A water drop of diameter \(cm\) is broken into \(64\) equal droplets. The surface tension of water is \(0.075\,N / m\). In this process the gain in surface energy will be ...........\(J\)

\(\Delta SE = T \Delta A\)

\(=0.075\left( A _{ f }- A _{1}\right)\)

\(A _{ i }=4 \pi r ^{2}\)

\(A _{ t }=4 \pi r _{0}^{2} \times 64\)

By volume conservation

\(\frac{4}{3} \pi r ^{3}=64 \cdot \frac{4}{3} \pi r_{0}^{3}\)

\(I _{0}=\frac{ r }{4}\)

\(A _{ f }=4 \pi\left(\frac{ r }{4}\right)^{2} \cdot 64=16 \pi r ^{2}\)

\(\Delta SE =0.075\left(16 \pi r ^{2}-4 \pi r ^{2}\right)\)

\(=0.075\left(12 \pi(0.01)^{2}\right)\)

\(=2.8 \times 10^{-4}\,J\)