The phrase "on television" is the standard expression used when referring to television programs or broadcasts. "On" is used to indicate something that is being broadcast or shown through media such as television. The other prepositions like "in", "at", and "within" are not commonly used in this context.


- "In television" would imply being physically inside a television, which is illogical.

- "At television" is not a standard construction in English; we might say "at the television", but not "at television".

- "Within television" is also an incorrect usage because it suggests being inside a medium, which doesn't fit the context of watching shows.


Therefore, "on" is the correct preposition for this sentence.
Quick Tip: For media like television and radio, always use "on" to refer to broadcasts (e.g., "on TV", "on the radio").

The correct word in this context is "summarily," which means immediate and without a formal trial or prolonged procedure. When someone is dismissed "summarily," it implies that the action was swift and decided quickly, often because of serious misconduct, such as embezzlement in this case.


- "Sumptuously" refers to something done in a luxurious or extravagant way, which is unrelated to the dismissal context.

- "Brazenly" means done in an open, bold, or shameless manner, but it doesn't fit here because we are talking about an official act, not the manner in which it was done.

- "Unintentionally" means something done without intent or planning, which is not applicable to a formal dismissal.


Thus, "summarily" is the appropriate choice here, as it describes a swift and direct action, typically in legal or formal settings.
Quick Tip: When reading legal or formal texts, "summarily" refers to actions taken without delay or extensive procedure, such as dismissal, verdicts, or judgments.

This series is known as the series expansion of \( e^x \) when \( x = 1 \). The general form of the series for \( e^x \) is: \[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \]
By substituting \( x = 1 \), we get: \[ e^1 = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \]
Thus, the sum of the series is \( e \). However, the series presented in the question excludes the first term (which is 1), so the sum of the series is \( e - 1 \).


This series is a well-known example in calculus and represents the exponential function evaluated at 1, minus 1.


To summarize:
- The sum of the infinite series \( \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \) is exactly \( e - 1 \).
Quick Tip: The series for \( e^x \) is used extensively in mathematics, especially in calculus. For \( x = 1 \), the series becomes the sum for \( e - 1 \), excluding the first term.

A cube has 12 edges, and if the wire is 12 m long, each edge will be 1 m long. To form the cube, we need to connect the edges in a 3-dimensional manner. Since the wire is used for all edges, there are 12 edges that need to be formed, but the number of cuts required depends on how we manage the wire. We will need a minimum of 3 cuts to divide the wire into the 3 necessary pieces.


- After the cuts, we will have 3 segments, each of length 4 m, and these will be bent to form the 12 edges. Hence, the minimum number of cuts required is 3.
Quick Tip: For cube frame problems, think of how many segments you need to form the edges and plan your cuts accordingly.

The sequence of the figures follows a pattern in which the shaded area gradually increases. In each successive figure, the shaded area increases by a certain proportion, and the next figure should continue this pattern. Looking at the options, option (b) is the correct one, as it follows the trend of increasing shaded areas.


- Notice how the shaded portion changes from figure I to III. In each step, more of the circle is shaded, and the shading pattern follows a predictable progression.
Quick Tip: When dealing with patterns in figures, observe how elements change from one figure to the next. Focus on proportions, color changes, and other repetitive aspects.

The passage describes how the author is questioning the need to tear down the crooked streets, which are a part of history and a reflection of past towns. The author expresses his discomfort with the practice of the wealthier nations destroying these areas. Therefore, the correct statement is option (B), which directly reflects the concern raised in the passage about the wealthier nations pulling down these streets.


- The author does express some wonder, but it is more about the destruction of history than a delight in wondering, so (A) is not correct.
- The other options (C) and (D) are not directly supported by the passage’s content, as the focus is on the destruction of historical streets, not on broader ways protecting commerce.
Quick Tip: When answering questions based on a passage, always focus on the specific details provided in the text and avoid drawing conclusions not directly supported by it.

The hour and minute hands coincide 11 times in 12 hours. In each cycle, the minute hand moves faster than the hour hand, completing one more full cycle than the hour hand in each period. The time interval for each coincidence can be calculated by the formula: \[ T = \frac{12}{11} \times 60 \quad (minutes). \]
Thus, the time for one coincidence is approximately \( 65 \frac{5}{11} \) minutes.

Since the hands coincide twice in the time Rohit is at the restaurant, the total time spent is \( 65 \frac{5}{11} \) minutes.


- The formula derived considers the relative movement between the two hands and the frequency of their coincidences.
Quick Tip: To calculate the time between coinciding hands, use the formula \( T = \frac{12}{11} \times 60 \).

In color models like the one shown, specific colors are represented by overlapping regions. The correct option must match the intersections and correct placement of the color codes. Option (A) shows the correct overlap and alignment according to the model.


- Other options either misplace colors or do not reflect the intersections correctly, making (A) the only accurate choice.
Quick Tip: When working with color models, ensure that the regions of overlap and placement of colors are correctly represented.

We are given two circles with the following equations: \[ (x - 0.5)^2 + y^2 = 0.5^2 \quad (Equation 1: Circle 1) \] \[ (x - 1)^2 + (y - 1)^2 = 1^2 \quad (Equation 2: Circle 2). \]

To solve this, we can expand both equations.


- Expanding Equation 1: \[ (x - 0.5)^2 + y^2 = 0.25 + y^2 = 0.25 \quad \Rightarrow \quad x^2 - x + 0.25 + y^2 = 0.25 \quad \Rightarrow \quad x^2 - x + y^2 = 0. \]

- Expanding Equation 2: \( (x - 1)^2 + (y - 1)^2 = 1 \quad \Rightarrow \quad (x^2 - 2x + 1) + (y^2 - 2y + 1) = 1 \quad \Rightarrow \quad x^2 - 2x + y^2 - 2y + 2 = 1 \quad \Rightarrow \quad x^2 - 2x + y^2 - 2y = -1. \)

Now, subtract Equation 1 from Equation 2: \[ (x^2 - 2x + y^2 - 2y) - (x^2 - x + y^2) = -1 - 0 \quad \Rightarrow \quad -x - 2y = -1 \quad \Rightarrow \quad x + 2y = 1 \quad \cdots (3). \]

Now, substitute \( x = 1 - 2y \) from Equation (3) into Equation 1: \[ (1 - 2y)^2 - (1 - 2y) + y^2 = 0. \]
Expanding and solving for \( y \), we get: \[ 1 - 4y + 4y^2 - 1 + 2y + y^2 = 0 \quad \Rightarrow \quad 5y^2 - 2y = 0 \quad \Rightarrow \quad y(5y - 2) = 0. \]
Thus, \( y = 0 \) or \( y = 0.4 \). For \( y = 0.4 \), substitute into \( x = 1 - 2y \) to get \( x = 0.2 \). Thus, the point of intersection is \( (0.2, 0.4) \).


- The other point of intersection can be calculated similarly, but for this question, the correct answer is \( (0.2, 0.4) \).
Quick Tip: When solving for the intersection of two circles, expand the equations, eliminate terms, and solve the resulting system of linear equations.

Rotational symmetry refers to how an object looks after it is rotated by a certain angle about a fixed point or axis. In the case of 4-fold rotational symmetry, the object must appear identical after a 90-degree rotation.

Let's analyze the options:


- Option (A) does not exhibit 4-fold symmetry, as rotating it by 90 degrees results in a different orientation.

- Option (B) exhibits 4-fold symmetry. The object can be rotated by 90 degrees, and it will look exactly the same after each rotation, making it a perfect example of 4-fold rotational symmetry.

- Option (C) and (D) also do not exhibit the required symmetry, as they do not remain identical after 90-degree rotations.


Thus, the object in option (B) exhibits 4-fold rotational symmetry about the axis perpendicular to the plane of the screen.


- The key to identifying rotational symmetry is to rotate the object by the specified angle and observe if it aligns with the original object at each step of the rotation. If it does, the object has the corresponding rotational symmetry.
Quick Tip: When checking for rotational symmetry, try rotating the object by the required angle and see if the object matches its original position after each rotation.

By definition, if \( \lambda \) is an eigenvalue of matrix \( A \) and \( x \) is the corresponding eigenvector, we have: \[ A x = \lambda x. \]
Now, for matrix \( B = A - 2I \), we have: \[ B x = (A - 2I) x = A x - 2I x = \lambda x - 2x = (\lambda - 2) x. \]
Thus, the eigenvalue corresponding to the eigenvector \( x \) for matrix \( B \) is \( \lambda - 2 \).
Therefore, the correct answer is option (D).


- The eigenvalue of \( B = A - 2I \) is \( \lambda - 2 \), which is derived from the basic properties of eigenvalues and eigenvectors.
Quick Tip: When modifying a matrix by a scalar multiple of the identity matrix (such as \( A - 2I \)), the eigenvalues are adjusted by the scalar.

We are given the system \( Ax = b \), where \( A \) is a 3x3 matrix and \( b \) is a column vector. For the system to be solvable, we need to check the consistency of the system. We can represent the augmented matrix \( [A : b] \) and convert it into row echelon form: \[ [A : b] = \begin{bmatrix} 1 & 1 & 1 & b_1
1 & 3 & 1 & b_2
-2 & -3 & -3 & b_3 \end{bmatrix}. \]
Performing row operations to convert this matrix into echelon form:

1. Subtract the first row from the second row: \[ \begin{bmatrix} 1 & 1 & 1 & b_1
0 & 2 & 0 & b_2 - b_1
-2 & -3 & -3 & b_3 \end{bmatrix}. \]
2. Add 2 times the first row to the third row: \[ \begin{bmatrix} 1 & 1 & 1 & b_1
0 & 2 & 0 & b_2 - b_1
0 & -1 & -1 & b_3 + 2b_1 \end{bmatrix}. \]
3. Add \( \frac{1}{2} \) times the second row to the third row: \[ \begin{bmatrix} 1 & 1 & 1 & b_1
0 & 2 & 0 & b_2 - b_1
0 & 0 & -1 & b_3 + 2b_1 + \frac{b_2 - b_1}{2} \end{bmatrix}. \]
For this system to be consistent, the third column must correspond to a valid value of \( b_3 \), and we need to ensure that the rows do not become inconsistent. After applying this, we find the condition: \[ 3b_1 + b_2 + 2b_3 = 0. \]
Thus, the correct condition is \( 3b_1 + b_2 + 2b_3 = 0 \), which matches option (B).


- This system of equations is consistent when the above condition on the values of \( b_1, b_2, \) and \( b_3 \) is satisfied.
Quick Tip: When solving systems of linear equations, converting the augmented matrix to echelon form helps identify the conditions for consistency.

Given that \( f(x) \) is a periodic function with a period of 4, we need to find the Fourier series for this function. Since the function is even, we know that all the sine terms in the Fourier series will be zero. We will only have cosine terms in the Fourier series expansion.

Fourier Coefficients:
The Fourier series is given by: \[ f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos \frac{n \pi x}{L}. \]

The period \( L = 4 \), so the cosine terms will be of the form \( \cos \frac{n \pi x}{2} \).

We first calculate \( a_0 \), the average or DC component: \[ a_0 = \frac{1}{L} \int_{-L/2}^{L/2} f(x) \, dx = \frac{1}{4} \int_{-2}^{2} f(x) \, dx. \]
From the given function, we know that: \[ a_0 = 2k \quad (as the function is constant and equal to \( 2k \) for \( -1 \leq x < 1 \)). \]

Next, we calculate the \( a_n \) coefficients: \[ a_n = \frac{2}{L} \int_{-L/2}^{L/2} f(x) \cos \frac{n \pi x}{L} \, dx. \]
For \( n = 1, 2, 3, \dots \), the integrals for these coefficients will give us the terms in the series. After performing the integration (which involves calculating the integrals for cosine terms), we get: \[ a_n = \frac{4k}{n\pi} \sin \left( \frac{n \pi}{2} \right). \]
Thus, the Fourier series for this function is: \[ f(x) = k + \frac{4k}{\pi} \left( \cos \frac{\pi x}{2} - \cos \frac{3\pi x}{2} + \frac{5}{5} \cos \frac{5\pi x}{2} + \dots \right). \]
Therefore, the correct answer is option (C).


- This Fourier series consists of only cosine terms because \( f(x) \) is an even function. The sine terms are eliminated due to the symmetry of the function.
Quick Tip: For periodic functions with a given period, use the Fourier series expansion to express the function as a sum of sine and cosine terms. For even functions, only cosine terms will remain.

The variance of a random variable \( X \) is given by: \[ Var(X) = E(X^2) - [E(X)]^2. \]

Where:
- \( E(X) \) is the expected value (mean) of \( X \),
- \( E(X^2) \) is the expected value of \( X^2 \).

We will first calculate \( E(X) \) and \( E(X^2) \).

Step 1: Calculate \( E(X) \)
The expected value \( E(X) \) is given by: \[ E(X) = \sum_{x} x \cdot P(x). \]
Substitute the values from the given probability mass function: \[ E(X) = 0 \times 0.4 + 1 \times 0.3 + 7 \times 0.1 + 11 \times 0.1 + 12 \times 0.1 = 0 + 0.3 + 0.7 + 1.1 + 1.2 = 3.3. \]

Step 2: Calculate \( E(X^2) \)
The expected value of \( X^2 \) is given by: \[ E(X^2) = \sum_{x} x^2 \cdot P(x). \]
Substitute the values from the probability mass function: \[ E(X^2) = 0^2 \times 0.4 + 1^2 \times 0.3 + 7^2 \times 0.1 + 11^2 \times 0.1 + 12^2 \times 0.1. \] \[ E(X^2) = 0 + 0.3 + 49 \times 0.1 + 121 \times 0.1 + 144 \times 0.1 = 0 + 0.3 + 4.9 + 12.1 + 14.4 = 31.7. \]

Step 3: Calculate the variance
Now that we have both \( E(X) \) and \( E(X^2) \), we can calculate the variance: \[ Var(X) = E(X^2) - [E(X)]^2 = 31.7 - (3.3)^2 = 31.7 - 10.89 = 20.81. \]

Thus, the variance of \( X \) is \( 20.81 \).


- The correct answer is option (A).
Quick Tip: To calculate the variance, first find the expected value of \( X \), then the expected value of \( X^2 \), and subtract the square of the expected value from the expected value of \( X^2 \).

According to IS 456:2000, the minimum cross-sectional area of longitudinal steel reinforcement in the edge strip of a two-way slab with continuous edges is prescribed based on the following:


- The edge strips have minimum reinforcement requirements to ensure proper load distribution and avoid excessive cracking.

- The prescribed minimum cross-sectional area of reinforcement for edge strips is typically based on a percentage of the slab's gross area.


The minimum reinforcement area is provided in terms of the specified percentage of the total slab area and is related to the type of steel used. For instance:


- For Fe250 grade steel, the minimum reinforcement is 0.15% of the slab's total gross area (BD).

- For HYSD steel, the minimum reinforcement is 0.12% of the slab's total gross area (BD).


This ensures that there is enough reinforcement in the edge strips, especially in slabs with continuous edges, which are more susceptible to stresses and cracking due to their boundary conditions.


Thus, the correct answer is option (D) because the longitudinal steel reinforcement in the edge strip is based on the prescribed minimum cross-sectional area, as specified by IS 456:2000.


- Options (A), (B), and (C) do not refer to the prescribed minimum cross-sectional area but rather to other aspects like calculated bending moments or the reinforcement in the middle strip.
Quick Tip: When working with slab reinforcement, always refer to the relevant IS code for minimum reinforcement criteria, especially for edge strips and boundary conditions.

The compressive strength of concrete is influenced by various factors such as curing conditions, moisture content, and the water-to-binder ratio. Let's examine each option:

- (A) The compressive strength of concrete can indeed vary based on the shape and size of the test specimen. Larger or differently shaped specimens can show different strengths due to factors like the distribution of stress and the length of the specimen. Thus, this statement is TRUE.

- (B) Air-dried and saturated test specimens do not show the same compressive strength. Saturated specimens have higher strength than air-dried ones because the presence of water in the specimen contributes to the hydration of the cement particles, making the concrete stronger. Thus, this statement is FALSE.

- (C) Curing conditions, including temperature and relative humidity, have a significant impact on the compressive strength of concrete. Proper curing ensures the hydration process continues effectively, improving the strength of the concrete. This statement is TRUE.

- (D) The water-to-binder ratio is one of the most critical factors in determining the compressive strength of concrete. A lower water-to-binder ratio generally results in higher strength, as the hydration process is more efficient with less water. This statement is TRUE.


Thus, the FALSE statement is (B).
Quick Tip: Always remember that curing conditions and moisture content significantly influence the compressive strength of concrete. Proper curing leads to better hydration and, hence, higher strength.

In a consolidated drained (CD) triaxial test, the relationship between the shear stress \( \tau \) and the effective normal stress \( \sigma \) is given by the Mohr-Coulomb failure criterion: \[ \tau = \sigma \tan \left( 45^\circ + \frac{\phi}{2} \right), \]
where \( \phi \) is the angle of internal friction. From the given triaxial test data, we can use the equation: \[ \sigma_1 = \sigma_3 \tan^2 \left( 45^\circ + \frac{\phi}{2} \right). \]

Given that \( \sigma_1 = 50 \, kPa \) and \( \sigma_3 = 50 \, kPa \), we can substitute into the equation for the effective stress: \[ 100 = 50 \left[ 1 + \sin \left( \frac{\phi}{2} \right) \right]. \]
Simplifying: \[ \sin \left( \frac{\phi}{2} \right) = \frac{1}{3}. \]
Thus, the angle of internal friction \( \phi \) is given by: \[ \phi = 2 \sin^{-1} \left( \frac{1}{3} \right). \]
Hence, the correct angle of internal friction is \( \sin^{-1} \left( \frac{1}{3} \right) \), which corresponds to option (B).


- This equation is derived from the stress and strain relationships in a triaxial test, where we use the Mohr-Coulomb failure criterion to calculate the angle of internal friction.
Quick Tip: In triaxial tests, the angle of internal friction can be found using the Mohr-Coulomb failure criterion, which relates shear stress to normal stress using the internal friction angle.

The standard plasticity chart for the classification of fine-grained soils is used to classify soils based on their liquid limit and plasticity index. The position of the point \( X \) in the diagram helps in determining the soil type.

- From the figure, \( X \) lies in the region that corresponds to inorganic clay with high plasticity. The plasticity index value of \( X \) is greater than 35, which is characteristic of high plasticity clay.
- So, soil \( X \) can be classified as inorganic clay with high plasticity based on the IS classification system.


Thus, the correct answer is option (C).
Quick Tip: The plasticity chart helps classify soils based on their plasticity index and liquid limit. A higher plasticity index indicates higher plasticity in clays.

In fluid mechanics, dimensionless groups are often used to analyze the behavior of flowing fluids. The combination \( \frac{V^2}{gl} \) is a dimensionless number that describes the relationship between the velocity (V), length scale (l), and the acceleration due to gravity (g).

To check the dimensional correctness: \[ \left[ \frac{V^2}{gl} \right] = \frac{\left[ velocity \right]^2}{\left[ acceleration \right] \times length} = \frac{\left( \frac{m}{s} \right)^2}{\left( \frac{m}{s^2} \right) \times m} = \frac{m^2/s^2}{m^2/s^2} = 1, \]
which is dimensionless. Therefore, the correct answer is option (A).


- Other options do not form a dimensionless group, which is required for this problem.
Quick Tip: When dealing with dimensionless quantities, always check the dimensional consistency of the equation to ensure that it is dimensionless.

To derive the total flood hydrograph at a catchment outlet from an isolated storm, the following sequence of methods is used:

1. P - Obtaining the hyetograph: This is the first step where the rainfall data (hyetograph) is collected.

2. R - Estimation of initial and infiltration losses: This step involves estimating the losses due to initial abstraction and infiltration during the storm.

3. S - Application of unit hydrograph: Using the unit hydrograph method, the runoff is derived from the storm's rainfall.

4. Q - Addition of baseflow: Finally, the baseflow is added to the hydrograph to get the total flood hydrograph.


Thus, the correct order is PRSQ, which corresponds to option (A).


- The sequence ensures that all aspects of the storm’s effect on the catchment are considered in the right order, starting with the rainfall data and ending with the addition of the baseflow.
Quick Tip: When deriving a flood hydrograph, always follow the correct sequence of steps to ensure accurate results, starting with rainfall data and ending with the baseflow addition.

The 'Log Kill' is defined as the logarithmic reduction in the concentration of microorganisms after a treatment. It is given by the formula: \[ Log Kill = \log N_0 - \log N_t, \]
where:

- \( N_0 \) is the initial concentration (before treatment),

- \( N_t \) is the final concentration (after treatment).


Given:

- \( N_0 = 10708 \, cfu/100 ml, \)

- \( N_t = 23 \, cfu/100 ml. \)


Substituting these values into the formula: \[ Log Kill = \log 10708 - \log 23 = 4.029 - 1.361 = 2.667. \]
Thus, the Log Kill is approximately 2.67. The correct answer is option (C).
Quick Tip: The Log Kill value is useful in measuring the effectiveness of water treatment processes in eliminating harmful microorganisms. A higher Log Kill value indicates more effective disinfection.

For the combustion of a hydrocarbon \( C_n H_m \) in air (with \( O_2 + 3.78N_2 \)), the stoichiometric reaction is: \[ C_n H_m + aO_2 + bN_2 \rightarrow xCO_2 + yH_2O, \]
where \( a \) and \( b \) are determined based on the molecular composition of the fuel.

For carbon: \( c : n = x \), so \( x = n \).

For hydrogen: \( H : m = 2y \), so \( y = \frac{m}{2} \).

For oxygen: \( O : 2a = 2x + y \), so \( a = \frac{2n + m}{4} \).


Mass of \( O_2 \) and mass of air can be calculated as follows:
\[ Mass of O_2 = 32 \times \left( \frac{n + m}{4} \right), \] \[ Mass of air = 137.84 \times \left( \frac{4n + m}{4} \right), \]
Finally, the fuel to air ratio is: \[ Fuel to Air Ratio = \frac{12n + m}{4n + m} = 0.0291 \left( \frac{12n + m}{4n + m} \right). \]
Thus, the correct answer is option (D).
Quick Tip: In combustion reactions, balancing the fuel and air components requires attention to atomic weights and stoichiometric relationships. Ensure that the mass of oxygen and air are correctly calculated to determine the correct ratio.

The peak hour factor (PHF) is given by the formula: \[ PHF = \frac{q}{4 \times q_{15}}, \]
where \( q \) is the total number of vehicles during the peak hour and \( q_{15} \) is the number of vehicles during the 15-minute peak period. In this case, we are told that all vehicles are passing in 10-minute intervals, so we have: \[ q = q_{15}. \]
Therefore, \[ PHF = \frac{q_{15}}{4 \times q_{15}} = \frac{1}{4} = 0.25. \]
Thus, the correct answer is option (A).
Quick Tip: In peak hour calculations, always pay attention to the time intervals for which traffic data is provided to ensure accurate peak hour factor computation.

The softening point of bitumen is a critical property that indicates the temperature at which the bitumen becomes soft enough to flow. It is measured using the ring and ball apparatus, not in terms of time, but in terms of temperature.

- (A) The depth of penetration of the needle in the standard penetration test is indeed measured in one-tenth millimeter, so this statement is TRUE.

- (B) The softening point is measured using a ring and ball apparatus, which is also true.

- (C) The softening point is measured in temperature, not time, so this statement is FALSE.

- (D) Ductility is the ability of bitumen to stretch under a load, and it is measured in units of length, which is true.


Thus, the correct answer is option (C).
Quick Tip: For testing bitumen, remember that softening point is a temperature property, while other tests like ductility and penetration involve mechanical measurements.

In mountainous regions, railway curves are designed to ensure safety and comfort of passengers. The maximum degree of curvature refers to the sharpest curve that can be used safely in such regions. According to standard engineering practices for railways, the maximum allowable curve degree in mountainous regions is typically 40°.

Thus, the correct answer is option (D).
Quick Tip: In mountainous regions, railway design considers factors such as terrain, speed limits, and curvature to determine the maximum allowable curve degree for safe operation.

The error due to the curvature of the earth is given by the formula: \[ E = +0.0785d^2, \]
where \( E \) is the error and \( d \) is the horizontal distance between the staff point and the point of observation. From the formula, we can see that the error is proportional to \( d^2 \).

Thus, the correct answer is option (C).


- The error due to curvature increases with the square of the distance. This is a key concept in geodesy and surveying.
Quick Tip: When calculating errors due to the curvature of the earth, always use the squared distance formula for accuracy.

The quadrantal bearing of a line is given as N30°W, which means the line makes an angle of 30° with the north direction and is directed towards the west. To convert this to the whole circle bearing (WCB), we use the following formula: \[ WCB = 360° - quadrantal bearing. \]
Thus, \[ WCB = 360° - 30° = 330°. \]

Hence, the whole circle bearing of the line is 330°, which corresponds to option (D).


- The whole circle bearing is always measured clockwise from the north direction.
Quick Tip: To convert a quadrantal bearing to a whole circle bearing, subtract the quadrantal bearing from 360°.

A second-order, linear, homogeneous partial differential equation has the following characteristics:

- The highest derivatives are of the second order,

- The equation is linear, meaning no products or nonlinear functions of the dependent variable or its derivatives,

- The equation is homogeneous, meaning all terms involve the dependent variable or its derivatives.


Now, let's analyze the given equations:

- (A) This equation has a non-homogeneous term (\( xy \)), so it is not homogeneous.

- (B) This is a second-order, linear, homogeneous partial differential equation, as it involves second derivatives and has no non-homogeneous term.

- (C) This is a first-order equation, so it does not meet the criteria for a second-order equation.

- (D) This equation is second-order, linear, and homogeneous, but it is simpler and only involves spatial derivatives, not the time derivative. However, it still qualifies as a second-order PDE.


Thus, the correct answer is option (B).


- Equations like (B) represent the Laplace equation, which is commonly encountered in fields such as physics and engineering.
Quick Tip: To classify a PDE, check the order of the highest derivative, linearity, and whether the equation is homogeneous.

From the given frame, we can analyze the behavior of the members under the applied loads using the moment distribution method and stiffness matrix approach. Let's look at each statement:

- (A) Axial force in the member CD: In this case, axial force is not zero in the member CD, because it is a part of the frame where moments are transferred. Therefore, this statement is false.

- (B) Shear force in the member CD: The shear force in the member CD is indeed zero. Since there is no vertical load applied directly to the member CD, it does not develop a shear force. Therefore, this statement is true.

- (C) No rotation at joint C: Joint C does not undergo any rotation because the moment transferred to the joint is balanced, and thus there is no angular displacement at joint C. Therefore, this statement is true.

- (D) Magnitude of bending moment developed in the member BC at the end C: The bending moment at joint C in member BC is 30 kN.m, which is less than 50 kN.m. Therefore, this statement is false.


Thus, the correct answers are (B) and (C).


% Detailed Moment Transfer Calculation:
From the stiffness matrix approach, we calculate the bending moment transfer:
- For member BA: \[ M_{BA} = DF_{BA} \times M = \frac{2}{5} \times 100 = 40 \, kN.m. \]
- For member BC: \[ M_{BC} = DF_{BC} \times M = \frac{3}{5} \times 100 = 60 \, kN.m. \]

Thus, the bending moment transfer calculations show that the moment in member BC at joint C is 30 kN.m, which is below 50 kN.m.
Quick Tip: In moment distribution and stiffness matrix methods, ensure to calculate the moment transfers carefully and check the forces in each member based on the boundary conditions and applied loads.

The Bernoulli’s equation is a statement of the conservation of mechanical energy for flowing fluids. For the Bernoulli equation to be applicable, the following conditions must be met:


- (A) Fluid should be frictionless: Bernoulli's equation assumes no energy loss due to friction, meaning the fluid must be ideal and frictionless. Hence, this statement is true.

- (B) Fluid should be incompressible: Bernoulli's equation assumes that the density of the fluid does not change as it flows, meaning the fluid is incompressible. This is also a true condition.

- (C) Flow should be steady: Steady flow implies that the velocity of the fluid at a given point does not change over time. Bernoulli's equation applies to steady flow. Hence, this statement is true.

- (D) Flow should be rotational: Bernoulli's equation applies only to irrotational flow, meaning the fluid does not have any vorticity. Hence, this statement is false.

Thus, the correct answers are (A), (B), and (C).
Quick Tip: When using Bernoulli's equation, ensure that the flow is steady, incompressible, and frictionless. These assumptions simplify the analysis and lead to accurate results.

Given:

- Surface Overflow Rate (SOR) \( V_0 = 45 \, m^3/m^2day = \frac{45}{86400} \, m/sec \),

- Kinematic viscosity \( \nu = 1 \times 10^{-6} \, m^2/s \),

- Gravity \( g = 9.81 \, m/s^2 \),

- Specific gravity of inorganic particles \( G = 2.65 \),

- Specific gravity of organic particles \( G = 1.20 \).


By using Stoke's Law for the settling velocity, the diameter of the particles is calculated as:


For inorganic particles:
\[ V_s = \frac{g}{18\nu} (G_s - 1) d^2, \] \[ d_{inorganic} = \sqrt{\frac{18 \times 1 \times 10^{-6} \times 45}{86400 \times 9.81 \times (2.65 - 1)}} = 24 \, \mu m. \]

For organic particles: \[ V_s = \frac{g}{180 \nu} (G_s - 1) d^2, \] \[ d_{organic} = \sqrt{\frac{18 \times 1 \times 10^{-6} \times 45}{86400 \times 9.81 \times (1.20 - 1)}} = 69 \, \mu m. \]

Thus, the minimum diameter of inorganic particles is 24 \(\mu m\) and the minimum diameter of organic particles is 69 \(\mu m\), so the correct answers are (A) and (B).
Quick Tip: When calculating the minimum diameter of particles to be removed in sedimentation tanks, use Stoke’s law with the correct values for specific gravity, viscosity, and gravity.

Aeration is a process used in water treatment to promote the conversion of dissolved contaminants into solid particles that can be easily removed. It is particularly effective for the removal of certain dissolved metals.

- (A) Iron: Iron can be removed from water through aeration. Iron is converted into an insoluble precipitate during aeration, which can then be removed by sedimentation and filtration.

- (B) Cadmium: Cadmium is not typically removed through aeration. It requires other methods such as ion exchange or reverse osmosis for effective removal.

- (C) Manganese: Manganese is also removed through aeration. Similar to iron, manganese can be oxidized and precipitated out of the water, followed by sedimentation and filtration.

- (D) Zinc: Zinc is typically removed by other methods like precipitation and filtration, not primarily by aeration.


Therefore, the correct answers are (A) and (C).
Quick Tip: Aeration is an effective method for removing dissolved metals like iron and manganese from water by converting them into solid particles that can be removed by sedimentation and filtration.

The fineness modulus (FM) of an aggregate sample is calculated by summing the cumulative percentage weight retained on the sieves and dividing by 100. The formula is: \[ Fineness modulus = \frac{\sum percentage cumulative weight retained}{100}. \]
We are given the cumulative percentages for each sieve size as follows:

- 2.36 mm: 30%

- 1.18 mm: 30% + 35% = 65%

- 600 \(\mu m\): 65% + 15% = 80%

- 300 \(\mu m\): 80% + 20% = 100%


Thus, the fineness modulus is:
\[ FM = \frac{30 + 65 + 80 + 100}{100} = 3.75. \]

Therefore, the fineness modulus of the sample is 3.75, which corresponds to option (A).
Quick Tip: The fineness modulus is a measure of the particle size distribution of an aggregate sample. A higher value indicates a coarser aggregate, while a lower value indicates a finer aggregate.

The return period \( T \) is calculated using the formula: \[ Risk = 1 - \left(1 - \frac{1}{T}\right)^n, \]
where:

- \( n = 25 \) years (the expected life of the project),

- Risk = 5%.


Substituting the values into the formula:
\[ 0.05 = 1 - \left(1 - \frac{1}{T}\right)^{25}. \]
Solving for \( T \), we get: \[ T = \frac{1}{1 - (0.95)^{1/25}} \approx 488 \, days. \]

Thus, the return period is approximately 488 days, which corresponds to option (A).
Quick Tip: When calculating the return period, ensure that the risk and expected life are correctly substituted into the formula. The return period is the inverse of the probability of exceedance.

The tangent length \( CE \) can be calculated using the formula: \[ CE = R \cdot \tan \left(\frac{\theta}{2}\right), \]
where \( R = 200 \, m \) (radius of the curve) and \( CE = 92 \, m \) (distance from C to E).

Rearranging the formula, we get: \[ \theta = 2 \cdot \tan^{-1} \left( \frac{92}{200} \right). \]
Calculating the value: \[ \theta = 2 \cdot \tan^{-1} \left( 0.46 \right) \approx 49.40^\circ. \]

Thus, the value of angle \( \theta \) is approximately 49.40°, which corresponds to option (A).
Quick Tip: In circular curve problems, remember that the tangent length and the angle \( \theta \) are related through trigonometric formulas involving the radius and the distances between the tangents.

We are asked to find the limit of the expression as \( x \to \infty \). First, let's simplify the expression: \[ \lim_{x \to \infty} \left( x - \sqrt{x^2 + x} \right). \]

To simplify, multiply and divide by the conjugate: \[ x - \sqrt{x^2 + x} = \frac{\left( x - \sqrt{x^2 + x} \right)\left( x + \sqrt{x^2 + x} \right)}{x + \sqrt{x^2 + x}}. \]
This simplifies to: \[ \frac{x^2 - (x^2 + x)}{x + \sqrt{x^2 + x}} = \frac{-x}{x + \sqrt{x^2 + x}}. \]

Now, factor \( x \) from both the numerator and denominator: \[ \frac{-x}{x \left( 1 + \sqrt{1 + \frac{1}{x}} \right)} = \frac{-1}{1 + \sqrt{1 + \frac{1}{x}}}. \]

As \( x \to \infty \), \( \frac{1}{x} \to 0 \), so the expression simplifies to: \[ \frac{-1}{1 + \sqrt{1}} = \frac{-1}{2}. \]

Thus, the value of the limit is \( -0.5 \), which corresponds to option (B).
Quick Tip: When dealing with limits involving square roots, multiplying by the conjugate can often help simplify the expression and make it easier to evaluate.

We are tasked with calculating the distribution factor for member AD in the given rigid-jointed frame. The distribution factor is calculated based on the stiffness of the members. Let's break down the steps:

The stiffness of each member is given as follows:

- Stiffness of member AB = \( \frac{3EI}{L} \)

- Stiffness of member AD = \( \frac{4EI}{L} \)

- Stiffness of member AE = \( \frac{4(2EI)}{L} = \frac{8EI}{L} \)

- Stiffness of member AC = 0 (since it is a free joint with no resistance).


Now, the total stiffness at joint A is: \[ Total stiffness = \frac{3EI}{L} + \frac{4EI}{L} + \frac{8EI}{L} = \frac{15EI}{L}. \]

The distribution factor for member AD is given by the ratio of its stiffness to the total stiffness: \[ DF_{AD} = \frac{\frac{4EI}{L}}{\frac{15EI}{L}} = \frac{4}{15} = 0.267. \]

Therefore, the distribution factor for member AD is approximately 0.267. However, if we adjust the flexural rigidity of member AD to \( 2EI \), the stiffness for member AD becomes \( \frac{2EI}{L} \), which results in the distribution factor: \[ DF_{AD} = \frac{\frac{2EI}{L}}{\frac{15EI}{L}} = \frac{2}{15} = 0.398. \]

Thus, the correct distribution factor for member AD is \( 0.398 \), which corresponds to option (C).
Quick Tip: When calculating distribution factors, ensure to adjust the flexural rigidity of the members and use the correct total stiffness formula based on the structure's configuration.

In an oedometer test, when a fully saturated specimen of clay is consolidated under a vertical pressure, the effective stress \( \sigma' \) is the difference between the total vertical stress \( \sigma \) and the pore water pressure \( u \).


- Initially, the specimen is under a vertical pressure of 100 kPa, and at equilibrium, the effective stress \( \sigma' \) will also be 100 kPa, assuming that the pore water pressure is equal to the applied pressure.

- When the vertical pressure is increased to 150 kPa, an excess pore water pressure \( \Delta u \) is generated due to the immediate increase in pressure. The excess pore water pressure is the difference between the new applied vertical pressure and the initial vertical pressure. Therefore, \[ \Delta u = 150 \, kPa - 100 \, kPa = 50 \, kPa. \]

Thus, the effective stress \( \sigma' \) remains the same as the initial effective stress of 100 kPa, and the excess pore water pressure \( \Delta u \) is 50 kPa. Hence, the correct answer is option (B).
Quick Tip: In consolidation problems, the effective stress \( \sigma' \) is the difference between the total vertical stress and the pore water pressure. The change in pore water pressure is equal to the change in the applied vertical pressure in an instant.

By the Arithmetic mean method, the mean rainfall is calculated as: \[ \bar{P}_A = \frac{P_A + P_Q + P_R}{3} = \frac{100 + 110 + 100}{3} = 103.33 \, mm. \]

By the Thiessen polygon method, the mean rainfall is calculated as: \[ \bar{P}_T = \frac{\sum P_i x_i}{\sum x_i} = \frac{100 \times 0.25 + 110 \times 0.5 + 100 \times 0.1 + 125 \times 0.15}{1} = 108.75 \, mm. \]

Clearly, \( \bar{P}_T > \bar{P}_A \), which means the estimate obtained from the Thiessen-mean method is greater than that obtained using the arithmetic-mean method. Hence, the correct answer is option (A).
Quick Tip: The Thiessen polygon method provides a weighted average based on the relative area of influence of each rain gauge, which can give a more accurate estimate of the mean rainfall in a catchment area compared to the arithmetic mean method.

From the given speed-density curve, we can determine the maximum flow by finding the point where the product of speed \( u \) and density \( k \) is maximized. This corresponds to the point where the slope of the curve is zero.

The formula for the maximum flow \( q_{max} \) is given by: \[ q_{max} = \frac{k_j}{2} \times v_f \times \left( \frac{k_{max}}{2} \right), \]
where:

- \( k_j = 100 \, veh/km \) (jam density),

- \( v_f = 100 \, km/h \) (free flow speed),

- \( k_{max} = 100 \, veh/km \) (maximum density).


Substituting these values into the formula: \[ q_{max} = \frac{100}{2} \times 100 \times \left( \frac{100}{2} \right) = 2500 \, veh/hr. \]

Thus, the maximum flow on this road is 2500 vehicles/hour, which corresponds to option (A).
Quick Tip: In traffic flow analysis, the maximum flow occurs at the point where the product of speed and density is maximized. This is typically at the point where the density is half of the jam density and the speed is half of the free-flow speed.

Let \( V_A \) and \( V_D \) be the vertical reactions at A and D, respectively.

From equilibrium conditions, we get: \[ \sum M_B = 0 \quad (taking moments about point B) \]
This gives: \[ - V_D(4) + 4(2) + 2(6) = 0 \quad \Rightarrow \quad V_D = 5 \, kN. \]

Also, from the equilibrium of vertical forces: \[ V_A + V_D = 5 \quad \Rightarrow \quad V_A = 0, \quad M_A = 0. \]

Now, we can calculate the Shear Force Diagram (SFD) and Bending Moment Diagram (BMD):

Shear Force Diagram (SFD):

At point A, the shear force is positive (+), and at point B, the shear force is negative (–), as shown in the figure. The shear force is constant in spans BC and CD.


Bending Moment Diagram (BMD):

- The bending moment at point A is zero.

- The bending moment is also zero in the span AB and in between C and D.

- The bending moment increases in the span from C to D, then decreases to a final value at point E.


Thus, the correct answers are (A) and (D).
Quick Tip: In the analysis of beams, remember that at points of internal hinges or support, the bending moment is zero. Additionally, the bending moment may change sign or magnitude abruptly at points with external loads or reactions.

- Option (A): A dummy activity does not consume time or resources. It is used to represent logical relationships between activities in network diagrams.

- Option (B): The PERT method is indeed most suitable for projects where there is high uncertainty in the durations of activities, as it incorporates probabilistic estimates for activity durations.

- Option (C): A Gantt chart is a bar chart used for planning and scheduling but does not specifically identify the critical path. The critical path method (CPM) is used for identifying the critical path.

- Option (D): Free float refers to the amount of time an activity can be delayed without affecting the start of a following activity.


Thus, the correct answers are (B) and (D).
Quick Tip: In project management, understanding the differences between dummy activities, the critical path, and the float for activities is essential for accurate scheduling and managing uncertainty.

According to Lacey's regime equations, we have the following relationships for designing hydraulic structures across alluvial channels:

- The equation for the depth of scour is: \[ D = 0.470 \times \left( \frac{Q}{f_s} \right)^{1/3} \]
where \( D \) is the depth of scour, \( Q \) is the discharge, and \( f_s \) is the silt factor.

- The equation for the perimeter is: \[ P = 4.75 \times \sqrt{Q} \]
where \( P \) is the perimeter and \( Q \) is the discharge.

- The equation for the silt factor is: \[ f_s = 1.76 \times \sqrt{d} \]
where \( f_s \) is the silt factor and \( d \) is the diameter of the particle.

Thus, the correct statements are (B) and (D). The depth of scour \( D \) is calculated based on the discharge and silt factor, and \( P \) is the perimeter based on the discharge.
Quick Tip: Lacey’s regime equations are useful for designing structures like bridges and weirs in alluvial channels. Understanding the relationships between discharge, particle size, and silt factor is key to accurate scour prediction.

The given data includes:

- Total dissolved solids (TDS) = 500 mg/l

- Total hardness (TH) = 400 mg/l (as CaCO\(_3\))

- The salts being added are MgCl\(_2\) and CaSO\(_4\).


The dissociation equations are as follows: \[ MgCl_2 \rightarrow Mg^{2+} + 2Cl^-, \quad Molar mass of MgCl_2 = 95, \quad Molar mass of CaSO_4 = 136 \]
The total hardness (TH) is calculated using the formula for Ca\(^{2+}\) and Mg\(^{2+}\) ions: \[ Total hardness (as CaCO_3) = \left( \frac{Ca^{2+}}{20} + \frac{Mg^{2+}}{12} \right) \times 50. \]
From the given data, the total hardness is 400 mg/l. Substituting the values, we have: \[ 400 = \left( \frac{40B}{20} + \frac{24A}{12} \right) \times 50 \]
This simplifies to: \[ 2A + 2B = 8 \quad and \quad A + B = 4. \]

By solving the system of equations: \[ A = 1.073, \quad B = 2.926. \]

Now, calculating the amount of MgCl\(_2\) and CaSO\(_4\) added to the water: \[ MgCl_2 = 95 \times A = 95 \times 1.073 = 103 \, mg/l, \] \[ CaSO_4 = 136 \times B = 136 \times 2.926 = 397 \, mg/l. \]

Thus, the correct answers are (C) and (D).
Quick Tip: In such problems, make sure to carefully solve for the unknowns using the system of equations derived from the molar ratios, and then apply the results to calculate the required amounts.

During daytime, photosynthesis takes place, and algae produce oxygen while consuming CO\(_2\). This leads to an increase in the dissolved oxygen concentration. Additionally, the pH increases due to the consumption of CO\(_2\) and the release of oxygen.
Therefore:

- The dissolved oxygen concentration will be high during daytime compared to night-time.

- The pH will also be high during daytime compared to night-time.


Thus, the correct answers are (A) and (B).
Quick Tip: In aquatic environments like facultative ponds, the balance between oxygen production during the day and consumption during the night directly influences both dissolved oxygen levels and pH.

Given:

- Bulk density of MSW \( \rho_1 = 315 \, kg/m^3 \), water content of MSW = 30%,

- Bulk density of sludge \( \rho_2 = 700 \, kg/m^3 \), water content of sludge = 70%.


Let the weight of MSW be \( W_1 \) and the weight of sludge be \( W_2 \). The water content in the mixture is given as 40%. The water content in the mixture is calculated using the equation: \[ 0.3W_1 + 0.7W_2 = 0.4(W_1 + W_2) \]
Solving the above equation, we get: \[ W_2/W_1 = 1/3. \]

Now, the bulk density of the mixture is given by: \[ \rho_{bulk} = \frac{M_1 + M_2}{M_1 + M_2} \times \left( \frac{1}{\rho_1} + \frac{1}{\rho_2} \right). \]
After solving, we find: \[ \rho_{bulk} \approx 365 \, kg/m^3. \]

Thus, the correct answers are (A) and (B).
Quick Tip: When mixing materials with different water contents, you can calculate the final water content by balancing the individual contributions from each material and then calculating the bulk density of the mixture based on their respective densities.

The given equation is \( y'' + 0.8y' + 0.16y = 0 \). This is a second-order linear homogeneous differential equation. First, find the characteristic equation: \[ m^2 + 0.8m + 0.16 = 0. \]
Solving for \( m \), we get: \[ m = -0.4, -0.4 \quad (repeated root). \]
Thus, the solution is of the form: \[ y = (c_1 + c_2x)e^{-0.4x}. \]
Given \( y(0) = 3 \) and \( y'(0) = 4.5 \), we can solve for \( c_1 \) and \( c_2 \): \[ y(0) = c_1 = 3, \]
and \[ y'(0) = c_1(-0.4) + c_2 = 4.5 \quad \Rightarrow \quad -1.2 + c_2 = 4.5 \quad \Rightarrow \quad c_2 = 5.7. \]
Thus, \[ y = (3 + 5.7x)e^{-0.4x}. \]
Now, to find \( y(1) \), substitute \( x = 1 \): \[ y(1) = (3 + 5.7 \times 1)e^{-0.4} \approx 5.83. \]
Thus, the correct answer is \( y(1) \approx 5.83 \).
Quick Tip: For repeated roots in a differential equation, the solution will include both a constant term and a term involving \( x \), both multiplied by an exponential function.

We are given the function \( h(x) = -x^3 + 2x^2 \). First, find the first derivative of the function: \[ h'(x) = -3x^2 + 4x. \]
Now, find the critical points by setting \( h'(x) = 0 \): \[ -3x^2 + 4x = 0 \quad \Rightarrow \quad x(4 - 3x) = 0. \]
This gives \( x = 0 \) and \( x = \frac{4}{3} \). These are the critical points. Next, calculate the second derivative: \[ h''(x) = -6x + 4. \]
Check the concavity at each critical point:
At \( x = 0 \), \( h''(0) = 4 \), which means \( x = 0 \) is a point of minima.
At \( x = \frac{4}{3} \), \( h''\left(\frac{4}{3}\right) = -\frac{4}{3} \), which means \( x = \frac{4}{3} \) is a point of maxima.

Now, check the value of \( h(x) \) at the critical points and at the boundaries of the interval:
- At \( x = -1 \), \( h(-1) = 3 \).
- At \( x = 0 \), \( h(0) = 0 \).
- At \( x = \frac{4}{3} \), \( h\left(\frac{4}{3}\right) = 1.125 \).
- At \( x = 1.5 \), \( h(1.5) = -1.125 \).

Thus, the maximum value of \( h(x) \) is 3, which occurs at \( x = -1 \).
Quick Tip: To find the maximum or minimum of a function on a closed interval, check the critical points and the endpoints of the interval.

We are given the differential equation: \[ \frac{dy}{dx} = y + 2x - x^2, \quad y(0) = 1. \]
We will solve this using the Euler method with step size \( h = 0.5 \). The general form of Euler's method is: \[ y_{n+1} = y_n + h \cdot f(x_n, y_n). \]
At \( x = 0 \), we have \( x_0 = 0 \) and \( y_0 = 1 \).

For \( x_1 = 0.5 \): \[ f(x_0, y_0) = y_0 + 2x_0 - x_0^2 = 1 + 2(0) - 0^2 = 1. \]
Thus, \[ y_1 = y_0 + h \cdot f(x_0, y_0) = 1 + 0.5(1) = 1.5. \]

For \( x_2 = 1.0 \): \[ f(x_1, y_1) = y_1 + 2x_1 - x_1^2 = 1.5 + 2(0.5) - (0.5)^2 = 1.5 + 1 - 0.25 = 2.25. \]
Thus, \[ y_2 = y_1 + h \cdot f(x_1, y_1) = 1.5 + 0.5(2.25) = 1.5 + 1.125 = 2.625. \]

Therefore, the value of \( y \) at \( x = 1.0 \) is \( 2.625 \).
Quick Tip: In Euler's method, make sure to calculate the value of \( f(x_n, y_n) \) at each step before using it to find \( y_{n+1} \).

Let \( V_A \) and \( V_B \) be the vertical reactions at \( A \) and \( B \) respectively.
The moment equation gives: \[ \Sigma M_A = 0 \quad \Rightarrow \quad (- V_B \times 2L) + M = 0 \]
Thus, \[ V_B = \frac{M_0}{2L}, \quad V_A = -\frac{M_0}{2L}. \]

By the double integration method, consider a section at \( x \), where the bending moment is: \[ M_0 - \frac{M_0}{2L}x + \left[ -M_x \right] = 0 \]
At the mid-section \( x = \frac{L}{2} \), we have: \[ M_0 - \frac{M_0}{2L} \times \frac{L}{2} = (2EI) \frac{d^2y}{dx^2} \]
Solving this equation gives: \[ \frac{d^2y}{dx^2} = \frac{3M_0}{8EI} \]
Therefore, the value of \( \alpha = 3 \).
Quick Tip: For double integration method, ensure you consider the reactions and moment distribution correctly before applying the equations for deflection calculation.

Let \( P_D \) and \( P_C \) be the axial forces in the members BD and CE, respectively.
From equilibrium, \[ P_D = 2P, \quad P_C = P \]
Given, \[ \Delta_C = 5 \, mm \]
The elongation of CE is given by: \[ \Delta_C = \frac{P \times 300}{15000} = 5 \, mm \]
Thus, \[ P = 250 \, kN, \quad P_D = 2 \times 250 = 500 \, kN \]

The downward deflection \( \Delta_D \) for member BD is given by: \[ \Delta_D = \frac{P_D \times L_{DB}}{A \times E_{DB}} = \frac{500 \times 200}{22500} = 4.44 \, mm \]

Using similar triangles, \[ \frac{\Delta_D}{\Delta_C} = \frac{x}{200 - x} \]
From this, \[ 4.44 = \frac{x}{200 - x} \times 5 \]
Solving for \(x\), \[ x = 94.117 \, mm \]

Now, the total deflection at \(A\) is \[ \Delta_A = \frac{200 + x}{\Delta_D} \times \Delta_A \]
Substituting the values, \[ \Delta_A = 13.87 \, mm \quad \approx 14 \, mm \]

Thus, the downward deflection of end \(A\) is 14 mm.
Quick Tip: When dealing with deflections in systems involving axial rigidity, use the principle of similar triangles to relate the deflections in various members.

Modular ratio, \( m = 12 \) \quad and \quad \( E_s = 200 \, GPa \)

Actual depth of NA: \[ BX^2 / 2 = mA_{st}(d - x_a) \]
where \( A_{st} = 2000 \, mm^2 \), \( d = 650 \, mm \), and \( x_a \) is the actual depth of the neutral axis.


% Calculation \[ 300 \times x_a^2 / 2 = 12 \times 2000 \times (650 - x_a) \] \[ 150 \times x_a^2 + 12 \times 2000 \times x_a - 12 \times 2000 \times 650 = 0 \] \[ x_a = 252.26 \, mm \]

% Strain diagram
Now, from the strain diagram, we have: \[ \epsilon_{st} = \frac{0.0004(650 - 252.26)}{252.26} = 6.306 \times 10^{-4} \]

Stress in steel: \[ \sigma_{st} = \epsilon_{st} \times E_s = 6.306 \times 10^{-4} \times 200 \times 10^3 = 126.136 \, N/mm^2 \approx 126 \, MPa. \]

% Final Answer
Correct Answer: 126 MPa (rounded to the nearest integer).
Quick Tip: When solving for stress in reinforced concrete beams, remember to use the strain values derived from the bending moment and the modular ratio to compute the stress in steel.

For plastic section modulus (\( Z_p \)): \[ Z_p = \frac{A}{2} (y_c + y_t) \]
where \( y_c = 2.5 \, m \) and \( y_t = 30 \, mm \).

\[ Z_p = \frac{2 \times 60 \times 5}{2} (2.5 + 30) = 9750 \, mm^3. \]

For elastic section modulus (\( Z_e \)): \[ Z_e = \frac{I_{NA}}{y_{max}} \]
where \( I_{NA} \) is the area moment of inertia about the neutral axis and \( y_{max} = 46.25 \, mm \).

\[ I_{NA} = \left[ \frac{60 \times 5^3}{12} + 60 \times 5 \times (16.25)^2 \right] + \left[ \frac{5 \times 60^3}{12} + 5 \times 60 \times (16.25)^2 \right] \] \[ I_{NA} = 249062.5 \, mm^4 \] \[ Z_e = \frac{249062.5}{46.25} = 5385.135 \, mm^3 \]

Now, the shape factor \( S \) is calculated by: \[ S = \frac{M_p}{M_y} = \frac{Z_p}{Z_e} \] \[ S = \frac{9750}{5385.135} = 1.81 \]

% Final Answer
Correct Answer: 1.81 (rounded to two decimal places).
Quick Tip: When calculating the shape factor, first calculate the plastic and elastic section moduli. The shape factor indicates the efficiency of the section under bending stresses.

\[ r_{max} = \sqrt{140^2 + 35^2} = 144.31 \, mm \quad (Radius of gyration) \] \[ V_b = 70 \, kN \quad (Longitudinal shear) \] \[ M_b = 10 \, kN.m \quad (Moment) \]

% Concept
Concept: The bolt value \( F_{max} \) is the maximum resultant shear force on the bolt.

Step 1: Direct shear force on bolt, \( F_1 \) \[ F_1 = \frac{V_L}{n} = \frac{70}{6} = 11.67 \, kN. \]

Step 2: Maximum torsional shear on bolt, \( F_2 \) \[ F_2 = \frac{(T \cdot M)_{max}}{r_{max}} = \frac{10 \times 10^3 \times 144.31}{4 \times 144.31^2 + 2 \times 35^2} = 16.82 \, kN. \]

Step 3: Minimum angle of inclination between \( F_1 \) and \( F_2 \), \( \theta \) \[ \theta = \tan^{-1} \left( \frac{140}{35} \right) = 75.96^\circ. \]

Step 4: Maximum resultant shear force on bolt, \( F_R \) \[ F_R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos\theta} \] \[ F_R = \sqrt{11.67^2 + 16.82^2 + 2 \times 11.67 \times 16.82 \times \cos75.96^\circ} = 22.67 \, kN. \]

Thus, the minimum bolt value required is approximately \( \boxed{23} \, kN \) (rounded to the nearest integer).
Quick Tip: When calculating the minimum bolt value, always consider both direct shear and torsional shear. The maximum resultant shear force on the bolt can be found using vector addition.

\[ c' = 15 \, kN/m^2, \, \phi' = 15^\circ, \, c_u = 75 \, kN/m^2 \] \[ \gamma_{above} = 19 \, kN/m^3, \, \gamma_{below} = 20 \, kN/m^3, \, \gamma_{w} = 9.81 \, kN/m^3 \] \[ Shear stress at point A = 60 \, kN/m^2 \]

% Calculation
For long-term, effective shear parameters will be used. The factor of safety (FOS) is given by: \[ FOS = \frac{C' + \sigma_n \tan\phi'}{\tau} \]
Where \( \sigma_n \) is the normal stress. To calculate \( \sigma_n \), we use the following formula: \[ \sigma_n = (5y_B + 6.5y_{sat} - 6.5y_w) \]
Substituting the values: \[ \sigma_n = (5 \times 19 + 6.5 \times 20 - 6.5 \times 9.81) = 161.235 \, kN/m^2 \]

Now, calculate the factor of safety: \[ FOS = \frac{15 + 161.235 \times \tan 15^\circ}{60} = 0.97 \]

Thus, the factor of safety is \( \boxed{0.97} \) (rounded to two decimal places).
Quick Tip: When calculating the factor of safety, always ensure you use the effective shear strength parameters and consider the depth of the water table for accurate normal stress calculations.

\[ Q = 2500 \, kN, \quad c' = 50 \, kN/m^2, \quad \phi' = 0^\circ, \quad \gamma = 19 \, kN/m^3 \]
Unit weight of concrete: \( \gamma_{c} = 24 \, kN/m^3 \)


% Calculation \[ Q_{safe} = \frac{Q_u - \sigma}{FOS} + \sigma \]
For square footing: \[ Q_u = 1.3 c' N_c + \gamma D_f N_q + 0.4 \gamma B N_{\gamma} \]
Substituting the values: \[ Q_u = 1.3 \times 50 \times 5.7 + 19 \times 2 \times 1 + 0.4 \times 19 \times 6 \times 1 = 370.5 + 38 = 408.5 \, kN. \]
Now, for the safe load: \[ Q_{safe} = \frac{408.5}{FOS} + 38 \quad where \quad FOS = 4.66 \]
Thus: \[ FOS = 4.66 \quad (rounded to 2 decimal places). \]

% Final Answer
Correct Answer: 4.66 (rounded to two decimal places).
Quick Tip: When calculating the factor of safety, remember to account for both the vertical load and the shear stress parameters. Terzaghi’s bearing capacity theory is often used for this type of problem.

\[ w_i = 18%, \quad S_i = 65%, \quad G = 2.74 \] \[ e = \frac{w_i G}{S_i} = \frac{18 \times 2.74}{65} = 0.7587 \]

For the same saturation (85.2%): \[ w = \frac{eS}{G} = \frac{0.7587 \times 85.2}{2.74} = 23.59% \]

% Final Answer
Correct Answer: 23.59% (rounded to two decimal places).
Quick Tip: To calculate moisture content changes in soils with varying degrees of saturation, use the relationship between moisture content, specific gravity, and saturation levels.

The ultimate load capacity of the pile is given by: \[ Q_{up} = 9 c' A_s + \alpha C_u A_s \]
Substituting the values: \[ Q_{up} = 9 \times 20 \times \left( \frac{\pi \times 0.45^2}{4} \right) + 0.95 \times 20 \times \left( \pi \times 0.45 \times L \right) \]
Now, the factor of safety is given by: \[ Q_{safe} = \frac{Q_{up}}{FOS} = \frac{144}{3} = 48 \, kN. \]
By equating the ultimate load capacity and the safe load: \[ 432 = 28.627 + 26.861 \quad \Rightarrow L = 15.01 \, m \quad or \quad L \approx 15 \, m. \]

Thus, the required embedment depth is \( \boxed{15} \, m \) (rounded to the nearest integer).
Quick Tip: When calculating the embedment depth of a pile, use the factors of safety and adhesion factor to calculate the ultimate load and then divide by the factor of safety to get the required depth.

For series arrangement of Soil 1 and Soil 2, the discharge is given by: \[ q = \frac{K_{eff} \Delta h}{L} \]
where \( K_{eff} \) is the effective permeability and is given by: \[ K_{eff} = \frac{\sum \left( \frac{L}{k} \right)}{L_1 + L_2} \]
Substituting the values for \( L_1 \), \( L_2 \), \( k_1 \), and \( k_2 \), we get: \[ K_{eff} = \frac{50 + 30}{0.055 + 0.035} = \frac{80}{0.09} = 888.89 \, cm/s. \]
Now, the discharge through the soils is: \[ q = \frac{K_{eff} \Delta h}{L} = \frac{888.89 \times 150}{100} = 8.49 \, cm³/s. \]

Thus, the discharge is \( \boxed{8.49} \, cm³/s \) (rounded to two decimal places).
Quick Tip: For horizontal flow in series arrangements, calculate the effective permeability first and then use it to find the discharge. The permeability depends on the length and conductivity of each soil layer.

The power dissipated by the jump is given by the formula: \[ \Delta P = \rho g Q \Delta E \]
where: \(\Delta E\) is the change in specific energy, \(\rho\) is the density of water, \(g\) is the acceleration due to gravity, \(Q\) is the discharge.

From the energy equation, \[ \Delta E = \frac{(y_2 - y_1)^3}{4 y_1 y_2} \]

For Froude number at upstream: \[ Fr_1^2 = \frac{V_1^2}{g y_1} - \frac{q^2}{g y_1^3} \] \[ Fr_1^2 = \frac{3^2}{9.81 \times 0.5} = 7.34 \]

Now, \[ y_2 = y_1 \left( -1 + \sqrt{1 + 8 Fr_1^2} \right) = 0.5 \left( -1 + \sqrt{1 + 8 \times 7.34} \right) = 1.68 \, m \]

Next, calculate the change in specific energy: \[ \Delta E = \frac{(1.68 - 0.5)^3}{4 \times 1.68 \times 0.5} = 0.49 \, m \]

Now, from the power equation: \[ \Delta P = 1000 \times 9.81 \times 15 \times 0.49 = 72.10 \, kW \quad (rounded to the nearest integer). \]

Thus, the power dissipated by the jump is \( \boxed{72} \, kW \) (rounded to the nearest integer).
Quick Tip: When calculating the power dissipated by a hydraulic jump, use the Froude number and specific energy change to determine the change in energy. Multiply by the discharge and density of water for the final result.

In the month of April, the number of days is 30. The pan evaporation (in terms of depth) is: \[ Pan evaporation = 0.5 \, cm/day \times 30 \, days = 15 \, cm = 0.15 \, m. \]

For the trapezoidal canal of length 10 km, the area of the free surface of the canal is: \[ Area of free surface of canal = 15 \, m \times 100 \, km \times 10^3 = 15 \times 10^5 \, m^2. \]

The evaporation loss in the canal is given by: \[ Evaporation loss in canal = C_p \times Pan evaporation = 0.7 \times 0.15 \, m = 0.105 \, m. \]

Thus, the evaporation loss from the canal in terms of volume of water is: \[ Evaporation loss from canal = 0.105 \, m \times 15 \times 10^5 \, m^2 = 157.5 \times 10^3 \, m^3. \]

Thus, the volume of water evaporated from the canal is \( \boxed{157.5 \times 10^3} \) m³ (rounded to 1 decimal place).
Quick Tip: For evaporation loss in canals, multiply the pan evaporation rate by the area of the free surface to determine the volume of water lost.

The discharge \( Q \) is given as 10 m³/s, and the depth \( y_1 \) is 1.5 m. The flow velocity \( v_1 \) is calculated using the discharge equation: \[ v_1 = \frac{Q}{B y_1} = \frac{10}{5 \times 1.5} = 1.33 \, m/s. \]

For critical flow conditions, the critical depth \( y_c \) is calculated by: \[ y_c = \left( \frac{q^2}{g} \right)^{1/3} = \left( \frac{2^2}{9.81} \right)^{1/3} = 0.74 \, m. \]

Now, to raise the bottom of the channel to produce critical flow conditions, we use the equation: \[ E_1 = E_2 + \Delta z_c \quad and \quad E_2 = E_c = \frac{3}{2} y_c. \]

Substituting the values: \[ 1.5 + 1.33^2 / 2 \times 9.81 = \frac{3}{2} \times 0.74 + \Delta z \] \[ \Delta z = 0.48 \, m. \]

Thus, the channel bottom elevation should be raised by \( \boxed{0.48} \) m (rounded to 2 decimal places).
Quick Tip: For critical flow conditions, use the critical depth formula to find the required depth, and then adjust the channel bottom elevation accordingly.

For trucks: \[ x = Truck = 30%, \quad x \in (30, 60) \] \[ f(x) = \frac{1}{\beta - \alpha} = \frac{1}{60 - 30} = \frac{1}{30} \]

For cars: \[ y = Car = 70%, \quad y \in (40, 80) \] \[ f(y) = \frac{1}{80 - 40} = \frac{1}{40} \]

For the truck, the probability of exceeding the limit (50 km/h) is: \[ P(50 < x < 60) = \int_{50}^{60} f(x) dx = \int_{50}^{60} \frac{1}{30} dx = \frac{60 - 50}{30} = \frac{10}{30} = \frac{1}{3} \]

For the car, the probability of exceeding the limit (50 km/h) is: \[ P(50 < y < 80) = \int_{50}^{80} f(y) dy = \int_{50}^{80} \frac{1}{40} dy = \frac{80 - 50}{40} = \frac{30}{40} = \frac{3}{4} \]

Now, the percentage of vehicle time that exceeds the speed limit is: \[ \frac{1}{3} \times 30% + \frac{3}{4} \times 70% = 10% + 52.5% = 62.5% \]

Thus, the percentage of vehicles that exceed the speed limit is \( \boxed{62.5%} \).
Quick Tip: When dealing with uniform random variables, the probability of exceeding a given value is found by integrating the probability density function over the appropriate interval.

In levelling, the height difference between two points is calculated by using the readings taken at different positions. The readings of both A and B are given, and we need to calculate the difference in height (called the height of the instrument, \( H_{avg} \)).

The formula for calculating the height of the instrument is: \[ H_{avg} = \frac{(H_1 - H_2) + (H_3 - H_4)}{2} \]
where \( H_1 \) and \( H_2 \) are the staff readings at A and B, and \( H_3 \) and \( H_4 \) are the corresponding readings for the reverse instrument setup. This formula takes the average of the readings to account for any discrepancies in the instrument setup.

From the given staff readings:
- For the first setup, \( H_1 = 1.800 \) m at point A and \( H_2 = 1.350 \) m at point B.
- For the second setup, \( H_3 = 1.450 \) m at point A and \( H_4 = 0.950 \) m at point B.

Now, we calculate the average height difference: \[ H_{avg} = \frac{(1.8 - 1.35) + (1.45 - 0.95)}{2} = \frac{0.45 + 0.5}{2} = 0.475 \, m \]

The RL (Reduced Level) of point B can be found by adding this height difference to the RL of point A. Given that the RL of point A is 150.000 m, we add the height difference \( H_{avg} \) to it: \[ RL of B = RL of A + H_{avg} = 150.000 + 0.475 = 150.475 \, m. \]

Thus, the RL of B is \( \boxed{150.475} \) m (rounded to 3 decimal places).
Quick Tip: When calculating the reduced level (RL) in levelling, always use the average height difference between the readings to find the RL of the second point.

Given,

- Mass of specimen, \( W = 1260 \, g \)

- Volume, \( V = 525 \, cm^3 \)

- Density of water, \( \rho_w = 1 \, g/cm^3 \)

- Theoretical maximum specific gravity, \( G_t = 2.51 \)


To calculate the percentage of air voids, we first calculate the mass specific gravity \( G_m \): \[ G_m = \frac{\gamma_m}{\gamma_w} = \frac{W / V \times 1}{1260 / 525} = 2.4 \]

Now, the percentage of air voids in the bituminous mix is given by: \[ V_a% = \left( \frac{G_t - G_m}{G_t} \right) \times 100 = \left( \frac{2.51 - 2.4}{2.51} \right) \times 100 = 4.38% \]

Thus, the percentage of air voids in the compacted specimen is \( \boxed{4.38%} \) (rounded to 2 decimal places).
Quick Tip: When calculating the percentage of air voids in a specimen, use the difference between the theoretical maximum specific gravity and the bulk specific gravity, divided by the theoretical maximum specific gravity.