Question 1

Inventory control (EOQ model): If the annual demand for an item is trebled (multiplied by 3) and the ordering/setup cost is reduced to one third (S/3), while the holding cost per unit per year remains unchanged, how does the Economic Order Quantity …

Accepted Answer

Introduction: The Economic Order Quantity (EOQ) model is a foundational result in operations and supply chain management. It balances the trade-off between ordering costs (or setup costs) and inventory holding costs to minimize total annual cost. This problem tests your understanding of how EOQ scales when demand and ordering cost change simultaneously. Given Data / Assumptions: Classical EOQ applies with constant, deterministic demand and instantaneous replenishment. Original parameters: demand D, ordering cost S, holding cost H. Changes: D becomes 3D (trebled), S becomes S/3 (one third), H is unchanged. No quantity discounts, no stockouts, and lead time effects are ignored. Concept / Approach: The EOQ formula is EOQ = sqrt(2 * D * S / H). The key insight is that EOQ depends on the square root of the product D * S. If two adjustments offset each other so that D * S remains the same, then EOQ remains identical. We will substitute the changed values and compare to the original EOQ. Step-by-Step Solution: Start with the standard formula: EOQ = sqrt(2 * D * S / H).Apply the changes: D' = 3D and S' = S / 3.Compute the new EOQ: EOQ' = sqrt(2 * (3D) * (S/3) / H).Algebraic simplification: EOQ' = sqrt(2 * D * S / H).Conclusion: EOQ' equals the original EOQ; therefore, EOQ remains unchanged. Verification / Alternative check: Use proportional (sensitivity) reasoning. Since EOQ ∝ (D * S)^(1/2) * H^(−1/2), the combined change factor for D * S is (3) * (1/3) = 1. Thus (D * S)^(1/2) is multiplied by 1^(1/2) = 1, and EOQ is unaffected. This confirms the algebraic result without numbers. Why Other Options Are Wrong: is trebled. EOQ would triple only if D * S increased by a factor of 9 (since EOQ scales with the square root), which it does not. decreases by a factor of 3. This would require D * S to drop by a factor of 9; not the case here. decreases by a factor of 1/3. Would imply D * S decreased to 1/9 of its value; incorrect. increases by a factor of √3. Would require D * S to triple; instead, it is unchanged. Common Pitfalls: Students often forget the square-root relationship and attempt linear reasoning (e.g., “demand tripled, so EOQ triples”). Always inspect how parameters enter the EOQ formula—EOQ scales with the square root of the product D * S and inversely with the square root of H. Final Answer: remains unchanged.

Question 2

Radiation sources: cobalt-60 is widely used in medical therapy and industrial radiography as a sealed source emitting which dominant type of penetrating ionizing radiation?

Accepted Answer

Introduction: Cobalt-60 (Co-60) is a man-made radioisotope produced by neutron activation of cobalt-59. Its decay scheme makes it extremely useful as a high-energy, high-intensity radiation source for teletherapy units and for non-destructive testing (NDT) in industry. Identifying its principal emission is fundamental to shielding design and safety protocols. Given Data / Assumptions: Co-60 decays by beta emission to excited states of nickel-60. The excited daughter de-excites by emitting characteristic photons. Applications require deeply penetrating radiation. Concept / Approach: While Co-60 undergoes beta decay, the practically important radiation for therapy and radiography is the cascade of two high-energy gamma photons (~1.17 MeV and ~1.33 MeV). These γ-rays are highly penetrating, enabling treatment of tumors at depth and radiographic inspection of thick components. X-rays are produced by electron transitions or Bremsstrahlung in X-ray tubes, not by Co-60 decay directly. Step-by-Step Solution: Recognize Co-60 decay: β- emission to excited Ni-60.Identify emitted photons: two γ-rays near 1.17 and 1.33 MeV.Select γ-rays as the dominant useful radiation from sealed Co-60 sources. Verification / Alternative check: Databases and shielding charts list Co-60 as a gamma emitter with average energies around 1.25 MeV, which dictate lead or concrete thicknesses for safe operation. Why Other Options Are Wrong: X-rays: Generated by tubes or secondary processes, not the primary emission of Co-60. α-rays: Not part of Co-60 decay. β-rays: Emitted internally but are fully absorbed within sealed capsules and are not the principal useful radiation. Neutrons: Not produced by Co-60 decay. Common Pitfalls: Equating the decay mode (beta) with the useful external radiation (gamma). In sealed sources, betas are shielded by the capsule; gammas escape and are used. Final Answer: γ-rays

Question 3

Magnetic materials: choose the incorrect statement about magnetic behavior and material selection for permanent magnets versus electromagnets.

Accepted Answer

Introduction: Different applications demand different magnetic properties. Electromagnets used for transformers and motors prioritize low losses and high permeability, while permanent magnets require high coercivity and remanence to retain magnetization without an external field. Understanding which statements align with these principles helps in material selection and design. Given Data / Assumptions: Terminology: 'soft' versus 'hard' magnetic refers to coercivity and hysteresis behavior, not mechanical hardness. Curie temperature marks the transition from ferromagnetic to paramagnetic behavior. Examples: pure iron (soft), tungsten steel and alnico (hard). Concept / Approach: Soft magnetic materials exhibit high initial permeability, low coercive force, and low hysteresis loss—ideal for alternating fields. Hard magnetic materials have high coercivity and retain magnetization, suitable for permanent magnets. Above the Curie point, even ferromagnets lose ordered domains and behave paramagnetically, confirming option (a). Alnico and tungsten steel are classic hard magnet materials, contradicting any statement that claims they are not hard magnetic. Step-by-Step Solution: Evaluate each statement against standard definitions and examples.Identify the outlier: the claim that tungsten steel and alnico are not hard magnetic is false.Mark this as the incorrect ('wrong') statement. Verification / Alternative check: Magnetics references classify tungsten steel and alnico as traditional permanent-magnet materials with substantial coercive force and remanence. Why Other Options Are Wrong: (a): Correct: Curie transition removes ferromagnetism. (b): Correct distinction between permanent magnets and electromagnet cores. (c): Correct properties of soft magnets such as low loss and high permeability. (e): Correct by definition for hard magnets. Common Pitfalls: Confusing mechanical 'hardness' with magnetic 'hardness'; they are different properties influenced by microstructure and composition. Final Answer: Tungsten steel and alnico are not hard magnetic materials.

Question 4

Plant utilities design: for compressed air service in industrial pipelines, what is a typical maximum permissible flow velocity (order-of-magnitude value) used to limit pressure drop and noise while maintaining reasonable line sizes?

Accepted Answer

Introduction: Compressed air distribution must balance capital cost (pipe size), operating cost (pressure drop and compressor power), and practical concerns such as noise and erosion at fittings. A rule-of-thumb maximum velocity helps designers choose economical line sizes without detailed iteration at early stages. Given Data / Assumptions: General industrial compressed air networks (not high-pressure gas transmission). Acceptable pressure drops over reasonable distances. Target is a widely cited maximum permissible velocity. Concept / Approach: As velocity increases, frictional pressure drop and acoustic noise rise sharply. Many design guides suggest keeping main line velocities around 6–10 m/s for efficiency, with branches and short runs up to about 20 m/s. The question asks for a typical maximum permissible value—an upper bound used where line lengths are limited or when cost pressure favors smaller pipes. Step-by-Step Solution: Relate velocity to pressure drop using Darcy–Weisbach: ΔP ∝ f * (L/D) * (ρ * v^2 / 2).Recognize noise and efficiency concerns above ~20 m/s in plant air lines.Select 20 m/s as a representative maximum often tolerated in practice. Verification / Alternative check: Compressed-air handbooks frequently present tables with recommended velocities around 6–10 m/s for mains and up to ~20 m/s for branches; beyond this, pressure drops and noise become problematic. Why Other Options Are Wrong: 5 or 10 m/s: Conservative and common for mains, but the question seeks a maximum permissible order-of-magnitude value. 40 or 60 m/s: Excessive for plant air; would cause large pressure losses and noise. Common Pitfalls: Applying a single velocity criterion without checking length, roughness, and allowable pressure drop; detailed design should verify with friction calculations. Final Answer: 20

Question 5

Photovoltaics and energy conversion: A solar cell in a photovoltaic module converts incident sunlight directly into which form of usable energy (without any intermediate thermal or mechanical stage)?

Accepted Answer

Introduction / Context: Solar cells are semiconductor devices that transform solar radiation into electricity by means of the photovoltaic effect. Unlike solar-thermal systems, they do not first heat a working fluid; rather, they produce electric power directly at the junction of specially doped materials. Given Data / Assumptions: Device is a solid-state photovoltaic (PV) cell. Illumination is by sunlight within the device's spectral response. No moving parts or heat engines are involved. Concept / Approach: In a PN junction under illumination, photons with energy greater than the bandgap create electron–hole pairs. The built-in electric field separates these charge carriers, generating a photovoltage and photocurrent. The net result is direct conversion of radiant energy into electrical energy at the terminals of the cell. Step-by-Step Solution: Recognize that PV cells rely on the photovoltaic effect, not on heating a fluid.Identify the output form of energy: electron flow (current) at a measurable voltage.Conclude that the direct product is electrical power (P = V * I).Therefore select the option describing electrical energy. Verification / Alternative check: Current–voltage (I–V) curves under standard test conditions (STC) quantify electrical output without any mechanical stage, confirming direct conversion to electricity. Why Other Options Are Wrong: Thermal: solar-thermal collectors convert sunlight to heat, not PV cells. Mechanical: requires turbines or engines, which PV systems do not employ. Chemical: while solar energy can drive electrochemical storage (batteries), the cell itself outputs electricity. Common Pitfalls: Confusing PV panels with solar water heaters. Assuming energy storage is inherent to PV; storage is external. Final Answer: Electrical

Question 6

Bearing materials and service conditions: Tin-based white metals (Babbitt alloys) are typically selected for bearings operating under which combination of load and pressure conditions?

Accepted Answer

Introduction / Context: Bearing material choice balances conformability, embeddability, fatigue resistance, and friction. Tin-based white metals (Babbitts) are classic bearing linings known for their soft, conformable matrix with hard intermetallic phases that support load over a hydrodynamic oil film. Given Data / Assumptions: Application is a hydrodynamic journal bearing with adequate lubrication. Tin-based Babbitt lining over a stronger backing (e.g., steel/bronze shell). Normal operating temperatures and good oil cleanliness. Concept / Approach: White metals provide excellent embeddability and conformability but limited load-carrying capacity compared with bronzes or high-strength overlays. They are therefore favored where unit pressures and loads are modest, misalignment must be tolerated, and seizure resistance is important. Step-by-Step Solution: Identify key properties: soft matrix, good seizure resistance, embeddability.Map properties to service: suitable for light-duty bearings with low P and low W.Avoid high-temperature or severe load applications where fatigue life is limiting.Select the condition 'Low pressure and low load'. Verification / Alternative check: Manufacturer handbooks rate Babbitt linings for relatively low specific pressures; heavier-duty bearings often use copper–lead or aluminum–tin alloys with higher fatigue strength. Why Other Options Are Wrong: High pressure/high load: exceeds Babbitt fatigue capacity. High temperature: risks softening of the matrix and reduced film strength. Large surface wear: while conformable, Babbitt is not a wear-resistant hard-facing material. Common Pitfalls: Assuming Babbitt alone carries the load; it is a lining supported by a stronger shell. Ignoring lubrication; boundary conditions can rapidly damage soft linings. Final Answer: Low pressure and low load

Question 7

Bonding in solids and directionality: Which type of interatomic bonding is strongly directional in crystalline solids, leading to specific bond angles and directional properties?

Accepted Answer

Introduction / Context: The nature of chemical bonding affects mechanical, electronic, and thermal properties of solids. Directionality refers to a bond's preference for particular angles and orientations in space, which strongly influences crystal structure and anisotropy. Given Data / Assumptions: Comparison among common bonding types: ionic, metallic, covalent, and Van der Waals. Crystalline solid contexts, not molecular gases. Focus on angular preferences and spatial orientation. Concept / Approach: Covalent bonds arise from shared electron pairs between atoms and favor discrete directional orbitals (e.g., sp^3 tetrahedral in diamond), imposing strict bond angles and directional rigidity. Ionic and metallic bonds are largely non-directional due to electrostatic attractions or delocalized electron clouds, while Van der Waals forces are weak and non-directional dispersion interactions. Step-by-Step Solution: Identify bonding types and assess directionality.Covalent: strong directionality from orbital hybridization (e.g., sp, sp^2, sp^3).Ionic/metallic: bond strength arises from omnidirectional fields or electron seas.Therefore, select 'Covalent bonding'. Verification / Alternative check: Materials with covalent networks (diamond, silicon) exhibit high anisotropy and specific bond angles; metals with metallic bonding tend to have close-packed structures with slip systems reflecting non-directionality. Why Other Options Are Wrong: Van der Waals: weak, non-directional interactions. Ionic: electrostatic, largely isotropic in solids. Metallic: delocalized electrons lead to non-directional bonding and ductility. Common Pitfalls: Conflating molecular geometry (directional) with bulk crystalline packing in non-covalent solids. Assuming hydrogen bonding here; it is not one of the listed fundamental solid-state bonds. Final Answer: Covalent bonding

Question 8

Heat loss from a bare steam pipe in a still room: Which listed parameter does NOT affect the total heat loss by combined natural convection and thermal radiation from a hot, bare pipe (no fans present)?

Accepted Answer

Introduction / Context: In quiescent indoor air, a hot bare pipe loses heat primarily via natural convection and thermal radiation. The total heat loss depends on surface properties, geometry, and temperature differences that drive both mechanisms. Given Data / Assumptions: No forced convection (no fans). Pipe is horizontal, bare, and surrounded by still air. Radiation exchange is with the room surfaces approximated as a large enclosure. Concept / Approach: Natural convection heat transfer coefficient depends on Grashof/Prandtl numbers, which are functions of geometry (diameter, length) and temperature difference. Radiative heat exchange depends strongly on surface emissivity and the fourth-power temperature difference. Therefore, emissivity, size, and temperatures all affect the heat loss—none can be dismissed. Step-by-Step Solution: Consider natural convection: h_nc ∝ (ΔT)^n * f(geometry, properties).Consider radiation: q_rad = ε * σ * (T_s^4 − T_sur^4) * A.Emissivity ε clearly influences radiative loss; pipe dimensions set area A and characteristic length.Therefore all listed parameters matter; choose 'None of these'. Verification / Alternative check: Empirical correlations (e.g., Churchill–Chu for cylinders) include diameter and ΔT explicitly; radiative exchange is proportional to emissivity and area, confirming full dependence. Why Other Options Are Wrong: Selecting any one parameter as irrelevant contradicts the governing equations for convection and radiation. Common Pitfalls: Neglecting radiation in high-emissivity paints; it can rival convection. Assuming only surface temperature matters and ignoring size effects on h_nc. Final Answer: None of these (all influence heat loss)

Question 9

Atmospheric lapse rate in the troposphere: If the sea-level air temperature is 15 °C, the temperature t (in °C) at height h (in metres) within the troposphere follows which standard relation?

Accepted Answer

Introduction / Context: The International Standard Atmosphere (ISA) adopts a constant temperature lapse rate in the troposphere, the weather-bearing layer of Earth's atmosphere. This linear relation provides a handy approximation for many engineering calculations, aviation, and environmental assessments. Given Data / Assumptions: Sea-level temperature taken as 15 °C. Standard tropospheric lapse rate: 6.5 °C per 1000 m. Valid only within the troposphere (up to the tropopause). Concept / Approach: A lapse rate of 6.5 °C per 1000 m means temperature decreases with altitude. Converting to per metre gives 0.0065 °C/m. Thus, at height h metres, subtract 0.0065 h from 15 °C to obtain the approximate air temperature. Step-by-Step Solution: Standard lapse: 6.5 °C per 1000 m → 0.0065 °C per m.Form relation: t(h) = 15 − 0.0065 h.Check units and sign: negative slope because temperature falls with height.Select the option matching this equation. Verification / Alternative check: At h = 2000 m, t = 15 − 13 = 2 °C, a realistic tropospheric estimate under standard conditions. Why Other Options Are Wrong: Positive slope implies warming with height, which is non-standard in the troposphere. Alternate slopes (0.0035) do not match ISA values. Common Pitfalls: Applying the relation above the tropopause where the lapse rate changes. Ignoring local weather inversions that deviate from the standard. Final Answer: t = 15 − 0.0065 h

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