CUET UG Physics Booster Test 3-Technological Evolution and Scientists -

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CUET UG Physics Booster Test 3-Technological Evolution and Scientists

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3-Technological Evolution and Scientists

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QUESTION 1 OF 20

James Clerk Maxwell estimated molecular parameters from measurable quantities like ________, before moving on to unify the laws of electricity and ________.

Answer Review

1 James Clerk Maxwell estimated molecular parameters from measurable quantities like ________, before moving on to unify the laws of electricity and ________.

A Temperature, Light

B Viscosity, Magnetism

C Friction, Gravity

D Mass, Optics

💡 Short Explanation

�� Maxwell studied gases using measurable macroscopic quantities. �� Viscosity helped estimate molecular properties. �� He later unified electricity and magnetism.

✅ Correct Answer Explanation

Maxwell's early work involved deriving molecular parameters from measurable quantities such as the viscosity of gases. This demonstrated how microscopic molecular behavior could be inferred from macroscopic observations. Later, Maxwell unified the laws of electricity and magnetism into a single framework through his electromagnetic theory. Option B correctly identifies both viscosity and magnetism. Temperature alone was not the specific quantity emphasized in this context. Friction and gravity were not central to this research, and optics was not the phenomenon unified with electricity.

❓ Why Other Options Are Incorrect

�� Option A (Temperature, Light) → Temperature is measurable, but viscosity was the key quantity used in Maxwell's molecular studies.
�� Option C (Friction, Gravity) → Neither represents Maxwell's molecular parameter research or electromagnetic unification.
�� Option D (Mass, Optics) → Optics was explained through electromagnetism but was not unified with electricity in the same way as magnetism.

🎯 Strategy

Used

Elimination

Application: Eliminate options containing unrelated concepts such as gravity, friction, and optics.

Final Logic: Maxwell's famous achievement was the unification of electricity and magnetism, making Option B the only valid choice.

🧠 Memory Trick

"Viscosity → Molecules, Magnetism → Maxwell"

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Maxwell's Contributions and Electromagnetic Theory

2 Choose the correct statements about the implications of Maxwell's equations,
1. They mathematically express all the basic laws of electromagnetism when combined with the Lorentz force formula.
2. They predict the existence of coupled time-varying electric and magnetic fields propagating in space.
3. They proved that electricity must be particulate in nature.
4. They resolved an inconsistency in Ampere's circuital law by introducing the conduction current.

A 1 and 2

B 1, 2, and 4

C 2 and 3

D 1, 3, and 4

💡 Short Explanation

�� Maxwell's equations describe electromagnetism. �� They predict electromagnetic waves. �� Statement 4 is incorrect because Maxwell introduced displacement current.

✅ Correct Answer Explanation

Statement 1 is correct because Maxwell's equations, together with the Lorentz force law, form the foundation of classical electromagnetism. Statement 2 is correct because Maxwell predicted self-sustaining electromagnetic waves consisting of oscillating electric and magnetic fields. Statement 3 is incorrect because Maxwell's equations do not prove that electricity is particulate. The electron was discovered later. Statement 4 is incorrect because Maxwell corrected Ampere's law by introducing displacement current, not conduction current. Hence only Statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes Statement 4, which is incorrect.
�� Option C → Includes Statement 3, which is incorrect.
�� Option D → Includes Statements 3 and 4, both incorrect.

🎯 Strategy

Used

Elimination

Application: Check each statement individually and eliminate options containing incorrect statements.

Final Logic: Only Statements 1 and 2 are scientifically valid.

🧠 Memory Trick

"Maxwell → Waves, not electrons."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Maxwell's Equations and Electromagnetic Waves

3 In demonstrating the existence of electromagnetic waves, Hertz,
1. used an AC circuit oscillating at the frequency of visible light.
2. established the identity of their reflection and refraction with that of light.
3. produced waves in the high frequency region like X-rays.
4. relied exclusively on the displacement current inside a capacitor.

A 2 only

B 1 and 2

C 2 and 3

D 1 and 4

💡 Short Explanation

�� Hertz experimentally verified Maxwell's prediction. �� He demonstrated reflection and refraction of radio waves. �� He did not generate visible light or X-rays.

✅ Correct Answer Explanation

Statement 2 is correct because Hertz showed that electromagnetic waves exhibit reflection and refraction similar to light. Statement 1 is incorrect because his apparatus generated radio-frequency waves, not visible-light frequencies. Statement 3 is incorrect because he did not produce X-rays. Statement 4 is incorrect because his experiment involved spark-gap oscillators and was not based exclusively on displacement current. Therefore only Statement 2 is correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 1 is incorrect.
�� Option C → Statement 3 is incorrect.
�� Option D → Statements 1 and 4 are both incorrect.

🎯 Strategy

Used

Elimination

Application: Remove statements inconsistent with Hertz's experimental setup.

Final Logic: Only Statement 2 accurately describes Hertz's work.

🧠 Memory Trick

"Hertz proved radio waves behave like light."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Hertz's Experimental Verification

4 Identify the incorrect statement about Heinrich Hertz's contributions,

A He was the first to broadcast and receive radio waves.

B He showed that radio waves share reflection and refraction properties with heat waves.

C He formulated the theoretical set of equations governing all electromagnetism.

D He discovered the photoelectric effect while pioneering research on gas discharges.

💡 Short Explanation

�� Maxwell formulated the equations. �� Hertz experimentally verified them. �� Hertz observed the photoelectric effect.

✅ Correct Answer Explanation

The theoretical equations governing electromagnetism were formulated by James Clerk Maxwell, not Hertz. Hertz's contribution was experimental verification of electromagnetic waves and observation of the photoelectric effect. Therefore Option C is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of Hertz's radio-wave experiments.
�� Option B → Hertz demonstrated wave properties such as reflection and refraction.
�� Option D → Hertz observed the photoelectric effect during discharge studies.

🎯 Strategy

Used

Odd One Out

Application: Identify the statement describing Maxwell rather than Hertz.

Final Logic: Only Option C belongs to Maxwell's achievements.

🧠 Memory Trick

"Maxwell wrote, Hertz proved."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Hertz and Maxwell

5 Bose produced waves with lengths from λmax = 25 mm to λmin = 5 mm. If the frequency corresponding to λmin is νmax and to λmax is νmin, the ratio νmax / νmin in free space is

A 5

B 0.2

C 25

D 1

💡 Short Explanation

�� Frequency is inversely proportional to wavelength. �� ν = c/λ. �� Smaller wavelength means larger frequency.

✅ Correct Answer Explanation

νmax = c/λmin νmin = c/λmax νmax/νmin = λmax/λmin = 25/5 = 5 Therefore the ratio is 5.

❓ Why Other Options Are Incorrect

�� Option B → Reciprocal ratio.
�� Option C → Incorrect numerical calculation.
�� Option D → Frequencies are not equal.

🎯 Strategy

Used

Substitution

Application: Directly substitute wavelengths into ν = c/λ.

Final Logic: 25 ÷ 5 = 5.

🧠 Memory Trick

"Frequency flips wavelength."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Relation Between Frequency and Wavelength

6 Consider the statements regarding analysis of Bose's vs Marconi's historical impact. Choose the correct statements.
1. Bose generated millimeter-scale waves but his experiments remained confined to the laboratory.
2. Marconi's transmission over many kilometers directly utilized Bose's 5 mm waves.
3. Both built upon Hertz's successful experimental test of Maxwell's theory.
4. Marconi's success marked the beginning of commercial communication using EM waves.

A 1, 3, 4

B 2, 3, 4

C 1, 2, 3

D 1, 4

💡 Short Explanation

�� Bose worked mainly in laboratories. �� Hertz inspired both researchers. �� Marconi enabled practical communication.

✅ Correct Answer Explanation

Statement 1 is correct. Bose generated millimeter waves but did not develop commercial communication systems. Statement 2 is incorrect because Marconi used much longer radio wavelengths for long-distance communication. Statement 3 is correct because both researchers built upon Hertz's verification of Maxwell's theory. Statement 4 is correct because Marconi's work initiated practical wireless communication. Hence Statements 1, 3, and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 2.
�� Option C → Includes incorrect Statement 2.
�� Option D → Omits correct Statement 3.

🎯 Strategy

Used

Elimination

Application: Evaluate each historical statement independently.

Final Logic: Statement 2 is the only incorrect statement.

🧠 Memory Trick

"Bose Lab, Marconi Communication."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Bose and Marconi Contributions

7 If a Marconi-type long wave transmitter emits a radio signal at a frequency of 500 kHz, what is the wave vector k (magnitude) of the propagating wave in vacuum? (c = 3 × 10⁸ m/s, π ≈ 3.14)

A 1.05 × 10⁻² rad/m

B 600 rad/m

C 1.50 × 10¹⁴ rad/m

D 3.14 × 10⁻² rad/m

💡 Short Explanation

�� k = 2π/λ. �� λ = c/f. �� Use both relations sequentially.

✅ Correct Answer Explanation

f = 500 × 10^3 Hz λ = (3 × 10^8)/(5 × 10^5) = 600 m k = (2π)/600 = 6.28/600 ≈ 1.05 × 10^-2 rad/m Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Confuses wavelength with wave number.
�� Option C → Unrealistically large value.
�� Option D → Calculation error.

🎯 Strategy

Used

Substitution

Application: Use λ = c/f and then k = 2π/λ.

Final Logic: k ≈ 0.0105 rad/m.

🧠 Memory Trick

"Find λ first, then k."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Wave Number and Wavelength

8 The transition from Hertz's laboratory verification to Marconi's long-distance transmission demonstrated that electromagnetic waves,

A require a physical medium to propagate over many kilometres.

B inherently lose their magnetic component when transmitted beyond a laboratory.

C can be utilized to carry energy and information from one place to another over vast distances.

D become longitudinal waves when used for communication.

💡 Short Explanation

�� EM waves travel through vacuum. �� They carry energy and information. �� Wireless communication depends on this property.

✅ Correct Answer Explanation

Marconi demonstrated the practical use of electromagnetic waves for transmitting signals over large distances. This showed that EM waves can carry information and energy effectively.

❓ Why Other Options Are Incorrect

�� Option A → EM waves do not require a material medium.
�� Option B → Electric and magnetic fields remain coupled.
�� Option D → EM waves remain transverse.

🎯 Strategy

Used

Elimination

Application: Remove statements contradicting EM-wave properties.

Final Logic: Only Option C reflects the significance of Marconi's achievement.

🧠 Memory Trick

"Marconi = Message Carrier."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Applications of Radio Waves

9 Match List I with List II

List I	List II
1. Cellular phones	a. Ultrahigh frequency (UHF)
2. FM radio	b. 88 MHz to 108 MHz
3. TV waves	c. 54 MHz to 890 MHz
4. Short wave bands	d. Higher frequencies up to 54 MHz

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

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

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

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

💡 Short Explanation

�� Cellular phones use UHF. �� FM radio uses 88–108 MHz. �� TV waves span a wide MHz range.

✅ Correct Answer Explanation

1 → a : Cellular phones operate in UHF bands. 2 → b : FM radio occupies 88–108 MHz. 3 → c : TV waves span 54–890 MHz. 4 → d : Short-wave related transmissions extend up to about 54 MHz. Thus Option A is the correct matching sequence.

❓ Why Other Options Are Incorrect

�� Option B → 1 and 2 are interchanged.
�� Option C → FM and TV assignments are incorrect.
�� Option D → Multiple mismatches occur.

🎯 Strategy

Used

Option Grouping

Application: Match known frequency bands first and eliminate remaining options.

Final Logic: FM = 88–108 MHz immediately confirms Option A.

🧠 Memory Trick

"FM = 88–108."

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

10 Identify the incorrect statement regarding the transmission ranges of TV and AM radio

A AM bands span from 530 kHz to 1710 kHz.

B TV waves encompass frequencies lower than the standard AM radio band.

C TV waves span from 54 MHz to 890 MHz.

D Cellular phones utilize higher frequencies (UHF) compared to standard AM radio.

💡 Short Explanation

�� TV frequencies are much higher than AM frequencies. �� AM operates in kHz range. �� TV operates in MHz range.

✅ Correct Answer Explanation

TV waves occupy approximately 54 MHz to 890 MHz, whereas AM radio operates around 530–1710 kHz. Since MHz frequencies are much higher than kHz frequencies, the statement that TV frequencies are lower than AM frequencies is incorrect. Therefore Option B is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct AM frequency range.
�� Option C → Correct TV frequency range.
�� Option D → Cellular phones use higher UHF frequencies than AM radio.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application: Compare MHz and kHz units.

Final Logic: MHz > kHz, so Option B is false.

🧠 Memory Trick

"TV in MHz, AM in kHz."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Frequency Bands and Communication Systems

11 Gamma rays and X-rays in medicine statements,
1. Both lie at the highest frequency range of the electromagnetic spectrum.
2. Gamma rays are emitted by radioactive nuclei whereas X-rays are typically generated by bombarding a metal target with electrons.
3. Both are strictly used for physical therapy of muscle tissue.
4. Both have wavelengths generally less than 1 nm.

A 1, 2, 4

B 2, 3, 4

C 1, 3

D 1, 2, 3, 4

💡 Short Explanation

�� Gamma rays and X-rays occupy the highest frequency region. �� Their origins are different. �� Both possess very short wavelengths.

✅ Correct Answer Explanation

Statement 1 is correct because X-rays and gamma rays belong to the highest-frequency region of the electromagnetic spectrum. Statement 2 is correct because gamma rays originate from radioactive nuclei, whereas X-rays are generally produced when high-speed electrons strike a metal target. Statement 3 is incorrect because these radiations are mainly used for imaging, diagnosis, sterilization, and cancer treatment, not strictly for muscle therapy. Statement 4 is correct because both typically have wavelengths shorter than 1 nm. Hence Statements 1, 2, and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 3.
�� Option C → Omits correct Statements 2 and 4.
�� Option D → Includes incorrect Statement 3.

🎯 Strategy

Used

Elimination

Application: Evaluate each statement using known properties and applications.

Final Logic: Statement 3 is the only incorrect statement.

🧠 Memory Trick

"Gamma-Nucleus, X-ray-Target."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: X-rays and Gamma Rays

12 For high precision applications like LASIK surgery, the beam spread must be minimized. The ability to focus a beam narrowly is inversely related to its wavelength λ. If an infrared laser has wavelength λir and a UV laser has wavelength λuv, the focusing precision factor ratio (UV compared to IR), assuming it is proportional to (λir / λuv), will be,

A greater than 1, since λir > λuv

B less than 1, since λir > λuv

C exactly 1, since all EM waves travel at speed c

D negative, due to the different speeds in a medium

💡 Short Explanation

�� UV wavelengths are shorter than IR wavelengths. �� Shorter wavelengths allow better focusing. �� The ratio λir/λuv exceeds 1.

✅ Correct Answer Explanation

Since infrared radiation has a larger wavelength than ultraviolet radiation, λir > λuv Therefore, λir/λuv > 1 Because focusing precision is proportional to this ratio, the UV beam can be focused more precisely than the IR beam. Hence the precision factor is greater than 1.

❓ Why Other Options Are Incorrect

�� Option B → Opposite conclusion of the wavelength relationship.
�� Option C → Speed of propagation does not determine focusing precision.
�� Option D → The ratio cannot be negative.

🎯 Strategy

Used

Substitution

Application: Directly compare the wavelength relationship.

Final Logic: Larger numerator and smaller denominator give a ratio greater than 1.

🧠 Memory Trick

"Shorter λ → Sharper Focus."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Wavelength Dependence of Applications

13 Infrared radiation in household devices,
1. relies on the absorption of radiation by greenhouse gases.
2. is emitted by semiconductor light emitting diodes.
3. involves waves shorter than 400 nm.
4. detects changes in atmospheric ozone.
Choose the correct statements:

A 2 only

B 1 and 3

C 2 and 4

D 4 only

💡 Short Explanation

�� TV remotes use infrared LEDs. �� Infrared wavelengths are longer than visible red light. �� Ozone detection mainly uses ultraviolet radiation.

✅ Correct Answer Explanation

Statement 2 is correct because infrared radiation used in remote controls is produced by semiconductor light-emitting diodes (LEDs). Statement 1 is incorrect because greenhouse gas absorption is unrelated to the operation of household infrared devices. Statement 3 is incorrect because wavelengths below 400 nm belong to ultraviolet radiation, not infrared radiation. Statement 4 is incorrect because atmospheric ozone monitoring primarily involves ultraviolet radiation. Therefore only Statement 2 is correct.

❓ Why Other Options Are Incorrect

�� Option B → Statements 1 and 3 are incorrect.
�� Option C → Statement 4 is incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Elimination

Application: Identify the only statement directly related to household IR devices.

Final Logic: Infrared remotes operate using IR LEDs.

🧠 Memory Trick

"Remote = IR LED."

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

14 If a microwave oven's frequency is not properly matched to the resonant frequency of water molecules, the result would most likely be that

A the food's water molecules would freeze due to inverse resonance.

B the energy transfer to the kinetic energy of the water molecules would be highly inefficient.

C the microwaves would convert into high-energy gamma rays.

D the klystron valve would generate visible light instead.

💡 Short Explanation

�� Microwave heating depends on resonance. �� Efficient absorption requires frequency matching. �� Mismatch reduces heating efficiency.

✅ Correct Answer Explanation

Microwave ovens heat food by causing water molecules to absorb electromagnetic energy and increase their molecular motion. If the microwave frequency is far from the effective absorption range, energy transfer becomes inefficient and heating decreases significantly. Thus Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → No such inverse-resonance freezing process occurs.
�� Option C → Microwaves cannot spontaneously convert into gamma rays.
�� Option D → Frequency mismatch does not cause visible-light generation.

🎯 Strategy

Used

Elimination

Application: Reject physically impossible outcomes.

Final Logic: Frequency mismatch mainly reduces energy absorption.

🧠 Memory Trick

"Mismatch = Less Heating."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Microwave Applications

15 If an electric charge oscillates harmonically with a frequency of 10⁹ Hz, and the corresponding electromagnetic wave travels in a vacuum, what is the wavelength of the emitted wave? (c = 3 × 10⁸ m/s)

A 3.0 m

B 0.3 m

C 30 m

D 0.03 m

💡 Short Explanation

�� Use λ = c/f. �� Frequency and wavelength are inversely related. �� Vacuum speed is 3 × 10⁸ m/s.

✅ Correct Answer Explanation

λ = c/f λ = (3 × 10^8)/(10^9) λ = 0.3 m Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ten times larger.
�� Option C → Hundred times larger.
�� Option D → Ten times smaller.

🎯 Strategy

Used

Substitution

Application: Directly apply λ = c/f.

Final Logic: 3 × 10⁸ ÷ 10⁹ = 0.3 m.

🧠 Memory Trick

"GHz gives decimeter waves."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Frequency-Wavelength Relation

List I	List II
1. Gamma radiation	a. Characteristic size of an atomic nucleus (10⁻¹⁴ m to 10⁻¹⁵ m)
2. Radio waves	b. Same size as the macroscopic transmitting antenna
3. Visible radiation	c. Much longer in wavelength than atomic size, but emitted by atoms
4. X-rays	d. Wavelength comparable to interatomic spacing in crystals

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

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

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

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

💡 Short Explanation

Gamma rays originate at nuclear scales. Radio wavelengths match antenna dimensions. Visible light is emitted by atoms. X-rays have wavelengths comparable to interatomic crystal spacing.

✅ Correct Answer Explanation

1 → a: Gamma rays have wavelengths comparable to nuclear dimensions. 2 → b: Radio-wave transmission requires antennas with dimensions related to the wavelength. 3 → c: Visible light is emitted by atoms although its wavelength is much larger than atomic size. 4 → d: X-rays have wavelengths comparable to the spacing between atoms in crystals. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B: Gamma rays are not antenna-sized waves.
Option C: Radio waves are not associated with atomic emission wavelengths.
Option D: Gamma rays are not associated with crystal spacings.

🎯 Strategy

Used

Option Grouping

Application: Match the most obvious pair first (Gamma ↔ Nucleus).

Final Logic: The remaining radiations match uniquely with their characteristic length scales.

🧠 Memory Trick

"Gamma–Nucleus, Radio–Antenna, Visible–Atom, X-ray–Crystal."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Wavelength Scales.

17 Choose the correct statements about the detection of light and UV rays,
1. Both can be detected using photocells.
2. Both can be recorded using photographic film.
3. Both rely on the principle of nuclear radioactive decay for detection.
4. Detection by photocells involves electrons in atoms moving between energy levels.

A 1, 2, 4

B 2, 3, 4

C 1, 3

D 1, 2, 3, 4

💡 Short Explanation

�� Photocells detect light and UV radiation. �� Photographic films can record both. �� Radioactive decay is not involved.

✅ Correct Answer Explanation

Statement 1 is correct because photocells respond to incident electromagnetic radiation. Statement 2 is correct because photographic films can record both visible and ultraviolet radiation. Statement 3 is incorrect because their detection does not depend on radioactive decay. Statement 4 is correct because photocell operation involves electron excitation and movement between energy states. Thus Statements 1, 2, and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 3.
�� Option C → Omits correct Statements 2 and 4.
�� Option D → Includes incorrect Statement 3.

🎯 Strategy

Used

Elimination

Application: Identify the statement unrelated to optical detection.

Final Logic: Radioactive decay has no role in photocell detection.

🧠 Memory Trick

"Photocell = Electron Motion."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Detection of Light and UV Radiation

18 Radio waves are detected by receiver's ________, whereas Gamma rays are detected by ________ tubes.

A Aerials, Geiger

B Photocells, Klystron

C Diodes, Magnetron

D Bolometers, Photographic

💡 Short Explanation

�� Radio signals are received by antennas. �� Gamma rays ionize gases. �� Geiger tubes detect ionizing radiation.

✅ Correct Answer Explanation

Radio waves are commonly detected using receiving aerials or antennas that convert electromagnetic signals into electrical signals. Gamma rays are detected using Geiger-Müller tubes because they ionize the gas present inside the detector. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Klystrons generate microwaves rather than detect gamma rays.
�� Option C → Magnetrons generate microwaves.
�� Option D → Bolometers detect infrared radiation.

🎯 Strategy

Used

Odd One Out

Application: Match each detector with its associated radiation type.

Final Logic: Antenna ↔ Radio, Geiger ↔ Gamma.

🧠 Memory Trick

"Gamma → Geiger."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Detection Methods

19 Satellites equipped with infrared detectors can monitor the growth of crops largely because,

A crops reflect high amounts of ultraviolet radiation from the sun.

B infrared sensors can detect the heat signatures and thermal radiation characteristics of vegetation.

C Earth's ozone layer completely amplifies the infrared signal from plants.

D plants emit strong microwaves during photosynthesis.

💡 Short Explanation

�� Healthy vegetation has characteristic thermal signatures. �� Infrared sensing is widely used in agriculture. �� Satellites monitor plant growth remotely.

✅ Correct Answer Explanation

Plants absorb, reflect, and emit radiation differently depending on their health and growth conditions. Infrared detectors can measure these thermal characteristics and provide valuable information about crop growth, stress, and vegetation cover. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Crop monitoring is not mainly based on UV reflection.
�� Option C → Ozone does not completely amplify infrared signals.
�� Option D → Plants do not emit strong microwaves during photosynthesis.

🎯 Strategy

Used

Contextual/Tonal Matching

Application: Connect infrared sensing with agricultural monitoring applications.

Final Logic: Thermal signatures provide useful crop information.

🧠 Memory Trick

"IR sees plant heat."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Infrared Applications

20 Identify the incorrect statement regarding the interaction between electromagnetic waves and the atmosphere,

A The ozone layer absorbs most of the sun's ultraviolet light.

B Incoming visible light passes relatively easily through the atmosphere.

C The earth re-radiates absorbed visible light as shorter wavelength ultraviolet radiation.

D Infrared radiation is trapped by greenhouse gases like carbon dioxide.

💡 Short Explanation

�� Earth absorbs solar energy and re-emits infrared radiation. �� Ozone absorbs UV radiation. �� Greenhouse gases trap infrared radiation.

✅ Correct Answer Explanation

Earth absorbs incoming solar radiation and re-radiates the energy primarily as longer-wavelength infrared radiation, not as shorter-wavelength ultraviolet radiation. Therefore Statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct; ozone absorbs most harmful UV radiation.
�� Option B → Correct; visible light largely passes through the atmosphere.
�� Option D → Correct; greenhouse gases absorb and trap infrared radiation.

🎯 Strategy

Used

Elimination

Application: Compare atmospheric processes with known greenhouse-effect concepts.

Final Logic: Earth emits infrared, not ultraviolet, radiation.

🧠 Memory Trick

"Sun gives visible, Earth gives IR."

📚 Class XII – Physics Part II – Chapter: Electromagnetic Waves – Topic: Interaction of EM Waves with the Atmosphere

CUET UG Physics Booster Test 3-Technological Evolution and Scientists

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