GATE PI 2019

Q11GATE 2019MCQ1MEngineering Mathematics

For any real, square and non-singular matrix B , the det B -1 is

📖 Explanation

This question tests a fundamental property relating the determinant of a matrix to the determinant of its inverse.

Let $B$ be a real, square, and non-singular matrix. Since $B$ is non-singular, its determinant, $\det B$ , must not be zero ( $\det B \neq 0$ ). This ensures that the inverse matrix, $B^{-1}$ , exists.

We use two key properties to derive the relationship:

Matrix Inverse Definition: A matrix multiplied by its inverse yields the identity matrix $I$ .
$B \times B^{-1} = I$
Determinant Product Property: The determinant of a product of matrices is the product of their individual determinants.
$\det(AB) = (\det A) \times (\det B)$

Applying the determinant property to the inverse definition:
$\det(B \times B^{-1}) = \det(I)$
$(\det B) \times (\det B^{-1}) = \det(I)$

The determinant of the identity matrix $I$ is always 1:
$\det(I) = 1$

Substituting this back into our equation:
$(\det B) \times (\det B^{-1}) = 1$
Since $\det B \neq 0$ , we can divide by $\det B$ to solve for $\det B^{-1}$ :
$\det B^{-1} = \frac{1}{\det B}$
This can also be written using exponent notation as:
$\det B^{-1} = (\det B)^{-1}$

Therefore, the determinant of the inverse of a matrix $B$ is $(\det B)^{-1}$ .

Q12GATE 2019MCQ1MEngineering Mathematics

For a complex number z = 1 - 4i with i = √-1, the value of $\left|\dfrac{z+3}{z-1}\right|$ is

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📖 Explanation

To find the value of $\left|\dfrac{z+3}{z-1}\right|$ for $z = 1 - 4i$ , we first calculate the numerator and the denominator separately.

For the numerator, $z+3$ :
$z+3 = (1 - 4i) + 3 = (1 + 3) - 4i = 4 - 4i$ .

For the denominator, $z-1$ :
$z-1 = (1 - 4i) - 1 = (1 - 1) - 4i = 0 - 4i = -4i$ .

Now, substitute these back into the expression:
$\dfrac{z+3}{z-1} = \dfrac{4 - 4i}{-4i}$ .
We can simplify this by dividing both numerator and denominator by 4:
$\dfrac{4 - 4i}{-4i} = \dfrac{1 - i}{-i}$ .

To eliminate the complex number in the denominator, multiply both the numerator and denominator by the conjugate of $-i$ , which is $i$ :
$\dfrac{1 - i}{-i} \times \dfrac{i}{i} = \dfrac{(1 - i) \times i}{(-i) \times i} = \dfrac{1 \times i - i \times i}{-i^2}$ .
Recall that $i^2 = -1$ :
$\dfrac{i - (-1)}{-(-1)} = \dfrac{i + 1}{1} = 1 + i$ .

Finally, we need to find the modulus of $1 + i$ . The modulus of a complex number $a + bi$ is given by $|a + bi| = \sqrt{a^2 + b^2}$ .
For $1 + i$ , we have $a=1$ and $b=1$ .
$|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$ .

Therefore, the value of $\left|\dfrac{z+3}{z-1}\right|$ is $\sqrt{2}$ .

Q13GATE 2019MCQ1MEngineering Mathematics

The vector that is normal to the surface 2xz 2 - 3xy - 4x = 7 at the point (1, -1, 2) is

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📖 Explanation

To find the vector normal (perpendicular) to a surface given by an equation $F(x, y, z) = c$ , we use the gradient of the function $F$ . The gradient vector, $\nabla F$ , is always normal to the surface at any point $(x, y, z)$ .

First, define the function $F(x, y, z)$ from the given surface equation $2xz^2 - 3xy - 4x = 7$ :
$F(x, y, z) = 2xz^2 - 3xy - 4x$
Next, calculate the partial derivatives of $F$ with respect to $x$ , $y$ , and $z$ to form the gradient vector $\nabla F = \frac{\partial F}{\partial x}\mathbf{i} + \frac{\partial F}{\partial y}\mathbf{j} + \frac{\partial F}{\partial z}\mathbf{k}$ .
$\frac{\partial F}{\partial x} = \frac{\partial}{\partial x}(2xz^2 - 3xy - 4x) = 2z^2 - 3y - 4$
$\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(2xz^2 - 3xy - 4x) = -3x$
$\frac{\partial F}{\partial z} = \frac{\partial}{\partial z}(2xz^2 - 3xy - 4x) = 4xz$
So, the gradient vector is:
$\nabla F(x, y, z) = (2z^2 - 3y - 4)\mathbf{i} + (-3x)\mathbf{j} + (4xz)\mathbf{k}$
Finally, evaluate $\nabla F$ at the given point $(1, -1, 2)$ by substituting $x = 1$ , $y = -1$ , and $z = 2$ :
$(\frac{\partial F}{\partial x})_{(1, -1, 2)} = 2(2)^2 - 3(-1) - 4 = 2(4) + 3 - 4 = 8 + 3 - 4 = 7$
$(\frac{\partial F}{\partial y})_{(1, -1, 2)} = -3(1) = -3$
$(\frac{\partial F}{\partial z})_{(1, -1, 2)} = 4(1)(2) = 8$
Therefore, the normal vector at $(1, -1, 2)$ is:
$\nabla F(1, -1, 2) = 7\mathbf{i} - 3\mathbf{j} + 8\mathbf{k}$

Q14GATE 2019MCQ1MEngineering Mathematics

If roots of the auxiliary equation of $\dfrac{d^2y}{dx^2}\;+\;a\dfrac{dy}{dx}\;+\;by=0$ are real and equal, the general solution of the differential equation is

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📖 Explanation

We need to find the general solution of the differential equation $\dfrac{d^2y}{dx^2} + a\dfrac{dy}{dx} + by = 0$ when its auxiliary equation has real and equal roots.

1. Form the Auxiliary Equation:
For a differential equation of the form $A\dfrac{d^2y}{dx^2} + B\dfrac{dy}{dx} + Cy = 0$ , the auxiliary equation is $Am^2 + Bm + C = 0$ .
Here, the given equation is $\dfrac{d^2y}{dx^2} + a\dfrac{dy}{dx} + by = 0$ .
So, the auxiliary equation is $m^2 + am + b = 0$ .

2. Condition for Real and Equal Roots:
For a quadratic equation $Am^2 + Bm + C = 0$ , the roots are real and equal if the discriminant $B^2 - 4AC = 0$ .
In our auxiliary equation $m^2 + am + b = 0$ :
$A = 1$ , $B = a$ , $C = b$ .
Setting the discriminant to zero: $a^2 - 4(1)(b) = 0 \implies a^2 - 4b = 0 \implies b = \dfrac{a^2}{4}$ .

3. Find the Repeated Root:
When roots are real and equal, the repeated root $m$ is given by $m = \dfrac{-B}{2A}$ .
Using our auxiliary equation coefficients: $m = \dfrac{-a}{2(1)} = -\dfrac{a}{2}$ .
So, the repeated root is $m = -\dfrac{a}{2}$ .

4. General Solution for Real and Equal Roots:
For a second-order linear homogeneous differential equation with constant coefficients, if the auxiliary equation has real and equal roots $m_1 = m_2 = m$ , the general solution is $y = (c_1 + c_2x)e^{mx}$ .
Substitute $m = -\dfrac{a}{2}$ into this general solution form:
$y = (c_1 + c_2x)e^{\left(-\frac{a}{2}\right)x}$
$y = (c_1 + c_2x)e^{-\frac{ax}{2}}$

Q15GATE 2019MCQ1MEngineering Mathematics

The solution of $\int_1^a {\int_1^b {\frac{{dxdy}}{{x\,y}}} }$ is

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📖 Explanation

This problem involves evaluating a double integral. The integral given is $\int_1^a {\int_1^b {\frac{{dxdy}}{{x\,y}}} }$
The integrand $\frac{1}{x\,y}$ can be written as a product of a function of $x$ and a function of $y$ , i.e., $\frac{1}{x} \cdot \frac{1}{y}$ . Since the limits of integration are constant for both $x$ and $y$ , we can separate this double integral into a product of two single integrals:
$\left( \int_1^a {\frac{dx}{x}} \right) \times \left( \int_1^b {\frac{dy}{y}} \right)$
First, evaluate the integral with respect to $x$ :
$\int_1^a {\frac{dx}{x}} = [\ln|x|]_1^a = \ln|a| - \ln|1|$
Assuming $a>1$ , $\ln|a| = \ln(a)$ and $\ln(1) = 0$ . So, the $x$ -integral evaluates to $\ln(a) - 0 = \ln(a)$ .

Next, evaluate the integral with respect to $y$ :
$\int_1^b {\frac{dy}{y}} = [\ln|y|]_1^b = \ln|b| - \ln|1|$
Assuming $b>1$ , $\ln|b| = \ln(b)$ and $\ln(1) = 0$ . So, the $y$ -integral evaluates to $\ln(b) - 0 = \ln(b)$ .

Finally, multiply the results of the two integrals:
$\text{Result} = \ln(a) \times \ln(b) = \ln(a) \ln(b)$

Q16GATE 2019MCQ1MGeneral Engineering

Match the crystal structure in Column A with the corresponding packing fractions in Column B of the table Column A Column B 1. Simple cubic P. 0.74 2. Hexagonal close-packed Q. 068 3. Body-centered cubic R. 0.52 4. Face-centered cubic

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📖 Explanation

To match the crystal structures with their packing fractions, we first understand that the packing fraction (or Atomic Packing Factor, APF) is the ratio of the volume occupied by atoms in a unit cell to the total volume of the unit cell: $\text{Packing Fraction} = \frac{\text{Volume of atoms \in unit cell}}{\text{Volume of unit cell}}$

Simple Cubic (SC):
- Number of atoms per unit cell = 1 (atoms at 8 corners, each shared by 8 unit cells).
- Relationship: $a = 2r$ .
- Volume of atoms: $1 \times \frac{4}{3}\pi r^3$ .
- Volume of unit cell: $a^3 = (2r)^3 = 8r^3$ .
- Packing Fraction: $\frac{\frac{4}{3}\pi r^3}{8r^3} = \frac{\pi}{6} \approx 0.52$ . This corresponds to R.
Hexagonal Close-Packed (HCP):
- HCP structures have a high packing efficiency.
- Packing Fraction: $\approx 0.74$ . The theoretical value is $\frac{\pi}{3\sqrt{2}} \approx 0.7405$ . This corresponds to P.
Body-Centered Cubic (BCC):
- Number of atoms per unit cell = 2 (1 from corners + 1 in the center).
- Relationship: $a = \frac{4r}{\sqrt{3}}$ .
- Volume of atoms: $2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3$ .
- Volume of unit cell: $a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}}$ .
- Packing Fraction: $\frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} = \frac{8\pi}{64} \times \sqrt{3} = \frac{\pi\sqrt{3}}{8} \approx 0.68$ . This corresponds to Q.
Face-Centered Cubic (FCC):
- Number of atoms per unit cell = 4 (1 from corners + 3 from faces).
- FCC structures also have high packing efficiency.
- Packing Fraction: $\approx 0.74$ . The theoretical value is $\frac{\pi}{3\sqrt{2}} \approx 0.7405$ . This corresponds to P.

Summarizing the matches:

Simple Cubic (1) $\rightarrow$ R (0.52)
Hexagonal Close-Packed (2) $\rightarrow$ P (0.74)
Body-Centered Cubic (3) $\rightarrow$ Q (0.68)
Face-Centered Cubic (4) $\rightarrow$ P (0.74)

This leads to the matching: 1 - R, 2 - P, 3 - Q, 4- P.

Q17GATE 2019MCQ1MGeneral Engineering

The link lengths of a planar four bar mechanism are AB = 100 mm, BC = 25 mm, CD = 75 mm and DA = 90 mm. For achieving the full rotation of both the input (crank) as well as the output (follower) links, the link that needs to be fixed is

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📖 Explanation

To achieve full rotation for both the input (crank) and output (follower) links in a four-bar mechanism, we must first verify if it's a Grashof mechanism. The link lengths are given as AB = 100 mm, BC = 25 mm, CD = 75 mm, and DA = 90 mm.
Let's identify the shortest ( $s$ ) and longest ( $l$ ) links, and the other two links ( $p$ , $q$ ):
$s = 25 \text{ mm (BC)}$
$l = 100 \text{ mm (AB)}$
$p = 75 \text{ mm (CD)}$
$q = 90 \text{ mm (DA)}$

According to Grashof's Law, for a mechanism to be Grashofian (allowing at least one link to rotate fully), the sum of the shortest and longest link lengths must be less than the sum of the lengths of the other two links:
$s + l < p + q$
Let's check this condition:
$s + l = 25 \text{ mm} + 100 \text{ mm} = 125 \text{ mm}$
$p + q = 75 \text{ mm} + 90 \text{ mm} = 165 \text{ mm}$
Since $125 \text{ mm} < 165 \text{ mm}$ , the Grashof condition is satisfied, meaning this is a Grashof mechanism.

For a Grashof mechanism, to get a double-crank (where both input and output links can make full 360-degree rotations), the shortest link must be fixed. If the shortest link ( $s$ ) is fixed, then the two links connected to it (the two adjacent links) will act as cranks. In this case, the shortest link is BC (25 mm).
If BC (shortest link $s$ ) is fixed, the links connected to it are AB (100 mm) and CD (75 mm). Also, the link opposite to the fixed shortest link is DA (90 mm).
In a Grashof mechanism with the shortest link fixed:

The link opposite the shortest link ( $DA = 90 \text{ mm}$ ) will fully rotate (crank).
The longest link ( $AB = 100 \text{ mm}$ ) will also fully rotate (crank).
The links adjacent to the fixed shortest link ( $AB = 100 \text{ mm}$ and $CD = 75 \text{ mm}$ ) are the ones that rotate, but specifically, the link opposite the shortest (DA) and the longest link (AB) will achieve full rotation.
Therefore, if BC is fixed, we get a double-crank mechanism where both DA and AB can achieve full 360-degree rotation. This fulfills the requirement for both input and output links to have full rotation.

Q18GATE 2019MCQ1MManufacturing Processes II

The process used for producing continuous insulation coating on an electrical wire is

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Q19GATE 2019MCQ1MManufacturing Processes II

The correct statement pertaining to the friction welding process is

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Q20GATE 2019MCQ1MManufacturing Processes II

The end product obtained using spinning process is shown in the figure. The initial blank thickness is 2.5 mm. The blank diameter (in mm) is

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📖 Explanation

In the spinning process, the total developed length of the initial blank metal remains constant. This means the length of the blank's outer edge will form the entire profile of the spun part.

From the figure, we have:
$\sin 30^\circ = \frac{45}{\text{Hypotenuse}}$
Since $\sin 30^\circ = 0.5$ , we get:
$\text{Hypotenuse} = \frac{45}{0.5} = 45 \times 2 = 90 \text{ mm}$
The total initial blank diameter ( $D$ ) is calculated by summing the two slanted lengths (hypotenuses) and the flat bottom section:
$D = (2 \times \text{Hypotenuse}) + 30 \text{ mm}$
$D = (2 \times 90) + 30 = 180 + 30 = 210 \text{ mm}$

Q21GATE 2019MCQ1MIndustrial Engineering

For a classical (Wilson) model of determining economic order quantity (EOQ), the carrying and ordering costs are C r and C o , respectively. For an annual demand D, the minimum yearly total inventory cost is

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📖 Explanation

The Wilson Economic Order Quantity (EOQ) model aims to find the optimal order quantity that minimizes total inventory costs, which include both ordering and carrying costs. The total annual inventory cost, $TC(Q)$ , is given by the sum of annual ordering cost $\left( \frac{D}{Q} \right) C_o$ and annual carrying cost $\left( \frac{Q}{2} \right) C_r$ :
$TC(Q) = \left( \frac{D}{Q} \right) C_o + \left( \frac{Q}{2} \right) C_r$
To find the minimum cost, we differentiate $TC(Q)$ with respect to $Q$ and set it to zero:
$\frac{dTC}{dQ} = -\frac{DC_o}{Q^2} + \frac{C_r}{2} = 0$
Solving for $Q$ gives the optimal order quantity, $Q^*$ :
$Q^* = \sqrt{\frac{2DC_o}{C_r}}$
Substitute this $Q^*$ back into the $TC(Q)$ formula to find the minimum yearly total inventory cost, $TC_{min}$ :
$TC_{min} = \left( \frac{D}{\sqrt{\frac{2DC_o}{C_r}}} \right) C_o + \left( \frac{\sqrt{\frac{2DC_o}{C_r}}}{2} \right) C_r$
Simplifying both terms:
First term: $D C_o \sqrt{\frac{C_r}{2DC_o}} = \sqrt{\frac{D^2 C_o^2 C_r}{2DC_o}} = \sqrt{\frac{DC_o C_r}{2}} = \frac{1}{2} \sqrt{2DC_o C_r}$
Second term: $\frac{1}{2} \sqrt{\frac{2DC_o}{C_r}} C_r = \frac{1}{2} \sqrt{\frac{2DC_o C_r^2}{C_r}} = \frac{1}{2} \sqrt{2DC_o C_r}$
Adding these two simplified terms yields the minimum total inventory cost:
$TC_{min} = \frac{1}{2} \sqrt{2DC_o C_r} + \frac{1}{2} \sqrt{2DC_o C_r} = \sqrt{2DC_o C_r}$

Q22GATE 2019MCQ1MIndustrial Engineering

A company has purchased an asset by investing Rs. 30,000. The useful life of the asset is 5 years and it has no salvage value at the end of its useful life. The depreciation cost (in Rs.) for the 2 nd year using sum-of-years-digit (SYD) method is

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📖 Explanation

To calculate depreciation for the 2nd year using the Sum-of-Years'-Digits (SYD) method, we first need the sum of the digits representing the asset's useful life. The useful life ( $n$ ) is 5 years, so the sum ( $S$ ) is calculated as $S = \frac{n(n+1)}{2} = \frac{5(5+1)}{2} = \frac{5 \times 6}{2} = 15$ . The depreciable amount is the asset cost minus the salvage value, which is Rs. $30,000 - 0 = 30,000$ . For the 2nd year, the depreciation fraction is $\frac{\text{Remaining Useful Life at start of Year 2}}{S}$ . The remaining useful life at the start of year 2 is $5 - 1 = 4$ years. So, the depreciation for the 2nd year is Rs. $30,000 \times \frac{4}{15} = \text{Rs. } 2,000 \times 4 = 8,000$ .

Q23GATE 2019MCQ1MManufacturing Processes II

In an NC milling operation, the tool path is generated using absolute programming for the trajectory shown in the figure The corresponding block of the NC program is

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📖 Explanation

This problem involves understanding G-codes for NC milling. The trajectory shown is a circular arc. First, we need to identify the direction of the arc from the start point to the end point. Observing the figure (which is implied to be given in the question context), the trajectory of the point is in the clockwise direction. For clockwise circular interpolation, the specific G-code used is G02.

Next, we need to determine the coordinates of the endpoint of this arc and its radius. According to the problem setup, the final point (endpoint) has coordinates X = 120 and Y = 60. The radius of this circular arc is given as R = 60.

In absolute programming, the block for a circular interpolation specifies the G-code, the absolute X and Y coordinates of the endpoint, and the radius. Combining these, the corresponding NC program block will be:

$G02 \text{ X 120.0 Y 60.0 R 60.0}$

For reference:

G00: Rapid traverse (point-to-point positioning).
G01: Linear interpolation (straight cut at cutting feed rate).
G02: Circular Interpolation - Clockwise (CW).
G03: Circular Interpolation - Counter-Clockwise (CCW).

Q24GATE 2019MCQ1MIndustrial Engineering

The SQC chart based on Binomial distribution is

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📖 Explanation

SQC charts monitor quality. The Binomial distribution models scenarios with two outcomes (e.g., conforming/nonconforming), where the probability of success is constant across independent trials.

p chart: Monitors the proportion of nonconforming items in a sample. Since it deals with the proportion of 'successes' (nonconforming items) in a fixed number of trials (sample size), it is fundamentally based on the Binomial distribution.
c chart: Monitors the number of defects per unit and typically follows a Poisson distribution.
$\bar{X}$ chart: Monitors the average of measurements and, by the Central Limit Theorem, relies on the Normal distribution for sample means.
R chart: Monitors the range of measurements within samples and typically assumes the underlying process data is normally distributed.
Therefore, the p chart is based on the Binomial distribution.

Q25GATE 2019MCQ1MIndustrial Engineering

The capacity of a passenger airline is expressed in terms of

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📖 Explanation

The capacity of a passenger airline is best expressed by Available Seat Miles (ASM). ASM is a crucial metric that combines the number of seats offered with the distance those seats travel. It is calculated using the formula:

$\text{ASM} = \text{Number of Seats} \times \text{Distance (\in miles)}$

For instance, if an airline operates a flight with 100 seats over a distance of 500 miles, the ASM for that flight would be:

$\text{ASM} = 100 \text{ seats} \times 500 \text{ miles} = 50,000 \text{ ASM}$

This metric accounts for both the quantity of seats and the distance covered, making it a comprehensive and standardized measure of an airline's passenger transport potential, unlike merely counting available seats, miles, or sectors.

Q26GATE 2019MCQ1MIndustrial Engineering

REL chart is used in

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Q27GATE 2019NAT1MGeneral Engineering

A metallic rod of diameter d 0 is subjected to the tensile test. The engineering stress and the true stress at fracture are 800 MPa and 900 MPa, respectively. The ratio of the rod diameter at fracture d f to the initial diameter d 0 is (round off to 2 decimal places)

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📖 Explanation

Let's break down this problem on stress and strain in a tensile test. We know the relationship between true stress ( $\sigma_T$ ) and engineering stress ( $\sigma$ ) is given by $\sigma_T = \sigma(1 + \varepsilon)$ , where $\varepsilon$ is the engineering strain.

For an incompressible material under tension, the volume remains constant. This means $A_0 L_0 = A_f L_f$ , where $A$ is the cross-sectional area and $L$ is the length. From this, we can derive the relation:
$1 + \varepsilon = \frac{L_f}{L_0} = \frac{A_0}{A_f} = \frac{d_0^2}{d_f^2}$
Here, $d_0$ is the initial diameter and $d_f$ is the diameter at fracture.

Given:

True stress at fracture ( $\sigma_T$ ) = 900 MPa
Engineering stress at fracture ( $\sigma$ ) = 800 MPa

Using the relation $\sigma_T = \sigma(1 + \varepsilon)$ :
$900 = 800(1 + \varepsilon)$
$1 + \varepsilon = \frac{900}{800} = 1.125$
Now, substitute this into the diameter relationship:
$1.125 = \left(\frac{d_0}{d_f}\right)^2$
Solving for the ratio $\frac{d_0}{d_f}$ :
$\frac{d_0}{d_f} = \sqrt{1.125} \approx 1.0606$
Finally, we need the ratio $\frac{d_f}{d_0}$ :
$\frac{d_f}{d_0} = \frac{1}{1.0606} \approx 0.942$
Rounding off to two decimal places, the ratio $\frac{d_f}{d_0}$ is $0.94$ .

Q28GATE 2019NAT1MGeneral Engineering

A heat pump is to supply heat at the rate of 10 kW to a building to be maintained at 22°C. The outside temperature is 2°C. The minimum power (in kW) required to run the heat pump is (round off to 2 decimal places)

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📖 Explanation

To find the minimum power required, we need to consider the heat pump operating at its maximum possible Coefficient of Performance (COP). The COP of a heat pump is defined as $COP = \frac{\text{Desired effect}}{\text{Energy input}}$ , which for a heat pump is $COP_{HP} = \frac{Q_H}{W}$ . The maximum theoretical COP for a reversible heat pump is given by $COP_{max} = \frac{T_H}{T_H - T_L}$ .

Given:

Heat supplied, $Q_H = 10 \text{ kW}$
Building temperature, $T_H = 22^\circ\text{C} = 273 + 22 \text{ K} = 295 \text{ K}$
Outside temperature, $T_L = 2^\circ\text{C} = 273 + 2 \text{ K} = 275 \text{ K}$

First, calculate the maximum COP:
$COP_{max} = \frac{T_H}{T_H - T_L} = \frac{295}{295 - 275} = \frac{295}{20} = 14.75$
Since $W_{min} = \frac{Q_H}{COP_{max}}$ , the minimum power required is:
$W_{min} = \frac{10}{14.75} = 0.68 \text{ kW}$

Q29GATE 2019NAT1MGeneral Engineering

One kilogram of air is compressed at a constant temperature of 150°C until its volume is halved. Considering gas constant R = 0.287 kJ/kg-K for air, the magnitude of heat rejected (in kJ) in the compression process is ______ (round off to 2 decimal places)

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📖 Explanation

This problem involves an isothermal compression process, where the temperature remains constant. For an isothermal process, the change in internal energy ( $dU$ ) is zero. According to the First Law of Thermodynamics, $\delta Q = dU + \delta W$ , which simplifies to $\delta Q = \delta W$ when $dU = 0$ . This means the heat transfer ( $Q$ ) is equal to the work done ( $W$ ). The work done for an isothermal process is given by $W = P_1V_1 \ln \left(\frac{V_2}{V_1}\right)$ . Using the ideal gas equation, $P_1V_1 = mRT_1$ , we can substitute this into the work equation: $W = Q = mRT_1 \ln \left(\frac{V_2}{V_1}\right)$ .

Given values: mass ( $m$ ) = 1 kg, initial temperature ( $T_1$ ) = 150°C = 423 K (since $T_1 = T_2$ for isothermal), volume is halved ( $V_1 = 2V_2$ , or $\frac{V_2}{V_1} = \frac{1}{2}$ ), and gas constant ( $R$ ) = 0.287 kJ/kgK.

Substitute these values into the formula:
$W = Q = 1 \times 0.287 \times 423 \times \ln\left(\frac{V_2}{2V_2}\right)$
$W = Q = 1 \times 0.287 \times 423 \times \ln\left(\frac{1}{2}\right)$
$W = Q = -84.148 \text{ kJ}$
The negative sign indicates that heat is rejected from the system during compression. Therefore, the magnitude of heat rejected is 84.148 kJ.

Q30GATE 2019NAT1MManufacturing Processes II

For the abrasive jet machining process, the ratio of abrasive volume to carrier gas volume is 0.25. Further, the ratio of abrasive density to carrier gas density is 25. The mass ratio of abrasive to the mixture of abrasive and carrier gas is (round off to 2 decimal places)

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📖 Explanation

In Abrasive Jet Machining (AJM), we define the Mixing Ratio (MR) as the ratio of abrasive volume to carrier gas volume, given by $MR = \frac{V_a}{V_g}$ . We also need to find the Mass Ratio ( $\alpha$ ), which is the mass of abrasive particles to the total mass of the mixture (abrasive + carrier gas):
$\alpha = \frac{M_a}{M_{a+g}} = \frac{M_a}{M_a + M_g}$
We know that mass ( $M$ ) is density ( $\rho$ ) multiplied by volume ( $V$ ), so $M_a = \rho_a V_a$ and $M_g = \rho_g V_g$ . Substituting these into the formula for $\alpha$ :
$\alpha = \frac{\rho_a V_a}{\rho_a V_a + \rho_g V_g}$
To simplify calculations, it's often easier to work with the reciprocal $\frac{1}{\alpha}$ :
$\frac{1}{\alpha} = \frac{\rho_a V_a + \rho_g V_g}{\rho_a V_a} = 1 + \frac{\rho_g V_g}{\rho_a V_a} = 1 + \left(\frac{\rho_g}{\rho_a}\right) \times \left(\frac{V_g}{V_a}\right)$
Given: $\frac{\rho_a}{\rho_g} = 25$ (so $\frac{\rho_g}{\rho_a} = \frac{1}{25}$ ) and $\frac{V_a}{V_g} = 0.25$ (so $\frac{V_g}{V_a} = \frac{1}{0.25}$ ).
Substituting these values:
$\frac{1}{\alpha} = 1 + \left(\frac{1}{25} \times \frac{1}{0.25}\right) = 1 + \left(0.04 \times 4\right) = 1 + 0.16 = 1.16$
Therefore, $\alpha = \frac{1}{1.16} \approx 0.862$ . Rounding off to two decimal places, the mass ratio is $0.86$ .

Q31GATE 2019NAT1MManufacturing Processes II

In a typical turning tool life test, the following data are generated for tools A and B: Tool name Cutting speed (m/min) Tool life (min) A 200 20 B 150 58 Assuming the same tool life exponent for the tools, the value of constant in the Taylor’s tool life equation (with cutting speed in m/min and tool life in min) is ______ (round off to 2 decimal places)

🚀 Solve in Practice Mode

📖 Explanation

The core concept here is Taylor's Tool Life Equation, which states $V T^n = C$ . Here, $V$ is the cutting speed, $T$ is the tool life, $n$ is the tool life exponent, and $C$ is a constant.

Given the data for Tool A ( $V_A = 200 \text{ m/min}$ , $T_A = 20 \text{ min}$ ) and Tool B ( $V_B = 150 \text{ m/min}$ , $T_B = 58 \text{ min}$ ), and assuming the same tool life exponent $n$ for both, we can write:
$200 \times 20^n = C \quad \text{(1)}$
$150 \times 58^n = C \quad \text{(2)}$
Equating (1) and (2) to find $n$ :
$200 \times 20^n = 150 \times 58^n$
$\frac{200}{150} = \left(\frac{58}{20}\right)^n$
$1.33 = (2.9)^n$
Taking the natural logarithm on both sides:
$\ln(1.33) = n \times \ln(2.9)$
$n = \frac{\ln(1.33)}{\ln(2.9)} \approx \frac{0.285}{1.065} \approx 0.267 \approx 0.27$
Now, substitute the value of $n$ back into equation (1) to find $C$ :
$C = 200 \times 20^{0.27}$
$C \approx 200 \times 2.2466$
$C \approx 449.33$

Q32GATE 2019NAT1MIndustrial Engineering

The average proportion non-conforming of 20 samples each of size 100 items is 0.12. The upper control limit for the relevant chart is ______ (round off to 2 decimal places)

🚀 Solve in Practice Mode

📖 Explanation

This problem involves calculating the Upper Control Limit (UCL) for a P-chart, which is used to monitor the proportion of non-conforming (defective) items in a process. The P-chart is particularly useful when items are classified as either accepted or rejected.

The formula for the control limits of a P-chart is given by:
$\text{Control Limits} = \bar p \pm 3 \times \sqrt {\frac{{\bar p\;\left( {1 - \bar p} \right)}}{n}}$
Given:

Average proportion non-conforming ( $\bar p$ ) = 0.12
Sample size (n) = 100

To find the Upper Control Limit (UCL):
$\text{UCL} = 0.12 + 3 \times \sqrt {\frac{{0.12\;\left( {1 - 0.12} \right)}}{100}}$
$\text{UCL} = 0.12 + 3 \times \sqrt {\frac{{0.12\;\left( 0.88 \right)}}{100}}$
Therefore, UCL = 0.21

Q33GATE 2019NAT1MIndustrial Engineering

For a process which is in a state of statistical control (within ± 3σ), the estimated process standard deviation (σ) is 3 mm. The specification limits for the corresponding product are 100 ± 7 mm. The capability ratio C r is ______ (round off to 3 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

The Process Capability Ratio ( $C_p$ ) is a measure of how well a process can meet its specification limits. It's calculated by comparing the width of the specification limits to the natural variability of the process (which is typically considered as $6\sigma$ ). The formula for $C_p$ is:

$C_p = \frac{\text{Upper Specification Limit (USL)} - \text{Lower Specification Limit (LSL)}}{6 \times \text{Process Standard Deviation } (\sigma)}$

Given:
USL = $100 + 7 = 107$ mm
LSL = $100 - 7 = 93$ mm
Process standard deviation ( $\sigma$ ) = $3$ mm

Substituting these values into the formula:
$C_p = \frac{107 - 93}{6 \times 3} = \frac{14}{18}$
$C_p = 0.7777...$

Rounding off to 3 decimal places, the capability ratio $C_p$ is $0.778$ .

Q34GATE 2019NAT1MIndustrial Engineering

In a work study experiment, normal time was recorded as 140 sec with a rating of 100%. Considering 2% allowance, the standard time (in s) is ______ (round off to 1 decimal place)

🚀 Solve in Practice Mode

📖 Explanation

The concept of standard time is crucial in work study. It accounts for personal needs, fatigue, and unavoidable delays.

Concept:
Standard Time is calculated using the formula:
$\text{Standard time} = \text{Normal time} \times \left(1 + \frac{\text{Allowance}(\%)}{100}\right)$

Calculation:
Given:

Normal Time = $140$ seconds
Allowance = $2\%$

Substitute these values into the formula:
$\text{Standard time} = 140 \times \left(1 + \frac{2}{100}\right)$
$\text{Standard time} = 140 \times (1 + 0.02)$
$\text{Standard time} = 140 \times 1.02$
$\text{Standard time} = 142.8 \text{ sec}$

Important Point:
Remember that Normal time is also derived from observed time and rating factor:
$\text{Normal time} = \frac{\text{Rating factor}(\%)}{100} \times \text{Observed time}$

Q35GATE 2019NAT1MIndustrial Engineering

A warehouse has a loading dock and 3 persons for loading operations. The arrival rate of trucks follows a Poisson distribution with a mean of 4 trucks/hour. The average loading time (by three persons together) per truck is exponentially distributed with a mean of 10 minutes. The charge of the trucks per hour and loading charges per person per hour are Rs.20 and Rs.6, respectively. The total cost (in Rs./hour) is ______.

🚀 Solve in Practice Mode

📖 Explanation

The total cost per hour is the sum of the loading crew cost and the cost incurred by trucks waiting.
The loading crew cost is calculated as: $\text{Number of loaders} \times \text{Hourly wage rate}$ .
The cost of waiting time is calculated as: $W_s \times \lambda \times (\text{truck charge per hour})$ , where $W_s = \frac{1}{\mu - \lambda}$ is the expected waiting time per truck, and $\lambda$ is the expected arrival rate per hour.

Given:
$\lambda = 4 \text{ trucks/hr}$
Service rate by three persons, $\mu = \frac{1 \text{ truck}}{10 \text{ minutes}} \times \frac{60 \text{ minutes}}{1 \text{ hour}} = 6 \text{ trucks/hr}$
Loading cost = Rs. $6 \text{/person/hr}$
Truck charge = Rs. $20 \text{/hr}$

Calculation:

Loading crew cost: $3 \text{ loaders} \times \text{Rs. } 6 \text{/person/hr} = \text{Rs. } 18 \text{/hr}$ .
Cost of waiting time:
$W_s = \frac{1}{\mu - \lambda} = \frac{1}{6 - 4} = \frac{1}{2} \text{ hours}$
$\text{Cost of waiting} = W_s \times \lambda \times (\text{truck charge per hour}) = \frac{1}{2} \times 4 \times 20 = \text{Rs. } 40 \text{/hr}$ .
Total cost per hour: $\text{Loading crew cost} + \text{Cost of waiting time} = 18 + 40 = \text{Rs. } 58 \text{/hr}$ .

Q36GATE 2019MCQ2MEngineering Mathematics

If the Laplace transform of e ωt is $\dfrac{1}{s-\omega}$ , the Laplace transform of t cosh t is

🚀 Solve in Practice Mode

📖 Explanation

Alright, let's break down how to find the Laplace transform of $t \cosh t$ .

First, remember the fundamental Laplace transform of an exponential function: $L\{e^{\omega t}\} = \dfrac{1}{s-\omega}$ .

Step 1: Find the Laplace Transform of $\cosh t$

We know that $\cosh t$ can be expressed in terms of exponentials: $\cosh t = \dfrac{e^t + e^{-t}}{2}$ .
Using the linearity property of Laplace transforms, which states $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ , we can write:
$L\{\cosh t\} = L\left\{\frac{e^t + e^{-t}}{2}\right\} = \frac{1}{2} \left( L\{e^t\} + L\{e^{-t}\} \right)$
Now, apply the given Laplace transform $L\{e^{\omega t}\} = \dfrac{1}{s-\omega}$ :
For $e^t$ , $\omega = 1$ , so $L\{e^t\} = \dfrac{1}{s-1}$ .
For $e^{-t}$ , $\omega = -1$ , so $L\{e^{-t}\} = \dfrac{1}{s-(-1)} = \dfrac{1}{s+1}$ .

Substitute these back:
$L\{\cosh t\} = \frac{1}{2} \left( \frac{1}{s-1} + \frac{1}{s+1} \right)$
Combine the fractions inside the parenthesis by finding a common denominator $(s-1)(s+1)$ :
$L\{\cosh t\} = \frac{1}{2} \left( \frac{(s+1) + (s-1)}{(s-1)(s+1)} \right) = \frac{1}{2} \left( \frac{2s}{s^{2-}1} \right)$
Simplifying this, we get:
$L\{\cosh t\} = \frac{s}{s^{2-}1}$

Step 2: Apply the Multiplication by $t$ Property

A crucial property of Laplace transforms is that if $L\{f(t)\} = F(s)$ , then $L\{t f(t)\} = -\dfrac{d}{ds} F(s)$ .
Here, $f(t) = \cosh t$ and we just found $F(s) = L\{\cosh t\} = \dfrac{s}{s^{2-}1}$ .
Now, we need to find the derivative of $F(s)$ with respect to $s$ . We'll use the quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ .
Let $u = s$ and $v = s^{2-}1$ .
Then, $u' = \dfrac{d}{ds}(s) = 1$ .
And $v' = \dfrac{d}{ds}(s^{2-}1) = 2s$ .

Applying the quotient rule:
$\dfrac{d}{ds} F(s) = \dfrac{d}{ds} \left( \frac{s}{s^{2-}1} \right) = \frac{(1)(s^{2-}1) - (s)(2s)}{(s^{2-}1)^2}$
$\dfrac{d}{ds} F(s) = \frac{s^{2-}1 - 2s^2}{(s^{2-}1)^2} = \frac{-s^{2-}1}{(s^{2-}1)^2}$

Step 3: Final Calculation

Now, using the property $L\{t f(t)\} = -\dfrac{d}{ds} F(s)$ :
$L\{t \cosh t\} = - \left( \frac{-s^{2-}1}{(s^{2-}1)^2} \right)$
Distributing the negative sign, we get:
$L\{t \cosh t\} = \frac{s^{2+}1}{(s^{2-}1)^2}$
This result matches option A.

Q37GATE 2019MCQ2MEngineering Mathematics

General solution of the Cauchy-Euler equation $x^2\frac{d^2y}{dx^2}-7x\dfrac{dy}{dx}+16y=0$ is

🚀 Solve in Practice Mode

📖 Explanation

This problem involves solving a Cauchy-Euler differential equation. These equations have a special structure, allowing us to assume a solution of the form $y = x^m$ .

Given the differential equation: $x^2\frac{d^2y}{dx^2}-7x\dfrac{dy}{dx}+16y=0$ .
We assume a solution $y = x^m$ .
Then, the first derivative is $\dfrac{dy}{dx} = mx^{m-1}$ , and the second derivative is $\dfrac{d^2y}{dx^2} = m(m-1)x^{m-2}$ .

Substitute these into the original equation:
$x^2\left[m(m-1)x^{m-2}\right]\$ - 7x\left[mx^{m-1}\right]$ + 16\left[x^m\right]$ = 0 $This simplifies to:$ m(m-1)x^m - 7mx^m + 16x^m = 0 $Factor out$ x^m $(assuming$ x \neq 0 $):$ x^m[m(m-1) - 7m + 16] = 0$

Since $x^m \neq 0$ , we set the expression in the brackets to zero to form the auxiliary equation:
$m(m-1) - 7m + 16 = 0$
$m^2 - m - 7m + 16 = 0$
$m^2 - 8m + 16 = 0$

This is a quadratic equation. We can recognize it as a perfect square:
$(m-4)^2 = 0$
Solving for $m$ , we find a repeated root:
$m_1 = m_2 = 4$

For a Cauchy-Euler equation with repeated real roots $m_1 = m_2 = m$ , the general solution is given by the formula:
$y = (c_1 + c_2 \ln x) x^m$

Substituting our repeated root $m=4$ :
$y = (c_1 + c_2 \ln x) x^4$

Q38GATE 2019MCQ2MGeneral Engineering

A uniform cantilever beam ABC of length L is subjected to a point load P at point B and a concentrated moment M at point C (as shown in figure). Let E be the Young’s modulus of the beam material and / be the area moment of inertia of the beam’s cross-section. Assuming the validity of the Euler-Bernoulli theory of slender beams, the downward deflection at point C is

🚀 Solve in Practice Mode

📖 Explanation

To find the total downward deflection at point C, we can use the principle of superposition, which means we'll add the deflection caused by the point load P and the deflection caused by the moment M.

Key Concept Recall:

For a cantilever beam with a point load $P$ at its free end (length $L_c$ ), the deflection at the free end is $\frac{PL_c^3}{3EI}$ and the slope is $\frac{PL_c^2}{2EI}$ .
For a cantilever beam with a moment $M$ at its free end (length $L_c$ ), the deflection at the free end is $\frac{ML_c^2}{2EI}$ and the slope is $\frac{ML_c}{EI}$ .

Calculation:
The deflection at C due to load P acting at B and the moment M at C is calculated as follows:
Deflection at C = (Deflection at B due to load P) + (Slope at B due to load P $\times$ Length BC) + (Deflection at C due to moment M).

Here, the length of the beam section from A to B is $L/2$ , and from B to C is also $L/2$ . So, we consider the beam from A to B as a cantilever of length $L/2$ subjected to load P at B.
$\text{Deflection at C} = \frac{P\left ( \frac{L}{2} \right )^3}{3EI}+\frac{P\left ( \frac{L}{2} \right )^2}{2EI}\times \frac{L}{2}+\frac{ML^2}{2EI}$
$\text{Deflection at C} = \frac{PL^3}{24EI}+\frac{PL^3}{16EI}+\frac{ML^2}{2EI}$
To combine the terms with $PL^3$ :
$\text{Deflection at C} = \frac{2PL^3}{48EI}+\frac{3PL^3}{48EI}+\frac{ML^2}{2EI}$
$\text{Deflection at C} = \frac{5PL^3}{48EI}+\frac{ML^2}{2EI}$
Thus, the downward deflection at point C is $\frac{5PL^3}{48EI}+\frac{ML^2}{2EI}$ .

Q39GATE 2019MCQ2MGeneral Engineering

Three Carnot engines E 1 E 2 , E 3 ; operate as shown in the figure (T 1 > T 2 > T 3 ). The efficiency of the engine E 3 in terms of the efficiencies η 1 and η 2 of the engines E 1 and E 2 respectively, is

🚀 Solve in Practice Mode

📖 Explanation

The thermal efficiency ( $\eta$ ) of a Carnot heat engine is defined as $\eta =1-\frac{T_C}{T_H}$ , where $T_H$ is the source temperature and $T_C$ is the sink temperature.

Given the efficiencies for engines $E_1$ , $E_2$ , and $E_3$ :
For engine $E_1$ : $\eta_1 = 1 - \frac{T_2}{T_1} \Rightarrow \frac{T_2}{T_1} = 1 - \eta_1$ ...... (1)
For engine $E_2$ : $\eta_2 = 1 - \frac{T_3}{T_2} \Rightarrow \frac{T_3}{T_2} = 1 - \eta_2$ ...... (2)
For engine $E_3$ : $\eta_3 = 1 - \frac{T_3}{T_1} \Rightarrow \frac{T_3}{T_1} = 1 - \eta_3$ ...... (3)

Now, multiply equations (1) and (2):
$\frac{T_2}{T_1} \times \frac{T_3}{T_2} = \frac{T_3}{T_1} = (1 - \eta_1)(1 - \eta_2)$
Substitute $\frac{T_3}{T_1}$ from equation (3) into this result:
$1 - \eta_3 = (1 - \eta_1)(1 - \eta_2)$
Expand the right side:
$1 - \eta_3 = 1 - \eta_2 - \eta_1 + \eta_1 \eta_2$
Rearrange the terms to solve for $\eta_3$ :
$\eta_3 = 1 - (1 - \eta_1 - \eta_2 + \eta_1 \eta_2)$
$\eta_3 = \eta_1 + \eta_2 - \eta_1 \eta_2$

Q40GATE 2019MCQ2MManufacturing Processes II

True centrifugal casting process in horizontal configuration is to be used for casting a metallic cylinder with outside diameter 0.275 m and inside diameter 0.250 m. If G-factor (ratio of centrifugal force experienced by the rotating cast metal to its weight) is 65 and acceleration due to gravity is 9.8 m/s 2 , the minimum rotational speed (in rpm) required is closest to

🚀 Solve in Practice Mode

📖 Explanation

This problem asks us to find the minimum rotational speed (in rpm) for a true centrifugal casting process. The core idea is that the centrifugal force must be strong enough to hold the molten metal against the mold wall, which is quantified by the G-factor.

The G-factor ( $G$ ) is the ratio of centrifugal acceleration ( $a_c$ ) to the acceleration due to gravity ( $g$ ):
$G = \frac{a_c}{g}$
The centrifugal acceleration at the outer surface (which determines the minimum speed required) is $a_c = \omega^2 r$ , where $\omega$ is the angular velocity and $r$ is the radius. Here, the outer diameter is $D_o = 0.275$ m, so the radius $r = D_o/2 = 0.275/2$ m.
Thus, we have:
$G = \frac{\omega^2 (D_o/2)}{g}$
We need to find the rotational speed $N$ in rpm. The relationship between angular velocity $\omega$ (in rad/s) and rotational speed $N$ (in rpm) is $\omega = \frac{2 \pi N}{60}$ .
Substituting $\omega$ into the G-factor equation:
$G = \frac{\left(\frac{2 \pi N}{60}\right)^2 (D_o/2)}{g}$
Rearranging to solve for $N$ :
$N = \frac{30}{\pi} \sqrt{\frac{2 G g}{D_o}}$
Given $G = 65$ , $g = 9.8$ m/s $^2$ , and $D_o = 0.275$ m:
$N = \frac{30}{\pi} \sqrt{\frac{2 \times 65 \times 9.8}{0.275}}$
$N = \frac{30}{\pi} \sqrt{\frac{1274}{0.275}}$
$N = \frac{30}{\pi} \sqrt{4632.727...}$
$N \approx 9.5493 \times 68.064$
$N \approx 649.75 \text{ rpm}$
Comparing this to the options, the closest value is 650 rpm.

Q41GATE 2019MCQ2MManufacturing Processes II

Ina sine bar, let h denote height of slip gauge and l be the distance between the rollers. The relationship between error in angular measurement (dθ) and errors in the slip gauge combination (dh) and in the spacing of the rollers (dl) is

🚀 Solve in Practice Mode

📖 Explanation

A sine bar uses the principle of trigonometry to measure angles. The fundamental relationship is derived from the right-angled triangle formed by the sine bar, slip gauges, and reference surface:
$\sin\ \theta = \frac{h}{l}$
Here, $h$ is the total height set by the slip gauges, $l$ is the distance between the roller centers, and $\theta$ is the angle.

To find the relationship between small errors in angle measurement ( $d\theta$ ), slip gauge height ( $dh$ ), and roller spacing ( $dl$ ), we use differential calculus. Taking the natural logarithm of both sides of the basic formula:
$\ln(\sin\ \theta) = \ln(h) - \ln(l)$
Now, we differentiate both sides: $d(\ln(\sin\ \theta)) = \frac{\cos\ \theta}{\sin\ \theta} d\theta = \cot\ \theta \ d\theta$ . Similarly, $d(\ln(h)) = \frac{dh}{h}$ and $d(\ln(l)) = \frac{dl}{l}$ . Applying these, we get:
$\cot\ \theta \ d\theta = \frac{dh}{h} - \frac{dl}{l}$
Solving for $d\theta$ :
$d\theta = \frac{1}{\cot\ \theta} \left( \frac{dh}{h} - \frac{dl}{l} \right)$
Since $\frac{1}{\cot\ \theta} = \tan\ \theta$ , the final relationship is:
$d\theta = \tan\ \theta \left( \frac{dh}{h} - \frac{dl}{l} \right)$

Q42GATE 2019MCQ2MManufacturing Processes II

A 100 mm long cylindrical workpiece of diameter 50 mm is reduced to 25 mm diameter using extrusion process. The flow curve for the metal has strength coefficient as K = 750 MPa and the strain hardening co-efficient is 0.15. Assuming no friction and no redundant work, the required ram pressure (in MPa) is closest to

🚀 Solve in Practice Mode

📖 Explanation

This problem asks us to calculate the ram pressure needed for an extrusion process, assuming ideal conditions (no friction, no redundant work). We're given the initial and final dimensions of a cylindrical workpiece and the material's flow curve characteristics: strength coefficient $K$ and strain hardening coefficient $n$ .

First, let's list the given parameters:

Initial workpiece length, $L_0 = 100$ mm
Initial workpiece diameter, $D_0 = 50$ mm
Final workpiece diameter, $D_f = 25$ mm
Strength coefficient, $K = 750$ MPa
Strain hardening coefficient, $n = 0.15$
Assumptions: No friction, no redundant work

The core idea here is that the ram pressure ( $P$ ) is directly related to the work done per unit volume ( $W_v$ ) during the plastic deformation. This work is calculated by integrating the material's flow stress ( $\sigma = K\epsilon^n$ ) over the true strain ( $\epsilon_t$ ) experienced during the process.

Step 1: Calculate True Strain ( $\epsilon_t$ )

For an extrusion process involving a change in cross-sectional area, the true strain is given by:
$\epsilon_t = \ln\left(\frac{A_0}{A_f}\right)$
Where $A_0$ is the initial area and $A_f$ is the final area. For a cylindrical workpiece, $A = \frac{\pi D^2}{4}$ . So, the area ratio simplifies to:
$\frac{A_0}{A_f} = \frac{\frac{\pi D_0^2}{4}}{\frac{\pi D_f^2}{4}} = \left(\frac{D_0}{D_f}\right)^2$
Substituting the given diameters:
$\frac{A_0}{A_f} = \left(\frac{50 \text{ mm}}{25 \text{ mm}}\right)^2 = (2)^2 = 4$
Now, calculate the true strain:
$\epsilon_t = \ln(4)$
Using the property $\ln(a^b) = b \ln(a)$ :
$\epsilon_t = 2 \ln(2)$
Given $\ln(2) \approx 0.693147$ :
$\epsilon_t \approx 2 \times 0.693147 = 1.386294$

Step 2: Determine Ram Pressure ( $P$ )

Under ideal conditions (no friction, no redundant work), the ram pressure ( $P$ ) is equal to the average flow stress multiplied by the true strain, or more precisely, the work done per unit volume ( $W_v$ ). This is found by integrating the flow stress over the true strain:
$W_v = P = \int_0^{\epsilon_t} \sigma d\epsilon = \int_0^{\epsilon_t} K \epsilon^n d\epsilon$
Integrating this expression gives:
$P = \frac{K \epsilon_t^{n+1}}{n+1}$
Now, substitute the values:

$K = 750$ MPa
$n = 0.15$
$\epsilon_t \approx 1.386294$
$n+1 = 0.15 + 1 = 1.15$

First, calculate $\epsilon_t^{n+1}$ :
$\epsilon_t^{n+1} = (1.386294)^{1.15} \approx 1.45754$
Now, calculate the pressure $P$ :
$P = \frac{750 \text{ MPa} \times 1.45754}{1.15}$
$P = \frac{1093.155 \text{ MPa}}{1.15}$
$P \approx 950.57 \text{ MPa}$
Comparing this value with the given options, $950.57$ MPa is closest to $950$ MPa.

Q43GATE 2019MCQ2MIndustrial Engineering

An LPP is defined as Minimize z = 15x 1 + 12x 2 subject to, x 1 + 2x 2 ≤ 3 2x 1 - 4x 2 ≤ 5 x 1 , x 2 ≥ 0 The objective function of the dual of this LPP is

🚀 Solve in Practice Mode

📖 Explanation

To find the objective function of the dual LPP, recall the duality rules. For a minimization primal problem, the dual will be a maximization problem. The number of dual variables equals the number of primal constraints. Crucially, the Right Hand Side (RHS) values of the primal constraints become the coefficients of the dual objective function.

Given the primal LPP:
Minimize $z = 15x_1 + 12x_2$
Subject to:
$x_1 + 2x_2 \le 3$
$2x_1 - 4x_2 \le 5$
$x_1, x_2 \ge 0$

Here, the primal problem has two constraints with RHS values $3$ and $5$ . Since there are two primal constraints, there will be two dual variables, let's call them $y_1$ and $y_2$ . The dual problem will be a maximization problem. Therefore, the coefficients of the dual objective function will be $3$ and $5$ , corresponding to $y_1$ and $y_2$ respectively.

Thus, the dual objective function is:
Maximize $w = 3y_1 + 5y_2$

Q44GATE 2019MCQ2MManufacturing Processes II

A 20 mm HSS drill with a point angle of 118° is used for drilling a through hole on a metallic plate of thickness 100 mm with a cutting speed of 333.33 mm/s and feed of 0.22 mm/rev. Assuming that the drill is touching the surface of the plate at the start, the drilling time (in s) is closest to

🚀 Solve in Practice Mode

📖 Explanation

This problem asks us to calculate the time taken to drill a through hole. We'll first find the drill's rotational speed, then determine the total depth the drill needs to travel, and finally calculate the drilling time using the feed rate.

First, let's find the rotational speed ( $N_{rev/sec}$ ) of the drill using the cutting speed ( $V$ ) and drill diameter ( $D$ ):
$N_{rev/sec} = \frac{V}{\pi D}$
Given $V = 333.33 \text{ mm/s}$ and $D = 20 \text{ mm}$ :
$N_{rev/sec} = \frac{333.33 \text{ mm/s}}{\pi \times 20 \text{ mm}} \approx \frac{333.33}{62.8318} \approx 5.3051 \text{ rev/s}$
Next, we calculate the total length ( $L_{total}$ ) the drill must travel. This includes the plate thickness ( $T$ ) and the extra penetration required for the drill's conical point ( $L_{point}$ ). The drill point's axial length is given by:
$L_{point} = \frac{r}{\tan(\alpha/2)}$
Given drill radius ( $r = D/2 = 10 \text{ mm}$ ) and half point angle ( $\alpha/2 = 118\degree/2 = 59\degree$ ):
$L_{point} = \frac{10 \text{ mm}}{\tan(59\degree)} \approx \frac{10 \text{ mm}}{1.66428} \approx 6.0088 \text{ mm}$
The total drilling length is:
$L_{total} = T + L_{point} = 100 \text{ mm} + 6.0088 \text{ mm} \approx 106.0088 \text{ mm}$
Finally, we calculate the drilling time ( $T_{drill}$ ). First, find the axial feed rate ( $F_{axial}$ ) by multiplying the feed rate ( $f$ ) by the rotational speed:
$F_{axial} = f \times N_{rev/sec} = 0.22 \text{ mm/rev} \times 5.3051 \text{ rev/s} \approx 1.1671 \text{ mm/s}$
Then, the drilling time is:
$T_{drill} = \frac{L_{total}}{F_{axial}}$
$T_{drill} = \frac{106.0088 \text{ mm}}{1.1671 \text{ mm/s}} \approx 90.828 \text{ seconds}$
The calculated drilling time is approximately 90.83 seconds, which is closest to 90 seconds.

Q45GATE 2019MCQ2MIndustrial Engineering

An acceptance sampling plan is selected with sample size n = 80, acceptance number c = 2 for a lot size of 10,000 units. The probability of accepting the lot is based on Poisson distribution. Assuming rectification inspection, if incoming lot quality p is 0.03 and mean (λ) is 2.4, the average outgoing quality (AOQ) is closest to

🚀 Solve in Practice Mode

📖 Explanation

We need to calculate the Average Outgoing Quality (AOQ) for a given acceptance sampling plan with rectification inspection. The core formula for AOQ is $\text{AOQ} = p \times P(a)$ , where $p$ is the incoming proportion of defectives and $P(a)$ is the probability of accepting the lot.

Given: Sample size ( $n$ ) = 80, Acceptance number ( $c$ ) = 2, Incoming lot quality ( $p$ ) = 0.03, Mean ( $\lambda$ ) = 2.4.

The probability of acceptance, $P(a)$ , using the Poisson distribution with $c=2$ and $\lambda=2.4$ is calculated as $P(a) = P(X \le c) = \sum_{x=0}^{c} \frac{e^{-\lambda} \lambda^x}{x!}$ .
$P(a) = P(X=0) + P(X=1) + P(X=2)$
$P(a) = \frac{e^{-2.4} (2.4)^0}{0!} + \frac{e^{-2.4} (2.4)^1}{1!} + \frac{e^{-2.4} (2.4)^2}{2!}$
$P(a) = \frac{0.090718 \times 1}{1} + \frac{0.090718 \times 2.4}{1} + \frac{0.090718 \times 5.76}{2}$
$P(a) = 0.090718 + 0.217723 + 0.261268 = 0.57019$

Now, substitute $p$ and $P(a)$ into the AOQ formula:
$\text{AOQ} = 0.03 \times 0.57019 = 0.0171057$

Comparing this value with the given options, $0.0171057$ is closest to $0.0170$ .

Q46GATE 2019MCQ2MIndustrial Engineering

The mean time to repair (MTTR) for a repairable system is 30 minutes. When maintenance time changes from 20 minutes to 40 minutes, the net increase in maintainability is closest to

🚀 Solve in Practice Mode

📖 Explanation

Let's break down how changing the available maintenance time impacts a system's "maintainability."

Understanding Maintainability

Maintainability ( $M$ ) tells us the probability of successfully repairing a system within a given time ( $t$ ). A key concept here is Mean Time To Repair (MTTR), which is the average time it takes to fix a system. We're given an MTTR of 30 minutes.

For repair calculations, we usually assume an exponential distribution. The formula for maintainability in this case is:
$M(t) = 1 - e^{-\lambda t}$
Here, $\lambda$ is the repair rate, which is the inverse of the MTTR ( $\lambda = \frac{1}{\text{MTTR}}$ ).

Step-by-Step Calculation

Calculate the Repair Rate ( $\lambda$ ):
Given MTTR = 30 minutes,
$\lambda = \frac{1}{30} \text{ repairs/minute}$
Calculate Initial Maintainability ( $M_1$ ) at $t_1 = 20$ minutes:
$M_1 = 1 - e^{-(\frac{1}{30}) \times 20} = 1 - e^{-2/3}$
$M_1 \approx 1 - 0.5134 = 0.4866$
Calculate New Maintainability ( $M_2$ ) at $t_2 = 40$ minutes:
$M_2 = 1 - e^{-(\frac{1}{30}) \times 40} = 1 - e^{-4/3}$
$M_2 \approx 1 - 0.2466 = 0.7534$
Determine the Net Increase in Maintainability:
The net increase is $M_2 - M_1$ .
$\text{Net Increase} = 0.7534 - 0.4866 = 0.2668$
Alternatively, using the exact exponential terms:
$\Delta M = e^{-2/3} - e^{-4/3} \approx 0.513417 - 0.246604 \approx 0.266813$

Comparing with Options

The calculated net increase is approximately 0.2668. Let's compare this with the given options:

A. 0.15: $|0.2668 - 0.15| = 0.1168$
B. 0.25: $|0.2668 - 0.25| = 0.0168$
C. 0.45: $|0.2668 - 0.45| = 0.1832$
D. 0.60: $|0.2668 - 0.60| = 0.3332$

The closest option is B, with a difference of 0.0168. Therefore, the net increase in maintainability is closest to 0.25.

Q47GATE 2019MCQ2MIndustrial Engineering

A company invests Rs. 50 thousand in assets. The initial investment is Rs. 30 thousand with two subsequent investments of Rs.10 thousand each at the end of 1 st year and 2 nd year. The useful life of the assets is 10 years with no salvage value at the end. If the interest rate is 10% and the minimum attractive rate of return (MARR) is 12%, the annual capital recovery and return (CRR) in thousands of Rs. is

🚀 Solve in Practice Mode

📖 Explanation

To calculate the Capital Recovery and Return (CRR), we first need to find the Present Worth (PW) of all investments using the given interest rate, and then convert this PW into an equivalent annual amount using the Minimum Attractive Rate of Return (MARR).

Step 1: Calculate the Present Worth (PW) of all investments.
The total investment includes an initial payment of Rs. 30,000, Rs. 10,000 at the end of year 1, and Rs. 10,000 at the end of year 2. We use an interest rate ( $i$ ) of 10% (0.10) to bring all these future amounts to their present value (Year 0).
The formula for Present Worth is $PW = \sum_{t=0}^{n} \frac{CF_t}{(1+i)^t}$ .

PW of initial investment (Year 0): Rs. 30,000
PW of investment at Year 1: $\frac{10,000}{(1+0.10)^1} = \frac{10,000}{1.10} \approx$ Rs. 9,090.91
PW of investment at Year 2: $\frac{10,000}{(1+0.10)^2} = \frac{10,000}{1.21} \approx$ Rs. 8,264.46

Total Present Worth (PW) = Rs. 30,000 + Rs. 9,090.91 + Rs. 8,264.46 = Rs. 47,355.37.
In thousands, this is Rs. 47.355 thousand.

Step 2: Convert the total Present Worth (PW) to an Equivalent Uniform Annual Series (CRR).
We use the Minimum Attractive Rate of Return (MARR, $i'$ ) of 12% (0.12) and the useful life ( $n$ ) of 10 years to find the annual capital recovery and return. This is done using the Capital Recovery Factor (CRF).
The formula for CRR is $CRR = PW \times CRF(i', n)$ .
The Capital Recovery Factor (CRF) formula is $CRF(i', n) = \frac{i'(1+i')^n}{(1+i')^n - 1}$ .

First, calculate CRF for $i' = 12\%$ and $n = 10$ years:
$CRF(12\%, 10) = \frac{0.12(1+0.12)^{10}}{(1+0.12)^{10} - 1}$
$CRF(12\%, 10) = \frac{0.12(1.12)^{10}}{(1.12)^{10} - 1}$
Since $(1.12)^{10} \approx 3.105848$ ,
$CRF(12\%, 10) = \frac{0.12 \times 3.105848}{3.105848 - 1} = \frac{0.37270176}{2.105848} \approx 0.176984$

Now, calculate CRR:
$CRR = \text{Rs. } 47,355.37 \times 0.176984 \approx \text{Rs. } 8,381.15$

Converting to thousands of Rupees, the annual capital recovery and return (CRR) is approximately Rs. 8.38 thousand.

Q48GATE 2019MCQ2MIndustrial Engineering

The man-hours required (T n ) to manufacture the n th unit in a plant is given by T n = T 1 n b , where b = -0.322 at the 80% learning rate. If the manufacturing time for the first unit (T 1 ) is 80 man-hours, the total time (in man-hours) required to manufacture the first 4 units, at 80% learning rate, is

🚀 Solve in Practice Mode

📖 Explanation

This problem involves the learning curve concept, which states that the time required to produce a unit decreases as more units are manufactured. This is due to improved efficiency, skill development, and better methods over time.

The formula provided for the man-hours required for the $n^{th}$ unit is:
$T_n = T_1 \cdot n^b$
Here, $T_n$ is the time for the $n^{th}$ unit, $T_1$ is the time for the first unit, $n$ is the unit number, and $b$ is the learning curve exponent.

We are given:

$T_1 = 80$ man-hours (time for the first unit).
$b = -0.322$ (learning curve exponent, corresponding to an 80% learning rate).

To find the total time for the first 4 units, we need to calculate $T_1, T_2, T_3,$ and $T_4$ individually and then sum them up.

For the 1st Unit ( $T_1$ ):
This is given directly:
$T_1 = 80 \text{ man-hours}$
For the 2nd Unit ( $T_2$ ):
Using the formula with $n=2$ :
$T_2 = T_1 \times 2^b = 80 \times 2^{-0.322}$
$T_2 = 80 \times 0.79997 \approx 63.998 \text{ man-hours}$
For the 3rd Unit ( $T_3$ ):
Using the formula with $n=3$ :
$T_3 = T_1 \times 3^b = 80 \times 3^{-0.322}$
$T_3 = 80 \times 0.70453 \approx 56.362 \text{ man-hours}$
For the 4th Unit ( $T_4$ ):
Using the formula with $n=4$ :
$T_4 = T_1 \times 4^b = 80 \times 4^{-0.322}$
$T_4 = 80 \times 0.64000 \approx 51.200 \text{ man-hours}$

Now, sum these individual times to get the total time for the first 4 units:
$\text{Total Time} = T_1 + T_2 + T_3 + T_4$
$\text{Total Time} = 80.000 + 63.998 + 56.362 + 51.200$
$\text{Total Time} \approx 251.560 \text{ man-hours}$

Q49GATE 2019MCQ2MIndustrial Engineering

A firm, with a production target of 50,000 units/year, has the following data for the selection of a new location for its plant Location Fixed cost (Rs.) Variable costs per unit (Rs.) P 110,000 2 Q 95,000 2.5 R 80,000 3 S 75,000 3.5 The most economical location for the firm is

🚀 Solve in Practice Mode

📖 Explanation

To find the most economical location, we need to calculate the total cost for each location at the production target of 50,000 units/year. The total cost is given by the formula:
$\text{Total Cost} (TC) = \text{Fixed Cost} (FC) + (\text{Variable Cost per Unit} (VC) \times \text{Production Volume})$
Let's calculate the total cost for each location:

For Location P:
$\text{Total Cost}_P = 10,000 + (2 \times 50,000) = 10,000 + 100,000 = 110,000 \, \text{Rs.}$

For Location Q:
$\text{Total Cost}_Q = 95,000 + (2.5 \times 50,000) = 95,000 + 125,000 = 220,000 \, \text{Rs.}$

For Location R:
$\text{Total Cost}_R = 80,000 + (3 \times 50,000) = 80,000 + 150,000 = 230,000 \, \text{Rs.}$

For Location S:
$\text{Total Cost}_S = 75,000 + (3.5 \times 50,000) = 75,000 + 175,000 = 250,000 \, \text{Rs.}$
Comparing the total costs: Location P has a total cost of Rs. 110,000, Location Q has Rs. 220,000, Location R has Rs. 230,000, and Location S has Rs. 250,000. Location P has the lowest total cost.

Q50GATE 2019MCQ2MManufacturing Processes II

Considering included angle θ of the thread to be 60° using the Best-Wire method, the difference between the effective diameter (E) and the dimension under the wire (T) for M 10 x 1.0 mm is closest to

🚀 Solve in Practice Mode

📖 Explanation

This question asks for the difference between the effective diameter (E) and the dimension under the wire (T) for an M 10 x 1.0 mm thread, using the Best-Wire method. We are given the thread's nominal major diameter ( $D$ ) as 10 mm, pitch ( $P$ ) as 1.0 mm, and included thread angle ( $\theta$ ) as 60°.

First, we calculate the best wire diameter ( $w$ ) using the formula:
$w = \frac{P}{2 \cos(\theta/2)}$
Given $P = 1.0$ mm and $\theta/2 = 60^\circ / 2 = 30^\circ$ , we substitute these values:
$w = \frac{1.0}{2 \times \cos(30^\circ)} = \frac{1.0}{2 \times \frac{\sqrt{3}}{2}} = \frac{1.0}{\sqrt{3}} \approx 0.57735 \text{ mm}$
The difference between the effective diameter (E) and the dimension under the wire (T) is given by $E - T = \frac{w}{2}$ .
$E - T = \frac{0.57735}{2} \approx 0.288675 \text{ mm}$
Rounding to three decimal places, the difference is $0.289$ mm.

Q51GATE 2019NAT2MEngineering Mathematics

If z is a complex variable with i = √-1, the length of the minor axis of an ellipse defined by |z - (1 + i) | + |z - (9 + i) | = 10 is ______

🚀 Solve in Practice Mode

📖 Explanation

The given equation for the ellipse is $|z - (1 + i)| + |z - (9 + i)| = 10$ . This is in the standard complex form for an ellipse: $|Z - Z_1| + |Z - Z_2| = 2a$ .
Comparing the given equation with the standard form, we identify the length of the major axis as $2a = 10$ , so $a = 5$ . The foci are $Z_1 = (1, 1)$ and $Z_2 = (9, 1)$ .
The distance between the center and each focus is $c$ . The distance between the two foci is $|Z_1 - Z_2| = |(1+i) - (9+i)| = |-8| = 8$ . Since the distance between foci is $2c$ , we have $2c = 8$ , which means $c = 4$ .
For an ellipse, the relationship between $a$ , $b$ (half of the minor axis length), and $c$ is $a^2 - b^2 = c^2$ .
Substituting the values of $a$ and $c$ :
$5^2 - b^2 = 4^2$
$25 - b^2 = 16$
$b^2 = 25 - 16 = 9$
$b = 3$
The length of the minor axis is $2b = 2 \times 3 = 6$ .

Q52GATE 2019NAT2MEngineering Mathematics

The numerical value of the definite integral $\int^1_0{e^{-x}}dx$ using trapezoidal rule with function evaluations at points x = 0, 0.5 and 1 is ______ (round off to 3 decimal places)

🚀 Solve in Practice Mode

📖 Explanation

The Trapezoidal Rule approximates a definite integral using trapezoids. The general formula is:
$\int^b_a{f(x)dx} = \frac{h}{2}\left[ y_0 + y_n + 2(y_1 + y_2 + y_3 + \ldots) \right]$
Here, $h$ is the step size, $a$ is the lower limit, and $b$ is the upper limit. The number of intervals $n = \frac{b-a}{h}$ .

Given $x$ values are $0, 0.5, 1$ . This means the step size $h = 0.5$ .
The function is $f(x) = e^{-x}$ . We have:
$y_0 = f(0) = e^0 = 1$
$y_1 = f(0.5) = e^{-0.5} = 0.6065$
$y_2 = f(1) = e^{-1} = 0.3678$ (This corresponds to $y_n$ since it's the last point).

Applying the trapezoidal rule for two intervals ( $n=2$ ):
$\int_0^1e^{-x}dx = \frac{h}{2}[y_0 + y_2 + 2y_1]$
Substituting the values:
$\int_0^1e^{-x}dx = \frac{0.5}{2}[1 + 0.3678 + (2 \times 0.6065)]$
$\int_0^1e^{-x}dx = 0.25[1.3678 + 1.213]$
$\int_0^1e^{-x}dx = 0.25[2.5808] = 0.6452$
Rounding off to 3 decimal places, the value is $0.645$ .

Q53GATE 2019NAT2MGeneral Engineering

A thin walled cylindrical pressure vessel with an inside diameter of 300 mm and wall thickness of 3 mm is subjected to an internal gauge pressure of 1.5 MPa. The maximum shear stress (in MPa) at a point located on the inner surface of the pressure vessel is

🚀 Solve in Practice Mode

📖 Explanation

Let's break down how to find the maximum shear stress in this thin-walled cylindrical pressure vessel.

First, we need to understand the stresses acting on the vessel. In a thin cylindrical pressure vessel, we have two primary stresses:

Circumferential (Hoop) Stress ( $\sigma_c$ ): This acts around the circumference of the cylinder, like a hoop. It's given by $\sigma_c = \frac{Pr}{t} = \frac{Pd}{2t}$ .
Longitudinal Stress ( $\sigma_l$ ): This acts along the length of the cylinder. It's given by $\sigma_l = \frac{Pr}{2t} = \frac{Pd}{4t}$ .

Here, $P$ is the internal gauge pressure, $d$ is the inside diameter, and $t$ is the wall thickness.

Now, let's talk about maximum shear stress ( $\tau_{max}$ ). Shear stress tends to cause material to fail by shearing, and engineers are often interested in the maximum shear stress because it's a critical factor in design. The general formula for maximum shear stress is $\tau_{max} = \frac{\sigma_{max} - \sigma_{min}}{2}$ .

We consider two scenarios for shear stress:

In-plane shear stress: This is the shear stress acting within the plane defined by the circumferential and longitudinal stresses. Here, $\sigma_{max} = \frac{Pd}{2t}$ and $\sigma_{min} = \frac{Pd}{4t}$ . So, the in-plane shear stress is $\tau_{max} = \frac{Pd}{8t}$ .
Out-of-plane shear stress: This considers the stresses in a plane perpendicular to the surface. Here, the principal stresses are $\sigma_1 = \frac{Pd}{2t}$ , $\sigma_2 = \frac{Pd}{4t}$ , and $\sigma_3 = 0$ (since the outer surface is exposed to atmospheric pressure, we consider the stress perpendicular to the surface to be zero). Comparing these, the true maximum and minimum principal stresses are $\sigma_{max} = \frac{Pd}{2t}$ and $\sigma_{min} = 0$ . Therefore, the out-of-plane maximum shear stress is $\tau_{max} = \frac{Pd}{4t}$ .

Important Note for GATE: When a question asks for the "maximum shear stress" in a thin-walled pressure vessel without further clarification, it refers to the absolute maximum shear stress, which is almost always the out-of-plane shear stress because it's higher than the in-plane shear stress and often dictates failure.

Let's plug in the given values:
$d = 300 \, \text{mm}$
$t = 3 \, \text{mm}$
$P = 1.5 \, \text{MPa}$

Using the formula for out-of-plane maximum shear stress:
$\tau_{max} = \frac{Pd}{4t}$
$\tau_{max} = \frac{1.5 \times 300}{4 \times 3}$
$\tau_{max} = \frac{450}{12}$
$\tau_{max} = 37.5 \, \text{MPa}$

Q54GATE 2019NAT2MGeneral Engineering

A cam is designed to achieve a simple harmonic motion of a flat-faced follower. Starting from the rest, the follower rises to the maximum height of 50 mm at 180° of cam rotation as shown in the figure. If the cam rotates at a uniform angular speed of 100 rpm, the speed of the follower (in mm/s) at the instance when the cam rotates 45° from the initial position is ______

🚀 Solve in Practice Mode

📖 Explanation

For a follower undergoing Simple Harmonic Motion (SHM), the instantaneous velocity ( $V_0$ ) is given by:

$V_0 = \frac{S}{2} \times \frac{\pi \omega}{\theta_0} \sin\left(\frac{\pi \theta}{\theta_0}\right)$

First, we need to convert the given angular speed from RPM to rad/s: $\omega = \frac{2\pi N}{60}$ .
Given values are: Stroke length ( $S$ ) = $50 \text{ mm} = 0.05 \text{ m}$ , Outstroke angle ( $\theta_0$ ) = $180^\circ = \pi \text{ rad}$ , Cam angular speed ( $N$ ) = $100 \text{ rpm}$ , and Instantaneous angle ( $\theta$ ) = $45^\circ = \frac{\pi}{4} \text{ rad}$ .

Let's calculate $\omega$ :
$\omega = \frac{2\pi \times 100}{60} = 10.4719 \text{ rad/s}$

Now, substitute all values into the velocity formula:
$V_0 = \frac{50}{2} \times \left(\frac{\pi \times 10.4719}{\pi}\right) \times \sin\left(\frac{\pi \times \frac{\pi}{4}}{\pi}\right)$
$V_0 = 25 \times 10.4719 \times \sin\left(\frac{\pi}{4}\right)$
$V_0 = 261.7975 \times 0.7071$
$V_0 = 185.118 \text{ mm/s}$

Q55GATE 2019NAT2MGeneral Engineering

An open tank of 2 m x 2 m x 2 m is filled with layers of two fluids. Depth of each layer is one meter. The top layer is that of an oil of specific gravity 0.8. The bottom layer is of water. Consider the density of water ρ w = 1000 kg/m 3 and acceleration due to gravity g = 9.8 m/s 2 . Neglecting the effect of atmospheric pressure, the force (in N) exerted by the fluids on one of the side walls of the tank is ______

🚀 Solve in Practice Mode

📖 Explanation

To calculate the force exerted by the fluids on a side wall, we can use the concept of a pressure prism. The tank dimensions are given as: Depth $2$ m, width $2$ m, Length $2$ m.
The specific weight of water is $\rho_w g = 1000 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 = 9800 \text{ N/m}^3$ . Let's denote specific weight generally as $w$ .
For oil, the specific gravity is $0.8$ , so its specific weight is $0.8w$ .

The pressure at the interface between oil and water (at $h_1 = 1$ m from the top) is:
$P_1 = S_{oil} \times w \times h_1 = 0.8w \times 1 = 0.8w$

The pressure at the bottom of the tank (at $h_2 = 1$ m from the interface, or $2$ m from the top) is:
$P_2 = P_1 + w \times h_2 = 0.8w + w \times 1 = 1.8w$

The force $F$ on one side wall is the area of the pressure diagram multiplied by the width of the wall. The pressure diagram consists of a triangle (for the oil layer) and a trapezoid (for the water layer).
$F = \text{width} \times \left( \frac{1}{2} (0.8w) \times 1 + \frac{1}{2} (1.8w + 0.8w) \times 1 \right)$
$F = 2 \times \left( 0.4w + \frac{1}{2} (2.6w) \right)$
$F = 2 \times (0.4w + 1.3w)$
$F = 2 \times (1.7w)$
$F = 3.4w$

Now, substitute the value of $w = 9800 \text{ N/m}^3$ :
$F = 3.4 \times 9800 = 33320 \text{ N}$

Q56GATE 2019NAT2MGeneral Engineering

During a storm, the wind speed is 90 km/hr. In a high-rise building, there is a window of size 1.2 m x 1.8 m facing the storm on an upper floor. Neglecting the ground effects on wind speed and considering the density of air ρ air = 1.2 kg/m 3 , the force (in N) acting on the window due to the storm is ______

🚀 Solve in Practice Mode

📖 Explanation

The force ( $F$ ) acting on the window due to the storm is determined by the pressure difference across the window multiplied by its area. When wind hits the window, the pressure on the windward side becomes the stagnation pressure, given by $P_o = P_\infty + \frac{1}{2}\rho_{air}V^2$ . Thus, the force is $F = (P_o - P_\infty ) \times A_{Window} = \frac{1}{2}\rho_{air}V^2 \times A_{Window}$ .

Given: $\rho_{air} = 1.2 \text{ kg/m}^3$ , $A_{Window} = 1.2 \text{ m} \times 1.8 \text{ m}$ , and $V = 90 \text{ km/hr} = 25 \text{ m/sec}$ .

Substituting these values into the formula:
$F = \frac{1}{2} \times 1.2 \times (25)^2 \times (1.2 \times 1.8)$
$F = 810 \text{ N}$

Q57GATE 2019NAT2MManufacturing Processes II

The heat transfer efficiency in arc welding of a plate using a current of 250 A at 20 V is 90%. The heat required to melt the material is 10 J/mm 3 . If the cross-sectional area of the weld joint is 30 mm 2 and the travel speed is 5 mm/s, the melting efficiency (in %) is (round off to 2 decimal places)

🚀 Solve in Practice Mode

📖 Explanation

The melting efficiency ( $\eta_m$ ) in welding is the ratio of the heat required to melt the material ( $H_m$ ) to the total heat supplied ( $H_s$ ).
${\eta _m} = \frac{{{H_m}}}{{{H_s}}}$
The heat supplied per unit volume ( $H_s$ ) can be calculated using the formula:
${H_s} = \frac{{VI}}{{{A} \times v}} \times {\eta _h}\;\;J/m{m^3}$
Given: $H_m = 10 \text{ J/mm}^3$ , Arc Voltage $V = 20 \text{ V}$ , Current $I = 250 \text{ A}$ , Arc efficiency $\eta_h = 0.90$ , Travel speed $v = 5 \text{ mm/s}$ , and Weld bead area $A = 30 \text{ mm}^2$ .
Substitute these values into the formula for $H_s$ :
${H_s} = \frac{{20~\times ~250}}{{{30} ~\times ~ 5}} \times {0.90} = 30 \text{ J/mm}^3$
Now, calculate the melting efficiency $\eta_m$ :
${\eta _m} = \frac{{{H_m}}}{{{H_s}}}~=~\frac{10}{30} = 0.3333$
Therefore, the melting efficiency is $33.33\%$ .

Q58GATE 2019NAT2MManufacturing Processes II

A sand casting process has a mold constant of 2 s/mm 2 and solidification exponent of 2. If the solidification time is to be doubled for a given unit volume of material, the corresponding reduction in the cast surface area (in %) is ______

🚀 Solve in Practice Mode

📖 Explanation

The solidification time ( $t_s$ ) in sand casting is described by Chvorinov's Rule: $t_s = C{\left( {\frac{V}{A}} \right)^n}$ , where $C$ is the mold constant, $V$ is the volume of the casting, $A$ is the surface area of the casting, and $n$ is the solidification exponent (given as 2).

Given that $n=2$ , the formula becomes $t_s = C{\left( {\frac{V}{A}} \right)^2}$ .
For a given unit volume of material, $V$ is constant. The mold constant $C$ is also constant.
Therefore, $t_s \propto \left(\frac{1}{A}\right)^2$ , which implies $A^2 \propto \frac{1}{t_s}$ , or $A \propto \frac{1}{\sqrt{t_s}}$ .

We are given that the solidification time is to be doubled, so $t_2 = 2t_1$ .
Using the proportionality $A \propto \frac{1}{\sqrt{t_s}}$ :
$\frac{A_2}{A_1} = \frac{\sqrt{t_1}}{\sqrt{t_2}} = \frac{\sqrt{t_1}}{\sqrt{2t_1}} = \frac{1}{\sqrt{2}} \approx 0.707$ .

The reduction in cast surface area is calculated as:
$\frac{A_2 - A_1}{A_1} = \frac{A_2}{A_1} - 1 = 0.707 - 1 = -0.2928$ .
The negative sign indicates a reduction.
Therefore, the percentage reduction in the cast surface area is $0.2928 \times 100 = 29.28\% \approx 29.3\%$ .

Q59GATE 2019NAT2MManufacturing Processes II

During a turning operation of a specific work material having a shear strength of 220 MPa under orthogonal cutting condition, the process parameters are Feed 0.2 mm/rev Depth of cut 1 mm Rake angle -5° Given chip thickness ratio as 0.5, friction angle as 49.2° and shear angle as 25.4°, the feed force (in N) is ________

🚀 Solve in Practice Mode

📖 Explanation

To determine the feed force ( $F_t$ ), we first need to calculate the shear force ( $F_s$ ).
The shear force ( $F_s$ ) is given by the formula: $F_s = \frac{t_0 b}{\sin \phi} \times \tau_s$
Given values are: Rake angle ( $\alpha$ ) = $-5^\circ$ , friction angle ( $\beta$ ) = $49.2^\circ$ , shear angle ( $\phi$ ) = $25.4^\circ$ , shear strength ( $\tau_s$ ) = $220 \text{ MPa}$ .
The feed ( $f$ ) is the uncut chip thickness ( $t_0$ ) = $0.2 \text{ mm/rev}$ , and the depth of cut ( $d$ ) is the width of chip ( $b$ ) = $1 \text{ mm}$ .
Substituting these values: $F_s = \frac{0.2 \times 1}{\sin 25.4^\circ} \times 220 = 102.58 \text{ N}$
Now, we can calculate the feed force ( $F_t$ ) using the formula: $F_t = \frac{F_s \sin (\beta - \alpha)}{\cos (\phi + \beta - \alpha)}$
Plugging in the values: $F_t = \frac{102.58 \times \sin (49.2^\circ - (-5^\circ))}{\cos (25.4^\circ + 49.2^\circ - (-5^\circ))} = \frac{102.58 \times \sin (54.2^\circ)}{\cos (79.6^\circ)}$
Therefore, $F_t = 460.88 \text{ N}$ .

Q60GATE 2019NAT2MManufacturing Processes II

A CO 2 laser in continuous mode is used for drilling a plate. The process parameters and their values are Laser power intensity 1 × 10 8 W/mm 2 Vaporization energy 5 × 10 6 J/mm 3 Efficiency of the process 15% Laser spot diameter 200 micrometer The drilled depth (in mm) after 2 seconds is ______

🚀 Solve in Practice Mode

📖 Explanation

The core concept here is that the volume of material vaporized depends on the effective energy supplied and the energy required to vaporize a unit volume.

First, let's calculate the laser power being delivered. The laser power ( $P$ ) is the laser power intensity multiplied by the spot area:
$P = \text{Laser power intensity} \times \frac{\pi d^2}{4}$
$P = 1 \times 10^8 \text{ W/mm}^2 \times \frac{\pi \times (0.2 \text{ mm})^2}{4}$
$P = 3.14 \text{ MW}$

Next, we find the net energy effectively used for vaporization over 2 seconds, considering the process efficiency:
$\text{Net energy provided} = P \times \eta_{\text{process}} \times \text{time}$
$\text{Net energy provided} = 3.14 \times 10^6 \text{ W} \times 0.15 \times 2 \text{ s} = 942.477 \text{ kJ}$

Now, we calculate the volume of metal vaporized using the net energy and the vaporization energy:
$\text{Volume of metal vaporized} = \frac{\text{Net energy provided}}{\text{Vaporization energy}}$
(Note: Vaporization energy $= 5 \times 10^6 \text{ J/mm}^3 = 5000 \text{ kJ/mm}^3$ )
$\text{Volume of metal vaporized} = \frac{942.477 \text{ kJ}}{5000 \text{ kJ/mm}^3} = 0.188 \text{ mm}^3$

Finally, assuming a cylindrical drilled hole, the volume of the vaporized metal is $\frac{\pi}{4} \times d^2 \times H$ , where $H$ is the drilled depth.
$\frac{\pi}{4} \times (0.2 \text{ mm})^2 \times H = 0.188 \text{ mm}^3$
$H = 6 \text{ mm}$

Q61GATE 2019NAT2MIndustrial Engineering

A process which is in a state of statistical control (within ± 3σ) has an estimate of standard deviation (σ) 2 mm. The specification limits for the corresponding product are 120 ± 8 mm. When process mean shifts from 118 mm to 122 mm with no change in process standard deviation, the difference in process capability index C pk is ______

🚀 Solve in Practice Mode

📖 Explanation

The Process Capability Index ( $C_{pk}$ ) helps us understand how well a process is performing relative to its design specifications. It's calculated using the formula: $C_{pk} = \min\left[ \frac{\text{USL} - \mu}{3\sigma}, \frac{\mu - \text{LSL}}{3\sigma} \right]$ Here, $\text{USL}$ is the Upper Specification Limit, $\text{LSL}$ is the Lower Specification Limit, $\mu$ is the process mean, and $\sigma$ is the standard deviation.

Given $\sigma = 2$ mm, and specification limits $120 \pm 8$ mm, we have $\text{USL} = 120 + 8 = 128$ mm and $\text{LSL} = 120 - 8 = 112$ mm.

(i) When the process mean $\mu = 118$ mm:
$C_{pk} = \min\left[ \frac{128 - 118}{3 \times 2}, \frac{118 - 112}{3 \times 2} \right] = \min\left[ \frac{10}{6}, \frac{6}{6} \right] = \min[1.67, 1] = 1$

(ii) When the process mean $\mu = 122$ mm:
$C_{pk} = \min\left[ \frac{128 - 122}{3 \times 2}, \frac{122 - 112}{3 \times 2} \right] = \min\left[ \frac{6}{6}, \frac{10}{6} \right] = \min[1, 1.6671] = 1$

Therefore, the difference in $C_{pk}$ values is $1 - 1 = 0$ .

Q62GATE 2019NAT2MIndustrial Engineering

A monitoring system has seven components. The reliability of each component is shown in the figure. The system reliability is ______ (round off to 2 decimal places)

🚀 Solve in Practice Mode

📖 Explanation

Let's break down how to calculate the reliability of this system. We need to remember two key formulas:

For components connected in parallel: The system reliability, $R_p(t)$ , is given by:
$R_p(t) = 1 - \left[ \left( 1 - R_1(t) \right) \left( 1 - R_2(t) \right) \ldots \left( 1 - R_n(t) \right) \right]$
This formula calculates the probability that at least one parallel component works.
For components connected in series: The system reliability, $R_s(t)$ , is given by:
$R_s(t) = R_1(t) \times R_2(t) \times \ldots \times R_n(t)$
This formula means all series components must work for the system to work.

Now, let's apply these to our problem. We'll calculate the reliability step-by-step for different sections of the system:

Calculation:

First, let's find the reliability of the two series blocks:

For components $R_2$ and $R_3$ (in series):
$R_{s1} = R_2 \times R_3 = 0.5 \times 0.7 = 0.35$
For components $R_5$ and $R_6$ (in series):
$R_{s2} = R_5 \times R_6 = 0.8 \times 0.75 = 0.6$

Next, we calculate the reliability of the parallel arrangement consisting of $R_{s1}$ , $R_4$ , and $R_{s2}$ . Let's call this $R_{s3}$ :

For $R_{s1}$ , $R_4$ , and $R_{s2}$ (in parallel, where $R_4$ has a reliability of 0.5):
$R_{s3} = 1 - (1 - R_{s1})(1 - R_4)(1 - R_{s2})$
$R_{s3} = 1 - (1 - 0.35)(1 - 0.5)(1 - 0.6)$
$R_{s3} = 1 - (0.65)(0.5)(0.4) = 1 - 0.13 = 0.87$

Finally, the overall system reliability ( $R_s$ ) is determined by the series connection of $R_1$ , $R_{s3}$ , and $R_7$ :

For $R_1$ , $R_{s3}$ , and $R_7$ (in series):
$R_s = R_1 \times R_{s3} \times R_7$
$R_s = 0.5 \times 0.87 \times 0.4 = 0.174$

Rounding off to two decimal places, the system reliability is $\mathbf{0.17}$ .

Q63GATE 2019NAT2MIndustrial Engineering

A PERT project network consists of 5 activities A to E. The time estimates of these activities follow Beta-distribution. The predecessor—successor (P-S) relationships between the nodes and time estimates of activities are given in table. Activity P – S Optimistic time (days) Most likely time (days) Pessimistic time (days) A 1 – 2 2 4 6 B 2 – 3 4 5 12 C 2 – 4 5 8 11 D 3 – 5 2 5 08 E 4 – 5 4 6 14 The variance (in day 2 ) of the critical path is ______ (round off to 2 decimal places)

🚀 Solve in Practice Mode

📖 Explanation

To find the variance of the critical path in a PERT network, we first need to calculate the expected completion time ( $T_e$ ), standard deviation ( $\sigma$ ), and variance ( $\sigma^2$ ) for each activity using the Beta-distribution formulas:

$T_e = \frac{T_o + 4T_m + T_p}{6}$
$\sigma = \frac{T_p - T_o}{6}$
$Variance = \sigma^2$

Where $T_o$ is the optimistic time, $T_m$ is the most likely time, and $T_p$ is the pessimistic time.

Now, let's calculate these values for each activity:

| Activity | $T_o$ | $T_m$ | $T_p$ | $T_e$ | $\sigma$ | Variance |
|---|---|---|---|---|---|---|
| A | 2 | 4 | 6 | 4 | $2/3$ | $4/9$ |
| B | 4 | 5 | 12 | 6 | $4/3$ | $16/9$ |
| C | 5 | 8 | 11 | 8 | 1 | 1 |
| D | 2 | 5 | 8 | 5 | 1 | 1 |
| E | 4 | 6 | 14 | 7 | $5/3$ | $25/9$ |

Next, we identify all possible paths through the network and calculate their total expected completion times:
Path 1: $1 \to 2 \to 3 \to 5$ has a total expected time of $T_e(A) + T_e(B) + T_e(D) = 4 + 6 + 5 = 15$ days.
Path 2: $1 \to 2 \to 4 \to 5$ has a total expected time of $T_e(A) + T_e(C) + T_e(E) = 4 + 8 + 7 = 19$ days.

Comparing the total expected times, the critical path is $1 \to 2 \to 4 \to 5$ with an expected completion time of 19 days.

Finally, the variance of the critical path is the sum of the variances of the activities on that path:
Variance of Critical Path = Variance(A) + Variance(C) + Variance(E) = $\frac{4}{9} + 1 + \frac{25}{9} = \frac{29}{9} + 1 = \frac{38}{9} = 4.222...$

Rounding off to two decimal places, the variance of the critical path is $\textbf{4.22}$ .

Q64GATE 2019NAT2MIndustrial Engineering

The sales data of a product for 5 years are Year 2014 2015 2016 2017 2018 Sales (units) 280 268 259 270 287 Assume the forecast for the year 2014 as 260 units. Using an exponential smoothing method with smoothing constant α = 0.5, the sales forecast (units) for the year 2019, is ______

🚀 Solve in Practice Mode

📖 Explanation

The Exponential Smoothing Method helps us predict future values based on past data. The core idea is that the forecast for the next period is an adjustment of the old forecast, taking into account the most recent actual data. The formula is given by:
$F_t = \alpha D_{t-1} + (1 - \alpha) F_{t-1}$
This can also be written as:
$F_t = F_{t-1} + \alpha(D_{t-1} - F_{t-1})$
Where:

$F_t$ is the forecast for the current period (the one we're calculating).
$D_{t-1}$ is the actual sales (demand) from the previous period.
$F_{t-1}$ is the forecast that was made for the previous period.
$\alpha$ (alpha) is the smoothing constant, which determines how much weight we give to the most recent actual data.

We are given the sales data for several years and an initial forecast for 2014 ( $F_{2014} = 260$ units) with a smoothing constant $\alpha = 0.5$ . We need to find the forecast for 2019 ( $F_{2019}$ ).

Let's calculate the forecasts step-by-step:

Forecast for 2015 ( $F_{2015}$ ):
$F_{2015} = F_{2014} + \alpha(D_{2014} - F_{2014})$
$F_{2015} = 260 + 0.5(280 - 260) = 260 + 0.5(20) = 260 + 10 = 270$ units.
Forecast for 2016 ( $F_{2016}$ ):
$F_{2016} = F_{2015} + \alpha(D_{2015} - F_{2015})$
$F_{2016} = 270 + 0.5(268 - 270) = 270 + 0.5(-2) = 270 - 1 = 269$ units.
Forecast for 2017 ( $F_{2017}$ ):
$F_{2017} = F_{2016} + \alpha(D_{2016} - F_{2016})$
$F_{2017} = 269 + 0.5(259 - 269) = 269 + 0.5(-10) = 269 - 5 = 264$ units.
Forecast for 2018 ( $F_{2018}$ ):
$F_{2018} = F_{2017} + \alpha(D_{2017} - F_{2017})$
$F_{2018} = 264 + 0.5(270 - 264) = 264 + 0.5(6) = 264 + 3 = 267$ units.
Forecast for 2019 ( $F_{2019}$ ):
$F_{2019} = F_{2018} + \alpha(D_{2018} - F_{2018})$
$F_{2019} = 267 + 0.5(287 - 267) = 267 + 0.5(20) = 267 + 10 = 277$ units.

Therefore, the sales forecast for the year 2019 is $277$ units.

Q65GATE 2019NAT2MIndustrial Engineering

The layout for an AGV system is shown in figure. The loading time is 0.5 minutes and the unloading time is also 0.5 minutes. All distances are in meters. Considering a vehicle velocity of 50 m/min, availability of 0.95 and traffic factor of 0.9, the number of vehicles required to satisfy a demand = of 50 delivery/hour is ______

🚀 Solve in Practice Mode

📖 Explanation

To find the number of AGVs required, we first calculate the ideal cycle time ( $T_i$ ). This includes loading, delivery, unloading, and return travel times.
$T_i = 0.5 + \frac{(35 + 40 + 35)}{50} + 0.5 + \frac{(20 + 40 + 20)}{50} = 4.8 \text{ min}$
Next, we adjust this ideal cycle time for real-world factors like traffic and AGV availability to get the actual cycle time ( $T_a$ ).
$T_a = \frac{4.8}{0.9 \times 0.95} = 5.61 \text{ min}$
Now, we determine how many deliveries one AGV can make per hour ( $d$ ).
$d = \frac{60}{5.614} = 10.68 \text{ deliveries/hour}$
Finally, to meet a demand of 50 deliveries per hour, we calculate the number of AGVs ( $n$ ) needed.
$n = \frac{50}{10.68} = 4.67 \approx 5 \text{ vehicles}$
Therefore, the number of vehicles required is 5.

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