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A soap bubble A of radius 0.03 m and another bubble B of radius 0.04 m are brought together so that the combined bubble has a common interface of radius r, then the value of r is

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0.48m

0.36m

0.24m

0.12m

answer is C.

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Detailed Solution

$r = \frac{r_{1} r_{2}}{r_{1} - r_{2}} = 0.03 \times 0.04$

$= 0.12 m$

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