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When two soap bubbles of radius r_{1} and r_{2} (r_{2}>r{1}) c

Question

When two soap bubbles of radius $r_{1}$ and $r_{2} (r_{2} > r 1)$ coalesce, the radius of curvature of common surface is

r_{2} - r_{1}

\frac{r _{2} - r _{1}}{r _{1} r _{2}}

\frac{r _{1} r _{2}}{r _{2} - r _{1}}

r_{2} + r_{1}

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Benjamin Discussed

When two soap bubbles of radius

r_{1}

and

r_{2} (r_{2} > r 1)

coalesce, the radius of curvature of common surface is

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Lets consider, two soap bubbles with radius r1 nad r2 such that r1 > r2 .

For bubble 1 -

P_{1} - P_{0} = 4 T / r_{1}

P_{1} = P_{0} + 4 T / r_{1}

...(1)

For bubble 2 -

P_{2} - P_{0} = 4 T / r_{2}

P_{2} = P_{0} + 4 T / r_{2}

...(2)

Let's, consider two bubbles merge so that-

P_{2} - P_{1} = 4 T / r

P_{0} + 4 T / r_{2} - P_{0} + 4 T / r_{1} = 4 T / r

4 T (1/ r_{2} - 1/ r_{1}) = 4 T / r

1/ r = 1/ r_{2} - 1/ r_{1}

r = \frac{r _{1} r _{2}}{r _{1} + r _{2}}

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Question Text	When two soap bubbles of radius $r_{1}$ and $r_{2} (r_{2} > r 1)$ coalesce, the radius of curvature of common surface is
Answer Type	Text solution:1
Upvotes	152