Surface Energy

Question

A single large mercury drop of radius R is broken up into 1000 identical tiny droplets without any loss of mass. If the surface tension of mercury is S, what is the increase in the total surface energy of the system?

RankGuru · Accepted Answer

Surface energy equals surface tension multiplied by the total exposed area, because creating new surface requires work against the cohesive forces between molecules. When the parent drop splits, volume is conserved, so $\frac{4}{3}\pi R^3 = 1000	imes\frac{4}{3}\pi r^3$, giving $r = R/10$. The original surface area is $4\pi R^2$, while the combined area of the droplets is $1000	imes4\pi r^2 = 1000	imes4\pi (R/10)^2 = 1000	imes\frac{4\pi R^2}{100} = 40\pi R^2$. The increase in area is $40\pi R^2 - 4\pi R^2 = 36\pi R^2$, so the rise in surface energy is $\Delta E = 36\pi R^2 S$. The option $4\pi R^2 S$ is merely the original drop's energy. The option $40\pi R^2 S$ forgets to subtract the parent surface. The option $1000	imes4\pi R^2 S$ wrongly uses the parent radius for every droplet. This follows the NCERT principle that subdivision increases surface area and therefore stores extra energy, with the general result $\Delta E = 4\pi R^2 S\,(n^{1/3} - 1)$ reducing to $36\pi R^2 S$ when $n = 1000$. The energy must be supplied externally, which is why the fragmented droplets are cooler if the energy is drawn from their internal thermal store. As a sanity check, breaking one drop into many must increase area and require energy input, so a positive, large multiple of $\pi R^2 S$ is exactly what we expect.

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