Ultrasonic Testing (UT) Level II Practice Questions

30 original questions in 6 topics, each with the correct answer, an explanation and its basis. Last reviewed October 16, 2026.

About this UT Level II practice test

These 30 multiple-choice questions are original study questions written by Atlantis NDT for ultrasonic technicians preparing for a Level II examination. They are not ASNT exam questions. They cover the work a Level II is expected to do: set up and calibrate the equipment, carry out straight-beam and angle-beam examinations, interpret and evaluate indications against a procedure, and report the results.

Under ASNT SNT-TC-1A (2024 edition), the employer certifies Level II technicians after general, specific and practical examinations set out in its written practice. ASNT also runs central ASNT NDT Level II examinations (general and method exams); where those are used, the employer still carries out the visual acuity, practical and job-specific examinations. Use this set to find weak areas before either route.

Atlantis NDT runs NDT training year-round, with theory online and hands-on practice on its Practical NDT simulator or in person at the client's own facility. Simulator practice supports training; it does not replace the practical examination or the experience hours the employer's written practice requires.

How to use this practice test

  • Answer each question before opening the explanation, and write down your choice.
  • Work the calculations on paper with units; most errors come from mixing millimeters, meters and microseconds.
  • Note the topics where you miss more than one question and study those first.
  • Re-test after a few days of study and compare your results by topic, not just overall.

Topics in this set:

Practice questions, not exam questions

Atlantis NDT wrote these questions as original study material. They are not taken from ASNT or from any employer's examination, and real examinations differ in wording, emphasis and format.

Certification is a separate step: under ASNT SNT-TC-1A the employer certifies its technicians after the examinations its written practice sets, and ASNT runs central Level II examinations for several methods. A good score here is a study signal, not a qualification.

Principles and physics

Question 1. What is the approximate wavelength of a 5 MHz longitudinal wave in steel (velocity 5,900 m/s)?

  1. A. 0.59 mm
  2. B. 1.18 mm
  3. C. 2.36 mm
  4. D. 11.8 mm
Show answer

Correct answer: B. 1.18 mm

λ = v/f = 5,900 m/s ÷ 5,000,000 Hz = 0.00118 m, or about 1.18 mm. 2.36 mm would be the wavelength at 2.5 MHz.

Basis: General UT principle (λ = v/f)

Question 2. Using N = D²·f/(4v), what is the approximate near-field length of a 10 mm diameter, 5 MHz straight-beam probe in steel (longitudinal velocity 5,900 m/s)?

  1. A. 42.4 mm
  2. B. 10.6 mm
  3. C. 84.7 mm
  4. D. 21.2 mm
Show answer

Correct answer: D. 21.2 mm

N = (10 mm)² × 5,000,000 Hz ÷ (4 × 5,900,000 mm/s) = 5×10⁸ ÷ 2.36×10⁷ ≈ 21.2 mm. Doubling the diameter would quadruple N, not double it.

Basis: General UT principle (near-field approximation)

Question 3. What is the main reason a liquid or gel couplant is used between a contact transducer and a steel part?

  1. A. It displaces air, whose very low acoustic impedance would reflect almost all the sound
  2. B. It raises the frequency of the transmitted pulse so that smaller flaws can be resolved
  3. C. It converts the longitudinal wave into a shear wave as the sound enters the steel
  4. D. It shortens the near field of the transducer
Show answer

Correct answer: A. It displaces air, whose very low acoustic impedance would reflect almost all the sound

Acoustic impedance (density × velocity) of air is several orders of magnitude lower than that of steel, so an air gap gives almost total reflection. Couplant displaces the air; it does not change the frequency or the wave mode.

Basis: General UT principle (acoustic impedance)

Question 4. An acrylic wedge (longitudinal velocity 2,730 m/s) is to produce a 45° refracted shear wave in steel (shear velocity 3,230 m/s). What incident angle is required?

  1. A. 45.0°
  2. B. 57.7°
  3. C. 36.7°
  4. D. 22.1°
Show answer

Correct answer: C. 36.7°

Snell's law: sin θi = (2,730 ÷ 3,230) × sin 45° = 0.845 × 0.707 ≈ 0.598, so θi ≈ 36.7°. 57.7° is the second critical angle for this wedge and steel combination.

Basis: Snell's law

Question 5. For an acrylic wedge on steel, what happens when the incident angle exceeds the second critical angle?

  1. A. Only a refracted longitudinal wave exists in the steel
  2. B. Both longitudinal and shear waves are refracted into the steel
  3. C. Neither a longitudinal nor a shear bulk wave is refracted into the steel
  4. D. The refracted shear wave travels at exactly 45°
Show answer

Correct answer: C. Neither a longitudinal nor a shear bulk wave is refracted into the steel

Beyond the first critical angle the refracted longitudinal wave is lost; beyond the second critical angle the refracted shear wave is lost too (near the second critical angle a surface wave is generated). Shear-wave probes therefore use incident angles between the two critical angles.

Basis: Snell's law; critical angles

Equipment, probes and calibration

Question 6. On an IIW-type calibration block (ISO 2400 block No. 1), what is the 100 mm radius mainly used for with an angle-beam probe?

  1. A. Checking the near-surface resolution of a straight-beam probe
  2. B. Measuring the refracted angle of the beam in steel
  3. C. Measuring the beam spread and focal distance of a dual-element probe
  4. D. Locating the probe index point and calibrating the sound-path range
Show answer

Correct answer: D. Locating the probe index point and calibrating the sound-path range

Maximizing the echo from the 100 mm radius places the probe index point at the center of the arc, and the radius echo and its repeat set the time base in sound path. The refracted angle is checked separately, using the holes and angle scales on the block.

Basis: ISO 2400 calibration block No. 1

Question 7. Compared with a lightly damped transducer of the same frequency, what does a heavily damped (broadband) transducer generally give?

  1. A. Better depth resolution but lower sensitivity
  2. B. Poorer depth resolution but higher sensitivity
  3. C. A longer pulse containing more cycles
  4. D. A narrower bandwidth around the center frequency
Show answer

Correct answer: A. Better depth resolution but lower sensitivity

Heavy backing damping shortens the pulse to fewer cycles, which broadens the bandwidth and improves the separation of closely spaced reflectors in depth, at the cost of some sensitivity.

Basis: General UT principle (transducer damping)

Question 8. What does a screen height (vertical) linearity check confirm?

  1. A. That the time base is accurate across the full screen width
  2. B. That two echoes keep their amplitude ratio as the gain is changed
  3. C. That the probe frequency matches its marked value
  4. D. That the pulse repetition rate is low enough to avoid ghost echoes
Show answer

Correct answer: B. That two echoes keep their amplitude ratio as the gain is changed

Screen height linearity is checked by setting two echoes at a known amplitude ratio (commonly 2:1) and confirming the ratio holds as the gain is varied, so displayed amplitude is proportional to the received signal. Time-base accuracy is horizontal linearity.

Basis: ASME BPVC Section V, Article 4 (instrument linearity checks)

Question 9. How is a distance-amplitude correction (DAC) curve normally constructed for angle-beam weld examination?

  1. A. By plotting the back-wall echo heights from a set of plates of different thickness
  2. B. By plotting the noise level at several gain settings across the screen
  3. C. By plotting peak echoes from same-size reference reflectors at several sound paths
  4. D. By plotting echoes from natural flaws in a production weld of the same material
Show answer

Correct answer: C. By plotting peak echoes from same-size reference reflectors at several sound paths

A DAC joins the maximized responses from same-size reference reflectors, such as side-drilled holes, at increasing sound paths in a block of material similar to the part. It shows how the response from a given reflector falls with distance in that material, and indications are judged against it.

Basis: General UT practice; ASME BPVC Section V, Article 4

Question 10. An echo peaks at 20% full screen height (FSH). Approximately how much gain must be added to bring it to 80% FSH on a linear display?

  1. A. 6 dB
  2. B. 12 dB
  3. C. 18 dB
  4. D. 24 dB
Show answer

Correct answer: B. 12 dB

The amplitude ratio is 80 ÷ 20 = 4, and 20·log₁₀(4) ≈ 12 dB. Each 6 dB doubles the amplitude: 20% to 40% to 80%.

Basis: General UT principle (decibel ratios)

Question 11. Why do many UT procedures require the reject (suppression) control to be off during calibration and evaluation?

  1. A. It increases the pulse repetition frequency, which can cause ghost echoes
  2. B. It changes the refracted angle of the probe in the test material
  3. C. It reduces the near-field length of the transducer
  4. D. It can hide low-level signals and make the amplitude display non-linear
Show answer

Correct answer: D. It can hide low-level signals and make the amplitude display non-linear

Reject suppresses everything below a threshold and, on many instruments, makes the amplitude response non-linear, so low-level flaw echoes can be hidden and comparisons with a DAC become unreliable.

Basis: General UT practice

Angle-beam technique and scanning

Question 12. What is the full skip distance for a 60° shear-wave probe on a 20 mm thick plate?

  1. A. 34.6 mm
  2. B. 40.0 mm
  3. C. 69.3 mm
  4. D. 80.0 mm
Show answer

Correct answer: C. 69.3 mm

Full skip = 2 × T × tan θ = 2 × 20 × tan 60° = 40 × 1.732 ≈ 69.3 mm. 34.6 mm is the half skip.

Basis: Angle-beam geometry

Question 13. A 45° probe on a 25 mm plate gives an echo at a sound path of 50 mm. What is the depth of the reflector below the scanning surface?

  1. A. 14.6 mm
  2. B. 25.0 mm
  3. C. 35.4 mm
  4. D. 10.4 mm
Show answer

Correct answer: A. 14.6 mm

Apparent depth = 50 × cos 45° ≈ 35.4 mm, which exceeds the thickness, so the reflector is on the second leg after a bounce off the back wall. True depth = 2T − 35.4 = 50 − 35.4 ≈ 14.6 mm. The surface distance is 50 × sin 45° ≈ 35.4 mm.

Basis: Angle-beam geometry

Question 14. To detect transverse cracks in a butt weld, in which direction should the angle beam be directed?

  1. A. Perpendicular to the weld axis from both sides
  2. B. Essentially parallel to the weld axis, on the weld or alongside it
  3. C. Straight down at 0° through the weld cap only
  4. D. Toward the heat-affected zone using a surface-wave probe only
Show answer

Correct answer: B. Essentially parallel to the weld axis, on the weld or alongside it

A transverse crack lies across the weld, so the beam must travel roughly along the weld axis to strike its face. Beams directed perpendicular to the weld axis are used for longitudinal flaws.

Basis: General UT practice; ASME BPVC Section V, Article 4

Question 15. Why do UT procedures require each scan pass to overlap the previous pass?

  1. A. To reduce couplant consumption
  2. B. To allow the time base to be checked and recalibrated during scanning
  3. C. To increase the effective refracted angle at the weld fusion faces
  4. D. To avoid gaps in coverage, as the beam is less sensitive at its edges
Show answer

Correct answer: D. To avoid gaps in coverage, as the beam is less sensitive at its edges

The beam is not uniformly sensitive across its width, so passes are indexed with a specified overlap of the transducer dimension to avoid unexamined strips between passes.

Basis: ASME BPVC Section V, Article 4 (scanning)

Question 16. Why is manual scanning usually performed at a gain higher than the reference level, with evaluation then carried out at reference sensitivity?

  1. A. To make the near field shorter
  2. B. Because the reference level is used only for thickness measurement, not flaw detection
  3. C. So that weak or poorly oriented reflectors are less likely to be missed
  4. D. To calibrate the time base more accurately on the component
Show answer

Correct answer: C. So that weak or poorly oriented reflectors are less likely to be missed

Extra scanning gain helps detect weak or poorly oriented reflectors during the search. Once an indication is found, its amplitude is evaluated at the reference sensitivity the acceptance criteria are based on.

Basis: General UT practice; ASME BPVC Section V, Article 4

Question 17. What is a transfer correction (transfer loss measurement) used to compensate for?

  1. A. Differences in surface condition, coupling and attenuation between block and part
  2. B. Drift in the instrument's time base during a long shift
  3. C. The wedge delay of a dual-element probe
  4. D. The difference between longitudinal and shear velocities in the component material
Show answer

Correct answer: A. Differences in surface condition, coupling and attenuation between block and part

Comparing the same type of reference signal on the block and on the part shows how much extra loss the part introduces. The gain is adjusted by that amount so the DAC remains valid on the component.

Basis: General UT practice

Interpretation and evaluation

Question 18. During straight-beam testing of a 30 mm steel plate, the back-wall echo disappears over an area and a strong echo appears at about 15 mm depth with its own repeats. What is the most likely cause?

  1. A. Loss of coupling under the probe
  2. B. A transverse crack open to the scanning surface
  3. C. Scattered fine porosity through the thickness
  4. D. A lamination lying parallel to the surface
Show answer

Correct answer: D. A lamination lying parallel to the surface

A planar reflector parallel to the surface blocks the sound from reaching the back wall and gives a strong echo with multiple repeats. Loss of coupling would remove all echoes, including the mid-wall one.

Basis: General UT interpretation (straight-beam plate testing)

Question 19. Which angle-beam echo behavior is most typical of a planar crack?

  1. A. An echo that looks the same, and has similar amplitude, from both sides of the weld and from every direction
  2. B. An echo at a fixed position on the time base that does not move when the probe is moved
  3. C. Amplitude that varies strongly with beam direction, with a ragged, multi-peaked echo
  4. D. A cluster of many small, low-amplitude echoes that appear and disappear independently
Show answer

Correct answer: C. Amplitude that varies strongly with beam direction, with a ragged, multi-peaked echo

A crack is planar and often faceted, so its response is strongly directional with a ragged echo dynamic. A rounded reflector responds similarly from all directions, a cluster of small echoes suggests porosity, and an echo that never moves comes from the probe or wedge, not the part.

Basis: General UT interpretation (echo dynamics)

Question 20. What is the most reliable way to decide whether an echo from the weld root area is a geometric indication from the root bead rather than a flaw?

  1. A. Raise the gain until the echo saturates, then judge it by its width on the screen
  2. B. Plot its position on the measured weld profile and confirm it from the opposite side
  3. C. Accept it as geometric if its peak stays below 100% of full screen height
  4. D. Change to a higher-frequency probe and treat it as geometric if the echo fades
Show answer

Correct answer: B. Plot its position on the measured weld profile and confirm it from the opposite side

Geometric echoes are identified by plotting. A reflector that plots on the root bead or counterbore position, and behaves consistently from the opposite side, is geometric. Amplitude or a change of frequency cannot prove where an echo comes from.

Basis: General UT interpretation (plotting indications)

Question 21. In the 6 dB drop method for sizing a reflector that is larger than the beam, the reflector edge is taken where the echo amplitude does what?

  1. A. Falls to half of its maximum value
  2. B. Falls to 10% FSH
  3. C. Falls by 20 dB from its maximum
  4. D. Rises above the DAC curve
Show answer

Correct answer: A. Falls to half of its maximum value

A 6 dB drop halves the amplitude, which corresponds to the beam axis being on the reflector edge. The method is only valid for reflectors larger than the beam; for smaller reflectors it measures the beam width rather than the flaw.

Basis: General UT sizing practice

Question 22. Why is echo amplitude alone an unreliable measure of the size of a planar flaw?

  1. A. Because amplitude does not change with distance from the probe
  2. B. Because amplitude also depends on orientation, roughness and position in the beam
  3. C. Because larger flaws always give smaller echoes
  4. D. Because echo amplitude can only be measured accurately with a dual-element contact probe
Show answer

Correct answer: B. Because amplitude also depends on orientation, roughness and position in the beam

A large crack tilted away from the beam, or with a rough face, can give a lower echo than a small, well-oriented reflector. This is why length is sized by probe movement and height is often sized by tip-diffraction techniques rather than by amplitude.

Basis: General UT sizing practice

Codes, procedures and documentation

Question 23. During a periodic calibration check, the reference sensitivity is found to have dropped by more than the procedure allows. What is normally required?

  1. A. Increase the gain by the amount lost and carry on scanning from the current position
  2. B. Record the change in the report and take no further action
  3. C. Average the old and new sensitivities for all indications
  4. D. Correct the calibration and re-examine everything done since the last valid check
Show answer

Correct answer: D. Correct the calibration and re-examine everything done since the last valid check

If sensitivity fell beyond the tolerance, work done since the last good check may have missed indications, so the calibration is corrected and that work is repeated. ASME Section V Article 4 contains this type of requirement; the procedure sets the tolerance.

Basis: ASME BPVC Section V, Article 4 (calibration confirmation)

Question 24. Why do procedures require the calibration block to be close in temperature to the component being examined?

  1. A. Because velocity and attenuation, especially in the wedge, change with temperature
  2. B. Because the block expands enough to change its nominal dimensions and hole positions significantly
  3. C. Because couplant only works at room temperature, whatever its type
  4. D. Because the probe frequency doubles when the temperature rises by a few degrees
Show answer

Correct answer: A. Because velocity and attenuation, especially in the wedge, change with temperature

Wedge velocity is quite temperature-sensitive, so a temperature difference shifts the refracted angle, the timing and the attenuation. ASME Section V Article 4 limits the block-to-component temperature difference for this reason.

Basis: ASME BPVC Section V, Article 4 (calibration block temperature)

Question 25. Under ASNT SNT-TC-1A, which task is normally reserved for a Level III rather than a Level II?

  1. A. Setting up and calibrating the equipment
  2. B. Interpreting and evaluating results against the applicable code
  3. C. Approving NDT procedures and establishing techniques
  4. D. Organizing and reporting examination results
Show answer

Correct answer: C. Approving NDT procedures and establishing techniques

SNT-TC-1A describes Level II personnel as able to set up and calibrate equipment, interpret and evaluate results against codes, and organize and report results. Developing, qualifying and approving procedures and establishing techniques are Level III responsibilities.

Basis: ASNT SNT-TC-1A (levels of qualification)

Question 26. Which set of information should a UT report contain so that the examination can be repeated?

  1. A. The accept or reject decision, the date and the technician's signature
  2. B. Equipment, search units, couplant, calibration block, reference level, procedure and coverage
  3. C. The part number, the drawing revision and the technician's signature
  4. D. Screen images of every rejected indication, with the decision for each
Show answer

Correct answer: B. Equipment, search units, couplant, calibration block, reference level, procedure and coverage

Another qualified technician must be able to reproduce the set-up and coverage from the report. The decision, signature and images are needed too, but on their own they do not allow the examination to be repeated.

Basis: General NDT reporting practice; ASME BPVC Section V, Article 4

Question 27. For an ASME Section V Article 4 angle-beam examination of a plate (non-piping) weld, which reference reflectors in the basic calibration block are used to build the DAC?

  1. A. Side-drilled holes
  2. B. Flat-bottom holes drilled from the back face
  3. C. Spherical voids
  4. D. Natural cracks from a previous weld
Show answer

Correct answer: A. Side-drilled holes

The non-piping basic calibration block in Article 4 uses side-drilled holes at known depths. Piping calibration blocks use notches, and flat-bottom holes are typical of straight-beam reference blocks.

Basis: ASME BPVC Section V, Article 4 (calibration blocks)

Method limitations

Question 28. An austenitic stainless steel weld gives high background noise and poor penetration with a 5 MHz shear-wave probe. Which change is most likely to improve the examination?

  1. A. Increase the frequency to 10 MHz to obtain a sharper and better-resolved beam
  2. B. Use a lower frequency, and consider refracted longitudinal-wave dual probes
  3. C. Increase the pulse repetition frequency to average out the noise
  4. D. Use a smaller crystal at the same frequency and the same shear-wave angle
Show answer

Correct answer: B. Use a lower frequency, and consider refracted longitudinal-wave dual probes

Coarse, columnar austenitic grains scatter and attenuate sound, more so at higher frequencies and for shear waves. Lower frequency and refracted longitudinal waves reduce scatter and beam skewing.

Basis: General UT principle (attenuation in coarse-grained materials)

Question 29. With a single-element contact straight-beam probe, why are reflectors very close to the scanning surface hard to detect?

  1. A. The beam diverges too quickly in the first few millimeters below the probe
  2. B. Shear waves are generated near the surface and swamp the longitudinal echoes
  3. C. The back-wall echo arrives first
  4. D. The initial pulse and ring-down mask echoes from the first few millimeters
Show answer

Correct answer: D. The initial pulse and ring-down mask echoes from the first few millimeters

While the transducer is still ringing after excitation it cannot display separate echoes, so the first few millimeters below the surface form a dead zone. Dual-element or delay-line probes move the interface signal away from the near-surface region and reduce this problem.

Basis: General UT principle (dead zone)

Question 30. Which flaw is most likely to be missed by a straight-beam (0°) examination from the plate surface?

  1. A. A lamination in the parent plate lying parallel to the surface
  2. B. A slag inclusion lying flat and parallel to the plate surface
  3. C. A large, rounded gas pore near the middle of the plate thickness
  4. D. A vertical planar flaw, such as sidewall lack of fusion in a square-edge joint
Show answer

Correct answer: D. A vertical planar flaw, such as sidewall lack of fusion in a square-edge joint

A 0° beam travels parallel to a vertical planar flaw, so almost no energy is reflected back to the probe. Angle beams are used to strike such flaws at a favorable angle.

Basis: General UT principle (flaw orientation)

Practise the geometry behind the questions

Many Level II questions are angle-beam and time-of-flight arithmetic. Check your working with the UT angle beam calculator and the TOFD calculator, see the principle on the simplified A-scan demo, and read how training, exams and certification fit together.

Next step: UT Level 2 course

Practice questions check recall. Structured training adds the classroom and hands-on hours your certification route requires. The UT Level 2 course covers angle-beam weld examination on flawed samples, calibration and reference sensitivity, and evaluating indications against ASME, AWS and API acceptance criteria. Completing a course is not certification: under an employer-based scheme such as SNT-TC-1A, your employer certifies you after its own examinations.

For employers: ASNT Level III consulting covers procedures, the written practice, qualification examinations and audit support for your technicians.

Where to go next

Atlantis NDT courses run year-round, with theory online and hands-on practice on the Practical NDT simulator or in person at the client's own facility. From here:

Enquire about UT Level II training for yourself or for a team. Tell us the method, the level and where the trainees are based.

Practise before the exam

Written papers test what you know; employers and practical examinations test what you can do with a probe in your hand. Use the free practice tests to find your gaps, then practise the technique.

Free downloads

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  • ExcelUT weld inspection reportExcel workbook, written as Atlantis Engineering Consultants with no logo, so you can add your own.Download .xlsxAbout this template
  • ExcelUT thickness survey reportExcel workbook, written as Atlantis Engineering Consultants with no logo, so you can add your own.Download .xlsxAbout this template
  • PDFUT Level II General practice questions40 questions. PDF with an answer key and explanations.Download PDF
  • PDFUT Level II Specific practice questions20 questions. PDF with an answer key and explanations.Download PDF

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