the volume is conserved during the splitting process, the volume of the original drop must equal the total volume of the smaller drops:

$$ \frac{4}{3} \pi R^3 = 1000 \cdot \frac{4}{3} \pi r^3 $$

Simplifying gives:

$$ R^3 = 1000 r^3 $$

$$ R = 10r $$

The surface energy $ E $ is proportional to the surface area. Thus, we calculate the surface energy for the original and smaller drops:

Surface area of the original drop:

$$ A_1 = 4 \pi R^2 $$

Surface energy of the original drop:

$$ E_1 = \sigma \cdot 4 \pi R^2 $$

where $ \sigma $ is the surface tension.

For the 1000 smaller droplets:

Surface area of one small drop:

$$ A_2 = 4 \pi r^2 $$

Total surface area for 1000 drops:

$$ 1000 \cdot 4 \pi r^2 = 4000 \pi r^2 $$

Total surface energy for 1000 drops:

$$ E_2 = \sigma \cdot 4000 \pi r^2 $$

Substitute $ R = 10r $ into the expression for $ E_1 $ and $ E_2 $:

$ E_1 = \sigma \cdot 4 \pi (10r)^2 = \sigma \cdot 4 \pi \cdot 100r^2 = \sigma \cdot 400 \pi r^2 $

$ E_2 = \sigma \cdot 4000 \pi r^2 $

Compute the ratio $ \frac{E_1}{E_2} $:

$$ \frac{E_1}{E_2} = \frac{\sigma \cdot 400 \pi r^2}{\sigma \cdot 4000 \pi r^2} = \frac{400}{4000} = \frac{1}{10} $$

From $\frac{E_1}{E_2} = \frac{x}{10}$, we find that $ x = 1 $.

Therefore, the value of $ x $ is:

Option D: 1

Fluid Mechanics

Continue this paper