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If 1000 droplets of water of surface tension 0.07 $\mathrm{N} / \mathrm{m}$. having same radius $1 \mathrm{~mm}$ each, combine to from a single drop. In the process the released surface energy is( Take $\left.\pi=\frac{22}{7}\right)$
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The correct answer is:
$7.92 \times 10^{-4} \mathrm{~J}$
We have $V_f=V_i$
$\Rightarrow \frac{4}{3} \pi r_{\mathrm{f}}^3=1000 \times \frac{4}{3} \pi \mathrm{r}_{\mathrm{i}}^3 \Rightarrow \mathrm{r}_{\mathrm{f}}^3=1000 \mathrm{r}_{\mathrm{i}}^3$
$\Rightarrow r_f=10 r_i$
So, released energy $=$ Initial surface energy - final surface energy
$=1000 \times \mathrm{T} \times 4 \pi \mathrm{r}_{\mathrm{i}}^2-\mathrm{T} \times 4 \pi \mathrm{r}_{\mathrm{f}}^2$
$=4 \pi \mathrm{T}\left(1000 \mathrm{r}_{\mathrm{i}}^2-\mathrm{r}_{\mathrm{f}}^2\right)$
$=4 \pi \times 0.07\left(1000 \mathrm{r}_{\mathrm{i}}^2-100 \mathrm{r}_{\mathrm{i}}^2\right)$
$=4 \pi \times 0.07 \times 900 r_i^2$
$=4 \pi \times 63 \times 10^{-6}=7.92 \times 10^{-4} \mathrm{~J}$
$\Rightarrow \frac{4}{3} \pi r_{\mathrm{f}}^3=1000 \times \frac{4}{3} \pi \mathrm{r}_{\mathrm{i}}^3 \Rightarrow \mathrm{r}_{\mathrm{f}}^3=1000 \mathrm{r}_{\mathrm{i}}^3$
$\Rightarrow r_f=10 r_i$
So, released energy $=$ Initial surface energy - final surface energy
$=1000 \times \mathrm{T} \times 4 \pi \mathrm{r}_{\mathrm{i}}^2-\mathrm{T} \times 4 \pi \mathrm{r}_{\mathrm{f}}^2$
$=4 \pi \mathrm{T}\left(1000 \mathrm{r}_{\mathrm{i}}^2-\mathrm{r}_{\mathrm{f}}^2\right)$
$=4 \pi \times 0.07\left(1000 \mathrm{r}_{\mathrm{i}}^2-100 \mathrm{r}_{\mathrm{i}}^2\right)$
$=4 \pi \times 0.07 \times 900 r_i^2$
$=4 \pi \times 63 \times 10^{-6}=7.92 \times 10^{-4} \mathrm{~J}$
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