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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 thecommon interface is

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a

0.003 m

b

0.133 m

c

1.2 m

d

8.9 m

answer is B.

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Detailed Solution

P0+4TR2−P0+4TR1=4TR⇒1R=1R2−1R1⇒R=R1R2R1−R2=50×8030mm=4003mm =4003×1000m=430m=0.133m
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