Surface Tension And Surface Energy

As described in NCERT Class 11, Chapter 10 (Mechanical Properties of Fluids), surface tension makes a liquid behave as though its surface is a stretched membrane, and the surface energy is proportional to the total surface area. Splitting one big drop into many small drops conserves volume but increases surface area. Volume conservation gives $\frac{4}{3}\pi R^3=n\cdot \frac{4}{3}\pi r^3$, so $r=R/n^{1/3}$. With $n=1000$, $n^{1/3}=10$, hence $r=R/10$. The original area is $4\pi R^2$, and the total new area is $n\cdot 4\pi r^2=1000\cdot 4\pi (R/10{)}^2=1000\cdot 4\pi R^2/100=10\cdot 4\pi R^2$. The area therefore grows by a factor of 10. The option 100 confuses the radius ratio squared with the wrong power. The option 1000 mistakes the number of drops for the area factor. The option 30 has no physical basis. The general result worth remembering is that splitting one drop into $n$ equal drops increases the surface area by a factor of $n^{1/3}$, since radius scales as $n^{-1/3}$ while area scales as the square of radius times the number of drops. A sanity check: more surface area means greater surface energy, so energy must be supplied to create the spray, which is why atomisers and nozzles do work on the liquid. The factor of 10 increase, equal to $1000^{1/3}$, is therefore both mathematically and physically consistent with the energy that must be expended.