A liquid drop having diameter 2mm and surface tension 0.08N/m. This drop splits into 512 identical small drops. Find change in surface energy.

The correct option is 7.034×10⁻⁶.

To find the change in surface energy when a larger drop splits into smaller drops, we can follow these steps:

1. Identify the given values:
Diameter of the initial liquid drop, $D=2\text{mm}$
Radius of the initial liquid drop, $R=\frac{D}{2}=1\text{mm}={10}^{-3}\text{m}$
Surface tension of the liquid, $T=0.08\text{N/m}$
Number of identical smaller drops, $n=512$

2. Relate the radii of the drops using volume conservation:
Since volume remains conserved when the large drop splits, the volume of the single large drop equals the total volume of the 512 smaller drops. Let $r$ be the radius of each small drop:
$\frac{4}{3}\pi R^3=n\cdot \frac{4}{3}\pi r^3$
$R^3=512r^3$
Taking the cube root on both sides:
$R=8r\Rightarrow r=\frac{R}{8}$

3. Calculate the change in surface area:
Initial surface area of the single large drop:
$A_i=4\pi R^2$
Final total surface area of all small drops:
$A_f=n\cdot 4\pi r^2=512\cdot 4\pi {\left(\frac{R}{8}\right)}^2=512\cdot 4\pi \frac{R^2}{64}=8\cdot 4\pi R^2$
The change in surface area ($\mathrm{\Delta A}$) is:
$\mathrm{\Delta A}=A_f-A_i=8\cdot 4\pi R^2-4\pi R^2=7\cdot 4\pi R^2=28\pi R^2$

4. Calculate the change in surface energy:
The change in surface energy ($\mathrm{\Delta E}$) is given by the product of surface tension and the change in surface area:
$\mathrm{\Delta E}=T\cdot \mathrm{\Delta A}$
$\mathrm{\Delta E}=T\cdot 28\pi R^2$
Substitute the given values:
$\mathrm{\Delta E}=0.08\cdot 28\cdot \pi \cdot {\left({10}^{-3}\right)}^2$
$\mathrm{\Delta E}=2.24\cdot 3.14159\cdot {10}^{-6}$
$\mathrm{\Delta E}\approx 7.034\times {10}^{-6}\text{J}$