d
(d) Volume of liquid remain same i.e. volume of \(1000\) small drops will be equal to volume of one big drop \(n\frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {R^3}\) \(⇒\) \(1000{r^3} = {R^3}\) \(⇒\) \(R = 10r\)

\(\therefore \frac{r}{R} = \frac{1}{{10}}\)

\(\frac{{{\rm{surface \,energy\, of\, one \,small\, drop}}}}{{{\rm{surface\, energy\, of\, one \,big \,drop}}}} = \frac{{4\pi {r^2}T}}{{4\pi {R^2}T}} = \frac{1}{{100}}\)