Question 1

Vehicle Dynamics – Overturning on a Level Circular Path
To avoid overturning while negotiating a level (unbanked) circular path, the speed of the vehicle must be __________ a certain critical value.

Accepted Answer

Introduction / Context: While turning on a level (unbanked) road, a vehicle experiences lateral acceleration requiring a centripetal force. If speed is too high, the lateral load transfer can lead to wheel lift and overturning. Given Data / Assumptions: Level (unbanked) circular path of radius r. Vehicle with track width b, center-of-gravity height h, and mass m. Static friction provides lateral force; rollover is considered based on tipping about the outer wheels. Concept / Approach: Overturning begins when the resultant of weight and inertial (centrifugal) effect passes through the outer tire contact edge. The critical speed v_crit can be estimated from moment balance about the outer wheel contact line. Step-by-Step Solution: Centrifugal effect (in vehicle frame) = m v^2 / r acting laterally at height h. Weight W = m g acting downward through CG; half-track b/2 is the lever arm to the outer wheel line. At impending overturn: (m v^2 / r) * h = W * (b/2) = m g * (b/2). Solve: v_crit = sqrt( (g r b) / (2 h) ). To avoid overturning, the operating speed v must be less than v_crit. Verification / Alternative check: The same criterion is obtained via resultant force line passing within the base of support. Banking raises the allowable speed; level path is the limiting case here. Why Other Options Are Wrong: 'Greater than' or 'equal to': at or above the critical speed, overturning risk exists. 'Independent of': speed clearly governs lateral acceleration. 'Randomly varying': physically meaningless. Common Pitfalls: Confusing skid (friction limit) with roll-over (geometric + inertial limit). Both impose speed limits but via different criteria. Final Answer: less than

Question 2

Physics — Definition Check Centripetal force is the force that acts along the radius of a circle and is directed toward the centre of the circle.

Accepted Answer

Introduction / Context: The item tests your understanding of uniform circular motion from introductory mechanics. Specifically, it asks whether the definition of centripetal force has been stated accurately. Centripetal force is essential for keeping an object moving in a circular path and always points toward the circle’s centre. Given Data / Assumptions: An object moves along a circular path of radius r. The motion may be uniform (constant speed) or non-uniform; in either case, the direction of the required inward force is radial toward the centre. Concept / Approach: In circular motion, velocity constantly changes direction. A radial inward acceleration, called centripetal acceleration, is required to change this direction. The net inward force providing this acceleration is the centripetal force. For uniform circular motion, magnitude a_c = v^2 / r and F_c = m * v^2 / r. The direction is always toward the centre (radially inward). Step-by-Step Solution: Identify what centripetal means: 'centre-seeking'. State the force direction: along the radius, inward to the centre. Match with standard formula in uniform motion: F_c = m * v^2 / r points inward. Conclude that the given sentence matches the standard definition. Verification / Alternative check: In the absence of sufficient inward force, the object would not curve inward; it would move tangentially by inertia. Observing conical pendulum, banked curves, or planets orbiting confirms the inward nature of the required force. Why Other Options Are Wrong: 'False' would imply the force is outward or tangential, which contradicts mechanics. The outward sensation (often called centrifugal) is a pseudo-effect in a rotating frame, not a real force in an inertial frame. Common Pitfalls: Confusing centripetal (real, inward) with centrifugal (apparent, outward in a rotating frame). Also, assuming the force is a new kind of force; it is simply the net inward force (tension, gravity, friction, etc.). Final Answer: True.

Question 3

Machines — Differential Pulley Block A differential pulley block has effort pulley diameters of 100 mm (larger) and 80 mm (smaller). What is its velocity ratio (VR)?

Accepted Answer

Introduction / Context: Differential (Weston) pulley blocks are simple lifting machines used to gain mechanical advantage. The velocity ratio (VR) links the distance moved by the effort to the distance moved by the load, and it depends on the diameters (or radii) of the two effort pulleys. Given Data / Assumptions: Larger pulley diameter D = 100 mm. Smaller pulley diameter d = 80 mm. Standard ideal relation for Weston differential pulley: VR = 2D / (D - d) when diameters are used. Concept / Approach: For a Weston differential pulley, the chain passes over two co-axial pulleys of slightly different diameters. One revolution advances one side and retracts the other, creating a small net movement of the load. Thus VR grows as the difference (D - d) becomes small, per the formula VR = 2D / (D - d). Step-by-Step Solution: Identify formula: VR = 2D / (D - d). Substitute values: VR = 2 * 100 / (100 - 80). Compute denominator: 100 - 80 = 20. Compute VR: 200 / 20 = 10. Verification / Alternative check: Using radii R and r (R = 50 mm, r = 40 mm) gives VR = 2R / (R - r) = 100 / 10 = 10, consistent with the diameter-based calculation. Why Other Options Are Wrong: 5, 20, and 40 do not satisfy the standard relation for the given diameters; they would require different D and d values or a different machine type. Common Pitfalls: Mixing up the formula with simple pulley systems; forgetting the factor of 2; using (D + d) instead of (D - d). Final Answer: 10.

Question 4

Centroid of a Trapezium For a trapezium of height h with parallel sides a (top) and b (base), the centroid lies at a distance y from the base b. What is y?

Accepted Answer

Introduction / Context: Locating the centroid (centre of gravity for a lamina of uniform density) is fundamental in mechanics and structural analysis. For trapezoids (trapezia), the centroid position along the height depends on the relative sizes of the parallel sides. Given Data / Assumptions: Trapezium with parallel sides a (top) and b (base), separated by height h. Uniform thickness and density (so centroid = centre of area). Distance y is measured upward from the base b toward the top side a. Concept / Approach: The centroid of a trapezium can be obtained using the first-moment (area * distance) method or standard compiled results. When measured from the base b toward the smaller side a, the centroid lies closer to the larger base and is given by a weighted ratio favouring the larger side length in the numerator. Step-by-Step Solution: Area of trapezium: A = (a + b) * h / 2. Using first moment of area about the base and simplifying leads to y = h * (2a + b) / [3 * (a + b)]. This places the centroid between base and mid-height, closer to the larger base when b > a. Verification / Alternative check: Special case a = b (rectangle): y = h * (2a + a) / [3 * (2a)] = h * 3a / (6a) = h / 2, which agrees with a rectangle’s centroid. If a → 0 (triangle), y → h * b / (3b) = h / 3 from the base, matching the triangle result. Why Other Options Are Wrong: y = h * (a + 2b) / [3(a + b)] is the expression measured from the top when roles of a and b swap; not correct for distance from base b as posed. y = (h * (a + b)) / 2 is just mid-height, valid only for a = b. y = h * (b - a) / [3(a + b)] has incorrect dependence and can be negative. Common Pitfalls: Confusing which side is the reference (top vs base) and swapping a, b; forgetting that centroid shifts toward the larger parallel side. Final Answer: y = [h * (2a + b)] / [3 * (a + b)].

Question 5

Kinematics — Identify the Scalar Quantity

Accepted Answer

Introduction / Context: Physical quantities are classified as scalars (magnitude only) or vectors (magnitude and direction). Correctly identifying which is which is fundamental to solving mechanics problems correctly. Given Data / Assumptions: Options include common motion-related quantities: force, speed, velocity, acceleration. Assume standard physics definitions in SI units. Concept / Approach: A scalar quantity is fully described by a number and appropriate unit, with no directional attribute. A vector quantity requires both magnitude and direction and follows vector addition rules. Step-by-Step Solution: Force requires magnitude and direction and adds vectorially ⇒ vector. Velocity describes rate of change of displacement with direction ⇒ vector. Acceleration describes rate of change of velocity with direction ⇒ vector. Speed is the rate at which distance is covered, with no directional component ⇒ scalar. Verification / Alternative check: Speed is the magnitude of velocity. Magnitude alone is scalar; only when direction is attached (velocity) does it become a vector. Why Other Options Are Wrong: Force, velocity, and acceleration inherently require direction to be fully specified and obey vector laws (e.g., parallelogram rule). Hence they are vectors, not scalars. Common Pitfalls: Confusing speed with velocity; thinking that “fast” implies a direction; treating acceleration as “just speeding up” rather than a vector that can also indicate slowing or turning. Final Answer: Speed.

Question 6

Simple Harmonic Motion (SHM) The periodic time T of a particle in SHM is ________ proportional to the angular frequency ω.

Accepted Answer

Introduction / Context: In SHM, the time for one complete oscillation is called the period T, and the angular frequency is ω. Their relationship is a fundamental property that appears across mechanical and electrical oscillators. Given Data / Assumptions: Standard SHM with displacement x(t) = A sin(ωt + φ). Angular frequency ω is constant for the system being considered. Concept / Approach: By definition, the period is the time taken for the phase to increase by 2π. Thus T and ω are tied by the identity T = 2π / ω, regardless of amplitude (for linear SHM). Step-by-Step Solution: Start with x(t) periodicity: x(t + T) = x(t) ⇒ ωT = 2π. Solve for T: T = 2π / ω. Interpret: T varies inversely with ω; if ω doubles, T halves. Verification / Alternative check: For a mass–spring oscillator, ω = sqrt(k / m), so T = 2π * sqrt(m / k); increasing stiffness increases ω and decreases T. For a simple pendulum (small angle), ω = sqrt(g / l), so T = 2π * sqrt(l / g), again showing inverse proportionality to ω. Why Other Options Are Wrong: 'Directly' contradicts T = 2π / ω. 'Depends on amplitude' is false for ideal linear SHM (period is independent of amplitude). 'Neither' is incorrect because there is a clear inverse relationship. Common Pitfalls: Confusing frequency f with angular frequency ω, and forgetting the factor 2π (i.e., ω = 2πf). Final Answer: inversely.

Question 7

Rotational Dynamics — SI Unit of Mass Moment of Inertia In SI units, the mass moment of inertia has the unit kg·m^2.

Accepted Answer

Introduction / Context: Mass moment of inertia measures an object’s resistance to angular acceleration about an axis. Correct units are essential for dimensional consistency in torque and angular acceleration relationships. Given Data / Assumptions: SI base unit of mass is kilogram (kg) and of length is metre (m). Moment of inertia involves mass multiplied by squared distance from an axis. Concept / Approach: The defining expression is I = ∫ r^2 dm for continuous bodies or I = Σ m_i r_i^2 for discrete masses. Thus, the dimension is [mass] * [length]^2, which in SI is kg·m^2. Step-by-Step Solution: Start with I = Σ m r^2. Replace m with kg and r with m, giving units kg·m^2. Check against rotational dynamics: τ = I * α has N·m on the left; α is rad/s^2 (dimensionless radians), leaving N·m = (kg·m^2) * (1/s^2), which is consistent since N = kg·m/s^2. Verification / Alternative check: For a solid disk, I = (1/2) m R^2; substituting SI units yields kg·m^2, confirming the unit again. Why Other Options Are Wrong: 'False' would imply a different unit, contradicting dimensional analysis and standard definitions in mechanics. Common Pitfalls: Confusing mass moment of inertia (kg·m^2) with area moment of inertia (m^4) used in beam bending; mixing CGS (g·cm^2) with SI (kg·m^2). Final Answer: True.

Question 8

Equilibrium — Triangle Law of Forces If forces acting at a point can be represented in magnitude and direction by the three sides of a triangle taken in order, then the forces are not in equilibrium.

Accepted Answer

Introduction / Context: The triangle law of forces is a fundamental condition for equilibrium of three concurrent forces. This question deliberately states the condition incorrectly to test conceptual clarity. Given Data / Assumptions: Three coplanar forces act at a point. They are representable head-to-tail by the sides of a triangle taken in order. Concept / Approach: Triangle law of forces: If three concurrent forces can be represented (in order) by the three sides of a triangle, they are in equilibrium; the resultant is zero because the vector polygon closes (returns to the start). Step-by-Step Solution: Represent F1, F2, F3 as vectors drawn head-to-tail. If the final head coincides with the initial tail, the vector sum F1 + F2 + F3 = 0. Zero resultant means equilibrium for a particle (no net force). Therefore, the given statement claiming 'not in equilibrium' is false. Verification / Alternative check: Equivalent polygon law: If the vector polygon closes, the system is in equilibrium. The triangle is just the special case for three forces. Why Other Options Are Wrong: 'Agree' supports a false statement; it contradicts standard statics results. Common Pitfalls: Confusing the order (vectors must be taken in order, not arbitrarily); mixing up triangle law of forces with parallelogram law of vector addition (two-force case). Final Answer: Disagree.

Question 9

Dynamics — Newton’s Laws “The rate of change of momentum of a body is directly proportional to the impressed force and takes place in the same direction as the force.” This statement refers to:

Accepted Answer

Introduction / Context: Newton’s three laws form the basis of classical mechanics. Each law has a distinct wording and meaning. This question checks recognition of the precise wording of Newton’s second law in its momentum form. Given Data / Assumptions: Momentum p = m * v. Force is related to the time rate of change of momentum. Concept / Approach: Newton’s second law: F ∝ dp/dt and acts in the same direction as dp/dt. In SI units, proportionality becomes equality with F = dp/dt. For constant mass, this reduces to F = m * a. Step-by-Step Solution: Identify wording: 'rate of change of momentum ... proportional to force'. Recall second law: F = dp/dt, direction same as dp/dt. Match to options: This is Newton’s second law. Verification / Alternative check: First law is inertia (no net force ⇒ velocity constant). Third law describes action–reaction pairs equal and opposite, not rate of change of momentum. Why Other Options Are Wrong: First law lacks momentum-rate wording; third law relates two interacting bodies, not a single body’s dp/dt; 'None' is unnecessary because the second law fits perfectly. Common Pitfalls: Confusing momentum form with constant-mass form; forgetting that direction of force equals direction of acceleration and dp/dt. Final Answer: Newton's second law of motion.

Question 10

Oscillations — Simple Pendulum Period What is the periodic time T (time for one oscillation) of a simple pendulum of length l (small-angle assumption)?

Accepted Answer

Introduction / Context: The simple pendulum is a classic oscillator used to illustrate periodic motion. Under small angular displacements, its motion approximates simple harmonic motion (SHM) with a well-known expression for the period T in terms of length l and gravitational acceleration g. Given Data / Assumptions: Point mass at the end of a massless string of length l. Small-angle approximation so that sinθ ≈ θ (in radians). Local gravitational acceleration g is constant. Concept / Approach: Linearizing the pendulum’s equation leads to angular frequency ω = sqrt(g / l). Using the identity T = 2π / ω gives T = 2π * sqrt(l / g). The period is independent of amplitude for small angles and independent of mass. Step-by-Step Solution: Start: ω = sqrt(g / l) for small angles. Use T = 2π / ω. Compute: T = 2π / sqrt(g / l) = 2π * sqrt(l / g). Verification / Alternative check: Dimensional analysis: sqrt(l / g) has units sqrt(m / (m/s^2)) = sqrt(s^2) = s; multiplied by 2π gives time (seconds). Why Other Options Are Wrong: Any expression with sqrt(g / l) in the numerator or missing the factor 2π does not match SHM theory. Options (b), (c), and (d) are dimensionally or numerically incorrect for a simple pendulum under small angles. Common Pitfalls: Forgetting the small-angle condition; mixing up g and l; dropping the 2π factor when converting from angular frequency to period. Final Answer: T = 2π * sqrt(l / g).

Question 11

Work and Energy — Concept Check Energy may be defined as the capacity to do work.

Accepted Answer

Introduction / Context: The statement tests the foundational definition of energy in physics. Energy is an abstract scalar that quantifies the ability of a system to perform work or produce change, appearing in many forms such as kinetic, potential, thermal, and electrical energy. Given Data / Assumptions: Work is defined as W = F * s * cosθ (in mechanics) and measured in joules (J). Energy, measured in joules, is transferable and conserved in isolated systems. Concept / Approach: Operationally, if a system can exert a force through a distance (or cause equivalent change), it possesses energy. Different forms of energy interchange under conservation laws, but their sum remains constant (ignoring losses to surroundings where applicable). Step-by-Step Solution: Relate energy to work: performing work requires energy; the amount of work done can reduce or transform stored energy. Examples: A raised mass (gravitational potential) can do work while descending; a moving body (kinetic) can do work in a collision. Unit check: Both work and energy share the unit joule in SI, strengthening the conceptual linkage. Verification / Alternative check: In electrical contexts, energy = power * time; in thermal contexts, heat transfer is energy in transit. All instances reflect capacity to produce change or work. Why Other Options Are Wrong: Labeling the definition 'Incorrect' would contradict standard physics terminology used across mechanics, thermodynamics, and electromagnetism. Common Pitfalls: Treating energy as substance rather than a quantitative state property; confusing energy with power (rate of doing work, measured in watts). Final Answer: Correct.

Question 12

Fundamental kinematics

What is the physical quantity defined as the rate of change of displacement of a body with respect to time?

Accepted Answer

Introduction / Context: This question checks core kinematics vocabulary used in physics and engineering. Displacement, velocity, acceleration, and momentum are often confused, yet each has a precise definition critical for motion analysis, control, and design. Given Data / Assumptions: A moving body has a displacement s(t) measured from a reference point. Time t is continuous and differentiable for ordinary motions considered in mechanics. Standard SI units are used: displacement in metres (m) and time in seconds (s). Concept / Approach: Velocity is defined as the rate of change of displacement with respect to time. In calculus terms, instantaneous velocity v is the first time derivative of displacement: v = ds/dt. Acceleration a is the rate of change of velocity: a = dv/dt = d^2s/dt^2. Momentum p is mass * velocity: p = m * v. Step-by-Step Solution: Identify the quantity linked directly to displacement by a first derivative.Use the definitions: velocity v = ds/dt, acceleration a = dv/dt, momentum p = m * v.Therefore, the correct match to “rate of change of displacement of a body” is velocity. Verification / Alternative check: Consider uniform motion with s = k * t. Then ds/dt = k is constant (velocity), while dv/dt = 0 (acceleration). This confirms that the first derivative corresponds to velocity and not acceleration. Why Other Options Are Wrong: Acceleration: Rate of change of velocity, not displacement. Momentum: Mass multiplied by velocity; not a rate of change. None of these: Incorrect because “velocity” precisely matches the definition. Speed: Magnitude of velocity only; the question refers to the vector rate (velocity) associated with displacement. Common Pitfalls: Using “speed” for any rate of motion. Speed lacks direction, whereas velocity incorporates direction and directly derives from displacement, which is a vector from a reference. Final Answer: velocity

Question 13

Free fall convention in kinematics

If a body falls freely under gravity, which value (with sign) is typically taken for gravitational acceleration g when the downward direction is chosen as positive?

Accepted Answer

Introduction / Context: Sign conventions in dynamics are crucial for error-free calculations. This problem asks which signed value of the gravitational acceleration g is taken in free fall when we deliberately choose the downward direction as positive. Given Data / Assumptions: Standard magnitude of gravitational acceleration near Earth's surface is about 9.8 m/s^2. Coordinate system is user-defined; here, the downward direction is specified as positive. Air resistance is ignored (ideal free fall). Concept / Approach: The sign of acceleration depends on the chosen coordinate axis. If downward is positive, gravitational acceleration is +9.8 m/s^2. If upward were taken as positive, the same acceleration would be written as −9.8 m/s^2. The physics does not change; only the algebraic signs do. Step-by-Step Solution: Choose axis: down is + (given).Magnitude of g ≈ 9.8 m/s^2.Therefore, the signed value is +9.8 m/s^2 for free fall. Verification / Alternative check: Take the position equation s = (1/2) g t^2 with s measured positive downward. With g = +9.8, s grows positively as time increases, matching physical intuition for a falling object. Why Other Options Are Wrong: ±8.9 m/s^2: Incorrect magnitude. −9.8 m/s^2: Would be correct only if upward were taken as positive. +10 m/s^2 (approx.): A useful approximation in mental math, but the question asks for the typical precise value. Common Pitfalls: Mixing up sign convention; always define the axis first. The vector points downward physically; assign the sign consistent with your axis. Final Answer: +9.8 m/s^2

Question 14

Neutral equilibrium of a smooth cylinder

A smooth circular cylinder lying on which of the following surfaces is in neutral equilibrium (its potential energy does not change for small displacements)?

Accepted Answer

Introduction / Context: Equilibrium classification (stable, unstable, neutral) is widely used in mechanics and machine design. A body is in neutral equilibrium if its potential energy remains unchanged for small displacements; it neither returns to the original position nor moves away. Given Data / Assumptions: The cylinder is perfectly smooth (no friction), so only normal reactions and weight act. We consider small displacements/rolls on the contacting surface. Gravitational potential energy depends on the height of the mass centre. Concept / Approach: If the height of the cylinder’s centre remains the same for small movements, the gravitational potential energy remains constant. That is the definition of neutral equilibrium. On a truly horizontal plane, when a smooth cylinder rolls or slides slightly, the centre stays at the same height. Step-by-Step Solution: Model the cylinder of radius R on a flat (horizontal) surface.The centre is at height R above the surface.For any small horizontal displacement, the centre height remains R → potential energy unchanged.Therefore, the cylinder is in neutral equilibrium on a horizontal surface. Verification / Alternative check: Contrast with other surfaces: on a convex support, the centre height changes with small rolls, usually increasing initially (unstable). In a concave cradle, height decreases and then increases, creating a restoring tendency (stable). Why Other Options Are Wrong: Concave groove: Produces a restoring torque (stable equilibrium), not neutral. Convex surface: Tends to topple or roll off (unstable). Inclined plane: Net component of weight causes motion downhill (not equilibrium). Rough stepped surface: Discrete geometry breaks neutrality; height generally changes on moving. Common Pitfalls: Assuming “smooth” implies stability. Smoothness removes friction but does not by itself decide stability; the geometry and centre-height variation do. Final Answer: horizontal surface

Question 15

Definition check: radius of gyration

True or False — “The radius of gyration is the distance where the whole mass (or area) of a body is assumed to be concentrated.”

Accepted Answer

Introduction / Context: Radius of gyration is fundamental in strength of materials and dynamics because it links mass distribution to rotational inertia. Misstatements often omit the crucial reference to an axis, which this question highlights. Given Data / Assumptions: Standard engineering definition is intended. Moment of inertia I and total mass m (or area A) exist for a chosen axis. Concept / Approach: The radius of gyration k about a given axis is defined by I = m * k^2 (or I = A * k^2 for area moments). It is the distance from that axis at which the entire mass (or area) could be imagined concentrated so that it would have the same moment of inertia as the actual distribution. Reference to an axis is essential. Step-by-Step Solution: Write the defining relation: I = m * k^2 (mass) or I = A * k^2 (area).Solve for k: k = sqrt(I / m) or k = sqrt(I / A).Note that I is always taken about a specified axis; hence k is axis-dependent.The given statement omits the axis; therefore it is incomplete/incorrect. Verification / Alternative check: For a thin ring of radius R about its central axis, I = m * R^2, so k = R. If you change the axis (e.g., a diameter), I changes and so does k, confirming the axis dependence. Why Other Options Are Wrong: Correct / Partly correct: Both fail because they ignore the axis requirement. Correct only for point masses: Trivializes the concept; definition applies to distributed mass. Meaningless without density: Density is not required for the definition; I and m (or A) suffice. Common Pitfalls: Quoting radius of gyration without specifying the axis, leading to misuse in column buckling (k = r) and shaft design formulas. Final Answer: Incorrect

Question 16

Velocity in simple harmonic motion (SHM)

For a particle executing SHM with amplitude r and instantaneous displacement y from the mean position, which expression gives its speed v at that instant?

Accepted Answer

Introduction / Context: Simple harmonic motion (SHM) models oscillations of springs, pendulums (small angles), and many vibrating systems. The velocity–displacement relation is essential for predicting speeds at different positions and for energy-based design. Given Data / Assumptions: Amplitude r (maximum magnitude of displacement). Instantaneous displacement y from the mean (equilibrium) position. Angular frequency ω (rad/s) characterizes the system. Concept / Approach: For SHM: y = r * sin(ω t + φ). Differentiate to get velocity: v = dy/dt = r * ω * cos(ω t + φ). Using the identity sin^2θ + cos^2θ = 1 and eliminating time gives v^2 = ω^2 (r^2 − y^2), hence v = ω * sqrt(r^2 − y^2). This highlights maximum speed at y = 0 and zero speed at y = ±r. Step-by-Step Solution: Start with y = r * sin(ω t + φ).Differentiate: v = dy/dt = r * ω * cos(ω t + φ).Use cos^2 = 1 − sin^2 and substitute sin = y / r to eliminate time.Obtain v^2 = r^2 * ω^2 * (1 − (y^2 / r^2)) = ω^2 (r^2 − y^2).Therefore, v = ω * sqrt(r^2 − y^2). Verification / Alternative check: Energy method: total energy E = (1/2) k r^2 is constant. At displacement y, elastic potential is (1/2) k y^2, so kinetic K = E − U = (1/2) k (r^2 − y^2). Using k = m ω^2 and K = (1/2) m v^2 gives v = ω * sqrt(r^2 − y^2). Why Other Options Are Wrong: ω^2(r − y), ω(r + y), ω y: Do not satisfy boundary conditions (e.g., v = 0 at y = ±r). sqrt(ω^2 r^2 + y^2): Speed would increase with |y|, contradicting SHM behaviour. Common Pitfalls: Forgetting that maximum speed occurs at the mean position and misusing signs when eliminating time. Final Answer: v = ω * sqrt(r^2 - y^2)

Question 17

Friction fundamentals

Which of the following statements about static friction are true under typical dry-contact conditions?

Accepted Answer

Introduction / Context: Static friction prevents relative motion until a limiting value is reached. Understanding its direction, proportionality to normal reaction, and weak dependence on apparent contact area is essential in machine design, braking, and robotics. Given Data / Assumptions: Dry, clean surfaces with no lubrication (Coulomb friction model). Normal reaction N acts perpendicular to the interface. Body is on the verge of motion (impending slip) or at rest. Concept / Approach: The Coulomb model states: maximum static friction F_s,max = μ_s * N. Below this limit, static friction adjusts to match the needed tangential force to prevent slip. Direction of static friction opposes the tendency (impending direction) of motion. The model is largely insensitive to the apparent area of contact; true micro-contact area scales with load and material properties. Step-by-Step Solution: Check proportionality: F_s,max / N = μ_s (constant for the material pair) → constant ratio at the limit.Check area dependence: experiments show weak correlation with apparent area for many dry contacts → statement holds as a practical rule.Check direction: static friction acts to oppose impending motion, maintaining equilibrium until the limit is reached. Verification / Alternative check: Place a block on a rough plane. Increase the applied tangential force slowly; friction matches it until F = μ_s N, then motion starts. Varying the block’s footprint at the same weight gives nearly the same limiting force, confirming area insensitivity. Why Other Options Are Wrong: Any single choice: In isolation they are each true, but the option “all of the above” correctly captures the full behaviour described by the Coulomb model. None of the above: Contradicted by standard friction laws. Common Pitfalls: Confusing static with kinetic friction, and assuming friction always equals μ_s N even below the limit. Static friction takes whatever value is necessary up to μ_s N. Final Answer: all of the above

Question 18

Effect of a pure couple in planar mechanics

A free couple acting on a rigid body produces which type of motion (neglecting constraints and other forces)?

Accepted Answer

Introduction / Context: A couple is a system of two equal, opposite, and parallel forces separated by a perpendicular distance. Its net force is zero, but it produces a pure moment. Recognizing the motion caused by a couple is fundamental in statics and dynamics. Given Data / Assumptions: Rigid body, unconstrained (free) unless otherwise stated. Only a couple (net moment M) acts; net resultant force is zero. Planar motion considered for simplicity. Concept / Approach: Because the vector sum of forces in a couple is zero, there is no net translational effect on the center of mass. However, the net moment is non-zero, which produces angular acceleration about the mass centre according to M = I_c * α, where I_c is the mass moment of inertia about the centre. Step-by-Step Solution: Compute resultant force: F_resultant = 0 (forces cancel).Compute resultant moment: M_resultant = F * d (non-zero).Equations of motion: ΣF = m * a = 0 → a = 0; ΣM_c = I_c * α = M_resultant → α ≠ 0.Hence, the body experiences pure rotation about its mass centre with no translation. Verification / Alternative check: Real-world examples: turning a steering wheel with both hands in opposite directions, or applying a torque with a wrench while the nut is free to rotate. In both cases, the object spins without translating. Why Other Options Are Wrong: Translatory / Combined: Require a non-zero resultant force. None of the above / SHM: A free couple does not cause zero motion nor oscillation; it produces rotation with angular acceleration proportional to the couple. Common Pitfalls: Confusing a couple with two non-parallel forces (which would create both a net force and a moment). A true couple is a free vector producing only rotational effect. Final Answer: rotational motion

Question 19

Conservation of mechanical energy (ideal case)

For an isolated system of moving bodies where only conservative internal forces act (no external work or dissipation), the total mechanical energy of the system:

Accepted Answer

Introduction / Context: Conservation laws are pillars of mechanics. The total mechanical energy (kinetic + potential) of an isolated system remains constant when only conservative forces (like gravity or ideal springs) act and when non-conservative effects (like friction) are absent. Given Data / Assumptions: System is isolated (no external work input/output). Only conservative internal forces are present. No energy losses due to friction, air drag, plastic deformation, or heat. Concept / Approach: Total mechanical energy E is defined as E = K + U, where K is total kinetic energy and U is total potential energy. For conservative forces, work done equals the negative change in potential energy. The work–energy principle and path independence lead to constant E for the entire system. Step-by-Step Solution: Write E = K + U for the system.For conservative forces, ΔE = ΔK + ΔU = 0 because work by internal conservative forces only redistributes energy between K and U.Therefore, dE/dt = 0 → E is constant with time. Verification / Alternative check: Consider two masses connected by an ideal spring on a frictionless surface. As one mass speeds up and the other slows down, K changes, the spring stores/releases U, yet K + U remains constant, validating the principle. Why Other Options Are Wrong: Varies from point to point / Max–min claims: These describe non-conservative scenarios or arbitrary stages; in the ideal conservative case, total energy does not systematically increase or decrease. Depends only on mass: Energy depends on both configuration (potential) and motion (kinetic), not mass alone. Common Pitfalls: Applying energy conservation without checking for non-conservative forces. Always verify isolation and absence of losses before using E = constant. Final Answer: is constant at every instant

Question 20

In a screw jack, the effort required to lower the load is __________ the effort required to raise the same load.

Accepted Answer

less than

Engineering Mechanics Questions