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A liquid drop having surface energy ' $E$ ' is spread into 216 droplets of the same size. The final surface energy of the droplets is
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Verified Answer
The correct answer is:
$6 \mathrm{E}$
The correct option is (D).
Concept: Surface Energy is proportional to the surface area $\mathrm{E} \propto 4 \pi R^2$.
Considering volume conservation, the size of small droplets $r$ is:
$V=\frac{4}{3} \pi R^3=216 \times \frac{4}{3} \pi r^3$
Therefore, $\mathrm{R}=6 \mathrm{r}$
The total surface area of the new 216 droplets is:
$216 \times\left(4 \pi r^2\right)$
or $6 \times\left(4 \pi R^2\right)$
Given $E \propto 4 \pi R^2$
Therefore, the total surface energy of the new droplets would be $6 \mathrm{E}$
Concept: Surface Energy is proportional to the surface area $\mathrm{E} \propto 4 \pi R^2$.
Considering volume conservation, the size of small droplets $r$ is:
$V=\frac{4}{3} \pi R^3=216 \times \frac{4}{3} \pi r^3$
Therefore, $\mathrm{R}=6 \mathrm{r}$
The total surface area of the new 216 droplets is:
$216 \times\left(4 \pi r^2\right)$
or $6 \times\left(4 \pi R^2\right)$
Given $E \propto 4 \pi R^2$
Therefore, the total surface energy of the new droplets would be $6 \mathrm{E}$
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