Question 1

Atmospheric science – How does air pressure vary with increasing altitude above the Earth's surface?

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Introduction:Understanding how atmospheric pressure varies with height is fundamental in meteorology, aviation, and fluid statics. Pressure is the weight of the air column above a given point; as you move higher, less air remains above you, so the pressure changes in a predictable way.Given Data / Assumptions:Standard atmosphere assumptions (well-mixed troposphere, no severe weather anomalies).Altitude measured upward from mean sea level.Temperature lapse rate treated as typical; qualitative trend sought.Concept / Approach:The hydrostatic relation dp/dz = − rho * g shows that pressure decreases with increasing altitude z because density rho and gravity g are positive. While the decrease is approximately exponential (barometric formula) rather than linear, the monotonic trend is unambiguous: higher altitude means lower pressure. This behavior affects boiling point, aircraft performance, and human physiology at high elevations.Step-by-Step Solution:Step 1: Write the hydrostatic equation: dp/dz = − rho * g.Step 2: Recognize rho > 0, g > 0, hence dp/dz Step 3: Conclude pressure decreases as altitude increases.Verification / Alternative check:At sea level P ≈ 101.3 kPa; at ~5.5 km, it is roughly half that value under standard conditions. Barometric pressure sensors and altimeters use this principle to infer altitude from pressure readings.Why Other Options Are Wrong:Does not change / increases: Contradicts hydrostatic balance and observation.Oscillatory or non-monotonic trends: Do not represent the standard atmospheric profile; local inversions affect temperature, not the monotonic pressure decrease.Common Pitfalls:Assuming a linear decrease; the actual dependence follows the barometric formula and depends on temperature profile. For most qualitative engineering problems, “decreases with height” is sufficient.Final Answer:decreases

Question 2

Floating bodies – Stability criterion using metacentre A floating body is said to be not in (stable) equilibrium if its metacentre M lies below its centre of gravity G. Do you agree with this statement?

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Introduction:For floating bodies, small-angle stability is evaluated using the metacentric height GM, where G is the centre of gravity and M is the metacentre. The sign of GM determines the nature of equilibrium, which is critical for ship design, buoys, and pontoon platforms.Given Data / Assumptions:Small angular displacement (initial stability).Homogeneous liquid; body floats partially submerged.Quasi-static conditions (no waves or dynamic effects).Concept / Approach:For small heel angles, the restoring moment per unit heel angle is proportional to GM. If GM > 0 (M above G), a restoring moment returns the body to its original position (stable). If GM = 0, the body exhibits neutral equilibrium. If GM Step-by-Step Solution:Step 1: Define GM = distance from G to M measured along the vertical through G.Step 2: Relate sign of GM to stability: positive → stable, zero → neutral, negative → unstable.Step 3: Given M below G implies GM negative; hence the body is not in stable equilibrium.Verification / Alternative check:Ship stability textbooks use the righting arm GZ ≈ GM * sin(theta) for small theta. If GM is negative, the righting arm reverses sign, indicating capsizing tendency, confirming instability.Why Other Options Are Wrong:Disagree: Contradicts the metacentric height criterion.Angle or density caveats: Initial stability result holds without those extra restrictions.Common Pitfalls:Confusing small-angle (initial) stability with large-angle behavior; GM is strictly an initial stability parameter but suffices for the statement posed.Final Answer:Agree

Question 3

Open-channel flow – Manning’s discharge form According to Manning's formula, the discharge Q through an open channel can be written in terms of area A, hydraulic mean depth m (i.e., R), bed slope i, and Manning's constant M = 1/n. Which expression i…

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Introduction:Manning's equation is a cornerstone for uniform open-channel flow. It connects discharge to channel geometry, roughness, and slope, enabling practical sizing of canals, sewers, and natural streams. The question asks for the correct symbolic form when using Manning's constant M = 1/n and hydraulic mean depth m (also written R) for hydraulic radius.Given Data / Assumptions:Uniform, steady open-channel flow.Cross-sectional area A; hydraulic radius m = A / P.Bed slope (energy grade) i.Manning roughness n with M = 1 / n.Concept / Approach:Velocity in Manning form is V = (1/n) * m^(2/3) * i^(1/2) = M * m^(2/3) * i^(1/2). Thus discharge is Q = A * V = A * M * m^(2/3) * i^(1/2). Any alteration of the exponents 2/3 and 1/2 misrepresents the empirical scaling embedded in Manning's relation.Step-by-Step Solution:Step 1: Write V = M * m^(2/3) * i^(1/2).Step 2: Multiply by area: Q = A * V.Step 3: Conclude Q = A * M * m^(2/3) * i^(1/2).Verification / Alternative check:Dimensional consistency: V has dimensions of L/T; m^(2/3) carries L^(2/3) and i^(1/2) is dimensionless, with M carrying appropriate empirical scaling. Multiplying by A (L^2) yields Q (L^3/T).Why Other Options Are Wrong:Other exponent patterns (1/2 and 2/3 swapped, powers on A or M) do not match the accepted Manning exponents.A * M * m * i: Omits the essential empirical exponents.Common Pitfalls:Confusing hydraulic mean depth m (R) with flow depth y in non-rectangular sections; always compute m = A / P first.Final Answer:A * M * m^(2/3) * i^(1/2)

Question 4

Hydrostatics – Pressure variation with depth in a liquid at rest At any point in a static liquid, the intensity of pressure is __________ to the depth measured below the free surface.

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Introduction:Hydrostatic pressure is foundational for manometers, dam design, and submerged structural analysis. In incompressible fluids at rest, pressure increases with depth due to the weight of the overlying fluid column, a relationship sometimes called Pascal's hydrostatic law.Given Data / Assumptions:Liquid at rest; density rho approximately constant.Uniform gravitational field g.Depth h is measured vertically below the free surface.Concept / Approach:The hydrostatic equilibrium equation is dp/dz = − rho * g (z upward). Integrating between the free surface (pressure P0) and a point at depth h gives p = P0 + rho * g * h. Therefore, gauge pressure p − P0 is directly proportional to depth h, independent of the vessel shape (hydrostatic paradox).Step-by-Step Solution:Step 1: Write dp/dz = − rho * g.Step 2: Integrate to obtain p − P0 = rho * g * h.Step 3: Conclude proportionality to depth.Verification / Alternative check:Manometer readings confirm that equal increments in depth produce equal increments in pressure; changing vessel shape does not alter the local pressure at a given depth.Why Other Options Are Wrong:Equal / inverse / independent: Contradict the derived linear relation.Proportional to container volume: Pressure is intensive and does not depend on total volume.Common Pitfalls:Assuming wider containers give higher pressure; only depth and density matter (for given P0 and g).Final Answer:directly proportional

Question 5

Orifices & mouthpieces – Relative coefficient of discharge Compared to a short external (projecting) mouthpiece, the coefficient of discharge Cd for a short internal (re-entrant, Borda’s) mouthpiece is generally:

Accepted Answer

Introduction:Mouthpieces modify the flow leaving an orifice. Their geometry affects contraction, velocity distribution, and thus the discharge coefficient Cd. A common comparison is between a short internal (re-entrant) mouthpiece, often called Borda's mouthpiece, and a short external projecting mouthpiece.Given Data / Assumptions:Sharp-edged openings; steady flow from a large reservoir.Short internal re-entrant mouthpiece vs short external mouthpiece.Neglect significant viscous losses beyond those embedded in Cd.Concept / Approach:In a re-entrant (internal) mouthpiece, the jet contracts more severely at the vena contracta located approximately at the mouthpiece plane, producing a smaller effective area and lower Cd (around 0.5). In contrast, short external mouthpieces that run full typically have Cd nearer 0.82–0.85. Hence, Cd(internal) Step-by-Step Solution:Step 1: Identify stronger contraction in the re-entrant geometry.Step 2: Recall typical Cd values: internal ≈ 0.5; external short ≈ 0.82.Step 3: Conclude “less than.”Verification / Alternative check:Classic hydraulics experiments (Borda) document reduced discharge for re-entrant openings compared to plain or externally projecting mouthpieces at the same head.Why Other Options Are Wrong:More than / equal / unrelated: Inconsistent with observed contraction behavior and standard data.Head-dependent reversal: While Cd varies slightly with head, the ordering does not invert in the usual operating range.Common Pitfalls:Confusing coefficients of contraction, velocity, and discharge. Remember Cd = Cc * Cv and strong contraction lowers Cc and thus Cd.Final Answer:less than

Question 6

Rheology – Shape of shear stress vs rate-of-shear curve For a non-Newtonian fluid, the graph of shear stress (tau) versus rate of shear (du/dy) is generally:

Accepted Answer

Introduction:Rheology studies how materials flow and deform. Newtonian fluids obey a linear relation tau = mu * (du/dy), yielding a straight-line plot through the origin. Non-Newtonian fluids deviate from this linearity, producing a curve on a tau versus shear-rate diagram, reflecting shear-thinning, shear-thickening, or yield-stress behavior.Given Data / Assumptions:Isothermal, steady simple shear.Non-Newtonian fluid (e.g., polymer solutions, slurries, paints, blood).Plot of shear stress against shear rate.Concept / Approach:Non-Newtonian classes include: pseudoplastic (shear-thinning), dilatant (shear-thickening), Bingham plastic (yield stress then linear), and Herschel–Bulkley (yield plus power-law). Their constitutive laws lead to nonlinear tau–(du/dy) relationships, hence a curve rather than a straight line. The exact curvature depends on microstructure alignment, particle interactions, and polymer chain dynamics under shear.Step-by-Step Solution:Step 1: Define the axes: tau on vertical, du/dy on horizontal.Step 2: Recognize that for Newtonian fluids, the plot is a straight line of slope mu.Step 3: For non-Newtonian fluids, mu_app = tau / (du/dy) varies with shear rate, bending the curve.Step 4: Conclude the general shape is curved (nonlinear), not a straight line.Verification / Alternative check:Rheometer data for paints and polymer melts show decreasing apparent viscosity with shear rate (concave-down curves), while cornstarch–water mixtures show increasing stress with shear rate more than linearly (concave-up), confirming curvature in both cases.Why Other Options Are Wrong:Straight or horizontal/vertical lines represent special limiting cases (Newtonian, perfectly plastic), not generic non-Newtonian behavior.Common Pitfalls:Assuming a single viscosity can describe all shear rates; for non-Newtonian fluids, report viscosity together with the shear rate or provide a flow curve.Final Answer:Correct

Question 7

Bernoulli principle – What stays constant under ideal assumptions? According to Bernoulli's equation for steady, incompressible, inviscid flow with no shaft work or heat transfer, which quantity remains constant along a streamline?

Accepted Answer

Introduction:Bernoulli's equation provides an energy balance for fluid flow, forming the basis for pitot tubes, venturimeters, and many engineering estimates. It defines a conserved sum of energy heads along a streamline under specific ideal conditions.Given Data / Assumptions:Steady flow along a streamline.Incompressible fluid, negligible viscosity (no head loss).No pumps or turbines adding or extracting shaft work; negligible heat interaction.Concept / Approach:The Bernoulli head (total head) is H = p/(rho * g) + v^2/(2 * g) + z, the sum of pressure head, velocity head, and elevation head. Under the stated assumptions, H is constant along a streamline. In real systems, additional terms account for pump head, turbine head, and head loss.Step-by-Step Solution:Step 1: Identify the three mechanical energy contributions per unit weight.Step 2: State the conservation along a streamline for the ideal case.Step 3: Recognize deviations in real flows (friction, work interactions) alter constancy.Verification / Alternative check:Venturimeter analysis directly applies Bernoulli between inlet and throat, confirming that changes in one head term are offset by the others when losses are negligible.Why Other Options Are Wrong:p + rho * v, p * v, or p/rho − g z: Do not represent the conserved mechanical energy per unit weight.v + z: Lacks pressure term and wrong dimensions for energy conservation.Common Pitfalls:Applying Bernoulli across a pump/turbine without adding the corresponding head term, or across regions with significant viscous dissipation without head-loss correction.Final Answer:p/(rho * g) + v^2/(2 * g) + z (total head)

Question 8

Flow measurement devices – Terminology for an opening in a tank wall An opening in the side of a tank or vessel such that the tank's liquid surface is below the top edge of the opening is called a:

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Introduction:Weirs, notches, and orifices are common flow-measurement and discharge-control devices. Their definitions are tied to geometry and installation location, and exam questions often test the precise terminology distinguishing them.Given Data / Assumptions:Opening is made in the side wall of a tank or vessel.The tank's free surface lies below the top edge of the opening.Free discharge over the opening is intended.Concept / Approach:A notch is an opening in the side of a tank or channel with the liquid surface below the top edge of the opening; it is typically used for flow measurement (e.g., V-notch, rectangular notch). A weir is similar in function but built across an open channel to raise water level and measure discharge over a crest spanning the channel. An orifice is an opening in a wall or floor where the liquid level is above the top edge, causing submerged discharge through the opening rather than free overfall.Step-by-Step Solution:Step 1: Identify relationship of free surface to the opening: below the top edge.Step 2: Map to definitions: that condition corresponds to a notch.Step 3: Exclude orifice (liquid level above top edge) and weir (across the width of a channel).Verification / Alternative check:Standard hydraulics texts define notches and weirs in these terms; device drawings confirm the geometric distinctions used for classification.Why Other Options Are Wrong:Weir: Built across a channel; not merely an opening in a tank wall.Orifice: Requires liquid level above the opening for discharge through the plate.Sluice gate / venturi: Different devices serving control/measurement by other principles.Common Pitfalls:Confusing notches with orifices because both are plate openings; the key cue is the location of the liquid surface relative to the opening top edge.Final Answer:notch

Question 9

Centre of pressure – Vertical plane surface in a liquid For a vertically immersed plane surface, the centre of pressure lies, relative to the surface's centre of gravity (centroid), at a position:

Accepted Answer

Introduction:The centre of pressure is the point of action of the resultant hydrostatic force on a submerged plane surface. Due to the linear increase of pressure with depth, the resultant acts below the geometric centroid when the surface is vertical or inclined, a fact vital for the design of gates and retaining walls.Given Data / Assumptions:Static liquid, constant density.Plane surface is vertical (or inclined but not horizontal).Free surface pressure is uniform (often atmospheric).Concept / Approach:Because pressure increases with depth, the lower part of the surface experiences greater pressure than the upper part, pulling the resultant downward relative to the centroid. Quantitatively, for an inclined surface the depth of the centre of pressure is y_cp = ȳ + I_G / (A * ȳ), where ȳ is the depth of the centroid, A is area, and I_G is the second moment of area about the centroidal axis parallel to the free surface. Since I_G / (A * ȳ) is positive, y_cp exceeds ȳ (deeper), i.e., the centre of pressure lies below the centroid.Step-by-Step Solution:Step 1: Express linear pressure distribution p = rho * g * h.Step 2: Integrate to find the resultant force R = rho * g * A * ȳ.Step 3: Use moment balance to locate the line of action: y_cp = ȳ + I_G / (A * ȳ).Step 4: Conclude y_cp > ȳ, i.e., centre of pressure is below the centroid for vertical surfaces.Verification / Alternative check:For a horizontal plane surface, pressure is uniform and the centre of pressure coincides with the centroid; this special case confirms that the downward shift for vertical surfaces arises from the pressure gradient with depth.Why Other Options Are Wrong:Above / same level: Would require an inverse pressure gradient, which does not occur in normal hydrostatics.Depends only on density or needs flow velocity: Location also depends on geometry (I_G, A, ȳ) and does not require flow.Common Pitfalls:Forgetting that the centre of pressure moves closer to the deeper edge as the surface becomes more inclined, and that for horizontal surfaces it coincides with the centroid.Final Answer:below

Question 10

Water hammer in pipelines – factors affecting magnitude The magnitude (severity) of water hammer produced by sudden valve operations in a pipeline depends primarily upon:

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Introduction:Water hammer is the rapid pressure rise or surge that occurs when flowing liquid is forced to stop or change direction suddenly. This question tests which physical factors govern the magnitude of that surge, a crucial topic in civil and mechanical pipeline design.Given Data / Assumptions:Liquid initially flowing steadily in a long pipe.Valve operation imposes a rapid change in velocity.Pipe wall is elastic; liquid is slightly compressible.Concept / Approach:The classical Joukowsky relation gives the peak pressure rise: Δp = ρ * a * ΔV where a is the wave speed. The wave speed depends on both fluid bulk modulus and pipe-wall elasticity. Closure time relative to wave travel time sets how fully the Joukowsky rise develops.Step-by-Step Solution:1) Recognize that stiffer walls and less compressible liquids yield larger wave speed a.2) Larger a makes Δp greater for the same velocity change.3) Faster valve closure (small closure time) increases ΔV per wave reflection, approaching the full Joukowsky surge.4) Hence all three—pipe elasticity, fluid compressibility, and closure speed—affect magnitude.Verification / Alternative check:Surge analysis software and field records show that installing air vessels or using slow-closure valves reduces peak surge by altering ΔV and the effective compressibility of the system.Why Other Options Are Wrong:Only one factor (a, b, or c) is incomplete; surge depends on combined properties and operation.Pipe diameter alone cannot explain wave speed or closure timing effects.Common Pitfalls:Ignoring pipe supports and entrapped air; both modify effective wave speed and observed surge.Final Answer:all of the above

Question 11

Lock gates – reaction between the two leaves In a lock gate, the resultant hydrostatic pressure on the gate is P, and each gate leaf is inclined at angle α to the normal to the lock side. The reaction between the two gates equals:

Accepted Answer

Introduction:Mitre gates in navigation locks split the hydrostatic resultant into reactions along each leaf. This item checks your ability to resolve forces based on the gate geometry.Given Data / Assumptions:Total hydrostatic resultant on the mitre is P acting normal to the lock wall plane.Two identical leaves meet at the mitre with each leaf at angle α to the wall normal.Friction and weight components are neglected for the force-resolution concept.Concept / Approach:The reaction R between the leaves acts along the bisector at the mitre. By symmetry, each leaf carries the same axial thrust. Resolving P into components along the two leaves gives the relation for R in terms of P and α.Step-by-Step Solution:1) The axial force along each leaf is R.2) Horizontal components of the two leaf forces must balance P.3) Each contributes R * 2 * sin α in total toward balancing P.4) Therefore P = 2 * R * sin α → R = P / (2 * sin α).Verification / Alternative check:As α → 90°, sin α → 1 and R → P / 2, matching intuition that when gates are nearly perpendicular, each carries half the load axially.Why Other Options Are Wrong:P * sin α or P * cos α: incorrect dimensionally for axial reaction.P / (2 * cos α): arises from wrong component resolution.P / 2: valid only for α = 90°, not general.Common Pitfalls:Confusing angles measured to the gate face vs the lock wall; a wrong reference angle flips sine/cosine.Final Answer:P / (2 * sin α)

Question 12

Newton’s law of viscosity – correct functional relationship Newton’s law of viscosity expresses the relationship between which two quantities?

Accepted Answer

Introduction:Newton’s law of viscosity defines the constitutive behaviour of Newtonian fluids and links internal friction to deformation rate. This law underpins laminar flow solutions and many turbulence models’ near-wall treatments.Given Data / Assumptions:Newtonian fluid assumption.Simple shear with velocity gradient perpendicular to flow.Continuum hypothesis and steady conditions.Concept / Approach:The law states τ = μ * (du/dy), where τ is shear stress, μ is dynamic viscosity, and du/dy is the rate of shear strain. The proportionality μ characterizes resistance to layer-by-layer motion.Step-by-Step Solution:1) Consider two adjacent fluid layers moving with different velocities.2) Define the velocity gradient du/dy between them.3) The required tangential force per unit area to sustain this gradient is τ.4) For a Newtonian fluid, τ is directly proportional to du/dy with constant μ.Verification / Alternative check:Poiseuille flow in a circular pipe yields a linear τ–du/dy relation consistent with Newton’s law, producing a parabolic velocity profile.Why Other Options Are Wrong:Pressure/temperature terms do not define the shear law.Bulk velocity alone does not determine τ without the gradient.Common Pitfalls:Confusing dynamic viscosity μ with kinematic viscosity ν = μ / ρ.Final Answer:shear stress and rate of shear strain

Question 13

Fluid property – resistance to relative layer motion The property of a liquid that offers resistance to the movement of one layer over another adjacent layer is called:

Accepted Answer

Introduction:Several fluid properties influence different phenomena. This question seeks the term for internal friction that resists sliding motion between neighbouring fluid layers.Given Data / Assumptions:Layers experience relative motion under shear.Liquid is Newtonian for simplicity.External factors like temperature may affect property magnitude but not the definition.Concept / Approach:Viscosity quantifies resistance to deformation in shear. The higher the viscosity, the larger the shear stress required to maintain a given velocity gradient across layers.Step-by-Step Solution:1) Identify the phenomenon: resistance to sliding between layers.2) Recall the constitutive law τ = μ * (du/dy).3) The property μ (dynamic viscosity) is therefore the correct descriptor.Verification / Alternative check:Everyday comparison—water vs. honey—illustrates that higher μ liquids flow more slowly for the same driving head.Why Other Options Are Wrong:Surface tension governs interfaces, not bulk shear.Compressibility relates to volume change with pressure.Capillarity concerns surface tension/meniscus effects in small tubes.Elasticity describes solids or transient liquid compressibility, not steady shear resistance.Common Pitfalls:Mixing up ν (m^2/s) with μ (Pa·s); both increase resistance, but units and meaning differ.Final Answer:viscosity

Question 14

Sharp-crested (narrow-crested) weir – thumb rule for thickness In a sharp-crested weir, the thickness of the crest is kept less than half of the head of water above the crest. State whether this design guideline is correct:

Accepted Answer

Introduction:Weir discharge equations assume a sharp, effectively thin crest so the nappe springs clear and the flow approximates an orifice sheet. The thickness guideline helps maintain sharp-crested behaviour.Given Data / Assumptions:Free overfall with aerated nappe.Head H measured upstream above the crest elevation.Negligible scale effects for standard laboratory/field conditions.Concept / Approach:Keeping crest thickness less than about H/2 limits streamline curvature within the crest and avoids broad-crested behaviour, which would alter coefficients and calibration.Step-by-Step Solution:1) Begin with objective: produce a sharp overfall with minimal crest interference.2) Use the empirical rule t 3) This ensures discharge coefficient correlations for sharp-crested weirs remain valid.Verification / Alternative check:Laboratory calibrations show that as t/H increases, coefficients shift toward broad-crested behaviour, increasing sensitivity to approach velocity and crest geometry.Why Other Options Are Wrong:False or restrictive cases: standard rectangular and triangular sharp crests follow this guidance.Very thick crests behave differently and need different formulae.Common Pitfalls:Ignoring aeration under the nappe; air-cushion failure changes calibration even with a thin crest.Final Answer:True

Question 15

Compressible flow regimes – definition of supersonic A flow is called “supersonic” if which of the following conditions holds?

Accepted Answer

Introduction:Compressible-flow classification uses the Mach number M = V / a, where a is the local speed of sound. This question asks you to identify the condition for supersonic flow.Given Data / Assumptions:Local thermodynamic state defines a (e.g., a = sqrt(gamma * R * T) for an ideal gas).Flow speed V may vary spatially.Standard regime naming: subsonic, sonic, supersonic, hypersonic.Concept / Approach:By convention, subsonic: M 1 up to about 5 (some texts extend to ~6); hypersonic: typically M ≥ 5. The key distinguishing feature is M > 1 for supersonic, where compressibility shocks and expansion fans dominate behaviour.Step-by-Step Solution:1) Compute M from local V and a.2) If M > 1, classify as supersonic; if M ≈ 1, sonic; M 3) The option that encodes M > 1 is “between 1 and 6.”Verification / Alternative check:In a converging–diverging nozzle, the diverging section supports supersonic flow only after the throat reaches M = 1 (choked) and backpressure is low enough; measured Mach numbers then exceed unity.Why Other Options Are Wrong:“Very high velocity” is vague and medium-dependent.Discharge difficulty is not a defining criterion.Reynolds number compares inertia to viscosity, not sound speed.“None” is false because M-based definition exists.Common Pitfalls:Using a fixed velocity threshold in m/s without considering the medium’s sound speed.Final Answer:the Mach number is between 1 and 6 (i.e., M > 1)

Question 16

Venturimeter installation – effect of inclination on reading When a venturimeter is installed in an inclined position, then for a given discharge the instrument will indicate __________ differential head reading (between its pressure taps):

Accepted Answer

Introduction:Venturimeters measure flow rate from the pressure drop between an upstream section and a throat. This question probes whether instrument inclination alters the indicated differential for the same flow.Given Data / Assumptions:Same two tap locations relative to the converging section and throat, regardless of pipe orientation.Elevation difference between taps is properly accounted for in static head (as usual in Bernoulli).Flow, fluid, and temperature are unchanged.Concept / Approach:The venturimeter reading corresponds to the pressure difference arising from velocity change, not from instrument orientation. Bernoulli’s equation includes elevation head z; when taps move together with the pipe, their relative elevation difference remains the same. Thus, for the same Q, the pressure differential (and manometer reading) is unchanged.Step-by-Step Solution:1) Apply Bernoulli between the upstream tap and the throat.2) For identical geometry, Δz between taps is the same whether the meter is horizontal or inclined along its axis.3) The velocity head difference for a fixed Q is the same.4) Therefore the measured Δp (converted to differential head) is the same.Verification / Alternative check:Manufacturers’ calibration curves specify discharge vs differential head independent of installation angle, provided taps are at the specified axial positions and manometer connections are correct.Why Other Options Are Wrong:More or less: would imply geometry-dependent Δz changed, which it does not for the same tap spacing along the meter.Indeterminate or zero: contradicts standard venturimeter principles.Common Pitfalls:Confusing “inclined manometers” (used to increase reading sensitivity) with an inclined venturimeter body; these are different ideas.Final Answer:the same

Question 17

Terminology – specific weight (weight density) “The specific weight of a fluid is also called weight density.” Decide whether this statement is:

Accepted Answer

Introduction:Fluid properties have multiple names across unit systems and industries. This question checks the equivalence of two common terms for the same quantity.Given Data / Assumptions:Specific weight γ is defined as weight per unit volume.Gravity is treated as uniform.Units may be N/m^3 (S.I.) or lbf/ft^3 (U.S. customary).Concept / Approach:Specific weight γ = ρ * g expresses how much weight a given volume of fluid possesses in a gravitational field. The term “weight density” is a synonym that emphasizes “weight per volume,” in contrast to mass density ρ (mass per volume).Step-by-Step Solution:1) Write γ = W / V.2) Relate W to mass m via W = m * g and m = ρ * V.3) Substitute to obtain γ = ρ * g.4) Therefore, “specific weight” and “weight density” refer to the same physical quantity.Verification / Alternative check:For fresh water near room temperature, ρ ≈ 1000 kg/m^3, so γ ≈ 9.81 kN/m^3; both terms are used interchangeably in handbooks.Why Other Options Are Wrong:“Incorrect” and conditional versions contradict the definition; the equality is unit-agnostic.Claiming weight is not used is incorrect; engineering often uses weight-based units and heads.Common Pitfalls:Confusing mass density (ρ) with weight density (γ); they differ by g.Final Answer:Correct

Question 18

Compressible fluids – variability of specific weight “The specific weight of compressible fluids does not remain constant.” Is this statement:

Accepted Answer

Introduction:Unlike liquids (often approximated as incompressible), gases are compressible and their density varies significantly with pressure and temperature. This affects specific weight, which depends directly on density.Given Data / Assumptions:Specific weight γ = ρ * g.Compressible fluid: ρ is a function of p and T.Gravity g is approximately constant for terrestrial applications.Concept / Approach:Because γ is proportional to ρ, any change in ρ due to compression/expansion or heating/cooling changes γ. For gases obeying the ideal-gas relation, ρ = p / (R * T), so γ varies with p and T even at modest ranges.Step-by-Step Solution:1) Write γ = ρ * g.2) For a gas, substitute ρ from its equation of state (e.g., ideal gas).3) Observe that p↑ or T↓ increases ρ and thus γ; p↓ or T↑ decreases γ.4) Therefore γ is not constant for compressible fluids.Verification / Alternative check:Atmospheric γ decreases with altitude as p and ρ drop; engineering tables list air density and weight density at different T and p.Why Other Options Are Wrong:False statements ignore the p–T dependence of ρ.Restricting to “very high pressure” or “cryogenic” is unnecessary; variability exists over wide conditions.g alone does not fix γ without ρ.Common Pitfalls:Applying an incompressible approximation to gases outside narrow ranges.Final Answer:True

Question 19

Composite notches – discharge over a trapezoidal notch The discharge over a trapezoidal (Cipolletti-type) notch can be expressed as the discharge over an equivalent rectangular notch __________ the discharge over a triangular (V) notch:

Accepted Answer

Introduction:A trapezoidal notch combines a rectangular section and sloping sides that behave like a triangular notch. Conceptually, its flow is the superposition of the two components.Given Data / Assumptions:Standard notch coefficients apply, with the trapezoid composed of a central rectangle and two equal triangles.Head H is measured above the crest.Coefficient variations with H are modest over the calibrated range.Concept / Approach:For small viscosity and surface-tension effects, discharges add linearly: Q_total = Q_rectangular + Q_triangular. The rectangular part scales roughly with H^(3/2), while the triangular side portions scale with H^(5/2).Step-by-Step Solution:1) Split area into a rectangle (width b) and two triangles (side slope m:1).2) Use standard formulae: Q_rect proportional to b * H^(3/2); Q_tri proportional to m * H^(5/2).3) Superpose to get Q_trap = Q_rect + Q_tri.Verification / Alternative check:Cipolletti notch (with specific side slope) is designed so that the triangular side discharge compensates for end contractions of the rectangular portion, reinforcing the additive concept.Why Other Options Are Wrong:Minus, multiply, divide, or exponentiate: physically incorrect for independent parallel flow paths combining at the nappe.Common Pitfalls:Forgetting coefficient-of-discharge corrections; while the conceptual addition holds, coefficients must be applied appropriately in calculations.Final Answer:plus

Question 20

Flow Classification – Time Variation of Discharge A flow in which the quantity of liquid flowing per second (discharge) remains constant with time at a given section is called steady flow.

Accepted Answer

Introduction:Fluid flows can vary with time and/or space. This question checks the definition of steady flow, where discharge at a fixed cross section does not change with time, a key assumption in many engineering calculations.Given Data / Assumptions:Single-phase liquid in a conduit or open channel.Observation made at a fixed section.Flow properties may vary across the section but we consider the section-mean values.Concept / Approach:Steady flow means that any property at a point (or section-mean discharge) is constant with time: dQ/dt = 0. This is independent of whether velocity changes along the length (uniform vs non-uniform) and independent of flow regime (laminar vs turbulent).Step-by-Step Solution:1) Let Q be discharge through a chosen section.2) If Q is constant in time, then the flow is steady by definition.3) Therefore the correct term for the described flow is steady flow.Verification / Alternative check:Many pipelines operated at fixed pump settings display nearly constant Q over long intervals, qualifying as steady even though velocity profiles across the section are non-uniform.Why Other Options Are Wrong:Streamline: Describes particle paths, not time variation of discharge. Turbulent: Refers to regime, not time variation. Unsteady: Means Q varies with time. Uniform: Means velocity is the same at all sections along the length, which is a spatial criterion.Common Pitfalls:Confusing steady/unsteady (time) with uniform/non-uniform (space) and with laminar/turbulent (regime).Final Answer:steady

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