Question

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

Answer 1

Energy due to surface Tension $E=\sigma \Delta A$

By law of conservation of mass, volume of drop $V_1+V_2=V$

$r_1=0.lcm=0.1\times {10}^{-3}m={10}^{-3}m$

$r_2=0.2cm=2\times {10}^{-3}m$

$\Delta A=4\pi r_1^2-4\pi R^2=4\pi [r_1^2+r_2^2-R^2]$

$R$ is the radius of new drop formed by the combination of two smaller drops.

$\frac{4}{3}\pi R^3=\frac{4}{3}\pi r_1^3+\frac{4}{3}\pi r_2^3$

$\frac{4}{3}\pi R^3=\frac{4}{3}\pi [r_1^3+r_2^3]⟹R^3=r_1^3+r_2^3$

$R^3=\left[\right(1\times {10}^{-3}{)}^3+(2\times {10}^{-3}{)}^3]=[{10}^{-9}+8\times {10}^{-9}]=9\times {10}^{-9}{m}^3$

$\therefore R=2.1\times {10}^{-3}m$

$E=\Delta A\sigma$

$=4\times 3.14\times \left[\right({10}^{-3}{)}^2+(2.0\times {10}^{-3}{)}^2-(2.1\times {10}^{-3}{)}^2]\times 435.5\times {10}^{-3}$

$E=4\times 3.14\times 435.5\times {10}^{-9}[5-41]$

$E=1742.0\times 3.14\times {10}^{-9}\left[0.59\right]$

$E=32.2723\times {10}^{-7}J$

Energy is released due to formation of bigger drop from smaller drops and finally area will be smaller.