Saved Bookmarks
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.
| 301. |
The highest frequency of radio waves which the when sent at some angle towards the ionosphere , gets reflected from ionosphere and returns to earth is called ……. . |
| Answer» Correct Answer - maximum usable frequency (MUF) | |
| 302. |
The ionosphere is used for the propagation ofA. sky wavesB. space wavesC. ground wavesD. sound waves |
| Answer» Correct Answer - A | |
| 303. |
Because of tilting, which waves finally disappear?A. Space wavesB. MicrowavesC. Sky wavesD. Surface waves |
| Answer» Correct Answer - D | |
| 304. |
The Ionosphere reflects theA. ground wavesB. sky wavesC. space wavesD. very high or ultra high frequency waves |
| Answer» Correct Answer - B | |
| 305. |
What is the factor which limits the covering range of T.V. transmission ? |
|
Answer» The T.V. Transmission by TV towers uses space wave. The covering range of space wave is limited by the curvature of the earth. |
|
| 306. |
When microwave signals follow the curvature of earth, this is known as:A. ionosphere reflectionB. ductingC. critical velocityD. modulation |
| Answer» Correct Answer - B | |
| 307. |
Electromagnetic waves of frequency ……. are reflected from ionosphere.A. 100 MHzB. 2 MHz to 30 MHzC. upto `1.5` MHzD. less than `1.5` MHz |
| Answer» Correct Answer - B | |
| 308. |
Name the two types of orbits (other than the geostationary orbit) used in satellite communication. |
| Answer» (i) Polar orbit (ii) Highly elliptical orbit. | |
| 309. |
At night time. the following ionospheric layers do not exist.A. D and `F_2`B. E and `F_1`C. D and `F_1`D. D and E |
| Answer» Correct Answer - C | |
| 310. |
At daytime, the ionosphere oonsists of the following layersA. 1. D,E and `F_1`B. 2. D, `F_1` and `F_2`C. 3. D,E,`F_1` and `F_2`D. 4. E,`F_1` and `F_2` |
| Answer» Correct Answer - C | |
| 311. |
What fraction of the surface area of earth can be covered to establish communication by one geostationary satellite ?A. 1B. 0.5C. 0.33D. 0.2 |
| Answer» Correct Answer - C | |
| 312. |
Why can moon be not used as a communication satellite? |
|
Answer» Moon is a natural satellite of earth, but it is not an active satellite as it carries no electornic equipments for receiving. Amplifying and transmitting the signal back to earth. Moon can not be used as communication satellite because ltbr. (i) the distance between Earth and Moon is not proper (ii) the period of revolution of Moon is not 24 hours. (iii) the moon is not revolving in the equatorial plane of the earth. |
|
| 313. |
STATEMENT-1. The transmission using FM is more efficient as compared to AM STATEMENT-2: In FM, amplitude and therefore power handled is constantA. Statement-1 is True. Statement-2 is True Statement-2 is a correct explanation for statement-8B. Statement is True Statement-2 is True Statement-2 is NOT a correct explanation for Statement-8C. Statement-1 Is True Statement-2 is FalseD. Statement-1 is False. Statement 2 is True |
| Answer» Correct Answer - A | |
| 314. |
STATEMENT-1 : A photodiode is used in receivers in optical communication STATEMENT-2 The photodiode has a resistance that varies according to intensity of light incident on itA. Statement-1 is True. Statement-2 is True Statement-2 is a correct explanation for statement-9B. Statement is True Statement-2 is True Statement-2 is NOT a correct explanation for Statement-9C. Statement-1 Is True Statement-2 is FalseD. Statement-1 is False. Statement 2 is True |
| Answer» Correct Answer - B | |
| 315. |
Rohan, a Science student purchased a few cells from a shop keeper for his torch. He noticed that the torch was not working will with these new cells. He thought of checking the emf of each cell. Using a potentiometer he measured their emfs and noticed that the emf of each cell was less than the value of oblige him. He then lodged a complaint with consumer forum and got theredressal. Read the above passage and answer the following question: (i) What are the basic values displayed by Rohan? (ii) Why rohan used potentiometer instead of voltmeter to measure emf of the cells? (iii) If the lenght of potentiometer wire is (a) doubled and (b) halved, what will be the effect on the position of zero deflection in a potentiometer. |
|
Answer» (i) Rohan used a potentionmeter for checking the emf of cells. It shows that his knowledge in this respect is good. Instead of fighting with the shopkeeper, he approached the consumer forum for redressel of his (ii) EMF is the maximum potential differnce potential difference between two electrodes two electrodes of a cell in open circuit. while measuring the emf of a cell using potentiometer method, a potentiometer does not draw any current from the cell whose emf is to be determined, whereas a voltmeter always draws some little current. Therefore, emf measured by voltmeter is slightly less than actual value of emf measured by potentionmeter. (iii) (a) when the lenght of wire is doubled, the potential gradient across the potentiometer wire will decrease. Now the position of zero deflection will occur at longer lenght. (b) When lenght of wire is halved, the reverse will true. |
|
| 316. |
In which frequency range, space waves are normally propagated?A. HFB. VHFC. UHFD. SHF |
| Answer» Correct Answer - C | |
| 317. |
The frequency of signals up to ` 1.5` MHz can be propagated usingA. ground wave propagationB. space wave propagationC. sky wave propagationD. satellite communication |
| Answer» Correct Answer - A | |
| 318. |
Broadcasting antennas are generallyA. omnidirectional typeB. vertical typeC. horizontal typeD. none of these |
| Answer» Correct Answer - B | |
| 319. |
Give the reason why transmission of T.V. signals via sky waves is not possible. |
|
Answer» The T.V. signals lie in the frequency range of 54 MHz to 890 Mhz, which cannot be reflected by ionosphere. That is why the transmission of T.V. signals is not possible via sky waves. |
|
| 320. |
What are T.V. signals? How can they be received? |
|
Answer» T.V. signals are those signals which have frequency range from 54 MHz to 890 MHz. Since these waves neither follow the curvature of the earth nor get reflected by ionosphere, their reception is possible either by using communication geostationary satellite which reflects the T.V. signals back to earth or by using tall antennas which may directly intercept the signals. |
|
| 321. |
Antennas are used to receive and transmitA. mechanical wavesB. electromagnetic wavesC. ultrasonic wavesD. de Broglie waves |
| Answer» Correct Answer - B | |
| 322. |
Assertion: Only microwaves are used in radar. Reason: Because microwaves have very small wavelength. |
|
Answer» In a radar, a beam signals is needed in particular direction which is possible if wavelength of signal wave is very small. Since the wavelength of microwaves is a few millimetre, hence they are used in radar. |
|
| 323. |
Microwaves are the electromagnetic waves with frequency, in the range ofA. Micro hertzB. Giga hertzC. Mega hertzD. Hertz |
| Answer» Correct Answer - B | |
| 324. |
Explain that microwaves are better carriers of signals than radio waves? |
|
Answer» Microwaves are the electromagnetic waves of wavelength of the order of a few millimeters, which is less than those of T.V. signals. On account of smaller wavelength, the microwaves can be transmitted as beam signals in a particular direction, much better than radiowaves because microwaves do not spread or bend around the corners of any obstacle coming in their way. Therefore, microwaves are better carriers of signals than radio waves. |
|
| 325. |
In communication system, the process of superimposing a low frequency signal on a high frequency wave is known asA. RepeaterB. AttenuationC. ModulationD. Demodulation |
|
Answer» Correct Answer - C This process is known as modulation |
|
| 326. |
The sky wave propagation is suitable for radiowaves of frequencyA. upto 2 MHzB. from 2 MHz to 20 MHzC. from 2 MHz to 30 mHzD. none of these |
| Answer» Correct Answer - C | |
| 327. |
Statement-1: Television signals are received through skywave propagation. Statement-2: The ionosphere reflects electromagnetic waves of frequencies greater than a certain critical requency.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 | |
| 328. |
When electromagnetic waves enter the ionised layer of ionosphere, then the relative permittivity i.e. dielectric constant of the ionised layerA. does not changeB. appears to increaseC. appears to decreaseD. sometimes appears to increase and sometimes to decrease |
| Answer» Correct Answer - C | |
| 329. |
A sinusoidal carrier voltage of 120 V is amplitude modulated by a sinusoidal voltage of frequency 10KHz. What is the modulation index if the maximum modulated carrier amplitude is 150 V ?A. A. 0.2B. B. 0.25C. C. 0.3D. D. 0.45 |
|
Answer» Correct Answer - B `A_C`=120 V , `A_"max"`= 150 `therefore A_m=A_"max"-A_C=150-120`= 30 V `therefore mu=A_m/A_C=30/120`=0.25 or 25% |
|
| 330. |
Television signal on earth cannot be recevied at distances greater than `100 km` from the transmission station. The reasion behind this is thatA. the receiver antenna is unable to detect the signal at a distance greater then 100 kmB. the TV programme consists of both audio and video signalsC. The TV signals are less powerful than radio signals.D. the surface of earth is curved like a sphere |
| Answer» Correct Answer - D | |
| 331. |
Television signal on earth cannot be recevied at distances greater than `100 km` from the transmission station. The reasion behind this is thatA. the receiver antenna is unable to detect the signal at a distance greater then 100 kmB. the TV programme consists of both audio and video signalsC. Te TV signals are less powerful than radio signalsD. the surface of earth is curved like a sphere |
| Answer» Correct Answer - D | |
| 332. |
A tansmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode ? Given radius of earth is `6.4 xx 10^(6) m`.A. 45.5 kmB. 40.5 kmC. 35 kmD. 32 km |
|
Answer» Correct Answer - A `h_T`=32 m , `h_"Receiver"`=100 m `d_"max"=sqrt(2Rh_T) + sqrt(2Rh_"Receiver")` `=sqrt(2xx64xx10^5xx32)+sqrt(2xx64xx10^5xx50)` `=64xx10^2xxsqrt10+8xx10^3xxsqrt10` `=144xx10^2 sqrt10` =`45.5 xx 10^3` = 45.5 Km |
|
| 333. |
Why should transmitters broadcasting programmes use different carrier frequencies? |
|
Answer» If all transmitters were to use same carrier frequency, it will not be possible to tune the receiver to a particular transmission. As frequency range of all message signals are identical, they will get mixed up almost inseparably. |
|
| 334. |
What is a transducer? Give two examples. |
|
Answer» A transducer is a device, which converts one form of energy into another. A mocrophone at the transmitting station and a loudspeaker at the receiving station are two examples of transducers. Infact a transducer provides output in electrical form or it has input in electrical form. |
|
| 335. |
What are the limitaions of amplitude modulation? |
|
Answer» (i) Efficiency of AM transmission is low. (ii) Amplitude modulation suffers from noise. (iii) Quality of audio signal is often poor. |
|
| 336. |
What are the three frequencies in an amplitude modulated wave? What are LSB and USB? |
|
Answer» In an amplitude modulated wave, the three frequencies are `v_c, v_c-v_m:v_c+v_m`. LSB stands for lower side band of frequency `(v_c-v_m)`. USB stands for upper side band of frequency `(v_c + v_m)`. |
|
| 337. |
What type of modulation is required for television broadcasts? |
|
Answer» Amplitude modulation for video signals and frequency modulation for voice signals. |
|
| 338. |
A `60^@` prism has a refractive index of `1.5`. Calculate (a) the angle of incidence for minimum deviation (b) angle of minimum deviation ( c) the angle of emergence of light at maximum deviation (d) angle of maximum deviation. |
|
Answer» Here, `A=60^(@), mu=1.5` When deviation is minimum , `r=(A)/(2)=(60)/(2)=30^(@)` As `mu=(sin i)/(sin i)` `sini=mu sin r=1.5 sin 30^(@)=0.75` `i=sin^(-1)(0.75)=49^(@)` (b) `del_(m)=2i-A=2xx49^(@)-60^(@)=38^(@)` (c) For maximum deviation, `i_(1)=90^(@)` and `r_(1)=C=sin^(=1)((1)/(mu))=sin^(-1)((1)/(3//2))=sin^(-1)(0.6667)=42^(@)` As `r_(1)+r_(2)=A` `r_(2)=A-r_(1)=60-42^(@)=18^(@)` From `mu=(sini_(2))/(sinr_(2)),` `sini_(2)=mu sin r_(2)=1.5 sin 18^(@)=1.5xx0.31=0.4650` `i_(2)=sin^(-1)(0.4650)=28^(@)` This is the angle of emergent (d) `del_("max")=i_(1)+i_(2)-A=90^(@)+28^(@)-60^(@)=58^(@)` |
|
| 339. |
Two wires of resistance `R_(1)` and `R_(2)` have temperature coefficient of resistance `alpha_(1)` and `alpha_(2)` respectively. These are joined in series. The effective temperature coefficient of resistance is |
|
Answer» Let `R_(1),R_(2)` be the resistances of two wires at `0^(@)C`. `R_(t_(1)),R_(t_(2))` be the resistance of two wires at `t^(@)C`. Then `R_(t_(1))=R_(1)(1+alpha_(1)t)` and `R_(1_(2))=R_(2)(1+alpha_(2)t)` When wires are in series, then total resistance is `R_(t_(s))=R_(t_(1))+R_(t_(2))=R_(1)(1+alpha_(1)t)+R_(2)(1+alpha_(2)l)=(R_(1)+R_(2))+(R_(1)alpha_(1)+R_(2)alpha_(2))t` `=R_(s)+(R_(1)alpha_(1)+R_(2)alpha_(2))t` or `R_(t_(s))-T_(s)=(R_(1)alpha_(1)+R_(2)alpha_(2))t` where `R_(1)+R_(2)=R_(s)=` effective resistance of two wires in series at `0^(@)C` If `alpha_(s)` is the effective temperature coefficient of resistance of the two wires in series, then `alpha_(s)=(R_(t_(s))-T_(s))/(R_(s)xxt)=((R_(1)alpha_(1)+R_(2)alpha_(2))t)/((R_(1)+R_(2))t)=(R_(1)alpha_(1)+R_(2)alpha_(2))/(R_(1)+R_(2))` |
|
| 340. |
A digital signal-A. is less reliable than analog signalB. is more reliable than analog signalC. is equally reliable as the analog signalD. none of above |
| Answer» Correct Answer - B | |
| 341. |
The message signal is usually of:A. audio frequency rangeB. radio frequency rangeC. audio or radio frequency rangeD. mixture of both |
| Answer» Correct Answer - A | |
| 342. |
Modulation is the phenomenon of :A. superimposing the audio frequency signal over a carrier waveB. separting the audio frequency signal from the carrier waveC. separting carrier wave from the modulated waveD. any of (1), (2), (3) above |
| Answer» Correct Answer - A | |
| 343. |
Give one example each of a system that uses the (i) sky wave (ii) Space wave mode of propagation. |
|
Answer» (i) Short wave broadcast. (ii) Television transmission. |
|
| 344. |
The phenomenon of loss of power of a ground wave is called …….. |
| Answer» Correct Answer - attenuation | |
| 345. |
A mode of wave propagation is also known as ……… |
| Answer» Correct Answer - ground or surface wave propagation | |
| 346. |
Line of slight propagation is also called ______propagationA. sky wavesB. Ground waveC. Sound waveD. Space wave |
| Answer» Correct Answer - D | |
| 347. |
Line of sight propagation isA. ground wave propagationB. sky wave propagationC. microwave propagationD. none of these |
| Answer» Correct Answer - C | |
| 348. |
A sinusoidal carrier wave of amplitude 40 V is amplitude modulated by a sinusoidal signal voltage. What is the amplitude of each side band if the modulation index is 25%.A. A. 5 VB. B. 4 VC. C. 6 VD. D. 8 V |
|
Answer» Correct Answer - A `A_C` = 40 V and `mu`=25% = 0.25 `therefore` Amplitude of each side band `=(muA_C)/2=(0.25xx40)/2`=5 V |
|
| 349. |
Write two factors justifying the need of modulating a signal. A carrier wave of peak voltage `12 V` is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75% ? |
|
Answer» Modulation of a signal is necessary to transmit a signal in the audio frequency range over a long distance on account of (i) suitable size of the antenna, (ii) suitable effective power radiated by the antenna , (iii) avoiding mixing up of signals from different transmitters. here, `A_c = 12 V, A_m = ?` `mu = 75% = 0.75` As ` mu = A_m/A_c` `:. A_m = muA_c = 0.75 xx 12 = 9V. ` |
|
| 350. |
Frequencies higher than 10 Mhz are found not to be reflected by the ionosphere on a particular day at a place. Calculate the maximum electron density of the ionosphere. |
|
Answer» Here, `v_c = 10 MHz = 10^(7) Hz,` `N_(max) = ?` Now, `v_c = 9 (N_(max))^(1//2)` or `N_(max) = v_(c)^(2)/ 81 = (10^7)^(2)/81 = 1.23 xx 10^12 m^(-3)` |
|