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)
$\frac{4}{3}\pi R^{3}=n.\frac{4}{3}\pi r^{3}$
$R^{3}=nr^{3}$
$R=n^{\frac{1}{3}}.r$
$E_{2}=surface \, energy \, of \, n \, drops=n\times 4\pi r^{2}\times T$
$E_{1}=surface \, of \, energy \, of \, big \, drop=4\pi R^{2}T$
$\therefore \frac{E_{2}}{E_{1}}=\frac{n r^{2}}{R^{2}}=\frac{n r^{2}}{\left(\right. n^{\frac{1}{3}} . r \left.\right)^{2}}=\frac{n r^{2}}{n^{\frac{2}{3}} . r^{2}} \\ \therefore \frac{E_{2}}{E_{1}}=n^{\frac{1}{3}}=\sqrt[3]{n}:1$