Small liquid drops assume a spherical shape because

The surface tension of a liquid is $70 \, dyne/cm$. In $MKS$ system its value is

When water is filled carefully in a glass,one can fill it to a height $h$ above the rim of the glass due to the surface tension of water. To calculate $h$ just before water starts flowing,model the shape of the water above the rim as a disc of thickness $h$ having semicircular edges,as shown schematically in the figure. When the pressure of water at the bottom of this disc exceeds what can be withstood due to the surface tension,the water surface breaks near the rim and water starts flowing from there. If the density of water,its surface tension and the acceleration due to gravity are $10^3 \ kg \ m^{-3}$,$0.07 \ N \ m^{-1}$ and $10 \ m \ s^{-2}$,respectively,the value of $h$ (in $mm$) is:

$A$ metal wire of density $\rho$ floats on the water surface horizontally. If it is not to sink in water,then the maximum radius of the wire is ($T$ = surface tension of water,$g$ = gravitational acceleration).

Consider a spherical drop of radius $R$. The surface tension of the liquid is $\sigma$. The force of surface tension on the shaded sub-hemisphere due to the remaining drop is $0.5\,\pi \sigma R$. The value of the angle $\theta$ (angle subtended by the sub-hemisphere at the center of the drop) is ......$^o$.

If a drop of liquid breaks into smaller droplets,it results in the lowering of the temperature of the droplets. Let a drop of radius $R$ break into $N$ small droplets,each of radius $r$. Estimate the drop in temperature.