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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
- A particle of mass 1.96×10⁻¹⁵ kg is kept in equilibrium between two horizontal metal
- A current is flowing through resistance. A potential difference of 120 V
- A satellite of mass m revolves around the earth of radius R at a height x from
- Two discs of moment of inertia I₁ and I₂ and angular speeds ?￢ﾂﾁ and ?￢ﾂﾂ are rotating
- A particle moves in a circle of radius 5 cm with constant speed and time period
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.