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Speed of sound using a resonance tube: NEET MCQs with answers

49 questions on this topic of Experimental Skills (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In a resonance tube apparatus, a tuning fork of frequency 320 Hz produces first resonance at an air column length of 24.5 cm and second resonance at 76.5 cm. The speed of sound in air calculated from this experiment is:

  1. 332.8 m/s
  2. 326.4 m/s
  3. 340.0 m/s
  4. 346.2 m/s
Show answer and explanation

Answer: (A) 332.8 m/s

The difference between two successive resonance lengths is (l₂ - l₁) = λ/2. Here, λ/2 = 76.5 - 24.5 = 52.0 cm = 0.52 m, which gives λ = 1.04 m. Therefore, the speed of sound is v = f × λ = 320 × 1.04 = 332.8 m/s.

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2. Using first and second resonance lengths l₁ and l₂ with a tuning fork of frequency ν, the speed of sound is:

  1. v = 4νl₁
  2. v = ν(l₁ + l₂)
  3. v = 2ν(l₂ - l₁)
  4. v = ν(l₂ - l₁)
Show answer and explanation

Answer: (C) v = 2ν(l₂ - l₁)

l₂ + e = 3λ/4 and l₁ + e = λ/4, so l₂ - l₁ = λ/2 and v = νλ = 2ν(l₂ - l₁). The end correction cancels.

NCERT Class 11

hard

3. In a resonance tube experiment, moist air has accumulated above the water surface. Relative to dry air at the same temperature and atmospheric pressure, the resonant lengths will:

  1. Remain unchanged because frequency remains constant
  2. Decrease because moist air has lower density, decreasing sound speed
  3. Increase because moist air has higher density, increasing sound speed
  4. Increase because moist air has lower density, increasing sound speed
Show answer and explanation

Answer: (D) Increase because moist air has lower density, increasing sound speed

Moist air has water vapor (molecular mass 18 g/mol) mixed with air (effective molecular mass ~29 g/mol). Consequently, the density of moist air is lower than that of dry air at the same temperature and pressure. Since v = √(γP/ρ), the speed of sound in moist air is greater. For a given tuning fork frequency f, λ = v/f increases, which means the resonance lengths must increase.

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4. In a resonance tube experiment, a student measures the first resonance length as l₁ = (16.0 ± 0.1) cm and the second resonance length as l₂ = (49.0 ± 0.1) cm using a tuning fork of marked frequency (500 ± 1) Hz. The maximum percentage error in the calculated speed of sound is:

  1. 0.61%
  2. 0.81%
  3. 1.21%
  4. 0.41%
Show answer and explanation

Answer: (B) 0.81%

Speed of sound is v = 2f(l₂ - l₁). Let L = l₂ - l₁ = 49.0 - 16.0 = 33.0 cm. The absolute error ΔL = Δl₁ + Δl₂ = 0.1 + 0.1 = 0.2 cm. Fractional error Δv/v = Δf/f + ΔL/L = (1/500) + (0.2/33.0) = 0.0020 + 0.00606 = 0.00806. In percentage, it is 0.81%.

easy

5. Resonance in the tube is detected by:

  1. Vibration stopping
  2. A maximum loudness of sound
  3. A minimum loudness
  4. A change in colour
Show answer and explanation

Answer: (B) A maximum loudness of sound

At resonance, the natural frequency of the air column matches the tuning fork, and the sound heard is loudest.

NCERT Class 11

hard

6. In an experiment to measure the speed of sound, a student measures the first and second resonance lengths as l₁ = (16.0 ± 0.1) cm and l₂ = (50.0 ± 0.1) cm using a tuning fork of known frequency ν = (500 ± 1) Hz. The maximum percentage error in the calculated speed of sound is approximately:

  1. 0.79%
  2. 0.59%
  3. 1.25%
  4. 1.45%
Show answer and explanation

Answer: (A) 0.79%

The formula for speed of sound is v = 2ν(l₂ - l₁). The difference is Δl = l₂ - l₁ = 34.0 cm with absolute error δ(Δl) = 0.1 + 0.1 = 0.2 cm. The fractional error is δv/v = δν/ν + δ(Δl)/Δl = 1/500 + 0.2/34.0 = 0.0020 + 0.00588 = 0.00788, which corresponds to 0.79%.

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7. In a resonance tube experiment, a tuning fork of frequency 500 Hz produces successive resonances at air column lengths of 17.0 cm and 51.0 cm. The speed of sound in air calculated from this observation is:

  1. 340 m/s
  2. 330 m/s
  3. 350 m/s
  4. 360 m/s
Show answer and explanation

Answer: (A) 340 m/s

The distance between two consecutive resonances in a resonance tube corresponds to half a wavelength: λ/2 = l₂ - l₁ = 51.0 - 17.0 = 34.0 cm = 0.34 m. Thus, λ = 0.68 m. The speed of sound is v = νλ = 500 × 0.68 = 340 m/s.

easy

8. In a resonance tube experiment, stationary longitudinal waves are formed in the air column. What are the natures of the boundaries at the open end and at the water surface, respectively?

  1. Both ends act as displacement nodes
  2. Displacement node at the open end, displacement antinode at the water surface
  3. Pressure antinode at the open end, pressure node at the water surface
  4. Pressure node at the open end, pressure antinode at the water surface
Show answer and explanation

Answer: (D) Pressure node at the open end, pressure antinode at the water surface

At the open end, the air particles are free to oscillate, creating a displacement antinode and therefore a pressure node. At the closed end (water surface), the air particles cannot move, forming a displacement node and consequently a pressure antinode.

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9. The end correction of a tube of internal diameter d is approximately:

  1. 0.3d
  2. 3d
  3. d/2
  4. Zero
Show answer and explanation

Answer: (A) 0.3d

The antinode forms slightly outside the open end; the end correction is about 0.3 times the internal diameter (0.6 times the radius).

NCERT Class 11

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10. Given below are two statements regarding the resonance tube experiment: Statement I: The air column inside the resonance tube vibrates in forced oscillations when excited by a vibrating tuning fork held above it. Statement II: Maximum sound intensity is heard when the natural frequency of the vibrating air column matches the frequency of the tuning fork. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

The tuning fork acts as an external periodic driver that drives the air column into forced oscillations. When the length of the column is adjusted such that one of its natural frequencies equals the driver's frequency, acoustic resonance occurs and the amplitude of oscillation, and hence sound intensity, reaches a maximum. Both statements are scientifically accurate.

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11. A tuning fork of frequency 340 Hz excites a resonance tube. The first resonance is obtained at 24.0 cm and the second resonance at 74.0 cm. What is the inner radius of the resonance tube?

  1. 2.50 cm
  2. 3.33 cm
  3. 1.67 cm
  4. 5.00 cm
Show answer and explanation

Answer: (C) 1.67 cm

From the two resonance lengths, end correction e = (l₂ - 3l₁) / 2 = (74.0 - 3 × 24.0) / 2 = (74.0 - 72.0) / 2 = 1.0 cm. Since e = 0.6 × r (where r is the inner radius), r = e / 0.6 = 1.0 / 0.6 = 1.67 cm.

easy

12. A student performs the resonance column experiment using a tuning fork of frequency 480 Hz. If the room temperature increases from 15 °C to 30 °C during the day, how will the length of the first resonance position change?

  1. It will decrease slightly
  2. It will remain exactly the same
  3. It will increase slightly
  4. It will first decrease and then become zero
Show answer and explanation

Answer: (C) It will increase slightly

The speed of sound in air increases with temperature according to v ∝ √T. Since v = 4(l₁ + e)ν, an increase in sound velocity v for a constant fork frequency ν results in an increase in the resonance length l₁.

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