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 $\mu m$)

$64$ small mercury droplets,each of radius $r$ and charge $q$,coalesce to form a single large drop. What is the ratio of the surface charge density of a small droplet to that of the large drop?

$1000$ droplets of water,each having a diameter of $2 \ mm$,coalesce to form a single drop. Given the surface tension of water is $0.072 \ N/m$. The energy loss in the process is

The radius of a soap bubble is $r$ and the surface tension of the soap solution is $S$. The electric potential to which the soap bubble must be raised by charging it so that the pressure inside the bubble becomes equal to the pressure outside the bubble is $(\varepsilon_0 = \text{permittivity of the free space})$

The excess pressure in a soap bubble is thrice that in another one. Then the ratio of their volume is

If the work done in blowing a bubble of volume $V$ is $W$,then the work done in blowing the bubble of volume $2V$ from the same soap solution will be