GATE CE 2021 Official Paper: Shift 2 (Previous Year Paper)

Explanation:

According to Rule 1 and Rule 3, (i) and (iv) are correct.

Rules of subject-verb agreement,

Rule 1: Collective nouns such as group, jury, family, audience, population are usually regarded as singular subjects and the verb mostly used is singular except in some cases. 

Rule 2: Nouns connected by the conjunction and in the subject work as the plural subject and take a plural verb.

Rule 3: When the prepositional phrases separate the subjects from the verbs, they have no effect on the verbs.

Rule 4: If the conjunction ‘and’  is replaced by together with/ along with/ accompanied by/ as well as, the verb will have no effect for the later part of these expressions. The words prior to these expressions are the subjects.

Rule 5: The verb in an or, either/or, or neither/nor sentence agrees with the noun or pronoun closest to it.

Explanation:

Given,

Mirror image on X-axis, it should look like the image below

Concept:

\(Probability = ;\frac{{Probable;Outcomes}}{{Total;Outcomes}}\)

\(P(A \cap B) = P(A) \times P(B)\)

Calculation:

Given,

Two cubes numbered 1 to 6.

So, Total outcomes = {1, 2, 3, 4, 5, 6} = 6

Possibility of even number = Favourable outcomes = {2, 4, 6} = 3

The probability that an even number is rolled out  for Cube 1 and Cube 2,

\(\begin{array}{l} P(A) = \frac{3}{6} = \frac{1}{2}\ P(B) = \frac{3}{6} = \frac{1}{2} \end{array}\)

The probability that an even number is rolled out on each dice is,

 \(P(A \cap B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)

Explanation:

given,

p ⊙ q = p - q, and p ⊕ q = p × q, that means

⊙ = - (subtraction) and ⊕ = × (multiplication)

Then,

(9 ⊙ (6⊕ 7)) ⊙ (7 ⊕ (6 ⊙ 5)) = (9 - (6 × 7)) - (7 × (6 - 5)) = (9 - 42) - 7 = - 40

(9 ⊙ (6⊕ 7)) ⊙ (7 ⊕ (6 ⊙ 5)) = - 40

Explanation:

Given,

Four people = P, Q, R, S

R shouldn't be seated at the second position from the left end of the row

Possible arrangements:

The total number of ways = 3 × 3! = 3 × 3 × 2 = 18

Calculation:

\(OQ = \sqrt {{{(3)}^2} + {{(4)}^2}} = 5units\)

Explanation:

According to the above passage,

• The author has basically shared his opinion about the importance of rehearsals.
• He said Actors rehearse for their new play but public speakers just go and speak without any prior practice.
• But according to his opinion, rehearsals are equally important for actors as well as public speakers.

Explanation:

For statement 1, Some football players play cricket, the possible Venn diagram can be, 

For statement 2, Some football players play cricket, the possible Venn diagram can be, 

According to both of them, Option 1, Some football players play hockey can be the possibility.

Concept:

\(Required;Probability = ;\left[ {\frac{{Shaded;Area}}{{Total;Area}}} \right]\)

Total Area = Area of square = (side)²

Shaded area = 2 ×  [Area of sector - (Area of half square or Area of triangle)]

Area of sector = \({\frac{{\pi {r^2}}}{4}}\)

Area of half square or Area of triangle = \({\frac{1}{2}{r^2}}\)

Calculation:

Given,

Side of square = r

Radius of sector = r

Total Area = Area of square = (side)2 = r²

Shaded area = 2 ×  [Area of sector - (Area of half-square or Area of triangle)] = \(2 \times \left( {\frac{{π {r^2}}}{4} - \frac{1}{2}{r^2}} \right)\)

\(Required;Probability = ;\left[ {\frac{{2 \times \left( {\frac{{\pi {r^2}}}{4} - \frac{1}{2}{r^2}} \right)}}{{{r^2}}}} \right] = \left[ {\frac{{{r^2} \times \left( {\frac{\pi }{2} - 1} \right)}}{{{r^2}}}} \right] = \left( {\frac{\pi }{2} - 1} \right)\)

Concept:

\(Area;of;equilateral;triangle = ;\frac{{\sqrt 3 }}{4} \times sid{e^2}\)

Calculation:

Assume the side of ΔPQR = s 

Given that,

\(\left( {\frac{a}{4},\frac{a}{6},\frac{a}{8}} \right)\) is an integer

Minimum area of ΔPQR when s is the LCM (Least Common Multiple of (4, 6, 8)

s = LCM (4, 6, 8) = 24

\(Area;of;\Delta PQR = ;\frac{{\sqrt 3 }}{4} \times {24^2}; = 144\sqrt 3 sq.units\)

Concept:

L’ Hospital’s Rule:

\(\mathop {\lim }\limits_{x \to a} \left{ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right} = \mathop {\lim }\limits_{x \to a} \left{ {\frac{{f'\left( x \right)}}{{g'\left( x \right)}}} \right}\)

This rule is only applicable for \(\frac{0}{0}\) and \(\frac{\infty }{\infty }\) indeterminate forms.

Calculation

As 

\(\mathop {\lim }\limits_{x \to \infty } \frac{{xln;x}}{{1 + {x^2}}} = \frac{\infty }{\infty }\)

Which is an indeterminate form.

Using L hospitality rule for \(\mathop {\lim }\limits_{x \to \infty } \frac{{xln;x}}{{1 + {x^2}}}\)

We get

\(\mathop {\lim }\limits_{x \to \infty } \frac{{x \times \frac{1}{x} + lnx}}{{2x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{1 + lnx}}{{2x}} = \frac{\infty }{\infty };Indeterminate;form\)

Again applying the L hospitality rule, we get 

\(\mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{x}}}{2} = \frac{{\frac{1}{\infty }}}{2} = 0\)

∴ Value of 

\(\mathop {\lim }\limits_{x \to \infty } \frac{{xln;x}}{{1 + {x^2}}} = 0\)

Concept:

Rearrange the rows with leading zeroes on the left and convert them into echelon form

The rank of the matrix = number of non-zeroes row in the echelon form

Calculation:

Given:

 matrix = \(\begin{bmatrix} 5 & 0 & -5 & 0 \\ 0 & 2 & 0 & 1 \\ -5 & 0 & 5 & 0 \\ 0 & 1 & 0 & 2 \end{bmatrix}\)

Replace R2 with R4 matrix becomes

  \(\begin{bmatrix} 5 & 0 & -5 & 0 \\ 0 & 1 & 0 & 2 \\ -5 & 0 & 5 & 0 \\ 0 & 2 & 0 & 1 \end{bmatrix}\)

Applying row transformations R₃ → R₃ + R₁ Now the matrix becomes.

  \(\begin{bmatrix} 5 & 0 & -5 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 1 \end{bmatrix}\)

Applying Row transformation for row 4: R₄ → R₄ - 2 R₂

 \(\begin{bmatrix} 5 & 0 & -5 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -3 \end{bmatrix}\)

 Replace R3 with R4, the matrix becomes

 \(\begin{bmatrix} 5 & 0 & -5 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & -3 \\ 0 & 0 & 0 & 0 \end{bmatrix}\)

Therefore Number of non zero rows is 3 

∴ The rank of the matrix is 3

Concept:

Gradient / Normal vector of surface / Slope of surface / Rate of maximum increase or decrease:

Let ϕ(x, y, z) = C represents any scalar point function, then its gradient is defined as

\(grad;\phi = \nabla \phi = \hat i\frac{{\partial \phi }}{{\partial x}} + \hat j\frac{{\partial \phi }}{{\partial y}} + \hat k\frac{{\partial \phi }}{{\partial z}}\) = Normal vector \( = \sum \hat i\frac{{\partial \phi }}{{\partial x}}\) 

1. Direction of (grad ϕ) = Direction of Normal vector = Direction of surface

2. Rate of maximum increase = |grad ϕ|

3. Unit normal on \('\phi ' = \hat n = \frac{{grad;\phi }}{{\left| {grad;\phi } \right|}}\) 

Calculation:

Given,

Surface S : x2 + y2 + z2 - 48 = 0

We know that; \(grad;\phi = \nabla \phi = \hat i\frac{{\partial \phi }}{{\partial x}} + \hat j\frac{{\partial \phi }}{{\partial y}} + \hat k\frac{{\partial \phi }}{{\partial z}}\)

∴ grad (S) = 2x î + 2y ĵ + 2z k̂

At point (4,4,4) we get:

grad (S) = 8 î + 8 ĵ + 8 k̂

∴ Unit normal on the sphere will be:

\( \therefore;\frac{{grad;S}}{{\left| {grad;S} \right|}} \Rightarrow\frac{{8\hat i;+;8\hat j;+;8\hat k}}{{\sqrt {{8^2};+; {8^2};+;{8^2}} }} \)

\(= \frac{{\ \hat i \ +\ \hat j \ +\ \hat k}}{{\sqrt{3}}}\)

Concept:

Orthogonal matrix: When the product of a matrix to its transpose gives identity matrix.

Suppose A is a square matrix with real elements and of n x n order and AT or A’ is the transpose of A.

AAT = I

Important Points

• A square matrix such that A2 = I is called the involuntary matrix.
• A square matrix such that A2 = A is called the Idempotent matrix.
• Any square matrix A is said to be non-singular if |A| ≠ 0, and a square matrix A is said to be singular if |A| = 0.
• A square matrix A = [aij] is said to be Hermitian matrix if aij = a̅ij

Explanation:

(i) Surveying in the science of determining the relative positions on ground, along with their elevations.

(ii) The relative positions are determined by measuring the horizontal distances, horizontal angles, vertical distances and the vertical angles. for this, various instrument are needed.

(iii) The entire work stages of survey operation in civil engineering may be divided into the following three stages:

• Field work
• Office work
• Care and adjustment of instruments

Field work:

(i) The field work consists of the measurement of all the necessary horizontal and vertical distances, horizontal and vertical angles, elevations, etc. and keeping a systematic record of what has been done in a field book. Field work is further subdivided into;

• Reconnaissance
• Field measurements
• Observationsfield record.

Office Work:

(i) The office work of a surveyor consists of three types based on the field records.

• Drafting
• Computing
• Designing.

​Care and adjustment of instruments:

(i) A great care is required to handle the survey instruments both in field and office. A beginner should always be made familiar with care and adjustment of the instruments and their limitations. Precise instruments like theodolite, level, prismatic compass, etc. need more care than the equipment such as chains, arrows, ranging rods, etc.

So. from the above observation correct sequence of surveying operation is

Reconnaissance  → Field observations → Data analysis → Map making

Explanation:

(i) A is Limit of Proportionality: Beyond which linear variation ceases

(ii) B is elastic limit: The maximum stress up to which a specimen regains its original length on the removal of the applied load.

(iii) C is upper yield point: The magnitude of the stress corresponding to C depends on the cross-sectional area, shape of the specimen and type of equipment used to perform the test. It has no practical significance

(iv) D is lower yield point: This is also called actual yield point

(v) DE Represent strain hardening: In this range, further addition of stress (more than yield stress) gives additional strain. However, strain increases with faster rate in this region. The material in this range changes its atomic and crystalline structure, resulting in increased to further deformation.

(vi) E is ultimate point: The stress corresponding to this point is ultimate stress and the corresponding strain is about 20% for mild steel.

(vii) F is fracture point: Stress corresponding to this is called breaking stress and strain is called fracture strain.

(viii) Region between E and F is the necking region in which the area of cross-section is drastically decreased.

Explanation

Natural circular frequency (ω) = \(\sqrt {\frac{K}{m}} \) rad/second

Or, Circular frequency (f) = \(\frac{ω }{{2\pi }}\) cycle per second

Springs in series and parallel:

Combined stiffness (K) i.e. resistance per unit deflection.

If springs are in series: \(\frac{1}{K} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}} + \frac{1}{{{K_3}}}...... + \frac{1}{{{K_n}}}\)

If springs are in parallel: K = K1 + K2 + K3 +..........+ Kn

Given,

It is given that for any single column the deflection is

  \(δ = \dfrac{PL^3}{12 EI}\)

We know 

The stiffness factor is defined as k = P/δ 

 The building Model has 3 columns.

 It is similar to the system of springs in parallel connection. 

 Stiffness (k) = 12EI/L³

So For the combined system Stiffness factor is 

As they are connected parallel. 

K_ef = k₁ + k₂ + k₃ = 3 k = 3 × 12EI/L3 = 36EI/L³ 

Natural circular frequency (ω) = \(\sqrt {\frac{K}{m}} \) rad/second

 ω = \(\sqrt {\frac{36EI}{L^3m}} \)

 ω = \(6\sqrt {\frac{EI}{L^3m}} \)

∴ The natural frequency of the system for horizontal oscillation is ω = \(6\sqrt {\frac{EI}{L^3m}} \)

Explanation:

Seasoning of timber increases strength, durability and stiffness but reduces its self-weight due to the reduction in moisture content.

Seasoning of timbers:

A newly fell tree consists of more than 50 % of water by weight, hence cannot be put into service when used for engineering purposes. This process of drying the timber section up to the optimum moisture content is called seasoning.

Seasoning of timber section is done:

1. To make it more workable.
2. To increase its resistance against cracking twisting and warping
3. To increase its resistance against ole forming agencies like termites, insects, fungi, etc.
4. To maintain its uniform shape and size.
5. To permit the application of decorative treatment over it.
6. To reduce its transportation cost.
7. To ensure its uniform burning when used as fuel.

• Water in the timber section is present either in the form of sap or in the form of moisture.
• It is comparatively easy to remove moisture in comparison to the sap.
• When the timber section is subjected to drying it is the cavity water that gets vaporize first and followed by the wall water.
• The stage at which entire cavity water is lost and cell water is intact are termed the fibre saturation point.
• Seasoning of the timber section can be done either naturally or artificially.
• In natural seasoning there is no control over the factors which control the evaporation thereby tendency of development of cracks is comparatively more.

Explanation:

Two Envelop System:

(i) A Two Envelope System separates vendor responses to Formal Solicitations into two sealed bids, or envelopes: a technical proposal and a price proposal. 

(ii) This process is followed in cases where technical specifications, TOR (Terms of Reference) or the technology/process etc are pre-determined. Now only the two things namely: Eligibility criterion of the bidders and the financial offer needs to be evaluated.

(iii) As far as opening the is concerned, obviously the envelop one containing eligibility criterion is opened first and the financial proposal of only those who meet the qualification criterion are opened. And the tender is awarded to the most economical bidder.

Three Envelop System:

(i) In this case the bidder are asked to submit the documents as below:

• Envelope 1 : Documents related to eligibility criteria.
• Envelope 2 : Technical bid.
• Envelope 3 : Financial bid.

(ii) The envelops are opened in sequence, that means first, second and then third. Through the documents submitted in envelop one, the eligibility of the bidder is ascertained.

(iii) Envelop 2 will be opened of only those bidders who meets the qualification criterion. At this stage technical offer of all the Eligible bidders is opened. generally at pre-defined date a conference is arranged to discuss the various offer submitted by various bidders, to discuss the merits and demerits of the technical offers. The bidders who offer the most viable and technologically superior technical solution are declared. All successful bidders at this stage are now asked to modify their financial bids, if required, as the technical specification has changed.

(iv) Envelope 3 will be opened of only those successful bidders at stage 2. And the contract is awarded to the most economical bidder.

Explanation:

Types of Triaxial test on the basis of Drainage:

1. Unconsolidated Undrained test (UU test):

• It is quick test and may complete in 5-10 minutes.
• In this test water is not allowed to leave the soil either during the consolidation stage (confining stage) nor shear stage (deviator stage).
• Such tests are suitable for low permeable soil such as clays with fast loading.
• UU test is carried out for evaluation of short-term stability of the structure.

2. Consolidated Undrained Test (CU test):

• During the first stage of confining pressure, drainage is allowed from the soil sample. Hence consolidation will take place. But during vertical shear loading, drainage is not permitted.
• Example: stability analysis of earthen dam during the sudden drawdown.

3. Consolidation Drain test (CD test):

• Drainage is permitted during the self-pressure (confining pressure) stage and shear stage both.
• This test is the most time taking and for some soils may take several weeks. So, it is also called ‘slow test;
• CD test is carried out for evaluation of long-term stability of the structure.

Explanation:

Classification of fine-grained soil:

(i) In USSC, fine-grained soils are classified on the basis of plasticity chart and compressibility.

(ii) generally, soils are considered as fine soils, when 50% or more of the total material by weight pass 75 μ sieve.

(iii) Liquid limit and plastic limit are determined for 425 μ sieve fraction and corresponding plasticity index is find out

CASE 1: Low plastic soil (Low compressibility) (LL < 35%)

CL → Low plastic inorganic soil

ML → Low plastic inorganic silt

OL → Low plastic organic clay

CASE 2: Medium Plastic soil (medium compressibility) (35% < LL < 50%)

CI → Medium plastic inorganic soil

MI → Medium plastic inorganic silt

OI → Medium plastic organic clay

CASE 3: Highly plastic soils (High compressibility) (LL > 50%)

CH → High plastic inorganic soil

MH → High plastic inorganic silt

OH → High plastic organic clay

Explanation:

a) Momentum correction factor (β):

The momentum correction factor is defined as the ratio of momentum of the flow per second based on actual velocity to the momentum of the flow per second based on average velocity across a section.

\(β = ;\frac{{Momentum;per;second;based;on;actual;velocity}}{{Momentum;per;second;based;on;average;velocity}}\)

⇒ \(β = \frac{1}{{A{V^2}}}\smallint {u^2}.dA\)

For turbulent flow, the momentum correction factor is slightly higher than one near to 1.2 and

 for laminar, its value is 1.33 = 4/3  --- (1)

b) Kinetic energy correction factor(α): 

• It is defined as the ratio of kinetic energy/second based on actual velocity to the kinetic energy/second based on average velocity.
• \(α = \frac{1}{A}\mathop \smallint \limits_A^{} {\left( {\frac{u}{V}} \right)^3}dA\)
• where A = area, V= average velocity, u= local velocity at distance r.
• For laminar flow in a circular pipe α = 2  --- (2)

​ (1) divided by (2)

​∴ Ratio of β/α = (4/3)/2 = 2/3

Concept:

Free Mean Speed:

From the speed-density diagram, it can be defined as the maximum speed at which the number of vehicles in a unit length is zero i.e. density is zero.

Density:: It can be defined as the number of vehicles per unit length. The unit of measurement is vehicles/km.

Jam Density: From the speed-density relation and flow-density curve, it can also be seen that it is the maximum density at which there is no flow on the road.

      

Traffic Flow: The number of vehicles passing through a particular point in certain time interval is defined as traffic flow. Also, number of vehicles counted in one hour is called traffic flow (q).

The relation between k, u and q is given below-

\({\bf{q}} = {\bf{k}} \times {\bf{u}};{\bf{veh}}/{\bf{hour}}\)

From the curves above,

As long as the relation between density and speed is linear, it can be seen that maximum flow (or flow capacity) occurs at kj/2 and Vf/2.

∴ Maximum flow occurs at speed V_f/2

Maximum flow is given by

\({{\bf{q}}_{{\bf{max}}}} = \left( {\frac{{{{\bf{k}}_{\bf{j}}}}}{2}} \right)\left( {\frac{{{{\bf{u}}_{\bf{f}}}}}{2}} \right);{\bf{veh}}/{\bf{hour}}\)

Explanation

a) Reducing solid waste is reducing the amount of garbage that goes into our landfills. These are items we use each day, and then get rid of by putting them into the trash. Solid waste comes from homes, businesses, and industries.

We can reduce the amount of solid waste by reducing the consumption of raw materials: 

Ex:

1) Use a reusable bottle/cup for beverages on-the-go

2) Use reusable grocery bags, and not just for groceries

3) Buy secondhand items and donate used goods

4) Curb your use of paper: mail, receipts, magazines

b) 

• Both technological processes and consumptive processes result in the formation of solid wastes.
• Solid waste is generated, in the beginning, with the recovery of raw materials and thereafter at every step in the technological process as the raw material is converted to a product for consumption generation of solid waste during technological processes. Ex: mining, manufacturing, and packaging.

c) The objective of solid waste management is to reduce the quantity of solid waste disposed off on land by the recovery of materials and energy from solid waste. This in turn results in the lesser requirement of raw material and energy as inputs for technological processes.

Explanation

The hardness of water is defined as the concentration of multivalent metallic cations which destroys the surfaced property of soap (Concentration of ions) which reacts with Soap and leads to the formation of a precipitate.

Ca ²+, Mg²⁺ are major constituents

Al+3, Sr3+, Fe2+ Cu2+ are minor constituents of hardness.

Hardness is classified into two categories:

Carbonate hardness and non-carbonate hardness

Total Hardness (TH) = Carbonate Hardness (CH) + Non-Carbonate Hardness (NCH)

• Hardness in water can be determined by the determination of amounts of calcium and magnesium ions present in water by filtration
• Amount of Ca 2+, Mg2+ is determined by titration with versanate solution ( EDTA Method)
• In the EDTA method water is titrated with ethylene diamine tetraacetic acid using Erichrome Black T (EBT) as an indicator.
• EBT forms wine red colour and titration changes the colour to blue.

Explanation

Softening Point

• The softening point is measured by the ‘Ring and Ball’ test.

• The softening point is the temperature at which bitumen attains a particular degree of softness under standardized test conditions.

• The temperature at which a standard steel ball placed on a layer of bitumen kept in a standard ring passes through the bitumen layer and touches the bottom plate kept at a distance of 2.54 cm is the softening point.

• As a result of oxidation, the hydrogen atom in a bituminous hydrocarbon is combined with atmospheric oxygen, and polymerization takes place along with the formation of water.

• It has a higher softening point and lower susceptibilities to temperature.

Important Points

Bitumen: Some of the physical properties possess by Bitumen are-

• It is a viscous liquid; black or brown in color.
• It consists predominantly of hydrocarbons derived from petroleum crude.
• It is soluble in carbon disulfide.
• It is not soluble in water and the specific gravity is less than 1.
• It is thermoplastic, oxidizes slowly, and is chemically inert.

Explanation:

Photochemical Smog :

(i) This type of smog is also called as Los Angeles type smog (observed in Los Angeles in 1950s)

(ii) It does not require any smoke or fog. The word “smog” is misnoere here.

(iii) Photochemical smog is the chemical reaction of sunlight, nitrogen oxides, and volatile organic compounds in the atmosphere, which leaves airborne particles and ground-level ozone.

(iv) It is formed in the month of summer during the afternoon when there is bright sunlight so that photochemical reactions take place.

(v) Effects of photochemical smog: There are the various effect of photochemical smog

• Eye irritation: caused by aldehyde, PAN
• Vegetation damage-caused by O3, NO2, and PAN
• Visibility reduction
• Cracking of rubber
• Fading of eye

Note: Increased levels of carbon monoxide in the indoor environment result in the formation of carboxyhemoglobin and the long-term exposure becomes a cause of cardiovascular diseases and even leads to death.

Explanation

Given,

 Function f(x) = x e^|x|

 Integral is -1 to 1.

If f(-x) = f(x) then the function is said to be even function

 If f(-x) = - f(x) then the function is said to be odd function.

 f(-x) = -x e^|-x| = -x e|x| = - f(x)

 ∴ The given function is an odd function.

 For an odd function:

 \(\mathop \smallint \nolimits_{ - a}^a x;{f(x)}dx\) = 0

For a even function

 \(\mathop \smallint \nolimits_{ - a}^a x;{f(x)}dx\) = 2 × \(\mathop \smallint \nolimits_{ 0}^a x;{f(x)}dx\)

 Now, as the function is odd

 \(\mathop \smallint \nolimits_{ - 1}^1 x;{e^{\left| x \right|}}dx\) = 0

Concept:

∵ we know that, torsional moment equation

 \(\frac{\tau }{r} = \frac{T}{J} = \frac{{G\theta }}{L}\)

The above equation can be written as, \(\tau = \frac{{Tr}}{J}\)……….(i)

Where, J = Polar moment of inertia

For the circular section, J = (π/32 )× D⁴

And r = radius = D/2

Put the value of J and r in equation (i)

 \(\tau = \frac{{16T}}{{\pi {D^3}}}\)

Calculation:

For OP member:

T_OP = 3 kN-m

Dia (D) = 0.1 m

So, τ_max for OP, \({\tau _{\max }} = \frac{{16T}}{{\pi {D^3}}}\)

 \({\tau _{\max }} = \frac{{16 \times 3 \times {{10}^6}}}{{\pi \times {{100}^3}}} = 15.278\ MPa\)

For PQ member:

T_PQ = 1 kN-m

Dia (D) = 0.08 m

So, τmax for PQ member, \({\tau _{\max }} = \frac{{16T}}{{\pi {D^3}}}\)

 \({\tau _{\max }} = \frac{{16 \times 1 \times {{10}^6}}}{{\pi \times {{80}^3}}} = 9.94\ MPa\)

∴ Absolute maximum shear stress in the member = Maximum of [τ_OP, τ_PQ]

τ_abs = 15.278 MPa

Calculation:

ΣFy = 0, R_A + R_C = 10 × 4

R_A + R_C = 40…………..(i)

Take moment at internal hinge point B (From right);

ΣM_B = 0

R_c × 2 = 10 × 2 × 1

R_c = 10 kN

From (i), R_A = 30 kN

Concept:

Area ratio:

(i) Area ratio can be expressed as

 \(% {A_r} = \frac{{{A_o} - {A_i}}}{{{A_i}}} \times 100\)

(ii) Area ratio should be kept as less as possible, lesser area ratio means cutting edge is sharp. It should be less than 10% but in case it should exceed 20%.

Calculation:

Given, internal diameter (D_i ) = 48 mm, Outer diameter (D₀ ) = 52 mm

Area ratio is given by, \(% {A_r} = \frac{{{A_o} - {A_i}}}{{{A_i}}} \times 100\)

\(% {A_r} = \frac{{\frac{\pi }{4} \times D_o^2 - \frac{\pi }{4} \times D_i^2}}{{\frac{\pi }{4} \times D_i^2}} \times 100\)

\(% {A_r} = \frac{{D_o^2 - D_i^2}}{{D_i^2}} \times 100\)

\(% {A_r} = \frac{{{{52}^2} - {{48}^2}}}{{{{48}^2}}} \times 100\)

A_r = 17.36%

Calculation:

Let, catchment area = A km²

We know that,

Area of hydrograph = Total direct runoff volume

Area of hydrograph = Area of catchment × Runoff depth

1/2 × 144 × 3600 × 23 = (1/100) × A × 10⁶

A = 596.16 km²

Concept:

Absolute pressure: The pressure measured above absolute zero (or complete vacuum) is known as Absolute pressure.

Absolute Pressure = Atmospheric pressure + Gauge pressure

Absolute Pressure = Atmospheric pressure – Vaccum pressure

Calculation:

Given,

Atmospheric pressure = 91 kPa

Depth of lake = 60 m

Unit weight of lake water (γ) = 9790 N/m³

Gauge pressure at the bottom of lake = ρgh = γh = 9790 × 60 × 10^-3 = 587.4 kPa

Absolute pressure at the bottom of the lake (P_abs) = Atmospheric pressure + Gauge pressure

P_abs = 91 + 587.4 = 678.4 kPa

Concept:

Webster method:

In this method, the optimum cycle length (C_o) is calculated on the basis of the least total delay to the vehicles at the signalized intersection.

∴ Optimum cycle length, \({C_o} = \frac{{1.5L + 5}}{{1 - Y}}\)

Where,

L = total time lost per cycle

Y - sum of critical flow ratios of each phase 

Y = Y_A + Y_B = q_A/S_A + q_B/S_B

Calculation:

Given,

Y_A = 0.18, Y_B = 0.32, Y_C = 0.22

Y = 0.18 + 0.32 + 0.22 = 0.72

Total lost time in each phase = 2 sec

And number of phase = 3, so, total lost time = 2 × 3 = 6 seconds

∴ Optimum cycle length, \({C_o} = \frac{{1.5L + 5}}{{1 - Y}}\)

⇒ \({C_o} = \frac{{1.5 \times 6 + 5}}{{1 - 0.72}}\)

C_o = 50 seconds

Explanation

The most probable value of an angle is equal to its weighted arithmetic mean.

Given

 Angles and their weight as 

 

 The most Probable value is 

\(M.P.V = \frac{{36^\circ {{30}'} \times 4\ + \ 36^\circ {{00}'} \times 3 \ + \ 35^\circ {{30}'} \times 8\ + \ 36^\circ {{30}'} \times 4 }}{{4 + 3 + 8 + 4 }}\)

 MPV = 36°00'

Explanation

Given,

\(\dfrac{dy}{dx} - \dfrac{y}{x}=1\)

Comparing with

\(\frac{{dy}}{{dx}} + py = q\)

We get,

 p = -1/x, q = 1

We know Integrating Factor IF is given by 

\(IF = {e^{\smallint pdx}} = {e^{\smallint \frac{-1}{x}dx}} = {e^{-\ln x}} = 1/x\)

Now,

General solution can be written as,

y × IF = ∫ q × IF dx + k

y × 1/x = ∫1 × 1/x dx + k

 y/x = ∫ (1/x) dx + k

 y/x = lnx + lnk

 y = x ln(kx)

∴ The general Solution is x ln(kx)

Explanation:

Eigenvector (X) that corresponding to Eigenvalue (λ) satisfies the equation AX = λX.

Determining eigenvalues first by characteristic Equation 

Given matrix A = \(\begin{bmatrix} 2 & -2 \\ -1 & 6 \end{bmatrix}\)

The characteristic equation for the given matrix is

  |A - λI| = 0

 \(M = \left| {\begin{array}{*{20}{c}} {2 - λ }&-2\ -1&{6 - λ } \end{array}} \right| = 0\)

⇒ (2 - λ) × (6 - λ) - (-2)(-1) = 0 

⇒ 12 - 2 λ - 6 λ + λ² - 2 = 0 

⇒  λ2 - 8λ + 10 = 0

 \(λ = \frac{{ - 8 \pm √ {64 - 40} }}{2} = \frac{{ - 8 \pm √ {24} }}{2} = 4 \pm √ 6 \)

⇒ λ = 4 + √6 , 4 - √6 

∴ Eigen values are 4 + √6 & 4 - √6

Smallest Eigen Value = 4 - √6 = 1.55 

Now Determining Eigenvectors corresponding to the eigenvalue λ = 1.55.

For λ = 1.55 The corresponding Eigenvector is 

  \(\left[ {\begin{array}{*{20}{c}} {2 - 1.55}&-2\ -1&{6 - 1.55} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right] = ;\left[ {\begin{array}{*{20}{c}} 0\ 0 \end{array}} \right]\)

 \(\left[ {\begin{array}{*{20}{c}} {0.45}&-2\ -1&{4.45} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right] = ;\left[ {\begin{array}{*{20}{c}} 0\ 0 \end{array}} \right]\)

 0.45 x₁ - 2 x₂ = 0  -- (1)

  -x₁ + 4.45 x₂ = 0  -- (2)

 Solving (1) and (2) we get.

 x₁ = 2 & x₂ = 0.45 

 ∴ The eigenvector corresponding to 1.55 is \(\left[ {\begin{array}{*{20}{c}} { 2}\ 0.45 \end{array}} \right]\)

Calculation:

Plastic moment (M_p) using the static method:

ΣF_y = 0, R_A + R_C = P

ΣM_C = 0, R_A × L = P × 2L/3

R_A = 2P/3

So, moment at appoint B, M_P = 2P/3 × L/3 = 2PL/9

Plstic moment using the kinematic method:

From the comptability condition

\(\frac{L}{3} \times \theta = \frac{{2L}}{3} \times \varphi \)

⇒  θ = 2ϕ

Now, External work done = Internal work done 

\(P \times \frac{L}{3} \times \theta = {M_p}\left( {\theta + \phi } \right)\)

Put the value of θ in the above equation

\(P \times \frac{L}{3} \times 2\varphi = {M_p}\left( {2\varphi + \phi } \right)\)

\(\frac{{2PL}}{3} \times \varphi = 3{M_p} \times \varphi \)

\({M_p} = \frac{{2PL}}{9}\)

Concept:

Application of minimum potential energy:

"Among the all geometrically compatible state of structure which satisfy deflection boundary condition and force equilibrium requirement will have final stable condition when its total potential energy is minimum"

If U is the total strain energy stored in frame. The total strain energy will be minimum, when,

\(\frac{{\partial U}}{{\partial R}} = 0\)

\(U = \int {\frac{{{M^2}dx}}{{2EI}}} \)

It is an application of castigliano's theorem and based on principal of least work.

Calculation:

Let R be the propped reaction at G.

∵ We know that, the total strain energy stored in the beam

\(U = \int {\frac{{{M^2}dx}}{{2EI}}} \)

Total strain energy stored in beam, U = U_GF + U_FE

Portion GF:

M_x (x from G) = \(Rx - \frac{{w{x^2}}}{2}\)

\(U = \int {\frac{{M_x^2dx}}{{2EI}}} \)

⇒ \({U_{GF}} = \int\limits_0^L {\frac{{{{\left( {Rx - \frac{{w{x^2}}}{2}} \right)}^2}dx}}{{2EI}}} \)

⇒ \(\frac{{\partial {U_{GF}}}}{{\partial R}} = \int\limits_0^L {\frac{{2\left( {Rx - \frac{{w{x^2}}}{2}} \right)(x)dx}}{{2EI}}} \)

⇒ \(\frac{{\partial {U_{GF}}}}{{\partial R}} = \int\limits_0^L {\frac{{\left( {R{x^2} - \frac{{w{x^3}}}{2}} \right)dx}}{{EI}}} \)

\(\frac{{\partial {U_{GF}}}}{{\partial R}} = \frac{{R{L^3}}}{3} - \frac{w}{2} \times \frac{{{L^4}}}{4} = \frac{{R{L^3}}}{3} - \frac{{w{L^4}}}{8}\)..............(i)

Portion FE:

M_x ( x from F) = \(RL - \frac{{w{L^2}}}{2}\)

\(U = \int {\frac{{M_x^2dx}}{{2EI}}} \)

\({U_{FE}} = \int\limits_0^{2L} {\frac{{{{\left( {RL - \frac{{w{L^2}}}{2}} \right)}^2}dx}}{{2EI}}} \)

\(\frac{{\partial {U_{FE}}}}{{\partial R}} = \int\limits_0^{2L} {\frac{{2\left( {RL - \frac{{w{L^2}}}{2}} \right)(L)dx}}{{2EI}}} \)

\(\frac{{\partial {U_{FE}}}}{{\partial R}} = \int\limits_0^{2L} {\frac{{\left( {R{L^2} - \frac{{w{L^3}}}{2}} \right)dx}}{{EI}}} \)

\(\frac{{\partial {U_{FE}}}}{{\partial R}} = R{L^2} \times 2L - \frac{{w{L^3}}}{2} \times 2L = 2R{L^3} - w{L^4}\) ....................(ii)

According to the principal of minimum potential energy, the true value of redundant will be when, total potential energy in a frame is minimum

∴  \(\frac{{\partial U}}{{\partial R}} = 0\) 

⇒ \(\frac{{\partial {U_{GF}}}}{{\partial R}} + \frac{{\partial {U_{FE}}}}{{\partial R}} = 0\)

From equation (i) and (ii),

\(\frac{{R{L^3}}}{3} - \frac{{w{L^4}}}{8} + 2R{L^3} - w{L^4} = 0\)

\(\frac{{7R{L^3}}}{3} = \frac{9}{8}w{L^4}\)

⇒ \(R = \frac{{27}}{{56}}wL = 0.482wL\)

Explanation:

If pre-consolidation stress p_c is greater than effective overburden pressure po but less than p_o + Δp i.e., p_c < p_o + Δp, the settlement is computed in two parts.

(i) settlement for pressure p_o to p_c

(ii) settlement for pressure p_c to (p_o + Δp)

For the first part, the recompression index (C_c) is applicable, whereas, for the second part, the compression index (C_s) is used.

 \({S_c} = \frac{{{H_o}{C_s}}}{{1 + {e_o}}}\log \frac{{{p_c}}}{{{p_o}}} + {S_c} = \frac{{{H_o}{C_c}}}{{1 + {e_o}}}\log \frac{{{p_o} + \Delta p}}{{{p_c}}}\)

⇒ \({S_c} = \frac{{{H_o}}}{{1 + {e_o}}}\left[ {{C_s}\log \frac{{{p_c}}}{{{p_o}}} + {C_c}\log \frac{{{p_o} + \Delta p}}{{{p_c}}}} \right]\)

Hint: In Question, it is asked depth of flow where Specific energy is minimum for the given discharge (Q) which means critical flow condition.

Concept:

Critical flow analysis:

1. Specific energy is minimum for a given discharge.

2. Specific force is minimum for a given discharge.

3. Discharge is maximum for a given specific energy.

4. Discharge is maximum for a given specific force.

5. Froude number = Fr = 1

6. Velocity head is equal to half of the hydraulic depth, \(\frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}} = \frac{{\rm{D}}}{2}\)

Relations for critical flow:

General equation valid for the critical flow of any shape of the channel

\(\frac{{{{\rm{Q}}^2}{\rm{T}}}}{{{\rm{g}}{{\rm{A}}^3}}} = {\rm{F}}_{\rm{r}}^2\)

For critical flow in a rectangular channel

\({{\rm{Y}}_{\rm{c}}} = {\rm{critical;depth}} = {\left( {\frac{{{{\rm{Q}}^2}}}{{{\rm{g}}{{\rm{b}}^2}}}} \right)^{\frac{1}{3}}}\)

 We know Q = q × b 

 \({{\rm{Y}}_{\rm{c}}} = {\left( {\frac{{{{\rm{q}}^2}}}{{{\rm{g}}{}}}} \right)^{\frac{1}{3}}}\)

\({{\rm{E}}_{\rm{c}}} = {\rm{Specific;energy}} = \frac{3}{2} × {{\rm{Y}}_{\rm{c}}}\)

Calculation:

Given,

 Width of channel (b) = 6 m

 Discharge Q = 20 m³/sec, g = 9.81 m²/sec

 Discharge per unit width q = Q/b = 20/6 = 3.33 m³/sec/m

Critical Depth is: 

 \({{\rm{Y}}_{\rm{c}}} = {\left( {\frac{{{{\rm{q}}^2}}}{{{\rm{g}}{}}}} \right)^{\frac{1}{3}}}\)

 \({{\rm{Y}}_{\rm{c}}} = {\left( {\frac{{{{\rm{3.33}}^2}}}{{{\rm{9.81}}{}}}} \right)^{\frac{1}{3}}}\)

 y_c = 1.04 m

∴ Depth of flow in the channel at which the specific energy of the flowing water is minimum for the given discharge is 1.04 m

Explanation

a) For statement 4 & statement 2

According to Fick's law of diffusion

It is assumed that the plume has a Gaussian distribution in both z ( vertical ) and y (horizontal ) direction

The concentration of a gas or aerosol ( < 20 μ ) calculated at ground level for a distance downward x is given by 

 \({C_{x,y}} = \frac{Q}{{\pi \times U{\sigma _z} \times {\sigma _y};}}{e^{\left( { - \frac{1}{2}{{\left( {\frac{H}{{{\sigma _z}}}} \right)}^2}} \right)}} \times {e^{\left( { - \frac{1}{2}{{\left( {\frac{y}{{{\sigma _y}}}} \right)}^2}} \right)}} \)

Where,

C is the concentration of pollutant

U is mean wind speed

σ_y is the standard deviation of horizontal plume concentration, in the crosswind direction

σ_x is the standard deviation of horizontal plume concentration, in the vertical direction

H is effective stack height (m)

X is the downward distance along the plume mean centerline from a point source (m)

Y – crosswind distance from the centerline of the plume (m)    

 From the relation, we can see that the Downwind concentration of air pollutants due to an elevated point source is not directly proportional to the wind speed.

whereas it is Inversely proportional to Concentration.

Statement 4 is false. Statement 2 is true.

b) 

Category

A – Extremely unstable

B – Moderately unstable

C – Slightly unstable

D – Neutral

E – Slightly stable

F – Moderately Stable

 

For Neutral (D) Velocity >  For Very unstable (A)

∴ The value of the wind profile exponent for the 'very unstable' atmosphere is smaller than the wind profile exponent for the 'neutral' atmosphere

Statement 4 is false whereas Statement 1 is true.

Concept:

The head loss during the backwash (H_e) is given by

Hb = z × ( 1 – n ) × (S – 1) = ze × ( 1 – ne ) × ( S – 1 )

Where,

ze is the thickness of the expanded sand bed

ne is the porosity of the expanded sand bed

s is the specific gravity of sand

z is the thickness of sand bed and

n is the porosity of the sand bed before expansion

The rate of backwash VB is given by

VB = Vs × ne4.5,

where Vs is the settling velocity of sand

Given by 

\({V_s} = ;\frac{{{\rm{g}}\left( {{\rm{s}} - 1} \right){{\rm{d}}^2}}}{{18{\rm{v}}}}\)

for laminar condition

where,

d is the size of sand particles and, v is the kinematic viscosity of water

Calculation

Given,

Shape factor = 0.85 , particle diameter (d) = 0.4 mm, Specific Gravity (S) = 2.55

 Depth of filter Bed (z) = 0.75 m , Porosity (η ) = 0.35

 The porosity of expanded bed  η_e = 0.65

 Terminal velocity V_s = 4.5 cm/sec = 4.5 × 10^-2 m/sec

 The rate of backwash VB is:

VB = 4.5 × 10^-2 × (0.65)4.5

  V_B = 6.47 × 10^-3 m/sec

Concept: 

The equation for closing error and bearing of closing error is given by,

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

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

Where,

e = closing error, θ = bearing of closing error

ΣL = Algebraic sum of latitudes of all lines  

ΣD = Algebraic sum of departures of all lines

Calculation

Given,

Sum of latitude ΣL = + 2.1 m , Sum of Departure ΣD = - 2.8 m

 As the departure is -ve and Latitude is +ve Line lies in IV th Quadrant i.e. in N-W direction

 a) Length of closing error e

   \({\rm{e}} = \sqrt {{{\left( {{\rm{2.1}}} \right)}^2} + {{\left( {{\rm{-2.8}}} \right)}^2}} \) = 3.5 m

 b) Closing error

  \({\rm{\theta }} = {\tan ^{ - 1}}\left[ {\frac{{{\rm{Σ D}}}}{{{\rm{Σ L}}}}} \right]\) = \({\rm{\theta }} = {\tan ^{ - 1}}\left[ {\frac{{{\rm{-2.8}}}}{{{\rm{\ 2.1}}}}} \right]\) = - 53.13 ° 

 = 53° 7'48'' NW

Explanation

we know, 

a) Toughness Index: Toughness index is defined as the ratio of plasticity index (I_P) of the soil to the flow index (I_F) of the soil.

The toughness index varies between 0 to 3.

Given,

Liquid Limit (LL) = 60 %, Plastic Limit (PL) = 27%, Water content (w) = 32 %, and Flow indx (If) = 27

 Plasticity Index (PI) = LL - PL = 60 - 27 = 33% 

\(\begin{array}{l} Toughess;index = \frac{{Plasticity;index}}{{Flow;index}}\ \end{array}\)

Toughness index = 33/27

∴ Toughness index = 1.22

b) Liquidity Index

The Liquidity Index, IL­ is given as:

\({I_L} = \frac{{{W_n} - ;{PL}}}{{{LL} - ;{PL}}}\)

Where,

 LL - PL = Plasticity Index

\({I_L} = \frac{{{32}\ - ;{27}}}{{{60} - ;{27}}}\) = 5/33 = 0.1515

∴  Liquidity Index is 0.15

Concept

Note

i) Gradient (∇) converts the scalar function into vector function

ii) Divergence (∇.\({\rm{\bar F}}\)) Converts vector function to scalar (Dot product)

iii) Curl (∇ × \({\rm{\bar F}}\)) converts vector function to vector function (Cross product)

Calculation

Given, f(x, y) = xey

Let Point P is ( 2, 0), point Q is (1/2,2)

Line \(\overline {PQ} = \left( {1/2- 2} \right)\bar i + \left( {2 - 0} \right)\bar j \)

 \(\overline {PQ} = -\ 3/2\ \bar i + 2\bar j\)

 Unit vector along PQ is  \(\frac{{\frac{{ - 3}}{2}i + 2j}}{{\sqrt {\frac{9}{4} + 4} }}\)

The gradient of f (x, y) is given by ∇f

\(\nabla f = ;\frac{{\partial f}}{{\partial x}}\bar i + ;\frac{{\partial f}}{{\partial y}}\bar j \)

 \(= ;\frac{{\partial \left( {xe^y} \right)}}{{\partial x}};\bar i + ;\frac{{\partial \left( {xe^y} \right)}}{{\partial y}};\bar j \)

  \( = {e^y}i + x{e^y}j\)

 Directional derivative at point P (2,0) is

 = i + 2 j 

 The directional derivative of f (x, y) at P (2, 0) along the line \(\overline {PQ} \) is

 \(D.D = {\left( {\nabla f} \right)_p}.\frac{{\overline {PQ} }}{{\left| {\overline {PQ} } \right|}}\)

  \( = \frac{{\left( {i + 2j} \right).\left( {\frac{{ - 3}}{2}i + 2j} \right)}}{{\frac{5}{2}}}\)

   \( = \frac{{\left( {\frac{{ - 3}}{2} + 4} \right)}}{{\frac{5}{2}}} = \frac{{5/2}}{{5/2}} = 1\)

Concept:

This is a similar condition to the column subjected to a load P.

Here buckling happens due to the weight of water in the tank.

It is mentioned to ignore the self-weight so only the weight of water is considered

Weight of water W = ρ × g × Volume of tank = Buckling Load = P_cr

We know Buckling load is given by

where Le = effective length of a column

Here It is a case of one end is fixed and the other is free ∴ Effective length = 2 L

Calculation:

Given:

Height of water tank = h

The diameter of water tank = 1.5 m

E of steel = 20GPa, Length of steel column L = 4 m , Effective Length (L_e) = 2 × 4 = 8 m

Diameter of column = 75 mm = 75 × 10^-3 m

Density of water = 1000 kg/m³, g = 10 m/s²

Weight of water W = ρ × g × Volume of tank = ρ × g × Area of tank × h

Area of Tank = π × (1.5)²/4

Inertia of column = π D⁴/64 = π × (75 × 10-3 )⁴/64

Buckling load = weight of water

h = 2.7 m

∴ Maximum depth of water tank = 2.7 m

Additional Information

The value of the effective length varies according to the support condition.

Explanation

Degree of freedom:

The degrees of freedom (DOF) of a rigid body is defined as the number of independent movements.

To determine the DOF of this body we must consider how many distinct ways the bar can be moved.

Single degree of freedom system says it has one degree of freedom

Given,

Flexural Stiffness of beam = 4 π² kN/m

Mass = 10 kg

Given it is SDOF (i.e. it has 1 degree of freedom)

We know natural frequency is given by 

\(f = \frac{1}{{2\pi }} \times \sqrt {\left( {\frac{k}{m}} \right)} \)

Where k is stiffness, m is mass

\(f = \frac{1}{{2\pi }} \times \sqrt {\frac{{4 \times {\pi ^2} \times {{10}^3};N/m;}}{{10;kg}}} = \frac{{2\pi }}{{2\pi }} \times 10 = 10;Hz\)

∴ Frequency is 10 Hz

Concepts:

The cable can carry on tension force i.e. Bending moment at every point on cable is zero.

Using this concept along with equilibrium equation in X and Y direction support reactions developed at A and B can be evaluated easily.

Calculation:

Take moment at F is zero, ΣM_F = 0 (take left of F portion)

H_E × 3 + V_E × 2 = 0

H_E = (2/3) × V_E …………..(i)

Take moment at H = 0

ΣM_H = 0

V_E × 6 – H_E × 1 -10 × 4 – 12 × 2 = 0

6V_E – H_E = 64…………(ii)

Put the value of HE in equation (ii)

6V_E - (2/3) × V_E = 64

V_E = 12 kN

From (i), H_E = 8 kN

Tension in the cable segment EF

\({F_{EF}} = \sqrt {V_E^2 + H_E^2} = \sqrt {{{12}^2} + {8^2}} \)

F_EF = 4√13 kN

Free body diagram of cable segment EF

ΣF_H = 0, F_FG × cosθ₁ = 8 …………..(iii)

ΣF_V = 0, F_FG × sinθ₁ + 10 = 12

F_FG sinθ₁ = 2…………..(iv)

From (iii) and (iv), (F_FG)² = 8² + 2² = 68

F_FG = 8.25 kN

Explanation

The water jet performs projectile motion

We know in a projectile motion horizontal component of velocity remains constant

∴ V₂ = V₁ cos θ

Given velocity of water V = 50 m/s, θ = 45°

V₂ = 50 cos 45° = 35.36 m/s

As the energy is conserved

Applying Bernoulli’s theorem 

 \(\frac{P_1}{\rho_1 g}+\frac{V_1^2}{2g}+Z_1=\frac{P_2}{\rho_2 g}+\frac{V_2^2}{2g}+Z_2 + h_L\)

In both sections

Pressure remains the same as it is exposed to the atmosphere i.e. P₁ = P₂

Datum is same

The equation now becomes 

\(\frac{{V_1^2}}{{2g}} = \frac{{V_2^2}}{{2g}} + h\)

\(h = \frac{{V_1^2}}{{2g}} - \frac{{V_2^2}}{{2g}} = \frac{{{{50}^2} - {{35.36}^2}}}{{2 \times 9.81}} = 63.71;m\)

∴ The maximum height reached by stream above hose nozzle will be 63.71 m

Explanation

 To determine the centroidal distance from the center of reinforcement we need to make the body of uniform material.

 i.e. Normal section to Equivalent Section.

 

Nominal to Equivalent Section

For the Equivalent Concrete section

Equivalent area = Gross Area – Area of steel + m × Area of steel

Equivalent area = Gross area + (m - 1) × Area of steel

Where m is modular ratio

As

Steel Section is converted to concrete.(E_s > E_c) Width of concrete increases.

Change in width = \(\left( {m - 1} \right) × \frac{{{A_{st}}}}{{Depth;of;steel;section}}\)

Given,

The diameter of reinforcing bar = 16 mm,

Effective depth d = 350 - 35 = 315 mm

 Es = 2.1 × 105 MPa

 Ec = 2 × 104 MPa

Area of steel A_st = 3×π×16²/4 = 603.186 mm²

Modular ratio (m) = E_s/E_c = (2.1 × 105)/(2 × 104) = 10.5

Calculation of Centroid of the equivalent section from Top fiber:

\(NA = \frac{{{A_1} \times {y_1} + {A_2} \times {y_2}}}{{{A_1} + {A_2}}}\)

\(NA = \frac{{\left( {200 \times 350} \right) \times \frac{{350}}{2} + \left( {m - 1} \right) \times {A_{st}} \times 315}}{{\left( {200 \times 350} \right) + \left( {m - 1} \right) \times {A_{st}}}}\)

\(NA = \frac{{\left( {200 \times 350} \right) \times \frac{{350}}{2} + 9.5 \times 603.186 \times 315}}{{\left( {200 \times 350} \right) + 9.5 \times 603.186}}\)

NA = 185.6 mm

Value of X = d - NA = 315 - 185.6 

X = 129.407 mm

Explanation

Given,

 

 Following the given data if we draw the network we get this.

 

 ∴ The total time of the project is 56 days

Explanation

Book Value by

a) By Straight line method 

Book value (Bm) at the end of 'm ' years life is given by 

\({B_m} = {C_i} - m × {D_m}\)

Where Dm is depreciation 

Ci is the initial cost of an asset

Cs is Salvage or scrap value (Estimated at the end of utility period)

Depreciation (D_m) is given by 

\({D_m} = \frac{{{C_i} - {C_s}}}{n}\)

Given,

 Initial Cost (C_i) = 1,60,000 Salvage Value (C_s) = 10000

 Estimated Life (n) = 5 years

 Asked, at 4^th year of life (m) = 4

Book Value is 

\(Book;Value = 1,60,000 - 4 \times \frac{{1,60,000 - 10,000;}}{5}\)

Book Value = 1,60,000 – 1,20,000

Book Value = 40,000/-

b) By Sum of years Digit Method:

\(BV = {C_i} - \frac{{\left( {{C_i} - {C_s}} \right)}}{{n \times \frac{{n + 1}}{2}}} \times \left( {mn - \frac{{m\left( {m - 1} \right)}}{2}} \right)\)

\(BV = 160000 - \frac{{\left( {160000 - 10000} \right)}}{{5 \times \frac{{5 + 1}}{2}}} \times \left( {4 \times 5 - \frac{{4\left( {4 - 1} \right)}}{2}} \right)\)

\(BV = 160000 - \frac{{150000}}{{15}} \times 14\)

Book Value = 20,000

∴ The difference between the book values (in Rs., in integer) obtained at the end of 4^th year using the straight-line method and the sum of years digit method of depreciation is 40,000 - 20,000 = 20,000/-

Concept:

Meyerhoff's Method:

(i) Mayerhoff's theory takes into account depths, shape, and inclination factors together in the bearing capacity equation.

(ii) It can be applied to shallow, deep and foundations to all shapes and loading conditions, except foundations on built-up shapes.

(iii) The ultimate bearing capacity is given by

q_u = c N_c S_c d_c i_c + γD_f N_q S_q d_q i_q + 0.5Bγ N_γ S_γ d_γ i_γ

Where,

S_c, S_q, S_γ are shape correction factor

d_c, d_q, d_γ are depth correction factor

i_c, i_q, i_γ are inclination correction factor

Calculation:

Given,

Soil is clay for which, γ = 18.2 kN/m³, ϕ = 0°, c = 40 kN/m²

Footing size = (2.8 × 3.5) m, D_f = 1.5 m

For ϕ = 0°, N_c = 5.14, N_q = 1 and N_γ = 0

Shape factors: S_c = 1.16, S_q = 1.0 and S_γ = 1.0

Depth factors: d_c = 1.1, d_q = 1.0 and d_γ = 1.0

Inclination factors: i_c = 1.0, i_q = 1.0, i_γ = 1.0

Given loading is eccentric for which, e = 0.8 m

B' = 2(B/2 - e) = (B - 2e) = 2.8 - 2 × 0.8 = 1.2 m

L' = L = 3.5 m

The ultimate bearing capacity  is given by,

qu = c Nc Sc dc ic + γDf Nq Sq dq iq + 0.5B'γ Nγ Sγ dγ i_γ 

∵ N_γ = 0, so

qu = c Nc Sc dc ic + γDf Nq Sq dq i_q 

q_u = (40 × 5.14) × (1.16 × 1.1 × 1) + (18.2 × 1.5) × (1) × (1 × 1 × 1) = 289.64 kN/m²

Net ultimate bearing capacity (q_nu) = q_u - γD_f = 289.64 - 18.2 × 1.5 = 262.35 kN/m²

Net safe bearing capacity (q_ns) = (qu - γDf)/Fos = 262.35/2.5 = 104.94 kN/m²

Net safe Load that can be applied on the footing is the (Net safe bearing capacity) × L'B'

So, Net safe load = 104.94 × 3.5 × 1.2 = 440.75 kN

Concept:

 

Primary settlement is given by when C_c is known:

 \(Δ H = \frac{{{H_o}{C_c}}}{{1 + e}}\log \left( {\frac{{\bar \sigma + Δ \bar \sigma }}{{\bar \sigma }}} \right)\)

Where, σ̅ = Initial effective overburden pressure at the Centre of layer C-C

σ̅ + Δσ̅ = Effective overburden pressure after loading

C_c = Compression index

e = Initial void ratio

Calculation:

Given,

σ̅ + Δσ̅ = Eff. stress at the Centre of clay layer after embankment loading = 180 kN/m²

γ_w = 9.81 kN/m³

H = 6 m

γ_sub of clay layer, \({\gamma _{sub}} = \frac{{G - 1}}{{1 + e}}{\gamma _w}\) ...............(i)

∵ We know that, Se = wG

e = wG {∵ S = 1 for consolidation)

e = 0.45 × 2.65 = 1.1925

From (i), \({\gamma _{sub}} = \frac{{G - 1}}{{1 + e}}{\gamma _w} = \frac{{2.65 - 1}}{{1 + 1.1925}} \times 9.81\)

γ_sub = 7.383 kN/m³

σ̅  = Eff. stress at the centre of clay layer before embankment loading

σ̅   = 18.5 × 2 + 7.383 × 3 = 59.149 kN/m²

Primary settelement is given by, \(Δ H = \frac{{{H_o}{C_c}}}{{1 + e}}\log \left( {\frac{{\bar \sigma + Δ \bar \sigma }}{{\bar \sigma }}} \right)\)

\(Δ H = \frac{{6 \times 0.25}}{{1 + 1.1925}}\log \left( {\frac{{180}}{{59.149}}} \right)\)

ΔH = 0.33 m

Concept:

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

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

where,

b is the upper limit, a is the lower limit, h is the step size

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, Upper limit b = 0.5,

Lower limit a = 0, divided into 5 divisions 

Step size (h) = (0.5 - 0)/5 = 0.1 

 f(x) = 10x - 20x2 

 

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{5}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}+ {{\rm{y}}_4}{\rm{;}}} \right)} \right]\)

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{5}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}+ {{\rm{y}}_4}{\rm{;}}} \right)} \right]\)

\(I = \frac{0.1}{2} \times \left[ {\left( {0 + 0} \right) + 2\left( {0.8 + 1.2 + 1.2 + 0.8} \right)} \right]\)

 I = 0.4

Concept:

Coefficient of Consolidation:

• It is defined as the parameter used to measure the rate at which the saturated clay or soil undergoes consolidation when subjected to an increase in pressure. It is usually measured in terms of area per unit time.
• It is closely related to the liquid limit of the soil and decreases as the liquid limit of soil increases.
• The coefficient of consolidation is given by, \({{\rm{c}}_{\rm{v}}} = \frac{{\rm{k}}}{{{{\rm{m}}_{\rm{v}}}{{\rm{γ }}_{\rm{w}}}}}\)

where C_V is coefficient of consolidation,

 k is the coefficient of permeability of the soil, 

 m_v is the coefficient of volume compressibility and γ_w is the unit weight of water. 

We know, \({{\rm{m}}_{\rm{v}}} = \frac{{{{\rm{a}}_{\rm{v}}}}}{{1 + {{\rm{e}}_{\rm{o}}}}}\) 

  \({m_v} = \frac{{ - \left( {\frac{{{\rm{\Delta }}e}}{{{\rm{\Delta }}\bar \sigma }}} \right)}}{{1 + {e_o}}}\)

Δe = Change in void ratio of the soil

eo = initial void ratio

 \({\rm{\Delta }}\bar \sigma =Change \ in:effective;stress\)

 The ratio of the coefficient of consolidation of different samples is given by 

  \(\frac{{{C_{v1}}}}{{{C_{v2}}}} = \frac{{{k_1}}}{{{k_2}}} \times \frac{{{m_{v2}}}}{{{m_{v1}}}}\)

Calculation

Given,

There are 2 samples 

For M sample 

 Initial Void ratio = 0.575 

 Final void ratio = 0.510

 Change in effective stress = 180- 120 = 60 kN/m²

 \({m_v} = ;\frac{{{\rm{\Delta }}e}}{{1 + {e_0}}} \times \frac{1}{{{\rm{\Delta }}\sigma '}}\)

 \(= ;\frac{{0.575 - 0.51}}{{1 + 0.575}} \times \frac{1}{{60}}\)

 m_v = 6.878 × 10^-4 m²/kN

 For N sample 

Change in effective stress = 60 kN/m²

 Initial Void Ratio = 0.600

 Final Void Ratio = 0.550 

\({m_v} = ;\frac{{{\rm{\Delta }}e}}{{1 + {e_0}}} \times \frac{1}{{{\rm{\Delta }}\sigma '}}\)

\( = ;\frac{{0.6 - 0.55}}{{1 + 0.60}} \times \frac{1}{{60}}\)

 m_v = 5.208 × 10^-4 m²/kN

The ratio of Hydraulic Conductivity of M sample to N sample = 0.125 

 So the coefficient of consolidation ratios is 

 \(\frac{{{C_{vM}}}}{{{C_{vN}}}} = \frac{{{k_M}}}{{{k_N}}} \times \frac{{{m_{vN}}}}{{{m_{vM}}}}\)

 \( = 0.125 \times \frac{{5.20 \times {{10}^{ - 4}}}}{{6.878 \times {{10}^{ - 4}}}}\)

 = 0.0946 

∴ The ratio of the coefficient of consolidation is 0.0946

Concept:

Φ-Index:

(i) The ϕ index is the average rainfall above which the rainfall volume is equal to the runoff volume. The ϕ-index is derived from the rainfall hyetograph with the edge of the resulting run-off volume.

 \(\phi = \frac{{P - R}}{t}\)

Where

P = Total rainfall

R = Total runoff

t = Duration of rainfall excess

Calculation:

Given,

Total rainfall = 3 cm = 30 mm

Direct runoff = 1.6 cm = 16 mm

Time of rainfall = 3.5 hr

Considering the whole period of rainfall to be the period of rainfall excess.

 ∴ \(\phi = \frac{{30 - 16}}{{3.5}} = 4\ mm/hr\)

Since the time period (2.0 to 2.5 hr) and (3.0 to 3.5 hr) are not the period of rainfall excess, these periods are not considered in the calculation of ϕ - index.

 \(\phi = \frac{{30 - 0.5 \times 3 - 0.5 \times 4 - 16}}{{3.5 - 2 \times 0.5}} = 4.2\ mm/hr\)

Concept:

Venturimeter:

(i) It is a device used for measuring the rate of flow of fluid through pipe.

(ii) The actual discharge is given by;

\(Q = {C_d}\frac{{{a_1}{a_2}\sqrt {2gh} }}{{\sqrt {a_1^2 - a_2^2} }}\)

where, h = Difference between pressure head knows as venturi head = \(x\left( {\frac{{{ρ _m}}}{ρ } - 1} \right)\)

x = Reading of venturimeter or deflection in the manometer

ρ_m = Manometric fluid

ρ = Standard fluid

a₁ = Area of pipe, a₂ = Area of throat

Calculation:

Given,

Diameter of pipe = 20 cm and Diameter of throat = 10 cm

x = 15 cm, ρ_m = 13.6, ρ = 1

Piezometric head difference, \(h = x\left( {\frac{{{\rho _m}}}{\rho } - 1} \right)\)

⇒ \(h = 0.15\left( {\frac{{13.6}}{1} - 1} \right) = 1.89\ m\)

Area of pipe, \(a1 = \frac{\pi }{4} \times d_1^2 = \frac{\pi }{4} \times {0.2^2}\)

Area of throat, \({a_2} = \frac{\pi }{4} \times d_2^2 = \frac{\pi }{4} \times {0.1^2}\)

For no loss, coefficient of discharge, C_d = 1

The actual discharge is given by; \(Q = {C_d}\frac{{{a_1}{a_2}\sqrt {2gh} }}{{\sqrt {a_1^2 - a_2^2} }}\)

\(Q = 1 \times \frac{{\frac{\pi }{4} \times {{0.2}^2} \times \frac{\pi }{4} \times {{0.1}^2} \times \sqrt {2 \times 9.81 \times 1.89} }}{{\sqrt {{{\left( {\frac{\pi }{4} \times {{0.2}^2}} \right)}^2} - {{\left( {\frac{\pi }{4} \times {{0.1}^2}} \right)}^2}} }}\)

Q = 0.049395 m³/s = 49.39 liter/sec

Concept:

Duty: The duty of water is the relationship between the volume of water and the area of the crop it matures. It is defined as the area irrigated per cumec of discharge running for base period B. The duty is generally represented by D.

 \(D = \frac{{8.64B}}{\Delta }\)

Where,

Δ = Total depth of water required (in meter)

D = Duty in ha/cumec

B = Base period (in days)

Calculation:

Given:

B = 120 days

Live storage = 300 Mm³

Crop Area = 40000 hectares

Since the tank gets filled twice a year.

∴ Total Volume of water required for the crop area = 2 × 300 = 600 Mm³

⇒ Total depth of water required (Δ), \(\Delta = \frac{{600 \times {{10}^6}}}{{400 \times {{10}^4}}} = 1.5\ m\)

⇒ \(\rm D = \frac{{8.64 \times 120}}{{1.5}} = 691.2\ ha/cumec\)

Concept: 

Activated Sludge Process (ASP)

ASP works on a suspended culture system. In this process, food and microorganisms both are in suspension and microorganisms chase food.

ASP consists of two units

1) Aeration tank: Food and oxygen supply to organisms

2) Secondary clarifier: Microorganisms settle as biomass

Yi =  B.O.D of influent, Ye = B.O.D of effluent

Xe = Mixed Liquor Volatile Suspended Solids (MLSS) in the effluent

Qe = Flow rate of influent, Qu = Under flow rate

V = Volume of the aeration tank

Xu = MLSS in under flow

Qr = Rate of return sludge

1) Food to microorganisms ratio [F/M ratio]:

\(\frac{{\rm{F}}}{{\rm{M}}} = \frac{{{\rm{Q}}\left( {{{\rm{Y}}_{\rm{i}}} - {{\rm{Y}}_{\rm{o}}}} \right)}}{{{\rm{VX}}}}\)

Generally, F/M ranges from 0.2 to 0.4 Day-1

2) Mean Cell Residence Time (M.C.R.T or θC):

 \({\rm{M}}.{\rm{C}}.{\rm{R}}.{\rm{T}} = \frac{{{\rm{VX}}}}{{{{\rm{Q}}_{\rm{w}}}{{\rm{X}}_{\rm{u}}} + \left( {{\rm{Q}} - {{\rm{Q}}_{\rm{w}}}} \right){{\rm{X}}_{\rm{e}}}}}\)

Calculation

Given,

Discharge of municipalwaste water (Q) = 7500 m³/day

Ultimate BOD of influent (Y_i) = 200 mg/l  &  Ultimate BOD of effluent (Y_e) = 20 mg/l       

X = 3000 mg/l, HRT = 0.2 day

 V = Volume of the aeration tank = Q × HRT = 7500 × 0.2

X_U = 12000 mg/l

MCRT (θ_C) = 10 days

 \({\theta _c} = \frac{{VX}}{{{Q_w}{X_U} + \left( {Q - {Q_w}} \right){X_e}}}\)

 As X_e is not mentioned it is assumed as 0

Equation Becomes 

 \( {\theta _c} = \frac{{VX}}{{{Q_w}{X_u}}}\)

 \({Q_w} = \frac{{VX}}{{{\theta _c}× X_U}}\)

 \(\Rightarrow {Q_w} = ;\frac{{\left( {7500 × 0.2} \right) × 3000}}{{10 × 12000}}\)

  Q_w = 37.5 m³/day

∴ The underflow volume of sludge wastage is 37.5 m3/day

Explanation:

Grit chambers are a basin to remove the inorganic particles to prevent damage to the pumps, and to prevent their accumulation in sludge digesters.

Grit chambers are nothing but sedimentation tanks, designed to separate the intended heavier inorganic materials (specific gravity about 2.65) and to pass forward the lighter organic materials.

 For 100 % removal

The settling velocity can be determined through the equation,

\( \frac{H}{{{V_S}}} = ;\frac{L}{{{V_H}}}\)

where H = depth of the tank, L = length of the tank,

Vs  = Settling velocity, and VH = Flow-through velocity

Given,

Grit chamber is designed to remove particle of 0.25 mm dia 

 Terminal velocity Vs = 2.5 cm/sec = 0.025 m/sec

 Width of chamber = 0.50 m, Depth of flow H = 0.75 m 

 Flow through velocity V_H = 0.3 m/sec, L = ? 

 As it is asked to remove 0.25 mm particle completely efficiency is 100 % 

 From the equation 

 \( \frac{H}{{{V_S}}} = ;\frac{L}{{{V_H}}}\)

 \( \frac{0.75}{{{0.025}}} = ;\frac{L}{{{0.3}}}\)

 L = 9

 ∴ Length of chamber = 9 m

Concept:

Theoretical Specific gravity or apparent specific gravity:

It is the specific gravity without considering air voids is given as

 \({G_t} = \frac{{{W_1} + {W_2} + {W_3} + {W_4}}}{{\frac{{{W_1}}}{{{G_1}}} + \frac{{{W_2}}}{{{G_2}}} + \frac{{{W_3}}}{{{G_3}}} + \frac{{{W_4}}}{{{G_4}}}}}\)

If weight of constituents are given in percent of total mix, then

 \({G_t} = \frac{{100}}{{\frac{{{W_1}(% )}}{{{G_1}}} + \frac{{{W_2}(% )}}{{{G_2}}} + \frac{{{W_3}(% )}}{{{G_3}}} + \frac{{{W_4}(% )}}{{{G_4}}}}}\)

Calculation:

Given,

Proportions of coarse aggregate (W₁%) = 55%

Proportions of fine aggregate (W₂%) = 40%

Proportions of mineral filler (W₃%) = 5%

G₁ = 2.55, G₂ = 2.65, G₃ = 2.70

Theoretical specific gravity is given by, \({G_t} = \frac{{100}}{{\frac{{{W_1}(% )}}{{{G_1}}} + \frac{{{W_2}(% )}}{{{G_2}}} + \frac{{{W_3}(% )}}{{{G_3}}}}}\)

 \({G_t} = \frac{{100}}{{\frac{{55}}{{2.55}} + \frac{{40}}{{2.65}} + \frac{5}{{2.70}}}}\)

G_t = 2.60

Explanation

We know.

The minimum distance available on a highway at any spot should be of sufficient length to stop a vehicle travelling at design speed safely without collision with any other obstruction is called SSD

Stopping sight distance (SSD) = Lag distance + Braking distance

 SSD =  \(vt + \frac{{{v^2}}}{{2gf}}\)

Where, v =  design speed of vehicle in m/s

f = Coefficient of friction

We know retarding acceleration (a) = f  × g

Given

SSD = 140 m, Design speed V = 90 km/h = 25 m/sec

 Acceleration due to gravity (g) = 9.81 m/s2, Deceleration rate (a) = 3.5 m/s2

 \(SSD = vt + \frac{{{v^2}}}{{2gf}}\) 

  Substituting f = a/g

  \(140 = vt + \frac{{{v^2}}}{{2g\left( {\frac{a}{g}} \right)}}\)

  \(140 = 25 \times t + \frac{{{{25}^2}}}{{2 \times 3.5}}\)

 140 = 25 × t + 89.29 

 t = 2.028

 ∴ Perception time is 2.028 sec

Concept

For the Indian railway's track

Theoretical Cant is given by:

\({e_{th}} = \frac{{G \times V_{max}^2}}{{127;R}}\)

Actual cant or equilibrium cant provided is given by 

\({e_{act}} = \frac{{G \times V_{avg}^2}}{{127 \times R;}}\)

Where,

eth is the maximum allowable super-elevation in (m) 

Vmax is the maximum speed on the curve in kmph,

V_avg is the average speed on the curve

G is the gauge length of the track in m and

R is the radius of the curve in m where

D is the curvature of the curve in degree.

,\({\rm{R}} = \frac{{1750}}{{\rm{D}}}\), 

\({e_{th}} = {e_{act}} + CD\)

Where,

eact is the actual or equilibrium super-elevation 

CD is Cant deficiency

Calculation:

Given,

Degree of curve = 2° 

Type of Gauge is Broad Gauge (BG) G = 1750 mm

 V_max = 100 kmph, V_avg = 80 kmph

Radius of curve  \(R = \frac{{1750}}{D}\) = 1750/2 = 875 m

Taking V_max theoretical cant is 

\({e_{th}} = \frac{{G \times V_{max}^2}}{{127;R}}\)

\({e_{th}} = \frac{{1.750 \times {{100}^2}}}{{127 \times 875;}}\) = 157.480 mm

Taking Veq equilibrium cant is 

\({e_{act}} = \frac{{G \times V_{avg}^2}}{{127 \times R;}}\)

 \({e_{eq}} = \frac{{1.750 \times {{80}^2}}}{{127 \times 875;}}\) = 100.787 mm

∴ Cant Deficiency = 157.480 - 100.787 = 56.693 mm

 ∴ Cant Deficiency is 57 mm as in Question it is asked to roundoff to integer.