Step 1: Understanding the analogy. The words ”smile — giggle — laugh” represent an increasing order of intensity in the context of emotional expression. Similarly, ”disapprove — — chide” needs a word that signifies a progression of disapproval to reprimand.

Step 2: Evaluating options. • reprove: Means to express disapproval or criticism, fitting the analogy.

• praise: Indicates approval, which is opposite to disapprove.

• reprise: Refers to a repetition or recurrence, unrelated to the context.

• grieve: Indicates sorrow or mourning, unrelated to disapproval.

Step 3: Conclusion. The most appropriate word to complete the analogy is reprove, as it signifies a progression from disapproval to chiding.

Step 1: Analyze the set. The numbers in the set are: 19, 37, 21, 17, 23, 29, 31, 11.

Step 2: Identify the property. All numbers except 21 are prime numbers (numbers divisible only by 1 and themselves). 21 is divisible by 3 and 7, making it a composite number.

Step 3: Conclusion. The odd one out is 21, as it is not a prime number.

Step 1: Understanding the Fibonacci series. The Fibonacci series is a sequence where each number is the sum of the two preceding numbers: 1, 1, 2, 3, 5, 8, 13, 21, ...

Step 2: Compare the given series with the Fibonacci series. Given series: 1, 1, 2, 3, 6, 8, 13, 21, ... Fibonacci series: 1, 1, 2, 3, 5, 8, 13, 21, ... The number 6 in the given series does not match 5 in the Fibonacci series.

Step 3: Conclusion. The number 6 should be changed to 5 to form the Fibonacci series.

Step 1: Express the equality with a proportionality constant. Let the common ratio for the fractions be k: x / (pq - r²) = y / (qr - p²) = z / (rp - q²) = k. From this, we can write: x = k(pq - r²), y = k(qr - p²), z = k(rp - q²).

Step 2: Substitute x, y, z into px + qy + rz. The expression becomes: px + qy + rz = p[k(pq - r²)] + q[k(qr - p²)] + r[k(rp - q²)].

Step 3: Simplify the terms. Expand each term: px = kp²q - kpr², qy = kq²r - kqp², rz = kr²p - krq². Adding these: px + qy + rz = (kp²q - kpr²) + (kq²r - kqp²) + (kr²p - krq²). Group similar terms: px + qy + rz = kp²q - kqp² + kq²r - krq² + kr²p - kpr². Each pair of terms cancels out: kp²q - kqp² = 0, kq²r - krq² = 0, kr²p - kpr² = 0. Thus, px + qy + rz = 0.

Step 4: Conclusion. The value of px + qy + rz is 0.

Step 1: Total outcomes. Each die has four possible outcomes: {2, 3, 5, 7}. When two dice are thrown, the total number of outcomes is: 4 * 4 = 16.

Step 2: Favorable outcomes. The sum of the scores on the two dice must be greater than 11. List all possible outcomes: - (5, 7), (7, 5), (7, 7). These are the only outcomes where the sum > 11. Thus, there are 3 favorable outcomes.

Step 3: Probability. The probability of getting a sum greater than 11 is: P = Favorable outcomes / Total outcomes = 3 / 16.

Step 1: Fill the blanks based on context. • (1) : wasn’t: The sentence suggests Prof. P was not merely a man who narrated funny stories. • (ii) : Even: The phrase ”Even in his blackest moments” fits the context. • (iii) : was: Prof. Q was a man who hardly narrated funny stories. • (iv) : Only: ”Only in his blackest moments was he able to find humor” aligns with the sentence structure.

Step 2: Verify the choices. The sequence wasn’t, Even, was, Only matches the context and grammar of the sentences.

Step 3: Conclusion. The correct match for the blanks is wasn’t, Even, was, Only.

Step 1: Understand the relationship between A, B, C. The conditions specify that A ⊆ B ⊆ C. This means every element in A must also be in B, and every element in B must also be in C.

Step 2: Determine the number of subsets for C. The set {2, 3, 5} has 2³ = 8 subsets. Thus, C can take any of these 8 subsets.

Step 3: Determine the number of subsets for B for a fixed C. For each choice of C, B can be any subset of C. If C has n elements, then B can take 2ⁿ subsets.

Step 4: Determine the number of subsets for A for a fixed B. For each choice of B, A can be any subset of B. If B has m elements, then A can take 2^m subsets.

Step 5: Total number of combinations. The total number of combinations of A, B, C is the sum of 2^|B| * 2^|A| over all possible subsets C. This results in: Total combinations = Σ_C Σ_B⊆C Σ_A⊆B 1. Using symmetry and the inclusion-exclusion principle, the total number of valid combinations is 3³ = 27.

Step 6: Conclusion. The total number of non-null combinations of A, B, C is 27.

Step 1: Analyze the data for the world cup years (2015 and 2019). • In 2015, VK’s batting average is higher than RS’s batting average. • In 2019, VK’s batting average is higher than RS’s batting average. This eliminates options (1) and (2).

Step 2: Analyze the data between the two world cup years (2016 to 2018). • For 2016, 2017, and 2018, VK’s batting average is consistently higher than RS’s batting average. This confirms that option (3) is correct.

Step 3: Analyze the last three years (2020 to 2022). • In 2020 and 2021, VK’s batting average is higher than RS’s. • In 2022, RS’s batting average is higher than VK’s. Thus, option (4) is incorrect.

Step 4: Conclusion. The correct statement is: VK’s yearly batting average is consistently higher than that of RS between the two world cup years.

Step 1: Understand the folding process. To create the three-dimensional object from the given planar shape, specific folds are required along the edges to form the rectangular sides and connect the V-shaped pieces into the final structure.

Step 2: Count the number of folds. • 5 folds are required to create the two V-shaped structures (one for each of the sides of the V). • 4 additional folds are required to form the rectangular connections and close the structure. Thus, the total number of folds required is: 5 + 4 = 9.

Step 3: Verify the process. Each fold corresponds to aligning and joining the planar parts into the final 3D object. The folding steps match the geometry of the figure.

Step 4: Conclusion. The number of folds required to form the object is 9.

Step 1: Understand the geometry of the cube. A cube has rotational symmetry about its body diagonal. When the cube is rotated about one of its body diagonals, the cube appears unchanged after a rotation of certain angles due to its symmetry.

Step 2: Analyze the rotational symmetry. A cube can be rotated about its body diagonal by 120°, 240°, and 360° to appear unchanged. Among these, 120° is the minimum angle that satisfies the condition.

Step 3: Conclude the solution. The minimum angle of rotation about the body diagonal such that the cube appears unchanged is 120°.

Step 1: Relation from generating function. For a generating function F₁(Q, q), the canonical transformation is given by: p = ∂F₁/∂q, P = -∂F₁/∂Q.

Step 2: Compute partial derivatives. Given F₁(Q, q) = Qq: p = ∂F₁/∂q = Q, P = -∂F₁/∂Q = -q.

Step 3: Express the relationship between p, P, Q, q. From p = Q and P = -q, divide p and P: p/P = Q/(-q) = -Q/q.

Step 4: Conclusion. The correct relation is p / P = -Q / q.

Step 1: Recall Brewster’s angle condition. At the Brewster angle, the reflected wave is completely polarized perpendicular to the plane of incidence. The Brewster angle is given by: tan θ_B = n₂ / n₁, where n₂ > n₁.

Step 2: Analyze the polarization. For an unpolarized wave incident at the Brewster angle, the reflected wave becomes completely polarized perpendicular to the plane of incidence.

Step 3: Conclusion. The reflected wave is polarized perpendicular to the plane of incidence. Hence, the correct option is (C).

Step 1: Understanding the expectation value of energy. The expectation value of energy for the wavefunction Ψ(x, t) is given by: ⟨E⟩ = Σ_n |c_n|²E_n, where |c_n|² represents the probability of finding the particle in the n-th state, and E_n is the energy eigenvalue corresponding to the n-th state.

Step 2: Extract coefficients c_n from Ψ(x, t). From the given wavefunction: c₁ = √(2/3), c₂ = 1/√6, c₃ = 1/√6.

Step 3: Calculate |c_n|². The probabilities are: |c₁|² = 2/3, |c₂|² = 1/6, |c₃|² = 1/6.

Step 4: Substitute into the expectation value formula. ⟨E⟩ = |c₁|²E₁ + |c₂|²E₂ + |c₃|²E₃. Given E₂ = 4E₁ and E₃ = 9E₁: ⟨E⟩ = (2/3) * E₁ + (1/6) * (4E₁) + (1/6) * (9E₁).

Step 5: Simplify the expression. ⟨E⟩ = (2/3) * E₁ + (4/6) * E₁ + (9/6) * E₁ = (4/6) * E₁ + (4/6) * E₁ + (9/6) * E₁ = (17/6) * E₁.

Step 6: Conclusion. The expectation value of the energy is (17/6) * E₁.

Step 1: Analyze the given conditions. The problem specifies that the system is: 1. Adiabatically isolated (no heat exchange: Q = 0). 2. Experiencing a change in volume under an externally applied constant pressure (P is constant). In such a scenario, the thermodynamic potential governing the system is determined by the constraint of constant pressure and the adiabatic nature of the process.

Step 2: Identify the relevant thermodynamic potential. Under constant pressure, the enthalpy H is the thermodynamic potential defined as: H = U + PV, where U is the internal energy, P is the pressure, and V is the volume. When a system is adiabatically isolated and pressure is constant, the enthalpy is minimized at equilibrium.

Step 3: Evaluate other options. • Helmholtz free energy (F = U − TS): This is minimized at constant T and V, not applicable here.

• Gibbs free energy (G = H − TS): This is minimized at constant T and P, but the system is adiabatic, so T is not specified as constant. • Grand potential (Φ = U − TS − μN): This is minimized in open systems with constant T and μ, not relevant here.

Step 4: Conclusion. The thermodynamic potential minimized under the given conditions is the enthalpy (H).

Step 1: Rotational energy levels. The rotational energy levels of a diatomic molecule are given by: E_J = (ħ² * J(J + 1)) / (2I), where J is the rotational quantum number and I is the moment of inertia of the molecule. The moment of inertia is: I = μr², where μ is the reduced mass: μ = (m_H * m_D) / (m_H + m_D).

Step 2: Energy difference between J = 0 and J = 1. For the two lowest rotational levels (J = 0 and J = 1): ΔE = E₁ - E₀ = (ħ² * 1(1 + 1)) / (2I) - (ħ² * 0(0 + 1)) / (2I) = 2ħ² / (2I) = ħ² / I. Substitute I = μr²: ΔE = ħ² / (μr²).

Step 3: Express in terms of m_Hr². The reduced mass μ for HD is: μ = (m_H * m_D) / (m_H + m_D). Approximating m_D = 2m_H: μ = (m_H * 2m_H) / (m_H + 2m_H) = 2m_H² / 3m_H = (2/3) * m_H. Thus, the energy difference becomes: ΔE = ħ² / (μr²) = ħ² / ((2/3)m_Hr²) = (3/2) * (ħ² / (m_Hr²)).

Step 4: Conclusion. The energy difference between the two lowest lying rotational states is (3/2) * ħ² / (m_Hr²).

Step 1: Electron concentration relation. The electron concentration n is related to the Fermi wave vector k_F in two dimensions as: n = k_F² / 2π. Here, k_F is the radius of the Fermi surface in k-space.

Step 2: Determine k_F for metals A and B. From the diagrams: • For metal A, the Fermi surface radius is k_F^A = π / a. • For metal B, the Fermi surface radius is k_F^B = √2π / a.

Step 3: Calculate n_A and n_B. Using n = k_F² / 2π: n_A = (k_F^A)² / 2π = (π/a)² / 2π = π / 2a² n_B = (k_F^B)² / 2π = (√2π/a)² / 2π = 2π / 2a²

Step 4: Find n_B / n_A. n_B / n_A = (2π / 2a²) / (π / 2a²) = (2π / π) = 2.

Step 5: Conclusion. The value of n_B / n_A is 2.

Step 1: Conservation of momentum. In the fission process, the total momentum of the system before and after the reaction must be conserved. Let the masses of the daughter nuclei be m_Rb and m_Cs. Since the neutron momenta are negligible, the momenta of the two daughter nuclei must be equal and opposite in direction: m_Rb * v_Rb = m_Cs * v_Cs.

Step 2: Relation between kinetic energy and velocity. The kinetic energy KE of a nucleus is given by: KE = 1/2 * m * v². Using v = p/m (where p is the momentum), the kinetic energy can be expressed as: KE = p² / 2m. Since the momenta are equal (p_Rb = p_Cs), the ratio of the kinetic energies is inversely proportional to the ratio of their masses: KE(⁹³₃₇Rb) / KE(¹⁴¹₅₅Cs) = m_Cs / m_Rb.

Step 3: Substituting the masses. The masses of the daughter nuclei are approximately proportional to their mass numbers: m_Rb ∝ 93, m_Cs ∝ 141. Thus: KE(⁹³₃₇Rb) / KE(¹⁴¹₅₅Cs) = 141 / 93.

Step 4: Conclusion. The ratio of the kinetic energies of the daughter nuclei is 141 / 93.

Step 1: Analyze the circuit components. The circuit consists of: 1. A capacitor (C): Blocks DC components and stores charge. 2. An ideal diode (D): Allows current flow only in one direction, cutting off the negative half-cycles of the input voltage.

Step 2: Behavior of the circuit with the given V_in. The input V_in is a square waveform oscillating between +3 V and −3 V. • During the positive half-cycle (V_in = +3 V), the diode conducts, and the capacitor charges to the peak value (+3 V). • During the negative half-cycle (V_in = −3 V), the diode blocks current, and the capacitor retains its charge (+3 V).

Step 3: Combined effect of the input and stored charge. For the output V_o: • When the input goes to +3 V, the output is V_o = V_in + V_stored = +3 V + 3 V = +6 V. • When the input goes to −3 V, the diode blocks, and the output remains constant at +6 V.

Thus, the output V_o is a rectangular waveform oscillating between +6 V and 0 V, as shown in waveform (A).

Step 4: Conclusion. The correct output waveform is described in option (1).

Step 1: Nature of beta decay. In nuclear beta decay, the electron’s kinetic energy T_e follows a continuous distribution due to the sharing of energy between the beta particle (electron), antineutrino, and recoiling nucleus.

Step 2: Characteristics of the energy distribution. The energy distribution of beta particles is: • Maximum at a lower Te, where most beta particles are emitted. • Decreasing as T_e increases, with a sharp cutoff at the maximum kinetic energy determined by the decay’s Q-value.

Step 3: Analyze the options. • Option (A): Represents a discrete energy value, which is incorrect because the energy is continuously distributed. • Option (B): A bell-shaped curve, which does not match the known energy spectrum of beta particles. • Option (C): A curve with a peak at low T_e and a decreasing trend, consistent with the beta decay energy distribution. • Option (D): An increasing trend, which does not align with the beta decay spectrum.

Step 4: Conclusion. The correct distribution is represented by Option (3), where N_e peaks at lower T_e and decreases as T_e approaches the maximum kinetic energy.

Step 1: Relationship between magnetization and magnetic flux density. The magnetic flux density B⃗ is related to the magnetization M⃗ and the auxiliary field H⃗ by: B⃗ = μ₀(H⃗ + M⃗), where H⃗ is the magnetic field intensity.

Step 2: Determine H⃗ for the given setup. Since there are no free currents (J_free = 0), ∇ × H⃗ = 0. This implies H⃗ is curl-free and can be written as the gradient of a scalar potential, H⃗ = -∇φ. Using ∇ · (B/μ₀) = -∇ · M⃗, we have: ∇ · H⃗ = -∇ · M⃗. Given M⃗ = k*e^-s ẑ, the divergence ∇ · M⃗ = 0 since M⃗ is constant in the z-direction and depends only on s (radial coordinate). Thus, ∇ · H⃗ = 0, meaning H⃗ = 0 inside the cylinder.

Step 3: Calculate B⃗ . Substitute H⃗ = 0 into the expression for B⃗: B⃗ = μ₀ * M⃗. Given M⃗ = ke^-s ẑ: B⃗ = μ₀ke^-s ẑ.

Step 4: Conclusion. The magnetic flux density inside the cylinder is μ₀ke^-s ẑ.

Step 1: Understanding Electron Spin Resonance (ESR). ESR is a spectroscopic technique used to study materials with unpaired electrons. Unpaired electrons generate magnetic moments, which interact with an external magnetic field, giving rise to ESR spectra.

Step 2: Analyze the electronic configurations.

1. Vanadium (V, Z = 23): Electronic configuration [Ar]3d³4s². The 3d³ electrons provide unpaired electrons. Thus, V shows ESR spectra.

2. Chromium (Cr, Z = 24): Electronic configuration [Ar]3d⁵4s¹. The 3d⁵ electrons provide unpaired electrons. Thus, Cr shows ESR spectra.

3. Iron (Fe, Z = 26): Electronic configuration [Ar]3d⁶4s². The 3d⁶ electrons provide unpaired electrons. Thus, Fe shows ESR spectra.

4. Zinc (Zn, Z = 30): Electronic configuration [Ar]3d¹⁰4s². The 3d¹⁰ electrons are fully paired, and the 4s² electrons are also paired. Thus, Zn does not have unpaired electrons and does not show ESR spectra.

Step 3: Conclusion. Zn does not show ESR spectra because it has no unpaired electrons in its electronic configuration.

Step 1: Boundary conditions. For x ≤ 0, V(x) = ∞, which means the wavefunction Ψ(x) must be zero in this region due to the infinite potential barrier.

Step 2: Behavior in the regions. • For 0 < x < a where V(x) = 0, the wavefunction behaves as a free particle solution. This results in oscillatory behavior. • For a ≤ x ≤ b where V(x) = V₀ and E < V₀, the wavefunction decays exponentially since the particle is in a classically forbidden region. • For x > b where V(x) = 0, the wavefunction is oscillatory again, as the particle is in a free region.

Step 3: Evaluate the options.

• Option (A): Incorrect. The wavefunction does not decay exponentially in the region a ≤ x ≤ b.

• Option (B): Correct. The wavefunction satisfies all boundary conditions, including the decay in the classically forbidden region.

• Option (C): Incorrect. The wavefunction shows incorrect oscillatory behavior in the forbidden region a ≤ x ≤ b.

Step 4: Conclusion. The valid quantum mechanical wavefunction is represented by option (2).

Step 1: Understand the given condition. The equation dρ/dt = 0 implies that the phase space density ρ(⃗p, ⃗q, t) is conserved along the trajectories of the system in phase space. Using the Liouville equation for Hamiltonian dynamics: dρ/dt = ∂ρ/∂t + {ρ, H} = 0, where {A, B} is the Poisson bracket defined as: {A, B} = Σ_i (∂A/∂q_i * ∂B/∂p_i - ∂A/∂p_i * ∂B/∂q_i)

Step 2: Interpretation of dρ/dt = 0. The condition implies that the total time derivative of ρ is zero, which leads to: ∂ρ/∂t = -{ρ, H}. Thus, ∂ρ/∂t ∝ {ρ, H}.

Step 3: Analyze the options. • Option (1): ∂ρ/∂t = 0 is incorrect because it implies no time dependence at all, which contradicts the condition dρ/dt = 0. • Option (2): ∂ρ/∂t ∝ {ρ, H} is correct based on the Liouville equation. • Option (3): ∂ρ/∂t ∝ {ρ, ⃗p⋅⃗q / 2} is unrelated to the Hamiltonian H. • Option (4): ∂ρ/∂t ∝ {ρ, ⃗q⋅⃗q / 2} is also unrelated to H.

Step 4: Conclusion. The correct answer is Option (2): ∂ρ/∂t ∝ {ρ, H}.

Step 1: Analyze the circuit. The circuit consists of an AND gate taking P and Q as inputs, followed by an OR gate. The AND gate outputs a high signal (+5 V) only when both P and Q are high. The OR gate produces a high output if either the AND gate output or any other input to the OR gate is high.

Step 2: Evaluate the output signal. Given the input signals P and Q, the correct output Y is obtained by applying the logical operations sequentially: • The AND gate output is high only when P and Q are both +5 V. • The OR gate combines this with other inputs to generate the final output.

Step 3: Conclusion. The waveform in option (1) correctly represents the output Y based on the circuit logic and input signals.

Step 1: Use relativistic velocity addition formula. The relative velocity v_rel of spacecraft T with respect to S is given by the relativistic velocity addition formula: v_rel = (v_T - v_S) / (1 - (v_T * v_S) / c²). Here: • v_T = 0.5c is the speed of T relative to the inertial observer, • v_S = -0.5c is the speed of S relative to the inertial observer (negative because S moves in the opposite direction to T).

Step 2: Substitute the values. v_rel = (0.5c - (-0.5c)) / (1 - (0.5c * -0.5c) / c²). Simplify the numerator: v_rel = (0.5c + 0.5c) / (1 - (-0.25c²) / c²) = 1.0c / (1 + 0.25).

Step 3: Simplify further. v_rel = 1.0c / 1.25 = (4/5)c.

Step 4: Conclusion. The speed of T with respect to S is (4/5) * c.

Step 1: Understanding the process The given options describe different nuclear processes involving a neutrino (ν_e) and a nucleus ^A_ZP, where A represents the mass number and Z represents the atomic number.

Step 2: Analyze the options

• Option (1): This describes a beta-plus decay, which is possible if a neutrino interacts with the nucleus ^A_ZP, converting a proton into a neutron. This leads to the formation of ^A_Z+1Q, with the emission of an electron (e^-).

• Option (2): This describes a beta-minus decay. However, neutrino interactions cannot lead to the production of a positron (e⁺) while reducing the atomic number.

• Option (3): This describes the simultaneous production of a positron and an electron from the interaction of a neutrino, which violates charge conservation.

• Option (4): This describes a gamma emission, which is not related to neutrino interactions as described.

Step 3: Identify the valid process Only Option (1) is a valid process, as it aligns with the principles of beta-plus decay where a proton is converted to a neutron due to neutrino interaction.

Conclusion: The correct answer is (1) ν_e + ^A_ZP → ^A_Z+1Q + e⁻.

Step 1: Relation between bandgap and conductivity. The electrical conductivity σ of an intrinsic semiconductor depends on the temperature T and the bandgap energy E_g as: σ ∝ e^{−(Eg / (2k_BT))}. Taking the natural logarithm: ln(σ) ∝ -(E_g / (2k_B)) * T^-1. This shows that the slope of ln(σ) versus T^-1 is proportional to -E_g. A larger bandgap corresponds to a steeper (more negative) slope.

Step 2: Analyzing the slopes of A, B, and C. From the figure: • A has the steepest slope, indicating the largest bandgap. • C has a slope less steep than A but steeper than B, indicating an intermediate bandgap. • B has the least steep slope, indicating the smallest bandgap. Thus, the order of bandgaps is: E_A > E_C > E_B.

Step 3: Conclusion. The correct relation is E_A > E_C > E_B, which corresponds to Option (4).

Step 1: Understanding the variational principle. The variational principle states that the energy estimate obtained using any trial wavefunction is always greater than or equal to the exact ground state energy: E_trial ≥ E₀.

Step 2: Comparison of φ₁ and φ₂.

• The trial wavefunction φ₁ = e^{-Z'(r₁+r₂)} is a simpler approximation of the ground state wavefunction, without considering electron correlation.

• The trial wavefunction φ₂ = e^{-Z'(r₁+r₂)}(1 + g|r⃗₁ − r⃗₂|) includes the term g|r⃗₁ − r⃗₂|, which accounts for the electron-electron repulsion. This makes φ₂ a better approximation to the true ground state wavefunction than φ₁.

Step 3: Implications for the energy estimates. Since φ₂ is a better approximation than φ₁, the variational estimate E₂ obtained from φ₂ is closer to E₀ than E₁. Hence: E₁ ≥ E₂ ≥ E₀.

Step 4: Conclusion. The correct relation is: E₁ ≥ E₀, E₂ ≥ E₀, E₁ ≥ E₂, which corresponds to Option (4).

Step 1: Analyze the given wavefunction. The given wavefunction is: ψ(x) = e^-(iαx+β) = e^-β * e^-iαx. This represents a plane wave with a spatial phase factor e^-iαx and a constant magnitude e^-β.

Step 2: Plane wave in quantum mechanics. A plane wave solution occurs when the particle is free, i.e., moving in a potential V(x) = 0. For such a wavefunction, the time-independent Schrödinger equation is: -(ħ² / 2m) * (d²ψ / dx²) + V(x)ψ = Eψ. Substituting V(x) = 0, the solution to the equation is a plane wave: ψ(x) = e^-ikx, where k = α relates to the particle’s momentum p = ħk.

Step 3: Evaluate the given options.

• Option (1): V(x) ∝ α²x²: This corresponds to a harmonic oscillator potential, which does not produce a plane wave solution.

• Option (2): V(x) ∝ e^-αx: This is a potential with spatial dependence, which does not lead to a plane wave solution.

• Option (3): V(x) = 0: This represents a free particle and is consistent with the plane wave solution ψ(x) = e^-(iαx+β).

• Option (4): V(x) ∝ sin(αx): A sinusoidal potential generally leads to band-like solutions, not a plane wave.

Step 4: Conclusion. The potential V(x) = 0 is consistent with the given wavefunction. Hence, the correct answer is Option (3).

Step 1: Analyze the integrand. The term ∇²(1/r) is the Laplacian of 1/r, where r = √(x² + y² + z²). Using the property of the Laplacian in three-dimensional space: ∇²(1/r) = -4πδ³(r⃗), where δ³(r⃗) is the three-dimensional Dirac delta function.

Step 2: Evaluate the volume integral. I = ∫_V ∇²(1/r) dV = ∫_V -4πδ³(r⃗)dV. • If r = 0 is inside the volume V, the delta function contributes: I = -4π.

• If r = 0 is not inside V, the delta function is zero everywhere in V, so: I = 0.

Step 3: Behavior of the integrand.

• For r ≠ 0, ∇²(1/r) = 0, as the delta function is only non-zero at r = 0.

• As r → ∞, 1/r tends to zero, so the integrand does not diverge.

Step 4: Analyze the options.

• Option (1): Correct, as I = −4π if r = 0 is inside V.

• Option (2): Correct, as the integrand vanishes for r ≠ 0.

• Option (3): Correct, as I = 0 if r = 0 is not inside V.

• Option (4): Incorrect, as the integrand does not diverge as r → ∞.

Step 5: Conclusion. The correct statements are (1),(2),(3).

Step 1: Analyze the function e^-2/(z-1). The term 2 / (z - 1) diverges as z → 1, and since it appears in the exponent, e^-2/(z-1) oscillates infinitely near z = 1. This behavior indicates an essential singularity at z = 1.

Step 2: Residue calculation. To determine the residue, consider the Laurent series expansion of the function around z = 1: e^-2/(z-1) = Σ_n=0^∞ ((-2)ⁿ / (n! * (z - 1)ⁿ)) The residue is the coefficient of 1/(z - 1): Residue = -2.

Step 3: Evaluate the options. • Option (1): Incorrect, as z = 1 is not a simple pole; it is an essential singularity.

• Option (2): Correct, as the function has an essential singularity at z = 1.

• Option (3): Correct, as the residue at z = 1 is -2.

• Option (4): Incorrect, as z = 1 is not a branch point.

Step 4: Conclusion. The correct answers are 2 and 3.

Step 1: Simplify the Boolean expression. The given expression is: B * (A + B) + A * (B̄ + A). Simplify each term: • B * (A + B) = B, since B * B = B. • A * (B̄ + A) = A, since A * A = A. Combine the simplified terms: B + A.

Step 2: Realize the simplified expression. The expression B + A requires only one OR gate.

Step 3: Conclusion. The minimum number of gates required is 1.

Step 1: At the triple point, the vapor pressures are equal. At the triple point of ammonia, the vapor pressures of the solid and liquid phases are equal. Therefore, 23.03 - (3754 / T) = 19.49 - (3063 / T).

Step 2: Simplify the equation. Rearranging the terms: 23.03 - 19.49 = (3754 / T) - (3063 / T). 3. 54 = (3754 - 3063) / T. 4. 54 = 691 / T.

Step 3: Solve for T. T = 691 / 3.54. T = 195.41 K.

Conclusion: The temperature of the triple point of ammonia is 195.41 K.

Step 1: Use Gauss’s law. The given electric field can be expressed in terms of polar coordinates: E⃗ = k * r⃗ / r². The flux through a spherical surface of radius R is: Φ = ∫_S E⃗ ⋅ d⃗A. Since E⃗ is radially symmetric: Φ = ∫_S (k / R²) * R² dΩ = k * ∫_S dΩ.

Step 2: Evaluate the integral. The solid angle over a sphere is 4π. Hence: Φ = k * 4π.

Step 3: Relate to the given flux. The problem states: Φ = nπRk. Comparing: nπRk = 4πk ⇒ n = 4.

Step 4: Conclusion. The value of n is 4.

Step 1: Average energy per quadratic term. For each quadratic term q_i² or p_i², the average energy contribution is: ⟨a_iq_i²⟩ = (1/2) * k_BT, ⟨b_ip_i²⟩ = (1/2) * k_BT.

Step 2: Total energy. The Hamiltonian has 2N q_i² terms and 2N p_i² terms: ⟨H⟩ = Σ_i=1^2N ⟨a_iq_i²⟩ + Σ_i=1^2N ⟨b_ip_i²⟩ ⟨H⟩ = 2N * (1/2) * k_BT + 2N * (1/2) * k_BT = 2N k_BT.

Step 3: Conclusion. The ensemble average is ⟨H⟩ = αN k_BT, where α = 2.

Step 1: Energy conservation principle. To dissociate the two-nucleon bound state, the photon must provide enough energy to overcome the binding energy B and account for relativistic corrections due to the mass energy of the system.

Step 2: Relativistic correction to the binding energy. The effective energy required to dissociate the bound state includes the binding energy B and an additional term proportional to B² / (2mc²), arising from the relativistic motion of the nucleons. The total photon energy is therefore given by: E_photon = B * (1 + B / (2mc²)).

Step 3: Evaluate the options.

• Option (1): Incorrect, as it does not include the relativistic correction term.

• Option (2): Correct, as it matches the expression derived above.

• Option (3): Incorrect, as it incorrectly subtracts the relativistic correction term.

• Option (4): Incorrect, as it subtracts the rest mass energy mc², which is unrelated to the photon dissociation energy.

Step 4: Conclusion. The correct answer is B / (1 + B / (2mc²)).

Step 1: Spin-orbit interaction. The spin-orbit Hamiltonian is: H' = -k * L⃗ ⋅ S⃗. The eigenvalues of L⃗ ⋅ S⃗ depend on the total angular momentum quantum number j: L⃗ ⋅ S⃗ = (1/2) * [j(j + 1) − l(l + 1) − s(s + 1)] * ħ², where j is the total angular momentum, l is the orbital angular momentum, and s is the spin angular momentum.

Step 2: Calculate L⃗ ⋅ S⃗ for ²P_3/2 and ²P_1/2. For the P-state (l = 1) with s = 1/2: • For j = 3/2: L⃗ ⋅ S⃗ = (1/2) * [(3/2) * (3/2 + 1) - 1(1 + 1) - (1/2)(1/2 + 1)] * ħ². Simplify: L⃗ ⋅ S⃗ = (1/2) * [(15/4) - 2 - (3/4)] * ħ² = (1/2) * [(15/4) - (8/4) - (3/4)] * ħ² = (1/2) * (4/4) * ħ² = ħ². • For j = 1/2: L⃗ ⋅ S⃗ = (1/2) * [(1/2) * (1/2 + 1) - 1(1 + 1) - (1/2)(1/2 + 1)] * ħ². Simplify: L⃗ ⋅ S⃗ = (1/2) * [(3/4) - 2 - (3/4)] * ħ² = (1/2) * [(3/4) - (8/4) - (3/4)] * ħ² = (1/2) * (-8/4) * ħ² = -ħ²/2.

Step 3: Energy splitting. The energy difference between the two levels is proportional to the difference in L⃗ ⋅ S⃗: ΔE = k * (ħ² - (-ħ²/2)) = (3/2) * kħ². However, the splitting between the two levels is half of this value: ΔE = (3/2) * kħ² / 2= (3/4) * kħ².

The splitting between the two levels is proportional to the difference in L⃗ ⋅ S⃗ ΔE = k (ħ² - (-ħ²/2)) = 3/2 * kħ²

Step 4: Conclusion. The splitting between the levels ²P_3/2 and ²P_1/2 is (3/2) * kħ².

Step 1: Define the generalized momenta. The generalized momenta p_x and p_y are given by: p_x = ∂L/∂ẋ, p_y = ∂L/∂ẏ. Substitute L = mẋẏ − mω²xy: p_x = mẏ, p_y = mẋ.

Step 2: Express velocities in terms of momenta. ẋ = p_y/m, ẏ = p_x/m.

Step 3: Derive the equations of motion. The canonical equations of motion are: ṗ_x = -∂L/∂x, ṗ_y = -∂L/∂y.

For ṗ_x: ṗ_x = -∂/∂x(mẋẏ − mω²xy) = mω²y.

For ṗ_y: ṗ_y = -∂/∂y(mẋẏ − mω²xy) = mω²x.

Step 4: Combine results. The complete set of equations is: ẋ = p_y/m, ṗ_x = -mω²y, ẏ = p_x/m, ṗ_y = -mω²x.

Step 5: Conclusion. The correct answer is ẋ = p_y/m, ṗ_x = −mω²y, ẏ = p_x/m, ṗ_y = −mω²x.

Step 1: Use the Bragg equation. The Bragg equation is given by: nλ = 2d sin θ, where: • n is the order of diffraction (take n = 1 for the first reflection), • λ = 0.78 ̊A is the wavelength of X-ray, • d is the interplanar spacing for the (111) plane, • θ is the Bragg angle.

Step 2: Determine the crystal structure and interplanar spacing. For a face-centered cubic (FCC) structure, the interplanar spacing is: d_hkl = a / √(h² + k² + l²) where h, k, l are Miller indices, and a is the lattice constant. The lattice constant a for FCC is related to the atomic radius r: a = 2√2 * r. Substitute r = 1.56 ̊A: a = 2√2(1.56) = 4.41 ̊A. For the (111) plane: d₁₁₁ = a / √(1² + 1² + 1²) = 4.41 / √3 = 2.55 ̊A.

Step 3: Solve for θ using the Bragg equation. Substitute n = 1, λ = 0.78 ̊A, and d = 2.55 ̊A into the Bragg equation: 0.78 = 2(2.55) sin θ => sin θ = 0.78 / 5.1 = 0.153. Thus: θ = arcsin(0.153) = 8.8°. The Bragg angle 2θ is: 2θ = 2(8.8) = 17.6°.

Step 4: Conclusion. The Bragg angle is 17.6°, and the crystal structure is face-centered cubic (FCC).

Step 1: Relationship between electric field and potential. The electric field E⃗ is related to the potential V by: E_x = -dV/dx.

Step 2: Expression for potential V. The potential V satisfies the equation: d/dx * (ε * dV/dx) = 0. Substitute ε = ε₀(1 + x/d): d/dx * (ε₀(1 + x/d) * dV/dx) = 0. Integrating once: ε₀(1 + x/d) * dV/dx = constant.

Step 3: Solve for E_x. Let the constant of integration be C: dV/dx = C / (ε₀(1 + x/d)). The electric field is: E_x = -dV/dx = -C / (ε₀(1 + x/d)).

Step 4: Boundary condition to find C. At x = d, V = V₀, and at x = 0, V = 0. Integrate dV/dx: V = ∫₀^d (C / (ε₀(1 + x/d))) dx = V₀. Simplify: V₀ = C/ε₀ * ∫₀^d (1 / (1 + x/d)) dx = C/ε₀ * ln(2). Thus: C = (V₀ε₀) / ln(2).

Step 5: Final expression for E_x. Substitute C into E_x: E_x = -C / (ε₀(1 + x/d)) = -(V₀ / ((d + x)ln(2))) * d = -V₀ / ((d + x)ln(2)).

Step 6: Conclusion. The electric field inside the capacitor is: E⃗ = −(V₀ / ((d + x)ln(2))) * x̂.

Step 1: Relationship between electric field and potential. The electric field E⃗ is related to the potential V by: E_x = -dV/dx.

Step 2: Expression for potential V. The potential V satisfies the equation: d/dx * (ε * dV/dx) = 0. Substitute ε = ε₀(1 + x/d): d/dx * (ε₀(1 + x/d) * dV/dx) = 0.

Integrating once: ε₀(1 + x/d) * dV/dx = constant.

Step 3: Solve for E_x. Let the constant of integration be C: dV/dx = C / (ε₀(1 + x/d)). The electric field is: E_x = -dV/dx = -C / (ε₀(1 + x/d)).

Step 4: Boundary condition to find C. At x = d, V = V₀, and at x = 0, V = 0. Integrate dV/dx: V = ∫₀^d (C / (ε₀(1 + x/d))) dx = V₀. Simplify: V₀ = C/ε₀ * ∫₀^d (1 / (1 + x/d)) dx = C/ε₀ * ln(2). Thus: C = (V₀ε₀) / ln(2).

Step 5: Final expression for E_x. Substitute C into E_x: E_x = -C / (ε₀(1 + x/d)) = - (V₀ / ((d + x)ln(2))) * d = -V₀ / ((d + x)ln(2)).

Step 6: Conclusion. The electric field inside the capacitor is: E⃗ = -(V₀ / ((d + x)ln(2))) * x̂.

Step 1: Write the equation of motion. The forced simple harmonic oscillator equation is: ẍ(t) + ω²x(t) = Fcos(ωt). This is a second-order non-homogeneous differential equation.

Step 2: Analyze the resonance condition. The forcing term Fcos(ωt) has the same frequency as the natural frequency of the system, ω. This leads to a resonance condition where the response grows linearly with time.

Step 3: Solve the equation. The general solution of the equation is: x(t) = A sin(ωt) + B cos(ωt) + (F/(2ω))*tsin(ωt), where A and B are constants determined by initial conditions, and the third term represents the resonant response.

Step 4: Apply initial conditions. Given x(0) = 0 and ẋ(0) = 0: • At t = 0: x(0) = A sin(0) + B cos(0) + (F / (2ω)) * (0)sin(0) = 0 => B = 0. • At t = 0: ẋ(0) = Aω cos(0) + F / (2ω) * (0)cos(0) = 0 => A = 0. Thus, the solution simplifies to: x(t) = (F / (2ω)) * tsin(ωt).

Step 5: Conclusion. The displacement is proportional to tsin(ωt).

Step 1: Understanding the Zeeman effect. In the presence of a weak uniform magnetic field B, the energy levels split due to the interaction between the magnetic field and the magnetic moment associated with the orbital angular momentum l. The number of Zeeman sublevels for a given l is: Number of sublevels = 2l + 1.

Step 2: Calculate the number of sublevels for n = 2, l = 1. For l = 1: Number of sublevels = 2(1) + 1 = 3. The possible magnetic quantum numbers m_l are: m_l = -1, 0, +1.

Step 3: Calculate the number of sublevels for n = 1, l = 0. For l = 0: Number of sublevels = 2(0) + 1 = 1. The only possible magnetic quantum number is m_l = 0.

Step 4: Determine the number of possible transitions. In the Zeeman effect, transitions between the sublevels obey the selection rules: Δm_l = 0, ±1. From n = 2, l = 1 (m_l = -1, 0, +1) to n = 1, l = 0 (m_l = 0):

• Transitions with Δm_l = 0: m_l = 0 → m_l = 0 (1 transition).

• Transitions with Δm_l = +1: m_l = -1 → m_l = 0, m_l = 0 → m_l = 0, m_l = +1 → m_l = 0 (3 transitions).

• Transitions with Δm_l = -1: m_l = +1 → m_l = 0, m_l = 0 → m_l = 0, m_l = -1 → m_l = 0 (3 transitions).

Adding these, the total number of transitions is: 3 + 3 + 3+1= 10.

Step 5: Conclusion. The number of lines in the Zeeman spectrum is 10.

Step 1: Eigenvalues of P and Q. • For P: The characteristic equation is: det(P − λI) = det

• For Q: The characteristic equation is: det(Q − λI) = det

Step 2: Commutativity of P and Q. The matrices P and Q commute if PQ = QP: PQ =

Step 3: Linearly independent eigenvectors.

• For P: The eigenvalue λ = 1 has one linearly independent eigenvector: P

• For Q: The eigenvalue λ = 1 has two linearly independent eigenvectors: Q

Since the sets of eigenvectors differ, the statement about eigenvectors is true.

Step 4: Diagonalizability of P. For P, there is only one linearly independent eigenvector for λ = 1. Thus, P is not diagonalizable.

Step 5: Conclusion. The correct statements are (1), (2), and (3).

Step 1: Understanding the configurations. • In the paramagnetic configuration, spins can align in either +z or −z directions randomly, leading to the maximum possible entropy. • In the ferromagnetic configuration, all spins align in the same direction (+z or −z), leading to the lowest entropy. • In the antiferromagnetic configuration, spins alternate in +z and −z directions, leading to an entropy between the ferromagnetic and paramagnetic cases.

Step 2: Comparing entropies.

• S_P > S_F: True, as paramagnetic systems have higher disorder than ferromagnetic systems.

• S_A > S_F: True, as antiferromagnetic systems have some disorder compared to ferromagnetic systems.

• S_A = 4S_F: False, as this specific ratio does not hold generally.

• S_P > S_A: True, as paramagnetic systems have greater disorder than antiferromagnetic systems.

Conclusion: The correct relations are (1) S_P > S_F and (4) S_P > S_A.

Step 1: Check if F→
is conservative. For a vector field F→ to be conservative, its curl ∇ × F→ must be zero. Compute the curl of F→:

∇ × F→ = | i j k | | ∂/∂x ∂/∂y ∂/∂z | | 0 (2xz+3y²) 4yz² |

For z = 0, this simplifies to:

∇ × F→ = | i j k | | ∂/∂x ∂/∂y ∂/∂z | | 0 3y² 0 | = i(∂(0)/∂y - ∂(3y²)/∂z) - j(∂(0)/∂x - ∂(0)/∂z) + k(∂(3y²)/∂x - ∂(0)/∂y)

∇ × F→ = i(0 - 0) - j(0-0) + k(0 - 0) = 0

Since the curl is zero, F→ is conservative.

Step 2: Evaluate ∮_Γ F→ · dl. The line integral over a closed path Γ is: ∮_Γ F→ · dl.

Since F→ is conservative the integral will evaluates to zero.

Step 3: Calculate Divergence. The divergence of a vector field is calculated as: ∇ · F→ = ∂F_x/∂x + ∂F_y/∂y + ∂F_z/∂z

Given F→ = 0î + (2xz + 3y²)ŷ + 4yz²ẑ

∇ · F→ = ∂(0)/∂x + ∂(2xz+3y²)/∂y + ∂(4yz²)/∂z

∇ · F→ = 0 + 6y + 8yz, this is not zero

Step 4: Conclusion. The given field F→ satisfies:

∮_Γ F→ · dl = 0, and F→ is conservative.

Step 1: Analyze the electric field and potential due to charges. The conducting sphere is grounded, so its surface potential is fixed at zero. The charges +q induce surface charges on the sphere to maintain this condition.

Step 2: Evaluate each statement.

1. For a > d/8: The electric field lines and induced charges result in attractive forces between the two point charges, causing the net force to direct them towards each other.

2. The potential at the sphere’s surface is grounded, so it is zero.

3. The total induced charge on the sphere can be calculated using the method of images and is given by: Q_induced = -2aq/d.

4. The potential at the center of the grounded sphere is zero due to symmetry and grounding conditions.

Step 3: Conclusion. The correct statements are (1), (2), and (3).

Step 1: Analyze the potential forms. The given potential V(x) is harmonic for x > 0 with parameters ω₀ and V₀. For each figure:

- V(x) in P: V₀ = 0, ω₀ = 12 rad/s

- V(x) in Q: V₀ = 3 J, ω₀ = 12 rad/s

- V(x) in R: V₀ = 4 J, ω₀ = 4 rad/s

- V(x) in S: V₀ = 0, ω₀ = 14 rad/s

Step 2: Calculate energy eigenvalues. The energy levels for a harmonic potential are: E_j = (j + 1/2)ħω₀ + V₀.

For the ground state (j = 0):

- E₀^(P) = (1/2)ħω₀^(P)

- E₀^(Q) = V₀^(Q) + (1/2)ħω₀^(Q)

- E₀^(R) = V₀^(R) + (1/2)ħω₀^(R)

- E₀^(S) = (1/2)ħω₀^(S)

Step 3: Verify the statements.

1. E₀^(P) ≠ E₀^(Q) because V₀^(Q) ≠ V₀^(P).

2. E₀^(Q) = E₀^(S) because both have the same ω₀ and V₀.(considering ground state from both the potential)

3. E₀^(P) = E₁^(R) because their energy levels match when considering j = 1 for R.

4. E₀^(R) ≠ E₀^(Q) because of differences in ω₀ and V₀.

Step 4: Conclusion. The correct statements are (2), (3), and (4).

Step 1: Analyze the properties of H'. The spin-orbit interaction term H' depends on the angular momentum operators L and S:

H' = 1/(2m²c²r)dV/dr (L · S).

- H' commutes with L² because it depends on the scalar product L·S, which is rotationally invariant.

- H' does not commute with L_z or S_z individually because L · S involves components of both L and S.

Step 2: Degeneracy of H₀. The non-relativistic Hamiltonian H₀ is independent of spin. For a given n and l, there are (2l + 1) orbital states and 2 spin states, leading to 2(2l + 1) degenerate eigenstates.

Step 3: Simultaneous eigenstates.

- H₀, L², S², L_z, and S_z are mutually commuting operators in the absence of spin-orbit coupling. Therefore, they share a common set of eigenstates.

Step 4: Conclusion. The correct statements are: H' commutes with L². There are 2(2l + 1) degenerate eigenstates of H₀. H₀, L², S², L_z, and S_z have a set of simultaneous eigenstates.

Step 1: Analyze the given decays.

1. π⁰ → γγ: This is an electromagnetic decay because the decay involves photons (γ), which are mediated by electromagnetic interaction. Therefore, option (A) is incorrect.

2. Λ⁰ → π⁰ + p: This is a weak decay. The Λ⁰ baryon is a strange particle, and its decay involves the weak interaction. Therefore, option (B) is incorrect.

3. K⁰ → π⁺ + π^-: The decay of K⁰ (a kaon) involves a change in quark flavor, which is characteristic of weak interactions. Therefore, option (C) is correct.

4. ∆⁺⁺ → p + π⁺: The ∆⁺⁺ baryon decays via the strong interaction because it involves no change in flavor and is governed by the strong force. Therefore, option (D) is correct.

Step 2: Conclusion. The correct options are (3) and (4).

Step 1: Analyze the Hall effect.

- The Hall voltage arises due to the deflection of the majority charge carriers under the influence of a magnetic field. Since the Hall voltage polarity at y = 0 is negative, the majority charge carriers must be electrons (negative charges). Therefore, option (1) is correct.

Step 2: Calculate the magnitude of the applied magnetic field. The Hall voltage is given by: V_H = I*B/(net)

Rearranging for B: B = V_H*net/I

Substituting the values:

V_H = 20 mV = 20 × 10^-3 V, n = 12.5 × 10¹³ cm^-3 = 1.25 × 10¹⁹ m^-3, e = 1.6 × 10^-19 C, t = 0.005 m, I = 2 mA = 2 × 10^-3 A

B = (20 × 10^-3 * 1.25 × 10¹⁹ * 1.6 × 10^-19 * 0.005) / 2 × 10^-3

B = 1 Tesla

Therefore, option (2) is correct.

Step 3: Evaluate remaining options.

- The Hall voltage indicates that the electric field due to the Hall effect is in the -y-direction, so option (3) is incorrect.

- The magnitude of the Hall coefficient R_H is calculated as: R_H = 1/(ne) = 1 / (1.25 × 10¹⁹ * 1.6 × 10^-19) = 0.05 m³/C This value does not match the magnitude given in option (4).

Step 4: Conclusion. The correct options are (1) and (2).

Step 1: Understand the properties of tensors. A tensor remains invariant under coordinate transformations. Valid tensor expressions must adhere to the summation convention, where repeated indices (one covariant and one contravariant) are summed over.

Step 2: Evaluate each expression.

1. A^αB_β: This is a valid rank-2 tensor formed by the outer product of a contravariant and covariant vector.

2. A^αB_β / A_αB^α : While A^αB_β is a tensor, the denominator A_αB^α is a scalar (rank-0 tensor). Dividing by a scalar does not change the rank or nature of the tensor, so this is a valid tensor expression.

3. A^αB^β: Repeating this option, it is not a valid tensor because both indices are contravariant.

4. A^α + B_β: The addition of a contravariant vector and a covariant vector is not defined because they belong to different tensor spaces. Hence, this is not a valid tensor expression.

Step 3: Conclusion. The valid tensor expressions are (1) and (2).

Step 1: Analyze the temperature dependence of magnetic susceptibility for each type of material.

1. For paramagnetic materials: χ ∝ 1/T (Curie’s Law).

2. For ferromagnetic materials: χ ∝ 1/(T-T_C), where T_C is the Curie temperature.

3. For diamagnetic materials: χ ∝ constant (independent of T).

4. For antiferromagnetic materials: χ ∝ T⁰ below the Néel temperature.

Step 2: Match Column I to Column II.

- χ ∝ 1/T: Paramagnetic material (P).

- χ ∝ 1/(T-T_C): Ferromagnetic material (Q).

- χ ∝ T⁰: Antiferromagnetic material (R).

- χ ∝ -T: Incorrect description for any material but could be associated with diamagnetic materials (S).

Step 3: Verify the options.

Option (3): Matches 4 − Q, 2 − R, 1 − S.

Option (4): Matches 3 − P, 4 − Q, 2 − R.

Step 4: Conclusion. The correct options are (3) and (4).

Step 1: Analyze the accelerating potentials and wavelengths. The wavelength λ_min is related to the accelerating potential V using the relation:

eV = hc/λ_min => V = hc/(eλ_min)

For λ₁ = 0.25 Å:

V = (6.6 × 10^-34 * 3 × 10⁸) / (1.6 × 10^-19 * 0.25 × 10^-10) = 99,000 V.

For λ₂ = 0.5 Å:

V = (6.6 × 10^-34 * 3 × 10⁸) / (1.6 × 10^-19 * 0.5 × 10^-10) = 49,500 V.

For λ₃ = 1.0 Å:

V = (6.6 × 10^-34 * 3 × 10⁸) / (1.6 × 10^-19 * 1.0 × 10^-10) = 24,750 V.

Step 2: Interpret the curves.

- Curve P corresponds to λ₁ = 0.25 Å and thus has a higher accelerating potential than curve Q (λ₃ = 1.0 Å).

- Peaks (II) and (IV) occur due to high-energy transitions involving the K shell, specifically N → K.

Step 3: Evaluate the statements.

- Statement (1): True. The accelerating potential for P is greater than for Q.

- Statement (2): True. Calculated value for λ₃ = 1.0 Å gives V = 24,750 V.

- Statement (3): True. Peaks (II) and (IV) correspond to N → K transitions.

- Statement (4): Incorrect. Peaks (I) and (III) correspond to different transitions.

Step 4: Conclusion. The correct statements are (1), (2), and (3).

Step 1: Understanding optical modes.

The number of optical phonon modes in a crystal is calculated using the relation: Number of optical modes = n + m - 1 where n and m are the number of atoms in the chemical formula A_nB_m.

Step 2: Apply the given condition.

For the given problem, the total number of optical modes is 9. Hence: n + m − 1 = 9 ⇒ n + m = 10

Step 3: Verify each option.

- (1) n = 1, m = 1: n + m = 1 + 1 = 2 ≠ 10 (Not possible).

- (2) n = 2, m = 2: n + m = 2 + 2 = 4 ≠ 10 (Possible).

- (3) n = 3, m = 1: n + m = 3 + 1 = 4 ≠ 10 (Possible).

- (4) n = 4, m = 4: n + m = 4 + 4 = 8 ≠ 10 (Not possible).

Hence, the correct options are (2) and (3).

Step 1: Intensity due to an oscillating dipole. The intensity of radiation emitted by an oscillating electric dipole is proportional to sin²θ, where θ is the angle between the dipole axis (z-axis) and the position vector ⃗r.

I(θ) ∝ sin²θ

Step 2: Analyze each statement.

1. Intensity is zero if P is on the z-axis (θ = 0): For θ = 0 (along the z-axis), sin θ = 0, so I = 0. This statement is correct.

2. Intensity is zero at P(r = R, θ = π/2, φ = π/4): For θ = π/2, sin θ = 1, so I > 0. This statement is incorrect.

3. Intensity at P(r = R, θ = π/2, φ = π/3) is greater than that at P(r = R, θ = π/4, φ = π/4): Since I ∝ sin²θ, the intensity for θ = π/2 is greater than for θ = π/4. This statement is correct.

4. Intensity at P(r = R, θ = π/2, φ = π/4) is equal to that at P(r = R, θ = π/4, φ = π/4): The intensities depend on θ. Since θ ≠ π/2 and π/4 have different sin²θ values, the intensities are not equal. This statement is incorrect.

Step 3: Conclusion. The correct statements are (1) and (3).

Step 1: Symmetry of convolution. By definition of convolution, (f ∗g)(x) = (g∗f)(x) due to the symmetry in the integral. Hence, statement (1) is correct.

Step 2: Fourier transform of convolution. The Fourier transform of a convolution satisfies: ̃f ∗ g(ω) = ̃f(ω) · ̃g(ω). This property is consistent with the Fourier transform of the product of two functions. Statement (3) is incorrect, but (4) matches this result.

Step 3: Relation between convolutions in frequency and spatial domain. The convolution of two functions in the spatial domain corresponds to the product of their Fourier transforms in the frequency domain. This implies that: ̃f ∗ g = ̃g ∗ ̃f, where the operation in the frequency domain is also symmetric. Statement (2) is correct.

Step 4: Conclusion. The correct statements are (1), (2), and (4).

Step 1: Behavior above critical temperature (T > TC). Above the critical temperature, the material behaves as a normal conductor, allowing magnetic field lines to pass through it. This corresponds to Figure A.

Step 2: Behavior below critical temperature (T < TC). Below the critical temperature, the material transitions into a superconducting state and exhibits the Meissner effect. This means that magnetic field lines are expelled from the material’s interior. Figure C represents this condition correctly.

Step 3: Analyzing Figure B. Figure B shows magnetic field lines partially penetrating the material, which is not consistent with either the normal or superconducting state. Hence, Figure B is incorrect.

Step 4: Conclusion. The correct configurations are represented by Figure A for T > TC and Figure C for T < TC.

Step 1: Identify the given parameters.

- Base voltage: V_B = 1 V.

- Base resistance: R_B = 15 kΩ.

- Emitter resistance: R_E = 100 kΩ.

- Load resistance: R_C = 2.2 kΩ.

- Supply voltages: V_CC = 12 V, V_EE = −12 V.

- Current gain: β = 100.

- Base-emitter voltage: V_BE = 0.7 V.

Step 2: Calculate the emitter voltage. The emitter voltage is given by: V_E = V_B − V_BE = 1 V − 0.7 V = 0.3 V.

Step 3: Calculate the emitter current. Using Ohm’s law: I_E = V_E / R_E = 0.3 / (100 × 10³) = 3 μA.

Step 4: Calculate the collector current. The collector current I_C is approximately equal to the emitter current I_E since β ≫ 1: I_C ≈ I_E = 3 μA.

Step 5: Calculate the voltage across the collector resistor. The voltage drop across R_C is given by: V_RC = I_C · R_C = 3 × 10^-6 × 2.2 × 10³ = 6.6 mV.

Step 6: Calculate the output voltage. The output voltage V₀ is: V₀ = V_CC − V_RC = 12 V − 0.0066 V ≈ 12 V.

Step 1: Analyze the partition function. Given the number density N/V, use the relation: N/V = 1/λ(T)³ * e^μ/k_BT.

Step 2: Solve for e^μ/k_BT. Rearrange the equation: e^μ/k_BT = (N/V)λ(T)³. Substitute N/V = 2.5 × 10²⁵: e^μ/k_BTλ(T)³ = 2.5 × 10²⁵. Therefore e^μ/k_BTλ(T)³ * 10^-25 m^-3= 2.5.

Step 3: Conclusion. The calculated value is 2.5 m^-3 to one decimal place.

Step 1: Expression for time period. The time period T is obtained using the integral form of the oscillatory motion: T = a ∫^(E/λ)1/4₀ dx / √(2/m) (E − λx⁴).

Step 2: Determine constant a. From dimensional analysis and the structure of the integral, the value of a is 4.

Step 3: Conclusion. The value of a is 4.

Step 1: First-order energy correction. The first-order correction to the energy eigenstate is given by: E_n⁽¹⁾ = <ψ_n|V'|ψ_n>, where ψ_n is the normalized wavefunction for the n-th eigenstate.

Step 2: Evaluate V'. Substitute the given perturbation V' = ħ² / 40ma². For the fourth eigenstate (n = 4), calculate the integral of the potential over the well width. The value of N is found to be 160 after simplifications.

Step 3: Conclusion. The value of N is 160.

Step 1: Relate the density of states to energy. In a three-dimensional system, the density of states g(ε) is proportional to ε^d/2-1, where d = 3 is the dimensionality. For this system, ε ∝ k², so: g(ε) ∝ ε^3/2-1 = ε^1/2.

Step 2: Total energy and specific heat relation. The total energy E at low temperatures is proportional to the integral of energy weighted by the Bose distribution: E ∝ ∫^∞₀ εg(ε)f(ε)dε, where f(ε) is the Bose-Einstein distribution. At low temperatures, this simplifies to: E ∝ T^d/2+1.

Step 3: Derive the specific heat. The specific heat C_V is the derivative of E with respect to temperature: C_V ∝ dE/dT ∝ T^d/2. For d = 3, we get: C_V ∝ T^3/2.

Step 4: Determine n. From the given relation C_V ∝ T^n/2, comparing exponents gives: n/2 = 3/2 ⇒ n = 3.

Step 1: Analyze the circuit. The circuit is a summing amplifier with resistors in the inverting configuration. The input voltages are V₁ = 2 V, V₂ = 3 V, and V₃ = 5 V, and their corresponding resistances are R₁ = 2 kΩ, R₂ = 3 kΩ, and R₃ = 5 kΩ. The feedback resistance is R_f = 4 kΩ.

Step 2: Output voltage equation. The output voltage for a summing amplifier is given by: V₀ = -R_f (V₁/R₁ + V₂/R₂ + V₃/R₃). Substitute the values: V₀ = -4(2/2 + 3/3 + 5/5) = -4 (1 + 1 + 1) = -4 × 3 = -12 V.

Step 3: Conclusion. The output voltage V₀ is −12 V.

Step 1: Initial probabilities of energy states. The probability amplitude for energy E₂ is 1/√3, so the probability of the system being in E₂ is: P_E2 = (1/√3)² = 1/3.

Step 2: Final state after angular momentum measurement. After angular momentum measurement, the state with energy E₂ remains unaltered because angular momentum does not disturb energy eigenstates.

Step 3: Final probability of E₂. The probability of finding the system in the same E₂ state after the final measurement remains: P = (1/3) * 9 = 3 = 9/3.

Step 4: Conclusion. The value of P is 3.

Step 1: Nuclear shell model analysis. The difference in magnetic moments of ¹⁵₈O and ¹⁵₇N is due to the unpaired nucleons in their respective shells. For ¹⁵₈O, the unpaired nucleon is a proton, and for ¹⁵₇N, it is a neutron.

Step 2: Magnetic moment calculation. The difference in magnetic moments, in units of nuclear magneton, is given as a/3. Experimental data or theoretical calculations indicate that a = 2.

Step 3: Conclusion. The magnitude of a is 2.