Question

A big drop is formed by coalescing 1000 small identical drops of water. If E₁ be the total surface energy of 1000 small drops of water and E₂ be the surface energy of single big drop of water, the E₁ : E₂ is x : 1 where x = ________.

Answer 1

Correct Answer: 10

The surface energy of a droplet is given by:

\[ E = \sigma \times A, \]

where \( \sigma \) is the surface tension and \( A \) is the surface area.

Let the radius of each small drop be \( r \). The surface area of one small drop is:

\[ A_1 = 4 \pi r^2. \]

For 1000 small drops, the total surface area is:

\[ A_1(\text{total}) = 1000 \times 4 \pi r^2 = 4000 \pi r^2. \]

The total surface energy of the small drops is:

\[ E_1 = \sigma \times A_1(\text{total}) = \sigma \times 4000 \pi r^2. \]

When the 1000 small drops coalesce, the total volume remains the same. The volume of one small drop is:

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

The total volume of 1000 small drops is:

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

If the radius of the large drop is \( R \), the volume of the large drop is:

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

Equating the volumes:

\[ \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (1000 r^3), \]

which simplifies to:

\[ R^3 = 1000 r^3 \Rightarrow R = 10r. \]

The surface area of the large drop is:

\[ A_2 = 4 \pi R^2 = 4 \pi (10r)^2 = 4 \pi \times 100r^2 = 400 \pi r^2. \]

The surface energy of the large drop is:

\[ E_2 = \sigma \times A_2 = \sigma \times 400 \pi r^2. \]

The ratio of surface energies is:

\[ \frac{E_1}{E_2} = \frac{\sigma \times 4000 \pi r^2}{\sigma \times 400 \pi r^2}. \]

Simplifying:

\[ \frac{E_1}{E_2} = \frac{4000}{400} = 10. \]

Final Answer: The ratio of surface energies is: 10.