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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2007

  • question_answer
    A radioactive substance contains 10000 nuclei  and its half-life period is 20 days. The number of nuclei present at the end of 10 days is

    A)  7070                                     

    B)  9000

    C)  8000                                     

    D)  7500

    Correct Answer: A

    Solution :

    We know,           \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{t/T}}\] \[\Rightarrow \]               \[\frac{N}{100000}={{\left( \frac{1}{2} \right)}^{10/20}}\] \[\Rightarrow \]               \[N=\frac{10000}{\sqrt{2}}=\frac{10000}{1.414}=\frac{10000}{1.414}=7070\]


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