Official Paper

GATE CE 2023 Official Paper: Shift 2 (Previous Year Paper)

65 questions · 180 minutes · with answers · free

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General Aptitude (10 questions)

The line ran _______ the page, right through the centre, and divided the page into two.

((a))
across
((b))
of
((c))
between
((d))
about

Show Answer

Answer: ((a))

across

The line ran ACROSS the page, right through the centre, and divided the page into two.

Kind : ________ :: Often : Seldom

(By word meaning)

((a))
Cruel
((b))
Variety
((c))
Type
((d))
Kindred

Show Answer

Answer: ((a))

Cruel

Explanation

Antonyms of “often” - infrequently, rarely, seldom

likewise Antonyms of “kind” - aloof, cruel, hateful, harsh etc

In how many ways can cells in a 3 × 3 grid be shaded, such that each row and each column have exactly one shaded cell? An example of one valid shading is shown.

((a))
2
((b))
9
((c))
3
((d))
6

Show Answer

Answer: ((d))

Solution:

Number of such case can be = ³C₁ x ²C₁ x ¹C₁

= 6 (ans)

There are 4 red, 5 green, and 6 blue balls inside a box. If 𝑁 number of balls are picked simultaneously, what is the smallest value of 𝑁 that guarantees there will be at least two balls of the same colour?

One cannot see the colour of the balls until they are picked.

((a))
4
((b))
15
((c))
5
((d))
2

Show Answer

Answer: ((a))

Given:

4 Red, 5 Green and 6 Blue

We select three balls in worst case

1 Red, 1 Green and 1 Blue

If we select fourth ball then we found two balls are of same colour.

Consider a circle with its centre at the origin (O), as shown. Two operations are allowed on the circle.

Operation 1: Scale independently along the x and y axes.

Operation 2: Rotation in any direction about the origin.

Which figure among the options can be achieved through a combination of these two operations on the given circle?

((a))
((b))
((c))
((d))

Show Answer

Answer: ((a))

Operation 1:

Operation 2:

Elvesland is a country that has peculiar beliefs and practices. They express almost all their emotions by gifting flowers. For instance, if anyone gifts a white flower to someone, then it is always taken to be a declaration of one’s love for that person. In a similar manner, the gifting of a yellow flower to someone often means that one is angry with that person.

Based only on the information provided above, which one of the following sets of statement(s) can be logically inferred with certainty?

(i) In Elvesland, one always declares one’s love by gifting a white flower.

(ii) In Elvesland, all emotions are declared by gifting flowers.

(iii) In Elvesland, sometimes one expresses one’s anger by gifting a flower that is not yellow.

(iv) In Elvesland, sometimes one expresses one’s love by gifting a white flower.

((a))
only (ii)
((b))
(i), (ii) and (iii)
((c))
(i), (iii) and (iv)
((d))
only (iv)

Show Answer

Answer: ((d))

only (iv)

Explanation:

In statement 1, “always” is incorrect word as in paragraph it said “if someone gifts” not always with everyone

In statement 2, “all emotions” is incorrect word as in paragraph it is not concluded that for each and every emotions they gift flowers, where it said “almost”.

In statement 3 , “not yellow” is incorrect, as in paragraph it is clearly mentioned that to show anger they gift yellow flowers.

In statement 4, “sometimes” is correct.

Therefore, only 4th statement is correct

Three husband-wife pairs are to be seated at a circular table that has six identical chairs. Seating arrangements are defined only by the relative position of the people. How many seating arrangements are possible such that every husband sits next to his wife?

((a))
16
((b))
4
((c))
120
((d))
720

Show Answer

Answer: ((a))

H₁ W₁, H₂ W₂, H₃ W₃

Treat every husband-wife pairs as one block

There are three pairs

So all three pairs can be arranged by (3 – 1)! ways and in all pairs husband and wife can be arranged by 2! ways.

So required number of ways are 2! × 2³ = 16

Based only on the following passage, which one of the options can be inferred with certainty?

When the congregation sang together, Apenyo would also join, though her little screams were not quite audible because of the group singing. But whenever there was a special number, trouble would begin; Apenyo would try singing along, much to the embarrassment of her mother. After two or three such mortifying Sunday evenings, the mother stopped going to church altogether until Apenyo became older and learnt to behave.

At home too, Apenyo never kept quiet; she hummed or made up silly songs to sing by herself, which annoyed her mother at times but most often made her become pensive. She was by now convinced that her daughter had inherited her love of singing from her father who had died unexpectedly away from home.

((a))
The mother was embarrassed about her daughter’s singing at home.
((b))
The mother’s feelings about her daughter’s singing at home were only of annoyance.
((c))
The mother was not sure if Apenyo had inherited her love of singing from her father.
((d))
When Apenyo hummed at home, her mother tended to become thoughtful.

Show Answer

Answer: ((d))

When Apenyo hummed at home, her mother tended to become thoughtful.

Explanation:

As per the statement in paragraph “she hummed or made up silly songs to sing by herself, which annoyed her mother at times but most often made her become pensive”.

the word Pensive means thoughtful , thinking.

Therefore, statement 4 can be inferred with certainty.

If x satisfies the equation $\rm 4^{8^x}=256$ , then x is equal to ______.

((a))
$\frac{1}{2}$
((b))
log₁₆ 8
((c))
$\frac{2}{3}$
((d))
log₄ 8

Show Answer

Answer: ((c))

$\frac{2}{3}$

$4^{8^x}$ = 256

⇒ $4^{8^x}$ = 4⁴

⇒ 8^x= 4

⇒ $\left(2^3\right)^x$ = 2²

So, x = $\frac{2}{3}$

Consider a spherical globe rotating about an axis passing through its poles. There are three points P, Q, and R situated respectively on the equator, the north pole, and midway between the equator and the north pole in the northern hemisphere. Let P, Q, and R move with speeds v_p, v_Q, and V_R, respectively.

Which one of the following options is CORRECT?

((a))
v_P < v_R < v_Q
((b))
v_P < v_Q < v_R
((c))
v_P > v_R > v_Q
((d))
v_P = v_R ≠ v_Q

Show Answer

Answer: ((c))

v_P > v_R > v_Q

Solution:

velocity = ŵ.r

here , ŵ = angular rotation = constant

hence, more is the distance away from the axis of rotation, more will be the velocity

therefore, V_P > V_R > V_Q

Civil Engineering (55 questions)

Let 𝜙 be a scalar field, and 𝒖 be a vector field. Which of the following identities is true for div(𝜙𝒖)?

((a))
div(𝜙𝒖) = 𝜙div(𝒖) + 𝒖 ⋅ grad(𝜙)
((b))
div(𝜙𝒖) = 𝜙div(𝒖) + 𝒖 × grad(𝜙)
((c))
div(𝜙𝒖) = 𝜙grad(𝒖) + 𝒖 ⋅ grad(𝜙)
((d))
div(𝜙𝒖) = 𝜙grad(𝒖) + 𝒖 × grad(𝜙)

Show Answer

Answer: ((a))

div(𝜙𝒖) = 𝜙div(𝒖) + 𝒖 ⋅ grad(𝜙)

Concept:

The divergence of a vector field

The divergence computes a scalar quantity from a vector field by differentiation.

If a(x, y, z) is a vector function of position in 3 dimensions, that is a = a1ˆı + a2ˆ + a3kˆ

then its divergence at any point is defined in Cartesian co-ordinates by,

$div:a = \frac{\delta a_{1}}{\delta x} +\frac{\delta a_{2}}{\delta y}+\frac{\delta a_{3}}{\delta z}$

We can write this in a simplified notation using a scalar product with the ∇ vector differential operator:

$div:a = (\hat{i}\tfrac{\delta }{\delta x}+\hat{j}\tfrac{\delta }{\delta y}+\hat{k}\tfrac{\delta }{\delta z}).a = \triangledown.a$

Notice that the divergence of a vector field is a scalar field.

The gradient of a scalar field

Recall the discussion of temperature distribution throughout a room in the overview, where we wondered how a scalar would vary as we moved off in an arbitrary direction.

Here we find out how.

If U(r) = U(x, y, z) is a scalar field, i.e a scalar function of position r = [x, y, z] in 3 dimensions, then its gradient at any point is defined in Cartesian co-ordinates by,

$grad:U = \frac{\delta U}{\delta x}\hat{i}+\frac{\delta U}{\delta y}\hat{j}+\frac{\delta U}{\delta z}\hat{k}\equiv \triangledown U$

div of Ua

Suppose that U(r) is a scalar field and that a(r) is a vector field and we are interested in the product Ua. This is a vector field, so we can compute its divergence.

For example the density ρ(r) of a fluid is a scalar field, and the instantaneous velocity of the fluid v(r) is a vector field, and we are probably interested in mass flow rates for which we will be interested in ρ(r)v(r).

The divergence (a scalar) of the product Ua is given by:

∇ · (Ua) = U(∇ · a) + (∇U) · a

∇ · (Ua) = Udiva + (gradU) · a

Substituting the notations as per the questions:

∇(𝜙𝒖) = 𝜙(∇.𝒖) + 𝒖 ⋅ grad(𝜙)

div(𝜙𝒖) = 𝜙div(𝒖) + 𝒖 ⋅ grad(𝜙) (ans)

Which of the following probability distribution functions (PDFs) has the mean greater than the median?

((a))
Function 1
((b))
Function 2
((c))
Function 3
((d))
Function 4

Show Answer

Answer: ((b))

Function 2

Explanation

A positively skewed distribution is one in which the data is distributed more on one side of the scale with a long tail on the right, and in which the mean, median, and mode are all positive values rather than negative or zero.

The mean is typically to the right of the data median in this distribution, which is sometimes referred to as the right-skewed distribution.

Mean > Median > Mode

A distribution type that is negatively skewed is one in which the mean is smaller than the median and mode, the distribution tail is longer on the left side of the graph, and more values plot on the right side of the graph.

Mean < Median < Mode

A remote village has exactly 1000 vehicles with sequential registration numbers starting from 1000. Out of the total vehicles, 30% are without pollution clearance certificate. Further, even- and odd-numbered vehicles are operated on even- and odd-numbered dates, respectively.

If 100 vehicles are chosen at random on an even-numbered date, the number of vehicles expected without pollution clearance certificate is ________.

((a))
15
((b))
30
((c))
50
((d))
70

Show Answer

Answer: ((b))

Solution:

Probability of selecting even vehicles on even numbered dates = 1

Since 30% of the total vehicles are without pollution clearance certificate.

∴ The no. of vehicles expected without pollution clearance certificate

= 100 × 1 × 0.3

= 30

30 vehicle are expected to be without pollution clearance certificate.

A circular solid shaft of span L = 5 m is fixed at one end and free at the other end. A torque T = 100 kN.m is applied at the free end. The shear modulus and polar moment of inertia of the section are denoted as G and J, respectively. The torsional rigidity GJ is 50,000 kN.m² /rad. The following are reported for this shaft:

Statement i) The rotation at the free end is 0.01 rad

Statement ii) The torsional strain energy is 1.0 kN.m

With reference to the above statements, which of the following is true?

((a))
Both the statements are correct
((b))
Statement i) is correct, but Statement ii) is wrong
((c))
Statement i) is wrong, but Statement ii) is correct
((d))
Both the statements are wrong

Show Answer

Answer: ((b))

Statement i) is correct, but Statement ii) is wrong

Solution:

Given:

Circular solid shaft of span L = 5 m

Torque T = 100 kNm

Torsional rigidity GJ = 50,000 kNm2 /rad

Statement (i):

Rotation at free end (θ) = $\rm \frac{\tau L}{GJ}=\frac{100\times 5(kNm^2)}{50000(kNm^2/rad)}$ = 0.01rad

Therefore, statement (i) is correct.

Statement (ii):

Torsional strain energy = $\frac{1}{2}\times T \times \theta=\frac{1}{2}\times 100\times 0.01$

= 0.5 kN-m

Therefore, statement (ii) is incorrect.

M20 concrete as per IS 456: 2000 refers to concrete with a design mix having _______.

((a))
an average cube strength of 20 MPa
((b))
an average cylinder strength of 20 MPa
((c))
a 5-percentile cube strength of 20 MPa
((d))
a 5-percentile cylinder strength of 20 MPa

Show Answer

Answer: ((c))

a 5-percentile cube strength of 20 MPa

Explanation

In M20, M refers to mix and 20 to characteristic cube strength.

As per clause no. 6.1.1, IS456 : 2000 characteristic strength is defined as the strength below which not more than 5 percent of the test results are expected to fall.

Additional Information

Concrete mix design

Concrete mix design is classified as nominal mix design and the design mix

Nominal mix design:

This is the very rough method of concrete mix design which gives wrong translations of concrete grades.
IS 456:2000 provides a more precise nominal mix proportion for M5, M7.5, M10, M15, and M20 grades of concrete in terms of the total mass of aggregates, proportions of fines to coarse aggregates, and volume of water to be used per 50 kg of cement.
Nominal mix concrete can only be used in ordinary concrete constructions involving concrete
grades not higher than M20.
For higher grades of concrete, design mix concrete is adopted.

Design mix concrete:

Design mix concrete is based on the principle of "mix design" and always preferred over the nominal mix of concrete.
It yields concrete of desired quality and is more economical than the nominal mix.
The IS recommendations of the mix design are given in IS 10262:1982 and SP 23:1982.

When a simply-supported elastic beam of span L and flexural rigidity EI (E is the modulus of elasticity and I is the moment of inertia of the section) is loaded with a uniformly distributed load w per unit length, the deflection at the mid-span is

$\rm \Delta_0=\frac{5}{384}\frac{wL^4}{El}$

If the load on one half of the span is now removed, the mid-span deflection _______.

((a))
reduces to Δ₀/2
((b))
reduces to a value less than Δ₀/2
((c))
reduces to a value greater than Δ₀/2
((d))
remains unchanged at Δ₀

Show Answer

Answer: ((a))

reduces to Δ₀/2

Explanation:

As per the super- position law

$\Delta_0=\frac{5}{384}\frac{WL^4}{EI}$ $\Delta^{\prime},=,{\frac{\Delta_{0}}{2}}={\frac{1}{2}}!\left({\frac{5}{,384}},{\frac{W L^{4}}{E{I}}}\right)$

So, midspan deflection reduced to ∆₀/2

Muller-Breslau principle is used in analysis of structures for ________.

((a))
drawing an influence line diagram for any force response in the structure
((b))
writing the virtual work expression to get the equilibrium equation
((c))
superposing the load effects to get the total force response in the structure
((d))
relating the deflection between two points in a member with the curvature diagram in-between

Show Answer

Answer: ((a))

drawing an influence line diagram for any force response in the structure

Explanation

Influence line diagrams for both determinate and indeterminate structures are created using the Muller-Breslau principle.

"It says that any stress function's impact line may be found by taking away the function's constraint and adding a directly associated generalized unit displacement at that point in the stress function's direction."

"The ordinate of influence line diagram for a reaction is given by the ordinate of elastic curve if a unit deflection is applied in the direction of reaction"

Note:

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

A standard penetration test (SPT) was carried out at a location by using a manually operated hammer dropping system with 50% efficiency. The recorded SPT value at a particular depth is 28. If an automatic hammer dropping system with 70% efficiency is used at the same location, the recorded SPT value will be _________.

((a))
28
((b))
20
((c))
40
((d))
25

Show Answer

Answer: ((b))

Concept:

Standard Penetration Test (SPT)

The standard penetration test is an in-situ test that is coming under the category of penetrometer tests.
The standard penetration tests are carried out in borehole.
The test will measure the resistance of the soil strata to the penetration undergone.
A penetration emphirical correlation is derived between the soil properties and the penetration resistance.
The test is extremely useful for determining the relative density and the angle of shearing resistance of cohesionless soils.
It can also be used to determine the unconfined compressive strength of cohesive soils.

Procedure for Standard Penetration Test

The test is conducted in a bore hole by means of a standard split spoon sampler.
Once the drilling is done to the desired depth, the drilling tool is removed and the sampler is placed inside the bore hole.
By means of a drop hammer of 63.5kg mass falling through a height of 750mm at the rate of 30 blows per minute, the sampler is driven into the soil.
The number of blows of hammer required to drive a depth of 150mm is counted.
Further it is driven by 150 mm and the blows are counted.
Similarly, the sampler is once again further driven by 150mm and the number of blows recorded.
The number of blows recorded for the first 150mm not taken into consideration.
The number of blows recorded for last two 150mm intervals are added to give the standard penetration number (N).
N = No: of blows required for 150mm penetration beyond seating drive of 150mm
If the number of blows for 150mm drive exceeds 50, it is taken as refusal and the test is discontinued.
The standard penetration number is corrected for dilatancy correction and overburden correction.

The requirements to conduct an SPT soil test are:

Split Spoon Sampler
Drop Hammer weighing 63.5kg
Guiding rod
Driving head (anvil).

Split Spoon Sampler

This is a thick-walled sample tube having an outer diameter of 50.8 mm and an inner diameter of 35 mm. The length of the sampler is around 65 cm.
The split spoon sampler consists of a steel tube, driving shoes, a check valve, coupling, and vent ports.
Notably, the thick split spoon sampler is used for the collection of disturbed soil samples.

Drop Hammer

The drop hammer is used to drive the split spoon sampler into the borehole.
The weight of the hammer in the SPT test is 63.5 kg.

Driving Head (Anvil)

The driving head stops the hammer at a certain point.

Guiding Rod

The Guiding rod serves the purpose of guiding the hammer to the anvil.

Corrections in Standard Penetration Test

Before the SPT values can be used in empirical correlations and design charts, the standard penetration value N needs to be corrected as per IS 2131. The corrections are:

Correction for Dilatancy
Overburden Pressure Correction

Overburden Pressure Correction

Overburden pressure has a profound effect on penetration resistance.

When two granular soils have the same relative density, soil with a higher confining pressure will have a higher 'N' value.

As the depth of the soil increases, the confining pressure also gets increased.

As such, the value of N is underestimated at shallow depths and overestimated at greater depths.

In order to take into account the effect of overburden pressure, the obtained value of N is altered to incorporate the effects of the overburden pressure. The right answer for 'N' is

N1 = N (350/(S + 70))

Here, N1 is the correction made for the overburden pressure and N is the recorded value

S = Overburden pressure less than 280KN/m2

Dilatancy Correction

Silty fine sands and fine sands below the water table are subjected to the pore water pressure. The pore pressure makes the soil harder to break through, which increases the penetration number (N).

When the observed value of N is more than 15, Terzaghi and Peck (1967) suggested the following correction in the case of the silty fine sands.

N_C = 15 + 0.5 (N₁ -15),

Where, N₁ is the recorded value after Overburden Pressure Correction, and N_C is the final corrected value.

If N₁ is less than or equal to 15, then N_C = N.

we know that efficiency of blows is inversely proportional to number of blows and SPT number, i.e.

$\rm Efficiency ∝ \frac{1}{Number \ of \ blows}∝ \frac{1}{SPT\ value}$

⇒ η₁N₁ = η₂N₂

⇒ 0.5 × 28 = 0.7 × N₂

⇒ $N_2=\frac{0.5 \times 28}{0.7}=20$

Therefore, the recorded SPT value = 20

A vertical sheet pile wall is installed in an anisotropic soil having coefficient of horizontal permeability, 𝑘_𝐻 and coefficient of vertical permeability, 𝑘_𝑉. In order to draw the flow net for the isotropic condition, the embedment depth of the wall should be scaled by a factor of ______, without changing the horizontal scale.

((a))
$\rm \sqrt{\frac{k_H}{k_V}}$
((b))
$\rm \sqrt{\frac{k_V}{k_H}}$
((c))
1.0
((d))
$\rm {\frac{k_H}{k_V}}$

Show Answer

Answer: ((a))

$\rm \sqrt{\frac{k_H}{k_V}}$

A steady-state, homogeneous, anisotropic sysem can be mathematically transformed into an isotropic system by coordinate transformation, creating what is sometimes called a transformed section.

The coordinates in the true anisotropic sysem are x and z. In the tranformed isotropic system the coordinates are X and Z, where

X = x

$\rm Z =Z\sqrt{\frac{k_x}{k_z}}$

$\rm Z =Z\sqrt{\frac{k_H}{k_V}}$

So, the embedment depth of wall should be scaled by factor of $\rm \sqrt{\frac{k_H}{k_V}}$ .

Identify the cross-drainage work in the figure.

((a))
Super passage
((b))
Aqueduct
((c))
Siphon aqueduct
((d))
Level crossing

Show Answer

Answer: ((a))

Super passage

Explanation

cross drainage works

When a canal and a natural drain cross, a cross drainage works structure is built to keep the drain water from combining with the canal water. Because it is more expensive, this kind of building should be avoided at all costs. There are two approaches to prevent cross-drainage works:

By rearranging the canal waterway's orientation
The construction of a single cross drainage work is made more cheap by combining two or three streams into one.

There are three types of cross drainage works structures:

Type - 1: Cross drainage work carrying canal over the drain

Aqueduct - The canal water level is referred as full supply level (FSL) and drainage water level is referred as high flood level (HFL). The HFL is below the canal bed level.The water in drainage flows under gravity and possess the atmospheric pressure
Syphon Aqueduct - canal water is carried above the drainage but the high flood level (HFL) of drainage is above the canal trough. The drainage water flows under syphonic action and there is no presence of atmospheric pressure in the natural drain.

Type - 2: Cross Drainage work carrying Drainage over the canal

Super passage - The full supply level of canal is below the drainage trough in this structure. The water in canal flows under gravity and possess the atmospheric pressure. This is simply a reverse of Aqueduct structure
Canal Syphon - drainage is carried over canal similar to a super passage but the full supply level of canal is above than the drainage trough so the canal water flows under syphonic action and there is no presence of atmospheric pressure in canal.

Type - 3: Cross drainage works admitting canal water into the canal

Level Crossing

Canal inlets

The river in the specified Cross drainage work is above the canal, and the bottom trough is adequately above the canal's full supply level. Thus, it's a super passage.

Which one of the following options provides the correct match of the terms listed in Column-1 and Column-2?

Column-1	Column-2
P.	Horton equation	I.	Precipitation
Q.	Muskingum method	II.	Flood frequency
R.	Penman method	III.	Evapotranspiration
		IV.	Infiltration
		V.	Channel routing

((a))
P - IV, Q - V, R - III
((b))
P - III, Q - IV, R - I
((c))
P - IV, Q - III, R - II
((d))
P - III, Q - I, R - IV

Show Answer

Answer: ((a))

P - IV, Q - V, R - III

Explanation

Precipitation is the process of condensation of atmospheric water either in liquid or frozen form,which falls back to earth due to gravitational forces of attraction.

In the process of precipitation, a part of the atmosphere saturates itself with water vapour, and at the right temperature, it condenses and precipitates onto the surface of the earth. Air becomes saturated by the cooling of molecules present in the air, and the addition of water vapour to these molecules results in the saturation of the air.

Precipitation is measured using rain gauge.

Evapo-transpiration is a kind of process in which water is transferred from the Earth's surface to the atmosphere through the combined processes of plant transpiration and evaporation. Evaporation happens when liquid water converts to vapor, while transpiration refers to water loss from plant leaves. This process plays a important role in the Earth's water cycle, influencing climate, energy balance, and ecosystem dynamics

Can be directly measured using various methods such as lysimeters, eddy covariance towers, soil moisture sensors and Penman's equation.

Infiltration is the flow of water through the soil surface into a porous medium under gravitational action and pressure effects. The factors that influence the infiltration are the soil type such as texture, structure, hydrodynamic characteristics which influence capillary forces and adsorption, and the soil coverage

measurement of infiltration -ring infiltromenter, rainfall simulator, hydrograph analysis, Horton's equation, philip's equation, green ampt equation.

A common method of Hydrologic channel routing is Muskingum method, in which the channel storage in a reach is expressed as a function of both inflow and outflow discharge.

S = k [x.I^m – (1 - x)Q^m]

Where, s = Storage in a channel reach

I = Inflow Discharge

Q = Outflow Discharge

k → Coefficient known as storage time coefficient. It was unit of time, it is approx. equal to time

of travel of flood wave through the channel reach.

x → Weightage factor. Its range series from 0 to 0.5

In channels, the Muskingum approach is applied to flood response analysis.

In the context of Municipal Solid Waste Management, ‘Haul’ in ‘Hauled Container System operated in conventional mode’ includes the _________.

((a))
time spent by the transport truck at the disposal site
((b))
time spent by the transport truck in traveling between a pickup point and the disposal site with a loaded container
((c))
time spent by the transport truck in picking up a loaded container at a pickup point
((d))
time spent by the transport truck in driving from the depot to the first pickup point

Show Answer

Answer: ((b))

time spent by the transport truck in traveling between a pickup point and the disposal site with a loaded container

Explanation

"Haul" time in the hauled container system comprises the amount of time needed to go to the disposal site, beginning when the container to be emptied has been loaded onto the track, as well as the amount of time the truck needs to travel after leaving the disposal site to get to the spot where the empty container needs to be placed again.

There is no accounting for time spent at the disposal location.

Which of the following is equal to the stopping sight distance?

((a))
(braking distance required to come to stop) + (distance travelled during the perception-reaction time)
((b))
(braking distance required to come to stop) – (distance travelled during the perception-reaction time)
((c))
(braking distance required to come to stop)
((d))
(distance travelled during the perception-reaction time)

Show Answer

Answer: ((a))

(braking distance required to come to stop) + (distance travelled during the perception-reaction time)

Concept:

Stopping Sight Distance:-

Stopping sight distance (SSD) is the minimum sight distance available on a highway at any spot having sufficient length to enable the driver to stop a vehicle travelling at design safely without collision with any other obstruction.

The stopping sight distance is the sum of the lag distance and the braking distance.

Lag distance is the distance the vehicle traveled during the reaction time ‘t’ and is given by vt, where v is the velocity in m/sec2.
Braking distance is the distance traveled by the vehicle during braking operation.

If F is the maximum frictional force developed and the braking is l, then work done against friction in stopping the vehicle is Fl = fWl, where W is the total weight of the vehicle. The kinetic energy at the design speed is,

$\frac{1}{2}m{v^2} = \frac{1}{2}\frac{{W{v^2}}}{g};;fWl = \frac{{W{v^2}}}{{2g}}$

$l = \frac{{{v^2}}}{{2gf}}$

SSD = Lag distance + Braking distance

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

Where, v is the design speed in m/sec2,

t is the reaction time in sec,

g is the acceleration due to gravity

f is the coefficient of friction.

Coefficient of longitudinal friction:-

Speed (kmph)	< 30	40	50	60	> 80
f	0.40	0.38	0.37	0.36	0.35

Important Points

Intermediate Sight Distance (ISD) = SSD for 2-way 2- lane road
ISD = 2 × SSD -- for single lane 2 way
ISD = SSD --- for single lane one - way

The magnetic bearing of the sun for a location at noon is 183˚ 30ˊ. If the sun is exactly on the geographic meridian at noon, the magnetic declination of the location is ________.

((a))
3˚ 30ʹ W
((b))
3˚ 30ʹ E
((c))
93˚ 30ʹ W
((d))
93˚ 30ʹ E

Show Answer

Answer: ((a))

3˚ 30ʹ W

Explanation

At any place 12 :00 PM sun will be exactly over the true meridian of that place.

Declination = True bearing – Magnetic bearing

= 180° – 183°30′

= - 3°30′

NEGATIVE DECLINATION = WEST DECLINATION

POSITIVE DECLINATION = EAST DECLINATION

= 3°30′ W (ANS)

With regard to the shear design of RCC beams, which of the following statements is/are TRUE?

((a))
Excessive shear reinforcement can lead to compression failure in concrete
((b))
Beams without shear reinforcement, even if adequately designed for flexure, can have brittle failure
((c))
The main (longitudinal) reinforcement plays no role in the shear resistance of beam
((d))
As per IS456:2000, the nominal shear stress in the beams of varying depth depends on both the design shear force as well as the design bending moment

Show Answer

Answer: ((a))

Excessive shear reinforcement can lead to compression failure in concrete

Explanation

Concrete becomes stronger in diagonal tension upon failure compared to diagonal compression failure when the area of shear reinforcement is high, i.e., in case of excessive shear reinforcement, and compression failure may occur prior to the shear reinforcement yielding.

Therefore, option 1 is correct

When a flexural fracture in beams lacking shear reinforcement reaches the longitudinal reinforcement, it will propagate suddenly and may result in brittle failure.

Therefore, option 2 is correct

Main reinforcement raises the concrete's depth, restricts the width of cracks, and acts as a dowel to boost shear resistance. The proportion of tensile reinforcement and the concrete's grade determine its design shear strength.

Therefore, option 3 is incorrect

for beams with varying depth,nominal shear stress, as per clause 40.1.1, IS 456 : 2000

$\tau_{v} = \frac{V_{u}\pm \frac{M_{u}tan\beta }{d}}{bd}$

Therefore, option 4 is correct

For the matrix

$\rm [A]=\begin{bmatrix}1&-1&0\\ -1&2&-1\\ 0&-1&1\end{bmatrix}$

which of the following statements is/are TRUE?

((a))
[𝐴]{𝑥} = {𝑏} has a unique solution
((b))
[𝐴]{𝑥} = {𝑏} does not have a unique solution
((c))
[𝐴] has three linearly independent eigenvectors
((d))
[𝐴] is a positive definite matrix

Show Answer

Answer: ((a))

[𝐴]{𝑥} = {𝑏} has a unique solution

As |A| = 0

So, one of the eigen value is zero.

|A – λI| = 0

$\begin{bmatrix}1-\lambda&-1&0\\ -1&2-\lambda&-1\\ 0&1&1-\lambda\end{bmatrix}=0$

(1 – λ)[(2 – λ)(1 – λ) – 1] + 1 [λ - 1] = 0

(1 – λ)[(2 – λ)(1 – λ) – 1 – 1] = 0

λ(1 – λ)(3 – λ)=0

λ = 0, 1, 3

As there are three eigen values so number of linearly independent eigen vector are 3.

Since, |A| = 0

So, [A]{x} = {b} does not have a unique solution

For the positive definite matrix, all the eigen values must be positive.

But, here one eigen value i.e. λ = 0, so, A is not a positive definite matrix.

In the frame shown in the figure (not to scale), all four members (AB, BC, CD, and AD) have the same length and same constant flexural rigidity. All the joints A, B, C, and D are rigid joints. The midpoints of AB, BC, CD, and AD, are denoted by E, F, G, and H, respectively. The frame is in unstable equilibrium under the shown forces of magnitude 𝑃 acting at E and G. Which of the following statements is/are TRUE?

((a))
Shear forces at H and F are zero
((b))
Horizontal displacements at H and F are zero
((c))
Vertical displacements at H and F are zero
((d))
Slopes at E, F, G, and H are zero

Show Answer

Answer: ((a))

Shear forces at H and F are zero

Explanation:

Since AD and BC are subjected to pure bending, Hence shear force at H & F are zero

Therefore, option (a) correct

since axial deformations are neglected and due to symmetry horizontal displacement of H & F = zero

Therefore, option (b) correct

Vertical displacement at H & F is not equal to 0

Therefore, option (c) is incorrect

Also due to symmetry slopes at E, F, G, H = 0

Therefore, option (d) is correct

The reason(s) of the nonuniform elastic settlement profile below a flexible footing, resting on a cohesionless soil while subjected to uniform loading, is/are:

((a))
Variation of friction angle along the width of the footing
((b))
Variation of soil stiffness along the width of the footing
((c))
Variation of friction angle along the depth of the footing
((d))
Variation of soil stiffness along the depth of the footing

Show Answer

Answer: ((a))

Variation of friction angle along the width of the footing

Explanation

The non-uniform elastic settlement profile below a flexible footing resting on cohesionless soil subjected to uniform loading is due to the non-linear behaviour of the soil stiffness along the width of the footing.
When a flexible footing is placed on cohesionless soil, the soil deforms non-linearly due to the soil's low shear strength. This leads to a differential settlement, where the soil settles more near the edges of the footing than in the centre.
The non-uniform settlement profile is also influenced by the size and shape of the footing, as well as the intensity and distribution of the applied load.
Additionally, the soil's compressibility and deformation characteristics play a crucial role in determining the magnitude and distribution of the settlement.
The modulus of elasticity varies with width of footing so there is a variation in the stiffness along the width of footing.

Therefore, we can conclude that the nonuniform elastic settlement profile below a flexible footing, resting on a cohesionless soil while subjected to uniform loading is due to variation of soil stiffness along the width of the footing

Which of the following is/are NOT active disinfectant(s) in water treatment?

((a))
OH (hydroxyl radical)
((b))
O3 (ozone)
((c))
OCl− (hypochlorite ion)
((d))
Cl− (chloride ion)

Show Answer

Answer: ((a))

OH (hydroxyl radical)

Explanation

In water treatment O3, OH (hydroxyl radical), and OCl- are regarded as the active disinfectants.

(OH-) Hydroxyl ion is not a disinfectant that is active.
The disinfection effectiveness of free chlorine (OCl- and HOCl) is decreased by the (Cl-) chloride ion.

Therefore, (Cl-) chloride ion in not active disinfectant from the above options

As per the Indian Roads Congress guidelines (IRC 86: 2018), extra widening depends on which of the following parameters?

((a))
Horizontal curve radius
((b))
Superelevation
((c))
Number of lanes
((d))
Longitudinal gradient

Show Answer

Answer: ((a))

Horizontal curve radius

Explanation

As per IRC 86 : 2018 Clause 8.6 Widening of Carriageway on Curves

To allow for the safe passage of vehicles, the roadway must be widened at steep horizontal curves.

There are two parts to the necessary widening:

(i) mechanical widening to accommodate the extra width occupied by a vehicle on a curve due to tracking of the rare wheels, and

(ii) psychological widening to allow vehicles to cross easily because lane-bound vehicles have a tendency to wander more on curves than on straightaways.

$W_E=\frac{nl^2}{2R}+\frac{V}{9.5\sqrt R}$

Where,

n = Number of lanes

l = length of wheel base

R = Horizontal curve radius

V = velocity of vehicle

WE ∝ n

$W_E∝\frac{1}{R}$

From the above formula of extra widening, it depends on Number of lanes, length of wheel base, Horizontal curve radius, and velocity of vehicle.

The steady-state temperature distribution in a square plate ABCD is governed by the 2-dimensional Laplace equation. The side AB is kept at a temperature of 100°C and the other three sides are kept at a temperature of 0°C. Ignoring the effect of discontinuities in the boundary conditions at the corners, the steady-state temperature at the center of the plate is obtained as T₀°C. Due to symmetry, the steady-state temperature at the center will be same (T₀°C), when any one side of the square is kept at a temperature of 100°C and the remaining three sides are kept at a temperature of 0°C. Using the principle of superposition, the value of T₀ is _________ (rounded off to two decimal places).

An unconfined compression strength test was conducted on a cohesive soil. The test specimen failed at an axial stress of 76 kPa. The undrained cohesion (in kPa, in integer) of the soil is _____

The pressure in a pipe at X is to be measured by an open manometer as shown in figure. Fluid A is oil with a specific gravity of 0.8 and Fluid B is mercury with a specific gravity of 13.6. The absolute pressure at X is ________ kN/m2 (round off to one decimal place).

[Assume density of water as 1000 kg/m3 and acceleration due to gravity as 9.81 m/s2 and atmospheric pressure as 101.3 kN/m2]

For the elevation and temperature data given in the table, the existing lapse rate in the environment is _____ ℃/100 m (round off to two decimal places).

Elevation from ground level (m)	Temperature (°C)
5	14.2
325	16.9

If the size of the ground area is 6 km × 3 km and the corresponding photo size in the aerial photograph is 30 cm × 15 cm, then the scale of the photograph is 1 : ________ (in integer).

The solution of the differential equation

$\rm \frac{d^3y}{dx^3}-5.5\frac{d^2y}{dx^2}+9.5\frac{dy}{dx}-5y=0$

is expressed as 𝑦 = 𝐶₁𝑒^2.5𝑥 + 𝐶₂𝑒^𝛼𝑥 + 𝐶₃𝑒^𝛽𝑥 , where 𝐶₁, 𝐶₂, 𝐶₃, 𝛼, and 𝛽 are constants, with α and β being distinct and not equal to 2.5. Which of the following options is correct for the values of 𝛼 and 𝛽?

((a))
1 and 2
((b))
−1 and −2
((c))
2 and 3
((d))
−2 and −3

Show Answer

Answer: ((a))

1 and 2

Auxillary equation is,

m³ – 5.5 m² + 9.5 m – 5 = 0

By solving above equation, we get m = 2.5, 1, 2

So, m₁ and m₂ are 1 and 2.

Two vectors [2 1 0 3]^𝑇 and [1 0 1 2]^𝑇 belong to the null space of a 4 × 4 matrix of rank 2. Which one of the following vectors also belongs to the null space?

((a))
[1 1 −1 1]^T
((b))
[2 0 1 2]^T
((c))
[0 −2 1 −1]^T
((d))
[3 1 1 2]^T

Show Answer

Answer: ((a))

[1 1 −1 1]^T

ρ(A_{4 × 4}) = 2

N (A) = Number of column – Rank

= 4 – 2 = 2

i.e. Null space of A will consist only two linearly independent vectors which is given as x and y.

Eigen vectors of matrix A, $\begin{bmatrix}2\\ 1\\ 0 \\ 3 \end{bmatrix}\rm and \begin{bmatrix}1\\ 0 \\ 1 \\ 2 \end{bmatrix}$

As these are linearly independent eigen vectors so remaining eigen vectors of null space must be linearly dependent.

Hence, $\rm X - Y=\begin{bmatrix}1\\ 1\\ -1 \\ 1 \end{bmatrix}$

Cholesky decomposition is carried out on the following square matrix [𝐴].

$\rm [A]=\begin{bmatrix}8&-5\\ -5&a_{22}\end{bmatrix}$

Let 𝑙_ij and 𝑎ij be the (i, j)^th elements of matrices [𝐿] and [𝐴], respectively. If the element 𝑙₂₂ of the decomposed lower triangular matrix [𝐿] is 1.968, what is the value (rounded off to the nearest integer) of the element 𝑎₂₂?

((a))
5
((b))
7
((c))
9
((d))
11

Show Answer

Answer: ((b))

LL^T = A

$\begin{bmatrix}L_{11}&0\\ L_{21}&L_{22} \end{bmatrix} \begin{bmatrix} L_{11}&L_{21}\\ 0&L_{22} \end{bmatrix}= \begin{bmatrix} 8&-5\\ -5&a_{22} \end{bmatrix}$

L₁₁ = 2√2 , L₂₁= $-\frac{5}{2\sqrt2}$ and $L^2_{21}+L^2_{22}=a_{22}$

$a_{22}=\left( -\frac{5}{2\sqrt2} \right)^2$ + 1.968 = 6.99 ≃ 7

In a two-dimensional stress analysis, the state of stress at a point is shown in the figure. The values of length of PQ, QR, and RP are 4, 3, and 5 units, respectively. The principal stresses are ________. (round off to one decimal place)

((a))
σ_x = 26.7 MPa, σ_y = 172.5 MPa
((b))
σ_x = 54.0 MPa, σ_y = 128.5 MPa
((c))
σ_x = 67.5 MPa, σ_y = 213.3 MPa
((d))
σ_x = 16.0 MPa, σ_y = 138.5 MPa

Show Answer

Answer: ((c))

σ_x = 67.5 MPa, σ_y = 213.3 MPa

$\rm \cos \theta=\frac{4}{5}\ and \ \sin \theta=\frac{3}{5};\theta = 36.87^\circ$

σ_n = σ_x cos²θ + σ_ysin²θ

$120=\sigma_x\times\left(\frac{4}{5} \right)^2+\sigma_y \left(\frac{3}{5} \right)^2$

25 × 120 = 16σ_x + 9σ_y

τ_xy = (σ_y – σ_x) sinθ cosθ

$70=(\sigma_y-\sigma_x)\left(\frac{4}{5} \right) \left(\frac{3}{5} \right)$

70 × 25 = 12σ_y + 12σ_x

Solving (i) and (ii), σ_x = 67.5 MPa and σ_y = 213.3 MPa

Two plates are connected by fillet welds of size 10 mm and subjected to tension, as shown in the figure. The thickness of each plate is 12 mm. The yield stress and the ultimate stress of steel under tension are 250 MPa and 410 MPa, respectively. The welding is done in the workshop (partial safety factor, 𝛾_𝑚𝑤= 1.25). As per the Limit State Method of IS 800: 2007, what is the minimum length (in mm, rounded off to the nearest higher multiple of 5 mm) required of each weld to transmit a factored force P equal to 275 kN?

((a))
100
((b))
105
((c))
110
((d))
115

Show Answer

Answer: ((b))

105

$\rm P=L_w(ks)\frac{f_u}{\sqrt3\gamma_{mw}}$

275 × 10³ = L_w × 0.7 × 10 × $\frac{410}{\sqrt3 \times 1.25}$

L_w = 207.45 mm

Length of each weld = $\frac{207.45}{2}=103.73$ mm ≃105 mm

In the given figure, Point O indicates the stress point of a soil element at initial non-hydrostatic stress condition. For the stress path (OP), which of the following loading conditions is correct?

((a))
𝜎_𝑣 is increasing and 𝜎_ℎ is constant
((b))
𝜎_𝑣 is constant and 𝜎_ℎ is increasing
((c))
𝜎_𝑣 is increasing and 𝜎_ℎ is decreasing
((d))
𝜎_𝑣 is decreasing and 𝜎_ℎ is increasing

Show Answer

Answer: ((a))

𝜎_𝑣 is increasing and 𝜎_ℎ is constant

σ_v = Major principal stress

σ_h = Minor principal stress

For 1:1 slope, ${dq \over dp} = 1$

${d\sigma v - d\sigma h \over d\sigma v + d\sigma h} =1$

d𝜎v – d𝜎h = d𝜎v + d𝜎h

2d𝜎h = 0

d𝜎h = 0

Hence, if σh is constant then increasing σv or

decreasing σv will lead to 1:1 slope.

So, σ_h is constant and σ_v is increasing.

The figure shows a vertical retaining wall with backfill consisting of cohesive-frictional soil and a failure plane developed due to passive earth pressure. The forces acting on the failure wedge are: P as the reaction force between the wall and the soil, R as the reaction force on the failure plane, C as the cohesive force along the failure plane and W as the weight of the failure wedge. Assuming that there is no adhesion between the wall and the wedge, identify the most appropriate force polygon for the wedge.

((a))
((b))
((c))
((d))

Show Answer

Answer: ((c))

Explanation:

The forces acting on the failure wedge are:

P as the reaction force between the wall and the soil,

R as the reaction force on the failure plane,

C as the cohesive force along the failure plane and

W as the weight of the failure wedge

.For given passive condition we can draw the FBD diagram of failure wedge. Assuming that there is no adhesion between the wall and the wedge

From the above Free Body Diagram of failure wedge, we can draw force polygon as shown below:

Therefore, option 3 is correct match for the given conditions.

A compound symmetrical open channel section as shown in the figure has a maximum of _______ critical depth(s).

B_m – Bottom width of main channel

B_f – Bottom width of flood channel

y_m – Depth of main channel

y – Total depth of the channel

n_m – Manning’s roughness of the main channel

n_f – Manning’s roughness of the flood channel

((a))
3
((b))
2
((c))
1
((d))
4

Show Answer

Answer: ((a))

Case 1:

For Q = Q₁ and depth of flow (y) < y_m

then, y_c1 is the critical depth.

Case 2:

For Q = Q₂ and depth of flow (y) > y_m

then, y_c1 , y_c2 and y_c3 be the critical depth.

Case 3:

Only y_c3 will exist at any discharge Q₃

such that y_c3 > y₃

So, the maximum number of critical depth would be three.

The critical flow condition in a channel is given by _______.

[Note: 𝛼 – kinetic energy correction factor; 𝑄 – discharge; A_c – cross-sectional area of flow at critical flow condition; T_c – top width of flow at critical flow condition; 𝑔 – acceleration due to gravity]

((a))
$\rm\frac{\alpha Q^2}{g}=\frac{A_c^3}{T_c}$
((b))
$\rm\frac{\alpha Q}{g}=\frac{A_c^3}{T_c^2}$
((c))
$\rm\frac{\alpha Q^2}{g}=\frac{A_c^3}{T_c^2}$
((d))
$\rm\frac{\alpha Q}{g}=\frac{A_c^3}{T_c}$

Show Answer

Answer: ((a))

$\rm\frac{\alpha Q^2}{g}=\frac{A_c^3}{T_c}$

Froude number, $F_{r}=;\frac{V}{\sqrt{g D}}$ ; where D = hydraulic depth = $\frac{A}{T}$

⇒ ${ F}_{r}^{2};=;\frac{V^{2}}{g D}$

⇒ $F_{r}^{2};=;\frac{Q^{2}T}{g A^{2}A}=\frac{Q^{2}T}{g A^{3}}$

For critical condition, F_r = 1

⇒ $\frac{Q^{2}T_{c}}{g A^{3}},=,1$

⇒ $\frac{Q^{2}}{g}=\frac{A_{c}^{3}}{T_{c}}$

Now, taking into account, kinetic energy correction factor,

$\frac{\alpha Q^{2}}{g};=;\frac{A_{c}^{3}}{T_{c}}$

Match the following air pollutants with the most appropriate adverse health effects:

Air pollutant	Health effect to human and/or test animal
P.	Aromatic hydrocarbons	I.	Reduce the capability of the blood to carry oxygen
Q.	Carbon monoxide	II.	Bronchitis and pulmonary emphysema
R.	Sulfur oxides	III.	Damage of chromosomes
S.	Ozone	IV.	Carcinogenic effect

((a))
(P) – (II), (Q) – (I), (R) – (IV), (S) – (III)
((b))
(P) – (IV), (Q) – (I), (R) – (III), (S) – (II)
((c))
(P) – (III), (Q) – (I), (R) – (II), (S) – (IV)
((d))
(P) – (IV), (Q) – (I), (R) – (II), (S) – (III)

Show Answer

Answer: ((d))

(P) – (IV), (Q) – (I), (R) – (II), (S) – (III)

Explanation:

Carcinogenic agents include aromatic hydrocarbons like 3, 4-benzpyrene and other polycyclic organic chemicals that result from incomplete combustion of hydrocarbons. These are the ones that cause cancer.
When inhaled, sulfur dioxide (SO2), an irritating gas, affects mucosal membranes. Bronchitis and pulmonary emphysema are the result. Patients with asthma are severely impacted.
Carbon Monoxide (CO): CO and blood hemoglobin have a significant affinity for one another, forming carboxyhemoglobin, or COHb. As a result, haemoglobin's capacity to deliver oxygen to body tissues is diminished. Because CO binds to hemoglobin around 200 times more strongly than oxygen does, modest quantities of CO can nevertheless lead to high levels of COHb. The central nervous system is also affected by carbon monoxide.
Ozone (O3) can harm the respiratory tract's tissue, resulting in irritation and inflammation and it damages the chromosomes in plants. This can aggravate asthma symptoms and cause symptoms like coughing and tightness in the chest.

Therefore, the correct match from above explanation is option 4

A delivery agent is at a location R. To deliver the order, she is instructed to travel to location P along straight-line paths of RC, CA, AB and BP of 5 km each. The direction of each path is given in the table below as whole circle bearings. Assume that the latitude (L) and departure (D) of R is (0, 0) km. What is the latitude and departure of P (in km, rounded off to one decimal place)?

Paths	RC	CA	AB	BP
Direction (in degrees)	120	0	90	240

((a))
L = 2.5; D = 5.0
((b))
L = 0.0; D = 5.0
((c))
L = 5.0; D = 2.5
((d))
L = 0.0; D = 0.0

Show Answer

Answer: ((b))

L = 0.0; D = 5.0

For location R,

Latitude, L_R = 0

Departure, D_R = 0

For location P,

Latitude, L_P = L_R + Σ∆_L

= 0 + 5cos120° + 5cos0° + 5cos90° + 5cos240° = 0

Departure, D_P= D_R + Σ∆_D

= 0 + 5sin120° + 5sin0° + 5sin90° + 5sin240° = 5

So, L_P = 0

D_P = 5

Alternatively,

Path (l)	Direction (θ)	Latitude (l cos θ)	Departure (l sinθ)
RC	120°	–2.5	4.33
CA	0°	5	0
AB	90°	0	5
BP	240°	–2.5	–4.33

Length of each path (l) = 5 km

Latitude of P in km = Σlatitude = 0

Departure of P in km = Σdeparture = 5

Which of the following statements is/are TRUE?

((a))
The thickness of a turbulent boundary layer on a flat plate kept parallel to the flow direction is proportional to the square root of the distance from the leading edge
((b))
If the streamlines and equipotential lines of a source are interchanged with each other, the resulting flow will be a sink
((c))
For a curved surface immersed in a stationary liquid, the vertical component of the force on the curved surface is equal to the weight of the liquid above it
((d))
For flow through circular pipes, the momentum correction factor for laminar flow is larger than that for turbulent flow

Show Answer

Answer: ((a))

The thickness of a turbulent boundary layer on a flat plate kept parallel to the flow direction is proportional to the square root of the distance from the leading edge

Explanation:

Thickness of boundary layer in turbulent flow:

$\rm \frac{\delta}{x}=\frac{0.376}{R\mathrm{e}{x}^{1/5}}=\frac{0.376}{\left(\frac{\rho u{\infty}x}{\mu}\right)^{1/5}}$

$\rm{\delta}=\frac{0.376x^{4/5}}{\left(\frac{\rho u_{\infty}}{\mu}\right)^{1/5}}$

δ ∝ x4/5

The thickness of a turbulent boundary layer on a flat plate kept parallel to the flow direction is not proportional to the square root of the distance from the leading edge

Therefore, option 1 is incorrect

If the streamlines and equi-potential lines of a source are reversed(change the direction), the resulting flow will be a sink

Therefore, option 2 is incorrect

PH = the total pressure on the vertical plane's projected area of the curved surface.

PH acts at the center of pressure of the lane surface and is equal to the entire pressure that the liquid exerts on an imaginary vertically submerged plane surface, which is the vertical projection of the curved surface.

Pv = the weight of the liquid inside the section that rises vertically above the curved surface and reaches the liquid's free surface.

PV will operate through the liquid's center of gravity in the section that extends above the curved surface and up to the liquid's free surface (in this case, represented by the profile ABCDEFA).

Therefore, option 3 is correct

For flow through circular pipes, the momentum correction factor for laminar flow is always larger than that for turbulent flow.

Therefore, option 4 is correct

In the context of water and wastewater treatments, the correct statements are:

((a))
particulate matter may shield microorganisms during disinfection
((b))
ammonia decreases chlorine demand
((c))
phosphorous stimulates algal and aquatic growth
((d))
calcium and magnesium increase hardness and total dissolved solids

Show Answer

Answer: ((a))

particulate matter may shield microorganisms during disinfection

Explanation

Particulate matter reduces the efficacy of UV or chlorine disinfection by shielding the targeted microorganisms.

Therefore option 1 is correct.

Ammonia generates a high demand for chlorine because it is converted to nitrogen gas during the chlorination process, which turns chlorine into chloride.

Therefore option 2 is incorrect.

Excessive levels of phosphorus can lead to the growth of big aquatic plants and algae. It is known as "eutrophication" and can lead to lower dissolved oxygen levels.

Therefore option 3 is correct.

Total dissolved solids are made up of dissolved organic matter and inorganic salts, primarily calcium, magnesium, potassium, sodium bicarbonates, chlorides, and sulfates.
The existence of multivalent cations is what causes hardness. The majority of the time, soluble bicarbonates, chlorides, and calcium and magnesium sulfates are to blame.

Therefore option 4 is correct.

Which of the following statements is/are TRUE for the aerobic composting of sewage sludge?

((a))
Bulking agent is added during the composting process to reduce the porosity of the solid mixture
((b))
Leachate can be generated during composting
((c))
Actinomycetes are involved in the process
((d))
In-vessel composting systems cannot be operated in the plug-flow mode

Show Answer

Answer: ((a))

Bulking agent is added during the composting process to reduce the porosity of the solid mixture

Explanation:

A bulking agent is added to support the sludge by increasing its porosity to increase effective aeration.

Therefore, option 1 is incorrect

Leachates grows during aerobic composting of sewage sludge.

Therefore, option 2 is correct

Actinomycetes are aerobic spore bacteria.These species are thermophilic, meaning that temperature, water content, and aerobic conditions all affect how they develop. These organisms are supported in their proliferation by the heat generated during aerobic composting.

Therefore, option 3 is correct

There are two types of in-vessel composting systems: agitated bed and plug flow. When the system is in plug flow mode, the first-in, first-out principle is used and the interaction between the particles in the composting mass remains constant throughout the process. While, composting material is mechanically combined in an agitated bed system.

Therefore, option 4 is incorrect

The figure presents the time-space diagram for when the traffic on a highway is suddenly stopped for a certain time and then released. Which of the following statements are true?

((a))
Speed is higher in Region R than in Region P
((b))
Volume is lower in Region Q than in Region P
((c))
Volume is higher in Region R than in Region P
((d))
Density is higher in Region Q than in Region R

Show Answer

Answer: ((a))

Speed is higher in Region R than in Region P

Explanation:

Time-space diagram for when the traffic on a highway.

Any vertical line in the graph that intersects arrow lines indicates the whereabouts of the cars at a given moment in various regions.
The speed of cars in a certain region is represented by the slope of a line in that area.

From the curve we can infer:

Option (1) is incorrect because, slope of distance time graph is greater in region P as compared to region R.
Volume in region Q is lower than in region P because in region Q, vehicles are at halted conditions. So, in region Q, velocity of vehicle is zero and hence, volume is zero. Option (2) is correct.
Option (3) is correct because slope of shock wave is much higher than Q to R as compared to P to Q.
Density is higher in region Q as compared with region R and region P, because in region Q vehicles are practically at Jam density that is maximum. So, option (4) is correct.

Consider the Marshall method of mix design for bituminous mix. With the increase in bitumen content, which of the following statements is/are TRUE?

((a))
the Stability decreases initially and then increases
((b))
the Flow increases monotonically
((c))
the air voids (VA) increases initially and then decreases
((d))
the voids filled with bitumen (VFB) increases monotonically

Show Answer

Answer: ((a))

the Stability decreases initially and then increases

Concept:

Marshall stability and flow:

Marshall stability of a test specimen is the maximum load required to produce failure when the specimen is preheated to a prescribed temperature placed in a special test head and the load is applied at a constant strain (5 cm per minute). during the stability test, a dial gauge is used to measure the vertical deformation of the specimen. The deformation at the failure point expressed in units of 0.25 mm is called the marshall flow value of the specimen.

Graphical plot:

The average value of each of the above properties is found for each mix with the different bitumen contents. Graphs are plotted with the bitumen content on the x-axis and the following value on the y-axis.

(i) Marshall stability value

(ii) Flow value

(iii) Unit weight

(iv)Percent air voids in the total mix

(v) Percent voids filled with bitumen(VFB)

From the above graphs we can infer

Stability increases initially and then decreases

Hence option (a) is INCORRECT

Flow increases monotonically

Hence option (b) is CORRECT

Air voids decreases continously

Hence option (c) is INCORRECT

VFB increases monotonically

Hence option (d) is CORRECT

A 5 cm long metal rod AB was initially at a uniform temperature of T0°C. Thereafter, temperature at both the ends are maintained at 0°C. Neglecting the heat transfer from the lateral surface of the rod, the heat transfer in the rod is governed by the one-dimensional diffusion equation $\rm\frac{\partial T}{\partial t}=D\frac{\partial^2T}{\partial x^2}$ , where D is the thermal diffusivity of the metal, given as 1.0 cm2/s.

The temperature distribution in the rod is obtained as

$\rm T(x,t)=\Sigma_{n=1,3,5...}^{\infty}C_n\sin\frac{n\pi x}{5}e^{-\beta n^2t}$ ,

where x is in cm measured from A to B with 𝑥 = 0 at A, t is in s, 𝐶𝑛 are constants in °C, T is in °C, and β is in s−1 .

The value of β (in 𝑠−1 , rounded off to three decimal places) is ________.

A beam is subjected to a system of coplanar forces as shown in the figure. The magnitude of vertical reaction at Support P is ______ N (round off to one decimal place).

For the frame shown in the figure (not to scale), all members (AB, BC, CD, GB, and CH) have the same length, 𝐿 and flexural rigidity, 𝐸𝐼. The joints at B and C are rigid joints, and the supports A and D are fixed supports. Beams GB and CH carry uniformly distributed loads of 𝑤 per unit length. The magnitude of the moment reaction at A is 𝑤𝐿2⁄𝑘. What is the value of 𝑘 (in integer)? ____________

Consider the singly reinforced section of a cantilever concrete beam under bending, as shown in the figure (M25 grade concrete, Fe415 grade steel). The stress block parameters for the section at ultimate limit state, as per IS 456: 2000 notations, are given. The ultimate moment of resistance for the section by the Limit State Method is ________ kN.m (round off to one decimal place).

[Note: Here, As is the total area of tension steel bars, b is the width of the section, d is the effective depth of the bars, f_ck is the characteristic compressive cube strength of concrete, f_y is the yield stress of steel, and x_u is the depth of neutral axis.]

A 2D thin plate with modulus of elasticity, 𝐸 = 1.0 N/m², and Poisson’s ratio, 𝜇 = 0.5, is in plane stress condition. The displacement field in the plate is given by 𝑢 = 𝐶𝑥²𝑦 and 𝑣 = 0, where 𝑢 and 𝑣 are displacements (in m) along the 𝑋 and 𝑌 directions, respectively, and 𝐶 is a constant (in m⁻²). The distances x and y along X and Y, respectively, are in m. The stress in the 𝑋 direction is 𝜎_𝑋𝑋 = 40𝑥𝑦 N/m², and the shear stress is 𝜏_𝑋𝑌 = 𝛼𝑥² N/m². What is the value of 𝛼 (in N/m⁴, in integer)? ________

An idealised frame supports a load as shown in the figure. The horizontal component of the force transferred from the horizontal member PQ to the vertical member RS at P is _______ N (round off to one decimal place).

A square footing is to be designed to carry a column load of 500 kN which is resting on a soil stratum having the following average properties: bulk unit weight = 19 kN/m³; angle of internal friction = 0° and cohesion = 25 kPa. Considering the depth of the footing as 1 m and adopting Meyerhof’s bearing capacity theory with a factor of safety of 3, the width of the footing (in m) is _________ (round off to one decimal place)

[Assume the applicable shape and depth factor values as unity; ground water level at greater depth.]

A circular pile of diameter 0.6 m and length 8 m was constructed in a cohesive soil stratum having the following properties: bulk unit weight = 19 kN/m³; angle of internal friction = 0° and cohesion = 25 kPa.

The allowable load the pile can carry with a factor of safety of 3 is __________ kN (round off to one decimal place).

[Adopt: Adhesion factor, α = 1.0 and Bearing capacity factor, N_c = 9.0]

For the flow setup shown in the figure (not to scale), the hydraulic conductivities of the two soil samples, Soil 1 and Soil 2, are 10 mm/s and 1 mm/s, respectively. Assume the unit weight of water as 10 kN/m³ and ignore the velocity head. At steady state, what is the total head (in m, rounded off to two decimal places) at any point located at the junction of the two samples? _____

A consolidated drained (CD) triaxial test was carried out on a sand sample with the known effective shear strength parameters, 𝑐′ = 0 and 𝜙′ = 30°. In the test, prior to the failure, when the sample was undergoing axial compression under constant cell pressure, the drainage valve was accidentally closed. At the failure, 360 kPa deviatoric stress was recorded along with 70 kPa pore water pressure. If the test is repeated without such error, and no back pressure is applied in either of the tests, what is the deviatoric stress (in kPa, in integer) at the failure? ___________

A catchment may be idealized as a circle of radius 30 km. There are five rain gauges, one at the center of the catchment and four on the boundary (equi-spaced), as shown in the figure (not to scale).

The annual rainfall recorded at these gauges in a particular year are given below.

Gauge	G₁	G₂	G₃	G₄	G₅
Rainfall (mm)	910	930	925	895	905

Using the Thiessen polygon method, what is the average rainfall (in mm, rounded off to two decimal places) over the catchment in that year? ________

The cross-section of a small river is sub-divided into seven segments of width 1.5 m each. The average depth, and velocity at different depths were measured during a field campaign at the middle of each segment width. The discharge computed by the velocity area method for the given data is _____ m³/s (round off to one decimal place).

Segment	Average depth (D) (m)	Velocity (m/s) at different depths
0.2 D	0.6 D	0.8 D
1	0.40	--	0.40	--
2	0.70	0.76	--	0.70
3	1.20	1.19	--	1.13
4	1.40	1.25	--	1.10
5	1.10	1.13	--	1.09
6	0.80	0.69	--	0.65
7	0.45	--	0.42	--

The theoretical aerobic oxidation of biomass (C₅H₇O₂N) is given below:

C₅H₇O₂N + 5O₂ → 5CO₂ + NH₃ + 2H₂O

The biochemical oxidation of biomass is assumed as a first-order reaction with a rate constant of 0.23/d at 20ºC (logarithm to base e). Neglecting the second-stage oxygen demand from its biochemical oxidation, the ratio of BOD₅ at 20ºC to total organic carbon (TOC) of biomass is _______ (round off to two decimal places).

[Consider the atomic weights of C, H, O and N as 12 g/mol, 1 g/mol, 16 g/mol and 14 g/mol, respectively

A system of seven river segments is shown in the schematic diagram. The R_i’s, Q_i’s, and C_i’s (i = 1 to 7) are the river segments, their corresponding flow rates, and concentrations of a conservative pollutant, respectively. Assume complete mixing at the intersections, no additional water loss or gain in the system, and steady state condition. Given: Q₁ = 5 m³/s ; Q₂ = 15 m³/s ; Q₄ = 3 m³/s ; Q₆ = 8 m³/s ; C₁ = 8 kg/m³; C₂ = 12 kg/m³; C₆ = 10 kg/m³. What is the steady state concentration (in kg/m³, rounded off to two decimal place) of the pollutant in the river segment 7 ? ______

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