A drop of liquid of radius $R = 10^{-2}\ \mathrm{m}$ having surface tension $S = 0.1/(4\pi)\ \mathrm{Nm^{-1}}$ divides itself into $K$ identical drops. In this process the total change in the surface energy $\Delta U = 10^{-3}\ \mathrm{J}$. If $K = 10^{\alpha}$ then the value of $\alpha$ is ______.

Step 1: Apply conservation of volume.

The total volume of the $K$ identical smaller drops equals the volume of the original drop.

$$K \cdot \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi R^{3}$$ $$r = R K^{-1/3}$$

Step 2: Determine the initial and final surface areas.

Calculate the surface area before and after the division.

$$A_i = 4\pi R^{2}$$ $$A_f = K \cdot 4\pi r^{2}$$ $$A_f = 4\pi R^{2} K^{1/3}$$

Step 3: Formulate the change in surface energy.

The change in surface energy is the product of surface tension and the change in surface area.

$$\Delta U = S \left(A_f - A_i\right)$$ $$\Delta U = S \cdot 4\pi R^{2} \left(K^{1/3} - 1\right)$$

Step 4: Calculate the constant factor.

Substitute the given values $S = \frac{0.1}{4\pi}\,\text{Nm}^{-1}$ and $R = 10^{-2}\,\text{m}$.

$$S \cdot 4\pi R^{2} = \frac{0.1}{4\pi} \cdot 4\pi \left(10^{-2}\right)^{2}$$ $$S \cdot 4\pi R^{2} = 10^{-5}\,\text{J}$$

Step 5: Solve for $K$.

Substitute the calculated factor and the given $\Delta U = 10^{-3}\,\text{J}$ into the energy equation.

$$10^{-5} \left(K^{1/3} - 1\right) = 10^{-3}$$ $$K^{1/3} - 1 = 100$$ $$K^{1/3} = 101$$ $$K = 101^{3}$$ $$K \approx 10^{6}$$

Step 6: Find the value of $\alpha$.

Compare the value of $K$ with $10^{\alpha}$.

$$K = 10^{\alpha}$$ $$10^{\alpha} \approx 10^{6}$$ $$\alpha = 6$$

Answer: $6$