GATE Mechanical Engineering 2022 SET-2

Question 1
F(t) is a periodic square wave function as shown. It takes only two values, 4 and 0, and stays at each of these values for 1 second before changing. What is the constant term in the Fourier series expansion of F(t)?

A
1
B
2
C
3
D
4
Engineering Mathematics   Calculus
Question 1 Explanation: 
The constant term in the Fourier series expansion of F(t) is the average value of F(t) in one fundamental period i.e.,
\frac{\int_{0}^{1}4dt+\int_{1}^{2}0dt}{2}=\frac{4}{2}=2
Question 2
Consider a cube of unit edge length and sides parallel to co-ordinate axes, with its centroid at the point (1, 2, 3). The surface integral \int_{A}^{}\vec{F}.d\vec{A} of a vector field \vec{F}=3x\hat{i}+5y\hat{j}+6z\hat{k} over the entire surface A of the cube is ______.
A
14
B
27
C
28
D
31
Engineering Mathematics   Calculus
Question 2 Explanation: 
Given,
\begin{aligned} \bar{F} &=3x\bar{i}+5y\bar{j}+6z\bar{k} \\ \triangledown \cdot \bar{F}&= \frac{\partial }{\partial x}(3x)+\frac{\partial }{\partial x}(5y)+\frac{\partial }{\partial x}(6z)\\ &= 3+5+6=14 \end{aligned}
By gauss divers once Theorem
\begin{aligned} \int_{A}^{}\bar{F}\cdot dA &=\int \int \int (\triangledown \cdot F)dV =\int \int \int 14\; dv\\ &=14 \times \text{volume of a cube of side 1 unit } \\ &=14 \times (1)^3=14\end{aligned}
Question 3
Consider the definite integral
\int_{1}^{2}(4x^2+2x+6)dx
Let I_e be the exact value of the integral. If the same integral is estimated using Simpson's rule with 10 equal subintervals, the value is I_s . The percentage error is defined as e=100 \times (I_e-I_s)/I_e . The value of e is
A
2.5
B
3.5
C
1.2
D
0
Engineering Mathematics   Numerical Methods
Question 3 Explanation: 
Exact value
\begin{aligned} &=\int_{1}^{2} (4x^2+2x+6)dx\\ &=\frac{4x^3}{3}+\frac{2x^2}{2}+6x\\ &=\frac{4}{3}(\pi) \times (3)+6\\ &=\frac{28}{3}+9=\frac{55}{3} \end{aligned}
Approximate value
Here f(x) is a polynomial of degree 2, so Simpsons rule gives exact value with zero error
\begin{aligned} \therefore \;\; \text{Approx value}&=\frac{55}{3}\\ \frac{I_e-I_s}{I_e}&=0\\ \therefore \;\;e=\left (\frac{I_e-I_s}{I_e} \right )\times 100&=0 \end{aligned}
Question 4
Given \int_{-\infty }^{\infty }e^{-x^2}dx=\sqrt{\pi}
If a and b are positive integers, the value of \int_{-\infty }^{\infty }e^{-a(x+b)^2}dx is ___.
A
\sqrt{\pi a}
B
\sqrt{\frac{\pi}{a}}
C
b \sqrt{\pi a}
D
b \sqrt{\frac{\pi}{a}}
Engineering Mathematics   Calculus
Question 4 Explanation: 
\begin{aligned} &\text{ Let }(x+b)=t\\ &\Rightarrow \; dx=dt\\ &\text{When ,} x=-\infty ;t=-\infty \\ &\int_{-\infty }^{-\infty }e^{-n(x+b)^2}dx=\int_{-\infty }^{-\infty }e^{-at^2}dt\\ &\text{Let, }at^2=y^2\Rightarrow t=\frac{y}{\sqrt{a}}\\ &2at\;dt=3y\;dy\\ &dt=\frac{ydy}{at}=\frac{ydy}{a\frac{y}{\sqrt{a}}}=\frac{y}{\sqrt{a}}\\ &\int_{-\infty }^{-\infty }e^{-at^2}dt=\int_{-\infty }^{-\infty }e^{-y^2}\cdot \frac{dy}{\sqrt{a}}=\sqrt{\frac{\pi}{a}} \end{aligned}
Question 5
A polynomial \phi (s)=a_{n}s^{n}+a_{n-1}s^{n-1}+...+a_{1}s+a_0 of degree n \gt 3 with constant real coefficients a_n, a_{n-1},...a_0 has triple roots at s=-\sigma . Which one of the following conditions must be satisfied?
A
\phi (s)=0 at all the three values of s satisfying s^3+\sigma ^3=0
B
\phi (s)=0, \frac{d \phi (s)}{ds}=0, \text{ and }\frac{d^2 \phi (s)}{ds^2}=0 \text{ at }s=-\sigma
C
\phi (s)=0, \frac{d \phi (s)}{ds}=0, \text{ and }\frac{d^4 \phi (s)}{ds^4}=0 \text{ at }s=-\sigma
D
\phi (s)=0, \text{ and }\frac{d^3 \phi (s)}{ds^3}=0 \text{ at }s=-\sigma
Engineering Mathematics   Differential Equations
Question 5 Explanation: 
Since \varphi (s) has a triple roots at s=-\sigma
Therefore, \varphi (s)=(s+\sigma )^3\psi (s)
It satisfies all the conditions in option (B) is correct.
Question 6
Which one of the following is the definition of ultimate tensile strength (UTS) obtained from a stress-strain test on a metal specimen?
A
Stress value where the stress-strain curve transitions from elastic to plastic behavior
B
The maximum load attained divided by the original cross-sectional area
C
The maximum load attained divided by the corresponding instantaneous crosssectional area
D
Stress where the specimen fractures
Strength of Materials   Stress-strain Relationship and Elastic Constants
Question 6 Explanation: 
Tensile Strength: The tensile strength, or ultimate tensile strength (UTS), is the maximum load obtained in a tensile test, divided by the original cross-sectional area of the specimen.
\sigma _u=\frac{P_{max}}{A_o}
where, \sigma _u= Ultimate tensile strength, kg/mm^2
P_{max}= Maximum load obtained in a tensile test, kg
A_o= Original cross-sectional area of gauge length of the test piece, mm^2
The tensile strength is a very familiar property and widely used for identification of a material. It is very easy to determine and is a quite reproducible property. It is used for the purposes of specifications and for quality control of a product. Tensile strength can be empirically correlated to other properties such as hardness and fatigue strength. For brittle materials, the tensile strength is a valid criterion for design.
Question 7
A massive uniform rigid circular disc is mounted on a frictionless bearing at the end E of a massive uniform rigid shaft AE which is suspended horizontally in a uniform gravitational field by two identical light inextensible strings AB and CD as shown, where G is the center of mass of the shaft-disc assembly and g is the acceleration due to gravity. The disc is then given a rapid spin \omega about its axis in the positive x-axis direction as shown, while the shaft remains at rest. The direction of rotation is defined by using the right-hand thumb rule. If the string AB is suddenly cut, assuming negligible energy dissipation, the shaft AE will

A
rotate slowly (compared to \omega ) about the negative z-axis direction
B
rotate slowly (compared to \omega ) about the positive z-axis direction
C
rotate slowly (compared to \omega ) about the negative y-axis direction
D
rotate slowly (compared to \omega ) about the positive y-axis direction
Theory of Machine   Gyroscope
Question 7 Explanation: 


The spin vector will chase the couple on torque vector and produce precision in system.
Hence precision will be -y direction. Rotate slowly (compared to \omega ) about negative z-axis direction.
Question 8
A structural member under loading has a uniform state of plane stress which in usual notations is given by \sigma _x=3P,\sigma _y=-2P,\tau _{xy}=\sqrt{2}P , where P \gt 0. The yield strength of the material is 350 MPa. If the member is designed using the maximum distortion energy theory, then the value of P at which yielding starts (according to the maximum distortion energy theory) is
A
70 Mpa
B
90 Mpa
C
120 Mpa
D
75 Mpa
Machine Design   Satic Dynamic Loading and Failure Theories
Question 8 Explanation: 
Given,
\sigma _x=3P
\sigma _y=-2P
\tau =\sqrt{2}P
According to maximum distortion energy theory,
\begin{aligned} \sqrt{\sigma _x^2-\sigma _x\sigma _y+\sigma _y^2+3\tau _{xy}^2} &=\frac{S_{yt}}{FOS} \\ P\sqrt{3^2-3(-2)+(-2)^2+3(\sqrt{2})^2}&= \frac{350}{1}\\ P \times 5&=350 \\ P&=70\; MPa \end{aligned}
Question 9
Fluidity of a molten alloy during sand casting depends on its solidification range. The phase diagram of a hypothetical binary alloy of components A and B is shown in the figure with its eutectic composition and temperature. All the lines in this phase diagram, including the solidus and liquidus lines, are straight lines. If this binary alloy with 15 weight % of B is poured into a mould at a pouring temperature of 800^{\circ}C , then the solidification range is

A
400 \; ^{\circ}C
B
250 \; ^{\circ}C
C
800 \; ^{\circ}C
D
150 \; ^{\circ}C
Manufacturing Engineering   Engineering Materials
Question 9 Explanation: 


Solidification range =A'B
\triangle ABC \text{ and } \triangle MB'C is similar
\begin{aligned} \frac{MA}{MB'}&=\frac{BC}{BC'}\\ \frac{700-T_A}{700-400}&=\frac{15}{30}\\ 700-T_A&=\frac{1}{2} \times 300\\ T_A&=550^{\circ}C \end{aligned}
Solidification range =T_A-T_B=550-400=150 ^{\circ}C
Question 10
A shaft of diameter 25^{^{-0.04}}_{-0.07} mm is assembled in a hole of diameter 25^{^{+0.02}}_{0.00} mm.
Match the allowance and limit parameter in Column I with its corresponding quantitative value in Column II for this shaft-hole assembly.

Allowance and limit parameter (Column I)
P. Allowance
Q. Maximum clearance
R. Maximum material limit for hole

Quantitative value (Column II)
1. 0.09 mm
2. 24.96 mm
3. 0.04 mm
4. 25.0 mm
A
P-3, Q-1, R-4
B
P-1, Q-3, R-2
C
P-1, Q-3, R-4
D
P-3, Q-1, R-2
Manufacturing Engineering   Metrology and Inspection
Question 10 Explanation: 


(1) Allowance = Lower limit of hole - upper limit of shaft
Allowance = 25.00 - 24.96 = 0.04 mm

(2) Maximum clearance C_{max} = Upper limit of hole - lower limit of shaft
C_{max} = 25.02 - 24.93 = 0.09 mm

(3) Maximum material limit for hole = minimum size of hole = 25.00
There are 10 questions to complete.

GATE Mechanical Engineering 2022 SET-1

Question 1
The limit
p=\lim_{x \to \pi}\left ( \frac{x^2+\alpha x+2\pi ^2}{x-\pi+2 \sin x } \right )
has a finite value for a real \alpha . The value of \alpha and the corresponding limit p are
A
\alpha =-3\pi, \text{ and }p= \pi
B
\alpha =-2\pi, \text{ and }p= 2\pi
C
\alpha =\pi, \text{ and }p= \pi
D
\alpha =2\pi, \text{ and }p= 3\pi
Engineering Mathematics   Calculus
Question 1 Explanation: 
\begin{aligned} p&=\lim_{x \to \pi}\left ( \frac{x^2+\alpha x+2 \pi ^2}{x-\pi+2 \sin x} \right ) \\ p&=\left ( \frac{\pi ^2+\alpha \pi +2 \pi ^2}{\pi-\pi+2 \sin \pi } \right ) \\ &= \frac{2 \pi ^2+\alpha \pi}{0}\\ \therefore \;\; \alpha &= -3 \pi\\ p&=\lim_{x \to \pi}\left ( \frac{x^2- 3 \pi x+2 \pi ^2}{x-\pi+2 \sin x} \right ) \\ &=\lim_{x \to \pi}\left ( \frac{2x- 3 \pi }{1+2 \cos \pi} \right ) \\ &= \frac{2 \pi-3 \pi}{1-2}=\frac{-\pi}{-1}=\pi\\ \therefore \; \alpha &=-3 \pi \text{ and }p= \pi \end{aligned}
Question 2
Solution of \triangledown^2T=0 in a square domain (0 \lt x \lt 1 and 0 \lt y \lt 1) with boundary conditions:
T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1, y) = 1 + y is
A
T(x,y)=x-xy+y
B
T(x,y)=x+y
C
T(x,y)=-x+y
D
T(x,y)=x+xy+y
Engineering Mathematics   Differential Equations
Question 2 Explanation: 
T(x, 0) = x \Rightarrow option (c) is not correct.
T(0, y) = y \Rightarrow all options satisfied.
T(x, 1) = 1 + x; \Rightarrow only option (b) is satisfied.
T(1, y) = 1 + y is \Rightarrow only option (b) is satisfied.
Question 3
Given a function \varphi =\frac{1}{2}(x^2+y^2+z^2) y in threedimensional Cartesian space, the value of the surface integral
\oiint_{S}{\hat{n}.\triangledown \varphi dS}
where S is the surface of a sphere of unit radius and \hat{n} is the outward unit normal vector on S, is
A
4 \pi
B
3 \pi
C
4 \pi/3
D
0
Engineering Mathematics   Calculus
Question 3 Explanation: 
\begin{aligned} \varphi &=\frac{1}{2}(x^2+y^2+z^2)\\ \triangledown \varphi &=(x\hat{i}+y\hat{j}+z\hat{k})=\bar{F}\\ \oiint_{S}(\triangledown \varphi\cdot \bar{n})dS&=\int \int _v\int Div\; \bar{F} dv\\ &=\int \int \int 3dv\\ &=3v\\ &=3\left ( \frac{4}{3} \pi \right )=4\pi \end{aligned}
Question 4
The Fourier series expansion of x^3 in the interval -1\leq x\leq 1 with periodic continuation has
A
only sine terms
B
only cosine terms
C
both sine and cosine terms
D
only sine terms and a non-zero constant
Engineering Mathematics   Calculus
Question 4 Explanation: 
f(x)=x^3, \;\; -1 \leq x \leq 1
It is an odd function
Fourier series contains only sine terms.
Question 5
If A=\begin{bmatrix} 10 &2k+5 \\ 3k-3 & k+5 \end{bmatrix} is a symmetric matrix, the value of k is ___________.
A
8
B
5
C
-0.4
D
\frac{1+\sqrt{1561}}{12}
Engineering Mathematics   Linear Algebra
Question 5 Explanation: 
A=\begin{bmatrix} 10 & 2k+5\\ 3k-3 & k+5 \end{bmatrix}
(2k + 5) = (3k - 3)
k=8
Question 6
A uniform light slender beam AB of section modulus EI is pinned by a frictionless joint A to the ground and supported by a light inextensible cable CB to hang a weight W as shown. If the maximum value of W to avoid buckling of the beam AB is obtained as \beta \pi ^2 EI, where \pi is the ratio of circumference to diameter of a circle, then the value of \beta is

A
0.0924\; m^{-2}
B
0.0713\; m^{-2}
C
0.1261\; m^{-2}
D
0.1417\; m^{-2}
Strength of Materials   Euler's Theory of Column
Question 6 Explanation: 
Draw FBD of AB

\Sigma M_A=0
W \times 2.5=T \sin 30^{\circ} \times 2.5
T=2W
Compressive load acting on AB =T \cos 30^{\circ}=2W \times \frac{\sqrt{3}}{2}=\sqrt{3}W
Buckling happens when \sqrt{3}W=P_{cr}=\frac{\pi ^2 EI}{L_e^2}
\sqrt{3}W=\frac{\pi ^2 EI}{L^2} \;\;(\because L_e=L \text{as both ends hinged})
W=\frac{1 \times \pi ^2 EI}{\sqrt{3} \times (2.5)^2}=0.0924 \pi^2 EI
W0.0924 \pi^2 EI=\beta \pi ^2 EI
\beta =0.0924 m^{-2}
Question 7
The figure shows a schematic of a simple Watt governor mechanism with the spindle O_1O_2 rotating at an angular velocity \omega about a vertical axis. The balls at P and S have equal mass. Assume that there is no friction anywhere and all other components are massless and rigid. The vertical distance between the horizontal plane of rotation of the balls and the pivot O_1 is denoted by h . The value of h=400 mm at a certain \omega . If \omega is doubled, the value of h will be _________ mm.

A
50
B
100
C
150
D
200
Theory of Machine   Gyroscope
Question 7 Explanation: 
h_1 = 400 mm, h_2 = ?
\omega _1=\omega \;\;\;\omega _2=2\omega
For Watt governor,
\begin{aligned} h&=\frac{g}{\omega ^2}\\ h\propto \frac{1}{\omega ^2}\\ \Rightarrow h_1\omega _1^2&=h_2\omega _2^2\\ 400 \times \omega ^2&=h_2 \times (2\omega )^2\\ h_2&=100mm \end{aligned}
Question 8
A square threaded screw is used to lift a load W by applying a force F. Efficiency of square threaded screw is expressed as
A
The ratio of work done by W per revolution to work done by F per revolution
B
W/F
C
F/W
D
The ratio of work done by F per revolution to work done by W per revolution
Machine Design   Bolted, Riveted and Welded Joint
Question 8 Explanation: 
\text{Screw efficiency}=\frac{\text{Work done by the applied force/rev}}{\text{Work done in lifting the load/rev}}
Efficiency of screw jack \eta =\frac{\tan \alpha }{\tan(\alpha +\phi )}
Efficiency depends on helix angle and friction angle.
Question 9
A CNC worktable is driven in a linear direction by a lead screw connected directly to a stepper motor. The pitch of the lead screw is 5 mm. The stepper motor completes one full revolution upon receiving 600 pulses. If the worktable speed is 5 m/minute and there is no missed pulse, then the pulse rate being received by the stepper motor is
A
20 KHz
B
10 kHz
C
3 kHz
D
15 kHz
Manufacturing Engineering   Machining and Machine Tool Operation
Question 9 Explanation: 
No. of steps required for one full revolution of stepper motor shaft or lead screws n_S= 600
Pitch (p) = 5 mm
Linear table speed V_{table}= 5 m/min = 5000 mm/min
RPM of lead Screw (N_S)= \frac{V_{table}}{p}=1000 rpm
We have equation of frequency of pulse generator
\begin{aligned} f_p&= N_s \times n_S\\ f_p&= 1000 \times 600=600,000 pulses/min\\ f_p&=\frac{600000}{60}pulses/sec\\ f_p&=10000 pulses/sec \text{ or }Hz\\ f_p&=10kHz \end{aligned}
Question 10
The type of fit between a mating shaft of diameter 25.0^{\begin{matrix} +0.010\\ -0.010 \end{matrix}}mm and a hole of diameter 25.015^{\begin{matrix} +0.015\\ -0.015 \end{matrix}}mm is __________.
A
Clearance
B
Transition
C
Interference
D
Linear
Manufacturing Engineering   Metrology and Inspection
Question 10 Explanation: 


If,
D_{hole} = 25.00 mm,
D_{shaft} = 25.01 mm ( Interference fit.)
D_{hole} = 25.03 mm,
D_{shaft} = 24.99 mm
(Clearance fit)
Some of the assemblies provide clearance fit and some provides interference fit.
Hence, It is transition fit
There are 10 questions to complete.

Verbal Ability

Question 1
Fish belonging to species S in the deep sea have skins that are extremely black (ultra-black skin). This helps them not only to avoid predators but also sneakily attack their prey. However, having this extra layer of black pigment results in lower collagen on their skin, making their skin more fragile.
Which one of the following is the CORRECT logical inference based on the information in the above passage?
A
Having ultra-black skin is only advantageous to species S
B
Species S with lower collagen in their skin are at an advantage because it helps them avoid predators
C
Having ultra-black skin has both advantages and disadvantages to species S
D
Having ultra-black skin is only disadvantageous to species S but advantageous only to their predators
GATE ME 2022 SET-2   General Aptitude
Question 1 Explanation: 
Ultra-black skin advantages that avoids predators and attacks their prey.
Ultra-black skin disadvantage is that the skin is more fragile means easily broken or destroyed.
Question 2
A person was born on the fifth Monday of February in a particular year.
Which one of the following statements is correct based on the above information?
A
The 2nd February of that year is a Tuesday
B
There will be five Sundays in the month of February in that year
C
The 1st February of that year is a Sunday
D
All Mondays of February in that year have even dates
GATE ME 2022 SET-2   General Aptitude
Question 2 Explanation: 
February month have 5 Mondays.
February - 29 days = 4 weeks 1 day -> Monday
29th Feb is Monday
1st | 8th | 15th | 22nd | 29th Feb is Mondays.
So, 2nd Feb is Tuesday.
Question 3
Writing too many things on the ________ while teaching could make the students get _________.
A
bored / board
B
board / bored
C
board / board
D
bored / bored
GATE ME 2022 SET-2   General Aptitude
Question 3 Explanation: 
Board means a surface, frame, or device for posing notices or writing on the blackboard. Bored means filled with or characterized by boredom.
Question 4
Given below are three conclusions drawn based on the following three statements.

Statement 1: All teachers are professors.
Statement 2: No professor is a male.
Statement 3: Some males are engineers.

Conclusion I: No engineer is a professor.
Conclusion II: Some engineers are professors.
Conclusion III: No male is a teacher.

Which one of the following options can be logically inferred?
A
Only conclusion III is correct
B
Only conclusion I and conclusion II are correct
C
Only conclusion II and conclusion III are correct
D
Only conclusion I and conclusion III are correct
GATE ME 2022 SET-1   General Aptitude
Question 4 Explanation: 


No male is teacher is correct.
Engineer and Professor relation not given.
Question 5
A rhombus is formed by joining the midpoints of the sides of a unit square.
What is the diameter of the largest circle that can be inscribed within the rhombus?
A
\frac{1}{\sqrt{2}}
B
\frac{1}{2\sqrt{2}}
C
\sqrt{2}
D
2\sqrt{2}
GATE ME 2022 SET-1   General Aptitude
Question 5 Explanation: 


From above diagram,
Clearly diameter of circle = side of Rhombus = a

a=\sqrt{\left ( \frac{1}{2} \right )^2+\left ( \frac{1}{2} \right )^2}=\frac{1}{\sqrt{2}}
Question 6
Humans are naturally compassionate and honest. In a study using strategically placed wallets that appear "lost", it was found that wallets with money are more likely to be returned than wallets without money. Similarly, wallets that had a key and money are more likely to be returned than wallets with the same amount of money alone. This suggests that the primary reason for this behavior is compassion and empathy.
Which one of the following is the CORRECT logical inference based on the information in the above passage?
A
Wallets with a key are more likely to be returned because people do not care about money
B
Wallets with a key are more likely to be returned because people relate to suffering of others
C
Wallets used in experiments are more likely to be returned than wallets that are really lost
D
Money is always more important than keys
GATE ME 2022 SET-1   General Aptitude
Question 6 Explanation: 
Compassion and empathy refers to suffering of others.
Question 7
Four girls P, Q, R and S are studying languages in a University. P is learning French and Dutch. Q is learning Chinese and Japanese. R is learning Spanish and French. S is learning Dutch and Japanese.
Given that: French is easier than Dutch; Chinese is harder than Japanese; Dutch is easier than Japanese, and Spanish is easier than French.
Based on the above information, which girl is learning the most difficult pair of languages?
A
P
B
Q
C
R
D
S
GATE ME 2022 SET-1   General Aptitude
Question 7 Explanation: 
Difficulty level as per given statements

French < Dutch
Japanese < Chinese
Dutch < Japanese
Spanish < French


From above relation, we can conclude following relation
French < Dutch < Japanese < Chinese
Q leaving most difficult pair of languages.
Question 8
After playing ________ hours of tennis, I am feeling _______ tired to walk back.
A
too / too
B
too / two
C
two / two
D
two / too
GATE ME 2022 SET-1   General Aptitude
Question 8 Explanation: 
Two hours of tennis is time,
too tired means to an excessive degree.
Question 9
The world is going through the worst pandemic in the past hundred years. The air travel industry is facing a crisis, as the resulting quarantine requirement for travelers led to weak demand.
In relation to the first sentence above, what does the second sentence do?
A
Restates an idea from the first sentence.
B
Second sentence entirely contradicts the first sentence.
C
The two statements are unrelated.
D
States an effect of the first sentence.
GATE ME 2021 SET-2   General Aptitude
Question 9 Explanation: 
First option is wrong because second sentence does not contradict the first sentence. Third option is wrong because two sentences are related. Fourth option is wrong because the second sentence does not repeat the first one.
Hence second option is correct which shows the result of the cause.
Question 10
Given below are two statements 1 and 2, and two conclusions I and II.

Statement 1: All entrepreneurs are wealthy.
Statement 2: All wealthy are risk seekers.

Conclusion I: All risk seekers are wealthy.
Conclusion II: Only some entrepreneurs are risk seekers.

Based on the above statements and conclusions, which one of the following options is CORRECT?
A
Only conclusion I is correct
B
Only conclusion II is correct
C
Neither conclusion I nor II is correct
D
Both conclusions I and II are correct
GATE ME 2021 SET-2   General Aptitude
Question 10 Explanation: 
Possible cases are:

Conclusion-I is incorrect becaue some risk seeker are wealthy.
Conclusion-II is also incorrect because all the entrepreneurs are risk seeker as well as wealthy


There are 10 questions to complete.

Numerical Ability

Question 1
Equal sized circular regions are shaded in a square sheet of paper of 1 cm side length. Two cases, case M and case N, are considered as shown in the figures below. In the case M, four circles are shaded in the square sheet and in the case N, nine circles are shaded in the square sheet as shown.


What is the ratio of the areas of unshaded regions of case M to that of case N?
A
2:3
B
1:1
C
3:2
D
2:1
GATE ME 2022 SET-2   General Aptitude
Question 1 Explanation: 


2d=1\Rightarrow d=\frac{1}{2}
Area of circle =\frac{\pi d^2}{4}=\frac{\pi}{16}
Total circle area = 4 \times \frac{\pi}{16}=\frac{\pi}{4}


3d=1\Rightarrow d=\frac{1}{3}
Area of circle =\frac{\pi d^2}{4}=\frac{\pi}{36}
Total circle area = 9 \times \frac{\pi}{36}=\frac{\pi}{4}
\frac{\text{Unshaded area M}}{\text{Unshaded area N}}=\frac{1^2-\frac{\pi}{4}}{1-\frac{\pi}{4}}=\frac{1}{1}=1:1
Question 2
Four cities P, Q, R and S are connected through oneway routes as shown in the figure. The travel time between any two connected cities is one hour. The boxes beside each city name describe the starting time of first train of the day and their frequency of operation. For example, from city P, the first trains of the day start at 8 AM with a frequency of 90 minutes to each of R and S. A person does not spend additional time at any city other than the waiting time for the next connecting train. If the person starts from R at 7 AM and is required to visit S and return to R, what is the minimum time required?

A
6 hours 30 minutes
B
3 hours 45 minutes
C
4 hours 30 minutes
D
5 hours 15 minutes
GATE ME 2022 SET-2   General Aptitude
Question 2 Explanation: 
R at 7 am
1 hour journey
Reached Q at 8 am
At Q buses available timings are 5 am, 7am, 9 am ..
Person started at Q at 9 am
1 hour journey reached P at 10 am.
Buses timings at P are 8 am, 9:30am, 11 am, 12:30 pm...
Person started at P at 11 am
1 hour journey
Reached S at 12 noon.
Buses timings at S are
8am, 8:45 am, 9:30 am, 10:15 am, 11am, 11:45 am, 12:30pm ....
Person started at S at 12:30 pm
1 hour journey
Reached R at 1:30 pm
Minimum Total time = 1:30 pm - 7 am
= 6 hrs 30 min
Question 3
Consider the following functions for non-zero positive integers, p and q

f(p,q)=\underbrace{p \times p \times p \times ...p}_{q \;\;times} =p^q;\;\; f(p,1)=p
g(p,q)=p^{p^{p^{\vdots ^{q \;\;times}}}};\;\; g(p,1)=p

Which one of the following options is correct based on the above?
A
f(2,2)=g(2,2)
B
f(g(2,2),2) \lt f(2,g(2,2))
C
g(2,1) \neq f(2,1)
D
f(3,2) \gt g(3,2)
GATE ME 2022 SET-2   General Aptitude
Question 3 Explanation: 
f(2,2) = 2 x 2 = 4
g(2,2) = 4
f(2,2) = g(2,2)
Question 4
For the past m days, the average daily production at a company was 100 units per day.
If today's production of 180 units changes the average to 110 units per day, what is the value of m?
A
18
B
10
C
7
D
5
GATE ME 2022 SET-2   General Aptitude
Question 4 Explanation: 
Total production = Average x No. of days
P= 100 m
and
P + 180 = 110 (m+1)
Solving, m = 7
Question 5


Which one of the groups given below can be assembled to get the shape that is shown above using each piece only once without overlapping with each other?
(rotation and translation operations may be used).

A
A
B
B
C
C
D
D
GATE ME 2022 SET-2   General Aptitude
Question 5 Explanation: 
Priority: Shape, position and size.
As per size point of view parallelogram is not suitable.
Question 6
If f(x)=2 \ln\sqrt{e^x} what is the area bounded by f(x) for the interval [0,2] on the x- axis?
A
\frac{1}{2}
B
1
C
2
D
4
GATE ME 2022 SET-2   General Aptitude
Question 6 Explanation: 
\begin{aligned} f(x)&=2 \ln ( \sqrt{e^x})\\ &=2 \ln (e^{x/2})=2 \log _e e^{x/2}\\ f(x)&=2\left ( \frac{x}{2} \right )=x \end{aligned}


Area =(1/2) x 2 x 2=2
Question 7
Which one of the following is a representation (not to scale and in bold) of all values of x satisfying the inequality 2-5x\leq -\frac{6x-5}{3} on the real number line?

A
A
B
B
C
C
D
D
GATE ME 2022 SET-2   General Aptitude
Question 7 Explanation: 
\begin{aligned} 2-5x \leq -\frac{6x-5}{3}\\ 1-9x\leq 0\\ \Rightarrow 1\leq 9x\Rightarrow x\geq \frac{1}{9} \end{aligned}
Question 8
In a 12-hour clock that runs correctly, how many times do the second, minute, and hour hands of the clock coincide, in a 12-hour duration from 3 PM in a day to 3 AM the next day?
A
11
B
12
C
144
D
2
GATE ME 2022 SET-1   General Aptitude
Question 8 Explanation: 
3 pm - 3am = 12 hrs
Coincide is
11 times in 12 hours
Question 9
An equilateral triangle, a square and a circle have equal areas.
What is the ratio of the perimeters of the equilateral triangle to square to circle?
A
3\sqrt{3}: 2: \sqrt{\pi}
B
\sqrt{3\sqrt{3}}: 2: \sqrt{\pi}
C
\sqrt{3\sqrt{3}}: 4: 2\sqrt{\pi}
D
\sqrt{3\sqrt{3}}: 2: 2\sqrt{\pi}
GATE ME 2022 SET-1   General Aptitude
Question 9 Explanation: 


\begin{aligned} \frac{\sqrt{3}}{4}a^2 &=s^2 &=\pi r^2&=k\\ a&=\sqrt{\frac{4k}{\sqrt{3}}}&s=\sqrt{k} &\;\;\; r=\sqrt{\frac{\pi}{k}}\\ &\text{Perimeters:}\\ 3a&:&4a&:&2 \pi ^2\\ 3\frac{2\sqrt{k}}{3^{1/4}}&:&4\sqrt{k}&:&2 \pi \frac{\sqrt{k}}{\sqrt{\pi}}\\ 3^{3/4}&:&2&:&\sqrt{\pi} \end{aligned}
Question 10


A block with a trapezoidal cross-section is placed over a block with rectangular cross section as shown above.
Which one of the following is the correct drawing of the view of the 3D object as viewed in the direction indicated by an arrow in the above figure?

A
A
B
B
C
C
D
D
GATE ME 2022 SET-1   General Aptitude
Question 10 Explanation: 
View from arrow direction is


There are 10 questions to complete.

General Aptitude

Question 1
Equal sized circular regions are shaded in a square sheet of paper of 1 cm side length. Two cases, case M and case N, are considered as shown in the figures below. In the case M, four circles are shaded in the square sheet and in the case N, nine circles are shaded in the square sheet as shown.


What is the ratio of the areas of unshaded regions of case M to that of case N?
A
2:3
B
1:1
C
3:2
D
2:1
GATE ME 2022 SET-2      Numerical Ability
Question 1 Explanation: 


2d=1\Rightarrow d=\frac{1}{2}
Area of circle =\frac{\pi d^2}{4}=\frac{\pi}{16}
Total circle area = 4 \times \frac{\pi}{16}=\frac{\pi}{4}


3d=1\Rightarrow d=\frac{1}{3}
Area of circle =\frac{\pi d^2}{4}=\frac{\pi}{36}
Total circle area = 9 \times \frac{\pi}{36}=\frac{\pi}{4}
\frac{\text{Unshaded area M}}{\text{Unshaded area N}}=\frac{1^2-\frac{\pi}{4}}{1-\frac{\pi}{4}}=\frac{1}{1}=1:1
Question 2
Four cities P, Q, R and S are connected through oneway routes as shown in the figure. The travel time between any two connected cities is one hour. The boxes beside each city name describe the starting time of first train of the day and their frequency of operation. For example, from city P, the first trains of the day start at 8 AM with a frequency of 90 minutes to each of R and S. A person does not spend additional time at any city other than the waiting time for the next connecting train. If the person starts from R at 7 AM and is required to visit S and return to R, what is the minimum time required?

A
6 hours 30 minutes
B
3 hours 45 minutes
C
4 hours 30 minutes
D
5 hours 15 minutes
GATE ME 2022 SET-2      Numerical Ability
Question 2 Explanation: 
R at 7 am
1 hour journey
Reached Q at 8 am
At Q buses available timings are 5 am, 7am, 9 am ..
Person started at Q at 9 am
1 hour journey reached P at 10 am.
Buses timings at P are 8 am, 9:30am, 11 am, 12:30 pm...
Person started at P at 11 am
1 hour journey
Reached S at 12 noon.
Buses timings at S are
8am, 8:45 am, 9:30 am, 10:15 am, 11am, 11:45 am, 12:30pm ....
Person started at S at 12:30 pm
1 hour journey
Reached R at 1:30 pm
Minimum Total time = 1:30 pm - 7 am
= 6 hrs 30 min
Question 3
Consider the following functions for non-zero positive integers, p and q

f(p,q)=\underbrace{p \times p \times p \times ...p}_{q \;\;times} =p^q;\;\; f(p,1)=p
g(p,q)=p^{p^{p^{\vdots ^{q \;\;times}}}};\;\; g(p,1)=p

Which one of the following options is correct based on the above?
A
f(2,2)=g(2,2)
B
f(g(2,2),2) \lt f(2,g(2,2))
C
g(2,1) \neq f(2,1)
D
f(3,2) \gt g(3,2)
GATE ME 2022 SET-2      Numerical Ability
Question 3 Explanation: 
f(2,2) = 2 x 2 = 4
g(2,2) = 4
f(2,2) = g(2,2)
Question 4
For the past m days, the average daily production at a company was 100 units per day.
If today's production of 180 units changes the average to 110 units per day, what is the value of m?
A
18
B
10
C
7
D
5
GATE ME 2022 SET-2      Numerical Ability
Question 4 Explanation: 
Total production = Average x No. of days
P= 100 m
and
P + 180 = 110 (m+1)
Solving, m = 7
Question 5
Fish belonging to species S in the deep sea have skins that are extremely black (ultra-black skin). This helps them not only to avoid predators but also sneakily attack their prey. However, having this extra layer of black pigment results in lower collagen on their skin, making their skin more fragile.
Which one of the following is the CORRECT logical inference based on the information in the above passage?
A
Having ultra-black skin is only advantageous to species S
B
Species S with lower collagen in their skin are at an advantage because it helps them avoid predators
C
Having ultra-black skin has both advantages and disadvantages to species S
D
Having ultra-black skin is only disadvantageous to species S but advantageous only to their predators
GATE ME 2022 SET-2      Verbal Ability
Question 5 Explanation: 
Ultra-black skin advantages that avoids predators and attacks their prey.
Ultra-black skin disadvantage is that the skin is more fragile means easily broken or destroyed.
Question 6


Which one of the groups given below can be assembled to get the shape that is shown above using each piece only once without overlapping with each other?
(rotation and translation operations may be used).

A
A
B
B
C
C
D
D
GATE ME 2022 SET-2      Numerical Ability
Question 6 Explanation: 
Priority: Shape, position and size.
As per size point of view parallelogram is not suitable.
Question 7
A person was born on the fifth Monday of February in a particular year.
Which one of the following statements is correct based on the above information?
A
The 2nd February of that year is a Tuesday
B
There will be five Sundays in the month of February in that year
C
The 1st February of that year is a Sunday
D
All Mondays of February in that year have even dates
GATE ME 2022 SET-2      Verbal Ability
Question 7 Explanation: 
February month have 5 Mondays.
February - 29 days = 4 weeks 1 day -> Monday
29th Feb is Monday
1st | 8th | 15th | 22nd | 29th Feb is Mondays.
So, 2nd Feb is Tuesday.
Question 8
If f(x)=2 \ln\sqrt{e^x} what is the area bounded by f(x) for the interval [0,2] on the x- axis?
A
\frac{1}{2}
B
1
C
2
D
4
GATE ME 2022 SET-2      Numerical Ability
Question 8 Explanation: 
\begin{aligned} f(x)&=2 \ln ( \sqrt{e^x})\\ &=2 \ln (e^{x/2})=2 \log _e e^{x/2}\\ f(x)&=2\left ( \frac{x}{2} \right )=x \end{aligned}


Area =(1/2) x 2 x 2=2
Question 9
Which one of the following is a representation (not to scale and in bold) of all values of x satisfying the inequality 2-5x\leq -\frac{6x-5}{3} on the real number line?

A
A
B
B
C
C
D
D
GATE ME 2022 SET-2      Numerical Ability
Question 9 Explanation: 
\begin{aligned} 2-5x \leq -\frac{6x-5}{3}\\ 1-9x\leq 0\\ \Rightarrow 1\leq 9x\Rightarrow x\geq \frac{1}{9} \end{aligned}
Question 10
Writing too many things on the ________ while teaching could make the students get _________.
A
bored / board
B
board / bored
C
board / board
D
bored / bored
GATE ME 2022 SET-2      Verbal Ability
Question 10 Explanation: 
Board means a surface, frame, or device for posing notices or writing on the blackboard. Bored means filled with or characterized by boredom.


There are 10 questions to complete.

GATE Mechanical Engineering 2021 SET-2

Question 1
Consider an n x n matrix A and a non-zero n x 1 vector p. Their product Ap=\alpha ^2p, where \alpha \in \mathbb{R} and \alpha \notin \{-1,0,1\}. Based on the given information, the eigen value of A^2 is:
A
\alpha
B
\alpha ^2
C
\sqrt{\alpha }
D
\alpha ^4
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
Given, A P=\alpha^{2} P
By comparison with A X=\lambda X \Rightarrow
\Rightarrow \quad \lambda=\alpha^{2}
Hence, eigen value of A is \alpha^{2}, so eigen value of A^{2} is \alpha^{4}.
Question 2
If the Laplace transform of a function f(t) is given by \frac{s+3}{(s+1)(s+2)} , then f(0) is
A
0
B
\frac{1}{2}
C
1
D
\frac{3}{2}
Engineering Mathematics   Differential Equations
Question 2 Explanation: 
By using partial fraction concept.
\begin{aligned} f(t) &=L^{-1}\left[\frac{s+3}{(s+1)(s+2)}\right] \\ &=L^{-1}\left[\frac{2}{s+1}-\frac{1}{s+2}\right] \\ \Rightarrow \qquad f(t) &=2 e^{-t}-e^{-2 t} \\ \text { So, } \qquad f(c)&=2 e^{0}-e^{0}=2-1=1 \end{aligned}
Question 3
The mean and variance, respectively, of a binomial distribution for n independent trials with the probability of success as p, are
A
\sqrt{np},np(1-2p)
B
\sqrt{np}, \sqrt{np(1-p)}
C
np,np
D
np,np(1-p)
Engineering Mathematics   Probability and Statistics
Question 3 Explanation: 
Mean= np
Variance = npq = np(1 - p)
Question 4
The Cast Iron which possesses all the carbon in the combined form as cementite is known as
A
Grey Cast Iron
B
Spheroidal Cast Iron
C
Malleable Cast Iron
D
White Cast Iron
Manufacturing Engineering   Engineering Materials
Question 4 Explanation: 
On the basis of nature of carbon present in cast iron, it may be divided into white cast iron and gray cast iron.
In the gray cast iron, carbon is present in free form as graphite. Under very slow rate of cooling during solidification, carbon atoms get sufficient time to separate out in pure form as graphite. In addition, certain elements promote decomposition of cementite. Silicon and nickel are two commonly used graphitizing elements.
In white cast iron, carbon is present in the form of combined form as cementite. In normal conditions, carbon has a tendency to combine with iron to form cementite.
Question 5
The size distribution of the powder particles used in Powder Metallurgy process can be determined by
A
Laser scattering
B
Laser reflection
C
Laser absorption
D
Laser penetration
Manufacturing Engineering   Forming Process
Question 5 Explanation: 
Particle Size, Shape, and Distribution:
Particle size is generally controlled by screening, that is, by passing the metal powder through screens (sieves) of various mesh sizes. Several other methods also are available for particle-size analysis:
1. Sedimentation, which involves measuring the rate at which particles settle in a fluid.
2. Microscopic analysis, which may include the use of transmission and scanning- electron microscopy.
3. Light scattering from a laser that illuminates a sample, consisting of particles suspended in a liquid medium; the particles cause the light to be scattered, and a detector then digitizes the signals and computes the particle-size distribution.
4. Optical methods, such as particles blocking a beam of light, in which the particle is sensed by a photocell.
5. Suspending particles in a liquid and detecting particle size and distribution by electrical sensors.
Question 6
In a CNC machine tool, the function of an interpolator is to generate
A
signal for the lubrication pump during machining
B
error signal for tool radius compensation during machining
C
NC code from the part drawing during post processing
D
reference signal prescribing the shape of the part to be machined
Manufacturing Engineering   Computer Integrated Manufacturing
Question 6 Explanation: 
In contouring systems the machining path is usually constructed from a combination of linear and circular segments. It is only necessary to specify the coordinates of the initial and final points of each segment, and the feed rate. The operation of producing the required shape based on this information is termed interpolation and the corresponding unit is the "interpolator". The interpolator coordinates the motion along the machine axes, which are separately driven, by providing reference positions instant by instant for the position-and velocity control loops, to generate the required machining path. Typical interpolators are capable of generating linear and circular paths.
Question 7
The machining process that involves ablation is
A
Abrasive Jet Machining
B
Chemical Machining
C
Electrochemical Machining
D
Laser Beam Machining
Manufacturing Engineering   Machining and Machine Tool Operation
Question 7 Explanation: 
Laser beam machining (LBM) is a nonconventional machining process, which broadly refers to the process of material removal, accomplished through the interactions between the laser and target materials. The processes can include laser drilling, cutting, grooving, writing, scribing, ablation, welding, cladding, milling, and so on. LBM is a thermal process, and unlike conventional mechanical processes, LBM removes material without mechanical engagement. In general, the workpiece is heated to melting or boiling point and removed by melt ejection, vaporization, or ablation.
Question 8
A PERT network has 9 activities on its critical path. The standard deviation of each activity on the critical path is 3. The standard deviation of the critical path is
A
3
B
9
C
27
D
81
Industrial Engineering   PERT and CPM
Question 8 Explanation: 
In CPM,
\begin{array}{l} \sigma=\sqrt{\text { sum of variance along critical path }} \\ \sigma=\sqrt{\sigma^{2}+\sigma^{2}+\ldots .+\sigma^{2}} \\ \sigma=\sqrt{9 \sigma^{2}}=\sqrt{9 \times 9}=9 \end{array}
Question 9
The allowance provided in between a hole and a shaft is calculated from the difference between
A
lower limit of the shaft and the upper limit of the hole
B
upper limit of the shaft and the upper limit of the hole
C
upper limit of the shaft and the lower limit of the hole
D
lower limit of the shaft and the lower limit of the hole
Manufacturing Engineering   Metrology and Inspection
Question 9 Explanation: 
It is minimum clearance or maximum interference. It is the intentional difference between the basic dimensions of the mating parts. The allowance may be positive or negative.

Question 10
In forced convective heat transfer, Stanton number (St), Nusselt number (Nu), Reynolds number (Re) and Prandtl number (Pr) are related as
A
\text{St}=\frac{\text{Nu}}{\text{Re Pr}}
B
\text{St}=\frac{\text{Nu Pr}}{\text{Re}}
C
\text{St}=\text{Nu Pr Re}
D
\text{St}=\frac{\text{Nu Re}}{\text{Pr}}
Heat Transfer   Free and Forced Convection
Question 10 Explanation: 
S t=\frac{N u}{R e \times P r}
There are 10 questions to complete.

GATE Mechanical Engineering 2021 SET-1

Question 1
If y(x) satisfies the differential equation

(\sin x)\frac{dy}{dx}+y \cos x =1

subject to the condition y(\pi /2)=\pi /2, then y(\pi /6) is
A
0
B
\frac{\pi}{6}
C
\frac{\pi}{3}
D
\frac{\pi}{2}
Engineering Mathematics   Differential Equations
Question 1 Explanation: 
\begin{aligned} \frac{d y}{d x}+y \cot x&=\text{cosec} x\\ 1.F. \qquad&=e^{\int \cot x d x}=e^{\log \sin x}=\sin x\\ \Rightarrow \quad y(\sin x)&=\int \text{cosec} x \sin x d x+c\\ \Rightarrow \qquad y \sin x&=x+c\\ \Rightarrow \qquad \frac{\pi}{2} \sin \frac{\pi}{2} & =\frac{\pi}{2}+c \\ \Rightarrow \qquad \frac{\pi}{2} & =\frac{\pi}{2}+c \quad \Rightarrow c=0 \\ \Rightarrow \qquad y \sin x & =x\\ \Rightarrow \qquad y \sin \frac{\pi}{6}&=\frac{\pi}{6}\\ \Rightarrow \qquad y\left(\frac{1}{2}\right) &=\frac{\pi}{6} \\ \Rightarrow y &=\frac{\pi}{3} \end{aligned}
Question 2
The value of \lim_{x \to 0}\left ( \frac{1- \cos x}{x^2} \right ) is
A
\frac{1}{4}
B
\frac{1}{3}
C
\frac{1}{2}
D
1
Engineering Mathematics   Calculus
Question 2 Explanation: 
\begin{aligned} \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)&=? \;\;\;\;\;\;\left(\frac{0}{0} \text { form }\right) \\ \text { Applying } L \cdot H \text { rule } & =\lim _{x \rightarrow 0} \frac{\sin x}{2 x}\left(\frac{0}{0}\right)=\lim _{x \rightarrow 0} \frac{\cos x}{2}=\frac{1}{2} \end{aligned}
Question 3
The Dirac-delta function (\delta (t-t_0)) \text{ for }t,t_0 \in \mathbb{R}, has the following property

\int_{a}^{b}\varphi (t)\delta (t-t_0)dt=\left\{\begin{matrix} \varphi (t_0) & a \lt t_0 \lt b\\ 0 &\text{otherwise} \end{matrix}\right.

The Laplace transform of the Dirac-delta function \delta (t-a) for a \gt 0;
\mathcal{L} (\delta (t-a))=F(s) is
A
0
B
\infty
C
e^{sa}
D
e^{-sa}
Engineering Mathematics   Differential Equations
Question 3 Explanation: 
\begin{aligned} \because \qquad \int_{0}^{-} f(t) \delta(t-a) d t&=f(a) \\ \therefore \qquad L\{\delta(t-a)\}&=\int_{0}^{-} e^{-s t} \delta(t-a) d t=e^{-a s} \end{aligned}
Question 4
The ordinary differential equation \frac{dy}{dt}=-\pi y subject to an initial condition y(0)=1 is solved numerically using the following scheme:

\frac{y(t_{n+1})-y(t_n)}{h}=-\pi y(t_n)

where h is the time step, t_n=nh, and n=0,1,2,.... This numerical scheme is stable for all values of h in the interval.
A
0 \lt h \lt \frac{2}{\pi}
B
0 \lt h \lt 1
C
0 \lt h \lt \frac{\pi}{2}
D
for all h \gt 0
Engineering Mathematics   Differential Equations
Question 4 Explanation: 
\begin{aligned} \frac{y\left(t_{n+1}\right)-y\left(t_{n}\right)}{h} &=-\pi y\left(t_{n}\right) \\ y_{n+1} &=-\pi / y_{n}+y_{n}=(-\pi h+1) y_{n} \end{aligned}
It is recursion relation between y_{n+1} and y_{n}
So solution will be stable if
\begin{aligned} |-\pi h+1| & \lt 1 \\ -1 \lt -\pi h+1 & \lt 1 \\ -2 \lt -\pi h & \lt 0 \\ 0 & \lt \pi h \lt 2 \\ 0 & \lt h \lt \frac{2}{\pi} \end{aligned}
Therefore option (A) is correct.
Question 5
Consider a binomial random variable X. If X_1,X_2,..., X_n are independent and identically distributed samples from the distribution of X with sum Y=\sum_{i=1}^{n}X_i, then the distribution of Y as n\rightarrow \infty can be approximated as
A
Exponential
B
Bernoulli
C
Binomial
D
Normal
Engineering Mathematics   Probability and Statistics
Question 6
The loading and unloading response of a metal is shown in the figure. The elastic and plastic strains corresponding to 200 MPa stress, respectively, are

A
0.01 and 0.01
B
0.02 and 0.01
C
0.01 and 0.02
D
0.02 and 0.02
Strength of Materials   Stress and Strain
Question 6 Explanation: 
Elastic strain : Which can be recovered = 0.03 - 0.01 = 0.02
Plastic strain : Permanent strain = 0.01
Question 7
In a machining operation, if a cutting tool traces the workpiece such that the directrix is perpendicular to the plane of the generatrix as shown in figure, the surface generated is

A
plane
B
cylindrical
C
spherical
D
a surface of revolution
Manufacturing Engineering   Machining and Machine Tool Operation
Question 7 Explanation: 


Question 8
The correct sequence of machining operations to be performed to finish a large diameter through hole is
A
drilling, boring, reaming
B
boring, drilling, reaming
C
drilling, reaming, boring
D
boring, reaming, drilling
Manufacturing Engineering   Machining and Machine Tool Operation
Question 8 Explanation: 
Drilling: to produce a hole, which then may be followed by boring it to improve its dimensional accuracy and surface finish.

Boring: to enlarge a hole or cylindrical cavity made by a previous process or to produce circular internal grooves.

Reaming: is an operation used to (a) make an existing hole dimensionally more accurate than can br achived by drilling alone and (b) improve its surface finish. The most accurate holes in workpieces generally are produced by the following sequence of operation.

Centering -> Drilling -> Boring -> Reaming.
Question 9
In modern CNC machine tools, the backlash has been eliminated by
A
preloaded ballscrews
B
rack and pinion
C
ratchet and pinion
D
slider crank mechanism
Manufacturing Engineering   Computer Integrated Manufacturing
Question 9 Explanation: 


Question 10
Consider the surface roughness profile as shown in the figure.

The center line average roughness (R_a \text{ in }\mu m) of the measured length (L) is
A
0
B
1
C
2
D
4
Manufacturing Engineering   Metrology and Inspection
Question 10 Explanation: 
R_{G}=\frac{\sum_{i=1}^{n} y}{n}=\frac{4}{4}=1
There are 10 questions to complete.

GATE Mechanical Engineering-Topic wise Previous Year Questions

GATE 2022 Mechanical Engineering Syllabus

Revised syllabus of GATE 2022 Mechanical Engineering by IIT.

Practice GATE Mechanical Engineering previous year questions

Year wise | Subject wise | Topic wise

Section 1: Engineering Mathematics

Linear Algebra: Matrix algebra, systems of linear equations, eigenvalues and eigenvectors.
Calculus: Functions of single variable, limit, continuity and differentiability, mean value theorems, indeterminate forms; evaluation of definite and improper integrals; double and triple integrals; partial derivatives, total derivative, Taylor series (in one and two variables), maxima and minima, Fourier series; gradient, divergence and curl, vector identities, directional derivatives, line, surface and volume integrals, applications of Gauss, Stokes and Green’s theorems.
Differential equations: First order equations (linear and nonlinear); higher order linear differential equations with constant coefficients; Euler-Cauchy equation; initial and boundary value problems; Laplace transforms; solutions of heat, wave and Laplace’s equations.
Complex variables: Analytic functions; Cauchy-Riemann equations; Cauchy’s integral theorem and integral formula; Taylor and Laurent series.
Probability and Statistics: Definitions of probability, sampling theorems, conditional probability; mean, median, mode and standard deviation; random variables, binomial, Poisson and normal distributions.
Numerical Methods: Numerical solutions of linear and non-linear algebraic equations; integration by trapezoidal and Simpson’s rules; single and multi-step methods for differential equations.
Section 2: Applied Mechanics and Design

Engineering Mechanics: Free-body diagrams and equilibrium; friction and its applications including rolling friction, belt-pulley, brakes, clutches, screw jack, wedge, vehicles, etc.; trusses and frames; virtual work; kinematics and dynamics of rigid bodies in plane motion; impulse and momentum (linear and angular) and energy formulations; Lagrange’s equation.
Mechanics of Materials: Stress and strain, elastic constants, Poisson’s ratio; Mohr’s circle for plane stress and plane strain; thin cylinders; shear force and bending moment diagrams; bending and shear stresses; concept of shear centre; deflection of beams; torsion of circular shafts; Euler’s theory of columns; energy methods; thermal stresses; strain gauges and rosettes; testing of materials with universal testing machine; testing of hardness and impact strength.
Theory of Machines: Displacement, velocity and acceleration analysis of plane mechanisms; dynamic analysis of linkages; cams; gears and gear trains; flywheels and governors; balancing of reciprocating and rotating masses; gyroscope.
Vibrations: Free and forced vibration of single degree of freedom systems, effect of damping; vibration isolation; resonance; critical speeds of shafts.
Machine Design: Design for static and dynamic loading; failure theories; fatigue strength and the S-N diagram; principles of the design of machine elements such as bolted, riveted and welded joints; shafts, gears, rolling and sliding contact bearings, brakes and clutches, springs.
Section 3: Fluid Mechanics and Thermal Sciences

Fluid Mechanics: Fluid properties; fluid statics, forces on submerged bodies, stability of floating bodies; controlvolume analysis of mass, momentum and energy; fluid acceleration; differential equations of continuity and momentum; Bernoulli’s equation; dimensional analysis; viscous flow of incompressible fluids, boundary layer, elementary turbulent flow, flow through pipes, head losses in pipes, bends and fittings; basics of compressible fluid flow.
Heat-Transfer: Modes of heat transfer; one dimensional heat conduction, resistance concept and electrical analogy, heat transfer through fins; unsteady heat conduction, lumped parameter system, Heisler’s charts; thermal boundary layer, dimensionless parameters in free and forced convective heat transfer, heat transfer correlations for flow over flat plates and through pipes, effect of turbulence; heat exchanger performance, LMTD and NTU methods; radiative heat transfer, Stefan- Boltzmann law, Wien’s displacement law, black and grey surfaces, view factors, radiation network analysis.
Thermodynamics: Thermodynamic systems and processes; properties of pure substances, behavior of ideal and real gases; zeroth and first laws of thermodynamics, calculation of work and heat in various processes; second law of thermodynamics; thermodynamic property charts and tables, availability and irreversibility; thermodynamic relations.
Applications: Power Engineering: Air and gas compressors; vapour and gas power cycles, concepts of regeneration and reheat. I.C. Engines: Air-standard Otto, Diesel and dual cycles.
Refrigeration and airconditioning: Vapour and gas refrigeration and heat pump cycles; properties of moist air, psychrometric chart,
basic psychrometric processes. Turbomachinery: Impulse and reaction principles, velocity diagrams, Peltonwheel, Francis and Kaplan turbines; steam and gas turbines.
Section 4: Materials, Manufacturing and Industrial Engineering

Engineering Materials: Structure and properties of engineering materials, phase diagrams, heat treatment, stressstrain diagrams for engineering materials.
Casting, Forming and Joining Processes: Different types of castings, design of patterns, moulds and cores; solidification and cooling; riser and gating design. Plastic deformation and yield criteria; fundamentals of hot and cold working processes; load estimation for bulk (forging, rolling, extrusion, drawing) and sheet (shearing, deep drawing, bending) metal forming processes; principles of powder metallurgy. Principles of welding, brazing, soldering and adhesive bonding.
Machining and Machine Tool Operations: Mechanics of machining; basic machine tools; single and multi-point cutting tools, tool geometry and materials, tool life and wear; economics of machining; principles of nontraditional machining processes; principles of work holding, jigs and fixtures; abrasive machining processes; NC/CNC machines and CNC programming.
Metrology and Inspection: Limits, fits and tolerances; linear and angular measurements; comparators; interferometry; form and finish measurement; alignment and testing methods; tolerance analysis in manufacturing and assembly; concepts of coordinate-measuring machine (CMM).
Computer Integrated Manufacturing: Basic concepts of CAD/CAM and their integration tools; additive manufacturing.
Production Planning and Control: Forecasting models, aggregate production planning, scheduling, materials requirement planning; lean manufacturing. Inventory Control: Deterministic models; safety stock inventory control systems.
Operations Research: Linear programming, simplex method, transportation, assignment, network flow models, simple queuing models, PERT and CPM.

Download the GATE 2022 Mechanical Engineering Syllabus pdf from the official site of IIT Bombay. Analyze the GATE 2022 revised syllabus for Mechanical Engineering.

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GATE Mechanical Engineering 2019 SET-1

Question 1
Consider the matrix
P=\begin{bmatrix} 1 & 1 &0 \\ 0&1 &1 \\ 0& 0 & 1 \end{bmatrix}
The number of distinct eigenvalues of P is
A
0
B
1
C
2
D
3
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
\text { Given: } A=\left[\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]
It is an upper triangular matrix. It's diagonal elements are eigen values.
The eigen values of the matrix are 1, 1, 1.
\therefore Number of distinct eigen values = 1
Hence, option (B) is correct.
Question 2
A parabola x=y^2 \; with \; 0\leq x\leq 1 is shown in the figure. The volume of the solid of rotation obtained by rotating the shaded area by 360^{\circ} around the x-axis is
A
\frac{\pi}{4}
B
\frac{\pi}{2}
C
\pi
D
2 \pi
Engineering Mathematics   Calculus
Question 2 Explanation: 
\text { Given: } y^{2}=x, 0 \leq x \leq 1
The value of solid obtained by rotating the area bounded by the curve
\begin{aligned} \mathrm{y}^{2}&=\mathrm{x}, 0 \leq \mathrm{x} \leq 1 \text{ about x-axis is}\\ V &=\int_{a}^{b} \pi y^{2} d x \\ V &=\int_{0}^{1} \pi x d x \\ &=\left(\frac{\pi \mathrm{x}^{2}}{2}\right)_{0}^{1} \\ &=\frac{\pi}{2} \end{aligned}
Question 3
For the equation \frac{dy}{dx} + 7x^2 y=0, if y(0)=3/7, then the value of y(1)is
A
\frac{7}{3}e^{-7/3}
B
\frac{7}{3}e^{-3/7}
C
\frac{3}{7}e^{-7/3}
D
\frac{3}{7}e^{-3/7}
Engineering Mathematics   Differential Equations
Question 3 Explanation: 
Given \frac{d y}{d x}+7 x^{2} y=0 \ldots(1)
With y(0)=\frac{3}{7} \ldots(2)
Now,(1) is written as
\Rightarrow \int \frac{1}{y} d y+\int 7 x^{2} d x=C
\Rightarrow \log y+\frac{7 x^{3}}{3}=C
\Rightarrow y=e^{\frac{7 x^{3}}{3}+c} \; \; ...(3)
Using (2),(3) becomes \frac{3}{7}=\mathrm{e}^{0} \cdot \mathrm{e}^{\mathrm{C}}(\mathrm{or}) \mathrm{e}^{\mathrm{C}}=\frac{3}{7} \; \; ...(4)
\therefore The solution of (1) with (3) &(4) is given by
y=y(x)=e^{\frac{-7 x^{3}}{3}+c}=e^{\frac{-7 x^{3}}{3}} \cdot e^{c}=\frac{3}{7} \cdot e^{\frac{-7 x^{3}}{3}}
Hence, y(1)=y=\frac{3}{7} \cdot e^{-\frac{7}{3}}
Question 4
The lengths of a large stock of titanium rods follow a normal distribution with a mean (\mu) of 440 mm and a standard deviation (\sigma) of 1mm. What is the percentage of rods whose lengths lie between 438 mm and 441 mm?
A
81.85%
B
68.40%
C
99.75%
D
86.64%
Engineering Mathematics   Probability and Statistics
Question 4 Explanation: 



\begin{array}{l} \mathrm{Z}=\frac{x-\mu}{\sigma} \\ \mathrm{Z}(\mathrm{x}=438)=\frac{438-440}{1}=-2 \\ \mathrm{P}(\mathrm{Z}=-2)=2.28 \% \\ \mathrm{Z}(\mathrm{x}=441)=\frac{441-440}{1}=1 \\ \mathrm{P}(\mathrm{Z}=1)=84.13 \% \\ \end{array}
The percentage of rods whose lengths lie between 438 mm and 441 mm =
\begin{array}{c} =\mathrm{P}(\mathrm{Z}=1)-\mathrm{P}(\mathrm{Z}=-2) \\ =84.13 \%-2.28 \%=81.85 \% \end{array}
Question 5
A flat-faced follower is driven using a circular eccentric cam rotating at a constant angular velocity \omega. At time t=0, the vertical position of the followeris y(0)=0, and the system is in the configuration shown below.

The vertical position of the follower face, y(t) is given by
A
e \sin \omega t
B
e(1+ \cos 2\omega t)
C
e(1- \cos \omega t)
D
e \sin 2 \omega t
Theory of Machine   Cams
Question 5 Explanation: 


\begin{aligned} \mathrm{AA}_{1}=\mathrm{y} &=\mathrm{e}(1-\cos \theta) \\ &=\mathrm{e}(1-\cos \omega \mathrm{t}) \end{aligned}
Question 6
The natural frequencies corresponding to the spring-mass systems I and II are \omega _I \; and \; \omega _{II}, respectively. The ratio \frac{\omega _I}{\omega _{II}} is
A
\frac{1}{4}
B
\frac{1}{2}
C
2
D
4
Theory of Machine   Vibration
Question 6 Explanation: 
System: I
k_{e q}=\frac{k \cdot k}{k+k}=\frac{k}{2}
\omega_{\mathrm{n}_{1}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}}
System - II
\mathrm{k}_{\mathrm{eq}}=2 \mathrm{k} \quad \omega_{\mathrm{n}_{2}}=\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}
\frac{\omega_{\mathrm{n}_{1}}}{\omega_{\mathrm{n}_{2}}}=\frac{\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}}}{\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}} \times \frac{\mathrm{m}}{2 \mathrm{k}}}=\frac{1}{2}
Question 7
A spur gear with 20^{\circ} full depth teeth is transmitting 20 kW at 200 rad/s. The pitch circle diameter of the gear is 100 mm. The magnitude of the force applied on the gear in the radial direction is
A
0.36 kN
B
0.73 kN
C
1.39 kN
D
2.78kN
Theory of Machine   Gear and Gear Train
Question 7 Explanation: 
\begin{array}{l} \phi=20^{\circ}, P=20 k W, \omega=200 \mathrm{rad} / \mathrm{s}, d=100 \mathrm{mm}=0.1 \mathrm{m} \\ \text { Torque }=\text { Power } / \omega \\ \mathrm{T}=\frac{20000}{200}=100 \mathrm{Nm} \\ \text { Now, } \mathrm{T}=\mathrm{F}_{\mathrm{T}} \times \frac{\mathrm{d}}{2} \\ \Rightarrow 100=\mathrm{F}_{\mathrm{T}} \times \frac{0.1}{2} \\ \Rightarrow \mathrm{F}_{\mathrm{T}}=2000 \mathrm{N}\\ \frac{F_{R}}{F_{T}}=\tan \phi \\ \Rightarrow \mathrm{F}_{\mathrm{R}}=2000 \times \tan 20^{\circ} \\ \Rightarrow \mathrm{F}_{\mathrm{R}}=727.94 \mathrm{N}=0.73 \mathrm{kN} \end{array}
Question 8
During a non-flow thermodynamic process(1-2) executed by a perfect gas, the heat interaction is equal to the work interaction (Q_{1-2}=W_{1-2}) when the process is
A
Isentropic
B
Polytropic
C
Isothermal
D
Adiabatic
Thermodynamics   Thermodynamic System and Processes
Question 8 Explanation: 
Given, \mathrm{Q}_{1-2}=\mathrm{W}_{1-2}
\therefore \Delta \mathrm{U}_{1-2}=0
\Rightarrow \mathrm{c}_{\mathrm{v}}\left[\mathrm{T}_{2}-\mathrm{T}_{1}\right]=0
\Rightarrow \mathrm{T}_{1}=\mathrm{T}_{2}
So, the process is isothermal.
Question 9
For a hydrodynamically and thermally fully developed laminar flow through a circular pipe of constant cross-section, the Nusselt number at constant wall heat flux (Nu_q) and that at constant wall temperature (Nu_T) are related as
A
Nu_q \gt Nu_T
B
Nu_q \lt Nu_T
C
Nu_q = Nu_T
D
Nu_q = ( Nu_T)^2
Heat Transfer   Heat Transfer in Flow Over Plates and Pipes
Question 9 Explanation: 
For laminar flow through circular tube:
\mathrm{Nu}_{\mathrm{q}}=4.36 (For constant heat flux)
\mathrm{Nu}_{\mathrm{T}}=3.66 (For constant wall temperature)
\mathrm{Nu}_{\mathrm{q}} \gt \mathrm{Nu}_{\mathrm{T}}
Question 10
As per common design practice, the three types of hydraulic turbines, in descending order of flow rate, are
A
Kaplan, Francis, Pelton
B
Pelton, Francis, Kaplan
C
Francis, Kaplan, Pelton
D
Pelton, Kaplan, Francis
Fluid Mechanics   Turbines and Pumps
Question 10 Explanation: 
Kaplan turbine has highest flow area hence it can handle highest discharge. On the other hand, Pelton turbine has lowest flow area hence it works on low discharge.
\therefore \mathrm{Q}_{\mathrm{Kaplan}} \gt \mathrm{Q}_{\mathrm{Francis}}>\mathrm{Q}_{\mathrm{Pclton}}
There are 10 questions to complete.