Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

Multiple choice question for engineering

Set 1

1. Stress intensity factor is the critical value of stress at which crack extension occurs.
a) True
b) False

View Answer

Answer: b [Reason:] Stress intensity specifies the stress intensity at the tip of the crack.

2. The critical value at which crack extension occurs is called
a) Stress Intensity Factor
b) Toughness
c) Fracture Toughness
d) None of the mentioned

View Answer

Answer: c [Reason:] Fracture toughness is the critical value of stress intensity at which crack extension occurs.

3. Fracture toughness does not depend upon geometry of the part containing crack
a) True
b) False

View Answer

Answer: b [Reason:] Fracture toughness is directly proportional to a factor Y that depends upon geometry of the part having crack.

4. How many modes are there for crack propagation?
a) 2
b) 3
c) 4
d) 5

View Answer

Answer: b [Reason:] Opening, sliding and tearing are the 3 modes.

5. A curved beam has neutral axis is curved while loaded and straight when unloaded.
a) True
b) False

View Answer

Answer: b [Reason:] Curved beam’s neutral axis is always curved irrespective of the loading.

6. The bending stress in a straight beam varies linearly with the distance from neural axis like that in a curved beam.
a) True
b) False

View Answer

Answer: b [Reason:] Bending stress in a curved beam varies hyperbolically with the distance from neutral axis.

7. If for a curved beam of trapezoidal cross section, radius of neutral axis is 89.1816mm and radius of centroidal axis is 100mm, then find the bending stress at inner fibre whose radius is 50mm. Area of cross section of beam is 7200mm² and the beam is loaded with 100kN of load.
a) 97.3
b) 95.8
c) 100.6
d) None of the mentioned

View Answer

Answer: c [Reason:] e=100-89.816=10.8184mm, h=89.1816-50=39.1816mm, M=100 x 100 N-m Therefore σ=Mh/AeR or σ=10000 x 39.1816/ [7200 x 10.8184 x 50] or σ=100.6N/mm².

8. A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm².Considering only bending stress.
a) 14.80mm
b) 13.95mm
c) 16.5mm
d) 17.2mm

View Answer

Answer: b [Reason:] σ(b)=Mh/ AeR or σ(b)=1000x4Dx(4D-0.2D-3.5D)/ 0.8D²x0.02Dx3.5D , σ(b)=21428.6/D² 21428.6/D² = 110 or D=13.95mm.

9. A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm² and considering only direct tensile stress.
a) 4.7mm
b) 6.8mm
c) 13.95mm
d) 3.4mm

View Answer

Answer: d [Reason:] Direct Tensile Stress=1000/0.8D² or σ (t) =1250/D² 1250/D²=110 or D=3.4mm.

10. A curved beam with eccentricity 0.02D is loaded with 1kN.Centroidal radius=4D and inner and outer radii are 3.5D and 4.5D respectively. Area of cross section is 0.8D². Find the dimension D if allowable stress is 110N/mm² and considering combined effect of direct stress and bending stress.
a) 15.8mm
b) 14.35mm
c) 17.9mm
d) 18.1mm

View Answer

Answer: b [Reason:] σ(b)=Mh/ AeR or σ(b)=1000x4Dx(4D-0.2D-3.5D)/ 0.8D²x0.02Dx3.5D , σ(b)=21428.6/D² DirectTensile Stress=1000/0.8D² or σ (t) =1250/D² Total stress=22678.6/D² N/mm²= 110 or D=14.35mm.

11. If a hollow steel tube is heated from a temperature of 25’C to 250’C then fid the expansion of tube if area of the cross section is 300mm²,length of tube=200mm and coefficient of thermal expansion is 10.8 x 10⁻⁶ per ⁰C.
a) 1.22mm
b) 0.486mm
c) 0.878mm
d) 1.52mm

View Answer

Answer: b [Reason:] Expansion=ἀxlx∆T= 10.8 x 10⁻⁶ x 200 x 225=0.486mm.

12. All type of stresses vanishes after as soon as the applied load is removed.
a) True
b) False

View Answer

Answer: b [Reason:] Residual stresses are independent of the load.

13. Residual stresses are always added in the load stresses and hence are always harmful.
a) True
b) False

View Answer

Answer: b [Reason:] Residual stresses may be beneficial when they are opposite to load stresses and hence are subtracted from load stresses.

Set 2

1. The stress represented by sin (t) + 1 belongs to which category?
a) Fluctuating Stresses
b) Alternating stresses
c) Repeated Stresses
d) Reversed Stresses

View Answer

Answer: c [Reason:] The minimum stress value is zero and hence is belongs to Repeated Stress category.

2. The stress represented by sin (t) + 2 belongs to which category?
a) Fluctuating Stresses
b) None of the mentioned
c) Repeated Stresses
d) Reversed Stresses

View Answer

Answer:a [Reason:] The mean as well as the amplitude value is non zero hence it belongs to Fluctuation Stress Category.

3. The stress represented by sin (t) + 4 belongs to which category?
a) Alternating Stresses
b) None of the mentioned
c) Repeated Stresses
d) Reversed Stresses

View Answer

Answer: a [Reason:] The mean as well as the amplitude value is non zero hence it belongs to Alternation Stresses category which is the other name of Fluctuation Stresses.

4. The stress represented by cos (t) belongs to which category?
a) Fluctuating Stresses
b) Alternating Stresses
c) Repeated Stresses
d) Reversed Stresses

View Answer

Answer: d [Reason:] Half cycle is in tensile stress and other half in compressive stress, hence it belongs to Reversed Stresses Category.

5. If the mean stress value for a sinusoidal stress function is zero, then this type of stress falls in which category?
a) Fluctuating Stresses
b) Alternating Stresses
c) Repeated Stresses
d) Reversed Stresses

View Answer

Answer: d [Reason:] If mean is to be zero, then there must be compressive as well as tensile stresses and hence belongs to reversed stresses category.

6. The phenomenon of decreased resistance of the materials to fluctuating stresses is the main characteristic of _____ failure.
a) Fracture
b) Fatigue
c) Yielding
d) None of the mentioned

View Answer

Answer: b [Reason:] Fatigue failure of the material is the failure at low stress levels under fluctuating syresses.

7. Fatigue failure is time dependent failure.
a) True
b) False

View Answer

Answer: a [Reason:] Fatigue failure is defined as time delayed fracture under cyclic loading.

8. There is sufficient plastic deformation prior to fatigue failure, which gives a warning well in advance.
a) True
b) False

View Answer

Answer: b [Reason:] Fatigue cracks are not visible until they reach the surface and by that time the failure has already taken place. Material nevers enters in the plastic range.

Set 3

1. Which of the following property is affected by heat treatment?
a) Hardness
b) Strength
c) Ductility
d) All of the mentioned

View Answer

Answer: d [Reason:] Heat treatment involves changes in the micro structure and hence all the internal properties are effected.

2. Annealing involves heating the component to a temperature
a) Slightly above the critical temperature
b) Equal to critical temperature
c) Slightly less than critical temperature
d) None of the mentioned

View Answer

Answer: a [Reason:] In annealing, component is heated to a temperature above than critical temperature.

3. Which of the following is true?
a) Rate of cooling in normalising is faster then in annealing
b) Annealing improves ductility
c) Normalising improves grain structure
d) All of the mentioned

View Answer

Answer: d [Reason:] In annealing, the furnace is switched off and component cools slowly. In normalising, component is air cooled.

4. Quenching
a) Consists of heating the component to critical temperature
b) Cooling rapidly
c) Increases hardness
d) All of the mentioned

View Answer

Answer: d [Reason:] During quenching, component is rapidly cooled which leads to formation of martensite. Hence hardness increases.

5. Tempering involves
a) Reheating the quenched component to a temperature greater than critical temperature
b) Increases the brittleness
c) Reheating the quenched component to a temperature equal to critical temperature
d) None of the mentioned

View Answer

Answer: d [Reason:] Tempering involves reheating the quenched product to a temperature less than transformation range. It improves ductility and reduces brittleness.

6. Silicon addition in spring steel increases its toughness.
a) True
b) False

View Answer

Answer: a [Reason:] Silicon addition increases strength without lowering the ductility.

7. Nickel addition in alloys
a) Increases toughness
b) Increases hardenability and impact resistance
c) Limit grain growth during heat treatment process
d) All of the mentioned

View Answer

Answer: d [Reason:] Nickel addition increases toughness by limiting grain growth.

8. Flame hardening involves
a) Heating the surface above the trAnswerformation range
b) Quenching after heating
c) Minimum case depth is 1mm
d) All of the mentioned

View Answer

Answer: d [Reason:] Flame Hardening is a process of heating the surface with a flame above critical temperature and then quenching it.

9. Induction hardening process involves
a) Heating surface by induction in field of invariable current
b) Case depth minimum of 2mm are produced
c) Heating surface by induction in field of alternating current
d) None of the mentioned

View Answer

Answer: c [Reason:] Heating can only be done in presence of alternating current and not constant current.

10. Case carburising involves
a) Introducing carbon at surface layer
b) Heating range 880 to 980’C
c) Case depths up to 2mm are possible
d) All of the mentioned

View Answer

Answer: d [Reason:] Case carburising involves introducing carbon at surface layer. Medium can be liquid, solid or gas and high case depths are possible.

11. Which of the following are not true for carbo nitriding?
a) Introducing carbon and nitrogen at surface layer
b) Component is heated in range of 650 to 920’C
c) Cyaniding is similar to carbo nitriding except that the medium is liquid
d) This process gives a lower wear resistance compared to case carburising process

View Answer

Answer: d [Reason:] Carbo nitriding gives wear resistance greater than compared to case carburising.

12. Which of the following are true for nitriding?
a) Nascent oxygen is involved
b) Temperature range 490 to 590’C
c) Gaseous or liquid medium
d) All of the mentioned

View Answer

Answer: d [Reason:] In nitriding nascent oxygen is acted on the surface of the product at a temperature of 490::590’C in a gaseous or liquid medium.

Set 4

1. A force 2P is acting on the double transverse fillet weld. Leg of weld is h and length l. Determine the shear stress in a plane inclined at θ with horizontal.
a) PSinθ(Sinθ+Cosθ)/hl
b) P(Sinθ+Cosθ)/hl
c) Pcosθ(Sinθ+Cosθ)/hl
d) None of the listed

View Answer

Answer: a [Reason:] F=PSinθ and width=h/(Sinθ+Cosθ).

2. Maximum shear stress in transverse fillet weld of leg h and length l is
a) P/hl
b) 1.21P/hl
c) P/1.21hl
d) None of the listed

View Answer

Answer: b [Reason:] τ= PSinθ(Sinθ+Cosθ)/hl, by maximising it θ=67.5’ and hence find corresponding τ.

3. A sunk key fits in the keyway of the _____ only.
a) Hub
b) Sleeve
c) Both hub and sleeve
d) Neither hub nor sleeve

View Answer

Answer: a [Reason:] Sunk key fits halfway in the hub and halfway in the shaft.

4. Hollow saddle key is superior to flat saddle key as far as power transmitting capability is concerned.
a) True
b) False

View Answer

Answer: b [Reason:] The resistance to slip in case of flat key is more.

5. Saddle key is more suitable than sunk key for heavy duty applications.
a) True
b) False

View Answer

Answer: b [Reason:] In sunk key, relative motion is also prevented by shear resistance of sunk key and hence sunk key is recommended.

6. The main advantage of sunk key is that it is a _____ drive.
a) Positive
b) Negative
c) Neutral
d) None of the listed

View Answer

Answer: a [Reason:] Sunk key is a positive drive and no slip occurs.

7. Woodruff key permits _____ movement b/w shaft and the hub.
a) Axial
b) Radial
c) Eccentric
d) None of the listed

View Answer

Answer: b [Reason:] Woodruff key is a sunk key and doesn’t permit axial moment.

8. Determine the length of kennedy key required to transmit 1200N-m and allowable shear in the key is 40N/mm². The diameter of shaft and width of key can be taken as 40mm and 10mm respectively.
a) 49mm
b) 36mm
c) 46mm
d) 53mm

View Answer

Answer: d [Reason:] l=M/[dbτ√2].

9. Splines are keys.
a) True
b) False

View Answer

Answer: a [Reason:] Splines are keys made with shafts.

10. Involute splines have stub teeth with a pressure angle of ___
a) 30
b) 45
c) 60
d) Can’t be determined

View Answer

Answer: b Explannation: Pressure angle is 30’ and not 60’.

Set 5

1. Knuckle Joint can’t be used to connect two intersecting rods.
a) Yes
b) No, it can’t be used
c) It can be used with some modificatios
d) It is expensive and hence isn’t used

View Answer

Answer: b [Reason:] Knuckle Joint is used to connect two rods whose axes coincide or intersect and lie in a same plane.

2. A knuckle joint is unsuitable for two rotating shafts, which transmit torque
a) True
b) False

View Answer

Answer: a [Reason:] Knuckle joint can’t be used for torque transmission.

3. A maximum of how many roads may be connected using a knuckle joint?
a) 2
b) 3
c) 4
d) 5

View Answer

Answer: b [Reason:] In rare explanation, two rods with forks and one rod with eye is connected.

4. A knuckle joint is also called socket pin joint.
a) True
b) False

View Answer

Answer: b [Reason:] A knuckle joint is also called a Forked Pin Joint.

5. Which of the following are important parts of knuckle joint?
a) Eye
b) Pin
c) Fork
d) Each of the mentioned

View Answer

Answer: d [Reason:] All the mentioned parts are important components of knuckle joint.

6. Calculate the diameter of pin from shear consideration with maximum shear stress allowed is 40NN/mm² and an axial tensile force of 50kN is acting on the rod.
a) 39mm
b) 44mm
c) 49mm
d) 52mm

View Answer

Answer: a [Reason:] As the pin is subjected to double shear diameter (D) = √(2P/π x τ) = 38.80mm.

7. If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.
a) 40mm
b) 50mm
c) 60mm
d) 70mm

View Answer

Answer: a [Reason:] d=P/2aσ = 40mm.

8. If any cross section is subjected to direct tensile stress and bending stress, then find the dimension of cross section. Given length & breadth are t and 2t respectively. F=25kN acts on the top fibre of the cross section, M=F x t . Also maximum allowable tensile stress =100N/mm².
a) 25.5mm
b) 30.2mm
c) 27.55mm
d) None of the mentioned

View Answer

Answer: a [Reason:] σ= [P/A] + [My/I], where y=t & I=t(2t)ᴲ/12.

9. A knuckle joint can be used in valve mechanism of a reciprocating engine.
a) Yes
b) No
c) Yes but there are stress probles
d) No as it is very dangerous to use

View Answer

Answer: a [Reason:] Knuckle joint can be used till the rods coincide or intersect in a plane.

.woocommerce-message { background-color: #98C391 !important; }