8000 identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is

To find the ratio of the final surface energy to the initial total surface energy, let us analyze the physical process of combining the drops step-by-step.

Step 1: Relate the radii of the drops using volume conservation
Let $r$ be the radius of each identical small water drop and $R$ be the radius of the combined big drop.
Since the volume of the water remains constant, the volume of the single big drop is equal to the sum of the volumes of the $n=8000$ small drops:
$\frac{4}{3}\pi R^3=n\cdot \left(\frac{4}{3}\pi r^3\right)$

By canceling out the common terms, we get:
$R^3=n{r}^3$
Taking the cube root of both sides:
$R=n^{\frac{1}{3}}r$

Given that $n=8000$:
$R={\left(8000\right)}^{\frac{1}{3}}r=20r$

Step 2: Express the initial and final surface energies
The surface energy of a spherical liquid drop with surface tension $T$ and surface area $A=4\pi a^2$ is given by $E=T\cdot A$.

The initial surface energy ($E_i$) of all the 8000 small drops together is:
$E_i=n\cdot \left(T\cdot 4\pi r^2\right)=8000\cdot T\cdot 4\pi r^2$

The final surface energy ($E_f$) of the single big drop is:
$E_f=T\cdot 4\pi R^2$

Substitute $R=20r$ into the equation for final surface energy:
$E_f=T\cdot 4\pi {\left(20r\right)}^2=400\cdot T\cdot 4\pi r^2$

Step 3: Calculate the ratio of the final to initial surface energy
Now, find the ratio $\frac{E_f}{E_i}$:
$\frac{E_f}{E_i}=\frac{400\cdot T\cdot 4\pi r^2}{8000\cdot T\cdot 4\pi r^2}=\frac{400}{8000}=\frac{1}{20}$

Thus, the ratio of the final surface energy to the initial surface energy of all the drops together is 1:20.