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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
                  Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At \[t=0\] it was 1600 counts per second and \[t=8\] seconds it was 100 counts per second. The count rate observed, as counts per second, at \[t=6\] seconds is close to- [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) 200                                          

    B) 150  

    C) 400                              

    D)   360

    Correct Answer: A

    Solution :

    \[{{A}_{0}}=1600\text{ }cps\] at \[t=8\text{ }second\] \[100=\frac{1600}{{{2}^{n}}}\] \[{{2}^{n}}\,=\,16\,=\,{{2}^{4}}\] \[n=4\,=\,\frac{8}{{{T}_{1/2}}}\] \[{{T}_{1/2}}\text{  }2\] second In \[t=6\] sec number of half life?s = 3 \[\therefore \,\,\,\,A\,=\,\frac{1600}{{{2}^{3}}}\,=\,200\,cps\]


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