Gate Ce 1

GATE 2024
Civil Engineering
(Volume - I)
TOPIC WISE GATE SOLUTIONS

2013-2023
unacademy
GATE SYLLABUS
Section 1 : Engineering Mathematics
Linear Algebra : Matrix algebra; Systems of linear equations; Eigen values and Eigen
vectors.
Calculus : Functions of single variable; Limit, continuity and differentiability; Mean value
theorems, local maxima and minima; Taylor series; Evaluation of definite and indefinite
integrals, application of definite integral to obtain area and volume; Partial derivatives;
Total derivative; Gradient, Divergence and Curl, Vector identities; Directional
derivatives; Line, Surface and Volume integrals.
Ordinary Differential Equation (ODE) : First order (linear and non-linear) equations;
higher order linear equations with constant coefficients; Euler-Cauchy equations; initial
and boundary value problems.
Partial Differential Equation (PDE) : Fourier series; separation of variables; solutions of
one-dimensional diffusion equation; first and second order one-dimensional wave
equation and two-dimensional Laplace equation.
Probability and Statistics : Sampling theorems; Conditional probability; Descriptive
statistics – Mean, median, mode and standard deviation; Random Variables – Discrete
and Continuous, Poisson and Normal Distribution; Linear regression.
Numerical Methods : Error analysis. Numerical solutions of linear and non-linear
algebraic equations; Newton’s and Lagrange polynomials; numerical differentiation;
Integration by trapezoidal and Simpson’s rule; Single and multi-step methods for first
order differential equations.
Section 2 : Structural Engineering
Engineering Mechanics : System of forces, free-body diagrams, equilibrium equations;
Internal forces in structures; Frictions and its applications; Centre of mass; Free
Vibrations of undamped SDOF system.
Solid Mechanics : Bending moment and shear force in statically determinate beams;
Simple stress and strain relationships; Simple bending theory, flexural and shear
stresses, shear centre; Uniform torsion, Transformation of stress; buckling of column,
combined and direct bending stresses.
Structural Analysis : Statically determinate and indeterminate structures by force/
energy methods; Method of superposition; Analysis of trusses, arches, beams, cables
and frames; Displacement methods: Slope deflection and moment distribution
methods; Influence lines; Stiffness and flexibility methods of structural analysis.
Construction Materials and Management : Construction Materials: Structural Steel –
Composition, material properties and behaviour; Concrete - Constituents, mix design,
short-term and long-term properties. Construction Management: Types of construction
projects; Project planning and network analysis - PERT and CPM; Cost estimation.
Concrete Structures: Working stress and Limit state design concepts; Design of beams,
slabs, columns; Bond and development length; Prestressed concrete beams.
Steel Structures : Working stress and Limit state design concepts; Design of tension and
compression members, beams and beam- columns, column bases; Connections - simple
and eccentric, beam-column connections, plate girders and trusses; Concept of plastic
analysis –beams and frames.
Section 3 : Geotechnical Engineering
Soil Mechanics : Three-phase system and phase relationships, index properties; Unified
and Indian standard soil classification system; Permeability - one dimensional flow,
Seepage through soils - two - dimensional flow, flow nets, uplift pressure, piping,
capillarity, seepage force; Principle of effective stress and quicksand condition;
Compaction of soils; One- dimensional consolidation, time rate of consolidation; Shear
Strength, Mohr’s circle, effective and total shear strength parameters, Stress-Strain
characteristics of clays and sand; Stress paths.
Foundation Engineering : Sub-surface investigations - Drilling bore holes, sampling,
plate load test, standard penetration and cone penetration tests; Earth pressure
theories - Rankine and Coulomb; Stability of slopes - Finite and infinite slopes, Bishop’s
method; Stress distribution in soils - Boussinesq’s theory; Pressure bulbs, Shallow
foundations - Terzaghi’s and Meyerhoff’s bearing capacity theories, effect of water
table; Combined footing and raft foundation; Contact pressure; Settlement analysis in
sands and clays; Deep foundations - dynamic and static formulae, Axial load capacity of
piles in sands and clays, pile load test, pile under lateral loading, pile group efficiency,
negative skin friction.
Section 4 : Water Resources Engineering
Fluid Mechanics : Properties of fluids, fluid statics; Continuity, momentum and energy
equations and their applications; Potential flow, Laminar and turbulent flow; Flow in
pipes, pipe networks; Concept of boundary layer and its growth; Concept of lift and
drag.
Hydraulics : Forces on immersed bodies; Flow measurement in channels and pipes;
Dimensional analysis and hydraulic similitude; Channel Hydraulics - Energy-depth
relationships, specific energy, critical flow, hydraulic jump, uniform flow, gradually
varied flow and water surface profiles.
Hydrology : Hydrologic cycle, precipitation, evaporation, evapo-transpiration,
watershed, infiltration, unit hydrographs, hydrograph analysis, reservoir capacity, flood
estimation and routing, surface run-off models, ground water hydrology - steady state
well hydraulics and aquifers; Application of Darcy’s Law.
Irrigation : Types of irrigation systems and methods; Crop water requirements - Duty,
delta, evapo-transpiration; Gravity Dams and Spillways; Lined and unlined canals,
Design of weirs on permeable foundation; cross drainage structures.
Section 5 : Environmental Engineering
Water and Waste Water Quality and Treatment : Basics of water quality standards -
Physical, chemical and biological parameters; Water quality index; Unit processes and
operations; Water requirement; Water distribution system; Drinking water treatment.
Sewerage system design, quantity of domestic wastewater, primary and secondary
treatment. Effluent discharge standards; Sludge disposal; Reuse of treated sewage for
different applications.
Air Pollution : Types of pollutants, their sources and impacts, air pollution control, air
quality standards, Air quality Index and limits.
Municipal Solid Wastes : Characteristics, generation, collection and transportation of
solid wastes, engineered systems for solid waste management (reuse/ recycle, energy
recovery, treatment and disposal).
Section 6 : Transportation Engineering
Transportation Infrastructure : Geometric design of highways - cross-sectional
elements, sight distances, horizontal and vertical alignments.
Geometric design of railway Track - Speed and Cant.
Concept of airport runway length, calculations and corrections; taxiway and exit taxiway
design.
Highway Pavements : Highway materials - desirable properties and tests; Desirable
properties of bituminous paving mixes; Design factors for flexible and rigid pavements;
Design of flexible and rigid pavement using IRC codes
Traffic Engineering : Traffic studies on flow and speed, peak hour factor, accident study,
statistical analysis of traffic data; Microscopic and macroscopic parameters of traffic
flow, fundamental relationships; Traffic signs; Signal design by Webster’s method;
Types of intersections; Highway capacity.
Section 7 : Geomatics Engineering
Principles of surveying; Errors and their adjustment; Maps - scale, coordinate system;
Distance and angle measurement - Levelling and trigonometric levelling; Traversing and
triangulation survey; Total station; Horizontal and vertical curves.
Photogrammetry and Remote Sensing - Scale, flying height; Basics of remote sensing
and GIS.
General Aptitude (GA)
Verbal Aptitude : Basic English grammar: tenses, articles, adjectives, prepositions,
conjunctions, verb-noun agreement, and other parts of speech Basic vocabulary: words,
idioms, and phrases in context Reading and comprehension Narrative sequencing.
Quantitative Aptitude : Data interpretation: data graphs (bar graphs, pie charts, and
other graphs representing data), 2- and 3-dimensional plots, maps, and tables
Numerical computation and estimation: ratios, percentages, powers, exponents and
logarithms, permutations and combinations, and series Mensuration and geometry
Elementary statistics and probability.
Analytical Aptitude : Logic: deduction and induction Analogy, Numerical relations and
reasoning.
Spatial Aptitude : Transformation of shapes: translation, rotation, scaling, mirroring,
assembling, and grouping Paper folding, cutting, and patterns in 2 and 3 dimensions.
CONTENTS
S. No. Topics
1. Strength of Materials
1. Properties of Metals, Stress and Strain
2. Shear Force and Bending Moment
3. Principle Stress and Principle Strain
4. Bending and Shear Stress
5. Deflection of Beams
6. Torsion of Shaft and Pressure Vessels
7. Theory of Columns and Shear Centre
2. Structural Analysis
1. Determinacy and Indeterminacy (Static & Kinematic)
2. Force and Energy Methods
3. Displacement Method of Analysis
4. Analysis of Trusses
5. Influence Line Diagram and Rolling Loads
6. Analysis of Arches
7. Matrix and Stiffness Method
8. Structural Dynamics
3. RCC Structure & Pre-Stress Concrete
1. IS Code Recommendations and Fundamentals
2. Design & Analysis of Beam and Slab
3. Shear, Bond, Torsion, Anchorage & Development Length
4. Column & Footing
5. Prestressed Concrete
4. Design of Steel Structure
1. General Design Requirements
2. Connections
3. Design of Tension Members
4. Design of Compression Members
5. Design of Beams
6. Design of Plate Girders, Industrial Roof Truss and
Gantry Girders
7. Plastic Analysis
5. Engineering Mechanics
6. Construction Materials & Management
7. Geomatics Engineering
1. Fundamental Concepts
2. Levelling
3. Traversing
4. Theodolites and Plane Table Surveying
5. Tacheometric & Triangulation Surveying
6. Measurement of Area, Volume & Theory of Errors and
Survey Adjustment
7. Curves
8. Field Astronomy & Photogrammetric Surveying
9. Basics of GIS, GPS & Remote Sensing
8. Engineering Mathematics
1. Linear Algebra
2. Differential Equations
3. Integral and Differential Calculus
4. Vector Calculus
5. Maxima and Minima
6. Mean Value Theorem
7. Complex Variables
8. Limit & Series Expansion
9. Probability & Statistics
10. Numerical Methods
11. Transform Theory
9. General Aptitude
1. Numerical Ability
2. Logical Reasoning
3. Verbal Ability
STRENGTH
OF MATERIALS
Syllabus : Strength of Materials
Bending moment and shear force in statically determinate beams; Simple stress and strain
relationships; Simple bending theory, flexural and shear stresses, shear centre; Uniform torsion,
Transformation of stress; buckling of column, combined and direct bending stresses.
Contents : Strength of Materials
S. No. Topics
1. Properties of Metals, Stress and Strain
2. Shear Force and Bending Moment
3. Principal Stress and Principal Strain
4. Bending and Shear Stress
5. Deflection of Beams
6. Torsion of Shaft and Pressure Vessels
7. Theory of Columns and Shear Centre
1 Properties of Metals, Stress and Strain
2013 IIT Bombay 2015 IIT Kanpur

1.1 Creep strains are [1 Mark] 1.4 A tapered circular rod of diameter varying
(A) caused due to dead load only. from 20 mm to 10 mm is connected to
(B) caused due to live load only. another uniform circular rod of diameter 10
(C) caused due to cyclic load only. mm as shown in the following figure. Both
(D) independent of load. bars are made of same material with modulus
2014 IIT Kharagpur of elasticity, E = 2 × 105 MPa. When
1.2 A box of weight 100 kN shown in the figure subjected to a load P = 30π kN , the
is to be lifted without swinging. If all forces deflection at point A is _______ mm.
are coplanar, the magnitude and direction [2 Marks]
(θ) of the forces ( F ) with respect to x-axis d1 = 20 mm
should be [2 Marks]
y
C
40 kN
90 kN
F 2m
450
300 q x
100 kN B
d 2 = 10 mm
(A) F = 56.389 kN and θ = 28.280 1.5 m
(B) F = −56.389 kN and θ = −28.280 A

0
(C) F = 9.055 kN and θ = 1.414
P = 30 p kN
(D) F = −9.055 kN and θ = −1.4140
1.3 The values of axial stress (σ) in kN/m2 , 2016 IISc Bangalore
bending moment (M) in kNm, and shear 1.5 An elastic isotropic body is in a hydrostatic
force (V) in kN acting at point P for the state of stress as shown in the figure. For no
arrangement shown in the figure are change in the volume to occur, what should
respectively. [2 Marks] be its Poisson’s ratio? [2 Marks]
Cable
Frictionless sy
P Beam Q pulley
(0.2 m ´ 0.2 m)
3m
sx
50 kN
(A) 1000, 75 and 25 sz
(B) 1250, 150 and 50
(C) 1500, 225 and 75 (A) 0.00 (B) 0.25
(D) 1750, 300 and 100 (C) 0.50 (D) 1.00
Strength of Materials 1
2017 IIT Roorkee 1.8 In a material under a state of plane strain, a
10 × 10 mm square centered at a point gets
1.6 An elastic bar of length L, uniform cross-
sectional area A, coefficient of thermal deformed as shown in the figure.
y
expansion α , and Young’s modulus E is
fixed at the two ends. The temperature of the 10 mm
bar is increased by T, resulting in an axial

stress σ . Keeping all, other parameters 10 mm
unchanged, if the length of the bar is

p x
doubled, the axial stress would be + 0.0005 rad
2
[1 Mark] 0.004
mm
(A) σ (B) 2σ If the shear strain γ xy at this point is
(C) 0.5 σ (D) 0.25σ expressed as 0.001 k (in rad), the value of
k is [1 Mark]
1.7 Consider the stepped bar made with a linear
(A) 0.50 (B) 0.25
elastic material and subjected to an axial load (C) – 0.25 (D) – 0.50
of 1 kN, as shown in the figure. [2 1.9 A 2 m long, axially loaded mild steel rod of
Marks] 8 mm diameter exhibits the load-
displacement ( P -δ) behavior as shown in
the figure. [2 Marks]
L1 = 400 mm
A1 = 100 mm 2 14000
1 12000
E1 = 2 ´ 105 MPa
Axial load, P (kN)
10000
8000
6000
L2 = 900 mm
4000
A2 = 60 mm 2
2 2000
E2 = 3 ´ 105 MPa
0
0 1 2 3 4 5 6 7 8 9 10
Displacement, d (mm)
P = 1 kN Assume the yield stress of steel as 250 MPa.

The complementary strain energy (in Nmm)
Segments 1 and 2 have cross-sectional area stored in the bar up to its linear elastic
behavior will be _______.
of 100 mm 2 and 60 mm2 . Young’s modulus
2018 IIT Guwahati
of 2 ×105 MPa and 3 ×105 MPa, and length
1.10 A plate in equilibrium is subjected to
of 400 mm and 900 mm, respectively. The
uniform stresses along its edges with
strain energy (in Nmm up to one decimal magnitude σ xx = 30 MPa and σ yy = 50 MPa
place) in the bar due to the axial load is as shown in the figure. [2 Marks]
________.
2 Strength of Materials
s yy = 50 MPa subjected to axial tensile force as shown in
the figures. When the bar is subjected to
y axial tensile forces P1 and P2 , the strain
s xx = 30 MPa
energies stored in the bar are U1 and U 2
x
respectively. [2 Marks]
P1
The Young’s modulus of the material is

2 ×1011 N/m 2 and the Poisson’s ratio is 0.3. P2
If σ zz is negligibly small and assumed to be
zero, then the strain ε zz is
(P1+P2)
(A) −120 × 10−6 (B) −60 × 10−6
(C) 0.0 (D) 120 × 10−6 If U is the strain energy stored in the same
bar when subjected to an axial tensile force
2020 IIT Delhi
( P1 + P2 ) the correct relationship is
1.11 A rigid, uniform, weightless, horizontal bar
is connected to three vertical members P, Q (A) U = U1 + U 2 (B) U < U1 + U 2
and R as shown in the figure (not drawn to (C) U = U1 − U 2 (D) U > U1 + U 2
the scale). All three members have identical
axial stiffness of 10 kN/mm. The lower end 2021 IIT Bombay
of bars P and R rest on a rigid horizontal
1.13 Strain hardening of structural steel means
surface. When NO load is applied, a gap of
[1 Mark]
2 mm exists between the lower end of the bar
Q and the rigid horizontal surface. When a (A) Strengthening steel member externally
vertical load W is placed on the horizontal for reducing strain experienced
bar in the downward direction, the bar still (B) Experiencing higher stress than yield
remains horizontal and gets displaced by 5 stress with increased deformation
mm in the vertically downward direction. (C) Decreases in the stress experienced with
Rigid Uniform Weightless
Horizontal Bar
increasing strain
(D) Strain occurring before plastic flow of
steel material
2022 IIT Kharagpur
P Q R
1.14 Consider two linearly elastic rods HI and IJ,
each of length b, as shown in the figure. The
Gap
rods are co-linear, and confined between two
Rigid Horizontal
fixed supports at H and J. Both the rods are
Surface initially stress free. The coefficient of linear
The magnitude of the load W (in kN, round thermal expansion is 𝛼 for both the rods. The
off to the nearest integer), is _______. temperature of the rod IJ is raised by ∆𝑇,
[2 Marks] whereas the temperature of rod HI remains
1.12 A prismatic linearly elastic bar of length L, unchanged. An external horizontal force P is
cross-sectional area A, and made up of a now applied at node I. It is given that
material with Young’s modulus E, is α = 10−6 0 C−1 , ΔT = 50 0 C, b = 2m,
AE = 106 N. The axial rigidities of the rods field in the plate is given by u = Cx 2 y and
HI and IJ are 2AE and AE, respectively. v = 0, where u and v are displacements (in
b m) along the X and Y directions,
b
P
respectively, and C is a constant (in m−2).
AE The distances x and y along X and Y,
2AE I
H J respectively, are in m. The stress in the 𝑋
To make the axial force in rod HI equal to direction is σ XX = 40 xy N/m 2 , and the
zero, the value of the external force P (in N) shear stress is τ XY = αx 2 N/m 2 . What is the
is _______. (Round off to the nearest
integer) [2 Marks] value of α (in N/m4, in integer)?
___________
1.15 For linear elastic and isotropic material, the
correct relationship among young’s modulus 1.18 A hanger is made of two bars of different
of elasticity (E), Poisson ratio (ν ) and shear sizes. Each bar has a square cross-section.
modulus (G). [1 Mark] The hanger is loaded by three-point loads in
the mid vertical plane as shown in the figure.
G E
(A) E = (B) G = Ignore the self-weight of the hanger. What is
1 + 2ν 2(1 + ν)
the maximum tensile stress in N/mm2
E G anywhere in the hanger without considering
(C) G = (D) E =
1 + 2ν 2 (1 + ν ) stress concentration effects?
2023 IIT Kanpur
100 mm
1.16 A 5 cm long metal rod AB was initially at a
uniform temperature of T0 °C. Thereafter,
temperature at both the ends are maintained
at 0°C. Neglecting the heat transfer from the 100 kN 100 kN
lateral surface of the rod, the heat transfer in 50 mm
the rod is governed by the one-dimensional
∂T ∂ 2T
diffusion equation = D 2 , where D is 50 kN
∂t ∂x
(A) 15.0 (B) 25.0
the thermal diffusivity of the metal, given as
(C) 35.0 (D) 45.0
1.0 cm2/s.
The temperature distribution in the rod is
obtained as
nπx −βn2t
T ( x, T ) =  n =1,3,5... Cn sin
∞
e ,
5
Where, x is in cm measured from A to B with
𝑥 = 0 at A, t is in s, 𝐶𝑛 are constants in °C, T
is in °C, and β is in s−1.
The value of β (in 𝑠−1, rounded off to three
decimal places) is _________.
1.17 A 2D thin plate with modulus of elasticity, E
= 1.0 N/m2, and Poisson’s ratio, μ = 0.5 , is
in plane stress condition. The displacement
1.1 (A) Tertiary creep : Slope increase with time
rapidly accelerating strain rate upto failure.
Creep is a time dependent process where a
It is associated with formation of internal
material under an applied stress exhibits
cracks, voids, grain, boundary, separation,
deformation at high temperature. So, creep strain
necking, etc.
occur due to constant load i.e., dead load.
Hence, the correct option is (A)  Effects of stress and temperature on
creep
 Key Point T3 > T2 > T1
 Creep (Sometimes known as cold flow) s3 > s 2 > s1
can occur as a result of long term

exposure of high levels of stress that are T3 or s3
still below the yield strength of the
Creep strain
material. T2 or s 2
 Creep is also temperature dependent T1 or s1

process, and always increases with
temperature. T < 0.4 Tm
 Threshold for creep : The critical
Time
temperature for creep is 40% of the
With increase in stress or temperature :
melting temperature.
If T > 0.4 Tm → creep likely to occur 1. The instantaneous strain increases
Here, Tm = Melting temperature which is mainly because of
 The rate of deformation is known as elasticity.
creep rate. It is slope of the line in a 2. The steady-state creep rate increases.
creep strain vs time curve.
Rupture 3. The time to rupture decreases.
1.2 (A)
Primary
It is asked in the question that the box should be
Tertiary
lifted without swinging, it means that there should
Strain
Secondary
be no moment and the forces should be balanced.
The FBD of given question is,
Instantaneous
F sin q0
Time tr
 Creep stages : 40sin 450

Primary creep : Slope of curve (creep rate) 90sin 300
decreases with time work hardening. 40 cos 450
90 cos 300
Secondary creep : Constant slope balance Fcosq
of work hardening and recovery.
100
Basics of Signals 200

 FH = 0 , Bending moment = 50 × 3 = 150 kN.m
Shear force = 50 kN
90 cos 300 = 40 cos 450 + F cos θ
Hence, the correct option is (B).
 F cos θ = 49.66 kN …(i)
 Key Point
 FV = 0 , While solving the problem, care should be taken
that point P lies on the beam, not on the cable. So,
90sin 300 + F sin θ + 40sin 450 = 100 at point P, there are axial stress, bending moment
 F sin θ = 26.72 kN …(ii) and shear force. If point P lie on the cable, then
there will be only axial stress and no shear force
Dividing equation (ii) by equation (i), we get and bending moment. This is because cable can
26.72 with stand only axial tension and no other forces
tan θ = = 0.5380 i.e., cable is unable to carry bending moment,
49.66
shear force, torsion or axial compression.
θ = 28.280
1.4 15
Substituting value of θ in equation (ii), we
Given : Diameters of tapered circular rod,
get
( D1 ) = 20 mm and ( D2 ) = 10 mm
F sin (28.28) = 26.72
Modulus of elasticity ( E ) = 2 ×105 MPa
F = 56.397 kN Load ( P ) = 30πkN
D1 = 20 mm
Hence, the correct option is (A).
1.3 (B) C
Cable 2m
Frictionless
P Beam Q pulley
(0.2 m ´ 0.2 m) B
3m D2 = 10 mm
1.5 m
P = 30 p kN
50 kN
Total deflection at point A is sum of elongation of
According to question, the FBD of given problem
uniform circular rod due to axial load P and the
after removing the cable can be drawn as,
elongation of tapered circular rod due to axial load
Beam Frictionless
pulley P.
P
T T
Q ∴ Deflection of point A = Elongation of
3m uniform circular rod AB + elongation of
T tapered circular rod BC.
T
Δ A = Δ AB + Δ BC
PL PL
= +
50 kN
AE π D D E
1 2
So, T = 50 kN 4
(30π×103 ) × (1.5 ×103 ) (30π× 103 ) × (2 ×103 )
50 kN = +
Axial stress = = 1250 kN/m2 π 2 π
0.2 × 0.2  × 10  × (2 × 10 5
) × 20 × 10 × (2 × 105 )
4  4
= (9 + 6) mm 1.6 (A)
BA = 15 mm
The given problem can be depicted in figure as
Hence, the deflection at point A is 15 mm. shown below,
 Key Point a
A E
Extension of a tapered bar under axial loading
L
:
4PL The axial stress resulting from rise in temperature
1. In circular tapered bar, δL = of the bar is given by
π D1 D2 E
σ = αΔTE
2. In rectangular tapered bar,
From above equation we can see that length have
PL B
δL = ln 2 no effect on thermal stress. Hence, the axial
Et ( B2 − B1 ) B1 stress will remain same (σ) .
1.5 (C) Hence, the correct option is (A).
1.7 35
Change in volume of an elastic isotropic body
subjected to triaxial stress is given by, Given : Axial load, ( P ) = 1kN
 σ + σ y + σz  Segment 1 :
dV =  x  (1 − 2μ) × Volume
 3  Length of segment 1, ( L1 ) = 400 mm,
sy Area of segment 1, ( A1 ) = 100 mm 2 ,

Young’s modulus ( E1 ) = 2 ×105 MPa
Segment 2 :
sx Length of segment 2, ( L2 ) = 900 mm
Area of segment 2, ( A2 ) = 60 mm 2
sz
Young’s modulus ( E2 ) = 3 ×105 MPa
For hydrostatic stress, σ x = σ y = σ z = σ
A
L1 = 400 mm
Now, for no change in volume, dV = 0 A1 = 100 mm 2

1
 3σ  E1 = 2 ´ 105 MPa
So, 0 =   (1 − 2μ) × Volume
 3 
B
∴ 1 − 2μ = 0
L2 = 900 mm
1 A2 = 60 mm 2
μ = = 0.5 2
E2 = 3 ´ 105 MPa
2
Hence, the correct option is (C)
C
 Key Point P = 1 kN
 When an object is submerged in water, it
FBD of the given question can be shown as,
experiences uniform compression because of
1 kN 1 kN
uniform pressure in all directions i.e. the
stress is equal in all direction. This state of
stress often known as hydrostatic stress. 1 2
 Mohr’s circle becomes a point in case of
hydrostatic stress. 1 kN 1 kN
:Method 1:
Elongation of bar = Elongation of segment 1 g xy
+ Elongation of segment 2 p
+ g xy
Δ AC = Δ AB + Δ BC 2
-ve Shear strain
PL1 PL2
= + From the figure we can see that because of shear
A1 E1 A2 E2
strain, the angle has been increased.
(1× 103 ) × (400) (1× 103 ) × (900) Therefore, shear strain should be negative.
= +
(100) × (2 × 105 ) (60) × (3 × 105 ) ∴ γ xy = −0.0005 rad
= 0.02 + 0.05 = 0.07 mm
 −0.0005 = 0.001 k
1
So, strain energy, U = × P × Δ AC k = −0.50
2
1 Hence, the correct option is (D).
= × (1×103 ) × 0.07 = 35 N-mm
2  Key Point
Hence, the strain energy stored in bar due to axial Sign convention for shear stress :
loading is 35 N-mm. 1. The shear stress will be considered positive
:Method 2: when a pair of shear stress acting on opposite
Strain energy of stepped bar under an axial loading sides of the element produce a counter
is given by, clockwise torque (couple).
P 2 Li (- t)
U =
2 Ai Ei
(+ t) (+ t)
So, in given problem strain energy is given as,
P 2 L1 P 2 L2 (- t)
U= +
2 A1 E1 2 A2 E2 2. The shear stress will be considered negative
3 2 3 2
(1× 10 ) × (400) (1×10 ) × 900 when a pair of shear stress acting on opposite
= +
2 × (100) × (2 × 10 ) 2 × (60) × (3 × 105 )
5
sides of the element produce a clockwise
= 10 + 25 = 35 N-mm torque (couple)
Hence, the strain energy stored in bar due to axial Sign convention for shear strain :
Positive y-face
loading is 35 N-mm.
y
1.8 (D) g t Positive x -face
p q
2
According to question, x
t
y t
z
10 mm t r
Negative x-face
o
Negative y -face
p g p
-g +g
10 mm 2 2 2
1. If the angle between two positive faces (or

p x two negative faces) are reduced, the shear
+ 0.0005 rad
2 strain will be considered positive.
0.004
mm 2. If the angle between two positive forces (or
The given figure is equivalent to the figure shown two negative faces) are increased, the shear
below, strain will be considered negative.
1.9 15625000
Given : Length of rod ( L) = 2 m = 2000 mm dC
Diameter of rod ( D ) = 8 mm dC = δdP  = δ ← Differential of
dP
Yield stress (σ y ) = 250 MPa complementary energy is deflection.
P
It has
The area of the region above the force-deflection Load Complementary energy C C = ò ddP no physical
B
curve is called complementary energy. A 0
meaning and
P
is just for
Stress mathematical
Complementary strain dP convenience
energy per unit volume Strain energyU
sy
d
U = ò Pd d
D 0
O d Deflection Strain energy

Strain energy dd produced by
per unit volume load P
 For linear elastic material the value of

eE Strain
complementary energy is equal to that of
Load (P)
strain energy.
 The concept of total complementary energy
of an elastic system is particularly useful in
12500 kN the solution of statically indeterminate
structures, in which infinite number of
Complementary stresses distribution and reactive forces may
strain energy
be found to satisfy the requirement of
equilibrium.
2.5 mm deflection (d)
1.10 (A)
Complementary strain energy
Given : Stress in x-direction (σ xx ) = 30 MPa
1
= Strain energy = Pδ Stress in y-direction (σ yy ) = 50 MPa
2
1 Young’s modulus ( E ) = 2 ×1011 N/m 2
= × (12500 ×103 ) × 2.5 = 2 × 105 N/mm 2
2
= 15625000 N-mm Poisson’s ratio (μ) = 0.3
Hence, the complementary strain energy is Stress in z-direction (σ zz ) = 0
15625000 N -mm . σ zz μ(σ xx + σ yy )
Strain, ε zz = −
 Key Point E E
δ 0.3(30 + 50)
 Strain energy, U =  Pd δ = 0− = −120 × 10−6
2 ×105
0
(Negative sign represents contraction)
dU Hence, the correct option is (A).
dU = Pd δ  = P ← Differential of
dδ
1.11 130
strain energy is load.
P1 L
 Complementary strain energy, Given : δ1 = 5 mm =
P AE
C =  δdP AE
0
= 10 kN/mm
L
PL
δ2 = 3 mm = 2
AE
We conclude that, U > U1 + U 2
Hence, the correct option is (D).
1 2 3
1.13 (B)
P1 P2 P3
1.14 50
Given :
W
b
b
P
P1 P2 P3 = P1
AE
H 2 AE I J
P1 + P2 + P1 = W
P1 = 10 × 5 = 50 kN α = 10−6 C −1
b=2 m
P2 = 10 × 3 = 30 kN
W = 2 × (50) + 30 = 130 kN ΔT = 500 C
Hence, the magnitude of the load W is 130 kN. AE = 106 N
1.12 (D) PHI = 0 then P = ?
According to question, Axial deformation should be equal to zero
L δ1 + δ2 = 0
P1
1 2
L H I I J
P2 δ2 = 0 = δad + δTH
− PL
L 0= + αΔTL
AE
( P1 + P2 )
PL
When bar is subjected to load P1 , the strain energy = αΔTL
AE
stored, P = 10−6 × 50 × 106
2
U1 =
P L 1
… (i) P = 50 N
2 AE
1.15 (B)
Where, L = length of bar
When bar is subjected to load P2 , the strain energy We know that,
stored, E = 2G (1 + v )
P22 L E
U2 = … (ii) G=
2 AE 2 (1 + v)
Now when bar is subjected to load ( P1 + P2 ) , the Hence, the correct option is (B).
strain energy stored, 1.16 0.394
2
( P1 + P2 ) L ∞
 nπx  −βn2t
U=
2 AE
T ( x, t ) = 
n =1,3,5
cn sin 
 5 
e
P12 L P22 L 2 P1 P2 L − n 2 π2 α 2 t
U= + + … (iii) ∞
 nπx 
2 AE 2 AE 2 AE T ( x, t ) =  βn sin  e
l2
By equation (i), (ii) and (iii), n =1  l 
α2 = D = 1 ( l = 5) 15 x 2 ×1
τ xy = = 5x2
2 (1 + 0.5 )
π2α 2 π (1) π2 ( 3.14 )
2 2
∴ β= 2 = = = = 0.394 By comparing τ xy , we get α = 5

l 25 25 25
Hence, the correct answer is 0.394. Hence, the correct answer is 5.
1.17 5 1.18 (B)
Given : E = 1.0 N/m 2 , μ = 0.5

A 50 kN
Displacement field in the plate is
u = Cx y2 100 mm
B
v=0
∂u B
∴ εx = = 2Cxy
∂x
∂v
εy = =0 100 kN 100 kN
∂y
50 mm
Stress in x direction , σ xx = 40 xy  2 
N C
. C
m 
Shear stress, τ xy = α x 2  2 
N 50 kN 50 kN
250 kN
m 
We know, shear modulus
A
τ
G = xy ... ( A )
γ xy
Stress in x direction;
 E 
2 ( x
σx =  ε + με y ) ... ( i )
1− μ 
B
Substitute all known value in equation (i)
we get,
250 kN
1
40 xy = ( 2Cxy + 0.5 × 0 ) Fig. (i)
(1 − 0.52 ) Fig. (ii) FBD of member BC and AB
∴ C = 15 Stress in member BC,
∂v ∂u
Shear strain γ xy = + = 0 + Cx 2
∂x ∂y PB 50 ×103
σB = = = 20 N/mm2
γ xy = Cx2 AB 50 × 50
Substitute γ xy and C in equation (A), Stress in member AB,
τ xy
G= PA 250 ×103
γ xy σA = = = 25 N/mm2
AA 100 ×100
E τ
= xy2 So, maximum tensile stress = 25 N/mm2
2 (1 + μ ) Cx Hence, the correct option is (B).
2 Shear Force and Bending Moment
2015 IIT Kanpur respectively. A force P acts at C as shown.

Let RAh and RBh be the horizontal reactions
2.1 A horizontal beam ABC is loaded as shown
in the figure below. The distance of the at supports A and B, respectively, and RAv
point of contra flexure from end A (in m) is be the vertical reaction at support A. Self-
________. [1 Mark] weight of the member may be ignored.
10 kN
B
0.75 m P
A B C 2m
1m C
2016 IISc Bangalore

6m
2.2 The portal frame shown in the figure is
subjected to a uniformly distributed vertical
load w (per unit length).
w kN/m A
Q R
1.5 m
L/2 1.5 m
Which one of the following sets gives the
P S correct magnitudes of RAv , RBh and RAh ?
L [2 Marks]
The bending moment in the beam at the 1 2
joint Q is [2 Marks] (A) RAv = 0, RBh = P & RAh = P
3 3
(A) Zero 2 1
(B) RAv = 0, RBh = P & RAh = P
wL2 3 3
(B) (hogging)
24 3 1.5
(C) RAv = P, RBh = P & RAh = P
wL2 8 8
(C) (hogging)
12 1.5 1.5
(D) RAv = P, RBh = P & RAh = P
8 8
wL2
(D) (sagging)
8 2017 IIT Roorkee
2.3 A rigid member ACB is shown in the figure. 2.4 Consider the three prismatic beams with
The member is supported at A and B by the clamped supports P, Q and R as shown
pinned and guided roller supports,
in the figures. [2 Marks, Set- (D) Maximum bending moment and
II] maximum shear force occur at the same
80 N section.
2018 IIT Guwahati

P
E, I 2.6 A vertical load of 10 kN acts on a hinge
8m located at a distance of L/4 from the roller
20 N/m support Q of a beam of length L (see
figure).
Q
10 kN
E, I
8m P Q
640 Nm
3L/4 L/4
R
E, I The vertical reaction at support Q is
8m [1 Mark]
Given that the modulus of elasticity, E is (A) 0.0 kN (B) 2.5 kN
2.5 × 104 MPa, and the moment of inertia, I (C) 7.5 kN (D) 10.0 kN
is 8 ×108 mm 4 , the correct comparison of 2020 IIT Delhi
the magnitudes of the shear force S and the 2.7 A weightless cantilever beam of span L is
bending moment M developed at the loaded as shown in the figure. For the entire
support is span of the beam, the material properties
are identical and the cross-section is
(A) S P < SQ < S R , M P = M Q = M R rectangular with constant width.
P
(B) S P = SQ < S R , M P = M Q > M R PL
L
(C) S P < SQ > S R , M P = M Q = M R
Form the flexure-critical perspective, the
(D) S P < SQ < S R , M P < M Q < M R most economical longitudinal profile of the
beam to carry the given loads amongst the
2.5 A simply supported beam is subjected to a options given below, is.
uniformly distributed load. Which one of [1 Mark]
the following statements is true? (A)
[1 Mark]
(A) Maximum or minimum shear force (B)
occurs where the curvature is zero.
(B) Maximum or minimum bending (C)
moments occurs where the shear force
is zero. (D)
(C) Maximum or minimum bending
moment occurs where the curvature is 2021 IIT Bombay
zero.
2.8 A propped cantilever beam EF is subjected 2.10 A propped cantilever beam xy , with an
to a unit moving load as shown in the figure internal hinge at the middle, is carrying a
(not to scale). The sign convention for uniformly distributed load of 10 kN/m, as
positive shear force at the left and right shown in the figure.
sides of any section is also shown 10 kN/m
Unit moving load
x y
E G
F
2m 2m
The vertical reaction at support x (in kN,
in integer) is _______. [2
Marks]
Sign convention for 2022 IIT Kharagpur
positive shear force
2.11 Consider a simply supported beam PQ as
The CORRECT qualitative nature of the shown in the figure. A truck having 100 kN
influence line diagram for shear force at G on the front axle and 200 kN on the rear
is [1 Mark] axle moves from left to right. The spacing
(A) between the axis is 3 m. The maximum
+ bending moment at point R is _______
E
– G
F kN/m (in integer) [2 Marks]
R
(B) P Q
1m 4m
+
E F
– G 2023 IIT Kanpur
(C) +
2.12 In the frame shown in the figure (not to
E F
G scale), all four members (AB, BC, CD, and
–
AD) have the same length and same
constant flexural rigidity. All the joints A,
(D) B, C, and D are rigid joints. The midpoints
+
of AB, BC, CD, and AD, are denoted by E,
E F
– G F, G, and H, respectively. The frame is in
2.9 The equation of deformation is derived to unstable equilibrium under the shown
be y = x 2 − xL for a beam shown in the forces of magnitude 𝑃 acting at E and G.
figure. Which of the following statements is/are
y TRUE?
A H D
x
P P
E G
L
The curvature of the beam at the mid-span
(in units, in integer) will be ________. B C
F
[1 Mark]
(A) Shear forces at H and F are zero
(B) Horizontal displacements at H and F magnitude 𝑀 at one-third spans, as
are zero shown in the figure. Which of the
(C) Vertical displacements at H and F are following statements is/are TRUE?
zero A B
(D) Slopes at E, F, G, and H are zero
M M
2.13 A beam is subjected to a system of coplanar
L L L
forces as shown in the figure. The
(A) Support reactions are
magnitude of vertical reaction at Support P
zero
is ________ N (Round off to one decimal
(B) Shear force is zero everywhere
place).
500 N
(C) Bending moment is zero
100 N everywhere
0.5 m
P
60 0
(D) Deflection is zero everywhere
Q
200 N
2.0 1.5 1.0 1.5 m
2.14 Consider the beam shown in the figure

(not to scale), on a hinge support at end A
and a roller support at end B. The beam
has a constant flexural rigidity, and is
subjected to the external moments of
2.1 0.25 10l 3

Deflection due to applied load 10 kN, δ2 =
3EI
:Method 1:
Deflection due to moment M,
According to question,
10 kN Ml 2 (0.25 ×10) × l 2
δ3 = =
0.75 m
2 EI 2 EI
A B C Since, it is propped at point B, the deflection at
1m point B should be zero. So, from
Given problem can be converted into a cantilever compatibility condition,
beam by shifting the effect of load applied at 10 × 0.753 2.5 × 0.752 RB × 0.753
+ − =0
point C to point B. It can be depicted as shown 3EI 2 EI 3EI
below, ∴ RB = 15 kN
10 kN
0.75 m BM at a distance x from free end,
A B
2.5 kN
RB
Here, RB is the reaction force at B due to prop.
RBl 3
∴ Deflection due to RB , δ1 =
3EI
10 kN M x = −10(1 − x) + RB (0.75 − x)
x
B At point of contraflexure, M x = 0
A C
∴ − 10 × (1 − x) + 15 × (0.75 − x ) = 0
x
x = 0.25 m
0.75 m 15 kN
Hence, the distance of point of contraflexure
x
from end A is 0.25 m.
BM x = 10 x − 15 × ( x − 0.25) = 0
∴ x = 0.75 m Scan for
So, from end A distance is 0.25 m. Video Solution
:Method 2:
1m 2.2 (A)
x 10 kN
w kN/m
B
A C Q R
x
0.75 m RB L/2
x
HP P S HS = 0
0.75RB
(+)
B’ VP L VS
Conjugate beam As there is no horizontal force.
2.5
(–) Hence, H p = H s = 0
10 L
ΣM Q = H p × =0
2
1.25 kNm
∴ BM at Q = 0
x
2.3 (D)
2.5 kNm RBh
By conjugate beam method, P
2m
Deflection at B = 0
C
1 2 
 2 × 0.75 RB × 0.75 × 3 × 0.75 −
6m
 0.75 1 
 2.5 × 0.75 × 2 + 2 
 =0 A
 × 7.5 × 0.75 × 2 × 0.75 RAh
 3  RAv
1.5 m
1.5 m
0.14 RB − 2.1 = 0
RB = 15 kN Resolving vertical forces,
Now, taking bending moment about section x-x ∴ ΣFV = 0
at a distance x from end A, RAv = P
Resolving horizontal forces, 20 N/m
∴ ΣFH = 0 FBD
E, I
RBh = RAh 8m
R
Taking moments about A equal to zero. 160 kN
ΣM A = 0 ,
SFD (+)
RBh × 8 + P ×1.5 = 0
1.5 1.5
RBh = − P= P (Magnitude)
8 8
BMD (–)
1.5
∴ RAh = P ( RBh = RAh )
8 20
Hence, the correct option is (D). 640 kNm
 Fv = 0 ,
Scan for
Video Solution R − 20 × 8 = 0
R = 160 N
2.4 (C) So, shear force at support, SQ = 160 N
Let R be the reaction force at the support. And, moment at support,
For P : M Q = 20 × 8 × 4 = 640 Nm
80 N
E, I For R :
FBD P 640 Nm
8m FBD R
R E, I
80 kN R 8m
SFD (+) Zero
SFD
640 kNm
BMD (+)
BMD (–)
 Fv = 0 ,
640 kNm
 Fv = 0, R=0
R − 80 = 0 So, shear force at support, S R = 0
R = 80 N
and, moment at support, M R = 640 Nm
So, shear force at support, S p = 80 N
Thus, form the above calculations,
and, moment at support, M p = 80 × 8 = 640 Nm
S P < SQ > S R and M P = M Q = M R
For Q :
Hence, the correct option is (C).
2.5 (B)
In case of simply supported beam which is

carrying a uniformly distributed load, the
maximum or minimum bending moment occurs
where the shear force is zero.
w
 Internal hinge reduces indeterminacy by
FBD P
one.
l
R R
l/2 l/2 2.7 (D)
For most economical longitudinal profile of
SFD V (+) cantilever, the depth of beam should vary along
Shear (–) V the length as per the shape of BMD.
P P
BMD PL PL
Mmax (+)
L L L
Moment Bending moment diagram

(+) (-)
Hence, the correct option is (C). PL ML
2.6 (A) Hence, the correct option is (D).

Internal hinge in a beam has zero bending 2.8 (A)
moment resistance.
10 kN 2.9 2
P Q Given :
A
3L/4 L/4
The FBD of given problem can be shown as
below. y = x 2 − xL , to find curvature of beam
10 kN + R We know, y = Deflection
dy
P
A Q = Slope
A dx
MP RQ
RP R
3L L d2y M 1
4 4 2
= Curvature = =
dx EI R
Considering the right hand section of beam,
dy
Taking moment about hinge,  M = 0 ∴ = 2x − l
dx
L
− RQ × = 0 d2y
4 =2
dx 2
RQ = 0
2.10 30
Given beam,
Scan for 10 kN/m
Video Solution
x y
 Key Point Hinge
 Internal hinges are provided in a structure
2m 2m
to reduce statical indeterminacy of the
structure. It makes structure more flexible x y
and allows structure to move which reduces
2m 2m
the reactive stresses.
Rx Ry
 Bending moment at internal hinge is
always zero. We know, sum of moment at internal hinge is 0.
M 0 = 0, Taking moment about hinge support Q,
Ry × 2 − 10 × 2 × 1 = 0 ΣM Q = 0
Ry = 10 kN 
f y = 0, P × (2 + 1.5 + 1 + 1.5) + (100 × 0.5) + (200 × 2.5)

Rx + Ry = 10 × 4 −500sin 60 × 4 = 0
Rx = 40 − 10 ∴ P = 197 N
∴ Rx = 30 kN Hence, the correct answer is 197.
Vertical reaction at support x is 30 kN. 2.14 A&B
M M
2.11 180 C D
A B
R
Q L L L
P
1m 4m RA RB
Equilibrium equations :
Maximum bending moment
= 200 × 0.8 + 100 × 0.2 M A =0
= 180 kNm  −M + M − RB × (3L) = 0

 RB = 0
2.12 A,B,D
F V = 0, RA + RB = 0
Member AD and BC are subjected to pure
 RA = 0
bending, so shear force at H and F are zero.
SFD
dM
V= A B
dx 3L
V = 0 (Zero) No shear force throughout the span.
Horizontal displacement of H and F are zero. BMD
C
Vertical displacement at H and F are not zero. L L L
Slopes at E, F, G and H are zero because at axis A

C D
B
−ve
of symmetry slope must be zero for symmetrical
M M
deflection profile.
Here, bending moment and deflection is not
Hence, correct options are (A), (B) and (D).
zero everywhere.
2.13 197 Hence, the correct options are (A), (B).
0
500 sin 60
500 N
100 N
60 0 0.5 m
500 cos60 0 200 N
2.0 m 1.5 m 1.0 m 1.5 m
P Q
3 Principal Stress and Principal Strain
2013 IIT Bombay s y = 50 MPa
3.1 The state of 2D stress at a point is given

y
by a [2 Marks]
s x = 50 MPa s x = 50 MPa
 σ xx τ xy  100 30  x
 =   MPa
 τ yx σ yy  30 20 
The maximum shear stress in MPa is s y = 50 MPa
(A) 50 (B) 75
(C) 100 (D) 110 (A) –50 MPa, on a plane 450 clockwise
w.r.t. x-axis.
2014 IIT Kharagpur
(B) –50 MPa, on a plane 450 anti-
3.2 For the state of stresses (in MPa) shown in clockwise w.r.t. x-axis.
the figure below, the maximum shear stress (C) 50 MPa, at all orientations.
(in MPa) is _____. [2 Marks] (D) Zero, at all orientations.
4
2016 IISc Bangalore
4
3.5 For the state of stress (in MPa) shown in the
2 2 figure, the major principal stress is 10 MPa.
[2 Marks]
5
4
2015 IIT Kanpur

3.3 Two triangular wedges are glued together
as shown in the following figure. The stress 5 5
acting normal to the interface, σ n is
______ MPa. [1 Mark] t
100 MPa
5
The shear stress τ is
sn (A) 10.0 MPa (B) 5.0 MPa
100 MPa 100 MPa (C) 2.5 MPa (D) 0.0 MPa
2019 IIT Madras
450
3.6 A element is subjected to biaxial normal
tensile strains of 0.0030 and 0.0020. The
100 MPa normal strain in the plane of maximum
3.4 For the plane stress situation shown in the shear strain is [1 Mark]
figure, the maximum shear stress and the (A) Zero (B) 0.0050
plane on which it acts are [1 Mark]
(C) 0.0010 (D) 0.0025
3.7 For a plane stress problem, the state of coordinate system to the X − Y coordinate
stress at a point P is represented by the system. The angle θ , locating the X-axis, is
stress element as shown in figure. assumed to be positive when measured
20 MPa from the x-axis in counter-clockwise
direction.
25 MPa y
Y X
80 MPa P 80 MPa σxx = 120 MPa
σ xy = 50 MPa
σ xy
25 MPa t
θ 300
x
20 MPa
σ yy = 35.6 MPa
By how much angle (θ) in degrees the
The absolute magnitude of the shear stress
stress element should be rotated in order to
component σ xy (in MPa, round off to one
get the planes of maximum shear stress?
[2 decimal place) in x − y coordinate system
Marks]
is ______. [1 Mark]
2022 IIT Kharagpur
3.10 Stresses acting on an infinitesimal soil
P
element are shown in the figure (with
σ z > σ x ). The major and minor principal
q stresses are σ1 and σ3 , respectively.
(A) 31.7 (B) 13.3 Considering the compressive stresses as
(C) 26.6 (D) 48.3 positive, which one of the following
2020 IIT Delhi expressions correctly represents the angle
between the major principal stress plane
3.8 The state of stress represented by Mohr’s
and the horizontal plane? [1 Mark]
circle shown in the figure is
[1 Mark]
Shear stress Stress plane
σz
τzx
Normal
(0, 0) stress τzx
σx
σx τzx
(A) biaxial tension of equal magnitude
(B) hydrostatic stress τzx
(C) pure shear σz
(D) uniaxial tension  τ zx  −1  τ zx 
(A) tan −1   (B) tan  
2021 IIT Bombay  σ1 − σ x   σ1 + σ3 
3.9 The state of stress in a deformable body is  τ zx  −1  τ zx 
(C) tan −1   (D) tan  
shown in the figure. Consider  σ1 + σ x   σ3 − σ x 
transformation of the stress from the x − y
2023 IIT Kanpur 3.12 The infinitesimal element shown in the
figure (not to scale) represents the state of
3.11 In a two-dimensional stress analysis, the
stress at a point in a body. What is the
state of stress at a point is shown in the
magnitude of the maximum principal stress
figure. The values of length of PQ, QR,
(in N/mm2, in integer) at the point?
and RP are 4, 3, and 5 units, respectively.
6 N/mm 2
The principal stresses are_________.
(Round off to one decimal place)
y 3 N/mm2
4 N/mm 2 5 N/mm 2
P σ = 120 MPa 450
τ = 70 MPa
σx
5 N/mm2 4 N/mm2
3 N/mm2
x
Q R 6 N/mm2
σy
(A) σ x = 26.7 MPa, σ y = 172.5 MPa

(B) σ x = 54.0 MPa, σ y = 128.5 MPa
(C) σ x = 67.5 MPa, σ y = 213.3MPa
(D) σ x = 16.0 MPa, σ y = 138.5 MPa
3.1 (A) ∴ σ1 = 110 MPa and σ2 = 10 MPa

According to the question, σ1 − σ 2 110 − 10
and, τ max = =
σ xx τ xy  100 30  2 2
Stress tensor,  =  MPa τmax = 50 MPa
 τ yx σ yy   30 20 
So, σ xx = 100 MPa, σ yy = 20 MPa, τ xy = 30 MPa Hence, the correct option is (A).
In case of 2-D state of stress principal stresses are  Key Point

given by,  Maximum shear stress in case of 2-D
2
stress can be directly calculated by,
σx + σ y  σ − σy  2
σ1/2 = ±  x  + τ xy  σ − σy 
2
2  2  τmax =  x 2
 + τ xy
2
 2 
100 + 20  100 − 20  2
σ1/ 2 = ±   + 30  It is given in question that state of stress
2  2  is 2-D. If it is asked in 3-D, then
σ1/2 = 60 ± 50 σ − σ3 110 − 0
τ max absolute = 1 = = 55 MPa
2 2
Where, σ3 = 0 MPa . 2
 σ − σy  2
τmax =  x  + τ xy
3.2 5  2 
Given : Stress in x-direction (σ x ) = −2MPa , σ1 − σ 2
= = radius of Mohr’s circle
2
Stress in y-direction (σ y ) = 4MPa ,
4. Plane of maximum shear stresses (θ s ) ,
Shear stress ( τ xy ) = 4 MPa .
 σ − σy 
2 tan 2θ s = −  x 
σ − σ2  σ − σy   2τ
So, τmax = 1 =  x 2
 + τ xy  xy 
2  2 
2
3.3 0
σx + σ y  σ − σy  2
Where, σ1 = +  x  + τ xy Given :
2  2 
Normal stress in x direction (σ x ) = 100 MPa
2
σx + σ y  σ − σy  2 Normal stress in y direction (σ y ) = −100 MPa
σ2 = −  x  + τ xy
2  2  Shear stress ( τ xy ) = 0
2
 −2 − 4  2 2 2
100 MPa
So, τ max =   + 4 = ( − 3) + 4
 2 
τmax = 5 MPa sn
100 MPa 100 MPa
4 450
2
4 100 MPa
s2 Method 1..
s1
Normal stress at any plane at an angle θ is
Hence, the value of maximum shear stress is 5 given by
MPa. σx + σ y
σn = σθ =
 Key Point 2
Member subjected to Biaxial stresses ( σ x , σ y )  σ x − σ y  
±   cos 2θ + τ xy 2 sin θ 
combined with shear stresses ( τ xy ) :  2  
1. Principal stresses (σ1 , σ 2 ) , So, normal stress at θ = 450
100 + ( −100)
2 σ 450 =
 σ + σy   σx − σ y  2 2
(σ1 , σ2 ) =  x ±   + τ xy
 2   2  ±
100 − (−100) 
cos(900 ) + 0 
2. Principal plane ( θ p ) ,  2 
σ 450 = 0 MPa
2τ xy
tan 2θ p = Hence, the stress acting normal to the interface,
σx − σ y
σ n is 0 MPa.
3. Maximum shear stresses (τmax ) , Method 2..
Since, no shear stress is acting along the plane x s y = 50 MPa
and y, then x and y planes are treated as principal
planes and σ x and σ y are equal to the principal y
stresses. s x = 50 MPa x
s x = 50 MPa
So, Mohr’s circle for the given condition is,
(0,100)
s y = 50 MPa
2q = 90 0
:Method 1:
(-100, 0) (100, 0) Maximum shear stress,
2
 σ − σy  2
( 0, -100) τmax =  x  + τ xy
 2 
From Mohr’s circle, normal stress is zero at 450 2
to the principal plane.  50 − 50  2
τ max =   +0
Hence, the stress acting normal to the interface is  2 
zero. τmax =0
Hence, the stress acting normal to the interface, ∴ Shear stress in all the directions is zero.
σ n is 0 MPa. Hence, the correct option is (D).
 Key Point :Method 2:
 Principal planes (Planes on which Mohr circle for the state of stress
Point circle
principal stresses act) :
2τ xy
tan(2θ p ) = 50 MPa
σx − σ y Thus, there is no shear stress at any point
 Plane of maximum shear stress : Hence, the correct option is (D).
− (σ x − σ y )  Key Point
tan(2θs ) =
2 τ xy The given stress condition is of hydrostatic
 Average stress (Shear stress is maximum) loading.
3.5 (B)
σx + σ y
σavg = Given :
2
0 Normal stress in x direction (σ x ) = 5 MPa
 θ = 45 , is the plane of maximum shear
stress. Normal stress in y direction (σ y ) = 5 MPa
According to given question
3.4 (D) 5
Given :
Normal stress in x direction (σ x ) = 50 MPa
y
Normal stress in y direction (σ y ) = 50 MPa
5 x 5
Shear stress ( τ xy ) = 0 MPa
According to the question,
t
Major principal stress, σ1 = 10 MPa :Method 1:
2 Angle of principal plane ( θ p ) is given by,
σx + σ y  σ − σy  2
σ1 = +  x  + τ xy 2τ xy −2 × 25
2  2  tan 2θ p = =
σx − σ y 80 − (−20)
2
5+5  5−5 2
10 = +   + τ xy θ p = −13.280
2  2 
10 = 5 + τ xy Note : Here, θ is measured from the plane at
which σ y is acting. So, τ xy is taken as negative,
∴ τ xy = 5 MPa
because shear stress is anticlockwise in direction
in the plane in which σ y act.
3.6 (D)
Now, angle of plane of maximum shear stress,
Given : Normal tensile strain in x direction
(∈x ) = 0.0030
θ = θ p + 450 = −13.280 + 450 = 31.710
Normal tensile strain in y direction Hence, the correct option is (A).
(∈y ) = 0.0020 :Method 2:
According to the question, element is subjected Angle of plane of maximum shear stress (θ s ) is
to biaxial normal tensile strains. calculated by,
Let x and y be the plane. (σ x − σ y ) [80 − (−20)]
tan 2θs = − =−
The plane of maximum shear strain occur at 450 τ xy 2 × (−25)
to the principal planes.
θ s = 31.710
So, normal strain in the plane of maximum shear
strain Hence, the correct option is (A).
∈ + ∈y 0.0030 + 0.0020
= x = Scan for
2 2 Video Solution
= 0.0025
Hence, the correct option is (D). 3.8 (C)
Scan for In case of pure shear state of stress condition. The

Video Solution centre of Mohr’s circle lies at origin.
 Key Point  Key Point
Normal strain (or stress) in the plane of  In case of hydrostatic stress, Mohr circle
maximum shear strain (or stress) is equal to the reduces to a point and zero shear.
average of normal strain in x and y direction. +t
3.7 (A)
s1 = s 2 = s
20 MPa -s +s
25 MPa
y
-t
80 MPa x 80 MPa
 In case of bi-axial tension of equal
25 MPa t magnitude
20 MPa
+t ∴ On substituting value in formula,
 40 + 35.6 
s1 = s 2 = s 120 =  
 2 
-s +s
 40 − 35.6 
+  cos(120) + z xy sin(120)
 2 
-t
∴ z xy = 96.186 MPa
 In case of uniaxial tension
+t
3.10 (A)
σz
+s
-s æs ö (s,0) τzx
ç ,0 ÷
è2 ø
A B
-t
−
3.9 96.186 σx
Given : θ τzx
120 MPa σ1
40 MPa
C
50 MPa
ΣFx = 0,
τxy 30 0
BC
sin θ =
35.6 MPa AC
σy (−) σ x ( BC ) − τ z × ( AB) + σ1 sin θ( AC ) = 0
τxy
 AC sin θ  AC cos θ σ1 sin θ( AC )
σx' σx   + τz × =
 cos θ  cos θ cos θ
σx τx 'y 0 σx
60
σ x tan θ + τ zx = σ1 tan θ
30 0
tan θ (σ1 − σ x ) = τ zx
σy  τ zx 
θ = tan −1  
σ x' = σ x cos 2 θ + σ y sin 2 θ  σ1 − σ x 
+ 2 z xy sin θ cos θ
3.11 (C)
 σ + σy   σx − σ y  τ xy = 0
Or σ x' =  x +  cos 2θ
 2   2  4
cos θ =
+ z xy sin 2θ 5
 σ x = 40 MPa , σ y = 35.6 MPa , 3
sin θ =
5
Q = 600 (From vertical)
σn = σx cos2 θ + σ y sin 2 θ + 2τxy sin θ cos θ
∴ σ x' = 120 MPa
...(i)
z x ' y ' = − 50
y
= 2 ± 5  −3& 7 MPa
P
70 MPa So, maximum principle stress = 7 MPa
θ 120 MPa
Hence, the correct answer is 7.
σx 4 Method-2 : (Mohr Circle)
5
θ τ
Q 3 x
x
R
R
3
σy
O 5
σ
 σ − σx  (2, 0) 90 0 = 2θ 6
τ= y  sin(2θ) + τ xy cos 2θ
 2  R 4
...(ii)
From equation (i),
2 2
4 3 R = x 2 + 32 = ( x − 1) 2 + 42
120 = σ x   + σ y   ...(iii)
5 5
From equation (ii), x 2 + 32 = x2 − 2 x + 1 + 42
 σ − σx  4 3 x=4
70 =  y  2 × × …(iv)
 2  5 5
σmax = 2 + R = 2 + x 2 + 32
By solving equation (iii) and (iv),
= 2 + 4 2 + 32 = 7 MPa
We get, σ x = 67.5 MPa, σ y = 213.33 MPa
3.12 7
Method-1 :
Given : σ y = 6 MPa , τ xy = 3 MPa ,
σ45 = σ2X cos 2 450 + σ2y sin 2 450

+ 2τ XY sin 450.cos 450
1 1 1
5 = σX × + 6 × + 2 × 3×
2 2 2
σ
5 = X +3+3
2
σ X = −2 MPa
2
σ + σY  σ − σY  2
σ p1 / σ p2 = X ±  X  + τ XY
2  2 
2
−2 + 6  −2 − 6 
 + 3 = 2 ± 16 + 9
2
= ± 
2  2 
4 Bending and Shear Stresses
2013 IIT Bombay 2018 IIT Guwahati

4.1 The ‘Plane section remains plane’ 4.5 A cantilever beam of length 2 m with a
assumption in bending theory implies square section of side length 0.1 m is loaded
[1 Mark] vertically at the free end. The vertical
(A) Strain profile is linear. displacement at the free end is 5 mm. The
(B) Stress profile is linear. beam is made of steel with Young’s modulus
(C) Both profiles are linear. of 2.0 ×1011 N/m 2 . The maximum bending
(D) Shear deformation is neglected. stress at the fixed end of the cantilever is
4.2 A symmetric I-section (with width of each [2 Marks]
flange = 50 mm, thickness of each flange 10
(A) 20.0 MPa (B) 37.5 MPa
mm, depth of web 100 mm, thickness of web
= 10 mm) of steel is subjected to a shear (C) 60.0 MPa (D) 75.0 MPa
force of 100 kN. Find the magnitude of the 4.6 An 8 m long simply-supported elastic beam
shear stress (in N/mm 2 ) in the web at its of rectangular cross-section (100 × 200) mm
junction with the top flange _______. is subjected to a uniformly distributed load
[1 Mark] of 10 kN/m over its entire span. The
2014 IIT Kharagpur maximum principal stress (in MPa, up to two
4.3 The first moment of area about the axis of decimal places) at a point located at the
bending for a beam cross section is extreme compression edge of a cross-section
and at 2 m from the support is _____.
[1 Mark]
[2 Marks]
(A) moment of inertia
(B) section modulus 2019 IIT Madras
(C) shape factor
(D) polar moment of inertia 4.7 For a given loading on a rectangular plain
2016 IISc Bangalore concrete beam with an overall depth of 500
mm, the compressive strain and tensile strain
4.4 A 450 mm long plain concrete prism is developed at the extreme fibers are of the
subjected to the concentrated vertical loads
same magnitude of 2.5 ×10−4 . The curvature
as shown in the figure. Cross-section of the
in the beam cross-section (in m-1, round off
prism is given as 150 mm ×150 mm .
to 3 decimal places) is ________.
Considering linear stress distribution across
the cross-section, the modulus of rupture [1 Mark]
(expressed in MPa) is ________. 4.8 For a channel section subjected to a
11.25 kN 11.25 kN downward vertical shear force at its centroid,
which one of the following represents the
P S
Q R correct distribution of shear stress in flange
and web?
150 mm 150 mm 150 mm
[2 Marks] [1 Mark]
(A) 50 50
300
100 100
50
N.A.
400
(B)
All dimensions are in mm
(A) 240 N (B) 480 N
(C) (C) 60 N (D) 120 N
2023 IIT Kanpur
4.10 The cross-section of a girder is shown in the

figure (not to scale). The section is
symmetric about a vertical axis (Y-Y). The
moment of inertia of the section about the
horizontal axis (X-X) passing through the
(D) centroid is________cm4 (round off to
nearest integer).
40 cm
Y
10 cm
4.9 Cross section of a built-up wooden beam as

shown in figure (not drawn to scale) is X X
subjected to a vertical shear force of 8 kN. 50 cm
The beam is symmetrical about the neutral
axis (NA), shown and the moment of inertia
about NA is 1.5 ×109 mm4 . Considering that
the nails at the location P are spaced
Y
longitudinally (along the length of the beam)
at 60 mm, each of the nails at P will be 20 cm
subjected to the shear force of
[2 Marks]
4.1 (A) 50 × (100 + 10 + 10)3 (50 − 10) × 1003
= −
“Plane sections remains plane after bending”. By 12 12
6 4
this assumption the strain will become directly I = 3.866 ×10 mm
proportional to the deflection i.e. ε ∝ δy 100 × 103 × 50 × 10 × 55
So, τ =
Hence, the correct option is (A). 3.866 × 106 × 10
 Key Point τ = 71.12 N/mm 2
dy Hence, magnitude of the shear stress in the web at
Initial plane Final plane
(Before bending) (After bending) its junction with the top flange 71.12 N /mm2 .
4.3 (B)
Linear displacement
variation Moment of inertia is given by,
I = Ay 2
i.e. moment of inertia is moment of first moment
dy
(moment of area) about axis of bending.
It is also called second moment of area but section
From figure. It is clear that the strain is directly modulus is given by,
proportional to deflection and strain profile
I Ay 2
obtained will be linear. Z= = = Ay
y y
Scan for So, the first moment of area about the axis of
Video Solution bending for a beam cross section is section
modulus.
4.2 71.12 Hence, the correct option is (B).
Given : Width of each flange (b) = 50 mm Scan for
Thickness of flange (t ) = 10 mm Video Solution
Depth of web (d ) = 100 mm

4.4 3
Thickness of web (tw ) = 10 mm
Given : Length of concrete prism ( L) = 450 mm
Shear force ( F ) = 100 kN
Cross section of the prism ( A) = 150 ×150 mm 2
According to question, 11.25 kN 11.25 kN
50 mm P S
10mm Q R
y = 55 mm RP RS
NA
10 mm 100 mm
150mm 150mm 150mm
Q R
10mm
BMD
P
Now, shear stress is given by, S
FA y
τ= Considering vertical forces,
Ib  Fy = 0 ,
BD 3 bd 3
and I = − RP + RS = 11.25 + 11.25
12 12
RP + RS = 22.5
Taking moment about point P, 0.1
6250 ×
 MP = 0, M 2
σb = × y =
11.25 ×150 + 11.25 × 300 = RS × 450 I 0.1× 0.13
RS = 11.25 kN 12
σb = 37.5 MPa
RP = 22.5 − RS
RP = 11.25 kN
4.6 90
Now, bending moment at point Q is given by,
BM Q = RP × 150 = 11.25 × 150 = 1687.5 kNmm Given : Length of beam ( L) = 8m
and, section modulus in given by, Cross section of beam (A) = 100 × 200 mm 2
I bd 3 bd 2 150 × 1502
z= = = = Uniformly distributed load (U.D.L) = 10 kN/m
y 12 × d 6 6 10 kN/m
C
2 A B
z = 562500 mm3 8m
RA RB
So, modulus of rupture (i.e. bending stress) is
2m
given by,
RA + RB = 10 × 8 = 80
M BM Q 1687.5 × 103
σ= = = 80
Z Z 562500 Due to symmetry, RA = RB = = 40 kN
2
σ = 3 N/mm = 3 MPa 2
Hence, the modulus of rupture is 3 MPa. Now, bending moment at point C is given by,
BM C = RA × 2 − 10 × 2 × 1
4.5 (B)
= 40 × 2 − 20 = 60 kNm
Given : Length of beam (l) = 2m,
And moment of inertia is given by,
Side of square beam (a) = 0.1m
Deflection (δ) = 5mm bd 3 0.1× 0.23
I= =
12 12
Young’s modulus ( E ) = 2 × 1011 N/m 2 −5
P I = 6.67 ×10 m4
So, bending stress is given by,
0.1m BM C 60 × 103 0.2
σb = ×y= −5
×
I 6.67 × 10 2
0.1m
Now, deflection in cantilever beam when σb = 89.96 MPa ≈ 90 MPa
subjected to point load at free end is given Principal stress,
by, σ + σy 1
σ p1 , σ p2 = x ± × (σ x − σ y ) 2 + 4τ2xy
Pl 3 2 2
δ=
3EI 90 + 0 1
σ p1 , σ p2 = ± × (90 − 0) 2
P × 23 2 2
5 × 10−3 =
0.1× 0.13 σ p1 = 90 MPa and σ p2 = 0 MPa
3 × 2 × 1011 ×
12
So, principal stress = 90 MPa.
P = 3125 N
Since there is no shear stress, so bending stress
So, bending moment, M = P × l = 3125 × 2 will be maximum principle stress.
M = 6250 Nm Hence, the maximum principal stress at a point
and, from bending equation bending stress is given located at the extreme compression edge of a
by,
cross-section and at 2 m from the support is 90
4.9 (A)
MPa.
Given : Shear force (V ) = 8kN
4.7 0.001 Moment of inertia about neutral axis,
Given : ( I NA ) = 1.5 × 109 mm 4
Depth of beam ( D) = 500 mm = 0.5m
Distance between two nails (d) = 60 mm
Strain developed (ε) = 2.5 ×10−4
Now, from bending equation,
σ E
=
y R
1 σ 1
= ×
R E y 150
1  1   σ
= ε×  ε = 
R  D/2  E 50
2 50
= 2.5 ×10−4 × = 1× 10−3 = 0.001m -1
0.5 Now, shear force per unit length is given by,
-4
e = 2.5 ´ 10 VAy
F=
I
y 8 × 103 × 50 × 100 × 150
D NA F= = 4 N/mm
1.5 × 109
So, shear force per unit length at each nail,
F = 4 × 60 = 240 N
e = 2.5 ´ 10-4
Hence, the curvature in the beam cross-section is Hence, the correct option is (A).
0.001 m–1. 4.10 468810
4.8 (C)
Y
Shear force in flange (horizontal member) is linear
with zero at free end and in web (vertical member) 40 cm
it is parabolic in nature. C1 10 cm
Shear force distribution for channel, 5 cm
tmax
X X
y1 C2 50 cm
y
tmax y2
20 cm
tmax
Y
A1 = 40 × 10 = 400 cm 2
A2 = 20 × 50 = 1000 cm 2
y1 = 50 + 5 = 55cm
50
y2 = = 25cm
2
Centroid of composite shape from bottom
fibre:-
A1 y1 + A2 y2
y=
A1 + A2
(400 × 55) + (1000 × 25)
=
400 + 1000
= 33.5714 cm
 103 
I xx =  40 × + 400(55 − 33.5714)2 
 12 
 50 3

+  20 × + 1000(33.5714 − 25)2 
 12 
I xx = 187007.29 + 281802.23
I xx = 468809.52 cm 4 ≅ 468810 cm 4
5 Deflection of Beams
2013 IIT Bombay 1 2πx 

(A) y = 1 − a cos 
P L 
5.1 A uniform beam (EI = constant) PQ in the
form of a quarter – circle of radius R is fixed 1 2πx 
(B) y = 1 − a sin 
at end P and free at the end Q, where a load P L 
W is applied as shown. The vertical nπx
(C) y = a sin
downward displacement, δ q , at the L
loaded point Q is given by nπx
(D) y = a cos
 WR 3  L
δq = β   5.3 The beam of an overall depth 250 mm
 EI  (shown below) is used in a building
Find the value of β (correct to 4 decimal subjected to two different thermal
places) ________. environments. The temperatures at the top
[2 Marks] and bottom surfaces of the beam are 360 C
Q and 720 C respectively. Considering
coefficient of thermal expansion (α) as
W
1.50 ×10−5 per 0 C , the vertical deflection of
the beam (in mm) at its mid-span due to
temperature gradient is ________.
R [2 Marks]
360 C
900
P 250
æ WR 3 ö mm
dq = b ç ÷ 720 C
è EI ø
1.5 m 1.5 m
2014 IIT Kharagpur 5.4 For the cantilever beam of span 3 m (shown
below), a concentrated load of 20 kN applied
5.2 If the following equation establishes
at the free end causes a vertical displacement
equilibrium in slightly bent position, the
of 2 mm at a section located at a distance of
mid-span deflection of a member shown in
1 m from the fixed end. If a concentrated
the figure is
vertically downward load of 10 kN is applied
d2y P at the section located at a distance of 1 m
+ y=0
dx 2 EI from the fixed end (with no other load on the
If a is amplitude constant for y, then beam), the maximum vertical displacement
[2 Marks] in the same beam (in mm) is _____.
y [2 Marks]
20 kN
P EI P 2mm
x
y
M N
L
1m 2m
5.5 The axial load (in kN) in the member PQ 2016 IISc Bangalore
for the arrangement/assembly shown in the
5.9 A 3m long simply supported beam of
figure given below is _______.
uniform cross section is subjected to a
[2 Marks]
P uniformly distributed load of
w = 20kN/m in the central 1m as shown
in the figure.
2m
160 kN
S
Q R
Hinge Beam
S If the flexural rigidity (EI) of the beam is
2m 2m
30 ×106 N-m 2 , the maximum slope
2015 IIT Kanpur
(expressed in radians) of the deformed beam
5.6 A steel strip of length, L = 200 mm is fixed is [2 Marks]
at end A and rests at B on a vertical spring of (A) 0.681×10 −7
(B) 0.943 ×10−7
stiffness, k = 2 N/mm. The steel strip is 5 mm
(C) 4.310 ×10−7 (D) 5.910 ×10−7
wide and 10 mm thick. A vertical load, P =
50 N is applied at B, as shown in the figure. 2017 IIT Roorkee
Considering E = 200 GPa, the force (in N) 5.10 Two prismatic beams having the same
developed in the spring is ________. flexural rigidity of 1000 kNm 2 are shown in
P
the figures. [2 Marks]
6 kN/m
A B
k d1
4m
120 kN
L
[2 Marks, Set-II] d2
5.7 Two beams are connected by a linear spring 1m 1m
as shown in the following figure. For a load If the mid-span deflections of these beams
P as shown in the figure, the percentage of are denoted by δ1 and δ2 (as indicated in the
the applied load P carried by the spring is
figures), the correct option is
_______. [2 Marks]
P (A) δ1 = δ2 (B) δ1 < δ2
L
(C) δ1 > δ2 (D) δ1 >> δ2
EI 3EI
kS = 2018 IIT Guwahati
(2 L3 )
5.11 The figure shows a simply supported
EI
beam PQ of uniform flexural rigidity EI
5.8 A simply supported reinforced concrete carrying two moments M and 2M.
beam of length 10 m sags while undergoing M 2M
shrinkage. Assuming a uniform curvature of P Q
0.004 m −1 along the span, the maximum L/3 L/3 L/3
deflection (in m) of the beam at mid-span is The slope at P will be [2 Marks]
________.
(A) 0 (B) ML/(9EI)
[2 Marks]
(C) ML/(6EI) (D) ML/(3EI)
2020 IIT Delhi = 5 m. Force Q is applied at the center of
beam WX such that the force in the spring
5.12 A cantilever beam PQ of uniform flexural VW becomes zero.
rigidity (EI) is subjected to a concentrated P
moment M at R as shown in the figure U 2a
V
[2 Marks] EI Q
R a a
P
Q W 2EI X
M L/2
L
The magnitude of force Q (in kN) is
________. (round off to the nearest integer)
The deflection at the free end Q is [2 Marks]
3ML2 ML2 2023 IIT Kanpur
(A) (B)
4 EI 6 EI
2
5.15 When a simply-supported elastic beam of
3ML ML2 span L and flexural rigidity EI (E is the
(C) (D)
8 EI 4 EI modulus of elasticity and I is the moment of
2021 IIT Bombay inertia of the section) is loaded with a
5.13 Employ stiffness matrix approach for the uniformly distributed load w per unit length,
simply supported beam as shown in the the deflection at the mid-span is
figure to calculate unknown displacement 5 wL4
/rotations. Take length, L = 8 m; modulus of Δ0 =
384 EI
elasticity, E = 3 ×104 N/mm 2 ; moment of If the load on one half of the span is now
inertia, I = 225 × 106 mm 4 . removed, the mid-span deflection_____.
E, 2I
P
(A) Reduces to Δ0 /2
A B E, I
C
(B) Reduces to a value less than Δ0 /2
(C) Reduces to a value greater than Δ0 /2
L L (D) Remains unchanged at Δ0
2 2
The mid-span deflection of the beam (in mm,
round off to integer) under P = 100 kN in
downward direction will be _______.
[2 Marks]
2022 IIT Kharagpur
5.14 The linearly elastic planar structure shown in
the figure is acted upon by two vertical
concentrated forces. The horizontal beams
UV and WX are connected with the help of
the vertical linear spring with spring constant
k = 20 kN/m. The fixed supports are
provided at U and X. It is given that flexural
rigidity EI = 105 kN-m2, P = 100 kN, and a
5.1 0.7854 5.2 (C)
B
X d2y P
2
= − . y = −ω2 x
R sin q dx EI
W
P
Where, ω = .
EI
R
dq Solution of difference equation,
q
d2y
2
= −ω2 x is given as,
A dx
X
y = A sin ωx + B cos ωx
Moment at any section X-X,
M xx = −WR sin θ From boundary condition, at x = 0, y = 0
∂M xx  B=0
∴ = − R sin θ Also, at x = l , y = 0
∂W
From strain energy method,  0 = A sin ω l
Deflection at point B , But A can not be zero, otherwise for all values
π /2
∂U (∂M xx / ∂W ) of x, y will be zero.
δB = =  M xx dx
∂W 0 EI Hence, ω l = nπ
π /2
(−WR sin θ)(− R sin θ) nπ n πx
= 
0
EI
dx ω=
l
 y = A sin
l
π/ 2
(WR 2 sin 2 θ) Rd θ Hence, the correct option is (C).
= 
0
EI
dx (dx = Rd θ)
5.3 2.43
π/ 2
WR 3 360 C

360
= 2sin 2 θd θ 250
2 EI 0 mm 720
3 π /2 720 C Temperature
WR
 (1 − cos 2θ)d θ
gradient
= 1.5 m 1.5 m
2 EI 0
Due to temperature gradient, bottom section get
3 π/ 2
WR  sin 2θ  expanded while upper section remains of same
= θ− 
2 EI  2 0 length.
π WR3
= …(i)
4 EI
R R
 WR 3 
Now, according to the question, δ B = β  
 EI  L/2 L/2
So, comparing it with equation (i), we have
π d
β = = 0.7854 L + LaDT1
4
Hence, the value of β is 0.7854 . L + LaDT2
5.4 1
2R – d Using Maxwell – Betti reciprocal theorem

20 kN
1m A 2m B
L/2 L/2 d A = 2 mm
d
10 kN
1m 2m
From circle,
dB = ?
L L
(2 R − δ) × δ = ×
2 2 20 × Δ B = 10 × (2)
(By the property of chord) ∴ Δ B = 1mm
L2 Hence, Maximum vertical displacement in the
2
2 Rδ − δ = beam under the load of 10 kN at a distance of 1 m
4
from fixed end is 1 mm.
L2  Key Point
2 Rδ = (δ2 <<<< 2 Rδ)
4 Maxwell – Betti Reciprocal Theorem :
The theorem states that the workdone by forces
L2
δ= …(i) acting through displacement of the second system
8R is same as the work done by the second system of
Now from the circular arc forces acting through the displacements of the first
system.
R R+h
= Thus, P.Δ1 = QΔ 2
L + LαΔT1 L + LαΔT2
P Q
(Since angle included between them is same)
α(ΔT1 − ΔT1 ) h D1 D2
=
1 + αΔT1 R This is also valid even when the first system of
As α × ΔT1 <<< 1 forces is P1 , P2 .......Pn and the second system of
forces is given by Q1 , Q2 ........Qn by the forces
h
R= …(ii) P1 , P2 ..........Pn only δ1 , δ2 ........δn be the
α(ΔT2 − ΔT1 )
displacements due to system of forces
Putting value of R from equation (ii) to equation Q1 , Q2 ........Qn only acting on the beam as shown
(i), we get
in fig.
2
L α(ΔT2 − ΔT1 ) P1 P2 Pi
∴ δ=
8h
d1 d2 di
32 ×1.5 ×10−5 × 36
=
8 × 250 ×10−3 Q1 Q2 Qi
= 2.43mm
D1 D2 Di
Hence, the mid span deflection due to temperature
gradient is 2.43 mm. So, in general Pi Δi = Qi δi
to compression.
5.5 50
Now, since end B of beam is attached to
FBD of given problem is,
spring,
P Deflection of point B = Compression in spring
So, from FBD of beam,
( P − R) L3
RQ Deflection of point B = …(i)
RQ 160 kN 3EI
R
Q
S
R and, compression in spring =
2m 2m k
…(ii)
By compatibility condition,
From equation (i) and (ii), we have
(Deflection at point Q in beam QR ) ( P − R) L3 R
=
= (Deflection at point Q in axial member PQ ) 3EI k
(50 − R )(200)3 R
RQ L3 RQ L  =
∴ δ S + θS × QS − = 5 × 10 3
2
3EI AE 3 × 200 × 103 ×
12
160 × 23 160 × 22 RQ × 43  R = 3.0075 N
 + ×2− =0
3 × EI 2 EI 3 × EI Hence, the force developed in the spring is
(Deflection due to axial force will be very less 3.0075 N .
as compared to bending forces)
Scan for
 6400 = RQ ×128 Video Solution
 RQ = 50 kN 5.7 25
P
Hence, the axial load in the member PQ is 50 kN. L
5.6 3.0075 EI 3EI

kS =
(2 L3 )
Given : Length, L = 200 mm
EI
Stiffness of spring, k = 2 N/mm
The FBD of given problem is,
Vertical load, P = 50 N P
E = 200 GPa. A B
The FBD for given problem is, R
R
P
B Beam ks
A
R
R R
R
k Spring C D
L The spring attached to the two beams is going to

Here R = Force developed in the spring due be in compression due to applied load and this
compression is equal to the difference of OA2 + AQ 2 = OQ 2
deflection caused in two beams. 2
Let R be the force developed in spring due to  10 
OA +   = 2502
2
compression.  2
∴ Compression in spring  OA = 2502 − 52 = 249.95 m
= (Deflection in upper beam
∴ Deflection at mid span
– Deflection in lower beam)
= OA '− OA = 250 − 249.95 = 0.05 m
 Δ spring = δ B − δ D
Hence, the maximum deflection of the beam at
R ( P − R) L3 RL3 mid span is 0.05 m.
= −
ks 3EI 3EI 5.9 (*)
3 3
R ( P − R) L RL :Method 1:
= −
3EI 3EI 3EI Moment area method :
3
2L
R(2 L3 ) ( P − R) L3 RL3
= −
3EI 3EI 3EI
2 R = ( P − R) − R
P Due to symmetrical loading
 R=
4 20 ×1
∴ Percentage of the applied load carried by Rp = RQ = = 10 kN
2
the spring
Now, M A = R p × 1 = 10 ×1 = 10 kN-m
R P/4
= ×100 = ×100 = 25%
P P 20 × 0.52
and, M B = 10 ×1.5 − = 12.5kN - m
Hence, the percentage of the applied load carried 2
by the spring is 25%. 12.5
10
EI
EI III
Scan for
Video Solution
I II
5.8 0.05
P Q
Given problem can be depicted in figure as A
0.5 m
B C
shown below, 1m 1m 1m
O So, maximum slope in beam occurs at supports P
and Q,
R = 250 m M
∴ Maximum slope in beam = Area of
10 m EI
P
A 900 5m Q diagram between point A and B
= AI + AII + AIII
A' 1 10   10 
Given : Length, L = 10 m =  × 1×  +  0.5 × 
2 EI   EI 
1 2 2.5 
Curvature = = 0.004 m−1 +  × 0.5 ×
R 
3 EI 
1 10.833 10.833
∴ R= = 250 m = = ×103
0.004 EI 30 ×10 6
Now, from the geometry, in ΔOAQ −4

= 3.611× 10 radians
:Method 2: 5 wl 4
δ1 =
Macaulay’s method : 384 EI
w/m X
5 6 × 44
δ1 = ×
384 1000
w/m
RP x X RQ  δ1 = 0.02 m …(i)
120 kN
w w
M x = Rp x − ( x − 1) 2 + ( x − 2) 2
2 2 d2
According to Macaulay method 1m 1m
d2y w w Mid - span deflection of simply supported
EI 2 = R p x − ( x − 1) 2 + ( x − 2) 2
dx 2 2 with point load, δ2
After integrating once Pl 3 120 × 23
dy R p x
2
w w δ2 = =
EI = + C1 − ( x − 1)3 + ( x − 2)3 48EI 48 ×1000
dx 2 6 6  δ2 = 0.02 m …(ii)
Due to symmetrical loading slope will be zero at
So, from equation (i) and (ii),
mid section ( x = 1.5 m)
δ1 = δ2
dy
∴ at x = 15m, = 0 Hence, the correct option is (A).
dx
R p (1.5) 2
w 5.11 (C)
0= + C1 − (1.5 − 1)3
2 6 :Method 1:
9R w 9w w RP 3M
= p + C1 − = + C1 −
8 48 16 48 L
M 2M
w 9w 26w 13w
∴ C1 = − =− =− P Q
48 16 48 24
∴ Equation of slope,
L/3 L/3 L/3 RQ
2
dy R p x 13w w w( x − 2)3
EI . = − − ( x − 1)3 + M
dx 2 24 6 6
The slope will be maximum at the support,
M M
−13w −13 × 20 ×103
∴ θmax ( x =0) = = BMD
24 EI 24 × 30 ×106 M/EI
= − 3.611×10− 4 radian P Q
5.10 (A) M/EI M/EI

R1 R2
Given : EI = 1000 kNm 2 Conjugate beam
Also, flexural rigidity of both beam is same M
6 kN/m R1 + R2 = (Area of diagram in conjugate
EI
beam)
d1
1 L M ML
4m Now, R1 + R2 = × × =
Mid span deflection of simply supported with 2 3 EI 6 EI
UDL, δ1 ΣM Q = 0 ,
1 L M  2L L  1 L M 5.12 (C)
R1 L − × ×  + − × ×
2 3 EI  3 9  2 3 EI R
P
Q
 L L  1 L M 2L L/2
 + + × × × =0 M
 3 9  2 3 EI 9 L
7 ML 4ML ML ML Deflection diagram,

R1 = + − =
54EI 54 EI 27 EI 6 EI P R Q
∴ Slope at P, d1
θ p = Shear force at point a in conjugate beam
ML d2
θ p = SFp =
6EI δQ = δ1 + δ 2
Hence, the correct option is (D). 2
l l
:Method 2: M  M 
2 2 l
Moment area method =   +   
2 EI EI  2 
dQ / P Ml 2 Ml 2
= +
P Q 8 EI 4 EI
M 2M
Ml 2 + 2 Ml 2 3Ml 2
L/3 L/3 L/3 = =
M 8 EI 8EI
EI + Hence, the correct option is (A).
– –
M M 5.13 119
EI EI Given : Length = 8 m
BMD/ EI E = 3 ×104 N/mm 2
δQ / P = Deflection of point Q w.r.t. to tangent at I = 225 ×106 mm 4
point P P = 100 kN
δQ / P P
θP = E, 2 I B
L A E, I
C
δQ / P = Moment of area of BMD/EI
 1 M L   2L L 
δQ / P =  − × ×   + 
 2 EI 3   3 9  L L
2 2
 1 M L  L L 
+ − × ×  + 
 2 EI 3   3 9 
PL
 1 M L   2L  4
+ × ×   BMD
 2 EI 3   9 
ML2
= PL
6 EI 4 EI PL
ML 2 8EI
M
δQ / P ML
θP = = 6 EI = L B L EI
L L 6 EI 2 2
Now, convert it into conjugate beam, since its K = 20 kN/m
simply supported beam then its conjugate beam EI = 105 kN-m 2
M P = 100 kN
remain same and diagram becomes our
EI a=5 m
loading diagram.
PL
Free body diagram,
P
4 EI ( y1 )
PL 2a
8EI
R=0
A C R=0
L B L
2 2
R=0
Taking moment about B.
ΣM B = 0
Q
 1 PL L  L 1  l   
Then VA ( L) −  × × ×  +   
 2 8 EI 2  2 3  2    a a
 1 Pl 2 l  2 l   2 EI
− × × ×  = 0
 2 4 EI 2  3 2  
3
PL
Deflection, y1 =
Pl 2 Pl 2 Pl 2 3EI
VA = + = P(2a)3
48 EI 48 EI 24 EI y1 =
L  1 PL L 1 L  3EI
∴ M B = VA × − × × × × =0
2  2 8EI 2 3 2  y1 =
100 ×103 (10)3
3 ×105 ×103
Pl 2 l Pl 3
MB = × − y1 = 0.333 m
24 EI 2 192 EI
R=0 Q
Pl 3
MB =
64 EI a a
 In conjugate beam where we want to find
2 EI
deflection, take out moment about that point.
Pl 3 100 × 83 ×103 ×109
∴ δB = =
64 EI 64 × 3 ×104 × 225 ×106 Q
Qa
= 118.5 mm
a a
5.14 640
Given : Q
P
U a a
2a y2
V Qa
EI
Q
By using moment area method
a a
1 1 5a 
2 EI y2 =  aQa × 
2 EI  2 3
X
5Qa 3
y2 = = 0.33
12 EI
5Q53
0.33 =
12 ×105 ×103
Q = 640000 N
Q = 640 kN
5.15 A
5 wL4
Given : Δ 0 =
384 EI
From above, when load on one half of span is

Δ0
removed, the mid span deflection reduced to .
2
6 Torsion of Shafts and Pressure Vessels
2017 IIT Roorkee 2 kN.m 1 kN.m
6.1 A hollow circular shaft has an outer diameter

0.1 m 0.8 m
of 100 mm and inner diameter of 50 mm. If
O Q
the allowable shear stress is 125 MPa, the P
1m 0.5 m
maximum torque (in kN-m) that the shaft can
The absolute maximum shear stress in the
resist is ________. member (in MPa, round off to one decimal
[2 Marks] place) is _____.
2018 IIT Guwahati 2022 IIT Kharagpur
6.2 A solid circular beam with radius of 0.25 m 6.5 The hoop stress at point on the surface of a
and length of 2 m is subjected to a twisting thin cylinder pressure vessel is computed to
moment of 20 kNm about the z-axis at the be 30 MPa . The value of Maximum shear
free end, which is the only load acting as stress at this point is
shown in the figure. The shear stress (A) 15 MPa (B) 30 MPa
component τ xy at point M in the cross (C) 7.5 MPa (D) 22.5 MPa
section of the beam at a distance of 1 m from [1 Mark]
the fixed end is 2023 IIT Kanpur
[1 Mark]
x 6.6 A circular solid shaft of span L = 5 m is fixed
20 kNm
M M at one end and free at the other end. A torque
z
y T = 100 kN.m is applied at the free end. The
2m shear modulus and polar moment of inertia
(A) 0.0 MPa (B) 0.51 MPa of the section are denoted as G and J,
(C) 0.815 MPa (D) 2.0 MPa respectively. The torsional rigidity GJ is
6.3 A closed thin-walled tube has thickness t,
50,000 kN.m 2 /rad . The following are
mean enclosed area within the boundary of
reported for this shaft:
the centerline of tube's thickness is Am and
Statement i) The rotation at the free end is
shear stress τ . Torsional moment of 0.01 rad
resistance T, of the section would be
Statement ii) The torsional strain energy is
(A) 2τAm t (B) 4τAm t 1.0 kN.m
(C) τAmt (D) 0.5τAmt With reference to the above statements,
2021 IIT Bombay which of the following is true?
(A) Both the statements are correct
6.4 A solid circular torsional member OPQ is
(B) Statement i) is correct, but Statement ii)
subjected to torsional moments as shown in
is wrong
the figure (not to scale). The yield shear
(C) Statement i) is wrong, but Statement ii)
strength of the constituent material is 160
MPa. [1 Mark] is correct
(D) Both the statements are wrong
A B
6.1 23
Z
Given : Outer diameter of the shaft T
( D ) = 100 mm A B
Inner diameter of the shaft (d ) = 50 mm
Allowable shear stress (τ) = 125 MPa Scan for
Video Solution
Let the maximum torque be (Tmax )
6.3 (A)
π   D4 − d 4 
Tmax = τ  
16   D   Torsional moment of resistance T is given by,
τ× J  TR 
π   100 4 − 50 4   T=  τ = 
= 125   R  J 
16   100 
τ× 2πR3t
= 23009711.82 Nmm T= = τ× 2πR 2t
R
Tmax = 23009 Nm = 23kNm T = 2τAmt ( πR 2
= Am )
Hence, the maximum torque that the shaft can
resist is 23 kN-m.
6.4 15.279
6.2 (A)
2 kNm
According to question, d 1 = 0.1 m
x 3 kNm 1 kNm
T Material element aligned with
O P
cylindrical coordinate axes 5m
1 kNm
y Radial d 2 = 0.08 m
direction p 1 kNm
p P Q
T 2m
All other stress
16TOP 16 × 3 ×106
components are (τmax )OP = =
zero in longitudinal π× d13 π× (100)3
direction l
The only non-zero stresses are τθz = τ zθ = τ, , if θ (τmax )OP = 15.279 MPa
is 900 then θ = y . 16TpQ 16 ×1×106
(τmax ) PQ = =
16 T π× d 23 π× (80)3
∴ τ zy = τ yz = τmax =
πd 3
(τmax ) PQ = 9.947 MPa
3
16 × 20 ×10
= = 0.815 MPa ∴ Absolute maximum shear stress in member
π(2 × 0.25)3
= max  ( τ max )OP , ( τ max ) PQ 
But in rest of the planes shear stresses are zero.
So, τ xy = τ yx = 0 = 15.279 MPa
6.5 (A)
Thin cylindrical pressure vessel

pd
Hoop stress, σ hoop = = 30 MPa
2t
Longitudinal stress,
pd 30
σlongitudinal = = = 15 MPa
2t 2
Maximum shear stress
 σ σ σ − σ2 
= max  1 , 2 , 1 
2 2 2 
 30 15 30 − 15 
= max  , , 
2 2 2 
= 15 MPa
6.6 (B)
T Gθ
(i) We know, =
J L
Rotation at the free end is given as
TL 100 ×103 × 5 500
θ= = 3
= = 0.01
GJ 50000 ×10 50000
∴ θ = 0.01 rad
1 T 2L
(ii) Strain energy = × T × θ =
2 2GJ
(100) 2 × 5
=
2 × 50000 = 0.5 kN-m
Thus, statement I is true and Statement II is False.
7 Theory of Columns and Shear Centre
2013 IIT Bombay PL PL

(A) (B)
4k 2k
7.1 Two steel column P (length, L and yield P Pk
strength, f y = 250 MPa ) and Q (length, 2 (C) (D)
4k 4L
L and yield strength, f y = 500 MPa ) have 2017 IIT Roorkee
the same cross-section and end condition 7.4 Consider two axially loaded columns,
the ratio of buckling load of column P to namely 1 and 2, made of a linear elastic
that of column Q is material with Young’s modulus
5
[1 Mark] 2 ×10 MPa , square cross-section with
(A) 0.5 (B) 1.0 side 10 mm, and length 1 m. For column 1,
(C) 2.0 (D) 4.0 one end is fixed and the other end is free.
For column 2, one end is fixed and the other
2014 IIT Kharagpur end is pinned. Based on the Euler’s theory,
the ratio (up to one decimal place) of the
7.2 The possible location of shear centre of the
buckling load of column 2 to the buckling
channel section, shown below is
load of column 1 is ________.
[1 Mark]
[2 Marks, Set-I]
2018 IIT Guwahati
7.5 A column of height h with a rectangular

P Q R S
cross-section of size a × 2 a has a buckling
load of P. If the cross-section is changed to
0.5a × 3a and its height changed to 1.5h,
the buckling load of the redesigned column
(A) P (B) Q will be [1 Mark]
P P
(C) R (D) S (A) (B)
12 4
2015 IIT Kanpur
P 3P
(C) (D)
7.3 In a system two connected rigid bars AC 2 4
and BC are of identical length, L with pin 2021 IIT Bombay
supports at A and B. The bars are
interconnected at C by a frictionless hinge. 7.6 An elevated cylindrical water storage tank
The rotation of the hinge is restrained by a is shown in the figure. The tank has inner
rotational spring of stiffness, k. The system diameter of 1.5 m. It is supported on a solid
initially assumes a straight line steel circular column of diameter 75 mm
configuration, ACB. Assuming both the and total height (L) of 4 m. Take water
bars as weightless, the rotation at supports, density 1000 kg/m3 and acceleration due
A and B, due to a transverse load, P applied to gravity 10 m/s 2 .
at C is [2 Marks]
6 EI
h (B) 2 rad/s
mL3
EI
(C) 6 rad/s
mL3
1 2 EI
E, I (D) rad/s
L L m
7.8 Consider the cross-section of a beam made
up of thin uniform elements having
thickness t (t  a) shown in the figure. The
If elastic modulus (E) of steel is 200 GPa, (x, y) coordinates of the points along the
ignoring self-weight of the tank. For the center-line of the cross-section are given in
supporting steel column to remain the figure.
unbuckled. The maximum depth (h) of the y
2a a
water permissible (in m, round off to one K (a , 3a )
H
decimal place) is _______.
( − 2 a , 3 a) t
[2 Marks]
7.7 A single story building model is shown in t
3a
the figure. The rigid bar of mass ‘m’ is
supported by three massless elastic
columns whose ends are fixed against
rotation. For each of the columns, the x
J (0, 0)
applied lateral force (P) and corresponding
moment (M) are also shown in the figure. The coordinates of the shear center of this
The lateral deflection (δ ) of the bar is cross-section are : [1 Mark]
(A) x = 0, y = 3 a
PL3
given by δ= , where L is the (B) x = 2 a , y = 2 a
12 EI
effective length of the column, E is the (C) x = − a , y = 2 a
Young’s modulus of elasticity and l is the (D) x = − 2 a , y = a
area moment of inertia of the column cross-
section with respect to its neutral axis.
Rigid bar
δ
M
P
Column
Lateral deflection
Building model profile of any column
For the lateral deflection profile of the
columns as shown in the figure, the natural
frequency of the system for horizontal
oscillation is [1 Mark]
2 EI
(A) rad/s
L m
7.1 (D) 7.3 (A)
Let, PP and PQ be the buckling load for column P According to question,
P
and Q respectively. For any loading condition, all L L
the factors i.e., E, π and I will be same and cancel A C B
q q
out, except length L.
Since, buckling load is independent of yield
2q
strength, we can write D
PP (2 L) 2 Deflection under the load, P = θL

= 2 =4 Now, work done by the load P is given by,
PQ L
1
Hence, the correct option is (D). (W ) = P(θL)
2
7.2 (A)
( θ = Rotation due to load P)
According to question, Strain energy stored in spring is given by,
F
1
U = [ k (2θ)] (2θ) = 2k θ2
V
h 2
P
e
Q R S Energy stored in spring will be equal to the work
done.
F
:Method 1: 1
∴ P(θL) = 2k θ2
When resultant passes through shear centre then 2
there will be no twisting moment. PL
θ=
So, for no twisting moment condition, 4k
Ve = Fh Hence, the correct option is (A).
Fh 7.4 8
e=
V Given : Young’s modulus of material
i.e., location of shear centre of given section will
be at point P. ( E ) = 2 ×105 MPa
Hence, the correct option is (A). Cross-section area of each column
:Method 2: ( A) = 10 ×10 = 100 mm 2
The given channel is un-symetrical about loading Length of each column (l ) = 1m
axis. If transverse load is applied through centroid
Let, P1 be the buckling load for column 1 and P2
(R) then torsion will also develop in section.
To avoid torsion, transverse load must be applied be the buckling load for column 2.
through shear centre P. Now, buckling load for column 1 (one end is fixed
Hence, the correct option is (A). and the other end is free) is given by,
 Key Point π2 EI
P1 = …(i)
The principle involved in loading the shear 4 L2
centre for a cross section of a beam is that the Buckling load for column 2 (one end is fixed and
loads acting on beam must lie in a plane which the other end is pinned) is given by,
contains the resultant shear force on each cross
2π2 EI
section of beam, as computed from the shearing P2 = …(ii)
stresses produced in the beam when it is loaded L2
so that it does not twist at its ends. Now, the ratio of P1 and P2
P1 2π2 EI 4 L2 π
= × 2 =8 FBottom = 1000 ×10 × h × ×1.52
P2 L2 π EI 4
Hence, the ratio (up to one decimal place) of the FBottom = 17671.45868h N
buckling load of column 2 to the buckling load of To avoid buckling, FBottom ≤ PE
column 1 is 8.0. 17621.45868 h ≤ 47903.22
7.5 (A) h ≤ 2.71 m
Given : Buckling load of column =P ∴ hmax = 2.7 m
Height of the column = h 7.7 (C)
Cross-section = a × 2 a
1. For one column
Load on the column, M
δ
 2a × a 3  P
π2 E   2 4
P=  12  = π Ea …(i) P 3
h2 6h 2 δ=
12 EI
Let, P0 be the buckling load of the redesigned
column then,
2. For combined system
 3a × (0.5a )3  δ
π2 E  
P0 =  12 P
(1.5h) 2
π2 Ea 4 1 π2 Ea 4
P0 = = ×
72h 2 12 6h 2
…(ii)
From equation (i) and (ii) we have, P3 P
δ= =
1 P 36 EI K
P0 = × P =
12 12 36EI
K= 3
Hence, the correct option is (A). 
7.6 2.7 K EI
w= =6 rad/sec
Given : Diameter of solid steel column, m m3
D = 0.075 m Hence, the correct option is (C).
Height of column, L = 4 m 7.8 (A)
Effective length of column, Le = 2 L = 8 m Y
2a a
By Euler’s formula,
X
π2 EI
PE =
L2e
π 3a
π2 × 200 ×109 × × 0.0754
PE = 64
82
PE = 47903.22 N
(0, 0)
π
FBottom = ρgh × ×1.52 Co-ordinates of shear centre = (0,3a )
4
STRUCTURAL
ANALYSIS
Syllabus : Structural Analysis
Statically determinate and indeterminate structures by force/energy methods; Method of

superposition; Analysis of trusses, arches, beams, cables and frames; Displacement methods:
Slope deflection and moment distribution methods; Influence lines; Stiffness and flexibility
methods of structural analysis.
Contents : Structural Analysis
S. No. Topics
1. Determinacy & Indeterminacy (Static & Kinematic)
2. Force and Energy Methods
3. Displacement Method of Analysis
4. Analysis of Trusses
5. Influence Line Diagram and Rolling Loads
6. Analysis of Arches
7. Matrix and Stiffness Method
8. Structural Dynamics
1 Determinacy & Indeterminacy (Static & Kinematic)
2014 IIT Kharagpur

1.1 The static indeterminacy of the two-span
continuous beam with an internal hinge,
shown below is ________. [1 Mark]
Internal hinge
(A) 11 (B) 8
(C) 3 (D) 0
2017 IIT Roorkee
1.2 The degree of static indeterminacy of a rigid
jointed frame PQR supported as shown in 1.5 A planar truss tower structure is shown in the
figure is [1 Mark] figure. [2 Marks]
S
Cable
y
0
45
R
EI
900
Q x
EI P
(A) Zero (B) One Consider the following statements about the
(C) Two (D) Unstable external and internal determinacies of the
truss.
2015 IIT Kharagpur
P. Externally determinate
1.3 A guided support as shown in the figure Q. External static indeterminacy = 1
below is represented by three springs R. External static indeterminacy = 2
(horizontal, vertical and rotational) with S. Internally determinate
stiffness k x , k y and kθ respectively. The T. Internal static indeterminacy = 1
limiting values of k x , k y and kθ are U. Internal static indeterminacy = 2
Which one of the following options is
[1 Mark] correct?
(A) P-False, Q-True; R-False; S-False; T-
(A) ∞, 0, ∞ (B) ∞, ∞, ∞ False; U-True
(B) P-False, Q-True; R-False; S-False; T-
(C) 0, ∞, ∞ (D) ∞, ∞, 0
True; U-False
2016 IISc Bangalore (C) P-False, Q-False; R-True; S-False; T-
1.4 The kinematic indeterminacy of the plane False; U-True
truss shown in the figure is (D) P-True, Q-True; R-False; S-True; T-
[1 Mark] False; U-True
Structural Analysis 1
1.6 Consider the frame shown in the figure Hinge
2022 IIT Kharagpur

If the axial and shear deformations in
different members of the frame are assumed 1.8 Consider a beam PQ fixed at P hinged at Q
to be negligible, the reduction in the degree and subjected to a load F as shown. The
of kinematic indeterminacy would be equal static and kinematic degree of indeterminacy
to [1 Mark] respectively are
(A) 5 (B) 6 F
P
(C) 7 (D) 8 θ Q
2019 IIT Madras
(A) 1 and 2 (B) 2 and 0
1.7 The degree of static indeterminacy of the
(C) 2 and 2 (D) 2 and 1
plane frame as shown in the figure is _____
[1 Mark]
[1 Mark]
1.1 0 1.2 (A)
According to question, According to question,

Internal hinge S
Cable
0
Note : The beam given is having only vertical 45
R
load hence, there will be no horizontal reaction.
∴ The number of equilibrium equation = 2
EI
Number of external reaction, R = 3
Degree of external indeterminacy,
900 P
Dse = R − 2 = 3 − 2 = 1 Q
EI
Degree of internal indeterminacy,
Dsi = 3C − rr Ds = Dse + Dsi
Where, C = Number of closed loop = 0. Dse = R − 3
rr = (m '− 1) = (2 − 1) = 1 Number of external reaction = 4
Dsi = 3 × 0 − 1 = −1 Dse = 4 − 3 = 1
Dsi = 3C − rr
Hence, Ds = Dse + Dsi = 1 − 1 = 0
Where, C = Closed loop = 0
Note : Stable for vertical loading and unstable
rr = Σ(m '− 1) = 2 − 1 = 1
for horizontal loading.
Dsi = 3 × 0 − 1 = −1
Hence, the static indeterminacy of the two-span
continuous beam with an internal hinge, shown Ds = Dse + Dsi = 1 − 1 = 0
below is 0. Hence, the correct option is (A).
2 Structural Analysis
We know that, for plane truss, kinematic
Scan for
Video Solution
indeterminacy,
Dk = 2 j − R
1.3 (A) Where, total number of joint, j = 7
According to question, Total number of external reaction, R = 3
∴ Dk = 2 × 7 − 3 = 14 − 3 = 11
The give support is guided roller support which Hence, the correct option is (A).
has 2 reaction.
1.5 (A)
1. Horizontal and
2. Moment i.e. θ = 0 According to question,
Stiffness in x-direction,
Load R
Kx = = x
Deflection Δ
∴ There is restriction in x-direction Δ = 0.
R
Kx = x = ∞
0
Stiffness in y-direction,
Load R R
Ky = = y = y
Deflection Δ ∞
 There is no restriction in y-direction Δ → ∞
.
Ry Degree of external static indeterminacy,
Ky = =0 Dse = R − Equilibrium equation
∞
Similarly, stiffness in θ direction, Where, R = Number of support reaction = 4
Moment M θ Equilibrium equation = 3
Kθ = =
Rotation θ Dse = 4 − 3 = 1
∴ θ is restricted i.e. θ = 0 .
M (Externally indeterminacy)
Kθ = θ = ∞ ∴ External static indeterminacy is equal to 1.
0
Hence, K x , K y , K θ are respectively ∞, 0, ∞ . Degree of internal static indeterminacy,
Dsi = m − (2 j − 3) (for plane truss)
Scan for
Video Solution Where, Number of members in the structures
m = 15
1.4 (A) Number of joints in the structure j = 8
According to question, Dsi = 15 − (2 × 8 − 3) = 2
(Internally indeterminate)
And internal static indeterminacy is 2.
Scan for
Video Solution
1.6 (B) rr = 2 − 1 = 1
According to question, Dsi = 3 × 4 − 1 = 11
Ds = Dse + Dsi = 4 + 11 = 15
Hence, the degree of static indeterminacy of the
plane frame as shown in the figure is 15.
..Method 2..
By formula, Ds = 3m − (3 j − R ) − rr
Given that, the frame shown in figure is rigid.
(for plane frame)
Hence, the degree of kinematic indeterminacy,
Where, number of member in structure, m = 22
Dk = (3 j − R) − m " Number of joints in structure, j = 19
Where, number of joints in the structure, j = 6 Total reaction acting at support, R = 7
Total support reaction, R = 4 rr = Σ(m '− 1)
Number of inextensible (rigid) members m " = 6 Where, m ' = Number of member meeting at
Note : In question reduction of kinematic hybrid joint.
indeterminacy is asked when member are rr = 2 − 1 = 1
assumed to be rigid.
Hence, Ds = 3 × 22 − (3 ×19 − 7) − 1 = 15
Case 1 : When member are extensible,
Hence, the degree of static indeterminacy of the
Dk1 = 3 j − R = 3 × 6 − 4 = 14
plane frame as shown in the figure is 15.
(members are extensible)
Scan for
Case 2 : When member are inextensible (rigid), Video Solution
then
Dk2 = 3 j − R − m " 1.8 (D)
Dk 2 = 3 × 6 − 4 − 6 = 8 Given :
Reduction in Dk = Dk − Dk = 14 − 8 = 6
1 2 θ
Q
1.7 15
..Method 1..
Degree of external indeterminacy,
Dse = R − Available equilibrium equation
Total external reaction at support, R = 5 (reactions)
R = 3+ 2+ 2 = 7 Equilibrium reaction = 3
Degree of static indeterminacy,
Dse = 7 − 3 = 4
DSI = R − 3  2
Degree of internal indeterminacy, Kinematic indeterminacy, KI = ?
Dsi = 3C − rr  At simply supported only one chances of
Where, C = Number of closed loop = 4. rotation or of is free to rotate.
rr = m '− 1 So, KI = 1
Where, m ' = Number of member meeting at

hybrid joint
2 Force and Energy Methods
2013 IIT Bombay Under the action of a concentrated load P at

C as shown, the magnitude of tension
2.1 A uniform beam (EI = constant) PQ in the developed in the rope is
form of a quarter circle of radius R is fixed at
3P P
end P and free at the end Q, where a load W (A) (B)
is applied as shown. The vertical downward 2 2
displacement δQ at the loaded point Q is 3P
(C) (D) 2P
 WR3  8
given by δQ = β   . Find the value of β 2.4 Two beams PQ (fixed at P and with a roller
 EI  support at Q, as shown in figure I, which
correct to 4-decimal places. [2 Marks] allows vertical movement) and XZ (with a
W
Q hinge as Y) are shown in the figures I and II
respectively. The spans of PQ and XZ and L
and 2L respectively. Both the beams are
R under the action of uniformly distributed
load (w) and have the same flexural stiffness,
P
EI (where, E and I respectively denote
2014 IIT Kharagpur modulus of elasticity and moment of inertia
2.2 The axial load (in kN) in the member PQ for about axis of bending). Let the maximum
the arrangement/assembly shown in the deflection and maximum rotation be δ max1
figure given below is _______. [2 Marks]
P and θ max1 , respectively, in the case of beam
PQ and the corresponding quantities for the
beam XZ be δ max 2 and θ max 2 , respectively.
2m 160 kN
[2 Marks]
w
Q R Q
2m 2m P
2016 IISc Bangalore L
2.3 An assembly made of a rigid arm ABC Figure - I
w
hinged at end A and supported by an elastic
rope CD at end C is shown in the figure. The X Z
Hinge Y
members may be assumed to be weightless L L
and the lengths of the respective members
are as shown in the figure. [1 Mark] Figure - II
A D Which one of the following relationship is
true?
(A) δ max1 ≠ δ max 2 and θ max1 ≠ θ max 2
L
Rigid arm
P L (B) δ max1 = δ max 2 and θ max1 ≠ θ max 2
Rope
(C) δ max1 ≠ δ max 2 and θ max1 = θ max 2
B C (D) δ max1 = δ max 2 and θ max1 = θ max 2
L L
2019 IIT Madras the loads acting at joints L and M are along
the positive X direction. [2 Marks]
2.5 A portal frame shown in figure (not drawn to Y
10 kN 10 kN 10 kN
scale) has a hinge support at joint P and a
roller support at joint R. A point load of 50 I J K
kN is acting at joint R in horizontal 1m

R N P S
direction. The flexural rigidity EI, of each X
member is 106 kNm2 , under the applied 1m
load, the horizontal displacement (in mm, L 10 kN Q M 10 kN

1m 1m 1m 1m 1m 1m
round off to 1 decimal place) of joint R
would be ______. The magnitude of the horizontal component
P
5m
Q of reaction (in kN) at S, is
EI (A) 10 (B) 5
(C) 20 (D) 15
EI 10 m 2023 IIT Kanpur
2.7 An idealised frame supports a load as shown
R 50 kN in the figure. The horizontal component of
the force transferred from the horizontal
member PQ to the vertical member RS at P
2020 IIT Delhi
is ______ N (round off to one decimal
2.6 Joints I, J, K, L, Q and M of the frame shown place).
in the figure (not drawn to the scale) are 1.2 m 0.6 m
pins. Continuous members IQ and LJ are
S
connected through a pin at N. Continuous 0.3 m U
members JM and KQ are connected through P Q
θ
a pin at P. The frame has hinge supports at 1.0 m
joints R and S. The loads acting at joints I, J
and K are along the negative Y direction and 1.0 m T 10 N
R
2.1 0.7854 m1 = Bending moment at any section due to

W unit load applied at the location of desired
deflection
Q
x = R sin q M = − Wx (due to given loading)
dx y
x m = −1× x (due to unit load)
R
R R cos q
q x = R sin θ , y = R − R cos θ
q dqr
O and dx = Rd θ
P R
Then, M = −WR sin θ and m = − R sin θ
Using unit load method,
π/2 ( − WR sin θ)( − R sin θ) R
Mm1 δQ =  dθ
Δ= dx 0 EI
EI
WR 3 π /2
Where, M = Bending moment at any section due
to given loading,
δQ =
EI 
0
sin 2 θd θ
WR 3 π /2 (1 − cos 2θ) Note :
δQ =
EI 0 2
dθ P
B C
 2 1 − cos 2θ  A
 sin θ =
SB S1
2 
qB qB L
3 π /2
WR  sin 2θ  L L
δQ = θ − 2 
2EI 0 δC = δ B + θ B L
π  wR3  Hence, the axial load 50 kN.
δQ =  
4  EI  2.3 (B)
π A D
On comparing, β = = 0.7854
4
Hence, the value of β is 0.7854.
2.2 50
L L
Rigid arm P
According to question, Rope
P
C
B
L L
In above problem under the action of concentrated
2m 160 kN
load P at C, the tensile force (T ) will develop in
the rope which is inclined at 450 to the horizontal.
Q R
2m 2m A
From the given problem axial load in the member
PQ can be found out by considering PQ to be T sin450
T
rigid. Hence deflection at Q will be zero. L
By drawing free body diagram,
For point Q, 450
T cos45
0
B C
L
P
R  L
160 kN 0
R  tan 45 = 1, tan θ = L 
Q Taking moment about hinge A,

2m 2m ΣM A = 0 ,
At Q,
T cos 450 × L + T sin 450 × L − P × L = 0
Upward deflection due to reaction R
TL TL
= Downward deflection due to 160 kN load + = PL
2 2
R (4)3 160(2)3 160(2) 2
= + ×2 P
3EI 3EI 2 EI T=
64 R 160 × 8 2
= + 160 × 4
3 3 ∴ The magnitude of tension developed in the
R = 50 kN P
rope is .
∴ Axial load in PQ = 50 kN 2
2.4 (D)  Key Point
M
According to question, P
w qP qQ
Q
Q
P ML ML
θP = and θQ =
L 6 EI 3 EI
L P
Figure - I
Q
w and R
3 2
X Z PL PL
Hinge Y ΔR = and θ R =
L L 3EI 2 EI
Figure - II Hence, the horizontal displacement of joint R

would be 25 mm.
In figure-I, at point Q there is a roller with a hinge
in the vertical plane, which will provide both ..Method 2..
rotation and vertical deflection at the support. Using strain energy method,
P 5m
Similarly in figure-II, at point Y, which is an HP Q
internal hinge and vertical deflection are possible.
Due to symmetry of loading and boundary VP
10 m
conditions figure-I is the free body diagram of left
part of figure-II.
Hence, the correct option is (D). R HR = 50 kN
2.5 25
VR
..Method 1..
dM x
50 ´ 10 Mx ×
δU dH R
qQ ΔR = = Σ dx = 0
qQ δH R EI
Þ ΣFx = 0 , H P = 50 kN
Taking moment about joint ' R ' ,
R 50 kN ΣM R = 0 ,
qQ ´ 10 50 ´ 103
3EI VP × 5 − H P ×10 = 0
Total horizontal deflection of joint R, 50 × 10
VP = = 100 kN
5
= ΔR + θ Q × L
Considering member PQ ,
x
PL3  ML '  50
= + × L
3EI  3EI 
50 × (10)3 (50 ×10) × 5 100
= + × 10
3EI 3EI M x = 100 x
dM x 100 x
50 ×103 500 × 5 = = 2x
= + ×10 dH R 50
3 ×106 3 ×106
Limit 0 to 5 m.
= 0.025 m = 25 mm
Considering member QR ,
Q
50
P 5m xs = 6 2 cos θ
Virtual displacements of all the points.
100
x δyI = 2 cos θ d θ, δy j = 2 cos θ d θ,
10 m
δyk = 2 cos θ d θ
δxL = − 2 sin θ d θ,
δxm = −5 2 sin θ d θ,
M x = 100 × 5 − 50 x δxs = −6 2 sin θ d θ
dM x
= 10 − x Principle of virtual work to find unknown
dH R horizontal force H S .
Limit 0 to 10 m. δU = 0
δU
ΔR = = −10 × 2 cos θ d θ × 3 + 10 ×− 2 sin θ d θ
δH R
5 (100 x )(2 x ) 10 (500 − 50 x )(10 − x ) + 10 ×− 5 2 sin θ d θ −  Hs ×− 6 2 sin θ d θ
ΔR =  dx +  dx
0 EI 0 EI
30 2 cos θ + 10 2 sin θ + 50 2 sin θ
50 x 3
10 Hs =
5 5000 x + 500 x + 2 6 2 sin θ
200( x)3 3
ΔR = + 0
Put, θ = 45 ,
3EI 0 EI 0
90
Δ R = 0.025 m = 25 mm Hs = = 15 kN
6
Hence, the horizontal displacement of joint R Hence, the correct option is (D).
would be 25 mm.
2.7 18
Scan for
1
Video Solution tan θ =
1.2
2.6 (D)  1 
θ = tan −1   = 39.80°
Degree of indeterminacy = 4 − 3 = 1  1.2 
Y
10 kN 10 kN 10 kN FBD :
VS
P F cos θ U
I J K Q
2m HP θ
1m
R q N P S F
HR HS X
10 N
1m
F sin θ
L 10 kN Q M 10 kN
1m 1m 1m 1m 1m 1m Taking moment about P = 0
Hence, for determinate structure hinge support at  ΣM p = 0
S is replace with roller support as shown in the  F sin θ×1.2 = 10 × (1.2 + 0.6)
figure.  F = 23.43 N
Coordinates of all the points where forces are
Horizontal component of force transferred from
acting.
PQ to member RS atP
yI = 2 sin θ, y j = 2 sin θ,
= F cos θ = 23.43 cos(39.805) = 18 N
yk = 2 sin θ Hence, the correct answer is 18.
xI = 2 cos θ, xm = 5 2 cos θ, 
3 Displacement Method of Analysis
20 kN
2013 IIT Bombay
3m 1m 1m 3m
3.1 All members in the rigid-jointed frame M
shown are prismatic and have the same EI EI EI
flexural stiffness EI. Find the magnitude of Internal hinge
the B.M. at Q (in kNm) due to given loading.
3.4 Considering the symmetry of a rigid frame
[2 Marks] as shown below, the magnitude of the
3m 4m bending moment (in kNm) at P (preferably
S
using the moment distribution method) is
2m [2 Marks]
P T R 24 kN/m
100 kNm
4Ic P 4Ic
2m
6m I Ic Ic
c
Q
2014 IIT Kharagpur 8m 8m

3.2 For the cantilever beam of span 3 m (shown (A) 170 (B) 172
below), a concentrated load of 20 kN applied (C) 176 (D) 178
at the free end causes a vertical displacement
of 2 mm at a section located at a distance of 2017 IIT Roorkee
1 m from the fixed end. If a concentrated
3.5 The value of M in the beam ABC shown in
vertically downward load of 10 kN is applied
the figure is such that the joint B does not
at the section located at a distance of 1 m
rotate.
from the fixed end (with no other load on the
beam), the maximum vertical displacement [2 Marks]
in the same beam (in mm) is ______. 30 kN/m M
[2 Marks] A B C
20 kN
2 mm
4m 6m
The value of support reaction (in kN) at B
should be equal to ______.
1m 2m 3.6 Consider the portal frame shown in the
figure and assume the modulus of elasticity,
2015 IIT Kanpur
E = 2.5 × 10 4 MPa and the moment of
3.3 For the beam shown below, the value of the
inertia, I = 8 × 108 mm 4 for all the members
support moment M is _____ kNm.
of the frame.
[1 Mark]
P [2 Marks]
2000 kN
1650 L P
(E, I) 2m Q
kN/m R
EI
Q (E, I) S
(E, I) 4m EI L
R
4m S
The rotation (in degrees) up to one decimal (A) 7.5 (B) 3.0
place at the rigid joint Q would be ______.
(C) 48.0 (D) 0.1
[2 Marks]
2020 IIT Delhi
2018 IIT Guwahati
3.7 A vertical load of 10 kN acts on a hinge 3.10 The planar structure RST shown in the figure
located at a distance of L / 4 from the roller is roller-supported at S and pin-supported at
support Q of a beam of length L (see figure) R. Members RS and ST have uniform
[1 Mark] flexural rigidity (EI) and S is a rigid joint.
10 kN Consider only bending deformation and
neglect effects of self-weight and axial
P Q stiffening.
[2 Marks]
3L/4 L/4 P T
The vertical reaction at support Q is
L/2
(A) 0.0 kN (B) 2.5 kN
(C) 7.5 kN (D) 10.0 kN R S
3.8 A prismatic beam PQR of flexural rigidity
EI = 1 × 10 4 kNm 2 is subjected to a moment L
When the structure is subjected to a
of 180 kNm at Q as shown in the figure.
concentrated horizontal load P and the end
180 kNm
P R
T, the magnitude of rotation at the support R,
Q is
5m 4m PL2 PL
(A) (B)
The rotation at Q (in radian) up to two 12 EI 6 EI
decimal places is _______. [2 Marks] PL 3
PL2
(C) (D)
2019 IIT Madras 12 EI 6 EI
2021 IIT Bombay
3.9 The rigid jointed plane frame QRS shown in
figure is subjected to load P at the joint R. 3.11 A square plate O-P-Q-R of a linear elastic
Let the axial deformations in the frame be material with sides 1.0 m is loaded in a state
neglected. If the support S undergoes a of plane stress. Under a given stress
PL3 condition, the plate deforms to a new
settlement of Δ = , the vertical reaction configuration O-P’-Q’-R’ as shown in the
β EI
figure (not to scale). Under the given
at the supports S will be become zero when deformation the edges of the plate remain
β is equal to straight.
y 10 mm
H
Q'
R'
10 mm
R 5m
Q EI
P' F
10 mm M
O x
P
20 mm EI EI 4m
The horizontal displacement of the point (0.5
m, 0.5 m) in the plate O-P-Q-R (in mm, G
K
round off to one decimal place) is _______.
3m 3m
[2 Marks]
If the magnitude (absolute value) of the
3.12 A frame EFG is shown in the figure. All support moment at H is 10 kN-m, the
members are prismatic and have equal magnitude (absolute value) of the applied
flexural rigidity. The member FG carries a moment M (in kN-m) to maintain static
uniformly distributed load w per unit length. equilibrium is _____. (round off to the
Axial deformation of any member is nearest integer)
neglected. [2 Marks]
L 3.14 Which of the following statement(s) is/are
w
correct? [2 Marks]
F (A) If a linearly elastic structure is subjected
G
to a set of loads, the partial derivative of
the total strain energy with respect to the
deflection at any point is equal to the
2L load applied at that point.
(B) If a linearly elastic structure is subjected
to a set of loads, the partial derivative of
E the total strain energy with respect to the
load at any point is equal to the
Considering the joint F being rigid, the
deflection at that point.
support reaction at G is
(C) If a structure is acted upon by two force
[2 Marks] system Pa and Pb , in equilibrium
(A) 0.500 wL (B) 0.375 wL separately, the external virtual work
(C) 0.453 wL (D) 0.482 wL done by a system of forces Pb during the
2022 IIT Kharagpur deformations caused by another system
of forces Pa is equal to the external
3.13 Consider the linearly elastic plane frame
shown in the figure. Members HF, FK and virtual work done by the Pa system
FG are welded together at joint F. Joints K, during the deformation caused by the Pb
G and H are fixed supports. A counter- system.
clockwise moment M is applied at joint F. (D) The shear force in a conjugate beam
Consider flexural rigidity EI = 105 kN-m 2 loaded by the M/EI diagram of the real
for each member and neglect axial beam is equal to the corresponding
deformations. deflection of the real beam.
2023 IIT Kanpur reaction at A is wL2 /k . What is the value
3.15 For the frame shown in the figure (not to of 𝑘 (in integer)? _________.
scale), all members (AB, BC, CD, GB, and
w w
CH) have the same length, 𝐿 and flexural
rigidity, 𝐸𝐼. The joints at B and C are rigid
G B C H
joints, and the supports A and D are fixed
supports. Beams GB and CH carry L
uniformly distributed loads of 𝑤 per unit
A D
length. The magnitude of the moment
L L L
3.1 25 Hence, the magnitude of the B.M. at Q due to

given loading 25 kN-m.
3m 4m
S 3.2 1
20 kN
2m 2 mm
P T R
100 kNm
2m 1m 2m
Using Betti’s law,
Q 10 × 2 = 20 × Δ
Δ = 1mm
3EI
Stiffness of member TP, ( KTP ) = Hence, the maximum vertical displacement at the
L
end when load is placed at 1 m from the fixed
Stiffness of member TS , ( K TS ) = 0
support is 1 mm.
4 EI 4 EI Note : Betti’s theorem, the virtual work done by
Stiffness of member TQ, ( KTQ ) = =
L 2 a P-force system is equal to the virtual work done
= 2EI by Q-force system in going through the
4 EI 4 EI deformation of P-force system.
Stiffness of member TR, ( KTR ) = = P1
L 4
= EI y1
ΣK = EI + 0 + 2EI + EI = 4EI P2
Distribution factor for member, y2

KTQ 2 EI 1
TQ = = = Then, P1 y 2 = P2 y1
ΣK 4 EI 2
∴ M TQ = ( DF )TQ × M 3.3 5
20 kN
1 3m 1m 1m 3m
M TQ = × 100 = 50 kN-m
2 M
EI EI EI
Moment at ' Q ' is carry over moment of M TQ ,
Internal hinge
1 Due to symmetry the load of 20 kN acting at a
M Q = COF × M TQ = × 50 = 25 kN-m
2 hinge will be distributed equally to both side of
magnitude 10 kN. BM releases when there is 2
EI C
internal hinge. 1
Distribution factor for BA = 3 =
Then, 8EI C 4
10 kN
3
3m 1m
2 EI C 3
M Distribution factor for BP = =
8 EI C 4
M
3
10 kN-m Fixed end moment, M BA = 0 and M AB = 0 .
1  1 wL2 24 × 82
M = × 10 = 5 kN-m COF =  M BP = − =− = −128kN-m
2  2 12 12
Hence, the value of the support moment M is 5 wL2 24 × 82
kN-m. M PB = = = 128kN-m
12 12
A B P
Scan for 1/4 3/4
Video Solution
FEM 0 0 – 128 + 128
32 96
3.4 (C) Carry 16 48
According to question, over
24 kN/m Final 16 32 – 32 176
4Ic P 4Ic
Scan for
6m I Ic Ic
c Video Solution
8m 8m 3.5 60
From given problem we can see that, axis of According to question,
symmetry is passing through a column, hence, it 30 kN/m M
can be treated as, B C
A
P
B
4Ic 4m 6m
8m Since, due to the moment ' M ' the joint B does
6m I
c not rotate, it will behave like a fixed support and
A member BC shall be treated as a null beam since,
there is no loading.
Stiffness of member BA, 30 kN/m
4 EI 4 EI C 2
K BA = = = EI C A B
L 6 3
Stiffness of member BP , RA 4m RB
4 EI 4 E × 4 I C R A + RB = 30 × 4
K BP = = = 2 EI C
L 8 Due to symmetry R A = RB
2 8 EI C 2 RB = 30 × 4
ΣK = EI C + 2 EI C =
3 3
RB = 60 kN
 Ki 
 D.F . =  The value of support reaction at B should be equal
 ΣK i 
to 60 kN.
Scan for 3.7 (A)
Video Solution According to question,
10 kN
3.6 1.0
P RP 3L/4 L/4
2000 kN RQ
1650 2m Taking moment about the internal hinge from right
(E, I)
kN/m side,
Q (E, I) S L
RQ × = 0
4
(E, I) 4m RQ = 0
R
Scan for
4m Video Solution
Moment at ' Q ' due to given loading,
2 3.8 0.01
M Q = 2000 × 2 − 1650 × 2 ×
2 According to question,
M Q = 700 kN-m 180 kNm
P R
EI = 2.5 × 10 4 × 8 × 108 Q
EI = 20 × 1012 Nmm 2 = 20000 kNm 2 5m 4m

4 EI : Method 1 :
Stiffness of member QR, ( K QR ) =
L 4 EI 4 EI
Stiffness of member QP, ( K QP ) = =
4 EI L 5
= = EI
4 4 EI 4 EI
Stiffness of member QR, ( K QR ) = =
4 EI 4 EI L 4
Stiffness of member QS , ( K QS ) = =
L 4 = EI
= EI Then, stiffness of joint’s ' Q ' ,
ΣK = 2 EI = Stiffness at joint ' Q ' K Q = ΣK = K QP + K QR
Rotation at rigid joint, 4 9 EI
KQ = EI + EI =
Moment at rigid joint 5 5
θ=
Stiffness at that joint Rotation at joint ' Q ' ,
MQ 700 M 180
θ= = = 0.0175 rad θQ = Q = = 0.01rad
KQ 2 × 20000 K Q 9 × 1× 104
5
0.0175 ×1800 Hence, the rotation at Q up to two decimal places
θ= = 1.002670
π is 0.01 rad.
θ = 1.00 : Method 2 :
Hence, the rotation up to one decimal place at the 2 EI
rigid joint Q would be 1.00. M QP = M FQP + (2θQ + θ P )
L
Scan for ( M FQP = 0, θ P = 0)
Video Solution
2 ×104 VQ = VR
= 0+ × (2θQ + 0) = 8000 × θQ
5 Taking moment about Q.
2 EI ΣM Q = 0 ,
M QP = M FQP + (2θQ + θ R )
L 3EI Δ 4.5 EI Δ
( M FQR = 0, θ R = 0) − 2
+ VR × L − =0
L L2
2 ×104 7.5 EI Δ
= 0+ (2θQ ) = 10000 θQ P = VR =
4 L3
P
R 7.5EI Δ
Hence, sway force, P =
M QP M QP
Q
M QR M QR
L3
ΣM Q = 0 , PL3
Δ= …(i)
− M QP − M QR − 180 = 0 7.5EI
18000 × θQ = −180 PL3
On comparing with, Δ =
−180 βEI
θQ = = 0.01 radian β = 7.5
18000
[(– ve) = Anticlockwise] Hence, the vertical reaction at the supports S will
3.9 (A) be become zero when β is equal to 7.5.
Given that the support S undergoes a settlement of : Method 2 :
Δ , then R will also sinks by ' Δ ' . Slop deflection method
M QR P
: Method 1 :
6 EI Δ H
M QR = −
L2
M RS
6 EI Δ
M RQ =− 2 P H
L
and M RS = M SR = 0
Due to symmetry distribution factor at ' R ' is 0.5
H
and 0.5 using moment distribution method.
Q R S M SR
0.5 0.5
6 EI D 6EI D 0
Fixed end
- -
0 2 EI  3Δ 
moments L2 L2 M QR =  θR − 
Balancing 3EI D 3EI D L  L 
moment L2 L2 2 EI  3Δ 
Carry over 1.5 EI D 1.5 EI D M RQ =  2θR − 
moment L2 L2 L  L 
Find end
-
4.5EI D
-
3EI D 3EI D 1.5 EI D If reaction at S is equal to zero,
moment L2 L2 L2 L2 M RQ + M QR + P.L = 0
Considering QR member, 6 EI θ R 12 EI Δ
− + PL = 0
4.5EI D Q L2
R 3EI D
L
L2 L2 6 EI θ R 12 EI PL3
− 2 × + PL = 0
L L βEI
VQ VR = P 6 EI θR 12 PL
− + PL = 0 …(i)
ΣFy = 0 , L β
From equilibrium of joint, 3.11 2.5
M RQ + M RS = 0 y
8 EI θ R 6 EI Δ 10 mm
− 2 =0
L L
10 mm
6 EI θ R 6  6 EI PL3 
=  2 ×  R
L 8 L β EI 
6 EI θ R 36 PL
− …(ii) c
L 8β
From equation (i) and (ii),
36 PL 96 PL 10 mm
− + PL = 0 5 mm
8β 8β O
x
20 mm
60 PL
+ PL = 0 Horizontal displacement :
8β
10 mm
8β = 60
60
β= = 7.5 1
8 x=
3
Hence, the vertical reaction at the supports S will
be become zero when β is equal to 7.5.
c
Scan for (0.5, 0.5) m
Video Solution 2
1− x =
0.5 m 3
3.10 (A)
20 mm
10 20
− =−
x 1− x
1 x
=
2 1− x
1 − x = 2x
1
x=
3
AB 20
=
2 2
− 0.5
3 3
Neglecting axial deformations,
AB = 5 mm
 PL 
×L
ML  2  PL2 So, horizontal displacement of
θR = = =
6 EI 6 EI 12 EI 5
C= = 2.5 mm
(Anticlockwise) 2
Hence, the correct option is (A). Moving in horizontal direction,
3.13 60
10 mm
Given: Flexural rigidity ( EI ) = 10 5 kN-m 2
Magnitude of the support
e Moment at H = 10 kN-m
The magnitude (absolute value) of the applied
moment M is,
10 mm M
5 mm = 10 kNm
6
10 + 5 M = 6 ×10
Vertical displacement of C = = 7.5 mm
2 M = 60 kNm
3.12 (D)
3.14 (A), (B), (C)
∂U  As per betti’s theorem
Compatibility condition, =0
∂R P1 P2
∂  M 
2
  dx  = 0
∂R  2 EI  δ1 δ2
PA System
  wx 2 
2
 wL2 
2

  Rx −  2L 
RL −  
∂   2  2  P3 P4
  dx +   dx  = 0
∂R 2 EI 2 EI
 0
 PB System
  δ3 δ4
 wx 2   2 wL3  P1 S 3 + P2 S 4 = P3 S1 + P4 S 2
2
L 
Rx −  2L 
RL − 
 2  ( x ) dx +  2 
0 2 EI 0 2 EI ( L)dx = 0  As per Castigliano’s theorem-1
P
 2 wx  3
 2 wL  3
 Rx −  2 L  RL −  dx ∂U
L
 2   2  δ=
0 EI
+
0
EI
=0 ∂P
U = Total strain energy
3 4 3
RL w L
− × + RL2 (2 L) −
wL
(2 L) = 0 P = Load applied
3 2 4 2 ∂U
 =P
RL3 wL4 ∂S
− + 2 RL3 − wL4 = 0
3 8 As per Castigliano theorem.
RL3 wL4 9 3.15 6
2 RL3 + = + wL4 = wL4
2 8 8 E
w kN/m B C w kN/m
H
7 RL3 9 4 L L
= wL
3 8
L L
Axis of
27 symmetry
R= wL = 0.482wL
56 A D
Distribution factors of member BA and BC
2
wL
(Applied moment)
2 L
B 2
C'
k BC
2
k BA Imaginary frame
A
k
Join Membe D.F =
t r
k k k
B BA I 2
L 3 I  3
 
1 I  2L 1
BC’   3
2 L
Distribution of end moment →
From above, moment at A

wL2
= ... ( i )
6
Given moment at A
wL2
= ... ( ii )
k
From equation (i) and (ii), α = 6

4 Analysis of Trusses
P (0, 4)
2013 IIT Bombay 22.840
4.1 The pin-jointed 2D truss is loaded with a

80 kN
horizontal force of 15 kN at joint S and
another 15 kN vertical force at joint U as
shown in figure. Find the force in member y
RS (in kN) and. Report your answer taking
tension as +ve and compression as –ve. 104.030
x 53.130
[1 Mark]
4m 4m 4m Q (1, 0) R (3, 0)
R S (A) 30 kN compressive
15 kN
(B) 30 kN tensile
4m
(C) 50 kN compressive
V U T (D) 50 kN tensile
Q
4.4 For the truss shown below, the member PQ
4m 15 kN is short by 3 mm. The magnitude of vertical
P
displacement of joint R (in mm) is _______.
[2 Marks]
R
4.2 A uniform beam weighing 1800 N is
supported at E and F by cable ABCD.
3m
Determine the tension force in segment AB P Q
at this cable (correct to 1 decimal place). 4m 4m
Assume the cable ABCD, BE and CF are
weightless. [2 Marks] 2015 IIT Kanpur
A D 4.5 For the 2D truss with the applied loads
1m shown below, the strain energy in the
C member XY is _____ kN-m. For member XY,
B
assume AE = 30 kN, where A is cross-section
area and E is the modulus of elasticity.
E F [2 Marks]
5 kN
0.5 m 1m 2m 0.5 m 0.5 m 10 kN
2014 IIT Kharagpur 3m
4.3 Mathematical idealization of a crane has

three bars with their vertices arranged as 3m
shown in the figure with a load of 80 kN
X Y
hanging vertically. The coordinates of the
vertices are given in parentheses. The force 3m
in the member QR, FQR will be

[2 Marks]
3m
2016 IISc Bangalore Which one of the following sets gives the
correct values of the force, stress and change
4.6 Consider the plane truss with load P as
in length of the horizontal member QR?
shown in the figure. Let the horizontal and
[2 Marks]
vertical reactions at the joint B be H B and (A) Compressive force = 25 kN;
VB , respectively and VC be the vertical Stress = 250 kN/m2 ;
reaction at the joint C . [1 Mark] Shortening = 0.0118 m
L
A G (B) Compressive force = 14.14 kN;
0
L
60
0
60
Stress = 141.4 kN/m2 ;
P Extension = 0.0118 m
E F
L (C) Compressive force = 100 kN;
600 600
Stress = 1000 kN/m2 ;
B C
D Shortening = 0.0417 m
L L (D) Compressive force = 100 kN;
Which one of the following sets gives the Stress = 1000 kN/m2 ;
correct values of VB , H B and VC ? Extension = 0.0417 m
(A) VB = 0, H B = 0, VC = P 4.8 A plane truss with applied loads is shown in
the figure. [2 Marks]
P P
(B) VB = , H B = 0, VC = 20 kN
2 2 10 kN
10 kN J
P P 1m
(C) VB = , H B = P (sin 600 ), VC = H K
1m
2 2 G U V
L
F M 1m
0
(D) VB = P, H B = P (cos 60 ), VC = 0 E N 1m
T S R Q P
4.7 Consider the structural system shown in the
figure under the action of weight W. All the 2m 2m 2m 2m 2m 2m
joints are hinged. The properties of the The members which do not carry any force
members in terms of length (L), area (A) and are
the modulus of elasticity (E) are also given (A) FT, TG, HU, MP, PL
in the figure. Let L, A and E be 1 m, 0.05 m2 (B) ET, GS, UR, VR, QL
(C) FT, GS, HU, MP, QL
and 30 × 106 N/m2 , respectively, and W be (D) MP, PL, HU, FT, UR
100 kN.
2018 IIT Guwahati
P 4.9 Consider the deformable pin-jointed truss

L with loading, geometry and section
900 properties as shown in the figure.
A,
E
A,
2P
Joint C
450 2A, E 450
Q R P
450 450
E
A,
2AE AE L
A,
900
E
S
AE
W
L
Given that E = 2 ×1011 N/m2 , A = 10 mm2 , RR denote the vertical reactions (upward
L = 1 m and P = 1 kN. The horizontal positive) applied by the supports at P , Q and
displacement of joint C (in mm up to one R . Respectively on the truss. The correct
decimal place) is ______. [2 Marks] combination of ( RP , RQ , RR ) is represented
4.10 All the members of the planar truss (see
by
figure), have the same properties in terms of
area of cross-section (A) and modulus of [1 Mark]
elasticity (E).
2m
P 3m
P 30 kN
L Q R
3m 3m 3m
P
(A) (10,30, −10) kN (B) (30, −30,30) kN
L
(C) (20, 0,10) kN (D) (0, 60, −30) kN
For the loads shown on the truss, the
statement that correctly represents the nature 2020 IIT Delhi
of forces in the members of the truss is 4.13 Consider the planar truss shown in the figure
[1 Mark] (not drawn to the scale)
(A) There are 3 members in tension and 2 L L L
members in compression
(B) There are 2 members in tension 2
member in compression and 1 zero-force L
members P
(C) There are 2 members in tension 1
member in compression and 2 zero-force Neglecting self-weight of the members, the
members number of zero-force members in the truss
(D) There are 2 members in tension and 3 under the action of the load P, is [1 Mark]
zero force members (A) 7 (B) 6
2019 IIT Madras (C) 8 (D) 9
4.14 The plane truss has hinge supports at P and
4.11 A plane truss is shown in figure not drawn to
W and is subjected to the horizontal forces as
scale.
L
shown in the figure (not drawn to the scale)
20 kN 20 kN 2m 2m
1m
S V
20 kN K N 20 kN 10 kN Z
1m
I P
G S 1m
4m
1m
E T R U
F H J M O R 10 kN Y
2m 2m 2m 2m 2m 2m 2m
Which one of the following options contains 4m
ONLY zero force members in the truss? Q T X
10 kN
(A) FG , FI , HI , RS (B) FI , FG , RS , PR
(C) FG , FH , HI , RS (D) FI , HI , PR, RS 4m
4.12 Consider the pin-jointed truss shown in P W

figure (not drawn to scale). Let RP , RQ and
Q
Representing the tensile force with ‘+’ sign R
E, 2 A
and the compressive force with ‘–’ sign, the
force in member XW (in kN, round off to the E, A E, A E, A
nearest integer) is _______. E, A
[2 Marks] T
P S
2021 IIT Bombay E, A E, A
L L
4.15 A truss EFGH is shown in the figure, in
which all the members have the same axial F
rigidity R. In the figure, P is the magnitude If load, F = 10 3 kN, moment of inertia,
of external horizontal forces acting at joints
I = 8.33 × 106 mm 4 , area of cross-section,
F and G.
F G A = 104 mm 2 , and length, L = 2 m for all
P P the members of the truss, the compressive
stress (in kN/m2, in integer) carried by the
L
member Q - R is ________. [2 Marks]
2022 IIT Kharagpur
E H
4.17 The plane truss shown in the figure is
subjected to external force (P). It is given
L that P = 70 kN, a = 2 m and b = 3 m .
3
If R = 500 × 10 kN , P = 150 kN and L = 3
m, the magnitude of the horizontal
displacement of joint G (in mm round off to
one decimal place) is ________.
[1 Mark]
4.16 Refer the truss a shown in the figure (not to
scale). The magnitude (absolute value) of force (in
kN) in the member EF is ______. (Round off
to nearest integer).
[2 Marks]
2023 IIT Kanpur
4.18 An idealised bridge truss is shown in the figure. The force in Member U2L3 is ___________ kN
(round off to one decimal place).
20 kN 20 kN 20 kN 20 kN 20 kN
U0 U1 U2 U3 U4 U5 U6
3m
L1 L2 L3 L4 L5
6 @3 m = 18 m
4.1 0 4.2 1311.96
4m 4m 4m R2
R1
R S 15 kN A D
4m q
1m
C
V U T
Q q
B
4m 15 kN
P R1 E F
H2
0.5 m 1m 2m 0.5 m 0.5 m

R2
From concept of zero force member 1800
FQR and FQV are zero force member. Weight of bar i.e.1800 N will act at centre of bar
i.e. at a distance of 2 m from either end.
Taking moment about point ‘V’ from right side.
 MV = 0,  Fy = 0 ,
R1 × 8 − 15 × 4 − 15 × 4 = 0 R1 + R2 = 1800N …(i)
R1 = 15kN Taking moment about point ‘D’,
Now from method of section,  M D = 0,
X
S
R 15 kN R1 × 4 − 1800 × 2.5 = 0
R1 = 1125 N
V
U
H1 and R2 = 675 N [From equation (i)]
X 15 kN
R1 Considering joint ‘D’,
H2
R2
R2
Considering right side of section X-X,

FRS S D
15 kN TCD
450
FRU
U 450
FUV H1
TCD cos 450
15 kN R1 = 15 kN
Taking moment about point ‘U’,  Fy = 0 ,

FRS × 4 − 15 × 4 + R1 × 4 = 0 R2 − TCD cos 450 = 0
FRS = 15 × 4 −15 × 4 = 0 R2
 TCD = = 675 2N
Hence, FRS = 0 cos 450
Now, drawing free body diagram that is  RQ = 120 kN
TAB
675 2 From equation (i),
C 450
q B
RR = −40kN
Considering point ‘P’,
E F FPQ 14.030
2m 2m FPR cos 53.130

1800 N FPQ sin14.030 53.130
 Fy = 0 ,
14.030 FPR
TAB sin θ + 675 2 sin 450 = 1800
FPQ
 TAB sin θ = 1125 …(ii) 80 kN
and  Fx = 0 ,  Fx = 0 ,
− FPQ sin14.030 + FPR cos53.130 = 0
TAB cos θ − 675 2 cos 450 = 0
FPR = 0.404 FPQ …(ii)
 TAB cosθ = 675 …(iii)
 Fy = 0 ,
By dividing equation (ii) to equation (iii),
5 FPQ cos14.030 − FPR sin 53.130 − 80 = 0
tan θ =
3 FPQ × cos14.030 −(0.404 FPQ )sin 53.130 = 80
θ = 59.0360 [From equation (ii)]
By putting the value of θ in equation (ii), we get 0.647 FPQ = 80
1125 FPQ = 123.65kN
TAB = = 1311.96 N
sin(59.036)
Now, considering joint ‘Q’,
4.3 (A) FPQ
FPQ cos14.03
P (0, 4)
53.130 14.030
q q = 14.030
FQR
80 kN 22.840 FPQ sin14.03
4m
120
 Fx = 0 ,
104.030
53.130
Q R (3, 0) FQR + FPQ sin14.03 = 0
(0, 0)
RQ RR FQR = −30.00 = −30 kN (Compressive)
1m 2m
3m Hence, the correct option is (A).
 Fy = 0 , 4.4 2
Since PQ is short by 3 mm joint R will displace
RQ + RR = 80 kN …(i)
upward, so we have to find the vertical
Taking moment about point ‘R’, displacement of joint R,
80 × 3 − RQ × 2 = 0 Δ R = U λ
Where, λ = Lack of fit of any member. 4.5 5
Let apply a unit load at R as shown below,
Given : AE = 30 kN
1
5 kN
R
10 kN
U PR U RQ
3m
q = 36.87 0 Q
P
U PQ
X 3m
1 1 X Y
VP = VQ =
2 2
3m
1
Due to symmetry, VP = VQ = A B
2 H1
X
Considering joint ‘P’,
3m
R1 R2
U PR
 Fx = 0 ,
q U PQ H1 = 10kN
 Fy = 0 ,
1
2 R1 + R2 = 5
 Fy = 0 , Taking moment about ‘A’,
R2 × 3 − 5 × 3 − 10 × 9 = 0
1
U PR sin θ = R2 = 35kN (Upward)
2
1 And R1 = 5 − R2 = 5 − 35 = − 30kN
U PR = (Downward)
2sin 36.87 0
By method of section, cutting section X-X.
U PR = 0.833 …(i)
For the calculation of force in member X-Y
 Fx = 0 , consider downward portion of section X-X.
U PQ + U pr cosθ = 0 Fxy
4
U PQ = −U PR cos36.870 = −0.833 ×
5 450
B
H A = 10 kN
U PQ = −0.67
RA = 30 kN RB = 35 kN
Δ R = U PQ (λ PQ ) = −0.67 × (−3)
 Fx = 0,
Δ R = 2mm (upwards) Fxy = 10 kN
Hence, the magnitude of vertical displacement of Strain energy in member xy,
joint R is 2 mm.
P 2 L (10) 2 × 3
U= = = 5 kN-m
Scan for 2 AE 2 × 30
Video Solution Hence, the strain energy in the member XY is 5
kN-m.
Considering joint ‘S’
Scan for
FSQ FSR
Video Solution
4.6 (A) 450 450
L
A G1
L
100
P
E F  Key Point
L
We can also apply Lami’s rule,
100 FSR F
HB B C 0
= 0
= SQ
D sin 90 sin135 sin135
L L
 Fx = 0,
VB VC
FSQ = FSR
 Fy = 0,
 Fx = 0 ,
FSQ sin 450 + FSR sin 450 = 100
HB = 0
1
 Fy = 0 , 2 FSQ × = 100
2
VB + VC = P …(i) FSQ = 50 2
Taking moment about ‘B’ And FSR = 50 2
Vc × 2L − P × 2L = 0 Considering joint ‘Q’
FQP
Vc = P
And VB = 0 [From equation (i)] 450
FQR
Hence, the correct option is (A). 450
4.7 (C)
FSQ
Given : A = 0.05m2 , L = 1m
 Fy = 0,
E = 30 × 106 N/m 2 and W = 100 kN
FQP sin 450 − FSQ sin 450 = 0
P
FQP = FSQ = 50 2
L  Fx = 0,
FQR + FQP cos 450 + FSQ cos 450 = 0
0 0
45 45
Q R 1 1
450 450 FQR = −50 2 × − 50 2 ×
2 2
900 FQR = −100kN (compressive)
FQR100
S Stress σQR = = 1000 kN/m2
=
(2 A) 2 × 0.05
W Due to compressive force member QR will go
Under shortening RB × L = 2 × L + 1× L
PL FQR × LQR RB = 3kN and RA = 1kN
Δ QR = =
AE (2 A) E (Downward)
100 × ( 2 ×1) ×10 3
Consider joint ' B ' ,
= = 0.0471m
2 × 0.05 × 30 ×106 FAB = 1 (Compressive)
Hence, the correct option is (C). FBC = 3 (Compressive)
4.8 (A) Consider joint ' A ' ,
20 kN FAC = 2 (Tension)
10 kN
10 kN J Now, force in each member due to unit load at C
H K
G L in horizontal direction.
U V C
F M 1
E N
T S R Q P
From the concept of zero force member FT , TG

, HU, MP, PL are the members which do not carry
450 B
any force. A HB = 1
VA = 1 VB = 1
 Key Point
 Fx = 0 ,
If three members meet at a joint out of which two
are collinear and the third member which is non- HB = 1
linear will carry zero force, provided that there is  Fy = 0 ,
no external loading at that joint.
VA = VB
4.9 2.7 MA = 0,
11 2 2
Given : E = 2 ×10 N/m , A = 10mm VB = 1
L = 1m and P = 1kN . and VA = 1 (Downward)
2P=2 kN
Consider joint B ,
1kN=P FAB = 1 (Compressive)
C
FBC = 1 (Compressive)
2L FBC L Consider joint ' A ' ,
FAC FAC = 2 (Tensile)
0 B
A 45
P [Assuming tensile force as + ve and compressive
force as −ve ]
L
PKL
RA = 1kN RB = 3kN Member P K L AE
AE
 Fx = 0 , AB −1 −1 1 AE 1 / AE
HB = P = 1
BC −3 −1 1 AE 3 / AE
CA 2 2 2 2AE 2 2
 Fy = 0 ,
2AE
RA + RB = 2P = 2 PKL 5.414
MA = 0,
 AE = AE
The horizontal deflection, 4.12 (B)
 PKL 5.414 ×1000
Δ HC = =
AE (10 ×10−6 ) (2 ×1011 ) 2m
−3
3m
Δ HC = 2.707 ×10 m = 2.7 mm
The horizontal displacement of joint C is 2.7 mm. P Q R
30 kN
Scan for
3m 3m 3m RR
Video Solution
RP RQ
4.10 (D) From force equilibrium  Fx = 0 ,

D P C HP = 0
P P
 Fy = 0 ,
RP + RQ + RR = 30 …(i)
A B Taking moment about joint ' P ' ,
P P
FAB  MP = 0 ,
At joint ' B ' , 9 RR + 6 RQ − 30 × 3 = 0
FAB = P and FBC = 0
3RR + 2RQ = 30 …(ii)
At joint ' D ' ,
Using method of section, cutting section ' X -X '
FDC = P (Tensile)
and considering left portion of section X -X .
and FDA = 0 X
There is no component of forces in AC direction
hence
FAC = 0
X
Hence, there are 2 members in tension and 3 zero
30 kN
force members.
4.11 (B) RP
From the concept of zero force member ΣFy = 0 ,

L
20 kN 20 kN RP = 30kN (upward)
20 kN K 20 kN
N By putting the value of RP in equation (i),
I
G
P
S 30 + RQ + RR = 30
RQ = − RR
E T
F H J M O R
From equation (ii),
If three members meet at a joint out of which two 3RR + 2(− RR ) = 30 [RQ = − RR ]
are collinear and the third member which is non-
collinear will carry zero force, provided that there RR = 30kN (upward)
is no external loading at that joint. and RQ = −30 kN (downward)
Hence, FG , FI , RS and PR are zero force
Hence, RP = 30kN
members.
Hence, the correct option is (B). RQ = −30 kN and RR = 30kN .
G E
Scan for
Video Solution
4.13 (C) 1
Given the planner truss,

G E C F
A
B
D 1
ΣPUl P × 1× l
F P X =− 2
=− 2
ΣU l 1 ×l
At joint A,
FAC
X = −P
Force in member GF = FGF + Ux
Force in member GF = FGF + Ux
FAB
= P − (− P × 1) = 0
ΣFx = 0, ΣFy = 0
FAC = 0, FAB = 0 0 0
0 0 0
Similarly, at joint B, FBD = FBC = 0
0 0
Similarly, at joint C, FCE = FCD = 0
0
At joint D,
FDE
No. of zero force member is 8.
FDF 4.14 – 30
P Method of sections,
ΣFy = 0, ΣFy = 0 10 kN
S V Z 10 kN
S V Z
FDE = P, FDF = 0 4m
R
Truss reduce (it is 1 degree indeterminate) 10 kN
R U Y 10 kN
U Y
G l E 4m
Q Q
X 10 kN X
10 kN
P
l (x)
T
(x)
P W
F QP F XW
F
4m
Take redundant as vertical support,
P P E M Q =0
FXW × 4 + 10 × 4 + 10 × 8 = 0
P
P P 2 120
FXW = − = −30 kN
4
F Hence, the force in member XW is – 30 kN.
Applied unit load at redundant,
4.15 0.9 P (1)(3) 150(1)(3)
∴ δG = =
AE 500 × 103
Given : Axial rigidity, R = 500 ×103 kN
= 9 ×10−4 m = 0.9 mm
P = 150 kN
4.16 500
Pkl (1)
δG = Σ =?
AE Q R
P G E, 2 A
P P
E, A
3a E, A E, A
E, A
3m a 2 a
T
P S
E, A a E, A
(1)
VP F = 10 3 kN VS
3m
V p = Vs = 5 3 kN
Consider joint G :
Consider equilibrium of LHS of section (1) - (1),
P G Q
FQR
FQT
3a
ΣFx = 0 2
T
ΣFy = 0 P
a
FPT
To final K, applied unit load at G

1 kN G 5 3
1 kN
Taking moment about ( T ),
2 kN ΣMT (CW) = 0
1 kN
 3a 
(5 3 × a) + fQR   = 0
 2 
1 kN
FQR = −10 kN or 10 kN (C)
PkL Compressive stress in member QR ( σc )
Members P k L
AE
FQR
σc =
AB 0 1 3 0 2A
0 − 2 0 10 kN
BC 3 2 σc = 4−6 2
= 500 kN/m2
2(10 ×10 ) m
AC 0 1 3 0
4.17 30
BG P 1 3 P(1)(3)
 M E = 0,
AE
RA × 4 − H A × 4 = 0
GC 0 0 3 0
RA = H A
 M j = 0,
RA × 8 − 70 × 4 − H A ×1 = 0
RA = 40kN
H A = 40 kN
At joint E
 Fy = 0,
F = 70 − 40 = 30 kN
4.18 14.14
Vertical reaction at supports
100
= = 50 kN ( Due to symmetry of structure )
2
20 20 20 20 20
U0 U1 U 2 U3 U4 U5 U6
50 50
L1 L2 L3 L4 L5
3.5 m@6 span

By using method section:
20 kN 20 kN
U0 U1 U2
Fu2 u3
Fu 2 l3
50 kN
Fl2 l3
L1 L2
V = 0
Fu2l3 cos 45° + 20 + 20 − 50 = 0
Fu2l3 = 10 2 = 14.14 kN
Hence, the correct answer is 14.14.

5 Influence Line Diagram and Rolling Loads
2013 IIT Bombay R. 0.5
5.1 Beam PQRS has internal hinges in spans PQ

and RS as shown. The beam may be
0.5
subjected to a moving distributed vertical L 3L
load of maximum intensity 4 kN/m of any 4 4
length anywhere on the beam. The maximum S. 1.0
absolute value of the shear force (in kN) that
can occur due to this loading just to the right
of support Q shall be [2 Marks]
P Q R S L 3L
4 4
5m 5m 20 m 5m 5m (A) P (B) Q
(A) 30 kN (B) 40 kN (C) R (D) S
(C) 45 kN (D) 55 kN 2015 IIT Kanpur

5.3 A simply supported beam AB of span, L = 24
2014 IIT Kharagpur m is subjected to two wheel loads acting at a
5.2 In a beam of length L, four possible influence distance, d = 5 m apart as shown in the figure
line diagrams for shear force at a section below. Each wheel transmits a load P = 3 kN
L and may occupy any position along the
located at a distance of from the left end beam. If the beam is an I-section having
4
support (marked as P, Q, R and S) are shown section modulus, S = 16.2 cm3 , the
below. The correct influence line diagram is maximum bending stress (in GPa) due to the
[1 Mark] wheel loads is ______. [2 Marks, Set - II]
P P
P. 0.75
A B
d
L
0.25
L 3L 2017 IIT Roorkee
4 4
5.4 Consider the beam ABCD shown in the
Q. 0.6
figure
A B C D
Internal hinge
0.6 4m 4m 10 m
L 3L
4 4 For a moving concentrated load of 50 kN on
the beam, the magnitude of the maximum
bending moment (in kNm) obtained at the P Q R S T
support C will be equal to _______.
[2 Marks]
1.5m 3m 2m 1.5m
2019 IIT Madras The maximum negative (hogging) bending
5.5 A long uniformly distributed load of 10 moment (in kN-m) that occurs at point R, is
kN/m and q concentrated load of 60 kN are [2 Marks]
moving together on the beam ABCD shows (A) 150.00 (B) 56.25
in the figure (not drawn to scale). The (C) 22.50 (D) 93.75
relative positions of the two loads are not
fixed. The maximum shear force (in kN 2023 IIT Kanpur
round off to the nearest integer) caused at the 5.7 Muller-Breslau principle is used in analysis
internal hinge B due to the two loads is of structures for _________.
______.
Internal hinge
(A) Drawing an influence line diagram for
any force response in the structure
A B C D
(B) Writing the virtual work expression to
2m 2m 4m get the equilibrium equation
(C) Superposing the load effects to get the
2020 IIT Delhi
total force response in the structure
5.6 Distributed load (s) of 50kN/m may occupy
(D) Relating the deflection between two
any position (s) (either continuously or in
patches) on the girder PQRST as shown in points in a member with the curvature
the figure (not drawn to the scale) diagram in-between
5.1 (C) 1
Hence, ordinate at A = × 5 = 0.25
P A Q R B S 20
To get the maximum absolute value of S.F. UDL
5m 5m 20 m 5m 5m should be placed in spans ' PQ ' and ' QR '
4 kN/m 1 1
Hence, SFmax = ×10 × 0.25 × 4 + ×1× 20 × 4
2 2
Q’ SFmax = 45kN
0.25 A’ 1 Hence, the correct option is (C).
+
+ 5.2 (A)
Q –
0.25 A C B
B’
Ordinate at Q = 1 ( unit displacement)
L/3 3L/4
3L / 4
1 0.75 =
Ordinate at B = × 5 = 0.25 L
20
(By similar triangle rule) L/4
While releasing the S.F. just right of ' Q ' by = 0.25
L
making a cut and giving unit displacement slope at Hence, the correct option is (A).
right of ' Q ' and left of ' Q ' should be same.
 Key Point 28.89 ×103
(σb )max =
b 16.2 ×10− 6
( a + b)
a
(σb )max = 1.783 ×109 N/m2 = 1.783GPa
b
Hence, the maximum bending stress due to the
C wheel loads is 1.783 GPa.
a Scan for
( a + b) Video Solution
5.3 1.783 5.4 200

The bending stress in beam is given by,
M M A B C D
σb = y=
I z
To get maximum bending stress, the bending
4m 4m 10 m
moment in beam should also be maximum.
Using Muller Breslau principle, ILD for moment
Maximum bending moment will occur under load
at ‘C’.
P, by placing the load P in such a way that centre
of span lies in between C.G. of load system and Giving unit rotation at C,
load under consideration ( P ) . 50 kN
Centre of load system, B C D
A
3 R 3 q=1
2.5 m 2.5 m 4m
5m B’
Due to symmetry it will act at 2.5 m from either 4m 4m 10 m
side.
3 kN 3 kN
 Key Point
2.5 m 2.5 m
Influence line for CD will not occur because it
5m
P1
will be a curve and since the beam is
5.935 P2 Center of
beam determinate, curve cannot take place.
C
θ = tan 450
a = 10.75 m 2.5 8.25 BB '
1.25 1.25
b = 13.25 m tan 450 = 1 =
12 m 12 m
BC
ab 10.75 ×13.25 BB ' = BC = 4
Ordinate at P1 = = = 5.935 m
L 24 Hence, ordinate at ‘B’ = 4 m.
5.935 × 8.25 For maximum bending moment load should be
Ordinate at P2 = = 3.695 m
13.25 placed at ‘B’.
(By similar triangle rule) Hence, M max = Load × Ordinate
Hence, maximum bending moment
M max = ΣLoad × Ordinate M max = 50 × 4 = 200 kN-m
M max = 3 × 5.935 + 3 × 3.695 = 28.89 kN-m Hence, the magnitude of the maximum bending
moment obtained at the support C will be equal to
M max
Hence, (σb )max = 200 kN-m.
z
5.5 70 5.7 A
According to question, 

Internal hinge
A B C D
2m 2m 4m
From Muller’s Breslau’s principle
2m 2m 4m
B C D
A
1
B’
For maximum shear force the UDL should be
placed in portion AB and the concentrated load
should be placed at B.
Hence, VB (max) = UDL × Area of ABB '
+ Concentrated load × Ordinate

1 
VB (max) = 10 ×  × 2 × 1 + 60 × 1
2 
VB (max) = 70kN
Hence, the maximum shear force caused at the
internal hinge B due to the two loads is 70 kN.
Scan for
Video Solution
5.6 (B)
Given : Distributed load = 50 kN/m.

ILD for BM at the centre of span ‘QS’
50 kN/m
P Q R S T
1.5m 3m 2m 1.5m
(+) 1.20
( -) ( -)
0.6 0.9
1  1 
M R = −  × 1.5 × 50 × 0.6  −  × 1.5 × 50 × 0.9 
2  2 
= −56.25 kN-m or 56.25 kN-m (hogging)
6 Analysis of Arches
2017 IIT Roorkee Neglecting self-weight, the maximum

bending moment generated in the structure
6.1 The figure shows a two-hinged parabolic (in kN.m, round off to the nearest integer), is
arch of span L subjected to a uniformly _______.
distributed load of intensity q per unit length.
[1 Mark]
[1 Mark]
q per unit length 2022 IIT Kharagpur
6.3 A semi-circular bar of radius R m, in a
vertical plane, is fixed at the end G, as shown
in the figure. A horizontal load of magnitude
P kN is applied at the end H. The magnitude
of the axial force, shear force, and bending
L moment at point Q for θ = 450 , respectively,
The maximum bending moment in the arch are [2 Marks]
is equal to
Q
qL2 qL2
(A) (B)
8 12 R
2
qL θ
(C) Zero (D) G
10 P
H O
2020 IIT Delhi P P PR

(A) kN, kN, and kNm
6.2 A planar elastic structure is subjected to 2 2 2
uniformly distributed load, as shown in the P P
(B) kN, kN, and 0 kNm
figure (not drawn to the scale) 2 2
12 kN/m P PR
(C) 0 kN, kN, and kNm
2 2
2.4 m P PR
(D) kN, 0 kN, and kNm
2 2
Parabolic profile
8.0 m
6.1 (C)  Key Point

For a given loading if the free BMD correspond
For a two hinged and three hinged parabolic arch to the actual shape of the structure, then such
bending moment at every point is zero if they are structure are called funicular polygon.
subjected to UDL trough out its length. For a funicular profile the BM and S.F. at any
Hence, the correct option is (C). section is zero.
Scan for
 Key Point
Video Solution
6.2 96
12 kN/m
2.4 m
HA = 0 Parabolic profile
VA VB
8.0 m
ΣFy = 0 ,
VA + VB = wL
 VA = VB
∴ VA + VA = wL
6.3 (A)
2VA = wL
Q
12 × 8
VA = = 48 kN
2 P cosθ R
R sinθ
VA = VB = 48 kN P
θ
θ 450
As horizontal thrust is zero so it behaves like a

P sinθ
beam (curved beam)
wL2 P
M max = (At crown) ( FQ ) = P sin θ = (at θ = 450 )
8 2
P
12 × 82 ( SQ ) = P cos θ = (at θ = 450 )
M max = = 96 kNm 2
8
PR
Hence, the maximum bending moment generated M Q = PR sin θ = (at θ = 450 )
2
in the structure is 96 kN-m.

7 Matrix and Stiffness Method
2015 IIT Kanpur 6EI 12EI

(A) (B)
L2 L3
7.1 For the beam shown below, the stiffness
3EI EI
coefficient k22 can be written as (C) (D)
L 6 L2
[1 Mark]
Note : 1, 2 and 3 are 2
the degree of freedom
L 3
A, E, I 1
7.1 (B) 12 EI Δ
RB =
2 L3
12 EI
L 3
So, k22 = RB = 3 [ Δ = 1]
L
A, E, I 1
Scan for
k22  Force at 2 due to a unit deformation at 2. Video Solution
To kind k22 given unit deflection in the direction
2 only and measure force developed in the same

direction.
6EI D
L2
6EI D
L2 D =1
B
A
L
12 EI D 12 EI D
RA = RB =
L3 L3
Taking moment about ‘A’,
 M A = 0,
6 EI Δ 6 EI Δ
RB × L − − 2 =0
L2 L
12 EI Δ
RB × L =
L2
8 Structural Dynamics
2019 IIT Madras The natural frequency of the system is

[1 Mark]
8.1 A simple mass-spring oscillatory system
m k
consists of a mass m, suspended from a (A) (B)
k m
spring stiffness k. Considering z as the
displacement of the system at any time ‘t’ m k
(C) (D)
the equation of motion for the free k m
vibration of the of system is mz + kz = 0.
8.1 (D)
Given : Equation of motion for the free vibration,
mz + kz = 0
k
z+ z=0

m
z + ω2n z = 0
Comparing with 
k
We get, ωn =
m

RCC STRUCTURE AND
PRE-STRESS CONCRETE
Syllabus : RCC Structure & Pre-Stress Concrete
Working stress and Limit state design concepts; Design of beams, slabs, columns; Bond and
development length; Prestressed concrete beams.
Contents : RCC Structure & Pre-Stress Concrete
S. No. Topics
1. IS Code Recommendations & Fundamentals
2. Design & Analysis of Beam and Slab
3. Shear, Bond, Torsion, Anchorage & Development Length
4. Column & Footing
5. Prestressed Concrete
1 General Design Requirements
1995 IIT Kanpur 1.4 Factor of safety adopted by IS 800 : 1984

while arriving at the permissible stress in
1.1 As per IS 800 : 1984, the minimum pitch of axial compression is [1 Mark]
rivets in a row is recommended as the (A) 2.00 (B) 1.00
diameter of the rivet times
(C) 1.67 (D) 1.50
[1 Mark]
2005 IIT Bombay
(A) 2.0 (B) 2.5
1.5 Which one of the following is not correct for
(C) 3.0 (D) 4.0
steel sections as per IS 800 : 1984?
1996 IISc Bangalore [2 Marks]
1.2 Generally the maximum deflection/span (A) The maximum bending stress in tension
ratio of a steel member should not exceed. or in compression in extreme fiber
calculated on the effective section of a
(A) 1/750 (B) 1/500
beam shall not exceed 0.66 f y
(C) 1/325 (D) 1/250
(B) The bearing stress in any part of a beam
1.3 Generally, fatigue life of welded steel when calculated on the net area shall not
structure to fatigue life of riveted steel
exceed 0.75 f y
structure ratio is [1 Mark]
(A) Smaller than 1 (C) The direct stress in compression on the
cross-sectional area of axially loaded
(B) Equal to 1 compressive member shall not exceed
(C) Greater than 1 0.6 f y
(D) Greater than 2.1 (D) None of the above
1997 IIT Madras
1.1 (B) Span

δ max (in mm) > (in mm)
325
The minimum spacing between centres of
fasteners shall not be less than 2.5 time the
nominal diameter of the fasteners. 1.3 (A)
Hence, the correct option is (B). Welded connection are weak under fatigue loading
1.2 (C) and prone to cracking. So fatigue life of welded
structure is always less than the fatigue life of
Maximum deflection in the beam should not
riveted structure.
Span
exceed ,
325
Steel Structures 1
Fatigue life of welded structure
<1
Fatigue life of riveted structure
1.4 (C)
Permissible stress in axial compression as per IS

800 : 1984
i.e., σ allowable = 0.6 f y
 1 
∴ Factor of safety =   = 1.67
 0.6 
1.5 (D)
All statements (a), (b) and (c) are correct

according to IS 800 : 2007.

2 Steel Structures
2 Connections
2014 IIT Kharagpur (D) Forces due to the friction between

connected parts
2.1 The tension and shear force (both in kN) in
2016 IISc Bangalore
each bolt of the joint, as shown below,
respectively are 2.4 Two plates are connected by fillet welds of
[2 Marks, Set-1] size 10 mm and subjected to tension, as
shown in the figure. The thickness of each
plate is 12 mm. The yield stress and the
ultimate tensile stress of steel are 250 MPa
4 and 410 MPa, respectively. The welding is
5
3 Pu = 250 kN done in the workshop ( γ mw = 1.25) .
P
(A) 30.33 and 20.00 100 mm
(B) 30.33 and 25.00
(C) 33.33 and 20.00
(D) 33.33 and 25.00
2015 IIT Kanpur
2.2 A bracket plate connected to a column flange
transmits a load of 100 kN as shown in the
following figure. The maximum force for
which the bolts should be designed is
_______ kN. 150 mm
[2 Marks, Set-1]
P
100 kN As per the Limit State Method of IS 800:
600 2007, the minimum length (rounded off to
the nearest higher multiple of 5 mm) of each
75 weld to transmit a force P equal to 270 kN
75
(factored) is [2 Marks]
75 75 (A) 90 mm (B) 105 mm
All dimension
are in mm (C) 110 mm (D) 115 mm
2.5 Two bolted plates under tension with
2.3 Prying forces are [1 Mark, Set-2] alternative arrangement of bolt holes are
(A) Shearing forces on the bolts because of shown in figures 1 and 2. The hole diameter,
the joints pitch and gauge length are d , p and g ,
(B) Tensile forces due to the flexibility of respectively.
connected parts
P P P g P
(C) Bending forces on the bolts because of d
p
the joints
Steel Structures 1
Which one of the following conditions must 2.9 A fillet weld is simultaneously subjected to
be ensured to have higher net tensile capacity factored normal and shear stresses of 120
of configuration shown in figure 2 than that MPa and 50 MPa, respectively. As per IS
shown in figure 1? 800 : 2007 the equivalent stress (in MPa, up
[2 Marks, Set-2] to two decimal places) is ______.
(A) p 2 > 2 gd (B) p 2 > 4 gd [1 Mark, Set-2]
2.10 Four bolts P, Q, R and S of equal diameter
(C) p 2 > 4 gd (D) p > 4 gd are used for a bracket subjected to a load of
2017 IIT Roorkee 130 kN as shown in the figure.
[2 Marks, Set-2]
2.6 Two plates of 8 mm thickness each are Centerline
connected by a fillet weld of 6 mm thickness 200 mm
as shown in the figure.
6 130 kN
Q P
240 mm
P 50 mm 100 mm P
R S
100 mm
The permissible stresses in the plate and
the weld are 150 MPa and 110 MPa,
100 mm
respectively. Assuming the length of the
The force in bolt P is
weld shown in the figure to be the effective
length, the permissible load P (in kN) is (A) 32.50 kN (B) 69.32 kN
_______. [2 Marks] (C) 82.50 kN (D) 119.32 kN
2.7 A column is subjected to a load through a 2019 IIT Madras
bracket as shown in the figure.
2.11 A 16 mm thick gusset plate is connected to
15 cm P = 10 kN the 12 mm thick flange plate of an I – section
cm
10 using fillet welds on both sides as shown in
1 figure (not drawn to scale). The gusset plate
900 is subjected to point of 350 kN acting at a
distance of 100 mm from the flange plate
size of filet weld is 10 mm.
[2 Marks]
10
cm
350 kN
100 mm
The resultant force (in kN) up to one decimal
place in the bolt 1 is ________. 16 mm thick
gusset plate
[2 Marks] 500 mm
500 mm
2018 IIT Guwahati Fillet weld
I-section Fillet weld

2.8 In a fillet weld, the direct shear stress and Flange (12 mm thick)
bending tensile stress are 50 MPa and 150 (Front view) (Side view)
MPa respectively. As per IS 800 : 2007, the The maximum resultant stress (in MPa,
equivalent stress (in MPa, up to two decimal round off to 1 decimal place) in the fillet
places) will be _______. weld along the vertical plane would be
[1 Mark, Set-1] _______.
2 Steel Structures
2020 IIT Delhi weld is 108 MPa. The location of the
centroid of the angle is represented by C yy
2.12 Two steel plates are lap jointed in a
workshop using 6 mm thick fillet weld as in the figure, where C yy = 28.4 mm. The
shown in the figure (not drawn to the scale), area of cross-section of the angle is
The ultimate strength of the weld is 410 1903 mm 2 . Assuming the effective throat
MPa. thickness of the weld to be 0.7 times the
given weld size, the lengths L1 and L2
(rounded-off to the nearest integer) of the
120 mm
Fillet weld required to transmit a load equal to the
weld full strength of the tension member are,
respectively
200 mm
5mm weld
As per Limit State Design of IS 800 : 2007, L1
the design capacity (in kN, round off to three Cyy

decimal places) of the welded connection, is 100 mm
_____. [2 Marks]
L2
5mm weld
2021 IIT Bombay
15 mm thick gusset plate
2.13 A column is subjected to a total load (P) of
60 kN supported through a bracket (A) 541 mm and 214 mm
connection as shown in the figure (not to (B) 214 mm and 541 mm
scale).
40 40 (C) 380 mm and 151 mm
mm mm
(D) 151 mm and 380 mm
60 kN
[2 Marks]
2023 IIT Kanpur
30 mm
R 2.15 Consider the fillet-welded lap joint shown in
30 mm
the figure (not to scale). The length of the
weld shown is the effective length. The
welded surfaces meet at right angle. The
100 mm weld size is 8 mm, and the permissible stress
in the weld is 120 MPa. What is the safe load
The resultant force in bolt R (in kN, round P (in kN, rounded off to one decimal place)
off to one decimal place) is ________. that can be transmitted by this welded joint?
[2 Marks] 75 mm
2022 IIT Kharagpur
50 mm
P P
2.14 A weld is used for joining an angle section
ISA 100 mm ×100 mm ×10 mm to a gusset
plate of thickness 15 mm to transmit a tensile
load. The permissible stress in the angle is
150 MPa and the permissible shear stress on
the section through the throat of the fillet
Steel Structures 3
2.1 (D) 75
tan θ =
75
tan θ = 1 (∴ θ = 450 )
FR = F12 + F22 + 2F1F2 cos θ
4
5
3 Pu = 250 kN FR = (20)2 + (141.42) 2 + 2 × 20 ×141.42cos 450
FR = 156.2 kN
Hence, the bolts should be designed is 156.2 kN.
Pu = 250 kN
4 Scan for
cos θ= Video Solution
5
3
Sin θ = 2.3 (B)
5
Pu cos θ Prying force is additional tensile force develop in
Tension in each bolt = bolt due to flexibility of connection.
6 2P 2P
250 4
= × = 33.33kN
6 5
P sin θ
Shear in each bolt = u
6
250 3
= × = 25 kN
6 5
Hence, the correct option is (D). 2P 2P
2P
Scan for
Video Solution
2.2 156.2
Q P+Q P+Q Q
75 F2
q P+Q P+Q
Q Q
F1
75
2P
75 75
100
Q = Prying force
We know, F1 = = 20 kN Hence, the correct option is (B).
5
Per (100 × 0.6) × (75 2) × 10−3 2.4 (B)
F2 = =
 r2 (75 2) 2 × 4 ×10−6 Refer Solution No. 2.19
F2 = 141.42kN Hence, the correct option is (B).
4 Steel Structures
Always Remember Pplate = σat × Anet
For safe weld design = 150 × (50 × 8)
Vdw ≥ Tdg = 60 kN
Vdw = Design strength of weld So permissible load = 60 kN
Tdg = Gross strength of plate Hence, the permissible load P is 60 kN.
Scan for
2.5 (C)
Video Solution
P P P g P
d
p 2.7 6
Tensile strength of plate in arrangement 2 will be Given : Number of bolts, n = 4

greater than in arrangement 1 Applied load, P = 10 kN
As per IS 800 : 2007 clause 6.3, P
Pe
10 kN
 Fup   Fup  F2
 0.9 Anet  >  0.9 Anet 
 γ m1 2  γ m1 1 1 1350
F1 r = 5cm
( Anet )2 > ( Anet )1
 P2  
 B − 2 d +  t  > [ ( B − d )t ]1
 4g  2
15cm
P2 P 10
B − 2d + > B−d F1 = = = 2.5 kN
4g n 4
P2 Pe 10 × 15
>d F2 = × r1 = × 5 = 7.5 kN
4g  r12
4 × 52
P2 > 4 gd FR = F12 + F22 + 2F1F2 cos θ

Hence, the correct option is (C). FR = 2.52 + 7.52 + 2 × 2.5 × 7.5cos1350
2.6 60 FR = 6 kN
Given : Permissible stress in plate σat = Hence, the resultant force up to one decimal place
in the bolt 1 is 6 kN.
150 MPa
Scan for
Permissible stress in weld Sb = 110 MPa
Video Solution
6
2.8 173.20
Given : Direct bending tensile stress
P 50 mm 100 mm P f a = 150 MPa
100 mm fu = 410 MPa
Pw = Fb × Effective area Direct shear stress q = 50 MPa
According to IS 800 : 2007, clause 10.5.10.1.1,
= 110 × (leff × tt )
the equivalent stress,
= 110(100 + 100 + 50) × 0.7 × 6 fu
f e = f a 2 + 3q 2 ≤
= 115.5 kN 3 γ mw
Steel Structures 5
410 FR = F12 + F22 + 2F1F2 cos θ
f e = 1502 + 3 × 502 ≤
3 ×1.25
f e = 173.20 ≤ 189.37 (Hence ok) FR = 32.52 + 502 + 2 × 32.5 × 50 × 0.384
So equivalent stress = 173.20 MPa . FR = 69.33kN
Hence, the equivalent stress will be 173.20 MPa. Hence, the correct option is (B).
Scan for 2.11 78.10
Video Solution
t1 t2
2.9 147.99 350 kN
P
Given : Factored normal stress, f a = 120 MPa 100mm
500 mm Pe = m
fu = 410 MPa 500mm
Factored shear stress, q = 50 MPa

According to IS 800 : 2007, clause 10.5.10. 10.1.1
the equivalent stress, Given : P = 350 kN, e = 100 mm
f e = f a2 + 3q 2 ≤
fu S = Size of weld = 10 mm
3 γ mw tt = Effective throat thickness
410
f e = 1202 + 3 × 502 ≤ tt = 0.7S
3 ×1.25
f e = 147.99 ≤ 189.37 tt = 0.7 ×10
So equivalent stress = 147.99 MPa tt = 7 mm
Hence, the equivalent stress is 147.99 MPa. Shear stress due to direct force
2.10 (B) P
q=
200 mm 2 × dt
130 KN
350 ×103
q= = 50 MPa
Q P
2 × 500 × 7
240 mm q F2 Normal stress due to bending
R S F1
My Pe × d / 2
Fb = =
I tt d 3
2×
P 130 12
F1 = = = 32.5 kN
n 4 350 × 103 × 100 × 500
Fb = = 60 MPa
Pe 5003
F2 = × rp 2×7× ×2
 r12 12
{ rp = rQ = rR = rs = 502 + 1202 = 130 mm } Resultant stress = Fb2 + q 2
(130 × 200) × 130
F2 = Fr = 602 + 502
4 × 130 2
200 Fr = 78.10 MPa
F2 = = 50 kN
4 Hence, the maximum resultant stress in the fillet
50 weld along the vertical plane would be 78.10
And cos θ = = 0.384
130 MPa.
6 Steel Structures
Pe.r1
Scan for F2 =
Video Solution Σr12
60 × 100 × 40
2.12 413.59 F2 = = 18.18 kN
Σ 4 × 50 2 + 2 × 40 2
Given : Thickness of weld, t = 6 mm ,
F2
Throat thickness, tt = 0.7 × t ,
F2
tt = 0.7 × 6
180 0
θ = 180
0
F1
Ultimate strength of weld = 410 MPa
Design capacity of welded connection,
F1
Ps = fb × leff × tt
R = 102 + 18.182 + 2 ×10 ×18.18cos1800
F1 θ = 0
F2
R = 102 + 18.182 + 2 ×10 ×18.18cos 00
R = 28.18 kN
2.14 (A)
410
P= × 520 × 0.7 × 6 Case 1 : (Ignoring shear lag)
3 × 1.25
Full strength of tension member,
716352
= = 413598 N P = σsat × Ag
3
= 413.586 kN P = 150 ×1903
P = 285450 N
Hence, the design capacity of the welded
connection, is 413.59 kN. Weld strength of L1 ,
2.13 28.18 P1 = L1 × (0.7 × s) ×108
40 40 P1 = L1 × 0.7 × 5 ×108
mm mm
60 kN P1 = 378L1 …(i)
Weld strength of L2 ,
30 mm P2 = L2 × (0.7 × s) ×108
R
30 mm
P2 = L2 × 0.7 × 5 ×108
P2 = 378L2 …(ii)
Applying moment CG equal to zero,
100 mm
P1 × 28.4 = P2 (100 − 28.4)
Resultant force, R = F12 + F22 + 2F1F2 cos θ P1 71.6
=
Total load ( P) P2 28.4
F1 =
Number of bolt (n) L1 × 378 71.6
= = 2.521
60 L2 × 378 28.4
F1 = = 10 kN
6 L1 = 2.521 L2 …(iii)
Steel Structures 7
Total weld strength = Tensile strength
( L1 + L2 ) × 3.5 ×108 = 285450
L1 + L2 = 755.158
Using equation (iii),
2.521 L2 + L2 = 7585.158
L2 = 214.472 mm
L1 = 540.685mm
Case 2 : (Consider shear lag)
Anet = A1 + K1. A2
Equal angle section, K = 0.75 Anet
Anet = 1.75 × ( A1 )
1903
= 1.75 × = 1665.125 mm 2
2
Full strength of tension member
= 150 ×1665.125s
= 249768.75 N
( L1 + L2 ) × 3.5 ×108 = 249768.75
L1 + L2 = 660.763 mm
Let us check the options (C) and (D).
None of the options full fill
L1 + L2 = 660.763 mm
2.15 134.4
Given : Weld size (S) = 8 mm
Permissible stress ( σ ps ) = 120 MPa
Psafe = K × S × leff × σ ps
= 0.7 × 8 × (75 + 75 + 50) ×120
= 134.40 kN

8 Steel Structures
3 Shear, Bond, Torsion, Anchorage & Development Length
2013 IIT Bombay 2016 IISc Bangalore
3.1 As per IS 456 : 2000, for M 20 grade 3.4 In shear design of an RC beam, other than
concrete and plain bars in tension, the design the allowable shear strength of concrete (τc )
bond stress τbd = 1.2 MPa . Further, IS 456 : , there is also an additional check suggested
in IS 456 : 2000 with respect to the
2000 permits this design bond stress value to
maximum permissible shear stress ( τ c max ) .
be increased by 60% for HYSD bars. The
stress in the HYSD reinforcing steel bars in The check for τc max is required to take care
tension, σ s = 360 MPa . Find the required of
[1 Mark]
development length, Ld for HYSD bars in
(A) Additional shear resistance from
terms of the bar diameter, φ ______. reinforcing steel.
[1 Mark] (B) Additional shear stress that comes from
accidental loading.
2014 IIT Kharagpur (C) Possibility of failure of concrete by
diagonal tension.
3.2 A rectangular beam of width 230 mm and
(D) Possibility of crushing of concrete by
effective depth 450 mm is reinforced with
diagonal compression.
four bars of 12 mm diameter. The grade of
3.5 As per IS 456 : 2000 for the design of
concrete is M 20 and grade of steel is Fe 500.
reinforced concrete beam, the maximum
Given that for M 20 grade of concrete the
allowable shear stress ( τ c max ) depends on
ultimate shear strength, τuc = 0.36 N/mm 2
the [1 Mark]
for steel percentage, p = 0.25 , and τuc =
(A) Grade of concrete and grade of steel.
0.48 N/mm2 for p = 0.50. For a factored (B) Grade of concrete only.
shear force of 45 kN, the diameter (in mm) (C) Grade of steel only.
of Fe 500 steel two-legged stirrups to be used (D) Grade of concrete and percentage of
at spacing of 375 mm, should be reinforcement.
[1 Mark] 3.6 A haunched (varying depth) reinforced
(A) 8 (B) 10 concrete beam is simply supported at both
ends, as shown in the figure. The beam is
(C) 12 (D) 16
subjected to a uniformly distributed factored
2015 IIT Kanpur load of intensity 10 kN/m. The design shear
force (expressed in kN) at the section X-X of
3.3 The development length of a deformed
the beam is ________.
reinforcement bar can be expressed as
[2 Marks]
 1   φσs  5m X
   . From the IS 456 : 2000, the 10 kN/m
 k   τbd 
400 mm
value of k can be calculated as _______. 600 mm
[1 Mark] X
Effective span = 20 m
RCC Structure & Pre-stress Concrete 1

2018 IIT Guwahati 2022 IIT Kharagpur
3.7 An RCC beam of rectangular cross-section 3.10 A reinforced concrete beam with rectangular
has factored shear of 200 kN at its critical cross section (width = 300 mm, effective
section. Its width b is 250 mm and effective depth = 580 mm) is made of M30 grade
depth d is 350 mm. Assume design shear concrete. It has 1 % longitudinal tension
strength, τc of concrete as 0.62 N/mm 2 and reinforcement of Fe 415 grade steel. The
design shear strength for this beam is
maximum allowable shear stress, τc ,max in
0.66 N/mm2 . The beam has to resist a
concrete as 2.8 N/mm2 . If two-legged 10 factored shear force of 440 kN. The spacing
mm diameter vertical stirrups of Fe 250 of two-legged, 10 mm diameter vertical
grade steel are used, then the required stirrups of Fe415 grade steel is _______
spacing (in cm, up to one decimal place) as (mm).
per limit state method will be _________. [2 Marks]
[2 Marks]
2023 IIT Kanpur
2019 IIT Madras
3.11 With regard to the shear design of RCC
3.8 Tie bars of 12 mm diameter are to be beams, which of the following statements
provided in a concrete pavement slab. The is/are TRUE?
working tensile stress of the tie bars is 230 (A) Excessive shear reinforcement can
MPa, the average bond strength between a lead to compression failure in
tie bar and concrete is 2 MPa, and the joint concrete
gap between the slabs is 10 mm. Ignoring the
(B) Beams without shear reinforcement,
loss of bond and the tolerance factor, the
design length of the tie bars (in mm, round even if adequately designed for
off to the nearest integer) is ________. flexure, can have brittle failure
[2 Marks] (C) The main (longitudinal)
reinforcement plays no role in the
2020 IIT Delhi
shear resistance of beam
3.9 A reinforcing steel bar, partially embedded (D) As per IS456:2000, the nominal
in concrete, is subjected to a tensile force P.
shear stress in the beams of varying
The figure that appropriately represents the
depth depends on both the design
distribution of the magnitude of bond stress
shear force as well as the design
(represented as hatched region), along the
bending moment
embedded length of the bar, is
[1 Mark]
(A)
P
(B)
P
(C)
P
(D)
P
2 RCC Structure & Pre-stress Concrete

3.1 46.875 Given :
Width of beam, b = 230 mm
Given : Design bond stress, τbd = 1.2 MPa
Effective depth of beam, d = 450 mm
Stress in bar, σ s = 360 MPa
Diameter of main bars, φ = 12 mm
To find : Development length for HYSD bars, Ld
Grade of concrete = M 20
=?
Grade of steel = Fe500
According to question and as per clause 26.2.1,
Percentages of steel = 0.25 and 0.50
mentioned in IS 456 : 2000,
For p = 0.25 ,
The design bond stress value to be increased by
60% for HYSD bars .i.e., τ bd will become Ultimate shear strength, τuc = 0.36 N/mm 2
1.6τbd For p = 0.50 ,
∴ Development length, Ultimate shear strength, τuc = 0.48 N/mm 2

φσ s φ × 360 φ × 360 Factored shear force, Vu = 45 kN
Ld = = =
4τbd 4(1.6τbd ) 4 × 1.6 × 1.2 Spacing of 2 legged Fe500 stirrups,
Ld = 46.875φ Sv = 375 mm
Hence, the required development length 46.875. To find : Diameter of stirrups φ s = ?
As per clause 40.1 mentioned in IS 456 : 2000,
Scan for
Video Solution Nominal shear stress in beam,
V 45 × 103
τV = u = = 0.435 N/mm 2
bd 230 × 450
 Key Point Percentage of steel,
1. As per clause 26.2.1 of IS 456 : 2000 for π 
deformed bars conforming to IS 1786. 4 ×  × φ2 
100 As 4  ×100
Design bond stress values τ bd , should be p= =
bd 230 × 450
increased by 60 percent. π 
4 ×  ×122 
2. For bars in compression, these values of τ bd 4  ×100
=
shall be increased by 25 percent. 230 × 450
3. For fusion bonded epoxy coated deformed p = 0.437%
bars, design bond stress values shall be taken Design shear strength of concrete,
as 80 percent of τ bd given in 26.2.1. 0.12
τc = 0.36 + × (0.437 − 0.25)
0.25
3.2 (A)
τc = 0.45 N/mm 2
∴ τ v < τc
Hence, there is a requirement of providing of
minimum shear reinforcement.
As per clause 26.5.1.6, IS 456 : 2000,
Minimum shear reinforcement in the form of
stirrups shall be provided such that,

Asv 0.4 As per clause 26.2.1, mentioned in IS 456 : 2000,
=
bSv 0.87 f y For deformed bars conforming to IS1786, the
value of design bond stress ( τbd ) in limit state
 0.4   bS v 
∴ Asv =  ×
 0.87   f y  method shall be increased by 60%.i.e., τ bd will
become 1.6τbd .
 0.4   230 × 375 
Asv =  × 
 0.87   500  φσ s φσ s φσ s
Ld = = = …(ii)
Asv = 79.31 mm 2 4τbd 4(1.6τbd ) 6.4τbd
So, diameter of stirrups will be calculated by using Comparing equation (i) and (ii),
Fe500, 2-legged stirrups as, 1  φσ s  φσ s
k  τbd  = 6.4τ
π  
Asv =  × φs 2  × 2
bd
4  1 1
=
φ s = 7.12 mm  8 mm k 6.4
Hence, providing 8φs Fe500 two legged stirrups ∴ k = 6.4
@ centre to centre spacing of 375 mm. Hence, the IS 456 : 2000, the value of k can be
calculated as 6.4.
Scan for
Scan for Video Solution
Video Solution
3.4 (D)
 Key Point In shear design of a RCC beam as per IS 456 :
(a) Shear reinforcement 2000, there is a provision of two checks for the
1. τv < 0.5τc no shear reinforcement is value of nominal shear stress (τv ) . They are :
provided (i) Check for allowable shear strength of
2. 0.5τc ≤ τv < τc minimum shear concrete (τc )
reinforcement is provided (ii) Check for maximum permissible shear stress
3. τv > τc ( τ c max )
Then shear reinforcement (vertical or As per clause 40.2.3, IS 456 : 2000,
inclined) are designed for shear force of Under no circumstances even with shear
Vus = [τv − τc ] bd reinforcement, shall the value of nominal shear
(b) As per clause 40.4 of IS 456 : 2000. The stress, τv exceeds maximum permissible shear
characteristic strength of the stirrup or bent stress, τc max .
up reinforcement shall not be taken greater
This clause is provided to check the beam for the
than 415 N/mm 2 .
diagonal compression failure.
3.3 6.4 If in any case τv exceeds τc max .i.e., τv > τc max , then
Given : Development length of a deformed the beam should be redesigned because it is weak
reinforcement bar, in shear and may fail due to crushing by diagonal
1  φσ s  compression.
Ld = …(i)
k  τbd  Hence, the correct option is (D).
To find : Value of k=?

3.5 (B) Hence, the design shear force at the section X-X of
the beam is 65 kN.
As per Table No. 20 mentioned in IS 456 : 2000
for limit state of collapse in shear, the maximum Scan for
Video Solution
allowable shear stress ( τ c max ) depends on the
grade of concrete only.
3.7 8.20
Maximum allowable
Grade of Given :
shear ( τ c max ) in
concrete Factored shear force, Vu = 200 kN = 200 × 103 N
N/mm 2
Width of beam, b = 250 mm = 0.25 m
M15 2.5
Effective depth of beam, d = 350 mm = 0.35 m
M20 2.8
M25 3.1 Design shear strength of concrete,
τc = 0.62 N/mm 2
M30 3.5
M35 3.7 Maximum allowable shear stress in concrete,
M40 and above 4.0 τc max = 2.8 N/mm2
Two legged vertical stirrups having diameter,
 Key Point
φ = 10 mm
Also, τ c max = 0.63 f ck To find: Required spacing of shear reinforcement,
Hence, the correct option is (B). Sv = ?
As per clause 40.1, mentioned in IS 456 : 2000,
3.6 65
Nominal shear stress in beam,
X
5m V
10 kN/m τV = u
bd
600 mm d
400 mm
200 ×103
b τV = = 2.286 N/mm 2
250 × 350
X
Effective span = 20 m ∴ τV < τc max
Shear force at section X-X, Hence, OK (as per clause 40.2.3, IS46: 2000)
Vu = 100 − 5 × 10 = 50 kN But, τV > τc , therefore, from clause 40.4, IS 456 :
2000,
Depth at section X-X,
Shear reinforcement would carry a shear force,
200
d = 400 + × 5 = 500 mm = 0.5 m Vus = (Vu − τc ) bd
10
Vus = τV bd − τcbd = (τV − τc )bd
Moment at section X-X,
Vus = (2.286 − 0.62) × 250 × 350
M u = 100 × 5 − 10 × 2.5 × 5 = 375kNm
Vus = 145775 N = 145.775 kN
Design shear force at section X-X,
∴ For vertical stirrups, strength of shear
IS clause 40.1.1 beams of varying depth
reinforcement,
Mu
Vu ,design = Vu +
tan β 0.87 f y Asv d
d Vus =
SV
β = Angle between top & bottom edges
0.87 f y Asv d
375 200 Sv =
Vu ,design = 50 + × = 65 kN Vus
0.5 10000
 π 
0.87 × 250 ×  2 × × (10)2  × 350 3.10 101.63
Sv =  4 
Given : Width, B = 300 mm
145775
Sv = 82.03 mm  8.20 cm Effective depth, d = 580 mm
Hence, the required spacing (up to one decimal Used M30 grade concrete and Fe415 steel
place) as per limit state method will be 8.20 cm. pt = 1%
3.8 700 τc = 0.66 Mpa
Given : Diameter of tie bar (d ) = 12 mm π
Asv = 2 × ×102 = 157.08 mm 2
Working tensile stress (σ st ) = 230 N/mm 2 4
Average bond strength between a tie bar and Vus = 400 kN
concrete Sb = 2 N/mm 2 440 ×103
τv = = 2.52 N/mm 2
Thickness of slab (t ) = 10 mm 300 × 580
d σ st 0.87 f y Asv d
Length of tie bar = t + Vus = (τv − τs ) bd =
2 × Sb Sv
12 × 230 = (2.52 − 0.66) × 300 × 580
= 10 + = 10 + 690
2× 2
0.87 × 415 × 157.08 × 580
= 700 mm =
Sv
3.9 (D)
Sv = 101.63mm
Reinforcing steel bar, partially embedded in
Check : 0.75d = 0.75 × 580 = 435 mm
concrete, is subjected to a tensile force P, bond
stress distribution shown in figure, 0.87 f y σ sv
( Sv ) min = = 472 mm
L 0.46
S v = 101.63 mm
P
3.11 A, B, D
umin
uavg In case of excessive shear reinforcement,
umax concrete become stronger in diagonal on failure
Bond stress distribution and compression Failure may occur before the
shear reinforcement has yielded.
The main reinforcement increase shear resistance
P
of beam.
Nominal shear stress for beam with varying
P P M
Vu ± u ( tan β)
depth, τv = d
 Key Point bd
● For bars in compression, the values of bond Where, τv depend on both design shear force as
stress for bars in tension shall be increased well as the design bending moment.
by 25 %.
Hence, correct options are (A), (B) and (D).
● The deformed bars conforming to IS : 1786

these values shall be increased by 60 %.

4 Column & Footing
2014 IIT Kharagpur 4.4 An RCC short column (with lateral ties) of
rectangular cross-section of 250 mm × 300
4.1 While designing, for a steel column of Fe mm is reinforced with four numbers of 16
250 grade, a base plate resting on a concrete
mm diameter longitudinal bars. The grades
pedestal of M 20 grade, the bearing strength
of steel and concrete are Fe 415 and M 20
of concrete (in N/mm2) in limit state method respectively. Neglect eccentricity effect.
of design as per IS 456 : 2000 is ______.
Considering limit state of collapse in
[1 Mark, Set-1] compression (IS 456 : 2000), the axial load
2015 IIT Kanpur carrying capacity of the column (in kN) up
4.2 A column of size 450 mm × 600 mm has to one decimal place is _______.
unsupported length of 3 m and is braced [2 Marks, Set-1]
against side sway in both directions. 2021 IIT Bombay
According to IS 456 : 2000, the minimum 4.5 A combined trapezoidal footing of length L
eccentricities (in mm) with respect to major
supports two identical square columns ( P1
and minor principle axes are
[1 Mark, Set-2] and P2 ) of size 0.5 m × 0.5m , as shown in
(A) 20 and 20 (B) 26 and 21 the figure. The column P1 and P2 carry loads
(C) 26 and 20 (D) 21 and 15 of 2000 kN and 1500 kN respectively.
2018 IIT Guwahati L
4.3 A structural member subjected to

compression has both translation and
rotation restrained at one end, while only 5m 1.5m
translation is restrained at the other end. As P1 P2
per IS 456 : 2000, the effective length factor
recommended for design is 5m
[1 Mark, Set-2]
If the stress beneath the footing is uniform,
(A) 0.50 (B) 0.65
the length of combined footing L (in m,
(C) 0.70 (D) 0.80 round off to two decimal places) is
________. [2 Marks]
4.1 9 Hence, the bearing strength of concrete in limit

state method of design as per IS 456 : 2000 is 9
Grade of concrete = M30 (∴ f ck = 30 N/mm 2 ) N/mm2.
Grade of steel = Fe500 (∴ f y = 500 N/mm 2 )  Key Point
As per IS 456 : 2000, According to clause 38.1, IS 456 : 2000,
Permissible bearing stress in concrete The relationship between compressive stress
= 0.446 f ck = 0.446 × 20 = 9N/mm 2 distribution in concrete and strain in concrete may
be assumed to be rectangle, trapezoid, parabola or
RCC Structure & Pre-Stress Concrete 1

any other shape which results in prediction of
strength in substantial agreement with the results Let us consider x-x be the major axis and y-y be
of tests. An acceptable stress-strain curve. For minor axis. Then,
design purposes, the compressive strength of S. No. Eccentricity calculation
concrete in structure is assumed to be 0.67 times 1. Minimum eccentricity with respect
the characteristic strength. The partial safety factor to minor axis (x-x) :
γ m = 1.5 is applied in addition to this. Thus the L D
∴ emin yy = +
permissible design strength of concrete is taken as 500 30
0.67 f ck 3000 450
= 0.446 f ck emin yy = + = 21 mm
1.5 500 30
For short axially loaded column : 2. Minimum eccentricity with respect
When a short column is axially loaded, the strain to major axis (y-y) :
distribution across the section will be rectangular. L B
emin xx = +
At failure, the strain in the concrete will be 500 30
uniform at 0.002. The stress in concrete will be, 3000 600
emin xx = + = 26 mm
0.67 f ck 0.67 f ck 500 30
= = 0.446 f ck
γ mc 1.5
As per clause 25.4, IS 456 : 2000,
The steel may develop full design stress in the case The minimum value of eccentricity is the
of mild steel reinforcement when the concrete maximum of the following :
attains a limiting strain of 0.002. However, with
L D
cold twisted reinforcements ( f y = 415 N/mm 2 or (a) emin = +
500 30
f y = 500 N/mm2 ), the full design stress will not (b) emin = 20 mm
develop at a strain of 0.002. Hence the correct option is (B).
The maximum stresses in the materials, at a strain 4.3 (D)
of 0.002 is tabulated as:
Let, L = Unsupported length of compression
Material Stresses
member
Concrete 0.446 f ck
As per ANNEX-E, Table No.-28, and IS 456 :
Mild steel 0.87 f y 2000,
Fe 415 steel 0.796 f y For a structural member subjected to compression
has both translation and rotation restrained at one
Fe 500 steel 0.75 f y
end, while only translation is restrained at the
other end, the effective length factor
4.2 (B) recommended for design, Leff = 0.80 L .
y
600 mm
x x
Refer Solution 4.5 table
4.4 918.1
y
Given :
450 mm Grade of concrete = M20 (∴ f ck = 20 N/mm 2 )

Grade of steel = Fe415 (∴ f y = 415 N/mm 2 ) = 2.393 m
h
Dimension of column = 250 mm × 300 mm
Number of bars in longitudinal reinforcement, n
=4
Diameter of bars, φ = 16 mm a
b
Neglecting eccentricity effect.
Gross area of the column section,
Ag = 250 × 300 = 75000 mm 2
Reference
Area of steel,
π C.G. of trapezium
Asc = 4 × × 16 2 = 804.25 mm 2 2a + b h 1.5 × 2 + 5 L
4 = × = ×
∴ Area of concrete, a+b 3 1.5 + 5 3
Ac = Ag − Asc = 0.410 L
2.393 = 0.410 L
Ac = 75000 − 804.25 = 74195.75 mm 2
L = 5.833 m
Here, the eccentricity effect is neglected for the
column. Hence, as per clause 39.6, IS 456 : 2000, 
Axial load carrying capacity on the column,
Puz = 0.45 f ck Ac + 0.75 f y Asc
Puz = 0.45 × 20 × 74195.75
+ 0.75 × 415 × 804.25
Puz = 918084.563 N  918.1 kN
Hence, the axial load carrying capacity of the
column up to one decimal place is 918.1 kN.
4.5 5.833
Reference
L
2000 kN 1500 kN
5m 1.5 m
P1 P2
5m
0.25
A1Y1 + A2Y2
C.G. of load =
A1 + A2
C.G. of load
(2000 × 0.25) + [1500 × (5 + 0.25)]
=
2000 + 1500

5 Prestressed Concrete
(Based on 1343:1980)
2013 IIT Bombay 2017 IIT Roorkee

5.1 A rectangular concrete beam 250 mm wide 5.3 A pre-tensioned rectangular concrete beam
150 mm wide and 300 mm depth is
and 600 mm deep is prestressed by means of
prestressed with three straight tendons, each
16 high tensile wires, each of 7 mm
having a cross-sectional area of 50 mm2 , to
diameter, located at 200 mm from the bottom an initial stress of 1200 N / mm2 . The
face of the beam at a given section. If the tendons arc located at 100 mm from the
effective prestress in the wires is 700 MPa, soffit of the beam. If the modular ratio is 6,
what is the maximum sagging bending the loss of prestressing force (in kN), up to
moment (in kNm) correct to one decimal one decimal place due to the elastic
place due to live load which this section of deformation of concrete only is ______.
[2 Marks]
the beam can withstand without causing
5.4 A simply supported rectangular concrete
tensile stress at the bottom face of the beam?
beam of span 8 m has to be prestressed with
Neglect the effect of dead load of beam
a force of 1600 kN. The tendon is of
_______. parabolic profile having zero eccentricity at
[2 Marks] the supports. The beam has to carry an
external uniformly distributed load of
2015 IIT Kanpur
intensity 30 kN/m. Neglecting the self-
5.2 In a prestressed concrete beam section weight of the beam, the maximum dip (in m)
(shown in the figure), the net loss is 10% and up to two decimal places of the tendon at the
the final prestressing force applied at X is mid-span to balance the external load should
750 kN. The initial fiber stresses (in N/mm2) be ________.
at the top and bottom of the beam are [2 Marks]
[2 Marks] 2018 IIT Guwahati
b 5.5 A 6 m long simply supported beam is
b = 250 mm prestressed as shown in the figure.
d1 = 200 mm Neutral
d1
d 2 = 100 mm axis
Prestressing cable
X (straight profile)
d1
The beam carries a uniformly distributed
d2
load of 6 kN/m over its entire span. If the
(A) 4.166 and 20.833 effective flexural rigidity EI = 2 ×104 kNm 2
(B) – 4.166 and – 20.833 and the effective prestressing force is 200
kN, the net increase in length of the
(C) 4.166 and – 20.833
prestressing cable (in mm) up to two decimal
(D) – 4.166 and 20.833
places is ________. [2 Marks]
2020 IIT Delhi experienced at point ‘Q’ will be _______.
[2 Marks]
5.6 A simply supported prismatic concrete beam
of rectangular cross- section, having a span 2022 IIT Kharagpur
of 8 m, is prestressed with an effective 5.9 A post tension concrete member 15 m and
prestressing force of 600 kN. The cross-section 450 mm × 450 mm (3 steel
eccentricity of the prestressing tendon is zero tendons) each of cross-section area
at supports and varies linearly to a value of e
200 mm2 , The tendons are tensioned one
at the mid-span. In order to balance an
external concentrated load of 12 kN applied after another to a stress of 1500 MPa. All
at the mid-span, the required value of e (in tendons are straight located at 125 mm from
mm, round off to the nearest integer) of the bottom of the member Assume prestress to
tendon, is ______. [2 Marks] be same in all tendon, Modular ratio 6. The
average loss of prestress due to elastic
5.7 A concrete beam of span 15 m, 150 mm wide
deformation of concrete considering all three
and 350 mm deep is prestressed with a
tendons. [2 Marks]
parabolic cable as shown in the figure (not
drawn to the scale). Coefficient of friction (A) 28.32 MPa (B) 14.16 MPa
for the cable is 0.35, and coefficient of wave (C) 42.48 MPa (D) 7.08 MPa
effect is 0.0015 per metre.
Centroidal axis
50 mm
350 mm
50 mm
70 mm
7.5 m
Cable
15 m
If the cable is tensioned from one end only,
the percentage loss (round off to one decimal
place) in the cable force due to friction, is
______. [2 Marks]
2021 IIT Bombay
5.8 A prismatic cantilever prestressed concrete
beam of span length, L = 1.5 m has one
straight tendon placed in the cross-section as
shown in the following figure (not to scale).
The total prestressing force of 50 kN in the
tendon is applied at dc = 50 mm from the top
in the cross-section of width, b = 200 mm
and depth, d = 300 mm.
Prestressing tendon
P
dc
X
Q d
X'
L b
Section X- X'
If the concentrated load, P = 5 kN, the
resultant stress (in MPa, in integer)

5.1 86.206 Hence, the effect of dead load of beam 86.206
kNm.
Given : Width of beam, b = 250 mm
5.2 (D)
Depth of beam, d = 600 mm
Given : Net loss = 10%
Number of tensile wires, n = 16 Final prestressing force applied at X after loss =
Diameter of tensile wires, φ = 7 mm 750 kN
Effective prestress in wires, P = 700MPa Initial prestressing force,
250 750
P=
+ 0.9
P = 833.33kN = 833.33×103 N
600
e = 100 mm Eccentricity, e = d1 − d2 = 200 − 100 = 100 mm
200 mm
+ Initial stresses at top and bottom fiber of beam is

P Pe M given by,
y y
A I I P Pe P Pe
f =  = 
Since the prestressing force is located at 200 mm A Z A  bd 2 
 
from the bottom face of the beam.  6 
∴ Eccentricity, Then, calculating the initial stresses at top and
600 bottom fiber of beam as :
e = 600 − − 200 = 100 mm (i) Initial stresses in top fiber of beam,
2
Prestress force in all wire,  833.33 ×103 
ftop = 
π  250 × 400 
P = 700 × (7)2 ×16 = 431.03 ×103 N
4  833.33 ×103 ×100 × 6 
Since, the tensile stress at bottom face of the beam − 
 250 × 4002 
is zero i.e., ( f = 0) . Then,
f top = 8.33 − 12.50
Stress at bottom fiber
P Pe M P Pe M ftop = −4.17 N/mm2 (tensile)
f = + y− y= + −
A I I A Z Z (ii) Initial stresses in bottom fiber of beam,
P
f = +
Pe
−
M  833.33 ×103 
A  bd   bd 2 
2 fbottom =  
     250 × 400 
 6   6 
 833.33 ×103 ×100 × 6 
3
431.03 ×10 431.03 × 10 ×100 × 6 3 + 
0= +  250 × 4002 
250 × 600 250 × 6002
fbottom = 8.33 + 12.50
M ×6
−
250 × 6002 f bottom = 20.83 N/mm 2 (Compressive)
 2.87 + 2.87 = 6.66 ×10−8 M Hence, the correct option is (D).
∴ Maximum sagging bending moment, Scan for
M = 86206000 Nmm = 86.206 kNm Video Solution

5.3 4.8 5.4 0.15
30 kN/m
Given : Width of beam, b = 150 mm
Depth of beam, d = 300 mm
P P
Number of tendons, n = 3
e
Area of cross-section of tendons, Ast = 50 mm 2
Initial stress, f initial = 1200 N/mm 2
Distance of tendons from soffit of beam
We know maximum bending moment ( BM max )
= 100 mm
due to UDL on simply supported beam,
Modular ratio, m = 6
wl 2
( BM ) max =
8
P
e = 50 mm
P ( BM )max = Pe
100 mm
BM max
Central dip e =
P
Prestressing force, wl 2 30 × 82
e= =
P = n × Ast × finiitial 8P 8 ×1600
e = 0.15 m
P = 3 × 50 ×1200 = 180000 N
Hence, the external load should be 0.15 m.
Eccentricity,
5.5 0.12
D   300 
e =  − 100  =  − 100  = 50 mm
2   2  Given : Span of beam = 6 m=6000mm
Stress in concrete at the location of steel, Flexural rigidity,
P P×e EI = 2 ×104 kNm2
fc = + ×e
A I EI = 2 ×104 ×103 ×106 = 2 ×1013 Nmm2
180000 180000 × 502 Prestressing force, P = 200 kN
fc = +
150 × 300  150 × 3003  Total load = 6 kN/m
  Eccentricity, e = 50 mm
 12 
f c = 4.0 + 1.33 = 5.333 N/mm 2 Step 1 : Slope of beam due to P-force is given by,
Loss of prestress due to elastic deformation in q1 q1

prestressing,
Δf = m × f c = 6 × 5.3333 = 32 N/mm 2 L/2 L/2
P.e
Δf = 32 ×10−3 kN/mm2
Loss of prestresing force, PeL
θ1 =
ΔP = n × Ast × Δf 2 EI
ΔP = 3 × 50 × 32 ×10−3 = 4.8kN 200 ×103 × 50 × 6000
θ1 =
2 × 2 × 1013
Hence, the elastic deformation of concrete only is
4.8 kN. θ1 = 1.5 × 10 −3 (upward)

Step 2 : Slope of beam due to UDL is given by, 12 × 103 × 8 × 103
600 × 103 × 1× e =
wL3 4
θ2 =
24 EI 12 ×103 × 8 ×103
q2 q2 e= = 40 mm
4 × 600 ×103
Hence, the required value of e of the tendon is 40
2
mm.
wl /8
5.7 4.5
L/2 L/2
6 × (6000)3 Given : Span of concrete beam ( L) = 15 m
θ2 = = (+) 2.7 ×10−3
24 × 2 ×10 13 Width of beam = 150 mm
(Downward) Depth of beam = 350 mm
Step 3 : Net slope of beam, Coefficient of friction (μ) = 0.35
θ = θ1 + θ2 Coefficient of wave effect (k ) = 0.0015 per metre
θ = (−)1.5 ×10−3 + (+) 2.7 ×10−3
θ = 1.2 ×10−3
Step 4 :
e q q
eq eq Assume initial pre-stress force = P0

Total net increase in length of cable
Loss of pre-stress due to friction = P0 (kx + μα)
= 2e θ = 2 × 50 ×1.2 ×10−3 = 0.12 mm
For tensioning one end only x = L = 15 m
Hence, the net increase in length of the
prestressing cable up to two decimal places is 0.12
8h 8 × (70 + 50)
α= = = 0.064
mm. L 15000
So, loss = P0 (0.0015 ×15 + 0.35 × 0.064)
Scan for
Video Solution = 0.0449P0
Lossin force
5.6 40 So, percentage loss = ×100
Initial force
Given : Span of beam = 8 m 0.0449 P0
Effective prestressing force = 600 kN = ×100 = 4.49 ≈ 4.5%
P
External concentrated load = 12 kN
12 kN Hence, the percentage loss in the cable force due
to friction is 4.5 %.
P P = 600 kN
5.8 0
Given :
e
5 kN
50 mm
Q P 300 mm
8m
1.5 m
wl
P cos θ× e = 200 mm
4
θ=0 0 Let us calculate,
cos θ = 1 Bending moment = 5 × 1.5 = 7.5 kN-m

P 50, 000 300 ×103 (300 × 103 ) × 1002
= = 0.833 fc = +
A 200 × 300 450 × 450 450 × 4503
Pe 50, 000 ×100 12
= = 1.67
z 200 × 3002
 bd 3 
6  I = 12 
 
M 7.5 ×106
= = 2.5 f c = 2.36 MPa
z 200 × 3002
6 Δσ = m × f c
Stress at top = 0.833 + 1.67 − 2.5 = 6 × 2.36 = 14.16 MPa
−3 2
= 3 ×10 N/mm Here three tendons so average loss of prestress
(tendons)
+ +
Zero stress 14.16 MPa stress 2 × 14.16 MPa stress
0 + 14.16 + 2(14.16)
(Δσ) avg =
+ + 3
P Pe M ( Δσ) avg = 14.16 MPa
A τ τ
Resultant stress is Q = 0. 
5.9 (B)
Given : Length L = 15 m
b × d = 450 ∗ 450
Modular ratio = 6
Section area = 200 mm2
Tendons stress = 1500 MPa
450 mm
e = 100 mm
125 mm
450
Total load P = stress * Area
1500 × 200
=
1000
= 300 kN
( Δσ) average = ? (due to elastic shorting)
(Δσ) elastic shorting = m.7c
P Pe2
fc = +
A I

DESIGN OF STEEL
STRUCTURE
Syllabus : Design of Steel Structure
Working stress and Limit state design concepts; Design of tension and compression members,
beams and beam- columns, column bases; Connections - simple and eccentric, beam-column
connections, plate girders and trusses; Concept of plastic analysis -beams and frames.
Contents : Design of Steel Structure
S. No. Topics
1. General Design Requirements
2. Connections
3. Design of Tension Members
4. Design of Compression Members
5. Design of Beams
6. Design of Plate Girders, Industrial Roof Truss and
Gantry Girders
7. Plastic Analysis
1 General Design Requirements
1995 IIT Kanpur 1.4 Factor of safety adopted by IS 800 : 1984

while arriving at the permissible stress in
1.1 As per IS 800 : 1984, the minimum pitch of axial compression is [1 Mark]
rivets in a row is recommended as the
(A) 2.00 (B) 1.00
diameter of the rivet times
(C) 1.67 (D) 1.50
[1 Mark]
(A) 2.0 (B) 2.5 2005 IIT Bombay
(C) 3.0 (D) 4.0 1.5 Which one of the following is not correct for
steel sections as per IS 800 : 1984?
1996 IISc Bangalore
[2 Marks]
1.2 Generally the maximum deflection/span (A) The maximum bending stress in tension
ratio of a steel member should not exceed. or in compression in extreme fiber
(A) 1/750 (B) 1/500 calculated on the effective section of a
(C) 1/325 (D) 1/250 beam shall not exceed 0.66 f y
1.3 Generally, fatigue life of welded steel (B) The bearing stress in any part of a beam
structure to fatigue life of riveted steel when calculated on the net area shall not
structure ratio is [1 Mark] exceed 0.75 f y
(A) Smaller than 1
(C) The direct stress in compression on the
(B) Equal to 1 cross-sectional area of axially loaded
(C) Greater than 1 compressive member shall not exceed
(D) Greater than 2.1 0.6 f y
1997 IIT Madras (D) None of the above
1.1 (B) Span

δmax (in mm)  (in mm)
325
The minimum spacing between centres of
fasteners shall not be less than 2.5 time the
1.3 (A)
nominal diameter of the fasteners.
Hence, the correct option is (B). Welded connection are weak under fatigue loading
1.2 (C)
and prone to cracking. So fatigue life of welded
structure is always less than the fatigue life of
Maximum deflection in the beam should not
riveted structure.
Span
exceed , Fatiguelifeof weldedstructure
325 <1
Fatiguelifeof rivetedstructure
Steel Structures 1
1.4 (C)
Permissible stress in axial compression as per IS

800 : 1984
i.e., allowable  0.6 f y
 1 
 Factor of safety     1.67
 0.6 
1.5 (D)
All statements (a), (b) and (c) are correct

according to IS 800 : 2007.

2 Steel Structures
2 Connections
2014 IIT Kharagpur (D) Forces due to the friction between

connected parts
2.1 The tension and shear force (both in kN) in
2016 IISc Bangalore
each bolt of the joint, as shown below,
respectively are 2.4 Two plates are connected by fillet welds of
[2 Marks, Set-1] size 10 mm and subjected to tension, as
ultimate tensile stress of steel are 250 MPa
4 and 410 MPa, respectively. The welding is
5
3 Pu = 250 kN done in the workshop ( γ mw = 1.25) .
P
(A) 30.33 and 20.00 100 mm
(B) 30.33 and 25.00
(C) 33.33 and 20.00
(D) 33.33 and 25.00
2015 IIT Kanpur
2.2 A bracket plate connected to a column flange
transmits a load of 100 kN as shown in the
following figure. The maximum force for
which the bolts should be designed is
_______ kN. 150 mm
[2 Marks, Set-1]
P
100 kN As per the Limit State Method of IS 800:
600 2007, the minimum length (rounded off to
the nearest higher multiple of 5 mm) of each
75 weld to transmit a force P equal to 270 kN
75
(factored) is [2 Marks]
75 75 (A) 90 mm (B) 105 mm
All dimension
are in mm
(C) 110 mm (D) 115 mm
2.5 Two bolted plates under tension with
2.3 Prying forces are [1 Mark, Set-2] alternative arrangement of bolt holes are
(A) Shearing forces on the bolts because of shown in figures 1 and 2. The hole diameter,
the joints pitch and gauge length are d , p and g ,
(B) Tensile forces due to the flexibility of respectively.
connected parts
P P P g P
(C) Bending forces on the bolts because of d
p
the joints
Steel Structures 1
Which one of the following conditions must 2.9 A fillet weld is simultaneously subjected to
be ensured to have higher net tensile capacity factored normal and shear stresses of 120
of configuration shown in figure 2 than that MPa and 50 MPa, respectively. As per IS
shown in figure 1? 800 : 2007 the equivalent stress (in MPa, up
[2 Marks, Set-2] to two decimal places) is ______.
(A) p 2 > 2 gd (B) p 2 > 4 gd [1 Mark, Set-2]
2.10 Four bolts P, Q, R and S of equal diameter
(C) p 2 > 4 gd (D) p > 4 gd are used for a bracket subjected to a load of
2017 IIT Roorkee 130 kN as shown in the figure.
[2 Marks, Set-2]
2.6 Two plates of 8 mm thickness each are Centerline
connected by a fillet weld of 6 mm thickness 200 mm
as shown in the figure.
6 130 kN
Q P
240 mm
P 50 mm 100 mm P
R S
100 mm
The permissible stresses in the plate and
the weld are 150 MPa and 110 MPa,
100 mm
respectively. Assuming the length of the
The force in bolt P is
weld shown in the figure to be the effective
length, the permissible load P (in kN) is (A) 32.50 kN (B) 69.32 kN
_______. [2 Marks] (C) 82.50 kN (D) 119.32 kN
2.7 A column is subjected to a load through a 2019 IIT Madras
bracket as shown in the figure.
2.11 A 16 mm thick gusset plate is connected to
15 cm P = 10 kN
cm the 12 mm thick flange plate of an I – section
10 using fillet welds on both sides as shown in
1 figure (not drawn to scale). The gusset plate
90
0 is subjected to point of 350 kN acting at a
distance of 100 mm from the flange plate
size of filet weld is 10 mm.
[2 Marks]
10
cm
350 kN
100 mm
The resultant force (in kN) up to one decimal
place in the bolt 1 is ________. 16 mm thick
gusset plate
[2 Marks] 500 mm
500 mm
2018 IIT Guwahati Fillet weld
I-section Fillet weld

2.8 In a fillet weld, the direct shear stress and Flange (12 mm thick)
bending tensile stress are 50 MPa and 150 (Front view) (Side view)
MPa respectively. As per IS 800 : 2007, the The maximum resultant stress (in MPa,
equivalent stress (in MPa, up to two decimal round off to 1 decimal place) in the fillet
places) will be _______. weld along the vertical plane would be
[1 Mark, Set-1] _______.
2 Steel Structures
2020 IIT Delhi weld is 108 MPa. The location of the
centroid of the angle is represented by C yy
2.12 Two steel plates are lap jointed in a
workshop using 6 mm thick fillet weld as in the figure, where C yy = 28.4 mm. The
shown in the figure (not drawn to the scale), area of cross-section of the angle is
The ultimate strength of the weld is 410 1903 mm 2 . Assuming the effective throat
MPa. thickness of the weld to be 0.7 times the
given weld size, the lengths L1 and L2
(rounded-off to the nearest integer) of the
120 mm
Fillet weld required to transmit a load equal to the
weld full strength of the tension member are,
respectively
200 mm
5mm weld
L1
As per Limit State Design of IS 800 : 2007,
the design capacity (in kN, round off to three Cyy
decimal places) of the welded connection, is 100 mm
_____. [2 Marks]
L2
5mm weld
2021 IIT Bombay
15 mm thick gusset plate
2.13 A column is subjected to a total load (P) of
60 kN supported through a bracket (A) 541 mm and 214 mm
connection as shown in the figure (not to (B) 214 mm and 541 mm
scale).
40 40
(C) 380 mm and 151 mm
mm mm
(D) 151 mm and 380 mm
60 kN
[2 Marks]
2023 IIT Kanpur
30 mm
30 mm
R 2.15 Consider the fillet-welded lap joint shown in
the figure (not to scale). The length of the
weld shown is the effective length. The
welded surfaces meet at right angle. The
100 mm weld size is 8 mm, and the permissible stress
in the weld is 120 MPa. What is the safe load
The resultant force in bolt R (in kN, round P (in kN, rounded off to one decimal place)
off to one decimal place) is ________. that can be transmitted by this welded joint?
[2 Marks] 75 mm
2022 IIT Kharagpur
50 mm
P P
2.14 A weld is used for joining an angle section
ISA 100 mm ×100 mm ×10 mm to a gusset
plate of thickness 15 mm to transmit a tensile
load. The permissible stress in the angle is
150 MPa and the permissible shear stress on
the section through the throat of the fillet
Steel Structures 3
2.1 (D) 75
tan θ =
75
tan θ = 1 (∴ θ = 450 )
FR = F12 + F22 + 2F1F2 cos θ
4
5
3 Pu = 250 kN FR = (20)2 + (141.42)2 + 2 × 20 ×141.42cos 450
FR = 156.2 kN
Hence, the bolts should be designed is 156.2 kN.
Pu = 250 kN
4 Scan for
cos θ= Video Solution
5
3
Sin θ = 2.3 (B)
5
Pu cos θ Prying force is additional tensile force develop in
Tension in each bolt = bolt due to flexibility of connection.
6 2P 2P
250 4
= × = 33.33kN
6 5
P sin θ
Shear in each bolt = u
6
250 3
= × = 25 kN
6 5
Hence, the correct option is (D). 2P 2P
2P
Scan for
Video Solution
2.2 156.2
Q P+Q P+Q Q
75 F2
q P+Q P+Q
F1 Q Q
75
75 2P
75
100 Q = Prying force
We know, F1 = = 20 kN
5 Hence, the correct option is (B).
Per (100 × 0.6) × (75 2) ×10−3 2.4 (B)
F2 = =
 r2 (75 2)2 × 4 ×10−6 Refer Solution No. 2.19
F2 = 141.42 kN Hence, the correct option is (B).
4 Steel Structures
Always Remember Pplate = σat × Anet
For safe weld design = 150 × (50 × 8)
Vdw ≥ Tdg = 60 kN
Vdw = Design strength of weld So permissible load = 60 kN
Tdg = Gross strength of plate Hence, the permissible load P is 60 kN.
Scan for
2.5 (C)
Video Solution
P P P g P
d
p 2.7 6
Tensile strength of plate in arrangement 2 will be Given : Number of bolts, n = 4

greater than in arrangement 1 Applied load, P = 10 kN
As per IS 800 : 2007 clause 6.3, P
Pe
10 kN
 Fup   Fup  F2
 0.9 Anet  >  0.9 Anet 
 γ m1   γ m1  1 1350
2 1
F1 r = 5cm
( Anet )2 > ( Anet )1
 P2  
 B − 2 d +  t  > [ ( B − d )t ]1
 4g  2
15cm
P2 P 10
B − 2d + > B−d F1 = = = 2.5 kN
4g n 4
P2 Pe 10 × 15
>d F2 = × r1 = × 5 = 7.5kN
4g  r12
4 × 52
P 2 > 4 gd FR = F12 + F22 + 2F1F2 cos θ
Hence, the correct option is (C). FR = 2.52 + 7.52 + 2 × 2.5 × 7.5cos1350
2.6 60 FR = 6 kN
Given : Permissible stress in plate σat = Hence, the resultant force up to one decimal place
in the bolt 1 is 6 kN.
150 MPa
Permissible stress in weld Sb = 110 MPa Scan for
Video Solution
6
2.8 173.20
Given : Direct bending tensile stress
P 50 mm 100 mm P
f a = 150 MPa
100 mm fu = 410 MPa
Pw = Fb × Effective area Direct shear stress q = 50 MPa
According to IS 800 : 2007, clause 10.5.10.1.1,
= 110 × (leff × tt )
the equivalent stress,
= 110(100 + 100 + 50) × 0.7 × 6 fu
f e = f a 2 + 3q 2 ≤
= 115.5kN 3 γ mw
Steel Structures 5
410 FR = F12 + F22 + 2F1F2 cos θ
f e = 1502 + 3 × 502 ≤
3 ×1.25
f e = 173.20 ≤ 189.37 (Hence ok) FR = 32.52 + 50 2 + 2 × 32.5 × 50 × 0.384
So equivalent stress = 173.20 MPa . FR = 69.33kN
Hence, the equivalent stress will be 173.20 MPa. Hence, the correct option is (B).
Scan for 2.11 78.10
Video Solution
t1 t2
2.9 147.99 350 kN
P
Given : Factored normal stress, f a = 120 MPa 100mm
500 mm Pe = m
fu = 410 MPa 500mm
Factored shear stress, q = 50 MPa

According to IS 800 : 2007, clause 10.5.10. 10.1.1
the equivalent stress, Given : P = 350 kN, e = 100 mm
f e = f a2 + 3q 2 ≤
fu S = Size of weld = 10 mm
3 γ mw tt = Effective throat thickness
410
f e = 1202 + 3 × 502 ≤ tt = 0.7 S
3 ×1.25
tt = 0.7 ×10
f e = 147.99 ≤ 189.37
So equivalent stress = 147.99 MPa tt = 7 mm
Hence, the equivalent stress is 147.99 MPa. Shear stress due to direct force
2.10 (B) P
q=
200 mm 2 × dt
130 KN
350 ×103
q= = 50 MPa
Q
2 × 500 × 7
P
240 mm q F2 Normal stress due to bending
R S F1
My Pe × d / 2
Fb = =
I t d3
2× t
P 130 12
F1 = = = 32.5 kN
n 4 350 × 103 × 100 × 500
Fb = = 60 MPa
Pe 5003
F2 = × rp 2×7× ×2
 r12 12
{ rp = rQ = rR = rs = 502 + 1202 = 130 mm } Resultant stress = Fb2 + q 2

(130 × 200) × 130 Fr = 602 + 502
F2 =
4 × 130 2
200 Fr = 78.10 MPa
F2 = = 50 kN
4 Hence, the maximum resultant stress in the fillet
50 weld along the vertical plane would be 78.10
And cos θ = = 0.384
130 MPa.
6 Steel Structures
Pe.r1
Scan for F2 =
Video Solution Σr12
60 × 100 × 40
2.12 413.59 F2 = = 18.18 kN
Σ 4 × 50 2 + 2 × 40 2
Given : Thickness of weld, t = 6 mm ,
F2
Throat thickness, tt = 0.7 × t ,
F2
tt = 0.7 × 6
180 0
θ = 180
0
F1
Ultimate strength of weld = 410 MPa
Design capacity of welded connection,
F1
Ps = fb × leff × tt
R = 102 + 18.182 + 2 × 10 × 18.18 cos1800
F1 θ = 0
F2
R = 102 + 18.182 + 2 ×10 ×18.18cos 00
R = 28.18 kN
2.14 (A)
410
P= × 520 × 0.7 × 6 Case 1 : (Ignoring shear lag)
3 ×1.25
Full strength of tension member,
716352
= = 413598 N P = σsat × Ag
3
= 413.586 kN P = 150 ×1903
P = 285450 N
Hence, the design capacity of the welded
connection, is 413.59 kN. Weld strength of L1 ,
2.13 28.18 P1 = L1 × (0.7 × s) ×108
40 40 P1 = L1 × 0.7 × 5 ×108
mm mm
60 kN P1 = 378L1 …(i)
Weld strength of L2 ,
30 mm P2 = L2 × (0.7 × s) ×108
R
30 mm P2 = L2 × 0.7 × 5 ×108
P2 = 378 L2 …(ii)
Applying moment CG equal to zero,
100 mm
P1 × 28.4 = P2 (100 − 28.4)
Resultant force, R = F12 + F22 + 2 F1F2 cos θ P1 71.6
=
Total load ( P) P2 28.4
F1 =
Number of bolt (n) L1 × 378 71.6
= = 2.521
60 L2 × 378 28.4
F1 = = 10 kN
6 L1 = 2.521 L2 …(iii)
Steel Structures 7
( L1 + L2 ) × 3.5 ×108 = 285450
L1 + L2 = 755.158
Using equation (iii),
2.521 L2 + L2 = 7585.158
L2 = 214.472 mm
L1 = 540.685mm
Case 2 : (Consider shear lag)
Anet = A1 + K1. A2
Equal angle section, K = 0.75 Anet
Anet = 1.75 × ( A1 )
1903
= 1.75 × = 1665.125 mm 2
2
Full strength of tension member
= 150 ×1665.125s
= 249768.75 N
( L1 + L2 ) × 3.5 ×108 = 249768.75
L1 + L2 = 660.763 mm
Let us check the options (C) and (D).
None of the options full fill
L1 + L2 = 660.763 mm
2.15 134.4
Given : Weld size (S) = 8 mm
Permissible stress ( σ ps ) = 120 MPa
Psafe = K × S × leff × σ ps
= 0.7 × 8 × (75 + 75 + 50) ×120
= 134.40 kN

8 Steel Structures
3 Design of Tension Member
1995 IIT Kanpur angles are connected, one each on either side
of a 10 mm thick gusset plate, by 18 mm
3.1 The net effective cross-sectional area
diameter rivets arranged in one row. The
calculated in the steel angle tension
allowable stresses in rivet are
member design accounts for
f s = 90 N/mm 2 and fbr = 250 N/mm2 .
(A) The tensile force and bolt holes
(B) The eccentricity of the end connections 3.3 Maximum tensile stress in the tie in N/mm2
and the bolt holes is [2 Marks]
(C) The effectiveness of the tack connection (A) 93.6 (B) 87.5
along the length (C) 77.2 (D) 66.0
(D) The effectiveness of the end connection 3.4 Minimum number of rivets required at each
ends is [2 Marks]
[1 Mark]
(A) 2 (B) 3
2002 IISc Bangalore (C) 4 (D) 5
3.2 ISA 100 ×100 ×10 mm (cross-sectional area 2004 IIT Delhi
2
= 1908 mm ) serves as tensile member. This 3.5 Two equal angles ISA 100 mm × 100 mm of
angle is welded to a gusset plate along A and thickness 10 mm are placed back-to-back
B appropriately as shown in figure. and connected to the either side of a gusset
Assuming the yield strength of the steel to be plate through a single row of 16 mm
260 N/mm2 the tensile strength of the diameter rivets in double shear. The effective
member can be taken to be approximately areas of the connected and unconnected legs
[2 Marks] of each of these angles are 775 mm 2 and
950 mm 2 , respecti-vely. If these angles are
NOT tack riveted, the net effective area of
ISA 100 ´ 100 ´ 10 the is pair of angles is riveted, the net
effective area of this pair of angles is
[2 Marks]
2
(A) 3650 mm (B) 3450 mm 2
Gusset plate
(C) 3076 mm 2 (D) 2899 mm 2
(A) 500 kN (B) 300 kN
(C) 225 kN (D) 375 kN 2005 IIT Bombay
2003 IIT Madras 3.6 The permissible stress in axial tension σ st in

steel member on the net effective area of the
Common Data for section shall not exceed ( f y is the yield
Questions 3.3 & 3.4
stress) [1 Mark]
A truss tie consisting of 2 ISA 75 × 75 × 8 mm (A) 0.80 f y (B) 0.75 f y
carries a pull of 150 kN. At ends the two (C) 0.60 f y (D) 0.50 f y
Steel Structures 1
2006 IIT Kharagpur safety factor, γ mw = 1.25 ). As per the Limit
3.7 In the design of welded tension members, State Method of IS 800: 2007, what is the
consider the following statements : minimum length (in mm, rounded off to
[2 Marks] the nearest higher multiple of 5 mm)
A. The entire cross-sectional area of the required of each weld to transmit a
connected leg is assumed to contribute to factored force P equal to 275 kN?
100 mm
the effective area in case of angles.
P
B. Two angles back-to-back and tack-

welded as per the codal requirements
may be assumed to behave as a tee
section. P
C. A check on slenderness ratio may be 150 mm
necessary in some cases. (A) 100 (B) 105
The true statements are : (C) 110 (D) 115
(A) A and B (B) B and C
(C) A and C (D) A, B and C
2007 IIT Kanpur
3.8 A steel flat of rectangular section of size
70 × 6 mm is connected to a gusset plate by
three bolts each having a shear capacity of
15 kN in holes having diameter 11.5 mm. If
the allowable tensile stress in the flat is 150
MPa, the maximum tension that can be
applied to the flat is
[2 Marks]
15
20
T
20
15
35
(A) 42.3 kN (B) 52.65 kN

(C) 59.5 kN (D) 63.0 kN
2023 IIT Kanpur
3.9 Two plates are connected by fillet welds of
size 10 mm and subjected to tension, as
ultimate stress of steel under tension are
250 MPa and 410 MPa, respectively. The
welding is done in the workshop (partial
2 Steel Structures
3.1 (A) ISA size = 75 × 75 × 6 mm
Assuming both the angle between sections are not
In tension number,
tack riveted. (Maximum tensile stress is obtained
Net effective cross-sectional area
when angles are not tack bolted/ riveted)
= Gross area – Area of bolt holes
3 A1 3 × 412
Whereas in compression member, ∴ k1 = = = 0.685
3 A1 + A2 3 × 412 + 568
Net-effective cross-sectional area = Gross area
Ae = ( A1 + K1 A2 )2
3.2 (C)
= (412 + 0.685 × 568) × 2 = 1602.2 mm 2
Given : ISA 100 ×100 ×100 mm
P
Cross sectional area = 1908 mm 2 ∴ Maximum tensile stress =
Ae
Yield strength of weld f y = 260 N/mm 2 150 ×103
= = 93.6 N/mm2
1602.3
3.4 (C)
ISA 100 ´ 100 ´ 10
Given :
Diameter of rivet = 18 mm
Diameter of hole d 0 = 18 + 1.5 = 19.5
Gusset plate
Strength of rivets in shearing
A1 = Net area of connected leg
2 × π(19.5)2 × 90
= (100 − 5) ×10 = 950 mm 2 = = 53.76 kN
4
A2 = Net area of outstand leg Strength of rivets in bearing
= (100 − 5) ×10 = 950 mm 2 = 19.5 × 10 × 250 = 48.75kN
∴ Rivet value = 48.75kN
3 A1 3 × 950
K= = = 0.75 150
3 A1 + A2 3 × 950 + 950 No. of rivets = = 3.08 ≈ 4
48.75
Hence Anet = A1 + KA2
= 950 + 0.75 × 950
3.5 (D)
= 1662.5 mm 2
When angles are not tack riveted they will be
Tensile strength of member = 0.6 × f y × Anet considered as single angles connected on one side
= 0.6 × 260 × 1662.5 of gusset plate.
= 259.4 kN Ae = ( A1 + kA2 ) × 2
So on the safe side, tensile strength can be taken 3 A1 3 × 775
k= = = 0.71
as 225 kN. 3 A1 + A2 3 × 775 + 950
3.3 (A) Ae = (775 + 0.71× 950) × 2
Given : Ae = 2899 mm 2
Pull applied P = 150 kW Hence, the correct option is (D).
Steel Structures 3
3.6 (C)  P2 
Anet = ( B − 3 × d 0 ) + × 2 t
The permissible stress is axial tension. σ st shall  4g 
not exceed 0.60 f y .  2 × 352 
= (70 − 3 ×11.5) + ×6
S. Allowable  4 × 20 
Type of stresses FOS
No. stress Anet = 396.75 mm 2
1. Axial tensile (σat ) 0.6 f y 1.67
∴ Maximum tension allowed
2. Maximum axial 0.6 f y 1.67 = Anet × σat = 396.75 × 150 = 59.5 kN
compressive (σat )
Case 3 : By failure along (4)-(5)-(6)-(7)
3. Bending tensile 0.66 f y 1.51
Anet = ( B − 2 × d 0 ) × t
(σbt )
4. Maximum bending 1.51 = (70 − 2 ×11.5) × 6 = 282 mm 2
0.66 f y
compressive (σbc ) ∴ Maximum tension allowed = Anet × σat
= 282 × 150 = 42.3kN
3.7 (D) ∴ Minimum value among all four cases
1. Correct because correction factor k is > applied to A2 i.e. area of outstanding leg. Hence, the correct option is (A).
2. Correct because correction factor k = 1,
Scan for
when angles are connected back to back and Video Solution
tacked. Hence Anet = A1 + A2 as in case of T-
3.9 B
section.
3. Correct because, when slenderness ratio Given : fu = 410 MPa , f y = 250 MPa ,
increases beyond 400, it is called as cable. γ mw = 1.25 and t = 12 mm , Size of weld,
S = 10 mm .
3.8 (A) Factored force P = 275 kN
fu
P= × Lw × 0.7 × S ... ( A )
t = 6 mm 1 4 3γ mw
5 15
20 Assume minimum length of weld required on
70 mm 2 20
6 each size = l
15
3 7 ∴ Lw = 2 × l
35
Substitute P and Lw in equation (A),
2
Given : d 0 = 11.5 m and σat = 150 N/mm 410
275 × 103 = × 2l × 0.7 × 10
Shear capacity of each bolt = 15 kN 3 × 1.25
Case 1 : Let us suppose all bolt fail in shear l = 103.25 mm ≈ 105 mm
∴ Max force allowed = 3 ×15 = 45 kN Hence, the correct option is (B).
Hence only option below 45 kN i.e 42.5kN

Case 2 : By failure along section (4)-(5)-(2)-(6)-
(7)
4 Steel Structures
4 Design of Compression
2015 IIT Kanpur 4.4 The critical bending compressive stress in

the extreme fibre of a structural steel section
4.1 A steel member ‘M’ has reversal of stress is 1000 MPa. It is given that the yield
due to live loads, whereas member ‘N’ has strength of the steel is 250 MPa, width of
reversal of stress due to wind load. As per IS flange is 250 mm and thickness of flange is
800 : 2007, the maximum slenderness ratio 15 mm. As per the provisions of IS 800 :
permitted is 2007. The non-dimensional slenderness ratio
[1 Mark] of the steel cross-section is
(A) Less for member M than that of member [2 Marks]
N. (A) 0.50 (B) 0.75
(B) More for member M than for member N. (C) 0.25 (D) 2.00
(C) Same for both the members.
(D) Not specified in the code.
2018 IIT Guwahati
4.2 A steel column of ISHB 350 @ 72.4 kg/m is

subjected to a factored axial compressive
load of 2000 kN. The load is transferred to a
concrete pedestal of grade M 20 through a
square base plate. Consider the bearing
strength of concrete as 0.45 f ck where, f ck
is the characteristic strength of concrete.
Using limit state method and neglecting the
self-weight of base plate and steel column,
the length of a side of the base plate to be
provided is [1 Mark]
(A) 39 cm (B) 42 cm
(C) 45 cm (D) 48 cm
2019 IIT Madras
4.3 A steel column is restrained against both

translation and rotation at one end and is
restrained only against rotation but free to
translate at the other end. Theoretical and
design (IS 800 : 2007) values, respectively,
of effective length factor of the column are
[1 Mark]
(A) 1.0 and 1.2 (B) 1.2 and 1.0
(C) 1.0 and 1.0 (D) 1.2 and 1.2
Steel Structures 1
4.1 (A) (Self-weight of base plate and steel column is
neglected)
Less for member M than that of member N.
Total load on pedestral = 2000 kN
Bearing capacity of concrete
As per IS 800 : 2007
= 0.45 fck = 0.45 × 20 = 9 MPa
 Key Point
Area of steel base plate = Factored axial load
bearing capacity of concrete
(2000 ×103 )
= = 222222.22 mm2
9
Side of base plate = Area
= 222222.22 = 471.4 mm
= 47.14 cm
The length of a side of the base plate to be
provided = 47.14 = 48cm
4.3 (A)
Degree of end
restraint of Values of effective length
compression
members
Effectively held
in position and
restrained against
rotation in both
ends.
Theoretical value = 0.50l
Recommended
value = 0.65l
Effectively held
in position at both
ends, restrained
Scan for against rotation at
Video Solution one end.
4.2 (D) Recommended
Given : Factored axial load = 2000 kN value = 0.80l
2 Steel Structures
Effectively held 4.4 (A)
in position at both
ends, but not Given :
restrained against Yield strength of steel = 250 MPa
rotation. Width of flange = 250 mm
Thickness of flange = 15 mm
Recommended
value = 1.00l fy 250
λ= = = 0.5
f cr 1000
Effectively held
in position and Hence, the correct option is (A).
restrained against  Key Point
rotation at one
end and at the The strength curve of an axially loaded and
other restrained initially straight hinged column. This plot is
against rotation sometimes represented in non-dimensional form
Recommended
but not held in f cr fy
position. value = 1.20l versus shown in figure.
fy f cr
Effectively held f cr Shown in figure
in position and fy
restrained against Plastic yield
rotation in one 1.0
end, and at the
other partially Theoretical value = Nil Elastic Buckling
restrained against Recommended
rotation but not
value = 1.50l 0 1.0
held in position. fy
λ=
Effectively held f cr
in position at one
end but not Scan for
restrained against Video Solution
rotation, and at
the other end Theoretical value = 2.00l
restrained against 
Recommended
rotation but not
held in position. value = 2.00l
Effectively held
in position and
restrained against
rotation at one
end but not held Theoretical value = 2.00l
in position nor
restrained against Recommended
rotation at the value = 2.00l
other end.
Note : Here, l = Unsupported length of column
Steel Structures 3
5 Design of Beams
2014 IIT Kharagpur

5.1 A steel section is subjected to a combination Beam (I-section)
of shear and bending actions. The applied
shear force is V and the shear capacity of the Bearing plate
section is Vs . For such a section, high shear 75 mm
force (as per IS 800 : 2007) is defined as As per IS 800 : 2007, the web bearing
(A) V > 0.6Vs (B) V > 0.4Vs strength (in kN, round off to 2 decimal
places) of the beam is ________.
(C) V > 0.8Vs (D) V > 0.9Vs
[2 Marks]
[1 Mark, Set-1]
2020 IIT Delhi
2016 IISc Bangalore
5.5 The flange and web plates of the doubly
5.2 The semi-compact section of a laterally symmetric built-up section are connected by
unsupported steel beam has an elastic section continuous 10 mm thick fillet welds as
modulus, plastic section modulus and design
shown in the figure (not drawn to the scale).
bending compressive stress of 500 cm3 , 650 The moment of inertia of the section about
cm3 and 200 MPa, respectively. The design its principal axis X-X is 7.73 ×106 mm 4 . The
flexural capacity (expressed in kNm) of the permissible shear stress in the fillet welds is
section is ________.
100 N/mm2 . The design shear strength of the
[1 Mark, Set-1]
section is governed by the capacity of the
2019 IIT Madras fillet welds.
5.3 Assuming that there is no possibility of shear 100 mm
buckling in the web, the maximum reduction 10 mm
permitted by IS 800 : 2007 in the (low shear)

design bending strength of a semi - compact
120 mm
steel section due to high shear is X X
[1 Mark]
10 mm fillet weld (typical)
(A) Zero
(B) Governed by the area of the flange
(C) 50%
(D) 25% The maximum shear force (in kN, round off
5.4 A rolled I-section beam is supported on a 75 to one decimal place) that can be carried by
mm wide bearing plate as shown in the the section, is______.
figure. Thicknesses of flange and web of the [2 Marks]
I-section are 20 mm and 8 mm, respectively.
Root radius of the I-section is 10 mm.
Assume : material yield stress,
f y = 250 MPa and partial safety factor for
material, γ mo = 1.10 .
Steel Structures 1
5.1 (A)
Scan for
According to IS 800 : 2007 clause 9.2.1. no. Video Solution
Reduction in moment capacity of the section is
necessary as long as the cross section is subjected 5.3 (A)
to high shear force (factored value of shear force
is greater than 60% of the shear strength of the As per IS 800 : 2007
section). For semi-compact section
1. In low shear case (V ≤ 0.6Vd )
5.2 100
Ze f y
Given : Md =
Elastic section modulus Z e = 500 cm3 γ m0
Plastic section modulus Z P = 650 cm3 2. In high shear case (V ≥ 0.6Vd )

Design bending compressive stress, Ze f y
f d = 200 MPa Md =
γ mo
Design bending strength of laterally unsupported
beam, So reduction is zero.
M d = βb Z p fbd Hence, the correct option is (A).
Ze 5.4 272.73
βb = for semi compact section
Zp
Given : Stiff bearing length (b) = 75 mm
Z
M d = e × Z p × fbd = Z e × fbd
Zp Thickness of flange (t f ) = 20 mm
= 500 ×103 × 200 ×10−6 = 100 kN-m Thickness of web (tw ) = 8 mm
Hence, the flexural capacity of the section is
Root radius of I-section ( R) = 10 mm
= 100 kN-m
 Key Point Material yield stress ( f g ) = 250 MPa
We know design flexural capacity Web bearing strength
βZ f
Md = b p y fy
γ m0 = [b + 2.5(t f + R)] × tw ×
γ m0
βb = 1.0 (for plastic & compact section)
250
βb =
Ze
(For semi-compact section) = [75 + 2.5(20 + 10)] × 8 ×
Zp 1.1
Where, Z p = Plastic section modulus = 272.727 kN
Z e = Elastic section modulus Hence, the web bearing strength of the beam is
272.727 kN.
γ m 0 = Material factor of safety of steel
against yielding Scan for
f y = Yield stress of the material Video Solution
2 Steel Structures
5.5 393.5
10 mm
120 mm
x x
10 mm
100 mm
FAy
q=
I ×b
Where, q = Shear stress
F = Shear force
A = Area of cross section
I = Moment of inertia of the section
b = 4t = 4 × 0.7 ×10 = 28
F × (100 × 10) × 55
100 =
7.73 × 106 × 28
F = 393.527 kN
The maximum shear force that can be carried by
the section, is 393.5 kN.

Steel Structures 3
6 Design of Plate Girders, Industrial Roof Truss & Gantry
Girders
1993 IIT Bombay 6.5 List-I contains some elements in design of a

simply supported plate girder and List-II
6.1 A plate girder I-section is made by groove gives some qualitative locations on the
welding stress free web plate to two stress- girder. Match the items of two lists as per
free flange plates. After cooling of the welds good design practice and relevant codal
to the room temperature, the residual stress provisions. [2 Marks]
would be List-I
[1 Mark] (A)Flange splice
(A) Tensile at the free edges of the flanges. (B) Web splice
(B) Compressive at the free edges of the (C) Bearing stiffeners
flanges. (D) Horizontal stiffeners
(C) Compressive at the flange web List-II
intersection. 1. At supports (minimum)
(D) Zero everywhere. 2. Away from center of span
3. Away from support
1997 IIT Madras 4. In the middle of span
6.2 Effective length of a rafter member between 5. Longitudinally somewhere in the
two nodes at a distance L, perpendicular to compression flange
the plane of the truss is Codes :
[1 Mark] A B C D
(A) 2.00 L (B) 0.85 L (A) 2 3 1 5
(C) 1.50 L (D) 1.00 L (B) 4 2 1 3
(C) 3 4 2 1
1998 IIT Delhi (D) 1 5 2 3
6.3 The maximum permissible deflection for a 2004 IIT Delhi
gantry girder, spanning over 6 m, on which
6.6 In a plate girder, the web plate is connected
an EOT (Electric Overhead Travelling)
to the flange plates by fillet welding. The
crane of capacity 200 kN is operating, is
size of the fillet welds is designed to safely
[1 Mark]
resist [1 Mark]
(A) 8 mm (B) 10 mm (A) The bending stresses in the flanges.
(C) 12 mm (D) 18 mm (B) The vertical shear force at the section.
2003 IIT Madras (C) The horizontal shear force between the
flanges and the web plate.
6.4 Which of the following elements of a pitched (D) The forces causing buckling in the web.
roof industrial steel building primarily
resists lateral load parallel to the ridge? 2011 IIT Madras
[1 Mark] 6.7 The adjoining figure shows a schematic
(A) Bracings (B) Purlins representation of a steel plate girder to be
(C) Truss (D) Columns used as a simply supported beam with a
Steel Structures 1
concentrated load. For stiffeners, PQ ii.PQ should be placed in the tension side
(running along the beam axis) and RS of the flange.
(running between the top and bottom (B) i. RS helps to prevent local buckling of
flanges) which of the following pairs of the web.
statements will be TRUE? [2 Marks] ii.PQ should be placed in the compression
side of the flange.
(C) i. RS should be provided at supports.
R
P Q ii.PQ should be placed along the neutral
S axis.
(D) i. RS should be provided away from
L L points of action of concentrated loads.
(A) i. RS should be provided under the ii.PQ should be provided on the
concentrated load only. compression side of the flange.
6.1 (B) 6.3 (A)

Residual stress has a significant role in many The deflection of gantry girders under dead and
failure mechanisms, stress corrosion cracking imposed loads should not exceed the following
fatigue and fracture. values as per IS 800 : 1984
Tensile residual stresses decrease the reliability (i) Where cranes are manually operated L/500
and useful life of a component. (ii) Where electric overhead travelling L/750
Compressive residual stresses at the surface crane are operated upto 50t or 500 kN
increase the performance. (iii) Where electric overhead travelling L/1000
cranes are operated over 50t or 500 kN.
(iv) Other moving loads such as charging cars etc
L/600
Where, L = Span of crane runway girder
∴ For EOT crane with load capacity of 200 kN
maximum deflection
L 6000
= = = 8 mm
750 750
6.4 (A)
Truss are quite stable in their own plane. However
in the out of plane, these are very weak. To make
the trusses stable to carry out of plane loads, along
the ridge and lateral loads (wind and earthquake)
Fig. Residual stress distribution (+ indicates the truss must be braced.
tension and - compression) 6.5 (A)
Hence, the correct option is (B). (A) Flange splice : It resist most of the moment.
6.2 (B) The moment will be generally maximum at
centre for simply supported beam, flange
Effective length of rafter member = 0.85 L spice should be away from the centre of span
Hence, the correct option is (B). (where bending moment is less).
2 Steel Structures
(B) Web splice : It resist most of the shear. For
simply supported beam, maximum shear
force usually occurs at support thus web
splice should be away from support (where
shear force is less).
(C) Bearing stiffeners : It is provided at
supports (where reaction or concentrated
load act.)
(D) Horizontal stiffeners : It should be
provided near to the compressive flange
longitudinally to web plate to avoid
compression bucking of web.
Scan for
Video Solution
6.6 (C)
The size of the fillet welds in design to safety resist
the horizontal shear force between the flanges and
the web.
6.7 (B)
Horizontal stiffener or longitudinal stiffener
( PQ ) is provided to prevent the buckling of web
due to bending stress (compressive stress), so it
should be placed in the compression side of the
flange.
The first horizontal stiffener is provided at one-
fifth of the distance from the compression flange
to the tension flange. If required, another stiffener
is provided at the neutral axis. Horizontal
stiffeners are not continuous and are provided
between vertical stiffeners.
Vertical stiffener ( RS ) are provided to prevent
local buckling of web due to shearing buckling
stress.
It is assumed that the vertical stiffener is not
subjected to any load and is selected to provide
necessary lateral stiffness only.

Steel Structures 3
7 Plastic Analysis
2013 IIT Bombay R. Statical method of ultimate load analysis

S. Kinematical mechanism method of
7.1 As per IS 800 : 2007, the cross-section in
ultimate
which the extreme fiber in compression can
reach the yield stress, but cannot develop the Group-II
plastic moment of resistance due to local 1. Upper bound on ultimate load
bucking is classified as [1 2. Lower bound on ultimate load
Mark] 3. Material partial safety factor
(A) Plastic section 4. Load factor
(B) Compact section (A) P – 1 ; Q – 2 ; R – 3 ; S – 4
(C) Semi-compact section (B) P – 2 ; Q – 1 ; R – 4 ; S – 3
(D) Slender section
(C) P – 3 ; Q – 4 ; R – 2 ; S – 1
Common Data for (D) P – 4 ; Q – 3 ; R – 2 ; S – 1
Questions 7.23 & 7.24 7.5 The ultimate collapse load (P) in term of
plastic moment M P by kinematic approach
A propped cantilever made of a prismatic
steel beam is subjected to a concentrated for a propped cantilever of length L with P
load P at mid span as shown in the figure. acting at its mid-span as shown in the figure
P would be [1 Mark]
P
A B
X Y
1.5 m 1.5 m
L L
R 2 2
7.2 If load P = 80 kN, find the reaction R (in kN) 2M P 4M P
(correct to one decimal place) using elastic (A) P = (B) P =
L L
analysis. ________. [2 Marks]
7.3 If the magnitude of load P is increased till 6M P 8M P
(C) P = (D) P =
collapse and the plastic moment carrying L L
capacity of steel beam section is 90 kNm, 7.6 A prismatic beam (as shown below) has
determine reaction R (in kN) (correct to one plastic moment capacity M P , then the
decimal place) using plastic analysis collapse load P of the beam is
________. [2 Marks] P
P
2014 IIT Kharagpur 2
7.4 Match the information given in Group-I with

those in Group-II [1 Mark, Set-1] L L L
Group-I 2 2 3
P. Factor to decrease ultimate strength to 2M P 4M P
(A) (B)
design strength L L
Q. Factor to increase working load to 6M P 8M P
ultimate load for design (C) (D)
L L
Steel Structures 1
2015 IIT Kanpur having different plastic moment capacities
( M P , 2M P ) as shown in the figure.
7.7 A fixed end beam is subjected to a load, W at
1 [2 Marks, Set-2]
rd span from the left support as shown in 2L / 3 P
3
the figure. The collapse load of the beam is
P MP
L/3 2M P
X Z
Y MP L L
2MP
L L The minimum value of load ( P) at which the
2 2 beam would collapse (ultimate load) is
(A) 16.5 M P / L (B) 15.5 M P / L 7.5 M P 5.0 M P
(A) (B)
L L
(C) 15.0 M P / L (D) 16.0 M P / L
4.5 M P 2.5 M P
[2 Marks, Set-2] (C) (D)
L L
7.8 For formation of collapse mechanism in the
2018 IIT Guwahati
following figure, the minimum value of PU
is cM P /L M P and 3M P denote the plastic 7.11 A prismatic propped cantilever beam of span
moment capacities of beam sections as L and plastic moment capacity M, is
shown in this figure. The value of c is subjected to a concentrated load at its mid-
________. span. If the collapse load of the beam is
Pu M M
1m 1m α p , the value of α p is _______.
L L
[2 Marks, Set-2]
3MP MP
7.12 The dimension of a symmetrical welded I-
section are shown in the fig.
2m 140
9
[2 Marks, Set-1]
2016 IISc Bangalore
6.1 200
7.9 A propped cantilever of span L carries a
vertical concentrated load at the mid-span. If 9
the plastic moment capacity of the section is
140
M P , the magnitude of the collapse load is
(All dimensions are in mm)
8M P 6M P The plastic section modulus about the
(A) (B)
L L weakest axis (in cm3 , upto one decimal
4M P 2M P place) is ______. [2 Marks]
(C) (D)
L L
2019 IIT Madras
[2 Marks, Set-1]
7.10 A fixed end beam is subjected to a 7.13 If the section shown in the figure turns from
concentrated load ( P) as shown in the fully elastic to fully plastic, the depth of
figure. The beam has two different segments neutral axis (NA), y decreases by
2 Steel Structures
60 mm K
5 mm
y a a
60 mm
Z Z
H J
a a
5 mm
2020 IIT Delhi I
7.14 The ratio of the plastic moment capacity of a The values of S and M P (rounded-off to one
beam section to its yield moment capacity is decimal place) are
termed as [1 Mark] [2 Marks]
(A) shape factor (B) aspect ratio (A) S = 1.5, M p = 58.9 kN-m
(C) slenderness ratio (D) load factor
(B) S = 2, M p = 100.2 kN-m
2021 IIT Bombay
(C) S = 1.5, M p = 100.2 kN-m
7.15 A prismatic steel beam is shown in the
(D) S = 2.0, M p = 58.9 kN-m
figure. The plastic moment, Mp calculated
for collapse mechanism using static method 2023 IIT Kanpur
and kinematic method is,
7.17 Consider the horizontal axis passing through
[2 Marks]
P the centroid of the steel beam cross- section
shown in the figure. What is the shape factor
A C (rounded off to one decimal place) for the
B E, I
cross-section?
L 2L
3 3 b
2 PL
(A) M p , static > = M p , kinematic
9
b
2 PL
(B) M p , static = = M p ,kinematic
9
2 PL b
(C) M p , static = ≠ M p ,kinematic
9
2 PL b b b
(D) M p , static < = M p ,kinematic
9 (A) 1.5 (B) 1.7
2022 IIT Kharagpur (C) 1.3 (D) 2.0
7.16 For the square steel beam cross-section
shown in the figure, the shape factor about
Z-Z axis is S and the plastic moment capacity
is M P . Consider yield stress f y 250 MPa
and a = 100 mm.
Steel Structures 3
7.1 (C) 7.3 60
Cross section in which extreme fibre in P
compression can reach yield stress but cannot A B
develop plastic moment of resistance due to local R
buckling is called semi compact section. 1.5 m 1.5 m
q q
 Key Point Mp D
Plastic Section : Cross section which can
develop plastic hinges and have the rotation q q
capacity required for the failure of structure by Mp Mp
formation of plastic mechanism.
Compact Section : Cross section which can Lθ
Δ= , L=3 m
develop plastic moment of resistance but have 2
inadequate plastic hinge rotation capacity for External work done = Internal work done,
formation of plastic mechanism before buckling.
P × Δ = M P θ + M P (θ + θ)
Slender Section : Cross section in which the
elements buckle locally even before attainment of L
P× θ = 3M P θ
yield stress are called as slender section. 2
6 M P 6 × 90
7.2 25 Pu = = = 180 kN
L 3
P
At collapse condition  M A = 0 ,
A B
R × 3 + M P − P ×1.5 − M P + M P = 0
1.5 m 1.5 m
R× 3 + 90 − 180 ×1.5 = 0
From compatibility equation,
R = 60 kN
Net deflection at B = 0
Hence, reaction R using plastic analysis 60 kN.
RL3 P (1.5)3 P(1.5) 2
= + × 1.5
3EI 3EI 3EI Scan for
3 Video Solution
RL 1.125P 1.687 P
= +
3EI EI EI
7.4 (C)
9 R 1.125 P 1.687 P
= +
EI EI EI  Factor which increases working load to
5P 5 ultimate load is load factor. Therefore ‘Q’
R= = × 80 should be matched with ‘4’.
16 16
R = 25kN  Factor which decreases ultimate strength of
Hence, the reaction R using elastic analysis 25 kN. a material to design strength is material
partial safety factor. Therefore ‘P’ should be
Scan for matched with ‘3’.
Video Solution
4 Steel Structures
7.5 (C) P/2
P
P D2
q
L/2 L/2
D1
q q
q q
Mp D Mp
Lθ Lθ
q q Δ1 = and Δ 2 =
2 3
P
Mp Mp PΔ1 − Δ 2 = M P (θ + θ)
2
From principle of virtual work, PL P l
θ − × θ = M P (θ + θ)
External work done = Internal work done 2 2 3
P × Δ = M P θ + M P (θ + θ) PL
θ = 2M P θ
L 3
From figure, Δ = θ 6M P
2 P=
L
l
Pu × θ = 3M P θ General minimum collapse load is taken here in
z
6M P
6M P both mechanism collapse load ( Pu ) = .
Pu = L
L Hence, the correct option is (C).
7.7 (C)
7.6 (C)
No. of plastic hinges formed at collapse
Degree of static inderminacy = 0 = DS + 1 = 2 + 1 = 3
No of plastic hinges required for mechanism Mechanism 1 :
= DS + 1 = 0 + 1 = 1 W
L /3 2 L /3
Mechanism 1 : q a
P/2 2M p D Mp
P Mp
q a
q
D
2M p 2M p
L/2 L/2 L/3 L 2L

Δ= θ= α
From principle of virtual work 3 3
θ = 2α
External work done = Internal work done
From principle of virtual work internal work done
Pu = External work done
× Δ = M Pθ
2 L
2 M P θ + 20 M P θ + 2 M P α + M P α = W × θ
Pu L 3
× θ = M Pθ
2 3 θ WL
4 M P θ + 3M P = θ
6M P 2 3
Pu = 11M P θ WL
L = θ
Mechanism 2 : 2 3
Steel Structures 5
16.5 Pu
W= MP 3L / 4 L/4
L
q f
Mechanism 2 :
3M p D Mp
L/2 L/2
L/3 W f
q
q q Mp Mp
D
2M p Mp 3L L
θ= φ
q q
4 4
φ = 3θ
Mp Mp PL
3M P θ + M p θ + M p φ + M p φ = φ
From principal of virtual work 4
External work done = Internal work done PL
4M P θ + 2M P φ = φ
WL 4
= θ = 2M P θ + M P θ + M P θ + M P θ 4 PL
3 M Pφ + 2M P φ = φ
15 M P 3 4
W= 10 M P φ PL
L = φ
15 M P 3 4
Hence minimum collapse load is 40 M P
L P=
Hence, the correct option is (C). 3L
Out of two mechanisms smaller collapse load is
Scan for 40
Video Solution taken, so C = = 13.33
3
7.8 13.33 Scan for
Video Solution
Mechanism 1 :
Pu
7.9 (B)
L/2 L/4 L/4
P
q q
3M p D Mp L /2 L /2
q q q q
D
Mp
Mp Mp q q
L = 4m Mp Mp
From principle of virtual work, L
External work done = Internal work done Δ= θ
2
Pu × Δ = 3M P θ + M P θ + M P θ + M P θ
L
Pu × θ = 6 M P θ P × Δ = M P θ + M P (θ + θ)
4
24 M P 1
Pu = P × θ = 3M P θ
L 2
Mechanism 2 :
6 Steel Structures
6M P 2L
P= P× θ = 5M P θ
L 3
Hence, the correct option is (B). 15M P
P=
2
7.10 (A)
7.5M P
So minimum collapse load is
Degree of static indeterminacy DS = 2 L
∴ Number of plastic hinges required for
7.11 *
complete collapse = DS + 1 = 2 + 1 = 3
P
Mechanism 1 :
2L / 3 P 4L / 3
L
P
q f
D Mp
2M p q q
Mp D
q f
q q
2M p 2M p
2L 4L Mp Mp
Δ= θ= φ
3 3 L
Δ= θ
θ = 2φ 2
External work done = Internal work done From principle of virtual work
 2L 
P× θ  = 2 M P θ + 2 M P (θ + φ) + M P φ P × Δ = M Pθ + M Pθ + M Pθ
 3 
L
 2L  P× θ = 3M P θ
P× θ  = 4 M P θ + 3M P φ 2
 3  6M P
P= …(i)
 2L  θ L
P θ  = 4 M P θ + 3M P  
 3  2 αM P
Comparing equation (i) to we get α = 6
 2 L  11M P θ L
P θ =
 3  2 7.12 89.89
33M P 9 mm
P=
4L
Mechanism 2 : 6.1mm 182mm
2L / 3 P
9 mm
Mp 140
q q
2M p D A
We know, Z p = ( y1 + y 2 )
2
q q
140 × 9 182 × 6.1  6.1 6.1 
= 2× (35 + 35) +  + 
Mp Mp 2 2  4 4 
External work done = Internal work done = 89893.055mm3 = 89.89cm3
P × Δ = 2M P θ + M P θ + M P θ + M P θ
Steel Structures 7
A
Scan for Zp Mp ( y1 + y2 )
Video Solution Shape factor = = = 2
Ze My I/y
7.13 (D) ● Load factor is the ratio of actual amount of
60 mm load and the maximum permissible load.
5 mm P
Load factor = c
y P
Load factor = Factor of safety
NA
y1
× Shape factor
● Slenderness ratio is the ratio between length
5 mm of a column and the least radius of gyration
A1 y1 + A2 y 2 of its cross section.
y= L
A1 + A2 λ= e
γ min
60  5
60 × 5 × + 60 × 5 ×  60 +  ● Aspect ratio-For any geometric shape aspect
y=
2  2
ratio is found by dividing the value of one
(60 × 5) + (60 × 5)
dimension by another dimension in a
y = 46.25 mm structure. Generally, it is used for design of
The section is unsymmetrical about the neutral any structure like beam, for which the ratio
axis and hence the equal area axis has to be of width (b) to depth of cross section of beam
located. (d) is to be fixed.
60 mm Hence, the correct option is (A).
EA 5 mm  Key Point
NA 60 mm
5 mm
Let x ≤ 5 mm
(60 × 5) + (60 × 5)
60 × x =
2
x = 5 mm
∴ NA shifts by,
7.15 (B)
60 − y = 60 − 46.25 = 13.75mm
Hence, the correct option is (D). Arc
We know that, Angle =
Radius
Scan for ∴ Deflection is same at B from both A and C
Video Solution
L/3 2L/3
G
7.14 (A) α β
2L
● Shape factor is the ratio of plastic section L / 3α ×β
3
modulus ( Z p ) and section modulus ( Z e )
8 Steel Structures
L 2L bh
Now, α= β [δ = θ × L ] 2×
3 3 = 2 (y + y )× f
1 2 y
α = 2β 2
Ds = 0
h
No. of plastic hinges = Ds + 1 = 0 + 1 = 1
 2L 
P β  = M P (α ) + M P (β)
 3  b
 2L 
P  β  = M P (2β) + M P (β) h
 3 
2L
β P = 3M P β 100
3 100 2 ×
2L MP = 2 ×  h × 2  × 250
MP =  
P 2 3 
9
2 PL  100 
M P static = = M p kinetic M P = 5000 ×  × 2  × 250
9  2 ×3 
Hence, the correct option is (B). M P = 58.92 kN-m
7.16 (D) Hence, the correct option is (D).
Given : Yield stress, f y = 250 MPa 7.17 (B)
Dimension, a = 100 mm I 3b
Ze = , Ymax =
Ymax 2
a a b
h y
3b
2
a a
x b x
b
a = 100 m
y
b = 100 2 mm
3b × b3  b × b3 
100 2 100 I xx = + 2×  × (b × b) × b 2 
h= = mm 12  12 
2 2
1 1   29b 4
We know that, = b4  + 2  + b   =
Shape factor for diamond section = 2 4 12   12
Now plastic moment = Section plastic modulus 29b 4 2 29 3

Ze = × = b
× Yield stress 12 36 18
M P = ZP × f y A
Z P = [ y1 + y2 ]
2
A
MP = × ( y1 + y2 ) × f y
2 A = 3 b × b + 2 × b2 = 5 b2
Steel Structures 9
 b b 2 3 
 3b ×  × + b × b b  + 1
y1 = 
2 4
= 
8 
b
3b × × b 2 3 
2  2 + 1
11
y1 = b
20
A 5b 2  11 11 
Zp = ( y1 + y2 ) =  ×b + ×b
2 2  20 20 
11 3
= b
4
11 3
b 11×18
S .F = 4 = = 1.706 = 1.71
29 3 4 × 29
b
18

10 Steel Structures
ENGINEERING
5 MECHANICS
Syllabus : Engineering Mechanics
System of forces, free-body diagrams, equilibrium equations; Internal forces in structures;

Frictions and its applications; Centre of mass; Free Vibrations of undamped SDOF system.
5 ENGINEERING MECHANICS
Syllabus : Engineering Mechanics
System of forces, free-body diagrams, equilibrium equations; Internal forces in structures;

Frictions and its applications; Centre of mass; Free Vibrations of undamped SDOF system.
Q
2014 IIT Kharagpur
P
4
5.1 Polar moment of inertia ( I p ) , in cm , of a
450
rectangular section having width, b = 2 cm 60 0
x
and depth, d = 6 cm is _____. R
[1 Mark] The respective values of the magnitude (in
kN) and the direction (with respect to the x -
2016 IISc Bangalore
axis) of the resultant vector are
5.2 An assembly made of a rigid arm ABC [2 Marks]
hinged at end A and supported by an elastic (A) 290.9 and 96.00
rope CD at end C as shown in the figure. The
members may be assumed to be weightless (B) 368.1 and 94.70
and the lengths of the respective members (C) 330.4 and 118.90
are as shown in the figure.
D
(D) 400.1 and 113.50
A 2017 IIT Roorkee
5.4 A particle of mass 2 kg is travelling at a

L L velocity of 1.5 m/s . A force f (t ) = 3t 2 (in N)
P Rope
Rigid arm is applied to it in the direction of motion for
C
a duration of 2 seconds, where t denotes time
B
L L in seconds. The velocity (in m/s, up to one
Under the action of a concentrated load P at decimal place) of the particle immediately
C as shown, the magnitude of tension after the removal of the force is _____.
developed in the rope is [2 Marks]
[1 Mark]
2018 IIT Guwahati
3P P
(A) (B)
2 2 5.5 An aircraft approaches the threshold of a
runway strip at a speed of 200 km/h . The
3P
(C) (D) 2P pilot decelerates the aircraft at a rate of
8
5.3 The magnitude of vectors P, Q and R are 100 1.697 m/s2 and takes 18 sec to exit the
kN, 250 kN and 150 kN, respectively as runway strip. If the deceleration after exiting
shown in the figure
the runway is 1m/s2 , then the distance (in m,
up to one decimal place) of the gate position Neglecting the self-weight of the cable, the
from the location of exit on the runway is tension (in kN, integer value) in the cable
______m. due to the applied load will be ______.
[2 Marks] [2 Marks]
5.6 A cylinder of radius 250 mm and weight, 2020 IIT Delhi
W = 10 kN is rolled up an obstacle of height
5.9 A rigid weightless platform PQRS shown in
50 mm by applying a horizontal force P at its the figure (not drawn to the scale) can slide
centre as shown in the figure. freely in the vertical direction. The platform
is held in position by the weightless member
OJ and four weightless, frictionless rollers,
P Points O and J are pin connections. A block
of 90 kN rests on the platform as shown in
50mm the figure. [2 Marks]
W
1m 4m
All interfaces are assumed frictionless. The Roller (typical) W = 90kN

minimum value of P is [2 Marks]
P Q
(A) 4.5 kN (B) 5.0 kN
(C) 6.0 kN (D) 7.5 kN
1.5m
Rigid
5.7 Two rigid bodies of mass 5 kg and 4 kg are J
platform
at rest on a frictionless surface until acted S R
upon by a force of 36 N as shown in the 2m 3m
figure. The contact force generated between
the two bodies is O
[2 Marks] 4m
5 kg
36 N 4 kg
The magnitude of horizontal component of
(A) 4.0 N (B) 7.2 N the reaction (in kN) at pin O, is
(A) 150 (B) 120
(C) 9.0 N (D) 16.0 N
(C) 90 (D) 180
5.8 A cable PQ of length 25 m is supported at
two ends at the same level as shown in the 2021 IIT Bombay
figure. The horizontal distance between the
5.10 A wedge M and a block N are subjected to
supports is 20 m. A point load of 150 kN is
forces P and Q as shown in the figure. If
applied at point R which divides it into two
force P is sufficiently large, then the block N
equal parts
can be raised. The weights of the wedge and
20 m
the block are negligible compared to the
forces P and Q. The coefficient of friction (
P Q
μ ) along the inclined surface between the
wedges and the block is 0.2. All other
R
surface are frictionless. The wedge angle is
150 kN
300 .
2 Analog Electronics
Q
2022 IIT Kharagpur
Surface 3
Surface 2 ( μ 0.2)
5.13 A horizontal force mechanics of P kN is
Block N
applied to a homogeneous body of weight 25
kN, as shown in the figure. The coefficient
P
Wedge M of friction between the body and the floor is
30 0 0.3. Which of the following statement is/are
Surface 1 correct.
1m
The limiting force P, in terms of Q, required
for impending motion of block N to just P
move it in the upward direction is given as
P = αQ . The value of the coefficient ‘ α ’ 2m
(round off to one decimal place) is
[2 Marks]
(A) The motion of the body will occur by
(A) 2 (B) 0.5
overturning.
(C) 0.6 (D) 0.9
(B) Sliding of the body never occurs
5.11 A perfectly flexible and inextensible cable is
(C) No motion occurs for P ≤ 6 kN
shown in the figure (not to scale). The
external loads at F and G are acting (D) The motion of body will occur by sliding
vertically. only.
E [2 Marks]
1m
5.14 A uniform rod KJ of weight w shown in the
H
figure rests against a frictionless vertical
3m wall at the point K and a rough horizontal
surface at point J. It is given that w = 10 kN,
F a = 4 m and b = 3 m.
G
10 kN K
12 kN
2m 2m 2m
The magnitude of tension in the cable
segment FG (in kN, round off to two decimal
a
places) is _______. [2 Marks]
5.12 A prismatic fixed-fixed beam, modelled with
a total lumped mass of 10 kg as a single
degree of freedom system (SDOF) as shown J
in the figure.
A b
B C
m
The minimum coefficient of static friction
EI EI that is required at the point J to hold the rod
L L in equilibrium is ______. (round off to three
2 2 decimal places)
SDOF
[2 Marks]
If the flexural stiffness of beam is 4π2 kN/m
5.15 An undamped spring-mass system with mass
, its natural frequency of vibration (in Hz, in
m and spring stiffness k is shown in the
integer) in the flexural mode will be
figure. The natural frequency and natural
_______.
period of this system are ω rad/s and T s,
[2 Marks]
respectively. If the stiffness of the spring is
doubled and the mass is halved, then the (A) 2ω rad/s and T / 2 s
natural frequency and the natural period of (B) ω / 2 rad/s and 2T s
the modified system, respectively, are (C) 4ω rad/s and T / 4 s
[2 Marks] (D) ω rad/s and T s
y
5.1 40 A
According to the question,
b = 2 cm x dA
Y
r
y
X X d = 6 cm x
O
z
Y The above theorem can be easily proved. Let us
Now, polar moment of inertia is given by, consider an elemental area dA at a distance r from
bd 3 db3 O. Let the coordinates of dA be x and y. Then
I P = I X + IY = +
12 12 from definition,
(from perpendicular axis theorem) I zz =  r 2 dA = ( x 2 + y 2 )dA
2 × 6 6 × 23 432 + 48
3
I zz =  x 2 dA +  y 2 dA = I xx + I yy
IP = + =
12 12 12
480 5.2 (B)
IP = = 40 cm 4
12 Let, tension in rope be T
Hence, the correct answer is 40 cm4.
A
Scan for
Video Solution
2L
 Key Point L
P
Perpendicular axis theorem : The moment of T
inertia of an area about an axis perpendicular to
its plane (polar moment of inertia) at any point O C
B
is equal to the sum of moments of inertia about L
any two mutually perpendicular axes through the
same point O and lying in the plane of the area. Considering triangle ABC,
Referring to figure, if z-axis is normal to the plane AC 2 = AB 2 + BC 2
of paper passing through point O, as per theorem,
I zz = I xx + I yy AC = L2 + L2 = 2 L
Taking moments about A equal to zero.
∴ ΣM A = 0 , (w.r.t. – x-axis in clockwise direction)
− P × L + T × 2L = 0 ∴ θ ' = 180 − 61.114 = 118.8860
PL P θ ' ≈ 118.90
T= = (w.r.t. + x-axis in counter-clockwise direction).
2L 2
5.3 (C) Scan for
Video Solution
Let resultant of forces P, Q, R be Rres .
+ y 5.4 5.5
Q = 250 N
15
0
P = 100 N Given : Mass of the particle (m) = 2 kg
Initial velocity (u ) = 1.5 m/s
0
45
60 0 Applied force F (t ) = 3t 2 N
-x +x
150 Force duration (t ) = 2sec
R = 150 N
-y Let, final velocity be v.
Now, we know that,
Now, net horizontal force,
F = ma
ΣFH = P cos 60 − Q sin15 − R cos15 t
ΣFH = 100cos 60 − 250sin15 − 150cos15  F (t )dt = m(v − u)

0
2
ΣFH = −159.59 kN
0
3t 2 dt = 2(v − 1.5)
∴ Net vertical force, 2
t 3  = 2(v − 1.5)
ΣFV = P sin 60 + Q cos15 − R sin15 0
ΣFV = 100sin 60 + 250cos15 − 150sin15 (8 − 0) = 2(v − 1.5)

ΣFV = 289.26 kN v = 5.5 m/s
Now, the resultant is given by, Hence, the velocity of the particle immediately
after the removal of the force is 5.5 m/s.
Rres = (ΣFH ) 2 + (ΣFV ) 2
Scan for
Rres = (−159.59) 2 + (289.26) 2 Video Solution
Rres = 330.4 kN (Magnitude)
 Key Point
Since, ΣFH is negative and ΣFV is positive then
According to Newton’s second law of motion,
Rres is lying in second quadrant. Applied force on body
+y = Rate of change of linear momentum of
body
SFV d (mV )
Rres F=
dt
dV
-x q q' If m = constant, then F = m
+x dt
SFH
t v
 ΣF
θ = tan −1  V
 −1  289.26 
∴  F dt = m
0
dV = m[v − u ]
 = tan  −159.59 
u
 ΣFH   
5.5 312.5
θ = − 61.1140
Given :
Initial speed of the aircraft (u ) = 200 km/h AB = 2502 − 2002 = 150 mm
De-acceleration upto the runway exit At the instant when the cylinder just starts rolling
2
(a1 ) = −1.697 m/s over obstacle, it will be lifted from point D. the
contact will only be there with point B and no
Time taken upto the runway exit (T ) = 18sec
reaction force takes place at contact point B.
De-acceleration after exiting runway Now, taking the moments about B equal to zero,
(a2 ) = −1m/s 2 ΣM B = 0 ,
Let, v1 be the velocity at runway exit (after 18 sec), − P × AC + W × AB = 0
S be the distance travelled by the aircraft from − P × 200 + 10 × 150 = 0
runway exit to the gate position with an
1500
acceleration a2 and final velocity be v2 , where P= = 7.5 kN
200
v2 = 0 .
Now, velocity at the runway exit,
Scan for
v1 = u + at Video Solution
200 × 1000
v1 = + ( −1.697 × 18) = 25 m/s
3600 5.7 (D)
Distance traveled from runway exit to the gate a
position is, ma 5 kg
a
v22 = v12 + 2aS 36 N FC

ma 4 kg
FC
0 = 252 + 2 × (−1) × S
2 S = 625 W1 W2
625 The applied load 36 N will cause both the blocks
S= = 312.5 m
2 to move together. Let a be the acceleration and FC
Hence, the distance of the gate position from the be the contact force for both the blocks.
location of exit on the runway is 312.5 m. Now, from Newton’s second law, F = ma
F 36
Scan for a= = = 4 m/s 2
Video Solution m 9
Similarly, contact force,
5.6 (D) FC = ma = 4 × 4 = 16 N
Given : Radius of the cylinder ( R) = 250 mm (for smaller block)
Weight of the cylinder (W ) = 10 kN Hence, the correct option is (D).
Obstacle height ( h) = 50 mm Note :
FC = 36 − ma can also be used to determine the
C value of FC (for bigger block) and value will be
P
B same.
A
W h = 50 mm Checking, FC = 36 − 5 × 4 = 36 − 20 = 16 N
D
AC = R − h = 250 − 50 = 200 mm Scan for
Video Solution
AB = CB 2 − AC 2
5.8 125 12.52 = x 2 + 102
Given : Length of cable PQ = 25 m x = 7.5 m
Horizental distance between PQ = 20 m
Bending moment at R = 0 ,
:Method 1:
10 m 10 m H P × 7.5 = 75 ×10
P TP S TQ Q H P = 100 kN
12
.5 0 0 m
m 53.13 53.13 2.5
0
1
0
Tension in cable = H P2 + VP2
36.87 36.87
R
150 kN = 752 + 1002 = 125 kN
Let, tension in cable PR and QR be TP and TQ
respectively, Now, in triangle PQR, Hence, the tension in the cable due to the applied
PS = SQ = 10 m load will be 125 kN.
PR = RQ = 12.5 m 5.9 (B)

 10 
∠PRS = SRQ = sin −1   = 53.13
0
4m
 12.5  0.5m
90 kN
1m
Using Lami’s theorem, HP
TP 150 HP 1.5m
= 1.5m
sin(90 + 36.87) sin(53.13 + 53.13) HR
R0
HR
TQ q = 36.87 0
= 2m
R0
sin(90 + 36.87)
H0
150 × sin(90 + 36.87)
TP = = 124.98 kN
sin(53.13 + 53.13)
TP = TQ = 124.99 ≈ 125kN
Hence, the tension in the cable due to the applied Σy = 0 ,
load will be 125 kN.
R0 sin 36.87 − 90 = 0
:Method 2:
10 m 10 m 90
VP VQ R0 = = 150 kN
HP HQ sin 3687 0
P Q
12
.5
m
x
.5
m Horizontal reaction at O = H 0
12
R = R0 cos36.87 = 150 × cos36.87
150 kN = 120 kN
VP + VQ = 150 kN
VP = VQ = 75 kN
5.10 (D)
Given : Friction factor of surface 2μ = 0.2 P = αQ
Q α = 0.878
α  0.9
5.11 8.25
Block N
N1 μN2
VE VH
N2 μN2
HE 1m
E H
N2 HH
3m
4m
P θ
Wedge M 2m
F G
30 0
2m
N3 10 KN
12 KN
Free body diagram,
Left of F, H E × 3 + VE × 2 = 0
N 2 cos300 = 0.2 N 2 sin 300 + Q
2
Q H E = VE …(i)
N2 = 3
0
cos 30 − 0.2sin(30) At support,
N2 = 1.305Q VE × 6 − H E ×1 − 10 × 4 − 12 × 2 = 0
N2 6VE − H E = 64 …(ii)
Put value of H E in equation (ii),
2
30 0 6VE − VE = 64
3
300
N1 3
VE = × 64 = 12 KN
16
0.2N2 2
H E = × 12 = 8 KN
3
Tension, in EF,
Q
FEF = VE 2 + H E 2
FEF = (12) 2 + (8) 2 = 4 13 KN

12 KN
μN2
8 KN E
0
P 30
30 0
F
θ
N2 FFG
10 KN
P = 0.2 ×1.305Q cos (30) ΣFH = 0 , FFG cos θ1 = 8 …(iii)
0
+ 1.305Q sin(30 ) ΣFV = 0 , FFG sin θ1 + 10 = 12
P = 0.878Q
FFG sin θ1 = 2 …(iv)
Given condition,
( FFG )2 = (8)2 + (2)2 1m
FFG = 8.2462 KN ( Pmin )overt urning
FFG = 8.25 KN
2m
5.12 10
0.5m
Given : Total lumped mass, m = 10 kg
We know that,
Aedge
1 k 25 kN
Frequency, f n =
2π m At the verge of overturning
where, k = Flexural stiffness and ( Pmin )overturning × 2 = mg × 0.5
m = Total lumped mass 25 × 0.5

( Pmin ) overturning =
2
1 4000π2 12.5
fn = ( Pmin ) overturning = = 6.25 kN
2π 10 2
1
Here, ( Pmin )overturning < ( Pmin ) sliding
fn = × 20 × π = 10 Hz
2π  First overturning will take place.
5.14 0.375
5.13 (A), (B), (C)
Given :
(Pmin ) slidi ng Smooth
K
J
Rough
b
w = 10 kN , a = 4 m , b = 3 m
25 kN = mg N1
Q
( Pmin ) sliding = (maximum static friction)max
W N2
= μN
fs
Normal reaction (N) = mg = 25 kN
Static equilibrium equation
( Pmin ) sliding = 0.3 × 25 = 7.5 kN ΣFH = 0
Maximum force for overturning, N1 = f s
ΣFv = 0
N2 = w
ΣM k = 0
w × 2.5cos θ + f s × 4 = N2 × 3
3
10 × 2.5 × + f s × 4 = 10 × 3
5
f s = 8.75 kN
f s ≤ f max
3.75 ≤ μs × N2
3.75
μs ≥
10
μs ≥ 0.375
μmin = 0.375
5.15 (A)
Natural frequency,
k 2π
ω= , T1 =
m ω1
Case : If stiffness is doubled & the mass is halved
k2 = 2 k & m2 = m / 2
2k
ω2 =
m/2
4k k
ω2 = =2
m m
ω2 = 2ω1
2π 2π
T2 = =
ω2 2ω
T1
T2 =
2

6 CONSTRUCTION MATERIALS
AND MANAGEMENT
Syllabus : Construction Materials & Management
Construction Materials: Structural Steel – Composition, material properties and behavior;

Concrete - Constituents, mix design, short-term and long-term properties. Construction
Management: Types of construction projects; Project planning and network analysis - PERT
and CPM; Cost estimation.
6 CONSTRUCTION MATERIALS AND MANAEMENT
Syllabus : Construction Materials & Management
Construction Materials: Structural Steel – Composition, material properties and behavior; Concrete -
Constituents, mix design, short-term and long-term properties. Construction Management: Types of
construction projects; Project planning and network analysis - PERT and CPM; Cost estimation.
(C) 3 1 2
Construction Materials
(D) 3 2 1
1. Maximum possible value of compaction 3. Consider the following statements for air-
factor for fresh (green) concrete is entrained concrete :
[1 Mark] (i) Air-entrainment reduces the water
(A) 0.5 (B) 1.0 demand for a given level of workability
(C) 1.5 (D) 2.0 (ii) Use of air-entrained concrete is required
2014 IIT Kharagpur in environments where cyclic freezing
2. Group-I contains representative stress- and thawing is expected.
strain curves as shown in the figure, while Which of the following is TRUE?
Group-II gives the list of materials. Match [1 Mark]
the stress strain curves with the (A) Both (i) and (ii) are True
corresponding materials. [1 Mark] (B) Both (i) and (ii) are False
(C) (i) is True and (ii) is False
(D) (i) is False and (ii) is True
4. Workability of concrete can be measured
using slump, compaction factor and Vee-bee
time. Consider the following statements for
workability of concrete : [1 Mark]
(i) As the slump increases, the Vee-bee time
Group-I increases
P. Curve J
(ii) As the slump increases, the compaction
Q. Curve K
factor increases
R. Curve L
Which of the following is TRUE?
Group-II
(A) Both (i) and (ii) are True
1. Cement paste
2. Coarse aggregate (B) Both (i) and (ii) are False
3. Concrete (C) (i) is True and (ii) is False
P Q R (D) (i) is False and (ii) is True
(A) 1 3 2 2016 IISc Bangalore
(B) 2 3 1
Construction Materials & Management 1

5. The compound which is largely responsible P. Walls of one brick thick are measured in
for initial setting and early strength gain of square meters
Ordinary Portland Cement is [1 Mark] Q. Walls of one brick thick are measured in
(A) C3 A (B) C3 S cubic meters.
R. No deduction in the brickwork quantity
(C) C2 S (D) C4 AF
if made for openings in walls up to
6. Bull’s trench kiln is used in the
0.1m2 area.
manufacturing of [1 Mark]
S. For the measurement of excavation from
(A) Lime (B) Cement
the borrow pit in a fairly uniform ground,
(C) Bricks (D) None of these
deadman are left at suitable intervals.
7. As per Indian standards for bricks, minimum For the above statements, the correct option
acceptable compressive strength of any class is
of burnt clay bricks in dry state is
(A) P-false; Q-True; R-false; S-True
[1 Mark]
(B) P-false; Q-True; R-false; S-False
(A) 10.0 MPa (B) 7.5 MPa
(C) P-True; Q-False; R-True; S-False
(C) 5.0 MPa (D) 3.5 MPa
(D) P-True; Q-False; R-True; S-True
2017 IIT Roorkee
2018 IIT Guwahati
8. Group I gives a list of test methods and test
10. The Le-Chatelier apparatus is used to
apparatus for evaluating some of the
determine [1 Mark]
properties of Ordinary Portland Cement
(OPC) and concrete. Group II gives the list (A) Compressive strength of cement
of these properties. [2 Marks] (B) Fineness of cement
Group I (C) Setting time of cement
P. Le Chatelier test (D) Soundness of cement
Q. Vee-Bee test 11. The setting time of cement is determined
R. Blaine air permeability test using [1 Mark]
S. The Vicat apparatus (A) Le-Chateller apparatus
Group II (B) Briquette testing apparatus
1. Soundness of OPC (C) Vicat apparatus
2. Consistency and setting time of OPC (D) Casagrande’s apparatus
3. Consistency or workability of concrete 2019 IIT Madras
4. Fineness of OPC
12. A box measuring 50 cm×50 cm  50 cm is
The correct match of the items in Group I
filled to the top with dry coarse aggregate of
with the items in Group II is
mass 187.5 kg . The water absorption and
(A) P-1, Q-3, R-4, S-2
specific gravity of the aggregate are 0.5 %
(B) P-2, Q-3, R-1, S-4
2.5, respectively. The maximum quantity of
(C) P-4, Q-2, R-4, S-1
water (in kg. round off to 2 decimal places)
(D) P-1, Q-4, R-2, S-3 required to fill the box completely is ______.
9. Consider the following statements : [2 Marks]
[2 Marks]
2 Construction Materials & Management

2020 IIT Delhi 2021 IIT Bombay
13. During the process of hydration of cement, 16. Gypsum is typically added in cement to
due to increase in Dicalcium Silicate (C2S) [1 Mark]
content in cement clinker, the heat of (A) enhance hardening
hydration [1 Mark] (B) increase workability
(A) increases (C) decrease heat of hydration
(B) initially decreases and then increases (D) prevent quick setting
(C) does not change 17. The Softening point of bitumen has the same
(D) decreases unit as that of [1 Mark]
14. The Los Angles test for stone aggregates is (A) Time (B) Temperature
used to examine [1 Mark] (C) Distance (D) Viscosity
(A) soundness 18. Seasoning of timber for use in construction
(B) abrasion resistance is done essentially to :
(C) specific gravity (A) Smoothen timber surfaces
(B) Increases strength and durability
(D) crushing strength
(C) Remove knots from timber logs
15. Group-I gives a list of test methods for
(D) Cut timber in right season and geometry
evaluating properties of aggregates. Group-
[1 Mark]
II gives the list of properties to be evaluated.
2022 IIT Kharagpur
[2 Marks]
Group - I Group - II 19. Which of the following equations is correct
(Test methods) (Properties) for the Pozzolanic reaction?
[1 Mark]
P. Soundness 1 Strength
(A) Ca(OH)2  Reactive superplasticiser
test .
Q Crushing test 2 Resistance to  H2O  C-S-H
. . weatherin (B) Ca(OH)2  Reactive silicon dioxide
g  H2O  C-S-H
R. Los Angeles 3 Adhesion (C) Ca(OH)2  Reactive sulphates  H2O
abrasio .
 C-S-H
n test
(D) Ca(OH)2  Reactive sulphur  H2O
S. Stripping 4 Hardness
value .  C-S-H
test 20. Match : [1 Mark]
The correct match of test methods under (i) C3 S

Group-I to properties Group-II, is (ii) C2 S
(A) P-3, Q-4, R-1, S-2
(iii) C3 A
(B) P-4, Q-1, R-2, S-3
(C) P-2, Q-4, R-3, S-1 (P) Early age strength
(D) P-2, Q-1, R-4, S-3 (Q) Later age strength
(R) Flash setting
(S) Highest heat of hydration
(T) Lowest heat of hydration Total floats (in days) for the activities 5-7
(A) (i) Q & T and 11-12 for the project are, respectively,
(ii) P & S [1 Mark]
(A) 25 and 1 (B) 1 and 1
(iii) R
(C) 0 and 0 (D) 81 and 0
(B) (i) P 2. The optimistic time (O), Most likely time
(ii)Q & T (M) and pessimistic Time (P) (in days) of the
(iii) R & S activities in the critical path are given below
(C) (i) T in the format O-M-P.
(ii) S
(iii) P & Q The expected completion time (in days) of
(D) (i) P the project is _____. [2 Marks]
(ii) Q & R 3. The activity on arrow network of activities
(iii) T for a construction project is shown in the
figure. The durations (expressed in days) of
Construction Planning the activities are mentioned below the
& Management
arrows.
2016 IISc Bangalore
1. A construction project consists of twelve
activities. The estimated duration (in days)
required to complete each of the activities
along with the corresponding network
diagram is shown below,
The critical duration for this construction

project is [2 Marks]
(A) 13 days (B) 14 days
(C) 15 days (D) 16 days
2017 IIT Roorkee
4. The activity details of a project are given
below
Depends Duration
Activity
on (in days)
P - 6
Q P 15
R Q,T 12
S R 16
T P 10
U Q, T 14
V U 16

The estimated minimum time (in days) for activities are shown in the table. The indirect
the completion of the project will be ______. cost incurred by the contractor in INR 5000
[2 Marks] per day.
5. Given that the scope of the construction
work is well defined with all its drawing,
specifications, quantities and estimates,
which one of the following types of contract
would be most preferred? [1 Mark]
(A) EPC contract
(B) Percentage rate contract
(C) Item rate contract
(D) Lump sum contract
6. For a construction project, the mean and
standard deviation of the completion time
are 200 days and 6.1 days, respectively. If the project is target for completion in 16
Assume normal distribution and use the days, the total cost (in INR) to be incurred by
value of standard normal deviate, z  1.64 the contractor would be _____. [2
for the 95% confidence level. The maximum Marks]
time required (in days) for the completion of 2021 IIT Bombay
the project would be _______. [1 Mark]
9. A small project has 12 activities – N, P, Q,
2018 IIT Guwahati R, S, T, U, V, W, X, Y and Z. The
7. A probability distribution with right skew is relationship among these activities and the
shown in figure [1 Mark] duration of these activities are given in the
Table.
Duration Depends
Activity
(in weeks) upon
N 2 -
P 5 N
Q 3 N
The correct statement for the probability R 4 P
distribution is S 5 Q
(A) Mean is equal to mode T 8 R
(B) Mean is greater than median but less than U 7 R, S
mode
V 2 U
(C) Mean is greater than median and mode
W 3 U
(D) Mode is greater than median
X 5 T, V
2019 IIT Madras Y 1 W
8. The network of a small construction project Z 3 X, Y
awarded to a contractors is shown in the
The total float of the activity “V” (in weeks,
following figure. The normal duration, crash
in integer) is ________.
duration, normal cost and crash cost of all the
[2 Marks]

10. Contractor X is developing his bidding integer) obtained at the end of 4th year using
strategy against contractor Y. The ratio of straight line method and sum of years digit
Y ' s bid price to X ' s cost for the 30 method of depreciation is _______.
previous bids in which Contractor X has [2 Marks]
competed against Contractor Y is given in 14. The activity details for a small project are
the table given in the table.
Ratio of Y ' s bid Number of Duration Depends
Activity
price to X ' s cost bids (days) on
1.02 6 A 6 -
1.04 12 B 10 A
1.06 3 C 14 A
1.10 6 D 8 B
1.12 3 E 12 C
Based on the bidding behavior of the F 8 C
Contractor Y, the probability of winning G 16 D, E
against Contractor Y at a markup of 8% for H 8 F, G
the next project is [2 Marks] K 2 B
(A) 0% L 5 G, K
(B) 2100 % The total time (in days, in integer) for project
(C) More than 0% but less than 50% completion is _______. [2 Marks]
(D) More than 50% but less than 100% 2022 IIT Kharagpur
11. The direct and indirect costs estimated by a
contractor for bidding a project is Rs. 15. The project activities are given in the
following table along with the duration and
160000 and Rs. 20000 respectively. If the
dependency.
mark up applied is 10% of the bid price, the
Duration Depends
quoted price (in Rs.) of the contractor is Activities
(days) on
[1 Mark]
P 10 -
(A) 200000 (B) 198000
Q 12 -
(C) 196000 (D) 182000
12. In case of bids in two-envelop system, the R 5 P
correct option is : [1 Mark] S 10 Q
(A) Either of the two (Technical and T 10 P, Q
Financial) bids can be opened first Which one of the following combinations is
(B) Financial bid is opened 1st correct? [2 Marks]
(C) Both (Financial and Technical) bids are (A) Total duration of the project  22 days,
opened simultaneously Critical path is Q  S
(D) Technical bid is opened 1st (B) Total duration of the project  20 days,
13. An equipment has been purchased at initial Critical path is Q  T
cost of Rs. 160000 and has an estimated (C) Total duration of the project  22 days,
salvage value of Rs. 10000. The equipment Critical path is P  T
has an estimated life of 5 years. The (D) Total duration of the project  20 days,
difference between the book values (in Rs. in
Critical path is P  R

16. The activity of a project are given in the (B) an average cylinder strength of 20 MPa
following table along with their duration and (C) a 5-percentile cube strength of 20 MPa
dependency
(D) a 5-percentile cylinder strength of 20
Duration Depend MPa
Activities
(Days) on
18. In cement concrete mix design, with the
A 10 - increase in water-cement ratio, which one of
B 12 - the following statements is TRUE?
C 5 A (A) Compressive strength decreases but
D 14 B workability increases
E 10 B, C (B) Compressive strength increases but
Total float of the activity E (in days) is workability decreases
_______ (in integer). [2 Marks] (C) Both compressive strength and
2023 IIT Kanpur workability decrease
(D) Both compressive strength and
17. M20 concrete as per IS 456: 2000 refers to
workability increase
concrete with a design mix having______.
(A) an average cube strength of 20 MPa
:Construction Materials:
2. (B)
1. (B) Stress
In the compaction factor test, the degree of Aggregate
workability of concrete is measured in terms of
internal energy required to compact the concrete Concrete
thoroughly.
Cement paste
Compaction factor
Weight of partially compacted Strain
concrete Hence, the correct option is (B).
=
Weight of fully compacted concrete 3. (A)
The maximum possible value of compaction factor An air entraining agent introduces air in the form
for fresh concrete can be 1. of bubbles that occupy upto 5 % of the volume of
Hence, the correct option is (B). concrete distributed uniformly throughout cement
 Key Point paste. Thus for the same amount of water/cement
ratio, we get higher workability.
Degree of workability Compacting factor
An air entraining agent improves freezing and
Very low 0.78 thawing resistance since cracks in the concrete
Low 0.85 during freezing and melting of water is avoided
because water can take up the air voids present is
Medium 0.92
concrete.
High 0.95 Hence, the correct option is (A).

4. (D) 3. Tricalcium Aluminate (C3 A) :
Slump test : Slumps is the subsidence of the  It rapidly reacts with water and is responsible
specimen recorded in mm. for flash set of finely grounded clinker.
The difference in levels between the height of the  It contributes in 24 hours strength after
mould and that of the highest point of the subsided addition of water but it contribute less.
concrete is measured. 4. Tetra-calcium Aluminate ferrite
The difference in height in mm is taken as slump (C4 AF ) :
of concrete. As water content increases, slump
increases.  It is also responsible for high heat of
Compaction factor test : In the compaction factor hydration as compare to C2 S and but less
test, the degree of workability of concrete is than C3 A .
measured in terms of internal energy required to  Its contribution in strength is very less.
compact the concrete thoroughly.
The compound which is largely responsible
Weight of patially compacted concrete
C.F .  for initial setting and early strength gain of
Weight of fully compacted concrete
OPC is C3 S .
Vee-bee test : This test is preferred for finding
workability of stiff concrete mix having very low
workability. The time required for the shape of  Key Point
concrete fully assumes a cylindrical shape, in Composition of cement clinker
second is known as vee-bee degree.
 As the slump increases, the vee-bee time
decreases
Hence first statement false.
 As the slump increases, the compaction
factor increases
5. (B) 6. (C)
1. Tricalcium silicate (C3 S ) : Kilns : A kiln is a large oven which is used to burn
 It is considered as very good strength bricks. The kilns which are used in the
compound it enables the clinker to grind manufacture of bricks are of the following two
easily. types :
 It hydrates rapidly generates high heat and (1) Intermittent kilns :
develop early hardness and strength. (a) Intermittent up-drought kilns
 The hydration of C3 S is mainly responsible (b) Intermittent down-drought kilns
for 7 days strength and hardness. (2) Continuous kiln :
2. Dicalcium silicate (C2 S ) : (a) Bull’s trench kiln
 It hydrates and hardens slowly and takes (b) Hoffman’s kiln
long time to add to the strength. It is (c) Tunnel kiln
responsible for ultimate strength. Hence, the correct option is (C).
 After 28 days the gain of strength is due to
C2 S

7. (D)  Key Point
As per Indian standards for bricks, minimum Apparatus Determine
acceptable compressive strength of any class of 1. Vicat apparatus (i) Consistency,
burnt clay bricks in dry state is 3.5 MPa. Initial and final
Hence, the correct option is (D) setting of cement
 Key Point 2. Briquette (ii) Tensile strength
Class of brick Minimum crushing strength testing of cement
apparatus
First class  10 N/mm2
3. Le-Chatelier (iii) Soundness of
Second class  7.5 N/mm2 apparatus cement
4. Slump cone (iv) Workability of
8. (A)
apparatus concrete
Le-Chatelier test Soundness of OPC 5. Permeability (v) Fineness of
Vee-Bee test Consistency of workability apparatus with cement
of concrete manometer and
Blaine air Fineness of OPC flow meter
permeability test 6. Le-Chatelier (vi) Specific gravity
Vicat apparatus Consistency and setting time flask of cement
of OPC 12. 50.94
Volume of the box = 0.5  0.5  0.5  0.125m3
9. (D)
Mass of aggregate  187.5 kg
1. Walls of one brick thick are measured in
Specific gravity of aggregate (Gagg ) = 2.5
square meters.
2. No deduction in the brickwork quantity is 187.5
Volume of aggregate   0.075m3
made for openings in walls up to 0.1m2 area. 2.5 1000
3. For the measurement of excavation from the Volume of empty space  0.125  0.075
borrowpit in a fairly uniform ground,  0.05m3
deadman are left at suitable intervals. Water absorption  0.5%
Hence, the correct option is (D). 0.5 187.5
Volume of water absorbed  
10. (D) 100 1000
The soundness of cement is determined either by  9.375  104
Le-Chatelier’s method of by means of autoclave Total volume of water that can be filled
test.  9.375 104  0.05  0.0509 m3
Le-Chatelier’s method detects unsoundness due to
Mass of water  50.94 kg
free lime only.
Hence, the correct option is (D). Hence, the maximum quantity of water required to
11. (C) fill the box completely is 50.94 kg .
The setting time of cement is determined using

vicat apparatus.
13. (D) 17. (B)
Due to increase in C2 S heat of hydration 18. (B)

decreases. Seasoning of timber increases strength and
Hence, the correct option is (D). durability of timber which are used for different
 Key Point engineering purposes.
(i) Rate of hydration,
C4 AF  C3 A  C3S  C2 S 19. (B)
(ii) Dicalcium silicate, 20. (B)

(2CaO  SiO2 or C2S)
Column-I Column-II
 Last compound formed during
1 C3 S P Early age strength
hydration of cement.
 Responsible for progressive later stage 2 C2 S Q Later age strength
strength.
3 C3 A R Flash setting
 Structure requires later stage strength
proportion of this component increase e.g. S Highest heat of hydration
hydraulic structure, bridges. T Lowest heat of Hydration
14. (B) (1)  (P)

(2)  (Q) and (T)
 Abrasion resistance is determined by Los
Angles test. (3)  (R) and (S)
 The soundness of cement is determined :Construction Planning & Management:
either by ‘Le-Chatelier’s method’ or by
means of a ‘Auto clave test’. 1. (C)
 The specific gravity of cement is obtained by
using ‘Le-Chatelier’s’ flask.
15. (D)
1. Soundness test is done to evaluate resistance Now, for activity 5 – 7,
against weathering of aggregate. Early Start Time ( EST )  38
2. Crushing test is done to evaluate strength of
Late Start Time ( LST )  38
aggregate.
3. Loss angles abrasion test is to evaluate Total float ( FT )  Late start time (LST)
hardness of aggregate. Early start time (EST )
4. Stripping value test is to evaluate ratio of ( FT )  38  38  0
uncovered area observed to the total area of
Similarly, for activity 11 – 12,
aggregate. The test is conducted to determine
the effects of moisture upon the adhesion of Early Start Time ( EST )  80
binding material. Late Start Time ( LST )  80
Hence, the correct option is (D). Total float ( FT )  Late start time (LST)
16. (D) Early start time (EST)

( FT )  80  80  0 4. 51
 Key Point
Total float can be determined using following
expressions/methods
1. Total float,
( FT )  Late Start Time (LST )
Early Start Time (EST )
Time taken by the critical path is considered as the
2. Total float shortest possible time to finish the project as the
( FT )  Late Finish Time (LFT ) delay critical activity tends to delay in overall
Early Finish Time (EFT ) project. Here, the time taken by the critical path
3. For any activity, if there is no margin left for (P-Q-U-V) is
the occurrence of events (the slack of all events are 6  15  14  16  51days
zero), the total float will be zero.
5. (C)
2. 37.84
In case, when the scope of the construction work
is well defined with all its drawing, specifications,
quantities and estimates, Item rate contract also
called a schedule contract/unit price contract is
mostly preferred.
The item rate contract is most commonly used for
Now, the expected completion time is given by, all types of engineering works financed by public
O  4M  P or government bodies. This type of contract, is
T suitable for works which can be split into various
6
8  4 10  14 6  4  8  11 items and quantities under each item can be
  estimated with accuracy.
6 6
5  4  7  10 7  4 12  18  Key Point
 
6 6 1. EPC (Engineering Procurement
 10.33  8.17  7.17  12.17 Construction) contract : The process of
T  37.84 days tendering, engineering and construction may have
the detailed design of the project, procure all the
equipment and materials necessary. The Front End
Engineering Design phase (FEED) packages are
3. (C) used as the basis for bidding on when the client
offers the EPC work to the market.
It can be observed that path P-Q-T-W-X consumes
2. Percentage rate contract : It is similar to
maximum amount of time, hence it is the critical
the item rate contract in almost all respects as the
path of the project.
contract documents are the same as in item rate
Critical time  2  3  5  3  2  15days
contract, making payments to the contractor, the
engineer is similar. The method of tendering the
unit rates are different in this case. While The given case is of unsymmetrical distribution
tendering, the contractors do not have to write the (positive skewed or right skewed distribution). In
rate or estimated cost of each item, but a this case, mean is greater than median and mode (
percentage figure by which the estimate unit rates Mean  median  mode ) as shown in figure.
are to be increased or decreased, the same
percentage figure being applicable to all the items.
3. Lumpsum contract : In this type of
contract, the contractor offers to do the whole
work as shown in drawings and described by
specifications, for a total stipulated sum of money.
There are no individual rates quoted, thus it
becomes difficult to make adjustments in the
contract value if any changes are to be made in the  Key Point
work later on. Mode : It is the most frequently occurring data
A lump sum contract is more suitable for value in a series of numbers. The terms mode,
works for which contractors have prior median and mean are used to identify the
construction experience. This experience enables midpoint in a given skewed distribution data set.
the contractors to submit a more realistic bid. This For the symmetric distribution, the mode, median
type of contract is not suitable for difficult and the mean is same i.e., the axis of symmetry.
foundations, excavations of uncertain character Median : It is the mid data point in a data series
and projects susceptible to unpredictable hazards organized in sequence.
and variations.
Mean : It is the sum of all the data divided by the
6. 210 number of data sets.
Given : Mean deviation of the completion time 8. 149500
(TM )  200days
Standard deviation of the completion time
()  6.1days
Standard normal deviate ( z )  1.64
Let, the maximum time required for the
completion of the project be Tmax
Now, we know that
Tmax  TM
z

Tmax  200
1.64 
6.1
Tmax  1.64  6.1  200  210days Considering normal durations, the total cost of the
project
TC  DC  IC
7. (C)

 INR 66000  INR 5000 18 Stage 2 : Crash activity Q and R (or U and R) by
 INR156000 1 day
Stage 1 : Crash activity R by 1 day  Project becomes
 Project becomes
TC  (INR 66750  INR 2000 + INR750)

TC  (INR 6600  INR 750) +(INR 5000 16)
 INR 5000 17  149500
 151750
9. 0
7 11
R T 20
3 5 25 28
4 X
Earliest P 8 8 Z
5 10 11
2 5 20 3
0 V 25 28
N D1 2 Y
1 2 1
2 18 22
0 2 Q W
3 5 U 7 9
S 11 3
Latest
4 6 18 24
7
5 11
6
Critical Path = N-P-R-D1-U-V-X-Z = 28 weeks

For any critical activity all floats are 0
So for V activity total float = 0
By formula,
Total float  L j  Ei  tij  20 18  2 = 0
10. (C) 3
5. 1.12 3
30
Ratio of Probability of
At mark up of 8%
S. Y’s bid Number type of
Bid price of contractor X  1.08
No. price to of bid bid from previous
X’s cost Contractor X will with if quoted bid price of
30 bids
contractor Y is greater than X bid price,
6 For type 4 and type 5, Y bid is higher.
1. 1.02 6
30 So, Probability of win of contractor X is
12 6 3 9
2. 1.04 12     0.3
30 30 30 30
3 Option 3 more than 0% but less than 50%.
3. 1.06 3 Hence, the correct option is (C).
30
6 11. (A)
4. 1.10 6
30 Given : Direct cost = 160000 Rs
Indirect cost = 20000 Rs.

Mark up rate = 10% of bid price 3
D3  (160000  1000)   30000
Total estimated cost by contractor 15
 160000  20000 B3  70000  30000  40000
 180000 Rs
2
Mark up cost = 10% D4  (160000  10000)   20000
15
0.9  Quoted price = Total estimate cost
B4  40000  20000 = 20000
180000
Quoted price   200000  Difference between value of straight line
0.9
method and sum of years digit method
 40000  20000  20000
12. (D)
14. 56
In case of bids in two envelop system technical bid
is opened first. From the above given data we can draw different
Hence, the correct option is (D). path of completion in following way.
7
13. 20000 K
D1 L
2 5
Given : Initial cost (Ci )  160000

A B D G
1 2 3 5 6 9
6 10 8 16 D2
C H
Salvage value (CS )  10000
E
14 12 8
8
F
n  Life of asset 5 year
4
16
Book value after 4th year by straight line method Here, D1, D2 are dummy activities
Critical path = longest duration path
 160000  10000 
 160000  4    40000  Total paths :
 5 
(i) A - B - K - L = 23 days
Now, depreciation after mth year by sum of year
(ii) A - B - D - G - D1 - L = 45 days
digit method
(iii) A - B - D - G - D2 - H = 48 days
 n  m 1 
 (Ci  Cs )  (iv) A - C - E - G - D1 - L = 53 days
n(n  1) 
  (v) A - C - E - G - D2 - H = 56 days
 2  (vi) A - C - E - F - H = 36 days
Depreciation after 1 year ( D1 )  5th path is longest path which takes 56 days
(5  1  1) to complete project hence it is critical.
D1  (160000  10000)
5  (5  1) 15. (A)
2
Given,
 50000
Activity Duration Depend on
Book value after 1 year ( B1 )
 160000  50000  110000 P 10 -
Similarly we can calculate other depreciation and Q 12 -
booked values R 5 P
4
D2  (160000  10000)   40000 S 10 Q
15
B2  110000  40000  70000 T 10 P, Q

FT  (26  15)  10
FT  26  25
FT  1 day
17. (C)
18. (A)
Path :
(i) Q and T  22 (i) Increase in water cement ratio, decrease in
compressive strength of concrete because
(ii) Q and S  22
higher volume of water occupies the volume
16. 1 of concrete and introduces voids.
C (ii) Increase in w/c ratio, increase in workability
E
of concrete due to availability of water,
A 5
16 reduce the friction between the particles.
10 10
26 As per Abram’s law :
12 D
Compressive Strength
B 14
15 26
E
10
16 26
Water Cement Ratio
Total float, Hence, the correct option is (A).
FT = Maximum available Time
– Activity duration


GEOMATICS
7 ENGINEERING
Syllabus : Geomatics Engineering
Principles of surveying; Errors and their adjustment; Maps - scale, coordinate system; Distance
and angle measurement - Levelling and trigonometric levelling; Traversing and triangulation
survey; Total station; Horizontal and vertical curves. Photogrammetry and Remote Sensing -
Scale, flying height; Basics of remote sensing and GIS.
Contents : Geomatics Engineering
S. No. Topics
1. Fundamental Concepts
2. Levelling
3. Traversing
4. Theodolites and Plane Table Surveying
5. Tacheometric & Triangulation Surveying
6. Measurement of Area, Volume & Theory of Errors and Survey Adjustment
7. Curves
8. Field Astronomy & Photogrammetric Surveying
9. Basics of GIS, GPS & Remote Sensing
1 Fundamental Concepts
2014 IIT Kharagpur 2021 IIT Bombay

1.1 The survey carried out to delineate natural 1.3 In general, the CORRECT sequence of
features, such as hills, rivers, forests and surveying operations is [1 Mark]
man-made features, such as towns, villages, (A) Field observation  Reconnaissance 
buildings, roads, transmission lines and Data analysis  Map making
canals is classified as [1 Mark]
(B) Reconnaissance  Field observation
(A) Engineering survey  Data analysis  Map making
(B) Geological survey (C) Data analysis  Reconnaissance 
(C) Land survey Field observation  Map making
(D) Topographic survey (D) Reconnaissance  Data analysis 
2019 IIT Madras Field observation  Map making
1.2 A survey line was measured to be 285.5 m 2022 IIT Kharagpur
with a tape having a nominal length of the 1.4 A line between stations P and Q laid on a
type was found to be 0.05 m too short. If the slope of 1 in 5 was measured as 350 m using
line lay on a slope of 1 in 10. The reduced a 50 m tape. The tape is known to be short
length (horizontal length) of the line for by 0.1 m.
plotting of survey work would be
The corrected horizontal length (in m) of the
(A) 285.6 mm (B) 284.5 mm line PQ will be [1 Mark]
(C) 283.6 mm (D) 285.0 mm (A) 356.20 (B) 350.70
(C) 349.3 (D) 342.52
1.1 (D) 1.2 (C)

Engineering survey : To prepare detail drawing
Given :
of project involving roads, railways etc. and to
determine the quantities on datas for designing of Length of measured survey line ( L ')  285.5
engineering works. Nominal length of of tape = 30 m
Land survey : It is a survey being done on land. Gradient = 1 in 10
It involves running of survey lines and
Actual tape length = 30 – 0.05 = 29.95 m
determining their length.
Geological survey : The type of survey in which, Standardization correction :
information about both the surface and subsurface Actual tape length < Nominal length of tape
is acquired. 29.95 m  30 m
Correction per tape length = 29.95 – 30
= – 0.05m
Geomatics Engineering 1
Standardization correction, 1.4 (D)
Measured length
Cs   Correction Given : Tape length  50 m
Nominal length of tape
Too short = 0.1 m
285.5
  0.05 Q
30 350m
 0.4758 m (negative) 1
P 5
Slope correction :
Slope correction,
l'
h2 Correct sloping length  L
Cslope   l
2L ' 49.9
28.552   350
Cslope   1.4275 (negative) 50
2  285.5  349.3
1
  tan 1    11.309
5
Horizontal distance between P and Q,
L
cos  
349.3
L  342.51m
Total correction,
CTotal  CS  CSlope
 0.4758 1.4275
 (0.4758  1.4275)  1.9033
Correct length L = Measured length – Correction
Correct length L  285.5 1.9033  283.596 m
Hence the correct length (L) of measured survey
line is 283.596 m.
 Key Point
Correction due to slope always negative.
Correction for standardization.
 Length of tape too long positive.
 Length of tape too short negative.
1.3 (B)
Correct sequence :
Reconnaissance  Field observation  Data
analysis  Map making
2 Geomatics Engineering
2 Levelling
2013 IIT Bombay (B) Mean Sea Level (MSL) is used as a

reference surface for establishing the
2.1 A theodolite is set up at station A and 3 m horizontal control
long staff is held vertically at station B. The (C) Mean Sea Level (MSL) is a
depression angle reading at 2.5 m marking simplification of the Geoid
on the staff is 6010' . The horizontal distance (D) Geoid is an equipotential surface of
between A and B is 2200 m. Height of gravity
instrument at station A is 1.1 m and R.L. of 2.5 In a levelling work, sum of the Back Sight
A is 880.88 m. Apply the curvature and (B.S.) and Fore Sight (F.S.) have been found
refraction correction, and determine the R.L to be 3.085 m and 5.645 m respectively. If
of B (in m) ________. [2 Marks] the Reduced Level (R. L.) of the starting
2014 IIT Kharagpur station is 100.000 m, the R.L. (in m) of the
last station is _______. [1 Mark, Set-2]
2.2 The Reduced Levels (RLs) of the points P
2.6 The combined correction due to curvature
and Q are + 49.600 m and + 51.870 m
and refraction (in m) for a distance of 1 km
respectively. Distance PQ is 20 m. The
on the surface of Earth is [1 Mark, Set-2]
distance (in m from P) at which the + 51.000
(A) 0.0673 (B) 0.673
m contour cuts the line PQ is [1 Mark]
(C) 7.63 (D) 0.763
(A) 15.00 (B) 12.33
2.7 Two pegs A and B were fixed on opposite
(C) 3.52 (D) 2.27
banks of a 50 m wide river. The level was set
2.3 A levelling is carried out to establish the
up at A and the staff readings on Pegs A and
Reduced Levels (RL) of point R with respect
B were observed as 1.350 m and 1.550 m,
to the Bench Mark (BM) at P. The staff
respectively. Thereafter, the instrument was
readings taken are given below :
shifted and set up at B. The staff readings on
Staff BS IS FS RL
Pegs B and A were observed as 0.750 m and
station (m) (m) (m) (m)
0.550 m, respectively. If the R.L. of Peg A is
P 1.655 -- -- 100.000
100.200 m, the R.L. of Peg B (in m) is
Q – 0.950 -- – 1.500 --
_______. [2 Marks]
R -- -- 0.750 ?
If RL of P is 100.000 m, then RL of R (in m) 2016 IISc Bangalore
is [2 Marks, Set-1] 2.8 The staff reading taken on a workshop floor
(A) 103.355 (B) 103.155 using a level is 0.645 m. The inverted staff
(C) 101.455 (D) 100.355 reading taken to the bottom of a beam is
2015 IIT Kanpur 2.960 m. The reduced level of the floor is
40.500 m. The reduced level (expressed in
2.4 Which of the following statements is
FALSE? [1 Mark] m) of the bottom of the beam is [1 Mark]
(A) Plumb line is along the direction of (A) 44.105 (B) 43.460
gravity (C) 42.815 (D) 41.145
2.9 The vertical angles subtended by the top of a
tower T at two instrument stations set up at
P and Q, are shown in the figure. The two
stations are in line with the tower and spaced 2021 IIT Bombay
at a distance of 60 m. Readings taken from
2.13 Which of the following is NOT a correct
these two stations on a leveling staff placed
statement? [1 Mark]
at the benchmark (BM = 450.000 m) are also
shown in the figure. The reduced level of the (A) The first reading from a level station is a
top of the tower T (expressed in m) is ‘Fore Sight’.
______. [2 Marks] (B) Planimeter is used for measuring ‘area’.
(C) Contours of different elevations may
intersect each other in case of an
overhanging cliff.
(D) Basic principle of surveying is to work
from whole to parts.
2017 IIT Roorkee
2023 IIT Kanpur
2.10 An observer standing on the deck of a ship
just sees the top of a lighthouse. The top of 2.14 Trigonometric levelling was carried out
the lighthouse is 40 m above the sea level from two stations P and Q to find the reduced
and the height of the observer’s eye is 5 m level (R. L.) of the top of hillock, as shown
above the sea level. The distance (in km, up in the table. The distance between Stations P
to one decimal place) of the observer from and Q is 55 m. Assume Stations P and Q, and
the lighthouse is ______. [2 Marks, Set-2] the hillock are in the same vertical plane. The
2018 IIT Guwahati R. L. of the top of the hillock (in m) is
_______. (round off to three decimal places).
2.11 A level instrument at a height of 1.320 m has
been placed at a station having a Reduced Vertical
Level (RL) of 112.565 m. The instrument angle of Staff
R.L. of
reads- 2.835 m on a levelling staff held at the Station the top reading on
benchmark
bottom of a bridge deck. The RL of the of benchmark
bottom of the bridge deck (in m) is hillock
[2 Marks] P 1845' 2.340 m 100.000 m
(A) 116.720 (B) 116.080 Q 1245' 1.660 m
(C) 114.080 (D) 111.050
2019 IIT Madras
2.12 A staff is placed on a benchmark (BM) of
reduced level (RL) 100.000 m and a
theodolite is placed at a horizontal distance
of 50 m from the BM to measure the vertical
angles. The measured vertical angles from
the horizontal at the staff readings of 0.400
m and 2.400 m are found to be the same.
Taking the height of the instrument as 1.400
m, the RL (in m) of the theodolite station is
______. [2 Marks]
2.1 642.106 Last (RL) 100  0.705  0.75
( RL) R  100  1.455  101.455
Given : ( RL) A  880.88 m
2.4 (B)
Explanation : Mean Sea level (MSL) is not used
R
tan 6 10 ' 
0
as a reference surface for establishing the
2200 m horizontal control. It is used for establishing the
x  237.70m horizontal vertical control.
True staff reading at station B Hence, the correct option is (B).
 2.5  0.06728 d 2  2.5  0.06728 (2.2)2 2.5 97.44
 2.5  0.326  2.174 m
Given :  BS  3.085 m
So, ( RL) B = ( RL) A  HI  x  2.174
 FS  5.645 m
 880.88  1.1  237.70  2.174 First RL  100 m
 642.106 m
Last RL  ?
 FS   FS  Last RL  First RL
3.085  5.645  Last RL  100
Last RL  97.44 m
2.2 (B)
Hence, the R.L. of the last station is 97.44 m.
Given : Distance between P and Q is  20m
2.6 (A)
Given : d  1km
Correction due to curvature, Cc   0.07853d 2
PRS ~ PQN ,
Correction due to refraction, CR   0.01121d 2
20 x
  Combine correction, C  CC  CR
51.870  49.600 51.00  49.600
 x  12.33m C   0.07853d 2  0.01121d 2
Hence, the correct option is (B). Combined correction, Ccombined  0.0673d 2
Where d = Distance in kms
Cc   0.0673 12  0.0673m
2.3 (C) Hence, the correct option is (A).

We know that,
Last (RL) – First (RL)   BS   FS
2.9 476.915
Given : Distance between P and Q  60 m
Reduced level of BM  450 m
2.7 100
Given : ( RL) A  100.200 m

ha  1.350 m , hb  1.550 m
h 'a  0.550 m , h 'b  0.750 m
Difference in elevation,
(hA  hB )  (hA'  hB' )
H  In ATO ,
2
2 y
(1.350  1.550)  (0.550  0.750) tan10.50  …(i)
H  60  x
2 In BO ' T
H   0.2 m y
tan 16.50 
Since point A is more elevated than point B x
( RL) B  ( RL) A  H y  x tan16.50
 100.200  0.2  100 m From equation (i),
2  x tan16.50
tan10.50 
60  x
60 tan10.50  x tan10.50  2  x tan16.50
2.8 (A) 60 tan10.50  2  x(tan16.50  tan10.50 )
x  82.26 m
Given : RL of workshop floor = 40.500 m
y  x tan16.50  24.36 m
Staff reading from floor = 0.645 m
RL of point T  ( RL) BM  2.555  24.36
Height of instrument (H.I)
 450  2.555  24.36  476.915 m
 ( R.L.) floor  Staff reading from floor
Hence, the reduced level of the top of the tower T
 40.500  0.645  41.145 m is 476.915 m.
2.10 33.001
Given : Height of observer’s eye  h1  5m
Height of light house  h2  40 m
R.L. of bottom of beam

= H.I. + Inverted staff reading from
bottom of beam
 41.145  2.960  44.105 m
Hence, the correct option is (A). Distance of observer from the light house,
d  d1  d 2
d  3.855( h1  h2 )
d  3.855( 5  40)  33.0012 km  101.4 1.4  100 m
Hence, the distance of the observer from the Hence, the RL of the theodolite station is 100 m.
lighthouse is 33.0012 km.
2.13 (A)
2.11 (A)
2.14 137.682
Given : HI (Height of instrument)  1.320 m
RL (Reduced level of BM)  112.565 m
2.34 18 45'
12 45 '
1.66 P x
55
Q
Height of instrument RL = 100 m Datum
 RL  HI level When reading taking from P,
 112.565  1.320  113.885 m RL of hillock = HI P  x tan1845'
RL of bottom of bridge deck  100  2.34  x tan1845' ...  i 
 HI  2.835
When reading taking from Q,
 113.885  2.835  116.720 m
Hence, the correct option is (A). RL of hillock  HI Q   x  55  tan1245 '
 100  1.66   x  55  tan1245 ' ...  ii 
By Equating equation (i) and (ii),
2.12 100
100  2.34  x tan1845'  100  1.66   x  55  tan1245'
Given : Reduced level (RL) of bench mark (BM)
= 100.000 m
Horizontal distance between theodolite and staff = 2.34  0.339x  1.66  12.445  0.226x
50 m  x  104.115
 RL of the top of the hillock
 100  2.34  104.115tan1845'  137.682 m
Height of instrument = R.L. of B.M. + Back sight

 100  (0.4  1)  101.4 m
R.L. of theodolite station
 Height of instrument – Height
3 Traversing
2013 IIT Bombay attraction. The magnetic Fore bearing of line

AO and OA were observed to be S 520 40' E
3.1 The latitude and departure of a line AB are +
78 m and – 45.1 m respectively. The whole and N 500 20'W , respecti-vely. What is the
circle bearing of the line AB is true Fore bearing of line AB? [2 Marks]
[1 Mark] (A) N 81050' E (B) N 82010' E
(A) 300 (B) 1500 (C) N 84010' E (D) N 77050' E
(C) 2100 (D) 3300 3.5 The bearings of two inaccessible stations, S1
3.2 Following bearing are observed while (Easting 500 m, Northing 500 m) and S 2
traversing with a compass :
(Easting 600 m, Northing 450 m) from a
Fore Back
Line station S3 were observed as 2250 and
bearing bearing
AB 0 1530 26 ' respectively. The independent
126 45 ' 3080 00 '
Easting of station S3 (in m) is [2 Marks]
BC 45015 ' 227 030 '
CD (A) 450.000 (B) 570.710
340030 ' 1610 45 '
(C) 550.000 (D) 650.000
DE 258030 ' 78030 '
EA 2016 IISc Bangalore
212030 ' 310 45'
Applying correction due to local attraction, 3.6 The reduced bearing of a 10 m long line is N
the correct fore bearing of line BC will be 300 E . The departure of the line is
[2 Marks] [1 Mark]
0 0
(A) 48 15 ' (B) 50 15 ' (A) 10.00 m (B) 8.66 m
(C) 49 0 45 ' (D) 480 45 ' (C) 7.52 m (D) 5.00 m
2015 IIT Kanpur 2017 IIT Roorkee
3.3 In a closed loop traverse of 1 km total length, 3.7 The observed bearings of a traverse are given
the closing errors in departure and latitude below :
are 0.3 m and 0.4 m, respectively. The Line Bearing Line Bearing
relative precision of this traverse will be : PQ 0
46 15' QP 226015 '
[1 Mark]
QR 108015' RQ 286015 '
(A) 1 : 5000 (B) 1 : 4000
RS 201030 ' SR 20 030 '
(C) 1 : 3000 (D) 1 : 2000
3.4 In a region with magnetic declination of ST 3210 45 ' TS 1410 45 '
20 E , the magnetic Fore bearing (FB) of a The station(s) most likely to be affected by
line AB was measured as N 79050' E . There the local attraction is/are [2 Marks]
was local attraction at A. To determine the (A) Only R (B) Only S
correct magnetic bearing of the line, a point (C) R and S (D) P and Q
O was selected at which there was no local
3.8 The accuracy of an Electronic Distance Which of the triangles are ill-conditioned
Measuring Instrument (EDMI) is specified and should be avoided in Triangulation
as (a mm  b ppm). Which one of the surveys? [1 Mark]
following statements is correct? (A) Both Q and S (B) Both P and S
[1 Mark] (C) Both Q and R (D) Both P and R
(A) Both a and b remain constant, 3.11 The data from a closed traverse survey PQRS
irrespective of the distance being (run in the clockwise direction) are given in
measured. the table [1 Mark]
(B) a remains constant and b varies in Line Included angle (in degrees)
proportion to the distance being PQ 88
measured. QR 92
(C) a varies in proportion to the distance RS 94
being measured and b remains constant. SP 89
The closing error for the traverse PQRS (in
(D) Both a and b vary in proportion to the degrees) is _______.
distance being measured.
2020 IIT Delhi
2018 IIT Guwahati
3.12 An open traverse PQRST is surveyed using
3.9 The following details refer to a closed theodolite and the consecutive coordinates
traverse : [2 Marks] obtained are given in the table [2 Marks]
Consecutive coordinate
Consecutive Coordinates
Line Northing Southing Easting Westing
(m) (m) (m) (m) Northi Eastin Westin
Southin
Line ng g g
PQ - 437 173 - g
( ( (
QR 101 - 558 - (m
m m m
RS 419 - - 96 )
) ) )
SP - 83 - 634 PQ 110.2 - 45.5 -
The length and direction (whole circle QR 80.6 - - 60.1
bearing) of closure, respectively are RS - 90.7 - 70.8
(A) 1m and 900 ST - 105.4 55.5 -
(B) 2 m and 900 If the independent coordinates (Northing,
(C) 1m and 2700 Easting) of station P are (400 m, 200 m), the
(D) 2 m and 2700 independent coordinates (in m) of station T,
are
2019 IIT Madras
(A) 405.3, 229.9 (B) 205.3, 429.9
3.10 The interior angles of four triangles are given (C) 194.7, 370.1 (D) 394.7, 170.1
below, 3.13 The lengths and bearings of a traverse PQRS
Triangle Interior Angles are :
P Segment Length (m) Bearing
850 , 500 , 450
PQ 40 800
Q 1000 , 550 , 250 QR 50 10 0
R 1000 , 450 , 350 RS 30 2100
S 1300 , 300 , 200 The length of line segment SP (in m, round
off to two decimal places), is _____.
[2 Marks]
2021 IIT Bombay 3.18 If the magnetic bearing of sun at a place at
noon is S 20 E the magnetic declination (in
3.14 Traversing is carried out for a closed traverse
degree) at that place is.
PQRS. The internal angles at vertices P, Q,
(A) 20 W (B) 40 W
R and S are measured as 920 , 68, 1230 and
(C) 20 E (D) 40 E
770 respectively. If fore bearing of a line PQ [1 Mark]
is 270 , fore bearing of line RS (in degrees,
2023 IIT Kanpur
in integer) is ______. [2 Marks]
3.15 Which of the following is/are correct 3.19 The magnetic bearing of the sun for a
statement (s)? [1 Mark] location at noon is 183˚ 30ˊ. If the sun is
(A) The boundary of water of a calm water exactly on the geographic meridian at noon,
pond will represent contour line. the magnetic declination of the location
(B) If the whole circle bearing of a line is is_______ .
2700 , its reduced bearing is 900 NW. (A) 3˚ 30ʹ W
(C) In the case of fixed hair stadia (B) 3˚ 30ʹ E
tachometry, the staff intercept will be larger, (C) 93˚ 30ʹ W
when the staff is held nearer to the (D) 93˚ 30ʹ E
observation point. 3.20 A delivery agent is at a location R. To deliver
(D) Back bearing of line is equal to Fore
the order, she is instructed to travel to
Bearing 1800 location P along straight-line paths of RC,
3.16 For a given traverse, latitudes and departures
CA, AB and BP of 5 km each. The direction
are calculated and it is found that sum of
of each path is given in the table below as
latitudes is equal to +2.1 and the sum of
whole circle bearings. Assume that the
departures is equal to  2.8 m. The length
latitude (L) and departure (D) of R is (0, 0)
and bearing of the closing error,
km. What is the latitude and departure of P
respectively, are :
(in km, rounded off to one decimal place)?
(A) 2.45 m and 530 7'48'' NW
Paths RC CA AB BP
(B) 3.50 m and 530 7'48'' NW Directions (in
120 0 90 240
(C) 0.35 m and 53.130 SE degrees)
(D) 3.50 and 53.130 SE
(A) L = 2.5; D = 5.0
[2 Marks] (B) L = 0.0; D = 5.0
2022 IIT Kharagpur (C) L = 5.0; D = 2.5
3.17 The bearing of a survey is N 31017 'W . It’s a (D) L = 0.0; D = 0.0
azimuth observed from north is _______
degree. (rounded-off to two decimal places)
[1 Mark]
3.1 (D)
Given : Latitude = L cos   78 m
Departure = L sin    45.1m
Since latitude is positive and departure is negative
the line in 4th quadrant.
 45.1  Key Point
tan  
78 In a closed traverse the algebraic sum of the
Whole circle bearing (WCB)  3600  300 latitudes (L ) should be zero and the algebraic
 330 0 sum of departure (D ) should be zero.
3.4 (C)
3.2 (D)
Given : Station A is subjected to local attraction
Given : Fore bearing of DE – Back bearing of DE but station O is free from local attraction hence
 1800 fore bearing of OA will be correct.
Station D and E are free from local attraction, Correct FB of OA  3600  500 20 '  3090 40 '
CD – DC  340030 ' 1600 45' Correct back bearing of OA  3090 40 ' 180 '
Error at station ‘C’  (1800  1780 45')  1290 40 '
 ()1015' Given FB of AO  S 520 40'E

 1800  520 40 '
Correction at station ‘C’  ()1015'
 1270 20 '
Correction bearing CB  227 30 ' 1 15'
0 0
So local attraction of point A
 2280 45 '  1290 40 ' 127 0 20 '  20 20 '
BC  CB  45015' 2280 45'  180030 ' True bearing of line AB
Error at station B  183030 ' 1800  ()3030'  79050 ' 20 20 ' 20  N 84010'E
Correction at station B  3030 ' Hence, the correct option is (C).
Corrected bearing of BC  45015' 3030 '
 480 45'
3.5 (C)
Given :
3.3 (D)
Given : Perimeter of closed loop traverse  1km
Total error
Relative precision 
Perimeter of closed loop
e
Rp 
p
1
Total error  eL2  eD2  0.4 2  0.32  m
2 Let distance of S1S3  L1
Rp 
1/2 m

1 and distance of S2 S3  L2
1 10 m 2000
3
Easting of S3 ,
500  L1 sin 450  600  L2 sin 26034 '
L1 sin 450  L2 sin 26034 '  100 …(i) local attraction, then station R is most likely to be
Northing of S3 , affected.
500  L1 cos 450  450  L2 cos 26034 '
L1 cos 450  L2 cos 26034 '  50 …(ii)
Solving equation (i) and (ii), we get
L1  70.707 3.8 (B)
L2  111.800 The accuracy of an electronic distance measuring

instrument (EDMI) is
 Easting of S3  500  70.707sin 450   (a mm  b ppm)
= 550 m The value of a is constant instrument and the value
Hence, the correct option is (C). of b is directly proportional to the distance.
3.6 (D)
3.9 (A)
Given : Length of line L  10 m
All north reading are latitude,
 L  (419  101)  (437  83)  0
All east west reading are departure,
 D  (173  558)  (96  634)  1
 Length of closure
 ( L)2  ( D) 2  02  12  1
And direction of closure,
The departure of a line is, D 1
tan      tan 1    900
1 L 0
 L sin 300  10   5m
2 Hence, the correct option is (A).
3.10 (A)
3.7 (A) For a well-conditioned triangle, the interior angle
Given : should not be less than 300 .
In this way, triangle ‘Q’ and ‘S’ having less angles
Line Bearing Line Bearing
(acute angle).
PQ 46015' QP 226015 ' Q and S are ill conditioned.
QR 108015' RQ 286015 ' Hence, the correct option is (A).
RS 201030 ' SR 20 030 '
ST 3210 45 ' TS 1410 45 '
Difference of fore bearing and back bearing of PQ
and ST are 180 0 . So Station P, Q, S, T are free from 3.11 3
Assuming it as anticlockwise traverse.
Note : Mathematically sum of interior angle for a l cos   30.215
closed traverse l sin   33.081
 (2n  4)  900  [(2  4  4)]  900  3600 Length of line segment SP,
Where, n  Number of sides. l  30.215   33.081  44.802 m
2 2
Given sum of interior angles,

Hence, the length of line segment SP is 44.80 m.
 880  920  94  89  3630
3.14 196
The error in interior angle  3600  3630  30 Q
Hence, the closing error for the traverse PQRS is
30. 68 0
0
27
P 0
92
123 0 R
77 0
3.12 (D)
Given independent coordinates (Northing, S
Easting) of station P (400, 200) ( FB) RS  ?
N
( FB) Pq  27
400 m, 200 m
( BB ) PQ  27 0  1800  207 0
( FB )QR  207 0  680  1390
( BB )QR  1390  1800  319 0
W E
( FB) RS  319  123  1960
3.15 A, B, D
3.16 (B)
S Given : Sum of latitude, L  2.1 m
L  190.8 196.1  5.3 Sum of departure, D  2.8 m
D  101 130.9  29.9
Closing error (e)  (L)2  (D)2
Co-ordinate of T are  (400  5.3, 200  29.9)
 (394.7, 170.1) e  (2.1)2  (2.8)2  3.5 m
N
3.13 44.80 53 07' 4.37''
Segme Beari Lat. Dep. W E

Lengt ( l cos ) ( l sin )
n n
h
t g
PQ 40 m 800 6.945 39.392
QR 50 m 49.250 8.682 S
10 0
 D  1  2.8 
RS 30 m 2100
- 25.980 - 15 Bearing,   tan 1    tan  
SP 1 l cos  l sin   L   2.1 

  530 7 ' 48.37"
L  l cos   30.215  0 Hence, the correct option is (B).
D  l sin   33.081  0
3.17 328.71
0
N 31 17 ' W
WCB  360  31017 '  328.7160

3.18 (C)
Magnetic bearing of sun  S 20 E
Magnetic declination  
Magnetic bearing of sun in WCB  1780
True bearing of sun must be at noon is 180 0
Declination  TB  MB
 1800  1780
 20
Declination is positive it means declination is
eastward.
3.19 (A)
Given : Magnetic Bearing of sun = 183030'
We know, true bearing of sun at noon = 180o
True Bearing = magnetic Bearing  declination
Declination  180  183030'  330'  3030' W
3.20 (B)
Given each path length (l) = 5 km
Path Direction Latitude Departure
(l) ( )  l cos    l sin  
RC 120 0 -2.5 4.33
CA 00 5 0
0
AB 90 0 5
BP 2400 -2.5 -4.33
 Latitude of p in km =
latitude  2.5  5  2.5  0 and Departure
of p in km = departure  4.33  5  4.33  5
Hence, Latitude (L) = 0 and Departure (D) = 5
4 Theodolites & Plane Table Surveying
2014 IIT Kharagpur 450 . The horizontal distance between

stations P and Q is 20 m. The staff reading at
4.1 List-I contains tool/instrument while List-II a benchmark S of RL 433.050 m is 2.905 m.
contains the method of surveying. Match the Neglecting the errors due to curvature and
tool/instrument with the corresponding refraction, the RL of the station Q (in m), is
method of surveying. [2 Marks] [2 Marks]
List-I List-II (A) 413.050 (B) 413.955
A. Alidade 1. Chain surveying (C) 435.955 (D) 431.050
B. Arrow 2. Levelling
4.4 A theodolite is set up at station A. The RL of
C. Bubble tube 3. Plain table surveying
instrument axis is 212.250 m. The angle of
D. Stadia hair 4. Theodolite surveying
elevation to the top of a 4 m long staff, held
Codes : vertical at station B, is 7 0 . The horizontal
A B C D distance between stations A and B is 400 m.
(A) 3 2 1 4 Neglecting the errors due to curvature of
(B) 2 4 3 1 earth and refraction, the RL (in m, round off
(C) 1 2 4 3 to three decimal places) of station B is
(D) 3 1 2 4 _______.
[2 Marks]
2017 IIT Roorkee
2023 IIT Kanpur
4.2 The method of orientation used, when the
plane table occupies a position not yet 4.5 The direct and reversed zenith angles
located on the map, is called as observed by a theodolite are 56o00'00" and
[1 Mark] 303o00'00", respectively. What is the
(A) Traversing (B) Radiation vertical collimation correction?
(C) Levelling (D) Resection (A) + o00'00"
2020 IIT Delhi (B) − 1o00'00"
(C) − 0o30'00"
4.3 A theodolite was set up at a station P. The (D) + 0o30'00"
angle of depression to a vane 2 m above the
foot of a staff held at another station Q was
4.1 (D) A. Alidade : It is a straight edge rular used in

plane table surveying which is used for
List-I List-II sighting the objects and drawing the lines in
A. Alidade 1. Plate table survey the drawing sheet
B. Arrow 2. Chain survey B. Arrow : Arrows are used to mark the
C. Bubble tube 3. Levelling position of end of the chain or tape on the
D. Stadia hair 4. Theodolite ground.
C. Bubble tube : It is used to check the level of x  20 m
the instrument when the bubble of the tube
RL of Q  433.05  2.905  x  2
comes in the centre, then the instrument is
levelled.  433.05  2.905  20  2
D. Stadia hair : Line on diaphragm, the  413.955 m
intercept ‘S’ between two stadia hairs of a
vertically held rod gives the distance Hence, the correct option is (C).
between tacheometer and the rod as 4.4 257.36
D  KS (k = stadia constant)
Given : R.L. of instrument axis = 212.250 m
Horizontal distance between A and B = 400 m
4.2 (D)
Traversing : Traverse means to pass across.

Traversing refer to the framework of series of lines
forming an open or closed polygon.
Radiation : It is a method of locating a point by
drawing a radial line from the plane table station
to the station under consideration.
Levelling : It is the method of determining the
V  400 tan 7 0  49.113
difference of elevations or levels of different
points on the earth’s surface. x  49.113  4  45.113
Resection : It is the method of locating a station RLB  212.25  45.113  257.363 m
occupied by the plane table when the position of
that station has not been plotted earlier on the sheet Hence, the RL of station B is 257.36 m.
when the plane table occupied other station. 4.5 (D)
Given : Observed value of direct zenith angle
 1   5600 '00"
Observed value of reversed zenith angle
4.3 (B)  2   30300 '00"
Given : Error 
 1  2   360
2
Distance between stations P and Q = 20 m
Staff 20 m

 56  303  360  30 '
2.905 m
0
45 2
Bench mark S P Correction  30'
RL = 433.05 m x
2m
x
 tan 450
20
5 Tacheometric & Triangulation Surveying
2014 IIT Kharagpur distance (in m) between the points Q and R

is _______.
5.1 A tacheometer was placed at point P to
[2 Marks]
estimate the horizontal distances PQ and PR.
The corresponding stadia intercepts with the
telescope kept horizontal, are 0.320 m and
0.210 m, respectively. The  QPR is
measured to be 610 30' 30". If the stadia
multiplication constant = 100 and the stadia
addition constant = 0.10 m. The horizontal
5.1 28.8
Distance of PQ  KS  C
 1 0 0 ( 0 . 3 5 ) 0 . 1
3 2 . 1 0 m
Distance of PR  KS  C
 100  (0.21)  0.1
 21.10 m
Applying the cosine rule,
QR  PQ2  PR 2  2( PQ)( PR) cos 
Where   61030 '30 ''
(32.10)2  (21.10)2
QR 
 2(32.10)(21.10) cos(61030'30'')
QR  28.799 m  28.80 m
Hence, the horizontal distance between the points
Q and R is 28.80 m.
6 Measurement of Area, Volume & Theory of Errors and Survey
Adjustment
2015 IIT Kanpur 2021 IIT Bombay

6.1 In a survey work, three independent angles, 6.3 A horizontal angle  is measured by four
X, Y and Z were observed with weights different surveyors multiple times and the
WX , WY , WZ respectively. The weight of the values reported are given below. [1 Mark]
sum of angles X, Y and Z is given by, Angle Number of
Surveyor
[2 Marks] observation
0
1 1 36 30 ' 4
(A) 0
 1 1 1  2 36 00 ' 3
W  W  W  3 0
8
 X Y Z 
35 30 '
0
4 36 30 ' 4
 1 1 1 
(B)     The most probable value of the angle  (in
 WX WY WZ 
degree, round off to two decimal places) is
(C) WX  WY  WZ
________.
(D) WX2  WY2  WZ2 2022 IIT Kharagpur
2019 IIT Madras 6.4 The error in measuring the radius of a 5 cm
6.2 A series of perpendicular off sets taken from circular rod was 0.2%. If the cross-sectional
a curved boundary wall to a straight 1.67, area of the rod was calculated using this
2.04, 2.34, 1.87 and 1.15 m. The area (in m 2 measurement, then the resulting absolute
, sound off to 2 decimal places) bounded by percentage error in the computed area
the survey line curved boundary wall, the is______. (round off to two decimal places)
first and the last offset, determined using [1 Mark]
Simpson’s rule, is _______. [2 Marks]
6.1 (A) Wa
K.A(K=constant)
K2
Various laws of weight.
A
Let there be quantities A and B with weight Wa and Wa K 2
K
Wb respectively, Hence, the correct option is (A).
Operation Resultant weight
1
A B 1

1
Wa Wb
1
A B 1

1
Wa Wb
6.2 68.54 A 2rr

A A
A 2rr

A r 2
A  r   r 
 2    0.2 % given 
A  r   r 
A
 2  0.2  0.40%
Area by Simpson’s rule, A
d
A [(h0  hn )  4(h1  h3  ....)
3
 2(h2  h4  ....)]
6
 [(1.22  1.15)  4(1.67  2.34  1.87)
3
2(2.04  2.14)]
 68.50 m 2
Hence, the first and the last offset, using
Simpson’s rule, is 68.50 m2.
 Key Point
Area by trapezoidal rule,
d
A [(h0  hn )  2(h1  h2  ....hn1 )]
2
Area by Simpson’s one third rule,
d
A [(h0  hn )  4(h1  h3  ....hn1 )
3
2(h2  h4  hn2 )]
6.3 360
Σ (angle × No. of observation)

MPV 
No. of observation
36030' 4  360  3  35030' 8  36030' 4

MPV 
19
MPV  360
6.4 0.4
A  r 2
For percentage error A  2rr
7 Curves
2014 IIT Kharagpur (A) 0.577 R (B) 1.155 R

(C) 1.732 R (D) 3.464 R
7.1 The chainage of the intersection point of two
straight line is 1585.60 m and the angle of 2017 IIT Roorkee
intersection is 140°. If the radius of a circular 7.3 The VPI (vertical point of intersection) is
curve is 600.00 m, the tangent distance (in 100 m away (when measured along the
m) and length of the curve (in m), horizontal) from the VPC (vertical point of
respectively are [2 Marks] curvature). If the vertical curve is parabolic,
(A) 418.88 and 1466.08 the length of the curve (in meters and
(B) 218.38 and 1648.49 measured along the horizontal) is ______.
(C) 218.38 and 418.88 [1 Mark]
(D) 418.88 and 218.38
2016 IISc Bangalore
7.2 A circular curve of radius R connects two
straights with a deflection angle of 600 . The
tangent length is [1 Mark]
7.1 (C) 2R l


3600 
Given : Angle of intersection  1400 2R   2 600  40
Radius of curve R  600 m l   418.88m
3600 3600
7.2 (A)
Given : Deflection angle   600

Tangent length  R tan
2
600
 R tan
2
  1800  1400  400
 0.577 R
Tangent distance T1P or T2 P ,

 R tan  600 tan 200  218.38 m
2
Length of the curve T1CT2 ,
7.3 200
Given : VPI is 100 m away (When measured
along horizontal) from VPL.
VPL is horizontally midway between VPC and

VPT,
L
VPI  VPC 
2
L
VPC  VPI 
2
L
0  100 
2
L
 100
2
L  200 m
Hence, the length of the curve is 200 m.
8 Field Astronomy & Photogrammetric Surveying
2016 IISc Bangalore 2018 IIT Guwahati

8.1 Optimal flight planning for a 8.4 A square area (on the surface of the earth)
photogrammetric survey should be carried with side 100 m and uniform height, appears
out considering [1 Mark] as 1cm 2 on a vertical aerial photograph. The
topographic map shows that a contour of 650
(A) Only side-lap
m passes through the area. If focal length of
(B) Only end-lap the camera lens is 150 mm, the height from
(C) Either side-lap or end-lap which the aerial photograph was taken, is
[2 Marks]
(D) Both side-lap as well as end-lap (A) 800 m (B) 1500 m
8.2 A tall tower was photographed from an (C) 2150 m (D) 3150 m
elevation of 700 m above the datum. The 8.5 An aerial photograph of a terrain having an
radial distance of the top and bottom of the average elevation of 1400 m is taken at a
tower from the principal points are 112.50 scale of 1 : 7500. The focal length of the
mm and 82.40 mm, respectively. If the camera is 15 cm. The altitude of the flight
bottom of the tower is at an elevation of 250 above mean sea level (in m, up to one
m above the datum, then the height decimal place) is ______. [2 Marks]
(expressed in m) of the tower (expressed in
2019 IIT Guwahati
m) is _____. [1 Mark]
8.6 A camera with a focal length of 20 cm fitted
2017 IIT Roorkee
in an air craft is used for taking vertical
8.3 Two towers, A and B, standing vertically on photographs of a torsion. The average
a horizontal ground, appear in a vertical elevation of the torsion is 1200 m above
aerial photograph as shown in the figure.
mean sea level (MSL). What is the height
[2 Marks]
above MSL at which an air craft must fly in
order to get the aerial photographs at a scale
of 1/8000? [2 Marks]
(A) 3200 m (B) 2600 m
(C) 3000 m (D) 2800 m
The length of the image of the tower A on the
photograph is 1.5 cm and of the tower B is 2022 IIT Kharagpur
2.0 cm. The distance of the top of the tower 8.7 An aerial photograph is taken from a flight at
A (as shown by the arrow head) is 4.0 cm and
a height of 3.5 km above mean sea level,
the distance of the top of the tower B is 6.0
cm, as measured from the principal point p using a camera of focal length 152 mm. If the
of the photograph. If the height of the tower average ground elevation is 460 m above
B is 80 m, the height (in meters) of the tower mean sea level, then the scale of photograph
A is _______. is [1 Mark]
(A) 1 : 100000 (B) 1 : 20
(C) 1 : 20,000 (D) 1 : 2800
2023 IIT Kanpur
8.8 If the size of the ground area is 6 km × 3 km

and the corresponding photo size in the aerial
photograph is 30 cm × 15 cm, then the scale
of the photograph is 1: _______. (in integer).
8.1 (D) Hence, the height of the tower is 120.4 m.
Optional flight planning for a photogrammetric

survey should be carried out by considering both
side lap as well as end lap.
8.3 90
Note : It is describe to have about 60% Given :
longitudinal over lap and 30% side overlap there For tower A :
by covering the entire area to be surveyed. Relief displacement, d A  1.5cm
8.2 120.4
rA  4cm
For tower B :
Given : Flying height H  700 m
Relief displacement,
h0  250 m , r  112.50 mm, r1  82.40 mm . d B  2.0cm, rB  6cm hB  80m
hr
Relief displacement d 
H  havg
hB rB
For B, dB 
H  h0
80  6
2
H  h0
H  h0  240 m
hA rA
For A, dA 
H  h0
hA  4
1.5 
240
Relief displacement,
240 1.5
rh hA   90 m
d 4
H  h0 Hence, the height of the tower A is 90 m.
d  r  r1
d  112.50  82.40  30.1mm
30.1 (700  250)
h  120.4 m
112.50
H  2800 m
8.4 (C)
Given : Focal length f  150 mm
f
Scale, S 
H h
Elevation of ground (h)  650 m
RF (Representation fraction)
1cm 2 1
 2 2
 Hence, the correct option is (D).
(100) cm 10000
1 0.15

10000 H  650
H  650  1500
8.7 (C)
H  1500  650  2150 m
Hence, the correct option is (C). Given : f  152 mm
O
8.5 2525 3.5 km
1 h0  460 m
Given : Scale  , h  1400 m , f  15cm
7500 MSL
f
Scale  f 0.152 1
H h Scale =  
H  h0 (3.5  1000)  460 20000
1 0.15

7500 H  1400 8.8 20000
H 1400  0.15  7500  1125 Given : Ground area = 6 km  3 km
H  2525 m
Map area = 30 cm  15 cm
Hence, the altitude of the flight above mean sea
level is 2525 m. Map area
We know, scale =
Ground area
30 15 1
 
6 10  3 10
5 5
20000
8.6 (D)  Scale of photograph is 1:20000.
Given : Focal length f  20 cm
Hence, the correct answer is 1 : 20000.
As we know scale of vertical photograph
f

H  havg
1
Scale 
8000
1 20

8000 ( H  1200) 100
9 Basics of GIS, GPS & Remote Sensing
2016 IISc Bangalore 2017 IIT Roorkee
9.1 The minimum number of satellites needed 9.3 The number of spectral bands in the
for a GPS to determine its position Enhanced Thematic Mapper sensor on the
precisely is [1 Mark] remote sensing satellite LANDSAT-7 is
(A) 2 (B) 3 [1 Mark]
(C) 4 (D) 24 (A) 64 (B) 10
9.2 The system that uses the Sun as a source of (C) 8 (D) 15
electromagnetic energy and records the
naturally radiated and reflected energy
from the object is called [1 Mark]
(A) Geographical information system
(B) Global positioning system
(C) Passive remote sensing
(D) Active remote sensing
9.1 (C) Passive remote sensing systems record EMR that

Minimum four satellite must be required from is reflected or emitted from the surface of earth
which three are position coordinate and one for passive sensors can only be used to defect when
clock deviation from satellite time. the energy naturally occurring energy is available.
Hence, the correct option is (C). Hence, the correct option is (C).
9.2 (C) 9.3 (C)

Remote sensing system which measure energy i.e. The landsat enhanced thematic mapper plus (ETM
naturally available are called as passive remote +) sensor is carried on landsat-7 and images
sensing. The sun provide a very convenient source consist of 8 spectral bands with a spatical
of energy for remote sensing. resolution of 30 m for bands 1 to 7. The reduction
for band 8. (Panchromatic is 15 m).
Total number of spectrum band in enhanced.
Thematic mapper Wave length ( m)

sensor  8
Band 1 - Blue 0.45  0.52
Band 2 - Blue 0.42  0.60
Band 3 - Blue 0.63  0.69
Band 4 - Blue 0.76  0.90
Band 5 - Blue 1.55 1.75
Band 6 - Blue 3.0  5.0
Band 7 - Blue 2.08  2.35
Band 8 - Blue 10.4 12.5
ENGINEERING
8 MATHEMATICS
Syllabus : Engineering Mathematics

Linear Algebra : Matrix algebra; Systems of linear equations; Eigen values and Eigen vectors.
Calculus : Functions of single variable; Limit, continuity and differentiability; Mean value
theorems, local maxima and minima; Taylor series; Evaluation of definite and indefinite
integrals, application of definite integral to obtain area and volume; Partial derivatives; Total
derivative; Gradient, Divergence and Curl, Vector identities; Directional derivatives; Line,
Surface and Volume integrals. Ordinary Differential Equation (ODE) : First order (linear and
non-linear) equations; higher order linear equations with constant coefficients; Euler-Cauchy
equations; initial and boundary value problems. Partial Differential Equation (PDE) : Fourier
series; separation of variables; solutions of one- dimensional diffusion equation; first and
second order one-dimensional wave equation and two-dimensional Laplace equation.
Probability and Statistics : Sampling theorems; Conditional probability; Descriptive statistics –
Mean, median, mode and standard deviation; Random Variables – Discrete and Continuous,
Poisson and Normal Distribution; Linear regression. Numerical Methods : Error analysis.
Numerical solutions of linear and non-linear algebraic equations; Newton’s and Lagrange
polynomials; numerical differentiation; Integration by trapezoidal and Simpson’s rule; Single
and multi-step methods for first order differential equations.
Contents : Engineering Mathematics
S. No. Topics
1. Linear Algebra
2. Differential Equations
3. Integral and Differential Calculus
4. Vector Calculus
5. Maxima and Minima
6. Mean Value Theorem
7. Complex Variables
8. Limit & Series Expansion
10. Numerical Methods
11. Transform Theory
1 Linear Algebra
(A) If the trace of the matrix is positive and

2013 IIT Bombay
the determinant of the matrix is negative,
1.1 There are three matrixes  P 42 , Q 24 and at least one of its Eigen values is
 R 41 . The minimum number of
negative.
(B) If the trace of the matrix is positive, all
multiplication required to compute the
its Eigen values are positive.
matrix PQR is
(C) If the determinant of the matrix is
[1 Mark]
positive, all its Eigen values are positive.
2014 IIT Kharagpur (D) If the product of the trace and
determinant of the matrix is positive, all
3 2 1 its Eigen values are positive.
1.2 Given the matrices J   2 4 2  and
1.6 The rank of matrix
1 2 6 
[2 Marks]
 1
6 0 4 4 
K   2  , the product K T JK is _____.   2 14 8 18 
  1   is ______.
 14  14 0 10 
[1 Mark]
1.7 With reference to the conventional Cartesian
1.3 The sum of the Eigen value of the matrix,
( x, y ) coordinate system , the vertices of a
[ M ] is
triangle have the following coordinates :
[1 Mark]
( x1 , y1 )  (1,0) ; ( x2 , y2 )  (2, 2) ;
 215 650 795
( x3 , y3 )  (4,3)
where, [ M ]   655 150 835 
 485 355 550  The area of triangle is equal to
[1 Mark]
(A) 915 (B) 1355
3 3
(C) 1640 (D) 2180 (A) (B)
2 4
1.4 The determinant of matrix
4 5
[1 Mark] (C) (D)
5 2
0 1 2 3
1 0 3 0  2015 IIT Kanpur
 is ______.
2 3 0 1 1.8 For what value of p the following set of
  equation will have no solution?
3 0 1 2
1.5 Which one of the following statements is [1 Mark]
TRUE about every n × n matrix with only 2x  3y  5
real Eigen values? 3 x  py  10
[1 Mark]
Engineering Mathematics 1
1.9 The smallest and largest Eigen values of the 1.14 The matrix P is the inverse of matrix Q. If
following matrix are I denotes the identify matrix, which one of
[2 Marks] the following option is correct?
 3 2 2  [1 Mark]
 4 4 6  (A) PQ  I but QP  I
 
 2 3 5  (B) QP  I but PQ  I
(C) PQ  I and QP  I
(A) 1.5 and 2.5 (B) 0.5 and 2.5
(D) PQ  QP  I
(C) 1.0 and 3.0 (D) 1.0 and 2.0
1.10 Let A  [aij ], 1  i, j  n with n  3 and  5 1
1.15 Consider the matrix   . Which one of
aij  i. j. The rank of A is 4 1 
the following statements is TRUE for the
[1 Mark] Eigen values and Eigen vectors of this
(A) 0 (B) 1 matrix?
(C) n – 1 (D) n [2 Marks]
2 1  (A) Eigen value 3 has a multiplicity of 2, and
1.11 The two Eigen Values of the matrix  
1 p  only one independent Eigen vector
have a ratio of 3 : 1 for p = 2. What is another exists.
value of p for which the Eigen values have (B) Eigen value 3 has a multiplicity of 2, and
the same ratio of 3 : 1 two independent Eigen vectors exists.
(C) Eigen value 3 has a multiplicity of 2, and
[2 Marks]
no independent Eigen vector exists.
(A) 2 (B) 1 (D) Eigen value are 3 and – 3, and two
7 14 independent Eigen vectors exists.
(C) (D)
3 3 1.16 Consider the following simultaneous
equations (with c1 and c2 being constants) :
2016 IISc Bangalore
3x1  2 x2  c1
1.12 If the entries in each column of a square
4x1  x2  c2
matrix M add up to 1, then an Eigenvalue of
The characteristics equation for these
M is
simultaneous equation is
[1 Mark] [1 Mark]
(A) 4 (B) 3
(A)   4  5  0
2
(C) 2 (D) 1
(B)  2  4  5  0
1.13 Consider the following linear system
(C)  2  4  5  0
x  2 y  3z  a
(D)  2  4  5  0
2 x  3 y  3z  b
1 5  3 7 
5x  9 y  6 z  c 1.17 If A    and B   T
 , AB is
 6 2   8 4 
This system is consistent if a, b and c equal to
satisfy the equation [2 Marks]
[2 Marks] 38 28  3 40 
(A) 7 a  b  c  0 (B) 3a  b  c  0 (A)   (B)  
32 56   42 8 
(C) 3a  b  c  0 (D) 7 a  b  c  0  43 27  38 32
(C)   (D)  
2017 IIT Roorkee 34 50   28 56
1.18 For the given orthogonal matrix Q,
2 Engineering Mathematics
 3 2 6   2  4
 7 7 1.20 The matrix   has
7   4  2
 
Q  
6 3 2  [2 Marks]
 7 7 7 
  (A) real Eigen values and Eigen vectors.
 2 6 3
  (B) real Eigen value but complex Eigen
 7 7 7  vectors.
The inverse is (C) complex Eigen value but real Eigen
[1 Mark] vectors.
 3 2 6  (D) complex Eigen value and Eigen vectors.
 7 7 7  1.21 The rank of the following matrix is
 
1 1 0  2 
(A)  
6 3 2 
 7 7 7  2 0 2 2 
   
 2 6 3  4 1 3 1 

 7 7 7  [2 Marks]
 3 2 6 (A) 1 (B) 2
 7 
7
 
7 (C) 3 (D) 4
 
2019 IIT Madras
(B) 
6 3 2
 
 7 7 7 1.22 Consider the hemispherical tank of radius 13
  m as shown in the figure (not drawn to
 2 
6 3 
 7 7 7  scale). What is the volume of water (in m 3
3 6 2  ) when the depth of water at the centre of
7 
7 
the tank is 6 m ?
7
  [2 Marks]
(C) 
2 3 6 
7 7 7 
 
6 2

3
 7 7 7 
 3 6 2
 7 7
 
7
(A) 156  (B) 396 
  (C) 468  (D) 78 
(D)  
2 3 6
 
 7 7 7 2 3 4
  1.23 The inverse of the matrix  4 3 1  is
 6 
2 3 
 7 7 7  1 2 4 
2018 IIT Guwahati [2 Marks]
 10 4 9 
1.19 Which one of the following matrices is
singular? (A)  15 4 14 
[1 Mark]  5 1 6 
 2 5  3 2  10 4 9 
(B)  15 4 14 

(A)   (B)  2 3
 1 3  
 5 1 6 
 2 4  4 3
(C)   (D)  
3 6 6 2
 4 9 1 2 2 3
 2  5  5 3 4 2 5 
  1.27 The Rank of matrix  is

(C) 3
4 14  5 6 2 7
 5 5   
  7 8 2 9
 1  1  6 [1 Mark]
 5 5  (A) 4 (B) 3
 4 9 
 2 5
(C) 2 (D) 1
5 
   5 0 5 0
(D)  3 
14  0 2 0 1 
4

 5 5 1.28 The rank of matrix  is
   5 0 5 0
 1 1 6   
 5 5  0 1 0 2
2020 IIT Delhi [1 Mark]
(A) 2 (B) 3
1.24 Consider the system of equations
(C) 4 (D) 1
1 3 2  1 
 2 2 3  x1  1  1.29 If A is a square matrix then orthogonality
   x2     property mandates
 4 4 6     2  [1 Mark]
   x3   
2 5 2  1  (A) AA  A
T 2
(B) AA  I
T
The value of x3 (round off to the nearest (C) AAT  A1 (D) AAT  0
integer), is ______. 1.30 The smallest eigen value and corresponding
[2 Marks]  2 2 
1.25 A 4  4 matrix [ P ] is given below eigen vector of the matrix   is
 1 6 
[2 Marks]
[2 Marks]
0 1 0
3
 2 2.55
3 4 
0 (A) 1.55 and  
[ P]   0.45
0 0 1
6
   2.00 
0 0 6
1 (B) 1.55 and  
The eigenvalues of [ P ] are 0.45
(A) 1,2,3,4 (B) 1,2,5,7 1.00
(C) 2.00 and  
(C) 0,3,6,6 (D) 3,4,5,7 1.00
2021 IIT Bombay  2.00 
(D) 1.55 and  
1 2  0 1   0.45 
1.26 If P    and Q   T T
 then Q P
 3 4   1 0  2022 IIT Kharagpur
is 1.31 The Cartesian co-ordinates of a point P in
[1 Mark] right - handed co-ordinate system are (1, 1,
1 2  1 3 1). The transformed co-ordinates of P due to
(A)   (B)   a 450 clockwise rotation of the co-ordinate
3 4   2 4
system about the positive x-axis are
 2 4  2 1
(C)   (D)   [2 Marks]
1 3  4 3 (A) (1, 0,  2) (B) (1, 0, 2)
(C) (1, 0, 2) (D) (1, 0,  2) [1 Mark]
1.32 The matrix M is defined as 2023 IIT Kanpur
[1 Mark] 1.36 For the matrix
1 3   1 1 0 
M  
4 2 [ A]   1 2 1
and has eigen value 5 and – 2.  0 1 1 
The matrix Q is formed as
Which of the following statements is/are
Q  M 3  4M 2  2M
TRUE?
Which of the following is/are the eigen (A) [𝐴]{𝑥} = {𝑏} has a unique
value(s) of matrix Q?
solution
(A) – 20 (B) 25
(B) [𝐴]{𝑥} = {𝑏} does not have a unique
(C) – 30 (D) 15
1.33 Let y be a non zero vector of size 2022  1. solution
(C) [𝐴] has three linearly independent
Which of the following statement(s) is/are
true? eigenvectors
[2 Marks] (D) [𝐴] is a positive definite matrix
2 1 0 3
T T T
(A) y y is an eigen value of yy 1.37 Two vectors and
(B) yy T has a rank of 2022 1 0 1 2
T
belong to the null space of
T
(C) yy is a symmetric matrix a 4 × 4 matrix of rank 2. Which one of the
T
(D) yy is invertible following vectors also belongs to the null
1.34 P and Q are two square matrices of the same space?
order. Which of the following statement(s) (A) 1 1 1 1
T
is/are correct?
(B)  2 0 1 2
T
[1 Mark]
(A) If P and Q are not invertible, then
(C)  0 2 1 1
T
 PQ
1 1 1
Q P
(D) 3 1 1 2
T
(B) If P and Q are invertible, then
 PQ
1
 P 1Q 1 1.38 Cholesky decomposition is carried out on the
following square matrix [𝐴].
(C) If P and Q are invertible, then
8 5 
 PQ
1
 Q 1 P 1  A    5
 a22 
(D) If P and Q are invertible, then
QP Let iij and aij be the  i, j 
1
 P 1Q 1
th
elements of
1.35 The component of pure shear strain in a matrices [𝐿] and [𝐴], respectively. If the
sheared material are given in the matrix element l22 of the decomposed lower
1 1 
E 
triangular matrix [𝐿] is 1.968, what is the
1 1 value (rounded off to the nearest integer)
Here, trace ( E )  0. Given P  Trace ( E 8 ) of the element a22 ?
and Q  Trace ( E "). (A) 5 (B) 7
Numerical value ( P  Q ) is equal to
(C) 9 (D) 11
_______ (in integer).
1.39 If M is an arbitrary real 𝑛×𝑛 matrix, then (A) The eigenvalues of [𝐴] are same as the
which of the following matrices will have eigenvalues of [𝐴]
non-negative eigenvalues? (B) The eigenvalues of [𝐴]−1 are the
(A) M 2 reciprocals of the eigenvalues of [𝐴]
(B) MM 𝑇 (C) The eigenvectors of [𝐴] are same as the
(C) M 𝑇M eigenvectors of [𝐴]
(D) (M 𝑇)2 (D) The eigenvectors of [𝐴]−1 are same as the
eigenvectors of [𝐴]
1 2 3 
1.40 For the matrix  A  3 2 1  , which of
 
3 1 2 
the following statements is/are TRUE?
1.1 16 3 2 1  1 
Total number of multiplication performed in K JK  1 2 113  2 4 2   2 
T
calculating Pmn  Qnl is  m  n  l  . 1 2 6  33  1 31
If we multiply PQ first : 1

P42  Q24 will have multiplications K JK   6 8 113  2 
T
 1 31
 4  2  4  32
( PQ)44  R41 will have multiplications K T JK   6  1  8  2  1  111  23
 4  4 1  16 Hence, the product of K T JK is 23.
Therefore, total number of multiplications
 16  32  48
If we multiply QR first :
Q24  R41 will have multiplications 1.3 (A)
 2  4 1  8
 215 650 795
P42  (QR)21 will have multiplications
Given : [ M ]   655 150 835 
 4  2 1  8
 485 355 550 
Therefore, total number of multiplications
 8  8  16 For any square matrix,
Therefore, minimum number of multiplications Sum of the Eigen values is equal to trace of the
performed = Min (48, 16) = 16. matrix (Sum of leading diagonal elements).
Hence, the minimum number of multiplications Therefore, sum of Eigen values
required to compute the matrix PQR is 16.  215  150  550  915
1.2 23
1.4 88
3 2 1  1
0 3
Given : J   2 4 2  and K   2 
  1 2
1 0 3 0 
1 2 6    1 Given : A  
2 3 0 1
 
3 0 1 2
Expanding about R1 , For a 2  2 matrix, determinant will be positive if
0 1 2 3 (i) Both Eigen values are positive.
(ii) Both Eigen values are negative.
1 0 3 0
A Therefore, the given statement is not true.
2 3 0 1 Option (D) : If the product of the trace and
3 0 1 2 determinant of the matrix is positive, all its Eigen
0 3 0 1 2 3 values are positive.
Let determinant of a 2  2 matrix be  15 and trace
 0 3 0 1 1 3 0 1
be  2 . Then, product of determinant and trace is
0 1 2 0 1 2
positive.
1 2 3 1 2 3 1  2   2 …(i)
2 0 3 0 3 0 3 0 12  15 …(ii)
0 1 2 3 0 1 From equation (i) and (ii),
A  0  1[1  12  9]  12  3(24) 1  3,  2   5
A  4  12  72  88 Therefore, the given statement is not true.
Hence, the determinant of the matrix is 88.
1.6 2
1.5 (A)
Option (A) : If the trace of the matrix is positive 6 0 4 4 
  2 14 8 18 
and the determinant of the matrix is negative, at Given : A =  
least one of its Eigen values is negative.  14  14 0 10 
Let determinant and trace of a 2  2 matrix be
. Method 1 :
 30 and 1 then,
By elementary transformations
Determinant = Product of Eigen values
R2  3R2  R1 , R3  6R3  14R1
 1 2   30
6 0 4 4 
Trace = Sum of Eigen values  1   2  1 
A  0 42 28 58 
 30 
1      1 0  84  56 116 
 1 
R3  R3  2 R2 ,
1  30  1  0
2
1  6,  5 6 0 4 4 
A  0 42 28 58
and 2  1 1
0 0 0 0 
 2   5, 6
There exist two non-zero rows.
Since, one of the Eigen value is negative, given
statement is true. So, ( A)  2
Option (B) : If the trace of the matrix is positive, Hence, the rank of the matrix is 2.
all its Eigen values are positive. . Method 2 :
From above example, trace = 1, then 1   5 , In the given matrix A, there exists linear
2  6 . relationship is given below,
Therefore, given statement is not true. R2  2R1  R3
Option (C) : If the determinant of the matrix is So, there are only linearly independent rows R1
positive, all its Eigen values are positive.
and R3
Determinant = Product of Eigen values
( A)  2 Number of unknowns ( n)  2
Hence, the rank of the matrix is 2. The augmented matrix is given by,
2 3 : 5 
[ A : B]   
 3 p : 10
By elementary transformation,
1.7 (A) 3
R2  R2  R1
Given : ( x1 , y1 )  (1,0) ; ( x2 , y2 )  (2, 2) ; 2
2 3 : 5
( x3 , y3 )  (4,3) .  
[ A : B] 
0 p  9 : 5 
 2 2
Since, ( A : B)  2
If no solution exists then ( A)  ( A : B) .
9
It is possible only, if p  0
2
p  4.5
Hence, the value of p is 4.5.
1.9 (D)
x1 y1 1
1 3  2 2
Area of triangle  x2 y2 1
2 Given : A =  4  4 6 
x3 y3 1
 2  3 5 
Area of triangle
The characteristic equation of matrix A is given
1
  x1 ( y2  y3 )  y1 ( x2  x3 )  1( x2 y3  y2 x3 ) by,
2
Area of triangle A  I  0 [where,   Eigen values]
1 3 2
 1(2  3)  0(2  4)  1(2  3  2  4)
2
2 4 4   6 0
1 3
Area of triangle   1  0  2  . 2 3 5
2 2
(3  )[( 4  )(5  )  18]
Area cannot be negative. Therefore area of triangle
3  2[4(5  )  12]
is equal to .
2 2[12  2( 4  )]  0
Hence, the correct option is (A). (3  ) (20  4  5   2  18)
1.8 4.5 2(20  4  12)
Given : The system of equation are,  2(12  8  2)  0
2x  3y  5  3  4 2  5  2  0
3 x  py  10 1  1,  2  1, 3  2
It is in the form of non-homogenous equation and
Therefore,   1, 1, 2
is given by,
Hence, smallest Eigen value is 1 and largest Eigen
AX  B
value is 2.
2 3  5
where, A    , B  Hence, the correct option is (D).
3 p  10
1.10 (B) 1 3

2 1
Given : A matrix is defined as, A  [aij ]
1  32
where, 1  i, j  n and n  3 and aij  i. j
Put the value of 1 in equation (i),
For n  3  j  3 ,
3 2   2  2  p
Let order of matrix be 2  3 .
4 2  2  p
 1 1 1 2 1 3 
Then, [ A]23    p2
 2  1 2  2 2  3 2 
4
1 2 3 Put the value of 1 in equation (ii),
[ A]23   
2 4 6 3 22  2 p  1
In the above matrix, there exist a linear
 p2
2
relationship as given below, 3   2 p 1
R2  2 R1  4 
Thus, only row R1 is an independent row.  p2  4 p  4 
3   2 p 1
So, ( A)  1  16 
For n  4  j  4 , 3 p 2  12 p  12  32 p  16
Now, let order of matrix be 3  4 . 3 p2  20 p  28  0
1 2 3 4  14
p
Then, [ A]34   2 4 6 8 
,2
3
 3 6 9 12  34 Hence, the correct option is (D).
In the above matrix, there exist a linear . Method 2 :
relationship as given below, 2 1 
Given : A   
R2  2 R1 and R3  3R1 1 p 
Thus, only row R1 is an independent row. The characteristics equation of the matrix A is
So, ( A)  1 given by,
Therefore, for any order of given matrix, rank of A A  I  0 [where,   Eigen values]
is always equal to 1. 2 1
Hence, the correct option is (B). 0
1 p
1.11 (D)
(2  ) ( p  )  1  0
2 1 
Given : A    2  ( p  2)  (2 p  1)  0
1 p 
From option (D) :
. Method 1 :
14
Let 1 and  2 be the Eigen value of matrix A p
3
1 3
Therefore,  for p  2 14   14 
2 1 So,  2    2    2   1  0
3   3 
Trace  1   2  2  p …(i) 20 25
2    0
Determinant  1 2  A  2 p  1 3 3
…(ii)   5,
5
3
5 1.13 (B)
So, ratio of two Eigen values   3:1
5/3 Given : The system of equations are,
Hence, the correct option is (D). x  2 y  3z  a
1.12 (D) 2 x  3 y  3z  b
Given : Entries in each column of a square matrix 5x  9 y  6 z  c
M add up to 1. It is in the form of a non-homogenous equation
Let us consider 2  2 matrix and generalize the and is given by,
result for all other like 3  3 , 4  4, 5  5 and so on AX  B
for M order square matrix. 1 2 3 a 
1 0 where, A   2 3 3  , B   b 
 
Example : A  
0 1   5 9 6   c 
The characteristics equation of the matrix A is The augmented matrix is given by,
given by,
1 2 3 : a 
A  I  0 [where,   Eigen values]
 A : B    2 3 3 : b 
1 0    0   5 9 6 : c 
0 1    0    0
    By elementary transformation,
1  0 R2  R2  2R1 , R3  R3  5R1
0
0 1  1 2 3 : a 
(1  )  0  A : B   0 1 9 : b  2a 

2
  1, 1 0 1 9 : c  5a 
1 1 3 R3  R3  R2
Example :  2 1  2 

1 2 3 : a 
  2 1 0 
 A : B   0 1 9 : b  5a 

The characteristics equation is given by, 0 0 0 : c  b  3a 
A  I  0
( A : B)  2
1  1 3 For system to be consistent,
2 1    2  0 ( A : B)  ( A)
2 1 0 The above condition is only possible if
(c  b  3a)  0
(1  )  (1  )  2
3a  b  c  0
1 (2  4)  3[2  2(1  )]  0
(1  )    2  2  2  4  6  0 1.14 (C)
3  5  6  0 Given : P  Q 1 …(i)
  1,  0.5  2.39i Multiplying Q on both side of equation (i),
From both examples, it is clear that atleast one of QP  QQ1 [Since, QQ 1  I ]
the Eigen value is unity when the sum of column
elements is unity. QP  I …(ii)
Hence, the correct option is (D). Post multiplying Q on both side of equation (i),
PQ  Q 1Q [Since, Q 1Q  I ]
PQ  I …(iii) s  2 1  1
From equation (ii) and (iii), Hence, the given matrix consist only one
PQ  I and QP  I independent Eigen vector.
Hence, the correct option is (C). Hence, the correct option is (A).
1.15 (A) 1.16 (A)
5 1 Given : The system of equations are,
Given : A   
4 1  3x1  2 x2  c1
The characteristic equation of the matrix A is given
4x1  x2  c2
by,
A  I  0 [where,   Eigen values] It is in form of non-homogeneous equation and is
given by,
5   1
0 AX=B
4 1 
3 2 c 
 (5  )(1  )  (4  1)  0 where, [ A]    ,B   1
4 1 c2 
 2  6  9  0
The characteristic equation of the matrix A is given
(  3)2  0
by,
  3, 3
A  I  0 [where,   Eigen values]
If an Eigen value  of a square matrix of order n
is repeated p times then the number p is called 3 2
algebraic multiplicity of Eigen value. 0
4 1 
Here, Eigen values are 3, 3.
Therefore Eigen value 3 has a multiplicity of 2. (3  )(1  )  (4  2)  0
Calculation of independent Eigen vector :  2  4  5  0
Geometric multiplicity of Eigen value is the Hence, the correct option is (A).
number of linearly independent Eigen vectors
associated with it. 1.17 (A)
Let s be a number of linearly independent vectors. 1 5  3 7 
Then s = [Number of variables  ( A  I ) ] Given : A    , B 
6 2  8 4
Here, number of variables  2
T
For   3 , 3 7  3 8
[ B]  
T
  
5  3 1  8 4 7 4
 A  I    
 4 1  3 1 5   3 8 
Hence, ABT    
 2 1 6 2  7 4 
 A  I    
 4 2  3  35 8  20 38 28
ABT     
18  14 48  8 32 56 
By elementary transformation,
R2  R2  2R1
 2 1
 A  I     1.18 (C)
0 0 
It has one non-zero row. Given :
Hence, ( A  I )  1
Thus, number of independent Eigen vector,
3 2 6 Hence, the correct option is (C).
7 7 7 1.19 (C)
 
6
Q
3 2
is an orthogonal matrix. If A  0 , A is said to be singular matrix.
7 7 7
  From option (A) :
2 6 3 
 2 5
 7 7 7  a 
1 3
. Method 1 :
a  6 5 1 0
If Qnn is an orthogonal matrix then, QQT  I From option (B) :
The inverse of the matrix, Q 1  QT  3 2
b 
 3 6 2   2 3
7 7 7 b  94  5  0
 
1
Q Q  T  2 3 6 From option (C) :
7 7 7  2 4
  c 
 6 2 3  3 6
 7 7 7  c  12 12  0
Hence, the correct option is (C). From option (D) :
. Method 2 : 4 3
d  
Inverse of a matrix is given by, Q 1 
[Adj Q ] 6 2
Q d  8 18  10  0
 21 42 14  Hence matrix (c) is singular.
 49 49 49  Hence, the correct option is (C).
 
14 21 42  1.20 (D)
Adj Q  
 49 49 49 
   2  4
Given : P  
 42 14 21 
 4  2

 49 49 49 
The characteristics equation of matrix P is given
3  9 12  2  18 4  by,
Q       
7  49 49  7  49 49  P  I  0 [where,   Eigen values]
6  36 6 
    2 4
7  49 49  0
4 2
3  21  2  14  6   42 
Q          (2  ) ( 2  )  ( 4) (4)  0
7  49  7  49  7  49 
 4  2  2   2  16  0
 63  28  252  343
Q   1  2  12  0
49  7 343
3 6 2 1   2 3 i and  2   2 3 i
7 7 7 These are the two complex Eigen values of the
 
Q 
1 2 3 6 matrix.
7 7 7 For Eigen vector [ X ] of a matrix (P)
 
6 2 3  corresponding to Eigen value  , the following
 7 7 7  equation satisfies.
[ P  I ] [ X ]  0 . 3
R3  R3  R2 ,
For 1  2 3 i , 2
1 1 0  2 
2  2 3 i 4   x1  0
      P  0  2 2 6 
 4  2  2 3 i   x2  0 0 0 0 0 
2  2 3 ix1  4 x2  0 There exists two non-zero rows.
2  2 3 ix1  4 x2 ( P )  2
x1 x2 Hence, the correct option is (B).

4 22 3i 1.22 (B)
 x1   4  Given : r  13 m , h  6 m
x   
 2  2  2 3 i  1
Volume of water  h2 (3r  h)
For  2   2 3 i , 3
1
2  2 3 i 4   x1  0   62  (3 13  6)
      3
 4  2  2 3 i   x2  0  396
2  2 3 ix1  4 x2  0 Hence, the correct option is (B).
1.23 (D)
2  2 3 ix1  4 x2
x1 x2  2 3 4

4 2 2 3i Let A   4 3 1 
1 2 4 
 x1   4 
x     Det ( A)  2(12  2)  3(16  1)  4(8  3)
 2  2  2 3 i 
 2(10)  3(15)  4(5)
Therefore, the given matrix has complex Eigen
values and complex Eigen vectors. Det ( A)  5
Hence, the correct option is (D). Minor of 2  12  2  10
1.21 (B) Minor of 3  16  1  15
Minor of 4  8  3  5
1 1 0  2 
Minor of 4  12  8  4
Given : P   2 0 2 2 
Minor of 3  8  4  4
 4 1 3 1 
Minor of 1  4  3  1
By elementary transformation, Minor of 1  3 12  9
R2  R2  2R1 and Minor of 2  2 16  14
R3  R3  4R1 Minor of 4  6 12  6
1 1 0  2  10 15 5 
P  0  2 2 6  Cofactors of A   4 4 1
0  3 3 9   9 14 6 
adj A  (Cofactors of A)T
 10 4 9  0 1 3 0
 2 4 
adj A   15 4 14  3 0
[ P]  
 5 1 6  0 0 6 1
 
Inverse of the matrix is given by, 0 0 1 6
 10 4 9  0 1 3 0
adjA 1 
1
A    15 4 14  2 3 0 4
A 5 Det ([ P]) 
 5 1 6  0 0 6 1
0 0 1 6
 4 9 
 2 5 5  1 3 0
  6 1
1  4 14   (2) 0 6 1  2(1)  70
A  3 1 6
 5 5  0 1 6
 
 1 1 6 
Trace of [ P]  0  3  6  6  15
 5 5 
Consider option (B),
1  2  5  7  15  Trace of [ P ]
1.24 3 1 2  5  7  70  Det [ P ]
Given : System of equations, So, the eigen values of P are 1, 2, 5 and 7.
1 3 2 1  Hence, the correct option is (B).
2   x1   
 2 3   1  1.26 (C)
x2 
4 4 6     2 
   x3    ( PQ)T  QT PT
2 5 2    1 
1 2 0 1   2 1
Echelon form is given by, PQ     
3 4 1 0  4 3
1 3 2 : 1
 2 2 3 : 1   2 4
 A : B    QT PT  ( PQ)T   
4 4 6 : 2 1 3 
  Hence, the correct option is (C).
3 5 2 : 1
1 3 2 : 1 1.27 (C)
0 1 2 : 1
 R2  R2  3R1
0 0 1 : 0
  R3  R3  5R1
0 0 0 : 0
1 2 2 3 
1 3 2  1 0 2 4 4 
0 1 2   x1   1  
  x     0 4 8 8 
0 0 1   2   3   
   x3    0 6 12 12
 0 0 0  0
R3  R3  2R2
 x3  3
R4  R4  3R2
Hence the value of x3 is 3.
1.25 (B)
Given :
1 2 2 3  0.45 2   x  0
0 2 4 4   1 4.45  y   0
      
0 0 0 0  45 x  2 y  0
 
0 0 0 0  45 x  2 y
p ( A)  2
x 2
Hence, the correct option is (C). 
y 0.45
1.28 (B)
5 0 1 0 5 0 1 0 1.31 (B)
0 2 0 1  R1  R1  R2 0 2 0 1 
   . Method 1 :
 5 0 1 0 0 0 0 0
    Given : (1, 1, 1) right handed coordinate system.
0 1 0 2 0 1 0 2
  450 clockwise rotation of coordinate about x-
5 0 1 0
0 2 0 1 
axis.
1
R4  R4  R2  z
 2
0 0 0 0 (1,1,1)
 
0 3 45 0
0 0 y
 2 
5 0 1 0 x
0 2 0 0 
 By rotation of yz plane, new coordinates will be
R3 
 R4  3
0 0 0
2  y '  y cos   z sin   1
1
 1
1
0
 2 2
0 0 0 0 
1 1
Rank ( A)  3 z '   y sin   z cos   1  1  2
2 2
So, new coordinate (1, 0, 2).
1.29 (B) Hence, the correct option is (B).
. Method 2 :
If, AAT  I or A1  AT
In a cartesian plane a point p (1,1,1) after rotating
The matrix is orthogonal.
450 on positive direction of x-axis.
 x '  1 0 0   x
1.30 (D)  y '  0 cos   sin   y 
    
 2 2  z '  0 sin  cos    z 
A   A  I  0
 1 6  Here   450 , at point p (1,1,1)
2  tr  [ A]  0  x '  1 0 0  1
 y '  0 cos 450 0
 sin 45  1
 2  8  10  0    
 z '  0 sin 450 cos 450  1
  1.55
[ A] X  [ X ]
[ A  I ] X ]  0
 x '  1 0 0  1 1 2 
 y '  0 1/ 2 1/ 2  1
  X    AAT    
    2 4
 z '  0 1/ 2 1/ 2  1
 Finding eigen values,
 
Characteristics equation is given by,
 x '  1   1 
 y '  1/ 2  (1/ 2)    0  X  I  0
     
 z '   1/ 2  1/ 2   2  1 2  1 0 
   2 4    0 1   0
   
Then new coordinate is (1, 0, 2) about positive x-
1  2
axis. 0
Hence, the correct option is (B). 2 4
1.32 (A), (D) (1  ) (4  )  4  0
1 3  4    4   2  4  0
Given : M     2  5  0
4 2
  5,  2 (  5)  0
Q  M 3  4M 2  2M   0,5 Eigen value of  AAT 
By Cayley Hamilton’s theorem every square  yT y    yT   y 

matrix satisfies its own characteristics equation.
1 
i.e., M    yT y   1 221    5
 2  21
Then the eigen value of Q  Eigen value of
We can see that  yT y  is the eigen value of
(M 3  4M 2  2M ).
Putting eigen values,  yT y  .
For M    5, So, option (A) is correct.
E1 Q   53  4  52  2  5 Option (B) :
Rank of a matrix,
E1 Q   15
p ( y )  min (rows, columns)
E2 Q   ( 2)3  4  ( 2) 2  2  ( 2) Given matrix  y 20221
E2 Q    20 p( y )  min (2022,1)
Hence, the correct option are (A) and (D). p ( y )  1 also y  non-zero vector that mean
1.33 (A), (C)
p( y)  1
So, option (B) is not correct.
Given : y  non-zero vector of order (2022 1)
Option (C) :
i.e.,  y 20221 Let’s take an example,
Checking options : 1 
Option (A) :
 A   
 2  21
Let’s take an example,
 AT   1 212
1 
 A    1  1 2
 2  21  AAT     1 221   
 2  21  2 4  22
 AT   1 212
We can see that  AAT  becomes a symmetric From equation (i) and (ii),
matrix. 1  2,  2   2
So, option (C) is correct. Eigen values of matrix ()
Option (D) : E   2,  2
 y 20221 ,  y T

12022 E 8   ( 2)8 , ( 2)8
 y 20221  yT 12022   yyT  20222022 E 8   16,16
We know that,
E 11   ( 2)11 , ( 2)11
p  AB   min  p ( A), p ( B ) 
E 11   32 2,  32 2
p  yyT   min  p( y ), p( yT ) 
p  yyT   min 1, 1 P  Trace(8 )  16  16  32
Q  Trace (11 )  32 2  ( 32 2)  0
p  yyT   1
Numerical values,
It is clear that  yyT  is not invertible because its
P  Q  32  0  32
rank is 1.
1.36 (B), (C)
So, option (D) is not correct.
Hence, the correct option are (A) and (C).  1 1 0
1.34 (C), (D) Given : [A]   1 2 1
Given : P and Q are two square matrices.  0 1 1  33
We know that, Characteristic equation, | A  I |  0
( AB)1  B 1 A1  1   1 0 
Also P and Q are invertible  1 2    1   0
 
then,  PQ   Q 1 P 1
1
 0 1 1   
QP
1
 P 1Q 1 (1  )[ 2  3  2  1]  1(  1)  0
Hence, the correct option are (C) and (D). (1  )[ 2  3  2  1  1]  0
1.35 32 (1  )( 2  3)  0
1 1   2  3   3  3 2  0
Given : Matrix   
1 1  3  4 2  3  0
Trace ()  1  (1)  0 (as given) ()(2  4  3)  0
1 1    
Det ()   1 (1)  11   2
1 1 Since there are three Eigen values so number of
We know that, linearly independent Eigen vectors are 3.
Sum of eigen values, Here one Eigen value   0 ,so [A] is not a
1   2  Trace () positive definite matrix.
1   2  0 …(i) Also, [A]{x}  {b} , does not have a unique
Product of eigen values, solution.
1 2  Det () Hence, correct options are (B) and (C).
1  2   2 …(ii) 1.37 (A)
Given : Matrix is 4  4 M 2 X  X 
Rank of matrix = 2
M 2 X  2 X ...(ii)
N  A   Number of column – Rank of matrix = 4
[  (non negative) is Eigen value of M2]
2
–2=2
We can say that null space of A will consists only MX  X
two linearly independent vectors which is given as M T X  X .... (iii)
x and y. [M, MT have same Eigen values]
2 1 
1  0 MM T X  MX
Eigen vectors of Matrix A,   and   MM T X  X 
0 1 
   
3 2 MM T X   2 X ....(iv)
Therefore, remaining eigen vectors must be [  2 (none negative) is Eigen value of MMT]
linearly dependent. MX  X
1 M T X  X
1
Hence, X Y    M T MX  MX
 1
  M T MX  (X )
1
Hence, the correct option is (A). M T MX   2 X ...(v)
1.38 (B) [  2 (non negative) is Eigen value of M T M ]
Cholesky decomposition, A  LLT MX  X
 8 5   L11 0   L11 L21  MMx  MX
 5 a    L L22   0 L22 
 22   21 M 2 X  X )
On compare both side (M T )   2 X
We get, L11  2 2
[ M , M T have same Eigen value]
5
L21  ( M T )2 X   2 X …(vi)
2 2
L221  L222  a22 [  2 is Eigen value of (M T )2 which non negative]
2 Hence, the correct options are (B) & (C).
 5 
   1.968  a22 1.40 (A), (B), (D)
2 2
Characteristics equation of A and AT are same.
a22  6.99
 Characteristics roots of A and AT are same.
a22  7
AX  X
Hence, the correct option is (B). A1 AX  A1X
1.39 (B), (C) IX  A1X
MX  X ...(i) X  A1 X
[  is Eigen value of (M)] 1
MMx  MX
X  A1 X

1
 is Eigen value of A1 .

Hence, the correct options are (A), (B) &
(C).

2 Differential Equations
2013 IIT Bombay 2.5 Consider the following difference equation

y y
2.1 Laplace equation for water flow in soil is x( y dx  x dy) cos  y ( x dy  y dx)sin
x x
2 H 2 H 2 H
given by  2  2  0. Which of the following is the solution of the
x 2 y z above equation (c is an arbitrary constant)?
Head H does not vary in y and z- direction. [2 Marks]
Boundary condition x y x y
(A) cos  C (B) sin  C
dH y x y x
x  0, H  5 m ,  1 .
dx y y
What is value of H (in m) at x  1.2 ? (C) xy cos C (D) xy sin C
x x
2014 IIT Kharagpur
2016 IISc Bangalore
2.2 The integrating factor for the differential
2.6 The type of partial differential equation
dP
equation  k2 P  k1L0e k1t is 2 P 2 P 2 P P P
dt   3 2   0 is
x 2
y 2
xy x y
[1 Mark]
 k1t  k 2t [1 Mark]
(A) e (B) e
(A) Elliptic (B) Parabolic
(C) e k1t (D) e k2t
(C) Hyperbolic (D) None of these
2.3 Water is flowing at a steady rate through a
2.7 The respective expressions for
homogenous and saturated horizontal soil
complementary function and particular
strip of 10 m length. The strip is being
integral part of the solution of the differential
subjected to a constant water head (H) of 5
equation
m at the beginning and 1 m at the end. If the
d4y d2y
governing equations of flow in the soil strip
4
 3 2  108 x 2 are
d 2H dx dx
is  0 (where, x is the distance along [2 Marks]
dx 2
the soil strip), the value of H (in m) at the (A) [c1  c2 x  c3 sin 3 x  c4 cos 3 x] and
middle of the strip is _______. [3x  12 x  c]
4 2
2015 IIT Kanpur (B) [c2 x  c3 sin 3 x  c4 cos 3 x] and

2.4 Consider the following second order linear [5x 4  12 x 2  c]
differential equation
[2 Marks] (C) [c1  c3 sin 3 x  c4 cos 3x] and
2
d y [3x 4  12 x 2  c]
 12 x 2  24 x  20
dx 2 (D) [c1  c2 x  c3 sin 3 x  c4 cos 3 x] and
The boundary conditions are : [5x 4  12 x 2  c]
x  0, y  5 and at x  2, y  21 .
The value of y at x  1 is______.
2.8 The solution of the partial differential (A) (1, 0) (B) (0,1)
u  2u
equation   2 is of the form (C) (1,1) (D) ( 2, 2)
t x
2018 IIT Guwahati
[1 Mark]
2.13 Variation of water depth (y) in a gradually
(A) C cos(kt ) C1e( k / ) x
 C2e ( k / ) x 
 varied open channel flow is given by the first
(B) Cekt C1e( k / ) x
 C2 e  ( k / ) x 

order differential equation
(C) Cekt C1 cos( k /  ) x 

10
ln( y )
dy 1 e 3

C2 sin( k /  ) x  dx 250  45e 3ln( y )
(D) C sin(kt ) C1 cos( k /  ) x Given initial conditions : y ( x  0) = 0.8 m.
The depth (in m, up to three decimal places)
C2 sin( k /  ) x  of flow at a downstream section at x  1 m
2017 IIT Roorkee from one calculation step of Single Step
Euler Method is ______.
2.9 Consider the following partial differential 2.14 The solution (up to three decimal places) at
equation x 1 of the differential equation
 2  2  2 2
d y dy
3  B  3  4  0  2  y  0 subject to boundary
x 2 xy y 2 dx 2
dx
For this equation to be classified as dy
conditions y (0)  1 and (0)  1 is
parabolic, the value of B 2 must be _____. dx
[1 Mark] _____.
dQ [2 Marks]
2.10 The solution of the equation  Q 1
dt 2.15 The solution at x  1, t  1 of the partial
with Q  0 at t  0 is  2u  2u
differential equation,  25 subject
[2 Marks] x 2 dt 2
t
(A) Q(t )  e  1 (B) Q(t )  1  et u
to initial condition of u (0)  3 x, (0)  3
(C) Q(t )  1  et (D) Q(t )  1  et t
is,
2.11 Consider the following second-order
[2 Marks]
differential equation
(A) 1 (B) 2
y " 4 y ' 3 y  2t  3t 2 (C) 4 (D) 6
The particular solution of the differential dy
2.16 The solution of the equation x  y  0
equation is dx
[2 Marks] passing through the point (1, 1) is
(A) 2  2t  t 2
(B) 2t  t 2
[1 Mark]
(C) 2t  3t 2 (D) 2  2t  3t 2
2
(A) x (B) x
2.12 The tangent to the curve represented by (C) x 1
(D) x 2
y  x ln x is required to have 450 inclination 2019 IIT Madras
with the x-axis. The coordinates of the 2.17 Consider a two dimensional flow through
tangent point would be isotropic soil along x-direction and z-
[2 Marks] direction. If h is the hydraulic head, the
Laplace’s equation of continuity is Vi ( n 1)  Vi ( n ) V ( n )  2Vi ( n )  Vi ( n1) 
expressed as (B)    i 1 
t  (x) 2 
h h h h
(A)   0 Vi ( n )  Vi ( n 1) V ( n )  2Vi ( n )  Vi ( n1) 
x x z z (C)    i 1 
h h t  (x) 2 
(B)  0
x z Vi ( n )  Vi ( n1) V ( n )  2Vi ( n )  Vi ( n1) 
(D)    i 1 
2h 2h 2h 2t  2x 
(C) 2   0
x xz z 2 2020 IIT Delhi
2h 2h
(D) 2  2  0 2.22 In the following partial differential equation,
x z  is a function of t and z , and D and K are
2.18 Consider the ordinary differential equation
functions of 
d2y dy
x2  2 x  2 y  0 . Given the values  2 K () 
dx 2
dx D ()   0
z 2 z t
of y (1)  0 and y (2)  2 , the value of y (3)
The above equation is
(round off to 1 decimal place), is _____. [1 Mark]
[2 Marks] (A) a second order non-linear equation
2.19 An ordinary differential equation is given (B) a second degree linear equation
 dy  (C) a second degree non-linear equation
below   ( x ln x)  y . The solution for the
 dx  (D) a second order linear equation
above equation is 2.23 For the ordinary differential equation
[2 Marks] d 2x dx
 5  6x  0 ,
(Note : K denotes a constant in the options) dt 2
dt
(A) y  Kxe x (B) y  Kxe x with initial conditions x(0)  0 and
(C) y  K ln x (D) y  Kx ln x dx
(0)  10 , the solution is
2.20 Euclidean norm (length) of the vector dt
[4 2 6]T is [2 Marks]
[1 Mark] (A) 10e  10e
2t 3t
(A) 48 (B) 56 (B) 5e 2t  6e3t

(C) 24 (D) 12 (C) 10e 2t  10e3t
2.21 A one-dimensional domain is discretized (D) 5e 2t  6e3t
into N sub-domain of width x with node 2.24 The following partial differential equation is
numbers i  0,1, 2,3,......, N . If the time scale defined for u : u ( x, y )
is discretized in steps of t , the forward- u  2u
 ; y  0; x1  x  x2
time and centred-space finite difference y x 2
approximation at i th node and nth time step, The set of auxiliary conditions necessary to
for the partial differential equation solve the equation uniquely, is
V  2V [1 Mark]
  2 is (A) two initial conditions and one boundary
t x
[2 Marks] condition
(B) three initial conditions
V ( n 1)
 Vi (n)
V ( n)
 2Vi  Vi ( n1) 
(n)
(A) i 1
 i 1
 (C) three boundary conditions
t  2x 
(D) one initial condition and two boundary (A) y  xk ln(k ) (B) y  x ln(kx)
conditions (C) y  k ln(kx) (D) y  x ln( x )
2.25 The ordinary differential equation
2
d u 2022 IIT Kharagpur
2
 2 x 2u  sin x  0 is 3/2
dx d3y  dy 
2.29  x    x2 y  0
[1 Mark] dx 3
 dx 
(A) linear and homogeneous The correct description is
(B) linear and non-homogeneous [1 Mark]
(C) non-linear and non-homogeneous (A) An ordinary differential equation of
(D) non-linear and homogeneous order 3 and degree 3
2.26 An ordinary differential equation is given (B) An ordinary differential equation of
below order 3 and degree 3/2
d 2 y dy (C) An ordinary differential equation of
6  y0 order 3 and degree 2
dx 2 dx
(D) An ordinary differential equation of
The general solution of the above equation
order 2 and degree 3
(with constants C1 and C2 ), is
2.30 Consider the differential equation
[2 Marks] dy
x x  4( x  2)  y
 dx
(A) y ( x)  C1e  C2e
3 2
For the initial condition y  3 at x  1, the
x x

(B) y ( x)  C1 xe 3
 C2 e 2 value of y at x  1.4 obtained using Euler’s
x x
method with a step-size of 0.2 is ________.

(C) y ( x)  C1e 3
 C2 e 2 [2 Marks]
x x
2.31 Consider the following expression

(D) y ( x)  C1e 3
 C2 xe 2 z  sin ( y  it )  cos ( y  it )
2021 IIT Bombay Where z, y and t are variable, and i  1 is

a complex number. The partial differential
2.27 The solution of the second-order differential equation derived from the above expression
d2y dy is
equation 2
 2  y  0 with boundary
dx dx [1 Mark]
conditions y  0   1 and y 1  3 is  z  z
2 2
(A) 2  2  0
[2 Marks] t y
  x   2 z 2 z
(A) e x  3e sin    1 xe x (B)  0
  2   t 2 y 2
 z z
 x   (C) i  0
(B) e x  3e sin    1 xe x t y
  2  
z z
(C) e  x   3e  1 xe  x (D)  i  0
t y
(D) e  x   3e  1 xe  x 2.32 The function F ( x, y ) satisfies the Laplace
2.28 If k is a constant, the general solution of equation,
dy y 2 F ( x, y)  0
  1 will be in form of :
dx x on a circular domain of radius r  1 with its
[2 Marks] center at point P with coordinates
x  0, y  0. The value of this function on dy
2.34 In the differential equation  xy  0 ,
the circular boundary of this domain is equal dx
to 3. The numerical value of f (0, 0) is  is a positive constant. If 𝑦 = 1.0 at
[1 Mark] 𝑥 = 0.0, and 𝑦 = 0.8 at 𝑥 = 1.0, the value of
(A) 1 (B) 0  is________ (rounded off to three
(C) 2 (D) 3 decimal places).
2023 IIT Kanpur du
2.35 The differential equation,  2tu 2  1
2.33 The solution of the differential equation dt
is solved by employing a backward
d3y d2y dy
3
 5.5 2
 9.5  5 y  0 difference scheme within the finite
dx dx dx
difference framework. The value of 𝑢 at
is expressed as y  C1e2.5 x  C2ex  C3ex , the (𝑛 − 1)th time-step, for some 𝑛, is 1.75.
where C1 , C2 , C3 ,  and  are constants, with The corresponding time (t) is 3.14 s. Each
time step is 0.01 s long. Then, the value of
 and  being distinct and not equal to 2.5.
Which of the following options is correct for
 un  un1  is____________.(round off to
the values of 𝛼 and 𝛽? three decimal places).
(A) 1 and 2 (B) −1 and −2
(C) 2 and 3 (D) −2 and −3
2.1 3.8 At x  1.2 m ,

H  5  1.2  3.8 m
2 H 2 H 2 H
Given :  2  2 0 Hence, the value of H is 3.8.
x 2 y z
Since, head H does not vary in y and z-direction. 2.2 (D)
2 H dP
Therefore, 0 Given :  k2 P  k1L0e k1t
x 2 dt
Integrating both sides, Above differential equation is in the form of
H Leibniz linear equation.
 C1
x dP
 KP  Q
H  C1 x  C2 …(i) dt
H '  C1 where, K  k2 , Q  k1 L0e  k1t
Using boundary conditions : The integrating factor is given by,

(i) When x  0, H  5 I.F.  e  e 2  ek2t
K dt k dt
5  C1  0  C2  C2  5 Hence, the correct option is (D).

2.3 3
dH
(ii) When x  0,  1
dx d 2H
Given : 0
1  C1  C1  1 dx 2
Integrating on both sides,
Put the values of C1 and C2 in equation (i),
dH
H  x  5  C1 …(i)
dx
H  C1 x  C2 …(ii) y  18
Using boundary conditions : Hence, the value of y is 18.
(i) When x  0, H  5
5  C1  0  C2
C2  5
2.5 (C)
(ii) When x  10, H  1
Given :
1  C1 10  5
 y  y
2 x( y dx  x dy ) cos    y ( xdy  ydx) sin  
C1  x x
5
Put the values of C1 and C2 in equation (ii),  y
sin  
x( y dx  x dy ) x
2 
H   x5 y ( xdy  ydx)  y
5 cos  
x
At the middle of strip i.e. x  5
y dx  x dy y  y
2
H    5  5  3m  tan  
xdy  ydx x x
5
Hence, the value of H is 3. Let y  vx
2.4 18 Differentiating both sides with respect to x,
dy xdv
d2y v  dy  vdx  xdv
Given : 2
 12 x 2  24 x  20 dx dx
dx
vxdx  x(vdx  xdv) vx  vx 
Integrating both sides with respect to x,  tan  
x(vdx  xdv)  vxdx x  x
dy
  4 x3  12 x2  20 x  c1
dx 2vxdx  x 2 dv
 v tan v
y   x 4  4 x3  10 x 2  c1 x  c2 x 2 dv
…(i) 2v dx
 1  v tan v
Using boundary conditions : x dv
(i) When x  0, y  5 2v dx
 v tan v  1
x dv
5   0  4  0 10  0  c1  0  c2
2dx (v tan v  1)
c2  5  dv
x v
(ii) When x  2, y  21 2dx  1
  tan v   dv
21   24  4  23  10  22  c1  2  c2 x  v
21  16  32  40  2c1  5 Integrating both sides,
2dx  1
2c1  21  16  32  40  5  x    tan v  v  dv
2c1  40 2 ln x  ln(sec v)  ln v  ln C
c1  20  sec v 
ln x 2  ln  C 
Put the values of c1 and c2 in equation (i),  v 
y   x 4  4 x3  10 x 2  20 x  5 x2  C
sec v
At x  1 , v
y  1  4  10  20  5
y C.F.  (c1  c2 x)e0 x
Put, v  on above equation,
x  e 0 x [c3 sin 3 x  c4 cos 3 x]
 y
sec   C.F.  c1  c2 x  c3 sin 3 x  c4 cos 3 x
x2  C x
y The particular integral is given by,
x 1
P.I.  108 x 2
C  y f ( D)
x  sec  
y x P.I.  4
1
108 x 2
 y C ( D  3D 2 )
xy  C sec   
 x  cos  y  P.I. 
1
108 x 2
   D 
4
x 3D 2 1  2
 y  3D 
xy cos    C
x 1
P.I.  108 x 2
Hence, the correct option is (C).  D  2
3D 2 1 
2.6 (C)  3 
1
2 P 2 P 2 P P P 1  D2 
Given :   3 2  0 P.I.  1  3  108 x
2
x 2
y 2
xy x y 3D 2  
Comparing the above equation with general form  By binomial expansion: 
of a linear second order partial differential  
equation, (1  x)  n  1  nx  n(n  1) x 2  ....
 2! 
 2u  2u  2u u u
a 2 b c 2 d  e  fu  g 1  D2 
x xy y x y P.I.  1  3  ....  108 x
2
where, a  1, b  3, c  1 3D 2  
b 2  4ac  9  4  5  0 1  D2 
P.I.  108 x 2
 [108 x 2 ]
Hence, Partial differential equation is hyperbolic. 3D 2  3 
Hence, the correct option is (C). 1  108 
P.I.  108 x  3  2 
2
2.7 (A) 3D 2
d4y d2y 1
Given : 4
 3 2
 108 x 2 P.I.  108x2  72
2 
dx dx 3D
This is in the form of a non-homogenous linear 1
P.I.  9 x4  36 x2   c
differential equation. 3
[ f ( D)] y   P.I.  3 x 4  12 x 2  c
The auxiliary equation is given by, The complete solution is given by,
f ( m)  0 y  C.F. + P.I.
m 4  3m 2  0 y  c1  c2 x  c3 sin 3 x
m1  0, m2  0, m3   3 i, m4  3 i  c4 cos 3 x  3 x 4  12 x 2  c
The complementary function is given by, Hence, the correct option is (A).
C.F.  (c1  c2 x)e m1x 2.8 (B)
e [c3 sin bx  c4 cos bx]
ax
u  2u
Given :  2
t x
The above given equation is heat equation.
Therefore, solution to the above equation is, 2.10 (D)
u ( x, t )  ( A cos px  B sin px)  Ce  p t
2
dQ
Given :  Q 1
K K dt
Put p   i ,
  ( D  1)Q  1
Now, put the value of p in given equation, This is in the form of a non-homogenous linear
 K K  Kt differential equation.
u ( x, t )   A cos ix  B sin ix  Ce [ f ( D )] Q  
   
The auxiliary equation is given by,
 K K  Kt
u ( x, t )   A cosh x  iB sinh x  Ce f ( m)  0
   
m 1  0
  K
x 
K
x  m  1
Kt   e e 
 
u ( x, t )  Ce  A   The complementary function is given by,
  2 
C.F.  C1emt  C1e  t
 K
x 
K
x 
e 
e 
 Particular integral is given by,
iB  
 2  1
P.I.   (1)
 f ( D)
 A  iB 
K
x
u ( x, t )  Ce e
Kt 
  1
  2  P.I.   1(1  D)1
D 1
 A  iB  
K
 x
e 
  P.I.  1(1  D  D2  D3.......)
 2 
 By binomial expansion: 
 K

K   
(1  x)  n  1  nx  n(n  1) x 2  ....
x x
u( x, t )  Ce C1e   C2e
Kt 

 2! 
P.I.  1  0  0.......  1
2.9 36
The complete solution is given by,
 2  2  2 Q (t )  C.F.  P.I.
Given : 3  B  3  4  0
x 2 xy y 2 Q(t )  C1et  1 …(i)
Comparing the above equation with general form
Using boundary condition :
of a linear second order partial differential
equation, When t  0 , Q  0
 2u  2u  2u u u 0  C1e  0  1
a 2 b c 2 d  e  fu  g
x xy y x y 0  C1  1
where, a  3 , b  B, c  3 C1  1
For the equation to be parabolic, Put the value of C1 in equation (i),
b 2  4ac  0
Q(t )  1  et
B  4  3 3  0
2

B 2  36
2.11 (A)
Hence, the value of B 2 is 36.
Given : y " 4 y ' 3 y  2t  3t 2 1
P.I.  3t 2  6t  6
( D2  4D  3) y  2t  3t 2 3
P.I.  t 2  2t  2
The particular integral is given by,
1 P.I.  (2  2t  t 2 )
P.I.  (2t  3t 2 )
f ( D) Hence, the correct option is (A).
1 2.12 (A)
P.I.  (2t  3t 2 )
D  4D  3
2
Given : Tangent to the curve is,
P.I. 
1
(2t  3t 2 ) y  x ln x
 D  4D 
2
Co-ordinates of the tangent point is given by,
3 1  
 3  tan  
dy
 ln x  1
1 dx
1  D2  4D 
P.I.  1   (2t  3t 2 ) tan 450  ln x  1
3 3 
1  ln x  1
 By binomial expansion:  ln x  0
 
 (1  x)  n  1  nx  n(n  1) x 2   x 1
 2!  Putting x  1 in the equation of curve, we get
 n(n  1)(n  2) 3  y  0.
  x  ....
 3!  Hence, the correct option is (A).
1   D2  4D   D2  4D  2.13 0.793
2
P.I.  1    
3   3   3  Given : Differential equation,
10


ln( y )
1 e 3
3
 D2  4D  dy
   ..... (2t  3t 2 ) 
 3   dx 250  45e 3ln( y )
According to Euler method,
1 1
P.I.  (2t  3t 2 )  {D 2 (2t  3t 2 ) y1  y0  h f ( x0 , y0 )
3 3
At x0  0, y0  0.8m
1
4D(2t  3t 2 )}  {D4 (2t  3t 2 ) 10
ln(0.8)
9 dy 1 e 3

 16 D 2 (2t  3t 2 )  8D3 (2t  3t 2 )}..... dx 250  45e 3ln(0.8)
1 dy
1
P.I.  (2t  3t 2 )  {6  4(2  6t )}  6.8 10 3  0.0068  f ( x0 , y0 )
3 3 dx
y1  y0  f ( x0 , y0 ) 1 y1  0.8  ( 0.0068)
1 
 {0  16(6)  0}  0.7932 m
9 
1 1 Hence, the depth of the flow is 0.793.
P.I.   2t  3t 2  ( 6  8  24t )
3 3 2.14 0.368
1  d2y dy
 (96)  Given : 2  y 0
9  dx 2
dx
1 14 96  ( D  2D  1) y  0
2
P.I.   2t  3t 2  8t   
3 3 9 This is in the form of a homogenous linear
differential equation.
 f ( D) y  0 Satisfying the conditions
u( x,0)
The auxiliary equation is given by, u ( x, 0)  f ( x) and  g ( x)
f ( m)  0 t
where, f ( x) is the initial displacement
m 2  2m  1  0
and g ( x ) is the initial velocity.
m  1, 1
The general solution is given by,
The complementary function is given by,
1
So, C.F. = (C1  C2 x)e  x u( x, t )  [ f ( x  ct )  f ( x  ct )]
2
Particular integral (P.I) is 0 because it is 1 x ct
homogenous equation.   g (s) ds
2c x ct
The complete solution is given by,
 2u  2u
y  (C1  C2 x)e  x …(i)  25 [Given]
x 2 t 2
Differentiating above equation with respect to x, Initial conditions:
dy u (0)  3 x  f ( x)
 (C1  C2 x)e x  C2e x
dx u
(0)  3  g ( x)
Using boundary conditions : t
(i) When x  0, y  1 Comparing the given equation with standard
From equation (i), equation,
1  (C1  C2  0)e0  2u 1  2u 1
 i.e. c 
1  C1 t 2
25 x 2
5
Therefore, by D-alembert’s solution is,
dy
(ii) When  1, x  0 1  1   1 
dx u( x, t )   f  x  t   f  x  t 
2  5   5 
1   (1  C2  0)e 0  C2e 0 1
x t
5
1 11  C2 1
1 1
 g ( s) ds
C2  0 2 x t
5 5
Put values of C1 and C2 in equation (i), At x  1, t  1
y  (1  0  x) e x 1   1   1 
u (1, 1)  3 1    3 1  
y  e x 2   5   5  
 1
At x  1 , 1 
 5
5
ye  
1
1 1
 0.368

2 
1
3 ds
e 2.718 1 
 5
Hence, the solution of the differential equation is 1 4 6 5
0.368. u (1, 1)   3   3     3[5]6/5
2 5 5 2
4/5
2.15 (D) 1 12 18  15  6 4 

u (1, 1)        
 2u  2u 2  5 5  2 5 5
Given :  25
x 2 t 2 1  30  15 2
u (1, 1)     
D-alembert’s formula for the equation, 2 5  2 5
 2u 2 u
2 u (1, 1)  3  3  6
 c ,   x  , t  0
t 2 x 2 Hence, the correct option is (D).
and c  0
2.16 (C)  D [ D  1] y  2 D y  2  0
dy [ D2  3D  2] y  0 …(ii)
Given : x  y  0
dx Equation is homogenous differential equation,
dy To solve auxiliary equation,
x  y
dx D 2  3D  2  0
dy dx  D  1, 2

y x Hence solution :
Integrating on both sides, y  C1et  C2e 2t
dy dx y  C1 x  C2 x 2 …(iii)
 y   x Or
Now substituting the given conditions y (1)  0 ,
ln ( y )   ln ( x)  ln C y (2)  2 in equation (iii),
ln y  ln( x1 )  ln C  y(1)  0  C1  C2  0 …(iv)
1
y  x C …(i) y(2)  2  2C1  4C2  2 …(v)
At x  1 and y  1 , Solving equations (iv) and (v),
1
1 1 C C1  1
C 1 C2  1
Put the value of C in equation (i),
Putting the value of C1 and C2 in equation (iii),
y  x 1 1
1
y  x2  x
Therefore, y  x
 y(3)  32  3  9  3  6
Hence, the value of y (3) is 6.
2.17 (D)
2.19 (C)
For function h  f ( x, y ) the Laplace equation of
continuity is  dy 
Given : x ln x    y
2h 2h  dx 
 0 Rearranging the above equation,
x 2 y 2
dy dx
Hence, the correct option is (D). 
y x ln x
2.18 6
Integrating both sides,
Given : Cauchy’s differential equation : dy dx
d2y dy  y   x ln x
x2 2
 2x  2 y  0
dx dx dx
Conditions : y (1)  0 ln y   …(i)
x ln x
y (2)  2
Let ln x  t
y (3)  ? dx
 dt
d2y dy x
x 2
2
 2x  2 y  0 …(i)
dx dx Substituting the values in equation (i),
dt
Putting x  et , we get x
dy
 Dy ln y  
dx t
2  ln y  ln t  ln K
2 d y d
x 2
 D( D  1) y , where D  ln y  ln Kt
dx dt
y  Kt …(ii) 2.22 (A)
Putting t  ln x in equation (ii),
 2 K () 
y  K ln x Given : D ()   0
z 2 z t
In first term of given PDE,
2.20 (B)
D  f ()
Given : Equation contains product of dependent variable
x  [4 2 6] T with it’s derivative, so it is non-linear equation.
4 The given PDE is 2nd order derivative, so it is a
x   2  second order non-linear equation.
 6  Hence, the correct option is (A).
Euclidean norm length 2.23 (C)
 16  4  36  56 d 2x dx
Given :  5  6x  0 ,
Hence, the correct option is (B). dt 2
dt
dx
where x(0)  0 and (0)  10
dt
Auxilary equation is m 2  5m  6  0
2.21 (B) m  2, 3
Given : So, complimentary function is given by,
V  2V C.F.  c1e 2t  c2e3t
 2 …(i)
t x
and PI = 0
Approximate time derivative in equation (i) with
Solution is x  CF  PI
forward difference,
V Vi n1  Vi n  C1e2t  C2e3t
  0(t ) …(ii)
t t And
dx
 2C1e2t  3C2e3t
Note that the right hand side only in value V at dt
x  x1 Here, C1  10 , C2  10
Use the central difference approximation to x  10e 2t  10e3t
  2V  Hence, the correct option is (C).
 2  and evaluate all the terms at time n .
 x  xi
2.24 (D)
 2V V n  2Vi n  Vi n1
 i 1  0( x 2 ) u  2u
x 2 xi  x2 Given : PDE,  , y  0, x1  x  x2
y x 2
…(iii)
Putting the values from equation (ii) and (iii) in As y  0 , PDE is parabolic which requires two
equation (i), we get boundary conditions (corresponding to x ) and one
Vi ( n 1)  Vi ( n ) Vi ( n1)  2Vi ( n )  Vi ( n1)  initial condition (corresponding to y ) to solve.
  
t  (x) 2  Hence, the correct option is (D).
0(t )  0(x 2 ) 2.25 (B)
Vi ( n 1)  Vi ( n ) Vi ( n1)  2Vi ( n )  Vi ( n1)  Given : Differential equation

   
t  (x) 2  d 2u
2
 2 x 2u  sin x  0
Hence, the correct option is (B). dx
d 2u  P
1
and Q  1
 2 x 2u   sin x
dx 2 x
1
Integration factor (I.F.)  e   e x
The above equation is in the form of linear non- Pdx dx 1

homogenous differential equation of second order. x
d 2u General solution is given by,
 P( x)u  Q( x) y ( IF )   Q( IF )dx  c
dx 2
Hence, the correct option is (B). 1 1
y     1. dx  ln k
2.26 (A) x x
y  x ln (kx)
d 2 y dy
Given : 6  y0 Hence, the correct option is (B).
dx 2 dx
2.29 (C)
Auxiliary equation is given by, 3/2
6m 2  m  1  0
d3y  dy 
 x   x2 y  0
 dx 
3
dx
1 1
m , m 3/2
d3y  dy 
3 2  x2 y   x  
 dx 
3
So, general solution is given by, dx
2
 d3y 2 
3
1 1
 dy 
 3  x y   x  
x x
y  C1e  C2 e
3 2
 dx   dx 
Order 3 degree 2.
2.27 (C) Hence, the correct option is (C).
d 2 y 2dy 2.30 6.4
Given :  y0
dx 2 dx dy
y (0)  1 Given :  4( x  2)  y  f ( x, y)
dx
y (1)  3 With initial condition y  3 at x  1, i.e., y (1)  3
( D2  2D  1) y  0 The value of y at x  1.4, i.e., y (1.4)  ?
Auxiliary equation, m  2m  1  0
2
By Euler’s method with step size h  0.2.
(m  1)2  0 yn1  yn  hf ( xn , yn )
m  1
y (1.2)  y (1)  hf 1, y (1) 
So, CF  PI  (C1  C2 x)e x
y (1.2)  3  0.2  4 (1  2)  3
y (0)  1  C1  1
y (1)  3   C2  3e  1 y (1.2)  3  0.2(12  3)
y (1.2)  3  0.2  9
So, y  [1  (3e  1) x]e x
Hence, the correct option is (C). y (1.2)  3  1.8  4.8
2.28 (B) y (1.4)  y (1.2)  hf 1.2, y (1.2) 
dy y y (1.4)  4.8  0.2  4 (1.2  2)  4.8 
Given :  1
dx x y (1.4)  4.8  0.2(4  3.2  4.8)
We know that,
y (1.4)  4.8  0.2(12.8  4.8)
dy
 Py  0 y (1.4)  4.8  0.2  8
dx
y (1.4)  4.8  1.6  6.4
2.31 (A) From equation (i),
C2  C4  3
Given : z  sin ( y  it )  cos ( y  it ), z, y and t are
So, f (0, 0)  3
variables.
i  1
2.33 (A)
z
 i cos ( y  it )   sin ( y  it )  ( i) d3y d2y
t Given :  5.5
dy
 9.5  5 y  0
3 2
2 z dx dx dx
  i sin ( y  it )  i  i cos ( y  it )  ( i) The differential equation can be written as
t 2
2 z D 3  5.5 D 2  9.5 D  5  0
 sin ( y  it )  cos ( y  it ) …(i) By solving above equation we get,
t 2
z D = 2, 1, 2.5
 cos ( y  it )  sin ( y  it )
y Hence the solution is given as
2 z y  c1e2.5 x  c2e1x  c3e 2 x
  sin ( y  it )  cos ( y  it ) …(ii)
y 2 So,  and  are 1 and 2 respectively.
2 z 2 z Hence, the correct option is (A).
 0
t 2 y 2 2.34 0.446
Hence, the correct option is (A). dy
2.32 (D)  xy y (0)=1
dx
Given : The function f ( x, y ) satisfies the Laplace dy
 xdx y(1)= 0.8
equation, y
2 f ( x, y)  0 dy
f ( x, y)  (C1 x  C2 ) (C3 y  C4 ) …(i)  y  xdx
As given circular domain of radius r  1 with its x2
center at point P with coordinates x  0, y  0. ln y   c …(i)
2
y
y (0)  1 ln1  0  c c0
(0, 1)
x2
ln y   …(ii)
2
P x 
(–1, 0) (0, 0) (1, 0) y(1)  0.8 ln (0.8) 
2
  2 ln (0.8)
(0, –1)
  ln (0.8)2
From figure,    ln (0.8)2  0.446

f (0,0)  C2  C4
f (0,1)  (C2 ) (C3  C4 )  3
2.35 – 0.151
f (1,0)  (C1  C2 ) (C4 )  3
Given : h = 0.01, un1  1.75 , tn1  3.14
f (1,0)  ( C1  C2 ) (C4 )  3
 f  t , u   1  2t u 2 
du
f (0, 1)  (C2 ) ( C3  C4 )  3
dt
From all above equations, we can conclude
From Euler’s backward method,
un  u n 1  hf  tn , un 
un  u n1  h 1  2tn un2 
un  1.75  0.011  2  3.15 un2 

un  1.75  0.01  0.63un2
0.63un2  un  1.76  0
By solving, un  1.599 , un1  1.75
So, un  un1  1.599  1.75  0.151
Hence, the correct answer is – 0.151.

3 Integral & Differential Equations
 1  sin x
2013 IIT Bombay

3.5 The value of 
0 1 x 2
dx  
0 x
dx is
3.1 The value of  6

cos4 3 sin 3 6 d  [2 Marks]

0
[2 Marks] (A) (B) 

2
1
(A) 0 (B) 3
15 (C) (D) 1
2
8 3.6 The area of the region bounded by the
(C) 1 (D)
3 parabola y  x 2  1 and the straight line
2015 IIT Kanpur x  y  3 is [2 Marks]
3.2 Given i  1, the value of the definite (A)
59
(B)
9
 6 2
2
cos x  i sin x
integral, I   dx is [1 Mark] (C)
10
(D)
7
0
cos x  i sin x 3 6
(A) 1 (B) – 1 3.7 The angle of intersection of the curves
(C) i (D) – i x 2  4 y and y 2  4 x at point (0, 0) is
3.3 Choose the most appropriate equation for the [2 Marks]
function drawn as a thick line, in the plot (A) 0 0 (B) 30
0
below.
(C) 450 (D) 900
3.8 The area between the parabola x 2  8 y and
the straight line y  8 is ______.
[2 Marks]
3.9 If f ( x)  2 x  3x  5 , which of the
7
following is a factor of f ( x) ?
(A) x  y  y (B) x  ( y  y ) (A) ( x3  8) (B) ( x  1)
(C) x  y  y (D) x  ( y  y ) (C) (2 x  5) (D) ( x  1)
2016 IISc Bangalore 2017 IIT Roorkee
3.4 In a process, the number of cycles to failure 3.10 Let x be a continuous variable defined over
x
decreases exponentially with an increase in the interval ( , ) and f ( x)  e  x  e .
load. At a load of 80 units, it takes 100 cycles
for failure. When the load is halved, it takes The integral g ( x)   f ( x)dx is equal to
10000 cycles for failure. The load for which [2 Marks]
the failure will happen in 5000 cycles is x x
(A) ee (B) e  e
(A) 40.00 (B) 46.02
(C) ee (D) e  x
x
(C) 60.01 (D) 92.02
3.11 Consider the following definite integral 2021 IIT Bombay
(sin 1 x) 2
1
I  dx 3.15 The volume determined from
1  x2 forV   2,3  1, 2   0,1
0
The value of the integral is,

 8xyz dV
v
will
[2 Marks] be (in integer) _______.

 3
 3 [1 Mark]
(A) (B) 3.16 The value (round off to one decimal places)
24 12
1
3 3
 xe
x
(C) (D) of dx is
48 64 1
[1 Mark]
2018 IIT Guwahati

2022 IIT Kharagpur
3.12 The value of the integral 
0
x cos 2 x.dx is
 x 2 x3 x 4 
[2 Marks] 3.17   2  3  4  ......  dx is equal to
x 
 2
 2
1
(A) (B) (A)   constant
8 4 1 x
2 1
(C) (D)  2 (B)   constant
2 1  x2
2020 IIT Delhi 1
(C)  constant
3.13 The area of an ellipse represented by an 1  x2
1
x2 y 2 (D)  constant
equation 2  2  1 is 1 x
a b
[1 Mark]
[1 Mark]
ab 2023 IIT Kanpur
(A) ab (B)
4 3.18 For the integral
4ab ab 1 1
(C) (D) I   2 dx
3 2 1 x
3.14 If C represents a line segment between which of the following statements is TRUE?
(0,0,0) and (1,1,1) in Cartesian coordinate (A) 𝐼 = 0
system, the value (expressed as integer) of
(B) 𝐼 = 2
the line integral
(C) 𝐼 = −2
 C ( y  z)dx  ( x  z)dy  ( x  y)dz  is (D) The integral does not converge
______.
[2 Marks]

3.1 (B) 6
I   cos4 3  2sin 3 cos3  d 
3

6 0
Given : I   cos 4 3 sin 3 6  d 
0
sin 2 A  2 sin A cos A

6 3.3 (B)
I   8cos 3 sin 3 d 
7 3
Given :
0
Let 3  t , 3d   dt
When   0, t  0
 
When   ,t
6 2

2
dt From figure,
I  8 cos7 t sin 3 t
0
3 x  0, y  0
 x  2, y   1
2
8
3 0
I sin 3 t cos7 t dt From option (A) :
x y y
Here, m  3, n  7 Put y   1 ,
8  3  1   7  1 7  3 7  5  x   1   1   2  (  2,  1)
I  
3   3  7  3  7  2  3  7  4 3  7  6 3  7  8  
From option (B) :
8 2 6 4 2 1 x  (y  y )
I  
3 10  8  6  4  2 15 Put y   1 ,
x   ( 1   1)  2  (2,  1)
 Key Point
From option (C) :
By the property of integrals,
 x y y
Put y   1 ,
2
 sin
m
x cos n xdx
0 x  1  1  0  (0,  1)
(m  1)(m  3).....(n  1) (n  3) (n  5)
 From option (D) :
(m  n) (m  n  2)...... x   y  y 
3.2 (C) Put y   1 ,
 /2
cos x  i sin x x    1  1    (0)  0  (0, 1)
Given : I  
0
cos x  i sin x
dx
Only option (B) is valid option for given figure.
 /2  /2 Hence, the correct option is (B).
eix
I  dx  e
2 ix
dx 3.4 (B)
0
e ix 0
Since, the number of cycles to failure decreases
1  /2 1
I  e2ix   ei  1 exponentially with an increase in load. The general
2i 0 2i equation is given by,
1
I  cos   i sin   1 y  ne mx
2i
where, y  number of cycle failure and x is load
1 2
I   1  0  1  i Given : y  100 and x  80
2i 2i
Hence, the correct option is (C). Therefore, 100  n e80 m …(i)
When load is halved, it take 10000 cycle for
failure.
10000  ne 40 m …(ii) Put s  0,

From equation (i) and (ii) sin x
e
0x
dx  cot 1 (0)
100  e 40 m 0
x
ln100 

m …(iii) I2  
sin x
dx 
40 x 2
0
For 5000 cycles to failure,
 
5000  ne xm …(iv) Hence, I  I1  I 2   
2 2
From equation (ii) and (iv),
2  e m ( x  40)
3.6 (B)
m( x  40)  ln 2
From equation (iii), Given : y  x 2  1 …(i)
ln 2 x y 3 …(ii)
x  40 
m From equation (ii),
ln 2 y  3 x
x  40   40
ln100 Put the above value in equation (i),
0.693 3  x  x2  1
x  40   40
4.605 x2  x  2  0
x  46.02 ( x  2)( x  1)  0
Hence, the correct option is (B). x  2, 1
3.5 (B) When x   2, y  5
Given : I  
 1  sin x When x  1, y  2
0 1  x2
dx  0 x dx Thus, the common points are ( 2, 5) and (1, 2) .
Let I  I1  I 2 ,
 1
where, I1   dx  [tan 1 x]0
0 1  x2
I1  tan 1   tan 1 0
 
I1  0 
2 2
 sin x
I2   dx
0 x
1
L[sin x]  2
s 1 1 3 x
 Area    dydx    dydx

 sin x  1
L   2 dx  [tan 1 s]s x 2 y  x 2 1
 x  s s 1 1 1
 sin x 
L  tan 1   tan 1 s
Area   [ y ]3yxx2 1dx   (2  x  x 2 )dx
 x 

x 2 x 2
1
 x 2 x3 
 sin x   Area   2 x   
L    tan 1 s  cot 1 s
 x  2  2 3  2
 sin x 
  1 1  8
L   e sx
sin x
dx  cot 1s Area   2      4  2  
 x  0  x  2 3  3
 12  3  2   10  64  64 
Area     Area  64    64  
 6   3  3  3
7 10 27 9 128
Area     Area  128   85.33
6 3 6 2 3
Hence, the correct option is (B). Hence, the area between the curves is 85.33.
3.7 (D) 3.9 (B)
Given : Curves x 2  4 y and y 2  4 x Given : f ( x)  2 x7  3x  5
Checking from the options,
From option (B) :
Put x  1 ,
f (1)  2 17  3 1  5  0
Since, x  1 satisfies the given function f ( x) .
Therefore, ( x  1) is a factor of given function
Angle between the curves is angle between the f ( x) .
tangents at the point of intersection. i.e. 900 Hence, the correct option is (B).
Hence, the correct option is (D). 3.10 (B)
3.8 85.33 x
Given : f ( x)  e  x e
Given : x 2  8 y …(i)
g ( x)   f ( x)dx   e x e dx
x
y 8 …(ii)
g ( x)   e  e e  x dx
x
From equation (i) and equation (ii), …(i)
x 2  8  8  64
Put e  x  t ,
x  8
Differentiating on both sides,
When x  8 , y  8
e x (1)dx  dt
When x   8 , y  8
e  x dx   dt
Thus, common points are (8,8) and (  8,8) .
From equation (i),
g (t )   et (1) dt
g (t )   et dt  (et )(1)

g (t )  et
8 8 8
Put t  e  x in above equation,
Area    dy dx   [ y ]8
x
x2
dx g ( x)  e e
y
8 x2 8 8
y Hence, the correct option is (B).
8
8
 x2   1 x3 
8
3.11 (A)
Area    8   dx  8 x   
(sin 1 x) 2
1
8 
8   8 3  8
Given : I   dx …(i)
1 8  1 ( 8) 
3 3 0 1  x2
Area  8  8    8  (8)  
8 3  8 3  Let sin 1 x  t
1
dx  dt
1  x2
When x  0 , t  sin 1 0  0 
I   cos2 xdx  I
 0
When x  1 , t  sin 1 1  
2 2 I   cos 2 xdx
From equation (i), 0
  /2
2 I  2  cos 2 xdx
2
I   t dt
2
0
0
  (2  1)   
2
 /2
t  1     2 I  2    2   
3 3
I       03  2 2  4 2
 3 0 3  2  
2
1   3 3 I
I   4
3 8 24 Hence, the correct option is (B).
 Key Point
3.12 (B) By the property of integrals,

Given : I   x cos 2 x.dx 2a  a
20 f ( x)dx, if f (2a  x)  f ( x)
0 (i)  f ( x)dx  
 0, if f (2a  x)   f ( x)
0
. Method 1 : 
  1  cos 2 x    /2  (n  1) (n  3)...... 
I   x .dx (ii)   cos n x dx   
0
 2  0 2 n(n  2)...... 
 x x  [if n  even]
I     cos 2 x .dx
0 2 2 
 x  x 3.13 (A)
I   dx   cos 2 x dx
0 2 0 2
x2 y 2
1  x2  1 
 Given : 2  2  1
2  2  0 2  0 (I) (II)
I  x cos 2 x dx a b
2 1  sin 2 x d  
I  x     ( x)   cos 2 x  dx 
4 2 2  dx  

2 1  x  sin 2 x cos 2 x 
I  
4 2  2 4  0
2 1  1 1  2 b
I      a

s
 x2  a2
4 2 4 4 4 Area    (1)dydx 
 
x  a y  b  x 2  a 2
(1)dydx
a
. Method 2 : b 2 2 b 2 2
 a x a x
a a a a
I   x cos 2 xdx Area  4 
0 x 0

y 0
(1) dydx  4 
x 0

y 0
dx

Area  ab
I   (  x) cos 2 (  x)dx
0
[By definite integral property]
 
I   cos xdx   x cos 2 xdx
2
0 0
3.14 3  d 
 (uv) dx  u  v dx    dx u  v dx  dx
Given : Line integral
I  ln (1  x)  1 dx
 ( y  z )dx  ( x  z )dy  ( x  y)dz 
C
 d 
Where, C is a line segment between (0, 0, 0) and    (ln (1  x))  1 dx  dx  C
(1, 1, 1)  dx 
I    ( ydx  xdy )  ( zdx  xdz )  ( zdy  ydz ) 
x
I  ln (1  x) x   dx  C
C 1 x
x  1 1
I    d ( xy )  d ( xz )  d ( yz )  I  ln (1  x) x   dx  C
C x 1
I  ( xy  yz  zx) (1,1,1)  1 
(0,0,0) I  ln (1  x) x   1   dx  C
I  (1  1  1)  (0  0  0) = 3  x 1 
I  x ln (1  x)  x  ln (1  x)  C
Hence, the value of the line integral is 3.
I  ln (1  x) ( x  1)  x  C
3.15 15
No option is correct.
Volume =    8 xyz dx dy dz 3.18 (D)
v
y
1 2 3 
V  8   z  y  x dx dy dz 
0 1 2 
 1 2 3
 Infinite area
V  8   zdz  ydy  x dx 
0 1 2 
1 2 3
 z 2   y 2   x2 
V  8        1  3 5  15 x
 2 0  2 1  2 2 -1 1
1 1
3.16 0 Given : I   dx
1 x2
1
 xe
x 1
dx  Odd function f  x  is not defined at x = 0
1 x2
f ( x)   f ( x) 0 1 1 1
1
So, I   2 dx   2 dx
1 x 0 x
 xe
x
dx  0 0 1
1  1  1
      
3.17 MTA  x  1  x 0
 x 2 x3 x 4  1
Here, is also not defined at x = 0. So, the integral
Given : I    x     ......  dx x
 2 3 4 
does not converge.
We know that,
Logarithmic series,
x 2 x3 x 4
ln (1  x)  x     ......
2 3 4 
I   ln (1  x) dx   ln (1  x)  1 dx
 
I function II function
Law of integration,
4 Vector Calculus
2014 IIT Kharagpur 2021 IIT Bombay

4.1 A particle moves along a curve whose 4.6 The shape of cumulative distribution
parametric equations are : function of Gaussian distribution is
x  t 3  2t , y  3e2t and z = 2 sin (5t) (A) Horizontal line
(B) Bell shaped
where x, y and z show variation of the
(C) Straight line at 45-degree angle
distance covered by the particle (in cm) with
(D) S-shaped.
time t (in s). The magnetic of the acceleration
[1 Mark]
of the particle (in cm/s 2 ) at t = 0 is ______.
4.7 A function is defined in Cartesian coordinate
2015 IIT Kanpur system as f ( x, y)  xe y . The value of
4.2 The directional derivative of the field directional derivative of the function (in
u( x, y, z )  x 2  3 yz in the direction of the integer) at the point (2, 0) along the direction
vector (iˆ  ˆj  2kˆ) at point (2, 1, 4) is of straight line segment from point (2, 0) to
________. [2 Marks] point (1 / 2, 2) is _______.
2017 IIT Roorkee [2 Marks]
4.8 The unit normal vector to the surface
4.3 The divergence of the vector field
x2  y 2  z 2  48  0 at the point (4, 4, 4) is :
V  x 2iˆ  2 y 3 ˆj  z 4 kˆ at x  1, y  2, z  3 is
[1 Mark]
_______. [1 Mark] 2 2 2 1 1 1
(A) , , (B) , ,
2018 IIT Guwahati 2 2 2 5 5 5
4.4 The value (up to two decimal places) of line 1 1 1 1 1 1
integral (C) , , (D) , ,
2 2 2 3 3 3
c
F (r )  dr , for F (r )  x 2iˆ  y 2 ˆj 2023 IIT Kanpur
along C which is a straight line joining (0, 0) 4.9 Let 𝜙 be a scalar field, and 𝒖 be a vector
to (1, 1) is ______. [2 Marks] field. Which of the following identities is
2019 IIT Madras true for div(𝜙𝒖)?
4.5 What is curl of the vector field (A) div(𝜙𝒖) = 𝜙div(𝒖) + 𝒖 ⋅ grad(𝜙)
2 x 2 yiˆ  5 z 2 ˆj  4 yzkˆ ? [1 Mark] (B) div(𝜙𝒖) = 𝜙div(𝒖) + 𝒖 × grad(𝜙)
(C) div(𝜙𝒖) = 𝜙grad(𝒖) + 𝒖 ⋅ grad(𝜙)
(A) 14ziˆ  2x2k
(D) div(𝜙𝒖) = 𝜙grad(𝒖) + 𝒖 × grad(𝜙)
(B) 6 zi  4 x 2 j  2 x 2 k
(C) 14 zi  6 yj  2 x 2 k
(D) 6 zi  8 xyj  2 x 2 yk
4.1 12 Directional derivative in direction of vector
(iˆ  ˆj  2kˆ) is given by,
Given : x  t 3  2t , y  3e2t , z = 2 sin (5t)
a iˆ  ˆj  2kˆ
Let position vector, r  xiˆ  yjˆ  zkˆ u   (4iˆ  12 ˆj  3kˆ) 
a 12  12  22
r  (t 3  2t )iˆ  3e2t ˆj  2sin(5t )kˆ a 4  12  6 14
u   
Velocity of particle is given by, a 6 6
dr
v a 7 2
dt u     5.715
a 3
v  (3t 2  2)iˆ  6e2t ˆj  10cos(5t )kˆ
Hence, the directional derivative is – 5.715.
Acceleration of particle is given by,
4.3 134
d 2r
a 2 Given : V  x 2iˆ  2 y 3 ˆj  z 4 kˆ
dt
a   6t  iˆ  12e2t  ˆj   50sin 5t  kˆ Divergence of the vector V is given by,
  ˆ  ˆ
At t  0 ,  V   iˆ  j  k  V
 x y z 
d 2r
 (6  0)iˆ  12(e 20 ) ˆj   ˆ  ˆ 2 ˆ
dt 2
 V   iˆ  j  k   ( x i  2 y 3 ˆj  z 4 kˆ)
 x y z 
[50sin(5  0)]kˆ
  
d 2r  V  ( x 2 )  (2 y 3 )  ( z 4 )
 12 ˆj x y z
dt 2
 V  2 x  6 y 2  4 z 3
2
d r
 12 ˆj  12 V  2 1  6  22  4  33  134
dt 2 (1,2,3)
Hence, the magnitude of acceleration of the Hence, the divergence is 134.

particle is 12. 4.4 0.67
4.2 – 5.715 Given : F (r )  x 2iˆ  y 2 ˆj
Given : u( x, y, z )  x 2  3 yz Coordinates of straight line = (0, 0) to (1, 1)
Gradient of a scalar function is given by, F (r )  x 2iˆ  y 2 ˆj
u u u dr  dx iˆ  dy ˆj
u  iˆ  ˆj  kˆ
x y z F (r )  dr   x 2iˆ  y 2 ˆj    dx iˆ  dy ˆj 
 
u  iˆ ( x 2  3 yz )  ˆj ( x 2  3 yz ) F (r )  dr  x 2 dx  y 2 dy
x y
Integrating both sides,

kˆ ( x 2  3 yz ) 1
z C
F (r )  dr   ( x 2 dx  y 2 dy)
0
ˆ ˆ
u  2 xi  3zj  3 yk ˆ 1 1
At point (2,  1, 4 ) , C
F (r )  dr   x 2 dx   y 2 dy
0 0
 4iˆ  12 ˆj  3kˆ
1 1
u  x3   y 3 
(2, 1, 4)
C F ( r )  dr   3   3 
 0  0
1 1 2 3  3iˆ 4
F (r )  dr  
C 3 3
nˆ    iˆ  2 ˆj     ˆj
5 2  5 5
2
C F (r )  dr  3  0.67 3 4
DD  f  nˆ    2   1
5 5
Hence, the value of line integral is 0.67.
4.8 (D)
4.5 (A)
  x2  y 2  z 2  48 , P (4, 4, 4)
Given : V  2 x y iˆ  5 z ˆj  4 yz kˆ
2 2
  
Grad     iˆ  ˆj  kˆ
The curl of vector field V is given by, x y z
iˆ ˆj kˆ iˆ ˆj kˆ  (2 x)iˆ  (2 y) ˆj  (2 z )kˆ
      n  (grad ) P  8iˆ  8 ˆj  8kˆ
 V  
x y z x y z n 8iˆ  8 ˆj  8kˆ iˆ  ˆj  kˆ
vx v y vz 2 x y 5 z 4 yz
2 2 nˆ   
n 64  64  64 3
    1 1 1 
  V  iˆ  ( 4 yz )  (5 z 2 )  nˆ 
 y z 
, , 
 3 3 3
   Hence, the correct option is (D).
 ˆj  ( 4 yz )  (2 x 2 y ) 
 x z  4.9 (A)
  
 kˆ  (5 z 2 )  (2 x 2 y )  div (u )   div(u )  u.grad ()
 x y  Hence, the correct option is (A).
  V  iˆ  4 z  10 z   ˆj  0  0 

 kˆ [0  2 x 2 ]
   V  14 ziˆ  2 x 2 kˆ
4.6 (B)
4.7 1
f ( x, y)  xe y
  f
f f ˆ f ˆ
  iˆ  j k
x y z
  e y iˆ  xe y ˆj  0kˆ
(2,0)  iˆ  2 ˆj
3
a  ( x2  x1 )iˆ  ( y2  y1 ) ˆj   iˆ  2 ˆj
2
3
 iˆ  2 ˆj
a 2
nˆ  
a  3
2
    (2)
2
 2
5 Maxima & Minima
2015 IIT Kanpur (A) The function x x , ( x  0) , has the global

5.1 While minimizing the function f ( x) , minima at x  e
necessary and sufficient conditions for a (B) The function x x , ( x  0) , has the global
point, x0 to be a minima are [1 Mark] maxima at x  e
(A) f '( x0 )  0 and f ''( x0 )  0 (C) The function x 3 has neither global
minima nor global maxima
(B) f '( x0 )  0 and f ''( x0 )  0
(D) The function x has the global minima at
(C) f '( x0 )  0 and f ''( x0 )  0
x0
(D) f '( x0 )  0 and f ''( x0 )  0
2022 IIT Kharagpur
2016 IISc Bangalore 5.6 Let max {a, b} denote the maximum of two
5.2 The optimum value of the function real number a and b. Which of the following
f ( x)  x2  4 x  2 is [1 Mark] statement(s) is/are TRUE about the function
[2 Marks]
(A) 2 (maximum)
(B) 2 (minimum) f ( x)  max 3  x, x  1 ?
(C) – 2 (maximum) (A) It is continuous on its domain
(D) – 2 (minimum) (B) It has a local maximum at x  2
5.3 The quadratic approximation of (C) It is differentiable on its domain
f ( x)  x  3x  5 at the point x  0 is
3 2
(D) It has a local minimum at x  2
[2 Marks] 5.7 Consider the polynomial
(A) 3x 2  6 x  5 (B)  3x 2  5 f ( x)  x  6 x  11x  6 on the domain S
3 2
given by 1  x  3. The first and second

(C)  3x2  6 x  5 (D) 3x 2  5
derivatives are f '( x) and f "( x). Consider
2018 IIT Guwahati the following statements :
5.4 At the point x  0 , the function f ( x)  x3 I. The given polynomial is zero at the
boundary points x  1 and x  3.
has [1 Mark]
II. There exists one local maxima of f ( x)
(A) Local maximum
with in the domain S.
(B) Local minimum
III. The second derivative f "( x )  0
(C) Both local maximum and minimum
throughout the domain S.
(D) Neither local maximum nor local
minimum IV. There exists one local minima of f ( x)
with in the domain S.
2019 IIT Madras
The correct option is :
5.5 Which one of the following is NOT a correct (A) Only statement II and IV are correct
statement? [2 Marks]
(B) Only statement I and IV are correct
(C) Only statement I, II and IV are correct (A) The function is continuous at all x
(D) Only statement I, II and III are correct (B) The function is differentiable at all 𝑥
[2 Marks] (C) The function is periodic
2023 IIT Kanpur (D) The function is bounded
5.8 For the function f  x   e x sin x ; x  ,

which of the following statements is/are
TRUE?
5.1 (D) f (0)  5
For stationary points, f '( x0 )  0

f '( x)  3x 2  6 x
f '(0)  0
For minima condition, f "( x0 )  0
f "( x)  6 x  6
The necessary and sufficient condition for a point
f "(0)  6
x0 to be minima are,
x2
f '( x0 )  0, f ''( x0 )  0 Quadratic approximation  5  x  0  ( 6)
2
 3x  5
2
5.2 (D)
Given : f ( x)  x 2  4 x  2 5.4 (D)
Differentiating both sides with respect to x,
Given : f ( x)  x3
f '( x)  2 x  4 …(i)
. Method 1 :
The stationary points are given by,
f '( x)  0 Differentiating function f ( x) with respect to x,
2x  4  0 f '( x)  3x 2
x2 The stationary point is given by,
Differentiating both sides of equation (i) with f '( x)  0
respect to x, 3x 2  0
f "( x)  2  0 [Minimum]
x0
Hence, f ( x) is minimum at x  2 . Differentiating function f '( x) with respect to x,
Therefore minimum value of f ( x) is, f ''( x)  6 x
f (2)  4  8  2   2 f ''(0)  0
Hence, the correct option is (D). [Neither local minima nor maxima]
5.3 (B) Hence, the correct option is (D).
Given : f ( x)  x3  3x 2  5 . Method 2 :
Quadratic approximation is given by, Graph for f ( x)  x3 will be drawn as,
( x  0)2
f ( x )  f (0)  ( x  0) f '(0)  f "(0)
2!
f ( x)  x 3  3 x 2  5
f ( x)  f ''( x)   f '( x) 
2
(3  2 ln x)

 f ( x)  2
x3
 3  2 ln x 
   f ( x)2   f '( x) 2
 f ''( x)  x 3

f ( x)
…(i)
At x  0 , Substituting x  e in equation (i),
The function y  x3 has neither local minima nor 1 1e 1
f "(e)  (e )   1
maxima (i.e., it is a point of inflection). e3  2 
 e
e
5.5 (A) ee
1 1e
f "(e)  (e )  e 2  0.063
1
Given : f ( x)  x x  ( x) x
e 3
1 f "(e)  0
ln f ( x)  ln x
x  x  e is point of minima.
Differentiating w.r.t x,
For option (C), f ( x)  x3 has global minima at
1 1  ln x
 f '( x)  x  0.
f ( x) x2
For option (D), f ( x)  x is not differentiable at
1  ln x 1  ln x 1x
 f '( x)   f ( x )  x x  0.
x2 x2
1  ln x Hence, the correct option is (A).
  1
 2 
 x
x
Now, f '( x)  0
1  ln x  0 5.6 (A), (D)
ln x  1
xe f ( x)  max 3  x, x  1
Now, we calculate f ''( x) at x  e
x 1 ; x  2
f '( x) 1  ln x f ( x)  
 3  x ; x  2
f ( x) x2
Differentiating above equation, we get Here, f (2)  f (2 )  f (2 )  2
y
 1
x     (1  ln x)  2 x
2
f ( x)  f ''( x)   f '( x)
2
  x 4
 f ( x)2
(0,3)
x Intersection
point
f ( x)  f ''( x)   f '( x)  y  x 1
2
 x  2 x(1  ln x)

 f ( x)  2 (0,1) (2,1)
x4
(1,0) (3,0)
f ( x)  f ''( x)   f '( x) 
2 x
 x  2 x  2 x ln x 0 (2,0) y  3 x

 f ( x) 2
x4 (0,–1)
f ( x)  f ''( x)   f '( x) 
2
3x  2 x ln x

 f ( x)2 x4 It is continuous on its domain.
1 ; x  2 f "( x  1.423)  (1.423)3  6  (1.423) 2
f '( x)  
3 ; x  2  111.423  6
Here, f '(2 )  f '(2 )   0.384  0 (+ve)
At x  1.423, it shows minima.
It is not differentiable at x  2.
That means, there exists one local maxima and one
Clearly it is visible from figure that at x  2 it local minima in the domain S.
shows local minima. At x  1,
Hence, the correct option are (A) and (D). f ( x  1)  13  6 12  111  6  0
5.7 (C) At x  3,
Given : Polynomial f ( x  3)  33  6  32  11 3  6  0
f ( x)  x3  6 x2  11x  6; 1  x  3 So at x  1 and x  3, the given polynomial is
Differentiation w.r.t x, zero at the boundary points.
f '( x)  3x 2  12 x  11 Statements I, II and IV are correct.
Again differentiation w.r.t x, Hence, the correct option is (C).
f "( x)  6 x  12 5.8 (A)
For maxima or minima,
f '( x)  0 y
3 x 2  12 x  11  0
Stationary points, x  2.577, 1.423 8
x  1,3
6
2.577, 1.423  1, 3
At x  2.577, 4
f "( x  2.577)  (2.577)  6  (2.577)
3 2
2
 11 2.577  6
x
  0.384  0 (–ve) ( , 0) (0, 0) (, C)
At x  2.577, it shows maxima. Hence, the correct option is (A).
At x  1.423,

6 Mean Value Theorem
2019 IIT Madras x ln 

(D)
ln  ln   1
6.1 Consider the functions : x   ln  and
y   ln  . Which one of the following the 2022 IIT Kharagpur
 6.2 The Fourier cosine series of a function is
correct expression for ? [2 Marks]
x given by,

ln 
(A) f ( x)   f n cos nx
ln  ln   1 n 0
x ln  For f ( x)  cos4 x, the numerical value of

(B)
ln  ln   1 ( f 4  f5 ) is ________. (round off to three
ln  decimal places) [1 Mark]
(C)
ln  ln   1
6.1 (A) 1 

 1 ln  x
Given : x   ln  and y   ln   
x  x  x
   
Then ?    
x
x   1
x y   1 
 and    x x ln 
ln  ln 
  x 1 
 
2

2  1
 y  ln y  ln(ln ) x   ln  

ln  
 ln     x ln    
1
Now, differentiating wrt x, x   ln  
1  
ln y  ln(ln )  x 0     ln  
1
  

  ln   x  x  x ln    
x ln y  ln(ln )2 Putting x   ln 
1 x  ln 
   
ln y  ln(ln ) ln y  ln(ln y)2 x ln  ln   1
x
But ln y  ln(ln )  Hence, the correct option is (A).

6.2 0.125
 1
x  
 1 x  ln  Given : f ( x)   f n cos nx
  
x  x  x
2
n 0
 
    For f ( x)  cos 4 x, ( f 4  f5 )  ?
f ( x)  cos4 x  cos2 x  cos2 x
 1  cos 2 x   1  cos 2 x 
f ( x)    
 2  2 
1
f ( x)  1  cos 2 x  cos 2 x  cos 2 2 x 
4
1
f ( x)  1  2cos 2 x  cos2 2 x 
4
1 1  cos 4 x 
f ( x)  1  2 cos 2 x  
4 2
1
f ( x)  2  4cos 2 x  1  cos 4 x
8
1 1 1 1
f ( x)   cos 2 x   cos 4 x
4 2 8 8
3 1 1
f ( x)   cos 2 x  cos 4 x …(i)
8 2 8
Expansion of F.S. in given function, for 4th and 5th
harmonic is
f ( x)  .......  f 4 cos 4 x  f5 cos5x  .......
…(ii)
So, from the equation (i) and (ii),
1
f 4  and f5  0
8
1
Then, f 4  f5   0
8
f 4  f5  0.125

7 Complex Variables
2014 IIT Kharagpur Which of the following is one of the residues

of the above function? [2 Marks]
2  3i
7.1 Z can be expressed as [1 Mark] (A) 1
9
5  i (B)
16
(A)  0.5  0.5i (B)  0.5  0.5i (C) 2 (D) 9
(C) 0.5  0.5i (D) 0.5  0.5i
2015 IIT Kanpur
7.2 Consider the following complex function,
9
f ( z) 
( z  1)( z  2)2
7.1 (B) R( z )  ( z  1)  f ( z )z 1
2  3i  9 
Given : Z  R( z )  ( z  1)  2
5  i  ( z  1) ( z  2)  z 1
1
(2  3i ) (  5  i)
Z  R( z ) 
9
1
( 5  i) ( 5  i) (1  2)2
10  2i  15i  3i 2 10  13i  3 For multiple poles, the residue of f ( z ) at z  a
Z 
52  i 2 25  1 with order m is given by,
13i 13 (i 1) i 1 1  d m1 m 
Z  13  R( z ) z  a   m1  f ( z ) ( z  a)  
26 26 2 (m  1)!  dz z  a
i 1
Z    0.5  0.5i Residue at z   2 is,
2 2
1  d  9  
Hence, the correct option is (B). R( z ) z   2     ( z  2)2  
(2  1)!  dz  ( z  1)( z  2) 2
 z   2
7.2 (A)
 d  9 
R( z ) z   2    
Given : f ( z ) 
9  dz  z  1   z   2
( z  1) ( z  2) 2
 9  9
Poles of f ( z ) are given by, R( z ) z   2   2
  1
 ( z  1)  z   2 9
( z  1) ( z  2)2  0
So, –1 is the required residue.
z  1,  2,  2 Hence, the correct option is (A).
f ( z ) has poles at z  1,  2
Residue of z  1 is,
8 Limit & Series Expansion
2014 IIT Kharagpur  sin 4 x   tan x 

(C) lim    2 and lim  

x  0 sin 2 x
 x 0
 x 
 x  sin x 
8.1 lim   equal to [1 Mark]  sin 4 x   tan x 
x 
 x  (D) lim    2 and lim   1
(A)  (B) 0 
x 0 sin 2 x
 x 0
 x 
8.7 For a small value of h, the Taylor series
(C) 1 (D) 
expansion for f ( x  h) is [1 Mark]
xa 1
8.2 The expression lim is equal to h2
a 0 a (A) f ( x)  hf '( x)  f ''( x)
2
[2 Marks]
h3
(A) log x (B) 0  f ''( x)  .....
3
(C) x log x (D) 
h2
(B) f ( x)  hf '( x)  f ''( x)
2015 IIT Kanpur 2!
 1
2x
h3
8.3 lim 1   is equal to [1 Mark]  f ''( x)  .....
x 
 x 3!
2
(A) e 2 h
(B) e (C) f ( x)  hf '( x)  f ''( x)
2!
(C) 1 (D) e 2
h3
2016 IISc Bangalore  f ''( x)  .....
3!
xy h2
8.4 What is the value of lim ? (D) f ( x)  hf '( x)  f ''( x)
x 0
y 0
x  y2
2
2
[1 Mark] h3
 f ''( x)  .....
(A) 1 3
(B) – 1 8.8 The following inequality is true for all x
(C) 0 close to 0. [1 Mark]
(D) Limit does not exist x 2
x sin x
2  2
 tan x  3 1  cos x
8.5 lim  2  is equal to _____. [1 Mark]
x 0 x  x
  x sin x
What is the value of lim ?
x 0 1  cos x
2019 IIT Madras
(A) 1 (B) 0
8.6 Which of the following is correct?
1
[1 Mark] (C) (D) 2
2
 sin 4 x   tan x 
(A) lim    1 and lim   1 2020 IIT Delhi

x  0 sin 2 x
 x 0
 x 
 sin 4 x   tan x  x2  5x  4
(B) lim     and lim   1
8.9 The value of lim
x  4 x 2  2 x
is [1 Mark]

x 0 sin 2 x
 x 0
 x 
1 The limit (correct up to one decimal place) is
(A) (B) 1
4 _______. [1 Mark]
1 x ln( x)
(C) (D) 0 8.13 The value of lim is [2 Marks]
2 x 1  x 2
(A) 0.5 (B) 

9 x 2  2020
8.10 The value of lim is (C) 0 (D) 1
x  x7
[1 Mark] 2022 IIT Kharagpur
7 8.14 A set of observations of independent
(A) 3 (B)
9 variable (x) and the corresponding
(C) indeterminable (D) 1 dependent variable (y) is given below.
8.11 The Fourier series to represent x  x 2 for x 5 2 4 3
  x   is given by y 16 10 13 12
 
a
x  x 2  0   an cos nx   bn sin nx Based on the data, the coefficient a of the
2 n 1 n 1 linear regression model
The value of a0 (round off to two decimal y  a  bx
places), is _______. [2 Marks] is estimated as 6.1. The coefficient b is
2021 IIT Bombay ______. (round off to one decimal place)
[2 Marks]
8.12 Consider the limit, 
 1 1 
lim   

x 1 ln x x 1 
8.1 (C) d a
( x  1)
da x a log x
f (a )  lim 
 x  sin x  a 0 d 1
Given : f ( x)  lim   (a)
x 
 x  da
 sin x  x a log x
f ( x)  lim 1   f (a)  lim  1 log x  log x
x 
 x  a 0 1
sin x sin Hence, the correct option is (A).
f ( x)  1  lim  1
x  x  . Method 2 :
f ( x)  1  0  1 [ 1  sin   1 ] a mx  1
lim  m log a
Hence, the correct option is (C). x 0 x
8.2 (A) Substitute m  1 and x  a ,
xa 1
xa 1 0 lim  1 log x  log x
Given : f (a)  lim  x 0 a
a 0 a 0
. Method 1 :
 Key Point
0 d x
For form, applying L-Hospital’s rule, (i) (a )  a x log a
0 dx
a mx  1
(ii) lim  m log a
x 0 x
d
8.3 (D) tan x
2x f ( x)  lim dx
 1 x 0 d
Given : y  lim 1   ( x 2  x)
x 
 x dx
Taking log on both sides, sec 2 x sec 2 0
 1 f ( x)  lim 
log y  lim 2 x log 1   x 0 2x 1 2  0 1
x 
 x 1
f ( x)   1
 1 1
2 log 1  
log y  lim  x  0  tan x 
x  1/ x 0 Hence, lim  2
x 0 x  x
 is equal to – 1.
 
0
For form, applying L’ Hospital’s rule, 8.6 (D)
0
2  1   sin 4 x   tan x 
  2  and lim 
 1  x 
To find : lim   
1   
x  0 sin 2 x
 x 0
 x 
log y  lim 
x
 sin 4 x   2sin 2 x cos 2 x 
1   lim
x 
 2 So, lim   
x
x  0 sin 2 x
  x 0
 sin 2 x 
log y  lim
2
2 sin 2  2sin .cos 
x  1
1  sin 4 x 
  lim(2cos
x lim  2 x)  2 1  2
y  e2 x  0 sin 2 x
  x0
Hence, the correct option is (D).  tan x 
and lim  
8.4 (D) x 0
 x 
xy 0 0
Given : f ( x, y )  lim  It is of form. So applying L’ Hospital rule
x 0
y 0
x y
2 2
0 0
Let, y  mx d
tan x  sec 2 x 
y  0, mx  0  lim dx  lim  
x 0 d x 0
 1 
x2m m x
f ( x, y)  lim  dx
x 0 x 2 (1  m2 ) 1  m2
Putting limit x  0
For different values of m we get different limits
 sin 4 x 
So, limit is not unique therefore limit does not  lim  2
x 0  sin 2 x 
exist.
Hence, the correct option is (D).  tan x 
lim 
x 0 
 1
8.5 –1 x 
 tan x  tan 0 0
Let, f ( x )  lim  2   8.7 (C)
x 0 x  x
  00 0
: Method 1 :
0
For form, applying L’ Hospital’s rule, Taylor series expansion of f ( x) around point x0
0
is given as
f ( x0 )  f ( x0 )  ( x  x0 ) f '( x0 )
f "( x0 ) f "( x0 ) Hence, the correct option is (D).
( x  x0 )2 .  ( x  x0 )3
2! 3! : Method 2 :
Taylor series expansion of f ( x) around x0  h ,  x sin x   x sin x 
lim    lim  
f ( x0  h)  f ( x0 )  ( x  h  x0 ) f '( x0 ) x 0 1  cos x  x 0 2 x
 2sin 
f "( x0 )  2
( x  h  x0 )2 .
2!
 f ( x0  h)  f ( x0 )  h f '( x0 )  x x
 x  2sin 2 cos 2 
h2  lim  

f "( x0 )  ...... x 0
 2sin 2
x

2!  2 
Transforming the variable, is x0  x
 x
h2  x cos 2 
f ( x  h)  f ( x)  h f '( x)  f "( x)  ....  lim 
2! x 0 x 
 sin 
Hence, the correct option is (C).  2 
: Method 2 :  x
Going by options.  2 cos 2 
 lim 
x 0 x 
(i) Taylor series expansion has terms with  sin 
positive sign. So options (B) and (D) are  2 
eliminated.  x 
 2 
(ii) Taylor series expansion contains terms with
factorials. So option (A) is eliminated.  sin  
 lim  1
Hence, the correct option is (C). 0  
8.8 (D) Applying limits,
 x sin x  2 1
Given : lim   =2
x 0  1  cos x 
 1
: Method 1 :
0 8.9 (A)
Above limit is of form
0 x2  5x  4
Given : lim
 Applying L’ Hospital Rule, x  4 x 2  2 x
d Applying L’ Hospital rule,

[ x sin x ]
lim dx  5 4  5 4
x 0 d x 2 1   2  1   2 
[1  cos x] lim  x x   lim  x x 
dx x   2 x   2
x2  4   4 
 x cos x  sin x   x  x
 lim  
x 0  sin x 1 0  0 1
 
Again it is of form 40 4
 Applying L’ Hospital Rule Hence, the correct option is (A).
  x sin x  cos x  cos x  8.10 (A)
 lim  
x 0  cos x
9 x 2  2020
Taking limits, Given : lim
x  x7
0 11
 2
1
2020 x y x2 xy
9
x2 5 16 25 80
 lim
x   7 2 10 4 20
1   4 13 16 52
 x
3 12 9 36
On putting the limit, we get  x  14  y  51  xy  188
 x  54
2
90
3 We know that, normal equation for fitting of
1 0 straight lines are
Hence, the correct option is (A).  y  na  b  x
8.11 – 6.58 and  xy  a  x  b  x 2
a0  
Here, n  4
Given : x  x 2    an cos nx   bn sin nx
2 n 1 n 1
So, 51  4a  b(14)
Let f ( x)  x  x 2 188  a (14)  b(54)
 After solving, a  6.1 and b  1.9
Fourier series expansion of f ( x) from  to 
is given by,

1 
a0   f ( x)dx
 
1 
a0   ( x  x 2 )dx
 
a0    x dx  x2 dx 
1  
    
a0  0  2 x 2 dx 
1 
 0 
x is odd and x 2 is even

2  x 3  2  3  2 2
   
  3  0   3  
 6.5797  6.58
Hence, the value of a0 is 6.58 .
8.12 1/2
 1 1   ( x  1)  ln x  1
lim    lim  

x 1 ln( x) x  1 x1  ( x  1) ln x  2

8.13 (C)
 1 
 x    ln x 
 x ln x    
 lim   
x
lim  2    
x  x  2
    x  2x   
 1
0  x   1  1
lim    lim    0
x   2  x   2 x  2  
8.14 1.9
9 Probability & Statistics
2013 IIT Bombay 4 1

(C) (D)
9.1 Find the value of  such that function f ( x) 5 5
9.6 If X is a continuous, real valued random
is valid probability density function
variable defined over the interval (,  )
and its occurrence is defined by the density
f ( x)    x  1 2  x  ; for 1  x  2
function given as,
=0 ; otherwise 1 x  a 
2
1  
[2 Marks] f ( x)  e2 b 
9.2 A 1 hour rainfall of 10 cm has return 2  b
period of 50 year. The probability that 1 where ‘a’ and ‘b’ are the statistical attributes
hour of rainfall of 10 cm or more will occur of the random variable X. The value of the
in each of two successive years is integral
2
(A) 0.04 (B) 0.2 a 1 x  a 
1  
 e2 b 
(C) 0.2 (D) 0.0004 dx is [1 Mark]
 2  b
2014 IIT Kharagpur (A) 1 (B) 0.5
9.3 The probability density function of 
evaporation E on any day during a year in a (C)  (D)
2
watershed is given by 9.7 An observer counts 240 veh/h at a specific
1 highway location. Assume that the vehicle
 ; 0  E  5 mm/day
f (E)  5 arrival at the locations is Poisson distributed,
 0; otherwise
the probability of having one vehicle
The probability that E lies in between 2 and arriving over a 30-second time interval is -
4 mm/day in the watershed is (in decimal) _______.
_____. [1 Mark] 2015 IIT Kanpur
9.4 A traffic officer imposes on an average 5 9.8 Consider the following probability mass
number of penalties daily on traffic violators. function (p.m.f) of a random variable X
Assume that the number of penalties on  q, if X  0
different days is independent and follows a 
p ( X , q )  1  q, if X  1
Poisson distribution. The probability that
 0,
there will be less than 4 penalties in a day is  otherwise
______. [2 Marks] If q  0.4 , the variance of X is _______.
9.5 A fair (unbiased) coin was tossed four times [1 Mark]
in succession and resulted in the following 9.9 For probability density function of a random
outcomes : variable, x is
(i) Head (ii) Head (iii) Head (iv) Head. x
f ( x)  (4  x 2 ) ; for 0  x  2
The probability of obtaining a ‘Tail’ when 4
the coin is tossed again is [1 Mark] 0 ; otherwise
1 The mean  x of the random variable is
(A) 0 (B)
2 ______. [2 Marks]
2016 IISc Bangalore (B) Mean of f ( x) and g ( x) are same;
Variance of f ( x) and g ( x) are
9.10 Probability density function of a random
variable X is given below different.
0.25; if 1  x  5 (C) Mean of f ( x) and g ( x) are different;
f ( x)   Variance of f ( x) and g ( x) are same.
0 ; otherwise
P ( X  4) is [2 Marks] (D) Mean of f ( x) and g ( x) are different;
3 1 Variance of f ( x) and g ( x) are
(A) (B)
4 2 different.
1 1 9.14 Type II error in hypothesis testing is
(C) (D)
4 8 [1 Mark]
9.11 The spot speeds (expressed in km/hr) (A) acceptance of the null hypothesis when it
observed at a road section are 66, 62, 45, 79,
is false and should be rejected
32, 51, 56, 60, 53, and 49. The median speed
(expressed in km/hr) is ______. (B) rejection of the null hypothesis when it is
[1 Mark] true and should be accepted
[Note : Answer with one decimal accuracy]. (C) rejection of the null hypothesis when it is
9.12 X and Y are two random independent events. false and should be rejected
It is known that P( X )  0.40 and (D) acceptance of the null hypothesis when it
P( X  Y )  0.7. Which one of
C
the is true and should be accepted
following is the value of P ( X  Y ) ?
2017 IIT Roorkee
[1 Mark]
(A) 0.7 (B) 0.5 9.15 A two-faced fair coin has its faces designated
(C) 0.4 (D) 0.3 as head (H) and tail (T). This coin is tossed
9.13 If f ( x) and g ( x) are two probability three times in succession to record the
density functions, following outcomes : H, H, H. If the coin is
tossed one more time, the probability (up to
 x
 a 1 ;  a  x  0 one decimal place) of obtaining H again,
 given the previous realizations of H, H and
 x
f ( x)    1 ; 0  x  a H, would be _______. [1 Mark]
 a
 0 ; otherwise 9.16 The number of parameters in the univariate
 exponential and Gaussian distributions,

respectively are [1 Mark]
 x
 a ;  a  x  0 (A) 2 and 2 (B) 1 and 2

 x (C) 2 and 1 (D) 1 and 1
g ( x)   ; 0 xa
 a 9.17 For the function f ( x)  a  bx , 0  x  1 to
 0 ; otherwise be a valid probability density function,

 which one of the following statements is
Which one of the following statement is correct? [2 Marks]
true? [2 Marks] (A) a  1 , b  4 (B) a  0.5 , b  1
(A) Mean of f ( x) and g ( x) are same;
(C) a  0, b  1 (D) a  1, b  1
Variance of f ( x) and g ( x) are same.
2018 IIT Guwahati (B) Mean is greater than median but less than
mode.
9.18 The frequency distribution of the
(C) Mean is greater than median and mode.
compressive strength of 20 concrete cube
(D) Mode is greater than median.
specimens is given in the table.
Number of specimens with 2019 IIT Madras
f (MPa)
compressive strength equal to f 9.22 The probability that the annual maximum
23 4 flood discharge will exceed 25000 m3 /s , at
28 2 least once in next 5 years is found to be 0.25.
22.5 5
The return period of this flood event (in
31 5
29 4 years, round off to 1 decimal place) is
If  is the mean strength of the specimens ______.
9.23 The probability density function of a
and  is the standard deviation, the number
continuous random variable distributed
of specimens (out of 20) with compressive
uniformly between x and y (for y  x ) is
strength less than   3 is ______.
[2 Marks]
9.19 The graph of a function f ( x) is shown in the 1
figure (A) y  x (B)
yx
1
(C) x  y (D)
x y
2020 IIT Delhi
9.24 A fair (unbiased) coin is tossed 15 times. The
probability of getting exactly 8 Heads (round
off to three decimal places), is _______.
For f ( x) to be a valid probability density [1 Mark]
function, the value of h is [1 Mark] 2022 IIT Kharagpur
(A) 1/3 (B) 2/3
9.25 A pair of six-faced dice is rolled thrice. The
(C) 1 (D) 3 probability that the sum of the outcomes in
9.20 Probability (up to one decimal place) of each roll equals 4 in exactly two of the three
consecutively picking 3 red balls without attempts is _______. (round off to three
replacement from a box containing 5 red decimal places) [2 Marks]
balls and 1 white ball is ______.
2023 IIT Kanpur
[1 Mark]
9.21 A probability distribution with right skew is 9.26 A remote village has exactly 1000 vehicles
shown in the figure. with sequential registration numbers starting
from 1000. Out of the total vehicles, 30% are
without pollution clearance certificate.
Further, even- and odd-numbered vehicles
are operated on even- and odd-numbered
dates, respectively.
If 100 vehicles are chosen at random on an
The correct statement for the probability even-numbered date, the number of vehicles
distribution is [1 Mark] expected without pollution clearance
(A) Mean is equal to mode. certificate is _________.
(A) 15 (B) 30 respectively. What is the probability of
(C) 50 (D) 70 occurrence of at least A or B (rounded off to
9.27 The probabilities of occurrences of two one decimal place)? ____________
independent events A and B are 0.5 and 0.8,
9.1 6 9.2 (D)
   x  3x  2  ; 1  x  2
 2
Return period of rainfall, T = 50 years.
Given : f ( x)  
Probability of occurrence once in 50 years,

0 ; otherwise
1
By the property of probability density function P  0.02
(P.D.F.), 50
 Probability of occurrence in each of 2 successive
 f ( x) dx  1 years,
 P  0.02  0.02  0.0004
2
  x  3x  2  dx  1
2 Hence, the correct option is (D).
1 9.3 0.4
2
 x3 x2  1
   3  2 x   1  ; 0  E  5 mm/day
 3 2 1 Given : f ( E )   5
 0; otherwise
 8 1 3 
        4  1  2  2  1   1 Hence, the required probability is,
 3 3 2  4
 7 9 
    2  1 P(2  E  4)   f ( E ) dx
 3 2  2
4
 14  27  12  
1 1 4
dx   x 2
   1
 6 2
5 5
1 1
 (4  2)
   1
6 5
6 2
P(2  E  4)   0.4
Hence, the value of  is 6. 5
 Key Point Hence, the probability that E lies in between 2 and
4 mm/day in the watershed is 0.4.
Probability distribution or probability density
function of X :
(i) f  x  0

(ii)  f  x  dx  1

9.4 0.265
Given :
a
(iii) P  a  X  b    f  x  dx The mean of poisson distribution is m  5
b By poisson distribution probability is given by,
e m mr
P(r ) 
r!
Let X = Number of penalties per day
P( X  4)  P( X  0)  P( X  1) a 1 x  a 
2
 1 x  a 
2
1   1  
 e2   0.5
b 
 P( X  2)  P( X  3) e2 b 
 2  b a 2  b
m m m2  m m3  m
P( X  4)  e  me  e  e Hence, the correct option is (B).
2! 3!
 m 2 m3  9.7 0.27
P ( X  4)  e  m 1  m   
 2 6  Given : m  240 veh/h
 25 125 
 e 5 1  5    m
240
veh/min  4 veh/min
 2 6  60
e 5 4
P( X  4)  (6  30  75  125) m veh/sec  2 veh/30sec
6 60
P ( X  4)  0.265 On average, two vehicles arrive in 30 sec time
Hence, the probability that there will be less than interval.
4 penalties in a day is 0.265. So, m  2
9.5 (B) Hence, according to Poisson distribution, the
1 probability of having one vehicle arriving over 30
P (4 successively head in 4 tosses) 
24 sec. time interval is given by,
P (4 successively head and 1 tail in last toss during e m mr
1 P(r )  [where, r  1 ]
5 toss)  P( H H H H T )  5 r!
2
P (4 successively head and head in last toss during The required probability,
1 e2 21
5 toss)  5  P( X  1)   2e2  0.27
2 1!
P (4H, T) Hence, the probability of having one vehicle
Required probability 
P (4H, H)  P(4 H, T) arriving over a 30-second time interval is 0.27.
4
1 1 9.8 0.24
  
Required probability  2 2 
1
if X  0
4 4  q,
1 1 1 1 2 
      Given : p ( X , q )  1  q, if X  1
2 2 2 2  0,
Hence, the correct option is (B).  otherwise
9.6 (B) q  0.4
2
1 x  a  X 0 1
1  
Given : f ( x)  e2 b 
p(X) 0.4 0.6
2  b
Variance of X is given by,

Var ( X )  E ( X 2 )   E ( X )
2
…(i)
where, E[ X ]   X P( X )
E[ X ]  (0  0.4  1 0.6)  0.6
Similarly, E[ X 2 ]  X 2  P( X )
By the property of normal distribution, the E[ X 2 ]  (02  0.4  12  0.6)  0.6
distribution curve is divided into two parts which From equation (i),
are equally probable. So,
Var ( X )  E ( X 2 )   E ( X )
2
53  56
Median = = 54.5 kmph
2
Var ( X )  0.6  (0.6)2
Hence, the median speed is 54.5.
Var ( X )  0.6  0.36  0.24
Hence, the variance of X is 0.24.
9.9 1.066
9.12 (A)
x
 (4  x ); for 0  x  2
2
Given : f ( x)   4 Given : P( X )  0.40
0 ; otherwise P( X  Y C )  0.7
Mean  x of random variable is given by, Hence, P( X  Y C )  P( X )  P(Y C )  P( X  Y C )
 2
x Since, X and Y are two independent events,
x    xf ( x) dx  0 x  4 (4  x ) dx
2
Therefore, P( X  Y C )  P( X )  P(Y C )
2 Hence, P( X  Y C )  P( X )  P(Y C )
2
 2 x4   x3 1 x5 
x    x   dx   3  4 5 
 4   0  P( X )  P(Y C )
0
23 1 25 8 32 0.7  0.4  P(Y C )  0.4  P(Y C )

x       1.066
3 4 5 3 20 0.3  0.6  P(Y C )
Hence, the mean  x of the random variable is P(Y C )  0.5
1.066. 1  P (Y )  0.5
9.10 (A) P (Y )  0.5
0.25; if 1  x  5 P( X  Y )  P( X )  P(Y )  P( X  Y )
Given : f ( x)  
0 ; otherwise P( X  Y )  0.4  0.5  0.4  0.5  0.7
P( X  4)  

f ( x)dx
9.13 (B)
1 4 Given : f ( x) and g ( x) are two probability
P( X  4)  

f ( x)dx   f ( x)dx
1
density functions,
 x
 a 1 ; a  x  0
4
1 1 3
P( X  4)   dx  ( x)14 
4 4 4 
1
 x
Hence, the correct option is (A). f ( x)    1 ; 0 xa
 a
9.11 54.5  0 ; otherwise

Given : Increasing order of spot speeds are, 
32, 45, 49, 51, 53, 56, 60, 62, 66, 79  x
Median = Middle value of observed data  a ;  a  x  0

Since, the number of observation are 10 i.e. even  x
number. and g ( x)   ; 0 xa
 a
Therefore, we have to take average of 5th and 6th  0 ; otherwise
observation. 

Therefore,
Mean of f ( x) is given by,

 6a3  8a3 2a3 a3
E  f ( x)   xf ( x ) dx E( X 2 )   
 12 12 6
 x2    x2  a3 a3
Var  f ( x)   0 
0 a
E  f ( x)     x  dx     x  dx 6 6
a   0 
a a
0 a
Variance of g ( x ) is given by,
 x3 x 2    x3 x 2 
E  f ( x)          Var  g ( x)   E ( X 2 )  [ E ( X )]2
 3a 2  a  3a 2 0

(  a )3 (  a ) 2 (  a 3 ) a 2 E( X 2 )   x g ( x) dx
2
E  f ( x)      
3a 2 3a 2
 x3
0 a
a2 a2 a2 a2 x3
E  f ( x)      E( X 2 )   a a dx  0 a dx
3 2 3 2
E  f ( x)  0  x4 
0
 x4 
a
E( X )       
2
Mean of g ( x ) is given by,  4a  a  4a 0


 a3  a3
E  g ( x)    xg ( x) dx E ( X 2 )  (0)     
  4 4
 x2
0 a
x2 2a 3 a 3
E  g ( x)   a dx  0 a dx E( X 2 )  
a 4 2
a3 a3
Var  g ( x)   0 
0 a
 x3   x3 
E  g ( x)         2 2
 3a  a  3a 0
 a2  a2 9.14 (A)
E[ g ( x)]      0
 3  3 Type II Errors means acceptance of the null
Variance of f ( x) is given by, hypothesis when it is false and should be rejected.
Var  f ( x)   E ( X 2 )  [ E ( X )]2
9.15 0.5

E( X 2 )  x
2
f ( x) dx 1 1 1
Probability of getting [H, H, H]   
 2 2 2
 x30
2
a
  x3  and the probability of getting [H, H, H, H]
E ( X )     x  dx   
2
 x 2  dx 1 1 1 1 1
a 
a  0
a      
2 2 2 2 16
0 a
 x 4 x3    x 4 x3  According to given question, (H, H, H) has already
E( X )      
2
  come in toss.
 4a 3   a  4 a 3 0
Condition is that to toss a coin one more time to
 a3 a3    a3 a3  obtain H again,
E( X 2 )         
 4 3  4 3 Hence, probability of getting ‘H’ again after (H,
H, H)
 2a 3 2a 3 Probability of getting ( H , H , H , H )
E( X 2 )   
4 3 Probability of getting ( H , H , H )
1/16 8 1 31 5
    0.5
1/ 8 16 2 29 4
Hence, the conditional probability according to the n  20
question is 0.5. Mean strength is given by,
9.16 (B) (23  4)  (28  2)  (22.5  5)  (31 5)  (29  4)

For the univariate exponential function, 20
Probability density function is given by,   26.575 MPa
 a e  ax ; x  0 Standard deviation is given by,
f ( x)  
 0 ; x0  ( f  ) 2

where, a is a parameter n 1
For Gaussian distribution, probability density 4(23  26.575) 2  2(28  26.575) 2
function is given by,
( x  )2 5(22.5  26.575) 2  5(31  26.575) 2
1 
f ( x)  e 2
2
4(29  26.575) 2
 2 
where,  and  are parameters. 19
Hence, the number of parameters in univariate 51.1225  4.06125  83.028
exponential distribution is 1 and for Gaussian 97.903  23.5225
distribution the parameters are 2. 
19
Hence, the correct option is (B).   3.697 MPa
9.17 (B) Thus,   3  26.575  3(3.697)  15.485 MPa
Given : f ( x)  a  bx, 0  x  1 Hence, the number of specimen with compressive
By probability density function, strength less than   3 is 0.
 9.19 (A)

 f ( x) dx  1
Given :
1
Hence,  (a  bx) dx  1
0
1
 bx 2 
 ax  2   1
 0
b
a  1
2 According to the given graph, h lies between 0 and
Only option (B) i.e. a  0.5 and b  1 satisfies 1, 2h lies between 1 and 2 and 3h between 2 and
the above equation. 3.
1 2 3
0
f ( x) dx  f ( x) dx  f ( x) dx  1
1 2
9.18 0 1st i nterval 2nd i nterval 3rd i nterval
Given : n  20 1 1
 h 1   2h  (2  1)
Number of specimens with 2 2
f (MPa) 1
compressive strength equal to f   3h  (3  2)  1
23 4 2
h 2h 3h
28 2   1
22.5 5 2 2 2
0.5h  1h  1.5h  1
3h  1
1
h
3
9.20 0.5
Given : Number of red balls in the box = 5 For a left skew curve (from the graph) l A  lB
Number of white balls in the box = 1
Mean < Mode
5
Probability of first ball to be red  Mean is less than mode and median.
6
Probability of second ball to be red (out of 9.22 17.9
4 The probability is of binominal distribution.
remaining 5 balls) 
5 n  number of times of occurrence of event = 5
Probability of third ball to be red (out of remaining r  number of times of favorable case
3 Probability of flood discharge
4 balls) 
4  P(r  1)  1  P(r  0)
Hence, the probability of consecutively picking 3  1  5 C0 p 0 q 5
red balls without replacement
 0.25
5 4 3 1
     0.5 So, q  0.944  1  p
6 5 4 2
Hence, the probability is 0.5. So, p  1  0.944  0.056
Hence, return period
9.21 (C)
1 1
In a right skew distribution curve, T   17.85  17.9 years
p 0.056
9.23 (B)
The probability density function of a continuous
random variable distributed uniformly between x
1
and y (for y  x ) is .
yx
From the above curve, mean is always right of the
peak. Mode comes at the peak and median comes
in between mean and mode.
Therefore, for a right skew curve (from the graph)
lB  l A
So, Mean > Mode Total area of P.D.F. is 1
Hence, the correct option is (C). F ( X )  ( y  x)  1
 Key Point 1
F(X ) 
Similarly, in a left skew distribution curve, mean yx
is always left of the peak. Mode comes at the peak Hence, the correct option is (B).
and median comes in between mean and mode.
9.24 0.196
1 1
Given : n  15, p  , q 
2 2
Let x  Number of heads 2
 3  33
Required probability is given by, p (sum  4)  C2   
3
 36  36
P( x  8)  n Cx p x q n  x 2
 3  33
8 7 p(sum  4)  3    
1 1  36  36
P ( x  8)  15C8    
2 2 p (sum  4)  0.019
15
C8 6435 9.26 B
P ( x  8)  15
  0.196
(2) 32768
Given : 30% of total vehicles are without
Hence, the probability of getting exactly 8 heads
pollution clearance certificate.
is 0.196.
9.25 0.019
Randomly choosen vehicle = 100
Given : A pair of dice is rolled thrice. Probability of selecting even vehicles on even
Sample space n( s )  36 (pair of dice). numbered dates =1
Event : Sum of the outcomes in each roll equals to  The number of vehicles expected without
4 in exactly two of the three attempts. pollution clearance certificate =
For pair of dice, 1000 1 0.3  30
Sum 2 3 4 5 6 7 8 9 10 11 12
Total Hence, the correct option is (B).
number of 1 2 3 4 5 6 5 4 3 2 1 9.27 0.9
outcomes
Given : P ( A)  0.5 and P ( B )  0.8
For sum  4, there are 3 outcomes.
P (at least A (or) B)  1  P(None of A and B)
By Binomial distribution,
p( s)  nCr ( p) r (q) n r , p  q  1  1  P( A) P( B )
3 3 33  1  (0.5) (0.2)  1  0.10  0.9
p , q  1 p  1 
36 36 36 Hence, the correct answer is 0.9.
2 3 2
 3   33 
p(sum  4)  3C2     
 36   36  
10 Numerical Methods
2013 IIT Bombay zero while finding the roots of the

equation,
10.1 The magnitude as the error (correct to two
decimal places) in the estimation of f ( x)  2  6 x  4 x 2  0.5x3
 x  10  dx using Simpson
integral
4
4 The correction, x to be added to xini the
0
first iteration is_____. [1 Mark]
1/3rd rule is _____. [Take the step length as
1] [2 Marks] 10.5 For step-size, x  0.4, the value of
2015 IIT Kanpur 1

following integral using Simpson’s rule
x2
3
10.2 The integral 
x1
x 2 dx with x2  x1  0 is is _____.
0.8
evaluated analytically as well as
 (0.2  25x  200 x
2
numerically using a single application of 0
the trapezoidal rule. If I is the exact value  675x3  900 x 4  400 x5 ) dx

of the integral obtained analytically and J [2 Marks]
is the approximate value obtained using
2016 IISc Bangalore
the trapezoidal rule, which of the following
statements is correct about their 10.6 Newton-Raphson method is to be used to
relationship? [1 Mark] find root of equation 3 x  e x  sin x  0 . If
the initial trial value for the root is taken as
(A) J  I
0.333, the next approximation for the root
(B) J  I
would be _____. [Note : Answer up to
(C) J  I three decimal] [1 Mark]
(D) Insufficient data to determine the 2017 IIT Roorkee
relationship
du
10.3 The quadratic equation x 2  4 x  4  0 is to 10.7 Consider the equation  3t 2  1 with
dt
be solved numerically, starting with the u  0 at t  0. This is numerically solved
initial guess x0  3 . The Newton - by using the forward Euler method with a
Raphson method is applied once to get a step size, t  2. The absolute error in the
new estimate and then the Secant method solution at the end of the first time step is
is applied once using in the initial guess ______. [2 Marks]
and this new estimate. The estimated value 2018 IIT Guwahati
of the root after the application of the 10.8 The quadratic equation 2 x 2  3x  3  0 is
Secant method is _______. [2 Marks] to be solved numerically starting with an
10.4 In Newton-Raphson iterative method, the initial guess as x0  2. The new estimate
initial guess value ( xini ) is considered as of x after the first iteration using Newton-
Raphson method is _____. [1 Mark]
2019 IIT Madras 10.12 A continuous function f ( x) is defined. If
10.9 The value of the function f ( x) is given at the third derivative at xi is to be computed
n distinct values of x and its value is to be by using the fourth order central finite-
interpolated at the point x * , using all the n divided-difference scheme (with step
points. The estimate is obtained first by the length = h), the correct formula is
Lagrange polynomial, denoted by I L and [2 Marks]
(A) f ( xi 3 )  8 f ( xi  2 )  13 f ( xi 1 )
then by the Newton polynomial, denoted
by I N . Which one of the following  13 f ( xi 1 )  8 f ( xi 2 )  f ( xi 3 )
8 h3
statements is correct? [1 Mark]
(B) f ( xi 3 )  8 f ( xi  2 )  13 f ( xi 1 )
(A) I L is always greater than I N
 13 f ( xi 1 )  8 f ( xi 2 )  f ( xi 3 )
(B) No definite relation exists between I L
8 h3
and I N (C)  f ( xi 3 )  8 f ( xi  2 )  13 f ( xi 1 )
(C) I L and I N are always equal  13 f ( xi 1 )  8 f ( xi 2 )  f ( xi 3 )
(D) I L is always less than I N . 8 h3
2020 IIT Delhi (D)  f ( xi 3 )  8 f ( xi  2 )  13 f ( xi 1 )
10.10 The true value of ln(2) is 0.69. If the value  13 f ( xi 1 )  8 f ( xi 2 )  f ( xi 3 )

8 h3
of ln (2) is obtained by linear interpolation
between ln (1) and ln (6) , the percentage of 2021 IIT Bombay
absolute error (round off to the nearest 10.13 The value of abscissa (x) and ordinate (y)
integer), is [1 Mark] of curve are as follows : [2 Marks]
(A) 35 (B) 69 x y
(C) 48 (D) 84 2.0 5.00
1 2.5 7.25
 (5x  4 x 2  3x  2) dx is
3
10.11 The integral 3.0 10.00
0 3.5 13.25
estimated numerically using three 4.0 17.00
alternative methods namely the By Simpsons 1/3rd rule, the area under the
rectangular, trapezoidal and Simpson’s curve (round off to two decimal places) is
rules with a common step size. In this _______.
1
 e dx using
context, which one of the following 10.14The value of x
the trapezoidal
0
statements is TRUE? [1 Mark]
(A) Simpson’s rule as well as rectangular rule with four equal subintervals is
rule of estimation will give NON-zero (A) 2.192 (B) 2.718
(C) 1.727 (D) 1.718
error.
[2 Marks]
(B) Only Simpson’s rule of estimation will
give zero error. 10.15 Numerically integrate, f ( x)  10 x  20 x 2
(C) Only the rectangular rule of estimation from lower limit a  0 to upper limit
will give zero error. b  0.5 . Use trapezoidal rule with five
(D) Simpson’s rule, rectangular rule as well equal subdivisions. The value (in units,
as trapezoidal rule of estimation will give round off to two decimal places) obtained
NON-zero error. is _______. [2 Marks]
f ( x)
2022 IIT Kharagpur Actual function
Approximated function
10.16Consider the following recursive interaction
scheme for different values of variable P
with the initial guess x1  1
(0,0)
1 P x
xn 1   xn   , n  1, 2,3, 4,5 xl xu
2 xn 
(A) the numerical value of the area obtained
For P  2, x5 is obtained to be 1.414, using the trapezoidal rule will be less
rounded-off to three decimal places. than the actual
For P  3, x5 is obtained to be 1.732, (B) the numerical value of the area obtained
rounded-off to three decimal places. using the trapezoidal rule will be more
If P  10, the numerical value of x5 is than the actual
________. (rounded-off to three decimal (C) the numerical value of the area obtained
places) [1 Mark] using the trapezoidal rule will be exactly
equal to the actual
2023 IIT Kanpur
(D) with the given details, the numerical
10.17A function f  x  , that is smooth and value of area cannot be obtained using
convex-shaped between interval  xl , xu  is trapezoidal rule
shown in the figure. This function is
observed at odd number of regularly spaced
points. If the area under the function is
computed numerically, then _________.
10.1 0.53 40

h 1
4
x  10  dx
4
Given : I   4
Estimated value,
0
 x  10  dx  10  266 
1 4
4 1
From Simpson’s rule,
3 rd 0 3
x0  nh
h  4 11  91  2  26  
 f ( x)dx  ( y0  yn )
 x  10  dx  245.33
x0
3 4
4
0
4( y1  y3  ...  yn1 ) 4
 x5 
2( y2  y4  .....  yn  2 )  Exact value = 0  x  10  dx    10 x 
4
4
5 0
h
 y0  y4  4( y1  y3 )  2( y2 )
a
 b
f ( x) 
3 45

 10  4  244.8
x 0 1 2 3 4 5
Magnitude of error
y  x 4  10 10 11 26 91 266
= Estimated value – Exact value
y0 y1 y2 y3 y4
= 245.33 – 244.8 = 0.53
10.2 (A)  5  25 1
f    10  4 
2 4 4
By Secant method,
 ( x1  x0 ) 
x2  x1    f ( x1 )
 f ( x1 )  f ( x0 ) 
5   1 
3
5 2 1   2  1
x2        
2  1 1  4 2   3  4
4   4 
5 1 15  1 7
Exact value is computed by integration which x2      2.334
2 6 6 3
follows the exact shape of graph while computing
Hence, the estimated value of the root is 2.334.
the area.
Whereas, in Trapezoidal rule, the lines joining  Key Point
each points are considered straight lines which is Secant method :
not the exact variation of graph all the time like as  xn  xn 1 
xn 1  xn    f ( xn )
shown in figure.  f ( xn )  f ( xn 1 ) 
Therefore, J  I
Hence, the correct option is (A). 10.4 0.33
 Key Point Given : f ( x)   2  6 x  4 x 2  0.5x3

(i) For the calculation of integral, trapezoidal Differentiating function f ( x) with respect to x,
rule is more exact if the function is
f '( x)  6  8x  1.5x2
polynomial of maximum degree 1. i.e. (0,
1). xini  0 [Initial value]
(ii) By Simpson rule we get more exact value of f ( xini )   2  0  0  0   2
integral if the polynomial is upto degree 3 By Newton-Raphson method,
 1 
 3rd is more exact upto degree 2  f ( xini )  2  1
x1  xini   0  
  f '( xini )  6  3
 
th
3
 and is more exact up to degree 3  1
x  x1  xini   0  0.33
 8 
3
10.3 2.334 Hence, the correction to be added to first iteration
is 0.33.
Given : f ( x)  x 2  4 x  4
10.5 1.367
Initial guess, x0  3
Given : a  0, b  0.8, x  0.4
Differentiating function f ( x) with respect to x,
f '( x)  2 x  4 f ( x)  0.2  25 x  200 x 2
f ( x0 )  f (3)  32  4  3  4  1 675 x3  900 x 4  400 x5
x 0 0.4 0.8
f '( x0 )  f '(3)  2
f ( x) 0.2 2.456 0.232
By Newton - Rapshon method,
y0 y1 y2
f ( x0 ) 1 5
x1  x0   3 
f '( x0 ) 2 2
1
From Simpson’s rule,
3 rd
x0  nh t t1
h
 f ( x)dx  ( y0  yn )  du   (3t 2  1)dt
x0
3 0 t0
4( y1  y3  ...  yn1 ) 3t 3

2
u  t   23  2  0  10
2( y2  y4  .....  yn  2 ) 3 0
0.8 For approximate value,

0.4
 f ( x) dx  ( y0  y2 )  4 y1  By Euler’s method,
3
0
un1  un  h f (tn , un )
0.8
0.4
 f ( x) dx  (0.2  0.232)  4  2.456 Put n  0,
3
0
u1  u0  2(3t02  1)
0.8

0
f ( x) dx  1.367 u1  0  2(3  02  1)  2
Thus after first iteration,
Hence, the value of the given integral is 1.367.
u  2 at t  t1  2
10.6 0.36
Hence, Absolute error
Given : f ( x)  3x  e  sin x x
= Exact value – Approximate value
Differentiating function f ( x) with respect to x,  10  2  8
f '( x)  3  e x  cos x Hence, the absolute error is 8.
Initial guess : 0.333 10.8 1
f (0.333)  3  0.333  e0.333
Given : f ( x)  2 x 2  3x  3 …(i)
 180 
 sin  0.333   x0  2
  
From Newton-Raphson formula,
f (0.333)   0.06926
f ( x0 )
 180  x1  x0  …(ii)
f '(0.333)  3  e0.333  cos  0.333   f '( x0 )
  
f '(0.333)  2.549 Differentiating function f ( x) with respect to x,
By Newton Raphson method, f '( x)  4 x  3 …(iii)
x1  x0 
f ( x0 ) Put the value of x0  2 in equation (i) and (iii),
f '( x0 )
f ( x0 )  2(2) 2  3(2)  3
( 0.06926)
x1  0.333   0.360 f ( x0 )  8  6  3  5
2.549
Hence, the next approximation for the root would and f '( x0 )  4(2)  3
be 0.36.
f '( x0 )  8  3  5
10.7 8
Put the values of f ( x0 ) and f '( x0 ) in equation
du
Given :  3t 2  1 with u  0 at t  0 (ii),
dt
f ( x0 )
and step size, t  2 x1  x0 
Absolute error f '( x0 )
= Exact value – Approximate value 5
x1  2   1
du 5
For exact value,  3t 2  1
dt Hence, the new estimate of x is 1.
Integrating both sides with respect to t,
10.9 (C) 10.13 20.67
4
For a given set of values of ( xi , f ( xi )) , the inter-
Area   f ( x)dx
polating polynomials obtained by Lagrange and 2
Newton interpolation may look different. But theta Numerical Integrations, by Simson’s 1/3rd rule
re identical. So, I L and I N are always equal. h
 ( y0  y4 )  4( y1  y3 )  2 y2 
Hence, the correct option is (C). 3
10.10 (C) 1
 [(5  17)  4(7.25  13.25)  2 10]
6
Given : True value ln(2) = 0.69
 20.67
True value ln(1) = 0
10.14 (C)
True value ln(6) = 1.79
1
Using method of interpolation, According to trapezoidal rule,  e x dx
1.79  0 0
Actual value   (2  1)  0.358
(6  1) ba
Given : n  4 , a  0 , b  1, h   0.25
True value  Actual value x
% error  100
True value x 0 1/4 1/2 3/4 1
0.69  0.358 y 1 e1/ 4 e1/ 2 e 3/ 4
e1
 100
0.69
 48.11%
1
h
I   e x  dx   y0  y4  2( y1  y2  y3 )
Hence, the correct option is (C). 0
2
10.11 (B) I  1.726
1
Given :  (5 x3  4 x 2  3x  2) dx 10.15 0.4
0
Given : a  0, b  0.5, n  5
3
Numerical integration using Simpson th gives Intercept,
8
 b  a  0.5  0
exact result for the polynomial upto degree three. h   0.1
Hence, the correct option is (B).  n  5
10.12 (D) x0  0 , y0  0
From the fourth order central finite-divided- x1  0.1 , y1  0.8

difference scheme with step size h, the third x2  0.2 , y2  1.2
derivative is given by,
x3  0.3 , y3  1.2
 yi 3  8 yi  2  13 yi 1
x4  0.4 , y4  0.8
3 y 13 yi 1  8 yi  2  yi 3
 x5  0.5 , y5  0
x 3 xi
8h3
Using trapezoidal formula numerical integration
  f ( xi 3 )  8 f ( xi  2 )  13 f ( xi 1 ) 
 13 f ( x )  8 f ( x )  f ( x  3)  h
 [( y0  y5 )  2( y1  y2  y3  y4 )]
f '''( xi )   i 1
3
i 2 i  2
8h
0.1
Hence, the correct option is (D).  [(0  0)  2(0.8  1.2  1.2  0.8)]
2
0.1 8
  0.4
2
10.16 3.162
x1  1
1 p
xn 1   xn   , n  1, 2,3, 4,5
2 xn 
For p  2, x5 is to be 1.414
For p  3, x5 is to be 1.732
If p  10, x5  ?
x1  1 for p  10
1 10 
x2   x1  
2 x1 
1  10 
x2  1  
2 1
x2  5.5
1 10 
x3   5.5  
2 5.5 
x3  3.6590
1 10 
x4   3.6590  
2 3.6590 
x4  3.19600
1 10 
x5   3.196  
2 3.196 
x5  3.162
10.17 (A)
Approximated function has under estimation so
numerical value of the area obtained using
trapezoidal rule will be less than the actual.
f ( x)
(0,0)
x
xl xu

11 Transform Theory
2016 IISc Bangalore 2019 IIT Madras

11.1 The Fourier series of the function, 11.2 The Laplace transfer of sin h(at ) is
f ( x)  0,   x  0 [1 Mark]
f ( x )    x, 0 x s s
(A) (B)
in the interval [  , ] is s  a2
2
s  a2
2
 2  cos x cos 3 x  a a
f ( x)    2   .... (C) 2 2 (D) 2
4  1 32
 s a s  a2
 sin x sin 2 x sin 3 x  2023 IIT Kanpur
    ....
 1 2 3  11.3 The following function is defined over the
interval [−𝐿, 𝐿] : f  x   px 4  qx 5
The convergence of the above Fourier series
at x  0 gives [1 Mark]
If it is expressed as a Fourier series,

1 2
(A)   
  x   x  
f  x   a0   an sin    bn cos   
2
n 1 n 6
n 1   L  L 

(1)n1 2
(B) 
n 1 n2

12
which options amongst the following are
true?

1 2 (A) an , n  1, 2,...,  depend on 𝑝
(C)  (2n 1) 2

8
(B) an , n  1, 2,...,  depend on 𝑞
n 1
n 1

(1) 
(D)   (C) bn , n  1, 2,...,  depend on 𝑝
n 1 (2n  1) 4
(D) bn , n  1, 2,...,  depend on 𝑞
11.1 (C)  cos (0)  1 and sin (0)  0

Given : f ( x)  0,   x  0 f (0 )  f (0 )  2  1 1 
  1  2  2  ... 
f ( x )    x, 0 x 2 4  3 5 
The Fourier series of the function is,   2 1 1 
  1  2  2  ... 
 2  cos x cos 3 x  2 4  3 5 
f ( x)    2   ....
4  1 32
  f (0 )    0   and f (0 )  0 
 sin x sin 2 x sin 3 x 
    .... 1 1 2
 1 2 3  1   ... 
32 52 8
Put x  0 in the given series, 
1  2
 2 1 1 1
f (0)    2  2  2  ... 
 
n 1 (2n  1)
2

8
4  1 3 5 
11.2 (C)
Formula of Laplace transform
a
L sin h(at )  2
s  a2
11.3 (B),(C)
 nx 
l
1
an  
l l
f ( x).sin 
 l 
 dx
1  4   nx 
l
   px  2 x  sin 
5
 dx = Depends on q
l l  even   l 
odd
 nx 
l
1
bn  
l l
f ( x).cos 
 l 
 dx
 nx 
l
1
  ( px 4  qx5 ).cos   dx = Depends on p
l l odd .  l 
even.
Hence, the correct options are (B) & (C).

GENERAL
APTITUDE
Syllabus : General Aptitude
Verbal Aptitude : Basic English grammar: tenses, articles, adjectives, prepositions,

conjunctions, verb-noun agreement, and other parts of speech Basic vocabulary: words,
idioms, and phrases in context Reading and comprehension Narrative sequencing.
Quantitative Aptitude : Data interpretation: data graphs (bar graphs, pie charts, and
other graphs representing data), 2 and 3 dimensional plots, maps, and tables Numerical
computation and estimation: ratios, percentages, powers, exponents and logarithms,
permutations and combinations, and series Mensuration and geometry Elementary
statistics and probability.
Analytical Aptitude : Logic: deduction and induction Analogy Numerical relations and
reasoning.
Spatial Aptitude : Transformation of shapes: translation, rotation, scaling, mirroring,
assembling, and grouping Paper folding, cutting, and patterns in 2 and 3 dimensions.
Contents : General Aptitude
S. No. Topics
1. Numerical Ability
2. Logical Reasoning
3. Verbal Ability
1 NUMERIC AL
CHAPTER
ABILITY
Contents :
S. No. Topics
1. Number System & Series
2. Average, Percentage & Ratio
3. Mixture Allegation & Direction
4. Permutation & Combination
6. Time, Speed & Distance
7. Work & Time
1 Number System & Series
2013 IIT Bombay (D) Cannot be determined

1.6 The last digit of (2171) 7 + (2172) 9
1.1 A number is as much greater than 75 as it is
smaller than 117. The number is + (2173)11 + (2174)13 is [2 Marks]
[1 Mark] (A) 2 (B) 4
(A) 91 (B) 93 (C) 6 (D) 8
(C) 89 (D) 96 2018 IIT Guwahati
2014 IIT Kharagpur
1.7 Consider a sequence of numbers a1 , a2 , a3 ,
1.2 Find the odd one in the following group 1 1
…. an where an = − , for each integer
ALRVX, EPVZB, ITZDF, OYEIK n n+2
[2 Marks] (n > 0). What is the sum of the first 50 terms?
(A) ALRVX (B) EPVZB [2 Marks]
(C) ITZDF (D) OYEIK  1 1
(A)  1 +  −
2015 IIT Kanpur  2  50
 1 1
1.3 How many four digit numbers can be formed (B)  1 +  +
with the 10 digits 0, 1, 2, ……..9 if no  2  50
number can start with 0 and if repetitions are  1  1 1 
(C) 1 +  −  + 
not allowed? [2 Marks]  2   51 52 
2016 IISc Bangalore  1 1 
(D) 1 −  + 
 51 52 
1.4 The sum of the digits of a two-digit number
1.8 For non-negative integers, a, b, c, what
is 12. If the new number formed by reversing
would be the value of a + b + c if
the digits is greater than the original number
log a + log b + log c = 0 ? [1 Mark]
by 54, find the original number.
(A) 3 (B) 1
[1 Mark]
(C) 0 (D) −1
(A) 39 (B) 57
+ a + a + ... + a
a
(C) 66 (D) 93 1.9 = a 2b and
n times
2017 IIT Roorkee + b + b + ... + a
b
= ab 2 , where a, b, n and m
1.5 What is the value of x when [1 Mark] m times
x+2 2 x+4 are natural numbers. What is the value of
 16  3
81×   ÷  = 144 m + m + m + ... + m   n
+ n + n + ... + n 
 25  5 
  ?
 n times  m times 
(A) 1
[1 Mark]
(B) −1 2 2 4 4
(A) 2a b (B) a b
(C) −2
Numerical Ability 1
(C) ab (a + b) (D) a 2 + b 2 [1 Mark]
3
1.10 A faulty wall clock is known to gain 15 (A) ( f ( x )) (B) ( f ( x )) 2 (C)
minutes every 24 hours. It is synchronized to ( f ( x )) 4 (D) f ( x)
the correct time at 9 AM on 11th July. What 2021 IIT Bombay
will be the correct time to the nearest minute
when the clock shows 2 PM on 15th July of 1.16 ⊕ and  are two operators on numbers p
the same year? [2 Marks] p2 + q2
and q such that p⊕q = &
(A) 12:45 PM (B) 12:58 PM pq
(C) 1:00 PM (D) 2:00 PM p2
pq = ; if x ⊕ y = 2  2, then x =
2019 IIT Madras q
1.11 On a horizontal ground, the base of a straight (A) y (B) 2y
ladder is 6 m away from the base of a vertical (C) y/2 (D) 3y/2
pole. The ladder makes an angle of 450 to [1 Mark]
the horizontal. If the ladder is resting at a 1.17 A function λ is defined by
point located at one – fifth of the height of ( p − q) 2 , if p ≥ q
λ ( p, q ) = 
the pole from the bottom, the height of the  p + q, if p < q
pole is _____ meters. [1 Mark] The value of expression
(A) 15 (B) 25 λ(−(−3 + 2),(−2 + 3))
(C) 35 (D) 30 is : [2 Marks]
(−(−2 + 1))
2020 IIT Delhi 16
(A) 0 (B)
1.12 Insert seven numbers between 2 and 34, such 3
that the resulting sequence including 2 and (C) 16 (D) −1
34 is an arithmetic progression. The sum of 2022 IIT Kharagpur
these inserted seven numbers is _______.
1.18 Both the numerator and the denominator of
[1 Mark]
(A) 130 (B) 120 3
are increased by a positive integer x and
(C) 124 (D) 126 4
1.13 The sum of two positive numbers is 100. 15
those of are decreased by the same
After subtracting 5 from each number, the 17
product of the resulting numbers is 0. One of integer. This operation results in the same
the original numbers is _______. value for both the fractions.
[1 Mark] What is the value of x?
(A) 90 (B) 80 (A) 3 (B) 4
(C) 95 (D) 85 (C) 1 (D) 2
1.14 The unit’s place in 26591749110016 is [1 Mark]
_______. [2 Marks] 2023 IIT Kanpur
(A) 9 (B) 3 1.19 Let a = 30!, b = 50! &c = 100! Consider
(C) 1 (D) 6 the following numbers :
1.15 If f ( x ) = x for each x ∈ (−∞, ∞), then
2
loga c, logc a, logb a, loga b
f ( f ( f ( x)))
is equal to _______. Which one of the following inequalities is
f ( x)
CORRECT?
2 Numerical Ability
(A) logca < logb a < loga b < loga c
(B) logca < loga b < logb a < logb c
(C) logca < logb a < logac < loga b

(D) logba < logc a < loga b < loga c
1.1 (D) (ii) Repetition is not allowed.

Let the unknown number be y. In thousandth place, 9 digits excepts 0 can be
According to question, placed.
y = 75 + x … (i) In hundredth place, 9 digits can be placed
y = 117 − x … (ii) (including 0, excluding the one used in thousandth
From equation (i) and (ii), place).
y = 96 In tenth place, 8 digit can be placed (excluding the
Hence, the correct option is (D). ones used in thousandth and hundredth place)
In ones place, 7 digits can be placed (excluding the
1.2 (D)
ones used in thousandth, hundredth and tenth
Given : ALRVX, EPVZB, ITZDF, OYEIK place)
. Method 1 : Total number of combinations,
Option (A) : It consists of only one vowel. = 9 × 9 × 8 × 7 = 4536
Option (B) : It consists of only one vowel. Hence, 4 digit numbers that can be formed are
Option (C) : It consists of only one vowel. 4536.
Option (D) : It consists of three vowels.
1.4 (A)
. Method 2 : Given :
(i) Sum of digits of two digit number is 12
Option (A) : A – L – R – V – X
A – L : 10 alphabets between A and L. (ii) Reverse of the two-digit number is greater
than original number by 54.
L – R : 5 alphabets between L and R.
R – V : 3 alphabets between R and V. . Method 1 :
V – X : 1 alphabet between V and X. The new number formed by reversing the digits is
Similar for option (B) and (C). greater than the original number is possible in
Option (D) : OYEIK options A and B only because reverse of 93 will
O – Y : 9 alphabets between O and Y. be a smaller number and reverse of 66 will be
Y – E : 5 alphabets between Y and E. similar to itself.
E – I : 3 alphabets between E and I. From option (A) :
I – K : 1 alphabet between I and K. Sum of two digits in the number = 3 + 9 = 12
Hence, the correct option is (D). After reversing the two digits number = 93
1.3 4536 The difference between the new number formed
and original number = 93 – 39 = 54
Given : Hence, the correct option is (A).
(i) No number can start with zero. . Method 2 :
Numerical Ability 3
Let original number be xy. Cyclicity of 1 is always 1.
Then, (10 x + y ) + 54 = 10 y + x Cyclicity of 2 is,
9 x + 54 = 9 y 21 → 2
9 x − 9 y = −54 22 → 4
x − y = −6 … (i) 23 → 8
Also, x + y = 12 … (ii) 24 → 6
From equations (i) and (ii), 25 → 2
x = 3, y = 9 Cyclicity of 3 is,
Hence, the correct option is (A). 31 → 3
. Method 3 : 32 → 9
Let the digit at one’s place will be x and ten’s place
33 → 7
will be (12 – x).
Hence, the original number = 10 (12 − x) + x 34 → 1
Cyclicity of 4 is,
= 120 − 10x + x
= 120 − 9x … (i) 41 → 4
The number obtained by interchanging the digits 42 → 6
= 10 x + 12 − x Unit digit of 21717 = 1
= 9 x + 12 Unit digit of 21729 = 2
As per given condition,
Unit digit of 217311 = 7
9 x + 12 = 54 + 120 − 9 x
9 x + 9 x = 174 − 12 Unit digit of 217413 = 4
18 x = 162 Therefore, the last digit of (2171) 7 + (2172) 9
x=9 + (2173)11 + (2174)13 = 1 + 2 + 7 + 4 = 14
From equation (i), Hence, the correct option is (B).
Original number = 120 − 9 × 9 = 120 − 81 = 39
1.7 (C)
Given :
1.5 (B)
Sequence of numbers a1 , a2 .....an
x+2 2 x+4
 16  3 1 1
Given : 81×   ÷  = 144 Where an = −
 25  5 n n+2
x+2
 4  2  3
2 x+4
1 1 1 1 1
 9 ×   ÷  = 144 So, a1 = 1 − , a2 = − , a3 = −
 5   5 3 2 4 3 5
2 2  1 1 1 1 1
 36  x + 2   3  x + 2  Sum =  1 −  +  −  + ...... + −
   ÷    = 144  3  2 4 50 52
  5    5  
 1 1 1  1 1 1 1 
 36 5 
x+ 2 Sum =  1 + + .....  −  + + ..... 
 2 3 50   3 4 5 52 
 5 × 3  = 12
1 1 1
12 x+ 2 = 12 Sum = 1 + − −
2 51 52
So, x + 2 = 1
1  1 1 
x = −1 Sum = 1 + −  + 
2  51 52 
1.6 (B)
4 Numerical Ability
1.8 (A) From 15th July 9 am to 15th July 2 pm, total
minutes gain =5 × 0.625 min = 3.125 min .
Given : log a + log b + log c = 0
log abc = 0 So, from 11th July at 9 am to 15th July 2 pm, clock
has gain 63.125 minutes, which means the correct
abc = 1
time is 12.57 pm or approximately 12:58 pm.
Since, a, b and c are non-negative numbers and
their product is 1, their sum can not be negative or
1.11 (D)
0.
So, option (C) and (D) are incorrect. According to given data, the figure can be drawn
Since, a, b and c are greater than equal to zero, as shown below,
their sum can not be 1.
So, if a + b + c = 1 + 1 + 1 = 3 [option (A)]
l
1.9 (B)
5
Given :
450
(i) a+ +  +a = a 2b
a 
6m
n times
n × a = a 2b Let length of pole be l (in meters).

l /5
n 2 a 2 = a 4b 2 … (i) Then, tan 450 =
6
(ii) +
b +  +b = ab 2
b 
l
m times
= 6  l = 30 m
m × b = ab 2 5
m 2b 2 = a 2b 4 … (ii)
1.12 (D)
m
+ +  +
m m × n
+ +  +n = m × n × n × m
n 
n times m times Let the sequence after inserting 7 terms is given
2 2
m
++  +
m m × n
+ +  +n = m n
n  by,
n times m times
2, A1 , A2 , A3 ,......, A7 , 34
Multiplying equation (i) and (ii),
So, the total number of terms of arithmetic
m 2 n 2 a 2b 2 = a 6b 6 progression, n = 9
m 2 n 2 = a 4b 4 T9 = 34
The nth term of the arithmetic progression is given
1.10 (B) by,
Given : Faulty clock gains 15 minutes every 24 a + (n − 1)d = 34
hours 2 + (9 − 1)d = 34
On 11th July at 9 am, clock is synchronized to the 8d = 32
correct time. d =4
From 11th July, 9 am to 15th July 9 am = 4 days Inserted sequence is as shown below,
Total minutes gain = 15 × 4 = 60 min = 1 hour . 6, 10, 14, 18, 22, 26, 30
15  6 + 30 
Every hour the clock gains min = 0.625 min Sum =   × 7 = 126
24  2 
Numerical Ability 5
1.13 (C) ∴ x 2 + y 2 − 2 xy = 0
Assume, two positive numbers are a and b, ( x − y )2 = 0
As per question, ∴ x = y satisfy the condition.
a + b = 100 … (i) Hence, the correct option is (A).
(a − 5)(b − 5) = 0 … (ii) 1.17 (A)
(a − 5)(b − 5) is zero only when (a − 5) or (b − 5) Given expression easily solved as,
is zero λ[−(− 3 + 2), (− 2 + 3)] λ (1, 1)
= = λ (1, 1)
Assuming, (a − 5) = 0 − (− 2 + 1) 1
a=5  p = q =1
b = 95
∴ λ(1, 1) = (1 − 1)2 = 0
1.14 (C)
1.18 (A)
110016 2k
(26591749) = (26591749) Number 1
Unit place = 9 = 9 2 even
= 81 3 3+ x
operation =
4 4+ x
So, the unit digit is 1 Number 2
Hence, the correct option is (C). 15 15 − x
operation =
Note : Cyclicity of 9 is (9, 1, 9, 1, ….) 17 17 − x
1.15 (A) 3 + x 15 − x
=
4 + x 17 − x
Given : f ( x) = x 2 (3 + x)(17 − x) = (15 − x)(4 + x)
f ( f ( x)) = ( x 2 )2 = x 4 ( x − 17)( x + 3) = ( x − 15)( x + 4)
f ( f ( f ( x)) = ( x 4 )2 = x 8 x 2 − 14 x − 51 = x 2 − 11x − 60
f ( f ( f ( x))) x8 9 = 3x
= 2 = x6 x=3
f ( x) x
= ( f ( x))3
1.19 (A)
logca ,logca , logba , logba
1.16 (A)
p2 + q2
Given : p ⊕ q =  m log m 
pq log n = log n 
 
x2 + y 2
∴ x⊕ y = …(i) log100! log 30! log 50! log 30!
xy , , ,
log 30! log100! log 30! log 50!
p2
and pq = logca < logb a < loga b < loga c
q
∴ 22 = =2 …(ii)
2
x2 + y2
=2
xy
6 Numerical Ability
2 Average, Percentage & Ratio
2013 IIT Bombay (A) 3-4 years (B) 4-5 years

(C) 5-6 years (D) 6-7 years
2.1 Following table provided figures (in rupees)
2.4 A foundry has a fixed daily cost of Rs 50,000
on annual expenditure of a firm for two years
whenever it operates and a variable cost of
2010 and 2011.
Rs 800Q, where Q is the daily production in
Category 2010 2011 tonnes. What is the cost of production in Rs
Raw material 5200 6240 per tonne for a daily production of 100
Power and fuel 7000 9450 tonnes?
Salary and wages 9000 12600 [1 Mark]
Plant and machinery 20000 25000 2.5 One percent of the people of country X are
Advertising 15000 19500 taller than 6 ft. Two percent of the people of
Research and country Y are taller than 6 ft. There are thrice
22000 26400 as many people in country X as in country Y.
development
Taking both countries together, what is the
percentage of people taller than 6 ft?
In 2011, which of the following two
[2 Marks]
categories have registered by same
(A) 3.0 (B) 2.5
percentage? [2 Marks]
(C) 1.5 (D) 1.2
(A) Raw material and Salary & wages
2015 IIT Kanpur
(B) Salary & wages and Advertising
(C) Power & fuel and Advertising 2.6 The exports and imports (in crores of Rs.) of
(D) Raw material and research and a country from the year 2000 to 2007 are
development given in the following bar chart. In which
2.2 A firm is selling its product at Rs. 60 per year is the combined percentage increase in
unit. The total cost of production is Rs. 100 imports and exports the highest?
and firm is earning total profit of Rs. 500. [2 Marks]
120
Later, the total cost increased by 30%. By 110 Exports Imports
what percentage the price should be 100

90
increased to maintain the same profit level. 80
70
[2 Marks] 60
50
(A) 5 (B) 10 40
(C) 15 (D) 30 30
20
100
2014 IIT Kharagpur 0
2000 2001 2002 2003 2004 2005 2006 2007
2.3 The population of a new city is 5 million and
2.7 Read the following table giving sales data of
is growing at 20% annually. How many
five types of batteries for years 2006 to 2012
years would it take to double at this growth
rate? [1 Mark]
Numerical Ability 1
Type Type Type Type Type [2 Marks]
Year
I II III IV V (A) 2 : 3 (B) 3 : 4
2006 75 144 114 102 108 (C) 6 : 7 (D) 4 : 3
2017 IIT Roorkee
2007 90 126 102 84 126
2008 96 114 75 105 135 2.10 If the radius of a right circular cone is
increased by 50% its volume increases by
2009 105 90 150 90 75
[1 Mark]
2010 90 75 135 75 90 (A) 75% (B) 100%
2011 105 60 165 45 120 (C) 125% (D) 237.5%
2012 115 85 160 100 145 2018 IIT Guwahati
2.11 A fruit seller sold a basket of fruits at 12.5%

Out of the following, which type of battery loss. Had he sold it for Rs.108 more, he
achieved growth between the years 2006 and would have made a 10% gain. What is the
2012? [2 Marks] loss in Rupees incurred by the fruit seller?
(A) Type V (B) Type III [2 Marks]
(C) Type II (D) Type I (A) 48 (B) 52
2016 IISc Bangalore (C) 60 (D) 108
2.12 The price of a wire made of a super alloy
2.8 (x % of y) + (y % of x) is equivalent to material is proportional to the square of its
_______. [1 Mark] length. The price of 10 m length of wire is
 xy  Rs. 1600. What would be the total price (in
(A) 2% of xy (B) 2% of  
 100  Rs.) of two wires of lengths 4 m and 6 m?
(C) xy % of 100 (D) 100% of xy [2 Marks]
2.9 Two finance companies, P and Q, declared (A) 768 (B) 832
fixed annual rates of interest on the amounts (C) 1440 (D) 1600
invested with them. The rates of interest 2019 IIT Madras
offered by these companies may differ from
2.13 Population of state X increased by x% and
year to year. Year-wise annual rates of
interest offered by these companies are the population of state Y increased by y%
from 2001 to 2011. Assume that x is greater
shown by the line graph provided below.
10 than y. Let P be the ratio of the population of
9.5
9 9 state X to state Y in a given year. The
8
8 8 8 percentage increase in P from 2001 to 2011
7 7.5
6.5 6.5 6 is ______.
4
[2 Marks]
2000 2001 2002 2003 2004 2005 2006 x 100 ( x − y )
(A) (B)
y 100 + y
If the amounts invested in the companies, P and Q, 100( x − y)
in 2006 are in the ratio 8 : 9, then the amount (C) x − y (D)
100 + x
s received after one year as interests from
companies P and Q would be in the ratio :
2 Numerical Ability
2020 IIT Delhi 2023 IIT Kanpur
2.14 The total expenditure of a family, on 2.18 In the given figure, PQRSTV is a regular
different activities in a month, is shown in hexagon with each side of length 5 cm. A
the pie-chart. The extra money spent on circle is drawn with its centre at V such
education as compared to transport (in that it passes through P. What is the area
percent) is _______. [1 Mark] (in cm2) of the shaded region? (The
diagram is representative)
V
P
Q T
(A) 100 (B) 50

R S
(C) 33.3 (D) 5 25π 20π
(A) (B)
2.15 The ratio of ‘the sum of the odd positive 3 3
integers from 1 to 100’ to ‘the sum of the (C) 6π (D) 7π
even positive integers from 150 to 200’ is 2.19 A square of side length 4 cm is given.
_______. [2 Marks] The boundary of the shaded region is
(A) 1 : 1 (B) 45 : 95 defined by one semi-circle on the top and
(C) 50 : 91 (D) 1 : 2 two circular arcs at the bottom, each of
2021 IIT Bombay radius 2 cm, as shown.
The area of the shaded region is
2.16 In a company, 35% of the employees drink
___________ cm2.
coffee, 40% of the employees drink tea and 4 cm
10% of the employees drink both tea and
coffee. What % of employees drink neither
tea nor coffee? [1 Mark]
(A) 25 (B) 35
(C) 40 (D) 15
2022 IIT Kharagpur
2.17 If p : q = 1: 2, q : r = 4 : 3, r : s = 4 : 5 and u 2 cm
is 50% more than s, what is the ratio p : u = ? (A) 8 (B) 4
[1 Mark] (C)12 (D) 10
(A) 2 : 15 (B) 16 : 45
(C) 1 : 5 (D) 4 : 25
Numerical Ability 3
2.1 (D) (iii) Initial total profit, PT = 500
According to given data, Total selling price = Total cost price + Profit T
Increase in raw material, SPT = CPT + PT
6240 − 5200
= ×100 = 20% SPT = 100 + 500 = 600
5200
Increase in power and fuel If total number of unit produced is x, then
9450 − 7000 x × 60 = 600
= ×100 = 35%
7000 x = 10
Increase in salary and wages 30
12600 − 9000 New cost price, CPT ' = 100 + 100 × = 130
= ×100 = 40% 100
9000
New profit, PT ' = 500
Increase in plant and machinery
25000 − 20000 SPT ' = CPT ' + PT '
= ×100 = 25%
20000 10 × SP ' = 130 + 500 = 630
Increase in advertising SP ' = 63
19500 − 15000 63 − 60
= ×100 = 30% Increase in selling price = ×100% = 5%
15000 60
Increase in research and development Hence, the correct option is (A).
26400 − 22000
= ×100 = 20% 2.3 (A)
22000
Category 2010 2011 % increase Given : Population of the city = 5 million
Raw material 5200 6240 20 Annual growth in population = 20%
Power and . Method 1 .
7000 9450 35
fuel After one year, population,
Salary and 20
9000 12600 40 x1 = 5 + 5 × = 6 million
wages
100
Plant and
20000 25000 25 After two years, population,
machinery
20 6
Advertising 15000 19500 30 x2 = x1 + x1 × = 6 + = 7.2 million
100 5
Research and
22000 26400 20 After three years, population,
development
Therefore, the two categories that have registered 20
x3 = x2 + x2 ×
by same percentage are raw material and research 100
and development. 7.2
x3 = 7.2 + = 8.64 million
2.2 (A) After four years, population,
20
Given : x4 = x3 + x3 ×
100
(i) Selling price per unit, SP = 60/unit
8.64
(ii) Total cost price, CPT = 100 x4 = 8.64 + = 10.368 million
5
4 Numerical Ability
Between third and fourth year, the population 2.4 1300
doubles i.e. becomes 10 million.
Given :
. Method 2 . (i) Fixed daily cost = 50, 000
The law of natural growth and decay is given by, (ii) Variable cost = 800Q
dx where, Q = daily production in tonnes.
∝x [For growth] …(i)
dt For daily production of 100 tonnes,
dx Variable cost = 800 ×100 = 80, 000
∝−x [For decay]
dt
Total cost
From equation (i),
= Fixed daily cost + Variable daily cost
dx dx
= kx  = k dt CT = 50, 000 + 80, 000
dt x
Integrating both sides, CT = 1,30, 000
ln x = kt + C
Total cost for 100 tonnes = 1,30, 000
…(ii)
where, k and C are constants. So, cost of production per tonne
x represents the population of the city and t is 1,30, 000
= = 1300 Rs.
time. 100
Initially ( t = 0 ), population is 5 million. Hence, the cost of production in Rs. Per tonne for
ln(5 ×106 ) = 0 + C a daily production of 100 tonnes is 1300 Rs.
C = 15.42 2.5 (D)
…(iii)
Given :
At t = 1 , population grows 20%.
(i) Country X has 1% of people taller than 6ft.
20
x ' = 5 + 5× = 6 million (ii) Country Y has 2% of people taller than 6ft.
100
(iii) Country X has thrice as many people in
From equation (ii), country Y.
ln x ' = kt + C Let population of Y be 100. Then X’s population is
6
ln(6 ×10 ) = k ×1 + 15.42 300.
k = 15.60 − 15.42 = 0.18 …(iv) Total number of people in both country = 400
Now, for population to double i.e. x = 10 million Total number of people taller than 6ft in both
, countries,
From equations (ii), (iii) and (iv), 1 2
X× +Y ×
T= 100 100
ln(10 ×106 ) = 0.18 × t + 15.42
400
0.18t = 0.69 1 2
300 × + 100 ×
t = 3.84 100 100
T=
Therefore, the population will double in between 400
three to four years. 5
%T = ×100% = 1.25%
Hence, the correct option is (A). 400
Numerical Ability 5
2.6 2006 % Increase in imports,
According to the given graph shown table can be 120 − 90

×100% = 33.34%
drawn, 90
In the year 2001, % Increase in exports,
% Increase in imports, 100 − 70
×100% = 42.8%
50 − 40 70
×100% = 25%
40 In the year 2007,
% Increase in exports, % Increase in imports = 0%
60 − 50 % Increase in exports,
×100% = 20%
50 110 − 100
×100% = 10%
In the year 2002, 100
% Increase in imports, % %
Year Total %
60 − 50 Increase Increase
×100% = 20% % increase
50 in imports in exports
% Increase in exports, 2001 25% 20% 45%
70 − 60 2002 20% 16.67% 36.67%
×100% = 16.67% 2003 16.67% −14.2% 2.47%
60
In the year 2003, 2004 14.2% 16.67% 30.87%
2005 12.5% 0% 12.5%
% Increase in imports,
2006 33.34% 42.8% 76.14%
70 − 60
×100% = 16.67% 2007 0% 10% 10%
60
% Increase in exports, From above table, it is clear that years 2006 is the
60 − 70 year with highest combined % increase in import
×100% = −14.2%
70 and export.
In the year 2004, 2.7 (D)
% Increase in imports, Final − Initial
Growth rate is given by = ×100%
80 − 70 Initial
×100% = 14.2%
70 From the given table,
% Increase in exports, Type I batteries,
70 − 60 115 − 75
×100% = 16.67% Growth rate = ×100% = 55.3%
60 75
Type II batteries,
In the year 2005,
85 − 144
% Increase in imports, Growth rate = ×100% = −40.97%
144
90 − 80 (negative sign indicates decay)
×100% = 12.5%
80 Type III batteries,
% Increase in exports = 0% 160 − 114
Growth rate = ×100% = 40.35%
In the year 2006, 114
6 Numerical Ability
Type IV batteries, 2.10 (C)
100 − 102 Given : Radius of cone is increased by 50%.
Growth rate = ×100% = −1.96%
102 Volume of cone is given by,
(negative sign indicates decay) 1
V = πr 2 h
Type V batteries, 3
Where, r = radius of circular surface of cone
145 − 108
Growth rate = ×100% = 34.25% h = height
108
Now, r is increased by 50%.
Therefore, Type I batteries has the highest growth So r ' = r + 0.5r = 1.5r
between years 2006 and 2012.
1 1
V ' = πr '2 h = π(1.5r ) 2 h
Hence, the correct option is (D). 3 3
2.8 (A) 1
V ' = 2.25 × πr 2 h
3
x y 1 1
( x % of y) + ( y % of x) = ×y+ ×x V ' = πr 2 h + 1.25 × ρr 2 h
100 100 3 3
xy + xy 2 xy V ' = V + 1.25 V
= =
100 100 Therefore, volume increases by 125%.
= 2 % of xy Hence, the correct option is (C).
Hence, the correct option is (A). 2.11 (C)

Let the cost price be x. The seller sold the fruits at
2.9 (D)
12.5 % loss.
From given graph, in year 2006, 12.5
So, Selling price = x − x ×
Rate of interest offered by P = 6% 100
Rate of interest offered by Q = 4% SP = 0.875 x
Let new selling price be SP' with 10 % Gain.
Ratio of amount invested in the companies is
So, SP ' = 0.875 x + 108
8 : 9 i.e. let 8x and 9x be amount invested by P
and Q. SP '− CP
Gain = ×100%
CP
Interest offered by P in one year,
0.875 x + 108 − x
8 x × 6 ×1 10 = ×100
IP = … (i) x
100 x
= −0.125 x + 108
Interest offered by Q in one year, 10
9 x × 4 ×1 2.25 x = 1080
IQ = … (ii)
100 x = 480 Rs
From equation (i) and (ii), 12.5% loss when cost price is 480.
I P 48 4 12.5
= = loss = 480 × = 60 Rs
I Q 36 3 100
Numerical Ability 7
2.12 (B) 2.14 (B)
Given : From the pie chart,

(i) Price is proportional to square of length of Money spent on education = 15 %
wire i.e. P ∝ L2 . Money spent on transport = 10 %
Money spent on eduction 15
(ii) P1 = 1600, L1 = 10 m, L2 = 4 m, L3 = 6 m = = 1.5
Money spent on transport 10
P1 L12 Money spent on education = 1.5 (Money spent on
=
P2 L22 transport)
1600 102 So, money spent on education is 50 % more than
= 2 money spent on transport.
P2 4
1600 ×16
P2 = = 256 2.15 (C)
100
Odd series : 1, 3, 5…99
P1 L12 1600 102
= 2  = 2 99 = 1 + (n − 1)2
P3 L3 P3 6
n = 50
1600 × 36
P3 = = 576  1 + 99 
100 Sodd =   × 50 = 2500
 2 
P2 + P3 = 256 + 576 = 832 Even series : 150, 152, 154,…200
Hence, the correct option is (B). 200 = 150 + (n − 1) 2
2.13 (B) n = 26
 150 + 200 
Let population of X in 2001 be 100% and Seven =   × 26 = 4550
 2 
population of Y in 2001 be 100%.
Therefore, the ratio is given by,
In 2011 population of X increased by x% i.e.
Sodd 2500 50
population of X = 100 + x. = =
Seven 4550 91
In 2011 population of Y increased by y% i.e.
population of Y = 100 + y.
Ratio of population of X and Y in 2001, 2.16 (B)
P=1 Given : Employees drink coffee = 35%

Ratio of population of X and Y in 2011, Employees drink tea = 40%
Employees drink both tea and coffee = 10%
100 + x
P' = So from above data we can easily sketch Venn
100 + y
diagram,
Percentage increase in
100 + x
−1 Coffee Tea
100 + y 100 ( x − y )
P= ×100 =
1 100 + y
Both
8 Numerical Ability
Each angle of regular hexagon,
Coffee Tea (n − 2)1800
= 1200
n
25 10 30
θ 1200 2 25π
Re quired Area = πR 2
= π5 =
3600 3600 3
Both Hence, the correct option is (A).
25 + 10 + 30 = 65% employees are those who
2.19 (A)
either takes coffee or tea or both.
∴ (100 − 65)% = 35% are those who neither 1
Bottom shaded area ( A ) = 2 × π(2) 2 = 2π
take coffee nor tea. 4
Hence, the correct option is (B). π× 22
Top shaded area = 4 × 2 − = 8 − 2π
2
2.17 (B)
Total shaded area = 8 − 2 π + 2 π = 8
p 1 Hence, the correct option is (A).
Given : =
q 2
q 4 
=
r 3
r 4
=
s 5
s 2
=
u 3
p=x
q = 2x
3x
r=
2
3x 4
=
2× s 5
15 x
s=
8
15 x 2
=
8u 3
45 x
u=
16
x
p:u = = 16:45
45 x
16
2.18 (A)
0
Sum of interior angles = (n− 2)180
Numerical Ability 9
3 Mixture Alligation & Direction
2014 IIT Kharagpur facing east. The ant first turns 720
anticlockwise at P, and then does the
3.1 X is 1 km North-East of Y. Y is 1 km South-
following two steps in sequence exactly
East of Z. W is 1 km West of Z. P is 1 km
South of W. Q is 1 km East of P. What is five times before halting. [2 Marks]
the distance between X and Q in km. 1. Moves forward for 10 cm
[2 Marks] 2. Turns 1440 clockwise
North
(A) 1 (B) 2
(C) 3 (D) 2
2015 IIT Kanpur East
3.2 Mr. Vivek walks 6 meters North-East, then

turns and walks 6 meters South-East, both
at 60 degrees to East. He further moves 2 The pattern made by the trace left behind
meters South and 4 meters West. What is by the ant is.
the straight distance in meters between the (A) T S
point he started from and the point he
finally reached? [1 Mark] U R
(A) 2 2 (B) 2
P Q
(C) 2 (D) 1/ 2
PQ = QR = RS = ST = TU = UP = 10 cm
2021 IIT Bombay
(B) U
3.3 On a planer field, you travelled 3 units east W T
from a point O. Next you travelled 4 units
P R
south to arrive at point P . Then you
travelled from P in north-east direction Q S
such that you arrive at a point that is 6 units SW = WR = RP = PT = TQ = QU = US =
east of point O. Next, you travelled in the 10 cm
North-West direction, so that you arrive at (C) S
point Q that is 8 units north of point P .
T R
The distance of point Q to point O, in
same units, should be ________.
P Q
[2 Marks]
PQ = QR =RS = ST =TP = 10 cm
(A) 6 (B) 4
(C) 5 (D) 3 (D) S
2022 IIT Kharagpur T R

3.4 An ant walks in a straight line on a plane
leaving behind a trace of its movement. P Q
The initial position of the ant is at point P SQ = QT = TR = RP = PS = 10 cm
General Aptitude 1
3.1 (C) From figure,
N ∠AOB = ∠OBA = 600
NE
Triangle OAB is an equilateral triangle because
W E ∠OAB + ∠ABO + ∠BOA = 1800
∠OAB + 600 + 600 = 1800
SE
S ∠OAB = 600
According to the given statements in the question,
Hence, OB = 6 m, MD = BC = 2 m
the figure is shown below,
W 1 km Z O
X OM = OB − MB = 6 − 4 = 2
900 900
1k
m 450 450 1 km From figure,
0
45
1 km
Y
1 km OD 2 = OM 2 + MD 2
OD2 = 22 + 22
P Q P’
According to Pythagoras theorem, OD = 2 2 m
In triangle XYZ, Hence, the correct option is (A).
ZX 2 = ZY 2 + YX 2 = 12 + 12 = 2 3.3 (C)
ZX = 2
From above figure,
ZX = P ' Q = 2
In triangle QXP ' ,
QX 2 = P ' X 2 + P ' Q 2 = 12 + ( 2)2
X
QX = 3
3.2 (A)
According to given data, the figure is shown
below,
N
The distance between point Q to point O is,

A
6m
OQ = OX 2 + XQ2
6m
W
O
600M 600 B
E OQ = 32 + 42 = 9 + 16
2m
D 4m C OQ = 25
OQ = 5 units
S Hence, the correct option is (C).
2 General Aptitude
3.4 (D)
S
144 0
0 R
36
720 108 0
720
1080
108 0
T
1080
1080
920
Q
P
1440
PS = SQ = QT = TR = RP = 10 cm

General Aptitude 3
4 Permutation & Combination
2018 IIT Guwahati
4.1 A three-member committee has to be

formed a group of 9 people. How many
such distinct committees can be formed?
(A) 2 (B) 9
[1 Mark] (C) 3 (D) 6
(A) 27 (B) 72 4.4 There are 4 red, 5 green, and 6 blue balls
(C) 81 (D) 84 inside a box. If N number of balls are
picked simultaneously, what is the
2020 IIT Delhi
smallest value of N that guarantees there
4.2 In a school of 1000 students, 300 students will be at least two balls of the same
play chess and 600 students play football. colour?
If 50 students play both chess and football, One cannot see the colour of the balls until
the number of students who play neither is they are picked.
_________. [2 Marks] (A) 4 (B) 15
(A) 200 (B) 150 (C) 5 (D) 2
(C) 50 (D) 100 4.5 Three husband-wife pairs are to be seated
at a circular table that has six identical
2023 IIT Kanpur
chairs. Seating arrangements are defined
4.3 In how many ways can cells in a 3× 3 only by the relative position of the people.
grid be shaded, such that each row and How many seating arrangements are
each column have exactly one shaded possible such that every husband sits next
cell? An example of one valid shading is to his wife?
shown. (A) 16
(B) 4
(C) 120
(D) 720
4.1 (D) Hence, the correct option is (D).
Given : 4.2 (B)
(i) Number of member in a committee = 3 Given : Total numbers of student = 1000

(ii) Number of people in group = 9 Number of students play chess = 300
Out of 9 people, number of ways to form three Number of students play football = 600
9
member committee = Cr = C3 = 84
n
Let x = number of students play neither chess
nor foot ball
General Aptitude 1
4.5 (A)
Given : n = 3
( n − 1)! = 2!
2
C1C2
From Venn diagram, 2 2

C5 C6 C3 C4
250 + 50 + 550 + x = 1000
x = 150 students
Hence, the correct option is (B). = 2 × 2 × 2 × 2 = 16
4.3 (B)
Given : 
Cells of 3 × 3 grid are to be shaded such that each
row and each column have exactly one shaded
cell. An example of one valid shading is given
below,
On the basis of the example, let’s take a 3 × 3 grid.

C1 C2 C3
In column C1 there are 3 ways to shaded a cell and

with reference to the condition, for column C2
there will be 2 ways to shaded a cell and in column
C3 there will be only 1 way to shaded a cell.
Hence, total number of ways will be 3 × 2 ×1 = 6 .
Such that each row and each column have exactly
one shaded cell.
4.4 (A)
Given : 4 Red, 5 Green and 6 Blue

We select three balls in worst case
1 Red, 1 Green and 1 Blue
If we select fourth ball then we found two balls are
of same colour.
2 General Aptitude
5 Probability & Statistics
2014 IIT Kharagpur 1 2

(A) (B)
9 9
5.1 In any given year, the probability of an 1 4
earthquake greater than Magnitude 6 (C) (D)
3 9
occurring in the Garhwal Himalayas is
0.04. The average time between successive 2021 IIT Bombay
occurrences of such earth quakes is 5.5 2 identical cube shaped dice each with
________ years. [1 Mark] faces numbered 1 to 6 are rolled
5.2 10% of the population in a town is HIV+. simultaneously. The probability that an
A new diagnostic kit for HIV detection is even number is rolled out on each dice is
available; this kit correctly identifies HIV+ [1 Mark]
individuals 95% of the time and HIV– 1 1
(A) (B)
individuals 89% of the time. A particular 4 12
patient is tested using this kit and is found 1 1
(C) (D)
to be positive. The probability that the 8 36
individual is actually positive is ________. 2023 IIT Kanpur
[2 Marks]
5.6 Which one of the options can be inferred
2015 IIT Kanpur about the mean, median, and mode for the
given probability distribution (i.e.
5.3 Four cards are randomly selected form a
probability mass function), P(𝑥), of a
pack of 52 cards. If the first two cards are
variable x?
kings, what is the probability that the third P( x)
card is a king? [1 Mark]
4
(A)
52
2
(B)
50
1  1 
(C) × 
52  52 
1  1   1 
(D) × ×  −17 −13 −9 −5 −1 1 5 9 13 17
52  52   50 
x
2017 IIT Roorkee (A) mean = median ≠ mode
(B) mean = median = mode
5.4 Two dices are thrown simultaneously. The
(C) mean ≠ median = mode
probability that the product of the numbers
(D) mean ≠ mode = median
appearing on the top faces of the dice is a
perfect square is [1 Mark]
General Aptitude 1
5.1 25 Number of kings left = 2
Given : Probability of an earthquake greater than 2

Probability that third card is a king =
magnitude 6 in any given year = 0.04 50
So, the average time between successive Hence, the correct option is (B).
occurrences of such earthquakes is given by,
5.4 (B)
1 1
T= = = 25 years
P 0.04 When two dices are thrown simultaneously, the
Hence, the average time between successive total number of combinations of the number that
occurrences of such earth quakes is 25 years. will be shown on faces of both the dices is given
by,
5.2 0.4896
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
Given :
Probability of being HIV+ in the town = 0.1. (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
Probability of being HIV– in the town = 0.9. (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
Probability that the kit identifies HIV+ correctly = (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
0.95.
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
Probability that the kit identifies HIV– correctly =
0.89. (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
Probability that the kit identifies HIV– incorrectly The combinations that are bold shows the numbers
= 0.11. whose product will give perfect square.
Probability that a person is found HIV+ using the Therefore, the required probability is given by,
diagnostic kit is given by,
8 2
P( B) = 0.1× 0.95 + 0.11× 0.9 P ( A) = =
36 9
P( B) = 0.194
According to Bayes’ theorem, the probability that
the individual is actually HIV+, when the kit gives 5.5 (A)
positive results,
E1 → occurrence of even number on first dice
 A  P ( A ∩ B ) 0.1× 0.95
P  = =
B P( B) 0.194 E2 → occurrence of even number on second dice
 A 3 1
P   = 0.4896 P ( E1 ) = =
B 6 2
Hence, the probability that the individual is 3 1
actually positive is 0.4896. P ( E1 ) = =
6 2
5.3 (B)
1 1 1
P( E1 ∩ E2 ) = P(E1 ) × P( E2 ) = × =
Given : Four cards are selected randomly, two of 2 2 4
which are kings.
Number of cards left = 50
2 General Aptitude
5.6 (A)
−17 −13 −9 −5 −1 1 5 9 13 17
Modian Mode
Mode
−1 + 1
Median = =0
2
Mode = –13, 13
So, Median ≠ Mode

General Aptitude 3
6 Time, Speed & Distance
2017 IIT Roorkee 2018 IIT Guwahati
6.1 Budhan covers a distance of 19 km in 2 hours 6.2 Tower A is 90 m tall and tower B is 140 m
by cycling one fourth of the time and tall. They are 100 m apart. A horizontal
walking the rest. The next day he cycles (at skywalk connects the floors at 70 m in both
the same speed as before) for half the time the towers. If a taut rope connects the top of
and walks the rest (at the same speed as tower A to the bottom of tower B, at what
before) and covers 26 km in 2 hours. The distance (in meters) from tower A will the
speed in km/h at which Budhan walks is rope intersect the skywalk?
[2 Marks] [1 Mark]
(A) 1 (B) 4
(C) 5 (D) 6
Tower
6.1 (D) B
Tower
Given : Budhan covers a total distance of 19 km A a
in 120 minutes by cycling for 30 minutes and
walking for 90 minutes. b q c
140 m
So, 30C + 90W = 19 km …(i) x
90 m
The next day, Budhan covers 26 km in 120

70 m
minutes by cycling for 60 minutes and walking for 70 m

q
60 minutes.
d e
So, 60C + 60W = 26 km …(ii) 100 m
Both days the speed of cycling and walking is the In above figure,
same. AD AB
tan θ = =
From equations (i) and (ii), DE BC
C = 0.33 km/min 90 20
=
W = 0.1 km/min 100 x
W = 0.1× 60 = 6 km/h 2000
x= = 22.22 m
6.2 22.22
Hence, the distance between tower A and rope-
skywalk intersect is 22.22 m.
According to given data figure is shown below,

General Aptitude 1
7 Work & Time
2016 IISc Bangalore 4 objects. What is the minimum time

needed to complete all the job?
7.1 Ananth takes 6 hours and Bharath takes 4
[2 Marks]
hours to read a book. Both started reading
(A) 2 hours (B) 2.5 hours
copies of the book at the same time. After
how many hours is the number of pages to (C) 3 hours (D) 3.5 hours
be read by Ananth, twice that to be read by 2019 IIT Madras
Bharath? Assume Ananth and Bharath
read all the pages with constant pace. 7.3 An oil tank can be filled by pipe X in 5
hours and pipe Y in 4 hours, each pump
[2 Marks]
working on its own. When the oil tank is
(A) 1 (B) 2 full and the drainage hole is open, the oil is
(C) 3 (D) 4 drained in 20 hours. If initially the tank
2017 IIT Roorkee was empty and someone started the two
pumps together but left the drainage hole
7.2 Two machines M1 and M2 are able to open, how many hours will it take for the
execute any of four jobs P, Q, R, and S. tank to be filled?
The machines can perform one job on one
[2 Marks]
object at a time. Jobs P, Q, R and S take 30
(Assume that the rate of drainage is
minutes, 20 minutes, 60 minutes and 15
independent of the Head)
minutes each respectively. There are 10
objects each requiring exactly 1 job. Job P (A) 4.00 (B) 1.50
is to be performed on 2 objects, Job Q on 3 (C) 2.00 (D) 2.50
objects, Job R on 1 objected and Job S on
7.1 (C) 12
Then, 1 hr = = 3 pages
4
Given :
From option (A) :
(i) Ananth takes 6 hours to read a book 1hrs
(ii) Bharath takes 4 hours to read a book
Ananth Bharath
L.C.M. = 12
Read Not read Read Not read
The number of pages read by Ananth and Bharath 2 pages 10 pages 3 pages 9 pages
must be 12 (or) multiple of 12 only. 10 : 9
If Ananth read 12 number of pages in 6 hrs, From option (B) :

12
Then, 1 hr = = 2 pages
6
If Bharath read 12 number of pages in 4 hrs,
General Aptitude 1
2hrs
(ii) Pipe Y fills the oil tank in 4 hours i.e. in one
Ananth Bharath 1
hour it fills th of tank.
Read Not read Read Not read 4
4 pages 8 pages 6 pages 6 pages
(iii) The drainage drains the tank in 20 hours i.e.
8:6
4:3 1
it drains th of tank in 1 hours.
From option (C) : 20
3hrs
(iv) If both pipes and drainage are open then,
Ananth Bharath 1 X Y 1
+ +
Read Not read Read Not read 5 4
6 pages 6 pages 9 pages 3 pages
2:1
Oil tank
Therefore, after 3 hours is the number of pages to
be read by Ananth, twice that to be read by
1
Bharath. -
20
In one hour, the tank will be will be filled by
7.2 (A)
1 1 1 8
+ − = part
Given : 5 4 20 20
(i) Number of machines = 02 i.e. M1 and M2 20
So, it will take hours i.e. 2.5 hours to fill the
(ii) Number of jobs = 04 i.e. P, Q, R and S 8
(iii) Number of objects = 10 tank.
(iv) Two objects need job P, 3 objects need job Hence, the correct option is (D).
Q, 1 objects needs job R and 4 objects need
job S. 
(v) Machines can perform one job at one object
at a time.
In the first hour, machine M1 will perform job P
on two objects (30 minutes each) and M2 will
perform job R on one object.
In the second hour, M1 will perform job Q on three
objects (20 minutes each) and M1 will perform job
S on four objects (15 minutes each). Therefore, all
jobs can be performed on all 10 objects within 2
hours (one job on each object).
7.3 (D)
Given :
(i) Pipe X fills the oil tank in 5 hours i.e. in one
1
hour it fills th of tank.
5
2 General Aptitude
Contents :
S. No. Topics
1. Data Interpretation
2. Syllogism
3. Numerical Computation
4. Miscellaneous
1 Data Interpretation
2013 IIT Bombay tons and the total revenues are 250 crore
rupees. What is the ratio of the revenue
1.1 Abhishek is elder to Savar. Savar is younger
to Anshul. generated through export of item 1 per
Which of the given conclusions is logically kilogram to the revenue generated through
valid and is inferred from the above export of the item 4 per kilogram?
statements? [2 Marks] [2 Marks]
(A) Abhishek is elder to Anshul
(B) Anshul is elder to Abhishek
(C) Abhishek and Anshul are of the same age
(D) No conclusions follows

2014 IIT Kharagpur
1.2 In a group of four children, Som is younger (A) 1:2 (B) 2:1
to Riaz. Shiv is elder to Ansu. Ansu is
(C) 1:4 (D) 4:1
youngest in the group. Which of the
following statements is /are required to find 1.4 Anuj, Bhola, Chandan, Dilip, Eswar and
the eldest child in the group? Faisal live on different floors in a six-
storeyed building (the ground floor is
[2 Marks] numbered 1, the floor above it 2, and so on).
Statements : Anuj lives on an even-numbered floor.
1. Shiv is younger to Riaz. Bhola does not live on an odd numbered
2. Shiv is elder to Som. floor. Chandan does not live on any of the
(A) Statement 1 by itself determines the floors below Faisal's floor. Dilip does not
eldest child. live on floor number 2. Eswar does not live
(B) Statement 2 by itself determines the on a floor immediately above or immediately
eldest child. below Bhola. Faisal lives three floors above
(C) Statements 1 and 2 are both required to Dilip. Which of the following floor-person,
determine the eldest child. combinations is correct? [2 Marks]
(D) Statements 1 and 2 are not sufficient to
determine the eldest child.
1.3 The total exports and revenues from the
exports of a country are given in the pie
charts below. The pie chart for exports
1.5 The monthly rainfall chart based on 50 years
shows the quantity of each item as a
of rainfall in Agra is shown in the following
percentage of the total quantity of exports.
figure. Which of the following are true? (k
The pie chart for the revenues shows the
percentile is the value such that k percent of
percentage of the total revenue generated
the data fall below that value)
through export of each item. The total
[2 Marks]
quantity of exports of all the items is 5 lakh
General Aptitude 1
Q-Telecom, T-Finance
(C) P- Home, Q-Power, T-Defense,
S-Telecom, U-Finance
(D) Q- Home, U-Power, T-Defense,
R-Telecom, P-Finance
1.8 The given question is followed by two
statements: select the most appropriate
option that solves the question
Capacity of a solution tank A is 70% of the
capacity of tank B. How many gallons of
solution are in tank A and tank B?
[2 Marks]
(i) On average, it rains more in July than in Statements :
December I. Tank A is 80% full and tank B is 40%
(ii) Every year, the amount of rainfall in full
August is more than that in January II. Tank A if full contains 14,000 gallons of
(iii)July rainfall can be estimated with better solution
confidence than February rainfall (A) Statement I alone is sufficient
(iv) In August, there is at least 500 mm of (B) Statement II alone is sufficient
rainfall (C) Either statement I or II alone is sufficient
(A) (i) and (ii) (B) (i) and (iii) (D) Both the statements I and II together are
(C) (ii) and (iii) (D) (iii) and (iv) sufficient
2016 IISc Bangalore
2015 IIT Kanpur
1.9 A shaving set company sells 4 different types
1.6 If ROAD is written as URDG, then SWAN
of razors, Elegance, Smooth, Soft and
should be written as [1 Mark]
Executive. Elegance sells at Rs. 48, Smooth
(A) VXDQ (B) VZDQ at Rs. 63, Soft at Rs.78 and Executive at Rs.
(C) VZDP (D) UXDQ 173 per piece. The table below shows the
1.7 The head of a newly formed government numbers of each razor sold in each quarter of
desires to appoint five of the six selected a year.
members P, Q, R, S, T and U to portfolios of Quarter/
Elegance Smooth Soft Executive
Home, Power, Defense, telecom and Product
Finance. U does not want any portfolio if S Q1 27300 20009 17602 9999
gets one of the five. R wants either Home or Q2 25222 19392 18445 8942
Finance or no portfolio. Q says that if S gets Q3 28976 22429 19544 10234
either Power or Telecom, then she must get Q4 21012 18229 16595 10109
the other one. T insists on a portfolio if P gets
one. Which product contributes the greatest
Which is the valid distribution of portfolio? fraction to the revenue of the company in
[2 Marks] that year? [2 Marks]
(A) P-Home, Q-Power, R-Defense, (A) Elegance (B) Executive
S-Telecom, T-Finance (C) Smooth (D) Soft
(B) R- Home, S-Power, P-Defense,
2 General Aptitude
1.10 Consider the following statements relating Which of the following statements are
of the level of poker play of four players P, correct? [2 Marks]
Q, R and S. (i) The elevator never moves directly from
I. P always beats Q any non-ground floor to another non-
II. R always beats S ground floor over the one hour period.
III. S loses to P only sometimes (ii) The elevator stays on the fourth floor for
IV. R always loses to Q the longest duration over the one hour
Which of the following can be logically period.
inferred from the above statements? (A) Only (i)
[2 Marks] (B) Only (ii)
(i) P is likely to beat all the three other (C) Both (i) and (ii)
players (D) Neither (i) nor (ii)
(ii) S is the absolute worst player in the set 1.14 P, Q, R, S, T and U are seated around a
(A) (i) only circular table. R is seated two places to the
(B) (ii) only right of Q. P is seated three places to the left
(C) (i) and (ii) of R. S is seated opposite U. If P and U now
(D) neither (i) nor (ii) switch seats, which of the following must
1.11 If ‘relftaga’ means carefree ‘otaga’ means necessarily be true? [2 Marks]
careful and ‘fertaga’ means careless which (A) P is immediately to the right of R.
of the following could mean ‘aftercare’? (B) T is immediately to the left of P.
[1 Mark] (C) T is immediately to the left of P or P is
(A) Zentaga (B) Tagafer immediately to the right of Q.
(C) Tagazen (D) Relffer (D) U is immediately to the right of R or P is
immediately to the left of T.
2017 IIT Roorkee
1.15 Students applying for hostel rooms are
1.12 Four cards lie on a table. Each card has a allotted rooms in order of seniority. Students
number printed on one side and a colour on already staying in a room will move if they
the other. The faces visible on the cards are get a room in their preferred list. Preferences
2, 3, red and blue. [1 Mark] of lower ranked applicants are ignored
Proposition : If a card has an even value on during allocation.
one side, then its opposite face is red. Given the data below, which room will Ajit
The cards which must be turned over to stay in [2 Marks]
verify the above proposition are Room
(A) 2, red (B) 2, 3, red Student Current
Names preference
(C) 2, blue (D) 2, red, blue seniority room
list
1.13 The points in the graph below represent the Amar 1 P R, S, Q
halts of a lift for durations of 1 minute, over Akbar 2 None R, S
a period of 1 hour. Anthony 3 Q P
Ajit 4 S Q, P, R
(A) P (B) Q
(C) R (D) S
1.16 The bar graph below shows the output of five
carpenters over one month, each of whom
General Aptitude 3
made different items of furniture : chairs, letters are positioned in the figure such that
tablet, and beds. [2 Marks] ( A  B  C ), ( B  G  E ) and ( D  E  F ) are
equal. Which integer among the following
choices cannot be represented by the letters
A, B, C, D, E, F and G? [2 Marks]
A D
B G E
C F
(A) 4 (B) 5
Consider the following statements : (C) 6 (D) 9
(i) The number of beds made by carpenter 1.19 Each of the letters in the figure below
C2 is exactly the same as the number of represents a unique integer from 1 to 9. The
tables made by carpenter C3. letters are positioned in the figure such that
(ii) The total number of chairs made by all each of ( A  B  C ), (C  D  E ),
carpenters is less than total number of ( E  F  G ) and (G  H  K ) is equal to 13.
tables. Which integers does E represent?
Which one of the following is true? [2 Marks]
(A) Only i
(B) Only ii
(C) Both i and ii
(D) Neither i nor ii
1.17 The temperature T in a room varies as a
(A) 1 (B) 4
function of the outside temperature T0 and
(C) 6 (D) 7
the number of persons in the room p,
according to the relation T  K (p  T0 ), 1.20 The annual average rainfall in a tropical city
is 1000 mm. On a particular rainy day (24-
where  and K are constants. What would be
hour period), the cumulative rainfall
the value of  , which gives the following
experienced by the city is shown in the
data? [1 Mark]
graph. Over the 24-hour period, 50% of the
T0 p T rainfall falling on a rooftop, which had an
25 2 32.4 obstruction-free area of 50 m2 was
harvested into a tank. What is the total
30 5 42.0
volume of water collected in the tank in
liters? [2 Marks]
(A) 0.8 (B) 1.0
(C) 2.0 (D) 10.0
1.18 Each of the letters arranged as below
represents a unique integer from 1 to 9. The
4 General Aptitude
2020 IIT Delhi
1.24 If 0, 1, 2, , 7, 8 ,9 are coded as O, P, Q,

, V, W, X, then 45 will be coded as
_______. [1 Mark]
(A) SU (B) ST
(C) SS (D) TS
1.25 Five friends P, Q, R, S and T went camping.
At night, they had to sleep in a row inside the
(A) 25,000 (B) 18,750 tent. P, Q, and T refused to sleep next to R
(C) 7500 (D) 3125 since he snored loudly. P and S wanted to
avoid Q as he usually hugged people in
2019 IIT Madras
sleep.
1.21 If E = 10; J = 20; O = 30; T = 40 , what will Assuming everyone was satisfied with the
be P + E + S + T ? [1 Mark] sleeping arrangements, what is the order in
(A) 51 (B) 120 which they slept? [2 Marks]
(C) 82 (D) 164 (A) RSPTQ (B) QRSPT
1.22 P, Q, R, S and T are related and belong to the (C) QTSPR (D) SPRTQ
same family. P is the brother of S. Q is the 1.26 After the inauguration of the new building,
wife of P, R and T are the children of the the Head of the Department (HoD) collated
siblings P and S respectively. Which one of faculty preferences for office space. P
the following statements is necessarily wanted a room adjacent to the lab. Q wanted
FALSE? [2 Marks] to be close to the lift. R wanted a view of the
(A) S is the sister-in-law of Q playground and S wanted a corner office.
(B) S is the aunt of R Assuming that everyone was satisfied, which
among the following shows a possible
(C) S is the brother of P
allocation? [1 Mark]
(D) S is the aunt of T
(A)
1.23 Mohan, the manager, wants his four workers
to work in pairs. No pairs should work for
more than 5 hours. Ram and John have
worked together for 5 hours. Krishna and
Amir have worked as a team for 2 hours.
Krishna does not want to work with Ram.
Whom should Mohan allot to work with
John, if he wants all the workers to continue (B)
working? [2 Marks]
(A) Amir
(B) Krishna
(C) Ram
(D) None of the three
General Aptitude 5
(C) (A) 8250 (B) 9750
(C) 8750 (D) 11250
1.28 For the year 2019, which of the previous
year’s calendar can be used? [1 Mark]
(A) 2014 (B) 2012
(C) 2011 (D) 2013
2022 IIT Kharagpur
(D) 1.29 Given the statements :
* P is the sister of Q
* Q is the husband of R
* R is the mother of S
* T is the husband of P
Based on the above information, T is
_______ of S. [1 Mark]
(A) The grandfather (B) An uncle
1.27 The monthly distribution of 9 Watt LED
bulbs sold by two firms X and Y from (C) The father (D) A brother
January to June 2018 is shown in the pie- 1.30 Healthy eating is a critical component of
chart and the corresponding table. If the total healthy aging. When should one start eating
number of LED bulbs sold by two firms healthy? It turns out that it is never too early.
during April-June 2018 is 50000, then the For example babies who start eating healthy
number of LED bulbs sold by the firm Y in the first year are more likely to have better
during April-June 2018 is _________. overall health as they get older. [2 Marks]
[1 Mark] Which one of the following is correct logical
inference based on the information in the
above passage?
(A) Eating healthy and better overall health
are more correlated at a young age, but
not old age.
(B) Eating healthy can be started at any age,
earlier the better.
(C) Healthy eating in more important for
Ratio of LED bulbs adult than kid.
Month sold by two firms (D) Healthy eating is important for those
(X:Y) with good health conditions but not for
January 7:8 others.
February 2:3
March 2:1
April 3:2
May 1:4
June 9:11
6 General Aptitude
y
1.31
16
14
14
12 11 O x
Frequency
10 9
8 7
6
4 y
4 3
2
2
0
3 4 5 6 7 8 9 O x
Marks
The above frequency chart shows the
frequency distribution of marks obtained by (A)
a set of student in an exam.
y
From the data presented above, which one of
the following is CORRECT?
(A) Mode > Mean > Median O x
(B) Median > Mode > Mean
(C) Mean > Mode > Median
(B)
(D) Mode > Median > Mean
[2 Marks] y
2023 IIT Kanpur
1.32 Consider a circle with its centre at the

O x
origin (O), as shown. Two operations are
allowed on the circle.
Operation 1: Scale independently along (C)
the x and y axes.
y
Operation 2: Rotation in any direction
about the origin.
Which figure among the options can be
achieved through a combination of these O x
two operations on the given circle?
(D)
General Aptitude 7
1.1 (D) Given : Total quantity of exports of all the items
= 5 lakh tons.
According to the question, 1 ton = 907.185 kg
5 lakh tons  4.53 108 kg
Total revenue generated = 250 crore rupees.
11% of total export belongs to item 1. Therefore,
the quantity of item 1 is given by,
11
Q1  4.53 108   4.98 107 kg
100
Revenue generated by item 1 is given by,
12
R1  250 107   3 108 Rs.
Since, there is no information given about the 100
relationship between Anshul and Abhishek. So no Hence, the revenue generated through export of
conclusion can be made from the given item 1 per kg
information. 3 108 3
  10  6
Hence, the correct option is (D). 4.98 10 7
5
1.2 (A) 22% of total exports belongs to item 4. Therefore,
the quantity of item 4 is given by,
Given : A group has 4 children. 22
Q4  4.53 108   9.96 107 kg
Som is younger to Riaz and Shiv is elder to Ansu. 100
Ansu is youngest in the group. Revenue generated by item 4 is given by,
Riaz Shiv 6
R4  250 107   15 107 Rs.
  100
Som Ansu (youngest) Hence, the revenue generated through export of
item 4 per kg
Statement 1 : Shiv is younger to Riaz.
Riaz 15 107 15
   1.5
 9.96 10 10
7
Shiv, Som Therefore, the ratio of revenue generated through

export of item 1 per kg to the revenue generated

through export of item 4 per kg is given by,
Ansu
6 4
Hence, it is clear from statement 1 that Riaz is the Ratio  
1.5 1
eldest children without the requirement of
statement 2.
Hence, the correct option is (A). 1.4 (B)
1.3 (D) Given : Total number of floors  06

Revenue : It is the income that a business has from (i) Anuj : Lives on an even-numbered floor.
its normal business activities, usually from the (2, 4, 6)
sales of goods and services to customers; money (ii) Bhola : Lives on an even-numbered floor (2,
regularly received by a company. 4, 6)
8 General Aptitude
(iii) Chandan : Lives on any of the floor above Statement 4 : In August, there is at least 500 mm
Faisal’s. of rainfall.
(iv) Dilip : Does not live on floor number 2. This statement may or may not be correct. Since,
(v) Eshwar : Does not live on any of the floor the graph is average of 50 years.
above or below bhola. Hence, the correct option is (B).
(vi) Faisal : Lives 3 floor above Dilip 1.6 (B)
According to options, Dilip lives either on floor 1 Given : ROAD is written as URDG.
or 3 and Faisal lives either on floor 4 or 5. If Dilip By following A – Z alphabetic chart :
lives on 1st floor then Faisal lives on 4th floor and
R – U : In between two alphabets
if Dilip lives on 3rd floor then Faisal lives on 6th
O – R : In between two alphabets
floor (not possible because Chandan lives on any
A – D : In between two alphabets
of the floor above Faisal’s, but there is only 6th
D – G : In between two alphabets
floors).
So, S T U V
Hence, Dilip  1st floor
W X Y Z
Faisal  4th floor
A B C D
From options, option (C) and (D) are incorrect.
N O P Q
Now, Chandan  5th floor Hence, the correct option is (B).
[From option (A) and (B)]
1.7 (B)
Eshwar  3rd floor
Since, Eshwar cannot be immediately above or Given :
below, Bhola, So Bhola must be living on 6th floor (i) There are 5 portfolios, home, power,
which means Anuj lives on 2nd floor. defense, telecom and finance.
Hence, the correct option is (B). (ii) 6 selected members P, Q, R, S, T and U are
to be appointed in 5 portfolios.
1.5 (B)
(iii) U does not want any portfolio if S gets one
Given : The monthly rainfall chart based on 50 of the five.
years of rainfall in Agra is shown below, (iv) R wants either Home or Finance or no
Statement 1 : On average, it rains more in July portfolio.
than in December. (v) Q says that if S gets either Power or
According to the graph, above statement is correct. Telecom, then she must get the other one.
T insists on a portfolio if P gets one.
Statement 2 : Every year, the amount of rainfall R wants Home or Finance or no portfolio. In
in August is more than that in January. option (A) and (D), R has Defense and Telecom.
The graph shows the average rainfall so it is So, both options are incorrect.
impossible to predict if every year amount of U does not want a portfolio if S gets one of the
rainfall in August is more than that in January. five. In option (C), S and U both have portfolios.
Statement 3 : July rainfall can be estimated with So, option (C) is also incorrect.
better confidence than February rainfall. Hence, the correct option is (B).
Since, the graph shows that the gap between 5 1.8 (D)
percentile and 95 percentile from average is higher
in February than in July. Therefore, July rainfall Given : Capacity of a solution tank A is 70% of
can be better estimated than February. the capacity of tank B.
General Aptitude 9
Statement I : Tank A is 80% full and tank B is R v/s Q Q Always R
40% full.
From the given data, it can be clearly inferred that
Statement II : Tank A if full contains 14,000 P always beats Q. Since, Q always beats R, P can
gallons of solution. also beat R. P only sometimes beats S. Hence, P is
Statement I can be used to solve the question if likely to beat all the other three player. So,
capacity of both tanks is already known and statement (i) can be logically inferred.
statement II can be used if it is known how much S only sometimes loses to P i.e. sometimes S wins
each tank is full/empty. as well, meaning S beats the best player
Therefore, by using both statements, sometimes. Hence, S can’t be absolute worst
Let capacity of tank B is x, player in the set.
70 So, statement (ii) cannot be logically inferred.
Then, x  14000
x  20000 gallons 1.11 (C)
Since, tank A was 80% full,
Given :
80
A 14000  11200 gallons (i) ‘relftaga’ means carefree.
100 (ii) ‘otaga’ means careful.
Since, tank B was 40% full, (iii) ‘fertaga’ means careless.
40 Here, it is clear that ‘taga’ means care and if taga
B  20000  8000 gallons
100 comes at the end in code language then meaning
Total solution  11200  8000  19200 gallons has ‘care’ at the front.
Hence, the correct option is (D). So, option (A) and (D) are incorrect.
1.9 (B) In ‘fertaga’ ‘fes’ means less.
So, ‘Tagafer’ means less care.
Elegance Smooth Soft Executive
Therefore, ‘Tagazen’ means aftercare.
27300 20009 17602 9999
25222 19392 18445 8942
1.12 (C)
28976 22429 19544 10234
21012 18229 16595 10109 To verify the proposition the least number of cards
that must be turned will be two.
102510 80059 72186 39284
(i) Card with visible face of a colour.
Rs. 48 Rs. 63 Rs. 78 Rs. 173
(ii) Card with visible face of a number.
102510 80059 72186 39284
Case 1 : Blue card and ‘2’ card :
48 63 78 173
If the blue card has number 2 on the other side then
= 4920480 = 5043717 = 5630508 = 6796132 proposition is incorrect but if the other side has
From the above table it is clear that, more revenue number 3 then next card has to be turned.
is obtained from executive. Now, if card ‘2’ has colour red on the other side
Hence, the correct option is (B). then proposition is correct but if other side has
1.10 (A)
colour blue then the proposition is wrong.
Case 2 : Red card and ‘3’ card :
Given : If the red card has number 2 on the other side then
Winner Loses we need a confirmation whether a number card has
P v/s Q P Always Q a red colour on the other side or blue colour. For
R v/s S R Always S that we must turn card ‘3’. If card ‘3’ has red
S v/s P P/S Sometimes P/S colour on the other side then the proposition is
10 General Aptitude
wrong but if on the other side has blue colour then (iv) In both figure, U is not immediately to right
the proposition is correct. of R and P is not immediately to the left of
Case - 2 does not match with any of the options. T.
Hence, the correct option is (C). Hence, the correct option is (C).
1.13 (D)
1.15 (B)
From the given graph it is clear that,
Given table is shown below,
(i) The halt at fourth floor is of 19 minutes over
the period of one hour and at the ground
floor the halt is of 21 minutes. So, Room
Student Current
statement (ii) is wrong. Names preference
seniority room
(ii) It is clear from the graph that the left moves list
directly from any non-ground floor to Amar 1 P R, S, Q
another non-ground floor over the one
hour period multiple times such as, Akbar 2 None R, S
(a) At t = 5, from 1st to 5th floor Anthony 3 Q P
(b) At t = 10, from 1st to 4th floor and so on. Ajit 4 S Q, P, R
So, statement (i) is also wrong. Amar is the most senior student and his first
Hence, the correct option is (D). preference is room R. So, he will move to room R.
Akbar is second in the seniority list. Since, R is
1.14 (C)
allotted to Amar, Akbar will get room S. Next in
line is Anthony, who will move to room P. So, Ajit
will get room Q.
1.16 (C)
Given figure is shown below,
Now, P and U switch seats,
(i) Number of beds made by carpenter C2
 20 12  8
Number of tables made by carpenter C3
 13  5  8
So, statement (i) is true.
(ii) Total number of chairs made by all
Fig. (a) Fig. (b) components
(i) In both figure,  C1  C2  C3  C4  C5
P is not immediately to the right of R.  2  10  5  2  4  23
Total number of tables made by all carpenters
(ii) In figure (a), T is not immediately to the left
 C1  C2  C3  C4  C5
of P.
 7  2  8  3  10  30
(iii) In figure (a), P is immediately to the right of So, statement (ii) is also true.
Q. Hence, the correct option is (C).
In figure (b), T is immediately to the left of
1.17 (B)
P.
Given : T  K ( p  T0 )
Table is shown below,
General Aptitude 11
T0 p T (i) Each of the letters represent a unique integer
from 1 to 9.
25 2 32.4 So, A  B  C......  K  45 … (i)
30 5 42.0 (ii) A  B  C  13 … (ii)
C  D  E  13 … (iii)
According to the given data, E  F  G  13 … (iv)
32.4  K   2  25  … (i) G  H  K  13 … (v)
Adding all four equation,
42  K   5  30  … (ii)
A  B  2C  D  2E  F  2G  H  K  52
From equation (i) and (ii), From equation (i),
42 5  30 C  E  G  52  45

32.4 2  25 C  E G  7 … (vi)
84   1050  162   972
and E  F  G  13
78   78 From option (A),
 1 E 1
Hence, the correct option is (B). Then, C  G  6
1.18 (B) and F  G  12
Prime factorization of digit 1 to 9 can be written Now, C and G should be less than 6.
as, Since, E  1 , C  1
1 1 22 So, C  2 , G  4 or C  4 , G  2
33 4  2 2 When C  2 and E  1 , then D has to be 10 for the
55 6  3 2 sum 13 which is not possible.
77 8  2 2 2 When C  4 and G  2 , then F has to be 10,
9  3 3 which is also not possible.
If A G  E  B  G  E  D  E  F Hence, option (A) is incorrect.
Will be equal only when prime factorization of all From option (B),
3 term should be equal. E4
But in between 1 to 9 we can observe that 5 and 7 Then, C  G  3 and F  G  9
are not present as a prime factor of any term. Now, C and G should be less than 3.
To become 5 and 7 as a part of these multiplication Let C  1 , G  2 or C  2 , G  1
there should be a common alphabet, but we can When C  1 , G  2  F  7 [Possible]
observe there is no any alphabet common in these C  D  E  13
three. 1  D  4  13
Therefore, 5 and 7 can’t be as a part of these D 8
multiplication. So, C  1 , D  8 , E  4 , F  7 , G  2
Since, 7 is not present in option. Now, A  B  C  13
So, 5 will be correct option. So, A  9 , B  3 or A  3 , B  9
Hence, the correct option is (B). G  H  K  13
1.19 (B) So, H  6 , K  5 or H  5 , K  6
Given : It is clear that every alphabet has unique number.
12 General Aptitude
1.20 (C) Ram but since it is clearly mentioned that Krishna
does not want to work with Ram, Krishna has to
Given : Annual average rainfall = 1000 mm
work with John and Amir will then work with Ram
Rainfall falling on a rooftop = 50%
for all workers to continue working.
Obstruction-free area = 50 m2
Cumulative rainfall (from the graph) = 300 mm
50% of the cumulative rainfall 1.24 (B)
50
  300  150 mm  150 103 m Codes given are as shown below
100
Total volume of water collected in the tank =
Obstruction-free area  50% of cumulative
rainfall
Total volume of water collected in the tank
 50 150 103  7.5 m3  7500 liters
1.21 (B)
Given : E  10 , J  20 , O  30 , T  40 So, the codes for 45  ST
Here, E  2  Position of E (5 )  10
th
J  2  Position of J (10th )  20 and so on. 1.25 (A)
Therefore, P  2  Position of P (16 ) = 32
th
Going through the options
S  2  Position of S (19th )  38 Option (A) : RSPTQ is correct choice because
So, P  E  S  T  32  10  38  40  120 both conditions are fulfilled in option (A)
Hence, the correct option is (B). 1. P, Q and T are not next to R.
1.22 (D)
2. P and S are not next to Q.
Given : Hence, the correct option is (A).
(i) Q is wife of P .
1.26 (C)
(ii) R is child of P and T is child of S .
Since, T is child of S , S cannot be aunt of T . Statement 1 : P wanted a room adjacent to the lab.
Hence, the correct option is (D). Hence, option (A) is incorrect.
1.23 (B) Statement 2 : Q wanted to be close to the lift.
Given : Hence option (B) is incorrect.
(i) No pair should work for more than 5 hours. Statement 3 : S wanted a corner office. Hence
(ii) Ram and John have worked together for 5 option (D) is incorrect.
hours. Here, option (C) satisfies all the statements.
(iii) Krishna and Amir have worked as a team for Hence, the correct option is (C).
2 hours.
1.27 (B)
(iv) Krishna does not want to work with Ram.
John cannot work with Ram anymore since his 5 From the given pie chart and corresponding table
hours have been completed with Ram. Now, John Sales
can either work with Krishna or Amir. If John Month X:Y X Y
(X + Y)
works with Amir, Krishna will have to work with
General Aptitude 13
April 3:2 15% 9% 6% 1.30 (B)
May 1:4 10% 2% 8% Eating healthy can be started at any age, earlier is
9: 5.5 better.
June 10% 4.5%
11 % The given statement is directly related to given
Total LED bulbs  50000 information.
The number of LED bulbs sold by the firm Y Hence, the correct option is (B).
during April - June 2018 is given by,
1.31 (D)
Y  19.5% of 50000
16
19.5 14
Y  50000  9750 14
100
12 11
Frequency
10 9
1.28 (D)
8 7
. Method 1 : 6
4
Non-leap year repeats after 6 or 11 years 4 3
2
2
Note : Above statement is valid only if leap year
is repeated every 4 years. 0
3 4 5 6 7 8 9
2013  6  2019 Marks
According to above chart, clearly mode is 7 marks
So, 2013 and 2019 has same calendar.
as mode is the mark whose frequency is highest
Total marks
. Method 2 : Mean 
Total students
(3  3)  (4  9)  (5 11)  (6  7)
 (7 14)  (8  2)  (9  2)
Mean 
(3  9  11  7  14  2  4)
292
Mean   5.84
50
Mode = 3 median – 2 mean
So, 2013 and 2019 has same calendar. 7 = 3 median – 2  5.84
Hence, the correct option is (D). 7  11.68
= median
1.29 (B) 3
= 6.72 median
Family Tree diagram can be made as :
 mode > median > mean
+ – + –
T P Q R
1.32 (A)
S

Here, T is an uncle of S.
14 General Aptitude
2 Syllogism
2015 IIT Kanpur (D) Neither (i) nor (ii)

2019 IIT Madras
2.1 There are 16 teachers who can teach
Thermodynamics (TD), 11 who can teach 2.4 In a sports academy of 300 people, 105
Electrical Sciences (ES), and 5 who can play only cricket, 70 play only hockey, 50
teach both TD and Engineering mechanics play only football, 25 play both cricket and
(EM). There are a total of 40 teachers, 6 hockey, 15 play both hockey and football
can not teach any of the three subjects, i.e. and 30 play both cricket and football. The
EM, ES and TD. 4 can teach ES and TD. rest of them play all three sports. What is
How many can teach both ES and EM but the percentage of people who play at least
not TD? [2 Marks] two sports?
(A) 1 (B) 2 [2 Marks]
(C) 3 (D) 4 (A) 25.00 (B) 28.00
(C) 23.30 (D) 50.00
2016 IISc Bangalore
2021 IIT Bombay
2.2 Fact 1 : Humans are mammals.
2.5 1. Some football players play cricket
Fact 2 : Some humans are engineers.
2. All cricket players play hockey
Fact 3 : Engineers build house.
Among the options given below, the
If the above statements are facts, which of statement that logically follows from the
the following can be logically inferred? two statements 1 and 2 above, is :
[2 Marks] [2 Marks]
I. All mammals build houses. (A) All football players play hockey
II. Engineers are mammals. (B) Some football players play hockey
III. Some humans are not engineers. (C) No football player plays hockey
(A) II only (B) III only (D) All hockey players play football
(C) I, II and III (D) I only 2.6 Statements: Either P marries Q or X
marries Y [2
2017 IIT Roorkee
Marks]
2.3 Consider the following sentences : Among the options below, the logical
All benches are beds. No bed is a bulb. NEGATION of the above statement is :
Some bulbs are lamps. (A) Neither P marries Q nor X marries Y.
Which of the following can be inferred? (B) P does not marry Q and X marries Y.
[1 Mark] (C) P marries Q and X marries Y.
(i) Some beds are lamps. (D) X does not marry Y and P marries Q.
(ii) Some lamps are beds. 2022 IIT Kharagpur
(A) Only (i) 2.7 A survey of 450 students about their
(B) Only (ii) subject of interest resulted in the following
(C) Both (i) and (ii) outcome.
General Aptitude 1
 150 students are interested in (A) Either conclusion III or conclusion IV
mathematics is
 200 students are interested in physics (B) Either conclusion I or conclusion II is
 175 students are interested in correct
chemistry (C) Only conclusion I is correct
 50 students are interested in (D) Only conclusion I and conclusion III
mathematics and physics are correct
 60 students are interested in physics
[2 Marks]
and chemistry
 40 students are interested in 2023 IIT Kanpur
mathematics and chemistry 2.9 A duck named Donald Duck says “All
 30 students are interested in ducks always lie.”
mathematics physics and chemistry Based only on the information above,
 Remaining students are interested in which one of the following statements can
humanities be logically inferred with certainty?
Based on the about information the
(A) Donald Duck always
number of students interested in
lies.
humanities is. [1 Mark]
(A) 10 (B) 45 (B) Donald Duck always tells the truth.
(C) 30 (D) 40 (C) Donald Duck’s statement is true.
2.8 Given below are two statements and four (D) Donald Duck’s statement is false.
conclusion drawn based on the statements? 2.10 Based only on the truth of the statement
Statement 1 : Some soaps are clean 'Some humans are intelligent', which one
Statement 2 : All clean objects are wet. of the following options can be logically
Conclusion I : Some clean objects are inferred with certainty?
soaps. (A) No human is intelligent
Conclusion II : No clean objects is a soap (B) All humans are intelligent
Conclusion III : Some wet object are soaps. (C) Some non-humans are intelligent
Conclusion IV : All wet object are soaps. (D) Some intelligent beings are humans
Which one of the following option can be
logically inferred.
2.1 (A) So, number of teachers who can teach both ES and
EM is 1.
According to the given data, Venn diagram can be
drawn as shown below,
2.2 (B)
From given facts, the following Venn diagram can
be drawn,
H = Humans M = Mammals
E = Engineers BH = Build houses
2 General Aptitude
2.6 (A)
As we are directed to do logical negation, of given
statement, situation of “either or” will becomes
“Neither nor”
as : PQ XY
From above diagram, only statement III is true.
Negation : PQ XY
2.3 (D)
2.7 (B)
According to the given data, Venn diagram is
Total = 450
shown below,
50
150 200
Phy
hs
Ma t
90 20 120
30
10 30
None of the given statements can be inferred
according to the Venn diagram. 105 60
Chem
2.4 (A)
175
According to the given data, Venn diagram can be
The number of students interested in humanities
drawn as shown below,
will be find as
90+120 +105 +20+30+10+30 = 405
450  405  45
2.8 (D)
Venn diagram can be made as
Number of people who play at least two sports
 (15  25  30) (2 sports) + 5 (3 sports)
 75 S C Cw
75
% of people  100  25%
300
Hence, the correct option is (A). (A) True
2.5 (B) (B) False
(C) True
Hockey
(D) False
Only conclusion I and III are correct.
Football Cricket
2.9 (D)
Hence, the correct option is (B). Given :
General Aptitude 3
A duck named Donald Duck says “All ducks
always lie.”
According to the given information Donald
Duck’s statement is false.
2.10 (D)
Given :
Statement ‘Some humans are intelligent’.
Can be represented by given venn diagram as
shown below,
Humans Intelligent
Option (A) No human is intelligent, is wrong as

some human are intelligent.
Option (B) All humans are intelligent, is wrong as
we have no information about remaining part of
human.
Option (C) Some non-humans are intelligent, is
wrong as we have no information about non-
humans.
Option (D) Some intelligent beings are humans,
can be inferred as some intelligent being are
human can be seen in the diagram.

4 General Aptitude
3 Numerical Computation
3.1 X and Y are two positive real numbers such 3.5 A function f ( x) is linear and has a value
that 2 X  Y  6 and X  2Y  8 . For of 29 at x   2 and 39 at x  3 . Find its
which of the following values of ( X , Y ) , value at x  5 . [1 Mark]
the function f ( x, y )  3X  6Y will give (A) 59 (B) 45
maximum value? [2 Marks] (C) 43 (D) 35
 4 10   8 20  3.6 Choose the most appropriate equation for
(A)  ,  (B)  ,  the function drawn as a thick line, in the
3 3  3 3 
plot below. [2 Marks]
 8 10   4 20 
(C)  ,  (D)  , 
3 3  3 3 
3.2 If | 4 X  7 | 5 then the values of
2 | X |  |  X | is [2 Marks]
1 1
(A) 2, (B) ,3
3 2
3 2
(C) ,9 (D) ,9
2 3
2014 IIT Kharagpur
(A) x  y  y (B) x    y  y 
(D) x    y  y 
3.3 The smallest angle of a triangle is equal to
(C) x  y  y
two thirds of the smallest angle of a
quadrilateral. The ratio between the angles 2016 IISc Bangalore
of the quadrilateral is 3:4:5:6. The largest
3.7 A cube is built using 64 cubic blocks of
angle of the triangle is twice its smallest
side and unit. After it is built, one cubic
angle. What is the sum, in degrees, of the
block is removed from every corner of the
second largest angle of the triangle and the
cube. The resulting surface area of the
largest angle of the quadrilateral?
body (in square units) after the removal is
[2 Marks]
______. [2 Marks]
3.4 If y  5x2  3 , then the tangent at x  0 , (A) 56 (B) 64
y3 [1 Mark] (C) 72 (D) 96
3.8 If f ( x)  2 x7  3x  5 , which of the
(A) Passes through x  0, y  0
following is a factor of f ( x) ? [2 Marks]
(B) Has a slope of 1
(A) ( x3  8) (B) ( x  1)
(C) Is parallel to the x-axis
(C) (2 x  5) (D) ( x  1)
(D) Has a slope of 1
General Aptitude 1
3.9 In a process, the number of cycles of
failure decreases exponentially with an
increase in load. At a load of 80 units, it
takes 100 cycles of failure. When the load
is halved, it takes 10000 cycles for failure.
The load for which the failure will happen
in 5000 cycles is ________.
[2 Marks]
(A) 40.00 (B) 46.02
(C) 60.01 (D) 92.02
3.10 A square pyramid has a base perimeter x
and the slant height is half of the perimeter. (i) y  2 x  4 for – 3  x   1
What is the lateral surface area of the (ii) y  x  1 for –1  x  2
pyramid? [2 Marks]
(iii) y  x  1 for –1  x  2
(A) x 2 (B) 0.75 x 2
(iv) y  1 for 2  x  3
(C) 0.50 x 2 (D) 0.25 x 2
(A) (i), (ii) and (iii) only
2017 IIT Roorkee (B) (i), (ii) and (iv) only
3.11 The following sequences of number is (C) (i) and (iv) only
arranged in increasing order : (D) (ii) and (iv) only
1, x, x, x, y, y,9,16,18, given that the mean log P log Q log R
3.14 Given that    10 for
and median are equal, and are also equal to yz zx x y
twice the mode, the value of y is x  y  z, what is the value of the product
[1 Mark] PQR? [2 Marks]
(A) 5 (B) 6 (A) 0 (B) 1
(C) 7 (D) 8 (C) xyz (D) 10 xyz
2018 IIT Guwahati 3.15 In manufacturing industries, loss is usually
taken to be proportional to the square of the
3.12 Hema’s age is 5 years more than twice of deviation from a target. If the loss is Rs.
Hari’s age. Suresh’s age is 13 years less 4900 for a deviation of 7 units, what would
than 10 times Hari’s age. If Suresh is 3 be the loss in Rupees for a deviation of 4
times as old as Hema, how old is Hema? units from the target? [2 Marks]
[1 Mark] (A) 400 (B) 1200
(A) 14 (B) 17 (C) 1600 (D) 2800
(C) 18 (D) 19
2019 IIT Madras
3.13 Which of the following function(s) is an
accurate description of the graph for the 3.16 A square has sides 5 cm smaller than the
range(s) indicated? [2 Marks] sides of a second square. The area of the
larger square is four times the area of the
smaller square. The side of the larger
square is ______ cm. [2 Marks]
(A) 15.10 (B) 8.50
(C) 18.50 (D) 10.00
2 General Aptitude
3.17 A retaining wall with measurements (A) 40 (B) 40
30 m 12 m  6 m was constructed with (C) 33 (D) 26
bricks of dimensions 8cm  6 cm  6 cm . If 3.22 Four persons P, Q, R and S are to be seated
60% of the wall consists of bricks, the in row. R should not be seated at the second
number of bricks used for the construction position from the left end of row. The
is _______ lakhs. [1 Mark] number of distinct seating arrangements
(A) 30 (B) 40 possible is : [1 Mark]
(C) 75 (D) 45 (A) 6 (B) 18
3.18 Suresh wanted to lay a new carpet in his
(C) 24 (D) 9
new mansion with an area of 70  55
3.23
sq.mts. However an area of 550 sq.mts.,
had to be left out for flower pots. If the cost
of carpet is Rs.50 per sq.mts., how much
money (in Rs.) will be spent by Suresh for
the carpet now? [1 Mark]
(A) Rs.1,65,000 (B) Rs.2,75,000
(C) Rs.1,92,500 (D) Rs.1,27,500 In the figure shown above, PQRS is a
2021 IIT Bombay square. The shaded portion is formed by
the intersection of sectors of circles with
3.19 In an equilateral triangle PQR, side PQ is radius equal to the side of square and
divided into 4 equal parts, side QR is center at S and Q.
divided into 6 equal parts and side PR is The probability that any point picked
divided into 8 equal parts. The length of randomly within square falls in shaded
each sub-divided part in cm is an integer. area is _______. [2 Marks]
The minimum area of triangle PQR 1 
possible, in cm2, is [2 Marks] (A) (B) 4 
2 2
(A) 18 (B) 144 3  
(C)  1 (D)
2 4
(C) 48 3 (D) 24 3.24 Four persons P, Q, R and S are to be seated
3.20 Four persons P, Q, R and S are to be seated in a row, all facing the same direction, but
in row. R should not be seated at the second not necessarily in the same order. P and R
position from the left end of row. The can not sit adjacent to each other. S should
number of distinct seating arrangements be seated to the right of Q. The number of
possible is : [1 Mark] distinct seating arrangements possible is :
(A) 6 (B) 18 [1 Mark]
(C) 24 (D) 9 (A) 4 (B) 2
(C) 8 (D) 6
3.21  and are 2 operators on numbers p
3.25 Consider two rectangular sheets, sheet M
and q such that p q  p  q and
and sheet N of dimensions 6cm  4 cm
p  q  p  q, then
each.
9 (6  7)   7  (6 5)   Folding operation 1 : The sheet is folded
into half by joining the short edges of the
[1 Mark]
current shape.
General Aptitude 3
Folding operation 2 : The sheet is folded (C) Area of PXWR = Area of SPQT – Area
into half by joining the long edges of the of RUVZ
current shape. (D) Area of PXWR = Area of RUVZ –
Folding operation 1 is carried out on Sheet Area of SPQT
M three times. 3.28 Consider a cube made by folding a single
Folding operation 2 is carried out on Sheet sheet of paper of appropriate shape. The
N three times. interior faces of the cube are all blank, the
The ratio of perimeters of the final folded exterior faces that are not visible in the
shape of Sheet N to the final folded shape above view may not be blank.
of Sheet M is _______. [2 Marks]
(A) 3 : 2 (B) 7 : 5
(C) 13 : 7 (D) 5 : 13
3.26 Which one of the following represents a
P possible unfolding of the cube?

T Q (A) (B)
S R
Five line segments of equal lengths, PR, (C) (D)
PS, QS, QT and RT are used to form a star – –
–
as shown in the figure above.
The value of  , in degrees, is ______. [2 Marks]
[2 Marks]
3.29 P invested Rs. 5000 per month for 6
(A) 36 (B) 108
months of a year and Q invested x per
(C) 72 (D) 45
month for 8 month of the year in a
2022 IIT Kharagpur partnership business. The profit is shared
3.27 In the following diagram the point R is the in proportion to the total investment made
center of circle. The line PQ and ZV are in that year. If at the end of that investment
tangential to the circle. The relation among year, Q receives 4/9 of the total profit, what
the areas of the squares, PXWR, RUVZ is the value of x (in Rs.).
and SPQT is [1 Mark] (A) 8437 (B) 3000
X
(C) 2500 (D) 4687
[2 Marks]
P 3.30 Two straight lines pass through the origin
S W
( x0 , y0 )  (0, 0) one of them passes
T Z through the point ( x1 , y1 )  (1, 3) and the
Q R
other passes through the point
( x2 , y2 )  (1, 2) . What is the area enclosed
U V
between the straight line in the interval (0,
(A) Area of SPQT = Area of RUVZ = Area
1) on the x-axis? [1 Mark]
of PXWR
(A) 1.5 (B) 0.5
(B) Area of SPQT = Area of PXWR – Area
of RUVZ (C) 2.0 (D) 1.0
4 General Aptitude
3.31 In a partnership business the monthly Line L4 : 6 x  9 y  6
investment by three friends for the first six Which one among the following is the
months is in the ratio 3:4:5 after six correct statement?
months, they had to increase their monthly (A) L2 is parallel to L4 and L2 is
investments by 10%, 15% and 20% perpendicular to L1
respectively, of their initial monthly (B) L1 is parallel to L2 and L1 is
investment. The new investment ratio was perpendicular to L3
kept constant for the next six months. (C) L3 is parallel to L4 and L3 is
What is the ratio of their shares in the total perpendicular to L2
profit (in the same order) at the end of the (D) L4 is perpendicular to L2 and L4 is
year such that the share is proportional to parallel to L3
their individual total investment over the [2 Marks]
year? 1 1 1
(A) 22:23:24 (B) 63:86:110 3.33 x: y:z  : :
2 3 4
(C) 22:33:50 (D) 33:46:60 xz y
[2 Marks] What is the value of ?
y
3.32 Consider the following equations of
(A) 1.25 (B) 3.25
straight lines :
(C) 0.75 (D) 2.25
Line L1 : 2 x  3 y  5
[1 Mark]
Line L2 : 3x  2 y  8
Line L3 : 4 x  6 y  5
3.1 (A) considered only because at one of these points the

value of function f ( X , Y )  3 X  6Y is
Given : f ( X , Y )  3 X  6Y
maximum.
where, X and Y are positive real numbers with
At point A (3, 0) :
2X  Y  6 …(i)
f ( X ,Y )  3 3  6  0  9
X  2Y  8 …(ii)
According to the given data following plot is  4 10 
At point B  ,  :
drawn below, 3 3 
4 10
f ( X , Y )  3   6   24
3 3
At point C (0, 4) :
f ( X , Y )  3  0  6  4  24
 4 10 
The maximum value is 24, at the point B  , 
3 3 
and C (0, 4) .
3.2 (B)
Feasible region  (O  A  B  C  O )
Only the points lying on the boundary of the Given : | 4 X  7 | 5
feasible region (i.e. point A, B and C) are Here, due to modulus, two possibilities are there,
General Aptitude 5
(i) 4X  7  5 Z  6 K  6  200  1200
X 3 W  3K  3  200  600
(ii) 4 X  7  5 From equation (i) and (ii),
X  0.5 2
C   600  400
At X  3 , 3
2 3  3  2  3  3  3 B  2  C  2  400  800
Since, A  B  C  1800
At X  0.5,
A  800  400  1800
2 0.5  0.5  2  0.5  0.5  0.5
A  600
So, A  Z  600  1200  1800
3.3 1800 Hence, sum of the second largest angle of triangle
Given : and the largest angle of the quadrilateral is 180 0 .
(i) Smallest angle of triangle is equal to two 3.4 (C)
third of smallest angle of quadrilateral.
Given : y  5x2  3
(ii) Ratio between angles of quadrilateral is
3:4:5:6
(iii) Largest angle of triangle is twice its smallest
angle.
dy dy
Slope of tangent    10 x
dx xy0,3 dx
dy
 0
dx xy0,3
Slope of the tangent is 0.
Let us assume
Equation of tangent is given by,
(i) Largest angle of triangle is B and smallest
y  y1  m( x  x1 )
angle is C
y  3  0( x  0)
(ii) Largest angle of quadrilateral is Z and
smallest is W . y3
(iii) Angles of quadrilateral are 3K, 4K, 5K and
6K.
2
Then, C  W … (i)
3
B  2C … (ii)
Since, W  X  Y  Z  3600
3K  4 K  5 K  6 K  3600
K  200 Therefore, the tangent is parallel to the x-axis.
6 General Aptitude
Hence, the correct option is (C). 3.7 (D)
3.5 (C) Given :
Given : (i) Total number of cubes  64
(i) f ( x) is linear. (ii) Total number of cubes that are removed from
each corner  08
(ii) At x  2, f ( x)  29
At x  3, f ( x)  39
Let f ( x)  mx  C ,
Then, at x  2,
f ( x)  m  2  C
29  2m  C … (i)
and at x  3,
f ( x)  m  3  C
39  3m  C … (ii) . Method 1 :
From equation (i) and (ii), Area of first face of cube after removal,
m  2, C  33  12 1  12 m2 [Existing surface]
So, at x  5, For six faces, area  12  6  72 m 2
After removing 8 cubes, visible surface area at
f ( x)  mx  C  2  5  33
each corner  03  8 1  24 m 2
f ( x )  43
Total surface area  72  24  96 m 2
Hence, the correct option is (C). Hence, the correct option is (D).
3.6 (B) . Method 2 :
Given plot is shown below, Removed cube will leave same area as they
covered before so still area will remain same
From above figure,
Area of one surface  16 unit 2
when y  1  x  2
and total number of surfaces  6
From option (A) :
Total surface area  16  6  96 unit 2
x y y Hence, the correct option is (D).
x  1  1   2  2 [Incorrect] 3.8 (B)
From option (B) : Given : f ( x)  2 x7  3x  5
x   y  y  y  From option (A) :
x  1  x3  8  0
 1  1   2 [Correct]
x3   8
From option (C) :
x  2
x y y
For x   2,
x  1  1  0  2 [Incorrect]
f ( x)  2 x 7  3 x  5
From option (D) :
f ( 2)  2( 2)7  3( 2)  5
x    y  y    
 1  1  f ( 2)   256  6  5  267
x   (1  1)  0  2 [Incorrect]
So, this is not a factor of f ( x)
General Aptitude 7
From option (B) : x
Slant height  l
x 1  0 2
x 1
For x  1,
f ( x)  2 x 7  3 x  5
f (1)  2(1)7  3(1)  5  5  5  0
So, ( x  1) is a factor of f ( x) .
3.9 (B)
Since, the number of cycles to failure decreases
exponentially with an increase in load. The general
equation is given by, Lateral surface area of pyramid
y  n e mx 1
  Base perimeter (p)  slant height (l)
where, y  number of cycle failure and x is load 2
Given : y  100 and x  80 1
A   pl
Therefore, 100  n e80 m …(i) 2
1 x
When load is halved, it take 10000 cycle for A   x
failure. 2 2
10000  n e 40 m …(ii) x2
A
From equation (i) and (ii), 4
100  e 40 m A  0.25 x2
ln100 Hence, the correct option is (D).
m …(iii)
40 3.11 (D)
For 5000 cycles to failure,
Given :
5000  n e x m …(iv)
(i) Series is 1, x, x, x, y, y, 9, 16, 18.
From equation (ii) and (iv),
(ii) Mean and median are equal.
2  e m ( x  40) (iii) Mean and median are equal to twice the
m( x  40)  ln 2 mode.
From equation (iii), 1  x  x  x  y  y  9  16  18
Mean 
ln 2 9
x  40 
m 3x  2 y  44
M
ln 2 9
x  40   40
ln100 Median is defined as the number at the centre of
0.693 the series.
x  40   40 Hence, Median  y
4.605
x  46.02 Mode is defined as the number which is most
Hence, the correct option is (B). repeated in the series.
Mode  x
3.10 (D)
Now, Mean  median  2  Mode
Base perimeter of square pyramid  x  p
8 General Aptitude
3x  2 y  44 At x  2, y  1 (Accurate)
 y  2x
9 (iii) y  x  1 for  1  x  2
3x  2 y  44
(i)  2x At x  1, y  1  1  0 [Inaccurate]
9
15 x  2 y  44 … (i) (iv) y  1 for  1  x  2 [Accurate]
3x  2 y  44 Hence, the correct option is (B).
(ii) y
9 3.14 (B)
3 x  7 y  44 … (ii) log P log Q log R
Given :    10
From equation (i) and (ii), yz zx x y
y  8, x  4 From the above equation,
Hence, the correct option is (D). log P  10 ( y  z )
3.12 (D) log Q  10 ( z  x)
log R  10 ( x  y )
Given :
(i) Hema  2  Hari  5 log P  log Q  log R 10( y  z )  ( z  x)  ( x  y)
(ii) Suresh  10  Hari 13 log P  log Q  log R  0

(iii) Suresh  3  Hema log ( PQR)  log1
Suresh  10  Hari 13  3 Hema Value of the product PQR 1
3  Hema  13 Hence, the correct option is (B).
Hari  … (i)
10 3.15 (C)
Hema  2  Hari  5 … (ii)
Given :
(i) For loss, L1  4900, deviation D1  7
 3  Hema  13 
Hema  2    5 (ii) For loss, L2  ?, deviation D2  4
 10
 6  26 (iii) Loss is proportional to the square of
Hema 1    5 deviation of target i.e.
 10  10
4 76 L  D2
Hema    Hema  19
10 10 L1 D12
So, 
Hence, the correct option is (D). L2 D22
3.13 (B) 4900 (7)2
 2
Given : Graph is shown below, L2 (4)
(i) y  2 x  4 for  3  x  1 L2  1600
At x  3, y  2 (Accurate) Hence, the correct option is (C).
At x  2, y  0 (Accurate)
3.16 (D)
Given :
(ii) y  x  1 for  1  x  2
At x  0, y  1 (Accurate)
At x  1, y  0 (Accurate)
General Aptitude 9
and A2  4 A1 …(i)
Area of A1  (a  5) 2 , Area of A2  a 2
From equation (i), 6
8
a  4(a  5)
2 2
a 2  4a 2  100  40 a
3a 2  40a  100  0
a  10, 3.33 4
Since, a cannot be less than 5, So a  3.33 is a a a
For  , ,  to be integer, a must be LCM of 4,
incorrect. 4 6 8
Hence, the correct option is (D). 6 and 8. So a = 24
3.17 (D) 3 2 3
Area  a   242  144 3
Given : 4 4
(i) Dimension of wall  30 m 12 m  6 m Hence, the correct option is (B).
Volume  2160 m3 3.20 (B)
(ii) Dimension of brick Number of arrangements  3  3!  18

 8 cm  6 cm  6 cm Hence, the correct option is (B).
Volume of brick 3.21 (B)
[9  (6  7)]  [7 1]  33  7  40
6
 288 cm  288 10 m
3 3
(iii) 60% of the wall consists of bricks. 3.22 (B)
60 Number of arrangements  3 3!  18
So, 2160   1296 m3
100 Hence, the correct option is (B).
Number of bricks used for construction is given
3.23 (C)
by,
1296
N 6
 45 105  45 lakhs
288 10
3.18 (A)
Given :
fA
(i) Area of mansion  70  55  3850 m2 Probability 
TA
(ii) Area for flower plot  550 m2  r 2 r 2 
fA     2
(iii) Area for carpet  3850  550  3300 m2  4 2
So, total money spent on carpet  r 2 r 2 
   2
 3300  50  Rs.165000 fA  4 2  
    1
2 
2
Hence, the correct option is (A). TA r
3.19 (B) Hence, the correct option is (C).
10 General Aptitude
3.24 (D) N
Condition 1 : P and R can not sit adjacent to each 6
other.
Condition 2 : S should be seated to the right of Q, 4 4
According to this all possible cases will be
1. Q P S R
6
2. Q R S P st
1 folding: (Respective of long edges)
3. P Q S R
4. R Q S P
6
5. P Q R S
6. R Q P S 2 2
6
3.25 (C) 2nd folding: (Respective of long side)
According to given data, we can proceed step by 6
step as given below, 1 1
M 6
3rd folding: (Respective of long side)
6 6
0.5 0.5
6
4 4 Perimeter of folded shape N = 13
Ratio of perimeters of the final folded shape N to
6 the final folded shape of sheet M is 13:7.
1st folding: (Respective of short edges) Hence, the correct option is (C).
3
3.26 (A)
Here, start shape, will all sequent are equal is
4 4
shown below,
3
nd
2 folding: (Respective of short side)
3
2 2
3  P  T  S  R  Q  
3rd folding: (Respective of short side)
(Equal sides of triangle have equal angle)
1.5
In TBR,
2 2
1.5
Perimeter of folded shape M = 7
General Aptitude 11
ExtB  2 …(i) 4
Profit sharing of Q 
( Sum of opposite interior angle = 9
Exterior angle) Profit sharing of ‘P’ = Capital of ‘P’
In QAS ,  Investment of ‘Time’
Profit sharing of ‘Q’ = Capital of ‘Q’
 Investment Time of ‘Q’
30000 5

8x 4
x = ~3000
3.30 (B)
ExtA  2 …(ii)
Given : Straight line 1  A(0, 0), B(1, 3)
In PAB ,
Straight line 2  A(0, 0), C (1, 2)
R B
3
2 C
Sum of all interior angles in a   1800. 1

reqd
   2  2  1800
A 1P
  360
3.27 (B)
Given :
X
Area  Area (set APBR)  ar (ACP )
S a P  ar (ARB )
W
a a2  r 2 1 1
a  (3 1)  (1 2)   3 1
Z 2 2
T a Q r R
r r  0.5
U r V 3.31 (B)
Ar ( SPQT)  a 2
Given :
Ar ( PXWR)  a 2  r 2 A B C
Ar ( RUVZ)  r 2 300x 400x 500x 1st 6 month
Ar ( SPQT)  Ar ( PXWR) – Ar ( RUVZ) 330x 460x 600x 2nd 6 month
Hence, the correct option is (B). Net profit of A = (300x + 330x)
3.28 (D) Net profit of B = (400x + 460x) 6
3.29 (B) Net profit of C = (500x + 600x) 6
Given : Capital of P = ~5000 A:B:C
Investment time of P = 6 months 630 : 860 : 1100
Investment time of Q = 8 months 63 : 86 : 110
12 General Aptitude
3.32 (D) 5k  3 5
 
12k 4
L1 : 2 x  3 y  5
L2 : 3x  2 y  8

L3 : 4 x  6 y  5
L4 : 6 x  9 y  6
for lines to be parallel,
a1 x  b1 y  c1  0
a2 x  b2 y  c2  0
a1 b1 c1
 
a2 b2 c2
4 6 5
here in L3 & L4 :  
6 9 6
⸫ L3 & L4 are parallel to each other
For lines to be perpendicular, product of slope of
lines  1
a
Slope of line 
b
3
Slope of L2 
2
6
Slope of L4 
9
3 6
Slope of L2  slope of L4    1
2 9
⸫ L2 & L4 are perpendicular to each other.
3.33 (A)
1 1 1
x: y: z : :
2 3 4
1 1 1
x: k ; y: k , z : k
2 3 4
1 1 1
k k k
xz y 2 4 3

y 1
k
3
6k  3k  4k
 12
k
3
General Aptitude 13
4 Miscellaneous
2021 IIT Bombay Without occupying any of the shaded

squares at the end of each step, the
4.1 minimum number of steps required to go
from P2 to P5 is
1 2 3 4 5
P X X
X X
Q
The mirror image of the above text about R Y
the x-axis is [1 Mark] S X X
T X X
(A) Example : Allowed steps for a person at Y
(B) (A) 4 (B) 5

(C) (C) 6 (D) 7
(D) [2 Marks]
4.4 For the picture shown above which one of
4.2
the following is the correct picture
Y
representing reflection with respect to the
mirror shown as the dotted line?
[1 Marks]
TRIANGLE
X
The mirror image of the above text about
the X-axis is [1 Mark]
(A) (B)
(C) (D)
2022 IIT Kharagpur (A)
4.3 In the square grid shown on the left, a

person standing at P2 position is required
to move to P5 position.
The only movement allowed for a step
involves, “two moves along one direction (B)
followed by one move in a perpendicular
direction”. The permissible directions for
movement are shown as dotted arrows in
the right.
For example, a person at a given position
Y can move only to the positions marked
X on the right.
General Aptitude 1
(C) and midway between the equator and the
north pole in the northern hemisphere. Let
P, Q, and R move with speeds vP , vQ , and
vR , respectively.
Which one of the following options is
(D) CORRECT?
(A) vP  vR  vQ (B) vP  vQ  vR
(C) vP  vR  vQ (D) vP  vR  vQ
2023 IIT Kharagpur
4.5 Consider a spherical globe rotating about

an axis passing through its poles. There are
three points P, Q, and R situated
respectively on the equator, the north pole,
4.1 (C) 4.4 (D)

4.2 (B)
4.3 (B)
By option elimination only option (C) can be
Mirrar
made, as after opening the given closed dice,  As the mirrar is placed on the right side of figure.
this side will be attached to left side of this of side. The reflection will be shown as
Start A B End
J
I H C
G D
F E
Farthe piece to go from start to end. The minimum
number of moves are represented by the path
mention belay
(A) ABC (B) HCD
(C) DFF (D) GHI
Hence minimum 5 moves are required.
2 General Aptitude
4.5 (C)
Spherical globe
We know velocity (V)  r , where   Constant
 V r
Hence, more is the distance from the axis of
rotation more will be the velocity.
So, VP  VR  VQ

General Aptitude 3
Contents :
S. No. Topics
1. Synonyms & Antonyms
2. Analogies
3. Sentence Completion
4. Idioms & Phrases
5. Grammar
6. Case Studies
1 Synonyms & Antonyms
2013 IIT Bombay 2017 IIT Roorkee
1.1 Which of the following options is the 1.5 There was no doubt that their work was
closest in meaning to the word given below thorough.
[1 Mark] Which of the words below is closest in
Primeval meaning to the underlined word above?
(A) Modern (B) Historic [1 Mark]
(C) Primitive (D) Antique (A) Pretty (B) Complete

(C) Sloppy (D) Haphazard
2014 IIT Kharagpur
2020 IIT Delhi
1.2 A student is required to demonstrate a high
level of comprehension of the subject, 1.6 Select the most appropriate word that can
especially in the social sciences. replace the underlined word without
changing the meaning of the sentence :
The word closest in meaning to
comprehension is [1 Mark] Now-a-days, most children have a
tendency to belittle the legitimate
(A) Understanding
concerns of their parents. [2 Marks]
(B) Meaning
(A) disparage (B) begrudge
(C) Concentration (C) reduce (D) applaud
(D) Stability
2023 IIT Kanpur
2015 IIT Kanpur
1.7 Kind: ______ : : Often : Seldom
1.3 Which word is not a synonym for the word (By word meaning)
vernacular? [1 Mark] (A) Cruel (B) Variety
(A) regional (B) indigenous (C) Type (D) Kindred
(C) indigent (D) colloquial 1.8 Eject : Insert :: Advance : _______. (By
1.4 The word similar in meaning to ‘dreary’ is word meaning)
[2 Marks] (A) Advent
(A) cheerful (B) dreamy (B) Progress
(C) Retreat
(C) hard (D) dismal
(D) Loan
1.1 (C) Historic : Something that is famous or important

in history.
Primeval : Earliest time in history.
Primitive : Early stage of development in history.
Modern : Recent times; something of the present
Antique : Very old and valuable.
time.
General Aptitude 1
Its synonyms are Ancient, Prehistoric, antique, 1.6 (A)
primitive and primordial.
Disparage : To depreciate or to lower in rank or
But the closest meaning of primeval is primitive
reputation.
because primitive means early stage in the historic
development. Begrudge : Displeasure, disapproval or envy.
Hence, the correct option is (C). Reduce : Decrease or diminish in size, amount,
extend or number.
1.2 (A)
Applaud : To express approval, praise.
Comprehension : Ability to understand Belittle : To speak slightingly of.
something. To depreciate or to seem little or less.
Hence, the correct option is (A). Option (A) is closest in meaning to the word
1.3 (C) belittle.
Vernacular : A form of language spoken by Hence, the correct option is (A).
particular group of people or region or country. 1.7 (A)
Indigenous : Something that naturally exists in a Given : Often is related to seldom in special
place or country rather than coming from another manner as they are opposite to each other in
place. meaning. In the same way of relationship kind will
Indigent : Poor, needy. be related to cruel as they are opposite to each
Colloquial : Language, word or expression which other in meaning. Hence, the correct option is (A).
is familiar, not formal. 1.8 (C)
Regional : Something that is from a particular
region or place. Given :
Eject is related to insert in a special manner as they
Except indigent, all other options are synonym of
are opposite in meaning for each other. In the same
vernacular.
way of relationship advance will be related to
Hence, the correct option is (C). retreat, as advance means forward and retreat
1.4 (D) means backward.
Dreary : Dull, unattractive, nothing of interest.
Cheerful : Excited, joyful

Dreamy : Having qualities such as gorgeous,
fabulous, incredible.
Dismal : Sad or without hope, dull.
1.5 (B)
The underlined word thorough means complete

with regard to every detail.
Pretty : Attractive, Lovely.
Sloppy : Careless, Negligent.
Haphazard : Unsystematic.
2 General Aptitude
2 Analogies
2015 IIT Kanpur 2.2 Select the word that fits the analogy :
Partial : Impartial : : Popular : ________
2.1 Select the pair that best expresses a
[1 Mark]
relationship similar to that expressed in the
(A) Unpopular (B) Impopular
pair : [1 Mark]
(C) Dispopular (D) Mispopular
Children : Pediatrician
2.3 Select the word that fits the analogy :
(A) Adult : Orthopaedist
Fuse : Fusion :: Use : _______
(B) Females : Gynaecologist
[1 Mark]
(C) Kidney : Nephrologist
(A) Usage (B) User
(D) Skin : Dermatologist
(C) Uses (D) Usion
2020 IIT Delhi
2.1 (B) Only option (B) expresses relationship similar to

the given pair.
Children : Pediatrician
Pediatricians are the doctors who are specialist in
treatment and care of children only. 2.2 (A)
Option (A) : Adult : Orthopedist Partial and impartial are opposite, in the same way
Orthopedists are the doctors who are specialist in popular and unpopular are opposite words.
treatment and care of bones. Hence, the correct option is (A).
[Includes men, women and kids] 2.3 (A)
Option (B) : Females : Gynecologist
Fuse : Fusion is a verb-noun pair.
Gynecologist are the doctor who are specialist in
Similarly, use : usage.
treatment of the female reproductive systems
[Specific for women]
Option (C) : Kidney : Nephrologist
Nephrologists are the doctors who are specialist in 
treatment of kidney related problems. [Includes
men, women and kids]
Option (D) : Skin : Dermatologist
Dermatologist are the doctors who are specialist in
treatment of skin.
[Includes men, women and kids]
General Aptitude 1
3 Sentence Completion
2013 IIT Bombay (B) a, was, found, at

(C) the, was, found, on
3.1 Friendship, no matter how _________ it is,
(D) a, being, find at
has its limitations. [1 Mark]
(A) cordial (B) intimate 2016 IISc Bangalore
(C) secret (D) pleasant 3.6 A rewording of something written or spoken

2014 IIT Kharagpur is a __________. [1 Mark]
(A) Paraphrase (B) Paradox
3.2 Choose the most appropriate word from the
(C) Paradigm (D) Paraffin
options given below to complete the
3.7 Archimedes said, “Give me a lever long
following sentence. [1 Mark]
enough and a fulcrum on which to place it
_____ is the key to their happiness; they and I will move the world. “The sentence
are satisfied with what they have. above is an example of a _________
(A) Contentment (B) Ambition statement. [1 Mark]
(C) Perseverance (D) Hunger (A) Figurative (B) Collateral
3.3 Choose the most appropriate word from the
(C) Literal (D) Figurine
options given below to complete the
following sentence. One of his biggest 2017 IIT Roorkee
_______ was his ability to forgive. 3.8 _______ with someone else’s email account
[1 Mark] is now a very serious offence.
(A) Vice (B) Virtues [1 Mark]
(C) Choices (D) Strength (A) Involving (B) Assisting
2015 IIT Kanpur (C) Tampering (D) Incubating
3.4 Extreme focus on syllabus and studying for 2018 IIT Guwahati
tests has become such a dominant concern of 3.9 The driver applied the _____ as soon as she
Indian students that they close their minds to approached the hotel where she wanted to
anything _________ to the requirements of take a_______? [1 Mark]
the exam. [1 Mark]
(A) Brake, Break (B) Break, Break
(A) related (B) extraneous
(C) Brake, Brake (D) Break, Brake
(C) outside (D) useful
3.10 “It is no surprise that every society has had
3.5 The Tamil version of ____________ John
codes of behavior; however, the nature of
Abraham-starrer Madras Cafe _________
these codes is often __________.”
cleared by the Censor Board with no cuts last
[1 Mark]
week, but the film’s distributors __________
The word that best fills the blank in the
no takers among the exhibitors for a release
above sentence is
in Tamil Nadu _________ this Friday.
(A) Unpredictable (B) Simple
[1 Mark]
(C) Expected (D) Strict
(A) Mr., was, found, on
General Aptitude 1
3.11 “Although it does contain some pioneering 2021 IIT Bombay
ideas, one would hardly characterize the
3.18 The author said, “Musicians rehearse before
work as _______.”
their concerts. Actors rehearse there roles
The word that best fills the blank in the
before the opening of new play. On the other
above sentence is [1 Mark]
hand, I find it strange that many public
(A) Innovative (B) Simple
speakers think they can just walk on the
(C) Dull (D) Boring stage and start speaking. In my opinion, it is
3.12 “His face _______ with joy when the no less important for public speakers to
solution of the puzzle was _______ to him.” rehearse their talks.”
[1 Mark] Based on the above passage, which one the
The words that best fill the blanks in the following is TRUE? [2 Marks]
above sentence are (A) The author is of the opinion that
(A) Shone, shown (B) Shone, shone rehearsing is more important only for
(C) Shown, shone (D) Shown, shown musicians than public speakers.
2019 IIT Madras (B) The author is of the opinion that
rehearsing is less important for public
3.13 The lecture was attended by quite _______
speakers than for musicians and actors.
students, so the hall was not very ____.
(C) The author is of the opinion that
[1 Mark]
rehearsing is important for musicians, actors
(A) Few, quite (B) A few, quite
and public speakers.
(C) Few, quiet (D) A few, quiet
(D) The author is of the opinion that
3.14 Daytime temperatures in Delhi can _______
rehearsing is more important for actors than
40 0 C . [1 Mark]
musicians.
(A) get (B) peak
3.19 Human have the ability to construct worlds
(C) reach (D) stand
entirely in their minds, which don’t exist in
3.15 The growth rate of ABC Motors in 2017 was the physical world. So far as we known, no
the same _______ XYZ motors in 2016. other species possess this ability. This skill
[1 Mark] is so important that we have different words
(A) as that off (B) as those of to refer to its different flavors, such as
(C) as off (D) as that of imagination, invention and innovation.
2020 IIT Delhi Based on above passage, which one of the
following is TRUE? [2 Marks]
3.16 His hunger for reading is insatiable. He reads
indiscriminately. He is most certainly (A) The terms imagination, invention and
a/an_______ reader. [2 Marks] innovation refer to unrelated skills.
(A) voracious (B) precocious (B) No species possess the ability to
construct worlds in their mind.
(C) wise (D) all-round
(C) Imagination, invention and innovation
3.17 It is a common criticism that most of the
are unrelated to the ability to construct metal
academicians live in their _______, so, they
worlds.
are not aware of the real life challenges.
(D) We do not know of any species other
[2 Marks]
than humans who possess the ability to
(A) glass palaces (B) homes
construct mental worlds.
(C) big flats (D) ivory towers
2 General Aptitude
2023 IIT Kanpur (A) across (B) of
(C) between (D) about
3.20 The line ran ______ the page, right through
the centre, and divided the page into two.
3.1 (B) The given sentence states that due to extreme

focus on syllabus and tests, students have closed
Cordial : Warm and friendly
their minds.
Use of ‘no matter how’ in the sentence indicates
Now, option (A) is incorrect because students are
that even if the friendship has a very close bond or
focused on studies so they can’t close their minds
intimate bond, it still has its limitations. Therefore,
to anything related to the requirements of the
intimate is the appropriate word to complete the
exam.
sentence.
Hence, the correct option is (B). Extraneous means irrelevant or unrelated to the
subject being dealt with. It is perfectly suitable and
3.2 (A)
correct because students have closed their minds
Contentment : State of complete satisfaction. to anything that is unrelated to the requirements of
Ambition : Desire to achieve something. the exam.
Perseverance : Firmness, tenacity or the Hence, the correct option is (B).
persistence.
3.5 (C)
Hunger : Having a strong desire to eat.
Given statement is made of two sentences. The The Tamil version of the John Abraham starrer
blank space will be filled with the word which Madras cafe was cleared by the Censor board with
explains the reason of happiness in the first no cuts last week, but the film’s distributors found
sentence. The second sentence actually helps in no takers among the exhibitors for a release in
figuring out the missing word. Since, they are Tamil Nadu on this Friday.
satisfied with what they have which means they Hence, the correct option is (C).
feel content with whatever they have and
3.6 (A)
therefore, they are happy.
Hence, the correct option is (A). Paraphrase : A restatement of a text, passage or
a rewording of something written or spoken.
3.3 (B)
Paradox : A self-contradictory statement.
Vice : Immoral, wicked or bad behavior
Paradigm : Pattern or mode.
Virtues : Moral behavior or showing good
Paraffin : Soft solid derived from petroleum.
behavior.
One of his biggest virtues was his ability to
forgive. 3.7 (A)
Hence, the correct option is (B). Figurative : Representing by a figure or

resemblance or expressing one thing in terms,
3.4 (B)
normally denoting another with which it may be
regarded as analogous.
General Aptitude 3
Collateral : It is an asset which is offered to secure The given sentence states that the work contains
a loan. some pioneering ideas but it can hardly be
Literal : Taking words in their basic or usual characterized as innovative.
meaning. Hence, the correct option is (A).
Figurine : A small model of human. 3.12 (A)
Shone : Shone is the second form of verb ‘shine’.
3.8 (C) It means to glow, spark, or glitter.
Tampering : Modification in something which Shown : Shown is the second and third form of
would be harmful for a person. verb ‘show’. It means to display, exhibit or unveil.
Incubating : Siting on the eggs to hatch by Therefore, the correct flow of sentence will be,
keeping them warm. “His face shone with joy when the solution of the
Assisting : To help. puzzle was shown to him.”
Involving : Causing someone to take part in Hence, the correct option is (A).
something.
3.13 (D)
Tampering is the most suitable option.
Tampering with someone else’s email account is Quite : To a certain extent or to a significant
now a very serious offence. amount.
Hence, the correct option is (C). Quiet : Silence, hushed.
3.9 (A) Quite a few : A large number.
“The hall was not very quite”, does not make any
Brake : Device for slowing or stopping a moving
sense. Here, the correct flow will be, “The hall was
vehicle.
not very quiet”.
Break : Taking time off in between.
If Hall was not quiet, it means lecture was attended
So, the correct flow of sentence is, ‘The driver
by enough students to make noise which indicates
applied the brake as soon as she approached the
the use of ‘quite a few’.
hotel where she wanted to take a break.
3.14 (C)
3.10 (A)
The most suitable word here is unpredictable. Daytime temperatures in Delhi can reach 400 C
‘It is no surprise that every society has had codes Hence, the correct option is (C).
of behavior; however, the nature of these codes is 3.15 (D)
often unpredictable’.
In the given sentence, growth rate of ABC motors
and XYZ motors is said to be the same. The use of
3.11 (A) ‘off’ is incorrect therefore option (A) and (C) are
Word ‘Although’ means in spite of the fact or wrong. Since, growth rate is singular noun, ‘those’
even though. cannot be used. Therefore, the correct flow of the
sentence will be,
The word ‘innovative’ is most suitable because
pioneering ideas means exploring something for “The growth rate of ABC motors in 2017 was the
the first time, similar to innovation but not exactly same as that of XYZ motors in 2016”.
the same. Hence, the correct option is (D).
4 General Aptitude
3.16 (A)
Voracious : Wanting a lot of new information and

knowledge.
Precocious : Early, premature, untimely.
Therefore, voracious is the appropriate word to
complete the sentence.
3.17 (D)
To live or be in ivory tower means to be unknown

about or to want to be separated with the ordinary
and unpleasant things that happen in people’s real
life.
3.18 (C)
3.19 (D)
As given in the above passage “so far as we know,

no other species possess this ability”. By this we
can conclude option D is correct.
3.20 (A)
Given :
The line ran across the page, right through the
centre, and divided the page into two.

General Aptitude 5
4 Idioms & Phrases
2014 IIT Kharagpur 2016 IISc Bangalore

4.1 Which of the following options is the closest 4.2 He turned a deaf ear to my request.
in meaning to the sentence below? [1 Mark] What does the underlined phrasal verb
“As a woman, I have no country.” mean? [1 Mark]
(A) Women have no country. (A) Ignored (B) Appreciated
(B) Women are not citizens of any country. (C) Twisted (D) Returned
(C) Women’s solidarity knows no national
boundaries.
(D) Women of all countries have equal legal
rights.
4.1 (C)
Given : “As a woman, I have no country.”
Given statement is used as a metaphor to illustrate
that women of all countries are united and their
union is not affected by the fact that they belong
to different countries.
Option (C) relates to given statement perfectly.
[Solidarity means unity or harmony]
In option (A), literal meaning of given statement
is taken which is wrong.
In option (B), the statement is factually incorrect
as well as does not relate to anywhere near the
given statement because the statement in the
question does not comment about citizenship.
In option (D), it is stated that women of all
countries have equal legal rights.
Given statements in the question does not refer to
anything remotely related to legal rights of
women.
4.2 (A)
‘Turning a deaf ear’ means ignoring someone.

General Aptitude 1
5 Grammar
2013 IIT Bombay (D) The teacher insured students of good

results.
5.1 The professor ordered to
2016 IISc Bangalore
I II
the students to go out of the class. 5.5 Out of the following four sentences, select
III IV the most suitable sentence with respect to
Which of the above underlined parts of the grammar and usage. [1 Mark]
sentence is grammatically incorrect? (A) I will not leave the place until minister
[1 Mark] does not meet me.
(A) I (B) II (B) I will leave the place until the minister
(C) III (D) IV does not meet me.
2014 IIT Kharagpur (C) I will not leave the place until the
minister meet me.
5.2 Choose the most appropriate word from
(D) I will not leave the place until the
the options given below to complete the
minister meets me.
following sentence. [1 Mark]
5.6 If I were you, I ______ that laptop. It’s
A person suffering from Alzheimer’s
much too expensive. [1 Mark]
disease ______ short-term memory loss.
(A) Won’t buy (B) Shan’t buy
(A) Experienced (B) Unexperienced
(C) Wouldn’t buy (D) Would buy
(C) Is experiencing (D) Experiences
5.7 Choose the most appropriate set of words
2015 IIT Kanpur from the options given below to complete
5.3 Choose the most appropriate word from the following sentence. [1 Mark]
the options given below to complete the _______ is a will, ________ is a way.
following sentence. [1 Mark] (A) Wear, there, their
The official answered _________ that (B) Were, their, there
the complaints of the citizen would be (C) Where, there, there
looked into. (D) Where, their, their
(A) respectably (B) respectfully 2017 IIT Roorkee
(C) reputably (D) respectively
5.8 The event would have been successful if
5.4 Choose the statement where underlined
you _______ able to come. [1 Mark]
word is used correctly : [1 Mark]
(A) are
(A) The minister insured the victims that
(B) had been
everything would be all right.
(C) have been
(B) He ensured that the company will not
(D) would have been
have to bear any loss.
5.9 The bacteria in milk are destroyed when it
(C) The actor got himself ensured against
any accident. ______ heated to 80 0 C . [1 Mark]
(A) would be (B) will be
(C) is (D) was
General Aptitude 1
2019 IIT Madras 5.16 You should ______ when to say ______.
(A) know/no (B) no/no
5.10 They have come a long way in _______
(C) know/know (D) no/know
trust among the users. [1 Mark]
[1 Mark]
(A) Creating (B) Created
5.17 The movie was funny and I ______?
(C) Creation (D) Create
(A) Could helped laughed
5.11 The CEO’s decision to quit was as
shocking to Board as it was to _______. (B) Could help laughing
[1 Mark] (C) Couldn’t help laughing
(A) Myself (B) I (D) Couldn’t help laughed
(C) Me (D) My [1 Mark]
5.12 Hima Das was _______ only Indian athlete 5.18 In the last few years, several new shopping
to win _______ gold for India. [1 Mark] malls were opened in the city. The total
(A) the, a (B) an, a number of visitors in malls is impressive.
(C) an, the (D) the, many However the total revenue generated
through sales in the shops in these malls is
2020 IIT Delhi
generally low.
5.13 Rescue teams deployed _________ Which one of the following is the correct
disaster hit areas combat _________ a lot logical inference based on the information
of difficulties to save the people. [1 Mark] in the about passage?
(A) in, with (B) with, with (A) Fewer people are visiting the malls but
(C) to, to (D) with, at spending more.
2021 IIT Bombay (B) Fewer people are visiting the malls and
5.14 (i) Arun and Aparna are here not spending enough.
(ii) Arun and Aparna is here (C) More people are visiting the malls and
(iii)Arun’s families is here spending more.
(iv) Arun’s family is here (D) More people are visiting the malls but
Which of the above sentences are not spending enough.
grammatically CORRECT? [1 Mark] [1 Mark]
(A) 1 and 2 (B) 1 and 4
2023 IIT Kanpur
(C) 2 and 4 (D) 3 and 4
5.15 Getting to the top is ______ than staying 5.19 “I have not yet decided, what will I do this
on top [1 Mark] evening; I_______ visit a friend”.
(A) easier (B) more easier (A) Mite (B) Would
(C) much easier (D) easiest (C) Might (D) Didn’t
2022 IIT Kharagpur
5.1 (B) ‘to-infinitive’ is after the verb order which is

neither essential nor required.
In English grammar, we use ‘to-infinitive’ usually
to express purpose or after certain verbs of
thinking and feeling. In the given sentence, use of
2 General Aptitude
5.2 (D) then the main clause includes should have / would
have / could have + verb (III).
To express a fact, present simple tense is used in
Hence the correct assertion is,
general. Therefore, “experiences” is more
The event would have been successful if you had
appropriate.
been able to come.
5.3 (B)
5.9 (C)
The official answered respectfully that the
In case of factual sentences, the dependent and
complaints of the citizen would be looked into.
independent bot clauses are in present simple
tense.
5.4 (B) The bacteria in milk are destroyed when it is
Insure : Protection against possible contingency. heated to 800 C.
Ensure : Making sure or making certain that Hence, the correct option is (C).
something will occur. 5.10 (A)
The correct use of underlined word is done only in
option (B). The phrase “have come a long way” means
progress. The given sentence is in present perfect
continuous tense i.e. subject + has/have +
5.5 (D)
verb(ing)
Word ‘until’ itself negates the upcoming sentence, Therefore, in the given sentence, the correct flow
hence ‘does not’ is not required. According to will be, “They have come a long way in creating
rules of present simple tense, verb ‘meet’ should trust among the users”.
have ‘s’, since the object is noun. Hence, the correct option is (A).
5.11 (C)
5.6 (C)
‘I’, ‘me’ and ‘myself’ are subjective, objective and
According to the concept of second conditional reflective pronouns respectively.
sentences, if the ‘if clause’ is in past tense then the In the given sentence, CEO is subject and quitting
main clause includes should / would / could + verb is verb and object is missing from the sentence,
(I). which means we require an objective pronoun (i.e.
Hence the correct assertion is, ‘me’).
If I were you, I wouldn’t buy that laptop. It’s much Hence, the correct option is (C).
too expensive.
5.12 (A)
Article ‘the’ is definite article. It limits the
5.7 (C)
meaning of a noun to one particular thing. In the
Given sentence is a well-known quotation which given sentence, it is specified that Hima Das was
goes as, where there is a will, there is a way. ‘the’ only Indian athlete to win. Article ‘a’ is used
Hence, the correct option is (C). before words that start with consonants. Therefore,
5.8 (B) the correct form of sentence will be,
‘Hima Das was the only Indian athlete to win a
According to the concept of third conditional
sentences, if the ‘if clause’ is in past perfect tense gold for India’.
General Aptitude 3
5.13 (A)
The sentence mentions ‘areas’, hence use of “in”
will be grammatically correct. Latter, the correct
word would be “combat with”.
5.14 (B)
5.15 (A)
Getting to the top is easier than staying in top.
5.16 (A)
5.17 (C)
5.18 (D)
5.19 (C)
Given :
“I have not yet decided, what will I do this
evening; I might visit a friend”.

4 General Aptitude
6 Case Studies
2014 IIT Kharagpur 2015 IIT Kanpur
6.1 Moving into a world of big data will 6.3 Alexander turned his attention towards
require us to change our thinking about the India, since he had conquered Persia.
merits of exactitude. To apply the Which one of the statements below is
conventional mindset of measurement to logically valid and can be inferred from the
the digital connected world of the twenty- above sentence [2 Marks]
first century is to miss a crucial point. As (A) Alexander would not have turned his
mentioned earlier, the obsession with attention towards India had he not
exactness is an artefact of the information- conquered Persia.
deprived analog era. When data was (B) Alexander was not ready to rest on his
sparse, every data point was critical, and laurels, and wanted to march to India.
thus great care was taken to avoid letting (C) Alexander was completely in control
any point bias the analysis. From “BIG of his army and could command
DATA” Viktor Mayer-Schonberger and towards India.
Kenneth Cukier (D) Since Alexander’s kingdom extended
The main point of the paragraph is : to Indian borders after the conquest of
[2 Marks] Persia, he was keen to move further.
(A) The twenty-first century is a digital 6.4 Most experts feel that in spite of
world. possessing all the technical skills required
(B) Big data is obsessed with exactness. to be a batsman of the highest order, he is
(C) Exactitude is not critical in dealing unlikely to be so due to lack of requisite
with big data. temperament. He was guilty of throwing
(D) Sparse data leads to a bias in the away his wicket several times after
analysis. working hard to lay a strong foundation.
6.2 Rajan was not happy that Sajan decided to His critics pointed out that until he
do the project on his own. On observing his addresses to this problem, success at the
unhappiness, Sajan explained to Rajan that highest level will continue to elude him.
he preferred to work independently. [2 Marks]
Which one of the statements below is Which of the statement (s) below is/are
logically valid and can be inferred from the logically valid and can be inferred from the
above sentences? [1 Mark] above passage?
(A) Rajan has decided to work only in a (i) He was already a successful batsman at
group. the highest level
(B) Rajan and Sajan were formed into a (ii) He has to improve his temperament in
group against their wishes. order to become a great batsman
(C) Sajan had decided to give in to Rajan's (iii)He failed to make many of his good
request to work with him. starts count
(D) Rajan had believed that Sajan and he (iv) Improving his technical skills will
would be working together. success
General Aptitude 1
(A) (iii) and (iv) (B) (ii) and (iii) the order that they went in. However, the
(C) (i), (ii) and (iii) (D) (ii) only time they spent inside seemed to vary a lot
2016 IISc Bangalore some people came out in a matter of
minutes while for others it took much
6.5 Indian currency notes show the
longer. [2 Marks]
denomination indicated in at least
From this, what can one conclude?
seventeen languages. If this is not an
(A) The centre operates on a first-come-
indication of the nation’s diversity,
first-served basis, but with variable
nothing else is. [2 Marks]
service times, depending on specific
Which of the following can be logically
customer needs.
inferred from the above sentences?
(B) Customers were served in an arbitrary
(A) India is a country of exactly seventeen
order, since they tool varying amounts
languages.
of time for service completion in the
(B) Linguistic pluralism is the only
centre.
indicator of a nation’s diversity.
(C) Since some people came out within a
(C) Indian currency notes have sufficient
few minutes of entering the centre, the
space for all the Indian languages.
system is likely to operate on a last-
(D) Linguistic pluralism is strong evidence
come-first-served basis.
of India’s diversity.
(D) Entering the centre early ensured that
6.6 Today we consider, Ashoka as a great ruler
one would have shorter service times
because of the copious evidence he left
and most people attempted to do this.
behind in the form of stone carved edicts.
6.8 A map shows the elevations of Darjeeling,
Historians tend to correlate greatness of a
Gangtok, Kalimpong, Pelling and Siliguri.
king at his time with the availability of
Kalimpong is at a lower elevation than
evidence today. [2 Marks]
Gangtok. Pelling is at a lower elevation
Which of the following can be logically than Gangtok.
inferred from the above sentences? Pelling is at a higher elevation that Siliguri.
(A) Emperors who do not leave significant Darjeeling is at a higher elevation than
sculpted evidence are completely Gangtok.
forgotten. Which of the following statements can be
(B) Ashoka produced stone-carved edicts inferred from the paragraph above?
to ensure that later historians will [2 Marks]
respect him. (i) Pelling is at a higher elevation than
(C) Statues of kings are a reminder of their Kalimpong.
greatness. (ii) Kalimpong is at a lower elevation than
(D) A king's greatness, as we know him Darjeeling.
today, is interpreted by historians. (iii)Kalimpong is at a higher elevation than
2017 IIT Roorkee Siliguri.
(iv) Siliguri is at a lower elevation than
6.7 Bhaichung was observing the pattern of
Gangtok.
people entering and leaving a car service
(A) Only (i)
centre. There was a single window where
(B) Only (ii) and (iii)
customers were being served. He saw that
(C) Only (ii) and (iv)
people inevitably came out of the centre in
(D) Only (iii) and (iv)
2 General Aptitude
6.9 The old concert hall was demolished the commercial use of this technology is
because of fears that the foundation would not legal in India. Notwithstanding that,
be affected by the construction of the new reports indicate that the herbicide tolerant
metro line in the area. Modern technology cotton had been purchased by farmers at an
for underground metro construction tried average of Rs 200 more than the control
to mitigate the impact of pressurized air price of ordinary cotton and planted in
pockets created by the excavation of large 15% of the cotton growing area in the 2017
amounts of soil. But even with these Kharif season.
safeguards, it was feared that the soil Which one of the following statements can
below the concert. [2 Marks] be inferred from the given passage?
From this, one can infer that, [2 Marks]
(A) the foundations of old buildings create (A) Farmers want to access the new
pressurized air pockets underground, technology for experimental purposes.
which are difficult to handle during (B) Farmers want to access the new
metro construction. technology if India benefits from it.
(B) metro construction has to be done (C) Farmers want to access the new
carefully considering its impact on the technology even if it is not legal.
foundations of existing building. (D) Farmers want to access the new
(C) old building in an area form an technology by paying high price.
impossible hurdle to metro 6.12 “Popular Hindi fiction, despite - or perhaps
construction in that area. because of - its wide reach, often does not
(D) pressurized air can be used to excavate appear in our cinema. As ideals that
large amounts of soil from viewers are meant to look up to rather than
underground areas. identify with, Hindi film protagonists
2019 IIT Madras
usually read books of aspirational value :
textbooks, English books, or high value
6.10 “The increasing interest in tribal characters literature.”
might be a mere coincidence, but timing is Which one of the following CANNOT be
of interest. None of this, though, is to say inferred from the paragraph above?
that the tribal hero has arrived in Hindi [2 Marks]
cinema, or that the new crop of characters (A) Protagonists in Hindi movies, being
represent the acceptance of the tribal ideals for viewers, read only books of
character in the industry. The films and aspirational value.
characters are too few to be described as a (B) People do not look up to writers of
pattern”. textbooks, English books or high value
What does the word ‘arrived’ mean in the literature.
paragraph above? [2 Marks] (C) Though popular Hindi fiction has wide
(A) Came to a conclusion reach : it often does not appear in the
(B) Reached a terminus movies.
(C) Went to a place (D) Textbooks, English books or high
(D) Attained a status literature have aspirational value, but
6.11 The new cotton technology, Bollgard-II, not popular Hindi Fiction.
with herbicide tolerant traits has developed 6.13 The Newspaper reports that over 500
into a thriving business in India. However, hectares of tribal land spread across 28
General Aptitude 3
tribal settlements in Mohinitampuram specific period of time. Real interest rate
forest division have already been calculated on the basis of actual value
“alienated”. A top forest official said. (inflation-adjusted), is approximately
“First the tribal are duped out of their land equal to the difference between nominal
holdings. Second, the families thus rate and expected rate of inflation in the
rendered landless are often forced to economy. [2 Marks]
encroach further into the forests.” Which of the following assertions is best
On the basis of the information available in supported by the above information?
the paragraph, ______ is/are responsible (A) Under high inflation, real interest rate
for duping the tribal. [2 Marks] is low and borrowers get benefited.
(A) The Newspaper (B) Under low inflation, real interest rate is
(B) It cannot be inferred who low and borrowers get benefited.
(C) Forest officials (C) Under low inflation, real interest rate is
(D) Landless families low and borrowers get benefited.
2020 IIT Delhi (D) Under high inflation, real interest rate
is low and lenders get benefited.
6.14 The American psychologist Howard
Gardner expounds that human intelligence 2023 IIT Kanpur
can be sub-categorized into multiple kinds, 6.16 Elvesland is a country that has peculiar
in such a way that individuals differ with beliefs and practices. They express almost
respect to their relative competence in each all their emotions by gifting flowers. For
kind. Based on this theory, modern instance, if anyone gifts a white flower to
educationists insist on prescribing multi- someone, then it is always taken to be a
dimensional curriculum and evaluation declaration of one’s love for that person. In
parameters that enable development and a similar manner, the gifting of a yellow
assessment of multiple intelligences. flower to someone often means that one is
[2 Marks] angry with that person.
Which of the following statements can be Based only on the information provided
inferred from the given text? above, which one of the following sets of
(A) Howard Gardner insists that the statement(s) can be logically inferred with
teaching curriculum and evaluation certainty?
needs to be multi-dimensional. (i) In Elvesland, one always declares
(B) Modern educationists insist that the one’s love by gifting a white flower.
teaching curriculum and evaluation (ii) In Elvesland, all emotions are declared
needs to be multi-dimensional. by gifting flowers.
(C) Modern educationists want to develop (iii) In Elvesland, sometimes one
and assess the theory of multiple expresses one’s anger by gifting a
intelligences. flower that is not yellow.
(D) Howard Gardner wants to develop and (iv) In Elvesland, sometimes one
assess the theory of multiple expresses one’s love by gifting a
intelligences. white flower.
6.15 Nominal interest rate is defined as the (A) only (ii) (B) (i), (ii) and (iii)
amount paid by the borrower to the lender (C) (i), (iii) and (iv) (D) only (iv)
for using the borrowed amount for a
4 General Aptitude
6.17 Based only on the following passage, such that each part is a mirror image of the
which one of the options can be inferred other part about that line.
with certainty? The figure below consists of 20 unit
When the congregation sang together, squares arranged as shown. In addition to
Apenyo would also join, though her little the given black squares, upto 5 more may
screams were not quite audible because of be coloured black. Which one among the
the group singing. But whenever there was following options depicts the minimum
a special number, trouble would begin; number of boxes that must be coloured
Apenyo would try singing along, much to black to achieve two lines of symmetry?
the embarrassment of her mother. After (The figure is representative)
two or three such mortifying Sunday a b
evenings, the mother stopped going to c d
church altogether until Apenyo became g
e f h
older and learnt to behave.
i
At home too, Apenyo never kept quiet; she
j k
hummed or made up silly songs to sing by
herself, which annoyed her mother at times (A) d (B) c, d, i
but most often made her become pensive. (C) c, i (D) c, d, i, f, g
She was by now convinced that her 6.20 The James Webb telescope, recently
daughter had inherited her love of singing launched in space, is giving humankind
from her father who had died unexpectedly unprecedented access to the depths of time
away from home. by imaging very old stars formed almost
[Excerpt from These Hills Called Home by 13 billion years ago. Astrophysicists and
Temsula Ao] cosmologists believe that this odyssey in
(A) The mother was embarrassed about her space may even shed light on the existence
daughter’s singing at home. of dark matter. Dark matter is supposed to
(B) The mother’s feelings about her interact only via the gravitational
daughter’s singing at home were only interaction and not through the
of annoyance. electromagnetic-, the weak- or the strong-
(C) The mother was not sure if Apenyo had interaction. This may justify the epithet
inherited her love of singing from her “dark” in dark matter.
father. Based on the above paragraph, which one
(D) When Apenyo hummed at home, her of the following statements is FALSE?
mother tended to become thoughtful (A) No other telescope has captured
6.18 If x satisfies the equation 48  256, then
x
images of stars older than those
x is equal to ________. captured by the James Webb telescope.
(B) People other than astrophysicists and
1
(A) (B) log16 8 cosmologists may also believe in the
2
existence of dark matter.
2
(C) (D) log4 8 (C) The James Webb telescope could be of
3
use in the research on dark matter.
(D) If dark matter was known to interact
6.19 A line of symmetry is defined as a line that via the strong-interaction, then the
divides a figure into two parts in a way epithet “dark” would be justified.
General Aptitude 5
6.1 (C) away his wicket i.e. He failed to make many of his
good starts count.
It is stated in the passage that in the information-
deprived analog era, there was a obsession with Therefore, option (B) can be inferred from the
exactness because data was sparse and every data passage.
point was critical but in the era of big data, the Hence, the correct option is (B).
thinking about merits of exactitude has to be 6.5 (D)
changed which means big data is not obsessed
The word ‘at least’ gives away the fact that there
with exactness or exactitude is not critical in
are more than 17 languages. So, option (A) is
dealing with big data.
incorrect.
Option (C) is nowhere related to the passage.
6.2 (D)
The last statement of the passage shows that
According to given passage, Rajan was not happy linguistic pluralism is a strong evidence of India’s
that Sajan decided to do the project on his own, diversity.
which means that Rajan had thought that they will Hence, the correct option is (D).
be doing the project together as mentioned in
option (D). 6.6 (D)
Hence, the correct option is (D). The given passage states that the reason because
6.3 (A) of which we consider Ashoka as a great rules is
that he left behind evidence in form of stone
Given statement has no information about
carved edicts which were found by historians.
Alexander’s army or Indian borders. Hence,
Therefore, a king’s greatness, as we know him
option (C) and (D) are incorrect.
today, is interpreted by historians.
Option (B) states that Alexander was not ready to
rest on his laurels. Laurel means a type of ancient
crown which is an indication of victory. The given 6.7 (A)
statement does not comment on the state of mind The passage states that the people come out of the
of Alexander i.e. whether he wanted to conquer center in the order that they came in. So, the center
different countries or not. operates on a first-come-first-served basis and the
Therefore option (B) is in correct. time that the people stayed inside the center varies
Hence, the correct option is (A). according to whatever kind of service they
6.4 (B) required.
Option (A) clearly contradicts the essence of the Hence, the correct option is (A).
passage. Hence, it is incorrect. 6.8 (B)
Passage states that due to lack of requisite
Given : A map shows the elevations of Darjeeling,
temperament, it is highly unlikely that he will
Gangtok, Kalimpong, Pelling and Siliguri.
become a batsman of higher order. So, he has to
improve his temperament to become a great
batsman. (i) Darjeeling is at the highest elevation.
The passage states that after laying a strong (ii) Kalimpong and Pelling are at lower
foundation i.e. after a good head start, he throws elevation than Gangtok.
6 General Aptitude
(iii) Siliguri is at a lower elevation than Pelling. Hence, the correct option is (A).
From point (ii), statement-II can be inferred. 6.13 (B)
Siliguri is at a lower elevation than Pelling and In the given passage, nowhere it is mentioned that
Pelling is at a lower elevation than Gangtok. who is responsible for duping the tribal.
Therefore, statement-IV can be inferred from the Therefore, it can not be inferred.
paragraph. Hence, the correct option is (B).
6.14 (B)
6.9 (B)
Since, the theory is already given or formulated.
The passage states that pressurized air pockets are Therefore, option (D) is incorrect.
created underground due to the excavation of large Howard Gardner is not related to teaching
amount of soil. So, options (A) and (D) are curriculums. Therefore, option (A) is incorrect.
incorrect. The passage indicates that due to the Modern educationists have little to do with
construction of metro line, foundation of the old developing theories. Therefore, option (C) is also
concert hall are effected, not the other way around. incorrect.
So, option (C) is also incorrect. The given passage indicates that modern
Hence, the correct option is (B). educationists insist on prescribing multi-
6.10 (D) dimensional curriculum and evaluation
parameters that enable development and
The given passage indicates that Hindi cinema is
assessment of multiple intelligences.
taking an interest in tribal characters but it does not
mean that tribal characters have been completely
accepted in the industry. 6.15 (A)
In the sentence, “None of this, though, is to say Real interest = Nominal rate-inflation rate
that the tribal hero has arrived in Hindi cinema”, So, when inflation increases, real interest rate
use of ‘arrived’ indicates that tribal characters decreases and Borrower gets benefited.
have not attained a certain status yet.
6.16 (D)
Given :
6.11 (C)
Elvesland is a country that has peculiar beliefs and
The given passage indicates that even though use practices. They express almost all their emotions
of new cotton technology is not legal in India, by gifting flowers. For instance, if anyone gifts a
farmers are still buying herbicide tolerant cotton at white flower to someone, then it is always taken to
an average of Rs.200 more than ordinary cotton be a declaration of one’s love for that person. In a
which means farmers want to access the new similar manner, the gifting of a yellow flower to
technology even if it is not legal. someone often means that one is angry with that
Hence, the correct option is (C). person.
6.12 (A) (i) In Elvesland, one always declares one’s
love by gifting a white flower cannot be
As it is clearly mentioned in the given passage that
inferred with certainty, as gifting white
Hindi film protagonists usually read books of
flower is sign of declaration of ones love
aspirational value which means there is a
for the person. It is not declares by a
possibility that they read other books as well. So,
person.
statement mentioned in option (A) cannot be
inferred from the passage.
General Aptitude 7
(ii) In Elvesland, all emotions are declared by Option (C) cannot be inferred as she was
gifting flowers cannot be inferred with convinced that her daughter had inherited her love
certainty, as they express “almost all” their of singing from her father.
emotions by gifting flowers. Therefore Option (D) can be inferred as when open go
“All” (definite) cannot be true. hummed at home, her mother tended to become
(iii) In Elvesland, sometimes one expresses thought-full and then convinced.
one’s anger by gifting a flower that is not
yellow is definitely cannot inferred as the
gifting of a yellow flower to someone often 6.18 (C)
means that one is angry with that person. 48 x  256
(iv) In Elvesland, sometimes one expresses
 48 x  44
one’s love by gifting a white flower can be
inferred with certainty, as if anyone gifts a  8x  4
white flower to someone, then it is always
taken to be a declaration of one’s love for
 2  3 x
 22
that person. 2
So, x
6.17 (D) Hence, the correct option is (C).
Given : 6.19 (B)
When the congregation sang together, Apenyo Given :
would also join, though her little screams were not A line of symmetry is defined as a line that divides
quite audible because of the group singing. But a figure into two parts in a way such that each part
whenever there was a special number, trouble is a mirror image of the other part about that line.
would begin; Apenyo would try singing along,
The figure below consists of 20 unit squares
much to the embarrassment of her mother. After
arranged as shown. In addition to the given black
two or three such mortifying Sunday evenings, the
squares, upto 5 more may be coloured black.
mother stopped going to church altogether until
Apenyo became older and learnt to behave. a b
At home too, Apenyo never kept quiet; she c d
hummed or made up silly songs to sing by herself, g
e f h
which annoyed her mother at times but most often
made her become pensive. She was by now i
convinced that her daughter had inherited her love j k
of singing from her father who had died
In the figure given below, MN is the vertical line
unexpectedly away from home.
of symmetry for the figure, for which we have to
Option (A) cannot be inferred as her mother was colour the box ‘i’.
embarrassed about her daughter’s singing at
church.
Option (B) cannot be inferred as the mothers
feelings about her daughter’s singing at home
were not only of annoyance but also she become
pensive.
8 General Aptitude
M Hence, the correct option is (D).

a b
c d
e f g h
i
j k
N
In the figure given below, PQ is the horizontal line
of symmetry for the figure, for which we have to
colour box ‘c’ and ‘d’.
a b
c d
P e f g h Q
i
j k
Therefore, we have to colour 3 boxes c, d, i to

achieve two line of symmetry MN and PQ.
6.20 (D)
Given :
The James Webb telescope, recently launched in
space, is giving humankind unprecedented access
to the depths of time by imaging very old stars
formed almost 13 billion years ago.
Astrophysicists and cosmologists believe that this
odyssey in space may even shed light on the
existence of dark matter. Dark matter is supposed
to interact only via the gravitational interaction
and not through the electromagnetic-, the weak- or
the strong-interaction. This may justify the epithet
“dark” in dark matter.
According to the given paragraph statement in
option (D) is FALSE.
As, “Dark matter is supposed to interact only via
the gravitational interaction and not through the
electromagnetic-, the weak- or the strong-
interaction.”
General Aptitude 9

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GATE 2024

TOPIC WISE GATE SOLUTIONS

6. Construction Materials & Management

Syllabus : Strength of Materials

2013 IIT Bombay 2015 IIT Kanpur

(B) F = −56.389 kN and θ = −28.280 A

bar is increased by T, resulting in an axial

unchanged, if the length of the bar is

P = 1 kN Assume the yield stress of steel as 250 MPa.

The Young’s modulus of the material is

can occur as a result of long term

 Creep is also temperature dependent T1 or s1

 Creep stages : 40sin 450

Basics of Signals 200

sy Area of segment 1, ( A1 ) = 100 mm 2 ,

Now, for no change in volume, dV = 0 A1 = 100 mm 2

1. If the angle between two positive faces (or

O d Deflection Strain energy

 For linear elastic material the value of

By equation (i), (ii) and (iii), n =1  l 

∴ β= 2 = = = = 0.394 By comparing τ xy , we get α = 5

Given : E = 1.0 N/m 2 , μ = 0.5

Substitute γ xy and C in equation (A), Stress in member AB,

2015 IIT Kanpur respectively. A force P acts at C as shown.

2016 IISc Bangalore

2018 IIT Guwahati

2.0 1.5 1.0 1.5 m

2.14 Consider the beam shown in the figure

2.1 0.25 10l 3

In case of simply supported beam which is

Moment Bending moment diagram

2.6 (A) Hence, the correct option is (D).

f y = 0, P × (2 + 1.5 + 1 + 1.5) + (100 × 0.5) + (200 × 2.5)

= 180 kNm  −M + M − RB × (3L) = 0

Slopes at E, F, G and H are zero because at axis A

2013 IIT Bombay s y = 50 MPa

3.1 The state of 2D stress at a point is given

2015 IIT Kanpur

(A) σ x = 26.7 MPa, σ y = 172.5 MPa

3.1 (A) ∴ σ1 = 110 MPa and σ2 = 10 MPa

In case of 2-D state of stress principal stresses are  Key Point

100 MPa 100 MPa

Scan for In case of pure shear state of stress condition. The

σ45 = σ2X cos 2 450 + σ2y sin 2 450

2013 IIT Bombay 2018 IIT Guwahati

All dimensions are in mm

(A) 240 N (B) 480 N

(C) (C) 60 N (D) 120 N

2023 IIT Kanpur

4.10 The cross-section of a girder is shown in the

4.9 Cross section of a built-up wooden beam as

Depth of web (d ) = 100 mm

2013 IIT Bombay 1 2πx 

2R – d Using Maxwell – Betti reciprocal theorem

5.6 3.0075 EI 3EI

L The spring attached to the two beams is going to

Now, from the geometry, in ΔOAQ −4

5.10 (A) M/EI M/EI

7 ML 4ML ML ML Deflection diagram,