If the surface tension of a soap solution is $0.03 \, N/m$,then the excess pressure inside a soap bubble of diameter $6 \, mm$ over the atmospheric pressure will be:

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

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 $g = 10 \ m \ s^{-2}$)

If the work done in blowing a bubble of volume $V$ is $W$,then the work done in blowing the bubble of volume $2V$ from the same soap solution will be

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is . . . . . . . ($S$ is the surface tension of mercury). (in $\pi r^2 S$)

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