Rolling Contact Bearings, Spur Gears and Helical Gears in Machine Design:
Q. Which of the following are functions of bearings?
A. Ensure free rotation of shaft with minimum friction
B. Holding shaft in a correct position
C. Transmit the force of the shaft to the frame
D. All of the listed
.
Answer: D
Explanation: Bearings are used for all the above listed purposes.
Q. A radial bearing supports the load that acts along the axis of the shaft.
A. True
B. False
.
Answer: B
Explanation: Radial bearing supports the load acting perpendicular to the axis of the shaft.
Q. A_______ bearing supports the load acting along the axis of the shaft.
A. Thrust
B. Radial
C. Longitudinal
D. Transversal
.
Answer: A
Explanation: Thrust bearing supports load acting along axis of shaft.
Q. Sliding contact bearings, also called plain bearings have no problem of wear.
A. True
B. False
.
Answer: B
Explanation: Surface of shaft slide over surface of the bush resulting in friction and wear.
Q. In steam and gas turbines, rolling contact bearings are used.
A. True
B. False
.
Answer: B
Explanation: Sliding contact bearings are generally used.
Q. Which of the following are true about plasticity?
A. Permanent Deformation
B. Ability to retain deformation under load or after removal of load
C. Plastic deformation is greater than elastic deformation
D. All of the mentioned
.
Answer: D
Explanation: This is the basic definition of plasticity.
Q. Which of the following is measure of stiffness?
A. Modulus of elasticity
B. Modulus of plasticity
C. Resilience
D. Toughness
.
Answer: A
Explanation: Stiffness is the ability of material to resist deformation under external load. Hence it is measured by modulus of elasticity.
Q. Which of the following facts are true for resilience?
A. Ability of material to absorb energy when deformed elastically
B. Ability to retain deformation under the application of load or after removal of load
C. Ability of material to absorb energy when deformed plastically
D. None of the mentioned
.
Answer: A
Explanation: Toughness is ability to store energy till proportional limit during deformation and to release this energy when unloaded.
Q. Modulus of resilience is defined as
A. Strain energy per unit volume
B. Strain energy per unit area
C. Independent of strain energy
D. None of the mentioned
.
Answer: A
Explanation: Modulus of resilience is strain energy per unit volume.
Q. In gear boxes and small size motors, rolling contact bearings are used.
A. True
B. False
.
Answer: A
Explanation: In small size applications, rolling contact bearings are preferred.
Q. Deep groove ball bearings creates a lot of noise.
A. Yes
B. They create very less noise
C. Depends on the application
D. No reference frame for comparison is mentioned
.
Answer: B
Explanation: They create very less noise due to point contact.
Q. There is problem of alignment in deep groove ball bearings.
A. Yes
B. No, it is self-aligning
C. It aligns itself only in some particular cases
D. Can’t be determined
.
Answer: A
Explanation: It is not self-aligning.
Q. Deep groove ball bearing has immense rigidity.
A. True
B. No it has point contact and hence low rigidity
C. It has surface contact
D. It has line contact
.
Answer: B
Explanation: Due to point contact, rigidity is not so good.
Q. Cylindrical load bearing has lower load capacity as compared to deep groove ball bearing.
A. True
B. False
.
Answer: B
Explanation: Cylindrical load bearing has a line contact and hence higher load capacity.
Q. Angular contact bearing can take thrust as well as radial loads.
A. True
B. False
.
Answer: A
Explanation: The line of reaction at the contact surfaces makes an angle with axis of bearing and thus has two components, hence allowing it to take both type of loads.
Q. In angular contact bearings, ____ bearings are required to take thrust load in both directions.
A. 1
B. 4
C. 2
D. 3
.
Answer: C
Explanation: Angular contact bearing has this disadvantage of needing two bearings.
Q. The angular play bearing must be mounted without axial play.
A. Yes
B. Little tolerance is adjusted
C. Little tolerance is necessary
D. Can’t be stated
.
Answer: A
Explanation: There is no tolerance of misalignment.
Q. Taper rolling supports
A. Axial loads
B. Thrust loads
C. Both axial and thrust loads
D. None of the mentioned
.
Answer: C
Explanation: The line of reaction makes an angle with the axis of bearing and hence both type of loads can be carried.
Q. Which of the following isn’t the property of taper roller?
A. High rigidity
B. Easy dismantling
C. Take low radial and heavy loads
D. All are the properties of tapper roller
.
Answer: C
Explanation: Due to line contact it can take high radial and thrust loads.
Q. Which of the following cannot take radial load?
A. Cylindrical Roller bearing
B. Taper roller bearing
C. Thrust ball bearing
D. None of the listed
.
Answer: C
Explanation: There is no inclination in the line of reaction and hence only thrust loads can be carried.
Q. Which of the following cannot tolerate misalignment?
A. Angular contact bearing
B. Cylinder roller bearing
C. Thrust ball bearings
D. All of the listed
.
Answer: D
Explanation: All of these require précised alignment.
Q. Cylinder roller creates lesser noise than deep groove ball bearing.
A. True
B. False
.
Answer: B
Explanation: Due to line contact, cylinder roller creates far more noise
Q. Static load is defined as the load acting on the bearing when shaft is _____
A. Stationary
B. Rotating at rpm<10
C. Rotating at rpm<5
D. None of the listed
.
Answer: A
Explanation: As name suggests, static load means load during stationary position of the shaft.
Q. A total permanent deformation of 0.0001 of the ball diameter is taken for considering static load capacity of the shaft.
A. True
B. False
.
Answer: A
Explanation: Permissible magnitude of deformation is set as per the experimental data.
Q. Stribeck equation gives dynamic load capacity of the bearing.
A. True
B. False
.
Answer: B
Explanation: It gives static load capacity of the bearing.
Q. Which of the following is expression for stribeck equation?(Number of balls=z)?
A. C=kd²z/5
B. C=kd²z/15
C. C=kd²z/10
D. None of the listed
.
Answer: A
Explanation: C=(1/5)zP and P=kd².
Q. The life of an individual ball bearing is the time period for which it works without any signs of failures.
A. True
B. False
.
Answer: B
Explanation: It is measured by number of revolutions and not by time.
Q. For majority of bearings, actual life is considerably greater than rated life.
A. True
B. False
.
Answer: A
Explanation: Stastically it has been proved that life which 50% of a group of bearings will exceed is approximately five times rating of rated life.
Q. The dynamic load carrying capacity of a bearing is defined as the radial load in radial bearings that can be carried for a minimum life of 1000 revolutions.
A. True
B. False
.
Answer: B
Explanation: It is for 1 million revolutions.
Q. In the expression of dynamic load capacity P=XVF(r) + YF(A., V stands for ?
A. Race rotation factor
B. Radial factor
C. Thrust factor
D. None of the listed
.
Answer: A
Explanation: It represent race rotation factor which depends upon whether inner race is rotating or outer race.
Q. Calculate the bearing life if expected life for 90% bearings is 9000h and shaft is rotating at 1500rpm.
A. 850 million rev
B. 810 million rev
C. 810 h
D. 850h
.
Answer: B
Explanation: L=60x1500x9000/10⁶ million rev.
Q. The bearing is subjected to a radial load of 4000N. Expected life for 90% bearings is 9000h and shaft is rotating at 1500rpm. Calculate the dynamic load capacity.
A. 4Q.21kN
B. 3Q.29kN
C. 2Q.33kN
D. 3Q.22kN
.
Answer: B
Explanation: C=4000x(810)⅓.
Q. A rolling contact bearing is specified as X30Q. Calculate the bearing diameter.
A. 35mm
B. 28mm
C. 21mm
D. 7mm
.
Answer: A
Explanation: Diameter=5 times the last two digits.
Q. A rolling contact bearing is specified as X30Q. Determine the series.
A. Extra light series
B. Light series
C. Medium series
D. Heavy series
.
Answer: C
Explanation: 1- Extra light series,2-Light series,3-Medium series,4-heavy series.
Q. Needles bearing consist of rollers of small diameter and very small length.
A. True
B. False
.
Answer: B
Explanation: Also called as quill bearings, they consist of small diameter and comparatively longer b=rollers.
Q. Extreme pressure causes ______ wear in the bearing parts.
A. Abrasive
B. Corrosive
C. Pitting
D. All of the mentioned
.
Answer: B
Explanation: Extreme pressure elements in EP additives are added in lubricating oils.
Q. Scoring is a ________ phenomenon.
A. Stick-slip
B. Fracture
C. Fatigue
D. In-out
.
Answer: A
Explanation: Alternative welding and shearing takes place.
Q. Thick film lubrication describes a phenomenon where two surfaces are _______ separated.
A. Completely
B. Partially
C. Not
D. None of the mentioned
.
Answer: A
Explanation: There is no contact between the two surfaces.
Q. Hydrodynamic bearing is a self-acting bearing.
A. True
B. False
.
Answer: A
Explanation: The pressure in the bearing is created within the system due to rotation of the shaft.
Q. A journal bearing is a ______ contact bearing working on the hydrodynamic lubrication and which supports load in____ direction.
A. Sliding, Axial
B. Rolling, Radial
C. Sliding, Radial
D. Rolling, Axial
.
Answer: C
Explanation: This is how journal bearing works. It derives its name from the portion of the shaft inside the bearing.
Q. Partial bearing is preferred over journal bearing.
A. True
B. No
C. More friction losses
D. Can’t be determined
.
Answer: A
Explanation: Friction losses are lesser and construction is simple.
Q. Temperature rise in partial bearing is ____ than full bearing.
A. Lesser
B. Greater
C. Equal
D. Undeterminable
.
Answer: A
Explanation: Due to lesser friction, temperature rise is less.
Q. A clearance bearing is design accurately to keep the radius of journal and bearing equal.
A. Journal radius is kept larger
B. Journal radius is kept smaller
C. True
D. Can’t be determined
.
Answer: B
Explanation: Journal radius is kept smaller.
Q. Fitted bearing must be partial bearing.
A. True
B. No
C. No lubricating space is required
D. Can’t be stated
.
Answer: A
Explanation: To provide space for lubricating oil.
Q. Footstep bearing is an axial load bearing.
A. True
B. Thrust load
C. Shear load
D. None of the listed
.
Answer: B
Explanation: It is a thrust bearing in which shaft end is in contact with bearing surface.
Q. Hydrostatic and hydrodynamic lubrication are the same thing.
A. True
B. False
.
Answer: B
Explanation: In hydrodynamic, motion is provided by the shaft while in hydrostatic the motion is provided by an external source.
Q. If we exclude the cost factor, which bearing is preferred?
A. Hydrostatic
B. Hydrodynamic
C. Both are equally preferred
D. Cannot be determined
.
Answer: A
Explanation: High load capacity, no starting friction and no rubbing action.
Q. If fluid film pressure is high and surface rigidity is low than mode of lubrication is called as elastohydrodynamic lubrication.
A. True
B. False
.
Answer: A
Explanation: In such cases, elastic deflections of the parts causes development of the film.
Q. 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: B
Explanation: It is an internal resisting force.
Q. 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: B
Explanation: It is proportional to distance and not square of it.
Q. 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: A
Explanation: P/A proportional to U/h.
Q. Calculate the kinematic viscosity if Saybolt viscosity is 400cSt.
A. 400SUS
B. 40.25SUS
C. 8Q.32SUS
D. 8Q.55SUS
.
Answer: D
Explanation: z=0.22t-[180/t] where t=400.
Q. Viscosity of lubricating oil decrease with increasing temperature.
A. Yes
B. It increases linearly
C. It increases hyperbolically
D. it remains constant
.
Answer: A
Explanation: Intermolecular forces decrease on the increase of temperature.
Q. 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: D
Explanation: Greater the VI, lesser is the rate of change w.r.t temperature.
Q. 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: D
Explanation: M=fWr=f(2prl)r, W=projected area of bearing x pressure.
Q. 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: B
Explanation: 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.
Q. 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=0.8Q.
Calculate supply pressure
A. Q.2Pa
B. Q.01Pa
C. Q.01Mpa
D. Q.2Mpa
.
Answer: C
Explanation: P=2Wln(225/155)/[π(225²-155²)] N/mm².
Q. 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=0.8Q.
Calculate flow requirement
A. 0.89l/s
B. 3Q.94l/min
C. 2Q.8l/min
D. None of the mentioned
.
Answer: B
Explanation: Q=πPhᵌ/6µln(225/155) whereµ=z/10⁹ and z=0.86x[0.22×160-180/160]. µ=2Q.3 x 10¯⁹N-s/mm².
Q. 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=0.8Q.
Calculate power loss in pumping.
A. Q.68kW
B. Q.35kW
C. Q.6kW
D. Q.2kW
.
Answer: C
Explanation: kW=Q(P-0)x10¯⁶.
Q. 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=0.8Q.
Calculate frictional power loss.
A. None of the listed
B. Q.3kW
C. Q.56kW
D. Q.2kW
.
Answer: C
Explanation: kW=µn²(225⁴-155⁴)/hx5Q.05×10⁶.
Q. The fundamental equation for viscous flow is given by ∆pbhᵌ/6µl where symbols have their usual meanings.
A. 6 is to be replaced by 3
B. 6 is to be replaced by 12
C. 6 is to be replaced by 9
D. 6 is to be replaced by 15
.
Answer: B
Explanation: It is givenby ∆pbhᵌ/12µl.
Q. A single transverse weld is preferred over double transverse fillet weld.
A. True
B. False
.
Answer: B
Explanation: A single transverse weld is not preferred because the edge of the plate which is not welded can warp out of shape.
Q. Transverse fillet weld can be designed using the same equations as of parallel fillet weld.
A. True
B. False
.
Answer: B
Explanation: Vice versa is true as strength of transverse fillet weld is greater than that of parallel fillet weld.
Q. The common normal to the curves of the two teeth must not pass through the pitch point.
A. True
B. It must pass
C. It may or may not pass
D. None of the listed
.
Answer: B
Explanation: The common normal must pass through the point where two mating gears meet.
Q. Which of the following can be used for power transmission in intersecting shafts.
A. Spur Gear
B. Helical Gear
C. Bevel Gear
D. None of the listed
.
Answer: C
Explanation: Bevel gears are used for power transmission in case of intersecting shafts.
Q. Is it possible to transmit power between shafts lying in different planes using gears?
A. Yes
B. No
.
Answer: A
Explanation: Worm or crossed helical gears can be used in this case for power transmission.
Q. The two gears are said to have conjugate motion if
A. They have constant angular velocity ratio
B. Variable angular velocity ratio
C. Infinitely small angular velocity ratio
D. None of the mentioned
.
Answer: A
Explanation: Two gear are said to have conjugate motion and tooth profiles are said to have conjugate curves if they have constant angular velocity ratio.
Q. Which of the following is not true about gears?
A. Positive drive
B. Constant velocity ratio
C. Transmit large power
D. Bulky construction
.
Answer: D
Explanation: They have compact construction.
Q. Gear drive don’t require precise alignment of shafts.
A. True
B. False
.
Answer: B
Explanation: A minute level of misalignment isn’t tolerated in gears.
Q. Spur gears can be used only when the two shafts are parallel.
A. True
B. False
.
Answer: A
Explanation: The teeth are cut parallel to the axis of shaft.
Q. The teeth of the helical gears are cut parallel to the shaft axis.
A. True
B. False
.
Answer: B
Explanation: They are cut at an angle with the shaft axis.
Q. Herringbone gear can be used in
A. Intersecting shafts only
B. Parallel shafts only
C. Both intersection and parallel shafts
D. None of the mentioned
.
Answer: B
Explanation: It consist of two helical gears with the opposite hand of the helix.
Q. Bevel gears impose ____ loads on the shafts.
A. Radial and thrust
B. Radial
C. Thrust
D. Neither radial nor thrust
.
Answer: A
Explanation: Bevel gears have the shape of a truncated cone and tooth is cut straight or spiral.
Q. Which of the following are true for worm gears?
A. Worm is in the shape of threaded screw
B. Threads on the worm have small lead
C. Worm imposes high thrust loads
D. Characterised by low speed reduction ratio
.
Answer: D
Explanation: They are characterised by high speed reduction ratio.
Q. Greater the velocity ratio, smaller the gearbox.
A. True
B. Greater the gearbox
C. Size of gearbox remains unaffected
D. None of the listed
.
Answer: B
Explanation: Greater velocity leads to increase in size of gear wheel which results in size of gearbox.
Q. Required velocity ratio is 60:1, which of the following are recommended?
A. Worm
B. Spur
C. Bevel
D. None of the mentioned
.
Answer: A
Explanation: For high speed reduction ratio, worm gears are recommended.
Q. For a constant velocity ratio, the common normal to the tooth profile at point of contact must pass through a continuously variable point.
A. True
B. It pass through a fixed point
C. Constant velocity ratio isn’t required, hence variable point is preferred
D. None of the listed
.
Answer: B
Explanation: It must pass through a fixed point called pitch to maintain a constant velocity ratio.
Q. In cycloidal gears contact area is
A. Comparatively smaller
B. Comparatively larger
C. Can’t be determined
D. None of the listed
.
Answer: B
Explanation: Convex flank on one tooth meets with concave on the other thus increasing the contact area.
Q. Involute gears have greater contact area as compared to cycloidal gears.
A. True
B. False
.
Answer: B
Explanation: There is mating of two convex surfaces and hence lesser contact area.
Q. Cycloidal teeth consist of
A. Hypocycloid curve
B. Epicycloid gear
C. Both hypocycloid curve and epicycloid curve
D. None of the mentioned
.
Answer: C
Explanation: It consist of both and thus are hard to manufacture.
Q. Pressure angle remains constant in case of involute profile.
A. True
B. False
.
Answer: A
Explanation: The common normal always passes through the pitch point and thus maintain the constant inclination.
Q. Pressure angle is _____ in case of cycloidal teeth.
A. Constant
B. Variable
C. zero
D. None of the listed
.
Answer: B
Explanation: Cycloidal teeth consist of two profiles.
Q. Velocity ratio is the ratio angular velocity of driving gear to that of driven gear.
A. True
B. False
.
Answer: A
Explanation: Velocity ratio is simply the angular velocities ratio.
Q. Velocity ratio and transmission ratio are the same thing.
A. True
B. False
.
Answer: B
Explanation: Transmission ratio is measured between first and last gear.
Q. Contact ratio is always
A. =1
B. >1
C. <1
D. Can’t be determined
.
Answer: B
Explanation: Some overlapping is essential for continuous transfer of power.
Q. Product of diametric pitch and circular pitch is?
A. π
B. 1/π
C. None of the listed
D. 2
.
Answer: A
Explanation: CP=πd/z and circular pitch=z/d.
Q. Diameteral pitch is 5, then calculate module of the gear.
A. 0.2
B. 0.4
C. 5
D. 10
.
Answer: A
Explanation: Module is the inverse of diameteral pitch.
Q. If centre distance between the two gears on same shaft is unequal to the centre distance on the other two gears on the second shaft, then this gear train is called reverted gear train.
A. True
B. False
.
Answer: B
Explanation: The centre distance is equal in both the shafts.
Q. If one gear is fixed while the other gear has motion of two types i.e. rotary about its own axis and rotation about axis of fixed gear, than the gear train is _____
A. Epicyclic gear train
B. Reverted gear train
C. Kepler gear train
D. None of the mentioned
.
Answer: A
Explanation: Definition of epicyclic gear train.
Q. Which of the following are true about epicyclic gear train?
A. Fixed gear is called sun gear
B. Rotating gear are called earth gear
C. Crank is called sun carrier
D. None of the listed
.
Answer: A
Explanation: Rotating gears are called planet gears.
Q. Epicyclic gears are not generally recommended due to bulky construction.
A. True
B. False
.
Answer: B
Explanation: Epicyclc gears have compact construction.
Q. A compound gear train consists of at least 3 shafts connected to each other.
A. True
B. False
.
Answer: B
Explanation: A compound gear train is characterised by one shaft carrying two gears atleast.
Q. Interference is caused by?
A. Overlapping of tooth profiles
B. Large size of dedendum
C. Meshing of involute and no-involute profiles
D. All of the mentioned
.
Answer: D
Explanation: In some cases, portion of tooth below base circle is not involute due to large dedendum. This leads to overlapping of involute portion of teeth with the non-involute portion of the teeth.
Q. Interference can be removed by under cutting.
A. True
B. False
.
Answer: A
Explanation: Undercutting removes the interfering portion of the tooth.
Q. Interference is maximum when the largest pinion is in mesh with the smallest gear.
A. True
B. False
.
Answer: B
Explanation: It is maximum when smallest pinion is in mesh with the largest gear.
Q. Undercutting is consider healthy for the tooth as it eliminates undercutting.
A. True
B. False
.
Answer: B
Explanation: Although it eliminates undercutting but it also weakens the strength of the tooth and removes a small involute portion near to the base circle.
Q. Backlash is defined as
A. Difference in the width of tooth space and engaging tooth thickness
B. Amount by which engaging tooth thickness exceeds the tooth space
C. Other name for interference
D. None of the listed
.
Answer: A
Explanation: Backlash is the amount by which tooth space exceeds the engaging tooth thickness.
Q. The amount of backlash depends on
A. Diameteral pitch
B. Module
C. Centre distance
D. All of the mentioned
.
Answer: D
Explanation: Module and diameteral pitch are the same thing, and centre distance is also a factor while deciding the magnitude of the backlash.
Q. The two teeth have thickness t₁ and t₂. If backlash of t₁ is greater than that of t₂, than
A. t₁ > t₂
B. t₁ < t₂
C. t₁ = t₂
D. None of the mentioned
.
Answer: B
Explanation: Backlash=Tooth space-tooth thickness, greater the backlash lower is the tooth thickness.
Q. There are ____ standard systems for the shape of gear teeth.
A. 1
B. 2
C. 3
D. 4
.
Answer: C
Explanation: Q.5⁰ full depth involute system, 20⁰ full depth involute system and 20⁰ stub involute system.
Q. When the number of teeth reaches infinity, circle radius approaches infinity the gear becomes an infinite loop.
A. True
B. False
.
Answer: B
Explanation: It becomes a rack with straight sided teeth.
Q. 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 Q.5⁰
D. All of the statements are true
.
Answer: C
Explanation: Lower the pressure angle, quieter is the operation. Lower the pressure angle, lower is the breadth of the tooth at root.
Q. 20⁰ stub involute system have comparatively smaller interference.
A. True
B. False
.
Answer: A
Explanation: They have shorter addendum and shorter dedendum.
Q. 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: A
Explanation: Smaller moment arm of bending force leads to stronger stub teeth.
Q. As the module increases, index of size of gear decreases.
A. True
B. False
.
Answer: B
Explanation: Module is the measure of size of index of the gear tooth.
Q. Crowning is an abrasive process that debars the gear strength.
A. True
B. False
.
Answer: B
Explanation: Crowningis used to strengthen the tooth.
Q. Inaccuracies in tooth profile lead to concentration of pressure on the middle of tooth.
A. True
B. False
.
Answer: B
Explanation: Inaccuracies lead to shift of pressure at the end of tooth which can be improved by crowning.
Q. 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: C
Explanation: C = m(z(p)+z(g))/Q.
C = 5(25 + 115)/2 => 700/2 => 350mm.
Q. 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: D
Explanation: D = 5×25 = 125mm.
Q. 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: D
Explanation: D = 5×115 = 575mm.
Q. 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. Q.75mm
C. Q.25mm
D. 5mm
.
Answer: D
Explanation: H=m=5mm.
Q. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the dedendum.
A. Q.75mm
B. 5mm
C. Q.25mm
D. Q.68mm
.
Answer: C
Explanation: H = Q.25xm = Q.25x5mm = Q.25mm.
Q. 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. Q.23mm
B. Q.44mm
C. Q.854mm
D. Q.16mm
.
Answer: C
Explanation: T = Q.5708xm = Q.5708x5mm = Q.854mm.
Q. 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. Q.75mm
C. Q.5mm
D. Q.25mm
.
Answer: D
Explanation: C = 0.25m = 0.25x5mm = Q.25mm.
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the transverse module.
A. Q.3mm
B. Q.1mm
C. Q.9mm
D. Q.7mm
.
Answer: A
Explanation: m=4/Cos(22⁰).
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the transverse pressure angle in degrees.
A. 1Q.9
B. 20.4
C. 1Q.6
D. 1Q.4
.
Answer: B
Explanation: tanᾰ=tan(19⁰)/Cos(22⁰).
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the axial pitch.
A. None of the listed
B. 3Q.2mm
C. 3Q.4mm
D. 2Q.6mm
.
Answer: C
Explanation: p=πx(transverse module)/tan(22).
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the pitch circle diameter of pinion.
A. 6Q.7mm
B. 5Q.6mm
C. 5Q.6mm
D. 6Q.8mm
.
Answer: A
Explanation: d=zxm/Cos(22).
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the pitch circle diameter of the gear.
A. 17Q.6mm
B. 14Q.6mm
C. 180.3mm
D. 20Q.4mm
.
Answer: A
Explanation: d=zxm/Cos(22).
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the centre distance.
A. 12Q.4mm
B. 13Q.6mm
C. 11Q.65mm
D. 14Q.4mm
.
Answer: C
Explanation: C=Sum of diameter of pinion and gear/Q.
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate addendum circle diameter of the pinion.
A. 5Q.2mm
B. 7Q.7mm
C. 6Q.4mm
D. None of the listed
.
Answer: B
Explanation: D(A.=m[z/Cos(22) + 2].
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the dedendum circle diameter of the pinion.
A. 6Q.5mm
B. 5Q.7mm
C. 5Q.2mm
D. None of the listed
.
Answer: B
Explanation: D(f)=m[z/Cos(22) – Q.5].
Q. The direction of tangential component for a driving gear is same to the direction of rotation.
A. True
B. False
.
Answer: B
Explanation: The direction is opposite and not same.
Q. If tangential component of force on tooth is 200N and helix angle is 25⁰, calculate the axial component of the force.
A. 200N
B. 30Q.5N
C. 9Q.26N
D. 2Q.6N
.
Answer: C
Explanation: P(A.=200xtan(25).
Q. Among spur gear and helical gear, which has smooth engagement and thus lesser noise?
A. Helical Gears
B. Spur Gears
C. Both have equal noises
D. Can’t be determined
.
Answer: A
Explanation: There is a gradual pick up of load in helical gears and hence smooth operation.
Q. There is same type of tooth meshing in helical and spur gear.
A. True
B. False
.
Answer: B
Explanation: In spur gears contact occurs along entire face width which leads to impact condition while in helical contact begins from a single point and then there is gradual increase in load.
Q. Among the normal module and transverse module, which one has greater value?
A. Normal Module
B. Transverse Module
C. Both have equal module
D. Insufficient information
.
Answer: B
Explanation: Normal Module=Transverse modulexCos(helix angle).
Q. 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: B
Explanation: It is zero in both the cases.
Q. Helix angle of herringbone and double helical gears is relatively higher.
A. True
B. False
.
Answer: A
Explanation: There is no thrust force and hence higher angles are permitted.
Q. 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.Q. If permissible bending stress is 500N/mm², then calculate the beam strength.
A. 15000N
B. 12000N
C. 8000N
D. 10000N
.
Answer: B
Explanation: S=mbσY.
Q. 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.Q. Calculate the ratio factor Q.
A. Q.2
B. Q.4
C. Q.7
D. Q.4
.
Answer: C
Explanation: Q=2×100/100+20.
Q. 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.Q. Calculate the material constant K if surface hardness is 400BHN.
A. Q.25 N/mm²
B. Q.05 N/mm²
C. Q.25N/mm²
D. Q.56N/mm
.
Answer: C
Explanation: K=0.16x[BHN/100]².
Q. Helical gears mounted on parallel shafts are called crossed helical gears.
A. True
B. False
.
Answer: B
Explanation: Crossed helical gears are the helical gears mounted on non-parallel shafts.
Q. Crossed helical gears have very low load carrying capacity.
A. True
B. False
.
Answer: B
Explanation: There is point contact and hence very less area and thus wear is comparatively rapid.
Q. 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: C
Explanation: For same hand of helix, shaft angle=sum of helix angles of two gears.
Q. Lewis form factor is based on real number of teeth.
A. True
B. False
.
Answer: B
Explanation: It is based on virtual umber of teeth only.
Q. Beam strength indicates the maximum value of radial force that a tooth can transmit without fatigue failure.
A. True
B. False
.
Answer: B
Explanation: It indicates maximum force for bending failure and not fatigue failure.
Q. 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. Q.57
D. Q.44
.
Answer: C
Explanation: Ratio factor Q=2×90/90+2Q.
Q. 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. Q.08N/mm²
D. Q.66N/mm²
.
Answer: C
Explanation: K=0.16x[BHN/100]².
Q. 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. 44Q.5N
B. 112Q.6N
C. 797Q.9N
D. 10Q.2N
.
Answer: C
Explanation: S=bQdK/cos²Ɯ where Ɯ=25⁰, d=zm/cosƜ, Q=2×90/90+25, K=0.16x[BHN/100]².
