100+ Free IESL Part II Mechanical Practice Questions

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2026 Statistics

Key Facts: IESL Part II Mechanical Exam

40%

Passing Score

IESL exam regulations

5 Core

Subject Sections

IESL Part II mechanical syllabus

LKR 5,000+

Historical Fee (per paper)

IESL finance desk

AMIESL

Leads To (Academic Route)

Institution of Engineers, Sri Lanka

Dec 2022

Phased Out Date

IESL official council decision

The IESL Qualifying Examination Part II - Mechanical Engineering is a landmark credential for mechanical engineers in Sri Lanka seeking professional recognition. It comprises comprehensive examinations in thermodynamics, fluid mechanics, strength of materials, machine design, and manufacturing technology. Earning AMIESL eligibility requires passing all core sections with a minimum mark of 40%. Although the traditional qualifying exam was phased out in 2022, understanding these core competencies remains vital for the current IESL academic qualification assessment routes.

Sample IESL Part II Mechanical Practice Questions

Try these sample questions to test your IESL Part II Mechanical exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A closed thermodynamic system undergoes a process during which 50 kJ of heat is added to the system, and the system performs 30 kJ of work on its surroundings. According to the First Law of Thermodynamics, what is the change in the internal energy of the system?

A.80 kJ

B.20 kJ

C.-20 kJ

D.50 kJ

Explanation: The First Law of Thermodynamics for a closed system is expressed as Q = delta_U + W, where Q is heat added, delta_U is change in internal energy, and W is work done by the system. Rearranging the formula gives delta_U = Q - W. Substituting the given values: delta_U = 50 kJ - 30 kJ = 20 kJ.

2Air enters an insulated compressor operating at steady state at 100 kPa and 20°C (293 K) with negligible velocity, and exits at 1 MPa and 340°C (613 K) with a velocity of 80 m/s. The mass flow rate of air is 2 kg/s. Assuming air behaves as an ideal gas with Cp = 1.005 kJ/kg·K, calculate the power input required by the compressor.

A.321.6 kW

B.643.2 kW

C.649.6 kW

D.652.8 kW

Explanation: Applying the Steady Flow Energy Equation (SFEE) under adiabatic conditions gives the power input: W_dot = m_dot * [Cp * (T2 - T1) + (V2^2 - V1^2) / 2000]. The temperature difference is T2 - T1 = 340°C - 20°C = 320 K. Calculating the enthalpy change: Cp * delta_T = 1.005 * 320 = 321.6 kJ/kg. The kinetic energy change is 80^2 / 2000 = 3.2 kJ/kg. The total specific work is 321.6 + 3.2 = 324.8 kJ/kg. Multiplying by the mass flow rate of 2 kg/s yields W_dot = 2 * 324.8 = 649.6 kW.

3A piston-cylinder device contains 0.1 kg of air initially at a pressure of 2 MPa and temperature of 350°C (623 K). The air undergoes a polytropic expansion to a final pressure of 0.2 MPa. If the polytropic index is n = 1.3, calculate the boundary work done by the gas during this process. (Take the gas constant R = 0.287 kJ/kg·K).

A.18.5 kJ

B.24.6 kJ

C.32.1 kJ

D.41.2 kJ

Explanation: For a polytropic process, the final temperature T2 is calculated using T2 = T1 * (P2 / P1)^((n-1)/n) = 623 * (0.2 / 2)^((1.3-1)/1.3) = 623 * 0.1^0.2308 = 366.2 K. The boundary work done is W = m * R * (T1 - T2) / (n - 1). Substituting the values: W = 0.1 * 0.287 * (623 - 366.2) / (1.3 - 1) = 0.0287 * 256.8 / 0.3 = 24.57 kJ.

4A power plant heat engine operates between a high-temperature thermal reservoir at 600°C and a low-temperature sink reservoir at 25°C. What is the maximum theoretical thermal efficiency of this engine?

A.95.8%

B.65.9%

C.50.0%

D.34.1%

Explanation: The maximum theoretical efficiency of any heat engine operating between two temperatures is given by the Carnot efficiency: eta_max = 1 - TC / TH, where TC and TH are the absolute temperatures in Kelvin. Converting temperatures: TH = 600 + 273.15 = 873.15 K, and TC = 25 + 273.15 = 298.15 K. Calculating the efficiency: eta_max = 1 - 298.15 / 873.15 = 1 - 0.3415 = 0.6585, which is 65.9%.

5A Carnot refrigeration system operates in a room where the temperature is maintained at 27°C (300 K). If the refrigerator is required to maintain its cold space temperature at -13°C (260 K), calculate its Coefficient of Performance (COP).

A.1.5

B.5.2

C.6.5

D.7.5

Explanation: The Coefficient of Performance of a Carnot refrigerator is determined solely by the absolute temperatures of the cold space (TC) and the hot sink (TH): COP = TC / (TH - TC). Converting the temperatures to Kelvin: TC = -13 + 273 = 260 K, and TH = 27 + 273 = 300 K. Calculating COP: COP = 260 / (300 - 260) = 260 / 40 = 6.5.

6An inventor claims to have developed a cyclic heat engine that receives 800 kJ of heat from a thermal source at 400 K, produces 300 kJ of net mechanical work, and rejects 500 kJ of heat to a low-temperature sink at 290 K. Evaluate the feasibility of this engine.

A.Feasible, as it satisfies the First Law of Thermodynamics.

B.Feasible, because the entropy change of the universe is positive.

C.Infeasible, because it violates the Clausius Statement by rejecting heat to a lower temperature.

D.Infeasible, because its thermal efficiency exceeds the maximum theoretical limit (Carnot limit) for the given temperature boundaries.

Explanation: First, the claimed efficiency of the engine is eta = W / QH = 300 / 800 = 0.375 (37.5%). The maximum possible efficiency between these two reservoirs is the Carnot efficiency: eta_Carnot = 1 - TC / TH = 1 - 290 / 400 = 1 - 0.725 = 0.275 (27.5%). Since the claimed thermal efficiency (37.5%) is greater than the maximum theoretical efficiency (27.5%), it violates the Kelvin-Planck statement of the Second Law and is impossible.

7Two kilograms of nitrogen gas (assumed to be an ideal gas with gas constant R = 0.297 kJ/kg·K) undergo an isothermal expansion from an initial volume of 0.8 m³ to a final volume of 2.4 m³. Calculate the change in entropy of the nitrogen gas during this process.

A.0.326 kJ/K

B.0.653 kJ/K

C.0.980 kJ/K

D.1.306 kJ/K

Explanation: For an isothermal process of an ideal gas, the change in entropy is given by delta_S = m * R * ln(V2 / V1). Substituting the given values: delta_S = 2 * 0.297 * ln(2.4 / 0.8) = 0.594 * ln(3) = 0.594 * 1.0986 = 0.6526 kJ/K.

8A solid copper block of mass 5 kg initially at 100°C (373 K) is placed in an insulated tank containing 10 kg of liquid water initially at 20°C (293 K). Find the total entropy generated by this heat transfer process when the system reaches thermal equilibrium. (Take specific heat of copper CCu = 0.385 kJ/kg·K and water Cw = 4.184 kJ/kg·K).

A.12.4 J/K

B.34.2 J/K

C.57.9 J/K

D.92.1 J/K

Explanation: First, find the final equilibrium temperature Tf using an energy balance: m_Cu * C_Cu * (Ti_Cu - Tf) = m_w * C_w * (Tf - Ti_w) => 5 * 0.385 * (373 - Tf) = 10 * 4.184 * (Tf - 293) => 1.925 * (373 - Tf) = 41.84 * (Tf - 293) => Tf = 296.52 K. The entropy change of the copper block is delta_S_Cu = m_Cu * C_Cu * ln(Tf / Ti_Cu) = 5 * 0.385 * ln(296.52 / 373) = -0.4417 kJ/K. The entropy change of the water is delta_S_w = m_w * C_w * ln(Tf / Ti_w) = 10 * 4.184 * ln(296.52 / 293) = 0.4996 kJ/K. The total entropy generated is S_gen = delta_S_Cu + delta_S_w = -0.4417 + 0.4996 = +0.0579 kJ/K = 57.9 J/K.

9According to the Clausius inequality, for any thermodynamic cycle, what must be the mathematical value of the cyclic integral of the quantity dQ / T?

A.Always greater than zero

B.Less than or equal to zero

C.Always equal to zero

D.Greater than or equal to zero

Explanation: The Clausius inequality states that for any cycle, the cyclic integral of dQ / T is less than or equal to zero: oint(dQ / T) <= 0. The equality holds for internally reversible cycles, while the inequality holds for irreversible cycles. A cyclic integral greater than zero would violate the second law of thermodynamics.

10An ideal gas-turbine power plant operates on the air-standard Brayton cycle with a pressure ratio of 8. The air temperature at the compressor inlet is 300 K, and the temperature at the turbine inlet is 1300 K. Assuming constant specific heats with gamma = 1.4, calculate the thermal efficiency of this cycle.

A.32.5%

B.44.8%

C.56.2%

D.76.5%

Explanation: The thermal efficiency of an ideal Brayton cycle depends only on the pressure ratio rp and the specific heat ratio gamma: eta = 1 - (1 / rp)^((gamma-1)/gamma). Substituting the given values: eta = 1 - (1 / 8)^((1.4-1)/1.4) = 1 - 8^(-0.2857) = 1 - 0.552 = 0.448 (or 44.8%).

About the IESL Part II Mechanical Exam

The IESL Qualifying Examination Part II Mechanical Engineering is a comprehensive assessment covering thermodynamics, fluid mechanics, strength of materials, machine design, and manufacturing technology. It forms a key milestone for non-accredited degree holders aiming to qualify for Associate Membership (AMIESL) and the Chartered Engineer (C.Eng) pathway in Sri Lanka.

Questions

100 scored questions

Time Limit

3 hours per paper (historical)

Passing Score

40%

Exam Fee

Historic registration (contact IESL for modern equivalence paths) (Institution of Engineers, Sri Lanka (IESL))

IESL Part II Mechanical Exam Content Outline

20%

Thermodynamics

Laws of thermodynamics, entropy, gas/vapor power cycles, gas turbine and refrigeration cycles, and combustion principles.

20%

Fluid Mechanics

Fluid statics, kinematics, Bernoulli's equation, viscous internal/external flows, boundary layer theory, dimensional analysis, and turbomachinery.

20%

Strength of Materials

Stress-strain relationships, torsion of shafts, shear force and bending moment diagrams, deflection of beams, thin/thick cylinders, and column buckling.

20%

Machine Design

Design of machine elements (shafts, keys, gears, bearings, springs, clutches, brakes, and threaded joints) and fatigue failure theories.

20%

Manufacturing Technology

Metal casting, forming, machining operations, welding and joining, metrology and inspection, and CNC/non-traditional machining.

How to Pass the IESL Part II Mechanical Exam

What You Need to Know

Passing score: 40%
Exam length: 100 questions
Time limit: 3 hours per paper (historical)
Exam fee: Historic registration (contact IESL for modern equivalence paths)

Keys to Passing

Complete 500+ practice questions
Score 80%+ consistently before scheduling
Focus on highest-weighted sections
Use our AI tutor for tough concepts

IESL Part II Mechanical Study Tips from Top Performers

1For Thermodynamics, focus on cycle analysis (Rankine, Brayton, Otto, Diesel) and thermodynamic properties of pure substances.

2In Fluid Mechanics, master head loss calculations (Darcy-Weisbach) and boundary layer drag forces.

3Strength of Materials requires a strong grasp of principal stresses, Mohr's circle, and beam deflection techniques (e.g. Macaulay's method).

4For Machine Design, practice shaft design under combined bending and torsion, and bearing life calculations using dynamic load ratings.

5In Manufacturing, memorize Merchant's circle equations for orthogonal cutting, casting solidification time calculations (Chvorinov's rule), and limits and tolerances.

Frequently Asked Questions

What is the IESL Qualifying Examination Part II Mechanical Engineering?

Historically, the IESL Part II was a written examination administered by the Institution of Engineers, Sri Lanka. It served as a critical qualifying step for engineering diplomates and non-accredited degree holders to satisfy the academic requirements for Associate Membership (AMIESL).

Is the IESL Part II exam still active?

No. The IESL officially phased out the Part I, Part II, and Part III written examinations, concluding the final sessions in December 2022. Candidates seeking recognition now undergo the IESL Academic Qualification Assessment route or study through the accredited IESL College of Engineering.

What topics were covered in the Mechanical Engineering syllabus?

The syllabus required mastery of five core subjects: Thermodynamics (refrigeration, gas turbines, steam cycles, and laws), Fluid Mechanics (statics, pipe flow, boundary layer, and turbomachinery), Strength of Materials (shafts, SFD/BMD, deflection, and buckling), Machine Design (fatigue, joints, bearings, gears, and shafts), and Manufacturing Technology (casting, forming, cutting, welding, and metrology).

What was the passing mark for the IESL Part II exam?

The passing mark was 40% per subject paper. If a candidate passed a majority of the papers, they could be referred in the remaining papers and retake them in subsequent sessions.

Does practicing these questions help with the current IESL assessment?

Yes. The current Academic Qualification Assessment and the examinations conducted by the IESL College of Engineering cover the exact same engineering science fundamentals. Practicing these core mechanical engineering problems is highly beneficial.