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WB JEE Medical WB JEE Medical Solved Paper-2007

  • question_answer
    A set of 24 tuning forks are so arranged that each gives 6 beats/s with the previous one. If the frequency of the last tuning fork is double that of the first, frequency of the second  tuning fork is

    A)  138 Hz                                 

    B)  132 Hz

    C)  144 Hz                                 

    D)  276 Hz

    Correct Answer: C

    Solution :

                     The frequencies of tuning forks are the terms  of an AP whose common difference is 6. \[\therefore \]  \[l=a+(n-1)d\]                 \[2a=a+(24-1)\times 6\]                 \[a=23\times 6\] \[=138\] \[\therefore \] Second frequency = 138 + 6 =144 Hz


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