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 1Questions 1–10 of 80
  1. Q1medium

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

    1. Abody mass
    2. Bwater surface area
    3. Cpressure diagram on atmosphere
    4. Ddisplaced fluid volume
    💡 Explanation:

    Couple between CB and CG determines rotational stability.

  2. Q2hard

    For a fully submerged body, stable equilibrium requires

    1. Acentre of gravity below centre of buoyancy
    2. BCG above CB
    3. CCG coincident with free surface
    4. Dneutral metacentre at infinity
    💡 Explanation:

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

  3. Q3easy

    In fluid flow, a streamline is a line

    1. Aof constant pressure only
    2. Bparallel to channel bed only
    3. Cof zero shear stress
    4. Dtangent to velocity vector at every point
    💡 Explanation:

    Streamlines show flow direction; cannot cross in steady flow.

  4. Q4easy

    Steady flow means at a fixed point

    1. Aflow properties do not change with time
    2. Bvelocity is zero everywhere
    3. Cpressure is uniform in field
    4. DReynolds number is constant globally
    💡 Explanation:

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

  5. Q5easy

    Uniform flow in an open channel has

    1. Avarying depth along length
    2. Bdepth and velocity constant along the channel reach
    3. Czero discharge
    4. Dalways supercritical flow
    💡 Explanation:

    Uniform flow implies friction slope equals bed slope.

  6. Q6easy

    Turbulent flow is characterised by

    1. Arandom fluctuating velocity components and mixing
    2. Bsmooth layered motion only
    3. CReynolds number below 1
    4. Dzero wall shear
    💡 Explanation:

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

  7. Q7Past Paper · PPSC/FPSC/NTSeasy

    Continuity equation for incompressible steady flow is

    1. AP + ρgh = constant always
    2. Bτ = μ du/dy
    3. Ch_f = fLV²/(2gD)
    4. DQ = AV = constant
    💡 Explanation:

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

  8. Q8Past Paper · PPSC/FPSC/NTSeasy

    Bernoulli equation along a streamline for ideal flow states

    1. Ap = ρgh only
    2. BV = constant always
    3. Cp/ρg + V²/(2g) + z = constant
    4. DfL/D = constant
    💡 Explanation:

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

  9. Q9medium

    Application of Bernoulli between two points requires

    1. Aturbulent mixing always allowed
    2. Bcompressible supersonic flow always
    3. Cunsteady pulsating flow
    4. Dsteady, inviscid, incompressible flow along streamline
    💡 Explanation:

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

  10. Q10easy

    Velocity head is expressed as

    1. AV²/(2g)
    2. BV/g
    3. C2g/V²
    4. DρV²
    💡 Explanation:

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

  11. Q11medium

    Piezometric head at a point equals

    1. Ap/(ρg) + z
    2. BV²/(2g) only
    3. Cp/(ρg) only
    4. Dz − p/(ρg)
    💡 Explanation:

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

  12. Q12medium

    Total energy head H in open channel flow is

    1. Ay only
    2. BV²/(2g) only
    3. Cz − y
    4. Dz + y + V²/(2g) for datum at bed
    💡 Explanation:

    Elevation + depth + velocity head referenced to chosen datum.

  13. Q13Past Paper · PPSC/FPSC/NTSmedium

    Venturi meter measures flow rate using

    1. Atemperature change only
    2. Bmagnetic induction
    3. CManning equation
    4. Dpressure difference between converging and throat sections
    💡 Explanation:

    Continuity and Bernoulli relate throat velocity to upstream conditions.

  14. Q14medium

    Pitot tube measures

    1. Adynamic viscosity
    2. Bstagnation pressure to obtain velocity
    3. CManning roughness
    4. Dsediment concentration
    💡 Explanation:

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

  15. Q15medium

    Orifice meter causes permanent pressure loss because of

    1. ABernoulli being invalid
    2. Bzero velocity at throat
    3. Ccontraction and turbulence at the orifice
    4. Dlaminar flow requirement
    💡 Explanation:

    Higher head loss than Venturi but simpler and cheaper construction.

  16. Q16Past Paper · PPSC/FPSC/NTSmedium

    Momentum equation in fluid mechanics is based on

    1. Afirst law of thermodynamics only
    2. BFick law of diffusion
    3. CNewton second law for control volume
    4. DDarcy law for porous media
    💡 Explanation:

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

  17. Q17medium

    Force exerted by a jet on a flat plate is

    1. A½ρAV² only always
    2. BρgAH
    3. CρAV² for normal impact when plate is stationary
    4. Dzero for any angle
    💡 Explanation:

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

  18. Q18hard

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

    1. Aonly weight of vane
    2. Bchange in momentum in flow direction
    3. CManning n
    4. DReynolds number alone
    💡 Explanation:

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

  19. Q19Past Paper · PPSC/FPSC/NTSmedium

    Hydraulic jump occurs in open channel when

    1. Asubcritical becomes laminar pipe flow
    2. Bflow becomes uniform always
    3. Csupercritical flow transitions to subcritical flow
    4. Ddepth remains constant
    💡 Explanation:

    Rapid increase in depth dissipates excess kinetic energy.

  20. Q20hard

    Specific force in open channel momentum analysis is

    1. AQ/A only
    2. BQ²/(gA) + Aȳ
    3. CV²/(2g)
    4. DManning V
    💡 Explanation:

    Used to locate hydraulic jump and conjugate depths.

  21. Q21medium

    Impulse-momentum principle is preferred over Bernoulli when calculating

    1. Astatic pressure at a point only
    2. Bforces on bends, reducers and structures
    3. Ccapillary rise
    4. Dvapour pressure
    💡 Explanation:

    Momentum equation directly yields force on pipe fittings and gates.

  22. Q22Past Paper · PPSC/FPSC/NTSeasy

    Reynolds number Re is defined as

    1. AρVD/μ or VD/ν
    2. BμVD/ρ
    3. CρgD/μ
    4. DV²D/g
    💡 Explanation:

    Ratio of inertial to viscous forces; determines flow regime.

  23. Q23Past Paper · PPSC/FPSC/NTSeasy

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

    1. A100
    2. B50000
    3. C10
    4. D2300
    💡 Explanation:

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

  24. Q24medium

    Laminar pipe flow velocity profile is

    1. Auniform plug flow
    2. Blinear from wall to centre
    3. Czero at centre
    4. Dparabolic with maximum at centre
    💡 Explanation:

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

  25. Q25medium

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

    1. AV
    2. BV²
    3. CV³
    4. Dindependent of V
    💡 Explanation:

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

  26. Q26easy

    Hydraulic radius R for full circular pipe equals

    1. AD/2
    2. BD/4
    3. CπD²/4
    4. DD
    💡 Explanation:

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

  27. Q27hard

    Draft tube in reaction turbine serves to

    1. Aincrease atmospheric pressure
    2. Bmeasure discharge
    3. Cfilter sediment
    4. Drecover kinetic energy and maintain suction head at runner
    💡 Explanation:

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

  28. Q28Past Paper · PPSC/FPSC/NTSmedium

    Discharge over a rectangular sharp-crested weir is proportional to

    1. AH only
    2. BH^(3/2)
    3. CH²
    4. D√H for all weir types
    💡 Explanation:

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

  29. Q29easy

    V-notch weir is preferred for measuring

    1. Avery high floods only
    2. Bgroundwater seepage directly
    3. Clow discharges with better sensitivity
    4. Dpipe pressure only
    💡 Explanation:

    Triangular weir gives accurate small flow measurement in laboratories.

  30. Q30easy

    Coefficient of discharge for orifice is less than unity due to

    1. Agravity variation
    2. BManning n in pipe
    3. Ccontraction and energy losses
    4. DCoriolis effect
    💡 Explanation:

    Cd typically 0.6–0.65 for sharp-edged orifice.

  31. Q31Past Paper · PPSC/FPSC/NTSeasy

    Torricelli theorem gives jet velocity from orifice as

    1. Agh only
    2. B√(2gh)
    3. C2gh without root
    4. D√(gh/2)
    💡 Explanation:

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

  32. Q32Past Paper · PPSC/FPSC/NTSmedium

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

    1. An − k independent dimensionless groups
    2. Bn + k groups
    3. Ck − n groups
    4. Dalways one group
    💡 Explanation:

    Reduces variables for model testing and empirical correlation.

  33. Q33easy

    Reynolds number is a dimensionless group representing ratio of

    1. Agravity to viscosity
    2. Bpressure to shear only
    3. Csurface tension to gravity only
    4. Dinertial forces to viscous forces
    💡 Explanation:

    Ensures dynamic similarity in viscous flow model studies.

  34. Q34medium

    Froude number similarity is essential in modeling

    1. Afree-surface gravity-driven flows
    2. Bcreeping laminar pipe flow only
    3. CDarcy groundwater flow
    4. Dpure diffusion
    💡 Explanation:

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

  35. Q35hard

    Geometric, kinematic and dynamic similarities together ensure

    1. Aonly same colour of fluid
    2. Bcomplete model-prototype similitude for intended laws
    3. Conly same pipe material
    4. Donly same temperature
    💡 Explanation:

    Scale model tests in hydraulics require matching relevant dimensionless numbers.

  36. Q36easy

    Surface tension in a liquid is caused by

    1. Agravitational pull only
    2. Bviscous shear at walls
    3. CBernoulli pressure drop
    4. Dcohesive forces at the free surface
    💡 Explanation:

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

  37. Q37Past Paper · PPSC/FPSC/NTSeasy

    Dynamic viscosity of a fluid is defined as the ratio of

    1. Adensity to velocity
    2. Bshear stress to rate of shear strain
    3. Cpressure to temperature
    4. Dvelocity to area
    💡 Explanation:

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

  38. Q38Past Paper · PPSC/FPSC/NTSeasy

    Kinematic viscosity ν is related to dynamic viscosity μ by

    1. Aν = μρ
    2. Bν = ρ/μ²
    3. Cν = μ + ρ
    4. Dν = μ/ρ
    💡 Explanation:

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

  39. Q39medium

    The bulk modulus of elasticity of a liquid measures

    1. Asurface tension effects only
    2. Bresistance to uniform compression
    3. Cshear deformation only
    4. Dturbulent mixing rate
    💡 Explanation:

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

  40. Q40easy

    Vapour pressure of water increases with

    1. Adepth in a reservoir only
    2. Bpipe diameter only
    3. Cchannel slope only
    4. Dtemperature
    💡 Explanation:

    Higher temperature increases molecular escape tendency, lowering cavitation margin.

  41. Q41easy

    A Newtonian fluid is one in which

    1. Aviscosity changes with time always
    2. Bshear stress is directly proportional to velocity gradient
    3. Cdensity is constant only in gases
    4. DReynolds number exceeds 4000
    💡 Explanation:

    Water and air at normal conditions behave as Newtonian fluids.

  42. Q42easy

    Specific weight γ of a fluid equals

    1. Aρ/g
    2. Bμρ
    3. Cρg
    4. Dρ + g
    💡 Explanation:

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

  43. Q43medium

    Compressibility of water is generally considered negligible in

    1. Apipe flow and open channel problems at low Mach number
    2. Bcavitation analysis always
    3. Cwater hammer studies
    4. Dhigh-speed jet flows
    💡 Explanation:

    Water hammer and very high pressures require accounting for compressibility.

  44. Q44Past Paper · PPSC/FPSC/NTSeasy

    Hydrostatic pressure at a point in a static fluid depends on

    1. Adepth, fluid density and gravitational acceleration
    2. Bflow velocity only
    3. Cpipe roughness only
    4. DReynolds number
    💡 Explanation:

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

  45. Q45Past Paper · PPSC/FPSC/NTSeasy

    Pascal law states that pressure applied to a confined fluid

    1. Ais transmitted undiminished in all directions
    2. Bacts only vertically downward
    3. Cdecreases with depth always
    4. Ddepends on container shape for intensity
    💡 Explanation:

    Foundation of hydraulic jacks and pressure measurement principles.

  46. Q46Past Paper · PPSC/FPSC/NTSmedium

    The centre of pressure on a submerged plane surface is

    1. Aabove the centroid always
    2. Bcoincident with centroid always
    3. Cat the water surface
    4. Dbelow the centroid because pressure increases with depth
    💡 Explanation:

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

  47. Q47Past Paper · PPSC/FPSC/NTSmedium

    Total hydrostatic force on a vertical plane surface equals

    1. AγA only
    2. B½γH² only without area
    3. Cvelocity head times area
    4. Dγh̄A where h̄ is depth to centroid
    💡 Explanation:

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

  48. Q48easy

    Pressure head in fluid mechanics is expressed as

    1. Ap/(ρg)
    2. Bρgp
    3. Cpρg
    4. Dp + ρg
    💡 Explanation:

    Pressure head converts pressure to equivalent liquid column height.

  49. Q49easy

    Absolute pressure equals gauge pressure plus

    1. Avacuum pressure only
    2. Batmospheric pressure
    3. Cvelocity head
    4. Dvapour pressure always
    💡 Explanation:

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

  50. Q50easy

    A manometer reading measures

    1. Aflow rate directly
    2. BReynolds number
    3. Cdifference in pressure between two points or from atmosphere
    4. DManning n directly
    💡 Explanation:

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

  51. Q51medium

    Hydrostatic paradox demonstrates that

    1. Apressure depends on container shape at same depth
    2. Bdeeper vessels always have lower pressure
    3. Cforce on container bottom depends on depth and area, not total water volume shape
    4. Dhorizontal forces are always zero
    💡 Explanation:

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

  52. Q52Past Paper · PPSC/FPSC/NTSeasy

    Archimedes principle states buoyant force equals

    1. Aweight of the body in air
    2. Bweight of fluid displaced by the body
    3. Chalf the submerged volume only
    4. Dsurface tension force
    💡 Explanation:

    F_b = ρgV_displaced; basis for stability of floating structures.

  53. Q53Past Paper · PPSC/FPSC/NTSeasy

    A body floats when its average specific gravity is

    1. Aequal to zero only
    2. Bgreater than 2.0 always
    3. Cless than that of the fluid
    4. Dequal to fluid SG with no submergence
    💡 Explanation:

    Equilibrium requires displaced fluid weight to equal body weight.

  54. Q54medium

    Metacentric height GM is positive when a floating body is

    1. Aunstable and capsizing
    2. Bstable in small angular disturbances
    3. Cneutrally buoyant only
    4. Dfully submerged always
    💡 Explanation:

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

  55. Q55Past Paper · PPSC/FPSC/NTSeasy

    Darcy-Weisbach head loss formula is

    1. Ah_f = n²V²R^(4/3)
    2. Bh_f = f (L/D)(V²/2g)
    3. Ch_f = KL²
    4. Dh_f = μV only
    💡 Explanation:

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

  56. Q56Past Paper · PPSC/FPSC/NTSmedium

    Hazen-Williams formula is primarily used for

    1. Awater distribution pipes in turbulent regime
    2. Blaminar laboratory tubes only
    3. Copen channel Manning alternative only
    4. Dcompressible gas pipelines
    💡 Explanation:

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

  57. Q57medium

    Moody diagram relates friction factor f to

    1. AFroude number only
    2. BManning n only
    3. Cweir crest length
    4. DReynolds number and relative roughness
    💡 Explanation:

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

  58. Q58hard

    Colebrook-White equation is used to find

    1. ADarcy friction factor in turbulent pipe flow
    2. BManning n from grain size
    3. Ccritical depth directly
    4. Dpump NPSH
    💡 Explanation:

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

  59. Q59easy

    Minor losses in pipe systems are expressed as

    1. AK V²/(2g)
    2. BfL/D only
    3. CManning S
    4. DρgH pump head
    💡 Explanation:

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

  60. Q60easy

    Velocity in pipe for given discharge is maximum when

    1. Adiameter is largest
    2. Barea is minimum for same Q
    3. Croughness is highest
    4. Dpipe is horizontal
    💡 Explanation:

    V = Q/A; contraction increases local velocity.

  61. Q61medium

    Parallel pipes carrying same total head loss have

    1. Aequal discharge always
    2. Binverse relationship between resistance and discharge share
    3. Czero flow in smaller pipe
    4. Dsame velocity regardless of diameter
    💡 Explanation:

    1/Q_total² relationship via equivalent resistance network.

  62. Q62easy

    Series pipes have

    1. Asame head loss in each
    2. Bindependent discharges
    3. Csame discharge through each pipe section
    4. Dzero velocity in second pipe
    💡 Explanation:

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

  63. Q63hard

    Water hammer pressure rise is proportional to

    1. Achange in velocity and wave celerity
    2. Bpipe length only
    3. CManning n
    4. Dchannel width only
    💡 Explanation:

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

  64. Q64medium

    Cavitation in pumps occurs when local pressure falls below

    1. Aatmospheric pressure at all points
    2. Bstagnation pressure
    3. CManning normal depth
    4. Dvapour pressure of the liquid
    💡 Explanation:

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

  65. Q65Past Paper · PPSC/FPSC/NTSeasy

    Manning formula for open channel velocity is

    1. AV = √(gD)
    2. BV = fL/D
    3. CV = C√(RS) Hazen only
    4. DV = (1/n) R^(2/3) S^(1/2)
    💡 Explanation:

    Manning n is roughness coefficient; S is energy slope.

  66. Q66easy

    Manning n increases with

    1. Aflow depth decrease only
    2. Bhigher Reynolds always decreasing n
    3. Cchannel roughness and vegetation
    4. Dsubcritical flow only
    💡 Explanation:

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

  67. Q67medium

    Hydraulic radius for wide rectangular channel approximates

    1. Awidth b
    2. Bb + y
    3. Cy/2 only for all widths
    4. Ddepth y
    💡 Explanation:

    R = by/(b+2y) ≈ y when b >> y.

  68. Q68Past Paper · PPSC/FPSC/NTSmedium

    Specific energy E in open channel is

    1. Ay − V²/(2g)
    2. BQ²/(2g)
    3. Cy + V²/(2g)
    4. Dz + p/γ
    💡 Explanation:

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

  69. Q69Past Paper · PPSC/FPSC/NTSmedium

    Critical depth occurs when specific energy is

    1. Amaximum always
    2. Bzero
    3. Cminimum for given discharge
    4. Dequal to normal depth always
    💡 Explanation:

    At critical flow, Froude number equals unity.

  70. Q70Past Paper · PPSC/FPSC/NTSeasy

    Froude number Fr is defined as

    1. AV/gD
    2. B√(gD)/V
    3. CV/√(gD_h)
    4. DρV²D/μ
    💡 Explanation:

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

  71. Q71medium

    Subcritical flow in open channel is controlled from

    1. Aupstream only always
    2. Bdownstream conditions
    3. Cmid-channel only
    4. Datmosphere only
    💡 Explanation:

    Disturbances propagate upstream; downstream controls depth.

  72. Q72medium

    Supercritical flow is controlled from

    1. Aupstream
    2. Bdownstream
    3. Cboth ends equally always
    4. Dneither end
    💡 Explanation:

    Information cannot travel upstream; upstream boundary fixes conditions.

  73. Q73hard

    Gradually varied flow (GVF) profiles depend on

    1. Apipe friction factor only
    2. Bpump power only
    3. Crelation of actual depth to normal and critical depths
    4. DBOD loading
    💡 Explanation:

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

  74. Q74medium

    Rapidly varied flow includes

    1. Auniform flow in long canal
    2. Blaminar pipe flow
    3. CDarcy groundwater flow
    4. Dhydraulic jump and flow over weirs
    💡 Explanation:

    Curvature and non-hydrostatic pressure distribution dominate locally.

  75. Q75Past Paper · PPSC/FPSC/NTSeasy

    Centrifugal pump head is the energy imparted per unit

    1. Avolume only
    2. Bweight of liquid
    3. Cmass times g only without head concept
    4. Dtime only
    💡 Explanation:

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

  76. Q76Past Paper · PPSC/FPSC/NTSmedium

    NPSH available must exceed NPSH required to avoid

    1. Acavitation at pump suction
    2. Bwater hammer in delivery pipe
    3. CManning depth error
    4. Dsediment deposition
    💡 Explanation:

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

  77. Q77hard

    Specific speed N_s of a pump indicates

    1. Aonly motor RPM
    2. Bpipe diameter only
    3. Ctype and suitability for head-discharge duty
    4. DManning roughness
    💡 Explanation:

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

  78. Q78medium

    Characteristic curves of centrifugal pump relate

    1. Aonly pipe friction factor
    2. Bonly channel slope
    3. Chead, power and efficiency to discharge
    4. Donly rainfall intensity
    💡 Explanation:

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

  79. Q79Past Paper · PPSC/FPSC/NTSmedium

    Pelton turbine is best suited for

    1. Alow head high flow rivers
    2. Btidal barrages only
    3. Chigh head and low discharge
    4. Ddrainage pumps
    💡 Explanation:

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

  80. Q80Past Paper · PPSC/FPSC/NTSmedium

    Francis turbine is a

    1. Amixed-flow reaction turbine for medium head
    2. Bpure impulse axial only
    3. Cpositive displacement pump
    4. Dtrickling filter
    💡 Explanation:

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