Question

The surface tension of a soap solution is \(3.5 \times 10^{-2}\) N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is \(\alpha \times 10^{-6}\) J. The value of \(\alpha\) is ______. (\(\pi = 22/7\))

Answer 1

Correct Answer: 264

Step 1: Understanding the Concept:
The work done to expand a soap bubble is equal to the surface tension multiplied by the change in the total surface area. A soap bubble has two free surfaces (inner and outer), so the surface area is twice that of a spherical drop.
Step 2: Key Formula or Approach:
Work done, \(W = T \times \Delta A\)
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 \(r_1\) is the initial radius and \(r_2\) is the final radius.
Step 3: Detailed Explanation:
Given values:
Surface tension, \(T = 3.5 \times 10^{-2} \text{ N/m}\).
Initial radius, \(r_1 = 1 \text{ cm} = 0.01 \text{ m}\).
Final radius, \(r_2 = 2 \text{ cm} = 0.02 \text{ m}\).
\(\pi = \frac{22}{7}\).
Calculate the change in area:
\(\Delta A = 8 \pi ((0.02)^2 - (0.01)^2)\)
\(\Delta A = 8 \pi (0.0004 - 0.0001) = 8 \pi (0.0003) = 0.0024 \pi \text{ m}^2\).
Now calculate the work done:
\(W = 3.5 \times 10^{-2} \times 0.0024 \times \frac{22}{7}\)
Express decimals as fractions to simplify:
\(W = \frac{35}{1000} \times \frac{24}{10000} \times \frac{22}{7} \times 10 = \dots\)
Alternatively:
\(3.5 \times \frac{22}{7} = \frac{7}{2} \times \frac{22}{7} = 11\).
So, \(W = 11 \times 10^{-2} \times 0.0024 = 11 \times 10^{-2} \times 24 \times 10^{-4}\)
\(W = (11 \times 24) \times 10^{-6}\)
\(W = 264 \times 10^{-6} \text{ J}\).
Step 4: Final Answer:
Comparing this to \(\alpha \times 10^{-6} \text{ J}\), we find that \(\alpha = 264\).