Two small drops of mercury, each of radius R | vedclass

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Question

Let $\mathrm{r}$ be radius of common drop

$\frac{4}{3} \pi r^{3}=2 	imes \frac{4}{3} \pi R^{3}$

$\mathrm{r}=(2)^{\frac{1}{3}} \mathrm{R}$

Su

Vedclass · Accepted Answer

$2^{1/3} : 1$

Vedclass · Answer

Let $\mathrm{r}$ be radius of common drop

$\frac{4}{3} \pi r^{3}=2 	imes \frac{4}{3} \pi R^{3}$

$\mathrm{r}=(2)^{\frac{1}{3}} \mathrm{R}$

Surface energy before the coalesce

$=2 	imes 4 \pi \mathrm{R}^{2} \mathrm{T}$

Surface energy after the coalesce $=4 \pi r^{2} T$

 Ratio $=\frac{2 	imes 4 \pi \mathrm{R}^{2} \mathrm{T}}{4 \pi \mathrm{r}^{2} \mathrm{T}}=\frac{2 \mathrm{R}^{2}}{2^{2 / 3} \mathrm{R}^{2}}$

$=\frac{2^{\frac{1}{3}} \cdot 2^{\frac{2}{3}}}{2^{\frac{2}{3}}}=\frac{2^{\frac{1}{3}}}{1} $

Two small drops of mercury, each of radius $R$ coalesce to form a single large drop. The radio of the total surface energies before and after the change is

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