← MES Jr. Grade
Past Questions
In a single-point turning operation of steel with a cemented carbide tool, Taylor’s tool life exponent is 0.25. If the cutting speed is halved, the tool life will increase by
A)
two times
B)
four times
C)
eight times
D)
sixteen times
The relationship between cutting speed and tool life is governed by Taylor's Tool Life Equation: V * T^n = C, where V is the cutting speed, T is the tool life, n is the tool life exponent, and C is a constant.
Given:
- Tool life exponent (n) = 0.25
- Initial speed = V1
- Initial tool life = T1
- Final speed (V2) = V1 / 2
- Final tool life = T2
According to the equation:
V1 * (T1)^n = V2 * (T2)^n
Substitute the values:
V1 * (T1)^0.25 = (V1 / 2) * (T2)^0.25
Divide both sides by V1:
(T1)^0.25 = (1 / 2) * (T2)^0.25
Rearrange to solve for the ratio T2/T1:
(T2 / T1)^0.25 = 2
To solve for T2/T1, raise both sides to the power of (1 / 0.25), which is the power of 4:
(T2 / T1) = 2^(1 / 0.25)
(T2 / T1) = 2^4
(T2 / T1) = 16
Therefore, the tool life increases by 16 times.
Comparing with options:
- a. two times (Incorrect)
- b. four times (Incorrect)
- c. eight times (Incorrect)
- d. sixteen times (Correct)
The layout strategy that deals with low-volume, high-variety production is
A)
fixed-position layout
B)
retail/service layout
C)
all of the above
D)
none of the above
In Industrial Engineering, layout strategies are chosen based on the volume and variety of the product.
1. **Fixed-Position Layout**: This is used when the product is too large, bulky, or fragile to move (e.g., shipbuilding, aircraft assembly, bridge construction). Because these projects are typically unique, highly customized, and produced in very low volumes (often one at a time), this layout is the definitive strategy for 'low-volume, high-variety' production. The product remains stationary, and resources (labor, tools, materials) are brought to it.
2. **Retail/Service Layout**: This focuses on maximizing customer exposure to products and optimizing the flow of customers through a store or service center. It is not categorized by production volume/variety in the same way manufacturing layouts are.
3. **Process-Oriented Layout (Job Shop)**: While not listed as a primary option here, a process layout also handles low-volume/high-variety. However, among the given choices, the fixed-position layout is the only one that fits the description of low-volume, high-variety production (specifically for large-scale projects).
Since 'fixed-position layout' is the only applicable manufacturing strategy listed for this specific production profile, option (a) is the correct choice.
Which one of the following gating systems is best suited to obtain directional solidification?
A)
Top grating
B)
Part-line grating
C)
Bottom grating
D)
Stepped grating
Directional solidification is a process where the casting solidifies in a controlled direction, typically from the furthest point from the riser toward the riser, to prevent shrinkage cavities and porosity.
1. **Top Gating (Top grating):** In a top gating system, the molten metal enters from the top. The hottest metal remains at the top of the mold cavity, while the coldest metal settles at the bottom. Since the riser is also located at the top, the temperature gradient is established such that the bottom solidifies first and the top (near the riser) solidifies last. This creates an ideal condition for directional solidification.
2. **Bottom Gating (Bottom grating):** In bottom gating, the hottest metal ends up at the top of the casting, but the metal entering the bottom cools as it rises. While this reduces turbulence, it often leads to a temperature profile where the bottom may remain liquid longer or the gradient is not as steep as top gating, making it less ideal for strict directional solidification compared to top gating.
3. **Part-line Gating:** This is a compromise between top and bottom gating, typically used to reduce turbulence and dross, but it does not optimize the temperature gradient for directional solidification as effectively as top gating.
4. **Stepped Gating:** This is used for very thick sections to ensure the mold fills uniformly and to reduce turbulence, but it is not the primary choice for establishing a simple directional solidification gradient.
Therefore, Top Gating is the best suited for obtaining directional solidification because it ensures the highest temperature of the metal is maintained at the top, where the riser is located.
Light and intricate parts with close dimensional tolerances of the order of + 0.005 mm are produced by
A)
investment casting
B)
die casting
C)
centrifugal casting
D)
shell mould casting
Step-by-step reasoning:
1. **Analysis of Requirements**: The question specifies three key requirements: 'light and intricate parts', 'close dimensional tolerances', and a specific precision order of approximately ±0.005 mm (though typically investment casting is cited for tolerances in the range of ±0.05 to ±0.1 mm, in the context of these specific options, it is the most precise).
2. **Evaluating Investment Casting**: Also known as the 'lost-wax process', it is specifically designed for producing complex, intricate geometries with high surface finish and excellent dimensional accuracy. Because the mold is a ceramic shell created around a wax pattern, there is no parting line, allowing for extreme intricacy.
3. **Evaluating Die Casting**: While die casting offers good tolerances and high production rates, it is generally used for non-ferrous metals and cannot achieve the same level of intricate internal detail or the extreme precision associated with high-end investment casting for specialized parts.
4. **Evaluating Centrifugal Casting**: This is used primarily for cylindrical parts (like pipes or bushings) to ensure a dense, grain-refined structure. It is not used for 'intricate' parts.
5. **Evaluating Shell Mould Casting**: This provides better accuracy than sand casting but is significantly less precise than investment casting and cannot produce the same level of intricate detail.
6. **Conclusion**: Investment casting is the industry standard for high-precision, intricate components (such as turbine blades or jewelry) that require the tightest tolerances among the listed casting methods.
For 50% penetration of work material, a punch with single shear equal to thickness will
A)
reduce the punch load to half the value
B)
increase the punch load by half the value
C)
maintain the same punch load
D)
reduce the punch load to quarter load
In a punching or blanking operation, the force (punch load) required to shear the material is calculated by the formula: F = L × t × τ, where 'F' is the punch load, 'L' is the length of the cut (perimeter), 't' is the thickness of the material, and 'τ' is the shear strength of the material. In a standard single shear operation, the punch must penetrate the full thickness (t) of the material to complete the cut. If the operation is modified such that only 50% penetration of the work material is required (or if the effective thickness being sheared at any given moment is reduced by half), the value of 't' in the equation is effectively halved. Since the force F is directly proportional to the thickness t (F ∝ t), reducing the penetration/thickness by 50% results in the punch load being reduced to half of its original value. Therefore, option (a) is correct.
In grey cast iron, carbon is present in the form of-
A)
Cementite
B)
Free carbon
C)
Flakes
D)
Spheroids
Grey cast iron is characterized by the presence of graphite in the form of thin, interconnected flakes. This occurs because the carbon in the molten metal precipitates as graphite during slow cooling. These graphite flakes act as internal stress raisers, which gives grey cast iron its characteristic brittle nature and excellent vibration-damping properties.
Analysis of other options:
- a. Cementite (Fe3C) is a hard, brittle intermetallic compound found in white cast iron or steel, but in grey cast iron, most of the carbon is converted to graphite.
- b. 'Free carbon' is a general term for graphite, but 'Flakes' is the specific morphological form that defines 'Grey' cast iron.
- d. Spheroids (or nodules) are characteristic of Ductile Cast Iron (Nodular Cast Iron), where additives like magnesium or cerium are used to force the graphite into spherical shapes to improve toughness.
The ability of material to resist softening at high temperature is known as-
A)
Creep
B)
Hot tempering
C)
Hot hardness
D)
Fatigue
The ability of a material to maintain its hardness and resist softening (deformation) when subjected to high temperatures is defined as 'Hot Hardness'. This property is critical for tools used in high-speed machining (like High-Speed Steel - HSS), where friction generates significant heat.
Why other options are incorrect:
- a. Creep: This is the slow, progressive, permanent deformation of a material under a constant stress at high temperatures over a long period. It is a result of softening, not the resistance to it.
- b. Hot tempering: Tempering is a heat treatment process used to increase toughness and reduce brittleness in steel; it is a process, not a material property.
- d. Fatigue: This is the weakening of a material caused by repeatedly applied loads (cyclic stress), which eventually leads to structural failure, regardless of whether the temperature is high or low.
The linear programming techniques can be applied successfully to industries like
A)
iron and steel
B)
food processing
C)
banking
D)
all of the above
Linear Programming (LP) is a mathematical optimization technique used to achieve the best outcome (such as maximum profit or lowest cost) in a mathematical model whose requirements are represented by linear relationships. Its versatility allows it to be applied across virtually all sectors of industry:
1. Iron and Steel (Manufacturing): LP is used for 'blending problems' (mixing raw materials to achieve a specific grade of steel at minimum cost) and production scheduling to optimize furnace usage.
2. Food Processing: LP is used for 'least-cost feed formulation' (mixing ingredients to meet nutritional requirements at minimum cost) and optimizing the supply chain for perishable goods.
3. Banking (Finance): LP is used for portfolio optimization, managing asset allocation to maximize returns while minimizing risk, and optimizing loan approvals based on capital constraints.
Since LP is a general-purpose optimization tool applicable to any system with linear constraints and a linear objective function, it is successfully applied to all the listed industries.
A weight falls on a plunger fitted in a container filled with oil thereby producing a pressure of 1.5 N/mm² in the oil. The Bulk Modulus of oil is 2800 N/mm². Given this situation, the volumetric compressive strain produced in the oil will be:
A)
400 x 10⁻⁶
B)
800 x 10⁻⁶
C)
268 x 10⁻⁶
D)
535 x 10⁻⁶
To find the volumetric compressive strain, we use the definition of the Bulk Modulus of elasticity.
1. **Given Data:**
- Pressure produced (Change in pressure, ΔP) = 1.5 N/mm²
- Bulk Modulus of oil (K) = 2800 N/mm²
2. **Formula:**
The Bulk Modulus (K) is defined as the ratio of the change in pressure (ΔP) to the volumetric strain (εᵥ):
K = ΔP / εᵥ
3. **Calculation:**
Rearranging the formula to solve for volumetric strain (εᵥ):
εᵥ = ΔP / K
εᵥ = 1.5 / 2800
εᵥ ≈ 0.000535714
4. **Converting to Scientific Notation:**
εᵥ = 535.714 x 10⁻⁶
Comparing this result to the given options, it aligns most closely with option (d) 535 x 10⁻⁶.
- Option (a) 400 x 10⁻⁶ is incorrect as it would require a pressure of ~1.12 N/mm².
- Option (b) 800 x 10⁻⁶ is incorrect as it would require a pressure of ~2.24 N/mm².
- Option (c) 268 x 10⁻⁶ is incorrect as it would require a pressure of ~0.75 N/mm².
A solid circular shaft of diameter 100 mm is subjected to an axial stress of 50 MPa. It is further subjected to a torque of 10 kNm. The maximum principal stress experienced on the shaft is closest to
A)
41 MPa
B)
82 MPa
C)
164 MPa
D)
204 MPa
Step-by-step solution:
1. **Given Data:**
- Diameter of shaft (d) = 100 mm = 0.1 m
- Axial stress (σₓ) = 50 MPa = 50 × 10⁶ N/m²
- Torque (T) = 10 kNm = 10 × 10³ Nm
2. **Calculate Torsional Shear Stress (τ):**
The shear stress induced by torque in a solid circular shaft is given by:
τ = (16T) / (πd³)
τ = (16 × 10,000) / (π × 0.1³)
τ = 160,000 / (π × 0.001)
τ = 160,000,000 / π ≈ 50.93 MPa
3. **Calculate Maximum Principal Stress (σ₁):**
The shaft is subjected to a combined state of axial stress (σₓ) and shear stress (τ). The formula for the maximum principal stress is:
σ₁ = (σₓ / 2) + √((σₓ / 2)² + τ²)
σ₁ = (50 / 2) + √((50 / 2)² + 50.93²)
σ₁ = 25 + √(25² + 50.93²)
σ₁ = 25 + √(625 + 2593.86)
σ₁ = 25 + √3218.86
σ₁ = 25 + 56.73
σ₁ = 81.73 MPa
4. **Conclusion:**
The calculated maximum principal stress is approximately 81.73 MPa, which is closest to 82 MPa.
Comparing with options:
- (a) 41 MPa: Incorrect.
- (b) 82 MPa: Correct (closest value).
- (c) 164 MPa: Incorrect.
- (d) 204 MPa: Incorrect.
The shapes of the bending moment diagram for a uniform cantilever beam carrying a uniformly distributed load over its length is:
A)
A straight line
B)
A hyperbola
C)
An ellipse
D)
A parabola
To determine the shape of the Bending Moment (BM) diagram, we analyze the relationship between the bending moment and the position along the beam.
1. **Setup**: Consider a cantilever beam of length 'L' fixed at one end and free at the other, carrying a uniformly distributed load (UDL) of intensity 'w' per unit length.
2. **Moment Equation**: Let 'x' be the distance from the free end. The total load acting on the segment of length 'x' is (w * x). This load acts at the center of gravity of the segment, which is at a distance of (x/2) from the section being considered.
3. **Calculation**: The bending moment M at distance 'x' is given by:
M = (Load) * (Distance) = (w * x) * (x / 2) = (w / 2) * x².
4. **Analysis of the Equation**: The equation M = (w/2)x² is a second-degree polynomial equation. In mathematics, an equation where the dependent variable (M) is proportional to the square of the independent variable (x) represents a parabola.
**Comparison with other options:**
- **Straight line**: This occurs when the load is a point load (M is proportional to x).
- **Hyperbola/Ellipse**: These shapes do not occur in standard beam bending moment diagrams for common loading conditions like UDL or point loads.
A test specimen is stressed slightly beyond the yield point and then unloaded. Its yield strength will
A)
Decrease
B)
Increase
C)
Remains same
D)
Becomes equal to ultimate tensile strength
This phenomenon is known as 'Strain Hardening' or 'Work Hardening'. When a ductile material (like mild steel) is stressed beyond its yield point, it undergoes plastic deformation. During this process, dislocations within the crystal lattice of the metal multiply and tangle, creating barriers to further dislocation movement. When the specimen is unloaded and then re-loaded, the material exhibits a higher yield strength than it did originally. The new yield point is effectively the stress level at which the material was previously unloaded. Therefore, the yield strength increases.
- Option (a) is incorrect because the material becomes stronger (harder), not weaker.
- Option (c) is incorrect because the internal structure of the material has been permanently altered by plastic deformation.
- Option (d) is incorrect because while the yield strength increases, it does not automatically jump to the ultimate tensile strength unless the material was stressed up to that specific limit before unloading.
The advantage of the piston valve over D-slide valve is that in the former case
A)
Wear is less
B)
Power absorbed is less
C)
Both wear and power absorbed are low
D)
The pressure developed being high provides tight sealing
In steam engines and compressors, the D-slide valve is a common mechanism, but it suffers from significant friction. Because the steam pressure acts directly on the face of the D-slide valve, it presses the valve firmly against the valve seat, creating high frictional resistance. This leads to two main problems: high power absorption (to overcome friction) and rapid wear of the valve surfaces.
In contrast, a piston valve is cylindrical and is 'balanced' against the steam pressure. Since the pressure acts equally around the circumference of the piston, there is no resultant force pushing the valve against the walls of the chest. This 'balanced' nature significantly reduces the friction between the valve and its guide. Consequently, the power required to operate the valve is much lower, and the mechanical wear is greatly reduced compared to the D-slide valve. Therefore, both wear and power absorption are low in the piston valve.
Maximum angular velocity of the connecting rod with a crank to connecting rod ratio 1:5 for a crank speed of 3000 rpm is around:
A)
300 rad/s
B)
60 rad/s
C)
30 rad/s
D)
3000 rad/s
Step-by-step calculation:
1. **Given Data**:
- Crank speed (N) = 3000 rpm
- Ratio of crank length (r) to connecting rod length (l) = 1:5. Therefore, n = l/r = 5.
2. **Calculate Angular Velocity of Crank (ω)**:
ω = (2 * π * N) / 60
ω = (2 * π * 3000) / 60 = 100 * π ≈ 314.16 rad/s
3. **Formula for Maximum Angular Velocity of Connecting Rod (ω_con)**:
In a slider-crank mechanism, the angular velocity of the connecting rod is given by:
ω_con = ω * (tan θ) / (n / cos φ) ≈ ω * (sin θ) / n
The maximum angular velocity of the connecting rod occurs approximately when the crank is perpendicular to the connecting rod. A widely accepted approximation for the maximum angular velocity of the connecting rod is:
ω_con_max ≈ ω * (sin θ_max) / n
For a large n, the maximum value occurs when sin θ is maximum (approx 1) or more accurately using the formula:
ω_con_max = ω * (sin θ) / sqrt(n^2 - sin^2 θ)
At the point of maximum angular velocity, sin θ ≈ 1/n (approximately), but for practical engineering calculations in competitive exams, the simplified maximum is often taken as:
ω_con_max ≈ ω / n
4. **Calculation**:
ω_con_max ≈ 314.16 / 5
ω_con_max ≈ 62.83 rad/s
5. **Conclusion**:
Comparing 62.83 rad/s with the given options, the closest value is 60 rad/s.
- Option (a) 300 rad/s is too high (close to the crank speed).
- Option (c) 30 rad/s is too low.
- Option (d) 3000 rad/s is incorrect as it matches the RPM value, not rad/s.
What is the ratio of Youngs modulus E to shear modulus G in terms of Poisson ratio?
A)
2(1 + 4)
B)
2(1 + μ)
C)
172(1 - μ)
D)
12(1 + 4)
In the study of Strength of Materials and Elasticity, there are fundamental relationships between the elastic constants for an isotropic material. The three primary constants are Young's Modulus (E), the Shear Modulus (G), and the Bulk Modulus (K), linked by Poisson's ratio (μ or ν).
The relationship between Young's Modulus (E) and Shear Modulus (G) is given by the formula:
E = 2G(1 + μ)
To find the ratio of Young's modulus E to shear modulus G, we rearrange the formula:
E / G = 2(1 + μ)
Analyzing the options:
- Option (a) '2(1 + 4)' is mathematically incorrect and likely a typographical error for (1 + μ).
- Option (b) '2(1 + μ)' matches the derived formula exactly.
- Option (c) '172(1 - μ)' is incorrect and does not correspond to any standard elastic relationship.
- Option (d) '12(1 + 4)' is mathematically incorrect.
Therefore, the correct ratio is 2(1 + μ).
Deformation per unit length in the direction of force is known as
A)
Strain
B)
Lateral strain
C)
Linear strain
D)
Linear stress
Option A is correct: Strain.
In a journal bearing, the radius of the friction circle increases with the increase in
A)
load
B)
radius of the journal
C)
speed of the journal
D)
viscosity of the lubricant
In a journal bearing, the 'friction circle' is an imaginary circle on the surface of the journal where the total frictional force is assumed to act. The radius of this friction circle (r) is typically expressed as a fraction of the journal radius (R), often denoted as r = kR. According to the theory of lubrication and bearing design, the position of the friction circle depends on the load carried by the bearing. As the load (W) increases, the pressure distribution in the lubricant film changes, and the point of application of the resultant frictional force shifts further away from the center of the journal. Therefore, the radius of the friction circle increases as the load increases.
Analyzing other options:
- Radius of the journal (b): While the friction circle radius is proportional to the journal radius, the question asks what causes it to 'increase' (implying a change in the operating condition or design parameter for a given bearing). The ratio k is primarily influenced by the load.
- Speed of the journal (c): Increasing speed generally increases the thickness of the oil film (hydrodynamic lift), which typically reduces the coefficient of friction and does not directly increase the friction circle radius in the same manner as load.
- Viscosity of the lubricant (d): Viscosity affects the friction torque and film thickness, but the primary driver for the shift in the resultant force location (the friction circle radius) is the applied load.
In the case of an involute toothed gear, involute starts from
A)
addendum circle
B)
dedendum circle
C)
pitch circle
D)
base circle
An involute curve is defined as the path traced by a point on a straight line that rolls without slipping on a circle. In gear geometry, this 'rolling circle' is known as the base circle. Therefore, the involute profile of a gear tooth physically begins at the base circle and extends outward toward the addendum circle.
Analysis of options:
- a. Addendum circle: This is the outermost circle of the gear; the involute ends here.
- b. Dedendum circle: This is the bottom-most circle of the tooth space; the profile here is usually a radial fillet curve, not an involute.
- c. Pitch circle: This is the theoretical circle used to define the gear ratio and kinematics; while the involute passes through it, it does not start there.
- d. Base circle: This is the fundamental circle from which the involute is generated. The involute profile starts exactly at the base circle.
In a four stroke I.C. engine, the turning moment during the compression stroke is
A)
Negative throughout
B)
Positive throughout
C)
Negative during major portion of the stroke
D)
Positive during major portion of the stroke
In an Internal Combustion (IC) engine, the 'turning moment' (or torque) refers to the torque acting on the crankshaft.
1. **Positive Torque:** Occurs when the gas pressure in the cylinder pushes the piston down (during the power stroke), driving the crankshaft in its direction of rotation.
2. **Negative Torque:** Occurs when the crankshaft must do work to move the piston against the pressure of the gas (during the compression stroke). In this case, the gas pressure opposes the motion of the piston, and the energy is supplied by the flywheel.
During the compression stroke, the piston moves from Bottom Dead Center (BDC) to Top Dead Center (TDC), compressing the air-fuel mixture. Since the gas pressure is increasing and opposing the upward motion of the piston, the torque is negative for the vast majority of the stroke.
However, it is not 'negative throughout' because, near the dead centers (TDC and BDC), the geometry of the crank-slider mechanism results in a turning moment that approaches zero. Furthermore, in some specific engine geometries or due to inertia effects at very high speeds, there can be momentary fluctuations, but the dominant characteristic of the compression stroke is negative torque. Therefore, the most accurate description is that it is negative during the major portion of the stroke.
If the shear force acting at every section of a beam is of the same magnitude and of the same direction then it represents a
A)
Simply supported beam with a concentrated load at the centre.
B)
Overhung beam having equal overhang at both supports and carrying equal concentrated loads acting in the same direction at the free ends.
C)
Cantilever subjected to concentrated load at the free end.
D)
Simply supported beam having concentrated loads of equal magnitude and in the same direction acting at equal distances from the supports.
To determine the correct option, we must analyze the Shear Force (SF) distribution for each case. The condition given is that the shear force must be of the same magnitude and the same direction (constant SF) across every section of the beam.
1. **Option (a): Simply supported beam with a concentrated load at the center.** The SF is positive from the left support to the center and negative from the center to the right support. The magnitude is the same, but the direction changes. Thus, it is not constant.
2. **Option (b): Overhung beam with equal overhangs and equal loads at free ends.** The SF in the overhangs will be equal to the load, but the SF in the central span between the supports will be zero (since the loads at the ends cancel each other out in the internal section). Thus, it is not constant across the entire beam.
3. **Option (c): Cantilever subjected to a concentrated load at the free end.** For a cantilever of length L with a load P at the free end, the shear force at any section x (measured from the free end) is constant and equal to P. Since the load is acting in one direction, the SF remains the same in magnitude and direction throughout the entire length of the beam. This satisfies the condition.
4. **Option (d): Simply supported beam with equal loads at equal distances from supports.** The SF will be constant between the support and the load, but it will change magnitude (drop to zero or change sign) between the two loads. Thus, it is not constant across every section.
Therefore, only the cantilever beam with a point load at the free end maintains a constant shear force throughout its length.
The following is not a friction clutch
A)
Centrifugal clutch
B)
Cone clutch
C)
Disc clutch
D)
Fluid clutch
A friction clutch is a device that transmits torque from one shaft to another through the application of friction between two surfaces.
1. Centrifugal clutch (Option a): Uses centrifugal force to push friction pads against a drum to engage the drive. It is a friction clutch.
2. Cone clutch (Option b): Uses conical surfaces that press against each other to transmit torque via friction. It is a friction clutch.
3. Disc clutch (Option c): Uses flat friction plates (discs) pressed together to transmit torque. It is a friction clutch.
4. Fluid clutch (Option d): Also known as a fluid coupling, it transmits torque through the hydrodynamic action of a fluid (usually oil) between an impeller and a runner. There is no physical contact or sliding friction between the driving and driven members. Therefore, it is a hydraulic/fluid coupling, not a friction clutch.
During tensile-testing of a specimen using a Universal Testing Machine, the parameters actually measured include
A)
True stress and true strain
B)
Poisson’s ratio and Young’s modulus
C)
Engineering stress and engineering strain
D)
Load and elongation
In a tensile test conducted using a Universal Testing Machine (UTM), the machine is equipped with sensors (such as a load cell for force and an extensometer or crosshead displacement sensor for length change). Therefore, the raw data 'actually measured' by the hardware are the applied Load (P) and the resulting Elongation (ΔL).
All other parameters mentioned in the options are 'calculated' or 'derived' values:
- Engineering Stress (σ = P/A₀) and Engineering Strain (ε = ΔL/L₀) are calculated by dividing the measured load and elongation by the original cross-sectional area and original length, respectively.
- True Stress and True Strain are derived from engineering values by accounting for the instantaneous change in area and length during deformation.
- Young's Modulus (E = σ/ε) and Poisson's ratio are material properties calculated from the slope of the stress-strain curve and the ratio of lateral to longitudinal strain, respectively.
Since the question asks for parameters 'actually measured', the correct answer is Load and elongation.
Belt slip may occur due to
A)
Heavy load
B)
Loose belt
C)
Driving pulley too small
D)
All of these
Belt slip occurs when the frictional force between the belt and the pulley is insufficient to transmit the required torque. The condition for no-slip is governed by the belt tension ratio formula: (T1/T2) ≤ e^(μθ), where T1 is the tight side tension, T2 is the slack side tension, μ is the coefficient of friction, and θ is the angle of wrap.
1. Heavy load (Option a): An increase in load increases the required torque, which increases the difference between T1 and T2. If T1/T2 exceeds e^(μθ), the belt will slip.
2. Loose belt (Option b): A loose belt reduces the initial tension (T0), which directly lowers the normal force acting on the pulley surface. Since frictional force is proportional to the normal force (F = μN), a loose belt significantly reduces the grip, leading to slip.
3. Driving pulley too small (Option c): A smaller pulley diameter reduces the angle of wrap (θ) if the center distance remains constant, or it increases the bending stress and reduces the effective contact area/pressure distribution. A smaller θ decreases the value of e^(μθ), making the system more prone to slipping.
Since all three factors contribute to the occurrence of belt slip, the correct option is 'All of these'.
In V belt drive, the belt touches
A)
At the bottom
B)
At sides only
C)
Could touch anywhere
D)
None of these
A V-belt is designed with a trapezoidal cross-section and fits into a V-shaped groove on the pulley. Due to the wedge action, the belt is forced against the inclined sides of the pulley groove. Under normal operating tension, the belt makes contact with the side walls of the groove, but it does not touch the bottom of the groove. This 'wedge effect' increases the normal force between the belt and the pulley, which significantly increases the frictional grip and reduces the likelihood of slippage compared to flat belts. If the belt were to touch the bottom, it would indicate that the belt is too loose or worn out, leading to a loss of efficiency.
A reed type tachometer uses the principle of
A)
torsional vibration
B)
longitudinal vibration
C)
transverse vibration
D)
damped free vibration
A reed-type tachometer operates on the principle of mechanical resonance. It consists of a set of steel reeds (cantilever beams) of varying lengths, each having a specific natural frequency of transverse vibration. A cam or a gear attached to the rotating shaft strikes these reeds at a frequency equal to the shaft's rotational speed. When the frequency of the striking cam matches the natural frequency of a particular reed, that reed enters resonance and vibrates with maximum amplitude (transverse vibration). The operator identifies the rotating speed by observing which reed is vibrating most vigorously.
- Torsional vibration (a) involves twisting of the shaft, which is not the mechanism here.
- Longitudinal vibration (b) involves axial movement, whereas the reeds move perpendicular to their length.
- Damped free vibration (d) refers to a system returning to equilibrium without external force, whereas the tachometer relies on forced vibration and resonance.