The excess pressure inside a soap bubble is 1.5 times the excess pressure inside a second soap bubble. The volume of the second bubble is ' x ' times the volume of the first bubble. The valu

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The excess pressure inside a soap bubble is 1.5 times the excess pressure inside a second soap bubble. The volume of the second bubble is ' x ' times the volume of the first bubble. The value of ' x ' is

TestHub · Accepted Answer

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

Let the excess pressures of the first and second bubbles be $P_{1}$ and $P_{2}$ , and their radii be $R_{1}$ and $R_{2}$ respectively.

Given that $P_{1} = 1.5 P_{2} = \frac{3}{2} P_{2}$ .

$\frac{4 T}{R _{1}} = \frac{3}{2} (\frac{4 T}{R _{2}})$

$\frac{R _{2}}{R _{1}} = \frac{3}{2}$

The volume of a soap bubble is $V = \frac{4}{3} π R^{3}$ .

The ratio of the volume of the second bubble to the first bubble is:

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

$x = (\frac{3}{2})^{3} = \frac{27}{8}$

Answer: $\frac{27}{8}$

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