All questions of Engineering Materials for Mechanical Engineering Exam

Martensite is a super-saturated solution of carbon in 
  • a)
    Alpha iron
  • b)
    Beta iron
  • c)
    Gamma iron
  • d)
    Delta iron
Correct answer is option 'A'. Can you explain this answer?

Rajeev Sharma answered
Martensite is super saturated solution of carbon in a-iron. It is diffusion less process and marten site has BCT(body centered Tetragonal structure)

Cast iron is used for machine beds because of its high 
  • a)
    Tensile strength
  • b)
    Endurance strength
  • c)
    Damping capacity
  • d)
    Compressive strength
Correct answer is option 'C,D'. Can you explain this answer?

Cast iron is a popular material used for machine beds due to its unique properties. The following are the reasons why cast iron is preferred for machine beds:

Damping capacity:
Cast iron has a high damping capacity, making it an ideal material for machine beds. Damping capacity refers to the ability of a material to absorb and dissipate vibrations. This property is important in machine tools because it helps to reduce vibrations and improve accuracy.

Compressive strength:
Cast iron is also known for its high compressive strength, which is the ability to withstand crushing forces without deformation or failure. This property is important in machine tools because it ensures that the machine bed can support the weight of heavy workpieces and resist deformation under load.

Other properties:
Cast iron also has other properties that make it suitable for machine beds, including high wear resistance, good thermal conductivity, and low thermal expansion. These properties help to ensure that the machine bed can withstand the rigors of machining operations and maintain its accuracy over time.

Conclusion:
In conclusion, the high damping capacity and compressive strength of cast iron make it an ideal material for machine beds. Its other properties also contribute to its suitability for this application.

Which of the following metal shrinks most from molten state to solid state?
a) Cast iron
b) Admiralty metal
c) Brass
d) Cast steel
Correct answer is option 'B'. Can you explain this answer?

Shubham Mehra answered
Admiralty metal is an alloy of not less than 70 percent copper, about 1 percent tin, small amounts of other elements, and the balance zinc; tin brass.
Admiralty metal shrinks most from molten state to solid state

When the temperature of a solid metal increases,  
  • a)
    Strength of the metal decreases but ductility increases
  • b)
    Both strength and ductility of the metal decrease
  • c)
    Both strength and ductility of the metal increase
  • d)
    Strength of the metal increases but ductility decreases
Correct answer is option 'A'. Can you explain this answer?

When the temperature of a solid metal increases, its intramolecular bonds are brake and strength of solid metal decreases. Due to decrease its strength, the elongation of the metal increases, when we apply the load i.e. ductility increases.

Which one of the following pairs of axis lengths (a, b, c) and inter-axial angles (α,β, γ) represents the tetragonal crystal system?  
  • a)
    a = b = c;α = β = γ = 900
  • b)
    a = b ≠ c;α = β = γ = 900
  • c)
    a ≠ b ≠ c;α = β = γ = 900
  • d)
    a = b = c;α = β = γ ≠ 900
Correct answer is option 'B'. Can you explain this answer?

Sanskriti Basu answered
Understanding Tetragonal Crystal System
The tetragonal crystal system is one of the seven crystal systems in crystallography. It is characterized by specific relationships between the lengths of its axes and the angles between them.
Key Characteristics of Tetragonal System
- Axis Lengths: In a tetragonal crystal system, two axes are of equal length, while the third axis is of a different length. This can be summarized as:
- a = b ≠ c
- Inter-Axial Angles: All angles between the axes are right angles (90 degrees). Therefore, the angles can be defined as:
- α = β = γ = 90°
Analyzing the Options
- Option A: a = b = c; α = β = γ = 90°
- This describes a cubic system, not tetragonal.
- Option B: a = b ≠ c; α = β = γ = 90°
- This correctly represents the tetragonal system, as it has two equal axes and one different.
- Option C: a ≠ b ≠ c; α = β = γ = 90°
- This describes an orthorhombic system, where all axes are of different lengths.
- Option D: a = b = c; α = β = 90°
- This again describes a cubic system.
Conclusion
The correct answer is indeed Option B: a = b ≠ c; α = β = γ = 90°, accurately representing the characteristics of the tetragonal crystal system.

Major operations in the manufacture of steel balls used for Ball bearings are givenbelow 
1. Oil lapping
2. Cold heading
3. Annealing
4. Hardening
5. Rough grinding
The correct sequence of these operations is
  • a)
    3,2,4,1,5
  • b)
    3,2,1,4,5
  • c)
    2,3,4,5,1
  • d)
    2,3,5,4,1
Correct answer is option 'C'. Can you explain this answer?

Asha Nambiar answered
Understanding the Manufacturing Sequence of Steel Balls
The manufacture of steel balls for ball bearings involves several critical operations, each serving a specific purpose. The correct sequence of these operations is crucial to achieve the desired mechanical properties and surface finish.
1. Cold Heading
- This is the initial operation where steel wire is cut and shaped into rough ball forms using dies.
- It is essential for forming the basic shape and size of the balls.
2. Annealing
- After cold heading, the balls undergo annealing, which involves heating and then slowly cooling them.
- This process relieves internal stresses and improves the ductility of the material.
3. Rough Grinding
- Following annealing, rough grinding is performed to remove excess material and achieve a more precise size and shape.
- It prepares the balls for further refinement and improves their surface finish.
4. Hardening
- Hardening is a heat treatment process that significantly increases the hardness of the balls.
- This step ensures that the balls can withstand heavy loads and high friction during operation.
5. Oil Lapping
- The final operation is oil lapping, which involves polishing the balls with abrasive materials and oil.
- This step enhances the surface finish and reduces friction, ensuring smooth operation in ball bearings.
Conclusion
The correct sequence, therefore, is: Cold Heading (2), Annealing (3), Rough Grinding (5), Hardening (4), and Oil Lapping (1). This sequence (option 'C') ensures that each process effectively prepares the steel balls for the next, resulting in high-quality ball bearings.

Phenol formaldehyde is a/an  
  • a)
    Thermoplastic polymer
  • b)
    Thermoset polymer
  • c)
    Elastomer
  • d)
    Rubber
Correct answer is option 'B'. Can you explain this answer?

Pankaj Joshi answered
Phenol Formaldehyde as a Thermoset Polymer

Introduction:
Phenol formaldehyde, also known as phenolic resin, is a type of polymer that is formed by the reaction between phenol and formaldehyde. It is a versatile and widely used polymer due to its excellent thermal and mechanical properties. In this response, we will explain why phenol formaldehyde is classified as a thermoset polymer.

Definition of Thermoset Polymer:
Before delving into the classification of phenol formaldehyde, let's understand what a thermoset polymer is. A thermoset polymer is a type of polymer that undergoes a chemical reaction called crosslinking during its curing process. This crosslinking forms a three-dimensional network structure that cannot be melted or re-molded once it is set. Thermoset polymers are known for their high strength, rigidity, and resistance to heat and chemicals.

Explanation:
Phenol formaldehyde falls under the category of thermoset polymers due to the following reasons:

Crosslinking:
During the curing process, the phenol formaldehyde resin undergoes crosslinking reactions. The formaldehyde molecules react with phenol molecules to form methylene bridges. These bridges create strong covalent bonds between the polymer chains, resulting in a three-dimensional network structure. Once the crosslinking is complete, the polymer becomes rigid and cannot be reshaped.

Irreversible Curing:
The curing of phenol formaldehyde is an irreversible process. Once the resin is heated and cured, it solidifies into a hard and infusible material. Unlike thermoplastic polymers, which can be melted and reshaped multiple times, phenol formaldehyde cannot be melted without decomposing. This irreversibility is a characteristic feature of thermoset polymers.

Properties:
Phenol formaldehyde exhibits excellent mechanical properties, such as high strength, stiffness, and dimensional stability. It also possesses good thermal resistance, electrical insulation properties, and resistance to chemicals. These properties are desirable in various applications, including electrical components, adhesives, coatings, and molded products.

Conclusion:
In summary, phenol formaldehyde is classified as a thermoset polymer due to its crosslinking reactions during curing, irreversible curing process, and the resulting three-dimensional network structure. Its unique properties make it suitable for a wide range of industrial applications.

Match List I with List IT and select the correct answer using the codes given below the lists:  
List - I (Cutting tool Material)                 List - I I(Major characteristic constituent)
A. High speed steel                               1. Carbon
B. Stellite                                                2. Molybdenum
C. Diamond                                            3. Nitride
D. Coated carbide tool                           4. Columbium
                                                               5. Cobalt
Codes: A B C D                              A B C D
        (a) 2 1 3 5                           (b) 2 5 1 3
         (c) 5 2 4 3                           (d) 5 4 2 3
  • a)
    (a)
  • b)
    (b)
  • c)
    (c)
  • d)
    (d)
Correct answer is option 'B'. Can you explain this answer?

List I (Cutting tool Material) List II (Major characteristic constituent)
A. High speed steel 1. Carbon
B. Stellite 2. Molybdenum
C. Diamond 3. Nitride
D. Coated carbide tool 4. Columbium
5. Cobalt

Codes: A B C D A B C D
(a) 2 1 3 5 (b) 2 5 1 3
(c) 5 2 4 3 (d) 5 4 2 3

The correct answer is option 'B' (2 5 1 3). Let's understand why:

1. High speed steel: High speed steel is a cutting tool material that is known for its high hardness, toughness, and wear resistance. It is mainly composed of iron, carbon, tungsten, chromium, vanadium, and sometimes cobalt. The major characteristic constituent in high speed steel is carbon (Option 1).

2. Stellite: Stellite is a cobalt-based alloy that is commonly used as a cutting tool material. It is known for its excellent wear resistance, high temperature resistance, and corrosion resistance. The major characteristic constituent in Stellite is cobalt (Option 5).

3. Diamond: Diamond is an extremely hard material that is used in cutting tools for machining non-ferrous materials, ceramics, and composites. It has exceptional hardness and thermal conductivity. The major characteristic constituent in diamond is carbon (Option 1).

4. Coated carbide tool: Coated carbide tools are widely used in machining operations due to their high hardness, wear resistance, and thermal stability. The major characteristic constituent in coated carbide tools is tungsten carbide (Option 2). However, none of the given options match with tungsten carbide, so we can eliminate this option.

By eliminating option 4, we are left with option 'B' (2 5 1 3) as the correct answer.

Consider the following statements about medium carbon steel:  
1. It can be quench-hardened but not case-hardened.
2. It cannot be quench-hardened but case-hardening can be done.
3. It exhibits distinct yield point under tension test.
Which of the following statements given above are correct?
  • a)
    1 and 2
  • b)
    2 and 3
  • c)
    1 and 3
  • d)
    1, 2 and 3
Correct answer is option 'C'. Can you explain this answer?

Kritika Joshi answered
Statement 1: Medium carbon steel can be quench-hardened but not case-hardened.
Quench-hardening is a heat treatment process in which steel is heated to a high temperature and then rapidly cooled (quenched) in a liquid medium, typically oil or water. This process results in a hardened structure with increased strength and hardness. Medium carbon steel, typically containing around 0.3-0.6% carbon, can undergo quench-hardening. The high carbon content allows for the formation of a hard martensitic structure during the quenching process.

On the other hand, case-hardening is a heat treatment process used to increase the hardness of the outer surface of a steel component while retaining a relatively tough and ductile core. This is achieved by introducing carbon or nitrogen into the surface layer of the steel through a diffusion process. Medium carbon steel is not suitable for case-hardening as it already contains a sufficient amount of carbon to achieve the desired hardness through quench-hardening. Therefore, statement 1 is correct.

Statement 2: Medium carbon steel cannot be quench-hardened but case-hardening can be done.
This statement is incorrect. As mentioned earlier, medium carbon steel can be quench-hardened by heating it to a high temperature and then rapidly cooling it. Case-hardening, however, is not suitable for medium carbon steel.

Statement 3: Medium carbon steel exhibits a distinct yield point under tension test.
The yield point is the stress level at which a material begins to deform plastically, meaning it does not return to its original shape after the applied stress is removed. Medium carbon steel, like most steels, typically exhibits a distinct yield point during a tension test. This yield point is characterized by a sudden drop in stress after which the material undergoes plastic deformation. Therefore, statement 3 is correct.

In conclusion, statements 1 and 3 are correct. Medium carbon steel can be quench-hardened and exhibits a distinct yield point under tension test. Statement 2 is incorrect as medium carbon steel is not suitable for case-hardening. Therefore, the correct answer is option C.

Herbert cloudburst Hardness test is conducted to know 
  • a)
    Uniformity of hardness over a surface
  • b)
    Softness of non-metallic components
  • c)
    Hardness of non-metallic components
  • d)
    Hardness at specified depth inside the surface.
Correct answer is option 'A'. Can you explain this answer?

Ans. (a)
In this metal surface is subjected to rain of several hundred thousand 3 mm diameter hard steel balls talling from a known height. The impact of each ball produces an imprint varying inversely in size with the hardness of the material at the point of impact. The imprint-size pattern shows whether or not there is uniformity of hardness over the entire test are.

Coefficient of Expansion is practically nil in a particular alloy. What is this alloy?
  • a)
    Hadfield Manganese Steel
  • b)
    Invar 
  • c)
    Vitallium
  • d)
    Stellite
Correct answer is option 'B'. Can you explain this answer?

Divya Banerjee answered
Ans. (b) A nickel steel alloy containing 36% of nickel is known as invar. It has nearly zero coefficient of expansion. So it is in great demand for measuring instruments and standards of lengths for everyday use. (Invar - 64% Fe, 36% Ni)

Assuming atoms to be perfect spheres, what is the value of the highest possible atomic packing factor (APF) in metals?  
  • a)
    0.95
  • b)
    0.74
  • c)
    0.66
  • d)
    0.5
Correct answer is option 'B'. Can you explain this answer?

Rajat Khanna answered
Highest possible atomic packing factor (APF) in metals is 0.74.

Explanation:
Atomic packing factor (APF) is a measure of how efficiently atoms are arranged in a crystal structure. It is the ratio of the volume occupied by atoms to the total volume of the unit cell.

In metals, atoms are closely packed together in a regular pattern. The two most common types of arrangements are face-centered cubic (FCC) and body-centered cubic (BCC) structures.

- Face-centered cubic (FCC) structure:
In an FCC structure, each corner of the cube is occupied by an atom, and there is an additional atom at the center of each face. This gives a total of 4 atoms per unit cell.
The volume occupied by atoms in an FCC structure can be calculated as follows:
Volume occupied by atoms = (number of atoms) * (volume of each atom)
= 4 * (4/3 * π * (radius of atom)^3)
= 16/3 * π * (radius of atom)^3

The total volume of the unit cell in an FCC structure is given by:
Total volume of unit cell = (length of unit cell)^3

Therefore, the atomic packing factor (APF) for an FCC structure is:
APF = (volume occupied by atoms) / (total volume of unit cell)
= (16/3 * π * (radius of atom)^3) / ((length of unit cell)^3)

- Body-centered cubic (BCC) structure:
In a BCC structure, each corner of the cube is occupied by an atom, and there is an additional atom at the center of the cube. This gives a total of 2 atoms per unit cell.
The volume occupied by atoms in a BCC structure can be calculated as follows:
Volume occupied by atoms = (number of atoms) * (volume of each atom)
= 2 * (4/3 * π * (radius of atom)^3)
= 8/3 * π * (radius of atom)^3

The total volume of the unit cell in a BCC structure is given by:
Total volume of unit cell = (length of unit cell)^3

Therefore, the atomic packing factor (APF) for a BCC structure is:
APF = (volume occupied by atoms) / (total volume of unit cell)
= (8/3 * π * (radius of atom)^3) / ((length of unit cell)^3)

Comparing the APF values for FCC and BCC structures, it can be observed that the highest possible APF is 0.74, which corresponds to the FCC structure. Therefore, the correct answer is option 'B' (0.74).

Consider the following statements in respect of austenitic stainless steels:1. Austenitic stainless steels are hardened and strengthened by cold working.  
2. Austenitic stainless steels cannot be quenched and tempered.
Which of the statements given above is/are correct?
  • a)
    1 only
  • b)
    2 only
  • c)
    Both 1 and 2
  • d)
    Neither 1 nor 2
Correct answer is option 'C'. Can you explain this answer?

Ishaan Malik answered
Ans. (c)
• Austenitic stainless steels usually contain 18% Cr and 8% Ni in addition to other minor alloying elements. Ni stabilizes the austenitic phase assisted by C and N. Other alloying additions include Ti, Nb, Mo (prevent weld decay), Mn and Cu (helps in stabilizing austenite).
• These steels are very tough and can be forged and rolled but offer great difficulty in machining.
• These steels cannot be hardened by quenching, in fact they are softened by rapid cooling from about 1000°C.

What is the process by which two or more chemically different monomers are polymerised to form a cross link polymer together with a by-product such as water or ammonia, known as?  
  • a)
    Addition polymerization
  • b)
    Co-polymerisation
  • c)
    Linear polymerisation
  • d)
    Condensation polymerization
Correct answer is option 'D'. Can you explain this answer?

Sanskriti Basu answered
Condensation polymerization is the process by which two or more chemically different monomers are polymerized to form a cross-linked polymer along with the production of a by-product such as water or ammonia. This type of polymerization occurs through the formation of covalent bonds between the monomers, resulting in the removal of a small molecule as a by-product.

Here is a detailed explanation of the process of condensation polymerization:

1. Definition:
Condensation polymerization is a reaction where two different monomers react by forming covalent bonds and eliminating a small molecule as a by-product.

2. Monomers:
Condensation polymerization involves two or more chemically different monomers. These monomers have functional groups that are capable of reacting with each other, such as alcohol groups (-OH) and carboxylic acid groups (-COOH).

3. Reaction:
The process begins with the reaction between the functional groups of the monomers. For example, in the case of a polyester, the reaction occurs between an alcohol group of one monomer and a carboxylic acid group of another monomer. This reaction forms an ester linkage, resulting in the elimination of a water molecule as a by-product.

4. Polymerization:
As the reaction proceeds, more monomers join the growing polymer chain through similar condensation reactions. The polymerization continues until all the monomers are consumed or until the desired polymer length is achieved.

5. Cross-linking:
In some cases, the condensation polymerization process can lead to the formation of cross-linked polymers. Cross-linking occurs when the polymer chains are connected through additional covalent bonds, resulting in a three-dimensional network structure. This enhances the mechanical properties of the polymer, such as strength and toughness.

6. By-product:
During condensation polymerization, a small molecule, such as water or ammonia, is produced as a by-product. The formation of this by-product is a result of the elimination of functional groups from the monomers during the reaction.

7. Examples:
Condensation polymerization is commonly used in the production of various polymers, including polyesters, polyamides, and polyurethanes. For instance, the condensation reaction between ethylene glycol and terephthalic acid produces polyethylene terephthalate (PET), which is widely used in the manufacture of bottles and fibers.

In conclusion, condensation polymerization is a process in which two or more chemically different monomers react to form a cross-linked polymer, along with the generation of a by-product such as water or ammonia. This type of polymerization is important in the synthesis of various polymers, offering a wide range of applications in industries such as packaging, textiles, and coatings.

Which one of the following pairs is not correctly matched?  
  • a)
    Point defect in crystal lattice : Self interstitials
  • b)
    Linear defect in crystal lattice : Grain boundary
  • c)
    Planar defect in crystal lattice : External surface
  • d)
    Volume defect in crystal lattice : other phases
Correct answer is option 'B'. Can you explain this answer?

Vaibhav Khanna answered
Point defect in crystal lattice : Self interstitials
Linear defect in crystal lattice : Grain boundaries
Planar defect in crystal lattice : External surfaces
Volume defect in crystal lattice : Other phases

The correct answer is option B: Linear defect in crystal lattice - Grain boundaries.

Explanation:

Point Defects:
- Point defects refer to the defects that occur at a single or few lattice sites in a crystal lattice.
- These defects can be classified into two types: vacancy defects and interstitial defects.
- Vacancy defects occur when an atom is missing from its lattice site, leaving a vacant site.
- Interstitial defects occur when extra atoms occupy the interstitial sites between the regular lattice positions.

Linear Defects:
- Linear defects, also known as dislocations, occur when there is a deviation from the regular arrangement of atoms along a line or plane in the crystal lattice.
- They can be classified into two types: edge dislocations and screw dislocations.
- Edge dislocations occur when an extra half-plane of atoms is inserted into the crystal lattice, causing a misalignment of the atoms on either side of the dislocation line.
- Screw dislocations occur when the atoms are arranged in a helical pattern around the dislocation line.

Planar Defects:
- Planar defects refer to the defects that occur on a plane or surface within the crystal lattice.
- These defects can be classified into two types: grain boundaries and external surfaces.
- Grain boundaries occur when there is a discontinuity in the crystal lattice due to the presence of different crystal orientations or grains.
- External surfaces refer to the surfaces of the crystal lattice that are exposed to the external environment.

Volume Defects:
- Volume defects, also known as bulk defects, occur within the bulk of the crystal lattice.
- These defects can include impurities, second phases, or voids within the crystal structure.
- Impurities are foreign atoms or ions that are incorporated into the crystal lattice.
- Second phases refer to the presence of a different crystal phase within the same material.
- Voids are empty spaces or vacancies within the crystal lattice.

In the given options, the pair "Linear defect in crystal lattice: Grain boundaries" is not correctly matched because grain boundaries are planar defects, not linear defects. The correct pair for linear defects would be "Linear defect in crystal lattice: Dislocations" which include edge and screw dislocations.

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