Question

If a water drop of radius 2 cm breaks into \(10^9\) drops of equal size, the work done is about:
(Surface tension of water is \(72 \times 10^{-3}\) N/m)

• A. \(1500 \pi \times 10^{-4}\) J
• B. \(1800 \pi \times 10^{-4}\) J
• C. \(1150 \pi \times 10^{-4}\) J
• D. \(1000 \pi \times 10^{-4}\) J

Hint:

When a large drop breaks into many small drops, the work done is the increase in surface energy: \(W = \gamma ( \text{total area of small drops} - \text{area of large drop})\).

Answer 1

The Correct Option is C

Step 1: Radius of a small droplet.
Volume conservation: (4/3)π R³ = 10⁹ × (4/3)π r³ → r = R/10³ = 0.02/1000 = 2 × 10⁻⁵ m.

Step 2: Work as surface energy change.
W = ΔU = γ × (total final surface area - initial surface area).

Step 3: Computing surface energies.
Initial: U_i = 4πR²γ = 4π(0.02)² × 72×10⁻³. Final: U_f = 10⁹ × 4πr²γ = 10⁹ × 4π(2×10⁻⁵)² × 72×10⁻³.

Step 4: Difference.
W = U_f - U_i ≈ 1150π × 10⁻⁴ J.

Step 5: Conclusion.
The work done is 1150π × 10⁻⁴ J.