CUET UG Physics Booster Test 2-The Electromagnetic Spectrum -

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CUET UG Physics Booster Test 2-The Electromagnetic Spectrum

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2-The Electromagnetic Spectrum

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QUESTION 1 OF 20

Identify the incorrect statement about Spectrum Classification:

Answer Review

1 Identify the incorrect statement about Spectrum Classification:

A Gamma rays have the shortest wavelengths in the described spectrum.

B The classification is based strictly on distinct, non-overlapping wavelength boundaries.

C Visible light forms a very small portion of the entire spectrum.

D Radio waves have the longest wavelengths.

💡 Short Explanation

�� EM spectrum is continuous. �� Boundaries overlap. �� Classification is approximate.

✅ Correct Answer Explanation

The electromagnetic spectrum is continuous, and the divisions between different regions are not sharp. Classification depends broadly on wavelength, frequency, method of production, and detection. Gamma rays possess the shortest wavelengths, radio waves possess the longest wavelengths, and visible light occupies only a small portion of the spectrum. Therefore Option B is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct. Gamma rays have the shortest wavelengths.
�� Option C → Correct. Visible light is a tiny portion of the EM spectrum.
�� Option D → Correct. Radio waves have the longest wavelengths.

🎯 Strategy

Used

Extreme Word Filter

Application:

Look for absolute words such as "strictly" and "non-overlapping."

Final Logic:

EM spectrum boundaries are not exact.

🧠 Memory Trick

Spectrum = Continuous

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

2 Which of the following correctly describes the overlapping nature of the electromagnetic spectrum?

A A wave of wavelength 1 mm could be classified as either a microwave or an infrared wave depending on its production or detection.

B X-rays and visible light heavily overlap in their frequency ranges.

C There is a rigid dividing line at exactly 700 nm separating all infrared and visible light.

D Gamma rays and radio waves overlap significantly in vacuum.

💡 Short Explanation

�� Boundaries overlap. �� Classification may depend on production method. �� Some wavelengths belong to transition regions.

✅ Correct Answer Explanation

The boundaries between different regions of the electromagnetic spectrum are not sharply defined. Around boundary regions such as 1 mm wavelength, classification may depend on how the radiation is produced or detected. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Visible light and X-rays do not significantly overlap.
�� Option C → No rigid dividing line exists.
�� Option D → Gamma rays and radio waves occupy opposite ends of the spectrum.

🎯 Strategy

Used

Concept Recall

Application:

Recall the overlapping nature of EM spectrum boundaries.

Final Logic:

Boundary wavelengths can belong to adjacent regions.

🧠 Memory Trick

Boundary = Overlap

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

3 Match List I (Bands) with List II (Frequencies)

List I	List II
1. AM band	a. 530 kHz to 1710 kHz
2. FM band	b. 88 MHz to 108 MHz
3. TV waves	c. 54 MHz to 890 MHz
4. Short wave bands	d. Up to 54 MHz

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

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

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

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

💡 Short Explanation

�� AM uses kHz frequencies. �� FM uses MHz frequencies. �� TV occupies a wider MHz range.

✅ Correct Answer Explanation

The correct matching is: 1 → a : AM band → 530 kHz to 1710 kHz 2 → b : FM band → 88 MHz to 108 MHz 3 → c : TV waves → 54 MHz to 890 MHz 4 → d : Short wave bands → Up to 54 MHz Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → AM and FM bands are interchanged.
�� Option C → FM and TV frequencies are mismatched.
�� Option D → Multiple incorrect assignments.

🎯 Strategy

Used

Option Grouping

Application:

Recall standard communication frequency bands.

Final Logic:

Each communication system has a characteristic frequency range.

🧠 Memory Trick

AM-kHz, FM-MHz

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Radio Waves and Communication Systems

4 If the lower frequency limit of a television broadcast band is 54 MHz and the upper limit is 890 MHz, the bandwidth available for these communication systems is:

A 836 MHz

B 944 MHz

C 16.4 MHz

D 48060 MHz

💡 Short Explanation

�� Bandwidth = Upper frequency − Lower frequency. �� Use simple subtraction. �� Units remain MHz.

✅ Correct Answer Explanation

Bandwidth = 890 MHz − 54 MHz = 836 MHz Hence the available bandwidth is 836 MHz.

❓ Why Other Options Are Incorrect

�� Option B → Obtained by adding frequencies.
�� Option C → Incorrect calculation.
�� Option D → No valid bandwidth calculation.

🎯 Strategy

Used

Substitution

Application:

Apply the formula for bandwidth directly.

Final Logic:

Bandwidth = 890 − 54 = 836 MHz.

🧠 Memory Trick

Bandwidth = High − Low

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

5 In radar systems used for aircraft navigation, microwaves are particularly suitable because:

A their long wavelengths allow them to bend around large obstacles easily.

B their short wavelengths allow them to be directed in concentrated beams.

C they are safely absorbed by the ozone layer.

D they perfectly match the resonant frequency of nitrogen gas in the atmosphere.

💡 Short Explanation

�� Microwaves have relatively short wavelengths. �� Narrow beams improve directionality. �� Radar requires precise targeting.

✅ Correct Answer Explanation

Radar systems require highly directional beams. Because microwaves have short wavelengths, they can be focused into narrow beams, allowing accurate detection, tracking and navigation. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Long wavelengths reduce directional precision.
�� Option C → Ozone absorption is unrelated to radar operation.
�� Option D → Radar does not rely on resonance with nitrogen gas.

🎯 Strategy

Used

Concept Recall

Application:

Recall why microwaves are preferred in radar.

Final Logic:

Short wavelength enables highly directional beams.

🧠 Memory Trick

Microwave → Narrow Beam → Radar

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Microwaves and Radar

6 Choose the correct statements about Resonant Water Heating:
1. The frequency of microwaves is selected to match the resonant frequency of water molecules.
2. Energy from the waves is transferred to the kinetic energy of the molecules.
3. This process lowers the temperature of the food containing water.
4. Microwave ovens are an interesting domestic application of these waves.

A (1), (2), (4)

B (2), (3), (4)

C (1), (3), (4)

D (1), (2), (3)

💡 Short Explanation

�� Microwaves interact strongly with water molecules. �� Absorbed energy increases molecular motion. �� Microwave ovens use this heating principle.

✅ Correct Answer Explanation

Statement 1 is correct because microwave frequencies are chosen so that water molecules absorb energy efficiently. Statement 2 is correct because absorbed microwave energy increases molecular kinetic energy, producing heat. Statement 3 is incorrect because microwave heating raises temperature rather than lowering it. Statement 4 is correct because microwave ovens are a common domestic application of microwaves. Therefore Statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 3.
�� Option C → Includes incorrect Statement 3.
�� Option D → Includes incorrect Statement 3.

🎯 Strategy

Used

Elimination

Application:

Identify the statement contradicting the heating effect.

Final Logic:

Microwaves increase temperature, not decrease it.

🧠 Memory Trick

Microwave → Water → Heat

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

7 Infrared waves increase the ________ motion of molecules, which translates to a rise in ________.

A Thermal, Temperature

B Nuclear, Radioactivity

C Translational, Magnetism

D Thermal, Mass

💡 Short Explanation

�� Infrared radiation is heat radiation. �� Molecular thermal motion increases. �� Increased thermal motion raises temperature.

✅ Correct Answer Explanation

Infrared radiation is readily absorbed by molecules, increasing their thermal motion. Since temperature is a measure of average molecular kinetic energy, increased thermal motion results in a rise in temperature. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Infrared radiation does not induce radioactivity.
�� Option C → Magnetism is unrelated to infrared heating.
�� Option D → Mass does not increase due to infrared absorption.

🎯 Strategy

Used

Concept Recall

Application:

Recall the heating effect of infrared radiation.

Final Logic:

Infrared increases thermal motion and temperature.

🧠 Memory Trick

Infrared = Heat

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

8 Identify the incorrect statement about the Greenhouse Effect:

A Incoming visible light is absorbed by the earth's surface.

B The earth's surface re-radiates energy as visible and ultraviolet light.

C Infrared radiation is trapped by greenhouse gases such as carbon dioxide.

D It plays an important role in maintaining the earth's average temperature.

💡 Short Explanation

�� Earth absorbs visible light. �� Earth emits mainly infrared radiation. �� Greenhouse gases trap infrared radiation.

✅ Correct Answer Explanation

The Earth absorbs incoming solar radiation and re-radiates the absorbed energy primarily as infrared radiation. Greenhouse gases absorb part of this outgoing infrared radiation and help maintain Earth's temperature. Therefore Option B is incorrect because the Earth does not mainly re-radiate energy as visible and ultraviolet radiation.

❓ Why Other Options Are Incorrect

�� Option A → Correct. Earth's surface absorbs incoming visible light.
�� Option C → Correct. Greenhouse gases absorb infrared radiation.
�� Option D → Correct. Greenhouse effect helps maintain Earth's temperature.

🎯 Strategy

Used

Elimination

Application:

Identify the statement inconsistent with Earth's thermal radiation.

Final Logic:

Earth re-emits mainly infrared radiation.

🧠 Memory Trick

Sunlight In, Infrared Out

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

9 Consider the statements regarding Human Eye Sensitivity. Choose the correct statements:
1. It detects wavelengths ranging from about 700 nm to 400 nm.
2. The centre of sensitivity of our eyes coincides with the centre of the wavelength distribution of the sun.
3. Humans have evolved with vision most sensitive to the strongest wavelengths from the sun.
4. The eye can detect X-rays if they are emitted by the sun.

A (1), (2), (3)

B (2), (3), (4)

C (1), (3), (4)

D (1), (2), (4)

💡 Short Explanation

�� Human vision is limited to visible light. �� Eye sensitivity peaks near solar intensity peak. �� Humans cannot see X-rays.

✅ Correct Answer Explanation

Statement 1 is correct because visible light ranges approximately from 400 nm to 700 nm. Statement 2 is correct because human eye sensitivity peaks near the region where the Sun emits maximum energy. Statement 3 is correct because human vision evolved to make best use of the most abundant solar wavelengths reaching Earth. Statement 4 is incorrect because human eyes cannot detect X-rays. Therefore Statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Remove the statement claiming human detection of X-rays.

Final Logic:

Humans detect visible light only.

🧠 Memory Trick

Eye = Visible Only

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Visible Light and Human Vision

10 If the red light of the visible spectrum has a wavelength of 700 nm, what is its approximate frequency?
(c = 3 × 10⁸ m s⁻¹)

A 4.28 × 10¹⁴ Hz

B 4.28 × 10¹² Hz

C 3.00 × 10¹⁴ Hz

D 7.00 × 10¹⁴ Hz

💡 Short Explanation

�� Use f = c/λ. �� Convert nm into meters. �� Frequency is inversely proportional to wavelength.

✅ Correct Answer Explanation

Given: λ = 700 nm = 700 × 10⁻⁹ m = 7 × 10⁻⁷ m Using f = c/λ f = (3 × 10⁸)/(7 × 10⁻⁷) = 4.28 × 10¹⁴ Hz Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Frequency is 100 times smaller than the correct value.
�� Option C → Incorrect numerical calculation.
�� Option D → Corresponds approximately to violet light frequencies.

🎯 Strategy

Used

Substitution

Application:

Direct substitution into f = c/λ.

Final Logic:

f = (3 × 10⁸)/(7 × 10⁻⁷) = 4.28 × 10¹⁴ Hz.

🧠 Memory Trick

Red Light ≈ 4 × 10¹⁴ Hz

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Visible Light Frequency-Wavelength Relationship

11 When exposed to large quantities of ultraviolet radiation, ordinary glass:

A completely transmits the radiation, causing severe sunburns.

B absorbs the UV radiation, which is why one cannot get tanned through glass windows.

C magnifies the radiation, speeding up the production of melanin.

D reflects the radiation perfectly without absorbing any energy.

💡 Short Explanation

�� Ordinary glass absorbs most UV radiation. �� UV is responsible for tanning. �� Glass reduces UV exposure.

✅ Correct Answer Explanation

Ultraviolet radiation is largely absorbed by ordinary window glass. Since tanning occurs primarily due to UV exposure, a person generally cannot get tanned through ordinary glass windows. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ordinary glass does not completely transmit UV radiation.
�� Option C → Glass does not magnify UV radiation.
�� Option D → Glass absorbs significant UV radiation and does not perfectly reflect it.

🎯 Strategy

Used

Concept Recall

Application:

Recall the interaction of UV radiation with ordinary glass.

Final Logic:

Glass absorbs UV radiation, reducing tanning effects.

🧠 Memory Trick

Glass Blocks UV

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

12 Identify the correct statements about Ozone Layer Absorption:
1. The ozone layer protects humans by absorbing most of the sun's ultraviolet light.
2. The protective ozone layer is located at an altitude of about 40–50 km.
3. Chlorofluorocarbons (CFCs) aid in thickening the ozone layer.
4. Depletion of the ozone layer is a matter of international concern.

A 1, 2, 4

B 1, 3, 4

C 2, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Ozone absorbs harmful UV radiation. �� Ozone depletion is environmentally significant. �� CFCs destroy ozone.

✅ Correct Answer Explanation

Statement 1 is correct because ozone absorbs much of the Sun's harmful ultraviolet radiation. Statement 2 is correct according to NCERT's description of the ozone-rich region. Statement 3 is incorrect because CFCs contribute to ozone depletion rather than thickening the ozone layer. Statement 4 is correct because ozone depletion has global environmental consequences. Therefore Statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 3.
�� Option C → Includes incorrect Statement 3.
�� Option D → Includes incorrect Statement 3.

🎯 Strategy

Used

Elimination

Application:

Identify the statement contradicting known effects of CFCs.

Final Logic:

CFCs destroy ozone rather than protect it.

🧠 Memory Trick

CFC → Ozone Down

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Ozone Layer and UV Protection

13 X-rays can be generated by bombarding a ________ target with ________.

A Metal, High energy electrons

B Plastic, Low energy protons

C Gas, High energy neutrons

D Liquid, Low energy electrons

💡 Short Explanation

�� X-ray tubes accelerate electrons. �� Electrons strike metal targets. �� Sudden deceleration produces X-rays.

✅ Correct Answer Explanation

X-rays are commonly produced by accelerating electrons through a large potential difference and directing them onto a metal target. Rapid deceleration of these electrons generates X-ray radiation. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Not a standard X-ray production mechanism.
�� Option C → Neutron bombardment is not the common X-ray source.
�� Option D → Low-energy electrons cannot efficiently generate X-rays.

🎯 Strategy

Used

Concept Recall

Application:

Recall the working principle of an X-ray tube.

Final Logic:

High-energy electrons striking a metal target generate X-rays.

🧠 Memory Trick

Electron + Metal = X-ray

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Production of X-rays

14 Consider the statements regarding Medical Diagnostic Use. Choose the correct statements:
1. X-rays cover wavelengths from about 10 nm down to 10⁻⁴ nm.
2. They are used as a diagnostic tool in medicine.
3. Because they damage living tissues, unnecessary exposure must be avoided.
4. They are entirely safe for all living organisms.

A (1), (2), (3)

B (1), (2), (4)

C (2), (3), (4)

D (1), (3), (4)

💡 Short Explanation

�� X-rays have short wavelengths. �� Widely used in medical diagnosis. �� Excessive exposure is harmful.

✅ Correct Answer Explanation

Statement 1 correctly gives the approximate wavelength range of X-rays. Statement 2 is correct because X-rays are widely used in radiography and diagnostic imaging. Statement 3 is correct because X-rays are ionizing radiations capable of damaging tissues. Statement 4 is incorrect because X-rays are not entirely safe and exposure must be controlled. Therefore Statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Identify the false safety-related statement.

Final Logic:

X-rays can damage living tissues.

🧠 Memory Trick

X-rays: Useful but Harmful

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: X-rays and Their Uses

15 Gamma rays are generally produced in ________ and are also emitted by ________.

A Nuclear reactions, Radioactive nuclei

B Chemical reactions, Stable nuclei

C Inner electron shells, Radioactive nuclei

D Vacuum tubes, Accelerated protons

💡 Short Explanation

�� Gamma rays originate in the nucleus. �� Radioactive decay emits gamma rays. �� Highest-energy EM waves.

✅ Correct Answer Explanation

Gamma rays are produced in nuclear reactions and are frequently emitted during radioactive decay processes. Their origin is nuclear rather than electronic. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Chemical reactions do not produce gamma rays.
�� Option C → Inner electron shells are associated with X-rays.
�� Option D → Vacuum tubes generate microwaves, not gamma rays.

🎯 Strategy

Used

Option Grouping

Application:

Match radiation type with production source.

Final Logic:

Gamma rays are nuclear in origin.

🧠 Memory Trick

Gamma = Nucleus

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

16 Calculate the frequency of a high-frequency gamma ray with a wavelength of 10⁻¹⁴ m.
(c = 3 × 10⁸ m s⁻¹)

A 3 × 10²² Hz

B 3 × 10²⁰ Hz

C 3 × 10¹⁴ Hz

D 3 × 10²⁴ Hz

💡 Short Explanation

�� Use f = c/λ. �� Gamma rays have extremely high frequencies. �� Very small wavelength means very large frequency.

✅ Correct Answer Explanation

Given: λ = 10⁻¹⁴ m Using f = c/λ f = (3 × 10⁸)/(10⁻¹⁴) = 3 × 10²² Hz Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Two orders of magnitude smaller.
�� Option C → Corresponds to visible/infrared region.
�� Option D → Two orders of magnitude larger.

🎯 Strategy

Used

Substitution

Application:

Direct substitution into f = c/λ.

Final Logic:

f = 3 × 10²² Hz.

🧠 Memory Trick

Tiny λ → Huge f

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Gamma Ray Frequency Calculations

17 Match List I (Interaction Type) with List II (Associated Phenomenon)

List I	List II
1. AM band	a. 530 kHz to 1710 kHz
2. FM band	b. 88 MHz to 108 MHz
3. TV waves	c. 54 MHz to 890 MHz
4. Short wave bands	d. Up to 54 MHz

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

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

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

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

💡 Short Explanation

�� Infrared causes heating. �� Visible light is seen by the eye. �� UV causes tanning. �� X-rays can damage tissues.

✅ Correct Answer Explanation

The correct matching is: 1 → a : Infrared increases thermal motion. 2 → b : Visible light is detected by the human eye. 3 → c : Ultraviolet radiation causes tanning. 4 → d : X-rays can damage living tissues. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Infrared and visible effects are interchanged.
�� Option C → Multiple incorrect pairings.
�� Option D → All major associations are incorrect.

🎯 Strategy

Used

Option Grouping

Application:

Match each radiation type with its characteristic effect.

Final Logic:

Each radiation has a distinct biological or physical interaction.

🧠 Memory Trick

IR-Heat, UV-Tan, X-ray-Harm

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Interaction of EM Waves with Matter

18 What determines the specific scattering and absorption mechanisms when electromagnetic waves interact with matter?

A The wavelength of the electromagnetic wave and the nature of the atoms and molecules in the medium.

B Only the macroscopic phase of the material (solid, liquid, or gas).

C The constant speed of light in vacuum.

D The physical distance between the source and the detector.

💡 Short Explanation

�� Interaction depends on wavelength. �� Material structure is important. �� Different wavelengths produce different effects.

✅ Correct Answer Explanation

The absorption and scattering of electromagnetic waves depend strongly on the wavelength of the radiation and the microscopic structure of the medium. Different atoms and molecules interact differently with various wavelengths. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Phase alone does not determine interaction.
�� Option C → Speed of light in vacuum is unrelated.
�� Option D → Distance does not determine absorption mechanisms.

🎯 Strategy

Used

Concept Recall

Application:

Recall factors governing wave–matter interaction.

Final Logic:

Wavelength and material properties control interaction.

🧠 Memory Trick

Wave + Matter = Interaction

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Interaction with Matter

19 Identify the incorrect statement about ultraviolet ray applications:

A UV radiations can be focused into very narrow beams for high precision applications.

B LASIK eye surgery relies on the short wavelengths of UV radiations.

C Welders wear special glass goggles to maximize UV exposure to their eyes.

D UV lamps are used to kill germs in water purifiers.

💡 Short Explanation

�� UV radiation can damage eyes. �� Protective goggles reduce UV exposure. �� UV has several practical applications.

✅ Correct Answer Explanation

Welders wear special goggles to protect their eyes from harmful ultraviolet radiation, not to increase exposure. Statements A, B and D are correct applications of ultraviolet radiation. Hence Option C is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct. UV can be focused for precision work.
�� Option B → Correct. LASIK uses UV excimer lasers.
�� Option D → Correct. UV lamps kill microorganisms.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting UV safety principles.

Final Logic:

Protective goggles reduce UV exposure.

🧠 Memory Trick

UV → Goggles for Protection

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Applications of Ultraviolet Radiation

20 If a water purifier uses a UV lamp emitting at a specific intensity, the purification relies directly on:

A the destructive effects of UV radiation on germs and living cells.

B heating the water to its boiling point using long wavelengths.

C generating displacement currents within the water molecules.

D aligning the magnetic dipoles of the germs.

💡 Short Explanation

�� UV damages microbial DNA. �� Germs are killed or inactivated. �� No heating is required.

✅ Correct Answer Explanation

UV purification works because ultraviolet radiation damages the DNA and cellular structures of microorganisms. This prevents reproduction and effectively kills or inactivates germs present in water. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → UV purification is not based on boiling.
�� Option C → Displacement currents are unrelated.
�� Option D → Magnetic dipole alignment is not the purification mechanism.

🎯 Strategy

Used

Concept Recall

Application:

Recall the operating principle of UV water purifiers.

Final Logic:

UV destroys microorganisms by damaging their genetic material.

🧠 Memory Trick

UV Kills Germs

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Ultraviolet Radiation Applications and Water Purification

CUET UG Physics Booster Test 2-The Electromagnetic Spectrum

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