CUET UG Physics Booster Test 2-Electromagnetic Wave Theory -

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CUET UG Physics Booster Test 2-Electromagnetic Wave Theory

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 2-Electromagnetic Wave Theory

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QUESTION 1 OF 20

Consider the charge radiation statements. Choose the correct statements.
1. A proton moving at a constant velocity of 2 × 10⁵ m/s produces EM waves.
2. An electron oscillating harmonically at 1 MHz produces EM waves of 1 MHz.
3. A stationary charge produces a time-varying magnetic field.
4. The energy of the propagating wave comes at the expense of the source's energy.

Answer Review

1 Consider the charge radiation statements. Choose the correct statements.
1. A proton moving at a constant velocity of 2 × 10⁵ m/s produces EM waves.
2. An electron oscillating harmonically at 1 MHz produces EM waves of 1 MHz.
3. A stationary charge produces a time-varying magnetic field.
4. The energy of the propagating wave comes at the expense of the source's energy.

A 1, 2, 3

B 2, 4

C 1, 3, 4

D 2, 3, 4

💡 Short Explanation

�� Accelerated charges radiate. �� Radiation frequency equals source frequency. �� Energy carried by waves comes from the source.

✅ Correct Answer Explanation

Statement 2 is correct because an oscillating electron emits electromagnetic waves having the same frequency as its oscillation. Statement 4 is correct because electromagnetic waves transport energy away from the source, reducing the source's energy. Statement 1 is incorrect because a charge moving with constant velocity does not radiate. Statement 3 is incorrect because a stationary charge produces only a static electric field and no magnetic field.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statements 1 and 3.
�� Option C → Includes Statement 1.
�� Option D → Includes Statement 3.

🎯 Strategy

Used

Elimination

Application:

Remove statements involving non-accelerating charges.

Final Logic:

Only Statements 2 and 4 are correct.

🧠 Memory Trick

Acceleration → Radiation

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Accelerated Charge Radiation

2 An electric dipole forms a basic source of EM waves only when its charges are in ______ motion; the resulting radiation wavelength is typically correlated with the ______ of the radiating system.

A uniform, mass

B accelerated, characteristic size

C stationary, volume

D accelerated, temperature

💡 Short Explanation

�� Radiation requires acceleration. �� Oscillating dipoles are EM wave sources. �� Antenna size is linked to wavelength.

✅ Correct Answer Explanation

An oscillating dipole contains accelerating charges, which generate electromagnetic radiation. The wavelength of emitted radiation is commonly related to the characteristic dimensions of the radiating system, such as antenna length. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Uniform motion does not radiate.
�� Option C → Stationary charges do not radiate.
�� Option D → Temperature is not the relevant parameter here.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply the physical requirements for radiation.

Final Logic:

Accelerated charges and system size determine radiation characteristics.

🧠 Memory Trick

Dipole + Acceleration = EM Waves

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Oscillating Dipole Source

3 In the mechanism of wave propagation, the regenerating electric and magnetic fields

A rely on an external mechanical force to travel through space

B require a continuous presence of conduction current in free space

C propagate independently without influencing one another

D mutually induce one another, propagating the wave even in free space

💡 Short Explanation

�� Changing E produces B. �� Changing B produces E. �� No medium is required.

✅ Correct Answer Explanation

According to Maxwell's theory, a time-varying electric field generates a magnetic field, and a time-varying magnetic field generates an electric field. This mutual regeneration enables electromagnetic waves to propagate through vacuum without any material medium. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → No mechanical force is needed.
�� Option B → Free-space propagation requires no conduction current.
�� Option C → The fields continuously influence one another.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply Maxwell's field regeneration principle.

Final Logic:

Mutual induction sustains propagation.

🧠 Memory Trick

E → B → E → B

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Propagation Mechanism

4 Identify the incorrect statement about self-sustaining oscillations

A They do not require a material medium because they are self-sustaining.

B The time-varying electric field acts as a source for the magnetic field via displacement current.

C The time-varying magnetic field induces an electric field as per Faraday's law.

D The self-sustaining oscillations require constant conduction current in free space to prevent collapse.

💡 Short Explanation

�� EM waves are self-sustaining. �� No conduction current is needed in vacuum. �� Maxwell's equations explain propagation.

✅ Correct Answer Explanation

Electromagnetic waves propagate in free space because changing electric and magnetic fields continuously regenerate each other. No conduction current exists in vacuum, and none is required for propagation. Therefore Option D is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Correct role of displacement current.
�� Option C → Correct statement of Faraday's law.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting free-space propagation.

Final Logic:

Conduction current is unnecessary in vacuum.

🧠 Memory Trick

Vacuum Needs No Current

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Self-Sustaining Oscillations

5 If an electron in a linear antenna oscillates completely 500,000 times per second, the resulting radio wave's angular frequency ω in rad/s will be

A 5 × 10⁵ π

B 10⁶ π

C 2.5 × 10⁵ π

D 10⁶

💡 Short Explanation

�� Frequency = 5 × 10⁵ Hz. �� ω = 2πν. �� Substitute directly.

✅ Correct Answer Explanation

Given: ν = 500,000 Hz = 5 × 10⁵ Hz Angular frequency: ω = 2πν = 2π × 5 × 10⁵ = 10⁶π rad/s Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Missing factor of 2.
�� Option C → Half the correct value.
�� Option D → π omitted.

🎯 Strategy

Used

Substitution

Application:

Use ω = 2πν.

Final Logic:

ω = 10⁶π rad/s.

🧠 Memory Trick

ω = 2πν

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Frequency and Angular Frequency

6 Impracticality of generating visible light using a standard AC electronic circuit
1. Visible light has a frequency around 6 × 10¹⁴ Hz.
2. Modern electronic circuits cap out around 10¹¹ Hz.
3. Circuits can only generate magnetic waves, not electric waves.
4. Wavelength of yellow light is too large for an antenna.

A 1 and 4

B 2 and 3

C 1 and 2

D 3 and 4

💡 Short Explanation

�� Visible light frequency is extremely high. �� Electronic circuits cannot reach optical frequencies. �� Statements 3 and 4 are incorrect.

✅ Correct Answer Explanation

Statement 1 is correct because visible light frequencies are approximately 10¹⁴–10¹⁵ Hz. Statement 2 is correct because ordinary electronic circuits generally operate up to about 10¹¹ Hz. Statement 3 is incorrect because electromagnetic waves contain both electric and magnetic fields. Statement 4 is incorrect because visible-light wavelengths are very small, not large.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statement 4.
�� Option B → Includes Statement 3.
�� Option D → Both statements are incorrect.

🎯 Strategy

Used

Elimination

Application:

Select only physically correct reasons.

Final Logic:

Frequency limitation is the real reason.

🧠 Memory Trick

10¹¹ vs 10¹⁴

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Frequency Limitations of Electronic Circuits

7 Choose the correct statements about the speed of light in vacuum
1. It varies based on whether the wave is an X-ray or visible light.
2. It is used to define the standard of length because of its measured constancy.
3. It is independent of the wavelength of the electromagnetic wave.
4. Its value is exactly 1 / √(μ₀ε₀).

A 1 and 2

B 2, 3, 4

C 1, 3, 4

D All of the above

💡 Short Explanation

�� Speed in vacuum is constant. �� Independent of wavelength. �� Given by Maxwell's relation.

✅ Correct Answer Explanation

Statements 2, 3 and 4 are correct. The speed of light is constant in vacuum and equals: c = 1/√(μ₀ε₀) This value is used in defining the metre. Statement 1 is incorrect because all electromagnetic waves travel at the same speed in vacuum.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statement 1.
�� Option C → Includes Statement 1.
�� Option D → Statement 1 is false.

🎯 Strategy

Used

Elimination

Application:

Reject options containing Statement 1.

Final Logic:

Vacuum speed is independent of wavelength.

🧠 Memory Trick

All EM Waves → Same c

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Speed of Light in Vacuum

8 Match List I with List II

List I	List II
1. Speed of light in vacuum (c)	a. √(με/μ₀ε₀)
2. Speed of EM wave in a medium (v)	b. 1/√(μ₀ε₀)
3. Refractive index (c/v)	c. c
4. Ratio of amplitudes (E₀/B₀)	d. 1/√(με)

A 1-b, 2-d, 3-a, 4-c

B 1-d, 2-b, 3-a, 4-c

C 1-b, 2-d, 3-c, 4-a

D 1-a, 2-b, 3-d, 4-c

💡 Short Explanation

�� c = 1/√(μ₀ε₀) �� v = 1/√(με) �� E₀/B₀ = c

✅ Correct Answer Explanation

1 → b 2 → d 3 → a 4 → c Therefore: P-b, Q-d, R-a, S-c Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect c and v matches.
�� Option C → Incorrect refractive-index match.
�� Option D → Multiple mismatches.

🎯 Strategy

Used

Option Grouping

Application:

Match standard electromagnetic-wave formulas.

Final Logic:

Only Option A satisfies all relations.

🧠 Memory Trick

c, v, n, E/B

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Speed Relations and Refractive Index

9 If a plane electromagnetic wave of frequency 30 MHz is traveling in a vacuum, what will be the magnitude of the wave vector k?
(c = 3 × 10⁸ m/s)

A 0.2π rad/m

B 0.6π rad/m

C 1.5π rad/m

D 3.0π rad/m

💡 Short Explanation

�� λ = c/ν �� k = 2π/λ �� Use vacuum speed.

✅ Correct Answer Explanation

Given: ν = 30 MHz = 3 × 10⁷ Hz λ = c/ν = (3 × 10⁸)/(3 × 10⁷) = 10 m Therefore, k = 2π/λ = 2π/10 = 0.2π rad/m Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Three times larger than actual value.
�� Option C → Incorrect wavelength calculation.
�� Option D → Fifteen times larger than actual value.

🎯 Strategy

Used

Substitution

Application:

Compute λ first, then k.

Final Logic:

k = 0.2π rad/m.

🧠 Memory Trick

k = 2π/λ

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Wave Vector Magnitude

10 If the magnetic field of a wave is given by B = B₀ sin(kx + ωt) ĵ, the wave propagates in the direction of

A +x

B -x

C +y

D -y

💡 Short Explanation

�� Wave form determines direction. �� kx − ωt → +x direction. �� kx + ωt → −x direction.

✅ Correct Answer Explanation

A wave traveling in the negative x-direction is represented by sin(kx + ωt) whereas sin(kx − ωt) represents propagation in the positive x-direction. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to kx − ωt.
�� Option C → Wave propagates along x, not y.
�� Option D → Not along y-axis.

🎯 Strategy

Used

Formula Recall

Application:

Recall the standard traveling-wave forms.

Final Logic:

Plus sign with ωt implies motion along −x.

🧠 Memory Trick

(kx + ωt) → Backward

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Wave Propagation Direction

11 Identify the incorrect statement about the electric-magnetic ratio when E₀ = 60 V/m

A The amplitude of the magnetic field is B₀ = 2 × 10⁻⁷ T.

B The ratio of E₀ to B₀ equals c.

C The ratio B₀/E₀ equals 3 × 10⁸ m/s.

D They are related by E₀/B₀ = c.

💡 Short Explanation

�� E₀ = cB₀. �� E₀/B₀ = c. �� B₀/E₀ = 1/c.

✅ Correct Answer Explanation

For an electromagnetic wave: E₀ = cB₀ Therefore, B₀ = E₀/c = 60/(3 × 10⁸) = 2 × 10⁻⁷ T Hence Statement A is correct. Also, E₀/B₀ = c = 3 × 10⁸ m/s Therefore Statements B and D are correct. However, B₀/E₀ = 1/c = 3.33 × 10⁻⁹ s/m and not 3 × 10⁸ m/s. Hence Statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correctly calculates B₀.
�� Option B → Correct relation.
�� Option D → Standard EM wave relation.

🎯 Strategy

Used

Substitution

Application:

Apply E₀ = cB₀ and evaluate each statement.

Final Logic:

Only Statement C reverses the ratio incorrectly.

🧠 Memory Trick

E/B = c

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Electric-Magnetic Field Ratio

12 Consider the sinusoidal wave function statements. Choose the correct statements:
Ex = E₀ sin(kz − ωt)
1. The magnetic field must oscillate in the y-z plane.
2. The magnetic field is given by By = B₀ sin(kz − ωt).
3. The phase difference between the oscillating E and B fields is π.
4. The wave propagates along the +z direction.

A 2, 4

B 1, 2, 3

C 2, 3, 4

D 1, 4

💡 Short Explanation

�� E and B oscillate in phase. �� Wave travels along +z. �� B is along y-direction.

✅ Correct Answer Explanation

For Ex = E₀ sin(kz − ωt) the wave propagates in the +z direction. The magnetic field is perpendicular to both E and the propagation direction, so: By = B₀ sin(kz − ωt) Thus Statements 2 and 4 are correct. Statement 1 is incorrect because B oscillates along the y-axis, not in the y-z plane. Statement 3 is incorrect because E and B are in phase (phase difference = 0).

❓ Why Other Options Are Incorrect

�� Option B → Includes Statements 1 and 3.
�� Option C → Includes Statement 3.
�� Option D → Includes Statement 1.

🎯 Strategy

Used

Elimination

Application:

Remove options containing Statements 1 or 3.

Final Logic:

Only Statements 2 and 4 are correct.

🧠 Memory Trick

E and B: Same Phase

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Plane Electromagnetic Waves

13 The fundamental property that electric and magnetic fields in an EM wave are perpendicular to each other and to the direction of propagation proves that

A EM waves are longitudinal waves like sound

B EM waves can be deflected by steady magnetic fields

C EM waves are transverse in nature

D EM waves require a highly dense medium

💡 Short Explanation

�� E ⊥ B. �� E and B ⊥ propagation direction. �� This is the defining feature of transverse waves.

✅ Correct Answer Explanation

In electromagnetic waves: E ⊥ B E ⊥ propagation direction B ⊥ propagation direction Since oscillations occur perpendicular to the direction of propagation, EM waves are transverse waves. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes longitudinal waves.
�� Option B → Not implied by field geometry.
�� Option D → EM waves need no material medium.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the wave type from field orientation.

Final Logic:

Perpendicular oscillations imply transverse waves.

🧠 Memory Trick

Perpendicular = Transverse

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Transverse Nature

14 If B is along the +x-axis and the wave propagates along the +y-axis, E must be along the ______ axis, because E × B yields ______.

A +z, +y

B -z, +y

C -z, -y

D +z, -y

💡 Short Explanation

�� Propagation direction = E × B. �� Use right-hand rule. �� E × B must point along +y.

✅ Correct Answer Explanation

Given: B → +x Propagation → +y We require: E × (+x) = +y Using vector products: (-z) × (+x) = +y Therefore E must be along the -z direction. Hence the correct option is: B. -z, +y

❓ Why Other Options Are Incorrect

�� Option A → (+z) × (+x) = -y.
�� Option C → Gives wrong propagation direction.
�� Option D → Gives -y direction.

🎯 Strategy

Used

Vector Analysis

Application:

Apply right-hand-rule cross product.

Final Logic:

(-z) × (+x) = +y.

🧠 Memory Trick

E × B = Direction

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Vector Cross Product

15 If a medium has a magnetic permeability μ = μ₀ and an electric permittivity ε = 4ε₀, what will be the velocity of EM waves in this medium?
(c = 3 × 10⁸ m/s)

A 3.0 × 10⁸ m/s

B 1.5 × 10⁸ m/s

C 6.0 × 10⁸ m/s

D 0.75 × 10⁸ m/s

💡 Short Explanation

�� v = 1/√(με) �� Compare with vacuum speed. �� ε becomes four times larger.

✅ Correct Answer Explanation

v = 1/√(με) = 1/√(μ₀ × 4ε₀) = 1/(2√(μ₀ε₀)) = c/2 = (3 × 10⁸)/2 = 1.5 × 10⁸ m/s Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Vacuum speed.
�� Option C → Exceeds vacuum speed.
�� Option D → Extra factor of 2 error.

🎯 Strategy

Used

Substitution

Application:

Substitute μ and ε into the velocity formula.

Final Logic:

v = c/2.

🧠 Memory Trick

ε ×4 ⇒ v ÷2

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Velocity in Materials

16 When an EM wave passes from vacuum into a material medium with refractive index n = 1.5
1. Its velocity decreases to 2 × 10⁸ m/s.
2. Its frequency remains unchanged.
3. Its velocity increases to 4.5 × 10⁸ m/s.
4. Its wavelength in the medium is greater than in a vacuum.

A 1 and 4

B 2 and 3

C 1 and 2

D 3 and 4

💡 Short Explanation

�� v = c/n. �� Frequency remains constant. �� Wavelength decreases.

✅ Correct Answer Explanation

Given: n = 1.5 v = c/n = (3 × 10⁸)/1.5 = 2 × 10⁸ m/s Thus Statement 1 is correct. Frequency remains unchanged at a boundary, so Statement 2 is correct. Statement 3 is incorrect because velocity decreases. Statement 4 is incorrect because wavelength decreases in the medium.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statement 4.
�� Option B → Includes Statement 3.
�� Option D → Statements 3 and 4 are both incorrect.

🎯 Strategy

Used

Substitution

Application:

Use n = c/v and wave-boundary principles.

Final Logic:

Velocity decreases, frequency remains constant.

🧠 Memory Trick

Medium: v↓, λ↓, f Same

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Refractive Index Effects

17 Match List I with List II

List I	List II
1. Microwave oven	a. Raises temperature of molecules/substances
2. Infrared therapy lamps	b. Transfers energy to earth making life possible
3. X-ray machines	c. Matches resonant frequency of water molecules to transfer kinetic energy
4. Photosynthesis (Visible light)	d. Damages or destroys living tissues at high energy

A 1-c, 2-a, 3-d, 4-b

B 1-a, 2-c, 3-b, 4-d

C 1-c, 2-b, 3-d, 4-a

D 1-d, 2-a, 3-c, 4-b

💡 Short Explanation

�� Microwaves heat water. �� Infrared provides thermal effects. �� X-rays can damage tissue. �� Photosynthesis supports life.

✅ Correct Answer Explanation

1 → c : Microwave oven 2 → a : Infrared therapy lamps 3 → d : X-ray machines 4 → b : Photosynthesis Thus: 1-c, 2-a, 3-d, 4-b Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Multiple mismatches.
�� Option C → Incorrect therapy and photosynthesis pairing.
�� Option D → Incorrect microwave and X-ray pairing.

🎯 Strategy

Used

Option Grouping

Application:

Match each EM wave application with its effect.

Final Logic:

Only Option A correctly matches all pairs.

🧠 Memory Trick

Micro-Water, IR-Heat, X-Ray-Damage, Light-Life

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Energy Transport and Applications

18 Choose the correct statements about solar EM wave impact
1. The earth absorbs incoming visible light and re-radiates it as longer wavelength infrared radiation.
2. Greenhouse gases directly trap incoming visible radiation before it reaches the surface.
3. This absorption process plays an important role in maintaining the earth's average temperature.
4. Infrared radiation from the sun is entirely blocked by the ozone layer.

A 1 and 2

B 2 and 4

C 1 and 3

D All of the above

💡 Short Explanation

�� Earth reradiates absorbed energy as IR. �� Greenhouse effect regulates temperature. �� Statements 2 and 4 are incorrect.

✅ Correct Answer Explanation

Statement 1 is correct because Earth absorbs solar radiation and reradiates energy mainly in the infrared region. Statement 3 is correct because greenhouse warming helps maintain Earth's average temperature. Statement 2 is incorrect because greenhouse gases mainly absorb outgoing infrared radiation rather than incoming visible light. Statement 4 is incorrect because ozone primarily absorbs ultraviolet radiation.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statement 2.
�� Option B → Both statements are incorrect.
�� Option D → Includes incorrect statements.

🎯 Strategy

Used

Elimination

Application:

Reject scientifically incorrect atmospheric interactions.

Final Logic:

Only Statements 1 and 3 are correct.

🧠 Memory Trick

Sunlight In, IR Out

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Solar Radiation and Greenhouse Effect

19 Hertz's successful 1887 experiment was pivotal because it transferred theoretical physics into reality by verifying that

A charges in uniform motion can transmit energy

B oscillating electric and magnetic fields can detach from the source and propagate through space

C only low-frequency waves can exist in free space

D electricity is particulate in nature

💡 Short Explanation

�� Hertz verified Maxwell's prediction. �� Electromagnetic waves exist. �� Waves propagate independently through space.

✅ Correct Answer Explanation

Hertz experimentally generated and detected electromagnetic waves, confirming Maxwell's prediction that oscillating electric and magnetic fields can travel through space as self-propagating waves. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Uniform motion does not produce radiation.
�� Option C → Free space supports all EM frequencies.
�� Option D → Not the objective of Hertz's experiment.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the key outcome of Hertz's experiment.

Final Logic:

Hertz experimentally confirmed electromagnetic waves.

🧠 Memory Trick

Hertz = Proof of Maxwell

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Hertz's Experiment

20 Consider the Radio wave production statements. Choose the correct statements:
1. They are produced by accelerated motion of charges in conducting wires.
2. The FM radio band operates at higher frequencies than the AM band.
3. They are utilized by cellular phones in the UHF band.
4. They typically originate directly from atomic nuclei.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Radio waves come from accelerating charges. �� FM frequencies exceed AM frequencies. �� Cellular communication uses radio frequencies.

✅ Correct Answer Explanation

Statement 1 is correct because radio waves are generated by accelerating charges in antennas. Statement 2 is correct because FM broadcasting operates at frequencies much higher than AM broadcasting. Statement 3 is correct because cellular communication commonly uses frequencies within the UHF region. Statement 4 is incorrect because radio waves used in communication are not normally produced by atomic nuclei.

❓ Why Other Options Are Incorrect

�� Option B → Includes Statement 4.
�� Option C → Includes Statement 4.
�� Option D → Includes Statement 4.

🎯 Strategy

Used

Elimination

Application:

Remove all options containing Statement 4.

Final Logic:

Communication radio waves originate from accelerated charges, not nuclei.

🧠 Memory Trick

Antenna → Radio Waves

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Radio Wave Production and Applications

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