InterviewSolution
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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.
| 51. |
The electric field in an electromahetic wave is given by `E = 100sin omega (t-x//c)`. Find the energy contained in a cylinder of crross section `20cm^(2)` and length `1 m` along the `x`-axis. Also find intensity of wave. |
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Answer» The energy density `u_(av) = (1)/(2)in_(0)E_(0)^(2)` The volume of cylinder `V = Al` The enegry contained in this volume `U = u_(av) V = (1)/(2)in_(0)E_(0)^(2)Al` `= (1)/(2)(8.85 xx 10^(-12)) (100)^(2)(20 xx 10^(-4))(1)` `= 8.85 xx 10^(-11) J` The intensity of wave is `I = (1)/(2)in_(0)E_(0)^(2)c` `= (1)/(2)(8.85 xx 10^(-12))(100)^(2)(3xx 10^(8))` `= 13.3 W//m^(2)` |
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| 52. |
The electric field in an electromagnetic wave is given by `E=( 50N (C^-1)) sin omega (t-x/c).` Find the energy contained in a cylinder of cross section `10 cm^2` and length 50 cm along the x- axis. |
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Answer» Correct Answer - A::B `u_E = u_B = 1/2 epsilon_0 E^2` `= 1/2 epsilon_0 ((E_0)/(sqrt 2))^2` `=1/4 epsilon _0 E_0^2` `= 1/4 (8.86 xx 10^-12)(50)^2` `=5.54 xx 10^-9 J//m^3` Total energy = `(u_E + u_B)` (Given volume) `= 2 xx 5.54 xx 10^-9 xx (10 xx 10^-4)(0.50)` = `5.55 xx 10^-12 J`. |
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| 53. |
It is necessary ton use satellites for long distance TV transmission. Explain why? |
| Answer» TV waves (parts od radio waves)range from 54 MHz.Unlike short wave bands(used in radio boardcasts) which are reflected by ionosphere,TV waves are not properly reflected by ionosphere. This is why, satellities are used for distance TV transmission. | |
| 54. |
In a plane e.m. wave of frequency `1.5xx10^12Hz`, the amplitude fo the magnetic field is `6.0xx10^-6T`. (a) Calculate the amplitude of the electric field. (b) What is the total average energy density of the e.m. wave? |
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Answer» Correct Answer - (a) `1.8xx10^3 Vm^-1 (b) 1.4xx10^-5J//m^-3` Here, `v=1.5xx10^(12)Hz, B_0=6.0xx10^-6T` (a) Amplitude of electric field, `E_0=CB_0=(3xx10^8)xx(6.0xx10^-6)` `=1.8xx10^3Vm^-1` (b) Total average energy density is `U_(av)=1/2in_0E_0^2=1/2(8.85xx10^(-12))xx(1.8xx10^3)^2` `=1.4xx10^-5Jm^-3` |
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| 55. |
The magnetic field in a plane electromangnetic wave is given by `B=(300muT) sin (5.0xx10^-5x^-1)(t-x//c)` Find(i) the maximum electric field and (ii) The average energy density corresponding to the electric field. |
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Answer» Correct Answer - `9xx10^4Vm^-1 ; 1.79xx10^-2Jm^-3` Here, `B_0=300muT=3000xx10^-6T` `=3xx10^-4T` `E_0=cB_0 =(3xx10^8)xx(3xx10^-4)` `=9xx10^4Vm^-1` `u_E=1/4in_0E_0^2=1/4xx(8.85xx10^-12)xx(9xx10^4)^2` `=1.79xx10^-2Jm^-3` |
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| 56. |
How does a charge q oscillating at certain frequency produce electromangnetic waves? Skecth schematic diatram depiciting electric and magnetic field for an electromagnetic wave propagating along thex-direction. |
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Answer» Oscillating charge produces sinusoidal variation of electric and magnetic fields, which inturn produces electromagnetic wave. For the schematic diagram diagram of electromagnetic wave. |
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| 57. |
In an electomegnetic wave, the amplitude of electric field is `10V//m`. The frequency of wave is `5xx10^14Hz`. The wave is propagating along Z-axis, find (i) the average energy density of electric field (ii) the average energy density of magnetic field (iii) the total average energy density of e.m. wave. |
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Answer» Here, `E_0=10Vm^-1, v=5xx10^14Hz`. (i) Average energy density due to electric field is `u_E=1/2in_0E_(rms)^2=1/2in_0((E_0)/(sqrt2))^2 =1/4in_0E_0^2` `=1/4xx(8.85xx10^-12)xx(10)^2=2.21xx10^-10J//m^3` (ii) Average energy density due to magnetic field is `u_B=1/2 (B_(rms)^2)/(mu_0) =1/(2mu_0) ((B_0)/(sqrt2))^2=1/4 (B_0^2)/(mu_0)= 1/4((E_0//c)^2)/(mu_0)=1/(4mu_0) (E_0^2)/(c^2)=1/(4mu_0) (E_0^2)/((1//mu_0in_0))=1/4 in_0E_0^2=u_E` `=2.21xx10^-10J//m^3` (iii) Total average energy density of e.m. waves `u=u_E+u_B=2.21xx10^-10+2.21xx10^-10=4.42xx10^-10J//m^3` |
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| 58. |
Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitorA. is changing with timeB. decreasesC. does not changeD. decreases to zero |
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Answer» Correct Answer - A Displacement current aries when electric field in a region is changing with time. It will be so if the charge on a capacitor is changing with time. |
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| 59. |
A parallel plate capacitor has circular plates each of radius 6.0cm, It is charged such that the electric field in the gap between its plates rises constantly at the rate of `10^10V cm^-1s^-1`. What is the displacement current? |
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Answer» Here, `r=6xx10^-2m`, `A=pixx(6xx10^-2)^2=36pixx10^-4m^2` `(dE)/(dt)=10^10Vcm^-1s^-1=10^12vm^-1s^-1` Displacement current, `I_D=in_0(dphi_E)/(dt)=in_0A(dE)/(dt)` `=(8.85xx10^-12)xx(36pixx10^-14)xx10^12` `=01A` |
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| 60. |
Assertion : Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor does not charge Reason : The displacement current arises in the region in which the electric field is constant with time.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 Displacement current arises when electric field in a region is changing with time, which is given by `I_(D)=epsilon_(0)(dphi_(E))/(dt)`. It will be so if the charge on a capacitor is not constant but changing with time. |
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| 61. |
A TV tower has a height of 200m. How much population is covered by TV brood-cast ? Given radius of the earth `=6.4xx10^(6)` m and avarege density of population `=10^(3) km^(-2).` |
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Answer» Correct Answer - A |
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| 62. |
A TV transmitter is at a height of 100m. What is its coverage range ? By how much should its height be increased to double the coverage range. Radius of earth `=6400 "km".` |
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Answer» Correct Answer - A::C::D |
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| 63. |
The frequency band used for radar relay systems and television-A. UHFB. VLFC. VHFD. EHF |
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Answer» Correct Answer - a Reflective index of ionosphere is: `mu=sqrt(1(80.6N)/(v^(2)))` Where, N= Number density of `"electron"//m^(3)` `v=` frequency of wave is kHz `:. Mult1` |
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| 64. |
What is the amplitude of electric field produced by radiation coming from a `100` W bulb at a distance of `4` m? Assume efficiency of bulb to be `3.14%` and assume it to the point source:A. `2.42 V//m`B. `3.43 V//m`C. `4.2xx10^(4) V//m`D. `14xx10^(4) V.m` |
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Answer» Correct Answer - b Intensity at a distance `r=4` m is `I=(Ixx3.14%)/(4pi r^(2))=(100xx(3.14//100))/(4xx3.14xx(4)^(2))` `(1)/(64) W//m^(2)` Half of the intensity is provide by electric field and remaining half by magnetic field `:. (1)/(2)I=(1)/(2)(epsilon_(0)E_(rms)^(2)C)` `implies E_(rms)=sqrt((I)/(epsilon_(0)c))=sqrt({(1//64)/(8.85xx10^(-12)xx3xx10^(8))})` `sqrt(5.88)=2.42 V//m =sqrt(2)xx2.42=3.43 V//m` |
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| 65. |
Which of the following have zero average value un a plane electromagnetic wave? (i)electric field (ii)magnetic field (iii) electric energy (iv) magnetic enegryA. `(i),(ii)`B. `(ii),(iii)`C. `(i),(iii)`D. `(i),(iv)` |
| Answer» Correct Answer - A | |
| 66. |
Which of the following has/have zero average value in a plane electromagnetic wave?A. Both magnetic and electric fieldsB. Electric field onlyC. Magnetic field onlyD. None of these |
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Answer» Correct Answer - A Both magnetic and electric field have zero average value in a plane electromagnetic wave. |
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| 67. |
The maximum magnetic field in a plane electromagnetic wave is `4muT`. The wave is going in the `x`-direction and magnetic field is in `y`-direction. Find the maximum electric field in the wave and its direction. |
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Answer» `E_(0) = cB_(0) = 3 xx 10^(8) xx 4 xx 10^(-6) = 1200N//C` `vec(E), vec(B)` and direction of propagation are mutually perpendicualr. `vec(E)` is along `z`-axis. |
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| 68. |
An e.m. wave going through vacuum is denoted by `E=E_0sin(kx-omegat)`. Which of the following is /are independent of wavelength?A. `k`B. `omega`C. `k//omega`D. `komega` |
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Answer» Correct Answer - c Here, `k=2pi//lambda , omega=2piv=2pic//lambda` So, `k/omega=(2pi//lambda)/(2piv)=1/c=constant` So, it is independent of wavelength. |
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| 69. |
An em wave going through vacuum is described by `E=E_0sin(kx-omegat)` `B=B_0sin(kx-omegat)`A. `E_(0)k = B_(0)omega`B. `E_(0) omega = B_(0)k`C. `E_(0) B_(0) = omegak`D. None of these |
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Answer» Correct Answer - A From `k = (2pi)/(lambda)` and `omega = 2piv` we get, `k/(omega) = ((2pi)/(lambda))/(2piv) = v/(lambda) = 1/c` As, `(E_(0))/(B_(0)) = c rArr (k)/(omega) = (B_(0))/(E_(0))` `E_(0)k = B_(0)omega` |
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| 70. |
An electromagnetic wave going through vacuum is described by` E= E_0 sin(kx- omega t), B=B_0sin(kx-omega t)`. ThenA. `E_(0)k=B_(0)omega`B. `E_(0)omega=B_(0)k`C. `E_(0)B_(0)=omega k`D. `None of these |
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Answer» Correct Answer - a `E_(0)/(B_(0)=C`. Also `k=(2pi)/(lambda)` and `omega=2piv`. These relation gives `E_(0)K=B_(0)omega` |
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| 71. |
An em wave exerts pressure on the surface on which it is incident. Justify? |
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Answer» Since electromagnetic waves carry energy and momentum, therefore, they exerted pressure on the surface on which they are incident. |
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| 72. |
Even though an electric field `vecE` exerts a force `avecE` on a charged particle yet the electric field of an EM wave does not contribute to the radiation pressure (but transfer energy). Explain. |
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Answer» Electric field of an e.m. wave is an oscillating field. Due to it, the electric force caused by electric field in e.m. wave on a charged particle is an oscillating one. This electric force averaged over an integral number of cycle is zero, since its direction changes every half cycle. Hence, electric field is not responsible for radiation pressure. |
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| 73. |
In order to establish an instantaneous displacemet current of 1 mA in the space between the plates of 2`mu`F parallel plate capacitor, the potential difference need to apply isA. `100 V s^(-1)`B. `200V s^(-1)`C. `300Vs^(-1)`D. `500 V s^(-1)` |
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Answer» Correct Answer - D `I_(D)="1 mA" =10^(-3)A,C=2muF=2xx10^(-6)F` `I_(D)=I_(C)=(d)/(dt)(CV)=C(dV)/(dt)` Therefore, `(dV)/(dt)=(I_(D))/(C)=(10^(-3))/(2xx10^(-6))=500"V s"^(-1)` Therefore, applying a varying potential difference of 500 V `s^(-1)` would produce a displacement current of desired value. |
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| 74. |
How would you establish an instantaneou displacement current of 2.0A in the space between the parallel plates of `1muF` capacitor? |
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Answer» Correct Answer - A::B::C::D |
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| 75. |
Advantage of optical fibreA. High bandwidth and EM interferenceB. Low bandwidth and EM interferenceC. High band width, low transmission capacity and no EM interferenceD. High bandwidth, high data transmission capacity are no EM interference |
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Answer» Correct Answer - d Few advantages of optical fibres are that the number of signals carried by optional fibres is much more than that carried by the `Cu` wire or radio waves. Optical fibres are practically free from electromagnetic interence and problem of cross talks whereas ordinary cables and microwaves links suffer a lot from it. |
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| 76. |
You are given a `2mu F` parallel plate capacitor. How would you establish an instantaneous displacement current fo 1mA in the space between its plates? |
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Answer» Here, `I_D=1mA =10^-3A, c=2mu F=2xx10^-6F, I_D=I=d/(dt) (CV)=C (dV)/(dt)` Therefore, `(dV)/(dt) =(I_D)/C= (10^-3)/(2xx10^-6)=500V//s` Therefore, applying a varying potential difference of `500V//s` would produce a displacement current of desired value. |
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| 77. |
If `mu_(1)` and `mu_(2)` are the refractive indices of the materials of core and cladding of an optical fibre, then the loss of light due to its leakage can be minimized by havingA. `mu_(1)gtmu_(2)`B. `mu_(1)ltmu_(2)`C. `mu_(1)=mu_(2)`D. None of these |
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Answer» Correct Answer - a In optical fibre, the loss of light due to leakage can be minimized if critical angle is small. |
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| 78. |
An optical fibre communication system works on a wavelength of `1.3 mu_(m)`. The number of subscribers it can feed if a channel required `20kHz` areA. `2.3xx10^(10)`B. `1.15xx10^(10)`C. `1xx10^(5)`D. None of these |
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Answer» Correct Answer - b Optical source frequency `f=(c)/(lambda)=(3xx10^(8))/(1.3xx10^(-6))=2.3xx10^(14)Hz`. |
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| 79. |
A micro-wave and an ultrasonic sound wave have the same wavelength. Their frequencies are in the ratio (approximately)A. `10^(2)`B. `10^(4)`C. `10^(6)`D. `10^(8)` |
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Answer» Correct Answer - C Frequency of microwaves, `upsilon_(m) ~~ 10^(11)Hz` Frequency of ultrasonic sound waves, `upsilon_(u) ~~ 10^(5) Hz` `:. (upsilon_(m))/(upsilon_(u)) = (10^(11))/(10^(5)) = 10^(6)` |
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| 80. |
Statement-1: Short wave band are used for transmission fo radiowaves to a large distance. Statement-2: Short waves are reflected from ionosphere.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 - a Both Statement-1 and Statement and Statement-2 are correct. Here the Statement -2 is correct explanation of Statement -1. |
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| 81. |
Statement-1: Short wave band are used for transmission fo radiowaves to a large distance. Statement-2: Short waves are reflected from ionosphere.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 - b Short wave (wavelength `30 km` to `30 cm`). These waves are used for radio transmission and for general communication purpose to a longer distance from ionosphere. |
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| 82. |
Radiowaves of wavelength `360` m are transmitted from a transmitter. The inductance of the coil which must be connected with capacitor of capacity `3.6 muF` in a resonant citcuit to receive these waves will be appoximatelyA. `10^(3)H`B. `10^(2)H`C. `10^(-4)H`D. `10^(-8)H` |
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Answer» Correct Answer - d `lambda=(c)/(v)` and `v=(1)/(2pisqrt(LC))` `:. lambda=c.2pisqrt(LC)` `L=(lambda^(2))/(4pi^(2)c^(2)C)` `((360)^(2))/(4xx(3.14^(2))xx(3xx10^(8))xx3.6xx10^(-6))` `=10^(-8)` H |
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| 83. |
A capacitor of capacitance C, is being charged up by connecting it across a d.c. voltage source of voltage V. How do the conduction and displacement currents in this set up compare with each other (a) during the charging up process? (b) after the capacitor gets fully charged? |
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Answer» From the property of continuty, (a) Conduction current equals to displacement currents. During charging up process, there is a conduction current in connecting wires and equivalent displacement current in the region between the two plates of capacitor. (b) After the capacitor gets fully charged., the conduction current in the conneting wires becomes zero. Now displacement current in a region between the two plates is also zero due to property of continuty. |
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| 84. |
Consider the following types of electromagnetic radiation: radiowaves, infra-red, vissible light. Which of the following statements are correct? (i) Only radiowaves can be used to transmite audio information. (ii) Only infrared radiation is emitted by very hot object. (iii) Only visible light can be detect by humans.A. only (i) is correctB. only (ii) is correctC. only (iii) is correctD. None of the above is correct. |
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Answer» Correct Answer - d (i) Audio information can be transmitted in the from of the optical wave through optical fibres. (ii) very hot objects emit not only infrared radiation but also emit visible radiation. (iii) Humans can detect infrared radiations by feeling the heat on the portion of body exposed by radiation. Therefore, all the given statements are not completely correct. Hence the option(d) is correct. |
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| 85. |
Arrrange the following electromagnetic waves in the order of their increasing wavelength: (a) `gamma`-rays (b) microwaves. (c) x-rays (d) Radiowaves. |
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Answer» The electromagnetic waves in the orger of ltbr? Increasing wavelength are (a) `gamma`-rays, (b) x-rays, (c) microwaves and (d) Radiowave. |
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| 86. |
In the question 3, the ratio of conduction current and the displacement current isA. `((api)/(lamda))^(2)`B. `((a pi)/(lamda))`C. `((lamda)/(api))^(2)`D. `((lamda)/(2pi))` |
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Answer» Correct Answer - A The displacement current `I^(d)=intJ_(d)sdsd theta` `=(2pi)/(lambda^(2))I_(0)int_(theta=0)^(2pi)int_(s=0)^(a)ln.((a)/(s))sds sin(2piupsilont)d theta` `=((2pi)/(lambda))^(2)I_(0)int_(s=0)^(a)(1)/(2)ds^(2)ln((a)/(s))sin(2piupsilont)` `=(a^(2))/(4)((2pi)/(lambda))^(2)I_(0)int_(s=0)^(a)d((s)/(a))^(2)ln.((a)/(s))^(2)sin(2pi upsilont)` `=(a^(2))/(4)((2pi)/(lambda))^(2)I_(0)sin(2piupsilont)` `I^(d)=((api)/(lambda))^(2)I_(0)sin2pi upsilont=I_(0)^(d)sin2pi upsilontrArr(I_(0)^(d))/(I_(0))=((api)/(lambda))^(2)` |
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| 87. |
Give difference between displacement current and conduction current. |
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Answer» Conduction current is due to flow of electrons in the circuit. It exists even if the flow of electrons is at uniform rate. Displacement current is due to time varying electric field. It does not exist under steady condition. |
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| 88. |
Which of the following electromagnetic waves has smaller wavelength ?A. X-raysB. MicrowavesC. `gamma`-raysD. Radiowaves. |
| Answer» Correct Answer - C | |
| 89. |
The ultra high frequency band of radiowaves in electromagnetic wave is used as inA. television wavesB. cellular phone communicationC. commercial FM radioD. both (a) and (c) |
| Answer» Correct Answer - B | |
| 90. |
The conduction current is the same as displacement current when the source isA. ac onlyB. dc onlyC. either ac or dcD. neither dc nor ac |
| Answer» Correct Answer - C | |
| 91. |
X-rays and `gamma`-rays of same energies are distinguished by theirA. frequencyB. chargesC. ionising powerD. method of production |
| Answer» Correct Answer - D | |
| 92. |
The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma rays isA. microwave, infrared, ultraviolet, gamma raysB. infrared, microwave, ultraviolet, gamma raysC. gamma rays, ultraviolet, infrared, microwavesD. microwaves, gamma rays, infrarred, ultraviolet |
| Answer» Correct Answer - A | |
| 93. |
Which of the following is not an electromagnetic wave?A. X-raysB. `gamma`-raysC. `beta`-raysD. Heat rays |
| Answer» Correct Answer - C | |
| 94. |
The minimum frequency `v_(min)` of continuous `X`-rays is related to the applied pot. Diff `V` as:A. `V_(min) propV`B. `V_(min) propV^(1//2)`C. `V_(min)propV^(1//3)`D. `V_(min)prop V^(4)` |
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Answer» Correct Answer - a `v_(min)propV` |
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| 95. |
The wavelength of infrared rays is of the order of:A. `5xx10^(-7)m`B. `10^(-3)m`C. between `(a)` and `(b)`D. None of these |
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Answer» Correct Answer - c `lambda` for infra-red rays lies between `5xx10^(-7)m` and `10^(-3)m`. |
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| 96. |
The wavelength of ultraviolet rays is of the order of:A. `10^(-3)`B. `10^(-6)m`C. `10^(-8)m`D. between `(b)` and `(c)`. |
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Answer» Correct Answer - d `lambda` lies between `10^(-6)` `m` and `10^(-8)m`. |
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| 97. |
A plane electromagnetic wave is incident on a material surface. The wave delivers momentum p and energy E.A. `p=0, E=0`B. `p!=0, E!=0`C. `p!=0, E!=0`D. `p=0, E!=0` |
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Answer» Correct Answer - b `EM` waves carry momentum and hence can exert pressure on surfaces. They also transfer energy to the surface so `p != 0` and `E!=0`. |
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| 98. |
Which of the following rays has the maximum frequency?A. Gamma raysB. Blue lightC. Infrared raysD. Ultraviolet rays |
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Answer» Correct Answer - a `v_(gamma-"rays")gtv_("UV-rays")gtv_("Blue light")gtv_("infrared rays")` |
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| 99. |
A signal emitted by an antenna from a certain point can be received at another point of the surface in the form ofA. Sky waveB. Ground waveC. Sea waveD. Both `(a)` and `(b)` |
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Answer» Correct Answer - d Ground wave and sky wave both are amplitude modulated wave and the amplitude modulated signal is transmitted antenna at a distance place. |
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| 100. |
The oscillating electric and magnetic vectors of an electromagnetic wave are oriented alongA. The same direction but differ in phase by `90^(@)`B. The same direction and are in phaseC. Mtually perpendicular direction and are in phaseD. Mutually perpendicular and differ in phase by `90^(@)` |
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Answer» Correct Answer - c `vecE` and `vecB` are mutually perpendicular to each other and are in phase i.e, they become zero and minimum at the same place and at the same time. |
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