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STD 11 - 9.2 , surface tension — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 9.2 , surface tension4 Marks
MCQ
$n$ drops of $a$ liquid, each with surface energy $E$, joining to form a single drop
  • ✓
    some energy will be released in the process
  • B
    some energy will be absorbed in the process
  • C
    the energy released or absorbed will be$\frac{E}{n}(n-n^{2/3})$
  • D
    the energy released or absorbed will be $nE \  (2^{2/3}-1)$
✓

Answer

Correct option: A.
some energy will be released in the process
a
$\mathrm{E}=\mathrm{T} 4 \pi \mathrm{r}^{2} \Rightarrow \frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{n} \times \frac{4 \pi}{3} \mathrm{r}^{3}$

$\Rightarrow \mathrm{n}=\frac{\mathrm{R}^{3}}{\mathrm{r}^{3}} \Rightarrow \mathrm{R}=\mathrm{n}^{1 / 3} \mathrm{r}$

Surface energy of big drop

$\mathrm{E}^{\prime}=\mathrm{T} 4 \pi \mathrm{R}^{2}=\mathrm{T} 4 \pi \mathrm{n}^{2 / 3} \mathrm{r}^{2}=\mathrm{En}^{2 / 3}$

Energy released

$=n \mathrm{E}-\mathrm{E}^{\prime}=\mathrm{nE}-\mathrm{n}^{2 / 3} \mathrm{E}=\mathrm{E}\left(\mathrm{n}-\mathrm{n}^{2 / 3}\right)$

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