Question

# A drop of mercury of radius $$2\ mm$$ is split into $$8$$ identical drops. Find the increase in surface energy. Surface tension of mercury $$= 0.465\ Jm^{-2}$$

Solution

## $$R=2mm$$ $$\Rightarrow$$ $$2\times {10}^{3}m$$The volume of  drop is given as:$$\cfrac{4}{3}\pi{R}^{3}=8\times \cfrac{4}{3}\times {r}^{3}$$$$\therefore$$ $${R}^{3}=8{r}^{3}$$$$\cfrac{{R}^{3}}{8}={r}^{3}$$ $$\Rightarrow$$ $${(\cfrac{R}{2})}^{3}={r}^{3}$$$$\therefore$$ $$r=1mm={10}^{-3}m$$$${A}_{1}=4\pi {R}^{2}$$ $$\Rightarrow$$ $$4\pi {(2\times {10}^{-2})}^{2}=4\pi \times 4\times {10}^{-6}=16\pi \times {10}^{-6}$$$${A}_{2}=8\times 4\pi {r}^{2}=32\pi {({10}^{-2})}^{3}=32\pi {10}^{-6}$$$$\Delta A={A}_{1}-{A}_{2}$$$$(32\pi-16\pi){10}^{-6}=16\pi\times {10}^{-6}$$the increase in surface energy will be:$$E_s=T\Delta A=23.376$$Physics

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