| ⇦ | 
| ⇨ |
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: - When 1 kg of ice at 0⁰C melts to water at 0⁰C, the resulting change in its entropy
- When a charged oil drop moves upwards in an electric field,
- To demonstrate the phenomenon of interference, we require two sources which emit
- Two closed organ pipes when sounded simultaneously give 4 beats/second.
- The drift velocity of the electrons in a copper wire of length 2 m under
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When 1 kg of ice at 0⁰C melts to water at 0⁰C, the resulting change in its entropy
- When a charged oil drop moves upwards in an electric field,
- To demonstrate the phenomenon of interference, we require two sources which emit
- Two closed organ pipes when sounded simultaneously give 4 beats/second.
- The drift velocity of the electrons in a copper wire of length 2 m under
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.