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)?

1 | |

2 | |

3 | |

4 |

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

\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 ______.

14 | |

27 | |

28 | |

31 |

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}

\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

\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

2.5 | |

3.5 | |

1.2 | |

0 |

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}

\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 ___.

If a and b are positive integers, the value of \int_{-\infty }^{\infty }e^{-a(x+b)^2}dx is ___.

\sqrt{\pi a} | |

\sqrt{\frac{\pi}{a}} | |

b \sqrt{\pi a} | |

b \sqrt{\frac{\pi}{a}} |

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?

\phi (s)=0 at all the three values of s satisfying s^3+\sigma ^3=0 | |

\phi (s)=0, \frac{d \phi (s)}{ds}=0, \text{ and }\frac{d^2 \phi (s)}{ds^2}=0 \text{ at }s=-\sigma | |

\phi (s)=0, \frac{d \phi (s)}{ds}=0, \text{ and }\frac{d^4 \phi (s)}{ds^4}=0 \text{ at }s=-\sigma | |

\phi (s)=0, \text{ and }\frac{d^3 \phi (s)}{ds^3}=0 \text{ at }s=-\sigma |

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.

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?

Stress value where the stress-strain curve transitions from elastic to plastic behavior | |

The maximum load attained divided by the original cross-sectional area | |

The maximum load attained divided by the corresponding instantaneous crosssectional area | |

Stress where the specimen fractures |

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.

\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

rotate slowly (compared to \omega ) about the negative z-axis direction | |

rotate slowly (compared to \omega ) about the positive z-axis direction | |

rotate slowly (compared to \omega ) about the negative y-axis direction | |

rotate slowly (compared to \omega ) about the positive y-axis direction |

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

70 Mpa | |

90 Mpa | |

120 Mpa | |

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}

\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

400 \; ^{\circ}C | |

250 \; ^{\circ}C | |

800 \; ^{\circ}C | |

150 \; ^{\circ}C |

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

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

P-3, Q-1, R-4 | |

P-1, Q-3, R-2 | |

P-1, Q-3, R-4 | |

P-3, Q-1, R-2 |

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.