Question

Calculate the work done to break up a drop of mercury of diameter 10^-2 m into 8 drops all of the same size.

Answer 1

$$\begin{gathered} Now, \\ Volumeofonebigdroplet=8\times volumeofsmalllittledroplets \\ The, \\ \frac{4\pi }{3}\times {\left({10}^{-2}\right)}^3=8\times \frac{4\pi }{3}\times r^3 \\ \frac{1}{8}\times {\left({10}^{-2}\right)}^3=r^3 \\ r=\frac{1}{2}\times {10}^{-2}m \\ now, \\ surfaceareaofbigdoplet=4\pi {\left(R\right)}^2 \\ surfaceareaof8tinydroplets=8\times 4\pi {\left(r\right)}^2 \\ Thereforechangeinsurfacearea=8\times 4\pi {\left(r\right)}^2-4\pi {\left(R\right)}^2 \\ ∆A=8\times 4\pi \times \frac{1}{4}\times {10}^{-4}-4\pi \times {10}^{-4} \\ ∆A=4\pi \times {10}^{-4} \\ now, \\ workdone=∆A\times surfacetension \\ forwatersurfacetensio=7\times {10}^{-2} \\ therefore, \\ W=4\pi \times {10}^{-4}\times 7\times {10}^{-2} \\ W=28\pi \times {10}^{-6} \\ W=8.796\times {10}^{-5}J \end{gathered}$$