A) 60%
B) 65%
C) 70%
D) 75%
Correct Answer: B
Solution :
\[N={{N}_{0}}{{e}^{-\lambda t}}\] \[\therefore \ 0.9{{N}_{0}}={{N}_{0}}{{e}^{-\lambda \times 5}}\Rightarrow 5\lambda ={{\log }_{e}}\frac{1}{0.9}\] ...... (i) and \[x{{N}_{0}}={{N}_{0}}{{e}^{-\lambda \times 20}}\Rightarrow 20\lambda ={{\log }_{e}}\left( \frac{1}{x} \right)\] ...... (ii) Dividing (i) by (ii), we get \[\frac{1}{4}=\frac{{{\log }_{e}}(1/0.9)}{{{\log }_{e}}(1/x)}=\frac{{{\log }_{10}}(1/0.9)}{{{\log }_{10}}(1/x)}=\frac{{{\log }_{10}}0.9}{{{\log }_{10}}x}\] \[\Rightarrow {{\log }_{10}}x=4{{\log }_{10}}0.9\Rightarrow x=0.658=65.8%\]
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