When two soap bubbles of radii a and $\mathrm{b}(\mathrm{b}>\mathrm{a})$ coalesce, the radius of curvature of common surface is:

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Fluid Mechanics

When two soap bubbles of radii a and

b (b > a)

coalesce, the radius of curvature of common surface is:

(A) $\frac{ab}{a + b}$
(B) $\frac{ab}{b - a}$ ✓ Correct
(C) $\frac{b - a}{ab}$
(D) $\frac{a + b}{ab}$

Answer: (B)

Solution

The pressure inside a soap bubble of radius

r

exceeds the outside pressure by

\frac{4 T}{r}

, since the film has two surfaces. For the two bubbles,

P_{a} = P_{0} + \frac{4 T}{a}, P_{b} = P_{0} + \frac{4 T}{b}

and

P_{a} > P_{b}

because

a < b

. The common film separates the two interiors and bulges towards the larger bubble, so its radius of curvature

r

satisfies

P_{a} - P_{b} = \frac{4 T}{r}

\frac{4 T}{a} - \frac{4 T}{b} = \frac{4 T}{r} \Rightarrow \frac{1}{r} = \frac{b - a}{ab}

The radius of curvature of the common surface is

\frac{ab}{b - a}

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Single correct option

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