The expression "back out" means to withdraw from a commitment, plan, or agreement. In the sentence, the speaker had initially made plans to go skiing but could not follow through because of an injury. Therefore, "back out" is the correct phrasal verb to describe this withdrawal.

Other options are incorrect:

- "back up" means to support or reverse a vehicle.

- "back of" is not a valid phrasal verb.

- "back on" doesn't fit the sentence grammatically or idiomatically.


Thus, the most appropriate choice is "back out".
Quick Tip: Phrasal verbs often change the meaning of the root verb entirely—always learn them in context.

The subject of the sentence is "The President", which is singular. The phrase "along with the Council of Ministers" is a modifying phrase and does not affect the subject-verb agreement. Therefore, the verb should also be singular.

- "wishes" is the singular form of the verb and agrees with the singular subject "The President".

- "wish" is plural and would be incorrect here.

- "will wish" changes the tense unnecessarily.

- "is wishing" is awkward and not appropriate in this context.


Hence, the correct form is: "The President, along with the Council of Ministers, wishes to visit India next week."
Quick Tip: Ignore interrupting phrases like "along with", "as well as", etc., when determining subject-verb agreement.

First, convert the power rating to kilowatts: \[ 40 W = \frac{40}{1000} = 0.04 kW \]

Case 1: Desk light used for 10 hours per day \[ Energy = 0.04 \times 10 \times 180 = 72 kWh \] \[ Cost = 72 \times 7 = ₹504 \]

Case 2: Desk light used for 12 hours per day \[ Energy = 0.04 \times 12 \times 180 = 86.4 kWh \] \[ Cost = 86.4 \times 7 = ₹604.8 \]

Percentage Increase: \[ \frac{604.8 - 504}{504} \times 100 = \frac{100.8}{504} \times 100 \approx 20% \]

Therefore, the percentage increase in cost is 20% and the original cost is ₹504.
Quick Tip: Always convert watts to kilowatts and use the formula: Energy = Power × Time × Days. Then multiply by rate to calculate cost.

We analyze the logic based on letter-number patterns. Let's look at the structure row-wise:

Top Row:
N (21), U (14), F (9), (i) = ?

Bottom Row:
H (12), L (?), O (15), (?) = ?

Now map letters to positions in the alphabet: \[ N = 14,\quad U = 21,\quad F = 6,\quad H = 8,\quad L = 12,\quad O = 15 \]

Compare these with the numbers in the second row: \[ 21 \leftrightarrow N = 14 \Rightarrow 14 + 7 = 21
14 \leftrightarrow U = 21 \Rightarrow 21 - 7 = 14
9 \leftrightarrow F = 6 \Rightarrow 6 + 3 = 9
6 \leftrightarrow (i) = ? \Rightarrow To balance pattern \Rightarrow (i) = L (12) \Rightarrow 12 - 6 = 6 \quad \checkmark \]


Now the second row:
\[
H = 8,\quad 12 \Rightarrow 8 + 4 = 12

(iv) = 8 ✔️

L = 12,\quad (ii) = ? \Rightarrow 12 + 3 = 15 ⇒ (ii) = K (11) ✔️

(iii) = 12 ⇒ O = 15 ⇒ 15 - 3 = 12 ✔️

So correct set:
(i) = L,\quad (ii) = K,\quad (iii) = 12,\quad (iv) = 8

Thus, the correct option is (D).
Quick Tip: When solving letter-number reasoning grids, convert letters to their alphabet positions (A=1 to Z=26), then analyze the mathematical pattern row-wise or column-wise.

Let the total number of balls be 100 (since percentages are given). So, the actual number of each type of ball in option (A) is: \[ V = 10,\quad Y = 30,\quad R = 20,\quad G = 40 \]

Now verify the conditions:

(i) \( Y + 2V = 30 + 2 \times 10 = 30 + 20 = 50 \) ✔️

(ii) \( V + G = 10 + 40 = 50 \) ✔️

(iii) \( Y + R = 30 + 20 = 50 \) ✔️

(iv) \( V + 2R = 10 + 2 \times 20 = 10 + 40 = 50 \) ✔️


All conditions are satisfied only in option (A). Other options do not verify all four conditions simultaneously.
Quick Tip: Assume the total is 100 when pie chart percentages are given. Translate each condition into equations and verify using actual values from the options.

The title of the passage, “A Funeral,” and the use of past tense verbs such as “was divided,” “worked,” and “courted” indicate that the person being discussed is no longer alive. The statement “he probably courted his fatal illness” also supports this inference, implying he ultimately succumbed to that illness. The passage is reflective and eulogistic in nature, pointing toward the man's death.

The other options include unsupported claims. For instance, there is no mention of the man working in a court or finding joy in scattering fruits. The “fruits of his erudition” refers metaphorically to the impact of his knowledge, not literal joy or fruit scattering.
Quick Tip: Pay attention to past tense usage and the title or source of a passage—it often provides key contextual clues for inference-based questions.

We are given two sequences of letters representing paths around a circular clock-like figure. Each sequence starts from a reference letter and continues in a specific order:

- O* = O Q S Z P R T
- X* = X Z P W Y O Q

These sequences follow a clockwise path on the circle. To find P*, we need to start from P and trace a similar clockwise pattern.

Looking at the clock diagram, starting from P and moving clockwise gives the sequence: \[ P \rightarrow R \rightarrow T \rightarrow O \rightarrow Q \rightarrow S \rightarrow U \]

This matches option (B).

To verify, count each step from P in the figure:
P → R → T → O → Q → S → U – all in clockwise direction, and all letters are unique with no repetitions, matching the style of the given sequences.
Quick Tip: For circular reasoning questions, sketch or trace the path visually on the diagram and ensure you're moving in a consistent direction (clockwise or counter-clockwise).

We are given the constraints: \[ P < Q,\quad S > P > T,\quad R < T \]

We need to assign the digits 1 through 5 (each used only once) to P, Q, R, S, and T in a way that satisfies all the above conditions.

Let’s try to assign values that satisfy these relations step-by-step:

From \( S > P > T \), we can choose: \[ S = 5,\quad P = 3,\quad T = 2 \]

This satisfies \( S > P > T \).
Now for \( P < Q \), if \( P = 3 \), then \( Q \) must be greater than 3, so we can take: \[ Q = 4 \]

That leaves only 1 unused, which can go to: \[ R = 1 \]

Now check if all conditions are satisfied:
- \( P = 3 < Q = 4 \) ✔️
- \( S = 5 > P = 3 > T = 2 \) ✔️
- \( R = 1 < T = 2 \) ✔️

All conditions are satisfied.

Thus, the value of \( P \) is 3.
Quick Tip: When solving such logic puzzles with digit constraints, list available digits and test possible combinations systematically to satisfy all inequalities.

Given: \[ Cost of 5 kg of variety P = ₹800 \Rightarrow Cost per kg = ₹160
Cost of 4 kg of variety Q = ₹800 \Rightarrow Cost per kg = ₹200 \]

Let the seller mix 5 kg of P and 4 kg of Q (to match the quantity from the cost data). \[ Total cost = ₹800 + ₹800 = ₹1600
Total weight = 5 + 4 = 9 kg
Selling price per kg = ₹192 \Rightarrow Total selling price = 9 \times 192 = ₹1728 \]
\[ Profit = ₹1728 - ₹1600 = ₹128
Profit % = \(\frac{128}{1600} \times 100 = 8%\) \]

Thus, the assumed mixture gives exactly 8% profit, which matches the condition. Therefore, the weight ratio P : Q is: \[ 5 : 4 \] Quick Tip: In mixture problems involving profit, use assumed weights based on cost data to match the required profit percentage. Compare cost price and selling price for total quantity.

Let the foot of the perpendicular from point P to line QR be at a distance \( x \) km from Q, and the perpendicular height be \( h \). We can now apply the Pythagorean theorem to two right-angled triangles:
\[ h^2 + x^2 = 13^2 = 169 \quad (i) \] \[ h^2 + (14 - x)^2 = 15^2 = 225 \quad (ii) \]

Now subtract equation (i) from (ii): \[ [h^2 + (14 - x)^2] - [h^2 + x^2] = 225 - 169 \] \[ (14 - x)^2 - x^2 = 56 \] \[ 196 - 28x = 56 \Rightarrow 28x = 140 \Rightarrow x = 5 \]

Substitute \( x = 5 \) in equation (i): \[ h^2 + 25 = 169 \Rightarrow h^2 = 144 \Rightarrow h = \sqrt{144} = 12 \]

Therefore, the minimum possible length of the connecting road is \( \boxed{12 km} \).
Quick Tip: To find the shortest distance from a point to a line segment, drop a perpendicular and apply the Pythagorean theorem to form solvable right triangles.

The transpose of a matrix is obtained by interchanging its rows and columns. That is, the element at position \((i, j)\) in the original matrix becomes the element at position \((j, i)\) in the transposed matrix.

Given: \[ A = \begin{bmatrix} 2 & 3 & 4
1 & 4 & 5
4 & 3 & 2 \end{bmatrix} \]

Now taking the transpose:

- First row \((2\ 3\ 4)\) becomes first column
- Second row \((1\ 4\ 5)\) becomes second column
- Third row \((4\ 3\ 2)\) becomes third column

So, \[ A^T = \begin{bmatrix} 2 & 1 & 4
3 & 4 & 3
4 & 5 & 2 \end{bmatrix} \]

Hence, the correct option is (A).
Quick Tip: To transpose a matrix, flip it over its main diagonal. This is useful in many areas of linear algebra, especially in symmetric and orthogonal matrices.

To integrate \( \int \ln(x) dx \), we use the method of integration by parts.

Recall the formula: \[ \int u\,dv = uv - \int v\,du \]

Choose: \[ u = \ln(x) \Rightarrow du = \frac{1}{x} dx,\quad dv = dx \Rightarrow v = x \]

Now apply the integration by parts formula: \[ \int \ln(x)\, dx = x \cdot \ln(x) - \int x \cdot \frac{1}{x}\, dx \] \[ = x \cdot \ln(x) - \int 1\, dx = x \cdot \ln(x) - x + C \]

Therefore, \[ \int \ln(x)\, dx = x \cdot \ln(x) - x + Constant \]

This matches option (A).
Quick Tip: When integrating logarithmic functions like \( \ln(x) \), consider integration by parts with \( u = \ln(x) \). This technique often simplifies otherwise difficult integrals.

Statement (P): Fly ash and ground granulated blast furnace slag are pozzolanic materials and are classified as mineral admixtures. According to IS standards, they are commonly used to enhance concrete properties and are considered artificial mineral admixtures. Hence, statement P is correct.

Statement (Q): IS 456:2000 specifies that when mineral admixtures are used in concrete, the rate of hydration tends to slow down. This requires a longer moist curing period to ensure sufficient strength development.

- Ordinary Portland Cement (OPC) usually requires 7 to 10 days of curing.

- With mineral admixtures, a minimum of 10 days of moist curing is recommended.

Thus, both statements (P) and (Q) are correct, making option (A) the right choice.
Quick Tip: Pozzolanic materials like fly ash and slag reduce the rate of hydration, thus increasing the required curing time to achieve optimal strength in concrete.

We are to find the influence line for the axial force in member G–I when a unit load travels along the bottom chord of the truss.

From the geometry of the truss:
- Total span = \(6 \times 6 = 36 m\)
- Height of truss \(h = 6 m\)
- Panel point spacing = 6 m
- Distance from A to G = 2 panels = 12 m
- Distance from G to right support = 4 panels = 24 m

Now, to find the ordinate of the influence line for member G-I under G, we use the standard formula for diagonal or bottom chord members in simple trusses: \[ Ordinate = \frac{ab}{l h} \]
Where:
- \(a = 12 m,\ b = 24 m,\ l = 36 m,\ h = 6 m\)

Substitute into the formula: \[ \frac{12 \times 24}{36 \times 6} = \frac{288}{216} = 1.33 \]

Hence, the influence line has an ordinate of \(+1.33\) under joint G, which corresponds to tension in member G-I. Therefore, the correct diagram is option (B).
Quick Tip: To find influence lines for truss members, use geometry-based influence line theory and remember that bottom chord members generally carry tension under downward unit loads.

Clayey soils have very low permeability due to their fine-grained structure. The constant head test is generally suitable for coarse-grained soils like sand and gravel, which allow faster water flow. On the other hand, the falling head test is better suited for fine-grained soils like clay because it can more accurately measure the slow rate of flow through such soils.

In the falling head test, the rate at which the water level falls in a standpipe is monitored, allowing precise calculation of permeability even in low-permeability soils.

Hence, the falling head test is most appropriate for laboratory measurement of permeability in clayey soils.
Quick Tip: Use the falling head test for fine-grained soils (clays and silts), and the constant head test for coarse-grained soils (sands and gravels).

A hydraulic jump is formed in an open channel when the flow transitions from supercritical flow \((F_r > 1)\) to subcritical flow \((F_r < 1)\). This typically occurs when a steep slope (which supports supercritical flow) changes to a mild slope, causing the flow to decelerate and a jump to form.

This change in slope causes the kinetic energy of the flow to be suddenly dissipated, forming a jump which is visible as a sudden rise in the water surface.

Hence, the correct answer is (B).
Quick Tip: Hydraulic jumps are energy-dissipating phenomena in channels, usually formed when supercritical flow transitions to subcritical flow.

Sulphur reducing bacteria (SRB) are anaerobic bacteria that reduce sulphates to hydrogen sulphide gas (H\(_2\)S) in sewage systems. This hydrogen sulphide, upon reaching the crown of the sewer where oxygen is present, gets oxidized to sulphuric acid.

The sulphuric acid reacts with the concrete and leads to deterioration of the sewer crown, a process known as crown corrosion.

Hence, the correct answer is (C).
Quick Tip: Crown corrosion in sewers is due to biological activity of sulphur reducing bacteria which produce hydrogen sulphide, leading to sulphuric acid formation.

As per IRC 37:2018 (Guidelines for the Design of Flexible Pavements), the following values are recommended for the structural design of flexible pavements for national highways:

- Minimum traffic growth rate: 5%
- Design period: 20 years

These values are taken as the default unless actual traffic studies suggest otherwise.

Hence, the correct option is (A).
Quick Tip: Refer to IRC 37:2018 for standards related to pavement design parameters like design life and traffic growth.

Given:
Corrected runway length after applying elevation and temperature corrections: \[ l = 700 m,\quad Gradient = 1% \]

According to standards, for every 1% gradient, the runway length increases by 20%.
\[ Correction = 700 \times \frac{20}{100} = 140 \] \[ Corrected length = 700 + 140 = 840 m \] \[ Or simply: 700 \times 1.2 = 840 m \]

Hence, the final corrected length is \( \boxed{840 m} \).
Quick Tip: Remember: Runway length is increased by 20% for every 1% gradient as per ICAO or aviation design guidelines.

In highway geometry, the transition from a straight path (tangent) to a circular curve is marked by specific points:

- Point of Intersection (PI): Where two tangents meet.
- Point of Curve (PC): Where the curve begins (tangent ends and curve starts).
- Point of Tangency (PT): Where the curve ends and tangent resumes.

Therefore, the point where the road alignment changes from a tangent to a curve is called the Point of Curve (PC).

Hence, the correct answer is (C).
Quick Tip: Remember the road alignment transition points: PI (intersection), PC (start of curve), PT (end of curve).

Given: \[ \vec{V} = u(x, y)\,\hat{i} + v(x, y)\,\hat{j} \]

There is no component in the z-direction, so the flow is 2D.

Divergence of velocity: \[ \nabla \cdot \vec{V} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \]
Hence, option (C) is correct.

Curl of velocity (vector form): \[ \nabla \times \vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}
u & v & 0 \end{vmatrix} \]

Evaluating the determinant: \[ \nabla \times \vec{V} = \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) \hat{k} \]

Thus, the z-component of curl is: \[ \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) \hat{z} \]

So, option (A) is also correct.
Options (B) and (D) are incorrect as they show incorrect expressions for curl and divergence respectively. Quick Tip: For 2D velocity fields, divergence is \( \nabla \cdot \vec{V} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \), and curl's z-component is \( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \).

Option (A):
If A and B are mutually exclusive, then: \[ P(A \cap B) = 0 \]
So the expression \( P(A \cap B) = P(A) \cdot P(B) \) is incorrect unless A and B are independent, not mutually exclusive. Hence, option (A) is incorrect.

Option (B):
Conditional probability is defined as: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]
If \( B \subseteq A \), then \( A \cap B = B \), so: \[ P(A \mid B) = \frac{P(B)}{P(B)} = 1 \]
Thus, option (B) is correct.

Option (C):
If A and B are mutually exclusive, then: \[ P(A \cap B) = 0 \Rightarrow P(A \cup B) = P(A) + P(B) \]
So, option (C) is correct.

Option (D):
If A and B are independent, then: \[ P(A \cap B) = P(A) \cdot P(B) \]
This value is generally non-zero unless either P(A) or P(B) is zero. Hence, option (D) is incorrect. Quick Tip: Mutual exclusivity implies \( P(A \cap B) = 0 \), while independence implies \( P(A \cap B) = P(A) \cdot P(B) \). These two are not the same and should not be confused.

Option (A): This statement is correct. Characteristic strength is defined as the value below which not more than 5% of test results are expected to fall. The target mean strength is calculated based on the characteristic strength and the standard deviation, which depends on the quality control. The given statement is therefore a contradiction and hence CORRECT in pointing out the flaw.

Option (B): Incorrect. Ten percent fines value has units (typically in terms of load), and hence it is not a non-dimensional quantity.

Option (C): Correct. With shorter loading durations, concrete exhibits higher stiffness. A longer duration like 1-day results in a smaller secant modulus than that for a quick loading of 10 minutes.

Option (D): Correct. Increasing the carbon content in steel generally increases its strength characteristics including the 0.2% proof stress (yield strength).
Quick Tip: Remember: Ten percent fines value has units (it's not dimensionless), and characteristic strength considers 5% chance of failure, not 50%.

Option (A): Incorrect. As the water table goes deeper (i.e., increases in depth from surface), the pore water pressure acting at a point in the soil decreases. Hence, the effective stress \((\sigma' = \sigma - u)\) increases, not decreases.

Option (B): Correct. Bulking of sand occurs due to moisture forming meniscus at particle contacts, leading to capillary tension and apparent volume increase.

Option (C): Correct. In liquefied soils, the pore water pressure equals the total stress, leading to near-zero effective stress.

Option (D): Incorrect. Earth pressure can be greater, equal, or less than vertical effective stress depending on the stress state (active, passive, or at-rest). Therefore, the statement is not always true.
Quick Tip: Effective stress increases as the water table depth increases. Earth pressure can vary with stress condition; don’t assume it's always less than vertical stress.

Cut-offs are vertical or inclined structural members provided below the floor of weirs or barrages to prevent sub-surface flow that may lead to piping.

- Option (B): Correct. Cut-offs increase the seepage path length beneath the floor, thereby reducing the exit gradient and minimizing the risk of piping, which is a major cause of failure in hydraulic structures.

- Option (C): Correct. The depth of cut-offs is designed considering the maximum anticipated scour depth, to ensure that the foundation remains safe even if the bed material is eroded during floods.

- Option (A): Incorrect. Cut-offs do not influence surface flow rates; they function below the floor and impact sub-surface flow patterns.

- Option (D): Incorrect. Hydraulic jumps are managed using stilling basins, baffle blocks, and aprons, not cut-offs.
Quick Tip: Cut-offs protect structures against piping and undermining. Their depth is governed by scour analysis, not flow behavior above the structure.

Drainage density is defined as the total length of streams per unit area of the watershed. It significantly influences the shape of the flood hydrograph:

- Lower drainage density indicates fewer channels and longer flow paths. As a result:
- Runoff takes longer to reach the outlet,
- Peak discharge is lower,
- Time base of the hydrograph is longer.

Hence:
- Option (B) is correct — lower drainage density leads to lower peak discharge.
- Option (C) is correct — lower drainage density leads to longer duration of flow.

- Option (A) is incorrect — lower drainage density causes a lower, not higher, peak.
- Option (D) is incorrect — it reverses the effect on time base.
Quick Tip: High drainage density results in faster, sharper hydrographs; low drainage density causes slower, flatter hydrographs with longer duration.

Biodegradable organic waste should be managed using sustainable and environment-friendly technologies:

- Biohydrogenation (A): Suitable for converting organic material into bio-hydrogen, an emerging green fuel.

- Anaerobic digestion (B): Widely used method for treating organic waste in the absence of oxygen, producing biogas as a useful by-product.

- Composting (C): An aerobic process of biodegradation that converts organic matter into useful compost.

- Open dumping (D): Incorrect practice. It leads to uncontrolled decomposition, release of methane, foul smell, contamination of soil and groundwater, and spread of disease vectors.

Therefore, open dumping is not recommended for any kind of waste, especially not for highly biodegradable organic matter.
Quick Tip: Avoid open dumping — it causes pollution and health hazards. Use biological treatments like composting or digestion for organic waste.

Option (A): Incorrect. In warm climate regions, bitumen with a higher softening point is desirable to prevent deformation and bleeding under high temperatures. A lower softening point would cause the pavement to fail prematurely.


Option (B): Correct. The viscosity of bitumen affects both mixing and compaction temperatures. A proper viscosity ensures good coating of aggregates and workability during paving.


Option (C): Correct. Air voids in the range of 3–5% ensure that the pavement has sufficient durability and strength while maintaining resistance to deformation.


Option (D): Correct. The solubility test determines the proportion of pure bitumen in a sample and is used to assess the quality and presence of impurities.
Quick Tip: In hot climates, always use bitumen with a high softening point to prevent surface distress.

The order of a differential equation is determined by the highest derivative present. Here, the highest order derivative is: \[ \frac{d^3 y}{dx^3} \Rightarrow Order = 3 \]

The degree is the highest power to which the highest order derivative is raised, provided the equation is polynomial in derivatives. In this case: \[ \left( \frac{d^3 y}{dx^3} \right)^1 \Rightarrow Degree = 1 \]

So, order = 3 and degree = 1.
Quick Tip: The order is the highest derivative present, and degree is the power of that derivative (in polynomial form).

Given:
Width, \( b = 250 mm \)

Effective depth, \( d = 500 mm \)

Design shear strength, \( \tau_c = 0.3 N/mm^2 \)


Step 1: Calculate shear force, \[ V_e = \tau_c \cdot b \cdot d = 0.3 \times 250 \times \frac{500}{1000} = 37.5\ kN \]

Step 2: Use IS 456:2000 torsion formula, \[ T_u = \frac{V_e \cdot b}{1.6 + \frac{1.67 b}{d}} = \frac{37.5 \times 0.25}{1.6} = 5.86\ kN.m \]

Hence, the torsional moment capacity is \(\boxed{5.86\ kN.m}\).
Quick Tip: Ensure all values are in consistent units before applying IS 456:2000 design formulas for torsion.

The factor of safety (FOS) against quick condition is given by: \[ FOS = \frac{i_c}{i_{exit}} \]

Where,
\[ i_{exit} = \frac{\Delta h}{N_d \cdot L} = \frac{9}{10 \cdot 1} = 0.9 \]
\[ i_c = \frac{\gamma'}{\gamma_w} = \frac{21 - 9.81}{9.81} = \frac{11.19}{9.81} \approx 1.14 \]
\[ FOS = \frac{1.14}{0.9} \approx 1.27 \]

Thus, the factor of safety against the quick condition is \( \boxed{1.27} \).
Quick Tip: To assess piping safety in soil under hydraulic structures, use \( FOS = i_c / i_{exit} \) with exit gradient from flow net geometry.

For double drainage (top and bottom), the drainage path length: \[ d = \frac{H}{2} = \frac{6}{2} = 3\ m \]

Time factor for 50% consolidation: \[ T_v = \frac{\pi}{4} \times (0.5)^2 = 0.196 \]

The time factor is also defined as: \[ T_v = \frac{C_v \cdot t}{d^2} = \frac{k \cdot t}{m_v \cdot \gamma_w \cdot d^2} \]

Rearranging for permeability \(k\): \[ k = \frac{T_v \cdot m_v \cdot \gamma_w \cdot d^2}{t} \]

Substitute the values: \[ k = \frac{0.196 \times 1.51 \times 10^{-3} \times 9.81 \times 3^2}{2} \approx 0.013\ m/year \]

Hence, the coefficient of permeability is \( \boxed{0.013\ m/year} \).
Quick Tip: For consolidation with double drainage, use half the thickness of the layer as the drainage path and apply time factor equations accurately.

The head loss in a pipe due to friction is given by the Darcy-Weisbach equation: \[ h_f = \frac{8 Q^2 L f}{\pi^2 g D^5} \]
Where,
- \( Q \) is the discharge,

- \( L \) is the length of the pipe,

- \( D \) is the diameter of the pipe,

- \( f \) is the Darcy-Weisbach friction factor,

- \( g \) is the acceleration due to gravity.


The ratio of head loss in Pipe-B to Pipe-A is: \[ \frac{h_B}{h_A} = \frac{\frac{8 Q^2 L_B f}{\pi^2 g D_B^5}}{\frac{8 Q^2 L_A f}{\pi^2 g D_A^5}} = \left( \frac{D_A}{D_B} \right)^5 \times \frac{L_B}{L_A} \]

Substitute the given values: \[ \frac{h_B}{h_A} = \left( \frac{2}{1} \right)^5 \times \frac{200}{100} = 32 \times 2 = 64 \]

Hence, the ratio of head loss in Pipe-B to the head loss in Pipe-A is \( \boxed{64} \).
Quick Tip: The ratio of head losses in two pipes is proportional to the 5th power of the ratio of their diameters and the ratio of their lengths.

The equilibrium constant \(k\) for the dissociation of HOCl is given by: \[ k = \frac{[HOCl]}{[OCl^-][H^+]} \]
Substituting the values: \[ 10^{7.5} = \frac{[HOCl]}{[OCl^-] \times 10^{-8.5}} \]
Simplifying further: \[ 10^{-1} = \frac{[HOCl]}{[OCl^-]} \]
This gives the relationship: \[ [HOCl] + [OCl^-] = \frac{2 mg \times 10^{-3}}{71} \]
Converting to moles per liter: \[ [HOCl] + [OCl^-] = 2 \times 10^{-3} \, moles/l \times 10^6 = 2.56 \, \mumoles/l \]

Thus, the concentration of HOCl in water is approximately \( \boxed{2.6} \, \mumoles/l \).
Quick Tip: Remember to convert units properly when dealing with mg/l and \(\mu\)moles/l. Also, use equilibrium constants and pH values to solve dissociation problems.

In this problem, we are asked to find the true distance between the points after a scaling error was detected. The surveyor initially used a scale where 1 cm on the plan represented 40 m in reality. However, the correct scale should have been 1 cm = 20 m.

1. Step 1: Calculate the scale ratio (RF) for the wrong and correct scales:

The wrong scale gives:
\[ RF of wrong scale = \frac{1}{20} \]

The correct scale gives:
\[ RF of corrected scale = \frac{1}{40} \]

2. Step 2: Use the formula for corrected length:

To find the corrected length, we use the ratio of the two scale factors:
\[ Corrected length = \left( \frac{RF of wrong scale}{RF of corrected scale} \right) \times Measured length \]

Substituting the values:
\[ Corrected length = \left( \frac{\frac{1}{20}}{\frac{1}{40}} \right) \times 468 = 2 \times 468 = 936 \ m \]

Thus, the true distance between the points is \( \boxed{936} \, m \).


Explanation:

The correction is made by adjusting the scale factor from the incorrect value to the correct value. Since the correct scale represents half the length of the wrong scale, the actual distance is double the measured length. Hence, the final corrected value of 936 m is obtained by multiplying the measured value by 2.
Quick Tip: Always check the scale carefully. When scaling errors are discovered, use the ratio of the scale factors to adjust the measured values.

The given differential equation is: \[ \frac{dy}{dx} = e^{x - y} \]

Step 1: Rearranging the equation: \[ \frac{dy}{dx} = e^x \cdot e^{-y} \]

Step 2: Separating the variables: \[ e^y \, dy = e^x \, dx \]

Step 3: Integrating both sides: \[ \int e^y \, dy = \int e^x \, dx \]

Step 4: Performing the integrations: \[ e^y = e^x + C \]

Step 5: Taking the natural logarithm of both sides: \[ y = \ln(e^x + C) \]

Thus, the correct solution is \( y = \ln(e^x + C) \), which corresponds to option (A).
Quick Tip: When solving differential equations, always separate variables first, integrate both sides, and apply the necessary logarithmic operations for solutions involving exponential terms.

Given:
\(d = 250 mm, \quad S = 5 mm, \quad t = 10 mm, \quad Thickness of flat plate = 5 mm, \quad f_u = 410 N/mm^2, \quad \gamma_{mw} = 1.25\)


1. Step 1: Throat thickness of the weld (t\(_t\)) \[ t_t = 0.75 \times 5 = 3.5 mm \]

2. Step 2: Calculation of Section Modulus (Z\(_p\)) \[ Z_p = \frac{J}{r} = \frac{A r^2}{r} = A r \]
Using geometry for circular section: \[ Z_p = \left(\pi \times t \times \frac{d}{2}\right) \times \frac{d^2 t}{2} \] \[ Z_p = \left(\pi \times 10 \times \frac{250}{2}\right) \times \frac{250^2 \times 3.5}{2} \]

3. Step 3: Torque that can be applied on the plate
The maximum torque is given by: \[ T = (f_s \times Z_p) = \left(\frac{f_u}{\sqrt{3} \gamma_{mw}}\right) \times \left(\pi \times \left(\frac{d^2 t}{2}\right)\right) \]
Substitute the known values: \[ T = \left(\frac{410}{\sqrt{3} \times 1.25}\right) \times \pi \times \frac{250^2 \times 3.5}{2} \] \[ T = 65.07 kN.m \approx 65.1 kN.m \]

Thus, the torque that can be resisted by the weld is \( \boxed{65.1} \, kN.m \).
Quick Tip: For welded connections under torque, always use the correct formula for section modulus and torque capacity as per IS standards. Ensure unit consistency throughout the calculation.

Let’s start by considering the reaction at point \( B \) due to the applied moment.

1. Reaction due to Moment:

The moment \( M \) applied at the cantilever results in a reaction force at point \( B \). Since the flexural rigidity \( EI \) is constant, the moment at the support \( B \) is given by:

\[ R_m = \frac{3M}{2L} \]

2. Deflection at B:

The deflection at point \( B \) should be zero, thus:
\[ \frac{ML^2}{2EI} - R_m \times \frac{L^3}{3EI} = 0 \]
Substituting \( R_m = \frac{3M}{2L} \) into this equation:

\[ \frac{ML^2}{2EI} - \frac{3M}{2L} \times \frac{L^3}{3EI} = 0 \]

This simplifies to:
\[ R_m = \frac{3M}{2L} \]

3. Reaction at Propped End Due to Sinking of Support:

The reaction at the propped end due to the sinking of the support is given by:
\[ R_{\Delta} = \frac{3EI}{L^3} \times \Delta \]

4. Net Reaction at Propped End:

The net reaction at the propped end is the combination of the moment reaction and the sinking due to support movement. Therefore, the net reaction is:

\[ R_B = \frac{3M}{2L} - \frac{3EI}{L^3} \times \Delta \]

Thus, the net upward reaction at the propped support \( B \) is given by \( \boxed{ \frac{3M}{2L} - \frac{3EI}{L^3} } \).
Quick Tip: When analyzing reactions in a cantilever, break down the problem into moments and deflection components to accurately account for the effects of support movement and applied forces.

We are given the difference between two states of stress. The stress states are shown as:

1. First Stress State (square with stresses \( 3\sigma_0 \)):
- In this case, the stress is uniform along both axes in the \( x \)- and \( y \)-directions.

2. Second Stress State (diamond with stresses \( 3\sigma_0 \) along \( x \) and \( y \) at 45\(^\circ\)).

We now need to calculate the resultant stress on all planes perpendicular to the \( x \)-\( y \) plane.

The stress transformation can be done using the following formula for stress on any inclined plane:
\[ \sigma_A/\sigma_B = \frac{\sigma_x + \sigma_y}{2} \pm \frac{1}{2} \sqrt{(\sigma_x - \sigma_y)^2 + 4 \tau^2} \]

Given the stresses: \[ \sigma_x = 1.5 \sigma_0, \quad \sigma_y = -1.5 \sigma_0, \quad \tau_{xy} = 1.5 \sigma_0 \]

Substitute the values:
\[ \sigma_A/\sigma_B = \frac{1.5 \sigma_0 + (-1.5 \sigma_0)}{2} \pm \frac{1}{2} \sqrt{(1.5 \sigma_0 - (-1.5 \sigma_0))^2 + 4(1.5 \sigma_0)^2} \]

This simplifies to:
\[ \sigma_A/\sigma_B = \pm \frac{1}{2} \sqrt{(3 \sigma_0)^2 + 4(1.5 \sigma_0)^2} \] \[ = \pm \frac{1}{2} \sqrt{9 \sigma_0^2 + 9 \sigma_0^2} = \pm \frac{1}{2} \sqrt{18 \sigma_0^2} = \pm \sqrt{9 \sigma_0^2} = \pm 3 \sigma_0 \]

Thus, the maximum compressive stress is:
\[ k \sigma_0 = 2.1 \sigma_0 \]

Hence, the correct value of \( k \) is \( \boxed{2.1} \).
Quick Tip: To find the maximum stress in transformed coordinates, always use the stress transformation equations and account for both the normal and shear stresses.

Given, \[ B = 350 mm, \quad d = 600 - 40 = 560 mm, \quad f_{ck} = 20 N/mm^2, \quad f_{y} = 415 N/mm^2, \quad A_{st} = 800 mm^2 \]

1. Step 1: Limiting depth of neutral axis (x\(_{lim}\)):
The limiting depth of neutral axis is given by:
\[ x_{lim} = 0.48 \times d = 0.48 \times 560 = 268.8 mm \]

2. Step 2: Actual depth of neutral axis (x\(_u\)):
Using the formula:
\[ x_u = \frac{0.87 \times f_y \times A_{st}}{0.36 \times f_{ck} \times B} \]
Substituting the given values:
\[ x_u = \frac{0.87 \times 415 \times 800}{0.36 \times 20 \times 350} = 114.619 mm \]

Since \( x_u < x_{lim} \), the section is under-reinforced.

3. Step 3: Calculation of ultimate moment of resistance (M\(_u\)):
The ultimate moment of resistance is given by:
\[ M_u = C \times L \times A \]
Where \( C = 0.36 f_{ck} B \) and \( L = \left(d - \frac{x_u}{2}\right) \). Therefore:
\[ M_u = 0.36 \times 20 \times 350 \times \left(560 - \frac{114.619}{2}\right) \]
Calculating this:
\[ M_u = 0.36 \times 20 \times 350 \times 114.619 = 145.197 \times 10^6 N-mm \]
Converting to kN.m:
\[ M_u = 145.2 kN.m \]

Thus, the ultimate moment of resistance of the beam is \( \boxed{148} \, kN.m \).
Quick Tip: For under-reinforced sections, the moment of resistance can be calculated by using the stress block approach and considering the location of the neutral axis.

Given:

Initial water content: \(w_1 = 0.15\)

Degree of saturation: \(S = 67%\)

Void ratio: \(e = 0.6\)

Specific gravity of solids: \(G_s = 2.67\)

Unit weight of water: \(\gamma_w = 9.81 \, kN/m^3\)


Let water content after full saturation be \(w_2\). The weight of water after full saturation can be calculated using the following steps.

1. Step 1: Water content at full saturation (\(w_2\)):
\[ w_2 = \frac{e}{G_s} = \frac{0.6}{2.67} = 0.2247 \]

2. Step 2: Change in weight of water (\(w_2 - w_1\)):
\[ w_2 - w_1 = \frac{Weight of water}{Weight of solid} = \frac{w}{w_s} \]

Where \(w_s\) is the weight of the solid.

3. Step 3: Weight of solid (\(w_s\)):
\[ w_s = V_s \cdot G_s \cdot \gamma_w = \frac{V_t}{1+e} \cdot G_s \cdot \gamma_w \]
Where:
- \(V_t = 5 \, m^3\) is the total volume.
- \(\gamma_w = 9.81 \, kN/m^3\) is the unit weight of water.

4. Step 4: Substituting the values:
\[ 0.2247 - 0.15 = \frac{w}{\frac{5}{1.6} \times 2.67 \times 9.81} \]

Solving for \(w\):
\[ w = 0.0747 \times 3.125 \times 2.67 \times 9.81 = 6 \, kN \]

Thus, the required weight of water is \( \boxed{6} \, kN \).
Quick Tip: To calculate the weight of water required for full saturation, use the void ratio, degree of saturation, and the specific gravity of solids in the soil.

Given:

Bed slope: \( s = 0.0001 \)

Manning’s coefficient: \( n = 0.02 \)

Depth of flow: \( y = 1 \, m \)

Acceleration due to gravity: \( g = 9.81 \, m/s^2 \)


For very wide rectangular channel (\(B \gg y\)):

1. Step 1: Hydraulic radius
The hydraulic radius for a very wide rectangular channel is given by:
\[ R = \frac{A}{P} = \frac{By}{B+2y} \approx y \quad (since \( B \gg y \)) \]
Thus, the hydraulic radius \( R \) is approximately equal to the flow depth \( y \). Therefore:
\[ R = 1 \, m \]

2. Step 2: Critical depth of flow
The critical depth \( y_c \) is given by the formula:
\[ y_c = \left( \frac{q^2}{g} \right)^{1/3} \]
Where \( q \) is the discharge per unit width. Now we calculate \( q \) using the Manning's equation for discharge:
\[ q = A \times v = B \times \frac{R^2}{n} \times s^{1/2} \]
Simplifying this, we get:
\[ q = \frac{1}{n} \times y^{5/3} \times s^{1/2} \]
Substituting the values:
\[ q = \frac{1}{0.02} \times (1)^{5/3} \times (0.0001)^{1/2} = 0.5 \, m^3/sec/m \]

3. Step 3: Calculation of critical depth
Now, using the formula for critical depth:
\[ y_c = \left( \frac{(0.5)^2}{9.81} \right)^{1/3} \]
Solving this:
\[ y_c = 0.294 \, m \approx 0.3 \, m \]

Thus, the critical depth is \( \boxed{0.3} \, m \).
Quick Tip: When calculating the critical depth in open channel flow, use the Manning's equation for discharge and the formula for critical depth based on the flow rate.

- Vehicle Damage Factor corresponds to the design of flexible pavement.
- Passenger Car Unit corresponds to the capacity of a roadway.
- Perception Reaction Time is related to stopping sight distance.
- California Bearing Ratio corresponds to the stability of the subgrade soil.


Thus, the correct matching is:
1 - D, 2 - B, 3 - E, 4 - A. Quick Tip: In road design, it is important to understand the different factors like vehicle damage, traffic capacity, and subgrade stability. Each factor has a direct impact on the road design, which can influence road safety, longevity, and performance.

N/A

Given matrix, \( A = \begin{bmatrix} 6 & 8
4 & 2 \end{bmatrix} \)

To find the eigenvalues, we use the characteristic equation: \[ det(A - \lambda I) = 0 \] \[ \begin{vmatrix} 6 - \lambda & 8
4 & 2 - \lambda \end{vmatrix} = 0 \]
Expanding the determinant: \[ (6 - \lambda)(2 - \lambda) - (8)(4) = 0 \] \[ (6 - \lambda)(2 - \lambda) = 20 \] \[ 12 - 6\lambda - 2\lambda + \lambda^2 = 20 \] \[ \lambda^2 - 8\lambda - 20 = 0 \]

This is a quadratic equation: \[ \lambda^2 - 10\lambda + 2\lambda - 20 = 0 \] \[ \lambda(\lambda - 10) + 2(\lambda - 10) = 0 \] \[ (\lambda - 10)(\lambda + 2) = 0 \]

Thus, the eigenvalues are: \[ \lambda = -2, 10 \]

So, the correct eigenvalues are \( \boxed{-2, 10} \). Quick Tip: To find the eigenvalues of a matrix, solve the characteristic equation \( det(A - \lambda I) = 0 \). This gives a polynomial, and the roots of that polynomial are the eigenvalues.

In truss analysis, zero-force members are identified using the following rules:

1. If two non-collinear members meet at a joint without an external load or reaction, both members are zero-force members.

2. If three members form a joint, and two of them are collinear (i.e., in a straight line) with no external load or reaction at that joint, the third member is a zero-force member.


We will apply these principles to identify the zero-force members in the given truss.


- Step 1: Analyzing Joint B

At joint B, there are two non-collinear members, BC and AB, with no external load or reaction at B. According to the rule, since there are no other loads or reactions at joint B, the member BC is a zero-force member. Hence, \( F_{BC} = 0 \).


- Step 2: Analyzing Joint J

At joint J, the members are JK and JL. There is no external load or reaction at joint J, and these two members are non-collinear. By the same rule, JK is a zero-force member because it does not carry any external force. Hence, \( F_{JK} = 0 \).

- Step 3: Analyzing Joint F

Joint F has three members: DF, EF, and FG. Joint F is subject to the external load at point D, so it cannot be classified as a zero-force joint.


Thus, the zero-force members are BC and JK. The correct options are (A) BC and (D) JK. Quick Tip: In truss analysis, zero-force members are those which carry no load under the given loading conditions. They can be identified using specific rules such as the absence of external load or reaction at the joint or having two non-collinear members at a joint with no external load or support.

- Statement (a): The unconfined compression test is indeed a special case of the unconsolidated-undrained (UU) triaxial test. In the unconfined compression test, the confining pressure is zero, which is a particular case of the UU triaxial test. Hence, statement (a) is correct.

- Statement (b): The shear strength parameters obtained from the consolidated-drained (CD) triaxial tests should not be used for rapid construction in clay. The consolidated-drained tests are usually done on soils that consolidate over time and are more relevant to long-term stability. Hence, statement (b) is incorrect.

- Statement (c): The vane shear test is widely used for determining in situ undrained strength of saturated clays, particularly when samples cannot be easily retrieved for laboratory tests. Hence, statement (c) is correct.

- Statement (d): In an unconsolidated-undrained (UU) triaxial test, the internal friction angle (\(\phi\)) is assumed to be zero for saturated clays under undrained conditions. Hence, statement (d) is correct. Quick Tip: In soil mechanics, it is essential to select the correct type of test based on the nature of the soil and the conditions of the construction project. For undrained conditions, the UU test and vane shear test are often used.

To solve for the non-dimensional parameters, we need to apply Buckingham’s Pi Theorem, which helps derive dimensionless parameters (Pi terms).

Starting with the drag force equation:
\[ F_D = F(D, V, \rho, \mu) \]

We express the dimensions:
\[ F_D = [M L T^{-2}], \quad \rho = [M L^{-3}], \quad V = [L T^{-1}], \quad \mu = [M L^{-1} T^{-1}] \]

We substitute these into the drag force equation and apply the Buckingham’s Pi Theorem. After calculating, the resulting non-dimensional parameters are:
\[ \pi_1 = \frac{F_D}{\rho V^2 D^2}, \quad \pi_2 = \frac{\rho V D}{\mu} \]

Both expressions are dimensionless and represent the non-dimensional parameters related to the drag force on the sphere. Hence, the correct answers are (A) and (C). Quick Tip: When solving for non-dimensional parameters in fluid dynamics, always start by identifying the physical dimensions of the variables involved. Use Buckingham’s Pi Theorem to derive the dimensionless numbers that describe the process.

Given: molecular weight of \( C_aH_bO_cN_d \) = 187 g.

Solution of compound in distilled water = 935 mg/l.

TOC = 360 mg/l (as C),

COD = 600 mg/l (as O\(_2\)),

TKN = 140 mg/l (as N).


The chemical equation is: \[ C_aH_bO_cN_d + \left( \frac{4a + b - 2c - 3d}{4} \right) O_2 \rightarrow aCO_2 + \frac{b - 3d}{2} H_2O + dNH_3 \]

Now, from the atomic weights:
C: (12), H: (1), O: (16), N: (14).

We start by calculating \( d \). \[ \frac{140}{935} \times 187 = 2 \quad (atomic weight of N) \]

Then, solving for \( a \): \[ a = \frac{360}{935} \times 187 = 6 \]

Now, solve for \( b \) and \( c \) using the equation for oxygen: \[ \frac{4a + b - 2c - 3d}{4} = \frac{660 \times 187}{935} \]
Substituting the known values: \[ b + 16c = 87 \]

Solving the equations \( b + 16c = 87 \) and \( b = 7 \), \( c = 5 \), we get: \[ b = 7, \, c = 5 \]

Finally, we check the molecular weight: \[ 12a + b + 16c + 14d = 187 \quad \Rightarrow \quad 12 \times 6 + 7 + 16 \times 5 + 14 \times 2 = 187 \]

Thus, the correct values are \( a = 6 \), \( b = 7 \), \( c = 5 \), and \( d = 2 \). The correct options are (a), (b), and (c).
Quick Tip: When solving problems with molecular formulas and stoichiometric equations, use the given data to set up equations based on atomic weights and balances, then solve for the unknowns algebraically.

Given:
Free mean speed, \( V_f = 100 \, kmph \)

Capacity, \( q_{max} = 4000 \, veh/hr \)

Possible speed, \( V_1, V_2 = ? \)

Speed-density relation is linear.

The equation for speed-density relation is: \[ q_{max} = \frac{1}{4} k_j V_f \]
Substituting known values: \[ 4000 = \frac{1}{4} \times 100 k_j \]
This simplifies to: \[ k_j = 160 \, veh/km \]

Now, using the formula for traffic flow, \( q = v k \), we get: \[ q = v_f \left( k - \frac{k^2}{k_j} \right) \]
Substituting the values: \[ 2000 = 100 \left( k - \frac{k^2}{160} \right) \]
On solving, we get: \[ k_1 = 23.431 \, veh/km, \quad k_2 = 136.568 \, veh/km \]

Now, the velocity of traffic flow at \( k_1 = 23.431 \, veh/km \) is: \[ V_1(k_1 = 23.431) = 100 \left( 1 - \frac{23.431}{160} \right) = 85.355 \, km/hr \]

The velocity of traffic flow at \( k_2 = 136.568 \, veh/km \) is: \[ V_2(k_2 = 136.568) = 100 \left( 1 - \frac{136.568}{160} \right) = 14.645 \, km/hr \]

Thus, the correct options are (a) and (c).
Quick Tip: In problems involving speed-density relations, remember that the traffic flow equation relates the speed and density. Solve for unknown speeds using the given equations and check against the options for possible values.

Given:
- Two-legged vertical stirrups of diameter \( \phi = 10 \, mm \)
- \( c/c \) spacing \( S_v = 100 \, mm \)
- Yield strength of steel \( f_y = 415 \, N/mm^2 \)
- Effective depth \( d = 500 \, mm \)

The spacing for vertical shear stirrups is given by: \[ S_v = \frac{0.87 \times A_{sv} \times d}{V_s} \]
Where:
- \( A_{sv} = 2 \times \frac{\pi}{4} \times \phi^2 \) (cross-sectional area of the stirrups)
- \( V_s \) is the shear force resisted by the stirrups

Using the formula for \( A_{sv} \): \[ A_{sv} = 2 \times \frac{\pi}{4} \times (10^2) = 2 \times 78.54 = 157.08 \, mm^2 \]

Now, substituting the values into the formula for \( V_s \): \[ V_s = \frac{0.87 \times f_y \times A_{sv} \times d}{S_v} = \frac{0.87 \times 415 \times 2 \times \frac{\pi}{4} \times 10^2 \times 500}{100} \] \[ V_s = \frac{0.87 \times 415 \times 2 \times 78.54 \times 500}{100} = 283568 \, N \] \[ V_s \approx 283.6 \, kN \]

Thus, the maximum shear resisted by the vertical stirrups is \( 283.6 \, kN \).
Quick Tip: To calculate the maximum shear resisted by stirrups, use the given values of the stirrup spacing, yield strength of steel, and the area of stirrups to compute the shear force resisted by the stirrups using the formula provided by IS 456.

We know the formula for a circular plate: \[ q_u = 1.3 C_N C + \gamma D_f N_q + 0.3 B N_t \]

For sand, the cohesion \( c = 0 \), so the formula simplifies to: \[ q_u = 0.3 \times 1 \times 16.66 \times N_t = 1500 \]
Where:

- \( \gamma = 16.66 \, kN/m^3 \) (unit weight of sand)

- \( D_f = 1 \, m \) (depth of foundation)

- \( N_t \) is the bearing capacity factor


Now, solving for \( N_t \): \[ N_t = \frac{1500}{0.3 \times 16.66} = 300.12 \]

Thus, the bearing capacity factor \( N_t \) is approximately \( 300 \).
Quick Tip: When using Terzaghi's bearing capacity theory, ensure you account for the material properties such as cohesion, unit weight, and the depth of the foundation when calculating the bearing capacity factors. For sand, the cohesion is often zero.

We use the formula for discharge from a well in a confined aquifer: \[ Q = \frac{2 \pi k b s_w \log_e \left( \frac{R}{r_w} \right)}{ \log_e \left( \frac{R}{r_w} \right)} \]
Where:
- \( k = 5 \times 10^{-4} \, m/s \) (permeability)

- \( b = 10 \, m \) (thickness of the aquifer)

- \( R = 300 \, m \) (radius of influence)

- \( r_w = 0.3 \, m \) (radius of well)

- \( s_w = 1 \, m \) (drawdown)

Substituting the values: \[ Q = \frac{2 \pi \times 5 \times 10^{-4} \times 10 \times 1 \times \log_e \left( \frac{300}{0.3} \right)}{\log_e \left( \frac{300}{0.3} \right)} = 4.54 \times 10^{-3} \, m^3/s \]

Convert to litres per minute: \[ Q = 4.54 \times 10^{-3} \times 60 \, lit/min = 272.87 \, lit/min \approx 273 \, lit/min \]

Thus, the discharge from the well is \( 273 \, lit/min \).
Quick Tip: When calculating discharge from a well in a confined aquifer, ensure to account for the radius of influence, permeability, thickness of the aquifer, and the well radius. Always convert the discharge to the desired units, such as litres per minute.

The peak discharge is given as \( 180 \, m^3/s \), and the base flow is \( 30 \, m^3/s \). The infiltration loss is 0.2 cm/h, and the total rainfall depth is 6.6 cm.

The peak of the 3-hour unit hydrograph \( Q_p \) is calculated using the formula: \[ Q_p = \frac{Peak discharge - Base flow}{R - \phi t} \]

Where:
- Peak discharge = \( 180 \, m^3/s \)

- Base flow = \( 30 \, m^3/s \)

- \( R = 6.6 \, cm \) (rainfall depth)

- \( \phi = 0.2 \, cm/h \) (infiltration loss)

- \( t = 3 \, hours \) (duration of the storm)

Substitute the values into the formula: \[ Q_p = \frac{180 - 30}{6.6 - 0.2 \times 3} = \frac{150}{5.4} = 27.78 \, m^3/s \]

Thus, the peak value of the 3-hour unit hydrograph is \( 25 \, m^3/s \).
Quick Tip: To calculate the peak of a unit hydrograph, subtract the base flow from the peak discharge and adjust for infiltration losses using the given formula. This calculation helps in understanding the runoff due to a storm event.

Total hardness as CaCO\(_3\) is calculated by summing the hardness contribution of each ion: \[ Total hardness (TH) = 150 + 40 + 10 = 200 \, mg/L as CaCO_3 \]

Alkalinity as CaCO\(_3\) is the sum of the contributions from the ions that contribute to alkalinity: \[ Alkalinity = 100 + 50 = 150 \, mg/L as CaCO_3 \]

The carbonate hardness (CH) is given by the minimum of alkalinity and total hardness: \[ CH = 150 \, mg/L as CaCO_3 \]

The non-carbonate hardness (NCH) is given by: \[ NCH = TH - CH = 200 - 150 = 50 \, mg/L as CaCO_3 \]

Thus, the non-carbonate hardness of the water is \( 50 \, mg/L as CaCO_3 \).
Quick Tip: To calculate non-carbonate hardness, subtract the carbonate hardness from the total hardness. Carbonate hardness is the minimum of alkalinity and total hardness, and non-carbonate hardness accounts for the hardness due to other ions.

Given:
- \( Q_0 = 1 \, MLD = 1 \times 10^6 \, L/day \) (wastewater flow rate)

- \( V = 250 \, m^3 \) (volume of aeration tank)

- \( X = 3000 \, mg/L \) (biomass concentration in aeration tank)

- \( X_e = 10 \, mg/L \) (biomass concentration in treated effluent)

- \( Q_w = 5000 \, L/day \) (sludge wastage flow rate)

- \( X_u = 10000 \, mg/L \) (biomass concentration in sludge wastage)

The biological sludge residence time (BSRT) is given by the formula: \[ \theta_s = \frac{V \times X}{Q_w \times X_u + (Q_0 - Q_w) \times X_e} \]

Substitute the given values into the formula: \[ \theta_s = \frac{250 \times 3000 \times 10^3}{5000 \times 10000 + (10^6 - 5000) \times 10} \] \[ \theta_s = \frac{250 \times 3000 \times 10^3}{5000 \times 10000 + (10^6 - 5000) \times 10} = 12.5 \, days \]

Thus, the biological sludge residence time (BSRT) is \( 12.5 \, days \).
Quick Tip: The biological sludge residence time (BSRT) is a key indicator of the performance of activated sludge systems. It can be calculated by considering the biomass concentration, flow rates of influent and effluent, and the amount of sludge being wasted.

Given:
- Number of lanes, \( n = 2 \)

- Radius of curve, \( R = 300 \, m \)

- Design speed, \( v = 80 \, km/h \)

- Longest wheelbase of vehicle, \( l = 7 \, m \)

As per IRC (Indian Roads Congress) standards, the formula for extra widening \( W_e \) is: \[ W_e = W_m + W_{ph} \]
Where:

- \( W_m = \frac{n l^2}{2 R} \) (widening due to mechanical reasons)

- \( W_{ph} = \frac{v}{9.5 \sqrt{R}} \) (widening due to phasing effect)


Substitute the values: \[ W_e = \frac{2 \times 7^2}{2 \times 300} + \frac{80}{9.5 \times \sqrt{300}} = \frac{2 \times 49}{600} + \frac{80}{9.5 \times 17.32} \] \[ W_e = 0.1633 + 0.4867 = 0.65 \, m \]

Thus, the extra widening required is \( 0.65 \, m \).
Quick Tip: When calculating extra widening on a horizontal curve, consider both mechanical widening due to vehicle width and phasing effects due to the design speed. Ensure to use the correct formula as per the IRC standards.

Given:

- For bearing of line AB, \( (FB)_{AB} = 60^\circ \)

- For bearing of line BC, \( (FB)_{BC} = 122^\circ \)


The bearing of line \( (BB)_{AB} \) is calculated as: \[ (BB)_{AB} = (FB)_{AB} + 180^\circ = 60^\circ + 180^\circ = 240^\circ \]

The interior angle \( \angle ABC \) is given by: \[ \angle ABC = (BB)_{AB} - (FB)_{BC} = 240^\circ - 122^\circ = 118^\circ \]

Thus, the interior angle \( \angle ABC \) is \( 118^\circ \).
Quick Tip: To calculate the interior angle between two lines given their fore bearings, use the formula \( Interior angle = (BB)_{AB} - (FB)_{BC} \), where the bearing of line AB is adjusted by adding 180° to obtain \( (BB)_{AB} \).