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)

Let the radius of the initial drop be $R$ and the radius of each small droplet be $r$.

Given diameter $D = 2$ mm, so $R = 1$ mm $= 10^{-3}$ m.

By conservation of volume:

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

$R^3 = 512 r^3 \Rightarrow R = 8r \Rightarrow r = \dfrac{R}{8}$

Initial surface area, $A_i = 4\pi R^2$

Final surface area, $A_f = 512 \times 4\pi r^2 = 512 \times 4\pi \left(\dfrac{R}{8}\right)^2 = 32\pi R^2$

Change in surface area, $\Delta A = A_f - A_i = 32\pi R^2 - 4\pi R^2 = 28\pi R^2$

Change in surface energy, $\Delta U = T \Delta A$

$\Delta U = 0.08 \times 28\pi R^2$

Substituting $R = 10^{-3}$ m and $\pi \approx \dfrac{22}{7}$:

$\Delta U = 0.08 \times 28 \times \dfrac{22}{7} \times (10^{-3})^2$

$\Delta U = 0.08 \times 4 \times 22 \times 10^{-6}$

$\Delta U = 7.04 \times 10^{-6}$ J

Comparing with $\alpha \times 10^{-6}$ J, we get $\alpha \approx 7$.

Answer: $7$