Two mercury droplets of radii $0.1 \ cm$ and $0.2 \ cm$ coalesce into one single drop. What amount of energy is released? The surface tension of mercury is $T = 435.5 \times 10^{-3} \ N \ m^{-1}$.

(N/A) Given radii: $r_1 = 0.1 \ cm = 10^{-3} \ m$ and $r_2 = 0.2 \ cm = 2 \times 10^{-3} \ m$.
Surface tension $T = 435.5 \times 10^{-3} \ N \ m^{-1}$.
Let the radius of the larger drop be $R$.
Since the volume remains conserved: $\frac{4}{3} \pi R^3 = \frac{4}{3} \pi r_1^3 + \frac{4}{3} \pi r_2^3$.
$R^3 = r_1^3 + r_2^3 = (0.1)^3 + (0.2)^3 = 0.001 + 0.008 = 0.009 \ cm^3$.
$R = (0.009)^{1/3} \approx 0.208 \ cm = 2.08 \times 10^{-3} \ m$.
Energy released $\Delta E = T \times \Delta A = T \times (A_{initial} - A_{final})$.
$A_{initial} = 4 \pi (r_1^2 + r_2^2) = 4 \pi (0.01 + 0.04) \times 10^{-4} = 4 \pi (0.05) \times 10^{-4} = 0.2 \pi \times 10^{-4} \ m^2$.
$A_{final} = 4 \pi R^2 = 4 \pi (0.208 \times 10^{-2})^2 = 4 \pi (0.04326) \times 10^{-4} \approx 0.173 \pi \times 10^{-4} \ m^2$.
$\Delta A = (0.2 - 0.173) \pi \times 10^{-4} = 0.027 \pi \times 10^{-4} \ m^2$.
$\Delta E = 435.5 \times 10^{-3} \times 0.027 \times 3.14 \times 10^{-4} \approx 3.69 \times 10^{-6} \ J$.