MCQ 11 Mark
Range of frequencies allotted for commercial FM radio broadcast is
- ✓
$88$ to $108 \mathrm{MHz}$
- B
$88$ to $108 \mathrm{kHz}$
- C
$8$ to $88 \mathrm{MHz}$
- D
$88$ to $108 \mathrm{GHz}$
AnswerCorrect option: A. $88$ to $108 \mathrm{MHz}$
A maximum frequency deviation of $75 \mathrm{kHz}$ is permitted for commercial FM broadcast stations in the $88$ to $108 \mathrm{MHz} \ V \ \mathrm{HF}$ band.
View full question & answer→MCQ 21 Mark
A step index fibre has a relative refractive index of $0.88 \%$. What is the critical angle at the corecladding interface
- A
$60^{\circ}$
- B
$75^{\circ}$
- C
$45^{\circ}$
- ✓
AnswerHere $\frac{n_1-n_2}{n_1}=\frac{0.88}{100} \Rightarrow \frac{n_2}{n_1}=0.9912$
$\therefore$ Critical angle $\theta_c=\sin ^{-1}\left(\frac{n_2}{n_1}\right)=\sin ^{-1}(0.9912)=84^{\circ} 24^{\prime}$
View full question & answer→MCQ 31 Mark
Audio signal cannot be transmitted because
- A
The signal has more noise
- B
The signal cannot be amplified for distance communication
- C
The transmitting antenna length is very small to design
- ✓
The transmitting antenna length is very large and impracticable / The signal is not a radio signal
AnswerCorrect option: D. The transmitting antenna length is very large and impracticable / The signal is not a radio signal
Following are the problems which are faced while transmitting audio signals directly.
(i) These signals are relatively of short range.
(ii) If every body started transmitting these low frequency signals directly, mutual interference will render all of them ineffective.
(iii) Size of antenna required for their efficient radiation would be larger i.e. about $75 \mathrm{~km}$.
View full question & answer→MCQ 41 Mark
- A
The amplitude of modulated wave varies as frequency of carrier wave
- ✓
The frequency of modulated wave varies as amplitude of modulating wave
- C
The amplitude of modulated wave varies as amplitude of carrier wave
- D
The frequency of modulated wave varies as frequency of modulating wave/ The frequency of modulated wave varies as frequency of carrier wave
AnswerCorrect option: B. The frequency of modulated wave varies as amplitude of modulating wave
The process of changing the frequency of a carrier wave (modulated wave) in accordance with the audio frequency signal (modulating wave) is known as frequency modulation (FM).
View full question & answer→MCQ 51 Mark
A laser is a coherent source because it contains
- A
- B
Uncoordinated wave of a particular wavelength
- C
Coordinated wave of many wavelengths
- ✓
Coordinated waves of a particular wavelength
AnswerCorrect option: D. Coordinated waves of a particular wavelength
View full question & answer→MCQ 61 Mark
Through which mode of propagation, the radio waves can be sent from one place to another
AnswerRadio waves can be transmitted from one place to another as grand wave or sky wave or space wave propagation.
View full question & answer→MCQ 71 Mark
Maximum useable frequency (MUF) in $F$-region layer is $x$, when the critical frequency is $60 \mathrm{MHz}$ and the angle of incidence is $70^{\circ}$. Then $x$ is
- A
$150 \mathrm{MHz}$
- B
$170 \mathrm{MHz}$
- ✓
$175 \mathrm{MHz}$
- D
$190 \mathrm{MHz}$
AnswerCorrect option: C. $175 \mathrm{MHz}$
$M U F=\frac{f_c}{\cos \theta}=\frac{60}{\cos 70^{\circ}}=175 \mathrm{MHz}$
View full question & answer→MCQ 81 Mark
A sky wave with a frequency $55\ \mathrm{MHz}$ is incident on $D$-region of earth's atmosphere at $45.$ The angle of refraction is (electron density for $D$-region is $400$ electron/ $\mathrm{cm}$ )
- A
$60^{\circ}$
- ✓
$45^{\circ}$
- C
$30^{\circ}$
- D
$15^{\circ}$
AnswerCorrect option: B. $45^{\circ}$
$n_{\text {eff }}=n_0 \sqrt{1-\left(\frac{80.5 N}{v^2}\right)}=1 \sqrt{1-\frac{80.5\times\left(400 \times 10^6\right)}{\left(55 \times 10^6\right)^2}} \approx 1$
Also $n_{\text {eff }}=\frac{\sin i}{\sin r} \Rightarrow \sin r=\sin i \Rightarrow r=i=45^{\circ}$
View full question & answer→MCQ 91 Mark
In which frequency range, space waves are normally propagated
- A
$\mathrm{HF}$
- B
$\mathrm{VHF}$
- ✓
$\mathrm{UHF}$
- D
$\mathrm{SHF}$
AnswerCorrect option: C. $\mathrm{UHF}$
View full question & answer→MCQ 101 Mark
Advantage of optical fibre
- A
High bandwidth and EM interference
- B
Low bandwidth and EM interference
- C
High band width, low transmission capacity and no EM interference
- ✓
High bandwidth, high data transmission capacity and no EM interference
AnswerCorrect option: D. High bandwidth, high data transmission capacity and no EM interference
Few advantages of optical fibres are that the number of signals carried by optical fibres is much more than that carried by the $C u$ wire or radio waves. Optical fibres are practically free from electromagnetic interference and problem of cross talks whereas ordinary cables and microwave links suffer a lot from it.
View full question & answer→MCQ 111 Mark
An oscillator is producing FM waves of frequency $2 \mathrm{kHz}$ with a variation of $10 \mathrm{kHz}$. What is the modulating index
- A
$0.20$
- ✓
$5.0$
- C
$0.67$
- D
$1.5$
AnswerThe formula for modulating index is given by
$m_f=\frac{\delta}{v_m}=\frac{\text { Frequency variation }}{\text { Modulatingfrequency }}=\frac{10 \times 10^3}{2 \times 10^3}=5$
View full question & answer→MCQ 121 Mark
Laser beams are used to measure long distances because
- A
- B
They are highly polarised
- C
- ✓
They have high degree of parallelism
AnswerCorrect option: D. They have high degree of parallelism
Laser beams are perfectly parallel. So that they are very narrow and can travel a long distance without spreading. This is the feature of laser while they are monochromatic and coherent these are characteristics only.
View full question & answer→MCQ 131 Mark
Television signals on earth cannot be received at distances greater than $100 \mathrm{~km}$ from the transmission station. The reason behind this is that
AnswerCorrect option: D. The surface of earth is curved like a sphere
View full question & answer→MCQ 141 Mark
The characteristic impedance of a coaxial cable is of the order of
- A
$50 \Omega$
- B
$200 \Omega$
- ✓
$270 \Omega$
- D
AnswerCorrect option: C. $270 \Omega$
View full question & answer→MCQ 151 Mark
If $\mu$ and $\mu$ are the refractive indices of the materials of core and cladding of an optical fibre, then the loss of light due to its leakage can be minimised by having
- ✓
$\mu>\mu$
- B
$\mu<\mu$
- C
$\mu=\mu$
- D
AnswerCorrect option: A. $\mu>\mu$
View full question & answer→MCQ 161 Mark
In an amplitude modulated wave for audio frequency of $500 \mathrm{cycles} / \mathrm{second}$ the appropriate carrier frequency will be
- A
$50 \mathrm{cycles} / \mathrm{sec}$
- B
$100 \mathrm{cycles} / \mathrm{sec}$
- C
$500 \mathrm{cycles} / \mathrm{sec}$
- ✓
$50,000 \mathrm{cycles} / \mathrm{sec}$
AnswerCorrect option: D. $50,000 \mathrm{cycles} / \mathrm{sec}$
$50,000 \mathrm{cycles} / \mathrm{sec}$
View full question & answer→MCQ 171 Mark
For sky wave propagation of a $10 \ \mathrm{ MHz}$ signal, what should be the minimum electron density in ionosphere
AnswerCorrect option: A. $\simeq 1.2 \times 10^{12} \mathrm{~m}^{-3}$
The critical frequency of a sky wave for reflection from a layer of atmosphere is given by $f_c=9\left(N_{\max }\right)^{1 / 2}$
$\begin{aligned}& \Rightarrow 10 \times 10^6=9\left(N_{\max }\right)^{1 / 2}\end{aligned}$
$ \Rightarrow N_{\max }=\left(\frac{10 \times 10^6}{9}\right)^2 \simeq 1.2 \times 10^{12} \mathrm{~m}^{-3}$
View full question & answer→MCQ 181 Mark
A laser beam is used for carrying out surgery because it
AnswerSurgery needs sharply focused beam of light and laser can be sharply focused.
View full question & answer→MCQ 191 Mark
In short wave communication waves of which of the following frequencies will be reflected back by the ionospheric layer, having electron density $10$ per $m$
- ✓
$2 \mathrm{MHz}$
- B
$10 \mathrm{MHz}$
- C
$12 \mathrm{MHz}$
- D
$18 \mathrm{MHz}$
AnswerCorrect option: A. $2 \mathrm{MHz}$
By using $f_c \approx 9\left(N_{\max }\right)^{1 / 2} \Rightarrow f_c \approx 2 \mathrm{MHz}$
View full question & answer→MCQ 201 Mark
The process of superimposing signal frequency (i.e. audio wave) on the carrier wave is known as
AnswerCarrier + signal $\rightarrow$ modulation.
View full question & answer→MCQ 211 Mark
The maximum distance upto which TV transmission from a TV tower of height $h$ can be received is proportional to
- ✓
(a) $h$
- B
(b) $h$
- C
(c) $h$
- D
(d) $h$
AnswerCorrect option: A. (a) $h$
(a) $d=\sqrt{2 h R} \Rightarrow d \propto h^{1 / 2}$
View full question & answer→MCQ 221 Mark
What should be the maximum acceptance angle at the aircore interface of an optical fibre if $n$ and $n$ are the refractive indices of the core and the cladding, respectively
- A
$\sin ^{-1}\left(n_2 / n_1\right)$
- ✓
$\sin ^{-1} \sqrt{n_1^2-n_2^2}$
- C
$\left[\tan ^{-1} \frac{n_2}{n_1}\right]$
- D
$\left[\tan ^{-1} \frac{n_1}{n_2}\right]$
AnswerCorrect option: B. $\sin ^{-1} \sqrt{n_1^2-n_2^2}$
Core of acceptance angle $\theta=\sin ^{-1} \sqrt{n_1^2-n_2^2}$
View full question & answer→MCQ 231 Mark
Consider telecommunication through optical fibres. Which of the following statements is not true
- A
Optical fibres may have homogeneous core with a suitable cladding
- B
Optical fibres can be of graded refractive index
- ✓
Optical fibres are subject to electromagnetic interference from outside
- D
Optical fibres have extremely low transmission loss
AnswerCorrect option: C. Optical fibres are subject to electromagnetic interference from outside
Optical fibres are not subjected to electromagnetic interference from outside.
View full question & answer→MCQ 241 Mark
The attenuation in optical fibre is mainly due to
- A
- B
- C
Neither absorption nor scattering
- ✓
AnswerA very small part of light energy is lost from an optical fibre due to absorption or due to light leaving the fibre as a result of scattering of light sideways by impurities in the glass fibre.
View full question & answer→MCQ 251 Mark
Broadcasting antennas are generally
MCQ 261 Mark
The carrier frequency generated by a tank circuit containing $1 n F$ capacitor and $10 \mu \mathrm{H}$ inductor is
- A
$1592 \mathrm{~Hz}$
- B
$1592 \mathrm{MHz}$
- ✓
$1592 \mathrm{kHz}$
- D
$159.2 \mathrm{~Hz}$
AnswerCorrect option: C. $1592 \mathrm{kHz}$
$ v=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \times 3.14 \sqrt{10 \times 10^{-6} \times 1 \times 10^{-9}}}=1592 \mathrm{kHz}$
View full question & answer→MCQ 271 Mark
Long distance short-wave radio broadcasting uses
MCQ 281 Mark
The velocity factor of a transmission line $x$. If dielectric constant of the medium is $2.6,$ the value of $x$ is
- A
$ 0.26$
- ✓
$ 0.62$
- C
$2.6$
- D
$6.2$
AnswerCorrect option: B. $ 0.62$
$v . f .=\frac{1}{\sqrt{k}}=\frac{1}{\sqrt{2.6}}=0.62$
View full question & answer→MCQ 291 Mark
In AM, the centpercent modulation is achieved when
- ✓
Carrier amplitude $=$ signal amplitude
- B
Carrier amplitude $\neq$ signal amplitude
- C
Carrier frequency = signal frequency
- D
Carrier frequency $\neq$ signal frequency
AnswerCorrect option: A. Carrier amplitude $=$ signal amplitude
View full question & answer→MCQ 301 Mark
An amplitude modulated wave is modulated to $50 \%$. What is the saving in power if carrier as well as one of the side bands are suppressed
- A
$70 \%$
- B
$65.4 \%$
- ✓
$94.4 \%$
- D
$25.5 \%$
AnswerCorrect option: C. $94.4 \%$
$P_{s b}=P_c\left(\frac{m_a}{2}\right)^2=P_c \frac{(0.5)^2}{4}=0.0625 P_c$
Also $P=P_c\left(1+\frac{m_a^2}{2}\right)=P_c\left(1+\frac{(0.5)^2}{2}\right)=1.125 P_c$
$\therefore \%$ saving $=\frac{\left(1.125 P_c-0.0625 P_c\right)}{1.125 P_c} \times 100=94.4 \%$.
View full question & answer→MCQ 311 Mark
The bit rate for a signal, which has a sampling rate of $8 \mathrm{kHz}$ and where $16$ quantisation levels have been used is
- ✓
$32000\ \mathrm{bits} / \mathrm{sec}$
- B
$16000\ \mathrm{bits} / \mathrm{sec}$
- C
$64000\ \mathrm{bits} / \mathrm{sec}$
- D
$72000\ \mathrm{bits} / \mathrm{sec}$
AnswerCorrect option: A. $32000\ \mathrm{bits} / \mathrm{sec}$
If $n$ is the number of bits per sample, then number of quantisation level $=2$
Since the number of quantisation level is 16$\Rightarrow 2=16 \Rightarrow n=4$
$\therefore$ bit rate $=$ sampling rate $\times$ no. of bits per sample$=8000 \times 4=32,000 \mathrm{bits} / \mathrm{sec}$.
View full question & answer→MCQ 321 Mark
The audio signal used to modulate $60 \sin (2 \pi \times 10 t)$ is $15 \sin 300 \pi t$. The depth of modulation is
- A
$50 \%$
- B
$40 \%$
- ✓
$25 \%$
- D
$15 \%$
AnswerCorrect option: C. $25 \%$
$m_a=\frac{E_m}{E_c}=\frac{15}{60} \times 100=25 \%$
View full question & answer→MCQ 331 Mark
A $500 \mathrm{~Hz}$ modulating voltage fed into an FM generator produces a frequency deviation of $2.25 \mathrm{kHz}$. If amplitude of the voltage is kept constant but frequency is raised to $6 \mathrm{kHz}$ then the new deviation will be
- A
$4.5 \mathrm{kHz}$
- ✓
$54 \mathrm{kHz}$
- C
$27 \mathrm{kHz}$
- D
$15 \mathrm{kHz}$
AnswerCorrect option: B. $54 \mathrm{kHz}$
$m_f=\frac{\delta}{f_m}=\frac{2250}{500}=4.5$
$\therefore$ New deviation $=2\left(m_f f_m\right)=2 \times 4.5 \times 6=54 \mathrm{kHz}$.
View full question & answer→MCQ 341 Mark
The modulation index of an FM carrier having a carrier swing of $200 \mathrm{kHz}$ and a modulating signal $10 \mathrm{kHz}$ is
Answer$ \mathrm{CS}=2 \times \Delta f \text { or } \Delta f=C S / 2$
$ \therefore \Delta f=\frac{200}{2}=100 \mathrm{kHz}$
Now $m_f=\frac{\Delta f}{f_m}=\frac{100}{10}=10$
View full question & answer→MCQ 351 Mark
The total power content of an AM wave is $900 \mathrm{~W}$. For $100 \%$ modulation, the power transmitted by each side band is
- A
$50 \mathrm{~W}$
- B
$100 \mathrm{~W}$
- ✓
$150 \mathrm{~W}$
- D
$200 \mathrm{~W}$
AnswerCorrect option: C. $150 \mathrm{~W}$
$P_c=P_t\left[\frac{2}{2+m^2}\right]=900\left[\frac{2}{2+1}\right]=600 \mathrm{~W}$
Now, $P_{L S B}=\frac{m^2}{4} \times P_c=\frac{1}{4} \times 600=150 \mathrm{~W}$
View full question & answer→MCQ 361 Mark
The total power content of an AM wave is $1500 \mathrm{~W}$. For $100 \%$ modulation, the power transmitted by the carrier is
- A
$500 \mathrm{~W}$
- B
$700 \mathrm{~W}$
- C
$750 \mathrm{~W}$
- ✓
$1000 \mathrm{~W}$
AnswerCorrect option: D. $1000 \mathrm{~W}$
$ \frac{P_t}{P_c}=1+\frac{m^2}{2} \text { or } P_c=P_t\left[\frac{2}{2+m^2}\right] $
$ \therefore P_c=1500\left[\frac{2}{2+1}\right] \quad \because m=100 \%=1 $
$ =1000 \mathrm{~W}$
View full question & answer→MCQ 371 Mark
The antenna current of an AM transmitter is $8 \mathrm{~A}$ when only carrier is sent but increases to $8.96 \mathrm{~A}$ when the carrier is sinusoidally modulated. The percentage modulation is
- A
$50 \%$
- B
$60 \%$
- C
$65 \%$
- ✓
$71 \%$
AnswerCorrect option: D. $71 \%$
We know that $\left(\frac{I_t}{I_c}\right)^2=1+\frac{m^2}{2}$
Here, $I_t=8.96 \mathrm{~A}$ and $I_c=8 \mathrm{~A}$
$ \therefore\left(\frac{8.96}{8}\right)^2=1+\frac{m^2}{2} \text { or } 1.254=1+\frac{m^2}{2} $
$ \text { or } \frac{m^2}{2}=0.254 \text { or } m^2=0.508$ or $m=0.71=71 \%$
View full question & answer→MCQ 381 Mark
A transmitter supplies $9 k W$ to the aerial when unmodulated. The power radiated when modulated to $40 \%$ is
- A
$5 \mathrm{~kW}$
- ✓
$9.72 \mathrm{~kW}$
- C
$10 \mathrm{~kW}$
- D
$12 \mathrm{~kW}$
AnswerCorrect option: B. $9.72 \mathrm{~kW}$
$ P_t=P_c\left[1+\frac{m^2}{2}\right]=9\left[1+\frac{(0.4)^2}{2}\right] $
$ =9\left[1+\frac{0.16}{2}\right] \quad(\because m=40 \%=0.4) $
$ =9(1.08)=9.72 \mathrm{~kW}$
View full question & answer→MCQ 391 Mark
A photodetector is made from a semiconductor $\ln$ Ga As with $E=$ $0.73 \mathrm{eV}$. What is the maximum wavelength, which it can detect
- A
$1000 \mathrm{~nm}$
- ✓
$1703 \mathrm{~nm}$
- C
$500 \mathrm{~nm}$
- D
$173 \mathrm{~nm}$
AnswerCorrect option: B. $1703 \mathrm{~nm}$
Limiting value of $h v$ is $E$, such that $h v=\frac{h c}{\lambda}=E_g$
$ \text { or } \lambda=\frac{h c}{E_g}=\frac{6.63 \times 10^{-34} J-s \times 3 \times 10^8 \mathrm{~ms}^{-1}}{0.73 \times 1.6 \times 10^{-19} \mathrm{~J}} $
$ =1703 \mathrm{~nm}$
View full question & answer→MCQ 401 Mark
Consider an optical communication system operating at $\lambda 800 \mathrm{~nm}$. Suppose, only $1 \%$ of the optical source frequency is the available channel bandwidth for optical communication. How many channels can be accommodated for transmitting audio signals requiring a bandwidth of $8 \mathrm{kHz}$
- ✓
(a) $4.8 \times 10$
- B
- C
(c) $6.2 \times 10$
- D
(d) $4.8 \times 10$
AnswerCorrect option: A. (a) $4.8 \times 10$
(a) Optical source frequency $f=\frac{c}{\lambda}$$=3 \times 10 \%\left(800 \times 10^{-}\right)=3.8 \times 10^n \mathrm{~Hz}$Bandwidth of channel $(1 \%$ of above $)=3.8 \times 10 \mathrm{~Hz}$Number of channels =(Total bandwidth of channel $) /$ (Bandwidth needed per channel)
View full question & answer→MCQ 411 Mark
In a diode $AM-$detector, the output circuit consist of $R=1 \mathrm{k} \Omega$ and $C=10$ $p F$. A carrier signal of $100 \mathrm{kHz}$ is to be detected. Is it good
- ✓
- B
- C
Information is not sufficient
- D
View full question & answer→MCQ 421 Mark
In which of the following remote sensing technique is not used
AnswerRemote sensing is the technique to collect information about an object in respect of its size, colour, nature, location, temperature etc. without physically touching it. There are some areas or location which are inaccessible.
So to explore these areas or locations, a technique known as remote sensing is used. Remote sensing is done through a satellite.
View full question & answer→MCQ 431 Mark
The phenomenon by which light travels in an optical fibres is
- A
- B
- ✓
Total internal reflection
- D
AnswerCorrect option: C. Total internal reflection
In optical fibre, light travels inside it, due to total internal reflection.
View full question & answer→MCQ 441 Mark
In an FM system a $7 \mathrm{kHz}$ signal modulates $108 \mathrm{MHz}$ carrier so that frequency deviation is $50 \mathrm{kHz}$. The carrier swing is
- ✓
$7.143$
- B
$8$
- C
$ 0.71$
- D
$350$
AnswerCorrect option: A. $7.143$
Carrier swing $=\frac{\text { Frequency deviation }}{\text { Modulatingfrequency }}=\frac{50}{7}=7.143$
View full question & answer→MCQ 451 Mark
The waves used in telecommunication are
AnswerIn telecommunication, microwaves are used.
View full question & answer→MCQ 461 Mark
In a communication system, noise is most likely to affect the signal
- A
- ✓
In the channel or in the transmission line
- C
In the information source
- D
AnswerCorrect option: B. In the channel or in the transmission line
View full question & answer→MCQ 471 Mark
Indicate which one of the following system is digital
- A
Pulse position modulation
- ✓
- C
- D
Pulse amplitude modulation
AnswerPulse code modulation is a digital system.
View full question & answer→MCQ 481 Mark
While tuning in a certain broadcast station with a receiver, we are actually
- ✓
Varying the local oscillator frequency
- B
Varying the frequency of the radio signal to be picked up
- C
- D
AnswerCorrect option: A. Varying the local oscillator frequency
View full question & answer→MCQ 491 Mark
- A
That converts electromagnetic energy into radio frequency signal
- B
That converts radio frequency signal into electromagnetic energy
- ✓
That converts guided electromagnetic waves into free space electromagnetic waves and vice-versa
- D
AnswerCorrect option: C. That converts guided electromagnetic waves into free space electromagnetic waves and vice-versa
An antenna is a metallic structure used to radiate or receive EM waves.
View full question & answer→MCQ 501 Mark
When the modulating frequency is doubled, the modulation index is halved and the modulating voltage remains constant, the modulation system is