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Waves question

2014 · Shift 1 · Q51
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Waves question

2014 · Shift 1 · Q51

JEE AdvancedPhysicsWavesMultiple correct+3 / −1
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 He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas).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 : 167RT\sqrt {167RT}167RT​ = 640 J1/2 mol −-− 1/2; 140RT\sqrt {140RT}140RT​ = 590 J1/2 mol −-− 1/2. The molar mass M in grams is given in the options. Take the values of 10/M\sqrt {10/M}10/M​ for each gas as given there.)
  1. A
    Neon (M=20,1020=710)\left( {M = 20,\sqrt {{{10} \over {20}}} = {7 \over {10}}} \right)(M=20,2010​​=107​)
  2. B
    Nitrogen (M=28,1028=35)\left( {M = 28,\sqrt {{{10} \over {28}}} = {3 \over 5}} \right)(M=28,2810​​=53​)
  3. C
    Oxygen (M=32,1032=916)\left( {M = 32,\sqrt {{{10} \over {32}}} = {9 \over {16}}} \right)(M=32,3210​​=169​)
  4. D
    Argon (M=36,1036=1732)\left( {M = 36,\sqrt {{{10} \over {36}}} = {{17} \over {32}}} \right)(M=36,3610​​=3217​)
View written solutionFree

Correct answer: D

  1. Resonance condition in a closed tube

For a resonance column closed at one end, the first resonance occurs at

L=λ4L = \frac{\lambda}{4}L=4λ​

where LLL is the minimum resonating length.

Given:

L=(0.350±0.005) mL = (0.350 \pm 0.005)\,\text{m}L=(0.350±0.005)m

So,

λ=4L=4(0.350±0.005)=1.400±0.020 m\lambda = 4L = 4(0.350 \pm 0.005) = 1.400 \pm 0.020\,\text{m}λ=4L=4(0.350±0.005)=1.400±0.020m
  1. Find the speed of sound in the gas

Given frequency:

f=244 s−1f = 244\,\text{s}^{-1}f=244s−1

Using

v=fλv = f\lambdav=fλ

we get

v=244×1.400=341.6 m/sv = 244 \times 1.400 = 341.6\,\text{m/s}v=244×1.400=341.6m/s

Error in vvv comes from error in λ\lambdaλ:

Δv=244×0.020=4.88 m/s\Delta v = 244 \times 0.020 = 4.88\,\text{m/s}Δv=244×0.020=4.88m/s

Hence,

v=341.6±4.9 m/sv = 341.6 \pm 4.9\,\text{m/s}v=341.6±4.9m/s

So the possible range is approximately

336.7 to 346.5 m/s336.7 \text{ to } 346.5\,\text{m/s}336.7 to 346.5m/s
  1. Speed of sound formula

For a gas,

v=γRTMv = \sqrt{\frac{\gamma RT}{M}}v=MγRT​​

where MMM is in kg/mol.

If MMM is given in grams, then

v=1000γRTMv = \sqrt{\frac{1000\gamma RT}{M}}v=M1000γRT​​

Using the given form,

v=100γRT 10Mv = \sqrt{100\gamma RT}\,\sqrt{\frac{10}{M}}v=100γRT​M10​​

Now use the provided values.

  1. Check each option

(A) Neon

Neon is monoatomic, so γ=53\gamma = \frac{5}{3}γ=35​. Given:

167RT=640\sqrt{167RT} = 640167RT​=640

and

1020=710\sqrt{\frac{10}{20}} = \frac{7}{10}2010​​=107​

Thus,

v≈640×710=448 m/sv \approx 640 \times \frac{7}{10} = 448\,\text{m/s}v≈640×107​=448m/s

This is far from the observed range.

(B) Nitrogen

Nitrogen is diatomic, so γ=75\gamma = \frac{7}{5}γ=57​. Given:

140RT=590\sqrt{140RT} = 590140RT​=590

and

1028=35\sqrt{\frac{10}{28}} = \frac{3}{5}2810​​=53​

Thus,

v≈590×35=354 m/sv \approx 590 \times \frac{3}{5} = 354\,\text{m/s}v≈590×53​=354m/s

This is slightly above the upper limit 346.5 m/s346.5\,\text{m/s}346.5m/s, so not consistent.

(C) Oxygen

Also diatomic, so use

140RT=590\sqrt{140RT} = 590140RT​=590

and

1032=916\sqrt{\frac{10}{32}} = \frac{9}{16}3210​​=169​

Thus,

v≈590×916=331.875 m/sv \approx 590 \times \frac{9}{16} = 331.875\,\text{m/s}v≈590×169​=331.875m/s

This is below the lower limit 336.7 m/s336.7\,\text{m/s}336.7m/s, so not consistent.

(D) Argon

Argon is monoatomic, so use

167RT=640\sqrt{167RT} = 640167RT​=640

and

1036=1732\sqrt{\frac{10}{36}} = \frac{17}{32}3610​​=3217​

Thus,

v≈640×1732=340 m/sv \approx 640 \times \frac{17}{32} = 340\,\text{m/s}v≈640×3217​=340m/s

This lies well within the experimental range:

336.7 to 346.5 m/s336.7 \text{ to } 346.5\,\text{m/s}336.7 to 346.5m/s

So this matches.

  1. Conclusion

The gas is

Argon\boxed{\text{Argon}}Argon​

So the correct option is D.

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