MHT CET · Physics · Mechanical Properties of Fluids
A liquid drop of radius ' \(R\) ' is broken into ' \(n\) ' identical small droplets. The work done is \([\mathrm{T}=\) surface tension of the liquid]
- A \(4 \pi R^2\left(n^{\frac{2}{3}}-1\right) T\)
- B \(4 \pi R^2\left(n^{\frac{1}{3}}-1\right) T\)
- C \(4 \pi R^2\left(1-n^{\frac{1}{3}}\right) T\)
- D \(4 \pi R^2\left(1-n^{\frac{2}{3}}\right) T\)
Answer & Solution
Correct Answer
(B) \(4 \pi R^2\left(n^{\frac{1}{3}}-1\right) T\)
Step-by-step Solution
Detailed explanation
Volume of \(\mathrm{n}\) smaller droplets \(=\) Volume of bigger drop
\(\begin{aligned}
& \mathrm{n} \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3 \\
& \therefore \quad \mathrm{R}=\mathrm{n}^{\frac{1}{3}} \cdot \mathrm{r} \\
& \mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{\frac{1}{3}}}
\end{aligned}\)
\(\begin{aligned} \text {Work done } W & =\left[n \cdot 4 \pi r^2-4 \pi R^2\right] T \\ & =4 \pi\left[n \cdot \frac{R^2}{n^{\frac{2}{3}}}-R^2\right] T \\ & =4 \pi R^2\left[n^{\frac{1}{3}}-1\right] T\end{aligned}\)
\(\begin{aligned}
& \mathrm{n} \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3 \\
& \therefore \quad \mathrm{R}=\mathrm{n}^{\frac{1}{3}} \cdot \mathrm{r} \\
& \mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{\frac{1}{3}}}
\end{aligned}\)
\(\begin{aligned} \text {Work done } W & =\left[n \cdot 4 \pi r^2-4 \pi R^2\right] T \\ & =4 \pi\left[n \cdot \frac{R^2}{n^{\frac{2}{3}}}-R^2\right] T \\ & =4 \pi R^2\left[n^{\frac{1}{3}}-1\right] T\end{aligned}\)
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