A soap bubble of radius $r_{1}$ is placed on another soap bubble of radius $r_{2} (r_{1} < r_{2})$ . The radius $R$ of the soapy film separating the two bubbles is:

r_{1} + r_{2}

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\sqrt{r_{1}^{2} + r_{1}^{2}}

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(r_{1}^{2} + r_{1}^{2})

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\frac{r_{2} r_{1}}{{r_{2} - r}_{1}}

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Solution

The correct option is D $\frac{r_{2} r_{1}}{{r_{2} - r}_{1}}$
Two soap bubbles of radii and are placed on one another to form a double bubble.
Let T be the surface tension of the liquid and P be the atmospheric pressure.
Pressure inside the smaller bubble =
Pressure inside the larger bubble =
Therefore, Pressure difference =
Now pressure on the concave side of the soapy film separating the two bubbles is greater than that on the convex side by
Therefore,

Hence,

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A soap bubble of radius r1 is placed on another soap bubble of radius r2(r1<r2). The radius R of the soapy film separating the two bubbles is:

A soap bubble of radius $r_{1}$ is placed on another soap bubble of radius $r_{2} (r_{1} < r_{2})$ . The radius $R$ of the soapy film separating the two bubbles is: