The excess pressure inside a spherical drop of water is four times that of another drop. Their respective mass ratio is

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TestHub · Accepted Answer

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

Given that $Δ P_{1} = 4Δ P_{2}$ .

$\frac{2 T}{R _{1}} = 4 (\frac{2 T}{R _{2}})$

$\frac{R _{1}}{R _{2}} = \frac{1}{4}$

The mass of a spherical drop is $M = \frac{4}{3} π R^{3} ρ$ , where $ρ$ is the density of water.

The ratio of their masses is:

$\frac{M _{1}}{M _{2}} = (\frac{R _{1}}{R _{2}})^{3}$

$\frac{M _{1}}{M _{2}} = (\frac{1}{4})^{3} = \frac{1}{64}$

The respective mass ratio is $1 : 64$ .

Answer: $1 : 64$

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