Q74 / 90 · JEE Main 2019 (12 Jan Shift 2) · Physics

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 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to

  • A
    335 m s-1
  • B
    341 m s-1
  • C
    322 m s-1
  • Correct answer
    D
    328 m s-1

Solution

λ14=11+e ⇒v512 × 4=11+e    ...(1)

λ24=27+e ⇒v256 × 4=27+e    ...(2)

Subtracting the equations,

v256×4×12=16 cm=0.16 m

⇒v=0.16×2×4×256

         =328 m s-1

Hence, the correct answer is option: 'D'.

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