MCQOPTIONS
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MCQOPTIONS
This section includes 843 Mcqs, each offering curated multiple-choice questions to sharpen your Materials Science and Manufacturing Engineering knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Match the following: P. Feeler gauge Q. Fillet gaugeR. Snap gauge S. Cylindrical plug gauge (i) Radius of an object (ii) Diameter within limits by comparison (iii) Clearance or gap between components (iv) Inside diameter of straight hole |
| A. | P-(iii), Q-(i), R-(ii), S-(iv) |
| B. | P-(iii), Q-(ii), R-(i), S-(iv) |
| C. | P-(iv), Q-(ii), R-(i), S-(iii) |
| D. | P-(iv), Q-(i), R-(ii), S-(iii) |
| Answer» B. P-(iii), Q-(ii), R-(i), S-(iv) | |
| 152. |
Friction at the tool-chip interface can be reduced by |
| A. | decreasing the rake angle |
| B. | increasing the depth of cut |
| C. | decreasing the cutting speed |
| D. | increasing the cutting speed |
| Answer» E. | |
| 153. |
Match the Machine Tools (Group A) with the probable Operations (Group B) : Group A Group BP. Centre Lathe1. SlottingQ. Milling2. Counter-boringR. Grinding3. KnurlingS. Drilling4. Dressing |
| A. | P-1, Q-2, R-4, S-3 |
| B. | P-2, Q-1, R-4, S-3 |
| C. | P-3, Q-1, R-4, S-2 |
| D. | P-3, Q-4, R-2, S-1 |
| Answer» D. P-3, Q-4, R-2, S-1 | |
| 154. |
Assertion (A): In electron beam welding process, vacuum is an essential process parameter. Reason (R): Vacuum provides a highly efficient shield on weld zone. |
| A. | Both A and R are true and R is the correct explanation of A |
| B. | Both A and R are true but R is not a correct explanation of A |
| C. | A is true but R is false |
| D. | A is false but R is true |
| Answer» C. A is true but R is false | |
| 155. |
The main cutting force acting on a tool during the turning (orthogonal cutting) operation of a metal is 400 N. The turning was performed using 2 mm depth of cut and 0.1 mm/rev feed rate. The specific cutting pressure (in N/mm2) is |
| A. | 1000 |
| B. | 2000 |
| C. | 3000 |
| D. | 4000 |
| Answer» C. 3000 | |
| 156. |
Assertion (A) : A diamond tool is used for USM of glass workpiece. Reason (R) : Diamond is harder than glass |
| A. | Both A and R are true and R is the correct explanation of A |
| B. | Both A and R are true but R is not a correct explanation of A |
| C. | A is true but R is false |
| D. | A is false but R is true |
| Answer» E. | |
| 157. |
Orthogonal machining of a steel workpiece is done with a HSS tool of zero rake angle. The ratio of the cutting force and the thrust force on the tool is 1 : 0.372. The length of cut chip is 4.71 mm while the uncut chip length is 10 mm. What is the shear plane angle and friction angle in degree? Use Merchants theory |
| A. | 32.49, 10.22 |
| B. | 25.22, 20.41 |
| C. | 64.78 20.41 |
| D. | 25.22, 23.21 |
| Answer» C. 64.78 20.41 | |
| 158. |
Match List-I (Cutting tool materials) with ListII (Manufacturing methods) and select the correct answer using the codes given below : List-I List-IIA. HSS1. CastingB. Satellite2. ForgingC. Cemented carbide3. RollingD. UCON4. Extrusion5. Powder metallurgy |
| A. | A - 3 , B - 1 , C - 5 , D - 2 |
| B. | A - 2 , B - 5 , C - 4 , D - 3 |
| C. | A - 3 , B - 5 , C - 4 , D - 2 |
| D. | A - 2 , B - 1 , C - 5 , D - 3 |
| Answer» E. | |
| 159. |
Assertion (A) : The ratio of uncut chip thickness to actual chip thickness is always less than one and is termed as cutting ratio in orthogonal cutting. Reason (R) : The frictional force is very high due to the occurrence of sticking friction rather than sliding friction. |
| A. | Both A and R are true and R is the correct explanation of A |
| B. | Both A and R are true but R is not a correct explanation of A |
| C. | A is true but R is false |
| D. | A is false but R is true |
| Answer» D. A is false but R is true | |
| 160. |
The hole 40+0.020 +0.000 and shaft, 40+0.010 +0.000 when assembled will result in |
| A. | clearance fit |
| B. | interference fit |
| C. | transition fit |
| D. | drive fit |
| Answer» D. drive fit | |
| 161. |
A bush was turned after mounting the same on a mandrel. The mandrel diameter in millimeters is 40+0.00-0.05 and bore diameter of bush is 40+0.06-0.010. The maximum eccentricity of the bush in mm will be |
| A. | 0.01 |
| B. | 0.055 |
| C. | 0.1 |
| D. | 0.11 |
| Answer» C. 0.1 | |
| 162. |
Allowance in limits and fits refers to |
| A. | Maximum clearance between the shaft and hole |
| B. | Minimum clearance between the shaft and hole |
| C. | Difference between the maximum and minimum size of hole |
| D. | Difference between the maximum and minimum size of shaft |
| Answer» C. Difference between the maximum and minimum size of hole | |
| 163. |
In a 2 D CAD package, clockwise circular arc of radius 5, specified from P1 (15, 10) to P2 (10, 15) will have its center at |
| A. | (10, 10) |
| B. | (15, 10) |
| C. | (15, 15) |
| D. | (10, 15) |
| Answer» D. (10, 15) | |
| 164. |
A shaft has a dimension, 35-0.025-0.009 respective values of fundamental deviation and tolerance are |
| A. | 0.025, 0.008 |
| B. | 0.025, 0.016 |
| C. | 0.009, 0.008 |
| D. | 0.009, 0.016 |
| Answer» E. | |
| 165. |
What are the upper and lower limits of the shaft represented by 60 f8 ? Use the following data : Diameter 60 lies in the diameter step of 50-80 mm. Fundamental tolerance unit, i, in m = 0.45 D1/3 + 0.001 D, where D is the representative size in mm; Tolerance value for IT8 = 25i. Fundamental deviation for f shaft = 5.5 D 0.41 |
| A. | lower lim. = 59.924 mm, Upper lim. = 59.970 mm |
| B. | lower lim. = 59.954 mm, Upper lim. = 60.000 mm |
| C. | lower lim. = 59.970 mm, Upper lim. = 60.016 mm |
| D. | lower limit = 60.000 mm, Upper limit = 60.046 mm |
| Answer» B. lower lim. = 59.954 mm, Upper lim. = 60.000 mm | |
| 166. |
A hole is specified as 400.0000.050 mm. The mating shaft has a clearance fit with minimum clearance of 0.01 mm. The tolerance on the shaft is 0.04 mm. The maximum clearance in mm between the hole and the shaft is |
| A. | 0.04 |
| B. | 0.05 |
| C. | 0.10 |
| D. | 0.11 |
| Answer» D. 0.11 | |
| 167. |
GO and NO-GO plug gauges are to be designed for a hole 20.000+0.010+0.050 mm. Gauge tolerances can be taken as 10% of the hole tolerance. Following ISO system of gauge design, sizes of GO and NO-GO gauge will be respectively. |
| A. | 20.010 mm and 20.050 mm |
| B. | 20.014 mm and 20.046 mm |
| C. | 20.006 mm and 20.054 mm |
| D. | 20.014 mm and 20.054 mm |
| Answer» E. | |
| 168. |
In an interchangeable assembly, shafts of size 25.000+0.040-0.0100 mm mate with holes of size 25.000+0.020-0.000 mm. The maximum possible clearance in the assembly will be |
| A. | 10 microns |
| B. | 20 microns |
| C. | 30 microns |
| D. | 60 microns |
| Answer» E. | |
| 169. |
Match the following lists List-IList-IIA. Aluminium brake shoe 1. Deep drawingB. Plastic water bottles 2. Blow moulding C. Stainless steel cups 3. Centrifugal casting D. Soft drink can (aluminum) 4. Impact extrusion 5. Sand casting6.Upset forging |
| A. | A - 1 , B - 3, C - 2 , D - 5 |
| B. | A - 5 , B - 2, C - 4 , D - 3 |
| C. | A - 5 , B - 1, C - 3, D - 4 |
| D. | A - 5 , B - 2, C - 1 , D - 4 |
| Answer» E. | |
| 170. |
Match the following lists List-IList-IIA. ECM 1. Plastic shear B. EDM 2. Erosion/brittle fracture C. USM 3. Corrosive reaction D. LBM 4. Melting vaporization 5. lon displaeement6.Plastic shear and ion displacement |
| A. | A - 5 , B - 3, C - 2 , D - 4 |
| B. | A - 5 , B - 1, C - 4 , D - 3 |
| C. | A - 1 , B - 5, C - 2 , D - 4 |
| D. | A - 4 , B - 3, C - 2 , D - 1 |
| Answer» B. A - 5 , B - 1, C - 4 , D - 3 | |
| 171. |
During normalizing process of steel, the specimen is heated |
| A. | between the upper and lower critical temperature and cooled in still air |
| B. | above the upper critical temperature and cooled in furnace |
| C. | above the upper critical temperature and cooled in still air |
| D. | between the upper and lower critical temperature and cooled in furnace |
| Answer» D. between the upper and lower critical temperature and cooled in furnace | |
| 172. |
Match the items in columns I and II.Column-IColumn-IIP. Charpy test1. Fluidity Q. Knoop test 2. Microhardness R. Spiral test3. FormabilityS. Cupping test4. Toughness5. Permeability |
| A. | P-4, Q-5, R-3, S-2 |
| B. | P-3, Q-5, R-1, S-4 |
| C. | P-2, Q-4, R-3, S-5 |
| D. | P-4, Q-2, R-1, S-3 |
| Answer» E. | |
| 173. |
Match the following materials with their most appropriate application. Material Application1. Low carbon steelP. Machine tool base2. Stainless steelQ. Aircraft parts3. Gray cast ironR. Kitchen utensils4. Titanium alloysS. Car body panels |
| A. | P - 1 , Q - 3 , R - 2 , S - 4 |
| B. | P - 1 , Q - 4 , R - 2 , S - 3 |
| C. | P - 3 , Q - 2 , R - 4 , S - 1 |
| D. | P - 3 , Q - 4 , R - 2 , S - 1 |
| Answer» E. | |
| 174. |
A shaft of diameter 20+0.06-0.15 mm and a hole of diameter 20+0.20+0.10 mm when assembled would yield |
| A. | Transition fit |
| B. | Interference fit |
| C. | Clearance fit |
| D. | None of these |
| Answer» D. None of these | |
| 175. |
Match List-I with List-II and select the correct answer using the codes given below : List-I List-IIA. Plane approach angle1.Tool faceB. Rake angle2.Tool flankC. Clearances angle3.Tool face and flankD. Wedge angle4.Cutting edge5.Tool nose |
| A. | A - 1 , B - 4 , C - 2 , D - 5 |
| B. | A - 4 , B - 1 , C - 3 , D - 2 |
| C. | A - 4 , B - 1 , C - 2 , D - 3 |
| D. | A - 1 , B - 4 , C - 3 , D - 5 |
| Answer» D. A - 1 , B - 4 , C - 3 , D - 5 | |
| 176. |
Neglecting the contribution of the feed force towards cutting power, the specific cutting energy in J/mm3 is |
| A. | 0.2 |
| B. | 2 |
| C. | 200 |
| D. | 2000 |
| Answer» C. 200 | |
| 177. |
Consider the following statements: During the third stage of tool wear, rapid deterioration of tool edge takes place because 1. Flank wear is only marginal 2. Flank wear is large 3. Temperature of the tool increases gradually 4. Temperature of the tool increases drastically Which of the statements given above are correct? |
| A. | 1 and 3 |
| B. | 2 and 4 |
| C. | 1 and 4 |
| D. | 2 and 3 |
| Answer» C. 1 and 4 | |
| 178. |
Two cutting tools are being compared for a machining operation. The tool life equations are Carbide tool VT1.6 = 3000 HSS tool VT0.6 = 200 where V is the cutting speed in m/min and T is the tool life in min. The carbide tool will provide higher tool life if the cutting speed in m/min exceeds |
| A. | 15.0 |
| B. | 39.4 |
| C. | 49.3 |
| D. | 60.0 |
| Answer» C. 49.3 | |
| 179. |
A steel bar 200 mm in a diameter is turned at a feed of 0.25 mm/rev. with a depth of cut of 4 mm. The rotational speed of the workpiece is 160 rpm. The material removal rate in mm3 / s is |
| A. | 160 |
| B. | 167.6 |
| C. | 1600 |
| D. | 1675.5 |
| Answer» E. | |
| 180. |
What is the approximate% change in the life, f, of the tool with zero rake angle used in orthogonal cutting when its clearance angle, , is changed from 10 to 7 deg? (Hint: Flank wear rate is proportional to cot ) |
| A. | 30% increase |
| B. | 30% decrease |
| C. | 70% increase |
| D. | 70% decrease |
| Answer» C. 70% increase | |
| 181. |
Two plates of the same metal having equal thickness are to be butt welded with electric arc. When the plate thickness changes, welding is achieved by |
| A. | Adjusting the current |
| B. | Adjusting the duration of the current |
| C. | Changing the electrode size |
| D. | Changing the electrode coating |
| Answer» B. Adjusting the duration of the current | |
| 182. |
A welding operation is being performed with voltage = 30 V and current = 100 A. The crosssectional area of the weld bead is 20 mm2. The work-piece and filler are of titanium for which of the specific energy of melting is 14 J/mm . Assuming a thermal efficiency of the welding process 70%, the welding speed (in mm/s) is (correct to two decimal places). |
| A. | 5.7 |
| B. | 2.1 |
| C. | 33.9 |
| D. | 7.5 |
| Answer» E. | |
| 183. |
Match the following A. Abrasive jet machining1. HornB. EDM2. Wire as cutting toolC. LBM3. LightD. USM4. Vacuum 5. Nozzle6. Cladding |
| A. | A 5, B 2, C 3, D 4 |
| B. | A 5, B 2, C 4, D 1 |
| C. | A 5, B 4, C 3, D 1 |
| D. | A 5, B 2, C 3, D 1 |
| Answer» E. | |
| 184. |
A steel bar 200 mm in diameter is turned at a feed of 0.25 mm/rev with a depth of cut of 4 mm. The rotational speed of the workpiece is 160 rpm. The material removal rate in mm3/s is |
| A. | 160 |
| B. | 167.6 |
| C. | 1600 |
| D. | 1675.5 |
| Answer» E. | |
| 185. |
Circular arc on a part profile is being machined on a vertical CNC milling machine, CNC part program using metric units with absolute dimensions is listed below: ............................................ N60 G01 X 30 Y 55 Z 5 F50 N70 G02 X 50 Y 35 R 20 N80 G01 Z 5 ............................................ The coordinates of the centre of the circular arc are: |
| A. | (30, 55) |
| B. | (50, 55) |
| C. | (50, 35) |
| D. | (30, 35) |
| Answer» E. | |
| 186. |
Match the following part programming codes with their respective functions Part Programming Codes P. G01 Q. G03 R. M03 S. M05 Functions I. Spindle stop II. Spindle rotation, clockwise III.Circular interpolation, anticlockwise IV. Linear interpolation |
| A. | P-II, Q-I, R-IV, S-III |
| B. | P-IV, Q-II, R-III, S-I |
| C. | P-IV, O-III, R-II, S-I |
| D. | P-III, Q-IV, R-II, S-I |
| Answer» D. P-III, Q-IV, R-II, S-I | |
| 187. |
Match the following A. Stamping die plate B. Gas turbine blades C. Twist drill D. Carbide tool inserts E. Hole in a ceramic plate 1. EDM 2. LBM 3. ECM 4. EBM 5. ECG 6. Friction welding 7. Wire EDM 8. USM 9. Ultrasonic welding |
| A. | A 7, B 1, C 6, D 5, E 4 |
| B. | A 7, B 1, C 6, D 5, E 8 |
| C. | A 7, B 1, C 6, D 5, E 3 |
| D. | A 7, B 1, C 6, D 5, E 2 |
| Answer» C. A 7, B 1, C 6, D 5, E 3 | |
| 188. |
Which of the following process can be used for welding of Aluminum alloys? P. Submerged arc welding Q. Gas metal arc welding R. Electroslag welding S. Gas tungsten arc welding |
| A. | P and Q |
| B. | Q and S |
| C. | Q and R |
| D. | R and S |
| Answer» C. Q and R | |
| 189. |
Assertion (A): Basic hole system is advocated by Indian Standards for fit design. Explanation (E): Holes can be produced to any size |
| A. | Both A and E are correct and E explains A |
| B. | Both A and E are true but E does not explains A |
| C. | A is true but E is false |
| D. | A is false but E is true |
| Answer» B. Both A and E are true but E does not explains A | |
| 190. |
A mould has downsprue whose length is 20 cm and the cross sectional area at the base of the downsprue is 1 cm2. The downsprue feeds a horizontal runner leading into the mould cavity of volume 1000 cm3. The time required to fill the mould cavity will be |
| A. | 4.05 s |
| B. | 5.05 s |
| C. | 6.05 s |
| D. | 7.25 s |
| Answer» C. 6.05 s | |
| 191. |
Match the following A. Abrasive jet machining B. PAM C. Wire EDM D. Chemical machining 1. Masking 2. Stamping die 3. Engraving 4. Stainless steel profiling 5. Drilling fine holes |
| A. | A 4, B 3, C 2, D 1 |
| B. | A 5, B 4, C 2, D 1 |
| C. | A 5, B 3, C 2, D 1 |
| D. | A 5, B 3, C 2, D 4 |
| Answer» D. A 5, B 3, C 2, D 4 | |
| 192. |
Hardness of green sand mould increases with |
| A. | increase in moisture content beyond 6 percent |
| B. | increase in permeability |
| C. | decrease in permeability |
| D. | increases in both moisture content and permeability |
| Answer» D. increases in both moisture content and permeability | |
| 193. |
With a solidification factor of 0.97 106 s/m2, the solidification time (in seconds) for a spherical casting of 200 mm diameter is |
| A. | 539 |
| B. | 1078 |
| C. | 4311 |
| D. | 3233 |
| Answer» C. 4311 | |
| 194. |
The main purpose of spheroidising treatment is to improve |
| A. | hardenability of low carbon steels |
| B. | machinability of low carbon steels |
| C. | hardenability of high carbon steels |
| D. | machinability of high carbon steels |
| Answer» B. machinability of low carbon steels | |
| 195. |
From the lists given below, choose the most appropriate set of heat treatment process and the corresponding process characteristies Process P. Tempering Q. Austempering R. Martempering Characteristics 1. Austenite is converted into bainite 2. Austenite is converted into martensite 3. Cementite is converted into globular structure 4. Both hardness and brittleness are reduced 5. Carbon is absorbed into the metal |
| A. | P-3, Q-1, R-5 |
| B. | P-4, Q-3, R-2 |
| C. | P-4, Q-1, R-2 |
| D. | P-1, Q-5, R-4 |
| Answer» D. P-1, Q-5, R-4 | |
| 196. |
Match the items in Column I and Column II. Column I Column II P. Metallic Chills 1. Support for the core Q. Metallic Chaplets 2. Reservoir of the molten metal R. Riser 3. Control cooling of critical sections S. Exothermic Padding 4. Progressive solidification |
| A. | P 1, Q 3, R 2, S 4 |
| B. | P 1, Q 4, R 2, S 3 |
| C. | P 3, Q 4, R 2, S 1 |
| D. | P 4, Q 1, R 2, S 3 |
| Answer» E. | |
| 197. |
In a slab rolling operation, the maximum thickness reduction ( hmax) is given by hmax = . R , where R is the radius of the roll and is the coefficient of friction between the roll and the sheet. If = 0.1, the maximum angle subtended by the deformation zone at the centre of the roll (bite angle in degree) is _____. |
| A. | 5.7106 |
| B. | 32.7106 |
| C. | 51.7106 |
| D. | 0.7106 |
| Answer» B. 32.7106 | |
| 198. |
With respect to metal working, match Group A with Group B: Group A P. Defect in extrusion Q. Defect in rolling R. Product of skew rolling S. Product of rolling through cluster mill Group B I. alligatoring II. scab III.fish tail IV. seamless tube V. thin sheet with tight tolerance VI. semi-finished balls of ball bearing |
| A. | P-II, Q-III, R-VI, S-V |
| B. | P-III, Q-I, R-VI, S-V |
| C. | P-III, Q-I, R-IV, S-VI |
| D. | P-I, Q-II, R-V, S-VI |
| Answer» C. P-III, Q-I, R-IV, S-VI | |
| 199. |
Match the following A. Polyester resin B. Methyl methacrylate C. Polyurethane D. Polyvinyl chloride Product 1. Moulded luggage 2. Refrigerator insulation 3. FRP 4. Contact lenses 5. Floor tiles |
| A. | A 1, B 2, B 3, C 4 |
| B. | A 1, B 2, B 3, C 5 |
| C. | A 1, B 2, B 5, C 4 |
| D. | A 1, B 5, B 3, C 4 |
| Answer» B. A 1, B 2, B 3, C 5 | |
| 200. |
The maximum possible draft in cold rolling of sheet increases with the |
| A. | increase in coefficient of friction |
| B. | decrease in coefficient of friction |
| C. | decrease in roll radius |
| D. | increase in roll velocity |
| Answer» B. decrease in coefficient of friction | |