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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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Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Imagine a light planet revolving around a very massive star in a circular
- A circular disc of radius R is removed from a bigger circular disc of radius 2R
- A monoatomic gas at a pressure P, having a volume V expands isothermally
- Consider a uniform square plate of side ɑ and mass m. The moment of inertia
- A particle is kept at rest at the top of a sphere of diameter 42m.When disturbed slightly
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.