A) \[1.12\text{ }\mu g\]
B) \[0.1\text{ }\mu g\]
C) \[3.125\text{ }\mu g\]
D) \[125\text{ }\mu \text{g}.\]
Correct Answer: C
Solution :
[c] Using \[\frac{M}{{{M}_{0}}}={{\left( \frac{1}{2} \right)}^{t/T}}={{\left( \frac{1}{2} \right)}^{120/24}}={{\left( \frac{1}{2} \right)}^{5}}=\frac{1}{32}\] So \[M=\frac{{{M}_{0}}}{32}=\frac{0.1mg}{32}=3.125\,\mu g.\]
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