$A$ 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.

An air bubble of radius $r$ in water is at depth $h$ below the water surface. If $P$ is the atmospheric pressure, and $d$ and $T$ are the density and surface tension of water respectively, then the pressure inside the bubble will be:

$A$ capillary tube of radius $r$ is dipped in a liquid of density $\rho$ and surface tension $S$. If the angle of contact is $\theta$,what is the pressure difference between the two surfaces in the beaker and the capillary?

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

One large soap bubble of diameter '$D$' breaks into $64$ smaller bubbles of equal size. If the surface tension of the soap solution is '$T$',what is the change in surface energy (in $\pi T D^2$)?

One end of a uniform glass capillary tube of radius $r = 0.025 \ cm$ is immersed vertically in water to a depth $h = 1 \ cm$. The excess pressure in $N/m^2$ required to blow an air bubble out of the tube is: (Surface tension of water $T = 7 \times 10^{-2} \ N/m$, Density of water $\rho = 10^3 \ kg/m^3$, Acceleration due to gravity $g = 10 \ m/s^2$)