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JEE Advanced · Physics · 16. Waves & Sound

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 He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas).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 \((0.350 \pm 0.005) m\), the gas in the tube is (Useful information: \(\sqrt{167\text{ RT}}=640\text{ J}^{\frac{1}{2}}\text{ mol}^{-\frac{1}{2}} ;\) \(\sqrt{140\text{ RT}}=590\text{ J}^{\frac{1}{2}}\text{ mole}^{-\frac{1}{2}}\) . The molar masses M in grams are given in the options. Take the values of \(\sqrt{\frac{10}{\text M }}\) for each gas as given there.)

  1. A NeonM=20, 1020=710
  2. B Nitroge M=28, 1028=35
  3. C OxygenM=32, 1032=916
  4. D ArgonM=36, 1036=1732
Verified Solution

Answer & Solution

Correct Answer

(D) ArgonM=36, 1036=1732

Step-by-step Solution

Detailed explanation

v=0.35 , λ=4γ v=λf=0.35 ×4 ×244 340 ms2
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