Question 1

In the S.I. system, the quantity measured in candela (cd) is luminous intensity. Identify the correct application of the unit 'candela'.

Accepted Answer

Introduction / Context: S.I. assigns specific base and derived units to distinct physical quantities. Candela (cd) is one of the seven S.I. base units and plays a central role in lighting engineering, photometry, and visual ergonomics by quantifying the luminous intensity of a source in a given direction. Given Data / Assumptions: The unit under discussion is candela (cd). Options mix unrelated quantities such as power (W), frequency (Hz), and current (A). We must identify the quantity for which cd is the canonical S.I. unit. Concept / Approach: Luminous intensity measures perceived light power in a specified direction, weighted by the standard luminous efficacy curve (photopic vision). This distinguishes it from illuminance (lux), luminous flux (lumen), radiant intensity (W/sr), or electrical power (W). Step-by-Step Solution: Recall S.I. base units: cd is assigned to luminous intensity.Map distractors: current → ampere (A); power → watt (W); frequency → hertz (Hz); illuminance → lux (lx) = lumen/m^2.Therefore, select 'luminous intensity'. Verification / Alternative check: In photometric relations, luminous flux (lm) = luminous intensity (cd) * solid angle (sr). This confirms cd’s role as intensity per direction, not area-based reception (lux) or energy rate (W). Why Other Options Are Wrong: Electric current intensity: Measured in ampere (A). Power: Measured in watt (W). Frequency: Measured in hertz (Hz). Illuminance: Measured in lux (lx), not cd. Common Pitfalls: Mixing luminous intensity (cd) with luminous flux (lm) or illuminance (lx). Confusing radiometric units (W, W/sr) with photometric units (lm, cd) that include human eye response. Final Answer: luminous intensity

Question 2

In S.I. terminology, the joule (J) equals newton–metre (Nm). Identify the physical quantities for which these units are used in practice, noting the conventional use for torque as well.

Accepted Answer

Introduction / Context: In S.I., work and energy share the unit joule (J). Since 1 J = 1 Nm, the same dimensional product appears in torque calculations. However, conventionally engineers write torque in Nm (not J) to distinguish the vectorial, moment-like nature of torque from scalar energy or work, even though the dimensions match. Given Data / Assumptions: Work and energy are scalars measured in joules. Torque (moment of force) is a vector (or axial vector) commonly expressed as Nm. 1 J = 1 Nm in S.I. dimensional terms. Concept / Approach: Use the work definition W = F * s (component along displacement) and torque definition M = F * d (perpendicular distance). Both are force times distance, but they represent different physical concepts. Hence, the shared unit does not imply identical physical meaning. Step-by-Step Solution: Write work: W = F * s → unit Nm → called J in S.I.Write energy: Energy measured in J by definition (work-capacity equivalence).Write torque: M = F * d → unit Nm, conventionally not denoted as J to avoid confusion.Therefore, the units apply across work, energy, and torque, with naming conventions distinguishing torque. Verification / Alternative check: Dimensional analysis: N = kgm/s^2, so Nm = kgm^2/s^2. This dimension appears in both torque and energy computations, confirming the equivalence of units despite different physical interpretations. Why Other Options Are Wrong: Only work and energy: Excludes torque, which also uses Nm in practice. Common Pitfalls: Interchanging 'joule' with torque values in reports; best practice labels torque explicitly as Nm. Forgetting that torque’s direction depends on the right-hand rule, unlike scalar energy. Final Answer: All the above.

Question 3

In unit prefixes used across engineering and physics, the thickness called a 'micron' (micrometre) corresponds to which length in metres?

Accepted Answer

Introduction / Context: Engineers often encounter micro-scale dimensions in materials science, manufacturing tolerances, filtration, and biology. The term 'micron' is a common, older name for the micrometre, a standard S.I. submultiple of the metre that simplifies expression of very small lengths. Given Data / Assumptions: We must map 'micron' to a power-of-ten metre length. Common small-scale prefixes: milli (10^-3), micro (10^-6), nano (10^-9), pico (10^-12). Spelling note: micrometre (µm) is the S.I. term; micron is non-S.I. but widely used. Concept / Approach: Recall S.I. prefixes: micro (symbol µ) denotes 10^-6. Therefore, 1 micrometre = 1 µm = 10^-6 m. Many industrial specifications and scientific instruments reference µm for film thickness, surface roughness, and particle sizes. Step-by-Step Solution: Identify the relevant prefix: micro → 10^-6.Associate 'micron' with micrometre (µm).State equivalence: 1 µm = 10^-6 m → choose 10^-6 m. Verification / Alternative check: Cross-check common conversions: 1000 µm = 1 mm, since 10^-6 m * 1000 = 10^-3 m. This aligns with milli (10^-3), verifying internal consistency. Why Other Options Are Wrong: 10^-3 m: milli (mm), not micro. 10^-9 m: nano (nm). 10^-12 m: pico (pm). 10^-2 m: centi (cm) scale, far larger. Common Pitfalls: Writing 'micron' but using mm or nm values in calculations. Confusing micro (µ) with milli (m) in unit symbols, which can cause 1000× errors. Final Answer: 10^-6 m

Question 4

Identify the basic S.I. unit for thermodynamic temperature as used in modern engineering and science (historically referred to in some texts as 'degree kelvin').

Accepted Answer

Introduction / Context: Thermodynamic temperature is a fundamental base quantity in S.I. The modern S.I. unit name is 'kelvin' with symbol K (without the word 'degree'). Older textbooks and many exam banks still use the legacy phrasing 'degree kelvin' or 'absolute degree,' which can cause confusion during study or tests that expect the historical term. Given Data / Assumptions: The options reflect legacy naming conventions rather than strict modern S.I. usage. We must select the historically correct option that maps to the S.I. base unit 'kelvin (K)'. 'Centigrade' is an older name for Celsius; Fahrenheit and Rankine are non-S.I. Concept / Approach: Modern S.I.: base unit is kelvin (symbol K), not 'degree kelvin.' However, to remain consistent with the given options and many legacy question banks, the correct choice corresponds to that historical label. The physical concept is the same: absolute thermodynamic temperature measured from absolute zero. Step-by-Step Solution: Identify modern unit → kelvin (K).Map to given options → the intended equivalent is 'degree kelvin' (legacy phrasing).Select option that best matches S.I. base unit. Verification / Alternative check: Engineering references now write temperatures as, for example, T = 300 K (without the degree sign). Conversions: T(K) = T(°C) + 273.15; Rankine relates to Fahrenheit via R = °F + 459.67 (non-S.I.). Why Other Options Are Wrong: Degree fahrenheit/centigrade: Non-S.I. or secondary scales (Celsius is derived, not base). Degree absolute: Ambiguous legacy term. Rankine: Non-S.I. absolute scale based on Fahrenheit increments. Common Pitfalls: Using the degree symbol with kelvin (e.g., °K), which is incorrect in modern S.I. Confusing °C (relative to water’s properties) with absolute thermodynamic scale K. Final Answer: degree kelvin

Question 5

Select the correct set of conversion statements relating newton, kilogram-force, pressure in kgf/cm^2, and standard atmosphere (atm) to their S.I. equivalents.

Accepted Answer

Introduction / Context: Engineering calculations frequently involve converting between historical gravitational units (kgf, kgf/cm^2) and S.I. units (N, kN/m^2, Pa). Mastering a few canonical conversion constants prevents costly mistakes in structural, hydraulic, and machine design computations. Given Data / Assumptions: Standard gravity g0 = 9.80665 m/s^2. Force relation: 1 kgf = 9.80665 N; therefore 1 N ≈ 0.10197 kgf. Pressure: 1 kgf/cm^2 = 9.80665 * 10^4 N/m^2 = 98.0665 kN/m^2. Standard atmosphere: 1 atm = 101.325 kPa = 101.325 kN/m^2. Concept / Approach: Apply g0 to convert kgf ↔ N and scale area conversions carefully to move between cm^2 and m^2. Recognize 1 kPa = 1 kN/m^2 for pressure magnitudes in engineering usage. Step-by-Step Solution: From definition: 1 kgf = g0 N = 9.80665 N → invert to get 1 N ≈ 0.10197 kgf.For pressure: 1 cm^2 = 10^-4 m^2 → 1 kgf/cm^2 = 9.80665 N / 10^-4 m^2 = 98,066.5 N/m^2 = 98.0665 kN/m^2.Atmospheric pressure: 1 atm = 101,325 N/m^2 = 101.325 kN/m^2. Verification / Alternative check: Compare with common engineering rules of thumb: 1 kgf/cm^2 ≈ 0.980665 bar and 1 atm ≈ 1.01325 bar, showing consistency with the S.I. pascal scale (1 bar = 100 kPa). Why Other Options Are Wrong: Each A–D is individually correct; the most complete choice is E (All the above). Common Pitfalls: Forgetting the area conversion when moving from cm^2 to m^2, leading to 10^4 errors. Applying g0 to masses instead of forces when converting kgf to N. Final Answer: All the above.

Question 6

Identify the fundamental (base) mechanical units used in the S.I. system for length, mass, and time.

Accepted Answer

Introduction / Context: The S.I. system is built on seven base quantities; in mechanics, the core trio is length, mass, and time. Knowing their correct base units avoids confusion when checking dimensions, doing unit conversions, or comparing legacy literature (C.G.S., F.P.S.) to S.I. Given Data / Assumptions: We focus on mechanical bases only (L, M, T). Options mix metric and imperial magnitudes and different time scales. The correct S.I. base magnitudes are metre, kilogram, and second. Concept / Approach: Dimensional consistency relies on the base set. Any deviation (e.g., using centimetre or minute) leads to scaled systems (C.G.S.) or mixed units (non-S.I.). Step-by-Step Solution: List S.I. mechanical bases: m, kg, s.Screen options: C.G.S. → cm, g, s (not S.I. bases); F.P.S. → ft, lb, s (imperial); km, kg, min (scaled and mixed).Select the exact S.I. base trio → 'Metre, kilogram, second'. Verification / Alternative check: Check dimensional equations: force = mass * acceleration → N = kg * m/s^2, directly referencing the base trio. Why Other Options Are Wrong: Centimetre, gram, second: C.G.S. scaling, not S.I. Foot, pound, second: Imperial system, not S.I. Kilometre, kilogram, minute: Mixed/derived magnitudes; not base units. None of these: Incorrect because a correct S.I. base set is present. Common Pitfalls: Using minutes or hours in dynamics without converting to seconds, causing numerical errors. Confusing weight (force) with mass in base definitions. Final Answer: Metre, kilogram, second

Question 7

In S.I. units, the hertz (Hz) quantifies how many cycles occur per second. Identify the physical quantity measured in hertz.

Accepted Answer

Introduction / Context: The hertz (Hz) appears across many fields: alternating current systems, vibrations, acoustics, and wave phenomena. Correctly associating Hz with the underlying quantity ensures clear interpretation of time-dependent processes in engineering and physics. Given Data / Assumptions: Unit under consideration: hertz (Hz). By definition, 1 Hz = 1 cycle per second. Other options list different base or derived quantities with distinct units (A, W, cd, J). Concept / Approach: Frequency counts how often a periodic event repeats per unit time. This aligns precisely with cycles per second, the definition of hertz. Luminous intensity (cd), electric current (A), power (W), and energy (J) are unrelated quantities with different units. Step-by-Step Solution: Recall definition: frequency f = number of cycles / time.Unit mapping: f in s^-1, named hertz (Hz).Select the option 'frequency.' Verification / Alternative check: AC mains examples: 50 Hz or 60 Hz indicate 50 or 60 complete waveform cycles per second—textbook illustrations of frequency. Why Other Options Are Wrong: Luminous intensity: candela (cd). Electric current: ampere (A). Power: watt (W). Energy: joule (J). Common Pitfalls: Confusing angular frequency (rad/s) with cyclic frequency (Hz). Relation: omega = 2pif. Interpreting 'Hz' as amplitude rather than rate of repetition. Final Answer: frequency.

Question 8

By definition of the newton in S.I., one newton of force produces which acceleration on a body of specified mass?

Accepted Answer

Introduction / Context: The newton (N) is the S.I. unit of force. Its definition anchors many engineering calculations in statics and dynamics, ensuring that F = m * a has a clear numerical interpretation in base units. Selecting the correct pairing of mass and acceleration cements the meaning of 1 N in everyday calculations. Given Data / Assumptions: S.I. base units: kilogram (kg) for mass, metre (m) for length, second (s) for time. Newton's second law: F = m * a. 1 N corresponds to a particular acceleration produced on a 1 kg mass. Concept / Approach: From F = m * a, set F = 1 N and m = 1 kg to solve for a. This provides the definition of the newton in S.I. terms without resorting to gravitational units or centimetre–gram–second scales. Step-by-Step Solution: Start with F = m * a.Set F = 1 N and m = 1 kg.Compute a = F / m = 1 / 1 = 1 m/s^2.Therefore, 1 N produces 1 m/s^2 acceleration on a 1 kg mass. Verification / Alternative check: Dimensional analysis: N = kg*m/s^2, which rearranges to a = N/kg = m/s^2, confirming the numerical result for unit values. Why Other Options Are Wrong: 1 cm/s^2 cases: Use centimetre-based acceleration, not the S.I. metre per second squared; also mixes g and kg inconsistently. 1 m/s^2 on 1 g mass: Uses gram, a C.G.S. mass unit, not S.I. 10 m/s^2 on 1 kg: Would imply a 10 N force, not 1 N. Common Pitfalls: Switching to centimetres or grams mid-calculation, causing hidden 10× or 1000× errors. Confusing weight (in N) with mass (in kg) when applying F = m * a. Final Answer: 1 m/s^2 while acting on a body of 1 kg mass

Question 9

In the S.I. system used in engineering mechanics, identify the correct base unit for force (as used in F = m * a).

Accepted Answer

Introduction / Context: Force is one of the most fundamental quantities in mechanics. In the S.I. (International System of Units), consistent unit selection is crucial so that equations like F = m * a remain dimensionally correct and easy to apply across engineering problems ranging from structural analysis to fluid mechanics and machine design. Given Data / Assumptions: The S.I. base unit of mass is kilogram (kg) and of length is metre (m). Acceleration is measured in m/s^2. Force is defined by Newton’s second law: F = m * a. Energy and work use different units than force, even though they are related concepts. Concept / Approach: From Newton’s second law, force must carry units kg * m/s^2. The S.I. assigns a special name to this derived unit: the newton (symbol N). One newton is the force required to accelerate a mass of 1 kg at 1 m/s^2. Distinguishing force (newton) from energy (joule) and mass (kilogram) prevents common dimensional mistakes in calculations and design checks. Step-by-Step Solution: Write the governing relation: F = m * a.Insert S.I. units: m → kg; a → m/s^2; therefore F units → kg * m/s^2.Recognize that kg * m/s^2 is named “newton” (N) in the S.I. system. Verification / Alternative check: Check dimensional homogeneity in common formulas (e.g., work W = F * s has units N * m = joule), confirming that newton is a force unit distinct from joule (energy). Why Other Options Are Wrong: Kilograms: unit of mass, not force.Joule: unit of energy/work (N * m), not force.Erg: cgs unit of energy (1 erg = 10^-7 J), not S.I. force.Watt: unit of power (J/s), not force. Common Pitfalls: Confusing kilograms-force (kgf) with kilogram (mass) or with newton; in S.I., force should be expressed in newtons.Using energy units (J) when a force value (N) is required. Final Answer: Newton

Question 10

In the S.I. system, identify the physical quantity that uses m^2/s as its unit.

Accepted Answer

Introduction / Context: Recognizing units is a powerful shortcut in engineering exams. The unit m^2/s commonly appears in fluid mechanics when describing how momentum diffuses through a fluid, a property central to laminar versus turbulent behavior and to boundary-layer growth on surfaces. Given Data / Assumptions: Kinematic viscosity is denoted by nu and defined as dynamic viscosity mu divided by density rho. Dynamic viscosity mu in S.I. is Pa·s (or N·s/m^2), and density rho is kg/m^3. Other listed quantities have different standard S.I. units (e.g., pressure → N/m^2). Concept / Approach: Kinematic viscosity nu = mu / rho characterizes diffusivity of momentum. It determines Reynolds number Re = V * L / nu, which separates flow regimes and strongly influences drag, head loss, and heat/mass transfer coefficients. Because it is a ratio of viscosity to density, its S.I. unit simplifies to m^2/s, highlighting its “diffusivity-like” nature similar to thermal diffusivity (alpha) with units m^2/s. Step-by-Step Solution: Start with nu = mu / rho.Replace mu by N·s/m^2 and rho by kg/m^3: nu units = (N·s/m^2) / (kg/m^3).Use 1 N = 1 kg·m/s^2 to simplify: (kg·m/s^2 * s / m^2) / (kg/m^3) → (kg·m/(s·m^2)) / (kg/m^3) → (m/(s·m^2)) * (m^3/kg) * (kg) = m^2/s. Verification / Alternative check: Compare with water at 20°C: nu ≈ 1.0 × 10^-6 m^2/s (1 cSt), which matches the known magnitude and unit. Why Other Options Are Wrong: Surface tension: N/m.Pressure: N/m^2 (Pa).Power: watt (W = J/s = N·m/s).Specific weight: N/m^3. Common Pitfalls: Mixing dynamic viscosity (Pa·s) with kinematic viscosity (m^2/s).Confusing m^2/s with m/s (velocity) or m^2 (area). Final Answer: Kinematic viscosity

Question 11

Identify the correct S.I. statements: pressure, surface tension, force, and power expressed in standard units.

Accepted Answer

Introduction / Context: Unit literacy is essential for error-free engineering. This item cross-checks four standard S.I. expressions—pressure, surface tension, force, and power—that occur repeatedly in mechanics, fluids, and thermodynamics. Mastering them improves dimensional analysis and prevents formula misuse in longer numerical problems. Given Data / Assumptions: Pressure in S.I. is pascal (Pa) = N/m^2. Surface tension is force per unit length with unit N/m. Force is measured in newtons (N). Power is the rate of doing work or energy transfer per unit time: watt (W) = J/s. Concept / Approach: Each statement can be validated by definition: pressure = normal force / area; surface tension = tangential force along a line / length; force = mass * acceleration; power = work rate = (force * velocity) = J/s. Converting each to canonical S.I. form confirms correctness without needing numbers. Step-by-Step Solution: Pressure: P = F/A → N/m^2, hence Pa.Surface tension: T = F/length → N/m.Force: F = m * a → measured in newtons (kg·m/s^2).Power: P = dW/dt → W = J/s = N·m/s. Verification / Alternative check: Dimensional check: [Pressure] = [Force]/[Area] = (M·L/T^2)/(L^2) = M/(L·T^2); [Power] = [Energy]/T = (M·L^2/T^2)/T = M·L^2/T^3. Why Other Options Are Wrong: Any option that excludes one of the four correct statements is incomplete, since all four expressions are standard S.I. definitions. Common Pitfalls: Mixing N/m (surface tension) with N/m^2 (pressure).Confusing watt (power) with newton (force) or joule (energy). Final Answer: All the above statements are correct.

Question 12

Select the correct S.I. unit for surface tension (force per unit length at a liquid interface).

Accepted Answer

Introduction / Context: Surface tension quantifies the tangential force acting along a line at a liquid interface. It strongly affects capillarity, droplet formation, bubble stability, and wetting phenomena. Using the correct S.I. unit ensures formulas like rise in a capillary or pressure jump across a curved interface are applied consistently. Given Data / Assumptions: Surface tension is force per unit length. Force in S.I. is newton (N). Length in S.I. is metre (m). Concept / Approach: By definition, T = F / L. Therefore the unit must be N/m in S.I. This distinguishes surface tension from stress (N/m^2), density (kg/m^3), and dynamic viscosity (Pa·s). Remember this when decoding mixed-unit problem statements, especially those involving capillary tubes and contact angles. Step-by-Step Solution: State definition: T = F / L.Insert S.I. units: F → N; L → m; thus T → N/m.Confirm with typical values: water at room temperature ≈ 0.072 N/m. Verification / Alternative check: Check capillary-rise formula where surface tension multiplies a length and cosine term, yielding force units, consistent with T in N/m. Why Other Options Are Wrong: kg/m^3: density.kg/m^2 or kg/m: do not reduce to force/length.Pa·s: dynamic viscosity, not surface tension. Common Pitfalls: Confusing surface tension with pressure (N/m^2) due to both involving forces.Expecting a mass-based unit because many fluid properties use kg and m^3; surface tension is fundamentally force/length. Final Answer: N/m

Question 13

Electrical measurements: identify the correct S.I. unit for the intensity of electric current.

Accepted Answer

Introduction / Context: Electrical quantities have distinct meanings and S.I. units. Current (the rate of flow of electric charge) is foundational in circuit theory, power systems, and electronics. Using its correct unit keeps Ohm’s law, power equations, and continuity relations dimensionally consistent. Given Data / Assumptions: Current I is defined as charge flow per unit time: I = dQ/dt. Charge Q is measured in coulombs (C). Time is in seconds (s). Concept / Approach: From I = dQ/dt, the unit is C/s. The S.I. special name for C/s is ampere (A). Other electrical quantities—voltage (V), power (W), energy (J)—are related but not interchangeable. Correct unit selection simplifies analysis of DC/AC circuits, transient behavior, and instrumentation readings. Step-by-Step Solution: Start with definition: I = dQ/dt.Substitute units: Q → C; t → s; so I units → C/s.Recognize C/s as ampere (A): the S.I. base unit for electric current. Verification / Alternative check: Ohm’s law V = I * R: V (volt) = A * Ω. Power P = V * I = W, confirming consistent unit relations. Why Other Options Are Wrong: Volts: electric potential difference.Watts: power (rate of energy transfer).Joules: energy.Coulombs: quantity of charge, not rate of flow. Common Pitfalls: Confusing coulomb (charge) with ampere (current).Treating voltage as “strength of current,” which conflates potential difference with flow rate. Final Answer: Amperes

Question 14

Identify which fluid property in the S.I. system uses N·s/m^2 (Pa·s) as its unit.

Accepted Answer

Introduction / Context: Viscosity quantifies a fluid’s resistance to deformation. Two related but distinct properties appear in problems: dynamic viscosity (mu) and kinematic viscosity (nu). Distinguishing their units avoids errors in Reynolds number, head-loss predictions, and pump sizing calculations. Given Data / Assumptions: Dynamic viscosity mu relates shear stress to velocity gradient in Newtonian fluids: tau = mu * du/dy. Kinematic viscosity nu = mu / rho. Density rho is kg/m^3; specific weight gamma is N/m^3. Concept / Approach: From tau = mu * (du/dy), shear stress tau has units N/m^2 and du/dy has units 1/s. Therefore mu must have units (N/m^2) / (1/s) = N·s/m^2. The same unit is called pascal-second (Pa·s). Kinematic viscosity is a ratio of mu to rho and carries m^2/s, not N·s/m^2. Step-by-Step Solution: Write Newton’s law of viscosity: tau = mu * (du/dy).Rearrange units: mu = tau / (du/dy) → (N/m^2) / (1/s) = N·s/m^2 = Pa·s.Contrast with kinematic viscosity: nu = mu/rho → units m^2/s. Verification / Alternative check: Water at 20°C: mu ≈ 1.0 × 10^-3 Pa·s; nu ≈ 1.0 × 10^-6 m^2/s, confirming distinct magnitudes and units. Why Other Options Are Wrong: Kinematic viscosity: m^2/s.Mass density: kg/m^3.Specific weight: N/m^3.Surface tension: N/m. Common Pitfalls: Interchanging nu and mu; using Pa·s where m^2/s is required in Reynolds number.Dropping the “·s” in Pa·s, which changes the quantity entirely. Final Answer: Dynamic viscosity

Question 15

Stress in S.I. units: select the correct unit for normal or shear stress.

Accepted Answer

Introduction / Context: Stress is force per unit area and underpins the analysis of bars, beams, columns, shells, and pressure vessels. While many older texts used mass-based units (kgf/cm^2), the S.I. standardizes stress as pascals (Pa = N/m^2), enabling consistent integration with material properties like modulus (also in Pa). Given Data / Assumptions: Stress = Force / Area. Force in S.I. is the newton (N). Area uses square metres (m^2) in S.I. Concept / Approach: By definition, stress sigma or tau has the unit N/m^2 (pascal). Submultiples like MPa (10^6 Pa) and GPa (10^9 Pa) are common in structural design. Using “kg per square cm” is dimensionally inconsistent for stress in S.I. because kilogram is mass, not force; if using legacy gravitational units, “kgf/cm^2” would still not be S.I. Step-by-Step Solution: Write sigma = F / A.Insert S.I. units: F → N; A → m^2; therefore sigma unit = N/m^2.Relate to common design numbers: structural steels often have yield strengths ~250–500 MPa (i.e., 250 × 10^6 to 500 × 10^6 Pa). Verification / Alternative check: Check dimensional consistency with Hooke’s law: E (Pa) = stress/strain; strain is dimensionless, so stress must be in Pa as well. Why Other Options Are Wrong: kg per sq cm / kg per sq mm: mass/area, not force/area.N per sq mm / N per sq cm: although convertible to S.I., the question asks for S.I. unit, which is N/m^2 (Pa) in base form. Common Pitfalls: Confusing kgf with kg and mixing legacy units with S.I.Forgetting that MPa and GPa are standard multiples in engineering practice. Final Answer: Newton per sq m

SI Units Questions