1. Principle used

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

$W = T\,\Delta A$

where:

• $T$ = surface tension
• $\Delta A$ = increase in surface area

2. Find radius of each small drop

The original drop of radius $R$ is divided into $27$ identical drops. Since volume is conserved,

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

So,

$R^3 = 27r^3$ $r^3 = \frac{R^3}{27}$ $r = \frac{R}{3}$

3. Initial surface area

For the original drop,

$A_i = 4\pi R^2$

4. Final surface area

Each small drop has radius $r = \frac{R}{3}$. So surface area of one small drop is

$4\pi r^2 = 4\pi\left(\frac{R}{3}\right)^2 = \frac{4\pi R^2}{9}$

For $27$ such drops,

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

5. Increase in surface area

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

6. Work done

$W = T\,\Delta A = T \times 8\pi R^2$

Therefore,

$\boxed{W = 8\pi R^2 T}$

7. Option check

• A: $4\pi R^2T$ ❌
• B: $8\pi R^2T$ ✅
• C: $\frac{1}{8}\pi R^2T$ ❌
• D: $3\pi R^2T$ ❌

So the correct option is B.