Q147 / 180 · COMEDK 2014 · Physics

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

  • A
    $1: 2^{1 / 3}$
  • Correct answer
    B
    $2^{1 / 3}: 1$
  • C
    $2: 1$
  • D
    $1: 2$

Solution

As, radius of bigger drop, $R=n^{1 / 3} r=2^{1 / 3} r$ $$ \begin{aligned} &\Rightarrow R^{2}=2^{2 / 3} r^{2} \Rightarrow \frac{r^{2}}{R^{2}} \text { or } 2^{-2 / 3} \\ &\begin{aligned} \frac{\text { Initial surface energy }}{\text { Final surface energy }} &=\frac{2\left(4 \pi r^{2} T\right)}{\left(4 \pi R^{2} T\right)}=2\left(\frac{r^{2}}{R^{2}}\right) \\ &=2 \times 2^{-2 / 3}=2^{1 / 3} \text { or } 2^{1 / 3}: 1 \end{aligned} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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