Q100 / 125 · BITSAT 2010 · Physics

Two mercury drops (each of radius $\mathrm{r}$ ) merge to form a bigger drop. The surface energy of the bigger drop, if $\mathrm{T}$ is the surface tension, is

  • A
    $2^{5 / 3} \pi r^{2} T$
  • B
    $4 \pi \mathrm{r}^{2} \mathrm{~T}$
  • C
    $2 \pi^{2} T$
  • Correct answer
    D
    $2^{8 / 3} \pi \mathrm{r}^{2} \mathrm{~T}$

Solution

$\frac{4}{3} \pi R^{3}=2 \times \frac{4}{3} \pi r^{3} \Rightarrow \mathrm{R}=2^{1 / 3} \mathrm{r}$ Surface energy of bigger drop, $\mathrm{E}=4 \pi R^{2} T=4 \times 2^{2 / 3} \pi r^{2} T=2^{8 / 3} \pi r^{2} T$

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

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