Physics CSS Solved Papers

CSS Physics Paper-I 2023 Solved

Engr. Muhammad Yar Saqib

CSS Physics Paper-I 2023 Solved is a complete solved guide for aspirants who need real answers, not only topic hints. This post solves the subjective section question by question with definitions, derivations, formulas, numerical substitutions, final results and exam-ready explanations. It covers gradient, acceleration, curl, rotational kinetic energy, moment of inertia, angular speed of watch hands, relativity, length contraction, mass-energy equivalence, capillarity, fluid pressure, wave energy, longitudinal waves, molecular oscillator work, ideal gas work, intermolecular forces, Kepler’s law, Michelson interferometer and Young’s double slit experiment.

Central Argument: A CSS Physics solved paper should not only tell students which topic was asked. It should actually define, derive, calculate and explain the answer. Therefore, this CSS Physics Paper-I 2023 Solved post gives the complete route to the answer so students can reproduce the solution in the examination hall.

CSS Physics Paper-I 2023 Solved Study Scope

This post covers CSS Physics Paper-I 2023 Solved in a complete, structured and WordPress-ready 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: Gradient, Acceleration and Curl
  6. Question 3: Rotational Kinetic Energy and Inertia
  7. Question 4: Relativity, Length Contraction and Antenna Mass Equivalent
  8. Question 5: Capillary Rise, Fluids and Pool-Bottom Pressure
  9. Question 6: Wave Energy, Longitudinal Waves and Molecular Oscillator Work
  10. Question 7: Work by Gas, Intermolecular Forces and Final Temperature
  11. Question 8: Short Notes
  12. Revision Plan
  13. Internal and External Resources
  14. FAQs

Important Formula Sheet for CSS Physics Paper-I 2023 Solved

Vector Calculus

grad φ = ∇φ

∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

curl A = ∇ × A

a = dv/dt = d²r/dt²

Rotation and Inertia

Krot = 1/2 Iω²

I = Σmr²

ω = 2π/T

τ = Iα

Relativity, Fluids and Waves

E = mc²

L = L0/γ

h = 2Tcosθ/(ρgr)

P = Patm + ρgh

Gas Laws and Oscillations

PV = nRT

Wby = ∫P dV

W = 1/2 kx²

k = μω²

Question Map of CSS Physics Paper-I 2023 Solved

Question Main Area What Is Fully Solved
Q2 Vector calculus and kinematics Gradient definition, physical significance, acceleration components and curl at a point.
Q3 Rotational dynamics Rotational kinetic energy, formulas for disc, hoop and sphere, inertia, cylinder inertia and angular speed of clock hands.
Q4 Relativity Physics before relativity, Einstein’s theory, mass-energy equivalence, length contraction and antenna mass equivalent.
Q5 Fluid mechanics Capillary rise derivation, fluids and characteristics, pressure at the bottom of a swimming pool.
Q6 Waves and molecular oscillation Wave energy distribution, longitudinal waves in solids and work required to stretch CO molecule.
Q7 Thermodynamics and gas laws Work by gas, non-conservative pressure force, intermolecular forces, moles and final temperature of oxygen.
Q8 Orbital mechanics and optics Kepler’s law of periods, Michelson interferometer and Young’s double slit experiment.

Question 2: Gradient, Acceleration and Curl at a Point

Full Question

Q.2. (a) What is gradient of a scalar function? Give its physical significance and show that Grad φ = ∇φ.

(b) Define the term acceleration and find its Cartesian components.

(c) If A = xz³ i - 2x²z j + 2yz⁴ k, then find curl of A at the point (1, -1, 1).

Q2(a): Gradient of a Scalar Function

The gradient of a scalar function is a vector that gives the maximum rate of increase of the scalar function and points in the direction in which the scalar function increases most rapidly.

If φ(x,y,z) is a scalar field, then its gradient is:

Grad φ = ∇φ

In Cartesian coordinates:

∇ = i ∂/∂x + j ∂/∂y + k ∂/∂z

Therefore:

∇φ = i ∂φ/∂x + j ∂φ/∂y + k ∂φ/∂z

Hence:

Grad φ = ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

Physical Significance of Gradient

The physical significance of gradient is that it gives the direction and rate of the steepest increase of a scalar field. For example, if φ represents temperature, then ∇φ points in the direction in which temperature rises most rapidly. If V is electric potential, then the electric field is related to potential by:

E = -∇V

The negative sign shows that the electric field points in the direction of decreasing potential.

Answer: The gradient of a scalar function is ∇φ. It points in the direction of maximum increase of the scalar field, and its magnitude is the maximum rate of change per unit distance.

Q2(b): Acceleration and Its Cartesian Components

Acceleration is the time rate of change of velocity. If position vector is:

r = xi + yj + zk

Then velocity is:

v = dr/dt = (dx/dt)i + (dy/dt)j + (dz/dt)k

Acceleration is:

a = dv/dt = d²r/dt²

Therefore, in Cartesian components:

a = ax i + ay j + az k

where:

ax = d²x/dt²

ay = d²y/dt²

az = d²z/dt²

Answer: Acceleration is a = dv/dt. Its Cartesian components are ax=d²x/dt², ay=d²y/dt² and az=d²z/dt².

Q2(c): Curl of A at (1, -1, 1)

Given vector field:

A = xz³ i – 2x²z j + 2yz⁴ k

So:

Ax = xz³

Ay = -2x²z

Az = 2yz⁴

The curl of A is:

∇ × A =
(∂Az/∂y – ∂Ay/∂z)i
+ (∂Ax/∂z – ∂Az/∂x)j
+ (∂Ay/∂x – ∂Ax/∂y)k

Now calculate each component.

i-component

∂Az/∂y = ∂(2yz⁴)/∂y = 2z⁴

∂Ay/∂z = ∂(-2x²z)/∂z = -2x²

i-component = 2z⁴ – (-2x²)

i-component = 2z⁴ + 2x²

j-component

∂Ax/∂z = ∂(xz³)/∂z = 3xz²

∂Az/∂x = ∂(2yz⁴)/∂x = 0

j-component = 3xz² – 0

j-component = 3xz²

k-component

∂Ay/∂x = ∂(-2x²z)/∂x = -4xz

∂Ax/∂y = ∂(xz³)/∂y = 0

k-component = -4xz – 0

k-component = -4xz

Therefore:

∇ × A = (2z⁴ + 2x²)i + (3xz²)j – (4xz)k

At the point (1, -1, 1), we have x=1, y=-1, z=1.

i-component = 2(1)⁴ + 2(1)² = 2 + 2 = 4

j-component = 3(1)(1)² = 3

k-component = -4(1)(1) = -4

Final Answer for Q2(c): ∇ × A = 4i + 3j - 4k at (1,-1,1).

Question 3: Rotational Kinetic Energy and Inertia

Full Question

Q.3. (a) Explain the rotational kinetic energy and determine its formula for a disc, hoop and sphere.

(b) What do you mean by the term inertia in physics? Calculate respectively the rotational inertia of a solid cylinder and a hollow cylinder about an axis of symmetry.

(c) Calculate the angular speed of the second’s hand, minute hand and hour hand of a watch.

Q3(a): Rotational Kinetic Energy

Rotational kinetic energy is the energy possessed by a body due to its rotational motion. A rigid body rotating with angular speed ω can be treated as a collection of particles. Each particle has linear speed:

v = rω

The kinetic energy of one particle is:

K = 1/2 mv²

Substitute v=rω:

K = 1/2 m(rω)²

K = 1/2 mr²ω²

For the whole rigid body:

Krot = Σ 1/2 mi ri² ω²

Krot = 1/2 ω² Σ mi ri²

But:

I = Σ mi ri²

Therefore:

Krot = 1/2 Iω²
Rotational Kinetic Energy: Krot = 1/2 Iω².

Formulas for Disc, Hoop and Sphere

Body Moment of Inertia Rotational Kinetic Energy
Disc or solid cylinder about central symmetry axis I = 1/2 MR² K = 1/4 MR²ω²
Hoop or thin ring about central symmetry axis I = MR² K = 1/2 MR²ω²
Solid sphere about diameter I = 2/5 MR² K = 1/5 MR²ω²

Q3(b): Inertia in Physics

Inertia is the property of a body by which it resists any change in its state of rest or uniform motion. In linear motion, mass is the measure of inertia. In rotational motion, moment of inertia is the measure of resistance to change in rotational motion.

Moment of inertia depends on both mass and distribution of mass about the axis of rotation:

I = Σ mr²

Rotational Inertia of Solid Cylinder

For a solid cylinder of mass M and radius R about its central axis of symmetry:

I = 1/2 MR²

Rotational Inertia of Hollow Cylinder

For a thin hollow cylinder of mass M and radius R about its central axis:

I = MR²

If the hollow cylinder is thick with inner radius R1 and outer radius R2, then:

I = 1/2 M(R1² + R2²)
Final Answer for Q3(b): Solid cylinder: I=1/2MR². Thin hollow cylinder: I=MR². Thick hollow cylinder: I=1/2M(R1²+R2²).

Q3(c): Angular Speed of Watch Hands

Angular speed is:

ω = 2π/T

Second Hand

The second hand completes one revolution in 60 s.

ωs = 2π/60

ωs = 0.1047 rad/s

Minute Hand

The minute hand completes one revolution in 60 min = 3600 s.

ωm = 2π/3600

ωm = 1.745 × 10⁻³ rad/s

Hour Hand

The hour hand completes one revolution in 12 h = 43200 s.

ωh = 2π/43200

ωh = 1.454 × 10⁻⁴ rad/s

Final Answer for Q3(c): Second hand = 0.1047 rad/s, minute hand = 1.745 × 10⁻³ rad/s, hour hand = 1.454 × 10⁻⁴ rad/s.

Question 4: Relativity, Length Contraction and Antenna Mass Equivalent

Full Question

Q.4. (a) What was Physics like before relativity and how did Einstein come up with his theory? Mathematically explain how mass and energy is interchangeable.

(b) Discuss in detail the relativity of length using Einstein’s special theory of relativity.

(c) Calculate the mass equivalent of energy from an antenna radiating 10 kW for 24 hours.

Q4(a): Physics Before Relativity

Before relativity, physics was based mainly on Newtonian mechanics and Maxwell’s electromagnetism. Newtonian mechanics assumed absolute space and absolute time. This meant that lengths, time intervals and simultaneity were considered the same for all observers. The Galilean transformation was used to relate measurements between inertial frames.

However, Maxwell’s equations predicted that light travels with a fixed speed c. This created a problem because Galilean transformation suggested that measured speed should depend on the motion of the observer. The failure to detect ether and the results of experiments such as the Michelson-Morley experiment showed that the old view was incomplete.

How Einstein Developed Special Relativity

Einstein proposed special relativity in 1905 based on two postulates:

  1. The laws of physics are the same in all inertial frames.
  2. The speed of light in vacuum is constant for all inertial observers, independent of the motion of source or observer.

These postulates led to Lorentz transformation, time dilation, length contraction, relativity of simultaneity and mass-energy equivalence.

Mass-Energy Interchangeability

Einstein showed that mass and energy are not separate quantities. Mass is a form of energy. The rest energy of a body is:

E = mc²

This equation means that even a small mass corresponds to a very large amount of energy because is extremely large.

If mass decreases by Δm, the released energy is:

ΔE = Δmc²

This relation explains nuclear energy, binding energy, pair production and annihilation. For example, in nuclear reactions, a small mass defect appears as a large amount of energy.

Answer: Before relativity, physics assumed absolute space and time. Einstein replaced this with invariant light speed and Lorentz transformation. Mass and energy are interchangeable according to E=mc².

Q4(b): Relativity of Length

According to Einstein’s special theory of relativity, length is not absolute. The measured length of an object depends on the relative motion between the object and the observer.

The length measured in the frame in which the object is at rest is called proper length L0. If the object moves with speed v relative to an observer, then 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 less than or equal to L0. This effect is called length contraction.

Important Points About Length Contraction

  1. Length contraction occurs only along the direction of relative motion.
  2. Dimensions perpendicular to motion remain unchanged.
  3. The effect becomes significant only when speed is comparable to the speed of light.
  4. The proper length is always the longest length.
Length Relativity Result: L = L0√(1-v²/c²). This is equivalent to L=L0/γ.

Q4(c): Mass Equivalent of Energy from Antenna

Given:

Power, P = 10 kW = 10,000 W

Time, t = 24 h = 24 × 3600 s = 86,400 s

Energy radiated is:

E = Pt

E = 10,000 × 86,400

E = 8.64 × 10⁸ J

Using mass-energy relation:

E = mc²

m = E/c²

Take:

c = 3 × 10⁸ m/s

c² = 9 × 10¹⁶ m²/s²

Now:

m = (8.64 × 10⁸)/(9 × 10¹⁶)

m = 0.96 × 10⁻⁸ kg

m = 9.6 × 10⁻⁹ kg

Conversion:

9.6 × 10⁻⁹ kg = 9.6 × 10⁻⁶ g = 9.6 micrograms
Final Answer for Q4(c): Mass equivalent = 9.6 × 10⁻⁹ kg, or about 9.6 micrograms.

Question 5: Capillary Rise, Fluids and Pool-Bottom Pressure

Full Question

Q.5. (a) Define capillarity and derive an expression for the rise of liquid in a capillary tube to show that the height of the liquid column supported is inversely proportional to the radius of the tube.

(b) What are fluids? Write their important characteristics.

(c) A cylindrical swimming pool has radius 2 m and depth 1.3 m. It is filled completely with salt water. Given density of salt water = 1.03 × 10³ kgm⁻³, volume of water = 16.34 m³, and atmospheric pressure = 1.013 × 10⁵ Pa. Calculate the pressure at the bottom of the pool.

Q5(a): Capillarity

Capillarity is the rise or fall of a liquid in a narrow tube due to surface tension and adhesive/cohesive forces. If adhesive force between liquid and tube is greater than cohesive force within the liquid, the liquid rises. If cohesive force is greater, the liquid is depressed.

Derivation of Capillary Rise

Consider a capillary tube of radius r dipped in a liquid of density ρ and surface tension T. Let the liquid rise to height h. The angle of contact is θ.

The upward vertical component of surface tension around the circumference of the tube is:

Fup = 2πr T cosθ

The weight of the liquid column is:

W = volume × density × g

The volume of liquid column is:

V = πr²h

So weight is:

W = πr²h ρg

At equilibrium:

upward force = weight of liquid column

Therefore:

2πrTcosθ = πr²hρg

Cancel πr from both sides:

2Tcosθ = r h ρg

Solving for h:

h = 2Tcosθ/(ρgr)
Capillary Rise Formula: h = 2Tcosθ/(ρgr). Since h ∝ 1/r, the height of liquid column is inversely proportional to the radius of the capillary tube.

Q5(b): Fluids and Their Characteristics

A fluid is a substance that can flow and continuously deform under the action of shear stress. Liquids and gases are fluids.

Important Characteristics of Fluids

  1. No fixed shape: Fluids take the shape of the container.
  2. Ability to flow: Fluids cannot resist shear stress permanently.
  3. Pressure normal to surfaces: At rest, fluids exert pressure perpendicular to any surface.
  4. Density: Fluids have mass per unit volume, written as ρ=m/V.
  5. Viscosity: Real fluids resist relative motion between layers.
  6. Compressibility: Gases are highly compressible, while liquids are nearly incompressible.
  7. Surface tension: Liquids show surface tension due to molecular forces at the surface.
Answer: Fluids include liquids and gases. Their important properties are flow, pressure, density, viscosity, compressibility, surface tension and absence of fixed shape.

Q5(c): Pressure at the Bottom of the Pool

Given:

ρ = 1.03 × 10³ kg/m³ = 1030 kg/m³

h = 1.3 m

Patm = 1.013 × 10⁵ Pa

g = 9.8 m/s²

Pressure at the bottom is:

P = Patm + ρgh

Substitute values:

P = 1.013 × 10⁵ + (1030)(9.8)(1.3)

Calculate hydrostatic pressure:

ρgh = 1030 × 9.8 × 1.3

ρgh = 13122.2 Pa

Now:

P = 101300 + 13122.2

P = 114422.2 Pa

Final Answer for Q5(c): Pressure at the bottom of the pool is approximately 1.144 × 10⁵ Pa, or 114.4 kPa absolute. Gauge pressure is approximately 13.1 kPa.

Question 6: Wave Energy, Longitudinal Waves and Molecular Oscillator Work

Full Question

Q.6. (a) For a wave travelling through a medium, demonstrate that the total energy per unit volume is one half kinetic and one half potential energy.

(b) The longitudinal waves can pass through solids. How is it possible and on what parameters will the velocity of such waves depend?

(c) The angular vibrational frequency of CO molecule is 0.6 × 10¹⁵ s⁻¹. Calculate the amount of work required for stretching it by 0.5 Å from the equilibrium position.

Q6(a): Energy in a Travelling Wave

Consider a sinusoidal wave travelling through a medium:

y = A sin(kx – ωt)

The transverse velocity of a particle of the medium is:

∂y/∂t = -Aω cos(kx – ωt)

Kinetic energy per unit volume is proportional to the square of particle velocity:

uK = 1/2 ρ(∂y/∂t)²

So:

uK = 1/2 ρA²ω² cos²(kx – ωt)

The potential energy density for a harmonic wave is also proportional to strain squared. For a sinusoidal travelling wave, it has the same average value as kinetic energy density:

uP = 1/2 ρA²ω² cos²(kx – ωt)

Therefore:

uK = uP

Total energy density is:

u = uK + uP

u = ρA²ω² cos²(kx – ωt)

The time average of cos² is 1/2. Hence average total energy density is:

<u> = 1/2 ρA²ω²

Since average kinetic and potential parts are equal:

<uK> = <uP> = 1/4 ρA²ω²
Final Result for Q6(a): In a harmonic travelling wave, average energy is equally divided between kinetic energy and potential energy. Each contributes one half of the total average energy.

Q6(b): Longitudinal Waves Through Solids

Longitudinal waves can pass through solids because solids have elasticity. In a longitudinal wave, particles oscillate parallel to the direction of wave propagation. Solids can undergo compression and rarefaction because they have restoring forces between their particles.

The velocity of longitudinal waves depends mainly on elasticity and density. In a thin rod, the speed is:

v = √(Y/ρ)

where Y is Young’s modulus and ρ is density.

For a bulk longitudinal wave in a solid, the speed is:

v = √[(K + 4G/3)/ρ]

where K is bulk modulus, G is shear modulus and ρ is density.

Parameters Affecting Velocity

  1. Elastic modulus: Greater elasticity increases wave speed.
  2. Density: Greater density decreases wave speed.
  3. Nature of medium: Different solids have different elastic constants.
  4. Temperature and structure: These may affect elastic properties and therefore wave speed.
Answer: Longitudinal waves pass through solids because solids support compression and rarefaction. Their speed depends on elastic modulus and density.

Q6(c): Work Required to Stretch CO Molecule

The CO molecule may be treated as a harmonic oscillator. For a harmonic oscillator:

ω = √(k/μ)

So:

k = μω²

Work required to stretch the molecule by displacement x is stored as potential energy:

W = 1/2 kx²

Reduced Mass of CO Molecule

Carbon has atomic mass approximately 12 u, and oxygen has atomic mass approximately 16 u.

μ = (mC mO)/(mC + mO)

μ = (12 × 16)/(12 + 16) u

μ = 192/28 u

μ = 6.857 u

Since:

1 u = 1.6605 × 10⁻²⁷ kg

Therefore:

μ = 6.857 × 1.6605 × 10⁻²⁷

μ ≈ 1.138 × 10⁻²⁶ kg

Calculate Force Constant

Given:

ω = 0.6 × 10¹⁵ s⁻¹ = 6.0 × 10¹⁴ s⁻¹

Now:

k = μω²

k = (1.138 × 10⁻²⁶)(6.0 × 10¹⁴)²

k = (1.138 × 10⁻²⁶)(3.6 × 10²⁹)

k ≈ 4.10 × 10³ N/m

Calculate Work

Given stretch:

x = 0.5 Å = 0.5 × 10⁻¹⁰ m = 5.0 × 10⁻¹¹ m

Work is:

W = 1/2 kx²

W = 1/2 (4.10 × 10³)(5.0 × 10⁻¹¹)²

W = 0.5 × 4.10 × 10³ × 25 × 10⁻²²

W ≈ 5.12 × 10⁻¹⁸ J

In electron volts:

W = (5.12 × 10⁻¹⁸)/(1.602 × 10⁻¹⁹)

W ≈ 32 eV

Final Answer for Q6(c): Work required to stretch CO molecule by 0.5 Å is approximately 5.12 × 10⁻¹⁸ J, or about 32 eV.

Question 7: Work by Gas, Intermolecular Forces and Final Temperature

Full Question

Q.7. (a) An ideal gas is enclosed in a cylinder with movable piston. Calculate the work done on such gas and show that pressure force is non-conservative.

(b) State and briefly explain the intermolecular forces.

(c) Oxygen gas having a volume of 1130 cm³ at 42°C and a pressure of 101 kPa expanded until its volume is 1530 cm³ and its pressure is 106 kPa. Find the number of moles of oxygen in the system and its final temperature.

Q7(a): Work Done by Gas and on Gas

Consider an ideal gas enclosed in a cylinder with a movable piston. If the piston moves through a small distance dx under pressure P, and area of piston is A, then force exerted by gas is:

F = PA

Small work done by gas is:

dWby = F dx

Since:

dV = A dx

Therefore:

dWby = P dV

Total work done by gas is:

Wby = ∫ViVf P dV

Work done on gas is the negative of work done by gas:

Won = -∫ViVf P dV

Why Pressure Force Is Non-Conservative

A conservative force has work that depends only on initial and final states. But for gas expansion or compression, work depends on the path followed in the P-V diagram. For example, work in isothermal expansion differs from work in adiabatic expansion even between similar volume limits.

In a cyclic process:

∮ P dV ≠ 0

This means net work over a closed path can be non-zero. Therefore, pressure force in thermodynamics is non-conservative.

Final Result for Q7(a): Work done by gas is Wby=∫P dV, work done on gas is Won=-∫P dV, and pressure force is non-conservative because work depends on the thermodynamic path.

Q7(b): Intermolecular Forces

Intermolecular forces are forces of attraction or repulsion between molecules. They are weaker than chemical bonds but control many physical properties such as boiling point, viscosity, surface tension, melting point and real gas behavior.

Main Types of Intermolecular Forces

  1. London dispersion forces: Weak attractive forces caused by temporary dipoles. They exist in all molecules and atoms.
  2. Dipole-dipole forces: Attractive forces between permanent dipoles of polar molecules.
  3. Hydrogen bonding: A strong dipole-dipole interaction involving hydrogen bonded to highly electronegative atoms such as oxygen, nitrogen or fluorine.
  4. Ion-dipole forces: Attraction between an ion and a polar molecule, important in solutions.
  5. Short-range repulsive forces: At very small separations, electron clouds repel each other strongly.

In real gases, intermolecular attraction reduces effective pressure, while finite molecular volume reduces available volume. This leads to van der Waals equation:

(P + an²/V²)(V – nb) = nRT
Answer: Intermolecular forces include dispersion forces, dipole-dipole forces, hydrogen bonding, ion-dipole forces and short-range repulsive forces. They explain real gas behavior, surface tension, viscosity and phase changes.

Q7(c): Number of Moles and Final Temperature of Oxygen

Given initial data:

V1 = 1130 cm³ = 1130 × 10⁻⁶ m³ = 1.130 × 10⁻³ m³

T1 = 42°C = 42 + 273.15 = 315.15 K

P1 = 101 kPa = 101000 Pa

Use ideal gas equation:

P1V1 = nRT1

So:

n = P1V1/(RT1)

Substitute values:

n = (101000)(1.130 × 10⁻³)/(8.314 × 315.15)

n = 114.13/2620.7

n ≈ 0.0436 mol

Number of moles: n ≈ 0.0436 mol.

Now final data:

V2 = 1530 cm³ = 1.530 × 10⁻³ m³

P2 = 106 kPa = 106000 Pa

Using ideal gas equation again:

P2V2 = nRT2

So:

T2 = P2V2/(nR)

Substitute values:

T2 = (106000)(1.530 × 10⁻³)/(0.0436 × 8.314)

T2 = 162.18/0.3625

T2 ≈ 448 K

Convert to Celsius:

T2 = 448 – 273.15

T2 ≈ 175°C

Final Answer for Q7(c): Number of moles n ≈ 0.0436 mol. Final temperature T2 ≈ 448 K, or approximately 175°C.

Question 8: Short Notes

Full Question

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

(a) Kepler’s Law of Periods

(b) Michelson interferometer

(c) Young’s double slit experiment

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): Kepler’s Law of Periods

Kepler’s law of periods, also called Kepler’s third law, states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

T² ∝ a³

For a circular orbit, the gravitational force provides centripetal force:

GMm/r² = mv²/r

Cancel m:

GM/r² = v²/r

So:

v² = GM/r

For circular motion:

v = 2πr/T

Therefore:

(2πr/T)² = GM/r
4π²r²/T² = GM/r

Rearrange:

T² = 4π²r³/GM

Thus:

T² ∝ r³

For elliptical orbits, r is replaced by semi-major axis a:

T² = 4π²a³/GM
Note Summary: Kepler’s law of periods is T² ∝ a³. It follows from Newton’s law of gravitation and centripetal force.

Q8(b): Michelson Interferometer

The Michelson interferometer is an optical instrument used to split a light beam into two perpendicular beams and then recombine them to produce interference. It is used to measure small distances, wavelength differences, refractive indices and optical path changes.

Construction

A Michelson interferometer consists of:

  1. A monochromatic light source.
  2. A half-silvered glass plate or beam splitter.
  3. Two mirrors placed at right angles.
  4. A compensating plate.
  5. A telescope or screen for observing interference fringes.

Working

The beam splitter divides the incident light into two beams. One beam travels toward mirror M1, and the other travels toward mirror M2. After reflection, the two beams return and recombine. If there is a path difference between the two beams, interference fringes are formed.

The optical path difference is approximately:

Δ = 2(L1 – L2)

where L1 and L2 are the distances from beam splitter to the two mirrors.

If one mirror is moved by distance d, the optical path changes by 2d. If N fringes cross the field of view:

2d = Nλ

Therefore:

λ = 2d/N

Uses

  1. Measurement of wavelength of light.
  2. Measurement of very small length changes.
  3. Determination of refractive index.
  4. Study of spectral line separation.
  5. Historical role in the Michelson-Morley experiment.
Note Summary: Michelson interferometer splits and recombines two coherent beams to produce interference. A mirror displacement d gives path difference 2d, and wavelength can be found from λ=2d/N.

Q8(c): Young’s 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 slits act as coherent sources, and their waves overlap on a screen at distance D.

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

δ = d sinθ

For small angles:

sinθ ≈ tanθ = y/D

So:

δ = dy/D

Bright Fringes

Constructive interference occurs when:

δ = nλ

Therefore:

dy/D = nλ

y = nλD/d

Dark Fringes

Destructive interference occurs when:

δ = (n + 1/2)λ

Therefore:

y = (n + 1/2)λD/d

Fringe Width

The distance between two consecutive bright or dark fringes is:

β = λD/d

Importance

Young’s experiment proves that light has wave nature. It also provides a method for measuring wavelength of light and demonstrates the principle of superposition.

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

Revision Plan for CSS Physics Paper-I 2023 Solved

After reading this complete CSS Physics Paper-I 2023 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 gradient formula, acceleration components and curl calculation from memory.
Q3 Revise rotational kinetic energy and angular speeds of watch hands.
Q4 Explain relativity, length contraction and calculate antenna mass equivalent again.
Q5 Derive capillary rise and solve pressure at bottom of swimming pool.
Q6 Revise wave energy division, longitudinal wave velocity and CO oscillator work.
Q7 Derive gas work, explain non-conservative pressure force and solve oxygen final temperature.
Q8 Prepare any two notes but revise all three for safety.

Related Resources for CSS Physics Paper-I 2023 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 2023 Solved

What does CSS Physics Paper-I 2023 Solved include?

CSS Physics Paper-I 2023 Solved includes complete solved answers for Q2 to Q8 with definitions, derivations, numerical calculations, formulas and final answers.

What is the curl answer in Question 2?

For A = xz³i - 2x²zj + 2yz⁴k, the curl at (1,-1,1) is 4i + 3j - 4k.

What are the angular speeds of watch hands in Question 3?

The second hand has angular speed 0.1047 rad/s, the minute hand has 1.745 × 10⁻³ rad/s, and the hour hand has 1.454 × 10⁻⁴ rad/s.

What is the antenna mass equivalent in Question 4?

The mass equivalent of energy radiated by a 10 kW antenna for 24 hours is 9.6 × 10⁻⁹ kg, or about 9.6 micrograms.

What is the pressure at the bottom of the pool in Question 5?

The pressure at the bottom of the pool is approximately 1.144 × 10⁵ Pa, or 114.4 kPa absolute.

What is the work required to stretch CO molecule in Question 6?

The work required to stretch the CO molecule by 0.5 Å is approximately 5.12 × 10⁻¹⁸ J, or about 32 eV.

What are the number of moles and final temperature in Question 7?

The number of moles of oxygen is approximately 0.0436 mol, and the final temperature is approximately 448 K, or 175°C.

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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