Fluid Mechanics

JEE Main level, numerical answer, medium

103 of 192

Question 103 of 192

Two soap bubbles of radius 2 cm and 4 cm , respectively, are in contact with each other. The radius of curvature of the common surface, in cm , is ______ .

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Answer: 4

R=R2R1R2−R1=4×22=4 cmR=\frac{R_2 R_1}{R_2-R_1}=\frac{4 \times 2}{2}=4 \mathrm{~cm}

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