12 NEET previous-year questions on Electromagnetic Waves, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
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X-rays
Infra-red rays
Ultraviolet rays
γ-rays
Solution
Wavelength of the ray λ= hc E =0.826 Å Since λ< 100 Å So it is X- ray
A charge moving at constant velocity
A stationary charge
A chargeless particle
An accelerating charge
Solution
An accelerating charge can produce electromagnetic wave.
1.41 × 10 –8 T
2.83 × 10 –8 T
0.70 × 10 –8 T
4.23 × 10 –8 T
Solution
rms rms E cB = rms rms EB c= 8 6 31 0 = × Brms = 2 × 10 –8 Brms = 0 2 B 0r ms2BB =× = –822 1 0×× = 2.83 × 10 –8 T
– y direction
+ z direction
– z direction
– x direction
Solution
The direction of the magnetic field is perpendicular to both the electric field and the direction of propagation . Using the right-hand rule, if is along the +x axis and is along the +y axis, then must be along the +z axis. Therefore, option (b) is correct.
10 × 10 3 J
12 × 10 3 J
24 × 10 3 J
48 × 10 3 J
Solution
The energy received is given by . Substituting, , so option (c) is correct.
c : 1
1 : 1
1 : c
1 : c 2
Solution
The intensity of an electromagnetic wave is given by , where and are the electric and magnetic field amplitudes, respectively. Since , the contributions from the electric and magnetic fields are in the ratio . Therefore, option (a) is correct.
∧ ∧ ∧ ∧ + + j k, j k
∧ ∧ ∧ ∧ − + − − j k, j k
∧ ∧ ∧ ∧ + − − j k, j k
∧ ∧ ∧ ∧ − + − + j k, j k
Solution
For a plane electromagnetic wave propagating in the x-direction, the electric field (E) and magnetic field (B) must be perpendicular to each other and to the direction of propagation. The correct combination is and , ensuring orthogonality, so option (d) is correct.
219.3 m
219.2 m
2192 m
21.92 cm
Solution
The wavelength can be calculated using the formula . Substituting the values, , so option (b) is correct.
(a) - (iv), (b) - (iii), (c) - (ii), (d) - (i)
(a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)
(a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)
(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
Solution
Waves Wavelength AM radio waves 102 m Microwaves 10–2 m Infrared radiations 10–4 m X-rays 10–10 m (a) - (ii) (b) - (iii) (c) - (iv) (d) - (i)
36 × 107 J
36 × 104 J
36 × 105 J
1 × 105 J
Solution
Energy = Power × time E = 100 × 103 × 3600 = 36 × 107 J
1.6×10⁻⁶ T
1.6×10⁻⁹ T
1.6×10⁻⁸ T
1.6×10⁻⁷ T
Solution
The amplitude of the magnetic field is related to the electric field amplitude by . Substituting the given values, , so option (d) is correct.
Solution
In electromagnetic wave, and are in same phase and ; their planes are perpendicular to each other.
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