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Waves question

2012 · Shift 2 · Q44
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Waves question

2012 · Shift 2 · Q44

JEE AdvancedPhysicsWavesMCQ+3 / −0.75
A student is performing the experiment of resonance Column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38oC in which the speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the Resonance Column tube. When the first resonance occurs, the reading of the water level in the column is
  1. A
    14.0 cm
  2. B
    15.2 cm
  3. C
    16.4 cm
  4. D
    17.6 cm
View written solutionFree

Correct answer: B

Step-by-Step Solution:

  1. Identify the given information:

    • Diameter of the column tube, d = 4 cm.
    • Frequency of the tuning fork, f = 512 Hz.
    • Speed of sound in air at 38°C, v = 336 m/s.
    • The zero of the meter scale is at the top end of the tube.

  2. Calculate the wavelength of the sound wave: The relationship between the speed of sound (v), frequency (f), and wavelength (λ) is given by the formula: v=fλv = fλv=fλ We can rearrange this to solve for the wavelength: λ=vfλ = \frac{v}{f}λ=fv​ Substituting the given values: λ=336 m/s512 Hz=0.65625 mλ = \frac{336 \text{ m/s}}{512 \text{ Hz}} = 0.65625 \text{ m}λ=512 Hz336 m/s​=0.65625 m To work with the dimensions given for the tube, let's convert the wavelength to centimeters: λ=0.65625 m×100cmm=65.625 cmλ = 0.65625 \text{ m} \times 100 \frac{\text{cm}}{\text{m}} = 65.625 \text{ cm}λ=0.65625 m×100mcm​=65.625 cm

  3. Account for the end correction: In a resonance column, the antinode is not formed exactly at the open end of the tube but slightly above it. This phenomenon is known as end correction (e). For a cylindrical tube of diameter d, the end correction is approximated by: e=0.3de = 0.3de=0.3d Substituting the given diameter: e=0.3×4 cm=1.2 cme = 0.3 \times 4 \text{ cm} = 1.2 \text{ cm}e=0.3×4 cm=1.2 cm

  4. Apply the condition for first resonance: The first resonance in a column closed at one end occurs when the effective length of the air column is equal to one-quarter of the wavelength. The effective length is the actual length of the air column (L1L_1L1​) plus the end correction (e). L1+e=λ4L_1 + e = \frac{λ}{4}L1​+e=4λ​ The value of L1L_1L1​ represents the reading of the water level from the top end, which is what the question asks for.

  5. Solve for the first resonating length (L1L_1L1​): First, calculate λ/4: λ4=65.625 cm4=16.40625 cm\frac{λ}{4} = \frac{65.625 \text{ cm}}{4} = 16.40625 \text{ cm}4λ​=465.625 cm​=16.40625 cm Now, rearrange the resonance condition formula to solve for L1L_1L1​: L1=λ4−eL_1 = \frac{λ}{4} - eL1​=4λ​−e Substitute the calculated values: L1=16.40625 cm−1.2 cmL_1 = 16.40625 \text{ cm} - 1.2 \text{ cm}L1​=16.40625 cm−1.2 cm L1=15.20625 cmL_1 = 15.20625 \text{ cm}L1​=15.20625 cm

  6. Compare the result with the given options: The calculated value L1≈15.2L_1 ≈ 15.2L1​≈15.2 cm matches option B.

    • A: 14.0 cm
    • B: 15.2 cm
    • C: 16.4 cm (This would be the answer if end correction was ignored)
    • D: 17.6 cm

Therefore, the reading of the water level in the column for the first resonance is 15.2 cm.

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