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

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TestHub · Accepted Answer

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

Given $Δ P_{1} = 4Δ P_{2}$

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

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

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

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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