Question 1

Apparent weight in free fall When a body falls freely solely under gravity (ignoring air resistance), what apparent weight does it possess on a scale attached to it?

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Introduction / Context:Apparent weight is the normal reaction a supporting surface exerts on a body. It can differ from true weight depending on acceleration of the reference frame. A classic case is free fall, which underpins weightlessness sensations in elevators and spacecraft in orbit.Given Data / Assumptions:Body is in free fall under gravity g.No air resistance; body and attached scale accelerate together.Apparent weight is the scale reading (normal force N).Concept / Approach:For vertical motion with acceleration a, Newton’s second law on the body with upward positive gives: N − W = −m a (if downward acceleration). In free fall, a = g downward, so N = 0. The body is locally weightless although gravitational force W = m g still acts; what vanishes is the contact force that registers on a scale.Step-by-Step Solution:Write equilibrium with acceleration: N − m g = m a_y.For free fall: a_y = −g ⇒ N − m g = −m g ⇒ N = 0.Therefore, the apparent weight (scale reading) is zero—described here as “no weight”.True weight remains m g; it is just not felt or measured by the scale in free fall.Verification / Alternative check:In a falling elevator with broken cable (idealized), occupants and the floor accelerate together at g; hence no normal reaction is developed. Astronauts in orbit are in continuous free fall around Earth, experiencing microgravity (practically weightlessness).Why Other Options Are Wrong:minimum (but not zero): incorrect; it is exactly zero for ideal free fall.maximum / equal to true weight: apply to accelerating upward frames, not free fall.Common Pitfalls:Confusing true weight (m g) with apparent weight (normal force). Only the latter vanishes in free fall.Final Answer:no

Question 2

Resultant of two forces with angular separation – perpendicularity condition Two forces act at an angle of 120°. The larger force is 40 N. If the resultant is perpendicular to the smaller force, find the magnitude of the smaller force.

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Introduction / Context:Vector addition with angular separation often appears in statics and dynamics. A useful technique is to convert geometric constraints (like perpendicularity) into dot-product equations. Here, the resultant is perpendicular to one of the forces, which allows a straightforward algebraic solution.Given Data / Assumptions:Forces P (smaller, unknown) and Q (larger = 40 N).Angle between P and Q is 120°.Resultant R = P + Q is perpendicular to P.All vectors are coplanar; use standard Euclidean geometry.Concept / Approach:Use the dot product: If R ⟂ P, then R · P = 0. Since R = P + Q, we have P · P + Q · P = 0. Evaluate Q · P via magnitudes and the included angle: Q P cos 120°.Step-by-Step Solution:Let |P| = p and |Q| = 40.Condition: (P + Q) · P = 0 ⇒ P · P + Q · P = 0.Compute: P · P = p^2; Q · P = |Q||P| cos 120° = 40 p (−1/2) = −20 p.Equation: p^2 − 20 p = 0 ⇒ p(p − 20) = 0 ⇒ p = 20 N (reject p = 0).Verification / Alternative check:Law of cosines on triangle formed by vectors can also be used to find R and then check perpendicularity via Pythagoras on magnitudes; you will arrive at the same p = 20 N.Why Other Options Are Wrong:40 N or 80 N: do not satisfy the orthogonality condition when substituted.10√3 N: common distractor from mixing 60°/120° relations; algebra disproves it.Common Pitfalls:Using 60° instead of 120° (supplementary angle confusion) or forgetting the negative cosine for obtuse angles.Final Answer:20 N

Question 3

Elevator physics – apparent pressure while accelerating upward When a lift (elevator) moves upward with constant upward acceleration a, the pressure/normal force a person exerts on the floor is _____ proportional to a.

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Introduction / Context:The “heavier” feeling in an accelerating elevator is a classic application of Newton’s second law. The scale reading (or floor reaction) is often called the apparent weight or pressure. This item asks how that reaction varies with upward acceleration.Given Data / Assumptions:Person of mass m standing on the elevator floor.Elevator accelerates upward with acceleration a (constant).g is gravitational acceleration; neglect air drag and other forces.Concept / Approach:Apply Newton’s second law in the vertical direction. Take upward as positive. The floor exerts a normal force N on the person; weight acts downward.Step-by-Step Solution:Equation of motion: N − m g = m a.Solve for the normal force: N = m (g + a).Therefore, the apparent pressure/weight increases linearly with a (direct proportionality for fixed m and g).If a = 0 (constant speed), N = m g; if a > 0, N > m g; if a = −g (free fall), N = 0.Verification / Alternative check:A spring scale between the feet and the floor would read N; as a increases, the spring compresses more, confirming the direct proportionality.Why Other Options Are Wrong:inversely: would imply N decreases when a increases, contrary to N = m(g + a).independent of a: true only for a = 0 (constant velocity).quadratically: there is no a^2 term in the linear relation.Common Pitfalls:Mixing up the sign convention or interpreting “pressure” as atmospheric pressure rather than the normal reaction per unit area. Here “pressure” corresponds to the contact force effect.Final Answer:directly

Question 4

Law of the machine – maximum mechanical advantage For a lifting machine satisfying P = m W + C (law of the machine), what is the maximum possible mechanical advantage (MA) as the load W varies?

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Introduction / Context:The law of the machine expresses the required effort P to lift load W as P = m W + C, where m and C lump the kinematic and frictional characteristics. Mechanical advantage MA is defined as W / P. This question asks for the maximum MA achievable over the operating range.Given Data / Assumptions:Law of the machine: P = m W + C, with constants m > 0 and C ≥ 0.Mechanical advantage: MA = W / P.We consider W ≥ 0 and examine limiting behavior as W varies.Concept / Approach:Compute MA(W) = W / (m W + C). Treat C as the frictional (or lost effort) term. For fixed m and C, MA increases with W because the constant C becomes relatively less important at larger loads. The theoretical upper bound occurs as W → ∞.Step-by-Step Solution:Write MA(W) = W / (m W + C).Divide numerator and denominator by W (for W > 0): MA(W) = 1 / (m + C/W).As W increases without bound, C/W → 0.Limit as W → ∞: MA_max = 1 / m.Verification / Alternative check:Differentiate MA(W) with respect to W: d(MA)/dW = C / (m W + C)^2 > 0 for C > 0, confirming monotonic increase and the asymptotic limit 1/m. For an ideal machine (C = 0), MA is identically 1/m for all W, consistent with the same bound.Why Other Options Are Wrong:1 + m, 1 − m, m: have incorrect dimensions and do not follow from MA = W/(mW + C).Common Pitfalls:Confusing mechanical advantage with velocity ratio or efficiency. MA depends on load when C ≠ 0, whereas velocity ratio is purely kinematic.Final Answer:1 / m

Question 5

Triangle law of forces – equilibrium test If three forces acting at a point can be represented in magnitude and direction by the three sides of a triangle taken in order (head-to-tail), are the forces in equilibrium?

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Introduction / Context:One of the classical equilibrium criteria for concurrent coplanar forces is the triangle law. It provides a geometric way to check whether the vector sum of forces is zero without component arithmetic.Given Data / Assumptions:Three non-collinear forces acting simultaneously at the same point.Planar vectors arranged head-to-tail in a closed polygon (triangle).Magnitudes represented by side lengths; directions by side orientations.Concept / Approach:The condition for equilibrium is that the vector sum equals zero. If three vectors placed head-to-tail form a closed triangle, the final head coincides with the initial tail, which means their sum is the zero vector. This is the triangle law of forces—an immediate geometric proof of equilibrium.Step-by-Step Solution:Represent forces F1, F2, F3 as directed segments.Place them sequentially head-to-tail: F1 + F2 + F3 as a polygon.If the polygon closes (a triangle for three forces), then F1 + F2 + F3 = 0.Thus, the three forces can balance each other at the point of concurrency.Verification / Alternative check:Resolve into x and y components. If the triangle closes, each component sum equals zero: ΣFx = 0 and ΣFy = 0, confirming equilibrium.Why Other Options Are Wrong:False / conditional statements: Equilibrium does not require a right triangle or equal forces; any closed triangle suffices when vectors are drawn to scale and direction.Common Pitfalls:Taking magnitudes only and ignoring directions; the law requires both be represented faithfully and taken “in order”.Final Answer:True

Question 6

Kinematics sign convention – gravity during upward throw When a body is thrown upward, should gravitational acceleration be taken as zero, positive g, or negative g in the vertical equation of motion?

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Introduction / Context:Projectile and vertical throw problems hinge on consistent sign conventions. Gravity does not switch off during ascent; it constantly accelerates the body downward. The statement claims g is taken as zero on upward motion—this must be evaluated.Given Data / Assumptions:Vertical 1-D motion under uniform gravity g.Choose upward as the positive axis (common convention).Neglect air resistance for clarity.Concept / Approach:With upward positive, gravitational acceleration is a constant negative value: a = −g. Kinematic equations such as v = u + a t and s = u t + (1/2) a t^2 hold at all times, including ascent, apex, and descent. Setting g = 0 would eliminate the deceleration responsible for the body slowing down and turning.Step-by-Step Solution:Take upward positive; weight acts downward ⇒ a = −g.Use v = u + a t ⇒ at apex v = 0 occurs because a = −g reduces v from u to 0.Displacement: s = u t + (1/2)(−g) t^2 correctly gives a concave-down trajectory in time.Therefore, gravitational acceleration is not zero during ascent; it is constant and downward.Verification / Alternative check:Energy method: kinetic energy decreases on the way up because gravitational potential energy increases by m g h. If g were zero, there would be no energy exchange to halt the upward motion—contradicting observation.Why Other Options Are Wrong:Correct near the top / in vacuum: gravity acts everywhere near Earth; vacuum removes drag only.g = 0 while going up: physically incorrect; gravity does not switch off.Common Pitfalls:Confusing sign convention with magnitude. You always include g; the sign depends on chosen positive direction.Final Answer:Incorrect – use g with downward sign (−g if upward is positive)

Question 7

Energy due to position — what is it called? In classical mechanics, the energy a body possesses solely by virtue of its position or configuration (for example, a mass held at a height, a compressed spring, or a stretched elastic band) is known as:

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Introduction / Context:This item checks your understanding of the two fundamental mechanical energy forms—potential and kinetic—and asks you to name the energy stored due to position or configuration. Recognizing this helps in solving problems using the work–energy principle, conservation of mechanical energy, and in real-world systems such as dams, cranes, and springs.Given Data / Assumptions:No numbers are needed; this is a concept recall question.Examples include a mass at height h, a compressed spring, or a charged object in an electric field.Neglect dissipative effects (air resistance, friction) for the conceptual definition.Concept / Approach:Mechanical energy appears mainly as potential energy (energy stored due to position/configuration) and kinetic energy (energy of motion). Gravitational potential energy of a mass m at height h (relative to a reference level) is m * g * h. Elastic potential energy in a spring compressed or stretched by x is (1/2) * k * x^2. These forms are available to do work when the system is released and allowed to move toward a lower-energy configuration.Step-by-Step Solution:Identify what the stem asks: energy “by virtue of position.”Recall definitions: potential energy (position/configuration), kinetic energy (motion).Match to examples: mass at height, compressed spring ⇒ potential energy.Therefore, the correct term is potential energy.Verification / Alternative check:If the body is released, potential energy converts to kinetic energy as it moves (e.g., falling mass). This conversion validates that the stored energy was indeed potential energy poised to perform work.Why Other Options Are Wrong:Kinetic energy: depends on velocity, not position.Electrical energy: field-specific; the stem is generic mechanics.Chemical energy: stored in molecular bonds, not the body’s mechanical position.Common Pitfalls:Confusing “stored” energy in a moving object as potential; moving objects possess kinetic energy.Forgetting that configuration (elastic deformation) is also a “position” effect.Final Answer:potential energy

Question 8

Newton’s second law — correct equation form: If P is the net force on a body of mass m producing acceleration a, which one of the following equations correctly states Newton’s second law?

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Introduction / Context:Newton’s second law connects dynamics (forces) with kinematics (acceleration). It is foundational for solving virtually all engineering mechanics problems from particle motion to rigid-body dynamics.Given Data / Assumptions:P is the resultant (net) force acting on the body.m is the mass of the body (assumed constant).a is the linear acceleration in the direction of P.Concept / Approach:The standard vector form is P = m * a. Rearranged to one side for equation checking, P − m * a = 0 is the correct identity. Any other algebraic or dimensional manipulation in the options that violates this equality is incorrect.Step-by-Step Solution:Recall Newton’s second law: P = m * a.Move all terms to one side: P − m * a = 0.Compare with given choices; pick the expression that matches.Therefore, the correct option is P − m.a = 0.Verification / Alternative check:Dimensional consistency: [P] = N = kgm/s^2 and [ma] = kg*(m/s^2) = N; subtracting like dimensions is valid, confirming the identity structure P − ma = 0.Why Other Options Are Wrong:P + m.a = 0: implies P = −ma; sign is wrong for the general statement.P x m.a = 0: “x” suggests multiplication or a cross-product; not the law’s form and dimensions become N^2.P/m.a = 0: division gives a dimensionless number; cannot equal zero identically for all cases.Common Pitfalls:Forgetting that P is the net force (not a single applied force), i.e., include all forces to compute m*a.Final Answer:P - m.a = 0

Question 9

Machines — identify the incorrect statement: Which of the following statements about friction, efficiency, mechanical advantage, and velocity ratio is incorrect?

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Introduction / Context:Basic machine parameters—mechanical advantage (M.A.), velocity ratio (V.R.), and efficiency—are frequently confused in exams. This question asks you to spot the incorrect statement among commonly paired definitions.Given Data / Assumptions:M.A. = Load / Effort.V.R. = Distance moved by effort / Distance moved by load (for the same time interval).Efficiency η = M.A. / V.R.Limiting friction and normal reaction are standard in dry friction theory.Concept / Approach:Check each statement against standard definitions. Friction indeed opposes relative motion. Coefficient of friction μ = Limiting friction / Normal reaction. An ideal machine has η = 100% (no losses). V.R., however, is a kinematic ratio of distances or velocities, not a force ratio; the force ratio is M.A.Step-by-Step Solution:Evaluate (a): definition of friction — correct.Evaluate (b): μ = F_limiting / N — correct.Evaluate (c): ideal machine ⇒ η = 100% — correct.Evaluate (d): claims V.R. = Load / Effort (force ratio) — this is actually M.A. Hence (d) is incorrect.Verification / Alternative check:Consider a simple pulley: effort moves 2 m while load rises 1 m ⇒ V.R. = 2. If load is 100 N and effort is 60 N ⇒ M.A. = 100/60 ≈ 1.67 and η = 1.67/2 ≈ 83%. Distances vs forces clearly differ.Why Other Options Are Wrong (i.e., why they are correct statements):(a) and (b) align with standard friction definitions.(c) defines the idealized, lossless machine correctly.Common Pitfalls:Swapping M.A. and V.R. because both are ratios; remember V.R. is geometric/kinematic, M.A. is force-based.Final Answer:The velocity ratio of a machine is the ratio of load lifted to the effort applied.

Question 10

Elastic springs — name of energy stored on compression: The potential energy stored in a spring due to elastic deformation (compression or extension) is called strain energy. Is this statement correct?

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Introduction / Context:When elastic members deform under load, they store recoverable energy. In springs and many structural elements, this stored energy is central to vibration analysis, energy methods, and impact problems.Given Data / Assumptions:The material obeys linear elasticity within the working range.Load is removed slowly so that energy recovery is not dissipated as heat (idealization).Concept / Approach:Strain energy is the energy stored in a body due to strain (deformation). For a linear spring of stiffness k compressed or stretched by x, the strain energy U is U = (1/2) * k * x^2. This energy can be converted back into kinetic energy when the spring is released, which is why springs are used to do mechanical work (e.g., in watches, valves, toys).Step-by-Step Solution:Identify the system: a spring under compression or tension.Recognize the energy form: recoverable potential energy due to elastic deformation.Name of this energy: strain energy (a category of potential energy).Hence, the statement is correct.Verification / Alternative check:Area under the load–deflection curve equals strain energy. For a linear spring: area of triangle = (1/2) * Force_max * Deflection = (1/2) * (k * x) * x = (1/2) * k * x^2, confirming the naming and formula.Why Other Options Are Wrong:No: would deny the well-established term used across strength of materials and machine design.Common Pitfalls:Confusing “strain energy” (recoverable) with “plastic work” (mostly non-recoverable after yielding).Thinking springs only “store” energy in compression; extension also stores the same form of energy.Final Answer:Yes

Question 11

Classification of force systems — which term fits? Forces that are concurrent (meet at one point) and whose lines of action also lie in the same plane are called:

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Introduction / Context:Correctly classifying force systems helps you choose the right equilibrium equations. For particles (or pins) with forces meeting at a point, ΣF = 0 in two orthogonal directions is sufficient in 2D; for general rigid bodies, you must also consider moments.Given Data / Assumptions:The forces meet at a single point (concurrent).Their lines of action lie in the same geometric plane (coplanar).We are naming the type of system only.Concept / Approach:Terminology: “coplanar” refers to lying in the same plane; “non-coplanar” means 3D distribution. “Concurrent” means all lines of action intersect at a common point; “non-concurrent” means they do not. Thus, a set of forces meeting at one point in a single plane is a system of coplanar concurrent forces.Step-by-Step Solution:Identify planarity: all in one plane ⇒ coplanar.Identify concurrency: meet at one point ⇒ concurrent.Combine descriptors ⇒ coplanar concurrent forces.Select the option that states this explicitly.Verification / Alternative check:Particle equilibrium is a classic example: weight, normal reaction, tension at a point are coplanar and concurrent; ΣF_x = 0 and ΣF_y = 0 ensure equilibrium without needing a moment equation about that point.Why Other Options Are Wrong:Coplanar non-concurrent: lines of action do not meet—different class.Non-coplanar concurrent / non-concurrent: imply 3D, not the 2D case described.Common Pitfalls:Confusing “passing near a point” with true concurrency; the lines must intersect at exactly one point.Final Answer:coplaner concurrent forces

Question 12

Rotational kinematics — which of the following are vector quantities? Select the correct set when considering direction-sensitive rotational measures.

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Introduction / Context:In rotational motion, many quantities possess both magnitude and direction (sense about an axis), making them vectors (more precisely, axial vectors or pseudovectors). Distinguishing vector from scalar quantities is essential for applying vector algebra and sign conventions correctly.Given Data / Assumptions:Standard rigid-body rotation about a fixed axis through small angles.Right-hand rule used to assign vector directions.Concept / Approach:Angular displacement, angular velocity, and angular acceleration each require a direction specification (clockwise vs counter-clockwise mapped to an axis via the right-hand rule). Hence they are treated as vectors in engineering analyses. Their directions are along the axis of rotation: thumb points along +ω when fingers curl in the rotation sense.Step-by-Step Solution:Angular displacement: for small rotations, represented as a vector along the axis (direction matters in composition).Angular velocity ω: vector indicating rotation rate and axis direction; used in v = ω × r.Angular acceleration α: time rate of change of ω; direction indicates speeding up or slowing down and precessional effects.Therefore, all listed quantities are vectors.Verification / Alternative check:Use vector relationships: linear velocity v at a point is v = ω × r, which only makes sense if ω is a vector. Similarly, tangential acceleration a_t = α × r requires vector α.Why Other Options Are Wrong:Selecting only one or two ignores the consistent vector treatment used in kinematics and dynamics.Common Pitfalls:Believing angular displacement is always scalar; for finite large rotations, non-commutativity arises, but in engineering practice small-angle vector treatment is standard.Final Answer:all of these

Question 13

Ideal machine — efficiency value: By definition, the efficiency of an ideal (lossless) machine is:

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Introduction / Context:Efficiency compares the useful output to the input. In machines, losses due to friction, deformation, and slip reduce efficiency below 100%. Understanding the limiting “ideal” case is vital for benchmarking real designs.Given Data / Assumptions:Efficiency η = Output / Input = M.A. / V.R.An ideal machine has no energy losses.Concept / Approach:In an ideal machine, all input work converts to useful output work. Therefore Output = Input for any load–movement pair, so η = 1 or 100%. Real systems always have η Step-by-Step Solution:Recall definition: η = M.A. / V.R.Ideal condition: M.A. equals V.R.; thus η = 1.Convert to percent: 1 * 100% = 100%.Hence, the correct answer is 100%.Verification / Alternative check:Energy viewpoint: Input work W_in = Effort * effort distance; Output work W_out = Load * load distance. For an ideal machine W_out = W_in, confirming η = 100%.Why Other Options Are Wrong:Ranges below 100% describe realistic but not ideal machines.Common Pitfalls:Confusing “high efficiency” with “ideal”; ideal is a theoretical limit used for comparison, not typically achievable.Final Answer:100%

Question 14

Impulse–momentum — comparing rebounds and stops: A lead ball and a rubber ball of the same mass strike a rigid wall with the same speed. The lead ball falls down (almost stops), whereas the rubber ball rebounds. Which statement correctly compares th…

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Introduction / Context:This classic comparison illustrates impulse, coefficient of restitution, and how rebound affects momentum change. It links to collision analysis, sports physics, and materials selection for impact absorption.Given Data / Assumptions:Both balls have the same mass m and approach the wall with the same speed u.Lead ball: comes nearly to rest after impact (final speed ≈ 0).Rubber ball: rebounds with some speed v (v > 0) in the opposite direction.Concept / Approach:Change in momentum Δp = p_final − p_initial (vector). Take the approach direction as positive. Lead: initial momentum = +mu; final ≈ 0 ⇒ |Δp_lead| ≈ mu. Rubber: initial = +mu; final = −mv ⇒ Δp_rubber = −mv − (+mu) = −m(u + v) ⇒ magnitude |Δp_rubber| = m(u + v), which is greater than mu because v > 0.Step-by-Step Solution:Lead: |Δp| ≈ mu.Rubber: rebounds ⇒ |Δp| = m(u + v) > m*u.Therefore, the rubber ball experiences a larger magnitude of momentum change.Impulse equals change in momentum, so the wall delivers a larger impulse to the rubber ball.Verification / Alternative check:Coefficient of restitution e > 0 for rubber; e ≈ 0 for lead. Higher e implies a more negative final velocity relative to the incident direction, increasing |Δp|.Why Other Options Are Wrong:Equal change: only true if both stop dead or both rebound equally, which is not the case.Rubber less than lead: contradicts the rebound reversal which adds to the momentum change.None of the above: incorrect because a valid comparison exists.Common Pitfalls:Comparing forces rather than impulse; the relevant measure is the integral of force over contact time.Ignoring direction in momentum; the rubber ball’s reversal makes the change larger.Final Answer:the change in momentum suffered by rubber ball is more than the lead ball

Question 15

Lami’s theorem — precise statement: Choose the correct statement describing Lami’s theorem for three forces acting at a point.

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Introduction / Context:Lami’s theorem is a powerful tool for solving coplanar concurrent three-force problems without resolving components. It appears frequently in truss joints, suspended bodies, and cable–pulley pin analyses.Given Data / Assumptions:Exactly three forces act at a single point.The forces are coplanar and in equilibrium (net force zero).The included angles are the angles between the directions of the other two forces.Concept / Approach:Lami’s theorem states: For three concurrent forces in equilibrium, F1/sin(α1) = F2/sin(α2) = F3/sin(α3), where α1 is the angle between F2 and F3 (the other two), and similarly for α2 and α3. This arises from the geometry of the force triangle and the sine rule.Step-by-Step Solution:Check for conditions: concurrency and exactly three forces.Write proportionality: F1/sin(angle opposite F1) = F2/sin(angle opposite F2) = F3/sin(angle opposite F3).Recognize this is the precise wording in option (d).Hence, select (d) as the correct statement.Verification / Alternative check:Construct the closed force triangle: join force vectors head-to-tail; by the sine rule for this triangle, the sides (force magnitudes) are proportional to the sines of the opposite angles, which are the pairwise angles between the other two forces in the original system.Why Other Options Are Wrong:Equal magnitudes or 120° spacing are special cases, not the general theorem.“Must be in equilibrium” is necessary but not a full statement of Lami’s relation.Common Pitfalls:Using the angle adjacent to the force instead of the angle between the other two forces (the opposite angle in the force triangle).Final Answer:if the three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two

Question 16

Engineering Mechanics – Moment of a Force about a Point For the force P acting on a rigid body, what is the correct expression for the moment of P about point O (i.e., torque at O), considering OB is the perpendicular distance from O to the line of …

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Introduction / Context: In engineering mechanics, the moment (torque) of a force about a point measures the tendency of that force to rotate the body about the point. This concept is fundamental to solving support reactions, designing levers, and checking equilibrium.Given Data / Assumptions: A force P acts on a rigid body along a known line of action.Point O is the reference about which the moment is taken.OB denotes the shortest (perpendicular) distance from O to the line of action of P.Sign of the moment depends on rotation sense, but the magnitude uses perpendicular distance.Concept / Approach: The scalar magnitude of the moment of a force about a point is given by M = P * d, where d is the perpendicular distance from the point to the force’s line of action. Any non-perpendicular segment (for example OA or OC) is not appropriate unless its perpendicular component is taken.Step-by-Step Solution: Identify the line of action of P. Drop a perpendicular from O to that line; its length is OB. Compute moment magnitude: M_O = P * OB. Assign sign based on clockwise or counterclockwise tendency (not asked here).Verification / Alternative check: Using vector formulation: M_O = r × P. The magnitude is |r| * |P| * sin(θ) which equals P times the perpendicular distance. Hence the shortest distance OB is correct.Why Other Options Are Wrong: P × OA / P × OC / P × AC: OA, OC, AC are not guaranteed to be perpendicular distances; using them blindly can overstate or understate the torque. P × (OA + OC): adds unrelated lengths and has no basis in the moment definition.Common Pitfalls: Using any convenient segment instead of the true perpendicular distance. Forgetting the sign (direction) of the moment when doing equilibrium.Final Answer: P × OB

Question 17

Collision Theory – Coefficient of Restitution for Inelastic Impacts For inelastic bodies undergoing impact, what is the typical range of the coefficient of restitution e?

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Introduction / Context: The coefficient of restitution e quantifies how “bouncy” a collision is by comparing relative speeds after and before impact. Understanding its range distinguishes elastic, inelastic, and perfectly inelastic collisions in dynamics problems.Given Data / Assumptions: Two bodies collide along the line of impact.No external impulse acts during the very short impact interval.e is defined along the common normal at impact.Concept / Approach: By definition, e = (relative speed of separation) / (relative speed of approach). Its value categorizes collisions: e = 1 (perfectly elastic), 0 1 indicate impact with added energy (e.g., active systems) and are not typical for passive mechanical impacts.Step-by-Step Solution: Identify inelastic collision: some kinetic energy lost. Use definition: inelastic ⇒ 0 Perfectly inelastic special case: e = 0. Hence for inelastic bodies in general, e lies between 0 and 1.Verification / Alternative check: Examine energy loss: elastic collisions conserve kinetic energy; inelastic collisions do not. Since speed of separation is reduced relative to approach, the ratio e must be less than 1 but nonnegative.Why Other Options Are Wrong: zero only: corresponds to perfectly inelastic, not all inelastic. one only: elastic, not inelastic. more than one / less than zero: not applicable to passive mechanical collisions modeled without active energy input.Common Pitfalls: Confusing “inelastic” with “perfectly inelastic”. Misapplying e across tangential directions instead of the line of impact.Final Answer: between zero and one

Question 18

Nature of Impact – Lead Spheres Approximation The impact between two lead spheres is typically approximated as which type of impact?

Accepted Answer

Introduction / Context: Real materials dissipate energy during impact through plastic deformation, internal friction, and heat. Lead is soft and exhibits significant plasticity, so its collision behavior helps illustrate the difference between elastic and inelastic impacts.Given Data / Assumptions: Two lead spheres collide along a line of impact.Material properties of lead: low yield strength and high plasticity.No external impulses during the very brief collision.Concept / Approach: In elastic impact, kinetic energy is conserved (e = 1). In inelastic impact, some kinetic energy converts into deformation and heat, giving 0 ≤ e Step-by-Step Solution: Identify material behavior: lead deforms plastically on impact. Plastic deformation absorbs kinetic energy. Therefore, the relative speed of separation is much less than relative speed of approach. Hence e is significantly less than 1 ⇒ inelastic impact.Verification / Alternative check: Practical drop tests with lead balls show poor bounce heights compared with steel or glass, reflecting low restitution and significant energy loss at contact.Why Other Options Are Wrong: elastic / perfectly elastic: would require negligible energy loss, which contradicts lead’s plastic response. super-elastic / rebound amplification: requires energy input during impact; not applicable.Common Pitfalls: Assuming all metals behave elastically at impact; material hardness matters.Final Answer: inelastic

Question 19

Worm and Worm Wheel – Velocity Ratio Calculation (Single-Start Worm) A single-start worm drives a worm wheel with 50 teeth. The effort wheel diameter is 100 mm and the load drum diameter is 50 mm. What is the velocity ratio (VR) of the lifting machi…

Accepted Answer

Introduction / Context: Velocity ratio (VR) links the movement of effort to the movement of load in simple machines. For a worm and worm wheel hoist with effort wheel and load drum, VR combines gear ratio and pulley radii effects.Given Data / Assumptions: Single-start worm (starts = 1).Number of teeth on worm wheel = 50.Diameter of effort wheel = 100 mm.Diameter of load drum = 50 mm.No slip; ideal kinematics.Concept / Approach: For worm gear hoists, VR = (teeth on worm wheel / worm starts) * (diameter of effort wheel / diameter of load drum). The gear part converts one worm revolution into many drum revolutions depending on wheel teeth, and the pulley part converts rotation to rope travel.Step-by-Step Solution: Gear ratio = 50 / 1 = 50. Pulley ratio = 100 / 50 = 2. Overall VR = 50 * 2 = 100. Thus, the effort must move 100 units while the load rises 1 unit (ideally).Verification / Alternative check: Dimensional/logic check: More wheel teeth and a larger effort wheel both increase VR, so a value larger than 50 is expected; 100 is consistent.Why Other Options Are Wrong: 50: ignores the wheel/ drum diameters ratio. 150 / 200: overestimates by multiplying with an incorrect factor. 25: would imply a half gear ratio or reversed diameters, which is not the case.Common Pitfalls: Confusing the number of worm starts with wheel teeth. Using radii instead of diameters inconsistently (ratio is the same but stay consistent).Final Answer: 100

Question 20

Basic Definition – Matter Contained in a Body In classical mechanics, the term for the amount of matter contained in a body is:

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Introduction / Context: Precise terminology avoids confusion when solving engineering problems. “Mass,” “weight,” and “momentum” are often mixed up by beginners, but they represent different physical quantities.Given Data / Assumptions: Standard SI units are used: mass in kilograms (kg), force in newtons (N).Near Earth’s surface g ≈ 9.81 m/s^2 for weight computations.Concept / Approach: Mass is a scalar measure of the amount of matter and the inertia of a body. Weight is the gravitational force acting on that mass. Momentum is mass multiplied by velocity and characterizes motion. An impulsive force is a large force applied over a short interval causing a finite change in momentum.Step-by-Step Solution: Define mass: amount of matter; intrinsic; unit kg. Relate weight: W = m * g (a force; units N). Relate momentum: p = m * v (units N·s). Therefore, the correct term for “matter contained” is mass.Verification / Alternative check: A body taken to the Moon retains the same mass but has reduced weight due to lower lunar gravity. This confirms that mass measures matter content, not gravitational pull.Why Other Options Are Wrong: weight: depends on g; not an amount of matter. momentum: requires motion; not a property of matter content. impulsive force: not a property; a type of force application. density: mass per unit volume; not the amount itself.Common Pitfalls: Saying “my weight is 70 kg” instead of “my mass is 70 kg”.Final Answer: mass

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