Physics CSS Solved Papers

CSS Physics Paper-I 2024 Solved

Engr. Muhammad Yar Saqib

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CSS Physics Paper-I 2024 Solved is a complete solved guide for aspirants who need real answers, not only short hints. This post solves the subjective section question by question with definitions, derivations, formulas, numerical substitutions, final results and exam-ready explanation. It covers Stokes theorem, motion in the xy-plane, relativity, energy-momentum relation, rotational work, conservative gravitational force, direction cosines, Earth’s angular momentum, fluid pressure, Doppler effect, standing waves, grating resolving power, heat pump coefficient of performance, Fermi-Dirac statistics, damped harmonic motion, Young’s double slit experiment and coherence.

Central Argument: A CSS Physics solved paper should not merely tell students what topic was asked. It should actually define, derive, calculate and explain the answer. Therefore, this CSS Physics Paper-I 2024 Solved post gives the full route to the answer so students can reproduce the method in the examination hall.

CSS Physics Paper-I 2024 Solved Study Scope

This post covers the subjective questions of CSS Physics Paper-I 2024 Solved in a complete and structured format. Each question includes the printed question, part-wise solution, important 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 clarity, correct units, clean derivations and disciplined presentation.

Show Table of Contents
  1. Overview
  2. Study Scope
  3. Important Formula Sheet
  4. Question Map
  5. Question 2: Stokes Theorem and Motion in xy-plane
  6. Question 3: Relativity, Energy-Momentum Relation and Rotational Work
  7. Question 4: Conservative Field, Direction Cosines and Earth Angular Momentum
  8. Question 5: Fluid Pressure and Doppler Effect
  9. Question 6: Standing Waves and Grating Resolving Power
  10. Question 7: Real Gas, Heat Pump and Fermi-Dirac Statistics
  11. Question 8: Comprehensive Notes
  12. Revision Plan
  13. Internal and External Resources
  14. FAQs

Important Formula Sheet for CSS Physics Paper-I 2024 Solved

Vector Calculus and Mechanics

∮C A·dl = ∬S (∇×A)·n dS

v = dr/dt, a = dv/dt

F = -∇U

L = r × p, L = Iω

Relativity and Rotation

γ = 1/√(1 - v²/c²)

L = L0/γ

E² = p²c² + m0²c⁴

dW = τdθ, Krot = 1/2 Iω²

Fluids, Waves and Optics

dP/dz = -ρg

P = P0 + ρgh

f' = f(v - vo)/(v - vs)

d sinθ = mλ, R = λ/Δλ = mN

Thermodynamics and Quantum Statistics

PV = nRT

PVγ = constant

COPHP = TH/(TH - TC)

f(E)=1/[e^((E-EF)/kT)+1]

Question Map of CSS Physics Paper-I 2024 Solved

Question Main Area What Is Fully Solved
Q2 Vector calculus and kinematics Stokes theorem statement, proof, significance, position, velocity and acceleration.
Q3 Relativity and rotational dynamics Length contraction, Lorentz consequences, energy-momentum proof, rotational work and kinetic energy.
Q4 Conservative force and angular momentum Conservative field, gravitational force from potential energy, direction cosines and comparison of Earth’s angular momenta.
Q5 Fluid mechanics and sound Pressure variation in fluid and atmosphere, Doppler effect numerical.
Q6 Waves and physical optics Standing waves, energy transport, nodes, grating dispersion and resolving power.
Q7 Thermodynamics and statistical physics Real vs ideal gas, adiabatic work, heat pump COP and Fermi-Dirac statistics.
Q8 Oscillations, interference and coherence Damped harmonic motion, Young double slit experiment, longitudinal and transverse coherence.

Question 2: Stokes Theorem and Motion in xy-plane

Full Question

Q.2. (a) State and prove Stokes theorem. Also explain its significance.

(b) A particle moves in the xy-plane and its coordinates vary with time according to x(t)=At²+Bt and y(t)=Ct²+D, where A=1.00 m/s², B=-32 m/s, C=5 m/s² and D=12 m. Find the position, velocity and acceleration of the particle when t=3 s.

Q2(a): Statement of Stokes Theorem

Stokes theorem states that the line integral of a vector field around a closed curve is equal to the surface integral of the curl of that vector field over any open surface bounded by that curve.

∮C A · dl = ∬S (∇ × A) · n dS

Here, C is the closed boundary curve, S is any open surface bounded by C, A is a vector field, dl is the small line element along the boundary, n is the unit normal to the surface, and ∇ × A is the curl of the vector field.

Q2(a): Proof of Stokes Theorem

Consider a smooth surface S bounded by a closed curve C. Divide the surface into a large number of small surface elements. Each small element has its own small boundary. For one such small element, the circulation of the vector field around the small boundary is approximately equal to the normal component of the curl multiplied by the small area:

small circulation = (∇ × A) · n ΔS

If we add the circulation around all small surface elements, the line integrals along internal boundaries cancel each other. This happens because every internal boundary is traversed twice in opposite directions by adjacent small elements. Therefore, only the outer boundary curve C remains.

Σ small circulations = ∮C A · dl

At the same time, the sum of all curl fluxes over small elements becomes the surface integral over the whole surface:

Σ (∇ × A) · n ΔS = ∬S (∇ × A) · n dS

Therefore, after cancellation of internal boundaries, we obtain:

∮C A · dl = ∬S (∇ × A) · n dS

This proves Stokes theorem.

Q2(a): Significance of Stokes Theorem

Stokes theorem is important because it connects a local property of a vector field with a global property. The curl ∇ × A tells the local tendency of rotation at a point, while the line integral ∮ A · dl tells the total circulation around a closed curve.

In physics, Stokes theorem is used in electromagnetism, fluid dynamics and vector calculus. Maxwell’s equations in integral and differential forms are connected through theorems like Stokes theorem. For example, the circulation of an electric or magnetic field around a loop can be related to the curl of that field over a surface.

Q2(b): Position, Velocity and Acceleration

Given:

x(t)=At²+Bt
y(t)=Ct²+D

A = 1.00 m/s²
B = -32 m/s
C = 5 m/s²
D = 12 m
t = 3 s

Position at t = 3 s

x(3) = A(3)² + B(3)
x(3) = 1(9) + (-32)(3)
x(3) = 9 – 96
x(3) = -87 m

y(3) = C(3)² + D
y(3) = 5(9) + 12
y(3) = 45 + 12
y(3) = 57 m

Position: r = -87 i + 57 j m

Velocity at t = 3 s

Velocity is the time derivative of position.

vx = dx/dt = 2At + B
vy = dy/dt = 2Ct

vx(3) = 2(1)(3) – 32
vx(3) = 6 – 32
vx(3) = -26 m/s

vy(3) = 2(5)(3)
vy(3) = 30 m/s

Velocity: v = -26 i + 30 j m/s

The speed is:

|v| = √[(-26)² + (30)²]
|v| = √(676 + 900)
|v| = √1576
|v| ≈ 39.7 m/s

Acceleration at t = 3 s

Acceleration is the time derivative of velocity.

ax = d²x/dt² = 2A = 2(1) = 2 m/s²
ay = d²y/dt² = 2C = 2(5) = 10 m/s²
Acceleration: a = 2 i + 10 j m/s²

The magnitude of acceleration is:

|a| = √(2² + 10²)
|a| = √104
|a| ≈ 10.2 m/s²
Final Answer for Q2: Position = -87 i + 57 j m, velocity = -26 i + 30 j m/s, acceleration = 2 i + 10 j m/s².

Question 3: Relativity, Energy-Momentum Relation and Rotational Work

Full Question

Q.3. (a) What is relativity of length according to Einstein? Also discuss consequences of Lorentz transformation for relativity.

(b) Prove E² = m²c⁴ + p²c².

(c) Derive formula for work and kinetic energy in rotational motion.

Q3(a): Relativity of Length According to Einstein

According to Einstein’s special theory of relativity, length is not absolute. The length of an object measured by an observer depends on the relative motion between the object and the observer. The proper length L0 is the length measured in the frame in which the object is at rest. If the object moves with speed v relative to an observer, the length measured parallel to the direction of motion is contracted.

L = L0/γ

where

γ = 1/√(1 – v²/c²)

Since γ ≥ 1, the moving length L is always smaller than or equal to the proper length L0. This effect is called length contraction.

Consequences of Lorentz Transformation

The Lorentz transformations are:

x’ = γ(x – vt)

t’ = γ(t – vx/c²)

y’ = y

z’ = z

These equations replace the Galilean transformation at speeds close to the speed of light. Their major consequences are:

  1. Relativity of simultaneity: Two events simultaneous in one frame may not be simultaneous in another frame.
  2. Time dilation: A moving clock appears to run slow. The relation is Δt = γΔt0.
  3. Length contraction: A moving rod contracts along the direction of motion. The relation is L = L0/γ.
  4. No object can exceed light speed: As velocity approaches c, γ becomes very large.
  5. Mass-energy relation: Energy and mass are connected through E = mc².

Q3(b): Proof of Energy-Momentum Relation

The relativistic total energy of a particle is:

E = γm0c²

The relativistic momentum is:

p = γm0v

Now calculate E² - p²c²:

E² – p²c² = (γm0c²)² – (γm0v)²c²

E² – p²c² = γ²m0²c⁴ – γ²m0²v²c²

E² – p²c² = γ²m0²c²(c² – v²)

E² – p²c² = γ²m0²c⁴(1 – v²/c²)

But:

γ² = 1/(1 – v²/c²)

Therefore:

E² – p²c² = m0²c⁴

Rearranging:

E² = p²c² + m0²c⁴
Final Result: E² = p²c² + m0²c⁴. If the symbol m is used for rest mass in the question, the relation becomes E² = p²c² + m²c⁴.

Q3(c): Work and Kinetic Energy in Rotational Motion

In linear motion, work is:

dW = F ds

In rotational motion, force is replaced by torque and linear displacement is replaced by angular displacement:

dW = τ dθ

Therefore, total work is:

W = ∫ τ dθ

For a rigid body rotating about a fixed axis:

τ = Iα

Since:

α = dω/dt

and

ω = dθ/dt

We can write:

α = dω/dt = (dω/dθ)(dθ/dt) = ω dω/dθ

Now substitute into work:

dW = τ dθ = Iα dθ

dW = I(ω dω/dθ)dθ

dW = Iω dω

Integrating from rest to angular speed ω:

W = ∫0ω Iω dω

W = I[ω²/2]0ω

W = 1/2 Iω²

This work appears as rotational kinetic energy:

Krot = 1/2 Iω²
Final Answer for Q3: Length contracts according to L=L0/γ; Lorentz transformation gives time dilation, length contraction and relativity of simultaneity; the energy-momentum relation is E²=p²c²+m0²c⁴; rotational kinetic energy is Krot=1/2 Iω².

Question 4: Conservative Field, Direction Cosines and Earth Angular Momentum

Full Question

Q.4. (a) What is conservative field? Prove that gravitational force is the negative derivative of potential energy.

(b) Find the direction cosines of Cartesian coordinates (2, -1, 2).

(c) Calculate which is greater: angular momentum of Earth associated with rotation about its own axis or angular momentum of Earth associated with its orbital motion around the Sun. Radius of Earth = 6400 km and radius of orbit around Sun = 1.5 × 10⁸ km.

Q4(a): Conservative Field

A force field is called conservative if the work done by the force in moving a particle from one point to another is independent of path. In a conservative field, work depends only on initial and final positions.

For a conservative force:

∮ F · dr = 0

and the force can be written in terms of potential energy:

F = -∇U

Gravitational Force as Negative Derivative of Potential Energy

For two masses M and m separated by distance r, gravitational potential energy is:

U(r) = -GMm/r

For radial motion, the force is:

F = -dU/dr

Differentiate U(r):

dU/dr = d/dr (-GMm/r)

dU/dr = GMm/r²

Therefore:

F = -dU/dr

F = -GMm/r²

The negative sign shows that gravitational force is attractive and acts toward the center of the attracting mass.

Result: Gravitational force is the negative derivative of gravitational potential energy: F = -dU/dr = -GMm/r².

Q4(b): Direction Cosines of (2, -1, 2)

For a vector:

A = 2i – j + 2k

Magnitude is:

|A| = √(2² + (-1)² + 2²)

|A| = √(4 + 1 + 4)

|A| = √9

|A| = 3

Direction cosines are:

l = x/|A| = 2/3

m = y/|A| = -1/3

n = z/|A| = 2/3

Final Answer: Direction cosines are (2/3, -1/3, 2/3).

Q4(c): Earth’s Rotational and Orbital Angular Momentum

Angular momentum associated with Earth’s rotation about its own axis is:

Lspin = Iω

Approximating Earth as a solid sphere:

I = 2/5 MR²

So:

Lspin = (2/5)MRE²(2π/Tday)

Angular momentum of Earth in its orbital motion around the Sun is:

Lorb = MRorb²(2π/Tyear)

Now compare by taking the ratio:

Lorb/Lspin = [MRorb²(2π/Tyear)] / [(2/5)MRE²(2π/Tday)]

Lorb/Lspin = (5/2)(Rorb/RE)²(Tday/Tyear)

Given:

RE = 6400 km

Rorb = 1.5 × 10⁸ km

Tday/Tyear ≈ 1/365

Now:

Rorb/RE = (1.5 × 10⁸)/(6400)

Rorb/RE = 23437.5

Therefore:

Lorb/Lspin = (5/2)(23437.5)²(1/365)

Lorb/Lspin ≈ 3.76 × 10⁶

Final Answer: Earth’s orbital angular momentum around the Sun is much greater than its spin angular momentum. It is approximately 3.8 × 10⁶ times greater.

Exam Explanation: Even though Earth spins once per day, the radius of its orbit around the Sun is enormously larger than Earth’s own radius. Since angular momentum depends on , the orbital angular momentum dominates.

Question 5: Fluid Pressure and Doppler Effect

Full Question

Q.5. (a) Write in detail about the variation of pressure in a fluid at rest and in the atmosphere with relevant mathematical formulas.

(b) The siren of a police car emits a pure tone at frequency 1125 Hz. Find the frequency perceived in your car when you are moving at 9 m/s and the police car is chasing behind you at 38 m/s.

Q5(a): Variation of Pressure in a Fluid at Rest

In a fluid at rest, pressure increases with depth. Consider a small vertical fluid element of height dz, area A and density ρ. The pressure at the bottom is slightly greater than the pressure at the top because the lower layer supports the weight of the fluid above it.

For equilibrium:

upward force – downward force – weight = 0

If vertical height z is measured upward:

P(z)A – [P(z+dz)A] – ρgA dz = 0

Simplifying:

P(z+dz) – P(z) = -ρg dz

Therefore:

dP/dz = -ρg

This is the hydrostatic pressure equation. It means pressure decreases as height increases and increases as depth increases.

If density is constant, integration gives:

P = P0 + ρgh

Here, P0 is atmospheric pressure at the surface, h is depth below the surface and ρgh is gauge pressure.

Fluid Pressure Result: Pressure in a liquid at rest increases linearly with depth: P = P0 + ρgh.

Pressure Variation in the Atmosphere

In the atmosphere, pressure also follows:

dP/dz = -ρg

However, air density is not constant. Using the ideal gas relation:

P = ρRT/M

So:

ρ = MP/RT

Substitute into hydrostatic equation:

dP/dz = -(MPg)/(RT)

For an isothermal atmosphere, temperature T is constant:

dP/P = -(Mg/RT) dz

Integrating from sea level to height z:

ln(P/P0) = -Mgz/RT

Therefore:

P = P0 e^(-Mgz/RT)

This is the barometric formula. It shows that atmospheric pressure decreases exponentially with height.

Atmospheric Pressure Result: For an isothermal atmosphere, P = P0 e^(-Mgz/RT).

Q5(b): Doppler Effect Numerical

Given:

Source frequency, f = 1125 Hz

Observer speed, vo = 9 m/s

Source speed, vs = 38 m/s

Speed of sound, v ≈ 340 m/s

The police car is behind and chasing the observer. The observer is moving away from the source, while the source is moving toward the observer. Therefore:

f’ = f (v – vo)/(v – vs)

Substitute values:

f’ = 1125 × (340 – 9)/(340 – 38)

f’ = 1125 × 331/302

f’ = 1125 × 1.096

f’ ≈ 1233 Hz

Final Answer for Q5(b): The perceived frequency is approximately 1233 Hz, or 1.23 kHz.

Exam Note: If the speed of sound is taken as 343 m/s, the answer remains approximately 1.23 kHz. The result is higher than the emitted frequency because the source is approaching the observer faster than the observer is moving away.

Question 6: Standing Waves and Grating Resolving Power

Full Question

Q.6. (a) If two waves having the same amplitude are propagating in opposite directions through a string, will they produce standing waves? Is energy transported, and are there any nodes?

(b) How can dispersion and resolving power of a grating be calculated in terms of wavelength?

Q6(a): Formation of Standing Waves

Yes. If two waves of the same amplitude, same frequency and same wavelength travel in opposite directions through a string, they superpose to produce a standing wave.

Let the two waves be:

y1 = A sin(kx – ωt)

y2 = A sin(kx + ωt)

The resultant displacement is:

y = y1 + y2

y = A sin(kx – ωt) + A sin(kx + ωt)

Using the identity:

sin(a-b) + sin(a+b) = 2 sin a cos b

We get:

y = 2A sin(kx) cos(ωt)

This is the equation of a standing wave. It is not of the form f(x-vt) or f(x+vt), so it does not represent a travelling wave. Instead, each point of the string oscillates with amplitude 2A sin(kx).

Nodes and Antinodes

Nodes occur where displacement is always zero:

2A sin(kx) = 0

sin(kx) = 0

kx = nπ

Since k = 2π/λ:

x = nλ/2

Antinodes occur where amplitude is maximum:

sin(kx) = ±1

x = (2n+1)λ/4

Is Energy Transported?

In a standing wave, there is no net transport of energy along the string. The two travelling waves carry equal energy in opposite directions, so the average energy flow cancels. Energy remains localized and changes between kinetic and potential forms.

Final Answer for Q6(a): Yes, equal-amplitude waves travelling in opposite directions produce a standing wave. Nodes are formed at x=nλ/2. There is no net energy transport along the string.

Q6(b): Grating Dispersion

For a diffraction grating, the grating equation is:

d sinθ = mλ

Here, d is grating spacing, θ is diffraction angle, m is order of spectrum and λ is wavelength.

Differentiate with respect to wavelength:

d cosθ (dθ/dλ) = m

Therefore, angular dispersion is:

dθ/dλ = m/(d cosθ)
Angular Dispersion: dθ/dλ = m/(d cosθ).

If a lens of focal length f is used, linear dispersion is:

dy/dλ = f dθ/dλ

dy/dλ = fm/(d cosθ)

Resolving Power of a Grating

Resolving power is the ability of a grating to separate two closely spaced wavelengths. It is defined as:

R = λ/Δλ

For a diffraction grating:

R = mN

where m is order of spectrum and N is the total number of illuminated slits.

Final Answer for Q6(b): Grating angular dispersion is dθ/dλ = m/(d cosθ), and resolving power is R = λ/Δλ = mN.

Question 7: Real Gas, Heat Pump and Fermi-Dirac Statistics

Full Question

Q.7. (a) Differentiate between real and ideal gas. Describe work done on ideal gas in thermal isolation.

(b) A heat pump acting as a refrigerator is used to heat a house. The temperature outside the house is -9°C and inside is kept at 21°C. Find the maximum coefficient of performance of the heat pump.

(c) What do you know about Fermi-Dirac statistics?

Scan Note: Some scanned copies write the temperature as “-9 micro C.” In this solved answer, it is interpreted as -9°C, because the heat pump problem physically requires Celsius temperature.

Q7(a): Ideal Gas and Real Gas

Ideal Gas Real Gas
Obeys PV=nRT exactly under all conditions in the ideal model. Obeys gas laws approximately, especially at low pressure and high temperature.
Molecules are treated as point particles. Molecules have finite size.
Intermolecular forces are neglected. Intermolecular attractive and repulsive forces exist.
No liquefaction in the ideal model. Can liquefy at low temperature and high pressure.
Internal energy depends only on temperature. Internal energy may depend on temperature and intermolecular forces.

A real gas is often described by the van der Waals equation:

(P + an²/V²)(V – nb) = nRT

Here, a corrects for intermolecular attraction and b corrects for finite molecular volume.

Work Done on Ideal Gas in Thermal Isolation

Thermal isolation means no heat is exchanged with the surroundings:

Q = 0

Such a process is called an adiabatic process. From the first law of thermodynamics:

ΔU = Q – Wby

Since Q=0:

ΔU = -Wby

So work done by the gas is:

Wby = -ΔU

Work done on the gas is:

Won = ΔU

For an ideal gas:

ΔU = nCv(T2 – T1)

Therefore:

Won = nCv(T2 – T1)

For a reversible adiabatic process:

PVγ = constant

Work done by gas is:

Wby = (P1V1 – P2V2)/(γ – 1)

Therefore, work done on gas is:

Won = (P2V2 – P1V1)/(γ – 1)
Final Result for Q7(a): In thermal isolation, Q=0. Work done on an ideal gas is Won=ΔU=nCv(T2-T1). For a reversible adiabatic process, Won=(P2V2-P1V1)/(γ-1).

Q7(b): Maximum Coefficient of Performance of Heat Pump

A heat pump used to heat a house has maximum coefficient of performance:

COPHP = TH/(TH – TC)

Convert Celsius temperatures to Kelvin:

TH = 21 + 273 = 294 K

TC = -9 + 273 = 264 K

Now substitute:

COPHP = 294/(294 – 264)

COPHP = 294/30

COPHP = 9.8

Final Answer for Q7(b): Maximum coefficient of performance of the heat pump is 9.8.

Important Note: If the device were treated purely as a refrigerator for cooling, the refrigerator COP would be COPR = TC/(TH-TC)=264/30=8.8. But because the question says it is used to heat a house, the heat pump COP is 9.8.

Q7(c): Fermi-Dirac Statistics

Fermi-Dirac statistics describes the distribution of identical particles called fermions among available energy states. Fermions have half-integral spin, such as 1/2, 3/2, and so on. Electrons, protons and neutrons are common examples.

The most important condition is the Pauli exclusion principle. It states that no two identical fermions can occupy the same quantum state simultaneously.

The Fermi-Dirac distribution function is:

f(E) = 1/[e^((E – EF)/kT) + 1]

Here, f(E) is the probability that a state of energy E is occupied, EF is the Fermi energy, k is Boltzmann’s constant and T is absolute temperature.

Important Features of Fermi-Dirac Statistics

  1. It applies to fermions, such as electrons.
  2. It obeys Pauli exclusion principle.
  3. At absolute zero, all states below Fermi energy are filled and all states above it are empty.
  4. It explains electron behavior in metals, semiconductors and white dwarfs.
  5. It is different from Bose-Einstein statistics, which applies to bosons.
Final Answer for Q7(c): Fermi-Dirac statistics gives the occupation probability of fermions and is written as f(E)=1/[e^((E-EF)/kT)+1]. It is based on Pauli exclusion principle and is essential for understanding electrons in solids.

Question 8: Comprehensive Notes

Full Question

Q.8. Write comprehensive note on any two of the following:

(a) Damped harmonic motion.

(b) Young double slit experiment.

(c) Longitudinal and transverse coherence.

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): Damped Harmonic Motion

Damped harmonic motion is oscillatory motion in which the amplitude gradually decreases with time due to a resistive force. The resistive force is usually proportional to velocity and acts opposite to the motion.

The damping force is:

Fd = -b dx/dt

The equation of motion is:

m d²x/dt² + b dx/dt + kx = 0

Dividing by m:

d²x/dt² + (b/m) dx/dt + (k/m)x = 0

Let:

β = b/2m

ω0 = √(k/m)

For underdamped motion, the solution is:

x(t) = A e^(-βt) cos(ωd t + φ)

where:

ωd = √(ω0² – β²)

Types of Damping

  1. Underdamping: Oscillations continue but amplitude decreases exponentially.
  2. Critical damping: The system returns to equilibrium in the shortest time without oscillating.
  3. Overdamping: The system returns slowly to equilibrium without oscillating.

The energy of a damped oscillator decreases with time because the resistive force converts mechanical energy into heat.

Note Summary: Damped harmonic motion is described by m x'' + b x' + kx = 0. For light damping, amplitude decays as e^(-βt) and the oscillator continues with reduced frequency.

Q8(b): Young Double Slit Experiment

Young’s double slit experiment demonstrates the wave nature of light through interference. A monochromatic light source illuminates two narrow slits separated by distance d. The two slits act as coherent sources. Their waves overlap on a screen placed at distance D, producing bright and dark fringes.

The path difference between waves reaching a point on the screen is:

δ = d sinθ

For small angles:

sinθ ≈ tanθ = y/D

Therefore:

δ = dy/D

Condition for Bright Fringes

Constructive interference occurs when path difference is an integral multiple of wavelength:

δ = nλ

So:

dy/D = nλ

y = nλD/d

Condition for Dark Fringes

Destructive interference occurs when path difference is an odd half multiple of wavelength:

δ = (n + 1/2)λ

So:

y = (n + 1/2)λD/d

Fringe Width

The distance between two consecutive bright or dark fringes is:

β = λD/d
Note Summary: Young’s double slit experiment proves interference of light. Bright fringes occur at d sinθ=nλ, dark fringes occur at d sinθ=(n+1/2)λ, and fringe width is β=λD/d.

Q8(c): Longitudinal and Transverse Coherence

Coherence describes the fixed phase relationship between waves. Interference fringes are clear only when waves maintain a stable phase relation. There are two important types of coherence: longitudinal coherence and transverse coherence.

Longitudinal Coherence

Longitudinal coherence, also called temporal coherence, refers to phase correlation along the direction of propagation. It tells how far a wave remains phase-correlated with itself along its direction of travel.

The coherence time is approximately:

τc ≈ 1/Δf

The coherence length is:

Lc = cτc

In terms of wavelength spread:

Lc ≈ λ²/Δλ

A source with narrow spectral width has large coherence length. Laser light has high longitudinal coherence, while white light has small coherence length.

Transverse Coherence

Transverse coherence refers to phase correlation across a wavefront, perpendicular to the direction of propagation. It tells whether two points on the same wavefront can act as coherent sources.

For a source of angular size θ, transverse coherence length is approximately:

Lt ≈ λ/θ

A smaller source gives greater transverse coherence. This is why narrow slits and pinholes are used in interference experiments.

Difference Between Longitudinal and Transverse Coherence

Longitudinal Coherence Transverse Coherence
Coherence along direction of propagation. Coherence across the wavefront.
Related to spectral width. Related to source size.
Measured by coherence length Lc. Measured by transverse coherence width Lt.
Important for path difference. Important for slit separation and source size.
Note Summary: Longitudinal coherence depends on spectral purity and coherence length, while transverse coherence depends on source size and phase correlation across the wavefront.

Revision Plan for CSS Physics Paper-I 2024 Solved

After reading this complete CSS Physics Paper-I 2024 Solved guide, revise it in three rounds. In the first round, learn the definitions and formulas. In the second round, reproduce the derivations without looking. In the third round, solve the numerical questions with units and compare your answer with the final result.

Question Revision Task
Q2 Write Stokes theorem from memory and recalculate position, velocity and acceleration at t=3s.
Q3 Derive length contraction, energy-momentum relation and rotational kinetic energy.
Q4 Prove F=-dU/dr, calculate direction cosines and compare Earth’s angular momenta.
Q5 Derive hydrostatic pressure equation and solve Doppler effect numerical again.
Q6 Derive standing wave equation and grating resolving power formula.
Q7 Revise adiabatic work, heat pump COP and Fermi-Dirac distribution.
Q8 Prepare any two notes but read all three for safety.

Related Resources for CSS Physics Paper-I 2024 Solved

Exam Note: CSS Physics papers sometimes appear in scanned form with unclear symbols. Where a scan distorts a symbol, use the standard Physics interpretation, state your assumption briefly and then solve with clean units. This is especially important in numerical questions.

FAQs About CSS Physics Paper-I 2024 Solved

What does CSS Physics Paper-I 2024 Solved include?

CSS Physics Paper-I 2024 Solved includes complete solved answers for Q2 to Q8 with definitions, derivations, numerical calculations, formulas and final answers.

Is Stokes theorem fully solved in this post?

Yes. The post states Stokes theorem, proves it by surface subdivision and internal-boundary cancellation, and explains its physical significance.

What is the answer for the xy-plane motion numerical in Q2?

At t=3s, position is -87i + 57j m, velocity is -26i + 30j m/s, and acceleration is 2i + 10j m/s².

What is the Doppler effect answer in Q5?

The perceived siren frequency is approximately 1233 Hz, or 1.23 kHz, assuming speed of sound as about 340 m/s.

Which angular momentum of Earth is greater in Q4?

Earth’s orbital angular momentum around the Sun is greater than its rotational angular momentum about its own axis by approximately 3.8 × 10⁶ times.

What is the heat pump COP in Q7?

The maximum coefficient of performance of the heat pump is 9.8, using TH=294K and TC=264K.

Are all notes in Q8 solved?

Yes. Although the paper asks for any two notes, this post solves all three: damped harmonic motion, Young double slit experiment, and longitudinal and transverse coherence.

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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