Question

If \(W_1\) is the work done in increasing the radius of a soap bubble from 'r' to '2r' and \(W_2\) is the work done in increasing the radius of the soap bubble from '2r' to '3r', then \(W_1: W_2 =\)

• A. 3:5
• B. 1:1
• C. 2:3
• D. 3:4

Hint:

Remember that a soap bubble has two surfaces, while a liquid drop has only one. For a liquid drop, the work done formula would be \(W = 4\pi T (R_f^2 - R_i^2)\). Always check whether the problem involves a bubble or a drop.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The work done to change the surface area of a liquid film is equal to the product of the surface tension and the change in surface area. A soap bubble has two surfaces (an inner and an outer surface) in contact with air, so the total surface area is twice that of a single sphere.
Step 2: Key Formula or Approach:
Work Done, \(W = T \times \Delta A_{total}\), where \(T\) is the surface tension.
For a soap bubble of radius \(R\), the total surface area is \(A_{total} = 2 \times (4\pi R^2) = 8\pi R^2\).
The work done in changing the radius from \(R_i\) to \(R_f\) is:
\[ W = T \times (A_{final} - A_{initial}) = T \times (8\pi R_f^2 - 8\pi R_i^2) = 8\pi T (R_f^2 - R_i^2) \] Step 3: Detailed Explanation:
Calculate \(W_1\):
The radius is increased from \(r\) to \(2r\). So, \(R_i = r\) and \(R_f = 2r\).
\[ W_1 = 8\pi T ((2r)^2 - r^2) = 8\pi T (4r^2 - r^2) = 8\pi T (3r^2) \] Calculate \(W_2\):
The radius is increased from \(2r\) to \(3r\). So, \(R_i = 2r\) and \(R_f = 3r\).
\[ W_2 = 8\pi T ((3r)^2 - (2r)^2) = 8\pi T (9r^2 - 4r^2) = 8\pi T (5r^2) \] Find the ratio \(W_1 : W_2\):
\[ \frac{W_1}{W_2} = \frac{8\pi T (3r^2)}{8\pi T (5r^2)} = \frac{3}{5} \] So, the ratio \(W_1 : W_2\) is 3:5.
Step 4: Final Answer:
The ratio of the work done is 3:5. Therefore, option (A) is correct.