CUET UG Physics Booster Test 2-Technological Evolution and Scientists -

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CUET UG Physics Booster Test -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 -Technological Evolution and Scientists

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QUESTION 1 OF 20

Maxwell's derivation of molecular parameters from measurable macroscopic quantities
Statements
1. linked fluid viscosity to the thermal velocity distribution of gas molecules.
2. was heavily based on the particulate nature of electricity from electrolysis.
3. showed his capability to bridge microscopic and macroscopic physics.
4. proved that visible light is a mechanical wave.

Answer Review

1 Maxwell's derivation of molecular parameters from measurable macroscopic quantities
Statements
1. linked fluid viscosity to the thermal velocity distribution of gas molecules.
2. was heavily based on the particulate nature of electricity from electrolysis.
3. showed his capability to bridge microscopic and macroscopic physics.
4. proved that visible light is a mechanical wave.

A 1 and 3

B 2 and 4

C 1, 3, and 4

D 1 and 2

💡 Short Explanation

�� Maxwell developed the kinetic theory of gases. �� Viscosity measurements were used to estimate molecular properties. �� His work connected microscopic molecular behavior with macroscopic observables.

✅ Correct Answer Explanation

Statement 1 is correct because Maxwell related gas viscosity to the thermal motion and velocity distribution of molecules. Statement 3 is correct because this work successfully connected microscopic molecular properties with measurable macroscopic quantities. Statement 2 is incorrect because particulate electricity from electrolysis is unrelated to this research. Statement 4 is incorrect because Maxwell showed that light is an electromagnetic wave, not a mechanical wave. Therefore, Statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Statements 2 and 4 are both incorrect.
�� Option C → Includes incorrect Statement 4.
�� Option D → Includes incorrect Statement 2.

🎯 Strategy

Used

Elimination

Application:

Identify which statements genuinely describe Maxwell's kinetic theory work.

Final Logic:

Only Statements 1 and 3 correctly describe Maxwell's molecular parameter research.

🧠 Memory Trick

Maxwell Gas → Viscosity + Molecules

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: James Clerk Maxwell and Scientific Contributions

2 Maxwell unified electricity, magnetism, and light by adding a new term to Ampere's law. If an electric field E changes over an area A, creating a rate of change of flux dΦE/dt, the displacement current id required for consistency is mathematically given by

A iₙ = μ₀ (dΦE / dt)

B iₙ = ε₀ (dΦE / dt)

C iₙ = (1/ε₀) (dΦE / dt)

D iₙ = μ₀ε₀ (dΦE / dt)

💡 Short Explanation

�� Maxwell introduced displacement current. �� It depends on changing electric flux. �� It restored consistency in Ampere's law.

✅ Correct Answer Explanation

The displacement current introduced by Maxwell is i_d = ε_0 (dΦ_E/dt)

❓ Why Other Options Are Incorrect

�� Option A → μ₀ does not appear in the displacement current expression.
�� Option C → Incorrect reciprocal dependence on ε₀.
�� Option D → μ₀ε₀ does not define displacement current.

🎯 Strategy

Used

Formula Recall

Application:

Recall Maxwell's displacement current equation directly.

Final Logic:

Displacement current equals ε₀(dΦE/dt).

🧠 Memory Trick

Displacement Current = ε₀ × Flux Change

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

3 Choose the correct statement regarding Hertz's experimental setup:
1. He generated an alternating current oscillating at the frequency of visible light.
2. He successfully tested Maxwell's theory using radio waves.
3. He measured the wavelength and speed of the generated waves.
4. He transmitted waves in the low frequency region compared to visible light.

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

�� Hertz experimentally verified Maxwell's theory. �� He generated radio waves. �� Radio frequencies are much lower than visible-light frequencies.

✅ Correct Answer Explanation

Statement 2 is correct because Hertz experimentally confirmed Maxwell's electromagnetic wave theory. Statement 3 is correct because he measured wavelength and speed of the generated waves. Statement 4 is correct because radio waves have frequencies much lower than visible light. Statement 1 is incorrect because Hertz did not generate visible-light frequency oscillations.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement 1.
�� Option C → Includes incorrect Statement 1.
�� Option D → Includes incorrect Statement 1.

🎯 Strategy

Used

Elimination

Application:

Identify the statement inconsistent with Hertz's radio-wave experiments.

Final Logic:

Hertz generated radio waves, not visible-light-frequency waves.

🧠 Memory Trick

Hertz = Radio, Not Visible

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

4 Choose the correct statements about the scope of Hertz's work
Statements
1. He established the identity of light and heat waves with electromagnetic waves.
2. He discovered the photoelectric effect.
3. He demonstrated that the speed of his generated waves was exactly half the speed of light.
4. He pioneered research on the discharge of electricity through gases.

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Hertz confirmed Maxwell's predictions. �� He discovered the photoelectric effect. �� He worked on electrical discharge in gases.

✅ Correct Answer Explanation

Statement 1 is correct because Hertz's work established the electromagnetic nature of light and heat waves. Statement 2 is correct because he discovered the photoelectric effect during his experiments. Statement 4 is correct because he worked extensively on electrical discharge through gases. Statement 3 is incorrect because the measured wave speed was approximately equal to the speed of light, not half of it.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Check which statement contradicts Hertz's experimental findings.

Final Logic:

Hertz found EM-wave speed approximately equal to c.

🧠 Memory Trick

Hertz → Light, Photoelectric, Gases

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

5 By generating electromagnetic waves in the 25 mm to 5 mm wavelength range, J.C. Bose essentially operated in a region of the electromagnetic spectrum that is now commonly classified as

A ultraviolet radiation.

B microwaves and short radio waves.

C low-frequency AM radio.

D X-rays.

💡 Short Explanation

�� 25 mm–5 mm corresponds to microwave wavelengths. �� Bose pioneered short-wavelength EM-wave generation. �� His work was foundational for microwave research.

✅ Correct Answer Explanation

Wavelengths between 25 mm and 5 mm fall in the microwave region of the electromagnetic spectrum. Bose successfully generated such waves in laboratory experiments. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → UV wavelengths are much shorter.
�� Option C → AM radio wavelengths are much longer.
�� Option D → X-rays have extremely small wavelengths.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Compare wavelength ranges with the EM spectrum.

Final Logic:

Millimetre wavelengths belong to the microwave region.

🧠 Memory Trick

Bose = Millimetre Microwaves

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: J.C. Bose Experiments

6 While Marconi's experiments successfully transmitted waves over ________ distances, Bose's experiments with shorter wavelengths were primarily restricted to the ________.

A Kilometres, Laboratory

B Millimetres, Laboratory

C Kilometres, Atmosphere

D Metres, Vacuum

💡 Short Explanation

�� Marconi achieved long-distance transmission. �� Bose's work remained laboratory-based. �� Both contributed significantly to EM-wave research.

✅ Correct Answer Explanation

Marconi successfully transmitted electromagnetic waves over many kilometres, establishing practical wireless communication. Bose's microwave experiments were largely confined to laboratory demonstrations. Therefore the correct combination is kilometres and laboratory.

❓ Why Other Options Are Incorrect

�� Option B → Marconi transmitted over kilometres, not millimetres.
�� Option C → Bose's experiments were not primarily atmospheric.
�� Option D → Neither term accurately describes the historical experiments.

🎯 Strategy

Used

Concept Recall

Application:

Compare Bose's and Marconi's achievements.

Final Logic:

Marconi = kilometres; Bose = laboratory.

🧠 Memory Trick

Marconi Far, Bose Lab

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

7 Identify the incorrect statement about Marconi's technological leap

A It followed the theoretical and experimental groundwork laid by Maxwell and Hertz.

B It utilized strictly visible light waves for kilometre-long transmissions.

C It initiated the era of wireless electromagnetic communication.

D It involved transmitting electromagnetic waves over many kilometres.

💡 Short Explanation

�� Marconi used radio waves. �� Wireless communication began with his work. �� Visible light was not used.

✅ Correct Answer Explanation

Marconi's wireless communication system used radio-frequency electromagnetic waves, not visible light. Options A, C and D accurately describe Marconi's achievement. Hence Option B is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct historical progression.
�� Option C → Correct contribution.
�� Option D → Correct description of his experiments.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement inconsistent with wireless radio communication.

Final Logic:

Marconi used radio waves, not visible light.

🧠 Memory Trick

Marconi = Radio, Not Light

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Marconi's Communication System

8 Match List I with List II based on the sequence of technological evolution in communication

List I	List II
1. Theoretical prediction of EM waves	a. Hertz
2. First experimental laboratory detection	b. Marconi
3. Production of millimeter-range waves	c. Maxwell
4. Kilometer-scale transmission	d. Bose

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

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

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

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

💡 Short Explanation

�� Maxwell predicted EM waves. �� Hertz detected them experimentally. �� Bose produced millimetre waves. �� Marconi achieved long-distance transmission.

✅ Correct Answer Explanation

1 → c (Maxwell): Predicted the existence of electromagnetic waves theoretically. 2 → a (Hertz): First to generate and detect electromagnetic waves experimentally. 3 → d (Bose): Produced and studied millimeter-wave radiation. 4 → b (Marconi): Achieved long-distance wireless transmission over kilometers.

❓ Why Other Options Are Incorrect

�� Option B → Reverses Hertz and Bose.
�� Option C → Reverses Maxwell and Hertz.
�� Option D → Incorrect historical sequence.

🎯 Strategy

Used

Chronological Matching

Application:

Arrange scientists according to historical developments.

Final Logic:

Maxwell → Hertz → Bose → Marconi.

🧠 Memory Trick

M-H-B-Ma

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

9 If a cellular phone uses the UHF band with a frequency of approximately 900 MHz (9 × 10⁸ Hz), what is the approximate wavelength of the electromagnetic wave carrying its voice communication?
(c = 3 × 10⁸ m/s)

A 0.33 m

B 3.0 m

C 0.03 m

D 33 m

💡 Short Explanation

�� Use λ = c/ν. �� UHF frequencies correspond to short wavelengths. �� Cellular communication uses radio waves.

✅ Correct Answer Explanation

λ = c/ν λ = (3 × 10^8)/(9 × 10^8) λ = 1/3 m λ ≈ 0.33 m Hence Option A is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Substitution

Application:

Directly apply λ = c/ν.

Final Logic:

0.33 m is obtained from c divided by 900 MHz.

🧠 Memory Trick

900 MHz → One-third metre

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

10 Identify the incorrect statement about the transmission of television and FM signals

A TV waves range from 54 MHz to 890 MHz.

B FM radio band extends from 88 MHz to 108 MHz.

C TV signals propagate through the vacuum of space at speeds much slower than c.

D These waves are produced by the accelerated motion of charges in conducting wires.

💡 Short Explanation

�� All EM waves travel at c in vacuum. �� TV and FM waves are electromagnetic waves. �� Accelerated charges generate these waves.

✅ Correct Answer Explanation

Electromagnetic waves, including TV and FM signals, travel in vacuum at the speed of light: c = 3 × 10^8 m/s Therefore Statement C is incorrect. Statements A, B and D are correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct TV frequency range.
�� Option B → Correct FM frequency range.
�� Option D → Correct generation mechanism.

🎯 Strategy

Used

Concept Recall

Application:

Use the fundamental property of EM-wave propagation.

Final Logic:

All EM waves travel at speed c in vacuum.

🧠 Memory Trick

TV/FM = EM → Speed c

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

11 Choose the correct statements about high-frequency medical treatment tools
1. Gamma rays used to destroy cancer cells have wavelengths less than 10⁻³ nm.
2. X-rays used in treatment are generated by bombarding a metal target with high energy electrons.
3. X-rays can damage living tissues and organisms, requiring careful exposure control.
4. Infrared waves are commonly used as a direct diagnostic tool for destroying cancer tumors.

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Gamma rays have extremely short wavelengths. �� X-rays are generated using energetic electrons. �� Exposure to X-rays must be controlled.

✅ Correct Answer Explanation

Statement 1 is correct because gamma rays occupy the highest-frequency region of the EM spectrum and typically have wavelengths below 10⁻³ nm. Statement 2 is correct because X-rays are produced when high-energy electrons strike a metal target. Statement 3 is correct because X-rays are ionising radiations and can damage living tissues. Statement 4 is incorrect because infrared radiation is mainly associated with heating and physical therapy, not destruction of cancer tumors. 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 statement inconsistent with medical radiation applications.

Final Logic:

Infrared radiation is not used to destroy cancer tumors.

🧠 Memory Trick

Gamma-X Treat, IR Heat

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Medical Applications of EM Waves

12 In the context of LASIK eye surgery, the use of ultraviolet radiation is advantageous over longer-wavelength visible light because

A UV radiation is strongly absorbed by ordinary glass.

B shorter wavelengths allow the radiation to be focussed into very narrow, high-precision beams.

C UV radiation induces the production of melanin.

D the ozone layer shields the operating room from background UV.

💡 Short Explanation

�� UV wavelengths are shorter. �� Shorter wavelengths improve focusing. �� LASIK requires high precision.

✅ Correct Answer Explanation

Ultraviolet radiation has shorter wavelengths than visible light. Short wavelengths can be focused into extremely narrow beams, allowing precise reshaping of the cornea during LASIK surgery. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → True but not the reason for LASIK precision.
�� Option C → Related to tanning, not eye surgery.
�� Option D → Unrelated to surgical precision.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the option directly linked to precision surgery.

Final Logic:

Shorter wavelength means higher focusing precision.

🧠 Memory Trick

Short λ → Sharp Beam

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

13 Consider the statements regarding infrared waves in household systems. Choose the correct statements:
1. They are produced by hot bodies and molecules.
2. They are readily absorbed by water molecules causing internal heating.
3. They are utilized in the form of semiconductor light emitting diodes for TV remotes.
4. They fall adjacent to the high-frequency end of the visible spectrum.

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Infrared is thermal radiation. �� Water absorbs infrared effectively. �� TV remotes use infrared LEDs.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. Infrared radiation is emitted by hot bodies, absorbed by water molecules causing heating, and is widely used in remote-control systems through semiconductor LEDs. Statement 4 is incorrect because infrared lies adjacent to the low-frequency (long-wavelength) end of visible light, not the high-frequency end.

❓ 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 incorrect statement regarding spectrum position.

Final Logic:

Infrared lies beyond red light, not violet light.

🧠 Memory Trick

IR = Beyond Red

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

14 If the resonant frequency of water molecules is νw, and a microwave oven produces microwaves of frequency νm to efficiently transfer energy to the water's kinetic energy, the design must ensure that

A νm = 0.5 νw

B νm = νw

C νm = 2 νw

D νm is entirely independent of νw

💡 Short Explanation

�� Resonance maximizes energy transfer. �� Microwave ovens exploit resonance. �� Water molecules absorb efficiently at resonance.

✅ Correct Answer Explanation

Maximum energy transfer occurs when the frequency of the external electromagnetic wave matches the natural resonant frequency of the water molecules. Therefore: νm = νw Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not resonance.
�� Option C → Not resonance.
�� Option D → Heating efficiency depends strongly on resonance.

🎯 Strategy

Used

Concept Recall

Application:

Apply the principle of resonance.

Final Logic:

Maximum absorption occurs when frequencies match.

🧠 Memory Trick

Resonance = Equal Frequencies

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

15 When charges oscillate harmonically with frequency ν in a transmitting antenna,
Statements
1. They generate electromagnetic waves of frequency 2ν.
2. The energy of the propagating wave comes at the expense of the energy of the accelerated charge.
3. They produce purely electrostatic fields that propagate.
4. The resulting magnetic fields do not vary with time.

A 2 only

B 1 and 2

C 3 and 4

D 1 and 4

💡 Short Explanation

�� Accelerated charges radiate. �� Radiated energy comes from the source. �� EM fields vary with time.

✅ Correct Answer Explanation

Statement 2 is correct because the energy carried by the electromagnetic wave originates from the accelerating charges. Statement 1 is incorrect because the emitted EM wave has the same frequency ν as the oscillating charge. Statement 3 is incorrect because propagating EM waves are not purely electrostatic. Statement 4 is incorrect because both electric and magnetic fields vary with time. Therefore only Statement 2 is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Use basic properties of EM-wave generation.

Final Logic:

Radiated wave frequency equals source frequency.

🧠 Memory Trick

Source ν → Wave ν

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Generation of EM Waves

16 An antenna radiating waves of 1 MHz will need to be extremely ________ compared to an antenna designed to efficiently radiate waves of 1 GHz, because the wavelength is directly proportional to the characteristic ________ of the radiating system.

A Small, Frequency

B Large, Size

C Short, Mass

D Heavy, Charge

💡 Short Explanation

�� Lower frequency means larger wavelength. �� Antenna size scales with wavelength. �� 1 MHz waves have much longer wavelengths than 1 GHz waves.

✅ Correct Answer Explanation

Since: λ = c/ν a 1 MHz wave has a wavelength 1000 times larger than a 1 GHz wave. Efficient antennas have dimensions comparable to wavelength. Therefore the 1 MHz antenna must be much larger, and wavelength is related to the characteristic size of the radiating system.

❓ Why Other Options Are Incorrect

�� Option A → Size increases as frequency decreases.
�� Option C → Mass is irrelevant.
�� Option D → Charge is irrelevant.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Use λ = c/ν and antenna-size relation.

Final Logic:

Long wavelength requires large antenna size.

🧠 Memory Trick

Low f → Long λ → Big Antenna

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

17 Match List I with List II

List I	List II
1. Gamma rays	a. Ionisation of gas in a chamber
2. Visible light	b. Vibration of atoms and molecules in a bolometer
3. Infrared	c. Excitation of electrons in atoms (photocells)
4. Ultraviolet	d. Fluorescence and photoelectric effect

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

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

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

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

💡 Short Explanation

Gamma rays ionise gases. Visible light excites electrons in photocells. Infrared causes thermal vibration. Ultraviolet produces fluorescence and photoelectric effects.

❓ Why Other Options Are Incorrect

Any option matching visible light with bolometers is incorrect.
Any option matching infrared with photocells is incorrect.
Any option matching gamma rays with fluorescence is incorrect.
Any option matching ultraviolet with ionisation chambers is incorrect.

🎯 Strategy

Used

Option Grouping

Application:

Associate each electromagnetic radiation with its characteristic detection mechanism.

Final Logic:

Different regions of the electromagnetic spectrum interact with matter differently, leading to distinct detection methods.

🧠 Memory Trick

Gamma–Ionise, Visible–Excite, IR–Heat, UV–Fluoresce

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Detection of Electromagnetic Radiations.

18 Because Geiger tubes rely on ionising gas to detect radiation, they are most effective for detecting

A long radio waves from aerials.

B high-energy, short-wavelength radiation like X-rays and Gamma rays.

C heat waves from hot bodies.

D the constant magnetic field of a steady current.

💡 Short Explanation

�� Geiger tubes work through ionisation. �� X-rays and gamma rays are strongly ionising. �� Radio waves are non-ionising.

✅ Correct Answer Explanation

Geiger tubes detect radiation by measuring ionisation produced in a gas. High-energy X-rays and gamma rays produce sufficient ionisation and are therefore efficiently detected. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Radio waves do not ionise gases.
�� Option C → Infrared heat waves are not detected this way.
�� Option D → Geiger tubes do not detect static magnetic fields.

🎯 Strategy

Used

Concept Recall

Application:

Identify the type of radiation that causes ionisation.

Final Logic:

Ionising radiation is best detected by Geiger tubes.

🧠 Memory Trick

Geiger = Gamma/X-ray

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Geiger Tube Detection

19 Earth satellites detect infrared radiation re-radiated by the Earth. If the peak wavelength of this re-radiated wave is roughly 10 μm (10 × 10⁻⁶ m), its frequency is approximately:
(c = 3 × 10⁸ m/s)

A 3 × 10¹³ Hz

B 3 × 10¹⁴ Hz

C 3 × 10¹² Hz

D 3 × 10¹⁵ Hz

💡 Short Explanation

�� Use ν = c/λ. �� λ = 10 × 10⁻⁶ m. �� Infrared frequencies are around 10¹³ Hz.

✅ Correct Answer Explanation

ν = c/λ = (3 × 10⁸)/(10 × 10⁻⁶) = (3 × 10⁸)/(10⁻⁵) = 3 × 10¹³ Hz Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Ten times too large.
�� Option C → Ten times too small.
�� Option D → Hundred times too large.

🎯 Strategy

Used

Substitution

Application:

Apply ν = c/λ directly.

Final Logic:

ν = 3 × 10¹³ Hz.

🧠 Memory Trick

10 μm → 10¹³ Hz

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

20 Identify the incorrect statement concerning the ozone layer and UV radiation

A Most of the sun's UV radiation is absorbed by the ozone layer at an altitude of 40–50 km.

B Depletion of the ozone layer by chlorofluorocarbons is an international concern.

C The ozone layer converts incoming visible light directly into infrared heat waves.

D Loss of the ozone layer would increase human exposure to harmful UV radiation.

💡 Short Explanation

�� Ozone absorbs UV radiation. �� CFCs damage the ozone layer. �� Ozone does not convert visible light into infrared radiation.

✅ Correct Answer Explanation

The ozone layer's primary role is absorption of harmful ultraviolet radiation. It does not convert visible light directly into infrared heat waves. Therefore Option C is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of ozone protection.
�� Option B → Correct environmental concern.
�� Option D → Correct consequence of ozone depletion.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement unrelated to ozone's actual function.

Final Logic:

Ozone primarily absorbs UV radiation.

🧠 Memory Trick

Ozone = UV Shield

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

CUET UG Physics Booster Test -Technological Evolution and Scientists

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