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The second overtone of an open pipe is in resonance with the first overtone of a closed pipe of length 2 m. Length of the open pipe is
Options
(a) 4 m
(b) 2 m
(c) 8 m
(d) 1 m
Correct Answer:
4 m
Explanation:
Frequency of second overtone of an open is ʋ = 3v / 2l₀ Frequency of first overtone of a closed pipe is ʋ’ = 3v / 4lc Given : ʋ = ʋ’ ⇒ 3v / 2l₀ = 3v / 4lc l₀ / lc = 2 ⇒ l₀ = 2lc = 2 × 2 = 4 m.
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Topics: Waves
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Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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