| ⇦ |
| ⇨ |
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.
Related Questions: - A block of wood weighs 10 N and is resting on an inclined plane. The coefficient
- At what height h above earth, the value of g becomes g/2 (where R=radius of earth)?
- A person has a minimum distance of distinct vision as 50 cm. The power of lenses
- If sound wave travel from air to water, which of the following remains unchanged?
- Which one of the following is a vector?
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A block of wood weighs 10 N and is resting on an inclined plane. The coefficient
- At what height h above earth, the value of g becomes g/2 (where R=radius of earth)?
- A person has a minimum distance of distinct vision as 50 cm. The power of lenses
- If sound wave travel from air to water, which of the following remains unchanged?
- Which one of the following is a vector?
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply