[MCQ’s]Machine design 2

Module 01

1. 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
Explanation: The common normal must pass through the point where two mating gears meet.

2. 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
Explanation: Bevel gears are used for power transmission in case of intersecting shafts.

3. Is it possible to transmit power between shafts lying in different planes using gears?
a) Yes
b) No
Explanation: Worm or crossed helical gears can be used in this case for power transmission.

4. 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
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.

5. Which of the following is not true about gears?
a) Positive drive
b) Constant velocity ratio
c) Transmit large power
d) Bulky construction
Explanation: They have compact construction.

6. Gear drive don’t require precise alignment of shafts.
a) True
b) False
Explanation: A minute level of misalignment isn’t tolerated in gears.

7. Spur gears can be used only when the two shafts are parallel.
a) True
b) False
Explanation: The teeth are cut parallel to the axis of shaft.

8. The teeth of the helical gears are cut parallel to the shaft axis.
a) True
b) False
Explanation: They are cut at an angle with the shaft axis.

9. 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
Explanation: It consist of two helical gears with the opposite hand of the helix.

10. Bevel gears impose ____ loads on the shafts.
c) Thrust
Explanation: Bevel gears have the shape of a truncated cone and tooth is cut straight or spiral.

11. Which of the following are true for worm gears?
a) Worm is in the shape of threaded screw
c) Worm imposes high thrust loads
d) Characterised by low speed reduction ratio
Explanation: They are characterised by high speed reduction ratio.

12. Greater the velocity ratio, smaller the gearbox.
a) True
b) Greater the gearbox
c) Size of gearbox remains unaffected
d) None of the listed
Explanation: Greater velocity leads to increase in size of gear wheel which results in size of gearbox.

13. Required velocity ratio is 60:1, which of the following are recommended?
a) Worm
b) Spur
c) Bevel
d) None of the mentioned
Explanation: For high speed reduction ratio, worm gears are recommended.

14. 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
Explanation: It must pass through a fixed point called pitch to maintain a constant velocity ratio.

15. In cycloidal gears contact area is
a) Comparatively smaller
b) Comparatively larger
c) Can’t be determined
d) None of the listed
Explanation: Convex flank on one tooth meets with concave on the other thus increasing the contact area.

16. Involute gears have greater contact area as compared to cycloidal gears.
a) True
b) False
Explanation: There is mating of two convex surfaces and hence lesser contact area.

17. Cycloidal teeth consist of
a) Hypocycloid curve
b) Epicycloid gear
c) Both hypocycloid curve and epicycloid curve
d) None of the mentioned
Explanation: It consist of both and thus are hard to manufacture.

18. Pressure angle remains constant in case of involute profile.
a) True
b) False
Explanation: The common normal always passes through the pitch point and thus maintain the constant inclination.

19. Pressure angle is _____ in case of cycloidal teeth.
a) Constant
b) Variable
c) zero
d) None of the listed
Explanation: Cycloidal teeth consist of two profiles.

20. Velocity ratio is the ratio angular velocity of driving gear to that of driven gear.
a) True
b) False
Explanation: Velocity ratio is simply the angular velocities ratio.

21. Velocity ratio and transmission ratio are the same thing.
a) True
b) False
Explanation: Transmission ratio is measured between first and last gear.

22. Contact ratio is always
a) =1
b) >1
c) <1
d) Can’t be determined
Explanation: Some overlapping is essential for continuous transfer of power.

23. Product of diametric pitch and circular pitch is?
a) π
b) 1/π
c) None of the listed
d) 2
Explanation: CP=πd/z and circular pitch=z/d.

24. Diameteral pitch is 5, then calculate module of the gear.
a) 0.2
b) 0.4
c) 5
d) 10
Explanation: Module is the inverse of diameteral pitch.

25) Which of the following formula is used to calculate oil-film thickness in hydrodynamic bearings?

A. h = (c + Є cos θ)
b. h =(c – Є cos θ)
c. h = c (1+ Є sin θ)
d. h = c (1+ Є cos θ)
ANSWER: h = c (1+ Є cos θ)

Explanation:
No explanation is available for this question!

26) Lubricating oil used in hydrodynamic bearing has total flow rate of 0.340 l/min and side leakage of 0.1520 l/min. If mass density of oil is 600 kg/m3 and specific heat is 1.05 kJ/kg oC, what is the rise in temperature if power lost in friction is 0.05 kW?

A.11.23o C
b. 15.11o C
c. 18.03o C
d. 22.23o C

Explanation:
No explanation is available for this question!

27) Lubricating oil of mass density 800 kg/m3 used in 360o hydrodynamic bearing has a flow rate of 6000 mm3. Neglecting side leakage if temperature rises to 10 oC and specific heat is 1.55 kJ/kg oC, what is the rate of heat dissipated in the bearing?

A.7.4 W
b. 236 W
c. 0.236 kW
d. 0.0744 kW

Explanation:
No explanation is available for this question!

28) A journal of 120 mm diameter rotates in a bearing at a speed of 1000 rpm. What is the power lost during friction if 8 kN radial load acts on the journal and coefficient of friction is 2.525 x 10-3?

A.0.126 kW
b. 0.253 KW
c. 2.365 kW
d. 7.615 kW

Explanation:
No explanation is available for this question!

29) What does number 6 indicate in the class of wire rope 6 x 37 (18/12/6/1)?

A.Diameter
b. Strands
c. Wires
d. None of the above

Explanation:
No explanation is available for this question!

30) Lang lay ropes offer more resistance to __________

A.fatigue failure
b. abrasive failure
c. both a. and b.
d. none of the above

Explanation:
No explanation is available for this question!

31. A V-belt pulley has belt velocity 20 m/s and mass 0.7 kg per meter. If allowable tension in the belt is 600 N then what will be the power transmitting capacity of belt?
(Assume μ = 0.5 & θ = 2.5 rad)

A.3.75 kW
b. 3.2 kW
c. 4.5 kW
d. 5.23 kW

Explanation:
No explanation is available for this question!

32. A V-belt pulley transmits power to flat belt pulley. If the coefficient of friction is 0.25 then which pulley governs the design? (β = 21o & α = 0.2)

A.Flat pulley
b. V-belt pulley
c. Both a. and b.
d. None of the above

Explanation:
No explanation is available for this question!

33.Which of the following statements is/are true for V-belts?

A.V-belt drives have low reduction ratio
b. Due to wedging action slip is negligible in V-belts
c. V-belt transmits power over long centre distances
d. All of the above
ANSWER: Due to wedging action slip is negligible in V-belts

Explanation:
No explanation is available for this question!

Module 01

1. Thick film lubrication describes a phenomenon where two surfaces are _______ separated.
a) Completely
b) Partially
c) Not
d) None of the mentioned
Explanation: There is no contact between the two surfaces.

2. Hydrodynamic bearing is a self-acting bearing.
a) True
b) False
Explanation: The pressure in the bearing is created within the system due to rotation of the shaft.

3. A journal bearing is a ______ contact bearing working on the hydrodynamic lubrication and which supports load in____ direction.
a) Sliding, Axial
d) Rolling, Axial
Explanation: This is how journal bearing works. It derives its name from the portion of the shaft inside the bearing.

4. Partial bearing is preferred over journal bearing.
a) True
b) No
c) More friction losses
d) Can’t be determined
Explanation: Friction losses are lesser and construction is simple.

5. Temperature rise in partial bearing is ____ than full bearing.
a) Lesser
b) Greater
c) Equal
d) Undeterminable
Explanation: Due to lesser friction, temperature rise is less.

6. 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
Explanation: Journal radius is kept smaller.

7. Fitted bearing must be partial bearing.
a) True
b) No
c) No lubricating space is required
d) Can’t be stated
Explanation: To provide space for lubricating oil.

8. Footstep bearing is an axial load bearing.
a) True
d) None of the listed
Explanation: It is a thrust bearing in which shaft end is in contact with bearing surface.

9. Hydrostatic and hydrodynamic lubrication are the same thing.
a) True
b) False
Explanation: In hydrodynamic, motion is provided by the shaft while in hydrostatic the motion is provided by an external source.

10. If we exclude the cost factor, which bearing is preferred?
a) Hydrostatic
b) Hydrodynamic
c) Both are equally preferred
d) Cannot be determined
Explanation: High load capacity, no starting friction and no rubbing action.

11. If fluid film pressure is high and surface rigidity is low than mode of lubrication is called as elastohydrodynamic lubrication.
a) True
b) False
Explanation: In such cases, elastic deflections of the parts causes development of the film.

Module 02

1. Thick film lubrication describes a phenomenon where two surfaces are _______ separated.
a) Completely
b) Partially
c) Not
d) None of the mentioned
Explanation: There is no contact between the two surfaces.

2. Hydrodynamic bearing is a self-acting bearing.
a) True
b) False
Explanation: The pressure in the bearing is created within the system due to rotation of the shaft.

3. A journal bearing is a ______ contact bearing working on the hydrodynamic lubrication and which supports load in____ direction.
a) Sliding, Axial
d) Rolling, Axial
Explanation: This is how journal bearing works. It derives its name from the portion of the shaft inside the bearing.

4. Partial bearing is preferred over journal bearing.
a) True
b) No
c) More friction losses
d) Can’t be determined
Explanation: Friction losses are lesser and construction is simple.

5. Temperature rise in partial bearing is ____ than full bearing.
a) Lesser
b) Greater
c) Equal
d) Undeterminable
Explanation: Due to lesser friction, temperature rise is less.

6. 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
Explanation: Journal radius is kept smaller.

7. Fitted bearing must be partial bearing.
a) True
b) No
c) No lubricating space is required
d) Can’t be stated
Explanation: To provide space for lubricating oil.

8. Footstep bearing is an axial load bearing.
a) True
d) None of the listed
Explanation: It is a thrust bearing in which shaft end is in contact with bearing surface.

9. Hydrostatic and hydrodynamic lubrication are the same thing.
a) True
b) False
Explanation: In hydrodynamic, motion is provided by the shaft while in hydrostatic the motion is provided by an external source.

10. If we exclude the cost factor, which bearing is preferred?
a) Hydrostatic
b) Hydrodynamic
c) Both are equally preferred
d) Cannot be determined
Explanation: High load capacity, no starting friction and no rubbing action.

11. If fluid film pressure is high and surface rigidity is low than mode of lubrication is called as elastohydrodynamic lubrication.
a) True
b) False
Explanation: In such cases, elastic deflections of the parts causes development of the film. Answer the questions 12-18 with respect to this figure

12. The bearing characteristic number shown in the graph is directly proportional to the unit bearing pressure.
a) True
b) False
Explanation: BCN=µN/P.

13. In the graph shown, which regions points the partial metal to metal contact.
a) BC
b) CD
c) Both BC and CD
d) Can’t be determined
Explanatiion: BC is the condition of thin film and hence surfaces are not completely separated.

14. Hydrodynamic lubrication takes place in region?
a) BC
b) CD
c) Both BC and CD
d) None of the mentioned
Explanation: In CD thick film of lubrication has developed.

15. Coefficient of friction is minimum at
a) It is same at each point
b) B
c) C
d) D
Explanation: It is minimum at the transition zone.

16. Bearing modulus is the bearing characteristic number at
a) B
b) C
c) D
d) None of the listed
Explanation: BCN modulus is the the BCN number corresponding to the minimum coefficient of friction.

17. If viscosity of the lubricant is very low than BCN will lie in the region
a) BC
b) Point C
c) CD
d) Can’t be determined
Explanation: Due to low viscosity, lubricant will not separate the two surfaces resulting in partial contact.

18. 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
Explanation: It is givenby ∆pbhᵌ/12µl.

19. A single transverse weld is preferred over double transverse fillet weld.
a) True
b) False
Explanation: A single transverse weld is not preferred because the edge of the plate which is not welded can warp out of shape.

20. Transverse fillet weld can be designed using the same equations as of parallel fillet weld.
a) True
b) False
Explanation: Vice versa is true as strength of transverse fillet weld is greater than that of parallel fillet weld.

Module 03

1. The size of a cam depends upon
a) base circle
b) pitch circle
c) prime circle
d) pitch curve
Explanation: Base circle is the smallest circle that can be drawn to the cam profile.
Pitch circle is a circle drawn from the centre of the cam through the pitch points.
Prime circle is the smallest circle that can be drawn from the centre of the cam and
tangent to the pitch curve.

2. The angle between the direction of the follower motion and a normal to the pitch curve is called
a) pitch angle
b) prime angle
c) base angle
d) pressure angle
Explanation: It is the angle between the direction of the follower motion and a normal to the pitch curve. This angle is very important in designing a cam profile. If the pressure angle is too large, a reciprocating follower will jam in its bearings.

3. A circle drawn with centre as the cam centre and radius equal to the distance between the cam centre and the point on the pitch curve at which the pressure angle is maximum, is called
a) base circle
b) pitch circle
c) prime circle
d) none of the mentioned
Explanation: Base circle is the smallest circle that can be drawn to the cam profile.
Pitch circle is a circle drawn from the centre of the cam through the pitch points.
Prime circle is the smallest circle that can be drawn from the centre of the cam and
tangent to the pitch curve.

4. The cam follower generally used in automobile engines is
a) knife edge follower
b) flat faced follower
c) spherical faced follower
d) roller follower
Explanation: When the contacting end of the follower is of spherical shape, it is called a spherical faced follower. It may be noted that when a flat-faced follower is used in automobile engines, high surface stresses are produced. In order to minimise these stresses, the flat end of the follower is machined to a spherical shape.

5. The cam follower extensively used in air-craft engines is
a) knife edge follower
b) flat faced follower
c) spherical faced follower
d) roller follower
Explanation: When the contacting end of the follower is a roller, it is called a roller follower. Since the rolling motion takes place between the contacting surfaces (i.e. the roller and the cam), therefore the rate of wear is greatly reduced. In roller followers also the side thrust exists between the follower and the guide. The roller followers are extensively used where more space is available such as in stationary gas and oil engines and aircraft engines.

6. In a radial cam, the follower moves
a) in a direction perpendicular to the cam axis
b) in a direction parallel to the cam axis
c) in any direction irrespective of the cam axis
d) along the cam axis
Explanation: In radial cams, the follower reciprocates or oscillates in a direction perpendicular to the cam axis.

7. A radial follower is one
a) that reciprocates in the guides
b) that oscillates
c) in which the follower translates along an axis passing through the cam centre of rotation.
d) none of the mentioned
Explanation: When the motion of the follower is along an axis passing through the centre of the cam, it is known as radial follower.

8. Ofset is provided to a cam follower mechanism to
a) minimise the side thrust
b) accelerate
c) avoid jerk
d) none of the mentioned
Explanation: When the motion of the follower is along an axis away from the axis of the cam centre, it is called off-set follower.

9. For low and moderate speed engines, the cam follower should move with
a) uniform velocity
b) simple harmonic motion
c) uniform acceleration and retardation
d) cycloidal motion
Explanation: None

10. For high speed engines, the cam follower should move with
a) uniform velocity
b) simple harmonic motion
c) uniform acceleration and retardation
d) cycloidal motion
Explanation: Since with high speed engines, maximum acceleration is required and that is possible only through cycloidal motion.

11. Which of the following displacement diagrams should be chosen for better dynamic performance of a cam-follower mechanism ?
a) simple hormonic motion
b) parabolic motion
c) cycloidal motion
d) none of the mentioned
Explanation: Only cycloidal motion gives maximum acceletation. Hence, it is considered the most dynamic cam- follower mechanism.

12. The linear velocity of the reciprocating roller follower when it has contact with the straight flanks of the tangent cam, is given by
a) ω(r1-r2)sinθ
b) ω(r1-r2)cosθ
c) ωr1+r2)sinθsec2θ
d) ω(r1+r2)cosθcosec2θ
Explanation: Velocity = ωr1+r2)sinθsec2θ
where ω = Angular velocity of the cam shaft,
r1 = Minimum radius of the cam,
r2 = Radius of the roller, and
θ = Angle turned by the cam from the beginning of the displacement for contact of roller with the straight flanks.

13. The displacement of a flat faced follower when it has contact with the flank of a circular arc cam, is given by
a) R(1-cosθ)
b) R(1-sinθ)
c) (R-r1)(1-cosθ)
d) (R-r1)(1-sinθ)
Explanation: Displacement = (R-r1)(1-cosθ)
where R = Radius of the flank,
r1 = Minimum radius of the cam, and
θ = Angle turned by the cam for contact with the circular flank.

14. The retardation of a flat faced follower when it has contact at the apex of the nose of a circular arc cam, is given by
a) ω2×OQ
b) ω2×OQsinθ
c) ω2×OQcosθ
d) ω2×OQtanθ
Explanation: Retardation is maximum when α−θ = 0 or θ = α ,
Maximum retardation = ω2×OQ
where OQ = Distance between the centre of circular flank and centre of nose.

15) Calculate the limiting speed to avoid cam jump, if an eccentric plate circular cam has eccentricity of 30 mm and provides motion to the follower of 4 kg mass. Stiffness of the spring is 30 N/mm2. Preload in the spring is 600 N

d. None of the above

16) Which motion of follower is best for high speed cams?

a. SHM follower motion
b. Uniform acceleration and retardation of follower motion
c. Cycloidal motion follower
d. All of the above

17) Which of the following statements is false for SHM follower motion?

a. SHM can be used only for moderate speed purpose
b. The acceleration is zero at the beginning and the end of each stroke
c. The jerk is maximum at the mid of each stroke
d. Velocity of follower is maximum at the mid of each stroke
ANSWER: The acceleration is zero at the beginning and the end of each stroke

18) Which of the following conditions can be used to minimize undercutting in cam and follower mechanism?

a. By using larger roller diameter
b. By using internal cams
c. By decreasing the size of the cam
d. All of the above

19) Which of the following equation is used to measure pressure angle between direction of follower motion and force exerted by the cam on follower when eccentricity is zero?

Where,
y = displacement of follower

a. cot Φ = (dy /dθ) / ( rb + y)
b. tan Φ = (dy /dθ) / ( rb + y)
c. tan Φ = (dy /dθ) x ( rb + y)
d. cot Φ = (dy /dθ) x ( rb + y)
ANSWER: tan Φ = (dy /dθ) / ( rb + y)

20) To avoid jump phenomenon, which of the following condition should be true?

Where,
P = preload in spring, ω = cam speed, k = stiffness of spring, e = eccentricity

a. ω > √ (me) / ( P + 2 ke)
b. ω > √( P + 2 ke) / (me)
c. ω < √( P + 2 ke) / (me)
d. ω < √ (me) / ( P + 2 ke)
ANSWER: ω < √( P + 2 ke) / (me)

21) What is meant by jump phenomenon in cam and follower system?

a. Follower looses contact with cam surface when cam rotates beyond particular speed due to inertia forces
b. Follower looses contact with cam surface when follower rotates beyond particular speed due to gravitational force
c. Follower looses contact with cam surface when cam rotates beyond particular speed due to gravitational forces
d. None of the above
ANSWER: Follower looses contact with cam surface when cam rotates beyond particular speed due to inertia forces

22) A cam operating roller follower has the following dimensions, radius of base and nose circle as 15 mm and 10 mm respectively and distance between them is 8 mm. Determine lift made by the follower

a. 5 mm
b. 12.5 mm
c. 3 mm
d. 17 mm

23) Which of the following statements is/are true for cam profile?

a. Pitch point on the pitch curve has minimum pressure angle
b. In case of roller follower, trace point represents centre of the roller
c. Pitch circle is drawn through trace point from the center of cam
d. All of the above
ANSWER: In case of roller follower, trace point represents centre of the roller

24) Which type of cam does not require any external force to have contact between cam and follower?

b. Conjugate cam
c. Both a. and b.
d. None of the above

Module 04

1. Is use of a flywheel recommended when a large motor is required only for a small instant of time?
a) True
b) False
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.

2. Flywheels are used in punching and shear operation.
a) True
b) False
Explanation: Flywheels store the kinetic energy imparted during idle positions and delivers this energy during actual shearing or punching.

3. 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
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.

4. 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
Explanation: The arms are free in split flywheel to contract and hence are better for stress reduction.

5. Flywheel and governor can be interchanged.
a) True
b) False
Explanation: Flywheel influences cyclic speed fluctuations while governor control mean speed.

6. 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
Explanation: Governor is used to control the mean speed and if mean speed is held constant then the governor will not operate.

7. The operation of flywheel is continuous.
a) True
b) False
Explanation: Flywheel is used to control the fluctuations and thus is continuously working.

8. Which of the following doesn’t waste energy?
a) Flywheel
b) Governor
c) Both flywheel and governor
d) Neither flywheel nor governor
Explanation: Flywheel has kinetic energy which is 100% convertible while governor suffer from friction losses.

9. Which of the following is not true for cast iron flywheels?
a) Excellent damping
b) Cheap
c) Given complex shape
d) Sudden failure
Explanation: Due to low tensile strength failure is sudden.

10. When the driving torque is more than load torque, flywheel is ______
a) Accelerated
b) Decelerated
c) Constant velocity
d) Can’t be determined

Answer the following in context of this figure
The following is a work cycle graph for a flywheel. T₂ is the torque supplied by motor while T₁ and T₃ are the torques required by machine. 11. During the portion AB the flywheel is _____
a) Accelerated
b) Decelerated
c) Maintain constant speed
d) None of the listed
Explanation: Driving torque is more than the torque supplied by motor.

12. During the portion BC the flywheel is _____
a) Accelerated
b) Decelerated
c) Maintain constant speed
d) None of the listed
Explanation: Driving torque is lesser than the torque supplied by motor.

13. The flywheel is accelerated during the
a) AB
b) BC
c) CD
d) Both AB and CD
Explanation: Whenever the torque supplied by the motor is greater than the torque needed by machine, flywheel will accelerate.

14. Belt, chain and rope are called rigid drives.
a) True
b) Belt is a flexible drive only
c) All three are flexible drives
d) None of the listed
Explanation: They all are flexible drive examples.

15. In flexible drives, rotary motion of driving shaft is directly converted to rotary motion of driven shafts
a) Yes
b) No, translator motion is involved
c) No, Cylindrical motion is involved
d) No, Parabolic motion is involved
Explanation: It is first converted into translator and then into rotary.

16. Flexible drive can absorb shock loads and damp vibrations.
a) True
b) No
c) Depends on the load applied
d) Doesn’t damp vibrations
Explanation: Intermediate link is long and flexible.

17. Velocity ratio in both flexible and rigid drive is constant.
a) True
b) False
Explanation: In flexible drive, velocity ratio is not constant.

18. Is there any worry for overloading conditions in flexible belt drives?
a) Yes
b) No
c) Only after a particular threshold limit
d) Depends on external factrors
Explanation: In case of overloading, belt slips over the pulley and hence protect it from the overload.

19. V belts have v shaped cross section.
a) No, rectangular
b) No, trapezium
c) No, circular
d) No, spherical
Explanation: They have trapezoidal cross section.

20. The force of friction between belt and V grooved pulley is high.
a) Yes, supported by wedge action
b) No
c) There is no wedge action involved
d) None of the listed
Explanation: Due to wedge action, force of friction is high.

21. V belts result in smooth and quite operation even at high speeds.
a) Yes
b) No they are very noisy
c) They are not endless and hence not smooth motion
d) None of the listed
Explanation: V belts are endless.

22. The efficiency of flat belt drive is 35%. If all the parameters are same and flat belt is replaced by V belt, than the efficiency of V belt will be?
a) <35%
b) >35%
c) =35%
d) Cant be determined
Explanation: The efficiency of flat belt drive is more that V belt drive.

23. There is problem of bending stress in the V belt drive.
a) True
b) False
Explanation: The ratio of cross sectional height to pulley diameter is high.

24. The layer of a belt is generally called as
a) Ply
b) Layer
c) Segment
d) Sediment
Explanation: Terminology.

25. Velocity ratio for chain drive is lesser than that for belts.
a) Yes
b) No
c) In some cases
d) Can’t be determined
Explanation: Velocity ratio for chain drives is about 15:1 while for belts it is around 7:1.

26. Fabric rubber belts are not widely used as they can’t be operated at high speeds.
a) They can’t be used at velocities 50m/s
b) They can be used at high velocities
c) Limiting velocity is 20m/s
d) None of the listed
Explanation: These belts can be operated at 300m/s.

27. Power transmitting capacity of V belts is more than that of flat belt.
a) Yes
b) No
c) Only for V angle > 15
d) None of the listed
Explanation: Coefficient of friction is 2.92times of flat belts in V belts for identical materials.

28. The optimum velocity of the belt for maximum transmission is given by √P/6m.
a) Yes
b) No, by replacing 6 by 3
c) Replacing 6 by 9
d) None of the listed
Explanation: It is given by √P/3m.

29. Creep is the slight absolute motion of the belt as it passes over the pulley.
a) Yes
b) No, it is a relative motion
c) None of the listed
d) It is absolute motion
Explanation: It is a relative and not absolute motion.

30. In horizontal belt, the loose side is generally kept on the bottom.
a) True
b) False
Explanation: Loose side is kept on the top so that arc of contact increases and hence efficiency of the drive increases.

31. Vertical drives are preferred over horizontal.
a) True
b) False
Explanation: In vertical drive, gravitational force on the belt causes slip which reduces the efficiency.

32. The law of belting states that the centreline of the belt when it approaches a pulley must not lie in the midplane of the pulley.
a) True
b) False
Explanation: The centreline must lie in the midplane.

33. It is possible to use the belting reverse direction without violating the law of belting.
a) True
b) False
Explanation: Law of belting is violated if belt is used in reverse direction.

34. A shorter centre distance is always preferred in belt drives.
a) Yes due to stability
b) No due to stability
c) Yes due to instability
d) No due to instability
Explanation: It is more stable and compact.

35. If velocity ratio is less than 3, then centre distance is given by D+.5d.
a) True
b) False
Explanation: It is given by D+1.5d.

36. If velocity ratio is more than 3, then centre distance is given by 2D.
a) By 3D
b) By D
c) By 4D
d) By 5D
Explanation: It is given by D.

37. Is it possible to reduce the centre distance as much as we want?
a) Yes power transmission increases
b) No
c) Can’t be cited
d) None of the listed
Explanation: It depends on physical dimension and the minimum angle of wrap required to transfer the required power.

38. The diameter of the shorter pulley in leather belt drive is 265mm. It is rotating at 1440 rpm. Calculate the velocity of the belt.
a) 25m/s
b) 20m/s
c) 30m/s
d) None of the mentioned
Explanation: v=πdn/60×1000.

39. Calculate the angle of wrap if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.
a) 174.8⁰
b) 167.8⁰
c) 159.3⁰
d) None of the mentioned
Explanation: ὰ=180 – 2sin¯¹(D-d/2C).

40. Calculate the arc of contact if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.
a) 1.04
b) 1.03
c) 1.01
d) 1.02
Explanation: ὰ=180 – 2sin¯¹(D-d/2C). Factor=1+ (1.04-1)(180-174.8)/(180-170).

41. Calculate the belt length if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.
a) 6.5m
b) 4.66m
c) 6.94m
d) 5.26m
Explanation: L=2C + π(D+d)/2 + (D-d)²/4C.

42. Crowns are never mounted on the pulley.
a) True
b) False
Explanation: Crowns are used to avoid slip in case of misalignment or non-parallelism.

43. In a cast iron pulley minor axis is generally kept in the plane of rotation.
a) True
b) False
Explanation: Keeping minor axis in plane of rotation increases the cross section.

44. The number of V belts required for a given application are given by (ignoring correction factor for arc of contact and belt length) Transmitted power/kW rating of single belt x Industrial Service Factor.
a) True
b) False
Explanation: It is given by Transmitted power x Industrial Service Factor /kW rating of single belt.

45. The pitch diameter of bigger pulley D in terms of small diameter d is given by
a) dx[speed of smaller pulley/speed of bigger pulley].
b) dx[speed of bigger pulley/speed of smaller pulley].
c) d
d) None of the mentioned
Explanation: Product of diameter and speed of pulley is constant.

46. If maximum tension in the belt is 900N and allowable belt load is 500N. Calculate the number of belts required to transmit power.
a) 2
b) 3
c) 4
d) 5
Explanation: No of belts=900/500.

47. The belt tension is maximum when velocity of belt is 0.
a) It is max at v=infinity
b) True
c) It is velocity independent
d) It has a constant value
Explanation: P₁-mv²/P₂-mv²=e^(fa/sinθ/2). Hence belt tension is maximum when v=0.

48. If belt tension in the two sides is 730N and 140N and belt is moving with a velocity of 10m/s, calculate the power transmitted.
a) 4.5kW
b) 5.9kW
c) 6.2kW
d) None of the mentioned
Explanation: Power=(P₁-P₂)xv.

49. If tensions in the belt are P₁ and P₂, then find P₁-mv²/P₂-mv². Contact angle for smaller pulley is 156⁰, Groove angle is 36⁰ and coefficient of friction is 0.2.
a) 6.21
b) 5.83
c) 4.66
d) 5.36
Explanation: P₁-mv²/P₂-mv²=e^(fὰ/sinθ/2).

Module 04

1. Clutch and coupling perform the same action.
a) Both being permanent joints
b) No they are different type of joints
c) Both being temporary joints
d) None of the listed
Explanation: Clutch is a temporary joint while coupling is a permanent joint.

2. Eddy current clutch is a type of friction clutch.
a) Yes
b) No, it is an electromagnetic type clutch
c) It is a mechanical clutch
d) None of the listed
Explanation: Eddy current clutch is a type of electromagnetic clutch.

3. In positive contact clutches, power transmission is achieved by means of friction.
a) Yes
b) It is achieved by shear contact
c) Major part is achieved by friction
d) None of the listed
Explanation: Power transmission is achieved by interlocking of jaws or teeth.

4. The jaw clutches show great amount of slip.
a) Yes
b) Zero slip
c) Used in non-synchronous applications
Explanation: Jaw clutches do not slip and are used in synchronous applications.

5. For a new friction lining, uniform wear theory is used.
a) True
b) False
Explanation: Uniform pressure theory is used.

6. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is 100mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is 1.5N/mm². Assuming uniform wear theory, calculate the operating force in the clutch.
a) 15546N
b) 12344N
c) 23562N
d) 24543N
Explanation: P=πpd(D-d)/2.

7. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is 100mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is 1.5N/mm². Assuming uniform wear theory, calculate the torque transmitting capacity of the clutch.
a) 412.23N-m
b) 353.43N-m
c) 334.53N-m
d) 398.34N-m
Explanation: P=πpd(D-d)/2 and M=μP(D+d)/4.

8. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is 100mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is 1.5N/mm². Assuming uniform wear theory, calculate the power transmitting capacity of the clutch at 80rad/s.
a) 27.8kW
b) 32.4kW
c) 21.2kW
d) 34.5kW
Explanation: P=πpd(D-d)/2 and M=μP(D+d)/4.Power=Mxω.

9. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is 100mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is 1.5N/mm². Assuming uniform pressure theory, calculate the operating force in the clutch.
a) 15546N
b) 12344N
c) 23562N
d) 35343N
Explanation: P=πp(D²-d²)/4.

10. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is 100mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is 1.5N/mm². Assuming uniform pressure theory, calculate the torque transmitting capacity of the clutch.
a) 412.23N-m
b) 549.78N-m
c) 567.54N-m
d) 678.86N-m
Explanation: P= πp(D²-d²)/4 and M=μP(Dᵌ-dᵌ)/3(D²-d²).

11. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is 100mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is 1.5N/mm². Assuming uniform wear theory, calculate the power transmitting capacity of the clutch at 80rad/s.
a) 27kW
b) 32kW
c) 39kW
d) 44kW
Explanation: P= πp(D²-d²)/4 and M=μP(Dᵌ-dᵌ)/3(D²-d²).Power=Mxω.

12. If number of contacting surfaces are 5, then number of disks required in multi disk clutch are?
a) 4
b) 5
c) 6
d) Can’t be determined
Explanation: Disks=Contacting surfaces+1.

13. Multi disk clutches are dry clutches.
a) Plasma clutches
b) Wet clutches
c) Yes
d) Depends on the lubrication used
Explanation: They are wet clutches as a lot of heat is dissipated due to more contacting surfaces.

14. In scooters, generally single plate clutches are used.
a) True
b) False
Explanation: Multi disk plate clutches which are compact are used.

15. The coefficient of friction is high in multi disk plate clutch.
a) Yes
b) Coefficient of friction is less
c) Coeffficient of friction is high
d) None of the listed
Explanation: In multi disk clutch, due to oil cooling, coefficient of friction decrease.

16. A cone clutch consists of inner conical surface and outer cylindrical surface.
a) Both cylindrical
b) Both conical
c) Outer conical and inner cylindrical
d) True
Explanation: It consists of both inner and outer conical surfaces.

17. Power is transmitted only by key and friction in the cone clutch.
a) Only by spline
b) By key, spline and friction
c) By friction only
d) By key only
Explanation: Power is transmitted via key, friction and spline.

18. The torque transmitting capacity of cone clutch increases as its semi vertical angle increase.
a) True
b) Decreases
c) Remains constant
d) None of the listed
Explanation: Torque transmitted is inversely proportional to the sin of the semi vertical angle.

19. When semi vertical angle is greater than angle of static friction, clutch results in self-engagement.
a) True
b) False
Explanation: Self engagement takes place when semi vertical angle is reduced so low to increase torque capacity that it become smaller than static friction angle.

20. A cone clutch with a small semi cone angle requires a relatively large force to engage and small force to disengage.
a) True
b) False
Explanation: It requires large force to disengage.

21. A cone clutch transmits 24kW at 490rpm. The coefficient of friction is 0.2 and allowable intensity of pressure is 0.35N/mm². The semi cone angle is 12⁰. The outer diameter is fixed as 310mm.Assuming uniform wear theory; find the maximum torque which is transmitted.
a) 502.4N-m
b) 542.3N-m
c) 467.72N-m
d) 454.5N-m
Explanation: M=Powerx10ᵌx60/2πN.

22. A cone clutch transmits 24kW at 490rpm. The coefficient of friction is 0.2 and allowable intensity of pressure is 0.35N/mm². The semi cone angle is 12⁰. The outer diameter is fixed as 310mm.Assuming uniform wear theory; find the inner diameter.
a) 275mm
b) 300mm
c) 290mm
d) 280mm
Explanation: M=πpμd(D²-d²)/8Sinἀ.

23. A cone clutch transmits 24kW at 490rpm. The coefficient of friction is 0.2 and allowable intensity of pressure is 0.35N/mm². The semi cone angle is 12⁰. The outer diameter is fixed as 310mm.Assuming uniform wear theory; find the face width of friction lining.
a) None of the listed
b) 48.1mm
c) 52.2mm
d) 56.8mm
Explanation: b=D-d/2Sinἀ.

24. A cone clutch transmits 24kW at 490rpm. The coefficient of friction is 0.2 and allowable intensity of pressure is 0.35N/mm². The semi cone angle is 12⁰. The outer diameter is fixed as 310mm.Assuming uniform wear theory; calculate force required to engage clutch.
a) 3546.9N
b) 2231.5N
c) 3241.5N
d) 4354.5N
Explanation: P=4MSinἀ/μ(D+d).

25. Centrifugal clutches are not recommended for IC engine applications.
a) True
b) False
Explanation: IC engine is not able to start under load ad hence centrifugal clutches are used.

26. In a centrifugal clutch, when the centrifugal force is slightly more than the spring force, shoe begins to move in a radially inward direction.
Explanation: It moves in radially outward direction.

27. Chain saws, lawnmowers, golf carts etc. never use centrifugal clutches.
a) True
b) No, it is highly used
c) Depends on the load required
d) None of the listed
Explanation: The electric motor has time to accelerate to reach the operating speed before it takes the load and hence centrifugal clutches are used.

28. The centrifugal clutches are used in light duty vehicles and not in heavy duty vehicles.
a) True
b) False
Explanation: Centrifugal clutches are used in heavy uty applications as they provide a time delay that is sufficient to permit the prime mover to gain momentum before taking over the load.

29. Find number of contacting surfaces for a multi disk clutch plate transmitting torque of 10N-m and inner and outer diameters of friction lining are 70mm and 100mm respectively. The operating force is of magnitude 305N and coefficient of friction is 0.2.
a) 5
b) 2
c) 4
d) 3
Explanation: z=4M/μP(D+d).

30. The intensity of normal pressure between the friction lining and the brake drum at any point is proportional to square of the vertical distance from the pivot.
a) It is independent
b) Proportional to vertical distance linearly
c) Inversely proportion to vertical distance
d) None of the listed
Explanation: It is proportional to the vertical distance linearly.

31. The coefficient of friction in internally expanding brakes is constant.
a) True
b) No, it varies with increase in speed linearly
c) It decreases with increase in speed linearly
d) None of the listed
Explanation: Coefficient of friction is constant in these types of brakes.

32. The centrifugal force acting on the shoe can never be omitted.
a) True
b) Always omitted
c) Depends on the magnitude
d) There is no centrifugal force acting
Explanation: Centrifugal force on the internal expanding brakes is generally neglected as it is very small.

33. These kinds of brakes require little maintenance.
a) Yes
b) No
c) Very much maintenance is required
d) Depends on the working environment
Explanation: Very little maintenance is required as there is less wear.

34. In internally expanding brake, a large actuating force produces a small braking torque and hence this brake isn’t used generally.
a) True
b) No, small actuating braking torque is produces
c) Zero actuating force is required
d) None of the listed
Explanation: Small actuating force produce large braking torque.

35. Heat dissipating capacity of internally expanding brakes is a problem.
a) True
b) No
c) There is no heat generated
d) Heat is convected away on its own
Explanation: It has relatively poor dissipation of heat.

36. Internally expanding brakes never suffer the problem of self-locking.
a) True
b) Wear might lead to self-locking
c) Brakes are never self-locked
d) None of the listed
Explanation: Due to improper design, wear may be caused which might lead to self-locking of brakes.

37. Moment of normal force and frictional forces about the pivot axis are 640000N-mm and 250000N-mm respectively. If force acts at a distance of 190mm from the pivoted point, calculate the actuating force.
a) 1078.6N
b) 2052.6N
c) 3223.5N
d) 4454.5N
Explanation: P=M(n)-M(f)/C.

a) True
b) False

39. Needles bearings have not much load capacity as they are quite small.
a) True
b) False
Explanation: Needle bearings have quite large loading capacity as compared to their size.

40. The energy absorb by brake is always kinetic.
a) No, potential
b) Kinetic or potential
c) Potential
d) Strain Energy
Explanation: It can be either kinetic or potential.

41. Pneumatic brakes are same as electrical brakes.
a) Yes both are concerned with electricity
b) No, one deals with pressure and other with electricity
c) Yes both deals with pressure
d) None of the listed
Explanation: Pneumatic brakes are operated by fluid pressure.

42. Disc brakes are radial brakes.
a) True
b) False
Explanation: Disc brakes are axial brakes.

43. Internal shoe brakes are radial while external shoe brakes are axial brakes.
a) True
b) False
Explanation: Both internal and external shoe brakes are radial brakes.

44. A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Calculate energy absorbed by the brake if square of radius of gyration is 0.2.
a) None of the mentioned
b) 134.3kJ
c) 165.3kJ
d) 134.2kJ
Explanation: E=mk²ω²/2.

45. A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the average angular velocity during braking period.
d) None of the mentioned
Explanaton: ω(avg)=[ω()initial+ω(final)]/2.

46. A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the angle through which disk rotated during braking period.
Explanation: ω(avg)=[ω()initial+ω(final)]/2 and θ=ωt.

47. A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the torque capacity of the brake.
a) 812.4N-m
b) 4583.6N-m
c) 612.4N-m
d) Noe of the listed