The excess pressure inside the first soap bub | vedclass

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

Vedclass · Accepted Answer

$1: 8$

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(D) The excess pressure $P$ 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 twice that of the second bubble $(P_{2})$,we have $P_{1} = 2P_{2}$.Substituting the formula for excess pressure: $\frac{4T}{R_{1}} = 2 	imes \frac{4T}{R_{2}}$.Simplifying this,we get $\frac{1}{R_{1}} = \frac{2}{R_{2}}$,which implies $\frac{R_{2}}{R_{1}} = 2$ or $R_{2} = 2R_{1}$.The volume $V$ of a spherical bubble is given by $V = \frac{4}{3} \pi R^{3}$.The ratio of the volumes is $\frac{V_{1}}{V_{2}} = \frac{\frac{4}{3} \pi R_{1}^{3}}{\frac{4}{3} \pi R_{2}^{3}} = \left( \frac{R_{1}}{R_{2}} ight)^{3}$.Since $\frac{R_{1}}{R_{2}} = \frac{1}{2}$,the ratio of volumes is $\left( \frac{1}{2} ight)^{3} = \frac{1}{8}$.

The excess pressure inside the first soap bubble of radius $R_{1}$ is two times that inside the second soap bubble of radius $R_{2}$. The ratio of the volumes of the first bubble to that of the second bubble is:

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