Multiple choice question for engineering
1. What is the coefficient of thermal expansion of silicon carbide?
a) 9 * 10-6
b) 4.3 * 10-6
c) 3.2 * 10-6
d) 10.6 * 10-6
Answer: b [Reason:] Silicon carbide has a good coefficient of thermal expansion of 4.3 * 10-6 /oC and that of alumina is 9 * 10-6 /oC. The coefficients of thermal expansion of silicon nitride and sialon are 3.2 * 10-6/oC and 3.04 * 10-6/oC respectively. Among these common ceramics, partially stabilized zirconia (PSZ) has the highest value of 10.6 * 10-6/oC.
2. Silicon carbides resist oxidation up to ______
Answer: d [Reason:] Silicon carbide is the hardest of traditional abrasive materials and has semiconductor properties. It can resist oxidation temperature as high as 1500oC, which is higher than the melting point of steel.
3. What disadvantage does silicon carbide have?
a) Tensile strength
b) Oxidation resistance
c) Thermal conductivity
Answer: d [Reason:] Silicon carbide has excellent tensile strength, oxidation resistance, and the highest thermal conductivity among the common engineering ceramics. However, it is expensive and only available in limited shapes and sizes. It also has a low strength.
4. Which of the following is not a type of silicon nitride ceramics?
a) Reaction bonded silicon nitride
b) Hot pressed silicon nitride
c) Stiff silicon nitride
d) Pressure-less sintered silicon nitride
Answer: c [Reason:] Silicon nitride is a fully resistant ceramic when exposed to most acids. They are classified into different types based on manufacturing method. Their types are Reaction bonded silicon nitride (RBSN), Hot pressed silicon nitride (HPSN), Sintered silicon nitride (SSN), and Pressure-less sintered silicon nitride (PSSN).
5. Partially stabilized zirconia does not exist in which crystalline structure?
Answer: c [Reason:] Zirconium oxide initially exists in the cubic structure at elevated temperature. This changes into a tetragonal structure upon cooling. At room temperature, this becomes a monocyclic crystal structure, which causes cracks during cooling.
6. How much of a stabilizing oxide is present in partially stabilized zirconium?
Answer: a [Reason:] Zirconia is added with stabilizing oxides to avoid cracking. These stabilizing oxides may be MgO, CaO, or Y2O3. Partially stabilized zirconium contains about 5% stabilizing oxide, while fully stabilized oxides contain about 18% of the same.
7. Which ceramic material does Si3Al3O3N5 denote?
a) Silicon carbide
b) Silicon nitride
d) Silicon alumino nitrate
Answer: c [Reason:] Sialon is derived from its constituents, which has been formed into an acronym Si-Al-O-N. These ceramics have good hardness, toughness, strength, and mechanical properties. They are used for cutting tool materials, dies, nozzles etc.
8. On average, what is the maximum use temperature of engineering ceramics?
Answer: d [Reason:] Compared to most metals and plastics, ceramics have a very good limit of maximum use temperature. It lies around 2760oC for ceramics, while metals and plastics lay lower at 815oC and 260oC respectively.
9. How is the creep strength of ceramics when compared to other materials?
Answer: d [Reason:] Engineering ceramics have a good tensile strength, compressive strength, hardness, and excellent creep resistance. Metals usually have gold creep resistance, but plastics fair poorly.
1. Which of the following is not a natural abrasive?
c) Silicon carbide
Answer: c [Reason:] An abrasive is a common ceramic material which is hard and mechanically resistant. Synthetic abrasives are preferred as they can be made as per the desired properties and also exhibit more uniformity. Diamond, sandstone, quartz, and emery are a few examples of natural abrasives. Silicon carbide and aluminum oxide are examples of synthetic abrasives.
2. What is a common application of diamond?
a) Drilling of rocks
c) Wood finishing
d) Polishing of metals
Answer: a [Reason:] Diamond is natural abrasive and is known as the hardest material. It is used for wire drawing dies, drilling rocks, and dressing grinding wheels. It is also used for polishing of hard carbide metals, glass, and ceramics. Emery, on the other hand, is used for general purpose metal polishing.
3. Sandpaper is a common application of ______ abrasives.
a) Garnet and flint
b) Rottenstone and pumica
c) Walnut shells
d) Silicon carbide
Answer: a [Reason:] Garnet and flint are natural abrasives typically used for sandpaper. Walnut shells find their application for cleaning of aircraft engine parts, whereas rottenstone and pumica are used for wood finishing.
4. Silicic acid deposits are used to form _____
Answer: c [Reason:] Tripoli is an abrasive material obtained from deposits of silicic acid. It is used for fine grinding of brass, aluminum, and other precious metals. Quartz, emery, and diamond are, however, natural abrasives.
5. A high-temperature electric arc of _____ is used for the manufacture of silicon carbide.
Answer: c [Reason:] Silicon carbide is an artificially manufactured abrasive material. This is done by mixing sand, coke, and sawdust and passing it through an electric arc at 4500oF. This mixture is converted into silicon carbide and crushed.
6. Which abrasive has a common trade name as carborundum?
a) Aluminum oxide
b) Silicon carbide
c) Tungsten carbide
d) Boron carbide
Answer: b [Reason:] Silicon carbide (SiC) is an artificial abrasive used for making grinding wheels, pipes, and pumps. It is better known by its trade name of carborundum. Aluminum oxide is another artificial abrasive.
7. Which of the following is regarded as a modern abrasive?
b) Silicon carbide
d) Boron carbide
Answer: d [Reason:] Boron carbide and tungsten carbide are generally regarded as modern abrasives. They are close to the hardness of a diamond, which is a natural abrasive. Boron carbide is used for polishing pastes.
8. The polishing of cast iron and finishing of stainless steels are applications of ______
a) Silicon carbide
b) Boron carbide
c) Tungsten carbide
d) Aluminum oxide
Answer: d [Reason:] Aluminum oxide is an abrasive which is obtained by heating of aluminum salts or from ores of bauxite. It has a lighter color compared to silicon carbide and is not as hard, but is more resistant to impact. It is used for polishing cast irons, non-ferrous materials, and for giving a lustrous finish to stainless steels.
9. Which silicate structure does the following diagram represent?
Answer: a [Reason:] The given figure denotes the silicon-oxygen tetrahedron structure. It is the primary structural unit of silicates. In such a structure, one silicon atom is connected with four oxygen atoms.
10. What is the composition of a double tetrahedral structure?
Answer: b [Reason:] A double tetrahedral silicate structure is produced as a result of overcoming the deficiency of electrons in oxygen. Its composition is Si2O7 whereas that of a polyhedral unit is Si3O9.
11. Which of the following structures can be seen for the vitreous form of glass?
Answer: b [Reason:] Glass is a vitreous silicate having a vitreous structure. It consists of a three-dimensional structure of covalent bonds. The figure represents the vitreous nature of silica in two-dimensional form.
12. Conversion of a tetrahedral unit into a three-dimensional structure gives a ______ structure.
Answer: d [Reason:] When a silicate tetrahedral unit extends into three dimensions, it gives a framework structure. On the other hand, a sheet structure extends into a two-dimensional plane similar to how framework extends in three-dimensions.
13. Quartz and Feldspar are examples having ______ structure
c) Poly tetrahedral
Answer: d [Reason:] When a silicate tetrahedral unit extends into three dimensions, it gives a framework structure. It has a relatively low density, low atomic packing factors, and good hardness. Common examples of framework structures can be found in cristobalite, quartz, and Feldspar.
14. Which silicate structure does the following figure represent?
Answer: b [Reason:] A silicate sheet structure is found when a double chained structure extends into a two-dimensional plane. Clay and mica are examples of ceramics having a sheet structure.
15. The ability of a material to exist in more than one crystal structure is known as _______
c) Polyhedral phase
Answer: a [Reason:] Polymorphism is the property of a material which allows it to exist in more than one form of space lattice or crystal structure. When such a process is reversible, it is defined as allotropy.
1. What kind of steel requires definite amounts of other alloying elements?
a) Carbon steel
b) Alloying steel
c) Stainless steel
d) Tool steel
Answer: b [Reason:] Alloy steels are those steels which require a specific amount of the elements making up its composition. Alloy steels consist of manganese, silicon, and copper as primary elements whose quantities are equal to or more than 1.65%, 0.60%, and 0.60% respectively.
2. Which of these is not a function of alloy steels?
a) Increases strength
b) Improves ductility
c) Reduces cost
d) Improves machinability
Answer: c [Reason:] Alloy steels are used to improve properties such as strength, hardness, ductility, grain structure, and machinability, among others. This, however, results in increased costs due to multiple elements involved in the process.
3. Steels containing up to 3% to 4% of one or more alloying elements are known as ________
a) Low alloy steels
b) HSLA steels
c) High alloy steels
d) Stainless steels
Answer: a [Reason:] Low alloy steels consist of 3% to 4% of alloying elements making up its composition. They have similar microstructure and heat treatments as plain carbon steels. HSLA and AISI steels are the types of low alloy steels.
4. What does AISI steel stand for?
a) American-Indian Steel Institute
b) American-Indian Society of Iron
c) American Iron and Steel Institute
d) Alloys, Iron and Steel Institute
Answer: c [Reason:] American Iron and Steel Institute (AISI) is an association, established in America, which produces steel. AISI steels are used in machine construction. They are otherwise known as construction steels or structural steels.
5. Which of these is not an application of HSLA steels?
b) Automobiles and trains
c) Building columns
d) Leaf and coil springs
Answer: d [Reason:] HSLA steels are known as high-strength-low-alloy steels. These high strength steels are primarily used as structural materials or construction alloys. They are used to reduce weight on bridges, automobiles, pressure vessels, building columns, etc.
6. Steels containing more than 5% of one or more alloying elements are known as ________
a) HSLA steels
b) High alloy steels
c) Tool and die steels
d) Stainless steels
Answer: b [Reason:] High alloy steels are composed of more than 5% of alloying elements. They have different microstructure and heat treatments than those of plain carbon steels. Tool and die, and stainless steels are the types of high alloy steels.
7. Which of the following groups of alloying elements stabilize austenite?
a) Ni, Mn, Cu, and Co
b) Cr, W, Mo, V, and Si
c) Cr, W, Ti, Mo, Nb, V, and Mn
d) Co, Al, and Ni
Answer: a [Reason:] The alloying elements such as Ni, Mn, Cu, and Co have a tendency to alleviate austenite, whereas Cr, W, Mo, V, and Si tend to stabilize ferrites. Alloying elements such as Cr, W, Ti, Mo, Nb, V, and Mn tend to form carbides. Other elements like Co, Al, and Ni help to weaken carbides and thereby form graphite.
8. Which family of steels are referred to as chromoly?
Answer: b [Reason:] The family of 41xx steel is usually called as chromoly or chrome-moly due to its primary alloying elements, chromium and molybdenum. Steels such as 4130 and 4140 are generally used for bicycle frames, and as parts of firearms, flywheels etc.
9. What is the common name of COR-TEN steel?
a) Weathering steel
b) Control-rolled steel
c) Pearlite-reduced steel
d) Microalloyed steel
Answer: a [Reason:] Weathering steels are otherwise also known as COR-TEN or corten steels. These steels produce a corrosion resistance, which makes them ideal for eliminating the need to paint. All choices¸ including corten steels, are classifications of HSLA steels.
10. Alloy steels containing 0.05% to 0.15% of alloying elements are called _______
a) Weathering steel
b) Stainless steel
c) Tool and die steel
d) Microalloyed steel
Answer: d [Reason:] Microalloyed steels contain alloying elements in small quantities (0.05% to 0.15%). These elements include niobium, vanadium, titanium, molybdenum, rare earth metals, among others. They are used to refine the microstructure of the grain or for precipitate hardening process.
1. What is the melting point of pure aluminum?
Answer: b [Reason:] Aluminum is an element which is available in abundance on earth. Its melting point is 600oC. Pure aluminum has low strength, due to which it is added with alloying elements.
2. What is the tensile strength of aluminum?
a) 122.5 GPa
b) 220 MPa
c) 70.5 GPa
d) 45 MPa
Answer: d [Reason:] Pure aluminum has a low tensile strength of 45 MPa and Young’s modulus of 70.5 GPa. Copper has a tensile strength of 220 MPa and Young’s modulus of 122.5 GPa.
3. Compared to copper, how is the electrical conductivity of aluminum?
Answer: a [Reason:] When equal weights are compared, aluminum is a better conductor than copper. It was found that aluminum conducts 201% more current than copper.
4. _______ is coated onto aluminum to improve its soldering ability.
Answer: c [Reason:] To improve soldering or joining ability of aluminum, it is coated with tin and then plated with other metals. Magnesium, lithium, and copper are better suited to form alloys to serve various purposes.
5. Which aluminum alloy is known as aircraft aluminum?
Answer: d [Reason:] Aluminum alloy 7075 consists of zinc as its primary element, along with a bit of magnesium and copper. Due to its high strength and good corrosion resistance, it is often used in aircrafts, and also known as aerospace aluminum. However, the other alloys are also employed in a few aerospace functions and structures.
6. Which of the following is not a classification of aluminum alloys?
a) Crucible alloys
b) Wrought alloys
c) Cast alloys
d) Heat-treatable alloys
Answer: a [Reason:] Aluminum alloys are categorized into wrought alloys, cast alloys, heat-treatable, and non heat-treatable alloys. An example of each is 2.5% Mg, 0.25% Cr, and rest Al (wrought), 12% Si and rest Al (cast), 0.4-0.9% Mg, 0.3-0.7% Si, and rest Al (heat-treatable), and (0.8-1.5% Mn and rest Al (non heat-treatable).
7. Which of these is not a property of duralumin?
a) High strength
b) 1/3 the weight of steel
c) Excellent casting and forging abilities
d) Poor machinability
Answer: d [Reason:] Duralumin contains 4% Cu along with 0.4-0.7% of both Mn and Mg. This provides it excellent casting and forging abilities and high strength while maintaining only 1/3 of the weight. It also possesses high machinability.
8. Which among the following is an example of a non heat-treatable alloy?
Answer: d [Reason:] Non heat-treatable alloys can be strengthened by cold-working operations. Al-Mn alloy is an example of a non heat-treatable alloy. The leftover choices are examples of heat-treatable alloys. Duralumin and Y-alloy are the important Al-Cu alloys.
9. Artificial aging process takes place at a temperature range of ________
d) Room temperature
Answer: a [Reason:] Ageing of alloys at 190-260oC accelerates the aging process and reduces the overall time. This phenomenon is known as artificial aging. Natural aging occurs at room temperature.
10. What happens when the maximum strength is achieved by aging process?
a) Precipitate hardening
b) Age hardening
d) Natural ageing
Answer: c [Reason:] As the aging temperature decreases, the maximum strength increases and reaches its peak. At the peak strength, the strength starts to decrease. This is referred to as over-aging.
11. Which of these is not a stage in precipitation hardening treatment?
a) Solution treatment
Answer: b [Reason:] The alloy is heated to a certain temperature and treated at that temperature. Then it is quenched (rapid cooling) and finally heated below solvus temperature; this is known as aging.
12. How much copper is present in Y-alloys?
Answer: a [Reason:] Y-alloy is an important type of Al-Cu alloy. Similar to duralumin, it contains 4% copper. Addition of 2% nickel and 1.5% magnesium is done to make Y-alloys.
13. What is the melting point of zinc?
Answer: a [Reason:] Zinc has a relatively low melting point of 419. 5oC. Its boiling point, on the other hand, is at 907oC. The melting point of aluminum is 600oC.
14. With the addition of which element, does zinc create a resistance to creep?
Answer: c [Reason:] When magnesium is added to copper, it increases its resistance to creep. Lead affects intercrystalline corrosion, while cadmium improves hardening effect. Copper makes the Cu-Zu alloys more ductile.
15. The most common casting process for zinc alloys is _______
a) Sand casting
b) Die casting
c) Investment casting
d) Centrifugal casting
Answer: b [Reason:] Die casting is generally employed for zinc alloys. These die-castings range from a few grams to 20 kilograms. Zinc base die casting alloys find applications in hardware items like car body, grills etc.
16. What is the appearance of zinc?
Answer: a [Reason:] Zinc is a metallic element whose color looks blue and grey. It has a lustrous property and is diamagnetic in nature.
17. What is the temperature at which zinc become malleable
a) <100 oC
b) 100-150 oC
c) >210 oC
d) 419 oC
Answer: b [Reason:] Zinc is brittle at most temperatures. However, it becomes malleable between 100 and 150 oC. Above 210 oC, it becomes brittle again, reaching its melting point at 41 oC.
1. Which of the following factors does not influence the variety and quality of metal?
a) Rate of heating and cooling
b) Quenching medium
d) Grain size
Answer: d [Reason:] The variety of metal and various metallurgical processes depend upon the method and rate of heating and cooling, furnaces used, and quenching medium. Grain size is one of the effects of heat treatment, not a cause for any change.
2. How does the rate of cooling affect the hardness of the metal?
a) Slow cooling, hard material
b) Slow cooking, soft material
c) Rapid cooling, soft material
d) No effective change
Answer: b [Reason:] The rate of cooling is the controlling factor in developing either a hard or soft structure. Rapid cooling from critical range results in a hard structure, whereas very slow cooling gives a soft structure.
3. Which of the following is not a stage of annealing?
Answer: c [Reason:] Annealing is defined as the softening process which involves heating the material at an elevated temperature and then slowly cooling it. The annealing cycle consists of three stages of heating to the desired temperature, holding at that temperature (soaking), and then slowly cooling (quenching).
4. Full annealing is applied to which kind of materials?
a) Steel castings
b) Steel wires
c) High carbon steels
d) Sheet products
Answer: a [Reason:] When we refer to annealing, usually we only talk about full annealing. This method is used to soften the material, to refine crystalline structure, and relieve stresses. This method is applied on steel castings and steel ingots.
5. For full annealing of hypoeutectoid steels, they are heated in a range above __________
Answer: d [Reason:] Hypoeutectoid steels contain carbon content which is less than 0.77%. For full annealing of this steel, it is heated 30-60oC above the A3 line. It is held at this temperature for a period of time, and then slowly cooled to room temperature.
6. Cooling of hypoeutectoid steels in done is furnace by decreasing the temperature to at least ________ below the A1 line.
Answer: d [Reason:] For full annealing of this steel, it is heated 30-60oC above the A3 line. Then it is cooled within the furnace by decreasing the temperature 10-30oC per hour, to at least 30oC below A1 line.
7. What is the result of full annealing of hypoeutectoid steels?
a) Coarse pearlite
Answer: a [Reason:] Hypoeutectoid steels are heated above the A3 line and then cooled in the furnace. Then it removed from the furnace and then cooled at room temperature. This results in coarse pearlite with excess ferrite.
8. How does heating for hypereutectoid steels differ from hypoeutectoid steels?
a) Heated 30-60oC above A1 line
b) Heated 30-60oC below A1 line
c) Heated 30-60oC below A2 line
d) Heated 30-60oC below A3 line
Answer: a [Reason:] Hypoeutectoid steels are heated above the A3 line and then cooled in the furnace. On the other hand, hypereutectoid steels are heated 30-60oC above the A1 line and then cooled similar to hypoeutectoid steels.
9. Heating of hypereutectoid steels results in formation of coarse pearlite with excess __________
Answer: c [Reason:] Similar to hypoeutectoid steels, hypereutectoid steels are heated 30-60oC above the A1 line. This results in a coarse pearlite with excess cementite in a dispersed spheroidal form. This improves the mechanical properties, ductility, and toughness.