When two soap bubbles of radius and $r_{1}$ and $r_{2} (r_{2} > r_{1})$ coalesce, the radius of curvature of common surface is

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Solution

The correct option is C

If $r_{1}$ and $r_{2}$ are the radii of smaller and larger bubble and $P_{0}$ is the atmospheric pressure, then the pressure inside them will be $P_{1} = P_{0} + \frac{4 T}{r_{1}} a n d P_{2} = P_{0} + \frac{4 T}{r_{2}}$ .

Now as $r_{1} < r_{2} ∴ P_{1} > P_{2}$
So for interface
$△ P = P_{1} - P_{2} = 4 T [\frac{1}{r_{1}} - \frac{1}{r_{2}}]$ ....(1)

As excess pressure acts from concave to convex side, the interface will be concave towards the smaller bubble and convex towards larger bubble and if r is the radius of interface.
$△ P = \frac{4 T}{r}$ .....(ii)
From (i) and (ii) $\frac{1}{r} = \frac{1}{r_{1}} - \frac{1}{r_{2}}$
$∴$ Radius of the interface r = $\frac{r_{1} r_{2}}{r_{2} - r_{1}}$

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When two soap bubbles of radius $r_{1}$ and $r_{2}$ ( $r_{2}$ > $r_{1}$ ) coalesce, the radius of curvature of common surface is

r_{1}

r_{2}

r_{1} > r_{2}

r_{1}

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(r_{1} > r_{2})

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a

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(b > a)

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