Official Paper

GATE CE 2015 Official Paper: Shift 1 (Previous Year Paper)

65 questions · 180 minutes · with answers · free

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General Aptitude (10 questions)

1

Extreme focus on syllabus and studying for tests has become such a dominant concern of Indian students that they close their minds to anything __________ to the requirements of the exam.

  1. ((a))

    Related

  2. ((b))

    Extraneous

  3. ((c))

    Outside

  4. ((d))

    Useful

Show Answer
Answer: ((b))

Extraneous

Extraneous - meaning irrelevant/ unconnected / unrelated to the requirements of the exam. It is the right answer because a students desire to focus on aceing their tests makes them avoid anything unrelated to their exams.

Related - means connected

Outside - means a place on the exterior

Useful - is something that will help - this is incorrect to the context because if something is useful no student would avoid it or ‘close their minds’ so to say.

2

Select the pair that best express a relationship similar to that expressed in the pair:

Children: Pediatrician

  1. ((a))

    Adult: Orthopedist

  2. ((b))

    Females: Gynecologist

  3. ((c))

    Kidney: Nephrologist

  4. ((d))

    Skin: Dermatologist

Show Answer
Answer: ((b))

Females: Gynecologist

A child is treated from birth till the age of youth by mostly the same Pediatrician or a child specialist; similarly

A female is treated for practically all her ailments specially child bearing period by mostly the same Gynecologist., who is also called a women’s doctor, just like a Pediatrician is called a children’s doctor. So this is the right answer.

‘Orthopedist ‘ treats bone injuries and an adult will go to be treated by him for a very short time, that too in case of such an injury

‘Nephrologist’ treats diseases related to kidneys.

Dermatologist- a specialist in dermatology, especially a doctor who specializes in the treatment of diseases of the skin

3

The Tamil version of _________ John Abraham – starrer Madras Café _________ cleared by the Censor Board with no cuts last week, but the film’s distributors _________ no takers among the exhibitors for a release in Tamil Nadu _________ this Friday.

  1. ((a))

    Mr., was, found, on

  2. ((b))

    A, was, found, at

  3. ((c))

    The, was, found, on

  4. ((d))

    A, being, find at

Show Answer
Answer: ((c))

The, was, found, on

First blank - The definite article‘The’ as a particular film is named

Second blank - ‘Last week’ refers to a past tense as also does the word ‘cleared’ so the auxiliary ‘was’ is appropriate.

Third blank - To agree with the past tense in the first part this blank also needs the same so ‘found’ is appropriate.

Fourth blank - ‘On’ is appropriate here as a particular day is named.

So ‘C’ is the right answer

4

If ROAD is written as URDG, then SWAN should be written as:

  1. ((a))

    VXDQ

  2. ((b))

    VZDQ

  3. ((c))

    VZDP

  4. ((d))

    UXDQ

Show Answer
Answer: ((b))

VZDQ

Each letter is replaced by the letter that comes three places after it in alphabetical order.

R + 3 = U,

O + 3 = R,

A + 3 = D,

D + 3 = G

Therefore, SWAN will become,

S + 3 = V,

W + 3 = Z,

A + 3 = D,

N + 3 = Q

Therefore, the required word is VZDQ.

5

A function f(x) is linear and has a value of 29 at x = – 2 and 39 at x = 3. Find its value at x = 5.

  1. ((a))

    59

  2. ((b))

    45

  3. ((c))

    43

  4. ((d))

    35

Show Answer
Answer: ((c))

43

Given, function f(x) is linear

Let the linear function be f(x) = ax + b, where a and b are constants

Now, f(x) has a value of 29 at x = – 2 and 39 at x = 3

∴ -2a + b = 29 and 3a + b = 39

Solving the above two equations we get, a = 2 and b = 33

f(x) = 2x + 33

Now value of f(x) at x = 5

f(5) = 2 × 5 + 33 = 43

6

Alexander turned his attention towards India, since he had conquered Persia. Which one of the statements below is logically valid and can be inferred from the above sentence?

  1. ((a))

    Alexander would not have turned his attention towards India had he not conquered Persia.

  2. ((b))

    Alexander was not ready to rest on his laurels, and wanted to march to India

  3. ((c))

    Alexander was completely in control of his army and could command it to move towards India

  4. ((d))

    Since Alexander’s kingdom extended to Indian borders after the conquest of Persia, he was keen to move further

Show Answer
Answer: ((a))

Alexander would not have turned his attention towards India had he not conquered Persia.

Explanation:

'A' ‘Turned his attention’ means that since Persia was conquered now he felt the desire to conquer some other territory. So ‘A’ is valid and can be inferred from the above sentence.

‘B’ is ambiguous and can’t really be validated

‘C’ too seems not very clear, because being a king who had already marched against many kingdoms, he definitely must have had control of his army, though if history is read further his army had revolted after his fight with the Indian king Porus.

‘D’ This too is not logically valid as it may or may not have been the reason as Persia did not extend to the Indian borders if history is to be believed.

7

Most experts feel that in spite of possessing all the technical skills required to be a batsman of the highest order, he is unlikely to be so due to lack of requisite temperament. He was guilty of throwing away his wicket several times after working hard to lay a strong foundation. His critics pointed out that until he addressed this problem, success at the highest level will continue to elude him.

Which of the statements (s) below is/are logically valid and can be inferred from the above passage?

i) He was already a successful batsman at the highest level

ii) He has to improve his temperament in order to become a great batsman

iii) He failed to make many of his good starts count

iv) Improving his technical skills will guarantee success.

  1. ((a))

    iiii) and iv)

  2. ((b))

    ii) and iii)

  3. ((c))

    i), ii) and iii)

  4. ((d))

    ii) only

Show Answer
Answer: ((b))

ii) and iii)

Explanation:

The statements that are logically valid and can be inferred from the above passage are ii and iii 

Because in the given lines we are told that though ‘he had all the technical skills required to be a batsman of the highest order, he had thrown away his wicket several times, due to lack of requisite temperament’ telling us very clearly that he could become a great batsman and had lost out many times after ‘working hard to lay a strong foundation’ due to his ‘temperament’ hence both (ii and iii) are valid and can be inferred from the above passage.

8

The exports and imports (in crores of Rs.) of a country from the year 2000 to 2007 are given in the following bar chart. In which year is the combined percentage increase in imports and exports the highest?

9

Choose the most appropriate equation for the function drawn as a thick line, in the plot below.

  1. ((a))

    x = y - |y|

  2. ((b))

    x = - (y - |y|)

  3. ((c))

    x = y + |y|

  4. ((d))

    x = - (y + |y|)

Show Answer
Answer: ((b))

x = - (y - |y|)

For all values of y > 0, x = 0                        ------- (1)

The line below the x axis passes through origin (0, 0) and (2, -1)

Equation of line is given by

x/2 – y/1 = 0

∴ x = 2y for y < 0

|x| = x for both x > 0 and x < 0

Now from equation (1)

x = 0 for y > 0

∴ x = y – y for y > 0

Also x = y + y for y < 0

∴ x = |y| - y for all y.

The above equation can also be written as follows

x = -(y - |y|)

10

The head of newly formed government desires to appoint five of the six selected members P, Q, R, S, T and U to portfolios of Home, Power, Defense, Telecom and Finance. U does not want any portfolio if S gets one of the five. R wants either Home or finance or no portfolio. Q says that if S gets either power or Telecom, then she must get the other one. T insists on a portfolio if P gets one .

Which is the valid distribution of portfolios?

  1. ((a))

    P – Home, Q – Power, R – Defence, S – Telecom, T – Finance

  2. ((b))

    R – Home, S – Power, P – Defence, Q – Telecom, T – Finance

  3. ((c))

    P – Home, Q – Power, T – Defence, S – Telecom, U – Finance

  4. ((d))

    Q – Home, U – Power, T – Defence, R – Telecom, P – Finance

Show Answer
Answer: ((b))

R – Home, S – Power, P – Defence, Q – Telecom, T – Finance

Let us check each option,

a) R has received a portfolio (Power) that is neither home nor finance, therefore this does not fulfil the given conditions.

b) R has received Home, S and Q have power and telecom, and P and T both have portfolios while U does not, thereby satisfying all conditions.

c) S and U both have portfolios, therefore making this impossible.

d) R has received a portfolio that is neither home nor finance, therefore this does not fulfil the given conditions.

Therefore, b is the correct distribution.

Civil Engineering (55 questions)

11

For what value of p the following set of equations will have no solution?

2x+3y=5 3x+py=10 {2x + 3y = 5}\ {3x + py = 10}

12

The integral  \(\mathop \smallint \limits_{{x_1}}^{{x_2}} {x^2}dx\) with  x2>x1>0{x_2} > {x_1} > 0  is evaluated analytically as well as numerically using a single application of the trapezoidal rule. If I is the exact value of the integral obtained analytically and J is the approximate value obtained using the trapezoidal rule, which of the following statements is correct about their relationship?

  1. ((a))

    J > I

  2. ((b))

    J < I

  3. ((c))

    J = I

  4. ((d))

    Insufficient data to determine the relationship

Show Answer
Answer: ((a))

J > I

Exact value of integration is computed by integration which follows the exact shape of graph while computing the area.

Whereas, in Trapezoidal rule, the lines joining each points are considered straight line which in not the exact variation of graph all the time

Also we know that approximate value calculated by trapezoidal rule is always greater than the exact value calculated by integration

∴ J > I

Where, J = approximate value, I = Exact value

13

Consider the following probability mass function (p.m.f.) of a random variable X:

\(p\left( {x,q} \right) = \left{ {\begin{array}{{20}{c}} q\ {1 - q}\ 0 \end{array}} \right.\begin{array}{{20}{c}} {if;X = 0}\ {if;X = 1}\ {otherwise} \end{array}\)

If q = 0.4, the variance of X is___________.

14

Workability of concrete can be measured using slump, compaction factor and Vee-bee time. Consider the following statements for workability of concrete:

(i) As the slump increases, the Vee-bee time increases.

(ii) As the slump increases, the compaction factor increases.

Which of the following is TRUE?

  1. ((a))

    Both (i) and (ii) are True

  2. ((b))

    Both (i) and (ii) are False

  3. ((c))

    (i) is True and (ii) is False

  4. ((d))

    (i) is False and (ii) is True

Show Answer
Answer: ((d))

(i) is False and (ii) is True

Concept:

Workability of concrete:

Workability is the relative ease with which concrete can be mixed, transported, molded, and compacted. we can also say, it is the energy to overcome friction while compacting.

The workability of concrete can be measured using the following methods,

1) Slump test:

The slump test is a measure of consistency or the wetness of the concrete mix. It is suitable for field applications, suitable for concrete of high and medium workability. Slumps for various types of concrete is measured in mm.

2) Compacting factor test:

It is more accurate than a slump test. It is suitable for concrete mixes of medium and low workability. It is suitable for determining workability in the laboratory.

3) Vee-bee consistometer:

The time required for complete remoulding of concrete in seconds after placed in the mould with a slump cone. It is expressed in seconds. It is the method used to determine the workability of very dry mixes with low workability.

4) Flow table test:

Based on the percentage flow the workability of the concrete is determined. The values could range from 0 to 150 %

5) Kelly ball test:

It is a simple field testing machine that determines the depth to which a 15 cm diameter metal hemisphere weighing 13.6 kg, will sink under its own weight into the fresh concrete. Based on the depth of the penetration workability of concrete is determined

Explanation:

ConsistencySlump in mmCompaction FactorVee-Bee time in seconds
Very Dry0-250.7-0.820-10
Dry25-500.8-0.8510-5
Plastic50-1000.85-.955-2
Semi-fluid100-1750.95-12-0

We can see,

As slump increases, Vee bee time decreases, as it becomes easier for concrete to flow.

Also as the slump increases, the compaction factor increases.

Hence the correct answer is option 4

15

Consider the following statements for air-entrained concrete:

(i) Air-entrainment reduces the water demand for a given level of workability.

(ii) Use of air-entrained concrete is required in environments where cyclic freezing and thawing is expectedWhich of the following is TRUE?

  1. ((a))

    Both (i) and (ii) are True

  2. ((b))

    Both (i) and (ii) are False

  3. ((c))

    (i) is True and (ii) is False

  4. ((d))

    (i) is False and (ii) is True

Show Answer
Answer: ((a))

Both (i) and (ii) are True

Air Entraining agent:

  • These are the admixtures which when used in concrete, forms millions of air bubbles, which act as flexible ball bearings and gives a lubricating effect.
  • The entrained air increases the workability of concrete resulting in its greater uniformity.
  • The air bubbles formed due to the air-entraining agent act as a fine aggregate and enable the reduction of sand in concrete. The reduction of fine aggregate results in the reduction of water requirement without any ill effect on workability and slump.
  • Air-entrained concrete contains billions of microscopic air cells per cubic foot. These air pockets relieve internal pressure on the concrete by providing tiny chambers for water to expand into when it freezes and useful in freeze and thaw resistance.

Both the given statements are correct, hence option 1 is correct

16

Consider the singly reinforced beam shown in the figure below:

 

<br>

At cross-section XX, which of the following statements is TRUE at the limit state?

  1. ((a))

    The variation of stress is linear and that of strain is non-linear

  2. ((b))

    The variation of strain is linear and that of stress is non-linear

  3. ((c))

    The variation of both stress and strain is linear

  4. ((d))

    The variation of both stress and strain is non-linear

Show Answer
Answer: ((b))

The variation of strain is linear and that of stress is non-linear

Bassed on the first assumption of limit state method, the variation of strain is linear and the variation of stress diagram is parabolic i.e non-linear.

17

For the beam shown below, the stiffness coefficient K22 can be written as

 

  1. ((a))

    6EIL2\frac{{6EI}}{{{L^2}}}

  2. ((b))

    12EIL3\frac{{12EI}}{{{L^3}}}

  3. ((c))

    3EIL\frac{{3EI}}{L}

  4. ((d))

    EI6L2\frac{{EI}}{{6{L^2}}}

Show Answer
Answer: ((b))

12EIL3\frac{{12EI}}{{{L^3}}}

Concept:

<br>

By giving unit displacement in 2nd direction without giving displacement in any other direction, the force developed RB is

 K22=12EIL3{K_{22}} = \frac{{12EI}}{{{L^3}}}

18

The development length of a deformed reinforcement bar can be expressed as Ld=ϕ×σsk×τbd{L_d} = \frac{{\phi\times {\sigma _s}}}{{k\times{\tau _{bd}}}} From the IS : 456 - 2000, the value of k can be calculated as _______________.

19

For the beam shown below, the value of the support moment M is _______________ kN-m.

 

20

Two triangular wedges are glued together as shown in the following figure. The stress acting normal to the interface, σn is _______________ MPa.

 

21

A fine-grained soil has 60% (by weight) silt content. The soil behaves as semi-solid when water content is between 15% and 28%. The soil behaves fluid-like when the water content is more than 40%. The ‘Activity’ of the soil is

  1. ((a))

    3.33

  2. ((b))

    0.42

  3. ((c))

    0.30

  4. ((d))

    0.20

Show Answer
Answer: ((c))

0.30

The Activity of the soil (A)

The activity of the soil is given by the ratio of the plasticity index (Ip) and the percentage of clay fraction (C) in the soil.

It indicates water absorption capacity or indicates swelling and shrinkage characteristics

A=IPC{\bf{A}} = \frac{{{{\bf{I}}_{\bf{P}}}}}{{\bf{C}}}

Where,

C = percentage of clay particles less than 2-micron size

IP = plasticity index = WL - WP

WL = liquid limit of the soil, WP = plastic limit of the soil

Activity numberType of soil
A < 0.75Inactive
A = 0.75 to 1.25Normal
A > 1.25Active

Calculation:

Given,

Fine-grained soil has 60% (by weight) silt content

∴ % clay = 100 – silt content = 100 – 60 = 40%

Soil behaves as a semi solid between 15 % and 28 %

∴ WS = 15 %, WP = 28 %

Soil behaves fluid-like when the water content is more than 40%

∴ WL = 40 %

IP = 40 – 28 = 12 %

A=IP% CA = \frac{{{I_P}}}{{\%\ C}}

A=1240A = \frac{{{12}}}{{40}}

Activity = 0.3

22

Which of the following statements is TRUE for the relation between discharge velocity and seepage velocity?

  1. ((a))

    Seepage velocity is always smaller than discharge velocity

  2. ((b))

    Seepage velocity can never be smaller than discharge velocity

  3. ((c))

    Seepage velocity is equal to the discharge velocity

  4. ((d))

    No relation between seepage velocity and discharge velocity can be established

Show Answer
Answer: ((b))

Seepage velocity can never be smaller than discharge velocity

Concept:

Apparent velocity (V):

It is called apparent velocity because the actual flow is through pores in the cross-section and not through the entire cross-sectional area.

It is also called as discharge velocity

Actual flow velocity (Va):

The velocity of water through pores is called seepage velocity (VS). As the flow is continuous, discharge Q must be the same throughout the system

Q = A × V = AV × Va

Where,

A = total cross-sectional area, AV = area of voids in the total cross-sectional area

\({\rm{V}} = \frac{{{{\rm{A}}{\rm{V}}}}}{{\rm{A}}} \times {{\rm{V}}{\rm{a}}}\)

The porosity of soil is given by,

\({\rm{n}} = \frac{{{\rm{volume;of;void}}}}{{{\rm{total;volume}}}} = \frac{{{{\rm{A}}{\rm{V}}} \times {\rm{L}}}}{{{\rm{A}} \times {\rm{L}}}} = \frac{{{{\rm{A}}{\rm{V}}}}}{{\rm{A}}}{\rm{;}}\)

V = n × Va

∴Va=Vn\therefore {{\bf{V}}_{\bf{a}}} = \frac{{\bf{V}}}{{\bf{n}}}

Explanation:

Vs=Vn{V_s} = \frac{V}{n}

porosity (n) ranges between 0 < n < 1

VVS<1 \frac{V}{V_S} <1

Hence,  Vs>VV_s>V

Seepage velocity (Vs) is always greater than or equal to discharge velocity.

23

Which of the following statements is TRUE for degree of disturbance of collected soil sample?

  1. ((a))

    Thinner the sampler wall, lower the degree of disturbance of collected soil sample

  2. ((b))

    Thicker the sampler wall, lower the degree of disturbance of collected soil sample

  3. ((c))

    Thickness of the sampler wall and the degree of disturbance of collected soil sample are unrelated

  4. ((d))

    The degree of disturbance of collected soil sample is proportional to the inner diameter of the sampling tube

Show Answer
Answer: ((a))

Thinner the sampler wall, lower the degree of disturbance of collected soil sample

Concept:

There are two types of soil samples

1) Disturbed soil samples:

These are the samples in which the natural structure of the soil gets disturbed during sampling.

These samples can be used to determine the index properties of soil, such as grain size, plasticity characteristics, specific gravity.

The samplers used to get disturbed samples is split spoon sampler

2) Undisturbed soil samples:

These are the samples in which the natural structure of the soil does not get disturbed. But we should know that, it is impossible to get a truly undisturbed soil sample.

These are samples are used for determining engineering properties of the soil such as compressibility, shear strength, and permeability, and also shrinkage index.

the samplers used to get undisturbed samples are shelby tubes and thin walled sampler, piston sampler, and hand carved sampler

The smaller the degree of disturbance, the greater will be the reliability of the results.

Explanation:

The factors affecting the degree of disturbance are as follows,

Factors affecting degree of disturbanceValue to obtain undisturbed soil sample
Area ratio≤ 10 %
Inside clearance0.5 to 3.0 %
Outside clearance0 to 2 %
Recovery ratio96 to 98 %
Inside wall frictionThe inside wall of sampler should be smooth
Non-return valveShould have large orifice area
Method of apply forceThe sampler should be pushed and not driven

The sampler having thinner wall like Shelby tube have a lesser area ratio (i.e 6 to 9 %) hence thin wall samplers are used for getting undisturbed samples

The sampler having a thicker wall, like split spoon sampler has a larger area ratio (i.e up to 110 %), hence it is impossible to get the undisturbed sample with thicker wall sampler.

24

In an unconsolidated untrained triaxial test, it is observed that an increase in cell pressure from 150 kPa to 250 kPa leads to a pore pressure increase of 80 kPa. It is further observed that, an increase of 50 kPa in deviatoric stress results in an increase of 25 kPa in the pore pressure. The value of Skempton’s pore pressure parameter B is:

  1. ((a))

    0.5

  2. ((b))

    0.625

  3. ((c))

    0.8

  4. ((d))

    1.0

Show Answer
Answer: ((c))

0.8

Concept:

Skemptons pore pressure parameter (A and B)

These are the parameters that help to obtain effective stress from the total stresses. Skempton gave the pore pressure parameters which express the response of pore pressure due to change in the total stresses under drained conditions. The expressions for pore water pressure parameters are expressed separately for the following three conditions.

1) Pore pressure parameters under Isotropic consolidations

ΔU3 = B × Δσ3   

Where,

ΔU3 = Pore water pressure developed in  the first stage during the application of confining stress Δσ3

B = It is a function of degree of saturation,

Δσ3 = Difference in cell pressure  

i.e For saturated soil B = 1, and for dry soil B = 0

2) Pore pressure parameters under deviator stress

ΔUd = A× B × Δσd

Where,

ΔUd = Increase in porewater pressure

Δσd = increase in deviator stress  

A = It is a function that varies with Over-consolidation ratio of the soil and also with deviator stress Δσ3

3) Combined effect

ΔU = ΔU3 + ΔUd

ΔU = (B × Δσ3) +( A × B × Δσd)

ΔU = B × (Δσ3 +( A × Δσd) )    

Calculation:

​Δσ3 = Difference in cell pressure

∴ Δσ3 = 250 - 150 = 100 kPa

Δu3= 80 kPa

Δσd = 50 kPa, Δud = 25 kPa

we know that,

ΔU3 = B × Δσ3   

B=ΔU3Δσ3{\rm{B}} = \frac{{{\rm{Δ }}{{\rm{U}}_3}}}{{{\rm{Δ }}{{\rm{σ }}_3}}}

B=80100=0.8{\rm{B}} = \frac{{80}}{{100}} = 0.8

25

Which of the following statements is NOT correct?

  1. ((a))

    Loose sand exhibits contractive behavior upon shearing.

  2. ((b))

    Dense sand when sheared under undrained condition, may lead to generation of negative pore pressure.

  3. ((c))

    Black cotton soil exhibits expansive behaviour.

  4. ((d))

    Liquefaction is the phenomenon where cohesionless soil near the downstream side of dams or sheet-piles loses its shear strength due to high upward hydraulic gradient

Show Answer
Answer: ((d))

Liquefaction is the phenomenon where cohesionless soil near the downstream side of dams or sheet-piles loses its shear strength due to high upward hydraulic gradient

Liquefaction of soil:

In loose saturated sand, on shear disturbance due to dynamic forces (e.g. earthquake, large vibrations near large machinery, and railway site) there is a decrease in volume as soil molecules come closer. Hence pore pressure is set, due to which effective stress reduces suddenly and thus shear strength of soil reduces.

Hence the condition where soil will undergo continued deformation at constant low residual effective stress or with zero effective stress is known as liquefaction of soil

Important point:

Piping is the phenomenon where cohesionless soil near the downstream side of dams or sheet-piles loses its shear strength due to high upward hydraulic gradient.

26

In a two-dimensional steady flow field, in a certain region of the x-y plane, the velocity component in the x-direction is given by   vx=x2{v_x} = {x^2}and the density varies as ρ=1x\rho = \frac{1}{x}  Which of the following is a valid expression for the velocity component in the y-direction, vy?

  1. ((a))

    vy=−x/y{v_y} = - x/y

  2. ((b))

    vy=x/y{v_y} = x/y

  3. ((c))

    vy=−xy{v_y} = - xy

  4. ((d))

    vy=xy{v_y} = xy

Show Answer
Answer: ((c))

vy=−xy{v_y} = - xy

Concept:

Continutiy equation in Three-Dimension

∂ρ∂t+∂(ρU)∂X+∂(ρV)∂Y+∂(ρW)∂Z=0;\frac{{\partial {\rm{\rho }}}}{{\partial {\rm{t}}}} + \frac{{\partial \left( {{\rm{\rho U}}} \right)}}{{\partial {\rm{X}}}} + \frac{{\partial \left( {{\rm{\rho V}}} \right)}}{{\partial {\rm{Y}}}} + \frac{{\partial \left( {{\rm{\rho W}}} \right)}}{{\partial {\rm{Z}}}} = 0{\rm{;}}

Where,

U, V, and W are components of velocity in X, Y and Z direction respectively

When the flow is steady,;;∂ρ∂t=0,;;\frac{{\partial {\bf{\rho }}}}{{\partial {\bf{t}}}} = 0

∴∂(ρU)∂X+∂(ρV)∂Y+∂(ρW)∂Z=0\therefore \frac{{{\rm{\partial }}\left( {{\rm{\rho U}}} \right)}}{{{\rm{\partial X}}}} + \frac{{{\rm{\partial }}\left( {{\rm{\rho V}}} \right)}}{{{\rm{\partial Y}}}} + \frac{{{\rm{\partial }}\left( {{\rm{\rho W}}} \right)}}{{{\rm{\partial Z}}}} = 0

When flow is steady and incompressible, ρ = constant

∴∂U∂X+∂V∂Y+∂W∂Z=0;\therefore \frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} + \frac{{\partial {\rm{W}}}}{{\partial {\rm{Z}}}} = 0{\rm{;}}

When the flow is steady, incompressible and 2-D, ∂W∂Z=0\frac{{\partial {\bf{W}}}}{{\partial {\bf{Z}}}} = 0

∴∂U∂X+∂V∂Y=0;\therefore \frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} = 0{\rm{;}}

Calculation:

U = component of velocity in x direction = Vx = x2

ρ=1X\rho = \frac{1}{{{X}}}

As per continuity equation, for two dimensional steady flow,

∴∂(ρU)∂X+∂(ρV)∂Y=0\therefore \frac{{{\rm{\partial }}\left( {{\rm{\rho U}}} \right)}}{{{\rm{\partial X}}}} + \frac{{{\rm{\partial }}\left( {{\rm{\rho V}}} \right)}}{{{\rm{\partial Y}}}} = 0

∴∂(1X×X2)∂X+∂(VX)∂Y=0\therefore \frac{{\partial \left( {\frac{1}{X} \times {{\rm{X}}^2}} \right)}}{{\partial {\rm{X}}}} + \frac{{\partial \left( {\frac{{\rm{V}}}{X}} \right)}}{{\partial {\rm{Y}}}} = 0

∴∂(X)∂X+1X×∂(V)∂Y=0\therefore \frac{{\partial \left( X \right)}}{{\partial {\rm{X}}}} + \frac{1}{X} \times \frac{{\partial \left( {\rm{V}} \right)}}{{\partial {\rm{Y}}}} = 0

∴1X×∂(V)∂Y=−1\therefore \frac{1}{X} \times \frac{{\partial \left( {\rm{V}} \right)}}{{\partial {\rm{Y}}}} = - 1

∴∂(V)∂Y=−X\therefore \frac{{\partial \left( {\rm{V}} \right)}}{{\partial {\rm{Y}}}} = - {\rm{X}}

Integrating on both the side

∴ V = - XY + C

27

For steady incompressible flow through a closed-conduit of uniform cross-section, the direction of flow will always be:

  1. ((a))

    from higher to lower elevation

  2. ((b))

    from higher to lower pressure

  3. ((c))

    from higher to lower velocity

  4. ((d))

    from higher to lower piezometric head

Show Answer
Answer: ((d))

from higher to lower piezometric head

Concept:

Bernoulli’s equation in the form of energy per unit weight is given by,

Pγ+V22g+Z=H\frac{{\rm{P}}}{{\rm{\gamma }}} + \frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}} + {\rm{Z}} = {\rm{H}}

Where,

Pγ=Static;head,V22g=Dynamic;head,;Z=dattum;head,;H=total;head\frac{{\rm{P}}}{{\rm{\gamma }}} = {\rm{Static;head}},\frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}} = {\rm{Dynamic;head}},{\rm{;Z}} = {\rm{dattum;head}},{\rm{;H}} = {\rm{total;head}}

Explanation:

Let's assume point 1 is upstream from point 2

Given,

Steady flow, ∴ Z1 = Z2

Uniform cross-section, ∴ V1 = V2

[Pγ+V122g+Z1]−[Pγ+V222g+Z2]=H\left[ {\frac{{\rm{P}}}{{\rm{\gamma }}} + \frac{{V_1^2}}{{2{\rm{g}}}} + {{\rm{Z}}_1}} \right] - \left[ {\frac{{\rm{P}}}{{\rm{\gamma }}} + \frac{{V_2^2}}{{2{\rm{g}}}} + {{\rm{Z}}_2}} \right] = H

[P1γ]−[P2γ]=H\left[ {\frac{{{{\rm{P}}_1}}}{{\rm{\gamma }}}} \right] - \left[ {\frac{{{{\rm{P}}_2}}}{{\rm{\gamma }}}} \right] = H

h1 - h2 = H

Now head loss is always positive number 

∴ h1 > h2

So the fluid flows from the point with higher piezometric head(h1) to the point with lower piezometric head(h2)

28

A circular pipe has a diameter of 1 m, bed slope of 1 in 1000, and Manning’s roughness coefficient equal to 0.01. It may be treated as an open channel flow when it is flowing just full, i.e., the water level just touches the crest. The discharge in this condition is denoted by Qfull. Similarly, the discharge when the pipe is flowing half-full, i.e., with a flow depth of 0.5 m, is denoted by Qhalf. The ratio Qfull/Qhalf is:

  1. ((a))

    1

  2. ((b))

    2\sqrt 2

  3. ((c))

    2

  4. ((d))

    4

Show Answer
Answer: ((c))

2

Concept:

Manning’s formula

Manning’s formula gives an empirical formula according to which the mean velocity is expressed in terms of a coefficient of roughness (n), called Manning’s roughness coefficient.

V=1n×R23×S12;{\rm{V}} = \frac{1}{{\rm{n}}} \times {{\rm{R}}^{\frac{2}{3}}} \times {{\rm{S}}^{\frac{1}{2}}}{\rm{;}}

Where,

R = hydraulic radius of the channel section,S = bed slope of the channel

Geometrics of channel

  1. Hydraulic Radius or hydraulic mean depth (R):

R=wetted;areawetted;perimeter=AP{\rm{R}} = \frac{{{\rm{wetted;area}}}}{{{\rm{wetted;perimeter}}}} = \frac{{\rm{A}}}{{\rm{P}}}

  1. Hydraulic depth (D)

D=wetted;areaTop;width;of;channel=AT{\rm{D}} = \frac{{{\rm{wetted;area}}}}{{{\rm{Top;width;of;channel}}}} = \frac{{\rm{A}}}{{\rm{T}}}

Discharge in the channel

The flow of water through the channel is given as

Q = A × V

Where,

A = Area of channel section, V = Mean velocity of water flowing through the channel

Calculation:

Given,

S = 1 in 1000, n = 0.01

1) when the pipe is flowing just full

Qfull = Afull × V

Afull = (πr2), P = 2πr

R=AP=πr22πr=r2=d4;{\rm{R}} = \frac{{\rm{A}}}{{\rm{P}}} = \frac{{{\rm{\pi }}{{\rm{r}}^2}}}{{2{\rm{\pi r}}}} = \frac{{\rm{r}}}{2} = \frac{{\rm{d}}}{4}{\rm{;}}

Qfull = A × V

Qfull=;(πr2)×;1n×R23×S12{{\rm{Q}}_{{\rm{full}}}} = {\rm{;}}\left( {{\rm{\pi }}{{\rm{r}}^2}} \right){\rm{}} \times {\rm{;}}\frac{1}{{\rm{n}}} \times {{\rm{R}}^{\frac{2}{3}}} \times {{\rm{S}}^{\frac{1}{2}}}

Qfull=;(πr2)×;10.01×(d4)23×(11000)12{{\rm{Q}}_{{\rm{full}}}} = {\rm{;}}\left( {{\rm{\pi }}{{\rm{r}}^2}} \right){\rm{}} \times {\rm{;}}\frac{1}{{0.01}} \times {\left( {\frac{{\rm{d}}}{4}} \right)^{\frac{2}{3}}} \times {\left( {\frac{1}{{1000}}} \right)^{\frac{1}{2}}}

2) when the pipe is flowing half

Qhalf = Ahalf × V

Ahalf = (πr2)/2, Phalf = πr

R=AP=πr22πr=r2=d4;{\rm{R}} = \frac{{\rm{A}}}{{\rm{P}}} = \frac{{\frac{{{\rm{\pi }}{{\rm{r}}^2}}}{2}}}{{{\rm{\pi r}}}} = \frac{{\rm{r}}}{2} = \frac{{\rm{d}}}{4}{\rm{;}}

Qhalf = A × V

Qhalf=(πr22)×;1n×R23×S12{{\rm{Q}}_{{\rm{half}}}} = \left( {\frac{{{\rm{\pi }}{{\rm{r}}^2}}}{2}} \right){\rm{}} \times {\rm{;}}\frac{1}{{\rm{n}}} \times {{\rm{R}}^{\frac{2}{3}}} \times {{\rm{S}}^{\frac{1}{2}}}

Qhalf=;(πr22)×;10.01×(d4)23×(11000)12{{\rm{Q}}_{{\rm{half}}}} = {\rm{;}}\left( {\frac{{{\rm{\pi }}{{\rm{r}}^2}}}{2}} \right){\rm{}} \times {\rm{;}}\frac{1}{{0.01}} \times {\left( {\frac{{\rm{d}}}{4}} \right)^{\frac{2}{3}}} \times {\left( {\frac{1}{{1000}}} \right)^{\frac{1}{2}}}

3) The ratio of Qfull to Qhalf is

\(\frac{{{{\rm{Q}}{{\rm{full}}}}}}{{{{\rm{Q}}{{\rm{half}}}}}} = \frac{{{\rm{;}}\left( {{\rm{\pi }}{{\rm{r}}^2}} \right){\rm{;;}} \times {\rm{;}}\frac{1}{{0.01}} \times {{\left( {\frac{{\rm{d}}}{2}} \right)}^{\frac{2}{3}}} \times {{\left( {\frac{1}{{1000}}} \right)}^{\frac{1}{2}}}}}{{\left( {\frac{{{\rm{\pi }}{{\rm{r}}^2}}}{2}} \right){\rm{;;}} \times {\rm{;}}\frac{1}{{0.01}} \times {{\left( {\frac{{\rm{d}}}{2}} \right)}^{\frac{2}{3}}} \times {{\left( {\frac{1}{{1000}}} \right)}^{\frac{1}{2}}}}} = \frac{2}{1};\)

\(\frac{{{{\rm{Q}}{{\rm{full}}}}}}{{{{\rm{Q}}{{\rm{half}}}}}} = \frac{2}{1};\)

Important points:

The circular channel section running half-full on one day and running full on another day, the ratio of the velocity of flow of section running full to running half is 1. As the hydraulic mean depth (R) is same in both the condition.

\({\rm{i}}.{\rm{e}}\frac{{{{\rm{V}}{{\rm{full}}}}}}{{{{\rm{V}}{{\rm{half}}}}}} = 1\)

The circular channel section running half-full on one day and running full on another day, the ratio of the headloss due to friction of section running full to running half is 1. As the mean velocity is same in both the condition.

\({\rm{i}}.{\rm{e}}\frac{{{{\rm{(h_f)}}{{\rm{full}}}}}}{{{{\rm{(h_f)}}{{\rm{half}}}}}} = 1\)

29

The two columns below show some parameters and their possible values.

ParameterValues
P – Gross Command AreaI – 100 hectare/cumecs
Q – Permanent Wilting PointII – 6 °C
R – Duty of canal waterIII – 1000 hectares
S – Delta of wheatIV – 1000 cm
V – 40 cm
VI – 0.12

Which of the following options matches the parameters and the values correctly?

  1. ((a))

    P-I, Q-II, R-III, S-IV 

  2. ((b))

    P-III, Q-VI, R-I, S-V

  3. ((c))

    P-I, Q-V, R-VI, S-II

  4. ((d))

    P-III, Q-II, R-V, S-IV

Show Answer
Answer: ((b))

P-III, Q-VI, R-I, S-V

Concept:

Gross command area (GCA):

It is the total area that can be irrigated by a canal so that an unlimited quantity of water is available. A canal is usually aligned along with the watershed between two drainage valleys.

In other words, GCA is the total area lying between the drainage boundaries that can be irrigated by the canal system. It is measured in hectares.

GCA = CCA + UCA

Where,

CCA = culturable command area

UCA = unculturable command area

Permanent wilting point (PWP):

The water content of the soil at which the plant is no longer able to extract water from the soil for its growth.

At a permanent wilting point**, films of water are held very tightly** and hence plant roots are not able to extract sufficient water for their growth

\(PWP = \frac{{\rm{;Water;of;moisture;held;by;the;soil;when;plants;get;permanently;wilted}}}{{\rm Unitweightof~water}}\)

It has no unit

Duty of canal water (D)

Duty is defined as the area of the land irrigated by a unit discharge of water flowing continuously for the duration of the base period of the crop.

D=Area;of;land;being;irrigatedRate;of;water;supply=AQ{\rm{D}} = \frac{{{\rm{Area;of;land;being;irrigated}}}}{{{\rm{Rate;of;water;supply}}}} = \frac{A}{Q}

Duty is measured in hectares/cumec (ha/cumec)

Delta of a crop (Δ):

It is defined as the total thickness of water required for complete maturity of the crop within its base period. It is measured in cm

The sugarcane crop has maximum delta, i.e Δ = 120 cm. Other crops have a delta of less than 120 cm

30

Total Kjeldahl Nitrogen (TKN) concentration (mg/L as N) in domestic sewage is the sum of the concentrations of:

  1. ((a))

    Organic and inorganic nitrogen in sewage

  2. ((b))

    Organic nitrogen and nitrate in sewage

  3. ((c))

    Organic nitrogen and ammonia in sewage

  4. ((d))

    Ammonia and nitrate in sewage

Show Answer
Answer: ((c))

Organic nitrogen and ammonia in sewage

Explanation:

Nitrogen in water:

The nitrogen in water indicates organic contamination of water i.e when water is contaminated by sewage then nitrogen compounds are traced in water. We can trace the following compounds,

  1. Ammonia nitrogen

  2. Organic nitrogen or Albuminoid nitrogen

  3. Nitrites

  4. Nitrates

Nitrogen compounds are measured by colorometry

Total Kjeldahl nitrogen (TKN): 

Kjeldahl Nitrogen = Organic nitrogen + free ammonia

It is the sum of organic nitrogen, ammonia (NH3), and ammonium (NH4+) in the chemical analysis of soil, water and wastewater.

To calculate Total Nitrogen (TN), the concentrations of nitrate-N and nitrite-N are determined and added to the total Kjeldahl nitrogen

31

Solid waste generated from an industry contains only two components, X and Y as shown in the table below

ComponentComposition (% weight)Density (kg/m3)
Xc1ρ1
Yc2ρ2

Assuming (c1 + c2) = 100, the composite density of the solid waste (ρ) is given by:

  1. ((a))

    100(c1ρ1+c2ρ2)\frac{{100}}{{\left( {\frac{{{c_1}}}{{{\rho _1}}} + \frac{{{c_2}}}{{{\rho _2}}}} \right)}}

  2. ((b))

    100(ρ1c1+ρ2c2)100\left( {\frac{{{\rho _1}}}{{{c_1}}} + \frac{{{\rho _2}}}{{{c_2}}}} \right)

  3. ((c))

    100(c1ρ1+c2ρ2)100\left( {{c_1}{\rho _1} + {c_2}{\rho _2}} \right)

  4. ((d))

    100(ρ1ρ2c1ρ1+c2ρ2)100\left( {\frac{{{\rho _1}{\rho _2}}}{{{c_1}{\rho _1} + {c_2}{\rho _2}}}} \right)

Show Answer
Answer: ((a))

100(c1ρ1+c2ρ2)\frac{{100}}{{\left( {\frac{{{c_1}}}{{{\rho _1}}} + \frac{{{c_2}}}{{{\rho _2}}}} \right)}}

Concept:

The Specific gravity of municipal solid waste (M.S.W)  containing components 1 and 2 is calculated by,

\(\frac{{100}}{{{{\rm{S}}{{\rm{M}}.{\rm{S}}.{\rm{W}}}}}} = \frac{{{\rm{% ;}}{{\rm{C}}1}}}{{{{\rm{S}}{{\rm{C}}1}}}} + \frac{{{\rm{% ;}}{{\rm{C}}2}}}{{{{\rm{S}}{{\rm{C}}2}}}} + .{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.\frac{{{\rm{% ;}}{{\rm{C}}{\rm{n}}}}}{{{{\rm{S}}_{{\rm{Cn}}}}}}\)

The above formula in terms of density (ρ )

\(\frac{{100}}{{{{\rm{\rho }}{{\rm{M}}.{\rm{S}}.{\rm{W}}}}}} = \frac{{{\rm{% ;}}{{\rm{C}}1}}}{{{{\rm{\rho }}{{\rm{C}}1}}}} + \frac{{{\rm{% ;}}{{\rm{C}}2}}}{{{{\rm{\rho }}{{\rm{C}}2}}}} + .{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.{\rm{;}}.\frac{{{\rm{% ;}}{{\rm{C}}{\rm{n}}}}}{{{{\rm{\rho }}_{{\rm{Cn}}}}}}\)

Where,                                                   

SM.S.W, SC1, SC2  = Specific gravity of total municipal solid waste, component 1, and component 2, respectively

% C1  and % C2 = weight in percentage by weight of component  1 and component 2 , respectively.

ρ1 and ρ2 = density of component 1 and component 2 respectively.  

Calculation:

Let density of sludge is ρ

\(\frac{{100}}{{{{\rm{\rho }}_{{\rm{M}}.{\rm{S}}.{\rm{W}}}}}} = \frac{{{\rm{% ;}}{{\rm{C}}1}}}{{{{\rm{\rho }}{{\rm{C}}1}}}} + \frac{{{\rm{% ;}}{{\rm{C}}2}}}{{{{\rm{\rho }}{{\rm{C}}2}}}} \)

\(\therefore {{\rm{\rho }}_{{\rm{M}}.{\rm{S}}.{\rm{W}}}} = \frac{{100}}{{\frac{{{\rm{% ;}}{{\rm{C}}1}}}{{{{\rm{\rho }}{{\rm{C}}1}}}} + \frac{{{\rm{% ;}}{{\rm{C}}2}}}{{{{\rm{\rho }}{{\rm{C}}2}}}}}}\)

32

The penetration value of a bitumen sample tested at 25°C is 80. When this sample is heated to 60°C and tested again, the needle of the penetration test apparatus penetrates the bitumen sample by d mm. The value of d CANNOT be less than _______________ mm.

33

Which of the following statements CANNOT be used to describe free flow speed (uf) of a traffic stream?

  1. ((a))

    uf is the speed when flow is negligible.

  2. ((b))

    uf is the speed when density is negligible.

  3. ((c))

    uf is affected by geometry and surface conditions of the road.

  4. ((d))

    uf is the speed at which flow is maximum and density is optimum.

Show Answer
Answer: ((d))

uf is the speed at which flow is maximum and density is optimum.

For the option A:

For the speed - flow relationship:

It can be observed that the free flow speed (uf) is the speed when flow is negligible. 

At low speed, traffic volume would be low. With an increase in speed, volume increases up to a certain limit (approximately half of free-flow speed) then reduced similarly after that limit.

For option B: 

 

From the above diagrams, it can be observed that free flow speed *(*uf ) is the speed when density is negligible.

With the increase in density, traffic volume increases upto certain limit then decreases.

For option C: Traffic speed uf is affected by geometry and surface conditions of the road which is correct.

For option D: From the above statements we can conclude that uf is the speed at which flow is maximum and density is optimum is incorrect.

Mistake point:

In such questions we get confused, or we dont read the question carefully. As in this question we have to choose incorrect statement. 

We can see that all the options except Option D are correct, as in question they asked about incorrect statement, so our answer will be option D

34

Which of the following statements is FALSE?

  1. ((a))

    Plumb line is along the direction of gravity.

  2. ((b))

    Mean Sea Level (MSL) is used as a reference surface for establishing the horizontal control.

  3. ((c))

    Mean Sea Level (MSL) is a simplification of the Geoid.

  4. ((d))

    Geoid is an equi-potential surface of gravity.

Show Answer
Answer: ((b))

Mean Sea Level (MSL) is used as a reference surface for establishing the horizontal control.

Plumb line:

Plumb line is a string having a metal weight at the end when it is suspended freely, it points directly towards the center of gravity.

It denotes the verticality of the object and the depth of the water.

Mean sea level (M.S.L):

M.S.L is the datum to which elevation and contour intervals are generally referred. It is used as a reference surface for establishing vertical control.

Geoid:

Geoid serves as a reference surface from which topographic heights and ocean depths are measured. It coincides with M.S.L over the ocean and continues in continental areas as an imaginary sea level surface defined by a spirit level. Mathematically speaking geoid is an equipotential surface of gravity

We can say that M.S.L is the simplification of the geoid.

35

In a closed loop traverse of 1 km total length, the closing errors in departure and latitude are 0.3 m and 0.4 m, respectively. The relative precision of this traverse will be:

  1. ((a))

    1: 5000 

  2. ((b))

    1: 4000 

  3. ((c))

    1: 3000 

  4. ((d))

    1: 2000

Show Answer
Answer: ((d))

1: 2000

Concept:

The equation for closing error and angle fo misclosure is given by,

e=(ΣL)2+(ΣD)2{\rm{e}} = \sqrt {{{\left( {{\rm{Σ L}}} \right)}^2} + {{\left( {{\rm{Σ D}}} \right)}^2}}

θ=tan⁡−1[ΣDΣL]{\rm{\theta }} = {\tan ^{ - 1}}\left[ {\frac{{{\rm{Σ D}}}}{{{\rm{Σ L}}}}} \right]

Where,

e = closing error, θ = angle of misclosure

ΣL = Algebraic sum of latitudes of all lines  

ΣD = Algebraic sum of departures of all lines

The relative error of closure (r)

r=Closing;error;of;a;traverseperimeter;of;a;traverse=eP{\rm{r}} = \frac{{{\rm{Closing;error;of;a;traverse}}}}{{{\rm{perimeter;of;a;traverse}}}} = \frac{{\rm{e}}}{{\rm{P}}}

Where, P =perimeter of a traverse = Total length of a traverse

The degree of accuracy or the relative precision

Relative;precision=1Pe{\rm{Relative;precision}} = \frac{1}{{\frac{{\rm{P}}}{{\rm{e}}}}}

Calculation:

Given,

ΣL = 0.4m, ΣD = 0.3 m

P = 1 km = 1000 m

we know that

e=(ΣL)2+(ΣD)2{\rm{e}} = \sqrt {{{\left( {{\rm{Σ L}}} \right)}^2} + {{\left( {{\rm{Σ D}}} \right)}^2}}

e=(0.4)2+(0.3)2{\rm{e}} = \sqrt {{{\left( {{\rm{0.4}}} \right)}^2} + {{\left( {{\rm{0.3}}} \right)}^2}}

e = 0.5 m

Relative;precision=1Pe=110000.5=510000=12000{\rm{Relative;precision}} = \frac{1}{{\frac{{\rm{P}}}{{\rm{e}}}}}=\frac{1}{{\frac{{\rm{1000}}}{{\rm{0.5}}}}} =\frac{5}{{10000}} = \frac{1}{{2000}}

36

The smallest and largest Eigen values of the following matrix are:

\(\left[ {\begin{array}{*{20}{c}} 3&{ - 2}&2\ 4&{ - 4}&6\ 2&{ - 3}&5 \end{array}} \right]\)

  1. ((a))

    1.5 and 2.5

  2. ((b))

    0.5 and 2.5

  3. ((c))

    1.0 and 3.0

  4. ((d))

    1.0 and 2.0

Show Answer
Answer: ((d))

1.0 and 2.0

Concept:

Eigen Values

Let A be a square matrix of order ‘n’ and ‘λ’ be a scalar.

​∣A−λI∣=0\left| {A - \lambda I} \right| = 0 is called the characteristic equation of matrix A.

The roots of the characteristic equation are called Eigenvalues.

Corresponding to each eigen value ‘λ’, there exists a non-zero vector ‘X’ such that AX = λX or (A -λI)X = 0

Calculation:

Given matrix is,

\(\left[ {\begin{array}{*{20}{c}} 3&{ - 2}&2\ 4&{ - 4}&6\ 2&{ - 3}&5 \end{array}} \right]\)

The characteristic eqaution for the given matrix is as follows

 ∣A−λI∣=0\left| {A - \lambda I} \right| = 0

\( \Rightarrow \left| {\begin{array}{*{20}{c}} {3 - \lambda }&{ - 2}&2\ 4&{ - 4 - \lambda }&6\ 2&{ - 3}&{5 - \lambda } \end{array}} \right| = 0\)

⇒ (3 – λ) (– 20 + 4λ – 5λ + λ2 + 18) + 2 (20 – 4λ – 12) + 2 (-12 + 8 + 2λ) = 0

⇒ λ3 – 4λ2 + 5λ – 2 = 0

Now we can put values from the given options, and see which option satisfies the above equation

Only 1 and 2 satisfy this equation.

λ = 1, 1, 2

Hence, smallest Eigen value = 1 and Largest Eigen value = 2

37

The quadratic equation x2−4x+4=0{x^2} - 4x + 4 = 0 is to be solved numerically, starting with the initial guess

X0 = 3. The Newton-Raphson method is applied once to get a new estimate and then the Secant method is applied once using the initial guess and this new estimate. The estimated value of the root after the application of the Secant method is _______________.

38

Consider the following differential equation:

x(ydx+xdy)cos⁡yx=y(xdy−ydx)sin⁡yxx\left( {ydx + xdy} \right)\cos \frac{y}{x} = y\left( {xdy - ydx} \right)\sin \frac{y}{x}

Which of the following is the solution of the above equation (c is an arbitrary constant)?

  1. ((a))

    xycos⁡yx=c\frac{x}{y}\cos \frac{y}{x} = c

  2. ((b))

    xysin⁡yx=c\frac{x}{y}\sin \frac{y}{x} = c

  3. ((c))

    xycos⁡yx=cxy\cos \frac{y}{x} = c

  4. ((d))

    xysin⁡yx=cxy\sin \frac{y}{x} = c

Show Answer
Answer: ((c))

xycos⁡yx=cxy\cos \frac{y}{x} = c

Given differential eqaution is,

x(ydx+xdy)cos⁡yx=y(xyd−ydx)sin⁡yx ∴ydx+xdyxdy−ydx=yxtan⁡yx........(1)\begin{array}{l} x\left( {ydx + xdy} \right)\cos \frac{y}{x} = y\left( {xyd - ydx} \right)\sin \frac{y}{x}\ \therefore\frac{{ydx + xdy}}{{xdy - ydx}} = \frac{y}{x}\tan \frac{y}{x}........ (1) \end{array}

Let, y = v × x

dy = vdx + xdv

By substituting values of Y and dY in equation 1, we get

vxdx+vxdx+x2dvvxdx+x2dv−vxdx=vtan⁡v xdv+2vdxxdv=v;tan;v 1+2vxdxdv=v;tan;v; 2vxdxdv=vtan⁡v−1\begin{array}{l} \frac{{vxdx + vxdx + {x^2}dv}}{{vxdx + {x^2}dv - vxdx}} = v\tan v\ \frac{{xdv + 2vdx}}{{xdv}} = v;tan;v\ 1 + \frac{{2v}}{x}\frac{{dx}}{{dv}} = v;tan;v;\ \frac{{2v}}{x}\frac{{dx}}{{dv}} = v\tan v - 1 \end{array}

Integrating both sides

2 log x = log |sec v| - log v + log c

⇒x2=csec⁡vv ⇒x2yx=csec⁡yx ⇒x2yx=csec⁡yx xycos⁡yx=c\begin{array}{l} \Rightarrow {x^2} = \frac{{c\sec v}}{v}\ \Rightarrow \frac{{{x^2}y}}{x} = c\sec \frac{y}{x}\ \Rightarrow {x^2}\frac{y}{x} = c\sec \frac{y}{x}\ xy\cos \frac{y}{x} = c \end{array}

39

Consider the following complex function:

f(z)=9(z−1)(z+2)2f\left( z \right) = \frac{9}{{\left( {z - 1} \right){{\left( {z + 2} \right)}^2}}}

Which of the following is one of the residues of the above function?

  1. ((a))

    −1

  2. ((b))

    9/16

  3. ((c))

    2

  4. ((d))

    9

Show Answer
Answer: ((a))

−1

Concept:

The residue of f(z):

If Zo is an isolated singular point of f(Z) then the coefficient of 1Z−Zo\frac{1}{{{\rm{Z}} - {{\rm{Z}}_{\rm{o}}}}} in Laurent’s series expansion of f(z) about the point Zo is called the residue of f(Z) at Zo

Residue at poles:

a) If f(z)=P(z)q(z){\rm{f}}\left( {\rm{z}} \right) = \frac{{{\rm{P}}\left( {\rm{z}} \right)}}{{{\rm{q}}\left( {\rm{z}} \right)}} where P(z) and q(z) are polynomials and has a simple pole at Z = Zo then

\({\rm{Res;}}\left[ {{\rm{f}}\left( {\rm{z}} \right):{\rm{Z}} = {{\rm{Z}}{\rm{o}}}} \right] = \mathop {\lim }\limits{{\rm{z}} \to {{\rm{z}}{\rm{o}}}} \left[ {\left( {{\rm{Z}} - {{\rm{Z}}{\rm{o}}}} \right){\rm{f}}\left( {\rm{z}} \right)} \right]\)

b) If f(z) has poles of order m at Z = Zo then

\({\rm{Res;}}\left[ {{\rm{f}}\left( {\rm{z}} \right):{\rm{Z}} = {{\rm{Z}}{\rm{o}}}} \right] = \frac{1}{{\left( {m - 1} \right)!}}\mathop {\lim }\limits{{\rm{z}} \to {{\rm{z}}{\rm{o}}}} \left{ {\frac{{{d^{m - 1}}}}{{d{Z^{m - 1}}}}\left[ {\left( {{\rm{Z}} - {{\rm{Z}}{\rm{o}}}} \right){\rm{f}}\left( {\rm{z}} \right)} \right]} \right};\)

Calculation:

Given,

f(z)=9(Z−1)(Z+2)2{\rm{f}}\left( {\rm{z}} \right) = \frac{9}{{\left( {{\rm{Z}} - 1} \right){{\left( {{\rm{Z}} + 2} \right)}^2}}}

The function f(z) has singular points at Z = 1 and Z = -2, these are the poles of order 1 and order 2 respectively.

R1 = Res [f(z) : Z =1] 

\(\begin{array}{l} R_1= \mathop {\lim }\limits_{z \to 1} \left( {z - 1} \right)f\left( z \right)\ R_1= \mathop {\lim }\limits_{z \to 1} \begin{array}{*{20}{c}} {\frac{9}{{{{\left( {z + 2} \right)}^2}}}} \end{array} \end{array}\)

R1=9(1+2)2R_1= \mathop {\frac{9}{{{{\left( {1 + 2} \right)}^2}}}}

R1 = 1

R2 = Res [f(z) : Z =-2] 

R2=lim⁡z→−2ddz2−1[(z+2)2f(z)] R2=lim⁡z→−2ddz(9z−1) R2=lim⁡z→−2−9(z−1)2R2=−1\begin{array}{l} R_2= \mathop {\lim }\limits_{z \to - 2} \frac{d}{{d{z^{2 - 1}}}}\left[ {{{\left( {z + 2} \right)}^2}f\left( z \right)} \right]\ R_2= \mathop {\lim }\limits_{z \to - 2} \frac{d}{{dz}}\left( {\frac{9}{{z - 1}}} \right)\ R_2= \mathop {\lim }\limits_{z \to - 2} \frac{{ - 9}}{{{{\left( {z - 1} \right)}^2}}} \R_2= - 1 \end{array}

Hence the option 1 is corect

40

The directional derivative of the field u(x,y,z)=x2−3yzu\left( {x,y,z} \right) = {x^2} - 3yz in the direction of the vector (i^+j^−2k^)\left( {\hat i + \hat j - 2\hat k} \right) at point (2,−1,;4)\left( {2, - 1,;4} \right) is ______.

41

The composition of an air-entrained concrete is given below:

Water                                                    : 184 kg/m3

Ordinary Portland Cement (OPC)        : 368 kg/m3

Sand                                                     : 606 kg/m3

Coarse aggregate                                 : 1155 kg/m3

Assume the specific gravity of OPC, sand and coarse aggregate to be 3.14, 2.67 and 2.74,

respectively. The air content is _______________ liters/ m3.

42

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.

43

Consider the singly reinforced beam section given below (left figure). The stress block parameters for the cross - section from IS : 456 - 2000 are also given below (right figure). The moment of resistance for the given section by the limit state method is _______________ kN-m.

44

For the formation of collapse mechanism in the following figure, the minimum value of Pu is cMp/L. Mp and 3Mp denote the plastic moment capacities of beam sections as shown in this figure. The value of c is _______________.

45

A tapered circular rod of diameter varying from 20 mm to 10 mm is connected to another uniform circular rod of diameter 10 mm as shown in the following figure. Both bars are made of same material with the modulus of elasticity, E = 2x105 MPa. When subjected to a load P = 30π kN, the deflection at point A is _______________ mm.

46

Two beams are connected by a linear spring as shown in the following figure. For a load P as shown in the figure, the percentage of the applied load P carried by the spring is _______________.

 

47

For the 2-D truss with the applied loads shown below, the strain energy in the member XY is _______________ kN-m. For member XY, assume AE = 30 kN, where A is cross-section area and E is the modulus of elasticity.

48

An earth embankment is to be constructed with compacted cohesionless soil. The volume of the embankment is 5000 m3 and the target dry unit weight is 16.2 kN/m3. Three nearby sites (see figure below) have been identified from where the required soil can be transported to the construction site. The void ratios (e) of different sites are shown in the figure. Assume the specific gravity of soil to be 2.7 for all three sites. If the cost of transportation per km is twice the cost of excavation per m3 of borrow pits, which site would you choose as the most economic solution? (Use unit weight of water = 10 kN/m3)

  1. ((a))

    Site X

  2. ((b))

    Site Y

  3. ((c))

    Site Z

  4. ((d))

    Any of the sites

Show Answer
Answer: ((a))

Site X

Concept:

The dry unit weight of soil is given by,

\({{\rm{γ }}{\rm{d}}} = \frac{{{\rm{G}} × {{\rm{γ }}{\rm{w}}}}}{{1 + {\rm{e}}}}\)

where,

e = voids ratio, G = specific gravity of soil, Ws = weight of soil in kg, V = volume of soil in m3

γ­d = unit weight of dry soil in kN/m3, γ­w = unit weight of water in kN/m3

The relation between porosity (n) and voids ratio (e) is given by,

n=e1+e{\rm{n}} = \frac{{\rm{e}}}{{1 + {\rm{e}}}}

where,

\({\rm{n}} = \frac{{{{\rm{V}}{\rm{V}}}}}{{\rm{V}}},{\rm{;e}} = \frac{{{{\rm{V}}{\rm{V}}}}}{{{{\rm{V}}_{\rm{S}}}}}\)

\(\frac{{{{\rm{V}}{\rm{V}}}}}{{\rm{V}}} = \frac{{\frac{{{{\rm{V}}{\rm{V}}}}}{{{{\rm{V}}_{\rm{S}}}}}}}{{1 + {\rm{e}}}}\)

∴VS=V1+e\therefore {{\rm{V}}_{\rm{S}}} = \frac{{\rm{V}}}{{1 + {\rm{e}}}}

VV = volume of voids, VS = volume of solids, V = total volume of soil

V = VV + VS

Calculation:

Given,

Cost of transportation per km = 2 × cost of excavation per m3

Volume of embankment, V = 5000 m3

γd  = 16.2 kN/m3, γW = 10 kN/m3 and G = 2.7, 

Volume of solids remain constant.

For embankment

\({{\rm{γ }}{\rm{d}}} = \frac{{{\rm{G}} × {{\rm{γ }}{\rm{w}}}}}{{1 + {\rm{e}}}}\)

16.2=(2.71+e)×10\begin{array}{l} 16.2 = \left( {\frac{{2.7}}{{1 + e}}} \right) × 10 \end{array}

e = 0.667

V = 5000

Vs=V1+e=3000 m3{V_s} = \frac{V}{{1 + e}} = 3000\ {m^3}

Site X

Vs=Vx1+e=3000 m3 Vx=3000×1.6=4800 m3\begin{array}{l} {V_s} = \frac{V_x}{{1 + e}} = 3000\ {m^3}\ {V_x} = 3000 × 1.6 = 4800\ {m^3} \end{array}

Total cost =4800x+140x×2=5080x = 4800x + {{140x}}\times {2} = 5080x

Site Y

Vy=3000×1.7=5100 m3{V_y} = 3000 × 1.7 = 5100\ {m^3}

Total cost =5100x+80x×2=5260x = 5100x + 80{x}\times{2} = 5260x

Site Z

Vz=3000×1.64=4920;m3{V_z} = 3000 × 1.64 = 4920;{m^3}

Total cost =4920x+100x×2=5120x = 4920x + 100{x}\times{2} =5120x

Hence, choose site X which has minimum cost.

49

A water tank is to be constructed on the soil deposit shown in the figure below. A circular footing of diameter 3 m and depth of embedment 1 m has been designed to support the tank. The total vertical load to be taken by the footing is 1500 kN. Assume the unit weight of water as 10 kN/m3 and the load dispersion pattern as 2V:1H. The expected settlement of the tank due to primary consolidation of the clay layer is ___________ mm.

50

A 20 m thick clay layer is sandwiched between a silty sand layer and a gravelly sand layer. The layer experiences 30 mm settlement in 2 years.

Given:

\({T_v} = \left{ {\begin{array}{{20}{c}} {\frac{\pi }{4}{{\left( {\frac{U}{{100}}} \right)}^2}}\ {1.781 - 0.933{{\log }_{10}}\left( {100 - U} \right)} \end{array}\begin{array}{{20}{c}} {for;U \le 60% }\ {for;U > 60% } \end{array}} \right.;\)

Where Tv is the time factor and U is the degree of consolidation in %.

If the coefficient of consolidation of the layer is 0.003 cm2/s, the deposit will experience a total of 50 mm settlement in the next _______________ years.

51

A non-homogeneous soil deposit consists of a silt layer sandwiched between a fine-sand layer at top and a clay layer below. Permeability of the silt layer is 10 times the permeability of the clay layer and one tenth of the permeability of the sand layer. Thickness of the silt layer is 2 times the thickness of the sand layer and two-third of the thickness of the clay layer. The ratio of equivalent horizontal and equivalent vertical permeability of the deposit is ________.

52

A square footing (2 x 2) m is subjected to an inclined point load, P as shown in the figure below. The water table is located well below the base of the footing. Considering one-way eccentricity, the net safe load carrying capacity of the footing for a factor of safety of 3.0 is _______ kN.

The following factors may be used:

Bearing capacity factors: Nq = 33.3, Nγ = 37.16; Shape factors; Fqs = Fγs = 1.314; Depth factors: Fqd = Fγd = 1.113; Inclination factors: Fqi = 0.444, Fγi = 0.02

53

Two reservoirs are connected through a 930 m long, 0.3 m diameter pipe, which has a gate valve. The pipe entrance is sharp (loss coefficient = 0.5) and the valve is half-open (loss coefficient = 5.5). The head difference between the two reservoirs is 20 m. Assume the friction factor for the pipe as 0.03 and g =10 m/s2. The discharge in the pipe accounting for all minor and major losses is _______________ m3/s.

54

A hydraulic jump is formed in a 2 m wide rectangular channel which is horizontal and frictionless. The post-jump depth and velocity are 0.8 m and 1 m/s, respectively. The pre-jump velocity is _______________ m/s. (use g = 10 m/s2)

55

A short reach of a 2 m wide rectangular open channel has its bed level rising in the direction of flow at a slope of 1 in 10000. It carries a discharge of 4 m3/s and its Manning’s roughness coefficient is 0.01. The flow in this reach is gradually varying. At a certain section in this reach, the depth of flow was measured as 0.5 m. The rate of change of the water depth with distance, dy/dx, at this section is _______________ (use g = 10 m/s2).

56

The drag force, FD, on a sphere kept in a uniform flow field depends on the diameter of the sphere, D; flow velocity, V; fluid density, ρ; and dynamic viscosity, μ. Which of the following options represents the non-dimensional parameters which could be used to analyze this problem?

  1. ((a))

    FDVD\frac{{{F_D}}}{{{V_D}}} and μρVD\frac{\mu }{{\rho VD}}

  2. ((b))

    FDρVD2\frac{{{F_D}}}{{\rho V{D^2}}} and ρVDμ\frac{{\rho VD}}{\mu }

  3. ((c))

    FDρV2D2\frac{{{F_D}}}{{\rho {V^2}{D^2}}} and ρVDμ\frac{{\rho VD}}{\mu }

  4. ((d))

    FDρV3D3\frac{{{F_D}}}{{\rho {V^3}{D^3}}} and μρVD\frac{\mu }{{\rho VD}}

Show Answer
Answer: ((c))

FDρV2D2\frac{{{F_D}}}{{\rho {V^2}{D^2}}} and ρVDμ\frac{{\rho VD}}{\mu }

Explanation:

Given,

Drag force is the function of D, V, ρ, and μ 

i.e FD = fn (D, V, ρ, μ) 

We know that the formula for Drag force is,

FD=CD×12ρAV2{F_D} = {C_D} \times \frac{1}{2}ρ A{V^2}

 FD=CD12ρ(π4D2)V2\ {F_D} = {C_D}{\frac{1 }{2}} ρ \left( {\frac{\pi }{4}{D^2}} \right){V^2}

\({{\rm{C}}{\rm{D}}} = \frac{{{{\rm{F}}{\rm{D}}} \times 8}}{{{\rm{ρ }} \times {\rm{\pi }} \times {{\rm{D}}^2} \times {{\rm{V}}^2}}}\)

Eliminating constants

\({{\rm{C}}{\rm{D}}} = \frac{{{{\rm{F}}{\rm{D}}}}}{{{\rm{ρ }} \times {{\rm{D}}^2} \times {{\rm{V}}^2}}}\)

This can be solved by multiplying the Reynold's number 

Re=ρVDμ=dimensionless\begin{array}{l} {R_e} = \frac{{ρ VD}}{μ } = dimensionless \end{array}

Hence the non-dimensional parameters to analyze the Drag force is \({{\rm{C}}{\rm{D}}} = \frac{{{{\rm{F}}{\rm{D}}}}}{{{\rm{ρ }} \times {{\rm{D}}^2} \times {{\rm{V}}^2}}}\) and Re=ρVDμ\begin{array}{l} {R_e} = \frac{{ρ VD}}{μ }\end{array}

57

In a catchment, there are four rain-gauge stations, P, Q, R, and S. Normal annual precipitation values at these stations are 780 mm, 850 mm, 920 mm, and 980 mm, respectively. In the year 2013, stations Q, R, and S, were operative but P was not. Using the normal ratio method, the precipitation at station P for the year 2013 has been estimated as 860 mm. If the observed precipitation at stations Q and R for the year 2013 were 930 mm and 1010 mm, respectively; what was the observed precipitation (in mm) at station S for that year?

58

The 4-hr unit hydrograph for a catchment is given in the table below. What would be the maximum ordinate of the S-curve (in m3/s) derived from this hydrograph?

Time (hr)024681012141618202224
Unit hydrograph Ordinate (m3/s)00.63.110139520.70.30.20.10
59

The concentration of Sulfur Dioxide (SO2) in ambient atmosphere was measured as 30 μg/m3. Under the same conditions, the above SO2 concentration expressed in ppm is _______________. Given: P/(RT) = 41.6 mol/m3; where, P = Pressure; T = Temperature; R = universal gas constant; Molecular weight of SO2 = 64.

60

Consider a primary sedimentation tank (PST) in a water treatment plant with Surface Overflow Rate (SOR) of 40 m3/m2/d. The diameter of the spherical particle which will have 90 percent theoretical removal efficiency in this tank is _______________ μm. Assume that settling velocity of the particles in water is described by Stokes’s Law.

Given: Density of water = 1000 kg/m3; Density of particle = 2650 kg/m3; g = 9.81 m/s2; Kinematic viscosity of water= 1.1x 10-6 m2/s

61

The acceleration-time relationship for a vehicle subjected to non-uniform acceleration is,

dvdt=(α−βv0)e−βt\frac{{dv}}{{dt}} = \left( {α - β {v_0}} \right){e^{ - β t}}

where, v is the speed in m/s, t is the time in s, α and β are parameters, and v0 is the initial speed in m/s. If the accelerating behaviour of a vehicle, whose driver intends to overtake a slow moving vehicle ahead, is described as,

dvdt=(α−βv)\frac{{dv}}{{dt}} = \left( {α - β v} \right)

Considering α = 2 m/sec2, β = 0.05 s-1 and dvdt=1.3;m/s2\frac{{dv}}{{dt}} = 1.3;m/{s^2} at t = 3 s, the distance (in m) travelled by the vehicle in 35 s is ___. m

62

On a circular curve, the rate of super-elevation is e. While negotiating the curve a vehicle comes to a stop. It is seen that the stopped vehicle does not slide inwards (in the radial direction). The coefficient of side friction is f. Which of the following is true:

  1. ((a))

    e ≤ f

  2. ((b))

    f < e < 2f

  3. ((c))

    e ≥ 2f

  4. ((d))

    None of the above

Show Answer
Answer: ((a))

e ≤ f

From the above figure under equilibrium condition

fN ≥ mg sin θ

where N = Normal reaction = m g cosθ 

f × m × g × cos θ ≥ m × g × sin θ

f × cos θ ≥ sin θ

f ≥ tan θ 

where tanθ = superelevation = e

f ≥ e

63

A sign is required to be put up asking drivers to slow down to 30 km/h before entering Zone Y (see figure). On this road, vehicles require 174 m to slow down to 30 km/h (the distance of 174 m includes the distance travelled during the perception-reaction time of drivers). The sign can be read by 6/6 vision drivers from a distance of 48 m. The sign is placed at a distance of x m from the start of Zone Y so that even a 6/9 vision driver can slow down to 30 km/h before entering the zone. The minimum value of x is _______________ m.

64

In a survey work, three independent angles X, Y and Z were observed with weights WX, WY, WZ, respectively. The weight of the sum of angles X, Y and Z is given by:

  1. ((a))

    1(1WX+1WY+1WZ)\frac{1}{{\left( {\frac{1}{{{W_X}}} + \frac{1}{{{W_Y}}} + \frac{1}{{{W_Z}}}} \right)}}

  2. ((b))

    (1WX+1WY+1WZ)\left( {\frac{1}{{{W_X}}} + \frac{1}{{{W_Y}}} + \frac{1}{{{W_Z}}}} \right)

  3. ((c))

    WX+WY+WZ{W_X} + {W_Y} + {W_Z}

  4. ((d))

    WX2+WY2+WZ2W_X^2 + W_Y^2 + W_Z^2

Show Answer
Answer: ((a))

1(1WX+1WY+1WZ)\frac{1}{{\left( {\frac{1}{{{W_X}}} + \frac{1}{{{W_Y}}} + \frac{1}{{{W_Z}}}} \right)}}

From the method of least squares the following laws of weights are established:

  1. The weight of the arithmetic mean of the measurements of unit weight is equal to the number of observations.
  2. The weight of the weighted arithmetic means is equal to the sum of the individual weights.
  3. The weight of the algebraic sum of two or more quantities is equal to the reciprocals of the individual weights.
  4. If a quantity of given weight is multiplied by a factor, the weight of the result is obtained by dividing its given weight by the square of the factor.
  5. If a quantity of given weight is divided by a factor, the weight of the result is obtained by multiplying its given weight by the square of the factor.
  6. If an equation is multiplied by its weight, the weight of the resulting equation is equal to the reciprocal of the weight of the equation.
  7. The weight of the equation remains unchanged, if all the signs of the equation are changed or if the equation is added or subtracted from a constant.

Hence as per the third law of weights, correct option is option 1

65

In a region with magnetic declination of 2°E, the magnetic Fore bearing (FB) of a line AB was measured as N79°50’E. There was local attraction at A. TO determined the correct magnetic bearing of the line, a point O was selected at which there was no local attraction. The magnetic FB of line AO and OA were observed to be S52° 40’E and N50°20’W, respectively. What is the true FB of line AB?

  1. ((a))

    N81°50’E

  2. ((b))

    N82°10’E

  3. ((c))

    N84°10’E

  4. ((d))

    N77°50’E

Show Answer
Answer: ((c))

N84°10’E

Magnetic Declination, δ =2°E

Magnetic FB of AB = N 79°50’E

To find local attraction at station A

As station O is free from local attraction

Hence FB of OA will be correct

Correct FB of OA = N 50°20’W @ 309° 40’

∴ Correct BB of OA = @ 309° 40’ - 180° = 129° 40’

∵ Observed FB of AO= Observed BB of OA= S52° 40’E= 127°20’

Error = MB – TB = 127° 20’ – 129° 40’ = - 2° 20’

Correction = + 2° 20’

Local attraction at station A

= + 2°20’ @   2°20’ E

∴ Magnetic F.B of AB = N 79° 50’ E

δ = 2° E and local attraction = 2° 20’E

∴ True Bearing of FB of AB = 79° 50’ + 2° 20’ + 2° = N84° 10’E

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