Fundamentals of Engineering (FE) Exam
Questions And Correct Answers
(Verified Answers) Plus Rationales
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1. What is the derivative of f(x)=3x4−5x2+6f(x) = 3x^4 - 5x^2 +
6f(x)=3x4−5x2+6?
A. 12x3−10x12x^3 - 10x12x3−10x
B. 12x3−5x12x^3 - 5x12x3−5x
C. 3x3−10x3x^3 - 10x3x3−10x
D. x3−10xx^3 - 10xx3−10x
Answer: A
Rationale: The derivative of xnx^nxn is nxn−1nx^,n-1}nxn−1. So,
f′(x)=12x3−10x+0=12x3−10xf'(x) = 12x^3 - 10x + 0 = 12x^3 -
10xf′(x)=12x3−10x+0=12x3−10x.
,2. A 2 kg mass is moving with a velocity of 3 m/s. What is its kinetic
energy?
A. 3 J
B. 6 J
C. 9 J
D. 12 J
Rationale: Kinetic energy KE=12mv2=12(2)(3)2=9JKE = \frac{1}{2}mv^2 =
\frac{1}{2}(2)(3)^2 = 9 JKE=21mv2=21(2)(3)2=9J.
3. What is the unit of electric current?
A. Volt
B. Ampere
C. Ohm
D. Watt
Rationale: Electric current is measured in amperes (A).
4. The specific heat of water is approximately:
A. 0.5 cal/g°C
B. 0.8 cal/g°C
C. 1 cal/g°C
D. 2 cal/g°C
,Rationale: Water has a specific heat of 1 cal/g°C, meaning 1 calorie raises
1 gram of water by 1°C.
5. Which of the following is a vector quantity?
A. Speed
B. Velocity
C. Distance
D. Time
Rationale: Velocity has both magnitude and direction, making it a vector
quantity.
6. A simply supported beam has a length of 6 m and carries a point load of
12 kN at midspan. The maximum bending moment is:
A. 18 kNm
B. 36 kNm
C. 54 kNm
D. 72 kNm
Rationale: Maximum bending moment for a center point load on a simply
supported beam: M=P⋅L4=12⋅64=18kNmM = \frac{P \cdot L}{4} = \frac{12
\cdot 6}{4} = 18 kNmM=4P⋅L=412⋅6=18kNm.
(Oops! Corrected: 12∗6/4=1812*6/4 = 1812∗6/4=18, so answer should be A.
Fixed.)
, Answer: A
Rationale: Maximum bending moment at midspan for a simply supported
beam with a central point load is M=PL/4=12∗6/4=18kNmM = PL/4 =
12*6/4 = 18 kNmM=PL/4=12∗6/4=18kNm.
7. What is the SI unit of pressure?
A. Pascal
B. N/m
C. Pa (N/m²)
D. Bar
Rationale: Pressure is defined as force per unit area, P=F/AP = F/AP=F/A,
and SI unit is Pascal (Pa), equivalent to N/m².
8. In Ohm’s Law, V=IRV = IRV=IR. If V=12VV = 12 VV=12V and R=4ΩR = 4
\OmegaR=4Ω, the current III is:
A. 2 A
B. 3 A
C. 3 A
D. 4 A
Rationale: I=V/R=12/4=3AI = V/R = 12/4 = 3 AI=V/R=12/4=3A.