NEET Physics - Chapter 17
Quink's Tube – Complete Notes, Revision, Important Questions & Downloads
Quink's Tube is built around the subtopic Interference Apparatus for Sound Velocity Measurement, where two sound paths are superposed and the detector shifts between maxima as path difference changes. The core derivation on this page is constructive condition l2 - l1 = N lambda, followed by the displacement relation 2x = lambda when the sliding tube is moved from one maximum to the next. NEET usually tests this as a short numerical: identify why tube displacement changes path by 2x, extract lambda from x, then compute sound speed using v = n0 lambda = 2n0x. You must track geometry and interference condition together, because option traps mix x and 2x or treat successive maxima as lambda/2 path difference.
Interference SetupPath DifferenceNCERT-Aligned
Expected QuestionsQ
1
Typically appears as one direct numerical or concept check on successive maxima, path difference, and velocity relation in sound interference apparatus.
Time Required⏱
1 h
About 25 minutes to lock derivation and 35 minutes to solve mixed MCQs that combine interference condition, geometry of tube motion, and velocity computation.
Difficulty⚡
Medium
Formula count is small, but mistakes are frequent because students confuse path difference with tube displacement and miss the factor of 2 in 2x = lambda.
NRI USA Curriculum GapUS
Bridge Needed
Many US high-school wave labs stop at qualitative interference observation, while NEET expects rapid algebraic extraction of lambda and v from measured displacement between successive maxima.
1Subtopics
20Practice Questions
4Free Downloads
1 hPrep Time
Quink's Tube Weightage and Trend
Waves and Sound - Topic 14| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2020 | 1 | 4 | |
| 2021 | 0 | 0 | |
| 2022 | 1 | 4 | |
| 2023 | 0 | 0 | |
| 2024 | 1 | 4 | |
| 2025 | 0 | 0 | |
| Estimated topic-linked asks in recent NEET papers | 3 | 12 |
This topic is tested through interference maxima logic, not through long derivation; students must immediately write l2 - l1 = N lambda and next-maximum shift 2x = lambda.
The displacement x of the sliding tube changes path by 2x because the wave traverses the moved segment twice in the U-tube arm.
Velocity question stems are usually one-line: once lambda is obtained from x, use v = n0 lambda and keep source frequency n0 unchanged.
The displacement x of the sliding tube changes path by 2x because the wave traverses the moved segment twice in the U-tube arm.
Velocity question stems are usually one-line: once lambda is obtained from x, use v = n0 lambda and keep source frequency n0 unchanged.
0.5
Avg Questions / Year
12
Total Marks (6 yrs)
Irregular
Pattern
Medium
Difficulty
5-Step Quink's Tube Solve Routine
1
Write both constructive conditions first Start with l2 - l1 = N lambda at first maximum and l2' - l1 = (N + 1) lambda at the next maximum before touching numbers.
2
Insert geometry relation before subtraction Use l2' = l2 + 2x explicitly; this is the point where most errors occur if tube displacement is used as x instead of path increment 2x.
3
Extract wavelength cleanly Subtract the two maxima equations to get 2x = lambda and carry units so lambda comes out in meters.
4
Compute speed with source frequency Apply v = n0 lambda = 2n0x and check that n0 is in Hz; if frequency is in kHz, convert before multiplication.
5
Run a trap check on maxima condition Successive maxima differ by one lambda in path difference, not lambda/2; lambda/2 corresponds to displacement relation x = lambda/2 only after converting path change 2x.
Quink's Tube Download Kit
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Focused notes on Quink's tube setup, interference conditions for maxima, derivation of 2x = lambda, and speed calculation from measured displacement.
Download PDFFormula Sheet
Single-sheet formula map with l2 - l1 = N lambda, l2' = l2 + 2x, 2x = lambda, and v = n0 lambda = 2n0x plus unit checks.
Download PDFMCQ Practice
Practice set on path-difference updates, successive-maximum interpretation, and velocity extraction with timing-oriented elimination traps.
Download PDFPYQ Workbook
Year-tagged interference and sound-velocity workbook with Quink-style apparatus questions and short solving templates for exam-speed use.
Download PDFSubtopics in Quink's Tube
2-Column Table| Column A | Column B |
|---|---|
Rapid Revision Cards
Concept → Trap → Example1) Interference Apparatus for Sound Velocity Measurement
Successive maxima methodFor Quink's tube maxima: l2 - l1 = N lambda, l2' = l2 + 2x, and l2' - l1 = (N + 1) lambda, giving 2x = lambda and v = n0 lambda = 2n0x.
- At first maximum, enforce constructive path difference as an integer multiple of lambda.
- When the sliding arm is moved by x, the acoustic path in that arm changes by 2x because the wave traverses the shifted portion twice.
- Trap: many students write x = lambda and overestimate velocity by a factor of 2; always derive 2x = lambda from successive maxima equations.
Example (NEET-style)If n0 = 512 Hz and the tube shift from one maximum to the next is x = 0.33 m, then lambda = 2x = 0.66 m and sound speed is v = n0 lambda = 512 x 0.66 = 337.92 m/s, which is consistent with room-temperature air.
Curriculum Gap: India vs USA
Two concrete preparation gaps to bridge for NEET readinessAP Physics 1 wave units emphasize qualitative interference, NEET expects apparatus-based algebra
AP-level treatment often stops at observing constructive and destructive interference, while NEET asks for equation chaining from apparatus geometry to numerical sound speed.
- Practice converting the setup into l2 - l1, l2' - l1, and l2' = l2 + 2x in under 20 seconds.
- Drill two-step numericals where x is measured and n0 is given in different frequency units.
US lab reports focus uncertainty commentary; NEET rewards fast deterministic reduction
In many US labs, students discuss error bars and instrumentation, but NEET single-correct MCQs reward immediate formula reduction and option elimination under strict time limits.
- Train with one-minute questions that force direct use of 2x = lambda and v = n0 lambda.
- Maintain a trap list: confusing x with 2x, using lambda/2 for successive maxima path difference, and forgetting unit conversion.
Concept IQ Check
2 MCQs1In Quink's tube, the sliding arm is shifted by 0.25 m to move from one sound-intensity maximum to the next. If the source frequency is 680 Hz, what is the speed of sound in air?Interference Apparatus for Sound Velocity Measurement
170 m/s
340 m/s
510 m/s
680 m/s
For successive maxima in Quink's tube, the path-difference increment is one wavelength. Because tube displacement x changes path by 2x in the moved arm, we use 2x = lambda. With x = 0.25 m, lambda = 0.50 m. Then v = n0 lambda = 680 x 0.50 = 340 m/s. Option A comes from wrongly using lambda = x. Option C is from multiplying by an incorrect 0.75 m wavelength. Option D is from forgetting wavelength multiplication entirely and copying frequency as speed. The key physics step is recognizing why geometric displacement contributes double path length in this setup.
2A student writes x = lambda for the shift between two successive maxima in Quink's tube. Which correction is physically correct?Interference Apparatus for Sound Velocity Measurement
Correct relation is x = lambda/4 because two maxima are separated by quarter cycle
Correct relation is x = 2 lambda because both arms contribute equally
Correct relation is 2x = lambda because moving one arm by x changes that path length by 2x
Correct relation is x = lambda only when frequency is known
The detector moves from one constructive-interference maximum to the next when path difference increases by one lambda. In Quink's tube, only one branch is physically shifted, but acoustic travel in that branch includes two traversals of the shifted segment, so path difference increment is 2x. Hence 2x = lambda and x = lambda/2. Option A invents quarter-cycle logic not used for successive maxima in this geometry. Option B incorrectly doubles lambda without geometric basis. Option D mixes geometric relation with frequency information: x-lambda relation is purely geometric-interference, independent of whether frequency value is supplied. This trap appears frequently because students memorize the final x = lambda/2 without understanding the 2x path update.
Practice Questions
Click "Reveal Answer" after attempting1In a Quink's tube experiment, the source frequency is 500 Hz and the tube shift between consecutive maxima is 0.34 m. Find the speed of sound.
170 m/s
340 m/s
510 m/s
680 m/s
👁 Reveal Answer
Correct option: B. For successive maxima in Quink's tube, 2x = lambda. So lambda = 2 x 0.34 = 0.68 m. Then v = n0 lambda = 500 x 0.68 = 340 m/s. Option A results from taking lambda = x. Option C comes from arithmetic mismatch after getting lambda correctly. Option D doubles the computed velocity by applying factor 2 twice.
2If n0 is unchanged and measured shift x for successive maxima doubles, what happens to computed sound speed v in Quink's tube?
v becomes half
v remains same
v doubles
v becomes four times
👁 Reveal Answer
Correct option: C. In this setup, v = 2n0x. With frequency n0 fixed, speed is directly proportional to x. If x doubles, v doubles. Option A reverses the proportionality. Option B would be true only if x did not change. Option D would require quadratic dependence on x, which does not appear in v = 2n0x.
3A student gets x = 0.20 m and frequency n0 = 850 Hz. They compute lambda = 0.20 m and v = 170 m/s. Identify the error.
Frequency should be divided by 2
Path change should be x/2
For successive maxima lambda = 2x, not x
Velocity must always be 343 m/s
👁 Reveal Answer
Correct option: C. In Quink's tube, moving the branch by x introduces a path difference 2x, so for consecutive maxima lambda = 2x = 0.40 m. Then v = n0 lambda = 850 x 0.40 = 340 m/s. Option A is unsupported because n0 is source-given. Option B is opposite to the geometry of two traversals. Option D is incorrect because measured speed can vary with air conditions and experiment, although around 340 m/s is common near room temperature.
4In deriving 2x = lambda, which pair of equations is subtracted after using l2' = l2 + 2x?
l2 - l1 = N lambda and l2' - l1 = (N + 1) lambda
l2 - l1 = (N + 1) lambda and l2' - l1 = N lambda
l2 + l1 = N lambda and l2' + l1 = (N + 1) lambda
l2 - l1 = N lambda and l2' - l1 = (N + 2) lambda
👁 Reveal Answer
Correct option: A. The first maximum gives l2 - l1 = N lambda and the next maximum gives l2' - l1 = (N + 1) lambda. Subtracting the first from the second yields l2' - l2 = lambda. With l2' = l2 + 2x, this gives 2x = lambda. Option B reverses the sequence and changes sign. Option C uses incorrect sum relation. Option D assumes skipping one maximum and would correspond to a different displacement condition.
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Frequently Asked Questions
Notes · Downloads · Revision · Important QuestionsWhy does moving the tube by x create a path difference of 2x in Quink's tube?
The shifted arm contributes extra travel on both segments of that branch in the U-path, so acoustic distance increases by 2x rather than x. This is the geometric reason the successive-maxima condition gives 2x = lambda. If this doubling is missed, all derived wavelengths and speeds are underestimated by half.
How is Quink's tube different from a simple two-slit light-interference analogy?
The interference idea is common, but Quink's tube is a guided sound-path apparatus with adjustable geometric path length in one branch. The exam focus is not fringe spacing on a screen but extracting wavelength and speed from maxima shifts and known frequency. So geometric path accounting is central in this topic.
Can I use destructive-interference condition to solve Quink's tube numericals?
Yes, but most NEET-style problems on this page are framed through consecutive maxima because the derivation directly yields 2x = lambda. If a question is built on minima, keep the half-integer path difference condition carefully and do not mix maxima and minima equations in the same algebraic step.
Why is source frequency n0 treated as fixed while wavelength changes in calculations?
Frequency is set by the tuning fork or source and does not depend on detector position. In this apparatus, what you infer from geometry is wavelength in air, then speed follows from v = n0 lambda. Students lose marks when they treat frequency as an adjustable variable while processing maxima shifts.
What is the fastest check to catch x versus 2x mistakes in the final answer?
After solving, quickly compute lambda/x from your result. For consecutive maxima in Quink's tube, lambda/x must be 2. If your ratio is 1, you used x as wavelength. This one-step audit is especially useful in timed objective papers where arithmetic may look correct but geometry was applied incorrectly.
Does Quink's tube measure speed directly or indirectly?
It measures speed indirectly. You experimentally obtain displacement between maxima, convert that to wavelength using lambda = 2x, and then compute speed through v = n0 lambda using known source frequency. So speed emerges from interference geometry plus source data, not from direct stopwatch-style travel timing.
If two successive maxima are farther apart than expected, what experimental factor should I suspect first?
Suspect reading error in tube displacement or wrong identification of truly consecutive maxima. Jumping over one maximum doubles the measured shift and distorts wavelength. In exam-style data, this appears as unrealistic speeds. Always verify whether the stated shift is between adjacent maxima before substituting into formulas.
How should NRI students prepare this topic for NEET if school labs cover it only qualitatively?
Convert each lab observation into equations and solve a short set of timed numericals. Memorize the three core links: constructive condition, geometric update l2' = l2 + 2x, and v = 2n0x. This converts a descriptive lab demonstration into an objective-question skill, which is the NEET requirement for this topic.
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