Physics - Fluid Question with Solution | TestHub

When small droplets coelesce to form a bigger drop, energy released in this process is given by, $$4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]$$ Where, $R=$ radius of big drop $r=$ radius of small drop $T=$ surface tension According to question $$\begin{array}{rl}& \frac{1}{2}m{v}^2=4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]\\ \Rightarrow & \frac{1}{2}\left[\frac{4}{3}\pi R^3\rho \right]v^2=4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]\\ \Rightarrow V^2& =\frac{6T}{\rho }\left[\frac{1}{r}-\frac{1}{R}\right]\Rightarrow V={\left[\frac{6T}{\rho }\left(\frac{1}{r}-\frac{1}{R}\right)\right]}^{\frac{1}{2}}\end{array}$$