Figure $(a)$ shows a thin liquid film supporting a small weight $= 4.5 \times 10^{-2} \, N$. What is the weight supported by a film of the same liquid at the same temperature in Figure $(b)$ and $(c)$? Explain your answer physically.

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.

$A$ soap film of surface tension $3 \times 10^{-2} \, N/m$ is formed in a rectangular frame and can support a straw. The length of the film is $10 \, cm$. The mass of the straw the film can support is ........ $g$.

The force required to take away a flat circular plate of radius $2 \, cm$ from the surface of water will be (the surface tension of water is $70 \, dyne/cm$).

It is easier to spray water to which soap is added because the addition of soap to water:

Hairs of a shaving brush cling together when it is removed from water due to