A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be :
Solution
<p>$r' = {r \over 4}$</p>
<p>$\Rightarrow \Delta E = T(\Delta S)$</p>
<p>$= T \times 4\pi (nr{'^2} - {r^2}),\,n = 64$</p>
<p>$= T \times 4\pi \times (4 - 1){r^2}$</p>
<p>$\Rightarrow \Delta E = 0.075 \times 4 \times 3.142(3) \times {10^{ - 4}}\,$ J</p>
<p>$= 2.8 \times {10^{ - 4}}$ J</p>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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