Question

The work done in blowing a soap bubble of radius \(R\) is \(W_1\) at room temperature. Now, the soap solution is heated. From the heated solution another soap bubble of radius \(2R\) is blown and the work done is \(W_2\). Then

• A. \(W_2 = 0\)
• B. \(W_2 = 4W_1\)
• C. \(W_2 = W_1\)
• D. \(W_2<4W_1\)

Hint:

Surface tension always decreases with increase in temperature.

Answer 1

The Correct Option is D

Step 1: Work done in forming a soap bubble.
\[ W = 8\pi T R^2 \] where \(T\) is surface tension.
Step 2: Effect of heating.
On heating, surface tension decreases:
\[ T_{\text{heated}}<T_{\text{room}} \]
Step 3: Comparing works.
If surface tension were same:
\[ W_2 = 8\pi T (2R)^2 = 4W_1 \] But since \(T\) decreases on heating, actual \(W_2\) will be less than \(4W_1\).
Step 4: Conclusion.
\[ W_2<4W_1 \]