Question

A large number of droplets, each of radius, $r$ coalesce to form a bigger drop of radius, $R$. An engineer designs a machine so that the energy released in this process is converted into the kinetic energy of the drop. Velocity of the drop is ($T=$ surface tension, $\rho=$ density)

• A. $\left[\frac{T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1/2}$
• B. $\left[\frac{6T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1/2}$
• C. $\left[\frac{3T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1/2}$
• D. $\left[\frac{2T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1/2}$

Answer 1

Answer: (B)

$\left[\frac{6T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1/2}$

Answer 2

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 $\frac{1}{2}mv^{2}=4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]$ $\Rightarrow \frac{1}{2}\left[\frac{4}{3}\pi R^{3}\rho\right]v^{2}=4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]$ $\Rightarrow V^{2}=\frac{6T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right] \Rightarrow V=\left[\frac{6T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{\frac{1}{2}}$