InterviewSolution
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InterviewSolution
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.
| 201. |
A carrier wave of `1000W` is subjected to `100%` modulation. Calculate (`i`) Power of modulated wave, (`ii`) power is `USB`, (`iii`) power is `LSB` |
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Answer» (`i`) Total pressure of modulated wave `P_(T)=P_(C)(1+(m^(2))/(2))`, `=1000(1+(1^(2))/(2))=1500watt` (`ii`) Power in `USB=(1)/(2)P_(SB)` Where power carried by side bands is given by amplitude modulation and detection `P_(SB)=P_(C)((m^(2))/(2))`, `=1000((1^(2))/(2))=500watt` `P_(USB)=(1)/(2)P_(SB)=(1)/(2)xx500=250watt` (`iii`) Since power in `LSB=` Power in `USB` `P_(LSB)=P_(USB)=250watt` |
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| 202. |
In frequency modulationA. The amplitude of modulated wave varies as frequency of carrier waveB. The frequency of modulated wave varies as amplitude of modulating waveC. The amplitude of modulated wave varies as amplitude of carrier waveD. The frequency of modulated wave varies as frequency of modulating wave |
| Answer» Correct Answer - B | |
| 203. |
If `f_(c)` and `f_(m)` are the frequencies of carrier wave and signal, then the band width isA. `f_(m)`B. `2f_(m)`C. `f_(c)`D. `2f_(c)` |
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Answer» Correct Answer - B band width `=(f_(c)+f_(m))-(f_(c)-f_(m))=2f_(m)` |
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| 204. |
The process of superimposing signal frequency (i.e. audio wave) on the carrier wave is known asA. TransmissionB. ReceptionC. ModulationD. Detection |
| Answer» Correct Answer - C | |
| 205. |
The audio signal voltage is given by `V_(m)=2sin12pixx10^(3)t`. The band width and `LSB` if carrier wave has a frequency `3.14xx10^(6)rad//s`A. `12 KHz`, `494 KHz`B. `6 KHz`, `313 KHz`C. `6 KHz`, `494 KHz`D. `18KHz`, `494 KHz` |
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Answer» Correct Answer - A bandwidth `=2f_(m)` `LSB=f_(c)-f_(m)` |
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| 206. |
The velocity factor of a transmission line `x`. If dielectric constant of the medium is `1.8`, the value of `x` isA. `0.26`B. `0.62`C. `0.74`D. `6.2` |
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Answer» Correct Answer - C `V.f.=(1)/(sqrt(k))`, `=(1)/(sqrt(1.8))=0.74` |
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| 207. |
The modulation index of an `FM` carrier having a carrier swing of `300kHz` and a modulating signal `10kHz` isA. `15`B. `10`C. `20`D. `25` |
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Answer» Correct Answer - A `CS=2Deltaf` `Deltaf=(CS)/(2)=(300KHz)/(2)=150 KHz` `m_(f)=(Deltaf)/(f_(m))=(150KHz)/(10KHz)=15` |
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| 208. |
A ground reciever is receiving a signal at (a) 5 MHz, and (b) 100 MHz, transmitted from a ground transmitter at a height of 300 m located at a distance of 100 km, Identify whether it is coming via space wave or sky wave propagation or satellite transponder. [Radius of earth `~~ 6.4 xx 10^(6) m , N_(max)` of ionosphere `= 10^(12) m^(3)`]A. Space waveB. Sky wave propagationC. Satellite transponderD. All of these |
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Answer» Correct Answer - b |
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| 209. |
Why do we need a higher bandwidth for transmission of music compared to that for commercial telephone communicaton? |
| Answer» As compared to speech signals in telephone communication, the music signals are more complex and correspond to higher frequency range. | |
| 210. |
In a video signal for transmission of picture. What value of bandwidth is used in communication system?A. 2.4 MHzB. 4.2 MHzC. 24MHzD. 42MHz |
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Answer» Correct Answer - B 4.2 MHz bandwidth is required in a video signal for transmission of picture . |
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| 211. |
T.V. transmission tower at a particular station has a height of `160m`. (`a`) What is the coverage range? (`b`) How much population is covered by transmission, if the average population density around the tower is `1200 per km^(2)`? (`c`) What should be the height of tower to double the coverage range |
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Answer» (`a`) Coverage range `d=sqrt(2Rh)` `=sqrt(2xx6400xx10^(3)xx160m)` `=45.254 km` (`b`) Population covered `=("population density")xx("area covered")` `=(1200)xx(pi d^(2))` `=(2400piRh)=2400xx3.14xx6.4xx10^(3)xx0.16` `=77.17lac` (`c`) Coverage range `propsqrt(h)` Therefore coverage range can be doubled by making height of the lower four times to `640m`. So, height of the lower should be increased by `480 m`. |
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| 212. |
Through which mode of propagation, the radio waves can be sent from one place to anotherA. Ground wave propagationB. Sky wave propagationC. Space wave propagationD. All of them |
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Answer» Correct Answer - D |
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| 213. |
The carrier frequency generated by a tank circuit containing `1 nF` capacitor and `10 muH` inductor isA. `1592 Hz`B. `1592 MHz`C. `1592 kHz`D. `159.2 Hz` |
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Answer» Correct Answer - C |
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| 214. |
A laser beam is used for carrying our surgery because itA. Is highly monochromaticB. Is highly coherentC. s highly directionalD. Can be sharply focussed |
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Answer» Correct Answer - D |
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| 215. |
A laser beam of pluse power `10^(12)` watt is focussed on an object are `10^(-4) cm^(2)`. The energy flux in `watt//cm^(2)` at the point of focus isA. `10^(20)`B. `10^(16)`C. `10^(8)`D. `10^(4)` |
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Answer» Correct Answer - B |
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| 216. |
A modulating signal is a square wave as shown in figure. The carrier wave is given by `c(t)=2 sin(8pit)"volt"`. The modulation index isA. `2`B. `0.75`C. `0.5`D. `1.5` |
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Answer» Correct Answer - C `mu=(A_(m))/(A_(c))=(1)/(2)=0.5` |
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| 217. |
A wave is represented as `e=(10^(8)t+6 sin 1250t)sin(W_c)t` then the modulating index isA. 10B. 1250C. `10^(8)`D. 6 |
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Answer» Correct Answer - d |
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| 218. |
The load on an `Am` diode detector consists of a resistance of `50KOmega` in parallel with a capacitor of `0.001 muF`. Determine the maximum modulation index that the detector can handle without distortion when modulation frequency is (`i`) `1kHz` (`ii`) `5 kHz` |
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Answer» Here `R_(C)` and `C` are in parallel combination and phase difference between `i_(1)` and `i_(2)` is `(pi)/(2)` `i^(2)+i_(1)^(2)+i_(2)^(2)` `[(V)/(Z_(m))]^(2)=[(V)/(R_(c))]^(2)+[(V)/(X_(c))]^(2)` `Z_(m)=(1)/(sqrt((1)/(R_(c))^(2)+(1)/((X_(c))^(2))))` `M_(max)=(Z_(m))/(R_(c))` `=(1)/(R_(c)sqrt((1)/(R_(c))^(2)+(1)/((X_(c))^(2))))` `=(1)/(sqrt(1+(2pifCR_(c))^(2)))` For `f=1kHz` `M_(max)=(1)/(sqrt(1+0.098696))=0.945` For `f=5kHz` `M_(max) =(1)/(sqrt(1+24674))=0.537` |
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| 219. |
The range of ground wave transmission can be increased byA. increasing the power of transmitter with the use of `HF`B. increasing the power of transmitter with the use of `VLF`C. decreasing the power and increasing the frequency of radio wavesD. decreasing both power and frequency of radio waves |
| Answer» Correct Answer - B | |
| 220. |
The ground wave eventually disappears, as one moves away from the transmitter, because ofA. interference from the sky waveB. loss of line of signal conditionC. maximum single-hop distance limitationD. diffraction effect causing tilting of the wave |
| Answer» Correct Answer - D | |
| 221. |
Figure shows a block diagram of a transmitter. Identify the boxes `X` and `Y` ? A. Amlitude modulator, detectorB. Dectector, power amliferC. Amplitude modulator, power amplifierD. Capacitor, Detector |
| Answer» Correct Answer - C | |
| 222. |
In an amplitude modulated wave for audio frequency of `500 "cycle"//second`,the appropriate carrier frequency will beA. `50 "cycles"//sec`B. `100 "cycles"//sec`C. `500 "cycles"//sec`D. `50,000 "cycles"//sec` |
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Answer» Correct Answer - D Carrier frequency `gt` audio frequency |
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| 223. |
In an amplitude modulated wave for audio frequency of `500 "cycle"//second`,the appropriate carrier frequency will beA. 50 cycles/secB. 100 cycles/secC. 500 cycles/secD. 50000 cycle/sec |
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Answer» Correct Answer - D Carrier frequency value should be vary large compared to modulating signal frequency |
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| 224. |
Why is an AM signal likely to be more noisy than a FM signal upon transmission through a channel? |
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Answer» In case of AM, the instantaneous voltage of carrier waves is varied by the modulating wave voltage. So, during the transmission, noise signals can also be added and receiver assumes noise a part of the modulating signal. In case of FM, the frequency of carrier waves is changed as the change in the instantaneous voltage of modulating waves. This can be done by mixing and not while the signal is transmitting in channel. So, noise does not affect FM signal. |
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| 225. |
Audio signal cannot be transmitted becauseA. the signal has more noiseB. the signal cannot be amplified for distance communicationC. the transmitting atenna length is very small to designD. the transmitting antenna length is very large and impractiable |
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Answer» Correct Answer - D Audio signal cannot be transmitted , because the length of transmitting antenna is very large and impracticle. |
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| 226. |
Audio signal cannot be transmitted becauseA. The signal has more noiseB. The signal cannot be amplified for distance communicationC. The transmitting antenna length is very small to designD. The transmitting antenna length is very large and impracticable |
| Answer» Correct Answer - D | |
| 227. |
For television broadcasting, the frequency employed is normallyA. `30-300 MHz`B. `30-300 GHz`C. `30-300 KHz`D. `30-300 Hz` |
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Answer» Correct Answer - A `VHF` (Very High Frequency) band having frequency range `30 MHz` to `300 MHz` is typically used for `TV` and radar transmission. |
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| 228. |
Assertion: The television broadcasting becomes weaker with increasing distance. Reason: The power transmitted from T.V. transmitter varies inversely as the distance of the receiver.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - c |
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| 229. |
The electromagnetic waves used in the telecommunication areA. ultravioletB. infraredC. visibleD. microwave |
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Answer» Correct Answer - D Electromagnetic radiation relevant to telecommunication is microwave . |
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