In a resonance tube experiment when the tube is filled with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of 24.5 cm. If the velocity of sound in air is 330 m/s, the tuning fork frequency is:

Option 1 -

1100 Hz

Option 2 -

2200 Hz

Option 3 -

3300 Hz

Option 4 -

550 Hz

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago
    Correct Option - 2


    Detailed Solution:

    f = V (n)/ (4 (1 - 17/100) (as closed from end)
    f = (n - 2) (v)/ (4 (1 - 24.5/100)
    nV (100)/4 (83) = (n - 2) (v) (100)/ (4) (75.5)
    75.5n = 83 (n - 2)
    75.5n = 83 (M - 2)
    7.5n = 166
    n = 22 (approx)
    f = (330) (22) (100)/ (4 (83) = 2200 Hz

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