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 :-
$\left( {\mathrm{Take}\,\pi = {{22} \over 7}} \right)$
Solution
$1000 \times \frac{4 \pi}{3}(1)^{3}=\frac{4 \pi}{3} \mathrm{R}^{3}$
<br/><br/>$\mathrm{R}=10 \mathrm{~mm}$
<br/><br/>$\mathrm{T} \times 1000 \times 4 \pi\left(10^{-3}\right)^{2}-\mathrm{T} \times 4 \pi\left(10 \times 10^{-3}\right)^{2}=\Delta \mathrm{E}$
<br/><br/>$\Rightarrow$ $\Delta \mathrm{E}=4 \times \pi \times 7 \times 10^{-2}[1000-100] \times 10^{-6}$
<br/><br/>$\Rightarrow$ $\Delta \mathrm{E}=7.92 \times 10^{-4} \mathrm{~J}$
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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