The excess pressure inside the first soap bub | vedclass

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

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$1$:$27$

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(A) The excess pressure inside a soap bubble of radius $R$ is given by $P = \frac{4T}{R}$, where $T$ is the surface tension of the soap solution.Given that the excess pressure in the first bubble $(P_1)$ is three times that in the second bubble $(P_2)$:$P_1 = 3P_2$$\frac{4T}{R_1} = 3 	imes \frac{4T}{R_2}$$\frac{1}{R_1} = \frac{3}{R_2} \implies R_2 = 3R_1$The volume of a spherical bubble is given by $V = \frac{4}{3}\pi R^3$.The ratio of the volumes of the first to the second bubble is:$\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi R_1^3}{\frac{4}{3}\pi R_2^3} = \left(\frac{R_1}{R_2}ight)^3$Substituting $R_2 = 3R_1$:$\frac{V_1}{V_2} = \left(\frac{R_1}{3R_1}ight)^3 = \left(\frac{1}{3}ight)^3 = \frac{1}{27}$Thus, the ratio of the volumes is $1:27$.

The excess pressure inside the first soap bubble of radius $R_1$ is three times that inside the second soap bubble of radius $R_2$. The ratio of volumes of the first to second bubble is

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