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

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.

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

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

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.

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

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.

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.

Concept:

Combined imports and exports in the year 2000 = 50 + 40 = 90 crores

Combined imports and exports in the year 2001 = 60 + 50 = 110 crores

Combined imports and exports in the year 2002 = 70 + 60 = 130 crores

Combined imports and exports in the year 2003 = 60 + 70 = 130 crores

Combined imports and exports in the year 2004 = 70 + 80 = 150 crores

Combined imports and exports in the year 2005 = 70 + 90 = 160 crores

Combined imports and exports in the year 2006 = 100 + 120 = 220 crores

Combined imports and exports in the year 2007 = 110 + 120 = 230 crores

% increase in combined imports and exports in the year 2001 \(= \frac{{110 - 90}}{{90}} \times 100% = 22.22% \)

% increase in combined imports and exports in the year 2002 \(= \frac{{130 - 110}}{{110}} \times 100% = 18.18% \)

There is no increase in 2003.

% increase in combined imports and exports in the year 2004 \(= \frac{{150 - 130}}{{130}} \times 100% = 15.38% \)

% increase in combined imports and exports in the year 2005 \(= \frac{{160 - 150}}{{150}} \times 100% = 6.67%\)

% increase in combined imports and exports in the year 2006 \(= \frac{{220 - 160}}{{160}} \times 100% = 37.5%\)

% increase in combined imports and exports in the year 2007 \(= \frac{{230 - 220}}{{220}} \times 100% = 4.54% \)

Thus % increase in combined imports and exports in the year 2006 is highest.

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|)

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.

Concept:

A = [a_ij]_{m × n} = coefficient of matrix, X = Column matrix of the variables

B = column matrix of constants

The system AX = B has

1. A unique solution, If Rank of A = Rank [A|B] and is equal to the number of variables.

2. Infinitely many solutions, If Rank of A = Rank of [A|B] < number of variables

3. No solution, If Rank of A ≠ Rank of [A|B], i.e. Rank of A < Rank of [A|B].

Calculation:

Given,

AX = B

\(\left[ {\begin{array}{*{20}{c}} 2&3\ 3&P \end{array}} \right]\left[ {\begin{array}{*{20}{c}} X\ Y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5\ {10} \end{array}} \right]\)

Consider the augmented matrix [A|B]

\(\left[ {{\rm{A|B}}} \right] = \left[ {\begin{array}{*{20}{c}} 2&3\ 3&{\rm{P}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5\ {10} \end{array}} \right]\)

3R₁ - 2R₂

\(\left[ {{\rm{A|B}}} \right] = \left[ {\begin{array}{*{20}{c}} 2&3\ 0&{2{\rm{P}} - 9} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5\ {10} \end{array}} \right]\)

For  “No solution” condition, Rank of A ≠ Rank of [A|B],    

∴ If 2P – 9 = 0, then Rank of A will be 1, and Rank of [A|B] will be 2.

2P – 9 = 0

∴ p = 4.5

Shortcut trick:

Consider the eqautions

a₁x + b₁y = c₁

a₂x + b₂y = c₂

\(1);{\rm{If;}}\frac{{{{\rm{a}}_1}}}{{{{\rm{a}}_2}}} \ne \frac{{{{\rm{b}}_1}}}{{{{\rm{b}}_2}}},{\rm{;then;it;is;a;Unique;solution}}\)

\(2);{\rm{If;}}\frac{{{{\rm{a}}_1}}}{{{{\rm{a}}_2}}} = \frac{{{{\rm{b}}_1}}}{{{{\rm{b}}_2}}} = \frac{{{c_1}}}{{{c_2}}},{\rm{;then;it;is;an;Infinite;solution}}\)

\(3);{\rm{If;}}\frac{{{{\rm{a}}_1}}}{{{{\rm{a}}_2}}} = \frac{{{{\rm{b}}_1}}}{{{{\rm{b}}_2}}} \ne \frac{{{c_1}}}{{{c_2}}},{\rm{;then;it;is;a;No;solution}}\)

Given,

2x + 3y = 5

3x + Py = 10

 a₁ = 2, b₁ = 3, c_1­ = 5, a₂ = 3, b₂ = P, c₂ = 10

\(\frac{{{{\rm{a}}_1}}}{{{{\rm{a}}_2}}} = \frac{{{{\rm{b}}_1}}}{{{{\rm{b}}_2}}}\)

\(\frac{2}{3} = \frac{3}{p}\)

P = 4.5

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

Concept:

Mean:

Let X is a discrete random variable having the possible values x₁, x₂, ……x_n

If P(X = x_i) = f(x_i), where i = 1, 2……n, then

E(X) = mean = x₁ f(x₁) + x₂ f(x₂)+ …….x_n f(x_n)

\({\rm{i}}.{\rm{e;E}}\left( {\rm{x}} \right) = \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{X}}_{\rm{i}}}{\rm{;f}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\)

\({\rm{E}}\left( {{{\rm{x}}^2}} \right) = \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} X_i^2{\rm{;f}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\)

Variance:

V(X) = E(X²) – E(X)²

Calculation:

Given q = 0.4

Required value = V(X) = E(X²) – [E(X)]²

\(\begin{array}{l} E\left( X \right) = \mathop \sum \limits_i {X_i}{p_i} = 0 \times 0.4 + 1 \times 0.6 = 0.6\ E\left( {{X^2}} \right) = \mathop \sum \limits_i X_i^2{p_i} = {0^2} \times 0.4 + {1^2} \times 0.6 = 0.6\ \therefore V\left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2} = 0.6 - 0.36 = 0.24; \end{array}\)

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:

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

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

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.

Concept:

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

 \({K_{22}} = \frac{{12EI}}{{{L^3}}}\)

Concept:

The development length of bars is given by,

\({L_d} = \frac{{ϕ × {σ _s}}}{{4 × {τ _{bd}}}}\)

where,

ϕ = nominal diameter of the bar

σ_s = stress in bar at the section considered at design load

τ_bd = design bond stress 

Calculation:

Given,

\({L_d} = \frac{{ϕ × {σ _s}}}{{4 × {τ _{bd}}}}\)

For deformed bars τ_bd value is increased by 60%

∴ τ_bd = 1.6 × τ_bd 

\({L_d} = \frac{{ϕ {σ _s}}}{{4 × 1.6{τ _{bd}}}} = \frac{{ϕ {σ _s}}}{{6.4{τ _{bd}}}}\)

So the value of k will be 6.4

Concept:

Carry-over factor (C.O.F) is given by,

\({\rm{C}}.{\rm{O}}.{\rm{F}} = \frac{{{\rm{Carry;over;moment}}}}{{{\rm{Applied;moment}}}}\)

The applied moment is the moment due to external load at the far end of the beam.

Carry over moment will be moment acting at the fixed support of the beam.

Calculation:

By symmetry we can consider two parts of beam separately with half the load.

Moment at joint B = 10 × 1 = 10 kNm

∴ Applied moment = 10 kNm

Carry over factor (C.O.F) = 1/2 

∴ Carry over moment = C.O.F × Applied moment

Carry over moment = 1/2 × 10 = 5 kNm

Hence carry-over moment at joint A due to 10 kNm moment at the propped end will be 5 kNm.

Concept:

The normal stress (σ_n)acting on a plane which is inclined at an angle θ, with the plane of major principle stress (i.e major principal plane) is given by,

\({{\rm{σ }}_{\rm{n}}}{\rm{;or;}}{{\rm{σ }}_{\rm{θ }}} = \frac{{{{\rm{σ }}_{\rm{x}}} + {{\rm{σ }}_{\rm{y}}}}}{2} + \frac{{{{\rm{σ }}_{\rm{x}}} - {{\rm{σ }}_{\rm{y}}}}}{2} \times \cos 2{\rm{θ }} + {{\rm{\tau }}_{{\rm{xy}}}}\times \sin 2{\rm{θ }}\)

Where,

σ_x  = normal stress acting on a plane in the X-axis

σ_y  = normal stress acting on a plane in the y-axis

τ_xy = shear stress acting in XY plane  

θ = an angle made by a plane with major principal plane

Calculation:

Given,

σx  = 100 MPa

σy  = -100 MPa

τxy = 0

θ = 45° 

Now we have,

\({{\rm{σ }}_{\rm{n}}}{\rm{;or;}}{{\rm{σ }}_{\rm{θ }}} = \frac{{{{\rm{σ }}_{\rm{x}}} + {{\rm{σ }}_{\rm{y}}}}}{2} + \frac{{{{\rm{σ }}_{\rm{x}}} - {{\rm{σ }}_{\rm{y}}}}}{2} \times \cos 2{\rm{θ }} + {{\rm{\tau }}_{{\rm{xy}}}}\times \sin 2{\rm{θ }}\)

\({{\rm{σ }}_{\rm{n}}}{\rm{;or;}}{{\rm{σ }}_{\rm{θ }}} = \frac{{100 + \left( { - 100} \right)}}{2} + \frac{{100 - \left( { - 100} \right)}}{2} \times \cos 2 \times 45^\circ + 0 \times \sin 2 \times 45^\circ \)

\({{\rm{σ }}_{\rm{n}}}{\rm{;or;}}{{\rm{σ }}_{\rm{θ }}} = \frac{{100 - 100}}{2} + \frac{{100 + 100}}{2} \times \cos 90^\circ \)

σ_n or σ_θ = 0 MPa

Alternate method:  

As plane AB and BC are principle planes, therefore Mohr’s circle for the given condition is

Here, normal stress is zero at 45° to the principle plane.

The Activity of the soil (A)

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

It indicates water absorption capacity or indicates swelling and shrinkage characteristics

\({\bf{A}} = \frac{{{{\bf{I}}_{\bf{P}}}}}{{\bf{C}}}\)

Where,

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

I_P = plasticity index = W_L - W_P

W_L = liquid limit of the soil, W_P = plastic limit of the soil

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 %

∴ W_S = 15 %, WP = 28 %

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

∴ W_L = 40 %

I_P = 40 – 28 = 12 %

\(A = \frac{{{I_P}}}{{%\ C}}\)

\(A = \frac{{{12}}}{{40}}\)

Activity = 0.3

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 (V_a):

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

Q = A × V = A_V × V_a

Where,

A = total cross-sectional area, A_V = 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 × V_a

\(\therefore {{\bf{V}}_{\bf{a}}} = \frac{{\bf{V}}}{{\bf{n}}}\)

Explanation:

\({V_s} = \frac{V}{n}\)

porosity (n) ranges between 0 < n < 1

\( \frac{V}{V_S} <1 \)

Hence,  \(V_s>V\)

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

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,

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.

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

ΔU_d = A× B × Δσ_d

Where,

ΔU_d = Increase in porewater pressure

Δσ_d = 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 + ΔU_d

ΔU = (B × Δσ₃) +( A × B × Δσ_d)

ΔU = B × (Δσ₃ +( A × Δσ_d) )    

Calculation:

​Δσ3 = Difference in cell pressure

∴ Δσ₃ = 250 - 150 = 100 kPa

Δu₃= 80 kPa

Δσ_d = 50 kPa, Δu_d = 25 kPa

we know that,

ΔU3 = B × Δσ3   

\({\rm{B}} = \frac{{{\rm{Δ }}{{\rm{U}}_3}}}{{{\rm{Δ }}{{\rm{σ }}_3}}}\)

\({\rm{B}} = \frac{{80}}{{100}} = 0.8\)

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.

Concept:

Continutiy equation in Three-Dimension

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

Where,

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

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

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

When flow is steady and incompressible, ρ = constant

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

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

\(\therefore \frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} = 0{\rm{;}}\)

Calculation:

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

\(\rho = \frac{1}{{{X}}}\)

As per continuity equation, for two dimensional steady flow,

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

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

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

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

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

Integrating on both the side

∴ V = - XY + C

Concept:

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

\(\frac{{\rm{P}}}{{\rm{\gamma }}} + \frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}} + {\rm{Z}} = {\rm{H}}\)

Where,

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

Explanation:

Let's assume point 1 is upstream from point 2

Given,

Steady flow, ∴ Z₁ = Z₂

Uniform cross-section, ∴ V₁ = V₂

\(\left[ {\frac{{\rm{P}}}{{\rm{\gamma }}} + \frac{{V_1^2}}{{2{\rm{g}}}} + {{\rm{Z}}_1}} \right] - \left[ {\frac{{\rm{P}}}{{\rm{\gamma }}} + \frac{{V_2^2}}{{2{\rm{g}}}} + {{\rm{Z}}_2}} \right] = H\)

\(\left[ {\frac{{{{\rm{P}}_1}}}{{\rm{\gamma }}}} \right] - \left[ {\frac{{{{\rm{P}}_2}}}{{\rm{\gamma }}}} \right] = H\)

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)

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.

\({\rm{V}} = \frac{1}{{\rm{n}}} \times {{\rm{R}}^{\frac{2}{3}}} \times {{\rm{S}}^{\frac{1}{2}}}{\rm{;}}\)

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):

\({\rm{R}} = \frac{{{\rm{wetted;area}}}}{{{\rm{wetted;perimeter}}}} = \frac{{\rm{A}}}{{\rm{P}}}\)

2. Hydraulic depth (D)

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

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

Q_full = A_full × V

A_full = (πr²), P = 2πr

\({\rm{R}} = \frac{{\rm{A}}}{{\rm{P}}} = \frac{{{\rm{\pi }}{{\rm{r}}^2}}}{{2{\rm{\pi r}}}} = \frac{{\rm{r}}}{2} = \frac{{\rm{d}}}{4}{\rm{;}}\)

Q_full = A × V

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

\({{\rm{Q}}_{{\rm{full}}}} = {\rm{;}}\left( {{\rm{\pi }}{{\rm{r}}^2}} \right){\rm{}} \times {\rm{;}}\frac{1}{{0.01}} \times {\left( {\frac{{\rm{d}}}{4}} \right)^{\frac{2}{3}}} \times {\left( {\frac{1}{{1000}}} \right)^{\frac{1}{2}}}\)

2) when the pipe is flowing half

Q_half = A_half × V

A_half = (πr²)/2, P_half = πr

\({\rm{R}} = \frac{{\rm{A}}}{{\rm{P}}} = \frac{{\frac{{{\rm{\pi }}{{\rm{r}}^2}}}{2}}}{{{\rm{\pi r}}}} = \frac{{\rm{r}}}{2} = \frac{{\rm{d}}}{4}{\rm{;}}\)

Q_half = A × V

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

\({{\rm{Q}}_{{\rm{half}}}} = {\rm{;}}\left( {\frac{{{\rm{\pi }}{{\rm{r}}^2}}}{2}} \right){\rm{}} \times {\rm{;}}\frac{1}{{0.01}} \times {\left( {\frac{{\rm{d}}}{4}} \right)^{\frac{2}{3}}} \times {\left( {\frac{1}{{1000}}} \right)^{\frac{1}{2}}}\)

3) The ratio of Q_full to Q_half 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.

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

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

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

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.

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

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

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

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,                                                   

S_M.S.W, S_C1, S_C2  = 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

\(\frac{{100}}{{{{\rm{\rho }}_{{\rm{M}}.{\rm{S}}.{\rm{W}}}}}} = \frac{{{\rm{% ;}}{{\rm{C}}_1}}}{{{{\rm{\rho }}_{{\rm{C}}1}}}} + \frac{{{\rm{% ;}}{{\rm{C}}_2}}}{{{{\rm{\rho }}_{{\rm{C}}2}}}} \)

\(\therefore {{\rm{\rho }}_{{\rm{M}}.{\rm{S}}.{\rm{W}}}} = \frac{{100}}{{\frac{{{\rm{% ;}}{{\rm{C}}_1}}}{{{{\rm{\rho }}_{{\rm{C}}1}}}} + \frac{{{\rm{% ;}}{{\rm{C}}_2}}}{{{{\rm{\rho }}_{{\rm{C}}2}}}}}}\)

Concept:

Penetration test:

1. The penetration test on bitumen is done by using a penetrometer.

2. It determines the softening of bitumen, by measuring the depth in tenths of millimeters (1/10^th) to which a standard loaded needle will penetrate vertically in five seconds.  

3. The IRC recommended bitumen grade is 30/40, 60/70, and 80/100.

4. For hot climate lower penetration grade is used i.e 30/40.

5. For cold climate higher penetration value is used i.e 80/100.

The penetration value = 1/10^th of the bitumen grade.

Calculation:

Given,

Bitumen grade = 80

The penetration value = 1/10^th of the bitumen grade.

The penetration value = (1/10) × 80 = 8 mm

Hence the needle of the penetration test apparatus penetrates the bitumen sample by 8 mm

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

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.

Concept:

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

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

\({\rm{\theta }} = {\tan ^{ - 1}}\left[ {\frac{{{\rm{Σ D}}}}{{{\rm{Σ L}}}}} \right]\)

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)

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

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

The degree of accuracy or the relative precision

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

Calculation:

Given,

ΣL = 0.4m, ΣD = 0.3 m

P = 1 km = 1000 m

we know that

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

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

e = 0.5 m

\({\rm{Relative;precision}} = \frac{1}{{\frac{{\rm{P}}}{{\rm{e}}}}}=\frac{1}{{\frac{{\rm{1000}}}{{\rm{0.5}}}}} =\frac{5}{{10000}} = \frac{1}{{2000}}\)

Concept:

Eigen Values

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

​\(\left| {A - \lambda I} \right| = 0\) 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

 \(\left| {A - \lambda I} \right| = 0\)

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

⇒ (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

Concept:

The iteration formula given by Newton Raphson is,

\({{\rm{x}}_{{\rm{n}} + 1}} = {{\rm{x}}_{\rm{n}}} - \frac{{{\rm{f}}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}{{{\rm{f'}}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}\)

x_n+1 = (n+1)^th approximation to the root

xn = nth approximation to the root

f(x_n) = value of function by substituting x = x_n

f'(xn) = derivative of function f(x_n)

The iteration formula given by secant method is,

\({{\rm{x}}_{{\rm{n}} + 1}} = {{\rm{x}}_{\rm{n}}} - \frac{{{\rm{f}}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}{{{\rm{f'}}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}\)

\({\rm{f'}}\left( {{{\rm{x}}_{\rm{n}}}} \right) = \frac{{{\rm{f}}\left( {{{\rm{x}}_{\rm{n}}}} \right) - {\rm{f}}\left( {{{\rm{x}}_{{\rm{n}} - 1}}} \right)}}{{{{\rm{x}}_{\rm{n}}} - {{\rm{x}}_{{\rm{n}} - 1}}}}\)

Calculation:

Given,

f(x) = x² – 4x + 4

∴ f’(x) = 2x – 4

x₀ = 3

f(3) = 1, f’(3) = 2

By Newton Rapshon method

 \(\begin{array}{l} {x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\ x_1=3 - \frac{1}{2} = \frac{5}{2}\ f\left( {x_1} \right) =f\left( {\frac{5}{2}} \right) = \frac{{25}}{4} - 10 + 4 = \frac{1}{4} \end{array}\)

By secant method,

\(\begin{array}{l} {x_2} = {x_1} - \frac{{\left( {{x_1} - {x_0}} \right)f\left( {{x_1}} \right)}}{{f\left( {{x_1}} \right) - f\left( {{x_0}} \right)}}\ x_2= \frac{{{f_1}{x_0} - {f_0}{x_1}}}{{{f_1} - {f_0}}}\ x_2= \frac{{\left( {\frac{1}{4} \times 3} \right) - \left( {1 \times \frac{5}{2}} \right)}}{{\frac{1}{4} - 1}} \x_2= 2.33 \end{array}\)

Given differential eqaution is,

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

Let, y = v × x

dy = vdx + xdv

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

\(\begin{array}{l} \frac{{vxdx + vxdx + {x^2}dv}}{{vxdx + {x^2}dv - vxdx}} = v\tan v\ \frac{{xdv + 2vdx}}{{xdv}} = v;tan;v\ 1 + \frac{{2v}}{x}\frac{{dx}}{{dv}} = v;tan;v;\ \frac{{2v}}{x}\frac{{dx}}{{dv}} = v\tan v - 1 \end{array}\)

Integrating both sides

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

\(\begin{array}{l} \Rightarrow {x^2} = \frac{{c\sec v}}{v}\ \Rightarrow \frac{{{x^2}y}}{x} = c\sec \frac{y}{x}\ \Rightarrow {x^2}\frac{y}{x} = c\sec \frac{y}{x}\ xy\cos \frac{y}{x} = c \end{array}\)

Concept:

The residue of f(z):

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

Residue at poles:

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

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

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] 

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

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

R₁ = 1

R₂ = Res [f(z) : Z =-2] 

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

Hence the option 1 is corect

Concept:

Directional Derivative (D.D):

If ϕ(X, Y, Z) is a differential scalar function then the rate of change of ϕ at a point P in the direction of a given vector \(\bar a=\left( {x\hat i + y\hat j - z\hat k} \right)\)  is called a D.D of ϕ. It is given by,

 \({\rm{D}}.{\rm{D}} = {\rm{∇ }}ϕ .\frac{{\rm{\bar a}}}{{\left| {{\rm{\bar a}}} \right|}}\)

Where,

∇ϕ = derivative of function ϕ(X, Y, Z)  

\(\left| \bar a \right|; = \sqrt {{x^2} + {y^2} + {z^2}} \)

Calculation:

ϕ(x, y, z) = x² – 3yz

\(\begin{array}{l} ∇ {\rm{\phi}} = 2x\hat i - 3zj - 3y\hat k\ ∇ \phi_{{at\left( {2, - 1,4} \right)}} = 4\hat i -12\hat j + 3\hat k \end{array}\)

Directional derivative

\(\begin{array}{l} = \left( {4\hat i - 12\hat j + 3\hat k} \right).\frac{{\hat i + \hat j - 2k}}{{\sqrt 6 }}\ = \frac{{4 - 12 - 6}}{{\sqrt 6 }} \ = - \frac{{7\sqrt 6 }}{3} \= - 5.715; \end{array}\)

Concept:

The density (ρ) of a substance is given by,

\({\rm{ρ }} = \frac{{{\rm{Mass;}}}}{{{\rm{Volume}}}} = \frac{{\rm{M}}}{{\rm{V}}},{\rm{;in;}}\frac{{{\rm{kg}}}}{{{{\rm{m}}^3}}}\)

∴ Mass = ρ × V

\({\rm{Specific;gravity}} = \frac{{{{\rm{ρ }}_{{\rm{material}}}}}}{{{{\rm{ρ }}_{{\rm{water}}}}}}\)

where, ρwater = 1000 kg/m3 

The total volume of concrete is given by,

V_T = V_C + V_F.A + V_C.A + V_W + V_air

Where, V_C, V_F.A, V_C.A, V_W, and V_air are volume of cement, fine aggregate, coarse aggregate, water and air respectively.

To find the mass of any of the above materials in 1 m³ of concrete

\(1 = \frac{{{{\rm{M}}_{\rm{c}}}}}{{{{\rm{ρ }}_{\rm{c}}}}} + \frac{{{{\rm{M}}_{{\rm{F}}.{\rm{A;}}}}}}{{{{\rm{ρ }}_{{\rm{F}}.{\rm{A}}}}}} + \frac{{{{\rm{M}}_{{\rm{C}}.{\rm{A}}}}}}{{{{\rm{ρ }}_{{\rm{C}}.{\rm{A}}}}}} + \frac{{{{\rm{M}}_{\rm{W}}}}}{{{{\rm{ρ }}_{\rm{W}}}}} + \frac{{{{\rm{M}}_{{\rm{air}}}}}}{{{{\rm{ρ }}_{{\rm{air}}}}}}\)

M_C, M_F.A, M_C.A, M_W, and M_air are mass of cement, fine aggregate, coarse aggregate, water and air respectively.

ρ_C, ρ_F.A, ρ_C.A, ρ_W, and ρ_air are density of cement, fine aggregate, coarse aggregate, water and air respectively.

Calculation:

Given,

Masses of materials in 1 m³ of concrete

MC = 368 kg, MF.A = 606 kg, MC.A = 1155 kg, and MW = 184 kg 

Densities of materials 

ρC = 3.14× 1000, ρF.A = 2.67× 1000, ρC.A = 2.74 × 1000, and ρW = 1000r

For total volume of concrete of 1m³

^{\(1 = \frac{{{{\rm{M}}_{\rm{c}}}}}{{{{\rm{ρ }}_{\rm{c}}}}} + \frac{{{{\rm{M}}_{{\rm{F}}.{\rm{A;}}}}}}{{{{\rm{ρ }}_{{\rm{F}}.{\rm{A}}}}}} + \frac{{{{\rm{M}}_{{\rm{C}}.{\rm{A}}}}}}{{{{\rm{ρ }}_{{\rm{C}}.{\rm{A}}}}}} + \frac{{{{\rm{M}}_{\rm{W}}}}}{{{{\rm{ρ }}_{\rm{W}}}}} + \frac{{{{\rm{M}}_{{\rm{air}}}}}}{{{{\rm{ρ }}_{{\rm{air}}}}}}\)}

\(1=\frac{{368}}{{3.14 × 1000}} + \frac{{606}}{{2.67 × 1000}} + \frac{{1155}}{{2.74 × 1000}} + \frac{{184}}{{1000}} + {V_v} \)

V_v = 0.0503 m³ = 50.3 litre/m³

Concept:

The maximum resultant shear force between f_{1 ­}and f₂ is f_R given by,

\({{\rm{f}}_{\rm{R}}} = \sqrt{{{\rm{f}}_1^2 + {\rm{f}}_2^2 + 2{{\rm{f}}_1}{{\rm{f}}_2}{\rm{cosθ }}} }\)

Where,

f₁ = shear force in any rivet due to direct load ‘P’

\({{\rm{f}}_1} = \frac{{\rm{P}}}{{\rm{n}}},{\rm{;where;n}} = {\rm{number;of;rivets}}\)

f₂ = shear force in rivet due to twisting moment M

\({{\rm{f}}_2} = \frac{{{\rm{P}} \times {\rm{e}} \times {\rm{r}}}}{{{\rm{\Sigma }}{{\rm{r}}^2}}}{\rm{;}}\)

r = radial distance of rivet from C.G of rivet group

The critical rivet is the one for which r is maximum and θ is minimum 

Calculation:

Where, S = Direction of shear force and B = direction of bending force for bolts.

Given,

P = 100 kN, n = 5, e = 600 mm

\({{\rm{f}}_1} = \frac{{\rm{P}}}{{\rm{n}}} = \frac{{100}}{5} = 20{\rm{;kN}}\)

\({{\rm{f}}_2} = \frac{{{\rm{P}} \times {\rm{e}} \times {\rm{r}}}}{{{\rm{\Sigma }}{{\rm{r}}^2}}} = \frac{{100 \times 600 \times {{10}^{ - 3}} \times \left( {75\sqrt 2 \times {{10}^{ - 3}}} \right)}}{{{{\left( {75\sqrt 2 } \right)}^2} \times 4 \times {{10}^{ - 6}}}} = 141.42{\rm{;kN}}\)

f_r = \(\sqrt{20^2+141.42^2+(2\times 20\times 141.42\times cos45^0)}\)

\({{\rm{f}}_{\rm{R}}} = 156.2{\rm{;kN}}\)

\({A_{st}} = 4 \times \frac{\pi }{4}\left( {{{12}^2}} \right) = 452.39;m{m^2}\)

Finding actual depth of neutral axis (x_u),

C = T

0.36 f_ck Bx_u = 0.87 f_y A_st

\({x_u} = \frac{{0.87 \times 415 \times 452.39}}{{0.36 \times 25 \times 200}}\) = 90.274 mm

x_{u, max} = 0.48d = 0.48 × 300 =144 mm

x_u < x_u,max, ∴ it is an under reinforced section

MOR = 0.36 x_u f_ck. B (d – 0.42 x_u)

MOR = 0.36 × 90.74 × 25 × 200 (300 – 0.42 × 90.74)

MOR = 42.77 kN-m

Explanation:

Length of the Beam = L = 4 m

Mechanism 1:

3M_Pθ + M_Pθ + M_Pθ + M_Pθ = P_u × θ × (L/4)

6M_P = (L/4) × Pu 

\( {P_u} = \frac{{24{M_P}}}{L}\)

Mechanism 2 :

Lϕ = 3Lθ

ϕ = 3θ

3M_P θ + M_P (θ +ϕ) + M_P ϕ = P_u ×( L/4) × ϕ

3MP θ + MP (θ +3θ) + MP 3θ = Pu ×( L/4) × 3θ 

10 M_P θ = P_u × (L/4) × 3θ 

Comparing the P = 40 MP/3L with cMp/L

\(\begin{array}{l} P = \frac{{40{M_P}}}{{3L}}\ ∴ C = \frac{{40}}{3}=13.33 \end{array}\)

∴ The Value of C = 13.33

Concept:

The change in length of a prismatic bar with axial force P is given by,

\({\rm{δ l}} = \frac{{{\rm{PL}}}}{{{\rm{AE}}}}\)

The change in length of a tapered bar with diameter d₁ at one end and diameter d₂ at other end is given by,

\({\rm{δ l}} = \frac{{{\rm{PL}}}}{{\frac{{\rm{π }}}{4}\left( {{\rm{d_1}}.{\rm{d_2}}} \right){\rm{E}}}}\)

Where,

δl = change in length of the bar

P = Axial force on the bar

A = cross-sectional area of a bar

E = modulus of elasticity

Calculation:

P = 30π kN, d₁ = 20 mm, d₂ = 10 mm

L₁ = 2 m, L₂ = 1.5 m

Total elongation

δL_Total = δL_AB + δL_BC

\(\begin{array}{l} δL_{Total}= \frac{{PL}}{{AE}} + \frac{{4PL}}{{π {d_1}{d_2}E}}\δL_{Total} = \frac{{30π \times 1.5}}{{\frac{{π \times {{\left( {0.01} \right)}^2}}}{4} \times 2 \times {{10}^5}}} + \frac{{30π \times 2}}{{\frac{{π \times \left( {0.01} \right) \times \left( {0.02} \right)}}{4} \times 2 \times {{10}^5}}} \end{array}\)

δLTotal = 9 + 6 = 15 mm

Concept:

The stiffness of the spring ‘K’ is given by,

\({\rm{K}} = \frac{{\rm{P}}}{{\rm{\Delta }}}\)

Where,

P = load in N

Δ = deflection in the spring in cm

Calculation:

Given,

\({\rm{K_S}} = \frac{{3{\rm{EI}}}}{{{{\rm{2\times L}}^3}}}\)

From the above figure

The load at point B = P - R 

The load at point D = R

\({{\rm{\Delta }}_{spring}} = {{\rm{\Delta }}_B} - {{\rm{\Delta }}_D}\)

\(\frac{{\rm{R}}}{{{{\rm{K}}_{\rm{S}}}}} = \frac{{\left( {{\rm{P}} - {\rm{R}}} \right){{\rm{L}}^3}}}{{3{\rm{EI}}}} - \frac{{{\rm{R}}{{\rm{L}}^3}}}{{3{\rm{EI}}}}{\rm{;}}\)

\(\frac{{{\rm{R}}\left( {2{{\rm{L}}^3}} \right)}}{{3{\rm{EI}}}} = \frac{{\left( {{\rm{P}} - {\rm{R}}} \right){{\rm{L}}^3}}}{{3{\rm{EI}}}} - \frac{{{\rm{R}}{{\rm{L}}^3}}}{{3{\rm{EI}}}}{\rm{;}}\)

2R = (P - R) - R

4R = P

R = P/4

∴ Percentage of force carried by spring = 25%

Concept:

Analysis of truss by the method of section

1. A section is cut through truss such that it cuts at the maximum of three members in which forces are unknown.

2. Thus, if the truss is cut into two parts at the section, three equilibrium equations can be written for each part, thus three unknown forces can be found out.

Deflection of truss joint by Castigliano’s method

\({\rm{\Delta }} = \frac{{\partial {\rm{U}}}}{{\partial {\rm{F}}}}\)

Where,

Δ = deflection at the point of application of force ‘F’ in the direction of a force ‘F’

F = force

U = strain energy in the system

\({\rm{U}} = \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} \frac{{{{\rm{P}}^2}{\rm{L}}}}{{2{\rm{AE}}}}{\rm{;;}}\)

Where,

P = force in the member of a truss, A = sectional area of each member, E = modulus of elasticity

Calculation:

First calculating reactions

ΣM_A = 0

R_B × 3 – 10 × 9 – 5 × 3 = 0

∴ R_B = 35 kN

As there is only one horizontal reaction i.e 10 kN

∴ H_A = 10 kN

ΣM_B = 0

R_A × 3 + 10 × 9 = 0

∴ R_A = - 30 kN

Let us cut a section 1 – 1 as shown in figure and consider the lower part.

ΣF_horizontal = 0

F_XY - H_A = 0

F_XY = 10 kN

Strain energy

\({\rm{U}} = \frac{{{{\rm{P}}^2}{\rm{L}}}}{{2{\rm{AE}}}}{\rm{;;}}\)

Given AE = 30 kN and L = 3 ,

Calculating for force P in XY member,

\({\rm{U}} = \frac{{{{\rm{P}}^2}{\rm{L}}}}{{2{\rm{AE}}}} = \frac{{{{10}^2} \times 3}}{{2 \times 30}} = 5{\rm{;kN}} {\rm{m;}}\)

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, W_s = weight of soil in kg, V = volume of soil in m³

γ­_d = unit weight of dry soil in kN/m³_, γ­_w = unit weight of water in kN/m³

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

\({\rm{n}} = \frac{{\rm{e}}}{{1 + {\rm{e}}}}\)

where,

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

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

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

V_V = volume of voids, V_S = volume of solids, V = total volume of soil

V = V_V + V_S

Calculation:

Given,

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

Volume of embankment, V = 5000 m³

γ_d  = 16.2 kN/m³, γ_W = 10 kN/m³ and G = 2.7, 

Volume of solids remain constant.

For embankment

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

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

e = 0.667

V = 5000

\({V_s} = \frac{V}{{1 + e}} = 3000\ {m^3}\)

Site X

\(\begin{array}{l} {V_s} = \frac{V_x}{{1 + e}} = 3000\ {m^3}\ {V_x} = 3000 × 1.6 = 4800\ {m^3} \end{array}\)

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

Site Y

\({V_y} = 3000 × 1.7 = 5100\ {m^3}\)

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

Site Z

\({V_z} = 3000 × 1.64 = 4920;{m^3}\)

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

Hence, choose site X which has minimum cost.

Concept:

The total stress due to total weight of soil medium is given by,

σ = γ × h

Where,

γ = unit weight of soil in kN/m³

h = depth of soil medium

Consolidation settlement when compression index C_C is known is given by,

\({\rm{Δ H}} = \frac{{{{\rm{H}}_{\rm{o}}}{{\rm{C}}_{\rm{C}}}}}{{1 + {{\rm{e}}_{\rm{o}}}}} \times {\log _{10}}\left( {\frac{{{\rm{\ \barσ ;}} + {\rm{Δ \barσ }}}}{{\rm{\barσ }}}} \right)\)

Where,

e_o = initial void ratio

H_o = depth of the clay layer

By the load strip method

\({\rm{\Delta \sigma }} = \frac{{{\rm{Total;load;on;footing}}}}{{{\rm{Area;of;footing}}}} = \frac{{\rm{W}}}{{\frac{{\rm{\pi }}}{4}{{\left( {{\rm{D}} + {\rm{nZ}}} \right)}^2}}}\)

Where,

D = diameter of footing

n = coefficient of horizontal, i.e for 2V : 1H, n = 1

Z = depth of the soil below the footing

Calculation:

Given,

Total vertical load = 1500 kN

Diameter of footing = 3 m

C_C = 0.3, e_o = 0.7, and H_o = 10 m, Z = 12 m

Initial  at the centre of clay layer

\(\begin{array}{l} \bar \sigma= 2 \times 15 + 6 \times 18 + 5 \times 18 - 11 \times 10\ \bar\sigma= 118;kN/{m^2} \end{array}\)

\(\Delta\bar\sigma\) at the centre of clay layer

\(\begin{array}{l} {\rm{Δ }}\barσ = \frac{{1500}}{{\frac{\pi }{4}{{\left( {3 + 12} \right)}^2}}} = 8.488;kN/{m^2}\ \end{array}\)

\({\rm{Δ H}} = \frac{{{{\rm{H}}_{\rm{o}}}{{\rm{C}}_{\rm{C}}}}}{{1 + {{\rm{e}}_{\rm{o}}}}} \times {\log _{10}}\left( {\frac{{{\rm{\ \barσ ;}} + {\rm{Δ \barσ }}}}{{\rm{\barσ }}}} \right)\)

\( {\rm{Δ }}H = \frac{{0.3 \times 10}}{{1.7}}\log_{10} \left( {\frac{{118 + 8.488}}{{118}}} \right)\)

ΔH = 53.24 mm

Concept:

1) Time factor

\({{\rm{T}}_{\rm{V}}} = \frac{{{{\rm{C}}_{\rm{V}}} \times {\rm{t}}}}{{{{\rm{d}}^2}}}\)

Where,

C_V = coefficient of consolidation, T = time of consolidation

d = drainage path

d = H_o/2 for double drainage

d = H_o for single drainage

2) Degree of consolidation

\({\rm{U}} = \frac{{\rm{S}}}{{{{\rm{S}}_{\rm{f}}}}} \times 100\)

S = settlement occurred up to certain time, S_f = final settlement after consolidation

\({\rm{If;U}} \le 60{\rm{;% }},{\rm{;then;}}{{\rm{T}}_{\rm{V}}} = \frac{{\rm{\pi }}}{4} \times {\left[ {\frac{{{\rm{U;% }}}}{{100}}} \right]^2}\)

\({\rm{If;U}} > 60{\rm{;% }},{\rm{;then;}}{{\rm{T}}_{\rm{V}}} = 1.781 - 0.933\times{\log _{10}}\left[ {100 - {\rm{U;% }}} \right]\)

Calculation:

H_o = 20 m, d = H_o/2 = 20/2 = 10 m (double drainage)

S = 30 mm, t = 2 years, C_V = 0.003 cm2/s

\(\begin{array}{l} {{\rm{T}}_{\rm{V}}} = \frac{{{{\rm{C}}_{\rm{V}}} \times {\rm{t}}}}{{{{\rm{d}}^2}}} = \frac{{0.003 \times 2 \times 365 \times 24 \times 60 \times 60}}{{{{\left( {10 \times 100} \right)}^2}}} \end{array}\)

T_V = 0.1892

\(\frac{\pi }{4}{U^2} = 0.1892\)

\(U = 0.4908 < 0.6\) (o.k.)

\({\rm{U}} = \frac{{\rm{S}}}{{{{\rm{S}}_{\rm{f}}}}} \times 100\)

\(S_f = \frac{{30}}{{0.4908}} = 61.12;mm\)

Final settelment remains constant.

Now, U for 50 mm settlement \( = \frac{{50}}{{61.12}} \times 100 = 81.8% \)

\(\begin{array}{l} {T_v} = 1.781 - 0.933\log \left( {100 - U} \right)\ {T_v} = 1.781 - 0.933\log \left( {100 - 81.8} \right) = 0.6053\ {T_v} = \frac{{Cv.t}}{{{d^2}}}\ t = \frac{{0.6053 \times {{\left( {1000} \right)}^2}}}{{0.003 \times 365 \times 24 \times 3600}} = 6.397;years \end{array}\)

Additional number of years = 6.397 - 2 = 4.39 years

Coefficient of permeability can be expressed in two components based on the direction of flow of water

1. Average coeffecient of permeability for horizontal flow is given by,

\({K_H} = \frac{{{K_1}{H_1} + {K_2}{H_2} + {K_3}{H_3}}}{{{H_1} + {H_2} + H_3 }}\)

2. Average coeffecient of permeability for Vertical flow is given by,

\({K_V} = \frac{{{H_1} + {H_2} + {H_3}}}{{\frac{{{H_1}}}{{{K_1}}} + \frac{{{H_2}}}{{{K_2}}} + \frac{{{H_3}}}{{{K_3}}}}}\)

Where,

H₁, H₂, and H₃ are width of the sandwiched soil layers 1, 2 and 3 respectively

K₁, K₂, and K₃ are permiabilty of the sandwiched soil layers 1, 2 and 3 respectively

Calculation:

It is given that,

K₂ = 10 K3, ∴ K₃ = 0.1 K₂

K₂ = 0.1 K₃, ∴ K₁ = 10 K₂

H₂ = 2 H₁, ∴ H₁ = 0.5 H₂

H₂ = 2/3 H₃, ∴ H₃ = 1.5 H₂    

Average coeffecient of permeability for horizontal flow

\({K_H} = \frac{{{K_1}{H_1} + {K_2}{H_2} + {K_3}{H_3}}}{{{H_1} + {H_2} + H_3 }}\)

\(\begin{array}{l} {K_H} = \frac{{{K_1}{H_1} + {K_2}{H_2} + {K_3}{H_3}}}{{{H_1} + {H_2} + {H_3}}}\ K_H = \frac{{10{K_2} × 0.5{H_2} + {K_2}{H_2} + 0.1{K_2} × 1.5{H_2}}}{{0.5{H_2} + {H_2} + 1.5{H_2}}}\ K_H= \frac{{6.15{K_2}{H_2}}}{{3{H_2}}} = \frac{{2.05{K_2}{H_2}}}{{{H_2}}} \K_H= 2.05{K_2} \end{array}\)

Average coeffecient of permeability for Vertical flow 

\({K_V} = \frac{{{H_1} + {H_2} + {H_3}}}{{\frac{{{H_1}}}{{{K_1}}} + \frac{{{H_2}}}{{{K_2}}} + \frac{{{H_3}}}{{{K_3}}}}} = \frac{{3{H_2}}}{{\frac{{0.5{H_2}}}{{10{K_2}}} + \frac{{{H_2}}}{{{K_2}}} + \frac{{1.5{H_2}}}{{0.1{K_2}}}}}\)

K_V = 0.1869 K₂

\(\frac{{{K_H}}}{{{K_V}}} = \frac{{2.05}}{{0.1869}} = 10.97\)

Concept:

The ultimate bearing capacity of soil by Meyerhof’s theory:

\({{\rm{q}}_{\rm{U}}} = {\rm{C}}{{\rm{N}}_{\rm{c}}}{{\rm{S}}_{\rm{c}}}{{\rm{d}}_{\rm{c}}}{{\rm{i}}_{\rm{c}}} + {\rm{γ }}{{\rm{D}}_{\rm{f}}}{{\rm{N}}_{\rm{q}}}{{\rm{S}}_{\rm{q}}}{{\rm{d}}_{\rm{q}}}{{\rm{i}}_{\rm{q}}} + 0.5{\rm{γ B}}{{\rm{N}}_{\rm{γ }}}{{\rm{S}}_{\rm{γ }}}{{\rm{d}}_{\rm{γ }}}{{\rm{i}}_{\rm{γ }}}\)

Where,

C = cohesion of soil, D_f = depth of the footing, B = width of the footing

S = shape factor, d = depth factor, i = load inclination factor

N_c,N_q and N_γ are Terzaghi’s bearing capacity factors

If angle at which load is acting, α = 0° ,then  i_c = i­_q= i_γ = 1

If ϕ = 0° then N_c = 5.14, N_q = 1, and N_γ = 0

The net ultimate bearing capacity of soil:

\({{\rm{q}}_{{\rm{nU}}}} = {{\rm{q}}_{\rm{U}}} - {\rm{γ }}{{\rm{D}}_{\rm{f}}}\)

\({{\rm{q}}_{{\rm{nU}}}} = {\rm{C}}{{\rm{N}}_{\rm{c}}}{{\rm{S}}_{\rm{c}}}{{\rm{d}}_{\rm{c}}}{{\rm{i}}_{\rm{c}}} + {\rm{γ }}{{\rm{D}}_{\rm{f}}}({{\rm{N}}_{\rm{q}}} - 1){{\rm{S}}_{\rm{q}}}{{\rm{d}}_{\rm{q}}}{{\rm{i}}_{\rm{q}}} + 0.5{\rm{γ B}}{{\rm{N}}_{\rm{γ }}}{{\rm{S}}_{\rm{γ }}}{{\rm{d}}_{\rm{γ }}}{{\rm{i}}_{\rm{γ }}}\)

Factor of safety

\({\rm{F}}.{\rm{O}}.{\rm{S}} = \frac{{{\rm{net;ultimate;bearing;capacity}}}}{{{\rm{net;safe;bearing;capacity}}}} = \frac{{{{\rm{q}}_{{\rm{nu}}}}}}{{{{\rm{q}}_{{\rm{ns}}}}}}\)

Calculation:

Given,

Nq = 33.3, Nγ = 37.16, S_q = S_γ = 1.314, d_q = d_γ = 1.113, i_q = 0.444, i_γ = 0.02

Square footing 2m × 2m

FOS = 3 m

\({{\rm{q}}_{\rm{U}}} = {\rm{C}}{{\rm{N}}_{\rm{c}}}{{\rm{S}}_{\rm{c}}}{{\rm{d}}_{\rm{c}}}{{\rm{i}}_{\rm{c}}} + {\rm{γ }}{{\rm{D}}_{\rm{f}}}{{\rm{N}}_{\rm{q}}}{{\rm{S}}_{\rm{q}}}{{\rm{d}}_{\rm{q}}}{{\rm{i}}_{\rm{q}}} + 0.5{\rm{γ B}}{{\rm{N}}_{\rm{γ }}}{{\rm{S}}_{\rm{γ }}}{{\rm{d}}_{\rm{γ }}}{{\rm{i}}_{\rm{γ }}}\)

It is given that c = 0

\(\therefore {{\rm{q}}_{\rm{U}}} = {\rm{γ }}{{\rm{D}}_{\rm{f}}}{({\rm{N}}_{\rm{q}}}-1){{\rm{S}}_{\rm{q}}}{{\rm{d}}_{\rm{q}}}{{\rm{i}}_{\rm{q}}} + 0.5{\rm{γ B}}{{\rm{N}}_{\rm{γ }}}{{\rm{S}}_{\rm{γ }}}{{\rm{d}}_{\rm{γ }}}{{\rm{i}}_{\rm{γ }}}\)

\(\begin{array}{l} q_{nu}= \left( {18 × 1 × \left( {33.3 - 1} \right) × 1.314 × 0.444 × 1.113 + 0.5 × 18 × 2 × 37.16 × 1.314 × 0.02 × 1.113} \right) \end{array}\)

q_nu = 397.08 kN/m²

\({\rm{F}}.{\rm{O}}.{\rm{S}} = \frac{{{{\rm{q}}_{{\rm{nu}}}}}}{{{{\rm{q}}_{{\rm{ns}}}}}} = \frac{{397.08}}{3} = 132.36{\rm{;kN}}/{{\rm{m}}^2}\)

For one way eccentricity, we will have to take reduced area of footing

Reduced side = 2 × eccentricity = 2 × 0.85 = 1.70 m

Reduced Area = 2 × 1.70 = 3.4 m²

Load carrying capacity in kN = P = 132.36 × Area = 132.36 × 3.4 = 450.03 kN

Concept:

According to Darcy’s Weisbach equation:

\({\rm{Head;loss}} = {{\rm{h}}_{\rm{f}}} = \frac{{{\rm{fL}}{{\rm{V}}^2}}}{{2{\rm{gD}}}}\)

 Where,

f = friction factor of the pipe, L = Length of the pipe, V = velocity of water through the pipe

g = gravitational acceleration, D = diameter of the pipe

Losses in the pipe at bends and fittings

\({\rm{h}} = {\rm{k}} × \frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}}\)

where k =bend or fitting coefficient, depends on the angle of bend, type of bend, type of fitting and type of threading

Calculation:

d = 0.3 m

k₁ = loss co-efficient = 0.5 (sharp entrance)

k₂ = loss co-efficient for value = 5.5

H = 20 m, f = 0.03, g = 10 m/s², L = 930 m

Total head loss is given by

\(\begin{array}{l} {h_L} = \frac{{fL{V^2}}}{{2gD}} + \frac{{{k_1}{V^2}}}{{2g}} + \frac{{{k_2}{V^2}}}{{2g}} + \frac{{{V^2}}}{{2g}}\ 20 = \frac{{\left( {0.03} \right)930{V^2}}}{{2 × 10 × 0.3}} + \frac{{0.5{V^2}}}{{2 × 10}} + \frac{{5.5{V^2}}}{{2 × 10}} + \frac{{{V^2}}}{{2 × 10}} \end{array}\)

20 = 5V²

V = 2 m/sec

we know that Q = A × V 

\(Q = \frac{\pi }{4}\left( {{{0.3}^2}} \right) × 2 = 0.1414;{m^3}/sec\)

Concept:

The relation between Sequent depths and Froude number is given by

\(\frac{{{Y_2}}}{{{Y_1}}} = \frac{1}{2} \times \left[ { - 1 + \sqrt {1 + 8{{\left( {{F_{r_2}}} \right)}^2};} } \right]\)

Where,

Y1 = initial depth or prejump depth

Y2 = final depth or post jump depth or sequent depth

\(\frac{{{{\rm{Y}}_2}}}{{{{\rm{Y}}_1}}} = {\rm{Sequent;depth;ratio}}\)

Froude number is given by:

\({{\rm{F}}_{\rm{r}}} = \frac{{\rm{V}}}{{\sqrt {{\rm{gD}}} {\rm{;}}}}\)

Where,

D = Y = hydraulic depth

Calculation:

y₂ = 0.8 m, V₂ = 1 m/sec, and g = 10 m/sec²

\({{\rm{F}}_{\rm{r_2}}} = \frac{{\rm{V}}}{{\sqrt {{\rm{gY_2}}} {\rm{;}}}} = \frac{1}{{\sqrt {10 \times 0.8} }} = 0.3535\)

\(\frac{{{Y_1}}}{{{Y_2}}} = \frac{1}{2} \times \left[ { - 1 + \sqrt {1 + 8{{\left( {{F_{r_2}}} \right)}^2};} } \right]\)

\({Y_1} = \frac{{0.8}}{2}\left( { - 1 + \sqrt {1 + 8 \times {{\left( {0.3535} \right)}^2}} } \right)\)

Y₁ = 0.1656

v₁y₁ = v₂y₂

\({V_1} = \frac{{1 \times 0.8}}{{0.1656}} = 4.83 m/sec;\)

Concept:

Equation of Gradually Varied Flow(G.V.F):

The rate of change of water depth with distance is given by,

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{{\rm{S}}_{\rm{o}}} - {{\rm{S}}_{\rm{f}}}}}{{1 - {\rm{F}}_{\rm{r}}^2}}\)

Where,

S_o = slope of the channel bottom

S_f = slope of the Total Energy Line(T.E.L)

F_r = Froude's number

Froude number is given by:

\({{\rm{F}}_{\rm{r}}} = \frac{{\rm{V}}}{{\sqrt {{\rm{gD}}} {\rm{;}}}}\)

Where,

D = Y = hydraulic depth

Calculation:

B = 2m, S = - 1/10000 (- ve as bed level is rising),

 Q = 4 m³/sec, n = 0.01, g = 10 m/sec²

\(\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{{S_o} - {S_e}}}{{1 - F{r^2}}}\ Fr = \frac{V}{{\sqrt {gy} }} = \frac{Q}{{2y\sqrt {gy} }} = \frac{4}{{2 \times 0.5\sqrt {10 \times 0.5} }} = 1.789\ R = \frac{A}{P} = \frac{{2 \times 0.5}}{{2 + 2 \times 0.5}} = \frac{1}{3}\ Q = \frac{1}{n}{R^{\frac{2}{3}}}{S^{\frac{1}{2}}}.A\ 4 = \frac{1}{{0.01}} \times {\left( {\frac{1}{3}} \right)^{\frac{2}{3}}}\times S{_f^{\frac{1}{2}}} \times 1 \end{array}\)

S_f = 6.923 × 10^-3

\(\frac{{dy}}{{dx}} = \frac{{ - \frac{1}{{10000}} - 6.923 \times {{10}^{ - 3}}}}{{1 - {{\left( {1.789} \right)}^2}}} = 0.0031\)

Explanation:

Given,

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

i.e F_D = f_n (D, V, ρ, μ) 

We know that the formula for Drag force is,

\({F_D} = {C_D} \times \frac{1}{2}ρ A{V^2}\)

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

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

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 

\(\begin{array}{l} {R_e} = \frac{{ρ VD}}{μ } = dimensionless \end{array}\)

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 \(\begin{array}{l} {R_e} = \frac{{ρ VD}}{μ }\end{array}\)

Concept:

Estimation of missing rainfall data (P_x)

P₁, P₂, P₃ ……… P_n are respective annual precipitation at stations 1, 2, 3 ….. n

N₁, N₂, N₃………N_n are normal precipitation at stations 1, 2, 3 ….. n

N_x is normal precipitation where the data is missing.

The P_x is estimated by one of the following methods

1) Simple mean method

If N₁, N₂, N₃………N_n falls between (N_x – 10 % N_x) to (N_x + 10 % N_x) then we use a simple mean method

\({{\rm{P}}_{\rm{x}}} = \frac{{{{\rm{P}}_1} + {{\rm{P}}_2} + {{\rm{P}}_3} \ldots ..{{\rm{P}}_{\rm{n}}}}}{{\rm{n}}}\)

Where n = number of rain gauge stations working

2) Normal ratio method

If N₁, N₂, N₃………N_n does not fall between (N_x – 10 % N_x) to (N_x + 10 % N_x) then we use the Normal ratio method.

\({{\rm{P}}_{\rm{x}}} = \frac{{{{\rm{N}}_{\rm{x}}}}}{{\rm{n}}}\left[ {\frac{{{{\rm{P}}_1}}}{{{{\rm{N}}_1}}} + \frac{{{{\rm{P}}_2}}}{{{{\rm{N}}_2}}} + \ldots \frac{{{{\rm{P}}_{\rm{n}}}}}{{{{\rm{N}}_{\rm{n}}}}}} \right]\)

Calculation:

Given,

Normal annual precipitations for stations P,Q,R and S are 780 mm, 850 mm, 920 mm and 980 mm respectively

Observed precipitation for stations Q and R are 930 mm and 1010 mm

Estimated precipitation at station P i.e P_P = 860 mm

Let, P_x = Precipitation in 2013 at station X

N_x = Normal annual precipitation at station X

\(\begin{array}{l} \frac{{{P_P}}}{{{N_P}}} = \frac{1}{3}\left[ {\frac{{{P_Q}}}{{{N_Q}}} + \frac{{{P_R}}}{{{N_R}}} + \frac{{{P_S}}}{{{N_S}}}} \right]\ \frac{{860}}{{780}} = \frac{1}{3}\left[ {\frac{{930}}{{850}} + \frac{{1010}}{{920}} + \frac{{{P_S}}}{{980}}} \right] \end{array}\)

P_S = 1093.43 mm

Hence, maximum ordinate is 22 m³/s

Concept:

Ideal gas equation is given by

PV = nRT

P = pressure, T = temperature, R = universal gas constant, V = volume of gas, n = number of moles of gas

Calculation:

Given,

P/(RT) = 41.6 mol/m³

1 m³ of air has 30 μg SO₂

10⁶ m³ of air has 30g SO₂

From PV = nRT

Concentration of SO₂ in PPM = 0.0113 ppm

Concept:

The terminal settling velocity (V_S)

It is estimated with the help of stokes equation

\({{\rm{V}}_{\rm{S}}} = \frac{{{\rm{g}} \times \left( {{\rm{G}} - 1} \right){{\rm{d}}^2}}}{{18 \times {\rm{\nu }}}}\)

Where,

\({\rm{\nu }} = {\rm{Kinematic;viscosity}} = \frac{{\rm{\mu }}}{{{{\rm{\rho }}_{\rm{w}}}}}\)

μ = dynamic viscosity, ρ_W = density of water,

G = Specific gravity of particle, d = diameter of spherical particle

The particle removal efficiency of i^th particle (η_i)

\({{\rm{\eta }}_{\rm{i}}} = \frac{{{{\rm{V}}_{{\rm{Si}}}}}}{{{{\rm{V}}_{\rm{o}}}}} \times 100\)

Where V_o = Surface Overflow Rate

Calculation:

V_o = 40 m³/m²/day = 40/86400 m/sec

ρ_P = density of particle = 2650 kg/m³, η_i = 90 %

ρ_W = 1000 kg/m³, g = 9.81 m/sec², ν = 1.1 × 10^-6 m²/sec

\({\rm{G}} = \frac{{2650}}{{1000}} = 2.65\)

\(90 = \frac{{{{\rm{V}}_{{\rm{Si}}}}}}{{\frac{{40}}{{86400}}}} \times 100\)

\(\therefore {{\rm{V}}_{{\rm{Si}}}} = 0.9 \times \frac{{40}}{{86400}}\)

\(0.9 \times \frac{{40}}{{86400}} = \frac{{{\rm{g}} \times \left( {{\rm{G}} - 1} \right){{\rm{d}}^2}}}{{18 \times {\rm{\nu }}}}\)

\({{\rm{d}}^2} = \frac{{0.9 \times 40 \times 18 \times 1.1 \times {{10}^{ - 6}}}}{{86400 \times 9.81 \times 1.64}}\)

∴ d = 2.2576 × 10^-5 m

Hence diameter of a spherical particle in μm is d =22.58 μm

Explanation:

\(\begin{array}{l} \frac{{dv}}{{dt}} = α - β v\end{array}\)

Integrating on both the side

\( \smallint \frac{{dv}}{{\left( {α - β v} \right)}} = \smallint dt\)

\(\frac{{log_e\left( {α - β v} \right)}}{{ - β }} = t + C\)

Given,

α = 2 m/sec², β = 0.05 s^-1

at \(t = 3sec,\frac{{dv}}{{dt}} = α - β v = 1.3;m/{s^2}\)

\(∴ \frac{{log_e\left( {1.3} \right)}}{{ - 0.05}} = 3 + c\)

\( ∴ c = - 8.247\)

\(\frac{{log_e\left( {α - β v} \right)}}{{ - β }} = t + C\)

Putting values

∴ Iog_e (2 – 0.05 × v) = (t – 8.247) × (– 0.05)

∴ 2 – 0.05 × v = 1.51 e^-0.05t

\(\begin{array}{l} ∴ v = \frac{{ds}}{{dt}} = 40 - 30.21{e^{ - 0.05t}}\ ∴ s = 40t + 604.15{e^{ - 0.05t}} + C' \end{array}\)

∴ at t = 0, s = 0

∴  c’ = – 604.15

Hence, s = 40t + 604.15 e^-0.05t – 604.15

at t = 35 sec

s = 40 × 35 + 604.15 e^{-0.05 × 35} – 604.15

s = 900.84 m

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

Concept:

Visual acuity:

Visual acuity is the ability to see fine details of the object. It denotes the quality of an individual's sight along his/her direct line of vision. Visual acuity is usually measured using the Snellen eye chart, which is based on the usual standard length of the eye exam room (6 m). Normal visual acuity is taken as 6/6.

If the driver with a visual acuity of 6/6 can read upto distance of 'D' m, then the driver with visual acuity of 6/9 can read upto \(\left( {\frac{6}{9} \times {\rm{D}}} \right){\rm{m}}\)

Calculation:

For a driver having 6/6 vision can see the board from distance 48 m.

∴ For a driver having 6/9 vision can see the board from a of distance \(48 \times \frac{6}{9} = 32;m\) 

Given, Vehicle requires 174 m to slow down to 30 km/hr.

So minimum distance of X from the start of Y zone

= 174 – 32

= 142 m

Hence, the minimum value of X = 142 m

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

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