JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    A man measures time period of a pendulum \[(T)\] in stationary lift. If the lift moves upward with acceleration \[\frac{g}{4},\] then new time period will be                           [BHU 2001]

    A)             \[\frac{2T}{\sqrt{5}}\]

    B)             \[\frac{\sqrt{5}T}{2}\]

    C)             \[\frac{\sqrt{5}}{2T}\]           

    D)             \[\frac{2}{\sqrt{5}T}\]

    Correct Answer: A

    Solution :

                    \[T=2\pi \sqrt{\frac{l}{g}}\] Þ \[\frac{{{T}'}}{T}=\sqrt{\frac{g}{{{g}'}}}=\sqrt{\frac{g}{g+\frac{g}{4}}}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}\]           


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