Physics CSS Solved Papers

CSS Physics Paper-II 2020 Solved

Engr. Muhammad Yar Saqib

CSS Physics Paper-II 2020 Solved is a complete solved guide for CSS aspirants who need real answers, not short hints. This post solves the subjective section question by question with definitions, derivations, formulas, numerical substitutions, final answers, tables and exam-ready explanations. It covers Coulomb’s law, electric field of point charges, Poisson equation, Laplace equation, Poynting vector, Maxwell equations, magnetic vector potential, Heisenberg uncertainty principle, barrier tunneling, electron momentum, semiconductor doping, NPN and PNP transistors, MOSFET, natural radioactivity, radioactive decay law, alpha-decay energy, nuclear fission, nuclear reactors, radiation detection methods, dielectric polarization, Ampere’s law and particle accelerators.

Central Argument: A CSS Physics solved paper should not only name the topic. It should define the principle, derive the equation, show the numerical working and explain the physical meaning. Therefore, this CSS Physics Paper-II 2020 Solved post gives the complete route to each answer so students can reproduce the solution in the examination hall.

Scan Note: The uploaded text uses spellings such as “Dopping” and “MOFET.” In this solved post, they are corrected to doping and MOSFET, while the original question meaning is preserved.

CSS Physics Paper-II 2020 Solved Study Scope

This post covers CSS Physics Paper-II 2020 Solved in a complete, structured and WordPress-ready format. Each question includes the printed question, part-wise solution, formulas, mathematical working, final answer and examiner-friendly explanation.

Use this solved paper as a study document. First revise the formula sheet, then read Q2 to Q8 one by one. After reading, close the post and rewrite each solution from memory. CSS Physics rewards clean definitions, correct assumptions, dimensional accuracy, correct units and disciplined presentation.

Show Table of Contents
  1. Overview
  2. Study Scope
  3. Important Formula Sheet
  4. Question Map
  5. Question 2: Coulomb Law, Poisson Equation and Electric Field
  6. Question 3: Poynting Vector, Maxwell Equations and Vector Potential
  7. Question 4: Uncertainty Principle, Tunneling and Electron Momentum
  8. Question 5: Doping, NPN/PNP Transistors and MOSFET
  9. Question 6: Natural Radioactivity, Radioactive Decay and Alpha-Decay Energy
  10. Question 7: Nuclear Fission, Reactors and Radiation Detection
  11. Question 8: Dielectric Polarization, Ampere Law and Accelerators
  12. Revision Plan
  13. Internal and External Resources
  14. FAQs

Important Formula Sheet for CSS Physics Paper-II 2020 Solved

Electrostatics

F = (1/4πε0)(q1q2/r²)

E = F/q0

E = (1/4πε0)(q/r²)

∇·E = ρ/ε0

∇²V = -ρ/ε0

Electromagnetism

S = E × H

S = (1/μ0)E × B

B = ∇ × A

∮B·dl = μ0Ienc

∇×B = μ0J + μ0ε0∂E/∂t

Quantum and Semiconductor Physics

ΔxΔp ≥ ℏ/2

p = mv

λ = h/p

n-type: donor impurity

p-type: acceptor impurity

Nuclear Physics

N = N0e^(-λt)

A = λN

T1/2 = 0.693/λ

Q = Δmc²

1 u = 931.5 MeV/c²

Question Map of CSS Physics Paper-II 2020 Solved

Question Main Area What Is Fully Solved
Q2 Electrostatics Electric field of point charges, Poisson equation from Gauss law, Laplace equation and electric field numerical.
Q3 Electromagnetism Energy transport, Poynting vector, Maxwell equations in integral and differential forms and vector potential.
Q4 Quantum mechanics Heisenberg uncertainty principle, barrier tunneling and electron momentum numerical.
Q5 Semiconductor physics Doping, n-type and p-type semiconductors, NPN and PNP transistors and MOSFET.
Q6 Nuclear physics Natural radioactivity, radioactive decay law and alpha-decay energy of uranium-238.
Q7 Nuclear technology Nuclear fission, basic principles of nuclear reactors and radiation detection methods.
Q8 Short notes Dielectric medium and polarization, Ampere law and particle accelerators.

Question 2: Coulomb Law, Poisson Equation and Electric Field

Full Question

Q.2. (a) Discuss electric field of point charges, keeping in view the magnitude of force acting on test charge according to Coulomb’s law.

Q.2. (b) Derive Poisson’s equation from Gauss’s law. Also write the expression for Laplace’s equation.

Q.2. (c) Find out the electric field due to charge of 2e at a distance of 26.5 × 10⁻¹² m. Given ε0 = 8.85 × 10⁻¹² C²/Nm² and e = 1.60 × 10⁻¹⁹ C.

Q2(a): Electric Field of Point Charges

Electric field is the force experienced by a unit positive test charge placed at a point in space. If a test charge q0 experiences force F, then electric field is:

E = F/q0

According to Coulomb’s law, the force between a source charge q and a test charge q0 separated by distance r is:

F = (1/4πε0)(qq0/r²)

Therefore, electric field due to point charge q is:

E = F/q0
E = (1/4πε0)(q/r²)

In vector form:

E = (1/4πε0)(q/r²) r̂

The direction of electric field depends on the sign of the source charge. For a positive charge, electric field is radially outward. For a negative charge, electric field is radially inward.

Electric Field Due to Multiple Point Charges

Electric field obeys the principle of superposition. If several point charges are present, the net electric field at a point is the vector sum of electric fields due to all charges:

Enet = E1 + E2 + E3 + …
Final Answer for Q2(a): The electric field of a point charge is E=(1/4πε0)(q/r²)r̂. The force on a test charge is F=q0E.

Q2(b): Derivation of Poisson’s Equation from Gauss’s Law

Gauss’s law in differential form is:

∇ · E = ρ/ε0

where ρ is volume charge density.

Electric field is related to electric potential V by:

E = -∇V

Substitute this into Gauss’s law:

∇ · (-∇V) = ρ/ε0
-∇ · ∇V = ρ/ε0

But:

∇ · ∇V = ∇²V

Therefore:

-∇²V = ρ/ε0

So:

∇²V = -ρ/ε0
Poisson’s Equation: ∇²V = -ρ/ε0.

Laplace’s Equation

In a charge-free region:

ρ = 0

Therefore, Poisson’s equation becomes:

∇²V = 0
Laplace’s Equation: ∇²V = 0.

Q2(c): Electric Field Due to Charge 2e

Given:

q = 2e = 2(1.60 × 10⁻¹⁹ C)
q = 3.20 × 10⁻¹⁹ C
r = 26.5 × 10⁻¹² m
ε0 = 8.85 × 10⁻¹² C²/Nm²

The electric field due to a point charge is:

E = (1/4πε0)(q/r²)

Using:

1/4πε0 = 8.99 × 10⁹ Nm²/C²

Substitute:

E = (8.99 × 10⁹)(3.20 × 10⁻¹⁹)/(26.5 × 10⁻¹²)²

Calculate denominator:

(26.5 × 10⁻¹²)² = 7.0225 × 10⁻²²

Calculate numerator:

(8.99 × 10⁹)(3.20 × 10⁻¹⁹) = 2.8768 × 10⁻⁹

Now:

E = (2.8768 × 10⁻⁹)/(7.0225 × 10⁻²²)
E ≈ 4.10 × 10¹² N/C
Final Answer for Q2(c): The electric field is approximately 4.10 × 10¹² N/C.

Question 3: Poynting Vector, Maxwell Equations and Vector Potential

Full Question

Q.3. (a) Discuss in detail the energy transport and the Poynting vector.

Q.3. (b) Write the four Maxwell’s equations both in integral and differential forms.

Q.3. (c) Explain vector potential.

Q3(a): Energy Transport and Poynting Vector

Electromagnetic fields carry energy and momentum. When an electromagnetic wave travels through space, energy is transported by the electric and magnetic fields. The rate of energy flow per unit area is represented by the Poynting vector.

The Poynting vector is defined as:

S = E × H

In vacuum, since:

H = B/μ0

we can write:

S = (1/μ0) E × B

The direction of S is the direction of electromagnetic energy propagation. Its SI unit is:

watt per square metre = W/m²

Energy Density of Electromagnetic Field

The energy density stored in electric and magnetic fields is:

u = 1/2 ε0E² + B²/(2μ0)

For an electromagnetic wave in vacuum, electric and magnetic energy densities are equal on average.

Poynting Theorem

Poynting theorem is the energy conservation law of electromagnetism:

∂u/∂t + ∇·S + J·E = 0

This means that the decrease in electromagnetic field energy inside a volume equals the energy flowing out through the surface plus the work done on charges inside the volume.

Final Answer for Q3(a): The Poynting vector S=E×H=(1/μ0)E×B represents electromagnetic energy flow per unit area per second.

Q3(b): Maxwell’s Equations in Integral and Differential Forms

Law Integral Form Differential Form Meaning
Gauss’s law for electricity ∮ E·dA = Qenc/ε0 ∇·E = ρ/ε0 Electric charges are sources or sinks of electric field.
Gauss’s law for magnetism ∮ B·dA = 0 ∇·B = 0 There are no isolated magnetic monopoles.
Faraday’s law ∮ E·dl = -dΦB/dt ∇×E = -∂B/∂t Changing magnetic field produces electric field.
Ampere-Maxwell law ∮ B·dl = μ0Ienc + μ0ε0 dΦE/dt ∇×B = μ0J + μ0ε0∂E/∂t Current and changing electric field produce magnetic field.

Importance of Maxwell’s Equations

  1. They unify electricity and magnetism.
  2. They predict electromagnetic waves.
  3. They show that light is an electromagnetic wave.
  4. They provide the foundation of radio, optics, antennas and communication technology.
Final Answer for Q3(b): Maxwell’s equations consist of Gauss’s law for electricity, Gauss’s law for magnetism, Faraday’s law and Ampere-Maxwell law, written in both integral and differential forms above.

Q3(c): Vector Potential

The magnetic vector potential is a vector field A defined such that magnetic field is the curl of A:

B = ∇ × A

This definition automatically satisfies Gauss’s law for magnetism:

∇ · B = 0

because divergence of curl is always zero:

∇ · (∇ × A) = 0

Scalar and Vector Potentials

In electromagnetism, electric and magnetic fields can be expressed in terms of scalar potential V and vector potential A:

B = ∇ × A
E = -∇V – ∂A/∂t

Importance of Vector Potential

  1. It simplifies electromagnetic field calculations.
  2. It is used in radiation and antenna theory.
  3. It is important in quantum mechanics, especially the Aharonov-Bohm effect.
  4. It helps express Maxwell’s equations in potential form.
Final Answer for Q3(c): Vector potential A is defined by B=∇×A. It is useful because it automatically satisfies ∇·B=0 and simplifies electromagnetic theory.

Question 4: Uncertainty Principle, Tunneling and Electron Momentum

Full Question

Q.4. (a) State and explain Heisenberg’s uncertainty principle.

Q.4. (b) Discuss the phenomenon of barrier tunneling.

Q.4. (c) Find the momentum of an electron moving with a speed of 1.88 × 10⁶ m/s, where mass of electron is 9.11 × 10⁻³¹ kg.

Q4(a): Heisenberg’s Uncertainty Principle

Heisenberg’s uncertainty principle states that it is impossible to know simultaneously the exact position and exact momentum of a microscopic particle. The more accurately position is known, the less accurately momentum can be known, and vice versa.

The mathematical form is:

Δx Δp ≥ ℏ/2

Since:

ℏ = h/(2π)

we can also write:

Δx Δp ≥ h/(4π)

Energy-Time Uncertainty

Another form of uncertainty principle is:

ΔE Δt ≥ ℏ/2

Physical Meaning

The uncertainty principle is not due to defective instruments. It is a fundamental property of quantum systems. A particle localized sharply in space must be represented by a wave packet containing many wavelengths, which creates uncertainty in momentum.

Significance

  1. It explains why electrons cannot exist inside the nucleus as ordinary confined particles.
  2. It supports the wave nature of matter.
  3. It explains zero-point energy.
  4. It limits classical determinism at microscopic scales.
Final Answer for Q4(a): Heisenberg’s uncertainty principle is ΔxΔp ≥ ℏ/2. It shows that exact position and momentum cannot be known simultaneously.

Q4(b): Barrier Tunneling

Barrier tunneling is a quantum phenomenon in which a particle passes through a potential barrier even when its energy is less than the barrier height. In classical physics, this is impossible. In quantum mechanics, the particle has a wave function that can penetrate the barrier.

Classical View

If a particle has energy E and faces a barrier of height V0, then classically:

E < V0

means the particle cannot cross the barrier.

Quantum View

In quantum mechanics, the wave function inside the classically forbidden barrier region does not become zero immediately. Instead, it decays exponentially:

ψ(x) ∝ e^(-κx)

where:

κ = √[2m(V0-E)]/ℏ

If the barrier is thin enough, the wave function may still have non-zero value on the other side. Therefore, there is a finite probability that the particle will appear beyond the barrier.

Transmission Probability

For a rectangular barrier of width a, the tunneling probability roughly behaves like:

T ∝ e^(-2κa)

This shows that tunneling probability decreases rapidly as barrier height or barrier width increases.

Applications of Tunneling

  1. Alpha decay: Alpha particles escape nuclei by tunneling through nuclear potential barriers.
  2. Tunnel diode: Semiconductor device based on quantum tunneling.
  3. Scanning tunneling microscope: Uses tunneling current to image surfaces at atomic scale.
  4. Nuclear fusion in stars: Tunneling helps particles overcome Coulomb repulsion.
Final Answer for Q4(b): Barrier tunneling occurs when a quantum particle passes through a potential barrier even if E<V0. It is explained by the non-zero wave function inside and beyond the barrier.

Q4(c): Momentum of Electron

Given:

m = 9.11 × 10⁻³¹ kg
v = 1.88 × 10⁶ m/s

The speed is much less than the speed of light, so non-relativistic momentum is sufficient:

p = mv

Substitute values:

p = (9.11 × 10⁻³¹)(1.88 × 10⁶)
p = 17.1268 × 10⁻²⁵
p = 1.71 × 10⁻²⁴ kg m/s
Final Answer for Q4(c): Momentum of the electron is approximately 1.71 × 10⁻²⁴ kg m/s.

Question 5: Doping, NPN/PNP Transistors and MOSFET

Full Question

Q.5. (a) What do you understand by the term doping? How can we make semiconductors as n-type or p-type with doping?

Q.5. (b) Discuss in detail the NPN and PNP transistors.

Q.5. (c) Explain MOSFET.

Q5(a): Doping in Semiconductors

Doping is the deliberate addition of a small controlled amount of impurity atoms to a pure semiconductor to increase its electrical conductivity. Pure silicon or germanium is called an intrinsic semiconductor. After doping, it becomes an extrinsic semiconductor.

n-Type Semiconductor

An n-type semiconductor is made by adding pentavalent impurity atoms to silicon or germanium. Pentavalent atoms have five valence electrons. Four electrons form covalent bonds with neighbouring semiconductor atoms, while the fifth electron becomes free for conduction.

Examples of pentavalent donor impurities are:

  • Phosphorus
  • Arsenic
  • Antimony
n-Type Result: Majority carriers are electrons, and minority carriers are holes.

p-Type Semiconductor

A p-type semiconductor is made by adding trivalent impurity atoms. Trivalent atoms have three valence electrons, so one bond remains incomplete. This creates a hole, which behaves as a positive charge carrier.

Examples of trivalent acceptor impurities are:

  • Boron
  • Aluminium
  • Gallium
  • Indium
p-Type Result: Majority carriers are holes, and minority carriers are electrons.
Feature n-Type Semiconductor p-Type Semiconductor
Impurity Pentavalent donor Trivalent acceptor
Majority carriers Electrons Holes
Minority carriers Holes Electrons
Examples P, As, Sb B, Al, Ga, In
Final Answer for Q5(a): Doping increases semiconductor conductivity. Pentavalent impurities produce n-type material, while trivalent impurities produce p-type material.

Q5(b): NPN and PNP Transistors

A bipolar junction transistor is a three-layer semiconductor device used for amplification and switching. It has three terminals:

  • Emitter
  • Base
  • Collector

NPN Transistor

An NPN transistor consists of a thin p-type base between two n-type regions. Electrons are the majority carriers. In active mode, the emitter-base junction is forward biased and the collector-base junction is reverse biased.

NPN transistor:

Emitter Base Collector
N | P | N
electrons move from emitter toward collector

In an NPN transistor, a small base current controls a much larger collector current. Conventional current enters the collector and leaves through the emitter.

PNP Transistor

A PNP transistor consists of a thin n-type base between two p-type regions. Holes are the majority carriers. In active mode, the emitter-base junction is forward biased and the collector-base junction is reverse biased.

PNP transistor:

Emitter Base Collector
P | N | P
holes move from emitter toward collector

Comparison Between NPN and PNP Transistors

Feature NPN Transistor PNP Transistor
Structure N-P-N P-N-P
Majority carriers Electrons Holes
Emitter arrow Arrow points outward Arrow points inward
Speed Generally faster because electrons are more mobile Generally slower than NPN
Common use Amplifiers and switches Amplifiers and switches with opposite polarity bias

Transistor Action

The emitter is heavily doped to inject carriers. The base is very thin and lightly doped, so most carriers pass through it. The collector is moderately doped and collects the carriers. This allows a small base current to control a large collector current.

Final Answer for Q5(b): NPN and PNP transistors are three-layer devices. NPN uses electrons as majority carriers, PNP uses holes, and both can work as amplifiers or switches.

Q5(c): MOSFET

MOSFET stands for Metal-Oxide-Semiconductor Field-Effect Transistor. It is a voltage-controlled device in which gate voltage controls current between source and drain.

Main Terminals of MOSFET

  • Gate: Controls the channel.
  • Source: Supplies charge carriers.
  • Drain: Collects charge carriers.
  • Body/Substrate: Semiconductor base region.

Structure and Working

The gate is separated from the semiconductor channel by a thin insulating oxide layer, usually silicon dioxide. Because the gate is insulated, gate current is extremely small. When a suitable gate voltage is applied, a conducting channel is formed between source and drain.

Types of MOSFET

Type Meaning
n-channel MOSFET Current is mainly carried by electrons.
p-channel MOSFET Current is mainly carried by holes.
Enhancement MOSFET Normally off; channel forms when gate voltage is applied.
Depletion MOSFET Normally on; gate voltage can reduce channel conduction.

Applications of MOSFET

  1. Digital logic circuits.
  2. Microprocessors and memory chips.
  3. Power switching circuits.
  4. Amplifiers.
  5. Voltage regulators and motor control circuits.
Final Answer for Q5(c): A MOSFET is a voltage-controlled transistor in which gate voltage controls source-drain current through an insulated gate structure.

Question 6: Natural Radioactivity, Radioactive Decay and Alpha-Decay Energy

Full Question

Q.6. (a) Discuss in detail the process of natural radioactivity.

Q.6. (b) Discuss in detail the radioactive decay.

Q.6. (c) Find the energy released during the alpha decay of ²³⁸U. Atomic masses are: ²³⁸U = 238.050785 u, ²³⁴Th = 234.043539 u and ⁴He = 4.002603 u.

Q6(a): Natural Radioactivity

Natural radioactivity is the spontaneous disintegration of unstable atomic nuclei found in nature. During this process, nuclei emit radiation in order to become more stable. The phenomenon was discovered by Henri Becquerel and later studied by Marie Curie and Pierre Curie.

Types of Natural Radioactive Emissions

Radiation Nature Effect on Nucleus Penetrating Power
Alpha radiation Helium nucleus ⁴He Mass number decreases by 4 and atomic number decreases by 2. Low
Beta radiation Electron or positron Atomic number changes by 1, mass number remains nearly same. Medium
Gamma radiation High-energy photon No change in atomic number or mass number. High

Alpha Decay

In alpha decay, a heavy nucleus emits an alpha particle:

²³⁸U → ²³⁴Th + ⁴He + Q

Alpha decay occurs mainly in heavy nuclei because emission of an alpha particle increases nuclear stability.

Beta Decay

In beta-minus decay, a neutron changes into a proton:

n → p + e⁻ + ν̄

In beta-plus decay, a proton changes into a neutron:

p → n + e⁺ + ν

Gamma Decay

Gamma decay occurs when an excited nucleus releases energy as a gamma photon:

N* → N + γ
Final Answer for Q6(a): Natural radioactivity is spontaneous nuclear decay by emission of alpha, beta or gamma radiation from unstable nuclei.

Q6(b): Radioactive Decay

Radioactive decay is a random nuclear process. It is impossible to predict exactly when one particular nucleus will decay, but the decay of a large number of nuclei follows a definite statistical law.

Decay Law

The rate of decay is proportional to the number of undecayed nuclei:

dN/dt = -λN

where λ is decay constant.

Separate variables:

dN/N = -λdt

Integrate:

lnN = -λt + constant

At t=0, N=N0. Therefore:

N = N0e^(-λt)

Activity

Activity is the number of decays per second:

A = -dN/dt

Since dN/dt=-λN:

A = λN

Therefore:

A = A0e^(-λt)

Half-Life

Half-life is the time in which half of the radioactive nuclei decay:

T1/2 = 0.693/λ

Mean Life

Mean life is:

τ = 1/λ
Final Answer for Q6(b): Radioactive decay follows N=N0e^(-λt). Activity is A=λN, and half-life is T1/2=0.693/λ.

Q6(c): Energy Released in Alpha Decay of Uranium-238

The decay is:

²³⁸U → ²³⁴Th + ⁴He + Q

Given atomic masses:

m(²³⁸U) = 238.050785 u
m(²³⁴Th) = 234.043539 u
m(⁴He) = 4.002603 u

Step 1: Mass Defect

Δm = m(parent) – [m(daughter) + m(alpha)]
Δm = 238.050785 – (234.043539 + 4.002603)
Δm = 238.050785 – 238.046142
Δm = 0.004643 u

Step 2: Convert Mass Defect to Energy

Use:

1 u = 931.5 MeV/c²

Therefore:

Q = Δm × 931.5 MeV
Q = 0.004643 × 931.5
Q ≈ 4.33 MeV

Step 3: Energy in Joules

Since:

1 MeV = 1.602 × 10⁻¹³ J

Therefore:

Q = 4.33 × 1.602 × 10⁻¹³
Q ≈ 6.94 × 10⁻¹³ J
Final Answer for Q6(c): The energy released during alpha decay of ²³⁸U is approximately 4.33 MeV, or 6.94 × 10⁻¹³ J.

Question 7: Nuclear Fission, Reactors and Radiation Detection

Full Question

Q.7. (a) Discuss in detail the phenomenon of fission.

Q.7. (b) Explain the basic principles of nuclear reactors.

Q.7. (c) Briefly write about the methods of detection of nuclear radiation.

Q7(a): Nuclear Fission

Nuclear fission is the process in which a heavy nucleus splits into two medium-mass nuclei with the release of energy and neutrons. It usually occurs when a heavy nucleus such as uranium-235 absorbs a slow neutron.

A typical fission reaction is:

²³⁵U + ¹n → ²³⁶U* → ¹⁴¹Ba + ⁹²Kr + 3¹n + energy

Features of Fission

  1. A heavy nucleus splits into two lighter fragments.
  2. Two or three neutrons are usually released.
  3. A large amount of energy is released due to mass defect.
  4. The released neutrons can cause further fission events.
  5. This can lead to a chain reaction.

Chain Reaction

A chain reaction occurs when neutrons released in one fission event trigger further fission events. If controlled, it is used in nuclear reactors. If uncontrolled, it can produce a nuclear explosion.

Energy Released in Fission

The energy appears as kinetic energy of fragments, kinetic energy of neutrons, gamma rays and later heat. The average energy released per fission of uranium-235 is about 200 MeV.

Final Answer for Q7(a): Nuclear fission is splitting of a heavy nucleus into lighter nuclei with release of neutrons and energy. It can produce a chain reaction.

Q7(b): Basic Principles of Nuclear Reactors

A nuclear reactor is a device in which a controlled nuclear chain reaction is maintained. The heat produced by fission is used to generate steam, which can drive turbines for electricity production.

Main Parts of a Nuclear Reactor

┌──────────────────────────────┐
│ Biological Shield │
│ ┌────────────────────────┐ │
Control Rods ────┼──┤ || || || || │ │
Fuel Rods ───────┼──┤ [U] [U] [U] [U] │ │
Moderator ───────┼──┤ water / graphite │ │
Coolant In ─────┼──► coolant removes heat │ │
Coolant Out ─────┼──┤ hot coolant exits │ │
│ └────────────────────────┘ │
└──────────────────────────────┘

Heat Exchanger → Steam → Turbine → Generator → Electricity

Part Function
Fuel Contains fissile material such as uranium-235 or plutonium-239.
Moderator Slows down fast neutrons to thermal energies so they can cause further fission.
Control rods Absorb excess neutrons and control the chain reaction.
Coolant Removes heat from the reactor core.
Shielding Protects workers and surroundings from radiation.
Heat exchanger Transfers heat to water to produce steam.
Turbine and generator Convert thermal energy into electrical energy.

Principle of Operation

Fission releases neutrons and energy. The moderator slows neutrons. Control rods adjust neutron population. Coolant removes heat. The heat produces steam, and the steam drives a turbine connected to a generator.

Final Answer for Q7(b): A nuclear reactor maintains a controlled fission chain reaction using fuel, moderator, control rods, coolant and shielding.

Q7(c): Methods of Detection of Nuclear Radiation

Nuclear radiation is detected through ionization, excitation, photographic effect or semiconductor charge production. Common detectors are described below.

Detector Working Principle Use
Geiger-Muller counter Radiation ionizes gas inside a tube, producing electrical pulses. Counting radiation events.
Ionization chamber Measures ionization current produced by radiation. Radiation dose measurement.
Scintillation counter Radiation produces light flashes in a scintillator, converted into electrical signal. Gamma-ray and particle detection.
Cloud chamber Charged particles leave visible tracks by ionizing vapour. Visual study of particle paths.
Bubble chamber Particles produce trails of bubbles in superheated liquid. High-energy particle experiments.
Semiconductor detector Radiation creates electron-hole pairs in semiconductor material. High-resolution energy measurement.
Photographic plate Radiation blackens photographic emulsion. Radiation exposure record.
Final Answer for Q7(c): Radiation can be detected by GM counters, ionization chambers, scintillation counters, cloud chambers, bubble chambers, semiconductor detectors and photographic plates.

Question 8: Dielectric Polarization, Ampere Law and Accelerators

Full Question

Q.8. Write notes on any two of the following:

(a) Dielectric medium and electric polarization

(b) Ampere’s law

(c) Accelerators

Exam Strategy: The paper asks for any two notes, but all three are solved below so students can choose the two they understand best.

Q8(a): Dielectric Medium and Electric Polarization

A dielectric medium is an insulating material that does not conduct electricity easily but becomes polarized when placed in an electric field. Examples include glass, mica, rubber, plastic, paper, oil and ceramics.

Electric Polarization

Electric polarization is the dipole moment per unit volume of a dielectric material:

P = dipole moment / volume

In vector form:

P = Σp / ΔV

where p is electric dipole moment.

How Polarization Occurs

When a dielectric is placed in an electric field, positive and negative charges inside atoms or molecules shift slightly in opposite directions. This creates induced dipoles. In polar molecules, permanent dipoles tend to align with the applied field.

Types of Polarization

Type Explanation
Electronic polarization Displacement of electron cloud relative to nucleus.
Ionic polarization Relative displacement of positive and negative ions.
Orientational polarization Alignment of permanent dipoles with applied field.
Space-charge polarization Accumulation of charges at interfaces or defects.

Effect on Capacitor

When a dielectric is inserted between capacitor plates, capacitance increases:

C = κC0

where κ is dielectric constant.

Note Summary: A dielectric is an insulator that becomes polarized in an electric field. Polarization is dipole moment per unit volume and increases capacitance in capacitors.

Q8(b): Ampere’s Law

Ampere’s circuital law states that the line integral of magnetic field around a closed path is equal to μ0 times the current enclosed by that path.

∮ B · dl = μ0 Ienc

Proof for a Long Straight Wire

Consider a long straight wire carrying current I. Magnetic field at distance r is circular and constant in magnitude on a circular Amperian loop.

For a circular path:

∮ B · dl = B∮dl

The circumference of the circular path is:

∮dl = 2πr

Therefore:

∮ B · dl = B(2πr)

Ampere’s law gives:

B(2πr) = μ0I

So:

B = μ0I/(2πr)

Ampere-Maxwell Law

Maxwell corrected Ampere’s law by adding displacement current:

∮ B·dl = μ0Ienc + μ0ε0 dΦE/dt

In differential form:

∇×B = μ0J + μ0ε0∂E/∂t

Applications

  1. Magnetic field of a long straight wire.
  2. Magnetic field inside a solenoid.
  3. Magnetic field of a toroid.
  4. Foundation of electromagnetic wave theory.
Note Summary: Ampere’s law is ∮B·dl=μ0Ienc. With Maxwell’s correction, it includes displacement current and becomes essential for electromagnetic waves.

Q8(c): Accelerators

Particle accelerators are machines that increase the kinetic energy of charged particles using electric fields and guide them using magnetic fields. They are used in nuclear physics, particle physics, medicine, industry and materials research.

Basic Principle

A charged particle in an electric field experiences force:

F = qE

The electric field does work on the particle and increases its kinetic energy:

ΔK = qV

Magnetic fields are used to bend and focus charged particles:

F = qvB

Types of Accelerators

Accelerator Working Use
Linear accelerator Accelerates particles in a straight line using alternating electric fields. Medical radiation therapy and research.
Cyclotron Uses magnetic field and alternating voltage between D-shaped electrodes. Nuclear physics and isotope production.
Synchrotron Uses varying magnetic field and radiofrequency cavities to keep particles in circular orbit. High-energy physics and synchrotron radiation.
Betatron Accelerates electrons by electromagnetic induction. High-energy electron beams.
Collider Accelerates particles in opposite directions and collides them. Particle discovery and fundamental physics.

Applications of Accelerators

  1. Study of atomic nuclei and elementary particles.
  2. Cancer treatment using particle beams.
  3. Production of medical isotopes.
  4. Material analysis and ion implantation.
  5. Synchrotron radiation for biology, chemistry and materials science.
Note Summary: Accelerators use electric fields to accelerate charged particles and magnetic fields to guide them. They are central to nuclear physics, medicine and modern particle research.

Revision Plan for CSS Physics Paper-II 2020 Solved

After reading this complete CSS Physics Paper-II 2020 Solved guide, revise it in three rounds. In the first round, learn the definitions and formulas. In the second round, reproduce each derivation without looking. In the third round, solve the numerical questions again with units and compare your answer with the final result.

Question Revision Task
Q2 Derive electric field from Coulomb law, derive Poisson equation and recalculate 4.10×10¹² N/C.
Q3 Write Poynting vector, Poynting theorem, four Maxwell equations and vector potential definitions.
Q4 Explain uncertainty, tunneling and solve electron momentum numerical again.
Q5 Compare n-type and p-type semiconductors, NPN and PNP transistors, and explain MOSFET.
Q6 Write decay law, activity, half-life and alpha-decay energy calculation.
Q7 Explain fission chain reaction, draw reactor diagram and list radiation detectors.
Q8 Prepare any two notes but revise all three: dielectric polarization, Ampere law and accelerators.

Related Resources for CSS Physics Paper-II 2020 Solved

Exam Note: For CSS Physics Paper-II, many answers combine theory and equations. Write definitions first, then formula, then derivation or explanation, and close with one application. This makes answers clearer and more examiner-friendly.

FAQs About CSS Physics Paper-II 2020 Solved

What does CSS Physics Paper-II 2020 Solved include?

CSS Physics Paper-II 2020 Solved includes complete solved answers for Q2 to Q8 with definitions, derivations, numerical calculations, formulas, final answers and exam-ready notes.

What is the electric field answer in Question 2?

The electric field due to charge 2e at distance 26.5×10⁻¹² m is approximately 4.10×10¹² N/C.

What is Poisson’s equation?

Poisson’s equation is ∇²V = -ρ/ε0. In a charge-free region, it becomes Laplace’s equation: ∇²V = 0.

What is the Poynting vector?

The Poynting vector is S=E×H, or S=(1/μ0)E×B in vacuum. It gives electromagnetic energy flow per unit area per second.

What is the momentum of the electron in Question 4?

For m=9.11×10⁻³¹ kg and v=1.88×10⁶ m/s, electron momentum is approximately 1.71×10⁻²⁴ kg m/s.

What is the alpha-decay energy in Question 6?

The alpha decay of ²³⁸U releases approximately 4.33 MeV, or 6.94×10⁻¹³ J.

What is the difference between n-type and p-type semiconductors?

n-type semiconductors are formed by pentavalent donor doping and have electrons as majority carriers. p-type semiconductors are formed by trivalent acceptor doping and have holes as majority carriers.

What are the main parts of a nuclear reactor?

Main reactor parts include fuel, moderator, control rods, coolant, shielding, heat exchanger, turbine and generator.

Can I paste this HTML into WordPress?

Yes. The post avoids an H1 heading so your WordPress post title can remain the only H1. The internal structure begins with H2 and continues with H3/H4 headings.

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