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NEET AIPMT SOLVED PAPER MAINS 2012

  • question_answer
    A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an \[\text{ }\!\!\alpha\!\!\text{ -}\]particle to describe a circle of same radius in the same field?

    A)  2 MeV                 

    B)  1 MeV 

    C)  0.5 MeV             

    D)  4 MeV

    Correct Answer: B

    Solution :

    For proton \[r=\frac{\sqrt{2m(KE)}}{qB}\]                 So           \[q\propto \sqrt{m(KE)}\] Hence    \[\frac{e}{2e}=\sqrt{\frac{({{m}_{p}})(1MeV)}{(4{{m}_{p}})(KE)}}\]                 \[\frac{1}{4}=\frac{1MeV}{4KE}\]                 \[KE=1\,MeV\]


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