MCQ on Bevel Gears, Worm Gears and FlyWheel in Machine Design
1. There are ____ common types of bevel gears.
a) 1
b) 2
c) 3
d) 4
.
Answer: b
Explanation: Spiral and straight.
2. Straight bevel gears are easy to design and manufacture and give reasonably good service with quieter operation.
a) True
b) No, noise problem
c) No, vibration problem
d) None of the listed
.
Answer: b
Explanation: They produce a lot of noise while working.
3. Which of the following creates smoother motion?
a) Straight bevel gears
b) Spiral bevel gears
c) Equal for straight and spiral
d) None of the mentioned
.
Answer: b
Explanation: Spiral bevel gear has smooth engagement which results in quieter operation.
4. When the pitch angle is greater than 90⁰, it is called external bevel gear.
a) True
b) Internal
c) Helical
d) Herringbone
.
Answer: b
Explanation: For external bevel gear, pitch angle is less than 90⁰.
5. For crown bevel gear, pitch angle is
a) <90 b) =90 c) >90
d) None of the listed
.
Answer: b
Explanation: Crown gears are characterised by pitch angle of 90⁰.
6. If pitch angle is >90, the bevel gear is?
a) Internal
b) External
c) Crown
d) Static
.
Answer: a
Explanation: Nomenclature.
7. Miter gears are the bevel gears mounted on shaft which are intersecting at angle greater than 90.
a) True
b) False
.
Answer: b
Explanation: The shafts which are intersecting at 90.
8. Crown gears having a pitch angle of 90⁰ are mounted on shafts intersecting at an angle
a) =90
b) <90 c) >90
d) None of the listed
.
Answer: c
Explanation: Application of crown gear according to its structure.
9. Which of the following are characteristics of skew bevel gears?
a) Straight teeth
b) Mounted on parallel shafts
c) Mounted on intersecting shafts
d) All of the mentioned
.
Answer: a
Explanation: They are used in non-parallel and non-intersecting shafts.
10. Hypoid gears are based on surfaces that are paraboloids of revolution.
a) True
b) False
.
Answer: b
Explanation: They are hyperboloids of revolution.
11. When two hyperboloids are rotated, the resulting motion is
a) Sliding
b) Turning
c) Combination of turning and sliding
d) Rotary
.
Answer: c
Explanation: There is turning as well as sliding motion.
12. In hypoid bevel gears, shafts may continue past each other.
a) True
b) False
.
Answer: a
Explanation: Offset of the shaft is quite considerable in case of hypoid gears.
13. Hypoid gears have the following characteristics.
a) Curved teeth
b) Mounted on non -parallel non intersecting shafts
c) Efficiency of 97%
d) All of the mentioned
.
Answer: d
Explanation: It has curved teeth and are mounted on non-parallel non intersecting shafts with efficiency of 96-98%.
14. Zerol gears are straight bevel gears with zero spiral angle.
a) True
b) False
.
Answer: b
Explanation: They happen to fall in the category of spiral bevel gears.
15. Zerol gears give lesser contact ratio.
a) True
b) No, larger contact ratio
c) Zero contact
d) None of the listed
.
Answer: b
Explanation: Zerol gears have more gradual contact and slighter larger contact ratio.
16. 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: a
Explanation: Face angle=pitch angle+ addendum angle.
17. 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: b
Explanation: Root angle=pitch angle-dedendum angle.
18. 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: b
Explanation: Formative number=2r/m.
19. 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: d
Explanation: 12/z = 2×12/4.
20. 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: c
Explanation: A=√(20/2)²+(24/2)².
21. 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: b
Explanation: tanϒ=D(p)/D(g)=150/210.
22. 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: b
Explanation: r=(Dp/2)-(bsinϒ/2).
23. 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: b
Explanation: P=M/r. r=(Dp/2)-(bsinϒ/2) where ϒ is ptch angle and is calculated by tanϒ=D(p)/D(g)=150/210.
24. 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: d
Explanation: P radial=P tangential x[tan20 Cos ϒ].
25. 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: b
Explanation: P radial=P tangential x[tan20 Sin ϒ].
26. 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: c
Explanation: Face gear consist of a spur or helical gear mating with a conjugate gear of disk form.
27. Lewis equation is used to obtain ____ strength of bevel gears.
a) Beam
b) Abrasive
c) Wear
d) Corrosive
.
Answer: a
Explanation: According to Lewis equation, Beam strength=product of module, face width of elemental section, permissible bending stress and Lewis form factor.
28. Beam strength of the bevel gear is independent of the cone distance.
a) True
b) False
.
Answer: b
Explanation: In other form S=mbσY[1- b/A] where A= cone distance.
29. Beam strength indicates the maximum value of the ___________ force at the large end of the tooth that the tooth can transmit without bending.
a) Tangential
b) Radial
c) Axial
d) None of the listed
.
Answer: a
Explanation: Beam strength is analysed by using the pitch radius at the larger end of the tooth.
30. Wear strength of the bevel gear can be calculated by using ________ equation.
a) Buckingham
b) Lewis
c) Newtonian
d) Rayleigh
.
Answer: a
Explanation: Bevel gear is considered to be equivalent to a formative spur gear in a plane which is perpendicular to the large end and hence Buckingham equation is applied.
31. If back cone distance is 10mm, then pitch circle diameter of the formative pinion is given by?
a) 10mm
b) 20mm
c) 5mm
d) 10√2 mm
.
Answer: b
Explanation: PCD=2xr.
32. The wear strength indicates the maximum value of radial force at the large end of the tooth that the tooth can transmit without pitting failure.
a) True
b) False
.
Answer: b
Explanation: It indicates the maximum value of tangential force.
33. If velocity is 5m/s, then velocity factor for a cut teeth is
a) 0.55
b) 0.66
c) 1.55
d) 1.66
.
Answer: a
Explanation: C=6/6+v.
34. If velocity is 5m/s, then velocity factor for generated teeth is
a) 0.71
b) 1.1
c) 0.9
d) 1.71
.
Answer: a
Explanation: C=5.6/5.6+√v.
35. Calculate the ratio factor Q if a pair of bevel gear consist of 25 teeth pinion meshing with a 40 teeth gear.
a) 1.964
b) 1.438
c) 1.554
d) 0.998
.
Answer: b
Explanation: tanϒ=25/40 or ϒ=32⁰. Q=2×40/ [40+25xtan(32)].
36. If surface hardness for a par of bevel gears is 400BHN, then material constant is
a) 3N/mm²
b) 2.56N/mm²
c) 0.98N/mm²
d) 1.44N/mm²
.
Answer: b
Explanation: K=0.16x[BHN/100]².
37. Calculate the wear strength for a pair of bevel gears having face width=20mm, module=5mm, No of teeth on pinion and gear 25 and 40 respectively and PCD of pinion=75mm.
a) 3668.5N
b) 4884.5N
c) 5126.6N
d) 4117.3N
.
Answer: b
Explanation: S=0.75xbxQxDxK/Cos ϒ. Tan ϒ=25/40 or ϒ=32⁰. Q=2×40/ [40+25xtan(32)] K=0.16x[BHN/100]².
38. If service factor is 1.4 & tangential ad dynamic load is 1180N and 1500N respectively, then calculate the effective load.
a) 3662N
b) 2889N
c) 3152N
d) 2236N
.
Answer: c
Explanation: Effective load=1.4×1180 + 1500.
39. Worm gear drives are used to transmit power between two non-intersecting shafts which are generally at right angles to each other.
a) True
b) False
.
Answer: a
Explanation: This is how the worm gear design fits into use.
40. The worm and worm wheel both are threaded screw.
a) True
b) Worm wheel is a toothed gear
c) Worm is a toothed gear
d) None of the listed
.
Answer: b
Explanation: Worm wheel is a toothed gear.
41. Which of the following is not true about worm gears?
a) Compact
b) Smooth and silent operation
c) Low speed reduction
d) All the mentioned are true
.
Answer: c
Explanation: Speed reduction can be high up to 100:1.
42. Is it possible to use worm gears in cranes for lifting purpose?
a) True
b) No self-locking and hence not possible
c) Possible up to a threshold load
d) None of the listed
.
Answer: a
Explanation: Worm gears support self-locking operation and hence are advantageous to use in lifting operations.
43. The power transmitting capacity of worm gears is high although efficiency is low.
a) True
b) False
.
Answer: b
Explanation: Both power transmitting capacity and efficiency of worm gears are low.
44. Can worm gears be used in steering mechanism?
a) True
b) False
.
Answer: a
Explanation: In steering mechanism, efficiency is of little importance but major requirement is of large mechanical advantage.
45. The worm helix angle is the _____ of the worm lead angle.
a) Complement
b) Half
c) Double
d) Supplement
.
Answer: a
Explanation: Worm helix angle+worm lead angle=90⁰.
46. If worm helix angle is 30⁰, then worm should have at least ___ threads.
a) 5
b) 6
c) 7
d) 8
.
Answer: a
Explanation: The permissible helix angle is 6⁰ and hence there should be at least five threads i.e. 30/6.
47. A pair of worm gear is written as 2/40/12/6. Calculate the centre distance.
a) 40mm
b) 156mm
c) 200mm
d) 80mm
.
Answer: b
Explanation: C=m(q+z)/2 where m=6mm, q=12 and z=40.
48. A pair of worm gear is written as 2/40/12/6. Calculate the speed reduction.
a) 2
b) 20
c) 15
d) 6
.
Answer: b
Explanation: i=40/2.
49. A pair of worm gear is written as 2/40/12/6. Calculate the pitch circle diameter of worm wheel.
a) 72mm
b) 240mm
c) 260mm
d) 320mm
.
Answer: b
Explanation: d=mxz where m=6mm and z=40.
50. A pair of worm gear is written as 2/40/12/6. Calculate the throat diameter of the worm wheel.
a) 220.5mm
b) 246.4mm
c) 190.44mm
d) 251.7mm
.
Answer: d
Explanation: d(t)=m[z+4cosϒ-2] where ϒ=9.46⁰ is the lead angle. tanϒ=2/12, z=40 and m=6mm.
51. A pair of worm gear is written as 2/40/12/6. Calculate the root diameter of the worm wheel.
a) 186.22mm
b) 250.4mm
c) 225.6mm
d) 250.44mm
.
Answer: c
Explanation: d=m[z-2-0.4cosϒ] where ϒ=9.46⁰ is the lead angle. tanϒ=2/12, z=40 and m=6mm.
52. If tangential force on worm is 1500N, then axial force on worm wheel will be?
a) 1500N
b) 3000N
c) 1500√2 N
d) 750N
.
Answer: a
Explanation: P₂(axial)=P₁(tangential).
53. Is use of a flywheel recommended when a large motor is required only for a small instant of time?
a) True
b) False
.
Answer: a
Explanation: Fly wheel allows the large motor to be replaced by a smaller motor as the peak power is required only for a small instance of time and not through the entire operation.
54. Flywheels are used in punching and shear operation.
a) True
b) False
Answer: a
Explanation: Flywheels store the kinetic energy imparted during idle positions and delivers this energy during actual shearing or punching.
55. Which of the following are functions of flywheel?
a) Store and release energy during work cycle
b) Reduce power capacity of the electric motor
c) Reduce amplitude of speed fluctuations
d) All of the listed
.
Answer: d
Explanation: Flywheel provides uniform motion by storing energy during idle positions and using this at actual operational time and thus reducing the power capacity of the electric motor.
56. When comes down to stress reduction, which one is preferred?
a) Solid flywheel
b) Split flywheel
c) Both have equal stresses
d) Cannot be determined
.
Answer: b
Explanation: The arms are free in split flywheel to contract and hence are better for stress reduction.
57. Flywheel and governor can be interchanged.
a) True
b) False
.
Answer: b
Explanation: Flywheel influences cyclic speed fluctuations while governor control mean speed.
58. If load on the engine is constant, the mean speed will be constant and ___ will not operate.
a) Flywheel
b) Governor
c) Both flywheel and governor
d) None of the mentioned
.
Answer: b
Explanation: Governor is used to control the mean speed and if mean speed is held constant then the governor will not operate.
59. The operation of flywheel is continuous.
a) True
b) False
.
Answer: a
Explanation: Flywheel is used to control the fluctuations and thus is continuously working.
60. Which of the following doesn’t waste energy?
a) Flywheel
b) Governor
c) Both flywheel and governor
d) Neither flywheel nor governor
.
Answer: a
Explanation: Flywheel has kinetic energy which is 100% convertible while governor suffer from friction losses.
61. Which of the following is not true for cast iron flywheels?
a) Excellent damping
b) Cheap
c) Given complex shape
d) Sudden failure
.
Answer: d
Explanation: Due to low tensile strength failure is sudden.
62. When the driving torque is more than load torque, flywheel is ______
a) Accelerated
b) Decelerated
c) Constant velocity
d) Can’t be determined
.
Answer: a
Explanation: I[dω/dt]=Driving torque-load torque.
63. Feather key can be used to prevent axial motion between two elements.
a) True
b) False
.
Answer: b
Explanation: All types of key but feather key can be used to prevent axial motion between two machine elements.
64. Keyed joints never lead to stress concentration on shafts.
a) True
b) False
.
Answer: b
Explanation: Keyway results in stress concentration in the shaft and makes the part weak.
