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 area 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 in newton.

Answer

45

Explanation

We have discussed above,
`F=(2AT)/(d) = (2A^(2)T)/(Ad)`
But, `Ad =` volume
`:. F = (2A^(2)T)/(V)`
Substituting the values we get,
`F =(2xx(40xx10^(-4))^(2)xx(70xx10^(-3)))/(0.05 xx 10^(-6))`
`= 45 N`