Engineering Mechanics and Dynamics MCQs 2026
90 questions with detailed answers · 32 from past papers · 9 quiz batches available
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- 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
💡 Explanation:Equilibrium requires vector sum of forces and moments to vanish.
- 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
💡 Explanation:First law defines inertia and need for net external force to change motion.
- Q3 easy
Newton's second law in vector form is stated as
💡 Explanation:Net force equals mass times acceleration for a particle.
- Q4 easy
The unit of force in SI system is
💡 Explanation:1 N = 1 kg·m/s².
- Q5 Past Paper · PPSC/FPSC/NTS easy
For a rigid body in plane motion, the number of independent equations of equilibrium is
💡 Explanation:ΣFx=0, ΣFy=0, ΣM=0 for coplanar force systems.
- Q6 medium
The moment of a force about a point equals
💡 Explanation:M = F × d⊥.
- Q7 medium
Varignon's theorem states that the moment of a force about a point equals
💡 Explanation:Moment of F equals moment of Fx plus moment of Fy.
- Q8 Past Paper · PPSC/FPSC/NTS easy
The coefficient of static friction μs is defined as the ratio of
💡 Explanation:μs = Fmax / N at impending motion.
- Q9 medium
Angle of friction φ is related to coefficient of friction μ by
💡 Explanation:Friction cone half-angle satisfies tan φ = μ.
- Q10 medium
Wedge friction problems are solved using
💡 Explanation:Free-body diagrams with limiting friction on each contact.
- Q11 Past Paper · PPSC/FPSC/NTS easy
Work done by a constant force moving through displacement s along its line of action is
💡 Explanation:W = F·s when force and displacement are parallel.
- Q12 easy
Kinetic energy of a particle of mass m moving with speed v is
💡 Explanation:Translational KE = ½mv².
- Q13 medium
Work-energy principle states that work done by all forces equals
💡 Explanation:T₂ − T₁ = ΣU₁₋₂ for particle motion.
- Q14 Past Paper · PPSC/FPSC/NTS easy
Potential energy of a weight W raised through height h is
💡 Explanation:Gravitational PE increase = Wh near earth surface.
- Q15 medium
Conservation of mechanical energy applies when
💡 Explanation:KE + PE remains constant if non-conservative work is zero.
- Q16 Past Paper · PPSC/FPSC/NTS medium
Linear impulse is defined as
💡 Explanation:Impulse J = ∫F dt.
- Q17 medium
Impulse-momentum equation for a particle is
💡 Explanation:Net impulse equals change in linear momentum.
- Q18 hard
Coefficient of restitution e for direct impact is
💡 Explanation:e = (v₂ − u₂)/(u₁ − v₁) along line of impact.
- Q19 Past Paper · PPSC/FPSC/NTS easy
For a projectile neglecting air resistance, horizontal component of velocity
💡 Explanation:No horizontal force implies ax = 0.
- Q20 medium
Maximum height of a projectile launched vertically upward with speed u is
💡 Explanation:Using v² = u² − 2gh at apex v=0.
- Q21 medium
Time of flight for projectile launched and landing at same level with speed u at angle θ is
💡 Explanation:Total time = 2 × time to apex.
- Q22 Past Paper · PPSC/FPSC/NTS medium
Range of projectile on level ground is maximum when launch angle is
💡 Explanation:R = u² sin 2θ / g is maximum at θ = 45°.
- Q23 medium
Relative velocity of A with respect to B is given by
💡 Explanation:Relative velocity uses vector subtraction.
- Q24 hard
In river-boat problems, minimum time crossing requires boat velocity
💡 Explanation:Maximize component normal to banks in ground frame.
- Q25 Past Paper · PPSC/FPSC/NTS easy
Centripetal acceleration of a particle in circular motion of radius r at speed v is
💡 Explanation:a_c = v²/r = rω² toward center.
- Q26 medium
Centrifugal force in a rotating frame is
💡 Explanation:Appears in non-inertial analysis as −ma_fictitious.
- Q27 hard
Belt friction equation T₂/T₁ = e^(μθ) is valid when
💡 Explanation:Capstan equation relates tensions and wrap angle.
- Q28 Past Paper · PPSC/FPSC/NTS medium
Mass moment of inertia of a thin rod of mass M and length L about center is
💡 Explanation:Standard result from integration or parallel axis.
- Q29 medium
Parallel axis theorem states I = I_c +
💡 Explanation:Shift axis parallel to centroidal axis distance d.
- Q30 medium
Radius of gyration k is defined by
💡 Explanation:k represents equivalent distance for rotational inertia.
- Q31 Past Paper · PPSC/FPSC/NTS hard
D'Alembert's principle converts a dynamics problem into
💡 Explanation:Inertia force −ma is included in FBD for kinetics as statics.
- Q32 hard
Coriolis acceleration in a rotating frame is proportional to
💡 Explanation:Coriolis term 2Ω×v appears for motion relative to rotating frame.
- Q33 medium
Lami's theorem applies to
💡 Explanation:Each force proportional to sine of opposite angle.
- Q34 Past Paper · PPSC/FPSC/NTS medium
The resultant of two equal forces P each acting at angle θ between them has magnitude
💡 Explanation:Vector addition gives R = 2P cos(θ/2).
- Q35 easy
A couple produces
💡 Explanation:Couple moment is free vector same about any point.
- Q36 easy
Kinetic friction is generally
💡 Explanation:Once sliding starts, friction typically drops to μk N.
- Q37 Past Paper · PPSC/FPSC/NTS easy
Power is defined as
💡 Explanation:P = dW/dt = F·v for constant force.
- Q38 medium
Angular momentum of a particle about point O is
💡 Explanation:H_O = r × p.
- Q39 hard
For central force motion, which quantity is conserved
💡 Explanation:Torque about center is zero.
- Q40 Past Paper · PPSC/FPSC/NTS medium
Banked curve without friction at design speed relies on
💡 Explanation:N sin θ = mv²/r at equilibrium speed.
- Q41 hard
Instant center of rotation for a body in plane motion is
💡 Explanation:IC has v = 0 at that instant.
- Q42 medium
Rolling without slipping condition relates linear and angular speed by
💡 Explanation:Contact point has zero relative velocity.
- Q43 Past Paper · PPSC/FPSC/NTS medium
Moment of inertia of a solid sphere about diameter is
💡 Explanation:Standard solid sphere result.
- Q44 medium
Principle of transmissibility of force states
💡 Explanation:External effects unchanged if line of action preserved.
- Q45 easy
A truss with all two-force members analyzed by method of joints assumes
💡 Explanation:Standard ideal truss assumptions.
- Q46 Past Paper · PPSC/FPSC/NTS medium
Three-force member in equilibrium has lines of action that
💡 Explanation:Three coplanar forces in equilibrium intersect at one point.
- Q47 hard
Efficiency of a screw jack neglecting collar friction is approximately
💡 Explanation:Square-thread jack efficiency formula.
- Q48 Past Paper · PPSC/FPSC/NTS easy
Collision is perfectly elastic when coefficient of restitution is
💡 Explanation:Relative speed of separation equals approach speed.
- Q49 medium
A body sliding down rough incline of angle θ with coefficient μ accelerates if
💡 Explanation:Component mg sin θ exceeds friction μ mg cos θ.
- Q50 medium
Conical pendulum period depends on
💡 Explanation:T = 2π√(L cos θ / g) for conical pendulum.
- Q51 Past Paper · PPSC/FPSC/NTS easy
The impulse required to stop a moving body equals
💡 Explanation:Change in momentum equals impulse.
- Q52 medium
For a particle in SHM, acceleration is proportional to
💡 Explanation:a = −ω²x.
- Q53 easy
Work done by friction on a sliding block is generally
💡 Explanation:Friction opposes relative motion doing negative work.
- Q54 Past Paper · PPSC/FPSC/NTS easy
Mechanical advantage of ideal machine equals
💡 Explanation:MA = F_out / F_in for ideal machine.
- Q55 medium
Velocity ratio of belt drive equals
💡 Explanation:Speed inversely proportional to pulley diameter.
- Q56 hard
Creep in belt drive refers to
💡 Explanation:Belt stretches differently on tight and slack sides.
- Q57 Past Paper · PPSC/FPSC/NTS hard
Product of inertia I_xy for an area is zero when
💡 Explanation:Symmetry cancels first moment product.
- Q58 medium
Polar moment of inertia J for solid circular shaft equals
💡 Explanation:J used in torsion of circular members.
- Q59 easy
A body in limiting equilibrium on horizontal plane has friction force equal to
💡 Explanation:At impending motion F = μs N.
- Q60 Past Paper · PPSC/FPSC/NTS easy
Newton's third law states that action and reaction forces
💡 Explanation:Third law pairs act on different bodies simultaneously.
- Q61 easy
Resultant of concurrent forces found by polygon method uses
💡 Explanation:Closing side gives resultant vector.
- Q62 medium
For a lift accelerating upward, apparent weight of passenger equals
💡 Explanation:Normal reaction increases with upward acceleration.
- Q63 Past Paper · PPSC/FPSC/NTS hard
At the top of vertical loop, minimum speed to maintain contact is
💡 Explanation:N = 0 gives mv²/r = mg.
- Q64 hard
Relative acceleration in general plane motion includes
💡 Explanation:a_A = a_B + α × r + ω × (ω × r).
- Q65 hard
The principle of virtual work for ideal systems in equilibrium states Σ(δW)=0 for
💡 Explanation:Static equilibrium via virtual work.
- Q66 hard
A force system reducible to a couple and single force not through that couple's axis has
💡 Explanation:General spatial force system reduces to wrench.
- Q67 Past Paper · PPSC/FPSC/NTS hard
Dynamic equilibrium using D'Alembert includes inertia torque for rotation as
💡 Explanation:Rotational analog −Iα included in moment equilibrium.
- Q68 easy
For projectile at maximum range on level ground, sin 2θ equals
💡 Explanation:Maximum when 2θ = 90° so θ = 45°.
- Q69 medium
Sliding velocity in belt friction problem affects
💡 Explanation:T2 > T1 on side toward which belt tends to slip.
- Q70 Past Paper · PPSC/FPSC/NTS medium
The kinetic energy of a rigid body rotating about fixed axis is
💡 Explanation:Rotational KE uses moment of inertia about axis.
- Q71 medium
Linear momentum of a system is conserved when
💡 Explanation:Internal forces cancel in pairs; need zero net external force.
- Q72 medium
Friction angle on inclined plane at which body is on verge of sliding down equals inclination when
💡 Explanation:At limit mg sin θ = μ mg cos θ.
- Q73 Past Paper · PPSC/FPSC/NTS easy
Centrifugal force magnitude on mass m at radius r with angular speed ω is
💡 Explanation:Outward fictitious force in rotating frame.
- Q74 hard
Coriolis effect deflects moving objects on rotating earth because of
💡 Explanation:Deflection due to 2Ω×v term.
- Q75 Past Paper · PPSC/FPSC/NTS medium
Work done by internal forces in a rigid body is
💡 Explanation:Internal pairs cancel for rigid bodies.
- Q76 medium
Instantaneous center for a wheel rolling on ground lies at
💡 Explanation:Contact point has zero velocity in pure rolling.
- Q77 easy
A particle moving with constant speed in a circle has
💡 Explanation:Speed constant implies no tangential acceleration.
- Q78 Past Paper · PPSC/FPSC/NTS medium
The principle of conservation of angular momentum applies when
💡 Explanation:Zero net external torque required.
- Q79 medium
For two blocks connected by string over frictionless pulley (Atwood), acceleration equals
💡 Explanation:Standard Atwood machine result.
- Q80 medium
Limiting friction is independent of
💡 Explanation:For rigid surfaces area does not enter μN relation.
- Q81 Past Paper · PPSC/FPSC/NTS easy
The moment of a couple is measured in
💡 Explanation:Couple moment has same units as torque.
- Q82 easy
If resultant force on a particle is zero, particle
💡 Explanation:Newton's first law consequence.
- Q83 medium
Energy lost in partially inelastic impact appears mainly as
💡 Explanation:Inelastic collision dissipates mechanical energy.
- Q84 Past Paper · PPSC/FPSC/NTS medium
Belt tension on slack side is lower because
💡 Explanation:Capstan effect increases tension on tight side.
- Q85 easy
Mass moment of inertia of thin ring about central axis is
💡 Explanation:All mass at radius R.
- Q86 hard
D'Alembert inertia force acts
💡 Explanation:Inertia force = −m a in dynamic equilibrium.
- Q87 Past Paper · PPSC/FPSC/NTS medium
Relative motion analysis in mechanisms commonly uses
💡 Explanation:Kinematic analysis uses relative velocity/acceleration.
- Q88 easy
For a particle under uniform gravitational field, trajectory in vacuum is
💡 Explanation:Constant g gives parabolic path.
- Q89 Past Paper · PPSC/FPSC/NTS easy
Magnitude of cross product |a × b| equals
💡 Explanation:Cross product magnitude uses sine of included angle.
- Q90 easy
Dot product of two perpendicular vectors is
💡 Explanation:a·b = ab cos 90° = 0.