# GATE EE 2010

 Question 1
The value of the quantity P, where $P=\int_{0}^{1}xe^{x}dx$, is equal to
 A 0 B 1 C e D 1/e
Engineering Mathematics   Calculus
Question 1 Explanation:
$P=\int_{0}^{1}xe^x\; dx$
Integrating by parts:
\begin{aligned} \text{Let } \; u&= x,\\ dv&= e^x\; dx\\ du&= dx,\\ v&=\int e^x\; dx=e^x \\ \text{Now, } \int udv&= uv-\int v\; du\\ \therefore \;\;\int x e^x\; dx &=xe^x-\int e^x\; dx \\ &=xe^x-e^x+c \\ \int_{0}^{1} xe^x\; dx&= [xe^x-e^x]_0^1\\ &= (1 \times e^1 -e^1)-(0 \times e^0 -e^0)\\ &= 0-(-1)\\ &= 1 \end{aligned}
 Question 2
Divergence of the three-dimensional radial vector field $\vec{r}$ is
 A 3 B 1/r C $(\hat{i}+\hat{j}+\hat{k})$ D $3(\hat{i}+\hat{j}+\hat{k})$
Engineering Mathematics   Calculus
Question 2 Explanation:
\begin{aligned} \vec{r} &= x\hat{i}+y\hat{j}+z\hat{k}\\ \text{div } \vec{r}&= \bigtriangledown \cdot \vec{r}\\ &=\left ( \hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z} \right )(x\hat{i}+y\hat{j}+z\hat{k})\\ &=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}\\ &=1+1+1=3 \end{aligned}
 Question 3
The period of the signal $x(t)=8 sin(0.8 \pi t+ \pi/4)$ is
 A 0.4 $\pi$ s B 0.8 $\pi$ s C 1.25 s D 2.5 s
Engineering Mathematics   Complex Variables
Question 3 Explanation:
If period $= T,$ then $x(t+T)=x(t)$
Here, $x(t)= 8 \sin(0.8 \pi t+\pi /4)$
putting $x(t+T)=x(t)$ we get
$8 \sin(0.8\pi (t+T)+\pi /4)=8 \sin (0.8 \pi t+\pi /4)$
$\sin(0.8\pi (t+T)+\pi /4)= \sin (0.8 \pi t+ \pi /4)$
$\sin(\theta _1)=\sin(\theta _2)\Rightarrow \theta _1=\theta _2+2 \pi$
$\therefore \;\; 0.8\pi (t+T)+ \pi /4=0.8 \pi t+ \pi /4+2 \pi$
$0.8\pi T=2 \pi$
$T=2.5 s$
 Question 4
The system represented by the input-output relationship $y(t)=\int_{-\infty }^{5t}x(\tau )d\tau , \; t\gt0$
 A Linear and causal B Linear but not causal C Causal but not linear D Neither liner nor causal
Signals and Systems   Linear Time Invariant Systems
Question 4 Explanation:
Integrator is always a linear system.
\begin{aligned} \text{Since, }y(t)&=\int_{-\infty }^{5t}x(\tau )d\tau \;\;t \gt 0 \\ &\text{For t=1,} \\ y(t)&= \int_{-\infty }^{5}x(\tau )d\tau \end{aligned}
here value at t=1 depends on future values like at t=2,3,... of input x(t).
So, it is a noncausal systems.
 Question 5
The switch in the circuit has been closed for a long time. It is opened at t = 0. At t =$0^+$, the current through the 1 $\mu$F capacitor is
 A 0A B 1A C 1.25A D 5A
Electric Circuits   Transients and Steady State Response
Question 5 Explanation:
As the switch has been closed for a long time, the circuit is in steady state. At steadystate, capacitor is open circuit,

Using KVL,
$5-I-4I=0 I=1A$
$V_c(0^-)=4 \times 1=4V$
As the voltage across capacitorcan not change abruptly,
So, $V_c(0^+)=V_c(0^-)=4V$
Circuit at $t=0^+$

Current through capacitor at $t=0^+$
$I_c(0^+)=\frac{4}{4}=1A$
 Question 6
The second harmonic component of the periodic waveform given in the figure has an amplitude of
 A 0 B 1 C $2/\pi$ D $\sqrt{5}$
Signals and Systems   Fourier Series
Question 6 Explanation:
The given signal is odd as wel as having half wave symmetry.
So, it has only sine terms with odd harmonics. So, for second harmonic term amplitude =0.
 Question 7
As shown in the figure, a 1 $\Omega$ resistance is connected across a source that has a load line v+i=100. The current through the resistance is
 A 25A B 50A C 100A D 200A
Electric Circuits   Basics
Question 7 Explanation:
A resistor has linear characteristics
i.e. $V=Ri$
$\Rightarrow \;\; V=i$
Load line,
$V+i=100$
$i+i=100$
current through resistance,
$i=\frac{100}{2}=50A$
 Question 8
A wattmeter is connected as shown in figure. The wattmeter reads.
 A Zero always B Total power consumed by $Z_1 \; and \; Z_2$ C Power consumed by $Z_1$ D Power consumed by $Z_2$
Electrical and Electronic Measurements   Measurement of Energy and Power
Question 8 Explanation:

Potential coil draws negligible current. So current through $Z_1$ and $Z_2$ is same.
current through current coil $=I_{cc}=I$
Voltage across potential coil $=V_{pc}$
Voltage across $Z_1=V_{pc}=V$
Wattmeter reads power consumed by $Z_2$, as voltage across potential coil = Voltage across $Z_2$
Current through current coil = Current through $Z_2$.
 Question 9
An ammeter has a current range of 0-5 A, and its internal resistance is 0.2 $\Omega$. In order to change the range to 0-25 A, we need to add a resistance of
 A 0.8 $\Omega$ in series with the meter B 1.0 $\Omega$ in series with the meter C 0.04 $\Omega$ in parallel with the meter D 0.05 $\Omega$ in parallel with the meter
Electrical and Electronic Measurements   Galvanometers, Voltmeters and Ammeters
Question 9 Explanation:
To extend the range of ammeter, a resistance $R_{sh}$ is connected across the meter.

$I_m=$ full scale deflection current=5A
$I=25A$
Multiplying power $=m=\frac{I}{I_m}=\frac{25}{5}=5$
$I_m=\frac{IR_{sh}}{R_{sh}+R_m}$
$\frac{I}{I_m}=\frac{R_{sh}+R_m}{R_{sh}}$
$m=1+\frac{R_m}{R_{sh}}$
$R_{sh}=\frac{R_m}{m-1}=\frac{0.2}{5-1}=0.05$
 Question 10
As shown in the figure, a negative feedback system has an amplifier of gain 100 with $\pm$10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately :
 A $10\pm 1$% B $10\pm 2$% C $10\pm 5$% D $10\pm 10$%
Control Systems   Feedback Characteristics of Control Systems
Question 10 Explanation:
$G=100\pm 10$%
$\frac{\Delta G}{G}=10$% or 0.1
$H=\frac{9}{100}$
Overall gain,
$T=\frac{G}{1+GH}$
$T=\frac{100}{1+100 \times \frac{9}{100}}=10$
Sensitivity w.r.t. $'G'=\frac{1}{1+GH} \times 10$%
$\;\;\;=\frac{1}{1+10 \times 1}\times 10$%
$\;\;\;=\frac{10}{11}\cong 1$%
There are 10 questions to complete.