Question

A liquid drop of diameter \(2\) mm breaks into \(512\) droplets. The change in surface energy is \( \alpha \times 10^{-6} \) J. (Take surface tension of liquid = \(0.08\) N/m). The value of \(\alpha\) is ____.

• A. \(10\)
• B. \(7\)
• C. \(8\)
• D. \(11\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When a large drop breaks into smaller droplets, the total volume remains constant, but the total surface area increases. The energy required to increase this surface area is equal to the product of surface tension and the change in surface area.
Step 2: Key Formula or Approach:
1. Volume conservation: \(R = n^{1/3} r\).
2. Surface energy change: \(\Delta U = T \cdot \Delta A = T [n(4\pi r^2) - 4\pi R^2]\).
Step 3: Detailed Explanation:
Initial radius \(R = 1 \text{ mm} = 10^{-3} \text{ m}\).
Number of droplets \(n = 512\).
Using volume conservation: \(R = (512)^{1/3} r = 8r \implies r = \frac{R}{8}\).
Change in area \(\Delta A = 4\pi [n r^2 - R^2] = 4\pi [512 \cdot (\frac{R}{8})^2 - R^2]\)
\[ \Delta A = 4\pi [512 \cdot \frac{R^2}{64} - R^2] = 4\pi [8R^2 - R^2] = 4\pi(7R^2) = 28\pi R^2 \].
Substituting the values:
\[ \Delta U = T \times 28\pi R^2 = 0.08 \times 28 \times 3.14 \times (10^{-3})^2 \]
\[ \Delta U = 7.0336 \times 10^{-6} \text{ J} \].
Comparing with \(\alpha \times 10^{-6}\), we get \(\alpha \approx 7\).
Step 4: Final Answer:
The value of \(\alpha\) is 7.