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JEE Main & Advanced Physics Simple Harmonic Motion JEE PYQ-Simple Harmonic Motion

  • question_answer
    In an experiment to determine the period of a simple pendulum of length 1m, it is attached to different spherical bobs of radii \[{{r}_{1}}\] and \[{{r}_{2}}\]. The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be \[5\times {{10}^{-4}}s,\] the difference in radii, \[|{{r}_{1}}-{{r}_{2}}|\] is best given by –                                       [JEE Online 09-04-2017]

    A)  0.01 cm           

    B)      (2) 0.1 cm

    C)  0.5 cm

    D)       1 cm

    Correct Answer: B

    Solution :

    [b]  \[T\propto \sqrt{l}\]   \[l=1\]
    \[\frac{\Delta T}{l}\,=\frac{1}{2}\,\,\frac{\Delta l}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta l={{r}_{1}}-{{r}_{2}}\]
    \[5\times {{10}^{-4}}\,=\frac{1}{2}\,\frac{{{r}_{1}}-{{r}_{2}}}{1}\]
    \[{{r}_{1}}-{{r}_{2}}\,=10\times {{10}^{-4}}\]
    \[{{10}^{-3}}\,m={{10}^{-1}}\,cm\,=0.1\,cm\]


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