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Multiple choice question for engineering

Set 1

1. Which of following is not a process involved in glass production?
a) Extrusion
b) Forming and shaping
c) Heat treatment
d) Finishing

View Answer

Answer: a [Reason:] Glass is produced by high-temperature operation of raw materials in a furnace. These processes include melting and refining forming and shaping, heat treatment, and finishing.

2. At what temperature are raw materials inserted into the furnace?
a) 1000oC
b) 1500oC
c) 2000oC
d) 3000oC

View Answer

Answer: b [Reason:] A mixture of raw materials such as SiO2, Na2CO3, dolomite, and several other are inserted into a furnace at 1400-1500oC for melting. This results in molten glass, which must be drawn off the furnace at 1000o C.

3. Which forming method is used for the production of hollow glasses?
a) Blowing
b) Pressing
c) Drawing
d) Casting

View Answer

Answer: a [Reason:] Glass blowing is a forming technique used to produce hollow objects like bottles and jars. In these machines, molten glass in ribbon form blows holes while air is supplied from above.

4. Which method of forming cannot be used to produce sheet glass?
a) Floating
b) Rolling
c) Drawing
d) Casting

View Answer

Answer: d [Reason:] In drawing operation, the glass is derived from a molten mixture and is passed through the rollers, where it is passed between cooled rolls. In floating operation, the molten glass is formed into a sheet. Any of these three operations may be performed to manufacture sheet glass.

5. Which operation does the below figure represent?
engineering-materials-metallurgy-questions-answers-glass-production-processing-q5
a) Drawing
b) Pressing
c) Blowing
d) Rolling

View Answer

Answer: c [Reason:] Machine blowing is forming operation carried out to manufacture hollow products. In this machine, the molten glass in ribbon form blows holes while air is supplied from above.

6. Which forming operation does the below figure represent?
engineering-materials-metallurgy-questions-answers-glass-production-processing-q6
a) Drawing
b) Pressing
c) Floating
d) Rolling

View Answer

Answer: a [Reason:] In the drawing process, glass is formed into tubing and rods. Molten glass is supplied to a bushing containing orifices. Through this, continuous glass fibers and filaments are formed.

7. The temperature at which a non-crystalline material transforms from a supercooled liquid to rigid glass is __________
a) Melting point
b) Glass transition temperature
c) Boiling point
d) Crystalline temperature

View Answer

Answer: b [Reason:] Glass transition temperature is defined as that temperature at which a change from supercooled liquid state to stiff glass occurs. Below this temperature, the glass structure remains frozen while retaining characteristics of the liquid structure. This temperature is denoted as Tg.

8. Glasses show evidence of _____ fractures.
a) No
b) Brittle
c) Ductile
d) Oblique

View Answer

Answer: b [Reason:] Generally, materials experience either ductile or brittle type of fractures. Glasses experience a brittle fracture. In such types of fracture, the glass shows no signs of a fracture occurring until it occurs i.e. no indication or deformation.

Set 2

1. How is glass wool formed?
a) Hot working
b) Direct melting
c) Cold working
d) Extrusion

View Answer

Answer: b [Reason:] Glass wool is material that provides exceptional insulation against heat and cold. It is formed by the direct melting process. Air and steam jets are used in this process.

2. Short fibers of glass wool can be used up to _______
a) 500oC
b) 600oC
c) 700oC
d) 800oC

View Answer

Answer: a [Reason:] Short fibers of glass wool have a natural density of 24 kg per cu. m and can be used up to a temperature of 500oC. These can be bonded with a thermosetting resin, due to which they can only be used up to 200oC. Bonding with thermosetting resins is done to achieve any desired density and curing.

3. What are the applications of the unbounded type of glass wool?
a) Space suits
b) Sound insulation in airplanes
c) Air filters
d) House ceiling

View Answer

Answer: d [Reason:] The unbonded type of glass wool, with a vapor wall, is used in exterior walls and ceilings of buildings. These, along with bonded type, are used for furnaces, ovens, water heaters ecetera.

4. Which type of glass wool is used in space suits?
a) Bonded type
b) Low-density wool
c) Coarse wool
d) Unbonded type

View Answer

Answer: b [Reason:] Extremely low-density wool is used for thermal and sound insulation in airplanes. This is due to its ability to decrease dead load and also acts as a fire retardant. This is also used in space suits and capsule couches.

5. What is known as expandable polystyrene?
a) Glass wool
b) Fiberglass
c) Thermocole
d) Cement

View Answer

Answer: c [Reason:] Thermocole is a white-colored and odorless material produced during the Second World War. It is the trademark name for the material known as expandable polystyrene. It is used as heat insulation and for packing or stationary purposes.

6. What is the maximum heat insulation of thermocole?
a) 100oC
b) 200oC
c) 500oC
d) 750oC

View Answer

Answer: b [Reason:] Thermocole is a chemically stable and lightweight material. It is a good heat insulator up to 200oC. Although it is generally resistant to salts, alkalies, and acids, it can be attacked by benzene.

7. Which of these is not a standard thickness in which thermocole is available in?
a) 10 mm
b) 37.5 mm
c) 50 mm
d) 75 mm

View Answer

Answer: d [Reason:] Thermocole sheets are a lightweight material having air in their pores. Standard sheets are available in sizes of 10, 12.5, 20, 25, 37.5, and 50 mm thickness.

8. Which of these is not a property of thermocole?
a) Heat insulator
b) Lightweight
c) Receptive to fungus
d) Odorless

View Answer

Answer: c [Reason:] Thermocole is a white-colored odorless material that is chemically stable and resistant to fungus. It has a density of 10-25 kg/cu.m which makes it extremely light in weight. Thermocole is a good heat insulator up to 200oC.

Set 3

1. For hardening of steel by quenching, the steel is cooled in __________
a) Furnace
b) Still air
c) Oil bath
d) Cooling tower

View Answer

Answer: c [Reason:] After heating and soaking of the steel, it must be properly cooled. The steel is quenched to room temperature in a water or oil bath.

2. The cooling rate must be _________ the critical cooling rate for hardening of steel by quenching.
a) Higher than
b) Lower than
c) Equal to
d) Half of

View Answer

Answer: a [Reason:] Post heating and soaking, the steels must be cooled. The cooling rate must be higher than the critical cooling rate to get the completely martensitic structure. The steel is quenched to room temperature with the help of a water bath or oil bath.

3. Phase transformation during hardening transforms _________
a) BCC to FCC
b) FCC to BCT
c) BCT to HCP
d) FCC to HCP

View Answer

Answer: b [Reason:] Due to rapid cooking, austenite is supercooled by nearly 500oC. The large force helps convert the FCC into BCT structure. The resulting structure is called martensite.

4. The slip does not occur in martensite due to the presence of _______ in the lattice.
a) Silicon
b) Germanium
c) Carbon
d) Tin

View Answer

Explanation: Martensite is a super-saturated solution of carbon in α-iron. Due to the presence of carbon in the lattice, slip does not occur. As a result, martensite is strong, hard, and brittle.

5. The hardening process is carried out on ________ steel
a) No carbon
b) Low carbon
c) Medium carbon
d) High carbon

View Answer

Answer: d [Reason:] As the carbon content increases, the hardness also increases. Due to this, the hardening process is carried out on high carbon steels containing 0.35-0.50% C.

6. How does the rate of cooling affect the hardness of steel?
a) Faster cooling results in low hardness
b) Slow cooling results in high hardness
c) Fast cooling results in high hardness
d) No change is found

View Answer

Answer: c [Reason:] We know that hardness depends on the nature and properties of the quenching medium. It was found that faster cooling resulted in greater hardness of the steel, and slow cooling lowers the hardness.

7. How does the size of the specimen affect the hardness of steel?
a) Smaller size results is high hardness
b) Smaller size results in low hardness
c) Larger size results in high hardness
d) No change is found

View Answer

Answer: a [Reason:] The size of the specimen also greatly affects the hardness of the steel during the hardening process. As the size of the specimen increases, its hardness decreases. For example, hardness with a 50 mm diameter steel bar will be higher than a similar one of 100 mm diameter.

8. ___________ is defined as the ease of forming martensite.
a) Hardness
b) Hardenability
c) Toughness
d) Strength

View Answer

Answer: b [Reason:] Hardenability is defined as a measure or ease with which hardness is achieved. It can also be said that hardenability is the ease with which martensite is formed. Hardness, on the other hand, is the ability of a metal to resist abrasion, indentation, and scratching.

9. Hardenability of a material can be measured using __________ test.
a) Jominy end-quench
b) Charpy
c) Rockwell
d) Izod

View Answer

Answer: a [Reason:] Hardenability is the measure of inclination of a material to achieve hardness. It is affected by the alloying elements in the material and the grain sizes. The hardenability of the material can be measured using the Jominy end-quench test method. Charpy and Izod are impact testing methods, whereas Rockwell is a hardness testing scale.

10. Which of the following factors affect the hardenability of a material?
a) Composition of steel
b) Grain size
c) Temperature of specimen
d) Quenching medium

View Answer

Answer: c [Reason:] The hardenability of the material can be measured using the Jominy end-quench test method. The composition of steel, austenitic grain size, structure before quenching, and quenching medium and method affect the hardenability of the steel.

11. Quenching of the sample in Jominy end-quench method is done at _______
a) 0oC
b) 25oC
c) 125oC
d) 200oC

View Answer

Answer: b [Reason:] The given sample of material is heated above the upper critical temperature and dropped into position in the frame of the apparatus. It is then quenched on one end at 25oC to achieve different rates of cooling along the length of the sample.

12. Hardness readings are taken every ________ after quenching in Jominy end-quench test.
a) 0.01 mm
b) 0.1 mm
c) 1.6 mm
d) 2.5 mm

View Answer

Answer: c [Reason:] The given sample of material is heated above the upper critical temperature. It is then quenched at one end at 25oC. After cooling, a flat is ground along the length of the bar. Rockwell C hardness readings are then taken for every 1.6 mm along the length from the quenched end, which are then plotted in the form of a graph.

Set 4

1. What kind of indenter is used in a Brinell test?
a) Diamond cone
b) Steel ball
c) Pen dot
d) Long tube

View Answer

Answer: b [Reason:] Brinell Hardness Test is one of the three available tests for checking the hardness of a material. Here, a hardened steel or tungsten carbide ball indenter is forced into the surface of the metal to be tested.

2. For how long is the load maintained in a Brinell test?
a) 2-4 seconds
b) 10-15 seconds
c) 20-30 seconds
d) 45-55 seconds

View Answer

Answer: b [Reason:] In the Brinell hardness test, the indenter is forced into the surface of the metal and maintained there for 10-15 seconds. This can take a standard load of 500-3000 kg in 500 kg increments.

3. How does the Vicker’s hardness test differ from Brinell’s?
a) Duration of indentation
b) Type of indenter
c) Materials to be tested
d) Load applied

View Answer

Answer: b [Reason:] Similar to Brinell’s test, Vicker’s test also employs forcing of the indenter into the surface of metal for about 10-15 seconds. However, instead of a ball indenter, a pyramid-shaped indenter is used. This test is commonly also called the diamond-pyramid hardness test.

4. What is the angle of indenter in Vicker’s hardness test?
a) 96 degrees
b) 110 degrees
c) 136 degrees
d) 150 degrees

View Answer

Answer: c [Reason:] In Vicker’s hardness test, a square pyramid-shaped indenter is used for indentation. This indenter is placed at an angle of 136 degrees between opposite faces. This angle was chosen as it approximates the most desirable ratio of indentation diameter to ball diameter in Brinell hardness test.

5. How is the Vicker’s hardness number defined as?
a) 1.8544PD2
b) 1.8544P 1/D2
c) 8.544P 1/D2
d) 8.544PD2

View Answer

Answer: b [Reason:] The diamond-pyramid number (DPH) is defined as the applied load divided by the surface area of indentation. Vickers hardness number (VHN or VPN) is also known as diamond-pyramid hardness number. Mathematically, it is defined as 1.8544P 1/D2 . This can be calculated since angle of indenter is known as 136 degrees.

6. The indenter used in C-scale of Rockwell hardness test is named _________
a) CABAL
b) BRALE
c) BRALL
d) IDOL

View Answer

Answer: b [Reason:] There are many scales on which Rockwell hardness tests can be conducted. The B-scale uses a 1/16 inch diameter steel ball indenter. The C-scale uses a diamond cone indenter of 120 degrees, which is named as BRALE.

7. Which of the following is not a Rockwell hardness scale?
a) A
b) E
c) N
d) K

View Answer

Answer: c [Reason:] There are several scales on which Rockwell hardness tests may be conducted. The A-scale uses a diamond cone indenter, whereas E-scale and K-scale use a 1/8 inch diameter steel ball of 100 kg and 150 kg indenting loads respectively.

8. Which materials are to be tested using a F-scale?
a) Copper and brass
b) Case hardened steels
c) Bronze, gunmetal, and beryllium copper
d) Thermoplastics

View Answer

Answer: a [Reason:] The Rockwell hardness F-scale uses a 1/16 inch diameter steel ball indenter with a total indenting load of 60 kg. This is used for testing of copper and brass.

9. Which of the following scales are used for testing of very soft thermoplastics?
a) A
b) E
c) H
d) R

View Answer

Answer: d [Reason:] The A-scale is used for thin hardened steel strips, E-scale is used for cast iron, aluminum alloys, and magnesium alloys, and H-scale is used for soft aluminum and thermoplastics. The R-scale uses a ½ inch diameter steel ball indenter and is used for very soft thermoplastics. For other thermoplastics, the L-scale is used.

10. What is the load that the Vickers hardness test can carry?
a) 10 kg
b) 120 kg
c) 150 kg
d) 3000 kg

View Answer

Answer: b [Reason:] Brinell’s hardness tests can be used up to 3000 kg for a steel ball. Both B-scale and C-scale of Rockwell have a minor load capacity of 10 kg and a major load capacity of 100 kg and 150 kg respectively. The Vickers test has a load of 1-120 kg and can be used in all metal alloys and ceramics.

Set 5

1. Why are Hume Rothery’s rules followed?
a) Extensive solid solution
b) Liquid solution
c) Weak solid solution
d) Extensive liquid solution

View Answer

Answer: a [Reason:] To form an extensive solid solution, Hume Rothery’s rules are obeyed. An extensive solid solution is generally considered as one that is greater than 10 atomic percent soluble.

2. According to Hume Rothery’s rules, size of atoms must not differ by more than ________
a) 5%
b) 15%
c) 35%
d) 55%

View Answer

Answer: b [Reason:] Hume Rothery’s rules state that the atomic radius or size of solute and solvent must not differ by more than 15%. This must be in order to minimize the lattice strain.

3. What is formed when the electronegativities of atoms differ?
a) Solid solutions
b) Liquid solution
c) Intermetallic compound
d) Maraging steel

View Answer

Answer: c [Reason:] According to Hume Rothery’s rules, the solute and solvent mist have similar electronegativities. If this difference is too large, the metals are inclined to form intermetallic compounds in place of solid solutions.

4. For interstitial solid solutions, atomic radii difference must not differ by more than ________
a) 25%
b) 37%
c) 59%
d) 73%

View Answer

Answer: c [Reason:] For interstitial solid solutions, Hume Rothery’s rules state that the radius of solute atoms must not be larger than solvent atoms by more than 59%. Moreover, they should have similar electronegativities and valencies.

5. Dissolution of limited amount of solute in solvent, the solution is a __________
a) Saturated solution
b) Unsaturated solution
c) Supersaturated solution
d) Oversaturated solution

View Answer

Answer: a [Reason:] If the solvent is dissolving a limited quantity of solute, it is known as a saturated solution. An unsaturated solution is considered one where a small quantity of solute is dissolved in a solvent, whereas for a supersaturated solution, the amount of solute in a solvent is more.

6. A solution of exchange of impurities for solvent atoms is called a _________
a) Interstitial solid solution
b) Substitutional solid solution
c) Saturated solution
d) Unsaturated solution

View Answer

Answer: b [Reason:] When the solute atoms (impurities) are substituted for parent solvent atoms in a crystal lattice, the atoms are called substitutional atoms. Such a mixture of two elements is called a substitutional solid solution.

7. What kind of solid solution is found in a Cu-Ni crystal?
a) Interstitial solid solution
b) Substitutional solid solution
c) Supersaturated solution
d) Unsaturated solution

View Answer

Answer: b [Reason:] The Cu-Ni system obeys Hume Rothery’s laws of similar atomic radii (1.28 and 1.25), same FCC crystal structure, similar valencies (+1 and +2), and similar electronegativities (1.9 and 1.8). These elements are completely soluble in one another and form a substitutional solid solution.

8. Which of the following is a random substitutional solid solution?
a) Cu-Zn
b) Au-Cu
c) Cu2MnAl
d) Carbon in ϒ iron

View Answer

Answer: a [Reason:] A random substitutional solid solution is one in which the solute and solvent atoms occupy random positions in the crystal lattice. Cu-Zn, or brass, is such a random substitutional solid solution.

9. Which type of solid solution does this figure illustrate?
engineering-materials-metallurgy-questions-answers-types-solid-solutions-q5

a) Interstitial solid solution
b) Substitutional solid solution
c) Supersaturated solution
d) Unsaturated solution

View Answer

Answer: a [Reason:] An interstitial solid solution is one which solute atoms fit into spaces between solvent atoms. These spaces are known as interstices. In the figure, a system of carbon in FCC ϒ iron is shown.

10. Au-Cn is an example of _________ solid solution.
a) Interstitial solid solution
b) Random substitutional solid solution
c) Supersaturated solution
d) Ordered substitutional solid solution

View Answer

Answer: d [Reason:] An ordered substitutional solid solution is one in which the solute and solvent atoms occupy specific or preferred positions in the crystal lattice. Au-Cu and Cu2MnAl are examples of such ordered substitutional solid solution.

11. Interstitial solutions have a _________ distribution.
a) Random
b) Linear
c) Alternating
d) Dendritic

View Answer

Answer: a [Reason:] In interstitial solutions, the solute atoms are randomly distributed throughout the solvent. In interstitial compounds, however, the pattern is regular in the specific compound.

12. Cu3Al and NiAl are examples of __________
a) Interstitial solutions
b) Interstitial compounds
c) Electron compounds
d) Valency compounds

View Answer

Answer: c [Reason:] If two metals consist of atoms of similar valencies, then the compounds formed are known as electron compounds. Cu3Al, CuZn, Cu3Sn and NiAl are examples of such electron compounds.