Fluid Mechanics Viscosity Question 322
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)
Options:
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}} $
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Answer:
Correct Answer: B
Solution:
[b] When small droplets coalesce 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] $ 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}{2}\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}}} $
BETA
