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

Concept: Surface energy: \[ E = T \times A \] where \(T\) is surface tension and \(A\) is surface area. When a liquid drop breaks into smaller drops, surface area increases, hence surface energy increases.
Step 1:Find radius of original drop} Diameter \(=2\) mm \[ R = 1 \text{ mm} = 10^{-3} \text{ m} \]
Step 2:Relation of radii} Volume conserved: \[ \frac{4}{3}\pi R^3 = 512 \times \frac{4}{3}\pi r^3 \] \[ R^3 = 512 r^3 \] \[ r = \frac{R}{8} \]
Step 3:Surface areas} Initial surface area: \[ A_1 = 4\pi R^2 \] Final surface area: \[ A_2 = 512 \times 4\pi r^2 \] \[ =512 \times 4\pi \left(\frac{R}{8}\right)^2 \] \[ =512 \times 4\pi \frac{R^2}{64} \] \[ =32\pi R^2 \]
Step 4:Increase in surface area} \[ \Delta A = A_2 - A_1 \] \[ =32\pi R^2 - 4\pi R^2 \] \[ =28\pi R^2 \]
Step 5:Increase in surface energy} \[ \Delta E = T \Delta A \] \[ =0.08 \times 28\pi (10^{-3})^2 \] \[ =0.08 \times 28\pi \times 10^{-6} \] \[ \approx 7.04 \times 10^{-6} \text{ J} \] Thus \[ \alpha \approx 7 \] Closest option: \[ \boxed{8} \]