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FE — General FE Exam - General Practice Test 2025

FE — GeneralModel questions
110marks
360min
10questions

This is the official FE — General FE Exam - General question paper for 2025, as set in the Model questions examination. It carries 110 full marks and a time allowance of 360 minutes, across 10 questions. On Kekkei you can attempt this FE Exam - General past paper online with a timer, get instant AI feedback and step-by-step solutions, and track the topics where you lose marks — completely free. Whether you are revising for your FE — General FE Exam - General exam or solving previous years' question papers, this 2025 paper is a great way to practise under real exam conditions.

LevelFE — General
SubjectFE Exam - General
Year2025 BS
Exam sessionModel questions
Full marks110
Time allowed360 minutes
Questions10, all with step-by-step solutions
A

Fundamentals of Engineering

Select the best answer.

10 questions·1 mark each
1Multiple choice1 mark

What is the derivative of f(x)=3x4−5x2+7x−2f(x) = 3x^4 - 5x^2 + 7x - 2 with respect to xx?

  • a

    12x3−10x+712x^3 - 10x + 7

  • b

    12x3−5x+712x^3 - 5x + 7

  • c

    12x3−10x−212x^3 - 10x - 2

  • d

    3x3−10x+73x^3 - 10x + 7

Correct answer: a

12x3−10x+712x^3 - 10x + 7

Using the power rule ddx[xn]=nxn−1\dfrac{d}{dx}[x^n] = nx^{n-1}, we differentiate each term:

f′(x)=3⋅4x3−5⋅2x+7=12x3−10x+7.f'(x) = 3 \cdot 4x^3 - 5 \cdot 2x + 7 = 12x^3 - 10x + 7.

The constant term −2-2 differentiates to zero. Therefore f′(x)=12x3−10x+7f'(x) = 12x^3 - 10x + 7.

mathematicsderivatives
2Multiple choice1 mark

A manufacturing process produces bolts whose lengths are normally distributed with a mean of μ=50 mm\mu = 50\text{ mm} and a standard deviation of σ=2 mm\sigma = 2\text{ mm}. What is the approximate probability that a randomly selected bolt has a length between 46 mm46\text{ mm} and 54 mm54\text{ mm}?

  • a

    68.27%

  • b

    95.45%

  • c

    99.73%

  • d

    84.13%

Correct answer: b

95.45%

We need P(46<X<54)P(46 < X < 54). Converting to zz-scores:

z1=46−502=−2,z2=54−502=2.z_1 = \frac{46 - 50}{2} = -2, \quad z_2 = \frac{54 - 50}{2} = 2.

By the empirical rule (68-95-99.7 rule), approximately 95.45%95.45\% of values fall within ±2\pm 2 standard deviations of the mean.

probabilitystatisticsnormal-distribution
3Multiple choice1 mark

A project requires an initial investment of \10{,}000andwillgenerateannualreturnsofand will generate annual returns of$3{,}000attheendofeachyearfor5years.Iftheinterestrateisat the end of each year for 5 years. If the interest rate is8%peryear,whatisthenetpresentworth(NPW)ofthisproject?Usetheuniformseriespresentworthfactorper year, what is the net present worth (NPW) of this project? Use the uniform series present worth factor\left(\dfrac{(1+i)^n - 1}{i(1+i)^n}\right)$.

  • a

    \5{,}000$

  • b

    \1{,}978$

  • c

    -\1{,}200$

  • d

    \3{,}520$

Correct answer: b

\1{,}978$

The present worth of the annual returns is:

PW=A⋅(1+i)n−1i(1+i)n=3000⋅(1.08)5−10.08(1.08)5.PW = A \cdot \frac{(1+i)^n - 1}{i(1+i)^n} = 3000 \cdot \frac{(1.08)^5 - 1}{0.08(1.08)^5}.

Compute (1.08)5=1.46933(1.08)^5 = 1.46933. Then:

PW=3000⋅0.469330.11755=3000×3.9927=11,978.PW = 3000 \cdot \frac{0.46933}{0.11755} = 3000 \times 3.9927 = 11{,}978.

Net present worth: NPW = 11{,}978 - 10{,}000 = \1{,}978$.

engineering-economicspresent-worth
4Multiple choice1 mark

A particle is in static equilibrium under the action of three coplanar forces. Force F1=100 NF_1 = 100\text{ N} acts along the positive xx-axis, and force F2=80 NF_2 = 80\text{ N} acts along the positive yy-axis. What is the magnitude of the third force F3F_3 required to maintain equilibrium?

  • a

    180 N180\text{ N}

  • b

    128.1 N128.1\text{ N}

  • c

    90 N90\text{ N}

  • d

    20 N20\text{ N}

Correct answer: b

128.1 N128.1\text{ N}

For static equilibrium: F⃗1+F⃗2+F⃗3=0⃗\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}, so F⃗3=−(F⃗1+F⃗2)\vec{F}_3 = -(\vec{F}_1 + \vec{F}_2). The magnitude is ∣F3∣=1002+802=16400≈128.1 N|F_3| = \sqrt{100^2 + 80^2} = \sqrt{16400} \approx 128.1\text{ N}.

staticsforce-equilibrium
5Multiple choice1 mark

A 5 kg5\text{ kg} block rests on a frictionless horizontal surface. A horizontal force of F=20 NF = 20\text{ N} is applied to the block. What is the acceleration of the block?

  • a

    2 m/s22\text{ m/s}^2

  • b

    100 m/s2100\text{ m/s}^2

  • c

    4 m/s24\text{ m/s}^2

  • d

    0.25 m/s20.25\text{ m/s}^2

Correct answer: c

4 m/s24\text{ m/s}^2

By Newton's second law, F=maF = ma, so a=F/m=20/5=4 m/s2a = F/m = 20/5 = 4\text{ m/s}^2. Since the surface is frictionless, there are no other horizontal forces.

dynamicsnewtons-second-law
6Multiple choice1 mark

Water flows through a horizontal pipe that narrows from a cross-sectional area of A1=0.02 m2A_1 = 0.02\text{ m}^2 to A2=0.01 m2A_2 = 0.01\text{ m}^2. If the velocity at the wider section is v1=3 m/sv_1 = 3\text{ m/s} and the pressure there is P1=250 kPaP_1 = 250\text{ kPa}, what is the pressure P2P_2 at the narrower section? Assume the density of water is ρ=1000 kg/m3\rho = 1000\text{ kg/m}^3 and the flow is ideal (incompressible, inviscid, steady).

  • a

    236.5 kPa236.5\text{ kPa}

  • b

    263.5 kPa263.5\text{ kPa}

  • c

    250.0 kPa250.0\text{ kPa}

  • d

    223.0 kPa223.0\text{ kPa}

Correct answer: a

236.5 kPa236.5\text{ kPa}

Using the continuity equation: v2=A1v1/A2=0.02×3/0.01=6 m/sv_2 = A_1 v_1 / A_2 = 0.02 \times 3 / 0.01 = 6\text{ m/s}.

Apply Bernoulli's equation:

P2=P1+12ρ(v12−v22)=250,000+500(9−36)=250,000−13,500=236,500 Pa=236.5 kPa.P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) = 250{,}000 + 500(9 - 36) = 250{,}000 - 13{,}500 = 236{,}500\text{ Pa} = 236.5\text{ kPa}.
fluid-mechanicsbernoullis-equation
7Multiple choice1 mark

A closed system undergoes a process in which 150 kJ150\text{ kJ} of heat is added to the system and the system performs 90 kJ90\text{ kJ} of work on the surroundings. What is the change in internal energy ΔU\Delta U of the system?

  • a

    240 kJ240\text{ kJ}

  • b

    −60 kJ-60\text{ kJ}

  • c

    60 kJ60\text{ kJ}

  • d

    150 kJ150\text{ kJ}

Correct answer: c

60 kJ60\text{ kJ}

The first law of thermodynamics for a closed system: ΔU=Q−W=150−90=60 kJ\Delta U = Q - W = 150 - 90 = 60\text{ kJ}. The internal energy increases by 60 kJ60\text{ kJ}.

thermodynamicsfirst-lawenergy-balance
8Multiple choice1 mark

A steel rod with a cross-sectional area of A=200 mm2A = 200\text{ mm}^2 and a Young's modulus of E=200 GPaE = 200\text{ GPa} is subjected to a tensile load of P=40 kNP = 40\text{ kN}. What is the normal stress in the rod and the resulting strain?

  • a

    σ=200 MPa,ε=0.001\sigma = 200\text{ MPa}, \varepsilon = 0.001

  • b

    σ=20 MPa,ε=0.0001\sigma = 20\text{ MPa}, \varepsilon = 0.0001

  • c

    σ=200 MPa,ε=0.01\sigma = 200\text{ MPa}, \varepsilon = 0.01

  • d

    σ=2000 MPa,ε=0.01\sigma = 2000\text{ MPa}, \varepsilon = 0.01

Correct answer: a

σ=200 MPa,ε=0.001\sigma = 200\text{ MPa}, \varepsilon = 0.001

Normal stress: σ=P/A=40×103/(200×10−6)=200 MPa\sigma = P/A = 40 \times 10^3 / (200 \times 10^{-6}) = 200\text{ MPa}.

Strain via Hooke's law: ε=σ/E=200×106/(200×109)=0.001\varepsilon = \sigma/E = 200 \times 10^6 / (200 \times 10^9) = 0.001.

materials-sciencestress-strain
9Multiple choice1 mark

An engineer working for a consulting firm discovers that a client's proposed building design has a structural flaw that could pose a safety risk to occupants. The client insists on proceeding without modifications to save costs. According to the NSPE Code of Ethics, what should the engineer do?

  • a

    Comply with the client's wishes since the client is paying for the project

  • b

    Refuse to approve the design and notify appropriate authorities if the client will not correct the flaw

  • c

    Quietly resign from the project without informing anyone

  • d

    Approve the design but add a written disclaimer limiting personal liability

Correct answer: b

Refuse to approve the design and notify appropriate authorities if the client will not correct the flaw

The NSPE Code of Ethics holds that engineers shall hold paramount the safety, health, and welfare of the public (Fundamental Canon No. 1). The engineer must refuse to approve the design and, if necessary, notify the appropriate authorities. Complying with the client, resigning quietly, or adding a disclaimer would all fail to protect public safety.

ethicsnspe-code-of-ethics
10Multiple choice1 mark

In a simple DC circuit, a 12 V12\text{ V} battery is connected in series with three resistors: R1=2  ΩR_1 = 2\;\Omega, R2=4  ΩR_2 = 4\;\Omega, and R3=6  ΩR_3 = 6\;\Omega. What is the current flowing through the circuit and the voltage drop across R2R_2?

  • a

    I=1 A,V2=4 VI = 1\text{ A}, V_2 = 4\text{ V}

  • b

    I=3 A,V2=12 VI = 3\text{ A}, V_2 = 12\text{ V}

  • c

    I=1 A,V2=6 VI = 1\text{ A}, V_2 = 6\text{ V}

  • d

    I=2 A,V2=8 VI = 2\text{ A}, V_2 = 8\text{ V}

Correct answer: a

I=1 A,V2=4 VI = 1\text{ A}, V_2 = 4\text{ V}

Total resistance: Rtotal=2+4+6=12  ΩR_{\text{total}} = 2 + 4 + 6 = 12\;\Omega.

Current: I=V/Rtotal=12/12=1 AI = V/R_{\text{total}} = 12/12 = 1\text{ A}.

Voltage across R2R_2: V2=I×R2=1×4=4 VV_2 = I \times R_2 = 1 \times 4 = 4\text{ V}.

electricityohms-lawcircuit-analysis

Frequently asked questions

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The full FE — General FE Exam - General 2025 (Model questions) question paper is available free on Kekkei. You can read every question online and attempt the paper under timed exam conditions.
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Yes. Every question on this FE Exam - General past paper includes a step-by-step solution, plus instant AI feedback when you attempt it on Kekkei.
How many marks is the FE — General FE Exam - General 2025 paper?
The FE — General FE Exam - General 2025 paper carries 110 full marks and is meant to be completed in 360 minutes, across 10 questions.
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Yes — reading and attempting this FE Exam - General past paper on Kekkei is completely free.