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NEET • PHYSICS • ELECTROMAGNETIC WAVES

Electromagnetic waves practice · Mixed difficulty

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31 Medium
31 of 145
Remembering +4 | -1
The velocity of electromagnetic radiation in a medium of permittivity ε0{\varepsilon _0} and permeability μ0{\mu _0} is given by
View answer and explanation
Correct option: C
1μ0∈0\frac{1}{{\sqrt {{\mu _0}{ \in _0}} }}
Option C, 1μ0∈0\frac{1}{{\sqrt {{\mu _0}{ \in _0}} }}, is correct because it represents the inverse square root of the product of permittivity and permeability, which defines the speed of electromagnetic radiation.
32 Medium
32 of 145
Understanding +4 | -1
Which of the following relation is correct?
View answer and explanation
Correct option: B
μ0ε0E0=B0\sqrt {{\mu _0}{\varepsilon _0}} {E_0} = {B_0}
Option B (μ0ε0E0=B0\sqrt {{\mu _0}{\varepsilon _0}} {E_0} = {B_0}) is correct because the speed of light in vacuum is c=E0B0=1μ0ε0c = \frac{E_0}{B_0} = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}, therefore B0=E0μ0ε0B_0 = E_0 \sqrt{\mu_0 \varepsilon_0}.
33 Medium
33 of 145
Application +4 | -1
An electromagnetic wave propagates in a medium, whose electric field vector is given as E=200sin⁡(18×108t+8x) V m−1E = 200\sin \left( {18 \times {{10}^8}t + 8x} \right)\,V\,{m^{ - 1}}. Refractive index of medium and amplitude of magnetic field will be
View answer and explanation
Correct option: D
μ=43; B0=89×10−6T\mu = \frac{4}{3};\,{B_0} = \frac{8}{9} \times {10^{ - 6}}T
Option D (μ=43; B0=89×10−6T\mu = \frac{4}{3};\,{B_0} = \frac{8}{9} \times {10^{ - 6}}T) is correct because the refractive index is the ratio of the speed of light in vacuum to the speed of light in the medium, and the amplitude of the magnetic field is the amplitude of the electric field divided by the speed of light in the medium.
34 Medium
34 of 145
Application +4 | -1
Magnetic field in a plane electromagnetic wave is given by Expression for corresponding electric field will be where c is speed of light
View answer and explanation
Correct option: A
E→=B0csin⁡(kx+ωt)k^V/m\overrightarrow E = {B_0}c\sin \left( {kx + \omega t} \right)\hat kV/m
Option A (E→=B0csin⁡(kx+ωt)k^V/m\overrightarrow E = {B_0}c\sin \left( {kx + \omega t} \right)\hat kV/m) is correct because the electric field is perpendicular to the magnetic field and the direction of propagation, and its magnitude is related to the magnetic field by E=cBE = cB.
35 Medium
35 of 145
Remembering +4 | -1
Ozone layer blocks the radiations of wavelength
View answer and explanation
Correct option: B
Equal to 3×10−73 \times {10^{ - 7}}m
Option A is correct because the ozone layer primarily blocks ultraviolet radiations with wavelengths less than 3 × 10-7 m.
36 Medium
36 of 145
Application +4 | -1
A plane electromagnetic wave, has frequency of 2.0×1010Hz2.0 \times {10^{10}}Hz and its energy density is 1.02×10−8J/m31.02 \times {10^{ - 8}}J/{m^3} in vacuum. The amplitude of the magnetic field of the wave is close to (Take, 14πε0=9×109Nm2C2\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\frac{{N{m^2}}}{{{C^2}}} and speed of light =3×108ms−1= 3 \times {10^8}m{s^{ - 1}} )
View answer and explanation
Correct option: B
160 nT
Option B (160 nT) is correct because the energy density of an electromagnetic wave is related to the amplitude of the magnetic field by u=B22μ0u = \frac{B^2}{2\mu_0}.
37 Hard
37 of 145
Application +4 | -1
Assume that all the energy from a 1000 W1000 \mathrm{~W} lamp is radiated uniformly, then the amplitude of electric field of radiation at a distance of 2 m2 \mathrm{~m} from the lamp is
View answer and explanation
Correct option: A
245.01 V/m245.01 \mathrm{~V} / \mathrm{m}
Option A (245.01 V/m245.01 \mathrm{~V} / \mathrm{m}) is correct because it is calculated using the relationship between power, intensity, and electric field amplitude for electromagnetic waves.
38 Hard
38 of 145
Understanding +4 | -1
The condition under which a microwave oven heats up a food item containing water molecules most efficiently in
View answer and explanation
Correct option: A
the frequency of the microwave must match the resonant frequency of the water molecules
Option A (the frequency of the microwave must match the resonant frequency of the water molecules) is correct because maximum energy transfer occurs when the microwave frequency matches the resonant frequency of water molecules, leading to efficient heating.
39 Medium
39 of 145
Application +4 | -1
Light with an energy flux of 25×104 Wm−225 \times {10^4}\,W{m^{ - 2}} falls on a perfectly reflecting surface at normal incidence, If the surface area is 15  cm215\,\,c{m^2}, The average force exerted on the surface is
View answer and explanation
Correct option: B
2.50×10−6N2.50 \times {10^{ - 6}}N
Option B (2.50×10−6N2.50 \times {10^{ - 6}}N) is correct because it is the average force calculated using the formula F=(2IA)/cF = (2IA)/c for a perfectly reflecting surface.
40 Medium
40 of 145
Application +4 | -1
If 150 J of energy is incident on area 2 m2{m^2}. If Qr=  15  J,{Q_r} = \;15\;\mathrm{J}, coefficient of absorption is 0.6, then amount of energy transmitted is
View answer and explanation
Correct option: B
45 J
Option C is correct because the magnetic field outside the capacitor plates decreases with distance from the axis.
41 Medium
41 of 145
Application +4 | -1
The magnetic field between the plate of a capacitor where r>Rr > R is given by (where rr is the distance from the axis of plates and RR is the radius of each plate of capacitor)
View answer and explanation
Correct option: C
μ0iD2πr\frac{{{\mathrm{\mu }_0}{i_D}}}{{2\mathrm{\pi }r}}
Option B is correct because the expression for magnetic field strength B at the point between the capacitor plates is given by ε0μ0r2dE/dt\frac{{{\varepsilon _0}{\mathrm{\mu }_0}r}}{2}dE/dt.
42 Medium
42 of 145
Application +4 | -1
A radiation of 200 W is incident on a surface which is 60% reflecting and 40% absorbing. The total force on the surface is
View answer and explanation
Correct option: A
1.07×10−61.07 \times {10^{ - 6}} N
Option A is correct because the total force is the sum of the force due to absorption and reflection.
43 Medium
43 of 145
Remembering +4 | -1
Speed of EM wave in free space is given by (where symbols have their usual meaning)
View answer and explanation
Correct option: C
c=1μ0ε0c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}
Option C, c=1μ0ε0c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}, is correct because it is the standard formula for the speed of electromagnetic waves in free space.
44 Hard
44 of 145
Application +4 | -1
The electric field of a plane electromagnetic wave is given by E = E0(x^+y^) sin⁡ (kz−ωt)E\, = \,{E_0}\left( {\mathrm{\hat x} + \mathrm{\hat y}} \right)\,\sin \,\left( {kz - \omega t} \right) its magnetic field will be given by
View answer and explanation
Correct option: B
E0c(−x^+y^) sin⁡ (kz−ωt)\frac{{{E_0}}}{c}\left( { - \hat x + \hat y} \right)\,\sin \,\left( {kz - \omega t} \right)
Option B (E0c(−x^+y^) sin⁡ (kz−ωt)\frac{{{E_0}}}{c}\left( { - \hat x + \hat y} \right)\,\sin \,\left( {kz - \omega t} \right)) is correct because the magnetic field must be perpendicular to both the electric field and the direction of propagation, and their cross product must point in the direction of propagation.
45 Medium
45 of 145
Application +4 | -1
In a uniform magnetic field of 10−5T{10^{ - 5}}T in free space, the energy density is u. The electric field which will produce the same energy density in free space is
View answer and explanation
Correct option: B
3×103V/m3 \times {10^3}V/m
Option B (3×103V/m3 \times {10^3}V/m) is correct because it equates the energy density of the magnetic field to the energy density of the electric field and solves for the electric field strength.
46 Easy
46 of 145
Application +4 | -1
If frequency of electromagnetic wave is 60 MHz60 \mathrm{~MHz} and it travels in air along zz direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in m\mathrm{m}) is
View answer and explanation
Correct option: D
5
Option D (5) is correct because the wavelength is calculated using the formula λ=c/f\lambda = c/f, where cc is the speed of light and ff is the frequency.
47 Hard
47 of 145
Application +4 | -1
A parallel plate capacitor of plate separation 2 mm is connected in on electric circuit having source voltage 400 volts. It the plate area is 60 cm260\,c{m^2}, then the value of displacement current for 10−6{10^{ - 6}} sec. will be
View answer and explanation
Correct option: B
1.062×10−2amp1.062 \times {10^{ - 2}}\mathrm{ amp}
Option C is correct because k/ωk/\omega represents the inverse of the wave's phase velocity, which equals 1/c1/c in vacuum, and thus is independent of wavelength.
48 Medium
48 of 145
Analyzing +4 | -1
Assertion: The relative velocity of two photons travelling in opposite directions is the velocity of light.

Reason: The rest mass of photon is zero
View answer and explanation
Correct option: B
If both assertion and reason are true, but reason is not correct explanation of the assertion.
Option B is correct because the relative velocity of two photons traveling in opposite directions is c, and the rest mass of a photon is zero, but the latter doesn't explain the former.
49 Easy
49 of 145
Application +4 | -1
nn photons of wavelength λ\lambda are absorbed by a black body of mass 'mm'. The momentum gained by the body is
View answer and explanation
Correct option: A
nhλ\frac{nh}{\lambda}
Option A (nhλ\frac{nh}{\lambda}) is correct because the momentum gained by the body is the sum of the momenta of the absorbed photons.
50 Medium
50 of 145
Remembering +4 | -1
The relation between electric field vector E\mathbf{E}, the displacement vector D\mathbf{D} and the polarization vector P\mathbf{P} for a dielectric placed in electric field E\mathbf{E} is given by
View answer and explanation
Correct option: C
D=ε0E+P\mathbf{D} = {\mathrm{\varepsilon }_0}\mathbf{E} + \mathbf{P}
Option C (D=ε0E+P\mathbf{D} = {\mathrm{\varepsilon }_0}\mathbf{E} + \mathbf{P}) is correct because it represents the fundamental relationship between the electric displacement vector, electric field vector, and polarization vector in a dielectric material.
51 Medium
51 of 145
Application +4 | -1
A plane electromagnetic wave of frequency 100 MHz is travelling in vacuum along the xx-direction. At a particular point in space and time, B =2.0×10−8k^TB\, = 2.0 \times {10^{ - 8}}\widehat kT (where, k^\widehat k is unit vector along zz-direction). What is EE at this point?
View answer and explanation
Correct option: C
6.0 j^ V/m6.0\,\widehat j\,V/m
Option C (6.0 j^ V/m6.0\,\widehat j\,V/m) is correct because the electric field is related to the magnetic field by E=cBE = cB and the direction is given by the Poynting vector.
52 Medium
52 of 145
Understanding +4 | -1
Electromagnetic waves propagate in the direction parallel to the vector (E‾\overline E is electric field and B‾\overline B is magnetic field)
View answer and explanation
Correct option: C
E×BE \times B
Option C (E×BE \times B) is correct because the direction of propagation of an electromagnetic wave is given by the Poynting vector, which is proportional to the cross product of the electric and magnetic field vectors.
53 Medium
53 of 145
Remembering +4 | -1
If μ0{\mu _0} is permeability of free space and ε0{\varepsilon _0} is permittivity of free space, the speed of light in vacuum is given by
View answer and explanation
Correct option: C
1μ0ε0\sqrt {\frac{1}{{{\mu _0}{\varepsilon _0}}}}
Option C (1μ0ε0\sqrt {\frac{1}{{{\mu _0}{\varepsilon _0}}}}) is correct because it represents the speed of light in a vacuum, where μ₀ is the permeability of free space and ε₀ is the permittivity of free space.
54 Medium
54 of 145
Remembering +4 | -1
Solar radiation is
View answer and explanation
Correct option: A
Transverse Electromagnetic wave
Option A is correct because solar radiation consists of transverse electromagnetic waves.
55 Easy
55 of 145
Understanding +4 | -1
During the porpagation of electromagnetic waves in a medium,
View answer and explanation
Correct option: C
electric energy density is equal to the magnetic energy density
Option C (electric energy density is equal to the magnetic energy density) is correct because in an electromagnetic wave, the electric and magnetic energy densities are equal on average.
56 Medium
56 of 145
Remembering +4 | -1
The pressure exerted by an electromagnetic wave of intensity I (Wm−2)\left(\mathrm{Wm}^{-2}\right) on a non-reflecting surface is ( CC is the velocity of light)
View answer and explanation
Correct option: C
I/cI/c
Option C (I/cI/c) is correct because the pressure exerted by an electromagnetic wave on a non-reflecting surface is the intensity divided by the speed of light.
57 Easy
57 of 145
Application +4 | -1
In an electromagnetic wave, at any instant the electric field was found to oscillate with an amplitude of 18 v/m18\, \mathrm{v} / \mathrm{m}. The magnitude of oscillating magnetic field will be -
View answer and explanation
Correct option: B
6×10−8 T6 \times 10^{-8} \mathrm{~T}
Option B (6×10−8 T6 \times 10^{-8} \mathrm{~T}) is correct because the magnitude of the oscillating magnetic field is calculated using the formula B0=E0cB_0 = \frac{E_0}{c}, where c is the speed of light.
58 Easy
58 of 145
Application +4 | -1
The sun delivers 103 W/m2{10^3}\,W/{m^2} of electromagnetic flux to the earth’s surface. Find the total power that is incident on a roof of dimensions 8 m×20 m8\,m \times 20\,m and the radiation force on the roof is given by 5.33×10−4N5.33 \times {10^{ - 4}}N
View answer and explanation
Correct option: C
1.6×105watt1.6 \times {10^5}\mathrm{ watt}
Option C (1.6×105watt1.6 \times {10^5}\mathrm{ watt}) is correct because the total power is the product of the intensity and the area of the roof.
59 Medium
59 of 145
Understanding +4 | -1
The electric and the magnetic field, associated with an e.m. wave, propagating along the + z− axis,\mathrm{ + }\,\mathrm{z - }\,\mathrm{axis,} can be represented by
View answer and explanation
Correct option: A
[E→=E0i^,  B→=B0j^]\left[ {\overrightarrow E = {E_0}\hat i,\,\,\overrightarrow B = {B_0}\hat j} \right]
Option A [E→=E0i^,,B→=B0j^]\left[ {\overrightarrow E = {E_0}\hat i,\\,\overrightarrow B = {B_0}\hat j} \right] is correct because the electric and magnetic fields must be perpendicular to each other and to the direction of propagation (+z-axis).
60 Easy
60 of 145
Application +4 | -1
If equation of wave is given by y=0.05sin⁡(2x−4t)y=0.05 \sin (2 x-4 t). Find velocity of wave?
View answer and explanation
Correct option: A
2
Option A (2) is correct because the velocity of the wave is the coefficient of t divided by the coefficient of x.
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