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: \(7\)