The excess pressure inside a spherical water drop A is four times that of another water drop B. Then, the ratio of the mass of water drop A to that of drop B is

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TestHub · Accepted Answer

The excess pressure inside a spherical water drop of radius $R$ and surface tension $T$ is given by $Δ P = \frac{2 T}{R}$

Given that $Δ P_{A} = 4Δ P_{B}$

$\frac{2 T}{R _{A}} = 4 (\frac{2 T}{R _{B}})$

$R_{B} = 4 R_{A} \Rightarrow \frac{R _{A}}{R _{B}} = \frac{1}{4}$

The mass of a spherical drop is $M = \frac{4}{3} π R^{3} ρ$

The ratio of their masses is $\frac{M _{A}}{M _{B}} = (\frac{R _{A}}{R _{B}})^{3}$

$\frac{M _{A}}{M _{B}} = (\frac{1}{4})^{3} = \frac{1}{64}$

The ratio of the mass of water drop A to that of drop B is $1 : 64$

Answer: $1 : 64$

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