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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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- What is the time taken by moon light to reach earth?
- An amplifier has a voltage gain Av=1000. The voltage gain in dB is
- The de-Broglie wavelength of a proton(charge = 1.6 x 10⁻¹⁹ C, Mass = 1.6 x 10⁻²⁷ kg)
- One mole of an ideal gas at an initial temperature of T K, does 6R joule of work
- In the Young’s double slit experiment, a point P on the central bright fringe
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.