CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    Two particles of equal charges after being accelerated  through  the  same  potential difference enter a uniform transverse magnetic field and describe circular path of radii\[{{R}_{1}}\]and\[{{R}_{2}}\]respectively. Then the ratio of their masses\[({{M}_{1}}/{{M}_{2}})\]is

    A)  \[\frac{{{R}_{1}}}{{{R}_{2}}}\]   

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

    C)         \[\frac{{{R}_{2}}}{{{R}_{1}}}\]                   

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

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

    Correct Answer: B

    Solution :

    Radius of circular path \[R=\frac{mv}{qB}\] but        \[mv=\sqrt{2mqV}\] \[\therefore \]  \[R=\frac{\sqrt{2mqV}}{qB}\]or \[R\propto \sqrt{m}\] Or           \[\frac{R_{1}^{2}}{R_{2}^{2}}=\frac{{{M}_{1}}}{{{M}_{2}}}\]or \[\frac{{{M}_{1}}}{{{M}_{2}}}=\frac{R_{1}^{2}}{R_{2}^{2}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner