GATE CE 2023 Official Paper: Shift 1 (Previous Year Paper)

Explanation:

• In the sentence, the word "might" is used to convey uncertainty or a potential future action.
• The speaker hasn't decided on their evening plans but is considering the possibility of visiting a friend. "Might" is a modal verb indicating a conditional or tentative action.
• Options 1 (mite) is likely a typo or incorrect word, 2 (would) suggests a conditional action but may not fit the context, and 4 (didn't) is past tense and does not align with the uncertainty expressed in the sentence.
• Therefore, "might" best captures the intended meaning of potential but undecided action.

Explanation:

• The correct analogy is Advance: Retreat.
• In the given analogy, "eject" is the opposite action to "insert," just as "retreat" is the opposite of "advance."
• Ejecting is removing or pushing out, and inserting is placing or pushing in.
• Similarly, advancing is moving forward, and retreating is moving backward.
• Options 1 (Advent) and 2 (Progress) do not convey the opposite relationship, and option 4 (Loan) is unrelated to the directional relationship between advance and retreat, making option 3 (Retreat) the appropriate choice based on word meanings and logical pairing.

Explanation:

Interior angle sum = (n – 2) × 180°

Each angle of a regular Hexagon:

\(\rm \frac{(n-2)\times 180^{\circ}}{n}=\) 120°

Required area = \(\rm \frac{\theta}{360^{\circ}}\times \pi \times 25=\frac{25\pi}{3}\)

Explanation:

The statement "Donald Duck always lies" creates a logical paradox. If the statement is true, then Donald Duck must be lying about always lying, which implies he sometimes tells the truth. Conversely, if the statement is false, it means Donald Duck sometimes tells the truth, contradicting the claim. Therefore, the certainty of Donald Duck's truthfulness or falsehood is indeterminate, making options 1 and 2 impossible to ascertain. However, option 4, "Donald Duck’s statement is false," logically follows as it captures the inherent contradiction in the statement, regardless of Donald Duck's overall truthfulness.

Explanation:

If box d or i is coloured block we get one line of symmetry, i.e., horizontal or vertical. But for two lines of symmetry, we need to at least colour 3 boxes, i.e., c, d, and I.

In another way, 

Both c, d, i and c, d, i, f, g make the figure symmetrical about two lines (horizontal and vertical), but as the minimum number is asked, option (B) is correct.

Explanation:

No human is intelligent - This statement directly contradicts the given information. We know that at least some humans are intelligent; therefore, it cannot be true that no human is intelligent.

All humans are intelligent - The information given does not provide enough evidence to infer this. The fact that some humans are intelligent does not mean that all humans are.

Some non-humans are intelligent - The original statement doesn't provide any information about non-humans, so we can't make any inferences about them based on that statement.

Some intelligent beings are humans - This can be logically inferred from the original statement. If 'some humans are intelligent,' then it necessarily follows that 'some intelligent beings are humans.'

So, the only logical inference that can be made with absolute certainty from the statement 'Some humans are intelligent,' is 'Some intelligent beings are humans.'

Explanation:

Mean:

• In probability distribution, the mean, represents the average of a random variable's possible values, weighted by their respective probabilities.
• It is a measure central to the distribution and indicates the center or average outcome.
• The mean provides insights into the long-term behavior of the random variable under consideration.

Median:

• The median is the middle value that separates the higher half from the lower half of a dataset when arranged in ascending order.
• It is a measure of central tendency and is not affected by extreme values.
• The median is particularly useful in describing the central position in skewed or asymmetric distributions.

Mode:

• The mode is the value or values that occur with the highest frequency.
• It represents the most probable outcome in a set of data. Unlike mean and median.
• The mode is not necessarily unique and a distribution can be unimodal (one mode), bimodal (two modes), or multimodal (more than two modes).

From the graph 

• It is a perfectly symmetrical distribution. So, mean = median.
• It is symmetrical about the y-axis (positive data = negative data). So, mean = median = 0
• Mode = -13, 13 (Bimodal)
• Thus, mean = median ≠ mode

Explanation:

The false statement is option 4. If dark matter was known to interact via the strong-interaction, then the epithet “dark” would be justified. 

• The paragraph explicitly states that dark matter is supposed to interact only via gravitational interaction and not through electromagnetic-, weak-, or strong-interaction.
• Therefore, the epithet "dark" in dark matter is justified precisely because it does not interact through the strong force or other known interactions.
• The other options (a, b, c) are not addressed in the paragraph, and their truth or falsity cannot be determined based on the information provided.

Explanation:

Given, a = 30!, b = 50! and c = 100!

We know that 

\(\rm \log_n m=\frac{\log m}{\log n}\)

log_a c, log_c a, log_a b, log_b a

\(\rm \frac{\log 100!}{\log 30!},\rm \frac{\log 30!}{\log 100!},\rm \frac{\log 50!}{\log 30!},\rm \frac{\log 30!}{\log 50!}\)

\(\rm \frac{\log 30!}{\log 100!}<\rm \frac{\log 30!}{\log 50!}<\rm \frac{\log 50!}{\log 30!}<\rm \frac{\log 100!}{\log 30!}\)

log_c a < log_b a < log_a b < log_a c

Explanation:

Area of quadrant,

A = \(\rm \frac{1}{4}\pi (2)^2=\pi\) cm²

Area of sector, B = Area of square – Area of quadrant  = 4 – π

So, the total shaded area

= 2A + 2B =2 × π + 2 × (4 – π)

= 2π + 8 – 2π = 8 cm²

Explanation:

\(\rm f(x)=\frac{1}{x^2}\) is not defined at x = 0

So, \(\rm I=\displaystyle\int^0_{-1}\frac{1}{x^2}dx+\rm \displaystyle\int^1_{0}\frac{1}{x^2}dx\)

\(\rm =\left(-\frac{1}{x}\right)_{-1}^0+\rm =\left(-\frac{1}{x}\right)_{0}^1=\infty\)

\(\rm \frac{1}{x}\) is not defined at x = 0

Therefore, the integral does not converge.

Explanation:

Stress in member (1),

\(\rm (σ_1)=\frac{P_1}{A_1}=\frac{250\times10^3}{100\times 100}=25 \ N/mm^2\)

Stress in member (2),

\(\rm (σ_2)=\frac{P_2}{A_2}=\frac{50\times10^3}{50\times 50}=20 \ N/mm^2\)

σ_max = 25 N/mm²

Explanation:

• Creep under compression is defined as an increase in the magnitude of strain under constant stress.
• The creep of concrete depends on stress in concrete, the age of loading, and the loading duration (increases with time).
• But, the rate of creep rapidly decreases with time.
• Although the exact mechanism of creep is not clear, it is generally attributed to:

1. Internal movement of adsorbed water
2. Growth in micro-cracks
3. Sliding between gel particles (C – S – H gel)
4. Moisture loss

• Under constant load, the Creep strain in concrete is Time-dependent as \( \frac{Ec}{1+\Theta }\)

      Where, \(\Theta \) is age factor

Explanation:

In a balanced section, the strain diagram under flexure is

\(\rm \frac{0.87f_y}{E_s}+0.002\)

So, strain in concrete = 0.0035

Strain in steel = \(\rm \frac{0.87\times415}{2\times10^5}+0.002=0.0038\)

Explanation:

• The water–cement ratio is the ratio of the weight of water to the weight of cement used in a concrete mix. A lower ratio leads to higher strength and durability but may make the mix difficult to work with and form.

Workability:

• The workability of concrete refers to its ease of handling, placing, and compacting during construction.
• It is influenced by factors such as water content, mix proportions, and aggregate properties.
• Optimal workability ensures efficient construction processes, proper compaction, and the achievement of desired strength and durability in the final concrete structure.
• The higher the water-cement ratio weaker the mix and the more workability.

Compressive Strength:

• The compressive strength of concrete is a measure of its ability to withstand axial loads or forces pushing the material together.
• As the water-cement ratio increases, the compressive strength decreases.
• A concrete design mix with a low water/cement ratio and also using larger aggregates results in gains in concrete compressive strength.

Explanation:

Given: Specific gravity, G = 2.6

Saturation, S = 0.5

Water content, w = 0.15

As we know

Se = Gw

⇒ 0.5 × e = 2.6 × 0.15

⇒ \(\rm e=\frac{2.6\times0.15}{0.5}=0.78\)

Explanation:

By Converse – Labarre formula

\(\rm \eta_g=\left[1-\frac{θ}{90}\left(\frac{m(n-1)+n(m-1)}{mn}\right)\right]\times100\)

\(\rm ⇒ 0.6=1-\frac{θ}{90}\left(\frac{3\times2+3\times2}{3\times3}\right)\)

⇒ \(\rm \frac{θ}{90}\left(\frac{6+6}{9}\right)=0.4\)

⇒ θ = 27° 

Now, \(\rm \tan \theta=\frac{d}{S}\) where d is diameter and S is spacing.

So, Spacing, S = \(\rm \frac{d}{\tan \theta}=\frac{0.3}{\tan 27^{\circ}}=0.587\)

Alternatively,

Group efficiency = \(\rm \frac{Q_{ug}}{nQ_{us}}\)

⇒ \(\rm 0.6=\frac{\frac{1}{2}(\gamma l)\tan \delta(2S+d)\times l\times4}{9\times \frac{1}{2}(\gamma l)\tan \delta (pdl)}\)

⇒ \(\rm 0.6\frac{(2S+d)\times4}{9\times \pi d}\)

⇒ \(\rm S=\frac{1}{2}\left[\frac{0.6\times9\times \pi \times 0.3}{4}-0.3\right]=0.486\)

Note:

As the exact answer of spacing is not matching with the given options by E Converse-Labarre formula, therefore we will use alternative method to find spacing.

Explanation:

Triaxial extension Test:

• The triaxial extension test is a geotechnical laboratory test used to determine the deformation characteristics of soil under different confining pressures.
• It involves subjecting a cylindrical soil sample to axial tension while measuring axial and radial strains.
• This test provides valuable information for soil engineering and slope stability analysis.
• At location I, σH > σV 

 Triaxial extension (As depth from ground level is less, σv is less).

Direct shear or shear box test:

• The direct shear test, conducted using a shear box apparatus, evaluates the shear strength of soils.
• It involves applying horizontal forces to a divided soil sample, measuring deformation, and determining the soil's shear strength parameters, such as cohesion and internal friction angle.
• At location II, the Failure plane is the same as the direct shear box test i.e. perpendicular to the vertical plane.

Triaxial compression test:

• The triaxial compression test is a geotechnical laboratory test that assesses the strength and stress-strain behavior of soils under confined conditions.
• It involves subjecting a cylindrical soil sample to axial compression within a confining pressure chamber, providing insights into soil stability and deformation characteristics.
• At location III, σv > σh 

 Triaxial compression test.

Vane shear test:

• The vane shear test is a geotechnical method for measuring the undrained shear strength of cohesive soils.
• A vane is inserted into the soil and rotated to measure the torque required for shearing. This test helps assess soil stability and bearing capacity.

Explanation:

• When a supercritical stream enters a mild-sloped (M) channel section, the type of flow profile would become "subcritical."
• In mild-sloped channels, the velocity of the flow decreases, transitioning from supercritical to subcritical flow.
• This change in flow profile occurs as the channel slope counteracts the kinetic energy of the supercritical flow, causing the water to slow down and the flow to become subcritical.
• When a supercritical stream enters a mild sloped (M) channel section, the type of flow profile will be M3.
• The depth of flow is less than critical depth of flow in case of M₃.

Explanation:

• Greenhouse gases (GHGs) act as a natural insulation layer for the Earth.
• These gases, such as carbon dioxide and water vapor, permit short-wavelength solar radiation to penetrate the atmosphere and reach the Earth's surface.
• Once absorbed by the surface, the Earth emits long-wavelength infrared radiation.
• GHGs, however, selectively absorb and trap much of this outgoing radiation, preventing its immediate escape into space.
• This process, known as the greenhouse effect, warms the Earth's surface and plays a vital role in maintaining a habitable climate.
• Human activities, particularly the burning of fossil fuels, enhance this effect, leading to global warming.

Explanation:

Statements (1) and (3) are correct.

Condition for summit curve (crest vertical curve),

+ve to +ve; +ve to flat

(G₁ > G₂)

+ve to –ve; –ve to –ve

(G₁ < G₂)

Condition for valley curve (sag vertical curve),

–ve to –ve; –ve to flat

(G₁ > G₂)

–ve to +ve; +ve to +ve

(G₁ < G₂)

Explanation:

Vertical collimation correction:

• Vertical collimation correction, refers to the adjustment made to correct for any deviation in the optical line of sight from being perfectly vertical.
• It accounts for errors in the leveling instrument or the telescope's alignment.
• The correction is crucial in ensuring accurate measurements of elevations and vertical angles, particularly in tasks such as leveling and height determination in construction, topography, and geodetic surveys.
• Applying vertical collimation correction helps eliminate systematic errors and enhances the precision of vertical measurements.

Calculation:

Given,

ϕ₁ = Observed value of direct zenith angle = 56° 00′ 00′′

ϕ₂ = Observed value of reversed zenith angle = 303° 00′ 00′′

So, Vertical collimation correction =

=

= 0.5° = 30’

Concept:

• The concept here is related to the resolution of an image, measured in dots per inch (dpi).
• Resolution determines the number of pixels in each inch of the image.
• If the student scans a 10-inch × 10-inch certificate at 600 dpi, the total number of pixels is 600×10×600×10.
• If the same certificate is scanned at 300 dpi, the total number of pixels becomes 300×10×300×10.
• The reduction in pixels can be calculated by comparing these two pixel counts, and the percentage reduction is determined by expressing this reduction as a percentage of the original pixel count.

Calculation:

Number of dots as per 600 dpi = (10 × 600)² = 36000000

Number of dots as per 300 dpi = (10 × 300)² = 9000000

Percentage reduction in number of pixels

\(\rm =\frac{36000000-9000000}{36000000}\times 100=\frac{27}{36}\times 100=75%\)

(*) Marks to all Marks to all

Let us take \(\rm M=\begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\) then E. values are λ = ± i

So, Eigen values of M2 are i2 and (– i)2 = –1, – 1

Eigen values of (MT)2 are same i.e. –1, – 1

Now, MMT and MTM are symmetric matrix, so they have also negative eigen values

so no option is correct i.e. wrong options.

Hence, it is marks to all.

Explanation:

\(\rm b_n=\frac{1}{l}\int_{-l}^lf(x)\cos\frac{(n\pi x)}{l}dx=\frac{1}{l}\int_{-l}^l(px^4+qx^5)\cos\left(\frac{n\pi x}{l}\right)dx\)​

\(=\rm \frac{1}{l}\int_{-l}^lpx^4\cos\left(\frac{n\pi x}{l}\right)dx+0\)  (∵2^nd integral is an odd functions)

​Thus, bn depend on p​

Similarly, 

\(\rm a_n=\frac{1}{l}\int_{-l}^lf(x)\sin\frac{(n\pi x)}{l}dx=\frac{1}{l}\int_{-l}^l(px^4+qx^5)\sin\left(\frac{n\pi x}{l}\right)dx\)

\(=\rm 0+\frac{1}{l}\int_{-l}^lqx^5\sin\left(\frac{n\pi x}{l}\right)dx\) (∵1^st integral is an odd functions)

Thus, an depend on q.

Explanation:

Structure I

r = 3 (VA, VC, ME)

s = 4 (ΣFy = 0, ΣM = 0, BMB = 0, BMD = 0)

Ds = r – s = 3 – 4 = –1 ⇒ negative implies unstable

Structure II

Rigid body rotation is possible since all four reactions are concurrent at A. Unstable,

Structure III

Shear force at x-x = Panel shear force = VA

There is no member to resist panel shear force.

So unstable.

Explanation:

• The waterborne diseases caused by viral pathogens among the given options are

1. Acute anterior poliomyelitis (caused by the poliovirus)
2. Infectious hepatitis (commonly caused by the hepatitis A virus)

• Cholera and typhoid fever are typically caused by bacterial pathogens (Vibrio cholerae for cholera and Salmonella typhi for typhoid fever).
• Infectious hepatitis - This term is broad, but it commonly refers to hepatitis A, B, C, D, and E. Among these, hepatitis A is primarily associated with waterborne transmission.
• 4. Typhoid fever- Typhoid fever is caused by the bacterium Salmonella typhi, not a viral pathogen.

Explanation:

• Option 1) This is generally FALSE. Refuse-Derived Fuel (RDF) is processed from municipal solid waste (MSW) where recyclable and non-combustible materials are removed. Therefore, the Higher Heating Value (HHV) of RDF is typically higher than unprocessed MSW because it is more concentrated in combustible materials.
• Option 2) This is TRUE. Refuse-Derived Fuel (RDF) can be made in various forms, including pellets, fluff, and even powdered form, according to the requirement of the end-use system.
• Option 3) This is FALSE. The inorganic fraction of MSW is typically not converted to RDF. Instead, RDF is made from the organic, combustible fraction of MSW. The noncombustible materials like metals and glass are often removed during the processing.
• Option 4) This is FALSE. RDF can indeed be used in conjunction with various other types of fuel, including coal, gas, and oil. Multifuel systems can burn RDF along with conventional fuels to produce energy.

Thus, among your options, the second statement is true, while the rest are false.

Explanation:

Given, P(A) = 0.5 and P(B) = 0.8 and A and B are independent.

The probability of occurrence of at least A or B is

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

P(A ∪ B) = P(A) + P(B) - P(A)P(B)

P(A ∪ B) = 0.5 + 0.8 – 0.5 × 0.8 = 0.9

Explanation:

Given, \(\rm \frac{dy}{dx}+α xy=0\)

Using variable separable method.

\(\rm \frac{dy}{dx}=-dxy\)

\(\rm \int\frac{dy}{y}=-α\int xdx+ln\ C\)

ln y = \(\rm -α\frac{x^2}{2}+ln C\)

\(\rm ln\left(\frac{y}{c}\right)=-\frac{α}{2}x^2\)

\(\rm y=Ce^{\frac{-α}{2}x^2}\)

By putting y(0) = 1

we get C = 1

By putting y(1) = 0.8

\(\rm e^{\frac{-α}{2}}=0.8\)

\(\rm lne^{\frac{-α}{2}}=ln0.8\)

\(\rm \frac{-α}{2}=ln0.8\)

α = – 2 ln0.8 = 0.446

Explanation:

Given: S = 8 mm, τ_vw = 120 MPa, L_w = 2 × 75 + 50 = 200 mm 

The safe load is calculated as per WSM 

∴ P_safe = τ_vw × A_w = τ_vw × L_w × (0.7 × s) [∵ t_t = 0.7s]

P = 120 × [2 × 75 + 50] × 0.7 × 8 × 10^–3 kN

P = 134.4 kN

Explanation:

The shear strength is given by

τ = c + σn tan ϕ 

Calculation:

Normal stress, σ_n = 50 kPa

Shear stress, τ = 35 kPa

τ = c + σ_n tan ϕ 

⇒ 35 = 50 tan ϕ (∵ c = 0, sandy soil)

⇒ tan ϕ = 0.7

⇒ ϕ = 35°

Explanation:

Given data

Area = 100 ha

Total depth of water, ∆₂= 35 cm

Total time period, B₂ = 120 days

Kor depth, ∆₁ = 12 cm

Kor period, B₁ = 30 days

(i)  Kor duty, \(\rm \Delta_1=\frac{8.64\times B_1}{\Delta_1}=\frac{8.64\times30}{0.12}=2160\) ha/m³/sec

Corresponding discharge= \(\rm \frac{Area}{Duty}=\frac{100}{2160}=0.046\) m³/sec

(ii) Overall duty, \(\rm \Delta_2=\frac{8.64\times B_2}{\Delta_2}=\frac{8.64\times120}{0.35}=2962.28\) ha/m3/sec

Corresponding discharge= \(\rm \frac{Area}{Duty}=\frac{100}{2962.28}=0.034\) m3/sec

The minimum discharge required will be the maximum of the above two:

∴ Q_req. = 0.046 m³/sec

Explanation:

Firstly lag the UH by 1 hr. By lagging 1 hr. We obtained a 2 hr UH of ordinate 4.5 m³/sec. at 1 hr and 15 m3/sec at 2 hr.

So, further lag the UH by 1 more hr we obtained a 3 hr, UH of ordinates 3 m³/sec at 1 hr and 10 m³/sec at 2 hr. Therefore D = 3 hr.

Explanation:

Given:

V = 15 m/s

R = 300 m

Jerk, C = 0.75 m³/s

So, the minimum length of the transition curve,

\(\rm L_t=\frac{V^3}{RC}=\frac{15^3}{300\times 0.75}=15\ m\)

Explanation:

• The trapezoidal rule approximates the area under a curve by connecting data points with straight lines.
• Since these lines are segments of the curve, the rule tends to underestimate the actual area by assuming a linear connection between adjacent points.
• As a result, it approximates the curve's shape with a series of trapezoids, leading to a conservative estimate that is less than the actual area.
• The trapezoidal rule tends to overestimate the value if the function is concave and underestimate the value if the function is convex.
• Since, the given function is convex-shaped parabola, the numerical value of the area obtained using the trapezoidal rule will be less than the actual.

Explanation:

We know, the strain at the compression zone, ε_sc = 0.035 x \(\frac{Xu-dc}{dc}\)

εsc= 0.0035 × \( \frac{Xu-dc}{dc}\)

So,  εsc  < 0.0035

But we can consider εsc ≈ 0.0035.

For Fe250,

The yield strain is \( \frac{0.87fy}{Es}=\frac{0.87 *250}{200000}\) =1.0875 x10^-3  < 0.0035 

When  εsc= 0.0035  fsc = 0.87 fy

For Fe500

The yield strain  = \( \frac{0.87fy}{Es}+0.002\)

\( \frac{0.87*500}{200000}+0.002\)

= 4.175 × 10–3

Hence,  fsc < 0.87 fy

Fe250 will reach 0.87f_y

Thus only Fe- 250 is used.

Explanation:

The plastic section modulus is

\(\rm Z_p=\frac{A}{2}[\overline{y}_1+\overline{y}_2]\)

\(\rm =(b\times 1.5b)\left[\frac{1.5b}{2}+\frac{1.5b}{2}\right]+\left[2b\times \frac{b}{2}\right]\times \left[\frac{b}{4}+\frac{b}{4}\right]=2.75b^3\)

The elastic section modulus is

\(\rm Z=\frac{I_{XX}}{y}=\frac{\left[\frac{b\times(3b)^3}{12}+\frac{2b\times b^3}{12}\right]}{1.5b}=\frac{29}{18}b^3\)

Shape factor is given by

S.F. = \(\rm \frac{Z_p}{Z}=\frac{2.75b^3}{\frac{29}{18}b^3}=1.7\)

Explanation:

At joint T,

ΣF_y = 0

⇒ F_TS sin 60° = P

\(\rm F_{TS}=\frac{2P}{\sqrt3}(T)\)

ΣF_x = 0

F_TS cos 60° = F_DT

\(\rm F_{DT}=\frac{2P}{\sqrt3}\left(\frac{1}{2}\right)=\frac{P}{\sqrt3}(C)\)

F_TS = F_SR = F_RQ = \(\rm\frac{2P}{\sqrt3}(T)\)

F_DT = F_CD = F_BL = \(\rm\frac{P}{\sqrt3}(C)\)

Similarly,

K_ST = K_SR = K_RQ = \(\rm\frac{2}{\sqrt3}(T)\)

\(\rm k_{DT}=k_{CD}=k_{BC}=\frac{1}{\sqrt3}(T)\)

The vertical deflection of joint T

\(\rm\therefore \delta_{VT}=\Sigma\frac{PkL}{AE}=\frac{\left(\frac{2P}{\sqrt3}\right)\left(\frac{2}{\sqrt3}\right)L\times3}{AE}+\frac{\left(\frac{P}{\sqrt3}\right)\left(\frac{1}{\sqrt3}\right)\frac{L}{2}\times3}{AE}\)

\(\rm \delta_{VT}=\frac{4PL}{AE}+\frac{PL}{2AE}=\frac{4.5PL}{AE}\)

⇒ k = 4.5

Explanation:

• Peak point of the compaction curve gives the maximum dry unit weight and optimum moisture content.
• With increase in the compaction effort, the maximum dry unit weight increases.
• With increase in the compaction effort, the optimum moisture content decreases.
• Compaction curve does not cross the zero-air-voids line.

Explanation:

The expression for the vertical stress (σz) at any point (r, z), within the half-space is

\(\rm \sigma_z=\frac{3P}{2\pi}\frac{z^3}{(r^2+z^2)^{5/2}}\)

The locus of the maximum σ_z 

\(\rm \frac{d\sigma_z}{dz}=0\) ⇒ \(\rm \frac{d\sigma_z}{2\pi}=\frac{3P}{2\pi}\left[\frac{(r^2+z^2)^{5/2}3z^2-z^3\frac{5}{2}(r^2+z^2)^{3/2}.2z}{(r^2+z^2)^5}\right]=0\)

⇒ (r² + z²)^3/2 z²[3(r² + z²) - 5z²] = 0

⇒ 3r² + 3z² - 5z² = 0

⇒ 3r² = 2z²

⇒ \(\rm z^2=\frac{3r^2}{2}\)

Explanation:

The net ultimate bearing capacity is

q_nu = q_u - \(\rm \overline{\sigma}\) = q_u - γD_f

1706 = {1.3 CN_C + γD_fN_q + 0.4B_γN_γ} - γDf

1706 = 18 × 1 × 58 + 0.4 × 2.5 × 10 N_γ - 18γ_f

N_γ = 68

For Plate(Surcharge at the plate level is zero)

The ultimate bearing capacity is

q_u = 1.3 CN_c + γDfNq + 0.3BγNγ

q_u = 0 + 0.3 × 0.3 × 18 × 68

q_u = 110.16

Explanation:

For a wide Rectangular channel

\(\rm R = \frac{A}{P}=\frac{B_y}{B+2y}=\frac{B_y}{b}=y\)

Discharge per unit width, q is given by

⇒ \(\rm q = \frac{1}{n}y y^{2/3}S^{1/2}\)

⇒ \(\rm q = \frac{1}{n} y^{5/3}S^{1/2}\)

Calculation of S_c ,

\(\rm y_c=\left(\frac{q^2}{g}\right)^{1/3}\)

⇒ \(\rm q=\frac{1}{n}y_c^{5/3}S_c^{1/2}\)

⇒ \(\rm q=\frac{1}{n}\left(\frac{q^2}{g}\right)^{5/3}S_c^{1/2}\)

⇒ \(\rm q=\frac{1}{n}\frac{q^{10/9}}{g^{5/3}}S_S^{1/2}\)

⇒ \(\rm S_c^{1/2}=\frac{nqg^{5/9}}{q^{10/9}}\)

⇒ \(\rm S_c=\left(\frac{ng^{5/9}}{q^{1/9}}\right)^2=\left(\frac{0.01\times9.8^{5/9}}{70^{1/9}}\right)^2\)

= 0.00049 (> 0.0001 and < 0.0009)

Concept:

Efficiency of plates cn be calculated by work done by work done per second to the input water power. 

Given:

v₁ = 20 m/s

u = 15 m/s

[u₁ = u₂ = u]

Efficiency, η = Work done per second / Input water power

\(\rm =\frac{F_x.u}{\frac{1}{2}mV_1^2}=\frac{m[v_1-u]\times u}{\frac{1}{2}mv_1^2}\)

\(=\frac{[20-15]\times15}{\frac{1}{2}\times (20)^2}=0.375\) ≃ 37.5%

Another method:

Efficiency of plates can be calculated by, 

Where, u = velocity of plate = 15 m/s

V = velocity of jet = 20 m/s

Explanation: 

Reverse Osmosis:

• Reverse osmosis is a water purification process where a semi-permeable membrane removes impurities.
• Concentration polarization occurs when retained solutes accumulate on the membrane, hindering water flow.
• It impacts efficiency and requires measures to minimize fouling, ensuring effective reverse osmosis for water treatment applications

Trickling filter:

• A trickling filter is a wastewater treatment method using a bed of rocks or synthetic media for microbial treatment.
• It enhances biological degradation.
• Ponding refers to water accumulation, often causing surface flooding.
• In wastewater treatment, trickling filters mitigate ponding by efficiently processing and reducing water contaminants, preventing surface water issues.

Coagulation:

• Coagulation in water treatment refers to the process of clumping together fine particles to form larger, settleable flocs.
• Charge neutralization involves adding chemicals to neutralize the electrical charges on particles, aiding coagulation.
• This reduces repulsion between particles, allowing them to come closer and form larger aggregates for effective removal during water treatment.

Additional Information

• Adsorption and concentration polarization are related phenomena, particularly in membrane processes.
• Concentration polarization occurs when solutes accumulate near a membrane surface, leading to reduced mass transport.
• Adsorption can contribute to concentration polarization as solutes may adsorb onto the membrane surface, impacting transport across the membrane.

Explanation:

Now, the capacity of road A,

\(\rm q_{max\ A}=\frac{1}{4}k_AU_A\)

Similarly, the capacity of road B,

\(\rm q_{max\ B}=\frac{1}{4}k_BU_B\)

Now, 

\(\rm \frac{C_A}{C_B}=\frac{\frac{1}{4}\times k_A\times U_A}{\frac{1}{4}\times k_B\times U_B}=\frac{k_AU_A}{k_BU_B}\)

Using standard properties of eigen values eigen vectors,

\([A]= \begin{bmatrix}1&2&3\\ 3&2&1\\ 3&1&2 \end{bmatrix} \)

If A is a square matrix, then its eigenvalues are equal to the eigenvalues of its transpose, since they share the same characteristic polynomial.

So, Ax = λx and A^Tx = λx 

So, statement 1 is correct

Ax = λx

A^-1Ax = A^-1(λx)

x = λA^-1x

A^-1x = \(x \over \lambda\)

So, eigen value and eigon vector of A^-1 is \(x \over \lambda\)

So, statement 2 is correct

If matrix A is n × n and is invertible (meaning it has an inverse, A^-1), then A and its inverse A^-1  have the same eigenvectors. This is because if v is an eigenvector of A with eigenvalue λ, then A^-1v is an eigenvector of A^-1 with the same eigenvalue λ. This property holds for invertible matrices.

So, statement 4 is also correct

Explanation:

Given, 

f(x) = e^x |sin x|

Consider,  

At  x = 0,

LHL = –e0 sin0 = 0

RHL = e0 sin 0 = 0

LHL = RHL = f(0) → continuous

At  x = 0,

LHD = –e0 × cos 0 – e0 sin 0 = –1

RHD = e0 cos 0 + e0 sin 0 = +1

LHD ≠ RHD → Not differentiable.

Explanation:

ΣF_y = 0

∴ V_A + V_B = 0       ...(i)

ΣM_A = 0

⇒ V_B × L – M + M = 0

∴ V_B = 0

From equation (i), we get

V_A = 0

∴ shear force is zero everywhere in the beams, and reactions are also zero.

The bending moment diagram is shown below

So, bending moment is not zero everywhere. Also deflection is not zero everywhere.

Maximum Mixing Depth (MMD):

• The Maximum Mixing Depth (Dmax) in the atmosphere refers to the maximum height reached by vertical air movements, such as turbulence or convection, during a specific time period.
• It represents the boundary between the mixed layer and the stable layer above. In the context of air pollution, Dmax is crucial, as it determines the vertical extent to which pollutants disperse and mix.
• Understanding Dmax helps assess the potential impact of pollutants on air quality and human health.
• The vertical dispersion of pollutants in the atmosphere is influenced by the maximum mixing depth (Dmax), which represents the height to which vertical mixing and turbulent processes extend.
• Up to Dmax, pollutants experience enhanced dispersion through vertical movement and mixing within the atmospheric boundary layer.
• The depth of the convective mixing layer in which vertical movement of pollutants is possible, is called the maximum mixing depth (MMD).
• The ventilation coefficient (Vc) is a parameter used in air quality modeling to estimate the dispersion of pollutants in the atmosphere.
• It is influenced by factors such as wind speed, atmospheric stability, and the mixing height or maximum mixing depth (Dmax).
• The relationship between the ventilation coefficient and Dmax is typically positive.

Explanation:

Solubility of Bitumen:

• The solubility of bitumen refers to its ability to dissolve in specific solvents, typically organic compounds. Bitumen exhibits low solubility, often requiring elevated temperatures for dissolution, impacting its use in various industrial applications, such as asphalt production.
• It is represented in terms of the amount of bitumen that can dissolve in a given solvent at a specific temperature, expressed as a percentage or ratio, indicating the solubility characteristics of bitumen in different conditions.

Softening point of bitumen:

• The softening point of bitumen is the temperature at which it attains a specific level of softness and deformation.
• It is determined by the ring and ball method. In this test, a steel ball is placed on a bitumen sample, and the temperature is gradually increased until the bitumen softens enough for the ball to sink a specified distance, represented by temperature.

Los Angeles abrasion test:

• The Los Angeles Abrasion Test is a method to assess the resistance of coarse aggregates to abrasion and degradation.
• It involves subjecting aggregates to repeated abrasion, simulating the impact of traffic.
• The percentage of material lost during the test provides a measure of aggregate durability for use in construction.

Ductility of bitumen:

• The ductility test for bitumen measures its ability to undergo deformation before breaking.
• A briquette-shaped specimen of bitumen is stretched at a specific rate and temperature until it breaks.
• The distance the bitumen stretches before rupture, expressed in centimeters, indicates its ductility and suitability for various applications, such as road construction.

Specific gravity of bitumen:

• Specific gravity of bitumen is a measure of its density about the density of water.
• It quantifies the mass of a given volume of bitumen compared to an equal volume of water.
• Specific gravity is crucial for assessing bitumen quality and consistency in construction and industrial applications.

Thin film oven test:

• The Thin Film Oven Test (TFOT) is a standard procedure used to simulate the aging of asphalt binders during the mixing and laying process.
• Asphalt samples are exposed to elevated temperatures in a thin film, evaluating their susceptibility to oxidative aging, hardening, and other changes that may affect performance.

Explanation:

Given differential equation

\(\rm \frac{du}{dt}+2tu^2=1 \)

\(\rm \frac{du}{dt}=f(t, u)=1-2tu^2\)

By backward Euler Iterative scheme,

u_n = u_n–1 + hf (t_n, u_n)

u_n – u_{n –1} = hf (t_n , u_n)

u_n – u_{n –1} = h[1 – 2t_n \(\rm u_n^2\)]

Given, h = 0.01, u_n = 1.75, t_n = 3.14

Putting these values we get

u_n – u_{n –1} = 0.01[1 – 2 × 3.14 × (1.752)²] = – 0.182325 ≃ – 0.182

Explanation:

Now, \(\rm σ^{'}_x\) = σ_x cos² θ + σ_y sin² θ + τ_xy sin 2θ      .........(i)

here θ = 45° 

\(\rm σ^{'}_x\) = 5 MPa

σ_y = 6 MPa

τ_xy = 3 MPa

from eq. (i), we get σ_x = -2 MPa

Now, the magnitude of the principal stresses at the point are

\(\rm σ_{1/2}=\frac{σ_x+σ_y}{2}\pm\frac{1}{2}\sqrt{(σ_y-σ_x)^2+4\tau xy^2}\) \(=\frac{-2+6}{2}\pm\frac{1}{2}\sqrt{(6+2)^2+4\times3^2}\)

σ₁ = 7 MPa

σ₂ = -3 MPa

The magnitude of the maximum principal stress is 7 N/mm²

Explanation:

Reactions to support:

Reactions at supports: Because of symmetry vertical reaction at supports

\(\rm=\frac{100}{2}=50 kN\)

Applying the method of sections and taking the section as shown in the figure above,

ΣF_y = 0

⇒ Fu₂L₃ cos45° + 20 + 20 – 50 = 0

Fu₂L₃ cos45° = 10

Fu₂L₃ = 10√2 kN = 14.12 kN (Tension)

Explanation:

A₁ = 40 × 10 = 400 cm²

A₂ = 20 × 50 = 1000 cm²

y̅₁ = 50 + 5 = 55 cm 

\(\rm \overline{y}_2=\frac{50}{2}=25\) cm

The centroid of composite shape, from bottom fiber,

\(\rm \overline{y}=\frac{A_1\overline{y}_1+A_2\overline{y}_2}{A_1+A_2}\)

\(\rm \overline{y}=\frac{(400\times55)+(1000\times 25)}{400+1000}\)

y̅ = 33.5714 cm

The moment of inertia of the section about the horizontal axis (X - X) passing through the centroid is 

\(\rm I_{xx}=\left[\frac{40\times10^3}{12}+400(\overline{y}_1-\overline y)^2\right]+\left[\frac{20\times50^3}{12}+1000(\overline y-\overline y_2)^2\right]\)

\(=\left[\frac{40\times10^3}{12}+400(55-33.5714)^2\right]+\left[\frac{20\times50^3}{12}+1000(33.5714-25)^2\right]\)

= 187007.2925 + 281802.2313

I_xx = 468809.5238 cm⁴ ≃ 468810 cm⁴

Explanation:

The expression can calculate the critical height of soil formation, 

C + γHC cos2 β tan ϕ = γHC sin β cos β

Calculation:

Given, 

Bulk unit weight (γ) = 19 kN/m3

The angle of internal friction (ϕ) = 25°

Cohesion (C) = 15 kPa

The inclination of rock slope (β) = 35°

Now, The critical height (in m) of the soil formation up to which it would be stable without any failure is

C + γH_C cos² β tan ϕ = γH_C sin β cos β

15 + 19 × H_C × cos² 35° tan 25° = 19 × H_C × sin 35° cos 35°

⇒ H_C = 5.03 m

Explanation:

Active earth pressure at base,

∴ σ = k_a(γz + q) - 2C \(\rm \sqrt{ka}\)

Calculation:

\(\rm k_a=\frac{1-\sin \phi}{1+\sin \phi}=\frac{1-\sin 25^{\circ}}{1+\sin 25^{\circ}}=0.406\)

q = 20 + 18 × 3 = 74 kN/m² 

γz = 19 × 4 = 76 kN/m²

C = 20 kN/m² 

So, the lateral active earth pressure acting at the base of the wall is

\(\rm \sigma=0.406\times (76+74)-2\times 20\times \sqrt{0.406}\)

= 35.413 kN/m² ≃ 35.4 kN/m²

Explanation:

Now, active pressure at depth z.

\(\rm \sigma_z=k_a\gamma z-2C\sqrt{k_a}-12z+kq\)

At z = 0

[∵ k = 1 for ϕ = 0°, q = 30 kN/m²; C = 20 kN/m²]

σ_z = –2 × C + q = 2 × 20 + 30 = –40 + 30 = –10 kN/m²

At z₀, σ_z = 0

18z₀ – 40 – 12z₀ + 30 = 0

⇒ 6z₀ – 10 = 0 ⇒ z₀ = 1.66 m

At depth H, \(\rm \sigma_H=k_a\gamma H-2C\sqrt{k_a}-12H+kq\)

So, σ_H = 18H – 40 – 12H + 30 = 6H – 10

For unsupported depth of excavation, total active thrust must be zero

So, \(\rm \frac{1}{2}\times 10\times 1.66=\frac{1}{2}\times (H-1.66)\times (6H-10)\)

⇒ \(\rm H=\frac{10}{3}=3.33\ m\)

Explanation:

Given: Time = 12 hr

Runoff, R = 100 mm

The total depth of precipitation,

\(\rm P=\frac{1}{2}[20\times4]+2\times 20+\frac{1}{2}\times 20\times 6=140 \ mm\)

∴ Total infiltration, I = P – R

= 140 – 100 = 40 mm

Now, W-Index = \(\rm \frac{l}{t}=\frac{140-100}{12}=3.33\ mm/hr\)

Now, \(\rm \frac{20}{4}=\frac{3.33}{t_1}\)

⇒ t₁ = 0.67 hr

Similarly, \(\rm \frac{20}{6}=\frac{3.33}{t_2}\)

⇒ t₂ = 0.99 hr

Now, ϕ-Index = \(\rm \frac{Total\ infiltration \ in \ which\ rainfall\ excess \ occur}{Time\ period\ in\ rainfall\ which\ excess\ occur}\)

\(\rm =\frac{I-infiltration \ in\ which\ no\ runoff}{time\ excess}\)

ϕ-Index = \(\rm \frac{40-\left[\frac{1}{2}\times3.33\times 0.67+\frac{1}{2}\times 3.33\times 0.99\right]}{12-0.67-0.99}\) = 3.6 mm/hr

Check:

Runoff = Area of hatched portion

\(\rm =\frac{1}{2}\times(20-3.6)\times (4-0.67)+(20-3.6)\times 2+\frac{1}{2}\times (20-3.6)\times (6-0.99)\)

= 100 mm (Hence, OK)

∴ ϕ-Index = 3.60 mm/hr

Explanation:

Given, y₁ = 0.2 m, y₂ = 1 m 

As we know

\(\rm \frac{2q^2}{g}=y_1y_2(y_1+y_2) \)

⇒ \(\rm \frac{2q^2}{10}=0.2×1(0.2+1)\)

⇒ \(\rm q^2=\frac{10}{2}× 0.2× 1.2=1.2\)

q = 1.095 m³/s/m

The values of the specific force at the pre-jump and post-jump sections are the same and are equal to

= \(\rm \frac{Q^2}{Ag}+A\overline y\)

\(\rm =\frac{q^2B^2}{By_1g}+\frac{By_1y_1}{2}\)

=\(\rm \left(\frac{1.095^2}{0.2× 102}+\frac{0.2^2}{2}\right)\)× 1 = 0.6195 m³ ≃ 0.62 m³

So, Specific force per unit width = 0.62 m³/m

Explanation:

Initial infiltration capacity, f₀ = 10 mm/h

Final infiltration capacity, f_c = 5 mm/h

Horton’s Decay constant, k_h = 0.5/h

The total infiltration depth (in mm) from a uniform storm of duration 12 h is

\(\rm F_p=\int_0^T\left[f_c+[f_0-f_c]e^{-k_nt}\right]dt\)

F_P \(\rm=\int_0^{12}(5+(10-5)e^{-0.5t})dt\)

FP ​\(\rm =[5t]_0^{12}+\left[\frac{5e^{-0.5t}}{(-0.5)}\right]_0^{12}\)

FP ​\(\rm =[5\times12]+\left[-10e^{-0.5\times12}-[-10e^{-0.5\times0}]\right]\)

FP ​= 60 + [9.975] = 69.97 mm ≃ 70 mm

Explanation:

Total Energy content (as discarded basis):

[0.2 × 4.5 + 0.45 × 16 + 0.05 × 14 + 0.1 × 32 + 0.2 × 8] = 13.6 MJ/kg

Energy content (on dry basis) for moisture content of 26% and dry content of 74%.

The energy content of the solid waste on dry-weight basis is

\(\rm =\frac{13.6\ MJ/kg\times 100%}{(100-26)%}=18.38\ MJ/kg\) ≃ 18.4 MJ/kg

Explanation:

The temporal mean velocity gradient is

\(\rm G_1=\sqrt{\frac{P}{μ_1\times v}}\)     .........(i)

\(\rm G_2=\sqrt{\frac{P}{μ_2\times v}}\)      .........(ii)

(P and v are constant)Divide (i) by (ii), we get

\(\rm \frac{G_1^2}{G_2^2}=\frac{μ_2}{μ_1}\)

\(\rm G_2^2=\left[\frac{(100s^{-1})^2\times 1.139\times 10^{-3}\ N-s/m^2}{1.518\times 10^{-3}\ N-s/m^2}\right]^{1/2}=86.62\ S^{-1}\)

The decrease in the average velocity gradient (in %) due to the reduction in water temperature is

G = \(\rm \frac{G_1-G_2}{G_1}\times 100=\frac{100-86.62}{100}\times 100=13.38%\)

Explanation:

Given: Q = 0.5 m³/sec

The food-to-microorganism ratio is 

\(\rm \frac{F}{M}=0.2\) mg/mg-d

S₀ = 150 mg/L

X = 2000 mg/L

Now, the volume of the aeration tank required for the plant is

\(\rm \frac{F}{M}=\frac{Q_0S_0}{V.X}\)

\(\rm 0.2d^{-1}=\frac{0.5\frac{m^3}{sec}\times86400\frac{sec}{d}\times 150\frac{mg}{L}}{V\times2000\frac{mg}{L}}\)

V = 16200 m³

Explanation:

Using triangle properties

Now, tan12°45′ = \(\rm \frac{V+0.68}{D+55}\)     .........(i)

Also, tan18°45′ = \(\rm \frac{V}{D}\)      .........(ii)

On solving equations (i) and (ii), we get

V = 35.287 m

D = 103.954 m

Now, R.L. of Hilltop is 

= R.L. of BM + 2.34 + V

= 100 + 2.34 + 35.287 = 137.627 m