Due to surface tension, the excess pressure inside a smaller drop is $9 \text{ units}$. If $27$ smaller drops combine, then the excess pressure inside the bigger drop is: (in $\text{ units}$)

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:

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

The radius of a soap bubble is $r$ and the surface tension of the soap solution is $S$. The electric potential to which the soap bubble must be raised by charging it so that the pressure inside the bubble becomes equal to the pressure outside the bubble is $(\varepsilon_0 = \text{permittivity of the free space})$

Two soap bubbles of radius $2 \ cm$ and $4 \ cm$,respectively,are in contact with each other. The radius of curvature of the common surface,in $cm$,is . . . . . . .

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