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A spherical drop of radium r is divided in to 8 equal
droplets. If the surface tension is S, then the work done in
the process will be
Options:
droplets. If the surface tension is S, then the work done in
the process will be
Solution:
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Verified Answer
The correct answer is:
$4 \pi r^2 \mathrm{~S}$
$\begin{aligned}
& \text { } \mathrm{U}_{\mathrm{i}}=\mathrm{S} \times 4 \pi \mathrm{r}^2 \\
& \text { as, } \mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}} \\
& \Rightarrow \frac{4}{3} \pi \mathrm{r}^3=8 \times \frac{4}{3} \pi \mathrm{r}^{\prime 3} \\
& \Rightarrow \mathrm{r}=2 \mathrm{r}^{\prime} \Rightarrow \mathrm{r}^{\prime}=\mathrm{r} / 2
\end{aligned}$
So, $\mathrm{U}_{\mathrm{f}}=8 \times\left[\mathrm{S} \times 4 \pi\left(\frac{\mathrm{r}}{2}\right)^2\right]$
$$
=2\left(\mathrm{~S} \times 4 \pi \mathrm{r}^2\right)
$$
So, $\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=\mathrm{S} \times 4 \pi \mathrm{r}^2$
& \text { } \mathrm{U}_{\mathrm{i}}=\mathrm{S} \times 4 \pi \mathrm{r}^2 \\
& \text { as, } \mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}} \\
& \Rightarrow \frac{4}{3} \pi \mathrm{r}^3=8 \times \frac{4}{3} \pi \mathrm{r}^{\prime 3} \\
& \Rightarrow \mathrm{r}=2 \mathrm{r}^{\prime} \Rightarrow \mathrm{r}^{\prime}=\mathrm{r} / 2
\end{aligned}$
So, $\mathrm{U}_{\mathrm{f}}=8 \times\left[\mathrm{S} \times 4 \pi\left(\frac{\mathrm{r}}{2}\right)^2\right]$
$$
=2\left(\mathrm{~S} \times 4 \pi \mathrm{r}^2\right)
$$
So, $\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=\mathrm{S} \times 4 \pi \mathrm{r}^2$
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