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

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Surface tension energy, $E=\sigma \Delta A$

As we can infer from the law of conservation of mass, the volume of a drop $V=v_1+v_2$

$r_1 = 0.1 cm = 0.001 m$

$r_2 = 0.2 cm = 0.002 m$

$\Delta A = 4 \pi r_1^{2} + 4 \pi r_2^{2} - 4\pi R^{2}$

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

The new bigger drop has a radius of R,

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

$R^{3}= r_1^{3}+r_2^{3}$

$R^{3}= 9 \times 10^{-9}$

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

Now, $E=\sigma \Delta A$

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

$E=1742\times 3.14\times 10^{-9}\times [-0.59]=-32.27 \times 10^{-7}~ J$

$Hence,E=-32.27 \times 10^{-7}~ J$ [Negative sign shows absorption]

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