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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    Two samples X and Y contain equal amount of radioactive substances. If\[\frac{1}{16}\]th of the sample X and\[\frac{1}{256}\]th of the sample V, remain after 8 h, then the ratio of half periods of X and Y is

    A)  2 : 1                      

    B)         1 : 2

    C)  1 : 4                      

    D)         1 : 16

    E)  4 : 1

    Correct Answer: A

    Solution :

    For X radioactive substances \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/{{T}_{x}}}}\] \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{8/{{T}_{x}}}}\] \[\frac{1}{16}={{\left( \frac{1}{2} \right)}^{8/{{T}_{x}}}}\] \[\Rightarrow \]               \[{{\left( \frac{1}{2} \right)}^{4}}={{\left( \frac{1}{2} \right)}^{8/{{T}_{x}}}}\] \[{{T}_{X}}=2\] For Y radioactive substances \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{8/{{T}_{Y}}}}\] \[\frac{1}{256}={{\left( \frac{1}{2} \right)}^{8/{{T}_{Y}}}}\] \[{{\left( \frac{1}{2} \right)}^{8}}={{\left( \frac{1}{2} \right)}^{8/{{T}_{Y}}}}\] Or           \[{{T}_{Y}}=1\] \[\therefore \]  \[\frac{{{T}_{X}}}{{{T}_{Y}}}=\frac{2}{1}\]


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