GATE EE 2015 SET 1

Question 1
A random variable X has probability density function f(x) as given below:

f(x)=\left\{\begin{matrix} a+bx & for \; 0 \lt x \lt 1\\ 0& otherwise \end{matrix}\right.

If the expected value E[X]=2/3, then Pr[X \lt 0.5] is _____________.
A
0.25
B
0.5
C
0.75
D
1
Engineering Mathematics   Probability and Statistics
Question 1 Explanation: 
\begin{aligned} f(x)&=\left\{\begin{matrix} a+bx & \text{for }0 \lt x \lt 1\\ 0 & \text{otherwise} \end{matrix}\right.&\\ \text{Now given }E(X)&=2/3\\ \int_{0}^{1}xf(x)dx&=\frac{2}{3}\\ \int_{0}^{1}x(a+bx)dx&=\frac{2}{3}\\ a\left ( \frac{x^2}{2} \right )_0^1+b\left ( \frac{x^3}{3} \right )_0^1&=\frac{2}{3}\\ a\left ( \frac{1}{2} \right )+b\left ( \frac{1}{3} \right )&=\frac{2}{3}\\ 3a+2b&=4\;\;...(i)\\ \text{Now, }\int_{0}^{1}f(x)dx &=1\\ (\text{total probability} & \text{ is always equal to 1})\\ \int_{0}^{1}(a+bx)dx&=\left ( ax+\frac{bx^2}{2} \right )_0^1=1\\ a+\frac{b}{2}&=1\\ 2a+b&=2\;\;...(ii)\\ \text{Now solving } & \text{ (i) and (ii), we get}\\ a&=0,b=2\\ So,\; \;f(x)&=\left\{\begin{matrix} 2x & \text{for }0 \lt x lt 1\\ 0 & \text{otherwise} \end{matrix}\right.&\\ \text{Now we need}&\\ P(x \lt 0.5)&=\int_{0}^{0.5}2xdx=2\left ( \frac{x^2}{2} \right )_0^{0.5}\\ &=0.5^2-0^2=0.25 \end{aligned}
Question 2
If a continuous function f(x) does not have a root in the interval [a,b], then which one of the following statements is TRUE?
A
f(a)\cdot f(b)=0
B
f(a)\cdot f(b) \lt 0
C
f(a)\cdot f(b) \gt 0
D
f(a)/ f(b)\leq 0
Engineering Mathematics   Calculus
Question 2 Explanation: 
Intermediate value theorem states that if a function is continious and f(a) \cdot f(b) \lt 0, then surely there is a root in (a,b). The contrapositive of this theorem is that if a function is continious and has no root in (a,b) then surely f(a) \cdot f(b) \geq 0. But since it is given that there is no root in the closed interval [a,b] it means f(a) \cdot f(b) \neq 0.
So surely f(a) \cdot f(b) \gt 0 which is choise(C).
Question 3
If the sum of the diagonal elements of a 2x2 matrix is -6, then the maximum possible value of determinant of the matrix is ________.
A
6
B
8
C
9
D
12
Engineering Mathematics   Linear Algebra
Question 3 Explanation: 
Consider a symmetric matrix A =\begin{bmatrix} a & b\\ b & d \end{bmatrix}
Given a+d=-6
|A|=ad-b^2
Now since b^2 is always non-negative, maximum determinant will come when b^2=0.
So we need to maximize
\begin{aligned} |A|&=ad-0=ad=a \times -(6+a) \\ &= -a^2-6a\\ \frac{d|A|}{da} &=-2a-6=0 \\ \Rightarrow \;\; a &=-3 \text{ is the only stationary point} \end{aligned}
Since, \left [\frac{d^2|A|}{da} \right ]_{a=-3}=-2 \lt 0, we have a maximum at a=-3.
Since a+d=-6, corresponding value of d=-3.
|A|=ad=-3 \times -3=9
Question 4
Consider a function \vec{f}=\frac{1}{r^{2}}\hat{r}, where r is the distance from the origin and \hat{r} is the unit vector in the radial direction. The divergence of this function over a sphere of radius R, which includes the origin, is
A
0
B
2\pi
C
4\pi
D
R\pi
Electromagnetic Fields   Coordinate Systems and Vector Calculus
Question 4 Explanation: 
\bar{f}=\frac{1}{r^2}\hat{r}
From divergence theorem as we know,
\begin{aligned} \int _{vol.} (\triangledown \cdot \bar{f})dV &=\oint \oint _S \bar{f}\cdot d\bar{S}\\ \oint \oint _S \bar{f}\cdot d\bar{S} &= \oint \oint _S \left ( \frac{1}{r^2} \cdot \hat{r}\right ) \times r^2 \sin \theta \cdot d\theta \cdot d\phi \cdot \hat{r} \\ &= \oint \oint _S \sin \theta \cdot d\theta \cdot d\phi=4\pi \end{aligned}
Question 5
When the Wheatstone bridge shown in the figure is used to find the value of resistor R_X, the galvanometer G indicates zero current when R_1=50\Omega ,R_2=65\Omega and R_3=100\Omega. If R_3 is known with \pm 5% tolerance on its nominal value of 100 \Omega , what is the range of R_X in Ohms?
A
[123.50, 136.50]
B
[125.89, 134.12]
C
[117.00, 143.00]
D
[120.25, 139.75]
Electrical and Electronic Measurements   Characteristics of Instruments and Measurement Systems
Question 5 Explanation: 
R_1=50\Omega
R_2=60\Omega
R_3=100\pm 5%\Omega
The value of R_3 \text{with} \pm 5% of tollerance,
\begin{aligned} R_3&= 100\pm 5\% \\ &= 100+100 \times \frac{5}{100}=105 \Omega \\ &= 100-100 \times \frac{5}{100}=95 \Omega \end{aligned}
In both condition, the bridge is balance, so under balance condition,
\begin{aligned} \frac{R_1}{R_3}&=\frac{R_2}{R_x} \\ R_x&=\frac{R_2R_3}{R_1} \\ (i) \; \text{when}, \; R_3&=105\Omega \\ \therefore \;\;R_x &= \frac{105 \times 65}{50}=136.50\Omega \\ (ii)\; \text{when}, \; R_3&=95 \Omega \\ \therefore \;\;R_x &= \frac{95 \times 65}{50}=123.50\Omega \end{aligned}
Question 6
A (0-50 A) moving coil ammeter has a voltage drop of 0.1 V across its terminals at full scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (0-500 A) is _______.
A
0.11
B
0.22
C
0.45
D
0.68
Electrical and Electronic Measurements   Galvanometers, Voltmeters and Ammeters
Question 6 Explanation: 


I_{fs}=50A,
V_m=0.1V,
R_m=\frac{0.1}{50}=2 \times 10^{-3}\Omega
\because \; m=10
R_{sh}=\frac{R_m}{(m-1)}
\;\;\;=\frac{2 \times 10^{-3}}{9}=0.22\Omega
Question 7
Of the four characteristics given below, which are the major requirements for an instrumentation amplifier?

P. High common mode rejection ratio
Q. High input impedance
R. High linearity
S. High output impedance
A
P, Q and R only
B
P and R only
C
P, Q and S only
D
Q, R and S only
Analog Electronics   Operational Amplifiers
Question 8
In the following chopper, the duty ratio of switch S is 0.4. If the inductor and capacitor are sufficiently large to ensure continuous inductor current and ripple free capacitor voltage, the charging current (in Ampere) of the 5 V battery, under steady-state, is_____.
A
1
B
1.5
C
2.5
D
3
Power Electronics   Choppers
Question 8 Explanation: 


Step down chopper
\begin{aligned} \Rightarrow \; V_0&=\alpha V_s\\ V_0&=I_0R+E\\ I_0&=\frac{V_0-E}{R}\\ I_0&=\frac{\alpha V_s-E}{R}\\ &=\frac{(0.4)20-5}{3}=1A \end{aligned}
Question 9
A moving average function is given by y(t)=\frac{1}{T}\int_{t-T}^{t}u(\tau )d\tau. If the input u is a sinusoidal signal of frequency \frac{1}{2\tau } Hz, then in steady state, the output y will lag u (in degree) by ______ .
A
30
B
60
C
90
D
120
Signals and Systems   Linear Time Invariant Systems
Question 9 Explanation: 
System input: \sin \omega _0 t
\begin{aligned} f_0&=\frac{1}{2T}Hz\\ \text{Therefore,}\\ \omega _0&=2 \pi f_0=\frac{2 \pi}{2T}=\frac{\pi}{T}\; rad/sec\\ y(t)&=\frac{1}{T}\int_{t-T}^{t}\sin \omega _0\tau \; d\tau\\ &=\frac{1}{T}[-\cos\omega _0 t ]_{t-T}^t\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\omega _0 (t-T) ]\\ &\left [ \because \; \omega _0=\frac{t}{T} \right ]\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\left ( \frac{\pi t}{T}- \pi \right ) ]\\ \\ &=-\frac{1}{T}[\cos\omega _0 t +\cos\omega _0 t ]\\ &=-\frac{2}{T} \cos\omega _0 t \\ &=\frac{2}{T}\sin (\omega _0 t -90^{\circ}) \\ &=\text{Output is lagging by }90^{\circ} \text{ w.r.t. input.} \end{aligned}
Question 10
The impulse response g(t) of a system, G , is as shown in Figure (a). What is the maximum value attained by the impulse response of two cascaded blocks of G as shown in Figure (b)?
A
\frac{2}{3}
B
\frac{3}{4}
C
\frac{4}{5}
D
1
Signals and Systems   Linear Time Invariant Systems
Question 10 Explanation: 
\begin{aligned} g(t)&=u(t)-u(t-1)\\ G(s)&=\frac{1}{s}-\frac{e^{-s}}{s}\\ G(s) \times G(s)&=g(t)* g(t) \end{aligned}

Maximum value =1
There are 10 questions to complete.
Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.