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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.
| 251. |
If `mu_0` be the permeability and `k_0` , the dielectric constant of a medium, its refractive index is given byA. `1/(sqrt(mu_0k_0))`B. `1/(mu_0k_0)`C. `sqrt(mu_0k_0)`D. `mu_0k_0` |
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Answer» Correct Answer - c Refractive index of medium is `mu=c/v, where c=1/(sqrt(mu_0k_0))` and `v=1/(sqrt(mu_0k_0mu_rk_r))` `:. mu=(1//sqrt(mu_0k_0))/(1//sqrt(mu_0k_0mu_rk_r))=sqrt(mu_rk_r)` Given `mu_r=mu_0 and k_r=k_0` then `mu=sqrt(mu_0k_0)` |
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| 252. |
Dimension of `(1)/(mu_(0)epsilon_(0))`, where symbols have usual meaning, areA. `L^-1T`B. `L^2T^2`C. `L^2T^-2`D. `LT^-1` |
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Answer» Correct Answer - c As `c=1/(sqrt(in_0mu_0)), c^2=1/(mu_0in_0),` so `[LT^-1]^2=1/(mu_0in_0) or [(mu_0in_0)]=L^2T^-2` |
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| 253. |
The amplitude of em wave in vacuum is doubled with no other changes made to wave. As a result of this doubling of the amplitude, which of the following statement is correct.A. The speed of the wave propagation chages onlyB. The frequency of the wave changes onlyC. The wavelength of the wave change onlyD. None of the above is true. |
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Answer» Correct Answer - d velocity of em wave `c=1/(sqrt(mu_0in_0))` `=3xx10^8 m//s` is independent of amplitude of e.m. wave, frequency and wavelength of e.m. wave. |
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| 254. |
The amplitude of em wave in vacuum is doubled with no other changes made to wave. As a result of this doubling of the amplitude, which of the following statement is correct.A. The speed of wave propagation changes onlyB. The frequency of the wave change onlyC. The wavelength of the wave changes onlyD. None of these |
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Answer» Correct Answer - D Velocity of electromagnetic wave `c=(1)/(sqrt(mu_(0)epsilon_(0)))=3xx10^(8)"m s"^(-1)` It is independent of amplitude, frequency and wavelength of electromagnetic wave. |
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| 255. |
Show that the radiation pressure exerted by an EM wave of intensity I on a surface kept in vacuum is `I//c`. |
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Answer» Pressure `=("Force")/("Area") =(F)/(A)` Force is the rate of change of momentum `"i.e., "" "F=(dp)/(dt)` `"Energy in time dt"" " U=p.C " or " p =U/c` `:. " ""Pressure" = 1/A .(U)/(C.dt)` `" prssure" =I/C " "[":." I = "Intensity" =(u)/(A.dt)]` |
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| 256. |
Is it necessary to use satellite for long distance T.V. transmission? Give reasons.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reasion are true but reason is not the correct explanation of assertion.C. If assertion is true but reasion is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - c The television signal being of high frequency are not reflected by the ionosphere. So the T.V. signals are broadcasted by tall antenna to get lager coverage, but for transmission over large distance satellites are needed. That is way, satellites are used for long distance T.V. transmission. |
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| 257. |
A `100 Omega` resistance and a capacitor of `100 Omega` reactance are connected in series across a `220` V source. When the capacitor is `50%` charged, the peak value of the displacement current isA. `4.4 A`B. `11sqrt(2) A`C. `2.2 A`D. `11 A` |
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Answer» Correct Answer - d `(i_(d))_(max)=(i_(c))_(max)=i_(0)(epsilon_(0))/(Z)=(200sqrt(2))/(sqrt(100^(2)+100^(2)))` `implies (i_(d))_(max)=(200sqrt(2))/(100sqrt(2))=2.24` As we are asked amplitude of displacement current. So, need not worry about charge on capacitor. |
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| 258. |
In an electric circuit, there is a capacitor of reactance `100Omega` connected across teh source of 220V. Find the displacement current. |
| Answer» Displacemnt current=conduction current. So, `I_D=V/(X_C)=220/100=2.2A` | |
| 259. |
Identify the electromagnetic waves whose wavelength very as: (a) `10^-12m lt lambda lt 10^-8m` (b) `10^-3m lt lambda lt 10^-1m`. Write one use each. |
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Answer» (a) X-rays, used in surgery and detective department (b) Microwave, used in radar communication and microwave oven. |
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| 260. |
Which of the following electromagnetic wave is used in high precision application like LASIK eye surgery?A. MicrowaveB. Ultraviolet raysC. Gamma raysD. X-rays |
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Answer» Correct Answer - B Due to its shorter wavelengths, ultraviolet radiations can be focussed in to very narrow beam for high precision application such as LASIK eye surgery. |
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| 261. |
Statement-1: Magnetic field lines cannot start from a point nor end at a point. Statement-2: The line integral of magnetic field induction over a closed path is not zero.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not correct explanation of Statement-1.C. Statement-1 is true, Statement -2 is false.D. Statement-1 is false, Statement -2 is true. |
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Answer» Correct Answer - c The magnetic lines of force form a closed path, hence the line integral of `vecB` over the closed path is zero. Thus, Statement-1 is true and Statement-2 is false. |
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| 262. |
Statement-1: Gamma rays are more energetic than X-rays. Statement-2: Gamma rays are of nuclear origin but x-rays are produced due to sudden deceleration of high energy electrons while falling on metal of high atomic number.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not correct explanation of Statement-1.C. Statement-1 is true, Statement -2 is false.D. Statement-1 is false, Statement -2 is true. |
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Answer» Correct Answer - b The Statement-1 and Statement-2 are true but the Statement-2 is not correct explanation of Statement-1. In face the energy of gamma ray is more than X-rays because the frequency of gamma ray is higher than that of X-rays as `E=hv`. |
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| 263. |
Statement-1: Out of the following radiations, microwave, ulrtaviolet and x-rays, microwaves has the shortest wavelength. Statement-2: The microwave do not deviate from the obstacle in their path while going form one location to another.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not correct explanation of Statement-1.C. Statement-1 is true, Statement -2 is false.D. Statement-1 is false, Statement -2 is true. |
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Answer» Correct Answer - d The energy of X-rays is maximum as compared microwave and ultraviolet rays. Since energy `E=hhv=(hc)/lambda`, so `lambda` is least for X-rays. Therefore, Statement-1 is wrong. The statement-2 is correct |
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| 264. |
The process of superimposing signal frequency (i.e. audio wave) on the carrier wave is known asA. TransmissionB. RecceptionC. ModulationD. Detection |
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Answer» Correct Answer - c Carrier+signal `to` modulation. |
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| 265. |
Indicate which one of the following system is digitalA. Pulse position modulationB. Pulse code modulationC. Pulse width modulationD. Pulse amplitude modulation |
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Answer» Correct Answer - b Pulse code modulation is a digital system. |
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| 266. |
In an FM system a `7kHz` signal modulates `108 MHz` carrier so that frequency deviation is `50 kHz`. The carrier swing isA. `7.143`B. `8`C. `0.71`D. `350` |
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Answer» Correct Answer - a `"Carrier swing"=("Frequency deviation")/("Modulating frequency")=(50)/(7)=7.143` |
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| 267. |
The modulation index of an `FM` carrier having a carrier swing of `200kHz` and a modulating signal `10kHz` isA. `5`B. `10`C. `20`D. `25` |
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Answer» Correct Answer - b CS`=2xx Delta f` or `Delta f=CS//2` `:. Delta f=(200)/(2)=100kHz` Now `m_(f)=(Deltaf)/(f_(m))=(100)/(10)=10` |
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| 268. |
In the given detector circuit, the suitable value of carrier frequency is A. `ltlt10^(9)Hz`B. `ltlt10^(5)Hz`C. `gtgt10^(9)Hz`D. None of these |
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Answer» Correct Answer - a Using `(1)/(f_("carrier"))ltltRC` we get time constant, `RC=1000xx10^(-12)=10^(-9)s` Now `v=(1)/(T)=(1)/(10^(-9))=10^(9)Hz` Thus the value of carrier frequency should be much less than `10^(9)Hz`, say `100 kHz`. |
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| 269. |
If a source is transmiting electric wave of frequency `8.2xx10^(6)` Hz, then wavelength of the electromagnetic waves transmitted from the source will beA. `36.6m`B. `40.5m`C. `42.3m`D. `50.9m` |
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Answer» Correct Answer - a Here, `lambda=(c)/(v)=(3xx10^(8))/(8.2xx10^(6))=36.6m` |
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| 270. |
An electromagnetic wave consists of oscillating electric and magnetic fields. What is the phase relationship between these fields? |
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Answer» The oscillations in electric and magnetic fields of an electromagnetic waves are in same phase. |
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| 271. |
A plane electromagnetic wave is moving along x-direction . The frequency of the wave is `10^(15)`Hz and the electric field at any point is varying sinusoidally with time with an amplitude of `2Vm^(-1)`. Calculate the average density of the electric and magnetic fields. |
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Answer» Correct Answer - A::B::C |
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| 272. |
For a plane electromagnetic wave, propagating along the z-axis, write the two possible pair of its oscillating electric and magnetic fields. How are the peak values of these (oscillating) fields related to each other? |
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Answer» `vecE=E_yhatj=E_0sin omega(t-z//c)hatj` `vecB=B_xhati=B_0 sin omega(t-z//c)hati` `E_0//B_0=c=` velocity of electromagnetic wave. |
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| 273. |
An electromagnetic wave travel along `z`-axis. Which of the following pair of space and time varying fields would generate such a wave?A. `E_(x), B_(y)`B. `E_(y), B_(x)`C. `E_(z), B_(x)`D. `E_(y), B_(z)` |
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Answer» Correct Answer - A `E_(x), B_(y)` `hati xx hatj = hatk` |
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| 274. |
The electric and the magnetic field, associted with an electromagnetic wave, propagating along the `+x-axis`, can be represented byA. `[E=E_(0)hatk, B=B_(0)hati]`B. `[E=E_(0)hatj, B=B_(0)hatj]`C. `[E=E_(0)hatj, B=B_(0)hatk]`D. `[E=E_(0)hati, B=B_(0)hatj]` |
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Answer» `mu=ExxB` `=E_(0)hati+B_(0)hatj=E_(0)B_(0)hatk` `ExxB` point in the direction of wave propagation. |
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| 275. |
The electric and the magnetic field, associated with an e.m. wave propagating along the `+z`axis, can be represented byA. `(vec(E) = E_(0)hatj, vec(B) = B_(0)hatk)`B. `(vec(E) = E_(0)hatj, vec(B) = B_(0)hatj)`C. `(vec(E) = E_(0)hatk, vec(B) = B_(0)hati)`D. `(vec(E) = E_(0)hatj, vec(B) = B_(0)hati)` |
| Answer» Correct Answer - D | |
| 276. |
A long straight cable of length l is placed symmetrically along z-axis and has radius `a(ltlt l).` The cable consists of a thin wire and a co- axial conducting tube. An alternating current `I(t) = I_(0) " sin " (2pi vt)`. Flows down the central thin wire and returns along the co-axial conducting tube. the induced electric at a distance s from the wire inside the cable is `E(s ,t) =mu_(0) I_(0) v " cos "(2pivt) In ((s)/(a)) hatk`. (i) Calculate the displacement current density inside the cable. (ii) Integrate the displacement current density across the cross- section of the cable to find the total displacement current `I^(d)`. (iii) compare the conduction current `I_(0)` with the displacement current `I_(0)^(d)`. |
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Answer» (i) Given , the induced electric field at a distance r from the wire inside the cable is `E(s,t)=mu_(0) I_(0)v "cos"(2pi vt) In ((s)/(a))hatk` Now displacement current density , `J_(0) = epsilon_(0) (dE)/(dt) =epsilon_(0) (d)/(dt) [mu_(0) I_(0) v"cos" (2pi vt) In ((s)/(a)) hatk]` `=epsilon_(0)mu_(0)I_(0)v(d)/(dt) ["cos"2pivt]In ((s)/(a))hatk` `=(1)/(c^(2))I_(0)v^(2)pi[-"sin"2pivt]In ((s)/(a))hatk` `=(V^(2))/(C^(2)) 2piI_(0)"sin"2pivtIn ((a)/(s))hatk" "[:. l_(4)(s)/(a) =- l_(4)(a)/(s)]` `=(1)/(lambda^(2))2piI_(0) In ((a)/(s)) "sin"2pivthatk` `=(2piI_(0))/(lambda_(2))In (a)/(s) "sin"2pivthatk` `(ii)" "I_(d) = int J_(d) sdsd0 = underset(s=0)overset(a)(int) J_(d) sds underset(0)overset(2pi)(int)d0 =underset(s=0)overset(a)(int) J_(d) sds xx 2pi` `=((2pi)/(lambda))^(2)` `rArr" "=((2pi)/(lambda))^(2) I_(0) underset(s=0)overset(a)(int) In ((a)/(s)) 1/2d(s^(2))."sin" 2pivt` `=(a^(2))/(2) ((2pi)/(lambda))^(2)I_(0) "sin"2pivt underset(s=0)overset(a)(int) In ((a)/(s)).d((s)/(a))^(2)` `=(a^(2))/(4)((2pi)/(lambda))^(2) I_(0)"sin" 2pivt underset(s=0)overset(a)(int) In ((a)/(s))^(2).d((s)/(a))^(2)` `=-(a^(2))/(4)((2pi)/(lambda))^(2)I_(0) "sin" 2pivt underset(s=0)overset(a)(int) In ((s)/(a))^(2).d((s)/(a))^(2)` `=-(a^(2))/(4)((2pi)/(lambda))^(2) I_(0)"sin"2pivt xx(-1)" "[:.underset(s=0)overset(a)(int)In ((s)/(a))^(2)d((s)/(a))^(2)=-1]` `:. I_(d) (a^(2))/(4)((2pi)/(lambda))^(2)I_(0)"sin" 2pivt` `rArr" "=((2pia)/(2lambda))^(2)I_(0)"sin" 2pivt` (iii) the displacement current `I_(0) =((2pia)/(2lambda))^(2) I_(0)"sin" 2pivt =I_(0d)"sin" 2 pivt` `"Here"" "I_(0d) =((2pia)/(2lambda))^(2)I_(0)=((api)/(lambda))^(2)I_(0)` `(I_(0d))/(I_(0)) =((api)/(lambda))^(2)` |
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| 277. |
In an electromanetic wave, the average energy density associated with electric field isA. `CV^(2)//2`B. `Q^(2)//2C`C. `epsi_(0)^(2)//2E_(0)`D. `epsi_(0)E^(2)//2` |
| Answer» Correct Answer - D | |
| 278. |
A plane EM wave travelling along z direction is described by `E=E_0sin (kz-omegat) hati and B=B_0 sin (kz-omegat)hatj`. show that (i) The average energy density of the wave is given by `u_(av)=1/4 epsilon_0 E_0^2+1/4 (B_0^2)/(mu_0)`. (ii) The time averaged intensity of the wave is given by `I_(av)=1/2 cepsilon_0E_0^2`. |
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Answer» (i)The e.m. wave carry energy which is due to electric field vector and magnetic field vector. In e.m. wave E and B vary from point to point and from moment to moment. Let E and B be their time average. The energy density due to electric field E is `u_E=1/2in_0E^2` The energy density due to magnetic field B is `u_B=1/2 (B^2)/(mu_0)` Total average energy density of em wave `u_(av)=u_E+u_B=1/2in_0E^2+1/2 (B^2)/(mu_0).....(i)` Let the em wave be propagation along z-direction. The electric filed vector and magnetic field vectors be represented by `E=E_0sin (kz-omegat)` `B=B_0sin (kz-omegat)` The time average value of `E^2` over complete cycle `=E_0^2//2`. and time average value of `B^2` over complete cycle `=B_0^2//2`. `:. u_(av)=1/2 in_0 (E_0^2)/2+1/2 mu_0 ((B_0^2)/2)=1/4 in_0E_0^2+(B_0^2)/(4mu_0)` (ii) we know that `E_0=cB_0 and c=1/sqrt(mu_0in_0)`. `:. 1/4 (B_0^2)/(mu_0)=1/4 (E_0^2//c^2)/(mu_0)=(E_0^2)/(4mu_0)xxmu_0in_0=1/4in_0E_0^2` `:. u_B=u_E` Hence, `u_(av)=1/4 in_0E_0^2+1/4 (B_0^2)/(mu_0)=1/4in_0E_0^2+1/4in_0E_0^2=1/4in_0E_0^2=1/2 (B_0^2)/(mu_0)` Time average intensity of the wave `I_(av)=u_(av)c=1/2 in E_0^2c=1/2in_0cE_0^2` |
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| 279. |
In an electromagnetic wave, the average energy density associated with magnetic field is:A. `Li_(0)^(2)//2`B. `B^(2)//2mu_(0)`C. `mu_(0)B^(2)//2`D. `mu_(0)//2B^(2)` |
| Answer» Correct Answer - B | |
| 280. |
If the wave-front of e.m.wave travelling in vacuum is given by , `vecr=hati+hatj+hatk`, and find the angle made by the direction of propagation of e.m. wave with y-axis. |
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Answer» If `theta` is the angle which the direction of propagation of e.m. wave makes with y-axis, then `cos theta=(vecr.hatj)/r=((hati+hatj+hatk).hatj)/(sqrt((1)^2+(1)^2+(1)^2))=1/(sqrt3) or theta=cos^-1(1/(sqrt3))` |
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| 281. |
A point source of electromagnetic radiation has an average power output of 800W. (a) Find the maximum value of electric field at a distance 3.5 m from the source. (b) What will be the maximum value of magnetic field? (c) What will be the energy density at a distance at a distance 3.5m from the source? |
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Answer» (a) `I=P/(4pir^2)=u_(av)c=1/2 in_0E_0^2c` `:. E_0=sqrt(P/(2pir^2in_0c))=sqrt(800/(2pixx(3.5)^2xx8.854xx10^-12xx3xx10^8))=62.6 V//m` (b) Maximum value of magnetic field, `B_0=(E_0)/c=62.6/(3xx10^8)=2.09xx10^-7T`. (c) Total energy density at 3.5 m is, `U=1/2 in_0E_0^2=1/2xx(8.85xx10^-12)xx(62.6)^2=1.73xx10^-8Jm^-3`. |
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| 282. |
What features of e.m. wave led Maxwell to conclude that light itself is e.m. wave? |
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Answer» Light and e.m. wave are of transverse nature and they travel with the same velocity in vacuum. This led Maxwell to conclude that light itself is e.m. wave. |
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| 283. |
Total energy of density of electromagnetic waves in vacuum is given by the relationA. `1/2.(E^(2))/(epsi_(0)) + (B^(2))/(2mu_(0))`B. `1/(2)epsi_(0)E^(2) + 1/2mu_(0)B^(2)`C. `(E^(2) + B^(2))/(c)`D. `(1)/(2) epsi_(0)E^(2) + (B^(2))/(2mu_(0))` |
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Answer» Correct Answer - D The energy in EM waves is divided equally between the electric and magnetic fields. The total energy per unit volume is `u - u_(e) + u_(m)` `1/2 epsi_(0)E^(2) + 1/2 (B^(2))/(mu_(0))` |
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| 284. |
The magnetic field of a beam emerging from a filter facing a floodlight is given by `B=12xx10^-8sin(1.20xx10^7z-3.60xx10^14t)T`. What is the average intensity of the beam? |
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Answer» Magnetic field `B=B_(0) "sin"omegat` Given equation `B=12 xx 10^(-8) " sin"(120 xx 10^(7) z-3.60 xx 10^(15) t)T.` On comparing this equation with standard equation we get `B_(0) =12 xx 10^(-8)` The average intensity of the beam`I_(av) =(1)/(2)(B_(0)^(2))/(mu_(0)) .c =1/2 xx((12 xx 10^(-8))^(2)xx 3 xx 10^(8))/(4pixx 10^(-7))` `=1.71 W//m^(2)` |
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| 285. |
Find the amplitude of the electric field in a parallel beam of light of intensity `8.0W//m^2`. |
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Answer» The intensity of plane electro-magnetic wave is given by `I=u_(av)c=1/2 in_0 E_0^2 or E_0=((2I)/(inc))^(1//2) =[(2.80)/((8.85 xx10^(-12))xx(3xx10^(8)))]^(1//2) =77.6 NC^(-1)` |
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| 286. |
Maxwell from his studies concluded that an electric field and a magnetic field changing with time in a direction perpendicular to each other produce a diusturbance which propagets in a direction perpendicular to both the fields. This disturbance is called electromagnetic wave. The em wave is of transverse nature. The velocity of em wave in vacuum is `c=1/(sqrtmu_0in_0)` where `mu_0=` magnetic permiability of free space and `in_0` =electric permittivity of free space. The wavelength of e.m. wave varies over a wide range from `10^(-14)` to `10^3m`. The em wave of different wavelengths travel with same speed in vacuum but move with different speeds in any medium. (i) What are the various e.m wave invisible to eye whose wavelength is lower than the smallest wavelength of visible light? (ii) How do you conclude that white light travels in vacuume with a speed `3xx10^8ms^-1`? (iii) What do you learn from the above study? |
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Answer» (i) The invisible e.m. wave whose wavelength is less than the smallest wavelength of visible light are ultraviolet radiations, x-rays and `gamma`-rays. (ii) The white light is an e.m. wave. The speed of e.m. wave in vacuum is given by `c=1/(sqrtmu_0in_0)`, where mu_0 =4pixx10^-7TmA^-1 and in_0=1/(4pixx9xx10^9)C^2N^-1m^-2` `:. c=1/(sqrt((4pixx10^-7) 1/((4pixx9xx10^9))))=3xx10^8ms^-1` (iii) From the above study, we find that nature/God that created all human beings alike. When they are exposed to different environments of the world and family, their thinking/views/speeds become different. God wants us to live in peace treating everyone as equal. |
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| 287. |
The electromagnetic waves are the radiations of large of wavelength. What are their velocity (i) in vacuum and (ii) in a medium? |
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Answer» The wavelength of electromagnetic wave ranges form `6xx10^-14m` to `6xx10^6m`. These waves travel with the same velocity `(=3xx10^8` `ms^-1)` in vacuum but with different velocities in a medium. In fact, the velocity of an electromagnetic wave is less in a medium than vacuum and a medium provides different values of refractive index to the electromagnetic waves of different wavelengths. |
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| 288. |
A flood light will be covered with filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave `E_x=38.8sin(1.2xx10^7z-3.6xx10^(15)t)V//m`. The average intensity of beam in` "watt"//m^2` will be: |
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Answer» Correct Answer - `2` Average intensity, `I_(av)=1/2 in_0E_0^2xxc` `=((8.85xx10^(-12)xx(38.8)^2xx(3xx10^8))/2) ~~2W//m^2` |
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| 289. |
Electromagnetic wave travel in medium at a speed of `2.0xx10^8ms^-1`. The relative permeability of the medium is 1.0. Find the relative permitivity. |
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Answer» Here `v=2xx10^8ms^-1, mu_r=1, c=3xx10^8ms^-1` Speed of e.m. wave in a medium is given by `(1)/(sqrt(muin)) =(1)/(sqrt(mu_(0)mu_(r)(in_(0)in_(r)))) =(1)/(sqrt(mu_(0)in_(0))) xx (1)/(sqrt(mu_(r)mu_(r)))` or `in_(r)=(c^(2))/(v^(2)mu_(r))= ((3xx10^(8))^(2))/((2xx10^(8))^(2)xx1) =2.25` |
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| 290. |
Why are infrared radiations referred to as hear waves also ? Name the radiations which are next to these radiation in electromagnetic spectrum having (i) shorter wavelength (ii) longer wavelength. |
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Answer» Infrared radiations falling on material are easily absorbed by water molecules present in them. As a result of it, the thermal motion of the water molecules present in material increases. Due to it, the material gets heated. That is why infrared radiations are often referred to as heat waves. (i) The radiations, next to infrared radiations, having shorter wavelength will be visible light. (ii) The radiations next to infrared radiations, having longer wavelength will be microwaves. |
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| 291. |
The speed of em wave in the unit of `10^8m//s`, in a medium of dielectric constant 2.25 and relative permeablity 4 is: |
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Answer» Correct Answer - `1` The speed of em wave in medium is given by `v=1/(sqrt(mu in))=1/(sqrt(mu_0mu_r in_0in_r))=c/(sqrt(mu_r in_r))=(3xx10^8)/(sqrt(4xx2.25))` `1xx10^8ms^-1` |
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| 292. |
The dielectric constant for air is 1.006. The speed of em wave travelling in air is `axx10^8ms^-1`, where a is about: |
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Answer» Correct Answer - `3` The speed of em wave in vacuum is `c=1/(sqrt(mu0 in_0))=3xx10^8ms^-1` Air almost acts as vacuum, therefore `3xx10^8=axx10^8` or a=3approximately |
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| 293. |
Name the electromagnetic radiation to which waves of wavelength in the range of `10^-2m` belong. Give one use of this part of EM spectrum. |
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Answer» Microwaves, These waves are used in Radar system for aircraft navigation. |
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| 294. |
A linearly polarised electromagnetic wave given as `E=E_(0)haticos(kz-omegat)` is incident normally on a perfectly reflecting wall `z=a`. Assuming that the material of the optically inactive, the reflected wave will be give asA. `vec(E)_(r)=-E_(0)hat(i)cos(kx-omegat)`B. `vec(E)_(r)=E_(0)hat(i)cos(kz+omegat)`C. `vec(E)_(r)=-E_(0)hat(i)cos(kz+omegat)`D. `vec(E)_(r)=E_(0) hat(i)sin(kz-omegat)` |
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Answer» Correct Answer - B As the well is perfectly reflecting, there is no change in amplitude `E_(0)`. Also the wall is optically inactive, so, there is no phase change. After reflection, the wave travels along `-ve` z direction, `therefore" "vecE_(r)=E_(0)hati cos (-kz-omegat)` `=E_(0)haticos(kz+omegat)" "(because cos(-theta)=costheta)` |
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| 295. |
A perfectly reflecting mirror has an area of `1cm^(2)` Light enery is allowed to fall on it 1 h at the rate of `10 W cm^(2)`. The force that acts on the mirror isA. `3.35 xx 10^(-8) N`B. `6.7 xx 10^(-8) N`C. `1.34 xx 10^(-7)N`D. `2.4 xx 10^(-4) N` |
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Answer» Correct Answer - B Let, E = Energy falling on the surface per second `= 10 J` Momentum of photons, `p = (h)/(lambda) = (h)/((c//v)) = (hv)/(c) = (E)/(c)` On reflection, change in momentum per second `=` force `= 2p = (2E)/(c) = (2 xx 10)/(3 xx 10^(8)) = 6.7 xx 10^(-8) N` |
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| 296. |
Light with an enargy flux of `25xx10^(4) Wm^(-2)` falls on a perfectly reflecting surface at normal incidence. If the surface area is `15 cm^(2)`, the average force exerted on the surface isA. `1.25xx10^-6N`B. `2.50xx10^-6N`C. `1.20xx10^-6N`D. `3.0xx10^-6N` |
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Answer» Correct Answer - b Average force =average pressurexxarea `:. F_(av)=(2I)/cxxA=(2xx(25xx10^4)xx(15xx10^-4))/(3xx10^8)` `=2.5xx10^-6N` |
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| 297. |
An AM wave has `1800` watt of total power content, For `100%` modulation the carrier should have power content equal toA. `1000` wattB. `1200` wattC. `1500` wattD. `1600` watt |
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Answer» Correct Answer - b `P_(t)=P_(c)(1+(m_(a)^(2))/(2))`, Here `m_(a)=1` `implies 1800=P_(c)(1+((1)^(2))/(2))impliesP_(c)=1200W` |
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| 298. |
A parallel plate capacitor consists of two circular plates each of radius `12 cm` and separated by `5.0 mm`. The capacitor is being charge by an external source. The charging current is constant and is equal to `0.15` A. The rate of change of potential difference between the plate will beA. `8.173xx10^(7) Vs^(-1)`B. `7.817xx10^(8)Vs^(-1)`C. `1.173xx10^(9)Vc^(-1)`D. `3.781xx10^(10)Vs^(-1)` |
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Answer» Correct Answer - c `(dV)/(dt)=(I)/(c)=(Id)/(epsilon_(0)A)` By this waves will just pass without reflection taking place. |
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| 299. |
A parallel- plate capacitor with plate area A and separation between the plates d, is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn summetrically between the plates. Find the displacement current through this area.A. `i`B. `(i)/(2)`C. `(i)/(4)`D. None of these |
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Answer» Correct Answer - b Suppose the charge on the capacitor at time `t` is `Q`, the electric field between the plates of the capacitor is `E=(Q)/(epsilon_(0)A)`. The flux through the area considered is `phi_(E)=(Q)/(epsilon_(0)A).(A)/(2)=(Q)/(2epsilon_(0))` `:.` The displacement current `i_(d)=epsilon_(0)(dphi_(E))/(dt)=epsilon_(0)((1)/(2epsilon_(0)))(dQ)/(dt)=(i)/(2)`. |
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| 300. |
A parallel plate capacitor consists of two circular plates each of radius `2 cm`, separated by a distance of `0.1 mm`. Ifvoltage across the plates is varying at the rate of `5xx10^(13) V//s`, then the value of displacement current is :A. `5.50` AB. `5.56xx10^(2)` AC. `5.56xx10^(3)` AD. `2.28xx10^(4)` A |
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Answer» Correct Answer - c `I_(d)=epsilon_(0)A(dE)/(dt)=epsilon_(0)(A)/(d)(dV)/(dt)` `(8.86xx10^(-12)xx3.14xx(2xx10^(-2))^(2))/(0.1xx10^(-3))xx5xx10^(13)` `5.56xx10^(3) A` |
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