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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2004

  • question_answer
        An \[3\times {{10}^{-3}}\Omega \] particle is accelerated through a p.d of 106 volt then K.E. of particle will be:

    A)  8 MeV                                 

    B)  4 MeV

    C)  2 MeV                                 

    D)  1 MeV

    Correct Answer: C

    Solution :

                    We know that charge on \[\text{55}00\text{ { }\!\!\mathrm{\AA}\!\!\text{ }}\]-particle is \[\text{6}000\text{ { }\!\!\mathrm{\AA}\!\!\text{ }}\] K.E. of a particle \[\text{5}000\text{ { }\!\!\mathrm{\AA}\!\!\text{ }}\] \[\text{4000 { }\!\!\mathrm{\AA}\!\!\text{ }}\]


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