Question

Under isothermal condition, two soap bubbles of radii \(r_1\) and \(r_2\) combine to form a single soap bubble of radius R. The surface tension of the soap solution is (P = outside pressure)

• A. \(\frac{P(R^3-r_1^3-r_2^3)}{4(r_1^2+r_2^2-R^2)}\)
• B. \(\frac{P(R^3+r_1^3+r_2^3)}{2(r_1^2-r_2^2+R^2)}\)
• C. \(\frac{P(r_1^3-r_2^3-R^3)}{(R^2+r^2+r^2)}\)
• D. \(\frac{P(R^3-r_1^3+r_2^3)}{2(r_1^2+r_2^2-R^2)}\)

Hint:

Apply Boyle law to the total gas, using the excess pressure 4S/r inside a soap bubble.

Answer 1

The Correct Option is A

Step 1: Pressures
Inside a soap bubble of radius \(r\) the pressure is \(P + \frac{4S}{r}\).

Step 2: Boyle law at constant temperature
Total \(PV\) before equals after:
\[ \left(P+\frac{4S}{r_1}\right)\frac43\pi r_1^3 + \left(P+\frac{4S}{r_2}\right)\frac43\pi r_2^3 = \left(P+\frac{4S}{R}\right)\frac43\pi R^3 \]

Step 3: Simplify
Cancel \(\frac43\pi\): \(P r_1^3 + 4Sr_1^2 + Pr_2^3 + 4Sr_2^2 = PR^3 + 4SR^2\). So \(4S(r_1^2+r_2^2-R^2) = P(R^3-r_1^3-r_2^3)\).

Step 4: Result
\[ S = \frac{P(R^3-r_1^3-r_2^3)}{4(r_1^2+r_2^2-R^2)} \]
Option (A).

Final Answer:
The surface tension is given by option A. \[ \boxed{\text{(A)}\ S=\frac{P(R^3-r_1^3-r_2^3)}{4(r_1^2+r_2^2-R^2)}} \]