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MGIMS WARDHA MGIMS WARDHA Solved Paper-2006

  • question_answer
    Two simple pendulums whose lengths are 100   cm and 121 cm are suspended side by side.         Their bobs are pulled together and then released After how many minimum oscillations of the longer pendulum, will the two be in phase again?

    A) 11                          

    B) 10  

    C) 21                          

    D) 20

    Correct Answer: B

    Solution :

    Time period of simple pendulum \[T=2\pi \sqrt{\frac{l}{g}}\] \[\therefore \]  \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{l}_{1}}}{{{l}_{2}}}}\]                 \[=\sqrt{\frac{121}{100}}\]                 \[=\frac{11}{10}\] Let after n oscillations, the pendulums are in same phase \[\therefore \] \[(n)\times 11=(n+1)\times 10\] \[\therefore \]  \[n=10\]


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