Pressure inside two soap bubbles is 1.01 , at | vedclass

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(B) The excess pressure inside a soap bubble is given by $\Delta P = P_{i} - P_{0} = \frac{4T}{r}$.Given $P_{0} = 1 \, atm$, the excess pressures are:

Vedclass · Accepted Answer

$27$

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(B) The excess pressure inside a soap bubble is given by $\Delta P = P_{i} - P_{0} = \frac{4T}{r}$.Given $P_{0} = 1 \, atm$, the excess pressures are:$\Delta P_{1} = 1.01 \, atm - 1 \, atm = 0.01 \, atm$$\Delta P_{2} = 1.03 \, atm - 1 \, atm = 0.03 \, atm$Since $\Delta P \propto \frac{1}{r}$, we have $\frac{\Delta P_{1}}{\Delta P_{2}} = \frac{r_{2}}{r_{1}}$.Substituting the values: $\frac{r_{2}}{r_{1}} = \frac{0.03}{0.01} = 3$.The ratio of the volumes is $\frac{V_{1}}{V_{2}} = \frac{\frac{4}{3}\pi r_{1}^{3}}{\frac{4}{3}\pi r_{2}^{3}} = \left(\frac{r_{1}}{r_{2}}\right)^{3}$.Since $\frac{r_{2}}{r_{1}} = 3$, then $\frac{r_{1}}{r_{2}} = \frac{1}{3}$.Therefore, $\frac{V_{1}}{V_{2}} = \left(\frac{1}{3}\right)^{3} = \frac{1}{27}$.Thus, the ratio of their volumes $V_{1}:V_{2}$ is $1:27$, or $V_{2}:V_{1}$ is $27:1$.

Pressure inside two soap bubbles is $1.01 \, atm$ and $1.03 \, atm$. The ratio between their volumes is (Pressure outside the soap bubble is $1 \, atm$). (in $ : 1$)

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