A drop of mercury of radius 2 mm is split into 8 identical...

Let the radius of the large drop be $R = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$ and the radius of each small droplet be $r$.

From the conservation of volume:

$\dfrac{4}{3}\pi R^3 = 8 \times \dfrac{4}{3}\pi r^3$

$R^3 = 8r^3 \Rightarrow r = \dfrac{R}{2}$

Initial surface area of the drop is $A_1 = 4\pi R^2$.

Final surface area of the $8$ droplets is $A_2 = 8 \times 4\pi r^2 = 32\pi \left(\dfrac{R}{2}\right)^2 = 8\pi R^2$.

Increase in surface area is $\Delta A = A_2 - A_1 = 8\pi R^2 - 4\pi R^2 = 4\pi R^2$.

Increase in surface energy is given by $\Delta U = T \Delta A$, where $T$ is the surface tension.

Substituting the given values:

$\Delta U = 0.465 \times 4\pi (2 \times 10^{-3})^2$

$\Delta U = 0.465 \times 16\pi \times 10^{-6}$

$\Delta U = 7.44\pi \times 10^{-6}$

$\Delta U \approx 23.37 \times 10^{-6} \text{ J} = 2.34 \times 10^{-5} \text{ J}$