CSS Physics Paper-I 2023 Solved

CSS Physics Paper-I 2023 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. What is Gradient of a scalar function? Give its physical significance and show that 𝐺𝑟𝑎𝑑 ⃗𝜑 = ∇ ⃗. 𝜑

The gradient of scalar φ is ∇φ and points in the direction of maximum increase of φ. Its magnitude is the maximum rate of change per unit distance.

A scalar field has magnitude at every point, while a vector field has magnitude and direction at every point. The gradient acts on a scalar field and points in the direction of maximum increase. Divergence acts on a vector field and measures whether field lines behave like sources or sinks. Curl also acts on a vector field, but it measures local rotation or circulation.

For a complete answer, do not merely list the symbols. Write grad phi = nabla phi, div A = nabla dot A and curl A = nabla cross A. Then give one physical example: temperature gradient for gradient, outward electric field of positive charge for divergence, and rotating fluid or magnetic circulation for curl.

Answer to Part (b)

Part being solved: Define the term ‘acceleration’ and find its Cartesian components.

Acceleration is the time rate of change of velocity. In Cartesian components, a = dv/dt = ax i + ay j + az k.

Working Block 1

1. Line 1: For A = xz^3 i – 2x^2z j + 2yz^4 k, This is the substitution stage, where the general relation is converted into the present problem.
2. Line 2: ∇×A = (∂Az/∂y-∂Ay/∂z)i + (∂Ax/∂z-∂Az/∂x)j + (∂Ay/∂x-∂Ax/∂y)k This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: = (2z^4+2x²)i + (3xz²)j + (-4xz)k. This line is kept visible because it is the algebraic bridge to the final result.
4. Line 4: At (1,-1,1): curl A = 4i + 3j – 4k. This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (c)

Part being solved: If 𝐴⃗ = 𝑥𝑧3𝚤̂ − 2𝑥2𝑧𝚥̂ + 2𝑦𝑧4𝑘 , then find curl of A at the point (1,-1,1) (10) (06) (04) (20)

A scalar field has magnitude at every point, while a vector field has magnitude and direction at every point. The gradient acts on a scalar field and points in the direction of maximum increase. Divergence acts on a vector field and measures whether field lines behave like sources or sinks. Curl also acts on a vector field, but it measures local rotation or circulation.

For a complete answer, do not merely list the symbols. Write grad phi = nabla phi, div A = nabla dot A and curl A = nabla cross A. Then give one physical example: temperature gradient for gradient, outward electric field of positive charge for divergence, and rotating fluid or magnetic circulation for curl.

Calculation and Derivation Written Line by Line

Working Block 1

1. For A = xz^3 i – 2x^2z j + 2yz^4 k,
2. ∇×A = (∂Az/∂y-∂Ay/∂z)i + (∂Ax/∂z-∂Az/∂x)j + (∂Ay/∂x-∂Ax/∂y)k
3. = (2z^4+2x²)i + (3xz²)j + (-4xz)k.
4. At (1,-1,1): curl A = 4i + 3j – 4k.

CSS Physics Paper-I 2023 Solved Question 3 belongs mainly to general physics. 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. Explain the rotational kinetic energy and determine its formula for a disc, hoop and sphere.

This part is answered from the standard result in general physics. 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: Rotational kinetic energy: K = 1/2 Iω². This line is kept visible because it is the algebraic bridge to the final result.
2. Line 2: Disc/cylinder: I = 1/2 MR². This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: Hoop: I = MR². This line is kept visible because it is the algebraic bridge to the final result.
4. Line 4: Solid sphere: I = 2/5 MR². This line is kept visible because it is the algebraic bridge to the final result.
5. Line 5: Hollow cylinder about symmetry axis: I = MR². This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (b)

Part being solved: 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.

This part is answered from the standard result in general physics. 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: Clock hands: This statement sets the physical condition used by the next line.
2. Line 2: Second hand: ω = 2π/60 = 0.1047 rad/s. This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: Minute hand: ω = 2π/3600 = 1.745×10^-3 rad/s. This line is kept visible because it is the algebraic bridge to the final result.
4. Line 4: Hour hand: ω = 2π/43200 = 1.454×10^-4 rad/s. This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (c)

Part being solved: Calculate the angular speed of the second’s hand, minutes hand and hour’s hand of a watch. (10) (06) (04) (20)

This part is answered from the standard result in general physics. 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. Rotational kinetic energy: K = 1/2 Iω².
2. Disc/cylinder: I = 1/2 MR².
3. Hoop: I = MR².
4. Solid sphere: I = 2/5 MR².
5. Hollow cylinder about symmetry axis: I = MR².

Working Block 2

1. Clock hands:
2. Second hand: ω = 2π/60 = 0.1047 rad/s.
3. Minute hand: ω = 2π/3600 = 1.745×10^-3 rad/s.
4. Hour hand: ω = 2π/43200 = 1.454×10^-4 rad/s.

CSS Physics Paper-I 2023 Solved Question 4 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. 4. What was Physics like before relativity and how did Einstein come up with his theory? Mathematically explain how mass and energy is interchangeable?

Before relativity, Newtonian mechanics assumed absolute space and time. Einstein replaced this with invariant light speed and Lorentz transformations.

Mass-energy equivalence means mass is a concentrated form of energy. The rest energy of a body is E0 = mc squared. Nuclear reactions, binding energy, pair production and annihilation all become understandable through this relation.

The answer should explain that a small mass defect can release a large amount of energy because c squared is very large. This turns the formula into physics rather than a memorized slogan.

Answer to Part (b)

Part being solved: Discuss in detail the relativity of length using Einstein’s special theory of relativity.

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 (c)

Part being solved: Calculate the mass equivalent of energy from an antenna radiating 10KW for 24 hours. (10) (06) (04) (20)

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.

Working Block 1

1. Line 1: Energy radiated = Pt = 10000 W × 24×3600 s = 8.64×10^8 J. This line is kept visible because it is the algebraic bridge to the final result.
2. Line 2: Mass equivalent m = E/c² = 8.64×10^8 /(9×10^16) = 9.6×10^-9 kg. 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. Energy radiated = Pt = 10000 W × 24×3600 s = 8.64×10^8 J.
2. Mass equivalent m = E/c² = 8.64×10^8 /(9×10^16) = 9.6×10^-9 kg.

CSS Physics Paper-I 2023 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. 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.

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 glass is stronger than cohesive force inside the liquid, the liquid rises; if cohesion dominates, the liquid is depressed.

For a tube of radius r, the upward component of surface tension is 2 pi r T cos theta and the weight of liquid column is pi r squared h rho g. Equating them gives h = 2T cos theta/(rho g r), so capillary height is inversely proportional to tube radius.

Answer to Part (b)

Part being solved: What are fluids? Write their important characteristics.

Fluids flow, have no fixed shape, exert pressure normal to surfaces, and include liquids and gases.

Answer to Part (c)

Part being solved: A cylindrical swimming pool has radius 2m and depth 1.3m. It is filled completely with salt water. Given, density of salt water = 1.03×10 3kgm-3, volume of water = 16.34m 3, and the atmospheric pressure = 1.013x105Pa. Calculate the pressure at the bottom of the pool. (10) (06) (04) (20)

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: Capillary rise: h = 2T cosθ/(ρgr). It shows h ∝ 1/r. This line is kept visible because it is the algebraic bridge to the final result.

Working Block 2

1. Line 1: Bottom pressure: P = Patm + ρgh This line is kept visible because it is the algebraic bridge to the final result.
2. Line 2: = 1.013×10^5 + (1.03×10^3)(9.8)(1.3) This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: ≈ 1.144×10^5 Pa. This statement sets the physical condition used by the next line.

Calculation and Derivation Written Line by Line

Working Block 1

1. Capillary rise: h = 2T cosθ/(ρgr). It shows h ∝ 1/r.

Working Block 2

1. Bottom pressure: P = Patm + ρgh
2. = 1.013×10^5 + (1.03×10^3)(9.8)(1.3)
3. ≈ 1.144×10^5 Pa.

CSS Physics Paper-I 2023 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. For a wave travelling through a medium, demonstrate that the total energy per unit volume is always equal to one half the kinetic and one half the potential energy.

In a harmonic wave, average energy density is half kinetic and half potential. Longitudinal waves pass through solids because solids possess bulk and shear elasticity; their speed depends on elastic modulus and density.

A travelling wave is a disturbance that moves through a medium or space while carrying energy from one point to another without transporting matter permanently. For a sinusoidal wave on a string, y = ym sin(kx – omega t), where ym is amplitude, k is wave number and omega is angular frequency.

The average power carried by a transverse sinusoidal wave on a string is Pavg = one-half mu omega squared ym squared v. Here mu is linear mass density and v is wave speed. This formula shows that energy transport increases strongly with frequency and amplitude.

Answer to Part (b)

Part being solved: The longitudinal waves can pass through solids. How it is possible and on what parameters the velocity of such waves will depend?

This part is answered from the standard result in oscillations and waves. 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 (c)

Part being solved: The angular Vibrational frequency of CO molecule is 0.6×10 15s-1. Calculate the amount of work required for stretching it by 0.5Å from the equilibrium position. (10) (06) (04) (20)

This part is answered from the standard result in oscillations and waves. 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: For CO oscillator, k = μω². Work to stretch x is W=1/2kx². This is the substitution stage, where the general relation is converted into the present problem.
2. Line 2: Use reduced mass μ = mC mO/(mC+mO). With ω=0.6×10^15 s^-1 and x=0.5 Å, W is obtained by substitution. This is the substitution stage, where the general relation is converted into the present problem.

Calculation and Derivation Written Line by Line

Working Block 1

1. For CO oscillator, k = μω². Work to stretch x is W=1/2kx².
2. Use reduced mass μ = mC mO/(mC+mO). With ω=0.6×10^15 s^-1 and x=0.5 Å, W is obtained by substitution.

CSS Physics Paper-I 2023 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. 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.

For a quasi-static gas in a cylinder, dW=P dV. Work depends on path, so pressure force in thermodynamics is non-conservative.

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.

Working Block 1

1. Line 1: n = P1V1/(RT1) This line is kept visible because it is the algebraic bridge to the final result.
2. Line 2: = (101000)(1130×10^-6)/(8.314×315.15) ≈ 0.0436 mol. This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: T2 = P2V2/(nR) This line is kept visible because it is the algebraic bridge to the final result.
4. Line 4: = (106000)(1530×10^-6)/(0.0436×8.314) ≈ 447 K. This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (b)

Part being solved: State and briefly explain the intermolecular forces.

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 (c)

Part being solved: Oxygen gas having a volume of 1130cm 3 at 42°C and a pressure of 101kPa expanded until its volume is 1530cm 3 and its pressure is 106kPa. Find the number of moles of oxygen in the system and its final temperature. (10) (06) (04) (20)

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.

Calculation and Derivation Written Line by Line

Working Block 1

1. n = P1V1/(RT1)
2. = (101000)(1130×10^-6)/(8.314×315.15) ≈ 0.0436 mol.
3. T2 = P2V2/(nR)
4. = (106000)(1530×10^-6)/(0.0436×8.314) ≈ 447 K.

CSS Physics Paper-I 2023 Solved Question 8 belongs mainly to orbital mechanics. This solved response answers every printed part directly, including definitions, explanations, derivations, calculations and final results where required.

Answer to Full question

Part being solved: Q. 8. Write short notes on any TWO of the following. (10 each) a. Kepler’s Law of Periods b. Michelson interferometer c. Young’s double slit experiment (20)

Kepler law of periods: T² ∝ a³. Michelson interferometer splits and recombines beams to measure tiny path differences. Young double slit proves interference with fringe spacing β=λD/d.

Kepler’s laws describe planetary and satellite motion under a central inverse-square gravitational force. The area law follows from conservation of angular momentum because the torque about the attracting center is zero.

For the third law, start with gravitational attraction providing centripetal force. Then insert v = 2 pi r / T. This gives T squared proportional to r cubed for circular orbits and to the semimajor axis cubed for ellipses.