GATE CE 2021 Official Paper: Shift 1 (Previous Year Paper)

Explanation:

When the comparison between two things we use the second degree of the adjective.

The degree form of easy are;

• Easy
• Easier
• Easiest

Explanation:

Mirror image:

The image of an object as seen in a mirror is known as the mirror image. 

In the given figure, the mirror is placed about the x-axis then the upper part of the object is seen downward and the lower part of the object is seen upward. In the given word, Letter I and E will remain unchanged in the mirror image. 

Hence, The mirror image of the figure will appear-

           

Explanation:

Given-

Total employees n(∪) = 100% 

Employees drink coffee = 35% 

Employees drink tea = 40%​

Employees drink Both tea and coffee = 10%

Employees, who drink either tea or coffee n(T ∪ C) = 25% + 10% + 30% = 65%

Employees, who drink either tea or coffee

n(T ∩ C) = n(∪) - n(T ∪ C)

n(T ∩ C) = (100 - 65)%

n(T ∩ C) = 35%.

Explanation:

 x ⊕ y = 2 ⊙ 2

$\frac{{{x^2} + {y^2}}}{{xy}} = \frac{{{2^2}}}{2}$

x² + y² = 2xy

(x - y)² = 0

x = y

Explanation:

Given,

1). P and R cannot sit adjacent to each other

2). S should be seated to the right of Q

These are total distinct seating arrangements possible followed these two conditions:

PQRS, PQSR, QRSP, RQPS, RQSP, QPSR

Explanation:

Statement Either P marries Q or X marries Y

The given statement explains any of the actions or both the actions follow.

Option 1 follows the statement because one action follows (X marries Y).

Option 2 follows the statement because one action follows (P marries Q).

Option 3 follows the statement because both actions follow (P marries Q and X marries Y).

Option 4 does not follow the statement because any of the actions follow (Neither P marries Q nor X marries Y).

Explanation:

Given,

Sheet M and Sheet N of dimensions 6 cm × 4 cm each.

Three times operation 1 performed on sheet M and Three times operation 2 performed on sheet N.

After performing operations,

Perimeter of M:

P_M = 2 (2 + 1.5) = 7 cm

Perimeter of N:

PN = 2 (6 + 0.5) = 13 cm

The ratio of perimeters of the final folded shape of Sheet N to the final folded shape of Sheet M:

Ratio = $\frac{{{P_N} }}{{{P_M}}}$ = $\frac{{{13} }}{{{7}}}$

Explanation:

Given

PR = PS = QS = QT = RT

Then ABCDEF will be regular pentagon.

Sum of the angle formed at the pentagon = 540⁰

Each angle of the pentagon = $\frac{{{Sum \ of \ the \ angles\ of \ pentagon}}}{{{5}}}$

= $\frac{{{540}}}{{{5}}}$ = 108⁰

Then, ∠EAB = ∠ABC = ∠BCD = ∠CDE = ∠DEA = 108⁰

∠PAB = 180⁰ - 108⁰ = 72⁰

∠PBA = 1800 - 1080 = 72⁰

From ΔPAB,

θ + ∠PAB + ∠PBA = 180⁰

θ + 72⁰ + 72⁰ = 1800

θ = 180⁰ - 144⁰ = 36⁰.

Explanation:

Given,

 \(λ \left( {p,q} \right) = { \begin{array}{*{20}{c}} {{{\left( {p - q} \right)}^2}}&{if;p \ge q.}\ {p + q}&{if;p < q.} \end{array};\)

Calculation:

 $\frac{{λ \left( { - \left( { - 3 + 2} \right),\left( { - 2 + 3} \right)} \right)}}{{\left( { - \left( { - 2 + 1} \right)} \right)}}$ = $λ \frac{{\left( {1,1} \right)}}{1} = λ \left( {1,1} \right)$

So, 1st definition will be applicable as p= q;

Hence,

λ (1,1) = (1 - 1)² = 0

Explanation:

In Option 1 and option 2, the word unrelated is use that is wrong because Imagination, invention and innovation, and the ability to construct mental worlds are related so both options are incorrect.

In option 3 sentence uses no species in which humans are also included so this option is also incorrect.

So, the correct option is 4.

Explanation:

\(\left[ {\begin{array}{*{20}{c}} 1&2&2&3\ 3&4&2&5\ 5&6&2&7\ 7&8&2&9 \end{array}} \right]\)

Using,

R₂ → R₂ - 3R₁,

R3 → R₃ - 5R1,

R4 → R₄ - 7R1,

A = \(\left[ {\begin{array}{*{20}{c}} 1&2&2&3\ 0&-2&-4&-4\ 0&-4&-8&-8\ 0&-6&-12&-12 \end{array}} \right]\)

Using,

R3 → R3 - 2R2,

R4 → R4 - 3R2,

A = \(\left[ {\begin{array}{*{20}{c}} 1&2&2&3\ 0&-2&-4&-4\ 0&0&0&0\ 0&0&0&0 \end{array}} \right]\)

So, ρ (A) = Number of non zero rows = 2

Concept:

Transpose of a Matrix:

If A = [aij]m × n, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A, denoted by A′ or (AT). AT = [aji]n × m

Calculation:

\(PQ = \left[ {\begin{array}{{20}{c}} 1&3 \ 2&4 \end{array}} \right]\left[ {\begin{array}{{20}{c}} 0&1 \ 1&0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&4 \ 1&3 \end{array}} \right]\)

\({\left( {PQ} \right)^T} = \left[ {\begin{array}{*{20}{c}} 2&4 \ 1&3 \end{array}} \right]\)

(PQ)^T = Q^TP^T

QTP^T = \(\left[ {\begin{array}{*{20}{c}} 2&4 \ 1&3 \end{array}} \right]\)

Concept:

Gaussian distribution:   

• All Gaussian distribution can be standardized to the reference Gaussian distribution, which is called the standard Gaussian distribution. Standardization in general is accomplished by subtracting the center of the distribution from a given element in the distribution and dividing the result by the standard deviation of the distribution.
• The distribution of a standardized Gaussian distribution—that is, a Gaussian distribution that has its elements standardized in this form—has its center at zero and has a variance of unity.
• The shape of the cumulative distribution function of Gaussian distribution is S-shaped.
• If a distribution is normal, then the values of the mean, median, and mode are the same. However, the value of the mean, median, and mode may be different if the distribution is skewed (not Gaussian distribution).
• The standard deviation of the mean used to check the given data distribution is close to Gaussian distribution.

The probability density function of a zero-mean Gaussian variable is as shown:

The probability distribution function of a Gaussian Random Variable is defined as;

$f\left( x \right) = \frac{1}{{\sqrt {2\pi {\sigma ^2}} }};{e^{\frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}$

Given distribution has zero mean i.e. μ = 0, so the above distribution can be written as:

$f\left( x \right) = \frac{1}{{\sqrt {2\pi {\sigma ^2}} }};{e^{\frac{{{x^2}}}{{2{\sigma ^2}}}}}$

Concept:

Muller Breslau Principle:

If an internal stress component (shear force, bending moment etc) or reaction component is allowed to act through a small distance thereby causing deformation of the structure, the curve of the deformed shape represents to some scale, the influence lines for that stress or the reaction component. It is applicable to all structures, determinate or indeterminate.

Note:

• In the case of determinate systems, the ILD can be obtained directly by allowing unit deformation corresponding to the constraint.
• For indeterminate structure, it is applicable only when the material is within the elastic limit and obeys Hook's law so that the law of superposition holds good.

Explanation:

Explanation:

Gypsum: Gypsum is composed of calcium sulphate (CaSO4) and water (H2O). Its chemical name is calcium sulphate dihydrate (CaSO4.2H2O). Gypsum is mined from sedimentary rock formations around the world. It acts as a retarder in concrete, increases its setting time thus prevent quick setting. 

Additional Information

a) Accelerators are used to increase the rate of gain of strength in the concrete.

It is an admixture that causes an increase in the rate of hydration of the cement and thus shortens the time of setting, increases the rate of strength development, or both.

Calcium chloride is a common accelerator, used to accelerate the time of set and the rate of strength gain.

Example: Calcium Chloride, Silicates, Na2SO4, NaCl, Fluorosilicates, Triethanolamine, etc.

b) Retarder increases the initial setting time of cement while accelerators reduce the setting time of cement.

Some commonly used retards are: CaSO4, Organic retardants include unrefined calcium, sodium, NH4, salts of ligno-sulfonic acids, hydrocarboxylic acids, and carbohydrates.

Explanation:

Total cost = Direct cost + Indirect cost

Total cost = 160000 + 20000 = 180000

According to question, Mark up applied is 10% of bid price.

Quoted bid price = Total Cost + (10/100 ) ×  Quoted bid price

⇒ 0.9 ×  Quoted bid price = 1,80,000

So,

Quoted price = 1,80,000/0.9 = 2,00,000

Additional Information

Indirect project Cost: Indirect project cost are those expenditures which cannot be clearly allocated to the individual activities of the project, but are assessed as a whole.

The indirect cost rises with an increase in duration. 

Direct Project Cost: These include labour cost, material cost, equipment cost etc.

Direct cost decreases with an increase in duration.

Explanation

Total stress = Effective stress + Pore pressure

When loading is done initially Total pressure increases.

The decrease in soil volume by squeezing out of the pore water on account of gradual dissipation of excess hydrostatic pressure induced by an imposed total stress is called consolidation.

Total Consolidation of soil is divided into

i) Primary consolidation

• Primary consolidation begins when soil is fully saturated.
• (In a saturated soil mass if increase in effective stress is Δσ1 , then initially it will be taken by pore water, Hence increase in pore pressure (Excess Pore Pressure (u) = Δσ1  )
• Primary consolidation is complete when expulsion of pore water stops. (u = 0). (i.e. decrease in pore water pressure). At this stage primary consolidation is complete for Δσ1.
• If further load continues to act then effective stress increases causing secondary consolidation.

ii) Secondary consolidation

• After completion of primary consolidation when expulsion of pore water is stopped and load continues to act, then at very slow rate further volume change may be recorded which is due to plastic readjustment of solids.

∴Correct sequence is

• i) Increase in Total stress
• ii) Increase in pore water pressure
• iii) Decrease in pore water pressure (u = 0)
• iv) Increase in effective stress.

Calculation:

It is given sample is consolidated to 50 kN/m² and then suddenly the load has been increased to 150 kN/m²

∴ Change in Total stress Δσ = 100 kN/m²

We know increase in vertical stress will lead to equal increase in pore water pressure

 Δu = 100 kN/m² 

Initially stage of soil

Since soil is in equilibrium at 50 kN/m² , i.e. (End of the consolidation )

Initial pore water pressure = 0 & Effective stress = 50 kN/m²

∴ Final Pore water pressure u = 100 kN/m²

b) We know

 Total Stress = 150kN/m²

 Effective stress = Total stress – pore water pressure

 = 150 - 100

 = 50 kN/m2

 ∴ Effective Stress = 50 kN/m²

Concept:

Bulk Unit weight (γb): 

It is defined as the ratio of the total weight of soil to the total volume of the soil mass.

${γ _b} = \frac{W}{V} = \frac{{{W_s} + {W_w}}}{{{V_s} + {V_w} + {V_a}}}$

Water Content (w):​

Water content or moisture content of a soil mass is defined as the ratio of the weight of water to the weight of solids (dry weight) of the soil mass.

\({\rm{w = }}\frac{{{{\rm{W}}{\rm{w}}}}}{{{{\rm{W}}{\rm{s}}}}};{\rm{w}} \ge 0\)

It is denoted by the w and is commonly expressed as a percentage. The minimum value for water content is 0. There is no upper limit for water content.

Dry Unit Weight (γ­d):-​

Dry unit weight is defined as the weight of soil solids per unit volume of soil. It is denoted by the letter symbol γd it has the unit of kN/m3.

\({{\rm{γ }}{\rm{d}}}{\rm{ = }}\frac{{{{\rm{W}}{\rm{s}}}}}{{\rm{V}}}{\rm{ = }}\frac{{{{\rm{W}}_{\rm{d}}}}}{{\rm{V}}}\)

It is used as a measure of the denseness of soil. A high value of dry unit weight indicates that more solids are packed in a unit volume of soil hence a more compact soil.

The specific gravity of solids (G):-​

The specific gravity of solids is defined as the ratio of the unit weight of solids to the unit weight of water. It is denoted by the letter G and is a unitless quantity.

\({\rm{G = }}\frac{{{{\rm{γ }}{\rm{s}}}}}{{{{\rm{γ }}{\rm{w}}}}}\)

Note: The relationship between the degree of saturation, water content, specific gravity, and the void ratio is:

s × e = w × G

Calculation:

Given,

w = 25%, γ = 18.5 kN/m³, G = 2.65, γ_w = 9.81 kN/m3 ​​

We know, s × e = w × G 

​$\gamma = \frac{{\left( {G + se} \right){\gamma _w}}}{{1 + e}}$ = $\frac{{\left( {G + wG} \right){\gamma _w}}}{{1 + e}}$ = $\frac{{G}{\left( {1 + w} \right){\gamma _w}}}{{1 + e}}$​

$\gamma = \frac{{G}{\left( {1 + w} \right){\gamma _w}}}{{1 + e}}$

18.5 = $\frac{{2.65}\times {\left( {1 + 0.25} \right)\times {9.81}}}{{1 + e}}$

e = 0.756

For fully saturated soil sample(s=1)

$\gamma_s = \frac{{\left( {G + e} \right){\gamma _w}}}{{1 + e}}$

$\gamma_s = \frac{{\left( {2.65 + 0.756} \right)\times {9.81}}}{{1 + 0.756}}$

$\gamma_s = 19.03$ kN/m3

Important Points

Air content:

Air content is defined as the ratio of the volume of air to the volume of voids. It is denoted by ac.

\({{\bf{a}}{\bf{c}}} = \frac{{{{\bf{V}}{\bf{a}}}}}{{{{\bf{V}}_{\bf{v}}}}} = \frac{{{\bf{Volume\ of\ air}}}}{{{\bf{Volume\ of\ voids}}}}\)

Porosity:

Porosity is defined as the ratio of the volume of voids to the total volume of soil. It is denoted by n. It varies between 0 and 1.

${\bf{n}} = \frac{{{{\bf{V}}_{\bf{v}}}}}{{\bf{V}}} = \frac{{{\bf{Volume\ of\ voids}}}}{{{\bf{Total\ volume}}}}$

Percentage air voids:

Percentage air voids are defined as the ratio of the volume of air to the total volume of soil. It is denoted by n­­a.

\({{\bf{n}}{\bf{a}}} = \frac{{{{\bf{V}}{\bf{a}}}}}{{\bf{V}}} \times 100 = \frac{{{\bf{Volume\ of\ air}}}}{{{\bf{Total\ volume}}}} \times 100\)

Degree of Saturation: 

The degree of Saturation of a soil mass is defined as the ratio of the volume of water in the voids to the volume of voids. It is denoted by S.

$S = \frac{{{V_w}}}{{{V_v}}} \times 100;;;0 \le S \le 100$

• For a fully saturated soil mass Vv = Vw, hence S = 100%
• For fully dry soil mass Vw = 0, hence S = 0%

Concept:

Most efficient channel: A channel is said to be efficient if it carries the maximum discharge for the given cross-section which is achieved when the wetted perimeter is kept a minimum.

a) For a triangular channel,

 The channel with a side slope of 1:1 is the most efficient channel.

Area of triangle A = 1/2 × y × 2 y

Wetted Perimeter P = 2 √2y 

The hydraulic radius for the given section is the ratio of the cross-sectional area to the wetted perimeter of the section. 

${\rm{R}} = {\rm{}}\frac{{\rm{A}}}{{\rm{P}}} = \frac{{{{\rm{y}}^2}}}{{2√ 2 {\rm{y}}}} = \frac{{\rm{y}}}{{2√ 2 }}$ --- (1)

b) Rectangular Section

Area of the flow, A = By

Wetted Perimeter, P = B + 2y

For efficient Rectangular channel y = B/2 → B = 2y

Area A = (2y) × (y) = 2y²

Perimeter (P) = 4y

Hydraulic Radius, $R = \frac{A}{P} = \frac{{By}}{{B + 2y}} = \frac{{2{y^2}}}{{4y}} = \frac{y}{2}$ ---- (2)

Explanation

The ratio of Hydraulic radius of triangular section to rectangular section is 

$\frac{{{R_1}}}{{{R_2}}} = \frac{{\frac{y}{{2\sqrt 2 }}}}{{\frac{y}{2}}} = \frac{1}{{\sqrt 2 }}$

Explanation:

Kinematic viscosity:

Kinematic viscosity is defined as the ratio of dynamic viscosity and density of the fluid.

$kinematic~ viscosity= \frac {dynamic ~ viscocity}{density ~of ~fluid}$

$ν = \frac {μ}{ρ }$

• SI unit = m2 / s
• CGS unit = Stokes or cm2 / s
• 1 Stoke = 10-4 m2 / s
• Dimensional Formula: M⁰L²T^-1  or L²/T

Additional Information

Dynamic Viscosity ( µ )

• Viscosity can be defined as the measure of a fluid's resistance to deformation at a given flow.
• Caused by friction within a fluid.
• Result of intermolecular forces between the particles within a fluid.
• SI unit N-s/m2 or Pa-s.

Shear stress is given by,

$Shear;stress = ;\mu ;\frac{{du}}{{dy}}$

where,

µ = Dynamic Viscosity of fluid

$\frac{u}{y}$= Rate of shear deformation

Explanation:

Pyrolysis:

• It is the process in which most of the organic matter upon heating in an oxygen-free atmosphere splits through a combination of thermal cracking and condensation reactions into gaseous, liquid, and solid fractions.
• The process typically occurs at temperatures above 430 °C (800 °F) and under pressure.
• It simultaneously involves the change of physical phase and chemical composition and is an irreversible process. So it is an endothermic process.
• This method is suitable for the disposal of solid waste which poses high calorific value but is comparatively costlier than other methods of disposal.

Important Points

Other methods for disposal of solid waste: 

Sanitary landfilling method:

• This is a method of disposal of municipal solid waste.
• In this method disposal of refuse is being carried out over the low lying area in the layers of approximately 1.5 m thick.
• Each layer after being disposed of is properly compacted and left for at least 7 days before the application of another layer over it.
• This entire process is biological in approach.
• For optimum decomposition moisture content of the refuse must be greater than 60%.
• This entire process completed within 2-12 months after which the height of the landfill is reduced by 25-40%.

​Composting:

• This is also a biological method of disposal of solid waste in which decomposition of the organic matter can be carried out either in the presence or absence of oxygen
• If composting is carried out aerobically mixing of the solid waste is insured either manually or mechanically.
• Aerobic composting is completed normally within 2-3 months and it is also known as Indore process.
• If composting is carried out an-aerobically mixing is avoided.
• Anaerobic composting is completed in normally 5-6 months and is also known as Bangalore process.
• For optimum composting C/N ratio must be in the range of 30-50.

Pulverisation and Shredding:

• In real terms, these are not the methods of disposal of solid waste these are only used to convert the heavier solid into the lighter one either by cutting or tearing action (Shredding) or by crossing or grinding action (Pulverisation).

​Autoclave:

• It is a low heat thermal process that is used for the disposal of biomedical waste in which steam under controlled temperature and pressure conditions is passed over it so as to carry out its disinfection.

Explanation:

• Chemoheterotrophs are microorganism (like bacteria, fungi, protozoa) which use organic matter both as energy source & carbon source and are the most important organism used in ASP.
• Young lakes having low nutrient & low plant productivity are called oligotrophic lake.
• Partially treated effluent will lead to greater depletion of D.O. at some distance d/s of the point of disposal.
• For effluent sample of STP BOD5 /BODu will always be less than one.

Explanation;

Classification of pollutants:

Concept:

• The ideal vertical curve is a 2° Parabola
• The most preferred curve for vertical alignment is parabolic.

Important Points

1. Square parabola is generally preferred due to the best riding quality, simplicity of calculation, and uniform rate of change of gradient.
2. In a circular curve, sight distance is available throughout the curve is constant.
3. Cubic parabola is generally preferred in the valley curve.

Concept:

Superelevation(e): Superelevation inroads is basically provided on the horizontally curved portion of the roads in which the outer edge of the road pavement is raised with respect to the inner edge, thus providing a transverse slope throughout the length of the horizontal curve of the road.

$e+f = \frac{V^{2}}{127R}$

where,

f = lateral friction = 0.15, V = Designed speed,

 R = Radius of curve

Calculation

Given,

Design Speed (V) = 80 kmph

Radius of curve (R) = 250 m, 

Lateral friction is assumed to develop fully i.e. f = 0.15

As, $e+f = \frac{V^{2}}{127R}$

 $e+0.15 = \frac{80^{2}}{127\times 250}$

e = 0.052

Explanation:

1. Levelling

• Operation includes, the line commences with a backsight and closes with a foresight.
• The first staff reading taken after setting up of instrument is always a Backsight and last staff reading taken before changing the instrument to the other position is always a Foresight.
• So Statement 1 is False

2. Contours

• Vertical cliff: Contour lines of different elevations unite to form one line.
• Valley: V-shaped contours with convexity towards higher ground
• Ridge: U - shaped contours with convexity towards lower ground
• Overhanging cliff: Two contours of different elevation cross each other in this case.
• 

3. Principle of Surveying

• Whole to part: In working from whole to part, the error will localise and prevent the accumulation of error while working from part to whole, the error will accumulate. Hence more error in working from part to whole.
• Location w.r.t at least 2 well-defined control point: The location of a point should be respected to at least 2 well-defined control points.

4. Planimeter:

• It is an instrument used in surveying to compute the area of any given plan. Planimeter only needs plan drawn on the sheet to calculate area.
• Generally, it is very difficult to determine the area of the irregular plot. So, by using planimeter we can easily calculate the area of any shape.

Concept:

We know the relationship between FB and BB 

Back bearing (BB) = Fore bearing (FB) ± 180°. 

+ve sign is used if FB is less than 180° and –ve sign is used if FB is more than 180°.

Calculation:

Given,

Fore Bearing of line = 270 °

∴ Back bearing = 270°  – 180° = 90° 

Important Points

• Whenever there is any station affected by local attraction, then the difference between the fore bearing and back bearing is not equal to 180°
• If the difference is exactly 180°, the two stations may be considered as not affected by local attraction.
• If the difference is not 180°, better to go back to the previous station and check the fore bearing.
• If that reading is the same as earlier, it may be concluded that there is a local attraction at one or both stations.

​

Additional Information

• In the case of a fixed hair stadia tachometer, the staff intercept will be smaller, when the staff is held nearer to the observation point.
• The boundary of water of calm water pond will represent contour line because elevation is some at all point on the calm water surface.

Explanation:

(D² + 2D + 1)y = 0

y(0) = 1

y(1) = 3

Auxiliary equation is;

m² + 2m + 1 + 0

m = -1, -1

CF = (C₁+C₂x)e^-x

and PI = 0

Using boundary conditions;

y(0) = 1, C₁ = 1

y(1) = 3, C₂ = 3e - 1

So the solution of equation is;

y = CF + PI

y = (C1+C2x)e^-x + 0

y = {1 + (3e - 1 )x}e^-x

y = e^-x + (3e - 1) xe^-x

Concept:

Trapezoidal rule:

\(I = \mathop \smallint \limits_a^b f\left( x \right);dx\)

${\rm{Number;of;intervals}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{;}}$

where b is the upper limit, a is the lower limit, h is the step size or subinterval.

According to the trapezoidal rule

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}{\rm{o}}} + {{\rm{y}}{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{;}} \ldots } \right)} \right]\)

Here, the interval [a, b] is divided into n number of intervals of equal width h.

It fits for a 1-degree polynomial.

Calculation:

Given, No. of interval = 4

Here $h = \frac{{1 - 0}}{4} = 0.25$ 

Calculating all values of f(x), we get:

According to Trapezoidal’s rule

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}{\rm{o}}} + {{\rm{y}}{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{;}} \ldots } \right)} \right]\)

\(\mathop \smallint \limits_{\rm{0}}^{\rm{1}}\ e^x\ {\rm{dx}} = \frac{h}{2}\left[ {\left( {{y_0} + {y_4}} \right) + 2\left( {{y_1} + {y_2} + {y_3}} \right)} \right]\)

\(\mathop \smallint \limits_{\rm{0}}^{\rm{1}}\ e^x\ {\rm{dx}} = \frac{0.25}{2}\left[ {\left( {{1} + {e}} \right) + 2\left( {{e^.25} + {e^.50} + {e^.75}} \right)} \right]\)

\(\mathop \smallint \limits_{\rm{0}}^{\rm{1}}\ e^x\ {\rm{dx}} = \frac{0.25}{2}\left[ {\left( {{1} + {e}} \right) + 2\left( {{1.284} + {1.648} + {2.117}} \right)} \right]\)

\(\mathop \smallint \limits_{\rm{0}}^{\rm{1}}\ e^x\ {\rm{dx}} = 1.727\)

Concept:

Estimation of solids:

Total solids mainly can be divided into two suspended solids and dissolved solids-

(1) Suspended solid

Raw water when passed through Whatman filter paper no. 9, the filter paper collects suspended solids which when dried under a hot air oven gives total suspended solids. It is determined by a gravimetric technique.

(2) Dissolved solid

The filtered water when heated at 104° - 105°C for 4 to 5 hours gives dissolved solids, which when subjected to a mettle furnace (600°C - 1000°C for 4 to 5 hours) results in nonvolatile solids, and hence the difference obtained is volatile solid.

Volatile solids = Dissolved solids(heated at 104° - 105°C) - Nonvolatile solids(heated at 600°C - 1000°C)

Calculation:

Given,

Sample size = 50 ml = 50 × 10^-3 = 0.050 L

Empty weight of the crucible = 54.352 g

Weight of crucible with the residue sample after drying at 104⁰C = 55.129 g

Weight of crucible with the residue sample after drying at 6000C = 54.783 g

Volatile solids weight = Dissolved solids(heated at 104° - 105°C) - Nonvolatile solids(heated at 600°C - 1000°C) 

Volatile solids weight = 55.129 - 54.783 = 0.346 g = 346 mg

Concentration of total volatile solids = $\frac{346}{0.050}$ = 6920 mg/L.

Explanation:

 

ΣF_y = 0 on block

N₂ sin60° - 0.2 N₂ sin30° - Q = 0

Q = 0.766 N₂

Putting in wedge,

ΣF_x = 0

0.2 N₂ cos30° + N₂ cos60° - P = 0

P = 0.67 N₂

$P = 0.67 \times \frac{Q}{{0.766}}$

P = 0.875 Q = 0.9 Q

So, 

α = 0.9

Explanation:

Mean of the bid to total cost ratio;

$\mu = \left[ {\frac{{1.02 \times 6 + 1.04 \times 12 + 1.06 \times 3 + 1.10 \times 6 + 1.12 \times 3}}{{30}}} \right] = 1.058$

Standard deviation,

$\sigma = \sqrt {\frac{{{\text{Σ }}{{\left( {{x_i} - \mu } \right)}^2}}}{N}}$

$= \sqrt {\frac{{6{{\left( {1.02 - 1.058} \right)}^2} + 12{{\left( {1.04 - 1.058} \right)}^2} + 3{{\left( {1.06 - 1.058} \right)}^2} + 6{{\left( {1.10 - 1.058} \right)}^2} + 3{{\left( {1.12 - 1.058} \right)}^2}}}{{30}}}$

= 0.034

Thus, Z at 8% markup level = $\left( {\frac{{x - \mu }}{\sigma }} \right)$

= $\left( {\frac{{1.08 - 1.058}}{{0.034}}} \right)$ = 0.0647

From normal deviate table;

By interpolation, probability for Z = 0.647

$= 72.6 + \left( {\frac{{75.8 - 72.6}}{{0.70 - 0.60}}} \right)\left( {0.647 - 0.60} \right)$

= 74.104%

Hence, the probability of winning against the competitor Y at 8% mark-up

= (1 - 0.74104)

= 0.25896

= 25.896%

Explanation:

• In dense sand and overconsolidated clays, the volume of the sample increases on shearing after a small decrease in volume.
• In loose sand and normal consolidated clays, the volume of the sample continuously decreases on shearing.

Concept:

Given,

the velocity profile in the pipe as

${{{V_r}}} = U\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)$

Comparing it with the original equation,

${V_r} = - \frac{1}{{4\mu }}\left( {\frac{{\partial P}}{{\partial x}}} \right){R^2}\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)$

⇒ ${U} = - \frac{1}{{4\mu }}\left( {\frac{{\partial P}}{{\partial x}}} \right){R^2}$

Given maximum velocity occurs at pipe centerline, that is at r = 0,

⇒ V_r = ${U}\left( {1 - \frac{{{{\left( 0 \right)}^2}}}{{{R^2}}}} \right) = {U}$

For average velocity(V̅r),

V̅_r = \(\frac{{\mathop \smallint \nolimits_0^R {V_r}.dA}}{A} = \frac{{\mathop \smallint \nolimits_0^R {U}\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right);.2\pi r;dr}}{{\pi {R^2}}}\)

⇒  V̅r = \(\frac{{{U} \times\ 2\pi }}{{\pi {R^2}}};\mathop \smallint \limits_0^R \left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)rdr\)

⇒ V̅r = $\frac{{{U} \times\ 2\pi }}{{\pi {R^2}}};\left| {\frac{{{r^2}}}{2} - \frac{{{r^4}}}{{{4R^2}}}} \right|_0^R$

⇒ V̅r  = $\frac{{{U} \times \ 2\pi }}{{\pi {R^2}}}\left( {\frac{{{R^2}}}{2} - \frac{{{R^4}}}{{4{R^2}}}} \right)$

⇒ V̅r  = $\frac{{{U} \times\ 2\pi }}{{\pi {R^2}}}\left( {\frac{{{R^2}}}{4}} \right) = \frac{{{U}}}{2}$

Hence, The average velocity of the fluid in the pipe is $\frac{U}{2}$.

Concept:

Procedure to find MPN:

• Select a series where three tubes each have positive results. (not necessary but recommended).
• Use sample size corresponding to which the MPN value is larger.
• The MPN index and confidence limit are given corresponding to the sample size of 10 mL, 1 mL, and 0.1 mL.

So if we are selecting sample size as 1 mL, 0.1 mL, and 0.01 mL, then the series of the sample used is one-tenth of the 10, 1, and 0.1 sample size, therefore multiply the MPN index value and confidence limit by 10 and similarly so on.

Calculations:

Since only three sample dilution has been taken.

We will search for a combination of +ve tubes as 5-3-1 in 0.01, 0.001 & 0.0001 ml from the table.

Since the MPN chart given MPN value of 110 per 100 ml for 10 ml, 1 ml & 0.1 ml dilutions than the result so obtained is multiplied by $\frac {10\ ml}{0.01\ ml}$ = 1000, to obtain MPN per 100 ml for 0.01, 0.001, 0.0001 ml dilutions.

Hence, MPN per 100 ml = 110 × 1000 = 110000.

Additional Information

Most probable number (MPN):

MPN stands the foremost probable number obtained in multiple tube fermentation tests. In this test, the water sample of different dilution ratios is mixed with lactose broth (nutrient for coliforms) and is incubated for 48 hours. During incubation, coliforms consume lactose broth and reduce it to acids and gasses. The presence of acid or gases in a sample tube indicates a positive test result otherwise negative.

Hence, No of positive tubes = No of tubes in which acid or gasses is detected

Concept:

Nitrogen Contents in sewage: The presence of nitrogen in sewage indicates the presence of organic matter, and may occur in one or more of the following forms: Free ammonia (also called Ammonia nitrogen), Albuminoid nitrogen (also called Organic nitrogen), Nitrites and Nitrates.

 Free ammonia (or Ammonia nitrogen):

• Ammoniacal nitrogen (NH3-N), is a measure for the amount of ammonia present in sewage.
• Ammonia has chemical formula of NH3 in the unionized state and NH4+ in the ionized state. The ammonia nitrogen is the sum of nitrogen present in both NH3 and NH4+.
• The free ammonia indicates the very first stage of decomposition of organic matter and the amount will progressively decrease as sewage gets treated.

Ammonia exist in the form of NH₄⁺ & NH₃

1. · At pH < 8, all ammonia is in the form of NH₄⁺
2. · At pH = 9.5, there are 50% NH₃ & 50% NH₄⁺
3. · At pH > 11, all ammonia is in the form of NH₃

_​∴ Option 1 is correct.

Extra Points 

 Albuminoid nitrogen (or Organic nitrogen): 

• Organic nitrogen is that nitrogen which describes nitrogen compounds that had its origin in living material.
• The nitrogen in protein and urea is organic nitrogen. Organic nitrogen can enter sewage as bodily wastes, discarded food material, or as components of cleaning agents.
• This albuminoid nitrogen indicates quantity of nitrogen present in sewage before the decomposition of organic matter is started.
• If this organic nitrogen reacts anaerobically then it decomposes to ammonia and if this reacts aerobically it converts to nitrites or nitrates.

Concept:

Time headway:

It is defined as the time interval between the passage of successive vehicles moving in the same lane and measured from head to head as they pass a point on the road.

$Time;headway = \frac{1}{{traffic;volume}} = \frac{1}{q}hours/vehicle$

Where,

q = traffic volume in vehicles/hour

Time Headway is also given by H_t = $\frac{{3600 }}{q_{max}}$

where q_max is the Theoretical maximum capacity 

Calculation:

Given

Traffic stream u = 70 - 0.7 k 

 we Know 

Traffic volume = Traffic Density × u

 q = ku

 q = k × ( 70 - 0.7 k )

 q = 70k - 0.7 k² 

For maximum q 

dq/dk = 0

 $\frac{{dq}}{{dk}} = 70 - 1.4 \ k = 0$

 k = 70/1.4 = 50

 Maximum q = 70k - 0.7 k2 = 70 × 50 - 0.7 × 50²

 q = 1750 veh/hr

So Theoretical maximum capacity is 1750 veh/hr

Now Time Head way is 

Time Headway given by H_t = $\frac{{3600 }}{1750}$ = 2.057 sec

∴ Time Headway = 2.057 sec ≈ 2.1 sec