RRB JE CBT 2 Mechanical Mock Test 2026: Practice Questions with Answers and Explanations

RRB JE CBT 2 Mechanical Mock Test: Overview

RRB JE CBT 2 Exam Pattern

The reported CBT 2 examination structure consists of 150 questions carrying 150 marks, with a total duration of 120 minutes. The technical section accounts for 100 questions.

Candidates should verify the applicable recruitment notification for the exact examination rules and marking scheme.

RRB JE CBT 2 Mechanical Mock Test: 25 Questions

Instructions for candidates

1. Attempt each question before checking the answer.

2. Select only one option unless the question specifies otherwise.

3. Use rough calculations where necessary.

4. Award yourself one mark for a correct answer.

5. Deduct one-third of a mark for an incorrect answer when calculating your practice score.

6. Review the explanations for every incorrect answer.

Section A: Strength of Materials

Question 1. Stress Calculation

A steel rod of cross-sectional area 500 mm² is subjected to an axial tensile load of 50 kN. What is the tensile stress developed in the rod?

A. 50 MPa
B. 100 MPa
C. 150 MPa
D. 200 MPa

Correct Answer: B. 100 MPa

Explanation:

Stress is calculated using:

Stress = Load / Cross-sectional area

Given:

• Load = 50,000 N

• Area = 500 mm²

Stress = 50,000 / 500 = 100 N/mm²

Since 1 N/mm² = 1 MPa, the tensile stress is 100 MPa.

Question 2. Young’s Modulus

A material experiences a normal stress of 200 MPa and a corresponding longitudinal strain of 0.001. Calculate Young’s modulus.

A. 100 GPa
B. 150 GPa
C. 200 GPa
D. 250 GPa

Correct Answer: C. 200 GPa

Explanation:

Young’s modulus is:

E = Stress / Strain

E = 200 MPa / 0.001

E = 200,000 MPa = 200 GPa.

Young’s modulus indicates the stiffness of a material within its linear elastic range.

Question 3. Bending Moment

A simply supported beam of span 4 m carries a concentrated load of 20 kN at its centre. What is the maximum bending moment?

A. 10 kN·m
B. 15 kN·m
C. 20 kN·m
D. 40 kN·m

Correct Answer: C. 20 kN·m

Explanation:

For a simply supported beam carrying a central point load:

Maximum bending moment = WL / 4

Where:

• W = 20 kN

• L = 4 m

Maximum bending moment = (20 × 4) / 4 = 20 kN·m.

Question 4. Poisson’s Ratio

A material has a longitudinal strain of 0.002 and a lateral strain of −0.0006. What is its Poisson’s ratio?

A. 0.15
B. 0.20
C. 0.30
D. 0.40

Correct Answer: C. 0.30

Explanation:

Poisson’s ratio is the negative ratio of lateral strain to longitudinal strain.

ν = −(Lateral strain / Longitudinal strain)

ν = −(−0.0006 / 0.002)

ν = 0.30.

The negative sign of lateral strain indicates contraction perpendicular to the direction of tensile loading.

Question 5. Factor of Safety

A component has an ultimate strength of 600 MPa. If the factor of safety based on ultimate strength is 3, what is the allowable stress?

A. 100 MPa
B. 150 MPa
C. 200 MPa
D. 300 MPa

Correct Answer: C. 200 MPa

Explanation:

Allowable stress = Ultimate strength / Factor of safety

= 600 / 3

= 200 MPa.

In actual design, the relevant design code may specify the appropriate strength basis and allowable stress criteria.

Section B: Thermodynamics

Question 6. First Law of Thermodynamics

A closed system receives 500 kJ of heat and performs 200 kJ of work. What is the change in internal energy?

A. 200 kJ
B. 300 kJ
C. 500 kJ
D. 700 kJ

Correct Answer: B. 300 kJ

Explanation:

For a closed system, using the convention that heat supplied to the system and work done by the system are positive:

ΔU = Q − W

ΔU = 500 − 200

ΔU = 300 kJ.

Question 7. Carnot Engine Efficiency

A Carnot engine operates between a source temperature of 600 K and a sink temperature of 300 K. Calculate its thermal efficiency.

A. 25%
B. 40%
C. 50%
D. 75%

Correct Answer: C. 50%

Explanation:

Carnot efficiency is:

η = 1 − (T₂ / T₁)

Where temperatures are in kelvin.

η = 1 − (300 / 600)

η = 0.5 = 50%.

This is the theoretical maximum efficiency for a heat engine operating between these two temperatures.

Question 8. Ideal Gas Equation

A gas occupies 0.5 m³ at an absolute pressure of 200 kPa and a temperature of 300 K. Using the ideal gas equation, calculate its mass. Take the specific gas constant as 287 J/(kg·K).

A. 0.58 kg
B. 1.16 kg
C. 1.74 kg
D. 2.32 kg

Correct Answer: B. 1.16 kg

Explanation:

The ideal gas equation is:

PV = mRT

Therefore:

m = PV / RT

Using SI units:

• P = 200,000 Pa

• V = 0.5 m³

• R = 287 J/(kg·K)

• T = 300 K

m = (200,000 × 0.5) / (287 × 300)

m ≈ 1.16 kg.

Question 9. Entropy of an Isolated System

According to the second law of thermodynamics, the entropy of an isolated system undergoing a spontaneous process:

A. Always decreases
B. Remains constant in every process
C. Never decreases
D. Is always zero

Correct Answer: C. Never decreases

Explanation:

The entropy of an isolated system remains constant for an ideal reversible process and increases for an irreversible process.

Therefore, its entropy cannot decrease during a physically admissible spontaneous process.

Question 10. Refrigerator Coefficient of Performance

A refrigerator removes 300 kJ of heat from a cold space while consuming 100 kJ of work. Calculate its coefficient of performance (COP).

A. 1
B. 2
C. 3
D. 4

Correct Answer: C. 3

Explanation:

For a refrigerator:

COP = Heat removed / Work input

COP = 300 / 100

COP = 3.

Unlike thermal efficiency, the COP of a refrigerator can be greater than 1.

Section C: Fluid Mechanics

Question 11. Continuity Equation

Water flows through a horizontal pipe. The cross-sectional area decreases from 0.04 m² to 0.02 m². If the velocity in the larger section is 2 m/s, calculate the velocity in the smaller section.

A. 1 m/s
B. 2 m/s
C. 4 m/s
D. 8 m/s

Correct Answer: C. 4 m/s

Explanation:

For steady incompressible flow:

A₁V₁ = A₂V₂

V₂ = A₁V₁ / A₂

V₂ = (0.04 × 2) / 0.02

V₂ = 4 m/s.

The fluid velocity increases as the pipe’s cross-sectional area decreases.

Question 12. Reynolds Number

Water flows through a pipe of diameter 0.05 m at an average velocity of 2 m/s. If the kinematic viscosity is 1 × 10⁻⁶ m²/s, calculate the Reynolds number.

A. 1,000
B. 10,000
C. 50,000
D. 100,000

Correct Answer: C. 100,000

Explanation:

Reynolds number is:

Re = VD / ν

Re = (2 × 0.05) / (1 × 10⁻⁶)

Re = 100,000.

This value indicates turbulent flow in a conventional pipe-flow context.

Question 13. Bernoulli’s Principle

For steady, incompressible, inviscid flow along a streamline, Bernoulli’s equation represents conservation of:

A. Mass only
B. Mechanical energy
C. Temperature only
D. Entropy only

Correct Answer: B. Mechanical energy

Explanation:

Bernoulli’s equation relates pressure energy, kinetic energy and potential energy per unit weight:

P/(ρg) + V²/(2g) + z = constant.

The equation applies under its stated assumptions. Pumps, turbines and energy losses require additional terms.

Question 14. Hydraulic Power

A pump delivers water at a flow rate of 0.02 m³/s against a head of 20 m. Take water density as 1,000 kg/m³ and gravitational acceleration as 9.81 m/s². Calculate the hydraulic power.

A. 1.962 kW
B. 3.924 kW
C. 5.886 kW
D. 9.810 kW

Correct Answer: B. 3.924 kW

Explanation:

Hydraulic power is:

P = ρgQH

P = 1,000 × 9.81 × 0.02 × 20

P = 3,924 W

Therefore, hydraulic power = 3.924 kW.

This is the power transferred to the fluid. Actual pump input power will be greater when pump efficiency is below 100%.

Question 15. Cavitation in Pumps

Cavitation in a centrifugal pump occurs when:

A. The liquid pressure falls below its vapour pressure
B. The liquid pressure becomes infinitely high
C. The impeller stops rotating permanently
D. The liquid density becomes zero

Correct Answer: A. The liquid pressure falls below its vapour pressure

Explanation:

Cavitation occurs when local pressure falls sufficiently for vapour bubbles to form. These bubbles may collapse in higher-pressure regions, producing noise, vibration, surface damage and reduced pump performance.

Adequate net positive suction head (NPSH) helps prevent cavitation.

Section D: Manufacturing Engineering

Question 16. Lathe Cutting Speed

A workpiece of diameter 100 mm rotates at 300 rpm. Calculate the cutting speed in metres per minute.

A. 47.12 m/min
B. 62.83 m/min
C. 94.25 m/min
D. 125.66 m/min

Correct Answer: C. 94.25 m/min

Explanation:

Cutting speed is:

V = πDN / 1,000

Where:

• D = 100 mm

• N = 300 rpm

V = (π × 100 × 300) / 1,000

V ≈ 94.25 m/min.

Question 17. Welding Process

Which welding process uses a non-consumable tungsten electrode and an inert shielding gas?

A. Shielded Metal Arc Welding
B. Gas Tungsten Arc Welding
C. Submerged Arc Welding
D. Resistance Spot Welding

Correct Answer: B. Gas Tungsten Arc Welding

Explanation:

Gas Tungsten Arc Welding (GTAW), also known as TIG welding, uses a non-consumable tungsten electrode. An inert shielding gas, commonly argon or helium, protects the weld area from atmospheric contamination.

Question 18. Casting Defect

Which casting defect is most directly associated with insufficient molten metal feeding during solidification?

A. Shrinkage cavity
B. Grinding burn
C. Tool chatter
D. Excessive tool wear

Correct Answer: A. Shrinkage cavity

Explanation:

Metals generally contract during solidification. If additional molten metal is not supplied adequately, a shrinkage cavity or related shrinkage porosity may form.

Proper riser design and directional solidification help reduce this problem.

Question 19. CNC Machining

What does CNC stand for in manufacturing engineering?

A. Computer Numerical Control
B. Central Network Cutting
C. Controlled Newtonian Calculation
D. Computerised Normal Casting

Correct Answer: A. Computer Numerical Control

Explanation:

CNC machines operate through programmed numerical instructions that control machine movements and operations.

They support repeatability, accuracy and automation in manufacturing.

Question 20. Cutting Tool Wear

Which type of tool wear commonly occurs on the clearance face of a cutting tool?

A. Flank wear
B. Shrinkage wear
C. Cavitation wear
D. Creep rupture

Correct Answer: A. Flank wear

Explanation:

Flank wear occurs on the tool’s clearance face due to interaction with the newly machined surface. Abrasion, adhesion and high-temperature effects can contribute to tool wear.

Section E: Theory of Machines and Machine Design

Question 21. Gear Ratio

A driving gear has 20 teeth and rotates at 600 rpm. It drives a gear with 60 teeth. Assuming ideal external gears, what is the speed of the driven gear?

A. 100 rpm
B. 200 rpm
C. 400 rpm
D. 1,800 rpm

Correct Answer: B. 200 rpm

Explanation:

For a simple pair of external gears:

N₁T₁ = N₂T₂

N₂ = N₁T₁ / T₂

N₂ = (600 × 20) / 60

N₂ = 200 rpm.

The driven gear rotates in the opposite direction to the driving gear.

Question 22. Simple Harmonic Motion

A body executes simple harmonic motion with an angular frequency of 10 rad/s and an amplitude of 0.2 m. What is its maximum velocity?

A. 0.5 m/s
B. 1 m/s
C. 2 m/s
D. 5 m/s

Correct Answer: C. 2 m/s

Explanation:

The maximum velocity in simple harmonic motion is:

Vmax = ωA

Vmax = 10 × 0.2

Vmax = 2 m/s.

Maximum velocity occurs when the body passes through its mean position.

Question 23. Bearing Function

What is the primary function of a bearing in a rotating machine?

A. To increase friction deliberately
B. To support relative motion and reduce friction
C. To eliminate all mechanical loads
D. To convert heat directly into electricity

Correct Answer: B. To support relative motion and reduce friction

Explanation:

Bearings support and guide rotating or moving components while reducing friction and managing applied loads.

The type of bearing selected depends on load direction, speed, lubrication and operating conditions.

Question 24. Mechanical Advantage

A machine lifts a load of 600 N using an effort of 150 N. Calculate its mechanical advantage.

A. 2
B. 3
C. 4
D. 5

Correct Answer: C. 4

Explanation:

Mechanical advantage is:

MA = Load / Effort

MA = 600 / 150

MA = 4.

The machine lifts a load four times the magnitude of the applied effort under the stated conditions.

Question 25. Factor of Safety in Design

A component is designed to operate under a maximum allowable stress of 120 MPa. If its relevant failure strength is 360 MPa, what is the factor of safety based on that failure strength?

A. 1.5
B. 2
C. 3
D. 4

Correct Answer: C. 3

Explanation:

Factor of safety = Failure strength / Allowable stress

= 360 / 120

= 3.

The strength basis used for the calculation must match the applicable design criteria.

RRB JE CBT 2 Mechanical Mock Test: Answer Key

Use this table to review your responses after attempting all 25 questions.

How to Calculate Your Mock Test Score

For this practice test, use the following formula:

Score = Correct answers − (Incorrect answers ÷ 3)

Unattempted questions receive zero marks.

For example, if you answer 20 questions correctly, 3 incorrectly and leave 2 unanswered:

Score = 20 − (3 ÷ 3)

Final score = 19 out of 25.

This calculation assumes the examination’s one-third negative-marking rule applies to the practice set. Confirm the marking scheme in the relevant recruitment notification.

Score Interpretation

The following bands are informal practice benchmarks, not official RRB cut-offs.

A single mini mock test cannot predict your final examination result. Performance across multiple full-length tests is more informative.

Important Mechanical Engineering Topics for RRB JE CBT 2

Candidates should prepare according to the official syllabus for their engineering discipline.

This list is a revision guide, not a substitute for the complete official syllabus.

How to Prepare for RRB JE CBT 2 Mechanical Engineering

1. Follow the Official Syllabus

Divide the syllabus into individual subjects and topics. Track completed chapters, pending concepts and areas that need additional practice.

2. Practise Numerical Questions Daily

Give regular attention to Strength of Materials, Thermodynamics, Fluid Mechanics and Machine Design. Learn the underlying formulas and practise applying them to different situations.

3. Revise Engineering Fundamentals

Technical questions may test conceptual understanding as well as calculations. Review definitions, assumptions, diagrams, laws and standard relationships.

4. Attempt Timed Mock Tests

Begin with shorter practice sets and gradually move to full-length tests. Use the results to identify weak topics and improve time management.

5. Analyse Every Incorrect Answer

Classify errors into three categories: conceptual mistakes, calculation errors and incorrect question interpretation. Revise the relevant topic before attempting similar questions again.

6. Avoid Blind Guessing

If negative marking applies, consider the penalty before making uncertain selections. Prioritise questions you can solve accurately.

7. Practise the Computer-Based Format

Use official RRB mock-test facilities when available to understand the examination interface, question navigation and any permitted on-screen tools.

Official RRB JE Preparation Resources

Candidates should verify recruitment details and examination instructions through official Railway Recruitment Board websites.

• RRB official portal: Railway Recruitment Boards

• RRB Chandigarh: Official RRB Chandigarh website

Check the relevant CEN notification, corrigenda and official mock-test announcements for the examination cycle applicable to your application.

Frequently Asked Questions (FAQs)

1. What is the RRB JE CBT 2 Mechanical Mock Test?

It is a practice test designed to help Mechanical Engineering candidates revise technical concepts and prepare for the second computer-based examination for Junior Engineer recruitment.

2. How many questions are included in this RightsAim mock test?

This mini mock test contains 25 multiple-choice questions with answers and explanations. It is not a full-length 150-question paper.

3. Is this an official RRB JE question paper?

No. The questions are original practice material prepared for revision. They should not be represented as official RRB questions or guaranteed predictions of the examination.

4. How many questions are there in RRB JE CBT 2?

The reported CBT 2 pattern contains 150 questions carrying 150 marks, including 100 questions in Technical Abilities.

5. What is the duration of RRB JE CBT 2?

The reported examination duration is 120 minutes. Candidates should verify the applicable notification for their recruitment cycle.

6. Is there negative marking in RRB JE CBT 2?

The reported marking scheme deducts one-third of the marks assigned to a question for each incorrect answer. Verify the applicable official notification before the examination.

7. Which Mechanical Engineering subjects should I revise first?

Start by assessing your strengths and weaknesses across Strength of Materials, Thermodynamics, Fluid Mechanics, Manufacturing Engineering and Theory of Machines. Then prioritise topics based on the official syllabus and your practice-test performance.

8. Can diploma holders prepare using this mock test?

Yes. Diploma holders preparing for eligible RRB JE Mechanical posts can use this set to revise engineering fundamentals, subject to the eligibility requirements in the relevant recruitment notification.

9. How can I improve my mock test score?

Revise the concepts behind incorrect answers, practise numerical problems, take timed tests and maintain an error notebook. Review progress across multiple practice sessions.

10. Does a high mock test score guarantee RRB JE selection?

No. Mock tests are preparation tools. Actual selection depends on examination performance, applicable cut-offs, vacancies, eligibility, document verification, medical standards and the official selection rules.

Conclusion

The RRB JE CBT 2 Mechanical Mock Test is a useful revision resource for candidates who want to strengthen technical knowledge and practise exam-style MCQs.

This RightsAim practice set covers fundamental concepts from Strength of Materials, Thermodynamics, Fluid Mechanics, Manufacturing Engineering and Theory of Machines. Candidates should attempt the questions independently, review the explanations and use their scores to identify areas for improvement.

For the best preparation, combine topic-wise practice with previous-year papers, official mock tests and full-length timed examinations. Always verify the syllabus, examination pattern and selection rules through the relevant official RRB notification.

Disclaimer: This practice test is intended for educational purposes only. It is not an official RRB examination paper and does not guarantee qualification or selection.