CUET UG Physics Booster Test 3- Nuclear Reactions and Radioactivity -

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CUET UG Physics Booster Test 3- Nuclear Reactions and Radioactivity

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3- Nuclear Reactions and Radioactivity

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QUESTION 1 OF 20

Correct statements about radioactivity discovery:
1. It revealed an emission capable of penetrating black paper and silver.
2. It proved that atomic nuclei have a specific constant density independent of mass number.
3. It was a nuclear phenomenon causing photographic blackening.
4. It required the compound to be actively fluorescing at the time of exposure.

Answer Review

1 Correct statements about radioactivity discovery:
1. It revealed an emission capable of penetrating black paper and silver.
2. It proved that atomic nuclei have a specific constant density independent of mass number.
3. It was a nuclear phenomenon causing photographic blackening.
4. It required the compound to be actively fluorescing at the time of exposure.

A (1), (3)

B (2), (4)

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

D (3), (4)

💡 Short Explanation

�� Radioactive emissions penetrated barriers. �� Photographic plates were blackened. �� Fluorescence was not required.

✅ Correct Answer Explanation

Statement 1 is correct because the emitted radiation penetrated black paper and even metallic barriers. Statement 2 is incorrect because constant nuclear density was not established by Becquerel's experiment. Statement 3 is correct because the blackening was caused by radioactive emissions from the nucleus. Statement 4 is incorrect because radioactivity was found to be spontaneous and independent of fluorescence. Therefore, statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Statements 2 and 4 are incorrect.
�� Option C → Includes statement 2, which is incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using the historical discovery of radioactivity.

Final Logic:

Only statements 1 and 3 accurately describe the discovery.

🧠 Memory Trick

Black Plate = Nuclear Emission

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Discovery of Radioactivity

2 Incorrect statement regarding the photographic plate in Becquerel's experiment

A The blackening occurred due to emissions that penetrated a piece of silver.

B The blackening proved that fluorescence from visible light irradiation was directly stored and re-emitted as X-rays.

C The package was separated from the photographic plate.

D The emission was spontaneous and not dependent on prior visible light exposure.

💡 Short Explanation

�� Radioactivity is spontaneous. �� Visible light was not responsible. �� Penetrating radiation exposed the plate.

✅ Correct Answer Explanation

Becquerel initially suspected a connection with fluorescence, but the experiment ultimately demonstrated spontaneous radioactive emission. The blackening was not due to stored visible-light energy being re-emitted as X-rays. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of the observation.
�� Option C → Correct experimental arrangement.
�� Option D → Correct conclusion regarding spontaneity.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement inconsistent with the conclusions of the experiment.

Final Logic:

Radioactivity is spontaneous, not re-emitted fluorescence.

🧠 Memory Trick

No Light Needed

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Discovery of Radioactivity

3 If the mass defect associated with an alpha decay event is 0.0054 u, what is the approximate energy Q released? (Use 1 u = 931.5 MeV/c²)

A 5.03 MeV

B 50.3 MeV

C 9.315 MeV

D 0.54 MeV

💡 Short Explanation

�� Energy comes from mass defect. �� Use E = Δmc². �� 1 u corresponds to 931.5 MeV.

✅ Correct Answer Explanation

Q = Δm × 931.5 MeV = 0.0054 × 931.5 ≈ 5.03 MeV Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Ten times larger than the correct value.
�� Option C → Incorrect multiplication.
�� Option D → Ten times smaller than the correct value.

🎯 Strategy

Used

Substitution

Application:

Substitute the mass defect into E = Δmc².

Final Logic:

0.0054 × 931.5 ≈ 5.03 MeV.

🧠 Memory Trick

Mass Defect × 931

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Mass Defect and Energy Release

4 In successive beta emissions intended to stabilize intermediate fission fragments, the primary mechanism at play

A forces the nucleus to absorb large amounts of kinetic energy

B changes the neutron-to-proton ratio without changing the overall mass number A

C requires overcoming a Coulomb barrier of 400 keV

D synthesizes helium nuclei directly from the fragment

💡 Short Explanation

�� Beta decay converts neutrons and protons. �� Mass number remains unchanged. �� Stability improves.

✅ Correct Answer Explanation

During beta decay, a neutron converts into a proton (or vice versa), changing the neutron-to-proton ratio. The total number of nucleons remains constant, so the mass number A does not change. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Beta decay does not require external kinetic-energy absorption.
�� Option C → Coulomb barriers are associated with fusion.
�� Option D → Helium nuclei are emitted in alpha decay.

🎯 Strategy

Used

Elimination

Application:

Differentiate beta decay from alpha decay and fusion.

Final Logic:

Beta decay changes N/Z ratio while keeping A constant.

🧠 Memory Trick

β → N/Z Change

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Beta Decay

5 Regarding gamma decay and binding energy, when a nucleus drops to a lower energy state by emitting a photon,

A it undergoes a transmutation into a completely different chemical element

B it absorbs energy to overcome the Coulomb repulsive force

C it releases hundreds of keV or more of energy while the atomic number and mass number remain entirely unaffected

D the mass defect is converted entirely into a positron

💡 Short Explanation

�� Gamma rays are photons. �� Nuclear composition remains unchanged. �� Only energy decreases.

✅ Correct Answer Explanation

Gamma decay occurs when an excited nucleus releases excess energy as a photon. The atomic number and mass number remain unchanged because no particle is emitted from the nucleus. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → No transmutation occurs.
�� Option B → Energy is released, not absorbed.
�� Option D → Positrons are associated with beta-plus decay.

🎯 Strategy

Used

Elimination

Application:

Identify the unique properties of gamma decay.

Final Logic:

Gamma decay changes energy state only.

🧠 Memory Trick

γ = Same Nucleus, Less Energy

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Gamma Decay

6 In a theoretical particle interaction framework, if x and y represent the number of fundamental negative charges (electrons) ejected simultaneously during two cascading high-energy interactions, what is the net charge ejected?

A (x + y)e

B -(x + y)e

C (x - y)e

D (y - x)e

💡 Short Explanation

�� Each electron has charge -e. �� Total electrons = x + y. �� Charges add algebraically.

✅ Correct Answer Explanation

Since each electron carries charge -e, the charge associated with x + y electrons is: Q = -(x + y)e Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Wrong sign.
�� Option C → Incorrect expression.
�� Option D → Incorrect expression.

🎯 Strategy

Used

Substitution

Application:

Multiply total electrons by charge per electron.

Final Logic:

Q = Number of electrons × (-e).

🧠 Memory Trick

Electron = Negative

📚 Class XI – NCERT Physics Part I – Chapter: Electric Charges and Fields – Topic: Quantisation of Charge

7 The transformation of Uranium-235 into intermediate mass nuclear fragments and the subsequent stabilization respectively represent:

A Exothermic splitting, Successive beta decays

B Endothermic fusion, Alpha scattering

C Gravitational collapse, Helium burning

D Plasma confinement, Neutron multiplication

💡 Short Explanation

�� U-235 undergoes fission. �� Energy is released. �� Fragments stabilize through beta decay.

✅ Correct Answer Explanation

Fission is an exothermic splitting process in which a heavy nucleus breaks into intermediate-mass fragments. These fragments are usually neutron-rich and undergo successive beta decays to reach stable configurations. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Describes unrelated processes.
�� Option C → Not associated with fission products.
�� Option D → Does not describe stabilization.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Follow the sequence of events after fission.

Final Logic:

Fission → Radioactive Fragments → Beta Decay.

🧠 Memory Trick

Split → Beta → Stable

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Fission

8 Match the nuclear metrics (List I) with their conceptual value/impact (List II) in heavy nucleus fission.

List I	List II
1. E_bn of original heavy nucleus	a. Appears as kinetic energy of fragments and heat
2. E_bn of intermediate fragments	b. Approximately 7.6 MeV
3. Difference in E_bn	c. Approximately 8.5 MeV
4. Disintegration Energy	d. Accounts for energy release (0.9 MeV per nucleon)

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

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

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

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

💡 Short Explanation

�� Heavy nuclei have lower Ebn. �� Intermediate fragments have higher Ebn. �� The difference produces energy.

✅ Correct Answer Explanation

1 → b : Heavy nuclei have Ebn ≈ 7.6 MeV. 2 → c : Intermediate fragments have Ebn ≈ 8.5 MeV. 3 → d : The difference (~0.9 MeV per nucleon) explains energy release. 4 → a : Disintegration energy appears as kinetic energy and heat. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Ebn values are interchanged.
�� Option C → Difference in Ebn is incorrectly matched.
�� Option D → Disintegration energy is mismatched.

🎯 Strategy

Used

Option Grouping

Application:

Match each nuclear quantity with its physical meaning.

Final Logic:

7.6 → Heavy, 8.5 → Fragment, 0.9 → Release, Energy → Heat.

🧠 Memory Trick

7.6 → 8.5 → Energy

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Binding Energy and Fission

9 Fission intermediate fragment statements:
1. The fragments possess higher binding energy per nucleon than the parent nucleus.
2. They are perfectly stable immediately after fission.
3. They carry a large portion of the disintegration energy as kinetic energy.
4. Examples include Barium, Krypton, Xenon, and Strontium.

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

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

C (1), (2)

D (1), (2), (3), (4)

💡 Short Explanation

�� Fragments have higher Ebn. �� Fragments are radioactive. �� They carry kinetic energy.

✅ Correct Answer Explanation

Statement 1 is correct because intermediate fragments are more tightly bound than the parent nucleus. Statement 2 is incorrect because the fragments are initially radioactive. Statement 3 is correct because much of the released energy appears as fragment kinetic energy. Statement 4 is correct because Barium, Krypton, Xenon and Strontium are common fission products. Therefore, statements 1, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 2, which is incorrect.
�� Option C → Includes statement 2 and omits correct statements.
�� Option D → Includes statement 2, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using known properties of fission fragments.

Final Logic:

Fragments are radioactive, energetic and more tightly bound.

🧠 Memory Trick

Fragments = Fast + Radioactive

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Fission Fragments

10 Correct statements about the neutrons in Uranium-235 fission:
1. They are solely responsible for carrying the 200 MeV energy.
2. 2, 3, or 4 neutrons can be produced depending on the specific fragments.
3. They are required to induce the initial fission.
4. They allow for the possibility of a chain reaction.

A (2), (3), (4)

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

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

D (1), (2)

💡 Short Explanation

�� Neutrons initiate fission. �� Neutron yield varies. �� Chain reactions depend on neutrons.

✅ Correct Answer Explanation

Statement 1 is incorrect because most of the energy is carried by the fission fragments, not solely by neutrons. Statement 2 is correct because 2–4 neutrons may be emitted depending on the fission channel. Statement 3 is correct because neutron absorption initiates U-235 fission. Statement 4 is correct because emitted neutrons can trigger further fissions. Therefore, statements 2, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 1, which is incorrect.
�� Option C → Includes statement 1, which is incorrect.
�� Option D → Includes statement 1 and omits statement 3.

🎯 Strategy

Used

Elimination

Application:

Separate neutron roles from energy-carrying mechanisms.

Final Logic:

Neutrons initiate and sustain chain reactions but do not carry all the energy.

🧠 Memory Trick

Neutrons → Start & Sustain

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Chain Reaction and Neutron Multiplication

11 Incorrect statement about fusion of light nuclei

A The binding energy per nucleon of the final fused system is greater than that of the initial lighter nuclei.

B Energy is released as the final system is more tightly bound.

C The process occurs spontaneously at room temperature because the strong nuclear force is long-ranged.

D The fusion of two deuterons into helium is an example of an exothermic nuclear fusion reaction.

💡 Short Explanation

�� Fusion requires extremely high temperatures. �� Strong nuclear force is short-ranged. �� Fusion does not occur spontaneously at room temperature.

✅ Correct Answer Explanation

The strong nuclear force acts only over a few femtometres and cannot bring distant nuclei together. Light nuclei must overcome the Coulomb barrier before the strong force becomes effective. Therefore, fusion does not occur spontaneously at room temperature. Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct because fusion products generally have higher binding energy per nucleon.
�� Option B → Correct explanation for energy release in fusion.
�� Option D → Fusion of light nuclei is an exothermic process.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting the known conditions for fusion.

Final Logic:

Fusion requires overcoming the Coulomb barrier; it does not occur spontaneously at room temperature.

🧠 Memory Trick

Fusion Needs Extreme Heat

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Fusion

12 In addressing the discrepancy between the estimated temperature needed to overcome the Coulomb barrier and the actual temperature of the sun's interior, scientists concluded that

A the core consists of an entirely different fuel than hydrogen

B fusion in the sun must rely on protons possessing kinetic energies significantly above the average thermal energy

C the strong nuclear force in the sun has a dramatically longer range

D gravitational force entirely negates the Coulomb barrier

💡 Short Explanation

�� Solar core temperature is lower than the classical barrier temperature. �� Some particles possess energies above the average. �� Fusion becomes possible through high-energy tail particles and tunnelling.

✅ Correct Answer Explanation

The average thermal energy in the solar core is insufficient to classically overcome the Coulomb barrier. Fusion occurs because a fraction of protons possess energies above the average and because quantum tunnelling allows penetration through the barrier. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Hydrogen is the primary fuel.
�� Option C → The strong nuclear force does not change its range.
�� Option D → Gravity does not eliminate the Coulomb barrier.

🎯 Strategy

Used

Elimination

Application:

Evaluate each explanation for solar fusion.

Final Logic:

Fusion occurs due to energetic particles and tunnelling, not altered forces.

🧠 Memory Trick

Sun = Hot Tail + Tunnelling

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Thermonuclear Fusion in the Sun

13 If the Coulomb barrier for two protons is ~ 400 keV, what is the equivalent energy in Joules? (Given 1 eV = 1.6 × 10⁻¹⁹ J)

A 6.4 × 10⁻¹⁴ J

B 3.2 × 10⁻¹⁴ J

C 1.6 × 10⁻¹³ J

D 4.0 × 10⁻¹⁹ J

💡 Short Explanation

�� Convert keV to eV. �� Use 1 eV = 1.6 × 10⁻¹⁹ J. �� Apply unit conversion directly.

✅ Correct Answer Explanation

400 keV = 400 × 10³ eV = 4 × 10⁵ eV Energy in joules: = (4 × 10⁵)(1.6 × 10⁻¹⁹) = 6.4 × 10⁻¹⁴ J Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Half the correct value.
�� Option C → Larger than the correct value.
�� Option D → Corresponds to only a few eV.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Convert keV to joules systematically.

Final Logic:

400 keV = 6.4 × 10⁻¹⁴ J.

🧠 Memory Trick

400 × 1.6 = 640

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Coulomb Barrier

14 The requirement of raising the temperature to ~ 3 × 10⁹ K to achieve thermonuclear fusion

A ensures the fuel turns into a solid crystalline state

B neutralizes the electric charge of the protons

C provides particles with sufficient average kinetic energy to overcome the Coulomb repulsive force

D initiates the spontaneous decay of stable helium into hydrogen

💡 Short Explanation

�� Higher temperature means higher kinetic energy. �� Coulomb repulsion must be overcome. �� Fusion requires close nuclear approach.

✅ Correct Answer Explanation

At temperatures around 3 × 10⁹ K, particles acquire enough average kinetic energy to overcome the electrostatic repulsion between positively charged nuclei and come within the range of the strong nuclear force. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Plasma forms, not a solid.
�� Option B → Proton charge remains unchanged.
�� Option D → Helium does not spontaneously decay into hydrogen.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Connect temperature with particle kinetic energy.

Final Logic:

High temperature helps nuclei overcome the Coulomb barrier.

🧠 Memory Trick

Heat → Speed → Fusion

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Thermonuclear Conditions

15 Match the specific multi-step components of the proton-proton cycle (List I) with their respective roles/products (List II)

List I	List II
1. First step (¹H + ¹H)	a. Yields ordinary ⁴He and two protons
2. Positron-electron annihilation	b. Produces a deuteron, positron, and neutrino
3. Formation of ³He	c. Yields two gamma-ray photons
4. Final step combining two ³He	d. Fuses a deuteron with a proton

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

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

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

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

💡 Short Explanation

�� Proton-proton cycle powers stars. �� Intermediate products are formed stepwise. �� Helium is the final product.

✅ Correct Answer Explanation

1 → b : Two protons produce a deuteron, positron, and neutrino. 2 → c : Positron-electron annihilation produces gamma photons. 3 → d : ³He forms when a deuteron fuses with a proton. 4 → a : Two ³He nuclei combine to form ⁴He and two protons. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Multiple stages are mismatched.
�� Option C → Initial and final reactions are incorrectly matched.
�� Option D → Several products are assigned incorrectly.

🎯 Strategy

Used

Option Grouping

Application:

Match each stage with its corresponding product.

Final Logic:

p+p → d, e⁺ annihilation → γ, d+p → ³He, ³He+³He → ⁴He.

🧠 Memory Trick

p+p → d → ³He → ⁴He

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Proton-Proton Cycle

16 Stellar cycle statements:
1. The depletion of hydrogen causes the core to cool and collapse under gravity.
2. Gravitational collapse decreases the temperature of the core.
3. At a core temperature of about 10⁸ K, helium nuclei fuse into carbon.
4. Elements more massive than those near the peak of the binding energy curve cannot be produced by fusion alone.

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

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

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

D (1), (4)

💡 Short Explanation

�� Hydrogen depletion destabilizes the core. �� Helium burning begins near 10⁸ K. �� Very heavy elements are not formed by ordinary fusion alone.

✅ Correct Answer Explanation

Statement 1 is taken as correct in the stellar-evolution sequence because hydrogen depletion leads to gravitational contraction of the core. Statement 2 is incorrect because gravitational collapse increases, not decreases, the core temperature. Statement 3 is correct because helium fusion into carbon begins around 10⁸ K. Statement 4 is correct because fusion beyond the iron region is not energetically favorable. Therefore, statements 1, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 2, which is incorrect.
�� Option C → Includes statement 2, which is incorrect.
�� Option D → Omits statement 3, which is correct.

🎯 Strategy

Used

Elimination

Application:

Check each statement against stellar evolution principles.

Final Logic:

Hydrogen depletion leads to contraction, heating, and helium burning.

🧠 Memory Trick

H Ends → He Burns

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Stellar Evolution

17 In controlled thermonuclear fusion devices, the fuel state and the required heating temperature are respectively:

A Gaseous, 10³ K

B Plasma, 10⁸ K

C Liquid metal, 3 × 10⁹ K

D Solid, 1.5 × 10⁷ K

💡 Short Explanation

�� Fusion fuel exists as plasma. �� Extremely high temperatures are required. �� Magnetic confinement is used.

✅ Correct Answer Explanation

Controlled fusion reactors use a plasma consisting of ions and electrons. The plasma must be heated to approximately 10⁸ K to sustain fusion reactions. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Temperature is far too low.
�� Option C → Fuel is not a liquid metal.
�� Option D → Fusion fuel is not solid.

🎯 Strategy

Used

Memory-Based Recall

Application:

Recall standard fusion-reactor operating conditions.

Final Logic:

Fusion reactors use plasma at about 10⁸ K.

🧠 Memory Trick

Fusion = Plasma + 10⁸ K

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Controlled Thermonuclear Fusion

18 Incorrect statement regarding thermonuclear fusion reactors

A The objective is to generate steady power for humanity.

B The fuel is a plasma mixture of positive ions and electrons.

C Successful implementation mimics the energy production mechanism of stars.

D The plasma must be confined within physical iron containers to withstand the 10⁸ K temperatures.

💡 Short Explanation

�� No material can withstand 10⁸ K. �� Magnetic confinement is required. �� Fusion reactors imitate stars.

✅ Correct Answer Explanation

At temperatures around 10⁸ K, any physical container would melt or vaporize. Therefore, plasma is confined using magnetic fields or other advanced confinement methods. Hence, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct objective of fusion research.
�� Option B → Correct description of plasma.
�� Option C → Fusion reactors mimic stellar fusion.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement violating known plasma-confinement principles.

Final Logic:

No physical container can hold 10⁸ K plasma.

🧠 Memory Trick

Magnetic Bottle, Not Metal Bottle

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Fusion Reactors

19 If we measure the separated negative charge of x completely ionized uranium atoms and y completely ionized carbon atoms in a hypothetical isolated mass-energy interconversion thought experiment, what is the total negative charge collected?

A -(x + y)e

B (x + y)e

C (x - y)e

D (y - x)e

💡 Short Explanation

�� Ionized uranium contributes 92 electrons. �� Ionized carbon contributes 6 electrons. �� Atomic numbers must be included.

✅ Correct Answer Explanation

The total charge collected depends on the total number of electrons removed. For x uranium atoms: Charge = -92xe For y carbon atoms: Charge = -6ye Hence, Q = -(92x + 6y)e None of the given options match this expression.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect sign.
�� Option C → Incorrect algebraic form.
�� Option D → Incorrect algebraic form.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Count electrons using atomic numbers.

Final Logic:

Total charge depends on the number of electrons removed from each atom.

🧠 Memory Trick

Charge = Z × e

📚 Class XI – NCERT Physics Part I – Chapter: Electric Charges and Fields – Topic: Quantisation of Charge

20 Correct statements about comparing 1 kg of uranium and 1 kg of coal:
1. Burning coal produces 10⁷ J.
2. Fissioning uranium produces 10¹⁴ J.
3. The mass defect in the coal reaction is exactly zero.
4. The nuclear source produces a million times more energy due to the vastly larger mass defects involved in nuclear binding compared to chemical binding.

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

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

C (2), (3)

D (1), (2), (3), (4)

💡 Short Explanation

�� Coal produces about 10⁷ J/kg. �� Uranium produces about 10¹⁴ J/kg. �� Chemical mass defects are tiny but non-zero.

✅ Correct Answer Explanation

Statement 1 is correct because coal combustion releases about 10⁷ J per kilogram. Statement 2 is correct because uranium fission releases about 10¹⁴ J per kilogram. Statement 3 is incorrect because chemical reactions also involve tiny mass defects, though they are negligible. Statement 4 is correct because nuclear mass defects are vastly larger than chemical mass defects. Therefore, statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 3, which is incorrect.
�� Option C → Includes statement 3 and omits statement 1.
�� Option D → Includes statement 3, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Compare nuclear and chemical energy scales.

Final Logic:

Nuclear reactions release far more energy because of larger mass defects.

🧠 Memory Trick

10⁷ vs 10¹⁴

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear and Chemical Energy Comparison

CUET UG Physics Booster Test 3- Nuclear Reactions and Radioactivity

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