Soap Bubble Coalescence Radius

This problem concerns the radius ($r$) of a new soap bubble formed when two bubbles with radii $r_1$ and $r_2$ merge (coalesce) in a vacuum under isothermal conditions. The resulting formula depends on the physical principle governing the coalescence.

Surface Area Conservation Principle

The provided answer suggests a specific relationship ($r = \sqrt{r_1^2 + r_2^2}$). This relationship arises if we assume that the total surface area of the bubbles remains constant during coalescence. The surface area ($A$) of a sphere with radius $R$ is given by the formula $A = 4\pi R^2$.

• The initial surface area of the first bubble is $A_1 = 4\pi r_1^2$.
• The initial surface area of the second bubble is $A_2 = 4\pi r_2^2$.
• The total initial surface area is $A_{total\_initial} = A_1 + A_2 = 4\pi r_1^2 + 4\pi r_2^2$.
• The surface area of the newly formed bubble with radius $r$ is $A_{final} = 4\pi r^2$.

Radius Calculation Steps

By applying the principle of conservation of surface area, we equate the total initial surface area to the final surface area:

$ A_{final} = A_{total\_initial} $

Substituting the area formulas:

$ 4\pi r^2 = 4\pi r_1^2 + 4\pi r_2^2 $

Divide both sides of the equation by $4\pi$:

$ r^2 = r_1^2 + r_2^2 $

Taking the square root of both sides provides the final radius:

$ r = \sqrt{r_1^2 + r_2^2} $

This result matches Option 4.