$A$ water drop of $0.01 \ cm^3$ is squeezed between two glass plates and spreads into an area of $10 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,then the normal force required to separate the glass plates from each other will be: (in $N$)

The pressure inside a small air bubble of radius $0.1 \, mm$ situated just below the surface of water will be equal to. [Take surface tension of water $T = 70 \times 10^{-3} \, N/m$ and atmospheric pressure $P_0 = 1.013 \times 10^5 \, N/m^2$]

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

Two drops of equal radius coalesce to form a bigger drop. What is the ratio of the surface energy of the bigger drop to that of a smaller one?

$A$ water drop is divided into $8$ equal droplets. The pressure difference between the inner and outer side of the big drop will be

If two glass plates have water between them and are separated by a very small distance (see figure),it is very difficult to pull them apart. This is because the water in between forms a cylindrical surface on the side that gives rise to a lower pressure in the water in comparison to the atmosphere. If the radius of the cylindrical surface is $R$ and the surface tension of water is $T$,then the pressure in the water between the plates is lower by: