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A soap bubble has radius R and thickness d(d is very smaller as compared to the radius of the soap bubble{R}). It collapses into a spherical drop. find the ratio of excess pressure inside the drop as compared to surrounding to the excess pressure inside the bubble as compared to the surroundings.Obtain this ratio given that radius(r) is given by 64.8 cm and thickness(d) is given by 1mm

Dumb geeky nerd , 6 Years ago
Grade 12
anser 1 Answers
Arun

Last Activity: 4 Years ago

Let r  be the radius of the water drop formed.
Since the volume of the water forming bubble and drop is same,
34π(R3(Rd)3)=34πr3
r33R2d (neglecting d2  and d3)
Ratio of excess pressure in the drop to the excess pressure inside the bubble is given by,
Ratio=4σ/R2σ/r
Ratio=21(rR)
Substituting the value of r gives (24dR)1/3

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