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AIIMS AIIMS Solved Paper-2015

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

    A) 1                             

    B) \[\text{K}{{\text{O}}_{\text{2}}}\]  

    C) \[\text{C}{{\text{O}}_{\text{2}}}\]                          

    D) 1836

    Correct Answer: B

    Solution :

    \[\lambda \,\,\text{is smaller than 2d}\] \[a=\frac{{{C}^{2}}{{B}^{2}}l-F}{m}\]                 Or           \[a=\frac{F}{m+CBI}\] As S' is same \[a=\frac{F{{C}^{2}}{{B}^{2}}l}{m}\] \[a=\frac{F}{m+{{C}^{2}}{{B}^{2}}l}\] \[\left( \text{take g }=\text{ 1}0\text{ m}/{{\text{s}}^{\text{2}}} \right)\]    


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