Question

A big water drop is formed by the combination of $n$ small water drops of equal radii. The ratio of the surface energy of $n$ drops to the surface energy of the big drop is

• A. $n^2:1$
• B. $n:1$
• C. $\sqrt{n}:1$
• D. $\sqrt[3]{n}:1$

Answer 1

Answer: (D)

Volume of big drop = $n$ (Volume of small drop)
$\frac{4}{3}\pi R^3=n.\frac{4}{3}\pi r^3$
$R^3=n{r}^3$
$R=n^{\frac{1}{3}}.r$
$E_2=surfaceenergyofndrops=n\times 4\pi r^2\times T$
$E_1=surfaceofenergyofbigdrop=4\pi R^{2}T$
$$\begin{gathered} \therefore \frac{E_2}{E_1}=\frac{n{r}^2}{R^2}=\frac{n{r}^2}{(n^{\frac{1}{3}}.r{)}^2}=\frac{n{r}^2}{n^{\frac{2}{3}}.r^2} \\ \therefore \frac{E_2}{E_1}=n^{\frac{1}{3}}=\sqrt[3]{n}:1 \end{gathered}$$