A mercury drop shape as a round tablet of radius R and thickness h is located between two horizontal glass plates. Assuming that hltltR, find the mass m of a weight which has to be placed on the copper plate to diminish the distance between the plates by n-times the contact angle is equal to theta. calculate m if T is surface tension of the liquid.

A mercury drop shape as a round tablet of radius R and thickness h is

1.	A mercury drop shape as a round tablet of radius R and thickness h is located between two horizontal glass plates. Assuming that hltltR, find the mass m of a weight which has to be placed on the copper plate to diminish the distance between the plates by n-times the contact angle is equal to theta. calculate m if T is surface tension of the liquid.
Answer» `m=(2piRT^(2)\|costheta\|)/(gh)(n^(2)-1)` `m=(2piR^(2)T\|sintheta\|)/(gh)(n^(2)-1)` `m=(2piR^(2)T\|costheta\|)/(gh)(n^(2)+1)` `m=(2piR^(2)T\|costheta\|)/(gh)(n^(2)-1)` Solution :We know that pressure inside a FILM is greater than outside pressure by an amount `costheta=((H)/(2))/(r_(1))` or `r_(1)=(h)/(2\|costheta\|)` since, the tablet is between the PLATES, so `r_(2)=R` thus pressure Difference `P=T[(1)/(r_(1))+(1)/(r_(2))]=T[(1)/((h)/(2\|costheta\|))+(1)/(R)]` As h is small in comparison to R, so `(1)/(R)ltlt(1)/(h),:P=(2T\|costheta\|)/(h)` Let `R^(')` becomes the new radius of curvature when the distance between the plates is decreased by `n-` times ASSUMING mercury to be in compressible, `piR^(2)h=pi(R^('))^(2)(h)/(n),(R^('))^(2)=nR^(2)` The force EXERTED by the mercury drop now becomes `F^(')=(2pi(R^('))T\|costheta\|)/(((h)/(n)))=n^(2)F` If mg be the weight placed on the upper plate, then `F^(')=F+mg,m=(2piR^(2)T\|costheta\|)/(gh)(n^(2)-1)`