Strength of Materials MCQ Practice Set — 50 Questions with Answers
50 exam-oriented Strength of Materials multiple-choice questions with the correct answer and a clear explanation for each. Frequently asked in AE Level Civil Engineering, JE Level Civil Engineering. Solve the full set below for free — no login required.
- 1Strength of MaterialsMEDIUM
In a 'statically indeterminate' structure, 'compatibility' equations ensure:
AForces are in equilibrium onlyBDeformations are geometrically consistent with support conditions and member continuity — no gaps or overlapsCStresses are below yieldDThe structure is safe against bucklingAnswer: B. Deformations are geometrically consistent with support conditions and member continuity — no gaps or overlaps
Explanation: Compatibility (geometric) equations: ensure deformations are consistent — no gaps or overlaps at joints, supports are satisfied (zero displacement where constrained), continuous members remain connected. In force method (flexibility method): redundant forces chosen such that deformations at released points are compatible (closure conditions). Compatibility + equilibrium + constitutive = complete solution.
- 2Strength of MaterialsMEDIUM
The 'section classification' of steel sections in IS 800 (plastic/compact/semi-compact/slender) depends on:
AMaterial yield strength onlyBb/t ratios of compression elements relative to ε = √(250/fy) — governs whether plastic, compact, semi-compact, or slenderCOnly member lengthDConnection type (welded vs bolted)Answer: B. b/t ratios of compression elements relative to ε = √(250/fy) — governs whether plastic, compact, semi-compact, or slender
Explanation: IS 800 Table 2: section class depends on b/t ratios of compression elements (flanges, webs). Plastic (Class 1): can form plastic hinges with full rotation capacity (b/t ≤ 8.4ε, ε = √(250/fy)). Compact (Class 2): attain Mp but limited rotation. Semi-compact (Class 3): elastic moment only (Me). Slender (Class 4): local buckling before yield — effective section needed. Classification determines design method.
- 3Strength of MaterialsMEDIUM
The 'virtual work principle' states that for a system in equilibrium, if it undergoes a virtual displacement:
AExternal work = 0 alwaysBVirtual work of external forces = virtual internal strain energy for equilibrium (δW_ext = δW_int)CStresses must be zeroDDisplacement must be real, not virtualAnswer: B. Virtual work of external forces = virtual internal strain energy for equilibrium (δW_ext = δW_int)
Explanation: Principle of virtual work: for a system in equilibrium, the total virtual work done by all external forces through any compatible virtual displacement = virtual strain energy stored = 0 (for virtual work of external forces minus internal work). Used as: (1) equilibrium check; (2) computing deflections (unit load method). The virtual displacement must be kinematically admissible (compatible with constraints).
- 4Strength of MaterialsHARD
In the 'Muller-Breslau principle', the influence line for a function (reaction, shear, moment) is the:
AFree body diagram of the structureBDeflected shape when the restraint is removed and unit deformation applied — direct graphical construction of ILCThe loading pattern itselfDBending moment under full loadingAnswer: B. Deflected shape when the restraint is removed and unit deformation applied — direct graphical construction of IL
Explanation: Muller-Breslau: the influence line for any function X in a structure is the deflected shape of the structure when the restraint corresponding to X is removed and a unit displacement (for reactions/shears) or unit rotation (for moments) is applied in the direction of X. Extremely powerful: find IL shape without calculating ordinates individually. For propped reaction: remove prop, apply unit settlement = IL for reaction.
- 5Strength of MaterialsMEDIUM
The 'direct stiffness' global matrix after assembly is generally:
ADiagonal only — zero off-diagonal termsBSymmetric and banded (non-zeros near diagonal); singular until BCs applied — rigid body motion possibleCAlways upper triangular from assemblyDFull dense matrix with all non-zero termsAnswer: B. Symmetric and banded (non-zeros near diagonal); singular until BCs applied — rigid body motion possible
Explanation: Assembled global stiffness matrix [K]: (1) Symmetric: Kij = Kji (reciprocal theorem); (2) Banded: non-zero only near diagonal (elements near each other share DOF); (3) Singular before boundary conditions (rigid body modes possible); (4) Positive semi-definite. After applying boundary conditions (zeroing rows/columns of restrained DOF): becomes positive definite, non-singular → solvable. Bandwidth minimized by node numbering algorithms.
- 6Strength of MaterialsMEDIUM
The 'approximate analysis' of building frames under lateral loads (portal frame method) assumes:
AAll members have same stiffness (EI uniform)BInflection points at mid-height of columns and mid-span of beams; interior columns resist double shear of exteriorCOnly horizontal loads existDAll beam moments are zeroAnswer: B. Inflection points at mid-height of columns and mid-span of beams; interior columns resist double shear of exterior
Explanation: Portal method (lateral loads, low-rise frames): (1) Inflection point at mid-height of each column; (2) Inflection point at mid-span of each beam; (3) Internal columns carry twice the shear of exterior columns (assume each bay is a portal frame). Results in statically determinate sub-frames. Suitable for low-rise frames (< 10 storeys). Cantilever method: assumes column axial force proportional to distance from centroid (for tall frames).
- 7Strength of MaterialsHARD
The 'Timoshenko beam' theory differs from Euler-Bernoulli theory by including:
AOnly bending stress in the flangesBTransverse shear deformation and rotatory inertia — critical for deep beams or high-frequency vibration modesCAxial deformation onlyDTemperature effects along the beamAnswer: B. Transverse shear deformation and rotatory inertia — critical for deep beams or high-frequency vibration modes
Explanation: Timoshenko beam: includes shear deformation (transverse shear strain) AND rotatory inertia. E-B: assumes plane sections perpendicular to beam axis remain plane and perpendicular (no shear deformation). Timoshenko: sections remain plane but NOT necessarily perpendicular (shear angle φ). Important for: deep beams (span/depth < 10), composite beams with weak core, dynamic analysis (higher modes). Shear factor κ: 5/6 for rectangular, 0.9 for circular.
- 8Strength of MaterialsHARD
The 'plastic hinge rotation capacity' in steel beams depends on:
AApplied live load onlyBSection classification (Class 1 plastic), steel grade ductility, and lateral restraint — determines if plastic moment can be maintained during redistributionCOnly the span of the beamDConnection bolt grade onlyAnswer: B. Section classification (Class 1 plastic), steel grade ductility, and lateral restraint — determines if plastic moment can be maintained during redistribution
Explanation: Plastic hinge rotation capacity: needed when Pp (plastic collapse load) requires redistribution. Ductility depends on: (1) Steel grade (higher strength = lower ductility); (2) Section compactness (Class 1: can form and rotate plastic hinge; Class 2: can form but limited rotation; Class 3: reach My only); (3) Lateral restraint (prevent LTB); (4) Web-flange proportions. IS 800 limits Class 1 (plastic): b/t ≤ 8.4ε for flange, d/tw ≤ 84ε for web (ε = √(250/fy)).
- 9Strength of MaterialsMEDIUM
If the area of tensile steel reinforcement is doubled in a singly reinforced beam, the moment of resistance of the beam increases by about
A12%B22%C32%D42%Answer: C. 32%
Explanation: In a balanced or under-reinforced section, the moment of resistance is proportional to the area of steel (Ast) multiplied by the lever arm (d - n/3). While doubling Ast increases the force component, the neutral axis depth (n) also increases, which slightly reduces the lever arm. The net increase in moment of resistance is approximately 32%.
- 10Strength of MaterialsMEDIUM
A cantilever beam of length L is subjected to a uniformly distributed load w (load per unit length) over its whole length L and a concentrated load W (upward) at the free end. If the total downward load W_total = wL is equal to the upward concentrated load W, the deflection at the free end is:
AZeroBwL^4/8EI (downward)CWL^3/3EI (upward)DWL^3/8EI (downward)Answer: B. wL^4/8EI (downward)
Explanation: The downward deflection due to UDL is wL^4/8EI. The upward deflection due to point load W at the free end is WL^3/3EI. Since W = wL, the net deflection is (wL^4/8EI) - (wL^4/3EI), which is downward.
- 11Strength of MaterialsMEDIUM
With a decrease in lateral stiffness of a regular typical building, the design horizontal seismic coefficient will
AincreaseBdecreaseCremain unchangedDdepends on the heightAnswer: B. decrease
Explanation: According to IS 1893, the design horizontal seismic coefficient (Ah) is proportional to the spectral acceleration (Sa/g). A decrease in lateral stiffness increases the natural period (T). For most structures in the constant acceleration or velocity range, an increase in T leads to a decrease in Sa/g, thus decreasing Ah.
- 12Strength of MaterialsMEDIUM
The latest start of an activity is always
Aequal to the latest event time of the preceding nodeBless than the latest event time of the preceding nodeCgreater than or equal to the latest event time of the preceding nodeDindependent of the latest event time of the preceding nodeAnswer: C. greater than or equal to the latest event time of the preceding node
Explanation: For an activity i-j, the latest start time is LST = TL(j) - duration. The latest event time of the preceding node, TL(i), is the minimum of the latest start times of activities leaving that node. Hence an activity's LST is always greater than or equal to TL(i), with equality for at least one controlling outgoing activity.
- 13Strength of MaterialsMEDIUM
A beam AB of length 10 m having both ends fixed is acted upon by a concentrated load of 10 tonnes at C at a distance of 4 m from A. The B.M. at C is
A11.52 t-m (Hogging)B14.4 t-m (Hogging)C9.6 t-m (Sagging)D11.52 t-m (Sagging)Answer: D. 11.52 t-m (Sagging)
Explanation: For a fixed beam with load P at distance a from A and b from B (L=a+b), the moment at C is M_c = (P * a² * b²) / L³. Here P=10, a=4, b=6, L=10. M_c = (10 * 16 * 36) / 1000 = 5.76 t-m. The original options provided are inconsistent with standard structural analysis formulas for fixed beams.
- 14Strength of MaterialsMEDIUM
In case of mild steel, the value of the stress at elastic limit is
ASame as that at limit of proportionalityBMore than that at limit of proportionalityCLess than that at limit of proportionalityDIndependent and cannot be comparedAnswer: A. Same as that at limit of proportionality
Explanation: For mild steel, the limit of proportionality and the elastic limit are practically identical, though the elastic limit is theoretically slightly higher.
- 15Strength of MaterialsMEDIUM
The maximum deflection in a steel beam is limited to
AL/360BL/325CL/250DL/150Answer: A. L/360
Explanation: According to IS 800:2007, the maximum deflection for beams supporting brittle cladding is generally limited to L/300, and for general cases, it is often L/360.
- 16Strength of MaterialsMEDIUM
Which of the following bends will cause the maximum head loss?
A30° bendB60° bendC90° bendDU-bendAnswer: D. U-bend
Explanation: Head loss in a pipe bend increases with the angle of the bend. A U-bend (180°) causes more head loss than a 90° bend.
- 17TorsionEASY
Power transmitted by a shaft rotating at N rpm with torque T (N·m):
AP = 2πNT/60BP = NT/60CP = πNT/30DP = 2NT/60Answer: A. P = 2πNT/60
Explanation: P = T×ω = T×(2πN/60) = 2πNT/60 watts. Or P = 2πNT/60000 kW.
- 18Columns and StrutsEASY
Effective length L_e for a column fixed at one end and hinged at the other:
ALB0.5LC0.7LD2LAnswer: C. 0.7L
Explanation: Effective lengths: Pin-pin = L; Fixed-fixed = 0.5L; Fixed-free = 2L; Fixed-hinged = 0.7L (≈L/√2).
- 19TorsionMEDIUM
A hollow shaft (D_o=100mm, D_i=50mm) vs solid shaft (D=100mm), same material and torque. Hollow shaft weight saving:
A75%B25%C50%D12.5%Answer: B. 25%
Explanation: Weight ∝ cross-sectional area. Solid: π×100²/4. Hollow: π(100²−50²)/4 = π×7500/4. Saving = (100²−(100²−50²))/100² = 2500/10000 = 25%.
- 20Strength of MaterialsMEDIUM
The 'consistent deformation' (compatibility) method for redundant beams removes redundants and applies:
APrescribing displacement at all nodesBUnit redundant forces and compatibility equations: δ_load + Σfij×Xj = 0 → solve for redundantsCReplacing structure with equivalent platesDUsing finite elements onlyAnswer: B. Unit redundant forces and compatibility equations: δ_load + Σfij×Xj = 0 → solve for redundants
Explanation: Force method (flexibility/compatibility): (1) Remove redundants to get released (statically determinate) structure; (2) Find displacements at released points due to external loads (free body terms); (3) Apply unit values of redundants and find influence coefficients (flexibility coefficients fij); (4) Write compatibility equations: δ_released + Σ(fij × Xj) = 0; (5) Solve for redundants X. Then all internal forces by superposition.
- 21Strength of MaterialsMEDIUM
In 'combined bending and torsion' for a circular shaft, the 'equivalent twisting moment' Teq is:
ATeq = M + TBTeq = √(M² + T²) — combines bending and torsion effects for max shear stress checkCTeq = (M + T)/2DTeq = M × TAnswer: B. Teq = √(M² + T²) — combines bending and torsion effects for max shear stress check
Explanation: For shaft with bending M and torque T: Equivalent torque Teq = √(M²+T²) (used with τ formula: τ = Teq×r/J for max shear). Equivalent bending moment Meq = (M + √(M²+T²))/2 (used with bending stress σ = Meq/Z). Principal stress: σ1,2 = σ_b/2 ± √((σ_b/2)²+τ²). Design shafts for: Teq and τ ≤ τ_allow, or Meq and σ ≤ σ_allow.
- 22Strength of MaterialsHARD
In 'finite element analysis' (FEA), the 'isoparametric formulation' uses:
ADifferent functions for geometry and displacementBSame shape functions N_i for both geometric mapping and displacement interpolation — allows curved element boundariesCOnly triangular elementsDAnalytical integration onlyAnswer: B. Same shape functions N_i for both geometric mapping and displacement interpolation — allows curved element boundaries
Explanation: Isoparametric element: same shape functions N_i used for both geometry (x,y) and displacement (u,v): x=ΣN_i×x_i, u=ΣN_i×u_i. Allows curved element sides (essential for curved boundaries). Integration: numerical Gauss quadrature in natural coordinates (ξ,η). 4-node quadrilateral → bilinear; 8-node serendipity → quadratic. Jacobian J transforms from physical to natural coordinates.
- 23Strength of MaterialsMEDIUM
The 'slope' at the free end of a cantilever beam of span L, EI, under end point load P is:
Aθ = PL³/(3EI) (this is the deflection formula)Bθ = PL²/(2EI) — slope at free end of cantilever under end point loadCθ = PL/(EI)Dθ = 2PL²/(EI)Answer: B. θ = PL²/(2EI) — slope at free end of cantilever under end point load
Explanation: Cantilever deflection: slope at free end θ = PL²/(2EI); deflection at free end δ = PL³/(3EI). For cantilever under UDL w: θ = wL³/(6EI), δ = wL⁴/(8EI). Mohr''s theorems: (1) change in slope = area of M/EI diagram; (2) deflection = moment of M/EI area about reference. Standard results used in superposition (e.g. propped cantilever analysis).
- 24Strength of MaterialsMEDIUM
The 'temperature stress' in a bar fixed at both ends is:
Aσ = E×α×ΔT×L (includes length)Bσ = E×α×ΔT (compressive) — restrained expansion creates stress proportional to thermal strainCZero (temperature causes no stress)Dσ = α×ΔT only (no E needed)Answer: B. σ = E×α×ΔT (compressive) — restrained expansion creates stress proportional to thermal strain
Explanation: Temperature stress: bar fixed at both ends, temperature rises by ΔT. Free thermal strain = αΔT, but restrained → compressive stress developed = E×αΔT (compressive because bar wants to expand but cannot). No strain in bar. If one end is free: expansion ΔL = αΔTL, no stress. For prestress: temperature stress important in bridges where expansion joints are limited.
- 25Strength of MaterialsHARD
The 'collapse load' for a fixed-fixed beam of span L under central point load W is:
AW = 8Mp/LBW = 16Mp/L — three hinges (both ends + midspan) mechanism gives collapse loadCW = 4Mp/LDW = 12Mp/L (this applies to fixed-pinned)Answer: B. W = 16Mp/L — three hinges (both ends + midspan) mechanism gives collapse load
Explanation: Fixed-fixed beam under central W: collapse mechanism needs 3 plastic hinges (both fixed ends + midspan). At collapse: W×L/4 = Mp (from virtual work: W×δ = Mp×θ + Mp×2θ + Mp×θ = 4Mp×θ, δ = L/4×θ). So W_collapse = 16Mp/L. For fixed-pinned: W = 12Mp/L (2 hinges only). Plastic analysis always > elastic analysis because Mp > Me.
- 26Strength of MaterialsHARD
The 'funicular polygon' for a set of concentrated loads gives the shape of:
ABending moment diagram onlyBThe cable shape under those specific loads — pure tension in cable segments (or compression in equivalent arch)CDeflection of a beamDShear force diagramAnswer: B. The cable shape under those specific loads — pure tension in cable segments (or compression in equivalent arch)
Explanation: Funicular polygon: graphical method combining force polygon (vector polygon of loads + reactions) with space diagram to find the shape a cable or arch must take to carry those specific concentrated loads in pure tension or compression (no bending). For a cable: funicular polygon = cable shape. Rays from pole in force polygon correspond to cable segments. Useful for finding reactions and cable geometry.
- 27Propped Cantilever ReactionMEDIUM
A propped cantilever beam of span L with a uniformly distributed load w per unit length has a prop reaction at the free end of:
A3wL/8B5wL/8CwL/2DwL/4Answer: A. 3wL/8
Explanation: Propped cantilever with UDL w: prop reaction R = 3wL/8 (at the simply supported end). Fixed end reaction = 5wL/8, fixed end moment = wL^2/8. This is a standard one-degree statically indeterminate beam solved by compatibility (deflection at prop = 0). Bending moment at fixed end = wL^2/8 (hogging).
- 28Strength of Materials and Structural AnalysisMEDIUM
The core or kern of a section is the area within which load causes:
APlastic collapse onlyBNo tensile stress anywhere in sectionCMaximum shear at surfaceDZero compressive stressAnswer: B. No tensile stress anywhere in section
Explanation: If resultant passes through kern, stress remains compressive throughout.
- 29Strength of Materials and Structural AnalysisEASY
Effective length of a column fixed at one end and free at other is:
A2LBL/2CL/sqrt(2)DLAnswer: A. 2L
Explanation: A cantilever column has effective length twice its unsupported length.
- 30Strength of Materials and Structural AnalysisEASY
Effective length of a column hinged at both ends is:
A2LB0.5LCLD0.7LAnswer: C. L
Explanation: Pin-ended column has effective length equal to actual length.
- 31Strength of Materials and Structural AnalysisMEDIUM
Middle third rule applies to:
AOpen channel flowBRectangular section under eccentric compressionCFlexible pavementDTorsion of circular shaftAnswer: B. Rectangular section under eccentric compression
Explanation: Resultant within middle third avoids tension at base.
- 32Perry-Robertson FormulaMEDIUM
Perry-Robertson formula in column design accounts for:
AMaterial plasticityBInitial bow/imperfection of columnsCEnd fixity onlyDTemperature effectAnswer: B. Initial bow/imperfection of columns
Explanation: Perry-Robertson includes initial geometric imperfection (bow) of real columns; used in BS:449.
- 33Definition of Elastic ModulusEASY
The modulus of elasticity of a material is defined as:
AStress/strain within elastic limitBStrain/stress within elastic limitCUltimate stress/fracture strainDYield stress/plastic strainAnswer: A. Stress/strain within elastic limit
Explanation: Young's modulus E = σ/ε within the proportional (elastic) limit.
- 34Strength of MaterialsHARD
For ductile materials, failure theory commonly used under combined stress is:
Aminimum strain energy theoryBmaximum principal strain theory onlyCEuler theoryDmaximum shear stress theoryAnswer: D. maximum shear stress theory
Explanation: Tresca maximum shear stress theory is often applied to ductile materials.
- 35Strength of MaterialsHARD
The flexural rigidity of a beam is:
AEABM/ICGJDEIAnswer: D. EI
Explanation: Flexural rigidity is the product of modulus of elasticity and second moment of area.
- 36Strength of MaterialsHARD
Rankine's column formula combines:
Atorsion and bending onlyBcrushing and buckling effectsCfatigue and impact onlyDshear and creep onlyAnswer: B. crushing and buckling effects
Explanation: Rankine formula is an empirical interaction of crushing load and Euler buckling load.
- 37Strength of MaterialsHARD
The neutral axis of a homogeneous beam section under pure bending passes through:
Apoint of maximum shear onlyBcentroid of the sectionCtopmost fibreDbottommost fibreAnswer: B. centroid of the section
Explanation: For homogeneous prismatic beams, the neutral axis passes through the centroid.
- 38Theory of Structures and DesignHARD
The maximum shear stress in a rectangular beam section is:
A1.5 times average shear stressBzero at neutral axisCequal to average shear stressD2 times average shear stressAnswer: A. 1.5 times average shear stress
Explanation: For a rectangle, shear stress distribution is parabolic and maximum at the neutral axis is 1.5 times average shear stress.
- 39Theory of Structures and DesignHARD
For a simply supported beam carrying UDL w over full span, maximum deflection is:
AwL^3/(48EI)BwL^4/(8EI)CwL^4/(185EI)D5wL^4/(384EI)Answer: D. 5wL^4/(384EI)
Explanation: The maximum midspan deflection under full-span UDL is 5wL^4/(384EI).
- 40Theory of Structures and DesignHARD
For a rectangular section of width b and depth d, the plastic shape factor in bending about the centroidal axis parallel to b is:
A1.00B1.50C1.33D2.00Answer: B. 1.50
Explanation: For a rectangle, Zp = bd^2/4 and Ze = bd^2/6, hence shape factor = Zp/Ze = 1.5.
- 41Theory of Structures and DesignHARD
The member stiffness at end A of a prismatic beam AB, when end B is hinged, is:
A2EI/LBEI/LC3EI/LD4EI/LAnswer: C. 3EI/L
Explanation: The rotational stiffness is 4EI/L if the far end is fixed and 3EI/L if the far end is hinged.
- 42Strength of Materials and Structural AnalysisHARD
Shear stress at the top and bottom fibres of a rectangular beam section is:
AZeroBInfiniteCMaximumDEqual to average shearAnswer: A. Zero
Explanation: Shear stress distribution is parabolic and becomes zero at extreme fibres.
- 43Strength of Materials and Structural AnalysisHARD
In plastic analysis, shape factor is the ratio of:
ALoad to deflectionBPlastic moment to yield momentCShear to bending momentDYield stress to ultimate stressAnswer: B. Plastic moment to yield moment
Explanation: Shape factor measures reserve moment capacity beyond first yield.
- 44Strength of Materials and Structural AnalysisEASY
In a simply supported beam with central point load W, maximum bending moment is:
AWL/2BWL/4CWL/8DWLAnswer: B. WL/4
Explanation: Reactions are W/2 each and maximum moment occurs at midspan.
- 45Strength of Materials and Structural AnalysisHARD
Euler buckling load of a column is inversely proportional to:
AModulus of elasticity onlyBSquare of effective lengthCEffective length onlyDRadius of gyration onlyAnswer: B. Square of effective length
Explanation: Euler's critical buckling load: Pcr = π²EI / Le², where E is Young's modulus, I is moment of inertia, and Le is effective length of the column. Pcr is inversely proportional to Le² (square of effective length). Answer: Square of effective length. Effective length Le depends on end conditions: Le = L (pinned-pinned), Le = 0.5L (fixed-fixed), Le = 0.7L (fixed-pinned), Le = 2L (fixed-free/cantilever).
- 46Strength of Materials and Structural AnalysisHARD
In pure bending, shear force is:
AMaximumBInfiniteCEqual to bending momentDZeroAnswer: D. Zero
Explanation: Pure bending means constant bending moment and zero shear force.
- 47Strength of MaterialsMEDIUM
A rectangular beam is to be cut from a circular log of wood of diameter D. The depth of the strongest section is
AD/3BD/sqrt(3)C2D/3DD sqrt(2/3)Answer: D. D sqrt(2/3)
Explanation: For maximum section modulus of a rectangle inscribed in a circle, depth = D sqrt(2/3)
- 48Geotechnical EngineeringMEDIUM
The quantity of seepage of water through soil is proportional to: (I) coefficient of permeability of the soil; (II) total head loss through the soil.
AOnly IBOnly IICBoth I and IIDNeither I nor IIAnswer: C. Both I and II
Explanation: q = k i A = k (h/L) A, so the seepage is proportional to both the permeability and the total head loss.
- 49RCC DesignMEDIUM
A slab supported along two opposite sides is called a
AOverhanging slabBTwo way slabCOne way slabDCantilever slabAnswer: C. One way slab
Explanation: A slab spanning between two opposite supports bends in one direction and is a one-way slab.
- 50Irrigation & Water ResourcesMEDIUM
Which of the following methods of irrigation does NOT use open ditches for water delivery?
ASub-irrigationBTrickle irrigationCFurrow irrigationDCheck irrigationAnswer: B. Trickle irrigation
Explanation: Trickle (drip) irrigation delivers water through a closed network of pipes and emitters, not open ditches. [Self-solved — please verify.]