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A water drop of diameter $2 \mathrm{~cm}$ is broken into 64 equal droplets. The surface tension of water is $0.075 \mathrm{~N} / \mathrm{m}$ . In this process the gain in surface energy will be :

Option: 1
$2.8 \times 10^{-4} \mathrm{~J}$

Option: 2
$1.5 \times 10^{-3} \mathrm{~J}$

Option: 3
$1.9 \times 10^{-4} \mathrm{~J}$

Option: 4
$9.4 \times 10^{-5} \mathrm{~J}$

Answers (1)

best_answer

Let radius of small drops = r

Volume conservation of water

$\mathrm{\frac{4}{3} \pi R^{3} =64 \times \frac{4}{3} \pi r^{3}} \\$

$\mathrm{r^{3} =\frac{R^{3}}{64}, r=\frac{R}{4}}$

Initial surface energy

$\mathrm{U_{1}=S \times 4 \pi R^{2}}$

Final surface energy is system

$\mathrm{U_{2} =64 \times 5 \times 4 \pi r^{2}} \\$

$\mathrm{=64 \times 5 \times 4 \pi \times \frac{R^{2}}{16}} \\$

$\mathrm{=4 \times 5 \times 4 \pi R^{2}} \\$

$\mathrm{=4 U_{1}}$

Gain in energy

$\mathrm{\Delta U =U_{2}-U_{1}} \\$

$\mathrm{=4 U_{1}-U_{1}} \\$

$\mathrm{=3 U_{1} }\\$

$\mathrm{=3 \times 5 \times 4 \pi R^{2}} \\$

$\mathrm{=3 \times 0.075 \times 4 \pi \times 1 \times 10^{-4}} \\$

$\mathrm{=0.225 \times 4 \pi \times 10^{-4}} \\$

$\mathrm{=2.8 \times 10^{-4} \mathrm{~J}}$

Hence correct option is 1

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