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

  • question_answer A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centred at origin. A point charge q is moving towards the ring along the z-axis and has speed u at z = 4a. The minimum value of u such that it crosses the origin is : [JEE Main 10-4-2019 Morning]

    A) \[\sqrt{\frac{2}{m}}{{\left( \frac{1}{15}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]           

    B) \[\sqrt{\frac{2}{m}}{{\left( \frac{2}{15}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]

    C) \[\sqrt{\frac{2}{m}}{{\left( \frac{4}{15}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]

    D)   \[\sqrt{\frac{2}{m}}{{\left( \frac{1}{5}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]

    Correct Answer: B

    Solution :

    \[{{U}_{i}}+{{K}_{i}}={{U}_{f}}+{{K}_{f}}\] \[\frac{k{{q}^{2}}}{\sqrt{16{{a}^{2}}+9{{a}^{2}}}}+\frac{1}{2}m{{\text{v}}^{2}}=\frac{k{{q}^{2}}}{3a}\]           \[\frac{1}{2}m{{\text{v}}^{2}}=\frac{k{{q}^{2}}}{a}\left( \frac{1}{3}-\frac{1}{5} \right)=\frac{2k{{q}^{2}}}{15a}\] \[\text{v}=\sqrt{\frac{4k{{q}^{2}}}{15ma}}\]

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