Question

Two identical bubbles stick with each other and are in equilibrium, their radius is R. The area of common interface is equal to

Answer 1

Answer:

Area of common surface area of two bubbles is given as

$A = \pi(3R^2/4)$

Explanation:

Let the common surface radius is "y" and here system of two bubble will try to tend its surface area minimum

So here in order to minimize the free surface area the common surface must have maximum area of contact

So the two surface will cut orthogonal such that the distance between the two center will be equal to the radius of each sphere

So the radius of common surface is given as

$r = \sqrt{R^2 - (\frac{R}{2})^2}$

$r = \sqrt3 R/2$

so we have

$A = \pi y^2$

$A = \pi(3R^2/4)$

#Learn

Topic : Radius of soap bubble

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