VIT Engineering VIT Engineering Solved Paper-2010

  • question_answer
    Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii \[{{R}_{1}}\] and \[{{R}_{2}}\], respectively. The ratio of masses of X and Y is

    A)  \[\left( {{R}_{1}}/{{R}_{2}} \right)\]         

    B)  \[\left( {{R}_{2}}/{{R}_{1}} \right)\]

    C)  \[{{\left( {{R}_{1}}/{{R}_{2}} \right)}^{2}}\]        

    D)  \[\left( {{R}_{1}}/{{R}_{2}} \right)\]  

    Correct Answer: C

    Solution :

    \[\frac{1}{2}m{{v}^{2}}=qV\] or \[v=\sqrt{\frac{2qV}{m}}\] Centripetal force \[\frac{m{{v}^{2}}}{R}=q\times B\times v\] \[\therefore \] \[v=\frac{qBR}{m}\] Hence, \[\sqrt{\frac{2qV}{m}}=\frac{qBR}{m}\] or \[R={{\left( \frac{2mV}{q} \right)}^{1/2}}\times \frac{1}{B}\] Here, V, q and B are constants,                          Hence,           \[R\,\,\alpha \,\,m\] So,         \[\frac{{{m}_{1}}}{{{m}_{2}}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{2}}\]


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