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Physics

Electromagnetic Waves — JEE Main PYQs

34 previous year questions

Q1JEE Main 2026 Apr 2 Shift 2Medium

An electromagnetic wave travels in free space along the xx-direction. At a particular point in space and time, B=2×107j^\vec{B} = 2 \times 10^{-7}\,\hat{j} T is associated with this wave. The value of corresponding electric field E\vec{E} at this point is ________ V/m.

  1. A

    60k^60\,\hat{k}

  2. B

    60k^-60\,\hat{k}

  3. C

    30k^30\,\hat{k}

  4. D

    600k^-600\,\hat{k}

Show answer

Correct option: B

Q2JEE Main 2026 Apr 4 Shift 2Medium

A magnetic field vector in an electromagnetic wave is represented by B=B0sin(2πνt2πxλ)j^\vec{B} = B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{j}. Its associated electric field vector is _________.

  1. A

    E=νλB0sin(2πνt2πxλ)k^\vec{E} = -\nu\lambda\, B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{k}

  2. B

    E=νλB0sin(2πνt2πxλ)i^\vec{E} = -\nu\lambda\, B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{i}

  3. C

    E=νλB0sin(2πνt2πxλ)k^\vec{E} = \nu\lambda\, B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{k}

  4. D

    E=νλB0sin(2πνt2πxλ)i^\vec{E} = \nu\lambda\, B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{i}

Show answer

Correct option: A

Q3JEE Main 2026 Apr 5 Shift 1Medium

A displacement current of 4.0 A can be set up in the space between two parallel plates of 6 μF6 \text{ } \mu\text{F} capacitor. The rate of change of potential difference across the plates of the capacitor is nearly α×106\alpha \times 10^6 V/s. The value of α\alpha is ________.

  1. A

    0.58

  2. B

    0.67

  3. C

    0.82

  4. D

    0.75

Show answer

Correct option: B

Q4JEE Main 2026 Apr 5 Shift 2Easy

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : The electromagnetic wave exerts pressure on the surface on which they are allowed to fall.

Reason (R) : There is no mass associated with the electromagnetic waves.

In the light of the above statements, choose the correct answer from the options given below :

  1. A

    Both (A) and (R) are true and (R) is the correct explanation of (A)

  2. B

    Both (A) and (R) are true but (R) is not the correct explanation of (A)

  3. C

    (A) is true but (R) is false

  4. D

    (A) is false but (R) is true

Show answer

Correct option: B

Q5JEE Main 2026 Apr 6 Shift 1Easy

A point light source emits E.M. waves in free space. A detector, placed at a distance of LL m, measures the intensity as IoI_o. The detector is now shifted to another location on the same spherical surface ensuring the angle between original location and new location as 45°45°. The measured intensity at new location will be ________.

  1. A

    Io4\frac{I_o}{4}

  2. B

    IoI_o

  3. C

    Io2\frac{I_o}{\sqrt{2}}

  4. D

    Io2\frac{I_o}{2}

Show answer

Correct option: B

Q6JEE Main 2026 Apr 6 Shift 2Easy

For an electromagnetic wave propagating through vacuum, k\vec{k}, E\vec{E} and ω\omega represent propagation vector, electric field and angular frequency, respectively. The magnetic field associated with this wave is represented by :

  1. A

    E×kω\dfrac{\vec{E} \times \vec{k}}{\omega}

  2. B

    k×Eω\dfrac{\vec{k} \times \vec{E}}{\omega}

  3. C

    ω(E×k)\omega\left(\vec{E} \times \vec{k}\right)

  4. D

    ω(k×E)\omega\left(\vec{k} \times \vec{E}\right)

Show answer

Correct option: B

Q7JEE Main 2026 Apr 8 Shift 2Medium

A monochromatic source of light operating at 15kW15\,\text{kW} emits 2.5×10222.5 \times 10^{22} photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ________. (Take h=6.6×1034J.sh = 6.6 \times 10^{-34}\,\text{J.s} and c=3×108m/sc = 3 \times 10^8\,\text{m/s}).

  1. A

    Microwave

  2. B

    Infrared

  3. C

    Visible

  4. D

    Ultraviolet

Show answer

Correct option: D

Q8JEE Main 2025 Apr 3 Shift 1Medium

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2m away from it, is

  1. A

    3×1083 \times 10^{-8} Pascals

  2. B

    6×1086 \times 10^{-8} Pascals

  3. C

    1.5×1081.5 \times 10^{-8} Pascals

  4. D

    0

Show answer

Correct option: B

Q9JEE Main 2025 Apr 4 Shift 2MediumNumerical

If an optical medium possesses a relative permeability of 10π\dfrac{10}{\pi} and relative permittivity of 10.0885\dfrac{1}{0.0885}, then the velocity of light is greater in vacuum than that in this medium by __________ times.

(μ0=4π×107 H/m\mu_0 = 4\pi \times 10^{-7} \mathrm{~H/m}, ϵ0=8.85×1012 F/m\epsilon_0 = 8.85 \times 10^{-12} \mathrm{~F/m}, c=3×108 m/s\mathrm{c} = 3 \times 10^{8} \mathrm{~m/s})

Show answer

Answer: 6

Q10JEE Main 2025 Apr 7 Shift 2Medium

The unit of 2Iϵ0c\sqrt{\dfrac{2I}{\epsilon_0 c}} is :

(I = intensity of an electromagnetic wave, c : speed of light)

  1. A

    Vm\mathrm{Vm}

  2. B

    NC1\mathrm{NC^{-1}}

  3. C

    NC\mathrm{NC}

  4. D

    Nm\mathrm{Nm}

Show answer

Correct option: B

Q11JEE Main 2025 Jan 23 Shift 1Medium

The electric field of an electromagnetic wave in free space is E=57cos[7.5×106t5×103(3x+4y)](4i^3j^) N/C.\vec{E} = 57 \cos\left[7.5\times10^{6}\,\mathrm{t} - 5\times10^{-3}\left(3x+4y\right)\right]\left(4\hat{i}-3\hat{j}\right)\ \mathrm{N/C}. The associated magnetic field in Tesla is

  1. A

    B=573×108cos[7.5×106t5×103(3x+4y)](k^)\vec{B} = -\frac{57}{3\times10^{8}} \cos\left[7.5\times10^{6}\,\mathrm{t} - 5\times10^{-3}\left(3x+4y\right)\right]\left(\hat{k}\right)

  2. B

    B=573×108cos[7.5×106t5×103(3x+4y)](k^)\vec{B} = \frac{57}{3\times10^{8}} \cos\left[7.5\times10^{6}\,\mathrm{t} - 5\times10^{-3}\left(3x+4y\right)\right]\left(\hat{k}\right)

  3. C

    B=573×108cos[7.5×106t5×103(3x+4y)](5k^)\vec{B} = -\frac{57}{3\times10^{8}} \cos\left[7.5\times10^{6}\,\mathrm{t} - 5\times10^{-3}\left(3x+4y\right)\right]\left(5\hat{k}\right)

  4. D

    B=573×108cos[7.5×106t5×103(3x+4y)](5k^)\vec{B} = \frac{57}{3\times10^{8}} \cos\left[7.5\times10^{6}\,\mathrm{t} - 5\times10^{-3}\left(3x+4y\right)\right]\left(5\hat{k}\right)

Show answer

Correct option: C

Q12JEE Main 2025 Jan 23 Shift 2Easy

A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x+x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3 Vm1E_y = 9.3~\mathrm{Vm^{-1}}. Then, the magnetic field vector of the wave at that point is

  1. A

    Bz=6.2×108B_z = 6.2 \times 10^{-8} T

  2. B

    Bz=9.3×108B_z = 9.3 \times 10^{-8} T

  3. C

    Bz=3.1×108B_z = 3.1 \times 10^{-8} T

  4. D

    Bz=1.55×108B_z = 1.55 \times 10^{-8} T

Show answer

Correct option: C

Q13JEE Main 2025 Jan 23 Shift 2MediumNumerical

A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5 μF2.5~\mu\mathrm{F}. The dielectric constant of the medium between the capacitor plates is 1. It produces an instantaneous displacement current of 0.25 mA in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be __________ Vs1\mathrm{Vs^{-1}}.

Show answer

Answer: 100

Q14JEE Main 2025 Jan 24 Shift 2Easy

Arrange the following in the ascending order of wavelength (λ\lambda) :

(A) Microwaves (λ1\lambda_1)

(B) Ultraviolet rays (λ2\lambda_2)

(C) Infrared rays (λ3\lambda_3)

(D) X-rays (λ4\lambda_4)

Choose the most appropriate answer from the options given below :

  1. A

    λ3<λ4<λ2<λ1\lambda_3 < \lambda_4 < \lambda_2 < \lambda_1

  2. B

    λ4<λ3<λ1<λ2\lambda_4 < \lambda_3 < \lambda_1 < \lambda_2

  3. C

    λ4<λ3<λ2<λ1\lambda_4 < \lambda_3 < \lambda_2 < \lambda_1

  4. D

    λ4<λ2<λ3<λ1\lambda_4 < \lambda_2 < \lambda_3 < \lambda_1

Show answer

Correct option: D

Q15JEE Main 2025 Jan 28 Shift 1Medium

Due to presence of an em-wave whose electric component is given by E=100sin(ωtkx) NC1E = 100 \sin(\omega t - kx)\ \mathrm{NC^{-1}}, a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

  1. A

    50sin(ωtkx) NC150 \sin(\omega t - kx)\ \mathrm{NC^{-1}}

  2. B

    200sin(ωtkx) NC1200 \sin(\omega t - kx)\ \mathrm{NC^{-1}}

  3. C

    25sin(ωtkx) NC125 \sin(\omega t - kx)\ \mathrm{NC^{-1}}

  4. D

    400sin(ωtkx) NC1400 \sin(\omega t - kx)\ \mathrm{NC^{-1}}

Show answer

Correct option: B

Q16JEE Main 2025 Jan 28 Shift 2Hard

The magnetic field of an E.M. wave is given by B=(32i^+12j^)30sin[ω(tzc)]\vec{B} = \left(\dfrac{\sqrt{3}}{2}\hat{i} + \dfrac{1}{2}\hat{j}\right)30\sin\left[\omega\left(t - \dfrac{z}{c}\right)\right] (S.I. Units).

The corresponding electric field in S.I. units is :

  1. A

    E=(32i^12j^)30csin[ω(t+zc)]\vec{E} = \left(\dfrac{\sqrt{3}}{2}\hat{i} - \dfrac{1}{2}\hat{j}\right)30\,c\sin\left[\omega\left(t + \dfrac{z}{c}\right)\right]

  2. B

    E=(34i^+14j^)30ccos[ω(tzc)]\vec{E} = \left(\dfrac{3}{4}\hat{i} + \dfrac{1}{4}\hat{j}\right)30\,c\cos\left[\omega\left(t - \dfrac{z}{c}\right)\right]

  3. C

    E=(12i^32j^)30csin[ω(tzc)]\vec{E} = \left(\dfrac{1}{2}\hat{i} - \dfrac{\sqrt{3}}{2}\hat{j}\right)30\,c\sin\left[\omega\left(t - \dfrac{z}{c}\right)\right]

  4. D

    E=(12i^+32j^)30csin[ω(t+zc)]\vec{E} = \left(\dfrac{1}{2}\hat{i} + \dfrac{\sqrt{3}}{2}\hat{j}\right)30\,c\sin\left[\omega\left(t + \dfrac{z}{c}\right)\right]

Show answer

Correct option: C

Q17JEE Main 2024 Apr 4 Shift 1Easy

The electric field in an electromagnetic wave is given by E=i^ 40cosω(tzc) NC1\vec{\mathrm{E}} = \hat{i}~40 \cos\omega\left(\mathrm{t} - \frac{z}{c}\right)~\mathrm{N\,C^{-1}}. The magnetic field induction of this wave is (in SI unit) :

  1. A

    B=j^ 40cosω(tzc)\vec{\mathrm{B}} = \hat{j}~40 \cos\omega\left(\mathrm{t} - \frac{z}{c}\right)

  2. B

    B=k^ 40ccosω(tzc)\vec{\mathrm{B}} = \hat{k}~\frac{40}{c} \cos\omega\left(\mathrm{t} - \frac{z}{c}\right)

  3. C

    B=j^ 40ccosω(tzc)\vec{\mathrm{B}} = \hat{j}~\frac{40}{c} \cos\omega\left(\mathrm{t} - \frac{z}{c}\right)

  4. D

    B=i^ 40ccosω(tzc)\vec{\mathrm{B}} = \hat{i}~\frac{40}{c} \cos\omega\left(\mathrm{t} - \frac{z}{c}\right)

Show answer

Correct option: C

Q18JEE Main 2024 Apr 4 Shift 2Easy

Arrange the following in the ascending order of wavelength:

A. Gamma rays (λ1\lambda_1) B. x - rays (λ2\lambda_2) C. Infrared waves (λ3\lambda_3) D. Microwaves (λ4\lambda_4)

Choose the most appropriate answer from the options given below

  1. A

    λ4<λ3<λ2<λ1\lambda_4 < \lambda_3 < \lambda_2 < \lambda_1

  2. B

    λ4<λ3<λ1<λ2\lambda_4 < \lambda_3 < \lambda_1 < \lambda_2

  3. C

    λ2<λ1<λ4<λ3\lambda_2 < \lambda_1 < \lambda_4 < \lambda_3

  4. D

    λ1<λ2<λ3<λ4\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4

Show answer

Correct option: D

Q19JEE Main 2024 Apr 5 Shift 1Medium

An alternating voltage of amplitude 40 V and frequency 4 kHz is applied directly across the capacitor of 12 μF12\ \mu\mathrm{F}. The maximum displacement current between the plates of the capacitor is nearly :

  1. A

    10 A10\ \mathrm{A}

  2. B

    12 A12\ \mathrm{A}

  3. C

    13 A13\ \mathrm{A}

  4. D

    8 A8\ \mathrm{A}

Show answer

Correct option: B

Q20JEE Main 2024 Apr 5 Shift 2Medium

Match List-I with List-II :

List-I (EM-Wave) List-II (Wavelength Range)
(A) Infra-red (I) <103< 10^{-3} nm
(B) Ultraviolet (II) 400 nm to 1 nm
(C) X-rays (III) 1 mm to 700 nm
(D) Gamma rays (IV) 1 nm to 10310^{-3} nm

Choose the correct answer from the options given below :

  1. A

    (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

  2. B

    (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

  3. C

    (A)-(I), (B)-(III), (C)-(II), (D)-(IV)

  4. D

    (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Show answer

Correct option: B

Q21JEE Main 2024 Apr 6 Shift 1Easy

Electromagnetic waves travel in a medium with speed of 1.5×108 m s11.5\times10^{8}\ \mathrm{m\ s^{-1}}. The relative permeability of the medium is 2.0. The relative permittivity will be:

  1. A

    1

  2. B

    2

  3. C

    4

  4. D

    5

Show answer

Correct option: B

Q22JEE Main 2024 Apr 6 Shift 2Medium

In the given electromagnetic wave Ey=600sin(ωtkx) Vm1E_y = 600 \sin(\omega t - kx)\ \mathrm{Vm^{-1}}, intensity of the associated light beam is (in W/m2\mathrm{W/m^2}) : (Given ϵ0=9×1012 C2N1m2\epsilon_0 = 9 \times 10^{-12}\ \mathrm{C^2 N^{-1} m^{-2}})

  1. A

    729

  2. B

    972

  3. C

    243

  4. D

    486

Show answer

Correct option: D

Q23JEE Main 2024 Apr 8 Shift 1Medium

Average force exerted on a non-reflecting surface at normal incidence is 2.4×1042.4 \times 10^{-4} N. If 360 W/cm2360\ \mathrm{W/cm^2} is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:

  1. A

    0.2 m20.2\ \mathrm{m^2}

  2. B

    0.1 m20.1\ \mathrm{m^2}

  3. C

    0.02 m20.02\ \mathrm{m^2}

  4. D

    20 m220\ \mathrm{m^2}

Show answer

Correct option: C

Q24JEE Main 2024 Apr 9 Shift 1Medium

A plane EM wave is propagating along xx direction. It has a wavelength of 4 mm. If electric field is in yy direction with the maximum magnitude of 60 Vm160\ \mathrm{Vm^{-1}}, the equation for magnetic field is :

  1. A

    Bz=60sin[π2(x3×108t)]k^ TB_z = 60 \sin\left[\dfrac{\pi}{2}\left(x - 3 \times 10^{8}\, t\right)\right] \hat{k}\ \mathrm{T}

  2. B

    Bx=60sin[π2(x3×108t)]i^ TB_x = 60 \sin\left[\dfrac{\pi}{2}\left(x - 3 \times 10^{8}\, t\right)\right] \hat{i}\ \mathrm{T}

  3. C

    Bz=2×107sin[π2(x3×108t)]k^ TB_z = 2 \times 10^{-7} \sin\left[\dfrac{\pi}{2}\left(x - 3 \times 10^{8}\, t\right)\right] \hat{k}\ \mathrm{T}

  4. D

    Bz=2×107sin[π2×103(x3×108t)]k^ TB_z = 2 \times 10^{-7} \sin\left[\dfrac{\pi}{2} \times 10^{3}\left(x - 3 \times 10^{8}\, t\right)\right] \hat{k}\ \mathrm{T}

Show answer

Correct option: D

Q25JEE Main 2024 Apr 9 Shift 2Medium

The magnetic field in a plane electromagnetic wave is By=(3.5×107)sin(1.5×103x+0.5×1011t)\mathrm{B_y} = (3.5\times10^{-7})\sin(1.5\times10^3 x + 0.5\times10^{11} t) T. The corresponding electric field will be :

  1. A

    Ey=1.17sin(1.5×103x+0.5×1011t) Vm1E_y = 1.17\sin(1.5\times10^3 x + 0.5\times10^{11}t)~\mathrm{Vm^{-1}}

  2. B

    Ey=10.5sin(1.5×103x+0.5×1011t) Vm1E_y = 10.5\sin(1.5\times10^3 x + 0.5\times10^{11}t)~\mathrm{Vm^{-1}}

  3. C

    Ez=105sin(1.5×103x+0.5×1011t) Vm1E_z = 105\sin(1.5\times10^3 x + 0.5\times10^{11}t)~\mathrm{Vm^{-1}}

  4. D

    Ez=1.17sin(1.5×103x+0.5×1011t) Vm1E_z = 1.17\sin(1.5\times10^3 x + 0.5\times10^{11}t)~\mathrm{Vm^{-1}}

Show answer

Correct option: C

Q26JEE Main 2024 Feb 1 Shift 1Medium

A parallel plate capacitor has a capacitance C = 200 pF. It is connected to 230 V ac supply with an angular frequency 300 rad/s. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :

  1. A

    13.8 μ\muA and 13.8 μ\muA

  2. B

    1.38 μ\muA and 1.38 μ\muA

  3. C

    14.3 μ\muA and 143 μ\muA

  4. D

    13.8 μ\muA and 138 μ\muA

Show answer

Correct option: A

Q27JEE Main 2024 Feb 1 Shift 2Easy

If frequency of electromagnetic wave is 60 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) is :

  1. A

    10

  2. B

    5

  3. C

    2.5

  4. D

    2

Show answer

Correct option: B

Q28JEE Main 2024 Jan 27 Shift 1Medium

A plane electromagnetic wave propagating in xx-direction is described by

Ey=(200 Vm1) sin[1.5×107t0.05 x]E_y = (200\ \mathrm{Vm^{-1}})\ \sin[1.5 \times 10^7 t - 0.05\ x]; The intensity of the wave is :

(Use ϵ0=8.85×1012 C2N1m2\epsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2 N^{-1} m^{-2}})

  1. A

    53.1 Wm253.1\ \mathrm{Wm^{-2}}

  2. B

    106.2 Wm2106.2\ \mathrm{Wm^{-2}}

  3. C

    26.6 Wm226.6\ \mathrm{Wm^{-2}}

  4. D

    35.4 Wm235.4\ \mathrm{Wm^{-2}}

Show answer

Correct option: A

Q29JEE Main 2024 Jan 29 Shift 1Easy

Match List I with List II

LIST I LIST II
A. Bdl=μoic+μoεodϕEdt\oint \vec{B} \cdot \vec{dl} = \mu_o i_c + \mu_o \varepsilon_o \dfrac{d\phi_E}{dt} I. Gauss' law for electricity
B. Edl=dϕBdt\oint \vec{E} \cdot \vec{dl} = \dfrac{d\phi_B}{dt} II. Gauss' law for magnetism
C. EdA=Qεo\oint \vec{E} \cdot \vec{dA} = \dfrac{Q}{\varepsilon_o} III. Faraday law
D. BdA=0\oint \vec{B} \cdot \vec{dA} = 0 IV. Ampere - Maxwell law

Choose the correct answer from the options given below:

  1. A

    A-I, B-II, C-III, D-IV

  2. B

    A-IV, B-III, C-I, D-II

  3. C

    A-IV, B-I, C-III, D-II

  4. D

    A-II, B-III, C-I, D-IV

Show answer

Correct option: B

Q30JEE Main 2024 Jan 30 Shift 1Easy

The electric field of an electromagnetic wave in free space is represented as E=E0cos(ωtkz)i^\vec{\mathrm{E}} = \mathrm{E}_0 \cos(\omega t - kz)\,\hat{i}. The corresponding magnetic induction vector will be :

  1. A

    B=E0Ccos(ωtkz)j^\vec{\mathrm{B}} = \dfrac{\mathrm{E}_0}{\mathrm{C}} \cos(\omega t - kz)\,\hat{j}

  2. B

    B=E0Ccos(ωtkz)j^\vec{\mathrm{B}} = \mathrm{E}_0 \mathrm{C} \cos(\omega t - kz)\,\hat{j}

  3. C

    B=E0Ccos(ωt+kz)j^\vec{\mathrm{B}} = \dfrac{\mathrm{E}_0}{\mathrm{C}} \cos(\omega t + kz)\,\hat{j}

  4. D

    B=E0Ccos(ωt+kz)j^\vec{\mathrm{B}} = \mathrm{E}_0 \mathrm{C} \cos(\omega t + kz)\,\hat{j}

Show answer

Correct option: A

Q31JEE Main 2024 Jan 30 Shift 2Easy

Match List I with List II

List I List II
A. Gauss's law of magnetostatics I. Eda=1εoρdV\oint \vec{E} \cdot \vec{da} = \dfrac{1}{\varepsilon_o} \int \rho \, dV
B. Faraday's law of electro magnetic induction II. Bda=0\oint \vec{B} \cdot \vec{da} = 0
C. Ampere's law III. Edl=ddtBda\int \vec{E} \cdot \vec{dl} = \dfrac{-d}{dt} \int \vec{B} \cdot \vec{da}
D. Gauss's law of electrostatics IV. Bdl=μ0I\oint \vec{B} \cdot \vec{dl} = \mu_0 I

Choose the correct answer from the options given below:

  1. A

    A-IV, B-II, C-III, D-I

  2. B

    A-II, B-III, C-IV, D-I

  3. C

    A-I, B-III, C-IV, D-II

  4. D

    A-III, B-IV, C-I, D-II

Show answer

Correct option: B

Q32JEE Main 2024 Jan 30 Shift 2Medium

If the total energy transferred to a surface in time tt is 6.48×105 J6.48 \times 10^5\ J, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

  1. A

    2.16×103 kg m/s2.16 \times 10^{-3}\ kg\ m/s

  2. B

    4.32×103 kg m/s4.32 \times 10^{-3}\ kg\ m/s

  3. C

    1.58×103 kg m/s1.58 \times 10^{-3}\ kg\ m/s

  4. D

    2.46×103 kg m/s2.46 \times 10^{-3}\ kg\ m/s

Show answer

Correct option: A

Q33JEE Main 2024 Jan 31 Shift 1Medium

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5×10105 \times 10^{10} Hz and an amplitude of 50 Vm150\ \mathrm{Vm^{-1}}. The total average energy density of the electromagnetic field of the wave is : [Use ε0=8.85×1012 C2/Nm2\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2/Nm^2}]

  1. A

    1.106×108 Jm31.106 \times 10^{-8}\ \mathrm{Jm^{-3}}

  2. B

    2.212×1010 Jm32.212 \times 10^{-10}\ \mathrm{Jm^{-3}}

  3. C

    2.212×108 Jm32.212 \times 10^{-8}\ \mathrm{Jm^{-3}}

  4. D

    4.425×108 Jm34.425 \times 10^{-8}\ \mathrm{Jm^{-3}}

Show answer

Correct option: A

Q34JEE Main 2024 Jan 31 Shift 2Easy

Given below are two statements:

Statement I: Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields.

Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface.

In the light of the above statements, choose the most appropriate answer from the options given below:

  1. A

    Both Statement I and Statement II are correct.

  2. B

    Both Statement I and Statement II are incorrect.

  3. C

    Statement I is correct but Statement II is incorrect.

  4. D

    Statement I is incorrect but Statement II is correct.

Show answer

Correct option: A