A drop of water volume 0.05 ${\text{cm}}^3$ is pressed between two glass-plates, as a consequence of which, it spreads between the plates. The area of contact with each plate is ${\text{40\hspace{0.17em}cm}}^{\text{2}}$. If the surface tension of water is 70 dyne/cm, the minimum normal force required to separate out the two glass plates in newton is approximately (assuming angle of contact is zero):

Pressure inside the film is less than outside by an amount, $P=T[\frac{1}{r_1}+\frac{1}{r_2}]$ , where $r_{1}and{r}_2$ are the radii of curvature of the meniscus. Here $r_1=t/2and{r}_2=\infty$, then the force required to separate the two glass plates, between which a liquid film is enclosed (figure) is,

$F=P\times A=\frac{2AT}{t}$

Where t is the thickness of the film, A = area of film.

$F=\frac{2A^{2}T}{At}=\frac{2A^{2}T}{V}=\frac{2\times {(40\times {10}^{-4})}^2\times (70\times {10}^{-3})}{0.05\times {10}^{-6}}=45N$