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: Formula for Work Done in Expanding a Bubble: A soap bubble has two free surfaces (inner and outer). Work Done \( W = \text{Surface Tension} (T) \times \text{Change in Total Area} (\Delta A) \). \[ W = T \times 2 \times (4\pi R_f^2 - 4\pi R_i^2) = 8\pi T (R_f^2 - R_i^2) \] Thus, \( W \propto (R_f^2 - R_i^2) \).
Step 2: Calculate \( W_1 \) (r to 2r): \( R_i = r, R_f = 2r \). \[ W_1 \propto ((2r)^2 - r^2) = 4r^2 - r^2 = 3r^2 \]
Step 3: Calculate \( W_2 \) (2r to 3r): \( R_i = 2r, R_f = 3r \). \[ W_2 \propto ((3r)^2 - (2r)^2) = 9r^2 - 4r^2 = 5r^2 \]
Step 4: Calculate Ratio: \[ \frac{W_1}{W_2} = \frac{3r^2}{5r^2} = \frac{3}{5} \]