Fluid Mechanics and Hydraulics MCQs 2026

80 questions with detailed answers · 28 from past papers · 8 quiz batches available

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Page 1 of 1 Questions 110 of 80
  1. Q1 Past Paper · PPSC/FPSC/NTS easy

    Dynamic viscosity of a fluid is defined as the ratio of

    1. A density to velocity
    2. B shear stress to rate of shear strain
    3. C pressure to temperature
    4. D velocity to area
    💡 Explanation:

    Newton law of viscosity: τ = μ(du/dy); μ is dynamic viscosity in Pa·s.

  2. Q2 Past Paper · PPSC/FPSC/NTS easy

    Kinematic viscosity ν is related to dynamic viscosity μ by

    1. A ν = μρ
    2. B ν = ρ/μ²
    3. C ν = μ + ρ
    4. D ν = μ/ρ
    💡 Explanation:

    Kinematic viscosity has units m²/s and appears in Reynolds number.

  3. Q3 medium

    The bulk modulus of elasticity of a liquid measures

    1. A surface tension effects only
    2. B resistance to uniform compression
    3. C shear deformation only
    4. D turbulent mixing rate
    💡 Explanation:

    High bulk modulus makes liquids nearly incompressible for most hydraulic problems.

  4. Q4 easy

    Surface tension in a liquid is caused by

    1. A gravitational pull only
    2. B viscous shear at walls
    3. C Bernoulli pressure drop
    4. D cohesive forces at the free surface
    💡 Explanation:

    Surface tension creates capillary rise and affects small-scale flow phenomena.

  5. Q5 easy

    Vapour pressure of water increases with

    1. A depth in a reservoir only
    2. B pipe diameter only
    3. C channel slope only
    4. D temperature
    💡 Explanation:

    Higher temperature increases molecular escape tendency, lowering cavitation margin.

  6. Q6 easy

    A Newtonian fluid is one in which

    1. A viscosity changes with time always
    2. B shear stress is directly proportional to velocity gradient
    3. C density is constant only in gases
    4. D Reynolds number exceeds 4000
    💡 Explanation:

    Water and air at normal conditions behave as Newtonian fluids.

  7. Q7 easy

    Specific weight γ of a fluid equals

    1. A ρ/g
    2. B μρ
    3. C ρg
    4. D ρ + g
    💡 Explanation:

    γ = weight per unit volume; differs from mass density ρ by gravitational acceleration.

  8. Q8 medium

    Compressibility of water is generally considered negligible in

    1. A pipe flow and open channel problems at low Mach number
    2. B cavitation analysis always
    3. C water hammer studies
    4. D high-speed jet flows
    💡 Explanation:

    Water hammer and very high pressures require accounting for compressibility.

  9. Q9 Past Paper · PPSC/FPSC/NTS easy

    Hydrostatic pressure at a point in a static fluid depends on

    1. A depth, fluid density and gravitational acceleration
    2. B flow velocity only
    3. C pipe roughness only
    4. D Reynolds number
    💡 Explanation:

    p = ρgh for incompressible fluid; pressure increases linearly with depth.

  10. Q10 Past Paper · PPSC/FPSC/NTS easy

    Pascal law states that pressure applied to a confined fluid

    1. A is transmitted undiminished in all directions
    2. B acts only vertically downward
    3. C decreases with depth always
    4. D depends on container shape for intensity
    💡 Explanation:

    Foundation of hydraulic jacks and pressure measurement principles.

  11. Q11 Past Paper · PPSC/FPSC/NTS medium

    The centre of pressure on a submerged plane surface is

    1. A above the centroid always
    2. B coincident with centroid always
    3. C at the water surface
    4. D below the centroid because pressure increases with depth
    💡 Explanation:

    Moment of pressure forces shifts CP below centroid for inclined or vertical surfaces.

  12. Q12 Past Paper · PPSC/FPSC/NTS medium

    Total hydrostatic force on a vertical plane surface equals

    1. A γA only
    2. B ½γH² only without area
    3. C velocity head times area
    4. D γh̄A where h̄ is depth to centroid
    💡 Explanation:

    F = γh̄A; location of CP found using moment of inertia about water surface.

  13. Q13 easy

    Pressure head in fluid mechanics is expressed as

    1. A p/(ρg)
    2. B ρgp
    3. C pρg
    4. D p + ρg
    💡 Explanation:

    Pressure head converts pressure to equivalent liquid column height.

  14. Q14 easy

    Absolute pressure equals gauge pressure plus

    1. A vacuum pressure only
    2. B atmospheric pressure
    3. C velocity head
    4. D vapour pressure always
    💡 Explanation:

    p_abs = p_gauge + p_atm; vacuum is pressure below atmospheric.

  15. Q15 easy

    A manometer reading measures

    1. A flow rate directly
    2. B Reynolds number
    3. C difference in pressure between two points or from atmosphere
    4. D Manning n directly
    💡 Explanation:

    U-tube and inclined manometers are common in laboratory and field.

  16. Q16 medium

    Hydrostatic paradox demonstrates that

    1. A pressure depends on container shape at same depth
    2. B deeper vessels always have lower pressure
    3. C force on container bottom depends on depth and area, not total water volume shape
    4. D horizontal forces are always zero
    💡 Explanation:

    At equal depth, pressure is same regardless of container geometry.

  17. Q17 Past Paper · PPSC/FPSC/NTS easy

    Archimedes principle states buoyant force equals

    1. A weight of the body in air
    2. B weight of fluid displaced by the body
    3. C half the submerged volume only
    4. D surface tension force
    💡 Explanation:

    F_b = ρgV_displaced; basis for stability of floating structures.

  18. Q18 Past Paper · PPSC/FPSC/NTS easy

    A body floats when its average specific gravity is

    1. A equal to zero only
    2. B greater than 2.0 always
    3. C less than that of the fluid
    4. D equal to fluid SG with no submergence
    💡 Explanation:

    Equilibrium requires displaced fluid weight to equal body weight.

  19. Q19 medium

    Metacentric height GM is positive when a floating body is

    1. A unstable and capsizing
    2. B stable in small angular disturbances
    3. C neutrally buoyant only
    4. D fully submerged always
    💡 Explanation:

    Positive GM restores righting moment; GM = BM + BG − BG relations apply.

  20. Q20 Past Paper · PPSC/FPSC/NTS easy

    Darcy-Weisbach head loss formula is

    1. A h_f = n²V²R^(4/3)
    2. B h_f = f (L/D)(V²/2g)
    3. C h_f = KL²
    4. D h_f = μV only
    💡 Explanation:

    Friction factor f depends on Re and relative roughness ε/D.

  21. Q21 Past Paper · PPSC/FPSC/NTS medium

    Hazen-Williams formula is primarily used for

    1. A water distribution pipes in turbulent regime
    2. B laminar laboratory tubes only
    3. C open channel Manning alternative only
    4. D compressible gas pipelines
    💡 Explanation:

    V = C_RH^0.63 S^0.54 in SI form; C is Hazen-Williams coefficient.

  22. Q22 medium

    Moody diagram relates friction factor f to

    1. A Froude number only
    2. B Manning n only
    3. C weir crest length
    4. D Reynolds number and relative roughness
    💡 Explanation:

    Essential for estimating f in commercial pipes when Colebrook implicit solution not used.

  23. Q23 hard

    Colebrook-White equation is used to find

    1. A Darcy friction factor in turbulent pipe flow
    2. B Manning n from grain size
    3. C critical depth directly
    4. D pump NPSH
    💡 Explanation:

    Implicit relation between f, Re and ε/D; solved iteratively or by Swamee-Jain.

  24. Q24 easy

    Minor losses in pipe systems are expressed as

    1. A K V²/(2g)
    2. B fL/D only
    3. C Manning S
    4. D ρgH pump head
    💡 Explanation:

    K depends on fitting type: elbows, valves, sudden expansion.

  25. Q25 easy

    Velocity in pipe for given discharge is maximum when

    1. A diameter is largest
    2. B area is minimum for same Q
    3. C roughness is highest
    4. D pipe is horizontal
    💡 Explanation:

    V = Q/A; contraction increases local velocity.

  26. Q26 medium

    Parallel pipes carrying same total head loss have

    1. A equal discharge always
    2. B inverse relationship between resistance and discharge share
    3. C zero flow in smaller pipe
    4. D same velocity regardless of diameter
    💡 Explanation:

    1/Q_total² relationship via equivalent resistance network.

  27. Q27 easy

    Series pipes have

    1. A same head loss in each
    2. B independent discharges
    3. C same discharge through each pipe section
    4. D zero velocity in second pipe
    💡 Explanation:

    Q constant; total head loss is sum of individual losses.

  28. Q28 hard

    Water hammer pressure rise is proportional to

    1. A change in velocity and wave celerity
    2. B pipe length only
    3. C Manning n
    4. D channel width only
    💡 Explanation:

    Δp = ρaΔV; rapid valve closure causes surge in pressure pipes.

  29. Q29 medium

    Cavitation in pumps occurs when local pressure falls below

    1. A atmospheric pressure at all points
    2. B stagnation pressure
    3. C Manning normal depth
    4. D vapour pressure of the liquid
    💡 Explanation:

    Vapour bubbles form and collapse, damaging impellers and reducing efficiency.

  30. Q30 Past Paper · PPSC/FPSC/NTS easy

    Manning formula for open channel velocity is

    1. A V = √(gD)
    2. B V = fL/D
    3. C V = C√(RS) Hazen only
    4. D V = (1/n) R^(2/3) S^(1/2)
    💡 Explanation:

    Manning n is roughness coefficient; S is energy slope.

  31. Q31 easy

    Manning n increases with

    1. A flow depth decrease only
    2. B higher Reynolds always decreasing n
    3. C channel roughness and vegetation
    4. D subcritical flow only
    💡 Explanation:

    Smooth concrete has low n; vegetated floodplains have high n.

  32. Q32 medium

    Hydraulic radius for wide rectangular channel approximates

    1. A width b
    2. B b + y
    3. C y/2 only for all widths
    4. D depth y
    💡 Explanation:

    R = by/(b+2y) ≈ y when b >> y.

  33. Q33 Past Paper · PPSC/FPSC/NTS medium

    Specific energy E in open channel is

    1. A y − V²/(2g)
    2. B Q²/(2g)
    3. C y + V²/(2g)
    4. D z + p/γ
    💡 Explanation:

    Energy per unit weight relative to channel bottom for given discharge.

  34. Q34 Past Paper · PPSC/FPSC/NTS medium

    Critical depth occurs when specific energy is

    1. A maximum always
    2. B zero
    3. C minimum for given discharge
    4. D equal to normal depth always
    💡 Explanation:

    At critical flow, Froude number equals unity.

  35. Q35 Past Paper · PPSC/FPSC/NTS easy

    Froude number Fr is defined as

    1. A V/gD
    2. B √(gD)/V
    3. C V/√(gD_h)
    4. D ρV²D/μ
    💡 Explanation:

    D_h is hydraulic depth; Fr < 1 subcritical, Fr > 1 supercritical.

  36. Q36 medium

    Subcritical flow in open channel is controlled from

    1. A upstream only always
    2. B downstream conditions
    3. C mid-channel only
    4. D atmosphere only
    💡 Explanation:

    Disturbances propagate upstream; downstream controls depth.

  37. Q37 medium

    Supercritical flow is controlled from

    1. A upstream
    2. B downstream
    3. C both ends equally always
    4. D neither end
    💡 Explanation:

    Information cannot travel upstream; upstream boundary fixes conditions.

  38. Q38 hard

    Gradually varied flow (GVF) profiles depend on

    1. A pipe friction factor only
    2. B pump power only
    3. C relation of actual depth to normal and critical depths
    4. D BOD loading
    💡 Explanation:

    Mild, steep, horizontal and adverse beds yield M, S, H, A profile classes.

  39. Q39 medium

    Rapidly varied flow includes

    1. A uniform flow in long canal
    2. B laminar pipe flow
    3. C Darcy groundwater flow
    4. D hydraulic jump and flow over weirs
    💡 Explanation:

    Curvature and non-hydrostatic pressure distribution dominate locally.

  40. Q40 Past Paper · PPSC/FPSC/NTS easy

    Centrifugal pump head is the energy imparted per unit

    1. A volume only
    2. B weight of liquid
    3. C mass times g only without head concept
    4. D time only
    💡 Explanation:

    H = (p/ρg) + V²/(2g) + z increase across pump.

  41. Q41 Past Paper · PPSC/FPSC/NTS medium

    NPSH available must exceed NPSH required to avoid

    1. A cavitation at pump suction
    2. B water hammer in delivery pipe
    3. C Manning depth error
    4. D sediment deposition
    💡 Explanation:

    NPSHA depends on atmospheric pressure, vapour pressure and suction lift.

  42. Q42 hard

    Specific speed N_s of a pump indicates

    1. A only motor RPM
    2. B pipe diameter only
    3. C type and suitability for head-discharge duty
    4. D Manning roughness
    💡 Explanation:

    Low N_s: radial high-head; high N_s: axial low-head pumps.

  43. Q43 medium

    Characteristic curves of centrifugal pump relate

    1. A only pipe friction factor
    2. B only channel slope
    3. C head, power and efficiency to discharge
    4. D only rainfall intensity
    💡 Explanation:

    H-Q curve falls with increasing Q; efficiency has peak at BEP.

  44. Q44 Past Paper · PPSC/FPSC/NTS medium

    Pelton turbine is best suited for

    1. A low head high flow rivers
    2. B tidal barrages only
    3. C high head and low discharge
    4. D drainage pumps
    💡 Explanation:

    Impulse turbine with jet striking buckets; common in Himalayan projects.

  45. Q45 Past Paper · PPSC/FPSC/NTS medium

    Francis turbine is a

    1. A mixed-flow reaction turbine for medium head
    2. B pure impulse axial only
    3. C positive displacement pump
    4. D trickling filter
    💡 Explanation:

    Pressure drop occurs in runner; used widely in irrigation and hydropower.

  46. Q46 medium

    The centre of buoyancy of a submerged body is the centroid of

    1. A body mass
    2. B water surface area
    3. C pressure diagram on atmosphere
    4. D displaced fluid volume
    💡 Explanation:

    Couple between CB and CG determines rotational stability.

  47. Q47 hard

    For a fully submerged body, stable equilibrium requires

    1. A centre of gravity below centre of buoyancy
    2. B CG above CB
    3. C CG coincident with free surface
    4. D neutral metacentre at infinity
    💡 Explanation:

    Righting couple exists when CB moves more than CG during tilt.

  48. Q48 easy

    In fluid flow, a streamline is a line

    1. A of constant pressure only
    2. B parallel to channel bed only
    3. C of zero shear stress
    4. D tangent to velocity vector at every point
    💡 Explanation:

    Streamlines show flow direction; cannot cross in steady flow.

  49. Q49 easy

    Steady flow means at a fixed point

    1. A flow properties do not change with time
    2. B velocity is zero everywhere
    3. C pressure is uniform in field
    4. D Reynolds number is constant globally
    💡 Explanation:

    ∂/∂t = 0 at a point; unsteady flow varies with time.

  50. Q50 easy

    Uniform flow in an open channel has

    1. A varying depth along length
    2. B depth and velocity constant along the channel reach
    3. C zero discharge
    4. D always supercritical flow
    💡 Explanation:

    Uniform flow implies friction slope equals bed slope.

  51. Q51 easy

    Turbulent flow is characterised by

    1. A random fluctuating velocity components and mixing
    2. B smooth layered motion only
    3. C Reynolds number below 1
    4. D zero wall shear
    💡 Explanation:

    Re > critical value; eddies enhance momentum and energy dissipation.

  52. Q52 Past Paper · PPSC/FPSC/NTS easy

    Continuity equation for incompressible steady flow is

    1. A P + ρgh = constant always
    2. B τ = μ du/dy
    3. C h_f = fLV²/(2gD)
    4. D Q = AV = constant
    💡 Explanation:

    Conservation of mass: discharge Q equals area times mean velocity.

  53. Q53 Past Paper · PPSC/FPSC/NTS easy

    Bernoulli equation along a streamline for ideal flow states

    1. A p = ρgh only
    2. B V = constant always
    3. C p/ρg + V²/(2g) + z = constant
    4. D fL/D = constant
    💡 Explanation:

    Energy per unit weight: pressure + velocity + elevation heads sum constant.

  54. Q54 medium

    Application of Bernoulli between two points requires

    1. A turbulent mixing always allowed
    2. B compressible supersonic flow always
    3. C unsteady pulsating flow
    4. D steady, inviscid, incompressible flow along streamline
    💡 Explanation:

    Real flows need head loss term for viscous and minor losses.

  55. Q55 easy

    Velocity head is expressed as

    1. A V²/(2g)
    2. B V/g
    3. C 2g/V²
    4. D ρV²
    💡 Explanation:

    Kinetic energy per unit weight; converts velocity to equivalent head.

  56. Q56 medium

    Piezometric head at a point equals

    1. A p/(ρg) + z
    2. B V²/(2g) only
    3. C p/(ρg) only
    4. D z − p/(ρg)
    💡 Explanation:

    Sum of pressure head and elevation head; excludes velocity head.

  57. Q57 medium

    Total energy head H in open channel flow is

    1. A y only
    2. B V²/(2g) only
    3. C z − y
    4. D z + y + V²/(2g) for datum at bed
    💡 Explanation:

    Elevation + depth + velocity head referenced to chosen datum.

  58. Q58 Past Paper · PPSC/FPSC/NTS medium

    Venturi meter measures flow rate using

    1. A temperature change only
    2. B magnetic induction
    3. C Manning equation
    4. D pressure difference between converging and throat sections
    💡 Explanation:

    Continuity and Bernoulli relate throat velocity to upstream conditions.

  59. Q59 medium

    Pitot tube measures

    1. A dynamic viscosity
    2. B stagnation pressure to obtain velocity
    3. C Manning roughness
    4. D sediment concentration
    💡 Explanation:

    V = √(2gΔh) for ideal conditions from stagnation-static difference.

  60. Q60 medium

    Orifice meter causes permanent pressure loss because of

    1. A Bernoulli being invalid
    2. B zero velocity at throat
    3. C contraction and turbulence at the orifice
    4. D laminar flow requirement
    💡 Explanation:

    Higher head loss than Venturi but simpler and cheaper construction.

  61. Q61 Past Paper · PPSC/FPSC/NTS medium

    Momentum equation in fluid mechanics is based on

    1. A first law of thermodynamics only
    2. B Fick law of diffusion
    3. C Newton second law for control volume
    4. D Darcy law for porous media
    💡 Explanation:

    ΣF = ρQ(V_out − V_in) + pressure forces for steady control volume.

  62. Q62 medium

    Force exerted by a jet on a flat plate is

    1. A ½ρAV² only always
    2. B ρgAH
    3. C ρAV² for normal impact when plate is stationary
    4. D zero for any angle
    💡 Explanation:

    Momentum change Δ(mV) gives F = ρQ(V − 0) for normal jet.

  63. Q63 hard

    When a jet strikes a curved vane and deflects through angle θ, force component depends on

    1. A only weight of vane
    2. B change in momentum in flow direction
    3. C Manning n
    4. D Reynolds number alone
    💡 Explanation:

    F = ρQ(V_in cos α − V_out cos β) for curved vanes in turbines.

  64. Q64 Past Paper · PPSC/FPSC/NTS medium

    Hydraulic jump occurs in open channel when

    1. A subcritical becomes laminar pipe flow
    2. B flow becomes uniform always
    3. C supercritical flow transitions to subcritical flow
    4. D depth remains constant
    💡 Explanation:

    Rapid increase in depth dissipates excess kinetic energy.

  65. Q65 hard

    Specific force in open channel momentum analysis is

    1. A Q/A only
    2. B Q²/(gA) + Aȳ
    3. C V²/(2g)
    4. D Manning V
    💡 Explanation:

    Used to locate hydraulic jump and conjugate depths.

  66. Q66 medium

    Impulse-momentum principle is preferred over Bernoulli when calculating

    1. A static pressure at a point only
    2. B forces on bends, reducers and structures
    3. C capillary rise
    4. D vapour pressure
    💡 Explanation:

    Momentum equation directly yields force on pipe fittings and gates.

  67. Q67 Past Paper · PPSC/FPSC/NTS easy

    Reynolds number Re is defined as

    1. A ρVD/μ or VD/ν
    2. B μVD/ρ
    3. C ρgD/μ
    4. D V²D/g
    💡 Explanation:

    Ratio of inertial to viscous forces; determines flow regime.

  68. Q68 Past Paper · PPSC/FPSC/NTS easy

    For pipe flow, critical Reynolds number for transition to turbulence is approximately

    1. A 100
    2. B 50000
    3. C 10
    4. D 2300
    💡 Explanation:

    Below ~2300 flow is laminar; commercial pipes transition earlier due to disturbances.

  69. Q69 medium

    Laminar pipe flow velocity profile is

    1. A uniform plug flow
    2. B linear from wall to centre
    3. C zero at centre
    4. D parabolic with maximum at centre
    💡 Explanation:

    Hagen-Poiseuille flow: u = (g/4μ)(−dp/dx)(R² − r²).

  70. Q70 medium

    In laminar flow through circular pipe, head loss varies with velocity as

    1. A V
    2. B
    3. C
    4. D independent of V
    💡 Explanation:

    h_f ∝ V for laminar (Hagen-Poiseuille); h_f ∝ V² for turbulent.

  71. Q71 easy

    Hydraulic radius R for full circular pipe equals

    1. A D/2
    2. B D/4
    3. C πD²/4
    4. D D
    💡 Explanation:

    R = A/P = (πD²/4)/(πD) = D/4; used in Reynolds and Manning formulas.

  72. Q72 hard

    Draft tube in reaction turbine serves to

    1. A increase atmospheric pressure
    2. B measure discharge
    3. C filter sediment
    4. D recover kinetic energy and maintain suction head at runner
    💡 Explanation:

    Diffuser converts outlet kinetic head to pressure head, improving net head.

  73. Q73 Past Paper · PPSC/FPSC/NTS medium

    Discharge over a rectangular sharp-crested weir is proportional to

    1. A H only
    2. B H^(3/2)
    3. C
    4. D √H for all weir types
    💡 Explanation:

    Q = (2/3)Cd b √(2g) H^(3/2) for ideal; Cd is discharge coefficient.

  74. Q74 easy

    V-notch weir is preferred for measuring

    1. A very high floods only
    2. B groundwater seepage directly
    3. C low discharges with better sensitivity
    4. D pipe pressure only
    💡 Explanation:

    Triangular weir gives accurate small flow measurement in laboratories.

  75. Q75 easy

    Coefficient of discharge for orifice is less than unity due to

    1. A gravity variation
    2. B Manning n in pipe
    3. C contraction and energy losses
    4. D Coriolis effect
    💡 Explanation:

    Cd typically 0.6–0.65 for sharp-edged orifice.

  76. Q76 Past Paper · PPSC/FPSC/NTS easy

    Torricelli theorem gives jet velocity from orifice as

    1. A gh only
    2. B √(2gh)
    3. C 2gh without root
    4. D √(gh/2)
    💡 Explanation:

    Ideal velocity from head h; actual velocity reduced by coefficient of velocity.

  77. Q77 Past Paper · PPSC/FPSC/NTS medium

    Buckingham Pi theorem states that a problem with n variables and k fundamental dimensions has

    1. A n − k independent dimensionless groups
    2. B n + k groups
    3. C k − n groups
    4. D always one group
    💡 Explanation:

    Reduces variables for model testing and empirical correlation.

  78. Q78 easy

    Reynolds number is a dimensionless group representing ratio of

    1. A gravity to viscosity
    2. B pressure to shear only
    3. C surface tension to gravity only
    4. D inertial forces to viscous forces
    💡 Explanation:

    Ensures dynamic similarity in viscous flow model studies.

  79. Q79 medium

    Froude number similarity is essential in modeling

    1. A free-surface gravity-driven flows
    2. B creeping laminar pipe flow only
    3. C Darcy groundwater flow
    4. D pure diffusion
    💡 Explanation:

    Fr match required for open channel, spillway and ship model tests.

  80. Q80 hard

    Geometric, kinematic and dynamic similarities together ensure

    1. A only same colour of fluid
    2. B complete model-prototype similitude for intended laws
    3. C only same pipe material
    4. D only same temperature
    💡 Explanation:

    Scale model tests in hydraulics require matching relevant dimensionless numbers.