GATE PI 2024

The symbol '$\rightarrow$' indicates increasing intensity. The first sequence, sick $\rightarrow$ infirm $\rightarrow$ moribund, shows a progression from being slightly unwell (sick) to very weak (infirm) to nearly dying (moribund). We need to find a word that fits the blank in silly $\rightarrow$ ___ $\rightarrow$ daft, maintaining a similar increase in the intensity of foolishness. Silly means somewhat foolish, and daft means extremely foolish. We need a word that is more foolish than silly but less foolish than daft.

Let's evaluate the options:

• Frown: An expression of displeasure, unrelated to foolishness.
• Fawn: To excessively flatter, unrelated to foolishness.
• Vein: A blood vessel or a particular quality, unrelated to foolishness.
• Vain: Can mean producing no result (fruitless) or having excessive pride leading to foolish behavior. This represents a level of poor judgment or foolishness that is greater than silly but less extreme than daft.

Therefore, vain appropriately completes the analogy: silly $\rightarrow$ vain $\rightarrow$ daft.

This problem can be solved using graph coloring, a fundamental concept in graph theory. We represent each part of the figure as a vertex and draw an edge between two vertices if their corresponding parts share a boundary. The goal is to find the chromatic number of this graph, which is the minimum number of colors needed to color the vertices such that no two adjacent vertices (parts) have the same color. The Four-Color Theorem states that any planar graph (like our figure) can be colored using at most four colors. Therefore, by carefully examining the figure's structure, we can confirm that it can be colored with $4$ colors without any adjacent parts having the same color. The minimum number of colors required is $4$.

To form a 4-digit number divisible by 3 from the given digits $\{1, 3, 4, 6, 7\}$ without repetition, we first recall the divisibility rule for 3: a number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all five available digits is $1 + 3 + 4 + 6 + 7 = 21$. We need to select 4 digits from these 5. If we denote the sum of the four chosen digits as $S_4$, and the sum of all five digits as $S_5 = 21$, then $S_4 = S_5 - d$, where $d$ is the excluded digit. For $S_4$ to be divisible by 3, $21 - d$ must be divisible by 3. Since 21 is already divisible by 3, the excluded digit $d$ must also be divisible by 3.

From the set $\{1, 3, 4, 6, 7\}$, the digits divisible by 3 are 3 and 6. This gives us two cases for the set of 4 digits:

Case 1: Exclude digit 3.
The selected digits are $\{1, 4, 6, 7\}$. Their sum is $1 + 4 + 6 + 7 = 18$, which is divisible by 3.
The number of distinct 4-digit integers that can be formed using these 4 digits is $4! = 4 \times 3 \times 2 \times 1 = 24$.

Case 2: Exclude digit 6.
The selected digits are $\{1, 3, 4, 7\}$. Their sum is $1 + 3 + 4 + 7 = 15$, which is divisible by 3.
The number of distinct 4-digit integers that can be formed using these 4 digits is $4! = 4 \times 3 \times 2 \times 1 = 24$.

The total number of 4-digit positive integers divisible by 3 is the sum of the counts from these two valid cases:
Total Integers = $24 + 24 = 48$.

The given infinite series is $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \dots$. We can break this series down into a constant term and two separate infinite geometric series. The series can be written as:
$S = 2 + \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots\right) + \left(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots\right)$
The terms $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots$ form a geometric series $\sum_{k=1}^{\infty } \frac{1}{2^k}$.
The terms $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$ form a geometric series $\sum_{k=1}^{\infty } \frac{1}{3^k}$.
The sum of an infinite geometric series is given by the formula $S_{\infty } = \frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio (provided $|r| < 1$).

For the first geometric series (\sum_{k=1}^{∞} \frac{1}{2^k}$): First term$a = \frac{1}{2}$. Common ratio$r = \frac{1}{2}$. Sum$_2 = \frac{1/2}{1 - 1/2} = \frac{1/2}{1/2} = 1$.

For the second geometric series (\sum_{k=1}^{∞} \frac{1}{3^k}$): First term$a = \frac{1}{3}$. Common ratio$r = \frac{1}{3}$. Sum$_3 = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}$.

Now, combine the constant term and the sums of the two geometric series to find the total sum $S$:
$S = 2 + \text{Sum}_2 + \text{Sum}_3$
$S = 2 + 1 + \frac{1}{2}$
$S = 3 + \frac{1}{2}$
$S = \frac{6}{2} + \frac{1}{2} = \frac{7}{2}$.

First, find the total valid votes by subtracting invalid votes from the total votes cast:
$\text{Total valid votes} = 1,15,000 - 5,000 = 1,10,000$
Next, calculate the valid votes for candidates B and C individually based on their percentages from the pie chart:
$\text{Valid votes for B} = \frac{25}{100} \times 1,10,000 = 27,500$
$\text{Valid votes for C} = \frac{20}{100} \times 1,10,000 = 22,000$
Finally, sum these values to get the total valid votes for B and C:
$\text{Total valid votes for B and C} = 27,500 + 22,000 = 49,500$

The passage states that "mutations in a particular gene... resulted in these people developing the ability to digest dairy milk." This directly connects the digestion of dairy milk to a genetic mutation. Options A and B are incorrect because the passage specifies "some people" developed this ability, not all or none. Option C is incorrect as the passage discusses the "ability" to digest milk, not its necessity. Therefore, the inference that digestion of dairy milk in human beings resulted from a mutated gene (Option D) is directly supported by the text.

We want to find the probability of a family with three children having exactly two girls and one boy, given that the probability of having a boy (B) or a girl (G) is $P(B) = P(G) = 1/2$.

Method 1: Listing Outcomes

For three children, each birth has 2 possibilities (Boy or Girl). So, the total number of possible gender combinations is $2 \times 2 \times 2 = 2^3 = 8$.

The combinations with exactly two girls (G) and one boy (B) are: GGB, GBG, BGG. There are $3$ such favorable outcomes.

The probability is calculated as the ratio of favorable outcomes to total outcomes:
$P(\text{2 Girls, 1 Boy}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} = \frac{3}{8}$

Method 2: Binomial Probability Approach

This is a binomial probability problem where:

• Number of trials ($n$) = $3$ (three children)
• Number of successes ($k$) = $2$ (two girls)
• Probability of success ($p$) = $1/2$ (probability of having a girl)
• Probability of failure ($1-p$) = $1/2$ (probability of having a boy)

The binomial probability formula is $P(X=k) = C(n, k) \times p^k \times (1-p)^{n-k}$.

First, calculate the number of combinations $C(n, k)$:
$C(3, 2) = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{(2 \times 1)(1)} = 3$
Now, substitute these values into the formula:
$P(\text{2 Girls, 1 Boy}) = C(3, 2) \times (1/2)^2 \times (1/2)^{3-2}$
$P(\text{2 Girls, 1 Boy}) = 3 \times (1/4) \times (1/2) = \frac{3}{8}$

We need to find the annual rate of return for Mutual Fund A, let's call it $r_A$. We are given the investments made by Person 1 and Person 2 in three mutual funds (A, B, C) and the annual rate of return for funds B and C is $15\%$ or $0.15$.

First, let's calculate the profit (return) generated from funds B and C for each person:
For Person 1:
Fund B Profit: $₹20,000 \times 0.15 = ₹3,000$
Fund C Profit: $₹20,000 \times 0.15 = ₹3,000$
Total Profit from B & C (Person 1): $₹3,000 + ₹3,000 = ₹6,000$

For Person 2:
Fund B Profit: $₹15,000 \times 0.15 = ₹2,250$
Fund C Profit: $₹15,000 \times 0.15 = ₹2,250$
Total Profit from B & C (Person 2): $₹2,250 + ₹2,250 = ₹4,500$

Now, let's express the total return ($T$) for each person, which includes the profit from Fund A (using $r_A$) and the calculated profits from Funds B & C:
Person 1 Total Return ($T_1$): $T_1 = (₹10,000 \times r_A) + ₹6,000$
Person 2 Total Return ($T_2$): $T_2 = (₹20,000 \times r_A) + ₹4,500$

The problem states that Person 1's total amount is ₹500 more than Person 2's total amount: $T_1 = T_2 + ₹500$.
Substitute the expressions for $T_1$ and $T_2$ into this equation:
$(10,000 \times r_A) + 6,000 = (20,000 \times r_A) + 4,500 + 500$
Simplify the equation:
$10,000 \times r_A + 6,000 = 20,000 \times r_A + 5,000$
Rearrange the terms to solve for $r_A$:
$6,000 - 5,000 = 20,000 \times r_A - 10,000 \times r_A$
$1,000 = 10,000 \times r_A$
Solve for $r_A$:
$r_A = \frac{1,000}{10,000}$
$r_A = 0.10$
Converting this decimal to a percentage, we get:
$r_A = 0.10 \times 100\% = 10\%$
The annual rate of return for Mutual fund A is $10\%$.

To solve this problem, we need to determine the opposite faces of the dice using the provided views. In View 1, faces 5, 4, and 1 are visible. In View 2, faces 4, 6, and 3 are visible. In View 3, faces 2, 6, and 5 are visible. Since two opposite faces of a dice cannot be adjacent, we deduce the following:

• From View 1 (5, 4, 1 visible), 5 is adjacent to 4 and 1.
• From View 2 (4, 6, 3 visible), 6 is adjacent to 4 and 3. Since 1 is not visible with 6 and 4, and 6 is adjacent to 4, this implies that 6 must be opposite to 1.
• From View 3 (2, 6, 5 visible), 2 is adjacent to 6 and 5. Since 4 is not visible with 2, this implies that 2 must be opposite to 4.
  This leaves the remaining pair to be opposite: 3 is opposite to 5.
  Thus, the pairs of opposite faces are: 1 opposite 6, 2 opposite 4, and 3 opposite 5. Option A shows a net where 1 is opposite 6, 2 is opposite 4, and 3 is opposite 5, which matches our deductions.

Imagine two identical right circular cones, one placed upside down on top of the other, sharing the same circular base. This creates a symmetrical "double cone." Now, consider a plane that cuts through this combined shape. The key is that this cutting plane passes through both vertices (the pointy tops) of the cones.

Since the plane passes through both vertices ($V_1$ and $V_2$), it must contain the central axis that connects them. When a plane containing the axis cuts a cone, it slices along two "generators" (lines from the vertex to the base). In our double cone, the plane cuts the first cone along two generators ($V_1P$ and $V_1Q$) and the second cone along two generators ($V_2P$ and $V_2Q$). The common base is cut along its diameter ($PQ$).

The outer boundary of the cross-section is formed by these four segments: $V_1P$, $V_2P$, $V_1Q$, and $V_2Q$, forming a quadrilateral $V_1PV_2Q$. Because the cones are identical right circular cones, all generators have the same length, which is the slant height $L$. Therefore, $V_1P = V_1Q = V_2P = V_2Q = L$. A quadrilateral with all four sides of equal length is a rhombus. Thus, the outer boundary of the cross-section is a rhombus.

When we talk about Supply Chain Decisions, we're looking at the big-picture choices that shape how a product gets from its very beginning to the customer. These choices are strategic because they have a massive impact on costs, how efficient things are, and how happy customers are.

Let's break down the options:

• A. Location: Deciding where to put factories, warehouses, or stores is a huge supply chain decision. Your location directly influences transportation costs, how long it takes to deliver, and your access to markets.
• B. Inventory: Managing how much stock you have and where you keep it is critical. These decisions involve balancing the cost of holding inventory against the risk of running out of products, and also affect how much capital is tied up in goods.
• C. Distribution: This involves planning the entire network of how products move from suppliers all the way to customers through different channels. It directly impacts delivery speed and overall cost.

Now, let's look at D. Machine scheduling: While crucial for making production efficient, machine scheduling is typically an operational task within a factory, not a strategic, overarching supply chain decision. Although it affects capacity and lead time (which are supply chain concerns), the actual scheduling is managed at a very detailed, factory-level, rather than designing the entire network like location, inventory, or distribution do.

Therefore, Location, Inventory, and Distribution are core supply chain decisions, while machine scheduling falls under operational management.

To find the coefficient of $z^3$ in the Taylor series expansion of $\sin z$ around $z = 0$, we use the Taylor series formula: $f(z) = \sum_{n=0}^{\infty } \frac{f^{(n)}(a)}{n!}(z-a)^n$. For $f(z) = \sin z$ and $a = 0$, the expansion is $\sin z = f(0) + \frac{f'(0)}{1!}z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \dots$. We need to calculate the derivatives of $\sin z$ at $z = 0$:
$f(z) = \sin z \implies f(0) = \sin 0 = 0$
$f'(z) = \cos z \implies f'(0) = \cos 0 = 1$
$f''(z) = -\sin z \implies f''(0) = -\sin 0 = 0$
$f'''(z) = -\cos z \implies f'''(0) = -\cos 0 = -1$
Substituting these values into the series:
$\sin z = 0 + \frac{1}{1!}z + \frac{0}{2!}z^2 + \frac{-1}{3!}z^3 + \dots$
$\sin z = z - \frac{1}{3!}z^3 + \dots$
Since $3! = 3 \times 2 \times 1 = 6$, we have:
$\sin z = z - \frac{1}{6}z^3 + \dots$
Thus, the coefficient of the $z^3$ term is $-\frac{1}{6}$.

To evaluate the line integral $\oint_C \mathbf{F}(x,y) \cdot d\mathbf{l}$, we'll use Green's Theorem, which converts a line integral around a closed curve $C$ into a double integral over the region $D$ enclosed by $C$. Green's Theorem states:
$\oint_C \mathbf{F} \cdot d\mathbf{l} = \iint_D \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dx \, dy$
Given the vector field $\mathbf{F}(x, y) = (100x + 100y) \mathbf{i} + (-50x + 200y) \mathbf{j}$, we identify $M = 100x + 100y$ and $N = -50x + 200y$.

Now, let's calculate the partial derivatives:
$\frac{\partial N}{\partial x} = -50$
$\frac{\partial M}{\partial y} = 100$

Substitute these into Green's Theorem:
$\iint_D \left( -50 - 100 \right) \, dx \, dy = \iint_D (-150) \, dx \, dy$
The contour $C$ is an anticlockwise circle of radius $r = 2$. The area of this circular region $D$ is $A = \pi r^2 = \pi \cdot 2^2 = 4\pi$.
Therefore, the integral simplifies to:
$\iint_D (-150) \, dx \, dy = -150 \cdot (\text{Area of D}) = -150 \cdot 4\pi = -600\pi$
The value of the line integral $\oint_C \mathbf{F} \cdot d\mathbf{l}$ is $-600\pi$.

To find the static deflection at the free end of the cantilever beam, we use the Euler-Bernoulli beam theory. The deflection ($\delta$) of a cantilever beam at its free end due to a point load $F$ applied at a distance $a$ from the fixed end is given by the formula:
$\delta = \frac{F \cdot a^2 \cdot (3L-a)}{6EI}$
Here, the force $F$ is applied at a distance $a = 2L/3$ from the fixed end. Substituting this value of $a$ into the formula:
$\delta = \frac{F \cdot (2L/3)^2 \cdot (3L - 2L/3)}{6EI}$
Simplifying the expression:
$\delta = \frac{F \cdot (4L^2/9) \cdot (7L/3)}{6EI}$
$\delta = \frac{28FL^3}{162EI}$
Further simplification yields:
$\delta = \frac{14FL^3}{81EI}$
This is the magnitude of the static deflection at the free end of the beam.

The question investigates how the rate of heat loss ($Q$) from a current-carrying wire changes with the thickness ($t$) of its insulation sleeve. We're looking at steady-state conditions, where the heat generated in the wire is continuously dissipated.

The fundamental relationship for heat transfer is $Q = \frac{\Delta T}{R_{total}}$, where $\Delta T$ is the temperature difference between the wire surface ($T_w$) and the ambient ($T_{\infty }$), and $R_{total}$ is the total thermal resistance per unit length. Assuming $T_w - T_{\infty }$ is constant, $Q$ is inversely proportional to $R_{total}$.

The total thermal resistance ($R_{total}$) is the sum of two resistances:

1. Conductive Resistance of Insulation ($R_{cond}$): This resistance increases as the insulation thickness $t$ increases. For a cylindrical sleeve with inner radius $r_0$ (wire radius) and outer radius $r_1 = r_0 + t$, it's given by:
   $R_{cond}(t) = \frac{\ln((r_0+t)/r_0)}{2\pi k}$
   where $k$ is the thermal conductivity of the insulation.
2. Convective/Radiative Resistance from Outer Surface ($R_{conv+rad}$): This resistance decreases as $t$ increases because the outer surface area ($A = 2\pi r_1 = 2\pi(r_0+t)$) increases, allowing for more heat dissipation to the surroundings. It's given by:
   $R_{conv+rad}(t) = \frac{1}{h_{eff} A} = \frac{1}{h_{eff} 2\pi (r_0+t)}$
   where $h_{eff}$ is the effective heat transfer coefficient.

Combining these, the total resistance is:
$R_{total}(t) = \frac{\ln((r_0+t)/r_0)}{2\pi k} + \frac{1}{h_{eff} 2\pi (r_0+t)}$
To find how $R_{total}$ changes with $t$, we can analyze its derivative with respect to $r_1 = r_0+t$:
$\frac{dR_{total}}{dr_1} = \frac{1}{2\pi k r_1} - \frac{1}{h_{eff} 2\pi r_1^2}$
Setting the derivative to zero to find the minimum $R_{total}$ (and thus maximum $Q$):
$\frac{1}{k r_1} = \frac{1}{h_{eff} r_1^2} \implies r_1 = \frac{k}{h_{eff}}$
This critical radius, $r_c = k/h_{eff}$, corresponds to the insulation thickness $t_{min} = k/h_{eff} - r_0$ where $R_{total}$ is at its minimum.

• For $t < t_{min}$ (or $r_1 < k/h_{eff}$), the $R_{conv+rad}$ term dominates, and increasing $t$ decreases $R_{total}$.
• For $t > t_{min}$ (or $r_1 > k/h_{eff}$), the $R_{cond}$ term dominates, and increasing $t$ increases $R_{total}$.

Since $Q \propto 1/R_{total}(t)$:

• When $t$ increases from $0$ up to $t_{min}$, $R_{total}(t)$ decreases, causing $Q$ to increase.
• When $t$ increases beyond $t_{min}$, $R_{total}(t)$ increases, causing $Q$ to decrease.

Therefore, the rate of heat loss ($Q$) first increases with an increase in $t$, and then it decreases with further increase in $t$.

Chvorinov's Rule governs solidification time in casting, stating $t = C \left( \frac{V}{A} \right)^2$, where $t$ is solidification time, $V$ is volume, $A$ is surface area, and $C$ is a constant. For the cube, $V_{\text{cube}} = a^3$ and $A_{\text{cube}} = 6a^2$. For the cylinder with radius $r$ and height $4r$, $V_{\text{cylinder}} = \pi r^2 \times 4r = 4 \pi r^3$ and $A_{\text{cylinder}} = 2\pi r \times 4r + 2\pi r^2 = 10\pi r^2$.

Given that the solidification times are equal:
$C \left( \frac{a^3}{6a^2} \right)^2 = C \left( \frac{4\pi r^3}{10\pi r^2} \right)^2$
$\left( \frac{a}{6} \right)^2 = \left( \frac{2r}{5} \right)^2$
Taking the square root of both sides:
$\frac{a}{6} = \frac{2r}{5}$
Rearranging to find the ratio $r/a$:
$\frac{r}{a} = \frac{5}{12}$

Let's break down each deep drawing defect and its primary cause, as you'd need to know for your GATE exam.

P: Orange Peel on the surface of cup
Orange peel is a surface roughness that makes the cup look like an orange peel. This defect is directly linked to the internal structure of the material.

• Reason Match: 3 - Large grain size in the blank material. When the material has larger grains, the surface appears rougher after the deformation process of deep drawing.

Q: Wrinkling at the flange of cup
Wrinkling appears as folds or ripples, usually on the flange (the flat, untransformed part) of the cup. This happens because of compressive stresses acting on the material in this region.

• Reason Match: 1 - No blank holding force. If there isn't enough force holding the blank (sheet metal) down, the material in the flange can buckle under compression and move inwards, leading to wrinkles.

R: Tearing at the bottom corner of cup
Tearing, especially at the sharp corner where the punch starts to form the cup, means the material has stretched beyond its limit and fractured. This occurs due to very high localized stresses.

• Reason Match: 2 - Very small corner radius of the punch. A sharp punch corner creates a severe stress concentration point. This high stress can exceed the material's strength, causing it to tear at the base of the cup.

S: Earring at the top edge of the cup
Earring is when the top edge of the drawn cup forms an uneven, wavy shape, resembling "ears." This defect points to differences in the material's properties depending on the direction.

• Reason Match: 4 - Anisotropy of the blank material. Anisotropy means the material's properties (like how easily it deforms or strengthens) are not uniform in all directions. This directional variation causes uneven stretching during deep drawing, resulting in the formation of ears.

Therefore, the correct matching is P-3, Q-1, R-2, S-4.

To determine which pure metal has a hexagonal close-packed (HCP) crystal structure at room temperature, we need to recall the common crystal structures of the given metals. At room temperature, Magnesium (Mg) is known to have a hexagonal close-packed (HCP) structure. In contrast, Iron (Fe) is body-centered cubic (BCC), Aluminium (Al) is face-centered cubic (FCC), and Copper (Cu) is also face-centered cubic (FCC). Therefore, Magnesium is the correct answer.

To make 12 equal divisions on a disc using simple indexing on a milling machine, we need to determine how much to rotate the crank. The fundamental formula for simple indexing is:
$\text{Crank rotations} = \frac{\text{40}}{\text{Number of divisions required}}$
In this problem, the number of divisions required is 12. So, the total crank rotations needed for one division is:
$\text{Crank rotations} = \frac{40}{12} = \frac{10}{3} = 3 \frac{1}{3} \text{ turns}$
This means for each division, the crank needs to be rotated 3 full turns and an additional $\frac{1}{3}$ of a turn.

Now, let's check the given options to see which one provides this $\frac{1}{3}$ turn using a specific hole circle. We need to find a hole circle where $\frac{\text{number of holes to move}}{\text{number of holes on the circle}} = \frac{1}{3}$.

• Option A: "3 full rotations and 5 holes on a 15-hole circle".
  Here, the fractional part is $\frac{5}{15}$.
  $\frac{5}{15} = \frac{1}{3}$
  This matches the required $\frac{1}{3}$ turn. So, 3 full rotations and 5 holes on a 15-hole circle will give us the correct indexing for one division.

Let's briefly look at why other options are incorrect based on the fractional turn:

• Option B: "5 full rotations and 4 holes on a 16-hole circle". The fractional part is $\frac{4}{16} = \frac{1}{4}$. This is not $\frac{1}{3}$.
• Option C: "3 full rotations and 5 holes on a 18-hole circle". The fractional part is $\frac{5}{18}$. This is not $\frac{1}{3}$.
• Option D: "5 full rotations and 4 holes on a 20-hole circle". The fractional part is $\frac{4}{20} = \frac{1}{5}$. This is not $\frac{1}{3}$.

correctly provides the $\frac{1}{3}$ fractional turn needed for each of the 12 divisions.

CRAFT (Computerized Relative Allocation of Facilities Technique) optimizes facility layouts by exchanging department positions. A key constraint in CRAFT for department exchange is that departments must be either adjacent or maintain a feasible layout structure upon exchange. Let's analyze the given department pairs:

• P and Q: These are horizontally adjacent, so they are potentially exchangeable.
• R and S: These are horizontally adjacent, so they are potentially exchangeable.
• P and R: These are vertically aligned and thus adjacent, so they are potentially exchangeable.
• Q and R: These departments are not directly adjacent, neither horizontally nor vertically. Exchanging them would result in a non-contiguous or impractical layout structure, violating the basic principle of CRAFT for direct exchanges.

Therefore, the department pair that cannot be considered for exchange in CRAFT is Q and R.

Let's break down how "interchangeability" works in product design. When we talk about interchangeability, we mean that you can swap out any part of a product with another identical part without causing problems with how the product fits together or functions. Think of it like spare parts - any spare should work perfectly. This relies heavily on making parts with very precise measurements and following strict rules.

Now, let's see how each option relates to this idea:

• Standardization: This is super important for interchangeability. If components are standardized, it means they all meet the same specific requirements (like size, shape, and how they perform). This is the basic requirement to make them interchangeable.
• Simplification: When you simplify a product, you reduce the number of parts or make them less complex. Fewer, simpler parts are generally easier to standardize, which in turn makes them more interchangeable.
• Specialization: A specialized product or component is designed for a very particular job. However, even within that specialized product, the individual parts still benefit from and often need to be interchangeable for efficient manufacturing and easy repairs.
• Diversification: This is about offering a wider variety of products. When a company diversifies, it usually means creating many different products, which often require a broader range of unique components and systems. This can actually reduce the overall interchangeability across different product lines because the focus is on variety rather than commonality of parts.

Therefore, diversification is the concept that is least closely related to interchangeability.

THERBLIGS are fundamental motions in work study, and motion economy principles aim to enhance efficiency by eliminating or minimizing ineffective movements. Among the options, Search is the THERBLIG that does not directly advance the task; it represents unproductive time spent locating items. In contrast, Move, Grasp, and Preposition are typically necessary actions for interacting with or preparing objects for the next step. Therefore, applying motion economy principles, such as workplace organization and fixed locations for tools, can effectively eliminate the need for Search, thereby improving work progress.

When performing work sampling with a large number of observations, the sample size is estimated using the Normal distribution. This is because, according to the Central Limit Theorem, the distribution of sample proportions tends towards a Normal distribution as the sample size increases. This principle allows us to calculate the required sample size based on desired precision (margin of error) and confidence level, making the Normal distribution the standard for large-sample work sampling.

Linear Programming (LP) problems rely on certain core assumptions for their formulation and solution. The standard assumptions for an LP problem are:

• Certainty: All parameters (like profit per unit, resource availability, resource consumption per unit) are known precisely and remain constant.
• Proportionality: The contribution of each decision variable to the objective function and resource consumption is directly proportional to its level. For example, producing $2$ units requires exactly twice the resources and yields twice the profit as $1$ unit.
• Additivity: The total objective function value and total resource consumption are the sum of individual contributions from each decision variable. This implies no interaction effects.
• Divisibility: Decision variables can take any non-negative real (fractional) values. You can produce $2.5$ units of a product.

The question asks which option is NOT an assumption of a linear programming problem.

• Proportionality, Additivity, and Certainty are all fundamental assumptions of standard Linear Programming.
• Integrality is the assumption that decision variables must take on integer values. This is a characteristic of Integer Programming (IP) or Mixed-Integer Programming (MIP), which are extensions of LP, but not an assumption of standard Linear Programming itself. Standard LP assumes Divisibility, allowing fractional solutions.

Therefore, Integrality is not an assumption of a general linear programming problem.

In a single server Markovian queuing system, if customer arrivals follow a Poisson distribution, the time between consecutive arrivals (inter-arrival time) follows a specific probability distribution. A Poisson distribution describes the number of events (arrivals) in a fixed time interval, occurring randomly and independently at a constant average rate. A fundamental property of Poisson processes is that when the number of arrivals follows a Poisson distribution, the inter-arrival times are independent and identically distributed according to the Exponential distribution. Therefore, the inter-arrival time follows the Exponential distribution.

The Simple Exponential Smoothing (SES) method requires the least amount of historical data. SES primarily uses the most recent observation and the previous forecast to generate a new forecast, effectively needing only the last forecast value and the last actual value. In contrast, Econometric forecasting methods require significant historical data for estimating multiple variable relationships, Linear Regression methods need sufficient data points to establish reliable statistical relationships, and ARIMA methods require a substantial amount of historical time series data to identify model orders ($p, d, q$) and estimate coefficients. Therefore, for situations with limited data, SES is the most data-efficient option among those listed.

Total Productive Maintenance (TPM) is a strategy focused on maximizing equipment effectiveness by engaging all employees, particularly operators, in maintenance. A core principle of TPM is proactive and preventive action to eliminate breakdowns. Option A is true because 'Autonomous Maintenance' in TPM involves operators performing routine checks and minor adjustments, which are preventive tasks. Option B is false; reactive maintenance, which involves fixing equipment after it breaks down, is contrary to TPM's goal of preventing failures. TPM aims to prevent breakdowns, not just react to them. Option C is true as TPM ensures equipment reliability, which is crucial for the smooth flow required by Just-in-Time (JIT) systems. Option D is also true, as TPM is a fundamental part of Lean manufacturing, contributing to waste reduction by minimizing downtime and defects. Therefore, the statement that is not true about TPM is that it allows operators to perform reactive maintenance.

We are given a complex function $f(x, y) = u(x, y) + i v(x, y)$ that is analytic.

For $f(x, y)$ to be analytic, it must satisfy the Cauchy-Riemann equations:

1. $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$
2. $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

Additionally, if a function is analytic, its real part $u(x, y)$ and imaginary part $v(x, y)$ are harmonic functions. Harmonic functions satisfy Laplace's equation: $\frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2} = 0$.

Let's evaluate each option:

A. $\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0$
Since $u(x,y)$ is the real part of an analytic function, it must be harmonic. This equation represents Laplace's equation for $u(x,y)$. Therefore, this statement is TRUE.

B. $\frac{\partial^2v}{\partial x^2} + \frac{\partial^2v}{\partial y^2} = 0$
Similarly, since $v(x,y)$ is the imaginary part of an analytic function, it must be harmonic. This equation represents Laplace's equation for $v(x,y)$. Therefore, this statement is TRUE.

C. $\frac{\partial^2u}{\partial x^2} + \frac{\partial^2v}{\partial y^2} = 0$
Using the Cauchy-Riemann equation $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$, differentiating with respect to $y$ gives $\frac{\partial^2 u}{\partial y \partial x} = \frac{\partial^2 v}{\partial y^2}$. This does not directly imply the given equation.
Consider $f(z) = z^2 = (x+iy)^2 = x^2 - y^2 + i(2xy)$. Here, $u = x^2 - y^2$ and $v = 2xy$.
Then $\frac{\partial^2u}{\partial x^2} = 2$ and $\frac{\partial^2v}{\partial y^2} = 0$.
Their sum is $2 + 0 = 2 \neq 0$. Therefore, this statement is FALSE.

D. $\left(\frac{\partial u}{\partial x}\right) \left(\frac{\partial v}{\partial x}\right) + \left(\frac{\partial u}{\partial y}\right) \left(\frac{\partial v}{\partial y}\right) = 0$
Substitute the Cauchy-Riemann equations:
$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$
$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$
Substituting these into the given expression:
$\left(\frac{\partial v}{\partial y}\right) \left(\frac{\partial v}{\partial x}\right) + \left(-\frac{\partial v}{\partial x}\right) \left(\frac{\partial v}{\partial y}\right)$
$= \frac{\partial v}{\partial y} \frac{\partial v}{\partial x} - \frac{\partial v}{\partial x} \frac{\partial v}{\partial y}$
$= 0$
Therefore, this statement is TRUE.

The true equations are A, B, and D.

This question tests our understanding of mild steel's behavior under uniaxial tension, a fundamental concept for GATE. Let's break down each option:

Option A: The engineering stress-strain curve is linear within the elastic limit.
While the initial part of the curve (up to the proportional limit) is indeed linear, there can be slight non-linearity between the proportional limit and the yield point, even though this region is still elastic. So, stating it's always linear throughout the entire elastic limit isn't strictly true in all precise contexts.

Option B: The specimen fails in cup and cone type fracture.
Mild steel is a classic example of a ductile material. When ductile materials undergo necking (a localized reduction in cross-sectional area), they typically fracture in a "cup and cone" manner. This means one part of the fracture surface forms a cup, and the other forms a cone. This statement is TRUE.

Option C: The true stress is always more than the engineering stress at any finite strain.
Engineering stress ($\sigma_e$) is calculated using the original cross-sectional area ($A_0$), while true stress ($\sigma_t$) uses the instantaneous area ($A$).
$\sigma_t = \frac{P}{A} \quad \text{and} \quad \sigma_e = \frac{P}{A_0}$
During tensile testing, especially beyond the elastic limit, the cross-sectional area reduces ($A < A_0$). Therefore, for the same load $P$, $\frac{P}{A} > \frac{P}{A_0}$, implying $\sigma_t > \sigma_e$. This holds true for any finite strain where the area has reduced. This statement is TRUE.

Option D: The specimen does not regain its original dimensions after complete unloading from an initial stress above the yield stress.
The yield stress signifies the onset of plastic (permanent) deformation. If the material is loaded beyond this point, some of the deformation is plastic and non-recoverable. Upon unloading, only the elastic part of the deformation is recovered, leaving a permanent set. Thus, the specimen will not return to its original dimensions. This statement is TRUE.

Therefore, options B, C, and D are true statements describing the behavior of mild steel under uniaxial tensile load.

Let's break down each statement about Friction Stir Welding (FSW):

Statement A: "It can be used to produce lap, butt and tee joints."
Explanation: This is TRUE. FSW is a highly versatile solid-state welding process. It's capable of creating various joint types, including lap, butt, and tee joints, making it very adaptable for different applications.

Statement B: "A non-consumable rotating tool with shoulder and pin is used to melt the work-piece material."
Explanation: This is FALSE. While FSW does use a non-consumable rotating tool with a shoulder and pin, it is a solid-state joining process. This means it joins materials through plastic deformation, not by melting them. The friction generates heat that softens the material, allowing it to flow and mix mechanically.

Statement C: "Retreating side of the weld is where the linear velocity vector at a point on that side of the rotating tool and the welding direction are opposite."
Explanation: This is TRUE. Consider the rotating tool moving along the weld line. On the retreating side, the tangential velocity of the tool's surface (due to rotation) is in the direction opposite to the overall welding direction. This opposite motion helps in moving the plasticized material around the pin.

Statement D: "Advancing side of the weld is where the linear velocity vector at a point on that side of the rotating tool and the welding direction are opposite."
Explanation: This is FALSE. On the advancing side, the tangential velocity of the tool's surface is actually in the same direction as the welding direction. The condition where the velocity vector opposes the welding direction is what defines the retreating side, as explained in Statement C.

For a continuous random variable, the total probability over its entire range must be 1. This means the area under its Probability Density Function (PDF) curve must equal 1. We are given

.
To find $K$, we set the integral of the PDF over its domain equal to 1:
$\int_{-\infty }^{\infty } f(x) dx = 1$
Since $f(x)$ is non-zero only for $0 \le x \le 1$, the integral becomes:
$\int_{0}^{1} \frac{K}{4} dx = 1$
$\frac{K}{4} \int_{0}^{1} 1 dx = 1$
$\frac{K}{4} [x]_{0}^{1} = 1$
$\frac{K}{4} (1 - 0) = 1$
$\frac{K}{4} = 1$
Solving for $K$, we get:
$K = 4$
Thus, the value of $K$ is 4.

To solve $\lim_{x \to 1} \left(\frac{x^2 - 2ax + b}{x - 1}\right) = 8$, observe that as $x \to 1$, the denominator $(x-1)$ approaches $0$. For the limit to exist as a finite value (8), the numerator must also approach $0$, indicating a $\frac{0}{0}$ indeterminate form. Thus, setting the numerator to zero at $x=1$ gives $1^2 - 2a(1) + b = 0 \implies 1 - 2a + b = 0$, which means $b = 2a - 1$.

Now, since we have a $\frac{0}{0}$ indeterminate form, we can apply L'Hôpital's Rule. Differentiating the numerator and the denominator separately with respect to $x$:
$\lim_{x \to 1} \left(\frac{\frac{d}{dx}(x^2 - 2ax + b)}{\frac{d}{dx}(x - 1)}\right) = \lim_{x \to 1} \left(\frac{2x - 2a}{1}\right)$
Substituting $x=1$ into the differentiated expression and equating it to the given limit value:
$2(1) - 2a = 8$
$2 - 2a = 8$
$-2a = 6$
$a = -3$
Using the relation $b = 2a - 1$, we find $b = 2(-3) - 1 = -6 - 1 = -7$.
Finally, calculate $a - b = (-3) - (-7) = -3 + 7 = 4$.

Here's how to find the maximum force $F$ before the truss collapses:

1. Understand the string's behavior: Member AC is an inextensible string. This means it can only pull (be in tension, $T_{AC} \ge 0$). If it tries to push (go into compression), it will simply slacken, and the truss will lose its square shape and stability.

2. Analyze forces at Joint B:
   For horizontal equilibrium (\sum F_x = 0$) at joint B:$F_{BC} = -100 \text{ kN}$(This means member BC is in compression, pushing *away* from B with$100 \text{ kN}$, or pushing *towards* C with$100 \text{ kN}$).

3. Analyze horizontal forces at Joint C:
   The external force $F$ is at $60^\circ$. Its horizontal component is $F_x = F \cos 60^\circ = 0.5F$ (acting left).
   The force from member BC (compression) pushes Joint C to the right with $100 \text{ kN}$.
   The tension in string AC, $T_{AC}$, acts along the diagonal at $45^\circ$. Its horizontal component is $T_{AC} \cos 45^\circ$ (acting left).

   Applying horizontal equilibrium (\sum F_x = 0$) at Joint C:$100 - 0.5F - T_{AC} \cos 45^\circ = 0$

4. Determine the limiting condition for $F$:
   Rearrange the equation to solve for the term involving $T_{AC}$:
   $T_{AC} \cos 45^\circ = 100 - 0.5F$
   Since a string can only be in tension ($T_{AC} \ge 0$), and $\cos 45^\circ$ is positive, the right side of the equation must also be non-negative:
   $100 - 0.5F \ge 0$
   $100 \ge 0.5F$
   $F \le \frac{100}{0.5}$
   $F \le 200 \text{ kN}$

   This means the maximum force $F$ for which the string AC remains in tension (or just on the verge of slackening, i.e., $T_{AC}=0$) is $200 \text{ kN}$. If $F$ exceeds this value, the string goes slack, and the truss becomes unstable.
   The maximum value of force $F$ is $200.00 \text{ kN}$.

To find the stroke length of the slider, we need to determine its extreme positions.

1. First, identify the given dimensions: Crank length ($r$) is $l = 10\text{ cm}$. Connecting rod length ($L$) is $3l = 30\text{ cm}$. The offset ($e$) is $l = 10\text{ cm}$.

2. Next, calculate the horizontal position of the slider at its two dead centers. The stroke length ($S$) is the distance between these extreme points.

   • Outer Dead Center (Position 1): The crank and connecting rod are fully extended. The total length from the crank axis to the slider pin is $(L + r) = 30 + 10 = 40\text{ cm}$. Using the Pythagorean theorem to find the horizontal distance $x_1$ from the crank center to the slider:
     $x_1 = \sqrt{(L+r)^2 - e^2} = \sqrt{40^2 - 10^2} = \sqrt{1600 - 100} = \sqrt{1500} \approx 38.73\text{ cm}$
   • Inner Dead Center (Position 2): The crank and connecting rod are overlapping. The distance between their ends is $(L - r) = 30 - 10 = 20\text{ cm}$. Similarly, the horizontal distance $x_2$ from the crank center to the slider is:
     $x_2 = \sqrt{(L-r)^2 - e^2} = \sqrt{20^2 - 10^2} = \sqrt{400 - 100} = \sqrt{300} \approx 17.32\text{ cm}$

3. Finally, calculate the stroke length ($S$) as the difference between these two extreme horizontal positions:
   $S = x_1 - x_2 = 38.73 - 17.32 = 21.41\text{ cm}$
   Rounding off to one decimal place, the stroke length of the slider is $21.4\text{ cm}$.