A water drop of radius $1 \mathrm{~cm}$ is broken into 729 equal droplets. If surface tension of water is 75 dyne/ $\mathrm{cm}$, then the gain in surface energy upto first decimal place will be :
(Given $\pi=3.14$ )
Solution
<p>$729 \times {4 \over 3}\pi {r^3} = {4 \over 3}\pi {R^3}$</p>
<p>$\Rightarrow R = 9r$ ........ (1)</p>
<p>$\Delta U = S \times \Delta A$ ..... (2)</p>
<p>$\Rightarrow \Delta U = S \times \{ - 4\pi {R^2} + 729 \times 4\pi {r^2}\}$</p>
<p>$= S \times 4\pi \{ 729{r^2} - 81{r^2}\}$</p>
<p>$= 7.5 \times {10^{ - 4}}\,J$</p>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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