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1000 small water drops of equal size combine to form a big drop. The ratio of final surface energy to the total initial surface energy is
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The correct answer is:
$1:10$
Radii of the bigger and smaller drops are related as, $\mathrm{R} \propto \mathrm{n}^{\frac{1}{3}} \mathrm{r}$
Now, surface energy $E=T d A=T \times 4 \pi R^2$
$\begin{array}{ll}
\therefore \quad & \frac{\mathrm{E}_2}{\mathrm{E}_1}=\frac{4 \pi \mathrm{R}^2 \times \mathrm{T}}{\mathrm{n} \times 4 \pi \mathrm{r}^2 \mathrm{~T}} \\
\therefore \quad \frac{\mathrm{E}_2}{\mathrm{E}_1} & =\frac{\mathrm{R}^2}{\mathrm{nr}^2}=\frac{\mathrm{n}^{\frac{2}{3}} \times \mathrm{r}^2}{\mathrm{nr}^2} \\
& \frac{\mathrm{E}_2}{\mathrm{E}_1}=\frac{1}{\mathrm{n}^{\frac{1}{3}}}=\frac{1}{1000^{\frac{1}{3}}} \\
\therefore \quad \frac{\mathrm{E}_2}{\mathrm{E}_1} & =1: 10
\end{array}$
Now, surface energy $E=T d A=T \times 4 \pi R^2$
$\begin{array}{ll}
\therefore \quad & \frac{\mathrm{E}_2}{\mathrm{E}_1}=\frac{4 \pi \mathrm{R}^2 \times \mathrm{T}}{\mathrm{n} \times 4 \pi \mathrm{r}^2 \mathrm{~T}} \\
\therefore \quad \frac{\mathrm{E}_2}{\mathrm{E}_1} & =\frac{\mathrm{R}^2}{\mathrm{nr}^2}=\frac{\mathrm{n}^{\frac{2}{3}} \times \mathrm{r}^2}{\mathrm{nr}^2} \\
& \frac{\mathrm{E}_2}{\mathrm{E}_1}=\frac{1}{\mathrm{n}^{\frac{1}{3}}}=\frac{1}{1000^{\frac{1}{3}}} \\
\therefore \quad \frac{\mathrm{E}_2}{\mathrm{E}_1} & =1: 10
\end{array}$
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