$n$ number of liquid drops each of radius $r$ coalesce to form a single drop of radius $R$. The energy released in the process is converted into the kinetic energy of the big drop so formed. The speed of the big drop is [$T = \text{surface tension of liquid}, \rho = \text{density of liquid}$.]

The spherical shape of a rain-drop is due to

The surface tension of a liquid at its boiling point

The property of surface tension is obtained in

$A$ thread is tied slightly loose to a wire frame as shown in the figure. The frame is dipped into a soap solution and taken out,such that the frame is completely covered with a soap film. When the portion $A$ is punctured with a pin,the thread:

Surface tension is exhibited by liquids due to the force of attraction between the molecules of the liquid. The surface tension decreases with an increase in temperature and vanishes at the boiling point. Given that the latent heat of vaporization for water $L_v = 540 \text{ kcal/kg}$,the mechanical equivalent of heat $J = 4.2 \text{ J/cal}$,density of water $\rho_w = 10^3 \text{ kg/m}^3$,Avogadro's number $N_A = 6.0 \times 10^{26} \text{ molecules/kmol}$,and the molecular weight of water $M_A = 18 \text{ kg/kmol}$.
$(a)$ Estimate the energy required for one molecule of water to evaporate.
$(b)$ Show that the intermolecular distance for water is $d = \left( \frac{M_A}{N_A \rho_w} \right)^{1/3}$ and find its value.
$(c)$ $1 \text{ g}$ of water in the vapour state at $1 \text{ atm}$ occupies $1601 \text{ cm}^3$. Estimate the intermolecular distance at the boiling point in the vapour state.
$(d)$ During vaporisation,a molecule overcomes a force $F$,assumed constant,to go from an intermolecular distance $d$ to $d'$. Estimate the value of $F$.
$(e)$ Calculate $\frac{F}{d}$,which is a measure of the surface tension.