100+ Free PCN Infrared Thermography Category 2 Practice Questions

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Sample PCN Infrared Thermography Category 2 Practice Questions

Try these sample questions to test your PCN Infrared Thermography Category 2 exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1According to the Stefan-Boltzmann law, if the absolute temperature of a blackbody emitter increases from 300 K (26.85°C) to 600 K (326.85°C), by what factor does total radiant emittance increase?

A.2 times

B.4 times

C.8 times

D.16 times

Explanation: The Stefan-Boltzmann law states that radiant emittance E is directly proportional to absolute temperature raised to the fourth power (E = sigma * T^4). Doubling the absolute temperature from 300 K to 600 K increases the total radiant emittance by 2^4 = 16 times.

2Using Wien's Displacement Law (lambda_max * T = 2898 um*K), at what peak wavelength does an object at 500 K (226.85°C) emit its maximum spectral radiance?

A.2.898 um

B.5.796 um

C.9.660 um

D.14.49 um

Explanation: Wien's Displacement Law calculates peak spectral emission wavelength as lambda_max = 2898 um*K / T. For T = 500 K, lambda_max = 2898 / 500 = 5.796 um, which lies in the mid-wavelength infrared (MWIR) spectrum.

3For an opaque body (transmissivity tau = 0) in thermal equilibrium, what is the mathematical relationship between emissivity (epsilon) and reflectivity (rho) under Kirchhoff's Law?

A.epsilon + rho = 1

B.epsilon * rho = 1

C.epsilon - rho = 0

D.epsilon / rho = 1

Explanation: Kirchhoff's Law states that absorbivity equals emissivity (alpha = epsilon) for an object in thermal equilibrium. Conservation of radiant energy dictates alpha + rho + tau = 1. Since tau = 0 for an opaque body, epsilon + rho = 1, meaning high reflectivity directly implies low emissivity.

4Heat conduction through a flat wall is described by Fourier's Law (Q = k * A * delta_T / d). If wall thickness d is halved while surface temperature difference delta_T remains constant, how does conductive heat flux change?

A.Heat flux is halved

B.Heat flux remains unchanged

C.Heat flux doubles

D.Heat flux quadruples

Explanation: Fourier's Law shows conductive heat flow Q is inversely proportional to thickness d (Q/A = k * delta_T / d). Halving the wall thickness doubles the thermal gradient, thereby doubling conductive heat flux.

5During an outdoor electrical thermographic inspection, wind speed increases from 1 m/s (breeze) to 5 m/s. What effect does this have on observed electrical anomaly surface temperatures?

A.Apparent surface temperature of the hotspot decreases significantly due to forced convection

B.Apparent surface temperature of the hotspot increases due to friction heating

C.Surface temperature remains unchanged because internal resistance heating is constant

D.Hotspot emissivity increases, masking true temperature

Explanation: Higher wind speeds increase forced convection heat transfer coefficients (h_c), rapidly stripping heat from component surfaces. An electrical connection carrying constant current will display a much lower surface temperature rise under strong wind, potentially masking a critical fault.

6What is the primary characteristic of an ideal blackbody radiator in infrared thermography?

A.It absorbs 100% of incident radiation and emits maximum possible thermal energy at any given wavelength and temperature (epsilon = 1.0)

B.It reflects 100% of incident radiation and has zero thermal emission (epsilon = 0.0)

C.It transmits all infrared radiation without surface absorption or reflection (tau = 1.0)

D.It emits infrared energy only at a single discrete laser wavelength

Explanation: A blackbody is a theoretical ideal radiator with an emissivity (epsilon) and absorbivity (alpha) of 1.0 across all wavelengths. It reflects nothing (rho = 0) and transmits nothing (tau = 0), serving as the standard reference for thermal calibration.

7Thermal diffusivity (alpha = k / (rho * c_p)) measures a material's ability to conduct thermal energy relative to its volumetric heat capacity. Which material exhibits the highest thermal diffusivity?

A.Copper

B.Concrete

C.Polyethylene plastic

D.Water

Explanation: Copper has extremely high thermal conductivity (k ~ 400 W/m*K) and moderate heat capacity, giving it a thermal diffusivity of ~1.1 x 10^-4 m^2/s, enabling rapid thermal wave propagation compared to insulators like plastic or concrete.

8How does a 'graybody' differ from an ideal blackbody radiator?

A.A graybody has an emissivity less than 1.0 that is constant across all wavelengths

B.A graybody emits radiation only when illuminated by external visible light

C.A graybody has an emissivity that varies wildly with wavelength and angle

D.A graybody transmits all longwave infrared radiation while absorbing midwave radiation

Explanation: By definition, a graybody has a constant spectral emissivity (epsilon < 1.0) across the wavelength band of interest. Most industrial surfaces (painted metal, brick, wood) approximate graybody behavior over 8–14 um.

9What fundamental relationship is described by Planck's Law of Radiation?

A.The spectral distribution of radiant energy emitted by a blackbody as a function of absolute temperature and wavelength

B.The linear relationship between thermal expansion coefficient and component temperature

C.The rate of convective cooling under forced airflow conditions

D.The total conductive heat transfer across composite insulation materials

Explanation: Planck's Law gives the mathematical formula for spectral radiance emitted by a blackbody at temperature T across wavelength lambda. Integrating Planck's law over all wavelengths yields the Stefan-Boltzmann law, while differentiating it yields Wien's displacement law.

10When measuring temperature on a non-metallic surface with high emissivity, at what viewing angle (relative to the normal/perpendicular axis) does emissivity begin to drop rapidly?

A.Beyond 15 degrees from normal

B.Beyond 30 degrees from normal

C.Beyond 60 degrees from normal

D.Viewing angle has zero effect on emissivity for non-metals

Explanation: For non-metallic materials, directional emissivity remains relatively constant from 0 degrees (normal) up to about 45–60 degrees. Beyond 60 degrees from perpendicular, surface reflectivity increases rapidly and directional emissivity drops sharply towards zero.

About the PCN Infrared Thermography Category 2 Exam

PCN Infrared Thermography Category 2 is BINDT's ISO 18436-7 certification for thermographers who perform independent thermographic inspections, select measurement techniques, compensate for environmental and surface emissivity variables, analyze thermal anomalies, grade fault severity, and supervise Category 1 personnel.

Assessment

Multiple-choice examination aligned to ISO 18436-7 and BINDT CM/GEN Appendix B. Candidates are evaluated on heat transfer physics, camera radiometry, reflected temperature compensation, thermographic anomaly diagnosis, severity classification, and professional reporting.

Time Limit

2.0 hours

Passing Score

75%

Exam Fee

PSL/35-CM Issue 26 (2026): Category 1/2 English examination fee is £236 ex VAT (£283.20 inc VAT), inclusive of the PCN admin charge. ATO training course packages vary by provider. (British Institute of Non-Destructive Testing (BINDT) — PCN Scheme)

PCN Infrared Thermography Category 2 Exam Content Outline

20%

Heat Transfer & Thermography Principles

Conduction, convection, radiation physics, Stefan-Boltzmann law, Wien's displacement law, emissivity, reflectivity, transmissivity, and atmospheric attenuation.

20%

Infrared Equipment & Radiometry

Thermal imagers, detectors (cooled vs uncooled microbolometers), NETD, IFOV, MFOV, dynamic range, temperature measurement range, optics, and spectral bands.

15%

Data Acquisition & Measurement Techniques

Measurement procedures, background temperature compensation, reflected apparent temperature, distance/emissivity corrections, surface preparation, and environmental influences.

20%

Image Post-Processing & Diagnostic Analysis

Thermal anomaly detection, delta-T severity assessment, baseline comparison, image subtraction, statistical trending, and electrical/mechanical/building diagnosis.

10%

Severity Criteria & Condition Monitoring

Absolute temperature thresholds, relative/delta-T grading standards (NETA/NFPA/ISO), severity classification (minor, intermediate, critical), and PdM integration.

15%

Reporting, Safety & Standards

Report generation, corrective action recommendations, Category 1 supervision, electrical arc flash/thermal safety, ISO 18436-7 / BINDT CM/GEN compliance, and CP16-CM experience logging.

How to Pass the PCN Infrared Thermography Category 2 Exam

What You Need to Know

Passing score: 75%
Assessment: Multiple-choice examination aligned to ISO 18436-7 and BINDT CM/GEN Appendix B. Candidates are evaluated on heat transfer physics, camera radiometry, reflected temperature compensation, thermographic anomaly diagnosis, severity classification, and professional reporting.
Time limit: 2.0 hours
Exam fee: PSL/35-CM Issue 26 (2026): Category 1/2 English examination fee is £236 ex VAT (£283.20 inc VAT), inclusive of the PCN admin charge. ATO training course packages vary by provider.

Keys to Passing

Complete 500+ practice questions
Score 80%+ consistently before scheduling
Focus on highest-weighted sections
Use our AI tutor for tough concepts

PCN Infrared Thermography Category 2 Study Tips from Top Performers

1Master reflected apparent temperature (T_refl) compensation and emissivity measurement techniques (e.g. electrical tape / paint calibration method) — radiometry questions frequently test quantitative temperature corrections.

2Memorize Wien's Displacement Law (lambda_max * T = 2898 um*K) and Stefan-Boltzmann Law (E = epsilon * sigma * T^4 in Kelvin) for energy flux calculations.

3Practice calculating Measurement Field of View (MFOV = 3 * IFOV) to determine minimum target sizes at specific inspection distances.

4Study NETA / NFPA 70B / ISO delta-T severity classification matrices for electrical and mechanical component heating above reference baseline.

Frequently Asked Questions

What is BINDT PCN Infrared Thermography Category 2?

It is BINDT's PCN certification for ISO 18436-7 Category 2 thermographers. Certificated Category 2 thermographers select appropriate thermographic hardware and measurement techniques, perform quantitative thermal analysis, correct for emissivity and reflected background temperatures, evaluate fault severity, write comprehensive technical reports, and provide guidance to Category 1 personnel.

How many questions and how long is the official Cat 2 exam?

BINDT CM/GEN Appendix B specifies a multiple-choice examination totaling 60 questions with a 2.0-hour time limit (split into Part A General Theory 30 MCQs and Part B Sector Practical Application paper 30 MCQs). A minimum grade of 75% is required to pass.

What training and experience are required for Category 2 certification?

Candidates must hold Category 1 Thermography certification, complete 32 hours of approved Category 2 training (BINDT CM/GEN Appendix B Table 1), and document 24 months of cumulative verified thermography work experience logged on BINDT form CP16-CM.

How much does the PCN Category 2 examination cost?

Per BINDT PSL/35-CM (Issue 26, 2026), the initial examination fee for Category 1/2 English papers is £236 ex VAT (£283.20 inc VAT), which includes the PCN administrative fee. ATO course package fees are separate.

How does this practice bank prepare candidates for the Category 2 exam?

This bank provides 100 realistic multiple-choice questions with worked radiometry calculations, heat transfer physics, thermal image interpretation scenarios, delta-T severity grading, and BINDT CM/GEN procedural compliance questions.