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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.
| 151. |
For sky wave propagation of a 10 MHz signal, which of the following statements are true:A. The minimum electron density in the ionosphere is `1.2xx10^(12)m^(-3)`B. The minimum electron density in the ionosphere is `10^(6) m^(-3)`C. The reflection of the given sky wave signal takes place from E layer of ionosphereD. The reflection of the layer of ionosphere |
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Answer» Here, v=10MHz =`10^(7) Hz. N=?` `v=9sqrt(N)` or `10^(7)=9sqrt(N)` Squaring both sides, `10^(14)=81xxN` `N-(10^(14))/(81)=~1.2xx10^(12) m^(-3)` The 10 MHz signal is reflected from F-layer of ionosphere. |
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| 152. |
Assertion. Audio signal of frequencuy 10 KHz cannot be transmitted over long distance without modulation. Reason. Length of the antenna required `lambda//4`. Should have practical value.A. both, Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both, Assertion and Reason are false. |
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Answer» Both Assertion and Reason are true and Reson is the correct explanation of Assertion. If the audo signal of frequency 10 kHz is to be transmitted directly then wavelenght of signal is `lamdba=(c)/(v)=(3xx10^(8))/(10xx10^(3))=30xx10^(3)m=30km` Lenght of antenna used, `l=(lambda)/(4)=(30)/(4)=7.5km` which is impracticable. Hence audio signal cannot be trasmitted over long distance without modulation. in that situation the antenna used will be of lenght which is of practical value. |
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| 153. |
Audio signal cannot be transmitted becauseA. the signal has more noiseB. the signal can be amplified for long distance communicationC. energy of audio signal gets lost due to attenuationD. the transmitting antenna length is very large and impracticable |
| Answer» Energy of audio signal gets absorbed by the earth due to attenuation. Audio signals can not be transmitted over long distance. Also the transmitting antenna length is very large and impracticable. The signal can be amplified but cannot be trasmitted over long distances due to absorption of energy of energy by ground (being of low frequency) | |
| 154. |
In satellite communication 1. The frequenc used lies between 5 MHz and 10 MHz. 2. The uplink and downlink frequencies are different. 3. The orbit of geostationary satellite lies in the equatorial plane at an inclination of `0^(@)`. In the above statementsA. only 2 and 3 trueB. all are trueC. only 2 trueD. only 1 and 2 true |
| Answer» In satellite communication, the frequency used is more than 40 MHz. The uplink and downlink frequencies are different to avoid distortion of signal and the orbit of geostationary satellite lies in the equatorial plane at an inclination of `0^(@)`. | |
| 155. |
What should be the height of transmitting antenna if the T.V. telecast is to cover a radius of 128 km? `R_e = 6.4 xx 10^6`m. If the average population density around the tower is `1000//km^2`, how much population is covered? |
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Answer» Correct Answer - 1280m Here, d = 128 km `= 128 xx 10^(3) m, R = 6.4 xx 10^6 m` , population density `= 1000//km^2` Height of transmitting antenna ` h = d^2/(2R) = (128 xx 10^3)^2/(2 xx 6.4 xx 10^6) = 1280m.` Total population covered `= pid^2 xx` population density `= 3.14 xx (128)^2 xx 1000 = 5.14 xx 10^7`. |
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| 156. |
Statement-1 in a diode AM detector `R=1kOmega` and `C=10pF` circuit is good enough to detect a carrier signal of 100 kHz. Statement-2: the condition `(1)/(f_(c))ltltRC`A. Statement-1 is true, statement-2 is true, statement-2 is correct explanation for statement-1B. statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. statement-1 is true,statement-2 is falseD. statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - D | |
| 157. |
A diode AM detector with the output circuit consisting of `R=1 kOmega` and `C=1 muf` would be more suitable for detecting a carrier signal of:A. 10 KHzB. 0.5 KHzC. 1 KHzD. 0.75 KHz |
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Answer» Correct Answer - A `R=10^(3) Omega` and `C=10^(-6)F` For detection of a carrier signal (demodulation) `RC gt gt 1/f` In this case `RC = 10^3 xx 10^(-6) = 10^(-3) s` In the given options, when f=10 Kz = `10xx10^3` Hz `1/f=1/(10xx10^3)=0.1xx10^(-3) s therefore RC gt gt 1/f` Thus the frequency 10 KHz can be detected For `f=10^3` Hz, `1/f=10^(-3)` s and the other frequencies of 0.5 KHz and 0.75 KHz cannot be used for detection. |
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| 158. |
In a diode AM detector, the output circuit consists of `R = 1 k Omega and C = 10p F.` A carrier signal of 100 kHz is to be detected. Is it good? If yes, then explain why? If not, what value of C would you suggest ? |
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Answer» Correct Answer - `C = 1muF = 10^(-6)F` Here, `R = 1 k Omega = 10^3 Omega, C = 10pF` `= 10 xx 10^(-12) F = 10^(-11)F` `:. RC = 10^3 xx 10^(-11)s = 10^(-8)s and 1/v_c = 1/(100 xx 10^3) s = 10^(-5)s` We find that `1/v_c` is not less than RC, as is required for demodulation. Therefore, the arrangement is NO GOOD. For a satisfactory arrangement, let us try `C = 1muF = 10^(-6)F` `:. RC = 10^3 xx 10^(-6) s = 10^(-3)s ` Now, `1/v_c = (=10^(-5)s)lt lt RC(=10^(-3)s)` `:.` The condition is satisfied. This is good enough for demodulation. |
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| 159. |
A laser beam has intensity `3.0xx10^14Wm^-2`. Find the amplitudes of electric and magnetic fields in the beam. |
| Answer» `4.75xx10^(8) Vm^(-1), 1.58T` | |
| 160. |
In an amplitude modulated wave for audio frequency of `500 "cycle"//second`,the appropriate carrier frequency will beA. 50 cycles/sB. 100 cycle/sC. 500 cycle/sD. `50,000` cycles/s |
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Answer» Correct Answer - D Carrier frequency should be greater than audio frequency . |
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| 161. |
In an amplitude modulated wave for audio frequency of `500 "cycle"//second`,the appropriate carrier frequency will beA. 50 HzB. 100 HzC. 500 HzD. 50,000 Hz |
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Answer» Correct Answer - D Carrier frequency is very large as compared to the frequency of the audiosignal. `therefore` Carrier frequency is 50,000 Hz. |
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| 162. |
In an amplitude modulated wave for audio frequency of `500 "cycle"//second`,the appropriate carrier frequency will beA. 50 cycles/sB. 100 cycles/sC. 500 cycles/sD. 50,000 cycles/s |
| Answer» Correct Answer - D | |
| 163. |
In an amplitude modulated wave for audio frequency of `500 "cycle"//second`,the appropriate carrier frequency will beA. `50 c//s`B. `100 c//s`C. `500 c//s`D. `50000 c//s` |
| Answer» Correct Answer - D | |
| 164. |
State any two reason for necessity of modulation. |
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Answer» Modulation: The process of attaching i.e., superimposing the message signal to a high frequency carrier wave for the purpose of long distance transmission is called modulation. The resulted obtained is called modulated wave Necessity of modulation: (i). Size of antenna: Height of antenna`prop(1)/("frequency of signal")` Low frequency signal cannot be transmitted as such because height of antenna is large. By attaching the signal with high frequency carrier wave, we can reduce the height of antenna. (ii). Avoid mixing of signals: If two message signals of similar frequency are transmitted as such then they can get mixed with each other. Hence, different transmission channels are allocating different carrier wave frequencies. (iii). Power radiated: Power radiated by an antenna is directly proportional to square of frequency. power `propf` Hence, higher is the frequency more is the power transmitted and signal travel a larger distance. |
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| 165. |
An `AM` transmitter records an antenna current of `10.5 A`. The antenna current drops to `10 A` when only carrier is transmitted . What is the percentage modulation ? |
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Answer» Correct Answer - 0.453 Here , `I_(t) = 10.5 A and I_( c ) = 10 A , mu = ?` As ` sqrt ( 1 + (mu^(2))/(2)) = (I_(t))/(I_( c)) = ( 10.5)/( 10) = ( 21)/( 20)` `:. 1 + (mu^(2))/(2) = ( 441)/(400) =1 + (41)/( 400)` `mu^(2) = ( 82)/( 400) = 0.205` ` mu = sqrt( 0.205) = 0.453 = 45.3%` |
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| 166. |
When is a satellite said to be in sun-synchronous orbit? |
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Answer» A satellite is said to be in sun-synchronous orbit when it passes through a particular place of earth daily almost at the same local time. |
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| 167. |
State two factors by which the range of transmission of TV signal can be increased. |
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Answer» The range of transmission of T.V. signal can be increased (i) by increasing the heights of transmitting and receiving antennas and (ii) by increasing the power of transmitting signal from TV station. |
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| 168. |
GPS stands for ……. . |
| Answer» Correct Answer - Global Positioning System | |
| 169. |
Calculate the phase velocity of electromagnetic wavehaving electron density and frequency for D layer, N=400 electron/cc, v=300 kHzA. `3xx10^(8)m//s`B. `3-75xx10^(8) m//s`C. `6.8xx10^(8)m//s`D. `1.1xx10^(9)m//s` |
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Answer» `mu=sqrt(1-(81.45N)/(v^(2))=sqrt(1-(81.45xx400xx10^(6))/((300xx10^(3))^(2))` `=0.80` Phase velocity, `v=(c)/(mu)= (3xx10^(8))/(0.8)=3.75xx10^(8)m//s` |
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| 170. |
What is the refractive index of E-layer when the electromgnetic wave frequency is 50 MHz and the number density of electrons is `5xx10^(5)` electrons is `5xx10^(5)` electrons`//cm^(3)` ?A. 0-992B. 1-015C. 43481D. 1-000 |
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Answer» Hre, `N=5xx10^(5)` electrons//cm^(3)` `=5xx10^(5)xx10^(6)` electorns`//m^(3)` `=v50 MHz=50xx10^(6)`Hz `mu=sqrt(1-(81.45N)/v^(2))=sqrt(1-(81.45xx5xx10^(5)xx10^(6))/((50xx10^(6))^(2)))` `sqrt(1-(81.45xx10^(-3))/(5))` `=sqrt(1-0.01629)=sqrt(0.98371)=0.992` |
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| 171. |
Height of ionosphere is … km from ground surface.A. 12 kmB. 80 kmC. 400 kmD. 50 km |
| Answer» Correct Answer - C | |
| 172. |
The first Indian experimental satellite named …… was launched on ……… . |
| Answer» Correct Answer - Apple ; June `19 , 1981`. | |
| 173. |
The height of communication statellite from the surface of the earth is approximatelyA. `36 xx 10^(3) km`B. `36 km`C. `36 xx 10^(4) km`D. `36 xx 10^(2) km` |
| Answer» Correct Answer - A | |
| 174. |
The region that contains free electrons, negative ions and positive ions isA. isosphereB. mesosphereC. ionosphereD. stratosphere |
| Answer» Correct Answer - C | |
| 175. |
Maximum usable frequency (MUF) in F-region layer is x, when the critical frequency is 60 MHz and the angle of incidence is `70^(@)`, then x isA. 150 MHzB. 170 MHzC. 175 MHzD. 190 MHz |
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Answer» `MUF=v_(c)sec i=60xx10^(6)xxsec70^(@)` `=60xx10^(6)xx(1)/(0.342)` `=175.43xx10^(6)Hz=175.43 MHz` |
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| 176. |
One way communication is calledA. half duplexB. full duplexC. mono-communicationD. simplex |
| Answer» Correct Answer - D | |
| 177. |
The sky wave with a frequency of 20 MHz is incidnet on D-region at an angle of `45^(@)`, then angle of refraction isA. `12.5^(@)`B. `15^(@)`C. `30^(@)`D. `45^(@)` |
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Answer» For D-region, `N=10^(9)m^(-3)` Refractive index of a layer of atmosphere is `mu=[1-(81.45N)/(v^(2))]^(1//2)=[1-(81.45xx10^(9))/((20xx10^(9))^(2))]^(1//2)` approx 1` `mu=(sin i)/(sin r)approx1` or `sin r=sin i-=sin 45^(@)` or `r=45^(@)` |
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| 178. |
If sky wave with a frequency of 60 MHz is incident on D- region at an angle of `30^@`, then find the angle of refractive.A. `15^(@)`B. `30^(@)`C. `60^(@)`D. `45^(@)` |
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Answer» `v=50 MHz=50xx10^(6)Hz, i=30^(@)`, `N=10^(9)m^(-3)` `mu=sqrt(1-(81.45N)/(v^(2)))=sqrt(1-(81.45xx10^(9))/((50xx10^(6))))=1` `mu=(sin i)/(sin r)` or `1=(sin i)/(sin r)` or `sin i=sin r` or `r=i=30^(@)` |
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| 179. |
Sky wave propagation is also called |
| Answer» Correct Answer - ionospheric propagation | |
| 180. |
The distance between transmitting antenna and receiving antenna at which they can see each other is called ………. |
| Answer» Correct Answer - range of communication | |
| 181. |
What is a geostationary statellite? What are the basic requirements for such a satellite? |
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Answer» It is that satellite which appears to be at a fixed position at a definite height to an observer on earth. Essential conditions for geostationary satellites are as follows: 1. A geostationary satellite should be at a height nearly 36000 km above the equator of earth. Its period of revolution around the earth should be the same as that of earth about its axis i.e. exactly 24 hours. 3. It should revolve in an orbit concentric and coplaner with the equatorial plane. 4. Its sense of rotation should be the same as that of the earth about its own axis i.e. from west to east. Its orbital velocity is nearly 3-1 km/s. |
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| 182. |
If three wave no atmosphere, the average temperature on the surface of earth would be :-A. lowerB. same as nowC. higherD. zero |
| Answer» Correct Answer - A | |
| 183. |
What is a communicate satellite? Name the communication satellites of India. |
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Answer» A communication satellite is that satellite which can provide a communication link between two stations on earth at large distance apart. A geostationary satellite having electronic equipments by which the signals may be received, amplified and transmitted back to the earth can act as a communication satellite. The communication satellites of India are INSAT 2 B and INSAT 2 C. |
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| 184. |
What is the range of frequencies used for TV transmission? What is common between these waves and light waves? |
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Answer» Range of frequency used for TV transmission is `54 MHz "to" 890 MHz`. (VHF and UHF). These waves and light waves are electromagnetic waves. The ionosphere is unable to reflect back these waves to earth. |
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| 185. |
An electron oscillating with a frequency of `3xx10^6` Hz, would generate -A. X - raysB. ultraviolet raysC. radio wavesD. microwaves |
| Answer» Correct Answer - C | |
| 186. |
Temperature of troposphere decreases ….. `^(@)C//k`.m.A. `100`B. `50`C. `10`D. 6 |
| Answer» Correct Answer - D | |
| 187. |
Troposphere is used in the propagation ofA. Ground wavesB. Sky wavesC. Space wavesD. Sound waves |
| Answer» Correct Answer - A | |
| 188. |
Troposphere reflect the waves having frequency from ……A. 100 MHz to 200 MHzB. 500 KHz to 1500 KHzC. 0 Hz to 20 kHzD. 20 Hz to 20 kHz |
| Answer» Correct Answer - B | |
| 189. |
The distance between consecutive maxima and minima is given by-A. `lamda//2`B. `2lamda`C. `lamda`D. `lamda//4` |
| Answer» Correct Answer - D | |
| 190. |
An antenna behaves as resonant circuit only when its length isA. `(lamda)/(2)`B. `(lamda)/(4)`C. `lamda`D. `(lamda)/(2)` or integral multiple of `(lamda)/(2)` |
| Answer» Correct Answer - D | |
| 191. |
Long distance short-wave radio broadcasting usesA. ground waveB. lonospheric waveC. direct waveD. sky wave |
| Answer» Correct Answer - C | |
| 192. |
Which of the two - AM or FM is preferred for high fidelity reception? |
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Answer» Frequency modulation is preferred for high fidelity reception. |
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| 193. |
Give an expression for band-width in AM transmission. |
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Answer» In amplitude modulation, band width = ` 2 xx max`. frequency of audio signal. |
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| 194. |
What is the expressions for band width in FM transmission? |
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Answer» In frequency modulation, band width `= 2n xx ` frequency of audio modulating signal, where n is the number of significant side bands. |
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| 195. |
Perhaps, the most commonly used mode of communication involves use of free space as a communication channel between transmitter and receiver. The various possible modes of using of free space have been described below Ground wave communication : This mode is used for low frequency as at high frequency, losses become large. This is useful for distance of the order of 1 km. Space wave communication: The signals are transmitted directly from transmitter to receiver. There is no limit to upper frequency, but its range is limited to few hundred kilometers. The range depends on the height of transmitting antenna Sky wave communication : The signals are sent towards sky They are reflected back towards earth by ionosphere. Its range can be above 500 km. But, very high frequencies (above 10 MHz) are not reflected back. Satellite communication: The signals that cannot be reflected back by ionosphere are reflected back by satellite. Moon is a passive satellite of earth. The cordless phone installed in a house usesA. Ground wave communicationB. Space wave communicationC. Sky wave communicationD. Satellite communication |
| Answer» Correct Answer - A | |
| 196. |
Perhaps, the most commonly used mode of communication involves use of free space as a communication channel between transmitter and receiver. The various possible modes of using of free space have been described below Ground wave communication : This mode is used for low frequency as at high frequency, losses become large. This is useful for distance of the order of 1 km. Space wave communication: The signals are transmitted directly from transmitter to receiver. There is no limit to upper frequency, but its range is limited to few hundred kilometers. The range depends on the height of transmitting antenna Sky wave communication : The signals are sent towards sky They are reflected back towards earth by ionosphere. Its range can be above 500 km. But, very high frequencies (above 10 MHz) are not reflected back. Satellite communication: The signals that cannot be reflected back by ionosphere are reflected back by satellite. Moon is a passive satellite of earth. The frequency range allotted to FM radio broadcast is 88 MHz to 108 MHz. Which of the following communication channels can be used for its transmission?A. Sky wave communication onlyB. Space wave communication onlyC. Satellite communication onlyD. Either space wave communication or satellite communication |
| Answer» Correct Answer - D | |
| 197. |
Perhaps, the most commonly used mode of communication involves use of free space as a communication channel between transmitter and receiver. The various possible modes of using of free space have been described below Ground wave communication : This mode is used for low frequency as at high frequency, losses become large. This is useful for distance of the order of 1 km. Space wave communication: The signals are transmitted directly from transmitter to receiver. There is no limit to upper frequency, but its range is limited to few hundred kilometers. The range depends on the height of transmitting antenna Sky wave communication : The signals are sent towards sky They are reflected back towards earth by ionosphere. Its range can be above 500 km. But, very high frequencies (above 10 MHz) are not reflected back. Satellite communication: The signals that cannot be reflected back by ionosphere are reflected back by satellite. Moon is a passive satellite of earth. A receiver receives a signal of frequency 108 MHz from a transmitter 300 km away. It must be coming viaA. Ground wave communicationB. Space wave communicationC. Sky wave communicationD. Satellite communication |
| Answer» Correct Answer - D | |
| 198. |
What band width is required for transmission of T.V. signals? |
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Answer» For transmission of T.V. signals, band width required is 6 MHz. |
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| 199. |
A device X can convert one form to energy into another. Another device Y can be regarded as a combination of transmitter and receiver. Name of the devices X and Y. |
| Answer» X is transducer and y is a repeater. | |
| 200. |
The band width required for transmiting video signal is |
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Answer» 4.2 MHz is the band width for transmission of video signals. |
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