Theory of Machines and Vibrations MCQs 2026

83 questions with detailed answers · 40 from past papers · 9 quiz batches available

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Page 1 of 1 Questions 110 of 83
  1. Q1 Past Paper · PPSC/FPSC/NTS medium

    Helical gears differ from spur by

    1. A having no pitch circle
    2. B teeth cut at helix angle to axis
    3. C meshing only in vertical plane always
    4. D using only plastic always
    💡 Explanation:

    Gradual engagement; quieter and smoother at high speed.

  2. Q2 medium

    Helix angle in helical gear is measured at

    1. A root circle only
    2. B addendum circle only
    3. C lathe bed guide
    4. D pitch cylinder relative to gear axis
    💡 Explanation:

    Axial thrust generated; thrust bearings may be needed.

  3. Q3 hard

    Herringbone gear eliminates

    1. A net axial thrust of double helical arrangement
    2. B all tooth loading
    3. C need for lubrication
    4. D pitch diameter definition
    💡 Explanation:

    Two opposite helix halves on same gear.

  4. Q4 Past Paper · PPSC/FPSC/NTS medium

    Bevel gears transmit motion between

    1. A parallel axes only always
    2. B non-intersecting skew axes without hypoid
    3. C intersecting axes usually at 90 degrees
    4. D fixed ratio belt only
    💡 Explanation:

    Pitch cones roll without slip theoretically.

  5. Q5 Past Paper · PPSC/FPSC/NTS medium

    Worm gear pair has

    1. A high speed reduction and self-locking tendency possible
    2. B only parallel axis layout
    3. C equal size pinion and gear always
    4. D no sliding contact ever
    💡 Explanation:

    Worm is screw; wheel is partial helical gear.

  6. Q6 Past Paper · PPSC/FPSC/NTS easy

    Gear train velocity ratio is product of

    1. A modules only
    2. B pressure angles only
    3. C driven teeth divided by driver teeth for each stage
    4. D face widths only
    💡 Explanation:

    Overall ratio combines idler and compound stages.

  7. Q7 easy

    Idler gear in train changes

    1. A direction of rotation without affecting ratio magnitude
    2. B overall velocity ratio value
    3. C centre distance law only
    4. D module standard
    💡 Explanation:

    Teeth count cancels in ratio product.

  8. Q8 Past Paper · PPSC/FPSC/NTS medium

    Epicyclic gear train has

    1. A one or more gears rotating on moving arm
    2. B only fixed axis gears
    3. C no planet gears ever
    4. D only belt drives
    💡 Explanation:

    Automotive automatic transmission planetary sets.

  9. Q9 hard

    Reverted gear train has

    1. A input and output shafts coaxial
    2. B parallel offset shafts only
    3. C intersecting axes only
    4. D no idler allowed ever
    💡 Explanation:

    Compound arrangement with equal centre distances often.

  10. Q10 Past Paper · PPSC/FPSC/NTS easy

    Governor maintains

    1. A constant fuel tank level
    2. B constant tyre pressure
    3. C near-constant speed of engine despite load changes
    4. D constant sand moisture
    💡 Explanation:

    Centrifugal force of masses linked to throttle.

  11. Q11 Past Paper · PPSC/FPSC/NTS easy

    Watt governor is

    1. A hydraulic piston only
    2. B electronic ECU map only
    3. C pendulum type with flyballs on arms near spindle
    4. D turbocharger wastegate only
    💡 Explanation:

    Proportional controller; hunting at high speeds.

  12. Q12 Past Paper · PPSC/FPSC/NTS medium

    Porter governor has

    1. A no controlling force element
    2. B only hydraulic dashpot from start
    3. C central spring supporting flyball arms
    4. D fixed mass without arms
    💡 Explanation:

    Spring adds controlling force; improved sensitivity over Watt.

  13. Q13 medium

    Proell governor extends flyballs

    1. A inward toward spindle only
    2. B outward on extended links for greater speed rise effect
    3. C eliminates all sensitivity
    4. D replaces throttle completely
    💡 Explanation:

    Same balls; longer arms increase centrifugal moment.

  14. Q14 hard

    Sensitiveness of governor is ratio of

    1. A power output to fuel flow only
    2. B flywheel energy to torque only
    3. C range of speed to mean speed
    4. D gear ratio to module
    💡 Explanation:

    (N2-N1)/N where N mean speed.

  15. Q15 medium

    Hunting in governor means

    1. A stable immediate stop
    2. B zero response to load
    3. C continuous speed oscillation about mean
    4. D fixed throttle always
    💡 Explanation:

    Isochronous or damper needed to reduce hunting.

  16. Q16 Past Paper · PPSC/FPSC/NTS easy

    Flywheel stores energy as

    1. A chemical fuel energy only
    2. B rotational kinetic energy smoothing cyclic fluctuation
    3. C elastic spring potential in governor only
    4. D electrical capacitor charge only
    💡 Explanation:

    E = ½Iω²; reduces coefficient of speed fluctuation.

  17. Q17 Past Paper · PPSC/FPSC/NTS medium

    Coefficient of fluctuation of speed is

    1. A flywheel weight only
    2. B gear pressure angle
    3. C cam lift divided by base circle
    4. D (ω1 - ω2)/ω_mean
    💡 Explanation:

    Smaller value means tighter speed regulation.

  18. Q18 medium

    Flywheel rim has most mass at periphery because

    1. A centre hub stores more energy always
    2. B spokes alone sufficient always
    3. C bearing friction requires it
    4. D moment of inertia increases with radius squared
    💡 Explanation:

    I = mk²; k radius of gyration maximized at rim.

  19. Q19 Past Paper · PPSC/FPSC/NTS medium

    Turning moment diagram plots

    1. A velocity versus time only always
    2. B stress versus strain only
    3. C crank torque versus crank angle for one cycle
    4. D pressure versus volume only
    💡 Explanation:

    Area equals work per cycle; flywheel sizing from fluctuation.

  20. Q20 Past Paper · PPSC/FPSC/NTS easy

    Static balancing of rotating masses requires

    1. A couple in plane must be maximum
    2. B only single mass considered always
    3. C gear teeth must be involute
    4. D resultant centrifugal force in any plane is zero
    💡 Explanation:

    Σmrω² = 0 vector sum for coplanar masses.

  21. Q21 Past Paper · PPSC/FPSC/NTS medium

    Dynamic balancing requires

    1. A only force balance in one plane
    2. B resultant force and resultant couple both zero
    3. C no consideration of axial length
    4. D welding symmetry only
    💡 Explanation:

    Two-plane balance for long rotors.

  22. Q22 medium

    Balancing machine measures

    1. A only hardness HRC
    2. B only surface Ra
    3. C only cam lift
    4. D vibration or force due to unbalance at supports
    💡 Explanation:

    Determines amount and angular location of correction mass.

  23. Q23 hard

    Primary unbalance in reciprocating engine is

    1. A first-order force along line of stroke at crank speed
    2. B only constant torque always
    3. C only gear mesh frequency
    4. D zero at all speeds
    💡 Explanation:

    Partially balanced by counterweights on crank.

  24. Q24 hard

    Secondary unbalance occurs at

    1. A half crank speed only
    2. B twice crank speed in four-stroke inline engines
    3. C zero frequency always
    4. D cam shaft speed only for all orders
    💡 Explanation:

    Inherent in inline-four geometry.

  25. Q25 medium

    Gyroscope resists change in

    1. A linear speed along road only
    2. B fuel octane number
    3. C sand binder type
    4. D axis of rotation due to angular momentum
    💡 Explanation:

    Precession occurs when torque applied perpendicular to spin axis.

  26. Q26 hard

    Precession angular velocity of gyro is

    1. A Iω divided by T
    2. B T times ω only
    3. C zero for all applied torque
    4. D T / (Iω)
    💡 Explanation:

    ω_p = T/(Iω); slow precession for large Iω.

  27. Q27 medium

    Velocity diagram in relative motion shows

    1. A only scalar gear teeth count
    2. B only temperature entropy
    3. C only weld pool width
    4. D vector addition of absolute velocities of links
    💡 Explanation:

    Instant centre method or relative velocity polygons.

  28. Q28 hard

    Kennedy theorem states I-centres of three bodies

    1. A form equilateral triangle always
    2. B coincide at mass centre always
    3. C lie on a straight line
    4. D are unrelated
    💡 Explanation:

    Locates unknown I-centre using two known.

  29. Q29 hard

    Rubbing velocity at pin joint equals

    1. A angular velocity times pin radius if one link fixed locally
    2. B sum of link lengths
    3. C product of modules
    4. D zero always at all joints
    💡 Explanation:

    Used in wear and lubrication analysis.

  30. Q30 medium

    Instantaneous centre of rotation is

    1. A always at geometric centre of mass
    2. B fixed for all time always
    3. C point about which body appears to rotate at given instant
    4. D only for gears never links
    💡 Explanation:

    Kennedy theorem locates I-centres between links.

  31. Q31 Past Paper · PPSC/FPSC/NTS medium

    Velocity of point on link equals

    1. A angular velocity times radius to I-centre
    2. B only linear speed of slider always
    3. C zero at all pins always
    4. D product of masses only
    💡 Explanation:

    v = ωr in rotation about I-centre.

  32. Q32 hard

    Acceleration in mechanism analysis includes

    1. A only static weight
    2. B tangential and normal components relative to I-centre
    3. C only thermal expansion
    4. D only sand shrinkage
    💡 Explanation:

    Normal ω²r toward centre; tangential αr.

  33. Q33 hard

    Coriolis component appears in

    1. A pure rotation about fixed centre only
    2. B sliding motion on rotating link
    3. C static equilibrium only
    4. D heat conduction only
    💡 Explanation:

    2ωv relative in acceleration analysis of slider on crank.

  34. Q34 Past Paper · PPSC/FPSC/NTS easy

    Lower pair has

    1. A surface contact between elements
    2. B point or line contact only always
    3. C no contact ever
    4. D fluid film only in gears
    💡 Explanation:

    Revolute and prismatic joints are lower pairs.

  35. Q35 medium

    Higher pair example is

    1. A pin in hole revolute joint
    2. B gear teeth mesh or cam-follower point contact
    3. C slider in slot prismatic
    4. D welded fixed joint
    💡 Explanation:

    Line or point contact; one degree of freedom higher pair removes.

  36. Q36 Past Paper · PPSC/FPSC/NTS medium

    Kinematic inversion changes

    1. A tooth profile of gears only
    2. B material of links only
    3. C motor voltage only
    4. D which link is fixed without altering relative motions
    💡 Explanation:

    Same chain; different fixed link gives different applications.

  37. Q37 hard

    Double slider crank chain forms

    1. A Oldham coupling or Scotch yoke mechanisms
    2. B only simple pendulum
    3. C only flywheel energy store
    4. D only governor spring
    💡 Explanation:

    Two prismatic pairs with two revolute.

  38. Q38 Past Paper · PPSC/FPSC/NTS easy

    Cam-follower mechanism converts

    1. A cam rotation into prescribed follower translation or oscillation
    2. B electricity to refrigerant
    3. C sand to molten iron
    4. D rolling torque to billet only
    💡 Explanation:

    Profile determines displacement, velocity and acceleration laws.

  39. Q39 medium

    Disk cam with knife-edge follower has

    1. A point contact and high wear
    2. B surface contact lower pair
    3. C no pressure angle concern
    4. D constant velocity always
    💡 Explanation:

    Practical followers are roller or flat faced.

  40. Q40 Past Paper · PPSC/FPSC/NTS easy

    Roller follower reduces

    1. A cam lift amplitude to zero
    2. B need for cam profile
    3. C friction and wear compared to knife-edge follower
    4. D all dynamic forces to zero
    💡 Explanation:

    Rolling contact; offset affects pressure angle.

  41. Q41 Past Paper · PPSC/FPSC/NTS medium

    Pressure angle in cam is angle between

    1. A cam shaft and crank throw only
    2. B normal to profile and follower motion direction
    3. C gear helix and axis only
    4. D welding torch and plate only
    💡 Explanation:

    High pressure angle increases side thrust on bearings.

  42. Q42 easy

    Base circle of cam is

    1. A smallest circle centred on cam axis touching profile
    2. B pitch circle of gear
    3. C root circle of gear
    4. D lathe chuck diameter
    💡 Explanation:

    Lift measured radially outward from base circle.

  43. Q43 medium

    Simple harmonic motion cam profile gives

    1. A infinite jerk everywhere
    2. B constant velocity throughout dwell
    3. C random displacement
    4. D smooth acceleration at start and end of stroke
    💡 Explanation:

    Sinusoidal displacement law used for moderate speeds.

  44. Q44 easy

    Dwell in cam motion means

    1. A maximum velocity segment
    2. B follower oscillates rapidly
    3. C follower stationary while cam rotates
    4. D cam stops rotating
    💡 Explanation:

    Period with zero follower displacement.

  45. Q45 Past Paper · PPSC/FPSC/NTS easy

    Module m of spur gear equals

    1. A circular pitch divided by pi only
    2. B pitch circle diameter divided by number of teeth
    3. C addendum only
    4. D pressure angle in degrees
    💡 Explanation:

    m in mm; standardizes gear tooth size.

  46. Q46 Past Paper · PPSC/FPSC/NTS easy

    Circular pitch p of gear is

    1. A module divided by π
    2. B π times module
    3. C teeth divided by diameter
    4. D pressure angle cosine only
    💡 Explanation:

    p = πm; distance along pitch circle tooth to tooth.

  47. Q47 easy

    Addendum of gear tooth is

    1. A depth of dedendum only
    2. B whole face width
    3. C helix angle
    4. D radial height from pitch circle to tooth tip
    💡 Explanation:

    Typically equals one module for standard gears.

  48. Q48 easy

    Dedendum is measured from pitch circle to

    1. A tip circle only
    2. B base circle always as addendum
    3. C root of tooth
    4. D outside helix
    💡 Explanation:

    Clearance between tip of one gear and root of mate.

  49. Q49 Past Paper · PPSC/FPSC/NTS medium

    Law of gearing requires for constant velocity ratio

    1. A teeth must be rectangular
    2. B common normal at contact passes through pitch point
    3. C only helical gears qualify
    4. D pressure angle must be 90°
    💡 Explanation:

    Involute profile satisfies law when centres fixed.

  50. Q50 medium

    Involute tooth profile is generated by

    1. A extruding metal through die
    2. B sand casting pattern only
    3. C unwrapping string from base circle
    4. D resistance welding spot
    💡 Explanation:

    Conjugate action; centre distance tolerance permissible.

  51. Q51 Past Paper · PPSC/FPSC/NTS medium

    Contact ratio in gears should be

    1. A zero for quiet operation
    2. B negative for strength
    3. C exactly 0.5 always
    4. D greater than one for continuous smooth transmission
    💡 Explanation:

    Sum of lengths of path of contact over base pitch.

  52. Q52 hard

    Undercutting in spur gears occurs when

    1. A tip of mating gear cuts into root below base circle
    2. B pressure angle is too large only always
    3. C module is too large always
    4. D face width is excessive
    💡 Explanation:

    Small pinion teeth with low pressure angle prone; corrected by shift.

  53. Q53 Past Paper · PPSC/FPSC/NTS easy

    A kinematic chain becomes mechanism when

    1. A all links move freely without frame
    2. B no relative motion exists
    3. C one link is fixed to ground frame
    4. D only gears are present
    💡 Explanation:

    Mobility equation requires fixed link for constrained motion.

  54. Q54 Past Paper · PPSC/FPSC/NTS medium

    Degrees of freedom of plane mechanism given by Kutzbach equation is

    1. A F = L + j only
    2. B F = 3(L-1) - 2j - h
    3. C F = 2L always
    4. D F = j - h only
    💡 Explanation:

    L links, j lower pairs, h higher pairs; ground link included.

  55. Q55 Past Paper · PPSC/FPSC/NTS easy

    Four-bar linkage has

    1. A three links and five pairs
    2. B only prismatic pairs
    3. C no coupler link
    4. D four links connected by four revolute pairs
    💡 Explanation:

    Grashof condition determines crank existence.

  56. Q56 Past Paper · PPSC/FPSC/NTS medium

    Grashof law states for continuous relative rotation

    1. A all links equal length always
    2. B shortest plus longest link length ≤ sum of other two
    3. C only sliding pairs allowed
    4. D coupler must be longest always
    💡 Explanation:

    At least one link makes full revolution if satisfied.

  57. Q57 Past Paper · PPSC/FPSC/NTS easy

    Slider-crank mechanism converts

    1. A heat directly to electricity only
    2. B hydraulic pressure to sand mould
    3. C welding arc to lathe feed
    4. D reciprocating slider motion to rotary crank motion
    💡 Explanation:

    Basis of internal combustion engine piston-crank system.

  58. Q58 Past Paper · PPSC/FPSC/NTS medium

    Bevel gears transmit motion between

    1. A parallel axes only always
    2. B non-intersecting skew axes without hypoid
    3. C intersecting axes usually at 90 degrees
    4. D fixed ratio belt only
    💡 Explanation:

    Pitch cones roll without slip theoretically.

  59. Q59 Past Paper · PPSC/FPSC/NTS medium

    Worm gear pair has

    1. A high speed reduction and self-locking tendency possible
    2. B only parallel axis layout
    3. C equal size pinion and gear always
    4. D no sliding contact ever
    💡 Explanation:

    Worm is screw; wheel is partial helical gear.

  60. Q60 Past Paper · PPSC/FPSC/NTS easy

    Gear train velocity ratio is product of

    1. A modules only
    2. B pressure angles only
    3. C driven teeth divided by driver teeth for each stage
    4. D face widths only
    💡 Explanation:

    Overall ratio combines idler and compound stages.

  61. Q61 easy

    Idler gear in train changes

    1. A direction of rotation without affecting ratio magnitude
    2. B overall velocity ratio value
    3. C centre distance law only
    4. D module standard
    💡 Explanation:

    Teeth count cancels in ratio product.

  62. Q62 Past Paper · PPSC/FPSC/NTS medium

    Epicyclic gear train has

    1. A one or more gears rotating on moving arm
    2. B only fixed axis gears
    3. C no planet gears ever
    4. D only belt drives
    💡 Explanation:

    Automotive automatic transmission planetary sets.

  63. Q63 hard

    Reverted gear train has

    1. A input and output shafts coaxial
    2. B parallel offset shafts only
    3. C intersecting axes only
    4. D no idler allowed ever
    💡 Explanation:

    Compound arrangement with equal centre distances often.

  64. Q64 Past Paper · PPSC/FPSC/NTS easy

    Governor maintains

    1. A constant fuel tank level
    2. B constant tyre pressure
    3. C near-constant speed of engine despite load changes
    4. D constant sand moisture
    💡 Explanation:

    Centrifugal force of masses linked to throttle.

  65. Q65 Past Paper · PPSC/FPSC/NTS easy

    Watt governor is

    1. A hydraulic piston only
    2. B electronic ECU map only
    3. C pendulum type with flyballs on arms near spindle
    4. D turbocharger wastegate only
    💡 Explanation:

    Proportional controller; hunting at high speeds.

  66. Q66 Past Paper · PPSC/FPSC/NTS medium

    Porter governor has

    1. A no controlling force element
    2. B only hydraulic dashpot from start
    3. C central spring supporting flyball arms
    4. D fixed mass without arms
    💡 Explanation:

    Spring adds controlling force; improved sensitivity over Watt.

  67. Q67 medium

    Proell governor extends flyballs

    1. A inward toward spindle only
    2. B outward on extended links for greater speed rise effect
    3. C eliminates all sensitivity
    4. D replaces throttle completely
    💡 Explanation:

    Same balls; longer arms increase centrifugal moment.

  68. Q68 hard

    Sensitiveness of governor is ratio of

    1. A power output to fuel flow only
    2. B flywheel energy to torque only
    3. C range of speed to mean speed
    4. D gear ratio to module
    💡 Explanation:

    (N2-N1)/N where N mean speed.

  69. Q69 medium

    Hunting in governor means

    1. A stable immediate stop
    2. B zero response to load
    3. C continuous speed oscillation about mean
    4. D fixed throttle always
    💡 Explanation:

    Isochronous or damper needed to reduce hunting.

  70. Q70 Past Paper · PPSC/FPSC/NTS easy

    Flywheel stores energy as

    1. A chemical fuel energy only
    2. B rotational kinetic energy smoothing cyclic fluctuation
    3. C elastic spring potential in governor only
    4. D electrical capacitor charge only
    💡 Explanation:

    E = ½Iω²; reduces coefficient of speed fluctuation.

  71. Q71 Past Paper · PPSC/FPSC/NTS medium

    Coefficient of fluctuation of speed is

    1. A flywheel weight only
    2. B gear pressure angle
    3. C cam lift divided by base circle
    4. D (ω1 - ω2)/ω_mean
    💡 Explanation:

    Smaller value means tighter speed regulation.

  72. Q72 medium

    Flywheel rim has most mass at periphery because

    1. A centre hub stores more energy always
    2. B spokes alone sufficient always
    3. C bearing friction requires it
    4. D moment of inertia increases with radius squared
    💡 Explanation:

    I = mk²; k radius of gyration maximized at rim.

  73. Q73 Past Paper · PPSC/FPSC/NTS medium

    Turning moment diagram plots

    1. A velocity versus time only always
    2. B stress versus strain only
    3. C crank torque versus crank angle for one cycle
    4. D pressure versus volume only
    💡 Explanation:

    Area equals work per cycle; flywheel sizing from fluctuation.

  74. Q74 Past Paper · PPSC/FPSC/NTS easy

    Static balancing of rotating masses requires

    1. A couple in plane must be maximum
    2. B only single mass considered always
    3. C gear teeth must be involute
    4. D resultant centrifugal force in any plane is zero
    💡 Explanation:

    Σmrω² = 0 vector sum for coplanar masses.

  75. Q75 Past Paper · PPSC/FPSC/NTS medium

    Dynamic balancing requires

    1. A only force balance in one plane
    2. B resultant force and resultant couple both zero
    3. C no consideration of axial length
    4. D welding symmetry only
    💡 Explanation:

    Two-plane balance for long rotors.

  76. Q76 medium

    Balancing machine measures

    1. A only hardness HRC
    2. B only surface Ra
    3. C only cam lift
    4. D vibration or force due to unbalance at supports
    💡 Explanation:

    Determines amount and angular location of correction mass.

  77. Q77 hard

    Primary unbalance in reciprocating engine is

    1. A first-order force along line of stroke at crank speed
    2. B only constant torque always
    3. C only gear mesh frequency
    4. D zero at all speeds
    💡 Explanation:

    Partially balanced by counterweights on crank.

  78. Q78 hard

    Secondary unbalance occurs at

    1. A half crank speed only
    2. B twice crank speed in four-stroke inline engines
    3. C zero frequency always
    4. D cam shaft speed only for all orders
    💡 Explanation:

    Inherent in inline-four geometry.

  79. Q79 medium

    Gyroscope resists change in

    1. A linear speed along road only
    2. B fuel octane number
    3. C sand binder type
    4. D axis of rotation due to angular momentum
    💡 Explanation:

    Precession occurs when torque applied perpendicular to spin axis.

  80. Q80 hard

    Precession angular velocity of gyro is

    1. A Iω divided by T
    2. B T times ω only
    3. C zero for all applied torque
    4. D T / (Iω)
    💡 Explanation:

    ω_p = T/(Iω); slow precession for large Iω.

  81. Q81 medium

    Velocity diagram in relative motion shows

    1. A only scalar gear teeth count
    2. B only temperature entropy
    3. C only weld pool width
    4. D vector addition of absolute velocities of links
    💡 Explanation:

    Instant centre method or relative velocity polygons.

  82. Q82 hard

    Kennedy theorem states I-centres of three bodies

    1. A form equilateral triangle always
    2. B coincide at mass centre always
    3. C lie on a straight line
    4. D are unrelated
    💡 Explanation:

    Locates unknown I-centre using two known.

  83. Q83 hard

    Rubbing velocity at pin joint equals

    1. A angular velocity times pin radius if one link fixed locally
    2. B sum of link lengths
    3. C product of modules
    4. D zero always at all joints
    💡 Explanation:

    Used in wear and lubrication analysis.