SSRAO

Explanation:
For a Newtonian fluid, Newton's law of viscosity states: τ = μ(du/dy), where τ is shear stress (Pa), μ is dynamic viscosity (Pa·s), and du/dy is the velocity gradient (rate of shear strain, s⁻¹). This is a linear relationship — the defining characteristic of Newtonian fluids. Option B is for dilatant fluids (n>1), Option C is incorrect dimensionally, Option D has no physical basis.

Explanation:
Kinematic viscosity ν = μ/ρ, where μ is dynamic viscosity (Pa·s = kg/(m·s)) and ρ is density (kg/m³). Therefore ν = [kg/(m·s)] / [kg/m³] = m²/s. The CGS unit is Stoke (St) where 1 St = 10⁻⁴ m²/s. Options B, C, D are all units of dynamic viscosity μ, not kinematic viscosity.

Explanation:
Kinematic viscosity ν = μ/ρ = 0.048 / 800 = 6 × 10⁻⁵ m²/s. Step 1: ν = μ/ρ. Step 2: ν = 0.048 Pa·s ÷ 800 kg/m³. Step 3: ν = 0.000060 m²/s = 6 × 10⁻⁵ m²/s. Option B: 0.048 × 800 = 38.4 (wrong, multiplied). Option C: 800/0.048 (inverted). Option D: 0.048/1000 (wrong denominator).

Explanation:
Bulk modulus K = -V(dP/dV) = ρ(dP/dρ). The negative sign accounts for the fact that volume decreases when pressure increases (dV is negative when dP is positive), making K positive. For water at 20°C, K ≈ 2.2 GPa. For an ideal gas under isothermal conditions, K = P. Under adiabatic conditions, K = γP. Option B omits the negative sign. Option C omits V. Option D is the compressibility (1/K).

Explanation:
Surface tension arises from cohesive intermolecular forces. At the free surface, molecules experience a net inward force (unbalanced cohesive forces) since there are no liquid molecules above them. This creates a skin effect. Surface tension σ has units of N/m or J/m². It decreases with increasing temperature. Adhesive forces (Option B) cause capillary action but are not the cause of surface tension itself.

Explanation:
For a liquid droplet (single surface), the excess pressure is derived from force balance: ΔP × πr² = σ × 2πr, giving ΔP = 2σ/r. This is different from a soap bubble (two surfaces) where ΔP = 4σ/r. For a liquid jet (cylindrical): ΔP = σ/r. The formula ΔP = 2σ/r applies to spherical droplets. Option B (4σ/r) applies to soap bubbles in air.

Explanation:
Capillary rise formula: h = 2σcosθ/(ρgr). Derivation: Upward force = σ × 2πr × cosθ. Weight of liquid column = ρg × πr² × h. Equating: σ × 2πr × cosθ = ρg × πr² × h → h = 2σcosθ/(ρgr). For water in glass: θ ≈ 0°, cosθ = 1. For mercury in glass: θ ≈ 140°, cosθ < 0 (capillary depression). Option B misses factor of 2. Option C has extra factor of 2.

Explanation:
An ideal (perfect) fluid is defined as: (1) Incompressible (ρ = constant), (2) Inviscid (μ = 0, no viscosity), (3) No surface tension, (4) Flow is irrotational. Therefore, having viscosity is NOT a property of an ideal fluid — it is the opposite. Real fluids have viscosity. The concept of ideal fluid simplifies analysis using Bernoulli's equation without friction losses.

Explanation:
For liquids: Viscosity decreases with temperature because intermolecular cohesive forces weaken as temperature rises. For gases: Viscosity increases with temperature because viscosity in gases is due to molecular momentum transfer — higher temperature means more molecular activity and greater momentum exchange between layers. This is explained by Sutherland's law for gases.

Explanation:
Newton's law of viscosity: τ = μ(du/dy). For Newtonian fluids, shear stress τ is directly proportional to the rate of shear strain (velocity gradient du/dy), with dynamic viscosity μ as the constant of proportionality. The graph of τ vs du/dy is a straight line through the origin. Non-Newtonian fluids: Pseudoplastic (shear thinning), Dilatant (shear thickening), Bingham plastic (yield stress). Option B describes elastic solids. Option D describes dilatant fluids.

Explanation:
Compressibility β = 1/K, where K is the bulk modulus of elasticity. K = -V(dP/dV) = ρ(dP/dρ). Therefore β = 1/K = -dV/(V·dP). Compressibility has units of Pa⁻¹ or m²/N. For water: K ≈ 2.2 GPa, so β ≈ 4.55 × 10⁻¹⁰ Pa⁻¹. Liquids have very low compressibility (nearly incompressible). Gases are highly compressible. Option B is wrong (that is K itself). Option C is -dV/dP (missing V in denominator).

Explanation:
In a fluid at rest (static fluid), there is no relative motion between fluid layers, so the velocity gradient du/dy = 0. By Newton's law of viscosity: τ = μ(du/dy) = μ × 0 = 0. Therefore, shear stress is zero in a static fluid. This is the fundamental difference between fluids and solids — fluids cannot sustain shear stress at rest. Only normal (pressure) stresses exist in static fluids. This principle forms the basis of fluid statics.

Explanation:
Specific weight γ = ρg = 1000 kg/m³ × 9.81 m/s² = 9810 N/m³ ≈ 9.81 kN/m³. This is the weight per unit volume. Specific weight of mercury = 13600 × 9.81 = 133,416 N/m³. Note: Density ρ = 1000 kg/m³ (mass per unit volume). Specific weight γ = 9810 N/m³ (weight per unit volume). Specific gravity SG = 1.0 (dimensionless ratio). Option B is density in wrong units. Option C is g value.

Explanation:
Dilatant (shear-thickening) fluids: apparent viscosity increases with shear rate. Examples: cornstarch in water, wet sand. The flow curve is convex upward (n > 1 in power law τ = K(du/dy)^n). Pseudoplastic (shear-thinning): viscosity decreases with shear rate (n < 1). Examples: blood, paint, polymer solutions. Bingham plastic: requires yield stress before flowing (e.g., toothpaste, drilling mud). Thixotropic: viscosity decreases with time at constant shear rate.

Explanation:
Vapour pressure is the pressure exerted by the vapour of a liquid when it is in thermodynamic equilibrium with its liquid phase at a given temperature. At this pressure, the rate of evaporation equals the rate of condensation. Cavitation occurs when local pressure drops below vapour pressure. Vapour pressure increases with temperature. At boiling point, vapour pressure equals atmospheric pressure.

Explanation:
Cavitation occurs when the local pressure in a flowing liquid drops below the vapour pressure of the liquid at that temperature. This causes the liquid to vaporize locally, forming vapour bubbles (cavities). When these bubbles move to higher pressure regions, they collapse violently, causing noise, vibration, and erosion of surfaces. Cavitation is common in pumps, turbines, propellers, and control valves.

Explanation:
Newton's law of viscosity: τ = μ(du/dy), where τ = shear stress (Pa), μ = dynamic viscosity (Pa·s), du/dy = velocity gradient perpendicular to flow direction (s⁻¹). This law applies to Newtonian fluids. The velocity gradient du/dy represents the rate of angular deformation or rate of shear strain. Option B relates to pressure-driven flow. Option C relates to heat conduction (Fourier's law). Option D relates to mass diffusion (Fick's law).

Explanation:
Reynolds number Re = ρVL/μ = VL/ν = Inertia force / Viscous force. It determines whether flow is laminar (Re < 2300 for pipe flow) or turbulent (Re > 4000). Froude number Fr = V/√(gL) = Inertia/Gravity force. Mach number M = V/c = Inertia/Elastic force (compressibility). Weber number We = ρV²L/σ = Inertia/Surface tension force.

Explanation:
For a liquid droplet: ΔP = 2σ/r = 4σ/d. Given: σ = 0.073 N/m, d = 2 mm = 0.002 m, r = 0.001 m. ΔP = 2σ/r = 2 × 0.073 / 0.001 = 146 Pa. Alternatively: ΔP = 4σ/d = 4 × 0.073 / 0.002 = 146 Pa. Option B: 2σ/d = 73 Pa (used diameter instead of radius). Option C: 4σ/r = 292 Pa (soap bubble formula). Option D: σ/r = 73 Pa (missing factor of 2).

Explanation:
Dynamic viscosity of water at 20°C ≈ 1.002 × 10⁻³ Pa·s ≈ 1 mPa·s = 1 cP (centipoise). This is a standard reference value. At 0°C: μ ≈ 1.79 × 10⁻³ Pa·s. At 100°C: μ ≈ 0.28 × 10⁻³ Pa·s. Viscosity of water decreases significantly with temperature. Option B (10⁻⁶ Pa·s) is the kinematic viscosity of water (ν = μ/ρ = 10⁻³/1000 = 10⁻⁶ m²/s). Option C is 10 times too high. Option D is 1000 times too high.

Explanation:
Bingham plastic fluids require a minimum yield stress (τ₀) before they begin to flow. Once flowing, they behave like Newtonian fluids: τ = τ₀ + μ(du/dy). Examples: toothpaste, drilling mud, sewage sludge, some paints, mayonnaise. Water (Option B) is Newtonian. Air (Option C) is Newtonian gas. Honey (Option D) is Newtonian (viscous but no yield stress). The Bingham plastic model is important in drilling engineering and food processing.

Explanation:
Specific volume v = 1/ρ = Volume/Mass (m³/kg). It is the reciprocal of density. For water: v = 1/1000 = 0.001 m³/kg. Option B is density ρ (kg/m³). Option C is specific weight γ = ρg (N/m³). Option D is 1/γ = 1/(ρg) (m³/N). Specific volume is commonly used in thermodynamics for gases and steam tables.

Explanation:
Mercury on glass has a contact angle of approximately 135°-140°. Since θ > 90°, mercury does not wet glass — it shows capillary depression (mercury level inside tube is lower than outside). For water on glass: θ ≈ 0° (complete wetting, capillary rise). Contact angle < 90°: wetting liquid (capillary rise). Contact angle > 90°: non-wetting liquid (capillary depression). This explains why mercury thermometers show a convex meniscus.

Explanation:
Hydrostatic pressure equation: P = P₀ + ρgh, where P₀ = pressure at free surface (atmospheric), ρ = fluid density, g = gravitational acceleration, h = depth below free surface. This is derived from the hydrostatic equation: dP/dz = -ρg (pressure increases downward). The gauge pressure at depth h = ρgh. Absolute pressure = atmospheric + gauge = P₀ + ρgh. Option B would give pressure decreasing with depth (wrong). Option C ignores surface pressure.

Explanation:
Dynamic viscosity μ = ν × ρ. Density ρ = SG × ρ_water = 0.9 × 1000 = 900 kg/m³. μ = ν × ρ = 3 × 10⁻⁶ m²/s × 900 kg/m³ = 2.7 × 10⁻³ Pa·s. Step 1: ρ = 0.9 × 1000 = 900 kg/m³. Step 2: μ = 3 × 10⁻⁶ × 900 = 2700 × 10⁻⁶ = 2.7 × 10⁻³ Pa·s. Option B: ν/ρ (wrong, divided). Option C: used ρ=1 (wrong). Option D: used ρ=1000 (wrong density).

Explanation:
Capillary rise: h = 2σcosθ/(ρgr). From this formula: h ∝ 1/r (inversely proportional to radius). h ∝ σ (directly proportional to surface tension). h ∝ 1/ρ (inversely proportional to density). h ∝ cosθ (directly proportional to cosine of contact angle). Therefore, capillary rise is inversely proportional to the radius of the tube. This is why finer capillaries show greater rise — important in soil moisture movement and plant water transport.

Explanation:
For an ideal gas: PV = constant (isothermal). Differentiating: P·dV + V·dP = 0 → dP/dV = -P/V. Bulk modulus K = -V(dP/dV) = -V × (-P/V) = P. So K_isothermal = P. Under adiabatic conditions: PV^γ = constant → K_adiabatic = γP. For air at 1 atm: K_isothermal = 101325 Pa ≈ 0.1 MPa. Compare with water: K ≈ 2200 MPa — water is ~22,000 times less compressible than air at atmospheric pressure.

Explanation:
An incompressible fluid has infinite bulk modulus K → ∞. Since K = -V(dP/dV), if K → ∞, then dV → 0 for any finite dP, meaning volume does not change with pressure. Compressibility β = 1/K → 0 for incompressible fluids. Liquids are nearly incompressible (K ≈ 2.2 GPa for water). Gases are highly compressible (K = P ≈ 0.1 MPa at atmospheric pressure). In most hydraulic and liquid flow problems, liquids are treated as incompressible.

Explanation:
Surface tension of water at 20°C ≈ 0.0728 N/m ≈ 0.073 N/m. This is a standard reference value. Surface tension decreases with temperature: at 0°C ≈ 0.076 N/m, at 100°C ≈ 0.059 N/m. Surface tension of mercury at 20°C ≈ 0.480 N/m (much higher). Surface tension of alcohol ≈ 0.022 N/m (lower than water). Adding surfactants (soap) reduces surface tension significantly. Option B is 10× too high. Option C is 100× too high. Option D is 10× too low.

Explanation:
In gases, viscosity arises from molecular momentum transfer (diffusion of momentum) between adjacent layers moving at different velocities. Faster-moving molecules migrate to slower layers, transferring momentum and creating a drag force. This is why gas viscosity increases with temperature (more molecular activity). In liquids, viscosity is primarily due to cohesive intermolecular forces, which decrease with temperature. This fundamental difference explains opposite temperature dependence of viscosity in liquids vs gases.

Explanation:
Reynolds number Re = ρVD/μ = VD/ν, where ρ = density (kg/m³), V = mean velocity (m/s), D = pipe diameter (m), μ = dynamic viscosity (Pa·s), ν = kinematic viscosity (m²/s). Re is dimensionless: [kg/m³ × m/s × m] / [kg/(m·s)] = dimensionless. For pipe flow: Re < 2300 → laminar; 2300 < Re < 4000 → transition; Re > 4000 → turbulent. Option B has μ and ρ swapped. Option C has D in denominator. Option D has D².

Explanation:
Specific gravity (SG) or relative density = ρ_fluid / ρ_water at 4°C. The reference is water at 4°C because water has its maximum density (1000 kg/m³) at 4°C. SG is dimensionless. For mercury: SG = 13600/1000 = 13.6. For oil: SG ≈ 0.8-0.9. For seawater: SG ≈ 1.025. Using 4°C water as reference ensures a consistent, temperature-independent standard. Option B (20°C water) has density 998.2 kg/m³, not exactly 1000.

Explanation:
Shear stress τ = μ(du/dy) = μ × V/h (for linear velocity profile). τ = 0.02 × 0.5/0.002 = 0.02 × 250 = 5 Pa. Force F = τ × A = 5 × 2 = 10 N. Step 1: du/dy = V/h = 0.5/0.002 = 250 s⁻¹. Step 2: τ = μ × (du/dy) = 0.02 × 250 = 5 Pa. Step 3: F = τ × A = 5 × 2 = 10 N. Option B: F = τ only (forgot area). Option C: doubled (error in h). Option D: halved (error).

Explanation:
In Couette flow (one plate moving, one stationary, no pressure gradient), the velocity profile is linear: u(y) = V × y/h, where V is the velocity of the moving plate and h is the gap. This gives a constant velocity gradient du/dy = V/h throughout the gap, resulting in uniform shear stress τ = μV/h. Parabolic profile (Option B) occurs in Poiseuille flow (pressure-driven flow between stationary plates). Logarithmic profile occurs in turbulent boundary layers.

Explanation:
Surface tension σ has units of N/m (force per unit length) or equivalently J/m² (energy per unit area). These are equivalent: N/m = N·m/m² = J/m². Surface tension represents the energy required to increase the surface area by unit amount, or the force per unit length acting along the surface. Option B (N/m²) is pressure (Pa). Option C (N·m) is energy (J) or torque. Option D (N/m³) is specific weight. The CGS unit of surface tension is dyne/cm (1 N/m = 1000 dyne/cm).

Explanation:
Viscosity is the property of a fluid that measures its resistance to deformation or flow. It represents internal friction between fluid layers moving at different velocities. Dynamic viscosity μ (Pa·s) quantifies this resistance. High viscosity fluids (honey, glycerin) resist flow more than low viscosity fluids (water, air). Viscosity is the cause of energy loss in fluid flow. Surface tension (B) is a surface phenomenon. Compressibility (C) relates to volume change under pressure. Capillarity (D) is the rise/fall in tubes.

Explanation:
Dynamic viscosity μ = ν × ρ = 1.5 × 10⁻⁵ × 1.2 = 1.8 × 10⁻⁵ Pa·s. This corresponds approximately to air at 20°C (ν_air ≈ 1.5 × 10⁻⁵ m²/s, ρ_air ≈ 1.2 kg/m³, μ_air ≈ 1.8 × 10⁻⁵ Pa·s). Step 1: μ = ν × ρ. Step 2: μ = 1.5 × 10⁻⁵ × 1.2 = 1.8 × 10⁻⁵ Pa·s. Option B: ν/ρ = 1.5×10⁻⁵/1.2 = 1.25×10⁻⁵ (divided instead of multiplied). Option C: ν itself (forgot to multiply by ρ). Option D: wrong power of 10.

Explanation:
Capillary depression (liquid level inside tube is lower than outside) occurs when the contact angle θ > 90°. In this case, cosθ < 0, making h = 2σcosθ/(ρgr) negative, indicating depression. Example: mercury in glass (θ ≈ 140°). Capillary rise occurs when θ < 90° (cosθ > 0). Example: water in glass (θ ≈ 0°). At θ = 90°: cosθ = 0, no capillary effect. At θ = 0°: complete wetting, maximum capillary rise. The sign of cosθ determines rise or depression.

Explanation:
Mach number M = V/c, where V = flow velocity and c = speed of sound in the medium. c = √(γRT) for ideal gas, where γ = ratio of specific heats, R = gas constant, T = absolute temperature. M < 1: subsonic flow. M = 1: sonic flow. M > 1: supersonic flow. M > 5: hypersonic flow. Mach number represents the ratio of inertia force to elastic (compressibility) force. Option B is 1/M. Option C is incorrect (light speed is irrelevant).

Explanation:
A soap bubble has two surfaces (inner and outer film). For each surface: ΔP = 2σ/r. Total excess pressure = 2 × (2σ/r) = 4σ/r. Derivation: Work done by pressure = Work done against surface tension. ΔP × 4πr² × dr = σ × 2 × 8πr × dr (two surfaces). ΔP = 4σ/r. For a liquid droplet (one surface): ΔP = 2σ/r. For a liquid jet (cylindrical): ΔP = σ/r. The factor of 2 difference between droplet and soap bubble is due to the two surfaces of the soap film.

Explanation:
Boiling point is the temperature at which vapour pressure of a liquid equals the external (atmospheric) pressure. At this point, bubbles of vapour form throughout the liquid. Normal boiling point: temperature at which vapour pressure = 1 atm (101.325 kPa). For water: 100°C at 1 atm. At higher altitudes (lower pressure), water boils at lower temperatures. In pressure cookers (higher pressure), water boils at higher temperatures. Melting point (B) is solid-liquid transition. Critical temperature (C) is above which gas cannot be liquefied.

Explanation:
Specific weight γ = ρg = 850 × 9.81 = 8338.5 N/m³. Step 1: γ = ρ × g. Step 2: γ = 850 kg/m³ × 9.81 m/s² = 8338.5 N/m³ ≈ 8.34 kN/m³. Option B: density itself (forgot to multiply by g). Option C: 850/9.81 = 86.7 (divided instead of multiplied). Option D: 850 × 10 = 8500 (used g = 10 m/s² approximately). Specific weight of water = 1000 × 9.81 = 9810 N/m³. Specific weight of mercury = 13600 × 9.81 = 133,416 N/m³.

Explanation:
Surface tension enables a fluid to resist tensile stress at its free surface. The surface acts like a stretched elastic membrane that can support small loads. This is why small insects can walk on water and why a needle can float on water surface. Surface tension creates a tensile force along the surface. Bulk modulus (B) resists compression. Viscosity (C) resists shear deformation. Compressibility (D) is the inverse of bulk modulus. Surface tension is the only fluid property that provides tensile resistance.

Explanation:
Shear stress τ = μ × (du/dy) = 0.01 Pa·s × 100 s⁻¹ = 1 Pa. Step 1: τ = μ × (du/dy). Step 2: τ = 0.01 × 100 = 1 Pa. This is a direct application of Newton's law of viscosity. Option B: 0.01 × 100 × 10 = 10 (extra factor). Option C: 0.01 × 100 / 10 = 0.1 (divided by 10). Option D: 100 (used only velocity gradient, forgot μ). Units check: Pa·s × s⁻¹ = Pa.

Explanation:
Kinematic viscosity ν = μ/ρ, where μ = dynamic viscosity (Pa·s) and ρ = density (kg/m³). Units: ν = [Pa·s]/[kg/m³] = m²/s. CGS unit: Stoke (St), 1 St = 1 cm²/s = 10⁻⁴ m²/s. 1 centistoke (cSt) = 10⁻⁶ m²/s. Kinematic viscosity of water at 20°C ≈ 1 × 10⁻⁶ m²/s = 1 cSt. Kinematic viscosity of air at 20°C ≈ 15 × 10⁻⁶ m²/s. Option D (absolute viscosity) is another name for dynamic viscosity μ.

Explanation:
From Newton's law: τ = μ(du/dy) → μ = τ/(du/dy) = 5/50 = 0.1 Pa·s. Step 1: μ = τ/(du/dy). Step 2: μ = 5 Pa / 50 s⁻¹ = 0.1 Pa·s. This is a moderately viscous fluid (water is 0.001 Pa·s, glycerin is ~1.5 Pa·s). Option B: τ × (du/dy) = 5 × 50 = 250 (multiplied instead of divided). Option C: 50/5 = 10 (inverted ratio). Option D: 5/500 = 0.01 (wrong denominator). Units: Pa / s⁻¹ = Pa·s.

Explanation:
Gauge pressure P_gauge = ρgh = 1000 × 9.81 × 5 = 49050 Pa ≈ 49.05 kPa. Step 1: P = ρgh. Step 2: P = 1000 × 9.81 × 5 = 49050 Pa. This is the pressure above atmospheric. Absolute pressure = 49050 + 101325 = 150375 Pa ≈ 1.5 atm. Option B: ρ × h = 1000 × 5 = 5000 (forgot g). Option C: g × h = 9.81 × 5 = 49.05 (forgot ρ, wrong units). Option D: 1000 × 10 × 5 = 50000 (used g = 10).

Explanation:
Blood is a non-Newtonian fluid — it is pseudoplastic (shear-thinning) at low shear rates and shows Bingham plastic behavior at very low shear rates due to red blood cell aggregation. Blood viscosity decreases with increasing shear rate. Water (B), Air (C), and Mercury (D) are all Newtonian fluids with constant viscosity at given temperature and pressure. Other non-Newtonian fluids: paint, polymer solutions, ketchup, toothpaste, drilling mud, quicksand, cornstarch suspension.

Explanation:
Speed of sound c = √(γRT) = √(γP/ρ). These are equivalent expressions: Since P = ρRT (ideal gas law), γRT = γP/ρ. So both A and C are correct, making D the answer. Here γ = Cp/Cv (ratio of specific heats), R = specific gas constant (J/kg·K), T = absolute temperature (K), P = pressure (Pa), ρ = density (kg/m³). For air at 20°C: c = √(1.4 × 287 × 293) ≈ 343 m/s. Option B (√RT) is missing γ.

Explanation:
A perfectly wetting liquid has a contact angle θ = 0°. This means the liquid spreads completely over the solid surface. cosθ = cos0° = 1 (maximum value), giving maximum capillary rise: h = 2σ/(ρgr). Examples: water on clean glass (θ ≈ 0°), most liquids on high-energy surfaces. θ = 90°: neutral wetting (no capillary effect). θ = 180°: perfect non-wetting (theoretical). θ > 90°: non-wetting (capillary depression). The lotus effect uses superhydrophobic surfaces with θ > 150°.

Explanation:
Dynamic viscosity μ: τ = μ(du/dy). [μ] = [τ]/[du/dy] = [Force/Area] / [Velocity/Length] = [MLT⁻²/L²] / [LT⁻¹/L] = [ML⁻¹T⁻²] / [T⁻¹] = ML⁻¹T⁻¹. In SI: kg/(m·s) = Pa·s. Verification: Pa·s = N/m² × s = kg·m/s² / m² × s = kg/(m·s) = ML⁻¹T⁻¹. Option B (ML⁻¹T⁻²) is pressure [Pa]. Option C (M⁰L²T⁻¹) is kinematic viscosity [m²/s]. Option D (ML²T⁻²) is energy [J].

Explanation:
Kinematic viscosity ν = μ/ρ. [ν] = [ML⁻¹T⁻¹] / [ML⁻³] = L²T⁻¹. In SI: m²/s. Verification: m²/s = L²T⁻¹. CGS unit: Stoke (St) = cm²/s. 1 m²/s = 10⁴ St = 10⁶ cSt. Option B (ML⁻¹T⁻¹) is dynamic viscosity. Option C (ML⁻¹T⁻²) is pressure. Option D (LT⁻¹) is velocity. Kinematic viscosity is used in Reynolds number: Re = VL/ν (avoids need to know both μ and ρ separately).

Explanation:
For a cylindrical liquid jet (one curved surface, radius r = d/2): ΔP = σ/r = σ/(d/2) = 2σ/d. Derivation: Force balance on half-cylinder of length L: ΔP × 2rL = σ × 2L (two edges). ΔP = σ/r = 2σ/d. Compare: Spherical droplet: ΔP = 2σ/r = 4σ/d (two principal radii of curvature). Soap bubble: ΔP = 4σ/r = 8σ/d (two surfaces). Liquid jet has only one radius of curvature (infinite in axial direction), giving ΔP = σ/r = 2σ/d.

Explanation:
Viscosity is measured using a viscometer (or viscosimeter). Types: (1) Capillary viscometer (Ostwald) — measures flow time through capillary. (2) Falling sphere viscometer — measures terminal velocity of sphere. (3) Rotational viscometer (Brookfield) — measures torque on rotating cylinder. (4) Saybolt viscometer — measures time for fixed volume to flow. Manometer measures pressure. Barometer measures atmospheric pressure. Piezometer measures gauge pressure in liquids.

Explanation:
Archimedes' principle: A body immersed in a fluid experiences an upward buoyant force equal to the weight of fluid displaced. F_buoyancy = ρ_fluid × g × V_displaced. Apparent weight = Actual weight - Buoyant force = W - F_b. This principle is used in: ship design, submarine ballast, hydrometer (measures SG), hot air balloons. For a floating body: Buoyant force = Weight of body (equilibrium). For a submerged body: F_b = ρ_fluid × g × V_body.

Explanation:
Surface tension causes liquids to minimize their free surface area. This is because surface molecules have higher energy than interior molecules (unbalanced cohesive forces). Minimizing surface area minimizes surface energy. This explains: spherical shape of droplets (minimum surface area for given volume), spherical soap bubbles, beading of water on wax. Surface tension σ = dE/dA (energy per unit area = J/m² = N/m). Capillarity (C) is a consequence of surface tension, not the property itself.

Explanation:
ρ = SG × 1000 = 0.8 × 1000 = 800 kg/m³. μ = ν × ρ = 2 × 10⁻⁵ × 800 = 1.6 × 10⁻² Pa·s. Step 1: ρ = 0.8 × 1000 = 800 kg/m³. Step 2: μ = ν × ρ = 2 × 10⁻⁵ × 800 = 16 × 10⁻³ = 1.6 × 10⁻² Pa·s. Option B: ν/ρ = 2×10⁻⁵/800 = 2.5×10⁻⁸ (wrong). Option C: ν × 0.8 = 1.6×10⁻⁵ (used SG instead of density). Option D: ν × 1000 = 2×10⁻² (used water density).

Explanation:
Capillarity (capillary action) is the phenomenon of rise or fall of liquid in a narrow tube due to the combined effect of surface tension and adhesion/cohesion. Rise occurs when adhesion > cohesion (wetting liquid, θ < 90°). Fall (depression) occurs when cohesion > adhesion (non-wetting liquid, θ > 90°). Capillary rise h = 2σcosθ/(ρgr). Applications: rise of water in soil, movement of sap in plants, ink in pen nibs, oil in lamp wicks. Surface tension (C) is the cause; capillarity is the effect.

Explanation:
Absolute pressure = Gauge pressure + Atmospheric pressure. P_abs = P_gauge + P_atm. Gauge pressure is measured relative to atmospheric pressure. P_gauge > 0: pressure above atmospheric. P_gauge < 0: vacuum (below atmospheric), called vacuum pressure. Standard atmospheric pressure = 101.325 kPa = 1.01325 bar = 760 mmHg = 10.33 m of water. Absolute pressure is always positive. Vacuum pressure = P_atm - P_abs (positive value for pressures below atmospheric).

Explanation:
In gases, viscosity is due to molecular momentum transfer between layers. At higher temperatures, gas molecules have greater kinetic energy and move faster, increasing the rate of momentum exchange between adjacent layers. This increases the apparent viscosity. Sutherland's law: μ = μ₀(T/T₀)^(3/2) × (T₀+S)/(T+S), where S is Sutherland's constant. For air: μ increases from 1.71×10⁻⁵ Pa·s at 0°C to 2.17×10⁻⁵ Pa·s at 100°C. Option B is wrong — cohesive forces decrease with temperature.

Explanation:
Stokes' law: F_drag = 3πμDV = 6πμrV, where μ = dynamic viscosity, D = diameter, r = radius, V = velocity. This applies for Re < 1 (creeping flow). Derivation: F = CD × (π/4)D² × (ρV²/2), where CD = 24/Re for Stokes flow. F = (24μ/ρVD) × (π/4)D² × (ρV²/2) = 3πμDV. Applications: settling of particles, viscometer (falling sphere), sedimentation. Terminal velocity: V_t = D²(ρ_s - ρ_f)g/(18μ).

Explanation:
Hydrostatic equation: dP/dz = -ρg, where z is measured upward. The negative sign indicates pressure decreases as elevation increases. Integrating: P₂ - P₁ = -ρg(z₂ - z₁) = ρg(z₁ - z₂) = ρgh (where h = z₁ - z₂ = depth). This is derived from equilibrium of a fluid element: ΣF_z = 0 → (P - (P + dP))dA - ρg·dA·dz = 0 → dP/dz = -ρg. For incompressible fluid (ρ = constant): P = P₀ + ρgh.

Explanation:
From Stokes' law, terminal velocity: V_t = D²(ρ_s - ρ_f)g/(18μ). Therefore V_t ∝ D². Doubling the diameter increases terminal velocity by 4 times. This is valid for Re < 1 (Stokes regime). For larger particles (Newton's regime, 1000 < Re < 200000): V_t ∝ D^(1/2). For intermediate regime (Allen's regime): V_t ∝ D^(3/2). The D² dependence in Stokes regime is important in sedimentation, dust settling, and centrifugation.

Explanation:
Weber number We = ρV²L/σ = Inertia force / Surface tension force. It is important in problems involving free surface flows, droplet formation, bubble dynamics, and spray atomization. We >> 1: inertia dominates (surface tension negligible). We << 1: surface tension dominates. Reynolds number (B) = Inertia/Viscous. Froude number (C) = Inertia/Gravity. Bond number (D) = Gravity/Surface tension = ρgL²/σ. Weber number is named after Moritz Weber.

Explanation:
τ = μ(V/h) = 0.9 × (1.5/0.003) = 0.9 × 500 = 450 Pa. F = τ × A = 450 × 0.5 = 225 N. Step 1: du/dy = V/h = 1.5/0.003 = 500 s⁻¹. Step 2: τ = 0.9 × 500 = 450 Pa. Step 3: F = 450 × 0.5 = 225 N. Option B: τ × A = 450 × 1 = 450 (used A=1). Option C: 225/2 = 112.5 (halved). Option D: 225 × 3 = 675 (tripled). Units: N·s/m² × m/s / m × m² = N.

Explanation:
Pascal's law states that pressure at a point in a static fluid is equal in all directions (isotropic). This is proved by analyzing a triangular fluid element and showing P_x = P_y = P_z = P. Pascal's law is the basis of: hydraulic press, hydraulic brakes, hydraulic jack, hydraulic lift. Mechanical advantage: F₂/F₁ = A₂/A₁. Archimedes' principle (B) relates to buoyancy. Bernoulli's theorem (C) applies to flowing fluids. Newton's law (D) relates shear stress to velocity gradient.

Explanation:
Density of air at standard conditions (ISA: 15°C = 288.15 K, 101.325 kPa): ρ = P/(RT) = 101325/(287 × 288.15) = 101325/82699 = 1.225 kg/m³. This is the International Standard Atmosphere (ISA) value. At 0°C: ρ = 1.293 kg/m³. At 20°C: ρ = 1.204 kg/m³. At 100°C: ρ = 0.946 kg/m³. Air density decreases with temperature and altitude. At 10 km altitude: ρ ≈ 0.414 kg/m³. Option B (1.0) is too low. Option C (1.5) is too high.

Explanation:
Froude number Fr = V/√(gL) = √(Inertia force / Gravity force). It is important in free surface flows (open channels, ships, hydraulic jumps). Fr < 1: subcritical (tranquil) flow. Fr = 1: critical flow. Fr > 1: supercritical (rapid) flow. Named after William Froude. Reynolds number (B) = Inertia/Viscous. Weber number (C) = Inertia/Surface tension. Option D has no standard name. Froude number is used in ship model testing and hydraulic model studies.

Explanation:
Bulk modulus of water at 20°C ≈ 2.2 GPa = 2.2 × 10⁹ Pa. This high value confirms water is nearly incompressible. Compressibility β = 1/K = 1/(2.2×10⁹) ≈ 4.55 × 10⁻¹⁰ Pa⁻¹. For comparison: Steel K ≈ 160 GPa (much less compressible). Air at 1 atm (isothermal): K = P ≈ 0.1 MPa (22,000 times more compressible than water). The high bulk modulus of water is why water hammer (pressure surge) can be destructive in pipelines. Option B is 1000× too low. Option C is 10⁶× too low.

Explanation:
The no-slip condition states that the fluid velocity at a solid boundary equals the velocity of that boundary. If the boundary is stationary, fluid velocity = 0 at the wall. If the boundary moves at velocity V, fluid velocity = V at the wall. This is an empirical observation valid for most practical flows. It is the basis for: velocity profiles in pipes and channels, boundary layer theory, viscous drag calculations. Exception: rarefied gas flows (Knudsen number > 0.01) where slip occurs.

Explanation:
1 Poise (P) = 0.1 Pa·s = 1 g/(cm·s) = 1 dyne·s/cm². Therefore: 2 Poise = 2 × 0.1 = 0.2 Pa·s. Conversion: 1 Pa·s = 10 Poise. 1 centipoise (cP) = 0.001 Pa·s = 1 mPa·s. Water at 20°C: μ ≈ 1 cP = 0.001 Pa·s = 0.01 Poise. The Poise is named after Jean Léonard Marie Poiseuille. Option B: 2 Pa·s (forgot conversion factor). Option C: 20 Pa·s (multiplied by 10 instead of dividing). Option D: 0.02 Pa·s (divided by 100).

Explanation:
Density = SG × ρ_water = 7.8 × 1000 = 7800 kg/m³. This corresponds approximately to steel (SG ≈ 7.8-8.0). Other reference values: Aluminum: SG ≈ 2.7 (2700 kg/m³). Copper: SG ≈ 8.9 (8900 kg/m³). Lead: SG ≈ 11.3 (11300 kg/m³). Gold: SG ≈ 19.3 (19300 kg/m³). Ice: SG ≈ 0.917 (917 kg/m³). Wood (oak): SG ≈ 0.7 (700 kg/m³). Option B: 7800/10 = 780 (divided by 10). Option C: 7800/100 = 78. Option D: SG value itself.

Explanation:
τ = μ × (du/dy) = 0.5 × 200 = 100 Pa. Direct application of Newton's law of viscosity. Step 1: τ = μ × (du/dy). Step 2: τ = 0.5 Pa·s × 200 s⁻¹ = 100 Pa. Option B: 2 × 0.5 × 200 = 400 (extra factor of 2). Option C: 0.5/200 = 0.0025 (divided instead of multiplied). Option D: 0.5 × 200 × 2.5 = 250 (wrong). Units: Pa·s × s⁻¹ = Pa.

Explanation:
P = ρgh → h = P/(ρg). ρ = 0.85 × 1000 = 850 kg/m³. h = 101325 / (850 × 9.81) = 101325 / 8338.5 = 12.15 m. Step 1: ρ = 0.85 × 1000 = 850 kg/m³. Step 2: h = 101325/(850×9.81) = 101325/8338.5 = 12.15 m. For water (SG=1): h = 101325/9810 = 10.33 m. For mercury (SG=13.6): h = 101325/133416 = 0.76 m = 760 mm. Option C is water column height. Option B: 101325/(1000×9.81×1.2) = wrong.

Explanation:
Bingham plastic model: τ = τ₀ + μ_p(du/dy) when τ > τ₀ (yield stress). No flow when τ < τ₀. τ₀ = yield stress (Pa), μ_p = plastic viscosity (Pa·s). Examples: toothpaste, drilling mud, sewage sludge, some paints, blood at very low shear rates. Option B is Newtonian fluid (τ₀ = 0, n = 1). Option C is power-law (Ostwald-de Waele) model: n < 1 pseudoplastic, n > 1 dilatant, n = 1 Newtonian. Option D is dimensionally incorrect. Bingham plastic is named after Eugene Bingham (1878-1945).

Explanation:
h = 2σcosθ/(ρgr) = 2 × 0.073 × cos0° / (1000 × 9.81 × 0.0005). r = 0.5 mm = 0.0005 m. h = 2 × 0.073 × 1 / (1000 × 9.81 × 0.0005) = 0.146 / 4.905 = 0.02977 m ≈ 29.8 mm. Step 1: r = d/2 = 0.5 mm = 5×10⁻⁴ m. Step 2: h = 2×0.073×1/(1000×9.81×5×10⁻⁴) = 0.146/4.905 = 0.0298 m = 29.8 mm. Option B: h/2 = 14.9 (used d instead of r). Option C: 2h = 59.6 (doubled). Option D: h/4 = 7.45 (quartered).

Explanation:
1 Pa·s = 1000 cP (centipoise). Therefore: 1.5 × 10⁻³ Pa·s = 1.5 × 10⁻³ × 1000 cP = 1.5 cP. This is close to water viscosity at 20°C (≈ 1 cP). Conversion: 1 cP = 10⁻³ Pa·s = 1 mPa·s. 1 P (Poise) = 100 cP = 0.1 Pa·s. Option B: 1.5 × 10 = 15 (off by factor 10). Option C: 1.5/10 = 0.15 (off by factor 10). Option D: 1.5 × 100 = 150 (off by factor 100). The centipoise is a convenient unit because water ≈ 1 cP.

Explanation:
ρ_seawater = 1.025 × 1000 = 1025 kg/m³. P = ρgh = 1025 × 9.81 × 10 = 100552.5 Pa ≈ 100.5 kPa. Step 1: ρ = 1.025 × 1000 = 1025 kg/m³. Step 2: P = 1025 × 9.81 × 10 = 100,552.5 Pa ≈ 100.5 kPa. For fresh water: P = 1000 × 9.81 × 10 = 98.1 kPa. The difference is due to higher density of seawater. Option B: fresh water pressure (SG=1). Option C: atmospheric pressure (coincidental). Option D: 1.025 × 101.3 = 103.8 (wrong method).

Explanation:
τ = μ(V/h) = 0.6 × (3/0.004) = 0.6 × 750 = 450 Pa. F = τ × A = 450 × 0.4 = 180 N. Step 1: du/dy = V/h = 3/0.004 = 750 s⁻¹. Step 2: τ = 0.6 × 750 = 450 Pa. Step 3: F = 450 × 0.4 = 180 N. Option B: 180/4 = 45 (divided by 4). Option C: 180 × 4 = 720 (multiplied by 4). Option D: 180/2 = 90 (halved). Power = F × V = 180 × 3 = 540 W.

Explanation:
Mach number M = V/c, where V = flow speed and c = speed of sound. M < 1: subsonic. M = 1: sonic (transonic). 1 < M < 5: supersonic. M > 5: hypersonic. Speed of sound in air at 20°C: c = √(γRT) = √(1.4 × 287 × 293) ≈ 343 m/s. Mach number determines compressibility effects: for M < 0.3, flow can be treated as incompressible. Named after Ernst Mach (1838-1916). Reynolds number (B) = V·L/ν. Froude number (C) = V/√(gL). Strouhal number (D) = fL/V.

Explanation:
A hydrometer measures specific gravity (relative density) of liquids by measuring the depth to which it sinks in the liquid. By Archimedes' principle, the hydrometer floats at a depth where the weight of displaced liquid equals the weight of the hydrometer. Higher SG liquid → hydrometer sinks less (reads higher). Lower SG liquid → hydrometer sinks more (reads lower). Applications: battery acid (SG ≈ 1.28), alcohol content in beverages, milk quality testing. Manometer (B) measures pressure. Viscometer (C) measures viscosity. Barometer (D) measures atmospheric pressure.

Explanation:
A soap bubble has TWO surfaces (inner and outer), so the excess pressure = 4σ/r = 4σ/(d/2) = 8σ/d. For a soap bubble: ΔP = 4σ/r = 4σ/(d/2) = 8σ/d. For a liquid droplet (one surface): ΔP = 2σ/r = 4σ/d. The factor of 2 difference is because soap bubble has two surfaces. Option B (4σ/d) is for a liquid droplet. Option C (2σ/d) is for a liquid jet. Option D (σ/d) is incorrect.

Explanation:
ρ = γ/g = 8000/9.81 = 815.5 kg/m³. Step 1: γ = ρg → ρ = γ/g. Step 2: ρ = 8000/9.81 = 815.5 kg/m³. This corresponds to a light oil (SG ≈ 0.816). Option B: γ itself (forgot to divide by g). Option C: γ × g = 8000 × 9.81 = 78480 (multiplied instead of divided). Option D: 8000/10 = 800 (used g = 10 m/s²). Specific weight γ = ρg, so ρ = γ/g. Always divide by g to get density from specific weight.

Explanation:
ν = μ/ρ = 0.25/900 = 2.78 × 10⁻⁴ m²/s. Step 1: ν = μ/ρ. Step 2: ν = 0.25/900 = 0.000278 = 2.78 × 10⁻⁴ m²/s. This is about 278 cSt (centistokes) — a moderately viscous oil. Option B: μ × ρ = 0.25 × 900 = 225 (multiplied instead of divided). Option C: 0.25 × 900/62500 = 3.6×10⁻³ (wrong). Option D: 0.25/90 = 2.78×10⁻³ (wrong denominator). Units: Pa·s / (kg/m³) = m²/s.

Explanation:
In a static fluid, pressure difference depends only on the vertical distance (depth difference) between the two points: ΔP = ρg·Δh. Horizontal distance has no effect on pressure in a static fluid (pressure is constant along a horizontal plane). This is because the hydrostatic equation dP/dz = -ρg has no x or y component. Pascal's law confirms: pressure at the same horizontal level is equal throughout a connected static fluid. The shape of the container (D) does not affect pressure at a given depth.

Explanation:
τ = μ(V/h) = 0.1 × (2/0.001) = 0.1 × 2000 = 200 Pa. F = τ × A = 200 × 1 = 200 N. Power P = F × V = 200 × 2 = 400 W. Step 1: du/dy = V/h = 2/0.001 = 2000 s⁻¹. Step 2: τ = 0.1 × 2000 = 200 Pa. Step 3: F = 200 × 1 = 200 N. Step 4: P = F × V = 200 × 2 = 400 W. Option B: F only = 200 (forgot to multiply by V). Option C: 2 × 400 = 800 (doubled). Option D: 400/4 = 100 (quartered).

Explanation:
Surface tension decreases with increasing temperature because higher thermal energy weakens the cohesive intermolecular forces responsible for surface tension. At the critical temperature, surface tension becomes zero (liquid and vapor phases become identical). For water: σ = 0.076 N/m at 0°C, 0.073 N/m at 20°C, 0.059 N/m at 100°C. Adding surfactants (detergents) also reduces surface tension. Surface tension is relatively insensitive to pressure changes at moderate pressures. The Eötvös rule: σ ∝ (T_c - T)^(11/9).

Explanation:
Ideal gas law: PV = mRT → P = (m/V)RT = ρRT → ρ = P/(RT). Where P = absolute pressure (Pa), R = specific gas constant (J/kg·K), T = absolute temperature (K). For air: R = 287 J/kg·K. At standard conditions: ρ = 101325/(287 × 288.15) = 1.225 kg/m³. For any ideal gas: R = R_universal/M = 8314/M (J/kg·K), where M = molecular weight. Option B: RT/P (inverted). Option C: PRT (dimensionally wrong). Option D: P/(R²T) (wrong).

Explanation:
At the stagnation point, velocity = 0. By Bernoulli's equation: P + ½ρV² = P₀ + ½ρV₀² = constant. At stagnation: P_stag + 0 = P₀ + ½ρV². Therefore P_stag = P₀ + ½ρV², where P₀ is the free-stream static pressure and ½ρV² is the dynamic pressure. The stagnation pressure is always greater than static pressure. Pitot tube measures stagnation pressure. Velocity: V = √(2(P_stag - P_static)/ρ). Option B gives pressure less than static (wrong). Option C is static pressure only.

Explanation:
τ = μ(V/h) = 0.3 × (0.5/0.005) = 0.3 × 100 = 30 Pa. F = τ × A = 30 × 2 = 60 N. Power P = F × V = 60 × 0.5 = 30 W. Step 1: du/dy = 0.5/0.005 = 100 s⁻¹. Step 2: τ = 0.3 × 100 = 30 Pa. Step 3: F = 30 × 2 = 60 N. Step 4: P = 60 × 0.5 = 30 W. Option B: P/2 = 15 (halved). Option C: F only = 60 (forgot to multiply by V). Option D: P/4 = 7.5 (quartered).

Explanation:
For soap bubble: ΔP = 4σ/r = 4 × 0.04 / 0.005 = 0.16/0.005 = 32 Pa. Given: σ = 0.04 N/m, r = 5 mm = 0.005 m. ΔP = 4σ/r = 4 × 0.04 / 0.005 = 32 Pa. The factor 4 (not 2) is because soap bubble has two surfaces. Option B: 2σ/r = 2×0.04/0.005 = 16 Pa (liquid droplet formula, one surface). Option C: σ/r = 0.04/0.005 = 8 Pa (cylindrical jet formula). Option D: 8σ/r = 8×0.04/0.005 = 64 Pa (wrong formula).

Explanation:
γ = SG × γ_water = 0.75 × 9810 = 7357.5 N/m³. Step 1: ρ = SG × 1000 = 750 kg/m³. Step 2: γ = ρg = 750 × 9.81 = 7357.5 N/m³. Alternatively: γ = SG × γ_water = 0.75 × 9810 = 7357.5 N/m³. Option B: SG × 1000 = 750 (density, not specific weight). Option C: SG × 100 = 75 (wrong). Option D: SG × 10000 = 7500 (used g = 10). This corresponds to a light oil (SG = 0.75).

Explanation:
P_gauge = P_abs - P_atm = 200 - 101.325 = 98.675 kPa. Step 1: P_gauge = P_abs - P_atm. Step 2: P_gauge = 200 - 101.325 = 98.675 kPa. Since P_gauge > 0, this is a positive gauge pressure (above atmospheric). Option B: P_abs + P_atm = 301.325 (added instead of subtracted). Option C: P_abs itself (forgot to subtract atmospheric). Option D: P_atm itself. Vacuum pressure = P_atm - P_abs (used when P_abs < P_atm).

Explanation:
Hagen-Poiseuille equation: Q = πr⁴ΔP/(8μL) = πD⁴ΔP/(128μL). Derivation: Velocity profile u(r) = (ΔP/4μL)(R²-r²). Integrating: Q = ∫₀ᴿ u(r)·2πr·dr = πR⁴ΔP/(8μL) = πD⁴ΔP/(128μL). Valid for: laminar flow (Re < 2300), fully developed flow, incompressible Newtonian fluid, no-slip at wall. Mean velocity: V = Q/A = D²ΔP/(32μL). Maximum velocity: u_max = 2V (at centerline). Option B: D² instead of D⁴. Option C: factor 64 (off by 2). Option D: factor 32 (off by 4).

Explanation:
τ = μ × (du/dy) = 0.15 × 80 = 12 Pa. Step 1: τ = μ × (du/dy). Step 2: τ = 0.15 × 80 = 12 Pa. Option B: 0.15/80 = 0.001875 (divided). Option C: 80/0.15 = 533.3 (inverted). Option D: 0.15 × 8 = 1.2 (wrong gradient). Units: Pa·s × s⁻¹ = Pa. This is a moderately viscous fluid (between water and oil).

Explanation:
Surface tension arises from unbalanced (net inward) cohesive forces on molecules at the liquid-air interface. Interior molecules are surrounded by other liquid molecules on all sides (balanced forces, net force = 0). Surface molecules have liquid molecules below and on sides, but only air/vapor molecules above (fewer, weaker interactions). This creates a net inward force, pulling surface molecules inward and creating a tension in the surface. The surface acts like a stretched elastic membrane. Surface energy = σ × A (J).

Explanation:
τ = μ × (du/dy) = 0.04 × 25 = 1 Pa. Direct application of Newton's law of viscosity. Step 1: τ = μ × (du/dy). Step 2: τ = 0.04 Pa·s × 25 s⁻¹ = 1 Pa. Option B: 0.04/25 = 0.0016 (divided instead of multiplied). Option C: 25/0.04 = 625 (inverted). Option D: 0.04 × 16 = 0.64 (wrong multiplier). Units: Pa·s × s⁻¹ = Pa. This is a straightforward viscosity calculation.

Explanation:
Bulk modulus K = -V(dP/dV) = ρ(dP/dρ) determines resistance to compression. High K → difficult to compress (nearly incompressible). Low K → easy to compress. Water: K ≈ 2.2 GPa (nearly incompressible). Air at 1 atm: K ≈ 0.1 MPa (highly compressible). Compressibility β = 1/K. Viscosity (B) resists shear deformation. Surface tension (C) is a surface phenomenon. Specific gravity (D) is a density ratio. Bulk modulus is important in: water hammer analysis, acoustic wave propagation, hydraulic systems.

Explanation:
ν = μ/ρ = 5×10⁻⁴/800 = 6.25×10⁻⁷ m²/s. Convert to cSt: 1 m²/s = 10⁶ cSt. ν = 6.25×10⁻⁷ × 10⁶ = 0.625 cSt. Step 1: ν = 5×10⁻⁴/800 = 6.25×10⁻⁷ m²/s. Step 2: 6.25×10⁻⁷ m²/s × 10⁶ cSt/(m²/s) = 0.625 cSt. Option B: 6.25 (off by factor 10). Option C: 62.5 (off by factor 100). Option D: 625 (off by factor 1000). 1 cSt = 10⁻⁶ m²/s = 1 mm²/s.

Explanation:
1 centipoise (cP) = 10⁻³ Pa·s = 1 mPa·s. Therefore: 3 cP = 3 × 10⁻³ Pa·s. Conversion chain: 1 Poise (P) = 0.1 Pa·s. 1 cP = 0.01 P = 0.001 Pa·s = 10⁻³ Pa·s. Water at 20°C ≈ 1 cP = 10⁻³ Pa·s. Blood ≈ 3-4 cP. Glycerin ≈ 1500 cP. Honey ≈ 2000-10000 cP. Motor oil (SAE 30) ≈ 100-200 cP at 40°C. Option B: 3 × 10⁻² (off by factor 10). Option C: 3 × 10⁻¹ (off by factor 100). Option D: 3 (off by factor 1000).