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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 25

  • question_answer
    In the figure shown, A is a fixed charge. B (of mass m) is given a velocity v perpendicular to the line AB. At this moment the radius of curvature of the resultant path of B is

    A) \[0\]                 

    B)        \[\infty \] (Infinity)

    C) \[\left( \frac{4\pi {{\varepsilon }_{0}}m{{v}^{2}}}{{{q}^{2}}} \right){{r}^{2}}\]

    D)         r

    Correct Answer: C

    Solution :

    [c] Radius of curvature \[=\frac{{{v}^{2}}}{{{a}_{1}}}=\left( \frac{4\pi {{\varepsilon }_{0}}m{{v}^{2}}}{{{q}^{2}}} \right){{r}^{2}}\]


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