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

Vedclass · Accepted Answer

$1:27$

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(D) The excess pressure $\Delta P$ inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension of the soap solution.From this,we see that $\Delta P \propto \frac{1}{r}$.Let the excess pressure in the first bubble be $\Delta P_1$ and in the second bubble be $\Delta P_2$. Given $\Delta P_1 = 3 \Delta P_2$.Therefore,$\frac{\Delta P_1}{\Delta P_2} = 3$.Since $\frac{\Delta P_1}{\Delta P_2} = \frac{r_2}{r_1}$,we have $\frac{r_2}{r_1} = 3$,which implies $\frac{r_1}{r_2} = \frac{1}{3}$.The volume $V$ of a spherical bubble is given by $V = \frac{4}{3} \pi r^3$,so $V \propto r^3$.The ratio of the volumes is $\frac{V_1}{V_2} = \left( \frac{r_1}{r_2} \right)^3$.Substituting the ratio of radii,$\frac{V_1}{V_2} = \left( \frac{1}{3} \right)^3 = \frac{1}{27}$.

The excess pressure in a first soap bubble is three times that of another soap bubble. Then the ratio of the volume of the first bubble to the other is:

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