When two soap bubbles of radii a and b ( b a ) coalesce, the radius of curvature of common surface is:

Question

When two soap bubbles of radii a and b ( b > a ) coalesce, the radius of curvature of common surface is: (A) (a b/b-a) (B) (a+b/a b) (C) ( b - a / ab ) (D) (a b/a+b)

Tardigrade Admin · Accepted Answer

Excess pressure at common surface is given by P ex =4 T ((1/ a )-(1/ b ))=(4 T / r ) ∴ (1/ r )=(1/ a )-(1/ b ) r =( ab / b - a )

Q. When two soap bubbles of radii $a$ and $b (b > a)$ coalesce, the radius of curvature of common surface is:

$\frac{ab}{b - a}$

$\frac{a + b}{ab}$

$\frac{b - a}{ab}$

$\frac{ab}{a + b}$

Excess pressure at common surface is given by
$P_{e x} = 4 T (\frac{1}{a} - \frac{1}{b}) = \frac{4 T}{r}$
$∴ \frac{1}{r} = \frac{1}{a} - \frac{1}{b}$
$r = \frac{ab}{b - a}$

Q. When two soap bubbles of radii a and b(b>a) coalesce, the radius of curvature of common surface is:

b−aab​

aba+b​

abb−a​

a+bab​