GATE CE 2022 Official Paper: Shift 2 (Previous Year Paper)

The correct answer is 'couldn’t help laughing'.

Key Points

• From the given statement, it can be inferred that since the movie was funny, the speaker could not resist laughing out loud.
• The phrase 'couldn’t help laughing' is used when someone is unable to prevent themselves from laughing.

The complete sentence will be: The movie was funny and I couldn’t help laughing.

• Hence, option 3 is the correct answer.

Additional Information

• 'Could' is used to indicate the possibility
• For eg.- This could be possible.

**Explanation-**​

​\(x:y:z = \frac{1}{2}:\frac{1}{3}:\frac{1}{4}\)​

Let \(x = \frac{k}{2},y = \frac{k}{3},z = \frac{k}{4}\)

\(\frac{{x + z - y}}{y} = \frac{{\frac{k}{2} + \frac{k}{4} - \frac{k}{3}}}{{\frac{k}{3}}}\) = 1.25

∴ (x + z - y) / y = 1.25

Explanation-

Both the numerator and the denominator of \(\frac{3}{4}\) are increased by a positive integer, x

So the new number will be, \(\frac{{3 + x}}{{4 + x}}\)

Both the numerator and the denominator of \(\frac{15}{17}\) are decreased by a positive integer, x

So the new number will be, \(\frac{{15 - x}}{{17 - x}}\)

This operation results in the same value for both the fractions.

\(\frac{{3 + x}}{{4 + x}} = \frac{{15 - x}}{{17 - x}}\)

⇒ (3 + x)(17 -x) = (4 + x)(15 - x)

⇒ 14 x + 51 = 11x + 60

⇒ 3x = 9

⇒  x = 3

Explanation-

A survey of 450 students about their subjects of interest resulted in the following outcome.

• 150 students are interested in Mathematics. ⇒ n (M) = 150
• 200 students are interested in Physics. ⇒ n (P) = 200
• 175 students are interested in Chemistry. ⇒ n(C) = 175
• 50 students are interested in Mathematics and Physics. ⇒ n(M U P) = 50
• 60 students are interested in Physics and Chemistry. ⇒ n(P U C) = 60
• 40 students are interested in Mathematics and Chemistry. ⇒ n(M U C) = 40
• 30 students are interested in Mathematics, Physics, and Chemistry. ⇒ n(M U P U C) = 30

The remaining students are interested in Humanities is :

n(M U P U C) = n(M) + n(P) + n(C) - n(M U P) - n(P U C) - n(M U C) + n(M U P U C)

n(M U P U C) = 150 + 200 + 175 - 50 - 60 - 40 + 30 = 405

Total number of students interested in physics, math, and chemistry = 405

The number of students interested in Humanities = 450 - 405 = 45

The correct answer is 'More people are visiting the malls but not spending enough'.

Key Points

• Let's refer to the passage:
• 'The total number of visitors to the malls is impressive.'
• 'However, the total revenue generated through sales in the shops in these malls is generally low.'
• From the above-mentioned statements, it is evident that the correct logical inference based on the information in the above passage is that 'more people are visiting the malls but not spending enough'.
• Hence, option 2 is the correct answer.

Additional Information

• Revenue is income, especially when of an organization and of a substantial nature.
• For eg.- Traders have lost £10,000 in revenue since the traffic scheme was implemented.

Explanation:

In a partnership business,  the monthly investment by three friends for the first six months is in the ratio of 3: 4: 5.

Let capitals invested by three friends for six months are C₁ = 3x, C₂ = 4x, C₃ = 5x 

The new investment ratio was kept constant for the next six months.

Profit ratio for first six months = (3x × 6) : (4x × 6) : (5x × 6)

After six months, they had to increase their monthly investments by 10%, 15%, and 20%, respectively, of their initial monthly investment.

Profit ratio for next six months = (1.1 × 3x × 6) : (1.15 ×  4x × 6) : (1.20 × 5x × 6) 

The total ratio of profit = (3x × 6 + (1.1 × 3x × 6) : (4x × 6 + 1.15 ×  4x × 6) : (5x × 6 + 1.20 × 5x × 6) 

P1 ; P2 ; P3 = 6.3 : 8.6 : 11 = 63 : 86 : 110

Explanation-

When the multiplication of slopes of two lines is equal to -1, then the lines are said to be perpendicular.

 When the slope of two lines is equal, they are said to be parallel lines.

Given data and Calculation-

Line L1 : 2𝑥 − 3𝑦 = 5 , y = \(\frac{{2x - 5}}{3}\), slope = 2/3

Line L2 : 3𝑥 + 2𝑦 = 8 , y = \(\frac{{8 - 3x}}{2}\), slope = -3/2

Line L3 : 4𝑥 − 6𝑦 = 5 , y = \(\frac{{4x - 5}}{6}\), Slope = 4/6

Line L4 : 6𝑥 − 9𝑦 = 6 , y = \(\frac{{6x - 6}}{9}\), slope = 6/9

Multiplication of Slope L2 and L4 = \(\frac{{ - 3 \times 6}}{{2 \times 9}} = - 1\)

Slope of L1,L3 and L4 are equal.

So lines L4 and L3 are parallel.

So L4 is perpendicular to L2 and L4 is parallel to L3

Explanation-

Statement 1: Some soaps are clean.

Statement 2: All clean objects are wet.

From the above two statements, the Venn diagram can be drawn as-

Conclusion I: Some clean objects are soaps. - It is satisfied in both the probability diagram. So it can be logically inferred.

Conclusion II: No clean object is a soap.- It is not followed.

Conclusion III: Some wet objects are soaps.-It is satisfied in both the probability diagram. So it can be logically inferred.

Conclusion IV: All wet objects are soaps.-It is not followed.

So only conclusions I and III can be logically inferred.

Explanation-

The initial position of the ant is at point P facing east.

The ant first turns 72° anticlockwise at P

moves forward for 10 cm.

 

turns 144^o clockwise.

 

By repeating the above steps for 5 times, the final path traced by the ant will be

Explanation-

Given that,

The function f(x, y) satisfies the Laplace equation \(\rm \nabla ^2 f(x, y) = 0\)

on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. 

The value of this function on the circular boundary of this domain is equal to 3.

Here it is given that the value of the function is 3 for its domain, which signifies that it is a constant function whose value is 3.

So the value of the function at (0, 0) is 3.

The Given Question is Wrong, Marks are allotted to all.

Explanation-

Given that, \(I = \int {(x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + .....} \)

\(I = {\frac{x^2}{2}} - \frac{{{x^3}}}{6} + \frac{{{x^4}}}{{12}} - \frac{{{x^5}}}{{20}} + \)......

Expanding the options given,

option 1-

\(\frac{1}{{1 + x}} = {(1 + x)^{ - 1}} = 1 - x + {x^2} - {x^3}....... \)

option 2-

\(\frac{1}{{1 + {x^2}}} = {(1 + {x^2})^{ - 1}} = 1 - {x^2} + {x^4} - {x^6}....... \)

option 3-

\(\frac{{ - 1}}{{1 - x}} = - {(1 - x)^{ - 1}} = - (1 + x + {x^2} + {x^3}....... \)

option 4-

\(\frac{{ - 1}}{{1 - {x^2}}} = - {(1 - {x^2})^{ - 1}} = - (1 + {x^2} + {x^4} + {x^6}....... \)

So none of the options are matching with the correct answer.

Concept:

Elastic Modulus (E)

When the body is loaded within its elastic limit, the ratio of stress and strain is constant. This constant is known as Elastic modulus or Young's Modulus.

\({\rm{E}} = \frac{{{\rm{Stress}}}}{{{\rm{Strain}}}} = \frac{{\rm{\sigma }}}{\epsilon}\)

Rigidity modulus (G)

When a body is loaded within its elastic limit, the ratio of shear stress and shear strain is constant, this constant is known as the shear modulus.

\({\rm{G}} = \frac{{{\rm{Shear;stress;}}}}{{{\rm{Shear;strain}}}} = \frac{{\rm{\tau }}}{\phi }\)

Bulk modulus (K)

When a body is subjected to three mutually perpendicular like stresses of same intensity then the ratio of direct stress and the volumetric strain of the body is known as bulk modulus

\({\rm{K}} = \frac{{{\rm{Direct;stress}}}}{{{\rm{Volumetric;strain}}}} = \frac{{\rm{\sigma }}}{{\frac{{{\rm{\delta V}}}}{{\rm{V}}}}}\)

The relationship between E, K, G, and μ is:

\(E = 2G(1 + μ) \)

\(G ={ E\over 2(1 + μ) }\)

\(E = 3K (1 – 2μ) \)

\({\bf{E}} = \frac{{9{\bf{KG}}}}{{3{\bf{K}}\ + ;{\bf{G}}}} \)

\({\bf{\mu }} = \frac{{3{\bf{K}}\ -\ 2{\bf{G}}}}{{2{\bf{G}}\ +\ 6{\bf{K}}}}\)

Explanation:

Over reinforced section:

• An over reinforced section is a type of section in which we use the concrete for its ultimate compressive strength strength.
• The failure strain of concrete is reached before the yield strain of steel hence reinforced beam section undergoes a compressive failure.
• The percentage of tensile reinforcement is more than the amount of reinforcement provided for a balanced section. As in this type of section concrete fails first i.e it undergoes a brittle failure
• The main disadvantage of this section is it undergoes a sudden failure without warning.
• We will not advice to select this type of section for the structure as it adds an extra cost, by increasing the percentage of reinforcement.

Under reinforced beam:

• In these type of sections the area of reinforcement is less than limiting area of reinforcement.
• The failure is due to yielding of steel and hence called ductile failure.
• Gives sufficient warning before failure and hence preferred for general use.

Balanced section:

• The area of reinforcement is equal to limiting area of reinforcement.
• In this type of section the ultimate strength of concrete and steel are reached simultaneously.
• At the same point of time concrete and steel fails and ultimately structure fails by crushing of concrete.

Concept:

The following table shows different Bogue’s compounds and their properties.

Explanation-

Static indeterminacy-

• It is defined as the difference between a total number of unknowns (Total member forces+ reactions) and the total number of available equations from the conditions of equilibrium.
• Degree of static indeterminacy = Total number of unknown forces - Number of equilibrium equations available
• If the degree of static indeterminacy = 0, it is known as a statically determinate structure.
• If the degree of static indeterminacy > 0, it is known as statically indeterminate structure.

Kinematic indeterminacy-

• The degree of kinematic indeterminacy is the minimum number of movements (degrees of freedom, DOF) with which the kinematic configuration of the overall structure can be defined, that is, the number of unknown independent movements of the structure.

 

Given data and Calculation-

Total number of unknown forces, = (H_p, V_p, M_p , H_Q, V_Q) = 5

Number of equilibrium equations available = 3

Degree of static indeterminacy = 5 - 3 = 2

As the joint P is fixed, no movement is allowed at that joint, joint Q is hinged so only rotation is allowed at that joint.

So the degree of kinematic indeterminacy = 1 ( Rotation at joint Q)

Explanation:

Effect of water table on BC of soil:

• As the water table rises the moisture content of soil increases, and its BC decreases.
• This is due to the fact that when water gets into the pores of the soil, its load-carrying capacity decreases as water is nearly incompressible.
• Monsoon increases the water content of soil and hence is considered critical for designing shallow foundations on both sandy and clayey soil.

∴ Statement 1 is true

Conditions to be checked for the stability of slope of earthen dam:

• Stability of downstream slope during steady seepage.
• Stability of upstream slope during sudden drawdown.
• Stability of upstream and downstream slopes during and immediately after construction.

∴ Statement 2 is false

Quicksand condition:

• When upward seepage occurs it applies some seepage pressure on the soil and when this seepage pressure is equal to effective stress of soil, the quicksand condition occurs.
• In this soil loss, all its shear strength and the hydraulic gradient are equal to 1 and such hydraulic gradient is called critical hydraulic gradient.

∴ Statement 3 is false

Effect of water table on active thrust:

• When we decrease the water table, the BC of soil increases and it increases the effective stress of soil.
• With the increase, effective stress, active pressure, and hence active thrust also increase.

∴ Statement 4 is false

Explanation:

Taking the submission of horizontal forces as 0 ie Σ H = 0

σx × BC - τz × AB + σ1 × Sin θ = 0

\(σ_x × ({AC× Sin\ θ \over Cos\ θ})+ τ_{zx } × ({AC× Cos\ \ \theta \over Cos\ θ})={σ_1× AC × Sin\ \ θ \over Cos\ θ}\)

σx × Tan θ + τzx = σ1 × Tan θ 

Tan θ × (σ1 - σx) = τxy

 \(Tan\ \theta = {\tau_{xy} \over \sigma _1 -\sigma _x}\)

Explanation:

Horton equation:  It gives the infiltration of water as is given by

\(f_{b}=f_{c}+\left ( f_{o}-f_{c} \right )e^{-kt}\)

Where, f_b = Value of infiltration, f_c = equilibrium infiltration and f_o= Initial infiltration

Penman equation:  It gives the formula for evapotranspiration which is the sum of evaporation and transpiration

\(PET = \frac{AH_{n}+E_{a}\gamma }{A+\gamma }\)

Chezy formula:  It relates velocity through an open channel to its slope and hydraulic radius by the following formula

\(V = C\sqrt{RS}\)

Where V = Velocity, C = Chezy constant, R = Hydraulic radius and S = Slope

Dicken formula:  It gave a formula for discharge through the flood plain and is usually applicable for north and central India.

Q = C × A^3/4

Where,Q = Flood discharge, C = Constant, A = Area of catchment

Lacey theory: It is used to design of alluvial canal and is based on the following assumptions:

• The channel is flowing uniformly in the unlimited incoherent alluvial soil of the same character.
• The silt grade and silt charge are uniforms.
• The discharge remains constant

Concept:

Evapotranspiration:

• The sum of transpiration and evaporation is known as evapotranspiration.
• Which depends on various factors like type of soil and crop that is being grown on the soil.
• So, to counter these factors some constants like soil coefficient and crop coefficient are used and we modify the value of evapotranspiration we got from various formula like penmann's equation and the modified evapotranspiration is given by

E_o = E₁ × C₁ × C₂

Where, E₀ = Modified evapotranspiration, E₁ = Original evapotranspiration, C₁ = Crop coefficient, C₂ = Soil coeffiicient

Given:

E₁ = 6 mm/day, C₁ = 1.2, C₂ = 0.8

Calculation:

E_o = E₁ × C₁ × C₂

      = 6 × 1.2 × 0.8

E₀ = 5.76 mm/day

∴ The evapotranspiration is 5.76 mm/day

Explanation:

From Newton’s law of viscosity

\(\tau = \mu \cdot \frac{{du}}{{dy}}\)

μ = Proportionality constant/ coefficient of viscosity or viscosity

μ = dynamic viscosity (since it involves force)

\(\therefore \mu = \frac{\tau }{{du/dy}}\)

Dimensional formula for μ

∴ μ = M.L-1.T-1

Units:

Dynamic Viscosity (μ):

SI system:

μ = N.s/m2 or Pa.s = kg/m.s

CGS system:

μ = Dyne.sec/cm2 = 1 poise

1 N = 105 dynes

1 N.s/m2 = 105 dynes.sec/104 cm2

1 N.s/m2 = 10 dyne.sec/cm2

1 N.s/m2 = 10 poise = 1 Pa.s

Additional Information

Kinematic Viscosity (v): It is the ratio of dynamic viscosity to mass density.

\(v = \frac{{Dynamic;viscosity}}{{Mass;density}} = \frac{\mu }{\rho }\)

Dimensional formula for Kinematic viscosity (ν)

∴ ν = L2T-1

The dimensions of the kinematic viscosity show that they involves the magnitudes of length and time only.

The name kinematic viscosity has been given to the ratio (μ/ρ) because its unit (m2/s) is similar to the unit of kinematic quantities like velocity (m/s) and acceleration (m/s2).

Units

SI system:

m2/sec

CGS system:

cm2/sec or stoke

1 stoke = 1 cm2/sec = 10-4 m2/sec

Explanation:

Fugitive emission:

• Fugitive emissions are accidental emissions of vapours or gases from pressurised apparatus, either due to faulty equipment, leakage or other unforeseen mishaps.
• It can also occur through evaporation, in such sources as storage tanks or wastewater treatment facilities.

Total emission = Fugitive emission + Captured emission

Given is that Captured emission is 95 %. So, by using above formula, fugitive(escaped) emission is 100 - 95 = 5 %

And, Fugitive emission = percent of fugitive emission × Total emission

                                        = 0.05 × 5

        Fugitive emission = 0.25 kg/h

Important Points

• The best way to reduce fugitive emissions is through regular testing and maintenance.
• Leak detection should be scheduled at regular intervals using gas detection devices, which can measure the amount of vapours escaping, determine what they are and decide on an appropriate course of action.

Concept:

The nominal scale of measurement:

The nominal scale, sometimes called the qualitative type, places non-numerical data into categories or classifications.

For example:

• Placing cats into breed type. For example, a Persian is a breed of cat.
• Putting cities into states(Name of a city). Example: Jacksonville is a city in Florida.

The ordinal scale of measurement:

Ordinal scales are made up of ordinal data. Some examples of ordinal scales:

For example:

• High school class rankings: 1^st, 2^nd, 3^rd, etc.
• Social-economic class: working, middle, upper.
• Ranking of a city based on ease of business

The interval scale of measurement:

It has values of equal intervals that mean something.

For example:

• A thermometer might have intervals of ten degrees.

The ratio scale of measurement:

It is exactly the same as the interval scale except that the zero on the scale means: does not exist.

For example: 

• A weight of zero doesn’t exist;
• An age of zero doesn’t exist.
• Population density of a city

Hence, Match the following attributes of a city with the appropriate scale of measurements.

Explanation:

At Noon, the sun is exactly over the time median of the place. Thus true bearing of the line joining the sun and the place is either 0° and 360° if it is to the north.

and 180° if it is to the south of the place.

Magnetic bearing of sun at noon = 170°

True bearing = 180° 

True bearing = M.B + declination

180° - 178° = declination

So,

declination = 2° 

Declination is coming out as positive, so it will be in the eastward direction.

Explanation-

If A is a non-singular square matrix, there is an existence of n x n matrix A^-1, which is called the inverse of a matrix A such that it satisfies the property:

AA^-1 = A^-1A = I, where I is  the Identity matrix.

If A and B are the non-singular matrices, then the inverse matrix should have the following properties

• (A^-1)^-1 =A
• (AB)^-1 =A^-1B^-1
• (ABC)^-1 =C^-1B^-1A^-1
• (A1 A2….An)^-1 =A^n-1An-1^-1……A2^-1A1^-1
• (A^T)^-1 =(A^-1)^T

Given data and Analysis-

P and Q are two square matrices of the same order,

If P and Q are invertible, then [PQ]-1 = Q-1P-1

If P and Q are invertible, then [QP]-1 = P-1Q-1

Concept:

Plastic waste:

• Plastic is the general common term for a wide range of synthetic or semi-synthetic organic amorphous solid materials derived from oil and natural gas.
• Plastic waste, or plastic pollution, is ‘the accumulation of plastic objects (e.g.: plastic bottles and much more) in the Earth’s environment that adversely affects wildlife, wildlife habitat, and humans.’
• It also refers to the significant amount of plastic that isn’t recycled and ends up in landfills or, in the developing world, thrown into unregulated dumpsites.

Paper waste:

• It means newspapers, magazines, cardboard, and any other form of paper which is free of contaminated matter and is capable of being recycled.

Glass waste:

• It is discarded material from the glass manufacturing process or from used consumer products made of glass.

Food waste:

• Food waste refers to food such as plate waste (i.e., food that has been served but not eaten), spoiled food, or peels and rinds considered inedible that is sent to feed animals, to be composted or anaerobically digested, or to be landfilled or combusted with energy recovery.

Typical data of moisture content of municipal solid waste components are as follows:

Typical heat values of municipal solid waste components are as follows:

Hence, the decreasing order of moisture content in these wastes is MCf > MCp > MCg, where the moisture contents (MC) of food waste, paper waste, and glass waste were found to be MCf, MCp, and MCg, respectively

Concept:

Municipal solid waste collection route:

• Efficient routing of collection vehicles helps decrease costs by reducing the labor expended for collection.
• Proper planning of collection routes also helps conserve energy and minimize working hours and vehicle fuel consumption.
• It is necessary therefore to develop detailed route configurations and collection schedules for the selected collection system.

Design an optimum municipal solid waste collection route is depends on various rules are as follows:

• Routes should not be fragmented or overlapping.
• Each route should be compact, consisting of street segments clustered in the same geographical area.
• Total collection plus hauling time should be reasonably constant for each route in the community.
• The collection route should be started as close to the garage or motor pool as possible, taking into account heavily traveled and one-way streets.
• Heavily traveled streets should not be visited during rush hours.
• In the case of one-way streets, it is best to start the route near the upper end of the street, working down it through the looping process.
• Waste on a steep hill should be collected, when practical, on both sides of the street while the vehicle is moving downhill.
• Higher elevations should be at the start of the route.
• For collection from one side of the street at a time, it is generally best to route with many anti-clockwise turns around blocks.
• For collection from both sides of the street at the same time, it is generally best to route with long, straight paths across the grid before looping anticlockwise.
• For certain block configurations within the route, specific routing patterns should be applied.
• The last collection point on a route should be as close as possible to the waste disposal facility.

Explanation-

Macroscopic stream models represent how the behavior of one parameter of traffic flow changes with respect to another. Most important among them is the relation between speed and density.

The first and most simple relation between them is proposed by Greenshield. Greenshield assumed a linear speed-density relationship as illustrated in the figure to derive the model. The equation for this relationship is shown below.

• \(v = {v_f}\left( {1 - \frac{k}{{{k_j}}}} \right)\)
• where v is the mean speed at density k, v_f is the free speed and k_j is the jam density.

Traffic flow (q) = flow density (k) × Velocity(V)

Given data and Analysis -

\(v = {v_f}\left( {1 - \frac{k}{{{k_j}}}} \right)\)

Also it can be written as , \(k = {k_j}\left( {1 - \frac{v}{{{v_f}}}} \right)\)

q = k × v

\(q = k \times {v_f}\left( {1 - \frac{k}{{{k_j}}}} \right) \)

\(\rm q = \nu_f k - \left( \frac{\nu_f}{k_j} \right)k^2\)

\(q = v \times {k_j}\left( {1 - \frac{v}{{{v_f}}}} \right)\)

\(\rm q = k_j \nu - \left( \frac{k_j}{\nu_f} \right)\nu^2\)

So option 2 and 4 are correct.

Concept:

The absolute percentage error:

• It is a measure of how accurate a forecast system is.
• It measures this accuracy as a percentage.

The absolute percentage error in the computed area:

=\({e_A\over A} \times 100\) = \(2 \times {({e_r\over r}\times 100)}\)

 A = \(\pi r^2\) and e_A = \(2\pi re_r\)

Where A =  Area of cross-section, r = radius, e_r = error in radius

Calculation:

Given data

r = 5 cm

error in measuring the radius = 0.2%

Now

er = (0.2/100)5 = 0.01 cm

The absolute percentage error in the computed area:

=\({e_A\over A} \times 100\) = \(2 \times {({e_r\over r}\times 100)} = 2 \times {({0.01\over 5}\times 100)} \) = 0.4%

Explanation-

• The eigenvector of a matrix A is a vector represented by a matrix X such that when X is multiplied with matrix A, then the direction of the resultant matrix remains the same as vector X.
• AX = λX
• where A is any arbitrary matrix, λ are eigen values and X is an eigen vector corresponding to each eigen value.​Here, we can see that AX is parallel to X.So X is an eigen vector.

Some important properties of eigenvalues are-​

• If λ₁, λ₂…….λ_n are the eigenvalues of A, then \(\lambda _1^k,\lambda _2^k,......,\lambda _n^k\) are the eigenvalues of A^k
• Sum of Eigen Values = Trace of A (Sum of diagonal elements of A)

Given data and Calculation-

Given matrix = \(ε = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \)

\(A - \lambda I = \left( {\begin{array}{*{20}{c}} {1 - \lambda }&1\ 1&{ - 1 - \lambda } \end{array}} \right)\) = 0

\(\Rightarrow (1 - \lambda )( - 1 - \lambda ) - 1\) = 0

\( \Rightarrow {\lambda ^2} - 2 = 0\)

\( \Rightarrow {\lambda ^2} = 2\)

\( \Rightarrow \lambda = \pm \sqrt 2 \)

Eigenvalues of ε8 are \({\left( {\sqrt 2 } \right)^8} \) and \({\left( { - \sqrt 2 } \right)^8}\).

 Eigenvalues of ε¹¹ are \({\left( {\sqrt 2 } \right)^{11}}\) and \({\left( { - \sqrt 2 } \right)^{11}}\).

P = Trace (ε8) = 16 + 16 = 32

Q = Trace (ε11).= 0

P + Q = 32

Additional Information 

• Eigenvalues of real symmetric matrices are real.
• Eigenvalues of real skew-symmetric matrices are either pure imaginary or zero.
• Eigenvalues of unitary and orthogonal matrices are of unit modulus |λ| = 1
• If λ1, λ2…….λn are the eigenvalues of A, then kλ1, kλ2…….kλn are eigenvalues of kA.
• If λ1, λ2…….λn are the eigenvalues of A, then 1/λ1, 1/λ2…….1/λn are eigenvalues of A-1
• Eigenvalues of A = Eigen Values of AT (Transpose)
• Product of Eigen Values = |A|
• Maximum number of distinct eigenvalues of A = Size of A
• If A and B are two matrices of the same order then, Eigenvalues of AB = Eigenvalues of BA

Concept:

Soil Sampling:

The fundamental requirement of the sampling tool is that on being forced into the ground it should cause as little displacement, remolding, and disturbance as possible.

The degree of disturbance is controlled by three features of its design as follows:

• Cutting Edge
• Inside wall friction
• Non-return Valve

​

Cutting Edge:

A cutting edge embodies the following features:

a) Inside Clearance(Ci)

It is given by the following formula

\({C_i} = \frac{{{D_s} - {D_c}}}{{{D_c}}}\times 100\)

where Ds = inside diameter of sampling tube, Dc = inside diameter of cutting edge

For undisturbed sampling, it should lie between 1% and 3%.

b) Outside Clearance(Co)

It is given by the following formula

\({C_o} = \frac{{{D_w} - {D_T}}}{{{D_T}}}\times 100\)

where Dw = outside diameter of cutting edge, DT = outside diameter of the sampling tube.

The Outside clearance should not be greater than the inside clearance.

c) Area Ratio(Ar)

It is calculated by the following formula

\({A_r} = \frac{{D_w^2 - D_c^2}}{{D_c^2}} \times 100\)

where Dw = outside diameter of cutting edge, Dc = inside diameter of cutting edge

It should be kept as low as possible consistent with the strength requirements of the sample tube. Its value should not be greater than about 20% for stiff formations while for soft sensitive clays should not be greater than 10%.

Calculation:

Given data

Ds = inside diameter of sampling tube = 50 mm

Dc = inside diameter of cutting edge

Inside Clearance ( Ci) = 1%

So, Inside Clearance(Ci)

\({C_i} = \frac{{{D_s} - {D_c}}}{{{D_c}}} \times 100\)

\({1} = \frac{{{50} - {D_c}}}{{{D_c}}}\times 100\)

Dc = 49.5049 mm

Hence, The inside diameter (in mm) of the cutting edge is 49.5049 mm.

Explanation-

Stress due to a point load-

The stress due to a point load at any point at a depth Z m is given by,

\({σ _z} = {K_B}\frac{Q}{{{Z^2}}}\)

\({K_B} = \frac{3}{{2π }}{\left( {\frac{1}{{1 + \frac{{{r^2}}}{{{Z^2}}}}}} \right)^{\frac{5}{2}}} \)

• Where Q = Intensity of point load, Z vertical depth from the point load
• r - radial distance from the point load as shown in figure

Given data and Calculation-

The intensity of load(Q) = 1000 kN

Size of footing = 2 m × 2 m

Area of footing (A) = 4 m²

The stress due to a point load at any point at a depth Z m is given by,

\({σ _z} = {K_B}\frac{Q}{{{Z^2}}}\)

\({K_B} = \frac{3}{{2π }}{\left( {\frac{1}{{1 + \frac{{{r^2}}}{{{Z^2}}}}}} \right)^{\frac{5}{2}}} \)

r/z = 0 So K_B = 3\2π = 0.477

The stress due to a point load

 \({σ _z} = {0.477}\frac{Q}{{{Z^2}}}\)   ..........(i)

According to the given data in question

An isobar for which σ_z = 0.1 q = \(0.1 Q\over A\) = \(0.1 Q\over 4\)

σz  = \(0.1 Q\over 4\)  .........(ii)

Now equating equation (i) & (ii)

\({0.1 Q\over4} = {0.477}\frac{Q}{{{Z^2}}}\)

Z = 4.368 m

Explanation-

For the triaxial test, the equation will be \({σ _1} = {σ _3}{\tan ^2}(45 + \frac{ϕ }{2}) + 2c\tan (45 + \frac{ϕ }{2})\)

Where σ 1 = cell pressure + deviatoric stress

σ 3  = cell pressure

C = Cohesion of soil, ϕ = angle of internal friction 

Given data and Calculation-

cell pressure = σ ₃ = 100 kPa

deviatoric stress = 150 kPa

\({σ _1}\)= cell pressure + deviatoric stress = 100 + 150 = 250 kPa

\(250 = 100{\tan ^2}(45 + \frac{0}{2}) + 2c\tan (45 + \frac{0}{2})\)

\(C = \frac{{250 - 100}}{2} = 75\) kPa

Explanation-

Return period-

• This is the time period on average after which peak flood discharge is likely to be equal or exceeded. This is also called a recurrence interval.

Reliability and risk-

• It is the probability that the particular value of peak flood discharge is never equal or exceeded in the design life of the structure.
• Risk = 1-Reliability
• Risk is the probability of occurring of flood at least once in the design period.

Probability of occurrence-

P = \(\frac{1}{T}\) where T = Time period

q = 1 - p 

It follows the binomial probability distribution.

The probability of flood exceeding the design value r times in n years of the design period is given as,

P(r,n) = \({}^n{C_r}{p^r}{q^{n - r}}\)

Given data and Calculation-

The return period of flood = T

T = n, and n → ∞

Risk \(= 1 - {q^n} = 1 - {(1 - P)^n}\)

Risk = \(1 - {(1 - \frac{1}{T})^n}\)

For T = n, and n → ∞

Risk = \( = 1 - \frac{1}{e} = 0.632 = 63.2% \)

Explanation:

Basic runway length is based on some assumptions:-

a) Derived at Mean Sea level

b) Derived for a standard temperature of 15°C

c) The gradient must be zero.

The actual runway need not follow all these guidelines and hence we must apply some corrections.

The sequential correction is applied in the basic runway length.

The first correction is altitude correction, the second correction is temperature correction and the last correction is gradient correction.

Altitude Correction:

Basic runway length is increased by 7% for every 300 m rise above mean sea level.

This is done due to the decrease in drag force on the aircraft owing to the decrease in density of air as the altitude increases.

Given:

Height above MSL = 300 m, and Basic Runway length = 1500 m

After applying elevation correction,

\({\rm{Correction}} = 1500 \times \frac{7}{{100}} \times \frac{{300}}{{300}} = 105{\rm{;m}}\)

Corrected Runway Length = 1500 + 105 = 1605 m

Important Points

Temperature correction and gradient corrections are applied on the subsequently corrected runway length.

Explanation:

Given Data and Analysis-

Given polynomial is f(x) = x3 - 6x2 + 11x - 6. where 1 ≤ x ≤ 3.

Statement 1-The given polynomial is zero at the boundary points x = 1 and x = 3.

At x = 1, f(x) = 13 - 6 + 11 - 6=0

At x = 3, f(x) = 33 - 54 + 33 - 6=0

So statement 1 is correct.

f'(x) = 3x² - 12x + 11

f''(x) = 6x - 12

at x = 1, f''(x) = -6 < 0

for maximum or minimum, f'(x) = 0

⇒ 3x² - 12x + 11 = 0

⇒ x = 2.577 or 1.422

Both values of x exist within domain.

at x = 2.577, f''(x) ≥ 0, so f(x) will be minimum.

x = 1.492, f''(x) ≤ 0, so f(x) will be maximum.

At x = 1.422 the function will have a maximum value and at x = 2.577 it will have a minimum value.

So statements 2 and 4 are correct.

Statement 3 is wrong as for 1 ≤ x ≤ 2 , f''(x)  is not greater than zero.

Concept

Natural frequency (W_n) of any undamped single degree of freedom system is given by

W_n = √(k/m) 

W_n in rad/sec

k = Stiffness of system in N/m

m = mass of system in kg

Time period (T) = 2π / W_n

Given data and Calculation-

Given that, Initial mass (m₀) = m and stiffness (k₀) = k

Natural frequency = ω rad/s 

Natural time period = T s

Final mass m₁ = m/2 and Stiffness k₁ = 2k

Then, Wn1 = √(k₁/m₁) = √(4k/m) = 2 √(k/m) = 2 W_n  

T1 = 2π/W_n1 = 2π/(2 W_n) = T/2

Concept:

Shape factor SF =\(\frac{Z_p}{Z_e}\)

Z_p  is plastic section modulus,

Zp = \(\frac{A}{2}\) \(({\bar y_1} + {\bar y_2})\) 

Z_e = \(\frac{{{I_{NA}}}}{y}\)

\({\bar y_1}\) and \({\bar y_2}\) are centroids (from PNA) of the area above

PNA & below PNA respectively

INA = Moment of Inertia about ENA.

(2) For a square section

\({\bar y_1} = {\bar y_2} = \frac{a}{{3\sqrt 2 }}\) and

y = \(\frac{a}{{\sqrt 2 }}\)

I_NA = \(\frac{a^4}{{ 12 }}\)

calculation : given a = 100 mm,

Z_p \( = \frac{{100 × 100}}{2}\left( {\frac{{100}}{{2\sqrt 2 }} + \frac{{100}}{{3\sqrt 2 }}} \right)\) = 235.7 × 10³ mm³

Z_e = \(\frac{{a'1}}{{12 × \frac{a}{{\sqrt 2 }}}} = \frac{{{a^3}}}{{6\sqrt 2 }} = \frac{{{{(100)}^3}}}{{6\sqrt 2 }} = 117.85 × {10^3}m{m^3}\)

Shape factor Sf = \(\frac{Z_p}{Z_e}=\frac{235.7×10^3}{11785× 10^3}=2\)

M_P =Z_P f_y = (235.7 × 10³) mm³ × 250 N/mm²

= 58925 × 10³ N.mm

= 58.93140.m (Ans)

Concept:

1. Loss of prestress in any wire = m fcs where m is modular ratio & fcs is stress in concrete at the level of steel/wire.

2. The three wires are tensioned one-by-one. so, when wire 2 is tensioned mere is a loss of stress in wire (1) & also when wire (3) is tensioned, there is a loss of stress in wires (1) & (5). There is no loss of stress in the wire(3).

   

P_1, P₂, P₃ are forces in wires 1,2,3 respectively.

Loss in wire (1) due to P₂ and P₃ or (P₂+P₃)

Loss in wire (2) is due to P₃ only.

Loss in wire (3) is zero.

3) \({P_1} = {P_2} = {P_3} = P = \left( {\frac{{200 × 1500}}{{1000}}} \right)kN = 300,kN\) (given)

4. (P = Area of pre-stress × pre-stress)

Calculation

 

 

Total stress, \({f_{cs}}=\frac{P}{A} + {\frac{{Pe}}{I}^2}\)

1. Loss in wire (1) is due to P₂+P₃ = 2P; substitute 'P' in above with 'gp'

\({f_{c{s_1}}} = \frac{{2P}}{A} + \frac{{2P{e^2}}}{I}\)

\(= \left( {\frac{{2 × 300 × 100}}{{450 × 450}};} \right); + \frac{{2 × 300 × {{\left( {100} \right)}^2}}}{{\frac{{450 × {{\left( {450} \right)}^3}}}{{12}}}} - 3.96 + 1.76\)

= 4.72 N/mm²

Loss \({L_1} = m{f_{c{s_1}}} = 6 \times 4.72 = 28.32,N/m{m^2}\)

\({L_2} = m{f_{c{s_2}}} = 6 \times 2.36 = 14.16,N/m{m^2}\)

\({L_3} = m{f_{cs}}_{_3} = 6 \times 0 = 0\)

Ag Loss = \( = \frac{{28.32 + 14.16 + 0}}{3} = 14.16,N/m{m^2}\)

Loss in wire (2) is due to \(P3 = P = 300,kN.\)

\(fc{s_2}, = ,\frac{{300 \times 1000}}{{450 \times 450}}, + ,\frac{{300 \times 1000 \times {{\left( {100} \right)}^2}}}{{\frac{{450 \times {{\left( {450} \right)}^3}}}{{12}}}}\)

Explanation:

The relationship b/w volumetric strain and axial strain in case of loose sand and dense sand is given as:

From above, it can be said that volumetric strain decreases on increasing axial strain.

The relationship b/w pore water pressure and axial strain is given as:

From above, it can be said put initially for the same value of strain pore water pressure decreases thereafter increasing.

3. For reberated clay's the bearing capacity factor are given as:

N_c = 5.7

N_q = 1

N_γ = 0

This is because angle of internal friction be ϕ = 0

4. Skempton's pore pressure parameter 'B' is the ratio of change in pore pressure due to change in cell pressure

B = 0 for dry soils

B = 1 for saturated soils

Concept:

• The pressure of water in the pores of the soil is called pore water pressure (u).
• Under hydrostatic conditions, no water flow takes place, and the pore pressure at a given point is given by u =γWh ,where h = depth below water table or overlying water surface.
• At any point in a soil mass, the effective stress (represented by σ ') is related to total stress (σ) and pore water pressure (u) as σ ' = σ - u
• Both the total stress and pore water pressure can be measured at any point.

Given data and Calculation:

From the given figure,

Sum of the effective stress =

(γ_sat – γ_ω)h₂ + (γ_sat – γ_ω)[h₂ + h₃]

γ_sat – γ_ω = r’

so sum = 2γ ’h₂ + γ ’h₃

= (2h₂ + h₃)γ ’

Concept:

• pH, is the quantitative measure of the acidity or basicity of aqueous or other liquid solutions.
• pH is the negative logarithm of hydrogen ion concentration.
• P^H = -log₁₀ [H⁺]

Given data and Calculation:

Weight of HNO₃ = 100 mg

atomic weights of

H = 1 g/mol, N = 14 g/mol, O = 16 g/mol

Molecular weight of HNO₃ = 1 + 14 + 48 = 63 gm

No. of moles present = \(\frac{100 \times 10^{-3}}{63} = 0.00158\)

P^H = -log₁₀ [0.00158] = 2.8

Explanation:

Given,

The chemical formula of the biodegradable fraction of municipal solid waste (MSW) is C100H250O80N.

Oxygen in air (by weight) = 23 %

density of air = 1.3 kg/m3. 

Weight of the waste = 1 tonne 

2C₁₀₀H₂₅₀O₈₀N + 243.5O₂ → 200CO₂ + 247H₂O + 2NH₃

Molecular weight if MSW = (12 × 100) + (250 × 1) + (16 × 80) + (14)

= 2744 gm

5488 gm of MSW requires (243.5 × 32) gm of O₂

1000 kg of MSW requires = \(\frac{7792}{5488} \times 1000\)

= 1419.825 kg of O₂

Quantity of air required = \(\frac{1419.825}{0.23}\)

= 6173.15 kg

Vol. of air required = \(\frac{6173.15}{1.3}\)

= 4748.58

~ 4749 m³/tonne

Concept:

The relationship 6 / w traffic speed (V), Traffic density (K), and traffic (Q) is given as Q = K × V

Here (V) is space mean speed (SMS)

Also, Time headway (t_h) is given as :

\(\mathop Q\limits_{\begin{array}{*{20}{c}} \downarrow \ {veh/hr} \end{array}} = \frac{{3600}}{{th(rec)}}\)

Calculation:

Given : V = SMS = 30 Km/hr

K = 40 veh/ Km 

Traffic flow, Q = K × V = 40 × 30 = 1200 Veh/hr

Time headway is given as,

\(1200 = \frac{{3600}}{{{t_\lambda }}} \Rightarrow {t_\lambda } = 3\sec \)

Explanation-

Column matrix is a matrix in which all the elements are in a single column.

A column matrix has only one column and multiple rows. The order of a column matrix is n × 1, and it has n elements.

The elements are arranged in a vertical manner, with the number of elements equal to the number of rows in a column matrix.

The following properties of the column matrix, help in a deeper understanding of the column matrix.

• A column matrix has only one column.
• A column matrix has numerous rows.
• The number of elements in a column matrix is equal to the number of rows in the matrix.
• A column matrix is also a rectangular matrix.
• The rank of the column matrix is 1.
• The transpose of a column matrix is a row matrix.
• The column matrix can be added or subtracted to only a column matrix of the same order.
• A column matrix can be multiplied with only a row matrix
• The product of a column matrix with a row matrix gives a singleton matrix.
• Multiplication of column matrix and its transpose matrix gives a symmetric matrix.
• Multiplication of transpose and the given matrix is an eigenvalue of Multiplication of column matrix and its transpose matrix.
• Multiplication of column matrix and its transpose matrix gives a symmetric matrix that has determinant zero.

 

Given data and Analysis-

Let y be a non-zero vector of size 2022 × 1.

It is a column matrix.

So from the above properties explained,

 yyT is a symmetric matrix and yTy is an eigenvalue of yyT

As the determinant of yyT is zero, it is not invertible.

Alternate MethodConcept-We know that all nonzero eigenvalues of ABAB AB

 and BABA BA

 are the same.

Here, yyTyyT yy^T

 and yTyyTy y^T y

 have the same nonzero eigenvalue. Since yy y

 is nonzero, yyTyyT yy^T

 is a nonzero matrix. Additionally, a 1×11×1 1 $\times 1$

 matrix has its only entry as its eigenvalue. Hence, yyTyyT yy^T

 has a nonzero eigenvalue, which implies that yyTyyT yy^T

 is an eigenvalue of yTyyTy y^T y

Explanation:

Castigliano’s first theorem-

• For linearly elastic structure, where external forces only cause deformations, the complementary energy is equal to the strain energy.
• The first partial derivative of the total internal energy (strain energy) in a structure with respect to any particular deflection component at a point is equal to the force applied at that point and in the direction corresponding to that deflection component.
• This first theorem is applicable to linearly or nonlinearly elastic structures in which the temperature is constant and the supports are unyielding.

Castigliano’s second theorem-

• The first partial derivative of the total internal energy in a structure with respect to the force applied at any point is equal to the deflection at the point of application of that force in the direction of its line of action.
• The second theorem of Castigliano is applicable to linearly elastic (Hookean material) structures with constant temperature and unyielding supports.

 

Principle of Virtual work method-

• Relations between active forces can be determined directly without reference to the reactive forces.
• If a structure is acted upon by two force system Pa and Pb, in equilibrium separately, the external virtual work done by a system of forces Pb during the deformations caused by another system of forces Pa is equal to the external virtual work done by the Pa system during the deformation caused by the Pb system.

Conjugate beam method- 

• The shear force in a conjugate beam loaded by the M/EI diagram of the real beam is equal to the corresponding slope at that point of the real beam.
• The bending moment in a conjugate beam loaded by the M/EI diagram of the real beam is equal to the corresponding deflection at that point of the real beam.

​

So from the given options statement 1, 2 & 3 are correct.

Explanation:

Channel transition with hump:

 

At choked conditions, the flow will be critical.

From the graph shown above, it can be concluded that,

• In choked conditions, y1 decreases if the flow is supercritical and increases if the flow is subcritical.
• In choked conditions, y2 is equal to the critical depth if the flow is supercritical or subcritical.
• In unchoked conditions, y1 remains unaffected when the flow is supercritical or subcritical

Explanation:

Given data and Calculation:

Given that**,**

s0 = 5 μmol/m

D = 0.1 m2/s

The Total pollutants = area under the curve = 5 × 0.4 × 10 = 20 µmol.

x = 0.4L to x = L Concentration of pollutant = 0 Total concentration of pollutant in 10 m = 20 mol 

In the steady state condition i.e., in infinite time this concentration of pollutants will be uniformly distributed.

So, concentration at any point along the length is = 20/10 = 2 µ mol/m.

Steady state concentration at x = L/2 is 2 µ mol/m.

Explanation-

• Binomial distribution summarizes the number of trials, or observations when each trial has the same probability of attaining one particular value.
• The binomial distribution determines the probability of observing a specified number of successful outcomes in a specified number of trials.
• The binomial distribution formula is calculated as: \(P(n,x,p) = {}^n{C_x}{p^x}{(1 - p)^{n - x}}\)
• Where n is the number of trials (occurrences, X is the number of successful trials, p is the probability of success in a single trial, and nCx is the combination of n and x.

Given data and Calculation-

A pair of six-faced dice is rolled thrice.

The event for which probability is required is "the sum of the outcomes in each roll equals 4".

Total number of outcomes = 6 × 6 = 36

Outcome for event = {(1,3),(3,1),(2,2)}

Probability of occurance of  the event = \(\frac{3}{36}\) = \(\frac{1}{12}\)

Probability of non-occurence of the event = 1- \(\frac{1}{12}\) = \(​​\frac{11}{12}\)

The event has to occur exactly two of the three attempts, so \(P(3,2,\frac{1}{{12}}) = {}^3{C_2}{\left( {\frac{1}{{12}}} \right)^2}{(1 - \frac{1}{{12}})^{3 - 2}} \) 

p = 0.01909 = 0.02

1. It is given that here there is no axial force in HI so there is no reaction developed at H. Also, due to increase in temp. of rod IJ, there will be elongation in IJ, since ends are fixed this elongation has to be prevented by applying an external compressive force in member IJ.

Deflection due to 'p' = def. due to temp.

Δ_p = Δtemp.

\(\rm Δ_p=\frac{PL}{A E}\) and Δ_temp = L α Δ T

2. \(\rm Δ_p=\frac{PL}{A E}\) and Δtemp = L α Δ T

Calculation:- P = ?, L = 2m, Δ T = 50°C, α = 10^-6 °C^-1

AE = 10⁶ N

using Δ_p = Δ_temp

\(\rm \frac{PL}{AE}=L α Δ T\)

⇒ P = α ΔT AE

= 10^-6 × 50 × 10⁶

P = 50 N

Concept:

1. The deflection due to point Load in the Cantilever beam is given as:

 

\(Δ = \frac{PL^3}{3EI}\)

\(θ = \frac{PL^2}{2EI}\)

(2) Also

 

\(Δ_p = \frac{P x^3}{3EI} \)

\(θ = \frac{Px^2}{2EI}\)

Δ = Δ_p + Δ_p'

Δ_p' = y × θ = \(\left( \frac{Px^2}{2EI} \right) y\)

(3) The force is spring is developed only when there is a change in its length.

Calculation:

It is given that force in spring Vw is zero i.e., F_vw = 0; it is only possible when there is no change in length of spring and this can be ensured only if deflection at 'v' in beam uv is equal to deflection at w in beam xw.

(i.e.,) = \(\rm \Delta_v^{uv} = \Delta_w^{xw}\)

 

\(\rm \Delta_v^{uv} = \frac{P(2a)^3}{3EI} = \frac{8 Pa^3}{3EI}\)

\( \Delta_w^{xw} = \frac{θ a^3}{6EI} + \frac{θ a^3}{4EI}\)

 

\(= \frac{5}{12} \left( \frac{ θ a^3}{EI} \right)\)

Now, \(\rm \Delta_v^{uv} = \Delta_w^{xw}\)

\(\left( \frac{8}{3} \right) \left( \frac{Pa^3}{EI} \right) = \frac{5 }{12} \times \frac{θ a^3}{EI}\)

⇒ θ = \(\frac{32}{5} P\)

⇒ θ = \(\frac{32}{5} \times 100 = 640 \) KN

Concept:

Friction force = Coefficient of friction × Normal force

Calculation:

Given:

w = 10 kN, a = 4 m and b = 3 m

∵ KJ = \(\sqrt{4^2~+~3^2}~=~5~m\)

∴ KO = OJ = 2.5 m, OO' = 2.5 Cosθ, Cosθ = 3/5

Using equilibrium:

∑F_x = 0 ⇒ N₁ = f_s,

∑F_y = 0 ⇒ N₂ = w

∑M_k = 0 ⇒ w × (2.5 cos θ) + fs × 4 = N₂ × 3

⇒ \(10~\times~2.5~\times~\frac{3}{5}~+~4f_s~=~3~\times~10\)

⇒ f_s = 3.75 kN

∵ f_s ≤ (fs)_max

⇒  fs ≤ μ_sN₂

⇒ 3.75 ≤ μ_sw

⇒ μ_s ≥ \(\frac{3.75}{10}\)

⇒ μ ≥ 0.375

so, μ_min = 0.375

Concepts:

The total float of any activity is given as:

F_t = T_L^j – T_Eⁱ - t_ij

Where

T_L^j = latest allowable time of event j

T_Eⁱ = earliest expected time of event i

t_ij = duration of an activity

The AOA network for above is given as:

 

For Activity E,

T_L^j = 26 days

T_Eⁱ  =15 days

t_ij = 10 days

F_t = 26 – 15 – 10 = 1 day

concept:- The group efficiency of any pile group is given as

\(\rm n_g=\frac{θ_g}{θ_p\times N}\)

θ_g = Ultimate load carrying capacity of piles as group

θ_p = Ultimate load carrying capacity as single pile.

N = No. of piles.

Calculation:-

d = 1 m, L = 20 m

Design capacity of each pile θ_p = 1000 kN

\(\rm θ=\tan^{-1}\left(\frac{d}{r}\right)\)

\(\rm θ=\tan^{-1}\left(\frac{1}{3}\right)\)

= 18.43°

No. of rows = 4

No of column = 4

m = n = 4

Total = 16 piles

substituting these values in given equation.

\(\rm n_g=\left[1-\frac{18.43}{90}\left(\frac{4(4-1)+4(4-1)}{4\times4}\right)\right]=\frac{θ_g}{1000\times16}\)

θ_g = 11085 kN

Concepts:

The degree of Consolidation is given as:

\(U = \frac{\text{Dissipated excess pore water pressure (Δu)}}{\text{Initial excess pore water pressure }u^ e_i} \times 100\)

The dissipated excess pore water pressure at any time is given as:

Δu = u_i^e – u_t^e

u_t^e is excess pore water pressure at any time

Also, the relationship between total stress (σ), effective stress (σ^’) and pore water pressure is given as :

σ = σ^’ - u

Calculation:

Initial pore water pressure, u_o = 25 kPa

Initial total stress, σ_i = 150 kPa

Additional load applied = 100 kPa

Case1 : Just after application of additional load

At t =0 ; i.e. just application of load on soil strata, the initial excess pore water pressure will be given as:

u_i^e =  Additional  applied load =  100 kPa

Initial total pore water pressure, U_o^total = u_o  + u_i^e = 25 + 100 = 125 kPa

Initial total vertical stress just after application of load, σ_i = 150 + 100 = 250 kPa

The initial effective stress just after application of load, σ_i^’ =  σ_i  - U_o^total = 250 – 125  = 125 kPa

Case 2: Some time after application of additional load

At any time ‘t’; when effective stress is 175 kPa, then

Total stress will not change i.e. σ_t = 250 kPa

Total water pressure, U_t^total = 250 – 175 = 75 kPa

The excess pore water pressure, u_t^e = U_t^total – U_o = 75- 25 = 50 kPa

Now,  the degree of consolidation is given as:

\(U = \frac{100-50}{100} \times 100\)

U = 50 %

Concepts:

The energy loss due to hydraulic jump is given as:

\(E = \frac{{(y_2 - y_1)}^3}{4y_2y_1}\)

Where

y₁ is the pre jump depth

y₂ is the post jump depth

The relation between pre and post jump depth is given as

\(\frac{y_1}{y_2} = \frac{1}{2}[-1 +\sqrt{1+ 8F^2_2}]\)

Where

F₂ is the Froude number after = jump

Calculation:

Given:  y₁ = 0.5 m; E = 1.6 m

Using above equation:

\(1.6 = \frac{{(y_2 - 0.5)}^3}{4y_2 \times 0.5}\)

Solving the above equation, we get y₂ = 2.5 m

Now

\(\frac{0.5}{2.5} = \frac{1}{2}[-1 +\sqrt{1+ 8F^2_2}]\)

∴ F₂ = 0.35

Duty is given as:

\(D = \frac{8.64B}{\Delta }\)

Where

D is duty in ha/m³s.

B is base period in days

Δ  is amount of water required by plant in m

The power of pump is given as:

\(P = \frac{\gamma Q H}{\eta}\)

Where H is head and η is efficiency of pump

Calculation:

Given: Δ = 1.44 m , B = 120 Days

Duty, D = (8.64 × 120)/1.44

D = 720 ha/m³/s

The discharge required to irrigate the 108 ha area having water application efficient of 80 % is given as:

Q = 108/(720 × 0.8)

Q = 0.1875 m³/s

The power of pump is given as:

\(P = \frac{9810 \times 0.1875 \times 10 }{0.8}\)

 P = 22992 W

Power in h.p. = 22992/746 = 30.82 hp

The drag force on any particle is given as:

F = 3πμDV_S

Where

V_s is the settling velocity which is given as:

\(V_s = \frac{(ρ_b-ρ)D^2}{18μ}\)

D is the particle diameter

Based on the above, it can be inferred that for a  given mass the drag force is proportional to

F ∝ D³

Calculation:

Given:

For Particle P;  D_P = 0.5 mm

For Particle, Q; D_Q  = 1 mm

Now

\(\frac{F_Q}{F_P} = \frac{D^3_Q}{D^3_P}\)

Or

\(\frac{F_Q}{F_P} = \frac{1^3}{0.5^3}\)

∴ F_Q = 8 F_P

The drag force on Q is eight times the drag force on P.

Concepts:

The moisture content in MSW is given as:

\(MC = \frac{A- B}{A} \times 100\)

Where

A = Wet weight of MSW

B = dry weight of MSW

Calculations:

Initial wet weight of MSW, A = 30 MT @ Moisture content , MC = 10 %

\(0.1 = \frac{30 - B}{30} \)

∴ B = 27 MT

On adding water in MSW, dry weight will not change.

Let A^’ be the new wet weight of MSW when MC is increased to 50%, then using above formula, we get

\(0.5 = \frac{A^{'} - 27}{A^{'}}\)

A^’ = 54 MT

Extra amount of water added in MSW to bring its moisture content to ideal condition is

M_W­ = A^’ – A = 54- 30 = 24 MT

Concepts:

Let V_BT  and P_BT denote the volume of wet sludge before thickening and moisture content in wet sludge respectively.

Similarly, V_AT and P_AT denote the volume of wet sludge after thickening and moisture content in wet sludge respectively.

The relationship between these four parameters is given as:

V_AT (1- P_AT) = V_BT (1- P_BT)

Calculation:

Sewage flow rate, Q_s = 500 cum/day

TSS at inlet = 200 mg/L

Primary Sedimentation removes 60 % of total TSS i.e. 0.6 × 200 = 120 mg/l . These solids settle down at bottom of the tank and lead to the formation of sludge.

So, the total mass of sludge in wet sludge before thickening, M_BT = 120 × 5000 × 10³ mg/day

∴ M_BT  = 600 kg/day

Solid concentration in wet sludge before thickening (BT) is 2 %. It means that

100 kg of wet sludge (BT)  = 2 kg solids + 98 kg of water

Or

2 kg solids in Sludge (BT)  à 100 kg wet sludge (BT)

Or

600 kg/day solids  à (100/2) × 600 = 30,000 kg/day wet sludge (BT)

The volume of wet sludge (BT) = M_BT/ρ_BT = 30000/1000 = 30 cum/day

After thickening (AT), the solid concentration is 6 % i.e. moisture content is P_AT = 100 – 6 = 94 %

Using the above given relation, we get

30 × (1-0.98) = V_AT × (1-0.94)

∴ V_AT = 10 cum/day

Therefore, the volume of thickened sludge generated on daily basis is 10 cum/day

Concepts:

1. One mole of any gas occupies 22.4 liters. at 0⁰ C temperatures at 1 atm. pressure.
2.  The given temperature is 25 degrees C, therefore, we need to find out the what is the volume of SO₂ gas occupied by its one mole at 25⁰ C, This can be done using the ideal gas equation:

PV = nRT

Where

P is the pressure, V is the volume, n is the moles and T is the temperature

Calculation:

T₁ = 273 K and T₂ = 273 + 25 = 298 K

Since n and R are constant for a given gas, thus∴

\(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\)

\(\frac{1 \times 22.4}{273} = \frac{1 \times V_2}{298}\)

V = 24.45 ltr.

This means that 1 mole of SO₂ occupies 24.45 ltrs. at 25⁰ C and 1 atm.

{SO₂} = 0.04 ppm à this means that

10⁶ cum air has 0.04 cum SO₂

Or

10⁶ cum air  à 0.04 × 1000 = 40 ltrs. SO₂

1 mole of SO₂ occupies 24.45 liters. Therefore, 10⁶ cum air  à 40/24.45 = 1.636 moles of  SO₂

The molecular weight of SO₂ = 32 + 2 × 16 = 64 gm.

10⁶ cum air has 1.636 × 64 = 104.7 gm SO₂

Or  

1 cum air has  \(\frac{104.7 \times 10^6}{10^6}\)  μg/m3 SO₂

∴ SO₂ in μg/m³ = 104.7 μg/m³

Concepts:

The length of vertical crest curve is given as:

Case 1: If L > SSD; then

\(L = \frac{NS^2}{2(√H + √ h)^2}\)

Case 2: If L < SSD; then

\(L = 2S - \frac {2(√H + √ h)^2}{N}\)

Where

H is height of driver’s eye

h is height of obstacle

N is algebraic deviation of grade at entry and exit points.

Calculation:

Given:

Grades- n₁ = 1% and n₂ = -2%; H = 1.2m, h = 0.15; SSD = 200m

Assume: L > SSD i.e.  L > 200 m

Then, the length of vertical curve is given as:

\(L = \frac{(0.01-(-0.02)\times 200^2}{2 \times (\sqrt1.2 + \sqrt0.15)^2 }\)

On solving, we get L = 272.9 m

Since the calculated value of length of curve is found to be greater than SSD i.e. L >200 m so our assumption is right.

∴ L = 273 m

Concepts:

The shock wave speed is given as:

\(V_w = \frac{q_2-q_1}{k_2-k_1}\)

Where q₁ and q₂ are traffic flow for state 1 and state 2 respectively

K₁ and k₂ are traffic densities for state 1 and state 2 respectively

The relationship between traffic flow (q), traffic speed (v), and traffic density (k) is given as:

q = kV

Calculation:

State 1:

q₁ = 1200 Veh/hr and V₁ = 60 km/hr

Therefore, \(k_1 = \frac{q_1}{V_1} =\frac{1200}{60} = 20 :{Veh/km}\)

State 2:

q₂ = 1200 Veh/hr and V₂ = 60 km/hr

\(k_2 = \frac{q_2}{V_2} =\frac{1800}{30} = 60 ; \text {Veh/km}\)

Shock wave speed is calculated as:

\(v_w = \frac{1800-1200}{60-20}\)

v_w = 15 km/hr

Concepts:

The mass-specific gravity (G_m) is given as:

\(G_m = \frac{W_{mix}}{V_{CA} + V_{FA}}\)

Where

W_mix is the weight of the mixture of CA and FA

V_CA is the volume of Coarse Aggregate

V_FA is Volume of Fine Aggregate

The % of air Voids is given as:

\(V_a (\text{%}) = \frac{G_m-G_b}{G_m} × 100\)

Where

G_b is the bulk specific gravity

Calculation:

Given: W_CA= 260 gm; W_FA = 40 gm; G_CA = 2.6 and G_FA = 2.4

G_CA and G_FA are mass specific gravities.

The volume of Coarse aggregate and fine aggregates is given as:

V_CA = W_CA/G_CA; V_FA= W_FA/G_FA

Now,

\(G_m = \frac{240 + 260}{240/2.4 + 260/2.6}\)

Gm = 2.5

\(V_a = \frac{2.5-2.3}{2.5} \times 100\)

∴ the % of air voids, V_a = 8%

Concept:

As per Webster's Method, the optimum cycle length is given as:

\(C_0 = \frac{1.5L + 5 }{1- Y} \)

Where

L is the total Loss time, in all phases

\(Y = \Sigma \frac{q}{Q}\) = Sum of ratios of critical lane volume to saturation flow rate for all phases.

q = Critical lane Volume for each phase

Q = Saturation flow rate

Calculation:

Given: L = 9 sec, Q = 1800 veh/hr

Critical lane volume for any phase is the maximum design flow rate for that phase.

For phase I (East-West and West-East through Movements)

q₁ = Max ( 504, 460)

q₁ = 504 Veh/hr

For phase II (North-South and South-North through Movements)

q₂ = max(396, 360)

q₂ = 396 Veh/hr

Now

\(Y = \frac{q_1}{Q} + \frac{q_2}{Q} \)

 \(Y = \frac{504}{1800} + \frac{396}{1800} = 1/2 \)

Substituting the above values in the cycle length formula, we get

\(C_0 = \frac{1.5 \times 9 + 5}{1- 0.5}\)

C₀ = 37 sec