The pressures inside two soap bubbles A and B | vedclass

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Question

(A) The excess pressure inside a soap bubble is given by $\Delta P = P_{in} - P_{out} = \frac{4T}{r}$, where $T$ is the surface tension and $r$ is the

Vedclass · Accepted Answer

$8:1$

Vedclass · Answer

(A) The excess pressure inside a soap bubble is given by $\Delta P = P_{in} - P_{out} = \frac{4T}{r}$, where $T$ is the surface tension and $r$ is the radius of the bubble.For bubble $A$: $\Delta P_A = 1.02 - 1 = 0.02 	ext{ atm} = \frac{4T}{r_A}$.For bubble $B$: $\Delta P_B = 1.04 - 1 = 0.04 	ext{ atm} = \frac{4T}{r_B}$.Taking the ratio: $\frac{\Delta P_A}{\Delta P_B} = \frac{r_B}{r_A} = \frac{0.02}{0.04} = \frac{1}{2}$.Thus, $r_A = 2r_B$.The volume of a spherical bubble is $V = \frac{4}{3}\pi r^3$.The ratio of volumes is $\frac{V_A}{V_B} = \frac{r_A^3}{r_B^3} = (2)^3 = 8$.Therefore, the ratio is $8:1$.

The pressures inside two soap bubbles $A$ and $B$ are $1.02 \text{ atm}$ and $1.04 \text{ atm}$ respectively. The ratio of the volume of bubble $A$ to that of bubble $B$ is (outside pressure = $1 \text{ atm}$)

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