2. Solved Papers
  3. JCECE Engineering

JCECE Engineering JCECE Engineering Solved Paper-2009

  • question_answer
    A tuning fork \[A\] produces \[4\] beats/s with another tuning fork \[B\] of frequency\[320\,\,Hz\]. On filing one of the prongs of \[A,\,\,4\] beats/s 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: B

    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 then\[324\]. The beats between \[A\] and \[B\] will be 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\[320\,\,Hz).\]


You need to login to perform this action.
You will be redirected in 3 sec spinner