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  2. All Previous Year Questions List
S.No Question Option A Option B Option C Option D Answer Solution Comments Status Action
1 The point of contra flexure occurs in Cantilever beams Simply supported beams Overhanging beams Fixed beams c Point of contraflexure occur in overhanging beams at this point
Bending moment = 0		 Comments Active
2 TQM is related to Quality control Control chart Sampling Work study a TQM – Total quality management
* In this top management involves directly.
* Motto – longterm success through customer satisfaction.
* All member participates to improve processes, products, services, culture		 Comments Active
3 C-chart is based on one of the following Number of defects per unit of a product Fraction defectives in the sample Number of defectives in the sample None of the above a C – chart – no of defect per unit of a product
P – Chart – No of defectives in a sample		 Comments Active
4 There are ‘m’ rows and ‘n’ columns in a transportation problem. Degeneracy will occur if the number of allocations is Less than (m + n -1) Greater than (m + n -1) Equal to (m + n -1) Less than (m - n -1) a No. of Allocation < (m + n – 1) Comments Active
5 In sampling, AQL stands for Average quality level Acceptable quality level Asymmetric quality level Available quality level b AQL – Acceptable Quality level
(tell about now many defective components are considered acceptable during random sampling quality inspection)		 Comments Active
6 Which one of the following shows the percentage of the area in normal distribution curve for ± limits? \(2σ\) 99.73 % 95.45 % 68.26 % None of the above b \( ±1σ area=A+B\)
\(=34+34=68%\)
\(±2σ area=(A+B)+(C+D)\)
\(=68+(13.5+13.5)\)
\(=68+27=95%\)		 Comments Active
7 Term “Value” in value engineering refers to Total cost of the product Selling price of the product Utility of the product Manufacturing cost of the product c the term 'value' in value engineering refers to the utility of a product, service or system. Comments Active
8 Manufacturer’s risk is the probability of Rejecting a good lot which otherwise would have been accepted Defective batch being accepted which otherwise would have been rejected Bad components in a lot None of the above a Producer’s Risk: Probability of rejecting a good lot which otherwise would have been accepted
Consumer risk: Probability of defective lots being accepted. which otherwise would have been rejected.		 Comments Active
9 When ordering cost is increased to 16 times, the EOQ will be increased to 2 times 4 times 8 times None of the above b EOQ \(∝ordering cost\)
\(∝Co\)
\(\frac{EOQ_{1}}{EOQ_{2}}=\frac{Co}{16CO}=\frac{1}{4}\)
\(EOQ_{2}=4EOQ_{1} \)		 Comments Active
10 A production line is said to be balanced, if at each station There is equal number of machine There is equal number of operators Waiting time for service is same Operation time is same d Balance production line consists
* Operation time will be equal for each product.
* Equal no of worker on each assembly line.
* Equal no. of machines is allotted to each assembly line.		 Comments Active
11 Experts of same rank assemble for product development in Delphi technique Brain storming Direct expert comparison Morphological analysis b Brain stroming (Value analysis) Comments Active
12 Whirling speed of a shaft coincides with the natural frequency of its Longitudinal vibration Transverse vibration Torsional vibration Coupled bending torsional vibration b whirling of the shaft occur when the natural frequency of transvers vibration matches the frequency of a rotating shaft. Comments Active
13 The critical speed of a shaft is affected by its
1. Eccentricity
2. Span
3. Diameter
Which of the above are correct?		 1 and 2 1 and 3 2 and 3 1, 2 and 3 d Critical speed of shaft
\(ω_{n}=\frac{K}{m}=\frac{g}{δ} \)
\(K=Stiffness\)
\(m=mass\)
\(deflection (δ)=\frac{e}{(\frac{ω_{n}}{ω})^{2}}-1\)
\(y=deflection\)
\(e=eccentricity\)
\(ω=angular speed of shaft when shaft is loaded \)
\(centrally it acts like simply supported beam.\)
whiring of the shaft occur when the natural frequency of transvers vibration matches the frequency of a rotating shaft. \(∵ω_{n}=ω=\)
\(=\frac{K}{m}=\frac{g}{δ}\)
\(ω_{n}α1/δ\)
for simply supported beam. \(δ\)
\(δ=\frac{PL^{3}}{48EI}=Transverse deflection\)
depends on L \(ω_{n}\)
depends on \(-ω_{n}\) \(1=\frac{πd^{4}}{64}\)		 Comments Active
14 Standardization deals with the characteristics of product that include Its dimensions Method of testing the product Composition and properties of its material All of the above d Standardization provides inter changeability and availability. Comments Active
15 In a single point turning operation Taylor’s exponent is 0.25. If the cutting speed is halved then the tool life will become Half Two times Eight times Sixteen times d \(\) \(v_{1}=v_{1}, v_{2}=v/2\)
\(T_{1}=T, T_{2}=?, n=\frac{1}{4}\)
\(v_{1}T1n=v_{2}T2n⇒(\frac{T_{2}}{T_{1}})^{n}=\frac{v_{1}}{v_{2}}\)
\(=(\frac{T_{2}}{T_{1}})^{1/4}=(\frac{\frac{v}{v}}{2})=2^{4}=16\)
\(T_{2}=16T\)		 Comments Active
16 In an orthogonal cutting operation, the chip thickness and the uncut thickness are equal 0.45mm each. If the tool rake angle is 0°, the shear plane angle is 18° 30° 45° 60° c Chip thickness ratio
\(r_{c}=\frac{a_{1}}{a_{2}}=\frac{0.45}{0.45}=1\)
Shear angle
\(tan ∅=\frac{r_{c}cos α}{1-r_{c} sinα}\)
\(=\frac{1 cos 0}{1-1 sin 0}=\frac{1}{1}=1\)
Tan 45°
\(∅=45°\)		 Comments Active
17 Metal in electro-chemical machining process is removed by Migration of ions towards the tool Ionization and shearing Chemical action and abrasion Chemical etching a Machining Principle
ECM
Ion – displacement
AWJM, USM, AJM
Erosion
EDM, EBM, LBM, IBM
Fussion
CHM
Corrosion		 Comments Active
18 Lee and Shaffer equation showing relationship between rake angle (), shear angle () and friction angle () is expressed as \(α\) \(β\) \(θ\) \(α=\frac{π}{4}-Ø+θ\) \(∅=\frac{π}{4}+α-θ\) \( ∅=\frac{π}{4}-α+θ\) \( θ=\frac{π}{4}+∅-α\) b \( ∅=shear angle\)
\(α=rake angle\)
\(θ=Friction angle\)
\(∅=\frac{π}{4}+α-θ\)
\(∅+θ-α=π/4\)		 Comments Active
19 For the two shafts connected in parallel, which of the following in each shaft is same? Torque Shear stress Angle of twist Torsional stiffness c Shaft in parallel
\(-T=T_{1}+T_{2}\)
\(-Q_{1}=Q_{2}\)
\(\frac{T_{1}l_{1}}{J_{1}G}=\frac{T_{2}l_{2}}{I_{2}G}\)		 Comments Active
20 Use of jigs and fixtures leads to High operational cost High maintenance cost High Initial cost High manufacturing cost c Initial cost > Operational cost
After using jigs and fixture.
So manufactured cost also increases		 Comments Active
21 The tool life of a cutting tool mainly depends on Cutting speed Tool geometry Ambient temperature None of the above a According to modified Taylor’s tool life equation as.
\(T∝\frac{1}{VCa.f^{b}.d^{c}}\)
\(V_{c}=Cutting velocity\)
\(f=feed\)
\(d=depth of cut\)
\(a>b>c\)
Mainly \(T∝\frac{1}{v_{c}}\)		 Comments Active
22 The quality of machined surface depends on The material of the work piece Rigidity of machine work-tool system Cutting conditions All of the above d Quality of machined surface mainly depends on
- Feed
- Nose radius of cutting tool except these it also depends on
- Material to be machined
- Rigidity of machine
- Cutting velocity, depth of cut, coolant etc.		 Comments Active
23 Which of the following operations does not use a jig? Turning Drilling Reaming Tapping a Jig are specially used for drilling, taping, boaring, reaming etc.
In turning we need fixture to locate the work piece		 Comments Active
24 Which one of the following type of layout is used for the manufacturing of large aircrafts? Product layout Process layout Fixed position layout Combination layout c In fixed layout, facilities (man machine and material) are taken to product.
e.g: Racket, Rail coach, aircraft, ship etc.
Product layout: * Suitable for similar product.
* Similar operation are carried out at a single spot.		 Comments Active
25 In simplex method of linear programming the objective row of the matrix consists of Names of the variables Coefficient of the objective function Slack variables None of the above c Slack Variable: Variable which is added to inequality to make it equality. Comments Active
26 In Electro-chemical machining material removal is due to Corrosion Erosion Fusion Ion displacement d Machining Principle
ECM
Ion – displacement
AWJM, USM, AJM
Erosion
EDM, EBM, LBM, IBM
Fussion
CHM
Corrosion		 Comments Active
27 Which technique is utilized to find percent idle time for man or machine? Work sampling Time study Method study ABC analysis a Work sampling: To observe the workers and record the time of activity of them and extract the idle time.
ABC analysis: Method to categories the inventory according to values and usages.
Method study: Structured process of directly observing and measuring human work using time device.		 Comments Active
28 Which of the following is not the assumption in Merchant’s theory Tool is perfectly sharp Shear is occurring on a plane Uncut chip thickness is constant A continuous chip with built up edge (BUE) is produced d Assumption of merchant’s circle
- Sharp tool
- Orthogonal cutting
- Widthtool > Ww/P
- Shear occur along a plane
- Continous chip without built up edge
- Uncut chip thickness is constant
- Uniform cutting velocity
- Constt depth of cut		 Comments Active
29 The rate of work material removal in USM operation is proportional to the Volume of work material removed per impact Number of particles making impact per cycle Frequency of vibration All of the above d In USM
MRR∝ Volume of work material removal per impact
∝ Number of particles making impact per cycle.
∝ Frequency of vibration 20 KHz
∝ Amplitude (15 -20 μm)		 Comments Active
30 Which of the following is not a limitation for ECM process Very expensive Sharp corners are difficult to produce Surface finish is not good Use of corrosive media as electrolyte makes it difficult to handle c ECM provides excellent surface finish. Comments Active
31 For a two dimensional stress system, the coordinates of the centre of Mohr circles are? , 0 \(\frac{σ_{ x - }σ_{y}}{2}\) 0, \(\frac{σ_{x + σ_{y}}}{2}\) , 0 \(\frac{σ_{x + σ_{y}}}{2}\) 0, \(\frac{σ_{x - σ_{y}}}{2}\) c Co – ordinate of centre
\((\frac{σ_{x}+σ_{y}}{2}, 0)\)		 Comments Active
32 In machining processes, the percentage of heat generated in shear action is carried away by the chips to the extent of 10 % 25 % 50 % 90 % d Heat zone during machining
Max heat production occur in sear zone (near shear plane) around 80% of the total heat production, which is carried away by chip (mostly).		 Comments Active
33 Pneumatic comparators work on following theory Newton’s theory Bernouli’s theory Pascal’s theory Legendre’s theory c Pascal’s theory – In a fluid at rest in a closed container a pressure change in one part is transmitted without loss to every portion of the fluids and to the walls of container. Comments Active
34 Inter electrode gap in electro-chemical grinding is controlled by controlling the Pressure of electrolyte flow Applied static load Size of abrasives in the wheel Texture of the work piece c Inter electrode gap in ECG is controlled by controlling the size of diamond particles in the wheel. Comments Active
35 If the diameter of the hole is subjected to considerable variation, for locating in jigs and fixtures, the pressure type locator used is Conical locator – Adjustable type – For rough and uneven surface (casting) Spring loaded type– For unmachined surface. Equilizing support – For taper surface.
* Pin type locator – For small holes
* Conical locator – For shouldered and varying dia hole.
* Diamond pin locator – For round shape
* Cylindrical locator – For large holes
* Profile locator – For external unusual profile
* V – locator – For semi – circular or circular profile		 a Types of Locator
* Locating buttons: to locate flat surface
* locating pins: (a) Fixed type – For machined surface		 Comments Active
36 Strain in direction at right angle to the direction of applied force is known as Lateral strain Shear strain Volumetric strain None of the above a Lateral strain: poisson’s longitudinal strain ratio. Comments Active
37 The most suitable manufacturing process for machining a turbine blade made of nimonic alloy is Milling and lapping Electric discharge machining Ultrasonic machining Electro-chemical machining d ECM – By ECM, turbine blades of nimonic alloy is made. By investment casting turbine blade of Jet engine of air craft or of Gas turbine of titanium.
Nimonic alloy – Law creep super alloy.
50% Ni
18 – 21% Cr
15 – 21% Co
2 – 3% Ti
1 – 2% Al		 Comments Active
38 Which one of the following cannot be recycled? Thermoplastics Thermosets Elastomers Polymers b Thermosetting plastic can not be recycled. Comments Active
39 Addition of magnesium to cast iron increases its Hardness Corrosion resistance Creep resistance Ductility d Ductility – Mg is an innoculent, when added in CI (min 0.05%), it makes spheroidal form of graphite which improves tensile strength and ductility. Comments Active
40 Which one of the following is closest to the purest form of iron? Cast iron Wrought iron Grey cast iron Mild steel b Wrought iron – 99.99% Pure iron
Cast iron – C% (2% to 4.3%)
Mild steel – C% (upto 0.3%)		 Comments Active
41 Which one of the following is a point imperfection? Vacancy Frenkel defect Schottky defect All of the above d Point Imperfection
* Vacancy – Missing atom
* Substitution impurity – atom replacement
* Interstitial impurity – Foreign atoms sets on voids
* Self – interstitial impurity – misplace of same atom
* Frenkel – one kind of ion jumps from normal to interstitial position
* Schattley – Pair of ions missing
Line imperfection
* Edge dislocation
* Screw dislocation
Surface imperfection
* Grain boundary
* Stacking fault
* Twin boundary
Bulk imperfection
* Cracks
* Voids
* Pores
* Impurities		 Comments Active
42 In rolling process, the state of stress of the material undergoing deformation is Pure compression Pure shear Compression and shear Tension and shear c In rolling compression and shear both occur simultaneously.
In wire drawing: Tension.
In extrusion: Compression		 Comments Active
43 Toughness of steel is increased by adding Nickel Sulphur Chromium Tungsten a Alloying element
Effect
Ni
Increases tensile strength, ductility toughness
S
Improve machining properties
Cr
Improves corrosion resistance
W
Improves red hardness and wear resistance		 Comments Active
44 The temperature at which new stress free grains are formed in the metal is called Critical temperature Eutectic temperature Recrystallization temperature Yield temperature c Each material has a melting point. Let us suppose it is . \(T_{m}\)
Generally recrystallization temp is consider as to if working Temp is and \(\frac{1}{3} T_{m} \) \(\frac{3}{4} T_{m}\) \(T_{w}\)
(Cold working) \(T_{w}<\frac{T_{m}}{3}\)
\(\frac{T_{m}}{3}(Hot working) \(\frac{3}{4} T_{m}The temperature at which crystal structure take new from by destroying the older one is known as recrystallization temperature.		 Comments Active
45 Which one of the following materials is most elastic? Rubber Steel Aluminium Glass b Most elastic means which have lengthiest linear part where stress is directly proportional to strain. Comments Active
46 Cold working is the process of deforming a metal plastically At recrystallization temperature Below recrystallization temperature Above recrystallization temperature At annealing temperature b Each material has a melting point. Let us suppose it is . \(T_{m}\)
Generally recrystallization temp is consider as to if working Temp is and \(\frac{1}{3} T_{m} \) \(\frac{3}{4} T_{m}\) \(T_{w}\)
(Cold working) \(T_{w}<\frac{T_{m}}{3}\)
\(\frac{T_{m}}{3}(Hot working) \(\frac{3}{4} T_{m}The temperature at which crystal structure take new from by destroying the older one is known as recrystallization temperature.		 Comments Active
47 If there are bad effects on strain hardening on cold formed parts, the part must be Annealed Tampered Hardened Normalised a When cold working is done on a part, due to strain hardening internal stresses and brittleness increases and to remove these effects annealing is done. Comments Active
48 Principal stress at a point in a plane stressed element are: = = 500 Normal stress on the plane inclined at 45° to x-axis will be \(σ_{x }\) \(σ_{y}\) \(\frac{N}{m^{2}}\) 0 500 N/ \(m^{2}\) 707 N/ \(m^{2}\) 1000 N/ \(m^{2}\) b \( σ_{x}=σ_{y}=500 mPa\)
\(θ=45°\)
\(=σ_{n}=\frac{σ_{x}+σ_{y}}{2}+\frac{σ_{x}-σ_{y}}{2} cos 2 θ+τ_{xy} sin 2θ\)
\(=\frac{500+500}{2}+\frac{500-500}{2} cos 90°\)
\(σ_{n}=500 MPa\)		 Comments Active
49 Carbon content is highest in Mild steel Eutectoid steels Hypoeutectoid steels Hypereutectoid steels d Mild steel – Upto 0.25%
Eutectoid steel – 0.76%
Hypoeutectoid steel - < 0.76%
Hyper eutectoid steel - > 0.76% to <2.14%		 Comments Active
50 Important property requirements for tool materials employed for high speed machining are Impact strength, melting point and hardness Hot hardness, wear resistance and toughness Melting point, toughness and shear strength Shear strength, wear resistance and impact strength b Most important property of cutting tool
- Hot hardness (should be higher)
- Wear Resistance (Should be higher)
- Toughness (Should be higher)		 Comments Active
51 18/8 stainless steel contains 18% vanadium, 8% chromium 18% chromium, 8% nickel 18% tungsten, 8% nickel 18% tungsten, 8% chromium b 18/8 stainless steel (Austenitic steel)
18% - Chromium
8% - Nickel
< 0.8% - Carbon
50% - Iron \(≥\)		 Comments Active
52 The crystal structure of alpha iron is Body centered cubic Face centered cubic Hexagonal closed pack Simple cubic a There are 3 variants of steel
- Ferrite, α – iron – BCC
- Austenite, ν – iron – FCC
- \(δ-iron-BCC\)		 Comments Active
53 Which of the following statements is not true for austenitic stainless steels? They are hardened and strengthened by cold working They are most corrosion resistant amongst stainless steels Austenitic phase is extended to room temperature They are magnetic in nature d * Austenitic stainless steels cannot be hardened via heat treatment but can be hardened by cold working.
* Stainless steel are austenitic, so are corrosion resistant (mostly).
* In the pressure of Ni or Mn, Austenitic Steel can be occur at room temperature.
* But austenitic steel is non magnetic		 Comments Active
54 Slow plastic deformation in metals under a static load over a period of time is Fatigue Endurance Creep Dislocation c Time dependent deformation occur under static load is called creep. Under high temperature it accelerates. Comments Active
55 Austenite decomposes into ferrite and cementite at a temperature of 727 °C 1148 °C 1495 °C 1539 °C a When austenite is cooled below , eutectoid reaction occurs. \(727℃\)
\(Austenite-cool below (727℃, 0.8%)\)
\(Ferrite+Cementite=Pearlite\)		 Comments Active
56 Increase in ferrite phase in steel leads to increase in Strength Hardness Ductility Brittleness c Ferrite phase (It is considered as soft and ductile phase)
When it is added to steel, it increases ductility.		 Comments Active
57 Which one of the following is weaker than hydrogen bonds? Ionic bond Vander Waals bond Covalent bond Metallic bond b Strong bonds
Weak bond
Ionic (Nacl)
H – bonds
Metallic
Vander walls bonds (weakest)
Covalent (strongest) (H2O, CH4)		 Comments Active
58 Property of absorbing large amount of energy before fracture is known as Ductility Toughness Elasticity Hardness b \( Toughness= 0∈σdϵ\)
Area under the curve upto fracture point is known as toughness i.e. total energy absorbed upto fracture point under fracture load.		 Comments Active
59 Maximum principal strain theory of failure gives satisfactory result for Brittle materials only Brittle as well as ductile materials Ductile materials only None of the above a Max principal stress and max principal strain theory of failure give good result for brittle material. Comments Active
60 Eutectoid steel consists of Fully pearlite Fully Austenite Ferrite + Pearlite Cementite + Pearlite a Eutectoid – Fully pearlite \(α-iron(87%, white)+cementite (13%, black)\) Comments Active
61 Compressive test performed on cast iron will have fracture occurring Along an oblique plane Along the axis of load Perpendicular to the axis of load None of the above a Ductile material failure
Brittle material failure		 Comments Active
62 Which medium is used for fastest cooling during quenching of steel? Air Oil Water Brine (salt water) d Order according to cooling rate
\(Brine>Water>Oil>air>Furnace\)
Brine & Water = Martensite
Oil = Very fine pearlite
Air = Fine pearlite
Furnace = Coarse pearlite
\(Br.>Wa.>Oil>air>furn.\)		 Comments Active
63 In tensile test of mild steel, necking will start At lower yield stress At upper yield stress At ultimate tensile stress Just before fracture c Necking occur at ultimate point. Comments Active
64 Crystal lattice structure for mild steel is Single cubic BCC FCC HCP b Mild steel is ferrite or have BCC structure. \(α-iron\) Comments Active
65 If a material expands freely due to heating, it will develop Thermal stresses Tensile stresses Compressive stresses No stresses d During temperature increment if components are not constrained, there will be no thermal stress generation. Comments Active
66 Which one of the following has the highest value of specific stiffness? Steel Aluminium Fibre glass Carbon fibre composite d \( Specific stiffness= \frac{Young^{'}s Modulus }{Density (J/kg)}\) Comments Active
67 Which of the following statements is not true for diamond? It is hardest known material Diamond is non-metallic It has high thermal conductivity It has a very high electrical conductivity d Due to C – atom positioning in crystal, it have a unique structure called diamond cubic and have following properties like.
- Highest hardness
- Highest thermal conductivity
It is non – metallic material.		 Comments Active
68 Which statement is not true in case of martensite? Crystal structure is BCC Transformation does not involve diffusion Grains are plate like or needle like in appearance It is a non-equilibrium phase a Martensite have body centred tetragonal structure not cubic
- It is not shown on equilibrium phase diagram because it is an non – equilibrium phase.
- Martensite grain structure appears plate or needle like structure.
- Diffusion does not occur during martensite formation.		 Comments Active
69 Relationship between atomic radius ‘R’ and unit cell length ‘a’ for BCC crystal structure is a = \(\frac{4R}{3}\) a = 2R \(2\) a = \(\frac{2R}{3}\) a = 3R \(2\) a For BCC, relationship b/w a and R
- Choose the plane which have maximum density i.e. diagonal plane (110)
\(CD=CE^{2}+ED^{2}\)
\(CD=a^{2}+a^{2}\)
\(CD=a2\)
Now \(AD=R+2R+R\)
\(=a^{2}+2a^{2}=4R\)
\(4R=a3\)
\(a=\frac{4R}{3}\)		 Comments Active
70 Atomic packing factor for unit cell of HCP crystal structure is 0.68 0.52 0.74 0.82 c Atomic packing factor
\(APF=\frac{Volume of effective no of atoms in a unit cell}{Volume of unit cell}\)
Structure
APE
SC
0.52
BCC
0.68
FCC
0.74
DC
0.34
HCP
0.74		 Comments Active
71 Coordination number for FCC crystal structure is 4 6 No
Structure
6
Simple cubic
8
Body centred cubic
12
Face centred cubic		 12 d Co – ordination No – No of neighboring atom from a given atom at equal distance. Comments Active
72 In the graphical method of linear programming problem the optimum solution would lie in the feasible polygon at Its one corner Its center The middle of any side None of the above a Each point with in the region is feasible solution. But optimum solution will be always lie on one of the corner point because it uses all the resources upto an optimum level. Comments Active
73 A relatively large plate of glass is subjected to a tensile stress of 40 MP(a) If specific surface energy and Elastic modulus for glass are 0.3 J/m2 and 69 GPa, respectively, the maximum length of a surface crack that is possible without fracture is(a 4 µ m 8.2 µ m 41 µ m 82 µ m b \( Stress σ=40MPa\)
\(Sp. Energy=0.3 J/m^{2}=Ï‘\)
\(E=69 GPa\)
Griffet criterion for Brittle fracture
\(Critical Stress (σ_{c})\)
\(σ_{c}=\frac{2Eϑ}{πa} (Memorize it directly)\)
\(a=\frac{1}{2} of length of an crack fracture depends on stiffness\)
When energy decreases (Mechanical energy or potential energy) due to
- Strain energy
- Work done
Then a crack propagates or vise versa As the crack appears a new surface is introduced and then surface energy is release(d)
Now \(a=\frac{2×69×10^{9}×0.3}{π×(40×10^{6})^{2}}\)
\(=8.2×10^{-6}m=8.2μm\)		 Comments Active
74 Which one of the following is electrically most conductive? Copper Silver Aluminium Gold b Electrical conductivity comparison
\(Silver>Copper>Gold>Al\)		 Comments Active
75 Which one of the following correctively expresses the sensitivity of a governor? \( \frac{N_{1 + N_{2}}}{2N_{1 N_{2}}}\) \( (N_{1}-N_{2})/(N_{1}+N_{2})\) \( 2(N_{1}+N_{2})/(N_{1}-N_{2})\) \(\frac{2N_{1}N_{2}}{N_{1}+N_{2}}\) b \(Senstivity=\frac{Range of speed}{Mean speed}\)
\(N_{1}-max speed\)
\(N_{2}-Min speed\)
\(=\frac{N_{1}-N_{2}}{\frac{N_{1}+N_{2}}{2}}=\frac{2(N_{1}-N_{2})}{N_{1}+N_{2}}\)		 Comments Active
76 In a spring mass system if one spring of same stiffness is added in series, new frequency of vibration will be \(\frac{ω_{n}}{2}\) \(ω_{n}\) \(2 ω_{n}\) \(\frac{2}{ω_{n}}\) a Natural frequency
\(ω_{n}=\frac{K}{m}\)
\(K_{eq}=\frac{K_{1}K_{2}}{K_{1}+K_{2}}\)
\(=\frac{K.K}{K+K}\)
\(=\frac{K^{2}}{2K}\)
\(=\frac{K}{2}\)
\(ωn'=\frac{K_{eq}}{m}=\frac{K}{2m}=\frac{ω_{n}}{2}\)		 Comments Active
77 During the dwell period of the cam, the follower Remains at rest Moves in a straight line Moves with uniform speed Does simple harmonic motion a During the dwell period of the cam, the follower Remains at rest. Comments Active
78 For a 20° full depth involute gear teeth system, minimum number of teeth on a pinion is 12 14 16 18 d System of Gear teeth
Minimum no. of teeth on the pinion
\(14\frac{1}{2}^{°} full depth\)
32
20° full depth
18
20° Stabbed
14
oposite \(14\frac{1}{2}^{°} \)
12
Minimum no. of teeth on pinion
\(Z_{min}=\frac{2a}{1+\frac{1}{G}(\frac{1}{G}+2)sin^{2}∅-1}\)
\(a_{w}=addendum coefficient\)
\(G=Gear ratio\)
\(∅=Pressure angle\)
Or \(Z_{min}=\frac{2}{sin^{2}α}\)		 Comments Active
79 For a speed reduction of 50 : 1, which gear arrangement will be used? Spur gears Bevel gears Worm and worm wheel Herringbone gears c Gear
Speed reduction
Spur gear
Up to 5:1
Bevel gear
3:1 to 1:1
Worm and worm wheel
Upto 60:1
Helical gear
Upto 6:1		 Comments Active
80 The mathematical technique for finding the best use of limited resources in an optimum manner is called Linear programming Network analysis Queuing theory None of the above a Linear programming Comments Active
81 Primary unbalanced force due to inertia of reciprocating parts in a reciprocating engine is given by mr \(ω^{2}sinθ\) mr cos \(ω^{2}\) \(θ\) mr ( ) \(ω^{2}\) \(\frac{sin2θ}{n}\) m r () \(ω^{2}\) \(\frac{cos2θ}{n}\) b Inertial force on piston
\(F_{b}=mrω^{2}cosθ+mrω^{2}\frac{cos2θ}{n} \)
\(F_{b}=mrω^{2}cosθ=(primary unbalanced force) \)
\(mrω^{2}\frac{cos2θ}{n}= (Secondary unbalanced force)\)		 Comments Active
82 Magnification factor for a single degree of freedom vibration is expressed by = \(\frac{X}{X_{ST}}\) \(\frac{1}{(1-r^{2})^{2}+(2Ïšr)^{2}}\) = \(\frac{X}{X_{ST}}\) \(\frac{1}{(1+r)^{2}+(2Ïšr)^{2}}\) = \(\frac{X}{X_{ST}}\) \(\frac{1}{1-r^{2}}\) = \(\frac{X_{ST}}{X}\) \(\frac{1}{(1-r)^{2}-(2Ïšr)^{2}}\) a Magnification factor
\(M.F.=\frac{1}{(1-r^{2})^{2}+(2r)^{2}}\)
\(=\frac{Amplitude in forced motion}{Static deflection}\)
\(r=\frac{ω}{ω_{n}}\)
\(=damping coefficient option have\frac{X_{5}T}{X}.\)		 Comments Active
83 If is the actual coefficient of friction in a belt moving in grooved pulley and groove angle is .The virtual coefficient of friction will be \('μ^{'}\) \(2α\) \(μ/sinα\) \(\frac{µ}{COS α}\) \(μsinα\) \(μcosα\) a Virtual coefficient of friction
\(μ_{v}=\frac{μ}{sinα}\)
[Memorize directly]		 Comments Active
84 Which one of the following does not require a flywheel? Steam engine Engine driven press CI engine Gas turbine d Flywheel is used to store energy and releases when it is required Energy storage is necessary in case of all I.C engine, steam engine, punching press, shearing machine, riveting machine crushes etc
In these machine operations are intermittent.		 Comments Active
85 Which type of gears are used in connecting two coplanar and intersecting shafts? Spur gear Bevel gear Helical gear Worm and worm wheel b Spur gears
To connect 2 parallel shafts
Helical gears
To connect 2 parallel shafts
Worm & worm wheel
To connect 2 non – intersecting non – parallel shafts
Bevel gear
To connect 2 intersecting shafts		 Comments Active
86 Velocity of the belt for maximum power transmission by the belt and pulley arrangement is \(\frac{T_{max}}{3m}\) \(\frac{T_{max}}{4m}\) \(\frac{T_{max}}{5m}\) \(\frac{T_{max}}{m}\) a Condition for maximum power maximum tension in belt.
\(T_{max}=3 T_{c} (T_{c}=Centrifugal Tension)\)
Or linear velocity of belt
\(v=\frac{T_{max}}{3m}\)
\(m=mass of belt per unit length.\)		 Comments Active
87 Sensitivity of an isochronous governor is Zero One Two Infinity d \( Sensitiveness=\frac{Range of speee}{Mean speed}\)
For isochronous governor
Range of speed = 0
\(Sensitivity ∝\frac{1}{Sensitiveness}\)
\(∴Senstivity for isochroneous governor= ∞\)		 Comments Active
88 Porter governor is a Pendulum type governor Dead weight type governor Spring loaded governor Inertia type governor b Porter governor is dead weight loaded type of gravity controlled centrifugal governor.  Comments Active
89 If there is a gradual reduction in amplitude of vibration with time, the body is said to be in Free vibration Forced vibration Damped vibration Undamped vibration c Free Vibration: External force is used to excite initially the system and then is remove(d)
Forced Vibration: Body vibrated under influence of external force.
Periodically occurred vibration
Damped vibration: When there is reduction in amplitude over cycle. Vibration due to energy dissipation to overcome the friction.
Undamped vibration (Hypothetical): Vibration without friction.		 Comments Active
90 Which one of the following is the preferred mode of transmission of power from one shaft to another when distance between the shafts is relatively small Gears Belts Ropes Chains a Distance of Power transmission
Power Drive
> 15m
Rope drive
10 m to 15 m
Belt drive
2 m to 5 m
Chain drive
< 1 m
Gear drive		 Comments Active
91 If ratio of excitation and natural frequency of vibration ; the transmissibility of vibration will be \(\frac{ω}{ω_{n}}=2\) 0.5 1.0 1.5 2.0 b Transmissibity
\(β=(\frac{1}{1-(\frac{ω}{ω_{n}})^{2}})=(\frac{1}{1-(2)^{2}})\)
\(β=(\frac{1}{1-2})=(-1)=1\)		 Comments Active
92 A framed structure is said to be perfect if the following correlation is met between the number of joints ‘j’ and the number of the members ‘m’ m = 2j-3 m = 3j-3 m = 2j-1 m = j-2 a If n = no. of member
J = no of joints
If
\(n>2J-3 (Redundant truss)\)
\(n=2J-3 (Perfect truss)\)
\(n<2J-3(Deficient truss)\)		 Comments Active
93 Dimensional formula represents \(ML^{2}T^{-3}\) Work Force Momentum Power d \( ML^{2}T^{-3}=kg-m^{2}-s^{-3}\)
\(=\frac{Kg-m^{2}}{s^{3}}=\frac{Kg-m}{s^{2}}.\frac{m}{s}\)
\(=N.\frac{m}{sec}\)
\(Power=Force (N)×Velocity (m/sec)\)
\(=\frac{N-m}{sec}=\frac{J}{sec}=Watt\)		 Comments Active
94 A spring mass system shown in Figure is actuated by a load P = 0.75 sin2t. If mass of the block is 0.25 kg and stiffness of the spring is 4 N/m, displacement of the block will be P = 0.75 Sin 2t
 0.25 0.5 1.0 2.25 a \( =0.75sin2t, ω_{f}=\frac{2rad}{sec}, \)
\(x_{t}=(\frac{P_{o}}{k-mωf2})=(\frac{0.75}{4-0.25×2×2})=0.25\)		 Comments Active
95 Lewis equation in gears is used to find the Bending stress Tensile stress Centrifugal stress Fatigue stress a Lewis equation is used to calculate maximum load on tooth which it can bear without bending or failure.
\((P_{t}=m.(b)y.σ_{b})_{Permissible}\)
\(m-module \)
\(b-Width of tooth\)
\(y-Lweis from factor= πy\)
\(σ_{b}=Allowable bending stress.\)		 Comments Active
96 The equivalent bending moment under combined action of bending moment ‘M’ and torque ‘T’ is Principle stress
\(σ_{1,2}=\frac{σ_{d}+σ_{b}}{2}±(\frac{σ_{d}+σ_{b}}{2})^{2}+ζ^{2}\)
\(if σ_{d}=0\)
\(σ_{1,2}=\frac{32M}{2×πd^{3}}±(\frac{-32M}{2πd^{3}})^{2}+(\frac{16T}{πd^{5}})^{2}\)
\(=\frac{16M}{πd^{3}}±\frac{16}{πd^{3}}((M)^{2}+T^{2})\)
\(=\frac{16}{πd^{3}}(M±M^{2}+T^{2})\)
\(=\frac{32}{πd^{3}}(\frac{1}{2}(M±M^{2}+T^{2})) (Equivalent Bending Moment)\)		 \(\frac{1}{2}M^{2}+T^{2}\) M + \( M^{2}+T^{2}\) ) \(\frac{1}{2}(M+M^{2}+T^{2}\) d \( Direct stress σ_{d}=\frac{P}{A}\)
[Bending eqn] \(Bending stress σ_{b}=\frac{M.y}{I}=\frac{32M}{πd^{3}}\)
Torsional shear stress [Torsion eqn] \(ζ=\frac{T.r}{J}=\frac{16M}{πd^{3}}\)
Combined stress		 Comments Active
97 In EDM process the tool and workpiece are separated by Electrolyte A metal conductor Dielectric fluid None of the above c Common dielectric fluid
* Hydrocarbon oil (Kerosene)
* Deionized water		 Comments Active
98 An engine running at 150 r.p.m. drives a shaft with belt arrangement. If diameter of engine pulley is 55 cm and shaft pulley 33 cm, find the speed of shaft 100 r.p.m. 150 r.p.m. 200 r.p.m. 250 r.p.m. d RPM of engine (driver)
\(N_{d}=150 rpm\)
\(Engine pulley (d_{d})=55 cm\)
\(Shaft pulley (follower)\)
\(d_{f}=33 cm\)
Now
\(Velocity ratio=\frac{Nd}{Shaft speed (N_{f})}\)
\(=\frac{d_{f}}{d_{d}}=\frac{150}{N_{f}}\)
\(=\frac{150}{N_{f}}=\frac{33}{55}\)
\(n_{f}=250 rpm\)		 Comments Active
99 A spring-controlled governor is found unstable. It may be made stable by Increasing spring stiffness Decreasing spring stiffness Increasing ball weight Decreasing ball weight b \( Stiffness ∝\frac{1}{Movment of sleeve}\)
Unstable means as speed increases, lift decreases. It occurs when higher stiffness is used.
Lower stiffness higher the stability up to a certain point.		 Comments Active
100 Centre distance between two involute teeth gears of base radii R and r and pressure angle, is expressed by \( ∅\) (R + r) Sin \(∅\) (R + r) Cos \(∅\) (R + r)/Cos \(∅\) (R + r)/Sin \(∅\) c \( P=Pitch point\)
\(∅=Pressure angle\)
\(\frac{O_{2}A}{O_{2}P}=cos∅\)
\(=O_{2}A=O_{2}P. Cos ∅\)
\(\frac{O_{1}B}{O_{1}P}=cos∅\)
\(=O_{1}B=O_{1}Pcos ∅\)
Centre distance
\(C=O_{1}P+O_{2}P\)
\(=\frac{O_{1}B}{cos∅}+\frac{O_{2}A}{cos∅}\)
\(=\frac{R}{cos∅ }+\frac{r}{cos∅}\)
\(=\frac{R+r}{cos∅}\)		 Comments Active
101 A disc clutch has discs on driving shaft and discs on driven shaft. Number of pairs of contact surfaces will be \(n_{1}\) \(n_{2} \) n1+ n2 n1+ n2+1 n1+ n2-1 n1- n2 c \( n_{1}+n_{2}-1=no of pairs of contact\) Comments Active
102 If Hartnell governor uses a spring of greater stiffness, it will become Less sensitive More sensitive Remain unaffected Isochronous a \( Sensitivity ∝\frac{1}{Stiffness}\) Comments Active
103 Which one of the following in-line engine working on a four stroke cycle is completely balanced inherently? 2 cylinder engine 3 cylinder engine 4 cylinder engine 6 cylinder engine d Six cylinder engine firing order 1 – 5 – 3 – 6 – 2 – 4 Comments Active
104 Average tensions on the tight side and slack side of a flat belt drive are 700 N and 400 N respectively. If linear velocity of the belt is 5m/s, the power transmitted will be 1.5 kW 2.5 kW 2.8 kW 3.0 kW a Power transmission
\(P=(T_{1}-T_{2})ν\)
\(T_{1}=Tension on tight side (N)\)
\(T_{2}=Tension on slack side (N)\)
\(P=(T_{1}-T_{2})v=(700-400)×5\)
\(=300×5\)
\(=1500 Watt\)
\(=1.5 kW\)		 Comments Active
105 Which pair of gears usually has higher frictional losses Spur gears Helical gears Bevel gears Worm and worm wheel d Spur, helical and Bevel gears have line contact.
Worm – have threads so surface contact hence higher friction.		 Comments Active
106 In case of a flywheel, maximum fluctuation in energy is Sum of maximum and minimum energies Difference of maximum and minimum energies Ratio of maximum and minimum energies Ratio of minimum and maximum energies b \( \frac{Iω^{2}}{2})_{max}-\frac{Iω^{2}}{2})_{min}=∆E\) Comments Active
107 In a flat belt drive, slip between the driver and belt is 1% and that between belt and follower is 3%. If the pulley diameters are same, the velocity ratio of the drive is 0.99 0.98 0.97 0.96 d In case of slip
\(Velocity ratio=1-(\frac{S_{d}+S_{f}}{100}).\frac{d_{1}}{d_{2}}\)
\(1-(\frac{1+3}{100}).\frac{d}{d}\)
\(=1-\frac{4}{100}=\frac{96}{100}=0.96\)
\(S_{d}=% slip b/w driver & belt\)
\(S_{f}=% slip follower and belt\)
\(d_{1}=diameter of driver pulley\)
\(d_{2}=diameter of follower\)
\(d_{1}=d_{2} \)		 Comments Active
108 Axes of a pair of spur gears are 200 mm apart. The gear ratio is 3:1 and number of teeth on pinion is 20. The module of the gear is 4 mm 5 mm 8 mm 10 mm b \( Gear ratio=\frac{T_{G}}{T_{P}}=\frac{3}{1}\)
No of teeth on pinion
\(T_{P}=20\)
\(Centre distance (C)=200 mm.\)
\(C=r_{G}+r_{P}=\frac{d_{G}+d_{P}}{2}\)
\(=\frac{mT_{G}+mT_{P}}{2} (m=module)\)
\(=200=\frac{m(60+20)}{2}\)
\(=200=m.\frac{80}{2}\)
\(m=5 mm\)		 Comments Active
109 The outer circle of spur gear is called as Pitch circle Addendum circle Dedendum circle Base circle b Comments Active
110 Universal joint is an example of Lower pair Higher pair Rolling pair Sliding pair a In universal joint since surface contact occurs so it is a lower pair. Comments Active
111 The gear train usually employed in clocks is a Reverted gear train Simple gear train Sun and planet gear differential gear a 1st and 4th gears are coaxial in case of reverted gear train. Comments Active
112 The shearing area of a key of length ‘l’’, breadth ‘b’ and depth ‘d’ is equal to b x d l x d l x b l x \(\frac{d}{2}\) c \( (l×b)=Shearing Area\) Comments Active
113 A block resting on an inclined plane begins to slide down the plane when the angle of inclination is gradually increased to 30°. The coefficient of friction between the block and the plane is 0.50 0.578 0.72 0.866 b \( eaction R = W cos 30°\)
\(F=Wsin 30°\)
\(μR=mgsin30°\)
\(=μ×mgcos30°=mgsin30°\)
\(=μ=tan30°=\frac{1}{3}=\frac{1}{1.732}\)
\(=0.578\)		 Comments Active
114 A bullet of 0.03 kg mass moving with a speed of 400 m/s penetrates 12cm into a block of wood Force exerted by the wood block on the bullet is 10 kN 20 kN 25 kN 30 kN b \( m=0.03 kg, u=400 m/s\)
\(u=initial velocity\)
Final velocity (v) = 0
\(v^{2}=u^{2}=2as\)
\(0-400^{2}=2a×0.12\)
\(a=-\frac{160000}{0.24}\)
Now
\(F=ma\)
\(F=0.03×(\frac{160000}{0.24})\)
\(=20000 N\)
\(=2×10^{4}N=20kN\)		 Comments Active
115 A particle is projected at such an angle with the horizontal that the maximum height attained by the particle is one-fourth of the horizontal range. The angle of projection should be 30° 45° 60° 75° b Now according to question
\(H_{max}=\frac{R}{4}\)
\(=\frac{u^{2}sin 2θ}{g}=4×(\frac{u^{2}sin^{2}θ}{2g})\)
\(=2 sin θ cos θ=\frac{4}{2} sin^{2}θ\)
\(=θ=tan^{-1}(\frac{4}{4})=tan^{-1}(1)\)
\(=45°\)		 Comments Active
116 A body moving with a velocity of 1 m/s has kinetic energy of 1.5 Joules. Mass of the body is 0.75 kg 1.5 kg 3.0 kg 30 kg c \( .E=\frac{1}{2}mv^{2}\)
\(=1.5=\frac{1}{2}m×1^{2}\)
\(=m=3.0 kg\)		 Comments Active
117 An object falls from the top of a tower. If comes down half the height in 2 seconds. Time taken by the object to reach the ground is 2.8 s 3.2 s 4.0 s 4.5 s a \( h=ut+\frac{1}{2}gt^{2}\)
Here we can see \(h∝t^{2}\)
\(\frac{h_{1}}{h_{2}}=\frac{t12}{t22}\)
\(\frac{\frac{h}{2}}{h}=\frac{2^{2}}{t22}\)
\(t_{2}=8=22=2.828 sec\)		 Comments Active
118 A simply supported beam of length ‘l’ carries a point load ‘W’ at the mid span. Deflection in beam at the centre will be \(\frac{Wl^{3}}{3EI}\) \(\frac{Wl^{3}}{8EI}\) \(\frac{Wl^{3}}{48EI}\) \(\frac{5}{384}\frac{Wl^{3}}{EI}\) c Area of triangle
\(=\frac{1}{2}×\frac{wl}{4}.l=\frac{wl^{2}}{8}\)
\(γ_{max}=\frac{AX}{EI}=\frac{\frac{1}{2}×\frac{wl}{4}×\frac{l}{2}×\frac{l}{2}×\frac{2}{3}}{EI}\)
\(γ_{max}=\frac{wl^{3}}{48EI}\)		 Comments Active
119 A material has elastic modulus of 120 GPa and shear modulus of 50 GPa Poisson’s ratio for the material is 0.1 0.2 0.3 0.33 b E = 120 GPa
G = 50 GPa
E = 2G \((1+ν)\)
\(120=2×50(1+ν)\)
\(1.2=1+ν\)
\(Poisson^{'}s ratio (ν)=0.2\)		 Comments Active
120 Elastic constants E, G and K are related by the expression E = \(\frac{GK}{2K+G}\) E = \(\frac{2GK}{2K+G}\) E = \(\frac{3GK}{K+2G}\) E = \(\frac{9GK}{3K+G}\) d E, G, K relation
\(E=\frac{gKG}{3K+G}\)		 Comments Active
121 Uniformly distributed load ‘w’ act over per unit length of a cantilever beam of 3m length. If the shear force at the midpoint of beam is 6kN, what is the value of ‘w’ 2 kN/m 3 kN/m 4 kN/m 5 kN/m a Here
\(wl=6\)
\(w×3=6\)
\(w=2 KN/m\)		 Comments Active
122 A simply supported beam of length ‘l’ has uniformly distributed load ‘w’ kilogram acting per unit length. Bending moment at mid span is wl2/8 w l2/4 wl2/2 None of the above c \( M_{c}\)
\(\frac{wl}{2}.\frac{l}{2}-\frac{wl}{2}.\frac{l}{4}\)
\(\frac{wl^{z}}{4}-\frac{wl^{2}}{8}\)
\(BM_{c}=\frac{wl^{2}}{8}\)		 Comments Active
123 Elongation of bar under its own weight as compared to that when the bar is subjected to a direct axial load equal to its own weight will be The same One fourth A half Double c Elongation
\(∆=\frac{PL}{AE}\)
P = load
A = Cross – sectional area
E = image
L = length
Elongation due to self – weight
\(∆_{S}=\frac{PL}{2AE}\)
\(∆_{s}=\frac{∆}{2}\)		 Comments Active
124 The ratio of the compressive critical load for a long column fixed at both the ends and a column with one end fixed and the other end being free is 2 : 1 4 : 1 8 : 1 16 : 1 d Euler’s load
\(Pe=\frac{Ï€^{2}EI}{Leff2}\)
Both ends are fixed the
\(Le_{ff}=\frac{L}{2}=Pe_{1}=\frac{π^{2}EI}{L^{2}}×4\)
Some end is fixed and other end is free
\(L_{eff}=2L\)
\(=Pe_{2}=\frac{Ï€^{2}EI}{4L^{2}}\)
\(\frac{Pe_{1}}{Pe_{2}}=\frac{\frac{π^{2}EI×4}{L^{2}}}{\frac{π^{2}EI}{4L^{2}}}=16:1\)		 Comments Active
125 Maximum shear stress in a Mohr’s circle is Equal to the radius of Mohr’s circle Greater than the radius of Mohr’s circle times the radius of Mohr’s circle \(2\) Could be any of the above a Shear stress will be \((ζ)\)
Maximum when
\(ζ_{max}=radius of mohr^{'}s circle\)
\(ζ_{max}=(\frac{σ_{x}-σ_{y} }{2})^{2}+ζx^{2}y\)		 Comments Active
126 According to law of transmissibility of forces, effect of force acting on the body is Different at different points of the body Minimum when it acts at centre of gravity of the body Maximum when it acts at centre of gravity of the body Same at every point in its line of action d Law of transmissibility on same line of action the effect of force will be same at any point.
Overall of motion of the block will be same in forward direction.		 Comments Active
127 The electrolyte used in ECM process is Transformer oil White spirit Aqueous solution of common salt None of the above c NaCl (aq.) Comments Active
128 When a wire is stretched to double its original length, the longitudinal strain produced in it is 0.5 1.0 1.5 2.0 b Let \(l_{1}=l(Initial length), l_{2}=2l(Final length)\)
Longitudinal strain
\(∈_{L}=\frac{l_{2}-l_{1}}{l_{1}}=\frac{2l-l}{l}=1\)		 Comments Active
129 Turning a key into the lock is a case of Coplaner forces Non-coplaner forces Couple Moment c Couple: during turning a key, force couple acts. Comments Active
130 When the number of members ‘n’ in a truss is more than 2j-3, where ‘j’ is the number of joints, the frame is said to be Perfect truss Imperfect truss Deficient truss Redundant truss d If n = no. of member
J = no of joints
If
\(n>2J-3 (Redundant truss)\)
\(n=2J-3 (Perfect truss)\)
\(n<2J-3(Deficient truss)\)		 Comments Active
131 Which of the following equilibrium equation should be satisfied by the joints in truss , \(ΣH=0\) \(ΣM=0\) \(ΣH=0,ΣV=0\) \(ΣV=0,ΣM=0\) and \(ΣH=0,ΣV=0 \) \( ΣM=0\) b For truss equations for equilibrium are
Algebraic sum
\(Σ Horizontal forces=0\)
\(Σ Vertical forces=0\)		 Comments Active
132 Critical speed of a shaft depends on Diameter of disc Length of shaft Eccentricity All of the above d Critical speed of shaft
\(ω_{c}=\frac{K}{m}=\frac{g}{δ}=δ=\frac{e}{\frac{ω^{z}}{ωnz}-1}\)
\(K=stiffness \)
\(m=mass\)
\(δ=\frac{PL^{3}}{48EI}\)
\(I=\frac{Ï€d^{4}}{64} d=diameter\)
\(L=spam\)		 Comments Active
133 When the applied force is less than the limiting frictional force, the body will Start moving Remain at rest Slide backward Skid b Limiting frictional force is that minimum force which is needed to overcome the friction to start the motion. So if applied force is already less than limiting force, body will not move at all. For motion \(F>μN\) Comments Active
134 Coriolis component of acceleration exists whenever a point moves along a path that has Tangential acceleration Centripetal acceleration Linear motion Rotational motion d During rotation of path Comments Active
135 A slider on a link rotating with angular velocity have linear velocity The magnitude of Coriolis component of acceleration is \('ω'\) \('v'\) \(ω/2v\) \(2vω\) \(2v/ω\) ν/ 2 \(ω\) b \( a_{c}=Coriolis accelration \)
\(=2ωv\)		 Comments Active
136 The Coriolis component of acceleration acts Along the sliding surface Perpendicular to the sliding surface At 45° to the sliding surface None of the above b Coriolis component of acceleration can be obtained where sliding and rotational oscillation occur simultaneously.
e.g. Shaper (Quick return mechanism)
Coriolis acceleration
\(a_{c}=2ωv\)
Direction parallel to sliding velocity		 Comments Active
137 A column of length ‘l’ is fixed at both the ends. The equivalent length of the column is 2 l 0.5 l 4 l l b End condition of Beam
Leq
Both ends hunged
L
One end free, one end fixed
2L
Both ends fixed
L/2
One is fixed other is hinged
L/ \(2\)		 Comments Active
138 Ultrasonic machining is best suited for Amorphous material Brittle material Nonferrous material All of the above b Comments Active
139 If the diameter of a long column is reduced by 20 percent, the reduction in Euler buckling load in percentage is nearly 4 36 49 59 b Euler’s load
\(P_{e}=\frac{Ï€^{2}EI}{leff^{2}}\)
Here \(Pe∝I\)
\(Pe∝\frac{πd^{4}}{64}=Pe∝d^{4}\)
\(=\frac{Pe_{1}}{Pe_{2}}=\frac{d14}{d24}=Pe_{2}=Pe_{1}.\frac{d24}{d14}\)
\(=\frac{Pe_{2}}{Pe_{1}}=\frac{(0.8d)^{4}}{d^{4}}=0.8^{4}=0.4096\)
\(=1-\frac{Pe_{2}}{Pe_{1}}=(1-0.4096)\)
\(=\frac{Pe_{1}-Pe_{2}}{Pe_{1}}=0.5904\)
\(=59.04%\)		 Comments Active
140 Kinematic pair constituted by cam and follower mechanism is Higher and open type Lower and open type Lower and closed type Higher and closed type a Cam and follower have line or point contact so it is higher pair.
Generally, it is not constrained by a spring so it is open type. If spring controlled cam and follower is there that is closed type.		 Comments Active
141 If load at the free end of the cantilever beam is gradually increased, failure will occur at In the middle of beam At the fixed end Anywhere on the span None of the above b Failure of beam will occur there, where maximum bending moment appear, which is at fixed end.
\(M_{A}=W×l\)
\(M_{A}∝W\)
As W increases, will increase. \(M_{A}\)		 Comments Active
142 State of plane stress at a point is described by
and \(σ_{x}=σ_{y}=σ\) \(τ_{xy}=0\)
The normal stress on a plane inclined at 45° to the horizontal is		 \(σ/2\) σ \(√2\) σ \(√3\) σ d \( σ_{x}=σ_{y}=σ and τ_{xy}=0\)
Now
\(σ_{n}=\frac{σ_{x}+σ_{y}}{2}+\frac{σ_{x}-σ_{y}}{2} cos 2θ+τ_{xy}sin 2θ\)
\(\frac{σ_{ }+σ_{ }}{2}+\frac{σ_{ }-σ_{ }}{2} cos 90°+0×sin90°\)
\(=\frac{2σ}{2}=σ\)		 Comments Active
143 Mohr’s circle may be used to determine following stress on an inclined plane Normal stress Principal stress Tangential stress All of the above d Normal Stress
\(σ_{n}=\frac{σ_{x}+σ_{y}}{2}+\frac{σ_{x}-σ_{y}}{2} cos 2θ+τ_{xy}sin2θ\)
Tangential stress
\(ζ=\frac{σ_{x}+σ_{y}}{2} sin 2θ+τ_{xy} cos 2θ\)
Principal stresses
\(σ_{1}.σ_{2}=\frac{σ_{x}+σ_{y}}{2}+\frac{(σ_{x}-σ_{y})}{4}+τxy2\)		 Comments Active
144 A cable with uniformly distributed load per horizontal metre run will take the following shape Straight line Parabola Ellipse Hyperbola b the cable takes a parabolic shape. Comments Active
145 Impulse is Minimum momentum Maximum momentum Average momentum Final momentum - Initial momentum d (Final momentum) – (Initial momentum) = Impulse Comments Active
146 Kinetic energy of a solid cylinder of mass ‘m’, radius ‘r’ and angular velocity ‘’ is \(ω\) mr2 \(ω^{2}/2\) mr2 \(ω^{2}/4\) mr2 \(ω^{2}\) mr2 \(ω/2\) b K.E of rotating mass
\(K.E.=\frac{1}{2}Iω^{2}\)
\(I_{cylinder}=\frac{MR^{2}}{2}\)
\(∴KE=\frac{1}{2}\frac{MR^{2}}{2}.ω^{2}\)
\(=\frac{MR^{2}ω^{2}}{4}\)		 Comments Active
147 A ball of 2kg drops vertically onto the floor with a velocity of 20m/s. It rebounds with an initial velocity of 10m/s, impulse acting on the ball during contact will be 20 40 60 30 c Impulse = Force × time
\(=change in momentum\)
\(=m(v-u)\)
\(=2(-10-20)\)
\(=-60 kg-m/sec\)		 Comments Active
148 Two balls are dropped from a common point after an interval of 1 second. If acceleration due to gravity is 10m/, separation distance 3 second after the release of the first ball will be \(s^{2}\) 5m 15m 25m 30m c If both balls are dropped at the same time, so distance covered by the balls as \(\frac{1}{2} gt^{2}.\)
In 3 seconds, distance covered by first ball
\(=\frac{1}{2}×10×3^{2}=\frac{90}{2}=45m\)
In (3 – 102) seconds distance covered by seconds’ ball
\(=\frac{1}{2}gt^{2}=\frac{1}{2}×10×2^{2}=20 m\)
Separation distance \(=45-20=25m\)		 Comments Active
149 Two cars ‘A’ and ‘B’ move at 15m/s in the same direction. Car ‘B’ is 300m ahead of car ‘A’. If car ‘A’ accelerates at 6m/ while car ‘B’ continues to move with the same velocity, car ‘A’ will overtake car ‘B’ after \(s^{2}\) 7.5 s 10 s 12 s 15 s b \(\) \(U_{A}=15 m/s, a_{A}=6 m/s^{2}\)
\(U_{B}=15 m/s, a_{B}=0\)
Distance covered by car A in time t
\(S_{A}=u_{A}.t+\frac{1}{2}a_{A}t^{2} \)
\( =15t+3t^{2}\)
Distance covered by car B in time t
\(-S_{B}=u_{B}t+\frac{1}{2}a_{B}t^{2}\)
\( =15t+\frac{1}{2}×0×t^{2}\)
\( =15t\)
After time + cars meet (let suppose)
\(15t+3t^{2}=300+15t\)
\(t=10 sec\)		 Comments Active
150 A hollow shaft has external and internal diameters of 10cm and 5cm respectively. Torsional section modulus of shaft is 375 cm3 275 cm3 184 cm3 84 cm3 c Torsional Eqn
\(\frac{T}{J}=\frac{GQ}{l}=\frac{ζ}{R}\)
Now \(T.\frac{R}{J}=ζ=ζ=\frac{T}{(J/R)}\)
Torsional Rigidity = \(\frac{\frac{Ï€(Do4-D14)}{4}}{\frac{D_{o}}{2}}\)
J = Polar M.O.I. \(=\frac{π(10^{4}-5^{4})}{16×\frac{10}{2}}\)
\(=\frac{Ï€(10000-625)}{160}\)
\(=\frac{22}{7}×\frac{9375×2}{160}=84 cm^{3}\)		 Comments Active
151 A solid circular shaft is subjected to a maximum shear stress of 140MPa Magnitude of maximum normal stress developed in the shaft is 60 MPa 90 MPa 110 MPa 140 MPa d Normal stress
\(σ_{n}=\frac{σ_{x}+σ_{y}}{2}+\frac{σ_{x}-σ_{y}}{2} cos 2θ+τ_{xy}sin2θ\)
\(at θ=45°, τ_{xy} will be max.\)
\(σ_{x}=σ_{y}=0\)
\(σ_{n}=0+0+040.sin90°\)
\(σ_{n}=140 MPa\)		 Comments Active
152 In thick cylinder, the radial stresses in the wall thickness is zero negligible small varies from the inner face to outer face None of the above c For thick cylinder stress distribution
Lame’s Eqn
\(σ_{n}=\frac{b}{r^{2}}+a\)
\(σ_{r}=\frac{b}{r^{2}}-a\)		 Comments Active
153 The point of contra flexure occurs in Cantilever beams Simply supported beams Overhanging beams Fixed beams c Point of contra flexural
Bending moment = 0		 Comments Active
154 In a beam when shear force changes sign, the bending moment will be Zero Maximum Minimum Infinity b Where SF = 0 or changes sign.
BM will be max.		 Comments Active
155 Euler’s formula holds good for Short columns only Long columns only Both long and short columns Weak columns b Euler’s load \(P_{e}=\frac{π^{2}EI}{Le^{2}}\)
Equivalent length = Le
It is suitable for long column.
Rankine – Gorden’s formula is suitable for long or short column.
\(P=\frac{σ_{c}A}{1+a(\frac{L}{K})^{2}} K=I/A \)
\(K=radius of gyration \)		 Comments Active
156 The resultant deflection of a beam under unsymmetrical bending is Parallel to the neutral axis Perpendicular to the neutral axis Parallel to the axis of symmetry Perpendicular to the axis of symmetry b Try to remember this directly, to understand this concept, lots of derivations one required.
U – U and V – V are principle centroidal axis.
N – N – Neutral axis
W’ – N’ – deflection plane		 Comments Active
157 The shear stress at the centre of a circular shaft under torsion is Maximum Minimum Zero Unpredictable c Torsion Eqn
\(\frac{T}{τ}=\frac{τ}{R}=τ∝R=τ_{max} at r\)
\(Shear stress ∝Radius of shaft\)
Shear stress will be maximum at periphery and at centre it will be zero.		 Comments Active
158 A body is having a simple harmonic motion. Product of its frequency and time period is equal to Zero One Infinity 0.5 b In SHM time period
\(T=2π/ω\) \(∵ω=angular speed\)
And frequency \(f=\frac{1}{T}=\frac{ω}{2π}\)
So f.T \(=\frac{2π}{ω}.\frac{ω}{2π}=1\)		 Comments Active
159 For isotropic materials, shear and elastic moduli are related to each other and to Poisson’s ratio according to E = G ( 1 + 2 ) \( ν\) E = 2G (1 + ) \( ν\) E = G (2 + ) \(ν\) E = (2 + G) \(ν\) b For isotropic material elastic modulus for shear and elasticity
\(E=2G(1+ν)\)
\(E=Modulus of Elasticity\)
\(G=Modulus of rigidity \)
\(E=3K(1-2ν)\)
\(K=Bulk modulus\)
\(ν=Poisson^{'}s ratio.\)		 Comments Active
160 For a vibrating system with viscous damping, the characteristics equation is given as \(Mx+cx+kx=0.\)
If the roots of the characteristics equation are real and equal, the system is		 Over damped Critically damped Under damped Cannot be predicted b \( Eq^{n} mx+cx+kx=0\)
Roots of this will be equal and real
When \(C^{2}-4mK=0\)
Then system will be critically damped
If
\(C^{2}-4mK>0-Overdamped\)
\(C^{2}-4mK<0-underdamped \)
Over damped – system return to equilibrium by exponentially decaying to zero system and will not pass equilibrium position more than once.
Underdamped: System oscillates as it slowly return to equilibrium and the amplitude decreases overtime.
Critically damped: System returns to equilibrium very quickly without oscillating and without passing the equilibrium position at all.		 Comments Active
161 In operating characteristics curve, abscissa (x-axis) represents Number of defectives Percentage defectives Sample number Probability of acceptance b It describes the probability of accepting a lot as a function of lot’s quantity proportion. Comments Active
162 At breakeven point Sales revenue > total cost Sales revenue = total cost Sales revenue < total cost None of the above b Sales Revenue = Total cost at BEP
At BEP no profit and no loss occurs.		 Comments Active
163 Which one of the following is not the control chart for attributes p chart c chart R chart chart \(x\) * Both c and d
\(X, R & σ-Variability measurment chart\)
\(P, C u.np-Attribute charts \)		 Comments Active
164 If work station times are not same, the overall production rate of an assembly line is determined by the Fastest station time Slowest station time Average of all station times Average of slowest and fastest station times b Work station time – Time taken to complete the work assigned to a station. The station where maximum time is consumed is slowest station and for complete assembly, work at each station must be completed so overall production rate will be based on slowest station. Comments Active
165 Following is not a method of solving a transportation problem Northwest corner method Least cost method Vogel’s approximation method Dynamic method d Dynamic method is used to solve other type of multistage decision-making problem.
Eg. Assembly line programming		 Comments Active
166 Control limits of a X chart are \(X±σ\) \(X±2σ\) \(X±3σ\) \(X±6σ\) c Upper limit = \(X+3σ\)
Lower limit = \(X-3σ\)		 Comments Active
167 In ABC analysis items are classified in three categories namely A, B, and C in accordance with their Values Number Characteristics Priorities a A – 10% of total inventory
20% of total annual value
B – 20% of total inventory
20% of total annual value
C – 70% of total inventory
10% of total annual value		 Comments Active
168 Increase in economic order quantity results in Increase in inventory carrying cost Decrease in ordering cost Decrease in total cost Total cost first decreases and then increases d Comments Active
169 In ABC analysis of inventories, ‘A’ items usually constitute 10 % 20 % 30 % 70 % a A – 10% of total inventory
20% of total annual value
B – 20% of total inventory
20% of total annual value
C – 70% of total inventory
10% of total annual value
By ABC analysis we used to split the inventory according to their worth and quantity.		 Comments Active
170 In a queuing problem, if the arrivals are completely random then the probability distribution of number of arrivals in a given time follows Normal distribution Poisson distribution Binomial distribution Exponential distribution b Arrival shows poisson’s distribution as
\(P(x)=\frac{(t.λ)^{x}e^{-λt}}{x!}\)
\(x=no. of arrival in a given time\)
\(λ=mean arrival rate\)
\(t=certain time interval\)
Service rate follows exponential distribution function. When service time is variable and random.
\(P(t)=μ e^{-μt}\)
\(μ=avg. service rate\)		 Comments Active
171 Low helix angle drills are used for drilling holes in Plastics Copper Cast steel Carbon steel a Low helix angle is used for hard and brittle material like brass, plastic, C.I. etc
High helix angle is used for soft and ductile material like, steel, Cu, Al, etc
Thermosetting Plastic		 Comments Active
172 If in graphical solution of linear programming problem, the objective function line is parallel to the line representing constraint equation, then the solution of problem is Infeasible solution Unbound solution Multiple optimum solutions None of the above c Comments Active
173 In queuing theory, the ratio of mean arrival rate and the mean service rate is termed as Work factor Utilization factor Slack constant Production rate b Utilization factor for servers or traffic intensity
\(ρ=\frac{mean arrival rate(λ)}{mean service rate (μ)}\)
\(λ follows poisson^{'}s distributions\)
\(μ follows exponential distribution\)
This factor can also be written as
\(=\frac{Total engage time of server}{Total time available for the server}\)
\(=\frac{avg no. of busy server}{Total no of server}\)
It also defined productivity higher U.F, longer the weight time.		 Comments Active
174 The annual demand for an item is 3200 parts. Unit cost is Rs. 6 and the inventory carrying charges are estimated as 25% per annum. If the cost of one procurement is Rs. 150, what will be the number of orders per year? 4 6 8 10 a \(\) \(D=3200\)
\(C=Rs. 6/unit\)
\(C_{n}=\frac{25}{100}×6 Rs.\)
\(C_{o}=Rs. 150\)
\(EOQ=\frac{2DC_{o}}{C_{n}} \)
\(EOQ=\frac{2×3200×150}{\frac{25}{100}×6}=800\)
n \(no of order/year =\frac{3200}{800}=4\)		 Comments Active
175 Effect of stock out of a commodity is Loss of profit Loss of customers Loss of goodwill All of the above d Stock out means items is not available for manufacturing. Comments Active
176 If ‘m’ is the number of constraints in a linear programming problem with two variables ‘x’ and ‘y’ and non-negativity constraints x & y ≥ 0. The feasible region in the graphical solution will be surrounded by m lines m+1 lines m+2 lines m+4 lines c Form constraints and 2 variables (non negative) we need more than 2 lines as m + 2 other than and . \(x=0\) \(y=0\) Comments Active
177 An industry produces 300 spark plugs in one shift of 8 hours. If standard time per piece is 1.5 minute, the productivity would be 3/4 5/8 7/16 15/16 d Shift duration = 8 hrs
\( =8×60\)
\(=480 min\)
Standard time per piece = 1.5 min
No of spark plug in a shift = 300
Total standard production possible = \(\frac{480}{1.5}=320 pieces\)
Actual production = 300
\(Productivity=\frac{Actual Production}{Standard Production}\)
\(=\frac{300}{320}=\frac{15}{16}\)		 Comments Active
178 If the demand for an item is doubled and the ordering cost is halved, the economic order quantity for the item will be A half of the earlier quantity Double of the earlier quantity Increased by a factor of Will remain unchanged d Economic order quantity
\(EOQ=\frac{2×ordering cost×Demand}{Unit holding cost}\)
\(EOQ=\frac{2×C_{o}×D}{C_{n}}\)
\(A_{1}=A, A_{2}=A/2\)
\(D_{1}=D, D_{2}=2D\)
\(\frac{EOQ_{1}}{EOQ_{2}}=\frac{2AD/U}{2.A/2.2D/u}=1\)
(When unit holding cost is same in both case)		 Comments Active
179 Which control chart is used to measure variability of ‘Variability with in the sample’ chart \(x \) c chart u chart chart \(σ\) a \(\) \(σ:To check stability of process or operation over \)
\(time.\)
\(X chart-to control the mean valuable of a variable.\)
R chart – to control the variability of a variable based on a sample taken from process.
Note: It’s ans should be R – chart In option it is not given so choose chart. \(X\)		 Comments Active
180 In High velocity forming, the forming speed is greater than 3 m/s 5 m/s 8 m/s 15 m/s b HVF – 5 to 300 m/s
High velocity forming:
* Metals are formed by energy for a very short time with a velocity.
In high velocity forming velocity range 5 to 300 m/s. In conventional forming velocity range – 0.03 – 4 m/s.		 Comments Active