Question

A soap bubble of radius \(\frac{1}{\sqrt{π}}\) cm is expanded to radius \(\frac{3}{\sqrt{π}}\) cm. Surface tension of soap solution is 25 dyne/cm. The work done during expansion in erg is

• A. \(800\)
• B. \(1200\)
• C. \(1600\)
• D. \(2400\)

Hint:

A soap bubble has two surfaces, so \(W=T\times2\times\Delta A\).

Answer 1

The Correct Option is C

Step 1: Compute the area increase
One surface: $4\pi(r_2^2-r_1^2)=4\pi\cdot\dfrac8\pi=32$ cm$^2$. Two surfaces: 64 cm$^2$.

Step 2: Multiply by T
$W=25\times64=1600$ erg, option (C).

Final Answer:
The work is 1600 erg, option (C). \[ \boxed{1600\ \text{erg}} \]