A) 2.50 gm
B) 3.70 gm
C) 6.30 gm
D) 1.35 gm
Correct Answer: D
Solution :
\[M={{M}_{0}}{{e}^{-\lambda t}};\] given \[t=2\left( \frac{1}{\lambda } \right)\] \[\Rightarrow M=10{{e}^{-\lambda \left( \frac{2}{\lambda } \right)}}=10{{\left( \frac{1}{e} \right)}^{2}}\Rightarrow M=1.35\ gm\]
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