GATE CE 2012 Official Paper (Previous Year Paper)

The correct answer is option 2- setbacks

Explanation   

Setback: something that happens that delays or prevents a process from developing

From the given context, it is clear that the mission finally resolved the conflict after many failed attempts and obstacles. Thus, 'setback' is the correct word for the given blank.

   

The meaning of the other words:

• Attempt: an effort to achieve or complete a difficult task or action
• Meeting: an assembly of people for a particular purpose, especially for formal discussion
• Delegation: a group of people who have been sent somewhere to have talks with other people on behalf of a larger group of people.

You may be confused regarding the use of 'attempts' and 'setbacks'. Attempt does not have any negative connotation to it, whereas setback has one. As the sentence is specifically referring to fruitless attempts taken in the past, 'setbacks' would be a more apt choice.

As q = total number of quantities produced.

Total cost = 5q²

Total sales = 50q

Profit (P) = total cost – total sales

P = 5q² – 50q

Differentiating w.r.t q

$\frac{{dP}}{{dq}} = 10q - 50 = 0$

$q = \frac{{50}}{{10}} = 5$

$\frac{{{d^2}P}}{{d{q^2}}} = 10 > 0$

∴ At q = 5, profit (P) is maximum.

The correct answer is option 1- that

Explanation 

To answer this question, we need to understand the meaning of defining and non-defining clauses. 

• A defining clause (also called an essential clause or a restrictive clause) gives information essential to the meaning of the sentence. We always use 'that' with defining clauses.

Example: My bike that has a broken seat is in the garage.

• Unlike defining clauses, non-defining clauses (also called nonessential or nonrestrictive clauses) don’t limit the meaning of the sentence. You might lose interesting details if you remove them, but the meaning of the sentence wouldn’t change. We always use 'which' with non-defining clauses.

Example: The goat was standing on the roof of the house which was abandoned.

In the given sentence, if we consider the first clause, the information given is incomplete and it can only be completed via the introduction of the second clause. Hence, we have to use 'that' in this sentence.

The correct answer is option 4. 

Explanation 

As we can see, the question has been framed to test our grammatical knowledge regarding the use of the adjective 'less'. 

• 'Less' has been used correctly in option 1. Here, 'less' means 'minus'.
• 'Less' has been used correctly in option 2 as well. It has been used here to compare the country's expenditure to that of Bangladesh.
• 'Less' has been used correctly in option 3 as well. If we don't use the word 'than' after 'less', we have to use the comparative form of less- 'lesser'.
• The usage of 'less' is wrong in option 4. 'Less' is generally used as a way of comparing two quantities or amounts. Here, the word 'less' needs to be replaced with the words 'little' or 'low'.

The correct answer is 'Diminish'.

Key Points 

• The given word 'Mitigate' means make (something bad) less severe, serious, or painful.
• Example: "drainage schemes have helped to mitigate this problem"
• Synonyms are Diminish, lessen etc.
• Antonyms are increase, Intensify

Given:

y = 2x – 0.1x²

differentiating w.r.t x

$\frac{{dy}}{{dx}} = 2 - 0.2x = 0$

$x = \frac{2}{{0.2}} = 10$

${\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 10}} = - 0.2 < 0$

∴ At x = 10, y will have a maximum value.  

y_max = y(10) = (2 × 10) - 0.1(10²) = 10

Explanation:

• Clearly, in the given question there is nothing related to gender is mentioned in the advertisement, eliminating options 1) and 3)
• Xenophobic means having or showing a dislike of or prejudice against people from other countries. There, is no such thing related to people of other countries is mentioned in the advertisement, eliminating it too.
• Only general requirements are mentioned. So option 4 is the most suitable answer.

The given sequence is:

Explanation:

Let E(x) represent mean

Then by the property of mean

E(Cx) = cE(x) where c is constant

And E(x + c) = E(x) + E(c) = E(x) + c

∴ Options P and R always holds true.

Now,

Let Var (x) represent the variance

Then var (x + c) = var (x) + var (c)

 var (x + c) = var (x)    {∵ var (c) = 0}

∴ S.D (x + c) = S.D (x)

Now,

var (cx) =(c²)var (x)

S.D (cx) = (c)S.D (x)

∴ Options Q and S are incorrect.

Calculation: 

Let 100 shock absorbers are supplied

so, X supplies = 60 (60% of 100)

Y supplies = 40 (40% of 100)

Reliable supply by X = 96% of 60 = 57.6

Reliable supply by Y = 72% of 40 = 28.8

Required probability $P\left( Y \right) = \frac{{28.8}}{{57.6 + 28.8}} = 0.33$

Concept:

The Simpson’s rule is given by,

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right);dx = \dfrac{h}{3}[\left( {{y_0} + {y_n}} \right) + 2\left( {{y_2} + {y_4} + \ldots + {y_{n - 2}}} \right) + 4\left( {{y_1} + {y_3} + \ldots + {y_{n - 1}}} \right)\)

h –Width of interval / Step length

y₀, y₁, …y_n – Ordinates corresponding to x₀, x₁, ……x_n

Error =|Exact value – Approximate value|

Calculation:

Given:

\({\rm{I}} = \mathop \smallint \nolimits_{0.5}^{1.5} \dfrac{{{\rm{dx}}}}{x}\)

Three point function ∴ 0.5, 1, 1.5

h = 0.5

x = 0.5, y = 2

x = 1, y = 1

x = 1.5, y = 0.666

 

\(\mathop \smallint \nolimits_{0.5}^{1.5} \dfrac{{dx}}{x} = \dfrac{{0.5}}{3}\left[ {\left( {2 + 0.666} \right) + 4\left( 1 \right)} \right] = 1.111\)

The exact value of the given integral is

\(\mathop \smallint \nolimits_{0.5}^{1.5} \dfrac{{dx}}{x} = \left[ {\ln x} \right]_{0.5}^{1.5} = \ln 1.5 - \ln0.5 = 1.0986\)

Exact value = 1.0986

Error =|1.0986 –1.111| = 0.012

Concept:

• The normal distribution is a probability function that describes how the values of a variable are distributed.
• For a normally distributed variable x with mean μ and standard deviation σ, the normal variate z is given by the formula: $\rm z = \dfrac{x - \mu}{\sigma}$.

Calculation:

Given

Standard deviation σ = 200 mm, Mean μ = 1000 mm.

For x = 1200, $\rm z = \dfrac{x - \mu}{\sigma}=\dfrac{1200-1000}{200}=1$

P (X > 1200 mm) = P ( z > 1)

z is normal variate, 

We know (P ( - 1 < Z < 1 ) = 0.68 (i.e. 68% of data is within one standard deviation of mean)

 P ( 0 < Z < 1 ) = 0.68/2 = 0.34

So P (z >1) = 0.5 - 0.34 = 0.16 % < 50 %

Concept:

The Maclaurin’s series is given by

$f\left( x \right) = f\left( 0 \right) + x;f'\left( 0 \right) + \dfrac{{{x^2}}}{{2!}}f''\left( 0 \right) + \dfrac{{{x^3}}}{{3!}}f'''\left( 0 \right) + \ldots \infty$

Calculation:

Given:

$f\left( x \right) = 1 + x; + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \ldots$

Comparing the given series with the standard series, we get

f (0) = 0, f’(0) = 0, f’’(0) = 0

The only option satisfying this condition is e^x

Concept:

Poisson’s Ratio: Simon Poisson pointed out that within the elastic limit, lateral strain is directly proportional to the longitudinal strain. The ratio of the lateral strain to the longitudinal strain in a stretched wire is called Poisson’s ratio.

Here lateral strain the strain perpendicular to the applied force and its is given as (ΔR/R)

$\sigma = - \frac{{laterial;strain}}{{longitudinal;strain}} = ; - \frac{{\frac{{{\rm{\Delta }}R}}{R}}}{{\frac{{{\rm{\Delta }}L}}{L}}}; = - \frac{{{\rm{\Delta }}R}}{R} \times \frac{L}{{{\rm{\Delta }}L}}$

Here negative sign shows that as length of wire increases, the radius of wire will decrease.

The normal value of σ lies between -1 and +1/2

Whereas practical value of σ lies between 0 to +1/2

And just like strain Poisson’s ratio is also Dimensionless.

Note:

For a rigid body, the value of Poisson’s ratio is zero. A zero Poisson’s ratio means that there is no transverse deformation resulting from an axial strain.

• Most materials have Poisson's ratio values ranging between 0.0 and 0.5.
• A perfectly incompressible material deformed elastically at small strains would have a Poisson's ratio of exactly 0.5.
• Most steels and rigid polymers when used within their design limits (before yield) exhibit values of about 0.3, increasing to 0.5 for post-yield deformation which occurs largely at constant volume.
• Rubber has a Poisson ratio of nearly 0.5.
• Cork's Poisson ratio is close to 0, showing very little lateral expansion when compressed.

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Explanation:

From the above explanation we can see that Poisson’s ration is define as lateral strain upon longitudinal strain

i.e.,$;\sigma = - \frac{{laterial;strain}}{{longitudinal;strain}}$

Thus option 1 is correct among all

Explanation:

Relationship between Shear force, bending moment and Loading rate

• The slope of Shear force diagram at any section will be equal to the load intensity at that section

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$\dfrac{{{\rm{dS}}}}{{{\rm{;dx}}}} = {\rm{W}}$

Hence statement 2 is true.

• The slope of Bending moment diagram at any section of a loaded beam will be equal to Intensity of shear force at that section

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$\dfrac{{{\rm{dM}}}}{{{\rm{;dx}}}} = {\rm{S}}$

Hence statement 1 is true.

Curvature is given by

$\dfrac{{\rm{M}}}{{{\rm{EI}}}} = \dfrac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \dfrac{1}{{\rm{R}}}$

Slope of curvature is given by

$\dfrac{{{{\rm{d}}^3}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^3}}} = \dfrac{d}{{dx}}\dfrac{{\rm{M}}}{{{\rm{EI}}}} = \dfrac{V}{{EI}}$

Hence statement 3 is false.

y is the deflection. The second derivative is given by

$\dfrac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \dfrac{1}{{\rm{R}}}$

Hence statement 4 is true.

Concept:

As the Cube is submerged Deep in still water this indicates it is subjected to hydrostatic stress

Hydrostatic loading / Hydrostatic stress:

In the case of hydrostatic fluid, equal and alike normal stress acts on two mutually perpendicular planes without any shear i.e. σx = σy = σ and τxy = 0.

$Centre =\left ( \frac{σ_x;+;σ_y}{2} \right );and\;Radius=τ_{max}\Rightarrow\sqrt {{{\left({\frac{{{σ _{x}};-;{σ _{y}}}}{2}} \right)}^2} + τ _{xy}^2}$

∴ centre is at (σ, 0) and radius = 0, (Shear = 0)

which represents a point on x-axis / σ-axis or normal stress axis.

   

We know 

${\rm{;Maximum;shear;stress}} = \dfrac{{{\sigma _1} - {\sigma _2}}}{2}$

σ₁ - Major principal stress

σ₂ - Minor principal stress

Calculation:

Given:

The pressure exerted on all faces of the cube is p.

σ₁ = σ₂ = p

${\rm{Maximum;shear;stress}} = \dfrac{{{\rm{p}} - {\rm{p}}}}{2} = 0$

Explanation

The following are the assumptions of the design of flexural members employing limit state of collapse:

(i) Plane sections normal to the axis remain plane after bending.

(ii) As per IS 456: 2000, the maximum strain in outermost compression fiber in bending is taken as 0.0035.

(iii) As per IS 456: 2000, the maximum strain in outermost tension fiber in bending is taken as 0.002 + f_y/1.15E_s

Strain Diagram in Bending:

Loss due to Elastic Shortening: When the tendons are cut and the prestressing force is transferred to the member, the concrete undergoes immediate shortening due to the prestress.

Loss due to Friction: The friction generated at the interface of concrete and steel during the stretching of a curved tendon in a post-tensioned member. The loss due to friction does not occur in pre-tensioned members because there is no concrete during the stretching of the tendons.

Loss due to Relaxation: Relaxation is assumed to mean the loss of stress in steel under nearly constant strain at a constant temperature. It is similar to the creep of concrete. This loss is generally of the order of 2 to 8% of the initial stress.

Loss due to Anchorage Slip: In a post-tensioned member, the is loss of prestress due to the consequent reduction in the length of the tendon. The loss due to anchorage does not occur in pre-tensioned members.

∴ Loss due to the relaxation of strands is categorized as a long-term loss of pre-stress in a pre-stressed concrete member.

Concept:

Modes of failure in steel tension member:

a) Gross section yielding: 

Generally a tension member without bolt holes can resist loads up to the ultimate load without failure. But such a member will deform in the longitudinal direction considerably (nearly 10% - 15% of its original length) before fracture. At such a large deformation a structure becomes unserviceable.

b) Rupture Failure: 

When a tension member with a hole is loaded statically, the point adjacent to the hole reaches the yield stress first. On further loading, the stress at that point remains constant at yield stress and each fiber away from the hole progressively reaches the yield stress. Deformations continue with increasing load until finally rupture of the member occurs. 

As per IS 800: 2007, Clause 6.3.1

The design strength in tension of a plate, Tdn as governed by rupture of net cross-sectional area An at the hole is given by

$T_{dn} = \frac{0.9 A_nf_u}{\gamma_m}$

where,

γm = partial safety factor for failure at ultimate stress

fu = ultimate stress of the material

An = net effective area of the member

c) Block Shear Failure: 

The block shear failure becomes a possible mode of failure when the material bearing strength and bolt shear strength is higher. Block shear failure is the rupturing of the net tension plane(BC) and yielding on the gross shear plane(AB & CD), as shown in the figure, which results in rupturing of the shear plane as the connection length becomes shorter.

Hence, In bolted connections, the rupture of the net section is a mode of failure under tension.

Concept:

Theoretical critical buckling load is given by,

${P_{cr}} = \dfrac{{{\pi ^2}EI}}{{L_{eff}^2}}$

The above formula does not take into account the axial stress and the buckling load is given by this formula may be much more than the actual buckling load

Where Leff = effective length of column which depends on the end condition of the column.

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Calculation:

∵ Critical buckling load, ${P_{cr}} = \dfrac{{{\pi ^2}EI}}{{L_{eff}^2}}$

When both ends of the column are hinged then critical buckling load, ${P_{cr}} = \dfrac{{{\pi ^2}EI}}{{{L^2}}}$    ----(1)

When both ends of the column are fixed then critical buckling load ${P_{cr}} = \dfrac{{{\pi ^2}EI}}{{{(L/ 2)^2}}} = \dfrac{{{4\pi ^2}EI}}{{{(L)^2}}}$       ----(2)

The ratio of the theoretical critical buckling load for a column with fixed ends to that of another column with the same dimensions and material, but with pinned ends, is

(2)/(1)

Ratio = 4.

Concept:

we use the following equation to find out the applied shear stress;

τ = C + σ tanϕ 

Where, τ = Shear stress;  σ = normal stress

C = cohession of soil; ϕ = angle of internal friction or failure plane;

Calculation:

Given:

ϕ = 38∘ 

As the given soil is cohessionless, hence C = 0

Now, let's put the values into the equation.

∴ τ = 0 + σ × tan 38∘ 

⇒ τ/σ = tan 38∘

∴ τ/σ = 0.781

Explanation:

 

It can be seen from the above figure that an increase in the amount of compactive effort (i.e., compaction energy per unit volume) increases the maximum dry density and decreases the optimum water content.

Hence,

Only statement II is TRUE.

Note:

• The line of optimums shown in the given figure is joining the points indicating the maximum dry density. It is roughly parallel to the zero air void line.
• Zero air void line is a hypothetical line that indicates the theoretical maximum density plotted along with the compaction curve. It is also known as the 100% saturation line.

Concept: 

It is adopted by IS code. It was given by A-line Casagrande. It uses particle size distribution for Coarse soils and plasticity for fine Soils. 

​Equation of A-line I_P  = 0.73 (W_L- 20)

Soil classification as per Indian standards:

 

∴ According to IS Classification, there are 18 groups of soils:

8 groups of coarse-grained, 9 group of fine-grained, and 1 of peat

Calculations:

Given,  

LL =  40%  and   PI  =  28%    

Now it is used,

As per , PI = 0.73 ( LL-20) = 0.73(40 - 20)  = 14.6

I_P of soil  > I_P of  A –line

It will lie above A-line and also 35 < W_L < 50

So it is CI.

Additional Information

Equation of U- line IP  =  0.9 (WL - 8)  

Note:

 For IPs between 4,  and 7, dual Symbols are used.

This is a case of active earth pressure. Hence, σ₁ will be parallel to wall face and acting downwards.

Explanation:

Given,

bulk unit weight (γ_b) = 20 kN/m3

Undrained shear strength (S = C) = 25 kPa

Bearing Capacity N_c = 5.7, General Shear Failure

We know q_u = CN_c

q_u = γ D_f = 25 × (5.7) 

20 × D_f = 142.5

D_f = 7.124 m

Area of the trapezoidal cross-section = $\frac{1}{2}\left( {10 + \left( {4 + 10 + 4} \right)} \right) \times 3$

$= \frac{1}{2}\left( {28} \right) \times 3 = 42;{m^2}$

Side slope of the section = $\sqrt {{3^2} + {4^2}} = 5$

Wetted perimeter of the trapezoidal section = (10 + 5 + 5) = 20 m

Hydraulic radius = $\frac{{Wetted;Area}}{{Wetted;Perimeter}} = \frac{A}{P}$

$R = \frac{{42}}{{20}} = 2.1;m$

Concept

For a Rectangular channel

Area = B × y

Where, y is the depth of flow in the channel.

Froude number for rectangular channel is given by

${F_r} = \frac{V}{{\sqrt {gy} }}$

If

Fr = 1 then flow is critical

Fr < 1 then flow is Sub critical

Fr > 1 then flow is super critical

Calculation

Given,

B = 5 m, Q = 100 m3/sec, Fr = 0.8

We know for a rectangular channel

${F_r} = \frac{V}{{\sqrt {gy} }}$

$0.8 = \frac{V}{{\sqrt {9.81y} }}$      –--(1)

Q = AV

100 = B × y × V

$100 = 5 \times y \times V$

$V = \frac{{20}}{y}$      –-(2)

Equating 1 and 2

$0.8 = \frac{{20}}{{y; \times \sqrt {9.81y} }}$

y = 3.99 m ≈ 4 m

∴ The depth of the flow in the channel is 4 m

Explanation: 

Let,

 V1  =  6 m/ s ,V2 = 5 m/s 

 d1  = 4 m, d2  = 4 m

 V3  =   ?, d₃ = 2 m

From continuity Equation we know,

 Q = AV

Where, A is C/S area of flow, V is mean Velocity 

As section is circular, Area = πd²/4 ≈ 0.785× d²

Incoming Flow = Outgoing Flow

Q₁  = Q₂ + Q ₃

 A₁V₁ =  A₂V₂ + A₃V₃

0.785 × ( 4²)

6 =   0.785 × ( 4²) × 5 + 0.785 × ( 2²) × V₃

16 × 6 = 16 × 5  +  4 × V₃

16 = 4 × V₃

V₃ = 4 m/s

V =  4 m/ s.

Additional Information

Continuity Equation

• The equation is based on the principle of conservation of mass which means mass can neither be created nor be destroyed.
• For a fluid flowing through a pipe at all cross-sections, the quantity of fluid flowing per second is constant.
• For steady and compressible flow, the continuity equation is  ρ1A1V1 = ρ2A2V​2.
• For steady and incompressible flow, the continuity equation is A1V1 = A2V​2.

Explanation:

Actual Evapotranspiration is the quantity of water that is actually removed from a surface due to the processes of evaporation and transpiration.

Potential Evapotranspiration is a measure of the ability of the atmosphere to remove water from the surface through the processes of evaporation and transpiration assuming no control on water supply.

a) If the water supply to the plant is adequate, soil moisture will be at field capacity then, the ratio of AET to PET =1

b) If the water supply to the plant is inadequate, then the ratio of AET to PET less than 1

c) In clayey soils, AET/PET almost equal to 1.

d) When the soil moisture approaches the permanent wilting point, AET tends to 0.

Hence the range of the ratio of actual evapotranspiration to potential evapotranspiration is lies between 0 to 1.

Explanation:

(i) The COD (Chemical oxygen demand) test is widely used as an alternative to BOD test to estimate the strength of domestic and industrial wastes as COD test can give results in few hours as compared to BOD test which takes normally 5 days.

(ii) The COD test uses potassium dichromate (K2Cr2O7) in presence of concentrated sulfuric acid (H2SO4) solution that oxidizes both organic (predominate) and inorganic substances in a waste water sample.

(iii) The reagents for COD test are followings:

• Standard potassium dichromate solution (0.25N)
• Concentrated sulfuric acid
• Standard ferrous ammonium sulphate titrant (0.1N)
• Ferroin indicator solution.

Concept

• In transitions, additional factors must be considered, and when this transition is from a larger pipe to a smaller pipe the eddies formed are large and in long transitions, air entrapment may cause backing of flow. So, the transition from larger to smaller diameters shall not be made.
• The crowns of sewers are always kept continuous. In no case, the hydraulic flow line in the large sewers shall be higher than the incoming one. To avoid backing up, the crown of the outgoing sewer shall not be higher than the crown of the incoming sewer.

Explanation:

For the intersection of two-lane two-way roads from both sides:

Major conflicting points = 16, Minor conflicting points = 8 and thus total conflicting points (X) = 24

 

For the intersection of two-lane one-way rod and two-lane one-way road:

Major conflicting points = 4 Minor conflicting points = 2 and thus total conflicting points (Y) = 6

So, X/Y = 24/6 = 4

Additional Information

 

For the intersection of a two-lane one-way rod from both sides:

Major conflicting points = 7 Minor conflicting points = 4 and thus total conflicting points = 11

Important Points:

Concept

• Softening point denotes the temperature at which the bitumen attains a particular degree of softening under the specifications of the test. The test is conducted by using the Ring and Ball apparatus.
• Viscosity denotes the fluid property of bituminous material and is a measure of resistance to flow. At the application temperature, this characteristic greatly influences the strength of paving mixes.
• It measures the hardness or softness of bitumen by measuring the depth in tenths of a millimeter to which a standard loaded needle will penetrate vertically in 5 seconds.
• The softening point of sample X is 45 and for softening point for Y is 60

• The softening point of sample X is lower than Y so at the same temperature viscosity of X will be lower than that of Y, hence the penetration value of X will be more than that of Y.

Important Points

If the penetration value is 80/100, it means that the penetration value is 8 - 10 mm.

Bump Integrator is used to measure the roughness and mostly measure the longitudinal road profiles.

They are now most commonly installed on the floor of a vehicle with a cable connecting to the suspension to evaluate the roughness of the road.

Reciprocal levelling is adopted to accurately determine the level difference between two points which are separated by obstacles like a river, ponds, lakes, etc.

It eliminates the following errors:

i) error in instrument adjustments

ii) the combined effect of Earth's curvature and the refraction of the atmosphere

iii) variation in the average refraction

Important Points:

Reciprocal leveling eliminates the error due to collimation and error due to curvature of Earth completely, but as the refraction depends upon the atmosphere which may change every minute;

∴ Error due to refraction cannot be eliminated completely but it is reduced.

Concept:

Error in central difference formula is given by,

Error ∝ h²

h - Interval

Calculation:

Given:

When h = 0.03, Error = 2 × 10^-3

Error ∝ h²

⇒ Error = k × h²

2 × 10^-3 = k × 0.03²

k = 2.22

Error = 2.22 × (0.02)² = 8.88 × 10^-4 ≈ 9 × 10^-4

Concept:

It is given that positive and negative values are equally likely to occur, Binomial distribution can be adopted.

The probability of ‘r’ number of successes in ‘n’ trials is given by

${\rm{p}}\left( {\rm{x}} \right) = {\rm{n}}{{\rm{C}}_{\rm{r}}}{\rm{;}}.{{\rm{p}}^{\rm{r}}}.{{\rm{q}}^{{\rm{n}} - {\rm{r}}}}$

p - Probability of getting negative value

q - Probability of getting positive value

Calculation:

Given:

n = 5 trials

Positive and negative values are equally likely to occur,

p = 1/2 , q = 1/2

At most one negative value so it can be no negative value or 1 negative value

p (At most one negative) = p(r ≤ 1) =  p(r = 0) + p (r = 1)

${\rm{p}}\left( {{\rm{r}} = 0} \right) = 5{{\rm{C}}_0}{\rm{;}}.{\left( {\dfrac{1}{2}} \right)^0}.{\left( {\dfrac{1}{2}} \right)^{5 - 0}} = \dfrac{1}{{32}}$

${\rm{p}}\left( {{\rm{r}} = 1} \right) = 5{{\rm{C}}_1}{\rm{;}}.{\left( {\dfrac{1}{2}} \right)^1}.{\left( {\dfrac{1}{2}} \right)^{5 - 1}} = \dfrac{5}{{32}}$

p (At most one negative) = 6 / 32

Concept:

Characteristics polynomial of a matrix:

Let A be a square matrix of order n and λ be any scalar quantity. Then the polynomial in λ of degree n formed by solving the characteristic equation |A - λI| = 0 is called the characteristics polynomial of the matrix.

Eigenvalues:

The roots of the characteristic equation are called the eigenvalues or characteristic roots of latent roots of the matrix A.

Eigenvectors:

If λ is the eigenvalue of the matrix A then a non-zero vector X which satisfies AX = λX is called the eigenvector of the matrix corresponding to the eigenvalue λ.

Calculation:

\( \left[ {\begin{array}{*{20}{c}} 9&5\ 5&8 \end{array}} \right]\)

|A – λI| = 0

\(\left[ {\begin{array}{*{20}{c}} {9 - {\rm{\lambda }}}&5\ 5&{8 - {\rm{\lambda }}} \end{array}} \right] = 0\)

(9 – λ) (8 – λ) – 25 = 0

72 – 9λ – 8λ + λ2 – 25 = 0

λ2 – 17λ + 47 = 0

On solving the equation we get

λ = 3.48 and λ = 13.53

Concept:

The area of parallelogram is the magnitude of vector product (cross product) of its sides

The area of parallelogram whose sides are a and b is given by,

${\rm{Area;of;parallelogram}} = \left| {\vec a \times \vec b} \right|;$

Calculation:

Given:

$\overrightarrow {{\rm{OP}}} = a\hat i + b\hat j$

$\overrightarrow {{\rm{OR}}} = c\hat i + d\hat j$

\(\overrightarrow {{\rm{OP}}} \times \overrightarrow {OR} = \left| {\begin{array}{*{20}{c}} i&j&k\ a&b&0\ c&d&0 \end{array}} \right|\)

⇒$\left( {{\rm{ad}} - {\rm{bc}}} \right) {\rm{\hat k}}$

$\left| {\overrightarrow {OP} \times \overrightarrow {OR} } \right| = \sqrt {{0^2} + {0^2} + {{\left( {ad - bc} \right)}^2}}$

⇒ ad -bc

Concept:

$\frac{{dy}}{{dx}} + 2y= 0$

Since the variables are separable, let us adopt variable separable

In variable separable method, we have first variable on RHS and the second variable on LHS

Calculation:

Let us have y variable on LHS and x variable on RHS

$\frac{{dy}}{y} = - 2~;dx$

Let us integrate on both sides

$\smallint \frac{{dy}}{y} = - 2;\smallint dx$

ln y = -2x + c

y = e^-2x × e^c

Let e^c = k 

y = e-2x × k       ....(1)

At x =1 , y = 5

5= e- (2 × 1) × k

k = 5 × e²

Putting value of k in (1)

y = e-2x × 5 × e²

y = e-2x × 5 × 7.389 (∵ e² = 7.389)

y = 36.95 e-2x

The beam can be assumed as

${{\rm{\Delta }}_1} = \frac{{5\left( {\frac{w}{2}} \right){l^4}}}{{384EI}} + \frac{{P{L^3}}}{{48EI}}$

Now wl = P

${{\rm{\Delta }}_1} = \frac{{5P{l^3}}}{{768EI}} + \frac{{P{L^3}}}{{48EI}} \Rightarrow \frac{{21P{L^3}}}{{768EI}}$

Concept:

The three equilibrium conditions need to be satisfied are

∑ H = 0, H – Horizontal forces

∑ V =0, V – Vertical forces

∑ M_Z = 0, M_Z –Moment acting at a point

Calculation:

Free body diagram is drawn,

Sign Convention

Vertical forces -  ↑ - ( +ve ), ↓ - ( -ve ) 

Moments - clockwise moments are taken as positive, anticlockwise moments are taken as negative.

Since R is a hinged support it generates two reactions V_R and H_R

Applying Equilibrium condition ∑ V = 0, we get

V_R – P = 0

→ V_R = P

Applying Equilibrium condition ∑ M_Q = 0, we get

M – ( H_R × h ) = 0

${{\bf{H}}_{\bf{R}}} = \frac{{\bf{M}}}{{\bf{h}}}$

Concept:

In prestressed concrete, compressive stress is considered as positive, Tensile stress is considered as negative

When self weight (dead load) and live load is considered

In soffit of the beam, Eccentric load causes compressive stress, self weight and live load causes tensile stress

\({{\rm{\sigma }}{{\rm{soffit}}}} = \frac{{\rm{P}}}{{\rm{A}}} + \frac{{{\rm{P}} ;×; {\rm{e}}}}{{\rm{Z}}} - \frac{{{{\rm{M}}{\rm{D}}}}}{{\rm{Z}}} - \frac{{{{\rm{M}}_{\rm{L}}}}}{{\rm{Z}}}\)

In top of the beam Eccentric load causes tensile stress, self weight and live load causes compressive stress.

\({{\rm{\sigma }}{{\rm{top}}}} = \frac{{\rm{P}}}{{\rm{A}}} - \frac{{{\rm{P}} ;× ;{\rm{e}}}}{{\rm{Z}}} + \frac{{{{\rm{M}}{\rm{D}}}}}{{\rm{Z}}} + \frac{{{{\rm{M}}_{\rm{L}}}}}{{\rm{Z}}}\)

MD – Moment due to dead load

ML – Moment due to live load

P – Prestressing force

e – Eccentricity of prestressing force

Z – Section modulus of beam

A - Area of cross section of beam.

Calculation:

Given:

b = 500 mm,d = 750 mm

P = 1620 kN, e = 145 mm

l =7300 mm = 7.3 m

${\rm{Z}} = \frac{{b; × ;{d^2}}}{6}$

$Z = \frac{{b ;× ;{d^2}}}{6} = \frac{{500; × ;{{750}^2}}}{6}$

⇒ 46.875 × 10⁶ mm³

Moment due to live load and dead load is given by,

$M = \frac{{w; × ;{l^2}}}{8}$

$M = \frac{{45; × ;{7.3^2}}}{8}$

⇒ 299.76 kN - m

The stress at bottom (soffit) of the beam is given by

\({{\rm{\sigma }}{{\rm{soffit}}}} = \frac{{\rm{P}}}{{\rm{A}}} + \frac{{{\rm{P}}; ×; {\rm{e}}}}{{\rm{Z}}} - \frac{{{{\rm{M}}{\rm{}}}}}{{\rm{Z}}} \)

\({{\rm{\sigma }}{{\rm{soffit}}}} = \frac{{\rm{1620;× ;10^3}}}{{\rm{500;× ;750}}} + \frac{{{\rm{1620;× ;10^3}} ;× ;{\rm{145}}}}{{\rm{46.875;× ;10^6}}} - \frac{{{{\rm{299.76;× ;10^6}}{\rm{}}}}}{{\rm{46.875;×; 10^6}}} \)

⇒ 2.94 N/mm²

A positive sign indicates that the nature of force is compressive.

Explanation:

∑M_R = 0

V_P × l - W × l/2 = 0

${{\rm{V}}_{\rm{P}}} = \frac{W}{2}$

${{\rm{V}}_{\rm{R}}} = \frac{W}{2}$

By using the property of symmetry,

$\frac{y}{x} = \frac{h}{{\frac{l}{2}}}$

$y = \frac{{h; × ;x}}{{\frac{l}{2}}}$

$Bending;moment;at;Q ,~M_Q = \frac{W}{2} \times \frac{l}{2} - H \times h$

Horizontal reaction for any support is given by,

$H = \frac{{\smallint m ;\times; y;dx}}{{\smallint {y^2};dx}}$

Let us consider a section between P and R

\(H = \frac{{\mathop \smallint \nolimits_0^{l/2} \left( {\frac{W}{2} ;\times ;x} \right) ;\times; \left( {\frac{{h; \times; x}}{{\frac{l}{2}}}} \right)dx}}{{\mathop \smallint \nolimits_0^{l/2} {{\left( {\frac{{h; \times; x}}{{\frac{l}{2}}}} \right)}^2}dx}}\)

$H = \frac{{W ;\times; l}}{4;\times; h}$

$M_Q = \frac{W}{2} \times \frac{l}{2} - H \times h$

$M_Q = \frac{W}{2}; \times; \frac{l}{2} - \frac{{W ;\times; l}}{{4; \times; h}} \times h$

⇒ M_Q = 0

Concept:

The Load carried by weld is given by,

${P_{dw}} = {l_w} \times {t_t} \times \frac{{{f_u}}}{{\sqrt 3 \times {\gamma _{mw}}}}$

Where,

l_w = length of fillet weld

t_t = Throat thickness = K × S

K = Constant that depends upon the angle between weld faces

S = Size of the weld

f_U = Ultimate tensile stress

γ_mw = partial safety against weld strength = 1.25

Calculation:

P = load carried by both the welds = 270 kN

Load carried by each weld = 270/2 = 135 kN

K = 0.7, S = 10 mm

t_t = Throat thickness = K × S = 10 × 0.7 = 7

f_U = 410 kPa, γ_mw = 1.25

${P_{dw}} = {l_w} \times {t_t} \times \frac{{{f_u}}}{{\sqrt 3 \times {\gamma _{mw}}}}$

${l_w} = \frac{{{P_{dw}} \times \sqrt 3 \times {\gamma _{mw}}}}{{{f_u} \times {t_t}}} = \frac{{135 \times {{10}^3} \times \sqrt 3 \times 1.25}}{{410 \times 7}} = 101.84;mm$

l_w = 101.84 mm

Rounding off to the nearest higher multiple of 5 mm

l_w = 105 mm

Concept:

The permeability of soil specimen test in a falling head permeability test is given by

${\bf{k}} = \left( {\frac{{{\bf{a}} \times {\bf{L}}}}{{{\bf{A}} \times {\bf{t}}}}} \right) \times \ln \frac{{{{\bf{h}}_1}}}{{{{\bf{h}}_2}}}$

k – Permeability of soil

a – cross-sectional area of standpipe

A – Area of the cross-section of soil sample

h₁ – Initial head of the first specimen

h_2­ – Final head after time ‘ t ‘,

t – Time is taken for head to fall from h₁ to h₂

L – Length of sample

Calculation:

Given,

It is given that two specimens with identical geometric dimensions hence a, A and L will be same for both the specimens.

Also given fall of the head was measured after an identical time interval hence t will be the same for both the specimens

h₁^’ – Initial head of the second specimen

h₂^’ – Final head of the second specimen

k^’ – permeability of the second specimen

$\frac{{{{\rm{h}}_1}}}{{{{\rm{h}}_2}}} = 1.25$

k^’ = 5 × k ……. ( 1 )

${\rm{k}} = \left( {\frac{{{\rm{a}} \times {\rm{L}}}}{{{\rm{A}} \times {\rm{t}}}}} \right) \times \ln 1.25;\left( {\frac{{{h_1}}}{{{h_2}}} = 1.25;} \right)$

${\bf{k}}' = \left( {\frac{{{\bf{a}} \times {\bf{L}}}}{{{\bf{A}} \times {\bf{t}}}}} \right) \times \ln \frac{{{\bf{h}}'}}{{{\bf{h}}'}}$

$\begin{array}{l} \left( {\frac{{{\rm{a}} \times {\rm{L}}}}{{{\rm{A}} \times {\rm{t}}}}} \right) \times \ln \frac{{{\rm{h}}_1^{\rm{'}}}}{{{\rm{h}}_2^{\rm{'}}}} = 5 \times \left( {\frac{{{\rm{a}} \times {\rm{L}}}}{{{\rm{A}} \times {\rm{t}}}}} \right) \times \ln 1.25;\left( {{\bf{k'}} = 5 \times {\bf{k}};} \right) \ \end{array}$

$\frac{{{\rm{h}}_1^{\rm{'}}}}{{{\rm{h}}_2^{\rm{'}}}} = 3.05$

Concept:

The final consolidation settlement is given by,

${\rm{Δ H}} = \left( {\frac{{{C_c}; × ;{H_0}}}{{1; + ;{e_0}}}} \right) ;× ;{\log _{10}}\frac{{{σ _0}' ;+; {\rm{Δ }}σ }}{{{σ _0}'}}$

ΔH - ultimate consolidation settlement

C_c - Compression index, H₀ - Initial thickness of soil layer

σ₀^' - Initial effective stress, Δσ^'  - Increase in effective stress

Calculation:

Given:

σ0' = 100 kN/m², Δσ^' = 20 kN/m²

H₀ = `1 m, G = 2.7, W = 45%, C_c =0.45, C_r = 0.05

The relationship between w, G, S, and e is given by

w × G = S × e

0.45 × 2.7 = 1 × e

e = 1.215

${\rm{Δ H}} = \left( {\frac{{{C_c} ;× ;{H_0}}}{{1; + ;{e_0}}}} \right) × {\log _{10}}\frac{{{σ _0}'; + ;{\rm{Δ }}σ }}{{{σ _0}'}}$

 ${\rm{Δ H}} = \left( {\frac{{0.45; \times ;1}}{{1 ;+ ;1.215}}} \right) \times {\log _{10}}\frac{{100 ;+; 20}}{{100}}$

Δ H = 0.016 m = 16 mm.

Concept:

The given situation can be represented as

  

Effective Pressure = Total Pressure - Pore water pressure

Calculation:

Total Pressure (σ) = (1.5 + 0.5) γ_sat = 2 × 18 = 36 kN/m²

Since Q is present at the mid half of the P & R.

Pore water Pressure (u) at Q = 0.5 × (Pore water Pressure at P + Pore water Pressure at R)

u = 0.5 × (2 × γω + 2.1 ×γω) = 2.05 γ_ω

σ’ = σ – u = 36 – 2.05 × 9.81 = 15.89 kPa

Concept:

Two-point method velocity we don’t consider top surface maximum velocity. 

 

Mean velocity is given by

 V_avg $= \frac{{\left( {{V_{0.2y}} + {V_{0.8y}}} \right)}}{2}$

Where, 

Formula:

 Discharge in the channel  =   V_avg × area of Channel 

 V_avg = average mean velocity

Calculation

Given:

Top width of channel = 4 m

Depth of channel = 1 m

Velocity on surface ( V0y )  = 0.7 m/s

velocity on 0.2 m ( V_0.2y ) = 0.6 m/s

velocity on 0.8 m ( V_0.8y )  = 0.4 m/s

Using two-point method

Mean velocity

 V_avg $= \frac{{\left( {{V_{0.2y}} + {V_{0.8y}}} \right)}}{2}$

$= \frac{{0.6 + 0.4}}{2}$

= 0.5 m/s

Area of channel $= ;\frac{1}{2} × 4 × 1$

=2 m²

Discharge in the channel = 2 × 0.5

= 1.0 m³/sec

Explanation:

Important Points

Penman’s equation is based on energy balance and mass transfer.

Horton’s curve is a curve that represents the approximate infiltration capacity of soil which is used to find the total infiltration.

Horton’s infiltration equation

f(t) = f_c + (f_co - f_c) e^-kt 

Where, f_c = Final constant rate of infiltration at saturation.

f_co = Initial rate of infiltration capacity

k = Horton’s constant

Explanation:

Given:

Rainfall = 100 mm = 10 cm, Base period of rice = 120 days

water requirement by rice = 55 cm, irrigation efficiency = 60 %

Canal flow = 0.01 m3/sec

Now, the actual requirement of water through the canal for irrigation is

Δ = 55 cm – 10 cm = 45 cm = 0.45 m

we know that,

$\Delta = 8.64\frac{B}{D}$

Where, Δ = Delta of the crop in m

D = Duty, in hectares /cumec, B = Base period in days

$\therefore D = \frac{{8.64 ;\times; 120}}{{0.45}}$

D = 2304 hectare/cumec

∴ Area of land that can be irrigated with 0.01 m3/s is = 2304 × 0.01

= 23.04 hectares

Since the irrigation efficiency is 60% then the area of land that can be irrigated for rice crop is

$= 23.04 \times \frac{{60}}{{100}}$

= 13.824 hectares

Concept:

pH = - log10 [H+]

pOH = - log10 [OH-]

OH- → concentration of hydroxyl ions in mole/liter, 

H+ → concentration of hydrogen ions in mole/liter

pH+ pOH = 14

Mg/liter = mole/liters × Mx × 103

Where

Mx = Molecular weight of the ion/compound

Calculation:

Given, pH of sample = 9.25

pH+ pOH = 14

pOH = 14 – 9.25

pOH = 4.75

Now,

pOH = - log10[OH-]

∴ [OH-1] (mole/liters) = 10-[OH]

∴ [OH-] = 10-4.75 mole/liters

mg/l = mole/l × Mole weight × 103

mg/l = 10-4.75 × 17 × 103

mg/l = 0.01778 × 17

mg/l = 0.3023

∴ Concentration of hydroxyl ions is 0.3023 mg/liters.

Explanation:

Given,

Q = 4.2 m³/min = 0.07 m³/sec

Over flow Rate = 0.2 mm/s = 0.0002 m/sec

Depth of Basin = 3.5 m, Area of Basin = ?

Area = Q/Overflow rate

= 0.07/0.0002 = 350 m²

 Therefore required surface area = 350 m²

Concept:

The cumulative standard axles may be calculated as

${{\rm{N}}_{\rm{S}}} = \frac{{365 \times {\rm{A}} \times \left[ {{\rm{;}}{{\left( {{\rm{;}}1 + {\rm{r;}}} \right)}^{\rm{n}}} - 1{\rm{;}}} \right)] \times {\rm{F}} \times {\rm{D}}}}{{\rm{r}}}$

N_s – Cumulative standard axles in standard axles

A – Number of the commercial vehicle per day when pavement construction is completed

r – Rate of growth of traffic

n – Design life of the pavement

F – Vehicle damage factor

D – Lane distribution factor

Calculation:

Given

A = 4500 vehicles/ day

r = 8%

N_s = 100 × 10⁶

n = 15 years

$100 \times {10^6} = \frac{{365 \times 4500 \times \left[ {;{{\left( {;1 + 0.08} \right)}^{15}} - 1;} \right)] \times F}}{{0.08}}$

Solving we get F = 2.24

Important Points

When the construction of pavement is started A is given by

A = P × ( 1 + r )ⁿ

P – Initial traffic before construction is started

n – Period of construction

r -  Rate of growth of traffic.

Concept:

Setback Distance:

Setback distance m or the clearance distance is the distance required from the centerline of a horizontal curve to an obstruction on the inner side of the curve to provide adequate sight distance at a horizontal curve.

Let us assume the length of curve is more than the stopping sight distance

For two lanes the setback distance (m) from the inner lane is given by,

$m = (R - d) - \left( {R - d} \right) × \cos \frac{α }{2}$

m - Setback distance, R - Radius of the curve 

d - Distance between centerline of the road and the centerline of the inner lane

α - Angle subtended by the radius

$\frac{\alpha }{2} = \frac{{SSD; × ;180}}{{2; × ;{\rm{\pi }}; × ;\left( {{\rm{R}} - {\rm{d}}} \right)}}$

SSD - stopping sight distance

Calculation:

Given:

L = 200 m, SSD = 80 m, R = 300 m

$d = \frac{{7.5}}{4} = 1.875;m$

Length of curve > SSD, Hence we can use the formula

$\frac{\alpha }{2} = \frac{{80; × ;180}}{{2; ×; {\rm{\pi }}; ×; \left( {{\rm{300}}; - ;{\rm{1.875}}} \right)}}$

$\frac{\alpha }{2} = \frac{{SSD; × ;180}}{{2; ×; {\rm{\pi }}; × ;\left( {{\rm{R}} ;- ;{\rm{d}}} \right)}}$

$\frac{\alpha }{2} = 7.687$

m = (300 - 1.875) - (300 - 1.875) × cos (7.687)

⇒ 2.67 m

The set back distance (in m) required from the centre line of the inner lane of the pavement is 2.67 m. So the most appropriate answer is 2.54 m

Mistake Points For two lanes the setback distance (m) is given by,

$m = (R ) - \left( {R - d} \right) × \cos \frac{α }{2}$

Additional Information

The setback distance for a single lane is given by

Length of the curve is more than SSD

$m = R - (R \times \cos \frac{\alpha }{2})$

Concept:

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

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

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

k- Density of stream ( veh/km )

u_f - Free mean speed

k_j – Jam density

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

q = k × u

q – Flow given in veh/hr

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

${\bf{k}} = \frac{{1000}}{{{\bf{Space}};{\bf{headway}}}}$

At capacity or maximum flow,

\({{\rm{k}}0} = \frac{{{{\rm{k}}{\rm{j}}}}}{2}\)

${u_0} = \frac{{{{\rm{u}}_{\rm{f}}}}}{2}$

\({{\bf{q}}_{{\bf{max}}}} = {{\bf{k}}0} \times {{\bf{u}}0} = \frac{{{{\bf{k}}{\bf{j}}} \times {{\bf{u}}{\bf{f}}}}}{4}\)

q_max – Maximum flow or flow at capacity

k₀, u₀ – Density and speed at capacity

CALCULATION:

GIVEN:

Two-lane maximum capacity = 1800 veh/hr

At jam condition, space headway = 5m

${\rm{One;lane;capacity}} = {\rm{;}}{{\rm{q}}_{{\rm{max}}}} = \frac{{1800}}{2} = 900{\rm{veh}}/{\rm{hr}}$

${{\rm{k}}_{\rm{j}}} = \frac{{1000}}{5} = 200{\rm{;veh}}/{\rm{km}}$

$900 = \frac{{200 \times {{\rm{v}}_{\rm{f}}}}}{4}$

→ u_f = 18 km/hr

For two-lane flow = 1000 veh/hr

${\rm{For;one;lane;flow;q}} = \frac{{1000}}{2} = 500{\rm{veh}}/{\rm{hr}}$

${\rm{u}} = 18 - \left( {\frac{{18}}{{200}}} \right) \times {\rm{k}}$

$\frac{{500}}{{\rm{k}}} = 18 - \left( {\frac{{18}}{{200}}} \right) \times {\rm{k}};;\left( {;{\bf{u}} = \frac{{\bf{q}}}{{\bf{k}}};} \right)$

Solving we get,

k = 33.32 veh/km (for single lane )

For two-lane k = 2 × 33.32 = 66.64 ≈ 67

Explanation

Given:

The horizontal distance between stations P and Q = 100 m.

CALCULATION:

Assume distance between point and tower is ' x ' in the same plan  

From similar triangle height of tower:  

Height of tower = x tan5° + x tan0.5° = (100 + x) tan 3° + (100 + x) tan 0.1°

⇒ x ( tan5° + tan0.5° )=(100+x)(tan3° +tan0.1° )

$⇒ ;\frac{x}{{100 + x}} = \frac{{\left( {\tan 3^\circ + \tan 0.1^\circ } \right)}}{{\left( {\tan 5^\circ + \tan 0.5^\circ } \right)}}$

$⇒ \frac{x}{{100 + x}} = 0.563$

⇒ x=128.745 m

Putting the value of x in the equation

Height of tower = x(tan5° +tan0.5° )

=128.745 (tan5° +tan0.5° )

 = 12.387 m 

Important Points 

Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion.

In other words, similar triangles are the same shape, but not necessarily the same size. The triangles are congruent if, in addition to this, their corresponding sides are of equal length.

Concept:

Determination of seepage discharge when the medium is isotropic:

Seepage discharge is given by, $q = kh\frac{{{N_f}}}{{{N_d}}}$

Where h = Hydraulic head or head difference between upstream and downstream level or head loss through the soil

Nf = Total number of flow channels

Nd = Total number of equipotential drops

k = Coefficient of permeability

Calculations:

Data Given;

Permeability coefficient (k) = 0.09 m/day, specific gravity (G) = 2.7, void ratio (e) = 0.85

Seepage loss; $q = k \times h \times \frac{{{N_f}}}{{{N_d}}}$

Where, h = head difference, N_f = no. of flow channels, N_d = no. of potential drops

Now, k = 0.09 m/day, h = 10m - 1.5m = 8.5m, N_f = 4, N_d = 8

$q = 0.09 \times 8.5 \times \frac{4}{8}$= 0.3825 m³/day/m

So, Seepage loss (in m³ per day per unit length of the wall) = 0.3825 m³/day/m

Concept:

(i) The hydraulic gradient under which the quicksand or piping failure condition occurs is termed the critical hydraulic gradient. If void ratio and specific gravity of soil is known then ic may be given as

${i_{cr}} = \frac{{G - 1}}{{1 + e}} = (G - 1)(1 - n)$

n = Porosity, e = void ratio, G = Specific gradient

(ii) In order to prevent quicksand or piping failure, the hydraulic gradient should be less than the critical hydraulic gradient. Hence factor of safety against quicksand condition or piping failure is 

$FOS = \frac{{{i_{c}}}}{i}$

Calculation:

Data Given;

Permeability coefficient (k) = 0.09 m/day, Specific gravity (G) = 2.7, void ratio (e) = 0.85

F.O.S against piping = \(\frac{{{{\rm{i}}{\rm{c}}}}}{{{{\rm{i}}{\rm{e}}}}}\) = $\frac{{critical;hydraullic;gradient}}{{exit;gradient}}$

Where, ${i_c} = ;\frac{{G - 1}}{{1 + e}}$ , 

${i_e} = ;\frac{{head;difference}}{{no.;of;potential;drops; \times l}}$

h = head difference, N_f = no. of flow channels, N_d = no. of potential drop

h = 10m - 1.5m = 8.5m, N_f = 4, N_d = 8 

${i_c} = ;\frac{{2.7 - 1}}{{1 + 0.85}} = 0.9189$

${i_e} = ;\frac{{8.5}}{{8; \times ;3}} = 0.3542$

$F.O.S = \frac{{0.9189}}{{0.3542}} = 2.594$ = 2.6

Concept:

Food (F) to Micro - organsims (M) ratio is an important term related to activated sludge process particularly with aeration tank. F/M ratio is given by

$\rm \dfrac{F}{M} = \dfrac{{Daily;BOD;applied;to;the;aerator;system}}{{Total;microbial;mass;in;the;system}}$

$\dfrac{F}{M} = \dfrac{{Q × {Y_0}}}{{V × {X_t}}}$

Q - Sewage inflow per day 

Y0 - BOD of the influent sewage flow

V - Volume of the aeration tank

X_T - MLSS in the aeration tank

Calculation:

Given:

Q = 500 m3/hr = 500 × 24 m3/day

Y0 = 150 mg/l

V = 4000 m3

X_T = 2000 mg/l

 

$\dfrac{F}{M} = \dfrac{{500 \times 24× {150}}}{{4000 × {2000}}}$

⇒ 0.225 kg BOD per day/kg of MLSS

<br>

Important Points

1. Lower the F/M ratio, higher will be BOD removal
2. F/M ratio can be varied by varying MLSS concentration in the system.

Concept:

Sludge Age (θc):

The sludge age is an important parameter related to F/M ratio. Sludge age is also called mean cell residence time

$\rm Sludge;age\left( {{θ _c}} \right) = \dfrac{{Mass;of;suspended;solids;in;the;system;\left( M \right)}}{{Mass;of;solids;leaving;the;system;per;day}}$

• Mass of solids in the system = V× Xt
• Mass of waste sludge removed with the wasted sludge per day = QW × XR
• Mass of waste sludge removed with the effluent per day = (Q - QW) × XE

<br>

$\rm Sludge;age\left( {{θ _c}} \right) = \dfrac{{V\times X_T;\left( M \right)}}{{Q_W \times X_R~ +(Q - Q_W)\times X_E}}$

Q - Sewage inflow per day 

XT - MLSS in the aeration tank

QW - Flow of wasted sludge per day 

V - Volume of the aeration tank

XR - Concentration of solids in returned sludge

Calculation:

Given:

θC = 240 hr = 10 days

XT = 2000 mg/l = 2 kg/l

$\rm 10 = \dfrac{{4000 \times 2}}{{Mass;of;solids;leaving;the;system;per;day}}$

Mass of system leaving the system = 800 kg/day.

Concept:

The given section is a flanged section. For flanged sections we use IS 456:2000 ANNEX - G G-2 FLANGED SECTION

The given section can be taken as a combination of two sections

 

 

$Total;compressive;force = 0.36 × {f_{ck}} × b_w × {x_u} + 0.45 × {f_{ck}} × \left( {{b_f} - {b_w}} \right) × {D_f}$

f_ck - Characteristic compressive strength of concrete

b_w - breadth of web, b_f - breadth of flange

D_f - Depth of the flange, x_u - Depth of neutral axis

Total tensile force = 0.87 × fy × A_st

f_y - characteristic strength of reinforcement

A_st - Area of reinforcement

Calculation:

Given:

f_ck = 25 N/mm², b_w = 325 mm

b_f = 1000 mm, D_f = 100 mm

f_y = 415 N/mm², A_st = 4000 mm²

To find the depth of neutral axis we equate the total compressive force and total tensile force

$Total;compressive;force = 0.36 × {f_{ck}} × b_w × {x_u} + 0.45 × {f_{ck}} × \left( {{b_f} - {b_w}} \right) × {D_f}$

Total tensile force = 0.87 × f_y × A_st

0.36 × 25 × 325 × x_u + 0.45 × 25 × (1000 - 325) × 100 = 0.87 × 415 × 4000

x_u = 234.12 mm

The answer closely matching is option c) 236.3 mm