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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
Modern communication systems useA. analog circuitsB. digital circuitsC. combination of analog and digital circuitsD. none of the above |
| Answer» Correct Answer - B | |
| 252. |
One requires `11eV` of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in.A. visible regionB. infrared regionC. ultraviolet regionD. microwave region |
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Answer» Correct Answer - C Energy required to dissociate a carbon monoxide molecule into carbon and oxygen atoms ( E ) = 11 eV But E=hv `therefore v=E/h="11 eV"/(6.62xx10^(-34))` `=(11xx1.6xx10^(-19) J)/(6.62xx10^(-34)) =2.6xx10^15` Hz This frequency belongs to ultraviolet radiation |
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| 253. |
The wavelength of electromagnetic waves employed for space communication lie in the the range of-A. 1 mm to 30 mB. 1 mm to 300mC. 1 mm to 3 kmD. 1 mm to 30 km |
| Answer» The wavelenght of electromagnetic wave employed for space communication lie in the range of 1 mm to 30 km. | |
| 254. |
If sound waves of frequency 20 KHz are to be transmitted directly, then the length of the dipole antenna required isA. 3.75 kmB. 15 kmC. 7.5 kmD. 20 km |
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Answer» Correct Answer - C `lambda=c/f=(3xx10^8)/(20xx10^3) =15xx10^3` m = 15 km `therefore` the required length of the antenna `=lambda/2`=7.5 km |
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| 255. |
For television broadcasting, the frequency employed is normallyA. 30-300 M HzB. 30-300 GHzC. 30-300 kHzD. 30-300 Hz |
| Answer» Correct Answer - A | |
| 256. |
The sound waves after being converted into electrical waves are not transmitted as such because-A. They travel with the speed of soundB. The frequency is not constantC. They are highly absorbed by the atmosphereD. The height of antenna has to be increased several times |
| Answer» Correct Answer - C | |
| 257. |
For television broadcasting, the frequency employed is normallyA. 30-300 M HzB. 30 - 300 G HzC. 30 - 300 K HzD. 30 - 300 Hz |
| Answer» Correct Answer - A | |
| 258. |
Compute `LC` product of a tuned amplifer circuit required to generate a carrier `wave` of `1 MHz` for amplitude modulationA. `2.5xx10^(-14) s`B. `3xx10^(-14) s`C. `1.5xx10^(-14) s`D. `2xx10^(-14) s` |
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Answer» Correct Answer - A Frequency of the carrier wave = 1MHz = `10^6` Hz For tuned amplifier , the frequency f is `f=1/(2pisqrt(LC)) =10^6 therefore sqrt(LC)=1/(2pixx10^6)` `therefore LC = 1/(4pi^2xx10^12) =1/(4xx10xx10^12)` `=2.5xx10^(-14) s` (Take `pi^2`=10 ) |
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| 259. |
Modulation is done inA. TransmitterB. Radio receiverC. in communication channelD. None of these |
| Answer» Correct Answer - A | |
| 260. |
Beyond which frequency, the ionosphere bends any incidne electrognetic radiation but does not reflect if back towards the aarth?A. 20 MHzB. 30MHzC. 40MHzD. 50MHz |
| Answer» Ionosphere does not reflect electromagnetic waves of frequency more than 40 MHz. | |
| 261. |
Which of the following four alternatives is not correct, We need modulation :-A. To increase the3 selectiveityB. To reduce the time lag between transmission and reception of the information signalC. To reduce the zise of antennaD. To reduce the fractional hand waks that is the ratio of the signal hand width to the centre frequency. |
| Answer» We know that low frequencies signals cannot be transimitted to long distances. That is why, the low frequency signal is superimposed on a high frequency signal by a process known as modulation. In modulation process, the speed of electromagnetic wave will not change. Due to it, we can reduce the time lag between trasmission and reception of the information signals. | |
| 262. |
The need for doing modulation isA. to increase the intensity of audio signalB. to decrease the intensity of audio signalC. to transmit audio signal to large distancesD. none of the above |
| Answer» Correct Answer - C | |
| 263. |
"Need for modulation" arises due to which of the following reasons?A. Power radiated by an antenna `prop (1/lambda)^2`B. It allows multiple user-friendly communicationC. Height of antenna required `prop (lambda/4)`D. All of these |
| Answer» Correct Answer - D | |
| 264. |
Range of frequencies allotted for commercial `FM` radio broadcast isA. 88 MHz to 108 MHzB. 88 Hz to 88 kHzC. 800 kHz to 8000 MHzD. None of these |
| Answer» Correct Answer - A | |
| 265. |
Range of frequencies allotted for commercial `FM` radio broadcast isA. 88 to 108 MHzB. 88 to 108 kHzC. 8 to 88 MHz sD. 88 to 108 GHz |
| Answer» Correct Answer - A | |
| 266. |
An oscillator is producing FM waves of requency `2kHz` with a variation of `10kHz`. What is modulating index?A. `0.20`B. `5.0`C. `0.67`D. `1.5` |
| Answer» Correct Answer - B | |
| 267. |
The layer (or play) effective role in space communication at night is (or are) :A. D-layerB. E-layerC. `F_(1)`-layerD. `F_(2)-`layer |
| Answer» D-layer and `F_(1)` layer disappear at night. | |
| 268. |
For communication purpose of artifical satelliteA. micro waves are usedB. radio transponders are usedC. uplink and downlink have different frequenciesD. uplink and downlink have same frequencies |
| Answer» Microwaves are used for communication purpose of artificial satellite. Radio transponders is a receiving or transmitting equipment for satellite communication. Uplink and dwon link have different frequncies to avoid interference between uplink and downlink. | |
| 269. |
Name the type of radiowave propagation involved with T.V. signals broadcast by a tall antenna. |
| Answer» Space wave propagation or line of sight communication. | |
| 270. |
A 1000 kHz carrier wave is modulated by an audio signal of frequency range 100-5000 Hz. Then, the width of channel (in kHz) isA. 10 KHzB. 20 KHzC. 30 KHzD. 40 KHz |
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Answer» Correct Answer - A Band width or (width of the channel ) = 2 x (maximum frequency of the modulating signal or audio frequency signal) = 2 x 5000 Hz = 10,000 Hz = 10 KHz |
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| 271. |
The ratio waves of frequency `300 MHz` to `3000 MHz` belong toA. Super high frequency bandB. Ultra high frequency bandC. Very high frequency bandD. High frequency band |
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Answer» Correct Answer - B The radiowaves of frequency range 300 MHz to 3000 MHz belong to ultra high frequency band. |
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| 272. |
What is the frequency of a 20 m radiowave?A. a.15MHzB. b.12MHzC. c.1200 KHzD. d.25 MHz |
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Answer» Correct Answer - A `n=c/lambda=(3xx10^8)/20=15xx10^6`=15 MHz |
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| 273. |
A 1000 kHz carrier is modulated with 800 Hz audio signals. What are the frequencies of first pair of side bands: |
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Answer» Correct Answer - 1000.8 kHz , 999.2 kHz `v_( SB) = v_( c ) +- v_(m) = 1000 +- 0.8` `= 1000.8 kHz` and `999.2 kHz` |
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| 274. |
In a diode detector , output circuit consists of ` R = 1 M omega and C = 1 pF`. Calculate the carrier frequency it can detect. |
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Answer» Correct Answer - `v_(c) > > 1 MHz` Here , `RC = 10^(6) xx 10^(-12) = 10^(-6) s` For demodulation , `(1)/( v_( c )) lt lt RC or v_( c ) gt gt 1//RC` ` v_( c ) gt gt (1)/( 10^(-6)) = 10^(6) Hz = 1 MHz` ` :. v_( c ) gt gt 1 MHz` |
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| 275. |
A broadcast `AM` transmitter radiates ` 50 kW` of carrier power . What will be the radiated power at `50%` modulation ? |
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Answer» Correct Answer - 56.25 kW `P_(t) = P_( c ) ( 1 + (mu^(2))/(2)) = 50 ( 1 + (1)/(8))` `= (450)/(8) = 56.25 kW` |
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| 276. |
A message signal of frequency `12 kHz` and peak voltage 12 volt is used to modulate a carrier of frequency ` 1 MHz` and peak voltage 24 volt. What is the modulation index and the side bands produced ? |
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Answer» Correct Answer - 0.5 , 1012 kHz , 988 kHz Here , ` A_(m) = 12 "volt" , A_( c ) = 24 "volt," v_(m) = 12 kHz,` ` v_( c ) = 1 MHz = 1000 kHz` Modulation index , `mu = ( A_(m))/( A_( c )) = ( 12)/( 24) = 0.5` The upper side band frequency ` = v_( c ) + v _(m)` ` = 1000 + 12 = 1012 kHz` The lower side band frequency ` = v_( c ) - v _(m)` ` = 1000 - 12 = 988 kHz` |
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| 277. |
In a simple AM transmitter, the tuned circuit uses a coil of `40muH` and a shunt capacitor of 1 n F. If the oscillator output is modulated by audio frequencies upto 5kHz, what is the frequency range occupied by side bands? |
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Answer» Correct Answer - 791 to 801 kHz Here, `L = 40muH = 40 xx 10^(-6)H`, `C = 1 nF = 10^(-9) F, v_m = 5 kHz`. `v_c = 1/(2pi sqrt(LC)) = 1/(2pisqrt(40 xx 10^(-6) xx 10^(-9))) = 10^7/(4pi)` `= 796 kHz`. `:.` Frequency range of side bands `v_C +- v_m = (796 +-5) kHz = 791 kHz "to" 801kHz`. |
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| 278. |
Out of the following the radiation with lowest frequency isA. Visible lightB. X - raysC. MicrowavesD. Ultraviolet rays |
| Answer» Correct Answer - C | |
| 279. |
Which of the following is absorbed by the ozone layer ?A. Only radiowavesB. Only visible raysC. Ultraviolet radiationsD. `gamma` rays |
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Answer» Correct Answer - C The ozone layer absorbs the ultraviolet rays. |
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| 280. |
In short wave communication waves of which of the following frequeuncies will be reflected back by the ionospheric layer having electron density `10^(11)per m^(3)`?A. 2 MHzB. 10 MHzC. 12 MHzD. 18 MHz |
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Answer» Here, `N=10^(11)` per `m^(3)`, As `v=9sqrt(N)=9sqrt(10^(11))` `=9xx3.16xx10^(5)=28.3xx10^(5)=2.83xx10^(6)Hz` `-2.83 MHz` So waves of frequency 2 MHz will be reflected back by the ionosphere layer. |
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| 281. |
In short wave communication, waves of which of the following frequencies will be reflected back by the ionoshperic layer having electron density `10^11 " per " m^3`A. 2 MHzB. 10 MHzC. 12MHzD. 18 MHz |
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Answer» Correct Answer - A Critical frequency `f_c=9(N_"max")^(1//2)` where `N_"max"`=Electron density `therefore f_c=9(10^11)^(1//2) = 9(1/10xx10^12)^(1//2)` `=9/sqrt10xx10^6=2.8xx10^6` Hz = 2.8 MHz `therefore ` The frequency of 2 MHz which is less than the critical frequency will be reflected. |
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| 282. |
A message signal of 12 kHz and peak voltage 20 V is used to modulate a carrier wave of frequency 12 MHz and peak voltage 30 V. Calculate the (i) modulation index (ii) side- band frequencies. |
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Answer» Correct Answer - 0.67 ; 12.012 MHz ; 11.988MHz. Here, `v_m = 12kHz = 0.012 MHz, A_m = 20 V`. `v_C = 12 MHz, A_C = 30 V` `mu_a = A_m/A_c = 20/30 = 0.67` USB = `v_c + v_m = (12 + 0.012)MHz` `= 12.012 MHz` LSB `= v_c -v_m = (12 - 0.012)MHz` `= 11.988 MHz`. |
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| 283. |
The ozone layer in the atmosphere absorbsA. only the radiowavesB. only the visible lightC. only the Y raysD. X - rays and ultraviolet rays |
| Answer» Correct Answer - D | |
| 284. |
From which layer of the atmosphere,radio waves are reflected back?A. IonosphereB. ChromosphereC. MesophereD. None |
| Answer» Correct Answer - A | |
| 285. |
For a carrier wave of frequency 6 MHz, what should be the length of Marconi antenna? |
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Answer» Correct Answer - 12.5 m Here, `v = 6MHz = 6 xx 10^6 Hz` Length of Marconi antenna `lambda/4 = c/(4v) = (3xx10^8)/(4 xx 6 xx 10^6) m = 100/8 m ` `= 12.5m`. |
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| 286. |
Calculate the length of half wave dipole antenna at 30 MHz |
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Answer» v=30 MHz=`30xx10^6` Hz `lambda=c/v=(3xx10^8 ms^(-1))/(30xx10^6 s^(-1))`=10 m Hence length of antenna `l=lambda/2=10/2`=5 m |
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| 287. |
In which layer of the atmosphere, the water vapour is present ?A. TroposphereB. IonosphereC. StratosphereD. Mesosphere |
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Answer» Correct Answer - A Water vapour is present in the troposphere . |
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| 288. |
Calculate the length of half wave dipole antenna at 30 MHzA. 10 mB. 5 mC. 2.5 mD. 20 m |
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Answer» Correct Answer - B Length of a dipole antenna `l=lambda/2=c/(2f)` In this problem , `lambda=c/f =(3xx10^8)/(30xx10^6)` = 10 m `therefore l=lambda/2=5` m |
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| 289. |
If both the length of an antenna and the wavelength of the signal to be transmitted are doubled, the power radiated by the antennaA. is doubledB. is halvedC. remains the sameD. is quadrupled |
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Answer» Correct Answer - C The power radiated by an antenna `prop ("length"/"wavelength")^2` `therefore` If both l and `lambda` are doubled, P will remain the same. |
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| 290. |
Why do we require a satellite for long distance T.V. transmission? |
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Answer» The T. V. signal are of high frequency range 54 MHz to 890 MHz. They can not be reflected by ionosphere and their ground wave propagation is possible up to a limited distance due to heavily energy absorption while moving along the surface of earth. therefore, the propagation of T.V. signal is not possible by either sky waves or ground waves. The T.V. signal can be transmitted over a long distance through satellite communication. In satellite communication the transponder fitted in satellite receives the T.V. signals amplifies it and returns it towards the ground with different frequency waves. |
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| 291. |
Is it necessary to use satellite for long distance T.V. transmission? Give reasons. |
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Answer» The T.V. transmission involves the television signal waves having the frequency range 54 MHz to 890 MHz. These waves neither follow the curvature of earth nor they get reflected by ionosphere. Therefore, their communication via ground wave or sky wave is not possible. The reception of television signals is possible either (i) by using communication geostationary satellite which reflects the television signals back to earth or (ii) by using tall receiver antenna which may directly intercept the signals. |
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| 292. |
The Q of a resonant transmission line is:A. `Q=(omega)/(LR)`B. `Q=(omegaR)/(L)`C. `Q=(L)/(R)`D. `Q=(omegaL)/(R)` |
| Answer» Correct Answer - C | |
| 293. |
Line of sight (LOS) communication is also known asA. Ground wave communicationB. Space wave communicationC. Sky wave communicationD. ionospheric communication |
| Answer» Correct Answer - B | |
| 294. |
Polarization in electromagnetic wave is caused by-A. ReflectionB. RefractionC. transverse nature of e.m. wavesD. longitudinal nature of e.m. waves |
| Answer» Correct Answer - C | |
| 295. |
The characteristic impedance of los less transmission line is given by-A. `Z_(0)=sqrt(LC)`B. `Z_(0)=sqrt((L)/(C))`C. `Z_(0)=sqrt((C)/(L))`D. `Z_(0)=LC` |
| Answer» Correct Answer - B | |
| 296. |
The velocity of electromagnetic waves in a dielectric medium `(in_(r)=4)` isA. `3xx10^(6)` metre/secondB. `1.5xx10^(8)`metre/secondC. `6xx10^(6)` metre/secondD. `7.5xx10^(7)`metre/second |
| Answer» Correct Answer - B | |
| 297. |
The velocity of electromagnetic waves in a dielectric medium `(in_(r)=4)` isA. `3xx10^8` metre/secondB. `1.5xx10^8` metre/secondC. `6xx10^8` metre/secondD. `7.5xx10^7` metre/second |
| Answer» Correct Answer - D | |
| 298. |
The ionosphere is used for the propagation ofA. 1. sky wavesB. 2. space wavesC. 3. ground wavesD. 4. sound waves |
| Answer» Correct Answer - A | |
| 299. |
The velocity factor of a transmission line `x`. If dielectric constant of the medium is `2.6`,A. 0.26B. 0.62C. 2.6D. 6.2 |
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Answer» Correct Answer - B velocity factor `=(1)/(sqrt(k))=(1)/(sqrt(2.6))=0.62` |
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| 300. |
When microwave signals follow the curvature of earth, this is known as:A. space wave propagationB. ground wave propagationC. sky wave propagationD. tropospheric communication |
| Answer» Correct Answer - B | |