Question 1 |
In an electrostatic field, the electric displacement density vector, \vec{D} , is given by
\vec{D}(x,y,z)=(x^3\vec{i}+y^3\vec{j}+xy^2\vec{k})C/m^2
, where \vec{i},\vec{j},\vec{k} are the unit vectors along x-axis, y-axis, and z-axis, respectively. Consider a cubical region R centered at the origin with each side of length 1 m, and vertices at (\pm 0.5 m, \pm 0.5 m, \pm 0.5 m ). The electric charge enclosed within R is _________ C (rounded off to two decimal places).
\vec{D}(x,y,z)=(x^3\vec{i}+y^3\vec{j}+xy^2\vec{k})C/m^2
, where \vec{i},\vec{j},\vec{k} are the unit vectors along x-axis, y-axis, and z-axis, respectively. Consider a cubical region R centered at the origin with each side of length 1 m, and vertices at (\pm 0.5 m, \pm 0.5 m, \pm 0.5 m ). The electric charge enclosed within R is _________ C (rounded off to two decimal places).
0.25 | |
0.5 | |
0.75 | |
0.85 |
Question 1 Explanation:
\vec{D}(x,y,z)=(x^3\vec{i}+y^3\vec{j}+xy^2\vec{k})c/m^2
Q_{enc.}=\int _v \rho _v\cdot dV=\int (\triangledown \cdot \vec{D})dV
\triangledown \cdot \vec{D}=3x^2+3y^2
dV=dxdydz
\therefore \; Q_{enc.}=\int _v 3(x^2+y^2)dxdydz =3\left [ \int_{-0.5}^{0.5} x^2 dx \int_{-0.5}^{0.5}dy \int_{-0.5}^{0.5} dz+\int_{-0.5}^{0.5}dx\int_{-0.5}^{0.5}y^2dy\int_{-0.5}^{0.5}dz \right ]
=3\left [ \frac{x^3}{3}|_{-0.5}^{0.5} \times 1 \times 1 + \frac{y^3}{3}|_{-0.5}^{0.5} \times 1 \times 1\right ] =0.25+0.25=0.5C
Q_{enc.}=\int _v \rho _v\cdot dV=\int (\triangledown \cdot \vec{D})dV
\triangledown \cdot \vec{D}=3x^2+3y^2
dV=dxdydz
\therefore \; Q_{enc.}=\int _v 3(x^2+y^2)dxdydz =3\left [ \int_{-0.5}^{0.5} x^2 dx \int_{-0.5}^{0.5}dy \int_{-0.5}^{0.5} dz+\int_{-0.5}^{0.5}dx\int_{-0.5}^{0.5}y^2dy\int_{-0.5}^{0.5}dz \right ]
=3\left [ \frac{x^3}{3}|_{-0.5}^{0.5} \times 1 \times 1 + \frac{y^3}{3}|_{-0.5}^{0.5} \times 1 \times 1\right ] =0.25+0.25=0.5C
Question 2 |
A waveguide consists of two infinite parallel plates (perfect conductors) at a
separation of 10^{-4} cm, with air as the dielectric. Assume the speed of light in air to
be 3 \times 10^{8} m/s. The frequency/frequencies of TM waves which can propagate in
this waveguide is/are _______.
6 \times 10^{15}Hz | |
0.5 \times 10^{12}Hz | |
8 \times 10^{14}Hz | |
1 \times 10^{13}Hz |
Question 2 Explanation:
Cut-off frequency.
\begin{aligned} f_c&=\frac{c}{2a}\;\;\;(m=1) &=\frac{3 \times 10^8}{2 \times 10^{-4} \times 10^{-2}}\\ &=1.5 \times 10^{14}Hz \end{aligned}
f \gt f_c will only propagate
A and C will propage.
\begin{aligned} f_c&=\frac{c}{2a}\;\;\;(m=1) &=\frac{3 \times 10^8}{2 \times 10^{-4} \times 10^{-2}}\\ &=1.5 \times 10^{14}Hz \end{aligned}
f \gt f_c will only propagate
A and C will propage.
Question 3 |
Consider the following wave equation,
\frac{\partial^2 f(x,t)}{\partial t^2}=10000\frac{\partial^2 f(x,t)}{\partial x^2}
Which of the given options is/are solution(s) to the given wave equation?
\frac{\partial^2 f(x,t)}{\partial t^2}=10000\frac{\partial^2 f(x,t)}{\partial x^2}
Which of the given options is/are solution(s) to the given wave equation?
f(x,t)=e^{-(x-100t)^2}+e^{-(x+100t)^2} | |
f(x,t)=e^{-(x-100t)}+0.5e^{-(x+1000t)} | |
f(x,t)=e^{-(x-100t)}+\sin (x+100t) | |
f(x,t)=e^{j100 \pi(-100x+t)}+e^{j100 \pi(100x+t)} |
Question 3 Explanation:
As we know, wave equation is given by
\frac{\partial^2 f(x,t)}{\partial x^2}=\frac{C^2d^2f(x,y)}{dt^2}
Here, option (A) and (C) are satisfying the above standard wave equation.
\frac{\partial^2 f(x,t)}{\partial x^2}=\frac{C^2d^2f(x,y)}{dt^2}
Here, option (A) and (C) are satisfying the above standard wave equation.
Question 4 |
In a circuit, there is a series connection of an ideal resistor and an ideal capacitor.
The conduction current (in Amperes) through the resistor is 2\sin (t+\pi/2) .
The displacement current (in Amperes) through the capacitor is _________.
2 \sin (t) | |
2 \sin (t+\pi) | |
2 \sin (t+\pi /2) | |
0 |
Question 4 Explanation:
In series connection, current pass through each element remain same. Hence, i_c=i_d
So, i_d= 2\sin (t+\pi/2) .
Question 5 |
An antenna with a directive gain of 6\text{ dB}
is radiating a total power of \text{16 kW}. The amplitude of the electric field in free space at a distance of \text{8 km}
from the antenna in the direction of \text{6 dB}
gain (rounded off to three decimal places) is ______ \text{V/m}
.
0.244 | |
0.417 | |
1.254 | |
2.365 |
Question 5 Explanation:
\begin{aligned} G_{d}(\mathrm{~dB}) &=10 \log _{10}\left(G_{d}\right) \\ \Rightarrow\qquad 6 &=10 \log _{10} G_{d} \\ \Rightarrow\qquad G_{d} &=10^{0.6} \\ G_{d} &=3.981 \\ E_{m} &=\frac{\sqrt{60 G_{d} P_{\text {rad }}}}{r} \\ E_{m} &=\frac{\sqrt{60\left(16 \times 10^{3}\right)(3.981)}}{8 \times 10^{3}}=0.244(\mathrm{~V} / \mathrm{m}) \end{aligned}
Question 6 |
A standard air-filled rectangular waveguide with dimensions \text{a=8 cm, b=4 cm}, operates at \text{3.4 GHz}. For the dominant mode of wave propagation, the phase velocity of the signal in v_{p}. The value (rounded off to two decimal places) of v_{p}/c, where c denotes the velocity of light, is _____
2.24 | |
3.82 | |
1.2 | |
4.6 |
Question 6 Explanation:
f_{c 10}=\frac{c}{2 a}=\frac{3 \times 10^{8}}{2\left(8 \times 10^{-2}\right)}=1.875 \mathrm{GHz}
Guide phase velocity, V_{p}=\frac{C}{\sqrt{1-\left(\frac{f_{C 10}}{f}\right)^{2}}}
\frac{V_{p}}{C}=\frac{1}{\sqrt{1-\left(\frac{f_{C 10}}{f}\right)^{2}}}=\frac{1}{\sqrt{1-\left(\frac{1.875}{3.4}\right)^{2}}}=1.198
Guide phase velocity, V_{p}=\frac{C}{\sqrt{1-\left(\frac{f_{C 10}}{f}\right)^{2}}}
\frac{V_{p}}{C}=\frac{1}{\sqrt{1-\left(\frac{f_{C 10}}{f}\right)^{2}}}=\frac{1}{\sqrt{1-\left(\frac{1.875}{3.4}\right)^{2}}}=1.198
Question 7 |
For a vector field D=\rho\cos^{2}\:\varphi \:a_{\rho }+z^{2}\sin^{2}\:\varphi \:a_{\varphi }
in a cylindrical coordinate system \left ( \rho ,\varphi ,z \right )
with unit vectors a_{\rho },a_{\varphi }
and a_{z}
, the net flux of D leaving the closed surface of the cylinder \left ( \rho =3, 0\leq z\leq 2 \right )
(rounded off to two decimal places) is ________________
56.55 | |
22.12 | |
36.85 | |
76.34 |
Question 7 Explanation:
Method 1: \vec{D}=\rho \cos ^{2} \phi \hat{a}_{\rho}+z^{2} \sin ^{2} \phi \hat{a}_{\phi}
Electric flux crossing the closed surface is
\begin{aligned} \psi &=\oiint \vec{D} \cdot \overrightarrow{d S}=\iiint(\vec{\nabla} \cdot \vec{D}) d v \\ \vec{\nabla} \cdot \vec{D} &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho D_{\rho}\right)+\frac{1}{\rho} \frac{\partial D_{\phi}}{\partial \phi}+\frac{\partial D_{z}}{\partial z} \\ &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho \rho \cos ^{2} \phi\right)+\frac{1}{\rho} \frac{\partial}{\partial \phi}\left(z^{2} \sin ^{2} \phi\right)+0 \\ &=\frac{1}{\rho}(2 \rho) \cos ^{2} \phi+\frac{1}{\rho} z^{2} 2 \sin \phi \cos \phi=2 \cos ^{2} \phi+\frac{z^{2}}{\rho} \sin 2 \phi\\ \iiint(\vec{\nabla} \cdot \vec{D}) d v &=\iiint 2 \cos ^{2} \phi(\rho d \rho d \phi d z)+\iiint\left(\frac{z^{2}}{\rho} \sin 2 \phi\right) \rho d \rho d \phi d z \\ &=2 \int_{\rho=0}^{3} \rho d \rho \int_{\phi=0}^{2 \pi}\left(\frac{1+\cos 2 \phi}{2}\right) d \phi \int_{z=0}^{2} d z+\int_{\rho=0}^{2} d \rho \int_{\phi=0}^{2 \pi} \sin 2 \phi d \phi \int_{z=0}^{2} z^{2} d z \\ &=2\left(\frac{\rho^{2}}{-2}\right)_{\rho=0}^{3} \frac{1}{2}(2 \pi)(z)_{z=0}^{2}+0 \\ &=2\left(\frac{3^{2}}{2}\right) \pi(2)=18 \pi(\text { Coulomb })=56.55(\text { Coulomb }) \end{aligned}
Method 2: Electric flux crossing the closed surface is
\psi=\oiint \vec{D} \cdot \overrightarrow{d S}
Electric flux crossing \rho=3 cylindrical surface is
\begin{aligned} \left.\psi\right|_{\rho=3} &=\oiint\left(\rho \cos ^{2} \phi \hat{a}_{p}\right) \cdot(\rho d \phi d z) \hat{a}_{p} \\ &=3^{2} \int_{\phi=0}^{2 \pi} \cos ^{2} \phi d \phi \int_{z=0}^{2} d z \\ &=9 \frac{1}{2}(2 \pi)(2)=18 \pi(\text { coulomb })=56.55(\text { coulomb }) \end{aligned}
Electric flux crossing the closed surface is
\begin{aligned} \psi &=\oiint \vec{D} \cdot \overrightarrow{d S}=\iiint(\vec{\nabla} \cdot \vec{D}) d v \\ \vec{\nabla} \cdot \vec{D} &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho D_{\rho}\right)+\frac{1}{\rho} \frac{\partial D_{\phi}}{\partial \phi}+\frac{\partial D_{z}}{\partial z} \\ &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho \rho \cos ^{2} \phi\right)+\frac{1}{\rho} \frac{\partial}{\partial \phi}\left(z^{2} \sin ^{2} \phi\right)+0 \\ &=\frac{1}{\rho}(2 \rho) \cos ^{2} \phi+\frac{1}{\rho} z^{2} 2 \sin \phi \cos \phi=2 \cos ^{2} \phi+\frac{z^{2}}{\rho} \sin 2 \phi\\ \iiint(\vec{\nabla} \cdot \vec{D}) d v &=\iiint 2 \cos ^{2} \phi(\rho d \rho d \phi d z)+\iiint\left(\frac{z^{2}}{\rho} \sin 2 \phi\right) \rho d \rho d \phi d z \\ &=2 \int_{\rho=0}^{3} \rho d \rho \int_{\phi=0}^{2 \pi}\left(\frac{1+\cos 2 \phi}{2}\right) d \phi \int_{z=0}^{2} d z+\int_{\rho=0}^{2} d \rho \int_{\phi=0}^{2 \pi} \sin 2 \phi d \phi \int_{z=0}^{2} z^{2} d z \\ &=2\left(\frac{\rho^{2}}{-2}\right)_{\rho=0}^{3} \frac{1}{2}(2 \pi)(z)_{z=0}^{2}+0 \\ &=2\left(\frac{3^{2}}{2}\right) \pi(2)=18 \pi(\text { Coulomb })=56.55(\text { Coulomb }) \end{aligned}
Method 2: Electric flux crossing the closed surface is
\psi=\oiint \vec{D} \cdot \overrightarrow{d S}
Electric flux crossing \rho=3 cylindrical surface is
\begin{aligned} \left.\psi\right|_{\rho=3} &=\oiint\left(\rho \cos ^{2} \phi \hat{a}_{p}\right) \cdot(\rho d \phi d z) \hat{a}_{p} \\ &=3^{2} \int_{\phi=0}^{2 \pi} \cos ^{2} \phi d \phi \int_{z=0}^{2} d z \\ &=9 \frac{1}{2}(2 \pi)(2)=18 \pi(\text { coulomb })=56.55(\text { coulomb }) \end{aligned}
Question 8 |
The impedance matching network shown n the figure is to match a lossless line having characteristic impedance Z_{0}= 50 \:\Omega
with a load impedance Z_{L}. A quarter-wave line having a characteristic impedance Z_1=75\:\Omega
is connected to Z_{L}. Two stubs having characteristic impedance of 75\:\Omega
each are connected to this quarter-wave line. One is a short-circuited (\text{S.C}) stub of length 0.25\lambda
connected across PQ and the other one is an open-circuited (\text{O.C})
stub of length 0.5\lambda connected across RS.
The impedance matching in achieved when the real part of Z_{L} is
The impedance matching in achieved when the real part of Z_{L} is
112.5\:\Omega | |
75.0\:\Omega | |
50.0\:\Omega | |
33.3\:\Omega |
Question 8 Explanation:
\begin{aligned} Z_{\text {in } \lambda / 4}&=\frac{Z_{O}^{2}}{Z_{L}}=\frac{(75)^{2}}{0}=\infty \quad\left[\text { for } \frac{\lambda}{4} T_{X} \text { line }\right]\\ \text{Given }Z_{L}\text{ of }\frac{\lambda}{4}\text{ line is 0}(\mathrm{SC})\\ Z_{\text {in }_{\lambda / 2}}&=Z_{L}=\infty\left[\text { for } \frac{\lambda}{2} T_{X} \text { line }\right]\\ \text{Given }Z_{L}\text{ of }\frac{\lambda}{2}\text{ line is }\infty(O . C) \end{aligned}
The input impedance of \frac{\lambda}{4} transmission line, as well as \frac{\lambda}{2} transmission line is 00 , and they are in parallel with main transmission line, so they are not effective for main.
Transmission line.
Final configuration of given line is
For impedance matching, Z_{L}=\frac{Z_{1}^{2}}{Z_{0}}=\frac{(75)^{2}}{50}=112.5
The input impedance of \frac{\lambda}{4} transmission line, as well as \frac{\lambda}{2} transmission line is 00 , and they are in parallel with main transmission line, so they are not effective for main.
Transmission line.
Final configuration of given line is
For impedance matching, Z_{L}=\frac{Z_{1}^{2}}{Z_{0}}=\frac{(75)^{2}}{50}=112.5
Question 9 |
The refractive indices of the core and cladding of an optical fiber are 1.50 and 1.48, respectively. The critical propagation angle, which is defined as the maximum angle that the light beam makes with the axis of the optical fiber to achieve the total internal reflection, (rounded off to two decimal places) is _____ degree.
18.3 | |
3.45 | |
9.37 | |
5.82 |
Question 9 Explanation:
Given that
Refractive index of core \eta_{1}=1.50
Refractive index of clad \eta_{2}=1.48
Critical propagation angle \left(\theta_{P}\right)
\theta_{P}=\sin ^{-1}\left[ \frac{ \sqrt{\eta_{1}^{2}-\eta_{2}^{2}}}{\eta_{1}} \right]=\sin ^{-1} \left[ \frac{ \sqrt{1.5^{2}-1.48^{2}} }{1.5} \right] =9.37
Refractive index of core \eta_{1}=1.50
Refractive index of clad \eta_{2}=1.48
Critical propagation angle \left(\theta_{P}\right)
\theta_{P}=\sin ^{-1}\left[ \frac{ \sqrt{\eta_{1}^{2}-\eta_{2}^{2}}}{\eta_{1}} \right]=\sin ^{-1} \left[ \frac{ \sqrt{1.5^{2}-1.48^{2}} }{1.5} \right] =9.37
Question 10 |
Consider the vector field F\:=\:a_{x}\left ( 4y-c_{1}z \right )+a_y\left ( 4x + 2z\right )+a_{z}\left ( 2y +z\right ) in a rectangular coordinate system (x,y,z) with unit vectors a_{x},\:a_{y} and a_{z}. If the field F is irrotational (conservative), then the constant c_{1}
(in integer) is _________________
0 | |
1 | |
2 | |
3 |
Question 10 Explanation:
\begin{aligned} \nabla \times \vec{F}&=0\\ \left|\begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 4 y-c_{1} z & 4 x+2 z & 2 y+z \end{array}\right|&=0\\ =i(2-2)-j\left(0+c_{1}\right)+k(4-4)=0 \\ c_{1} &=0 \end{aligned}
There are 10 questions to complete.