A) \[2.67\]
B) \[5.3\]
C) \[0.53\]
D) \[0.267\]
Correct Answer: D
Solution :
\[\underset{2-0.8=12}{\mathop{\underset{2-\frac{40}{100}\times 2}{\mathop{\underset{2}{\mathop{PC{{l}_{5}}}}\,}}\,}}\,\underset{0.8}{\mathop{\underset{0}{\mathop{PC{{l}_{3}}}}\,}}\,+\underset{0.8}{\mathop{\underset{0}{\mathop{C{{l}_{2}}}}\,}}\,\] so, \[[PC{{l}_{5}}]=\frac{1.2}{2}=0.6\] \[[PC{{l}_{3}}]=\frac{0.8}{2}=0.4\] \[[C{{l}_{2}}]=\frac{0.8}{2}=0.4\] \[Kc=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}\] \[=\frac{0.4\times 0.4}{0.6}=\frac{1.6}{0.6}\] \[=0.267\]
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