A small liquid drop of radius $$R$$ is divided into $$27$$ identical liquid drops. If the surface tension is $$T$$, then the work done in the process will be:

A liquid drop of radius $$R$$ is divided into 27 identical smaller drops. By conserving volume, we have $$\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3$$, which simplifies to $$R^3 = 27r^3$$ and hence $$r = \frac{R}{3}$$.

The initial surface area of the drop is $$A_i = 4\pi R^2$$. After division into 27 drops of radius $$r$$, the total surface area becomes $$A_f = 27 \times 4\pi r^2 = 27 \times 4\pi \times \frac{R^2}{9} = 12\pi R^2$$. Therefore, the change in surface area is $$\Delta A = A_f - A_i = 12\pi R^2 - 4\pi R^2 = 8\pi R^2$$.

The work done in increasing the surface area against surface tension is given by $$W = T \times \Delta A = T \times 8\pi R^2 = 8\pi R^2 T$$.

The work done is $$8\pi R^2 T$$, which corresponds to Option 1.