Question

A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become :

• A. 100 times
• B. 10 times
• C. \(\frac{1}{100}th\)
• D. \(\frac{1}{10}th\)

Answer 1

The Correct Option is D

To determine the change in surface energy during droplet coalescence, we analyze the relationship between droplet size and surface area. Consider the scenario where 1000 small droplets merge into a single larger droplet.

The process can be described as follows:

1. Let \( r \) denote the radius of a single small droplet. The volume of one small droplet is \( V_{\text{small}} = \frac{4}{3}\pi r^3 \).
2. With 1000 small droplets, the total initial volume is \( V_{\text{total}} = 1000 \times \frac{4}{3}\pi r^3 \).
3. Upon coalescence, the volume of the resulting large droplet, with radius \( R \), remains constant: \( V_{\text{big}} = \frac{4}{3}\pi R^3 \).
4. Equating the initial total volume to the final volume yields \( \frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3 \).
5. Simplifying this equation, we get \( R^3 = 1000 \times r^3 \).
6. Taking the cube root of both sides, we find the radius of the large droplet to be \( R = 10r \).
7. Surface energy is directly proportional to surface area. The total surface area of the 1000 small droplets is \( A_{\text{small total}} = 1000 \times 4\pi r^2 \).
8. The surface area of the single large droplet is \( A_{\text{big}} = 4\pi R^2 \). Substituting \( R = 10r \), we get \( A_{\text{big}} = 4\pi (10r)^2 = 400\pi r^2 \).
9. The ratio of the surface area of the large droplet to the total surface area of the small droplets is \( \frac{A_{\text{big}}}{A_{\text{small total}}} = \frac{400\pi r^2}{1000 \times 4\pi r^2} = \frac{400}{4000} = \frac{1}{10} \).
10. Consequently, the surface energy of the large droplet is \(\frac{1}{10}\) times the total surface energy of the initial small droplets.

Therefore, the surface energy is reduced to \(\frac{1}{10}th\) of its original value.