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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.
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following pairs is wrong
- Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 eV.
- A ray of light passes from a medium A having refractive index 1.6 to the medium
- A mass m moves in a circle on a smooth horizontal plane with velocity v₀
- A body of mass m is placed on the earth’s surface. It is taken from the earth’s surface
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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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.