Fluid Mechanics and Hydraulics MCQs 2026
80 questions with detailed answers · 28 from past papers · 8 quiz batches available
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- Q1 Past Paper · PPSC/FPSC/NTS easy
Dynamic viscosity of a fluid is defined as the ratio of
💡 Explanation:Newton law of viscosity: τ = μ(du/dy); μ is dynamic viscosity in Pa·s.
- Q2 Past Paper · PPSC/FPSC/NTS easy
Kinematic viscosity ν is related to dynamic viscosity μ by
💡 Explanation:Kinematic viscosity has units m²/s and appears in Reynolds number.
- Q3 medium
The bulk modulus of elasticity of a liquid measures
💡 Explanation:High bulk modulus makes liquids nearly incompressible for most hydraulic problems.
- Q4 easy
Surface tension in a liquid is caused by
💡 Explanation:Surface tension creates capillary rise and affects small-scale flow phenomena.
- Q5 easy
Vapour pressure of water increases with
💡 Explanation:Higher temperature increases molecular escape tendency, lowering cavitation margin.
- Q6 easy
A Newtonian fluid is one in which
💡 Explanation:Water and air at normal conditions behave as Newtonian fluids.
- Q7 easy
Specific weight γ of a fluid equals
💡 Explanation:γ = weight per unit volume; differs from mass density ρ by gravitational acceleration.
- Q8 medium
Compressibility of water is generally considered negligible in
💡 Explanation:Water hammer and very high pressures require accounting for compressibility.
- Q9 Past Paper · PPSC/FPSC/NTS easy
Hydrostatic pressure at a point in a static fluid depends on
💡 Explanation:p = ρgh for incompressible fluid; pressure increases linearly with depth.
- Q10 Past Paper · PPSC/FPSC/NTS easy
Pascal law states that pressure applied to a confined fluid
💡 Explanation:Foundation of hydraulic jacks and pressure measurement principles.
- Q11 Past Paper · PPSC/FPSC/NTS medium
The centre of pressure on a submerged plane surface is
💡 Explanation:Moment of pressure forces shifts CP below centroid for inclined or vertical surfaces.
- Q12 Past Paper · PPSC/FPSC/NTS medium
Total hydrostatic force on a vertical plane surface equals
💡 Explanation:F = γh̄A; location of CP found using moment of inertia about water surface.
- Q13 easy
Pressure head in fluid mechanics is expressed as
💡 Explanation:Pressure head converts pressure to equivalent liquid column height.
- Q14 easy
Absolute pressure equals gauge pressure plus
💡 Explanation:p_abs = p_gauge + p_atm; vacuum is pressure below atmospheric.
- Q15 easy
A manometer reading measures
💡 Explanation:U-tube and inclined manometers are common in laboratory and field.
- Q16 medium
Hydrostatic paradox demonstrates that
💡 Explanation:At equal depth, pressure is same regardless of container geometry.
- Q17 Past Paper · PPSC/FPSC/NTS easy
Archimedes principle states buoyant force equals
💡 Explanation:F_b = ρgV_displaced; basis for stability of floating structures.
- Q18 Past Paper · PPSC/FPSC/NTS easy
A body floats when its average specific gravity is
💡 Explanation:Equilibrium requires displaced fluid weight to equal body weight.
- Q19 medium
Metacentric height GM is positive when a floating body is
💡 Explanation:Positive GM restores righting moment; GM = BM + BG − BG relations apply.
- Q20 Past Paper · PPSC/FPSC/NTS easy
Darcy-Weisbach head loss formula is
💡 Explanation:Friction factor f depends on Re and relative roughness ε/D.
- Q21 Past Paper · PPSC/FPSC/NTS medium
Hazen-Williams formula is primarily used for
💡 Explanation:V = C_RH^0.63 S^0.54 in SI form; C is Hazen-Williams coefficient.
- Q22 medium
Moody diagram relates friction factor f to
💡 Explanation:Essential for estimating f in commercial pipes when Colebrook implicit solution not used.
- Q23 hard
Colebrook-White equation is used to find
💡 Explanation:Implicit relation between f, Re and ε/D; solved iteratively or by Swamee-Jain.
- Q24 easy
Minor losses in pipe systems are expressed as
💡 Explanation:K depends on fitting type: elbows, valves, sudden expansion.
- Q25 easy
Velocity in pipe for given discharge is maximum when
💡 Explanation:V = Q/A; contraction increases local velocity.
- Q26 medium
Parallel pipes carrying same total head loss have
💡 Explanation:1/Q_total² relationship via equivalent resistance network.
- Q27 easy
Series pipes have
💡 Explanation:Q constant; total head loss is sum of individual losses.
- Q28 hard
Water hammer pressure rise is proportional to
💡 Explanation:Δp = ρaΔV; rapid valve closure causes surge in pressure pipes.
- Q29 medium
Cavitation in pumps occurs when local pressure falls below
💡 Explanation:Vapour bubbles form and collapse, damaging impellers and reducing efficiency.
- Q30 Past Paper · PPSC/FPSC/NTS easy
Manning formula for open channel velocity is
💡 Explanation:Manning n is roughness coefficient; S is energy slope.
- Q31 easy
Manning n increases with
💡 Explanation:Smooth concrete has low n; vegetated floodplains have high n.
- Q32 medium
Hydraulic radius for wide rectangular channel approximates
💡 Explanation:R = by/(b+2y) ≈ y when b >> y.
- Q33 Past Paper · PPSC/FPSC/NTS medium
Specific energy E in open channel is
💡 Explanation:Energy per unit weight relative to channel bottom for given discharge.
- Q34 Past Paper · PPSC/FPSC/NTS medium
Critical depth occurs when specific energy is
💡 Explanation:At critical flow, Froude number equals unity.
- Q35 Past Paper · PPSC/FPSC/NTS easy
Froude number Fr is defined as
💡 Explanation:D_h is hydraulic depth; Fr < 1 subcritical, Fr > 1 supercritical.
- Q36 medium
Subcritical flow in open channel is controlled from
💡 Explanation:Disturbances propagate upstream; downstream controls depth.
- Q37 medium
Supercritical flow is controlled from
💡 Explanation:Information cannot travel upstream; upstream boundary fixes conditions.
- Q38 hard
Gradually varied flow (GVF) profiles depend on
💡 Explanation:Mild, steep, horizontal and adverse beds yield M, S, H, A profile classes.
- Q39 medium
Rapidly varied flow includes
💡 Explanation:Curvature and non-hydrostatic pressure distribution dominate locally.
- Q40 Past Paper · PPSC/FPSC/NTS easy
Centrifugal pump head is the energy imparted per unit
💡 Explanation:H = (p/ρg) + V²/(2g) + z increase across pump.
- Q41 Past Paper · PPSC/FPSC/NTS medium
NPSH available must exceed NPSH required to avoid
💡 Explanation:NPSHA depends on atmospheric pressure, vapour pressure and suction lift.
- Q42 hard
Specific speed N_s of a pump indicates
💡 Explanation:Low N_s: radial high-head; high N_s: axial low-head pumps.
- Q43 medium
Characteristic curves of centrifugal pump relate
💡 Explanation:H-Q curve falls with increasing Q; efficiency has peak at BEP.
- Q44 Past Paper · PPSC/FPSC/NTS medium
Pelton turbine is best suited for
💡 Explanation:Impulse turbine with jet striking buckets; common in Himalayan projects.
- Q45 Past Paper · PPSC/FPSC/NTS medium
Francis turbine is a
💡 Explanation:Pressure drop occurs in runner; used widely in irrigation and hydropower.
- Q46 medium
The centre of buoyancy of a submerged body is the centroid of
💡 Explanation:Couple between CB and CG determines rotational stability.
- Q47 hard
For a fully submerged body, stable equilibrium requires
💡 Explanation:Righting couple exists when CB moves more than CG during tilt.
- Q48 easy
In fluid flow, a streamline is a line
💡 Explanation:Streamlines show flow direction; cannot cross in steady flow.
- Q49 easy
Steady flow means at a fixed point
💡 Explanation:∂/∂t = 0 at a point; unsteady flow varies with time.
- Q50 easy
Uniform flow in an open channel has
💡 Explanation:Uniform flow implies friction slope equals bed slope.
- Q51 easy
Turbulent flow is characterised by
💡 Explanation:Re > critical value; eddies enhance momentum and energy dissipation.
- Q52 Past Paper · PPSC/FPSC/NTS easy
Continuity equation for incompressible steady flow is
💡 Explanation:Conservation of mass: discharge Q equals area times mean velocity.
- Q53 Past Paper · PPSC/FPSC/NTS easy
Bernoulli equation along a streamline for ideal flow states
💡 Explanation:Energy per unit weight: pressure + velocity + elevation heads sum constant.
- Q54 medium
Application of Bernoulli between two points requires
💡 Explanation:Real flows need head loss term for viscous and minor losses.
- Q55 easy
Velocity head is expressed as
💡 Explanation:Kinetic energy per unit weight; converts velocity to equivalent head.
- Q56 medium
Piezometric head at a point equals
💡 Explanation:Sum of pressure head and elevation head; excludes velocity head.
- Q57 medium
Total energy head H in open channel flow is
💡 Explanation:Elevation + depth + velocity head referenced to chosen datum.
- Q58 Past Paper · PPSC/FPSC/NTS medium
Venturi meter measures flow rate using
💡 Explanation:Continuity and Bernoulli relate throat velocity to upstream conditions.
- Q59 medium
Pitot tube measures
💡 Explanation:V = √(2gΔh) for ideal conditions from stagnation-static difference.
- Q60 medium
Orifice meter causes permanent pressure loss because of
💡 Explanation:Higher head loss than Venturi but simpler and cheaper construction.
- Q61 Past Paper · PPSC/FPSC/NTS medium
Momentum equation in fluid mechanics is based on
💡 Explanation:ΣF = ρQ(V_out − V_in) + pressure forces for steady control volume.
- Q62 medium
Force exerted by a jet on a flat plate is
💡 Explanation:Momentum change Δ(mV) gives F = ρQ(V − 0) for normal jet.
- Q63 hard
When a jet strikes a curved vane and deflects through angle θ, force component depends on
💡 Explanation:F = ρQ(V_in cos α − V_out cos β) for curved vanes in turbines.
- Q64 Past Paper · PPSC/FPSC/NTS medium
Hydraulic jump occurs in open channel when
💡 Explanation:Rapid increase in depth dissipates excess kinetic energy.
- Q65 hard
Specific force in open channel momentum analysis is
💡 Explanation:Used to locate hydraulic jump and conjugate depths.
- Q66 medium
Impulse-momentum principle is preferred over Bernoulli when calculating
💡 Explanation:Momentum equation directly yields force on pipe fittings and gates.
- Q67 Past Paper · PPSC/FPSC/NTS easy
Reynolds number Re is defined as
💡 Explanation:Ratio of inertial to viscous forces; determines flow regime.
- Q68 Past Paper · PPSC/FPSC/NTS easy
For pipe flow, critical Reynolds number for transition to turbulence is approximately
💡 Explanation:Below ~2300 flow is laminar; commercial pipes transition earlier due to disturbances.
- Q69 medium
Laminar pipe flow velocity profile is
💡 Explanation:Hagen-Poiseuille flow: u = (g/4μ)(−dp/dx)(R² − r²).
- Q70 medium
In laminar flow through circular pipe, head loss varies with velocity as
💡 Explanation:h_f ∝ V for laminar (Hagen-Poiseuille); h_f ∝ V² for turbulent.
- Q71 easy
Hydraulic radius R for full circular pipe equals
💡 Explanation:R = A/P = (πD²/4)/(πD) = D/4; used in Reynolds and Manning formulas.
- Q72 hard
Draft tube in reaction turbine serves to
💡 Explanation:Diffuser converts outlet kinetic head to pressure head, improving net head.
- Q73 Past Paper · PPSC/FPSC/NTS medium
Discharge over a rectangular sharp-crested weir is proportional to
💡 Explanation:Q = (2/3)Cd b √(2g) H^(3/2) for ideal; Cd is discharge coefficient.
- Q74 easy
V-notch weir is preferred for measuring
💡 Explanation:Triangular weir gives accurate small flow measurement in laboratories.
- Q75 easy
Coefficient of discharge for orifice is less than unity due to
💡 Explanation:Cd typically 0.6–0.65 for sharp-edged orifice.
- Q76 Past Paper · PPSC/FPSC/NTS easy
Torricelli theorem gives jet velocity from orifice as
💡 Explanation:Ideal velocity from head h; actual velocity reduced by coefficient of velocity.
- Q77 Past Paper · PPSC/FPSC/NTS medium
Buckingham Pi theorem states that a problem with n variables and k fundamental dimensions has
💡 Explanation:Reduces variables for model testing and empirical correlation.
- Q78 easy
Reynolds number is a dimensionless group representing ratio of
💡 Explanation:Ensures dynamic similarity in viscous flow model studies.
- Q79 medium
Froude number similarity is essential in modeling
💡 Explanation:Fr match required for open channel, spillway and ship model tests.
- Q80 hard
Geometric, kinematic and dynamic similarities together ensure
💡 Explanation:Scale model tests in hydraulics require matching relevant dimensionless numbers.