A water drop of radius \(1\,cm\) is broken into \(729\) equal droplets. If surface tension of water is \(75\,dyne / cm\), then the gain in surface energy upto first decimal place will be\(...\times 10^{-4}\,J\)

Where \(T\) is surface tension and \(A\) is surface area

\(U _{ i }=\left(\frac{75 \times 10^{-5}}{10^{-2}} \frac{ N }{ m }\right) \times\left[4 \pi\left(1 \times 10^{-2}\right)^{2}\right]\)

\(=75 \times 10^{-3} \times 4 \pi \times 10^{-4}=942 \times 10^{-7}\,J\)

To get final radius of drops by volume conservation

\(\frac{4}{3} \pi R^{3}=729\left(\frac{4}{3} \pi r^{3}\right)\)

\(R=\text { Initial radius }\)

\(r=\text { final radius }\)

\(r=\frac{R}{(729)^{1 / 3}}=\frac{R}{9}=\frac{1}{9}\,cm\)

Final surface energy

\(U _{ f }=729[ TA ]\)

\(=729\left[\frac{75 \times 10^{-5}}{10^{-2}} \frac{ N }{ m }\right] \times\left[4 \pi\left(\frac{1}{9} \times 10^{-2}\right)^{2}\right]\)

\(=729\left[75 \times 10^{-3} \times \frac{4 \pi \times 10^{-4}}{81}\right]\)

\(=9\left[942 \times 10^{-7} J \right]\)

Gain in surface energy

\(\Delta U =9 \times 942 \times 10^{-7}-942 \times 10^{-7}\)

\(=8 \times 942 \times 10^{-7} J =7536 \times 10^{-7}\,J\)

\(=7.5 \times 10^{-4}\,J\)