GATE CE 2010 Official Paper (Previous Year Paper)

The correct answer is Indirect.

Key Points

• The word 'Circuitous' means not being forthright or direct in language or action.
• The synonyms of the word are "circular, indirect, roundabout".
• The antonyms of the word are "direct, straight, straightforward".
• From the Synonym of the given word 'Circuitous', we can say that the word 'Indirect' is its opposite meaning.
• The word 'Indirect' means not having a direct connection with something.
• Hence, we can say that 'Indirect' is the closest in meaning to the given word.
• Let's see the meaning of other given options-

Important Points

• The word 'Circuitous' is derived from the Medieval Latin "circuitosus", from Latin "circuitous".
• Its first known use was in the year 1664.
• Examples of 'Circuitous' in a sentence:
• "Such devices have an extra lead that uses a circuitous route to reach the otherwise hard-to-reach left ventricle."
• "So the agonised, circuitous discussions continue and continue, leading nowhere."

Unemployed → doesn't have a job to work 

Worker → Worker always works, and mostly has a job.

Workers are employed and Unemployed.

It implies a somewhat opposite relationship.

(1) Fallow → uncultivated land to improve its fertility.

Land → Land is usually cultivated.

Lands are cultivated and uncultivated.

It also implies a somewhat opposite relationship.

(2) Unaware and Sleeper (someone who sleeps is always unaware)

(3) Wit → A message whose ingenuity or verbal skill or incongruity has the power to evoke laughter.

Jester → A clown, joker

(4) Renovated → Make brighter or prettier or better

House → A place to lives humans

Hence, option (1) is the correct answer.

The correct answer is 'conserve'.

Key Points

• Let's explore the options:

 

• Thus, from the above-given explanation, it is clear that 'conserve' will be used in the blank.
• Therefore, the correct answer is option 4.

The correct answer is 'betrayed'.

Key Points

• Let's explore the options:

 

• Thus, from the above-given explanation, it is clear that 'betrayed' will be used in the blank.
• Therefore, the correct answer is option 3.

Given:

Total number of people in the room = 25

Number of people who play Hockey = 15

Number of people who play Football = 17

Number of people who play both = 10

Calculation:

Number of people who play only Hockey = 15 - 10 = 5

Number of people who play only Football = 17 - 10 = 7

⇒ 

⇒ The total number of person who play either of the two = 5 + 10 + 7 = 22

⇒  The number of persons playing neither hockey nor football = 25 - 22 = 3

∴ The required result will be 3.

The number of persons playing neither hockey nor football = ξ - n(A ⋃ B) - n(A) - n(B)

The number of persons playing neither hockey nor football = 25 - 10 - 7 - 5

The number of persons playing neither hockey nor football = 3

The correct answer is 'People in military establishments like to use chemical agents in war.'

Key Points

• Let's explore the options:
• In statement 1, let's take a look at the keyword 'civil strife' which means A war between factions or regions of the same country, which isn't correct in the context of passage, therefore option 1 is incorrect.
• In statement 2, it is mentioned that the chemical agents are useful in modern warfare but the passage doesn't have any facts that can justify the usefulness of the chemical agents, and hence incorrect.
• In statement 3, it is mentioned that chemical agents are undesirable, but in the passage it is mentioned that 'Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause.' which conveys that they are in fact desirable, thus the statement 3 is incorrect.
• In statement 4, it is mentioned that 'People in military establishments like to use chemical agents in war.' and in the passage, it is also mentioned that 'Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause', therefore statement 4 is correct.

By observation of summation, it can be seen that numbers are not in decimal.

The value of the answer is actually the remainder obtained by actually finding the sum of original numbers.

Let us proceed step by step.

First, take the unit digit position,

1. 7 + 6 = 13 ⇒ Rem (13/8) = 5

2. 3 + 7 + 1 (Carry from 13) = 11 ⇒ Rem (11/8) = 3

3. 1 + 2 + 1 (Carry from 11) = 4

Now, write in reverse order;

(137)8 + (276)8 = (435)8

Similarly;

1. 1 + 2 = 3

2. 3 + 7 = 10 ⇒ Rem (10/8) = 2

3. 7 + 6 + 1 (Carry from 10) = 14 ⇒ Rem (14/8) = 6

Leaves carry 1.

Write it reverse order;

(731)8 + (672)8 = (1623)8

Hence, "1623" is the correct answer.

5S₁ × 20 = 8 × S₂ × 2S = 10 × 0 × 30 = ……..

\(\Rightarrow \left( {2 \times \frac{a}{{100}} + 6 \times \frac{a}{{200}} + 5 \times \frac{a}{{300}}} \right) \times x = a\) 

⇒ x = 15 days

Given digits: 2, 2, 3, 3, 3, 4, 4, 4, 4

Case 1:

The first digit is 4.

4 _ _ _

Rest of the places are filled by 2, 2, 3, 3, 3, 4, 4, 4

\( \Rightarrow \begin{array}{*{20}{c}} 4&_&_&_\ \downarrow & \downarrow & \downarrow & \downarrow \ 1&3&3&3 \end{array}\)

The no. of cases = 1 × 3 × 3 × 3 – 1 = 26

Case 2:

The first digit is 3:

Rest of the places are filled by 2, 2, 3, 3, 4, 4, 4, 4

The exception cases are = 222, 333

Therefore:\(\begin{array}{*{20}{c}} 3&_&_&_\ \downarrow & \downarrow & \downarrow & \downarrow \ 1&3&3&3 \end{array}\)

No. of cases = 1 × 3 × 3 × 3 – 2

= 25

Case 3:

First digit is 2

No number can be possible which is greater than 3000

Total no. of cases = 26 + 25 = 51

1) H + G > I + S

2) |G - S| = 1

Meaning G and S will be next to each other in order.

So the option (1) is eliminated.

Because S and G are not next to each other, G is not the oldest, and S is not the youngest.

Assume ages are 6, 5, 4, and 2.

From option (2):

S = 6, G = 5, H = 4, I = 2

G + H > I + S

Option (2) satisfy all the conditions.

Now;

Options (3) and (4) are both impossible since in both those cases you would have I + S > H + G.

Hence, option (2) is the correct answer.

Explanation:

Given function f(x) = \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{2}{3}x }}{x }\)

as x → 0 f(x) → 0/0 form

Therefore, by applying L’ Hospital's rule....(differentiate both numerator and denominator)

We will get

\(\mathop {\lim }\limits_{x \to 0} \frac{{\cos \frac{2}{3}x }}{x } \times\frac{2}{3}\)

as x → 0 

\(\mathop {\lim }\limits_{x \to 0} \frac{{\cos \frac{2}{3}x }}{1 } \times\frac{2}{3} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos 0 }}{1 } \times\frac{2}{3}\) = 2/3

Explanation:

When a coin is tossed,

Probability of getting heads = 1/2

Probability of getting tails= 1/2

When two coins are tossed the probability of getting two heads simultaneously is getting heads in first trial and getting head in second trial

P (Getting heads in both trials) \(= \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\)

Concept:

The order of a differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the power of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned

Calculation:

Given:

Squaring on both sides to remove the radicals, we get

\({\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^2} = 16\left[ {{{\left( {\frac{{dy}}{{dx}}} \right)}^3} + {y^2}} \right] \)

The order of the highest derivative is 3

The power of the highest derivative is 2. Hence the degree of the equation is 2.

Concept:

The given information in question can be represented in the given figure:

The water will exert uniform pressure at the bottom plank in the upward direction.

Calculation:

Given:

Since it is given that Plank is floating on water. We get,

Upward force (buoyant force) exerted by water = Downward weight of pressure standing on Plank

(W')(L) = 2W

\(W' = \frac{{2W }}{L}\)

Consider midsection (left) of plank,

Consider Moment at the midpoint to be zero.

∑ Mmid = 0

\({M_{mid}} + W \times \frac{L}{4} = \frac{{{W'}\times L}}{2} \times \frac{L}{4}\)

\({M_{mid}} + \frac{{W \times L}}{4} = \left( {\frac{{2\times W }}{L}}\right)\times \left ( {\frac{L}{2}} \right)\times \left( {\frac{L}{4}} \right)\)

∴ Mmid = 0

Concept:

Let us use the method of joints 

Consider Joint S

∑F_y = 0

F_RS - P = 0

F_RS = P

Since F_RS is acting away from joint S it is tensile.

Consider Joint R

∑Fy = 0

F_TR sin45^° - F_RS = 0

⇒ F_TR = P × √ 2

Since F_TR is acting towards the joint R it is compressive

∑F_x = 0

F_TR cos 45^° - F_QR = 0

F_QR = P

Since F_QR is acting away from the joint it is tensile.

Concept:

When two stress are acting at a point, the maximum shear stress(τ_max) is given by

\({\tau _{max}} = \frac{{{σ _1} - {σ _3}}}{2}\)

σ₁ = Major principal stress

σ₃ = Minor principal stress

Calculation:

Given:

σ1 = 3 MPa

σ3 = -3 MPa

\({\tau _{max}} = \frac{{{3} - {(-3)}}}{2}\)

⇒ 3 MPa

Explanation:

Isotropic Material: 

• Isotropic materials have identical material properties in all directions at every given point.
• When a specific load is applied at any point of isotropic materials, it will exhibit the same strength, stress, strain, young’s modulus, and hardness in the x, y, or z-axis direction.

Examples: metals, glasses.

• Almost every metal exhibits isotropic behaviour at the micro-level because all single-crystal systems are isotropic with respect to mechanical properties.
• There are a few exceptional cases where metals exhibit anisotropic behaviour at the micro-level.
• Random localized distribution of anisotropic behaviour at the micro-level, nullify each other and the net effect at the macro level is isotropic behaviour.

For an isotropic, homogeneous, and elastic material obeying Hook's law, the number of independent elastic constants is 2.

It means to fully define the elastic behaviour of isotropic material only 2 elastic constants are good enough.

E = 2G (1 + μ) = 3K(1 - 2μ)

As seen in the above equations there are a total of 4 elastic constants.

If we know two of them, then we can find other constants also.

∴ there are 2 independent elastic constants required to define the stress-strain relationship for isotropic material.

Additional Information

Orthotropic material

• A material is said to be orthotropic if it has three different properties in three mutually perpendicular directions.
• There are 9 independent elastic constants required to define the stress-strain relationship for orthotropic material.

Anisotropic material

• A material is said to be anisotropic if it has different properties in each direction.
• There are 21 independent elastic constants required to define the stress-strain relationship for an anisotropic material.

Explanation:

Effective length is the distance between two consecutive points of contra-flexure or between points of zero bending moment.

The effective length of column used in calculating the Euler's crippling load are:

• Fixed end: Restrained in both translation and rotation
• Hinged end: Restrained in translation but free in rotation
• Free end: Free in both translation and rotation.

Euler's crippling load ( long columns only): \({P_{cr}} = \frac{{{n^2}{\pi ^2}EI}}{{{L_e}^2}}\) where,

n = no. of buckling loop

I = Minimum moment of inertia

Le = effective length of the column

E = elastic modulus

Concept:

A minimum grade of concrete to be used in the design of the prestressed concrete structure as per IS 1343 is as below:

1. For Post-tensioning minimum grade of concrete used is M-30.
2. For Pre-tensioning minimum grade of concrete used is M-40.

Additional Information

Cover to be used in the design of the prestressed concrete structure as per IS 1343 is as below:

1. For Post-tensioning minimum cover to be used is 30 mm.
2. For Pre-tensioning minimum cover to be used is 20 mm.

Concept:

The torque equation is given by

\(\dfrac{T}{{{I_P}}} = \dfrac{{\rm{\tau }}}{r} = \dfrac{{G\theta }}{L}\)

Where,T = torque, I_P = Polar moment of inertia, τ_max = Maximum shear stress, R = Radius of shaft,

G = Modulus of rigidity, θ = Angle of twist, L = Length of shaft

Calculation:

Given:

Diameter of shaft = d, Length of shaft =  L

 Torque = T, Shear modulus of the material = G.

Polar moment of inertia of the circular section is given by

\({{\rm{I}}_{\rm{P}}} = \dfrac{{{\rm{\pi }}{{\rm{d}}^4}}}{{32}}\)

From the equation: \(\dfrac{T}{{{I_P}}} = \dfrac{{\rm{\tau }}}{r} = \dfrac{{G\theta }}{L}\)

\(\theta = \dfrac{{{\rm{T}} \times {\rm{L}}}}{{{\rm{G}}{{\rm{I}}_{\rm{P}}}}}\)

\({\rm{\theta }} = \dfrac{{32 \times {\rm{T}} \times {\rm{L}}}}{{{\rm{G}} \times {\rm{\pi }}{{\rm{d}}^4}}}\)

Explanation:

The dry density (γ_d) of soil is given by,

\({\gamma _d} = \dfrac{{\left( {1 - {\eta _a}} \right) \times G \times {\gamma _w}}}{{1 + e}}\)

η_a - Percentage of air voids

G - Specific gravity of soil solids

e - Voids ratio

For zero air voids ⇒ volume of air is zero, hence η_a = 0, a_c = 0  

\({\gamma _d} = \dfrac{{G \times {\gamma _w}}}{{1 + e}}\)

\({\gamma _d} = \dfrac{{G \times {\gamma _w}}}{{1 + \left( {\dfrac{{w \times G}}{S}} \right)}}\)

 Also, S + a_c = 1. Hence, S = 1. So above equation becomes, 

\({\gamma _d} = \dfrac{{G \times {\gamma _w}}}{{1 + w \times G}}\)

Concept:

• Plasticity index of soil, I_P = W_L – W_P
• Plasticity index of A-line, I_P = 0.73 ( W_L - 20 )

Calculation:

Given,

Liquid limit ( W_L) = 60%

Plastic limit (W_p) = 20%

I_P of soil = W_L – W_P = 60 - 20 = 40%

I_P of A-line = 0.73 (W_L -20) = 0.73 (60 - 20) = 29.2%

I_P of soil > I_P of A-line, So it will lie above A-line and also W_L > 50%,

∴ Soil is CH.

Quick sand condition:

Quicksand is a condition and not a soil type. This condition is created in saturated thick layers of loose fine sandy soils when disturbed either due to vibration, such as, from pile driving in the neighborhood, or due to the pressure of flowing water (at the time of heavy pumping in excavation).

Explanation :

e – log p curve for normally consolidated clay is a straight line and for over consolidated clay is a curve.

e-log p curve for different clay -

Explanation:

Earth pressure at rest:

• It is the ratio of horizontal and vertical effective vertical stress when the retaining wall does not move at all.
• Sometimes it is also referred to as neural lateral earth pressure or lateral earth pressure
• Ko = \(1 - \sin \phi \)

The coefficient of earth pressure at rest (K0) is given by,

\(K_0=\dfrac{\sigma^{'}_h}{\sigma^{'}_v}\)

σh' – Effective horizontal stress

σv' –  Effective vertical stress

Active earth pressure:

• It is the ratio of horizontal to vertical stress when the retaining wall moves away from the retained soil.
• Ka= \( = \frac{{1 - \sin \phi }}{{1 + \sin \phi }}\)

​​Passive earth pressure:

• It is defined as the ratio of horizontal to vertical stress when the retaining wall moves towards the retaining wall.
• Kp= \(= \frac{{1 + \sin \phi }}{{1 - \sin \phi }}\)

​

Additional Information

Concept:

From \(\frac{{dy}}{{dx}} = \frac{{{S_0} - {S_f}}}{{1 - \frac{{{Q^2}T}}{{g{A^3}}}}} = \left[ {\frac{{1 - {{\left( {\frac{{{y_n}}}{y}} \right)}^{10/3}}}}{{1 - {{\left( {\frac{{{y_c}}}{y}} \right)}^3}}}} \right]{S_0}\)

Using Mannings equation for wide rectangular channel.

(i) As \(y \to {y_0};\frac{{dy}}{{dx}} \to 0;\) i.e, the water surface approaches normal depth line asymptotically

(ii) As \(y \to {y_c};\frac{{dy}}{{dx}} \to \infty ,\) i.e, the water surface meets the critical depth line vertically in region 2 and 3

(iii) As \(y \to \infty ,\frac{{dy}}{{dx}} \to 0;\) i.e, the water surface meets a very large depth as horizontal asymptote in region 1.

(iv) As \(y \to 0,\frac{{dy}}{{dx}} \to \infty\), means that the surface

• M1, M2, S2 and S3 meet y0 lines asymptotically
• M1, and S1 curve tend to horizontal as y → ∞
• M2, M3 and S2 meet yc line normally
• M3 and S3 meet channel bed normally.

So When A mild-sloped channel is followed by a steep-sloped channel

We Find the M2, S2 Profiles.

Concept:

When the width of the channel is reduced the term q (Discharge per unit width) increases. The no choke condition means critical depth of flow is not reached.

Case 1: Width of channel is B

The specific energy curve is drawn for the channel

Case 2: Width of channel is B₁ (B₁ ≤ B)

The specific energy curve corresponding to the width B₁ is drawn.

When we consider the subcritical portion the graph,

We find that for the same specific energy the depth decreases when there is reduction in width of the channel. Hence there is a drop at a downstream section.

Concept:

Horton's Formula:

Infiltration is the process in which water enters into the soil and the rate at which water enters into the soil is known as infiltration rate. Under the ponding condition, the infiltration rate decreases exponentially, and it can be represented by the Horton equation. According to Horton’s equation,

\({\rm{f}} = {{\rm{f}}_{\rm{c}}} + \left( {{{\rm{f}}_0} - {{\rm{f}}_{\rm{c}}}} \right) \times {{\rm{e}}^{ - {\rm{kt}}}}\)

Where,

f = infiltration capacity at time t, f0 = initial infiltration capacity,

fc = constant infiltration capacity after a long time, and k= Horton’s decay constant

Hydrological Channel Routing:

In channel routing, the storage is a function of both outflow and inflow discharge.

s = f (I, Q) 

The total volume stores in a channel reach can be considered under two categories

• Prism Storage
• Wedge storage

The total storage is given by,

s = K [xIm + (1 - x)Qm]

Where, m = a function depends on the type of a channel 

For a natural channel, m = 1

So, s = K [xI + (1 - x)Q]        ........(1)

So, Muskingum method is a type of hydrological channel routing.

​Synthetic Unit Hydrograph:

The hydrograph derived from basin characteristics is called Synthetic Unit Hydrograph. It was introduced by Synder.

 Empirical method of Evapotranspiration:

i) Blaney Criddle method

ii) Hargreaves pan evaporation method

iii) Modified penman method

Additional Information

Concept

Pitot tube:

Pitot tube is a pressure measurement instrument used to measure fluid velocity.

A pitot tube is used to measure the local velocity at a given point in the flow stream and not the average velocity in pipe or conduit.

It is an L shape tube whose one end is inserted in a pipe and the other end is open to the atmosphere. The point where the Pitot tube is inserted is called the ‘stagnation point’.

The stagnation point is the point at which the velocity head is converted into a pressure head.

Manometer:

Manometer is a device used to measure gauge and vacuum pressure, not absolute pressure.

Absolute pressure = Atmospheric pressure + Gauge pressure

Absolute pressure = Atmospheric pressure – Vacuum pressure. It is a measuring device used to measure the flow rate or discharge. It works on the principle of Bernoulli’s theorem.

Venturimeter:

Venturimeter consists of three-part

i) Converging cone

ii) Throat

iii) Diverging cone

Rate of flow is calculated by 

\({{\rm{Q}}_{{\rm{th}}}} = {C_d} \times \dfrac{{{A_1} \times {A_2}}}{{\sqrt {{A_1^2} - {A_2^2}} }} \times \sqrt {2 \times g \times h} \)

Where Cd is the coefficient of discharge of venturimeter.

A₁ is the cross-sectional area at the throat

A₂ is the cross-sectional area at enlarged section

h is the difference of pressure head between the throat and enlarged section 

Note: For venturimeter Cd = 0.94 – 0.98 

Anemometer:

An anemometer is an instrument that measures wind speed and wind pressure.

Anemometers are important tools for meteorologists, who study weather patterns

Explanation

Composting can be adopted when the MSW contains high organic content. Barging the MSW into the sea is now obsolete.

1) Landfill-

• In this method, refuse is dumped into the low-lying area into the layer of depths of approximately 1.5m. Each layer of refuse is termed a cell.
• Each layer is compacted properly and left for 7 days for its aerobic decomposition & it covered with a layer of mud in order to carry out the further process of decomposition. The entire process of decomposition in this method is biological and takes place in three stages.
• For the optimum decomposition of refuse, the moisture content in cells should not be kept below 60%. When decomposition of solids takes place height of the landfilling is reduced to 20% - 40% of its original height.
• In the initial stages decomposition takes place in the mesophilic range and in the final stages, it is shifted to the thermophilic range. The entire process of digestion is completed in 2 – 32 months.
• The only disadvantage associated with it is the release of gases and the formation of leachate. Leachate is a mixture of acids, alcohols, non–biodegradable waste, and water which may be carcinogenic. It is formed in the rainy season when water seeps through cells.

2) Incineration & thermal pyrolysis

• Incineration is a method in which refuse is burnt at a higher temperature. It is generally adopted for those wastes which pass high fuel value so that high heat can be produced which can be used for power generation. It is an exothermic process.
• Thermal pyrolysis is a method in which waste is burnt in a limited supply of O₂ to avoid its complete burning resulting in the formation of the charcoal-like product. It is done for waste that has a high calorific value and it is an endothermic process.

3) Dumping into the sea-

• Disposal of solid waste into sea is done only at the times of low tides for away from coasts (15-20 km away from shore) and 30 m below sea level. It comes back during high tide.
• Municipal Solid Waste (MSW) contains high moisture content, high organic materials, low calorific value, and low inorganic materials. The most effective and sustainable option for MSW management will be composting.

Key Points

• Incineration can be adopted when the calorific value of the MSW is high.
• Landfill can be adopted when the density of MSW is high. Since in the given MSW the quantity of inorganic material is low, its density is less.

Ambient air quality standards in respect of Noise as per Noise Pollution (Regulation and Control) Rules, 2000, of the Ministry of Environment & Forests, India are tabulated below:

Explanation:

Dry air cools at the rate of 9.8°C per km and it is called dry adiabatic lapse rate.

In saturated (wet) air, this rate is calculated to be 6°C per km and is known as wet adiabatic lapse rate.

Resulting temperature of air

\(= 40 - 9.8 \times \frac{{500}}{{1000}} = 40 - 4.9 = 35.1^\circ {\rm{C}}\)

Additional InformationEnvironmental Lapse Rate (ELR):

• In the troposphere, the temperature of the ambient air normally decreases with an increase in height. The rate of change of temperature of the ambient air is called environmental lapse rate (ELR) or ambient lapse rate.
• When dry air expands and cools adiabatically the lapse is about 9.8° C per km and known as the adiabatic lapse rate.
• When ELR is greater than the adiabatic lapse rate, the prevailing lapse rate is known as the super-adiabatic lapse rate. In this case, the environment is unstable.
• When ELR is less than the adiabatic lapse rate, the prevailing lapse rate is known as the sub-adiabatic lapse rate. In this case, the environment is stable.
• When the temperature of the ambient air increases with an increase in height, that situation is called an inversion in lapse rate.

Concept:

Aggregate impact value test gives an indication of the aggregate’s toughness property (i.e. property of a material to resist impact).

Aggregates to be used for wearing course, the impact value shouldn’t exceed 30 percent.

For bituminous macadam, the maximum permissible value is 35 percent.

For Water bound macadam base courses the maximum permissible value defined by IRC is 40 percent.

Additional Information

Explanation:

As per IRC: 67-2012, clause 3.2, Speed limit sign:

Speed limit signs are circular in shape with the red border and black symbols on white background.

Cautionary/ warning signs:

Cautionary/ warning signs are triangular in shape with the red border and black symbols on white background. For example, hairpin bend, narrow bridge, school ahead, etc.

Standard meridian = 82° 30’ E

Local meridian = 90° 40’ E

∴ Local meridian is ahead of standard meridian by 8°10'.

As 360°  corresponds to 24 hours

∴ 8°10' corresponds to \(\frac{{8^\circ 10'}}{{360}} \times 24 = 0.544;hours\) = 32 minutes 40 Second

∴ Local time = 6 hours 30 minutes + 0 hours 32 minutes 40 Second

∴ Local time = 7 hours, 02 minutes and 40 seconds

Explanation:

The given ordinary differential equation (ODE) is

\(\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}-6y=0\)

The given equation can be written as

D = d/dx

(D² +D - 6) y = 0

The auxiliary equation is D² + D - 6 = 0 ⇒ (D - 2) × (D + 3) = 0

The roots of auxiliary equation are -3 and 2

The complementary function C.F = c₁e^-3x + c₂e^2x

Since the RHS = 0, there in no particular integral

The solution is y = c₁e^-3x + c₂e^2x

Concept:

For a 2 x 2 matrix, there is a short-cut formula for inverse is given below

\(\rm {\begin{bmatrix} a & & b\ c& & d \end{bmatrix}}^{-1} = \frac{1}{(ad-bc)}\begin{bmatrix} d & & -b\ -c& & a \end{bmatrix}\)

Calculation:

We know that inverse of the matrix \(\rm {\begin{bmatrix} a & & b\ c& & d \end{bmatrix}}^{-1} = \frac{1}{(ad-bc)}\begin{bmatrix} d & & -b\ -c& & a \end{bmatrix}\)

a = 3 + 2i, b = -i

c = i, d = 3 - 2i

⇒ \({\begin{bmatrix} 3+2i & i\- i & 3-2i \end{bmatrix}}^{-1}= \frac{1}{(3+2i)(3-2i)-1}\begin{bmatrix} 3-2i & -i\ i & 3+2i \end{bmatrix}\)

⇒ \({\begin{bmatrix} 3+2i & i\- i & 3-2i \end{bmatrix}}^{-1}= \frac{1}{4+9-1}\begin{bmatrix} 3-2i & -i\ i & 3+2i \end{bmatrix}\)

⇒ \({\begin{bmatrix} 3+2i & i\- i & 3-2i \end{bmatrix}}^{-1}= \frac{1}{12}\begin{bmatrix} 3-2i & -i\ i & 3+2i \end{bmatrix}\)

Hence, option 2 is correct.

Concept:

The Simpson’s rule is given by,

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

h –Width of interval / Step length

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

Calculation:

Given:

h = 0.25

\(;{\rm{I}} = \dfrac{{0.25}}{3}[\left( {1 + 0.5} \right) + 4\left( {0.9412 + 0.64} \right) + 2\left( {0.8} \right) = 0.7854\)

Concept:

To find the partial differential equation differentiate the given equation with respect to x and y

Calculation:

Given:

z = ax + by + ab

\(\dfrac{{\partial {\rm{z}}}}{{\partial {\rm{x}}}} = {\rm{a}} = {\rm{p}}\)

\(\dfrac{{\partial {\rm{z}}}}{{\partial {\rm{y}}}} = {\rm{b}} = {\rm{q}}\)

Substituting the values of a and b in the given equation we get,

z = px + qy + pq

Concept: 

Length of the curve f(x) between x = a and x = b is given by

Length of curve f(x) \(= \mathop \displaystyle \int \nolimits_{\rm{a}}^{\rm{b}} \sqrt {1 + {{\left( {\dfrac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^2}} {\rm{dx}} \)

\(y=4h\dfrac{{{x^2}}}{{{L^2}}}\)

y - Vertical coordinate with the origin at the centre of the cable (Rise of cable)

h – Sag at mid span, x – Horizontal coordinate, L - Horizontal span between the supports

Calculation:

Given:

We know

\(y=4h\dfrac{{{x^2}}}{{{L^2}}}\)

\(\dfrac{{dy}}{{dx}} = 8h \times \dfrac{x}{{{L^2}}}\)

At supports x =0, y = 0

At mid span x = L/2, y =h

The cable is symmetrical about centre, we consider the left portion of cable

Length of the cable \(= 2 \times \mathop \displaystyle \int \nolimits_0^{\frac{{\rm{L}}}{2}} \sqrt {1 + {{\left( {\dfrac{{8 \times {\rm{h}} \times {\rm{x}}}}{{{{\rm{L}}^2}}}} \right)}^2}} {\rm{dx}} \)

⇒ Length of the cable \(= 2\mathop \displaystyle \int \nolimits_0^{{\rm{L}}/2} \sqrt {\begin{array}{*{20}{c}} {1 + 64\dfrac{{{{\rm{h}}^2}{{\rm{x}}^2}}}{{{{\rm{L}}^4}}}}\ ; \end{array}} {\rm{;dx}} \)

Concept:

\(\begin{array}{l} {\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = 4{{\rm{x}}^2} + 6{{\rm{y}}^2} - 8{\rm{x}} - 4{\rm{y}} + 8\ {{\rm{f}}_{\rm{x}}} = 8{\rm{x}} - 8,{\rm{;}}{{\rm{f}}_{\rm{y}}} = 12{\rm{y}} - 4\ {{\rm{f}}_{{\rm{xx}}}} = 8,{{\rm{f}}_{{\rm{xy}}}} = 0,{{\rm{f}}_{{\rm{yy}}}} = 12\ {{\rm{f}}_{\rm{x}}} = 0 \Rightarrow {\rm{x}} = 1,{{\rm{f}}_{\rm{y}}} = 0 \Rightarrow {\rm{y}} = \frac{1}{3} \end{array}\)

At \(\left( {1,\frac{1}{3}} \right);{{\rm{f}}_{{\rm{xx}}}}{\rm{;}}{{\rm{f}}_{{\rm{yy}}}} - {\rm{f}}_{{\rm{xy}}}^2 = 96 > 0\)

And \({{\rm{f}}_{{\rm{xx}}}} = 8 > 0 \Rightarrow {\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right)\) has a minimum at \(\left( {1,{\rm{;}}\frac{1}{3}} \right)\).

Minimum value \(= {\rm{f}}\left( {1,\frac{1}{3}} \right) = 4 + \frac{2}{3} - 8 - \frac{4}{3} + 8 = \frac{{10}}{3}\).

Explanation

As Nominal diameter of rivet is 20 mm, in the range of ( 12 mm - 24  mm )

Gross diameter = 20 + 1.5 = 21.5 mm

Strength of rivet in bearing = σb × d × t

Strength of rivet in single shear = τvf × π × d2/4

Strength of rivet in double shear = 2 × τvf × π × d2/4

Where,

d is gross dia of rivet

σb is allowable bearing stress in plate

τvf is allowable shear stress in rivets

For single shear

Rivet value = smaller (Bearing strength, single shear strength)

For Double shear

Rivet value = smaller (Bearing strength, Double shear strength)

Calculation:

From the figure it is clear that 

Pitch = 50 mm, Gauge = 50 mm, end distance = 50 mm

Permissible bearing stress = 300 MPa, permissible shearing stress = 100 MPa

t = 14 mm, number of bolts = 12

Thickness of cover plate = 10 m

Given,

d = 20 mm, Gross dia = 20 + 1.5

= 21.5 mm

σb = 300 Mpa, τvf = 300 MPa 

Thickness of plate (t) = 14 mm

Strength of rivet in bearing =  σb × d × t  (Double shear)

=  300 × 21.5 × 14

=  90300 N = 90 kN

Strength of rivet in double shear = 2 × τvf × π × d2/4

= 2 × 100 × π × 21.52/4 = 72.5 kN

Rivet value in single shear = min (90 kN, 72.5 kN)

= 72.5

12 number of bolts 

So strength = 12 × 72.5 kN = 871.32 kN

Concept:

When filet weld is applied to the square edge of the member, the maximum size of the weld should be less than the edge thickness by at least 1.5 mm.

Maximum size of the fillet weld = Thickness of the thinner plate – 1.5 mm

Strength of the thinner plate = σ × A_t

σ - Permissible stress in the plate, A_t - Area of thinner plate

Strength of the weld = t_t ×L× τ 

t_t - Throat thickness = 0.7 × size of weld

L - Length of weld, τ - Shear stress in the weld

Before failure, Strength of the thinner plate = Strength of the weld

Calculation:

Given:

Maximum size of plate = 10 - 1.5 = 8.5 mm

t_t  = 0.7 × 8.5 = 5.95 mm

A_t = 100 × 10 = 1000 mm²

Strength of the thinner plate = 150 × 1000 =150000 N/mm² ..... (1)

Strength of the weld = 110 × 5.95 × L = 654.5 × L N/mm² .....(2)

Equating 1 and 2 we get,

654.5 × L = 150000

⇒ L = 229.18 m

Concept:

Σ M_B = 0

R_A. L - M.L = 0

R_A = M

R_B = – M

BM at mid-span \(\Rightarrow {R_A} \times \frac{l}{2} - M \times \frac{l}{2} = 0\)

Hence bending moment is zero at the center.

Moreover, BM throughout the whole beam is zero

Confusion Points

Here 'M' is BM per unit length is given not the load per unit length.

Concept:

The distance of centroid of shaded region from centre ‘O’is given by

\({\rm{x}} = \dfrac{{{{\rm{A}}_1} \times {{\rm{x}}_1} - {{\rm{A}}_2} \times {{\rm{x}}_2}}}{{{{\rm{A}}_1} - {{\rm{A}}_2}}}\) (Negative sign because a some part of section is removed)

A₁ – Area of the disc

x₁ –   Radial distance of centroid of the disc from centre O

A₂ – Area of the hole

x₂ – Radial distance of centroid of hole from centre O

Calculation:

Given:

A₁ = π × r², x₁ = 0 (∵ centroid coincides with the centroid)

A₂ = π × (r/2)², x₂ = r/2

\({\rm{x}} = \dfrac{{\pi \times {r^2} \times 0 - {\rm{\pi }} \times {{\left( {\dfrac{r}{2}} \right)}^2} \times \dfrac{r}{2}}}{{\pi \times {r^2} - {\rm{\pi }} \times {{\left( {\dfrac{r}{2}} \right)}^2}}} = - \dfrac{r}{6}\)

Negative sign indicates the centroid is at a located at the third quadrant at a radial distance of r/6 from the centre O.

Explanation

Given 3 hinged arch

It is asked to determine resultant reaction at left support (i.e. A)

Σ F_y = 0

V_A + V_B = 10 kN

Σ M_B = 0

V_A × 20 – 10 × 15 = 0

V_A = 7.5 kN

Now taking moment about crown C

ΣM_C = 0

• H_A × 5 + V_A × 10 – 10 × 5 = 0

 7.5 × 10 – 10 × 5 = 5 × H_A

H_A = 5 kN

Now resultant Force at support A is 

\(\sqrt {V_A^2 + H_A^2} = \sqrt {{{7.5}^2} + {5^2}} = 9.01;\)

∴ Resultant Force is 9.01 kN

Asked angle of inclination with horizontal 

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{V_A}}}{{{H_A}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{7.5}}{5}} \right) = 56.3099^\circ \)

θ ≈ 56.31 °

Concept:

Vertical stress due to concentrated load is given as per Boussinesq’s equation.

\({{\rm{σ }}_{\rm{z}}}{\rm{;}} = {\rm{;}}\frac{3}{{2{\rm{\pi }}}}\frac{{\rm{Q}}}{{{{\rm{z}}^2}}}{\left( {\frac{1}{{1 + {{\left( {\frac{{\rm{r}}}{{\rm{Z}}}} \right)}^2}}}} \right)^{\frac{5}{2}}}\)

where,

σ_z = Stress magnitude, Q = Vertical point load, r = Radial distance, and z = Vertical distance.

Calculation:

Given:

For Point P₁  r = r, z = z

For Point P2  \(z = \frac{z}{2}\), \(r = \frac{r}{2}\)

Case 1: For Point P₁

 r = r, z = z

\({{\rm{σ }}_{\rm{z}}}{\rm{;}} = {\rm{;}}\frac{3}{{2{\rm{\pi }}}}\frac{{\rm{Q}}}{{{{\rm{z}}^2}}}{\left( {\frac{1}{{1 + {{\left( {\frac{{\rm{r}}}{{\rm{Z}}}} \right)}^2}}}} \right)^{\frac{5}{2}}}\)...... ( 1 )

Case 2: For Point P₂ 

\(r = \frac{r}{2}\),\(z = \frac{z}{2}\)

\({{\rm{σ }}_{\rm{z}}}{\rm{;}} = {\rm{;}}\frac{3}{{2{\rm{\pi }}}}\frac{{\rm{Q}}}{{{{\rm{(z/2)}}^2}}}{\left( {\frac{1}{{1 + {{\left( {\frac{{\rm{r/2}}}{{\rm{z/2}}}} \right)}^2}}}} \right)^{\frac{5}{2}}}\)

\({{\rm{σ' }}_{\rm{z}}}{\rm{;}} = {\rm{;}}4× \frac{3}{{2{\rm{\pi }}}}\frac{{\rm{Q}}}{{{{\rm{z}}^2}}}{\left( {\frac{1}{{1 + {{\left( {\frac{{\rm{r}}}{{\rm{Z}}}} \right)}^2}}}} \right)^{\frac{5}{2}}}\)

σ^'_z = 4 × σ_z

Concept:

The coefficient of permeability is given by Darcy's Law

v = k × i

v - Average discharge velocity

k - Coefficient of permeability

i - hydraulic gradient.

The coefficient of permeability can also be defined as the average velocity of flow under unit hydraulic gradient.

When i =1, v = k

\({\rm{Coefficient;of;permeability}} = \frac{{{\rm{Drop;in;heightof;water;level}}}}{{{\rm{Time;taken}}}}\)

Calculation:

Given:

Drop-in water level = 0.75 m

k = 10^-2 m/s

Using the equation: \({\rm{Coefficient;of;permeability}} = \frac{{{\rm{Drop;in;heightof;water;level}}}}{{{\rm{Time;taken}}}}\)

\({10^{ - 2}} = \frac{{0.75}}{{{\rm{Time;taken}}}}\)

⇒ Time taken = 75 s

Concept:

The ultimate strength pile is the sum of the total ultimate skin friction and total ultimate end bearing resistance. It is given by 

Qu = q Ab + q As = c_p × N_c ×  Ab + ∝ × c × A_s

∝ - Adhesion factor

c_p - the value of cohesion at the pile tip

c - value of cohesion along the length of the pile

For clay N_c = 9

Calculation:

Given:

Q_u = 700 kN, c_p= c = 40 kN/m²,

Size of pile (B × B) = 500 mm × 500 mm = 0.5 m × 0.5 m

Ab = 0.5 × 0.5,

As = 4 × Length of pile × B = 4 × 10 × 0.5

Length of pile = 10 m

Length of pile = 10 m

700 = 40 × 9 × 0.52 + α × 40 × (0.5 × 4 × 10)

∝ = 0.7625

The length of the pile is increased to 20 m & the cross-section of the pile is reduced to 250 mm × 250 mm

Ab = 0.25 × 0.25

As = 4 × Length of pile × B = 4 × 20 × 0.25

Q_u = 40 × 9 × 0.252 + 0.7625 × 40 × (0.25 × 4 × 20) --- (Using α = 0.7625)

Q_u = 632.5 kN.

Concept:

Most economical section is the one whose wetted perimeter is minimum for the given value of discharge.

For  a hydraulically efficient  rectangular channel of width b and depth d, the following conditions has to be satisfied

1. \({\rm{d}} = \frac{b}{2}\)..... (1)
2. \({\rm{m}} = \frac{d}{2}\)....(2)

Hydraulic radius (D) is defined as the ratio of the area of the channel to the top width.

\({\rm{D}} = \frac{A}{T}\)

Top width of channel = b = 2× d (from 1)

Hydraulic depth is given by

\({\rm{D}} = \frac{b × d}{b}\)

⇒Hydraulic depth, D = d

Perimeter of channel = b + d + d = b + 2 × d = 2 × d + 2 × d

∴ Perimeter of channel = 4 × d

Concept:

Froude number (F) for a rectangular channel is given by,

\(F= \frac{V}{{\sqrt {9.81 × y} }}\)

Where,

V = Velocity of flow through the rectangular channel

y = Depth of flow

Critical depth of flow y_c is given by

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

​q = discharge per unit width

\(q = \frac{Q}{b} = \frac{{V × A}}{b} = \frac{{V × y × b}}{b}\)

∴ q = V × y

​​Calculation:

Given:

F = 0.8, y = 1.5 m​​

\(0.8 = \frac{V}{{\sqrt {9.81 × 1.5} }}\)

V = 3.07 m/s

q = 3.07 × 1.5 = 4.6 m³/s/m

\({y_c} = {\left( {\frac{{{4.6^2}}}{9.81}} \right)^{\frac{1}{3}}}\)

∴ y_c = 1.29 m

Concept:

The discharge (Q) through a confined aquifer is given by,

\(Q = \dfrac{{2 \times \pi \times T \times \left( {{s_1} - {s_2}} \right)}}{{\ln \dfrac{{{r_1}}}{{{r_2}}};}}\)

Q – Discharge through the aquifer

T – Transmissivity

s₁, s₂ – Drawdown in observation wells 1 and 2 respectively

r₁, r₂ – radial distance of observation wells 1 and 2 respectively

Calculation:

Given:

r₁ = 10 m, r₂ = 100 m

s₁ = 3 m, s₂ = 0.5 m

Q = 2720 litres/minute = 2720 × 10^­-3 × 60 × 24 m³/day

\(2720 \times {10^{ - 3}} \times 24 \times 60 = \dfrac{{2 \times \pi \times T \times \left( {3 - 0.5} \right)}}{{\ln \dfrac{{100}}{{10}};}}\)

⇒ T = 576 m²/day

Concept:

The BOD consumed in ‘ t ’ days is given by,

Yt = L × [1 - e- (k × t)]

Yt – BOD consumed at t days

L – Ultimate BOD, t - Time in days, k - Rate constant. 

Y_t = L - L_t

Lt - BOD left after t days 

Calculation:

Given:

BOD3 = Y3 = 75 mg/L

t = 3 days, k = 0.345

\({{\rm{75}}} = {\rm{L}} \times \left( {;1 - {{\rm{e}}^{ - \left( {;{\rm{0.345}}; \times ;{\rm{3}};} \right)}};}\right)\)

L = 116.32 mg/L

The BOD consumed in 10 days Y10 is given by,

\({{\rm{Y}}_{\rm{10}}} = {\rm{116.32}} \times \left( {;1 - {{\rm{e}}^{ - \left( {;{\rm{0.345}}; \times ;{\rm{10}};} \right)}};} \right)\)

Y10 = 112.63 mg/L

BOD remaining after 10 days

112.63 = 116.32 - Lt

Lt = 3.69 mg/L

Concept:

Vertical curve:

Vertical alignment is the elevation of the centerline of the road. It may be in the form of a circular arc or a parabola

The common practice has been to use parabolic curves in summit curves. Parabolic curves with it can be laid out with ease as well as allowing a comfortable transition from one gradient to another. Hence statement I is true.

Although a circular curve offers equal sight distance at every point on the curve, for very small deviation angles a circular curve. and parabolic curves are almost congruent.

In valley curves the centrifugal force will be acting downwards along with the weight of the vehicle and hence impact to the vehicle will be more. It will result in jerking and cause discomfort to passengers. Hence comfort to passengers has to be considered in valley curves. Hence statement II is false.

The vertical curve is divided into two categories

1. Summit curve:

It is a crest curve with convexity upward.

The deviation angle between the two interacting gradients is equal o the algebraic difference between them.

When an ascending gradient meets with descending gradients the deviation angle will be maximum

N = n1 - (-n2) = n1 + n2.

Concept:

Off-tracking: 

The phenomenon of the rear wheel not following the front wheels while negotiating the curve is called off-tracking. It depends on

1. Length of wheelbase
2. Turning angle or radius of the curve.

\({ off -tracking}; = ;\frac{{{l^2}}}{{2R}}\)

The extra widening (W_e)of pavement is given by

We = Wm + Wp

Wm = Mechanical widening, Wp = physiological widening

Mechanical widening: The widening required to account for the off-tracking due to rigidity of wheelbase is known as Mechanical widening. It can be calculated as:

\({W_m} = \frac{{n{l^2}}}{{2R}}\)

n = Number of lanes

l = length of wheelbase = 6.1 m

R = Radius of circular curve

Psychological Widening: Extra width of pavement provided for psychological reasons such as overhangs of vehicles, greater clearance for crossing, etc is known as Psychological Widening.

\({W_{p}} = \frac{V}{{9.5\sqrt R }}\)

V = Design speed (kmph) 

Calculation:

Given - l = 6.6 m, n = 2, V = 80 km/hr

off-tracking = 0.096 m

\({0.096}; = ;\frac{{{6.6^2}}}{{2\times R}}\)

⇒ R = 226.875 m

\({{\rm{W}}_{\rm{e}}} = 2 \times 0.096 + \frac{{80}}{{9.5 \times \sqrt {226.875} }}\)

W_e = 0.752 m

Concept:

Warping stresses are due to daily variation of temperature and frictional stresses are due to seasonal variation in temperature.:

i) Warping stresses: Due to daily variation of temperature

• During the daytime the top surface of the pavement is at a higher temperature as it is directly exposed to the sun than the bottom surface, therefore there occurs a temperature difference between the top and bottom layer which causes warping of the pavement.
• As at top slab tries to expand to resist that internal compression stress develops on top and at the bottom, the slab tries to contract but to resist that internal tensile stress develops at the bottom.
• During night time the top surface of the pavement is at a lower temperature than the bottom surface, therefore there occurs a temperature difference between the top and bottom layer which causes warping of the pavement.
• As the top slab tries to contract but to resist it internal tensile stresses develop at the top and compressive stress at the bottom.

Hence statement 1 is false.

Tie bars 

• They used in longitudinal joints in concrete pavement.
• These are not load transfer devices but serve as a means to tie two slabs. Hence statement 2 is false.

Additional Information

Frictional stresses: Due to seasonal variation of temperature.

• Change in seasonal temperature induces frictional tension or frictional compression due to expansion and contraction.
• In an analysis, frictional stresses are assumed as constant throughout the length lx but actually, frictional stresses are zero as corners and maximum at center.

Concept:

The height of the instrument is first established by means of taking sight towards the benchmark of known elevation.

1. Height of instrument = R.L of bench mark + Back sight reading
2. Elevation of station = Height of instrument - Fore sight reading

Calculation:

Given:

R.L of benchmark = 100.0 m, Back sight reading = - 2.105 m (staff is inverted)

Fore sight reading = 1.105

Height of instrument = 100 + (-2.105) = 97.895 m

Elevation of plinth = 97.895 - 1.105 = 96.790 m.

Concept:

Total Hardness=\( \frac{{\left[ {C{a^{ + + }}} \right]mg/L}}{{Equivalent ~weight;of;{Ca}}} \times Equivalent;weight;of;CaC{O_3}\)(+ \frac{{\left[ {M{g^{ + + }}} \right]mg/L}}{{Equivalent~weight of;Mg}} \times Equivalent;weight;of;CaC{O_3}\)

Calculation:

Given:

[Ca²⁺] = 100 mg/l

[Mg²⁺] = 6 mg/l

Equivalent weight of calcium = 20 g

Equivalent weight of magnesium = 12 g

Equivalent weight of Calcium carbonate = 50 g

\({\rm{Total;hardness}} = \frac{{100}}{{20}} \times 50 + \frac{6}{{12}} \times 50\)

⇒ 275 mg/L

Concept:

Alkalinity in water is a measure of the concentration of carbonate, bicarbonate, and hydroxide ions in water

\(Alkalinity = \left[ {\frac{{\left[ {HCO_3^ - } \right]mg/L}}{{Eq.wt.of;HC{O_3}}} + \frac{{\left[ {O{H^ - }} \right]mg/L}}{{Eq.wt.of;OH}} + \frac{{\left[ {CO_3^2} \right]mg/L}}{{Eq.wt.;of;C{O_3}}}} \right] \times Equivalent;weight;of;CaC{O_3}.\)

Carbonate hardness is the minimum of total hardness and alkalinity.

1. When the alkalinity (as CaCO3) is greater than the total hardness, all the hardness is carbonate hardness, carbonate hardness  = Total hardness
2. When the alkalinity (as CaCO3) is less than the total hardness,

           carbonate hardness = alkalinity 

Calculation:

Given:

Total Hardness = 275 mg/L

[ HCO₃^-] = 250 mg/L

\({\rm{Alkalinity}} = \frac{{250}}{{61}} \times 50\)

⇒ 205 mg/L

Carbonate hardness is minimum of (275, 205)

Carbonate hardness = 205 mg/L.

Concept:

The capacity of the pump is given by,

\(Capacity~of ~the~pump= \frac{{{\rm{Volume;of;water;required}}}}{{Working;time;of;pump}}\)

The frequency of irrigation is given by,

\(f_w =\frac{{dw}}{{{C_u}}} \)

d_w - Depth of water to be supplied

c_u - Daily consumptive use rate

The water application efficiency(η ) is given by

\({\rm{η }} = \frac{{Depth;of;water;stored;in;root;zone}}{{Depth;of;water;applied~to ~soil}}\)

Calculation:

Given:

Area of the irrigation for crop 'X' = 36 hectare

cu = 0.5 cm/day, ηa - 75%

Moisture holding capacity of soil = 18 cm/m × 1 = 18 cm

Allowable depletion of moisture = 50% of available moisture

Allowable depletion of moisture = 0.5 × 18 = 9 cm

Frequency of irrigation crop, \(\frac{{dw}}{{{C_u}}} = \frac{{9}}{0.5} = 18;days\)

Hence water has to be supplied every 18 days

\({\rm{0.75}} = \frac{{9}}{{Depth;of;water;applied~ to~ soil}}\)

Depth of water applied = 12 cm

Quantity of water required = 0.12 ×  36 × 10⁴ = 43200 m³

\(Capacity ~of~ pump= \frac{{{\rm{43200\times 10^3}}}}{{18\times 10\times 60\times 60}} \)

⇒ 66.67 litres/sec

Concept:

The capacity of the pump is given by,

\(Capacity~of ~the~pump= \frac{{{\rm{Volume;of;water;required}}}}{{Working;time;of;pump}}\)

The frequency of irrigation is given by,

\(f_w =\frac{{dw}}{{{C_u}}} \)

dw - Depth of water to be supplied

cu - Daily consumptive use rate

The water application efficiency(η ) is given by

\({\rm{η }} = \frac{{Depth;of;water;stored;in;root;zone}}{{Depth;of;water;applied~to ~soil}}\)

Calculation:

Given:

cu = 4 mm/day = 0.4 cm/day, ηa - 75%

Moisture holding capacity of soil = 18 cm/m × 0.8 = 14.4 cm

Allowable depletion of moisture = 50% of available moisture

Allowable depletion of moisture = 0.5 × 14.4 = 7.2m

Frequency of irrigation crop, \(\frac{{dw}}{{{C_u}}} = \frac{{7.2}}{0.4} = 18;days\)

Hence water has to be supplied every 18 days

\({\rm{0.75}} = \frac{{7.2}}{{Depth;of;water;applied~ to~ soil}}\)

Depth of water applied = 9.6 cm

Quantity of water required = 0.096 × A 

\(\rm 40~litres/sec= \frac{{{\rm{0.096\times A}}}}{{18\times 10\times 60\times 60}}\)

A = 27 hectares.

Concept:

Resultant compressive force in concrete is given by,

\({C_C} = 0.36 \times {f_{ck}} \times b \times {x_u}\)

Resultant compressive force in the compressive steel is given by,

\({C_S} = {A_{sc}} \times \left( {{f_{sc}} - 0.45 \times {f_{ck}}} \right)\)

Given compression and tension steel yield

∴ fsc = 0.87 fy

\({C_S} = {A_{sc}} \times \left( {0.87 \times {f_y} - 0.45 \times {f_{ck}}} \right)\)

Total Tensile force T = 0.87 fy × Ast

To find the Actual Neutral Axis, Equate Tensile and Compressive forces

Total compressive force = Total tensile force

i.e. Cc + Cs = T

Calculation

Given:

b = 300 mm, d = 500 mm,

A_st = 2200 mm2, A_sc = 628 mm²

M 20,∴ f_ck = 20 N/mm², Fe 250, ∴ f_y = 250 N/mm²

Given compression and tension steel yield

∴ fsc = 0.87 fy

To find the neutral axis, we equate the Total compressive and tensile forces

\(0.36 \times {f_{ck}} \times b \times {x_u} + {A_{sc}} \times \left( {0.87 \times {f_y} - 0.45 \times {f_{ck}}} \right) = 0.87 \times {f_y} \times {A_{st}}\)

\({x_u} = \frac{{0.87 \times {f_y} \times {A_{st}} - ;{A_{sc}} \times \left( {0.87 \times {f_y} - 0.45 \times {f_{ck}}} \right)}}{{0.36 \times {f_{ck}} \times b}}\)

\({x_u} = \frac{{0.87 \times 250 \times 2200 - ;628 \times \left( {0.87 \times 250 - 0.45 \times 20} \right)}}{{0.36 \times 20 \times 300}};\)

= 160.91 mm

∴ The actual Neutral axis depth is 160.91 mm

Concept:

The moment of resistance(M_u) of the doubly reinforced beam is given by

M_u = C_C × (d - 0.42 × x_u) + C_S × (d - d')

Resultant compressive force in concrete (CC) is given by,

\({C_C} = 0.36 × {f_{ck}} × b × {x_u}\)

Resultant compressive force in the compressive steel(C_S) is given by,

\({C_S} = {A_{sc}} × \left( {{f_{sc}} - 0.45 × {f_{ck}}} \right)\)

Calculation:

Given:

b = 300 mm, d = 500 mm,

Ast = 2200 mm2, Asc = 628 mm2

M 20,∴ fck = 20 N/mm², Fe 250, ∴ fy = 250 N/mm2

Given compression and tension steel yield

∴ fsc = 0.87 fy

Resultant compressive force in concrete (CC) is given by,

\({C_C} = 0.36 × {{20}} × 300 × {160.91}\)(∵ x_u = 160.91 mm)

\({C_S} = {628} × \left( {{250×0.87} - 0.45 × {20}} \right)\)

Mu = 0.36 × 20 × 300 × 160.91 × (500 - 0.42 × 160.91) + 628 × (0.87× 250 - 0.45 × 20)  × (500 - 50)

M_u = 209.21 kN - m.

Concept:

The relationship between cohesion and unconfined compression strength is given by,

\(c = \frac{{{q _u}}}{2}\)

\(Cohesion = \frac{{Unconfined;compressive;strength}}{2}\)

Calculation:

Given:

q_u = 54 kPa

\(\rm c = \frac{{{54}}}{2}=\rm 27~ kPa\)

Additional Information

The relationship between major and minor principle stress in triaxial test is given by

\({σ _{1}} = {σ _{3}}{\tan ^2}\left( {45^\circ + \frac{\phi }{2}} \right) + 2c\tan \left( {45^\circ + \frac{\phi }{2}} \right)\)

σ1 - Major principle stress

σ3 - Confining pressure

 c - Cohesion of clay

In unconfined compression test confining pressure is zero ∴σ3 = 0

∴ \({σ _{1}} = 2c\tan \left( {45^\circ + \frac{\phi }{2}} \right)\)

For saturated clay ϕ =0

σ1 = 2 × c 

σ1 becomes qu (unconfined compressive strength)

 ∴ \(c = \frac{{{q _u}}}{2}\)

Concept:

As per Terzaghi’s theory, The ultimate bearing capacity of square footing is given by,

Ultimate bearing capacity (qu) = 1.3× c× Nc + q × Nq + 0.4 × γ × B × N_γ ....(1)

C - Cohesion of clay

Nc, Nq, N_γ - Bearing capacity factors, γ = unit weight of soil

q = overburden pressure = γ × D

D = depth of the footing, B = width of square footing

Calculation:

Given:

c = 27 kPa, Size of footing = 4 m × 4 m

For saturated clay ϕ = 0^°

For ϕ = 0^°, N_c = 5.7, N_q = 1, N_γ =0

∴ q_u = 1.3 × 27 × 5.7 + γ × D × 1

It is given that the footing rests on soil, D = 0

∴ qu = 1.3 × 27 × 5.7 = 200 kPa.