Free FE Other Disciplines Practice Questions

Sample Practice Questions

Try these 10 sample questions from the FE Other Disciplines practice bank. Each includes the answer and explanation. Continue with the interactive practice above to review the full question set.

1. 1For A = [[1, 2], [0, 3]] and B = [[2, 1], [4, 0]], what is the entry in row 1, column 1 of AB?

   1. A. 6
   2. B. 8
   3. C. 10
   4. D. 12

   Answer: C

   Explanation: NIST describes matrix multiplication as "This is the row times column rule." The first row of A dotted with the first column of B gives 1(2) + 2(4) = 10.

   Reference: NIST Digital Library of Mathematical Functions: Elementary Algebra — §1.2(v), equation 1.2.34

2. 2Four independent measurements from a population are 4, 5, 9, and 10. What is the sample-mean point estimate of the population mean?

   1. A. 2.5
   2. B. 6.0
   3. C. 7.0
   4. D. 28.0

   Answer: C

   Explanation: NIST states, "The sample mean is calculated by summing the values in the sample and dividing by the number of values in the sample." Thus the estimate is (4 + 5 + 9 + 10)/4 = 7.0.

   Reference: NIST/SEMATECH e-Handbook: Quantitative Techniques — §1.3.5, Interval Estimates

3. 3What property of a platinum resistance temperature detector is measured to infer temperature?

   1. A. A change in its electrical resistance
   2. B. A change in its dielectric capacitance
   3. C. A change in its junction-generated voltage
   4. D. A change in its electromagnetic inductance

   Answer: A

   Explanation: DOE states, "The change in temperature will cause the platinum wire to heat or cool, resulting in a proportional change in resistance." An RTD uses the calibrated resistance-temperature relation; junction-generated voltage instead characterizes a thermocouple.

   Reference: DOE Fundamentals Handbook: Instrumentation and Control, Volume 1 — Module IC-01, page 2, RTD Construction

4. 4An engineer following the NSPE Code is asked to perform an independent design in a field in which the engineer has neither relevant education nor experience. Which response meets the Code's competence requirement?

   1. A. Accept independently if the client agrees to bear the risk.
   2. B. Accept independently if professional liability insurance applies.
   3. C. Decline independent responsibility until appropriately qualified.
   4. D. Accept independently if the schedule allows extra calculations.

   Answer: C

   Explanation: The NSPE Code limits services to "areas of their competence." Client consent, insurance, and extra time do not themselves establish the technical qualifications needed to take independent responsibility for this design.

   Reference: NSPE Code of Ethics for Engineers — II.2 and II.2.a, Rules of Practice

5. 5A payment of $1,210 occurs at the end of year 2. At a 10% annual compound discount rate, what is its present worth?

   1. A. $980
   2. B. $1,000
   3. C. $1,100
   4. D. $1,464

   Answer: B

   Explanation: NIST explains that the single present value factor calculates "the present value (PV) of a future cash amount (Ft) occurring at the end of year t given discount rate d." The factor is 1/(1 + d)ᵗ, so P = 1210/(1.10)² = $1,000.

   Reference: Life Cycle Costing Manual for the Federal Energy Management Program — Chapter 3, page 23, Figure 3-1, PV formula for one-time amounts

6. 6Two concurrent planar forces are F₁ = 3i + 4j N and F₂ = −i + 2j N. What is their resultant?

   1. A. 2i + 6j N
   2. B. 4i + 2j N
   3. C. −3i + 8j N
   4. D. 3i + 6j N

   Answer: A

   Explanation: DOE states, "Components of a vector are vectors, which when added, yield the vector." Add corresponding components: Rₓ = 3 − 1 = 2 N and Rᵧ = 4 + 2 = 6 N. Multiplying components or subtracting the second vector does not give the resultant.

   Reference: DOE Fundamentals Handbook: Classical Physics — Module CP-02, page 10, Vector Components

7. 7A straight bar carries a concentric tensile load of 40 kN. Its cross-sectional area is 200 mm². What is the average normal stress?

   1. A. 20 MPa
   2. B. 80 MPa
   3. C. 200 MPa
   4. D. 800 MPa

   Answer: C

   Explanation: The handbook defines stress as "the force (F) applied per cross-sectional area (A) perpendicular to the force". Thus σ = 40,000/200 = 200 N/mm² = 200 MPa.

   Reference: DOE Fundamentals Handbook: Material Science, Volume 1 — MS-02, Stress, page 2, Equation (2-1)

8. 8Which material property describes the capacity to undergo substantial plastic deformation in tension before rupture?

   1. A. Hardness
   2. B. Stiffness
   3. C. Ductility
   4. D. Toughness

   Answer: C

   Explanation: DOE defines ductility as "the ability of a material to withstand plastic deformation without rupture." Hardness concerns indentation resistance, stiffness concerns the elastic stress-strain relation, and toughness concerns energy absorbed before fracture. Plastic tensile deformation capacity is ductility.

   Reference: DOE Fundamentals Handbook: Material Science, Volume 1 — MS-02, Physical Properties, Ductility, page 24

9. 9A constant current transfers 12 coulombs of charge through a conductor in 3 seconds. What is the current?

   1. A. 0.25 A
   2. B. 4.0 A
   3. C. 9.0 A
   4. D. 36 A

   Answer: B

   Explanation: DOE defines an ampere as "the movement of one coulomb of charge past any given point of a conductor during one second of time." Thus I = Q/t = 12/3 = 4 A; multiplying charge by time would give the wrong quantity.

   Reference: DOE Fundamentals Handbook: Electrical Science, Volume 1 — ES-01, Electrical Terminology, Current, page 10

10. 10A closed system receives 500 kJ of heat and performs 200 kJ of work on its surroundings. Changes in kinetic and potential energy are negligible. What is the change in internal energy?

    1. A. A decrease of seven hundred kilojoules
    2. B. A decrease of three hundred kilojoules
    3. C. An increase of three hundred kilojoules
    4. D. An increase of seven hundred kilojoules

    Answer: C

    Explanation: DOE states, "Energy can neither be created nor destroyed, only altered in form." For this closed system, heat received adds energy and work delivered removes energy. Thus ΔU = Qin − Wout = 500 − 200 = +300 kJ.

    Reference: DOE Fundamentals Handbook: Thermodynamics, Heat Transfer, and Fluid Flow, Volume 1 — HT-01, First Law of Thermodynamics, page 53