Surface Energy

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 $\frac{4}{3}\pi R^3=1000\times \frac{4}{3}\pi r^3$, giving $r=R/10$. The original surface area is $4\pi R^2$, while the combined area of the droplets is $1000\times 4\pi r^2=1000\times 4\pi (R/10{)}^2=1000\times \frac{4\pi R^2}{100}=40\pi R^2$. 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 $1000\times 4\pi R^{2}S$ 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.