GATE EE 2014 SET 2

Question 1
Which one of the following statements is true for all real symmetric matrices?
A
All the eigenvalues are real.
B
All the eigenvalues are positive.
C
All the eigenvalues are distinct.
D
Sum of all the eigenvalues is zero.
Engineering Mathematics   Linear Algebra
Question 2
Consider a dice with the property that the probability of a face with n dots showing up proportional to n. The probability of the face with three dots showing up is____.
A
0.1
B
0.33
C
0.14
D
0.66
Engineering Mathematics   Probability and Statistics
Question 2 Explanation: 
Let probability of occurence of one dot is P.
So, writing total probability
P+2P+3P+4P+5P+6P=1
P=\frac{1}{21}
Hence, problem of occurrence of 3 dot is =3P=\frac{3}{21}=\frac{1}{7}=0.142
Question 3
Minimum of the real valued function f(x)=(x-1)^{2/3} occurs at x equal to
A
-\infty
B
0
C
1
D
\infty
Engineering Mathematics   Calculus
Question 3 Explanation: 
f(x)=(x-1)^{2/3}=(\sqrt[3]{x-1})^2
As f(x) is square of \sqrt[3]{x-1}, hence its minimum value be 0 where at x=1.
Question 4
All the values of the multi-valued complex function 1^i, where i=\sqrt{-1}, are
A
purely imaginary
B
real and non-negative
C
on the unit circle
D
equal in real and imaginary parts
Engineering Mathematics   Complex Variables
Question 4 Explanation: 
Let z=1^i=1^{e^{i(4n+1)\pi/2}}\;\;\;n \in I
z=1 which is purly real and non negative.
Question 5
Consider the differential equation x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}-y=0. Which of the following is a solution to this differential equation for x \gt 0 ?
A
e^{x}
B
x^{2}
C
1/x
D
ln x
Engineering Mathematics   Differential Equations
Question 5 Explanation: 
\begin{aligned} x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y&=0\\ \text{Let, }x=e^z\leftrightarrow z&=\log x\\ s\frac{d}{dx}=xD=\theta &=\frac{d}{dz}\\ x^2D^2&=\theta (\theta -1)\\ (x^2D^2+xD-1)y&=0\\ [\theta (\theta -1)+\theta -1]y&=0\\ (\theta ^2-\theta +\theta -1)&=0\\ (\theta ^2-1)y&=0\\ \text{Auxiliary equation is }m^2-1&=0\\ m&=\pm 1\\ \text{CF is }C_1e^{-z}+C_2e^z&\\ \text{Solution is }y&=C_1e^{-z}+C_2e^z\\ y&=C_1x^{-1}+C_2x\\ y&=C_1\frac{1}{x}+C_2x \end{aligned}
One independent solution is \frac{1}{x}
Another independent solution is x.
Question 6
Two identical coupled inductors are connected in series. The measured inductances for the two possible series connections are 380 \muH and 240 \muH. Their mutual inductance in \muH is _____.
A
10
B
15
C
55
D
35
Electric Circuits   Magnetically Coupled Circuits, Network Topology and Filters
Question 6 Explanation: 
The two possible series connection are shown below:
Let the mutual inductance be M

(i) Additive connection,

(ii) Substractive connection,
L_{eqv}=L_1+L_2-2M=240\mu H
Thus, L_1+L_2+2M=380 \mu H\;\;...(i)
and L_1+L_2-2M=240 \mu H\;\;...(ii)
Solving equations (i) and (ii), we get:
4M=140\mu H
M=35\mu H
Therefore, mutual inductance M=35\mu H
Question 7
The switch SW shown in the circuit is kept at position '1' for a long duration. At t=0+, the switch is moved to position '2'. Assuming |V_{o2} | \gt |V_{o1}|, the voltage v_{c}(t) across the capacitor is
A
v_{c}(t)=-V_{o2}(1-e^{-t/2RC})-V_{o1}
B
v_{c}(t)=V_{o2}(1-e^{-t/2RC})+V_{o1}
C
v_{c}(t)=-(V_{o2}+V_{o1})(1-e^{-t/2RC})-V_{o1}
D
v_{c}(t)=(V_{o2}-V_{o1})(1-e^{-t/2RC})+V_{o1}
Electric Circuits   Transients and Steady State Response
Question 8
A parallel plate capacitor consisting two dielectric materials is shown in the figure. The middle dielectric slab is placed symmetrically with respect to the plates.

If the potential difference between one of the plates and the nearest surface of dielectric interface is 2 Volts, then the ratio \varepsilon _1:\varepsilon _2 is
A
1:4
B
2:3
C
3:2
D
4:1
Electromagnetic Fields   Electrostatic Fields
Question 8 Explanation: 
Let, A = Area of plates
Let C_1= C_3 be the capacitance formed with dielectric having dielectric constant \varepsilon _1.
C_{eqv} be the equivalent capacitance.
C_2 be the capacitance formed with dielectic having dielectric constant \varepsilon _{r_2}.
Then, C_1=C_3=\frac{\varepsilon _0\varepsilon _1A}{\frac{d}{4}}=\frac{4\varepsilon _0\varepsilon _1A}{d}
and C_2=\frac{\varepsilon _0\varepsilon _2A}{\frac{d}{2}}=\frac{2\varepsilon _0\varepsilon _2A}{d}
Also, equivalent capacitance = C_{eqv}
\begin{aligned} \therefore \;\;\frac{1}{C_{eqv}}&=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{2}{C_1}+\frac{1}{C_2} \\ &(\because \;\;C_1=C_3)(C_1,C_2,C_3 \;\text{all in series}) \\ \frac{1}{C_{eqv}} &= \frac{2d}{4\varepsilon _0\varepsilon _1d}+\frac{d}{2\varepsilon _0\varepsilon _2d}\\ C_{eqv}&=\frac{2\varepsilon _1\varepsilon _2\varepsilon _0A}{d(\varepsilon _1+\varepsilon _2)} \end{aligned}
Given, V_{eqv}= Total Voltage = 10 VOlt,
V_1=V_3=2 Volt
We know that : C\propto \frac{1}{V}
\begin{aligned} &\therefore \; \frac{C_{eqv}}{C_1}=\frac{V_1}{V_{eqv}}=\frac{1}{5} \\ &\frac{2\varepsilon _0\varepsilon _1\varepsilon _2 A}{d(\varepsilon _1+\varepsilon _2)}\times \frac{d}{4\varepsilon _0\varepsilon _1A}=\frac{1}{5} \\ &\frac{\varepsilon _2}{2(\varepsilon _1+\varepsilon _2)}=\frac{1}{5} \\ &5\varepsilon _2=2\varepsilon _1+2\varepsilon _2 \\ &2\varepsilon _1 =3\varepsilon _2\\ &\varepsilon _1:\varepsilon _2=3:2 \end{aligned}
Question 9
Consider an LTI system with transfer function
H(s)=\frac{1}{s(s+4)}
If the input to the system is cos(3t) and the steady state output is Asin(3t+\alpha) , then the value of A is
A
1/30
B
1/15
C
3/4
D
4/3
Signals and Systems   Laplace Transform
Question 9 Explanation: 
Given,
\begin{aligned} H(s)&=\frac{1}{s(s+4)}\\ r(t)&=\text{input}= \cos (3t)=\cos \omega t\\ \therefore \; \omega &=3 \text{ rad/s}\\ H(j\omega)&=\frac{1}{(j\omega)(j\omega+4)}\\ \text{Now, }|H(j\omega)|&=\frac{1}{\omega \sqrt{\omega^4+4^2}}\\ &=\frac{1}{3\sqrt{25}}=\frac{1}{15} \;\;\;(at\; \omega=3)\\ \angle H(j\omega)&=-90^{\circ}-\tan ^{-1}\frac{\omega}{4}\\ &=-90^{\circ}-\tan ^{-1}\frac{3}{4}=-126.86^{\circ}\\ \therefore \; c(t)&=\frac{1}{15} \cos (3t-126.86^{\circ})\\ &=\frac{1}{15} \sin (3t-36.86^{\circ})\;\;...(i)\\ c(t)&=A \sin (3t+\alpha )\;\;...(ii)\\ \text{Comparing }&\text{ eq. (i) and (ii), we have,}\\ A&=\frac{1}{15} \end{aligned}
Question 10
Consider an LTI system with impulse response h(t)=e^{-5t}u(t). If the output of the system is y(t)=e^{-3t}u(t)-e^{-5t}u(t) then the input, x(t), is given by
A
e^{-3t}u(t)
B
2e^{-3t}u(t)
C
e^{-5t}u(t)
D
2e^{-5t}u(t)
Signals and Systems   Linear Time Invariant Systems
Question 10 Explanation: 
Impulse response of an LTI system = transfer function =\frac{Y(s)}{X(s)}=H(s)
\begin{aligned} \text{where, } y(t)&=e^{-3t}u(t)-e^{-5t}u(t) \\ \therefore \; Y(s) &=\frac{1}{(s+3)}-\frac{1}{(s+5)} \\ &=\frac{2}{(s+3)(s+5)} \\ \text{Also, } h(t)&=e^{-5t}u(t) \\ \therefore \; H(s) &=\frac{1}{(s+5)} \\ \text{Therefore, } X(s)&=\frac{Y(s)}{H(s)} \\ &=\frac{2}{(s+3)(s+5)} \times (s+5) \\ &= \frac{2}{(s+3)}\\ \therefore \; \text{Input}&=x(t)=2e^{-3t}u(t) \end{aligned}
There are 10 questions to complete.
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