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

Given: - A big drop is formed by combining 1000 small droplets.

Step 1: Volume Conservation

Since the droplets coalesce to form one big drop, the total volume remains constant. Let \( r \) be the radius of each small droplet and \( R \) be the radius of the big drop.

The volume of one small droplet is:

\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]

The total volume of 1000 small droplets is:

\[ V_{\text{total}} = 1000 \times \frac{4}{3} \pi r^3 = \frac{4000}{3} \pi r^3 \]

The volume of the big drop is:

\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]

Equating the total volumes:

\[ \frac{4000}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

Simplifying:

\[ R^3 = 1000r^3 \]

Taking the cube root on both sides:

\[ R = 10r \]

Step 2: Surface Area Calculation

The surface area of one small droplet is:

\[ A_{\text{small}} = 4 \pi r^2 \]

The total surface area of 1000 small droplets is:

\[ A_{\text{total}} = 1000 \times 4 \pi r^2 = 4000 \pi r^2 \]

The surface area of the big drop is:

\[ A_{\text{big}} = 4 \pi R^2 = 4 \pi (10r)^2 = 4 \pi \times 100r^2 = 400 \pi r^2 \]

Step 3: Surface Energy Comparison

Surface energy is directly proportional to the surface area. Let \( E_{\text{small}} \) and \( E_{\text{big}} \) be the surface energies of the small droplets and the big drop, respectively. The ratio of the surface energies is:

\[ \frac{E_{\text{big}}}{E_{\text{total}}} = \frac{A_{\text{big}}}{A_{\text{total}}} = \frac{400 \pi r^2}{4000 \pi r^2} = \frac{1}{10} \]

Conclusion:

The surface energy will become \( \frac{1}{10} \)th of its original value.