Previous Year Paper

BPSC AE Paper 6 (Mechanical) 2019 Official Paper (Previous Year Paper)

Held in 2019

50 questions60 minutes
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Full Test (50 questions)

Question 1

The relation between Young's Modulus (E), shear modulus (G) and Poisson's ratio (v) is correctly given by the expression ________.

  1. AOption (a)

    E=G2(1+V)E = \frac{G}{{2\left( {1 + V} \right)}}

  2. BOption (b)

    G=E2(1+V)G = \frac{E}{{2\left( {1 + V} \right)}}

  3. COption (c)

    G=E(1+V)G = \frac{E}{{\left( {1 + V} \right)}}

  4. DOption (d)

    E=G(1+V)E = \frac{G}{{\left( {1 + V} \right)}}

View answer & explanation

Correct answer

Answer: (b)B.

G=E2(1+V)G = \frac{E}{{2\left( {1 + V} \right)}}

Why this answer is correct

Elastic Modulus (E)

When the body is loaded within its elastic limit, the ratio of stress and strain is constant. This constant is known as Elastic modulus

E=StressStrain=σϵ{\rm{E}} = \frac{{{\rm{Stress}}}}{{{\rm{Strain}}}} = \frac{{\rm{\sigma }}}{\epsilon}

Rigidity modulus (G)

When a body is loaded within its elastic limit, the ratio of shear stress and shear strain is constant, this constant is known as the shear modulus.

G=Shear;stress;Shear;strain=τϕ{\rm{G}} = \frac{{{\rm{Shear;stress;}}}}{{{\rm{Shear;strain}}}} = \frac{{\rm{\tau }}}{\phi }

Poisson’s ratio (ν)

The ratio of the lateral strain to the longitudinal strain is called Poisson’s ratio.

Relationship between E, G and ν is:

 E = 2G(1 + ν)  

G=E2(1+ν)G =\frac{E}{2(1+\nu)}

Additional Information

Other relationship between various elastic constants is:

E=9KG3K+;G{\bf{E}} = \frac{{9{\bf{KG}}}}{{3{\bf{K}} + ;{\bf{G}}}}

E = 3K (1 – 2μ) 

where K is bulk modulus

Question 2

The shearing yield strength (Ssy) is related to tensile yield strength (Sy) as

  1. AOption (a)

    Ssy = Sy

  2. BOption (b)

    Ssy = 0.414 Sy

  3. COption (c)

    Ssy = 0.577 Sy

  4. DOption (d)

    Ssy = 0.707 Sy

View answer & explanation

Correct answer

Answer: (c)C.

Ssy = 0.577 Sy

Why this answer is correct

Concept:

Maximum distortion energy theory (Von-mises theory)

  • According to this theory, the failure or yielding occurs at a point in a member when the distortion strain energy per unit volume reaches the limiting distortion energy (i.e. distortion energy at yield point) per unit volume as determined from simple tension test.
  • Equivalent stress under triaxial condition is given by:
  • \({{\left( {{\sigma }{1}}-{{\sigma }{2}} \right)}^{2}}+{{\left( {{\sigma }{2}}-{{\sigma }{3}} \right)}^{2}}+{{\left( {{\sigma }{1}}-{{\sigma }{3}} \right)}^{2}}\le 2\left(\frac{S_{yt}}{FOS}\right)^2\)
  • ⇒ τ=σy3\tau =\frac{\sigma _y}{\sqrt 3}
  • i.e. Ssy = 0.577Sy 

Maximum shear stress Theory (or) Guest & Tresca’s Theory

According to this theory, failure of the specimen subjected to any combination of the load when the maximum shearing stress at any point reaches the failure value equal to that developed at the yielding in an axial tensile or compressive test of the same material.

Graphical Representation

\({{\rm{\tau }}{{\rm{max}}}} \le \frac{{{{\rm{\sigma }}{\rm{y}}}}}{2}\) For no failure

\({{\rm{\sigma }}_1} - {{\rm{\sigma }}2} \le \left( {\frac{{{{\rm{\sigma }}{\rm{y}}}}}{{{\rm{FOS}}}}} \right)\) For design

τy≤σt2{\tau _y} \le \frac{{{\sigma _t}}}{2}

i.e. Ssy = 0.5Sy 

Where σt is the yield strength in tension and τy is the yield strength in shear.

Therefore, from both theory, the relation between the shearing yield strength (Ssy) and tensile yield strength (Sy) is, Ssy = 0.5Sy 

Question 3

When helical compression spring is cut into halves, the stiffness of the spring will be

  1. AOption (a)

    Same

  2. BOption (b)

    one-half

  3. COption (c)

    double

  4. DOption (d)

    one-fourth

View answer & explanation

Correct answer

Answer: (c)C.

double

Why this answer is correct

Concept:

When a spring of constant ko and length Lo is cut into pieces of length, say l1, l2, l3, ……. Ln, their respective spring constants can be calculated as,

k1l1 = k2l2 = k3l3 = ……. = knln = kolo

Here the spring is cut into two equal parts: l1 = l2 = L/2

k1l1 = k2l2 = KL

k1L2=k2L2=KL{k_1}\frac{L}{2} = {k_2}\frac{L}{2} = KL

k1 = k2 = 2K

  • The spring constant of spring may be defined as the force required per unit extension of the spring. K = F/x
  • If a spring of force constant K is cut into n equal parts, the spring constant of each part in nK
  • If a spring is cut into two pieces of equal size, each piece will have a force constant double the original
  • The spring constant of a spring is inversely proportional to the number of turns. (kn = constant) or k1n1 = k2n2

Question 4

For a ductile material, the limiting value of octahedral shear stress (τ0) is related to the yield stress (Sy) as

  1. AOption (a)

    τo=Sy23\tau_o=S_y \frac{\sqrt2}{3}

  2. BOption (b)

    τo=Sy32\tau_o=S_y 3{\sqrt2}

  3. COption (c)

    τo=Sy32\tau_o=S_y \frac{\sqrt3}{2}

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

τo=Sy23\tau_o=S_y \frac{\sqrt2}{3}

Why this answer is correct

Concept:

  • As per the Von-mises distortion energy theory.
  • \({{\left( {{σ }{1}}-{{σ }{2}} \right)}^{2}}+{{\left( {{σ }{2}}-{{σ }{3}} \right)}^{2}}+{{\left( {{σ }{1}}-{{σ }{3}} \right)}^{2}}\le 2\cdot {{\left( {{σ }_{per}} \right)}^{2}}\)..................(1)
  • The relation expressed in eq. (1) may also be obtained on the basis of what has come to be known a the Octahedral shearing stress theory.
  • This theory states that the failure of a member by yielding occurs when the Octahedral shearing stress in the member reaches the Octahedral shearing stress at yielding in the simple tension or compression test.
  • The Octaheddral plane is defined as one whose normal makes equal angles with the three principal directions.
  • For a triaxial state of stress, it can be shown that the Octahedral shearing stress, τoct, is given by,
  • τoct=13[(σ1−σ2)2+(σ2−σ3)2+(σ3−σ1)2]1/2\tau _{oct}=\frac{1}{3}[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2]^{1/2}.....................(2)
  • In the uniaxial tension test for which the only non-zero principal stress is σ1, failure by yielding occurs when this principal stress reaches σy.
  • Thus, the Octahedral shearing stress at the failure by yielding in the simple tension test is obtained from eq. (2) by setting σ2 = σ3 = 0 and σt = σy.
  • So, the equation becomes,
  • (τo)uniaxial=σy23(\tau_o)_{uniaxial}=\sigma_y \frac{\sqrt2}{3}

Question 5

Cementite contains Carbon to the tune of

  1. AOption (a)

    0.6%

  2. BOption (b)

    5%

  3. COption (c)

    3.6%

  4. DOption (d)

    6.67%

View answer & explanation

Correct answer

Answer: (d)D.

6.67%

Why this answer is correct

Concept:

  • Cementite or Iron Carbide has chemical formula Fe3C and contains 6.67% carbon by weight.
  • It is the hardest structure that appears in the iron-carbon equilibrium diagram.
  • Cementite has an orthorhombic crystal structure and is extremely hard and brittle as compared to austenite and ferrite.
  • Weight % of carbon in cementite =12×1003×56 + 12=6.67%=\frac{12\times100}{3\times56~+~12}=6.67 \%

Question 6

Which of the following does not aid in graphitization of cementite?

  1. AOption (a)

    Presence of Si

  2. BOption (b)

    Presence of Al

  3. COption (c)

    Presence of Ni

  4. DOption (d)

    Low temperature

View answer & explanation

Correct answer

Answer: (d)D.

Low temperature

Why this answer is correct

Concept:

Graphitization

  • is a microstructural change that sometimes occurs in carbon or low alloy steels subjected to moderate temperatures for long periods of time.
  • The microstructure and properties of Cast iron also differ from those of steel.
  • The excess carbon precipitates during solidification by a eutectic reaction either as the thermodynamically stable graphite phase (gray iron) and/or the metastable cementite phase (white iron). Whether the stable or metastable phase forms depends on nature and treatment given to the liquid, in particular, its graphitization potential, inoculation treatment, and the cooling rate.
  • Silicon increases the graphitization potential strongly and is always present in higher concentrations in grey irons.
  • Aluminium, Boron (< 0.15%), Copper, Nickel, and Titanium (< 0.25%) also  increases the graphitization potential
  • Bismuth, Boron (> 0.15%), Chromium, Manganese, Molybdenum, Tellurium, Vanadium, Titanium (> 0.25%) decreases the graphitization potential

Question 7

Which of the following phases exhibits highest solid solubility of carbon?

  1. AOption (a)

    Delta iron

  2. BOption (b)

    Gamma iron (austenite)

  3. COption (c)

    Alpha iron

  4. DOption (d)

    Ferrite

View answer & explanation

Correct answer

Answer: (b)B.

Gamma iron (austenite)

Why this answer is correct

Explanation:

Iron carbon equilibrium diagram: 

  • The phase diagram of Fe-Fe3C is not a true equilibrium because the iron carbide is an unstable phase that after prolonged heat treatment decomposes into iron and carbon( the graphite form).
  • The study is simplified by within assumption that sufficient time has been allowed at each new temperature for any necessary adjustment in phase compositions and relative amounts.

​Phase components:

Carbon is an interstitial impurity in iron and forms a solid solution in alpha and delta phases. Different phase components are as follows:

Ferrite: 

  • Pure iron contains 0.008% weight of carbon.
  • Upon heating, pure iron experiences a change in crystal structure before it melts.
  • The stable form at room temperature is called ferrite.
  • Alpha iron has a BCC crystal structure.
  • This phase is relatively soft can be made magnetic at a temperature below 768° C and has the maximum possible density of 7880 kg/m**3**.

Austenite: 

  • Ferrite experiences a  polymorphic transformation to FCC austenite (gamma iron) at 912°C.
  • The austinite persists up to 1394°C at which it becomes BCC-phase known as delta-iron.
  • Upon further heating, it melts at 1538°C.
  • The maximum solubility of carbon in gamma iron is 2.14% approximately 100 times the maximum solubility in the BCC ferrite phase because FCC interstitial positions are larger.
  • Austenite is a magnetic phase.

Cementite: 

  • A very hard and brittle phase is formed when carbon exceeds 6.67%.
  • It has two phases eutectic and eutectoid.

Pearlite: 

  • Microstructure for the eutectoid steel slowly cooled through the eutectoid temperature consist of alternating lamellae of the two phases alpha and Fe3C.
  • Carbon atoms diffuse away from the 0.22% ferrite, extending from the grain boundaries into unreacted austenite grain.
  • The resulting structure is called pearlite because it has the appearance of the mother of pearl.​​

Important Points

  • Ther are six important temperatures 727°C, 912°C, 1130°C, 1394°C, 1493°C, 1538°C at which phase transformations occur.

Question 8

The eutectic percentage of carbon in iron is 

  1. AOption (a)

    0.025%

  2. BOption (b)

    0.15%

  3. COption (c)

    2.03%

  4. DOption (d)

    4.33%

View answer & explanation

Correct answer

Answer: (d)D.

4.33%

Why this answer is correct

Concept:

Three important transformations of the iron-carbon equilibrium diagram

Eutectoid Reaction

  • At temperature 723° C and carbon composition 0.8%
  • Austenite (γ ) (S) ⇄ Ferrite (α-iron) (S) + Carbide (Fe3C) (S)
  • One solid converted into two solids.

Eutectic reaction

  • At temperature 1130° C and carbon composition 4.3%.
  • Liquid Iron (L) ⇄ Austenite (S) + Cementite (S)
  • One liquid converts in two solids.

Peritectic reaction

  • At temperature 1498°C and carbon composition 0.09%
  • δ –iron (S) + Liquid iron (L) ⇄  Austenite (S)
  • One liquid and one solid converts into another solid.

Question 9

The eutectic mixture of austenite and cementite in Iron-Carbon system is called

  1. AOption (a)

    Spheroidite

  2. BOption (b)

    Ledeburite

  3. COption (c)

    Pearlite

  4. DOption (d)

    Bainite

View answer & explanation

Correct answer

Answer: (b)B.

Ledeburite

Why this answer is correct

Concept:

  • Ledeburite is the eutectic mixture of austenite and cementite in Iron-Carbon system. Ledeburite consists of cementite as white matrix and austenite as grey particles at 1147°C. Ledeburite is not a type of steel as the carbon content is too high. It may exist as a separate constituent in some high carbon steels.
  • Spheroidite consist of sphere-like cementite particles is an alpha ferrite matrix. Spheroidite is formed when carbon steel is heated for more than 30 hours at a temperature of more than 690°C. Spheroidite is relatively soft and allows more formability. It finds wide application in railroad tracks, tyre cords, bridge cables, etc.
  • Pearlite is a two-phase, lamellar structure composed of 88% ferrite and 12% cementite. Pearlite is formed when austenite cools slowly below 727°C. It is hard as well as strong because of the layered structure. It finds wide application in cutting tools, knives, high strength wire, etc.
  • Bainite is composed of cementite and dislocation rich ferrite. Bainite forms as needles or plates, depending on the temperature of the transformation. Because of its high strength and formability, It is extensively used in automotive industries.

Components of the iron-carbon system

Phase or mixture of phasesName
Solid solution of carbon in α (δ) ironFerrite
Solid solution of carbon in γ - ironAustenite
Iron carbide (Fe3C)Cementite
Eutectic mixture of carbon solid solution in γ - iron with iron carbideLedeburite
Eutectoid mixture of carbon solid solution in α - iron with iron carbidePearlite

Question 10

The limitations of plain carbon steels include

  1. AOption (a)

    poor hardenability

  2. BOption (b)

    major loss of hardness on tempering

  3. COption (c)

    low corrosion resistance

  4. DOption (d)

    all of the above

View answer & explanation

Correct answer

Answer: (d)D.

all of the above

Why this answer is correct

Explanation:

Plain carbon steels:

Plain carbon steel is a type of steel having a maximum carbon content of 1.5% along with small percentages of silica, sulphur, phosphorus, and manganese. Carbon steel is made into a wide range of products, including structural beams, car bodies, kitchen appliances, and cans.

Classification of carbon steel

Low Carbon Steel (Mild Steel):

  • Low carbon steel is an alloy of iron and carbon. The carbon content varies from 0.05 to 0.15 % for dead mild steel and 0.15 to 0.3 % for mild steel.
  • The dead mild steel is very much ductile and is used for making various components like sheets, strips, wires, rods etc.
  • The mild steel is soft and ductile.
  • It is used for making various structural sections like angles, channels, beams, rods, pins, etc.

Medium carbon steel:

  • For medium carbon steel, the carbon percentage varies from 0.3 to 0.8%.
  • These steels have better strength and not very easy to shape as mild steel.
  • The hardness is slightly higher, and the ductility is poor.
  • The machine parts requiring strength like springs, automobile parts, shafts, dies for forging etc. are made from medium carbon steel.

High Carbon steel:

  • The carbon percentage varies from 0.8 to 1.5.
  • A high percentage of carbon gives hardness and strength to the metal.
  • It is mainly used for tools like chisels, hammers, dies, punches, broaches, reamers, drills, taps, etc. and the machine parts like springs, mandrels and similar parts requiring high strength and hardness.

Limitation of carbon steel

  • Low hardenability
  • low corrosion and oxygen resistance
  • Major loss of hardness on tempering

Question 11

The supersaturated interstitial solid solution of carbon in alpha iron is known as

  1. AOption (a)

    martensite

  2. BOption (b)

    pearlite

  3. COption (c)

    bainite

  4. DOption (d)

    ferrite

View answer & explanation

Correct answer

Answer: (a)A.

martensite

Why this answer is correct

Explanation:

Martensite: 

  • It is a metastable phase of steel formed by the transformation of austenite below 320°C.
  • Martensite is an interstitial supersaturated solid solution of carbon in α – Iron and has a body-centred tetragonal lattice.
  • It has a carbon content of up to 2% and is extremely hard and brittle.
  • It is a product of rapid cooling (quenching) and possesses an acicular or needle-like structure.
  • The rate of cooling and the amount of carbon percentage in steel are directly proportional to the amount of hardness achieved in martensitic transformation.

Additional Information

Bainite:

  • It is a plate-like microstructure that forms in steels at a temperature of 125-550° (depending on alloy content).
  • It forms by the decomposition of austenite at a temperature which is above MS but below that at which fine pearlite forms.

Ferrite (α – Iron):

  • The solid solution of carbon in α – Iron is called ferrite.
  • The carbon atoms are dissolved interstitially in the BCC structure.
  • The maximum solubility of carbon in α – Iron is 0.025% at 723°.
  • It is very soft and highly magnetic.
  • It is in-fact softest structure that appears in Iron – Carbon equilibrium diagram.

Pearlite:

  • It is formed by the decomposition of austenite at 723°C. The decomposition of austenite leads to the formation of a eutectoid mixture of 87% ferrite and 13% cementite called pearlite. The pearlite constituent consists of alternate lamellae of ferrite and cementite and contains 0.8% carbon.
  • For microstructure of a 0.2% carbon steel would consist of a quarter of pearlite and three- quarters of ferrite. As explained earlier, ferrite is soft and ductile, and pearlite is of medium hardness and it imparts mechanical strength to steel.
  • The higher the carbon content, the higher would be the pearlite content and hence higher mechanical strength.
  • Conversely, when the pearlite content increases, the ferrite content decreases and hence the ductility is reduced.

Austenite:

  • It is also known as Gamma-phase iron is a metallic, non-magnetic allotrope of iron or solid solution of iron, with an alloying element.
  • In plain carbon steel, austenite exists above the critical eutectoid temperature of 1000 K.
  • Austenite is of FCC crystal structure.

Ledeburite:

  • In iron and steel metallurgy, ledeburite is a mixture of 4.3 % Carbon in iron and is a eutectic mixture of austenite and cementite.
  • ​Ledeburite is not a type of steel as the carbon level is too high although it may occur as a separate constituent in some high carbon steel.

Question 12

The resultant microstructure after normalizing should be

  1. AOption (a)

    martensitic 

  2. BOption (b)

    pearlitic

  3. COption (c)

    bainitic

  4. DOption (d)

    ferritic

View answer & explanation

Correct answer

Answer: (b)B.

pearlitic

Why this answer is correct

Concept:

Normalizing: 

  • It is an annealing treatment in which steels are heated in the austenitic range followed by the air cooling (at higher cooling rates as compared to other annealing processes).
  • The heating is performed at least 55°C above the upper critical temperature that is above A3 for hypo eutectoid compositions and above Acm for hypereutectoid compositions.
  • It results in fine pearlitic steels are tougher than coarse pearlitic steels obtained by other annealing processes (as cooling rates are low due to furnace cooling).
  • It is performed to achieve any of the following purposes:
  • To eliminate coarse-grained microstructure obtained in previous working operations such as rolling and forging.
  • To modify and improve cast dendritic structure and reduce segregation by homogenizing the microstructure.
  • Main objective
  • Refine grain size in metal, improve strength and hardness, reduce ductility
  • Remove cold worked stress.
  • Remove dislocations due to hot working.

Question 13

Which of the following is the quench media with highest heat transfer equivalent?

  1. AOption (a)

    Brine 

  2. BOption (b)

    Air

  3. COption (c)

    Oil 

  4. DOption (d)

    Water

View answer & explanation

Correct answer

Answer: (a)A.

Brine 

Why this answer is correct

Concept:

  • The hardness of steel depends on the rate of cooling. Faster is the cooling rate, harder is the steel.
  • Decreasing cooling rate sequence
  1. Brine: Salt will be NaCl, NaOH, CaCl2
  2. Water + NaOH
  3. Water
  4. Oil
  5. Air

Question 14

Which of the following is a thermoplastic?

  1. AOption (a)

    Bakelite

  2. BOption (b)

    Epoxy resins (araldite)

  3. COption (c)

    Polypropylene

  4. DOption (d)

    Polyurethanes

View answer & explanation

Correct answer

Answer: (c)C.

Polypropylene

Why this answer is correct

Explanation:

Thermoplastics:

  • Thermoplastics are defined as polymers that can be melted and recast almost indefinitely.
  • They are molten when heated and harden upon cooling.
  • When frozen, however, thermoplastic becomes glass-like and subject to fracture.
  • These characteristics, which lend the material its name, are reversible, so the material can be reheated, reshaped, and frozen repeatedly. As a result, thermoplastics are mechanically recyclable.
  • Some of the most common types of thermoplastic are polypropylene, polyethylene, polyvinylchloride, polystyrene, Polyethylenetheraphthalate (PET), Teflon, Nylon, and polycarbonate.

Additional Information

Thermosetting plastic.

  • The plastic which keeps its shape and does not soften i.e the plastic which when molded once, can not be softened by heating is called Thermosetting plastic.
  • Thermosetting plastics are synthesized by condensation polymerization and have primary bonds between molecular chains and held together by strong cross-links.
  • Thermosetting plastics have a high melting point and tensile strength as compared to thermoplastics.
  • one of the most common types of thermoplastic are Bakelite, Melamine, Polyesters etc.

Question 15

Which of the following is not the function of cutting fluid?

  1. AOption (a)

    Removal of heat

  2. BOption (b)

    Lubrication

  3. COption (c)

    Protection against corrosion

  4. DOption (d)

    Abrasion of workpiece

View answer & explanation

Correct answer

Answer: (d)D.

Abrasion of workpiece

Why this answer is correct

Explanation:

Cutting fluid

Cutting fluid is a type of coolant and lubricant designed specifically for metalworking processes, such as machining and stamping. There are various kinds of cutting fluids, which include oils, oil-water emulsions, pastes, gels, aerosols (mists), and air or other gases.

Various functions of cutting fluids are

  1. Cutting fluid cools the workpiece and tool by carrying away the heat generated during machining
  2. It acts as a lubricant at the friction zones, hence tool life increases
  3. It prevents the corrosion of chips and machine
  4. As friction get reduced, the forces and electric power consumption decreases
  5. Using cutting fluids produces a better surface finish to the workpiece
  6. It causes to break the chips into small pieces
  7. It washes away the chips from the tool
  8. Improves dimensional accuracy and control on the workpiece

Question 16

Which of the following machining processes involves cutting of a hard workpiece by a soft material?

  1. AOption (a)

    Turning 

  2. BOption (b)

    Grinding

  3. COption (c)

    Electro-discharge machining

  4. DOption (d)

    Honing

View answer & explanation

Correct answer

Answer: (b)B.

Grinding

Why this answer is correct

Explanation:

​Grinding

  • Grinding is a metal cutting operation performed by means of a rotating abrasive tool, called “grinding wheel”.
  • Grinding is the most common form of abrasive machining.
  • It is a material cutting process that engages an abrasive tool whose cutting elements are grains of abrasive material known as grit.
  • These grits are characterized by sharp cutting points, high hot hardness, chemical stability, and wear resistance.
  • The grits are held together by a suitable bonding material to give the shape of an abrasive tool.

Designation of Grinding Wheel:

Grades of Grinding Wheel:

  • It indicates the hardness of the grinding wheel.
  • The grade of the grinding wheel is selected based on the mechanical properties of the workpiece material.
  • Soft wheels are used for grinding of hard workpiece because the rubbing forces induced by the blunt abrasive particle i.e. the self-sharpening is taking place and no dressing is required.
  • Hard wheels are used for grinding of the soft workpiece, the abrasive particle will be effectively utilized so that at the end of effective utilization the dressing will be carried for resharpening of grinding wheel.
  • A –H = Soft, I – P = Medium, Q – Z = Hard​

 

Electrical discharge machining:

  • Electrical discharge machining (EDM) is a manufacturing process whereby a desired shape is obtained by using electrical discharges (sparks).
  • Material is removed from the work-piece by a series of rapidly recurring current discharges between two electrodes, separated by a dielectric liquid and subject to an electric voltage.
  • EDM has the lowest specific power requirement and can achieve sufficient accuracy.
  • This method can be used with any material only it should be conductive in nature and hardness is not a constraint means any hardness material can be machined only it should be electrically conductive in nature.
  • In this process, a regulated  DC power supply used along with a liquid di-electric solution like paraffin or some light oil like transformer oil or kerosene.

Turning:

  • Turning is a form of machining, a material removal process, which is used to create rotational parts by cutting away unwanted material.
  • This may be classified as external or internal.

Honing:

  • Honing is a finishing process, in which a tool called hone carries out a combined rotary and reciprocating motion while the workpiece does not perform any working motion.
  • Most honing is done on an internal cylindrical surface, such as automobile cylindrical walls.

Question 17

Which of the following processes cannot be applied for machining of internal gears?

  1. AOption (a)

    Hobbing

  2. BOption (b)

    Shaping

  3. COption (c)

    Milling 

  4. DOption (d)

    Broaching

View answer & explanation

Correct answer

Answer: (a)A.

Hobbing

Why this answer is correct

Explanation:

Gear Hobbing Process

  • It is a gear generation process- In this process, the gear blank is rolled with a rotating cutter called a hob.
  • Gear hobbing is done by using a multipoint cutting tool called gear hob; It looks like a worm gear having a number of straight flutes all around its periphery parallel to its axis
  • These flutes are so shaped that by giving proper angles to them, these work as cutting edges
  • In gear hobbing operation, the hob is rotated at a suitable rpm and simultaneously fed to the gear blank; The gear blank is also kept as a revolving
  • Gear hobbing is used for making a wide variety of gears like spur gear, helical, hearing-bone, splines and gear sprockets, etc.

Advantages:

  • Gear hobbing is a fast and continuous process so it is realized as an economical process as compared to other gear generation processes
  • The method is versatile and can generate spur, helical, worm, and worm wheels

Limitation:

  • There is one glaring limitation of gear hobbing, that is the manufacturing of internal gears is not possible

 

​Shaping Process:

  • The operation of removing a layer of material from the surface of the workpiece or producing a slot on the workpiece by using a reciprocating single-point cutting tool is called a shaping operation.
  • In shaping operation, the workpiece is fixed and the cutting tool is movable i.e velocity is provided to the cutting tool.

​

Question 18

The purpose of applying flux coating on AC arc welding electrode includes

  1. AOption (a)

    maintaining continuity of arc

  2. BOption (b)

    protection against corrosion

  3. COption (c)

    removal of impurities

  4. DOption (d)

    all of the above

View answer & explanation

Correct answer

Answer: (d)D.

all of the above

Why this answer is correct

Explanation:

In arc welding, the flux is coated on the electrodes. The flux forms a protective coating of slag over the weld metal and creates a non-oxidising atmosphere.

Functions of the flux coating:

During welding, because of the heat of the arc the flux coating on the electrode melts and perform the following function:

  • It stabilizes the arc
  • Removes oxide, scale etc. and cleans the surfaces to be welded
  • It forms a gaseous shield around the arc which protects the molten weld pool from atmospheric contamination
  • It compensates the losses of certain elements which are burnt out during welding
  • It retards the rate of cooling of the deposited metal by covering with slags and improves its mechanical properties
  • It helps to give a good appearance to the weld and controls penetration
  • Both AC and DC can be used for the welding

Question 19

For producing more accurate holes, the sequence of operations to be followed is:

  1. AOption (a)

    Centering, drilling, boring, reaming

  2. BOption (b)

    Centering, boring, drilling, reaming

  3. COption (c)

    Drilling, centering, boring, reaming

  4. DOption (d)

    Drilling, reaming, boring, centering

View answer & explanation

Correct answer

Answer: (a)A.

Centering, drilling, boring, reaming

Why this answer is correct

Concept:

One typical series for hole making can be -

  1. Centering
  2. Drilling
  3. Boring
  4. Reaming
  5. Honing.

Centering is carried out to locate the hole for easy and accurate alignment of a drill.

Although it is optional, centering can improve precision.

Drilling is a process of originating a hole, while boring enlarges the diameter of an existing hole.

Reaming and honing are used to improve the surface finish and tolerance of the existing hole.

  • Drilling: Drilling is a cutting process that uses a drill bit to cut or enlarge a hole of circular cross-section in solid materials. The drill bit is a rotary cutting tool.
  • Boring: Boring is an operation to enlarging of an existing hole, which may have been made by a drill or maybe the result of a core in a casting.
  • Reaming: Reaming is a sizing operation that removes a small amount of metal from a hole already drilled. It is done for two purposes: to bring holes to a more exact size and to improve the finish of an existing hole.

Honing is a superfinishing process carried out using abrasive sticks for the removal of stock from metallic and non-metallic surfaces. This process produces a high surface finish, corrects the profiles of cylindrical surfaces, removes taper.

Question 20

The correct sequence of operations for making alloy steel gears is

  1. AOption (a)

    turning, hobbing, heat treatment, shaving

  2. BOption (b)

    heat treatment, turning, hobbing, shaving

  3. COption (c)

    shaving, hobbing, turning, heat treatment

  4. DOption (d)

    turning, hobbing, shaving, heat treatment

View answer & explanation

Correct answer

Answer: (d)D.

turning, hobbing, shaving, heat treatment

Why this answer is correct

Explanation:

For making gears, following sequence is used.

  1. Turning
  2. Hobbing
  3. Shaving
  4. Heat treatment

Turning:

  • It is performed on lathe machines. In this process, a single point cutting tool is attached to the tool post mounted on the cross slide. Cross slide is mounted on a carriage which can move over the guideways made on the bed of the machine.
  • The work is attached to a rotating chuck mounted on the headstock.
  • The work can also be supported through a dead center.
  • The feed is provided by the longitudinal motion of the carriage, while the depth of cut is provided by cross slide.

​

Hobbing: 

  • Gear hobbing is a multi-point machining process through which gear teeth are progressively generated with a helix cutting tool(hob) both the bob and the workpiece revolve constantly as the bob is fed across the face width of the gear blank.

       

​

  • Shaving: Gear shaving is the way of finishing the involute gear tooth that involves gear like tool rolling with the work on the crossed axis

​

  • Heat treatment: It is the process of heating and cooling of a material to change its physical and mechanical properties without changing the original shape and size.

Question 21

The correct sequence of operations for machining a precise rectangular window in a through-hardening die steel workpiece is

  1. AOption (a)

    heat treatment, surface grinding, drilling, wire EDM

  2. BOption (b)

    drilling, wire EDM, heat treatment, surface grinding

  3. COption (c)

    drilling, heat treatment, surface grinding, wire EDM

  4. DOption (d)

    surface grinding, heat treatment, wire EDM, drilling

View answer & explanation

Correct answer

Answer: (b)B.

drilling, wire EDM, heat treatment, surface grinding

Why this answer is correct

Concept:

The term hardened steel is often used for medium or high carbon steel that has been given heat treatment and then quenching followed by tempering.

  • Drilling Process after the Hardening is very difficult hence it should be done before Hardening Process
  • Wire EDM machining (Electrical Discharge Machining) is an electrothermal production process where a thin single-strand metal wire, along with de-ionized water (used to conduct electricity) allows the wire to cut through metal by the use of heat from electrical sparks while preventing rust.
  • Heat treatment is the process of heating the metal, holding it at that temperature, and then cooling it back.
  • During the process, the metal part will undergo changes in its mechanical properties. This is because the high temperature alters the microstructure of the metal.
  • Surface grinding is used to produce a smooth finish on flat surfaces. It is a widely used abrasive machining process in which a spinning wheel covered in rough particles **(**grinding wheel) cuts chips of metallic or nonmetallic substances from a workpiece, making a face of it flat or smooth.

Question 22

The correct sequence in increasing hardness of tool materials is

  1. AOption (a)

    HSS, tungsten carbide, cubic boron nitride, diamond

  2. BOption (b)

    HSS, cubic boron nitride, tungsten carbide, diamond

  3. COption (c)

    cubic boron nitride, HSS, tungsten carbide, diamond

  4. DOption (d)

    tungsten carbide, cubic boron nitride, HSS, diamond

View answer & explanation

Correct answer

Answer: (a)A.

HSS, tungsten carbide, cubic boron nitride, diamond

Why this answer is correct

Concept:

Hardness is a measure of the resistance to Plastic Deformation induced by either Mechanical Indentation or Abrasion.

Hardness measures in three different ways,

  1. Indentation Test
  2. Scratch Hardness Test
  3. Abrasion and Erosion Test​

​The hardest cutting tool materials are in descending order from hardest to toughest are:

  1. Diamond.
  2. Cubic boron nitride. (CBN)
  3. Ceramics (usually some form of aluminium oxide Al2O3)
  4. Cermets a hybrid between carbide and ceramic.
  5. Cemented carbides (tungsten carbide alloys)
  6. High-Speed Steel.​

Question 23

CNC machines are best suited, when

  1. AOption (a)

    production volumes are very high and there is no product variety

  2. BOption (b)

    holes are to be bored at precise locations

  3. COption (c)

    very small production volumes and product varieties are involved, while accuracy requirements are not very stringent

  4. DOption (d)

    a moderate production volume is combined with a moderate product variety and contoured shapes are to be cut

View answer & explanation

Correct answer

Answer: (a)A.

production volumes are very high and there is no product variety

Why this answer is correct

Concept:

  • Computer Numeric Control (CNC) is a computer-assisted process to control general-purpose machining from instruction generated by a processor and stored in a memory system or storage media.
  • CNC is a specific form of Control system where the position is Principle Control Variable.
  • CNC machine is best suited when:
  • There is batch Production because of their Highly Working
  • When there is no Precise Location
  • When Production Volume is High with a repeated job making
  • Advantages of CNC Machine
  • Improves machining accuracy.
  • Enables complex tasks, detail.
  • Creates flexibility in manufacturing.
  • Increases safety.
  • Boosts production volume.
  • Reduces setup-changeover time.
  • Precise Location Required more Jigs and Fixture which leads Increase in Operating cost

Question 24

Which of the following screw threads does not have a symmetric section?

  1. AOption (a)

    Buttress 

  2. BOption (b)

    Square

  3. COption (c)

    Whitworth

  4. DOption (d)

    Acme

View answer & explanation

Correct answer

Answer: (a)A.

Buttress 

Why this answer is correct

Explanation:

Square thread: In this thread, the flanks are perpendicular to the axis of the thread. This is used for transmitting motion or power. E.g. Fly presses, Screw jack, vice handles, cross-slide and compound slide, etc.

Trapezoidal threads: These threads have a profile that is neither square nor V thread form and have a form of a trapezoid. Different forms of trapezoidal threads are ACME thread, Buttress thread, Saw-tooth thread, Worm thread

  • ACME thread: It has an included angle of 29°. These are used in lathe lead screws.
  • Buttress Thread: In this thread, one flank is perpendicular to the axis of the thread and another flank is at 45°. These threads are used in the power press, carpentry vices, gun breeches, ratchets, etc.
  • Knuckle thread: The shape of the knuckle thread is not trapezoid, but it has a rounded shape. It has limited applications. It is used for valve spindles, railway carriage coupling, hose connections, etc.

Whitworth threads: The principal features of the British Standard Whitworth (BSW) thread form are that the angle between the thread flanks is 55 degrees and the thread has radii at both the roots and the crests of the thread.

Among all threads, only Buttress threads do not have a symmetric section.

Question 25

Which of the following is a wrong statement?

  1. AOption (a)

    Key is generally used for transmitting rotational motion

  2. BOption (b)

    Cotter is generally employed for transmitting axial motion

  3. COption (c)

    The keys are loaded in shear

  4. DOption (d)

    The cotters have to withstand tensile or compressive loads

View answer & explanation

Correct answer

Answer: (d)D.

The cotters have to withstand tensile or compressive loads

Why this answer is correct

Concept:

Keys :

  • Key is a Machinery Component that provides a Torque Transmitting link between Power Transmitting Elements
  • IC engines, Turbines, Rotating Shafts, Elements like Wheel, Gears, Pulleys, and Sprocket are connected with the help of Keys between the Shaft and Hub of the rotating member
  • Keys transfer power from shaft to Rotating Member. Hence it is used to Transfer rotational motion
  • Keys are Inserted into Keyways which are loaded in shear
  • Example of Keys: Rectangular Key, Round Key, Saddle Key, etc.

Cotter :

  • Cotter's joint is used to join two shafts which are in Rotation.
  • Cotter's joint is used when the members are subjected to Axial Tensile or Compressive Loading.
  • It is not used to connect the shafts that are rotating or transmitting torque. e.g. Joint between the piston rod and the crosshead of a steam engine.
  • There are three main parts of the Cotter joint
  1. Cotter
  2. Spigot
  3. Socket

Confusion Points

  • Cotter is generally employed for transmitting Tensile or Compressive axial motions and withstand the double shear and bending failure.

Question 26

When a belt happens to start slipping

  1. AOption (a)

    It can no longer transmit any power as kinetic friction is less than static friction

  2. BOption (b)

    It will still continue to transmit power with some slipping

  3. COption (c)

    Power transmission efficiency will increase

  4. DOption (d)

    Flat belts do not slip

View answer & explanation

Correct answer

Answer: (b)B.

It will still continue to transmit power with some slipping

Why this answer is correct

Explanation:

A firm frictional grip between belt and pulley or shaft is essential. But sometimes it becomes insufficient.

This may cause some forward motion of the belt without carrying the driven pulley with it, is called Slip of the belt.

  • Static friction is always greater than kinetic friction irrespective of slipping or non-slipping.
  • Power transmission (P) is a function of torque (T) and speed (N), i.e. P=2πNT60P=\frac{2\pi NT}{60}
  • Slipping reduces speed and torque
  • Slippage of the belt is less in the V-type belt with the respective pulley.
  • When the belt just starts slipping it will still continue to transmit power with some slip.

​

Additional Information

  • In pulley rotation certain portion of the belt will pass through the area where it will change from tight side to slack side, then the length of that specific portion of the belt will expand and contract subsequently and a relative motion will happen between the belt and pulley surface. This phenomenon is called creep of the belt.

Difference between slip and creep.

  • Creep is due to the elastic property of the belt whereas, the conventional slip is due to insufficient frictional grip between the belt and pulley.
  • However, the effect of the creep, as well as slip, is to reduce the speed ratio, and hence power transmission.

Question 27

The tight and slack sides of a belt connecting two pulleys are having tensions of 25 N and 15 N respectively, while the belt is running at 10 m/s. The power transmitted as

  1. AOption (a)

    250 W

  2. BOption (b)

    150 W

  3. COption (c)

    100 W

  4. DOption (d)

    Data provided is insufficient for calculating power

View answer & explanation

Correct answer

Answer: (c)C.

100 W

Why this answer is correct

Concept:

Power transmitted by a belt:

P = (Tt – Ts).v

where Tt the tension on the tight side (N), Ts is the tension on the slack side (N), V is the velocity of the belt in m/s

Calculation:

Given:

Tt = 25 N, Ts = 15 N, v = 10 m/s

Therefore, P = (25 - 15 ) × 10

P = 100 W

Question 28

Compared to a belt drive, a chain drive is generally characterized by

  1. AOption (a)

    constant velocity ratio 

  2. BOption (b)

    absence of slip

  3. COption (c)

    heavier construction 

  4. DOption (d)

    all of the above

View answer & explanation

Correct answer

Answer: (d)D.

all of the above

Why this answer is correct

Concept:

The power is transmitted from one shaft to the other by means of belts, ropes, chains, and gears.

Chain drives 

  • Chain drives are the positive types of the drive-in in which chain and sprocket mechanism are used to transmit the power by engaging the teeth of the sprocket with gaps in the chain.

  • As there will be no slip or loss while transmission so they are positive drives.

  • As the distance of transmission is more in the case of bicycle motorcycle and coal, chain drives are used for transmission of power.

  • The belts and ropes are flexible members which are used where the distance between the two shafts is large

  • The chains also have flexibility but they are preferred for intermediate distances

  • The gears are used when the shafts are very close to each other

  • This type of drive is also called a positive drive because there is no slip

  • If the distance is slightly larger, chain drive can be used for making it a positive drive

  • Belts and ropes transmit power due to the friction between the belt or rope and the pulley

  • There is a possibility of slip and creep and that is why this drive is not a positive drive

  • The gear drive and chain drive are positive drives

Question 29

Which of the following bearings are most suitable for supporting high axial thrust?

  1. AOption (a)

    Radial ball bearings

  2. BOption (b)

    Needle bearings

  3. COption (c)

    Cylindrical roller bearings 

  4. DOption (d)

    Tapered roller bearings

View answer & explanation

Correct answer

Answer: (d)D.

Tapered roller bearings

Why this answer is correct

Concept:

  • Tapered roller bearings are capable of carrying both radial and axial loads.
  • They are often used in pairs to take the thrust load in both directions.
  • The taper roller bearing consists of rolling elements in the form of a frustum of cone. They are arranged in such a way that the axes of individual rolling elements intersect in a common apex point on the axis of the bearing.
  • The taper roller bearing consists of rolling part and inner and outer raceways in the form of conical surfaces. The outer raceway or outer ring is called ‘cup’ and inner raceway is called ‘cone’.

Needle roller bearing

  • These bearings are relatively slender and completely fill the space so that neither a cage nor a retainer is needed.
  • These bearings are used when heavy loads are to be carried with an oscillatory motion.
  • For example, piston pin bearing in heavy-duty diesel engines where the reversal of motions tends to keep the roller in correct alignment.

Additional Information

The following Points are to be noted about the load-carrying capacity of different types of rolling contact bearings: 

  • Thrust Ball Bearing: Thrust ball bearing cannot take the radial load.
  • Taper Roller Bearing: can take heavy radial and thrust loads.
  • Angular Contact Bearing: can take both radial and thrust loads.
  • Cylindrical Roller Bearing: can take radial load only
  • Deep Groove Ball Bearing: takes load in radial as well as an axial direction

Question 30

A material store comprises of three 2-sided aisles, each having a length and height corresponding to 50 and 15 storage compartments. The total storage capacity will be

  1. AOption (a)

    9000 compartments

  2. BOption (b)

    4500 compartments

  3. COption (c)

    3500 compartments

  4. DOption (d)

    2500 compartments

View answer & explanation

Correct answer

Answer: (b)B.

4500 compartments

Why this answer is correct

Concept:

Total storage capacity = L × H × N × n

where, length of the aisle = L, Height of the aisle = H, Number of aisles = N, Number of sides of the aisle = n.

Calculation:

Given:

L = 50, H = 15, N = 3, n = 2.

Total storage capacity = L × H × N × n = 50 × 15 × 3 × 2 = 4500

Question 31

Which of the following is a wrong statement?

  1. AOption (a)

    PERT is suitable for projects having probabilistic time estimates

  2. BOption (b)

    CPM is suitable for projects having deterministic activities

  3. COption (c)

    Both PERT and CPM are event oriented

  4. DOption (d)

    PERT is event oriented, while CPM is activity oriented

View answer & explanation

Correct answer

Answer: (c)C.

Both PERT and CPM are event oriented

Why this answer is correct

Explanation:

A project may be defined as a combination of interrelated activities which must be executed in a certain order before the entire task can be completed.

The aim of planning is to develop a sequence of activities of the project so that the project completion time and cost are properly balanced.

To meet the objective of systematic planning, the management has evolved several techniques applying network strategy.

PERT (Programme Evaluation and Review Technique) and CPM (Critical Path Method) are network techniques which have been widely used for planning, scheduling and controlling the large and complex projects.

  • PERT (Project Evaluation and Review Technique) approach takes account of the uncertainties. In this approach, 3-time values are associated which each activity. So it is probabilistic.

  • CPM (Critical Path Method) involves the critical path which is the largest path in the network from starting to ending event and defines the minimum time required to complete the project.So it is deterministic.

Difference between PERT and CPM (Critical Path Method)

PERTCPM
1. Probabilistic approach1. Deterministic approach
2. Three-time estimate2. One - time estimate
3. Event oriented network model3. Activity-oriented network model
4. The slack concept is used4. Float concept is used
5. Project crashing is not possible5. Project crashing is possible
6. Deals with probabilistic time estimates6. Deals with deterministic time estimates

Question 32

While solving a linear programming model, if a redundant constraint is added, then what will be its effect on existing solution?

  1. AOption (a)

    There will be no effect

  2. BOption (b)

    The solution space will get further constrained

  3. COption (c)

    The solution space becomes concave

  4. DOption (d)

    The problem no longer remains solvable

View answer & explanation

Correct answer

Answer: (a)A.

There will be no effect

Why this answer is correct

Explanation:

  • Linear programming (LP) in industrial engineering is used for the optimization of our limited resources when there is a number of alternate solutions possible for the problem. The real-life problems can be written in the form of a linear equation by specifying the relation between its variables.
  • The general LP problem calls for optimizing a linear function for variables called the "objective function" subjected to a set of linear equations and inequalities called the constraints or restriction.
  • A redundant constraint is a constraint that can be removed from a system of linear constraints without changing the feasible region.
  • While solving a linear programming model, if a redundant constraint is added, then there will be no effect on the existing solution​.

​

Additional Information

  • Consider the following linear inequality constraints:

x = x1 and y = x2

  1. 2x +1y ≤ 8,
  2. 4x + 0y ≤ 15,
  3. 1x + 3y ≤ 9,
  4. 1x + 2y ≤ 14,
  5. 1x ≤ 4,
  6. 1x + 1y ≤ 5,

where, x + y ≤ 0

Methods for Identification of Redundant Constraints 

Many methods are available in the literature to identify the redundant constraints in linear programming problems. In this paper, the following five methods are discussed and compared

  1. Bounds method
  2. Linear programming method
  3. Deterministic method
  4. Stojkovic and Stanimirovic method
  5. Heuristic method

Question 33

Which of the following layout is useful when the product being processed is very big, heavy or difficult to move?

  1. AOption (a)

    Fixed position

  2. BOption (b)

    Process layout

  3. COption (c)

    Product layout

  4. DOption (d)

    Cellular layout

View answer & explanation

Correct answer

Answer: (a)A.

Fixed position

Why this answer is correct

Concept:

Layout planning in manufacturing and service organisations involves the physical arrangement of various resources available in the system to improve the performance of the operating system, thereby providing better customer services.

Layouts can be classified into the following four categories:

  1. Process layout
  2. Product layout
  3. Group layout (Combination layout)
  4. Fixed-position layout

Fixed-position layout:

In a fixed position layout, the service is performed around a customer that remains stationary while the work is being done.

For example, surgery is performed on a patient, where a patient remains stationary in operation theatre. Different doctors or specialist perform different activities in a sequence on a stationary patient.

Example: Manufacturers of aeroplanes, ships, locomotives, large turbines, heavy machinery, pressure vessels and others which involve heavy materials and sub-assemblies.

This type of layout is suitable:

  • When one or a few pieces of an identical product are to be manufactured
  • When the assembly consists of a large number of heavy parts, the cost of transportation of which is very high

 

Product Layout: In product layout, the workstations are located according to the processing sequence for the service.

Process Layout: In process layout, similar activities are grouped together according to the process or function they perform.

Question 34

The term 'jockeying' in queuing theory refers to

  1. AOption (a)

    not entering the long queue

  2. BOption (b)

    leaving the queue

  3. COption (c)

    shifting from one queue to another parallel queue

  4. DOption (d)

    none of the above

View answer & explanation

Correct answer

Answer: (c)C.

shifting from one queue to another parallel queue

Why this answer is correct

Concept:

There are four types of customer behaviours.

  1. Jockeying- When a customer keeps changing the queue just in hope to get fast service.
  2. Balking-Customer does not join the queue and leaves the system as the queue is very long
  3. Reneging- Customer joins the queue for a short period then leave the system as the queue is moving very slowly.
  4. Cheaters- Customer takes illegal means like bribing, fighting etc just in hope to get service faster.

Question 35

In Klein's construction for reciprocating engine mechanism, the scale of acceleration diagram will be

  1. AOption (a)

    linear scale of configuration diagram

  2. BOption (b)

    linear scale of configuration diagram multiplied by square of angular velocity of crank

  3. COption (c)

    square of the linear scale of configuration diagram

  4. DOption (d)

    square of the linear scale of configuration diagram multiplied by square of angular velocity of crank

View answer & explanation

Correct answer

Answer: (b)B.

linear scale of configuration diagram multiplied by square of angular velocity of crank

Why this answer is correct

Explanation:

Klein's Construction:

  • It is used to draw the velocity and acceleration diagrams for a single slider crank mechanism.
  • The velocity and acceleration of piston of a reciprocating engine mechanism can be determined by the figure given below.

Velocity diagram:

  • Draw the configuration diagram OAB for the slider-crank mechanism.
  • Draw OI perpendicular to OB and produce BA to meet OI at C. Then the triangle formed, is known as Klein's velocity diagram.
  • Velocity of connecting rod AB, VBA = ω × AC.
  • Velocity of piston B,VP = VBO = ω × OC.
  • Velocity of crankpin A, VAO = ω × OA.

Acceleration diagram:

  • With A as centre and AC as radius, draw a circle.
  • Locate D as the midpoint of AB.
  • With D as centre, and DA as radius, draw a circle to intersect the previously drawn circle at points H and E.
  • Join HE intersecting AB at F.
  • Produce HE to meet OB at G.
  • Then OAFG is Klein's acceleration diagram.
  • Acceleration of piston (Slider B), fBO = ω2 × OG.
  • Tangential acceleration of connecting rod, ftBA = ω2 × FG.
  • Normal acceleration of connecting rod, fnBA = ω2 × AF.
  • Total acceleration of connecting rod, fBA = ω2 × OG.
  • Angular acceleration of connecting rod, AB αAB=fBAt/AB=ω2FGAB;;;;(ccw)\alpha_{AB}=f^{t}_{BA}/AB=ω^{2}\frac{FG}{AB};;;;(ccw)
  • To find the acceleration of any point x in AB, draw a line X-X parallel to OB to intersect at X. Join OX.
  • Then, Acceleration of point x, fx = ω2 × OX.

Though, we can calculate both velocity and acceleration through Klein's construction but mainly it is used in calculating the linear acceleration of the piston.

Question 36

A piston having mass m = 3 kg executes pure simple harmonic motion, whose displacement in meters is given by x = 0.05 sin(10 t). Its inertia force at time t = 0 will be

  1. AOption (a)

    0 N

  2. BOption (b)

    – 0.15 N

  3. COption (c)

    – 1.5 N

  4. DOption (d)

    – 15 N

View answer & explanation

Correct answer

Answer: (a)A.

0 N

Why this answer is correct

Concept:

Inertia force:

"Inertial force" refers to the amount of force it would take to move a certain object or stop it from moving entirely.

This can be found using Newton's Second Law

F = ma  

where F = force of inertia, m = mass of object and a = acceleration

Acceleration is defined as:

Acceleration;(a)=Velocity;(v)Time;(t)=Displacement;(d)Time;(t)Time;(t)Acceleration;\left( a \right) = \frac{{Velocity;\left( v \right)}}{{Time;\left( t \right)}} = \frac{{\frac{{Displacement;\left( d \right)}}{{Time;\left( t \right)}}}}{{Time;\left( t \right)}}

Calculation:

Given:

mass of piston(m) = 3 kg, Displacement(x) =0.5 sin(10t), time =0, Acceleration(a), Inertia force(F) = ?

m = 3 kg

x = 0.05 sin(10t) meter

dxdt\frac{{dx}}{{dt}}​= 0.05 × 10cos(10t)

d2xdt2\frac{{{d^2}x}}{{d{t^2}}} = – 0.05 × 10 × 10sin(10t)

At t = 0, 

F = m × a = 3 × 0 = 0 Newton

Question 37

A cam displaces its follower by 20 mm through simple harmonic motion (SHM). The cam is rotating at 15 RPM and its angle of rise is 90°. The time period of the SHM will be

  1. AOption (a)

    4 seconds

  2. BOption (b)

    2 seconds

  3. COption (c)

    1 seconds

  4. DOption (d)

    0.5 second

View answer & explanation

Correct answer

Answer: (c)C.

1 seconds

Why this answer is correct

Explanation:

  • Cam is a mechanical element which is used to deliver the reciprocating or oscillating motion to another machine element known as a follower.
  • A simple harmonic motion of constant amplitude in which the acceleration is proportional and oppositely directed to the displacement of the body from a position of equilibrium.
  • A simple harmonic motion curve is widely used since it is simple to design. The curve is the projection of a circle about the cam rotation axis as shown in the figure.

where H = maximum height of follower in SHM motion

Calculation:          

Given:

Displacement of follower in SHM = 20 mm, angle of rise  θ = 90o = π /2, speed of cam rotation  N = 15 RPM,

Angular velocity of cam ω = 2πN60\frac{{2\pi N}}{{60}}

Now time period of the cam in SHM,

T = angle of rise in radian / angular velocity 

T=π22πN60;;T = \frac{{\frac{\pi }{2}}}{{\frac{{2\pi N}}{{60}}}};;

T=π×602×2πNT = \frac{{\pi \times 60}}{{2 \times 2\pi N}}

T = 1 sec

Key Points

The time period in SHM is simply angle turned upon angular velocity.

Question 38

If a flywheel having mass of 100 kg and radius of gyration 10 cm is rotating at 10 rad/s, its rotational kinetic energy will be

  1. AOption (a)

    10 J

  2. BOption (b)

    25 J

  3. COption (c)

    50 J

  4. DOption (d)

    100 J

View answer & explanation

Correct answer

Answer: (c)C.

50 J

Why this answer is correct

Concept:

  • Rotational kinetic energy: The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
  • A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
  • Mathematically rotational kinetic energy can be written as -

KE=12Iω2KE = \frac{1}{2}I{ω ^2}

where I = moment of inertia and ω = angular velocity, k = radius of gyration

The Moment of inertia of flywheel is given as, I = mk2 

Calculation:

Given:

m = 100 kg, ω = 10 rad/sec, k = 10 cm = 0.1 m

Therefore, KE=12(mk2)ω2KE = \frac{1}{2}(mk^2){ω ^2}

\(KE = \frac{1}{2}\times100~\times0.1^2\times~{10 ^2}=50~J\)

QuantityExpression
TorqueIα
Angular momentumIω
Rotational kinetic energy12Iω2\frac{1}{2}I{ω ^2}
Work doneτθ
Powerτω

Question 39

Calculate the critical speed of a lightweight, vertically mounted shaft carrying a rotor of mass 200 g at its mid-point. The stiffness of shaft at the location stiffness is 72000 N/m.

  1. AOption (a)

    360 rad/s

  2. BOption (b)

    19 rad/s

  3. COption (c)

    360000 rad/s

  4. DOption (d)

    600 rad /s

View answer & explanation

Correct answer

Answer: (d)D.

600 rad /s

Why this answer is correct

Concept:

The speed at which the shaft runs so that the additional deflection of the shaft from the axis of rotation becomes infinite is known as critical or whirling speed. And it happens when the critical speed of a shaft becomes equal to its natural frequency.

ωn=;ωc=km=;gδ{\omega _n} = ;{\omega _c} = \sqrt {\frac{k}{m}} = ;\sqrt {\frac{g}{δ }}

where k is the stiffness of the shaft, m is mass of the shaft and δ is the static deflection of the shaft.

Calculation:

Given:

m = 200 gm = 0.2 kg, k = 72000 N/m

Therefore, ωn=;ωc=km=;720000.2=600 rad/s{\omega _n} = ;{\omega _c} = \sqrt {\frac{k}{m}} = ;\sqrt {\frac{72000}{0.2 }} =600~rad/s

Question 40

Which of the following is a correct statements?

  1. AOption (a)

    A flywheel regulates speed during one cycle while the governor regulates speed from cycle to cycle

  2. BOption (b)

    The flywheel is an essential element of every prime mover

  3. COption (c)

    A flywheel controls the quantity or quality of the working agent

  4. DOption (d)

    A governor regulates speed by storing kinetic energy

View answer & explanation

Correct answer

Answer: (a)A.

A flywheel regulates speed during one cycle while the governor regulates speed from cycle to cycle

Why this answer is correct

Concept:

  • The flywheel does not maintain a constant speed, it simply reduces the fluctuation of speed.
  • It does not control the speed variations caused by the varying load. Governor regulates the mean speed of an engine when there are variations in the load.
  • As the force impulse is given by the change in momentum, i.e. ΔP = F.Δt. where P is momentum and t is time.
  • In the same way, the power impulse is given by the change or fluctuation in energy, i.e. ΔE = P.Δt. where P is power and t is time.
  • Thus, smooth out power impulses mean the smoothness in the fluctuation of energy, which is the work of a flywheel.

Important Points

  • A flywheel stores energy when the supply of energy is less than the required but the flywheel will not work, then the main supply will fail, i.e. when there is no supply of energy or zero supply of energy.
  • A governor, being an adjuster of supply with demand, is an essential element of every prime mover.
  • A flywheel is not an essential element of every prime mover. It is used only in case where there is undesirable cyclic fluctuations of energy output or input.

Question 41

The secondary force in a crank piston mechanism

  1. AOption (a)

    arises due to obliquity of connecting rod

  2. BOption (b)

    acts at double the frequency as that of the primary force

  3. COption (c)

    is smaller in magnitude than the primary force

  4. DOption (d)

    all of the above

View answer & explanation

Correct answer

Answer: (d)D.

all of the above

Why this answer is correct

Concept:

Consider a piston crank mechanism as shown;

m = mass of reciprocating part, l = length of connecting rod, r = radius of crank, ω  = angular speed of crank, θ = angle of inclination of crank with stroke lenth, n = ratio of length of connecting rod to radius of crank (l/r)

from acceleration analysis, acceleration of piston;

a = ω2 r.(cosθ + (cos2θn)\left( {\frac{{{\rm{cos}}2{\rm{\theta }}}}{{\rm{n}}}} \right))

This formula of acceleration is derived by considering the obliquity of connecting rod ( l > r) hence we can say that unbalance forces are created due to the obliquity of connecting rod.

now inertia force experienced by the mechanism due to this acceleration is 

Force = mass * acceleration

F=m.r.ω2(cosθ + (cos2θn)\left( {\frac{{{\rm{cos}}2{\rm{\theta }}}}{{\rm{n}}}} \right))

first part i.e. mrω2 cosθ of this inertia force is called primary force, and second part i.e. m.r.ω2(cos2θn){\rm{m}}.{\rm{r}}.{{\rm{\omega }}^2}\left( {\frac{{{\rm{cos}}2{\rm{\theta }}}}{{\rm{n}}}} \right) is called secondary force.

hence secondary force in piston crank mechanism

F2;=;m.r.ω2(cos2θn){\rm{F}}2{\rm{;}} = {\rm{;m}}.{\rm{r}}.{{\rm{\omega }}^2}\left( {\frac{{{\rm{cos}}2{\rm{\theta }}}}{{\rm{n}}}} \right)

NOW from this equation we can conclude that secondary force depends upon ;

  • obliquity of connecting rod (because n>1)
  • the secondary force acts as that of double the frequency of primary force
  • since the value of secondary force(mrω 2/n) is (1/n) times the maximum value of primary force(mrω 2) hence secondary force is smaller in magnitude as compared to primary force.

hence all the options are correct so d will be the answer.

Question 42

When a system is subjected to forced vibrations, then under steady-state conditions

  1. AOption (a)

    it vibrates at its natural frequency

  2. BOption (b)

    it vibrates at the imposed frequency

  3. COption (c)

    it vibrates at the mean of natural and imposed frequencies

  4. DOption (d)

    none of these

View answer & explanation

Correct answer

Answer: (b)B.

it vibrates at the imposed frequency

Why this answer is correct

Explanation:

Vibration: 

  • Vibration is a mechanical phenomenon, where oscillation takes place around an equilibrium point, it is a to and fro motion of an object.
  • There are two basic types of vibration. 
  • Free vibration
  • Forced vibration
  • Free vibration occurs when a mechanical system is set off with an initial input and then allowed to vibrate freely.
  • In the case of free vibration, the body will vibrate at its natural frequency of vibration.

Forced vibration:

  • If a system or body is subjected to periodic external forces then the resulting vibration is called Forced vibration.
  • when an external force is applied than the body starts vibration, and when the body vibrates with constant amplitude then we called it steady-state of vibration, now the body will not vibrate about its natural frequency of vibration, but vibrate with the frequency of applied external force which is called imposed frequency.

 

Steady state Amplitude

X=F0/k(1−r2)2+(2ξr)2X = \frac{{{F_0}/k}}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}

here, r=ωωhr=\dfrac{ω}{ω_h}

ω → inposed frequency

Important Points

In free vibration, object vibrate with their natural frequency & in forced vibration object with the frequency of excitation force i.e. imposed frequency.

Question 43

The problem of interference in an involute pinion, which has to come in external mesh with a gear, can be avoided by

  1. AOption (a)

    increasing the pressure angle

  2. BOption (b)

    positive profile correction on pinion and corresponding negative profile correction on gear

  3. COption (c)

    machining the pinion teeth through hobbing so that the interfering portion gets undercut

  4. DOption (d)

    all of the above

View answer & explanation

Correct answer

Answer: (d)D.

all of the above

Why this answer is correct

Concept:

  • When two gears are in mesh at one instant there is a chance to mate involute portion with the non-involute portion of the mating gear.
  • This phenomenon is known as the INTERFERENCE and occurs when the number of teeth on the smaller of the two meshing gears is less than a required minimum.
  • The phenomenon when the tip of the tooth undercuts the root on its mating gear is known as interference.
  • Methods of elimination of Gear tooth Interference:
  • Use of a larger pressure angle (having a larger pressure angle results in a smaller base circle. As a result, more of the tooth profiles become involute)
  • Under-cutting of a tooth (A portion of teeth below the base circle is cut off. When teeth are produced by this process, the tip of one tooth of gear will not contact the non-involute portion of the tooth of other gear)
  • Tooth stubbing (In this process a portion of the tip of the teeth is removed, thus preventing that portion of the tip of the tooth from contacting the non-involute portion of the other meshing tooth).
  • Increasing the number of teeth on the gear can also eliminate the chances of interference.
  • Increasing slightly the center distance between the meshing gears would also eliminate interference.
  • Tooth profile modification or profile shifting (Using profile shifted gears (gears with non-standard profile) can also be an option to eliminate interference. In profile shifted meshing gears, the addendum on the pinion is shorter compared with standard gears).

Question 44

The minimum recommended top land for hardened gears in terms of module m is

  1. AOption (a)

    0 m

  2. BOption (b)

    0.4 m

  3. COption (c)

    1 m

  4. DOption (d)

    1.2 m

View answer & explanation

Correct answer

Answer: (b)B.

0.4 m

Why this answer is correct

Concept:

Top land:

  • Top land is the top surface of gear teeth as shown in the figure below. The width of this top land is called the Face width of the gear teeth and the thickness of top land should be equal to the spacing between two consecutive gears to avoid meshing.
  • The minimum recommended Top land for hardened GEARS is  0.4m, where m refers to the module of the gear profile.

Question 45

Which of the following characteristics is generally exhibited by brittle materials?

  1. AOption (a)

    High impact strength

  2. BOption (b)

    Ultimate compressive strength higher than ultimate tensile strength

  3. COption (c)

    Good fracture toughness

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (b)B.

Ultimate compressive strength higher than ultimate tensile strength

Why this answer is correct

Explanation:

Brittleness

  • Brittleness is the opposite of ductility.
  • Brittle materials show little deformation before fracture and failure occurs suddenly without any warning i.e. it is the property of breaking without much permanent distortion.
  • Normally if the elongation is less than 5% the material is brittle. E.g. cast iron, glass, ceramics are typical brittle materials.
  • Brittle materials have the following characteristics:
  • High compressive strength [compressive strength > tensile strength]
  • Low tensile strength
  • Very low impact and impact strength and toughness

Important Points

Ductile material has high tensile strength, fatigue strength, impact strength, and toughness

Question 46

Consider a plane stress case, where σx = 3 Pa, σy = 1 Pa and τxy = 1 Pa. One of the principal directions w.r.t. x-axis would be

  1. AOption (a)

    0°

  2. BOption (b)

    15°

  3. COption (c)

    22.5°

  4. DOption (d)

    45°

View answer & explanation

Correct answer

Answer: (c)C.

22.5°

Why this answer is correct

Concept:

σ1,2=(σx+σy2)+(σx−σy2)cos⁡2θ+τxysin⁡2θ{\sigma _{1,2}} = \left( {\frac{{{\sigma _x} + {\sigma _y}}}{2}} \right) + \left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)\cos2\theta + {\tau _{xy}}\sin2\theta

τxy∗=−(σx−σy2)sin⁡2θ+τxycos⁡2θ\tau _{xy}^* = - \left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)\sin2\theta + {\tau _{xy}}\cos2\theta

At the principal plane, the value of shear stress is 0. i.e.τxy∗=0\tau _{xy}^* =0

Calculation:

Given:

τxy∗=0\tau_{xy}^*=0

−(σx−σy2)sin⁡2θ+τxycos⁡2θ=0 - \left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)\sin2\theta + {\tau _{xy}}\cos2\theta = 0

(σx−σy2)sin⁡2θ=τxycos⁡2θ\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)\sin2\theta = {\tau _{xy}}\cos2\theta

(3−12)sin⁡2θ=cos⁡2θ\left( {\frac{{3 - 1}}{2}} \right)\sin2\theta = \cos2\theta

sin⁡2θ=cos⁡2θ\sin2\theta = \cos2\theta

tan⁡2θ=1\tan2\theta=1

2θ=45∘2\theta=45^\circ

∴θ=22.5∘\therefore \theta = {22.5^\circ}

Question 47

A 3 m long beam, simply supported at both ends, carries two equal loads of 10 N each at a distance of 1 m and 2 m from one end. The shear force at the mid-point would be

  1. AOption (a)

    0 N

  2. BOption (b)

    5 N

  3. COption (c)

    10 N

  4. DOption (d)

    20 N

View answer & explanation

Correct answer

Answer: (a)A.

0 N

Why this answer is correct

Concept:

The figure shows the simply supported beam with the point loads.

Total vertical force:

∑Fy = RA + RB = 20

Calculation:

From figure,

RA = RB = 10 N and C is the midpoint of the beam AB.

From SFD, at point "C", the magnitude of Shear force is zero.

Question 48

Calculate the bending moment at the mid-point of a 6 m long simply supported beam carrying a 20 N point load at the mid-point.

  1. AOption (a)

    20 Nm

  2. BOption (b)

    30 Nm

  3. COption (c)

    45 Nm

  4. DOption (d)

    60 Nm

View answer & explanation

Correct answer

Answer: (b)B.

30 Nm

Why this answer is correct

Concept:

Bending moment:

A bending moment is the reaction induced in a structural element when an external force or moment is applied to the element causing the element to bend. Its SI unit is Nm

BM = F × L

Calculation:

Given

W(load) = 20 N, Total length = 6 m

So, RA + RB = 20 N,  RA = RB = 10 N (due to symmetrical loading)

BM=RA×(L2)BM=R_A × (\frac L2) = 10 × 3 = 30 Nm

Question 49

The stress-strain plot for ductile materials exhibits peak at ultimate strength

  1. AOption (a)

    because necking begins to occur, whereby engineering stress becomes less than the true stress

  2. BOption (b)

    because the material starts becoming weakening at the microstructural level

  3. COption (c)

    due to strain softening of the material

  4. DOption (d)

    none of the above

View answer & explanation

Correct answer

Answer: (a)A.

because necking begins to occur, whereby engineering stress becomes less than the true stress

Why this answer is correct

Concept:

  • The stress-strain curve for ductile material is shown below

  • So it is evident from the graph that the strain is proportional to stress or elongation is proportional to the load giving a straight line relationship. This law of proportionality is valid up to a point A. Point A is known as the limit of proportionality or the proportionality limit.

  • For a short period beyond point A, the material may still be elastic in the sense that the deformations are completely recovered when the load is removed. The limiting point B is termed as Elastic Limit.

  • Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus permanent deformation or permanent set when the load is removed. These two points are termed as upper and lower yield points respectively. The stress at the yield point is called the yield strength.

  • A further increase in the load will cause marked deformation in the whole volume of the metal. The maximum load which the specimen can withstand without failure is called the load at the ultimate strength. The highest point ‘E' of the diagram corresponds to the ultimate strength of a material.

  • Beyond point E, the bar begins to form the neck. The load falling from the maximum until a fracture occurs at F.

Question 50

For a round shaft subjected to pure torsion, the shear stress at the centre (axis) will be

  1. AOption (a)

    maximum 

  2. BOption (b)

    minimum

  3. COption (c)

    zero

  4. DOption (d)

    the information provided is insufficient

View answer & explanation

Correct answer

Answer: (c)C.

zero

Why this answer is correct

Concept:

The torsion equation for a circular member is

τr=Tj=C;θl\frac{{\rm{τ }}}{{\rm{r}}} = \frac{{\rm{T}}}{{\rm{j}}} = \frac{{{\rm{C;\theta }}}}{{\rm{l}}}

where τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress), r = Radius of the shaft, T = Twisting Moment or Torque J = Polar moment of inertia, C = Modulus of rigidity for the shaft material, l = Length of the shaft, θ = Angle of twist in radians on a length “l”

From the torsion equation,

τ=TJ×r;τ = \frac{T}{J} \times r;

At the center, the value of the radius is zero.

Therefore, τ = 0

The intensity of shear stress at any point in the cross-section of the shaft subjected to pure torsion is directly proportional to its distance from the center.

 

  • Torsional stress: When a machine member is under the twisting force then it is said to be the shaft is subjected to torsion. Due to this torsion in the shaft, the stresses induced in the shaft are known as torsional stress.
  • Twisting can be produced in the shaft when two equal and opposite couples acting in parallel planes.

​

50 questions in the full paper