A mercury drop shaped as a round tablet of radius R and thickness h is located between two horizontal plates, Assuming hltltR. Find the mass .m. of a weight which has to be placed on the copper plate to dimnish the distance between plates n-times (theta angle of contact)

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

1.	A mercury drop shaped as a round tablet of radius R and thickness h is located between two horizontal plates, Assuming hltltR. Find the mass .m. of a weight which has to be placed on the copper plate to dimnish the distance between plates n-times (theta angle of contact)
Answer» Solution :`Deltap=T{1/r_(1)+1/r_(2)}" now "Deltap=T{(2(costheta))/h}+1/R` `r_(1)=h/(2(costheta)),r_(2)=R" if "hltltRrArrDeltap=(2T(costheta))/h` then total force EXERTED by mercury on upper glass plate `F=(Deltap)` (A contact) = `({(2T(costheta))/h}piR^(2))/("if R is NEW radius of drop")` when `h^(1)=h/n` then CONSERVATION of volume `piR^(2)h=pi(R^(1))^(2)(h^(1))rArrR^(1)=sqrtnR" "f^(1)=n^(2)F` If .mg. is the weight placed on the upper plate, `F^(1)=f+mg` `rArrm^(1)=(F^(1)-f)/G=(2piR^(2)T(costheta))/(gh)(n^(2)-1)thereforem^(1)={(2piR^(2)T\|costheta\|)/(gh)}(n^(2)-i)`