A) 100%
B) 80%
C) 63.5%
D) 35.5%
Correct Answer: C
Solution :
\[=\int_{1}^{2}{{{\log }_{e}}1}\,dx+\int_{2}^{3}{{{\log }_{e}}2\,\,dx}++\int_{3}^{4}{{{\log }_{e}}}3\,\,dx\] \[=0+({{\log }_{e}}2)[x]_{2}^{3}+({{\log }_{e}}3)[x]_{3}^{4}\] \[=\int_{1}^{2}{{{\log }_{e}}1}\,dx+\int_{2}^{3}{{{\log }_{e}}2\,\,dx}++\int_{3}^{4}{{{\log }_{e}}}3\,\,dx\] \[=\int_{1}^{2}{{{\log }_{e}}}[x]dx+\int_{2}^{3}{{{\log }_{e}}[x]dx}++\int_{3}^{4}{{{\log }_{e}}}[x]dx\] Thus, \[=({{\log }_{e}}2)1+({{\log }_{e}}3)1\] Equivalent of \[={{\log }_{e}}6\] Thus, equivalent of \[y={{m}_{1}}x\] Equivalent of \[y={{m}_{2}}x\] Equivalent mass of \[4\sqrt{27}\] \[4\sqrt{18}\] Thus, percentage of \[2.0\times {{10}^{-5}}{{/}^{o}}C\]
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