`A` soap bubble has radius `R` and thickness `d(ltltR)` as shown. It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

Answer

`((R )/(24d))^(1//3)`

Explanation

`P_(1)=(4S)/(R ) " "implies " "P_(2)=(2S)/(r )`
where `P_(1)` is excess pressure is soap bubble and `P_(2)` is excess pressure in drop. Let `r` is the radius of spherical drop.
`(P_(2))/(P_(1))=(R )/(2r)" "….(1)`
From equation of volume
`4piR^(2)d=(4)/(3)" "," "r=(3R^(2)d)^(1//3)" "......(2)`
From `(1)` and `(2)`
`P_(2)/(P_(1))=(R)/(2(3R^(2)d)^(1//3))implies(P_(2))/(P_(1))=((R^(3))/(24R^(2)d))^(1//3)implies(P_(2))/(P_(1))=((R)/(24d))^(1//3)`