When two soap bubbles of radii r1 and r2(r2r1) coalesce, the radius of curvature of common surface is

Question

When two soap bubbles of radii r1 and r2(r2>r1) coalesce, the radius of curvature of common surface is (A) r2-r1 (B) (r2-r1/r1 r2) (C) (r1 r2/r2-r1) (D) r2+r1

Tardigrade Admin · Accepted Answer

For soap bubble of radius r1. P1-P0=(4 σ/r1), where P1 is the pressure inside it and P0 is the pressure of the atmosphere. Similarly, P2-P0=(4 σ/r2) When bubbles coalesce and form a common surface, P1-P2=(4 σ/r1)-(4 σ/r2)=(4 σ/R) ⇒ (1/R)=(1/r1)-(1/r2) or R=(r1 ⋅ r2/r2-r1)

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

$r_{2} - r_{1}$

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

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

$r_{2} + r_{1}$

Q. When two soap bubbles of radii r1​ and r2​(r2​>r1​) coalesce, the radius of curvature of common surface is

r2​−r1​

r1​r2​r2​−r1​​

r2​−r1​r1​r2​​

r2​+r1​

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

$r_{2} - r_{1}$

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

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

$r_{2} + r_{1}$