4λ1=11+e⇒512×4v=11+e...(1)
4λ2=27+e⇒256×4v=27+e...(2)
Subtracting the equations,
256×4v×21=16cm=0.16m
⇒v=0.16×2×4×256
=328ms−1
A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in air. A tuning fork of…
JEE Main 2019 — Physics Waves & Oscillations
2019mcqmedium
A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in air. A tuning fork of frequency 512Hz produces first resonance when the tube is filled with water to a mark 11cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256Hz which produces first resonance when water reaches a mark 27cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to
Official previous-year question
Held on 12 Jan 2019 · Verified 6 Jul 2026.
Options
- A
335ms−1
- B
341ms−1
- C
322ms−1
- D
328ms−1
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