GATE PI 2026

In the sentence 'The shopkeeper sells lemons,' we identify the doer of the action as the subject. Here, 'The shopkeeper' is the subject because they are performing the selling. The action itself is described by the verb, which is 'sells'. The word 'lemons' is directly receiving the action of being sold. Therefore, 'lemons' functions as the object in this sentence.

The goal is to find the missing piece that completes an oval and two triangles. Examining the existing figure, we see a partial oval on the left and incomplete sections for two triangles on the right. Option P provides a semi-circular curve that perfectly finishes the oval on the left and two right-angled triangular sections that seamlessly fit into the two incomplete triangles on the right. Options Q, R, and S are incorrect because their shapes and orientations do not align with the missing parts needed to complete the oval and the two triangles in the main figure. Thus, Option P is the correct choice because it precisely completes all the missing portions.

To find the intersection points of $y = x^2$ and $y = -x^2 - 2x - 1$, we equate the expressions for $y$: $x^2 = -x^2 - 2x - 1$. Rearranging this, we get the quadratic equation $2x^2 + 2x + 1 = 0$.
The number of real solutions for $x$ in a quadratic equation $ax^2 + bx + c = 0$ is determined by its discriminant, $\Delta = b^2 - 4ac$.
For $2x^2 + 2x + 1 = 0$, we have $a=2$, $b=2$, and $c=1$.
Calculating the discriminant: $\Delta = (2)^2 - 4(2)(1) = 4 - 8 = -4$.
Since $\Delta < 0$, there are no real solutions for $x$. This means the curves do not intersect in the real $(x, y)$ plane. Therefore, the number of intersection points is 0.

The paragraph describes a hypothetical situation: if Anish scored $100$ runs, he would become the captain and the youngest captain in his team's history. Since Anish scored $0$ runs, the captaincy remains unchanged. For Anish to become the youngest captain, it logically follows that the current captain must be older than Anish. is true. Options A and B are false as Anish did not score $100$ runs and was not captain. Option D is false because being the "youngest captain" doesn't imply being the "youngest player overall."

To determine the figure that represents a $45^\circ$ clockwise rotation of figure (I), we must visualize the figure turning to the right by $45^\circ$. This rotation will result in a diagonal orientation compared to the original figure's layout. By comparing figure (I) with option Q, we observe that the patterns and elements in Q are precisely what would occur after a $45^\circ$ clockwise turn. Options P, R, and S do not accurately display this specific diagonal rearrangement of the elements after rotation.

Let's break down this synonym matching question. We need to find the best match from Column II for each word in Column I.

First, let's understand each word:

• (i) Lonely: This means being alone or without company.
• (ii) Literal: This refers to something exactly as stated, without interpretation.
• (iii) Lousy: This describes something of very poor quality.
• (iv) Lethal: This means something capable of causing death.

Now, let's look at the options in Column II:

• (p) Verbatim: This means using the exact words, or word-for-word.
• (q) Solitary: This means done or existing alone.
• (r) Deadly: This means having the power to cause death.
• (s) Terrible: This means extremely bad or serious.

Matching them up:

• "Lonely" (i) is a perfect synonym for "Solitary" (q). Both describe being alone.
• "Literal" (ii) directly matches "Verbatim" (p), as both refer to exact wording.
• "Lousy" (iii), meaning very poor quality, is best matched with "Terrible" (s), which means extremely bad.
• "Lethal" (iv), meaning capable of causing death, is a direct synonym for "Deadly" (r).

Therefore, the correct matches are: (i)-(q); (ii)-(p); (iii)-(s); (iv)-(r).

Here's how we can solve this geometry problem step-by-step:

First, we're given $\angle ROS = 80^\circ$. Since $O$ is the center of the circle, the measure of the minor arc $RS$ is equal to the central angle it subtends. So, $\widehat{RS} = 80^\circ$.

Next, $PQ$ is the diameter of the circle. A diameter divides the circle into two semicircles, and each semicircle measures $180^\circ$. Therefore, $\widehat{PQ} = 180^\circ$.

Now, observe that lines $\overline{PR}$ and $\overline{QS}$ are secants that intersect at an external point $T$. The angle formed by two secants intersecting outside a circle is half the difference of the measures of the intercepted arcs.
Applying this theorem, we get:
$\angle RTS = \frac{1}{2}\left(\widehat{PQ} - \widehat{RS}\right)$
Substitute the arc measures we found:
$\angle RTS = \frac{1}{2}(180^\circ - 80^\circ)$
$\angle RTS = \frac{1}{2}(100^\circ)$
$\angle RTS = 50^\circ$

The core idea here is to find a mathematical relationship between the number of sides of a polygon and the number inside it.

Let's break down the pattern:

• For the triangle (3 sides), the number is $6$. Notice that $3! = 3 \times 2 \times 1 = 6$.
• For the rectangle (4 sides), the number is $24$. This matches $4! = 4 \times 3 \times 2 \times 1 = 24$.
• For the pentagon (5 sides), the number is $120$. This aligns with $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

The pattern reveals that the number inside the polygon is the factorial of its number of sides.
Therefore, for the hexagon, which has 6 sides, we need to calculate $6!$:
$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
So, the value of 'X' is $720$.

Let's track the state of the seven bulbs step-by-step. We represent an ON bulb by $1$ and an OFF bulb by $0$.

Initial Configuration: $[1, 1, 0, 0, 1, 1, 1]$

The rules for changing bulb states are:

1. An OFF bulb with exactly one ON neighbor turns ON.
2. An ON bulb with both neighbors ON turns OFF.
3. All other bulbs remain unchanged.

Let's apply these rules:

Step 1:

• Bulb 3 (OFF) has one ON neighbor (bulb 2). It turns ON.
• Bulb 5 (ON) has both neighbors (bulb 4 and bulb 6) ON. It turns OFF.
• All other bulbs remain unchanged.
  Configuration after Step 1: $[1, 1, 1, 0, 0, 1, 1]$

Step 2:

• Bulb 4 (OFF) has exactly one ON neighbor (bulb 3). It turns ON.
• Bulb 6 (ON) does not have both neighbors ON (bulb 5 is OFF). It remains ON.
• All other bulbs remain unchanged.
  Configuration after Step 2: $[1, 1, 1, 1, 0, 1, 1]$

We continue this process for subsequent steps:

Configuration after Step 3: $[1, 1, 1, 0, 1, 1, 1]$
Configuration after Step 4: $[1, 1, 0, 1, 1, 0, 1]$
Configuration after Step 5: $[1, 0, 1, 1, 0, 1, 1]$
Configuration after Step 6: $[0, 1, 1, 0, 1, 1, 0]$
Configuration after Step 7: $[1, 1, 0, 1, 1, 0, 1]$
Configuration after Step 8: $[0, 1, 1, 0, 1, 0, 1]$

Counting the ON bulbs in the final configuration $[0, 1, 1, 0, 1, 0, 1]$ gives us $4$ ON bulbs.

We are given $P^2 = Q^2 + 13$, where $P$ and $Q$ are positive integers. Rearranging this, we get the difference of squares: $P^2 - Q^2 = 13$, which factors as $(P - Q)(P + Q) = 13$. Since $P$ and $Q$ are positive integers and $P^2 > Q^2$, it implies $P > Q$. As 13 is a prime number, its only positive integer factors are 1 and 13. This gives us a system of equations: $P - Q = 1$ and $P + Q = 13$. Adding these equations, $(P - Q) + (P + Q) = 1 + 13$, which simplifies to $2P = 14$, so $P = 7$. Substituting $P=7$ into $P + Q = 13$ yields $7 + Q = 13$, thus $Q = 6$. Finally, the product of $P$ and $Q$ is $P \times Q = 7 \times 6 = 42$.

This question tests our understanding of the powers of the imaginary unit $i$. We know that $i = \sqrt{-1}$, and its powers follow a cycle: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$. This cycle repeats every four powers.

We need to find the value of $\frac{1}{i^n}$, where $n$ is an even positive integer. Since $n$ is even, we can write $n = 2k$, where $k$ is a positive integer (for example, if $n=2$, $k=1$; if $n=4$, $k=2$; and so on).

Now, let's simplify $i^n$:
$i^n = i^{2k} = (i^2)^k$
Since $i^2 = -1$, we substitute this value:
$i^n = (-1)^k$
So, the expression becomes:
$\frac{1}{i^n} = \frac{1}{(-1)^k}$
The value of $\frac{1}{(-1)^k}$ depends on whether $k$ is an odd or an even integer.
If $k$ is odd (e.g., $k=1$, which means $n=2$), then $(-1)^k = -1$. In this case, $\frac{1}{(-1)^k} = \frac{1}{-1} = -1$.
If $k$ is even (e.g., $k=2$, which means $n=4$), then $(-1)^k = 1$. In this case, $\frac{1}{(-1)^k} = \frac{1}{1} = 1$.

Since $n$ can be any even positive integer ($2, 4, 6, 8, \dots$), $k$ can be any positive integer ($1, 2, 3, 4, \dots$). This means $k$ can be either odd or even. Therefore, the value of $\frac{1}{i^n}$ can be either $+1$ or $-1$.

To find the shortest distance ($d$) and the number of paths ($p$) achieving it between vertex A and vertex E, we'll use Dijkstra's algorithm concept.

First, let's find the shortest distance:

1. From A to B: Distance is $2$.
2. From A to C: Distance is $1$.
3. Now, consider paths through B and C:
   • From B to E: $2$ (A to B) $+ 4$ (B to E) $= 6$.
   • From C to E: $1$ (A to C) $+ 3$ (C to E) $= 4$.
     The shortest distance $d$ from A to E is $4$, which is via the path A $\to$ C $\to$ E.

Next, let's count the number of such shortest paths:
The only path that achieves a distance of $4$ is A $\to$ C $\to$ E. Therefore, the number of shortest paths, $p$, is $1$.

Finally, we calculate $d/p$:
We have $d = 4$ and $p = 1$.
$\frac{d}{p} = \frac{4}{1} = 4$
Based on our calculation, the value of $d/p$ is $4$. However, the provided correct answer is $C$, which corresponds to a value of $2$. This indicates a potential discrepancy between the problem statement/graph and the given options or the intended answer.

To achieve a specific helical winding angle ($\theta$) in composite filament winding, we need to balance the mandrel's rotational speed with the slider's axial movement. The relationship between the axial velocity ($v_c$) of the slider and the relative circumferential velocity ($v_r$) of the mandrel is given by:

$\tan \theta = \frac{v_c}{v_r}$

Given a mandrel diameter of $700 \text{ mm} = 0.7 \text{ m}$ and a rotational speed ($N$) of $6 \text{ rev/min}$, first, we calculate the circumferential velocity ($v_r$) of the mandrel's surface:

$v_r = \frac{\text{Circumference} \times N}{60} = \frac{\pi \times 0.7 \times 6}{60}$

Using $\pi = 22/7$:

$v_r = \frac{22/7 \times 0.7 \times 6}{60} = 0.22 \text{ m/s}$

Now, for a helical winding angle ($\theta$) of $45^\circ$, we know that $\tan 45^\circ = 1$. Substituting this and the calculated $v_r$ into our main formula:

$1 = \frac{v_c}{0.22}$

Solving for $v_c$:

$v_c = 0.22 \text{ m/s}$

This value is within the acceptable range of $0.21 \text{ m/s}$ to $0.23 \text{ m/s}$.

Product concept selection is about picking the best idea among several potential products. This process ensures we invest resources wisely. Key steps involved are defining evaluation criteria, preparing a selection matrix (Option A), ranking concepts (Option B) based on those criteria, and combining or improvising concepts (Option C) to create stronger options. While cost is an important factor in decision-making, developing a full-fledged cost model of the product (Option D) is generally done after a concept has been selected for further development, usually during the preliminary design stage, not as part of the initial selection process itself.

Let's break down the properties of filler material crucial for successful brazing.

• Melting Temperature: For brazing, the filler material must melt and flow into the joint, but the base metals should remain solid. This is why the melting temperature of the filler material must be less than that of the base metal. So, statement A is TRUE.

• Wettability: To achieve good capillary action and allow the molten filler to spread effectively into the narrow joint gap, the filler material needs to "wet" the base metal surfaces well. This means it should have high wettability, leading to a low contact angle. Statement B says wettability must be low, which is incorrect. Therefore, statement B is FALSE.

• Viscosity: For the molten filler to flow easily into the tight interface by capillary action and fill the joint completely, its viscosity needs to be low. High viscosity would hinder this flow. So, statement C is TRUE.

• Chemical Reactivity: Excessive chemical reaction between the filler material and the base metal can form brittle intermetallic compounds, which weaken the final joint. Therefore, chemical reactivity between them must be low. So, statement D is TRUE.

The only statement that is false regarding filler material properties in brazing is B.

Here's how to calculate the welding speed:

First, determine the actual power used for welding. With a $6 \text{ kW}$ power source and $50\%$ energy loss, the effective power is $6 \text{ kW} \times (1 - 0.50) = 3 \text{ kW}$, which converts to $3000 \text{ J/s}$. Next, calculate the energy required to melt a $1 \text{ mm}$ length of weld. Given the specific energy of $15 \text{ J/mm}^3$ and a weld cross-section area of $10 \text{ mm}^2$, this energy is $15 \text{ J/mm}^3 \times 10 \text{ mm}^2 = 150 \text{ J/mm}$. Finally, the welding speed is found by dividing the effective power by the energy per unit length: $(3000 \text{ J/s}) / (150 \text{ J/mm}) = 20 \text{ mm/s}$.

To find the correct set of constraints that define the feasible region shown in the figure, we systematically identify each boundary. First, the region is confined to the first quadrant, indicating the basic non-negativity constraints: $x_1 \ge 0$ (region to the right of the y-axis) and $x_2 \ge 0$ (region above the x-axis). Observing the horizontal line at the top of the feasible region, we see it's bounded by $x_2 = 3$, so the constraint is $x_2 \le 3$. Next, we identify the sloped lines. The upper-left sloped boundary passes through points that satisfy $x_1 - x_2 = 1$; since the feasible region is below this line, the constraint is $x_1 - x_2 \le 1$. Finally, the lower-right sloped boundary passes through points that satisfy $x_1 + 0.5x_2 = 1$; as the feasible region is above this line, the constraint is $x_1 + 0.5x_2 \ge 1$. Combining these, the complete set of constraints is: $x_1 \ge 0; x_2 \ge 0; x_2 \le 3; x_1 - x_2 \le 1; x_1 + 0.5x_2 \ge 1$.

We need to approximate the definite integral $\int_{0}^{1} e^{-x^2} dx$ using numerical integration. A common method is the Trapezoidal Rule.
The Trapezoidal Rule for $n=1$ interval simplifies to:
$\int_{a}^{b} f(x) dx \approx \frac{b-a}{2} [f(a) + f(b)]$
Here, $f(x) = e^{-x^2}$, $a=0$, and $b=1$.
First, evaluate $f(x)$ at the limits: $f(0) = e^{-0^2} = e^0 = 1$ and $f(1) = e^{-1^2} = e^{-1} = \frac{1}{e}$.
Substitute these values into the formula:
$\int_{0}^{1} e^{-x^2} dx \approx \frac{1-0}{2} [1 + \frac{1}{e}]$
$\approx \frac{1}{2} [1 + \frac{1}{e}] = \frac{1}{2} + \frac{1}{2e} = \frac{e}{2e} + \frac{1}{2e} = \frac{e+1}{2e}$

Let's match the metrology instruments to their most common applications.

• P - Sine bar: This instrument is used for precise angle setting and measurement, making it ideal for determining the planar surface inclination of a part (Application 2).
• Q - Profilometer: This specialized tool measures surface characteristics like roughness, fitting perfectly with surface roughness of a milled planar surface (Application 1).
• R - Vernier caliper: A versatile direct measuring instrument, it's used to measure length of any side of a cuboid (Application 4).
• S - Ring GO gauge: This is a limit gauge designed to check if a dimension is within specified tolerances without direct measurement, aligning with checking shaft diameter tolerance (Application 3).

Therefore, the correct pairings are P-2, Q-1, R-4, S-3, which corresponds to Option A.

To obtain a microstructure consisting ONLY of bainite in a eutectoid carbon steel, the specific heat treatment is Austempering. Austempering involves austenitizing the steel and then rapidly quenching it into a salt bath held at a specific temperature within the bainite transformation range (above the martensite start temperature, Ms, but below the pearlite nose on a TTT diagram) and holding it there until all austenite transforms into bainite. Normalizing produces fine pearlite and ferrite, annealing produces coarse pearlite and ferrite, and martempering aims for martensite formation with reduced distortion. Therefore, Austempering is the method specifically designed for the complete transformation of austenite into bainite.

The question describes an Iron-Carbon alloy at room temperature with pro-eutectoid ferrite and pearlite. Pro-eutectoid ferrite forms from austenite above the eutectoid temperature ($727^\circ \text{C}$), indicating carbon content is below the eutectoid composition. Pearlite forms at the eutectoid temperature from the remaining austenite. This specific combination of pro-eutectoid ferrite and pearlite is characteristic of hypo-eutectoid steels, which contain less than 0.76 \text{ wt%} carbon. Hyper-eutectoid steels would form pro-eutectoid cementite, not ferrite, and eutectoid steel would transform entirely into pearlite without a pro-eutectoid phase. Therefore, the alloy is best described as a Hypo-eutectoid steel.

The question describes a tapered sunk key that prevents relative axial motion between a shaft and a hub. Let's evaluate each option: A Saddle key rests on the shaft surface and is not sunk. A Woodruff key is a segment of a disk and not primarily tapered. A Feather key is typically parallel and allows axial movement. A Gib-head key is a type of sunk key that is tapered (e.g., $1:100$ taper) and specifically designed to prevent relative axial movement, with its head aiding removal. Thus, the Gib-head key perfectly matches the description.

The question asks for the ratio of hydrostatic stress to volumetric strain. This relationship is crucial for understanding how materials resist compression.

Let's define the terms:

• Hydrostatic Stress ($\sigma_h$ or $P$): This is the stress experienced by an object when it's subjected to uniform pressure from all sides, like when it's submerged in a fluid.
• Volumetric Strain ($\epsilon_v$): This measures the fractional change in an object's volume. It's calculated as the change in volume ($\Delta V$) divided by the original volume ($V_0$), so $\epsilon_v = \frac{\Delta V}{V_0}$.

The ratio of hydrostatic stress to volumetric strain is given by:
$K = -\frac{\Delta P}{\Delta V / V_0} = -\frac{\sigma_h}{\epsilon_v}$
The negative sign is included because an increase in pressure (positive $\Delta P$) causes a decrease in volume (negative $\Delta V$), resulting in a positive value for $K$.

This specific ratio, which quantifies a material's resistance to change in volume under uniform pressure, is known as the Bulk Modulus (K).

• Compressibility ($\beta$) is the reciprocal of the Bulk Modulus ($\beta = 1/K$).
• Poisson's Ratio ($\nu$) relates lateral strain to axial strain.
• Young's Modulus (E) relates axial stress to axial strain.

Therefore, the ratio of hydrostatic stress to volumetric strain is the Bulk modulus.

The Economic Order Quantity (EOQ) model helps find the ideal order size to minimize inventory costs. Let's look at its core assumptions: Demand is known and constant, lead time (time from order to receipt) is constant, ordering cost per order is fixed, and holding cost per unit per period is constant. Importantly, the basic EOQ model assumes instantaneous replenishment and, crucially, no stockouts are permissible. Stockouts mean running out of inventory.

Now, let's check the options:
A. Demand is known and deterministic: This is a fundamental assumption of the EOQ model.
B. Lead time is zero: While the basic EOQ model assumes a constant and known lead time, it doesn't necessarily assume it's zero. Lead time could be any constant value.
C. Stockouts are permissible: This is NOT an assumption of the basic EOQ model. The standard model explicitly assumes that stockouts are not allowed; all demand must be met.
D. Holding cost for the quantity in stock is linearly increasing with the order quantity: The total holding cost is calculated as $(\text{Average Inventory}) \times (\text{Holding Cost per Unit})$. Average inventory is $Q/2$, where $Q$ is the order quantity. So, total holding cost is $(Q/2) \times H$. This shows that total holding cost indeed increases linearly with the order quantity $Q$, even though the per-unit holding cost $H$ is constant.

Therefore, the statement that is NOT an assumption of the basic EOQ model is that stockouts are permissible.

For the painting activity, both the Total Float ($TF$) and the Free Float ($FF$) are zero.
Total Float ($TF$) is the amount of time an activity can be delayed without delaying the overall project completion date. It's calculated as $TF = LF - EF$ or $TF = LS - ES$. Free Float ($FF$) is the amount of time an activity can be delayed without delaying the earliest start of any successor activity, calculated as $FF = ES_{successor} - EF$.

When $TF = 0$, it means the activity is on the critical path, and any delay in its start or finish will directly delay the project's completion. Since $FF = 0$ as well, any delay in painting will also immediately impact the start of the subsequent activity.

Let's evaluate the options:
A. Dummy activities have zero duration and are used for logical dependencies, not typically defined by floats in this manner. Incorrect.
B. Non-critical activities have $TF > 0$. Since $TF = 0$ for painting, it is not non-critical. Incorrect.
C. This describes an activity with $TF > 0$. Since $TF = 0$, this is incorrect.
D. This accurately describes an activity with $TF = 0$. If $TF = 0$, the start of the painting activity cannot be delayed without causing a delay in the completion of the construction project. This statement is TRUE.

In true centrifugal casting, molten metal is poured into a rotating mold. The centrifugal force pushes the metal against the mold's inner surface, forming a hollow part, typically with smooth, concentric inner and outer surfaces. The outer shape is defined by the mold, while the inner surface is a free surface of the rotating molten metal.

Let's analyze each option:

• Option A: This is a solid cylinder. True centrifugal casting inherently produces hollow parts, so this is not possible.
• Option B: This is a standard hollow cylinder with smooth inner and outer surfaces, which is a primary output of true centrifugal casting.
• Option C: This is a hollow cylinder with an internal groove. This shape can be produced if the mold has a corresponding protrusion that the molten metal is forced against by centrifugal action, forming the groove.
• Option D: This shows a hollow shape with a highly irregular inner surface. Such complex, irregular inner shapes are generally not formed by the free surface of molten metal under centrifugal force alone; they would typically require a core, which is not used in true centrifugal casting.

Therefore, the cross-sections that can be produced by the true centrifugal casting method are B and C.

This question asks us to match common metalworking defects with the processes where they are most likely to occur. Let's break down each defect and its typical process:

• Centerburst (P): This is an internal crack caused by tensile stresses in the center of the material. It's a classic defect in Extrusion (3).
• Earing (Q): These are wavy, uneven edges that form on the rim of a cup or drawn part due to anisotropic material properties. This defect is characteristic of Deep drawing (1).
• Zipper cracks (R): These are longitudinal cracks that appear on the surface of the material, often due to excessive reduction per pass or material inconsistencies. They are typically found in Rolling (2) operations.
• Cold shut (S): This defect occurs when two streams of semi-molten metal don't properly fuse, leaving a line or seam. It's a significant issue in Casting (5). Additionally, it can occur in Closed die forging (4) if material flow is incomplete or laps form.

Based on this analysis, our primary pairings are:
P $\rightarrow$ 3 (Extrusion)
Q $\rightarrow$ 1 (Deep drawing)
R $\rightarrow$ 2 (Rolling)
S $\rightarrow$ 4 (Closed die forging) AND S $\rightarrow$ 5 (Casting)

Now, let's examine the given options:

• Option A: P-3; Q-1; R-2; S-4. This option correctly matches P, Q, and R, and also correctly lists S with 4 (Closed die forging), which is a valid occurrence for cold shut.
• Option D: P-3; Q-1; R-2; S-5. This option also correctly matches P, Q, and R, and lists S with 5 (Casting), which is another valid occurrence for cold shut.

Since Cold shut (S) can occur in both Closed die forging (4) and Casting (5), both Option A and Option D provide valid combinations of defect-process matches.

For transferring geometrical models between different CAD systems, standard data transfer formats are crucial.

1. IGES (Initial Graphics Exchange Specification) is a neutral file format specifically designed for digital data exchange between CAD/CAM systems, supporting various geometric entities.
2. STEP (Standard for the Exchange of Product model data), defined by ISO 10303, is a comprehensive and widely adopted standard for robust CAD data interoperability and geometrical modeling.
3. NDC (Normalized Device Coordinates) is a coordinate system, typically $[0, 1]$ or $[-1, 1]$, for screen space in computer graphics, not a CAD model data transfer format.
4. STL (StereoLithography) is a standard format that represents 3D object surface geometry using connected triangles (a mesh), widely used for 3D printing and rapid prototyping.
   Therefore, IGES, STEP, and STL are well-established formats for geometrical modeling data transfer in CAD.

The correlation coefficient, denoted by $r$, quantifies the strength and direction of a linear relationship between two numerical variables. A crucial property to remember for GATE is that its value is always bounded between $-1$ and $+1$, inclusive. Mathematically, this means $-1 \le r \le 1$
Let's check the given options against this range:

• Option A: $0$ falls within $[-1, 1]$. This indicates no linear relationship.
• Option B: $10.3$ is outside the range, as $10.3 > 1$.
• Option C: $-0.9$ falls within $[-1, 1]$. This suggests a strong negative linear relationship.
• Option D: $2$ is outside the range, as $2 > 1$.
  Therefore, $0$ and $-0.9$ are the only possible values for the correlation coefficient.

The Carnot cycle is comprised of four reversible processes: two isothermal processes and two adiabatic processes. For a reversible adiabatic process, entropy ($S$) remains constant, meaning the process is isentropic ($\Delta S = 0$).

Let's examine each process in the Carnot cycle:

• Process 1-2: This is an isothermal expansion. Since temperature is constant but heat is exchanged, entropy changes ($\Delta S \neq 0$). Thus, it is not isentropic.
• Process 2-3: This is an adiabatic expansion. As it's a reversible adiabatic process, entropy remains constant ($\Delta S = 0$). Therefore, it is an isentropic process.
• Process 3-4: This is an isothermal compression. Similar to process 1-2, heat is exchanged at constant temperature, so entropy changes ($\Delta S \neq 0$). Thus, it is not isentropic.
• Process 4-1: This is an adiabatic compression. Being a reversible adiabatic process, entropy remains constant ($\Delta S = 0$). Therefore, it is an isentropic process.

Hence, the processes 2-3 and 4-1 are isentropic.

To find the minimum blanking force, we use the formula $F = \tau \times A$, where $F$ is the blanking force, $\tau$ is the shear strength, and $A$ is the shear area.

First, calculate the perimeter of the square:
$P = 4 \times \text{side length} = 4 \times 50 \text{ mm} = 200 \text{ mm}$

Next, determine the shear area, which is the perimeter multiplied by the sheet thickness:
$A = P \times \text{thickness} = 200 \text{ mm} \times 2 \text{ mm} = 400 \text{ mm}^2$

Now, substitute the values into the blanking force formula. Remember that $200 \text{ MPa}$ is equivalent to $200 \text{ N/mm}^2$:
$F = 200 \text{ N/mm}^2 \times 400 \text{ mm}^2 = 80000 \text{ N}$

Finally, express the force in the requested format, $\times 10^3 \text{ N}$:
$F = 80 \times 10^3 \text{ N}$

This problem asks for the probability of selecting only uncoated inserts when choosing 4 inserts randomly from a mixed box. We use combinations since the order of selection doesn't matter.

First, calculate the total number of ways to choose 4 inserts from the 15 available (5 coated + 10 uncoated):
Total combinations = $C(15, 4) = \frac{15!}{4!(15-4)!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$ ways.

Next, calculate the number of "favorable" ways to choose 4 uncoated inserts from the 10 available uncoated ones:
Favorable combinations = $C(10, 4) = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$ ways.

Finally, the probability is the ratio of favorable combinations to total combinations:
$\text{Probability} = \frac{\text{Favorable combinations}}{\text{Total combinations}} = \frac{C(10, 4)}{C(15, 4)} = \frac{210}{1365}$
Simplifying and rounding to two decimal places, we get $\frac{210}{1365} = \frac{2}{13} \approx 0.1538... \approx 0.15$.

To calculate the Net Present Value (NPV), we discount all future cash flows to their present value and then subtract the initial investment. The formula for NPV is: $NPV = \sum_{t=1}^{n} \frac{CF_t}{(1+r)^t} - \text{Initial Investment}$
Given an initial investment of INR $2,00,000$, annual payouts ($CF_t$) of INR $1,00,000$ for $n=3$ years, and a discount rate ($r$) of $15\%$ ($0.15$), we first find the present value of each payout:
Year 1 PV: $\frac{1,00,000}{(1 + 0.15)^1} = \frac{1,00,000}{1.15} \approx 86,956.52$
Year 2 PV: $\frac{1,00,000}{(1 + 0.15)^2} = \frac{1,00,000}{1.3225} \approx 75,614.37$
Year 3 PV: $\frac{1,00,000}{(1 + 0.15)^3} = \frac{1,00,000}{1.520875} \approx 65,751.61$
Summing these gives the Total Present Value of Payouts: $86,956.52 + 75,614.37 + 65,751.61 \approx 228,322.50$.
Finally, we calculate the NPV: $228,322.50 - 2,00,000 = 28,322.50$. Rounding to the nearest integer, the NPV is $28,323$.

The problem involves the modified Taylor tool life equation, which is a fundamental concept in machining. This equation is given as:
$V T^n f^m = \text{constant}$
Here, $V$ is the cutting speed (m/s), $T$ is the tool life in minutes, and $f$ is the feed in mm/rev. We are given the exponents $n = 0.25$ and $m = 0.5$.

We need to find the ratio of feeds ($f_1/f_2$) when the tool life is the same ($T_1 = T_2 = T$) under two different cutting conditions ($V_1, f_1$) and ($V_2, f_2$).
For condition 1: $V_1 T^n f_1^m = K$
For condition 2: $V_2 T^n f_2^m = K$

Since the constant $K$ is the same for both conditions, we can equate them:
$V_1 T^n f_1^m = V_2 T^n f_2^m$
As the tool life $T$ is constant, $T^n$ cancels out from both sides:
$V_1 f_1^m = V_2 f_2^m$
To find the ratio of feeds ($f_1/f_2$), rearrange the equation:
$\frac{f_1^m}{f_2^m} = \frac{V_2}{V_1}$
This can be written as:
$\left(\frac{f_1}{f_2}\right)^m = \frac{V_2}{V_1}$

We are given $V_1/V_2 = 2/3$, which means $V_2/V_1 = 3/2$. Substituting this value and $m = 0.5$ into the equation:
$\left(\frac{f_1}{f_2}\right)^{0.5} = \frac{3}{2}$
To solve for $f_1/f_2$, raise both sides to the power of $1/0.5$, which is 2:
$\frac{f_1}{f_2} = \left(\frac{3}{2}\right)^2$
$\frac{f_1}{f_2} = \frac{9}{4}$
$\frac{f_1}{f_2} = 2.25$
The ratio of corresponding feeds ($f_1/f_2$), rounded to two decimal places, is 2.25.

To evaluate the line integral $\int_C (2x dx + 2y dy + 2z dz)$, first we check if the vector field $\mathbf{F}(x, y, z) = \langle 2x, 2y, 2z \rangle$ is conservative.
The curl of $\mathbf{F}$ is calculated as:
$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2x & 2y & 2z \end{vmatrix}$
$= \mathbf{i} \left( \frac{\partial (2z)}{\partial y} - \frac{\partial (2y)}{\partial z} \right) - \mathbf{j} \left( \frac{\partial (2z)}{\partial x} - \frac{\partial (2x)}{\partial z} \right) + \mathbf{k} \left( \frac{\partial (2y)}{\partial x} - \frac{\partial (2x)}{\partial y} \right)$
$= \mathbf{i}(0 - 0) - \mathbf{j}(0 - 0) + \mathbf{k}(0 - 0) = \mathbf{0}$
Since $\nabla \times \mathbf{F} = \mathbf{0}$, the field $\mathbf{F}$ is conservative.

For a conservative field, we can find a scalar potential function $f(x, y, z)$ such that $\nabla f = \mathbf{F}$.
From $\frac{\partial f}{\partial x} = 2x$, we get $f(x, y, z) = x^2 + g(y, z)$.
Comparing $\frac{\partial f}{\partial y} = \frac{\partial g}{\partial y}$ with $2y$, we find $\frac{\partial g}{\partial y} = 2y$, so $g(y, z) = y^2 + h(z)$.
Thus, $f(x, y, z) = x^2 + y^2 + h(z)$.
Comparing $\frac{\partial f}{\partial z} = \frac{dh}{dz}$ with $2z$, we find $\frac{dh}{dz} = 2z$, so $h(z) = z^2 + K$.
Choosing $K=0$, the potential function is $f(x, y, z) = x^2 + y^2 + z^2$.

According to the Fundamental Theorem of Line Integrals, for a conservative field, the integral is the difference of the potential function evaluated at the end point and the start point.
The start point is $A = (0, 0, 0)$ and the end point is $B = (1, 0, 0)$.
Value at the end point: $f(1, 0, 0) = (1)^2 + (0)^2 + (0)^2 = 1$.
Value at the start point: $f(0, 0, 0) = (0)^2 + (0)^2 + (0)^2 = 0$.
The line integral is $f(\text{End Point}) - f(\text{Start Point}) = 1 - 0 = 1$.
The specific path (semi-circle) does not affect the result because the field is conservative.
The value of the line integral is 1.

To find the inventory at the end of November, we need to account for all units available and then subtract the demand for that month.

First, let's calculate the total supply available for November's demand:
$\text{Total Supply}_{\text{Nov}} = \text{Inventory}_{\text{Oct}} + \text{Production}_{\text{Nov}} + \text{Subcontracting}_{\text{Nov}}$
Given:

• Inventory at the end of October (which is the opening inventory for November) = $1500$ units
• Units produced in-house in November = $8000$ units
• Units subcontracted in November = $1000$ units

Substituting these values:
$\text{Total Supply}_{\text{Nov}} = 1500 + 8000 + 1000 = 10500 \text{ units}$
Now, to find the ending inventory for November, we subtract November's demand from this total supply:
$\text{Inventory}_{\text{Nov}} = \text{Total Supply}_{\text{Nov}} - \text{Demand}_{\text{Nov}}$
Given the demand for November is $10500$ units:
$\text{Inventory}_{\text{Nov}} = 10500 - 10500 = 0 \text{ units}$
Therefore, the inventory at the end of November is $0$ units.

In polymer extrusion, the final shape of the extruded product directly mirrors the shape of the die opening. For the given problem, extruded cross-section (P) is circular. Among the available die openings, die (1) is circular. Therefore, (P-1) is a correct match. Similarly, the extruded cross-section (Q) is square. From the die options, die (4) is square. Thus, (Q-4) is the correct match. The combination (P-1; Q-4) correctly identifies the die openings for the respective extruded shapes.

To determine the degree of densification, we compare the powder's bulk density to the density of the fully solid material. In this context, the apparent density ($\rho_{\text{app}}$) is used as the bulk density. The formula for the degree of densification is:

$\text{Degree of Densification} (\% ) = \frac{\text{Apparent Density}}{\text{Density of Solid Iron}} \times 100\%$

Given the apparent density ($\rho_{\text{app}}$) of $2.66 \text{ g/cm}^3$ and the density of solid iron ($\rho_{\text{solid}}$) as $7.85 \text{ g/cm}^3$, we substitute these values into the formula:

$\text{Degree of Densification} (\% ) = \frac{2.66 \text{ g/cm}^3}{7.85 \text{ g/cm}^3} \times 100\%$

$\text{Degree of Densification} (\% ) \approx 0.3388535 \times 100\% \approx 33.89\%$

To determine the correct matches, we'll check each matrix against the given properties.

Matrix P: Symmetric Property Check
A matrix is symmetric if it equals its transpose ($P = P^T$), meaning elements $P_{ij}$ and $P_{ji}$ are identical.

Here, $P_{12} = 17$ and $P_{21} = 17$; $P_{13} = 5$ and $P_{31} = 5$; $P_{23} = -12$ and $P_{32} = -12$. Since $P_{ij} = P_{ji}$ for all elements, Matrix P is symmetric (Property 3).