Manipal Engineering Manipal Engineering Solved Paper-2012

  • question_answer
    The length of the seconds pendulum is decreased by 0.3 cm when it is shifted to Chennai from London. If the acceleration due to gravity at London is \[981\,\,cm/{{s}^{2}}\], the acceleration due to gravity at Chennai is

    A) \[981\,\,cm/{{s}^{2}}\]       

    B) \[978\,\,cm/{{s}^{2}}\]

    C) \[984\,\,cm/{{s}^{2}}\]       

    D)        \[975\,\,cm/{{s}^{2}}\]

    Correct Answer: B

    Solution :

    \[{{L}_{1}}=\frac{{{g}_{1}}{{T}^{2}}}{4{{\pi }^{2}}}=\frac{{{g}_{1}}}{{{\pi }^{2}}};\,\,{{L}_{2}}=\frac{{{g}_{2}}{{T}^{2}}}{4{{\pi }^{2}}}=\frac{{{g}_{2}}}{{{\pi }^{2}}}\] Since, length is decreased,\[{{g}_{2}}\]is less than\[{{g}_{1}}\]. \[\therefore \]  \[{{L}_{1}}-{{L}_{2}}=\frac{{{g}_{1}}-{{g}_{2}}}{{{\pi }^{2}}}\] or            \[({{L}_{1}}-{{L}_{2}}){{\pi }^{2}}={{g}_{1}}-{{g}_{2}}\] or            \[0.3\times 10={{g}_{1}}-{{g}_{2}}\] \[\therefore \]  \[{{g}_{2}}=981-3=978\,\,cm/{{s}^{2}}\]


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