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Question

The excess pressure inside a spherical drop of water is 4 times that of another drop. Then their respective mass ratio is

A
1:4
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B
1:64
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C
1:16
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D
64:1
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Solution

The correct option is B 1:64
Excess pressure inside a liquid drop (ΔP)=2TR

Let the excess pressure in drop 1 is ΔP1 and drop 2 is ΔP2 and their radii are R1 and R2 respectively.
Given,
ΔP1=4×ΔP2
2TR1=4×2TR2
R2=4R1
R1R2=14

Mass ratio =mass of droplet 1 (m1)mass of droplet 2 (m2)
Here, m1=43πR31ρ;
m2=43πR32ρ
m1m2=R31R32=(R1R2)3=(14)3=164

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