| ⇦ | 
| ⇨ |
An open pipe resonates with a tuning fork of frequency 500 Hz. It is observed that two successive nodes are formed at distances 16 cm and 46 cm from the open end. The speed of sound in air in the pipe is
Options
(a) 230 m/s
(b) 300 m/s
(c) 320 m/s
(d) 360 m/s
Correct Answer:
300 m/s
Explanation:
Two successive modes are separated by λ/2. λ/2 = (46-16)=30 λ=60 cm=0.6 m v=nλ=500×0.6=300 m/s.
Related Questions: - The material suitable for making electromagnets should have
- When a Daniel cell is connected in the secondary circuit of a potentiometer,
- One mole of an ideal diatomic gas undergoes a transition from A to B along a path
- A thin prism of angle 15⁰ made of glass of refractive index µ₁ = 1.5 is combined
- When a ball is thrown up vertically with velocity vₒ,it reaches a maximum height
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The material suitable for making electromagnets should have
- When a Daniel cell is connected in the secondary circuit of a potentiometer,
- One mole of an ideal diatomic gas undergoes a transition from A to B along a path
- A thin prism of angle 15⁰ made of glass of refractive index µ₁ = 1.5 is combined
- When a ball is thrown up vertically with velocity vₒ,it reaches a maximum height
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
Question explain
Two successive modes are separated by λ/2. λ/2 = (46-16)=30 λ=60 cm=0.6 m v=nλ=500×0.6=300 m/s.