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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2005

  • question_answer
    A radioactive isotope has a half-life of 10 days. If today 125 mg is left over, what was its original weight 40 days earlier?

    A)  2 g                                        

    B)  600 mg

    C)  1 g                                        

    D)  1.5 g

    Correct Answer: A

    Solution :

    \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}},\,\,n=\frac{40}{10}=4\] \[\frac{125}{1000}=N{{\left( \frac{1}{2} \right)}^{4}},\,{{N}_{0}}=\frac{125\times {{2}^{4}}}{1000}=2\,g\]


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