A drop of water volume 0.05 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 40 cm 2 . If the surface tension

Question

A drop of water volume 0.05 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 40 cm 2 . If the surface tension of water is 70 dyne / cm , the minimum normal force required to seperate out the two glass plates in newton is approximately (assuming angle of contact is zero) :

TestHub · Accepted Answer

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 $r_{1} = t /2$ and $r_{2} = \infty$ , then the force required to separate the two glass-plates, between which a liquid film is enclosed (figure) is,

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

where t is the thickness of the film, $A =$ area of film.

$F = \frac{2 A ^{2} T}{At} = \frac{2 A ^{2} T}{V} = \frac{2 \times ( 40 \times 1 0 ^{- 4} ) ^{2} \times ( 70 \times 1 0 ^{- 3} )}{0.05 \times 1 0 ^{- 6}} = 45 N$

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