CUET UG Physics Booster Test 2-Mathematical Principles and Laws -

🚀

CUET UG Physics Booster Test 2-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 2-Mathematical Principles and Laws

1 / 20

QUESTION 1 OF 20

Match List I with List II

List I	List II
1. ∮ E · dA = Q/ε₀	a. Gauss's Law for magnetism
2. ∮ B · dA = 0	b. Gauss's Law for electricity
3. E between capacitor plates	c. ε₀(dΦE/dt)
4. Displacement current	d. (Q/A)/ε₀

Answer Review

1 Match List I with List II

List I	List II
1. ∮ E · dA = Q/ε₀	a. Gauss's Law for magnetism
2. ∮ B · dA = 0	b. Gauss's Law for electricity
3. E between capacitor plates	c. ε₀(dΦE/dt)
4. Displacement current	d. (Q/A)/ε₀

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

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

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

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

💡 Short Explanation

�� Gauss's law relates electric flux to enclosed charge. �� Magnetic flux through a closed surface is zero. �� Displacement current arises from changing electric flux.

✅ Correct Answer Explanation

The correct matching is: List I — List II 1 — b 2 — a 3 — d 4 — c ∮E·dA = Q/ε₀ represents Gauss's law for electricity. ∮B·dA = 0 represents Gauss's law for magnetism. Electric field between capacitor plates is E = (Q/A)/ε₀. Displacement current is ε₀(dΦE/dt). Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Electric and magnetic Gauss laws are interchanged.
�� Option C → Multiple mismatches occur.
�� Option D → Incorrect assignment of magnetic law and capacitor field.

🎯 Strategy

Used

Option Grouping

Application:

Associate each mathematical expression with its physical law.

Final Logic:

Each equation has a unique physical interpretation.

🧠 Memory Trick

Electric→Charge, Magnetic→Zero

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

2 Maxwell noticed an inconsistency in Ampere's circuital law when analyzing a charging capacitor. To fix it,

A he removed it by suggesting the existence of a displacement current

B he redefined the constant ε₀ to fix the enclosed charge ratio

C he proved that perfectly conducting media do not exist

D he proved conduction current equals zero everywhere

💡 Short Explanation

�� Charging capacitors revealed a contradiction. �� Maxwell introduced displacement current. �� Ampere's law became universally valid.

✅ Correct Answer Explanation

In a charging capacitor, conduction current exists in the wires but not across the gap between plates. This created an inconsistency in Ampere's law. Maxwell resolved this by introducing displacement current: Id = ε₀(dΦE/dt) This ensured continuity of current and completed Ampere-Maxwell law. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → ε₀ was not redefined.
�� Option C → Unrelated to Ampere's law.
�� Option D → Conduction current clearly exists in conductors.

🎯 Strategy

Used

Concept Recall

Application:

Recall Maxwell's correction to Ampere's circuital law.

Final Logic:

Displacement current solved the capacitor paradox.

🧠 Memory Trick

Capacitor Gap → Displacement Current

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

3 If the frequency of a plane electromagnetic wave is 30 MHz, what is its wavelength in vacuum?
(c = 3 × 10⁸ m/s)

A 10 m

B 1 m

C 0.1 m

D 100 m

💡 Short Explanation

�� Use λ = c/f. �� Convert MHz to Hz. �� EM waves travel at c in vacuum.

✅ Correct Answer Explanation

Given: f = 30 MHz = 30 × 10⁶ Hz Using: λ = c/f = (3 × 10⁸)/(30 × 10⁶) = 10 m Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Calculation error.
�� Option C → Requires much higher frequency.
�� Option D → Requires much lower frequency.

🎯 Strategy

Used

Substitution

Application:

Apply λ = c/f.

Final Logic:

30 MHz corresponds to 10 m.

🧠 Memory Trick

30 MHz → 10 m

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

4 Gauss's law for magnetism guarantees which of the following?
1. ∮ B · dA equals μ₀I
2. Magnetic monopoles act as analogous sources to electric charges
3. The right hand side of the equation is purely zero
4. Displacement current acts as the monopole source

A (1), (2)

B (3) only

C (2), (3)

D (1), (4)

💡 Short Explanation

�� Magnetic monopoles have not been observed. �� Net magnetic flux through a closed surface is zero. �� Magnetic field lines form closed loops.

✅ Correct Answer Explanation

Gauss's law for magnetism states: ∮ B·dA = 0 Therefore Statement 3 is correct. Statement 1 belongs to Ampere-Maxwell law. Statement 2 is incorrect because magnetic monopoles have not been experimentally detected. Statement 4 is incorrect because displacement current is not a monopole source. Hence only Statement 3 is correct.

❓ Why Other Options Are Incorrect

�� Option A → Both statements are incorrect.
�� Option C → Statement 2 is incorrect.
�� Option D → Both statements are incorrect.

🎯 Strategy

Used

Elimination

Application:

Select the statement directly expressing Gauss's magnetic law.

Final Logic:

Closed-surface magnetic flux is always zero.

🧠 Memory Trick

No Monopoles → Zero Flux

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Gauss's Law for Magnetism

5 Wave propagation directions are:

A Perpendicular, Perpendicular

B Parallel, Perpendicular

C Parallel, Parallel

D Perpendicular, Parallel

💡 Short Explanation

�� EM waves are transverse. �� E ⟂ B. �� Both fields are perpendicular to propagation.

✅ Correct Answer Explanation

In an electromagnetic wave: Electric field ⟂ Magnetic field Electric field ⟂ Direction of propagation Magnetic field ⟂ Direction of propagation Therefore both relationships are perpendicular. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → E and B are not parallel.
�� Option C → Neither relationship is parallel.
�� Option D → Contradicts EM-wave geometry.

🎯 Strategy

Used

Concept Recall

Application:

Recall the transverse nature of electromagnetic waves.

Final Logic:

E ⟂ B ⟂ Propagation.

🧠 Memory Trick

Three Mutual Perpendiculars

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

6 Consider the following statements on electromagnetic induction. Choose the correct statements:
1. Induced emf implies an induced electric field.
2. Magnetic fields changing with time create electric fields.
3. The emf equals the rate of change of magnetic flux.
4. The displacement current blocks the induced electric field.

A (1), (3), (4)

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

C (1), (2), (3), (4)

D (2), (3), (4)

💡 Short Explanation

�� Faraday's law links changing flux and emf. �� Changing B creates E. �� Displacement current does not block induction.

✅ Correct Answer Explanation

Statement 1 is correct because induced emf implies the presence of an induced electric field. Statement 2 is correct according to Faraday's law. Statement 3 is correct: emf = −dΦB/dt Statement 4 is incorrect because displacement current does not oppose or block induced electric fields. Therefore Statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Remove the statement contradicting electromagnetic induction.

Final Logic:

Displacement current does not suppress induced electric fields.

🧠 Memory Trick

Changing B → E

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

7 Incorrect statement about plane electromagnetic wave equations:

A The electric field Eₓ varies sinusoidally with z at a given time.

B The magnetic field Bᵧ oscillates perpendicular to Eₓ.

C The spatial variation is defined by the angular frequency ω.

D The wave propagates along the z-direction.

💡 Short Explanation

�� ω controls time variation. �� k controls spatial variation. �� Plane waves are sinusoidal.

✅ Correct Answer Explanation

A plane EM wave is represented as: E = E₀ sin(kz − ωt) Here: k determines spatial variation. ω determines temporal variation. Therefore Option C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct.
�� Option B → E and B are perpendicular.
�� Option D → Consistent with the given wave equation.

🎯 Strategy

Used

Formula Recall

Application:

Differentiate between k and ω.

Final Logic:

k → Space, ω → Time.

🧠 Memory Trick

k = Position, ω = Time

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

8 Identify the correct statements about frequency and wave numbers:
1. The angular frequency is ω = 2πν
2. The wave vector magnitude is k = 2π/λ
3. The speed of the wave is calculated by ω/k
4. k defines the energy density of the wave

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

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

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

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

💡 Short Explanation

�� ω = 2πν. �� k = 2π/λ. �� v = ω/k.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are standard wave relations. Statement 4 is incorrect because k measures spatial periodicity, not energy density. Therefore Option A is 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 assigning energy-density meaning to k.

Final Logic:

k describes wavelength, not energy density.

🧠 Memory Trick

ω = 2πν, k = 2π/λ

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

9 If an FM radio band transmits at a frequency of 100 MHz, what is its corresponding wavelength?
(c = 3 × 10⁸ m/s)

A 30 m

B 3 m

C 0.3 m

D 300 m

💡 Short Explanation

�� λ = c/f. �� Convert MHz to Hz. �� FM wavelengths are a few meters.

✅ Correct Answer Explanation

Given: f = 100 MHz = 100 × 10⁶ Hz Using: λ = c/f = (3 × 10⁸)/(100 × 10⁶) = 3 m Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to 10 MHz.
�� Option C → Corresponds to 1000 MHz.
�� Option D → Corresponds to 1 MHz.

🎯 Strategy

Used

Substitution

Application:

Apply λ = c/f.

Final Logic:

100 MHz corresponds to 3 m.

🧠 Memory Trick

100 MHz = 3 m

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Radio Wave Calculations

10 The refractive index of one medium with respect to another is equal to the ratio of the velocities of light in the two media. Choose the correct statements:
1. Velocity depends on electric permittivity and magnetic permeability.
2. Velocity is totally independent of the medium properties.
3. The formula v = 1/√(με) defines it.
4. Permeability μ directly replaces μ₀ in the material medium.

A (1), (3), (4)

B (1), (2)

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

D (3), (4)

💡 Short Explanation

�� Wave speed depends on ε and μ. �� Material properties affect velocity. �� Refractive index depends on speed ratio.

✅ Correct Answer Explanation

Statement 1 is correct because EM-wave speed depends on the medium's permittivity and permeability. Statement 3 is correct: v = 1/√(με) Statement 4 is correct because μ is the permeability of the material medium. Statement 2 is incorrect because velocity clearly depends on medium properties. Therefore 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 1.

🎯 Strategy

Used

Elimination

Application:

Identify the statement contradicting medium dependence.

Final Logic:

EM-wave speed varies with ε and μ.

🧠 Memory Trick

More ε or μ → Less Speed

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Velocity in Material Media and Refractive Index

11 Match List I with List II

List I	List II
1. B₀c	a. E₀
2. 2π/λ	b. k
3. 1/√(μ₀ε₀)	c. c
4. ω/k	d. Wave speed

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

💡 Short Explanation

�� E₀ = cB₀. �� k = 2π/λ. �� c = 1/√(μ₀ε₀).

✅ Correct Answer Explanation

The correct matching is: 1 → a : B₀c = E₀ 2 → b : 2π/λ = k 3 → c : 1/√(μ₀ε₀) = c 4 → d : ω/k = Wave speed Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Physical quantities mismatched.
�� Option C → Incorrect assignment of constants.
�� Option D → Multiple relations interchanged.

🎯 Strategy

Used

Option Grouping

Application:

Match standard EM-wave formulae with their corresponding quantities.

Final Logic:

Each expression directly represents a known EM-wave parameter.

🧠 Memory Trick

E = cB, k = 2π/λ

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Field Amplitudes and Wave Parameters

12 If the electric field component of a wave is, Eᶻ = 60 sin(0.5 × 10³x + 1.5 × 10¹¹t) V/m, what is the value of B₀?
(c = 3 × 10⁸ m/s)

A 2 × 10⁻⁷ T

B 1.8 × 10¹⁰ T

C 60 T

D 0.5 × 10³ T

💡 Short Explanation

�� E₀ = 60 V/m. �� E₀ = cB₀. �� Solve for B₀.

✅ Correct Answer Explanation

For electromagnetic waves: B₀ = E₀/c = 60/(3 × 10⁸) = 2 × 10⁻⁷ T Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Obtained using incorrect multiplication.
�� Option C → Ignores the E₀/c relation.
�� Option D → Unrelated to magnetic field amplitude.

🎯 Strategy

Used

Substitution

Application:

Apply B₀ = E₀/c directly.

Final Logic:

60/(3 × 10⁸) = 2 × 10⁻⁷ T.

🧠 Memory Trick

B = E/c

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

13 Average energy states are:

A Equal, Balanced

B Maximum, Minimum

C Zero, Infinite

D Static, Dynamic

💡 Short Explanation

�� Electric and magnetic energies share equally. �� Average energy densities are equal. �� Energy remains balanced.

✅ Correct Answer Explanation

In an electromagnetic wave: Average Electric Energy Density = Average Magnetic Energy Density Thus the energy distribution remains balanced between electric and magnetic components. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Average energies are not unequal.
�� Option C → Physically impossible.
�� Option D → Does not describe energy-density relation.

🎯 Strategy

Used

Concept Recall

Application:

Recall energy distribution in EM waves.

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

14 Identify the incorrect statement about energy densities:

A The energy associated with the propagating wave originates from stationary charges.

B The E field average energy density equals the B field average energy density.

C The oscillating fields regenerate each other during propagation.

D The wave can carry energy from one place to another.

💡 Short Explanation

�� EM waves originate from accelerated charges. �� Electric and magnetic energies are equal on average. �� Waves transport energy.

✅ Correct Answer Explanation

Electromagnetic waves are produced by accelerated charges, not stationary charges. Stationary charges produce static electric fields but do not radiate electromagnetic waves. Therefore Option A is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Correct property of EM waves.
�� Option C → Maxwell's theory explains mutual regeneration.
�� Option D → EM waves transport energy.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement inconsistent with wave generation.

Final Logic:

Radiation requires acceleration of charges.

🧠 Memory Trick

Accelerate → Radiate

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

15 Choose the correct statement regarding the direction of propagation:
1. E × B yields the z-direction if E is ĵ and B is k̂.
2. Since ĵ × k̂ = î, propagation is along the x-axis.
3. The fields are mutually parallel.
4. Vector algebra governs the field orientations.

A (1), (3), (4)

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

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

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

💡 Short Explanation

�� E × B gives propagation direction. �� ĵ × k̂ = î. �� E and B are perpendicular.

✅ Correct Answer Explanation

Statement 2 is correct because: ĵ × k̂ = î Hence propagation is along the x-axis. Statement 4 is correct because vector algebra determines field orientations. Statement 3 is incorrect since E and B are mutually perpendicular. The intended answer includes Statements 1, 2 and 4.

❓ 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:

Remove the statement violating orthogonality.

Final Logic:

E and B cannot be parallel.

🧠 Memory Trick

E × B = Direction

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

16 Identify the correct statements about spatial field orthogonality:
1. Eₓ and Bᵧ oscillate perfectly perpendicular to each other.
2. The direction of propagation is perpendicular to both fields.
3. The fields vibrate in phase with identical angular frequencies.
4. The magnetic field amplitude strictly equals the electric field amplitude.

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

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

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

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

💡 Short Explanation

�� E and B are perpendicular. �� Both are perpendicular to propagation. �� They oscillate in phase.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. Statement 4 is incorrect because: E₀ = cB₀ The amplitudes are related but not numerically equal. Hence Option A is 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:

Use E₀ = cB₀.

Final Logic:

Amplitude relation exists, but equality does not.

🧠 Memory Trick

In Phase, Not Equal

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

17 In vacuum, the fundamental speed constant c defines the relationship between the electric and magnetic fields.

A B₀ = E₀c

B c = E₀/B₀

C c = B₀/E₀

D c = E₀B₀

💡 Short Explanation

�� Electric and magnetic amplitudes are related. �� E₀ = cB₀. �� Rearranging gives c = E₀/B₀.

✅ Correct Answer Explanation

For electromagnetic waves: E₀ = cB₀ Therefore: c = E₀/B₀ Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct relation is E₀ = cB₀.
�� Option C → Inverse ratio.
�� Option D → Dimensionally incorrect.

🎯 Strategy

Used

Formula Recall

Application:

Apply the field amplitude relation.

Final Logic:

c equals the ratio E₀/B₀.

🧠 Memory Trick

c = E/B

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Field Amplitude Ratio

18 Because experiments show the velocity is independent of wavelength to within a few meters per second out of 3 × 10⁸ m/s, it is known that
1. the constancy of c dictates the standard of length definition
2. length standards vary based on light frequency
3. time-varying fields depend on the standard length
4. only optical measurements determine the length standard

A (1) only

B (1), (3)

C (2), (4)

D (1), (4)

💡 Short Explanation

�� c is invariant. �� Modern metre definition uses c. �� Frequency does not redefine length standards.

✅ Correct Answer Explanation

Statement 1 is correct because the SI metre is defined using the fixed value of c. Statements 2, 3 and 4 are incorrect. Therefore only Statement 1 is correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 3 is incorrect.
�� Option C → Statements 2 and 4 are incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Elimination

Application:

Retain only the statement directly linked to the SI definition.

Final Logic:

The fixed value of c defines the metre.

🧠 Memory Trick

Metre via Light

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

19 Match List I with List II

List I	List II
1. E = hν	a. Photon energy relation
2. ν = c/λ	b. Frequency formula
3. c	c. 3 × 10⁸ m/s
4. Visible light range	d. 700 nm – 400 nm

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

💡 Short Explanation

�� E = hν gives photon energy. �� ν = c/λ gives frequency. �� c = 3 × 10⁸ m/s.

✅ Correct Answer Explanation

Correct matching: 1 → a : E = hν → Photon energy relation 2 → b : ν = c/λ → Frequency formula 3 → c : c → 3 × 10⁸ m/s 4 → d : Visible range → 700 nm–400 nm Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Formulae incorrectly paired.
�� Option C → Energy and frequency relations interchanged.
�� Option D → Multiple mismatches.

🎯 Strategy

Used

Option Grouping

Application:

Associate each expression with its physical meaning.

Final Logic:

Standard definitions uniquely determine the matching.

🧠 Memory Trick

hν → Energy

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Planck Relation and EM Spectrum

20 Radio waves, possessing frequencies around 500 kHz to 1000 MHz, correlate with specific photon energies.

A Their photon energies are higher than those of X-rays.

B Their energies are lower than those of visible light photons.

C Their energies are defined by E = mc².

D They vibrate atoms directly through infrared absorption.

💡 Short Explanation

�� Photon energy depends on frequency. �� Radio waves have very low frequencies. �� Therefore radio photons have very low energies.

✅ Correct Answer Explanation

Using Planck's relation: E = hν Radio waves have frequencies much smaller than visible light, X-rays, and gamma rays. Hence their photon energies are significantly lower than visible-light photon energies. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → X-rays have much higher photon energies.
�� Option C → Photon energy is determined by E = hν, not E = mc².
�� Option D → Infrared absorption is unrelated to radio-wave photon energy.

🎯 Strategy

Used

Concept Recall

Application:

Apply Planck's relation.

Final Logic:

Lower frequency means lower photon energy.

🧠 Memory Trick

Lower ν → Lower E

📚 Class XII – Physics Part I – Chapter: Electromagnetic Waves – Topic: Photon Energy Scales and Planck Relation

CUET UG Physics Booster Test 2-Mathematical Principles and Laws

Test Complete!

Answer Review