# 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
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.