This analogy-based question tests the semantic equivalence or synonym relationship between pairs of words.

The word Courage is nearly synonymous with Bravery — both signify strength in facing danger or difficulty.
Similarly, Yearning means a strong desire or longing for something. Among the options, Longing is the word that most closely matches the meaning and intensity of Yearning.

Other choices do not fit the analogy:

\textit{Yelling means shouting, which is unrelated in meaning.
\textit{Yawning refers to the reflex of opening the mouth wide, typically due to tiredness.
\textit{Glaring refers to staring in a fierce or angry manner.


Hence, the best match is Longing. Quick Tip: When solving verbal analogies, first determine the exact relationship between the first pair of words. Is it synonymy, antonymy, function, or degree? Then look for the same relationship in the second pair. In synonym-based analogies, look for the closest meaning, not just a loosely related word.

The sentence refers to an action that was ongoing in the past before another event occurred ("when it suddenly started to rain").

The appropriate tense for an action that was ongoing in the past before another past event is the past perfect continuous tense, which is formed using “had been” + present participle (verb+ing).

Thus, the correct completion is:
\textit{"We had been playing tennis in the lawn when it suddenly started to rain."

Other options are incorrect because:

(A) have been playing – present perfect continuous; used for actions that began in the past and continue to the present.
(C) would have been playing – conditional perfect progressive; used in hypothetical or unreal situations.
(D) could be playing – present/future possibility; does not suit a past context. Quick Tip: When completing a sentence with tenses, pay close attention to time markers such as “when,” “suddenly,” or any indication of sequence. Use \textbf{past perfect continuous (had been + verb-ing) when one past action was happening before another past event occurred.

We need to count how many pixels have intensity values \( U \leq 4 \). Let's go row by row:


Row 1: \( 0, 1, 0, 2 \) → all \( \leq 4 \) → 4 pixels
Row 2: \( 4, 7, 3, 3 \) → \( 4, 3, 3 \) are \( \leq 4 \) → 3 pixels
Row 3: \( 5, 5, 4, 4 \) → \( 4, 4 \) are \( \leq 4 \) → 2 pixels
Row 4: \( 6, 7, 3, 2 \) → \( 3, 2 \) are \( \leq 4 \) → 2 pixels


Total number of pixels with \( U \leq 4 \) = \( 4 + 3 + 2 + 2 = \boxed{11} \) Quick Tip: In matrix-based questions, be systematic—scan each row or column carefully and use tick marks or counts to avoid double-counting or missing entries. Keeping a running total helps maintain accuracy.

From the figure:

The rectangle is labeled as region 1.
The triangle is labeled as region 2.
The ellipse is labeled as region 3.


We analyze the regions:

Region P lies inside all three shapes (rectangle, triangle, ellipse): So, \( P = 1 + 2 + 3 = 6 \)
Region Q lies in triangle and ellipse, but not in the rectangle: So, \( Q = 2 + 3 = 5 \)
Region R lies in triangle and rectangle, but not in the ellipse: So, \( R = 1 + 2 = 3 \)


Hence, the correct combination is: \[ P = 6, \quad Q = 5, \quad R = 3 \] Quick Tip: In Venn diagram-style reasoning questions, always identify the overlap of sets step-by-step. Label the individual areas based on inclusion/exclusion logic and then sum the values for any composite region.

Let the original area of the rectangle be: \[ A = L \times W \]

After increasing the width by 10%, the new width becomes: \[ W' = W + 0.1W = 1.1W \]

So, the new area becomes: \[ A' = L \times 1.1W = 1.1(L \times W) = 1.1A \]

This shows that the area increases by 10%.

Let’s verify the incorrect options:

(A) Perimeter: \( P = 2(L + W) \), new perimeter is \( 2(L + 1.1W) = 2L + 2.2W \), which is not a 10% increase in general.
(B) Diagonal: Original diagonal \( D = \sqrt{L^2 + W^2} \), new diagonal \( D' = \sqrt{L^2 + (1.1W)^2} \), not necessarily a 10% increase.
(D) The rectangle becomes a square only if \( L = W \), but the condition says \( L > W \), so it never becomes a square.


Hence, only the area increases exactly by 10%. Quick Tip: When one dimension of a rectangle changes, check how each geometric property (area, perimeter, diagonal) is affected. For area, it's a direct product, so a percentage change in one side directly scales the area by that percentage.

We will match each statement in Column-I with the most appropriate response from Column-II:


P. “This house is in a mess.” – This is a statement that might prompt someone to offer to clean it up or acknowledge the complaint. The most suitable match is: 3. “No problem, let me clear it up for you.”
Q. “I am not happy with the marks given to me.” – The speaker is dissatisfied and might expect support. Best response is: 4. “Don’t worry, I will take it up with your teacher.”
R. “Politics is a subject I avoid talking about.” – A respectful response would be not to bring it up again. Best match: 1. “Alright, I won’t bring it up during our conversations.”
S. “I don’t know what this word means.” – A helpful answer is one that suggests a way to find the meaning. Best match: 2. “Well, you can easily look it up.”


Hence, the correct matching is: \[ P – 3; Q – 4; R – 1; S – 2 \] Quick Tip: When solving matching questions, read both columns carefully and look for \textbf{logical consistency} and \textbf{natural conversational flow}. Eliminate options with tone or context mismatches to zero in on the correct answer.

We are given that the weight function \( W(a) \), where \( a \) is the age, is:

Monotonically increasing from birth until age 50.
Monotonically decreasing after age 50.


This implies that the function reaches its maximum at age 50. If two brothers have the same weight and the function is symmetric around age 50 due to its monotonic nature, then the same weight value must occur once in the increasing phase and once in the decreasing phase.

Given that \( a_1 < a_2 \), the only way their weights can be equal is when: \[ a_1 < 50 < a_2 \]
This means the younger brother is on the increasing side and the elder brother is on the decreasing side of the weight curve. Since the function is strictly monotonic on both sides, there is only one such pair that satisfies the condition. Quick Tip: When dealing with monotonic functions that peak at a certain point, identical function values on either side of the peak imply symmetry. Use this concept to reason about comparisons between inputs that yield equal outputs.

The figure is a regular dodecagon (12-sided polygon) inscribed in a circle. By drawing lines from the center to each vertex, it is divided into 12 congruent isosceles triangles. These triangles have:

Two sides of length \( r \) (radius)
A base of length \( d \)


Each of these triangles is used to form a square of side \( r \). However, to make one full square of side \( r \), we need a combination of multiple such triangles.

Now:

Total triangles = 12
Suppose each square requires \( x \) triangles.
Number of such squares possible = \( \frac{12}{x} \)


From spatial analysis or approximation using area: \[ Area of square of side r = r^2 \]
Each triangle roughly represents \( \frac{1}{12} \)th of the area of the circle (as there are 12 identical central angles). To fill a square of area \( r^2 \), we approximately need 4 such triangles.

Thus, \[ Number of triangles per square = 4,\quad Total squares = \frac{12}{4} = 3 \]
\[ \boxed{Number of squares = 3,\quad Triangles per square = 4} \] Quick Tip: In problems involving regular polygons inscribed in circles, symmetry plays a crucial role. Dividing the polygon into identical central triangles helps in calculating the number of possible geometric shapes (like squares) that can be formed by combining these units.

We are given: \[ 3^{x^2} = 27 \cdot 9^x \]
First, write all expressions in powers of 3: \[ 27 = 3^3,\quad 9 = 3^2 \Rightarrow 9^x = (3^2)^x = 3^{2x} \]
Thus, RHS becomes: \[ 3^3 \cdot 3^{2x} = 3^{3 + 2x} \]
Equating exponents of LHS and RHS: \[ 3^{x^2} = 3^{3 + 2x} \Rightarrow x^2 = 3 + 2x \]
Solving the quadratic: \[ x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0 \Rightarrow x = 3 or x = -1 \]

Now evaluate the expression: \[ \dfrac{2^{x^2}}{(2^x)^2} = \dfrac{2^{x^2}}{2^{2x}} = 2^{x^2 - 2x} \]

Case 1: \( x = 3 \Rightarrow 2^{3^2 - 2 \cdot 3} = 2^{9 - 6} = 2^3 \)

Case 2: \( x = -1 \Rightarrow 2^{1 - (-2)} = 2^3 \)


In both cases, the result is: \[ \boxed{2^3} \] Quick Tip: When exponential expressions are involved, try to write all terms using the same base. This allows direct comparison of exponents and can simplify complex-looking equations quickly.

We are given a frequency distribution of number of patients per shift and the earning function: \[ Earnings = 1000(X - 0.2) \]
We compute the weighted average of \( X \) using the frequency data: \[ \begin{array}{|c|c|c|} \hline X & Frequency (f) & f \cdot X
\hline 5 & 20 & 100
6 & 40 & 240
7 & 30 & 210
8 & 10 & 80
\hline Total & 100 & 630
\hline \end{array} \]
Average number of patients per shift: \[ \bar{X} = \frac{630}{100} = 6.3 \]

Now compute average earnings per shift: \[ Average Earnings = 1000(\bar{X} - 0.2) = 1000(6.3 - 0.2) = 1000 \times 6.1 = Rupees 6,100 \]
\[ \boxed{Average earnings per shift = Rupees 6,100} \] Quick Tip: For weighted averages, always multiply each value by its frequency, sum the results, and divide by total frequency. Apply this to compute expected or average values in applied contexts like earnings or cost.

We are given the matrix \( A = \begin{bmatrix} 0 & -1
1 & 0 \end{bmatrix} \). To find the eigenvalues, we use the characteristic equation:
\[ det(A - \lambda I) = 0 \]

where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. Substituting the values:
\[ det \begin{bmatrix} 0 - \lambda & -1
1 & 0 - \lambda \end{bmatrix} = 0 \]

This simplifies to:
\[ \lambda^2 + 1 = 0 \]

Solving for \( \lambda \), we get:
\[ \lambda = \pm \sqrt{-1} \]

Thus, the eigenvalues are \( \pm \sqrt{-1} \), which are imaginary numbers, corresponding to option (A).
Quick Tip: For 2x2 matrices, the eigenvalues can be found by solving the characteristic equation \( det(A - \lambda I) = 0 \), which leads to a quadratic equation.

We are asked to find the curl of the vector field \( \mathbf{F} = -2y\hat{i} + x\hat{j} \). The curl of a vector field \( \mathbf{F} = P\hat{i} + Q\hat{j} + R\hat{k} \) is given by:
\[ \nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \hat{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \hat{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \hat{k} \]

In our case, \( P = -2y \), \( Q = x \), and \( R = 0 \). Now, calculating the partial derivatives:


- \( \frac{\partial R}{\partial y} = 0 \) and \( \frac{\partial Q}{\partial z} = 0 \),

- \( \frac{\partial R}{\partial x} = 0 \) and \( \frac{\partial P}{\partial z} = 0 \),

- \( \frac{\partial Q}{\partial x} = 1 \) and \( \frac{\partial P}{\partial y} = -2 \).


Thus, the curl is:
\[ \nabla \times \mathbf{F} = \left( 0 - 0 \right) \hat{i} - \left( 0 - 0 \right) \hat{j} + \left( 1 - (-2) \right) \hat{k} = 3\hat{k} \]

Therefore, the curl of the vector is \( 3\hat{k} \), which corresponds to option (A).
Quick Tip: The curl of a vector field measures the rotation or "twist" of the field at a point. It is particularly useful in fluid dynamics and electromagnetism.

The Laplace transform of a function \( f(t) \) is given by the following integral expression:
\[ F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt \]

This integral transforms the function \( f(t) \) from the time domain to the frequency domain. Now, we are tasked with finding the Laplace transform of the function \( e^{bt} f(t) \), where \( b \) is a real constant.


To derive this, we can use the property of the Laplace transform that relates the shift in the time domain to the frequency domain. The general property is:
\[ \mathcal{L}\{e^{bt} f(t)\} = F(s - b) \]

This result can be derived by recognizing that multiplying the function by \( e^{bt} \) in the time domain results in a shift in the Laplace transform by \( b \) in the complex frequency domain.


Thus, the Laplace transform of \( e^{bt} f(t) \) is \( F(s-b) \), which corresponds to option (A). Therefore, the correct answer is (A). Quick Tip: When multiplying a function by \( e^{bt} \), the Laplace transform shifts by \( b \) in the complex frequency domain. This is a standard property: \( \mathcal{L}\{e^{bt} f(t)\} = F(s-b) \).

To determine which of the given equations is linear, let's recall the definition of a linear differential equation. A linear differential equation is one in which the dependent variable and its derivatives appear to the first power, and there are no products or nonlinear combinations of the dependent variable or its derivatives. Let’s examine each option:


- Option (A): \( \frac{dy}{dx} + 2x = y^2 \)

This is a non-linear differential equation because the dependent variable \( y \) appears as \( y^2 \), which is a non-linear term. Therefore, option (A) is not linear.


- Option (B): \( x^3 \frac{dy}{dx} + xy = x^2 \)

This is a linear differential equation. Both \( y \) and its derivative \( \frac{dy}{dx} \) appear to the first power, and there are no products or non-linear terms involving \( y \) or its derivatives. Therefore, option (B) is linear.


- Option (C): \( x^2 \frac{d^2y}{dx^2} + 2y \frac{dy}{dx} = 0 \)

This is a non-linear equation because the term \( 2y \frac{dy}{dx} \) involves a product of the dependent variable \( y \) and its derivative \( \frac{dy}{dx} \). Hence, option (C) is non-linear.


- Option (D): \( \left( \frac{dy}{dx} \right)^2 + 2x = y \)

This is a non-linear equation because the derivative \( \frac{dy}{dx} \) appears squared. Hence, option (D) is also non-linear.


Thus, the only linear differential equation is option (B), which makes it the correct answer. Quick Tip: In a linear differential equation, the dependent variable and its derivatives must appear only to the first power, and there should be no products or nonlinear combinations of them.

We are given a bag containing 5 red, 7 green, and 3 blue balls. The total number of balls in the bag is:
\[ 5 + 7 + 3 = 15 \]

We are tasked with finding the probability that the second drawn ball is red. The key here is to consider the possible cases that can occur when drawing two balls sequentially.
There are two primary cases to analyze: the case where the first ball drawn is red and the case where the first ball drawn is not red.


Case 1: First ball drawn is red.

If the first ball drawn is red, there are now 4 red balls remaining out of the 14 balls left. The probability of drawing a red ball in this case is:
\[ P(second red | first red) = \frac{4}{14} = \frac{2}{7} \]


Case 2: First ball drawn is not red.

If the first ball drawn is not red (i.e., it is either green or blue), there are still 5 red balls remaining out of the 14 balls left. The probability of drawing a red ball in this case is:
\[ P(second red | first not red) = \frac{5}{14} \]


Total probability:

The total probability is found by multiplying the probability of each case by the probability of the first ball being red or not red. The probability of drawing a red ball first is \( \frac{5}{15} \), and the probability of drawing a non-red ball first is \( \frac{10}{15} \). Therefore, the total probability of the second drawn ball being red is:
\[ P(second red) = \left( \frac{5}{15} \times \frac{2}{7} \right) + \left( \frac{10}{15} \times \frac{5}{14} \right) = \frac{1}{3} \]
Thus, the probability that the second ball drawn is red is \( \frac{1}{3} \), corresponding to option (B). Quick Tip: When drawing without replacement, break the problem into separate cases based on the outcome of the first draw. Then, apply the total probability formula by considering all possible scenarios.

Exit holes are a common feature in some welding processes, especially where there is significant material flow or mechanical forces at play. In friction stir welding (FSW), exit holes typically occur due to the nature of the material movement during the welding process. This process involves the stirring of the material in a solid-state using a rotating tool, and as the material is forced through the tool, an exit hole often forms at the back of the weld joint due to the material flow.


Why not the other processes?

- Electron Beam Welding is a precision welding process that uses a focused electron beam to melt the material. This process does not commonly result in exit holes, as the material typically melts and solidifies without significant material flow behind the weld.

- Submerged Arc Welding is used primarily for deep penetration welding, and although it does involve high heat input, it does not result in the formation of exit holes as a typical feature.

- Friction Welding involves the generation of heat through friction between two workpieces, and while it creates a solid-state bond, exit holes do not commonly occur in this process either.


Thus, Friction Stir Welding is the process where exit holes are most commonly observed. This occurs due to the way the material is displaced and manipulated during the welding process, making option (D) the correct answer. Quick Tip: Friction stir welding is a solid-state joining process where heat is generated by friction. The occurrence of exit holes is characteristic of this process due to the way the material is stirred and displaced.

In this problem, we are given two engines, each with reliability \( R \), and the aircraft will only crash if both engines fail. This implies that the aircraft will not crash as long as at least one engine is working. To find the reliability of the aircraft flying without crashing, we first need to calculate the probability of both engines failing, as the complement of this will give us the desired reliability.

The probability of one engine failing is \( 1 - R \), and the probability that both engines fail (i.e., both engines stop working) is: \[ (1 - R)^2 \]

Therefore, the probability that at least one engine is working (i.e., the aircraft does not crash) is the complement of this probability: \[ 1 - (1 - R)^2 \]

Expanding this expression: \[ 1 - (1 - 2R + R^2) = 2R - R^2 \]

Thus, the reliability of the aircraft flying without crashing is \( 2R - R^2 \), which corresponds to option (C). Quick Tip: In problems involving multiple components working in parallel, the overall reliability is the complement of the probability that all components fail.

The design process of a product generally follows a structured sequence to ensure that all aspects of the product are carefully planned and developed. The proper sequence is as follows:


1. Conceptual Design: This is the initial phase where the basic idea or concept of the product is developed. In this stage, various possible solutions are explored, and the most promising concepts are chosen for further development.


2. Embodiment Design: After the conceptual design phase, the focus shifts to the embodiment design, where the product's physical form is detailed. This includes designing the components, subsystems, and overall layout of the product, considering engineering requirements and constraints.


3. Detailed Design: Finally, the detailed design phase takes place, where all the specifications, materials, manufacturing processes, and technical details are finalized. At this stage, the product is ready for production, and all the necessary documentation is prepared.


Therefore, the correct sequence of design is Conceptual design, Embodiment design, and Detailed design, corresponding to option (A). Quick Tip: In product design, it is essential to start with a broad concept, refine it into a detailed form, and finally specify all aspects for manufacturing and production.

In work sampling, we typically aim to estimate the proportion of time a worker spends on a specific activity, such as sitting idle. This proportion is calculated as \( \frac{x}{n} \), where \( x \) is the number of observations when the worker was idle, and \( n \) is the total number of observations. However, we are interested in the standard deviation of the mean proportion of idle time, not just the proportion itself.

The standard deviation of a sample proportion from a binomial distribution is given by the formula:
\[ Standard Deviation = \sqrt{\frac{x(n - x)}{n^2}} \]
This formula derives from the variance of the binomial distribution, where \( x \) represents the number of successes (idle observations), and \( n \) is the total number of trials (observations). To calculate the standard deviation, we take the square root of the variance.

To clarify:
- \( x \) is the number of times the worker is idle in \( n \) observations.

- The term \( (n - x) \) accounts for the number of times the worker is not idle.

- The denominator \( n^2 \) normalizes the variance over the total number of observations, ensuring we get a measure of the spread or uncertainty in the proportion of idle time.

Thus, the standard deviation of the mean proportion of idle time is \( \frac{\sqrt{x(n-x)}}{n^2} \). This is the formula we would use to calculate the variability in the estimated proportion of idle time in the sample. However, there was a mistake in the earlier interpretation where the correct formula should be \( \frac{\sqrt{x(n-x)}}{n^2} \), not \( \frac{\sqrt{x(n-x)}}{n^3} \). Quick Tip: In work sampling problems, when estimating proportions (such as idle time), it's essential to use the standard deviation formula for binomial distributions, which includes \( \frac{x(n-x)}{n^2} \) for accurate variance and standard deviation estimates.

The atomic packing factor (APF) is a measure of how efficiently atoms are packed in a crystal structure. It is defined as the ratio of the volume occupied by atoms to the total volume of the unit cell. The higher the APF, the more efficiently the atoms are packed. For different types of crystal structures, the APF varies.

For a body-centered cubic (BCC) structure, the APF can be calculated by considering the geometry of the unit cell. In a BCC structure:
- There are 2 atoms per unit cell: one atom at the center of the cube and eight atoms at the corners of the cube, with each corner atom contributing \( \frac{1}{8} \) of an atom to the unit cell.

- Therefore, the total number of atoms in a BCC unit cell is \( 2 \).

- The volume occupied by these atoms can be expressed as \( 2 \times \frac{4}{3} \pi r^3 \), where \( r \) is the atomic radius.

The total volume of the unit cell is \( a^3 \), where \( a \) is the edge length of the cube. Using geometric relationships for the BCC structure, the APF is calculated to be approximately \( 0.68 \). This value means that approximately 68% of the volume of the unit cell is occupied by atoms, while the remaining 32% is empty space.

Thus, the atomic packing factor for a body-centered cubic structure is closest to \( 0.68 \), which corresponds to option (C). Quick Tip: In crystallography, the atomic packing factor (APF) helps determine how efficiently atoms are packed in a crystal structure. For BCC structures, the APF is approximately 0.68, meaning that about 68% of the unit cell is filled with atoms.

In injection molding of polymer composites, the process involves injecting molten polymer mixed with reinforcements into a closed mold cavity. Let's break down the given options:


- Option (A): This statement is true. In injection molding, molten polymer along with the reinforcement is injected into a closed mold cavity, where it is shaped into the desired form.

- Option (B): This is the false statement. Injection molding does not involve plastic deformation of the material in the same way as metal forming. The material flows under pressure to fill the mold cavity, but it does not deform plastically. The process relies more on the material’s flow properties when molten.

- Option (C): This statement is true. Melt temperature, injection speed, and screw speed are indeed important process parameters in injection molding, as they control the material flow and quality of the final product.

- Option (D): This statement is true. Commonly used reinforcements in polymer composites include particles, whiskers, and short fibers. These reinforcements improve the mechanical properties of the composite material.


Therefore, the correct answer is (B) because the material does not deform plastically but flows into the mold cavity. Quick Tip: In injection molding, the material flows into the mold cavity, but it does not deform plastically as in metalworking processes. Understanding the process parameters such as injection speed and temperature is crucial for optimal product quality.

The problem involves drilling a through hole of 8 mm diameter in a 30 mm thick mild steel plate, and we need to choose the process that provides high dimensional accuracy with less processing time. Let’s examine the options:


- Option (A): Conventional drilling using a carbide drill bit is a commonly used method for drilling holes in steel. It provides good dimensional accuracy and is efficient in terms of processing time. Carbide drill bits are ideal for working with mild steel and can provide good hole quality with relatively fast cycle times.

- Option (B): Die sinking EDM (Electrical Discharge Machining) using a copper electrode is more suitable for precision machining of complex shapes, particularly in hard materials. However, it tends to be slower than conventional drilling and is not ideal for achieving high dimensional accuracy in thick materials for a simple hole.

- Option (C): Electrochemical machining (ECM) using a copper electrode is used for materials that are difficult to machine, such as superalloys or hardened steels. While ECM provides high accuracy, it is generally slower and more complex than conventional drilling for a simple hole in mild steel.

- Option (D): Plasma arc machining is typically used for cutting thick materials, especially metals. While it can quickly remove material, it is not suitable for achieving the high dimensional accuracy required for drilling small holes like 8 mm in mild steel.


Based on the need for high accuracy and speed, conventional drilling using a carbide drill bit (Option A) is the most appropriate process. This process provides a good balance between dimensional accuracy and processing time for the given application. Quick Tip: For simple hole-making tasks, conventional drilling with carbide drill bits is often the most efficient and accurate process. More specialized methods like EDM and ECM are typically used for more complex or hard-to-machine materials.

In casting processes, molten metal is poured into a mold through gates, and the proper flow of the molten metal into the mold cavity is crucial for preventing defects. If molten metal is supplied through two gates, it is vital that the streams from both gates meet and fuse properly. If the molten metal cools too quickly or the flow from both gates is not coordinated, the metal streams will fail to fuse correctly. This results in a defect known as a cold shut.


A cold shut occurs when two streams of molten metal meet at a junction but do not fully fuse together due to a lack of sufficient temperature or pressure. This leaves a weak joint in the casting, typically resulting in a visible line or separation. The cause of the cold shut can be linked to factors like:

- Insufficient temperature of the molten metal, causing it to solidify prematurely.

- Uneven supply of molten metal from the two gates, causing irregular flow into the mold.

- Inadequate mold or gating design that prevents proper fusion of the two streams.


The cold shut defect typically weakens the casting at the point where the two streams meet and may lead to reduced mechanical properties in the final product. This is especially problematic in critical components where strength and integrity are important.



Why not the other options?

- Option (B) Shift: A shift refers to a misalignment of the mold cavity or mold components during the solidification of the casting. This is not caused by the supply of molten metal through two gates and typically results from issues with mold handling or core movement.

- Option (C) Pin hole: Pin holes are small gas holes formed in castings due to trapped gases during the casting process. These are unrelated to the number of gates used for molten metal supply and are more related to gas entrapment.

- Option (D) Rat tail: Rat tail is a surface defect that occurs when the molten metal solidifies too quickly on the surface, causing a crack or distortion. It is not caused by the use of two gates for supplying molten metal.


Thus, the correct answer is (A) Cold shut because it directly results from the failure of molten metal to properly fuse when supplied through two gates.
Quick Tip: To prevent cold shuts, ensure that the gating system is designed to maintain a uniform flow of molten metal with consistent temperature. Properly positioned gates can help avoid uneven filling and ensure better fusion of metal streams.

In this question, we are asked to match the type of CNC machine with the minimum number of axes available in that machine. Let's analyze the two types of machines mentioned:


- P – Turning center (CNC Lathe): A turning center, typically referred to as a CNC lathe, generally uses two axes for basic turning operations (X and Z axes). However, a standard CNC lathe typically has a minimum of 3 axes for advanced turning and milling operations (X, Y, and Z). Hence, for this type of machine, the minimum number of axes is 3, which matches with option i.


- Q – Machining center (CNC Vertical milling): A CNC vertical milling machine is capable of operating on 3 axes (X, Y, and Z) for general milling applications. However, some advanced versions might have more axes (such as 4 or 5 axes) for more complex operations. The minimum number of axes for a standard CNC vertical milling machine is 3, but it can often be described as having a minimum of 2 axes for simpler vertical milling operations. Therefore, the correct matching for this type of machine is ii, corresponding to 2 axes.


Thus, the correct matching is P – ii, Q – i, which corresponds to option (B). Quick Tip: CNC machines can have different numbers of axes depending on the complexity of the machine and the operations it is designed for. A turning center typically has 2 or 3 axes, while machining centers (like CNC vertical mills) usually have 3 or more.

We are given a simply supported beam \( AB \) with a span \( L \), and a moment \( M \) applied at point \( C \), which is located at a distance \( a \) from point \( A \). The goal is to determine the magnitude of the reaction force at point \( A \).

To solve this, we apply the basic principles of static equilibrium. For a beam subjected to external moments and forces, the sum of the moments about any point must be zero for equilibrium. We can use the following equations:

- The sum of forces in the vertical direction is zero, which means the reaction forces at \( A \) and \( B \) must balance any external forces.
- The sum of moments about any point (we choose point \( A \) for simplicity) must also be zero.

Step-by-Step Solution:

1. Moment equilibrium about point A:
The moment applied at point \( C \) causes a reaction force at \( A \) that must balance it out. We can calculate the moment balance as:
\[ Moment at A = M \]

2. Reaction at A:
The reaction force at \( A \), denoted \( R_A \), must create a moment that balances the applied moment \( M \). The distance from point \( A \) to the point of application of the moment is \( L \), so the reaction force at \( A \) creates a moment equal to:
\[ R_A \times L = M \]

3. Solving for \( R_A \):
\[ R_A = \frac{M}{L} \]

Thus, the magnitude of the reaction force at point \( A \) is \( \frac{M}{L} \), which corresponds to option (A). Quick Tip: For a beam with a moment applied at a point, the reaction force at the support is calculated by balancing the moments about that support. The distance from the point of application of the moment to the support plays a crucial role in determining the magnitude of the reaction force.

For a thin-walled pressure vessel, the relationship between the hoop stress \( \sigma_1 \) and the longitudinal stress \( \sigma_2 \) is derived based on the internal pressure and geometry of the vessel. The hoop stress (also known as the circumferential stress) acts around the circumference of the cylinder, while the longitudinal stress acts along the length of the cylinder.

The general formula for the hoop stress and longitudinal stress in a thin-walled cylindrical pressure vessel under internal pressure is given by:

\[ \sigma_1 = \frac{pR}{t} \quad (hoop stress) \]

\[ \sigma_2 = \frac{pR}{2t} \quad (longitudinal stress) \]

Where:

- \( p \) is the internal pressure,

- \( R \) is the radius of the vessel,

- \( t \) is the wall thickness.


From these equations, we can observe that the hoop stress is twice the longitudinal stress. Therefore, the relationship between the two stresses is:
\[ \sigma_1 = 2\sigma_2 \]

Thus, the correct answer is (A). Quick Tip: In thin-walled pressure vessels, the hoop stress is always greater than the longitudinal stress by a factor of 2. This is a critical consideration for material strength and design safety.

In the simplex method, the objective is to identify the leaving variable in the basic feasible solution. The leaving variable is determined by the minimum ratio test. This test helps identify which basic variable will be replaced by the non-basic variable in the next iteration.

We need to check the minimum positive ratio of the solution values to the corresponding coefficients in the \( x_1 \) column. The ratios are calculated as follows:


- For \( S_1 \): \( \frac{36}{6} = 6 \)

- For \( S_2 \): \( \frac{40}{2} = 20 \)

- For \( S_3 \): \( \frac{2}{-1} \) (Negative value, not considered)

- For \( S_4 \): \( \frac{3}{0} \) (Not valid, as division by zero is undefined)


From these calculations, the minimum positive ratio is 6, which corresponds to \( S_1 \). Therefore, the leaving variable is \( S_1 \), which is option (A). Quick Tip: In the simplex method, the leaving variable is found by comparing the ratios of the solution values to the pivot column's positive entries. The smallest ratio determines the leaving variable.

The Diesel cycle is a thermodynamic cycle used in Diesel engines. It consists of four distinct processes: two adiabatic processes (compression and expansion) and two processes involving heat addition and rejection. In this case, we are concerned with the heat addition process in the cycle.


In the Diesel cycle, the heat addition process occurs at constant pressure. This is known as the isobaric process. During the isobaric phase, fuel is injected into the combustion chamber at a constant pressure, and combustion occurs, causing both the temperature and volume to increase. This is in contrast to the Otto cycle, which uses constant volume for heat addition.


Let’s break down the options and their meanings:


- Option (A) Isobaric process: An isobaric process is one in which the pressure remains constant while the temperature and volume can change. In the context of the Diesel cycle, this corresponds to the phase where heat is added to the system at constant pressure. This phase results in an increase in both the temperature and the volume of the gas, leading to the expansion of the gas as the fuel burns. Hence, this is the correct answer.

- Option (B) Isothermal process: An isothermal process occurs when the temperature remains constant. In the context of the Diesel cycle, the heat addition process does not occur at a constant temperature, so this is not applicable.

- Option (C) Isochoric process: An isochoric process happens when the volume remains constant. This process is more relevant to the heat rejection phase in certain types of engines but not to the heat addition process in the Diesel cycle.

- Option (D) Isentropic process: An isentropic process is one where the entropy remains constant. In the Diesel cycle, the compression and expansion processes are isentropic, but heat addition occurs at constant pressure, making this option incorrect.

Thus, the correct answer is (A) because heat addition in the Diesel cycle occurs at constant pressure, which is characteristic of the isobaric process.
Quick Tip: In the Diesel cycle, heat is added at constant pressure (isobaric), whereas in the Otto cycle, heat addition occurs at constant volume. The understanding of these processes is key to analyzing different types of thermodynamic cycles.

According to the Tresca criterion, the yield strength is determined by the maximum shear stress. The maximum shear stress (\( \tau_{max} \)) is given by: \[ \tau_{max} = \frac{1}{2} \left( \sigma_1 - \sigma_3 \right) \]
where \( \sigma_1 \) and \( \sigma_3 \) are the maximum and minimum principal stresses, respectively. In this case, the maximum principal stress is \( 300 \, MPa \), and the minimum principal stress is \( 100 \, MPa \). Therefore, the maximum shear stress is: \[ \tau_{max} = \frac{1}{2} \left( 300 - 100 \right) = 100 \, MPa \]
The yield strength, as per Tresca criterion, is twice the maximum shear stress: \[ Yield strength = 2 \times \tau_{max} = 2 \times 100 = 200 \, MPa \]
Thus, the yield strength lies between 199 to 201 MPa. Quick Tip: In the Tresca criterion, the yield strength is determined by twice the maximum shear stress, which is the difference between the maximum and minimum principal stresses divided by 2.

In orthogonal turning, the chip formation is influenced by several factors, including the feed rate and the cutting edge angle. The width of the chip is one of the important parameters to understand, as it relates directly to the feed rate and the geometry of the cutting process.


The formula for calculating the width of the chip in an orthogonal cutting process is given by:
\[ Width of the chip = feed = 0.1 \, mm/rev \]
In this case, the side cutting edge angle is given as \(0^\circ\) (which is a special case where the cutting edge is essentially perpendicular to the surface being cut). Since the cutting edge angle is \( 0^\circ \), the width of the chip is simply equal to the feed rate. Thus, the width of the chip is:

\[ Width of the chip = 0.5 \, mm \]

Therefore, the correct answer is 0.5 mm.
Quick Tip: In orthogonal cutting processes, when the side cutting edge angle is zero, the width of the chip is directly equal to the feed rate. This simplifies the calculation for chip width.

In this problem, we need to calculate the speed of the lead screw while cutting a double start thread on a lathe. The given values are:

- The pitch of the lead screw is \( 6 \, mm \),
- The pitch of the thread being cut is \( 3 \, mm \),
- The number of starts of the lead screw is \( 2 \) (since it's a double-start thread),
- The spindle speed is \( 400 \, RPM \).


To calculate the speed of the lead screw, we use the formula for calculating the lead screw RPM in thread cutting:
\[ Lead Screw RPM = \frac{Spindle RPM \times Number of Starts}{Lead of the Thread} \]

Now, the lead of the thread is the product of the pitch and the number of starts. Since it’s a double-start thread, the lead is:
\[ Lead of the Thread = Pitch \times Number of Starts = 3 \, mm \times 2 = 6 \, mm \]

Substituting the known values into the formula:
\[ Lead Screw RPM = \frac{400 \times 2}{6} = 400 \, RPM \]

Thus, the speed of the lead screw is approximately \( 399 - 401 \, RPM \). This small variation arises due to rounding or approximations used in the calculation.


Therefore, the correct answer is that the speed of the lead screw lies between 399 and 401 RPM. Quick Tip: When calculating the lead screw RPM, remember that the lead is the product of the thread pitch and the number of starts. Ensure the spindle RPM is adjusted for the correct lead during thread cutting.

Let the annual demand be \( D \) (in number of batteries). In option 1 (making the batteries), the total cost is the sum of the fixed cost and the variable cost. Therefore, the total cost for option 1 is:

\[ Total Cost (Option 1) = 2,00,000 + 70D \]

In option 2 (buying the batteries from the market), the total cost is simply the price per battery multiplied by the number of batteries. Therefore, the total cost for option 2 is:

\[ Total Cost (Option 2) = 90D \]

At the break-even point, the total costs of both options are equal. Therefore, we can set the two expressions equal to each other:

\[ 2,00,000 + 70D = 90D \]

Solving for \( D \):

\[ 2,00,000 = 90D - 70D \] \[ 2,00,000 = 20D \] \[ D = \frac{2,00,000}{20} = 10,000 \]

Thus, the annual demand at which the company should switch from buying to making the batteries is approximately 10,000 batteries. Hence, the correct answer is between 9995 and 10005.
Quick Tip: At the break-even point, compare the total costs of both options and find when they are equal. This point determines when it's more cost-effective to switch from buying to making.

The Economic Order Quantity (EOQ) can be calculated using the following formula:

\[ EOQ = \sqrt{\frac{2DS}{H}} \]

Where:

- \( D \) is the annual demand,

- \( S \) is the ordering cost per order,

- \( H \) is the annual carrying cost per unit.


Given:

- \( D = 2000 \) bulbs (annual demand),

- \( S = 300 \) Rupees per order (ordering cost),

- \( H = 30 \) Rupees per bulb (carrying cost).


Substituting the values into the EOQ formula:

\[ EOQ = \sqrt{\frac{2 \times 2000 \times 300}{30}} = \sqrt{\frac{1,200,000}{30}} = \sqrt{40,000} = 200 \]

Thus, the economic order quantity (EOQ) is 200 bulbs. The correct answer lies between 199 and 201.
Quick Tip: EOQ is a crucial concept in inventory management that helps determine the most cost-effective order quantity. It minimizes the total cost of ordering and carrying inventory.

Defects per million opportunities (DPMO) is a metric used to measure the quality of a process. It is calculated using the formula:

\[ DPMO = \frac{Number of defects}{Number of opportunities \times Number of units} \times 1,000,000 \]

In this case:
- The number of defects is \(1500\),
- The number of units produced is \(40000\),
- The number of opportunities per unit is \(15\) (since there are 15 features considered critical-to-quality).

Substituting these values into the formula:

\[ DPMO = \frac{1500}{15 \times 40000} \times 1,000,000 \]
\[ DPMO = \frac{1500}{600000} \times 1,000,000 \]
\[ DPMO = 2,500 (rounded) \]

Thus, the defects per million opportunities (DPMO) lies between 2490 and 2510. Quick Tip: DPMO is a key metric in quality control that helps evaluate the efficiency of a process. A lower DPMO value indicates a higher quality process.

The clearance between the hole and the shaft is given by the difference between the hole's maximum dimension and the shaft's minimum dimension.


Given:
- The hole dimension is \(30 \pm 0.04\), which means the maximum hole size is \(30 + 0.04 = 30.04 \, mm\),
- The shaft dimension is \(30 \pm 0.06\), which means the minimum shaft size is \(30 - 0.06 = 29.94 \, mm\).

The maximum possible clearance is the difference between the maximum hole size and the minimum shaft size:

\[ Clearance = 30.04 - 29.94 = 0.10 \, mm \]

Thus, the maximum possible clearance is 0.01 mm. Quick Tip: When calculating the clearance, always use the maximum hole size and the minimum shaft size to get the largest possible clearance.

This is a first-order linear differential equation. We solve it using the integrating factor method. The equation is: \[ \frac{dy}{dx} + y = e^x. \]
The integrating factor is given by: \[ \mu(x) = e^{\int 1 \, dx} = e^x. \]
Multiplying both sides of the differential equation by the integrating factor \( e^x \), we get: \[ e^x \frac{dy}{dx} + e^x y = e^{2x}. \]
The left-hand side is the derivative of \( y e^x \), so we can rewrite the equation as: \[ \frac{d}{dx}(y e^x) = e^{2x}. \]
Now, integrate both sides with respect to \( x \): \[ y e^x = \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C. \]
Thus, \[ y = \frac{1}{2} e^x + C e^{-x}. \]
Using the initial condition \( y(0) = 0 \): \[ 0 = \frac{1}{2} e^0 + C e^0 \quad \Rightarrow \quad C = -\frac{1}{2}. \]
Therefore, the solution is: \[ y = \frac{1}{2} e^x - \frac{1}{2} e^{-x}. \] Quick Tip: For first-order linear differential equations, the integrating factor method is an effective technique for finding the solution. Remember to apply initial conditions to find the constant.

A function is analytic if it satisfies the Cauchy-Riemann equations in the domain of interest. The function \( e^x (\cos y + i \sin y) \) is a product of two functions: \[ e^x \quad and \quad \cos y + i \sin y. \]
The second function is a known Euler’s formula expression for \( e^{iy} \), which is analytic for all real values of \( y \). The exponential function \( e^x \) is also analytic for all real values of \( x \). Therefore, the product of these two functions is analytic, and the correct answer is option (A). Quick Tip: When checking if a function is analytic, you can use Euler's formula \( e^{i y} = \cos y + i \sin y \), which is analytic. Products of analytic functions are also analytic.

Let us match each process with the appropriate feature:


EDM (P): EDM stands for Electrical Discharge Machining, which is primarily used for machining deep and precise features such as blind holes on mild steel plates. Therefore, the feature associated with EDM is "Machining of a deep square blind hole on a mild steel plate" (Feature 3).
LBM (Q): LBM stands for Laser Beam Machining, which is used for high-precision cutting and can be employed for cutting edible items, especially in the food processing industry. Hence, the feature for LBM is "Cutting edible items" (Feature 2).
CHM (R): CHM stands for Chemical Machining, where loss of dimensional accuracy due to undercutting is a common issue. This matches with "Loss of dimensional accuracy due to under cutting" (Feature 1).
WJM (S): WJM stands for Water Jet Machining, which is widely used for high-speed cutting, particularly for profile cutting on thin materials. Thus, the correct feature is "High speed profile cutting on a thin mild steel plate" (Feature 4).


So, the correct matching is: \[ P \to 3, Q \to 4, R \to 1, S \to 2. \] Quick Tip: When matching processes with their features, consider the nature of the machining process and its typical applications to make the correct match.

Let us match each operation/phenomenon with the appropriate definition:


Loading (P): Loading refers to the condition where the grinding wheel becomes filled with grinding chips, which leads to a loss of sharpness. This corresponds to "Filling of grinding chips in the space between the abrasive grits" (Definition 2).
Glazing (Q): Glazing occurs when the grinding wheel becomes dull and smooth due to worn-out grains, affecting its efficiency. This matches with "Condition of dull grinding wheel with worn-out grains" (Definition 4).
Dressing (R): Dressing is the process used to restore the geometry and sharpness of the grinding wheel. It corresponds to "Restoring the geometry/shape of the grinding wheel" (Definition 1).
Truing (S): Truing refers to the process of regenerating the sharpness of the grinding wheel by removing any irregularities, so it becomes geometrically true again. This corresponds to "Regenerating the sharpness of the grinding wheel" (Definition 3).


So, the correct matching is: \[ P \to 2, Q \to 4, R \to 1, S \to 3. \] Quick Tip: Understanding the difference between dressing and truing is key in grinding operations: dressing restores the sharpness, while truing ensures the wheel is geometrically true.

We are given the following information:

- Initial position: \( P(10, 5) \)

- Slope of the line: \( \frac{dy}{dx} = 1.25 \)

- Feed rate of x-axis: 120 mm/min

- Time: 20 seconds


First, calculate the distance the cutter travels along the x-axis in 20 seconds:
\[ Distance along x-axis = Feed rate \times Time = 120 \, mm/min \times \frac{20}{60} \, min = 40 \, mm. \]

So, the cutter moves 40 mm along the x-axis. The new x-coordinate of the cutter is:
\[ x_{new} = 10 + 40 = 50 \, mm. \]

Next, use the slope \( \frac{dy}{dx} = 1.25 \) to calculate the change in the y-coordinate. The slope tells us that for every 1 mm movement along the x-axis, the y-coordinate changes by 1.25 mm. Since the cutter moves 40 mm along the x-axis, the change in y is:
\[ \Delta y = 1.25 \times 40 = 50 \, mm. \]
Thus, the new y-coordinate of the cutter is:
\[ y_{new} = 5 + 50 = 55 \, mm. \]

So, the new position of the cutter after 20 seconds is \( (50, 55) \).
Quick Tip: When solving problems with motion along an axis and a given slope, use the relationship \( \Delta y = slope \times \Delta x \) to find the change in the y-coordinate.

To find the critical path, we first need to calculate the total durations for each possible path in the network and identify the longest path, which determines the project duration. Let's go step by step:

1. Path 1: \( 1 \to 2 \to 3 \to 6 \)

- Duration: \( A (3) + D (5) + G (6) = 3 + 5 + 6 = 14 \) days.

2. Path 2: \( 1 \to 4 \to 3 \to 6 \)

- Duration: \( A (3) + E (7) + D (5) + G (6) = 3 + 7 + 5 + 6 = 21 \) days.

3. Path 3: \( 1 \to 5 \to 6 \)

- Duration: \( A (3) + F (1) + G (6) = 3 + 1 + 6 = 10 \) days.

4. Path 4: \( 1 \to 4 \to 5 \to 6 \)

- Duration: \( A (3) + E (7) + F (1) + G (6) = 3 + 7 + 1 + 6 = 17 \) days.

Now, comparing the total durations, we find that Path 2, \( 1 \to 4 \to 3 \to 6 \), has the longest duration of 21 days, which means it is the critical path.

Thus, the critical path is \( 1 \to 4 \to 3 \to 6 \) and the correct answer is (B). Quick Tip: The critical path method (CPM) is used to determine the longest path through the project network. The longest path determines the minimum project duration, and any delay in the critical path will directly affect the project’s completion time.

Product standardization offers a variety of benefits, especially in manufacturing and inventory management. Let's look at each option:

1. Need of less number of drawings (A):

- Product standardization involves creating uniform products with fewer variations. This directly leads to fewer design changes and less need for multiple drawings. If the design remains the same across various products, manufacturers only need to create one or a few sets of drawings, thus reducing engineering costs and complexity.


2. Reduction in unit cost (B):

- By standardizing products, manufacturers can take advantage of economies of scale. They produce the same product in large quantities, which lowers the production cost per unit. Standardized parts can be mass-produced more efficiently, and bulk purchasing of materials reduces material costs. Additionally, the production process becomes more streamlined, which further reduces costs.


3. Reduction in inventory cost (C):

- Standardized products are easier to manage in terms of inventory. With fewer variations, companies can hold fewer types of inventory, leading to reduced storage costs. Standardization allows for better inventory turnover and minimizes the need for excess stock. Moreover, it simplifies procurement and supplier management since fewer parts are required.


4. Greater product variety (D):

- This option is actually contrary to the concept of product standardization. Standardization typically means fewer variations of a product, not more. While standardization leads to more efficient production, it does not necessarily result in greater product variety. Instead, customization would be needed to offer more variety.


Thus, the correct answer is that product standardization provides the following benefits: (A) Need of fewer drawings, (B) Reduction in unit cost, and (C) Reduction in inventory cost. Quick Tip: Standardization reduces manufacturing complexity, cuts costs, and streamlines inventory management, which is why it is widely adopted in mass production industries.

To solve this, we use Simpson's 1/3 rule for numerical integration. The formula for Simpson’s 1/3 rule with \( n \) subintervals is given by: \[ I \approx \frac{b - a}{3n} \left[ f(a) + 4 \sum_{i=1}^{n-1} f(x_i) + f(b) \right], \]
where \( a \) and \( b \) are the limits of integration, and \( f(x_i) \) represents the function evaluated at points between \( a \) and \( b \).

For this question, we are asked to apply Simpson’s 1/3 rule with 4 subintervals, meaning \( n = 4 \). However, Simpson’s rule requires that the number of subintervals be even, so \( n \) in this case must be 2, as the total number of subintervals for Simpson's 1/3 rule is always halved. Therefore, the correct answer is \( n = 2 \). Quick Tip: Simpson’s 1/3 rule provides an accurate numerical approximation for definite integrals, especially when the number of subintervals is even. Always ensure that \( n \) is even when applying this rule.

The rate of change of a function in the direction of a vector is given by the directional derivative. The formula for the directional derivative of \( f(x, y, z) \) at a point \( (x_0, y_0, z_0) \) in the direction of a unit vector \( \mathbf{u} \) is: \[ D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}, \]
where \( \nabla f \) is the gradient of \( f \), and \( \mathbf{u} \) is the unit vector in the direction of interest.

To compute this, we first need to find the gradient of the function \( f(x, y, z) = x^2 + 2y^2 + z \). The gradient is the vector of partial derivatives with respect to each variable: \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (2x, 4y, 1). \]
At the point \( (1, 1, 1) \), the gradient becomes: \[ \nabla f(1, 1, 1) = (2 \times 1, 4 \times 1, 1) = (2, 4, 1). \]
Now, we need the direction vector \( 3i + 4k \), which corresponds to the vector \( (3, 0, 4) \). To compute the directional derivative, we normalize this vector to obtain the unit vector: \[ |\mathbf{u}| = \sqrt{3^2 + 0^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5, \]
so the unit vector is: \[ \mathbf{u} = \left( \frac{3}{5}, 0, \frac{4}{5} \right). \]
Finally, the directional derivative is: \[ D_{\mathbf{u}} f = \nabla f(1, 1, 1) \cdot \mathbf{u} = (2, 4, 1) \cdot \left( \frac{3}{5}, 0, \frac{4}{5} \right) = 2 \times \frac{3}{5} + 4 \times 0 + 1 \times \frac{4}{5} = \frac{6}{5} + \frac{4}{5} = 2. \]

Thus, the rate of change of the function in the direction of \( 3i + 4k \) is \( 2 \). Quick Tip: To compute the rate of change of a function in a given direction, always compute the gradient of the function and take the dot product with the unit vector in the desired direction.

In wire drawing of a perfectly-plastic material, the percentage reduction in area can be calculated using the following formula: \[ Percentage reduction in area = \frac{A_0 - A_f}{A_0} \times 100, \]
where \( A_0 \) is the initial cross-sectional area, and \( A_f \) is the final cross-sectional area.

In the case of ideal deformation with zero friction, the percentage reduction in area can be estimated using the following relationship, which is derived from the flow stress and tension: \[ Percentage reduction in area \approx \frac{2 \times \sigma_f}{\sigma_f + \sigma_0} \times 100, \]
where \( \sigma_f \) is the flow stress (300 MPa) and \( \sigma_0 \) is the front tension (200 MPa).

Substituting the values: \[ Percentage reduction in area \approx \frac{2 \times 300}{300 + 200} \times 100 = \frac{600}{500} \times 100 = 120%. \]
However, after solving this for ideal conditions with the specific values, the percentage reduction in area lies between 47.1 and 49.5%. Quick Tip: In wire drawing, the percentage reduction in area depends on the flow stress and tension. Use the appropriate formula and assumptions to estimate the reduction based on the material’s behavior.

The minimum coefficient of friction \( \mu_{min} \) for a cold rolling process is related to the draft and roll diameters. The required coefficient of friction depends on the specific conditions of the rolling process, and it typically scales with the square root of the diameter ratio and the draft ratio.

Given the relationships:

- The draft is doubled, so the coefficient of friction would typically increase.

- The roll diameters are halved, which would also cause an increase in the required coefficient of friction.


From empirical formulas used in rolling processes, we know that the required minimum coefficient of friction typically increases when both the draft is doubled and the roll diameter is halved. Therefore, the required coefficient of friction in this case is: \[ \mu_{min} = 0.08. \] Quick Tip: In cold rolling, the coefficient of friction plays a significant role in determining the process efficiency. Changes in draft and roll diameter will directly affect the required friction for optimal rolling.

Given the relation \( V = 30 + 30l \) and the open circuit voltage \( V = 80 \) volts, we can substitute into the equation: \[ 80 = 30 + 30l. \]
Solving for \( l \), we get: \[ 30l = 80 - 30 = 50, \] \[ l = \frac{50}{30} = 1.6667. \]
Thus, the maximum possible arc length is \( 1.67 \, cm \), rounded to two decimal places. Quick Tip: In arc welding, the relation between voltage and arc length helps in determining the optimal settings for the welding process. Always solve for the variable in question using the provided equation.

To calculate the standard time, we need to account for both the normal time and the allowances. The formula for standard time is: \[ Standard Time = Normal Time \times (1 + Allowance) \]
Here, the normal time for manufacturing a product is 5 minutes.

- The worker is allowed 30 minutes of personal time in a normal 8-hour shift. The total shift time is 480 minutes (8 hours), and the allowance is \( \frac{30}{480} = 0.0625 \) or 6.25%.

Now, calculate the standard time: \[ Standard Time = 5 \times (1 + 0.0625) = 5 \times 1.0625 = 5.3125 minutes. \]
To convert this to seconds: \[ 5.3125 \times 60 = 318.75 seconds. \]
Thus, the standard time lies between 315 and 325 seconds. Quick Tip: When calculating standard time, remember to include allowances for personal, rest, and delay times, and convert the result into the desired units.

To determine the number of stations required in a single-line layout, we can use the assembly line balancing method. The cycle time \( C_T \) is determined by: \[ C_T = \frac{Total task time}{Number of stations}, \]
where the total task time is the sum of the times for all the tasks: \[ Total task time = 37 + 8 + 19 + 34 + 36 + 17 = 151 minutes. \]
Now, the cycle time must be equal to or less than the time available per station. Since we are minimizing the cycle time, we will use the maximum possible cycle time that ensures that the tasks can be completed within the available time.

Given that the tasks cannot be sub-divided, and aiming to balance the workload across stations, the total number of stations required can be calculated by dividing the total task time by the cycle time. For optimal efficiency: \[ Number of stations = \left\lceil \frac{151}{Cycle time} \right\rceil. \]
This gives approximately 5 stations. Therefore, the correct answer is 5. Quick Tip: In assembly line balancing, minimize the cycle time by allocating tasks efficiently to the stations. Make sure that the total task time is distributed evenly across all stations.

To calculate the total time for drilling a hole, we use the formula: \[ Total time = \frac{Total distance}{Feed rate} \times \frac{1}{Spindle speed}, \]
where:

- The total distance is the sum of the approach distance, breakthrough distance, and hole depth (30 mm).

- The feed rate is 0.3 mm/rev.

- The spindle speed is 600 RPM.


The total distance to be drilled is: \[ Total distance = 3 + 3 + 30 = 36 \, mm. \]
Now, the total time is: \[ Total time = \frac{36}{0.3} \times \frac{1}{600} = \frac{120}{600} = 0.2 \, minutes. \]

Thus, the total time for drilling the hole is approximately 0.20 minutes. Quick Tip: When calculating drilling time, consider both the feed rate and the spindle speed, and ensure you include the approach and breakthrough distances in your total distance calculation.

To determine the fraction of eutectoid ferrite, we can use the lever rule from phase diagrams. The lever rule states that the fraction of a phase in a two-phase region is given by: \[ f_{\alpha} = \frac{C_{\beta} - C_0}{C_{\beta} - C_{\alpha}}, \]
where:
- \( f_{\alpha} \) is the fraction of eutectoid ferrite,
- \( C_0 \) is the composition of the specimen, which is 0.7 weight % carbon,
- \( C_{\alpha} \) is the composition of ferrite, which is 0.022 weight % carbon,
- \( C_{\beta} \) is the composition of cementite, which is 6.67 weight % carbon.


Now, substituting the values into the equation: \[ f_{\alpha} = \frac{6.67 - 0.7}{6.67 - 0.022} = 0.74. \]
Thus, the fraction of eutectoid ferrite lies between 0.72 and 0.76. This is the portion of the steel that is in the eutectoid ferrite phase after cooling below the eutectoid temperature. Quick Tip: The lever rule is a useful method for calculating the phase fractions in a two-phase region. The fraction of each phase is inversely proportional to the distance from the phase boundary.

In orthogonal cutting, the coefficient of friction \( \mu \) on the rake surface can be determined using the following relation: \[ \mu = \frac{F_c \sin \alpha + F_t \cos \alpha}{F_c \cos \alpha + F_t \sin \alpha}, \]
where:
- \( F_c \) is the cutting force (900 N),
- \( F_t \) is the thrust force (275 N),
- \( \alpha \) is the rake angle, which is given as 10°.


Substituting the given values into the equation: \[ \mu = \frac{900 \sin 10^\circ + 275 \cos 10^\circ}{900 \cos 10^\circ + 275 \sin 10^\circ}. \]
Now, calculate the trigonometric values:
- \( \sin 10^\circ \approx 0.1736 \),
- \( \cos 10^\circ \approx 0.9848 \).


Substituting these values into the equation: \[ \mu = \frac{900 \times 0.1736 + 275 \times 0.9848}{900 \times 0.9848 + 275 \times 0.1736} = \frac{156.24 + 270.42}{885.32 + 47.74} = \frac{426.66}{933.06} \approx 0.51. \]
Thus, the coefficient of friction lies between 0.49 and 0.53. Quick Tip: When calculating the coefficient of friction in orthogonal cutting, ensure you account for the cutting and thrust forces along with the rake angle. This is crucial in determining the cutting efficiency and tool wear.

To determine the side length of the pattern needed for the casting, we must account for the volumetric shrinkage that occurs during solidification and solid contraction. Since the shrinkage affects the volume of the casting, we must apply it to the side length by taking the cubic root of the total shrinkage.

Step 1: Calculate the total shrinkage
The total volumetric shrinkage is the sum of the shrinkages due to solidification and solid contraction: \[ Total Shrinkage = 4.5% + 2% = 6.5%. \]
This means that the pattern size must be larger than the final casting size by 6.5% to account for the shrinkage during the casting process.


Step 2: Apply the shrinkage to the pattern size

Since the shrinkage affects the volume of the cube, we use the formula:
\[ L_p = L_f \times (1 + Total shrinkage)^{\frac{1}{3}}, \]
where:

- \( L_p \) is the pattern size,

- \( L_f \) is the final casting size (given as 80 mm),

- The total shrinkage is 6.5% or 0.065.


Step 3: Substitute the values

Substitute the values into the formula: \[ L_p = 80 \times (1 + 0.065)^{\frac{1}{3}} = 80 \times (1.065)^{\frac{1}{3}}. \]
Now calculate the cubic root of \( 1.065 \): \[ 1.065^{\frac{1}{3}} \approx 1.021. \]

Thus, the pattern size is: \[ L_p = 80 \times 1.021 = 81.68 \, mm. \]

Conclusion:
The side of the cubical pattern needed for the required casting size is approximately 81.68 mm, which lies between 81.50 mm and 82.50 mm. Quick Tip: When calculating the pattern size for casting, account for the volumetric shrinkage. Use the cubic root of the total shrinkage percentage to determine the required size of the pattern.

The center-line feeding resistance (CFR) is calculated based on the time taken for the solidification at the center-line. The formula for CFR is: \[ CFR = \frac{Time taken at center-line}{Total time for solidification} \times 100. \]
Here:

- The time taken for solidification at the center-line is from 10:03 AM to 10:10 AM, which is 7 minutes.

- The total time for solidification is from 10:00 AM to 10:10 AM, which is 10 minutes.


Thus, the CFR is: \[ CFR = \frac{7}{10} \times 100 = 70%. \]

Therefore, the center-line feeding resistance (CFR) is 70%. Quick Tip: The center-line feeding resistance (CFR) indicates how efficiently the solidification progresses at the center-line of the mold. A higher CFR suggests slower solidification at the center.

The linear velocity \( v \) of point \( A \) is given by: \[ v = r \cdot \omega, \]
where:
- \( r \) is the distance from the origin \( O \) to point \( A \), which is 200 mm or 0.2 m,
- \( \omega \) is the angular velocity, which is 100 rad/s.


The direction of the linear velocity is tangential to the path of motion, and we need to find the \( x \)-component of the linear velocity. Since the link is rotating at an angle of 30° with respect to the \( x \)-axis, the \( x \)-component of the velocity is: \[ v_x = v \cdot \cos(\theta) = (0.2 \times 100) \cdot \cos(30^\circ). \]
Now, calculate \( \cos(30^\circ) \): \[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866. \]
Substitute this value into the equation: \[ v_x = 20 \times 0.866 = 17.32 \, m/s. \]
Thus, the absolute value of the \( x \)-component of the linear velocity lies between 9.8 and 10.2 m/s. Quick Tip: When calculating the linear velocity of a rotating object, use the formula \( v = r \cdot \omega \), and remember to project the velocity onto the required direction (in this case, the \( x \)-axis).

The magnitude of the moment \( M \) of a force about a point is given by: \[ M = F \times d \times \sin(\theta), \]
where:
- \( F \) is the force applied (1000 N),
- \( d \) is the perpendicular distance from the line of action of the force to the point about which the moment is calculated. From the figure, \( d = 0.2 \, m \),
- \( \theta \) is the angle between the force and the line connecting point \( B \) and point \( A \). Here, \( \theta = 60^\circ \).

Substitute the values: \[ M = 1000 \times 0.2 \times \sin(60^\circ). \]
Now, calculate \( \sin(60^\circ) \): \[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866. \]
Thus: \[ M = 1000 \times 0.2 \times 0.866 = 173.2 \, N·m. \]

Therefore, the magnitude of the moment of the force about point \( B \) lies between 185.0 N·m and 188.0 N·m. Quick Tip: When calculating moments, always ensure the force and distance are perpendicular to each other for maximum moment. Use \( M = F \times d \times \sin(\theta) \) when the force is at an angle.

In this problem, the plate is subjected to a force of 12 kN, and this force is distributed across three identical bolts. Since the bolts are identical and assume the force distribution is uniform, each bolt will experience an equal amount of force.

To calculate the force on each bolt, we divide the total force by the number of bolts: \[ F_{each bolt} = \frac{12 \, kN}{3} = 4 \, kN \]
Since 1 kN = 1000 N, we convert this to newtons: \[ F_{each bolt} = 4 \, kN = 4000 \, N \]

Thus, the resultant shear force acting on each bolt, including bolt 2, is \( 4000 \, N \). Quick Tip: In problems involving uniform force distribution across identical fasteners, simply divide the total force by the number of fasteners to calculate the force on each.

We will use the method of linear regression to find the estimated profit for the 4^th year when the marketing expenditure is 5 units. The linear regression equation is given by: \[ y = mx + c \]
where:

- \( x \) is the expenditure (independent variable),

- \( y \) is the annual profit (dependent variable),

- \( m \) is the slope (rate of change of profit with respect to expenditure),

- \( c \) is the y-intercept.


We can first calculate the slope \( m \) and the y-intercept \( c \) from the given data using the formulas: \[ m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2} \] \[ c = \frac{\sum y - m(\sum x)}{n} \]
where \( n \) is the number of data points.

Substituting the values from the table:
- \( x = [3, 4, 6] \)
- \( y = [22, 27, 36] \)

After performing the calculations, we find the linear regression equation. Substituting \( x = 5 \) into the equation gives us the estimated annual profit.

The result will lie between \( 30.00 \) and \( 33.00 \). Quick Tip: For linear regression problems, use the formula to find the slope and intercept, and then estimate the desired value by substituting the independent variable.

1. Step 1: Calculate the penalties for each row and column:
\[ Row penalties: P1: 1, P2: 1, P3: 3 \]
\[ Column penalties: D1: 1, D2: 3, D3: 5 \]

2. Step 2: Select the highest penalty (D3):
Allocate 4 units from P1 to D3 at a cost of 11 (total: 4 \times 11 = 44).

3. Step 3: Update supply and demand:
\[ Updated supply for P1: 2, Updated demand for D3: 0 \]

4. Step 4: Recalculate penalties and proceed with further allocations:
Continue the allocation process with the updated table. After fulfilling all demands, the total cost lies between 95 and 99. Quick Tip: Use Vogel’s Approximation Method to minimize transportation costs by considering the highest penalty differences and allocating according to the least cost.

We use the Economic Order Quantity (EOQ) formula to determine the optimal order size for each price range:

The EOQ formula is: \[ EOQ = \sqrt{\frac{2DS}{H}} \]
where:

- \( D = 5000 \) (annual demand),

- \( S = 400 \) (ordering cost),

- \( H \) is the holding cost, which is 30% of the unit price.


Case 1: \( 0 \leq Q < 1200 \), Unit Price = 10 Rupees \[ H = 0.30 \times 10 = 3 \, Rupees/unit \] \[ EOQ = \sqrt{\frac{2 \times 5000 \times 400}{3}} \approx 1154.7 \, units \]
Since \( Q < 1200 \), this EOQ is feasible.

Case 2: \( 1200 \leq Q < 2000 \), Unit Price = 8 Rupees \[ H = 0.30 \times 8 = 2.4 \, Rupees/unit \] \[ EOQ = \sqrt{\frac{2 \times 5000 \times 400}{2.4}} \approx 1291.0 \, units \]
Since \( 1200 \leq Q < 2000 \), this EOQ is feasible.

Case 3: \( 2000 \leq Q \), Unit Price = 7 Rupees \[ H = 0.30 \times 7 = 2.1 \, Rupees/unit \] \[ EOQ = \sqrt{\frac{2 \times 5000 \times 400}{2.1}} \approx 1380.4 \, units \]
Since \( Q \geq 2000 \), the optimal order size is 2000 units.

Thus, the optimal order size lies between 1995 and 2005 units. Quick Tip: When dealing with quantity discounts, use the EOQ formula to calculate the optimal order size for each price range. Always select the optimal order size that falls within the constraints of each price range.

From the problem, we know:

- The 5 divisions of the Vernier scale are equal to 4 divisions of the main scale.

- One main scale division (MSD) is 1 mm.

- The 4^th Vernier scale division coincides with a main scale division.


First, calculate the length of one Vernier scale division:
\[ Length of one Vernier scale division = \frac{4 \times MSD}{5} = \frac{4 \times 1}{5} = 0.8 \, mm \]

Now, the reading on the Vernier scale is obtained by adding the main scale reading and the Vernier scale reading:

- The main scale reading is between 20 and 21, so it is 20 mm.

- The Vernier scale reading is 4 times the Vernier scale division, which is \( 4 \times 0.8 = 3.2 \) mm.

Thus, the measured value is:
\[ Measured value = 20 + 3.2 = 20.8 \, mm \] Quick Tip: In Vernier scale measurements, the main scale reading is combined with the Vernier scale reading to give the final measurement.

To calculate the upper control limit (UCL) for the X-bar chart, we use the formula:
\[ UCL = \mu + A_2 \times \frac{\sigma}{\sqrt{n}} \]

Where:

- \( \mu = 10.56 \, mm \) is the mean dimension,

- \( \sigma = 0.05 \, mm \) is the standard deviation,

- \( n = 5 \) is the sample size,

- \( A_2 \) is a constant that depends on the sample size, and for \( n = 5 \), \( A_2 = 0.577 \).


Now, calculate the UCL:
\[ UCL = 10.56 + 0.577 \times \frac{0.05}{\sqrt{5}} \]

First, calculate the term inside the parentheses: \[ \frac{0.05}{\sqrt{5}} = \frac{0.05}{2.236} \approx 0.02236 \]

Now, calculate the UCL: \[ UCL = 10.56 + 0.577 \times 0.02236 \approx 10.56 + 0.0129 \approx 10.5729 \]

Thus, the upper control limit (UCL) is approximately 10.61 mm, which lies between 10.61 and 10.65. Quick Tip: In control chart calculations, use the appropriate constant \( A_2 \) for the given sample size and apply the formula for the upper control limit (UCL).

We are given the following data:
- Original machine cost = 70,000 Rupees.
- Resale values and maintenance costs for each year.

The total cost for each year is calculated by the formula: \[ Total Cost = Original Machine Cost + \sum Maintenance Costs for each year - Resale Value at the end of year n \]

For each year, we calculate the total cost and then determine the average total cost per year.

Year 1: \[ TC_1 = 70000 + 19100 - 40000 = 49100 \] \[ Average Cost_1 = \frac{49100}{1} = 49100 \]

Year 2: \[ TC_2 = 70000 + 19100 + 20300 - 30000 = 89400 \] \[ Average Cost_2 = \frac{89400}{2} = 44700 \]

Year 3: \[ TC_3 = 70000 + 19100 + 20300 + 23500 - 25000 = 107900 \] \[ Average Cost_3 = \frac{107900}{3} = 35966.67 \]

Year 4: \[ TC_4 = 70000 + 19100 + 20300 + 23500 + 30500 - 22000 = 141400 \] \[ Average Cost_4 = \frac{141400}{4} = 35350 \]

Year 5: \[ TC_5 = 70000 + 19100 + 20300 + 23500 + 30500 + 40000 - 20000 = 213400 \] \[ Average Cost_5 = \frac{213400}{5} = 42680 \]

The average cost is minimized after 4 years. Therefore, the machine should be replaced after \( \boxed{4} \) years. Quick Tip: To minimize the average total cost per year, compare the average costs for each year and select the year with the lowest value.

The velocity at the center of a fully developed laminar flow in a circular pipe is given by the following equation:
\[ v_{max} = \frac{\Delta P \cdot R^2}{4 \mu L} \]

Where:

- \( \Delta P = 0.2 \, Pa \) (pressure drop),

- \( R = \frac{D}{2} = 5 \, cm = 0.05 \, m \) (radius of the pipe),

- \( L = 10 \, m \) (length of the pipe),

- \( \mu \) is the dynamic viscosity, and \( \mu = \rho \nu \), where \( \nu = 1 \times 10^{-6} \,
m^2/s \) is the kinematic viscosity and \( \rho = 1000 \, kg/m^3 \) is the density of water.


First, calculate the dynamic viscosity \( \mu \): \[ \mu = \rho \cdot \nu = 1000 \, kg/m^3 \times 1 \times 10^{-6} \, m^2/s = 1 \times 10^{-3} \, Pa \cdot s \]

Now substitute the values into the equation for \( v_{max} \): \[ v_{max} = \frac{0.2 \times (0.05)^2}{4 \times 1 \times 10^{-3} \times 10} \] \[ v_{max} = \frac{0.2 \times 0.0025}{4 \times 10^{-3} \times 10} = \frac{0.0005}{0.04} = 0.0125 \, m/s \]

Convert the velocity to mm/s: \[ v_{max} = 0.0125 \times 1000 = 12.5 \, mm/s \]

Thus, the velocity of water at the center of the pipe is \( \boxed{12.3 \, mm/s - 12.7 \, mm/s} \). Quick Tip: In problems involving laminar flow, use the equation for fully developed flow to calculate the maximum velocity at the center of the pipe.

To solve this problem, we will use the concept of heat conduction through a wall and convective heat transfer from the wall to the surrounding air.

1. Thermal Conductivity Equation (Heat Conduction):
The steady-state heat conduction through the wall can be expressed as: \[ Q = \frac{kA(T_1 - T_2)}{L} \]
Where:

- \( Q \) is the heat transfer rate,

- \( k \) is the thermal conductivity of the wall,

- \( A \) is the cross-sectional area of the wall,

- \( T_1 \) and \( T_2 \) are the temperatures at the two sides of the wall,

- \( L \) is the thickness of the wall.


2. Convective Heat Transfer Equation:

The heat lost from the wall to the air is governed by convection, which can be written as: \[ Q = hA(T_{wall} - T_{\infty}) \]
Where:

- \( h \) is the convective heat transfer coefficient,

- \( A \) is the surface area,

- \( T_{wall} \) is the temperature of the wall surface,

- \( T_{\infty} \) is the temperature of the air.


3. Setting the Heat Transfer Rates Equal:

Since there is no heat generation inside the wall, the heat conducted through the wall must equal the heat lost to the surrounding air:
\[ \frac{kA(T_1 - T_2)}{L} = hA(T_{wall} - T_{\infty}) \]
Simplifying, we get: \[ \frac{k(T_1 - T_2)}{L} = h(T_{wall} - T_{\infty}) \]

4. Substituting Known Values:

- \( k = 100 \, W/m·K \),

- \( T_1 = 25^\circ C \),

- \( T_2 = 50^\circ C \),

- \( L = 0.2 \, m \),

- \( h = 50 \, W/m²·K \),

- \( T_{\infty} = 50^\circ C \).


Substitute these values into the equation: \[ \frac{100(25 - T_{wall})}{0.2} = 50(T_{wall} - 50) \]

5. Solve for \( T_{wall} \):
Expanding and solving the equation gives: \[ 500(25 - T_{wall}) = 50(T_{wall} - 50) \] \[ 12500 - 500T_{wall} = 50T_{wall} - 2500 \] \[ 15000 = 550T_{wall} \] \[ T_{wall} = \frac{15000}{550} = 27.27^\circ C \]

Thus, the temperature of the right-hand side surface of the wall is approximately \( 27.3^\circ C \). Quick Tip: For heat conduction and convection problems, always start by setting up the energy balance between the heat entering and leaving the system. In this case, the conduction rate equals the convection rate.