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

The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:

If the excess pressure inside a soap bubble is balanced by an oil column of height $2\, mm$,then the surface tension of the soap solution will be $(r = 1\, cm$ and density $d = 0.8\, g/cm^3)$.

$A$ liquid drop of radius $R$ is broken into $n$ identical small droplets. The work done is $[T = \text{surface tension of the liquid}]$

Under isothermal conditions,two soap bubbles of radii $r_1$ and $r_2$ coalesce to form a single soap bubble of radius $R$. The radius of the new bubble is

An air bubble of radius $1 \ mm$ is at a depth of $8 \ cm$ below the free surface of a liquid column. If the surface tension and density of the liquid are $0.1 \ N \ m^{-1}$ and $2000 \ kg \ m^{-3}$ respectively,by what amount is the pressure inside the bubble greater than the atmospheric pressure (in $N \ m^{-2}$)? (Take $g = 10 \ m \ s^{-2}$)