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VMMC VMMC Medical Solved Paper-2013

  • question_answer
    A simple pendulum has a time period \[{{T}_{1}}\]when on the earths surface and \[{{T}_{2}}\]when taken to a height R above the earths surface, where R is the radius of earth. The value of \[\frac{{{T}_{2}}}{{{T}_{1}}}\]is

    A)  1

    B)  \[\sqrt{2}\]

    C)  4

    D)  2

    Correct Answer: D

    Solution :

     \[{{g}_{1}}=\frac{GM}{{{R}^{2}}}\] \[{{g}_{2}}=\frac{GM}{{{(R+R)}^{2}}}=\frac{g}{4}\] At height R from surface of earth \[T=2\pi \sqrt{\frac{l}{g}}\] \[T\propto \frac{1}{\sqrt{g}}\] \[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{g}_{1}}}{{{g}_{2}}}}=\sqrt{\frac{g}{\frac{g}{4}}}\] \[=2\]


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