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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
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Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the equation y=asin(?t+kx),the dimensionals formula of ? is
- A glass flask weighing 390 g having internal volume 500 cc just floats
- A body of mass m is orbiting the earth at a radius r from the center of earth
- A bar magnet with magnetic moment 2.5×10³ JT⁻¹ is rotating in horizontal plane
- A hollow sphere of charge does not produce an electric field at any
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.