CUET UG Physics Booster Test 3-Mathematical Principles and Laws -

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CUET UG Physics Booster Test 3-Mathematical Principles and Laws

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3-Mathematical Principles and Laws

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QUESTION 1 OF 20

The missing term in Ampere's circuital law was found by analyzing the electric flux through surface S.

Answer Review

1 The missing term in Ampere's circuital law was found by analyzing the electric flux through surface S.

A By using Gauss's law, ΦE = (Q/ε₀) was utilized

B By using Faraday's law, ΦE was derived

C The surface integral of the magnetic field replaced the electric charge

D The surface charge integration directly yielded the conduction current

💡 Short Explanation

�� Maxwell analyzed charging capacitors. �� Electric flux was related to enclosed charge. �� This led to displacement current.

✅ Correct Answer Explanation

Maxwell used Gauss's law: ΦE = Q/ε₀ Differentiating with respect to time gives: dΦE/dt = (1/ε₀)dQ/dt Since dQ/dt = i, Maxwell obtained: Id = ε₀(dΦE/dt) This introduced displacement current into Ampere's law. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Faraday's law deals with magnetic flux, not electric flux.
�� Option C → Magnetic field integral does not replace charge.
�� Option D → Surface charge integration alone does not directly give conduction current.

🎯 Strategy

Used

Concept Recall

Application:

Recall Maxwell's derivation of displacement current.

Final Logic:

Gauss's law provided the electric-flux relation needed for Maxwell's correction.

🧠 Memory Trick

Gauss → Flux → Displacement Current

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Maxwell's Modification of Ampere's Law

2 If the charge Q on capacitor plates changes with time, a displacement current arises.
1. It relies on i = dQ/dt
2. It equals ε₀(dΦE/dt)
3. It maintains the continuity of total current across the capacitor
4. It nullifies the existence of Ex

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

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

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

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

💡 Short Explanation

�� Changing charge creates changing electric flux. �� Displacement current complements conduction current. �� Current continuity is preserved.

✅ Correct Answer Explanation

Statement 1 is correct because current is: i = dQ/dt Statement 2 is correct: Id = ε₀(dΦE/dt) Statement 3 is correct because displacement current ensures continuity across the capacitor gap. Statement 4 is incorrect because displacement current does not eliminate electric fields. Hence 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:

Reject the statement contradicting Maxwell's theory.

Final Logic:

Displacement current supports electric fields rather than removing them.

🧠 Memory Trick

Changing Flux = Displacement Current

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

3 If the magnetic field in a plane EM wave is
By = (2 × 10⁻⁷) sin(0.5 × 10³x + 1.5 × 10¹¹t) T,
what is its wavelength?

A 1.26 cm

B 2.5 cm

C 3.14 cm

D 0.5 m

💡 Short Explanation

�� Wave number k = 0.5 × 10³ rad/m. �� λ = 2π/k. �� Convert to centimeters.

✅ Correct Answer Explanation

Given: k = 0.5 × 10³ = 500 rad/m Using: λ = 2π/k = 6.283/500 = 0.01257 m = 1.26 cm Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Calculation error.
�� Option C → Incorrect substitution.
�� Option D → Much larger than actual value.

🎯 Strategy

Used

Substitution

Application:

Use λ = 2π/k directly.

Final Logic:

λ = 1.26 cm.

🧠 Memory Trick

λ = 2π/k

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

4 Field source states are:

A Symmetric, Asymmetric

B Existent, Non-existent

C Constant, Variable

D Straight, Curved

💡 Short Explanation

�� Electric charges exist. �� Magnetic monopoles have not been observed. �� Maxwell's equations reflect this asymmetry.

✅ Correct Answer Explanation

Electric fields originate from electric charges, which exist. Magnetic monopoles have never been experimentally detected. Thus source states are: Electric charge → Existent Magnetic monopole → Non-existent Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Does not describe source existence.
�� Option C → Not relevant to field sources.
�� Option D → Describes geometry, not sources.

🎯 Strategy

Used

Concept Recall

Application:

Recall Gauss's laws for electricity and magnetism.

Final Logic:

Charges exist; monopoles do not.

🧠 Memory Trick

Charge Exists, Monopole Missing

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Non-existence of Magnetic Monopoles

5 Match List I with List II

List I	List II
1. Time-varying B field	a. Induces E field
2. Time-varying E field	b. Induces B field
3. Accelerated charge	c. Radiates EM waves
4. Static charge	d. Produces electrostatic field only

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

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

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

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

💡 Short Explanation

�� Changing magnetic fields induce electric fields. �� Changing electric fields induce magnetic fields. �� Accelerated charges radiate.

✅ Correct Answer Explanation

A time-varying magnetic field induces an electric field according to Faraday's Law of Electromagnetic Induction. A time-varying electric field produces a magnetic field through the displacement current term in Maxwell's equations. An accelerated charge emits electromagnetic radiation and is therefore a source of EM waves. A static charge produces only an electrostatic field and does not radiate electromagnetic waves. Therefore: 1 → a 2 → b 3 → c 4 → d Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → First two relations are reversed.
�� Option C → Multiple incorrect pairings.
�� Option D → Accelerated charge incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Associate each phenomenon with Maxwell's theory.

Final Logic:

Changing fields generate each other; accelerated charges radiate.

🧠 Memory Trick

Changing B→E, Changing E→B

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

6 Faraday's law of electromagnetic induction states:
Statements
1. The induced emf is strictly zero if the magnetic flux is constant
2. The line integral ∮E·dl computes the induced emf
3. It provides the symmetrical counterpart to displacement current
4. It restricts electric fields to static charge origins only

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

B (1), (3), (4)

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

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

💡 Short Explanation

�� Faraday's law links changing flux and emf. �� Induced emf equals line integral of E. �� It is complementary to Maxwell's displacement current concept.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. If magnetic flux remains constant: emf = −dΦB/dt = 0 Also: ∮E·dl = −dΦB/dt Statement 4 is incorrect because electric fields can arise from changing magnetic fields.

❓ 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 contradicting Faraday's law.

Final Logic:

Electric fields need not originate only from charges.

🧠 Memory Trick

Changing B Creates E

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Faraday's Law of Induction

7 Identify the incorrect statement regarding sinusoidal wave equations:

A The phase term is represented by (kz − ωt).

B The spatial wave number k dictates the period of oscillation in time.

C The equations Ex = E₀ sin(kz − ωt) and By = B₀ sin(kz − ωt) demonstrate they are in phase.

D Substituting z and t values determines instantaneous field amplitudes.

💡 Short Explanation

�� k governs spatial variation. �� ω governs temporal variation. �� E and B remain in phase.

✅ Correct Answer Explanation

The period of oscillation in time depends on ω. k determines wavelength and spatial variation. Therefore Option B is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct phase expression.
�� Option C → E and B are in phase.
�� Option D → Substitution gives instantaneous field values.

🎯 Strategy

Used

Concept Recall

Application:

Separate spatial and temporal parameters.

Final Logic:

k controls space, ω controls time.

🧠 Memory Trick

k→Space, ω→Time

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

8 Choose the correct statements about ω and k:
1. ω = ck relates the temporal and spatial frequencies
2. k = 2π/λ connects wave number to wavelength
3. ω is defined as 2π/ν
4. The speed of the wave equals ω/k

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

B (1), (3), (4)

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

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

💡 Short Explanation

�� ω = ck. �� k = 2π/λ. �� v = ω/k.

✅ Correct Answer Explanation

Statements 1, 2 and 4 are correct. Statement 3 is incorrect. Correct relation: ω = 2πν not 2π/ν. Therefore Option A is 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

Formula Recall

Application:

Use standard wave equations.

Final Logic:

ω = 2πν and v = ω/k.

🧠 Memory Trick

ω = 2πν

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

9 If an electromagnetic wave's frequency ν is 2.0 × 10¹⁰ Hz, its wavelength λ in vacuum is:
(c = 3 × 10⁸ m/s)

A 1.5 × 10⁻² m

B 1.5 m

C 6 × 10¹⁸ m

D 0.66 × 10² m

💡 Short Explanation

�� λ = c/ν. �� Substitute given values. �� Convert carefully.

✅ Correct Answer Explanation

λ = c/ν = (3 × 10⁸)/(2 × 10¹⁰) = 1.5 × 10⁻² m Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Wrong power of ten.
�� Option C → Physically impossible.
�� Option D → Calculation error.

🎯 Strategy

Used

Substitution

Application:

Use λ = c/ν.

Final Logic:

λ = 1.5 × 10⁻² m.

🧠 Memory Trick

λ = c/ν

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

10 As electromagnetic waves propagate through glass rather than a vacuum, the changes observed are dictated by the medium's properties. Choose the correct statements:
1. The total fields differ from the external fields by factors ε and μ
2. The velocity reduces due to the material's permittivity and permeability
3. The speed equation becomes v = 1/√(μ₀ε₀)
4. Glass acts as a perfect insulator blocking all fields

A (1), (2)

B (3), (4)

C (1), (3)

D (2), (4)

💡 Short Explanation

�� Material properties influence EM waves. �� Wave speed decreases in glass. �� ε and μ determine behavior.

✅ Correct Answer Explanation

Statement 1 is correct because fields inside materials are affected by the medium's electric and magnetic properties. Statement 2 is correct because: v = 1/√(με) and typically v < c. Statement 3 is incorrect because 1/√(μ₀ε₀) applies only to vacuum. Statement 4 is incorrect because glass transmits light and does not block all fields. Therefore Statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Identify statements applicable specifically to material media.

Final Logic:

Medium properties reduce wave speed and modify field behavior.

🧠 Memory Trick

Glass: v < c

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Propagation in Material Media

11 Match List I with List II

List I	List II
1. E₀	a. Amplitude of electric field
2. B₀	b. Amplitude of magnetic field
3. c	c. E₀/B₀
4. Propagation direction	d. E × B

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

�� E₀ denotes electric field amplitude. �� B₀ denotes magnetic field amplitude. �� E × B gives propagation direction.

✅ Correct Answer Explanation

E₀ represents the maximum (amplitude) value of the electric field. B₀ represents the maximum (amplitude) value of the magnetic field. For an electromagnetic wave, the speed of light is given by: c = E₀/B₀ The direction of propagation of an electromagnetic wave is given by the vector product E × B (the direction of the Poynting vector). Therefore: 1 → a 2 → b 3 → c 4 → d Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Electric and magnetic amplitudes are interchanged.
�� Option C → Physical quantities are mismatched.
�� Option D → Multiple incorrect pairings.

🎯 Strategy

Used

Option Grouping

Application:

Match each physical quantity with its definition.

Final Logic:

Standard EM-wave relations uniquely determine the matching.

🧠 Memory Trick

E, B, c, E×B

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Field Amplitudes and Vector Relations

12 If the amplitude of the electric field E₀ = 48 V/m, what is the maximum magnetic field amplitude B₀?
(c = 3 × 10⁸ m/s)

A 1.6 × 10⁻⁷ T

B 1.44 × 10¹⁰ T

C 48 T

D 1.6 × 10⁻⁸ T

💡 Short Explanation

�� E₀ = cB₀. �� B₀ = E₀/c. �� Substitute the values.

✅ Correct Answer Explanation

Using: B₀ = E₀/c = 48/(3 × 10⁸) = 16 × 10⁻⁸ = 1.6 × 10⁻⁷ T Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Obtained by multiplying instead of dividing.
�� Option C → Ignores EM-wave relation.
�� Option D → Incorrect decimal placement.

🎯 Strategy

Used

Substitution

Application:

Apply B₀ = E₀/c directly.

Final Logic:

48 ÷ (3 × 10⁸) = 1.6 × 10⁻⁷ T.

🧠 Memory Trick

B = E/c

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Electric and Magnetic Field Amplitudes

13 Identify the incorrect statement regarding the energy associated with an electromagnetic wave:

A Energy is equally divided between the electric and magnetic fields.

B The average energy density of the electric field relies on stationary charges for propagation.

C The wave carries energy from the source accelerated charge to other locations.

D Electromagnetic waves transport energy without any physical mass transfer.

💡 Short Explanation

�� EM waves are self-sustaining. �� Accelerated charges generate waves. �� Stationary charges do not sustain propagation.

✅ Correct Answer Explanation

Electromagnetic waves propagate due to mutually regenerating electric and magnetic fields. Their propagation does not rely on stationary charges. Energy is transported through space by the electromagnetic fields themselves. Therefore Option B is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct; average electric and magnetic energy densities are equal.
�� Option C → Correct; accelerated charges are the source of radiation.
�� Option D → Correct; EM waves transport energy without transporting matter.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting EM-wave propagation.

Final Logic:

Propagation is maintained by oscillating fields, not stationary charges.

🧠 Memory Trick

Accelerated Charge → Radiation

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

14 Choose the correct statements about electromagnetic wave energy transfer:
1. Light carries energy from the sun to the earth.
2. Radio and TV signals carry energy via self-sustaining oscillations.
3. The magnetic field energy density completely outscales the electric field energy density.
4. The average energy densities of both fields are equal.

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

B (1), (2), (3)

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

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

💡 Short Explanation

�� EM waves transport energy. �� Electric and magnetic energy densities are equal on average. �� Radio signals are energy-carrying EM waves.

✅ Correct Answer Explanation

Statements 1 and 2 are correct because EM waves transport energy over large distances. Statement 4 is correct because: Average Electric Energy Density = Average Magnetic Energy Density Statement 3 is incorrect because neither field dominates the average energy density. Hence 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:

Remove the statement contradicting energy-density equality.

Final Logic:

Average electric and magnetic contributions are equal.

🧠 Memory Trick

Half Electric, Half Magnetic

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Energy Density and Energy Transfer

15 If the electric field is in the +y direction and the wave travels in the +x direction, the magnetic field is strictly defined.
1. The cross product requires B to be in the +z direction
2. The cross product ĵ × k̂ = î satisfies the condition
3. The magnetic field aligns with the x-axis
4. The electric field aligns with the z-axis

A (1), (2)

B (3), (4)

C (1), (3)

D (2), (4)

💡 Short Explanation

�� Propagation direction = E × B. �� ĵ × k̂ = î. �� B must point along +z.

✅ Correct Answer Explanation

Given: E along +y Propagation along +x Since: E × B = Propagation Direction ĵ × B = î Therefore: B = k̂ Statements 1 and 2 are correct. Statements 3 and 4 are incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Both statements are incorrect.
�� Option C → Statement 3 is incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Vector Analysis

Application:

Apply the right-hand rule to E × B.

Final Logic:

ĵ × k̂ = î.

🧠 Memory Trick

j × k = i

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

16 To ensure the logical consistency of Maxwell's equations, it was found that the fields inside a parallel plate capacitor demonstrate orthogonality.
1. The electric field is perpendicular to the plates.
2. The magnetic field circles parallel to the plates.
3. B and E are entirely parallel.
4. The displacement current induces a parallel electric field.

A (1), (2)

B (2), (3)

C (1), (3)

D (3), (4)

💡 Short Explanation

�� Electric field exists between plates. �� Magnetic field forms circular loops. �� E and B are not parallel.

✅ Correct Answer Explanation

Inside a charging capacitor: • Electric field is perpendicular to the plates. • Displacement current produces magnetic field lines that circle around the axis, parallel to the plates. Thus Statements 1 and 2 are correct. Statements 3 and 4 are incorrect.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Concept Recall

Application:

Use the capacitor model discussed by Maxwell.

Final Logic:

E and B remain mutually perpendicular.

🧠 Memory Trick

Capacitor: E Through, B Around

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

17 If a wave possesses a frequency of 10⁹ Hz, what is the wavelength of the electromagnetic wave produced?
(c = 3 × 10⁸ m/s)

A 0.3 m

B 3 m

C 30 m

D 0.03 m

💡 Short Explanation

�� λ = c/f. �� Use vacuum speed of light. �� Substitute values.

✅ Correct Answer Explanation

λ = c/f = (3 × 10⁸)/(10⁹) = 0.3 m Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Corresponds to 10⁸ Hz.
�� Option C → Corresponds to 10⁷ Hz.
�� Option D → Corresponds to 10¹⁰ Hz.

🎯 Strategy

Used

Substitution

Application:

Apply λ = c/f.

Final Logic:

λ = 0.3 m.

🧠 Memory Trick

10⁹ Hz → 0.3 m

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

18 Because of the strong experimental backing of the constant velocity of EM waves in vacuum, the scientific community

A defines the standard of length using this parameter

B rejects standard definitions of velocity

C relies strictly on optical illusions

D defines time intervals entirely via magnetic field amplitudes

💡 Short Explanation

�� c is a universal constant. �� The metre is defined using c. �� Modern SI units depend on fixed constants.

✅ Correct Answer Explanation

The modern definition of the metre is based on the distance traveled by light in vacuum during a specified fraction of a second. Thus the fixed value of c is used to define length standards. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Scientific standards rely on velocity definitions.
�� Option C → Not scientific.
�� Option D → Time is not defined using magnetic amplitudes.

🎯 Strategy

Used

Concept Recall

Application:

Recall the SI definition of the metre.

Final Logic:

Length standards are based on c.

🧠 Memory Trick

Metre Defined by Light

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Speed of Light and SI Standards

19 Photon scales:

A Energy, Frequency

B Mass, Velocity

C Volume, Density

D Static, Dynamic

💡 Short Explanation

�� Photon energy depends on frequency. �� E = hν. �� Higher frequency means higher energy.

✅ Correct Answer Explanation

Planck's relation states: E = hν Thus photon properties are fundamentally linked through energy and frequency. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Not Planck's relation.
�� Option C → Irrelevant photon properties.
�� Option D → Not physical photon scales.

🎯 Strategy

Used

Concept Recall

Application:

Recall Planck's quantum hypothesis.

Final Logic:

Energy is directly proportional to frequency.

🧠 Memory Trick

E = hν

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Planck's Relation

20 Match List I with List II

List I	List II
1. Ultraviolet rays	a. 400 nm to 1 nm
2. Infrared waves	b. 1 mm to 700 nm
3. Gamma rays	c. < 10⁻³ nm
4. Microwaves	d. 0.1 m to 1 mm

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

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

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

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

💡 Short Explanation

�� UV lies below visible wavelengths. �� Infrared lies above visible wavelengths. �� Gamma rays have the shortest wavelengths.

✅ Correct Answer Explanation

Ultraviolet rays have wavelengths shorter than visible light, typically from 400 nm to 1 nm. Infrared waves have wavelengths longer than visible light, ranging from 700 nm to 1 mm. Gamma rays possess the shortest wavelengths in the electromagnetic spectrum, generally less than 10⁻³ nm. Microwaves occupy the region between radio waves and infrared waves, with wavelengths from 0.1 m to 1 mm. Therefore: 1 → a 2 → b 3 → c 4 → d Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → All wavelength regions are mismatched.
�� Option C → UV and IR are interchanged.
�� Option D → Microwave and infrared regions are interchanged.

🎯 Strategy

Used

Option Grouping

Application:

Match each EM-wave type to its standard wavelength range.

Final Logic:

Standard electromagnetic spectrum ordering determines the answer.

🧠 Memory Trick

Gamma < UV < Visible < IR < Microwave

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

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