CSS Physics Paper-I 2022 Solved

Question 2: Small Oscillations, Charged Shell and Projectile Angle

Q2(a): Angular Frequency of Small Oscillations

The potential energy is:

For equilibrium, the first derivative of potential energy must be zero:

Differentiate:

At equilibrium x=x0:

For small oscillations near equilibrium, angular frequency is:

Here, the particle has unit mass, so m=1.

Differentiate again:

At equilibrium:

So:

Therefore:

Since m=1:

Q2(b): Electric Field of a Charged Hollow Spherical Shell

For a spherically symmetric charge distribution, use Gauss law:

On a spherical Gaussian surface of radius r, the electric field has constant magnitude and is radial:

Therefore:

Region I: r < a

Inside the hollow region, no charge is enclosed:

Region II: a < r < b

For a Gaussian sphere of radius r inside the charged material, only charge from radius a to radius r is enclosed:

Therefore:

Region III: r > b

Outside the shell, the Gaussian surface encloses the whole shell:

Therefore:

Common Special Case: ρ(r)=κ/r²

If the printed density is interpreted as ρ(r)=κ/r², then:

So:

For r > b:

Q2(c): Projectile Angle

For projectile motion, horizontal range is:

Maximum height is:

Given:

Substitute the formulas:

Cancel u²/g:

Use sin2θ = 2sinθcosθ:

Assuming sinθ ≠ 0, divide by sinθ:

Therefore:

Question 3: Star Pressure, Image Charge and Atwood Machine

Q3(a): Gravitational Pressure of a Uniform-Density Star

For a star in hydrostatic equilibrium:

For uniform density ρ:

Substitute into hydrostatic equation:

Integrate from radius r to surface R, where surface pressure is nearly zero:

At the center r=0:

For total mass M:

Substitute:

Since volume is:

Therefore:

So:

Q3(b): Point Charge Near Infinite Grounded Conducting Plane

Let a point charge +q be located at distance a from an infinite grounded conducting plane. Take the conducting plane as z=0 and the charge at (0,0,a).

By the method of images, the grounded conducting plane is replaced by an image charge -q at the mirror point (0,0,-a). The potential in the region z>0 is the sum of potentials due to the real charge and the image charge.

Potential

Let:

Then potential is:

On the grounded plane z=0, R1=R2, so:

This satisfies the boundary condition of a grounded conducting plane.

Electric Field

The electric field is:

Vector form of electric field in the region z>0 is:

where r1=(0,0,a) and r2=(0,0,-a).

Q3(c): Atwood Machine by Lagrangian Method

Consider an Atwood machine with masses m1 and m2 connected by a light inextensible string over a frictionless pulley. Let x be the downward displacement of mass m2. Then mass m1 moves upward by the same distance.

Both masses have speed ẋ. Therefore, kinetic energy is:

Taking gravitational potential energy such that increase in x lowers m2 and raises m1:

Lagrangian is:

Apply Euler-Lagrange equation:

Now:

Also:

Therefore:

Question 4: Poiseuille Flow, Quadrupole Potential and Soap Bubbles

Q4(a): Poiseuille Flow Through Parallel Tubes

For laminar flow through a tube, Poiseuille law gives:

For the same liquid and same pressure head:

For the first tube:

For the second tube:

Therefore:

So:

Total flow is:

Q4(b): Potential of Linear Quadrupole

The charge system has:

• +q at x=+a
• +q at x=-a
• -2q at the origin

Potential at a far point P(r,θ) is:

Here:

For r >> a, use expansion:

Adding the two terms:

Substitute into potential:

The monopole terms cancel:

Therefore:

Q4(c): Coalescence of Two Soap Bubbles

For a soap bubble, excess pressure is:

The surface energy of a soap bubble is:

A soap bubble has two surfaces, inner and outer. Therefore, total surface area is:

Surface energy is:

If two soap bubbles coalesce without loss of surface energy:

Cancel 8πT:

Therefore:

Question 5: Wave Function, Slot Potential and Two-Level Mean Energy

Q5(a): Wave Function

In quantum mechanics, a wave function ψ describes the state of a particle or system. It is generally complex and is not directly observable. However, the quantity |ψ|² gives probability density.

For a particle moving in one dimension, probability of finding it between x and x+dx is:

Conditions for Acceptable Wave Function

1. It must be finite.
2. It must be single-valued.
3. It must be continuous where required.
4. Its derivative must be continuous where potential is finite.
5. It must be normalizable.

Normalization means:

Wave Formula

A sinusoidal travelling wave can be written as:

or in complex form:

Here:

• A is amplitude
• k = 2π/λ is wave number
• ω = 2πf is angular frequency
• φ is phase constant

Phase Velocity

Phase velocity is the speed with which a point of constant phase moves.

Differentiate with respect to time:

Therefore:

Group Velocity

Group velocity is the speed with which the wave packet or energy travels.

For matter waves, group velocity corresponds to the velocity of the particle.

Q5(b): Potential Distribution in the Slot

The region between two grounded conducting plates satisfies Laplace equation:

Because the system is uniform in the z-direction, potential depends only on x and y:

Boundary conditions are:

Use separation of variables:

Substitute into Laplace equation:

Divide by XY:

Let separation constant be (nπ/a)². The solution satisfying grounded plates at y=0 and y=a is:

Since potential must vanish as x → ∞:

Thus general solution is:

Coefficients are found from the Fourier sine series of boundary potential f(y):

Special Case: Strip at Constant Potential V0

If the end strip is maintained at constant potential V0, then:

So:

Only odd values of n remain:

Q5(c): Mean Energy of Two-Level System

The two energy levels are:

For a classical particle in thermal equilibrium, the partition function is:

Therefore:

Mean energy is:

Substitute:

Question 6: Finite Potential Well, Spherical Boundary Potential and Linear Potential Mean

Q6(a): Particle in a Finite Potential Well

A finite potential well is a region where a particle has lower potential energy than outside, but the wall height is finite. Unlike an infinite well, the particle is not completely confined inside; its wave function penetrates slightly into the classically forbidden region.

Consider a one-dimensional finite square well:

Case 1: Bound States, E < V0

Inside the well, the Schrödinger equation gives oscillatory solutions:

where:

Outside the well, the wave function decays exponentially:

where:

The wave function and its derivative must be continuous at the boundaries. These boundary conditions give allowed discrete energy levels.

Case 2: Scattering States, E > V0

If the particle energy is greater than the well depth, the wave function is oscillatory both inside and outside the well. The particle is not bound; it can pass through the region with partial reflection and transmission.

Even and Odd Solutions

For a symmetric finite well, bound-state wave functions are either even or odd:

• Even states: ψ(x)=ψ(-x)
• Odd states: ψ(x)=-ψ(-x)

Comparison with Infinite Potential Well

Q6(b): Potential Inside a Hollow Sphere with Boundary Potential

If the potential is specified on the surface of a hollow sphere of radius R, the potential inside satisfies Laplace equation:

The general solution in spherical coordinates is expressed in spherical harmonics:

The terms r^l remain finite at the origin. Terms like r^(-l-1) are excluded inside the sphere because they diverge at r=0.

If the boundary potential is:

then expand it as:

Comparing at r=R:

Therefore:

So the potential inside is:

Q6(c): Mean Position in a Linear Potential

For a classical particle at temperature T, probability is proportional to the Boltzmann factor:

Given:

So:

Let:

Then:

Mean position is:

Now:

Therefore:

Substitute α=u0/kT:

Question 7: Adiabatic Expansion, Ampere Law and RMS Oxygen Speed

Q7(a): Degrees of Freedom from Adiabatic Expansion

For an adiabatic process:

Given:

Temperature is decreased by factor 1.32:

Apply adiabatic relation:

Substitute:

Cancel T1 and V1^(γ-1):

So:

Take natural logarithm:

Calculate:

For a gas molecule with f degrees of freedom:

Therefore:

Q7(b): Ampere’s Law

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

Proof for a Long Straight Current-Carrying Wire

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

Choose a circular path of radius r. Along the path, B is tangential and parallel to dl, so:

Since B is constant on the circular path:

The circumference is:

Therefore:

From Biot-Savart law for a long straight wire:

Substitute:

This proves Ampere’s law for this symmetric case.

Uses of Ampere’s Law

1. Magnetic field of a long straight wire.
2. Magnetic field inside a solenoid.
3. Magnetic field inside a toroid.
4. Magnetostatic field calculations with high symmetry.

Q7(c): RMS Speed of Oxygen Molecules at 0°C

The rms speed of gas molecules is:

For oxygen:

At 0°C:

Use:

Substitute:

Calculate numerator:

Divide by molar mass:

Take square root:

Question 8: Spin System, Electromagnetic Wave and Adiabatic Demagnetization

Q8(a): Free Energy of Spin-1/2 System

For one spin-1/2 particle, energy states are:

Partition function for one particle is:

Substitute:

Free energy per particle is:

Therefore:

The mean energy is:

where β=1/kT. Since:

we get:

Given:

Therefore:

Now:

So:

Free energy remains:

Since:

and:

Therefore:

Q8(b): Electromagnetic Wave Equation in Vacuum

Maxwell’s equations in vacuum are:

Wave Equation for Electric Field

Take curl of Faraday’s law:

Use vector identity:

In vacuum, ∇·E=0, so:

Right side becomes:

Therefore:

Wave Equation for Magnetic Field

Similarly:

The standard wave equation is:

Comparing:

This speed is the speed of light:

Properties of Monochromatic Electromagnetic Waves

1. They are transverse waves.
2. Electric field, magnetic field and direction of propagation are mutually perpendicular.
3. They travel in vacuum with speed c.
4. They do not require a material medium.
5. The fields are in phase.
6. The relation between field magnitudes is E/B = c.
7. They carry energy and momentum.
8. They can be polarized.

Q8(c): Adiabatic Demagnetization Using TdS Equations

Adiabatic demagnetization is a cooling method used for paramagnetic materials. The material is first magnetized isothermally in a strong magnetic field, so its magnetic dipoles become more ordered and entropy decreases. Heat is removed during this stage. Then the material is thermally isolated and the magnetic field is reduced adiabatically. The dipoles become disordered by absorbing internal energy, so the temperature falls.

Thermodynamic Basis

For a magnetic system, take Helmholtz free energy as a function of temperature and magnetic field:

The differential form may be written as:

Therefore:

From equality of mixed partial derivatives, the Maxwell relation is:

Now entropy differential is:

Using heat capacity at constant magnetic field:

So:

Using Maxwell relation:

Adiabatic Condition

For adiabatic demagnetization:

Therefore:

Rearrange:

This is the TdS-based cooling relation. For a paramagnet obeying Curie’s law:

Then:

Since this derivative is negative, reducing B adiabatically produces a fall in temperature.

Simple Ideal Result

For an ideal paramagnet where entropy depends mainly on the ratio B/T, adiabatic condition S=constant gives:

Therefore:

So: