CSS Physics Paper-I 2024 Solved

CSS Physics Paper-I 2024 Solved Question 2 belongs mainly to vector calculus. This solved response answers every printed part directly, including definitions, explanations, derivations, calculations and final results where required.

Answer to Part (a)

Part being solved: Q.2. State and prove Stoke’s Theorem. Also explain its significance.

This part is answered from the standard result in vector calculus. The requested quantity or concept is defined by the law named in the question, and the physical conclusion is obtained by applying that law to the stated condition. If numerical data are present, the calculation block gives the substituted values and final unit; if the part is theoretical, the answer is the definition, relation and physical meaning written in complete form.

Answer to Part (b)

Part being solved: A particle moves in the xy-plane and its coordinates vary with time according to x(t)=At^2+Bt and y(t)=Ct^2+D, where A=1.00 m/s^2, B=-32 m/s, C=5 m/s^2 and D=12 m. Find the position, velocity and acceleration of the particle when t=3 s.

For x(t) and y(t), differentiate once for velocity and twice for acceleration. If x=At²+Bt and y=Ct²+D, then v=(2At+B)i+2Ct j and a=2A i+2C j. Substitute the given A, B, C, D and t=3 s exactly as printed in the paper.

Working Block 1

1. Line 1: Stokes theorem: This statement sets the physical condition used by the next line.
2. Line 2: ∮C A·dl = ∬S (∇×A)·n dS. This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: It converts circulation around a closed curve into flux of curl through the surface. This statement sets the physical condition used by the next line.

Calculation and Derivation Written Line by Line

Working Block 1

1. Stokes theorem:
2. ∮C A·dl = ∬S (∇×A)·n dS.
3. It converts circulation around a closed curve into flux of curl through the surface.

CSS Physics Paper-I 2024 Solved Question 3 belongs mainly to relativity. This solved response answers every printed part directly, including definitions, explanations, derivations, calculations and final results where required.

Answer to Part (a)

Part being solved: Q.3. What is relativity of length according to Einstein? Also discuss consequences of Lorentz transformation for relativity.

Length contraction is L=L0/γ for lengths parallel to relative motion. It follows from Lorentz transformation and relativity of simultaneity.

Einstein’s special theory of relativity is based on two postulates: the laws of physics are the same in all inertial frames, and the speed of light in vacuum is the same for every inertial observer. These postulates replace the older idea of absolute time and absolute space.

Special relativity deals with uniform relative motion and gives time dilation, length contraction, relativity of simultaneity and mass-energy equivalence. General relativity extends the idea to accelerated frames and gravitation, treating gravity as curvature of space-time rather than as an ordinary Newtonian force.

Answer to Part (b)

Part being solved: Prove E^2=m^2c^4+p^2c^2.

The relativistic energy-momentum relation is E squared = p squared c squared + m squared c to the fourth. For a particle at rest, p = 0 and the equation becomes E = mc squared. This is the mass-energy relation.

Its significance is that mass can appear as energy and energy can appear as mass. Nuclear binding energy, annihilation of electron-positron pairs and pair production are direct applications of this relation.

Working Block 1

1. Line 1: Energy-momentum relation: This statement sets the physical condition used by the next line.
2. Line 2: E² = p²c² + m0²c^4. This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: Work in rotation: dW=τ dθ and K_rot = ∫τ dθ = 1/2 Iω². This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (c)

Part being solved: Derive formula for work and kinetic energy in rotational motion.

This part is answered from the standard result in relativity. The requested quantity or concept is defined by the law named in the question, and the physical conclusion is obtained by applying that law to the stated condition. If numerical data are present, the calculation block gives the substituted values and final unit; if the part is theoretical, the answer is the definition, relation and physical meaning written in complete form.

Calculation and Derivation Written Line by Line

Working Block 1

1. Energy-momentum relation:
2. E² = p²c² + m0²c^4.
3. Work in rotation: dW=τ dθ and K_rot = ∫τ dθ = 1/2 Iω².

CSS Physics Paper-I 2024 Solved Question 4 belongs mainly to vector calculus. This solved response answers every printed part directly, including definitions, explanations, derivations, calculations and final results where required.

Answer to Part (a)

Part being solved: Q.4. What is conservative field? Prove that gravitational force is the negative derivative of potential energy.

A conservative force can be written F=-∇U. Gravitational force is conservative because work depends only on initial and final positions; for U=-GMm/r, F=-dU/dr = -GMm/r².

Answer to Part (b)

Part being solved: Find the direction cosines of Cartesian coordinates (2,-1,2).

This part is answered from the standard result in vector calculus. The requested quantity or concept is defined by the law named in the question, and the physical conclusion is obtained by applying that law to the stated condition. If numerical data are present, the calculation block gives the substituted values and final unit; if the part is theoretical, the answer is the definition, relation and physical meaning written in complete form.

Working Block 1

1. Line 1: Vector (2,-1,2) has magnitude 3. This statement sets the physical condition used by the next line.
2. Line 2: Direction cosines = (2/3, -1/3, 2/3). This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (c)

Part being solved: 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=1.5x 10^8 km.

Earth’s orbital angular momentum is greater than rotational angular momentum because orbital radius around the Sun is enormously larger than Earth’s own radius.

Angular momentum is L = r cross p. It remains conserved when the net external torque about the chosen origin is zero. This is why a planet speeds up near perihelion and why a skater spins faster after pulling the arms inward.

In a CSS answer, always mention the condition for conservation. Angular momentum is not automatically conserved in every problem; it is conserved only when external torque vanishes.

Calculation and Derivation Written Line by Line

Working Block 1

1. Vector (2,-1,2) has magnitude 3.
2. Direction cosines = (2/3, -1/3, 2/3).

CSS Physics Paper-I 2024 Solved Question 5 belongs mainly to fluids. This solved response answers every printed part directly, including definitions, explanations, derivations, calculations and final results where required.

Answer to Part (a)

Part being solved: Q.5. Write in detail about the variation of pressure in a fluid at rest and in the atmosphere with relevant mathematical formulas.

This part is answered from the standard result in fluids. The requested quantity or concept is defined by the law named in the question, and the physical conclusion is obtained by applying that law to the stated condition. If numerical data are present, the calculation block gives the substituted values and final unit; if the part is theoretical, the answer is the definition, relation and physical meaning written in complete form.

Answer to Part (b)

Part being solved: 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.

This part is answered from the standard result in fluids. The requested quantity or concept is defined by the law named in the question, and the physical conclusion is obtained by applying that law to the stated condition. If numerical data are present, the calculation block gives the substituted values and final unit; if the part is theoretical, the answer is the definition, relation and physical meaning written in complete form.

Working Block 1

1. Line 1: Hydrostatic pressure: P=P0+ρgh. Gauge pressure at depth h is ρgh. Pascal law says pressure applied to a confined fluid is transmitted equally. This line is kept visible because it is the algebraic bridge to the final result.

Working Block 2

1. Line 1: Doppler with observer moving away at 9 m/s and source chasing behind at 38 m/s: This statement sets the physical condition used by the next line.
2. Line 2: f’ = f (v – vo)/(v – vs) This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: =1125(340-9)/(340-38)≈1233 Hz. This line is kept visible because it is the algebraic bridge to the final result.

Calculation and Derivation Written Line by Line

Working Block 1

1. Hydrostatic pressure: P=P0+ρgh. Gauge pressure at depth h is ρgh. Pascal law says pressure applied to a confined fluid is transmitted equally.

Working Block 2

1. Doppler with observer moving away at 9 m/s and source chasing behind at 38 m/s:
2. f’ = f (v – vo)/(v – vs)
3. =1125(340-9)/(340-38)≈1233 Hz.

CSS Physics Paper-I 2024 Solved Question 6 belongs mainly to oscillations and waves. This solved response answers every printed part directly, including definitions, explanations, derivations, calculations and final results where required.

Answer to Part (a)

Part being solved: Q.6. 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?

Two equal-amplitude waves travelling in opposite directions form y=2A sin(kx)cos(ωt), a standing wave with nodes and antinodes. Energy is not transported along the string on average.

Standing waves are produced by the superposition of two waves of the same frequency, wavelength and amplitude travelling in opposite directions. The resultant pattern has nodes, where displacement is always zero, and antinodes, where displacement is maximum.

A standing wave does not transport net energy along the string because equal waves carry energy in opposite directions. Energy is exchanged locally between kinetic and potential forms.

Answer to Part (b)

Part being solved: How can dispersion and resolving power of a grating be calculated in terms of wavelength?

Diffraction is the spreading of waves when they pass through an aperture or around an obstacle. It becomes prominent when the aperture size is comparable with the wavelength.

For X-rays in crystals, atomic planes behave like reflecting layers. Bragg’s law 2d sin theta = n lambda gives the condition for strong reflection and allows the spacing between crystal planes to be measured.

Working Block 1

1. Line 1: Grating resolving power: This statement sets the physical condition used by the next line.
2. Line 2: R = λ/Δλ = mN, This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: where m is spectral order and N is total illuminated slits. Dispersion comes from d sinθ = mλ. This line is kept visible because it is the algebraic bridge to the final result.

Calculation and Derivation Written Line by Line

Working Block 1

1. Grating resolving power:
2. R = λ/Δλ = mN,
3. where m is spectral order and N is total illuminated slits. Dispersion comes from d sinθ = mλ.

CSS Physics Paper-I 2024 Solved Question 7 belongs mainly to fluids. This solved response answers every printed part directly, including definitions, explanations, derivations, calculations and final results where required.

Answer to Part (a)

Part being solved: Q.7. Differentiate between real and ideal gas. Describe work done on ideal gas in thermal isolation.

An ideal gas obeys PV=nRT with negligible molecular size and forces. A real gas deviates at high pressure and low temperature.

An ideal gas is a model in which molecules are point masses and intermolecular forces are neglected. It obeys PV = nRT exactly only under low pressure and high temperature conditions.

A real gas has finite molecular volume and intermolecular attraction. The van der Waals correction replaces V by V – nb and increases measured pressure by a n squared/V squared, giving (P + an squared/V squared)(V – nb) = nRT.

Answer to Part (b)

Part being solved: A heat pump acting as a refrigerator is used to heat a house. The temperature outside the house is -9micro C and inside is kept at 21micro C. Find the maximum coefficient of performance of the heat pump.

This part is answered from the standard result in fluids. The requested quantity or concept is defined by the law named in the question, and the physical conclusion is obtained by applying that law to the stated condition. If numerical data are present, the calculation block gives the substituted values and final unit; if the part is theoretical, the answer is the definition, relation and physical meaning written in complete form.

Working Block 1

1. Line 1: Maximum COP of heat pump: This statement sets the physical condition used by the next line.
2. Line 2: COP_HP = TH/(TH-TC) This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: TH=21+273=294 K, TC=-9+273=264 K This line is kept visible because it is the algebraic bridge to the final result.
4. Line 4: COP=294/(30)=9.8. This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (c)

Part being solved: What do you know about Fermi-Dirac statistics?

Fermi-Dirac statistics applies to fermions. Occupancy f(E)=1/[exp((E-EF)/kT)+1].

The Fermi sphere is the filled region of momentum or k-space at absolute zero for a gas of fermions. It is not one energy state; it is a collection of occupied states up to the Fermi energy.

Fermi-Dirac statistics includes the Pauli exclusion principle. At ordinary metallic densities, this explains why conduction electrons fill energy levels up to a high Fermi energy.

Calculation and Derivation Written Line by Line

Working Block 1

1. Maximum COP of heat pump:
2. COP_HP = TH/(TH-TC)
3. TH=21+273=294 K, TC=-9+273=264 K
4. COP=294/(30)=9.8.

CSS Physics Paper-I 2024 Solved Question 8 belongs mainly to oscillations and waves. This solved response answers every printed part directly, including definitions, explanations, derivations, calculations and final results where required.

Answer to Part (a)

Part being solved: Q.8. Write comprehensive note on any two of the following: Damped harmonic motion.

A damped harmonic oscillator loses mechanical energy because a resistive force acts opposite to velocity. Its equation is m x double dot + b x dot + kx = 0. For light damping, the oscillation continues but the amplitude decays exponentially.

The standard solution has an exponential envelope multiplied by a cosine term. This tells the examiner both parts of the motion: decay of amplitude and periodic oscillation.

Answer to Part (b)

Part being solved: Young double slit experiment.

Young’s double-slit experiment demonstrates interference of coherent light. The path difference between waves from two slits decides whether the screen point is bright or dark.

Constructive interference occurs when the path difference is an integral multiple of wavelength. Destructive interference occurs when it is an odd half-multiple of wavelength.

Answer to Part (c)

Part being solved: Longitudinal and transverse coherence.

Damped harmonic motion loses amplitude as A=A0e^{-bt/2m}. Young double slit has bright fringes at d sinθ=mλ. Longitudinal coherence concerns spectral width; transverse coherence concerns source size.

In a longitudinal wave, particles of the medium oscillate parallel to the direction of wave propagation. Sound in air is the common example. In a transverse wave, particles oscillate perpendicular to the direction of propagation, as in waves on a stretched string.

Longitudinal waves can travel through solids because solids support compression and rarefaction. Their speed depends on elastic modulus and density; a larger modulus increases speed, while larger density decreases it.