Question 1

Units – Stoke (St) for Kinematic Viscosity Definition: One stoke equals 1 cm^2/s, which is numerically equal to 10^-4 m^2/s in SI units.

Accepted Answer

Introduction: This question assesses familiarity with unit conversions for kinematic viscosity, a parameter central to laminar–turbulent characterization and diffusion processes in fluids. Given Data / Assumptions: 1 stoke (St) = 1 cm^2/s by definition. SI base for kinematic viscosity is m^2/s. Concept / Approach: Convert square centimeters to square meters. Since 1 cm = 10^-2 m, 1 cm^2 = (10^-2 m)^2 = 10^-4 m^2. Therefore 1 St = 10^-4 m^2/s. Step-by-Step Solution: 1) Start with 1 St = 1 cm^2/s.2) Convert: 1 cm^2 = 10^-4 m^2.3) Thus 1 St = 10^-4 m^2/s. Verification / Alternative check: Compare with centistoke: 1 cSt = 10^-2 St = 10^-6 m^2/s, which is consistent with common values for light oils and water at moderate temperatures. Why Other Options Are Wrong: 10^-2, 10^-3, and 10^-6 m^2/s are inconsistent with the squared length conversion and are orders of magnitude off. Common Pitfalls: Forgetting that the conversion is for squared units; mixing stoke with centistoke; confusing dynamic and kinematic viscosity units. Final Answer: 10^-4 m2/s

Question 2

Viscous Force Definition – Relation to Shear Stress and Area In viscous flow, the resultant viscous force acting on a surface equals the shear stress due to viscosity multiplied by the area over which that stress acts.

Accepted Answer

Introduction: The item probes the basic relationship F = tau * A for shear forces in viscous flow, where tau is the shear stress and A is the area. This is foundational for wall shear calculations and drag estimation. Given Data / Assumptions: Continuum mechanics applies; Newtonian fluid. Shear stress tau is known or determined from velocity gradients and viscosity. Uniform average shear over area A for the purpose of this relation. Concept / Approach: Stress is force per unit area. Rearranging gives force as stress times area. In viscous flows near solid boundaries, tau = mu * (du/dy) for Newtonian fluids, and the net shear force follows directly. Step-by-Step Solution: 1) Start from definition: tau = F / A.2) Rearrange to F = tau * A.3) Recognize tau arises from viscosity via tau = mu * (du/dy). Verification / Alternative check: Dimensional consistency: [tau] = N/m^2, [A] = m^2, so F = N, confirming the product relation. Why Other Options Are Wrong: Sum, different, ratio: These words do not reflect the physics or dimensional analysis; only product yields correct force units. Common Pitfalls: Confusing wall shear stress with average shear; neglecting non-Newtonian behavior where tau may not be proportional to du/dy, though F = tau * A still holds once tau is known. Final Answer: product

Question 3

Power Transmission Through a Pipeline Condition for Maximum Power: The power delivered at the outlet is maximized when the head lost due to friction equals one-third of the total supply head.

Accepted Answer

Introduction: This item examines the condition for maximizing hydraulic power transmitted through a pipeline when head losses occur due to friction. It is a classic optimization result in fluid power systems. Given Data / Assumptions: Total supply head H at the pipeline inlet. Head loss due to friction h_f; remaining head at outlet H_d = H - h_f. Discharge Q depends on head loss via Darcy–Weisbach or equivalent; proportional reasoning suffices. Concept / Approach: Power transmitted P is proportional to Q * H_d. For a given pipeline, Q increases with head loss, but H_d decreases as h_f increases. Maximizing P with respect to h_f yields the optimal split. Step-by-Step Solution: 1) Let P = rho * g * Q * (H - h_f).2) For a given pipe, Q is proportional to sqrt(h_f) (for turbulent regimes with head loss proportional to V^2), so Q = k * h_f^(1/2).3) Then P = rho * g * k * h_f^(1/2) * (H - h_f).4) Differentiate with respect to h_f and set derivative to zero to maximize: dP/dh_f = 0 gives h_f = H / 3.5) Therefore maximum power occurs when friction loss is one-third of supply head. Verification / Alternative check: Substituting h_f = H/3 gives delivery head H_d = 2H/3 and confirms that small deviations reduce P on either side of this optimum. Why Other Options Are Wrong: One-fourth, one-half, two-third: These do not satisfy the first-order optimality condition and produce lower transmitted power. Common Pitfalls: Assuming maximum power at minimum losses; forgetting that Q depends on h_f; confusing efficiency optimization with power maximization. Final Answer: one-third of the total supply head

Question 4

Basic Property – Compressibility of Water Statement: For most engineering calculations at ordinary pressures, water is treated as an incompressible liquid.

Accepted Answer

Introduction: This question targets the standard modeling assumption regarding water compressibility in fluid mechanics and hydraulics. Recognizing when water can be idealized as incompressible simplifies analysis significantly. Given Data / Assumptions: Ambient temperatures and moderate pressures typical of water supply systems. Small density changes under pressure relative to gases. Concept / Approach: Compressibility describes density change with pressure. Water has a high bulk modulus, so density varies very little with pressure in common applications. Thus, continuity and momentum equations often assume constant density (incompressible flow). Step-by-Step Solution: 1) Identify property: liquids exhibit much smaller compressibility than gases.2) For water, bulk modulus is large, so delta rho / rho is negligible for moderate delta p.3) Therefore engineering analyses treat water as incompressible, except in phenomena like water hammer or very high-pressure systems. Verification / Alternative check: Compare typical density variation of water under several MPa to that of air; water changes are orders of magnitude smaller, supporting the incompressible idealization. Why Other Options Are Wrong: A compressible: Overly general for routine conditions. Weakly compressible only at very high pressure: Although qualitatively true, the statement in the stem asks for the standard classification used in typical calculations. Ideal gas: Physically incorrect. Common Pitfalls: Assuming perfect incompressibility even for rapid transients (e.g., water hammer) where compressibility matters; confusing thermodynamic compressibility with modeling simplifications. Final Answer: an incompressible

Question 5

Water Hammer in Pipes – Cause of Pressure Surge Statement: Water hammer occurs in pipelines due to a sudden change in the velocity of the flowing liquid (for example, rapid valve closure or pump trip).

Accepted Answer

Introduction: The question explores the physical cause of water hammer, a transient pressure surge that can damage pipes and fittings. Understanding its origin helps in designing protective measures such as surge tanks and slow-closing valves. Given Data / Assumptions: Pressurized pipe conveying a liquid (commonly water). Abrupt change in flow velocity imposed by system operation. Finite wave speed due to fluid compressibility and pipe wall elasticity. Concept / Approach: Sudden velocity changes create compression or expansion waves that travel at the elastic wave speed. The classic Joukowsky equation estimates the surge magnitude based on the instantaneous change in velocity. Step-by-Step Solution: 1) Root cause: rapid deceleration or acceleration of liquid mass.2) A pressure wave of magnitude delta p = rho * a * delta V propagates, where a is wave speed.3) Reflections at boundaries superimpose, producing high transient pressures (or vacuums).4) Therefore, sudden velocity change is the essential trigger for water hammer. Verification / Alternative check: Field observations show peak pressures immediately after quick valve closures; measurements agree with Joukowsky predictions for many systems. Why Other Options Are Wrong: Disagree: Ignores the fundamental mechanism. Occurs only due to temperature change: Thermal effects cause expansion but are not the classic water-hammer trigger. Occurs only in open channels: Water hammer is characteristic of pressurized conduits. Common Pitfalls: Assuming incompressibility eliminates water hammer; neglecting pipe wall elasticity; ignoring air pockets that can cushion or complicate transients. Final Answer: Agree

Question 6

Rectangular Weir – Discharge Formula with End Contractions Given Cd = 0.623, the standard SI form for discharge over a contracted rectangular weir is Q = 1.84 (L − 0.1 n H) H^(3/2), where L is crest length, H is head, and n is the number of end cont…

Accepted Answer

Introduction: The problem checks recognition of the widely used empirical discharge equation for a rectangular weir with end contractions, using a representative coefficient condensed into the constant 1.84 in SI units. Given Data / Assumptions: Rectangular sharp-crested weir with n end contractions. Head over crest H measured upstream under standard conditions. Composite constant 1.84 corresponds to Cd = 0.623 with g in SI. Concept / Approach: From the theoretical broad formula Q = (2/3) * Cd * b * sqrt(2 g) * H^(3/2), with end contractions the effective length is L_e = L − 0.1 n H. Collecting constants in SI yields the practical form Q = 1.84 (L − 0.1 n H) H^(3/2). Step-by-Step Solution: 1) Start with standard sharp-crested weir theory.2) Replace crest length by effective L_e = L − 0.1 n H.3) Use Cd = 0.623, combine constants to get 1.84.4) Final: Q = 1.84 (L − 0.1 n H) H^(3/2). Verification / Alternative check: Dimensional analysis confirms Q in m^3/s when L and H are in meters. Field calibration data support the exponent 3/2 for free overflow under nappe aeration. Why Other Options Are Wrong: H^2 and H^3 exponents are not supported by energy integration; H^(5/2) is not the classical sharp-crested relation. Replacing 0.1 n H with n H grossly overcorrects for contractions. Common Pitfalls: Measuring H too close to the crest; neglecting aeration or approach velocity corrections; miscounting contractions for sidewall abutments. Final Answer: 1.84(L - 0.1nH)H3/2

Question 7

Principle of Buoyancy – Dependence of Upthrust According to Archimedes’ principle, the buoyant force acting on a submerged or floating body equals the weight of the liquid displaced by the body.

Accepted Answer

Introduction: This item reviews Archimedes’ principle, a cornerstone of hydrostatics used to predict flotation, draft, and apparent weight reduction of submerged objects. Given Data / Assumptions: Object immersed in a static liquid. Hydrostatic pressure varies with depth but is at rest. Gravitational field is uniform. Concept / Approach: Net hydrostatic pressure on a submerged surface produces an upward resultant equal to the weight of the displaced liquid. For floating bodies, this upthrust balances the body weight at equilibrium. Step-by-Step Solution: 1) Compute pressure distribution p = rho * g * z on the surface.2) Integrate over wetted surface to get resultant upward force.3) Show that the resultant equals rho * g * V_displaced = weight of displaced liquid. Verification / Alternative check: Practical tests with hydrometers and floating vessels confirm that displacement weight equals vessel weight in equilibrium, validating the principle. Why Other Options Are Wrong: Pressure with which the liquid is displaced: Pressure distribution creates the force, but its integral equals the weight of displaced liquid, not the applied pressure value. Viscosity and compressibility do not determine static buoyant force in equilibrium. Common Pitfalls: Confusing buoyant force with drag; assuming it depends on object weight rather than displaced fluid; forgetting center of buoyancy location matters for stability but not magnitude. Final Answer: weight of the liquid displaced

Question 8

Fluid Mechanics — Energy Conservation for Ideal Incompressible Flow For a perfect (non-viscous) incompressible liquid flowing steadily along a continuous stream/streamline, the total head (pressure head + velocity head + datum head) of a fluid parti…

Accepted Answer

Introduction: Bernoulli's equation expresses conservation of mechanical energy for steady flow of an ideal (non-viscous, incompressible) fluid along a streamline. It relates pressure head, velocity head, and elevation head (datum head). Given Data / Assumptions: Fluid is incompressible. Flow is steady along a streamline. Viscous losses and shaft work are neglected (ideal conditions). No significant heat transfer; gravitational field is uniform. Concept / Approach: Bernoulli's statement: p/(rhog) + V^2/(2g) + z = constant along a streamline for ideal flow. Here p is pressure, rho is density, V is speed, g is gravitational acceleration, and z is elevation (datum). Step-by-Step Solution: Identify the heads: pressure head = p/(rhog), velocity head = V^2/(2g), datum head = z.For an ideal, steady, incompressible flow, the sum of these heads remains constant along a streamline.Therefore the statement in the question exactly matches Bernoulli's equation (or Bernoulli's theorem). Verification / Alternative check: If friction existed, the total head would drop; that situation is handled by the extended energy equation with head loss h_f. The question explicitly assumes a perfect incompressible liquid, so head loss is zero and Bernoulli applies directly. Why Other Options Are Wrong: Continuity equation: conserves mass (A*V constant) and does not state energy conservation. Pascal's law: pressure in a static fluid transmits equally in all directions; not an energy relation. Archimede's principle: buoyancy in static fluids; unrelated to flowing energy balance. Common Pitfalls: Confusing continuity (mass) with Bernoulli (energy); applying Bernoulli across pumps/turbines or across different streamlines without proper conditions; ignoring head losses in real fluids. Final Answer: Bernoulli's equation

Question 9

Manometer Fluid Selection — Required Properties The liquid used in manometers should have which characteristic so that pressure differences can be measured accurately and with short columns?

Accepted Answer

Introduction: Choosing a manometric fluid affects the sensitivity, size, and accuracy of pressure measurement devices. The key is to balance column height, capillary effects, vapor pressure, wetting, and chemical compatibility. Given Data / Assumptions: Static pressure measurement in U-tube or differential manometers. Small to moderate pressure differences typical of laboratory or industrial setups. Standard ambient conditions; clean, wettable tubes. Concept / Approach: Pressure difference is Delta p = rho * g * Delta h. For a given Delta p, the required column height is Delta h = Delta p / (rho * g). A larger density rho reduces the needed height, making instruments compact and less susceptible to reading error. Step-by-Step Solution: Start from Delta p = rhogDelta h.For fixed Delta p, increasing rho lowers Delta h and improves readability.High-density fluids (e.g., mercury) therefore allow shorter columns and higher stability. Verification / Alternative check: Compare water (rho about 1000 kg/m^3) to mercury (rho about 13,600 kg/m^3). For the same pressure, the mercury column is roughly 13.6 times shorter, confirming the benefit of high density. Why Other Options Are Wrong: Low density requires very tall columns, reducing practicality. Low or high surface tension alone does not guarantee accuracy; excessive capillarity can distort meniscus readings, but density is the primary requirement in this simplified list. Common Pitfalls: Ignoring vapor pressure (should be low), toxicity/compatibility, and temperature effects; reading the meniscus incorrectly; capillary rise in narrow tubes without correction. Final Answer: high density

Question 10

Hydraulic Gradient Line (HGL) vs Pipe Centre Line By how much does the hydraulic gradient line lie above the centre line of a flowing pipe, in terms of heads?

Accepted Answer

Introduction: The Hydraulic Gradient Line (HGL) and Energy Grade Line (EGL) are visual tools to interpret head distribution in pipe flow. Understanding what each represents helps diagnose pressure, cavitation risk, and available energy. Given Data / Assumptions: Incompressible flow in a full pipe. Steady conditions; elevation measured from a common datum. Neglecting minor losses for the conceptual definition. Concept / Approach: HGL represents piezometric head: p/(rhog) + z. The pipe centre line elevation is z. Therefore the vertical distance from the centre line up to HGL equals p/(rhog), which is the pressure head. The EGL is higher than the HGL by the velocity head V^2/(2g). Step-by-Step Solution: Write total head: H = p/(rhog) + V^2/(2g) + z.Define HGL = p/(rhog) + z.At the pipe centre line, elevation is z; distance to HGL is p/(rhog) = pressure head. Verification / Alternative check: Plot EGL and HGL for a uniform pipe: EGL - HGL = V^2/(2g) (velocity head). Thus HGL distance above the centre line is not velocity head but pressure head, confirming option A. Why Other Options Are Wrong: Velocity head corresponds to the vertical gap between EGL and HGL, not HGL and pipe centre line. Sum or difference of heads mixes concepts incorrectly for this definition. Common Pitfalls: Confusing EGL-HGL separation with HGL-to-centre-line spacing; ignoring that pressure can become negative (risk of cavitation) if HGL dips below vapour head. Final Answer: pressure head

Question 11

Effect of Viscosity on Orifice (Hole) Discharge The viscosity of a liquid ______ its rate of flow through a small hole (orifice) in a vessel under gravity.

Accepted Answer

Introduction: Real fluids experience internal friction (viscosity). When discharging through an orifice, viscous shear and boundary effects reduce the actual flow rate compared to the ideal inviscid prediction. Given Data / Assumptions: Small to moderate orifice, sharp-edged opening. Gravity-driven discharge from a tank. Steady conditions; no aeration or cavitation. Concept / Approach: Ideal discharge is given by Torricelli's formula Q_ideal = A * sqrt(2gh). Real discharge is Q = C_d * A * sqrt(2gh), where C_d (coefficient of discharge) accounts for contraction and viscosity. Higher viscosity reduces C_d and therefore reduces Q. Step-by-Step Solution: Start with Q = C_d * A * sqrt(2gh).As viscosity increases, energy losses rise and C_d decreases.Hence, for the same head h and area A, Q decreases. Verification / Alternative check: Compare water and glycerin under equal head: measured flow of glycerin is far smaller because its dynamic viscosity is orders of magnitude higher, confirming the decrease. Why Other Options Are Wrong: Increase: contradicts energy loss physics. No effect: true only in the inviscid idealization, which ignores real losses. First increases then decreases: not observed for ordinary ranges in simple orifices; monotonic reduction is typical as viscosity rises. Common Pitfalls: Using ideal Torricelli without discharge coefficients; ignoring temperature dependence of viscosity; misreading head levels causing apparent anomalies. Final Answer: decreases the rate of flow

Question 12

Definitions of Pressure — Comparing to Atmosphere When the pressure intensity at a point is greater than the local atmospheric pressure, the difference between them is called:

Accepted Answer

Introduction: Engineering pressure terms distinguish between measurements relative to absolute zero and measurements relative to the local atmosphere. Correct terminology avoids confusion in design and safety calculations. Given Data / Assumptions: Local atmospheric pressure p_atm is the reference for many instruments. Measured pressure p is greater than p_atm. Concept / Approach: Absolute pressure p_abs is measured from absolute vacuum. Gauge pressure p_g is measured relative to p_atm. If p > p_atm, then p_g = p - p_atm is positive and is called gauge pressure (often implicitly positive). Vacuum pressure typically refers to p_atm - p when p < p_atm. Step-by-Step Solution: Compute p_g = p - p_atm.Given p > p_atm, p_g > 0, named gauge pressure.Absolute pressure would be p_abs, not the difference; vacuum pressure applies when p < p_atm. Verification / Alternative check: Typical pressure gauges read zero at atmosphere. A reading of 200 kPa on a gauge implies 200 kPa gauge, not absolute; absolute would be approximately 200 + 101.3 ≈ 301.3 kPa (assuming standard atmosphere). Why Other Options Are Wrong: Absolute pressure: not a difference; it includes atmospheric baseline. Positive gauge pressure: a descriptive phrase, but the precise term is simply gauge pressure. Vacuum pressure: used when pressure is below atmospheric (negative gauge). Common Pitfalls: Confusing gauge and absolute scales; mixing units; forgetting local altitude and weather change p_atm slightly. Final Answer: gauge pressure

Question 13

Floating Condition — Balance of Forces When a body is placed on or in a liquid, it will float if which of the following is true?

Accepted Answer

Introduction: Floating and sinking are governed by Archimedes' principle and the force balance between weight and buoyant force (upthrust). Understanding this helps in ship design and stability assessments. Given Data / Assumptions: Body partially or fully submerged in a static liquid. Forces considered: weight W and buoyant force B. Equilibrium, negligible wave motion. Concept / Approach: Archimedes: Upthrust equals weight of displaced liquid. A body floats in static equilibrium when net vertical force is zero and restoring moments are adequate for stability. That requires B = W for neutral vertical equilibrium. Step-by-Step Solution: Write vertical force balance: Sum F_z = B - W.For floating without acceleration, Sum F_z = 0 ⇒ B = W.If B > W, the body accelerates upward until displacement reduces so that B = W; if B < W, it sinks until new displacement restores B = W or it rests on bottom. Verification / Alternative check: Consider a block of density rho_b in water of density rho_w. Fraction submerged f ≈ rho_b / rho_w so that displaced water weight equals block weight; this reproduces B = W at floatation. Why Other Options Are Wrong: B > W: transient upward motion, not static float condition. B < W: sinking occurs until support or new equilibrium is found. None of the above: incorrect because B = W is the correct float criterion. Common Pitfalls: Confusing stability (metacentric height) with simple float condition; ignoring density variations with temperature/salinity; overlooking trapped air effects. Final Answer: gravitational force is equal to the upthrust of the liquid

Question 14

Weir Discharge Formula — Ogee vs Sharp-Crested Rectangular Is the discharge formula for a sharp-crested weir and an ogee (ogee-crest) weir effectively the same as that of a rectangular sharp-crested weir, subject to appropriate coefficients?

Accepted Answer

Introduction: Weir equations relate discharge to head over the crest. Ogee weirs are shaped to follow the lower nappe of a sharp-crested weir, allowing high accuracy at design heads. Given Data / Assumptions: Free, fully aerated nappe conditions. Steady approach flow with appropriate approach velocity correction. Empirical coefficient of discharge C_d applied. Concept / Approach: For sharp-crested rectangular and ogee weirs, Q ≈ C_d * (2/3) * b * sqrt(2*g) * H^(3/2), with b as crest width and H as head above crest. Ogee weirs use similar head-discharge relation but with coefficients tuned to crest shape and approach conditions. Step-by-Step Solution: Start from the theoretical rectangular sharp-crest formula.Apply discharge coefficient C_d to account for contraction and viscosity.For an ogee crest, the same head exponent (3/2) applies; the profile mainly affects C_d and sensitivity to submergence. Verification / Alternative check: Standards (e.g., IS/USBPR/USACE) tabulate C_d for ogee and sharp crests; both retain Q proportional to H^(3/2) under free overflow, confirming functional sameness with adjusted coefficients. Why Other Options Are Wrong: Disagree: ignores the common formulation with different coefficients. Depends on notch angle only: angle pertains to V-notch (triangular) weirs, not rectangular/ogee. Only valid for triangular weirs: incorrect; the rectangular/ogee relation is independent of V-notch theory. Common Pitfalls: Using rectangular formula without velocity of approach correction; ignoring submergence; misapplying coefficients outside design head range. Final Answer: Agree

Question 15

Free Vortex — Speed–Radius Relationship For a free vortex (no external torque), the tangential velocity of a water particle varies with distance r from the centre as:

Accepted Answer

Introduction: Vortical motions are classified as free (irrotational, no forced swirl) or forced (solid-body rotation). The velocity–radius law distinguishes them and determines pressure distribution and surface shape. Given Data / Assumptions: Free vortex in incompressible, inviscid approximation. No external torque; angular momentum of each fluid particle is conserved. Steady, axisymmetric flow. Concept / Approach: For a free vortex, tangential momentum mv_thetar is constant along a streamline (neglecting viscosity). Therefore v_theta * r = constant ⇒ v_theta ∝ 1/r. The flow is irrotational except at the core. Step-by-Step Solution: Use angular momentum conservation: v_theta*r = K (constant).Rearrange to get v_theta = K/r.Hence as r increases, v_theta decreases proportionally to 1/r. Verification / Alternative check: Measured velocities in drain whirlpools and tornado outer regions follow approximately v ∝ 1/r, unlike forced vortices where v ∝ r. Why Other Options Are Wrong: v ∝ r: describes forced vortex (solid-body) rotation. v ∝ r^2 or v ∝ 1/r^2: do not conserve angular momentum for free vortices. Common Pitfalls: Confusing free with forced vortex; assuming solid-body rotation where none exists; ignoring viscous core where the ideal relation breaks down. Final Answer: inversely proportional to r

Question 16

Hydraulic Mean Depth (Hydraulic Radius) — Full Circular Pipe For a circular pipe of diameter d flowing full, the hydraulic mean depth R = A/P equals:

Accepted Answer

Introduction: The hydraulic mean depth (hydraulic radius) R = A/P is a key parameter in flow formulas such as Chezy and Manning. For common sections, closed-form expressions exist, simplifying design checks. Given Data / Assumptions: Circular conduit of diameter d flowing full. Area A = (pid^2)/4, Wetted perimeter P = pid. Steady uniform flow conceptually. Concept / Approach: Apply the definition R = A/P directly to the circular section values. Step-by-Step Solution: Compute A = (pid^2)/4.Compute P = pid.Then R = A/P = [(pid^2)/4]/(pid) = d/4. Verification / Alternative check: Dimensional check: R has dimensions of length; d/4 has length, consistent. Known tables list R = d/4 for a full circular pipe, confirming. Why Other Options Are Wrong: d/6 and d/2 and d are not obtained from A/P for the full circular case; they may appear in other hydraulic relations (e.g., depth to centroid), but not R. Common Pitfalls: Confusing hydraulic radius with radius of the pipe; using diameter in place of wetted perimeter for partially full conditions; mixing up A/P with perimeter/area. Final Answer: d/4

Question 17

Compressible Flow — Mach Number Classification When the Mach number M is less than 1 (M < 1), the flow regime is called:

Accepted Answer

Introduction: Mach number M = V/a (ratio of flow speed to local speed of sound) classifies compressible flow regimes and dictates which physical effects dominate, such as compressibility and shock formation. Given Data / Assumptions: Definition of Mach number M = V/a. Local thermodynamic state determines a (speed of sound). We consider standard aerodynamic regimes. Concept / Approach: Flow regimes: subsonic (M < 1), transonic (around M ≈ 1 with mixed zones), supersonic (M > 1 up to about 5), hypersonic (typically M ≥ 5). Sonic flow strictly refers to M = 1 at a point (e.g., throat of a choked nozzle). Step-by-Step Solution: Given M < 1, compressibility effects are mild to moderate; no detached shocks form in uniform external flow.Therefore the correct classification is subsonic. Verification / Alternative check: Textbook tables list standard ranges: subsonic M < 1; transonic ~0.8–1.2; supersonic >1; hypersonic ≥5. Our case falls clearly in subsonic. Why Other Options Are Wrong: Sonic: applies at M = 1, not M < 1. Super-sonic: requires M > 1. Hyper-sonic: extreme range M ≥ 5. Common Pitfalls: Confusing transonic with sonic; ignoring that local regions can reach M ≈ 1 even when freestream is below 1; using sea-level a for high-altitude cases without correction. Final Answer: sub-sonic flow

Question 18

A uniform rectangular body floats in water. Given: length = 3 m, width = 2 m, overall depth = 1 m. The observed depth of immersion is 0.6 m. Find the weight of the body (assume fresh water).

Accepted Answer

Introduction / Context: Buoyancy tells us that a floating body displaces a volume of liquid whose weight equals the body's weight. This question checks understanding of Archimedes' principle and how to compute displaced volume from plan area and immersion depth, then convert that displaced water into weight using density * gravity. Given Data / Assumptions: Length L = 3 m Width B = 2 m Overall depth (thickness) = 1 m (not directly needed for buoyancy at float) Immersion depth h = 0.6 m Fresh water density rho = 1000 kg/m^3 g = 9.81 m/s^2 Concept / Approach: For a floating body at rest, Weight of body W = Weight of displaced liquid. Displaced volume V = plan area * immersion depth. Then W = rho * g * V. Convert to kN by dividing by 1000. Step-by-Step Solution: Plan area A = L * B = 3 * 2 = 6 m^2Displaced volume V = A * h = 6 * 0.6 = 3.6 m^3Weight W = rho * g * V = 1000 * 9.81 * 3.6 NCompute: 9.81 * 3.6 = 35.316; so W = 35.316 kN ≈ 35.3 kN Verification / Alternative check: If g is rounded to 9.8, W ≈ 1000 * 9.8 * 3.6 = 35.28 kN, still ≈ 35.3 kN. Either way, the same option is selected. Why Other Options Are Wrong: 3.53 kN: Off by a factor of 10; likely confusion between meters and centimeters or forgetting g. 33.3 kN: Underestimates g (using 9.25) or rounding volume incorrectly. None of these: Incorrect because 35.3 kN is attainable and correct. Common Pitfalls: Using overall depth of 1 m instead of immersion depth; forgetting to multiply by g; using mass (kg) instead of weight (N); mixing units when converting to kN. Final Answer: 35.3 kN

Question 19

Critical depth in a rectangular channel in terms of unit discharge q (discharge per unit width): Select the correct expression for y_c.

Accepted Answer

Introduction / Context: Critical flow in open channels corresponds to Froude number Fr = 1. For a rectangular channel, the critical depth y_c is directly related to the unit discharge q (discharge per unit width). Correctly recalling this formula is fundamental in flow measurement and control structure design. Given Data / Assumptions: Rectangular channel of unit width for unit discharge definition Steady, prismatic, frictionless control section Acceleration due to gravity g is constant Concept / Approach: For a rectangular channel, specific energy E = y + v^2/(2g). At critical flow, dE/dy = 0 which leads to v_c^2 = g * y_c. Since q = v * y for unit width, combine to obtain y_c in terms of q and g. Step-by-Step Solution: At critical: v_c^2 = g * y_cUse q = v_c * y_c ⇒ v_c = q / y_cSubstitute: (q / y_c)^2 = g * y_cRearrange: q^2 = g * y_c^3Solve: y_c = (q^2 / g)^(1/3) Verification / Alternative check: Dimensional check: [q] = m^2/s for unit width; q^2/g has dimension m^3; cube root gives meters, consistent for depth. Why Other Options Are Wrong: (g / q^2)^(1/3): Inverts the relation; would have dimension 1/m, not a length. (q / g)^(1/2): Wrong power and units. q / g: Linear in q, dimension m^2/s^2, not a length. Common Pitfalls: Forgetting that q is per unit width; applying the same formula to non-rectangular sections without the appropriate geometric function; mixing up specific energy and specific force criteria. Final Answer: y_c = (q^2 / g)^(1/3)

Question 20

Weirs and End Contractions – Suppressed vs contracted crest When the end contractions of a rectangular weir are suppressed (i.e., the nappe is allowed to touch the side walls), the number of end contractions n used in the standard discharge formula …

Accepted Answer

Introduction: Rectangular weirs are categorized as suppressed or contracted depending on whether the nappe contracts at the side walls. In discharge calculations, a correction using the count of end contractions n is applied to the effective width. This question checks whether the learner knows that in a fully suppressed weir, the side contractions are absent and thus n is taken as zero in the standard formula. Given Data / Assumptions: Rectangular sharp-crested weir under free flow conditions. “Suppressed” means the weir crest spans the full channel width and the nappe touches the side walls. Standard empirical discharge correlations for small to moderate heads are assumed. Concept / Approach: The practical discharge expression for a rectangular weir includes side contraction adjustments. For a contracted weir, the effective width b_e is reduced from the physical width b by a term proportional to n * h, where h is the head over crest and n is the number of end contractions (0, 1, or 2). A “suppressed” weir has no lateral contraction, therefore n = 0, and the effective width equals the channel width. This increases discharge relative to an otherwise similar contracted weir at the same head. Step-by-Step Solution: Step 1: Identify the weir type: “end contractions suppressed” implies no side contraction.Step 2: Recall the effective width relation used in many handbooks: b_e = b − k * n * h, with k near unity for sharp crests.Step 3: For a suppressed weir, set n = 0, giving b_e = b.Step 4: Conclude that the statement “take n = 0” is correct for a fully suppressed rectangular weir. Verification / Alternative check: Laboratory correlations and design manuals consistently treat suppressed rectangular weirs with n = 0 because the side nappe is constrained by the channel walls and cannot contract laterally. Why Other Options Are Wrong: Disagree: Contradicts the standard definition of a suppressed weir.n = 1 for one side suppressed: That setting corresponds to exactly one contracted side, which is not “fully suppressed”.n = 2 for fully suppressed: Reverses the meaning; n = 2 is for two end contractions (both sides free).n depends only on head: Side contraction depends on geometry, not only on head. Common Pitfalls: Confusing “suppressed” with “submerged” or thinking the contraction count changes with head. Suppression refers to geometry (absence of side contraction), so n = 0 is a structural condition. Final Answer: Agree

Hydraulics and Fluid Mechanics Questions