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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.
| 351. |
A ground receiver station is receiving a signal at (a) 6.0 MHz and (b) 110 MHz, transmitted from ground transmitter at a hight 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 10^6m`, maximum number density of electrons in ionosphere = ` 10^(12)m^(-3)`. |
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Answer» Here, `h = 300 m, R = 6.4 xx 10^6m,` `N_(max) = 10^(12) m^(-3)` For space wave propagation, Maximum distance covered by space wave communication `d = sqrt (2Rh) = sqrt(2 xx 6.4 xx 10^6 xx 300)` `= 61.96 xx 10^3 m ~~ 62 km` Since the distance between transmitter and receiver is 100 km, hence for the given frequency signals of 6.0 MHz and 110 MHz, the propagation is not possible via space wave propagation. For sky wave propagation, the critical frequency, `v_c = 9 (N_(max))^(1//2) = 9 xx (10^12)^(1//2) = 9 xx 10^6 Hz` `= 9MHz` Since the frequency of signal (a) i.e. 6.0 MHz is less than `v_c`, so its propagation via sky wave is possible. Since the frequency of signal (b) i.e., 110 MHz is greater than `v_c`, hence its propagation is not possible via sky waves. For that, satellite communication is used. |
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| 352. |
A ground receiver station is receiving a signal at 500 MHz transmitted from a ground transmitter at a height of 300 m located at a distance of 500 km. It is coming viaA. Ground waveB. Space waveC. Sky waveD. Satellite communication |
| Answer» Correct Answer - D | |
| 353. |
A ground receiver station is receiving a signal at (a) `5 MHz and 100 MHz` , transmitted from a ground transmitter at a height of `400 m` located at a distance of ` 125 km`. Identify whether it is coming via space wave or sky wave propagation or statellite transponder. Radius of earth ` = 6.4 xx 10^(6) m` , maximum number density of electrons in ionsphere ` = 10^(12) m^(-3)`. |
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Answer» Correct Answer - (a) sky waves (b) satellite communication Maximum distance covered by space wave communication, ` d = sqrt (2 R h) = sqrt( 2 xx 6.4 xx 10^(6) xx 400)` `= 71.55 xx 10^(3) m = 71.55 km` Since the distance between transmitter and receiver is `125 km` , hence for the given frequency signals of ` 5 MHz and 100 MHz` , the propagation is not possible via space wave communication . For sky wave propagation , the critical frequency , `v_( c ) = 9 (N _(max))^(1//2) = 9 (10^(12))^(1//2) = 9 xx 10^(6) Hz` ` = 9 MHz` Since ` 5 MHz lt 9 MHz` , so the propagation of signal of frequency `5 MHz` is possible via sky wave . As `100 MHz gt 9 MHz`, hence the progation of signal of frequency `100 MHz` is not possible via sky waves . For that satellite communication is used . |
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| 354. |
A microwave telephone link operating at the central frequency of 10 GHz has been established. If 2% of this is available for microwave communication channerl and each telephone is alloted a bank width of 8 KHz. the number of telephone channels which cn be simultaneously granted is `2.5xx10^(a)`. What is the integer value of a? |
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Answer» Microwave freqeuncy used in telephone link `=10GHz=10xx10^(9)Hz=10^(10)Hz` Frequency available for microwave communication `2%` of 10 GHz `=(2)/(100)xx10^(10)=2xx10^(8)Hz` Bank width of each telephone channel =8 kHz `=8xx10^(3)` Hz Number of microwave telephone channels `=(2xx10^(8))/(8xx10^(3))=2.5xx10^(4)=2.5xx10^(a)` `therefore a =4` |
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| 355. |
If the sum of the heights of transmitting and receiving antennas in line of sight of communication is fixed at h, show that the range is maximum when the two antennas have a height `h//2` each. |
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Answer» The range (r) of line of sight communication between two antennas (during space wave propagation) is given by `r = sqrt(2Rh_(T)) + sqrt(2Rh_(R))` where `h_(T) and h_(R)` are the height of transmitting and receiving antennas. Here, `h_(T) + h_(R) = h. If h_(T) = H, then h_(R) = h-H` Then `r = sqrt(2R)[sqrt(H) + sqrt(h-H)]` For r to be maximum, `(dr)/(dH) = 0` `:. (dr)/(dH) = 0 = sqrt(2R) [ 1/(2sqrtH) + (1)/(2sqrt (h-H)xx(-1))]` or `1/(2sqrtH) - 1/(2sqrt(h-H)) = 0` or `2sqrt(H) = 2 sqrt(h-H)` or `H = h-H or h = 2H` or `H = h//2` Hence, for maximum range, `h_(T) = h_(R) = h//2`. |
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| 356. |
Heights of transmitting and receiving antennas from earth surface are `h_(T)` and `h_(R)` respectively. Write the relation for maximum line-of-sight (LOS) distance two antennas. |
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Answer» If the transmitting antenna is at a height `h_(T)`, then you can show that the distance to the horizontal `d_(T)` is given as `d_(T)=sqrt(2Rh_(T))`, where `R` is the radius of the earth (approximately 6400 km) `d_(T)` is also called the radio horizon of the transmitting antenna. With reference to figure above the maximum line-of-sight distance `d_(M)` between the two antennas having height `h_(T)` and `h_(R)` above the earth is given by `d_(M) = sqrt(2Rh_(T)) + sqrt(2Rh_(R))` where `h_(R)` is the height of receiving antenna. |
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| 357. |
In a communication system, operating at 1200 mm, only 2% of the source frequency is available for T.V. transmission , having a bandwidth of 5 MHz. The number of T.V. channels that can be transmitted isA. 1 millionB. 2 millionC. 0.5 millionD. 0.1 million |
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Answer» Correct Answer - A Frequency of the source `=c/lambda=(3xx10^8)/(1200xx10^(-9)) =(3xx10^8xx10^7)/12 = 25 xx 10^13` Hz Available band width for T.V. Transmission `=2/100xx25xx10^13=5xx10^12` Hz `therefore` No. of T.V. channels =` "Total bandwidth"/"Bandwidth per channel"=(5xx10^12)/(5xx10^6)=10^6`=1 million |
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| 358. |
For an optical communication system, operating at `lambda = 800 nm`, only `1 %` the optical source frequency is the available channel band width. How many channels can be accomodated for transmitting video T.V. signal requiring an approximate band width of 4.5 MHz? |
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Answer» Optical source frequency, `v = c/lambda = ( 3xx 10^8)/(800 xx 10^(-9)) = 3.75 xx 10^(14) Hz` Band width of channel = 1% of source frequency =` 1/100 xx 3.75 xx 10^(14) = 3.75 xx 10^(12) Hz` Number of channels for video T.V. signal `= (3.75 xx 10^(12))/(4.5 xx 10^(6)) = 8.3 xx 10^(5) `. |
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| 359. |
A communication channel having an operating wavelength of `lambda` metre can use only x% of its source frequency as its channel bandwidth . If the system is to be used for transmitting T.V. signals requiring a bandwidth of F Hertz . How many channels can this system transmit simultaneously ?A. `(xc)/(100lambdaF)`B. `(100lambdaF)/(xc)`C. `sqrt((xc)/(lambdaF))`D. `sqrt((100lambdaF)/(xc))` |
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Answer» Correct Answer - A Source frequency `f=c/lambda` `therefore` Bandwidth of the channel =x% of f `=x/100 xx c/lambda=(xc)/(100lambda)` `therefore` Number of channels = `"Total bandwidth of channel"/"Bandwidth of each channel"` `therefore N=(xc)/(100lambda)xx1/F` |
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| 360. |
In communication system, type of communication channel needed for given signal depends onA. phaseB. amplitudeC. powerD. bandwidth |
| Answer» Correct Answer - D | |
| 361. |
The carrier frequency generated by a tank circuit containing `1 nF` capacitor and `10 muH` inductor isA. 1592 HzB. 1592 MHzC. 1592 kHzD. `159.2` Hz |
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Answer» Correct Answer - C ` f = 1/(2 pi sqrt(LC))` ` = 1/(6.28 xx sqrt(1 xx 10^(-9) xx 10 xx 10^(-6)))` ` = 1592 ` kHz |
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| 362. |
Write an expression in a vector form for the Lorentz magnetic force `vecF` on a charge Q moving with velocity `vecV` in a magnetic field `vecB`. What is the direction of the magnetic force? |
| Answer» `vecF=Q(vecVxxvecB)`, The direction of `vecF` is perpendicular to the plane containing `vecV` and `vecB`, and is given by Right Hand Rule. | |
| 363. |
Ground/ surface wave propagation is a mode of propagation in which the signal wave glides over the surface of earth, while going from transmitter to receiver. In this mode, there is a loss of power (i.e., attenuation) of signal wave due to diffraction and absorption of signal wave energy by ground. The attenuation of ground / surface wave signal increases very rapidly with the increase in its frequency. Therefore, the ground/ surface wave propagation is not suitable for the propagation of high frequency signal wave and for very long range communication. Read the above passage and answer the following questions: (i) What is the frequency range of a signal wave employed for the ground wave propagation? (ii) For what purpose the ground wave propagation is used? (iii) What does the study of ground wave propagation imply in day to day life? |
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Answer» (i) For ground wave propagation, the frequency range of signal wave is `530 kHz "to" 1710 kHz.` (ii) The ground wave propagation is used for local broadcasting as a medium wave broadcast service. (iii) In day to day life, the ground wave propagation can be compared to the movement of a small child along the surface of ground. As child can go only upto certain distance on ground and gets exhausted, similarly the ground wave frequency signal loses its energy while gliding over the surface, of earth and hence can go up to limited distance. As ground wave propagation is for local broadcasting, similarly, the small child movement is confined with in home under the supervision of some elders. |
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| 364. |
The maximum range of ground or surface wave propagation depends on-A. the frequency of the radiowaves onlyB. power of the transmitter onlyC. both of themD. none of them |
| Answer» Correct Answer - C | |
| 365. |
On what factors, does the maximum range of ground wave propagation depend? How can the maximum range of ground wave propagation be increased? |
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Answer» The maximum range of ground wave propagation depends on : (i) the frequency of wave (ii) power of the transmitter. The maximum range of ground wave propagation can be increased by increasing the power of transmitter which is so only for very low frequency band (VLF) and not for medium frequency band (MF). |
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| 366. |
A steady current is flowing in a cylindrical conductor. Is there any electric field within the conductor ? If yes, what is its relation with current density? |
| Answer» Yes, an electric field existes within the conductor because it is the electric field which provides acceleration to electrons for the flow of current. | |
| 367. |
Compute `LC` product of a tuned amplifer circuit required to generate a carrier `wave` of `1 MHz` for amplitude modulation |
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Answer» Frequency of tuned amplifier is , ` v = 1/(2pisqrt(LC))` As per question, `v = 1MHz = 10^6Hz = 1/(2pisqrt(LC)) or LC = 1/(4pi^2 xx (10^6)^2) = (7 xx 7)/(4 xx (22 xx 22) xx 10^(12))` `= 2.5 xx 10^(-14)s`. |
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| 368. |
Which of the following is an advantage of FM over AM?A. Less noiseB. Larger bandwidthC. More circuit complexityD. Can be transmitted to longer distance |
| Answer» Correct Answer - A | |
| 369. |
For television broadcasting, the frequency employed is normallyA. 30 - 300 MHzB. 30-300 GHzC. 30-300 kHzD. 30-300 Hz |
| Answer» Correct Answer - A | |
| 370. |
The frequencies of electromagnetic waves employed in space communication lie in the range of -A. `10^(4)` Hz to `10^(7)Hz`B. `10^(4)` Hz to `10^(11)` HzC. 1 Hz to `10^(4)Hz`D. 1 Hz to `10^(11)` Hz |
| Answer» Correct Answer - B | |
| 371. |
The wavelength of electromagnetic waves employed for space communication lie in the the range of-A. 1mm to 30 mB. 1 mm to 300 mC. 1 mm to 3 kmD. 1mm to 30 km |
| Answer» Correct Answer - D | |