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Question

If a drop of radius $$r$$ breaks up into $$27$$ small drops then how much will be the change in the surface energy. The surface tension of liquid is $$T$$.


Solution

Let $$R=$$radius of small drops. Then, by equation the volumes we have.
$$\displaystyle \frac{4}{3}\pi r^{3}=27\left ( \frac{4}{3}\pi R^{3} \right )$$
or $$\displaystyle R=\frac{r}{3}$$
$$\Delta U=T\left ( \Delta A \right )$$
   $$=T\left ( A_{f}-A_{i} \right )$$
   $$=T\left [ 27\left ( 4\pi R^{2} \right )-4\pi r^{2} \right ]$$
   $$\displaystyle =4\pi T\left [ 27\left ( \frac{r}{2} \right )^{2}-r^{2} \right ]$$
   $$=8\pi r^{2}T$$

Physics

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