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Q.

Two soap bubbles, one of radius 50 mm and the other of radius 80 mm, are brought in contact so that they have a common interface. The radius of the curvature of the
common interface is

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a

1.2 m

b

0.003 m

c

8.9 m

d

0.133 m

answer is B.

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Detailed Solution

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$\begin{array}{l} [P_{0} + \frac{4 T}{R_{2}}] - [P_{0} + \frac{4 T}{R_{1}}] = \frac{4 T}{R} \\ ⇒ \frac{1}{R} = \frac{1}{R_{2}} - \frac{1}{R_{1}} \\ ⇒ R = \frac{R_{1} R_{2}}{R_{1} - R_{2}} = \frac{50 \times 80}{30} mm = \frac{400}{3} mm \\ = \frac{400}{3 \times 1000} m = \frac{4}{30} m = 0.133 m \end{array}$

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