Free GATE ME mock test — 65 questions, GATE marking
One GATE ME paper in the real exam format: 65 questions across one timed section, 3 hr on a server-controlled clock.
About the real exam India's most-attempted GATE branch — your gateway to M.Tech at the IITs/NITs and PSU recruitment in core mechanical engineering.
Free, reversible until you begin · one attempt per account · no retake, no card.
The paper, start to finish
65 question markers · Scored out of 100 marks; wrong answers can cost marks
Navigation: move freely between questions within a section.
Sample questions with worked answers
8 original practice questions written by Myndaq in the GATE ME style. They are not taken from this mock paper or from official past papers. Try each one, then open the answer.
- Question 1General AptitudeOne correct answer
Select the word that best fills the blank.
The inspection report was so thorough and so carefully documented that its conclusions were ______; not even the plant's harshest critics could find fault with them.
- tentative
- contentious
- unimpeachable
- ambiguous
Show answer and explanation
Answer: C
The second clause says nobody could find fault with the conclusions, so the blank needs a word meaning 'beyond challenge'. 'Unimpeachable' means exactly that.
'Tentative' (provisional) and 'ambiguous' (unclear) contradict 'thorough and carefully documented', and 'contentious' (likely to cause argument) contradicts the fact that even critics could not object.
- Question 2Engineering MathematicsType a numeric answer
The integral of x³ from x = 0 to x = 2 is estimated using the trapezoidal rule with 4 equal sub-intervals. The estimated value is ______ (rounded off to two decimal places).
Show answer and explanation
Answer: 4.25
Step size h = (2 - 0)/4 = 0.5, so the nodes are x = 0, 0.5, 1, 1.5, 2 with f(x) = x³ = 0, 0.125, 1, 3.375, 8.
Trapezoidal rule: I ≈ (h/2)[f0 + 2(f1 + f2 + f3) + f4] = 0.25 × [0 + 2(0.125 + 1 + 3.375) + 8] = 0.25 × [9 + 8] = 0.25 × 17 = 4.25.
Check: the exact value is 2⁴/4 = 4. The trapezoidal rule overestimates a function that is convex on the interval (f'' = 6x ≥ 0), so 4.25 > 4 is consistent.
- Question 3Applied Mechanics and DesignType a numeric answer
A simply supported beam of span 2 m carries a concentrated load of 10 kN at mid-span. The beam has a solid rectangular cross-section 50 mm wide and 100 mm deep. Neglecting self-weight, the maximum bending stress in the beam, in MPa, is ______.
Show answer and explanation
Answer: 60
Maximum bending moment for a central point load: M = PL/4 = 10 kN × 2 m / 4 = 5 kN·m = 5 × 10^6 N·mm.
Section modulus: Z = bh²/6 = 50 × 100² / 6 = 83 333.3 mm³.
σmax = M/Z = 5 × 10^6 / 83 333.3 = 60 N/mm² = 60 MPa.
Check with σ = My/I: I = bh³/12 = 50 × 100³ / 12 = 4.1667 × 10^6 mm⁴, y = 50 mm, so σ = 5 × 10^6 × 50 / 4.1667 × 10^6 = 60 MPa.
- Question 4Applied Mechanics and DesignOne correct answer
A rigid block of mass 2 kg is supported by two linear springs of stiffness 3000 N/m and 6000 N/m connected in series, and it vibrates freely along the spring axis without damping. The natural frequency of the system is closest to
- 67.1 rad/s
- 31.6 rad/s
- 5.03 rad/s
- 44.7 rad/s
Show answer and explanation
Answer: B
Springs in series: keq = k1k2/(k1 + k2) = (3000 × 6000)/(3000 + 6000) = 2000 N/m.
ωn = √(keq/m) = √(2000/2) = √1000 = 31.6 rad/s.
67.1 rad/s comes from adding the stiffnesses as if the springs were in parallel (√(9000/2)). 5.03 is the correct frequency expressed in hertz (31.6/2π), so it is wrong with the unit rad/s. 44.7 rad/s is √2000, which ignores the mass.
- Question 5Fluid Mechanics and Thermal SciencesType a numeric answer
Water (density 1000 kg/m³) flows with a mean velocity of 2 m/s through a horizontal pipe of internal diameter 100 mm and length 100 m. The Darcy friction factor is 0.02. Neglecting minor losses, the pressure drop over the length of the pipe, in kPa, is ______.
Show answer and explanation
Answer: 40
Darcy-Weisbach equation in pressure form: Δp = f (L/D) (ρV²/2).
L/D = 100/0.1 = 1000 and ρV²/2 = 1000 × 2² / 2 = 2000 Pa.
Δp = 0.02 × 1000 × 2000 = 40 000 Pa = 40 kPa.
Check via head loss: hf = fLV²/(2gD) = 0.02 × 100 × 4 / (2 × 9.81 × 0.1) = 4.08 m, and ρghf = 1000 × 9.81 × 4.08 ≈ 40 000 Pa. Because the pipe is horizontal and of uniform diameter, the whole head loss appears as pressure drop.
- Question 6Fluid Mechanics and Thermal SciencesOne or more correct answers
Which of the following statements is/are CORRECT?
- For the same compression ratio, an air-standard Otto cycle has a higher thermal efficiency than an air-standard Diesel cycle with a cut-off ratio greater than 1.
- The thermal efficiency of a reversible heat engine operating between two fixed-temperature reservoirs depends on the working fluid used.
- The entropy of an isolated system can never decrease.
- The specific enthalpy of an ideal gas depends on both its temperature and its pressure.
Show answer and explanation
Answer: A, C
A is correct: the Otto efficiency is 1 - 1/r^(γ-1), while the Diesel efficiency multiplies the second term by (ρ^γ - 1)/[γ(ρ - 1)], which is greater than 1 for any cut-off ratio ρ > 1, so the Diesel value is lower at the same r.
C is correct: for an isolated system, the second law gives dS ≥ 0 (increase of entropy principle).
B is wrong: by the Carnot principle, every reversible engine between the same two reservoirs has efficiency 1 - TL/TH, independent of the working fluid. D is wrong: for an ideal gas h = u + RT depends only on temperature (dh = cp dT).
- Question 7Materials, Manufacturing and Industrial EngineeringType a numeric answer
In a turning operation, the tool life follows Taylor's equation VT^n = C with n = 0.25. The tool lasts 81 min at a cutting speed of 36 m/min. The cutting speed, in m/min, at which the same tool will last 16 min is ______.
Show answer and explanation
Answer: 54
From V1T1^n = V2T2^n: V2 = V1 (T1/T2)^n = 36 × (81/16)^0.25.
(81/16)^0.25 = 3/2, because 81 = 3⁴ and 16 = 2⁴. So V2 = 36 × 1.5 = 54 m/min.
Check: 36 × 81^0.25 = 36 × 3 = 108 and 54 × 16^0.25 = 54 × 2 = 108, so C is the same. A shorter tool life must mean a higher speed. Inverting the ratio gives 24 m/min, which is wrong for that reason.
- Question 8Materials, Manufacturing and Industrial EngineeringOne correct answer
A component has a steady annual demand of 7200 units. The cost of placing one order is Rs 250 and the inventory holding cost is Rs 10 per unit per year. Using the basic economic order quantity (EOQ) model with no shortages, the EOQ and the number of orders placed per year are, respectively,
- 424 units and 17 orders
- 300 units and 24 orders
- 849 units and 8.5 orders
- 600 units and 12 orders
Show answer and explanation
Answer: D
EOQ = √(2DS/H) = √(2 × 7200 × 250 / 10) = √360 000 = 600 units. Orders per year = D/EOQ = 7200/600 = 12.
Check: at the EOQ the annual ordering cost (12 × 250 = Rs 3000) equals the annual holding cost (600/2 × 10 = Rs 3000), as the model requires.
424 units comes from leaving out the factor 2 (√180 000). 300 units is the average inventory (EOQ/2), not the order quantity. 849 units results from halving the holding cost (√720 000).
One attempt, explicitly
Adding this mock to your account is free and reversible - you can change the exam any time before you begin. Only the confirmed Begin action uses your one attempt. No retake, no reset, no card.
What you get back
You get your raw marks and per-section scores. No scaled score, band, rank or percentile is estimated for this paper.
- Scored out of 100 raw marks, with per-section totals.
- Wrong answers can cost marks where the paper defines it - the per-section marks in the paper map are the marks the scorer applies.
How the real GATE ME bands are read
GATE scores are reported as both raw marks (out of 100) and a normalised GATE score (out of 1000). IIT / IISc cutoffs are quoted in the 1000-scale; PSU cutoffs are quoted in the 100-scale. The General-category qualifying mark for ME in 2026 was 29.0 / 100 (OBC-NCL / EWS: 26.1; SC / ST / PwD: 19.3). The qualifying mark is only the floor — competitive M.Tech at the IITs and direct PSU calls require considerably higher.
- 850+ / 1000 (≈ 65+ / 100, AIR < 200)Strong shot at IIT Bombay / Delhi / Madras / Kanpur M.Tech in Mechanical, Thermal, Design, or Manufacturing. Direct interview calls at IOCL, ONGC, NTPC, BHEL, and HPCL.
- 700 – 850 / 1000M.Tech at IIT Roorkee, IIT Guwahati, IIT Hyderabad, IIT BHU, IIT (ISM) Dhanbad in core mechanical specialisations. Strong PSU written-round position.
- 550 – 700 / 1000M.Tech at lower-tier IITs (Patna, Bhilai, Jodhpur, Mandi, Goa), top NITs (Trichy, Surathkal, Warangal, Calicut), and most CFTIs. Useful for design / R&D roles in industry.
- Qualifying (29.0+ / 100, General)Clears the floor for M.Tech application eligibility at most CFTIs and several PSU written rounds. Many state-level PSUs also use GATE qualification as their primary recruitment filter.
Context only - this free mock returns: You get your raw marks and per-section scores. No scaled score, band, rank or percentile is estimated for this paper.
Everything about the GATE ME
GATE ME (Mechanical Engineering) is the most-attempted single paper in the Graduate Aptitude Test in Engineering — over 150,000 candidates write it every year. Conducted annually by an IIT (or IISc) on rotation, it tests undergraduate-level fundamentals across four core sections: Engineering Mathematics, Applied Mechanics & Design, Fluid Mechanics & Thermal Sciences, and Materials/Manufacturing/Industrial Engineering. A strong score (typically rank under 1,000) is the gateway to M.Tech programmes at the IITs, NITs, and CFTIs, and is the primary recruitment signal for PSUs like IOCL, ONGC, BHEL, NTPC, and HPCL. The exam follows the standard GATE pattern — 65 questions, 100 marks, 180 minutes, with a mix of MCQ, Multi-Select (MSQ), and Numerical Answer Type (NAT) questions.
Figures about the real exam, as recorded in the Myndaq course. They describe the real sitting, not this free mock paper - check current details with the exam body before you book.
What the real GATE ME covers
Who sits it
How people prepare for it
What it opens up
Figures above are as reported for the cycle named in each row and are refreshed each cycle. Sourced, dated facts for this exam (official links on every row) are on the GATE ME course page.
How this mirrors the real format
- The paper structure is the real format: a single timed section, 65 questions in total.
- Timing is server-controlled for the full 3 hr -- closing the tab does not pause the clock, exactly like our paid mock series.
- Marking follows the published paper: the per-section marks shown in the paper map are the marks the scorer applies.
- Navigation is free within a section: you can review and change answers before you submit.
- An on-screen calculator is provided in the sections where the exam allows one.
"Real format" means the paper shape, timing, marking and navigation above - not the official exam software, its visual identity, or any affiliation with the exam body.
What differs or is limited
- This free mock follows the same paper shape, timing, marking scheme and navigation as our paid GATE Mechanical Engineering mock series. It is scored automatically. It does not include tutor explanations, analysis dashboards or practice content.
- Includes the virtual calculator and free navigation, as in the real exam.
Device and setup checklist
- A stable internet connection for the full duration.
- An on-screen calculator is provided where the exam allows one.
- A laptop or desktop is recommended for the closest exam experience.