1. Surface energy formula

For a liquid drop, surface energy is $E = T \cdot A$ where $T$ is surface tension and $A$ is surface area.

For a spherical drop of radius $r$, $A = 4\pi r^2$ So, $E = T(4\pi r^2)$

2. Surface energy of 1000 small drops

Let radius of each small drop be $r$.

Surface energy of one small drop: $E_{\text{small}} = 4\pi r^2 T$

Therefore, for 1000 identical small drops, $E_1 = 1000 \cdot 4\pi r^2 T$

3. Radius of the big drop after coalescence

When 1000 drops coalesce, volume is conserved.

Volume of one small drop: $V_{\text{small}} = \frac{4}{3}\pi r^3$

Volume of 1000 small drops: $1000 \cdot \frac{4}{3}\pi r^3$

Let radius of big drop be $R$. Then $\frac{4}{3}\pi R^3 = 1000 \cdot \frac{4}{3}\pi r^3$

So, $R^3 = 1000r^3$ $R = 10r$

4. Surface energy of the big drop

$E_2 = 4\pi R^2 T = 4\pi (10r)^2 T = 100 \cdot 4\pi r^2 T$

5. Find the ratio $E_1 : E_2$

$E_1 : E_2 = 1000\cdot 4\pi r^2 T : 100\cdot 4\pi r^2 T$

Cancelling common terms, $E_1 : E_2 = 1000:100 = 10:1$

Hence, in the form $x:1$, $x=10$

6. Comparison with stored answer

Stored correct answer = 10.

Our derived answer also gives $x=10$, so it agrees.