Question 1

Dynamics – Effect of Scaling Both Masses in an Atwood-type System In the two-mass pulley system (both masses connected by a light inextensible string over a frictionless pulley), if both masses are doubled, how does the string acceleration change?

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Introduction / Context: The classic Atwood machine illustrates how acceleration depends on the difference and the sum of the two masses. This question checks understanding of proportionality when both masses are scaled by the same factor.Given Data / Assumptions: Two masses m1 and m2 on a light inextensible string over a frictionless pulley.Acceleration formula in magnitude: a = (m2 − m1) * g / (m1 + m2).Both masses are scaled: m1 → k m1 and m2 → k m2 with k = 2.Concept / Approach: Substitute the scaled masses into the expression for acceleration and simplify. Because both numerator and denominator scale by the same factor, the ratio remains unchanged.Step-by-Step Solution: Original acceleration: a = (m2 − m1) * g / (m1 + m2). After scaling by k: a′ = (k m2 − k m1) * g / (k m1 + k m2). Factor k: a′ = k (m2 − m1) * g / [k (m1 + m2)] = (m2 − m1) * g / (m1 + m2) = a.Verification / Alternative check: Choose numbers, e.g., m1 = 1, m2 = 3 ⇒ a = (2/4) g = 0.5 g. Doubling masses (2 and 6) ⇒ a = (4/8) g = 0.5 g (unchanged).Why Other Options Are Wrong: Half, double, or one-fourth would require changing the mass ratio, not uniform scaling. Adding g/2 is physically meaningless in this context.Common Pitfalls: Forgetting that both numerator and denominator scale by the same factor.Final Answer: same

Question 2

Lifting Machines – Mechanical Advantage vs Velocity Ratio For actual (real) machines with frictional losses, the mechanical advantage (MA) is __________ the velocity ratio (VR).

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Introduction / Context: In machine theory, the velocity ratio (VR) is a geometric/kinematic property, while the mechanical advantage (MA) depends on actual forces and is reduced by friction. Efficiency bridges the two.Given Data / Assumptions: Efficiency η = MA / VR.For real machines, 0 Concept / Approach: Rearranging efficiency gives MA = η * VR. Since η Step-by-Step Solution: Start: η = MA / VR. Rearrange: MA = η * VR. With η Verification / Alternative check: Ideal machine has η = 1 ⇒ MA = VR; adding friction reduces MA below VR.Why Other Options Are Wrong: Equal to: only for ideal machines. Greater than or reciprocal: contradicts η Unrelated: incorrect; η links them directly.Common Pitfalls: Confusing MA with VR or ignoring efficiency.Final Answer: less than

Question 3

Impact Mechanics – Loss of Kinetic Energy in a Perfectly Inelastic Collision Two bodies of masses m1 and m2 move along the same line with velocities u1 and u2, respectively, and stick together after impact (perfectly inelastic). What is the loss of …

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Introduction / Context: In one-dimensional inelastic impact where bodies coalesce, momentum is conserved but kinetic energy is not. The classic result expresses the energy loss in terms of masses and the initial relative speed.Given Data / Assumptions: Masses: m1 and m2.Initial velocities: u1 and u2 along the same line.Perfectly inelastic collision: common final velocity v after impact.No external impulse along line of motion.Concept / Approach: Use conservation of linear momentum to find v, then compute initial and final kinetic energies and subtract to get the loss ΔKE.Step-by-Step Solution: Momentum conservation: (m1 * u1 + m2 * u2) = (m1 + m2) * v ⇒ v = (m1 u1 + m2 u2) / (m1 + m2). Initial KE: KE_i = (1/2) m1 u1^2 + (1/2) m2 u2^2. Final KE: KE_f = (1/2) (m1 + m2) v^2. Subtract: ΔKE = KE_i − KE_f = (m1 * m2 * (u1 − u2)^2) / (2 * (m1 + m2)).Verification / Alternative check: The loss is proportional to the square of relative speed (u1 − u2); if u1 = u2, there is no impact and ΔKE = 0.Why Other Options Are Wrong: (b) misses the mass product and overestimates energy loss. (c) uses sum of speeds, not relative speed. (d) true only if u1 = u2; generally false. (e) dimensionally incorrect.Common Pitfalls: Forgetting to square the relative speed or misapplying momentum conservation.Final Answer: ΔKE = (m1 * m2 * (u1 − u2)^2) / (2 * (m1 + m2))

Question 4

Polygon Law of Forces – Definition Check If several forces acting simultaneously on a particle are represented in magnitude and direction by the sides of a polygon taken in order, then the resultant is represented by the closing side of the polygon …

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Introduction / Context: Graphical statics often uses the triangle or polygon law for vector addition. The polygon law generalizes the triangle law to more than two forces.Given Data / Assumptions: Concurrent force system acting on a particle.Vectors drawn to scale and direction head-to-tail (in order).Concept / Approach: The polygon law states that the vector sum of forces equals the closing side from the start to the final head. If the polygon closes, the resultant is zero (equilibrium).Step-by-Step Solution: Arrange force vectors head-to-tail in sequence. Draw the closing side from the tail of the first to the head of the last. This closing side equals the resultant in magnitude and direction (opposite if described tail-to-head).Verification / Alternative check: Component-wise addition (ΣFx, ΣFy) yields the same resultant as the graphical polygon construction.Why Other Options Are Wrong: It applies to any number of forces, not just two or equal forces; it is a planar vector addition rule (and extends to 3D via components).Common Pitfalls: Confusing order (head-to-tail) or drawing to inconsistent scales.Final Answer: Correct

Question 5

Circular Motion – Period and Angular Velocity For uniform circular motion with angular velocity ω (in rad/s), the period T (time for one revolution) is:

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Introduction / Context: Period and angular velocity are reciprocally related via the constant 2π because one complete revolution corresponds to an angular displacement of 2π radians.Given Data / Assumptions: Angular velocity ω in rad/s.Uniform circular motion (constant ω).Concept / Approach: Angular speed equals angle swept per unit time. One revolution is 2π radians, so time per revolution is 2π divided by the rate of angular change.Step-by-Step Solution: By definition: ω = dθ/dt. For one revolution: Δθ = 2π. Thus T = Δθ / ω = 2π / ω.Verification / Alternative check: Frequency f = 1/T, and ω = 2π f ⇒ T = 2π / ω (consistent).Why Other Options Are Wrong: (a) and (e) invert the relationship incorrectly. (c) multiplies instead of divides, wrong units. (d) misses a factor of 2.Common Pitfalls: Confusing radians with revolutions; always use 2π rad per revolution.Final Answer: 2π / ω

Question 6

Compound Pendulum – Centers of Suspension and Oscillation For a compound pendulum, the center of suspension and the center of oscillation are interchangeable. Is this statement true?

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Introduction / Context: A compound pendulum (physical pendulum) is any rigid body swinging about a horizontal axis. An important property, known as the reciprocity or interchangeability of centers, relates the center of suspension to the center of oscillation.Given Data / Assumptions: Small-angle oscillations (standard pendulum approximation).Rigid body with a fixed pivot (center of suspension).Concept / Approach: Bouguer’s theorem states that the center of suspension and the center of oscillation are interchangeable: if the pendulum is inverted and hung from the center of oscillation, the period remains the same and the original pivot becomes the new center of oscillation.Step-by-Step Solution: Define the equivalent simple pendulum length: L_eq = I_p / (m * h), where I_p is mass moment of inertia about pivot and h is distance from pivot to center of mass. The point along the line below the pivot at distance L_eq is the center of oscillation. By reciprocity, swapping the pivot to that point yields the same L_eq and hence the same period.Verification / Alternative check: Derivations from rotational dynamics about different axes using the parallel-axis theorem confirm the equality of equivalent lengths.Why Other Options Are Wrong: The property holds (within the small-angle model) without requiring point-mass concentration or drag-related assumptions.Common Pitfalls: Confusing physical pendulum with simple pendulum; the interchangeability is unique to the compound pendulum model.Final Answer: True

Question 7

Levers – Do “Leverage” and “Mechanical Advantage” Mean the Same for a Compound Lever? Assess the statement: “The terms leverage and mechanical advantage of a compound lever have the same meaning.”

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Introduction / Context: In machine elements, leverage usually refers to the ratio of effective arm lengths (effort arm to load arm), while mechanical advantage (MA) is the ratio of load to effort. For compound levers, multiple stages and friction complicate the relationship.Given Data / Assumptions: Compound lever with possible joints and pivots.Real (non-ideal) system may have friction at pins and links.Concept / Approach: For an ideal single lever without friction, MA equals leverage because moments balance: Effort * effort arm = Load * load arm ⇒ MA = load/effort = effort arm/load arm. However, with compound levers and real losses, the overall MA generally differs from the simple geometric leverage.Step-by-Step Solution: Define leverage: L = (effort arm) / (load arm) for a given lever stage. Define mechanical advantage: MA = Load / Effort across the entire compound system. In a compound lever, total MA is the product of stage MAs and is affected by friction; it is not necessarily equal to the simple arm-length ratio at any single pivot.Verification / Alternative check: If friction is neglected and geometry is cascaded ideally, MA can equal the product of lever arm ratios; but the blanket statement “same meaning” is still inaccurate.Why Other Options Are Wrong: Calling them the same conflates a geometric ratio with a force ratio across an entire system; only in special ideal cases do they coincide.Common Pitfalls: Assuming the simple lever relationship applies unchanged to multi-stage or real mechanisms.Final Answer: Incorrect

Question 8

Self-Locking Machines – Efficiency Criterion For a machine to be self-locking (i.e., it will not run backward under load when the effort is removed), the efficiency must be:

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Introduction / Context: Self-locking (non-reversible) behavior is crucial in lifting machines like screw jacks and worm gears to prevent back-driving. Efficiency provides a quick diagnostic.Given Data / Assumptions: Definitions: mechanical advantage MA, velocity ratio VR, efficiency η = MA / VR.Rule of thumb from machine theory relating reversibility to efficiency.Concept / Approach: A machine is reversible if it can be back-driven by the load when effort is removed. For many classical lifting mechanisms, reversibility occurs roughly when η > 50%. Conversely, self-locking generally requires η ≤ 50%, typically stated as less than 50%.Step-by-Step Solution: If η > 50% ⇒ losses are relatively small and back-driving is possible (reversible). If η Hence, for self-locking: η must be less than 50%.Verification / Alternative check: Detailed derivations for specific mechanisms (e.g., screw with friction angle > lead angle) also imply non-reversibility, which correlates with low efficiency.Why Other Options Are Wrong: Equal to or greater than 50% would suggest potential reversibility. 100% efficiency is ideal and certainly reversible. “Independent” is not supported; the 50% criterion is standard.Common Pitfalls: Confusing self-locking with braking; self-locking is a passive geometric/frictional effect.Final Answer: less than 50%

Question 9

Energy of Rebound – Ideal Elastic Bounce A rubber ball is dropped from a height of 2 m. Assuming no loss of velocity on rebound (perfectly elastic impact with the floor), to what maximum height will the ball rise?

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Introduction / Context: When a ball rebounds without energy loss (coefficient of restitution e = 1), mechanical energy is conserved between just-before-impact and just-after-rebound speeds, leading to symmetric rise and fall heights (ignoring air resistance).Given Data / Assumptions: Initial drop height h = 2 m.Perfectly elastic impact with the floor (e = 1).Neglect air drag and rolling effects.Concept / Approach: Use energy conservation or kinematics: the speed just before impact equals the speed just after rebound (upward) if e = 1. Therefore, the ball will climb back to the same height from which it was dropped.Step-by-Step Solution: Before impact: v_down = sqrt(2 g h). Elastic rebound: v_up = v_down. As it rises: v_up^2 = 2 g h′ ⇒ h′ = v_up^2 / (2 g) = (2 g h) / (2 g) = h = 2 m.Verification / Alternative check: Using coefficient of restitution: e = v_after / v_before = 1 ⇒ equal speeds; hence equal heights.Why Other Options Are Wrong: 1 m, 3 m, 4 m contradict energy conservation for e = 1; without losses the ball returns to its original height. “It will not rise” is obviously false for a bounce.Common Pitfalls: Assuming some loss automatically; here the problem explicitly states no loss.Final Answer: 2 m

Question 10

Machine Design — Screw Jack Efficiency (Condition for Maximum) For a square-threaded screw jack, the efficiency becomes maximum at a specific helix angle α relative to the angle of friction φ. Choose the correct relation for α at maximum efficiency.

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Introduction / Context: Screw jacks are classic simple machines used to lift loads through a threaded screw. Their efficiency depends on the helix angle α of the thread and the angle of friction φ. This question asks for the condition on α that maximizes efficiency for a square-threaded screw jack. Given Data / Assumptions: Square-thread screw jack raising a load under steady conditions. Helix angle α is measured on the screw mean radius helix. Angle of friction φ is defined by tan φ = μ, where μ is the friction coefficient. Idealized geometry; no collar friction considered in the core relation. Concept / Approach: For a screw jack, efficiency η during raising can be written in terms of α and φ. Maximizing η with respect to α gives a well-known design condition that balances mechanical advantage against frictional losses. Step-by-Step Solution: Start from the standard efficiency form (for raising): η = tan α / tan(α + φ). Differentiate η with respect to α and set dη/dα = 0 to find the maximum. Solving gives the optimum helix angle: α_opt = 45° - φ/2. Interpretation: a larger friction angle φ pushes the optimal α smaller than 45°. Verification / Alternative check: If φ = 0 (no friction), the condition yields α = 45°, which aligns with the symmetry of the tan-based expression and physical intuition that losses vanish at the geometric midpoint. Why Other Options Are Wrong: '45° + φ/2' would increase α with friction, worsening efficiency. Expressions with '90° ± φ' do not arise from maximizing η for screw threads and are dimensionally unrelated to the optimum condition. Common Pitfalls: Mixing definitions of α (lead vs helix angle), forgetting that the above result assumes square threads and neglects collar friction in the core optimization step. Final Answer: α = 45° - φ/2.

Question 11

Statics — Three Coplanar Forces in Equilibrium At a point O, three coplanar forces P, Q, and R act. If the system is in equilibrium, which statement(s) must hold true?

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Introduction / Context: Equilibrium of three concurrent coplanar forces is a foundational topic in engineering mechanics. Several equivalent characterizations exist, including the triangle law of forces and Lami’s theorem. This question asks which statements must hold when three forces are in equilibrium at a point. Given Data / Assumptions: Three forces P, Q, R acting at a single point O in one plane. System is in static equilibrium (no resultant force). Concept / Approach: For three concurrent forces to be in equilibrium, the vector polygon formed head-to-tail must close (triangle law), yielding a zero resultant. Lami’s theorem provides a scalar relation among magnitudes and the angles between forces when exactly three non-collinear forces are in equilibrium. Step-by-Step Solution: Triangle law: Draw P, then Q from the head of P, then R from the head of Q. If the tip of R returns to the start, P + Q + R = 0. Vector condition: Equilibrium requires the resultant to vanish, hence P + Q + R = 0. Lami’s theorem: For three forces in equilibrium, P/sin α = Q/sin β = R/sin γ, where α, β, γ are the angles opposite P, Q, R respectively. Verification / Alternative check: Resolving along any two perpendicular directions yields two independent equations. Solving gives a unique relation among magnitudes and angles equivalent to Lami’s theorem, confirming consistency of all three statements. Why Other Options Are Wrong: Each of (a), (b), and (c) is individually correct; picking only one omits complete characterization. Hence 'All of the above' is the best answer. Common Pitfalls: Misapplying Lami’s theorem to more than three forces; forgetting that angles in Lami’s theorem are those between the other two forces (opposite to the listed force). Final Answer: All of the above.

Question 12

Projectile on an Upward Inclined Plane A projectile is launched with initial speed u at an angle θ to the horizontal onto an upward inclined plane of angle β (β > 0). What is the range R measured along the plane?

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Introduction / Context: Range on an inclined plane is a classic extension of projectile motion. Instead of measuring horizontal range, we measure distance along a plane inclined at angle β to the horizontal. The result depends on both the launch angle θ and the slope β. Given Data / Assumptions: Initial speed u, launch angle θ to the horizontal. Inclined plane angle β (upward), no air resistance. Acceleration due to gravity g acts vertically downward. Concept / Approach: Resolve motion into components parallel and perpendicular to the incline or start from horizontal/vertical equations and impose the plane-impact condition y = x tan β at the point of contact. Algebra then gives range measured along the plane surface. Step-by-Step Solution: Parametric motion: x = u cos θ * t, y = u sin θ * t - (1/2) g t^2. Impact condition on the plane: y = x * tan β. Substitute x and y, solve for flight time t > 0 giving t = (2 u * sin(θ - β)) / (g * cos β). Distance along plane: R = x / cos β = (u cos θ * t) / cos β = (2 * u^2 * cos θ * sin(θ - β)) / (g * cos^2 β). Verification / Alternative check: For β = 0 (horizontal ground), formula reduces to R = (u^2 * sin 2θ) / g, which is the standard horizontal range, confirming correctness. Why Other Options Are Wrong: Using sin(θ + β) gives a larger angle and incorrect range sign for typical θ; the standard flat-ground result (option c) ignores the slope; the cos β factors in option d are not consistent with the plane geometry. Common Pitfalls: Forgetting to convert horizontal distance to along-plane distance (divide by cos β), mixing θ ± β signs, and algebraic slips solving the impact condition. Final Answer: R = (2 * u^2 * cos θ * sin(θ - β)) / (g * cos^2 β).

Question 13

Worm-Geared Pulley Block — Velocity Ratio Change In a worm-geared pulley block with a single-start worm, if the number of teeth on the worm wheel is doubled, what happens to the velocity ratio (VR)?

Accepted Answer

Introduction / Context: Worm-geared pulley blocks achieve high mechanical advantage via the gear ratio between a worm and a worm wheel. The velocity ratio (VR) is directly tied to the gear teeth count for a given worm start count. Given Data / Assumptions: Single-start worm (one thread), common in hoisting blocks. Other parts and diameters unchanged. VR is proportional to the worm wheel teeth count z for a single-start worm. Concept / Approach: For a single-start worm, each worm revolution advances the worm wheel by one tooth. Thus, the gear ratio equals z (number of teeth on the worm wheel). The overall VR for the block scales with that ratio (with additional factors from the sheave system if present, assumed constant here). Step-by-Step Solution: Initial VR ∝ z. Doubling z → VR_new ∝ 2z. Therefore, VR doubles if everything else remains unchanged. Verification / Alternative check: Compare rotations: one worm turn moves the wheel by 1 tooth; with twice the teeth, it takes twice the turns for one wheel revolution, doubling distance ratio between effort and load path, hence VR doubles. Why Other Options Are Wrong: 'Remains unchanged' contradicts the direct proportionality; 'half' or 'four times' would require changing worm starts or reeving pattern, which is not stated. Common Pitfalls: Confusing worm starts (single vs multi-start) with teeth count; ignoring additional pulley stages that also affect VR but are assumed constant here. Final Answer: True — the velocity ratio doubles.

Question 14

Area Moments — Solid Cylinder about Longitudinal (Polar) Axis Find the mass moment of inertia I of a solid circular cylinder (mass m, radius r, length l) about its longitudinal/polar axis (through the centre, along the axis).

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Introduction / Context: The mass moment of inertia quantifies resistance to angular acceleration about a specific axis. For a solid cylinder, the polar (longitudinal) axis is the central axis of symmetry; knowing I for this axis is essential in rotor dynamics and torsional vibration problems. Given Data / Assumptions: Solid cylinder: uniform density, radius r, length l, mass m. Axis of rotation: cylinder’s own axis (polar/longitudinal). Concept / Approach: For bodies of revolution, standard results give compact expressions for I. The solid cylinder about its own axis has a classical closed form derived from integrating r^2 over the cross-sectional disk and extruding along length l. Step-by-Step Solution: Definition: I = ∫ r^2 dm over the volume. For a disk element of thickness dz, I_disk = (1/2) * ρ * π * r^4 per unit thickness; integrating along l multiplies by l and relates to total mass m. Result simplifies to I = (1/2) * m * r^2. Verification / Alternative check: Compare with a thin ring of radius r: I_ring = m * r^2. A solid cylinder concentrates more mass near the centre than a ring, so I must be smaller; (1/2) m r^2 is consistent physically. Why Other Options Are Wrong: Fractions of 1/4, 1/6, 1/8 do not match the standard integral for a solid cylinder about its axis and yield values that are too small compared with established results. Common Pitfalls: Confusing polar axis (along the cylinder) with diametral axis (across the cylinder), which has a different I value: I_diametral = (1/4) m r^2 + (1/12) m l^2. Final Answer: I = (1/2) * m * r^2.

Question 15

Simple Harmonic Motion — Acceleration at Mean Position For a particle in SHM, what is the magnitude of acceleration at the mean (equilibrium) position?

Accepted Answer

Introduction / Context: In simple harmonic motion (SHM), displacement, velocity, and acceleration vary sinusoidally with time. Understanding where each quantity is maximum or zero is key to sketching motion and solving related dynamics problems. Given Data / Assumptions: Standard SHM: x(t) = A sin(ω t + φ). Mean position corresponds to x = 0 (equilibrium). Concept / Approach: Acceleration in SHM is proportional to displacement and directed toward the mean position: a = -ω^2 x. Therefore, acceleration depends directly on x, not on velocity. At x = 0, a must be zero. Step-by-Step Solution: Write the SHM relation: a = -ω^2 x. At mean position, x = 0. Hence a = -ω^2 * 0 = 0. Verification / Alternative check: Velocity at mean position is maximum (energy is all kinetic), which coexists with zero acceleration because the slope of velocity (change rate) crosses zero at the instant of peak speed in sinusoidal motion. Why Other Options Are Wrong: 'Maximum' acceleration occurs at extreme displacements x = ±A; 'minimum' is ambiguous but not appropriate here because acceleration actually becomes zero at the mean, not a small positive minimum. Common Pitfalls: Mixing up where velocity and acceleration are maximum; thinking acceleration must be large when the speed is large—this is not true for SHM at the instant of passing equilibrium. Final Answer: zero.

Question 16

Simple Harmonic Motion — Velocity at Mean Position For a particle in SHM, what is the velocity at the mean (equilibrium) position compared with elsewhere in the cycle?

Accepted Answer

Introduction / Context: SHM features an exchange between potential and kinetic energy. The mean position is where the restoring force (and hence potential energy for linear SHM) is zero, so kinetic energy — and velocity magnitude — are largest there. Given Data / Assumptions: Displacement x, velocity v, acceleration a in standard SHM. Total energy E is constant: E = (1/2) k A^2 = (1/2) k x^2 + (1/2) m v^2. Concept / Approach: At x = 0 (mean position), the potential energy part (1/2) k x^2 is zero, so kinetic energy is maximum; hence speed is maximum. Conversely, at extremes x = ±A, speed is zero and potential energy is maximum. Step-by-Step Solution: Use v(x) relation: v^2 = ω^2 (A^2 - x^2). At x = 0 ⇒ v^2 = ω^2 A^2 ⇒ v_max = ω A. At x = ±A ⇒ v = 0, confirming the contrast. Verification / Alternative check: From sinusoidal form x = A sin(ω t + φ), v = dx/dt = A ω cos(ω t + φ). Cosine reaches magnitude 1 at the mean, giving maximum |v| = A ω. Why Other Options Are Wrong: 'Zero' and 'minimum' occur at the extreme positions, not at the mean. Common Pitfalls: Assuming acceleration and velocity peak at the same place; in SHM, they are out of phase by 90 degrees in time (π/2 radians). Final Answer: maximum.

Question 17

Centres of Gravity — Solid Hemisphere A solid hemisphere of radius r has its centre of gravity located at a distance 3r/8 from the flat base along the symmetry axis. Is this statement correct?

Accepted Answer

Introduction / Context: Locating the centre of gravity (centroid for uniform density) of standard 3D solids is a common requirement in statics and strength of materials. For a solid hemisphere, there is a well-established distance of the centroid from the flat face (base). Given Data / Assumptions: Solid hemisphere of radius r, uniform density. Distance measured from the plane face (base) along the central vertical axis. Concept / Approach: The centroid of a solid hemisphere lies closer to the base than to the sphere centre. Standard integration (using disks or spherical coordinates) yields a compact result: ȳ = 3r/8 from the base plane toward the curved surface along the symmetry axis. Step-by-Step Solution: Model hemisphere as a stack of circular disks of radius ρ(z) with thickness dz. Compute first moment: ∫ z dV / ∫ dV and evaluate the bounds over the hemisphere. The integral simplifies to ȳ = 3r/8 measured from the base plane. Verification / Alternative check: Tabulated centroid locations in handbooks (solids of revolution) list ȳ = 3r/8 for the solid hemisphere, while the spherical surface (hemispherical shell) has a different value: ȳ = r/2 from the base. Why Other Options Are Wrong: 'Incorrect' would contradict established results derived by volume integration and universally referenced in engineering tables. Common Pitfalls: Confusing the solid hemisphere with a hollow hemispherical shell; mixing the reference point (from base versus from the sphere centre). Final Answer: Correct.

Question 18

Simple Pendulum — Changing the Period If the period T of a simple pendulum is to be doubled, by what factor should its length l be changed (small-angle assumption)?

Accepted Answer

Introduction / Context: The period of a simple pendulum depends on its length and the local acceleration due to gravity. Understanding how T changes with l is essential for timing devices and lab experiments using pendulums. Given Data / Assumptions: Small-angle approximation holds so that T = 2π * sqrt(l / g). Gravity g is constant. We desire T_new = 2 * T_old. Concept / Approach: Since T ∝ sqrt(l), scaling the period by a factor scales the length by the square of that factor. Therefore doubling T requires quadrupling l. Step-by-Step Solution: Start: T = 2π * sqrt(l / g). Let T_new = 2 * T_old ⇒ 2π * sqrt(l_new / g) = 2 * [2π * sqrt(l_old / g)]. Cancel 2π and square both sides: l_new / g = 4 * (l_old / g) ⇒ l_new = 4 * l_old. Verification / Alternative check: If l is quadrupled, T becomes 2π * sqrt(4 l_old / g) = 2 * 2π * sqrt(l_old / g), confirming the doubling. Why Other Options Are Wrong: Halving or doubling l changes T by factors of 1/√2 or √2 respectively, not 2. Common Pitfalls: Forgetting that T depends on sqrt(l), not linearly; overlooking that mass does not affect T for a simple pendulum. Final Answer: quadrupled.

Question 19

Classical mechanics — recoil of a gun  Which law of motion explains the recoil of a gun when a bullet is fired (the gun moves backward as the bullet moves forward)?

Accepted Answer

Introduction / Context:The recoil of a gun is a classic illustration used in physics to demonstrate action–reaction behavior. When a projectile is expelled at high speed, the gun experiences a backward motion. Identifying the correct Newtonian principle behind this phenomenon is a foundational skill for mechanics and engineering students.Given Data / Assumptions:A gun fires a bullet forward with high momentum.The gun and bullet form an isolated system for a very short interval.External horizontal forces during the impulse are negligible.Concept / Approach:Newton's third law states that for every action there is an equal and opposite reaction. Simultaneously, conservation of linear momentum applies to an isolated system: m_bullet * v_bullet + m_gun * v_gun = constant (typically zero before firing). The equal and opposite interaction forces during the short firing interval generate impulses that change the momenta of both bodies in opposite directions.Step-by-Step Solution:Action: expanding gases push the bullet forward.Reaction: the bullet exerts an equal and opposite force on the gun.Impulses are equal in magnitude and opposite in direction, so momenta are equal and opposite.Therefore, the gun recoils backward as a direct consequence of Newton's third law.Verification / Alternative check:Using momentum conservation: m_bullet * v_bullet + m_gun * v_gun = 0 (initially at rest). Thus v_gun = − (m_bullet / m_gun) * v_bullet, confirming backward motion with smaller speed due to higher gun mass.Why Other Options Are Wrong:First law: Inertia alone does not cause recoil.Second law: Relates force to acceleration but does not explain equal and opposite impulses here.None of these / energy only: Energy conversion occurs, but recoil specifically demonstrates action–reaction.Common Pitfalls:Attributing recoil solely to gas escape or energy without recognizing the paired interaction forces and momentum balance.Final Answer:Newton's third law of motion

Question 20

Friction and slopes — definition check  “The angle of an inclined plane at which a body just begins to slide down is called the helix angle.” Is this statement correct?

Accepted Answer

Introduction / Context:This item tests terminology in engineering mechanics and machine design. Two different angles are often confused: the angle of repose (slip threshold on an incline) and the helix angle (geometry of helical features like screw threads).Given Data / Assumptions:A rigid block rests on a rough inclined plane.We slowly increase the plane's angle until impending motion occurs.Basic Coulomb friction model applies.Concept / Approach:The angle of repose is the inclination at which a body is just about to slide. At this angle, tan(φ) = μ_s, where μ_s is the coefficient of static friction. The helix angle is a geometric angle formed by the helix (e.g., a screw thread) with respect to a plane perpendicular to the screw axis; it is unrelated to a block beginning to slip on a plane.Step-by-Step Solution:Define the phenomenon: onset of sliding on an incline.Recall definition: angle of repose φ satisfies tan(φ) = μ_s.Recognize that “helix angle” applies to helical paths (threads, springs), not to slip threshold.Therefore, the provided statement is incorrect.Verification / Alternative check:Practical test: pour granular material to form a conical pile; the slope angle at rest is the angle of repose, again demonstrating the correct term.Why Other Options Are Wrong:Only for screws / polished planes / friction zero: These do not change the definition; helix angle remains a geometric feature, not a slip threshold.Common Pitfalls:Mixing terms due to superficial similarity. Always connect “repose” with sliding onset and “helix” with helical geometry.Final Answer:False

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