A) 0.7
B) 0.81
C) 0.729
D) 0.6
Correct Answer: C
Solution :
\[A={{A}_{o}}{{e}^{-kt}}\] \[0.9{{A}_{o}}={{A}_{o}}{{e}^{-kt}}\] \[-kt=\ell n(0.9)\] \[-5\text{ }k=\ell n(0.9)\]\[\Rightarrow \]\[-15\text{ }k=3\ell n(0.9)\] \[A={{A}_{o}}{{e}^{-15k}}={{A}_{o}}{{e}^{-\ell n{{(0.9)}^{3}}}}\] \[={{(0.9)}^{3}}{{A}_{o}}=0.729\text{ }{{A}_{o}}\]You need to login to perform this action.
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