A mercury drop shaped as a round tablet of radius `R` and and thickness `h` is located between two horizontal glass plates. Assuming that `h lt lt R`, find the mass `m` of a weight which has to be placed on the upper plate to diminish the distance between the plates n-times. The contact angle equals `theta`. Calculate `m` if `R = 2.0 cm, h = 0.38 mm, n = 2.0`, and `theta = 135^@`.

Explanation

We must first calculate the pressure difference inside the film from that outside. This is
`p = alpha((1)/(r_1) + (1)/(r_2))`
Here `2 r_1|cos theta| = h` and `r_2 ~ - R` the radius of the tablet and can be neglected. Thus the total force exerted by mercury drop on the upper glass plate is
`(2 pi R^2 alpha|cos theta|)/(h)` typically
We should put `h//n` for `h` because the tablet is compresed `n` times. Then since `Hg` is nearly,
incompressible, `pi R^2 h = `constant so` R rarr R sqrt(n)`. Thus,
total force = `(2 pi R^2 alpha |cos theta|)/(h) n^2`
Part of the force is needed to keep the `Hg` in the shape of a table rather than in the shape of infinitely thin sheet. This part can be calculated being putting `n = 1` above. Thus
`mg + (2 pi R^2 alpha|cos theta|)/(h) = (2 pi R^2 alpha|cos theta|)/(h) n^2`
or `m = (2 pi R^2 alpha|cos theta|)/(hg) (n^2 - 1)= 0.7 kg`.
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