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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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