A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is $$\alpha \times 10^{-6}$$ J. The value of $$\alpha$$ is __________. (Take surface tension of liquid = 0.08 N/m)

The diameter of the original drop is $$2 \text{ mm}$$, hence its radius is
$$R = 1 \text{ mm} = 1 \times 10^{-3} \text{ m}$$

Let the radius of each of the $$512$$ identical smaller droplets be $$r$$.
Volume is conserved, so

$$\frac{4}{3}\pi R^{3}=512 \times \frac{4}{3}\pi r^{3}$$

$$\Rightarrow R^{3}=512\,r^{3} = 8^{3}\,r^{3}\; \Rightarrow\; r=\frac{R}{8}$$

Substituting $$R = 1\times10^{-3}\text{ m}$$:
$$r = \frac{1\times10^{-3}}{8}=1.25\times10^{-4}\text{ m}$$

Initial surface area of the single drop:
$$A_i = 4\pi R^{2}=4\pi (1\times10^{-3})^{2}=4\pi \times10^{-6}\; \text{m}^{2}$$

Surface area of one small droplet:
$$A_1 = 4\pi r^{2}=4\pi (1.25\times10^{-4})^{2}=4\pi (1.5625\times10^{-8})=6.25\pi\times10^{-8}\; \text{m}^{2}$$

Total surface area after breakup:
$$A_f = 512 \times A_1 = 512 \times 6.25\pi\times10^{-8} = 3200\pi\times10^{-8}=3.2\pi\times10^{-5}\; \text{m}^{2}$$

Increase in surface area:
$$\Delta A = A_f - A_i = 3.2\pi\times10^{-5} - 4\pi\times10^{-6} = 2.8\pi\times10^{-5}\; \text{m}^{2}$$

Surface tension of the liquid: $$T = 0.08 \text{ N m}^{-1}$$

Change in surface energy:
$$\Delta E = T\,\Delta A = 0.08 \times 2.8\pi\times10^{-5} = 2.24\pi\times10^{-6}\; \text{J}$$

Using $$\pi \approx 3.14$$:
$$\Delta E \approx 2.24 \times 3.14 \times10^{-6} \approx 7.03\times10^{-6}\; \text{J}$$

Thus $$\Delta E = \alpha \times 10^{-6}\text{ J}$$ with $$\alpha \approx 7$$.

Option B which is: $$7$$