Question 1

Siphon outlet velocity with head loss For a siphon of total centerline length l conveying between two reservoirs with total head difference H, the outlet velocity v (accounting for losses) is best expressed as:

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Introduction / Context:A siphon transfers liquid between elevations using atmospheric pressure and continuity. The actual discharge (and velocity) depends on the available head minus head losses along the siphon due to friction and fittings. Given Data / Assumptions: Total head difference between free surfaces: H. Head loss in siphon: h_f (includes friction and minor losses). Outlet discharges to atmosphere; velocities at large reservoirs are negligible. Concept / Approach: Apply Bernoulli between the upstream free surface and the outlet: z_1 + p_atm/γ = z_out + p_atm/γ + v^2/(2 g) + h_f. With z_1 − z_out = H, we get v^2/(2 g) = H − h_f, hence v = √(2 g (H − h_f)). Step-by-Step Solution: Write Bernoulli with losses: H = v^2/(2 g) + h_f.Rearrange: v = √(2 g (H − h_f)).Compute h_f using Darcy–Weisbach: h_f = f (L/D) v^2/(2 g) + ΣK v^2/(2 g) if needed. Verification / Alternative check: Neglecting losses (h_f ≈ 0) reduces to Torricelli form v = √(2 g H), which is an upper bound—consistent with the loss-inclusive expression. Why Other Options Are Wrong: (b) ignores losses and overestimates v; (c) adds losses incorrectly; (d) is dimensionally inconsistent; (e) is not a standard relation. Common Pitfalls: Omitting minor losses at bends and entrance; using reservoir velocities as non-negligible when they are typically small. Final Answer: v = √(2 g (H − h_f))

Question 2

Convergent–divergent mouthpiece characteristics Which statements correctly describe a convergent–divergent mouthpiece running full under suitable submergence?

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Introduction / Context:Mouthpieces are short tubes fitted to orifices to modify jet formation and recover pressure. A well-designed convergent–divergent mouthpiece can run full and deliver discharge close to the ideal by minimizing losses and recovering kinetic energy in the diffuser section. Given Data / Assumptions: Short convergent followed by a diffuser section. Steady incompressible flow; sufficient submergence to avoid air entrainment. Proper diffuser angle to limit separation and recover pressure. Concept / Approach: The convergent accelerates the flow to the throat (near vena contracta), while the diffuser converts velocity head back to pressure head. With good design, the discharge coefficient C_d approaches 1.0 and losses due to sudden enlargement are avoided because expansion is gradual, not abrupt. Step-by-Step Solution: Convergent → increases velocity to throat.Divergent → recovers part of dynamic head.Net effect → C_d ≈ 1.0 with minimal losses. Verification / Alternative check: Comparative tests show higher C_d for convergent–divergent mouthpieces than for sharp orifices or re-entrant mouthpieces under similar heads. Why Other Options Are Wrong: Each of (a), (b), and (c) is correct; hence (d) is the most complete answer, while (e) omits the key performance metric C_d ≈ 1.0. Common Pitfalls: Using too steep a diffuser angle causing separation; insufficient submergence that leads to cavitation or air ingestion. Final Answer: All the above

Question 3

Critical flow condition in open channels The flow in an open channel is termed “critical” when the Froude number Fr equals what value?

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Introduction / Context:The Froude number Fr = V / √(g D_h) compares flow inertia to gravity effects (D_h is hydraulic depth). It classifies free-surface flows as subcritical, critical, or supercritical—vital for control structures, spillways, and rapidly varied flow predictions. Given Data / Assumptions: Steady, one-dimensional open-channel flow. Hydraulic depth D_h = A / T (area over top width). Froude number Fr = V / √(g D_h). Concept / Approach: By definition, critical flow occurs when specific energy is minimized for a given discharge, and mathematically when Fr = 1. Subcritical flows have Fr < 1 (gravity-dominated), while supercritical flows have Fr > 1 (inertia-dominated). Step-by-Step Solution: Compute Fr from measured V and D_h.If Fr = 1 → critical; Fr < 1 → subcritical; Fr > 1 → supercritical.Design transitions (e.g., weirs, flumes) often target Fr ≈ 1 to create critical depth. Verification / Alternative check: At critical conditions, specific energy E has a stationary point with respect to depth, consistent with GVF theory. Why Other Options Are Wrong: (a) and (c) correspond to subcritical and supercritical regimes; (d) and (e) contradict the definition. Common Pitfalls: Confusing hydraulic depth with hydraulic radius; using local depth in irregular sections without proper definition of D_h. Final Answer: equal to 1.0

Question 4

Inclined jet kinematics – height vs horizontal span A water jet is issued at speed u making an angle θ with the horizontal. The maximum height H it reaches and the total horizontal span (range) R are measured. What is the ratio H / R expressed in te…

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Introduction / Context:Projectile motion under constant gravity provides useful analogies for free jets from nozzles. Relating the jet’s peak height to its horizontal reach helps in setting spray distances and shielding for hydraulic jets and fountains. Given Data / Assumptions: Initial speed u at launch angle θ above horizontal. Uniform gravitational acceleration g; air drag neglected. Ground at same elevation as launch point. Concept / Approach: Standard projectile relations: maximum height H = u^2 sin^2θ / (2 g) and range R = u^2 sin 2θ / g. Taking the ratio removes dependence on u and g, leaving a pure function of θ. Step-by-Step Solution: H = u^2 sin^2θ / (2 g).R = u^2 sin 2θ / g = u^2 (2 sinθ cosθ) / g.H / R = [u^2 sin^2θ / (2 g)] / [u^2 (2 sinθ cosθ) / g] = (sinθ) / (4 cosθ) = (1/4) tan θ. Verification / Alternative check: At θ = 45°, H/R = (1/4) tan 45° = 1/4, which is consistent since the 45° shot maximizes R and yields a moderate H relative to R. Why Other Options Are Wrong: sin θ, cos θ, tan θ, and cot θ ignore the factor 1/4 that arises from combining the height and range formulas. Common Pitfalls: Using degrees of trigonometric identities incorrectly; forgetting that sin 2θ = 2 sinθ cosθ, which introduces the factor 4 in the denominator. Final Answer: (1/4) tan θ

Question 5

Measuring very low (near-vacuum) pressures in fluid mechanics Among the following instruments, which is appropriate for measuring very low pressures (small positive/negative gauge values) in typical laboratory practice?

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Introduction / Context:Very low pressures (small departures from atmospheric) are commonly measured in labs using liquid columns because they can resolve tiny pressure differences with suitable geometry (e.g., inclined tubes). Selecting the proper instrument depends on whether absolute, gauge, or differential pressure is needed. Given Data / Assumptions: Non-volatile manometric fluids (e.g., water, mercury, alcohol) available. Small gauge pressures or slight vacuums to be measured. Steady conditions without rapid oscillations. Concept / Approach: A manometer translates pressure to a hydrostatic head difference h via p = ρ g h. For very small pressures, an inclined manometer increases the scale resolution. Piezometers only measure positive gauge pressure and cannot measure vacuum; barometers read atmospheric (absolute) pressure; differential manometers compare two points but are not specialized for near-zero absolute detection without careful setup; Bourdon gauges lack the fine resolution of an inclined liquid manometer for very low ranges. Step-by-Step Solution: Choose liquid column method for small Δp.Use vertical U-tube for moderate low pressures; switch to inclined manometer for very low Δp to magnify readings.Read head difference and convert to pressure using p = ρ g h. Verification / Alternative check: Instrument catalogs list micromanometers and inclined manometers specifically for low-pressure measurements, confirming suitability. Why Other Options Are Wrong: (a) Barometer measures absolute atmospheric pressure only. (b) Piezometer cannot register negative gauge pressure. (d) Differential manometers are comparative devices and not inherently optimized for very small single-point gauge pressures. (e) Bourdon gauges generally lack sufficient sensitivity at very low ranges without special design. Common Pitfalls: Using too small a-diameter tube causing capillarity errors; choosing a dense manometric fluid that reduces head sensitivity for tiny Δp. Final Answer: manometers

Question 6

Uniform flow in open canals: which statements correctly describe the geometric and energy characteristics?

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Introduction / Context:In open-channel hydraulics, “uniform flow” is a fundamental idealization used for canal design and analysis. When flow is uniform, depth and velocity remain constant along the channel, making energy and geometry relationships especially clear and testable in practice. Given Data / Assumptions: Steady, prismatic channel (constant cross-section and roughness). Uniform flow conditions: depth, velocity, and discharge remain constant with distance. No external work input (e.g., pumps) or withdrawals along the reach. Concept / Approach: Uniform flow implies a constant specific energy at each section apart from frictional dissipation balanced by the bed slope. The water surface slope, energy grade line slope, and bed slope are all equal and parallel. The hydraulic gradient coincides with the channel bed for uniform conditions, which is why the drop in water surface equals the head loss due to friction over a reach. Step-by-Step Solution: Uniform flow condition: d(depth)/dx = 0 and d(velocity)/dx = 0.Energy equation along the channel gives head loss due to friction equals fall of grade line over distance.For a prismatic canal at uniform flow, slopes satisfy S_f = S_w = S_0, where S_f is friction slope, S_w is water surface slope, and S_0 is bed slope.Hence the bed represents the hydraulic gradient and water surface is parallel to bed. Verification / Alternative check: Using Manning or Chezy for a reach with constant roughness and geometry, a constant normal depth y_n exists when S_f equals S_0, confirming the parallelism of bed and free surface and the energy balance. Why Other Options Are Wrong: Each of (a), (b), and (c) is individually true; selecting any single one omits the complete picture. Therefore “All the above” is the only fully correct choice. Common Pitfalls: Confusing gradually varied flow (non-uniform) with uniform flow; assuming parallel slopes in transitions or controls; ignoring that minor losses disturb exact equality over very short reaches. Final Answer: All the above.

Question 7

Classical units: one “metric slug” (hyl) is numerically equal to what quantity?

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Introduction / Context:Older engineering literature sometimes uses the term “metric slug” (also known as the hyl) as an analogue to the imperial “slug.” Understanding these legacy mass units helps interpret historic design tables and exam questions that mix gravitational and absolute unit systems. Given Data / Assumptions: Standard gravity g_0 ≈ 9.81 m/s^2. 1 kilogram-force (kgf) is the weight of 1 kg mass under g_0. “Metric slug” (hyl) is defined such that a force of 1 kgf produces an acceleration of 1 m/s^2 on it. Concept / Approach: From F = m * a, set F = 1 kgf and a = 1 m/s^2. But 1 kgf = 9.81 N. Therefore the mass m satisfying 9.81 N = m * 1 m/s^2 is m = 9.81 kg. Hence, one metric slug equals approximately 9.81 kilograms of mass, not kilogram-force (a force unit). Step-by-Step Solution: Define kgf: 1 kgf = 9.81 N.Apply F = m * a with F = 1 kgf and a = 1 m/s^2.Solve for mass: m = 9.81 N / (1 m/s^2) = 9.81 kg.Conclude: 1 metric slug (hyl) ≈ 9.81 kg mass. Verification / Alternative check: Compare with the imperial slug where 1 slug * 1 ft/s^2 = 1 lbf, yielding 1 slug ≈ 32.174 lbm. The analogy confirms the concept of defining mass via a chosen gravitational force unit and unit acceleration. Why Other Options Are Wrong: (a), (b), and (d) are weights (forces) expressed in kgf, not masses. (e) confuses the distinct imperial unit “slug” with the “metric slug”; they are not numerically equal. Common Pitfalls: Mixing mass and force units; overlooking that “kg wt” refers to kgf (a force), while the question asks for mass equivalence. Final Answer: 9.81 kg mass

Question 8

Real (practical) fluids: which intrinsic properties do they possess?

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Introduction / Context:Unlike the idealized “ideal fluid,” real or practical fluids exhibit a suite of physical properties that affect flow behaviour, measurement techniques, and design outcomes in pipelines, open channels, and equipment such as pumps and turbines. Given Data / Assumptions: Continuum assumption holds. Ambient temperature and pressure ranges common to engineering applications. Real fluids include liquids and gases with molecular interactions. Concept / Approach: Viscosity quantifies resistance to shear and is central in laminar and turbulent flow predictions (Reynolds number). Surface tension governs capillarity, droplet formation, and free-surface curvature at small scales. Compressibility, though small for liquids, is finite and important in water hammer, acoustics, and high-pressure systems; for gases, it is significant and modeled via equations of state. Step-by-Step Solution: Recognize viscosity affects shear stress: τ = μ (du/dy).Acknowledge surface tension effects in menisci, jets, and multiphase boundaries.Note compressibility via bulk modulus K = −V * dp/dV; even water has finite K.Therefore real fluids possess all three properties simultaneously. Verification / Alternative check: Empirical data: dynamic viscosity of water ~1 mPa·s at 20°C; surface tension ~0.072 N/m; bulk modulus ~2.2 GPa—none are zero, proving the point for a common fluid. Why Other Options Are Wrong: (a), (b), or (c) alone are incomplete. Only (d) captures the full set of inherent properties of practical fluids. Common Pitfalls: Assuming incompressibility equates to infinite bulk modulus; in reality, “incompressible” is an approximation where density change is negligible for the process in question. Final Answer: All the above.

Question 9

Definition check: an ideal fluid is best described as which of the following?

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Introduction / Context:Many fluid mechanics derivations—Bernoulli’s equation, potential flow, and vortex theory—use the abstraction of an “ideal fluid.” Correctly recalling this definition is crucial when deciding whether a model applies to a real-world situation or only provides a first approximation. Given Data / Assumptions: Ideal fluid is a theoretical construct. Used primarily for analytic tractability and conceptual understanding. Real fluids depart from this ideal to varying degrees. Concept / Approach: An ideal fluid has zero viscosity (inviscid) and is incompressible (density does not change with pressure). With no viscous stresses, shear stress is zero for any shear rate, so energy dissipation by viscosity is absent. Incompressibility eliminates density variations, simplifying continuity and momentum equations in many flows. Step-by-Step Solution: Identify the two defining attributes: μ = 0 and constant density.Infer that boundary layers and viscous drag are absent in this model.Apply inviscid momentum equations or potential flow theory when appropriate. Verification / Alternative check: Compare with Newtonian fluids (τ = μ du/dy). If μ ≠ 0, viscous effects exist; hence such a fluid is not ideal. Similarly, gases are compressible; liquids have finite viscosity—so neither are strictly ideal. Why Other Options Are Wrong: (b) is nonsensical; Newton’s law involves shear stress, not “velocity.” (c) Gases are typically compressible and viscous. (d) and (e) contradict inviscid, incompressible assumptions. Common Pitfalls: Over-applying ideal models to flows dominated by viscous effects or compressibility (e.g., near walls, high-speed gas flows). Final Answer: Frictionless (inviscid) and incompressible

Question 10

Sharp-crested weirs: for a clinging (depressed) nappe, the typical excess discharge over a free, ventilated nappe is approximately

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Introduction / Context:Weir calibration assumes a free, ventilated nappe. When air cannot freely access the nappe underside, sub-atmospheric pressure forms, causing the sheet to cling to the downstream face. This “clinging nappe” increases the effective discharge for a given head and invalidates standard coefficients unless corrected. Given Data / Assumptions: Sharp-crested rectangular or triangular weir. Comparison between ventilated (free) and non-ventilated (clinging) nappe at the same head. Similar approach conditions. Concept / Approach: When the nappe clings, pressure on its underside drops below atmospheric. The increased pressure differential accelerates the flow, raising discharge beyond the value predicted for a free nappe by roughly several percent—commonly cited about 6–7% for many practical cases. Step-by-Step Solution: Recognize that nappe ventilation controls pressure beneath the jet.If ventilation is blocked, the nappe clings; the discharge coefficient effectively rises.Typical observed excess discharge ≈ 6–7% over free-nappe calibration. Verification / Alternative check: Laboratory calibrations show consistent increases on the order of a few percent; restoring ventilation returns measured Q to free-nappe coefficients, confirming causality. Why Other Options Are Wrong: (b)–(e) greatly overstate the effect for standard weirs and heads; such large increases are not typical and would indicate other errors or geometric changes. Common Pitfalls: Failing to provide aeration vents; reading head too close to the crest; using free-nappe coefficients with a clinging nappe without correction. Final Answer: 6% to 7%

Question 11

Laminar flow in a circular pipe: what are the correction factors for momentum and energy?

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Introduction / Context:Velocity in pipe flow is not uniform across the section. The kinetic energy and momentum fluxes are therefore larger than what would be computed using the mean velocity alone. The energy (α) and momentum (β) correction factors account for this non-uniformity and are standard results for laminar flow in a round pipe. Given Data / Assumptions: Fully developed, steady, laminar flow in a circular pipe. Parabolic velocity profile u(r) with maximum at the centerline. Incompressible, Newtonian fluid. Concept / Approach: For a parabolic velocity profile, the derived factors are α = (∫A u^3 dA) / (A * U_m^3) = 2.0 and β = (∫A u^2 dA) / (A * U_m^2) = 4/3 ≈ 1.33, where U_m is the mean velocity. These increase the energy and momentum flux terms in Bernoulli and momentum equations, respectively. Step-by-Step Solution: Use the laminar profile u = 2 U_m (1 − r^2/R^2).Integrate u^3 and u^2 over the circular area to obtain α and β.Evaluate results to get α = 2.00 and β = 1.33. Verification / Alternative check: For turbulent flow with flatter profiles, α and β approach 1.0, matching intuition that uniform velocity fields need no correction. Why Other Options Are Wrong: (a) or (b) alone omit the other necessary factor. (d) and (e) contradict the well-known laminar results. Common Pitfalls: Using α = β = 1 for laminar conditions; forgetting these factors when computing head losses from measured velocity distributions. Final Answer: Both (a) and (b)

Question 12

Capillarity in a narrow glass tube dipped in a liquid: the rise or fall of the liquid column is

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Introduction / Context:Capillary rise or depression is essential in groundwater flow, soil mechanics, and microfluidics. It depends on interfacial forces, liquid properties, and the tube geometry—especially the diameter and contact angle with the wall. Given Data / Assumptions: Cylindrical, clean glass tube with small diameter. Static equilibrium of a liquid column in air. Surface tension σ and contact angle θ are well-defined. Concept / Approach: The standard relation for capillary rise is h = (4 σ cos θ) / (w * d), where w is specific weight of the liquid and d is tube diameter. Thus, h increases with surface tension and cos θ and decreases with specific weight and diameter. For water in clean glass, θ ≈ 0, so cos θ ≈ 1 and rise is pronounced in narrower tubes. Step-by-Step Solution: Start with vertical force balance: upward surface tension component = weight of liquid column.Write h = (4 σ cos θ)/(w d).Identify inverse dependence on d: halving d doubles h, all else equal. Verification / Alternative check: Check limiting cases: as d → large, h → 0 (no observable capillarity); as σ → 0 or cos θ → 0, rise vanishes, confirming the relationship. Why Other Options Are Wrong: (a) Ignores dependence on cos θ, w, and diameter. (b) Uses sine instead of cosine; the correct factor is cos θ. (c) Rise is inversely, not directly, proportional to specific weight. (e) Directly contradicts the formula. Common Pitfalls: Using radius instead of diameter without adjusting constants; neglecting contamination that changes θ; assuming the same rise for different liquids. Final Answer: Inversely proportional to the diameter of the glass tube

Question 13

Type of vortex motion: when the whole fluid mass rotates without external torque (due to gravity/pressure or imparted rotation), the motion is

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Introduction / Context:Vortical motions are categorized by the presence or absence of external torque. Classifying a flow as a free or forced vortex guides the expected velocity distribution, pressure field, and free-surface shape in rotating fluids (e.g., draining sinks, cyclone separators, rotating tanks). Given Data / Assumptions: No external torque input after initiation (no continuous stirring). Viscous dissipation is negligible over the short observation period. Axisymmetric swirl about a vertical axis. Concept / Approach: In a free vortex, angular momentum is conserved along streamlines, giving v_θ * r = constant and v_θ ∝ 1/r. The free surface assumes a depressed center (paraboloid under gravity), and pressure distribution satisfies radial equilibrium. In a forced vortex, continuous torque (e.g., a rotating container) imposes solid-body rotation, v_θ ∝ r, with a different free-surface shape and pressure field. Step-by-Step Solution: Identify absence of external torque: implies free vortex.Use v_θ r = constant to characterize velocity field.Relate to pressure/free surface via radial balance: dp/dr = ρ v_θ^2 / r. Verification / Alternative check: Everyday example: water draining from a basin forms a free vortex after initial stirring ceases. Contrastingly, a laboratory mixer creates a forced vortex while power is applied. Why Other Options Are Wrong: (b) Requires continuous torque; (c) is vague; (d) describes forced, not free, motion; (e) is not vortical. Common Pitfalls: Confusing transient spin-up (forced) with sustained swirl after torque removal (free); assuming v_θ is uniform across radius. Final Answer: Free vortex

Question 14

Hydrostatics: depth of centre of pressure for a vertical rectangular gate 4 m wide and 3 m high with its top at the free surface (water up to the top) is

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Introduction / Context:The centre of pressure is the point of action of the resultant hydrostatic force on a submerged plane surface. For vertical gates, correctly locating this point is crucial for hinge design, anchor sizing, and stress checks. Given Data / Assumptions: Vertical rectangular plate: width b = 4 m, height h = 3 m. Top edge at the free surface; hydrostatic pressure due to water only. Uniform gravitational acceleration and density. Concept / Approach: The depth to the centroid (centre of area) from the free surface is y_bar = h/2. The centre of pressure depth y_cp for a vertical plane surface with top at the free surface is y_cp = y_bar + (I_G / (A * y_bar)), where I_G is the second moment of area about the horizontal centroidal axis and A is the area. Step-by-Step Solution: Compute area: A = b * h = 4 * 3 = 12 m^2.Centroid depth: y_bar = h/2 = 1.5 m.Second moment about centroidal horizontal axis: I_G = (b * h^3)/12 = 4 * 27 / 12 = 9 m^4.Compute the term I_G/(A * y_bar) = 9 / (12 * 1.5) = 9 / 18 = 0.5 m.Therefore y_cp = 1.5 + 0.5 = 2.0 m from the free surface. Verification / Alternative check: Centre of pressure should lie below the centroid (because pressure increases with depth). The value 2.0 m > 1.5 m satisfies this physical expectation. Why Other Options Are Wrong: (a) and (b) place the resultant above the centroid, impossible here. (d) and (e) are deeper than warranted by the area moment ratio and overestimate lever arms and moments. Common Pitfalls: Using I about the free surface instead of about the centroid; forgetting that the top is at the surface, which simplifies the formula. Final Answer: 2.0 m

Question 15

Pressure terminology: a pressure value less than atmospheric pressure is referred to as

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Introduction / Context:Engineers use different reference frames for pressure: absolute, atmospheric, and gauge. Values below atmospheric occur in condensers, primed pump suctions, and vacuum systems, and are described by several equivalent terms in practice. Given Data / Assumptions: Atmospheric pressure p_atm as the reference for gauge readings. Gauge pressure p_gauge = p_absolute − p_atm. Negative gauge indicates p_absolute < p_atm. Concept / Approach: When p_absolute is less than atmospheric, p_gauge becomes negative. This condition is commonly described as “vacuum pressure,” “negative gauge pressure,” or colloquially as “suction,” acknowledging that devices do not pull fluids but create a pressure differential that ambient pressure uses to push fluids in. Step-by-Step Solution: Define p_gauge = p_abs − p_atm.If p_abs < p_atm, then p_gauge < 0 (negative gauge).Terminology mapping: suction pressure ≈ vacuum pressure ≈ negative gauge pressure. Verification / Alternative check: Vacuum gauges read in mmHg or kPa below atmospheric; converting these to gauge pressure yields negative values, confirming equivalence of terms. Why Other Options Are Wrong: (e) Absolute pressure is referenced to zero vacuum and is never negative; it cannot describe a state “less than atmospheric” without specifying the reference. Common Pitfalls: Assuming suction is a force; it is merely lower pressure. Confusing absolute and gauge scales when sizing equipment or interpreting specifications. Final Answer: All the above

Question 16

In fluid mechanics, for a totally submerged orifice discharging between two reservoirs, the discharge Q is directly proportional to which head measure? (Assume steady flow, sharp-edged orifice, and fully submerged upstream and downstream openings.)

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Introduction / Context:An orifice is an opening in a wall or plate through which fluid flows. When both the upstream and downstream openings are submerged, it is a totally submerged (drowned) orifice. Understanding how discharge depends on head is a core hydrostatics–hydrodynamics concept used in tanks, gates, and culverts. Given Data / Assumptions: Steady, incompressible flow of water. Sharp-edged orifice with coefficient of discharge Cd approximately constant in the considered range. Both sides submerged; head driving flow is the difference in water surface elevations (ΔH). Concept / Approach: From Bernoulli and continuity, the theoretical discharge through an orifice is Q_th = A * √(2 * g * ΔH), where A is the orifice area, g is gravitational acceleration, and ΔH is the difference in piezometric head. Real discharge is Q = Cd * A * √(2 * g * ΔH). Hence, for fixed A and Cd, Q varies with the square root of ΔH, not linearly with ΔH. Step-by-Step Solution: Identify the driving head for a drowned orifice: ΔH = H_up − H_down.Write discharge: Q = Cd * A * √(2 * g * ΔH).Conclude proportionality: Q ∝ √(ΔH) for constant Cd and A. Verification / Alternative check: Dimensional check: √(g * ΔH) has dimensions of velocity. Multiplying by area A gives volume flow per unit time—consistent with discharge units. Why Other Options Are Wrong: (a) Linear proportionality to ΔH is incorrect; the dependence is on the square root. (c) “Square root of the opening” is dimensionally inconsistent. (d) Reciprocal area contradicts continuity; Q increases with area. (e) is unnecessary since (b) is correct. Common Pitfalls: Confusing free orifice head (to atmosphere) with submerged head difference; assuming linear dependence on head without checking Bernoulli-based velocity relation. Final Answer: Square root of the difference in elevation of water surfaces

Question 17

The physical quantity with dimensions M L T^-2 (mass × length × time^-2) corresponds to which of the following?

Accepted Answer

Introduction / Context:Dimensional analysis helps engineers check equations, scale models, and identify governing non-dimensional parameters. Recognizing the base dimensions behind a symbol prevents unit mistakes in design and exams. Given Data / Assumptions: Base dimensions used: M (mass), L (length), T (time). Target dimension: M L T^-2. Classical mechanics definitions are assumed. Concept / Approach: From Newton’s second law, force F = m * a. Dimensions: [m] = M, [a] = L T^-2, so [F] = M L T^-2. Compare with other listed quantities to ensure no ambiguity. Step-by-Step Solution: Write dimensions of force: M L T^-2.Specific weight γ = weight/volume = (M L T^-2)/L^3 = M L^-2 T^-2.Discharge Q = volume/time = L^3 T^-1.Dynamic viscosity μ = shear stress / velocity gradient = (M L^-1 T^-2)/(T^-1) = M L^-1 T^-1. Verification / Alternative check: Check units in SI: Force in newtons (N) = kg·m/s^2, which maps directly to M L T^-2. Why Other Options Are Wrong: (a) Specific weight includes L^-2, not L^1. (c) Discharge lacks M. (e) Dynamic viscosity includes L^-1 T^-1, not L T^-2. (d) is unnecessary since a correct match exists. Common Pitfalls: Mixing specific weight with pressure dimensions, or confusing dynamic viscosity with kinematic viscosity (which is L^2 T^-1). Final Answer: Force

Question 18

Capillary rise of water in a clean glass capillary is governed by which proportionalities? (Assume contact angle near 0° for water–glass.)

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Introduction / Context:Capillary rise explains why water climbs in small tubes, soils, and porous media—vital for geotechnics, hydrology, and materials. It balances surface forces against weight of the liquid column. Given Data / Assumptions: Water in a clean glass tube (contact angle θ ≈ 0°, cos θ ≈ 1). Steady meniscus at equilibrium. Isothermal conditions; negligible evaporation and contamination. Concept / Approach: Equilibrium height h results from vertical surface-tension force around the tube circumference balancing the weight of the liquid column. The standard formula is h = 4 * σ * cos θ / (ρ * g * d), where σ is surface tension, ρ is density, g is gravity, and d is tube diameter. Step-by-Step Solution: Start from: upward force = perimeter * σ * cos θ = π d * σ * cos θ.Downward weight = volume * ρ * g = (π d^2 / 4) * h * ρ * g.Equate and solve: h = 4 * σ * cos θ / (ρ * g * d). Verification / Alternative check: Dimensional check: σ has N/m, dividing by ρ g d (N/m^3) yields meters—consistent. Empirical data for small diameters confirms large rises, illustrating inverse dependence on d. Why Other Options Are Wrong: Options (a), (b), and (c) are each individually true; therefore the most complete correct statement is (d) “All of these.” (e) contradicts the derived relation. Common Pitfalls: Forgetting cos θ; using wrong diameter (radius versus diameter); applying the formula to contaminated or non-wetting systems where θ differs significantly from 0°. Final Answer: All of these

Question 19

In open-channel flow theory, the specific energy at a section (per unit weight, measured relative to the channel bed) is equal to which expression?

Accepted Answer

Introduction / Context:Specific energy is central to analyzing critical flow, alternate depths, and transitions in open channels. It differs from total head by using the channel bed as the datum and excluding pressure head (since pressure distribution is approximately hydrostatic across the depth for subcritical flows). Given Data / Assumptions: Prismatic open channel with a free surface. Hydrostatic pressure distribution across depth. Depth of flow y and mean velocity V at the section are known. Concept / Approach: Specific energy E is defined per unit weight, measured above the channel bed: E = y + V^2/(2g). Here y is potential (elevation) above bed and V^2/(2g) is kinetic energy head. It is not the same as the total head H = z + p/γ + V^2/(2g) used in pipe flow (with z any external datum). Step-by-Step Solution: Write definition: E = potential above bed + velocity head.Potential above bed = y.Therefore E = y + V^2/(2g). Verification / Alternative check: At critical flow, dE/dy = 0 yields the classic relation Fr = 1 and minimum E for a given discharge, confirming consistency with open-channel critical-depth theory. Why Other Options Are Wrong: (b) is the total head, not specific energy relative to bed. (c) and (d) omit necessary terms. (e) incorrectly mixes hydrostatic head with y; p/γ inside a free-surface stream at the surface is atmospheric and not included in E. Common Pitfalls: Confusing “specific energy” with “specific head” in pressurized flow; ignoring that the datum for E is the channel bed. Final Answer: E = y + V^2/(2g)

Question 20

Choose the correct statement about fluid behavior and rheology:

Accepted Answer

Introduction / Context:Rheology classifies fluids by the relationship between shear stress and rate of deformation. Correctly identifying Newtonian versus non-Newtonian behavior is essential in pipeline design, food processing, drilling muds, and biomedical flows. Given Data / Assumptions: Shear flow between parallel plates (canonical reference). Steady shear; temperature constant. “Plastic” refers to materials exhibiting a yield stress (e.g., Bingham or Herschel–Bulkley types). Concept / Approach: Newtonian fluids obey τ = μ * (du/dy) where μ is constant at given conditions (e.g., water, air, light oils). Non-Newtonian fluids have a non-linear or rate-dependent relation (shear-thinning, shear-thickening, thixotropy, rheopexy). Viscoplastic materials (commonly referred to as “plastics” in older terminology) require a yield stress to initiate flow and then may behave Newtonian or otherwise. Step-by-Step Solution: Define Newtonian: linear proportionality, constant μ.Define non-Newtonian: τ is not linear in du/dy or μ is not constant.Define viscoplastic: τ < τ_y → no flow; τ ≥ τ_y → flow per constitutive law. Verification / Alternative check: Examples: Water (Newtonian), ketchup/paints (shear-thinning non-Newtonian), corn starch in water (shear-thickening), drilling muds (Bingham plastic with yield stress). Why Other Options Are Wrong: Options (a), (b), and (c) are each correct; therefore (d) “All of the above” is the best single choice. (e) is false. Common Pitfalls: Confusing yield stress with yield strain; assuming viscosity is constant for all fluids; ignoring temperature dependence even in Newtonian fluids. Final Answer: All of the above

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