GATE EE 2014 SET 3

Question 1
Two matrices A and B are given below:
A=\begin{bmatrix} p &q \\ r& s \end{bmatrix}; B=\begin{bmatrix} p^2+q^2 & pr +qs \\ pr+qs & r^2+s^2 \end{bmatrix}
If the rank of matrix A is N, then the rank of matrix B is
A
N/2
B
N-1
C
N
D
2N
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
\begin{aligned} A &=\begin{bmatrix} p & q\\ r & s \end{bmatrix} \\ A \times A=A^2&=\begin{bmatrix} p^2+q^2 & pr+qs\\ pr+qs & r^2+s^2 \end{bmatrix} =B \\ A^2 &=B \end{aligned}
Rank of amtrix does not change when we squaring the matrix, hence rank of B = rank of A=N.
Question 2
A particle, starting from origin at t=0s, is traveling along x-axis with velocity
v=\frac{\pi}{2}\cos (\frac{\pi}{2}t)m/s
At t=3s, the difference between the distance covered by the particle and the magnitude of displacement from the origin is_____
A
1
B
2
C
3
D
4
Engineering Mathematics   Calculus
Question 3
Let \triangledown \cdot (f v)=x^2y+y^2z+z^2x, where f and v are scalar and vector fields respectively. If v=yi+zj+xk , then v\cdot \triangledown f is
A
x^2y+y^2z+z^2x
B
2xy+2yz+2zx
C
x+y+z
D
0
Engineering Mathematics   Calculus
Question 3 Explanation: 
\begin{aligned} \vec{V}&=y\hat{i}+z\hat{j}+x\hat{k}\\ \hat{i}\frac{\partial (fV)}{\partial x}+\hat{j}\frac{\partial (fV)}{\partial y}+\hat{k}\frac{\partial (fV)}{\partial z}&=x^2y+y^2z+z^2x\\ y\frac{\partial f}{\partial x}+z\frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}&=x^2y+y^2z+z^2x\;\;...(i)\\ \vec{V}\cdot \Delta f&=y\frac{\partial f}{\partial x}+z\frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}\;\;...(ii)\\ \text{From equations (i) and (ii)}\\ \vec{V}\cdot \Delta f&=x^2y+y^2z+z^2x \end{aligned}
Question 4
Lifetime of an electric bulb is a random variable with density f(x)=kx^2, where x is measured in years. If the minimum and maximum lifetimes of bulb are 1 and 2 years respectively, then the value of k is _____
A
0.85
B
0.42
C
0.25
D
0.75
Engineering Mathematics   Probability and Statistics
Question 4 Explanation: 
Life time of an electric bulb with density
f(x)=Kx^2
If minimum and maximum lifetimes of bulb are 1 and 2 years respectively then
\begin{aligned} \int_{1}^{2}Kx^2dx &=1\\ \left.\begin{matrix} K\frac{x^3}{3} \end{matrix}\right|_1^2&=1\\ K\left ( \frac{8}{3}-\frac{1}{3} \right )&=1\\ \frac{7K}{3}&=1\\ K&=\frac{3}{7}=0.42 \end{aligned}
Question 5
A function f(t) is shown in the figure.

The Fourier transform F(\omega) of f(t) is
A
real and even function of w
B
real and odd function of w
C
imaginary and odd function of w
D
imaginary and even function of w
Signals and Systems   Fourier Transform
Question 5 Explanation: 
Fiven signal f(t) is an odd signal. Hence, F(\omega ) is imaginary and odd function of \omega .
Question 6
The line A to neutral voltage is 10 \angle 15^{\circ}V for a balanced three phase star connected load with phase sequence ABC . The voltage of line B with respect to line C is given by
A
10 \sqrt{3}\angle 105^{\circ} V
B
10 \angle 105^{\circ} V
C
10 \sqrt{3}\angle -75^{\circ} V
D
-10 \sqrt{3}\angle 90^{\circ} V
Electric Circuits   Three-Phase Circuits
Question 6 Explanation: 
Given,
V_{AN}=10\angle 15^{\circ}volt
As the system is balanced and phase sequence is ABC, therefore,
V_{AN}=10\angle 15^{\circ}
V_{BN}=10\angle 135^{\circ}V and V_{CN}=10\angle 255^{\circ}V
\therefore voltage of line w.r.t. line C is
V_{BC}=V_{BN}-V_{CN}
V_{BC}=10\angle 255^{\circ}-10\angle 135^{\circ}
V_{BC}=10\sqrt{3}\angle -75^{\circ} Volt
Question 7
A hollow metallic sphere of radius r is kept at potential of 1 Volt. The total electric flux coming out of the concentric spherical surface of radius R(\gt r) is
A
4\pi \varepsilon _{0}r
B
4\pi \varepsilon _{0}r^{2}
C
4\pi \varepsilon _{0}R
D
4\pi \varepsilon _{0}R^{2}
Electromagnetic Fields   Electrostatic Fields
Question 8
The driving point impedance Z(s) for the circuit shown below is
A
\frac{s^{4}+3s^{2}+1}{s^{3}+2s}
B
\frac{s^{4}+3s^{2}+4}{s^{2}+2}
C
\frac{s^{2}+1}{s^{4}+s^{2}+1}
D
\frac{s^{3}+1}{s^{4}+s^{2}+1}
Electric Circuits   Two Port Network and Network Functions
Question 8 Explanation: 


Driving point impedance, Z(s) is ,
Z(s)=s+\left ( \frac{\left ( s+\frac{1}{s} \right )\times \frac{1}{s}}{s+\frac{1}{s}+\frac{1}{s}} \right )
\;\;=s+\left ( \frac{s^2+1}{s^2} \right )\times \frac{s}{s^2+2}
\;\;=s+\frac{s^2+1}{s(s^2+2)}
\;\;=\frac{s^2(s^2+2)+^2+1}{s^3+2s}
Z(s)=\frac{s^4+3s^2+1}{s^3+2s}
Question 9
A signal is represented by
x(t)=\left\{\begin{matrix} 1 &|t| \lt 1 \\ 0& |t|\gt 1 \end{matrix}\right.
The Fourier transform of the convolved signal y(t)= x(2t)* x(t/2) is
A
\frac{4}{\omega ^{2}}sin(\frac{\omega }{2})sin(2 \omega )
B
\frac{4}{\omega ^{2}}sin(\frac{\omega }{2})
C
\frac{4}{\omega ^{2}}sin(2 \omega )
D
\frac{4}{\omega ^{2}}sin^{2} \omega
Signals and Systems   Fourier Transform
Question 9 Explanation: 
Given signal can be drawn as

Therefore,
\begin{aligned} &x(t)\leftrightarrow X(\omega )=2Sa(\omega ) \\ &\text{Now, } x(t)\leftrightarrow X(\omega )\\ &\text{then by time scaling,} \\ &x(at)\leftrightarrow \frac{1}{|a|}X(\omega /a) \\ &\therefore \; x(2t)\leftrightarrow Sa \left ( \frac{\omega }{2} \right )\;\;...(i) \\ &x\left ( \frac{t}{2} \right ) \leftrightarrow 4Sa(2\omega )\;\;...(ii)\\ &\text{Now, }y(t)=x(2t) \times x(t/2) \end{aligned}
Convolution in time domain multiplication in frequency domain
\begin{aligned} Y(\omega )&=4Sa\left ( \frac{\omega }{2}\right ) Sa(2\omega ) \\ Y(\omega )&=\frac{4\sin \left ( \frac{\omega }{2}\right ) }{\left ( \frac{\omega }{2}\right ) } \frac{\sin (2\omega )}{2\omega }\\ Y(\omega )&=\frac{4}{\omega ^2}\sin \left ( \frac{\omega }{2}\right ) \sin (2\omega ) \end{aligned}
Question 10
For the signal
f(t) = 3 \sin 8 \pi t + 6 \sin 12 \pi t + \sin 14 \pi t,
the minimum sampling frequency (in Hz) satisfying the Nyquist criterion is _____.
A
7
B
14
C
18
D
9
Signals and Systems   Sampling
Question 10 Explanation: 
\begin{aligned} f_{m_1} &=4Hz \\ f_{m_2} &=6Hz \\ f_{m_3} &=7Hz \end{aligned}
Then minimum sampling frequency satisfying the nyquist criterion is 7*2=14Hz.
There are 10 questions to complete.
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