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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    A disc of radius \[a\] and mass \[m\] is pivoted at the rim and is set in small oscillation. If a simple pendulum have the same period as that of the disc, then the length of the simple pendulum should be

    A) \[\frac{5}{2}a\]                                

    B) \[\frac{3}{2}a\]

    C) \[\frac{7}{2}a\]                                

    D)  \[\frac{1}{2}a\]

    Correct Answer: B

    Solution :

    We know that period of a physical pendulum \[T=2\pi \sqrt{\frac{{{I}_{0}}}{mgd}}=2\pi \sqrt{\frac{\frac{1}{2}m{{a}^{2}}+m{{a}^{2}}}{mga}}=2\pi \sqrt{\frac{3a}{2g}}\]                                                                     ... (i) \[T\]for simple pendulum            \[=2\pi \sqrt{\frac{l}{g}}\]            ... (ii) From Eqs. (i) and (ii), we get\[l=\frac{3}{2}a\]


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