Two soap bubbles with radii r_1 and r_2 (r_1 | vedclass

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(B) Let $p_0$ be the atmospheric pressure.Let $p_1$ and $p_2$ be the pressures inside the two bubbles of radii $r_1$ and $r_2$ respectively.The excess

Vedclass · Accepted Answer

$r = \frac{r_1 r_2}{r_1 - r_2}$

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(B) Let $p_0$ be the atmospheric pressure.Let $p_1$ and $p_2$ be the pressures inside the two bubbles of radii $r_1$ and $r_2$ respectively.The excess pressure inside a soap bubble is given by $\Delta p = \frac{4S}{r}$,where $S$ is the surface tension.Thus,the pressures inside the bubbles are:$p_1 = p_0 + \frac{4S}{r_1}$$p_2 = p_0 + \frac{4S}{r_2}$Since $r_1 > r_2$,we have $p_2 > p_1$.When they come in contact,the common surface acts as a membrane with a radius of curvature $r$. The pressure difference across this common surface is:$p_2 - p_1 = \frac{4S}{r}$Substituting the expressions for $p_1$ and $p_2$:$(p_0 + \frac{4S}{r_2}) - (p_0 + \frac{4S}{r_1}) = \frac{4S}{r}$$\frac{4S}{r_2} - \frac{4S}{r_1} = \frac{4S}{r}$Dividing by $4S$:$\frac{1}{r} = \frac{1}{r_2} - \frac{1}{r_1}$$\frac{1}{r} = \frac{r_1 - r_2}{r_1 r_2}$Therefore,$r = \frac{r_1 r_2}{r_1 - r_2}$.

Two soap bubbles with radii $r_1$ and $r_2$ $(r_1 > r_2)$ come in contact. Their common surface has a radius of curvature $r$. Find $r$.

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