Question 1 |
Consider a 3x3 real symmetric matrix S such that two of its eigenvalues are a\neq 0,\; b\neq 0 with respective eigenvectors \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix},\begin{bmatrix} y_{1}\\ y_{2}\\ y_{3} \end{bmatrix}. If a\neq b then x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3} equals
a | |
b | |
ab | |
0 |
Question 1 Explanation:
3\times 3 real symmetric matrix such that two of its
eigen value are a \neq 0 b \neq 0 with respective eigen
vectors \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]\left[\begin{array}{l}y_{1} \\ y_{2} \\ y_{3}\end{array}\right] \text{ if } a \neq b then
x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0 because they are orthogonal.
\therefore \qquad x^{T} y =0 \quad(\text { since } a \neq b)
\left[x_{1} x_{2} x_{3}\right]\left[\begin{array}{c}y_{1} \\y_{2} \\y_{3}\end{array}\right]=0
x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0
eigen value are a \neq 0 b \neq 0 with respective eigen
vectors \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]\left[\begin{array}{l}y_{1} \\ y_{2} \\ y_{3}\end{array}\right] \text{ if } a \neq b then
x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0 because they are orthogonal.
\therefore \qquad x^{T} y =0 \quad(\text { since } a \neq b)
\left[x_{1} x_{2} x_{3}\right]\left[\begin{array}{c}y_{1} \\y_{2} \\y_{3}\end{array}\right]=0
x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0
Question 2 |
If a function is continuous at a point,
the limit of the function may not exist at the point | |
the function must be derivable at the point | |
the limit of the function at the point tends to infinity | |
the limit must exist at the point and the value of limit should be same as the value of the function at that point |
Question 2 Explanation:
f(x) is continuous at any point
\text{if} \quad \lim_{x \rightarrow a^{-}} f(x)=\lim_{x \rightarrow a^{+}} f(x)=f(a)
\text{if} \quad \lim_{x \rightarrow a^{-}} f(x)=\lim_{x \rightarrow a^{+}} f(x)=f(a)
Question 3 |
Divergence of the vector field x^{2}z\hat{i}+xy\hat{j}-yz^{2}\hat{k} at (1,-1,1) is
0 | |
1 | |
5 | |
6 |
Question 3 Explanation:
\begin{aligned} \vec{F} &=x^{2} z \hat{i}+x y \hat{j}-y z^{2} \hat{k} \\ \nabla \cdot \vec{F} &=\frac{\partial}{\partial x}\left(x^{2} z\right)+\frac{\partial}{\partial y}(x y)-\frac{\partial}{\partial z}\left(y z^{2}\right) \\ \nabla \cdot \vec{F} &=2 x z+x-2 y z \\ \left.\therefore \nabla \cdot \vec{F}\right|_{1,-1,1} &=2 \times 1 \times 1+1-2 \times-1 \times 1 \\ &=2+1+2=5 \end{aligned}
Question 4 |
A group consists of equal number of men and women. Of this group 20% of the men and 50% of the women are unemployed. If a person is selected at random from this group, the probability of the selected person being employed is _______
0.25 | |
1 | |
4.5 | |
0.65 |
Question 4 Explanation:
Total \% of employed person
\begin{aligned} &=100-\frac{20+50}{2} \% \\ &=100-35 \%=65 \% \\ \text { Required prob } &=\frac{65}{100}=0.65 \end{aligned}
\begin{aligned} &=100-\frac{20+50}{2} \% \\ &=100-35 \%=65 \% \\ \text { Required prob } &=\frac{65}{100}=0.65 \end{aligned}
Question 5 |
The definite integral \int_{1}^{3}\frac{ 1}{x}\mathrm{d}x is evaluated using Trapezoidal rule with a step size of 1. The correct answer is _______
1.165 | |
2.365 | |
8.254 | |
9.548 |
Question 5 Explanation:

\begin{aligned} I &=\int_{1}^{3} \frac{1}{x} \mathrm{d} x=\frac{h}{2}\left[\left(y_{0}+y_{2}\right)+2 y_{1}\right] \\ &=\frac{1}{2}[1+0.33+2 \times 0.5]=\frac{2.33}{2}=1.165 \end{aligned}
Question 6 |
A rotating steel shaft is supported at the ends. It is subjected to a point load at the center. The maximum bending stress developed is 100 MPa. If the yield, ultimate and corrected endurance strength of the shaft material are 300 MPa, 500 MPa and 200 MPa, respectively, then the factor of safety for the shaft is _______
1 | |
3 | |
4 | |
2 |
Question 6 Explanation:
Same fiber under point load at the center of a rotating steel shaft supported at ends is subjected to reversal of stresses. Here only maximum stress is given, its sign (nature) is not given
\begin{aligned} \therefore \qquad \sigma_{\min }&=-100 \mathrm{MPa}\\ \sigma_{\max } &=100 \mathrm{MPa} \\ \sigma_{m} &=\frac{\sigma_{\max }+\sigma_{\min }}{2}=0 \\ \sigma_{a} &=\frac{\sigma_{\max }-\sigma_{\min }}{2} \\ &=\frac{100+100}{2}=100 \mathrm{MPa} \end{aligned}
As the steel is ductile material, Soderberg criteria will be applied.
\therefore \quad \frac{\sigma_{m}}{\sigma_{y}}+\frac{\sigma_{a}}{\sigma_{e}}=\frac{1}{N}
\text{or }\quad 0+\frac{100}{200}=\frac{1}{N} \Rightarrow N=2
\begin{aligned} \therefore \qquad \sigma_{\min }&=-100 \mathrm{MPa}\\ \sigma_{\max } &=100 \mathrm{MPa} \\ \sigma_{m} &=\frac{\sigma_{\max }+\sigma_{\min }}{2}=0 \\ \sigma_{a} &=\frac{\sigma_{\max }-\sigma_{\min }}{2} \\ &=\frac{100+100}{2}=100 \mathrm{MPa} \end{aligned}
As the steel is ductile material, Soderberg criteria will be applied.
\therefore \quad \frac{\sigma_{m}}{\sigma_{y}}+\frac{\sigma_{a}}{\sigma_{e}}=\frac{1}{N}
\text{or }\quad 0+\frac{100}{200}=\frac{1}{N} \Rightarrow N=2
Question 7 |
Two solid circular shafts of radii R_{1} and R_{2} are subjected to same torque. The maximum shear stresses developed in the two shafts are \tau_{1} and \tau_{2} . If R_{1} /R_{2}=2, then \tau_{2} /\tau_{1} is _______
4 | |
8 | |
10 | |
12 |
Question 7 Explanation:
\begin{aligned} \tau _{max}&=\frac{2T}{\pi R^3}\\ \frac{\tau _2}{\tau _1}&=\left ( \frac{R_1}{R_2} \right )^3\\ &=2^3=8 \;\;(\because \; T_1=T_2=T) \end{aligned}
Question 8 |
Consider a single degree-of-freedom system with viscous damping excited by a harmonic force. At resonance, the phase angle (in degree) of the displacement with respect to the exciting force is
0 | |
45 | |
90 | |
135 |
Question 8 Explanation:

X: Displacement
F_{0}: Excitation force
c \omega X: Damping force
k X: Spring force
m \omega^{2} X: Inertia force
Question 9 |
A mass m_{1} of 100 kg travelling with a uniform velocity of 5 m/s along a line collides with a stationary mass m_{2} of 1000 kg. After the collision, both the masses travel together with the same velocity. The coefficient of restitution is
0.6 | |
0.1 | |
0.01 | |
0 |
Question 9 Explanation:
Coefficient of restitution,
C_{R}=\frac{\text { Relative speed after collision }}{\text { Relative speed before collision }}
After collision, both the masses move with same velocity. So, there is zero relative velocity between them,
\therefore \quad C_{R}=\frac{0}{5}=0
C_{R}=\frac{\text { Relative speed after collision }}{\text { Relative speed before collision }}
After collision, both the masses move with same velocity. So, there is zero relative velocity between them,
\therefore \quad C_{R}=\frac{0}{5}=0
Question 10 |
Which one of following is NOT correct?
Intermediate principal stress is ignored when applying the maximum principal stress theory | |
The maximum shear stress theory gives the most accurate results amongst all the failure theories | |
As per the maximum strain energy theory, failure occurs when the strain energy per unit volume exceeds a critical value | |
As per the maximum distortion energy theory, failure occurs when the distortion energy per unit volume exceeds a critical value |
Question 10 Explanation:
The maximum shear stress theory predicts that
shear yield value \tau_{y} is 0.5 times the tensile yield value. This is about 15 % less than the value predicted by the distortion energy (or the
octahedral shear) theory. The maximum shear
stress theory gives values for design on the safe
side. Also, because of its simplicity, this theory
is widely used in machine design dealing with
ductile materials.
There are 10 questions to complete.