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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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(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The displacement ‘x’ (in meter) of a particle of mass ‘m’ (in kg) moving in one
- The equivalent resistance of two resistors connected in series is 6 Ω
- The rate of increase of thermo-e.m.f. with temperature at the neutral temperature
- A light whose frequency is equal to 6.0 x 10¹⁴ Hz is incident on a metal
- If X=3YZ², find dimensions of Y if X and Z are the dimensions of capacity
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.