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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. |
The ratio of `omega//k` for a travelling wave (where `omega` is the angular frequency and k is the angular wave number) is ......... |
| Answer» Correct Answer - speed | |
| 202. |
The orderly distrubution of electromagnetic radiations according to their wavelength or frequency is called the ............ |
| Answer» Correct Answer - electromagnetic spectrum | |
| 203. |
A plane electromagnetic wave of wave intensity `6 W//m^(2)` strikes a small mirror of area `9 cm^(2)`, held perpendicular to the approaching wave. The momentum transferred in kg-ms by the wave to the mirror each second will be:A. `1.2xx10^(-10)`B. `2.4xx10^(-9)`C. `3.6xx10^(-8)`D. `4.8xx10^(-7)` |
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Answer» Correct Answer - a Transferred momentum/secound to mirror is `P=(2AS)/(c)=1.21xx10_(10) kg ms ^(1)` |
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| 204. |
The bit rate for a signal, which has a sampling rate of `8 kHz` and whare `16` quantisation levels have been used isA. 32000 bits/secB. 16000 bits/secC. 64000 bits/secD. 72000 bits/sec |
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Answer» Correct Answer - a If n is the number of bits per sample, then number of quantisation level`=2^(n)` Since the number of quantisation level is `16` `implies 2^(n)=16impliesn=4` `:.` bit rate = sampling rate x no. of bits sample `=8000xx4=32,000 " bits"//"sec"`. |
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| 205. |
In previous question, the radiation froce on the roof will be:A. `8.53xx10^(-5) N`B. `2.3xx10^(-3) N`C. `1.33xx10^(-3) N`D. `5.33xx10^(-4) N` |
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Answer» Correct Answer - d Force of radiation=`("power")/(c)` `=(1.6xx10^(5))/(3xx10^(8))=5.32xx10^(-4) N` |
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| 206. |
Which of the following shows greenhouse effect?A. Ultraviolet raysB. Infrared raysC. `X`-raysD. None of these |
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Answer» Correct Answer - b Infrared radiations reflected by low lying clouds and keeps the earth warm. |
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| 207. |
Radiation of intensity `1 W//m^(2)` are striking on a metal plate. The pressure on the plate isA. `0.66 xx 10^(-8) N//m^(2)`B. `0.25 N//m^(2)`C. `2 N//m^(2)`D. `4 N//m^(2)` |
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Answer» Correct Answer - A As metal is reflecting surface and for reflecting surface radiation pressure, `p_(r) = (2S)/(c) = (2 xx 1)/(3 xx 10^(8)) = 0.667 xx 10^(-8) Nm^(-2)` |
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| 208. |
Radiations of intensity `0.5 W//m^(2)` are striking a metal plate. The pressure on the plate isA. `0.166xx10^(-8)"N m"^(-2)`B. `0.332xx10^(-8)"N m"^(-2)`C. `0.111xx10^(-8)"N m"^(-2)`D. `0.083xx10^(-8)"N m"^(-2)` |
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Answer» Correct Answer - A `P=(I)/(c)=(0.5)/(3xx10^(8))=0.166xx10^(-8)"N m"^(-2)` |
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| 209. |
A parallel plate capacitor is charged to `60 muC`. Due to a radioactive source, the plate losses charge at the rate of `1.8 xx 10^(-8) Cs^(-1)`. The magnitued of displacement current isA. `1.8 xx 10^(-8) Cs^(-1)`B. `3.6 xx 10^(-8) Cs^(-1)`C. `4.1 xx 10^(-11) Cs^(-1)`D. `5.7 xx 10^(-12) Cs^(-1)` |
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Answer» Correct Answer - A Displacement current is given by `I_(D) = (dq)/(dt) = 1.8 xx 10^(-5) Cs^(-1)` |
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| 210. |
Speed of electromagnetic waves is the sameA. for all wavelengthsB. for all intensitiesC. for all frequenciesD. in all media |
| Answer» Correct Answer - b | |
| 211. |
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the direction of electric and magnetic field vectors? |
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Answer» In e.m. wave the elctric and magnetic field vectors are perpendicular to each other and also perpendicular to the direction of propagation of e.m. wave. As plane e.m. wave is propagating along z-axis, so the electric field vector is along x-axis and magnetic field vector is along y-axis. |
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| 212. |
A plane electromagnetic wave travels in vacuum along z- direction. What can you say about the directions of ots electric and magnetic field vectors. If the frequency of the wave is `30 MHz`, what is its wavelength? |
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Answer» The electric and magnetic field vector `vecE and vecB` must be perpendicular to each other as well as perpendicular to the direction of propagation of e.m. wave. Thus `vecE and vecB` must lie in x-y plane and are mutually perpendicular to each other. wavelength, `lambda=c//v=3xx10^8//30xx10^6=10m` |
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| 213. |
The figure here gives the electric field of an EM wave at a certain point and a certain. The wave is transporting energy in the negative z direction. What is the direction of the magnetic field of the wave at the point and instant? A. Towards + X directionB. Towards - X directionC. Towards + directionD. Towards - Z direction |
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Answer» Correct Answer - a The direction of EM wave is fiven by the direction of `vecExxvecB`. |
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| 214. |
The velocity of electromagnetic radiatior in a medium of permittivity `epsilon_(0)` and permeability `mu_(0)` is given byA. `(1)/(sqrt(mu_(0)K_(0))`B. `(1)/(mu_(0)K_(0))`C. `sqrt(mu_(0)K_(0))`D. `mu_(0)K_(0)` |
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Answer» Correct Answer - c `mu=(c)/(v)` Now `c=(1)/(sqrt(mu_(0)inK_(0)))` and `v=(1)/(sqrt(mu_(0)K_(0)mu_(r)K_(r)))` `:. mu=sqrt(mu_(r)K_(r))` But `mu_(r)=u_(0)` and `K_(r)=K_(0)` `:.mu=sqrt(mu_(0)K_(0))` |
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| 215. |
The velocity of electromagnetic radiatior in a medium of permittivity `epsilon_(0)` and permeability `mu_(0)` is given byA. `sqrt(mu//E)`B. `sqrt(muepsi)`C. `1//sqrt(muepsi)`D. `sqrt(epsi//mu)` |
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Answer» Correct Answer - C For a medium with permittivity `epsi` and permeability `mu`, velocity of light is given by `c = (1)/(sqrt(muepsi))` |
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| 216. |
Which of the following wavelenght falls in `X`-rays region?A. `1Å`B. `10Å`C. `10^(-2)Å`D. `10^(-3)Å` |
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Answer» Correct Answer - a `X`-rays have wavelength of order of `1Å`. |
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| 217. |
The percentage power of `X`-ray increases with the increases in its:A. velocityB. intensityC. frequencyD. wavelength |
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Answer» Correct Answer - c `E=hv :. E prop v` |
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| 218. |
The range of wavelength of the visible light isA. `10Å` to `100Å`B. `4,000Å` to `8,000Å`C. `8,000Å` to `10,000Å`D. `10,000Å` to `15,000Å` |
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Answer» Correct Answer - b wavelength causes heating effect. |
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| 219. |
If the intensity of the incident radiowave of `1 "watt"//m^2` is reflected by the surface, find the pressure exerted on the surface. |
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Answer» Pressure exerted by reflected wave on the surface is, `P=(2I)/c=(2xx1)/(3xx10^8)=6.67xx10^-9N//m^2` |
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| 220. |
If a 100 watt bulb acts as a point source and has 2.5% efficiency, the intensity of this bulb at 3m isA. `2.9Wm^-1`B. `5.8Wm^-1`C. `1.5Wm^-1`D. `4.4Wm^-1` |
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Answer» Correct Answer - a In em wave, half of the intensity is due to electric field vector and half is due to magnetic field vector. Therefore, intensity due to electric field vector in e.m. wave is `=I/2=1/2 in_0 E_(rms)^2 c=1/2xx(0.022W//m^2)` or `E_(rms)= sqrt((0.022)/(in_(0)C)) =sqrt(0.022/(8.85xx10^(-12)xx3xx10^8))` `=2.9 V//m` |
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| 221. |
If a 100 watt bulb acts as a point source and has 2.5% efficiency, the intensity of this bulb at 3m isA. `0.002Wm^-2`B. `0.22Wm^-2`C. `4.4Wm^-2`D. `0.033Wm^-2` |
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Answer» Correct Answer - a Here, `P=100W, r=3m` Intensity, `I=("energy/time")/("area")=("Power")/(" area")` `=(100xx2.5//100)/(4pi(3)^2)=0.022W//m^2` |
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| 222. |
If the intensity of the incident radiowave is `1 "watt"//m^2` is absorbed by the surface, find the pressure exerted on the surface. |
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Answer» Pressure exerted by absorbed wave on the surface is, `P=I/c=2/(3xx10^8)=6.67xx10^-9N//m^2` |
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| 223. |
The frequency of `e.m` wave which is best suit to observe a particle of radius `3xx10^(-4)` is of order of:A. `10^(15)`B. `10^(14)`C. `10^(13)`D. `10^(12)` |
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Answer» Correct Answer - b Radius of particle is `lambda` `:.v=(c)/(lambda)=(3xx10^(8))/(3xx10^(-6))=10^(14)` For observation of particle the wave frequency should be more than `10^(4) Hz`, or wave length should be smaller. |
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| 224. |
Arrange the following electromagnetic radiations per quantum in the order of increasing energy: A Blue light B: Yellow light C:X-rays D:RadiowaveA. D,B,A,CB. A,B,D,CC. C,A,B,DD. B,A,D,C |
| Answer» Correct Answer - a | |
| 225. |
The energy of `X`-rays photon is `3.3xx10^(-16)` J.Its frequency is:A. `2xx10^(19)Hz`B. `5xx10^(18)Hz`C. `5xx10^(17)Hz`D. `5xx10^(16)Hz` |
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Answer» Correct Answer - c `E=hv` `V=E//h=5xx10^(17)Hz` |
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| 226. |
The wave associated with `2.7K` belong to:A. radio wavesB. micro wavesC. ultraviolet waysD. infrared waves |
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Answer» Correct Answer - b `lambda_(m)T=constant` `implieslambda_(m)xx(2.7K)=0.2896 cmK` `:. lambda_(m)=0.2896/(2.7)m` `=0.10cm=1mm` |
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| 227. |
A paprallel plate capacitor of capacitance `C=0.1 mu F` is connected across on a.c. source of (angular) frequency `500 "rad"s^(-1).` the value fo conduction current is `1m A.` What is the rms value of the voltage from the source ? What is the displacement current across the capacitor plates? |
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Answer» Correct Answer - A::B |
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| 228. |
If `E` and `B` represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of eletromagnetic wave is along.A. `vecE`B. `vecB`C. `vecExxvecB`D. None of these |
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Answer» Correct Answer - c `E.M`.wave travel with perpendicular to`E` and `B` which are also perpendicular to each other, `vecv =vecExxvecB` |
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| 229. |
What are the basic sources of electromagnetic waves? |
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Answer» The basic source of electromagnetic waves is the time varying electric field produces magnetic field and vice versa. |
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| 230. |
Name the scientists connected with history of electromagnetic wave. |
| Answer» Maxwell, Hertz, Bose and Marconi. | |
| 231. |
`X`-rays are produced by jumping of:A. electrons from lower to higher energy will of atomB. electrons from higher lower energy orbit of atomC. proton from lower to higher energy orbit of nuclenusD. proton from higher to lower energy orbit of nucleus |
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Answer» Correct Answer - b Electrons from higher energy jump to lower energy state. |
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| 232. |
About `6%` of the power of a 100 W light bulb is converted ot visible radiation. The average intensity of visible radiation at a distance of 8 m is (Assume that the radiation is emitted isotropically and neglect reflection.)A. `3.5xx10^(-3)"W m"^(-2)`B. `5.1xx10^(-3)"W m"^(-2)`C. `7.2xx10^(-3)"W m"^(-2)`D. `2.3xx10^(-3)"W m"^(-2)` |
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Answer» Correct Answer - C Here , power of bulb `=100 W` As intensity, `I = ("Power of visible light")/("area")` `= (100 xx 6//100)/(4pi(8)^(2)) = 7.2 xx 10^(-3) W m^(-2)` |
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| 233. |
The pressure exerted by an electromagnetic wave of intensity `I (Wm^(-2))` on a non-reflecting surface is ( c is the velocity of light)A. `Ic`B. `Ic^(2)`C. `I//c`D. `I//c^(2)` |
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Answer» Correct Answer - C When a surface intercepts electromagenetic radiation, a force and a pressure are exerted on the surface. As the surface is non-reflecting , so it is completely abosorbed and in such case the force is `F = (IA)/(c)` The radiation pressure is the force per unit area, `P = (F)/(A) = (I)/(c)` |
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| 234. |
A plane `E.M.` wave of frequency `40 MHz` travels along `X`-axis. At same point at same instant, the electric field `E` has maximum value of `750 N//C` in `Y`- direction. The magnitude and direction of magnetic field is |
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Answer» (i) Wavelength of the wave, `lambda = (c)/(v) = (3.0xx10^(8))/(40 xx 10^(6)) = 7.5 m` (ii) Given, maximum value of electric field, `E_(0) = 750 NC^(-1)` `:.` Magnetic field, `B_(0) = (E_(0))/(c) = (750)/(3xx10^(8)) = 2.5 xx 10^(-6) T` Since `E` and `B` are mutally perpendicular and they both are perpendicular to the propagation of wave. Thus, we concluded that magnetic field is in negative z-direction. |
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| 235. |
An L-C resonant circuit contains a `100 muF` capacitor an a `400 muH` inductor. It is not set into oscillation coupled to an antenna. Find the frequency of the radiated electromagnetic waves. |
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Answer» Given, `C = 100 pF = 100 xx 10^(-12) F, L = 400 muH` ` L = 400 xx 100^(-6) H` Frequency of radiated EM wave is `v = (1)/(2pisqrt(LC)) = (1)/(2 xx 3.14sqrt(400 xx 10^(-6) xx 100 xx 10^(-12)))` `= 0.0796 xx 10^(7) Hz` |
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| 236. |
An `L-c` resonant circuit contains a `200 pF` capacitor and a `200muH` inductor. It is into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves isA. `377 mm`B. `377 m`C. `377 cm`D. `3.77 cm` |
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Answer» Correct Answer - B Frequency of `L-C` oscilaltion is `v = (1)/(2pisqrt(LC)) = (1)/(2 xx 31.4sqrt(200 xx 10^(-6) xx 200 xx 10^(-12)))` `= 0.0796 xx 10^(7) Hz` So, wavelength `lambda = (c)/(v) = (3 xx 10^(8))/(0.0796 xx 10^(7)) = 377 m` |
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| 237. |
Which of the following statement is false for the properties of electromagnetic waves?A. Both electric and magnetic field vectors attain the maxima and minima at the same place and same timeB. The energy in electromagnetic wave is divided equally between electric and magnetic vectorsC. Both electric and magnetic filed vector are parallel to each other and perpendicular to the direction of propagation of waveD. These waves do not require any material medium for propagation |
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Answer» Correct Answer - C The time varying electric and magnetic fields are mutually perpendicular to each and also perpedicular to the direaction of propagation of the wave. |
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| 238. |
State any four properties of electromagnetic waves. |
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Answer» The electromagnetic waves consists of wide range of radiations, but all these radiations have the following common properties. 1. They are transverse in nature. (2) They do not require any material medium for propagation.(3) They travel with the same speed of `3xx10^8m//s` in vacuum.(4) They consists of mutually transverse varying electric and magnetic fields. |
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| 239. |
The transverse nature of electromagnetic waves is proved by which of the following?A. Interference phenomenaB. Diffraction phenomenaC. Dispersion phenomenaD. Polarisation phenomena |
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Answer» Correct Answer - D Phenomena of restricting the vibration of light in a particular direction is called polarisation, it helps to prove the transverse nature of elctromagnetic waves. |
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| 240. |
Assertion : The velocity of electromagnetic waves depends on electric and magnetic properties of the medium Reason : Velocity of electromagnetic waves in free space is constant.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - B `v=(1)/(sqrt(muepsilon))=(c)/(sqrt(mu_(r)epsilon_(r))` |
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| 241. |
The wavelength of the short radio waves, microwaves, ultraviolet waves are `lambda_(1),lambda_(2)` and `lambda_(3)`, respectively, Arrange them in decreasing order.A. `lambda_(1),lambda_(2),lambda_(3)`B. `lambda_(1),lambda_(3),lambda_(2)`C. `lambda_(3),lambda_(2),lambda_(1)`D. `lambda_(2),lambda_(1),lambda_(3)` |
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Answer» Correct Answer - A we know that , (a) Short radio waves i.e., wavelength range `= 1 xx 10^(-1) m` to `1 xx 10^(4) m` (ii) Microwaves wavelength range `= 1 xx 10^(-3) m` to `3 xx 10^(-1) m` (iii). Ultraviolet waves wavelength range `= 1 xx 10^(8) m` to `4 xx 10^(-8) m` So, decreasing order is `lambda_(1), lambda_(2), lambda_(3)`. |
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| 242. |
Assertion : Microwaves are heat waves which is used to heat up food in ovens. Reason : In ovens, microwaves heat up the vessel first and food inside gets heated up by the transfer of energy from the vessel.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - D Mecrowaves are not heat waves. Microwaves have frequency and energy smaller than visible light and wavelengths larger than it. In microwave ovens, the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the wave is transferred efficiently to the kinetic energy of the molecules. This raises the temperature of any food containing water. |
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| 243. |
Assertion : Microwaves are better carrier of signals than optical waves Reason : Microwaves move faster than optical waves.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - D The optical waves used in optical fibre communication are better carrier of signals than microwaves. The speed of microwave and optical wave is the same in vacuum. |
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| 244. |
The electric field associted with an electromagnetic wave in vacuum is given by `vecE=hati 40 cos(kz=6xx10^(8)t)`, when `E`, `z` and `t` are in volt/m metre and second respectivelyfind the wave vector.A. `2m^-1`B. `0.5m^-1`C. `6m^-1`D. `3m^-1` |
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Answer» Correct Answer - a Here, `E=40cos (kz-6xx10^8t)` Comparing it with the relation `E=E_0 cos .(2pi)/lambda(z-vt)=E_0cos ((2pi)/lambdaz-(2pi)/lambdavt)` we have `(2pi)/lambda=k` and `(2piv)/lambda=6xx10^8s^-1 or kv=6xx10^8` or `k=(6xx10^8)/v=(6xx10^8 s^-1)/(3xx10^8ms^-1)=2m^-1` |
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| 245. |
The electric field associted with an electromagnetic wave in vacuum is given by `vecE=hati 40 cos(kz=6xx10^(8)t)`, when `E`, `z` and `t` are in volt/m metre and second respectivelyfind the wave vector.A. `2m^(-1)`B. `0.5m^(-1)`C. `6m^(-1)`D. `3m^(-1)` |
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Answer» Correct Answer - A `vec(E) = hati 40 cos (kz - omegat)` `omega = 6xx10^(8) rad//sec` `k = (omega)/(c ) = (6 xx 10^(8))/(3xx10^(8)) = 2m^(-1)` |
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| 246. |
The electric field associated with an electromagnetic wave in vacuum is given by `vecE=40cos(kz-6xx10^(8)t)hati,` where E, z and t are in volt per meter, meter and second respectively. The value of wave vector k isA. 2 `m^(-1)`B. `0.5m^(-1)`C. `6m^(-1)`D. `3m^(-1)` |
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Answer» Correct Answer - A Compare the given equation with `E=E_(0)cos(kz-omegat)` We get, `omega=6xx10^(8)s^(-1)` `therefore" Wave vector, k"=(omega)/(c)=(6xx10^(8)s^(-1))/(3xx10^(8)"m s"^(-1))=2m^(-1)` |
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| 247. |
An e.m. wave travels in vacuum along z direction: `vecE=(E_1hati+E_2hatj)cos (kz-omega)`. Choose the correct option from the following :A. The associated magnetic field is given as `B=1/C (E_(1)hati -E_(2)hatj) " cos " (kz - omegat)`B. The associated magnetic field is given as `B=1/c (E_(1) hati -E_(2) hatj) " cos "(kz- omegat)`C. The given electromagnetic field is circulary polarisedD. The given electromagnetic wave is plane polarised |
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Answer» Correct Answer - D Here, in electromagnetic wave the electric field vector is given as , `E=(E_(1)hati +E_(2) hatj " cos " (kz -omegat)` in electromagnetic wave ,the associated magnetic field vector. `B =E/C = (E_(1)hati +E_(2)hatj)/(C)" cos "(kz-omegat)` Also E and B are perpendicular to each other and the propagation of electromagnetic wave is perpendicular to E as well as B so the given electromagnetic wave is plane polarised. |
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| 248. |
An e.m. wave trevels in vacuum along z direction: `vecE=(E_1hati+E_2hatj)cos (kz-omegat)`. Choose the correct option from the following :A. The associated magnetic field is given as `vecB=1/c(E_1hati+E_2hatj) cos (kz-omegat)`B. The associated magnetic field is given as `vecB=1/c(E_1hati-E_2hatj) cos (kz-omega)`C. The given electromagnetic field circular polarisedD. The given e.m. wave is plane polarised |
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Answer» Correct Answer - (a,d) In em wave, the electric field vector is given as, `vecE=(E_1hati+E_2hatj)cos (kz-omegat)` In em wave, the associated magnetic field vector, `vecB=(vecE)/c=(E_1hati+E_2hatj)/c cos (kz-omegat)` Thus, option (a) is correct. As `vecE and vecB` are perpendicular to each other and the propagation of em wave is `botr` to `vecE` as well as `vecB`, so the given em wave is plane polarised. Thus option (d) is true. |
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| 249. |
An electromagnetic wave of frequency `v=3.0 MHz` passes from vacuum into a dielectric medium with permittivity `epsilon=4.0`. ThenA. Wavelength is double and frequency become halfB. Wavelength is double and frequency becomes halfC. Wavelength is halved and frequency remains unchangedD. wavelengthand frequency remain unchanged |
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Answer» Correct Answer - c Refrance index `=sqrt((mu)/(mu_(0)epsilon_(0))` Here `mu` is not specified so we can consider `mu=mu_(0)` then refeactive index `=sqrt((epsilon)/(epsilon_(0)))=2` Hence, speed and wavelength of wave become half and frequency remain unchanged. |
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| 250. |
Electromagnetic waves travel in a medium which has relative permeability `1.3` and relative permittivity `2.14`. Then the speed of the electromagnetic wave in the medium will beA. `1.36xx10^(6)m//s`B. `1.8xx10^(2)m//s`C. `3.6xx10^(8)m//s`D. `1.8xx10^(8)m//s` |
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Answer» Correct Answer - d `v=(c)/(sqrt(mu_(e)epsilon_(r)))=(3xx10^(8))/(sqrt(1.3xx2.14))=1.8xx10^(8)m//sec` |
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