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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 10

  • question_answer
    A radioactive material of half-life T was produced in a nuclear reactor at different instants. The quantity produced second time was twice of that produced first time, if now their present activities are \[{{A}_{1}}\] and \[{{A}_{2}},\] respectively, then their age difference equals

    A) \[\frac{T}{In\,\,2}\left| In\,\frac{2{{A}_{1}}}{{{A}_{2}}} \right|\]         

    B) \[T\,\left| In\,\frac{{{A}_{1}}}{{{A}_{2}}} \right|\]

    C) \[\frac{T}{In\,2}\,\left| In\,\frac{{{A}_{2}}}{2{{A}_{1}}} \right|\]         

    D) \[T\,\left| In\,\frac{{{A}_{2}}}{2{{A}_{1}}} \right|\]

    Correct Answer: C

    Solution :

    [c] \[{{A}_{1}}=\lambda {{N}_{0}}{{e}^{-\lambda {{t}_{1}}}}\,\,\,\Rightarrow \,\,\,{{t}_{1}}=\frac{1}{\lambda }In\left( \frac{\lambda {{N}_{0}}}{{{A}_{1}}} \right)\] \[{{A}_{2}}=\lambda \left( 2{{N}_{0}} \right){{e}^{-\lambda {{t}_{2}}}}\] So,  \[{{t}_{1}}-{{t}_{2}}=\frac{T}{In\,\,2}\left| In\,\,\frac{{{A}_{2}}}{2{{A}_{1}}} \right|\]


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