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:
A soap bubble has two free surfaces (inner and outer). Therefore, the work done in increasing its surface area is twice the work done for a single surface. Work done is equal to Surface Tension multiplied by the total change in surface area.

Step 2: Key Formula or Approach:
1. Surface Area of a sphere \(A = 4\pi r^2\). 2. Change in area \(\Delta A = 2 \times (4\pi r_2^2 - 4\pi r_1^2) = 8\pi(r_2^2 - r_1^2)\). 3. Work \(W = T \times \Delta A\).

Step 3: Detailed Explanation:
1. Given: \(T = 3.5 \times 10^{-2}\) N/m, \(r_1 = 0.01\) m, \(r_2 = 0.02\) m.
2. \(\Delta A = 8 \times \frac{22}{7} \times (0.02^2 - 0.01^2) = \frac{176}{7} \times (0.0004 - 0.0001)\) \[ \Delta A = \frac{176}{7} \times 0.0003 = \frac{0.0528}{7} \text{ m}^2 \] 3. \(W = (3.5 \times 10^{-2}) \times \frac{0.0528}{7} = \frac{3.5}{7} \times 0.0528 \times 10^{-2}\) \[ W = 0.5 \times 0.0528 \times 10^{-2} = 0.0264 \times 10^{-2} = 264 \times 10^{-6} \text{ J} \] 4. Comparing with \(\alpha \times 10^{-6}\), \(\alpha = 264\).

Step 4: Final Answer:
The value of \(\alpha\) is 264.