JEE Advanced | 2014Sound WavesHard
Question
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 He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas).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 : 
= 640 J1/2 mole-1/2; 
= 590 J1/2mole-1/2. The molar masses M in grams are given in the options. Take the values of 
for each gas as given there.)
= 640 J1/2 mole-1/2; = 590 J1/2mole-1/2. The molar masses M in grams are given in the options. Take the values of for each gas as given there.)Options
A.Neon 
B.Nitrogen 
C.Oxygen 
D.Argon 
Solution
Minimum length = 
⇒ λ = 4l
Now, υ = fλ = (244) × 4 × l
as l = 0.350 ± 0.005
⇒ υ lies between 336.7 m/s to 346.5 m/s
Now, υ
, here M is molecular mass in gram
For monoatomic gas γ = 1.67 ⇒ υ = 640 ×
For diatomic gas, γ = 1.4 ⇒ υ = 590 ×
∴ υNe = 640 ×
= 448 m/s
υAr = 640 ×
= 340 m/s
υO2 = 590 ×
= 331.8 m/s
υN2 = 590 ×
= 354 m/s
∴ Only possible answer is Argon.
⇒ λ = 4l
Now, υ = fλ = (244) × 4 × l
as l = 0.350 ± 0.005
⇒ υ lies between 336.7 m/s to 346.5 m/s
Now, υ
, here M is molecular mass in gramFor monoatomic gas γ = 1.67 ⇒ υ = 640 ×
For diatomic gas, γ = 1.4 ⇒ υ = 590 ×
∴ υNe = 640 ×
= 448 m/s υAr = 640 ×
= 340 m/sυO2 = 590 ×
= 331.8 m/sυN2 = 590 ×
= 354 m/s∴ Only possible answer is Argon.
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