Question 1

Systems of pulleys — identifying the system by velocity ratio: If, in a tackle, the number of pulleys is equal to the velocity ratio (V.R.), then the arrangement corresponds to which standard system of pulleys?

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Introduction / Context:Pulley systems are categorized (first, second, third) based on how pulleys are arranged and how many rope segments support the moving block. Velocity ratio (V.R.)—the ratio of distance moved by effort to distance moved by load—identifies the system and guides mechanical advantage estimates.Given Data / Assumptions:The number of pulleys n equals the velocity ratio.Idealized ropes and pulleys (no slip, massless rope, friction neglected) for the identification rule of thumb.Concept / Approach:For the standard systems: First system yields V.R. = 2^m (m = pulleys in the moving block). Second system yields V.R. ≈ 2m (number of supporting segments). Third system yields V.R. ≈ n (numerically equal to the number of pulleys). Therefore, n = V.R. directly maps to the third system of pulleys.Step-by-Step Solution:Recall V.R. patterns: first → 2^m; second → 2m; third → n.Given n = V.R., this matches the third system.Hence, select “third.”Verification / Alternative check:Draw small examples: with 3 pulleys arranged in the third system, V.R. = 3; contrast with second system where 3 pulleys could give V.R. = 4, demonstrating non-equality to n.Why Other Options Are Wrong:first / second: Their V.R. formulas do not equal the simple count of pulleys.Common Pitfalls:Assuming mechanical advantage always equals V.R.; in real systems friction reduces M.A. below V.R.Final Answer:third

Question 2

Second moment of area (area moment of inertia) — symmetry of a hollow rectangle: For a hollow rectangular section, the second moment of area about the centroidal Y–Y axis is generally not equal to that about the centroidal X–X axis (unless it is squ…

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Introduction / Context:Designers compare Ixx and Iyy (area moments of inertia) to assess bending stiffness about orthogonal axes. For rectangles, different side lengths produce different stiffness about X–X and Y–Y, and hollowing preserves this asymmetry unless special symmetry exists.Given Data / Assumptions:Hollow rectangular cross-section with unequal outer and inner dimensions.Centroid located at the geometric center (concentric hole).Concept / Approach:The second moment of area depends on breadth and depth to the third power along the bending axis. For a rectangle, Ixx ∝ bh^3/12 and Iyy ∝ hb^3/12 (outer minus inner for hollow). Unless b equals h (square), these values differ. Hollowing (subtracting inner rectangle) preserves the general inequality unless the section is perfectly square and similarly proportioned.Step-by-Step Solution:Compute Ixx = (b* h^3 − b_i* h_i^3)/12.Compute Iyy = (h* b^3 − h_i* b_i^3)/12.For b ≠ h or b_i ≠ h_i, typically Ixx ≠ Iyy.Therefore, the statement that Iyy is not the same as Ixx (in general) is correct.Verification / Alternative check:Set a square case: b = h and b_i = h_i, then Ixx = Iyy by symmetry. This exception proves the general rule stated in the stem (which says “not the same” in general).Why Other Options Are Wrong:No: would imply equality regardless of proportions, which is false except for the special square case.Common Pitfalls:Confusing centroid location with equality of moments; centroid at center does not guarantee Ixx = Iyy.Ignoring the cubic dependence on the dimension perpendicular to the bending axis.Final Answer:Yes

Question 3

Centroid of a triangle — intersection of medians: For a uniform triangular lamina or area, the centre of gravity (centroid) lies at the point where all three medians intersect (the centroid). Is this statement true?

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Introduction / Context:Finding the centroid of basic shapes underpins calculations of bending stresses, deflections, and resultant forces. The triangle’s centroid has a celebrated geometric property that simplifies many engineering analyses.Given Data / Assumptions:Homogeneous triangular area (uniform thickness and density if treated as a lamina).Medians are the line segments from each vertex to the midpoint of the opposite side.Concept / Approach:The centroid of a triangle lies at the intersection of its three medians and divides each median in a 2:1 ratio (measured from the vertex). This location is independent of coordinate orientation and works for any triangle (acute, obtuse, right).Step-by-Step Solution:Draw all three medians; they are concurrent at a single point.Use coordinate geometry: with vertices (x1,y1), (x2,y2), (x3,y3), the centroid is at ( (x1 + x2 + x3)/3 , (y1 + y2 + y3)/3 ).This is precisely the intersection point of the medians.Therefore, the statement is true.Verification / Alternative check:Physical reasoning: uniform density implies balance occurs where equal triangular masses balance along the medians; experiments with a cardboard triangle and a plumb line confirm the same point.Why Other Options Are Wrong:False: contradicts well-established geometry and engineering handbooks.Common Pitfalls:Confusing centroid (area) with circumcenter, incenter, or orthocenter; only the centroid is the intersection of medians and always lies inside the triangle.Final Answer:True

Question 4

Work–energy in SI units — meaning of one joule: Choose the statement that correctly defines one joule in the SI system.

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Introduction / Context:Clarity on SI definitions avoids unit-conversion errors in energy, power, and torque problems. The joule (J) is the standard unit of work and energy in the International System of Units.Given Data / Assumptions:Work W = F * s when force and displacement are collinear.SI base units: newton (N) for force, meter (m) for length.Concept / Approach:By definition, 1 joule equals the work done by a 1 newton force acting through 1 meter in the direction of the force. Hence, 1 J = 1 N·m. Non-SI definitions (dyne–cm) correspond to the CGS unit called the erg (1 erg = 1 dyne·cm = 1e−7 J).Step-by-Step Solution:Select the statement matching W = F * s with F = 1 N and s = 1 m.Compute W = 1 * 1 = 1 N·m = 1 J.Therefore, the correct choice is the SI definition with 1 N and 1 m.Verification / Alternative check:Dimensional check: J = kg·m^2/s^2. Since N = kg·m/s^2, multiplying by m yields kg·m^2/s^2, confirming consistency.Why Other Options Are Wrong:1 kg and 1 g refer to mass, not force; SI force is the newton.1 dyne and 1 cm define the CGS unit (erg), not the SI joule.Common Pitfalls:Mixing energy (J) and torque (N·m). They share units but represent different physical quantities; context distinguishes them.Final Answer:work is done by a force of 1 N when it displaces a body through 1 m

Question 5

Railway curves — purpose of superelevation (cant): When a train negotiates a curve, raising the outer rail (providing superelevation) is intended to prevent excessive side thrust on the wheel flanges. Is this statement correct?

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Introduction / Context:On curves, railway vehicles experience lateral acceleration. Without correction, this causes high flange contact forces, wear, noise, and risk of derailment. Superelevation (cant) tilts the track to balance components of weight and centrifugal effects.Given Data / Assumptions:Train speed is within the designed speed range for the given cant.Curve of constant radius; track quality adequate.Concept / Approach:Tilting the outer rail raises the outer wheel, creating an inward component of the train’s weight. This component counteracts the outward lateral acceleration so that the resultant of weight and inertia passes closer to the rail midline, reducing flange thrust.Step-by-Step Solution:Centrifugal effect on a curve acts outward.Introduce cant so that a component of weight acts inward on the plane of the rails.Balance condition reduces the net lateral force at wheel–rail contact.Hence, side thrust on flanges is mitigated by raising the outer rail.Verification / Alternative check:Railway design uses equilibrium cant h such that (G * h / gauge) approximately balances lateral acceleration at a chosen speed. Empirical reductions in wear and noise confirm effectiveness.Why Other Options Are Wrong:No: Ignores fundamental dynamics of curved-track negotiation and standard practice in track design.Common Pitfalls:Assuming cant eliminates all lateral forces; it optimizes for a design speed, but over- or under-speed still produces residual thrust.Final Answer:Yes

Question 6

Projectile on an inclined plane — angle for maximum range: For a projectile launched up an inclined plane that makes an angle β with the horizontal, the range measured along the plane is maximum when the angle of projection α (measured from the hori…

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Introduction / Context:Optimizing projectile range on an incline appears in mining, ballistics, and sports engineering. Unlike the horizontal ground case (optimum 45°), the incline tilts the landing surface, shifting the best launch angle.Given Data / Assumptions:Inclined plane makes angle β with the horizontal.Launch speed is fixed; air resistance is neglected.Angle of projection α is measured from the horizontal.Concept / Approach:Resolve motion into components parallel and perpendicular to the plane or use standard x–y components and impose the landing condition y = x * tanβ (the plane line). Express range along the plane R_plane as a function of α and β, then maximize with respect to α for a fixed speed.Step-by-Step Solution:Equation of the plane: y = x * tanβ.Projectile: x(t) = u * cosα * t; y(t) = u * sinα * t − (1/2) * g * t^2.At impact: u * sinα * t − (1/2) * g * t^2 = u * cosα * t * tanβ.Solve for time t > 0, then get x and distance along plane: R_plane = x * secβ.Differentiating R_plane(α) with respect to α and setting derivative to zero yields optimum α = 45° + β/2.Verification / Alternative check:Check limiting cases: For β = 0° (horizontal ground), α = 45°, matching the well-known result. As β increases, the optimum α increases accordingly, which is physically sensible because the plane rises upward.Why Other Options Are Wrong:β/2, 30° + β/2, 60° + β/2: Do not satisfy the derived extremum condition and give suboptimal ranges except at special, coincidental β values.Common Pitfalls:Maximizing horizontal range instead of range along the plane; the objective function differs.Forgetting the secβ factor converting horizontal displacement to along-plane distance.Final Answer:45° + β/2

Question 7

Fundamentals of Mechanics – Definition Check What is the correct term for the force by which a body is attracted toward the center of the Earth due to gravity?

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Introduction / Context: In elementary mechanics, different physical quantities describe different aspects of motion and interaction. This question tests your ability to distinguish “weight” from related terms such as mass, momentum, and impulsive force, all of which have precise meanings in physics and engineering.Given Data / Assumptions: A body is near Earth’s surface.Gravitational field is approximately uniform (g ≈ 9.81 m/s^2).We use standard engineering definitions: mass in kilograms (kg), force in newtons (N).Concept / Approach: Weight is the gravitational force acting on a body. It depends on the local gravitational acceleration and the mass of the object. By contrast, mass is a measure of inertia, momentum relates mass and velocity, and an impulsive force acts over a short time interval to change momentum.Step-by-Step Solution: Define weight: W = m * g (units: newton). Mass is not a force: it is an intrinsic property (kg). Momentum: p = m * v (units: N·s), not a force. Impulsive force is a large force over short Δt; impulse J = ∫F dt changes momentum. Therefore, the force pulling a body toward Earth’s center is weight.Verification / Alternative check: Consider a 10 kg object: W = 10 * 9.81 ≈ 98.1 N directed toward Earth’s center. This matches everyday experience (scale reading) and is consistent with Newton’s law of universal gravitation in a near-surface approximation.Why Other Options Are Wrong: impulsive force: describes a short-duration force effect, not continuous gravity. mass: a scalar measure of matter, not a force. momentum: product of mass and velocity, not a force. inertia: tendency to resist acceleration; not a force and not listed in the original options but included here for contrast.Common Pitfalls: Confusing mass with weight; remember units (kg vs N). Ignoring that weight changes slightly with location (g varies), while mass does not.Final Answer: weight

Question 8

Systems of Pulleys – Second System (Weston-type) Velocity Ratio In the second system of pulleys comprising n pulleys, the velocity ratio (VR) is claimed to be n. Is this statement correct?

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Introduction / Context: Pulleys are classic simple machines used to gain mechanical advantage. There are three traditional “systems of pulleys.” Each system has a distinct expression for velocity ratio (VR), which relates effort movement to load movement. Misremembering which VR belongs to which system is a common exam error.Given Data / Assumptions: Second system of pulleys with n pulleys.Light, inextensible rope; ideal kinematics for VR (independent of friction).Concept / Approach: For the three classical systems: First system: VR = 2^n. Second system: VR = 2^n − 1. Third system: VR = n. The claim “VR = n” therefore pertains to the third system, not the second.Step-by-Step Solution: Identify the system: it is the second system. Recall standard formula: VR_second = 2^n − 1. Compare with claim VR = n ⇒ mismatch. Hence the statement is false.Verification / Alternative check: For n = 3: second system VR = 2^3 − 1 = 7. If one incorrectly used n, VR would be 3, which clearly contradicts textbook results and simple rope-segment counting.Why Other Options Are Wrong: “True”: incorrect; VR = n belongs to the third system. “True only if massless/frictionless”: VR formulas already assume ideal kinematics; friction affects efficiency, not VR. “True when n = 2”: still wrong; for n = 2, VR = 3 (not 2). “True for first system”: first system has 2^n, not n.Common Pitfalls: Mixing up the three systems’ formulas. Confusing velocity ratio (a kinematic quantity) with mechanical advantage (a force ratio affected by friction).Final Answer: False

Question 9

Screw Jack – Efficiency Formula in Terms of Helix Angle and Friction Angle Which of the following correctly gives the efficiency η of a screw jack in terms of helix angle α and friction angle φ (for raising the load)?

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Introduction / Context: Screw jacks convert a small effort over a long helical path into a large lifting force over a small linear distance. Their performance is captured by efficiency η, which depends on geometry (lead or helix angle α) and friction (expressed via friction angle φ).Given Data / Assumptions: Right-hand square/Acme thread model; helix angle α measured at mean radius.Friction angle φ where tan(φ) = μ.Raising the load (upward motion).Concept / Approach: For a screw jack, the efficiency in raising is the ratio of ideal mechanical advantage to the actual mechanical advantage. Using the force analysis on the screw thread (viewed as an inclined plane wrapped helically), the standard result is η = tan(α) / tan(α + φ) when lifting.Step-by-Step Solution: Relate friction to φ via tan(φ) = μ. Model thread as an inclined plane of angle α with friction φ. For raising: efficiency η = tan(α) / tan(α + φ). For lowering (for comparison): η_lower differs; self-locking depends on α and φ.Verification / Alternative check: If φ = 0 (no friction), η = tan(α)/tan(α) = 1 (100%), which is consistent with an ideal machine. As φ increases, denominator increases, efficiency drops logically.Why Other Options Are Wrong: tan(α − φ)/tan(α): not the raising-load expression; appears in other contexts. tan(α + φ)/tan(α): predicts η > 1 for φ > 0, impossible. tan(φ)/tan(α + φ) and product forms: not supported by thread equilibrium derivation.Common Pitfalls: Confusing raising vs lowering formulas. Using lead (l) and mean radius (r) inconsistently when deriving α via tan(α) = l / (2π r).Final Answer: η = tan(α) / tan(α + φ)

Question 10

Systems of Pulleys – First System with 4 Pulleys For the first system of pulleys having 4 pulleys, what is the velocity ratio (VR)?

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Introduction / Context: Velocity ratio (VR) in pulley systems measures how far the effort moves compared with the load. In the first system of pulleys, each additional pulley doubles the number of supporting rope segments of the load, doubling the VR.Given Data / Assumptions: First system of pulleys.Number of pulleys n = 4.Light, inextensible rope; ideal kinematics for VR.Concept / Approach: The well-known expression for the first system is VR = 2^n. The reason is that each pulley effectively doubles the number of load-supporting rope segments, thereby doubling how many units the effort must move to raise the load by 1 unit.Step-by-Step Solution: Write formula: VR = 2^n. Substitute n = 4. Compute: VR = 2^4 = 16.Verification / Alternative check: Count rope segments supporting the moving block in a standard diagram with 4 pulleys; you will find 16 units of effort travel per unit of load travel under ideal conditions.Why Other Options Are Wrong: 4 and 8 underestimate the doubling effect; 20 and 32 do not match 2^n.Common Pitfalls: Confusing first, second, and third pulley systems, which have distinct VR formulas.Final Answer: 16

Question 11

Parallel-Axis Theorem for Area Moments of Inertia According to the parallel-axis theorem, the moment of inertia IP of a plane area about any axis parallel to an axis through its centroid (C.G.) is:

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Introduction / Context: The parallel-axis theorem is a cornerstone of mechanics of materials and structural analysis. It allows engineers to compute the second moment of area (area moment of inertia) about any axis parallel to a centroidal axis without direct integration.Given Data / Assumptions: IG: centroidal area moment of inertia.A: area of the section.h: perpendicular distance between the centroidal axis and the new parallel axis.Concept / Approach: The theorem states that when shifting from the centroidal axis to a parallel axis a distance h away, the new moment equals the centroidal moment plus A times the square of that distance. This arises from the binomial expansion of (y + h)^2 during area integration.Step-by-Step Solution: Start from IP = ∫(y + h)^2 dA. Expand: ∫(y^2 + 2yh + h^2) dA = ∫y^2 dA + 2h ∫y dA + h^2 ∫dA. Because the centroid is on the axis, ∫y dA = 0. Thus IP = IG + h^2 A.Verification / Alternative check: Dimensional check: IG has units of length^4; A * h^2 also has length^4, so the sum is dimensionally consistent.Why Other Options Are Wrong: Subtracting A h^2 contradicts the derivation (moment increases as axis moves away). Division or inverse forms lack physical and dimensional meaning.Common Pitfalls: Confusing the area theorem with the mass moment version (form is analogous but quantities differ). Using wrong distance (must be perpendicular between parallel axes).Final Answer: IP = IG + A * h^2

Question 12

Rigid-Body Dynamics – Moment of Inertia of a Thin Spherical Shell For a thin spherical shell of mass m and radius r, what is the mass moment of inertia about any diameter?

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Introduction / Context: Standard mass moment of inertia formulas are essential in rotational dynamics. A thin spherical shell concentrates mass at a constant radius, changing its inertia compared with a solid sphere.Given Data / Assumptions: Rigid thin spherical shell (negligible thickness).Mass m uniformly distributed over the surface of radius r.Axis of rotation: any diameter of the sphere.Concept / Approach: Known results: Solid sphere about diameter: I = 2 m r^2 / 5. Thin spherical shell about diameter: I = 2 m r^2 / 3. The shell’s mass is farther from the center compared with a solid sphere, hence the larger coefficient.Step-by-Step Solution: Use standard result for a thin shell: I_diameter = 2 m r^2 / 3. No integration needed in an exam setting because it is a tabulated formula.Verification / Alternative check: Relative comparison: 2/3 ≈ 0.667 is greater than 2/5 = 0.4, reflecting the mass being distributed at radius r rather than throughout the volume.Why Other Options Are Wrong: m r^2 / 3: too small; not a standard result. 2 m r^2 / 5: that is for a solid sphere, not a thin shell. 3 m r^2 / 5 and m r^2: do not correspond to any standard spherical model about a diameter.Common Pitfalls: Mixing up solid vs thin-shell formulas; always recall shell has larger inertia for the same m and r.Final Answer: 2 m r^2 / 3

Question 13

Kinematics – Linear (Tangential) Speed in Circular Motion A body rotates with angular velocity ω (rad/s) along a circular path of radius r (m). What is its linear speed v?

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Introduction / Context: Uniform circular motion links angular quantities (radians, rad/s) with linear kinematics (meters, m/s). Converting between angular velocity and linear speed is fundamental in rotating machinery, gears, and belt drives.Given Data / Assumptions: Angular velocity ω in rad/s.Radius r in meters.Body moves along a circular path without slipping.Concept / Approach: Arc length s traveled in time t is s = r * θ. Differentiating with respect to time gives linear speed v = ds/dt = r * dθ/dt = r * ω.Step-by-Step Solution: Start from s = r * θ. Differentiate: v = ds/dt = r * dθ/dt. Recognize dθ/dt = ω. Thus v = r * ω.Verification / Alternative check: Units: ω (rad/s) is dimensionally 1/s; multiply by r (m) to get m/s, which is correct for linear speed.Why Other Options Are Wrong: ω / r and r / ω invert relationships incorrectly (wrong units). ω^2 / r and ω^2 * r are associated with centripetal acceleration (a_c = ω^2 r), not speed.Common Pitfalls: Confusing v = ω r with a_c = ω^2 r; remember speed vs acceleration.Final Answer: ω * r

Question 14

Highway Engineering – Ideal Banking Angle on Curves On a frictionless ideal banked road curve, the required banking angle depends primarily on which factor?

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Introduction / Context: Banking of roads allows vehicles to negotiate curves safely at a design speed without relying solely on tire-road friction. The ideal banking angle comes from balancing components of the normal reaction against centripetal requirements in a frictionless design scenario.Given Data / Assumptions: Frictionless idealization (design speed condition).Vehicle travels on a circular arc of radius R at speed v.Gravity g acts vertically downward.Concept / Approach: The standard relation for ideal banking is tan(θ) = v^2 / (g R). Thus θ depends on the square of speed and inversely on radius, independent of vehicle weight. Friction, surface type, and power are not part of the ideal balance condition (though they matter in real-world safety margins).Step-by-Step Solution: Equate horizontal component of normal reaction to centripetal requirement: N sinθ = m v^2 / R. Vertical equilibrium: N cosθ = m g. Divide: tanθ = (m v^2 / R) / (m g) = v^2 / (g R). Hence θ is determined by v^2 (and R), not by weight or engine power.Verification / Alternative check: Doubling speed quadruples v^2 and significantly increases θ; doubling vehicle mass cancels out of the ratio and does not change θ.Why Other Options Are Wrong: Weight: cancels from the equations. Nature of surface and coefficient of friction: relevant for non-ideal (with friction) conditions, not for ideal banking angle. Engine power: unrelated to the geometric equilibrium for banking.Common Pitfalls: Assuming heavier vehicles need a different banking angle; under ideal theory, they do not.Final Answer: (velocity)^2 of the vehicle

Question 15

Oscillations – Definition of Periodic Time “The time taken by a particle to complete one full oscillation is called the periodic time.” Evaluate this statement.

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Introduction / Context: Oscillations occur in mechanical, electrical, and structural systems. Consistent terminology helps compare behaviors such as frequency, period, and phase across domains. This item checks the definition of “periodic time.”Given Data / Assumptions: A system exhibiting repeated motion over time (oscillation/rotation/vibration).The motion is periodic (repeats identical states at regular intervals).Concept / Approach: Periodic time, usually denoted T, is the duration for one complete cycle after which the system repeats its state (position, velocity, etc.). The reciprocal quantity is frequency f, with the relation f = 1 / T, and for angular frequency ω, we have ω = 2π f = 2π / T.Step-by-Step Solution: Define: Periodic time T = time for one full oscillation. Frequency: f = number of oscillations per second; f = 1 / T. Angular frequency: ω = 2π f ⇒ T = 2π / ω. Hence the given statement matches the standard definition.Verification / Alternative check: In simple harmonic motion x(t) = A cos(ω t + φ), the motion repeats when ω t increases by 2π, so T = 2π / ω, which is the time for one complete oscillation.Why Other Options Are Wrong: Disagree: contradicts standard textbooks. Limiting to “small amplitudes” or “only for SHM” is unnecessary; the definition of period applies to any periodic motion. Requiring zero damping is incorrect; lightly damped systems can still be nearly periodic with a well-defined period (strictly, perfectly periodic if strictly undamped), but the definition of T remains the “time per cycle.”Common Pitfalls: Confusing period T with phase or amplitude; they are independent quantities.Final Answer: Agree

Question 16

Power in Rotational Motion – Torque–Speed Relation A body is acted upon by a constant torque T (N·m) and rotates with angular speed ω (rad/s). What is the power developed?

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Introduction / Context: Converting between torque, angular speed, and power is essential in rotating machinery (motors, gearboxes, turbines). Knowing the exact relation avoids unit mistakes when sizing drives and couplings.Given Data / Assumptions: Torque T in newton·metre (N·m).Angular speed ω in rad/s.Steady rotation (instantaneous power relation).Concept / Approach: Mechanical power is the rate of doing work. In rotation, the elemental work is dW = T dθ. Dividing by time gives power P = dW/dt = T (dθ/dt) = T ω. This yields power in watts when SI units are used.Step-by-Step Solution: Start: dW = T dθ. Differentiate: P = dW/dt = T * dθ/dt. Recognize dθ/dt = ω (rad/s). Therefore, P = T * ω (watts).Verification / Alternative check: Dimensional analysis: N·m × 1/s = N·m/s = watt, confirming SI consistency. When speed is given in revolutions per minute n (rpm), use ω = 2π n / 60 to obtain P = T * 2π n / 60 (watts) or P(kW) = (2π T n) / 60,000.Why Other Options Are Wrong: T * ω / 60: would be correct only if ω were in rpm; here ω is in rad/s already. T * ω / 75 or / 4500 in kW: arbitrary constants; not general. 2π * T * ω: double counts 2π since ω already includes radians.Common Pitfalls: Mixing rpm and rad/s; always convert rpm to rad/s before using P = T ω. Confusing torque units (N·m) with energy units (J); although numerically identical units appear, their physical meanings differ.Final Answer: T * ω (in watts)

Question 17

Engineering Mechanics — Rotational Inertia of a Thin Disc For a thin, solid circular disc of mass m and radius r, what is the mass moment of inertia about the axis through its center of gravity and perpendicular to the plane of the disc (the polar/s…

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Introduction / Context: In dynamics of rigid bodies, the mass moment of inertia measures resistance to angular acceleration about a chosen axis. For common shapes like discs and cylinders, standard closed-form results exist and are frequently used in design, vibration, and machine dynamics. Given Data / Assumptions: Body: thin, solid disc of uniform density. Mass m and radius r are known constants. Axis: through the disc’s center of gravity and perpendicular to its plane (polar axis). Concept / Approach: The polar mass moment of inertia I about the disc’s central axis is obtained by integrating r^2 over all elemental masses. For a thin disc, mass is distributed in the plane; symmetry simplifies the integral to a well-known formula. Step-by-Step Solution: Define elemental ring at radius ρ of thickness dρ: dA = 2πρ dρ. Mass element: dm = (m / (πr^2)) * 2πρ dρ = (2mρ / r^2) dρ. Moment of inertia: I = ∫ ρ^2 dm from 0 to r = ∫_0^r ρ^2 * (2mρ / r^2) dρ = (2m / r^2) ∫_0^r ρ^3 dρ. Compute integral: ∫_0^r ρ^3 dρ = r^4 / 4 ⇒ I = (2m / r^2) * (r^4 / 4) = (1/2) m r^2. Verification / Alternative check: Compare with a thin ring of radius r: I_ring = m r^2. Since a solid disc has more mass near the center, its I must be smaller; I = (1/2) m r^2 is consistent with this physical intuition. Why Other Options Are Wrong: The factors 1/4, 1/6, and 1/8 are not produced by the correct integration for a uniform disc and would under-predict rotational inertia. Common Pitfalls: Confusing the polar axis with a diametral axis. About a diametral centroidal axis, the inertia differs: I = (1/4) m r^2 + (1/12) m t^2 for a finite-thickness cylinder (t = length). Final Answer: mr^2/2.

Question 18

Machines — Terminology Check A non-reversible machine (no motion of the load when the effort is removed) is also called a self-locking machine. Is this statement correct?

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Introduction / Context: In the study of machines like screw jacks, hoists, and geared blocks, reversibility indicates whether a load can drive the machine in reverse when the effort is removed. This property determines safety and back-driving behavior. Given Data / Assumptions: Machine exhibits sufficient friction or geometry to prevent back-driving. Effort is removed after raising the load. Terminology per standard machine theory. Concept / Approach: A reversible machine allows the load to drive the effort side when the applied effort is removed (e.g., runs backward). A non-reversible machine does not back-drive; if it holds the load by itself due to friction or geometry, it is termed self-locking. Step-by-Step Reasoning: Define reversibility: machine back-drives under load without effort. If back-driving is impossible, the machine is non-reversible. Non-reversible machines that hold position are self-locking (e.g., screw jacks with small helix angles). Verification / Alternative check: For screw mechanisms, a < φ (helix angle less than friction angle) ensures self-locking; under this, the load cannot descend on its own, confirming non-reversibility equals self-locking in standard usage. Why Other Options Are Wrong: Disagreeing would contradict conventional definitions in machine theory and safety literature. Common Pitfalls: Confusing self-sustaining (holding momentarily) with self-locking (cannot back-drive at all); assuming all irreversible machines are efficient—they usually have higher friction losses. Final Answer: Agree — a non-reversible machine is called self-locking.

Question 19

Impact and Collisions — Behavior of Elastic Bodies When two elastic bodies collide, which of the following occurs during the impact process?

Accepted Answer

Introduction / Context: Collision of elastic bodies is described by phases of compression and restitution. Even “rigid” bodies deform locally at the contact region; elasticity allows them to store and release strain energy, which governs their post-impact velocities. Given Data / Assumptions: Elastic bodies with nonzero contact compliance. Short collision duration; external forces like weight are negligible during impact. Coefficient of restitution e ≤ 1; for perfectly elastic, e = 1. Concept / Approach: The impact has two stages: compression (bodies deform, relative approach decreases) and restitution (bodies recover shape, relative separation increases). At the instant of maximum compression, relative velocity at the contact becomes zero; then restitution begins. Step-by-Step Description: Compression: contact stresses cause local deformation; kinetic energy partially stores as strain energy. At peak compression: relative velocity at contact is zero; the bodies may be momentarily at rest in certain frames and configurations. Restitution: stored strain energy drives separation; shapes are largely recovered for elastic bodies. Verification / Alternative check: Newton’s impact law relates separation and approach speeds: v_separation = e * v_approach. For e close to 1, recovery is nearly complete, matching the described behavior. Why Other Options Are Wrong: Taken individually, (a) can be misread; however, together the options describe the full cycle (deform, pause at max compression, recover). Hence 'All of the above' best captures the correct sequence. Common Pitfalls: Assuming perfectly rigid contact (no deformation); confusing zero relative speed at maximum compression with zero absolute speed in every frame. Final Answer: All of the above — the deformation and recovery phases occur in elastic collisions, with a momentary stop at peak compression in the contact frame.

Question 20

Resultant of Two Equal Forces — Magnitude Formula Two equal forces each of magnitude P act at an angle θ to each other. What is the magnitude of their resultant?

Accepted Answer

Introduction / Context: Vector addition of two forces is a staple of statics and dynamics. For equal magnitudes with angle θ between them, the resultant has a compact expression that is widely used in engineering and physics. Given Data / Assumptions: Force magnitudes: P and P. Angle between forces: θ. Concept / Approach: Use the law of cosines for vectors: R^2 = P^2 + P^2 + 2 P P cos θ. Then take the square root to obtain R. Trigonometric half-angle identities lead to a simple form involving cos(θ/2). Step-by-Step Solution: R^2 = 2P^2 (1 + cos θ). Use identity: 1 + cos θ = 2 cos^2(θ/2). Hence R = √(2P^2 * 2 cos^2(θ/2)) = 2P cos(θ/2). Verification / Alternative check: Special cases: θ = 0 ⇒ R = 2P (collinear same direction). θ = 180° ⇒ R = 0 (equal and opposite), consistent with formula since cos 90° = 0. Why Other Options Are Wrong: Forms with sin, tan, or cot of θ/2 do not satisfy the cosine-law derivation and fail at boundary checks (e.g., θ = 0 or 180°). Common Pitfalls: Forgetting the half-angle identity or misapplying parallelogram/triangle construction with wrong included angle. Final Answer: 2 P cos(θ/2).

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