The pressure inside a soap bubble A is 1.01 t | vedclass

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(B) Outside pressure $P_0 = 1 	ext{ atm}$.Pressure inside soap bubble $A$,$P_A = 1.01 	ext{ atm}$.Pressure inside soap bubble $B$,$P_B = 1.02 	ext{

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$8: 1$

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(B) Outside pressure $P_0 = 1 	ext{ atm}$.Pressure inside soap bubble $A$,$P_A = 1.01 	ext{ atm}$.Pressure inside soap bubble $B$,$P_B = 1.02 	ext{ atm}$.Excess pressure in a soap bubble is given by $\Delta P = \frac{4T}{r}$,where $T$ is surface tension and $r$ is the radius.Excess pressure for bubble $A$: $\Delta P_A = P_A - P_0 = 1.01 - 1 = 0.01 	ext{ atm}$.Excess pressure for bubble $B$: $\Delta P_B = P_B - P_0 = 1.02 - 1 = 0.02 	ext{ atm}$.Since $\Delta P \propto \frac{1}{r}$,we have $r \propto \frac{1}{\Delta P}$.Therefore,$\frac{r_A}{r_B} = \frac{\Delta P_B}{\Delta P_A} = \frac{0.02}{0.01} = \frac{2}{1}$.The volume $V$ of a sphere is given by $V = \frac{4}{3}\pi r^3$,so $V \propto r^3$.Thus,the ratio of volumes is $\frac{V_A}{V_B} = \left(\frac{r_A}{r_B}ight)^3 = \left(\frac{2}{1}ight)^3 = \frac{8}{1}$.

The pressure inside a soap bubble $A$ is $1.01 \text{ atm}$ and that in a soap bubble $B$ is $1.02 \text{ atm}$. The ratio of the volume of $A$ to that of $B$ is

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