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JCECE Medical JCECE Medical Solved Paper-2014

  • question_answer
    A tuning fork A produces 4 beats \[{{s}^{-1}}\] with another tuning fork B of frequency 320 Hz. On filing one of the prongs of A, 4 beats \[{{s}^{-1}}\]are again heard when sounded with the same fork B. Then, the frequency of the fork A before filing is

    A)  328 Hz

    B)  316 Hz

    C)  324 Hz

    D)  320 Hz

    Correct Answer: D

    Solution :

     There are 4 beats between A and B therefore, the possible frequencies of/A are 316 or 324 that is \[(320\pm 4)\,Hz.\] When the prong of A is filled its frequency becomes greater than, the original frequency. If we assume that original frequency of A is 324, then on filing its frequency will be greater than 324. The beats between A and B will more than 4. But it is given that the beats are again 4, therefore 324 is not possible. Therefore, required frequency must be 316 Hz. This is true, because on filing the frequency may increase so as to give 4 beats with B of frequency 320 Hz.


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