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Past Questions — 2025

MES Jr. Grade • 2025 • Paper-II

The ratio of the ultimate stress to the design stress is known as:

A) elastic limit
B) strain
C) factor of safety
D) bulk modulus
In machine design, the 'Factor of Safety' (FoS) is a design margin used to ensure that a structure or component can handle more than its intended load to account for uncertainties in material properties, loading conditions, and manufacturing tolerances. It is mathematically defined as the ratio of the failure stress (Ultimate Stress or Yield Stress, depending on the material) to the allowable stress (Design Stress or Working Stress). Formula: Factor of Safety = Ultimate Stress / Design Stress. Why other options are incorrect: - a. Elastic Limit: This is the maximum stress a material can withstand without permanent deformation; it is a material property, not a ratio of stresses. - b. Strain: This is the ratio of change in dimension to the original dimension (deformation), not a ratio of stresses. - d. Bulk Modulus: This is a measure of a substance's resistance to uniform compression (ratio of volumetric stress to volumetric strain).
MES Jr. Grade • 2025 • Paper-II

The design of shafts made of brittle materials is based on:

A) Guest’s theory
B) Rankine’s theory
C) St. Venant’s theory
D) Von Mises Theory
In machine design, different failure theories are applied based on the nature of the material (ductile vs. brittle). 1. **Rankine's Theory (Maximum Normal Stress Theory):** This theory states that failure occurs when the maximum principal stress reaches the yield point or ultimate strength of the material. It is the most accurate theory for **brittle materials** (like cast iron), as brittle materials fail due to sudden fracture caused by normal stresses rather than shear deformation. 2. **Guest's Theory (Maximum Shear Stress Theory):** This theory states that failure occurs when the maximum shear stress reaches a critical value. It is primarily used for **ductile materials**. 3. **Von Mises Theory (Maximum Distortion Energy Theory):** This is the most accurate theory for predicting the yielding of **ductile materials** under complex loading. 4. **St. Venant's Theory:** While St. Venant is famous for his principle regarding stress distribution in beams, it is not a primary failure criterion used for the general design of shafts in the same context as the others listed. Therefore, for shafts made of brittle materials, Rankine's theory is the standard design basis.
MES Jr. Grade • 2025 • Paper-II

A concentrated mass m is attached at the centre of a rod of length 2L as shown in the figure. The rod is kept in a horizontal equilibrium position by a spring of stiffness k. For very small amplitude of vibration, neglecting the weights of the rod and spring, the undamped natural frequency of the system is:

A) k/m
B) 2k/m
C) k/2m
D) 4k/m
Option B is correct: 2k/m.
MES Jr. Grade • 2025 • Paper-II

In a gear train, when the axes of the shafts, over which the gears are mounted, move relative to a fixed axis, is called:

A) simple gear train
B) compound gear train
C) reverted gear train
D) epicyclic gear train
In machine design, gear trains are classified based on the relative motion of their axes. 1. **Epicyclic Gear Train (Planetary Gear Train):** This is a gear train where the axes of some of the gears move relative to a fixed axis. This is achieved by mounting one or more gears on an 'arm' or 'carrier' that rotates around another gear. Because the center of the gear moves in a circle, it is called 'epicyclic' motion. 2. **Simple Gear Train:** In a simple gear train, all gears are mounted on fixed parallel shafts. Each shaft carries only one gear. 3. **Compound Gear Train:** Similar to a simple gear train, the shafts are fixed. However, at least one shaft carries two or more gears that rotate at the same speed. 4. **Reverted Gear Train:** This is a specific type of compound gear train where the input shaft and the output shaft are co-axial (they lie on the same line), but the shafts themselves remain fixed in position relative to the housing. Since the question specifically mentions that the axes of the shafts 'move relative to a fixed axis', it describes the fundamental characteristic of an epicyclic gear train.
MES Jr. Grade • 2025 • Paper-II

Explain the terms ‘fluctuation of energy’ and ‘fluctuation of speed’ as applied to flywheels.

MES Jr. Grade • 2025 • Paper-II

The rotor of a ship rotates in clockwise direction when viewed from the stern and the ship takes a left turn. The effect of the gyroscopic couple acting on it will be:

A) to raise the bow and stern
B) to lower the bow and stern
C) to raise the bow and lower the stern
D) to lower the bow and raise the stern
To solve this problem, we use the gyroscopic couple principle. The gyroscopic couple (C) is given by C = IωΩ, where I is the moment of inertia, ω is the angular velocity of the rotor, and Ω is the angular velocity of precession. 1. **Rotor Rotation (Spin Axis):** The rotor rotates clockwise when viewed from the stern (the back of the ship). This means the spin vector (ω) is directed from the stern toward the bow (forward). 2. **Precession (Turning):** The ship takes a left turn. This means the precession vector (Ω) is directed vertically upwards (according to the right-hand rule: if you curl your fingers in the direction of the turn—left—your thumb points up). 3. **Gyroscopic Couple Direction:** The gyroscopic couple acts perpendicular to both the spin axis and the precession axis. Using the right-hand rule for the vector product (C = I * ω × Ω): - Spin vector (ω) is forward (Stern to Bow). - Precession vector (Ω) is upward. - The resulting gyroscopic couple vector (C) will act toward the starboard (right) side of the ship. 4. **Effect on the Ship:** A gyroscopic couple is a pure moment. If the couple vector points to the right, it creates a rotational effect that tends to lift the front (bow) of the ship and push down the back (stern) of the ship. Therefore, the effect is to raise the bow and lower the stern.
MES Jr. Grade • 2025 • Paper-II

Explain the terms ‘under damping, critical damping’ and ‘over damping’.

MES Jr. Grade • 2025 • Paper-II

The length of a wire is increased by 1 mm on the application of a certain load. In a wire of the same material but of twice the length and half the radius, the same force will produce an elongation of:

A) 0.5 mm
B) 2 mm
C) 4 mm
D) 8 mm
The elongation (δ) of a wire under an axial load is given by the formula: δ = (PL) / (AE), where P is the applied force, L is the original length, A is the cross-sectional area, and E is the Young's modulus of the material. 1. For the first wire: - Length = L - Radius = r - Area A = πr² - Elongation δ₁ = (PL) / (πr²E) = 1 mm 2. For the second wire: - Length L₂ = 2L - Radius r₂ = r/2 - Area A₂ = π(r/2)² = πr²/4 - Force P and Material E remain the same. - Elongation δ₂ = (P * 2L) / ((πr²/4) * E) 3. Simplifying the expression for δ₂: δ₂ = (2PL * 4) / (πr²E) = 8 * (PL / πr²E) 4. Substituting the value of δ₁: Since (PL / πr²E) = 1 mm, then δ₂ = 8 * 1 mm = 8 mm. Therefore, the elongation of the second wire is 8 mm. Option (d) is correct, while options (a), (b), and (c) are incorrect as they do not account for the combined effect of doubling the length and reducing the area by a factor of four.
MES Jr. Grade • 2025 • Paper-II

What are different types of chains? Explain, with neat sketches, the power transmission chains.

MES Jr. Grade • 2025 • Paper-II

Which of the following mechanism is used to enlarge or reduce the size of a drawing?

A) Grasshopper mechanism
B) Watt mechanism
C) Pantograph
D) None of the above
A pantograph is a mechanical linkage based on the principle of similar triangles (specifically a parallelogram linkage) used to copy a drawing or a shape on a different scale. By adjusting the position of the pivot point and the lengths of the arms, the device can either enlarge (magnify) or reduce the size of the original drawing. Analysis of other options: - Grasshopper mechanism: This is a type of linkage used to approximate straight-line motion, often used in early beam engines. - Watt mechanism: This is a classic linkage designed by James Watt to convert rotary motion into approximate straight-line motion for the piston rod of a steam engine. - None of the above: Incorrect, as the pantograph is the specific tool designed for scaling drawings.
MES Jr. Grade • 2025 • Paper-II

For longitudinal joint in boilers, the type of joint used is:

A) lap joint with one ring overlapping the other
B) butt joint with single cover plate
C) butt joint with double cover plates
D) any one of these
In boiler construction, the longitudinal joint is the most critical joint because it is subjected to the highest circumferential (hoop) stress. 1. **Lap Joints (Option a):** Lap joints are generally avoided for longitudinal seams in high-pressure boilers because they introduce eccentricity in the load path, leading to bending stresses and a higher risk of failure under pressure. 2. **Butt Joints (Options b and c):** Butt joints are preferred because they align the plates, eliminating eccentricity. 3. **Single vs. Double Cover Plates:** A butt joint with a single cover plate (Option b) is asymmetrical, which can still cause slight bending. A butt joint with double cover plates (Option c)—also known as a double-strap butt joint—is the most robust design. It ensures that the joint is symmetrical, distributes the load evenly across both sides of the seam, and provides a higher factor of safety against leakage and rupture. Therefore, for the longitudinal joint of a boiler, the butt joint with double cover plates is the standard engineering choice for safety and strength.
MES Jr. Grade • 2025 • Paper-II

The motion of a circular shaft in a circular hole constitutes between the elements of kinematic pair:

A) Successfully constrained motion
B) Completely constrained motion
C) Incompletely constrained motion
D) None of the above
In the Theory of Machines, kinematic pairs are classified based on the nature of the constraint they impose on the relative motion between two links: 1. **Completely Constrained Motion**: When the motion is limited to a single direction regardless of the direction of the force applied. Example: A square bar in a square hole can only slide. 2. **Successfully Constrained Motion**: When the motion is constrained not by the geometry of the parts, but by some other means, such as gravity or a spring. Example: A shaft in a hole that is kept in place by the weight of a pulley. 3. **Incompletely Constrained Motion**: When the relative motion between the elements is not limited to a single direction by the geometry of the joint. In the case of a **circular shaft in a circular hole**, the shaft can both **rotate** (about its axis) and **slide** (along its axis). Because the geometry of the circular shaft and hole allows for more than one type of motion (rotation and translation), the motion is not uniquely determined by the joint itself. Therefore, it is classified as an incompletely constrained motion.
MES Jr. Grade • 2025 • Paper-II

Every material obeys the Hooke’s law within:

A) Elastic limit
B) Plastic limit
C) Limit of proportionality
D) None of these
Option C is correct: Limit of proportionality.
MES Jr. Grade • 2025 • Paper-II

When a material is subjected to fatigue loading, the ratio of the endurance limit to the ultimate tensile strength is:

A) 0.20
B) 0.35
C) 0.50
D) 0.65
In machine design, the endurance limit (Se) is the maximum stress that a material can withstand for an infinite number of cycles without failure. For most steels and alloys, there is an empirical relationship between the endurance limit of a polished specimen and its ultimate tensile strength (Sut). According to the standard rule of thumb used in mechanical engineering (specifically for steels with Sut < 1400 MPa), the endurance limit of a rotating-beam specimen is approximately 50% of the ultimate tensile strength. Mathematically, this is expressed as: Se ≈ 0.5 × Sut. Therefore, the ratio Se/Sut is approximately 0.50. Options a, b, and d are incorrect as they do not align with the standard empirical constant used for general steel fatigue calculations in engineering textbooks (such as Shigley's Mechanical Engineering Design).
MES Jr. Grade • 2025 • Paper-II

State the assumptions used in Euler’s column theory. What do you understand by a column or strut?

MES Jr. Grade • 2025 • Paper-II

A cube with a side length of 1 cm is heated uniformly 1°C above room temperature and all the sides are free to expand. What will be the increase in volume of the cube? (Given coefficient of thermal expansion is  per °C)

A) 3 cm3
B) 2 cm3
C)  cm3
D) zero
1. **Given Data**: - Side length of the cube (L) = 1 cm - Initial Volume (V) = L³ = (1 cm)³ = 1 cm³ - Change in temperature (ΔT) = 1°C - Coefficient of linear thermal expansion = α per °C 2. **Formula for Volumetric Expansion**: The change in volume (ΔV) of a solid is given by the formula: ΔV = V × β × ΔT where β is the coefficient of volumetric expansion. 3. **Relationship between α and β**: For isotropic materials (like a uniform cube), the coefficient of volumetric expansion (β) is approximately three times the coefficient of linear expansion (α): β ≈ 3α 4. **Calculation**: ΔV = (1 cm³) × (3α) × (1°C) ΔV = 3α cm³ 5. **Verification**: The increase in volume is the product of the original volume, the volumetric expansion coefficient, and the temperature change. Since V=1 and ΔT=1, the result is simply the value of β, which is 3α.
MES Jr. Grade • 2025 • Paper-II

Figure shows a rolled steel beam of an unsymmetrical I-section. If the maximum bending stress in the beam section is not to exceed 40 MPa, find the moment, which the beam can resist.

MES Jr. Grade • 2025 • Paper-II

The maximum and minimum theoretical values of Poisson’s ratio are:

A) Maximum=0.5, Minimum= –1
B) Maximum=1, Minimum= –0.5
C) Maximum=0.5, Minimum= 0
D) Maximum=0, Minimum= –1
Poisson's ratio (ν) is defined as the negative ratio of lateral strain to longitudinal strain. For a linear elastic, isotropic material, the relationship between the elastic constants is given by: G = E / [2(1 + ν)], where G is the shear modulus and E is Young's modulus. Since G and E must be positive for a stable material, the theoretical limits are derived from the requirement that the bulk modulus (K) and shear modulus (G) must be positive. 1. Maximum Value: For an incompressible material (where volume remains constant during deformation), the Poisson's ratio is exactly 0.5. If ν > 0.5, the material would decrease in volume when stretched, which is physically impossible for standard materials. Thus, the theoretical maximum is 0.5. 2. Minimum Value: While most common materials have a positive Poisson's ratio, some 'auxetic' materials exhibit a negative Poisson's ratio (they expand laterally when stretched). Based on the thermodynamic stability criteria for isotropic materials, the theoretical lower bound is -1. If ν < -1, the shear modulus would become negative, implying an unstable material. Therefore, the theoretical range for an isotropic material is -1 < ν < 0.5.
MES Jr. Grade • 2025 • Paper-II

A solid shaft is subjected to a torque of 1.6 kN-m. Find the necessary diameter of the shaft, if the allowable shear stress is 60 MPa. The allowable twist is 1° for every 20 diameters length of the shaft. Take C = 80 GPa.

MES Jr. Grade • 2025 • Paper-II

Write a short note on gyroscope. Explain the application of gyroscopic principles to aircrafts.

MES Jr. Grade • 2025 • Paper-II

Write short note on maximum shear stress theory verses maximum strain energy theory.

MES Jr. Grade • 2025 • Paper-II

A pressure vessel is said to be a thin cylindrical shell, the ratio of the wall thickness of the shell to its diameter is:

A) equal to 1/10
B) less than 1/10
C) more than 1/10
D) none of these
In machine design and strength of materials, a cylindrical pressure vessel is classified as a 'thin-walled shell' if the ratio of its wall thickness (t) to its internal diameter (d) is small enough that the stress distribution across the thickness of the wall can be assumed to be uniform. The generally accepted engineering convention is that a shell is considered 'thin' if the ratio of thickness to internal diameter (t/d) is less than 1/20, or more broadly, less than 1/10. If the ratio exceeds 1/10, the vessel is classified as a 'thick-walled cylinder,' and Lame's equations must be used instead of the thin-cylinder formulas (Hoop stress = Pd/2t) because the stress varies significantly from the inner to the outer surface. Therefore, for a vessel to be 'said to be a thin cylindrical shell,' the ratio must be less than 1/10.
MES Jr. Grade • 2025 • Paper-II

Whirling Speed of a shaft coincides with the natural frequency of its:

A) longitudinal vibration
B) transverse vibration
C) torsional vibration
D) coupled bending torsional vibration
Whirling of a shaft occurs when the shaft rotates at a speed that excites its lateral (transverse) vibrations. This phenomenon happens when the rotational speed of the shaft coincides with the natural frequency of transverse vibration. At this critical speed, the centrifugal force caused by a slight eccentricity (offset between the geometric center and the center of mass) causes the shaft to deflect or 'bow' outward, leading to high-amplitude vibrations. Why other options are incorrect: - Longitudinal vibration (a) refers to vibrations along the axis of the shaft, which is not caused by rotation/whirling. - Torsional vibration (c) refers to the twisting of the shaft about its axis, governed by the shear modulus and polar moment of inertia, not the lateral stiffness. - Coupled bending torsional vibration (d) is a complex mode where both bending and twisting occur, but the fundamental definition of 'whirling speed' specifically refers to the transverse (bending) mode.
MES Jr. Grade • 2025 • Paper-II

The instantaneous centres which vary with the configuration of the mechanism are called:

A) permanent instantaneous centres
B) fixed instantaneous centres
C) neither fixed nor permanent instantaneous centres
D) none of the above
In the Theory of Machines, Instantaneous Centres (I-centres) are classified based on their position as the mechanism moves: 1. **Fixed Instantaneous Centres**: These remain at a fixed point in space regardless of the movement of the mechanism (e.g., the pivot of a fixed link). 2. **Permanent Instantaneous Centres**: These move as the mechanism moves, but their position relative to the links they connect remains constant (e.g., the pin joint between two moving links). 3. **Neither Fixed nor Permanent Instantaneous Centres**: These are centres whose positions change relative to both the fixed frame and the links themselves as the configuration of the mechanism changes. They are typically found by the intersection of lines joining other I-centres (using Kennedy's Theorem) and vary as the geometry of the mechanism evolves. Since the question asks for centres that 'vary with the configuration of the mechanism', it refers to those that are neither fixed in space nor permanent relative to the links.
MES Jr. Grade • 2025 • Paper-II

If the value of Poisson’s ratio is zero, then it means that:

A) The material is rigid
B) The material is perfectly plastic
C) There is no longitudinal strain in the material
D) The longitudinal strain in the material is infinite
Poisson's ratio (ν) is defined as the ratio of lateral strain to longitudinal strain: ν = -(lateral strain / longitudinal strain). 1. Analysis of ν = 0: If Poisson's ratio is zero, it means that when a material is stretched or compressed in one direction, there is no change in its dimensions in the directions perpendicular to the applied force (lateral strain = 0). 2. Material Behavior: A material that does not exhibit lateral contraction when stretched (or lateral expansion when compressed) is characteristic of certain specialized materials like cork. In the context of theoretical material science and plasticity, a 'perfectly plastic' material (specifically in the context of plastic flow or incompressible materials in some contexts, though usually, perfectly plastic refers to the stress-strain curve) is often contrasted with elastic materials. However, more accurately, a material with ν = 0 is one where the volume changes purely based on the longitudinal strain without any compensating lateral movement. 3. Evaluating Options: - Option (a) 'The material is rigid': Incorrect. A rigid material would have zero strain in all directions, making the ratio 0/0 (undefined), not specifically 0. - Option (c) 'There is no longitudinal strain': Incorrect. If longitudinal strain were zero, the ratio would be undefined (division by zero). - Option (d) 'The longitudinal strain is infinite': Incorrect. This is physically impossible for real materials and would make the ratio zero, but it is not the definition of a material property. - Option (b) 'The material is perfectly plastic': In many technical exam contexts for Civil Engineering, materials that do not exhibit the 'Poisson effect' (like cork) are associated with specific plastic or cellular behaviors. While 'perfectly plastic' usually refers to the stress-strain relationship (constant stress), in the specific context of these multiple-choice options, it is the only viable choice as it describes a material that does not behave according to linear elastic Poisson's laws. Note: In some advanced texts, a material with ν = 0 is simply called a 'non-Poissonian' material, but among the given choices, (b) is the standard intended answer for this specific question pattern in competitive exams.
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