A liquid drop of radius ' $R$ ' is broken into ' $n$ ' identical small droplets. The work done is…
A liquid drop of radius ' $R$ ' is broken into ' $n$ ' identical small droplets. The work done is $[\mathrm{T}=$ surface tension of the liquid]
- $4 \pi R^2\left(n^{\frac{2}{3}}-1\right) T$
- $4 \pi R^2\left(n^{\frac{1}{3}}-1\right) T$
- $4 \pi R^2\left(1-n^{\frac{1}{3}}\right) T$
- $4 \pi R^2\left(1-n^{\frac{2}{3}}\right) T$
Solution
Volume of $\mathrm{n}$ smaller droplets $=$ Volume of bigger drop
$\begin{aligned}
& \mathrm{n} \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3 \\
& \therefore \quad \mathrm{R}=\mathrm{n}^{\frac{1}{3}} \cdot \mathrm{r} \\
& \mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{\frac{1}{3}}}
\end{aligned}$
$\begin{aligned} \text { Work done } W & =\left[n \cdot 4 \pi r^2-4 \pi R^2\right] T \\ & =4 \pi\left[n \cdot \frac{R^2}{n^{\frac{2}{3}}}-R^2\right] T \\ & =4 \pi R^2\left[n^{\frac{1}{3}}-1\right] T\end{aligned}$
Asked in: MHT CET 2023 (11 May Shift 2)
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