GATE ME 2014 SET-1

Question 1
Given that the determinant of the matrix
\begin{bmatrix} 1 & 3 & 0\\ 2 & 6 & 4\\ -1 & 0 & 2 \end{bmatrix}
is -12, the determinant of the matrix
\begin{bmatrix} 2 & 6 & 0\\ 4 & 12 & 8\\ -2 & 0 & 4 \end{bmatrix}
A
-96
B
-24
C
24
D
96
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
Let D=-12 for the given matrix
|A|=\left|\begin{array}{ccc}2 & 6 & 0 \\4 & 12 & 8 \\-2 & 0 & 4\end{array}\right|=(2)^{3}\left|\begin{array}{ccc}1 & 3 & 0 \\2 & 6 & 4 \\-1 & 0 & 2\end{array}\right|
(Taking 2 common from each row)
\begin{aligned}\therefore \text{Det}(A) &=(2)^{3} \times D \\=& 8 x-12=-96\end{aligned}
Question 2
\lim_{x\rightarrow 0} \frac{x-sin x}{1-cos x} is
A
0
B
1
C
3
D
not define
Engineering Mathematics   Calculus
Question 2 Explanation: 
\lim_{x \rightarrow 0} \frac{x-\sin x}{1-\cos x}
Applying L ' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x-sinx)}{\frac{d}{dx}(1-cosx)}=\lim_{x \rightarrow 0} \frac{(1-\cos x)}{\sin x}
(It is still of \frac{0}{0} form)
Again applying L' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{d x}(1-\cos x)}{\frac{d}{d x}(\sin x)}=\lim_{x \rightarrow 0} \frac{\sin x}{\cos x}=\frac{0}{1}=0
Question 3
The argument of the complex number \frac{1+i}{1-i},where i=\sqrt{-1}, is
A
-\pi
B
-\pi/2
C
\pi/2
D
\pi
Engineering Mathematics   Complex Variables
Question 3 Explanation: 
\begin{aligned} Let z&=\frac{1+i}{1-i} \\ or z&=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{1+i^{2}+2 i}{1-i^{2}}=\frac{2 i}{2}=i \\ z&=x+i y=i \\ SO, x&=0 \\ y &=1 \end{aligned}
\begin{aligned} Arg(z) &=\tan ^{-1}\left(\frac{y}{x}\right) \\ &=\tan ^{-1}\left(\frac{1}{0}\right)=\tan ^{-1} \infty=\frac{\pi}{2} \end{aligned}
Question 4
The matrix form of the linear system \frac{dx}{dt}=3x-5y and \frac{dy}{dt}=4x+8y is
A
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 3 & -5\\ 4& 8 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix}
B
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 3 & 8\\ 4& -5 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix}
C
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 4 & -5\\ 3& 8 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix}
D
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 4 & 8\\ 3& -5 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix}
Engineering Mathematics   Linear Algebra
Question 4 Explanation: 
\begin{aligned} \frac{d x}{d t} &=3 x-5 y \\ \frac{d y}{d t} &=4 x+8 y \\ \frac{d}{d t}\left\{\begin{array}{l}x \\y\end{array}\right\}&=\left[\begin{array}{cc}3 & -5 \\4 & 8\end{array}\right]\left\{\begin{array}{l}x \\y\end{array}\right\}=\left[\begin{array}{cc} 3 x & -5 y \\4 x & +8 y\end{array}\right]\end{aligned}
Question 5
Which one of the following describes the relationship among the three vectors \hat{i}+\hat{j}+\hat{k}, 2\hat{i}+3\hat{j}+\hat{k} and 5\hat{i}+6\hat{j}+4\hat{k}
A
The vectors are mutually perpendicular
B
The vectors are linearly dependent
C
The vectors are linearly independent
D
The vectors are unit vectors
Engineering Mathematics   Calculus
Question 5 Explanation: 
For linear dependency, det \left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 1 \\ 5 & 6 & 4\end{array}\right| must be zero.
\therefore \Delta=1(12-6)-1(8-5)+1(12-15)
=6-3-3=0
\therefore There three vectors are linearly dependent.
Question 6
A circular rod of length 'L' and area of cross-section 'A' has a modulus of elasticity 'E' and coefficient of thermal expansion '\alpha'. One end of the rod is fixed and other end is free. If the temperature of the rod is increased by \Delta T, then
A
stress developed in the rod is E\alpha \Delta T and strain developed in the rod is \alpha \Delta T
B
both stress and strain developed in the rod are zero
C
stress developed in the rod is zero and strain developed in the rod is \alpha \Delta T
D
stress developed in the rod is E\alpha \Delta T and strain developed in the rod is zero
Strength of Materials   Stress and Strain
Question 6 Explanation: 
For thermal stress to be developed, there must be constraint in the system. So, although strain develops but there is no thermal stress.
Question 7
A mentalic rod of 500 mm length and 50mm diameter, when subjected to a tensile force of 100KN at the ends, experiences an increase in its length by 0.5 mm and a reduction in its diameter by 0.015 m. the poission's ratio of the rod material is_________
A
0.3
B
0.7
C
0.1
D
0.9
Strength of Materials   Stress-strain Relationship and Elastic Constants
Question 7 Explanation: 
\begin{array}{l} v=-\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=-\frac{\left(\frac{\Delta D}{D}\right)}{\left(\frac{\Delta L}{L}\right)} \\ v=\left(\frac{L}{D}\right)\left(-\frac{\Delta D}{\Delta L}\right) \\ \text{given,}L=500\mathrm{mm},D=50\mathrm{mm}\\ \Delta D=-0.015, \Delta L=+0.5 \mathrm{mm} \\ \qquad v=\frac{500}{50} \times \frac{0.015}{0.500}=0.3 \end{array}
Question 8
Critical damping is the
A
largest amount of damping for which no oscillation occurs in free vibration
B
smallest amount of damping for which no oscillation occurs in free vibration
C
largest amount of damping for which the motion is simple harmonic in free vibration
D
smallest amount of damping for which the motion is simple harmonic in free vibration
Theory of Machine   Vibration
Question 8 Explanation: 
The system with critical damping returns to equilibrium in fastest time without any oscillation. In critically damped free vibrations, the damping force is just sufficient to dissipate the energy within One cycle of motion. The system never executes a cycle, it approaches equilibrium with exponentially decaying displacement.
Question 9
A circular object of radius r rolls without slipping on a horizontal level floor with the center having velocity V. The velocity at the point of contact between the object and the floor is
A
zero
B
V in the direction of motion
C
V opposite to the direction of motion
D
V vertically upward from the floor
Engineering Mechanics   Plane Motion
Question 9 Explanation: 


\begin{aligned} V_{P}= & \omega \times \overline{O P}=2 \omega r \\ V_{C}= & \omega \times \overline{O C}=\omega r \\ V_{O}= & \omega \times 0=0 \end{aligned}
Question 10
For the given statements:

I. Mating spur gear teeth is an example of higher pair
II. A revolute joint is an example of lower pair
Indicate the correct answer
A
Both I and II are false
B
I is true and II is false
C
I is false and II is true
D
Both I an d II are true
Machine Design   Gears
Question 10 Explanation: 
Example of Lower Pair:
Revolute pair, Prismatic joint, Screw pair, cylindrical joint, spherical joint.
Example of higher pair:
Wheel rolling an surface, cam and follower contact, meshing teeth of two gears.
There are 10 questions to complete.

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