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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    41forks are so arranged that each produces 5 beats/s when sounded with its near fork. If the frequency of last fork is double the frequency of first fork, then the frequencies of the first and last fork are respectively

    A)  200, 400                              

    B)  205,410

    C)  195, 390                              

    D)  100, 200

    Correct Answer: A

    Solution :

    Each fork produces 5 beat with its near fork. So, total number of beats \[=40\times 5=200\] If \[{{n}_{1}}\] and \[{{n}_{2}}\]are the frequencies of first and last fork then \[{{n}_{2}}-{{n}_{1}}=200\]             ... (i) By the question   \[{{n}_{2}}=2{{n}_{1}}\]            ...(ii) Solving Eqs. (i) and (ii), we get                 \[{{n}_{1}}=200Hz,\]                 \[{{n}_{2}}=400Hz\]


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