Multiple choice question for engineering
Set 1
1. Lewis form factor is based on real number of teeth.
a) True
b) False
Answer
Answer: b [Reason:] It is based on virtual umber of teeth only.
2. Beam strength indicates the maximum value of radial force that a tooth can transmit without fatigue failure.
a) True
b) False
Answer
Answer: b [Reason:] It indicates maximum force for bending failure and not fatigue failure.
3. A pair of helical gears consist of 25 teeth pinion gear meshing with a 90 teeth gear. Calculate the ratio factor.
a) 0.74
b) 0.88
c) 1.57
d) 1.44
Answer
Answer: c [Reason:] Ratio factor Q=2×90/90+25.
4. A pair of helical gears consist of 25 teeth pinion gear meshing with a 90 teeth gear. Calculate the material constant k If surface hardness is 260BHN.
a) 0.64N/mm²
b) 0.88N/mm²
c) 1.08N/mm²
d) 2.66N/mm²
Answer
Answer: c [Reason:] K=0.16x[BHN/100]².
5. A pair of helical gears consist of 25 teeth pinion gear meshing with a 90 teeth gear. Calculate the wear strength If surface hardness is 260BHN. Also face width=35mm, module=4mm and helix angle=25⁰.
a) 443.5N
b) 1125.6N
c) 7971.9N
d) 1014.2N
Answer
Answer: c [Reason:] S=bQdK/cos²Ɯ where Ɯ=25⁰, d=zm/cosƜ, Q=2×90/90+25, K=0.16x[BHN/100]².
6. A pair of helical gears consist of 25 teeth pinion gear meshing with a 90 teeth gear. Calculate the tangential force If surface hardness is 260BHN. Also face width=35mm, module=4mm and helix angle=25⁰. The velocity of operation is 3.5m/s and service factor 1.5.
a) 1136.5N
b) 3983.7
c) 2012.6N
d) 3226.5N
Answer
Answer: b [Reason:] S=1.5xP/C where C=5.6/5.6+√v and S=bQdK/cos²Ɯ where Ɯ=25⁰, d=zm/cosƜ, Q=2×90/90+25, K=0.16x[BHN/100]².
7. Which of the following is herringbone gear?
a) Left one
b) Right one
c) Both
d) None
Answer
Answer: a [Reason:] Herringbone gear is without any groove.
8. The net axial force acting on bearing is zero in case of double helical gears while none zero in case of herringbone gears.
a) True
b) False
Answer
Answer: b [Reason:] It is zero in both the cases.
9. Helix angle of herringbone and double helical gears is relatively higher.
a) True
b) False
Answer
Answer: a [Reason:] There is no thrust force and hence higher angles are permitted.
10. A herringbone speed reducer consist of 20 teeth pinion driving a 100 teeth gear. The normal module of gear is 2mm. The face width of each half is 30mm and Lewis factor is 0.4. If permissible bending stress is 500N/mm², then calculate the beam strength.
a) 15000N
b) 12000N
c) 8000N
d) 10000N
Answer
Answer: b [Reason:] S=mbσY.
11. A herringbone speed reducer consist of 20 teeth pinion driving a 100 teeth gear. The normal module of gear is 2mm. The face width of each half is 30mm and Lewis factor is 0.4. Calculate the ratio factor Q.
a) 1.2
b) 1.4
c) 1.7
d) 1.4
Answer
Answer: c [Reason:] Q=2×100/100+20.
12. A herringbone speed reducer consist of 20 teeth pinion driving a 100 teeth gear. The normal module of gear is 2mm. The face width of each half is 30mm and Lewis factor is 0.4. Calculate the material constant K if surface hardness is 400BHN.
a) 1.25 N/mm²
b) 4.05 N/mm²
c) 3.25N/mm²
d) 2.56N/mm
Answer
Answer: c [Reason:] K=0.16x[BHN/100]².
13. Helical gears mounted on parallel shafts are called crossed helical gears.
a) True
b) False
Answer
Answer: b [Reason:] Crossed helical gears are the helical gears mounted on non-parallel shafts.
14. Crossed helical gears have very low load carrying capacity.
a) True
b) False
Answer
Answer: b [Reason:] There is point contact and hence very less area and thus wear is comparatively rapid.
15. Calculate the shaft angle for same hand of helix if helix angle of two gears are 20⁰ and 17⁰.
a) 17⁰
b) 20⁰
c) 37⁰
d) 3⁰
Answer
Answer: c [Reason:] For same hand of helix, shaft angle=sum of helix angles of two gears.
Set 2
1. The wire rope is subjected to
a) Tensile Stress
b) Bending Stress
c) Both bending and tensile stress
d) None of the listed
Answer
Answer: c [Reason:] There is direct tensile stress due to load being raised as well as bending stress.
2. In design of wire ropes, bending stress is converted into an equivalent bending load which is given by [d= diameter of individual wire and D= diameter of sheave].
a) AEd/D
b) 2AEd/D
c) AEd/2D
d) None of the listed
Answer
Answer: a [Reason:] P=σA and σ=Ed/D.
3. An elevator is assembled to raise equipment to a height of 22m. It is estimated that maximum weight of the material to be raised is 4.9kN. It is observed acceleration in such cases is 1.2m/s².10mm diameter wire ropes with fibre core are used. The tensile designation of the wire is 1500. Number of wire ropes can be taken 2. Calculate the weight of material raised by each load.
a) None of the listed
b) 2450N
c) 4900N
d) 9800N
Answer
Answer: b [Reason:] Weight of material raised by each rope= 4.9/2.
4. An elevator is assembled to raise equipment to a height of 22m. It is estimated that maximum weight of the material to be raised is 4.9kN. It is observed acceleration in such cases is 1.2m/s².10mm diameter wire ropes with fibre core are used. The tensile designation of the wire is 1500.Number of rope wires is assumed to be 2. Calculate the weight of wire if weight of 50m wire is 18kg.
a) 19.36N
b) 90.25N
c) 77.70N
d) 56.66N
Answer
Answer: c [Reason:] W=mass per unit length x length x g.
5. An elevator is assembled to raise equipment to a height of 22m. It is estimated that maximum weight of the material to be raised is 4.9kN. It is observed acceleration in such cases is 1.2m/s².10mm diameter wire ropes with fibre core are used. The tensile designation of the wire is 1500.Number of rope wires is assumed to be 2. Number of rope wires is assumed to be 2. Calculate the weight of wire if weight of 50m wire is 18kg. Calculate the force due to acceleration.
a) 1000N
b) 309.144N
c) 504.225N
d) 102.225N
Answer
Answer: b [Reason:] Mass of material raised by each wire rope= 2450/9.81 = 249.7kg. Mass of each wire rope= 77.7/9.81 =7.92kg. Force due to acceleration= [249.7+7.92] x 1.2.
6. An elevator is assembled to raise equipment to a height of 22m. It is estimated that maximum weight of the material to be raised is 4.9kN. It is observed acceleration in such cases is 1.2m/s².10mm diameter wire ropes with fibre core are used. The tensile designation of the wire is 1500.Number of rope wires is assumed to be 2. Number of rope wires is assumed to be 2. Calculate the total load on wire rope neglecting the bending load if weight of 50m wire is 18kg.
a) 1054.55N
b) 3504.55N
c) 2836.84N
d) 5678.6N
Answer
Answer: c [Reason:] F=4900/2 + 77.7 + 249.7×1.2 + 7.92×1.2.
7. Small sheaves are preferred over large sheaves on what parameters?
a) Increasingly centrifugal force in large sheaves
b) Higher cost of small sheaves
c) Preferring centrifugal force reduction even for more money
d) None of the listed
Answer
Answer: a [Reason:] Larger sheaves have higher centrifugal forces and higher cost because more material is used in their construction.
8. Rope operating over steel sheaves wear slowly than those used in conjunction with cast iron sheaves.
a) True
b) False
Answer
Answer: b [Reason:] Rope operating over steel sheaves wear faster. If wear on cast iron sheave is 100, than wear on steel sheave is 110.
9. Which type of rope drum is preferred?
a) Drums with helical grooves
b) Plain cylindrical drums
c) Both are equally preferred
d) None of the listed
Answer
Answer: a [Reason:] They have more bearing surface of the drum and prevent friction between adjacent turns of the rope.
10. If nominal diameter of the rope is 20mm, then pitch of the groove is around?
a) 2mm
b) 22mm
c) 40mm
d) 10mm
Answer
Answer: b [Reason:] p= d + 2 mm.
Set 3
1. There are ____ standard systems for the shape of gear teeth.
a) 1
b) 2
c) 3
d) 4
Answer
Answer: c [Reason:] 14.5⁰ full depth involute system, 20⁰ full depth involute system and 20⁰ stub involute system.
2. When the number of teeth reaches infinity, circle radius approaches infinity the gear becomes an infinite loop.
a) True
b) False
Answer
Answer: b [Reason:] It becomes a rack with straight sided teeth.
3. Which of the following statements are not true?
a) Increasing pressure angle improves the tooth strength
b) Contact duration is decreased with increase in pressure angle
c) 20⁰ pressure angle has quieter operation then 14.5⁰
d) All of the statements are true
Answer
Answer: c [Reason:] Lower the pressure angle, quieter is the operation. Lower the pressure angle, lower is the breadth of the tooth at root.
4. 20⁰ stub involute system have comparatively smaller interference.
a) True
b) False
Answer
Answer: a [Reason:] They have shorter addendum and shorter dedendum.
5. Which of the following have stronger teeth?
a) Stub teeth
b) Full depth teeth
c) Both have equal strength
d) Can’t be determined
Answer
Answer: a [Reason:] Smaller moment arm of bending force leads to stronger stub teeth.
6. As the module increases, index of size of gear decreases.
a) True
b) False
Answer
Answer: b [Reason:] Module is the measure of size of index of the gear tooth.
7. Crowning is an abrasive process that debars the gear strength.
a) True
b) False
Answer
Answer: b [Reason:] Crowningis used to strengthen the tooth.
8. Inaccuracies in tooth profile lead to concentration of pressure on the middle of tooth.
a) True
b) False
Answer
Answer: b [Reason:] Inaccuracies lead to shift of pressure at the end of tooth which can be improved by crowning.
9. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the centre distance.
a) 280mm
b) 269mm
c) 350mm
d) 305mm
Answer
Answer: a [Reason:] C=m(z(p)+z(g))/2.
10. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate pitch circle diameter of the pinion.
a) 95mm
b) 105mm
c) 115mm
d) 125mm
Answer
Answer: d [Reason:] D=5×25.
11. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the pitch circle diameter of the gear.
a) Cannot be determined
b) 31mm
c) 475mm
d) 575mm
Answer
Answer: d [Reason:] D=5×115.
12. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the addendum.
a) None of the listed
b) 4.75mm
c) 5.25mm
d) 5mm
Answer
Answer: d [Reason:] H=m=5mm.
13. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the dedendum.
a) 4.75mm
b) 5mm
c) 6.25mm
d) 6.68mm
Answer
Answer: c [Reason:] H=1.25xm=1.25x5mm.
14. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the tooth thickness.
a) 6.23mm
b) 5.44mm
c) 7.854mm
d) 8.16mm
Answer
Answer: c [Reason:] T=1.5708xm=1.5708x5mm.
15. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the bottom clearance.
a) None of the listed
b) 1.75mm
c) 2.5mm
d) 1.25mm
Answer
Answer: d [Reason:] C=0.25m=0.25x5mm.
Set 4
1. If pitch angle and addendum angles are 5⁰ and 12⁰ respectively, then face angle is equal to?
a) 17⁰
b) 7⁰
c) 5⁰
d) 12⁰
Answer
Answer: a [Reason:] Face angle=pitch angle+ addendum angle.
2. If pitch angle is 8⁰ and dedendum angle is 4⁰, then find root angle.
a) 12⁰
b) 4⁰
c) 8⁰
d) None of the listed
Answer
Answer: b [Reason:] Root angle=pitch angle-dedendum angle.
3. If back cone distance is 12mm and module at large end of the tooth is 4mm, then formative number of teeth will be?
a) 3
b) 6
c) 4
d) 12
Answer
Answer: b [Reason:] Formative number=2r/m.
4. If back cone distance is 12mm and module at large end of the tooth is 4mm, and virtual number of teeth is 12 then find the diameter of the tooth.
a) 5
b) 4
c) 3
d) 2
Answer
Answer: d [Reason:] 12/z = 2×12/4.
5. Calculate the cone distance of in a pair of bevel gears if pitch circle diameter of pinion and gear are 20mm and 24mm respectively.
a) 44mm
b) 22mm
c) 15.6mm
d) 20.2mm
Answer
Answer: c [Reason:] A=√(20/2)²+(24/2)².
6. Calculate the pitch angle if pitch circle diameter of the pinion and gear are 150mm and 210mm.
a) 28.14⁰
b) 35.54⁰
c) 36.22⁰
d) 63.22⁰
Answer
Answer: b [Reason:] tanϒ=D(p)/D(g)=150/210.
7. Calculate the radius of pinion at midpoint along the face width if PCD of pinion is 150mm and of gear is 210mm. Also face width of the tooth is 35mm.
a) 56.35mm
b) 64.83mm
c) 66.57mm
d) 58.69mm
Answer
Answer: b [Reason:] r=(Dp/2)-(bsinϒ/2).
8. Calculate the tangential component of gear tooth force if power transmitted is 6kW and diameters of pinion and gear are 150mm and 210 mm with face width of tooth being 35mm. Power is transmitted at 3000rpm.
a) 1668N
b) 2946N
c) 3000N
d) 3326N
Answer
Answer: b [Reason:] P=M/r. r=(Dp/2)-(bsinϒ/2) where ϒ is ptch angle and is calculated by tanϒ=D(p)/D(g)=150/210.
9. Calculate the radial component of gear tooth force if power transmitted is 6kW and diameters of pinion and gear are 150mm and 210 mm with face width of tooth being 35mm. Power is transmitted at 3000rpm. Also pressure angle is 20⁰.
a) 996.6N
b) 332.6N
c) 489.2N
d) 739.2N
Answer
Answer: d [Reason:] P radial=P tangential x[tan20 Cos ϒ].
10. Calculate the axial component of gear tooth force if power transmitted is 6kW and diameters of pinion and gear are 150mm and 210 mm with face width of tooth being 35mm. Power is transmitted at 3000rpm. Also pressure angle is 20⁰.
a) 660.05N
b) 528.06N
c) 448.21N
d) 886.6N
Answer
Answer: b
Explanaton: P radial=P tangential x[tan20 Sin ϒ].
11. The pinion of a face gear is a
a) Spur gear
b) Helical gear
c) Either spur or helical
d) None of the mentioned
Answer
Answer: c [Reason:] Face gear consist of a spur or helical gear mating with a conjugate gear of disk form.
Set 5
1. Viscosity is defined as the external resistance offered by a fluid to change its shape or relative motion of its parts.
a) Yes
b) It is internal resisting force
c) It is not offered but exerted on the fluids
d) None of the listed
Answer
Answer: b [Reason:] It is an internal resisting force.
2. Stream line flow happens when intermediate layers move with velocities proportional to the square of distance from the stationary plate.
a) True
b) False
Answer
Answer: b [Reason:] It is proportional to distance and not square of it.
3. Newton law of viscosity states that shear stress is proportional to rate of shear at any point in the fluid.
a) True
b) False
Answer
Answer: a [Reason:] P/A proportional to U/h.
4. Calculate the kinematic viscosity if Saybolt viscosity is 400cSt.
a) 400SUS
b) 40.25SUS
c) 86.32SUS
d) 87.55SUS
Answer
Answer: d [Reason:] z=0.22t-[180/t] where t=400.
5. Viscosity of lubricating oil decrease with increasing temperature.
a) Yes
b) It increases linearly
c) It increases hyperbolically
d) it remains constant
Answer
Answer: a [Reason:] Intermolecular forces decrease on the increase of temperature.
6. Which of the following lubricant has least rate of change of viscosity w.r.t temperature.
a) VI=20
b) VI=30
c) VI=40
d) VI=50
Answer
Answer: d [Reason:] Greater the VI, lesser is the rate of change w.r.t temperature.
7. Which of the following are not true for petroff’s equation?
a) Shaft is considered concentric with the bearing
b) Bearing is subjected to light load
c) Is used to find coefficient of friction
d) Frictional torque is given by fpr²l
Answer
Answer: d [Reason:] M=fWr=f(2prl)r, W=projected area of bearing x pressure.
8. In hydrodynamic lubrication, film thickness remains unaffected by change in speeds.
a) True
b) Increase with increase in speed
c) Decrease with increase in speed
d) Disappear as the speed tends to infinty
Answer
Answer: b [Reason:] As speed increases more and more lubricant is forces and pressure builds up thus separating the two surfaces. There is transition from thin film thick film.
9. For a hydrostatic thrust bearing,
Thrust load=450kN, shaft speed=730rpm, shaft diameter=450mm, recess diameter=310mm, film thickness=0.15mm, viscosity of lubricant=160SUS and specific gravity=0.86.
Calculate supply pressure
a) 10.2Pa
b) 4.01Pa
c) 4.01Mpa
d) 10.2Mpa
Answer
Answer: c [Reason:] P=2Wln(225/155)/[π(225²-155²)] N/mm².
10. For a hydrostatic thrust bearing,
Thrust load=450kN, shaft speed=730rpm, shaft diameter=450mm, recess diameter=310mm, film thickness=0.15mm, viscosity of lubricant=160SUS and specific gravity=0.86.
Calculate flow requirement
a) 0.89l/s
b) 38.94l/min
c) 28.8l/min
d) None of the mentioned
Answer
Answer: b [Reason:] Q=πPhᵌ/6µln(225/155) whereµ=z/10⁹ and z=0.86x[0.22×160-180/160]. µ=29.3 x 10¯⁹N-s/mm².
11. For a hydrostatic thrust bearing,
Thrust load=450kN, shaft speed=730rpm, shaft diameter=450mm, recess diameter=310mm,film thickness=0.15mm,viscosity of lubricant=160SUS and specific gravity=0.86.
Calculate power loss in pumping.
a) 2.68kW
b) 3.35kW
c) 2.6kW
d) 4.2kW
Answer
Answer: c [Reason:] kW=Q(P-0)x10¯⁶.
12. For a hydrostatic thrust bearing, Thrust load=450kN, shaft speed=730rpm, shaft diameter=450mm, recess diameter=310mm,film thickness=0.15mm,viscosity of lubricant=160SUS and specific gravity=0.86.
Calculate frictional power loss.
a) None of the listed
b) 2.3kW
c) 3.56kW
d) 4.2kW
Answer
Answer: c [Reason:] kW=µn²(225⁴-155⁴)/hx58.05×10⁶.