Category : JEE Main & Advanced
If n small drops of radius r coalesce to form a big drop of radius R then surface area of the liquid decreases.
Amount of surface energy released = Initial surface energy ? final surface energy
\[E=n4\pi {{r}^{2}}T-4\pi {{R}^{2}}T\]
Various formulae of released energy | ||||
\[4\pi T[n{{r}^{2}}-{{R}^{2}}]\] | \[4\pi {{R}^{2}}T({{n}^{1/3}}-1)\] | \[4\pi T{{r}^{2}}{{n}^{2/3}}({{n}^{1/3}}-1)\] | \[4\pi T{{R}^{3}}\left[ \frac{1}{r}-\frac{1}{R} \right]\] |
(i) If this released energy is absorbed by a big drop, its temperature increases and rise in temperature can be given by \[\Delta \theta =\frac{3T}{JSd}\left[ \frac{1}{r}-\frac{1}{R} \right]\]
(ii) If this released energy is converted into kinetic energy of a big drop without dissipation then by the law of conservation of energy.
\[\frac{1}{2}m{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\]
\[\Rightarrow \] \[\frac{1}{2}\left[ \frac{4}{3}\pi {{R}^{3}}d \right]{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\]
\[\Rightarrow \] \[{{v}^{2}}=\frac{6T}{d}\left[ \frac{1}{r}-\frac{1}{R} \right]\]
\[\therefore \] \[v=\sqrt{\frac{6T}{d}\left( \frac{1}{r}-\frac{1}{R} \right)}\]
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