A) \[1\]
B) \[\frac{4}{3}\]
C) \[\frac{2}{3}\]
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
[b] \[{{C}_{0}}=\frac{k{{\in }_{0}}A}{d}\] |
\[C=\frac{k{{\in }_{0}}2}{3d}+\frac{2k{{\in }_{0}}A}{3d}=\frac{4}{3}\frac{k{{\in }_{0}}A}{d}\] |
\[\therefore \]\[\frac{C}{{{C}_{0}}}=\frac{\frac{4}{3}\frac{k{{\in }_{0}}A}{d}}{\frac{k{{\in }_{0}}A}{d}}=\frac{4}{3}\] |
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