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).

$$E = \text{Surface Tension } (S) \times \text{Surface Area } (A)$$

$$\Delta E = E_{\text{initial}} - E_{\text{final}} = S \cdot (A_{\text{initial}} - A_{\text{final}})$$

$$\text{Volume of big drop} = 8 \times \text{Volume of one small drop}$$

$$\frac{4}{3}\pi R^3 = 8 \times \left(\frac{4}{3}\pi r^3\right)$$

$$R^3 = 8r^3 \implies R = \sqrt[3]{8r^3} = 2r$$

$$A_{\text{initial}} = 8 \times (4\pi r^2) = 32\pi r^2$$

$$A_{\text{final}} = 4\pi R^2 = 4\pi (2r)^2 = 4\pi (4r^2) = 16\pi r^2$$

$$\Delta E = S \cdot \left(A_{\text{initial}} - A_{\text{final}}\right)$$

$$\Delta E = S \cdot \left(32\pi r^2 - 16\pi r^2\right)$$

$$\Delta E = 16\pi r^2 S$$