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JEE • PHYSICS • ELECTROMAGNETIC WAVES

Electromagnetic waves practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 151-180 of 226 questions

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151 Medium
151 of 226
Understanding +4 | -1
Which one of the following rays (or wave) has
maximum speed in air?
View answer and explanation
Correct option: D
Heat radiations
Option D is correct because heat radiations, being electromagnetic waves, travel at the speed of light in air, which is the maximum speed among the given options.
152 Medium
152 of 226
Application +4 | -1
The electric field of an electromagnetic wave is given by E=100sin⁡ω(t−xv)N/C.E=100 \sin \omega\left(t-\frac{x}{v}\right) N / C.

Find the energy contained in a hypothetical space of volume 100 cm3100 \mathrm{~cm}^3

(Given ϵ0=9×10−12C2N−1m−2)\left.\epsilon_0=9 \times 10^{-12} C^2 N^{-1} m^{-2}\right)
View answer and explanation
Correct option: C
2.25×10−12 J2.25 \times 10^{-12} \mathrm{~J}
Option C (2.25×10−12 J2.25 \times 10^{-12} \mathrm{~J}) is correct because the energy density of an electromagnetic wave is proportional to the square of the electric field, and the energy is the energy density times the volume.
153 Medium
153 of 226
Remembering +4 | -1
The range of wavelengths for visible light is
View answer and explanation
Correct option: C
4000Ao4000 \mathop {\text{A}}\limits^{\text{o}} to 8000Ao8000 \mathop {\text{A}}\limits^{\text{o}}
Option C (4000Ao4000 \mathop {\text{A}}\limits^{\text{o}} to 8000Ao8000 \mathop {\text{A}}\limits^{\text{o}}) is correct because the visible light spectrum ranges approximately from 400 nm to 800 nm, which is 4000 to 8000 Angstroms.
154 Medium
154 of 226
Understanding +4 | -1
Which of the following Maxwell's equation is valid for time varying conditions but not valid for static conditions:
View answer and explanation
Correct option: D
∮E⃗⋅dl→=−∂ϕB∂t\oint \vec{E} \cdot \overrightarrow{d l}=-\frac{\partial \phi_{B}}{\partial t}
Option D (∮E⃗⋅dl→=−∂ϕB∂t\oint \vec{{E}} \cdot \overrightarrow{{dl}}=-\frac{{\partial \phi_{{B}}}}{{\partial t}}) is correct because this is Faraday's law of induction, which is only valid for time-varying magnetic fields.
155 Medium
155 of 226
Understanding +4 | -1
Assertion: Radar is used to detect an aeroplane in the sky

Reason: Radar works on the principle of reflection of waves.
View answer and explanation
Correct option: A
If both assertion and reason are true and the reason is the correct explanation of the assertion.
Option A is correct because radar uses the reflection of radio waves to detect objects like airplanes.
156 Medium
156 of 226
Understanding +4 | -1
Assertion: In electromagnetic waves electric field and magnetic field lines are perpendicular to each other.

Reason: E and B are self - sustaining
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
The correct answer is B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) because both statements are true, but the self-sustaining nature of E and B doesn't directly explain their perpendicularity.
157 Medium
157 of 226
Remembering +4 | -1
Which of the following is correct?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because Maxwell's equations describe that a time-varying magnetic field produces an electric field, a time-varying electric field produces a magnetic field, and E = Bc relates the magnitudes of the electric and magnetic fields in an electromagnetic wave.
158 Medium
158 of 226
Understanding +4 | -1
Statement I: If earth did not have atmosphere, its average surface temperature would be lower than the present average surface temperature

Statement II: Greehouse effect would be absent if earth did not have atmosphere
View answer and explanation
Correct option: A
Statement I and statement II are true
Option A is correct because the absence of an atmosphere would result in a lower average surface temperature due to the lack of the greenhouse effect.
159 Medium
159 of 226
Application +4 | -1
A plane electromagnetic wave of wave intensity 6Wm−26 {Wm}^{-2} strikes a small mirror of area 30 cm230 {~cm}^2, held perpendicular to the approaching wave. The the mirror each second will be
View answer and explanation
Correct option: A
1.2×10−101.2 \times 10^{-10}
Option A (1.2×10−101.2 \times 10^{-10}) is correct because the momentum transferred is calculated using the formula p = 2IA/c, where I is the intensity, A is the area, and c is the speed of light.
160 Medium
160 of 226
Analyzing +4 | -1
If E\mathrm{E} and B\mathrm{B} denote electric and magnetic fields respectively, which of the following is dimensionless?
View answer and explanation
Correct option: A
μ0ε0EB\sqrt{\mu_0 \varepsilon_0} \frac{\mathrm{E}}{\mathrm{B}}
Option A is correct because the expression μ0ε0\sqrt{μ_0 ε_0} EB\frac{\mathrm{E}}{\mathrm{B}} is dimensionless, as μ0ε0\sqrt{μ_0 ε_0} is the inverse of the speed of light and EB\frac{\mathrm{E}}{\mathrm{B}} is the speed of light.
161 Medium
161 of 226
Remembering +4 | -1
The intensity of electromagnetic wave is equal to (symbols have their usual meanings)
View answer and explanation
Correct option: A
ε0ErmsBrmsC2\varepsilon_0 E_{{rms}} B_{{rms}} C^2
Option B (ε0E2\varepsilon_0 E^2 rms) is incorrect. The correct formula is I=cϵ0Erms2=cBrms2μ0=cϵ0ErmsBrmsI = c \epsilon_0 E_{rms}^2 = c \frac{B_{rms}^2}{ \mu_0} = c \epsilon_0 E_{rms}B_{rms}
162 Medium
162 of 226
Understanding +4 | -1
If the total electromagnetic energy falling on a surface is UU, then the total momentum delivered (for complete absorption) is
View answer and explanation
Correct option: A
Uc\frac{U}{c}
Option A (Uc\frac{U}{c}) is correct because the momentum delivered for complete absorption is the energy divided by the speed of light.
163 Medium
163 of 226
Application +4 | -1
The amplitude of the sinusoidally oscillating electric field of plane wave is 60  Vm60\,\,\frac{V}{m}. Then the amplitude of the magnetic field is
View answer and explanation
Correct option: D
2×10−7T2 \times {10^{ - 7}}T
Option D (2×10−7T2 \times {10^{ - 7}}T) is correct because it is calculated using the relationship between the electric field amplitude, magnetic field amplitude, and the speed of light.
164 Medium
164 of 226
Understanding +4 | -1
An accelerated charge will not produce
View answer and explanation
Correct option: D
a Gravitational field
Option D is correct because an accelerated charge produces electromagnetic waves, electric fields, and magnetic fields, but not gravitational fields.
165 Medium
165 of 226
Analyzing +4 | -1
Statement I: Displacement current decreases with the increase in frequency of a.c. supplied to the capacitor

Statement II: Reactance due to capacitance is directly proprtional to the frequency of a.c.
View answer and explanation
Correct option: B
Statement I is true; statement II is false
Option B (Statement I is true; statement II is false) is correct because displacement current decreases with frequency, and capacitive reactance is inversely proportional to frequency.
166 Medium
166 of 226
Remembering +4 | -1
The energy density associated with electric field E⃗\vec{E} and magnetic field B⃗\vec{B} of an electromagnetic wave in free space is given by (ε0−\left(\varepsilon_{0}-\right. permittivity of free space, μ0\mu_{0} - permeability of free space)
View answer and explanation
Correct option: B
UE=ε0E22, UB=B22μ0U_{E}=\frac{\varepsilon_{0} E^{2}}{2}, \, U_{B}=\frac{B^{2}}{2 \mu_{0}}
Option B (UE=ε0E22, UB=B22μ0U_{E}=\frac{\varepsilon_{0} E^{2}}{2}, \, U_{B}=\frac{B^{2}}{2 \mu_{0}}) is correct because it represents the standard formulas for energy density of electric and magnetic fields in free space.
167 Medium
167 of 226
Remembering +4 | -1
Maxwell in his famous equation of electromagnetism introduced the concept of
View answer and explanation
Correct option: C
Displacement current
Option C (Displacement current) is correct because Maxwell introduced the concept of displacement current to modify Ampere's law and make it consistent with the continuity equation.
168 Medium
168 of 226
Remembering +4 | -1
Biological importance of Ozone layer is
View answer and explanation
Correct option: A
It stops ultraviolet rays
Option A (It stops ultraviolet rays) is correct because the ozone layer absorbs harmful ultraviolet radiation from the sun.
169 Medium
169 of 226
Application +4 | -1
If value of electric field E=6.3×1027volt/mE=6.3 \times 10^{27} \mathrm{volt} / \mathrm{m} for a electromagnetic wave. The value of magnetic field B will be:
View answer and explanation
Correct option: B
2.1×1019 T2.1 \times 10^{19} \mathrm{~T}
Option B (2.1×1019 T2.1 \times 10^{19} \mathrm{~T}) is correct because it is calculated using the relationship between the electric field, magnetic field, and speed of light in an electromagnetic wave.
170 Medium
170 of 226
Understanding +4 | -1
Which of the following can be deflected in magnetic field?
View answer and explanation
Correct option: B
Cathode rays
Option B (Cathode rays) is correct because cathode rays are streams of electrons, which are charged particles that are deflected by a magnetic field.
171 Medium
171 of 226
Application +4 | -1
A plane electromagnetic wave Ez=100cos⁡(6×108t+4x)V/m{E_z} = 100\cos \left( {6 \times {{10}^8}t + 4x} \right)V/m, where x is in metre and t is in second. Propagates in medium , what is refractive index of the medium ?
View answer and explanation
Correct option: B
2
Option B (2) is correct because the refractive index is the ratio of the speed of light in vacuum to the speed of light in the medium.
172 Medium
172 of 226
Remembering +4 | -1
Cascade theory of showers of Cosmic rays is invented by
View answer and explanation
Correct option: B
H. J. Bhabha and Heitler
Option B (H. J. Bhabha and Heitler) is correct because Bhabha and Heitler are credited with developing the cascade theory to explain the showers of cosmic rays.
173 Medium
173 of 226
Understanding +4 | -1
For a plane electromagnetic wave, average of which among the following is zero?
View answer and explanation
Correct option: A
Electric field
Option A (Electric field) is correct because the average value of the electric field over a complete cycle in a plane electromagnetic wave is zero.
174 Medium
174 of 226
Understanding +4 | -1
The magnetic field BB and electric field EE in an electromagnetic wave are:
View answer and explanation
Correct option: C
mutually perpendicular but with 0∘0^{\circ} phase difference
Option C is correct because in an electromagnetic wave, the electric and magnetic fields are mutually perpendicular and oscillate in phase (0° phase difference).
175 Medium
175 of 226
Understanding +4 | -1
In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size λ100,\frac{\lambda }{{100}}, where λ\lambda is the wavelength of the wave in free space. The phenomenon, which happens there will be:
View answer and explanation
Correct option: D
Scattering
Option D is correct because when the object size is much smaller than the wavelength, scattering occurs.
176 Medium
176 of 226
Application +4 | -1
An electromagnetic wave of intensity 50  Wm−250\,\,W{m^{ - 2}} enters in a medium of refractive index ‘n’ without any loss. The ratio of the magnitudes of electric fields and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by
View answer and explanation
Correct option: D
(n,1n)\left( {\sqrt n ,\frac{1}{{\sqrt n }}} \right)
Option D (n,1n)\left( {\sqrt n ,\frac{{1}}{{\sqrt n }}} \right) is correct because the ratio of electric fields is n\sqrt{n} and the ratio of magnetic fields is 1n\frac{1}{\sqrt{n}} after entering the medium.
177 Medium
177 of 226
Application +4 | -1
An earth orbiting satellite has solar energy collecting panel with total area 5  m25\mathrm{\;}{\mathrm{m}^2}. If solar radiations are perpendicular and completely absorbed, the average force associated with the radiation pressure is
(Solar constant=1.4  kWm−2= 1.4\mathrm{\;kW}{\mathrm{m}^{ - 2}})
View answer and explanation
Correct option: C
2.33×10−5  N2.33 \times {10^{ - 5}}\mathrm{\;N}
Option C (2.33×10−5  N2.33 \times {10^{ - 5}}\mathrm{\;N}) is correct because the force due to radiation pressure is calculated by dividing the power by the speed of light.
178 Medium
178 of 226
Understanding +4 | -1
For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric (Ue)({U_e}) and magnetic (Um)({U_m}) fields is:
View answer and explanation
Correct option: B
Ue=Um{U_e} = {U_m}
Option B (Ue=Um{U_e} = {U_m}) is correct because in an electromagnetic wave, the average energy density due to the electric field is equal to the average energy density due to the magnetic field.
179 Medium
179 of 226
Understanding +4 | -1
Statement I: A changing electric-field produces a magnetic field.

Statement II: A changing magnetic field produces an electric field.
View answer and explanation
Correct option: A
Statement I and statement II are true
Option A (Statement I and statement II are true) is correct because changing electric fields produce magnetic fields, and changing magnetic fields produce electric fields, as described by Maxwell's equations.
180 Medium
180 of 226
Application +4 | -1
Suppose that the electric field part of an electromagnetic wave in vacuum is E⃗=(3.1 N/C)cos⁡[(1.8 rad/m)y+(5.4×108 rad/s)t]i^\vec{E}=(3.1 \mathrm{~N} / \mathrm{C}) \cos \left[(1.8 \mathrm{~rad} / \mathrm{m}) y+\left(5.4 \times 10^8 \mathrm{~rad} / \mathrm{s}\right) {t}\right] \hat{i}. What is the frequency v{v}?
View answer and explanation
Correct option: A
86 MHz86 \mathrm{~MHz}
Option A (86 MHz) is correct because the frequency is calculated from the angular frequency using the formula v=ω2πv = \frac{\omega}{{2\pi}}.
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