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100+ Free MPPSC SES Mechanical Practice Questions

Pass your MPPSC State Engineering Service (Mechanical) Prelims exam on the first try — instant access, no signup required.

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Key Facts: MPPSC SES Mechanical Exam

150 MCQs

Total Written Exam Questions

MPPSC SES Notification

180 Mins

Exam Duration

MPPSC SES Exam Pattern

3:1

Marking Scheme (+3 for Correct, -1 for Incorrect)

MPPSC SES Marking Rules

40% / 30%

Sectional Qualifying Cutoff

MPPSC SES Syllabus & Guidelines

₹500 / ₹250

Application Fee for Gen / MP Domicile Reserved

MPPSC SES Advertisement

MPPSC SES Mechanical is a state-level exam for Assistant Engineer posts. It features 150 MCQs (50 GK, 100 Mechanical) in 3 hours with +3/-1 marking. This mock test provides 100 target Mechanical Engineering practice questions.

Sample MPPSC SES Mechanical Practice Questions

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

1A steel bar of 20 mm diameter and 1 m length is subjected to an axial tensile load of 50 kN. If the Young's modulus of steel is E = 200 GPa, what is the elongation of the bar?
A.0.398 mm
B.0.796 mm
C.1.592 mm
D.0.199 mm
Explanation: The elongation is calculated using the formula delta = P*L / (A*E). Here, P = 50,000 N, L = 1000 mm, A = (pi/4)*d^2 = (pi/4)*(20)^2 = 314.16 mm^2, and E = 200,000 MPa. Plugging in the values: delta = (50000 * 1000) / (314.16 * 200000) = 50000 / 62831.85 = 0.7958 mm, which rounds to 0.796 mm.
2For an isotropic material, what is the relation between the Young's Modulus (E), Shear Modulus (G), and Poisson's ratio (mu)?
A.E = 2G(1 - mu)
B.E = 2G(1 + mu)
C.E = 3G(1 - 2mu)
D.E = 3G(1 + 2mu)
Explanation: The relationship between Young's Modulus (E), Shear Modulus (G), and Poisson's ratio (mu) for an isotropic elastic material is E = 2G(1 + mu). For Bulk Modulus (K), the relationship is E = 3K(1 - 2mu).
3A cantilever beam of length L carries a uniformly distributed load of intensity w per unit length over its entire length. What is the maximum deflection of the beam at the free end? (EI is the flexural rigidity)
A.wL^4 / (8EI)
B.wL^4 / (3EI)
C.wL^4 / (48EI)
D.5wL^4 / (384EI)
Explanation: For a cantilever beam under a uniformly distributed load (w) over its entire length (L), the maximum deflection occurs at the free end and is given by delta = wL^4 / (8EI).
4At a point in a strained material, the principal stresses are 100 MPa (tensile) and 40 MPa (compressive). What is the maximum shear stress at this point?
A.30 MPa
B.70 MPa
C.60 MPa
D.140 MPa
Explanation: The maximum shear stress is given by tau_max = (sigma_1 - sigma_2) / 2. Since the second stress is compressive, sigma_2 = -40 MPa. Thus, tau_max = (100 - (-40)) / 2 = 140 / 2 = 70 MPa.
5What is the ratio of torque transmission capacity of a solid shaft of diameter D to a hollow shaft of outer diameter D and inner diameter d = D/2, if both are made of the same material and subjected to the same maximum shear stress?
A.16/15
B.15/16
C.8/7
D.2/1
Explanation: Torque transmission capacity T is proportional to the polar section modulus Zp. For solid shaft, Zp_solid = (pi/16)*D^3. For hollow shaft, Zp_hollow = (pi/16)*D^3 * (1 - (d/D)^4). Since d/D = 0.5, (d/D)^4 = 1/16. Thus, Zp_hollow = (pi/16)*D^3 * (15/16). The ratio T_solid / T_hollow = Zp_solid / Zp_hollow = 1 / (15/16) = 16/15.
6A column of length L has both ends pinned. Another column of the same material, length, and cross-section has both ends fixed. What is the ratio of the Euler's buckling load of the fixed-fixed column to that of the pinned-pinned column?
A.2
B.4
C.0.25
D.0.5
Explanation: Euler's buckling load is P_cr = pi^2 * EI / L_e^2. For pinned-pinned ends, the effective length L_e = L, so P_pinned = pi^2 * EI / L^2. For fixed-fixed ends, the effective length L_e = L/2, so P_fixed = pi^2 * EI / (L/2)^2 = 4 * pi^2 * EI / L^2. Thus, the ratio P_fixed / P_pinned is 4.
7A thin cylindrical pressure vessel of internal diameter d and thickness t is subjected to an internal pressure p. If E is the Young's modulus and mu is the Poisson's ratio, what is the volumetric strain (e_v) of the cylinder?
A.(pd / 4tE) * (5 - 4mu)
B.(pd / 2tE) * (5 - 4mu)
C.(pd / 4tE) * (1 - 2mu)
D.(pd / 2tE) * (1 - 2mu)
Explanation: The circumferential strain (e_1) is pd/(4tE)*(2 - mu) and longitudinal strain (e_2) is pd/(4tE)*(1 - 2mu). The volumetric strain for a thin cylinder is e_v = 2*e_1 + e_2 = 2 * [pd/(4tE)*(2 - mu)] + [pd/(4tE)*(1 - 2mu)] = (pd / 4tE) * (4 - 2mu + 1 - 2mu) = (pd / 4tE) * (5 - 4mu).
8A bar of cross-sectional area A and length L is subjected to an impact load. A weight W is dropped from a height h onto a collar at the end of the bar. What is the relation between the impact stress (sigma_i) and the static stress (sigma_s = W/A)?
A.sigma_i = sigma_s * [1 + sqrt(1 + 2h*E/(sigma_s*L))]
B.sigma_i = sigma_s * [1 + sqrt(1 + h*E/(2*sigma_s*L))]
C.sigma_i = sigma_s * [1 + sqrt(1 + h*E/(sigma_s*L))]
D.sigma_i = 2 * sigma_s
Explanation: By equating the potential energy lost by the falling weight to the strain energy stored in the bar at maximum deformation, we derive the impact factor: sigma_i / sigma_s = 1 + sqrt(1 + (2*h*E) / (sigma_s*L)).
9In a simply supported beam of length L subjected to a concentrated moment M at the center, what is the value of the bending moment at the point of application of the moment?
A.It is discontinuous, changing abruptly from -M/2 to +M/2
B.It is continuous and equals M/2
C.It is continuous and equals M
D.It is zero
Explanation: At any point where a concentrated external moment M is applied to a beam, the bending moment diagram (BMD) exhibits a vertical jump or discontinuity equal in magnitude to the applied moment M. In a simply supported beam of length L, the support reactions are R_A = -M/L and R_B = M/L. The bending moment just to the left of the center is -M/2, and just to the right of the center is +M/2, representing a total jump of M.
10Three coplanar forces are in equilibrium at a point. Force F1 is directed horizontally to the right. F2 acts at 120 degrees counterclockwise from F1. F3 acts at 240 degrees counterclockwise from F1. What is the ratio of the magnitudes |F1|:|F2|:|F3|?
A.1 : 1 : 1
B.1 : sqrt(3) : 2
C.1 : 2 : sqrt(3)
D.1 : 1 : sqrt(2)
Explanation: According to Lami's Theorem, for three coplanar forces in equilibrium: P / sin(alpha) = Q / sin(beta) = R / sin(gamma). The angles opposite to F1, F2, and F3 are: angle opposite F1 (between F2 and F3) = 120 degrees; angle opposite F2 (between F1 and F3) = 120 degrees; angle opposite F3 (between F1 and F2) = 120 degrees. Since all opposite angles are equal (120 degrees), sin(120) is equal for all. Thus, |F1| = |F2| = |F3|, which gives a ratio of 1:1:1.

About the MPPSC SES Mechanical Exam

The MPPSC State Engineering Service (SES) Mechanical Prelims is a state-level competitive examination for recruiting Assistant Engineers (AE) in various government departments of Madhya Pradesh. The exam consists of two parts: Part A (General Knowledge of MP, India, and World, plus ICT) with 50 questions, and Part B (Mechanical Engineering discipline) with 100 questions. The test is a 3-hour objective exam with negative marking (+3 for correct, -1 for incorrect). This practice bank offers 100 high-quality, exam-style technical questions designed to test your knowledge of core mechanical engineering topics.

Assessment

The written exam is conducted in a single session of 3 hours, carrying 150 questions (450 marks). Part A has 50 General Knowledge questions, and Part B has 100 Mechanical Engineering technical questions. There is a negative marking of -1 for each wrong answer. An interview of 50 marks follows for qualified candidates.

Time Limit

3 hours (180 minutes)

Passing Score

40% in each section for General, 30% for reserved categories of MP

Exam Fee

₹500 for General & Other States, ₹250 for MP Reserved (Madhya Pradesh Public Service Commission (MPPSC))

MPPSC SES Mechanical Exam Content Outline

10%

Engineering Mechanics & Mechanics of Solids

Equilibrium equations, force systems, stress-strain relationships, shear force and bending moment diagrams, torsion, and thin cylinders.

10%

Theory of Machines & Vibrations

Kinematic links, gears and gear trains, governors, flywheels, balancing of rotating masses, and single degree-of-freedom vibrations.

10%

Design of Machine Elements

Failure theories, fatigue strength, and design of keys, pins, joints, shafts, and bearings.

15%

Thermodynamics & IC Engines

Laws of thermodynamics, availability, gas power cycles, combustion, testing of IC engines, and emissions.

15%

Fluid Mechanics & Hydraulic Machinery

Fluid properties, kinematics, dynamics, boundary layer, turbines (Pelton, Francis, Kaplan), and pumps.

15%

Heat Transfer, Refrigeration & Air Conditioning

Conduction, convection, radiation, heat exchangers, VCR and VAR cycles, and psychrometrics.

15%

Production Engineering

Metal cutting, machine tools, casting, forging, rolling, extrusion, welding, and non-conventional machining.

10%

Industrial Management & CAD/CAM

Work study, plant layout, inventory control, PERT/CPM, and computer-aided design and manufacturing.

How to Pass the MPPSC SES Mechanical Exam

What You Need to Know

  • Passing score: 40% in each section for General, 30% for reserved categories of MP
  • Assessment: The written exam is conducted in a single session of 3 hours, carrying 150 questions (450 marks). Part A has 50 General Knowledge questions, and Part B has 100 Mechanical Engineering technical questions. There is a negative marking of -1 for each wrong answer. An interview of 50 marks follows for qualified candidates.
  • Time limit: 3 hours (180 minutes)
  • Exam fee: ₹500 for General & Other States, ₹250 for MP Reserved

Keys to Passing

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

MPPSC SES Mechanical Study Tips from Top Performers

1Focus heavily on mechanical core subjects like Thermodynamics, Fluid Mechanics, Strength of Materials, and Production, as they constitute 300 out of the 450 written exam marks.
2Get a solid grasp of Madhya Pradesh GK, geography, history, and current affairs (Part A), since a sectional score of at least 40% is mandatory here.
3Since each wrong answer costs 1 mark (while correct answers award 3 marks), avoid blind guessing. Aim for accuracy and skip highly uncertain questions.
4Practice numerical problems from Strength of Materials, Fluid Mechanics, and Heat Transfer, as MPPSC SES often includes simple to moderate calculation-based questions.
5Use previous years' question papers of MPPSC SES, UPSC ESE, and other state PSC AE exams to familiarize yourself with the question pattern and level of difficulty.

Frequently Asked Questions

What is the exam structure for MPPSC SES Mechanical?

The written exam is conducted in a single session of 3 hours, carrying 150 questions (450 marks). Part A has 50 General Knowledge questions, and Part B has 100 Mechanical Engineering technical questions. There is a negative marking of -1 for each wrong answer.

Is there a sectional cutoff in the MPPSC SES exam?

Yes, candidates must score at least 40% in Part A and Part B separately to qualify (30% for reserved categories of Madhya Pradesh).

What is the eligibility criteria for MPPSC SES Mechanical?

Candidates must have a B.E. or B.Tech degree in Mechanical Engineering from a recognized university. Additionally, candidate registration with the MP State Employment Exchange is required.

What is the age limit for the MPPSC SES exam?

Candidates must be between 21 and 40 years of age. Domiciles of Madhya Pradesh belonging to SC, ST, OBC, Women, or PwD categories receive a relaxation of up to 5 years (maximum age limit 45).

Is there an interview stage in MPPSC SES?

Yes. Candidates who clear the written exam are called for an interview worth 50 marks. The final merit list is prepared out of 500 total marks (450 written + 50 interview).