A) 0.06 mm
B) 0.4 mm
C) 0.6 mm
D) 0.32 mm
Correct Answer: A
Solution :
Drop evaporates if pressure inside drop is greater than its vapour pressure. So, to avoid evaporation, |
\[{{P}_{excess}}\le {{p}_{vapour}}\Rightarrow \frac{2S}{R}={{p}_{vapour}}\] |
\[R=\frac{2S}{P}\] |
\[=\frac{2\times 6.99\times {{10}^{-2}}}{2.33\times {{10}^{3}}}\] |
\[=6\times {{10}^{-5}}m=0.06mm\] |
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