Question 1

Resultant of two forces: angles for extremal magnitude  For two forces of fixed magnitudes, the resultant is maximum at which angle between them, and minimum at which angle (respectively)? Choose the correct ordered pair.

Accepted Answer

Given dataTwo fixed magnitudes P and Q with included angle θ.Concept/ApproachResultant magnitude: R = √(P² + Q² + 2PQ cosθ).Step-by-stepR is maximum when cosθ = 1 ⇒ θ = 0° (same direction).R is minimum when cosθ = −1 ⇒ θ = 180° (opposite directions).Common pitfallsThinking 90° gives minimum; at 90°, R = √(P² + Q²), which is intermediate.Final Answer0° and 180°

Question 2

Type of pendulum: small oscillations of a suspended rigid body under gravity  A rigid body is suspended vertically and oscillates with a small amplitude due to gravity. Identify the correct term for this system.

Accepted Answer

Concept/ApproachA compound (physical) pendulum is any rigid body oscillating about a horizontal axis under gravity. A simple pendulum is an idealized point mass on a massless string. A torsional pendulum oscillates by twisting a wire (restoring torque), not by gravity about a pivot. A second's pendulum specifically has a 2 s period.Final Answercompound pendulum

Question 3

Inclined plane with vertical effort: velocity ratio (VR)  An inclined plane makes an angle θ with the horizontal. The load is pulled up the plane by a vertical effort (idealized, no losses). Find the velocity ratio VR = (distance moved by effort) ÷ …

Accepted Answer

Concept/ApproachFor an ideal inclined plane, the mechanical advantage equals the velocity ratio. For a load displacement of s along the plane, the vertical rise is h = s sinθ.Step-by-stepIn the idealized vertical-effort arrangement, the effort point moves through the vertical rise h while the load moves a path length s along the plane.Therefore VR = (distance moved by effort) / (distance moved by load) = h / s = sinθ.But for an ideal machine, MA = W/E = 1/sinθ and MA = VR ⇒ VR = 1/sinθ = cosec θ.Common pitfallsConfusing VR with sinθ (a geometric ratio of rise to travel) without accounting for the ideal machine equivalence of MA and VR for the inclined plane model.Final Answercosec θ

Question 4

Resultant of two equal forces: find the angle between them  Two forces of equal magnitude act at a point. The resultant has the same magnitude as either force.  Choose the correct angle between the two forces.

Accepted Answer

Given dataTwo forces of equal magnitude F.Resultant R has magnitude equal to F.Concept/ApproachFor two equal forces separated by angle θ, the resultant is R = 2F cos(θ/2) (parallelogram law).Step-by-step calculationSet R = F → 2F cos(θ/2) = F.cos(θ/2) = 1/2 → θ/2 = 60°.Therefore θ = 120°.Common pitfallsUsing R = √(F2 + F2 + 2F·F cos θ) and forgetting to simplify for equal forces; both methods give the same result.Final Answer120°

Question 5

Triangle law of forces: verify the statement for the resultant  Statement: “If two forces acting simultaneously on a particle are represented (in magnitude and direction) by two sides of a triangle taken in order, then their resultant is represented…

Accepted Answer

Concept/ApproachThe triangle law of forces states exactly this: place the second force head-to-tail after the first; the side closing the triangle in the reverse sense gives the resultant in both magnitude and direction.Final AnswerAgree

Question 6

Torsional pendulum: frequency of oscillation formula  Let I be the mass moment of inertia of the suspended body and C be the torsional constant (restoring couple per radian).  Select the correct expression for the frequency of oscillation f.

Accepted Answer

Given dataTorsional pendulum with torsional constant C and moment of inertia I.Concept/ApproachEquation of motion: restoring couple = C × angular displacement. This is simple harmonic motion with angular frequency ω such that ω2 = C / I.Step-by-step relationω = √(C / I)f = ω / (2π) = (1 / 2π) × √(C / I)Common pitfallsUsing the simple pendulum formula √(g/L), which does not apply to torsional oscillations.Final Answerf = (1 / 2π) × √(C / I)

Question 7

Impact and energy loss in mechanics  Which one of the following statements about the kinetic energy of a body during an impact (collision) is generally correct?

Accepted Answer

Introduction / Context:When two bodies collide, some of the system's mechanical energy can be dissipated as sound, heat, and deformation. This question tests your understanding of how kinetic energy typically changes across an impact in engineering mechanics and physics.Given Data / Assumptions:Before impact: the body (or system) possesses a certain kinetic energy based on its mass and speed.During impact: internal forces act over very short time intervals; deformation and heat may occur.After impact: speeds generally change; some energy may be lost to non-recoverable modes unless the collision is perfectly elastic.Concept / Approach:Momentum is conserved for an isolated system during an impact, but kinetic energy is conserved only in a perfectly elastic collision. In most practical engineering situations (vehicle bumps, hammer blows, drop tests), the impact is partially or highly inelastic, so kinetic energy decreases.Step-by-Step Solution:Let KE_before be the total kinetic energy just before contact.Let KE_after be the total kinetic energy immediately after separation.For a perfectly elastic impact: KE_after = KE_before (rare in practice).For a perfectly plastic impact: KE_after For real impacts: energy loss occurs due to sound, heat, local yielding, and permanent set → KE_after is usually less than KE_before.Verification / Alternative check:Coefficient of restitution e characterizes elasticity. When 0 ≤ e Why Other Options Are Wrong:Remains constant / Equal before and after: True only for perfectly elastic collisions; not the general case.Before is less than after: Violates energy conservation without an external energy input.None of the above: Incorrect because the general case is a decrease in kinetic energy.Common Pitfalls:Confusing momentum conservation (almost always valid for isolated systems) with kinetic energy conservation (valid only for elastic impacts). Also ignoring dissipative effects like plastic deformation and sound.Final Answer:The kinetic energy of a body before impact is more than the kinetic energy of a body after impact.

Question 8

Center of gravity in engineering mechanics  The point through which the entire weight of a body may be considered to act, irrespective of its orientation, is known as the:

Accepted Answer

Introduction / Context:Designers and analysts must know where the weight of a body can be assumed to act to simplify calculations of support reactions, stability, and tipping. This conceptual question distinguishes multiple “centres” used in mechanics.Given Data / Assumptions:Weight is a force equal to mass * g acting downward toward Earth's centre.We assume a uniform gravitational field over the body's extent (a good approximation for most engineering applications).Concept / Approach:The centre of gravity (CG) is the point where the resultant weight of the body acts. For a uniform gravitational field, the centre of gravity coincides with the centre of mass, but the term CG emphasizes the action point of weight (a force).Step-by-Step Solution:Identify the definition that refers to the line of action of weight.CG is defined as the unique point at which the entire weight can be assumed to act for any orientation.Therefore, among the given choices, 'centre of gravity' matches the definition.Verification / Alternative check:In a non-uniform gravitational field, CG and centre of mass may differ slightly. In standard terrestrial problems, they effectively coincide, validating the selection.Why Other Options Are Wrong:Moment of inertia: A property measuring resistance to angular acceleration, not a point.Centre of percussion: A specific point on a swinging body where impact causes no reaction at the pivot.Centre of mass: Mass centroid; close in meaning but the question explicitly stresses weight action.Barycentre: General term for mass centroid of a system; used in orbital mechanics for two-body systems.Common Pitfalls:Using centre of mass and centre of gravity interchangeably without noting the emphasis on force (weight) in the definition.Final Answer:centre of gravity

Question 9

The maximum efficiency of a lifting machine is

Accepted Answer

1/(m x V.R.)

Question 10

Principle of transmissibility of forces  According to the principle of transmissibility, when a force acts upon a rigid body, its external effect (on overall equilibrium and motion) is:

Accepted Answer

Introduction / Context:The principle of transmissibility simplifies force systems in rigid-body mechanics. It allows engineers to shift a force along its line of action without changing the body's external effects.Given Data / Assumptions:The body is treated as rigid (no deformation effects considered).We consider only external effects such as overall equilibrium and motion, not internal stress distribution.Concept / Approach:The principle states: a force may be applied at any point along its line of action to a rigid body and will produce the same external effect. Internal stresses can differ, but the net force and moment about any point remain unchanged.Step-by-Step Solution:Identify the statement that preserves net force and net moment of the force system.Shifting the point of application along the same line of action does not create an additional couple.Hence, the correct description is 'same at every point on its line of action'.Verification / Alternative check:Compute moments about any reference: moving the force along its own line keeps the perpendicular distance to that line the same, so resultant moment contributions are unchanged.Why Other Options Are Wrong:Different at different points: Contradicts the principle for rigid bodies.Minimum/Maximum at CG: Not a general rule; effect depends on line of action, not simply proximity to CG.Zero on a parallel line: Shifting to a parallel but distinct line requires an added couple (equivalent force–couple system), changing external effects.Common Pitfalls:Confusing internal stresses (which do change) with external effects (which do not, for a rigid body).Final Answer:same at every point on its line of action

Question 11

Converting rotational speed to angular velocity  If a body rotates at N revolutions per minute (rpm), what is its angular velocity ω in radians per second?

Accepted Answer

Introduction / Context:Angular velocity is commonly required in radians per second for dynamics, power, and torque calculations. This question checks the fundamental conversion from rpm to rad/s.Given Data / Assumptions:N is given in revolutions per minute (rpm).1 revolution corresponds to 2π radians.1 minute equals 60 seconds.Concept / Approach:Use unit conversion: multiply by 2π radians per revolution and divide by 60 seconds per minute to convert rpm to rad/s.Step-by-Step Solution:Start with N rev/min.Convert revolutions to radians: N * 2π rad/min.Convert minutes to seconds: (N * 2π) / 60 rad/s.Therefore, ω = 2πN / 60 rad/s.Verification / Alternative check:For N = 60 rpm, ω should be 2π rad/s (one revolution per second). Substituting N = 60 gives ω = 2π, which matches.Why Other Options Are Wrong:πN/60 or πN/180: Miss the factor of 2 for a full revolution (2π).2πN/180: Divides by 180 instead of 60, underestimating by a factor of 3.N/2π: Inverts the correct relationship and has wrong units.Common Pitfalls:Forgetting that 1 rev = 2π rad, or failing to convert minutes to seconds.Final Answer:2πN/60

Question 12

Resultant of a force system  The process of determining a single force that can replace a given system of forces (i.e., the resultant) is called __________ of forces.

Accepted Answer

Introduction / Context:In engineering mechanics, two complementary operations are routinely used: finding a resultant (one force replacing many) and splitting a force into components. This question tests your terminology.Given Data / Assumptions:We are asked about the name of the process that yields a resultant force.No numerical data are required; this is a definition/terminology check.Concept / Approach:Composition of forces means combining multiple forces into a single resultant. By contrast, resolution of forces means decomposing a single force into components along chosen axes or directions.Step-by-Step Solution:Identify whether the operation goes from many → one or one → many.Many → one = composition (gives a resultant).One → many = resolution (gives components).Therefore, the blank must be filled with 'composition'.Verification / Alternative check:In textbooks, section titles typically read 'Composition of Forces' for resultants and 'Resolution of Forces' for components, confirming the terminology.Why Other Options Are Wrong:Resolution: Opposite operation (one force into components).Aggregation / Superposition: General words, not standard technical terms for the operation.Equilibration: Introducing equal and opposite forces to achieve equilibrium, not forming a resultant.Common Pitfalls:Mixing up composition and resolution due to their frequent paired use. Remember the direction of the operation.Final Answer:composition

Question 13

Skidding on a level circular path  A vehicle moves on a level circular track. Skidding can be avoided if the available frictional force between the tyres and the road is __________ the required centrifugal (outward) force for the turn.

Accepted Answer

Introduction / Context:Cornering safety depends on whether tyre–road friction can supply the lateral (centripetal) force needed for circular motion. If insufficient, the vehicle skids outward.Given Data / Assumptions:Level road (no banking), so lateral force is provided solely by friction.Centripetal force required: m * v^2 / r directed toward the centre of the curve.Maximum available static friction: F_max = μ_s * N = μ_s * m * g.Concept / Approach:For no skidding, the maximum available static friction must meet or exceed the required centripetal force. If the requirement exceeds what friction can provide, tyres will slide, producing skid.Step-by-Step Solution:Required: F_required = m * v^2 / r.Available: F_available = μ_s * m * g.Condition for no skidding: F_available ≥ F_required.Therefore, friction must be greater than or equal to the centrifugal tendency that must be countered.Verification / Alternative check:If μ_s * g = v^2 / r exactly, the tyres are at the limit of adhesion (impending skid). If μ_s * g is larger, there is a safety margin and no skidding occurs.Why Other Options Are Wrong:Less than / much less than: Guarantees skidding because friction cannot supply the needed centripetal force.Equal to: Only the limiting case; while it avoids skidding, the general safe condition is 'greater than or equal to'.Greater than: True but excludes the limiting (equality) case where skidding just does not occur.Common Pitfalls:Confusing 'centripetal' and 'centrifugal' descriptions. In vehicle dynamics we require sufficient inward (centripetal) frictional force; outward 'centrifugal' is an apparent effect in the rotating frame.Final Answer:greater than or equal to

Question 14

Type of motion: particle moving about a fixed axis  The motion of a particle moving around a fixed axis is best described as:

Accepted Answer

Introduction / Context:Classifying motion correctly is foundational in kinematics. This item distinguishes between translatory (linear), circular, and rotary (rotational) motion with respect to a fixed axis.Given Data / Assumptions:A particle traces a path about a fixed axis.The axis does not translate; the body's points rotate about it.Concept / Approach:When points of a body (or a particle constrained to a radius) move about a fixed axis, the motion is rotational (rotary). 'Circular' describes the path shape in a plane; 'rotary' emphasizes motion about an axis, which is the defining feature here.Step-by-Step Solution:Identify the defining characteristic: motion about a fixed axis → rotational.Translatory motion requires all points to have parallel, equal-velocity trajectories; that is not the case here.Therefore, 'rotary' is the most accurate descriptor among the options.Verification / Alternative check:A point on a rotating wheel follows a circular path, but the system motion is rotational about the axle. The axis criterion prioritizes the term 'rotary'.Why Other Options Are Wrong:Translatory: No fixed axis or angular description.Circular: Describes the trajectory shape but not the kinematic category; the question stresses a fixed axis.Translatory as well as rotatry: Misspelt and inapplicable to a single particle about a fixed axis.Oscillatory: Back-and-forth motion about an equilibrium, not continuous rotation.Common Pitfalls:Equating any circular path to the term 'circular' without considering whether the axis-based definition (rotational) is being tested.Final Answer:rotary

Question 15

Engineering mechanics – minimum external pull to initiate sliding on a rough horizontal plane A block of weight W rests on a rough, horizontal surface. The pull is applied at an angle θ to the horizontal. If θ equals the angle of friction, what is t…

Accepted Answer

Introduction / Context:This classical engineering mechanics question tests friction, impending motion, and the special case where the applied pull is inclined at the angle of friction. The goal is to find the least pull that will just overcome limiting friction and initiate sliding on a rough horizontal plane.Given Data / Assumptions:Block weight = W on a rough, horizontal plane.Coefficient of friction = μ, so angle of friction φ satisfies tan φ = μ.An external pull P is applied at an angle θ to the horizontal; for minimum P, θ = φ.Impending motion condition (limiting friction) is used.Concept / Approach:At impending motion, the frictional force reaches its maximum value F = μN. For an inclined pull P at angle θ above the horizontal, the normal reaction reduces due to the vertical component of P. Minimizing P with respect to θ leads to θ = φ, the angle of friction. Substituting θ = φ yields a compact expression for the least pull.Step-by-Step Solution:Resolve forces vertically: N + P sin θ = W ⇒ N = W − P sin θ.At impending motion: horizontal equilibrium gives P cos θ = F = μN = μ(W − P sin θ).Rearrange: P(cos θ + μ sin θ) = μW ⇒ P = μW / (cos θ + μ sin θ).For minimum P, set θ = φ where tan φ = μ. Then cos φ + μ sin φ = cos φ + (tan φ) sin φ = cos φ + (sin φ / cos φ) sin φ = (cos^2 φ + sin^2 φ)/cos φ = 1 / cos φ.Hence P_min = μW / (1 / cos φ) = μW cos φ = W (μ cos φ) = W sin φ (since sin φ = μ / √(1 + μ^2) and cos φ = 1 / √(1 + μ^2)).Verification / Alternative check:Using the identity sin φ = μ / √(1 + μ^2) and cos φ = 1 / √(1 + μ^2), the reduced form P_min = μW / √(1 + μ^2) equals W sin φ, confirming consistency.Why Other Options Are Wrong:W cos θ: corresponds to a vertical-balance component, not the least-pull result.W tan θ: dimensionally force, but not the friction-minimized pull.None of these: incorrect because W sin θ is correct when θ = φ.Common Pitfalls:Forgetting that the normal reaction decreases due to the vertical component of the pull, or not using θ = φ for minimum P.Final Answer:W sin θ (with θ equal to the angle of friction)

Question 16

Simple harmonic motion (SHM) – relation between acceleration and displacement For a particle executing SHM, the instantaneous acceleration a in terms of angular frequency ω and displacement y from the mean position is:

Accepted Answer

Introduction / Context:In simple harmonic motion (SHM), restoring forces lead to an acceleration proportional to displacement and directed toward the mean position. This item checks the fundamental proportionality between acceleration a, angular frequency ω, and displacement y.Given Data / Assumptions:Displacement from mean position = y.Angular frequency of oscillation = ω (in rad/s).Motion is ideal SHM (no damping, no driving force).Concept / Approach:The defining differential equation of SHM is a = d^2y/dt^2 = −ω^2 y. The magnitude of acceleration is ω^2 |y| and its direction is opposite to the displacement, producing the restoring tendency.Step-by-Step Solution:Start from the standard SHM form: y = A sin(ωt + φ).Differentiate twice: v = dy/dt = Aω cos(ωt + φ), a = dv/dt = −Aω^2 sin(ωt + φ) = −ω^2 y.Therefore, a is directly proportional to y with proportionality constant −ω^2. The negative sign indicates direction opposite to displacement.Verification / Alternative check:Dimensional analysis: [ω^2 y] has dimensions of acceleration since [ω] = s^−1 and [y] = m. Hence [ω^2 y] = m s^−2, consistent with acceleration.Why Other Options Are Wrong:ω · y: missing a factor of ω; dimensions are m s^−1, which is velocity, not acceleration.ω^2 / y: yields dimensions s^−2 m^−1, not acceleration.ω^3 · y: incorrect power of ω; dimensions m s^−3.Common Pitfalls:Ignoring the negative sign (direction). In many MCQs, the magnitude is asked implicitly; always remember the acceleration acts toward the mean position.Final Answer:a = −ω^2 y (magnitude ω^2 · y, opposite to displacement)

Question 17

Energy comparison – potential energy of a raised body vs. kinetic energy of a falling body For the same mass and height reference, the potential energy of a vertically raised body is ______ the kinetic energy gained by a body after freely falling th…

Accepted Answer

Introduction / Context:This question examines conservation of mechanical energy for vertical motion under gravity. When air resistance is negligible, potential energy (PE) and kinetic energy (KE) interconvert.Given Data / Assumptions:Mass m is the same in both comparisons.Height h is the same (reference taken from the lower level).Gravitational acceleration g is constant; air resistance is neglected.Concept / Approach:Potential energy at height h is PE = m g h. If a body falls freely from rest through height h, its loss in potential energy equals its gain in kinetic energy by energy conservation, so KE at the lower level becomes m g h.Step-by-Step Solution:Compute PE at height h: PE = m * g * h.Consider free fall from rest: using energy conservation, initial total energy = PE at top = m g h, final total energy = KE at bottom (taking PE at bottom as zero).Hence KE_bottom = m g h.Therefore, PE at height h equals KE gained after falling through height h.Verification / Alternative check:Use kinematics: v^2 = u^2 + 2 g h with u = 0 ⇒ v^2 = 2 g h ⇒ KE = 0.5 m v^2 = 0.5 m (2 g h) = m g h, agreeing with energy conservation.Why Other Options Are Wrong:Less than / greater than: contradict conservation of mechanical energy for ideal free fall.Cannot be compared: sufficient information is provided under standard assumptions.Common Pitfalls:Forgetting to neglect air resistance; with drag, some energy is dissipated as heat and sound, so KE would be less than m g h.Final Answer:equal to

Question 18

Triangle law of forces – equilibrium condition If three concurrent forces are represented in magnitude and direction by the three sides of a triangle taken in order (head-to-tail), are the forces in equilibrium?

Accepted Answer

Introduction / Context:The triangle law of forces provides a graphical equilibrium test for three concurrent coplanar forces. If the vector sum of forces is zero, a closed polygon (triangle for three forces) results.Given Data / Assumptions:Three forces act at a common point.Vectors are drawn to scale, in magnitude and direction.Sides are taken in order (head-to-tail construction).Concept / Approach:Equilibrium occurs when the resultant of all forces is zero. Graphically, placing force vectors head-to-tail to form a closed triangle ensures that the final head coincides with the initial tail, implying zero resultant.Step-by-Step Solution:Draw the first force vector to scale.From its head, draw the second force vector in correct direction and scale.From the second vector’s head, draw the third vector. If it ends at the starting point (closing the triangle), then F1 + F2 + F3 = 0.Thus, the three forces are in equilibrium.Verification / Alternative check:Analytically, resolve the three forces into perpendicular components. If the component sums are zero in both axes, the vector polygon closes; both methods are equivalent.Why Other Options Are Wrong:‘‘No’’: contradicts the triangle law of forces.Conditions about right angle or equal forces are unnecessary; any closed triangle suffices.Common Pitfalls:Forgetting to take vectors in order, or mis-scaling directions so the polygon does not represent the actual forces.Final Answer:Yes

Question 19

Machines – reversibility and efficiency criterion A machine whose efficiency is less than 50% is termed as:

Accepted Answer

Introduction / Context:In machine theory (e.g., screw jacks, hoists), reversibility refers to whether a machine, when loaded, can back-drive in reverse. Efficiency plays a decisive role in this classification.Given Data / Assumptions:Efficiency η = output / input (as a fraction or percentage).Reversible machine can be back-driven by load when input is removed.Self-locking (non-reversible) machine cannot be back-driven by the load.Concept / Approach:For several classical machines (notably screw mechanisms), the condition for self-locking is η Step-by-Step Solution:Identify the efficiency threshold: η = 50% is a key boundary for reversibility in screw-type mechanisms.If η Therefore, with efficiency less than 50%, the correct term is non-reversible machine.Verification / Alternative check:For a screw jack, self-locking condition is often derived using helix angle and friction angle relations. When friction angle exceeds helix angle, reverse motion under load is not possible; this aligns with low efficiency.Why Other Options Are Wrong:Reversible machine: requires sufficiently high efficiency and low friction.Neither of the above / ideal machine: not aligned with accepted definitions; an ideal machine would have η = 100%.Common Pitfalls:Assuming reversibility depends only on geometry; friction (hence efficiency) is crucial.Final Answer:non-reversible machine (self-locking)

Question 20

Impact mechanics – kinetic energy loss in a perfectly elastic collision In a perfectly elastic impact, the loss of kinetic energy is:

Accepted Answer

Introduction / Context:‘‘Elastic’’ and ‘‘inelastic’’ classify collisions based on kinetic energy behavior. In ideal elastic impacts, kinetic energy is conserved; in inelastic impacts, some kinetic energy transforms into other forms.Given Data / Assumptions:Coefficient of restitution e = 1 for a perfectly elastic impact.System is isolated along the line of impact; no external impulses.No deformation losses, heat, or sound energy losses in the idealization.Concept / Approach:For e = 1, both linear momentum (along the line of impact) and kinetic energy are conserved. Thus the total kinetic energy before equals that after impact.Step-by-Step Solution:State conservation: Σp_before = Σp_after and ΣKE_before = ΣKE_after.With e = 1, relative velocity of separation equals relative velocity of approach, a hallmark of elastic behavior.Therefore, loss of kinetic energy = 0.Verification / Alternative check:Classic 1D two-body formulas for elastic collision show KE_before − KE_after = 0; numerical examples with equal masses show a perfect velocity swap and identical total KE.Why Other Options Are Wrong:Saying some energy is always lost contradicts the ideal definition of ‘‘perfectly elastic’’.‘‘Depends only on masses’’ is incorrect; for e = 1, KE is conserved irrespective of specific masses.‘‘Cannot be determined’’ is false; it is determined by the definition of elastic impact.Common Pitfalls:Confusing real collisions (often slightly inelastic) with the idealized, perfectly elastic case used in theory.Final Answer:Agree – zero loss of kinetic energy

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