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JCECE Engineering JCECE Engineering 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 6 of frequency\[320\,\,Hz\].


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