A drop of mercury of radius 2mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 J/m²)

To find the increase in surface energy when a drop of mercury is split into multiple droplets, we need to calculate the change in the total surface area and then multiply it by the surface tension of mercury.

Let the radius of the initial single large mercury drop be $R$ and the radius of each of the $n$ identical smaller droplets be $r$.
Here, we are given:
Initial radius, $R=2\text{mm}=2\times {10}^{-3}\text{m}$
Number of droplets, $n=8$
Surface tension of mercury, $T=0.465{\text{J/m}}^2$

Since the total volume of mercury remains conserved during the splitting process, the volume of the large drop must equal the sum of the volumes of the $n$ droplets:
$\frac{4}{3}\pi R^3=n\times \frac{4}{3}\pi r^3$

Simplifying the volume conservation equation yields a relationship between $R$ and $r$:
$R^3=n{r}^3$
Taking the cube root on both sides:
$R=n^{\frac{1}{3}}$$r$

Substituting $n=8$ into the equation:
$R=8^{\frac{1}{3}}$$r=2r$
Therefore, the radius of each droplet is:
$r=\frac{R}{2}=\frac{2\times {10}^{-3}\text{m}}{2}$

Next, we determine the change in surface area ($\mathrm{\Delta A}$).
Initial surface area of the single large drop:
$A_i=4\pi R^2$
Final total surface area of the 8 smaller droplets:
$A_f=8\times 4\pi r^2$

The increase in surface area is:
$\mathrm{\Delta A}=A_f-A_i=8\times 4\pi r^2-4\pi R^2$
Substituting $r=\frac{R}{2}$:
$\mathrm{\Delta A}=32\pi {\left(\frac{R}{2}\right)}^2-4\pi R^2=32\pi \frac{R^2}{4}-4\pi R^2=8\pi R^2-4\pi R^2=4\pi R^2$

Now, we calculate the increase in surface energy ($\mathrm{\Delta E}$) using the relation:
$\mathrm{\Delta E}=T\times \mathrm{\Delta A}$
Substituting the values:
$\mathrm{\Delta E}=0.465\times 4\pi R^2$
$\mathrm{\Delta E}=0.465\times 4\times 3.14159\times {\left(2\times {10}^{-3}\right)}^2$
$\mathrm{\Delta E}=0.465\times 4\times 3.14159\times 4\times {10}^{-6}$
$\mathrm{\Delta E}=23.38\times {10}^{-6}\text{J}\approx 23.4\mu \text{J}$

Thus, the increase in surface energy is 23.4 μ J.