The work done to break a spherical drop of radius $R$ into $n$ drops of equal size is proportional to .............

Assume that a drop of liquid evaporates by a decrease in its surface energy,such that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is $T$,the density of the liquid is $\rho$,and $L$ is its latent heat of vaporization.

$A$ large number of water drops,each of radius $r$,combine to form a single drop of radius $R$. If the surface tension is $T$ and the mechanical equivalent of heat is $J$,the rise in heat energy per unit volume will be:

$A$ soap bubble, blown by a mechanical pump at the mouth of a tube, increases in volume with time at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by

Work done in splitting a drop of water of $1 \, mm$ radius into $10^6$ droplets is (Surface tension of water $= 72 \times 10^{-3} \, J/m^2$).

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