A drop of water of volume 0.05 $c m^{3}$ is pressed between two glass plates. as a consequence of which it spreads and occupies an area of 40 $c m^{2}$ . If the surface tension of water is 70 dyne/cm, find the normal force required to separate out the two glass plates in newton.

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Solution

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 $r_{2}$ and the radii of curvature of the meniscus. Here $r_{1} = t / 2$ and $r_{2} = \infty$ , then the force required to separate the two glass plates, between which the liquid film is enclosed, is
$F = P \times A = \frac{2 A T}{t}$
Where t is the thickness of the film, A = area of film
$r = t / 2$
$F = \frac{2 A^{2} T}{A t} = \frac{2 A^{2} T}{V} = \frac{2 \times (40 \times 10^{- 4})^{2} \times (70 \times 10^{- 3})}{0.05 \times 10^{- 6}} = 45 N$

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