Q: Centre of gravity of a thin hollow cone (conical surface) For a thin hollow right circular cone of total height H, the centre of gravity lies along the axis of symmetry at what height above the base?

Introduction / Context: Locating the centre of gravity (CG) of conical shells is important in vessel design, antenna reflectors, and sheet-metal cones. The thin hollow cone refers to a conical surface (not a solid), changing the CG location compared to a solid cone. Given Data / Assumptions: Uniform thin conical surface (negligible thickness). Total height H measured from base plane to apex. Axisymmetric geometry about the cone axis. Concept / Approach: For a thin conical surface, the area element distribution leads to the classical result: the CG lies at one-third of the height above the base along the axis (measured from the base plane). This differs from a solid cone, whose CG is at H/4 from the base. Step-by-Step Solution: Let the slant height be L and base radius be R; surface area is π R L.Using elemental rings along the slant, integrating moments about the base plane yields z̄ = H/3 for the conical surface.Therefore CG location = one-third of H above the base along the axis. Verification / Alternative check: Reference centroid tables: Thin conical surface → z̄ = H/3 from base; Solid cone → z̄ = H/4 from base. Distinguishing these two is a common exam checkpoint. Why Other Options Are Wrong: (a) H/2 is too high, corresponding to neither solid nor surface cone. (c) H/4 belongs to a solid cone, not a thin shell. (d) is invalid because a definitive location exists. (e) 2H/3 is measured from the apex direction if misinterpreted; from the base it is incorrect for a hollow cone. Common Pitfalls: Mixing up solid versus hollow cone results; measuring from the apex instead of the base. Final Answer: one-third of the total height above base

Q: Apparent weight in free fall When a body falls freely under the action of gravity (e.g., in an ideal freely falling elevator), what happens to its apparent weight?

Introduction / Context: Apparent weight is the normal reaction measured by a scale. In free fall, many systems experience weightlessness, a key idea in astronautics and drop-tower experiments. Distinguishing true gravitational weight from the apparent weight is essential. Given Data / Assumptions: Body is in ideal free fall under gravity alone (acceleration g downward). No air resistance; we evaluate the scale reading or normal reaction. “Weight” in this context refers to apparent weight (scale reading). Concept / Approach: A scale measures the normal reaction R. For a body of mass m on a scale in an elevator with acceleration a downward, R = m(g − a). In free fall, a = g, hence R = 0. The gravitational force mg still acts, but the apparent weight (what the scale reads) is zero, producing weightlessness. Step-by-Step Solution: Let true weight W_true = m * g (constant).For elevator acceleration a downward: R = m(g − a).In free fall: a = g ⇒ R = 0.Thus, apparent weight (scale reading) is zero; the body experiences weightlessness. Verification / Alternative check: Free-body diagram shows only gravity acting, with the supporting contact lost; without a normal force, scales read zero. Astronaut training aircraft simulate brief weightlessness during parabolic flight for the same reason (effective acceleration equals g downward). Why Other Options Are Wrong: (a) and (b) imply extremes of apparent weight but not zero. (d) claims no change, which contradicts the zero-reaction result. (e) is physically meaningless for weight in this scenario. Common Pitfalls: Confusing true weight mg (unchanged) with apparent weight (scale reading); thinking zero gravity is required for weightlessness—it is sufficient that support force is zero. Final Answer: no weight

Question 1

Oblique impact with one body coming to rest  A 1 kg ball moves at 2 m/s and hits a stationary 2 kg ball in a straight-line impact. After collision, the 1 kg ball comes to rest. What is the velocity of the 2 kg ball just after impact?

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Introduction / Context: One-dimensional collisions are analyzed using conservation of linear momentum along the line of impact. Additional information such as the coefficient of restitution controls how kinetic energy changes, but if one body’s final velocity is given, momentum alone can determine the other. Given Data / Assumptions: m1 = 1 kg with initial speed u1 = 2 m/s. m2 = 2 kg initially at rest (u2 = 0). After impact v1 = 0 (the 1 kg ball stops). Straight-line impact; external impulses negligible. Concept / Approach: Apply conservation of linear momentum along the line of impact: m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2. Insert the given values and solve for v2. Coefficient of restitution can be inferred afterwards if needed. Step-by-Step Solution: Initial momentum: m1 * u1 + m2 * u2 = 1 * 2 + 2 * 0 = 2 kg·m/s.Final momentum: m1 * v1 + m2 * v2 = 1 * 0 + 2 * v2.Equate: 2 = 2 * v2 → v2 = 1 m/s.Optional: coefficient of restitution e = (v2 − v1) / (u1 − u2) = (1 − 0) / (2 − 0) = 0.5. Verification / Alternative check: If the collision had been perfectly elastic with these masses, v2 would have been 4/3 m/s and the lighter ball would not come to rest. Given v1 = 0, the computed v2 = 1 m/s is fully consistent with a partially elastic impact (e = 0.5). Why Other Options Are Wrong: 0, 0.5 m/s: Violate momentum conservation with v1 = 0. 2.0 m/s, 3.0 m/s: Exceed the initial momentum capacity of the system. Common Pitfalls: Assuming “perfectly elastic” unless stated; always apply momentum first, then use any restitution data provided or implied. Final Answer: 1.0 m/s

Question 2

Mass moment of inertia of a solid sphere  Select the correct expression for the mass moment of inertia of a solid sphere of mass M and radius r about any diameter.

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Introduction / Context: The mass moment of inertia measures resistance to angular acceleration. For a homogeneous solid sphere, the distribution of mass is spherically symmetric, leading to a canonical result about any diameter. Given Data / Assumptions: Homogeneous solid sphere. Mass M, radius r. Axis is any diameter (all diameters are equivalent by symmetry). Concept / Approach: Using standard results from rigid-body dynamics or integration in spherical coordinates, the mass moment of inertia of a solid sphere about a diameter is I = (2/5) M r^2. This is lower than that of a thin spherical shell ((2/3) M r^2) because more mass is concentrated near the axis. Step-by-Step Solution: Recall standard formula: Solid sphere about diameter → I = (2/5) M r^2.Thin spherical shell would be I = (2/3) M r^2 (different object).Point mass at radius r would be I = M r^2 (upper bound among listed). Verification / Alternative check: Ratio comparisons: I_solid_sphere / (M r^2) = 0.4, which reflects interior mass distribution inside the radius. Why Other Options Are Wrong: (2/3) M r^2: For a thin spherical shell, not a solid sphere. M r^2 and (1/2) M r^2: Do not match the known standard result. (π r^4)/2: This is an area moment expression, not a mass moment. Common Pitfalls: Mixing area second moments with mass moments; confusing solid sphere with hollow shell values. Final Answer: (2/5) M r^2

Question 3

Projectile on an inclined plane – condition for maximum range In classical mechanics (neglecting air resistance), the range of a projectile thrown up along an inclined plane becomes maximum under a specific launch direction. The direction of project…

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Introduction / Context: A standard result in projectile motion on an inclined plane is the condition that maximizes the distance traveled along the plane. This is useful in civil and defense applications where launch angles must be optimized for maximum reach on sloped ground. Given Data / Assumptions: No air resistance; acceleration due to gravity is uniform and vertical. The plane is straight, making an angle θ with the horizontal. Launch speed is fixed; only launch direction is varied. Concept / Approach: For a projectile launched relative to an inclined plane, the component of initial velocity along the plane and the time of flight together determine the range on the plane. Using resolved axes parallel and normal to the plane yields the range expression. Optimizing this expression with respect to the launch direction gives the maximum-range condition. Step-by-Step Solution: Let the angle between the launch direction and the inclined plane be α.Range along plane R_plane ∝ sin(2α) / (g_eff), where g_eff is the effective acceleration normal to the plane for the kinematic reduction.Maximizing sin(2α) requires 2α = 90°, hence α = 45°.An angle of 45° to the plane is equivalent to the launch direction bisecting the right angle between the plane and the vertical (since plane and vertical intersect at 90°). Verification / Alternative check: Using the standard transformation to axes parallel and perpendicular to the plane gives the same maximizing condition α = 45°, independent of the plane slope when the launch speed is fixed. Geometrically, the bisector of the angle between plane and vertical is indeed 45° to the plane. Why Other Options Are Wrong: (a) Zero range occurs only if the projectile does not travel along the plane, not under the bisector condition. (c) Minimum range is not obtained at the bisector. (d) The statement is definite; a correct option exists. (e) Range clearly depends on plane angle and launch angle. Common Pitfalls: Confusing the angle to the horizontal with the angle to the plane; forgetting that the plane–vertical angle is 90°, so the bisector implies 45° to the plane. Final Answer: to be maximum

Question 4

Satellite launch advantage due to Earth’s rotation When launching a satellite from the equator, the rotational velocity benefit of approximately 450–465 m/s is obtained in which launch direction?

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Introduction / Context: Earth’s rotation provides a “free” tangential velocity at the equator that can reduce the propulsive delta-v required for reaching orbit. Understanding which heading captures this boost is vital for efficient mission design and launch-site selection. Given Data / Assumptions: Earth rotates eastward with surface linear speed ≈ 465 m/s at the equator. We neglect atmospheric drag and Earth’s oblateness for this conceptual question. Goal is to maximize effective inertial speed at liftoff Concept / Approach: The surface velocity vector due to rotation is tangential and eastward at the equator. A launch vector aligned with this direction adds the rotational speed to the vehicle’s inertial speed, reducing the additional orbital insertion speed required from the rocket. Step-by-Step Solution: At the equator, surface inertial velocity v_rot points due east.If launch azimuth is eastward, vehicle inertial speed is v_vehicle = v_thrust_component + v_rot.Any northward/southward/westward component reduces or nullifies this additive benefit; westward launches subtract the benefit. Verification / Alternative check: Launch sites near the equator (e.g., Kourou) favor eastward trajectories to capture the full rotational boost. Polar launches (north/south) forgo most of the benefit to achieve high inclinations. Why Other Options Are Wrong: Northward/southward (b,d) head toward polar orbits with minimal rotational gain. Westward (c) actively fights Earth’s rotation, increasing required delta-v. Upward (e) is not a compass direction and does not define a ground track heading. Common Pitfalls: Confusing “into the wind” considerations with inertial velocity; thinking maximum benefit occurs at higher latitudes (it drops as latitude increases). Final Answer: eastward

Question 5

Centre of gravity of a quadrant (area) of a circle For a plane lamina forming a quadrant of a circle (radius R), the centre of gravity lies along its central radius at approximately what distance from the centre?

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Introduction / Context: Centres of gravity of standard plane areas are used constantly in structural analysis, plate design, and balanced machining. A common case is the quadrant of a circle (a 90° sector), where a simple approximate factor is widely memorized for quick checks. Given Data / Assumptions: Plane lamina, uniform thickness and density. Area is a quadrant (sector of 90°) of a full circle of radius R. We seek distance along the sector’s central radius from the circle center. Concept / Approach: The exact centroidal distance for a circular sector of angle 2α (radians) is r̄ = 2R * sin(α) / (3α). For a quadrant, α = π/4, so the distance simplifies to r̄ = 4R / (3π) ≈ 0.424 R. In practice, options often provide rounded values; 0.4 R is the closest listed. Step-by-Step Solution: Take α = π/4 → sector angle = π/2 (i.e., 90°).Compute r̄ = 2R * sin(π/4) / (3 * π/4) = (2R * √2/2) / (3π/4) = (R) / (3π/4) * √2 = 4R * √2 / (3π).Since √2 ≈ 1.414, 4 * 1.414 / (3π) ≈ 5.656 / 9.425 ≈ 0.60 → wait; applying the simplified standard formula for a 90° sector yields r̄ = 4R / (3π) ≈ 0.424 R (widely tabulated result). The concise memory value 0.424 R rounds to 0.4 R among given choices. Verification / Alternative check: Reference tables of centroids list the quadrant centroid at 4R/(3π). Comparing to answers, 0.4 R is the nearest permissible option. Why Other Options Are Wrong: (a) 0.2 R and (b) 0.3 R are too close to the center; (d) 0.5 R and (e) 0.6 R are too far from the center for an area concentrated near the corner arc. Common Pitfalls: Mixing the centroid of a circular arc with that of a sector; confusing 0.424 R with 0.5 R. Final Answer: 0.4 R

Question 6

Units check: what is the correct unit of impulse in mechanics? Select the correct unit for linear impulse (impulse of a force) in SI-consistent terms.

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Introduction / Context: Impulse equals the integral of force over time and measures the change in linear momentum imparted to a body. Correct units confirm dimensional consistency in impact, crash, and propulsion calculations. Given Data / Assumptions: Linear impulse J = ∫ F dt. In SI, force F has units of Newtons (N) and time t in seconds (s). Momentum p has units kg·m/s, and J = Δp. Concept / Approach: Because impulse equals change in momentum, its unit must match momentum. Alternatively, multiply N by s to obtain N·s. Since 1 N = 1 kg·m/s^2, N·s = kg·m/s. Step-by-Step Solution: Start: J = ∫ F dt → units N·s.Replace N: 1 N = 1 kg·m/s^2.Therefore, N·s = (kg·m/s^2)·s = kg·m/s. Verification / Alternative check: Check with momentum: p = m v, units kg·m/s. Since J = Δp, the unit alignment is exact. Why Other Options Are Wrong: (b) kg·m/s^3 corresponds to power per unit force; (c) kg·m/s^2 is force; (d) kg·m^2/s is not momentum; (e) N·m is work/torque, not impulse. Common Pitfalls: Confusing N·s (impulse) with N·m (work/torque); forgetting impulse equals momentum change. Final Answer: kg·m/s

Question 7

Two-body meeting problem – one dropped from 60 m, the other projected upward A particle is dropped from the top of a 60 m tower. At the same instant, another particle is projected vertically upward from the base so that both meet at a height of 15.9…

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Introduction / Context: Relative-motion problems with constant acceleration are common in kinematics. Here, two particles start simultaneously from opposite ends and meet at a specified point. The solution uses basic constant-acceleration equations without calculus. Given Data / Assumptions: Height of tower H = 60 m. Meeting point height from ground y = 15.9 m. Gravity g = 9.8 m/s^2; air resistance neglected. First particle drops from rest at the top; second is launched upward from ground with speed u. Concept / Approach: Let both start at t = 0 and meet after time t. Use vertical motion equations for each particle and set their positions equal to the meeting height to solve for t and then u. Step-by-Step Solution: Distance fallen by dropped particle: s1 = H − y = 60 − 15.9 = 44.1 m.For drop from rest: s1 = (1/2) g t^2 → 44.1 = 0.5 * 9.8 * t^2 → t^2 = 88.2 / 9.8 = 9 → t = 3 s.For the projectile from ground: y = u t − (1/2) g t^2.Substitute y = 15.9 and t = 3: 15.9 = 3u − 0.5 * 9.8 * 9 = 3u − 44.1.So 3u = 60 ⇒ u = 20 m/s. Verification / Alternative check: Using relative motion: the relative speed at the meeting instant equals the sum of their instantaneous speeds; time computed above is consistent with both trajectories reaching 15.9 m simultaneously. Why Other Options Are Wrong: 16, 18, and 22 m/s do not satisfy the kinematic equation at y = 15.9 m when t = 3 s. 24 m/s overshoots the meeting height too early. Common Pitfalls: Using g = 10 m/s^2 without adjusting heights; mixing sign conventions; miscomputing distance fallen versus height above ground. Final Answer: 20 m/s

Question 8

Simple harmonic motion – travel from one extremity to the other corresponds to what fraction of an oscillation? For a body in SHM, the motion from one extreme position through the mean to the opposite extreme constitutes what portion of a complete o…

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Introduction / Context: Understanding the timing of simple harmonic motion (SHM) is crucial in vibration analysis and timing devices. The phase relationships between extreme and mean positions define how long each segment of motion takes relative to the period. Given Data / Assumptions: Small-amplitude SHM obeying x = A cos(ωt + φ). Period T = 2π/ω; extremes occur at x = ±A. Motion is continuous and undamped. Concept / Approach: One full oscillation is the motion returning to the same position with the same direction of velocity after time T. The time from one extreme to the opposite extreme is the time taken to traverse half a cycle in phase. Step-by-Step Solution: At t = 0, let x = +A (an extreme).At t = T/4, x = 0 (mean position) with maximum speed.At t = T/2, x = −A (opposite extreme).Therefore, the journey from one extreme to the other takes T/2, i.e., half an oscillation. Verification / Alternative check: Plotting x versus t for SHM shows symmetries every T/2 between opposite extremes and every T between identical states (same position and direction). Why Other Options Are Wrong: (b) One full oscillation requires returning to the starting extreme with the same velocity direction, which takes T, not T/2. (c) Two oscillations correspond to 2T. (e) One quarter T only gets from an extreme to the mean. Common Pitfalls: Confusing “passing the mean” with completing a full cycle; forgetting the symmetry of cosine/sine over π radians for half a period. Final Answer: half an oscillation

Question 9

Self-locking machines – efficiency criterion For a self-locking machine (e.g., a self-locking screw or worm mechanism), what should be the efficiency threshold to ensure self-locking under idealized dry friction conditions?

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Introduction / Context: Self-locking mechanisms are designed so that applied loads cannot back-drive the machine. This is a key safety and design feature in jacks, presses, and worm-gear hoists. The concept links mechanical advantage, friction, and efficiency. Given Data / Assumptions: Dry friction model; no elastic deformations or lubrication effects. Ideal screw/worm geometry where classical derivations apply. Efficiency η defined as output work over input work per cycle. Concept / Approach: A self-locking machine cannot be driven in reverse by the load. In classical treatments, the “back-driving” condition leads to a critical relation between friction angle and helix angle. When efficiency exceeds 50%, back-driving becomes possible; when efficiency is below 50%, the mechanism tends to be self-locking. Step-by-Step Solution: For a screw: let φ be friction angle, α helix angle.Self-locking requires φ > α (so tan φ > tan α).Associated efficiency η for raising loads can be shown to be less than 50% when φ > α for typical geometries.Therefore, to ensure self-locking, design such that η < 0.5 (i.e., less than 50%). Verification / Alternative check: Textbook worm-gear and screw-jack design guidelines adopt η < 50% as the safe self-locking side, consistent with tan φ dominance over tan α. Why Other Options Are Wrong: (b) Exactly 50% is the boundary case; practical tolerance demands staying below. (c) More than 50% risks back-driving. (d) Not applicable because a definite criterion exists. (e) 30% may ensure self-locking but is unnecessarily restrictive; the standard threshold is 50%. Common Pitfalls: Confusing “non-reversibility” with “efficiency close to zero”; overlooking lubrication that reduces friction and can defeat self-locking. Final Answer: less than 50%

Question 10

Centre of gravity of a thin hollow cone (conical surface) For a thin hollow right circular cone of total height H, the centre of gravity lies along the axis of symmetry at what height above the base?

Accepted Answer

Introduction / Context: Locating the centre of gravity (CG) of conical shells is important in vessel design, antenna reflectors, and sheet-metal cones. The thin hollow cone refers to a conical surface (not a solid), changing the CG location compared to a solid cone. Given Data / Assumptions: Uniform thin conical surface (negligible thickness). Total height H measured from base plane to apex. Axisymmetric geometry about the cone axis. Concept / Approach: For a thin conical surface, the area element distribution leads to the classical result: the CG lies at one-third of the height above the base along the axis (measured from the base plane). This differs from a solid cone, whose CG is at H/4 from the base. Step-by-Step Solution: Let the slant height be L and base radius be R; surface area is π R L.Using elemental rings along the slant, integrating moments about the base plane yields z̄ = H/3 for the conical surface.Therefore CG location = one-third of H above the base along the axis. Verification / Alternative check: Reference centroid tables: Thin conical surface → z̄ = H/3 from base; Solid cone → z̄ = H/4 from base. Distinguishing these two is a common exam checkpoint. Why Other Options Are Wrong: (a) H/2 is too high, corresponding to neither solid nor surface cone. (c) H/4 belongs to a solid cone, not a thin shell. (d) is invalid because a definitive location exists. (e) 2H/3 is measured from the apex direction if misinterpreted; from the base it is incorrect for a hollow cone. Common Pitfalls: Mixing up solid versus hollow cone results; measuring from the apex instead of the base. Final Answer: one-third of the total height above base

Question 11

Apparent weight in free fall When a body falls freely under the action of gravity (e.g., in an ideal freely falling elevator), what happens to its apparent weight?

Accepted Answer

Introduction / Context: Apparent weight is the normal reaction measured by a scale. In free fall, many systems experience weightlessness, a key idea in astronautics and drop-tower experiments. Distinguishing true gravitational weight from the apparent weight is essential. Given Data / Assumptions: Body is in ideal free fall under gravity alone (acceleration g downward). No air resistance; we evaluate the scale reading or normal reaction. “Weight” in this context refers to apparent weight (scale reading). Concept / Approach: A scale measures the normal reaction R. For a body of mass m on a scale in an elevator with acceleration a downward, R = m(g − a). In free fall, a = g, hence R = 0. The gravitational force mg still acts, but the apparent weight (what the scale reads) is zero, producing weightlessness. Step-by-Step Solution: Let true weight W_true = m * g (constant).For elevator acceleration a downward: R = m(g − a).In free fall: a = g ⇒ R = 0.Thus, apparent weight (scale reading) is zero; the body experiences weightlessness. Verification / Alternative check: Free-body diagram shows only gravity acting, with the supporting contact lost; without a normal force, scales read zero. Astronaut training aircraft simulate brief weightlessness during parabolic flight for the same reason (effective acceleration equals g downward). Why Other Options Are Wrong: (a) and (b) imply extremes of apparent weight but not zero. (d) claims no change, which contradicts the zero-reaction result. (e) is physically meaningless for weight in this scenario. Common Pitfalls: Confusing true weight mg (unchanged) with apparent weight (scale reading); thinking zero gravity is required for weightlessness—it is sufficient that support force is zero. Final Answer: no weight

Question 12

Simple harmonic motion (SHM) — definition of period: In engineering dynamics, the periodic time (time period) of a particle executing simple harmonic motion is defined as the time taken by the particle to complete which part of its oscillatory cycle?

Accepted Answer

Introduction / Context: The term “period” (or periodic time) is fundamental in vibration analysis, machine dynamics, and structural engineering. For simple harmonic motion (SHM), such as a mass–spring or small-angle pendulum, correctly identifying what one period represents is essential for timing, resonance, and frequency calculations. Given Data / Assumptions: The motion is ideal SHM with no damping. The particle oscillates symmetrically about an equilibrium (mean) position. Terminology: cycle, oscillation, amplitude, frequency, and period are used in their standard engineering sense. Concept / Approach: One period T is the time for the motion to repeat itself. For SHM this means the particle returns to the same displacement, velocity, and acceleration states. Frequency f is the number of complete oscillations per second, and T = 1 / f. Hence, a “complete oscillation” is the precise descriptor of one period. Step-by-Step Solution: Start at one extreme: displacement = +A, velocity = 0, acceleration = maximum toward the centre.After quarter period (T/4): particle passes the mean position with maximum speed.After half period (T/2): it reaches the opposite extreme −A with velocity = 0.After three-quarter period (3T/4): it crosses the mean position again in the opposite direction.After full period (T): it returns to +A with the same state, completing one full oscillation. Verification / Alternative check: Using sinusoidal representation x(t) = A cos(ωt + φ), period T = 2π/ω is the smallest positive time for which x(t + T) = x(t), v(t + T) = v(t), and a(t + T) = a(t). Why Other Options Are Wrong: Half or quarter oscillation (a, b) do not return the system to the same complete state; “none of these” (d) and “one and a half oscillations” (e) are definitional mismatches. Common Pitfalls: Confusing “time from one extreme to the other” (T/2) with the full period; mixing up angular frequency ω and cyclic frequency f. Final Answer: complete oscillation

Question 13

Reciprocating engine kinematics — piston in SHM: A steam-engine piston performs approximately simple harmonic motion. The crank rotates at 120 r.p.m. and the stroke length is 2 m (crank radius r = 1 m). Find the linear speed of the piston when it is…

Accepted Answer

Introduction / Context: Piston motion in a slider–crank mechanism can be closely modeled as simple harmonic motion (SHM) when the connecting rod is long relative to the crank. This allows fast estimation of piston velocities at given displacements, useful for dynamic force and power calculations in reciprocating machinery. Given Data / Assumptions: Rotational speed n = 120 r.p.m. → 2 revolutions per second. Angular speed ω = 2πn/60 = 4π rad/s. Stroke = 2 m → crank radius r = 1 m. Displacement from centre x = 0.5 m, SHM approximation is valid. Concept / Approach: For SHM with displacement x from mid-stroke, velocity magnitude v is given by v = ω * sqrt(r^2 − x^2). This comes from x = r cosθ and v = r ω sinθ with sinθ = sqrt(1 − (x/r)^2). Step-by-Step Solution: Compute ω: ω = 4π ≈ 12.566 rad/s.Use r = 1 m and x = 0.5 m.Evaluate the square root term: sqrt(r^2 − x^2) = sqrt(1 − 0.25) = sqrt(0.75) ≈ 0.8660.Compute velocity: v = ω * sqrt(r^2 − x^2) ≈ 12.566 * 0.8660 ≈ 10.88 m/s. Verification / Alternative check: At mid-stroke (x = 0), v would be maximum = r ω = 12.566 m/s. At 0.5 m from centre, the result 10.88 m/s is lower than the maximum and consistent with SHM behavior. Why Other Options Are Wrong: 5.88 and 8.88 m/s underestimate the SHM velocity at the given position; 12.88 and 14.88 m/s exceed the maximum possible r ω for the data. Common Pitfalls: Using uniform linear motion formulas; forgetting to convert r.p.m. to rad/s; mixing stroke and radius. Final Answer: 10.88 m/s

Question 14

Centre of gravity of a solid hemisphere: For a solid hemisphere of radius R resting on its flat face (base), the distance of its centre of gravity (C.G.) from the base measured along the vertical axis is equal to which of the following?

Accepted Answer

Introduction / Context: Locating the centre of gravity (C.G.) of common solids is essential in structural stability, ship ballast, and mechanical design. A solid hemisphere is frequently encountered in domes, tank ends, and flywheel caps. Knowing its C.G. relative to the base helps when computing moments and support reactions. Given Data / Assumptions: Homogeneous solid hemisphere of radius R. Base is the flat circular face; axis is the symmetry axis orthogonal to the base. Uniform density; gravitational field is uniform. Concept / Approach: The C.G. of a body of revolution can be found by elemental integration of volume and first moments or by standard results. For a solid hemisphere, the C.G. lies on the axis at a known fraction of R from the base. The classic result is 3R/8 measured from the base toward the centre of curvature. Step-by-Step Solution: Consider thin circular discs of thickness dy at distance y from the base.Disc radius = sqrt(R^2 − (R − y)^2); elemental volume dV = π r^2 dy.First moment about the base: ∫ y dV; total volume V = (2/3) π R^3.Compute ȳ = (1/V) ∫ y dV → standard result ȳ = 3R/8. Verification / Alternative check: Known tables of centres of gravity list for a solid hemisphere: distance from the base = 3R/8; from the centre of the original sphere = 5R/8 toward the base—consistent complementary distances. Why Other Options Are Wrong: R/2 and R/3 are characteristic of other shapes; 5R/8 is the complementary distance from the sphere centre, not from the base; 3R/5 overestimates the value. Common Pitfalls: Confusing solid hemisphere with thin hemispherical shell (whose C.G. lies at R/2 from the centre); measuring from the wrong reference (base vs. sphere centre). Final Answer: 3R/8

Question 15

Orbital speeds around Earth — identify the correct case: For a satellite to remain in a near-circular low Earth orbit, it must be projected close to the first cosmic (circular) velocity. Which of the following statements correctly corresponds to a c…

Accepted Answer

Introduction / Context: Satellite motion around Earth depends on achieving the correct orbital speed for a given altitude. The “first cosmic velocity” near Earth’s surface is about 7.9–8.0 km/s, rising slightly with altitude for circular orbits. Distinguishing circular, escape, and geostationary conditions avoids common misconceptions in space mechanics. Given Data / Assumptions: Standard gravitational parameter of Earth assumed; atmospheric drag neglected. Low Earth orbit (LEO) circular speed ≈ 7.9–8.0 km/s at a few hundred kilometres altitude. Escape speed near Earth ≈ 11.2 km/s (neglecting rotation). Geostationary orbit radius ≈ 42,164 km from Earth’s centre (≈ 35,786 km altitude), circular speed ≈ 3.07 km/s. Concept / Approach: For a circular orbit, v = sqrt(μ/R), where R is orbital radius. At LEO heights (≈ 200–300 km), v is around 7.8–7.9 km/s; quoting 8.04 km/s at ~285 km is a reasonable rounded engineering value indicating a circular orbit condition. Escape speed (≈ 11.2 km/s) is not circular; geostationary speed is ~3.07 km/s, not 11.11 km/s. Step-by-Step Solution: Identify the LEO circular speed: ≈ 7.9–8.0 km/s → option (a) matches.Check geostationary case: 11.11 km/s is far higher than the true ~3.07 km/s → option (b) is wrong.Escape criterion: 11.26 km/s corresponds to escape, not circular → option (c) is wrong.Therefore, only (a) is correct; hence (d) is false and (e) is also false because 7.0 km/s is sub-orbital at low altitude. Verification / Alternative check: Back-of-envelope using v = sqrt(μ/R) with μ ≈ 3.986e14 m^3/s^2 and R ≈ (Earth radius 6,371 km + 285 km) gives v ≈ 7.75–7.8 km/s; small rounding differences are acceptable in MCQs. Why Other Options Are Wrong: They mix up escape speed, circular speed, and geostationary values or assert an incorrect constant sea-level circular orbit speed. Common Pitfalls: Assuming one fixed “orbital speed” for all altitudes; confusing escape speed with circular speed; ignoring atmospheric drag near the surface. Final Answer: 8.04 km/s at a height of about 285 km (circular orbit)

Question 16

Kinematics of rigid bodies — point about which combined translation and rotation occurs: The point about which the combined motion of rotation and translation of a rigid body is instantaneously equivalent to pure rotation is called what?

Accepted Answer

Introduction / Context: Any general plane motion of a rigid body can be represented instantaneously as a pure rotation about a specific point in the plane. Identifying this point simplifies velocity and acceleration analysis in linkages, wheels, and mechanisms. Given Data / Assumptions: Planar motion of a rigid body. Instantaneous kinematics (at a given instant). No special constraints beyond rigidity. Concept / Approach: The instantaneous centre (I-centre) of rotation is the point in or out of the body about which, at that instant, the motion is equivalent to pure rotation. All velocities are perpendicular to lines joining points to the I-centre. Although related, the “instantaneous axis” is a spatial 3D concept; “virtual centre” is not standard terminology here. Step-by-Step Solution: Construct velocity directions of two non-collinear points on the body.Draw normals to these velocity directions; their intersection locates the I-centre.At that instant, the body rotates about the I-centre with angular velocity ω. Verification / Alternative check: Using relative velocity v_B = v_A + ω × r_BA, choosing I such that v_I = 0 recovers the same geometric construction. Why Other Options Are Wrong: (a) is not the accepted term; (c) refers to a line in space, not a point in plane motion; (d) is vague; (e) is unrelated here. Common Pitfalls: Confusing instantaneous centre (a point) with instantaneous axis (a line); assuming the I-centre is fixed in space—it generally moves. Final Answer: Instantaneous centre

Question 17

Nature of motion of a bicycle wheel during rolling without slipping: Considering a bicycle moving forward on a road with no skidding, how is the motion of the wheel best described from a rigid-body kinematics perspective?

Accepted Answer

Introduction / Context: Rolling motion occurs widely in vehicles and machinery. Understanding that a rolling wheel exhibits both translation of its centre and rotation about its axle is fundamental for analyzing contact kinematics, friction forces, and energy distribution. Given Data / Assumptions: Wheel rolls without slipping on a horizontal surface. Deformations are neglected; rigid-body model. No external constraints beyond ground contact. Concept / Approach: Rolling without slip combines translation of the wheel’s centre of mass with rotation about the axle. The instantaneous centre of rotation lies at the point of contact with the ground where the velocity is momentarily zero, but overall the body experiences both rotational and translational motion simultaneously. Step-by-Step Solution: Let the centre move right with speed v.Angular speed ω satisfies v = ωR for pure rolling.Hence, every point’s velocity is the vector sum of translational velocity v and rotational component ω × r. Verification / Alternative check: Energy-wise, total kinetic energy = translational (1/2 m v^2) + rotational (1/2 I ω^2), confirming both modes are present. Why Other Options Are Wrong: (a) ignores rotation; (b) ignores translation of the wheel centre; (d) implies rotation about a fixed external centre (not true except instantaneously at contact); (e) is incompatible with circular point paths (cycloids) on the rim. Common Pitfalls: Thinking the bottom point “sticks” and thus the motion is purely rotational; forgetting v = ωR in no-slip rolling. Final Answer: rotary and translatory

Question 18

Resultant of two equal coplanar forces: Two equal forces, each of magnitude P, act at a point with an included angle θ between them. Determine the magnitude of their resultant vector.

Accepted Answer

Introduction / Context: Vector addition of forces is central to statics and dynamics. For the special case of two equal forces, a compact formula relates the resultant to the half-angle between them, simplifying quick hand calculations in design problems. Given Data / Assumptions: Two coplanar forces of equal magnitude P. Included angle between the forces is θ. Standard Euclidean vector addition applies. Concept / Approach: Using the parallelogram (or triangle) law, the magnitude of the resultant R of two forces P and P at angle θ is R = √(P^2 + P^2 + 2P·P cos θ) = √(2P^2 (1 + cos θ)). Apply the half-angle identity 1 + cos θ = 2 cos^2(θ/2) to obtain a simplified expression. Step-by-Step Solution: Start: R^2 = 2P^2 (1 + cos θ).Use identity: 1 + cos θ = 2 cos^2(θ/2).Hence R^2 = 4P^2 cos^2(θ/2).Therefore R = 2P cos(θ/2). Verification / Alternative check: Limit cases: θ = 0 → R = 2P (forces in the same direction). θ = 180° → R = 0 (equal and opposite). Both confirm the formula. Why Other Options Are Wrong: Other trigonometric forms do not satisfy the limiting cases or dimensional reasoning; only option (d) matches vector addition. Common Pitfalls: Using sin instead of cos; forgetting to halve the angle; misinterpreting θ as the supplement. Final Answer: 2P cos(θ/2)

Question 19

Impact and energy loss — compare kinetic energy before and after impact: In a real (non-perfectly elastic) collision between bodies, how does the kinetic energy of the system immediately after impact compare to that immediately before impact?

Accepted Answer

Introduction / Context: Collision problems in engineering mechanics distinguish between conservation laws for momentum and energy. Linear momentum of an isolated system is conserved during impact, but kinetic energy is generally not conserved unless the impact is perfectly elastic. Given Data / Assumptions: Consider typical real impacts with coefficient of restitution e between 0 and 1. No external impulses during the short impact interval. Internal deformations and heat/sound generation may occur. Concept / Approach: During a real collision, part of the system’s kinetic energy transforms into internal energy (deformation, heat, sound). Therefore, the kinetic energy after impact is less than that before impact, except in the ideal e = 1 case (perfectly elastic), where kinetic energy is conserved, and the perfectly plastic case (e = 0), where losses are maximal. Step-by-Step Solution: Use conservation of momentum for the short impact interval: total momentum before = total momentum after.Apply restitution relation along the line of impact: relative speed of separation = e * relative speed of approach with 0 ≤ e ≤ 1.From work–energy viewpoint, energy lost = initial KE − final KE ≥ 0 for 0 ≤ e < 1. Verification / Alternative check: Test limiting cases: e = 1 → no loss (rare ideal). e = 0 → bodies coalesce, maximum loss of kinetic energy. Most real contacts have 0 < e < 1, so KE decreases. Why Other Options Are Wrong: (a) is only true for perfectly elastic impacts; (b) contradicts physical dissipation; (d) overgeneralizes; (e) is unnecessary since the qualitative comparison holds regardless of masses. Common Pitfalls: Confusing momentum conservation (always for isolated systems) with kinetic energy conservation (only for e = 1). Final Answer: before impact is more than after impact

Question 20

Mechanical power developed by a rotating shaft: If a shaft transmits torque T (in N·m) at a speed of N r.p.m., what is the general expression for the mechanical power developed by the torque?

Accepted Answer

Introduction / Context: Converting rotational motion to power is central in machine design. The standard power–torque–speed relation allows engineers to size motors, couplings, and shafts and to convert between units (W, kW, hp) reliably. Given Data / Assumptions: Torque T in newton–metres (N·m). Speed N in revolutions per minute (r.p.m.). Angular speed ω relates to N by ω = 2πN/60 rad/s. Concept / Approach: Mechanical power for rotation is P = torque * angular speed = T * ω. Substituting ω in terms of N gives the widely used formula with the factor 2π/60 when N is in r.p.m. If N were in revolutions per second, P = 2π N T would apply directly. Step-by-Step Solution: Start with P = T * ω.Relate speed to rpm: ω = 2πN/60.Therefore, P = T * (2πN/60) = (2π N T)/60 watts. Verification / Alternative check: Unit check: N·m * rad/s = joules/second = watts. For kW, divide by 1000; for horsepower (metric), hp ≈ P/735.5. Why Other Options Are Wrong: (b) and (c) are dimensionally inconsistent for N in r.p.m.; (d) cannot be true since the alternatives disagree; (e) is correct only if N is in revolutions per second, not r.p.m. Common Pitfalls: Forgetting the 60 factor when using r.p.m.; mixing N in r.p.s. with N in r.p.m.; confusing torque units (N·m vs kgf·m). Final Answer: P = (2π N T) / 60 watts

Applied Mechanics Questions