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

When $10^6$ small drops coalesce to make a new larger drop,then the drop:

The surface tension of a soap solution is $2 \times 10^{-2} \ N/m$. To blow a bubble of radius $1 \ cm$,the work done is:

$A$ water drop of radius '$r$' and volume '$V$' is kept between two identical glass plates such that it forms a thin layer of area '$A$' between the plates. $A$ force '$F$' is applied such that the two plates separate from each other. The surface tension '$T$' of the liquid is:

The lower end of a capillary tube of diameter $2.00 \; mm$ is dipped $8.00 \; cm$ below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at the temperature of the experiment is $7.30 \times 10^{-2} \; N m^{-1}$. Atmospheric pressure $= 1.01 \times 10^{5} \; Pa$,density of water $= 1000 \; kg m^{-3}$,$g = 9.80 \; m s^{-2}$. Also,calculate the excess pressure.

An air bubble of radius $1.0 \ mm$ is observed at a depth of $20 \ cm$ below the free surface of a liquid having surface tension $0.095 \ J/m^2$ and density $10^3 \ kg/m^3$. The difference between pressure inside the bubble and atmospheric pressure is . . . . . . $N/m^2$. (Take $g = 10 \ m/s^2$)