GATE CE 2019 Official Paper: Shift 1 (Previous Year Paper)

Correct answer: 3.

A few is the correct answer.

Few is a quantifier used in regards with plural countable nouns.Without the article “a,” few emphasizes a small number of something.Adding the article removes the emphasis—a few means some.

Quiet means silent and hence it is the correct answer.

Correct answer: 1.

The sentence requires the -ing form of create in order to make it correct. We generally use continuous aspect: for something happening before and after a given time.

A ladder is 6 m away from the base.

\(\tan 45^\circ = \frac{{{\rm{Opposite;Side}}}}{{{\rm{Adjacent;Side}}}} \Rightarrow 1 = \frac{{\frac{H}{5}}}{6} \Rightarrow 1 = \frac{H}{{30}}\)

∴ H = 30 m

∴ Height of Pole is 30 meters.

Sequence wise alphabetical code:

A = 1, B = 2, C = 3, D = 4, E = 5, F = 6, G = 7, H = 8, I = 9, J = 10, K = 11, L = 12, M = 13, N = 14, O = 15, P = 16, Q = 17, R = 18, S = 19, T = 20, U = 21, V = 22, W = 23, X = 24, Y = 25, and Z = 26

From above: E = 5, J = 10, O = 15, T = 20

But the given statement is: E = 10, J = 20, O = 30, T = 40

∴ Every value obtained must be multiplied by 2 to get the answer.

P + E + S + T = 2 × [16 + 5 + 19 + 20] = 120

The correct answer is 'me'

Key Points

• The correct answer is 'me'
• In the given sentence, the usage of 'to' means we need to use a prepositional object and that will be a noun/pronoun. According to the options we need to use a pronoun in an objective case.
• Let's see an example-
• To tell you the truth, I'm tired of listening to politicians tell us lies. ("politicians" is a noun and "to" is a preposition.)

Hence, the correct sentence is- The CEO’s decision to quit was as shocking to the Board as it was to me.

The paragraph mentions how farmers are trying to purchase the new technology despite its commercial use being illegal in India. Hence, option 2 is the correct answer. 

There is no mention of any benefits of the new technology to India or the farmers' desire to use the new technology experimentally. Therefore, options 1 and 3 cannot be inferred.

Option 4 is false as there is no mention of the willingness of farmers to pay a high price for the new technology.

Total players = 300

Count for players playing at least two sports = Total Player – (Player playing only one sport)

Count for players playing at least two sports = 300 – [Player playing only cricket + Player playing only hockey + Player playing only football]

Count for players playing at least two sports = 300 − [105 + 70 + 50] = 75 players

∴ % of players playing at least two sports = \(\frac{{75}}{{300}} \times 100 = 25%\)

Important Point:

Below formula can be used to compute any data for the given questions set.

Total players = Player playing only cricket + Player playing only hockey + Player playing only football + Player playing both hockey and cricket + Player playing both football and cricket + Player playing both hockey and football – 2 × (Player playing all three games)

Explanation:

Correct answer: 3

The sentence implies that the tribal hero, hitherto unrecognized in Hindi cinema, has now arrived on the big stage. He is now a specimen of inquiry, of wonder, and of discussion.

Hence the correct meaning of 'arrived' here is attained a status.

Given:

The side of small square is 5 cm less than larger one.

Area of large square is 4 times of area of small square.

Let,

Side of large square = x

side of small square = x - 5

∵ Area of large square = 4 × Area of small square

⇒ x² = 4 × (x - 5)²

⇒ x² = 4 (x² - 10x + 25)

⇒ 3x² - 40x + 100 = 0

⇒ 3x² - 30x - 10x + 100 = 0

⇒ 3 × (x - 10) – 10 × (x - 10) = 0

⇒ x - 10 = 0; 3x - 10 = 0

∴ x = 10 or x = 10/3 = 3.33

But x = 3.33 gives negative value of side of smaller square.

∴ x = 10 cm

From the given information,

Let's check each of the options:

1) S is the aunt of R → Cannot be determined (as the gender of S is not known, it is possible but not definite)

2) S is the aunt of T → Necessarily False (as S is the parent of T)

3) S is the sister-in-law of Q → Cannot be determined (as the gender of S is not known, it is possible but not definite)

4) S is the brother of P → Cannot be determined (as the gender of S is not known, it is possible but not definite).

Hence, the correct answer is S is the aunt of T''.

Concept:

We have following shortcuts for Limit’s Chapter:

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\ {x \to 0} \end{array};\left( {\frac{{\tan ;x}}{x}} \right) = 1;& ;\begin{array}{*{20}{c}} {{\rm{lim}}}\ {x \to 0} \end{array};\left( {\frac{{\sin ;x}}{x}} \right) = 1\)

Calculation:

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\ {x \to 0} \end{array};\frac{{\sin 4x}}{{\sin 2x}} = \left( {\frac{0}{0}} \right)form;\)

\( = \begin{array}{*{20}{c}} {{\rm{lim}}}\ {x \to 0} \end{array}\frac{{4.\cos 4x}}{{2.\cos 2x}};\left [{\because from;L'hospital;rule} \right]\)

\(= \frac{4}{2} = 2\)

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\ {x \to 0} \end{array}\frac{{\tan x}}{x};\left( {\frac{0}{0}} \right)\)

\(= \begin{array}{*{20}{c}} {{\rm{lim}}}\ {x \to 0} \end{array}\frac{{{{\sec }^2}x}}{1} = 1;\left[ {\because from;L'hospital;rule} \right]\)

\(\therefore \begin{array}{*{20}{c}} {{\rm{lim}}}\ {x \to 0} \end{array};\left( {\frac{{\sin 4x}}{{\sin 2x}}} \right) = 2;and;\begin{array}{*{20}{c}} {{\rm{lim}}}\ {x \to 0} \end{array};\left( {\frac{{\tan ;x}}{x}} \right) = 1\)

Explanation:

For fully saturated soil mass

By applying Darcy’s law

\({V_X} = {K_{X.}}{i_x} = {K_{X.}}\frac{{dh}}{{dx}};\)

\({V_Z} = {K_{Z.}}{i_Z} = {K_{X.}}\frac{{dh}}{{{d_Z}}}\)

Where, \({K_{X.}} = Permeability;in;X;direction\)

                \({K_z} = Permeability;in;Z;direction\)

From the continuity equation

\(\frac{{\delta U}}{{\delta x}} + \frac{{\delta V}}{{\delta Z}} = 0\)

\({K_{X.;}}\frac{{{\delta ^2}h}}{{\delta {X^2}}}; + {K_{Z.;}}\frac{{{\delta ^2}h}}{{\delta {Z^2}}}; = 0\)

For isotropic soil K_X = K_Z

\(\frac{{{\delta ^2}h}}{{\delta {X^2}}}; + \frac{{{\delta ^2}h}}{{\delta {Z^2}}} = 0\)

Explanation:

\(m.\ddot z + kz = 0\)

\(\ddot z + \frac{k}{m}.z = 0\)

Comparing with \(\dot x + \omega _n^2x = 0;\left( {Standard;Differential;Equation;for;Spring;} \right)\)

\( \Rightarrow \omega _n^2 = \frac{k}{m}\)

Natural frequency, \(\left( {{\omega _n}} \right) = \sqrt {\frac{k}{m}}\)

Explanation:

Taylor’s series method:

The Taylor series can be used to calculate the value of an entire function at every point, if the value of the function, and of all of its derivatives, are known at a single point.

Taylor's series expansion for f (x + h) is

\(f\left( {x + h} \right) = f\left( x \right) + hf'\left( x \right) + \frac{{{h^2}}}{{2!}}f''\left( x \right) + \frac{{{h^3}}}{{3!}} + f'''\left( x \right) + \ldots \infty \)

Concept:

Case 1:

Consider the truss given the figure below.

The two members at joint C are connected together at a right angle and there is no external load on the joint.

The free-body diagram of joint C, in the figure below, indicates that the force in each member must be zero in order to maintain equilibrium.

∑Fx = 0, FCB = 0

∑Fy = 0, FCD = 0

Furthermore, as in the case of joint A, in the figure below:

∑FY = 0, FAB sin θ = 0, F AB = 0 kN

∑Fx = 0, - FAE + 0 = 0, FAE = 0 kN

Case 2:

Consider the truss given the figure below.

Zero-force members also occur at joints having a geometry as joint D in the figure below.

Here no external load acts on the joint, so that a force summation in the y-direction, shown in the figure below, which is perpendicular to the two collinear members, requires that FDF = 0.

∑Fy = 0, FDF = 0

Using this result, FC is also a zero-force member, as indicated by the force analysis of joint F, in the figure below:

∑Fy = 0, FCF sin θ + 0 = 0

∴ FCF = 0

Calculation:

We know that, if three members are meeting at the joint, two of them are collinear and there is no point load acting on the joint, then the third member will carry zero force.

From the above statement,

We can say the GR, FI, RS, PR members will carry zero force

Concept:

For the plane strain condition:

\({ϵ_{{I}}} = \frac{{{ϵ_x};+ ;{ϵ_y}}}{2};+;\left( {\frac{{{ϵ_x};- ;{ϵ_y}}}{2}} \right)\cos 2\theta + \left( {\frac{{{γ _{xy}}}}{2}} \right) \sin;2\theta \)

\({ϵ_{{II}}} = \frac{{{ϵ_x};+ ;{ϵ_y}}}{2} - \left( {\frac{{{ϵ_x};- ;{ϵ_y}}}{2}} \right)\cos 2\theta - \left( {\frac{{{γ _{xy}}}}{2}} \right) \sin ;2\theta \)

\({γ _{x{y}}} = ; - \left( {\frac{{{ϵ_x};- ;{ϵ_y}}}{2}} \right)\sin 2\theta + \left( {\frac{{{γ _{xy}}}}{2}} \right) \cos2\theta \)

Calculation:

Given:

ϵ_x = 0.0030, ϵ_y = 0.0020, γ_xy = 0

Shear strain (\({γ _{x{y^I}}})\) for \({γ _{xy}} = 0\) would be maximum for θ = 45°

Maximum normal strain in plane of maximum shear strain i.e. θ = 45°

\({ϵ_{max}} = \frac{{{ϵ_x};+ ;{ϵ_y}}}{2} + \left( {\frac{{{ϵ_x};- ;{ϵ_y}}}{2}} \right)\cos;90^\circ + \left( {\frac{{{γ _{xy}}}}{2}} \right) \sin 90^\circ\)

\(\therefore;{ϵ_{max}} = \frac{{{ϵ_x};+ ;{ϵ_y}}}{2}\)

Maximum normal strain:

\(\frac{{{\varepsilon _x};+;{\varepsilon _y}}}{2} \Rightarrow \frac{{0.0030;+;0.0020}}{2} = \frac{{0.0050}}{2} \Rightarrow 0.0025\)

Explanation:

 

\(\sum {F_H} = 0\)

\({F_1} = 0 \ldots ..\left( i \right)\)

∑F_v = 0

P = 30 \(\ldots ..\left( {ii} \right)\)

∑F_H = 0

F₂ = F₃ ____(iii)

\(\sum {M_p} = 0\)

F₂ × 3 + 30 × 3 – F₃ × 1 = 0

F₂ × 3 – F₃ × 1 + 90 = 0

2F₂ = -90

⇒ F₂ = -45

\(\sum {M_Q} = 0\)

F₃ = 2 + R × 3 = 0

-45 × 2 + R × 3 = 0

R = 30

Q + R = 0

⇒ Q = - 30 kN

Concept:

We have for type sections are as follows:

Slender section:

The cross-section in which the elements buckle locally even before the attainment of yield stress is called slender sections.

Plastic section:

Cross-section, which can develop plastic hinge and have the rotation capacity required for the failure of the structure by formation of a plastic mechanism is called the plastic section.

Compact section:

Cross-section which can develop the plastic moment of resistance but have inadequate plastic hinge rotation capacity for the formation of a plastic mechanism before bucking are called a compact section.

Semi-compact section:

The cross-section in which the extreme fiber in compression can reach yield stress but cannot develop the plastic moment of resistance due to local buckling is called semi-compact or a non-compact section.

Low Shear Design:

V ≤ 0.6 × V_d

Design bending strength of a semi-compact section in case of low shear is

⇒ \(\rm{{M_{d1}} = \frac{{{Z_e}{f_y}}}{{{\gamma _{{m_O}}}}}}\)

High Shear Design

V > 0.6V_d

In the high shear design case, the whole portion of the web is completely involved in resisting against shear.

Design bending strength of a semi-compact section in case of high shear is

⇒ \(\rm{{M_{d2}} = \frac{{{Z_e}{f_y}}}{{{\gamma _{{m_O}}}}}}\)

\(\rm {For;Low;shear;Design,;No;Moment;reduction;is;recommended;by;IS;800 - 2007}\)

Explanation:

\(\rm Nominal~cover= \rm{{ Cover}} - \rm\frac{{Diameter; of; main; reinforcement;bar}}{2} - \rm{ Diameter;of;stirrup}\)

\(= 50;mm - \frac{{16}}{2}mm - 12\)

= 50 - 8 - 12 = 50 - 20 = 30 mm

Additional Information

Cover requirements as per IS 456:2000:

• In order to prevent the reinforcement from corrosion, a layer of concrete should be there in reinforced concrete structures.
• In actual construction practices, the specified minimum cover is very difficult to maintain. Thus IS 456:2000 specifies tolerance levels ranging from 0 mm to +10 mm i.e. no reduction in the clear cover is permitted but the nominal cover can be increased up to 10 mm.

Concept:

If any interior angle of a triangle is more than 120° or less than 30°; then the triangle is called ill-conditioned.

Explanation:

 

Additional Information

Triangulation surveying is the tracing and measurement of a series or network of triangles to determine distances and relative positions of points spread over an area, by measuring the length of one side of each triangle and deducing its angles and length of the other two sides by observation from this baseline.

Concept:

Braking distance is the distance traveled by the vehicle from the moment when the brake was applied to the moment when the vehicle stops.

Braking distance is also the work done by the frictional force on to the vehicle to stop it completely.

\(Braking;distance = \frac{{{V^2}}}{{2 \times g \times \left( {f + 0.01;G} \right) \times \eta }}\)

Where V = Speed of the vehicle before the brake is applied

                f = coefficient of lateral friction

                G = gradient in percentage

                η = braking efficiency

Calculation:

Braking distance for G% gradient

G = +G%

\(Braking;distance;\left( {{d_1}} \right) = \frac{{{V^2}}}{{2g\left( {{f_1} + 0.01G} \right)}}\)

Braking distance for 2G% gradient

G = + 2G%

\({\rm{Braking;distance;}}\left( {{{\rm{d}}_2}} \right) = \frac{{{V^2}}}{{2g\left( {f + 0.01 \times 2G} \right)}} = \frac{{{V^2}}}{{2g\left( {f + 0.02G} \right)}}\)

\({\rm{% ;change}} = \frac{{{d_1} - {d_2}}}{{{d_1}}} \times 100 = ;\frac{{\frac{{{V^2}}}{{2g\left( {f + 0.01G} \right)}} - \frac{{{V^2}}}{{2g\left( {f + 0.02G} \right)}}}}{{\frac{{{V^2}}}{{2g\left( {f + 0.01G} \right)}}}}\)

\(\frac{{\frac{1}{{f + 0.01G}} - \frac{1}{{f + 0.01G}}}}{{\frac{1}{{f + 0.01G}}}} = \frac{{\left( {f + 0.02G} \right) - ({f_r} + 0.01G)}}{{f + 0.02G}} = \frac{{0.01G}}{{f + 0.02G}} \times 100;\)

Explanation:

As the base of the slab is frictionless less, the Slab base is free to expand in all directions.

⇒ ∴ Thermal stress = zero

Additional Information

• When the temperature of material changes there will be a corresponding change in dimension.
• If a member is free to expand or contract due to the rise or fall of the temperature, no stress will be induced in the member.
• If the member is constrained (i.e. body is not allowed to expand or contract freely), change in length due to rise or fall of temperature is prevented and stresses are developed in the body which is known as thermal stress.

Explanation:

We know that velocity head \(= \frac{{{V^2}}}{{2g}}; \ldots \ldots \ldots \ldots \ldots ..equation;\left( i \right)\)

Area B × y

Top surface width (T) = B

Q = A × V

Q = B × y × V

For critical flow condition Froude number \(\left( {{F_r}} \right) = 1\)

\(\Rightarrow \frac{{{Q^2}T}}{{g{A^3}}} = 1\)

\({Q^2}T; = ;g \times {A^3}\)

\(\left( {B \times y \times V} \right)^2 \times 2T; = ;g \times {A^3}\)

\({B^2} \times {y^2} \times {V^2}\left( B \right) = g{\left( {B.y} \right)^3}\)

\( \Rightarrow y = \frac{{{V^2}}}{g}{\rm{;}} \ldots \ldots \ldots \ldots \ldots ..equation;\left( {ii} \right)\)

Dividing equation 1 & equation 2

\( \Rightarrow \frac{{Velocity;head}}{{depth}} = \frac{{\frac{{{V^2}}}{{2g}}}}{{\frac{{{V^2}}}{g}}} = ;\frac{1}{2}\)

Explanation:

Aqueduct:

When the HFL (Highest Flood Level) of the drain is sufficiently below the bottom of the canal such that the drainage water flows freely under gravity, the structure is known as Aqueduct.

Sluice gate: 

It is used to control the flow in a horizontal channel of unit width.

Super passage:

A super passage is just like a bridge in which the natural drain is carried over the canal. A super passage is constructed where the bed of the drain is well above the canal F.S.L (Full Supply Level).

Level Crossing:

The level crossing is an arrangement provided to regulate the flow of water through the drainage and the canal when they cross each other approximately at the same bed level.

Concept:

Average precipitation is calculated from any one of the following:

1. Arithmetic Mean Method
2. Theisen Polygon Method
3. Isohyetal Method
4. Inverse distance weighting (IDW) method

Arithmetic Mean Method for average precipitation calculation is the simplest method for determining the areal average.

\(\bar P = \frac{1}{N} \times \mathop \sum \limits_{i = 1}^N P\)

where P_i = rainfall at the i^th rain gauge station

N = total number of rain gauge stations

Theisen Polygon Method: This method assumes that any point in the watershed receives the same amount of rainfall as that measured at the nearest rain gauge station. Here, rainfall recorded at a gauge can be applied to any point at a distance halfway to the next station in any direction.

Any of the methods among two can be more accurate than other for the given specific and rain and areal characteristics.

Explanation:

For the retaining with supporting backfill of cohesionless soil having an angle of internal friction ϕ and inclined at an angle β with the horizontal.

Active Earth Pressure (P_a) acts at a height H/3 from the base of the wall and at an angle β with the horizontal.

Additional Information

It is assumed that the vertical stress and the lateral pressure acting on soil element are conjugate stresses i.e. the direction of forces are parallel to the plane on which the other acts.

The earth pressure coefficients are given as:

Concept:

For a structure constructed on cohesionless soil, and in contact with the water body. Due to the difference in the water level across the structure, the hydraulic gradient would be set up resulting in upward seepage. Water movement in upward direction would exert upward thrust on the soil and thereby reducing the effective stress.

If the head causing upward flow is increased, a stage is reached when the effective stress is reduced to zero. The condition so developed is known as quicksand condition. Quicksand is not any special type of sand, but it is a condition of zero effective stress in cohesionless soils.

For Quick sand condition, effective stress in the soil becomes zero (σ = 0)

At the bottom of the soil column,

\(\sigma = \gamma \times L\)

\(u = {\gamma _w};\left( {L + {\rm{\Delta }}H} \right)\)

During the quick sand condition, the effective stress is reduced to zero.

\(\gamma \times L = {\gamma _w} \times \left( {L + {\rm{\Delta }}H} \right)\)

\(L;\left( {\gamma - {\gamma _w}} \right) = {\gamma _w} \times {\rm{\Delta }}H\)

\(L \times {\gamma _b} = {\gamma _w} \times {\rm{\Delta }}H\)

\(\frac{{{\rm{\Delta }}H}}{L} = \frac{{{\gamma _b}}}{{{\gamma _w}}} = {i_c};\)

\({\rm{where;}}\)i_c = critical hydraulic gradient

Explanation:

Primary pollutant: 

• Pollutants that are emitted directly from identifiable sources, either from natural hazardous events like dust storms, volcanoes, etc, or from human activities like burning of wood, coal, oil in homes or industries or automobiles, etc.

Secondary pollutant:

• The primary pollutants often react with one another or with water vapor, aided and abetted by the sunlight, to form entirely a new set of pollutants, called the secondary pollutants.
• These are the chemical substances, which are produced from the chemical reactions of natural or anthropogenic pollutants or due to their oxidation, etc., caused by the energy of the sun.

Important Points

Concept:

Bending equation:

\(\frac{σ}{y} = \frac{M}{I} = \frac{E}{R}\)

where σ = bending stress, y = distance of fibre from the neutral axis,

M = bending moment on the cross-section of the beam

I = moment of inertia, E = modulus of elasticity,

R = radius of curvature during bending.

Calculation:

Given:

diameter, 

Overall Depth (d) = 500 mm,

Compressive and tensile strain = 2.5 × 10-4

From bending equation

\(\frac{σ}{y} = \frac{E}{R}\)

We know,

Stress ∝ strain

\(\sigma \propto \varepsilon \Rightarrow \sigma = E× \varepsilon \Rightarrow \frac{\sigma }{E} = \varepsilon \)

\(\frac{\sigma}{y} = \frac{E}{R} \Rightarrow \left( {\frac{{\rm{\sigma}}}{{\rm{E}}}} \right) = \frac{y}{R};\)

\(\Rightarrow 2.5 × {10^{ - 4;}} = y × \frac{1}{R}\)

\(y = \frac{d}{2} = \frac{{500}}{2} = 250;mm\)

\(\Rightarrow \frac{{250}}{{\rm{R}}} = 2.5 × {10^{ - 4}}\)

Curvature (1\R) = \( \frac{{2.5 × {{10}^{ - 4}}}}{{250}} \) = 1× 10^-6 mm^-1 = 1× 10-3 m-1 = 0.001 m^-1

Concept:

Surface overflow rate: It can be thought of as the settling velocity of that particle if introduced at the topmost point at the inlet of the tank gets settled at the bottom-most point just before the outlet of the tank.

Particles having sizes greater than the particles for which setting velocity is equal to overflow rate gets 100% removed.

Particles having settling velocity less than overflow rate will not get 100% removed.

The percentage removal of particles having settling velocity Vs' is given as \(=\frac{{{V}_{s}}'}{{{V}_{s}}}\times 100\)

Where, V_S’ = Settling velocity of smaller size particles.

Settling velocity is calculated by Stoke’s law for discrete particle settling. It is given as-

\({V_T} = \frac{{\left( {G - 1} \right)\gamma {d^2}}}{{18;\mu }} \Rightarrow {V_T} \propto {d^2}\)

Where Gs = Specific gravity of particles.

d = Particle size

ν = Kinematic Viscosity of fluid in which particle is settling.

Therefore, for different particles of the same density, a particle with a higher diameter will settle down faster.

Calculation:

For 30 second duration if 0.35 mm particle size settles completely then % removal of particle size 0.45 mm and 0.50 mm will be 100% respectively for each.

As settling velocity of particle size 0.45 mm and 0.50 mm will be greater than the settling velocity of size 0.35 mm (V_T ∝ d²).

Concept:

The five-minute peak hour factor for 5 minutes is defined as the ratio of peak hourly traffic for 5 minutes to the maximum possible traffic for minutes.

Calculation:

Given,

(maximum number of vehicle) in 5 minutes = 160 vehicles

Flow in the peak hour = 1000 vehicles

\(5;minute - peak;hour;factor; = \frac{{;peak;hour;flow}}{{12;\left( {5;minute;traffic} \right)}} = \frac{{1000}}{{12\left( {160} \right)}} = 0.52\)

Additional Information

The Peak Hour Factor (PHF) compares the traffic volume during the busiest 15 minutes of the peak hour with the total volume during the peak hour. It indicates how consistent traffic volume is during the peak hour.

\({\rm{PHF}} = \frac{{\left( {{\rm{total;hourly;volume}}} \right){\rm{;}}}}{{\left( {{\rm{peak;}}15{\rm{;minute;volume;within;the;hour;x;}}4} \right)}}\)

Explanation:

Given,

D₁ = 30 cm, D₂ = 15 cm

Differential manometer shows 8 cm of Hg

Discharge = Q_actual \(\rm{= {C_d} \times {Q_{therotical}} = {C_d} \times ;\frac{{\sqrt {2gh} ;{A_1}{A_2}}}{{\sqrt {A_1^2 - A_2^2} }}}\) 

\(\rm{{A_1} = \frac{\pi }{4}{\left( {0.3} \right)^2};;{A_2} = \frac{\pi }{4}{\left( {0.15} \right)^2}}\)

Where,  A₁ = Area of cross section (1), A₂ = Area of cross section (2)            

\(\rm{H = Pressure;Head = \left( {\frac{{{S_{Hg}}}}{{{S_w}}} - 1} \right) \times mercury;differential;head = \left( {\frac{{13.6}}{1} - 1} \right) \times 0.08 = 1.008;m}\)

\({\rm{Q}} = \frac{{\sqrt {2 \times 9.81 \times 1.008} \times \frac{{\rm{\pi }}}{4} \times {{\left( {0.3} \right)}^2} \times \frac{{\rm{\pi }}}{4}{{\left( {0.15} \right)}^2}}}{{\sqrt {{{\left[ {\frac{{\rm{\pi }}}{4} \times {{0.3}^2}} \right]}^2} - {{\left[ {\frac{{\rm{\pi }}}{4} \times {{0.15}^2}} \right]}^2}} }}\left[\because {{{\rm{C}}_{\rm{d}}} \approx 1;{\rm{Flow;is;assumed;as;frictionless;flow}}} \right]\)

∴ Q = 0.0811 m³/s

Concept:

The probability of the event occurring at least once in ‘n’ successive years, p = 1 – qⁿ

P = probability of occurrence = \(\frac{1}{T}\)

Q = probability of non-occurrence

n = no of years

Calculation:

Given,

Probability of flood discharge exceeds at least once in next 5 years = 0.25

\(p = 1 - {\left( {1 - \frac{1}{T}} \right)^n} = 0.025\)

\(⇒ 1 - 0.25 = {\left( {1 - \frac{1}{T}} \right)^5}\)

\({\left( {0.75} \right)^{\frac{1}{5}}} = 1 - \frac{1}{T}\)

\(\frac{1}{T} = 1 - {\left( {0.75} \right)^{\frac{1}{5}}}\)

\(⇒ \frac{1}{T} = 1 - 0.944 = 0.559\)

⇒ T = 17.89 years

Concept:

\(Dry;density;\left( {{\gamma _{dry}}} \right) = \frac{{{G_s}.{\gamma _w}}}{{1 + wG}}\)

Where G = Specific gravity of soil particle

                w = water content of the soil

                \({\gamma _w} =\)Density of water

Calculation:

Given,

Specific gravity (G) = 2.65

Mass density of water (γ_w) = 1000 kg/m³

Moisture content (w) = 10%

\(Dry;density\left( {{\gamma _{dry}}} \right) = \frac{{{G_s}.{\gamma _w}}}{{1 + wG}} = \frac{{2.65 \times 1000}}{{1 + 0.10 \times 2.65}} = \frac{{2650}}{{1.265}} = 2094.86;kg/{m^3}\)

Explanation:

Vertical stress increment at any depth (z) at a radial distance (r) as per Boussines’q theory is:

\({\rm{\Delta }}{\sigma _z} = {K_B}\frac{Q}{{{z^2}}}\)

Where,

KB = Boussinesq’s influence factor

\({K_B} = \frac{3}{{2\pi }}{\left[ {\frac{1}{{1 + {{\left( {\frac{r}{z}} \right)}^2}}}} \right]^{\frac{5}{2}}}\)

Given,

Q = 500 kN, 

\(Vertical;stress,;\left( {{\sigma _z}} \right) = \frac{Q}{{{Z^2}}};\frac{3}{{2\pi }}{\left[ {\frac{1}{{1 + {{\left( {\frac{r}{z}} \right)}^2}}}} \right]^{\frac{5}{2}}}\)

\({\sigma _z} = K.\frac{Q}{{{Z^2}}} \Rightarrow {{\rm{\sigma }}_z} \propto \frac{1}{{{Z^2}}}\)

Z₁ = 2 m, Z₂ = 4 m

\(\frac{{{\sigma _{z1}}}}{{{\sigma _{z2}}}} = {\left( {\frac{{{Z_2}}}{{{Z_1}}}} \right)^2} = {\left( {\frac{4}{2}} \right)^2} = 4{\rm{;}}\)

Explanation:

Option 1 & 4 are true, since \(\left| x \right|\) has global minimum at x = 0 and x³ has neither maximum nor minimum

Modulus of x has following graph and hence global minima is at x = 0

​

\({x^3};\) has following graph and hence it does not have global maxima or minima at any point

Let, \(y = \sqrt[x]{x} = {x^{\frac{1}{x}}}\)

\( \Rightarrow \ln y = \ln {x^{\frac{1}{x}}} = \frac{{\ln x}}{x} \Rightarrow y = {e^{\frac{{\ln x}}{x}}}\)

\(\therefore Consider;the;funciton;f\left( x \right) = \frac{{\ln x}}{x}\)

\(\Rightarrow f'\left( x \right) = 0 \Rightarrow \frac{{x\left( {\frac{1}{x}} \right) - \ln x}}{{{x^2}}} = 0 \Rightarrow \frac{{1 - \ln x}}{{{x^2}}} = 0\)

⇒ x = e → stationary point

\(f''\left( x \right) = \frac{{{x^2}\left[ {0 - \frac{1}{x}} \right] - \left[ {1 - \ln x} \right]2x}}{{{x^4}}}\)

\(\Rightarrow f''\left( x \right) = \frac{{ - x - 2x + 2x\ln x}}{{{x^4}}} = \frac{{ - 3x + 2x\ln x}}{{{x^4}}}\)

\(\therefore f''\left( e \right) = \frac{{ - 3e + 2e\left( 1 \right)}}{{{e^4}}} = \frac{{ - e}}{{{e^4}}} = - {e^{ - 3}} < 0\)

∴ x = e is a point of maximum

\(\therefore y = {e^{\frac{{lnx}}{x}}} \Rightarrow y = \sqrt[x]{X}\) has global maximum at x = e

Explanation:

\(\frac{{\partial v}}{{\partial t}} = \beta \frac{{{\partial ^2}v}}{{\partial {x^2}}}\)

\(\frac{{\partial v}}{{\partial t}} = \frac{{V_i^{\left( {n + 1} \right)}V_i^{\left( n \right)}}}{{{\rm{\Delta }}t}}\) (using forward time finite difference approximation)

Also, \({f^{11}}\left( x \right) = \frac{{{\partial ^2}f}}{{d{x^2}}} = \frac{{f\left( {x + h} \right) - 2f\left( x \right) + f\left( {x - h} \right)}}{{{h^2}}}\) 

(Using centered space finite difference approximation)

\(\Rightarrow \frac{{{\partial ^2}v}}{{\partial {x^2}}} = \frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{{{\left( {{\rm{\Delta }}x} \right)}^2}}}\)

\(\therefore \frac{{\partial v}}{{\partial t}} = \beta \frac{{{\partial ^2}v}}{{\partial {x^2}}}\) can be represented as

\(\frac{{V_i^{\left( {n + 1} \right)} - V_i^{\left( n \right)}}}{{{\rm{\Delta }}t}} = \beta \left[ {\frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{{{\left( {{\rm{\Delta }}x} \right)}^2}}}} \right]\)

Explanation:

Given,

width (B) = 5 m

depth (y) = 2 m

Area (B × y) = 5 × 2 = 10 m²

Perimeter (R) = B + 2y = 5 + 2 × 2 = 9 m

\(Hydraulic;radius;\left( R \right) = \frac{{Area}}{{Perimeter}}\)

\(R = \frac{{10}}{9} = 1.11\)

By the Manning's equation

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

\(2 = \frac{1}{n}{\left( {1.11} \right)^{\frac{2}{3}}}{\left( {0.001} \right)^{\frac{1}{2}}}\)

\(n = \frac{{1.072 \times 0.03162}}{2} = 0.0169; \cong 0.017\)

Manning’s roughness coefficient (n) = 0.017

Concept:

Statement - 1:

The Oedometer test

• The characteristics of soil during one-dimensional consolidation or swelling can be determined by means of the oedometer test.
• In the oedometer test, stress is applied to the soil specimen along the vertical axis, while strain in the horizontal directions is prevented.
• Therefore, the lateral stress in the soil while being tested in an oedometer is always at rest.

Statement - 2:

Pressure Distribution for different forms of footings:

For a perfectly rigid footing resting on the surface of a sand deposit, the contact pressure distribution is zero at the edges and maximum at the center.

However, for a very deep rigid footing on the sand, the contact pressure distribution may tend to become like that of rigid footing on clayey soil, with edge contact stress greater than at its center.

Statement - 3:

Overburden correction is applied to sandy as well as clayey soil but Dilatancy Correction is applied only for granular or sandy soil as sometimes it might not be able to dissipate pore pressure in case of dynamic loading, which results in a higher SPT-N value.

Thus, dilatancy correction is to be applied for granular soil only.

\(x; = ;\psi ln\phi ;& ;y; = \phi ln\psi\)

Partial differentiating both equations with respect to x

\(\Rightarrow 1 = {\psi _x}\ln \phi + \frac{\psi }{\phi } \times {\phi _x}\) …………..(i)

\( \Rightarrow 0 = {\phi _x}\ln \psi + \frac{\phi }{\psi } \times {\psi _x};\)

\(\Rightarrow \frac{\phi }{\psi } = \frac{{ - {\phi _x} \times \ln \psi }}{{{\psi _x}}}\)

\(\Rightarrow \frac{\phi }{{\psi \times {\phi _x}}} = \frac{{ - ;\ln \psi }}{{{\psi _x}}}\)

\(put;\left( {ii} \right)in;\left( i \right)\)

\(1 = {\psi _x}\ln \phi - \frac{{{\psi _x}}}{{ln\psi }} = ;{\psi _x} \times \left[ {\frac{{ln\phi ;ln\psi - 1}}{{\ln \psi }}} \right]\)

\({\psi _x} = ;\frac{{ln\psi }}{{ln\phi ln\psi - 1}}\)

Explanation

Shear stress acting at a section  \(;\left( \tau \right) = \frac{{FA\bar y}}{{Ib}}\)

F - Shear force acting at section

A - Area of the region above the section

Y̅ = Distance of centroid of the region from the N.A

I - Moment of inertia

Calculation:

Given:

F = shear force = 8 kN = 8000 N

A = Area of shaded portion = 50 × 100 = 5000 mm²

Y̅ = Distance of centroid of shaded portion form N.A

\(= 200 - \frac{{100}}{2} = 150;mm\)

I = 1.5 × 10⁹ mm⁴

Shear stress acting over the region

 \(\Rightarrow \tau = \frac{{8000 \times 50 \times 1000 \times 150}}{{1.5 \times {{10}^9} \times 50}} = 0.08;N/m{m^2}\)

⇒ Shear Force resisted by nails at P = shear stress × resisting area

⇒ Shear Force = 0.08 × 60 × 50 = 240 N

Explanation:

Method 1: Moment Distribution method

Assume ‘R’ sinks by Δ

Fixed End Moments are given by

\({M_{{F_{QR}}}} = \frac{{ - 6EI \times {\rm{\Delta ;}}}}{{{L^2}}};\)

\({M_{{F_{RQ}}}} = \frac{{ - 6EI \times {\rm{\Delta ;}}}}{{{L^2}}}\)

\({M_{{F_{RS}}}} = ;{M_{{F_{SR}}}} = 0\)

Distribution Factor at R is 0.5 & 0.5 for each end respectively due to symmetricity

By Analysis of beam QR,

\(\sum {M_Q} = 0\)

\(\frac{{4.5 \times EI \times {\rm{\Delta ;}}}}{{{L^2}}} + \frac{{3 \times EI \times {\rm{\Delta ;}}}}{{{L^2}}} = P \times L\)

\({\rm{\Delta ;}} = \frac{{{\rm{P}} \times {{\rm{L}}^3}}}{{7.5{\rm{;EI}}}}{\rm{;}}\)

Method 2: Slope deflection method

\({M_{QR}} = \frac{{2EI}}{l}\left( {{\theta _R} - \frac{{3{\rm{\Delta }}}}{l}} \right)\)

\({M_{RQ}} = \frac{{2EI}}{l}\left( {2{\theta _R} - \frac{{3{\rm{\Delta }}}}{l}} \right)\)

\({M_{RS}} = \frac{{2EI}}{l}\left( {2{\theta _R}} \right)\)

If reaction at S is equal to zero

M_RQ + M_QR + P × l = 0

\(\frac{{6EI{\theta _R}}}{l} - \frac{{12EI{\rm{\Delta }}}}{{{l^2}}} + Pl = 0\)

\(\frac{{6EI{\theta _R}}}{l} - \frac{{12EI}}{{{l^2}}} \times \frac{{P{l^3}}}{{\beta EI}} + Pl = 0\)

\(\frac{{6EI{\theta _R}}}{l} - \frac{{12Pl}}{\beta } + PL = 0\) -----(i)

From equilibrium of joint

M_RQ + M_RS = 0

\(\frac{{8EI{Q_R}}}{l} - \frac{{6EI{\rm{\Delta }}}}{{{l^2}}} = 0\)

\(\frac{{6EI{\theta _R}}}{l} = \frac{6}{8}\left( {\frac{{6EI}}{{{l^2}}} \times \frac{{P{l^3}}}{{\beta EI}}} \right)\)

\(\frac{{6EI{\theta _R}}}{l} = \frac{{36;Pl}}{{8\beta }}\) ----(ii)

⇒ From (i) & (ii)

\(\frac{{36Pl}}{{8\beta }} - \frac{{96Pl}}{{8\beta }}Pl = 0\)

\(\Rightarrow \frac{{60Pl}}{{8\beta }} + Pl = 0\)

\(\Rightarrow 8\beta = 60\)

\(\beta = \frac{{60}}{8} = 7.5\)

Explanation:

For plastic state

Area in compression = Area in tension

To locate the equal area axis, equate the area of both side

60 × 5 + (x - 5) × 5 = (65 - x) x5

300 + 5x – 25 = 325 – 5x

10x = 325 + 25 – 100 = 50

x = 5 mm from top

⇒ Neutral Axis lies at the intersection web and flange

To calculate the centroidal axis, taking the moment of area about bottom

\(\bar x = \frac{{60 \times 5 \times 30 + 60 \times 5 \times \left( {60 + \frac{5}{2}} \right)}}{{60 \times 5 + 60 \times 5}}\)

\(= \frac{{9000 + 18750}}{{600}} = 46.25mm;from;bottom\)

x̅ = 65 – 46.25 = 18.75 mm from top

Difference in NA = 18.75 – 5 = 13.75 mm

Explanation:

We know that

Settling velocity

\(\left( {{V_s}} \right) = \frac{{g; \times ;{\rho _w} \times \left( {G - 1} \right); \times ;{d^2}}}{{18;\mu }}\)

\({V_s} = \frac{{980 \times \left( {2.51 - 1} \right){{\left( {25 \times {{10}^{ - 4}}} \right)}^2} \times 1}}{{18 \times 0.01}} = 0.051;cm/sec\)

Given flow rate (Q) = 0.2 m²/sec

\(We;know;that;surface;overflow;rate;\left( {{V_0}} \right) = \frac{{Volume;/;time}}{{surface;area}};\)

\({V_0} = \frac{Q}{{surface;area}}\)

surface area = B × L = 8 × 32 = 256m²

\({V_0} = \frac{Q}{{BL}} = \frac{{\left( {0.2{m^3};/;sec} \right)}}{{256}} = 7.8125 \times {10^{ - 4}};m/sec\)

V₀ = 7.8125 × 10^-2 cm/sec

\(Particle;removal;efficiency;\left( \eta \right) = \frac{{{V_s}}}{{{V_0}}} \times 100;\)

\(\eta = \frac{{0.051}}{{0.078}} \times 100 = 65.38%\)

Concept:

\(True;length;line = \frac{{l'}}{l} \times measured;length\ \)

Calculation:

Given,

Measured distance (l) = 285.5 m

Length of tape = 30 m

True length of tape = 30 m - 0.05 = 29.95

\(Correct;length;\left( l \right) = \frac{{29.95}}{{30}} \times 285.5 = 285.024;m\)

The line measured on 1/10 slope

Correction due to slope

\( = \frac{{{h^2}}}{{2l}} = \frac{{{{\left( {\frac{{285.024}}{{10}}} \right)}^2}}}{{2 \times 285.024}} = 1.425\)

Corrected length = 285.024 – 1.425 = 283.6 m

Concept:

Fillet weld forms a rectangular shape on either side of gusset plate.

Taking this rectangular welds separately for analysis,

As per IS 800: 2007

Equivalent maximum resultant stress due to combination of normal & shear stress

\(= {f_C} = √ {f_a^2 + 3{q^2}}\)

f_a = normal stress due to bending

q = shear stress

\(q = \frac{{shear;force}}{{shear;area}} = \frac{{350;kN}}{{dt + dt}}\)

t = effective throat thickness = ks (s = size of weld)

⇒ t = 0.7 × 10 = 7 mm

\( ⇒ q = \frac{{350 \times 1000}}{{2 \times 10 \times 500}} = 50;N/m{m^2}\)

\({f_a} = \frac{{Pe}}{z} = \frac{{350 \times 1000 \times 100}}{{2 \times 7 \times \frac{{{{\left( {500} \right)}^2}}}{6}}} = 60;N/m{m^2}\) …………………….. \(\left( {\because Z = \frac{{\left( {2t} \right){d^2}}}{6}} \right)\) 

\(⇒ {f_c} = √ {{{60}^2} + 3 \times {{50}^2}} = 105.35;MPa\)

Alternate Solution:

\({\rm{Resultant;stress;}}~{{\rm{f}}_{\rm{e}}} = \sqrt {{\rm{f}}_{\rm{b}}^2 + {{\rm{q}}^2}} = \sqrt {{{60}^2} + {{50}^2}} \)

⇒ f_e = 78.10 MPa

Explanation:

 

 

Indirect cost = Rs. 500

Crashing possibility = 18 – 17 = 1 day

To reduce the project duration by 1 day, the following options available.

But R has least Cost Slop among all.

Therefore, Crashing ‘R’ by 1 day

After crashing ‘R’ by 1 day, the new network is as for follows.

 

Among all the above option, Q & R and U & R have least cost slope.

Let’s crash Q & R for the time being as crashing U will make the following activity reach critical level as due unavailability of any float period.

After crashing Q & R by 1 day, the new network is as follows.

Project duration = 16 days.

Total project cost = Total normal cost + crashing of ‘R’ by 1 day + crashing cost of Q & R by 1 day + indirect cost/day × Project duration

= 15000 + 6000 + 8000 + 7000 + 6000 + 4000 + 20000 + 750 × 1 + 2750 × 1 + 5000 × 16

= 1,49,500/-

Explanation:

Volume of box = 0.5 × 0.5 × 0.5 = 0.125 m³

Volume (V) = V_c + V_w

Volume of coarse aggregates \(\left( {{V_c}} \right) = \frac{{{W_c}}}{{G.{\gamma _w}}}\)

\(= \frac{{187.5;kg}}{{2.5 \times 1000;kg/{m^3}}} = 0.075;{m^3}\)

Volume of water = V – V_c = 0.125 – 0.075 = 0.05 m³

Water absorption is 0.5%

\(Water;absorbed = 0.5 \times \frac{{187.5}}{{100}} = 0.9375;kg\)

Total weight of water = volume × γ_w + weight of water absorbed

= 0.05 × 1000 + 0.9375 = 50.9375 kg

Explanation:

Σ F_x = 0

⇒ H_p = F = 0

⇒ H_p = -F

ΣM_p = 0

⇒ V_R × 5 + F × 10 = 0

⇒ V_R = -2F

ΣF_y = 0

⇒ V_p + V_R = 0

⇒ V_p = 2F

\({\Delta _{horizontal}}\) \(= \frac{{\partial U}}{{\partial F}} = \frac{{\partial {U_{PQ}}}}{{\partial F}} + \frac{{\partial {U_{QR}}}}{{\partial F}}\left( {Castigliano's;theorem} \right)\) 

\({U_{PQ}} = \smallint \frac{{{M^2}PQ}}{{2EI}}dx\)

\({M_{PQ}} = {V_R} \times x = 2{F_x};;;x\epsilon\left[ {0,5} \right]\)

\(\Rightarrow \frac{{\partial {U_{PQ}}}}{{\partial F}} = \frac{1}{{EI}}\smallint \frac{{\partial {M_{PQ}}}}{{\partial F}} \times {M_{PQ}}dx = \frac{1}{{EI}}\mathop \smallint \limits_0^5 2Fx \times 2x;dx\)

\(= \frac{{4F}}{{EI}}\left[ {\frac{{{x^3}}}{3}} \right]_0^5 = \frac{{4 \times 50}}{{{{10}^6}}} \times \frac{{125}}{3} = \frac{{25}}{{3000}}\)

\({U_{QR}} = \smallint \frac{{M_{QR}^2}}{{2EI}}dx;{M_{QR}} = {F_x};x;\epsilon\left[ {0,10} \right]\)

\(\Rightarrow \frac{{\partial {U_{QR}}}}{{\partial F}} = \frac{1}{{EI}}\smallint \frac{{\partial {M_{QR}}}}{{\partial F}}{M_{QR}}dx = \frac{1}{{EI}}\mathop \smallint \limits_0^{10} Fx \times xdx\)

\(= \frac{F}{{EI}} = \left[ {\frac{{{x^1}}}{3}} \right]_0^{10} = \frac{{50}}{{{{10}^6}}} \times \frac{{1000}}{3} = \frac{{50}}{{3000}}\)

\(\Rightarrow \frac{{\partial U}}{{\partial F}} = \frac{{25}}{{3000}} + \frac{{50}}{{3000}} = \frac{{25}}{{1000}} = 0.025m = 25mm\)

∴ \({\Delta _{horizontal}}{\rm{;}} = \frac{{\partial U}}{{\partial F}} = 25:mm\)

Explanation:

\(1;ppm;N{O_2} = \frac{{1;part;of;N{O_2}}}{{{{10}^6};part;of;air}} = \frac{{1;{m^3};of;N{O_2}}}{{{{10}^6};m;of;air}}\)

1 mole of NO₂ occupies 22.4 liters of volume at STP.

⇒ 1 mole of NO₂ occupies 0.0224 m³ volume at STP.

\(\Rightarrow 1;{m^3}of;N{O_2};contains\frac{1}{{0.0224}};moles = 44.64;moles\)

1 mole of NO₂ = 46 gms

⇒ 1 m³ of NO₂ = 44.64 × 46 gms = 2053.57 gms

⇒ 0.02 m³ of NO₂ = 41.07 gms of NO₂

\(\therefore 0.02;ppm;of;N{O_2} = \frac{{0.02;{m^3}of;N{O_2}}}{{{{10}^6}{m^3};of;air}} = 41.07;gms;per;{10^6};{m^3}\)

= 41.07 × 10^-6 gm/m³ = 41.07 μg/m³

Explanation:

Expanded bed depth \(= De = D\left[ {\frac{{1 - ;n}}{{1 - {n_e}}}} \right]\)

D = Depth of bed = 0.8m; n = Bed porosity = 0.4

n_c = porosity of expanded bed \( = {\left( {\frac{{{V_b}}}{{{V_s}}}} \right)^{0.22}}\)

V_s = terminal velocity of settling particles during back wasting = 0.05 m/s

V_b = backwash velocity

\({V_b} = \frac{Q}{{BL}}\left( {Q = Back;wash;flow;rate} \right)\)

[Q = 3.6 m³/min B = 3 m L = 4 m]

\(\Rightarrow {V_b} = \frac{{3.6}}{{60 \times 3 \times 4}} = 0.005;m/s\)

\(\Rightarrow {n_c} = {\left( {\frac{{0.005}}{{0.05}}} \right)^{0.22}} = 0.60255\)

\(\therefore De = 0.8\left[ {\frac{{1 - 0.4}}{{1 - 0.60255}}} \right] = 1.21;m/s\)

Explanation:

Given,

Waster water flow rate (Q) = 36 m³/hr

Model is N_t = N_0-e^-Kt

N_t = Number of micro-organism survived at time = t = 1% = 0.01 (∵ 99% efficiency is to be achieved)

N₀ = Number of micro-organism at time, t = 0

N_t = N_o × e^-Kt

0.01 = 1 × e^-0.23×1

e^-0.23t = 0.01

-0.23t = In.0.01

-0.23t = -4.605

t = 20 minutes

Volume = Q × t_d = 36 m³/hr × 20 minutes \(= 36\frac{{{m^3}}}{{hr}} \times \frac{{20}}{{60}}hrs\) 

Volume = 12 m³

B × L × H = 12

1 × L × 1.5 = 12

\(Length = \frac{{12}}{{1.5}} = 8m\)

Explanation:

\(\tan \theta = \frac{{2.4 - x}}{{50}} = \frac{{x - 0.4}}{{50}}\)

2x = 2.8

x = 1.4 m

Height Of Index = 100 + 1.4 = 101.4 m

RL of theodolite station = 101.4 – theodolite height = 101.4 – 1.4 = 100 m

Explanation:

Given DE

\({x^2}\frac{{{d^2}y}}{{d{x^2}}} - 2x\frac{{dy}}{{dx}} + 2y = 0\)

\(& ;y\left( 1 \right) = 0,;y\left( 2 \right) = 2\)

⇒ [x²D² – 2xD + 2] y = 0

\(Let;\theta = \frac{d}{{dz}};& ;x = {e^z}{\rm{;}}\)

Let xD = θ; x²D² = θ (θ - 1);

[θ × (θ - 1) - 2θ + 2]y = 0

⇒ [θ² - 3θ + 2] y = 0

The Auxiliary Equation is θ² - 3θ + 2 = 0

⇒ (θ - 2) × (θ - 1) = 0

⇒ θ = 1, 2

→ Obtained roots of Auxiliary Equation are real and distinct

∴ The solution of eq (1) is

y = C₁e^z + C₂e^2z

⇒ y = C₁x + C₂x² [∵ x = e^z]      ………..…(2)

Given ∴ y = 0 at x = 1

⇒ 0 = C₁ + C₂ ⇒ C₁ = -C₂

& y = 2 at x = 2

⇒ 2 = 2C₁ + 4C₂ ⇒ 2 = -2C₂ + 4C₂

C₂ = 1 and C₁ = -1

The solution is

y = -x + x² ⇒ y(3) = -3 + 3²

⇒ y(3) = 6.

CONCEPT:

The relationship between speed ( u )  and density ( k ) is given by the

\({\bf{u}} = {{\bf{u}}_{\bf{f}}} - \left( {\frac{{{{\bf{v}}_{\bf{f}}}}}{{{{\bf{k}}_{\bf{j}}}}}} \right) \times {\bf{k}}\)

u - Mean speed at density k ( m/s )

k- Density of stream ( veh/km )

uf - Free mean speed

kj – Jam density

Relation between flow, speed, and density is given by,

q = k × u

q – Flow given in veh/hr

Space headway is defined as the distance between corresponding points of two successive vehicles at any given time. It can also be taken as the average space occupied by each vehicle.

\({\bf{k}} = \frac{{1000}}{{{\bf{Space}};{\bf{headway}}}}\)

At capacity or maximum flow,

\({{\rm{k}}_0} = \frac{{{{\rm{k}}_{\rm{j}}}}}{2}\)

\({u_0} = \frac{{{{\rm{u}}_{\rm{f}}}}}{2}\)

\({{\bf{q}}_{{\bf{max}}}} = {{\bf{k}}_0} \times {{\bf{u}}_0} = \frac{{{{\bf{k}}_{\bf{j}}} \times {{\bf{u}}_{\bf{f}}}}}{4}\)

qmax – Maximum flow or flow at capacity

k0, u0 – Density and speed at capacity

Calculation:

Given,

Free-flow speed (V_f) = 60 km/hr

Jam density (K_J) = 120 vehicle/km

For a linear speed-density relationship

\(Maximum;flow;\left( {{q_{max}}} \right) = \frac{{{V_f} \times {K_J}}}{4}\)

\(= \frac{{60 \times 120}}{4}\left( {km/hr \times veh/km} \right)\)

q_max = 1800 veh/hr

Explanation:

Given,

λ(rate) = 360 vehicle/hour = 0.1 vehicle/hour

∴ f(x) = e^-λtx [probability density function of x] → Poisson's distribution

⇒ f(x) = 0.1e^-0.1x

∴ Required probability

\(P\left[ {6 < x < 10} \right] =e^{-\lambda{t_6}} - e^{-\lambda{t_{10}}}\) = \(e^{-0.1\times{6}} - e^{-0.1{\times 10}}\)

⇒ P [6 < x < 10] = 0.18

Explanation:

Vertical distance between VPC and VPI is

\(y = \frac{L}{2}{n_1} = \frac{{400}}{2} \times \frac{5}{{100}} = 10;m\)

Length of tie bars

∴ Length of tie bars = 2L_d + joint gap

∴ Length of tie bars = 2L_d + 10

where \(;{L_d} = \frac{{\phi {\sigma _{st}}}}{{4{\tau _{bd}}}}\)

\(Length; of;tie;bar= \frac{{2\phi {\sigma _{st}}}}{{4{\tau _{bd}}}} + 10 = \frac{{2 \times 12 \times 230}}{{4 \times 2}} + 10\)

∴ Length of tie bars = 700 mm

Explanation:

P = ( 4+ 8 + 15 + 10 + 8 + 3 + 1 ) ×(1/3) 

 = 16.3 mm

\(F; = \left[ {\left( {;4 + ;8; + ;3; + ;1;} \right) \times ;\frac{1}{3}} \right] + \left[ {4 \times \frac{{60}}{{60}}} \right] + \left[ {\frac{1}{2} \times \frac{{60}}{{60}} \times 6} \right]\)

F = 5.33 + 4 + 3 = 12.33 mm

R = P – F = 16.33 – 12.33 = 4 mm

 

i) Take t_e = 140 minutes = 2.33 hours

 P_e = 1.33 + 2.67 + 5 + 3.33 + 2.67 + 1 + 0.33 = 16.33 mm

   \(\phi ;-;index; = \frac{{{P_e} - R}}{{{t_e}}} = \frac{{16.33 - 4}}{{2.33}} = 5.292;mm/hr\)

ii) Take t_e = 80 mins = 1.3 hours

    P_e­ = 2.67 + 5 + 3.33 + 2.67 = 13.67

    \(\phi ;-;index; = \frac{{{P_e} - R}}{{{t_e}}} = \frac{{13.67 - 4}}{{1.33}} = 7.271;mm/hr\)

Explanation:

Given:

Applying Bernoulli’s equation at A & B:

\(\frac{{{P_A}}}{{{\gamma _w}}} + \frac{{V_A^2}}{{2g}} + {Z_A} = \frac{{{P_B}}}{{{\gamma _w}}} + \frac{{V_B^2}}{{2g}} + {Z_B} + {h_f}\)

P_A = P_B = P_atm = 10.3 meters of water;

V_A = V_B = 0

\(\Rightarrow {Z_A} - {Z_B} = {h_f};\left( {head;loss;due;to;friction} \right) = \frac{{fl{V^2}}}{{2gd}}\)

Where,

f = friction factor for the pipe, L = Length of pipe, V = Velocity of flow through pipe, d = diameter of pipe, g = acceleration due to gravity

\(l = 5000;m,;d = 0.1,;f = 0.02;and;{h_f} = {Z_A} - {Z_B} = 5\)

\(\Rightarrow 5 = \frac{{0.02 \times 5000 \times {V^2}}}{{2 \times 9.81 \times 0.1}}\)

\(\Rightarrow V = 0.313;m/s\)

Applying Bernoulli’s equation A & S:

\(\frac{{{P_A}}}{{{\gamma _w}}} + \frac{{V_A^2}}{{2g}} + {Z_A} = \frac{{{P_S}}}{{{\gamma _w}}} + \frac{{V_S^2}}{{2g}} + {Z_S} + {h_f}\)

\(\frac{{{P_s}}}{{{\gamma _w}}} = 2.5;m\) (Given)

[V_s = V, P_A = P_atm = 10.3 meters of water, V_A = 0]

\(\frac{{{P_{Ar}}}}{{{\gamma _w}}} = \frac{{{P_{atm}}}}{{{\gamma _w}}} = 10.326;m\)

[P_atm is equivalent to 10.326 m head of water]

\(10.326 + {Z_A} = 2.5 + \frac{{{V^2}}}{{2g}} + {Z_s} + {h_f}\)

⇒ Z_S - Z_A = h;

h_f = Friction head loss in pipe along inlet leg \(= \frac{{fl{V^2}}}{{2gd}}\)

\(l = 2000;m;and;v = 0.313;m/s\)

\(\Rightarrow 10.326 - 2.5 = \frac{{{{\left( {0.313} \right)}^2}}}{{2 \times 9.81}} + h + \frac{{0.02 \times 2000 \times {{\left( {0.313} \right)}^2}}}{{2 \times 9.81 \times 0.1}}\)

⇒ h = 5.823 m ≈ 5.8 m

Explanation:

Given data:

Velocity of plate, V = 1 cm/sec

Distance between the late = 5 mm

Dynamic viscosity of fluid = 0.01 N-S/m²

Shear stress at lower plate = 0

Pressure gradient: 

\(\frac{{\partial p}}{{\partial x}} = ?\)

We Know that, in case of coquette flow, shear stress (t)is given by

\(\tau = \frac{{\mu V}}{h} + \left( { - \frac{{\partial P}}{{\partial x}}} \right)\left( {\frac{B}{2} - y} \right)\)

At lower plate, y = 0; τ = 0 [Given]

\(0 = \frac{{0.01 \times 0.01}}{{0.005}}\left( {\frac{{\partial P}}{{\partial x}}} \right)\left[ {\frac{{0.005}}{2} - 0} \right]\)

\(\frac{{\partial P}}{{\partial x}} = 8;N/{m^2}per;m\)

Explanation:

For Clays

Pile Load capacity in the stiff clay is given by

Q = Q_{end bearing} + Q_{skin drag}

For calculating Pull out load, Only skin drag force would resist the external pull and thus end bearing strength is to neglected.

Q_u = Q_{skin drag}

Q_u = A_s.f_s = (πdL) × (∝ C)

= π × 0.6 × 12 × (0.5 × 110) = 1244.07 kN

Explanation:

FOS = 1.5

If there is seepage parallel to the slope

\(FOS = \frac{{C' + \gamma 'z \times {{\cos }^2}\beta \times ;\tan \phi }}{{{\gamma _{sat}}.z; \times \cos \beta ; \times \sin \beta }}\)

\(FOS = \frac{{\gamma 'z \times {{\cos }^2}\beta \times \tan \phi }}{{{\gamma _{sat}} \times z \times \cos \beta \times ;\sin \beta }}\)

\(FOS = \frac{{\gamma '}}{{{\gamma _{sat}}}}\frac{{\tan \phi }}{{\tan \beta }}\)

\(1.5 = \frac{{\left( {20 - 9.81} \right)}}{{20}}\frac{{\tan 30}}{{\tan \beta }}\)

β = 11.095°

Explanation:

Weight of Precast PCC Block, W = Volume × γ_c

  = 37.5 × 7.5 × 24

= 6750 kN

\(FOS;against;shear;failure;\left( F \right) = \frac{{Strength}}{{Applied;load}} = \frac{{\left( {c \times A; + ;W \times \tan \phi } \right)}}{T}\)

\(\Rightarrow FOS = \frac{{\left( {17 \times 37.5; + 6750 \times tan\phi } \right);}}{{1875}}\)

FOS = 2.0187

Explanation:

Given that soil is sand ⇒ C = 0

Also given ultimate bearing capacity (q_u)= 600 kPa

As per Terzaghi’s theory,

For square footing

Ultimate baring capacity (q_u) = 1.3 CN_c + γD.N_q + 0.4 γ × B × N_γ

600 kPa = 0 + 15 × 1 × N_q + 0.4 × 15 × 4 × 18.75

\(600 = 15;{N_q} + 450 \Rightarrow {N_q} = \frac{{600 - 450}}{{15}} = \frac{{150}}{{15}} = 10\)

∴ N_q = 10

Now, the depth of footing (D_f) = 2 m

 Net Ulimate bearing capacity = Ultimate bearing capacity - γ × D_f 

 q_nu = 1.3 CNc + γD_f(Nq - 1) + 0.4 γ × B × Nγ

Net ultimate baring capacity (q_nu) = 1.3 × CN_c + γD (N_q - 1) + 0.4 γB × N_γ

= 0 + 15 × 2 × (10 - 1) + 0.4 × 15 × 4 × 18.75

= 270 + 450 = 720 kPa

 Q _allowable or Q _safe is given by

\({{\rm{q}}_{{\rm{allowable}}}} = {{\rm{q}}_{{\rm{net;safe}}}} + {\rm{\gamma D}} = \frac{{{{\rm{q}}_{{\rm{net}}}}}}{{{\rm{FOS}}}} + {\rm{\gamma D}}\)

\({\rm{Safe;bearing;capacity;}}\left( {{q_{safe}}} \right) + γ {D_f} = \frac{{720}}{3} + 15 × 2 = 240 + 30 \Rightarrow {q_{safe}} = 270;kPa\)