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: Calculate the radius of one small drop.
Volume of large drop = volume of all small drops: \[ \frac{4}{3}\pi R^3 = 10^9 \cdot \frac{4}{3}\pi r^3 \Rightarrow r = \frac{R}{10^3} = \frac{0.02}{1000} = 2 \times 10^{-5}~\text{m} \]

Step 2: Work done is increase in surface energy.
Surface energy \(U = \gamma \cdot \text{surface area}\)

Step 3: Calculate surface energy.
- Original drop: \[ U_i = 4 \pi R^2 \gamma = 4 \pi (0.02)^2 \cdot 72 \times 10^{-3} = 0.00036288 \pi~\text{J} \] - Total small drops: \[ U_f = 10^9 \cdot 4 \pi r^2 \gamma = 10^9 \cdot 4 \pi (2 \times 10^{-5})^2 \cdot 72 \times 10^{-3} \approx 0.001152 \pi~\text{J} \]

Step 4: Work done.
\[ W = U_f - U_i \approx 0.001152 \pi - 0.000363 \pi \approx 0.000789 \pi \approx 1150 \pi \times 10^{-4}~\text{J} \]

Step 5: Conclusion.
The work done is \(1150 \pi \times 10^{-4}\) J.