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) \[{{\left[ \frac{2T}{\rho }\left( \frac{1}{r}-\frac{1}{R} \right) \right]}^{1/2}}\]
Correct Answer: B
Solution :
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 \[\frac{1}{2}m{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\] \[\Rightarrow \]\[\frac{1}{2}\left[ \frac{4}{3}\pi {{R}^{3}}\rho \right]{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\] \[\Rightarrow \]\[{{V}^{2}}=\frac{6T}{\rho }\left[ \frac{1}{r}-\frac{1}{R} \right]\Rightarrow \,V={{\left[ \frac{6T}{\rho }\left( \frac{1}{r}-\frac{1}{R} \right) \right]}^{\frac{1}{2}}}\]
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