Question

Under isothermal conditions, two soap bubbles of radii $r_1$ and $r_2$ coalesce to form a big drop. The radius of the big drop is

• A. $(r_1 - r_2)^{1/2}$
• B. $(r_1 + r_2)^{1/2}$
• C. $(r_1^2 + r_2^2)^{1/2}$
• D. $(r_1^2 - r_2^2)^{1/2}$

Hint:

For soap bubbles, surface area (not volume) is conserved during coalescence.

Answer 1

The Correct Option is C

Step 1: Consider surface area conservation.
For soap bubbles under isothermal conditions, total surface area remains conserved.

Step 2: Write surface area expressions.
Total initial surface area:
\[ 4\pi r_1^2 + 4\pi r_2^2 \] Final surface area of big bubble:
\[ 4\pi R^2 \]

Step 3: Equate areas.
\[ 4\pi R^2 = 4\pi (r_1^2 + r_2^2) \]

Step 4: Solve for $R$.
\[ R = \sqrt{r_1^2 + r_2^2} \]

Step 5: Conclusion.
The radius of the big bubble is $(r_1^2 + r_2^2)^{1/2}$.