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 are of `40 cm^(2)`. If the surface tension of water is `70 "dyne"//cm`, find the normal force required to separate out the two glass plates is newton.

Explanation

Pressure inside the film is less than outside by an amount, `P = T [ ( 1)/( r_(1)) + ( 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) = oo`, then the force required to separate the two glass plates, between which a liquid film is enclosed ( figure ) is , `F = P xx A = ( 2AT)/( t) `, where t is the thickness of the film, A = area of film.

`F= ( 2A^(2) T)/( At ) = ( 2A^(2) T )/( V)`
`= ( 2 xx ( 40 xx 10^(-4))^(2)xx ( 70- xx 10^(-3)))/(0.05 xx 10^(-6))= 45N`