Free FE Other Disciplines Practice Test: 55 Questions with Explanations
Work through 55 free FE Other Disciplines practice questions covering all 14 knowledge areas in the current NCEES specification, with a worked explanation under every question. This original, unofficial set is shorter than the real exam and does not predict a pass.
Practice questions
1. Mathematics (4 questions)
Question 1 of 55 · Mathematics · E. Single-variable calculus · Choose one answer
A manufacturer's average cost is C(q) = 0.04q + 900/q + 18 dollars per unit, where q is the batch size. Treat q > 0 as continuous. Which batch size minimizes average cost?
- A. 75 units
- B. 150 units
- C. 300 units
- D. 22,500 units
Answer and explanation
Answer: B. 150 units. Set the derivative to zero: C′(q) = 0.04 − 900/q² = 0, so q² = 22,500 and q = 150. The second derivative, C″(q) = 1,800/q³, is positive for every q > 0, so this is the minimum. Check: C(150) = $30 per unit, while C(75) and C(300) are both $33.
Why the other answers miss: D stops at q² and never takes the square root. A and C sit on either side of the minimum, and direct substitution gives the same higher cost for both. The constant 18 shifts the cost up but doesn't move the minimizing batch size.
Source for the principle: OpenStax Calculus Vol. 1, 4.7 Applied Optimization Problems
Go untimed, or give yourself 160 minutes to match the real exam's average pace. Keep an approved calculator and the free FE Reference Handbook beside you.
Question 2 of 55 · Mathematics · C. Numerical methods · Enter a number
Apply exactly one Newton–Raphson iteration to f(x) = x³ − 4x − 7, starting at x₀ = 3. What is x₁?
Enter your answer, rounded to three decimal places.
Answer and explanation
Answer: 2.652. Newton's update is x₁ = x₀ − f(x₀)/f′(x₀). Here f(3) = 27 − 12 − 7 = 8 and f′(x) = 3x² − 4, so f′(3) = 23. Then x₁ = 3 − 8/23 = 2.65217…, which rounds to 2.652.
We accept answers from 2.6515 up to, but not including, 2.6525.
Common mistakes: The question asks for one iteration, not the converged root. Writing the derivative of x³ as 3x instead of 3x², or adding the correction instead of subtracting it, gives a different number.
Source for the principle: OpenStax Calculus Vol. 1, 4.9 Newton's Method (Eq. 4.8)
Question 3 of 55 · Mathematics · D. Linear algebra · Choose one answer
Solve the system 2x + y − z = 2, x − y + 2z = 7, and 3x + 2y + z = 11. What is x?
- A. 3
- B. 2
- C. 7
- D. 1
Answer and explanation
Answer: B. x = 2. From the first equation, y = 2 − 2x + z. Substituting into the other two gives 3x + z = 9 and −x + 3z = 7. Solving: x = 2, z = 3, and then y = 1. All three original equations check. The coefficient determinant is −10, so this solution is the only one.
Why the other answers miss: D is the value of y, A is the value of z, and C just repeats a right-hand side. Always plug your answer back into every original equation.
Source for the principle: OpenStax College Algebra 2e, 7.6 Solving Systems with Gaussian Elimination
Question 4 of 55 · Mathematics · B. Differential equations · Choose one answer
Which is the general solution of y″ + 4y′ + 13y = 0?
- A. y = C₁e^(−2t) + C₂te^(−2t)
- B. y = e^(2t)(C₁ cos 3t + C₂ sin 3t)
- C. y = e^(−2t)(C₁ cos 3t + C₂ sin 3t)
- D. y = e^(−2t)(C₁ cos √13 t + C₂ sin √13 t)
Answer and explanation
Answer: C. y = e^(−2t)(C₁ cos 3t + C₂ sin 3t). The characteristic equation is r² + 4r + 13 = 0, so r = [−4 ± √(16 − 52)]/2 = −2 ± 3i. Complex roots λ ± μi give y = e^(λt)(C₁ cos μt + C₂ sin μt), with λ = −2 and μ = 3. The negative real part means the oscillation decays.
Why the other answers miss: B flips the sign of the real part, which would make the response grow. A is the form for a repeated real root, which needs b² − 4ac = 0. D uses √13 (the undamped natural frequency) instead of the imaginary part of the root.
Source for the principle: Paul's Online Notes (Lamar University), Differential Equations: Complex Roots
2. Probability and Statistics (3 questions)
Question 5 of 55 · Probability and Statistics · A. Estimation (confidence intervals) · Choose one answer
A random sample of 36 parts from a normally distributed process has a mean length of 20.04 mm. The population standard deviation is known to be 0.12 mm. Using z = 1.96, what is the two-sided 95% confidence interval for the population mean, rounded to 0.001 mm?
- A. 20.033 to 20.047 mm
- B. 19.805 to 20.275 mm
- C. 20.007 to 20.073 mm
- D. 20.001 to 20.079 mm
Answer and explanation
Answer: D. 20.001 to 20.079 mm. The standard error of the mean is σ/√n = 0.12/√36 = 0.0200 mm. The margin is 1.96 × 0.0200 = 0.0392 mm, so the interval is 20.0008 to 20.0792 mm.
Why the other answers miss: B uses the standard deviation of individual parts instead of the standard error. A divides by 36 instead of √36. C uses 1.645, the one-sided 95% (or two-sided 90%) factor. Note what the interval estimates: the process mean, not where 95% of individual parts fall.
Source for the principle: OpenStax Introductory Statistics 2e, 8.1 A Single Population Mean Using the Normal Distribution
Question 6 of 55 · Probability and Statistics · B. Expected value in decision making · Choose one answer
A day's equipment-interruption loss is $0 with probability 0.85, $500 with probability 0.10, or $6,000 with probability 0.05. These outcomes are mutually exclusive and cover every case. What is the expected daily loss?
- A. $350
- B. $300
- C. About $2,167
- D. $50
Answer and explanation
Answer: A. $350. Multiply each outcome by its probability and add: E[L] = 0(0.85) + 500(0.10) + 6,000(0.05) = 0 + 50 + 300 = $350.
Why the other answers miss: B and D each leave out one of the losses. C averages the three dollar amounts as if they were equally likely. The expected value doesn't have to match any single day's loss; it's the long-run average.
Source for the principle: OpenStax Introductory Statistics 2e, 4.2 Mean or Expected Value and Standard Deviation
Question 7 of 55 · Probability and Statistics · D. Goodness of fit · Select the three correct answers
A least-squares straight line fitted to data, including an intercept, has r² = 0.81 and a positive slope. Select the three true statements.
- A. About 81% of the variation in y is explained by the linear relationship with x.
- B. The correlation coefficient r is 0.90.
- C. About 19% of the variation in y is not explained by the line.
- D. 81% of the data points lie on the fitted line.
- E. The fit shows that changes in x cause changes in y.
Answer and explanation
Answer: A, B, and C. r² is the fraction of the variation in y explained by the regression line, so A is true and 1 − r² = 0.19 is unexplained (C). With a positive slope, r = +√0.81 = 0.90 (B).
Where people slip: D confuses explained variation with points landing exactly on the line. E fails because correlation alone doesn't show causation. You need all three correct choices and no wrong one to get this item.
Source for the principle: OpenStax Introductory Statistics 2e, 12.3 The Regression Equation (coefficient of determination)
3. Chemistry (3 questions)
Question 8 of 55 · Chemistry · B. Acids and bases · Choose one answer
Mix 25.0 mL of 0.200 M HCl with 40.0 mL of 0.100 M NaOH. Assume complete dissociation, additive volumes, and ideal dilute-solution behavior at 25 °C. What is the final pH, most nearly?
- A. 1.40
- B. 12.19
- C. 3.00
- D. 1.81
Answer and explanation
Answer: D. 1.81. The acid supplies 0.0250 L × 0.200 M = 0.00500 mol H⁺; the base neutralizes 0.0400 L × 0.100 M = 0.00400 mol. That leaves 0.00100 mol H⁺ in 0.0650 L, so [H⁺] = 0.01538 M and pH = −log(0.01538) = 1.81.
Why the other answers miss: A divides the excess by only the acid's original 25.0 mL. C takes the log of moles instead of concentration. B treats the leftover as excess base. Neutralize moles first, then divide by the total volume.
Source for the principle: OpenStax Chemistry 2e, 14.7 Acid-Base Titrations (Example 14.21)
Question 9 of 55 · Chemistry · C. Chemical reactions (stoichiometry) · Choose one answer
Completely burn 0.800 kmol of propane with oxygen supplied at 20% above the stoichiometric requirement. Use C₃H₈ + 5O₂ → 3CO₂ + 4H₂O and a molecular mass of 32.0 kg/kmol for O₂. What mass of oxygen is supplied?
- A. 76.8 kg
- B. 128.0 kg
- C. 153.6 kg
- D. 192.0 kg
Answer and explanation
Answer: C. 153.6 kg. Stoichiometric O₂ = 0.800 × 5 = 4.00 kmol. With 20% excess, 4.00 × 1.20 = 4.80 kmol. Mass = 4.80 kmol × 32.0 kg/kmol = 153.6 kg.
Why the other answers miss: A uses 16 kg/kmol (atomic oxygen) instead of 32 for O₂. B is the stoichiometric amount with no excess. D is the answer for 1 kmol of propane instead of 0.800 kmol.
Source for the principle: OpenStax Chemistry 2e, 4.3 Reaction Stoichiometry
Question 10 of 55 · Chemistry · A. Oxidation and reduction (corrosion control) · Choose one answer
A steel bracket is galvanized (coated with zinc). The coating gets scratched, exposing a small area of steel, and the part sits in a damp environment. What happens?
- A. The steel becomes the anode and the zinc becomes the cathode.
- B. The exposed steel corrodes faster than bare steel would, because zinc is more noble than iron.
- C. Neither metal corrodes, because the scratch is too small to matter.
- D. The zinc oxidizes preferentially as a sacrificial anode, protecting the exposed steel.
Answer and explanation
Answer: D. The zinc oxidizes preferentially as a sacrificial anode. Zinc has a lower reduction potential than iron, so it's more easily oxidized. In contact with steel in an electrolyte, zinc becomes the anode and corrodes, while the steel acts as the cathode and is protected. That's why galvanizing still protects steel at a scratch.
Why the other answers miss: B gets the reactivity backwards: zinc is more active than iron, not more noble. A reverses the electrode roles. C misses the point of cathodic protection: the small scratch exposes steel to the damp environment, but the zinc protects it by corroding preferentially.
Source for the principle: OpenStax Chemistry 2e, 17.6 Corrosion
4. Instrumentation and Controls (2 questions)
Question 11 of 55 · Instrumentation and Controls · A. Sensors (signal conversion) · Choose one answer
An ideal linear pressure transmitter maps 0 to 240 kPa gauge to 4 to 20 mA. The loop current passes through a 250 Ω resistor, and an ideal voltmeter across that resistor reads 3.50 V. What pressure does the transmitter indicate?
- A. 150 kPa gauge
- B. 210 kPa gauge
- C. 168 kPa gauge
- D. 240 kPa gauge
Answer and explanation
Answer: A. 150 kPa gauge. The loop current is I = V/R = 3.50 V / 250 Ω = 14.0 mA. Subtract the 4 mA zero and divide by the 16 mA span: P = 240 × (14 − 4)/(20 − 4) = 150 kPa gauge.
Why the other answers miss: C uses 14/20 and ignores the 4 mA offset. B divides 14 mA by the 16 mA span without subtracting the offset. D assumes full scale even though the signal is below 20 mA.
Source for the principle: Analog Devices Circuit Note CN0289 (4 mA-to-20 mA loop-powered transmitter); OpenStax University Physics Vol. 2, 10.2 Resistors in Series and Parallel
Question 12 of 55 · Instrumentation and Controls · B. Data acquisition (analog/digital) · Choose one answer
An ideal 12-bit analog-to-digital converter has a 0 to 10 V input range. Using 1 LSB = full-scale range / 2^N, what voltage does one least significant bit represent, most nearly?
- A. 39.1 mV
- B. 2.44 mV
- C. 4.88 mV
- D. 0.833 V
Answer and explanation
Answer: B. 2.44 mV. A 12-bit converter has 2¹² = 4,096 codes. 1 LSB = 10 V / 4,096 = 0.00244 V = 2.44 mV. That's the ideal voltage width of one quantization step.
Why the other answers miss: A is the 8-bit result (10 V / 256). C uses 2¹¹ instead of 2¹². D divides 10 V by 12, treating the bit count as the number of steps.
Source for the principle: Analog Devices Analog Dialogue StudentZone, Analog-to-Digital Conversion (Feb. 2022), 1 LSB = FS/2^N
5. Engineering Ethics and Societal Impacts (3 questions)
Question 13 of 55 · Engineering Ethics and Societal Impacts · A. Codes of ethics · Select the two correct answers
Apply the NSPE Code of Ethics for Engineers. An engineer determines that a set of drawings doesn't conform to applicable standards and endangers life. Management overrules the engineer and plans to proceed. Select the two actions the Code supports.
- A. Refuse to approve the drawings while they don't conform to applicable standards.
- B. Approve the drawings once management accepts the risk in writing.
- C. Notify the employer or client, and any other appropriate authority, of the danger.
- D. Treat confidentiality as forbidding any report outside the company.
- E. Wait until someone is injured before reporting.
Answer and explanation
Answer: A and C. Section II.1.b says engineers approve only documents that conform to applicable standards (A). Section II.1.a says that when an engineer's judgment is overruled under circumstances that endanger life or property, the engineer notifies the employer or client and other appropriate authority (C).
Where people slip: B: management accepting risk doesn't make nonconforming drawings conform. D: the confidentiality duty in II.1.c has exceptions for disclosures authorized or required by law or the Code. E: the duty to notify doesn't wait for an injury. This item applies the NSPE Code; your state board's own rules govern actual practice.
Source for the principle: NSPE Code of Ethics for Engineers, Section II.1.a–c
Question 14 of 55 · Engineering Ethics and Societal Impacts · B. Public protection issues (licensing boards) · Choose one answer
A candidate wants to know two things: whether they're eligible to sit for the FE exam, and how to become certified as an engineer intern (EI) after passing. Who decides these?
- A. NCEES
- B. The licensing board in the jurisdiction where they apply
- C. Pearson
- D. A professional engineering society such as NSPE
Answer and explanation
Answer: B. The licensing board in the jurisdiction where they apply. NCEES's Examinee Guide says eligibility to sit for an NCEES exam is determined by each licensing board, and it directs candidates to their licensing board for EI certification and licensure qualifications. NCEES develops and scores the exam; Pearson delivers it at test centers.
Why the other answers miss: A: NCEES runs the exam but doesn't grant EI certification or licenses. C: Pearson is the test-center provider. D: professional societies publish ethics codes but don't license engineers.
Source for the principle: NCEES Examinee Guide, May 2026, pp. 2 and 24
Question 15 of 55 · Engineering Ethics and Societal Impacts · C. Societal impacts (life-cycle analysis) · Choose one answer
A team compares two pipe materials using only the energy each factory uses to make the pipe. What change would turn this into a cradle-to-grave life-cycle comparison?
- A. Convert the manufacturing energy into dollars.
- B. Add more factories' manufacturing data.
- C. Add raw-material extraction, transport and delivery, the use phase, and disposal or recycling at end of life.
- D. Replace manufacturing energy with use-phase energy only.
Answer and explanation
Answer: C. Add raw-material extraction, transport and delivery, the use phase, and end of life. A product's life cycle begins when raw materials are extracted or harvested, continues through manufacturing and delivery, includes the product's use, and ends with disposal or recycling. A manufacturing-only comparison covers just one of those stages.
Why the other answers miss: B adds detail to the same single stage. A changes the units, not the scope. D swaps one stage for another instead of covering the whole life cycle.
Source for the principle: U.S. EPA, Green Engineering textbook, Ch. 13 Life-Cycle Concepts, Product Stewardship and Green Engineering
6. Safety, Health, and Environment (3 questions)
Question 16 of 55 · Safety, Health, and Environment · A. Industrial hygiene (exposure limits) · Choose one answer
A worker's exposure to one airborne contaminant is 80 ppm for 2 hours, 25 ppm for 4 hours, and 0 ppm for the remaining 2 hours of an 8-hour shift. What is the 8-hour time-weighted average concentration?
- A. 35.0 ppm
- B. 105 ppm
- C. 32.5 ppm
- D. 43.3 ppm
Answer and explanation
Answer: C. 32.5 ppm. OSHA's cumulative-exposure formula weights each concentration by its duration and divides by 8 hours: E = [80(2) + 25(4) + 0(2)] / 8 = 260/8 = 32.5 ppm.
Why the other answers miss: A averages the three readings as if they lasted equally long. D drops the two zero-exposure hours from the 8-hour denominator. B adds concentrations without time weighting. No exposure limit is given here, so this calculation alone doesn't say whether the exposure is acceptable.
Source for the principle: OSHA 29 CFR 1910.1000(d)(1)(i), cumulative exposure formula
Question 17 of 55 · Safety, Health, and Environment · A. Industrial hygiene (radiation, half-life) · Choose one answer
A sealed sample has a parent-nuclide activity of 12.0 MBq. Its half-life is 6.00 hours. What is the parent's activity after 15.0 hours? Count only the parent nuclide.
- A. 4.80 MBq
- B. 2.12 MBq
- C. 3.00 MBq
- D. 1.50 MBq
Answer and explanation
Answer: B. 2.12 MBq. 15.0 h is 15/6 = 2.5 half-lives, so A = 12.0 × (1/2)^2.5 = 2.12 MBq. Fractional half-lives are fine; decay is exponential.
Why the other answers miss: D rounds up to three half-lives and C rounds down to two. A scales activity linearly by 6/15 instead of applying exponential decay.
Source for the principle: OpenStax University Physics Vol. 3, 10.3 Radioactive Decay (Eqs. 10.15, 10.18)
Question 18 of 55 · Safety, Health, and Environment · E. Confined space entry · Put the three tests in order
Before employees enter a permit-required confined space under OSHA's general-industry rule, the atmosphere is tested. Put these tests in the required order, first to last.
- A. Toxic gases and vapors
- B. Oxygen
- C. Combustible gases and vapors
Your order (first to last): ___ → ___ → ___
Answer and explanation
Answer: Oxygen → combustible gases and vapors → toxic gases and vapors (B, C, A). 29 CFR 1910.146(d)(5)(iii) requires testing for oxygen first, then combustible gases and vapors, then toxic gases and vapors. OSHA's Appendix B explains why: most combustible-gas meters depend on oxygen and won't read reliably in an oxygen-deficient atmosphere, and fire or explosion is usually the more immediate threat than toxic exposure.
Where people slip: The usual slip is testing toxic gases first because they feel like the biggest health risk. Related numbers from the same rule: oxygen below 19.5% is oxygen-deficient and above 23.5% is oxygen-enriched, and a flammable gas, vapor, or mist above 10% of its lower flammable limit makes the atmosphere hazardous. You need the whole sequence in order to get this item.
Source for the principle: OSHA 29 CFR 1910.146(b) definitions and (d)(5)(iii); OSHA 1910.146 Appendix B, Procedures for Atmospheric Testing
7. Engineering Economics (3 questions)
Question 19 of 55 · Engineering Economics · A. Time value of money (present worth) · Enter a number
Equipment costs $18,000 now. Its operating cost is $3,200 at the end of each of years 1 through 5, and it returns $5,000 salvage at the end of year 5. At an 8% effective annual interest rate, ignoring taxes and inflation, what is the net present cost?
Enter your answer in dollars, rounded to the nearest whole dollar.
Answer and explanation
Answer: $27,374. Treat costs as positive. P = 18,000 + 3,200 × [1 − (1.08)^−5]/0.08 − 5,000 × (1.08)^−5 ≈ 18,000 + 12,776.672 − 3,402.916 = $27,373.756, which rounds to $27,374.
We accept answers from 27,373.5 up to, but not including, 27,374.5.
Common mistakes: The operating costs are an end-of-year annuity. Salvage is money coming back, so it reduces cost, and it must be discounted five years. Treating the payments as beginning-of-year, adding salvage as a cost, or leaving salvage undiscounted changes the answer.
Source for the principle: OpenStax Principles of Finance 2e, 9.1 Timing of Cash Flows (Eqs. 9.11–9.12)
Question 20 of 55 · Engineering Economics · C. Economic analyses (break-even) · Choose one answer
Process A has an annual fixed cost of $24,000 and a variable cost of $18 per unit. Process B has an annual fixed cost of $60,000 and a variable cost of $12 per unit. Both make acceptable output. At what annual volume are their total costs equal?
- A. 3,000 units/year
- B. 4,000 units/year
- C. 14,000 units/year
- D. 6,000 units/year
Answer and explanation
Answer: D. 6,000 units/year. Set 24,000 + 18q = 60,000 + 12q. Then q = (60,000 − 24,000)/(18 − 12) = 6,000 units. Both totals equal $132,000 there. Below 6,000 units, A costs less; above it, B does.
Why the other answers miss: A divides the fixed-cost difference by B's variable cost alone. B divides A's fixed cost by the variable-cost difference. C adds the two fixed costs before dividing.
Source for the principle: OpenStax Principles of Managerial Accounting, 3.2 Calculate a Break-Even Point in Units and Dollars
Question 21 of 55 · Engineering Economics · E. Project selection (unequal lives) · Choose one answer
Machine A costs $20,000, lasts 5 years, and costs $3,000 per year to operate. Machine B costs $32,000, lasts 10 years, and costs $2,000 per year to operate. Neither has salvage value, and each would be replaced with an identical machine. At an 8% effective annual interest rate, with operating costs paid at each year-end, which has the lower equivalent uniform annual cost (EUAC), and what is it?
- A. Machine A, $7,000 per year
- B. Machine B, $6,769 per year
- C. Machine B, $5,200 per year
- D. Machine A, $8,009 per year
Answer and explanation
Answer: B. Machine B, $6,769 per year. Annualize each first cost with the capital-recovery factor A/P = i(1 + i)ⁿ/[(1 + i)ⁿ − 1]. A: 20,000 × 0.25046 + 3,000 = $8,009 per year. B: 32,000 × 0.14903 + 2,000 = $6,769 per year. Annual costs compare machines with different lives directly, so B wins.
Why the other answers miss: A and C spread the first cost evenly with no interest ($20,000/5 + $3,000 and $32,000/10 + $2,000). D is A's correct EUAC, but it's the higher of the two.
Source for the principle: Penn State EME 460, Compound Interest Formulas III (capital-recovery factor A/P)
8. Statics (5 questions)
Question 22 of 55 · Statics · D. Equilibrium of rigid bodies (support reactions) · Choose one answer
A horizontal beam has a pin support A at x = 0 and a roller support B at x = 12 ft. It carries a downward uniform load of 60 lbf/ft over the full span plus a downward 180 lbf point load at x = 3 ft. What is the upward reaction at B?
- A. 405 lbf
- B. 495 lbf
- C. 360 lbf
- D. 45 lbf
Answer and explanation
Answer: A. 405 lbf. Replace the uniform load with its resultant: 60 × 12 = 720 lbf acting at 6 ft. Take moments about A: R_B(12) = 720(6) + 180(3) = 4,860, so R_B = 405 lbf. Vertical balance gives R_A = 900 − 405 = 495 lbf.
Why the other answers miss: B is the reaction at A. C leaves out the point load. D leaves out the distributed load.
Source for the principle: Engineering Statics (Baker & Haynes), 7.8 Distributed Loads; OpenStax University Physics Vol. 1, 12.1 Conditions for Static Equilibrium
Question 23 of 55 · Statics · B. Force systems (concurrent forces) · Choose one answer
A weightless joint carries a 900 N downward load. Its left cable runs up and to the left at 30° above horizontal; its right cable runs up and to the right at 60° above horizontal. What is the tension in the right cable, most nearly?
- A. 900 N
- B. 450 N
- C. 1,039 N
- D. 779 N
Answer and explanation
Answer: D. 779 N. Horizontal balance: T_R cos 60° = T_L cos 30°, so T_R = √3 T_L. Vertical balance: T_L sin 30° + T_R sin 60° = 900 N. Substituting gives T_L = 450 N and T_R = 779 N.
Why the other answers miss: B is the left cable's tension. A assumes one cable carries the whole load directly. C assumes the right cable alone supplies all 900 N vertically (900/sin 60°).
Source for the principle: OpenStax University Physics Vol. 1, 12.1 Conditions for Static Equilibrium
Question 24 of 55 · Statics · F. Area properties (centroids) · Choose one answer
Start with a rectangle covering 0 ≤ x ≤ 120 mm and 0 ≤ y ≤ 80 mm. Remove the upper-right rectangle covering 80 ≤ x ≤ 120 mm and 40 ≤ y ≤ 80 mm. Where is the centroid of the remaining area, measured from the original lower-left corner?
- A. (60, 40) mm
- B. (65.7, 42.9) mm
- C. (68, 44) mm
- D. (52, 36) mm
Answer and explanation
Answer: D. (52, 36) mm. Treat the cutout as negative area. A = 9,600 − 1,600 = 8,000 mm². x̄ = [9,600(60) − 1,600(100)]/8,000 = 52 mm. ȳ = [9,600(40) − 1,600(60)]/8,000 = 36 mm. Removing material from the upper right moves the centroid left and down, which matches.
Why the other answers miss: A ignores the cutout. B adds the cutout instead of subtracting it. C measures the right centroid from the wrong corner (120 − 52 and 80 − 36).
Source for the principle: Engineering Statics (Baker & Haynes), 7.5 Centroids Using Composite Parts
Question 25 of 55 · Statics · I. Weight and mass computations · Choose one answer
A component has a mass of 2,000 lbm. What is its mass in slugs, most nearly? (1 slug = 32.174 lbm.)
- A. 2,000 slug
- B. 62.2 slug
- C. 64,348 slug
- D. 907 slug
Answer and explanation
Answer: B. 62.2 slug. One slug is the mass that accelerates at 1 ft/s² under 1 lbf, which equals 32.174 lbm. So 2,000 lbm ÷ 32.174 lbm/slug = 62.2 slug. NIST's factors check this: 1 slug = 14.59390 kg and 1 lbm = 0.4535924 kg, a ratio of 32.174.
Why the other answers miss: A treats lbm and slug as the same unit. C multiplies by 32.174 instead of dividing. D converts to kilograms (907 kg) and keeps the slug label.
Source for the principle: NIST SP 811, Appendix B.8 (pound and slug to kilogram)
Question 26 of 55 · Statics · G. Static friction · Enter a number
A block rests on a plank. The coefficient of static friction between them is 0.35. One end of the plank is raised slowly. At what angle above horizontal does the block start to slip?
Enter your answer in degrees, rounded to one decimal place.
Answer and explanation
Answer: 19.3°. At the point of slipping, friction is at its maximum, f = μₛN. Along the incline: mg sin θ = μₛ mg cos θ, so tan θ = μₛ and θ = arctan(0.35) = 19.29°, which rounds to 19.3°. Mass cancels.
We accept answers from 19.25 up to, but not including, 19.35.
Common mistakes: Using arcsin(0.35) gives 20.5°, and arccos gives an unrelated angle. Friction balances the downslope component of weight, and the normal force is the cos θ component, so their ratio is tan θ.
Source for the principle: OpenStax University Physics Vol. 1, 6.2 Friction
9. Dynamics (5 questions)
Question 27 of 55 · Dynamics · F. Work, energy, and power (with friction) · Choose one answer
A 40 kg crate starts from rest and slides 10.0 m down a straight incline at 20° above horizontal. The kinetic friction coefficient is 0.120. Use g = 9.81 m/s² and ignore air resistance. What is its speed at the bottom, most nearly?
- A. 8.19 m/s
- B. 14.0 m/s
- C. 6.71 m/s
- D. 9.45 m/s
Answer and explanation
Answer: C. 6.71 m/s. Gravity does work mgL sin 20° and friction does −μₖ mgL cos 20°. Setting ½mv² equal to the net work: v = √[2gL(sin 20° − 0.120 cos 20°)] = 6.71 m/s. Mass cancels.
Why the other answers miss: A ignores friction. D adds friction's work instead of subtracting it. B treats the full 10 m slope as a vertical drop with no friction.
Source for the principle: OpenStax University Physics Vol. 1, 8.3 Conservation of Energy; OpenStax University Physics Vol. 1, 6.2 Friction
Question 28 of 55 · Dynamics · C. Angular motion · Choose one answer
A flywheel has a mass moment of inertia of 1.60 kg·m² and an initial angular speed of 15.0 rad/s in the motor's direction. A constant 24.0 N·m motor torque acts against a constant 8.0 N·m brake torque for 4.00 s. What is its angular displacement in that time?
- A. 140 rad
- B. 80 rad
- C. 180 rad
- D. 220 rad
Answer and explanation
Answer: A. 140 rad. Net torque is 24 − 8 = 16 N·m, so α = 16/1.60 = 10.0 rad/s². Then θ = ω₀t + ½αt² = 15(4) + ½(10)(16) = 60 + 80 = 140 rad.
Why the other answers miss: B drops the initial-speed term. C ignores the brake torque. D multiplies the final speed (55 rad/s) by the time, though the speed changes during the interval.
Source for the principle: OpenStax University Physics Vol. 1, 10.7 Newton's Second Law for Rotation (Eq. 10.25); OpenStax University Physics Vol. 1, 10.2 Rotation with Constant Angular Acceleration
Question 29 of 55 · Dynamics · H. Vibrations (natural frequency) · Choose one answer
A 50.0 kg mass slides without friction and is attached to two springs in parallel, with stiffnesses of 30.0 kN/m and 50.0 kN/m. Both springs stretch the same amount. Ignore damping. What is the natural frequency in hertz, most nearly?
- A. 40.0 Hz
- B. 6.37 Hz
- C. 5.03 Hz
- D. 3.08 Hz
Answer and explanation
Answer: B. 6.37 Hz. Parallel springs add: k = 80,000 N/m. ωₙ = √(k/m) = √(80,000/50) = 40.0 rad/s, and fₙ = ωₙ/(2π) = 6.37 Hz.
Why the other answers miss: A is ωₙ in rad/s labeled as hertz. C uses only the 50 kN/m spring. D combines the springs as if they were in series.
Source for the principle: OpenStax University Physics Vol. 1, 15.1 Simple Harmonic Motion
Question 30 of 55 · Dynamics · E. Impulse and momentum · Enter a number
A 0.150 kg ball hits a wall at 40.0 m/s and rebounds straight back at 30.0 m/s. Contact lasts 0.0100 s. What is the magnitude of the average net force on the ball?
Enter your answer in N, rounded to the nearest whole newton.
Answer and explanation
Answer: 1,050 N. Take the rebound direction as positive. Δp = m(v_f − v_i) = 0.150 × [30.0 − (−40.0)] = 10.5 kg·m/s. F_avg = Δp/Δt = 10.5/0.0100 = 1,050 N.
We accept answers from 1,049.5 up to, but not including, 1,050.5.
Common mistakes: Subtracting the speeds (40 − 30 = 10 m/s) ignores the reversal of direction and gives 150 N. Velocity is a vector; a rebound adds the two speeds.
Source for the principle: OpenStax University Physics Vol. 1, 9.2 Impulse and Collisions
Question 31 of 55 · Dynamics · A. Particle kinematics · Choose one answer
A vehicle traveling at 60.0 ft/s brakes with a constant deceleration of 15.0 ft/s². How far does it travel before stopping?
- A. 60.0 ft
- B. 240 ft
- C. 4.00 ft
- D. 120 ft
Answer and explanation
Answer: D. 120 ft. Use v² = v₀² + 2a(x − x₀) with v = 0 and a = −15.0 ft/s²: distance = −v₀²/(2a) = 60²/(2 × 15) = 3,600/30 = 120 ft. Check: it stops in 60/15 = 4 s at an average speed of 30 ft/s, and 30 × 4 = 120 ft.
Why the other answers miss: B leaves out the 2 (or multiplies the starting speed by the stopping time). C is the stopping time, 4.00 s, mislabeled as a distance. A multiplies the starting speed by an assumed one-second stopping time, ignoring the deceleration.
Source for the principle: OpenStax University Physics Vol. 1, 3.4 Motion with Constant Acceleration
10. Strength of Materials (5 questions)
Question 32 of 55 · Strength of Materials · F. Loads and deformations (axial) · Choose one answer
Two uniform bar segments are joined end to end with no load at the joint. Segment 1 is 18.0 in long with area 0.500 in²; segment 2 is 12.0 in long with area 0.250 in². Both have E = 29.0 × 10⁶ psi. The assembly carries 8.00 kip in tension (1 kip = 1,000 lbf). Assuming linear elastic behavior, what is the total elongation, most nearly?
- A. 0.00993 in
- B. 0.0132 in
- C. 0.0166 in
- D. 0.0232 in
Answer and explanation
Answer: D. 0.0232 in. Each segment carries the full 8,000 lbf, and their elongations add: δ = (P/E)(L₁/A₁ + L₂/A₂) = (8,000/(29.0 × 10⁶))(18/0.500 + 12/0.250) = (2.759 × 10⁻⁴)(36 + 48) = 0.0232 in.
Why the other answers miss: A and B are segment 1 and segment 2 alone. C uses the larger area for the whole 30 in.
Source for the principle: OpenStax University Physics Vol. 1, 12.3 Stress, Strain, and Elastic Modulus
Question 33 of 55 · Strength of Materials · G. Stress transformation and principal stresses · Choose one answer
At a point in plane stress, σx = 120 MPa (tension), σy = 40 MPa (tension), and τxy = 30 MPa. What is the maximum in-plane principal stress?
- A. 80 MPa
- B. 120 MPa
- C. 130 MPa
- D. 50 MPa
Answer and explanation
Answer: C. 130 MPa. σ₁,₂ = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²] = 80 ± √(40² + 30²) = 80 ± 50 MPa. So σ₁ = 130 MPa and σ₂ = 30 MPa.
Why the other answers miss: A is the center of Mohr's circle and D is its radius. B ignores the shear stress's effect on the principal directions.
Source for the principle: MIT OpenCourseWare 3.91 (Roylance), Transformation of Stresses and Strains, p. 6, Eq. 11
Question 34 of 55 · Strength of Materials · C. Stress from torsion · Choose one answer
A solid circular steel shaft 50 mm in diameter carries a torque of 1.00 kN·m. What is the maximum shear stress, most nearly?
- A. 20.4 MPa
- B. 81.5 MPa
- C. 40.7 MPa
- D. 163 MPa
Answer and explanation
Answer: C. 40.7 MPa. τ = Tc/J, with c = 25 mm and J = πd⁴/32 = π(0.050)⁴/32 = 6.136 × 10⁻⁷ m⁴. τ = (1,000 × 0.025)/(6.136 × 10⁻⁷) = 40.7 MPa.
Why the other answers miss: B uses the full diameter as c, or uses the area moment I = πd⁴/64 in place of J; either slip doubles the stress. A uses πd⁴/16, twice the correct J. D makes both slips at once.
Source for the principle: MIT OpenCourseWare 3.91 (Roylance), Torsion, p. 8, Eq. 14
Question 35 of 55 · Strength of Materials · C. Stress from bending · Choose one answer
A cantilever beam 2.00 m long carries a 5.00 kN downward point load at its free end. Its cross section is rectangular, 50 mm wide and 150 mm deep, loaded about its strong axis. What is the maximum bending stress, most nearly?
- A. 53.3 MPa
- B. 107 MPa
- C. 26.7 MPa
- D. 160 MPa
Answer and explanation
Answer: A. 53.3 MPa. The maximum moment is at the fixed end: M = PL = 5.00 × 2.00 = 10.0 kN·m. I = bh³/12 = 50(150)³/12 = 1.406 × 10⁷ mm⁴ and c = 75 mm. σ = Mc/I = (10.0 × 10⁶ N·mm)(75)/(1.406 × 10⁷) = 53.3 MPa.
Why the other answers miss: B uses the full depth, 150 mm, as c. C uses half the moment, PL/2. D swaps width and depth, bending the beam about its weak axis.
Source for the principle: MIT OpenCourseWare 3.91 (Roylance), Stresses in Beams, p. 4, Eqs. 5 and 7
Question 36 of 55 · Strength of Materials · H. Material failure (Euler buckling) · Choose one answer
A steel column with pinned ends is 3.00 m long. E = 200 GPa and its smallest area moment of inertia is I = 8.00 × 10⁻⁷ m⁴. What is its Euler critical buckling load, most nearly?
- A. 17.8 kN
- B. 43.9 kN
- C. 702 kN
- D. 175 kN
Answer and explanation
Answer: D. 175 kN. For pinned–pinned ends, the effective length equals the actual length. P_cr = π²EI/L_e² = π²(200 × 10⁹)(8.00 × 10⁻⁷)/(3.00)² = 175 kN.
Why the other answers miss: B uses L_e = 2L (fixed–free). C uses L_e = 0.5L (fixed–fixed). A leaves out π².
Source for the principle: MIT OpenCourseWare 3.91 (Roylance), Stresses in Beams, pp. 7–8, Eq. 11 and end conditions
11. Materials (3 questions)
Question 37 of 55 · Materials · F. Material selection · Select the two correct answers
A uniform tension rod is 2.00 m long with a 200 mm² cross section and carries 10.0 kN. It must stretch no more than 1.50 mm and have a mass no greater than 1.20 kg. All four candidates stay linearly elastic. Select the two candidates that meet both limits.
- A. Material A: E = 70 GPa, ρ = 2,700 kg/m³
- B. Material B: E = 210 GPa, ρ = 7,800 kg/m³
- C. Material C: E = 80 GPa, ρ = 2,900 kg/m³
- D. Material D: E = 45 GPa, ρ = 1,800 kg/m³
Answer and explanation
Answer: A and C. Use δ = PL/(AE) and m = ρAL, with A = 2.00 × 10⁻⁴ m². A: 1.43 mm and 1.08 kg (passes both). B: 0.476 mm but 3.12 kg (too heavy). C: 1.25 mm and 1.16 kg (passes both). D: 0.72 kg but 2.22 mm (stretches too much).
Where people slip: Check every requirement for every candidate. B and D each pass one limit and fail the other. The properties are illustrative, not specifications for named alloys.
Source for the principle: OpenStax University Physics Vol. 1, 12.3 Stress, Strain, and Elastic Modulus; OpenStax University Physics Vol. 1, 14.1 Fluids, Density, and Pressure
Question 38 of 55 · Materials · A. Phase diagrams (lever rule) · Enter a number
An alloy containing 35 wt% B is held in the two-phase α + liquid region. At that temperature, the tie line meets the α phase at 20 wt% B and the liquid at 60 wt% B. What mass fraction of the alloy is liquid?
Enter your answer, rounded to three decimal places.
Answer and explanation
Answer: 0.375. Lever rule: fraction liquid = (C₀ − C_α)/(C_L − C_α) = (35 − 20)/(60 − 20) = 15/40 = 0.375. The rest, 0.625, is α.
We accept answers from 0.3745 up to, but not including, 0.3755.
Common mistakes: 0.625 is the α fraction. The lever rule uses the opposite arm: the liquid fraction is the distance from the alloy to the α boundary, divided by the tie-line length.
Source for the principle: DoITPoMS (University of Cambridge), The Lever Rule, §12.7, credited LibreTexts republication
Question 39 of 55 · Materials · D. Thermal properties · Choose one answer
A metal rail 5.00 m long has a linear expansion coefficient of 23 × 10⁻⁶ per °C. Its temperature rises by 40.0 °C, and it's free to expand. How much longer does it get?
- A. 0.460 mm
- B. 4.60 mm
- C. 46.0 mm
- D. 9.20 mm
Answer and explanation
Answer: B. 4.60 mm. ΔL = αLΔT = (23 × 10⁻⁶ /°C)(5.00 m)(40.0 °C) = 0.00460 m = 4.60 mm.
Why the other answers miss: A and C are off by a factor of 10, usually from a meter-to-millimeter slip. D doubles the result.
Source for the principle: OpenStax University Physics Vol. 2, 1.3 Thermal Expansion
12. Fluid Mechanics (7 questions)
Question 40 of 55 · Fluid Mechanics · D. Fluid statics · Choose one answer
A vertical rectangular gate 1.50 m wide and 2.00 m tall has its top edge 1.00 m below an open water surface. Water is on one side and atmospheric air on the other. Use ρ = 1,000 kg/m³ and g = 9.81 m/s². What is the net hydrostatic force on the gate, most nearly?
- A. 58.9 kN
- B. 39.2 kN
- C. 88.3 kN
- D. 29.4 kN
Answer and explanation
Answer: A. 58.9 kN. Force on a plane surface = pressure at the centroid × area. The centroid is 1.00 + 1.00 = 2.00 m deep and A = 1.50 × 2.00 = 3.00 m². F = ρgh_c A = 1,000(9.81)(2.00)(3.00) = 58,860 N = 58.9 kN. Atmospheric pressure acts on both sides and cancels.
Why the other answers miss: D uses the pressure at the top edge over the whole gate. C uses the pressure at the bottom edge. B leaves out the 1.50 m width.
Source for the principle: Engineering Statics (Baker & Haynes), 7.9 Fluid Statics
Question 41 of 55 · Fluid Mechanics · E. Energy, impulse, and momentum equations (Bernoulli) · Enter a number
Water flows steadily from a 0.100 m diameter pipe into a 0.0500 m diameter pipe at the same elevation, at 0.0120 m³/s. The pressure in the larger pipe is 250 kPa gauge. Use ρ = 1,000 kg/m³, uniform velocity at each section, no pump or turbine, and negligible losses. What is the pressure in the smaller pipe?
Enter your answer in kPa gauge, rounded to one decimal place.
Answer and explanation
Answer: 232.5 kPa gauge. Continuity: v = Q/A gives v₁ = 0.0120/(π(0.100)²/4) = 1.528 m/s and v₂ = 0.0120/(π(0.0500)²/4) = 6.112 m/s. Bernoulli at equal elevation: p₂ = p₁ + ½ρ(v₁² − v₂²) = 250,000 + 500(2.334 − 37.35) = 232,492 Pa, or 232.5 kPa gauge.
We accept answers from 232.45 up to, but not including, 232.55.
Common mistakes: Halving the diameter quarters the area, so velocity rises by 4 and pressure falls. Using the diameter as a radius, or adding pressure as speed rises, gives the wrong value.
Source for the principle: OpenStax University Physics Vol. 1, 14.6 Bernoulli's Equation
Question 42 of 55 · Fluid Mechanics · C. Laminar and turbulent flow · Choose one answer
Newtonian oil (μ = 0.120 Pa·s, ρ = 900 kg/m³) flows steadily through a horizontal pipe 12.0 m long and 0.0400 m in diameter at an average velocity of 0.500 m/s. Assume fully developed flow and ignore entrance and fitting losses. What is the pressure drop?
- A. 3.60 kPa
- B. 14.4 kPa
- C. 28.8 kPa
- D. 57.6 kPa
Answer and explanation
Answer: B. 14.4 kPa. First check the regime: Re = ρVD/μ = 900(0.500)(0.0400)/0.120 = 150, well inside laminar flow. For laminar pipe flow, Δp = 32μLV/D² = 32(0.120)(12.0)(0.500)/(0.0400)² = 14,400 Pa = 14.4 kPa.
Why the other answers miss: A plugs the diameter into the radius form (8μLV/r²). C uses the centerline velocity, twice the average, in place of V. D plugs the radius into the diameter form.
Source for the principle: OpenStax University Physics Vol. 1, 14.7 Viscosity and Turbulence
Question 43 of 55 · Fluid Mechanics · F. Pipe flow and friction losses · Choose one answer
Water flows at 2.00 m/s through 100 m of 0.100 m diameter pipe. The Darcy friction factor is 0.020. Using g = 9.81 m/s², what is the head loss, most nearly?
- A. 0.408 m
- B. 8.15 m
- C. 4.08 m
- D. 40.0 m
Answer and explanation
Answer: C. 4.08 m. Darcy–Weisbach: h_f = f(L/D)(V²/2g) = 0.020 × (100/0.100) × (2.00²/(2 × 9.81)) = 0.020 × 1,000 × 0.2039 = 4.08 m.
Why the other answers miss: B divides by g instead of 2g. A uses L instead of L/D. D leaves out g entirely (V²/2 instead of V²/2g).
Source for the principle: Hydraulic Institute Data Tool, Fluid Flow: Darcy-Weisbach equation
Question 44 of 55 · Fluid Mechanics · J. Turbomachinery (pumps) · Choose one answer
A pump moves 0.0500 m³/s of water (ρ = 1,000 kg/m³) against a total head of 20.0 m. Pump efficiency is 75.0%. Using g = 9.81 m/s², what shaft power must the motor supply, most nearly?
- A. 13.1 kW
- B. 9.81 kW
- C. 7.36 kW
- D. 0.0131 kW
Answer and explanation
Answer: A. 13.1 kW. Total head H is the energy added per unit weight of fluid. Fluid power = ρgQH = 1,000(9.81)(0.0500)(20.0) = 9,810 W. Efficiency is useful output ÷ input, so shaft power = 9,810/0.750 = 13,080 W = 13.1 kW.
Why the other answers miss: B is the power delivered to the water, not the shaft input. C multiplies by efficiency instead of dividing; the motor must supply more than the fluid gets. D leaves out the density.
Source for the principle: OpenStax College Physics 2e, 12.3 The Most General Applications of Bernoulli's Equation (Power in Fluid Flow); OpenStax College Physics 2e, 7.6 Conservation of Energy (efficiency)
Question 45 of 55 · Fluid Mechanics · I. Flow measurement · Choose one answer
A pitot-static tube in low-speed airflow shows a difference of 800 Pa between total and static pressure. The air density is 1.20 kg/m³. Treating the flow as incompressible, what is the airspeed, most nearly?
- A. 1,333 m/s
- B. 25.8 m/s
- C. 667 m/s
- D. 36.5 m/s
Answer and explanation
Answer: D. 36.5 m/s. From Bernoulli's equation, V = √[2(p_t − p_s)/ρ] = √(2 × 800/1.20) = √1,333 = 36.5 m/s.
Why the other answers miss: B leaves out the factor of 2. C leaves out both the factor of 2 and the square root; A leaves out only the square root. Without the square root, the calculated quantity has units of m²/s², not m/s.
Source for the principle: NASA Glenn Research Center, Pitot-Static Tube Speedometer
Question 46 of 55 · Fluid Mechanics · K. Ideal gas law · Choose one answer
Air in a tank is at 200 kPa absolute and 300 K. Treat it as an ideal gas with R = 287 J/(kg·K). What is its density, most nearly?
- A. 3.50 kg/m³
- B. 1.18 kg/m³
- C. 2.32 kg/m³
- D. 80.2 kg/m³
Answer and explanation
Answer: C. 2.32 kg/m³. From the ideal gas law in mass form, ρ = p/(RT) = 200,000/(287 × 300) = 2.32 kg/m³. The ideal gas law needs absolute pressure and absolute temperature, both of which are given.
Why the other answers miss: B uses atmospheric pressure (101.3 kPa) instead of the tank pressure. A treats 200 kPa as gauge and adds atmospheric pressure, though the stem says absolute. D uses the universal constant 8.314 J/(mol·K), which gives moles per cubic meter, not kg/m³.
Source for the principle: OpenStax University Physics Vol. 2, 2.1 Molecular Model of an Ideal Gas
13. Basic Electrical Engineering (4 questions)
Question 47 of 55 · Basic Electrical Engineering · B. Current and voltage laws · Choose one answer
An ideal 24.0 V DC source supplies a 4.00 Ω resistor in series with a parallel pair of 12.0 Ω and 6.00 Ω resistors. What is the current in the 6.00 Ω resistor?
- A. 2.00 A
- B. 3.00 A
- C. 1.00 A
- D. 4.00 A
Answer and explanation
Answer: A. 2.00 A. The parallel pair is (12 × 6)/(12 + 6) = 4.00 Ω, so the total is 8.00 Ω and the source current is 24/8 = 3.00 A. The parallel pair has 3.00 A × 4.00 Ω = 12.0 V across it, so the 6.00 Ω branch carries 12/6 = 2.00 A. Check: the 12 Ω branch carries 1.00 A, and 2 + 1 = 3 A.
Why the other answers miss: C is the 12 Ω branch. B is the total source current. D puts the full 24 V across the 6 Ω resistor, ignoring the series resistor's voltage drop.
Source for the principle: OpenStax University Physics Vol. 2, 10.2 Resistors in Series and Parallel
Question 48 of 55 · Basic Electrical Engineering · C. AC circuits (power factor) · Choose one answer
A sinusoidal source supplies 200 V rms to a load with impedance Z = 30 + j40 Ω. What average (real) power does the load absorb?
- A. 800 W
- B. 240 W
- C. 480 W
- D. About 1,333 W
Answer and explanation
Answer: C. 480 W. |Z| = √(30² + 40²) = 50 Ω, so I = 200/50 = 4.00 A rms. Power factor = R/|Z| = 30/50 = 0.600. P = V I cos φ = 200 × 4.00 × 0.600 = 480 W. Check: I²R = 16 × 30 = 480 W.
Why the other answers miss: A is apparent power (VA), not real power. B applies a ½ factor that belongs only to peak values, not rms. D ignores the reactance when finding current (200²/30).
Source for the principle: OpenStax University Physics Vol. 2, 15.4 Power in an AC Circuit
Question 49 of 55 · Basic Electrical Engineering · E. Three-phase power · Choose one answer
A balanced three-phase load draws 30.0 A rms per line from a sinusoidal 480 V rms (line-to-line) supply at a power factor of 0.850 lagging. What real power does it consume, most nearly?
- A. 24.9 kW
- B. 12.2 kW
- C. 36.7 kW
- D. 21.2 kW
Answer and explanation
Answer: D. 21.2 kW. For a balanced three-phase load, P = √3 V_LL I_L cos θ, where V_LL and I_L are rms line values. P = √3 × 480 × 30.0 × 0.850 = 21,200 W = 21.2 kW.
Why the other answers miss: B treats it as single phase (no √3). C multiplies line-to-line voltage by 3 instead of √3. A leaves out the power factor, giving apparent power in kVA.
Source for the principle: University of Utah ECE 3600, Three-Phase Power notes, p. 3
Question 50 of 55 · Basic Electrical Engineering · D. Measuring devices · Choose one answer
You need to measure the current through a resistor and the voltage across it. How should an ammeter and a voltmeter be connected so they disturb the circuit as little as possible?
- A. Ammeter in series with very low resistance; voltmeter in parallel with very high resistance
- B. Ammeter in parallel with very high resistance; voltmeter in series with very low resistance
- C. Both meters in series with the resistor
- D. Ammeter in series with very high resistance, so it draws little current
Answer and explanation
Answer: A. Ammeter in series, low resistance; voltmeter in parallel, high resistance. An ammeter must carry the same current as the component, so it goes in series, and its resistance must be very low so it barely changes that current. A voltmeter must see the same voltage as the component, so it goes in parallel, and its resistance must be very high so it draws almost no current.
Why the other answers miss: B reverses both connections. C would put the voltmeter's high resistance in the current path. D gets the connection right but the resistance wrong; a high-resistance ammeter would cut the very current it's measuring.
Source for the principle: OpenStax University Physics Vol. 2, 10.4 Electrical Measuring Instruments
14. Thermodynamics and Heat Transfer (5 questions)
Question 51 of 55 · Thermodynamics and Heat Transfer · A. Thermodynamic laws (second law) · Choose one answer
A heat engine runs between reservoirs at 900 K and 300 K and takes in 600 kJ from the hot reservoir per cycle. What is the greatest net work it could possibly produce per cycle?
- A. 600 kJ
- B. 400 kJ
- C. 300 kJ
- D. 200 kJ
Answer and explanation
Answer: B. 400 kJ. No engine between two fixed-temperature reservoirs can beat a reversible (Carnot) engine. η_max = 1 − T_c/T_h = 1 − 300/900 = 2/3, so W_max = (2/3)(600) = 400 kJ per cycle.
Why the other answers miss: D is the heat rejected at that limit. C assumes an arbitrary 50% efficiency. A converts all the heat into work, which the second law forbids.
Source for the principle: OpenStax University Physics Vol. 2, 4.5 The Carnot Cycle
Question 52 of 55 · Thermodynamics and Heat Transfer · E. Heat transfer (conduction) · Enter a number
A plane wall with area 3.00 m² has two layers in series: layer 1 is 0.120 m thick with k = 0.600 W/(m·K), and layer 2 is 0.0800 m thick with k = 0.0400 W/(m·K). The outer surfaces are held at 60.0 °C and 20.0 °C. Assume steady one-dimensional conduction, no internal heat generation, perfect contact, and constant properties. What is the heat-transfer rate?
Enter your answer in W, rounded to one decimal place.
Answer and explanation
Answer: 54.5 W. Series layers add their resistances. Per unit area: L₁/k₁ + L₂/k₂ = 0.200 + 2.00 = 2.20 m²·K/W. Q = AΔT/(ΣL/k) = 3.00 × 40.0/2.20 = 54.5 W. A 40 °C difference is the same as 40 K.
We accept answers from 54.45 up to, but not including, 54.55.
Common mistakes: 18.2 W/m² is the heat flux, not the rate; multiply by the area. Averaging the two conductivities doesn't represent layers in series. Surface temperatures are given, so no convection coefficients are needed.
Source for the principle: OpenStax University Physics Vol. 2, 1.6 Mechanisms of Heat Transfer
Question 53 of 55 · Thermodynamics and Heat Transfer · F. Mass and energy balances · Choose one answer
Water flows steadily through a heater at 0.500 kg/s and is warmed from 15.0 °C to 45.0 °C. Take c = 4.186 kJ/(kg·K) and ignore changes in kinetic and potential energy and any work. What is the heating rate, most nearly?
- A. 62.8 kW
- B. 62,800 kW
- C. 2.09 kW
- D. 94.2 kW
Answer and explanation
Answer: A. 62.8 kW. Heat added per kilogram is cΔT, so the steady rate is Q = ṁcΔT = 0.500 kg/s × 4.186 kJ/(kg·K) × 30.0 K = 62.8 kJ/s = 62.8 kW.
Why the other answers miss: B converts 62.8 kJ/s to about 62,800 W but labels that value as kW. C leaves out the temperature rise. D uses the outlet temperature (45 °C) as the temperature change.
Source for the principle: OpenStax University Physics Vol. 2, 1.4 Heat Transfer, Specific Heat, and Calorimetry
Question 54 of 55 · Thermodynamics and Heat Transfer · E. Heat transfer (radiation) · Choose one answer
A small gray surface with emissivity 0.800 and area 1.00 m² is at 500 K. It is completely surrounded by a much larger isothermal enclosure at 300 K. Using σ = 5.67 × 10⁻⁸ W/(m²·K⁴), what is the net radiant heat loss, most nearly?
- A. 0.120 kW
- B. 2.84 kW
- C. 2.47 kW
- D. 0.0726 kW
Answer and explanation
Answer: C. 2.47 kW. Net radiation = εσA(T_s⁴ − T_surr⁴) = 0.800(5.67 × 10⁻⁸)(1.00)(500⁴ − 300⁴) = 0.800(5.67 × 10⁻⁸)(5.44 × 10¹⁰) = 2,468 W = 2.47 kW.
Why the other answers miss: B ignores the radiation coming back from the surroundings. A uses Celsius temperatures (227 °C and 27 °C); radiation needs kelvins. D raises the temperature difference to the fourth power, (500 − 300)⁴, instead of subtracting the fourth powers.
Source for the principle: OpenStax University Physics Vol. 2, 1.6 Mechanisms of Heat Transfer; ASHRAE Handbook—Fundamentals (2017), Ch. 4, Radiation: small surface in a large isothermal enclosure
Question 55 of 55 · Thermodynamics and Heat Transfer · I. Psychrometrics · Choose one answer
Air at 30 °C contains water vapor at a density of 12.8 g/m³. Saturation vapor densities: 10 °C, 9.40 g/m³; 15 °C, 12.8 g/m³; 20 °C, 17.2 g/m³; 25 °C, 23.0 g/m³; 30 °C, 30.4 g/m³. What is the initial relative humidity, and at about what temperature will condensation begin if the air is cooled uniformly in a rigid sealed container, with no change in water-vapor mass before condensation?
- A. 42.1%; about 15 °C
- B. 42.1%; about 30 °C
- C. 237.5%; about 15 °C
- D. 57.9%; about 20 °C
Answer and explanation
Answer: A. 42.1%; about 15 °C. Relative humidity = vapor density ÷ saturation vapor density at the air's temperature = 12.8/30.4 = 42.1%. Before condensation, the sealed container has a fixed water-vapor mass and volume, so the vapor density remains 12.8 g/m³. Condensation begins when the saturation density falls to 12.8 g/m³, which the table puts at 15 °C.
Why the other answers miss: B gives the right humidity but uses the air's current temperature as the condensation-onset temperature. C inverts the ratio. D reports the unsaturated fraction (100% − 42.1%) and picks the wrong row.
Source for the principle: OpenStax College Physics 2e, 13.6 Humidity, Evaporation, and Boiling (Table 13.5); ASHRAE Handbook—Fundamentals (2025), Ch. 1, §§4–5: vapor density and dew-point definitions
Check your answers
Here's the full key. Mark each answered question right or wrong, keep unanswered questions separate, then fill in the tally below it.
| Question | Answer | Knowledge area |
|---|---|---|
| 1 | B | Mathematics |
| 2 | 2.652 | Mathematics |
| 3 | B | Mathematics |
| 4 | C | Mathematics |
| 5 | D | Probability and Statistics |
| 6 | A | Probability and Statistics |
| 7 | A and B and C | Probability and Statistics |
| 8 | D | Chemistry |
| 9 | C | Chemistry |
| 10 | D | Chemistry |
| 11 | A | Instrumentation and Controls |
| 12 | B | Instrumentation and Controls |
| 13 | A and C | Engineering Ethics and Societal Impacts |
| 14 | B | Engineering Ethics and Societal Impacts |
| 15 | C | Engineering Ethics and Societal Impacts |
| 16 | C | Safety, Health, and Environment |
| 17 | B | Safety, Health, and Environment |
| 18 | B → C → A | Safety, Health, and Environment |
| 19 | $27,374 | Engineering Economics |
| 20 | D | Engineering Economics |
| 21 | B | Engineering Economics |
| 22 | A | Statics |
| 23 | D | Statics |
| 24 | D | Statics |
| 25 | B | Statics |
| 26 | 19.3° | Statics |
| 27 | C | Dynamics |
| 28 | A | Dynamics |
| 29 | B | Dynamics |
| 30 | 1,050 N | Dynamics |
| 31 | D | Dynamics |
| 32 | D | Strength of Materials |
| 33 | C | Strength of Materials |
| 34 | C | Strength of Materials |
| 35 | A | Strength of Materials |
| 36 | D | Strength of Materials |
| 37 | A and C | Materials |
| 38 | 0.375 | Materials |
| 39 | B | Materials |
| 40 | A | Fluid Mechanics |
| 41 | 232.5 kPa gauge | Fluid Mechanics |
| 42 | B | Fluid Mechanics |
| 43 | C | Fluid Mechanics |
| 44 | A | Fluid Mechanics |
| 45 | D | Fluid Mechanics |
| 46 | C | Fluid Mechanics |
| 47 | A | Basic Electrical Engineering |
| 48 | C | Basic Electrical Engineering |
| 49 | D | Basic Electrical Engineering |
| 50 | A | Basic Electrical Engineering |
| 51 | B | Thermodynamics and Heat Transfer |
| 52 | 54.5 W | Thermodynamics and Heat Transfer |
| 53 | A | Thermodynamics and Heat Transfer |
| 54 | C | Thermodynamics and Heat Transfer |
| 55 | A | Thermodynamics and Heat Transfer |
Every question is all-or-nothing, consistent with NCEES’s rule that individual items receive no partial credit. A multiple-select question counts only if you picked every correct choice and nothing else. The ordering question counts only if the whole sequence is right. A numeric answer counts if it lands in the accepted range shown under that question. Those ranges are this set’s grading rules, chosen for its stated rounding instructions; they do not claim to reproduce NCEES’s numerical tolerances. (NCEES Examinee Guide, May 2026, p. 11)
Your tally by knowledge area
| Knowledge area | Questions in this set | Your correct answers |
|---|---|---|
| 1. Mathematics | 4 (Questions 1–4) | ___ of 4 |
| 2. Probability and Statistics | 3 (Questions 5–7) | ___ of 3 |
| 3. Chemistry | 3 (Questions 8–10) | ___ of 3 |
| 4. Instrumentation and Controls | 2 (Questions 11–12) | ___ of 2 |
| 5. Engineering Ethics and Societal Impacts | 3 (Questions 13–15) | ___ of 3 |
| 6. Safety, Health, and Environment | 3 (Questions 16–18) | ___ of 3 |
| 7. Engineering Economics | 3 (Questions 19–21) | ___ of 3 |
| 8. Statics | 5 (Questions 22–26) | ___ of 5 |
| 9. Dynamics | 5 (Questions 27–31) | ___ of 5 |
| 10. Strength of Materials | 5 (Questions 32–36) | ___ of 5 |
| 11. Materials | 3 (Questions 37–39) | ___ of 3 |
| 12. Fluid Mechanics | 7 (Questions 40–46) | ___ of 7 |
| 13. Basic Electrical Engineering | 4 (Questions 47–50) | ___ of 4 |
| 14. Thermodynamics and Heat Transfer | 5 (Questions 51–55) | ___ of 5 |
| Total | 55 | ___ of 55 |
Session totals: ___ correct + ___ incorrect + ___ unanswered = 55. If you want a percentage, divide your correct count by 55 and multiply by 100; round to one decimal place. For example, 38 correct is 69.1% on this practice set.
What your score means
Your score tells you how you did on these 55 questions. It isn't an NCEES score, and it can't predict a pass.
NCEES counts your correct answers, converts that count to a scaled score that adjusts for differences between exam forms, and compares it with a standard set by subject-matter experts. It doesn't publish the passing score, and it doesn't take points off for wrong answers. (NCEES Examinee Guide, May 2026, p. 14; NCEES exam scoring) So there's no honest way to turn "38 of 55" into "ready" or "not ready." Two practical takeaways:
- Never leave a real exam question blank. A wrong answer costs nothing more than a blank.
- Treat small areas as clues, not verdicts. Instrumentation and Controls has two questions here. Missing one of them tells you to look closer, not that the whole area is weak.
If you've already taken the FE and failed, your diagnostic report shows a 0–15 scaled score for each knowledge area, relative to examinees who passed. (NCEES Examinee Guide, May 2026, pp. 20–21) Use that report to set your priorities and this set to practice the areas it flags.
Turn your misses into your next study session
For every question you missed, guessed, or needed the explanation to finish, write one line:
| Question | What went wrong | What I'll do next | Retry date and result |
|---|---|---|---|
| Example: Question 34 | Used the full diameter as c in τ = Tc/J | Redo it from a blank sheet; then solve one hollow-shaft torsion problem | Oct 10: correct, 3 min |
| Concept, model choice, units, arithmetic, reading, or handbook lookup | One specific action |
Then work through the log like this:
- Fix the cause, not just the question. A units slip needs a units routine (write every conversion as a chain of fractions). A concept gap needs the principle, which is linked under each explanation.
- Redo the missed question from a blank sheet a day or two later, without looking at the solution.
- Solve one new problem on the same principle, from your coursework or a textbook, so you're not just remembering this answer.
- Use the official ranges to help prioritize the misses in your log. Fluid Mechanics (12–18 questions) and Statics, Dynamics, Strength of Materials, and Thermodynamics and Heat Transfer (9–14 each) are the largest, per the NCEES specification. When two concepts need similar work, the one in a larger area is a reasonable starting point; your own error pattern still matters.
Need a week-by-week schedule? Our FE study plan and discipline guide groups the 14 Other Disciplines areas into four study passes across 12 weeks.
How this set compares with the real exam
| Row label | Real FE Other Disciplines exam | This practice set |
|---|---|---|
| Questions | 110, including a few unscored pretest questions you can't identify | 55 |
| Time | 5 hours 20 minutes of exam time, plus a 25-minute scheduled break | Untimed, or 160 minutes for the same average pace |
| Sections | Two. After about half the questions, you review and submit them, and they lock | One continuous set |
| Question types | Multiple choice plus multiple correct, point and click, drag and drop, and fill in the blank (numeric) | Multiple choice (44), multiple select (3), numeric entry (7), ordering (1); no point-and-click |
| Reference | The current FE Reference Handbook on screen as a searchable PDF | Your downloaded copy of the same handbook |
| Scoring | Scaled score, reported as pass or fail | A raw count on these questions |
Sources: NCEES FE exam page; NCEES Examinee Guide, May 2026, pp. 10–11 and 16.
Where 160 minutes comes from: 320 minutes ÷ 110 questions is about 2.9 minutes per question. This set is exactly half the length, so half the time is 160 minutes. That's an average pace, not a per-question limit: a definition question might take 30 seconds and a two-step fluids problem five minutes.
What this practice set covers
NCEES lists 14 knowledge areas for FE Other Disciplines, each with a range for how many of the 110 questions it gets. We sized this set roughly in proportion to those ranges.
| # | Knowledge area | Questions on the real exam | Questions in this set |
|---|---|---|---|
| 1 | Mathematics | 8–12 | 4 |
| 2 | Probability and Statistics | 6–9 | 3 |
| 3 | Chemistry | 5–8 | 3 |
| 4 | Instrumentation and Controls | 4–6 | 2 |
| 5 | Engineering Ethics and Societal Impacts | 5–8 | 3 |
| 6 | Safety, Health, and Environment | 6–9 | 3 |
| 7 | Engineering Economics | 6–9 | 3 |
| 8 | Statics | 9–14 | 5 |
| 9 | Dynamics | 9–14 | 5 |
| 10 | Strength of Materials | 9–14 | 5 |
| 11 | Materials | 6–9 | 3 |
| 12 | Fluid Mechanics | 12–18 | 7 |
| 13 | Basic Electrical Engineering | 6–9 | 4 |
| 14 | Thermodynamics and Heat Transfer | 9–14 | 5 |
| Total | 110 | 55 |
Real-exam ranges: NCEES FE Other Disciplines specification, effective with the July 2020 exams, pp. 1–3. We used the range midpoints as a starting point, then adjusted the allocation to include a useful mix of problems. These are editorial practice counts, not fixed NCEES weights.
Two limits to keep in mind. The ranges tell you what the exam covers, not how hard each area is or where you need work. And 55 questions can't reach every subtopic: this set doesn't test hypothesis testing, logic diagrams, electrical safety, open-channel flow, real-gas behavior, or combustion thermodynamics, among others. Open the official specification to see the full list before you plan your study.
Practice with the handbook and an approved calculator
Use the same handbook you'll get on exam day. NCEES supplies the current FE Reference Handbook on screen as a searchable PDF, and it's the only reference you can use. You search it with a search box on the left side of the screen; Ctrl+F doesn't work. (NCEES Examinee Guide, May 2026, p. 10) Download the current version free through your MyNCEES account (NCEES: exam reference handbooks), and keep it open while you work these questions. Practice finding each equation and reading its variable definitions, since that's where units and assumptions trip people up.
Bring one approved calculator. For 2026 NCEES exams, NCEES approved the HP 33s and HP 35s, Casio models beginning with "fx-115" or "fx-991," and Texas Instruments models beginning with "TI-30X" or "TI-36X." (NCEES notice dated October 20, 2025, p. 1) A TI-30XS is also available on screen during the exam. (NCEES Examinee Guide, May 2026, p. 8) Testing in a later year? Check the Calculator Policy link on the NCEES exams page. Practice on the exact model you'll bring.
Is the FE Other Disciplines exam changing?
As of October 7, 2026, NCEES's FE page still links the Other Disciplines specification that took effect with the July 2020 exams. (NCEES FE exam page; official Other Disciplines specification)
Before you finalize your study plan, check the NCEES FE exam page for the specification that applies to your test date.
Before you book
| Detail | FE exam |
|---|---|
| Fee | $225 paid to NCEES; your licensing board may charge its own application fee |
| Where and when | Computer-based, at NCEES-approved Pearson test centers, year-round |
| Results | Typically 7–10 days, reported as pass or fail |
| Retakes | One attempt per testing window (January–March, April–June, July–September, October–December) and no more than three in any 12 months; some boards are stricter |
Sources: NCEES Examinee Guide, May 2026, pp. 3, 5, and 14; NCEES FE exam page.
Your licensing board, not NCEES, decides whether you're eligible to sit for the exam. (NCEES Examinee Guide, May 2026, p. 2) For the step-by-step process, see our guides to NCEES registration and scheduling, exam accommodations, results, and retake rules.
Common questions
Can I switch to a different FE discipline after registering? Yes. You cancel your registration and register again for the new exam; NCEES refunds the registration fee minus a $50 administrative fee. If you already have an appointment, cancel it through Pearson at least 48 hours before its start; Pearson charges a $50 cancellation fee, and NCEES waives its own $50 fee when that Pearson fee has been paid. (NCEES Examinee Guide, May 2026, pp. 4 and 7) If you’re unsure which discipline to take, check with your licensing board first.
Does passing the FE make me an engineer intern? Not by itself. Engineer intern (EI) or engineer-in-training (EIT) certification comes from your licensing board, through its own application. (NCEES Examinee Guide, May 2026, p. 24)
Sources and verification
By Castleport Test Prep Editorial Team. Last verified: October 7, 2026.
On that date we checked the exam facts on this page against the NCEES FE exam page, the FE Other Disciplines specification, the May 2026 NCEES Examinee Guide, NCEES's exam scoring and handbook-access pages, and the NCEES 2026 calculator notice. We checked each question's underlying principle against the sources linked beneath it, independently solved all 55 items, and used arithmetic and text checks to verify numerical results, distractor explanations, and agreement with the question data. AI-assisted tools were used to develop and check this content. This is editorial source checking, not professional engineering review, and the questions have not been statistically validated as a readiness measure.
Official sources:
- NCEES: FE exam
- NCEES: FE Other Disciplines exam specification (PDF), effective with the July 2020 exams
- NCEES Examinee Guide, May 2026 (PDF) · current guide page
- NCEES: Exam scoring
- NCEES: Exam reference handbooks
- NCEES: Notice of future changes and 2026 approved calculators (PDF, October 20, 2025)
- OSHA 29 CFR 1910.146, permit-required confined spaces · OSHA 29 CFR 1910.1000, air contaminants
Teaching sources for each question are linked directly beneath it. How we check exam facts: our methodology. Spotted an error? Report it through our corrections page.
Castleport Test Prep is an independent exam prep publisher, not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES), Pearson, or any state licensing board. The practice questions on this page are original and unofficial; they are not NCEES exam questions. Exam and credential names identify their subjects; trademarks belong to their respective owners. This resource doesn't guarantee an exam result, eligibility, or licensure.