The pressures inside two soap bubbles A and B are 1.02 atmosphere and 1.04 atmosphere respectively. The ratio of volume of bubble A to that of bubble B is (outside pressure = 1 atmosphere)

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The excess pressure inside a soap bubble is given by $Δ P = \frac{4 T}{R}$ , where $T$ is the surface tension and $R$ is the radius of the bubble.

For bubble A, the excess pressure is $Δ P_{A} = 1.02 - 1 = 0.02$ atm.

For bubble B, the excess pressure is $Δ P_{B} = 1.04 - 1 = 0.04$ atm.

Taking the ratio of the excess pressures, we get $\frac{Δ P _{A}}{Δ P _{B}} = \frac{0.02}{0.04} = \frac{1}{2}$ .

Since $R \propto \frac{1}{Δ P}$ , the ratio of their radii is $\frac{R _{A}}{R _{B}} = \frac{Δ P _{B}}{Δ P _{A}} = \frac{2}{1}$ .

The volume of a spherical bubble is $V = \frac{4}{3} π R^{3}$ , which means $V \propto R^{3}$ .

The ratio of their volumes is $\frac{V _{A}}{V _{B}} = (\frac{R _{A}}{R _{B}})^{3} = (2)^{3} = 8$ .

Thus, the ratio of the volume of bubble A to that of bubble B is $8 : 1$ .

Answer: $8 : 1$

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