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 :
[b] When small droplets coelesce to form a bigger drop, energy released in this process is given by, |
Where, R = radius of big drop r = radius of small drop T = surface tension |
According to question |
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