text{ cm}$ is $\alpha \times 10^{-4} \text{ J}$. The value of $\alpha$ is . . . . . . (Take $\pi = 3.14$).

When two soap bubbles of radii $4 \, cm$ and $5 \, cm$ coalesce,what is the radius of the common interface surface in $cm$?

There are two liquid drops of different radii. The excess pressure inside over the outside is