JEE Main & Advanced Physics Fluid Mechanics, Surface Tension & Viscosity / द्रव यांत्रिकी, भूतल तनाव और चिपचिपापन Formation of Bigger Drop

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.

\[\frac{1}{2}m{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\]    

\[\Rightarrow \] \[\frac{1}{2}\left[ \frac{4}{3}\pi {{R}^{3}}d \right]{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\]    

\[\Rightarrow \] \[{{v}^{2}}=\frac{6T}{d}\left[ \frac{1}{r}-\frac{1}{R} \right]\] 

\[\therefore \]  \[v=\sqrt{\frac{6T}{d}\left( \frac{1}{r}-\frac{1}{R} \right)}\]