Let $R_1, R_2$ and $R_3$ be the radii of three mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

When a capillary tube is dipped into a tub containing mercury, the mercury level in the capillary goes down because the pressure just below the meniscus is

One end of a uniform glass capillary tube of radius $r = 0.025 \ cm$ is immersed vertically in water to a depth $h = 1 \ cm$. The excess pressure in $N/m^2$ required to blow an air bubble out of the tube is: (Surface tension of water $T = 7 \times 10^{-2} \ N/m$,Density of water $\rho = 10^3 \ kg/m^3$,Acceleration due to gravity $g = 10 \ m/s^2$)

Two small drops of mercury,each of radius $R$,coalesce to form a single large drop. The ratio of the total surface energies before and after the change is:

Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :

Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes is (in $ : 1$)