Q135 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Physics

Energy needed in breaking a liquid drop of radius $R$, into $n$ smaller drops each of radius $r$, is [T- Surface tension of the liquid]

  • Correct answer
    A
    $\left(4 \pi r^2 n-4 \pi R^2\right) T$
  • B
    $\left(\frac{4}{3} \pi r^3 n-\frac{4}{3} \pi R^3\right) T$
  • C
    $\left(4 \pi R^2-4 \pi r^2\right) n T$
  • D
    $\left(4 \pi R^2-n 4 \pi r^2\right) / T$

Solution

We have $\Delta \mathrm{U}=$ (surface energy of bigger drop) - (surface energy of lower drop) $\begin{aligned} & \Delta \mathrm{U}=\left(\mathrm{T} \times 4 \pi \mathrm{R}^2\right)-\left(\mathrm{nT} \times 4 \pi \mathrm{r}^2\right) \\ & |\Delta \mathrm{U}|=\left(4 \pi \mathrm{r}^2 \mathrm{n}-4 \pi \mathrm{R}^2\right) \mathrm{T}\end{aligned}$

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

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