Question 1

Weirs and Notches — Name of the Flowing Sheet In open-channel measurement, the thin sheet of water that flows over the crest of a notch or a weir is commonly known as:

Accepted Answer

Introduction: Weirs and notches are standard devices to measure discharge in open channels. The flowing sheet over the crest has a specific name because its behavior (aeration, contraction) affects the head–discharge relationship and calibration. Given Data / Assumptions: Sharp-crested notch or weir under free, fully aerated overflow. Steady approach conditions; head measured above the crest. Standard terminology in hydraulics texts. Concept / Approach: The thin, free-falling sheet that separates from the crest is called the nappe (also termed the vein). Its aeration and shape determine whether the standard formulas (with the 3/2 head exponent) apply without submergence corrections. Step-by-Step Solution: Identify crest (sill) as the physical top edge of the weir.Recognize the free-falling sheet of water as the nappe/vein.Therefore the correct term for the sheet is ‘‘nappe or vein’’. Verification / Alternative check: Diagrams for sharp-crested weirs label the thin falling sheet explicitly as the nappe; laboratory observations confirm this terminology. Why Other Options Are Wrong: Sill or crest refers to the edge/structure, not the water sheet. Orifice refers to flow through a small opening with enclosed boundaries, not free overfall. Common Pitfalls: Confusing the crest (structure) with the nappe (flow); ignoring the need for aeration to maintain a clean, separated nappe. Final Answer: nappe or vein

Question 2

Most Economical (Hydraulically Efficient) Channel Shape Among common prismatic open-channel sections used in practice, which cross-section is generally considered the most hydraulically efficient for conveying discharge (given similar area and rough…

Accepted Answer

Introduction: A channel section is termed “most economical” or “most hydraulically efficient” when it carries the maximum discharge for a given area, bed slope, and roughness—equivalently, when the wetted perimeter is minimized, maximizing the hydraulic radius R = A / P. Given Data / Assumptions: Comparison limited to practical prismatic sections used in fieldwork. Similar roughness and slope; steady uniform flow conditions. Feasible constructability and stability of banks are considered. Concept / Approach: The theoretically best (maximum R) is a semicircular channel, but it is rarely used at scale. Among common practical forms, the trapezoidal section—optimized with the condition that one-half of the top width equals one of the equal side lengths—yields the least wetted perimeter for a given area, providing superior efficiency over rectangular/triangular forms. Step-by-Step Solution: Define hydraulic radius R = A / P; larger R reduces frictional resistance.For a given area A, minimize P. The trapezoid allows both bottom width and side slopes to be tuned to minimize P.Optimal trapezoid (most economical) satisfies: hydraulic mean depth = half the flow depth, and side slopes chosen so one side equals half the top width; this increases R beyond rectangular/triangular shapes. Verification / Alternative check: Design handbooks list conditions for most economical trapezoidal and rectangular channels; the trapezoid consistently offers better efficiency and bank stability in practice. Why Other Options Are Wrong: Rectangular and square are special cases with higher wetted perimeter for the same area when not at their own “most economical” proportions; triangular channels are typically less efficient and less stable. Common Pitfalls: Confusing theoretical semicircular optimum with practical sections; ignoring side-slope stability and erosion criteria. Final Answer: trapezoidal

Question 3

Floating Bodies — Point About Which Small Oscillations Occur When a floating body in a liquid is given a small angular displacement, it oscillates about a specific point known as the:

Accepted Answer

Introduction: Stability of floating bodies is fundamental to naval architecture and hydraulic structures. Small angular disturbances lead to restoring or overturning moments depending on the relative positions of characteristic points within the body–fluid system. Given Data / Assumptions: Small angular displacement (linearized stability analysis). Homogeneous, incompressible liquid; body floating partially submerged. No waves or dynamic sloshing considered. Concept / Approach: For small heel angles, the line of action of buoyant force shifts because the submerged volume changes shape. The point about which the buoyant force line appears to rotate is the metacentre. If metacentre M lies above the centre of gravity G, the body experiences a restoring couple and is statically stable. Step-by-Step Solution: Identify key points: centre of gravity G (fixed in body), centre of buoyancy B (centroid of displaced liquid), and metacentre M.On heeling, B shifts; the new vertical through B intersects the original vertical at M.Oscillation for small disturbances occurs about M; stability requires GM > 0 (M above G). Verification / Alternative check: Model tests show righting moment proportional to GM * sin(theta) for small angles; oscillations centre on the metacentre under linear assumptions. Why Other Options Are Wrong: Centre of pressure applies to distributed pressure on surfaces; centre of buoyancy is the force application point but not the oscillation centre; centre of gravity is a body-fixed point, not the hydrodynamic pivot. Common Pitfalls: Confusing B with M; assuming large-angle behavior follows small-angle linear theory; neglecting free-surface effects on stability. Final Answer: metacentre

Question 4

Flow Classification — Time Variation of Discharge A flow in which the quantity of liquid flowing per second changes with time (i.e., not constant in time) is called:

Accepted Answer

Introduction: Classifying fluid motion helps engineers choose the right analysis tools. Besides spatial patterns (laminar vs turbulent), we also describe how flow variables change with time at a fixed point—steady versus unsteady. Given Data / Assumptions: Observation at a fixed location in space (Eulerian view). Quantity of flow per second (discharge) varies with time. Single-phase, incompressible liquid context. Concept / Approach: Steady flow means that properties such as velocity, pressure, and discharge at any fixed point do not change with time (∂/∂t = 0). When these quantities vary with time, the flow is unsteady. This definition is independent of whether the flow is laminar or turbulent. Step-by-Step Solution: Identify the given condition: discharge is not constant with time.Relate to definition: time variability at a point implies unsteady behavior.Conclude the correct classification: unsteady flow. Verification / Alternative check: Examples include pump start-up/shut-down, tidal variations in canals, and hydrograph responses where discharge varies with time—these are classic unsteady scenarios. Why Other Options Are Wrong: Steady flow requires time invariance; streamline flow is a dated synonym for laminar and does not address time change; turbulent flow describes disorderly fluctuations in space but can still be steady on average. Common Pitfalls: Equating turbulence with unsteadiness; forgetting that a statistically steady turbulent flow can have constant time-averaged discharge. Final Answer: unsteady flow

Question 5

Units and Definitions — What Quantity Has Units of Force per Unit Length? In fluid mechanics, which property has the dimension/unit of force per unit length (e.g., N/m)?

Accepted Answer

Introduction: Dimensional awareness helps identify fluid properties quickly. Recognizing which property is measured as force per unit length is useful in problems involving interfaces, droplets, jets, and capillary tubes. Given Data / Assumptions: Standard SI units; typical fluid-mechanics definitions. Interface between immiscible phases (e.g., water–air). Concept / Approach: Surface tension represents the tangential force per unit length acting along a line in the interface, equivalently the energy per unit area required to increase the surface. Its SI unit can be written as N/m (or J/m^2). Other listed properties have different units (e.g., viscosity is Pa·s). Step-by-Step Solution: Recall definition: surface tension sigma = force / length.Check dimensional equivalence: N/m equals J/m^2, matching the energetic interpretation.Therefore the correct choice is surface tension. Verification / Alternative check: Capillary-rise formulas include sigma with units N/m, reinforcing this identification. Why Other Options Are Wrong: Compressibility is fractional volume change per pressure (1/Pa). Capillarity is a phenomenon caused by surface tension and wetting; it is not a property with N/m units. Viscosity has units Pa·s (N·s/m^2). Common Pitfalls: Confusing the phenomenon (capillarity) with the property (surface tension); mixing dynamic viscosity with kinematic viscosity units. Final Answer: surface tension

Question 6

Application of Surface Tension — Lead Shot Manufacturing In the traditional production of small spherical lead shots by dropping molten lead from height into water, is the property of surface tension being utilized to form spheres?

Accepted Answer

Introduction: When small droplets of a liquid form in air and then solidify, their shapes reflect the interplay of interfacial forces and body forces. Lead shot production exploits this physics to achieve nearly spherical pellets without machining. Given Data / Assumptions: Molten lead droplets fall through air and quickly cool in water. Droplets are small enough for surface forces to dominate over gravity-driven distortion. No external fields deform the drops during flight. Concept / Approach: Surface tension minimizes surface area for a given volume, and the sphere provides the minimum surface area. Thus a freely falling droplet tends to a spherical shape before it solidifies. Rapid quenching in water “freezes” the spherical geometry into solid shot. Step-by-Step Solution: Molten lead separates into small droplets during fall.Surface tension acts isotropically along the interface, pulling the surface toward minimal area.The droplet adopts an approximately spherical shape, which is then preserved by solidification in water. Verification / Alternative check: Similar behavior is observed with water mist and molten glass beads; small droplets are nearly spherical due to surface tension dominance over aerodynamic forces. Why Other Options Are Wrong: Viscosity resists deformation but does not set the equilibrium minimal-area shape. Gravity initiates detachment and fall but tends to elongate drops at large sizes, not make them spherical. Common Pitfalls: Assuming large droplets remain spherical—at larger sizes, aerodynamic drag flattens and distorts the drop before solidification. Final Answer: Agree

Question 7

Convergent Mouthpiece — Pressure at Vena Contracta For a convergent mouthpiece discharging freely, is the absolute pressure head at the vena contracta equal to atmospheric pressure?

Accepted Answer

Introduction: Short-tube discharges (mouthpieces) produce a contracted jet. Understanding static pressure at the section of minimum area (vena contracta) is essential for applying Bernoulli and selecting the proper coefficients for discharge predictions. Given Data / Assumptions: Convergent mouthpiece discharging freely to the atmosphere (no submergence). Vena contracta forms at or just outside the mouthpiece exit. Steady, incompressible flow; negligible cavitation. Concept / Approach: At a free jet boundary exposed to the atmosphere, the static pressure equals atmospheric. For a convergent mouthpiece running free, the minimum jet section (vena contracta) is effectively at the free boundary; thus its static pressure equals atmospheric, even though velocity is maximum there. This is consistent with p_at_jet_surface = p_atm. Step-by-Step Solution: Apply Bernoulli from reservoir to jet: lossless ideal gives p/ gamma + V^2/(2g) + z = constant.At the free jet, static pressure equals p_atm; hence the pressure head term equals atmospheric head.The high speed at VC reflects conversion of pressure head to velocity head, not a sub-atmospheric static pressure at the free jet cross section. Verification / Alternative check: Experiments show pressure tapping at a free jet section reads atmospheric; sub-atmospheric readings occur inside sharp-edged orifices prior to the jet boundary, not at the exposed jet section. Why Other Options Are Wrong: False would imply sub- or super-atmospheric static pressure at a surface directly open to air; this contradicts boundary conditions. Common Pitfalls: Confusing interior throat pressure in enclosed nozzles (can be below atmospheric) with the static pressure of a free jet at the vena contracta. Final Answer: True

Question 8

Broad-Crested Weir — Condition for Maximum Discharge For a broad-crested weir, the discharge is maximum if the head of water on the downstream side of the weir is ______ the head of water on the upstream side.

Accepted Answer

Introduction: Broad-crested weirs often operate with free overflow, and their peak discharge condition relates to the energy distribution between upstream and downstream levels. Understanding this helps avoid submergence and ensures accurate rating curves. Given Data / Assumptions: Rectangular broad-crested weir under steady conditions. Negligible losses for the ideal condition; approach velocity small. Heads measured relative to the weir crest. Concept / Approach: Maximum discharge corresponds to critical flow over the crest. Energy and momentum considerations show a specific ratio between upstream and downstream heads when the crest section becomes critical, leading to the stated proportion. Step-by-Step Solution: Recognize that q = b * h * V and, at critical state over a broad crest, V^2 = g * h.The discharge reaches a maximum when the crest section runs at critical depth; energy distribution implies a fixed fraction relationship between downstream and upstream heads.This ideal analysis yields the practical rule: downstream head ≈ two-thirds of upstream head for maximum discharge. Verification / Alternative check: Textbook treatments of broad-crested weirs show crest-critical operation at maximum Q; empirical rating curves align with this condition in idealized settings. Why Other Options Are Wrong: Equal to or three-fourth do not correspond to the critical-flow proportion; one-third is associated with other hydraulic maxima conditions, not broad-crested weirs. Common Pitfalls: Applying this condition when the weir is submerged; ignoring approach velocity corrections; confusing critical depth criteria for different structures. Final Answer: two-third

Question 9

Rotating Liquid in a Cylindrical Vessel — Radial Pressure Variation When a cylindrical vessel containing liquid is rotated steadily about its vertical axis, how does the static pressure vary with radial distance r from the axis?

Accepted Answer

Introduction: Rigid-body rotation of a liquid about a vertical axis (no relative shear in the rotating frame) generates a parabolic free surface and a radial pressure gradient. This principle is important for centrifuges, rotating tanks, and free-vortex approximations. Given Data / Assumptions: Steady rotation at angular speed omega about a vertical axis. No radial/axial flow; liquid rotates as a solid body. Gravity acts downward with acceleration g. Concept / Approach: Radial equilibrium gives dp/dr = rho * omega^2 * r. Integrating from the axis (r = 0) yields p(r) - p(0) = (1/2) * rho * omega^2 * r^2. Hence pressure increases with r^2. The free surface forms a paraboloid: z = (omega^2 * r^2) / (2g) + constant. Step-by-Step Solution: Write force balance in radial direction: dp/dr = rho * omega^2 * r.Integrate: p(r) = p(0) + (rho * omega^2 / 2) * r^2.Conclude: pressure increases quadratically with radial distance. Verification / Alternative check: Observation of the free surface curvature in a rotating beaker matches the quadratic profile predicted by the pressure distribution. Why Other Options Are Wrong: Linear increase contradicts dp/dr proportional to r; decrease is physically impossible for outward centrifugal effect; “varies as square” is ambiguous about direction—pressure specifically increases with r^2. Common Pitfalls: Confusing free-vortex (v ∝ 1/r) with forced-vortex (rigid-body) behavior; neglecting hydrostatic vertical variations when comparing points at different elevations. Final Answer: increases as the square of the radial distance

Question 10

Forces acting in a moving fluid: Which characteristic forces are present simultaneously in a moving liquid under real (practical) flow conditions?

Accepted Answer

Introduction / Context: Any real liquid in motion experiences several fundamental forces at once. These include inertia (due to mass and acceleration), viscous (due to internal friction), gravity (weight), pressure, and sometimes surface tension and elastic effects. The question asks you to recognize that multiple forces act together rather than in isolation. Given Data / Assumptions: Continuum (fluid) mechanics framework. Real liquid with nonzero viscosity. Flow may be internal (pipes) or external (around bodies). Concept / Approach: Force balances in fluids are captured by the Navier–Stokes equations. The terms in these equations correspond to inertia, pressure, viscous, and body forces (e.g., gravity). In typical engineering flows, at least inertia, viscosity, and gravity are simultaneously present, though one may dominate depending on the regime (e.g., high Reynolds number → inertia dominance). Step-by-Step Solution: Identify forces: inertia (m * a), viscous (μ * velocity gradients), gravity (m * g).Recognize practical flows always have mass, viscosity, and weight → all act.Therefore, the correct choice aggregates these forces. Verification / Alternative check: Dimensionless groups confirm coexistence: Reynolds (inertia/viscous), Froude (inertia/gravity), Euler (pressure/inertia), Weber (inertia/surface tension). These ratios only make sense because each force is present to some degree. Why Other Options Are Wrong: Inertia force alone: Ignores friction and weight. Viscous force alone: Would imply a Stokes-like limit with no inertia/gravity—rare in general practice. Gravity force alone: Omits dynamics and frictional losses. Common Pitfalls: Thinking a single force “causes” the flow while others vanish. In reality, dominance varies by regime, but the other forces generally remain nonzero. Final Answer: all of these

Question 11

Venturimeter diffuser behavior: How does the flow velocity change in the divergent (diffuser) portion of a venturimeter under normal operation?

Accepted Answer

Introduction / Context: A venturimeter accelerates fluid in the convergent section to a high velocity at the throat, then decelerates it in the diffuser (divergent section) to recover static pressure. Understanding how velocity changes in each segment is essential for correct use and interpretation of differential-pressure readings. Given Data / Assumptions: Incompressible, steady flow. Well-designed diffuser with modest divergence angle to limit separation. Negligible elevation changes over the instrument length. Concept / Approach: Continuity requires v * A ≈ constant for incompressible flow. The diffuser has increasing cross-sectional area; therefore, velocity decreases. Bernoulli with losses shows the accompanying rise in static pressure (pressure recovery). Step-by-Step Solution: At the throat: A small → v high.In the diffuser: A increases → v decreases.Static pressure correspondingly increases (pressure recovery). Verification / Alternative check: Manometer connections at the throat and downstream of the diffuser typically show lower head at the throat and higher head downstream, consistent with decreased velocity in the diffuser. Why Other Options Are Wrong: Remains constant: Would require constant area; not the case. Increases: Opposite of diffuser action. Depends upon mass of liquid: Kinematics here depend on geometry and continuity, not system mass. Common Pitfalls: Thinking velocity keeps increasing through the meter; confusing diffuser with convergent section. Final Answer: decreases

Question 12

Wetting behavior of mercury on glass: Mercury does not wet glass primarily because of which liquid property?

Accepted Answer

Introduction / Context: Wetting depends on the balance between intermolecular attraction within the liquid (cohesion) and attraction between the liquid and a solid surface (adhesion). Whether a liquid spreads on a surface is determined by this balance along with surface tension and contact angle criteria. Given Data / Assumptions: Clean glass surface and pure mercury. Ambient conditions where mercury forms a convex meniscus. No chemical reactions at the interface. Concept / Approach: Mercury exhibits very strong cohesive forces between its own molecules compared with its adhesive attraction to glass. As a result, the contact angle exceeds 90 degrees (non-wetting), and mercury pulls into drops rather than spreading. While surface tension is high, the decisive comparison is cohesion > adhesion. Step-by-Step Solution: Assess intermolecular balance: cohesion (Hg–Hg) vs adhesion (Hg–glass).Since cohesion > adhesion → contact angle > 90° → non-wetting.Therefore, the key property explaining non-wetting is strong cohesion. Verification / Alternative check: Water on clean glass wets (concave meniscus) because adhesion (water–glass) > cohesion (water–water). Mercury is the opposite case. Why Other Options Are Wrong: Adhesion: If adhesion dominated, mercury would wet glass. Viscosity: Affects flow resistance, not equilibrium wetting. Surface tension alone: High surface tension contributes, but the wetting decision hinges on cohesion vs adhesion balance. Common Pitfalls: Attributing non-wetting solely to “surface tension” without comparing cohesion and adhesion; ignoring contamination that can alter contact angle. Final Answer: cohesion

Question 13

Closed cylindrical vessel completely filled and rotating about its vertical axis: What is the total resultant pressure on the top circular cover (radius r) due to rotation at angular speed ω, for a liquid of specific weight w?

Accepted Answer

Introduction / Context: When a completely filled cylindrical vessel rotates as a solid body about its vertical axis, the liquid develops a radial pressure variation. Even though there is no free surface at the top, the pressure at the top cover increases with radius, producing a net resultant force on the cover. Given Data / Assumptions: Liquid of specific weight w (so density ρ = w / g). Angular speed ω (rad/s), radius r. Solid-body rotation; neglect end effects. Concept / Approach: In solid-body rotation, radial equilibrium gives dp/dr = ρ * ω^2 * r. Integrate from the axis to radius r to obtain pressure rise relative to the axis: Δp(r) = (1/2) * ρ * ω^2 * r^2. The resultant on the circular top is the area integral of Δp over the disk area, giving a closed-form expression proportional to r^4. Step-by-Step Solution: Δp(r) = (1/2) * ρ * ω^2 * r^2 = (1/2) * (w/g) * ω^2 * r^2Total force F = ∬ Δp dA = ∫(0→r) Δp(ρ) * (2 * π * ρ) dρF = ∫(0→r) (1/2) * (w/g) * ω^2 * ρ^2 * (2 * π * ρ) dρ = (w * π * ω^2 / g) * ∫(0→r) ρ^3 dρF = (w * π * ω^2 / g) * (r^4 / 4) = (w * π * ω^2 * r^4) / (4 * g) Verification / Alternative check: Dimension check: w has units N/m^3, ω^2 r^4 / g gives m, multiplied by π yields N, consistent with force (total pressure resultant). Why Other Options Are Wrong: π * r^2 * (w * ω^2 * r^2) / (2 * g): Uses edge pressure as uniform—incorrect; average over the area is half of the edge value. (w * ω^2 * r^3) / g: Wrong power of r and missing π factor. (π * w * ω * r^4) / (2 * g): Uses ω instead of ω^2 and wrong constant. Common Pitfalls: Assuming uniform pressure equal to rim value; forgetting that pressure varies with r^2, making the average half the edge value and leading to the r^4 dependence after integration. Final Answer: (w * π * ω^2 * r^4) / (4 * g)

Question 14

Dimensional analysis: The ratio of inertia force to elastic (compressibility) force is known as which dimensionless number?

Accepted Answer

Introduction / Context: Dimensionless numbers classify flow regimes by comparing dominant forces. Inertia compared to viscous, gravity, surface tension, or elastic forces yields Reynolds, Froude, Weber, and Mach numbers respectively. Recognizing these pairings is core to similarity analysis and model testing. Given Data / Assumptions: Compressible flow context for elastic forces. Standard definitions of classical similitude groups. Concept / Approach: Mach number M = V / a compares flow speed V to the speed of sound a, which relates to fluid elasticity (bulk modulus). Hence, Mach represents inertia/elastic force. By contrast, Reynolds is inertia/viscous, Froude is inertia/gravity, and Weber is inertia/surface tension. Step-by-Step Solution: Identify force pair: inertia vs elastic → Mach.Reject the remaining options based on their canonical definitions. Verification / Alternative check: Compressibility effects (shocks, expansions) become important as Mach approaches or exceeds 1, which is tied to elastic propagation speed a, confirming the force pairing. Why Other Options Are Wrong: Reynolds: inertia/viscous. Froude: inertia/gravity. Weber: inertia/surface tension. Common Pitfalls: Memorization mistakes between Mach and Reynolds; forgetting that elastic forces matter only when compressibility is significant. Final Answer: Mach number

Question 15

Orifice flow theory: What is the theoretical jet velocity at a section having head H (ideal, no losses)?

Accepted Answer

Introduction / Context: Torricelli's theorem gives the ideal (lossless) efflux velocity under a head H as if a particle fell freely through that head. It underpins orifice and nozzle calculations before applying discharge and velocity coefficients for real flows. Given Data / Assumptions: Incompressible fluid. Negligible elevation differences aside from head H. No losses (ideal analysis). Concept / Approach: Bernoulli between the free surface and the jet section with atmospheric pressure on both locations reduces to v^2/(2g) = H, so v = sqrt(2 * g * H). Real jets use v = Cv * sqrt(2 * g * H) with Cv < 1. Step-by-Step Solution: Apply Bernoulli: (p/ρg + z + v^2/(2g)) constant.Atmospheric pressure cancels; surface velocity ~ 0.Thus v^2/(2g) = H ⇒ v = sqrt(2gH). Verification / Alternative check: Dimensional check: g has m/s^2, H has m; product gives m^2/s^2; square root gives m/s as required. Why Other Options Are Wrong: 2g*H: Missing square root; has wrong units. sqrt(H/(2g)): Inverts the relation. H/(2g): Again wrong units (seconds^2 per meter). Common Pitfalls: Forgetting to take the square root; confusing head at vena contracta with upstream head; mixing up Cv and Cd. Final Answer: v = sqrt(2 * g * H)

Question 16

Centre of pressure for a vertical plane surface: Select the correct formula for the depth h of the centre of pressure from the free surface (IG = second moment of area about the horizontal centroidal axis, A = area of immersed surface, x = depth of …

Accepted Answer

Introduction / Context: The centre of pressure is the point of application of the resultant hydrostatic force on a plane surface. For a vertical surface, pressure increases linearly with depth, shifting the resultant below the centroid. The exact depth ensures moment equivalence between the distributed pressure and a single resultant force. Given Data / Assumptions: Vertical plane surface fully or partially submerged. Hydrostatic condition (no motion). Free surface exposed to uniform atmospheric pressure. Concept / Approach: Resultant hydrostatic force F = ρ * g * A * x. Taking moments about the free surface and equating to the moment of the resultant at depth h yields h = x + IG / (A * x). Since IG/(A * x) is positive, h > x, meaning the centre of pressure lies below the centroid for a vertical surface. Step-by-Step Solution: Compute resultant: F = ρ * g * A * x.Hydrostatic moment: M = ρ * g * IG.Set F * h = ρ * g * IG ⇒ h = IG / (A * x) + x. Verification / Alternative check: For a thin vertical rectangle with centroid at x, the addition IG/(A*x) is nonzero, so h is always below the centroid, matching intuition from the linear pressure distribution. Why Other Options Are Wrong: x − IG/(A * x): Would place h above the centroid, contradicting physics. IG/(A * x) alone: Omits the centroidal depth term. (A * x)/IG: Inverts the relationship and has wrong dimensions. Common Pitfalls: Forgetting to include the centroidal term x; confusing IG about the centroid with second moment about the free surface. Final Answer: h = x + (IG / (A * x))

Question 17

Rotating liquid in a cylinder: The rise of liquid at the wall relative to the initial level is ________ the depression at the axis for solid-body rotation.

Accepted Answer

Introduction / Context: When a liquid in a cylinder rotates as a solid body, the free surface forms a paraboloid described by z = (ω^2 * r^2) / (2 * g) + constant. Volume conservation around the free surface implies that what rises at the walls must be balanced by a matching depression at the axis. Given Data / Assumptions: Open cylinder with a free surface. Steady, solid-body rotation with angular speed ω. Negligible evaporation and no inflow/outflow during rotation. Concept / Approach: The parabolic surface has its minimum at the axis and maximum at the wall. Integrating the free-surface elevation relative to the initial flat level over the cross-section yields zero net change in volume; thus, the peak rise at the wall equals the central depression in magnitude for the same reference level. Step-by-Step Solution: Free-surface shape: Δz(r) = (ω^2 * r^2) / (2 * g) − constant.Choose constant so the average elevation change over the disk is zero.Then Δz at r = 0 equals −Δz at r = R, making rise at wall = depression at axis. Verification / Alternative check: Compute mean of Δz over the radius; it is zero for the chosen constant. Therefore, positive volume at the rim equals negative volume at the center, confirming equality of magnitudes. Why Other Options Are Wrong: Less than / more than: Would violate volume conservation for the steady parabolic surface. Common Pitfalls: Confusing instantaneous sloshing (transient) with steady solid-body rotation; forgetting that the constant is chosen to conserve volume. Final Answer: same as

Question 18

Basic fluid types: A fluid with zero viscosity (no internal friction) is called a/an ________.

Accepted Answer

Introduction / Context: Idealizations simplify analysis and provide bounds for real behavior. An ideal fluid is one such construct: it has zero viscosity and is incompressible (in many treatments). Although no real fluid is perfectly ideal, the concept is useful for deriving basic relations like Bernoulli's equation. Given Data / Assumptions: Definition-focused conceptual question. Viscosity is the property of interest. Concept / Approach: A real fluid has finite viscosity; an ideal fluid is assumed to have none. Newtonian fluids obey τ = μ * du/dy with constant μ, while non-Newtonian fluids have μ that depends on shear rate; neither implies μ = 0. Therefore, a fluid with zero viscosity is an ideal fluid in the classical sense. Step-by-Step Solution: Set μ = 0 by definition.Conclude the classification: ideal fluid. Verification / Alternative check: Bernoulli's equation in its most basic form assumes inviscid flow—an idealization consistent with the notion of an ideal fluid. Why Other Options Are Wrong: Real fluid: Always has some viscosity. Newtonian fluid: μ constant but not zero. Non-Newtonian fluid: μ varies with shear rate; not zero in general. Common Pitfalls: Assuming “Newtonian” implies inviscid; it does not. Final Answer: ideal fluid

Question 19

Laminar pipe flow (Hagen–Poiseuille): Select the correct head loss due to viscosity for laminar flow in a circular pipe (d = diameter, l = length, v = mean velocity, μ = dynamic viscosity, w = specific weight).

Accepted Answer

Introduction / Context: In the laminar regime (Re < 2000 for pipes), head loss varies linearly with flow rate and viscosity. The Hagen–Poiseuille relation can be expressed in head form using specific weight, yielding a simple proportionality to length and velocity, and inversely to the square of diameter. Given Data / Assumptions: Steady, incompressible, fully developed laminar flow in a circular pipe. No slip at the wall; constant μ. v is mean (average) velocity. Concept / Approach: Darcy–Weisbach with laminar friction factor f = 64/Re leads to h_f = f * (l/d) * (v^2 / (2g)). Using Re = ρ * v * d / μ and w = ρ * g, rearrangement yields h_f = 32 * μ * v * l / (w * d^2). This is equivalent to the classical Hagen–Poiseuille pressure-drop formula divided by w to convert to head. Step-by-Step Solution: Start: h_f = f * (l/d) * (v^2 / (2g)), with f = 64 / Re.Re = ρ * v * d / μ ⇒ f = 64 * μ / (ρ * v * d).Substitute and simplify: h_f = (64 * μ / (ρ * v * d)) * (l/d) * (v^2 / (2g)) = (32 * μ * v * l) / (ρ * g * d^2).Use w = ρ * g ⇒ h_f = (32 * μ * v * l) / (w * d^2). Verification / Alternative check: For water at 20°C (μ ≈ 1.0e−3 Pa·s), small d and long l give appreciable head loss, matching practical experience in microchannels and viscous liquids. Why Other Options Are Wrong: 16*…/ (wd): Wrong constants and diameter power. 4…/(wd): Not derived from laminar friction factor; incorrect scaling. 64…/(w*d^2): Double the correct constant. Common Pitfalls: Using turbulent formulas in laminar range; forgetting v is mean velocity; mixing pressure drop and head loss without dividing by w. Final Answer: h_f = (32 * μ * v * l) / (w * d^2)

Question 20

Fluid properties – At standard lab conditions, what is the (dynamic vs kinematic) viscosity value commonly quoted for pure water at 20°C?

Accepted Answer

Introduction: Viscosity quantifies a fluid's internal resistance to motion. Two closely related measures are used in practice: dynamic viscosity (mu) and kinematic viscosity (nu). Many exam questions ask for the standard reference value for water at 20°C, a temperature often used for property tables and laboratory calibration. Given Data / Assumptions: Temperature is 20°C (approximately room temperature). Fluid is pure water at 1 atm. We distinguish dynamic viscosity mu (in Pa·s or poise) from kinematic viscosity nu (in m^2/s or stokes). Concept / Approach: Dynamic viscosity mu measures shear stress per unit shear rate, while kinematic viscosity nu = mu / rho measures momentum diffusivity. In SI, mu is in Pa·s (N·s/m^2) and nu in m^2/s; in CGS, mu is in poise and nu in stokes. For pure water at 20°C, tabulated values are mu ≈ 1.002 mPa·s ≈ 0.01002 poise = 1.002 centipoise (cP) and nu ≈ 1.004 × 10^−6 m^2/s = 1.004 centistokes (cSt) because rho ≈ 998 kg/m^3. Examinations typically expect the canonical statement “water has a viscosity of about one centipoise at 20°C.” Step-by-Step Solution: Step 1: Recall the widely memorized benchmark: mu(water, 20°C) ≈ 1 cP.Step 2: Map units: 1 cP = 0.01 poise and equals 1 mPa·s in SI.Step 3: Recognize that nu ≈ 1 cSt at 20°C because rho ≈ 1 g/cm^3; however, the default “viscosity” answer without qualifier is dynamic viscosity unless the question explicitly says “kinematic”.Step 4: Choose “one centipoise” as the best match. Verification / Alternative check: Property tables and handbooks list water at 20°C with mu ≈ 1.0 cP and nu ≈ 1.0 cSt. If the prompt explicitly asked for kinematic viscosity, “one centistroke” would be correct. Here the safest conventional choice is the dynamic value, one centipoise. Why Other Options Are Wrong: One stoke: 1 st = 10^−4 m^2/s; far larger than water's nu.One centistroke: This is nu, not mu; ambiguous unless specified.One poise or ten centipoise: Ten times too large for water at 20°C. Common Pitfalls: Mixing dynamic and kinematic viscosity units or forgetting that “poise” and “stoke” are CGS units. Also, quoting values at 25°C (slightly lower viscosity) instead of 20°C can cause small numeric drift. Final Answer: one centipoise

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