CUET UG Physics Booster Test 2-AC Generator Technology -

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CUET UG Physics Booster Test 2-AC Generator Technology

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2-AC Generator Technology

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QUESTION 1 OF 20

Identify the incorrect statement about mechanical to electrical conversion in generators

Answer Review

1 Identify the incorrect statement about mechanical to electrical conversion in generators

A It relies on the changing of effective area or orientation of a loop

B It requires an external means to mechanically rotate the armature or magnets

C It works by changing the magnetic flux enclosed by the circuit

D It creates electrical energy from purely static charges independent of magnetism

💡 Short Explanation

�� Generators operate on electromagnetic induction. �� Changing magnetic flux is essential. �� Static charges alone cannot generate AC power.

✅ Correct Answer Explanation

Electrical energy in generators is produced through electromagnetic induction. A changing magnetic flux linked with a circuit induces an EMF according to Faraday's law. This is achieved by rotating a coil or magnet. Options A, B, and C correctly describe the working principle of generators. Option D is incorrect because generator action fundamentally depends on magnetism and changing flux, not static charges.

❓ Why Other Options Are Incorrect

�� Option A → Correct; changing orientation changes magnetic flux.
�� Option B → Correct; mechanical rotation is required.
�� Option C → Correct; changing magnetic flux induces EMF.

🎯 Strategy

Used

Elimination

Application: Identify the statement inconsistent with Faraday's law.

Final Logic: Static charges alone cannot explain generator operation.

🧠 Memory Trick

Generator = Changing Flux, Not Static Charge

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Principle of AC Generator

2 Regarding Nicola Tesla's development of the highly evolved machine,

A it exploited the phenomenon of electromagnetic induction technologically

B it exploited the principles of purely static electricity

C it exploited the concept of steady unchanging magnetic flux

D it exploited the mechanics of constant direct currents

💡 Short Explanation

�� AC generators are based on induction. �� Tesla advanced AC technology. �� Changing flux is essential.

✅ Correct Answer Explanation

Tesla's contributions to AC systems utilized the phenomenon of electromagnetic induction. AC generators work because changing magnetic flux induces EMF in a conductor. Options B, C, and D do not describe the operating principle of AC generation.

❓ Why Other Options Are Incorrect

�� Option B → Static electricity is unrelated to generator operation.
�� Option C → Unchanging flux cannot induce EMF.
�� Option D → AC generators produce alternating, not constant DC currents.

🎯 Strategy

Used

Concept Recall

Application: Recall the fundamental principle behind generators.

Final Logic: Electromagnetic induction is the technological basis of AC generation.

🧠 Memory Trick

Tesla + AC = Induction

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Applications of Electromagnetic Induction

3 Consider statements regarding armature and rotor shaft setup. Which statements are correct?
1. The coil (armature) is mounted on a rotor shaft.
2. The axis of rotation of the coil is perpendicular to the direction of the magnetic field.
3. Rotation causes the effective area of the loop to change.
4. The axis of rotation of the coil is parallel to the direction of the magnetic field.
Which of the above statements are correct?

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Coil is mounted on the rotor shaft. �� Rotation changes flux linkage. �� Standard generator axis is perpendicular to the magnetic field.

✅ Correct Answer Explanation

The armature coil is mounted on the rotor shaft and rotates in a magnetic field. The rotation changes the effective area (A cos θ) exposed to the field, thereby changing magnetic flux. Statement 4 is incorrect because the axis of rotation is perpendicular, not parallel, to the magnetic field in the standard AC generator arrangement.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application: Identify the false statement.

Final Logic: Only statement 4 is incorrect.

🧠 Memory Trick

Perpendicular Axis → Changing Flux

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Construction of AC Generator

4 In the operation of an AC generator, the slip rings and brushes:

A connect the ends of the rotating coil to an external circuit

B provide the primary mechanical rotation to the armature

C generate the uniform magnetic field inside the stator

D prevent the alternating current from changing direction

💡 Short Explanation

�� Slip rings rotate with the coil. �� Brushes remain stationary. �� They transfer current to the external circuit.

✅ Correct Answer Explanation

Slip rings are attached to the rotating coil, while stationary brushes maintain contact with them. Together they provide a continuous electrical connection between the rotating armature and the external circuit.

❓ Why Other Options Are Incorrect

�� Option B → Rotation is provided by an external mechanical source.
�� Option C → Magnetic fields are produced by magnets/electromagnets.
�� Option D → AC naturally changes direction; slip rings do not prevent it.

🎯 Strategy

Used

Odd One Out

Application: Identify the actual function of slip rings and brushes.

Final Logic: Their role is electrical connection, not rotation or field production.

🧠 Memory Trick

Slip Rings + Brushes = Current Collection

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Slip Rings and Brushes

5 Consider the following statements regarding effective area and rotation of a coil in a magnetic field:
1. The effective area of the loop is the face perpendicular to the field.
2. The effective area is mathematically defined as A cos θ.
3. As the coil rotates, the angle θ changes with time as ωt.
4. The effective area remains strictly constant regardless of the angle θ.
Which of the above statements are correct?

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Effective area depends on orientation. �� Effective area = A cos θ. �� θ changes continuously during rotation.

✅ Correct Answer Explanation

The effective area exposed to the magnetic field is A cos θ. As the coil rotates, θ changes as ωt, causing periodic changes in magnetic flux. Therefore statements 1, 2 and 3 are correct. Statement 4 is incorrect because effective area varies with θ.

❓ 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.

Final Logic: Effective area changes continuously with angle.

🧠 Memory Trick

Effective Area = A cos θ

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Effective Area and Flux

6 Match List I with List II

List I	List II
1. Instantaneous Flux (ΦB)	a. ωt
2. Induced EMF (ε) formula	b. NBAω
3. Angle θ	c. −N(dΦB/dt)
4. Peak EMF (ε₀)	d. BA cos ωt

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

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

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

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

💡 Short Explanation

�� Flux = BA cos ωt. �� EMF = −N(dΦ/dt). �� θ = ωt. �� Peak EMF = NBAω.

✅ Correct Answer Explanation

The correct matching is: Instantaneous Flux → BA cos ωt Induced EMF → −N(dΦB/dt) Angle θ → ωt Peak EMF → NBAω This corresponds to Option A.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect assignment of angle and EMF.
�� Option C → Multiple mismatches.
�� Option D → Peak EMF and induced EMF are interchanged.

🎯 Strategy

Used

Option Grouping

Application: Match standard generator equations.

Final Logic: Only Option A satisfies all formula relationships.

🧠 Memory Trick

Flux → cos, EMF → derivative

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Mathematical Description of AC Generation

7 If an armature operates at 0.5 Hz with an area of 0.10 m² and 100 turns in a 0.01 T field, at what time t (in seconds) will the EMF first reach its absolute maximum value? (Assume θ = 0° at t = 0)

A 0.50 s

B 0.25 s

C 1.00 s

D 0.75 s

💡 Short Explanation

�� ε = ε₀ sin ωt. �� Maximum EMF when sin ωt = 1. �� First maximum occurs after one-quarter cycle.

✅ Correct Answer Explanation

Frequency: ν = 0.5 Hz Period: T = 1/ν = 2 s Maximum EMF occurs at: T/4 = 2/4 = 0.5 s Therefore, the first maximum occurs at 0.50 s, not 0.25 s. The provided answer is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to one-eighth of a cycle.
�� Option C → One complete half-cycle later.
�� Option D → Does not correspond to the first maximum.

🎯 Strategy

Used

Substitution

Application: Use T = 1/ν and first maximum at T/4.

Final Logic: T = 2 s ⇒ T/4 = 0.50 s.

🧠 Memory Trick

First Peak = T/4

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Sinusoidal EMF and Time Variation

8 If the instantaneous EMF equation is defined by ε = ε₀ sin ωt, the time dependence of the voltage output will be

A alternating and periodic

B direct and steady

C constant and unchanging

D exponential and continuously growing

💡 Short Explanation

�� Sine function oscillates. �� Voltage changes sign periodically. �� Output is AC.

✅ Correct Answer Explanation

The equation ε = ε₀ sin ωt represents a sinusoidal alternating voltage. The value repeatedly changes magnitude and polarity with time, making the output periodic and alternating.

❓ Why Other Options Are Incorrect

�� Option B → Describes DC.
�� Option C → Contradicts sinusoidal variation.
�� Option D → Sinusoidal functions do not grow exponentially.

🎯 Strategy

Used

Formula Recall

Application: Interpret the physical meaning of a sine function.

Final Logic: Sinusoidal output is alternating and periodic.

🧠 Memory Trick

Sine Wave = AC Wave

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Alternating EMF

9 Which of the following correctly matches a factor affecting peak EMF (ε₀) with its relationship to ε₀?

A Number of turns (N), Direct linear proportionality

B Area (A), Inverse proportionality

C Magnetic Field (B), Squared proportionality

D Angular Speed (ω), Inverse proportionality

💡 Short Explanation

�� ε₀ = NBAω. �� N appears directly. �� Peak EMF increases linearly with N.

✅ Correct Answer Explanation

Peak EMF is given by: ε₀ = NBAω Therefore ε₀ is directly proportional to N, B, A and ω. Option A correctly identifies the relationship.

❓ Why Other Options Are Incorrect

�� Option B → Area is directly proportional, not inverse.
�� Option C → Magnetic field is linear, not squared.
�� Option D → Angular speed is directly proportional.

🎯 Strategy

Used

Formula Recall

Application: Inspect proportionalities from ε₀ = NBAω.

Final Logic: Only N shows direct linear proportionality as stated.

🧠 Memory Trick

More Turns → More Voltage

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Peak EMF Dependence

10 Consider the following statements regarding peak values of induced EMF:
1. Peak voltage occurs when the angle θ is 90° or 270°.
2. Peak voltage occurs when sin ωt = ±1.
3. Peak voltage corresponds to the points where the change of flux is greatest.
4. Peak voltage occurs when the flux itself is absolutely maximized.
Which of the above statements are correct?

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� EMF depends on rate of flux change. �� Maximum at θ = 90° and 270°. �� Flux itself is zero there.

✅ Correct Answer Explanation

Peak EMF occurs when: sin ωt = ±1 which corresponds to θ = 90° or 270°. At these positions, the rate of change of flux is maximum, producing maximum EMF. Statement 4 is incorrect because maximum flux occurs at θ = 0° or 180°, where induced EMF is zero.

❓ 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: Distinguish between maximum flux and maximum EMF.

Final Logic: Peak EMF corresponds to maximum dΦ/dt, not maximum Φ.

🧠 Memory Trick

Max EMF = Max Flux Change

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Conditions for Maximum EMF

11 The equation relating instantaneous EMF, peak EMF, and frequency is

A ε = ε₀ sin 2πνt

B ε = ε₀ cos 2πνt

C ε = 2πν sin ε₀t

D ε = (ε₀ / 2πν) sin t

💡 Short Explanation

�� ω = 2πν. �� ε = ε₀ sin ωt. �� Substituting ω gives ε = ε₀ sin 2πνt.

✅ Correct Answer Explanation

The standard AC generator equation is: ε = ε₀ sin ωt Since: ω = 2πν Substituting, ε = ε₀ sin(2πνt) Hence Option A is correct. This equation directly relates instantaneous EMF, peak EMF, frequency, and time.

❓ Why Other Options Are Incorrect

�� Option B → Represents a phase-shifted form but is not the standard relation derived here.
�� Option C → Incorrect placement of ε₀ and ν.
�� Option D → Dimensionally incorrect and not obtained from AC generator equations.

🎯 Strategy

Used

Substitution

Application: Replace ω with 2πν in the standard EMF equation.

Final Logic: ε = ε₀ sin ωt ⇒ ε = ε₀ sin 2πνt.

🧠 Memory Trick

ω = 2πν

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Time Dependence of AC EMF

12 Location standard, Implication on mechanical rotation speed is:

A USA (60 Hz), Armature must rotate faster than in India

B India (50 Hz), Armature rotates at the same speed as USA

C Global (100 Hz), Armature must be extremely massive

D USA (0 Hz), Relies purely on constant magnetic fields

💡 Short Explanation

�� Frequency is proportional to rotational speed. �� USA uses 60 Hz. �� India uses 50 Hz.

✅ Correct Answer Explanation

The frequency of generated AC depends on the rotational speed of the generator. Since the USA standard frequency is 60 Hz while India's is 50 Hz, the generator must rotate faster to produce 60 Hz than to produce 50 Hz. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → 50 Hz and 60 Hz require different rotational speeds.
�� Option C → No global 100 Hz standard exists.
�� Option D → 0 Hz corresponds to no alternating current.

🎯 Strategy

Used

Conceptual Analysis

Application: Relate frequency directly to rotational speed.

Final Logic: Higher frequency requires higher rotational speed.

🧠 Memory Trick

60 Hz → Faster Spin

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Frequency of AC Generation

13 If an AC generator requires mechanical energy to rotate the armature, and this energy is specifically provided by the potential energy of water falling from a height, it is classified as a

A hydro-electric generator

B thermal generator

C nuclear power generator

D perpetual-motion machine

💡 Short Explanation

�� Falling water drives turbines. �� Turbines rotate generators. �� Gravitational potential energy is utilized.

✅ Correct Answer Explanation

Hydro-electric power stations use water stored at a height. As water falls, its gravitational potential energy converts into kinetic energy, rotating turbines connected to generators. Hence such generators are classified as hydro-electric generators.

❓ Why Other Options Are Incorrect

�� Option B → Uses steam generated by heating water.
�� Option C → Uses nuclear fuel as the energy source.
�� Option D → Violates conservation of energy.

🎯 Strategy

Used

Contextual Matching

Application: Identify the energy source described.

Final Logic: Falling water uniquely identifies hydro-electric generation.

🧠 Memory Trick

Hydro = Height + Water

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Hydroelectric Generation

14 Match List I with List II

List I	List II
1. Fuel in thermal plants	a. Coal or other sources
2. Fuel in nuclear plants	b. Nuclear fuel
3. Intermediate kinetic phase	c. Steam at high pressure
4. Final output	d. Electrical energy

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

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

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

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

💡 Short Explanation

�� Thermal plants use coal or other fuels. �� Nuclear plants use nuclear fuel. �� Steam drives turbines. �� Final output is electrical energy.

✅ Correct Answer Explanation

The correct matching is: Fuel in thermal plants → Coal or other sources Fuel in nuclear plants → Nuclear fuel Intermediate kinetic phase → Steam at high pressure Final output → Electrical energy This corresponds to Option A.

❓ Why Other Options Are Incorrect

�� Option B → Thermal and nuclear fuels are interchanged.
�� Option C → Multiple mismatches.
�� Option D → Incorrect energy pathway sequence.

🎯 Strategy

Used

Option Grouping

Application: Follow the energy conversion pathway.

Final Logic: Fuel → Steam → Turbine → Electrical Energy.

🧠 Memory Trick

Fuel → Steam → Electricity

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Sources of Mechanical Energy

15 A commercial generator produces 500 MW of power. If it supplies a city where each home consumes an average continuous power of 5 kW, how many homes can it completely support?

A 100,000

B 10,000

C 1,000,000

D 50,000

💡 Short Explanation

�� Total power = 500 MW. �� One home uses 5 kW. �� Divide total power by power per home.

✅ Correct Answer Explanation

500 MW = 500,000 kW Number of homes: = 500,000 / 5 = 100,000 Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Underestimates by a factor of 10.
�� Option C → Overestimates by a factor of 10.
�� Option D → Only half of the correct value.

🎯 Strategy

Used

Substitution

Application: Homes = Total Power ÷ Power per Home.

Final Logic: 500,000 kW ÷ 5 kW = 100,000 homes.

🧠 Memory Trick

500 ÷ 5 = 100

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Commercial Generator Output

16 Incorrect statement regarding the physical design of high-power commercial generators

A The coils are usually held stationary

B The electromagnets are rotated to induce EMF

C Modern day generators can light up to 5 million 100 W bulbs

D They rotate the coils exclusively while keeping the electromagnets perfectly stationary

💡 Short Explanation

�� Large generators use stationary armatures. �� Rotating electromagnets are preferred. �� Current collection becomes easier.

✅ Correct Answer Explanation

In commercial generators, heavy armature coils are usually kept stationary while electromagnets rotate. This avoids transferring large currents through rotating contacts. Therefore, Option D is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of commercial generators.
�� Option B → Correct commercial design.
�� Option C → Consistent with a 500 MW generator.

🎯 Strategy

Used

Elimination

Application: Compare simple and commercial generator designs.

Final Logic: Commercial generators rotate magnets rather than large coils.

🧠 Memory Trick

Big Generator → Rotate Magnet

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Commercial Generator Construction

17 Relative position of Area vector and Field vector, Rate of flux change is:

A Area vector parallel to Field (θ = 0°), Rate of flux change is zero

B Area vector perpendicular to Field (θ = 90°), Rate of flux change is zero

C Area vector anti-parallel to Field (θ = 180°), Rate of flux change is greatest

D Area vector parallel to Field (θ = 0°), Rate of flux change is greatest

💡 Short Explanation

�� Flux is maximum at θ = 0°. �� dΦ/dt becomes zero there. �� EMF is zero.

✅ Correct Answer Explanation

Flux is given by: Φ = BA cos θ At θ = 0°, flux is maximum. Since the flux curve is at an extreme value, its instantaneous rate of change is zero. Therefore induced EMF is also zero. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → At 90°, rate of flux change is maximum.
�� Option C → At 180°, rate of change is zero.
�� Option D → Contradicts the flux-rate relationship.

🎯 Strategy

Used

Conceptual Analysis

Application: Distinguish between maximum flux and maximum flux change.

Final Logic: Maximum flux does not mean maximum EMF.

🧠 Memory Trick

Max Flux → Zero EMF

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Flux and EMF Relationship

18 Consider the following statements regarding the current reversal mechanism in an AC generator:
1. The fundamental driver is the sine function varying from +1 to −1.
2. The periodic variation reverses the polarity of the induced EMF.
3. The direction of the current changes periodically, defining it as AC.
4. The slip rings mechanically force the current backwards every cycle.
Which of the above statements are correct?

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� AC changes direction periodically. �� Polarity reverses naturally. �� Slip rings only provide electrical contact.

✅ Correct Answer Explanation

The sinusoidal variation of induced EMF causes polarity reversal every half cycle. As a result, the current direction changes periodically, producing alternating current. Statement 4 is incorrect because slip rings do not force reversal; they merely transfer the generated AC to the external circuit.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application: Identify the false statement.

Final Logic: Current reversal occurs naturally from sinusoidal EMF.

🧠 Memory Trick

AC Reversal = Sine Wave

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Alternating Current Generation

19 The mathematical equation for power assumes P = V × I. If a typical highly evolved modern AC generator outputs 100 MW at a peak voltage of 10 kV, the peak current generated is

A 10,000 A

B 1,000 A

C 100,000 A

D 100 A

💡 Short Explanation

�� P = VI. �� I = P/V. �� Substitute the given values.

✅ Correct Answer Explanation

Given: P = 100 MW = 100 × 10⁶ W V = 10 kV = 10 × 10³ V Current: I = P/V = (100 × 10⁶)/(10 × 10³) = 10,000 A Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Ten times smaller than actual value.
�� Option C → Ten times larger than actual value.
�� Option D → Much smaller than actual value.

🎯 Strategy

Used

Substitution

Application: Direct application of P = VI.

Final Logic: 100 MW ÷ 10 kV = 10,000 A.

🧠 Memory Trick

MW ÷ kV = kA

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Power Output Calculations

20 The technological utility of Faraday and Henry's discovery:

A led directly to the development of modern day generators and transformers

B proved that only perfectly static magnetic fields can do mechanical work

C is merely of academic interest, restricted to theoretical physics laboratories

D bypasses the law of conservation of energy to create free electricity

💡 Short Explanation

�� Faraday and Henry discovered electromagnetic induction. �� Generators and transformers depend on induction. �� Modern power systems use this principle extensively.

✅ Correct Answer Explanation

The discovery of electromagnetic induction forms the foundation of generators, transformers, and modern electric power transmission systems. It enabled practical electricity generation and distribution, transforming modern civilization. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Static magnetic fields do not produce induction.
�� Option C → The discovery has enormous practical applications.
�� Option D → Conservation of energy is never violated.

🎯 Strategy

Used

Contextual/Tonal Matching

Application: Identify the statement reflecting the real impact of electromagnetic induction.

Final Logic: Modern generators and transformers are direct applications of induction.

🧠 Memory Trick

Faraday → Generator → Power Grid

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Applications and Significance of Electromagnetic Induction

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