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EAMCET Medical EAMCET Medical Solved Paper-1999

  • question_answer
    Two tuning forks A and B produce 6 beats per second. If the frequency of A is 512 Hz. The fork B is loaded with wax and they produce 4 beats per second- The frequency of B is:

    A)  506 Hz                                 

    B)  516 Hz

    C)  508 Hz                                 

    D)  518 Hz

    Correct Answer: D

    Solution :

                     Number of beats per second \[{{n}_{1}}-{{n}_{2}}=6\] Given: frequency of tuning fork A = 512 Hz Hence, frequency of B is \[512\text{ }\pm \text{ }6\] Hence, frequency of B is either \[=506Hz\]or \[518\,Hz\] After loading the fork 5, the frequency decreases Hence, frequency of fork B become \[512\pm B=4\] \[B=516\]or 508 Therefore, the frequency of fork before loading must be 518 Hz.


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