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 menicus. 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 is
`F=PxxA=(2AT)/t`
where `t` is the thickness of the film `A =` area of film.

`F=(2A^(2)T)/(At)=(2A^(2)T)/v=(2xx(40xx10^(-4))xx(70xx10^(-3)))/(0.05xx10^(-6))`
`=45N`