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

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TestHub · Accepted Answer

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

Given $Δ P_{A} = 4Δ P_{B}$ .

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

$R_{B} = 4 R_{A}$

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

The ratio of mass of drop A to that of drop B is

$\frac{M _{A}}{M _{B}} = (\frac{R _{A}}{R _{B}})^{3}$

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

Answer: $1 : 64$

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