Question 1 |
One of the eigenvectors of the matrix \begin{bmatrix} -5 & 2\\ -9 & 6 \end{bmatrix} is
\begin{Bmatrix} -1\\ 1 \end{Bmatrix} | |
\begin{Bmatrix} -2\\ 9 \end{Bmatrix} | |
\begin{Bmatrix} 2\\ -1 \end{Bmatrix} | |
\begin{Bmatrix} 1\\ 1 \end{Bmatrix} |
Question 1 Explanation:
The characteristic equation | A-\lambda D=0
\text{i.e.} \left|\begin{array}{cc}-5-\lambda & 2 \\ -9 & 6-\lambda\end{array}\right|=0
\text{or } (\lambda-6)(\lambda+5)+18=0
\text{or } \quad \lambda^{2}-6 \lambda+5 \lambda-30+18=0
\text{or } \quad \lambda^{2}-\lambda-12=0
\text{or } \lambda=\frac{1 \pm \sqrt{1+48}}{2}=\frac{1 \pm 7}{2}=4,-3
Corresponding to \lambda=4, we have
\; [A-\lambda I] x=\left[\begin{array}{cc}-5-\lambda & 2 \\-9 & 6-\lambda\end{array}\right]\left[\begin{array}{l}x \\y\end{array}\right]
=0
\text{Or, } \left[\begin{array}{ll}-9 & 2 \\ -9 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=0which gives only one independent equation,
-9 x+2 y=0
\therefore \frac{x}{2}=\frac{y}{9} gives eigen vector (2,9)
Corresponding to \lambda=-3,
=\left[\begin{array}{ll}-2 & 2 \\-9 & 9\end{array}\right]\left[\begin{array}{l}x \\y \end{array}\right]=0
which gives -x+y=0 (only one independent equation)
\therefore \quad \frac{x}{1}=\frac{y}{1} \quad which gives (1,1)
So, the eigen vectors are \left\{\begin{array}{l}2 \\ 9\end{array}\right\} and \left\{\begin{array}{l}1 \\ 1\end{array}\right\}
\text{i.e.} \left|\begin{array}{cc}-5-\lambda & 2 \\ -9 & 6-\lambda\end{array}\right|=0
\text{or } (\lambda-6)(\lambda+5)+18=0
\text{or } \quad \lambda^{2}-6 \lambda+5 \lambda-30+18=0
\text{or } \quad \lambda^{2}-\lambda-12=0
\text{or } \lambda=\frac{1 \pm \sqrt{1+48}}{2}=\frac{1 \pm 7}{2}=4,-3
Corresponding to \lambda=4, we have
\; [A-\lambda I] x=\left[\begin{array}{cc}-5-\lambda & 2 \\-9 & 6-\lambda\end{array}\right]\left[\begin{array}{l}x \\y\end{array}\right]
=0
\text{Or, } \left[\begin{array}{ll}-9 & 2 \\ -9 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=0which gives only one independent equation,
-9 x+2 y=0
\therefore \frac{x}{2}=\frac{y}{9} gives eigen vector (2,9)
Corresponding to \lambda=-3,
=\left[\begin{array}{ll}-2 & 2 \\-9 & 9\end{array}\right]\left[\begin{array}{l}x \\y \end{array}\right]=0
which gives -x+y=0 (only one independent equation)
\therefore \quad \frac{x}{1}=\frac{y}{1} \quad which gives (1,1)
So, the eigen vectors are \left\{\begin{array}{l}2 \\ 9\end{array}\right\} and \left\{\begin{array}{l}1 \\ 1\end{array}\right\}
Question 2 |
\lim_{x\rightarrow 0}\left ( \frac{e^{2x}-1}{\sin(4x)} \right ) is equal to
0 | |
0.5 | |
1 | |
2 |
Question 2 Explanation:
\lim_{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)}{\sin 4 x}, it is of\left(\frac{0}{0}\right)from
Applying L' Hospital's rule
\lim_{x \rightarrow 0} \frac{2 e^{2 x}}{4 \cos 4 x}=\frac{2 \times 1}{4 \times 1}=\frac{1}{2}
Applying L' Hospital's rule
\lim_{x \rightarrow 0} \frac{2 e^{2 x}}{4 \cos 4 x}=\frac{2 \times 1}{4 \times 1}=\frac{1}{2}
Question 3 |
Curl of vector \vec{F}=x^{2}z^{2}\hat{i}-2xy^{2}z\hat{j}+2y^{2}z^{3}\hat{k} is
(4yz^{3}+2xy^{2})\hat{i}+2x^{2}z\hat{j}-2y^{2}z\hat{k} | |
(4yz^{3}+2xy^{2})\hat{i}-2x^{2}z\hat{j}-2y^{2}z\hat{k} | |
2xz^{2}\hat{i}-4xyz\hat{j}+6y^{2}z^{2}\hat{k} | |
2xz^{2}\hat{i}+4xyz\hat{j}+6y^{2}z^{2}\hat{k} |
Question 3 Explanation:
\vec{F}=x^{2} z^{2} \hat{i}-2 x y^{2} z \hat{j}+2 y^{2} z^{3} \hat{k}
\nabla \times \vec{F}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^{2} z^{2}&-2 x y^{2} z^{2} & 2 y^{2} z^{3} \end{array}\right|
=\hat{i}\left[\frac{\partial}{\partial y}\left(2 y^{2} z^{3}\right)+\frac{\partial}{\partial z}\left(2 x y^{2} z\right)\right]
-\hat{j}\left[\frac{\partial}{\partial x}\left(2 y^{2} z^{3}\right)-\frac{\partial}{\partial z}\left(x^{2} z^{2}\right)\right]
+\hat{k}\left[\frac{\partial}{\partial x}\left(-2 x y^{2} z\right)-\frac{\partial}{\partial y}\left(x^{2} z^{2}\right)\right]
\nabla \times \vec{F}=\hat{i}\left[4 y z^{3}+2 x y^{2}\right]-\hat{j}\left[-2 z x^{2}\right]
=\left(4 y z^{3}+2 x y^{2}\right) \hat{i}+\left(2 x^{2} z\right) \hat{j}-\left(2 y^{2} z\right) \hat{k}
\nabla \times \vec{F}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^{2} z^{2}&-2 x y^{2} z^{2} & 2 y^{2} z^{3} \end{array}\right|
=\hat{i}\left[\frac{\partial}{\partial y}\left(2 y^{2} z^{3}\right)+\frac{\partial}{\partial z}\left(2 x y^{2} z\right)\right]
-\hat{j}\left[\frac{\partial}{\partial x}\left(2 y^{2} z^{3}\right)-\frac{\partial}{\partial z}\left(x^{2} z^{2}\right)\right]
+\hat{k}\left[\frac{\partial}{\partial x}\left(-2 x y^{2} z\right)-\frac{\partial}{\partial y}\left(x^{2} z^{2}\right)\right]
\nabla \times \vec{F}=\hat{i}\left[4 y z^{3}+2 x y^{2}\right]-\hat{j}\left[-2 z x^{2}\right]
=\left(4 y z^{3}+2 x y^{2}\right) \hat{i}+\left(2 x^{2} z\right) \hat{j}-\left(2 y^{2} z\right) \hat{k}
Question 4 |
A box contains 25 parts of which 10 are defective. Two parts are being drawn simultaneously in a random manner from the box. The probability of both the parts being good is
7/20 | |
42/125 | |
25/29 | |
5/9 |
Question 4 Explanation:
required prob =\frac{^{15} C_{2}}{^{25} C_{2}}=\frac{14 \times 15}{25 \times 24}=\frac{7}{20}
Question 5 |
The best approximation of the minimum value attained by e^{-x}sin(100x) for x\geq 0 is _______
-0.9844 | |
-1.2652 | |
-9.2658 | |
-8.2654 |
Question 5 Explanation:
f(x)=e^{-x} \sin (100 x)
P(x)=-e^{-x} \sin (100 x)+e^{-x} \cos (100 x) \times 100
for minima f^{\prime}(x)=0
from here \tan (100 x)=100
\begin{aligned} x &=\frac{1}{100} \tan ^{-1}(100)=0.0156 \\ f(x) &=e^{-0.0156} \sin (100 \times 0.0156)=-0.9844 \end{aligned}
P(x)=-e^{-x} \sin (100 x)+e^{-x} \cos (100 x) \times 100
for minima f^{\prime}(x)=0
from here \tan (100 x)=100
\begin{aligned} x &=\frac{1}{100} \tan ^{-1}(100)=0.0156 \\ f(x) &=e^{-0.0156} \sin (100 \times 0.0156)=-0.9844 \end{aligned}
Question 6 |
A steel cube, with all faces free to deform, has Young's modulus, E, Poisson's ratio, v, and coefficient of thermal expansion, \alpha. The pressure (hydrostatic stress) developed within the cube, when it is subjected to a uniform increase in temperature, \Delta T, is given by
0 | |
\frac{\alpha (\Delta T)E}{1-2v} | |
-\frac{\alpha (\Delta T)E}{1-2v} | |
\frac{\alpha (\Delta T)E}{3(1-2v)} |
Question 6 Explanation:
As all forces are free to deform, there is no thermal
stress.
Question 7 |
A two member truss ABC is shown in the figure. The force (in kN) transmitted in member AB is
15kN | |
60kN | |
80kN | |
20kN |
Question 7 Explanation:
\begin{aligned} A B &=1 \\ A C &=0.5 \mathrm{m} \\ B C &=\sqrt{1^{2}+0.5^{2}} \\ &=\sqrt{1.25}=1.118 \mathrm{m} \\ A_{x}+C_{x} &=0 \\ A_{y}+C_{y} &=10 \\ \;&\qquad\text{(from force equilibrium)}\\ \Sigma M_{A}&=0 \\ C_{x} \times 0.5&=10 \times 1 \\ \text{or }\quad C_{x}&=20 \mathrm{kN} \\ \text{and }\quad A_{x}&=-20 \mathrm{kN} \\ \end{aligned}
\begin{aligned} \Sigma M_{c}&=0 \\ \Rightarrow \quad F_{A B} \times 0.5&=20 \times 0.5 \\ \therefore \quad F_{A B}&=20 \mathrm{kN} \end{aligned}
Question 8 |
A 4-bar mechanism with all revolute pairs has link lengths l_f=20mm, l_{in}=40mm, l_{co}=50mm and l_{out}=60mm. The suffixes 'f', 'in', 'co' and 'out' denote the fixed link, the input link, the coupler and output link respectively. Which one of the following statements is true about the input and output links?
Both links can execute full circular motion | |
Both links cannot execute full circular motion | |
Only the output link cannot execute full circular motion | |
Only the input link cannot execute full circular motion |
Question 8 Explanation:
S+L \lt P+Q
In this case, shortest link is fixed and hence, the
resulting mechanism is double crank mechanism.
In this case, shortest link is fixed and hence, the
resulting mechanism is double crank mechanism.
Question 9 |
In vibration isolation, which one of the following statements is NOT correct regarding Transmissibility (T)?
T is nearly unity at small excitation frequencies | |
T can be always reduced by using higher damping at any excitation frequency | |
T is unity at the frequency ratio of \sqrt{2} | |
T is infinity at resonance for undamped systems |
Question 10 |
In a structure subjected to fatigue loading, the minimum and maximum stresses developed in a cycle are 200 MPa and 400 MPa respectively. The value of stress amplitude (in MPa) is
200 | |
100 | |
50 | |
150 |
Question 10 Explanation:
\begin{aligned} \sigma _{min}&=200 \; \text{MPa} \\ \sigma _{max}&=400 \; \text{MPa} \\ \sigma _a&= \frac{\sigma _{max}-\sigma _{min}}{2}\\ &= \frac{400-200}{2}\\ &=100\; \text{MPa} \end{aligned}
There are 10 questions to complete.
