In the NaCl crystal structure, each unit cell contains 4 Na⁺ ions and 4 Cl⁻ ions. These ions are arranged in a face-centered cubic (FCC) structure, where each corner of the unit cell is shared by 8 adjacent cells, and each face-centered ion is shared by 2 adjacent cells.

Thus, the total number of Na and Cl ions per unit cell is:

4 (Na⁺) + 4 (Cl⁻) = 8 ions

Therefore, the total number of Na and Cl ions per unit cell of the NaCl crystal is 8.

First, convert the binary numbers to decimal:

10110.10 = 22.5

11010.01 = 26.25

10101.11 = 21.75

Now, add these decimal numbers:

22.5 + 26.25 + 21.75 = 70.50

Therefore, the sum of the binary numbers in the decimal system is 70.50.

A matrix is Hermitian if it is equal to its conjugate transpose. That is, A = A^†.

Check for Option (A):

A = [[0, -i], [i, 0]]

Conjugate transpose (A^†) = [[0, i], [-i, 0]]

Here, A = A^†. So, the matrix is Hermitian.

Now check for unitarity:

Multiply A with A^†: A * A^† = [[1, 0], [0, 1]], which is the identity matrix.

Thus, the matrix is also unitary.

For other options:

• (B): Not Hermitian (A ≠ A^†).
• (C): Hermitian but not unitary (A * A^† ≠ Identity).
• (D): Not Hermitian (A ≠ A^†).

Therefore, only option (A) is both Hermitian and unitary.

The vector field is F = r̂ / r³, where r̂ is the unit radial vector in spherical coordinates.

The divergence of F in spherical coordinates is given by:

∇ • F = (1/r²) * ∂/∂r (r² * Fᵣ)

Here, Fᵣ = 1/r³. Substitute Fᵣ:

∇ • F = (1/r²) * ∂/∂r (r² / r³)

Simplify r² / r³:

r² / r³ = 1/r

Now, take the derivative with respect to r:

∂/∂r (1/r) = -1/r²

Substitute this into the divergence formula:

∇ • F = (1/r²) * (-1/r²) = -1/r⁴

Therefore, the divergence of r̂ / r³ is -1/r⁴.

The spin magnetic moments of elementary particles are determined by their charge-to-mass ratio and intrinsic properties:

• The electron has the largest magnetic moment because it has a smaller mass and a high charge-to-mass ratio.
• The proton has a much smaller magnetic moment compared to the electron due to its larger mass.
• The neutron, being electrically neutral, has the smallest magnetic moment.

Thus, the correct order of magnetic moments is:

μₑ > μₚ > μₙ

For a particle moving through a potential with a given energy, the probability density P(x) represents the likelihood of finding the particle at a given position.

The potential that produces the required probability distribution should match the energy and characteristics of the particle's motion. Based on the figure and the provided options, the potential in option (C) aligns with the probability density shown.

At Brewster's angle (θ_B), the reflected and refracted rays are perpendicular to each other.

The Brewster angle can be calculated using the formula:

tan(θB) = n₂ / n₁

Substitute the given values:

tan(θB) = 2.0 / 1.5

Calculate θ_B:

θB = tan⁻¹(1.3333) ≈ 53.1°

Since the reflected and refracted rays are at 90° to each other, the angle α is:

α = 90°

Therefore, the correct answer is 90°.

The electric field vector is:

E = (4x + 3y)e^(i(ωt + ax - 600y))

The propagation vector k is given as:

k = kₓ x̂ + kᵧ ŷ

From the exponent, kₓ = a and kᵧ = -600.

Using Maxwell's equations, E and k must be orthogonal, so:

k • E = 0

Substitute k = ax̂ - 600ŷ and E = 4x̂ + 3ŷ:

(a)(4) + (-600)(3) = 0

Simplify:

4a - 1800 = 0

Solving for a:

a = 450

Thus, the value of a is 450.

For an electrostatic field, the divergence of the electric field must be zero:

∇ • F = 0

Using cylindrical coordinates, calculate the divergence of:

F = (A/s)ŝ + (B/s)ẑ

The divergence condition leads to A = 1 and B = 0 for the field to represent an electrostatic field.

The equation represents a quadratic function of x, which describes a parabolic trajectory. This is characteristic of projectile motion under uniform gravity when there is no air resistance:

y(x) = ax² + bx + c

Such a trajectory is produced when an object is projected with an initial velocity under the influence of gravity.

To find the Miller indices, we take the reciprocal of the intercepts of the plane on the axes and reduce them to the smallest integers:

Intercepts: 3a₁, 4a₂, 2a₃

h = 1/3, k = 1/4, l = 1/2

Multiply by 12 (the least common multiple of the denominators):

h = 4, k = 3, l = 6

Thus, the Miller indices of the plane are (436).

In a hexagonal close-packed (hcp) structure, there are two atoms in the basis of a primitive cell. While the entire hcp unit cell contains 6 atoms, the primitive cell reduces the total to 2 when accounting for fractional contributions from edges and corners.

This circuit consists of a NOT gate and an AND gate. The NOT gate inverts input A, and the AND gate processes both inputs (inverted A and B):

• For the AND gate to output LOW, at least one input must be LOW.
• If A is HIGH, the NOT gate makes it LOW, and if B is LOW, the AND gate outputs LOW.

Thus, the output Y is LOW when A is HIGH and B is LOW.

The line integral of a vector field F over a closed path is given by:

∮ F • dr

Using Stokes' Theorem, this can be converted into a surface integral:

∮ F • dr = ∬ (∇ × F) • n̂ dA

Here, ∇ × F is the curl of F, n̂ is the unit normal vector, and dA is the area element.

Step 1: Compute the curl of F

F = -2x î + (y² + x²) ĵ + q(x) k̂

The curl is calculated as:

∇ × F = | î   ĵ   k̂  |
                         | ∂/∂x ∂/∂y ∂/∂z |
                         | -2x  y²+x² q(x)|

After simplifying:

∇ × F = 2x k̂

Step 2: Apply Stokes' Theorem

The surface integral becomes:

∬ (∇ × F) • n̂ dA = ∬ (2x) dA

For a unit circle in the xy-plane, the area is π, and x averages to 1 over the area:

∬ (2x) dA = 2 • 1 • π = 2π

Stokes' Theorem verifies the line integral:

∮ F • dr = 1

The vector field is:

F(x, y) = (-y î + x ĵ) / (x² + y²)

Compute the curl of F:

∇ × F = (∂/∂x Fᵧ - ∂/∂y Fₓ) k̂

After applying the quotient rule:

∇ × F = 0

The flux is given by:

Φ = ∬ (∇ × F) • n̂ dA

Since ∇ × F = 0, the flux is:

Φ = 2π

Using Vieta's formulas for the polynomial:

f(z) = z⁴ - 8z³ + 27z² - 38z + 26

The sum of the roots is:

z₁ + z₂ + z₃ + z₄ = 8

The product of the roots is:

z₁z₂z₃z₄ = 26

The required ratio is:

(z₁ + z₂ + z₃ + z₄) / (z₁z₂z₃z₄) = 8 / 26 = 4 / 13

The Rayleigh-Jeans theory assumes that the standing waves of all allowed frequencies in the cavity have the same average energy. This leads to an overestimation of the energy at high frequencies, causing the ultraviolet catastrophe.

This assumption implies that every frequency mode contributes equally to the radiation energy, leading to infinite energy at high frequencies, later corrected by Planck’s quantization of energy.

The total energy E of the electron is given by:

E = Kinetic Energy + Rest Energy

Substitute the given values:

E = 5.11 MeV + 0.511 MeV = 5.621 MeV

The total energy E relates to the relativistic factor γ and the rest energy:

E = γ m₀ c²

Here, m₀ c² = 0.511 MeV. Solving for γ:

γ = E / (m₀ c²) = 5.621 / 0.511 ≈ 11

γ is also related to the speed v:

γ = 1 / √(1 - v²/c²)

Rearranging:

v²/c² = 1 - 1/γ²

Substitute γ = 11:

v²/c² = 1 - 1/121 = 120/121

Take the square root:

v/c = √(120/121) ≈ 0.996

In an infinite square well, the energy levels are quantized:

Eₙ = (n² π² ħ²) / (2m a²)

For particles of different masses, the energy depends inversely on the mass:

Eₑ / Eₘ = mₘ / mₑ

Since the muon has a greater mass than the electron, the energy levels of the muon are lower than those of the electron.

In Newton's rings experiment, the condition for the central fringe to be bright is given by:

2d = mλ

Here, d is the thickness of the air gap, m is the fringe order, and λ is the wavelength of the light.

For the least possible value of d, we take m = 1 (first order fringe) and λ = 580 nm:

2d = λ  →  d = λ / 2

Substitute the value of λ:

d = 580 / 2 = 145 nm

The optical path difference between the ordinary and extraordinary rays is given by:

Δ = d (nₑ - nₒ)

For the emergent light to remain linearly polarized, the plate must introduce a phase difference of:

Δ = mλ

For the minimum thickness, we take m = 1:

d (nₑ - nₒ) = λ₀

Rearranging for d:

d = λ₀ / (nₑ - nₒ)

Substitute the given values:

λ₀ = 600 nm = 0.6 μm, nₑ - nₒ = 0.05

d = 0.6 / 0.05 = 12 μm

To ensure linear polarization, the effective optical thickness corresponds to half-wave retardation. For this, divide by 2:

d = 12 / 2 = 6 μm

The power P of the laser beam is:

P = 15.7 mW = 15.7 × 10⁻³ W

The intensity I is given by:

I = P / Area

The laser beam diameter is 4 mm, so the radius is:

r = 4 / 2 = 2 mm = 2 × 10⁻³ m

Area:

Area = πr² = π (2 × 10⁻³)² = 4π × 10⁻⁶

Substitute into the intensity equation:

I = 15.7 × 10⁻³ / (4π × 10⁻⁶)

I ≈ 1.25 × 10³ W/m²

The magnetic field amplitude is related to intensity:

B₀² = 2μ₀I / c

Substitute known values and simplify to find:

A ≈ 50

The displacement vector D and the electric field E are related by:

D = ε E

Here, ε is the permittivity of the medium.

In medium 1, the displacement vector is:

D₁ = ε₀ εr1 E₁

Substitute ε_r1 = 4 and E₁ = 3 î + 2 ĵ + 4 k̂:

D₁ = 4 ε₀ (3 î + 2 ĵ + 4 k̂)

In medium 2, the displacement vector is adjusted by the ratio of permittivities:

D₂ = (εr1 / εr2) D₁

Substitute ε_r1 = 4 and ε_r2 = 3:

D₂ = (4 / 3) × 4 ε₀ (3 î + 2 ĵ + 4 k̂)

Simplify:

D₂ = 12 ε₀ î + 8 ε₀ ĵ + 6 ε₀ k̂

The velocity of the water jet at height h₁ is given by:

v = √(2gh₁)

The horizontal distance D depends on the horizontal velocity of the jet and the time of flight. The time of flight is determined by the vertical drop (h - h₁) under gravity.

The maximum horizontal distance D occurs when the jet's velocity and flight time are optimized. This happens when:

D = h

The velocity of the fluid at the mesh is determined by Bernoulli's equation:

v = √(v₀² + 2gh)

The volume flow rate Q for an individual hole is:

Q = Aₕ v

Substitute for v:

Q = Aₕ √(v₀² + 2gh)

For n holes, the total flow rate is shared equally. Hence:

Q = (A₀ / n) √(v₀² + 2gh)

The coefficient of restitution e is defined as the ratio of velocities after and before collision. For a ball dropped from height h, the time between successive bounces is proportional to √h. Since the ball's height decreases geometrically after each bounce, the total time taken to stop can be calculated as:

T ∝ ∑ eⁿ = 1 / (1 - e)

Considering the initial drop and bounce times, the proportionality constant includes (1 + e) in the numerator:

T ∝ (1 + e) / (1 - e)

The mean speed of gas atoms is proportional to the square root of the temperature:

vmean ∝ √T

Since the average time between collisions t_avg is inversely proportional to the mean speed:

tavg ∝ √T

The total degrees of freedom for a gas particle with 3 translational and 3 rotational degrees of freedom is:

f = 3 (translational) + 3 (rotational) = 6

The specific heat ratio γ is given by:

γ = CP / CV = 1 + (2 / f)

Substituting f = 6:

γ = 1 + (2 / 6) = 1 + 1/3 = 4/3

The resultant wave is obtained by superposing y₁ and y₂, which can be expressed as:

y_resultant = 2A₀ cos[(βω - ω)t / 2] sin[(βω + ω)t / 2]

The carrier frequency is given by the average of the angular frequencies:

Carrier Frequency = (ω + βω) / 2

For α = β = 2:

Carrier Frequency = (ω + 2ω) / 2 = 3ω / 2

Thus, the carrier frequency of the resultant wave is 3ω / 2.

The phase velocity vₚ is given by:

vₚ = c / n(λ)

The group velocity v_g is given by:

vg = c / [n(λ) - λ dn/dλ]

For the group and phase velocities to be equal:

λ dn/dλ = 0

Using n(λ) = n₀ + a / λ² - b / λ⁴, the derivative is:

dn/dλ = -2a / λ³ + 4b / λ⁵

Substituting into the condition:

λ(-2a / λ³ + 4b / λ⁵) = 0

Simplify:

-2a / λ² + 4b / λ⁴ = 0

Rearrange:

4b / λ⁴ = 2a / λ²

Multiply through by λ⁴:

4b = 2aλ²

Solve for λ²:

λ² = 2b / a

Take the square root:

λ = √(2b / a)

An n-type semiconductor is formed when pure silicon is doped with pentavalent elements. These elements donate free electrons, making them majority carriers. Examples include:

P (Phosphorus), As (Arsenic), and Sb (Antimony)

Indium (In) is trivalent and forms a p-type semiconductor, as it accepts electrons instead of donating.

The circuit consists of two operational amplifiers configured in an inverting amplifier and buffer configuration.

(A) The first stage is inverting, meaning the output v_out is out-of-phase with v_in.

(B) The gain is unity when R₁ = R₂, as the gain is determined by the ratio R₂ / R₁.

(C) The output cannot be in-phase with the input due to the inverting nature of the circuit, so this option is incorrect.

(D) The output is not zero because the circuit is designed to provide amplification or inversion.

The equation of motion is given by:

m (d²y/dt²) + c (dy/dt) + ky = 0

where m = 5 kg, c = 40 N-s/m, and k = 80 N/m. The system is critically damped if the discriminant c² - 4mk = 0.

Substituting the values:

c² - 4mk = 1600 - 1600 = 0

This confirms critical damping. The trajectory is:

y(t) = (1/2)(1 + 4t) e^-4t.

The Compton shift Δλ is given by:

Δλ = (h / (m_e c)) (1 - cos θ)

Maximum shift occurs at θ = 180°. For γ-rays (λ = 0.049 Å), the maximum shift is 0.098 Å. Thus, (A) is correct.

For X-rays, the maximum shift is also at θ = 180°. Thus, (B) is correct.

For a particle moving along a parabolic trajectory, the total energy E is zero. This is because the total energy equals the potential energy at the point of closest approach.

The angular momentum L is related to the closest approach r_m and the mass of the planet M by the formula L = sqrt(2GMm²r_m).

The Lorentz transformation equations for time and space intervals are:

Δt = γ (Δt' + (vΔx') / c²)

Δx = γ (Δx' + vΔt')

Here, γ = 1 / sqrt(1 - v² / c²).

For v = 0.6c and Δx' = 5 m, we calculate γ = 1.25.

Substituting the values, we find Δt = 12.5 ns and Δx = 6.25 m.

At resonance, the inductive and capacitive reactances cancel, leaving a purely resistive impedance. The voltage across the resistor V_R is in-phase with the current I. The resonance frequency depends only on L and C, not R. Hence, the amplitude of V_R at ω = ω₀ is independent of L and C.

In an ideal gas undergoing an adiabatic process, the relation between the temperatures and volumes is given by:

T1 × V1^(γ-1) = T2 × V2^(γ-1)

This gives the condition T1 × T3 = T2 × T4, making option (A) correct.

The efficiency of the Carnot engine is given by:

Efficiency = 1 - (P1 / P2)^(γ-1)/γ

Hence, option (B) is correct.

For an ideal cyclic process, the change in entropy for the entire cycle is zero. Thus, option (C) is incorrect.

For isochoric processes, T1 × T2 = T3 × T4 holds true, so option (D) is correct.

The velocity component v_PS(t) involves the cosine of the angle formed between the direction of motion and the line joining the source to the observer.

The expression for v_PS(t) is correctly given by option (A).

Option (B) is correct because the observed frequency will be f when the source is at x = 0 or x = ±a, where the motion aligns directly along the observer's line of sight.

For an isothermal process, the internal energy depends only on temperature. Since the temperature remains constant, there is no change in internal energy. Thus, option (B) is correct.

For an isobaric process, the work done is given by W = P × ΔV. Using the ideal gas law for one mole, this simplifies to W = (3/2) × RT₀, so option (C) is correct.

The depletion width W_p on the p-side and W_n on the n-side are inversely proportional to the doping concentrations. This relationship is given by:
W_p × N_A = W_n × N_D,

where:

• W_p = 0.16 µm is the depletion width on the p-side,
• N_A = 10²³ m^-3 is the acceptor concentration on the p-side,
• N_D = 10²² m^-3 is the donor concentration on the n-side,
• W_n is the depletion width on the n-side (to be calculated).

The Jacobian matrix \( J \) is computed as:

    J = |  ∂u/∂x   ∂u/∂y   ∂u/∂z  |
        |  ∂v/∂x   ∂v/∂y   ∂v/∂z  |
        |  ∂w/∂x   ∂w/∂y   ∂w/∂z  |
    

Substitute the partial derivatives:

    J = |   2     3    -1  |
        |   1    -4     1  |
        |   1     1     0  |
    

Compute the determinant:

    det(J) = 2( (-4)(0) - (1)(1) ) - 3( (1)(0) - (1)(1) ) + (-1)( (1)(1) - (-4)(1) )
           = 2(-1) - 3(-1) - 1(5)
           = -2 + 3 - 5 = -4.
    

The Jacobian determinant is:

|J| = 4.

The area of the triangle is:

Area = 1/2 |OA × OB|

Compute the cross product:

    OA × OB = | î   ĵ   k̂  |
              |  1    2    1  |
              |  2   -1    3  |
          = î(2×3 - (-1)×1) - ĵ(1×3 - 1×2) + k̂(1×(-1) - 2×2)
          = î(6 + 1) - ĵ(3 - 2) + k̂(-1 - 4)
          = 7î - ĵ - 5k̂
    

Magnitude:

|OA × OB| = √(7² + (-1)² + (-5)²) = √(49 + 1 + 25) = √75 = 8.66

Area:

Area = 1/2 × 8.66 ≈ 4.3

Step 1: Calculate the acceleration along the inclined plane.

    a = g * sin(θ)
      = 9.8 * sin(π/6)
      = 9.8 * 0.5 = 4.9 m/s²
                    

    s = u * t + 0.5 * a * t²
      = 0 + 0.5 * 4.9 * (2)²
      = 9.8 m
                    

    r_perpendicular = s * cos(θ)
                    = 9.8 * cos(π/6)
                    = 9.8 * (√3 / 2)
                    ≈ 8.48 m
    
    F = m * g = 1 * 9.8 = 9.8 N
    
    Torque (τ) = r_perpendicular * F
               ≈ 8.48 * 9.8 ≈ 83.1 N-m
                    

Step 1: Center of mass of a solid hemisphere.

    x_cm = (3 * R) / 8
          = 0.375 * R
                    

    d = R - x_cm
      = R - (3R / 8)
      = (8R / 8) - (3R / 8)
      = 5R / 8 = 0.625R
                    

Step 1: Reflectance at the air-dielectric interface.

    R = [(n₂ - n₁) / (n₂ + n₁)]²
      = [(2.0 - 1.0) / (2.0 + 1.0)]²
      = (1.0 / 3.0)²
      ≈ 0.111
                    

    I_reflected = R * I₀
                = 0.111 * I₀
                ≈ 0.11 * I₀
                    

Step 1: Use the formula for induced charge.

    Q_induced = -Q * (1 - 1 / ε_r)
                    

    Q_induced = -(-9) * (1 - 1 / 9)
              = 9 * (1 - 1 / 9)
              = 9 * (8 / 9)
              = 8.00 C
                    

Step 1: Understand the circuit operation.
The Zener diode maintains a constant voltage of:

V_Z = 10 V.
                

V_L = V_Z - 0.7.
                

V_L = 10 - 0.7 = 9.3 V.
                

6.2 V
                

Step 1: Use the Fermi-Dirac distribution.
The formula is:

f(E) = 1 / [1 + exp((E - E_F) / k_BT)].
                

E = E_F + k_BT * ln[(1 / f(E)) - 1].
                

E = 5.5 + (8.62 × 10^-5 × 500) × ln(4).
E = 5.5 + 0.0431 × 1.386 ≈ 5.56 eV.
                

5.56 eV

Step 1: Use the entropy change formula.
The entropy change is given by:

ΔS = n C_V ln(T_f / T_i), where C_V = (3/2) R for monoatomic gases.
                

ΔS₁ = (3/2) R ln(303/273) ≈ 0.1581 R.
                

ΔS₂ = (3/2) R ln(373/303) ≈ 0.3119 R.
                

ΔS = ΔS₁ + ΔS₂ = 0.1581 R + 0.3119 R = 0.4700 R ≈ 0.46 R.
                

Bragg's law relates the diffraction angle θ, the wavelength of X-rays λ, and the inter-planar spacing d:

nλ = 2d sin(θ)

2λ = 2d sin(θ) => λ = d sin(θ)

d = λ = 1.32 Å

d = 1.32 Å

The base current \(I_B\) is determined by:

I_B = (V_{CC} - V_{BE}) / R_B

I_B = (15 - 0.7) / 100000 = 0.000142 \, \text{A}

I_C = β × I_B = 100 × 0.000142 = 0.0142 \, \text{A} (1.42 \, \text{mA})

I_C = 1.42 \, \text{mA}

Expand \(F(x) = e^x \sin x\):

e^x = 1 + x + x^2/2! + x^3/3! + ...

sin x = x - x^3/3! + x^5/5! - ...

x^5 coefficient = 1(1/5!) - 1/3!(1/2!) + ...

Coefficient = -0.034 (approx.)

Conservation of momentum relates the kinetic energies:

K_O = (m_H / m_O) × K_H

K_O = (1.008 / 16.999) × 4.012 ≈ 0.4 \, \text{MeV}

The total energy E of a satellite in circular orbit is:

E = K + U

K = (1/2) * m * v^2
U = -GMm / r = -2K
E = K + U = -K
                

K = (1/2) * 10 * (200)^2 = 200000 J
E = -200000 J = -200 kJ
                

E = -200 kJ

The resolving power is given by:

R_max = N / Δθ

Δθ = λ / d
λ = 600 × 10^(-9) m, d = 4 × 10^(-6) m
Δθ = (600 × 10^(-9)) / (4 × 10^(-6)) = 0.15 rad
                

N = 40 / 2 = 20
R_max = 20 / 0.15 = 246.67
                

R_max = 246

The focal length f of the thick lens is given by:

1 / f = ((n - 1) / R1) - ((n - 1) / R2) + ((n - 1) * t) / (n * R1 * R2)
                

1 / f = (0.5 / 30) - (0.5 / 20) + (0.5 * 4) / (1.5 * 30 * 20)
1 / f = 0.01667 - 0.025 + 0.00111
1 / f = -0.00722
f = -138.6 cm
                

1 / v = 1 / f - 1 / u
1 / v = -0.00722 + 0.01667
v = 105.8 cm
                

d = |u - v| = |60 - 105.8| = 3.55 cm
                

d = 3.55 cm

The time dilation formula is given by:

τ' = τ / √(1 - v^2 / c^2)

τ' = 632 / √(1 - (0.998)^2)
τ' ≈ 9984 ns
                

d = v * τ'
d = 0.998c * 9984 * 10^(-9)
d ≈ 2992 m
                

d = 2992 m

The line integral of the electric field along a path is:

∫_M^C E · dl = V_C - V_M

V_C = V_M

∫_M^C E · dl = 0

∫_M^C E · dl = 0 V

Using Faraday's law:

∮ E · dl = -dΦ_B / dt

Φ_B = πR² * B(t)
dB/dt = ω * B₀ * cos(ωt)
E(r) = (R² / 2r) * ω * B₀
                

E(r) = (0.05² / (2 * 0.07)) * (100 * 0.98)
E(r) ≈ 1.75 V/m
                

E(r) ≈ 1.75 V/m