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

Mechanical Engineering

Select the best answer.

10 questions·1 mark each
1Multiple choice1 mark

An ideal Carnot engine operates between TH=800KT_H = 800\,\text{K} and TC=300KT_C = 300\,\text{K}. The thermal efficiency is:

η=1TCTH\eta = 1 - \frac{T_C}{T_H}
  • a

    37.5%37.5\%

  • b

    50.0%50.0\%

  • c

    62.5%62.5\%

  • d

    75.0%75.0\%

Correct answer: c

62.5%62.5\%

η=1300800=0.625=62.5%\eta = 1 - \frac{300}{800} = 0.625 = 62.5\%
thermodynamics
2Multiple choice1 mark

A plane wall: L=0.2mL = 0.2\,\text{m}, k=50W/(m\cdotK)k = 50\,\text{W/(m\cdot K)}, T1=400KT_1 = 400\,\text{K}, T2=300KT_2 = 300\,\text{K}. Steady-state heat flux:

q=kT1T2Lq'' = k \frac{T_1 - T_2}{L}
  • a

    10,000W/m210{,}000\,\text{W/m}^2

  • b

    25,000W/m225{,}000\,\text{W/m}^2

  • c

    50,000W/m250{,}000\,\text{W/m}^2

  • d

    100,000W/m2100{,}000\,\text{W/m}^2

Correct answer: b

25,000W/m225{,}000\,\text{W/m}^2

q=50×1000.2=25,000W/m2q'' = 50 \times \frac{100}{0.2} = 25{,}000\,\text{W/m}^2
heat-transfer
3Multiple choice1 mark

Reynolds number Re=ρvDμRe = \frac{\rho v D}{\mu} for water (ρ=1000\rho = 1000, μ=103Pa\cdots\mu = 10^{-3}\,\text{Pa\cdot s}) at v=2m/sv = 2\,\text{m/s}, D=0.05mD = 0.05\,\text{m}. The flow regime is:

  • a

    Laminar (Re<2300Re < 2300)

  • b

    Transitional (2300<Re<40002300 < Re < 4000)

  • c

    Turbulent (Re>4000Re > 4000)

  • d

    Turbulent (Re=100,000Re = 100{,}000)

Correct answer: d

Turbulent (Re=100,000Re = 100{,}000)

Re=1000×2×0.05103=100,0004,000Re = \frac{1000 \times 2 \times 0.05}{10^{-3}} = 100{,}000 \gg 4{,}000

Fully turbulent flow.

fluid-mechanics
4Multiple choice1 mark

A truss member: F=50kNF = 50\,\text{kN}, A=500mm2A = 500\,\text{mm}^2. Normal stress:

σ=FA\sigma = \frac{F}{A}
  • a

    50MPa50\,\text{MPa}

  • b

    100MPa100\,\text{MPa}

  • c

    150MPa150\,\text{MPa}

  • d

    200MPa200\,\text{MPa}

Correct answer: b

100MPa100\,\text{MPa}

σ=50×103500×106=100MPa\sigma = \frac{50 \times 10^3}{500 \times 10^{-6}} = 100\,\text{MPa}
statics
5Multiple choice1 mark

A 10 kg block slides down a frictionless incline of height h=5mh = 5\,\text{m}. Velocity at bottom (g=9.81g = 9.81):

v=2ghv = \sqrt{2gh}
  • a

    7.0m/s7.0\,\text{m/s}

  • b

    9.9m/s9.9\,\text{m/s}

  • c

    14.0m/s14.0\,\text{m/s}

  • d

    19.6m/s19.6\,\text{m/s}

Correct answer: b

9.9m/s9.9\,\text{m/s}

v=2×9.81×5=98.19.9m/sv = \sqrt{2 \times 9.81 \times 5} = \sqrt{98.1} \approx 9.9\,\text{m/s}
dynamics
6Multiple choice1 mark

Steel rod: E=200GPaE = 200\,\text{GPa}, L=2mL = 2\,\text{m}, A=400mm2A = 400\,\text{mm}^2, P=80kNP = 80\,\text{kN}. Elongation:

δ=PLAE\delta = \frac{PL}{AE}
  • a

    1.0mm1.0\,\text{mm}

  • b

    2.0mm2.0\,\text{mm}

  • c

    4.0mm4.0\,\text{mm}

  • d

    8.0mm8.0\,\text{mm}

Correct answer: b

2.0mm2.0\,\text{mm}

δ=80,000×2400×106×200×109=0.002m=2.0mm\delta = \frac{80{,}000 \times 2}{400 \times 10^{-6} \times 200 \times 10^9} = 0.002\,\text{m} = 2.0\,\text{mm}
materials
7Multiple choice1 mark

Shaft transmits P=10kWP = 10\,\text{kW} at n=1500rpmn = 1500\,\text{rpm}. Torque:

T=P2πn/60T = \frac{P}{2\pi n / 60}
  • a

    31.8N\cdotm31.8\,\text{N\cdot m}

  • b

    63.7N\cdotm63.7\,\text{N\cdot m}

  • c

    95.5N\cdotm95.5\,\text{N\cdot m}

  • d

    127.3N\cdotm127.3\,\text{N\cdot m}

Correct answer: b

63.7N\cdotm63.7\,\text{N\cdot m}

T=10,0002π×1500/60=10,000157.08=63.7N\cdotmT = \frac{10{,}000}{2\pi \times 1500/60} = \frac{10{,}000}{157.08} = 63.7\,\text{N\cdot m}
machine-design
8Multiple choice1 mark

Ideal gas isothermal expansion: V1=0.1m3V_1 = 0.1\,\text{m}^3, V2=0.4m3V_2 = 0.4\,\text{m}^3, T=300KT = 300\,\text{K}, n=1moln = 1\,\text{mol}, R=8.314R = 8.314. Work:

W=nRTlnV2V1W = nRT \ln\frac{V_2}{V_1}
  • a

    1,729J1{,}729\,\text{J}

  • b

    3,458J3{,}458\,\text{J}

  • c

    4,988J4{,}988\,\text{J}

  • d

    6,915J6{,}915\,\text{J}

Correct answer: b

3,458J3{,}458\,\text{J}

W=1×8.314×300×ln4=2494.2×1.3863=3,458JW = 1 \times 8.314 \times 300 \times \ln 4 = 2494.2 \times 1.3863 = 3{,}458\,\text{J}
thermodynamics
9Multiple choice1 mark

Cylindrical fin: d=0.02md = 0.02\,\text{m}, k=400W/(m\cdotK)k = 400\,\text{W/(m\cdot K)}, h=20W/(m2\cdotK)h = 20\,\text{W/(m}^2\text{\cdot K)}. Fin parameter:

m=4hkdm = \sqrt{\frac{4h}{kd}}
  • a

    2.24m12.24\,\text{m}^{-1}

  • b

    3.16m13.16\,\text{m}^{-1}

  • c

    4.47m14.47\,\text{m}^{-1}

  • d

    6.32m16.32\,\text{m}^{-1}

Correct answer: b

3.16m13.16\,\text{m}^{-1}

m=4×20400×0.02=808=103.16m1m = \sqrt{\frac{4 \times 20}{400 \times 0.02}} = \sqrt{\frac{80}{8}} = \sqrt{10} \approx 3.16\,\text{m}^{-1}
heat-transfer
10Multiple choice1 mark

Mercury manometer (ρm=13,600\rho_m = 13{,}600), water pipe (ρw=1,000\rho_w = 1{,}000), Δh=0.15m\Delta h = 0.15\,\text{m}, g=9.81g = 9.81:

ΔP=(ρmρw)gΔh\Delta P = (\rho_m - \rho_w) g \Delta h
  • a

    10.4kPa10.4\,\text{kPa}

  • b

    18.5kPa18.5\,\text{kPa}

  • c

    20.0kPa20.0\,\text{kPa}

  • d

    25.1kPa25.1\,\text{kPa}

Correct answer: b

18.5kPa18.5\,\text{kPa}

ΔP=12,600×9.81×0.15=18,541Pa18.5kPa\Delta P = 12{,}600 \times 9.81 \times 0.15 = 18{,}541\,\text{Pa} \approx 18.5\,\text{kPa}
fluid-mechanics

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Yes. Every question on this FE — Mechanical past paper includes a step-by-step solution, plus instant AI feedback when you attempt it on Kekkei.
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The FE — Mechanical FE — Mechanical 2025 paper carries 110 full marks and is meant to be completed in 360 minutes, across 10 questions.
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