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

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.