JEE Main 2025 Jan 22 Shift 1, Physics Q21: Properties of Fluids

JEE Main2025Jan 22, Shift 1Physics

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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Solution

For two soap bubbles of radii $r_{1}$ and $r_{2}$ ( $r_{1} < r_{2}$ ) sharing a common interface, the radius of curvature of that interface is

$r = \frac{r _{1} r _{2}}{r _{2} - r _{1}}$ .

Substituting $r_{1} = 2$ cm and $r_{2} = 4$ cm:

$r = \frac{2 \times 4}{4 - 2} = \frac{8}{2} = 4$ cm.

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JEE Main 2025 Jan 22 Shift 1, Physics Q21: Properties of Fluids

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