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NEET Sample Paper NEET Sample Test Paper-71

  • question_answer
    Two tuning forks P and Q when set vibrating, give 4 beats/s. If a prong of the fork P is filed, the beats are reduced to 7.1s. What is frequency of P, if that of Q is 250 Hz?

    A) 246 Hz             

    B) 250 Hz

    C) 254 Hz             

    D) 252 Hz

    Correct Answer: A

    Solution :

    There are four beats between P and Q, therefore the possible frequencies of P are 246 or 254 (that is\[250 \pm  4\]) Hz. When the prong of P is filed, its frequency becomes greater than the original frequency. If we assume that the original frequency of P is 254, then on filing its frequency will be greater than 254. The beats between P and Q will be more than 4. But it is then that the beats are reduced to 2, therefore, 254 is not possible. Therefore, the required frequency must be 246 Hz. (This is true, because on filing the frequency may increase to 248, giving 2 beats with Q of frequency 250 Hz).


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