Eight mercury drops, each of radius \(r\), coalesce to form a bigger drop. The surface energy released in this process is _______. (\(S\) is the surface tension of mercury).

Let the radius of the bigger drop be \(R\). Since the total volume remains conserved during the process:

\(8 \times \dfrac{4}{3} \pi r^3 = \dfrac{4}{3} \pi R^3\)

\(R^3 = 8r^3 \Rightarrow R = 2r\)

Initial surface energy of the \(8\) drops is:

\(U_i = 8 \times (4 \pi r^2 S) = 32 \pi r^2 S\)

Final surface energy of the bigger drop is:

\(U_f = 4 \pi R^2 S = 4 \pi (2r)^2 S = 16 \pi r^2 S\)

The surface energy released in the process is:

\(\Delta U = U_i - U_f = 32 \pi r^2 S - 16 \pi r^2 S = 16 \pi r^2 S\)

Answer: \(16\pi r^2 S\)