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BHU PMT BHU PMT (Mains) Solved Paper-2006

  • question_answer
    An electron of mass rug, initially at rest, moves through a certain distance in a unifrom electric field in time\[{{t}_{1}}\] A proton of mass\[{{m}_{p}},\]also, initially at rest, takes time\[{{t}_{2}}\] to move through an equal distance in this unifrom electric field. Neglecting the effect of gravity, the ratio\[{{t}_{2}}/{{t}_{1}}\]is nearly equal to:

    A)  \[1\]                                    

    B)  \[{{({{m}_{p}}/{{m}_{e}})}^{1/2}}\]

    C)  \[{{({{m}_{e}}/{{m}_{p}})}^{1/2}}\]                       

    D)  \[1836\]

    Correct Answer: B

    Solution :

                     Electrostatic force,\[{{F}_{e}}=eE\] (for both the particles) But acceleration of electron                                 \[{{a}_{e}}=\frac{{{F}_{e}}}{{{m}_{e}}}\] and acceleration of proton,                                 \[{{a}_{p}}=\frac{{{F}_{e}}}{{{m}_{p}}}\] Distance travelled by both the particles \[s=\frac{1}{2}{{a}_{e}}t_{1}^{2}=\frac{1}{2}{{a}_{p}}t_{2}^{2}\] \[\therefore \]  \[\frac{{{t}_{2}}}{{{t}_{1}}}=\sqrt{\frac{{{a}_{e}}}{{{a}_{p}}}}=\sqrt{\frac{{{m}_{p}}}{{{m}_{e}}}}\]


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