1. Principle used

When a liquid drop is broken into smaller drops, the surface area increases.
The work required is equal to the increase in surface energy:

$W = T\,\Delta A$

where $T$ is the surface tension and $\Delta A$ is the increase in surface area.

2. Initial surface area of the big drop

If the big drop has radius $R$, then its surface area is

$A_i = 4\pi R^2$

3. Case 1: Big drop broken into 27 equal drops

Let the radius of each small drop be $r$.

By conservation of volume:

$\frac{4}{3}\pi R^3 = 27\cdot \frac{4}{3}\pi r^3$

$R^3 = 27r^3$

$r = \frac{R}{3}$

So total final surface area is

$A_f = 27\cdot 4\pi r^2 = 27\cdot 4\pi \left(\frac{R}{3}\right)^2$

$A_f = 27\cdot 4\pi \frac{R^2}{9} = 12\pi R^2$

Hence increase in area:

$\Delta A_1 = A_f - A_i = 12\pi R^2 - 4\pi R^2 = 8\pi R^2$

Thus,

$W_1 = T(8\pi R^2) = 10\text{ J}$

4. Case 2: Big drop broken into 64 equal drops

Let radius of each new small drop be $r'$.

Again by conservation of volume:

$\frac{4}{3}\pi R^3 = 64\cdot \frac{4}{3}\pi r'^3$

$R^3 = 64r'^3$

$r' = \frac{R}{4}$

Total final surface area:

$A_f' = 64\cdot 4\pi r'^2 = 64\cdot 4\pi \left(\frac{R}{4}\right)^2$

$A_f' = 64\cdot 4\pi \frac{R^2}{16} = 16\pi R^2$

Increase in area:

$\Delta A_2 = 16\pi R^2 - 4\pi R^2 = 12\pi R^2$

So,

$W_2 = T(12\pi R^2)$

Now,

$\frac{W_2}{W_1} = \frac{12\pi R^2}{8\pi R^2} = \frac{3}{2}$

$W_2 = \frac{3}{2}\times 10 = 15\text{ J}$

5. Option check

• A: $20\,$J $\rightarrow$ incorrect
• B: $10\,$J $\rightarrow$ incorrect
• C: $5\,$J $\rightarrow$ incorrect
• D: $15\,$J $\rightarrow$ correct

6. Comparison with stored answer

Derived answer = D: $15\,$J
Stored correct answer = D

They agree.