To solve this problem, we need to consider the change in surface energy when multiple small drops coalesce into a larger drop.$\\$Given:$\\$• Radius of each small drop $r. \\$• Radius of the large drop formed $R. \\$• Surface tension of the liquid $T. \\$The volume of a single small drop is given by:$\\ V_{small} = \frac{4}{3} \pi r^3 \\$Let $n$ be the number of small drops. The total volume of the small drops is:$\\ n \cdot V_{small} = n \cdot \frac{4}{3} \pi r^3 \\$The volume of the large drop is:$\\ V_{large} = \frac{4}{3} \pi R^3 \\$Since the total volume is conserved, we have:$\\ n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \\$Simplifying, we find:$\\ n \cdot r^3 = R^3 \\$The surface area of a single small drop is:$\\ A_{small} = 4 \pi r^2 \\$The total surface area of all small drops is:$\\ A_{total\_small} = n \cdot 4 \pi r^2 \\$The surface area of the large drop is:$\\ A_{large} = 4 \pi R^2 \\$The change in surface area when the drops coalesce is:$\\ \Delta A = A_{total\_small} - A_{large} \\ \Delta A = n \cdot 4 \pi r^2 - 4 \pi R^2 \\$Substituting $n = \frac{R^3}{r^3},$ we get:$\\ \Delta A = \frac{R^3}{r^3} \cdot 4 \pi r^2 - 4 \pi R^2 \\ \Delta A = 4 \pi \left( \frac{R^3}{r} - R^2 \right) \\$The change in surface energy is given by:$\\ \Delta E = T \cdot \Delta A \\ \Delta E = T \cdot 4 \pi \left( \frac{R^3}{r} - R^2 \right) \\$Simplifying further, we use $V = \frac{4}{3} \pi R^3$ to express $R^3: \\ R^3 = \frac{3V}{4\pi} \\$Substitute $R^3$ in $\Delta E: \\ \Delta E = T \cdot 4 \pi \left( \frac{\frac{3V}{4\pi}}{r} - R^2 \right) \\ \Delta E = 3VT \left( \frac{1}{r} - \frac{1}{R} \right) \\$Therefore, the energy released is $3VT \left( \frac{1}{r} - \frac{1}{R} \right). \\$This corresponds to Option 3.