Physics · chapter PYQs

Electromagnetic Waves, on the exam screen.

5 questions from NEET 2026, Re-NEET 2026, NEET 2020, 5 minutes, marked +4 / −1. See them with solutions.

Electromagnetic Waves · PYQs

Candidate: You

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Question 1 · NEET 2026 · Q45 · +4 / −1

Match List-I (Electromagnetic wave) with List-II (Production) and choose the correct answer from the options given below.

List I	List II
P) Microwave	I) Electrons in atoms emit light when they move from a higher energy level to a lower energy level
Q) Visible light	II) Radioactive decay of nucleus
R) Gamma rays	III) Vibration of atoms and molecules
S) Infra-red rays	IV) Klystron valve or magnetron valve

Answer and explanation

Answer: (3)

Microwaves are generated by klystron or magnetron valves (P-IV). Visible light is emitted when atomic electrons drop from higher to lower energy levels (Q-I). Gamma rays are produced in radioactive decay of the nucleus (R-II). Infra-red radiation arises from the vibration of atoms and molecules (S-III). This matches option (3).

Question 2 · Re-NEET 2026 · Q38 · +4 / −1
An electromagnetic wave travelling in a lossless dielectric medium having a dielectric constant, $\varepsilon_r = 9$ , has the electric field, $E_x = E_0\sin(kz - 2\pi \times 10^6 t)\ \text{Vm}^{-1}$ where $E_0$ is the amplitude and $k$ is the wave vector. Among the following options, the incorrect choice is
Answer and explanation
Answer: (2) or (3) (NTA accepted more than one option)
NTA accepted both option (2) and option (3) as answers. The refractive index is $n = \sqrt{\varepsilon_r} = 3$ , so $v = \frac{c}{n} = 10^8\ \text{m s}^{-1}$ , making option (1) correct. From $\omega = 2\pi \times 10^6$ , the frequency is $10^6\ \text{Hz}$ , hence $\lambda = \frac{v}{f} = 100\ \text{m}$ , not 300 m; therefore option (2) is incorrect. Option (3) is also incorrect as printed: the magnetic field should be $B_y = \frac{E_0}{v}\sin(kz-\omega t)$ , whereas the option uses $B_0/v$ . Option (4) is correct because $kz-\omega t$ represents propagation along $+z$ . NCERT Class 12, Electromagnetic Waves.
Question 3 · Re-NEET 2026 · Q43 · +4 / −1
The following table presents the part of the electromagnetic spectrum and their corresponding major applications. The correct option is:
List I
List II
P) Microwave
I) For purifying the water
Q) UV rays
II) For warming the food
R) Gamma rays
III) For AM and FM communication systems
S) Radio wave
IV) For treating the Cancer cells
Answer and explanation
Answer: (3)
Microwaves are absorbed by water molecules in food and heat it, so P-II. Ultraviolet radiation is germicidal and is used in water sterilisation, so Q-I. Gamma rays carry enough energy to destroy malignant tissue and are used in radiotherapy, so R-IV. Radio waves are the carriers in amplitude modulated (AM) and frequency modulated (FM) broadcasting, so S-III. NCERT Class 12, Electromagnetic Waves (electromagnetic spectrum).
Question 4 · NEET 2020 · Q142 · +4 / −1
Light with an average flux of 20 W cm $^{-2}$ falls on a non-reflecting surface at normal incidence having surface area 20 cm $^{2}$ . The energy received by the surface during a time span of 1 minute is
Answer and explanation
Answer: (2)
The energy received is given by E = flux × area × time. Substituting, $E = 20\ W/cm^2 \times 20\ cm^2 \times 60\ s = 24 \times 10^3\ J$ . Chapter: Electromagnetic Waves.
Question 5 · NEET 2020 · Q168 · +4 / −1
The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is (c = speed of electromagnetic waves)
Answer and explanation
Answer: (1)
In an electromagnetic wave the electric and magnetic energy densities are equal because E = cB and c $^{2}$ = 1/( $\mu$ $_{0}$ ε $_{0}$ ). Their contributions to intensity are therefore in the ratio 1:1. Chapter: Electromagnetic Waves.

List I	List II
P) Microwave	I) For purifying the water
Q) UV rays	II) For warming the food
R) Gamma rays	III) For AM and FM communication systems
S) Radio wave	IV) For treating the Cancer cells