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JEE Main & Advanced JEE Main Paper (Held On 9 April 2017)

  • question_answer
    A block of mass 0.1 kg is connected to an elastic spring of spring constant \[640\,\,N{{m}^{-1}}\] and oscillates in a damping medium of damping constant\[{{10}^{-2}}kg\,{{s}^{-1}}\]. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to - [JEE Online 09-04-2017]

    A)  2 s                                         

    B)  5 s

    C)  7 s                                         

    D)  3.5 s

    Correct Answer: D

    Solution :

                      \[E'=\frac{1}{2}\]              \[b=\frac{\lambda }{m}\]            \[A={{A}_{0}}{{e}^{-bt}}\] \[a'=\frac{a}{\sqrt{2}}\] \[\lambda =bm\] \[\frac{1}{\sqrt{2}}\,={{e}^{-\frac{t}{10}}}\] \[b=\frac{1}{10}\,\sqrt{2}\,={{e}^{\frac{1}{10}}}\] \[\ln \,\sqrt{2}\,=\frac{t}{20}\]


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