Q: Jet velocity from a tank orifice (Torricelli) A tank has a circular hole of diameter 2.5 cm in its vertical side. The water level is 50 m above the hole center. Neglecting losses, estimate the exit jet speed from the hole.

Introduction / Context: Torricelli’s theorem provides a quick estimate of exit speed from an orifice under a static head. It is frequently used for preliminary sizing of outlets and for sanity checks before applying detailed head-loss corrections. Given Data / Assumptions: Head h = 50 m measured to the orifice centerline. Neglect losses and velocity-of-approach; treat as a sharp orifice discharging to atmosphere. Gravity g = 9.81 m/s^2. Concept / Approach: Torricelli’s formula for ideal exit speed is v = sqrt(2 g h). The orifice diameter affects flow rate, not the ideal speed (under these assumptions). Step-by-Step Solution: Compute: v = sqrt(2 * 9.81 * 50) = sqrt(981) ≈ 31.3209 m/s.Round to one decimal place: ≈ 31.3 m/s.Hence, the correct choice is 31.3 m/s. Verification / Alternative check: Bernoulli between the free surface and the vena contracta with p_atm both sides and negligible losses reduces to the same expression. Including a discharge coefficient would reduce flow rate, not this ideal speed selection. Why Other Options Are Wrong: Values 31.1, 31.2, and 31.4 m/s are close but less accurate rounded forms; 31.5 m/s overshoots the ideal square-root value. Common Pitfalls: Using head to the bottom of the hole instead of its center; confusing speed (from head) with actual volumetric discharge (which also depends on area and C_d). Final Answer: 31.3 m/s

Q: Manometer capillarity avoidance – minimum tube diameter To minimize meniscus curvature and avoid significant capillary correction in pressure measurement, the manometer tube should have at least what internal diameter?

Introduction / Context: Capillary rise or depression in narrow tubes distorts liquid column readings in manometers, especially with water and clean glass. Selecting a sufficiently large internal diameter helps keep errors negligible without complex meniscus corrections. Given Data / Assumptions: Clean glass tube and water (or similar wetting fluid). Room temperature conditions where surface tension is typical. Objective: keep capillary rise/depression small relative to resolution. Concept / Approach: Capillary height is inversely proportional to tube radius: h_c ≈ 2 σ cos θ / (ρ g r). Increasing diameter (2r) reduces h_c rapidly. Practical laboratory and field practice often adopts ≥ 6 mm I.D. to make capillary corrections negligible for routine measurements. Step-by-Step Solution: Recognize h_c ∝ 1/r; doubling radius halves the capillary effect.Common guidance: for water–glass, I.D. ≥ 6 mm limits h_c to a few tenths of a millimeter, typically within reading uncertainty.Hence, select at least 6 mm internal diameter. Verification / Alternative check: Using typical values (σ ≈ 0.072 N/m, θ ≈ 0°, ρ ≈ 1000 kg/m^3) gives small h_c when r ≥ 3 mm. Many textbooks recommend 6–10 mm I.D.; 6 mm is the common minimum. Why Other Options Are Wrong: (a) and (b) are too small; capillarity is significant. (d) and (e) are acceptable but unnecessarily large relative to the “minimum” asked; the conventional cut-off is about 6 mm. Common Pitfalls: Ignoring contact-angle effects with dirty or coated tubes; forgetting that mercury behaves oppositely (depression instead of rise). Always read at consistent meniscus position if correction is unavoidable. Final Answer: 6 mm

Q: Capillarity in a glass tube (cohesion vs. adhesion): If the cohesion between fluid molecules exceeds the adhesion between the fluid and clean glass, what happens to the free liquid level inside a thin glass capillary dipped into the same liquid bath?

Introduction / Context: Capillarity describes how liquids rise or fall in narrow tubes due to the balance of surface tension forces and wetting behavior. Whether a liquid climbs (capillary rise) or depresses (capillary depression) depends on the relative magnitudes of cohesion (molecule–molecule attraction) and adhesion (liquid–solid attraction). Given Data / Assumptions: Clean, wettable glass tube of small internal diameter. Fluid with cohesion greater than adhesion to glass (non-wetting on glass). Atmospheric conditions; static equilibrium; gravity acts downward. Concept / Approach: When adhesion > cohesion (e.g., water on clean glass), the meniscus is concave and the liquid rises. When cohesion > adhesion (e.g., mercury on glass), the meniscus is convex and the liquid level inside the tube is depressed relative to the reservoir. The vertical displacement is governed by surface tension and contact angle via the classical capillary relation h = (2 σ cos θ) / (ρ g r). For non-wetting liquids, θ > 90°, cos θ is negative, leading to a negative h (depression). Step-by-Step Solution: Identify wetting: cohesion > adhesion ⇒ non-wetting behavior on glass.Non-wetting implies contact angle θ > 90° and a convex meniscus.Use h = (2 σ cos θ) / (ρ g r): with cos θ < 0, h is negative.Therefore, the capillary level inside the tube lies below the outside free surface. Verification / Alternative check: Mercury in glass is the canonical example: cohesion dominates, producing a convex meniscus and a depressed column inside capillaries. Why Other Options Are Wrong: (a) describes capillary rise (adhesion-dominant); (b) would be true only if the tube were very wide (negligible capillarity); (d) viscosity does not set the static height; (e) no oscillatory static profile occurs. Common Pitfalls: Confusing dynamic wetting/viscosity with static capillary height; mixing up concave vs. convex menisci. Final Answer: lower than the outside free surface

Q: Euler’s equation for inviscid flow — key assumptions: Which of the following is a core assumption used in deriving Euler’s equation of motion for liquids in continuum mechanics?

Introduction / Context: Euler’s equation provides the inviscid form of the linear momentum balance along a streamline (or in vector form in the field). It underpins Bernoulli’s equation and many first-approximation models in hydraulics and aerodynamics when viscous effects are negligible. Given Data / Assumptions: Continuum fluid with negligible viscosity (inviscid). Often treated as homogeneous and incompressible for liquids. Body forces (usually gravity) may act. No requirement that the flow be steady; Euler’s equation has both steady and unsteady forms. Concept / Approach: Euler’s equation is derived from Newton’s second law applied to a fluid element, neglecting viscous stresses (no shear stress terms). For many liquid-flow problems, additional simplification assumes a homogeneous, incompressible fluid. These assumptions lead to the familiar relationships that integrate to Bernoulli’s equation under steady flow. Step-by-Step Solution: State the inviscid momentum balance: pressure + inertia + body-force terms.Neglect viscous shear stresses entirely (ideal fluid).Assume fluid density is constant in space and time (homogeneous, incompressible).Conclude that option (b) captures a central assumption routinely invoked in the derivation for liquids. Verification / Alternative check: Bernoulli’s equation follows from steady, incompressible, inviscid Euler along a streamline, confirming the appropriateness of the assumption. Why Other Options Are Wrong: (a) contradicts the inviscid premise; (c) non-uniformity is permitted but not an assumption; (d) Euler allows steady or unsteady; (e) surface tension is generally negligible in bulk momentum balances. Common Pitfalls: Equating Euler’s equation with only steady flow; assuming viscosity can be small yet included—by definition Euler neglects it. Final Answer: The fluid is homogeneous and incompressible

Q: Definition of Mach number in fluid mechanics: Mach number (M) is defined as the ratio of inertial forces (or flow speed) to which fundamental resisting effect?

Introduction / Context: Mach number is central to compressible-flow analysis. It classifies regimes such as subsonic, transonic, supersonic, and hypersonic, determining whether compressibility and shock waves are important in aerodynamic and gas-dynamic problems. Given Data / Assumptions: Mach number M = V / a, where V is flow speed and a is speed of sound. Speed of sound encapsulates the medium’s elasticity via a = sqrt(∂p/∂ρ)_s. Continuum, single-phase, no phase change in the regime considered. Concept / Approach: Dimensionally, M compares inertial effects to elastic (compressibility) effects in the fluid. High M implies inertia dominates and compressibility cannot be ignored, while low M indicates quasi-incompressible behavior. Thus, Mach number represents a ratio of inertia to elasticity effects, not to viscosity, surface tension, or gravity. Step-by-Step Solution: Recognize M = V/a expresses inertia relative to elastic wave propagation.Relate a to elasticity through thermodynamic derivatives (e.g., isentropic bulk modulus).Conclude that the resisting “force” class is elasticity (compressibility), matching option (d). Verification / Alternative check: In Buckingham Pi terms, M arises from balancing inertia with compressibility, distinct from Reynolds (viscosity) and Weber (surface tension) numbers. Why Other Options Are Wrong: (a) relates to Reynolds number; (b) to Weber number; (c) to Froude number; (e) is irrelevant in standard fluid dynamics contexts. Common Pitfalls: Confusing Mach with Reynolds or Froude; assuming low M always means incompressible regardless of temperature effects—usually acceptable but context dependent. Final Answer: elasticity (compressibility)

Q: Jet trajectory from a free nozzle into air: When a liquid issues horizontally from a nozzle into the atmosphere (neglecting air resistance close to the nozzle), what is the ideal shape of the jet path under gravity?

Introduction / Context: Once a jet leaves a nozzle and is exposed to the atmosphere, it behaves like a continuous stream of fluid particles subjected mainly to gravity. Close to the nozzle (neglecting aerodynamic breakup), each fluid particle follows projectile motion principles. Given Data / Assumptions: Jet issues horizontally at speed u from a nozzle into still air. Air resistance is neglected over the short distance of interest. Gravity g acts downward; initial vertical velocity is zero. Concept / Approach: Projectile motion under constant gravitational acceleration produces a parabolic path. The horizontal velocity component remains constant (u), while the vertical component increases linearly with time (g t). Eliminating time yields y = (g x^2) / (2 u^2), the equation of a parabola in x–y coordinates. Step-by-Step Solution: Horizontal motion: x = u t ⇒ t = x / u.Vertical motion: y = (1/2) g t^2.Eliminate t: y = (1/2) g (x/u)^2 = (g x^2) / (2 u^2).Conclude geometric form is a parabola opening downward. Verification / Alternative check: Dimensional check: y has dimensions of length; expression (g/u^2) x^2 is consistent. Observationally, short horizontal jets (e.g., water taps) display a parabolic profile before breaking into droplets. Why Other Options Are Wrong: (a) would require zero gravity; (b) and (c) require centripetal constraints not present; (e) does not arise from uniform gravity with constant horizontal speed. Common Pitfalls: Including significant air drag near the nozzle; forgetting the vertical initial velocity is zero for a horizontal jet. Final Answer: parabolic trajectory

Q: Basic properties of liquids (continuum viewpoint): Which statement best characterizes liquids compared with solids and gases under ordinary conditions?

Introduction / Context: Understanding the macroscopic properties of liquids is foundational for hydraulics, process engineering, and materials handling. These properties differentiate liquids from solids (rigidity) and gases (high compressibility and expansibility). Given Data / Assumptions: Normal temperature and pressure, single-phase liquid. Continuum mechanics model (ignoring molecular-scale details). No phase change during observation. Concept / Approach: Liquids have very low compressibility, significant cohesion, and finite surface tension. They resist shear only transiently (flow under sustained shear) and thus do not maintain a fixed shape; instead they conform to the container, preserving essentially constant volume (neglecting slight compressibility). Step-by-Step Solution: Identify shape behavior: liquids flow and take the container’s shape.Recognize compressibility: small but nonzero (bulk modulus is large but finite).Acknowledge sensitivity to pressure/temperature: density and viscosity vary with both.Thus option (b) best captures the defining characteristic. Verification / Alternative check: Everyday observation (water in a glass) and standard property tables (bulk modulus, thermal expansion coefficient) support the description. Why Other Options Are Wrong: (a) Incorrect—compressibility is small but not zero; (c) false—properties change with T and p; (d) false—liquids do not expand to fill any volume; (e) false—surface tension is finite in equilibrium. Common Pitfalls: Assuming “incompressible” means mathematically zero compressibility; conflating liquid behavior with gas behavior. Final Answer: Liquids do not retain a definite shape but assume the container’s shape

Q: Error propagation for triangular weirs/notches: For a V-notch (triangular notch), discharge varies with head as Q ∝ H^(5/2). What is the ratio of the percentage error in discharge to the percentage error in measured head over the notch?

Introduction / Context: Flow measurement over notches relies on empirical–theoretical relations between head (H) and discharge (Q). Understanding how measurement errors in H propagate into errors in Q is crucial for instrumentation accuracy and calibration protocols. Given Data / Assumptions: Triangular (V-notch) weir with Q = K H^(5/2), where K includes notch angle and discharge coefficient. Small measurement errors relative to the nominal head (linearized error propagation). Steady flow; proper approach conditions. Concept / Approach: For Q = K H^n, fractional errors relate by dQ/Q = n dH/H. Thus, the percentage error in Q is n times the percentage error in H. For a V-notch, n = 5/2, significantly amplifying head measurement errors. Step-by-Step Solution: Start with Q = K H^(5/2).Take logarithms: ln Q = ln K + (5/2) ln H.Differentiate: dQ/Q = (5/2) dH/H.Therefore, %error(Q) / %error(H) = 5/2. Verification / Alternative check: Compare with rectangular notch where Q ∝ H^(3/2): the ratio would be 3/2, confirming the exponent-rule consistency. Why Other Options Are Wrong: (a), (b), (c), and (e) correspond to other exponents or are arbitrary; only 5/2 matches the V-notch exponent. Common Pitfalls: Using absolute errors instead of fractional; forgetting that the exponent multiplies the fractional error. Final Answer: 5/2

Question 1

Identifying steady, uniform flow – examples Which of the following most correctly represents a steady, uniform flow situation?

Accepted Answer

Introduction / Context: Textbook examples help cement the difference between steady/unsteady (time variation) and uniform/non-uniform (space variation). Correct classification supports valid use of Bernoulli and continuity in designs. Given Data / Assumptions: “Long pipe” implies constant cross-section and adequate entrance length. “Rate” refers to volumetric flow rate Q. Fluid is incompressible and properties are time-invariant for steady cases. Concept / Approach: Uniform flow means velocity magnitude is the same at successive cross-sections (averaged). Steady flow means velocity at each fixed point is time-invariant. A constant Q in a constant-area conduit satisfies both. Step-by-Step Solution: Constant-area pipe + constant Q ⇒ constant mean velocity V = Q/A.At each cross-section, mean V is the same ⇒ uniform.No time variation ⇒ steady. Verification / Alternative check: In an expanding tube at constant Q, velocity changes along the length (non-uniform), though it may be steady. Changing Q with time makes the flow unsteady regardless of geometry. Why Other Options Are Wrong: (a) and (e) vary with time → unsteady. (c) Steady but non-uniform. (d) Both non-uniform and unsteady. Common Pitfalls: Equating “steady” with “same everywhere”; assuming developing boundary layers violate uniformity—here we refer to section-averaged velocity and a sufficiently long, fully developed pipe. Final Answer: a long pipe at constant rate

Question 2

Gauge pressure – basic definition Which expression correctly defines gauge pressure in terms of absolute pressure and atmospheric pressure?

Accepted Answer

Introduction / Context: Pressure measurements in engineering commonly use gauge readings that exclude atmospheric pressure. Converting among absolute, gauge, and vacuum pressures is routine for pump sizing, vessel design, and HVAC systems. Given Data / Assumptions: Absolute pressure p_abs is measured relative to absolute vacuum. Atmospheric pressure p_atm is the local ambient pressure. Gauge pressure p_g is what a typical Bourdon gauge reports (relative to p_atm). Concept / Approach: By definition, p_g = p_abs − p_atm. If p_abs < p_atm, p_g becomes negative and is often reported as a vacuum pressure p_vac = p_atm − p_abs. Step-by-Step Solution: Start from absolute reference: p_abs = p_atm + p_g.Rearrange: p_g = p_abs − p_atm.For vacuum readings: p_vac = p_atm − p_abs = −p_g (when p_g negative). Verification / Alternative check: Instrumentation datasheets and standards consistently present p_abs = p_atm + p_g. Sample calculation: at sea level p_atm ≈ 101.3 kPa; if p_abs = 250 kPa, then p_g ≈ 149 kPa. Why Other Options Are Wrong: (b) Sum is incorrect; would double-count atmosphere. (c) This is vacuum pressure, not gauge pressure. (d) A correct relationship exists. (e) Mixing terms; vacuum pressure plus atmospheric equals absolute, not gauge. Common Pitfalls: Confusing negative gauge pressure with negative absolute pressure (which is impossible physically); forgetting that most gauges read zero at atmospheric conditions. Final Answer: absolute pressure - atmospheric pressure

Question 3

Jet velocity from a tank orifice (Torricelli) A tank has a circular hole of diameter 2.5 cm in its vertical side. The water level is 50 m above the hole center. Neglecting losses, estimate the exit jet speed from the hole.

Accepted Answer

Introduction / Context: Torricelli’s theorem provides a quick estimate of exit speed from an orifice under a static head. It is frequently used for preliminary sizing of outlets and for sanity checks before applying detailed head-loss corrections. Given Data / Assumptions: Head h = 50 m measured to the orifice centerline. Neglect losses and velocity-of-approach; treat as a sharp orifice discharging to atmosphere. Gravity g = 9.81 m/s^2. Concept / Approach: Torricelli’s formula for ideal exit speed is v = sqrt(2 g h). The orifice diameter affects flow rate, not the ideal speed (under these assumptions). Step-by-Step Solution: Compute: v = sqrt(2 * 9.81 * 50) = sqrt(981) ≈ 31.3209 m/s.Round to one decimal place: ≈ 31.3 m/s.Hence, the correct choice is 31.3 m/s. Verification / Alternative check: Bernoulli between the free surface and the vena contracta with p_atm both sides and negligible losses reduces to the same expression. Including a discharge coefficient would reduce flow rate, not this ideal speed selection. Why Other Options Are Wrong: Values 31.1, 31.2, and 31.4 m/s are close but less accurate rounded forms; 31.5 m/s overshoots the ideal square-root value. Common Pitfalls: Using head to the bottom of the hole instead of its center; confusing speed (from head) with actual volumetric discharge (which also depends on area and C_d). Final Answer: 31.3 m/s

Question 4

Manometer capillarity avoidance – minimum tube diameter To minimize meniscus curvature and avoid significant capillary correction in pressure measurement, the manometer tube should have at least what internal diameter?

Accepted Answer

Introduction / Context: Capillary rise or depression in narrow tubes distorts liquid column readings in manometers, especially with water and clean glass. Selecting a sufficiently large internal diameter helps keep errors negligible without complex meniscus corrections. Given Data / Assumptions: Clean glass tube and water (or similar wetting fluid). Room temperature conditions where surface tension is typical. Objective: keep capillary rise/depression small relative to resolution. Concept / Approach: Capillary height is inversely proportional to tube radius: h_c ≈ 2 σ cos θ / (ρ g r). Increasing diameter (2r) reduces h_c rapidly. Practical laboratory and field practice often adopts ≥ 6 mm I.D. to make capillary corrections negligible for routine measurements. Step-by-Step Solution: Recognize h_c ∝ 1/r; doubling radius halves the capillary effect.Common guidance: for water–glass, I.D. ≥ 6 mm limits h_c to a few tenths of a millimeter, typically within reading uncertainty.Hence, select at least 6 mm internal diameter. Verification / Alternative check: Using typical values (σ ≈ 0.072 N/m, θ ≈ 0°, ρ ≈ 1000 kg/m^3) gives small h_c when r ≥ 3 mm. Many textbooks recommend 6–10 mm I.D.; 6 mm is the common minimum. Why Other Options Are Wrong: (a) and (b) are too small; capillarity is significant. (d) and (e) are acceptable but unnecessarily large relative to the “minimum” asked; the conventional cut-off is about 6 mm. Common Pitfalls: Ignoring contact-angle effects with dirty or coated tubes; forgetting that mercury behaves oppositely (depression instead of rise). Always read at consistent meniscus position if correction is unavoidable. Final Answer: 6 mm

Question 5

Capillarity in a glass tube (cohesion vs. adhesion): If the cohesion between fluid molecules exceeds the adhesion between the fluid and clean glass, what happens to the free liquid level inside a thin glass capillary dipped into the same liquid bath?

Accepted Answer

Introduction / Context: Capillarity describes how liquids rise or fall in narrow tubes due to the balance of surface tension forces and wetting behavior. Whether a liquid climbs (capillary rise) or depresses (capillary depression) depends on the relative magnitudes of cohesion (molecule–molecule attraction) and adhesion (liquid–solid attraction). Given Data / Assumptions: Clean, wettable glass tube of small internal diameter. Fluid with cohesion greater than adhesion to glass (non-wetting on glass). Atmospheric conditions; static equilibrium; gravity acts downward. Concept / Approach: When adhesion > cohesion (e.g., water on clean glass), the meniscus is concave and the liquid rises. When cohesion > adhesion (e.g., mercury on glass), the meniscus is convex and the liquid level inside the tube is depressed relative to the reservoir. The vertical displacement is governed by surface tension and contact angle via the classical capillary relation h = (2 σ cos θ) / (ρ g r). For non-wetting liquids, θ > 90°, cos θ is negative, leading to a negative h (depression). Step-by-Step Solution: Identify wetting: cohesion > adhesion ⇒ non-wetting behavior on glass.Non-wetting implies contact angle θ > 90° and a convex meniscus.Use h = (2 σ cos θ) / (ρ g r): with cos θ < 0, h is negative.Therefore, the capillary level inside the tube lies below the outside free surface. Verification / Alternative check: Mercury in glass is the canonical example: cohesion dominates, producing a convex meniscus and a depressed column inside capillaries. Why Other Options Are Wrong: (a) describes capillary rise (adhesion-dominant); (b) would be true only if the tube were very wide (negligible capillarity); (d) viscosity does not set the static height; (e) no oscillatory static profile occurs. Common Pitfalls: Confusing dynamic wetting/viscosity with static capillary height; mixing up concave vs. convex menisci. Final Answer: lower than the outside free surface

Question 6

Euler’s equation for inviscid flow — key assumptions: Which of the following is a core assumption used in deriving Euler’s equation of motion for liquids in continuum mechanics?

Accepted Answer

Introduction / Context: Euler’s equation provides the inviscid form of the linear momentum balance along a streamline (or in vector form in the field). It underpins Bernoulli’s equation and many first-approximation models in hydraulics and aerodynamics when viscous effects are negligible. Given Data / Assumptions: Continuum fluid with negligible viscosity (inviscid). Often treated as homogeneous and incompressible for liquids. Body forces (usually gravity) may act. No requirement that the flow be steady; Euler’s equation has both steady and unsteady forms. Concept / Approach: Euler’s equation is derived from Newton’s second law applied to a fluid element, neglecting viscous stresses (no shear stress terms). For many liquid-flow problems, additional simplification assumes a homogeneous, incompressible fluid. These assumptions lead to the familiar relationships that integrate to Bernoulli’s equation under steady flow. Step-by-Step Solution: State the inviscid momentum balance: pressure + inertia + body-force terms.Neglect viscous shear stresses entirely (ideal fluid).Assume fluid density is constant in space and time (homogeneous, incompressible).Conclude that option (b) captures a central assumption routinely invoked in the derivation for liquids. Verification / Alternative check: Bernoulli’s equation follows from steady, incompressible, inviscid Euler along a streamline, confirming the appropriateness of the assumption. Why Other Options Are Wrong: (a) contradicts the inviscid premise; (c) non-uniformity is permitted but not an assumption; (d) Euler allows steady or unsteady; (e) surface tension is generally negligible in bulk momentum balances. Common Pitfalls: Equating Euler’s equation with only steady flow; assuming viscosity can be small yet included—by definition Euler neglects it. Final Answer: The fluid is homogeneous and incompressible

Question 7

Definition of Mach number in fluid mechanics: Mach number (M) is defined as the ratio of inertial forces (or flow speed) to which fundamental resisting effect?

Accepted Answer

Introduction / Context: Mach number is central to compressible-flow analysis. It classifies regimes such as subsonic, transonic, supersonic, and hypersonic, determining whether compressibility and shock waves are important in aerodynamic and gas-dynamic problems. Given Data / Assumptions: Mach number M = V / a, where V is flow speed and a is speed of sound. Speed of sound encapsulates the medium’s elasticity via a = sqrt(∂p/∂ρ)_s. Continuum, single-phase, no phase change in the regime considered. Concept / Approach: Dimensionally, M compares inertial effects to elastic (compressibility) effects in the fluid. High M implies inertia dominates and compressibility cannot be ignored, while low M indicates quasi-incompressible behavior. Thus, Mach number represents a ratio of inertia to elasticity effects, not to viscosity, surface tension, or gravity. Step-by-Step Solution: Recognize M = V/a expresses inertia relative to elastic wave propagation.Relate a to elasticity through thermodynamic derivatives (e.g., isentropic bulk modulus).Conclude that the resisting “force” class is elasticity (compressibility), matching option (d). Verification / Alternative check: In Buckingham Pi terms, M arises from balancing inertia with compressibility, distinct from Reynolds (viscosity) and Weber (surface tension) numbers. Why Other Options Are Wrong: (a) relates to Reynolds number; (b) to Weber number; (c) to Froude number; (e) is irrelevant in standard fluid dynamics contexts. Common Pitfalls: Confusing Mach with Reynolds or Froude; assuming low M always means incompressible regardless of temperature effects—usually acceptable but context dependent. Final Answer: elasticity (compressibility)

Question 8

Jet trajectory from a free nozzle into air: When a liquid issues horizontally from a nozzle into the atmosphere (neglecting air resistance close to the nozzle), what is the ideal shape of the jet path under gravity?

Accepted Answer

Introduction / Context: Once a jet leaves a nozzle and is exposed to the atmosphere, it behaves like a continuous stream of fluid particles subjected mainly to gravity. Close to the nozzle (neglecting aerodynamic breakup), each fluid particle follows projectile motion principles. Given Data / Assumptions: Jet issues horizontally at speed u from a nozzle into still air. Air resistance is neglected over the short distance of interest. Gravity g acts downward; initial vertical velocity is zero. Concept / Approach: Projectile motion under constant gravitational acceleration produces a parabolic path. The horizontal velocity component remains constant (u), while the vertical component increases linearly with time (g t). Eliminating time yields y = (g x^2) / (2 u^2), the equation of a parabola in x–y coordinates. Step-by-Step Solution: Horizontal motion: x = u t ⇒ t = x / u.Vertical motion: y = (1/2) g t^2.Eliminate t: y = (1/2) g (x/u)^2 = (g x^2) / (2 u^2).Conclude geometric form is a parabola opening downward. Verification / Alternative check: Dimensional check: y has dimensions of length; expression (g/u^2) x^2 is consistent. Observationally, short horizontal jets (e.g., water taps) display a parabolic profile before breaking into droplets. Why Other Options Are Wrong: (a) would require zero gravity; (b) and (c) require centripetal constraints not present; (e) does not arise from uniform gravity with constant horizontal speed. Common Pitfalls: Including significant air drag near the nozzle; forgetting the vertical initial velocity is zero for a horizontal jet. Final Answer: parabolic trajectory

Question 9

Basic properties of liquids (continuum viewpoint): Which statement best characterizes liquids compared with solids and gases under ordinary conditions?

Accepted Answer

Introduction / Context: Understanding the macroscopic properties of liquids is foundational for hydraulics, process engineering, and materials handling. These properties differentiate liquids from solids (rigidity) and gases (high compressibility and expansibility). Given Data / Assumptions: Normal temperature and pressure, single-phase liquid. Continuum mechanics model (ignoring molecular-scale details). No phase change during observation. Concept / Approach: Liquids have very low compressibility, significant cohesion, and finite surface tension. They resist shear only transiently (flow under sustained shear) and thus do not maintain a fixed shape; instead they conform to the container, preserving essentially constant volume (neglecting slight compressibility). Step-by-Step Solution: Identify shape behavior: liquids flow and take the container’s shape.Recognize compressibility: small but nonzero (bulk modulus is large but finite).Acknowledge sensitivity to pressure/temperature: density and viscosity vary with both.Thus option (b) best captures the defining characteristic. Verification / Alternative check: Everyday observation (water in a glass) and standard property tables (bulk modulus, thermal expansion coefficient) support the description. Why Other Options Are Wrong: (a) Incorrect—compressibility is small but not zero; (c) false—properties change with T and p; (d) false—liquids do not expand to fill any volume; (e) false—surface tension is finite in equilibrium. Common Pitfalls: Assuming “incompressible” means mathematically zero compressibility; conflating liquid behavior with gas behavior. Final Answer: Liquids do not retain a definite shape but assume the container’s shape

Question 10

Error propagation for triangular weirs/notches: For a V-notch (triangular notch), discharge varies with head as Q ∝ H^(5/2). What is the ratio of the percentage error in discharge to the percentage error in measured head over the notch?

Accepted Answer

Introduction / Context: Flow measurement over notches relies on empirical–theoretical relations between head (H) and discharge (Q). Understanding how measurement errors in H propagate into errors in Q is crucial for instrumentation accuracy and calibration protocols. Given Data / Assumptions: Triangular (V-notch) weir with Q = K H^(5/2), where K includes notch angle and discharge coefficient. Small measurement errors relative to the nominal head (linearized error propagation). Steady flow; proper approach conditions. Concept / Approach: For Q = K H^n, fractional errors relate by dQ/Q = n dH/H. Thus, the percentage error in Q is n times the percentage error in H. For a V-notch, n = 5/2, significantly amplifying head measurement errors. Step-by-Step Solution: Start with Q = K H^(5/2).Take logarithms: ln Q = ln K + (5/2) ln H.Differentiate: dQ/Q = (5/2) dH/H.Therefore, %error(Q) / %error(H) = 5/2. Verification / Alternative check: Compare with rectangular notch where Q ∝ H^(3/2): the ratio would be 3/2, confirming the exponent-rule consistency. Why Other Options Are Wrong: (a), (b), (c), and (e) correspond to other exponents or are arbitrary; only 5/2 matches the V-notch exponent. Common Pitfalls: Using absolute errors instead of fractional; forgetting that the exponent multiplies the fractional error. Final Answer: 5/2

Question 11

Why mercury is used in barometers: Select the best reason(s) that justify mercury as the working fluid in standard barometers.

Accepted Answer

Introduction / Context: Barometers measure atmospheric pressure via the height of a liquid column. The choice of liquid affects instrument size, sensitivity, and accuracy. Mercury has been the classical choice dating back to Torricelli’s experiment. Given Data / Assumptions: Ambient temperatures where mercury remains liquid. Standard gravity; vertical tube evacuated above the column. Clean, non-wetting glass conditions typical for barometers. Concept / Approach: Two key properties make mercury ideal: extremely low vapour pressure at room temperature (minimizing vapour above the column so pressure equals atmospheric) and high density (~13,600 kg/m^3), which keeps the column height manageable (~760 mm at 1 atm) and reduces instrument size compared with water (~10.3 m column). Step-by-Step Solution: Identify impact of vapour pressure: negligible vapour pressure prevents significant vapour cushion errors.Recognize effect of density: high density shortens the required column height to achieve hydrostatic balance p = ρ g h.Hence, statements (a) and (b) together justify mercury’s use. Verification / Alternative check: Compute water column for 1 atm: h = p/(ρ g) ≈ 101325/(1000*9.81) ≈ 10.3 m, impractical indoors; mercury reduces this to ~0.76 m. Why Other Options Are Wrong: (c) is vague and untrue universally; (e) includes (c), which is unjustified; only (d) provides the standard reasons. Common Pitfalls: Ignoring vapour pressure effects; assuming any dense liquid works equally well despite volatility and wetting issues. Final Answer: Both (a) and (b)

Question 12

Most economical trapezoidal channel section — identifying equivalent criteria: Which of the following conditions are satisfied by a hydraulically most economical trapezoidal open-channel section?

Accepted Answer

Introduction / Context: The “most economical” (best hydraulic) section minimizes wetted perimeter for a given area, thereby maximizing discharge for a given slope and roughness (or minimizing energy loss). For trapezoidal channels, several geometric conditions are equivalent indicators of this optimum shape. Given Data / Assumptions: Uniform, steady, open-channel flow. Trapezoidal cross-section with base width b, side slope m:1, and depth y. Optimization under constant area and slope (Manning/Chezy framework). Concept / Approach: For the most economical trapezoid, the condition is that the hydraulic radius R = A/P is maximized. This leads to canonical geometric results that can be expressed in multiple equivalent forms, including relationships among top width, side lengths, and inscribed semicircle tangency to the wetted boundary. Step-by-Step Solution: Start from optimum condition: d(P)/d(variable) = 0 at fixed A or use calculus of variations.Obtain standard results: half of top width equals sloping side length; side slopes meet the base so that the wetted perimeter is minimized.Show hydraulic mean depth R equals y/2 at optimum, implying efficient conveyance.Geometric construction: semicircle with diameter as top width tangent to wetted boundary confirms symmetry of distances. Verification / Alternative check: Textbook derivations using Chezy or Manning demonstrate identical optimum conditions; numerical examples confirm lower P for the same A at the optimum. Why Other Options Are Wrong: Here, (a), (b), and (c) are each true descriptors; thus selecting only one would be incomplete. The correct comprehensive choice is (d). Common Pitfalls: Confusing the most economical rectangle (b = 2y) with trapezoid conditions; overlooking that several geometric tests are equivalent at optimum. Final Answer: All of these

Question 13

Standard viscometers and kinematic viscosity correlation: Kinematic viscosity of liquids, when obtained via an empirical conversion equation from measured efflux time, is characteristically determined using which viscometer?

Accepted Answer

Introduction / Context: Several standardized cup viscometers measure efflux time of a fixed liquid volume through an orifice. Empirical equations then convert this “seconds” reading into kinematic viscosity (commonly in centistokes), enabling process and quality control for oils and fuels. Given Data / Assumptions: Efflux-time measurement at controlled temperature. Empirical calibration linking time and kinematic viscosity. Newtonian fluids in the applicable range. Concept / Approach: The Saybolt universal viscometer (SUV) reports Saybolt seconds (SSU). ASTM correlations relate SSU to kinematic viscosity ν (cSt) using a formula of the form ν = a t − b / t over a defined range. Redwood and Engler also use time scales, but the widely taught textbook “equation” mapping time to ν in cSt is associated with the Saybolt universal method. Step-by-Step Solution: Measure efflux time t in Saybolt seconds using the SUV apparatus.Apply the empirical conversion ν = a t − b / t (constants depend on standard range and calibration).Report ν as kinematic viscosity (e.g., cSt) at the test temperature. Verification / Alternative check: Standards (e.g., ASTM D445/D446 and older Saybolt methods) present tables/curves consistent with the above conversion. Why Other Options Are Wrong: Redwood and Engler scales exist but are not the classical source of the cited ν–t equation in many syllabi; “None” contradicts common lab practice; Ford cups are used mainly for paints with different correlations. Common Pitfalls: Interchanging absolute and kinematic viscosity; ignoring temperature control that strongly affects ν. Final Answer: Saybolt universal viscometer

Question 14

Velocity distribution in viscous flow through a circular pipe In internal laminar flow of a viscous fluid through a straight circular pipe (steady and fully developed), the axial velocity profile across the diameter is best described as:

Accepted Answer

Introduction / Context: Understanding the shape of the velocity profile in pipe flow is fundamental for computing discharge, wall shear, and energy losses. For steady, fully developed laminar flow of a Newtonian fluid in a circular pipe, the profile has a distinctive mathematical form. Given Data / Assumptions: Newtonian fluid with constant viscosity. Steady, incompressible, fully developed internal flow. Straight, circular pipe; no swirl; no acceleration along the axis. Concept / Approach: From the Navier–Stokes equations with the above assumptions, the solution is the Hagen–Poiseuille profile: velocity u decreases from a maximum at the centreline to zero at the wall due to the no-slip condition, following a quadratic (parabolic) dependence on radius. Step-by-Step Solution: Start with momentum balance in the axial direction for fully developed flow: pressure gradient is constant.Integrate twice with respect to radius and apply boundary conditions u = 0 at r = R (no slip) and finite du/dr at r = 0 (symmetry).Obtain u(r) = u_max * (1 − (r/R)^2), which is a parabola in r. Verification / Alternative check: The average velocity u_avg equals one-half of u_max in laminar pipe flow, which is consistent only with a parabolic profile. Why Other Options Are Wrong: Elliptical, hyperbolic, or circular are not the functional form of the solution; a uniform (plug) profile contradicts the no-slip condition in laminar regime. Common Pitfalls: Confusing turbulent (flatter) profiles with laminar ones; forgetting that the profile becomes parabolic only under fully developed conditions. Final Answer: Parabolic

Question 15

Buoyancy of a floating wooden block A wooden block (density = 0.75 relative to water) is 5 m long, 2 m wide, and 3 m high. When it floats in fresh water, what volume of water is displaced?

Accepted Answer

Introduction / Context: Archimedes’ principle governs floating bodies: the weight of the displaced fluid equals the weight of the body. For a wooden block with given relative density, the submerged volume is a fixed fraction of its total volume. Given Data / Assumptions: Block dimensions: 5 m × 2 m × 3 m. Relative density of wood = 0.75 (i.e., rho_wood = 0.75 * rho_water). Floating at equilibrium in fresh water; no additional loads. Concept / Approach: For floating equilibrium: Weight of block = Weight of displaced water. Since weights are proportional to densities times volumes, the submerged volume fraction equals the relative density. Thus, V_displaced = RD * V_block. Step-by-Step Solution: Compute total block volume: V_block = 5 * 2 * 3 = 30 m^3.Submerged fraction = 0.75.Displaced water volume: V_disp = 0.75 * 30 = 22.5 m^3. Verification / Alternative check: Because RD < 1, the block must float with part above water. The displaced volume (22.5 m^3) is less than the block volume (30 m^3), which is physically consistent. Why Other Options Are Wrong: 17.5, 20.0, and 25.0 m^3 do not equal RD * V_block; 30.0 m^3 implies full submergence (RD = 1), which is not the case. Common Pitfalls: Mixing density ratio with mass fraction; forgetting that volume fraction submerged equals relative density for floating bodies. Final Answer: 22.5 m^3

Question 16

Dynamic similarity for channels dominated by inertia and gravity When inertia and gravity are the primary forces and friction has a minor role, channel design and model–prototype comparison should be based on which dimensionless number?

Accepted Answer

Introduction / Context: Choice of the correct similarity parameter is crucial in hydraulic modeling. Free-surface flows such as rivers, spillways, and ship hydrodynamics are governed mainly by inertia and gravity, so the corresponding dimensionless group must capture that balance. Given Data / Assumptions: Free-surface flow conditions; wave and surface effects matter. Inertia and gravity dominate; viscous effects are relatively small. Geometric similarity between model and prototype is assumed. Concept / Approach: The Froude number Fr = V / sqrt(g * L) compares inertial forces to gravitational forces. Dynamic similarity requires Fr_model = Fr_prototype, ensuring homologous behavior of waves, hydraulic jumps, and critical transitions. Step-by-Step Solution: Identify dominant forces: inertia and gravity ⇒ use Froude scaling.Set Fr_model = Fr_prototype during design.Use scaled velocity: V_model = V_proto * sqrt(L_model / L_proto). Verification / Alternative check: Reynolds number governs inertia–viscosity; Weber number inertia–surface tension; Mach number compressibility; Euler number pressure–inertia. None of these match inertia–gravity dominance like Froude. Why Other Options Are Wrong: They address different force balances and cannot alone preserve wave-related phenomena in open channels. Common Pitfalls: Attempting to match both Reynolds and Froude simultaneously is often impossible; prioritize Froude for free-surface flows. Final Answer: Froude number

Question 17

Meniscus of mercury in an open tube In an open capillary or glass tube, the free surface of pure mercury (clean glass, air above) assumes which shape?

Accepted Answer

Introduction / Context: Surface tension and wetting determine the shape of a liquid's free surface near a solid boundary. Mercury and water show opposite wetting behavior on clean glass, producing different meniscus shapes that directly affect capillary rise or depression. Given Data / Assumptions: Open tube with air above the liquid. Clean glass surface. Liquid is mercury (high surface tension, non-wetting on glass). Concept / Approach: Contact angle greater than 90 degrees (non-wetting) yields a convex meniscus (bulging upward in the middle, depressed near the glass). For mercury on glass, the contact angle is obtuse, causing capillary depression and a convex meniscus. Step-by-Step Solution: Identify wetting: mercury does not wet glass ⇒ large contact angle.Large contact angle implies convex meniscus.Therefore, the free surface is convex, appearing curved downward near the walls relative to the centre. Verification / Alternative check: Water on clean glass (wetting) yields a concave meniscus and capillary rise, the opposite of mercury's behavior. Observations in manometers confirm mercury's capillary depression. Why Other Options Are Wrong: Concave is characteristic of wetting liquids like water; perfectly horizontal would require zero capillary effects; random curvature contradicts deterministic surface tension physics. Common Pitfalls: Confusing “curved upward” and “curved downward” descriptions; always tie the shape to the contact angle and wetting behavior. Final Answer: Convex meniscus (curved downward at the walls)

Question 18

Cause of a backwater curve in gradually varied flow A backwater curve (M-curve or S-curve rising upstream) in an open channel is typically produced when:

Accepted Answer

Introduction / Context: Backwater curves are part of gradually varied flow profiles, important for flood routing, afflux estimation, and design near controls. Recognizing the physical cause helps select the correct profile family (M1, M2, S1, etc.). Given Data / Assumptions: Prismatic, mild or steep channel as applicable. Steady flow with gradually varied depth changes. Presence or absence of a downstream control (e.g., weir, gate, dam). Concept / Approach: A backwater curve is a rising water surface upstream of a control that increases tailwater depth relative to normal depth. The obstruction forces subcritical approach flow to adjust gradually upstream, producing an elevated profile. Step-by-Step Solution: Identify control: a weir/barrage increases downstream depth.For subcritical flow (Fr < 1), disturbances propagate upstream, so depth rises upstream of the control.Resulting profile is a backwater curve (commonly M1 or S1 depending on slope and depths). Verification / Alternative check: GVF equation dy/dx = (S0 − Sf) / (1 − Fr^2) predicts positive dy/dx near controls where depth exceeds normal, consistent with backwater formation. Why Other Options Are Wrong: Friction vs bed slope balance alone does not create a control; narrowing typically lowers depth for subcritical flow; steepening bed increases energy slope and can lower water surface upstream. Common Pitfalls: Confusing drawdown (falling profile) with backwater; ignoring flow regime effects on information propagation. Final Answer: A downstream control such as a weir or barrage backs up the water surface

Question 19

Maximum efficiency of power transmission through a pipeline If the total head available at the pipe inlet is H and the head lost due to friction is h_f, what is the maximum possible efficiency of power transmission through the pipe?

Accepted Answer

Introduction / Context: When transmitting power by water through a long pipe to a turbine/nozzle, frictional losses reduce the head at the outlet. The efficiency depends on how much head is lost en route versus delivered for useful work. Given Data / Assumptions: Total head at inlet = H. Head loss due to friction along pipe = h_f. No other losses or shaft work until the outlet. Concept / Approach: Hydraulic power delivered is proportional to the product of discharge and available head at outlet. Optimization (with velocity related to head) gives a specific ratio of loss to available head at maximum efficiency: h_f = H/3, yielding η_max = 2/3. Step-by-Step Solution: Let head at outlet H_o = H − h_f.For a given pipe, h_f ∝ V^2; discharge Q ∝ V; delivered power P_d ∝ H_o * Q ∝ (H − k V^2) * V.Differentiate with respect to V, set dP_d/dV = 0 ⇒ optimal when h_f = H/3.Then η_max = H_o / H = (H − H/3) / H = 2/3. Verification / Alternative check: Classical derivation in fluid mechanics texts confirms η_max = 66.7% for power transmission by a pipeline without other losses. Why Other Options Are Wrong: Other percentages do not correspond to the optimal friction–head trade-off; they arise if h_f differs from H/3 and are suboptimal. Common Pitfalls: Maximizing outlet head alone (min V) or flow alone (max V) does not maximize power; the product must be optimized. Final Answer: 2/3 (66.7%)

Question 20

Primary purpose of manometers in fluid mechanics Manometers are most commonly used to measure which of the following?

Accepted Answer

Introduction / Context: Manometers are simple, reliable devices for pressure measurement using hydrostatics. They are widely used in laboratories and field applications for pipes, ducts, and vessels where accurate differential pressure readings are required. Given Data / Assumptions: Static fluids in the manometer legs. Known manometric fluid density (e.g., mercury, water). Hydrostatic equilibrium; negligible capillary effects in large tubes. Concept / Approach: Hydrostatic relation: p2 − p1 = ρ_m * g * Δh (adjusted for densities in each leg as needed). This makes manometers ideal for direct measurement of pressure differences. With one leg open to atmosphere, they measure gauge pressure; with both legs connected, they measure differential pressure between two points. Step-by-Step Solution: Set up connections to the two points whose pressure difference is required.Read level difference Δh between manometric fluid columns.Compute Δp using Δp = ρ_m * g * Δh (with appropriate corrections). Verification / Alternative check: Calibration with known pressure sources confirms the linear relation between Δh and Δp. Why Other Options Are Wrong: Atmospheric pressure is usually obtained via barometers; very low pressures often need McLeod or Pirani gauges; velocity is not directly measured by a manometer without additional devices like Pitot-static tubes. Common Pitfalls: Ignoring density differences and temperature effects; misreading Δh due to meniscus curvature in narrow tubes. Final Answer: Difference in pressure between two points

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