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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)
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- If the cold junction of a thermo-couple is kept at 0⁰C and the hot junction is kept
- 40 g of Argon is heated from 40⁰C to 100⁰C (R=2 cal/mole). Heat absorbed at constant
- Three capacitors each of capacity 4 μF are to be connected in such a way that the effective
- A bar magnet with magnetic moment 2.5×10³ JT⁻¹ is rotating in horizontal plane
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.