Question 1

Flow classification – definition of uniform flow in pipes or open channels A flow in a pipe or channel is said to be uniform when which of the following conditions holds true (at a given instant in time)?

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Introduction: Every flow field can be classified along two axes: spatial variation (uniform vs non-uniform) and temporal variation (steady vs unsteady). This question specifically tests your understanding of the spatial criterion for uniform flow in internal or open-channel systems. Given Data / Assumptions: Single-phase, incompressible fluid. Cross-section may be constant or varying; we focus on the definition, not a particular geometry. Comparison is made across positions along the flow at the same instant. Concept / Approach: Uniform flow means the mean velocity (and thus discharge per unit width, etc.) is the same at all locations along the streamline direction at that instant. In a constant-area pipe with no inflow/outflow along the length, uniformity implies v(x) is constant with x. Steady flow, by contrast, refers to invariance with time at a fixed point, i.e., ∂/∂t = 0; it is different from uniformity even though both often co-occur in simple cases. Step-by-Step Solution: Identify the key phrase: “at all sections have the same velocities” → spatial uniformity criterion.Note that “quantity flowing per second is constant” describes steadiness (time-invariance of Q), not spatial uniformity.A definite path describes laminarity, not uniformity; Reynolds threshold is irrelevant to the definition. Verification / Alternative check: A diffuser operating at constant Q is steady but non-uniform (velocity decreases along x). This counterexample confirms the distinction between steady and uniform. Why Other Options Are Wrong: Different velocities at sections: That is the definition of non-uniform flow.Constant discharge: Steady flow criterion.Definite path: Laminar flow idea; not the uniformity criterion.Reynolds number condition: Turbulence criterion, not uniformity. Common Pitfalls: Equating “steady” and “uniform”. They are orthogonal; a flow can be steady non-uniform or unsteady uniform. Final Answer: the liquid particles at all sections have the same velocities

Question 2

Hydrostatics – location of the centre of pressure for an inclined plane surface For a plane surface immersed in a liquid and inclined at an angle θ to the free surface, the depth of the centre of pressure, measured from the free surface, lies ______…

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Introduction: When a plane surface is submerged, hydrostatic pressure increases linearly with depth, producing a distributed force. The resultant acts at the centre of pressure, which is not at the centroid except in special cases. This question checks the qualitative position of that point for surfaces not horizontal. Given Data / Assumptions: Liquid is at rest; density is constant; gravity uniform. Plane surface fully or partially submerged and inclined by angle θ. Free-surface pressure uniform (e.g., atmospheric). Concept / Approach: The depth of the centre of pressure is y_cp = ȳ + I_G / (A * ȳ), where ȳ is centroid depth, A is area, and I_G is the second moment of area about an axis through the centroid parallel to the free surface. Since I_G, A, and ȳ are positive, the additional term is positive, making y_cp > ȳ. Thus the centre of pressure lies deeper than the centroid, i.e., below it when referenced to the free surface. Step-by-Step Solution: State linear pressure law: p = p0 + rho * g * y.Compute resultant R = rho * g * A * ȳ.Locate line of action using moment balance leading to y_cp = ȳ + I_G / (A * ȳ).Infer y_cp > ȳ → centre of pressure lies below centroid. Verification / Alternative check: For a horizontal plane (no depth gradient), I_G/(A*ȳ) → 0 for the relevant axis and centre of pressure coincides with the centroid, the expected special case. Why Other Options Are Wrong: Above / same depth: Would contradict the positive moment shift caused by increasing pressure with depth.Indeterminate/depends only on density: Location also depends on geometry via I_G and A. Common Pitfalls: Confusing centroid with centre of pressure; they coincide only when pressure is uniform across the surface. Final Answer: below

Question 3

Weirs – behavior of nappe at the ends of a rectangular crest As water flows over a rectangular weir (crest/sill), what happens to the jet stream along the ends of the sill due to side-edge effects?

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Introduction: Side contractions are a hallmark of sharp-crested rectangular weirs that do not span the entire channel width. The nappe narrows near the side edges, reducing the effective width and requiring a correction when computing discharge. This question asks for the qualitative change at the ends of the sill. Given Data / Assumptions: Free (not submerged) rectangular weir with end contractions present. Side walls are away from the nappe edges so lateral contraction can occur. Sharp crest; typical heads where standard formulas apply. Concept / Approach: Because flow lines near the side edges bend around the corners, the jet contracts laterally at each end, forming a reduced effective width. Empirical weir equations incorporate this via an “end contraction” correction, often expressed as L_e = L − k * n * H, where n is the number of contracting ends (usually 2) and k is an empirical factor of order unity. Step-by-Step Solution: Recognize geometry: finite crest length with free side edges.Identify effect: streamlines converge near edges causing a narrower nappe.Implication: use effective length L_e smaller than physical length L. Verification / Alternative check: Laboratory observations show visible narrowing and a drop in discharge relative to a suppressed weir (no side contraction), verifying the contraction phenomenon. Why Other Options Are Wrong: Expands/does not change: Opposite of observed end-contraction effect.Oscillation/thickening at center: Not the primary behavior explained by edge contraction. Common Pitfalls: Confusing end contraction with velocity-of-approach correction; they are separate effects. Final Answer: contracts

Question 4

Orifice coefficients – variation of Cv with head for a sharp-edged orifice How does the coefficient of velocity (Cv) of a sharp-edged orifice typically vary as the head of water (H) increases within normal operating ranges?

Accepted Answer

Introduction: Performance coefficients for orifices—coefficient of contraction Cc, coefficient of velocity Cv, and coefficient of discharge Cd—depend modestly on head. Understanding the trend helps in selecting appropriate Cd for design and in interpreting laboratory results. Given Data / Assumptions: Sharp-edged orifice discharging freely. Head H varied within a practical range where flow remains stable and non-cavitating. Fluid is water; temperature effects small. Concept / Approach: Cv represents the ratio of actual jet speed to ideal speed sqrt(2 * g * H). As H increases, entrance effects and relative influence of viscous losses diminish slightly; the jet accelerates more nearly ideally, so Cv tends to increase and approach a value near unity (often 0.97–0.99). The change is not dramatic but is monotonic over ordinary laboratory ranges, whereas Cc remains about 0.62 for sharp edges and Cd ≈ Cc * Cv rises slightly due to Cv's increase. Step-by-Step Solution: Recognize Cv = V_actual / V_ideal.Note that with larger H, entrance losses are a smaller fraction of available head, so V_actual approaches V_ideal.Therefore Cv shows a mild increasing trend with H. Verification / Alternative check: Textbook datasets typically report Cv ≈ 0.95 at low heads rising toward ≈ 0.98–0.99 at higher heads for sharp-edged plates, confirming the trend. Why Other Options Are Wrong: Decreases / constant at 1.0 / random: Do not match experimental behavior.Non-monotonic strong decrease: Not supported under normal conditions. Common Pitfalls: Assuming Cd changes dramatically with head; Cv’s increase is modest, and Cc remains roughly constant for sharp edges. Final Answer: increases

Question 5

Flow in varying-area conduits – classification for constant volumetric rate A liquid flows through an expanding tube (increasing cross-section) at a constant volumetric flow rate. How should this flow be classified?

Accepted Answer

Introduction: Correctly labeling flows is essential before applying momentum and energy equations. The two attributes are time variation (steady vs unsteady) and spatial variation along the conduit (uniform vs non-uniform). An expanding tube at constant flow rate provides a classic example to test these definitions. Given Data / Assumptions: Rigid tube with cross-sectional area increasing with downstream distance. Volumetric rate Q is constant in time. Incompressible liquid; no lateral inflow/outflow. Concept / Approach: Steady means properties at a fixed point do not change with time; constant Q satisfies that. Uniform refers to constancy along the flow direction at a given instant. Because area A(x) increases, continuity v(x) = Q / A(x) decreases with x, so velocity is not spatially constant—hence non-uniform. Therefore, the correct classification is steady non-uniform flow. Step-by-Step Solution: Apply continuity at each section: v(x) = Q/A(x).Note Q is time-invariant → steady.Observe v varies with x because A changes → non-uniform. Verification / Alternative check: A diffuser at constant throughput is the canonical steady non-uniform device; velocity and dynamic head decrease downstream while static pressure rises, consistent with Bernoulli plus losses. Why Other Options Are Wrong: Steady uniform: Would require constant area or special profile giving constant v; not the case.Unsteady options: Contradict the given constant rate condition.Periodic uniform: No periodic time variation is described. Common Pitfalls: Equating constant discharge with uniformity; it guarantees steadiness, not spatial uniformity. Final Answer: steady non-uniform flow

Question 6

Forces on a body resting on a liquid – weight vs upthrust When a body is placed over (or on) a liquid, it is subjected to two principal vertical forces: gravitational force (its weight) and the upward hydrostatic thrust (buoyant force). Is this stat…

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Introduction: Understanding the vertical force balance on a body interacting with a fluid underpins floatation, apparent weight, and stability problems. The question asks whether both weight and buoyant force act when a body is placed on or in contact with a liquid at rest. Given Data / Assumptions: Quasi-static conditions; fluid at rest (hydrostatics). Body may float partially submerged or rest on a support while touching the liquid. No dynamic lift or drag considered. Concept / Approach: The body’s weight W acts downward. The fluid exerts an upward pressure distribution over the wetted surface; the resultant is the buoyant (upthrust) force. If the body floats freely, equilibrium requires W = buoyant force. If it is supported (e.g., on a scale), the scale reading equals W − buoyant force, evidencing that both forces are present even without full submergence. Step-by-Step Solution: Identify forces: W downward; pressure-resultant upward (buoyancy) over immersed area.If floating: W = rho * g * V_displaced.If merely touching with partial wetting: The same hydrostatic mechanism creates an upward resultant on the wetted portion. Verification / Alternative check: Apparent weight experiments show a reading drop when an object is immersed partially—direct proof of a buoyant force acting concurrently with weight. Why Other Options Are Wrong: False or conditional statements are inconsistent with hydrostatics; buoyancy does not require full submergence, special fluids, or motion. Common Pitfalls: Equating “over” with “above but not touching.” The context here is contact with the liquid; once a wetted area exists, hydrostatic pressure generates an upthrust. Final Answer: True

Question 7

Open-channel instruments – what does a critical depth meter measure? A critical depth meter (a calibrated flume or control that establishes critical flow) is primarily used to measure which quantity in an open channel?

Accepted Answer

Introduction: Critical depth meters (such as flumes that force Froude number to unity at a throat) convert a difficult-to-measure discharge into an easily measured depth at a calibrated section. This question asks you to identify the primary hydraulic quantity obtained by using such a device. Given Data / Assumptions: Open-channel flow with a control that creates a critical section (Fr = 1) at a known geometry. Free-flow (modular) conditions without submergence at the control for proper calibration. Empirical rating relates depth at the control to discharge. Concept / Approach: At critical flow, specific energy is minimized, and for a given control geometry the depth at the throat (or at a gauge point) is a unique function of the discharge. Therefore, by measuring depth at that location, one infers the flow rate Q using a calibration curve or formula. Instruments like the Parshall flume and Venturi flume are examples of critical-depth meters used worldwide for flow gauging in irrigation canals and wastewater channels. Step-by-Step Solution: Introduce a control that enforces Fr = 1 at a specific section.Measure the stage (depth) at the designated gauge point.Convert measured depth to discharge using the device rating (Q = f(h)). Verification / Alternative check: Manufacturer calibration tables and standard hydraulics texts provide Q–h relationships for flumes; field use confirms accurate discharge measurement over wide ranges when submergence criteria are met. Why Other Options Are Wrong: Velocity only or depth only: The device’s purpose is to infer Q from depth under controlled conditions, not to report velocity or depth in isolation.Hydraulic jump detection and slope measurement are different diagnostic tasks. Common Pitfalls: Using the meter under submerged conditions without correction factors; submergence invalidates the simple Q–h relation for many flumes. Final Answer: rate of discharge (flow rate) in the open channel

Question 8

Unit conversions – litres contained in one cubic metre of water In basic engineering units, one cubic metre (1 m^3) of water corresponds to how many litres?

Accepted Answer

Introduction: Unit fluency prevents costly mistakes in design and billing. Converting between cubic metres and litres is especially common in water supply, irrigation, and laboratory settings. This question asks for the exact litre equivalent of a cubic metre. Given Data / Assumptions: SI base and derived units are used. 1 litre is defined as 1 cubic decimetre (1 dm^3). No temperature correction is required for volume units in this ideal conversion. Concept / Approach: By definition, 1 m = 10 dm. Therefore 1 m^3 = (10 dm)^3 = 1000 dm^3. Since 1 dm^3 = 1 L, it follows that 1 m^3 = 1000 L. This conversion is exact and independent of the substance occupying the volume (water, air, etc.). Step-by-Step Solution: Convert metres to decimetres: 1 m = 10 dm.Cube both sides: 1 m^3 = 10^3 dm^3 = 1000 dm^3.Recognize dm^3 as litres: 1000 dm^3 = 1000 L. Verification / Alternative check: Utility bills and water meters often list consumption in cubic metres; multiplying by 1000 gives litres directly (e.g., 2.5 m^3 = 2500 L), confirming the exact ratio. Why Other Options Are Wrong: 100, 250, 500: Underestimate by factors of 10–4.10,000: Overestimates by a factor of 10. Common Pitfalls: Confusing mass and volume equivalences (e.g., 1 L of water ≈ 1 kg near room temperature) and applying them when the question asks only about volume units. Final Answer: 1000 litres

Question 9

Centre of pressure vs centroid – vertical immersed surface For a vertically immersed plane surface in a static liquid, the centre of pressure (point of action of the resultant hydrostatic force) lies __________ the surface’s centre of gravity (centr…

Accepted Answer

Introduction: The distribution of hydrostatic pressure on submerged surfaces is linear with depth, pulling the resultant force toward the deeper parts of the surface. Recognizing the relative position of the centre of pressure to the centroid is essential for gate and wall design to ensure correct moment calculations. Given Data / Assumptions: Liquid at rest, constant density, uniform gravity. Plane surface is vertical and fully or partially submerged. Free-surface pressure is uniform (usually atmospheric). Concept / Approach: For a vertical plane, pressure increases linearly with depth. The resultant hydrostatic force acts at depth y_cp = ȳ + I_G / (A * ȳ), where ȳ is centroid depth, A is area, and I_G is the second moment of area about the centroidal axis parallel to the free surface. Because I_G/(Aȳ) is positive, y_cp is deeper than ȳ; thus, the centre of pressure lies below the centroid when referenced to the free surface. Step-by-Step Solution: Write p = p0 + rho * g * y.Integrate over the area to get R = rho * g * A * ȳ.Use moment equilibrium to obtain y_cp = ȳ + I_G/(Aȳ).Conclude y_cp > ȳ ⇒ centre of pressure is below the centroid. Verification / Alternative check: For a horizontal surface, the pressure is uniform and y_cp = ȳ; this limiting case reinforces that the downward shift arises from the depth gradient in a vertical surface. Why Other Options Are Wrong: At/above: Would contradict the increasing pressure with depth.Indeterminate/“midway”: Location depends on geometry (I_G, A, ȳ), not just fluid density. Common Pitfalls: Equating centroid and centre of pressure in all cases; they coincide only when pressure is uniform across the surface. Final Answer: below

Question 10

Open-channel hydraulics – section shape for maximum hydraulic efficiency For channels conveying open-channel flow on a given bed slope, the highest hydraulic efficiency (maximum discharge for a given cross-sectional area and roughness) is generally …

Accepted Answer

Introduction / Context: Hydraulic efficiency of an open channel section is typically assessed using the hydraulic radius R = A / P, where A is the flow area and P is the wetted perimeter. For a given area and roughness, a shape that minimizes wetted perimeter at a given area maximizes R and therefore discharge capacity on a given slope. Given Data / Assumptions: Uniform open-channel flow on a given slope with identical roughness characteristics. Comparison among standard geometric sections (circular run partly full, rectangular, trapezoidal, etc.). Friction formulae like Manning or Chezy apply; discharge increases with hydraulic radius. Concept / Approach: For a given area, the shape with the smallest wetted perimeter gives the largest hydraulic radius. Among common shapes, the circle has the smallest perimeter for a given area, hence the best perimeter-to-area ratio. In open channels, a semicircular trough is theoretically the most economical, and a circular conduit running at an appropriate depth closely approximates this optimum. Step-by-Step Solution: 1) Objective: maximize R = A / P for given A.2) Geometric isoperimetric principle: of all plane curves with a given perimeter, the circle encloses maximum area (equivalently, for a given area the circle has minimum perimeter).3) Therefore P is minimized and R maximized by a circular profile (or semicircular channel in open flow).4) Higher R reduces boundary shear effects relative to area, increasing discharge capacity for the same slope and roughness. Verification / Alternative check: Design texts derive the most economical proportions for rectangular and trapezoidal sections (e.g., b = 2y for best rectangular). Even at these best proportions, the achievable R for a given A is still lower than that of a circular/semicircular form, confirming the circular section's superior efficiency. Why Other Options Are Wrong: Square/rectangular: larger wetted perimeter for the same area than circular, hence smaller R. Trapezoidal: can be optimized but still does not beat the circular shape's isoperimetric advantage. Triangular: very large wetted perimeter for a given area; least efficient among those listed. Common Pitfalls: Confusing “most economical rectangular/trapezoidal” proportions (best within that family) with the globally best shape; the circle (or semicircle for open flow) remains superior in perimeter-to-area performance. Final Answer: circular

Question 11

Hydrostatics of dams – total hydrostatic force per metre length on a vertical wall For a vertical masonry dam wall retaining a liquid of specific weight w to depth H, what is the total hydrostatic force per metre length of the wall (resultant acting…

Accepted Answer

Introduction / Context: The design of retaining structures like dams requires evaluating the resultant hydrostatic force. On a vertical wall holding a liquid of depth H, the pressure varies linearly with depth, creating a triangular pressure distribution. Given Data / Assumptions: Liquid is static and homogeneous with specific weight w. Wall is vertical and of unit breadth (1 m) into the page. Free surface is at atmospheric pressure; gauge pressures are considered. Concept / Approach: Gauge pressure at depth y is p = w * y. The resultant horizontal force equals the area under the pressure diagram (a triangle with height wH at the base and width H). For unit width, the equivalent area gives the total force. Step-by-Step Solution: 1) Pressure at depth y: p(y) = w * y.2) Force on a horizontal strip of thickness dy per metre width: dF = p(y) * 1 * dy = w * y * dy.3) Integrate from y = 0 to H: F = ∫(0→H) w * y dy = w * H^2 / 2.4) Line of action occurs at 2H/3 below the free surface (useful for moment checks). Verification / Alternative check: The triangular pressure diagram has area (1/2) * base * height = (1/2) * H * (wH) = w * H^2 / 2, identical to the integration result, confirming consistency. Why Other Options Are Wrong: w * H or w * H / 2: these have incorrect dimensional dependence on H (linear rather than quadratic). w * H^2 / 4: underestimates the triangular area by a factor of 2. 2 * w * H^2: overestimates by a factor of 4. Common Pitfalls: Confusing force per metre length with average pressure; forgetting that pressure increases with depth leads to missing the H^2 dependence. Final Answer: w * H^2 / 2

Question 12

Floating stability – condition for neutral equilibrium using the metacentre A floating body is said to be in neutral equilibrium when which of the following conditions about the metacentre (M) relative to the centre of gravity (G) holds?

Accepted Answer

Introduction / Context: Stability of floating bodies is assessed by the relative positions of the metacentre (M), the centre of buoyancy (B), and the centre of gravity (G). Depending on these positions, the body may be stable, unstable, or in neutral equilibrium for small angles of heel. Given Data / Assumptions: Small angular displacements (initial stability analysis). Homogeneous fluid; hydrostatic conditions. The definitions of M, B, and G are standard. Concept / Approach: Initial stability is determined by metacentric height GM = distance between G and M. If GM > 0 (M above G), the body is stable. If GM < 0 (M below G), it is unstable. Neutral equilibrium corresponds to GM = 0, meaning M and G coincide so that the righting and overturning moments vanish for small inclinations. Step-by-Step Solution: 1) Consider a small heel angle: buoyancy shifts, locating a new B and defining M as the intersection of buoyancy normals.2) Compute GM; sign indicates stability type.3) GM = 0 ⇔ M coincides with G.4) Therefore, neutral equilibrium condition is M and G coincident. Verification / Alternative check: In neutral equilibrium, no restoring couple acts for small heels; experimentally, a body neither rights itself nor continues to overturn when slightly disturbed, matching GM = 0. Why Other Options Are Wrong: M above G: stable equilibrium (not neutral). M below G: unstable equilibrium. M between B and G (not coincident): implies nonzero GM, yielding stability or instability depending on ordering. M at water surface: incorrect generalization; M depends on geometry and heel. Common Pitfalls: Confusing centre of buoyancy B with metacentre M; B always lies below the waterline, while M may lie above or below G depending on geometry. Final Answer: M coincides with G

Question 13

Basic property definition – mass per unit volume at standard conditions The mass per unit volume of a fluid evaluated at a standard temperature and pressure is called:

Accepted Answer

Introduction / Context: Fluid properties are often tabulated at standard temperature and pressure (STP). Knowing basic definitions avoids confusion when converting between related quantities used in hydraulics and thermodynamics. Given Data / Assumptions: Mass density ρ is mass per unit volume. Specific weight γ is weight per unit volume. Specific gravity SG is a ratio relative to a reference density. Concept / Approach: Mass density is the fundamental property ρ = m / V. Specific weight uses the gravitational conversion γ = ρ * g. Specific gravity is dimensionless: SG = ρ / ρ_ref (for liquids, commonly water at 4 °C). The question explicitly asks for “mass per unit volume,” which is mass density. Step-by-Step Solution: 1) Identify the requested definition: mass per unit volume → ρ.2) Distinguish from γ: weight per unit volume = ρ * g.3) Distinguish from SG: ratio of ρ to a reference ρ_ref (dimensionless).4) Choose “mass density.” Verification / Alternative check: Dimensional consistency: ρ units are kg/m^3 (S.I.); γ units are N/m^3; SG has no units. This confirms that only “mass density” fits the statement exactly. Why Other Options Are Wrong: Specific weight / weight density: incorporate g and are not mass per volume. Specific gravity: a ratio, not an absolute property with units. None of these: incorrect because ρ is the standard term. Common Pitfalls: Using weight density data when a calculation requires mass density, leading to unit errors by a factor of g. Final Answer: mass density

Question 14

Internal mouthpieces (Borda type) – identifying flow regime In an internal mouthpiece fitted flush with a tank wall, if the jet after the vena contracta does not touch the sides of the mouthpiece along its length, the mouthpiece is said to be:

Accepted Answer

Introduction / Context: Short-tube or internal (Borda) mouthpieces can operate in two distinct regimes: running free and running full. Correctly identifying the regime is essential because coefficients of discharge and velocity differ between them. Given Data / Assumptions: Sharp-edged entrance causing a vena contracta immediately downstream of the orifice plane. Steady incompressible flow with modest heads. Tube length small (a few diameters). Concept / Approach: In the running free regime, the jet contracts and remains detached from the tube wall, exiting the mouthpiece without wetting the internal surface. If the tube is long enough (or pressure recovery causes expansion), the jet attaches to the wall and the mouthpiece runs full, behaving like a short pipe with additional friction and different coefficients. Step-by-Step Solution: 1) Observe the jet behaviour after contraction: does it contact the tube wall?2) If no contact occurs, the flow is detached; classify as running free.3) If contact occurs along the length, classify as running full.4) Recognize that length-to-diameter ratio strongly influences the regime (running free commonly when L/D < about 3). Verification / Alternative check: Laboratory visualizations show a clear annular space between the contracted jet and the tube wall when running free; discharge coefficients then approximate orifice flow values rather than short-pipe values. Why Other Options Are Wrong: Running full / partially running full: require wall contact and wetting. Partially running free: nonstandard term; regime is typically categorized as either free or full. Choked: not a standard term for incompressible mouthpiece flow. Common Pitfalls: Assuming any internal tube implies pipe-flow behaviour; in running free, friction inside the tube is negligible because the jet does not contact the wall. Final Answer: running free

Question 15

Newtonian shear – dependence of shear stress on layer spacing The shear stress transmitted between two parallel layers of a Newtonian liquid moving with a velocity difference across a gap is __________ proportional to the distance (gap) between the …

Accepted Answer

Introduction / Context: Newton's law of viscosity states that shear stress in a Newtonian fluid is proportional to the rate of shear strain (velocity gradient). Understanding how the gap between layers affects this gradient clarifies how shear stress scales in simple shear. Given Data / Assumptions: Two large parallel layers (or plates) with a relative velocity difference Δu. Gap (distance) between layers is y. Fluid behaves as Newtonian with dynamic viscosity μ. Concept / Approach: The constitutive relation is τ = μ * (du/dy). For a linear velocity profile between plates, du/dy ≈ Δu / y. Therefore, for fixed Δu and μ, shear stress varies inversely with the separation y: increasing the gap reduces the gradient and thus the shear stress. Step-by-Step Solution: 1) Write τ = μ * (du/dy).2) Replace du/dy by Δu / y for simple Couette flow.3) Obtain τ = μ * (Δu / y).4) Conclude τ ∝ 1 / y when Δu, μ are constants. Verification / Alternative check: Doubling the gap while holding plate speeds fixed halves du/dy and halves τ, which matches laboratory Couette apparatus observations. Why Other Options Are Wrong: Directly: would require τ ∝ y for constant Δu, contradicting τ = μ * Δu / y. Exponential/logarithmic/unrelated: not supported by Newtonian shear theory in this configuration. Common Pitfalls: Forgetting that if the velocity difference is not held constant—e.g., fixed wall shear—the scaling changes; here Δu is taken as fixed to isolate the effect of gap size. Final Answer: indirectly (inversely)

Hydraulics and Fluid Mechanics Questions