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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    In free space, a particle A of charge \[1\mu C\]is held fixed at a point P. Another particle B of the same charge and mass \[4\mu g\]is kept at a distance of 1 mm from P. if B is released, then its velocity at a distance of 9 mm from P is : \[\left[ \text{Take}\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} \right]\] [JEE Main 10-4-2019 Afternoon]

    A) \[2.0\times {{10}^{3}}m/s\]

    B) \[3.0\times {{10}^{4}}m/s\]

    C) \[1.5\times {{10}^{2}}m/s\]

    D) \[1.0\,m/s\]

    Correct Answer: A

    Solution :

    \[{{W}_{E}}=-[\Delta U]={{U}_{i}}-{{U}_{F}}=\frac{1}{2}m{{\text{v}}^{2}}\]           \[U=\frac{k{{q}_{1}}{{q}_{2}}}{r}\]       \[\frac{(9\times {{10}^{9}})\times {{10}^{-12}}}{{{10}^{-3}}}-\frac{(9\times {{10}^{9}})\times {{10}^{-12}}}{9\times {{10}^{-3}}}=\frac{1}{2}\times (4\times {{10}^{-6}}){{v}^{2}}\]           \[{{v}^{2}}=4\times {{10}^{6}}\]           \[v=2\times {{10}^{3}}m/s\]


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