Engineering Mechanics and Dynamics MCQs 2026

90 questions with detailed answers · 32 from past papers · 9 quiz batches available

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Page 1 of 1 Questions 110 of 90
  1. Q1 Past Paper · PPSC/FPSC/NTS easy

    A body is in static equilibrium when the resultant of all forces and all moments acting on it is

    1. A infinite
    2. B equal to its weight only
    3. C equal to friction force only
    4. D zero
    💡 Explanation:

    Equilibrium requires vector sum of forces and moments to vanish.

  2. Q2 Past Paper · PPSC/FPSC/NTS easy

    According to Newton's first law, a particle continues in its state of rest or uniform motion unless acted upon by

    1. A gravity alone always
    2. B internal stress only
    3. C an unbalanced external force
    4. D centripetal acceleration only
    💡 Explanation:

    First law defines inertia and need for net external force to change motion.

  3. Q3 easy

    Newton's second law in vector form is stated as

    1. A F = mv
    2. B F = m/a
    3. C F = m + a
    4. D F = ma
    💡 Explanation:

    Net force equals mass times acceleration for a particle.

  4. Q4 easy

    The unit of force in SI system is

    1. A joule
    2. B watt
    3. C newton
    4. D pascal
    💡 Explanation:

    1 N = 1 kg·m/s².

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

    For a rigid body in plane motion, the number of independent equations of equilibrium is

    1. A two
    2. B six
    3. C one
    4. D three
    💡 Explanation:

    ΣFx=0, ΣFy=0, ΣM=0 for coplanar force systems.

  6. Q6 medium

    The moment of a force about a point equals

    1. A force times total distance along action line
    2. B force divided by distance
    3. C force times angle only
    4. D force times perpendicular distance from point to line of action
    💡 Explanation:

    M = F × d⊥.

  7. Q7 medium

    Varignon's theorem states that the moment of a force about a point equals

    1. A the sum of moments of its components about the same point
    2. B the product of components only
    3. C zero always for concurrent forces
    4. D the moment of resultant only about another point
    💡 Explanation:

    Moment of F equals moment of Fx plus moment of Fy.

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

    The coefficient of static friction μs is defined as the ratio of

    1. A kinetic friction to velocity
    2. B normal force to weight always
    3. C limiting friction to normal reaction
    4. D applied force to acceleration
    💡 Explanation:

    μs = Fmax / N at impending motion.

  9. Q9 medium

    Angle of friction φ is related to coefficient of friction μ by

    1. A tan φ = μ
    2. B sin φ = μ
    3. C cos φ = μ
    4. D φ = μ in degrees
    💡 Explanation:

    Friction cone half-angle satisfies tan φ = μ.

  10. Q10 medium

    Wedge friction problems are solved using

    1. A only energy methods always
    2. B Bernoulli equation
    3. C Fourier law
    4. D equilibrium equations including friction on sliding surfaces
    💡 Explanation:

    Free-body diagrams with limiting friction on each contact.

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

    Work done by a constant force moving through displacement s along its line of action is

    1. A F/s
    2. B F s cos 90° only always
    3. C m g h only
    4. D F s
    💡 Explanation:

    W = F·s when force and displacement are parallel.

  12. Q12 easy

    Kinetic energy of a particle of mass m moving with speed v is

    1. A ½ m v²
    2. B m v
    3. C m v²
    4. D 2 m v²
    💡 Explanation:

    Translational KE = ½mv².

  13. Q13 medium

    Work-energy principle states that work done by all forces equals

    1. A change in kinetic energy
    2. B change in potential energy only
    3. C zero always
    4. D change in momentum only
    💡 Explanation:

    T₂ − T₁ = ΣU₁₋₂ for particle motion.

  14. Q14 Past Paper · PPSC/FPSC/NTS easy

    Potential energy of a weight W raised through height h is

    1. A W/h
    2. B ½ W h²
    3. C W h²
    4. D W h
    💡 Explanation:

    Gravitational PE increase = Wh near earth surface.

  15. Q15 medium

    Conservation of mechanical energy applies when

    1. A only conservative forces do work
    2. B friction is always present
    3. C external impulses act
    4. D mass varies with time
    💡 Explanation:

    KE + PE remains constant if non-conservative work is zero.

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

    Linear impulse is defined as

    1. A integral of force over time interval
    2. B force times distance only
    3. C mass times velocity squared
    4. D moment of inertia times angular speed
    💡 Explanation:

    Impulse J = ∫F dt.

  17. Q17 medium

    Impulse-momentum equation for a particle is

    1. A J = ½ m v²
    2. B J = m(v₂ − v₁)
    3. C J = F d only always
    4. D J = I ω
    💡 Explanation:

    Net impulse equals change in linear momentum.

  18. Q18 hard

    Coefficient of restitution e for direct impact is

    1. A ratio of masses
    2. B ratio of contact forces
    3. C ratio of relative speed of separation to relative speed of approach
    4. D always equal to 1
    💡 Explanation:

    e = (v₂ − u₂)/(u₁ − v₁) along line of impact.

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

    For a projectile neglecting air resistance, horizontal component of velocity

    1. A remains constant
    2. B increases linearly with time
    3. C decreases to zero at apex only
    4. D equals vertical component always
    💡 Explanation:

    No horizontal force implies ax = 0.

  20. Q20 medium

    Maximum height of a projectile launched vertically upward with speed u is

    1. A u/g
    2. B u²/(2g)
    3. C 2u/g
    4. D u²/g
    💡 Explanation:

    Using v² = u² − 2gh at apex v=0.

  21. Q21 medium

    Time of flight for projectile launched and landing at same level with speed u at angle θ is

    1. A u sin θ / g
    2. B u cos θ / g
    3. C 2u cos θ / g
    4. D 2u sin θ / g
    💡 Explanation:

    Total time = 2 × time to apex.

  22. Q22 Past Paper · PPSC/FPSC/NTS medium

    Range of projectile on level ground is maximum when launch angle is

    1. A 45°
    2. B 30°
    3. C 60°
    4. D 90°
    💡 Explanation:

    R = u² sin 2θ / g is maximum at θ = 45°.

  23. Q23 medium

    Relative velocity of A with respect to B is given by

    1. A v_A − v_B
    2. B v_A + v_B
    3. C v_B − v_A only when parallel
    4. D v_A × v_B
    💡 Explanation:

    Relative velocity uses vector subtraction.

  24. Q24 hard

    In river-boat problems, minimum time crossing requires boat velocity

    1. A parallel to river bank always
    2. B opposite to current only
    3. C perpendicular to river flow relative to water
    4. D zero relative to ground
    💡 Explanation:

    Maximize component normal to banks in ground frame.

  25. Q25 Past Paper · PPSC/FPSC/NTS easy

    Centripetal acceleration of a particle in circular motion of radius r at speed v is

    1. A v r toward center
    2. B r/v² outward
    3. C v²/r directed toward center
    4. D v/r tangential
    💡 Explanation:

    a_c = v²/r = rω² toward center.

  26. Q26 medium

    Centrifugal force in a rotating frame is

    1. A a fictitious outward force equal to m ω² r
    2. B a real force toward center
    3. C equal to gravitational force always
    4. D zero in inertial frame
    💡 Explanation:

    Appears in non-inertial analysis as −ma_fictitious.

  27. Q27 hard

    Belt friction equation T₂/T₁ = e^(μθ) is valid when

    1. A belt is on verge of slipping over a fixed drum
    2. B belt moves without friction
    3. C angle θ is in degrees without conversion
    4. D tension is zero
    💡 Explanation:

    Capstan equation relates tensions and wrap angle.

  28. Q28 Past Paper · PPSC/FPSC/NTS medium

    Mass moment of inertia of a thin rod of mass M and length L about center is

    1. A ML²/3
    2. B ML²/2
    3. C ML²/12
    4. D M L
    💡 Explanation:

    Standard result from integration or parallel axis.

  29. Q29 medium

    Parallel axis theorem states I = I_c +

    1. A M d
    2. B M d²
    3. C M d³
    4. D d² only
    💡 Explanation:

    Shift axis parallel to centroidal axis distance d.

  30. Q30 medium

    Radius of gyration k is defined by

    1. A I = k/M
    2. B I = M k²
    3. C k = I M
    4. D k = √M/I
    💡 Explanation:

    k represents equivalent distance for rotational inertia.

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

    D'Alembert's principle converts a dynamics problem into

    1. A an equivalent static equilibrium by adding inertia forces
    2. B a thermodynamic cycle analysis
    3. C a heat transfer problem
    4. D a fluid statics problem only
    💡 Explanation:

    Inertia force −ma is included in FBD for kinetics as statics.

  32. Q32 hard

    Coriolis acceleration in a rotating frame is proportional to

    1. A ω² r only
    2. B v_rel only
    3. C g always
    4. D 2 ω × v_rel
    💡 Explanation:

    Coriolis term 2Ω×v appears for motion relative to rotating frame.

  33. Q33 medium

    Lami's theorem applies to

    1. A any number of non-concurrent forces
    2. B only frictionless contacts
    3. C three concurrent coplanar forces in equilibrium
    4. D only dynamic impact problems
    💡 Explanation:

    Each force proportional to sine of opposite angle.

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

    The resultant of two equal forces P each acting at angle θ between them has magnitude

    1. A 2P sin(θ/2)
    2. B 2P cos(θ/2)
    3. C P cos θ
    4. D 2P
    💡 Explanation:

    Vector addition gives R = 2P cos(θ/2).

  35. Q35 easy

    A couple produces

    1. A translation without rotation
    2. B pure rotation without translation
    3. C zero moment about any point
    4. D only normal stress in bodies
    💡 Explanation:

    Couple moment is free vector same about any point.

  36. Q36 easy

    Kinetic friction is generally

    1. A less than limiting static friction
    2. B greater than static friction
    3. C equal to normal force
    4. D independent of surfaces
    💡 Explanation:

    Once sliding starts, friction typically drops to μk N.

  37. Q37 Past Paper · PPSC/FPSC/NTS easy

    Power is defined as

    1. A product of mass and velocity
    2. B force times time only
    3. C momentum per unit volume
    4. D rate of doing work
    💡 Explanation:

    P = dW/dt = F·v for constant force.

  38. Q38 medium

    Angular momentum of a particle about point O is

    1. A m v only
    2. B I ω only always
    3. C r × (m v)
    4. D F × t
    💡 Explanation:

    H_O = r × p.

  39. Q39 hard

    For central force motion, which quantity is conserved

    1. A angular momentum about force center
    2. B linear momentum always
    3. C kinetic energy always
    4. D potential energy always in all cases
    💡 Explanation:

    Torque about center is zero.

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

    Banked curve without friction at design speed relies on

    1. A weight component only
    2. B engine thrust
    3. C normal reaction component toward center providing centripetal force
    4. D Coriolis force
    💡 Explanation:

    N sin θ = mv²/r at equilibrium speed.

  41. Q41 hard

    Instant center of rotation for a body in plane motion is

    1. A center of mass always
    2. B center of gravity only for static bodies
    3. C any point on the body always
    4. D point with zero velocity in that instant
    💡 Explanation:

    IC has v = 0 at that instant.

  42. Q42 medium

    Rolling without slipping condition relates linear and angular speed by

    1. A v = ω r
    2. B v = ω/r
    3. C v = ω r²
    4. D v = ω only
    💡 Explanation:

    Contact point has zero relative velocity.

  43. Q43 Past Paper · PPSC/FPSC/NTS medium

    Moment of inertia of a solid sphere about diameter is

    1. A 2/3 M R²
    2. B M R²
    3. C 2/5 M R²
    4. D 4/3 M R²
    💡 Explanation:

    Standard solid sphere result.

  44. Q44 medium

    Principle of transmissibility of force states

    1. A force can be moved anywhere freely
    2. B force may be moved along its line of action without changing external effects on rigid body
    3. C force must always act at centroid
    4. D force direction may change freely
    💡 Explanation:

    External effects unchanged if line of action preserved.

  45. Q45 easy

    A truss with all two-force members analyzed by method of joints assumes

    1. A loads at joints and negligible member weight
    2. B distributed loading on each member always
    3. C plastic deformation in joints
    4. D dynamic impact loading
    💡 Explanation:

    Standard ideal truss assumptions.

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

    Three-force member in equilibrium has lines of action that

    1. A must always be perpendicular
    2. B must be concurrent or parallel
    3. C must pass through center of mass only
    4. D are unrelated
    💡 Explanation:

    Three coplanar forces in equilibrium intersect at one point.

  47. Q47 hard

    Efficiency of a screw jack neglecting collar friction is approximately

    1. A cos α only
    2. B μ only without angle
    3. C tan(α)/tan(α+φ) where α is thread lead angle
    4. D always 100%
    💡 Explanation:

    Square-thread jack efficiency formula.

  48. Q48 Past Paper · PPSC/FPSC/NTS easy

    Collision is perfectly elastic when coefficient of restitution is

    1. A 0
    2. B 0.5 always
    3. C infinity
    4. D 1
    💡 Explanation:

    Relative speed of separation equals approach speed.

  49. Q49 medium

    A body sliding down rough incline of angle θ with coefficient μ accelerates if

    1. A μ > tan θ
    2. B tan θ > μ
    3. C θ = 0 always
    4. D μ = 1 always
    💡 Explanation:

    Component mg sin θ exceeds friction μ mg cos θ.

  50. Q50 medium

    Conical pendulum period depends on

    1. A bob mass only
    2. B air density only
    3. C string thickness only
    4. D length and angle of string, not bob mass
    💡 Explanation:

    T = 2π√(L cos θ / g) for conical pendulum.

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

    The impulse required to stop a moving body equals

    1. A its kinetic energy
    2. B its initial momentum
    3. C its weight
    4. D half its momentum
    💡 Explanation:

    Change in momentum equals impulse.

  52. Q52 medium

    For a particle in SHM, acceleration is proportional to

    1. A velocity
    2. B time squared
    3. C negative displacement from mean position
    4. D constant value always
    💡 Explanation:

    a = −ω²x.

  53. Q53 easy

    Work done by friction on a sliding block is generally

    1. A always positive
    2. B negative
    3. C zero always
    4. D equal to normal force
    💡 Explanation:

    Friction opposes relative motion doing negative work.

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

    Mechanical advantage of ideal machine equals

    1. A input work divided by output work
    2. B velocity ratio only without efficiency
    3. C always less than 1
    4. D output force divided by input force
    💡 Explanation:

    MA = F_out / F_in for ideal machine.

  55. Q55 medium

    Velocity ratio of belt drive equals

    1. A always 1
    2. B tension ratio T2/T1
    3. C coefficient of friction only
    4. D ratio of driven pulley diameter to driver pulley diameter for speed of driven
    💡 Explanation:

    Speed inversely proportional to pulley diameter.

  56. Q56 hard

    Creep in belt drive refers to

    1. A elastic stretching only without slip
    2. B slight relative sliding due to tension difference over arc of contact
    3. C thermal expansion of belt only
    4. D centrifugal force only
    💡 Explanation:

    Belt stretches differently on tight and slack sides.

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

    Product of inertia I_xy for an area is zero when

    1. A area is any triangle always
    2. B area is unsymmetrical always
    3. C centroid is at origin only
    4. D x or y axis is axis of symmetry
    💡 Explanation:

    Symmetry cancels first moment product.

  58. Q58 medium

    Polar moment of inertia J for solid circular shaft equals

    1. A π d²/4
    2. B π d⁴/32
    3. C π d³/16
    4. D bh³/12
    💡 Explanation:

    J used in torsion of circular members.

  59. Q59 easy

    A body in limiting equilibrium on horizontal plane has friction force equal to

    1. A mg always regardless of applied force
    2. B zero
    3. C μ N
    4. D applied horizontal force always even below limit
    💡 Explanation:

    At impending motion F = μs N.

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

    Newton's third law states that action and reaction forces

    1. A cancel on same body always
    2. B act on same body
    3. C need not be collinear
    4. D are equal, opposite, and act on different bodies
    💡 Explanation:

    Third law pairs act on different bodies simultaneously.

  61. Q61 easy

    Resultant of concurrent forces found by polygon method uses

    1. A scalar addition of magnitudes only
    2. B head-to-tail vector addition
    3. C cross product only
    4. D only graphical moment method
    💡 Explanation:

    Closing side gives resultant vector.

  62. Q62 medium

    For a lift accelerating upward, apparent weight of passenger equals

    1. A m(g − a)
    2. B mg always
    3. C m(g + a)
    4. D zero
    💡 Explanation:

    Normal reaction increases with upward acceleration.

  63. Q63 Past Paper · PPSC/FPSC/NTS hard

    At the top of vertical loop, minimum speed to maintain contact is

    1. A gr
    2. B √(g r)
    3. C g/r
    4. D zero
    💡 Explanation:

    N = 0 gives mv²/r = mg.

  64. Q64 hard

    Relative acceleration in general plane motion includes

    1. A translation of reference plus rotational component
    2. B only centripetal term always
    3. C only Coriolis in inertial frame
    4. D zero if speeds equal
    💡 Explanation:

    a_A = a_B + α × r + ω × (ω × r).

  65. Q65 hard

    The principle of virtual work for ideal systems in equilibrium states Σ(δW)=0 for

    1. A virtual displacements consistent with constraints
    2. B actual displacements only with friction doing work
    3. C any displacement violating constraints
    4. D only dynamic motion
    💡 Explanation:

    Static equilibrium via virtual work.

  66. Q66 hard

    A force system reducible to a couple and single force not through that couple's axis has

    1. A pure couple only always
    2. B zero resultant always
    3. C only scalar resultant
    4. D a wrench (force-couple system)
    💡 Explanation:

    General spatial force system reduces to wrench.

  67. Q67 Past Paper · PPSC/FPSC/NTS hard

    Dynamic equilibrium using D'Alembert includes inertia torque for rotation as

    1. A I α added as real torque
    2. B zero always
    3. C m a only
    4. D −I α
    💡 Explanation:

    Rotational analog −Iα included in moment equilibrium.

  68. Q68 easy

    For projectile at maximum range on level ground, sin 2θ equals

    1. A 0
    2. B 0.5
    3. C 1
    4. D 2
    💡 Explanation:

    Maximum when 2θ = 90° so θ = 45°.

  69. Q69 medium

    Sliding velocity in belt friction problem affects

    1. A nothing in capstan equation
    2. B only mass moment of inertia
    3. C direction of impending slip determining which tension is larger
    4. D only gravitational potential
    💡 Explanation:

    T2 > T1 on side toward which belt tends to slip.

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

    The kinetic energy of a rigid body rotating about fixed axis is

    1. A ½ m v² only
    2. B ½ I ω²
    3. C I ω
    4. D m g h
    💡 Explanation:

    Rotational KE uses moment of inertia about axis.

  71. Q71 medium

    Linear momentum of a system is conserved when

    1. A internal forces are zero only
    2. B net external force is zero
    3. C friction always present
    4. D energy is not conserved
    💡 Explanation:

    Internal forces cancel in pairs; need zero net external force.

  72. Q72 medium

    Friction angle on inclined plane at which body is on verge of sliding down equals inclination when

    1. A always 45°
    2. B tan θ = μ
    3. C μ = 0 always
    4. D θ = 90° minus μ
    💡 Explanation:

    At limit mg sin θ = μ mg cos θ.

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

    Centrifugal force magnitude on mass m at radius r with angular speed ω is

    1. A m ω r
    2. B m ω/r
    3. C m g
    4. D m ω² r
    💡 Explanation:

    Outward fictitious force in rotating frame.

  74. Q74 hard

    Coriolis effect deflects moving objects on rotating earth because of

    1. A magnetic field only
    2. B rotation of reference frame
    3. C gravity variation only
    4. D air viscosity only
    💡 Explanation:

    Deflection due to 2Ω×v term.

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

    Work done by internal forces in a rigid body is

    1. A zero for rigid idealization
    2. B always equal to KE change alone
    3. C always positive
    4. D equal to impulse
    💡 Explanation:

    Internal pairs cancel for rigid bodies.

  76. Q76 medium

    Instantaneous center for a wheel rolling on ground lies at

    1. A center of wheel always
    2. B top of wheel
    3. C contact point with ground
    4. D infinity always
    💡 Explanation:

    Contact point has zero velocity in pure rolling.

  77. Q77 easy

    A particle moving with constant speed in a circle has

    1. A zero centripetal acceleration
    2. B zero tangential acceleration
    3. C zero velocity
    4. D zero angular speed
    💡 Explanation:

    Speed constant implies no tangential acceleration.

  78. Q78 Past Paper · PPSC/FPSC/NTS medium

    The principle of conservation of angular momentum applies when

    1. A any internal moment exists
    2. B linear momentum is not conserved
    3. C net external moment about axis is zero
    4. D only in statics
    💡 Explanation:

    Zero net external torque required.

  79. Q79 medium

    For two blocks connected by string over frictionless pulley (Atwood), acceleration equals

    1. A g(m1−m2)/(m1+m2)
    2. B g always
    3. C zero always
    4. D g(m1+m2)/(m1−m2)
    💡 Explanation:

    Standard Atwood machine result.

  80. Q80 medium

    Limiting friction is independent of

    1. A normal force
    2. B surface materials
    3. C apparent area of contact for rigid bodies per Amontons-Coulomb law
    4. D roughness through μ
    💡 Explanation:

    For rigid surfaces area does not enter μN relation.

  81. Q81 Past Paper · PPSC/FPSC/NTS easy

    The moment of a couple is measured in

    1. A N·m same as moment of force
    2. B N/m
    3. C N·m²
    4. D joule only never N·m
    💡 Explanation:

    Couple moment has same units as torque.

  82. Q82 easy

    If resultant force on a particle is zero, particle

    1. A must be at rest always
    2. B remains at rest or moves with constant velocity
    3. C must accelerate
    4. D must move in circle
    💡 Explanation:

    Newton's first law consequence.

  83. Q83 medium

    Energy lost in partially inelastic impact appears mainly as

    1. A deformation and heat
    2. B increase in total momentum
    3. C increase in angular momentum always
    4. D gravitational PE only
    💡 Explanation:

    Inelastic collision dissipates mechanical energy.

  84. Q84 Past Paper · PPSC/FPSC/NTS medium

    Belt tension on slack side is lower because

    1. A centrifugal force zero always
    2. B pulley mass reduces tension
    3. C friction over wrap angle transfers force from driver
    4. D gravity alone increases tight side
    💡 Explanation:

    Capstan effect increases tension on tight side.

  85. Q85 easy

    Mass moment of inertia of thin ring about central axis is

    1. A ½ M R²
    2. B M R²
    3. C 2/5 M R²
    4. D M R
    💡 Explanation:

    All mass at radius R.

  86. Q86 hard

    D'Alembert inertia force acts

    1. A same as acceleration
    2. B always vertical downward
    3. C opposite to acceleration direction through mass center
    4. D only in rotating frames
    💡 Explanation:

    Inertia force = −m a in dynamic equilibrium.

  87. Q87 Past Paper · PPSC/FPSC/NTS medium

    Relative motion analysis in mechanisms commonly uses

    1. A only thermodynamic charts
    2. B Mohr circle only
    3. C steam tables only
    4. D velocity and acceleration polygons or vector equations
    💡 Explanation:

    Kinematic analysis uses relative velocity/acceleration.

  88. Q88 easy

    For a particle under uniform gravitational field, trajectory in vacuum is

    1. A circular always
    2. B parabolic
    3. C straight line always
    4. D hyperbolic always
    💡 Explanation:

    Constant g gives parabolic path.

  89. Q89 Past Paper · PPSC/FPSC/NTS easy

    Magnitude of cross product |a × b| equals

    1. A a b sin θ
    2. B a b cos θ
    3. C a + b
    4. D a b
    💡 Explanation:

    Cross product magnitude uses sine of included angle.

  90. Q90 easy

    Dot product of two perpendicular vectors is

    1. A one
    2. B product of magnitudes
    3. C negative one always
    4. D zero
    💡 Explanation:

    a·b = ab cos 90° = 0.