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 big drop is

Physics | Physics

$n^{2} : 1$
$n : 1$
$\sqrt[3]{n} : 1$
$\sqrt{n} : 1$

Solution: The correct answer is C.

$ By\text{ }mass\text{ }conservation\text{ }new\text{ }radius\text{ }R\text{ }is~ $

$ R={{n}^{1/3}}r $

$ ratio\text{ }of\text{ }energy\text{ }is:\,~\frac{nT\times 4\pi {{r}^{2}}}{T\times 4\pi {{R}^{2}}}={{n}^{1/3}}~:1 $

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