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

  • question_answer
    A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by : [JEE Main 10-4-2019 Afternoon]

    A) \[2\pi \sqrt{\frac{L}{\sqrt{{{g}^{2}}+{{\left( \frac{qE}{m} \right)}^{2}}}}}\]

    B) \[2\pi \sqrt{\frac{L}{\sqrt{g+\left( \frac{qE}{m} \right)}}}\]

    C) \[2\pi \sqrt{\frac{L}{\left( g-\begin{matrix}    qE  \\    m  \\ \end{matrix} \right)}}\]          

    D) \[2\pi \sqrt{\frac{L}{\left( {{g}^{2}}-\begin{matrix}    {{q}^{2}}{{E}^{2}}  \\    {{m}^{2}}  \\ \end{matrix} \right)}}\]

    Correct Answer: A

    Solution :

    \[{{g}_{eff}}=\sqrt{{{g}^{2}}+{{\left( \frac{qE}{m} \right)}^{2}}}\]           \[T=2\pi \sqrt{\frac{\ell }{{{g}_{eff}}}}\]           \[=2\pi \sqrt{\frac{\ell }{\sqrt{{{g}^{2}}+{{\left( \frac{qE}{m} \right)}^{2}}}}}\]


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