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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2002

  • question_answer
    Forty one tuning forks are arranged in increasing order of frequencies such that every fork gives 5 beats with the next. The last fork has a frequency that is double the frequency of the first fork is:

    A)  210

    B)  400

    C)   205

    D)  200

    Correct Answer: D

    Solution :

     \[{{f}_{1}}=\] frequency of first fork. \[{{f}_{41}}=\] frequency of 41st fork. \[{{f}_{41}}=2{{f}_{1}}\] Between Ist and 41st there are 40 intervals each interval produces 5 beats. \[{{f}_{41}}-{{f}_{1}}=40\times 5\] \[2{{f}_{1}}-{{f}_{1}}=200\] \[{{f}_{1}}=200Hz\]


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