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JEE Advanced 2014Paper 1
Exam
JEE Advanced
Year
2014
Paper
Paper 1
Subject
Physics
Chapter
Waves and Sound

JEE Advanced 2014 · Paper 1 · Physics · Waves and Sound

A student is performing an experiment using a resonance column and...

A student is performing an experiment using a resonance column and a tuning fork of frequency 244s−1244{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 He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas).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( 0.350 \pm 0.005 ) m , the gas in the tube is (Useful information: 167RT=640J12mol−12;140RT=590J12mole−12\sqrt{167RT} = 640{J}^{\dfrac{1}{2}}mo{l}^{- \dfrac{1}{2}}; \sqrt{140RT} = 590{J}^{\dfrac{1}{2}}mol{e}^{- \dfrac{1}{2}} . The molar masses M in grams are given in the options. Take the values of 10M\sqrt{\dfrac{10}{M}} for each gas as given there.)

Options

  1. ANeon(M=20,1020=710)\left(M = 20, \sqrt{\dfrac{10}{20}} = \dfrac{7}{10}\right)
  2. BNitroge(M=28,1028=35)\left(M = 28, \sqrt{\dfrac{10}{28}} = \dfrac{3}{5}\right)
  3. COxygen(M=32,1032=916)\left(M = 32, \sqrt{\dfrac{10}{32}} = \dfrac{9}{16}\right)
  4. DArgon(M=36,1036=1732)\left(M = 36, \sqrt{\dfrac{10}{36}} = \dfrac{17}{32}\right)

Correct answer

D. Argon(M=36,1036=1732)\left(M = 36, \sqrt{\dfrac{10}{36}} = \dfrac{17}{32}\right)

Step-by-step solution

v=0.35,λ=4γv = 0.35, \lambda = 4\gamma

v=\lambdaf=0.35×4×244≅340ms2v = \lambdaf = 0.35 \times 4 \times 244 ≅ 340m{s}^{2}

v=14γ\gammaRTMv = \dfrac{1}{4\gamma}\sqrt{\dfrac{\gammaRT}{M}}

v=\gammaRTM⇒M=\gammaRTv2v = \sqrt{\dfrac{\gammaRT}{M}} ⇒ M = \dfrac{\gammaRT}{{v}^{2}}

For monoatomic gas : γ=1.67⇒M=6402100×10003402\gamma = 1.67 ⇒ M = \dfrac{{640}^{2}}{100} \times \dfrac{1000}{{340}^{2}}

≈(M2−117)2\approx {\left({M}^{2} - \dfrac{1}{17}\right)}^{2}

≈36\approx 36

M=36M = 36

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