Professional Engineer (PE) Exam Prep 2026 Comprehensive
Study Guide with 100 Practice Questions, Detailed
Explanations, and Verified Answers
SECTION 1: ENGINEERING FUNDAMENTALS (30 Questions)
Question 1
A simply supported beam of length L = 8 m carries a uniform load of w = 12
kN/m. What is the maximum bending moment?
A) 96 kN·m
B) 48 kN·m
C) 192 kN·m
D) 384 kN·m
Answer: A
Rationale: For a simply supported beam with a uniform load, the maximum
bending moment occurs at midspan and is calculated as: M_max = wL²/8.
Substituting values: M_max = (12 kN/m × 8² m²) / 8 = (12 × 64) / 8 = = 96
kN·m. Option B (48 kN·m) would be wL²/16. Option C (192 kN·m) would be
wL²/4 (fixed beam). Option D (384 kN·m) would be wL²/2 (cantilever beam with
uniform load).
Question 2
A steel rod with an original length of 2.0 m and a cross-sectional area of 200
mm² is subjected to a tensile load of 50 kN. The modulus of elasticity is 200 GPa.
What is the elongation of the rod?
A) 0.5 mm
B) 2.5 mm
C) 0.25 mm
D) 5.0 mm
Answer: B
,Rationale: Elongation is calculated using ΔL = (P × L) / (A × E) . Convert units:
A = 200 mm² = 200 × 10⁻⁶ m²; E = 200 GPa = 200 × 10⁹ Pa. ΔL = (50,000 N × 2.0
m) / (200 × 10⁻⁶ m² × 200 × 10⁹ Pa) = 100,,000,000 = 0.0025 m = 2.5 mm.
Option A (0.5 mm) would result from using E = 1,000 GPa. Option C (0.25 mm)
would result from using A = 2,000 mm². Option D (5.0 mm) would result from
using E = 100 GPa.
Question 3
The pressure at a depth of 10 m in water (density = 1,000 kg/m³) is most nearly:
A) 98.1 kPa
B) 9.81 kPa
C) 981 kPa
D) 1.0 kPa
Answer: A
Rationale: Hydrostatic pressure is calculated using P = ρgh. P = 1,000 kg/m³ ×
9.81 m/s² × 10 m = 98,100 Pa = 98.1 kPa. Option B (9.81 kPa) would be for a
depth of 1 m. Option C (981 kPa) would be for a depth of 100 m. Option D (1.0
kPa) is incorrect.
Question 4
A pump delivers 0.5 m³/s of water against a head of 40 m. The pump efficiency
is 80%. What is the power required by the pump? (γ_water = 9.81 kN/m³)
A) 196 kW
B) 245 kW
C) 157 kW
D) 392 kW
Answer: B
Rationale: Hydraulic power = γ × Q × H = 9.81 kN/m³ × 0.5 m³/s × 40 m = 196.2
kW. Pump power = Hydraulic power / Efficiency = 196..80 = 245.25 kW.
Option A (196 kW) is the hydraulic power without accounting for efficiency.
,Option C (157 kW) would result from dividing by 1.25. Option D (392 kW) would
result from dividing by 0.50.
Question 5
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300
K. What is the maximum possible thermal efficiency?
A) 50%
B) 33%
C) 67%
D) 100%
Answer: A
Rationale: The maximum possible thermal efficiency is given by the Carnot
efficiency: η_max = 1 - T_C / T_H = 1 - 300/600 = 0.50 = 50% . Option B (33%)
would be 1 - 400/600. Option C (67%) would be 1 - 200/600. Option D (100%) is
impossible.
Question 6
A project has an initial cost of $100,000** and generates annual net cash flows
of **$25,000 for 6 years. The discount rate is 10%. What is the approximate Net
Present Value (NPV)?
(Present value annuity factor for 10%, 6 years = 4.355)
A) $8,875
B) $50,000
C) $108,875
D) $150,000
Answer: A
Rationale: NPV = -Initial Cost + (Annual Cash Flow × PVAF). NPV = -$100,000
+ ($25,000 × 4.355) = -$100,000 + $108,875 = **$8,875**. Since NPV > 0, the
project is acceptable. Option B ($50,000) is the undiscounted total net cash flow.
Option C ($108,875) is the present value of cash flows without subtracting initial
cost. Option D ($150,000) is incorrect.
, Question 7
The first law of thermodynamics is best described as:
A) Energy cannot be created or destroyed, only converted from one form to
another
B) Entropy of an isolated system always increases
C) Heat flows from hot to cold
D) Pressure and volume are inversely proportional
Answer: A
Rationale: The first law of thermodynamics is the conservation of energy
principle: energy cannot be created or destroyed, only converted from one form to
another. Option B describes the second law of thermodynamics (entropy). Option
C is a consequence of the second law. Option D describes Boyle's law.
Question 8
A pipe with a diameter of 0.3 m carries water at a velocity of 2.5 m/s. What is the
flow rate?
A) 0.177 m³/s
B) 0.707 m³/s
C) 0.236 m³/s
D) 0.088 m³/s
Answer: A
Rationale: Flow rate Q = A × V. Area A = π × (D/2)² = π × (0.15)² = 0.0707 m². Q
= 0.0707 m² × 2.5 m/s = 0.177 m³/s. Option B (0.707 m³/s) would be for D = 0.6
m. Option C (0.236 m³/s) would be for V = 3.33 m/s. Option D (0.088 m³/s) would
be for V = 1.25 m/s.
Question 9
A 10 cm diameter solid steel shaft transmits 50 kW at 500 rpm. What is the
approximate torque?
Study Guide with 100 Practice Questions, Detailed
Explanations, and Verified Answers
SECTION 1: ENGINEERING FUNDAMENTALS (30 Questions)
Question 1
A simply supported beam of length L = 8 m carries a uniform load of w = 12
kN/m. What is the maximum bending moment?
A) 96 kN·m
B) 48 kN·m
C) 192 kN·m
D) 384 kN·m
Answer: A
Rationale: For a simply supported beam with a uniform load, the maximum
bending moment occurs at midspan and is calculated as: M_max = wL²/8.
Substituting values: M_max = (12 kN/m × 8² m²) / 8 = (12 × 64) / 8 = = 96
kN·m. Option B (48 kN·m) would be wL²/16. Option C (192 kN·m) would be
wL²/4 (fixed beam). Option D (384 kN·m) would be wL²/2 (cantilever beam with
uniform load).
Question 2
A steel rod with an original length of 2.0 m and a cross-sectional area of 200
mm² is subjected to a tensile load of 50 kN. The modulus of elasticity is 200 GPa.
What is the elongation of the rod?
A) 0.5 mm
B) 2.5 mm
C) 0.25 mm
D) 5.0 mm
Answer: B
,Rationale: Elongation is calculated using ΔL = (P × L) / (A × E) . Convert units:
A = 200 mm² = 200 × 10⁻⁶ m²; E = 200 GPa = 200 × 10⁹ Pa. ΔL = (50,000 N × 2.0
m) / (200 × 10⁻⁶ m² × 200 × 10⁹ Pa) = 100,,000,000 = 0.0025 m = 2.5 mm.
Option A (0.5 mm) would result from using E = 1,000 GPa. Option C (0.25 mm)
would result from using A = 2,000 mm². Option D (5.0 mm) would result from
using E = 100 GPa.
Question 3
The pressure at a depth of 10 m in water (density = 1,000 kg/m³) is most nearly:
A) 98.1 kPa
B) 9.81 kPa
C) 981 kPa
D) 1.0 kPa
Answer: A
Rationale: Hydrostatic pressure is calculated using P = ρgh. P = 1,000 kg/m³ ×
9.81 m/s² × 10 m = 98,100 Pa = 98.1 kPa. Option B (9.81 kPa) would be for a
depth of 1 m. Option C (981 kPa) would be for a depth of 100 m. Option D (1.0
kPa) is incorrect.
Question 4
A pump delivers 0.5 m³/s of water against a head of 40 m. The pump efficiency
is 80%. What is the power required by the pump? (γ_water = 9.81 kN/m³)
A) 196 kW
B) 245 kW
C) 157 kW
D) 392 kW
Answer: B
Rationale: Hydraulic power = γ × Q × H = 9.81 kN/m³ × 0.5 m³/s × 40 m = 196.2
kW. Pump power = Hydraulic power / Efficiency = 196..80 = 245.25 kW.
Option A (196 kW) is the hydraulic power without accounting for efficiency.
,Option C (157 kW) would result from dividing by 1.25. Option D (392 kW) would
result from dividing by 0.50.
Question 5
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300
K. What is the maximum possible thermal efficiency?
A) 50%
B) 33%
C) 67%
D) 100%
Answer: A
Rationale: The maximum possible thermal efficiency is given by the Carnot
efficiency: η_max = 1 - T_C / T_H = 1 - 300/600 = 0.50 = 50% . Option B (33%)
would be 1 - 400/600. Option C (67%) would be 1 - 200/600. Option D (100%) is
impossible.
Question 6
A project has an initial cost of $100,000** and generates annual net cash flows
of **$25,000 for 6 years. The discount rate is 10%. What is the approximate Net
Present Value (NPV)?
(Present value annuity factor for 10%, 6 years = 4.355)
A) $8,875
B) $50,000
C) $108,875
D) $150,000
Answer: A
Rationale: NPV = -Initial Cost + (Annual Cash Flow × PVAF). NPV = -$100,000
+ ($25,000 × 4.355) = -$100,000 + $108,875 = **$8,875**. Since NPV > 0, the
project is acceptable. Option B ($50,000) is the undiscounted total net cash flow.
Option C ($108,875) is the present value of cash flows without subtracting initial
cost. Option D ($150,000) is incorrect.
, Question 7
The first law of thermodynamics is best described as:
A) Energy cannot be created or destroyed, only converted from one form to
another
B) Entropy of an isolated system always increases
C) Heat flows from hot to cold
D) Pressure and volume are inversely proportional
Answer: A
Rationale: The first law of thermodynamics is the conservation of energy
principle: energy cannot be created or destroyed, only converted from one form to
another. Option B describes the second law of thermodynamics (entropy). Option
C is a consequence of the second law. Option D describes Boyle's law.
Question 8
A pipe with a diameter of 0.3 m carries water at a velocity of 2.5 m/s. What is the
flow rate?
A) 0.177 m³/s
B) 0.707 m³/s
C) 0.236 m³/s
D) 0.088 m³/s
Answer: A
Rationale: Flow rate Q = A × V. Area A = π × (D/2)² = π × (0.15)² = 0.0707 m². Q
= 0.0707 m² × 2.5 m/s = 0.177 m³/s. Option B (0.707 m³/s) would be for D = 0.6
m. Option C (0.236 m³/s) would be for V = 3.33 m/s. Option D (0.088 m³/s) would
be for V = 1.25 m/s.
Question 9
A 10 cm diameter solid steel shaft transmits 50 kW at 500 rpm. What is the
approximate torque?
