CSS Physics Paper-II 2026 Solved

Question 2: Gauss Law in Dielectrics and Charged Spherical Shell

Q2(a): Gauss Law in Dielectrics

Gauss law relates electric flux through a closed surface to the charge enclosed by that surface. In vacuum, it is written as:

∮E·dA = Qenc/ε0

In differential form:

∇·E = ρ/ε0

In a dielectric material, charges are of two types: free charges and bound charges. Bound charges appear because molecules become polarized. To handle this conveniently, we introduce electric displacement field D.

D = ε0E + P

Here P is polarization vector. Gauss law in dielectrics is written as:

∮D·dA = Qfree

In differential form:

∇·D = ρfree

For a linear, homogeneous and isotropic dielectric:

D = εE

where:

ε = εrε0

Significance of Gauss Law in Dielectrics

1. It separates free charge from bound polarization charge.
2. It simplifies electric-field calculations inside dielectric materials.
3. It is useful in capacitors filled with dielectric medium.
4. It explains why capacitance increases when a dielectric is inserted.
5. It provides the foundation for macroscopic electrostatics in materials.

Q2(b): Electric Field Outside a Uniformly Charged Spherical Shell

Consider a uniformly charged spherical shell of radius R and total charge q. We need the electric field at a point outside the shell at distance r from the center, where:

r > R

Because of spherical symmetry, the electric field is radial and has the same magnitude at every point on a spherical Gaussian surface of radius r.

Choose a spherical Gaussian surface of radius r. Its area is:

A = 4πr²

Gauss law gives:

∮E·dA = q/ε0

Since E is constant over the Gaussian surface and parallel to dA:

E∮dA = q/ε0

So:

E(4πr²) = q/ε0

Therefore:

E = q/(4πε0r²)

Using Coulomb constant:

k = 1/(4πε0)

we get:

E = kq/r²

This means that outside a uniformly charged spherical shell, the field is the same as if all charge were concentrated at the center.

Q2(c): Electric Field Numerical

Given:

q = 4 μC = 4×10⁻⁶ C

R = 0.3 m

r = 0.5 m

Since r=0.5 m is greater than the shell radius R=0.3 m, the point is outside the shell. Therefore:

E = kq/r²

Use:

k = 8.99×10⁹ N m²/C²

Substitute:

E = (8.99×10⁹)(4×10⁻⁶)/(0.5)²

E = (35.96×10³)/0.25

E = 1.4384×10⁵ N/C

Rounded answer:

E ≈ 1.44×10⁵ N/C

Question 3: Induced EMF and Faraday Law

Q3(a): Meaning of Induced EMF

Induced emf is the electromotive force produced in a circuit when the magnetic flux linked with the circuit changes. It is called “induced” because it is not produced by an ordinary battery or chemical source. It appears due to electromagnetic induction.

Magnetic flux through a surface is:

Φ = B·A = BA cosθ

Induced emf can be produced by:

1. Changing the magnetic field through a coil.
2. Changing the area of the coil.
3. Changing the angle between magnetic field and area vector.
4. Moving a conductor in a magnetic field.

Q3(b): Derivation of Induced EMF in a Coil

Faraday’s law states that the induced emf in a circuit is equal to the negative rate of change of magnetic flux linked with the circuit.

For one turn:

ε = -dΦ/dt

For a coil of N turns, total flux linkage is:

NΦ

Therefore, induced emf is:

ε = -d(NΦ)/dt

If N is constant:

ε = -N dΦ/dt

For a finite change in flux:

εavg = -N ΔΦ/Δt

The negative sign represents Lenz’s law. It shows that the induced emf opposes the change in magnetic flux that produces it.

Q3(c): Numerical Calculation of Induced EMF

Given:

N = 20

Φi = 0.5 Wb

Φf = 0.1 Wb

Δt = 0.2 s

Change in flux:

ΔΦ = Φf - Φi

ΔΦ = 0.1 - 0.5 = -0.4 Wb

Magnitude of induced emf:

|ε| = N|ΔΦ|/Δt

Substitute values:

|ε| = 20(0.4)/0.2

|ε| = 8/0.2

|ε| = 40 V

Question 4: Schrodinger Equation and One-Dimensional Potential Well

Q4(a): Time-Independent Schrodinger Wave Equation

The time-independent Schrodinger equation is an energy eigenvalue equation. It is used when the potential energy does not explicitly depend on time.

In one dimension, the equation is:

-(ℏ²/2m)d²ψ/dx² + V(x)ψ = Eψ

In operator form:

Hψ = Eψ

Here H is the Hamiltonian operator, ψ is the spatial wave function and E is the energy eigenvalue.

Why It Is Called Time Independent

The full wave function can be separated as:

Ψ(x,t) = ψ(x)e^(-iEt/ℏ)

The time-dependent factor is separated from the spatial part. The equation for ψ(x) contains no explicit time variable, so it is called time independent.

Q4(b): Particle in One-Dimensional Infinite Potential Well

Consider a particle of mass m confined in an infinite potential well from x=0 to x=L.

The potential is:

V(x)=0, 0<x<L

V(x)=∞, x≤0 and x≥L

Inside the well, the Schrodinger equation becomes:

-(ℏ²/2m)d²ψ/dx² = Eψ

Rearrange:

d²ψ/dx² + (2mE/ℏ²)ψ = 0

Let:

k² = 2mE/ℏ²

Then:

d²ψ/dx² + k²ψ = 0

The general solution is:

ψ(x) = A sin(kx) + B cos(kx)

Boundary condition at x=0:

ψ(0)=0

This gives:

B=0

So:

ψ(x)=A sin(kx)

Boundary condition at x=L:

ψ(L)=0

Therefore:

A sin(kL)=0

For non-zero A:

sin(kL)=0

Thus:

kL = nπ

k = nπ/L, where n=1,2,3,...

Energy is:

E = ℏ²k²/(2m)

Substitute k=nπ/L:

En = n²π²ℏ²/(2mL²)

Using ℏ=h/(2π):

En = n²h²/(8mL²)

Q4(c): Ground-State Energy Numerical

Given:

L = 1 nm = 1×10⁻⁹ m

m = 9.11×10⁻³¹ kg

h = 6.626×10⁻³⁴ J s

For ground state:

n = 1

Energy is:

E1 = h²/(8mL²)

Substitute:

E1 = (6.626×10⁻³⁴)²/[8(9.11×10⁻³¹)(1×10⁻⁹)²]

E1 ≈ 6.03×10⁻²⁰ J

Convert to electron-volts:

1 eV = 1.602×10⁻¹⁹ J

E1 = (6.03×10⁻²⁰)/(1.602×10⁻¹⁹) eV

E1 ≈ 0.376 eV

Question 5: Bravais Lattice, Centered Cubic Cells and Unit Cell Volume

Q5(a): Bravais Lattice

A Bravais lattice is an infinite periodic arrangement of points in space such that every lattice point has the same surroundings. The whole lattice can be generated by translating a point through primitive vectors.

In three dimensions, a lattice point is represented by:

R = n1a1 + n2a2 + n3a3

where a1, a2 and a3 are primitive translation vectors, and n1, n2, n3 are integers.

Important Features

1. It describes translational symmetry of a crystal.
2. Every lattice point is equivalent.
3. A crystal structure is formed by attaching a basis to each lattice point.
4. There are 14 Bravais lattices in three dimensions.

Q5(b): Volume of Centered Cubic Unit Cell

The side length is:

a = 0.4 nm

Volume of a cubic unit cell is:

V = a³

Substitute:

V = (0.4 nm)³

V = 0.064 nm³

Since:

1 nm = 10 Å

therefore:

0.4 nm = 4 Å

So, in cubic angstroms:

V = (4 Å)³

V = 64 ų

In cubic metres:

0.4 nm = 0.4×10⁻⁹ m

V = (0.4×10⁻⁹)³ m³

V = 6.4×10⁻²⁹ m³

Q5(c): Base-Centered and Face-Centered Cubic Cells

Base-Centered Cell

A base-centered cell has lattice points at all eight corners and at the centers of two opposite faces. If the centered faces are top and bottom faces, it is called C-centered or base-centered.

Effective number of lattice points:

8 corners × 1/8 = 1

2 face centers × 1/2 = 1

Total = 2 lattice points per conventional cell

Face-Centered Cubic Cell

A face-centered cubic cell has lattice points at all eight corners and at the centers of all six faces.

Effective number of lattice points:

8 corners × 1/8 = 1

6 face centers × 1/2 = 3

Total = 4 lattice points per conventional cell

Types of Base-Centered Cells

Base-centered cells are usually named according to the pair of faces that contain extra lattice points:

1. A-centered: Centering on faces perpendicular to the a-axis.
2. B-centered: Centering on faces perpendicular to the b-axis.
3. C-centered: Centering on faces perpendicular to the c-axis.

Question 6: Fermi Sphere, Fermi-Dirac Function and Copper Fermi Energy

Q6(a): Fermi Sphere

The Fermi sphere is the filled region in momentum space or k-space for a gas of fermions at absolute zero temperature. Electrons fill all available quantum states from the lowest energy upward until the highest occupied energy is reached.

The boundary of this filled region is called the Fermi surface. For free electrons in a metal, this surface is spherical in k-space, so it is called the Fermi sphere.

Is Fermi Sphere an Energy State?

No. The Fermi sphere is not a single energy state. It is a collection of occupied quantum states in k-space up to the Fermi energy.

At absolute zero:

• All states with E<EF are filled.
• All states with E>EF are empty.
• EF is the highest occupied energy at 0 K.

Q6(b): Fermi-Dirac Distribution Function

Fermi-Dirac distribution gives the probability that an energy state E is occupied by a fermion at temperature T.

The distribution function is:

f(E,T)=1/[exp((E-EF)/kBT)+1]

Here:

• f(E,T) is occupancy probability.
• E is energy of the state.
• EF is Fermi energy.
• kB is Boltzmann constant.
• T is absolute temperature.

Physical Interpretation

At T=0 K, the distribution becomes a step function. All states below EF are occupied and all states above EF are empty.

At finite temperature, the sharp edge becomes smooth. Some electrons occupy states above EF, and some states below EF become empty.

At E=EF:

f(EF,T)=1/[exp(0)+1]

f(EF,T)=1/2

Q6(c): Fermi Energy of Copper

Given:

n = 8.49×10²⁸ m⁻³

m = 9.11×10⁻³¹ kg

h = 6.62×10⁻³⁴ J s

Formula given in the question:

EF = h²/(2m)(3n/8π)^(2/3)

Substitute values:

EF = (6.62×10⁻³⁴)²/[2(9.11×10⁻³¹)] × [3(8.49×10²⁸)/(8π)]^(2/3)

First:

h²/(2m) = (6.62×10⁻³⁴)²/[2(9.11×10⁻³¹)]

h²/(2m) ≈ 2.405×10⁻³⁷

Next:

[3n/(8π)]^(2/3) = [3(8.49×10²⁸)/(8π)]^(2/3)

[3n/(8π)]^(2/3) ≈ 4.684×10¹⁸

Therefore:

EF ≈ (2.405×10⁻³⁷)(4.684×10¹⁸)

EF ≈ 1.126×10⁻¹⁸ J

Convert to electron-volts:

1 eV = 1.602×10⁻¹⁹ J

EF = (1.126×10⁻¹⁸)/(1.602×10⁻¹⁹) eV

EF ≈ 7.03 eV

Question 7: Alpha, Beta, Gamma Decay and Binding Energy

Q7(a): Difference Between Alpha, Beta and Gamma Decay

Radioactive decay is the spontaneous transformation of an unstable nucleus into a more stable state. The three common types are alpha decay, beta decay and gamma decay.

Alpha Decay

Alpha decay occurs mostly in heavy nuclei. A helium nucleus is emitted:

ᴬ_ZX → ᴬ⁻⁴_{Z-2}Y + ⁴₂He + Q

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 emits a high-energy photon and reaches a lower energy state:

X* → X + γ

Q7(b): Alpha Decay of Uranium

A common alpha decay example is uranium-238 decay:

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He + Q

Check mass number conservation:

238 = 234 + 4

Check atomic number conservation:

92 = 90 + 2

So the equation is balanced.

Q7(c): Binding Energy and Nuclear Mass

Binding energy is the energy required to separate a nucleus completely into its individual protons and neutrons. A nucleus with greater binding energy is more tightly bound and more stable.

Mass-energy relation is:

E = mc²

Binding energy is related to mass defect:

BE = Δmc²

The mass of a bound nucleus is less than the sum of masses of its separate nucleons because some mass is converted into binding energy.

If two nuclei have the same mass number and charge number, they have the same number of protons and neutrons. But if:

(B.E.)A > (B.E.)B

then nucleus A has a larger mass defect. Therefore, its actual mass is smaller.

So nucleus B has more mass.

Question 8: Fission, Elementary Particles, Composite Particles and Fusion

Q8(a): Nuclear Fission and Practical Application

Nuclear fission is the splitting of a heavy nucleus 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 neutron.

A typical fission reaction is:

²³⁵U + ¹n → fission fragments + 2 or 3 neutrons + energy

The released neutrons can cause further fission events. This produces a chain reaction.

Practical Application: Nuclear Reactor

In a nuclear reactor, fission is controlled to produce heat. This heat converts water into steam, and steam drives turbines to generate electricity.

Q8(b): Elementary and Composite Particles

Elementary particles are particles that have no known internal structure. In the Standard Model, quarks, leptons and gauge bosons are elementary particles.

Composite particles are made of smaller particles. Protons and neutrons are composite because they are made of quarks.

Families of Elementary Particles

Q8(c): Controlled Fusion and Why It Is Hard on Earth

Controlled fusion is the process of combining light nuclei under controlled conditions so that energy is released steadily and safely. The most commonly discussed fusion reaction for energy production is deuterium-tritium fusion:

²H + ³H → ⁴He + n + energy

Fusion releases energy because the final products have less mass than the initial nuclei. The mass difference appears as energy:

E = Δmc²

Why Fusion Is Difficult on Earth

1. Light nuclei repel each other due to Coulomb repulsion.
2. Extremely high temperature is needed to overcome this repulsion.
3. The fuel becomes plasma and must be confined.
4. Plasma must remain hot and dense for enough time.
5. Material walls cannot directly touch the hot plasma.
6. Energy losses through radiation and instabilities must be controlled.
7. Neutron damage and heat extraction are major engineering challenges.

Main Confinement Methods