There are two liquid drops of different radii. The excess pressure inside over the outside is

The excess pressure inside a spherical soap bubble of radius $1 \,cm$ is balanced by a column of oil (specific gravity $= 0.8$), $2 \,mm$ high. The surface tension of the bubble is: (in $\,N/m$)

$A$ spherical soap bubble has internal pressure $P_0$ and radius $r_0$ and is in equilibrium in an enclosure with pressure $P_1 = \frac{8P_0}{9}$. The enclosure is gradually evacuated. Assuming temperature and surface tension of the soap bubble to be fixed, find the value of $\frac{\text{final radius}}{\text{initial radius}}$ of the soap bubble.

$Assertion :$ Smaller drops of liquid resist deforming forces better than the larger drops.
$Reason :$ Excess pressure inside a drop is directly proportional to its surface area.

If $W$ amount of work is required to form a bubble of volume $V$,then how much work is required to form a bubble of volume $2V$?

Work done in splitting a drop of water of $1 \, mm$ radius into $10^6$ droplets is (Surface tension of water $= 72 \times 10^{-3} \, J/m^2$).