A tuning fork A, marked 512 Hz, produces 5 beats per second, where sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded?

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    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    This is a short answer type question as classified in NCERT Exemplar

    Frequency of tuning fork A

    v A = 512 Hz

    Probable frequency of tuning fork B

    v B = v A + 5 = 512 ± 5 = 517 Hz or 507Hz

    When B is loaded its frequency reduces .

    If it is 517Hz it might reduced to 507Hz given again a beat of 5Hz

    If it 507Hz reduction will always increase the beat frequency, hence v B = 517 Hz

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