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MGIMS WARDHA MGIMS WARDHA Solved Paper-2006

  • question_answer
    50 tuning forks are arranged in increasing order of their frequencies such that each gives 4 beats/s with its previous tuning fork. If the frequency of the last fork is octave of the first, then the frequency of the first tuning fork is :

    A) 200 Hz                                  

    B) 204 Hz

    C) 196 Hz                                  

    D) none of these

    Correct Answer: C

    Solution :

    Each tuning fork gives 4 beats/s with its previous tuning fork, i.e., the difference between frequency of any two consecutive tuning forks is 4. \[\therefore \]  \[{{f}_{n}}={{f}_{1}}+(n-1)d\] \[\Rightarrow \]               \[2f=f+(50-1)\times 4\] \[\Rightarrow \]               \[f=49-4\]                 \[f=196\,Hz\]


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