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 - <p>1100 Hz<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>2200 Hz</p>
Option 3 - <p>3300 Hz</p>
Option 4 - <p>550 Hz</p>
3 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
5 months 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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