CSS Physics Paper-I 2022 Solved

CSS Physics Paper-I 2022 Solved Question 2 belongs mainly to electromagnetism. 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. A particle of unit mass moves in potential V(x) = ax² + b/ x² where a & b are positive constants. Find the angular frequency o f small oscillations?

This part is answered from the standard result in electromagnetism. 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 V(x)=ax²+b/x², equilibrium: dV/dx=2ax-2b/x³=0 => x0=(b/a)^(1/4). This is the substitution stage, where the general relation is converted into the present problem.
2. Line 2: V”(x)=2a+6b/x^4; at x0, b/x0^4=a, so V”=8a. This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: For unit mass, omega = sqrt(8a). This is the substitution stage, where the general relation is converted into the present problem.

Answer to Part (b)

Part being solved: A hollow spherical shell carries charge density
=  in region ≤ ≤ . Find the electric field in three regions (i)  < (ii) < < (iii)  > .

For a charged spherical shell with volume charge density ρ between radii a and b plus a line charge λ on the axis, use Gauss law. In each region E(2πrL)=λ_enclosed/ε0 for cylindrical symmetry, and for a purely spherical charge use E4πr²=Q_enclosed/ε0.

Electric field is force per unit positive test charge: E = F/q0. For a point charge q, Coulomb’s law gives E = (1/4 pi epsilon0) q r-hat/r squared. The direction is away from a positive charge and toward a negative charge.

The field concept is useful because it separates the source charge from the test charge. Once E is known at a point, the force on any charge q0 placed there is simply F = q0 E.

Working Block 1

1. Line 1: Range R = u² sin2θ/g, maximum height H = u² sin²θ/(2g). This line is kept visible because it is the algebraic bridge to the final result.
2. Line 2: R = 3H => sin2θ = (3/2)sin²θ => 2sinθ cosθ = 1.5sin²θ => tanθ = 4/3. This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: θ = 53.13°. This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (c)

Part being solved: A projectile is fired in such a way that its hor izontal range is equal to three times its maximum height. Determine its angle of project ion. (08) (07) (05) (20)

This part is answered from the standard result in electromagnetism. 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. For V(x)=ax²+b/x², equilibrium: dV/dx=2ax-2b/x³=0 => x0=(b/a)^(1/4).
2. V”(x)=2a+6b/x^4; at x0, b/x0^4=a, so V”=8a.
3. For unit mass, omega = sqrt(8a).

Working Block 2

1. Range R = u² sin2θ/g, maximum height H = u² sin²θ/(2g).
2. R = 3H => sin2θ = (3/2)sin²θ => 2sinθ cosθ = 1.5sin²θ => tanθ = 4/3.
3. θ = 53.13°.

CSS Physics Paper-I 2022 Solved Question 3 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 Part (a)

Part being solved: Q. 3. Assume that a star has uniform density. Show that the gravitational pressure P is proportional to V-3/4 where V is volume.

For a uniform-density star, hydrostatic equilibrium dP/dr = -Gm(r)ρ/r² gives central pressure proportional to GM²/R⁴. Since V ∝ R³, pressure varies as R^-4 = V^-4/3; if the question expects gravitational pressure scaling under a different definition, state assumptions clearly.

Answer to Part (b)

Part being solved: Derive expressions for potential and electric field associated with point charge located near an infinite grounded conducting plane.

For a point charge near an infinite grounded conducting plane, replace the plane by an image charge -q at the mirror point. The potential is V = (1/4πε0)[q/r1 – q/r2].

Electric field is force per unit positive test charge: E = F/q0. For a point charge q, Coulomb’s law gives E = (1/4 pi epsilon0) q r-hat/r squared. The direction is away from a positive charge and toward a negative charge.

The field concept is useful because it separates the source charge from the test charge. Once E is known at a point, the force on any charge q0 placed there is simply F = q0 E.

Answer to Part (c)

Part being solved: Determine equation of motion of masses attached to the string of at-wood machine by Lagrangian methods. (08) (07) (05) (20)

For Atwood machine with masses m1 and m2, choose coordinate x. L = 0.5(m1+m2)x_dot² – (m2-m1)gx. Euler-Lagrange gives acceleration a=(m2-m1)g/(m1+m2).

CSS Physics Paper-I 2022 Solved Question 4 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. 4. Q cm 3 of water flows per second through a horizontal tub e of uniform bore of radius r & of length L. Another tube of half the l ength but radius 2r is connected in parallal to same pressure head. What will be the total quantity of water flowing / sec through these two tubes?

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: Poiseuille: Q ∝ r^4/L. This line is kept visible because it is the algebraic bridge to the final result.
2. Line 2: Tube 1: Q. This statement sets the physical condition used by the next line.
3. Line 3: Tube 2: length L/2 and radius 2r => Q2/Q = (2r)^4/(L/2) × L/r^4 = 16×2 = 32. This line is kept visible because it is the algebraic bridge to the final result.
4. Line 4: Total flow = 33Q. This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (b)

Part being solved: A linear quadruple is an arrangement of a system of charges which consist of −2 at the origin and + at the two point (±,0,0). Show that at distances much greater than (..  ≫ ), the potential may be written in the approximate form =  4 !(3#$% &−1 ),  ≫ 

For charges +q at ±a and -2q at origin, monopole and dipole terms vanish. The far potential is V = qa²(3cos²θ-1)/(4πε0 r³), the quadrupole term.

Answer to Part (c)

Part being solved: Two soap bubbles with radii r1 and r2 coalescs to f orm a bigger bubble of radii r. Show that r = ( r1² +r2² )1/2. (08) (07) (05) (20)

Soap-bubble excess pressure is 4T/r. On coalescence, surface energy conservation gives r² = r1² + r2², so r = sqrt(r1²+r2²).

Calculation and Derivation Written Line by Line

Working Block 1

1. Poiseuille: Q ∝ r^4/L.
2. Tube 1: Q.
3. Tube 2: length L/2 and radius 2r => Q2/Q = (2r)^4/(L/2) × L/r^4 = 16×2 = 32.
4. Total flow = 33Q.

CSS Physics Paper-I 2022 Solved Question 5 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. 5. Explain wave function. Derive wave formula and exp lain phase and group velocity.

Wave function ψ contains probability amplitude. The phase velocity is vp=ω/k, group velocity vg=dω/dk; for matter waves, group velocity equals particle velocity.

The wave function psi contains the quantum state of a particle. It is generally complex and is not itself directly observable, but |psi| squared gives probability density.

A physically acceptable wave function must be finite, single-valued, continuous where required and normalizable. Normalization means the total probability of finding the particle somewhere is one.

Answer to Part (b)

Part being solved: Two semi-infinite grounded metal plates parallel t o each other and to the xz- plane are located at ( =0 and ( = planes, respectively. The left ends of these two plates at ) =0, are closed off by a strip of width and extend to infinity in the z-direction. The strip is insulate d from both the plates and is maintained at a specific potential ((). Find the potential distribution in the slot.

For two grounded plates at y=0 and y=a with end strip potential V(y), solve Laplace equation by separation: Φ(x,y)=Σ An sin(nπy/a)e^{-nπx/a}. Coefficients come from the Fourier sine series of V(y).

Working Block 1

1. Line 1: Two-level system: E0=0, degeneracy g0=1; E1=E, degeneracy g1=3. This line is kept visible because it is the algebraic bridge to the final result.
2. Line 2: Z = 1 + 3e^{-E/kT} This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: < E > = 3E e^{-E/kT}/(1+3e^{-E/kT}). This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (c)

Part being solved: A two level system has energies 0& E. The level wi th zero energy is non- degenerate while the level with energy E is triply degenerate. Find the mean energy of a classical particle in this system at t emperature T. (08) (07) (05) (20)

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.

Calculation and Derivation Written Line by Line

Working Block 1

1. Two-level system: E0=0, degeneracy g0=1; E1=E, degeneracy g1=3.
2. Z = 1 + 3e^{-E/kT}
3. < E > = 3E e^{-E/kT}/(1+3e^{-E/kT}).

CSS Physics Paper-I 2022 Solved Question 6 belongs mainly to nuclear 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. 6. Explain the particle in finite potential well with all possible cases and solutions and make a comparison with infinite pote ntial well.

In a finite potential well, bound states have sinusoidal wave functions inside and exponentially decaying tails outside. Energies are discrete but fewer than in an infinite well.

The time-independent Schrodinger equation is an energy eigenvalue equation. It is called time independent because the potential does not explicitly depend on time and the time factor can be separated from the spatial wave function.

For a particle in a one-dimensional infinite well, the wave function must vanish at the walls. These boundary conditions allow only selected wavelengths and therefore selected energies.

Working Block 1

1. Line 1: For u(x)=u0 x, x>0, classical probability ∝ e^{-u0x/kT}. This is the substitution stage, where the general relation is converted into the present problem.
2. Line 2: Mean x = ∫x e^{-αx}dx / ∫e^{-αx}dx = 1/α = kT/u0. This line is kept visible because it is the algebraic bridge to the final result.

Answer to Part (b)

Part being solved: The potential (&) is specified on the surface of a hollow sphere, of radius

Potential inside a sphere with boundary V(R,θ,φ) is expanded in spherical harmonics: V(r,θ,φ)=Σ A_lm r^l Y_lm(θ,φ). Coefficients follow from boundary values.

Calculation and Derivation Written Line by Line

Working Block 1

1. For u(x)=u0 x, x>0, classical probability ∝ e^{-u0x/kT}.
2. Mean x = ∫x e^{-αx}dx / ∫e^{-αx}dx = 1/α = kT/u0.

CSS Physics Paper-I 2022 Solved Question 7 belongs mainly to nuclear 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. 7. When a gas expands adiabatically its volume is doub led while its absolute temperature is decreased by a factor 1.32. Compute number of degree of freedom of gas molecule?

This part is answered from the standard result in nuclear 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: Adiabatic: TV^(gamma-1)=constant. This line is kept visible because it is the algebraic bridge to the final result.
2. Line 2: When V2=2V1 and T2=T1/1.32: This line is kept visible because it is the algebraic bridge to the final result.
3. Line 3: 1/1.32 = 2^(1-gamma) => gamma-1 = ln(1.32)/ln2 ≈ 0.400. This line is kept visible because it is the algebraic bridge to the final result.
4. Line 4: For molecules f = 2/(gamma-1) ≈ 5. This is the substitution stage, where the general relation is converted into the present problem.

Answer to Part (b)

Part being solved: State and prove Ampere’s Law.

Ampere law: ∮B·dl = μ0 I_enclosed. It follows from magnetostatic symmetry and is used for long wires, solenoids and toroids.

Ampere’s law states that the line integral of magnetic field around a closed path equals mu0 times the current enclosed: integral B dot dl = mu0 I_enclosed. With Maxwell’s correction, displacement current is also included.

The law is most useful in symmetric situations such as long straight wires, solenoids and toroids, where B is constant on a suitable Amperian path.

Answer to Part (c)

Part being solved: Find the rms speed of oxygen molecules at O0c? (08) (07) (05) (20)

This part is answered from the standard result in nuclear 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: O2 rms speed at 0 C: This statement sets the physical condition used by the next line.
2. Line 2: v_rms = sqrt(3RT/M) = sqrt(3×8.314×273.15/0.032) ≈ 461 m/s. 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. Adiabatic: TV^(gamma-1)=constant.
2. When V2=2V1 and T2=T1/1.32:
3. 1/1.32 = 2^(1-gamma) => gamma-1 = ln(1.32)/ln2 ≈ 0.400.
4. For molecules f = 2/(gamma-1) ≈ 5.

Working Block 2

1. O2 rms speed at 0 C:
2. v_rms = sqrt(3RT/M) = sqrt(3×8.314×273.15/0.032) ≈ 461 m/s.

CSS Physics Paper-I 2022 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. An ensemble of non-interacting spin -1/2 particles is in contact with a heat bath at temperature T & is subjected to an externa l magnetic field. Each particle can be in one of the two quantum states o f energies ε0. If the mean energy per particle is –ε0/2, then find free energ y per particle?

For spin-1/2 particles with energies ±ε0, mean energy U = -ε0 tanh(ε0/kT). If U=-ε0/2, tanh(ε0/kT)=1/2. The free energy is F=-kT ln[2 cosh(ε0/kT)].

Answer to Part (b)

Part being solved: Derive the electromagnetic wave equation in vacuum and also describe the properties of monochromatic electromagnetic waves.

In vacuum, Maxwell equations give ∇²E = μ0ε0 ∂²E/∂t² and ∇²B = μ0ε0 ∂²B/∂t². Monochromatic waves are transverse, travel at c, and have E/B=c.

Answer to Part (c)

Part being solved: Discuss adiabatic demagnetization using TDS equatio ns mathematically in detail? (08) (07) (05) (20)

Adiabatic demagnetization cools paramagnetic salts by magnetizing isothermally and demagnetizing adiabatically, using the magnetic entropy relation dS=0.