Theory of Machines and Vibrations MCQs 2026

93 questions with detailed answers · 40 from past papers · 10 quiz batches available

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Page 1 of 1Questions 1–10 of 93
  1. Q1Past Paper · PPSC/FPSC/NTSeasy

    Governor maintains

    1. Aconstant fuel tank level
    2. Bconstant tyre pressure
    3. Cnear-constant speed of engine despite load changes
    4. Dconstant sand moisture
    💡 Explanation:

    Centrifugal force of masses linked to throttle.

  2. Q2Past Paper · PPSC/FPSC/NTSeasy

    A kinematic chain becomes mechanism when

    1. Aall links move freely without frame
    2. Bno relative motion exists
    3. Cone link is fixed to ground frame
    4. Donly gears are present
    💡 Explanation:

    Mobility equation requires fixed link for constrained motion.

  3. Q3Past Paper · PPSC/FPSC/NTSmedium

    Degrees of freedom of plane mechanism given by Kutzbach equation is

    1. AF = L + j only
    2. BF = 3(L-1) - 2j - h
    3. CF = 2L always
    4. DF = j - h only
    💡 Explanation:

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

  4. Q4Past Paper · PPSC/FPSC/NTSeasy

    Four-bar linkage has

    1. Athree links and five pairs
    2. Bonly prismatic pairs
    3. Cno coupler link
    4. Dfour links connected by four revolute pairs
    💡 Explanation:

    Grashof condition determines crank existence.

  5. Q5Past Paper · PPSC/FPSC/NTSmedium

    Grashof law states for continuous relative rotation

    1. Aall links equal length always
    2. Bshortest plus longest link length ≤ sum of other two
    3. Conly sliding pairs allowed
    4. Dcoupler must be longest always
    💡 Explanation:

    At least one link makes full revolution if satisfied.

  6. Q6Past Paper · PPSC/FPSC/NTSeasy

    Slider-crank mechanism converts

    1. Aheat directly to electricity only
    2. Bhydraulic pressure to sand mould
    3. Cwelding arc to lathe feed
    4. Dreciprocating slider motion to rotary crank motion
    💡 Explanation:

    Basis of internal combustion engine piston-crank system.

  7. Q7medium

    Instantaneous centre of rotation is

    1. Aalways at geometric centre of mass
    2. Bfixed for all time always
    3. Cpoint about which body appears to rotate at given instant
    4. Donly for gears never links
    💡 Explanation:

    Kennedy theorem locates I-centres between links.

  8. Q8Past Paper · PPSC/FPSC/NTSmedium

    Velocity of point on link equals

    1. Aangular velocity times radius to I-centre
    2. Bonly linear speed of slider always
    3. Czero at all pins always
    4. Dproduct of masses only
    💡 Explanation:

    v = ωr in rotation about I-centre.

  9. Q9hard

    Acceleration in mechanism analysis includes

    1. Aonly static weight
    2. Btangential and normal components relative to I-centre
    3. Conly thermal expansion
    4. Donly sand shrinkage
    💡 Explanation:

    Normal ω²r toward centre; tangential αr.

  10. Q10hard

    Coriolis component appears in

    1. Apure rotation about fixed centre only
    2. Bsliding motion on rotating link
    3. Cstatic equilibrium only
    4. Dheat conduction only
    💡 Explanation:

    2ωv relative in acceleration analysis of slider on crank.

  11. Q11Past Paper · PPSC/FPSC/NTSeasy

    Lower pair has

    1. Asurface contact between elements
    2. Bpoint or line contact only always
    3. Cno contact ever
    4. Dfluid film only in gears
    💡 Explanation:

    Revolute and prismatic joints are lower pairs.

  12. Q12medium

    Higher pair example is

    1. Apin in hole revolute joint
    2. Bgear teeth mesh or cam-follower point contact
    3. Cslider in slot prismatic
    4. D welded fixed joint
    💡 Explanation:

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

  13. Q13Past Paper · PPSC/FPSC/NTSmedium

    Kinematic inversion changes

    1. Atooth profile of gears only
    2. Bmaterial of links only
    3. Cmotor voltage only
    4. Dwhich link is fixed without altering relative motions
    💡 Explanation:

    Same chain; different fixed link gives different applications.

  14. Q14hard

    Double slider crank chain forms

    1. AOldham coupling or Scotch yoke mechanisms
    2. Bonly simple pendulum
    3. Conly flywheel energy store
    4. Donly governor spring
    💡 Explanation:

    Two prismatic pairs with two revolute.

  15. Q15Past Paper · PPSC/FPSC/NTSeasy

    Cam-follower mechanism converts

    1. Acam rotation into prescribed follower translation or oscillation
    2. Belectricity to refrigerant
    3. Csand to molten iron
    4. Drolling torque to billet only
    💡 Explanation:

    Profile determines displacement, velocity and acceleration laws.

  16. Q16medium

    Disk cam with knife-edge follower has

    1. Apoint contact and high wear
    2. Bsurface contact lower pair
    3. Cno pressure angle concern
    4. Dconstant velocity always
    💡 Explanation:

    Practical followers are roller or flat faced.

  17. Q17Past Paper · PPSC/FPSC/NTSeasy

    Roller follower reduces

    1. Acam lift amplitude to zero
    2. Bneed for cam profile
    3. Cfriction and wear compared to knife-edge follower
    4. Dall dynamic forces to zero
    💡 Explanation:

    Rolling contact; offset affects pressure angle.

  18. Q18Past Paper · PPSC/FPSC/NTSmedium

    Pressure angle in cam is angle between

    1. Acam shaft and crank throw only
    2. Bnormal to profile and follower motion direction
    3. Cgear helix and axis only
    4. Dwelding torch and plate only
    💡 Explanation:

    High pressure angle increases side thrust on bearings.

  19. Q19easy

    Base circle of cam is

    1. Asmallest circle centred on cam axis touching profile
    2. Bpitch circle of gear
    3. Croot circle of gear
    4. Dlathe chuck diameter
    💡 Explanation:

    Lift measured radially outward from base circle.

  20. Q20medium

    Simple harmonic motion cam profile gives

    1. Ainfinite jerk everywhere
    2. Bconstant velocity throughout dwell
    3. Crandom displacement
    4. Dsmooth acceleration at start and end of stroke
    💡 Explanation:

    Sinusoidal displacement law used for moderate speeds.

  21. Q21easy

    Dwell in cam motion means

    1. Amaximum velocity segment
    2. Bfollower oscillates rapidly
    3. Cfollower stationary while cam rotates
    4. Dcam stops rotating
    💡 Explanation:

    Period with zero follower displacement.

  22. Q22Past Paper · PPSC/FPSC/NTSeasy

    Module m of spur gear equals

    1. Acircular pitch divided by pi only
    2. Bpitch circle diameter divided by number of teeth
    3. Caddendum only
    4. Dpressure angle in degrees
    💡 Explanation:

    m in mm; standardizes gear tooth size.

  23. Q23Past Paper · PPSC/FPSC/NTSeasy

    Circular pitch p of gear is

    1. Amodule divided by π
    2. Bπ times module
    3. Cteeth divided by diameter
    4. Dpressure angle cosine only
    💡 Explanation:

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

  24. Q24easy

    Addendum of gear tooth is

    1. Adepth of dedendum only
    2. Bwhole face width
    3. Chelix angle
    4. Dradial height from pitch circle to tooth tip
    💡 Explanation:

    Typically equals one module for standard gears.

  25. Q25easy

    Dedendum is measured from pitch circle to

    1. Atip circle only
    2. Bbase circle always as addendum
    3. Croot of tooth
    4. Doutside helix
    💡 Explanation:

    Clearance between tip of one gear and root of mate.

  26. Q26Past Paper · PPSC/FPSC/NTSmedium

    Law of gearing requires for constant velocity ratio

    1. Ateeth must be rectangular
    2. Bcommon normal at contact passes through pitch point
    3. Conly helical gears qualify
    4. Dpressure angle must be 90°
    💡 Explanation:

    Involute profile satisfies law when centres fixed.

  27. Q27medium

    Involute tooth profile is generated by

    1. Aextruding metal through die
    2. Bsand casting pattern only
    3. Cunwrapping string from base circle
    4. Dresistance welding spot
    💡 Explanation:

    Conjugate action; centre distance tolerance permissible.

  28. Q28Past Paper · PPSC/FPSC/NTSmedium

    Contact ratio in gears should be

    1. Azero for quiet operation
    2. Bnegative for strength
    3. Cexactly 0.5 always
    4. Dgreater than one for continuous smooth transmission
    💡 Explanation:

    Sum of lengths of path of contact over base pitch.

  29. Q29hard

    Undercutting in spur gears occurs when

    1. Atip of mating gear cuts into root below base circle
    2. Bpressure angle is too large only always
    3. Cmodule is too large always
    4. Dface width is excessive
    💡 Explanation:

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

  30. Q30Past Paper · PPSC/FPSC/NTSmedium

    Helical gears differ from spur by

    1. Ahaving no pitch circle
    2. Bteeth cut at helix angle to axis
    3. Cmeshing only in vertical plane always
    4. Dusing only plastic always
    💡 Explanation:

    Gradual engagement; quieter and smoother at high speed.

  31. Q31medium

    Helix angle in helical gear is measured at

    1. Aroot circle only
    2. Baddendum circle only
    3. Clathe bed guide
    4. Dpitch cylinder relative to gear axis
    💡 Explanation:

    Axial thrust generated; thrust bearings may be needed.

  32. Q32hard

    Herringbone gear eliminates

    1. Anet axial thrust of double helical arrangement
    2. Ball tooth loading
    3. Cneed for lubrication
    4. Dpitch diameter definition
    💡 Explanation:

    Two opposite helix halves on same gear.

  33. Q33Past Paper · PPSC/FPSC/NTSmedium

    Bevel gears transmit motion between

    1. Aparallel axes only always
    2. Bnon-intersecting skew axes without hypoid
    3. Cintersecting axes usually at 90 degrees
    4. Dfixed ratio belt only
    💡 Explanation:

    Pitch cones roll without slip theoretically.

  34. Q34Past Paper · PPSC/FPSC/NTSmedium

    Worm gear pair has

    1. Ahigh speed reduction and self-locking tendency possible
    2. Bonly parallel axis layout
    3. Cequal size pinion and gear always
    4. Dno sliding contact ever
    💡 Explanation:

    Worm is screw; wheel is partial helical gear.

  35. Q35Past Paper · PPSC/FPSC/NTSeasy

    Gear train velocity ratio is product of

    1. Amodules only
    2. Bpressure angles only
    3. Cdriven teeth divided by driver teeth for each stage
    4. Dface widths only
    💡 Explanation:

    Overall ratio combines idler and compound stages.

  36. Q36easy

    Idler gear in train changes

    1. Adirection of rotation without affecting ratio magnitude
    2. Boverall velocity ratio value
    3. Ccentre distance law only
    4. Dmodule standard
    💡 Explanation:

    Teeth count cancels in ratio product.

  37. Q37Past Paper · PPSC/FPSC/NTSmedium

    Epicyclic gear train has

    1. Aone or more gears rotating on moving arm
    2. Bonly fixed axis gears
    3. Cno planet gears ever
    4. Donly belt drives
    💡 Explanation:

    Automotive automatic transmission planetary sets.

  38. Q38hard

    Reverted gear train has

    1. Ainput and output shafts coaxial
    2. Bparallel offset shafts only
    3. Cintersecting axes only
    4. Dno idler allowed ever
    💡 Explanation:

    Compound arrangement with equal centre distances often.

  39. Q39Past Paper · PPSC/FPSC/NTSeasy

    Watt governor is

    1. Ahydraulic piston only
    2. Belectronic ECU map only
    3. Cpendulum type with flyballs on arms near spindle
    4. Dturbocharger wastegate only
    💡 Explanation:

    Proportional controller; hunting at high speeds.

  40. Q40Past Paper · PPSC/FPSC/NTSmedium

    Porter governor has

    1. Ano controlling force element
    2. Bonly hydraulic dashpot from start
    3. Ccentral spring supporting flyball arms
    4. Dfixed mass without arms
    💡 Explanation:

    Spring adds controlling force; improved sensitivity over Watt.

  41. Q41medium

    Proell governor extends flyballs

    1. Ainward toward spindle only
    2. Boutward on extended links for greater speed rise effect
    3. Celiminates all sensitivity
    4. Dreplaces throttle completely
    💡 Explanation:

    Same balls; longer arms increase centrifugal moment.

  42. Q42hard

    Sensitiveness of governor is ratio of

    1. Apower output to fuel flow only
    2. Bflywheel energy to torque only
    3. Crange of speed to mean speed
    4. Dgear ratio to module
    💡 Explanation:

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

  43. Q43medium

    Hunting in governor means

    1. Astable immediate stop
    2. Bzero response to load
    3. Ccontinuous speed oscillation about mean
    4. Dfixed throttle always
    💡 Explanation:

    Isochronous or damper needed to reduce hunting.

  44. Q44Past Paper · PPSC/FPSC/NTSeasy

    Flywheel stores energy as

    1. Achemical fuel energy only
    2. Brotational kinetic energy smoothing cyclic fluctuation
    3. Celastic spring potential in governor only
    4. Delectrical capacitor charge only
    💡 Explanation:

    E = ½Iω²; reduces coefficient of speed fluctuation.

  45. Q45Past Paper · PPSC/FPSC/NTSmedium

    Coefficient of fluctuation of speed is

    1. Aflywheel weight only
    2. Bgear pressure angle
    3. Ccam lift divided by base circle
    4. D(ω1 - ω2)/ω_mean
    💡 Explanation:

    Smaller value means tighter speed regulation.

  46. Q46medium

    Flywheel rim has most mass at periphery because

    1. Acentre hub stores more energy always
    2. Bspokes alone sufficient always
    3. Cbearing friction requires it
    4. Dmoment of inertia increases with radius squared
    💡 Explanation:

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

  47. Q47Past Paper · PPSC/FPSC/NTSmedium

    Turning moment diagram plots

    1. Avelocity versus time only always
    2. Bstress versus strain only
    3. Ccrank torque versus crank angle for one cycle
    4. Dpressure versus volume only
    💡 Explanation:

    Area equals work per cycle; flywheel sizing from fluctuation.

  48. Q48Past Paper · PPSC/FPSC/NTSeasy

    Static balancing of rotating masses requires

    1. Acouple in plane must be maximum
    2. Bonly single mass considered always
    3. Cgear teeth must be involute
    4. Dresultant centrifugal force in any plane is zero
    💡 Explanation:

    Σmrω² = 0 vector sum for coplanar masses.

  49. Q49Past Paper · PPSC/FPSC/NTSmedium

    Dynamic balancing requires

    1. Aonly force balance in one plane
    2. Bresultant force and resultant couple both zero
    3. Cno consideration of axial length
    4. Dwelding symmetry only
    💡 Explanation:

    Two-plane balance for long rotors.

  50. Q50medium

    Balancing machine measures

    1. Aonly hardness HRC
    2. Bonly surface Ra
    3. Conly cam lift
    4. Dvibration or force due to unbalance at supports
    💡 Explanation:

    Determines amount and angular location of correction mass.

  51. Q51hard

    Primary unbalance in reciprocating engine is

    1. Afirst-order force along line of stroke at crank speed
    2. Bonly constant torque always
    3. Conly gear mesh frequency
    4. Dzero at all speeds
    💡 Explanation:

    Partially balanced by counterweights on crank.

  52. Q52hard

    Secondary unbalance occurs at

    1. Ahalf crank speed only
    2. Btwice crank speed in four-stroke inline engines
    3. Czero frequency always
    4. Dcam shaft speed only for all orders
    💡 Explanation:

    Inherent in inline-four geometry.

  53. Q53medium

    Gyroscope resists change in

    1. Alinear speed along road only
    2. Bfuel octane number
    3. Csand binder type
    4. Daxis of rotation due to angular momentum
    💡 Explanation:

    Precession occurs when torque applied perpendicular to spin axis.

  54. Q54hard

    Precession angular velocity of gyro is

    1. AIω divided by T
    2. BT times ω only
    3. Czero for all applied torque
    4. DT / (Iω)
    💡 Explanation:

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

  55. Q55medium

    Velocity diagram in relative motion shows

    1. Aonly scalar gear teeth count
    2. Bonly temperature entropy
    3. Conly weld pool width
    4. Dvector addition of absolute velocities of links
    💡 Explanation:

    Instant centre method or relative velocity polygons.

  56. Q56hard

    Kennedy theorem states I-centres of three bodies

    1. Aform equilateral triangle always
    2. Bcoincide at mass centre always
    3. Clie on a straight line
    4. Dare unrelated
    💡 Explanation:

    Locates unknown I-centre using two known.

  57. Q57hard

    Rubbing velocity at pin joint equals

    1. Aangular velocity times pin radius if one link fixed locally
    2. Bsum of link lengths
    3. Cproduct of modules
    4. Dzero always at all joints
    💡 Explanation:

    Used in wear and lubrication analysis.

  58. Q58medium

    Toggle mechanism at dead centre has

    1. Azero force transmission always
    2. Bmechanical advantage approaching infinity theoretically
    3. Cmaximum velocity ratio always
    4. Dno fixed link
    💡 Explanation:

    Used in presses near top dead centre.

  59. Q59hard

    Peaucellier linkage generates

    1. Aexact straight line motion from rotation
    2. Bparabolic cam profile only
    3. Chelical gear teeth only
    4. Drandom vibration path
    💡 Explanation:

    Inversor linkage of historical importance.

  60. Q60medium

    Whitworth quick return used in

    1. Alathe headstock only
    2. BCNC tool changer only
    3. Ccar differential only
    4. Dshaper and slotting machines
    💡 Explanation:

    Time ratio of cutting to return stroke > 1.

  61. Q61medium

    Geneva mechanism provides

    1. Aconstant smooth rotation only
    2. Blinear welding seam
    3. Cintermittent rotary motion from continuous input
    4. Dhydraulic pump flow
    💡 Explanation:

    Used in indexing tables and watches.

  62. Q62easy

    Ratchet and pawl allows

    1. Abidirectional equal torque
    2. Bvariable gear module
    3. Ccam dwell elimination
    4. Drotation in one direction only
    💡 Explanation:

    Prevents reverse rotation of wheel or shaft.

  63. Q63easy

    Belt drive velocity ratio depends on

    1. Aonly belt colour
    2. Bonly sand grain size
    3. Conly lathe feed rate
    4. Ddiameters of driver and driven pulleys if no slip
    💡 Explanation:

    N1/N2 = D2/D1 for flat belt ideal.

  64. Q64medium

    Creep in belt drive is

    1. Atotal belt fracture
    2. Bzero tension condition
    3. Cwelding arc instability
    4. Drelative slip due to elastic stretch difference tight and slack sides
    💡 Explanation:

    Small gradual slip even under load below breaking.

  65. Q65easy

    Chain drive compared to belt has

    1. Ano slip and suitable for short centre distance high load
    2. Bonly elastic stretch transmission
    3. Cmust always be vertical
    4. Dcannot work in oil
    💡 Explanation:

    Roller chain on sprockets; periodic lubrication.

  66. Q66medium

    Synchronizing mesh in manual gearbox uses

    1. Aonly open dog clutch always without sync
    2. Bfriction cones or blocker rings to match shaft speeds
    3. Chydraulic torque converter only
    4. DCVT rubber belt only
    💡 Explanation:

    Prevents gear clash before engagement.

  67. Q67easy

    Frictional clutch transmits torque by

    1. Agear tooth only always
    2. Bhydraulic fluid shear only in manual plate clutch
    3. Cnormal force times coefficient of friction on mating surfaces
    4. Delectromagnetic flux in plate clutch only always
    💡 Explanation:

    Single or multi-plate; pressure plate springs.

  68. Q68medium

    Universal joint transmits motion between

    1. Aperfectly constant velocity always alone
    2. Bparallel non-intersecting axes without CV joint
    3. Cinclined intersecting shafts with varying angular velocity
    4. Dfixed welding only
    💡 Explanation:

    Double Hooke joint restores constant velocity.

  69. Q69hard

    Primary unbalance in reciprocating engine is

    1. Afirst-order force along line of stroke at crank speed
    2. Bonly constant torque always
    3. Conly gear mesh frequency
    4. Dzero at all speeds
    💡 Explanation:

    Partially balanced by counterweights on crank.

  70. Q70medium

    Balancing machine measures

    1. Aonly hardness HRC
    2. Bonly surface Ra
    3. Conly cam lift
    4. Dvibration or force due to unbalance at supports
    💡 Explanation:

    Determines amount and angular location of correction mass.

  71. Q71Past Paper · PPSC/FPSC/NTSmedium

    Dynamic balancing requires

    1. Aonly force balance in one plane
    2. Bresultant force and resultant couple both zero
    3. Cno consideration of axial length
    4. Dwelding symmetry only
    💡 Explanation:

    Two-plane balance for long rotors.

  72. Q72Past Paper · PPSC/FPSC/NTSeasy

    Static balancing of rotating masses requires

    1. Acouple in plane must be maximum
    2. Bonly single mass considered always
    3. Cgear teeth must be involute
    4. Dresultant centrifugal force in any plane is zero
    💡 Explanation:

    Σmrω² = 0 vector sum for coplanar masses.

  73. Q73Past Paper · PPSC/FPSC/NTSmedium

    Turning moment diagram plots

    1. Avelocity versus time only always
    2. Bstress versus strain only
    3. Ccrank torque versus crank angle for one cycle
    4. Dpressure versus volume only
    💡 Explanation:

    Area equals work per cycle; flywheel sizing from fluctuation.

  74. Q74medium

    Flywheel rim has most mass at periphery because

    1. Acentre hub stores more energy always
    2. Bspokes alone sufficient always
    3. Cbearing friction requires it
    4. Dmoment of inertia increases with radius squared
    💡 Explanation:

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

  75. Q75Past Paper · PPSC/FPSC/NTSmedium

    Coefficient of fluctuation of speed is

    1. Aflywheel weight only
    2. Bgear pressure angle
    3. Ccam lift divided by base circle
    4. D(ω1 - ω2)/ω_mean
    💡 Explanation:

    Smaller value means tighter speed regulation.

  76. Q76Past Paper · PPSC/FPSC/NTSeasy

    Flywheel stores energy as

    1. Achemical fuel energy only
    2. Brotational kinetic energy smoothing cyclic fluctuation
    3. Celastic spring potential in governor only
    4. Delectrical capacitor charge only
    💡 Explanation:

    E = ½Iω²; reduces coefficient of speed fluctuation.

  77. Q77medium

    Hunting in governor means

    1. Astable immediate stop
    2. Bzero response to load
    3. Ccontinuous speed oscillation about mean
    4. Dfixed throttle always
    💡 Explanation:

    Isochronous or damper needed to reduce hunting.

  78. Q78hard

    Sensitiveness of governor is ratio of

    1. Apower output to fuel flow only
    2. Bflywheel energy to torque only
    3. Crange of speed to mean speed
    4. Dgear ratio to module
    💡 Explanation:

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

  79. Q79medium

    Proell governor extends flyballs

    1. Ainward toward spindle only
    2. Boutward on extended links for greater speed rise effect
    3. Celiminates all sensitivity
    4. Dreplaces throttle completely
    💡 Explanation:

    Same balls; longer arms increase centrifugal moment.

  80. Q80Past Paper · PPSC/FPSC/NTSmedium

    Porter governor has

    1. Ano controlling force element
    2. Bonly hydraulic dashpot from start
    3. Ccentral spring supporting flyball arms
    4. Dfixed mass without arms
    💡 Explanation:

    Spring adds controlling force; improved sensitivity over Watt.

  81. Q81Past Paper · PPSC/FPSC/NTSeasy

    Watt governor is

    1. Ahydraulic piston only
    2. Belectronic ECU map only
    3. Cpendulum type with flyballs on arms near spindle
    4. Dturbocharger wastegate only
    💡 Explanation:

    Proportional controller; hunting at high speeds.

  82. Q82Past Paper · PPSC/FPSC/NTSeasy

    Governor maintains

    1. Aconstant fuel tank level
    2. Bconstant tyre pressure
    3. Cnear-constant speed of engine despite load changes
    4. Dconstant sand moisture
    💡 Explanation:

    Centrifugal force of masses linked to throttle.

  83. Q83hard

    Reverted gear train has

    1. Ainput and output shafts coaxial
    2. Bparallel offset shafts only
    3. Cintersecting axes only
    4. Dno idler allowed ever
    💡 Explanation:

    Compound arrangement with equal centre distances often.

  84. Q84Past Paper · PPSC/FPSC/NTSmedium

    Epicyclic gear train has

    1. Aone or more gears rotating on moving arm
    2. Bonly fixed axis gears
    3. Cno planet gears ever
    4. Donly belt drives
    💡 Explanation:

    Automotive automatic transmission planetary sets.

  85. Q85easy

    Idler gear in train changes

    1. Adirection of rotation without affecting ratio magnitude
    2. Boverall velocity ratio value
    3. Ccentre distance law only
    4. Dmodule standard
    💡 Explanation:

    Teeth count cancels in ratio product.

  86. Q86Past Paper · PPSC/FPSC/NTSeasy

    Gear train velocity ratio is product of

    1. Amodules only
    2. Bpressure angles only
    3. Cdriven teeth divided by driver teeth for each stage
    4. Dface widths only
    💡 Explanation:

    Overall ratio combines idler and compound stages.

  87. Q87Past Paper · PPSC/FPSC/NTSmedium

    Worm gear pair has

    1. Ahigh speed reduction and self-locking tendency possible
    2. Bonly parallel axis layout
    3. Cequal size pinion and gear always
    4. Dno sliding contact ever
    💡 Explanation:

    Worm is screw; wheel is partial helical gear.

  88. Q88Past Paper · PPSC/FPSC/NTSmedium

    Bevel gears transmit motion between

    1. Aparallel axes only always
    2. Bnon-intersecting skew axes without hypoid
    3. Cintersecting axes usually at 90 degrees
    4. Dfixed ratio belt only
    💡 Explanation:

    Pitch cones roll without slip theoretically.

  89. Q89hard

    Rubbing velocity at pin joint equals

    1. Aangular velocity times pin radius if one link fixed locally
    2. Bsum of link lengths
    3. Cproduct of modules
    4. Dzero always at all joints
    💡 Explanation:

    Used in wear and lubrication analysis.

  90. Q90hard

    Kennedy theorem states I-centres of three bodies

    1. Aform equilateral triangle always
    2. Bcoincide at mass centre always
    3. Clie on a straight line
    4. Dare unrelated
    💡 Explanation:

    Locates unknown I-centre using two known.

  91. Q91medium

    Velocity diagram in relative motion shows

    1. Aonly scalar gear teeth count
    2. Bonly temperature entropy
    3. Conly weld pool width
    4. Dvector addition of absolute velocities of links
    💡 Explanation:

    Instant centre method or relative velocity polygons.

  92. Q92hard

    Precession angular velocity of gyro is

    1. AIω divided by T
    2. BT times ω only
    3. Czero for all applied torque
    4. DT / (Iω)
    💡 Explanation:

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

  93. Q93medium

    Gyroscope resists change in

    1. Alinear speed along road only
    2. Bfuel octane number
    3. Csand binder type
    4. Daxis of rotation due to angular momentum
    💡 Explanation:

    Precession occurs when torque applied perpendicular to spin axis.