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BHU PMT BHU PMT (Mains) Solved Paper-2011

  • question_answer
    Two pendulums of lengths 100 cm and 121 cm start vibrating. At some instant the two are at the mean position in the same phase. After how many vibrations of the longer pendulum will the two be in the same phase at the mean position again?

    A)  10    

    B)  11

    C)  20                                         

    D)  21

    Correct Answer: A

    Solution :

                     Let the two pendulums are in same phase, after n vibrations of the longer pendulum. In this time the shorter pendulum will complete \[(n+1)\]vibrations. \[\therefore \]  \[n2\pi \sqrt{\frac{{{l}_{2}}}{g}}=(n+1)2\pi \sqrt{\frac{{{l}_{1}}}{g}}\]                 \[n2\pi \sqrt{\frac{121}{g}}=(n+1)2\pi \sqrt{\frac{100}{g}}\]                 \[11n=10(n+1)\]                 \[n=10\]                              


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