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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The electromagnetic radiations are in descending order of wavelength in the following sequenceA. infra red waves, radio waves X-rays visible light raysB. radio-waves, infra-red waves, visible, X-raysC. radio waves, visible light, infra-red waves, X-raysD. X-rays, visible light, infra-red wave, radiowaves |
| Answer» Correct Answer - B | |
| 2. |
Radio waves and visible light in vacuum haveA. same velocity but different wavelengthB. continuous emission specturmC. band absorption spectrumD. line emission spectrum |
| Answer» Correct Answer - A | |
| 3. |
A man can take pictures of those objects which are not fully visible to the eye using camera films acceptable toA. ultraviolet raysB. sodium lightC. visible lightD. infra-red rays |
| Answer» Correct Answer - D | |
| 4. |
A parallel plate capacitor of plate separation `2 mm` is connected in an electric circuit having source voltage `400 V`. If the plate area is `60 cm^(2)`, then the value of displacement current for `10^(-6) sec` will beA. `1.062 amp`.B. `1.062xx10^(-2) amp`C. `1.062xx10^(-3) amp`D. `1.062xx10^(-4) amp` |
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Answer» Correct Answer - B `I_(D)=epsilon_(0)(delphi_(E))/(delt)=epsilon_(0)(EA)/(t)=(epsilon_(0)A)/(t)(v)/(d)` `I_(D)=(8.85xx10^(-12)xx400xx60xx10^(-4))/(2xx10^(-3)xx10^(-6))` `=1.062xx10^(-2)amp` Hence the corrent answer will be (`4`). |
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| 5. |
Which of the following are not electromagnetic waves ?A. Radio wavesB. gamma raysC. `beta-rays`D. `X-rays` |
| Answer» Correct Answer - C | |
| 6. |
Electrimagnetic waves are transverse is nature is evident byA. polarizationB. interferenceC. reflectionD. diffraction |
| Answer» Correct Answer - A | |
| 7. |
Suppose that the electric field amplitude of an electromagnetic wave us `E_0=120 N// C` and that its frequency is `50.0 MHz`. (a) Determine` B_0,omega,k` and `lambda`, (b) find expressions for `E` and `B`. |
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Answer» (`a`) `i` Using `C=(E_(0))/(B_(0))` we get `B_(0)=(E_(0))/(C )=(120)/(3xx10^(8))=4xx10^(-7)T=400nT` (`ii`) `omega=2piv=2xxpixx50xx10^(6)=3.14xx10^(8)rad//Sec` (`iii`) `c=ulambdaimplieslambda=(C )/(u)=(3xx10^(8))/(50xx10^(6))=6m` (`iv`) `K=(2pi)/(lambda)=(2pi)/(6)=(2xx3.14)/(6)=1.05m^(-1)` (`b`) `vecE=E_(0)sin(kx-omegat)` `=120sin(1.05x-3.14xx10^(8)t)` `B=B_(0)sin(kx-omegat)` `=400xx10^(-9)sin(1.05x-3.14xx10^(8)t)` |
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| 8. |
A plane electromagnetic wave of frequency `40 MHz` travels in free space in the `X`-direction. At some point and at some instant, the electric field `vecE` has its maximum value of `750 N//C` in `Y`- direction. The wavelength of the wave is-A. `3.5 m`B. `5.5 m`C. `7.5 m`D. `9.5 m` |
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Answer» Correct Answer - C `lambda=(C )/(f)=(3xx10^(8))/(4xx10^(7))=7.5 m` |
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| 9. |
An `AC rms` voltage of `2V` having a frequency of `50KHz` is applied to a condenser of capacity of `10muF`. The maximum value of the magnetic field between the plates of the condenser if the radius of plate is `10cm` isA. `0.4 p mu`B. `4pi mu T`C. `2 mu T`D. `40pi mu T` |
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Answer» Correct Answer - C `B=(mu_(0)C)/(2piR )(dv)/(dt)` |
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| 10. |
The velocity of all radiowaves in free space is `3xx10^(8),//s`, the frequency of a wave of wavelength `150 m` isA. 20 khzB. 2 kHzC. 2 MHzD. 1 MHz |
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Answer» Correct Answer - C As given `E=10cos(10^(7)t+kx)` Comparing it with standard equation of e.m. wave, `E=E_(0)cos(omegat+kx)` Amplitude `E_(0)=10 V//m` and `omega=10^(7) rad//s` `because c=vlamda=(omegalamda)/(2pi)` Or `lamda=(2pic)/(omega)=(2pixx3xx10^(8))/(10^(7))=188.4m` Also, `c=(omega)/(K)` or `k=(omega)/(c)=(10^(7))/(3xx10^(8))=0.033` The wave is propagating along y direction. |
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| 11. |
The velocity of all radiowaves in free space is `3xx10^(8),//s`, the frequency of a wave of wavelength `150 m` isA. `45 MHz`B. `2 MHz`C. `2 KHz`D. `20 KHz` |
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Answer» Correct Answer - B `C=nlambda` |
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| 12. |
The wave function (in `S.I unit`) for an electromagnetic wave is given as- `psi(x,t)=10^(3)sinpi(3xx10^(6)x-9xx10^(14)t)` The speed of the wave isA. `9xx10^(14)m//s`B. `3xx10^(8)m//s`C. `3xx10^(6)m//s`D. `3xx10^(7)m//s` |
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Answer» Correct Answer - B `C=(omega)/(K)` |
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| 13. |
The amplitude of magnetic field at a region carried by an electromagnetic wave is `0.1 muT`. The intensity of wave isA. `4 mu Wm^(-2)`B. `1.2 Wm^(-2)`C. `4 Wm^(-2)`D. `1.2 mu Wm^(-2)` |
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Answer» Correct Answer - B `I=(B_(0)^(2)C)/(2mu_(0))` |
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| 14. |
The sun radiates electromagnetic energy at the rate of `3.9xx10^(26)W`. Its radius is `6.96xx10^(8)m`. The intensity of sun light at the solar surface will be (in `W//m^(2)`)A. `1.4xx10^(4)`B. `2.8xx10^(5)`C. `64xx10^(6)`D. `5.6xx10^(7)` |
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Answer» Correct Answer - C `I=(P)/(4pir^(2))` |
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| 15. |
The intensity of electromagnetic wave at a distance of `1 Km` from a source of power `12.56 Kw`. IsA. `10^(-3)Wm^(-2)`B. `4xx10^(-3)Wm^(-2)`C. `12.56xx10^(-3)Wm^(-2)`D. `1.256xx10^(-3)Wm^(-2)` |
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Answer» Correct Answer - A `I=(P)/(4pir^(2))` |
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| 16. |
If the source of power` 4 kW` product`10^(20)` photons //second , the radiation belongs to a part spectrum calledA. `X-rays`B. Ultraviolet raysC. microwavesD. `gamma-rays` |
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Answer» Correct Answer - A `P=(nE)/(t)`, `E=(hc)/(lambda)` |
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| 17. |
A condenser is charged using a constant current. The ratio of the magnetic field at a distance of `(R )/(2)` and `R` from the axis of condenser (`R` is the radius of plate) while charging isA. `1:1`B. `2:1`C. `1:2`D. `1:4` |
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Answer» Correct Answer - C `r lt RB=(mu_(0)I_(0)r)/(2piR^(2))`, `r=R B=(mu_(0)I_(d))/(2piR )` |
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| 18. |
The frequency of a light wave in a material is `2xx10^(14)Hz` and wavelength is `5000 Å`. The refractive index of material will beA. `1.40`B. `1.50`C. `3.00`D. `1.33` |
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Answer» Correct Answer - C Velocity of light waves in material is `v=nlamda` ..(i) Refractive index of material is `mu=(c)/(v)` ..(ii) Where `c` is speed of light in vacuum or air. or `mu=(c)/(nlamda)` ..(iii) Given, `n=2xx10^(14)Hz` `lamda=5000 Å = 5000 xx 10^(-10) m`, `c = 3 xx 10^(8) m//s` Hence, from Eq (iii) we get `mu=(3xx10^(8))/(2xx10^(14)xx5000xx10^(-10))=3.00` |
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| 19. |
The frequency of a light wave in a material is `2xx10^(14)Hz` and wavelength is `5000 Å`. The refractive index of material will beA. 1.4B. 1.5C. 3D. 1.33 |
| Answer» Correct Answer - C | |
| 20. |
The maximum electric field of a plane electromagnetic wave is `88V//m`. The average energy density isA. `3.4xx10^(-8)Jm^(-3)`B. `13.7xx10^(-8)Jm^(-3)`C. `6.8xx10^(-8)Jm^(-3)`D. `1.7xx10^(-8)Jm^(-3)` |
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Answer» Correct Answer - C `U_(av)=epsilon_(0)E_(max)^(2)` |
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| 21. |
The `rms` value of electric field of a plane electromagnetic wave is `314V//m`.The average energy density of electric field and the average energy density areA. `4.3xx10^(-7) Jm^(-3)`, `2.15xx10^(-7) Jm^(-3)`B. `4.3xx10^(-7) Jm^(-3)`, `8.6xx10^(-7) Jm^(-3)`C. `2.15xx10^(-7) Jm^(-3)`, `4.3xx10^(-7) Jm^(-3)`D. `8.6xx10^(-7) Jm^(-3)`, `4.3xx10^(-7) Jm^(-3)` |
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Answer» Correct Answer - B `(U_(av))_(E)=(1)/(2)epsilon_(0)E_(max)^(2)`, `U_(av)=epsilon_(0)E_(max)^(2)` |
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| 22. |
The maxwells four equations are written as (`i`) `ointvecE.vec(dS)=(q_(0))/(epsilon_(0))` (`ii`) `ointvecB.vec(dS)=0` (`iii`) `ointvecE.vec(dl)=(d)/(dt)ointvecB.vec(dS)` (`iv`) `ointvecB.vec(dl)=mu_(0)epsilon_(0)(d)/(dt)ointvecE.vec(dS)` The equations which have sources of `vecE` and `vecB` areA. (`i`), (`ii`), (`iii`)B. (`i`), (`ii`)C. (`i`) and (`iii`) onlyD. (`i`) and (`iv`) only |
| Answer» Correct Answer - D | |
| 23. |
In an electromagnetic wave the average energy density is associated with-A. electric field onlyB. magnetic field onlyC. equally with electric and magnetic fields.D. average energy density is zero. |
| Answer» Correct Answer - C | |
| 24. |
Light with energy flux `18wcm^(-2)` is incident on a mirror of size `2cmxx2cm` normally. The force experienced by it and momentum delivered in one minute areA. `0.48 muN`, `28.8 mu kgms^(-1)`B. `48 muN`, `2.88 mu kgms^(-1)`C. `28.8 muN`, `4.8 mu kgms^(-1)`D. `0.24 muN`, `28.8 mu kgms^(-1)` |
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Answer» Correct Answer - A Momentum `P=(U)/(C )=(IA t)/(C )`, Force `F=(IA)/(C )` |
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| 25. |
Light with energy flux `36w//cm^(2)` is incident on a well polished metal square plate of side `2cm`. The force experienced by it isA. `0.96 muN`B. `0.24muN`C. `0.12 muN`D. `0.36 muN` |
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Answer» Correct Answer - A `F=(2U)/(Ct)=(2IA t)/(Ct)` , `=(2IA)/(C )` |
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| 26. |
The radiation force experienced by body exposed to radiation of intensity `I`, assuming surface of body to be perfectly absorbing is:A. `(piR^(2)I)/(c )`B. `(piRHI)/(c )`C. `(IRH)/(2c )`D. `(IRH)/(c )` |
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Answer» Correct Answer - D `F=(I)/(c )xx`effective area |
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| 27. |
Electromagnetic radiation with energy flux `50W cm^(-2)` is incident on a totally absorbing surface normally for `1 hour`: If the surface has an area of `0.05m^(2)`, then the avergae force due to the radiaton pressure, on it is,A. `8.3xx10^(-7)N`B. `8.3xx10^(-5)N`C. `1.2xx10^(-7)N`D. `1.2xx10^(-5)N` |
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Answer» Correct Answer - B Force `F=(IA)/(C )` |
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| 28. |
Light with energy flux of `24Wm^(-2)` is incident on a well polished disc of radius `3.5 cm` for one hour. The momentum transferred to the disc isA. `1.1mu kg ms^(-1)`B. `2.2mu kg ms^(-1)`C. `3.3mu kg ms^(-1)`D. `4.4mu kg ms^(-1)` |
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Answer» Correct Answer - B `P=(2U)/(C )=(2IAt)/(C )` |
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| 29. |
In a plane electromagnetic wave, the electric field oscillates sinnusoidally at a frequency of `2xx10^(10)Hz` and amplitude `48 V//m`. The amplitude of oscillating magnetic field will beA. `(1)/(16)xx10^(-8)Wb//m^(2)`B. `16xx10^(-8)Wb//m^(2)`C. `12xx10^(-7)Wb//m^(2)`D. `(1)/(12)xx10^(-7)Wb//m^(2)` |
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Answer» Correct Answer - B `B_(0)=(E_(0))/(C )` |
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| 30. |
Light with energy flux `36Wm^(-3)` is incident on a circular part of radius `1.4m` of a perfectly balck body. The force experienced by the body and the momentum delivered in `10 minutes` areA. `2.2 muN`, `7.2 mu kgms^(-1)`B. `3.5 muN`, `7.4 mu kgms^(-1)`C. `0.74 muN`, `444 mu kgms^(-1)`D. `7.4 muN`, `2.2 mu kgms^(-1)` |
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Answer» Correct Answer - C Momentum `P=(U)/(C )=(IA t)/(C )`, Force `F=(IA)/(C )` |
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| 31. |
The electrical field in the gap of a condenser charges as `10^(12)Vm^(-1)s^(-1)`. If the radius of each plate of the condenser is `3cm`, the magnetic field at the edge of plate in the gap isA. `1.67 mT`B. `0.167 muT`C. `0.5 muT`D. `5 muT` |
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Answer» Correct Answer - B `intvecB.vec(dl)=mu_(0)epsilon_(0)(dphi_(E))/(dt)` , `B(2pir)=(1)/(C^(2))(dE)/(dt)pir^(2)` |
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| 32. |
The diameter of the condenser plate is `4cm`. It is charged by an external current of `0.2 A`.The maximum magnetic field induced in the gapA. `2 muT`B. `4 muT`C. `6 muT`D. `8 muT` |
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Answer» Correct Answer - A `B=(mu_(0)i)/(2pir)` |
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| 33. |
The area of each plate of parallel plated condenser is `144 cm^(2)`. The electrical field in the gap between the plates changes at the rate of `10^(12)Vm^(-1)s^(-1)`. The displacement current isA. `(4)/(pi)A`B. `(0.4)/(pi)A`C. `(40)/(pi)A`D. `(1)/(10pi)A` |
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Answer» Correct Answer - B `I_(d)=epsilon_(0)A(dE)/(dt)` |
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| 34. |
A condenser has two conducting plates of radius `10cm` separated by a distance of `5 mm`. It is charged with a constant current of `0.15A`. The magnetic field at a point `2cm` from the axis in the gap isA. `1.5xx10^(6)T`B. `3xx10^(-8)T`C. `6xx10^(-8)T`D. `3xx10^(-6)T` |
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Answer» Correct Answer - C `B=(mu_(0)ir)/(2piR^(2))` |
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| 35. |
A parallel plate capacitor with circular plates of radius `1m` has a capacitor of `1nF`. At `t = 0`, it is connected for charging in series with a resistor `R = 1MOmega` across a `2V` battery. Calculate the magnetic field at a point `P`, halfway between the cnetre and the periphery of the plates, after `t = 10^(-3)sec`. |
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Answer» The time constant of the CR circuit is `tau=CR=10^(-3)`s. Then we have `q(t)=CV[1-exp(-t//tau)]` `=2xx10^(-8)[1-exp(-t//10^(-2)]` The electric field in between the plates at time t is `E=(q(t))/(epsilon_(0)A)=(q)/(piepsilon_(0)),A=pi(1)^(2)m^(2)=` area of the plates Consider now a circular loop of radius `(1//2)`m parallel tot he plates passing through P. the magnetic field B at all points on the loop is along the loop and of the same value. the flux `Phi_(E)` through this loop is `Phi_(E)=Exx` area of the loop `=Exxpixx((1)/(2))^(2)=(piE)/(4)=(q)/(4epsilon_(0))` The displacement current `i_(d)=epsilon_(0)=(edPhi_(E))/(dt)=(1)/(4)(dq)/(dt)=0.5xx10^(-6)exp(-1)` at `t=10^(-3)s`. Now, applying Ampere-maxwell law to the loop we get `Bxx2pixx((1)/(2))=pi_(0)(i_(c)+i_(d))=mu_(0)(0+i_(d))=0.5xx10^(-6)mu_(0)exp(-1)` or, `B=0.74xx10^(-3)T` |
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| 36. |
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. `1xx10^(9)Vs^(-1)`B. `2xx10^(10)Vs^(-1)`C. `3xx10^(12)Vs^(-1)`D. `2xx10^(9)Vs^(-1)` |
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Answer» Correct Answer - D `(dv)/(dt)=(dI_(d))/(epsilon_(0)A)` |
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| 37. |
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. `5.50A`B. `5.56xx10^(2)A`C. `5.56xx10^(3)A`D. `2.28xx10^(4)A` |
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Answer» Correct Answer - C `I_(d)=(epsilon_(0)A)/(d)xx(dv)/(dt)` |
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