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(A) The excess pressure inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension.Let the radii of the

Vedclass · Accepted Answer

$0.125$

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(A) The excess pressure inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension.Let the radii of the two bubbles be $r_1$ and $r_2$ respectively.Given that the excess pressure in the first bubble is twice that of the second:$\frac{4T}{r_1} = 2 	imes \frac{4T}{r_2}$This simplifies to $r_2 = 2r_1$.The volume of a spherical bubble is $V = \frac{4}{3}\pi r^3$.Given $V_1 = n 	imes V_2$,we have:$\frac{4}{3}\pi r_1^3 = n 	imes \frac{4}{3}\pi r_2^3$Substituting $r_2 = 2r_1$:$r_1^3 = n 	imes (2r_1)^3$$r_1^3 = n 	imes 8r_1^3$$n = \frac{1}{8} = 0.125$.

The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is $n$ times the volume of the second,where $n$ is:

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