The excess pressure inside one soap bubble is four times that of another. What is the ratio of the volume of the first bubble to the second bubble?

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T=$ Surface tension of water). (in $\pi TR^2$)

$A$ drop of water with a volume of $0.05 \ cm^3$ is pressed between two glass plates,causing it to spread between the plates. The area of contact with each plate is $40 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,the minimum normal force required to separate the two glass plates in newtons is approximately: (assuming the angle of contact is zero)

$8$ mercury drops coalesce to form one mercury drop. The energy changes by a factor of:

$A$ drop of water of volume $0.05\, cm^3$ is pressed between two glass plates,as a consequence of which it spreads and occupies an area of $40\, cm^2$. If the surface tension of water is $70\, dyne/cm$,then the normal force required to separate out the two glass plates will be in Newton.

$A$ liquid drop of diameter $D$ breaks into $27$ small drops of equal size. If the surface tension of the liquid is $\sigma$,then the change in surface energy is