Question

The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is \( \alpha \times 10^{-4} \) J. The value of \( \alpha \) is _______. (Take \( \pi = 3.14 \))}

• A. 0.86
• B. 0.64
• C. 0.62
• D. 0.30

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The work done in expanding a bubble is equal to the surface tension multiplied by the change in the total surface area.
Because a soap bubble has a thin film with two surfaces (an inner surface and an outer surface) in contact with air, the effective surface area is twice the geometric surface area of a sphere.
Step 2: Key Formula or Approach:
Work done \(W = T \cdot \Delta A\).
For a soap bubble, the change in area is \(\Delta A = 2 \times (4\pi R_2^2 - 4\pi R_1^2) = 8\pi(R_2^2 - R_1^2)\), where \(R_1\) and \(R_2\) are the initial and final radii.
Step 3: Detailed Explanation:
Given values:
Surface tension \(T = 0.03 \text{ N/m}\).
Initial diameter \(D_1 = 2 \text{ cm} \implies R_1 = 1 \text{ cm} = 0.01 \text{ m}\).
Final diameter \(D_2 = 6 \text{ cm} \implies R_2 = 3 \text{ cm} = 0.03 \text{ m}\).
Calculate the change in total surface area:
\[ \Delta A = 8\pi ((0.03)^2 - (0.01)^2) \] \[ \Delta A = 8\pi (0.0009 - 0.0001) \] \[ \Delta A = 8\pi (0.0008) = 0.0064\pi \text{ m}^2 \] Now calculate the work done:
\[ W = T \cdot \Delta A = 0.03 \times 0.0064\pi \] \[ W = 0.000192\pi \text{ J} \] Write this in the format \(\alpha \pi \times 10^{-4} \text{ J}\):
\[ W = 1.92 \pi \times 10^{-4} \text{ J} \] Comparing this to the given expression, we find \(\alpha = 1.92\).
Step 4: Final Answer:
The value of \(\alpha\) is \(1.92\).