Q135 / 160 · AP EAMCET 2024 (20 May Shift 2) · Physics

Two mercury drops, each with same radius $r$, merged to form a bigger drop. $T$ is the surface tension of mercury, then the surface energy of bigger drop is given by

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

Solution

$\mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}} \Rightarrow 2 \times \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3 \Rightarrow \mathrm{R}=(2)^{\frac{1}{3}} \mathrm{r}$ $\therefore \quad$ Surface energy of bigger drop, $\begin{aligned} & \mathrm{E}=\mathrm{TA}=\mathrm{T}\left(4 \pi \mathrm{R}^2\right)=4 \pi(2)^{2 / 3} \mathrm{r}^2 \mathrm{~T} \\ & =2^{8 / 3} \pi \mathrm{r}^2 \mathrm{~T} \end{aligned}$

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

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