| ⇦ | 
| ⇨ |
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 super conductor exhibits perfect
- The identical loops of copper and aluminium are moving with the same speed
- A long horizontal rod has a bead which can slide long its length
- Which of the following combinations should be selected for better tuning of an L-C-R
- A small signal voltage V(t)=V₀ sin ωt is applied across an ideal capacitor C
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A super conductor exhibits perfect
- The identical loops of copper and aluminium are moving with the same speed
- A long horizontal rod has a bead which can slide long its length
- Which of the following combinations should be selected for better tuning of an L-C-R
- A small signal voltage V(t)=V₀ sin ωt is applied across an ideal capacitor C
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