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JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    A tuning fork arrangement (pair) produces 4 beats/s with one fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is   AIEEE  Solved  Paper-2002

    A) 286 cps

    B)                           292 cps    

    C)                           294 cps                    

    D) 288 cps

    Correct Answer: B

    Solution :

    The tuning fork of frequency 288 Hz is producing 4 beats/s with the unknown tuning fork i.e., the frequency difference between them is 4. Therefore, the frequency of unknown tuning fork                                  \[=288\pm 4=292\] or 284              On placing a little wax of unknown tuning fork, its frequency decreases but now the number of beats produced per second is 2 i.e., the frequency difference now decreases. It is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of given tuning fork. Hence, the frequency of unknown tuning fork is 292 Hz.


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