Tuning fork above a partially-filled resonance tube. First resonance length L1 = λ/4 − e (end correction); second resonance length L2 = 3λ/4 − e. So λ = 2(L2 − L1) and speed v = f·λ. Avoid the end correction by using the difference. Apply temperature correction: v_T = v_0 √(T/T_0).
-- NCERT Physics Lab Manual Class 11, Part 7, p. 114Experiment 08 Speed of Sound Resonance
8 MCQs9-step worked example
Source: NCERT Experimental SkillsOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Experiment 08 Speed of Sound Resonance, explained for NEET
The trap in this experiment is finishing at the first resonance. A student lowers the water, hears the sound swell, measures l₁, and writes v = 4f l₁. That answer is always too small, and not by a random amount — the antinode does not sit at the rim of the tube. It sits a little above it, by an end correction e. The honest first-resonance relation is l₁ + e = λ/4, and e is unknown.
The second resonance is what rescues the measurement. Lower the water further and the same fork resonates again at l₂ + e = 3λ/4. Subtract: l₂ − l₁ = λ/2, with e gone. So
λ = 2(l₂ − l₁) and v = 2f(l₂ − l₁).
That subtraction is the whole design of the experiment, and it is why the procedure in the NCERT Physics Lab Manual Class 11, Part 7, page 114 insists on two resonance positions rather than one. If you do want e, the same two readings give it: e = (l₂ − 3l₁)/2, typically a centimetre or so for a standard tube.
Two structural points hold the rest together. The water surface is a rigid boundary, so it is a displacement node and a pressure antinode; the open top carries the displacement antinode. And f belongs to the fork, stamped on it — the tube supplies only lengths.
For NEET, questions are short: identify the node, pick the right combination of l₁ and l₂, or correct a reported speed to 0 °C using the ≈ 0.61 m s⁻¹ per °C rule. Watch-out: the speed you compute is the speed at the room's temperature, not a textbook constant. Quoting 331 m s⁻¹ as "the" answer at 27 °C is a lost mark.
Can you answer these Experiment 08 Speed of Sound Resonance MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
When the air column in a resonance tube is sounding at a resonance position, the water surface at the bottom of the column is a point of:
Show answer and why every option is right or wrong
Answer: C. C is correct. The water surface is a rigid boundary, so the air there cannot move — a displacement node — and displacement nodes are pressure antinodes, as set out for the closed air column in NCERT Physics Lab Manual Class 11, Part 7, page 114.
Why A is wrong: A is wrong because the air at a rigid surface cannot oscillate, so it cannot be a displacement antinode; that description belongs to the open top of the tube.
Why B is wrong: B is wrong on both counts: the surface is a node, not an antinode, and a pressure node sits where displacement is maximum, which is at the open end.
Why D is wrong: D is wrong because displacement node and pressure node cannot coincide — where displacement is zero the pressure variation is largest.
MCQ 2Easy RecallPractice
Acoustically, the vibrating air column above the water in a resonance tube behaves as a pipe that is:
Show answer and why every option is right or wrong
Answer: A. A is correct. The water closes the lower end and the tube mouth is open, so the shortest resonating column is about λ/4 — the closed-pipe arrangement described in NCERT Physics Lab Manual Class 11, Part 7, page 114.
Why B is wrong: B is wrong because a pipe open at both ends has its shortest resonance at about half a wavelength, not a quarter; also the water end is not open.
Why C is wrong: C is wrong because the top of the tube is open to the room, and a doubly closed pipe is not what the apparatus provides.
Why D is wrong: D is wrong on the boundary condition and on the length: no resonating column in this experiment is a full wavelength at the first position.
MCQ 3Easy RecallPractice
With a fork of frequency f and the first two resonance lengths l₁ and l₂, the speed of sound in the air column is calculated as:
Show answer and why every option is right or wrong
Answer: D. D is correct. Since l₂ − l₁ = λ/2, the wavelength is 2(l₂ − l₁) and v = fλ = 2f(l₂ − l₁) — the working relation of the experiment in NCERT Physics Lab Manual Class 11, Part 7, page 114.
Why A is wrong: A is wrong because it treats l₁ alone as exactly λ/4 and so silently ignores the end correction, returning a speed that is too small.
Why B is wrong: B is wrong because l₂ is about 3λ/4, not λ/4, so multiplying it by 4f overestimates the speed roughly threefold.
Why C is wrong: C is wrong because l₂ − l₁ is half a wavelength, not a whole one; the factor 2 is missing.
MCQ 4Direct ApplicationPractice
A 512 Hz fork gives resonance at column lengths l₁ = 0.160 m and l₂ = 0.500 m. The speed of sound in the air of the tube is:
Show answer and why every option is right or wrong
Answer: B. B is correct: l₂ − l₁ = 0.340 m, so v = 2 × 512 × 0.340 = 3.48 × 10² m s⁻¹, using the two-resonance relation of NCERT Physics Lab Manual Class 11, Part 7, page 114.
Why A is wrong: A is wrong because it uses f(l₂ − l₁) and drops the factor 2 that converts a half wavelength into a full one.
Why C is wrong: C is wrong because it comes from 4f l₁, which assumes the first column is exactly a quarter wavelength and so leaves the end correction uncorrected.
Why D is wrong: D is wrong because it adds the two lengths, 2f(l₂ + l₁), instead of subtracting them; only the difference removes the end correction.
MCQ 5Direct ApplicationPractice
In another trial the same apparatus resonates at l₁ = 0.170 m and l₂ = 0.530 m. The end correction for this tube is:
Show answer and why every option is right or wrong
Answer: A. A is correct: e = (l₂ − 3l₁)/2 = (0.530 − 0.510)/2 = 1.0 × 10⁻² m, the end correction obtained from the same pair of readings in NCERT Physics Lab Manual Class 11, Part 7, page 114.
Why B is wrong: B is wrong because it uses l₂ − 3l₁ without halving it; the relation carries a factor of one half.
Why C is wrong: C is wrong because 0.360 m is l₂ − l₁, which is half a wavelength, not the end correction.
Why D is wrong: D is wrong because subtraction removes e from the speed calculation, but e itself is still a real, non-zero length that these readings can report.
MCQ 6Direct ApplicationPractice
A resonance-tube run gives 3.48 × 10² m s⁻¹ in a room at 27 °C. Taking the speed of sound in air to change by 0.61 m s⁻¹ for each degree Celsius, the corresponding value at 0 °C is about:
Show answer and why every option is right or wrong
Answer: D. D is correct: 0.61 × 27 = 16.5 m s⁻¹, and 348 − 16.5 = 3.32 × 10² m s⁻¹, the temperature reduction noted with this experiment in NCERT Physics Lab Manual Class 11, Part 7, page 114.
Why A is wrong: A is wrong because it adds the correction; cooling the air from 27 °C to 0 °C lowers the speed, it does not raise it.
Why B is wrong: B is wrong because it treats the measured speed as temperature-independent, which is exactly what the room-temperature caveat warns against.
Why C is wrong: C is wrong because it applies roughly half the correction — about 0.3 m s⁻¹ per °C instead of 0.61.
MCQ 7CalculationPractice
From the same readings f = 512 Hz, l₁ = 0.160 m and l₂ = 0.500 m, student P computes the speed as 4f l₁ while student Q computes it as 2f(l₂ − l₁). P's value compared with Q's is:
Show answer and why every option is right or wrong
Answer: C. C is correct: P gets 4 × 512 × 0.160 = 327.7 m s⁻¹ against Q's 348.2 m s⁻¹, so P is low by about 2.0 × 10¹ m s⁻¹ — the size of the end-correction penalty described in NCERT Physics Lab Manual Class 11, Part 7, page 114.
Why A is wrong: A is wrong in direction: omitting the end correction shortens the assumed quarter wavelength, so the computed speed comes out below the two-length value, not above it.
Why B is wrong: B is wrong in size; 41 m s⁻¹ is roughly twice the actual gap, which is about 20 m s⁻¹ for these numbers.
Why D is wrong: D is wrong because the two formulas are not equivalent: 4l₁ equals λ only if the end correction is zero, and here it is 1.0 × 10⁻² m.
MCQ 8CalculationPractice
The same tube in the same room is used again with a 256 Hz fork in place of the 512 Hz fork. Compared with the earlier run, the quantity (l₂ − l₁) and the speed computed from it are:
Show answer and why every option is right or wrong
Answer: B. B is correct: the speed of sound is fixed by the air, so halving f doubles λ and therefore doubles l₂ − l₁ = λ/2, leaving v = 2f(l₂ − l₁) the same — the consistency check the procedure in NCERT Physics Lab Manual Class 11, Part 7, page 114 relies on.
Why A is wrong: A is wrong because the speed is a property of the air at that temperature and cannot double when only the fork is changed.
Why C is wrong: C is wrong in direction: a lower frequency means a longer wavelength, so the separation of the two resonance positions grows rather than shrinks.
Why D is wrong: D is wrong because the separation must change — it is λ/2 — and the speed must not; this option reverses both.
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How do you solve a Experiment 08 Speed of Sound Resonance question? A worked example
- 1
Given
Tuning fork frequency f = 512 Hz (exact — the stamped value on the fork). First resonance at l₁ = 16.0 cm, second at l₂ = 50.0 cm, read on a metre scale of least count 0.1 cm attached to the resonance tube. Room temperature 27 °C.
- 2
Required
The speed of sound in the air column, and the percentage uncertainty in it from the two length readings.
- 3
Concept
The air column closed by the water resonates when its effective length is an odd multiple of a quarter wavelength: l₁ + e = λ/4 and l₂ + e = 3λ/4. Subtracting eliminates the unknown end correction e, so the pair of lengths — not either one alone — carries the wavelength.
- 4
Formula
l₂ − l₁ = λ/2, hence λ = 2(l₂ − l₁) and v = fλ = 2f(l₂ − l₁). For the uncertainty, the difference of two scale readings carries the sum of their absolute uncertainties, and v is directly proportional to that difference, so Δv/v = Δ(l₂ − l₁)/(l₂ − l₁).
- 5
Substitution
l₂ − l₁ = 50.0 − 16.0 = 34.0 cm = 0.340 m. v = 2 × 512 × 0.340. Each length is uncertain by the least count, 0.1 cm, so Δ(l₂ − l₁) = 0.1 + 0.1 = 0.2 cm.
- 6
Calculation
v = 1024 × 0.340 = 348.16 m s⁻¹. Fractional uncertainty = 0.2/34.0 = 5.9 × 10⁻³, i.e. 0.59 %, giving Δv ≈ 0.0059 × 348 ≈ 2 m s⁻¹. The frequency 512 Hz is an exact stamped value and the 2 in 2f(l₂ − l₁) is a counting factor from the geometry; neither contributes to the significant-figure count, which is set by the three-figure lengths.
- 7
Final answer
v = (3.48 ± 0.02) × 10² m s⁻¹ at 27 °C — three significant figures, matching the three-figure length readings. Reduced to 0 °C using 0.61 m s⁻¹ per °C, this is about 3.32 × 10² m s⁻¹.
- 8
Common trap
Using l₁ by itself, v = 4f l₁ = 4 × 512 × 0.160 = 327.7 m s⁻¹, which is about 20 m s⁻¹ low. The shortfall is not experimental scatter; it is the end correction e = (l₂ − 3l₁)/2 = 1.0 × 10⁻² m being ignored. A second symptom is quoting the result as "the" speed of sound without naming the temperature it belongs to.
- 9
Similar NEET-style question
A resonance tube sounded with a 480 Hz fork resonates at 17.5 cm and 53.5 cm, both read on a scale of least count 0.1 cm. Find the speed of sound and the end correction, and state the percentage uncertainty in the speed. (Answer: v = 2 × 480 × 0.360 = 3.46 × 10² m s⁻¹; e = (53.5 − 52.5)/2 = 0.5 cm; uncertainty 0.2/36.0 = 0.56 %.)
What to remember before solving Experiment 08 Speed of Sound Resonance questions
More in Experimental Skills: 6 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
Experiment 08 Speed of Sound Resonance questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
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