Q41 / 60 · JEE Advanced 2014 (Paper 1) · Physics

A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s-1. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350± 0.005)m , the gas in the tube is (Useful information: 167RT=640 J12 mol-12;140RT=590 J12 mole-12 . The molar masses M in grams are given in the options. Take the values of 10M for each gas as given there.)

  • A
    NeonM=20, 1020=710
  • B
    Nitroge M=28, 1028=35
  • C
    OxygenM=32, 1032=916
  • Correct answer
    D
    ArgonM=36, 1036=1732

Solution

v=0.35 , λ=4γ

v=λf=0.35 ×4 ×244 ≅340 ms2

v=14γ  γRTM

v= γRTM ⇒  M=γRTv2

For monoatomic gas :   γ=1.67 ⇒  M=6402100 ×10003402

≈M2-1172

≈36

M=36

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

More Waves and Sound Questions