CSIR NET Physical Science July 2026 Memory Based Questions, Check Detailed Question & Solutions Here

CSIR NET Physical Science Exam 2026 Memory based questions is presented here and the initial reaction from candidates suggests that the paper was moderate in difficulty with a strong emphasis on conceptual understanding. While most aspirants found the overall paper manageable, several reported that the question distribution across units was uneven. Quantum Physics, Atomic & Molecular Physics, Electromagnetic Theory (EMT), and Numerical Analysis received significant weightage, whereas some traditionally important topics had fewer questions than expected.

Based on candidates' feedback immediately after the examination, here's a detailed analysis of the CSIR NET Physical Science 2026 exam.

CSIR NET Physical Science July 2026 Memory Based Questions 

Here we delves into memory-based questions from the CSIR NET Physical Science exam, providing in-depth solutions and conceptual clarity. It covers a range of topics from Quantum Mechanics, Statistical Physics, Electrodynamics, and Classical Mechanics, aiming to enhance understanding for competitive exam aspirants. Emphasis is placed on problem-solving techniques and core theoretical principles.

CSIR NET Physical Science Overall Difficulty Level

According to the majority of candidates, the CSIR NET Physical Science 2026 paper was not overly difficult. Students described it as a doable exam, but many emphasized that success depended more on conceptual clarity than lengthy calculations.

Several aspirants mentioned that:

• Part A was comparatively easy.

• Part B remained balanced and manageable.

• Part C required stronger conceptual understanding and careful interpretation.

Although some students completed almost all the questions, others felt that the conceptual nature of the paper made accuracy more challenging.

General Aptitude: Entropy in Processes

A question in the General Aptitude section concerned a process where entropy would decrease. While boiling water and melting ice exemplify entropy increase, boiling an egg in water demonstrates an entropy decrease. * (Memory Tip: When an egg boils, its globular proteins convert into fibrous proteins, and the liquid solidifies. This transformation from a more disordered liquid state to a more ordered solid state results in an entropy decrease.)*

Statistical Mechanics: Ising Model for Two Spins

This problem involved a two-spin system with the Hamiltonian: H = -jS₁S₂ - hS₁ - hS₂, where S₁ and S₂ are spins (+1 or -1), and h is a magnetic field. The objective was to find the average value of the spin ( or ).

Solution Steps:

1. Energy States (E_i):

• (S₁, S₂) = (+1, +1): E = -j - 2h

• (S₁, S₂) = (-1, -1): E = -j + 2h

• (S₁, S₂) = (+1, -1): E = +j

• (S₁, S₂) = (-1, +1): E = +j

2. Partition Function (Z):
   Z = 2e^(βj)cosh(2βh) + 2e^(-βj)

3. Expectation Value of S₁ ():
   = [ 2e^(βj)sinh(2βh) ] / [ 2e^(βj)cosh(2βh) + 2e^(-βj) ]
   = e^(βj)sinh(2βh) / [ e^(βj)cosh(2βh) + e^(-βj) ]

4. Approximation for Small h:
   For small h, using sinh(x) ≈ x and cosh(x) ≈ 1:
   ≈ βh e^(βj) / cosh(βj)
   Due to symmetry, and are identical.

Atomic & Molecular Physics: Ground State Term for Ar 3d³ 4s²

To determine the Ground State Term Symbol for Argon [Ar] 3d³ 4s²:

1. Relevant Electrons: Only 3d³ electrons contribute (4s² is filled).

2. L (Orbital Angular Momentum): For d-electrons, l = 2. Filling according to Hund's rules (first three with ms = +1/2):

• ml: +2, +1, 0, -1, -2

• Electrons: 1, 1, 1, 0, 0

• L = Σml = (+2) + (+1) + (0) = 3.

• L=3 corresponds to the F term.

3. S (Total Spin Angular Momentum): All three electrons have ms = +1/2.

• S = Σms = (1/2) + (1/2) + (1/2) = 3/2.

4. Multiplicity (2S+1): 2(3/2) + 1 = 4.

5. J (Total Angular Momentum): For less than half-filled subshells (d³), J = |L - S|.

• J = |3 - 3/2| = 3/2.

6. Term Symbol: ²S⁺¹L_J = 4F_3/2.

Statistical Mechanics: Microstates for Distinguishable Spin-1 Particles

Six distinguishable, non-interacting spin-1 particles are distributed into three energy levels: -e, 0, and +e. We need to find the number of microstates where the total energy is zero.

Conditions:

1. n_(-e) + n_0 + n_(+e) = 6 (Total 6 particles)

2. n_(-e)(-e) + n_0(0) + n_(+e)(+e) = 0 => n_(-e) = n_(+e)

Possible Distributions and Number of Ways:

1. (0, 6, 0): All 6 particles in the 0 energy level.

• Ways: 6C0 * 6C6 * 0C0 = 1

2. (1, 4, 1): One in -e, four in 0, one in +e.

• Ways: 6C1 * 5C4 * 1C1 = 6 * 5 * 1 = 30

3. (2, 2, 2): Two in -e, two in 0, two in +e.

• Ways: 6C2 * 4C2 * 2C2 = 15 * 6 * 1 = 90

4. (3, 0, 3): Three in -e, zero in 0, three in +e.

• Ways: 6C3 * 3C0 * 3C3 = 20 * 1 * 1 = 20

Total Number of Microstates: 1 + 30 + 90 + 20 = 141.

Solid State Physics: X-ray Diffraction Selection Rules for KCl and NaCl

The discussion involved Allowed Planes in KCl and NaCl for X-ray diffraction.

• NaCl Structure: A pure FCC (Face-Centered Cubic) structure. Its selection rule dictates that reflections are allowed only if all h, k, l indices are either all even or all odd (e.g., (111), (200) are allowed; (100), (110) are forbidden).

• KCl Structure: Has the same underlying structure as NaCl but behaves like a Simple Cubic (SC) lattice for X-ray diffraction due to the similar scattering factors of K⁺ and Cl⁻ ions. This implies different allowed reflection conditions compared to a standard FCC lattice, with specific planes being forbidden (e.g., (111) and (100) are generally not allowed for KCl when considered as a simple cubic). The lecturer further clarifies that it behaves as "twice of simple cubic."

Quantum Mechanics: Minimum Energy of a Hamiltonian

Given the Hamiltonian H = αL² + βħL_z, applying it to an |l, m> state yields energy E_lm = αl(l+1)ħ² + βmħ². The problem asked for conditions for minimum energy.

Analysis of Conditions:

• l = 0, m = 0: E = 0.

• l = 1, m = 1: E = 2αħ² + βħ². (Positive if α, β positive)

• l = 1, m = -1: E = 2αħ² - βħ².

• If α < β/2, then 2α < β, making 2α - β negative.

• This results in negative energy, which is the minimum energy value among the options, assuming α, β are generally positive constants.

Therefore, the minimum energy occurs when l = 1, m = -1, and α < β/2.

Numerical Methods: Numerical Integration (Trapezoidal and Simpson's 1/3 Rule)

To evaluate ∫ x⁴ dx from 0 to 4 with interval size h = 1:

1. Table of x and y (f(x)) values:

2. Trapezoidal Rule:
   ∫ f(x) dx ≈ (h/2) [y₀ + y_n + 2(y₁ + y₂ + … + y_n-1)]
   ∫ f(x) dx ≈ (1/2) [0 + 256 + 2(1 + 16 + 81)] = (1/2) [256 + 196] = 226.

3. Simpson's 1/3 Rule:
   ∫ f(x) dx ≈ (h/3) [y₀ + y_n + 4(y₁ + y₃ + …) + 2(y₂ + y₄ + …)]
   ∫ f(x) dx ≈ (1/3) [0 + 256 + 4(1 + 81) + 2(16)] = (1/3) [256 + 328 + 32] = (1/3) [616] ≈ 205.33.

Numerical Methods: Newton-Raphson Method

The Newton-Raphson method finds roots of f(x) = 0. Given f(x) = (x² - 1)(x - 2) = x³ - 2x² - x + 2, its roots are +1, -1, and +2. The derivative is f'(x) = 3x² - 4x - 1.

The iteration formula is xₙ₊₁ = xₙ - f(xₙ) / f'(xₙ).

Convergence of Trial Points:

Complex Analysis: Analytic Functions

Given f(z) = u + iv is an analytic function, and the real part u = x³ - 3xy². The task is to find the imaginary part v.

Using the Cauchy-Riemann (CR) equation ∂u/∂x = ∂v/∂y:

1. ∂u/∂x = 3x² - 3y².

2. We need to find v such that ∂v/∂y = 3x² - 3y². Integrating with respect to y, v = 3x²y - y³ + C(x). Comparing with options, the correct v would be of this form (e.g., v = 3x²y - y³).

Quantum Mechanics: Perturbation Theory

To calculate the first-order energy correction for the first excited state of a particle in a potential well (from 0 to 2L) with perturbation V' = V₀L δ(x - 3L/2):

1. First Excited State Wave Function (n=2):
   ψ₂(x) = √(1 / L) sin(πx / L)

2. First-Order Energy Correction (E₁'):
   E₁' = ⟨ψ₂|V'|ψ₂⟩ = ∫₀²ᴸ ψ₂(x) V'(x) ψ₂(x) dx
   E₁' = ∫₀²ᴸ (1/L) sin²(πx / L) [V₀L δ(x - 3L/2)] dx
   Using the Dirac delta property, evaluate at x = 3L/2:
   E₁' = (1/L)  V₀L * sin²(π(3L/2) / L) = V₀ * sin²(3π/2) = V₀ * (-1)² = V₀.
   The first-order energy correction is V₀.

Electromagnetism: Spherical Polar Coordinate Transformations

To determine ∂θ̂ / ∂φ in spherical polar coordinates, with r̂ = sinθ cosφ î + sinθ sinφ ĵ + cosθ k̂:

1. Derive θ̂: θ̂ = ∂r̂ / ∂θ = cosθ cosφ î + cosθ sinφ ĵ - sinθ k̂.

2. Calculate ∂θ̂ / ∂φ: ∂θ̂ / ∂φ = -cosθ sinφ î + cosθ cosφ ĵ.

• (Memory Tip: When calculating partial derivatives, terms without the variable of differentiation (e.g., φ in ∂/∂φ) become zero, so the k̂ component vanishes.)*

Quantum Mechanics: Expectation Value of Energy

Given E₁ = 1.2 eV and state |ψ⟩ = 3|s₁⟩ + 2|s₂⟩ + 3|s₄⟩.

The instructor states the sum of squares as 3² + 2² + 3² = 16, leading to a normalization factor of N = √16 = 4. (Note: Mathematically, 3²+2²+3² = 22, not 16. However, the lecture explicitly states 16 and proceeds with 4 as the normalization factor, which is preserved here as per instructions.)

1. Normalized Probabilities: P(s₁) = (3/4)² = 9/16, P(s₂) = (2/4)² = 4/16 = 1/4, P(s₄) = (3/4)² = 9/16. (Note: The lecturer's calculation for P(s4) was (3/16), which implies the initial coefficient for |s₄⟩ was √3, not 3, given the normalization factor of 4. This is preserved as part of the given information).

2. Energy of States (Eₙ = n²E₁): E₁ = 1.2 eV, E₂ = 4 * 1.2 eV, E₄ = 16 * 1.2 eV.

3. Expectation Value:
   ⟨E⟩ = (9/16) * (1.2) + (1/4) * (4 * 1.2) + (3/16) * (16 * 1.2)
   ⟨E⟩ = 0.675 + 1.2 + 3.6 = 5.475 eV, closest to 5.5 eV.

Electronics: JK Flip-Flop Asynchronous Counter

A circuit with three JK flip-flops (Q₂, Q₁, Q₀) as an asynchronous up counter (MOD 8), with J=K=1 (toggle mode) and clock triggered by inverted output (Q̄) of previous flip-flop. Initial data: 100.

Output Sequence:

The output after the 9th clock pulse is 101.

Electronics: Op-Amp Circuit with Zener Diode

For an Op-Amp circuit with a Zener diode, critical principles are:

• Virtual Ground / Virtual Short Concept: In an ideal Op-Amp with negative feedback, V- ≈ V+.

• Kirchhoff's Current Law (KCL): No current enters Op-Amp input terminals.

• Zener Diode Behavior:

• Forward Bias: Behaves like a normal diode (~0.7V drop).

• Reverse Bias (below Vz): Open circuit.

• Reverse Bias (at or above Vz): Maintains constant Zener voltage (Vz).

• (Memory Tip: Analyzing the Zener's state (forward, reverse below Vz, or breakdown) based on circuit voltages is key to determining V_out.)

Electronics: Modulation for Frequency Shifting

The primary benefit of using modulation to shift a signal to a higher frequency is to reduce the impact of low-frequency disturbances, specifically 1/F noise (flicker noise), which is inversely proportional to frequency.

Nuclear and Particle Physics: Mirror Nuclei

Given mirror nuclei with ΔE_c = 3.53 MeV, R = R₀A¹ᐟ³ (R₀ = 1.5 fm), and e² / (4πε₀) = 1.44 MeV·fm.

The Coulombic energy difference (ΔE_c) for mirror nuclei is:

ΔE_c = 2 * (3/5) * (e² / (4πε₀)) * (Z - 1) / (R₀A¹ᐟ³)

For mirror nuclei, A = 2Z - 1. Substituting this into the equation and solving for Z (by matching given options) yields Z = 8.

Nuclear and Particle Physics: Nuclear Magnetic Moment

To calculate the nuclear magnetic moment for a nucleus with Z=8 (protons) and N=9 (neutrons), which is ¹⁷O.

This is an odd-N nucleus, so the magnetic moment is determined by the last odd neutron.

1. Neutron Shell Model Configuration: The 9th neutron occupies the 1d₅ᐟ₂ state.

2. Quantum Numbers: From 1d₅ᐟ₂, l = 2 (d-shell) and j = 5/2. This is a j = l + 1/2 state.

3. Nuclear Magnetic Moment Formula (odd neutron, j = l + 1/2):
   μ = g_s * (1/2) + g_l * l
   Where g_s ≈ -3.826 (for neutron) and g_l = 0 (for neutron orbital).
   μ = (-3.826) * (1/2) + 0 * 2 = -1.913 nuclear magnetons (μ_N).

Magnetic Moment Formula Variations for Odd Nucleons:

Particle Physics: Reaction Analysis (Spin & Parity)

To determine if π⁺ → π⁰ + π⁰ is allowed or forbidden, we check conservation laws.

• Spin Analysis: Pions (π⁺, π⁰) all have spin 0. Initial total spin (π⁺) = 0. Final total spin (π⁰ + π⁰) = 0. Spin is conserved.

• Parity Analysis: Pions are pseudoscalar mesons, meaning they have negative intrinsic parity (-1).

• Initial state parity (π⁺) = -1.

• Final state parity (π⁰ + π⁰) = P(π⁰) * P(π⁰) = (-1) * (-1) = +1.

• Since parity changes from -1 to +1, Parity is not conserved.

Therefore, the reaction π⁺ → π⁰ + π⁰ is forbidden due to non-conservation of parity.

Electrodynamics: Rectangular Waveguide Cut-off Frequency

For a rectangular waveguide with dimensions A = 3 cm (0.03 m) and B = 2 cm (0.02 m), filled with air (εR=1, μR=1), the cut-off frequency is given by f_c = (C/2) * √[ (m/A)² + (n/B)² ].

For the lowest frequency, we consider the dominant TE₁₀ mode (m=1, n=0):

f_c(₁₀) = (3 × 10⁸ / 2) * √[ (1 / 0.03)² + (0 / 0.02)² ] = 1.5 × 10⁸ * (100/3) = 0.5 × 10¹⁰ Hz = 5 GHz.

Statistical Mechanics: Number of Microstates for N=2, G=10

Given N = 2 particles and G = 10 states, the number of microstates for different particle types:

• Fermions: Each state occupied by at most one particle.

• Combinations: ¹⁰C₂ = (10 * 9) / (2 * 1) = 45 microstates.

• Bosons: Multiple particles can occupy the same state.

• Formula: (N + G - 1) C N = (2 + 10 - 1) C 2 = ¹¹C₂ = (11 * 10) / (2 * 1) = 55 microstates.

• Classical / Distinguishable Particles (Maxwell-Boltzmann): Each particle is distinct.

• Formula: G^N = 10² = 100 microstates.

Quantum Mechanics: Commutator Bracket Evaluation

To evaluate [R, e^(αS)] given [R, [R,S]] = 0.

The condition [R, [R,S]] = 0 implies that [R,S] commutes with R.

Using the identity [A, f(B)] = f'(B) [A, B] when [A, [A,B]] = 0, we can directly state the result.

The commutator evaluates to α[R,S]e^(αS).

Quantum Mechanics: Time-Dependence of Operator Expectation Value

For an operator O and Hamiltonian H that are both time-independent, the time evolution of the expectation value is given by:

d/dt = (i/ħ) <[H, O]> + <∂O/∂t>

Since O is time-independent, <∂O/∂t> = 0.

Thus, d/dt = (i/ħ) <[H, O]>.

If O and H commute (i.e., [H,O] = 0), then d/dt = 0, meaning the expectation value of O is time-independent.

Quantum Mechanics: Time Evolution of a State and Probability

Given three energy eigenvalues: E₀, 2E₀, 3E₀ with corresponding eigenstates |Ψ₁⟩, |Ψ₂⟩, |Ψ₃⟩. Initial state at t=0: |Ψ(0)⟩ = (1, 0, 0)ᵀ.

To find the probability of finding the system in a specific new state at time t = πħ / (2E₀):

1. Express |Ψ(0)⟩ as a linear combination of normalized eigenstates: |Ψ(0)⟩ = c₁|Ψ₁⟩ + c₂|Ψ₂⟩ + c₃|Ψ₃⟩.

2. Apply time evolution: |Ψ(t)⟩ = c₁|Ψ₁⟩e^(-iE₁t/ħ) + c₂|Ψ₂⟩e^(-iE₂t/ħ) + c₃|Ψ₃⟩e^(-iE₃t/ħ).

3. Calculate the probability P = ||² for the target state |Φ⟩.

Linear Algebra: 4D Vector Space and Linear Independence

Given three vectors in a 4D vector space: v₁ = (1, 1, 0, 0)ᵀ, v₂ = (0, 1, 1, 0)ᵀ, v₃ = (0, 0, 1, 1)ᵀ. The task is to find a fourth vector (v₄) such that {v₁, v₂, v₃, v₄} form a basis. This requires linear independence.

One method is to check if v₄ can be written as a linear combination of v₁, v₂, v₃. For an option like v₄ = (1,0,-1,0)ᵀ, if we attempt to write it as a v₁ + b v₂ + c v₃, a system of equations arises. The lecturer concludes that this option results in an inconsistent system, implying v₄ is linearly independent. (Note: A direct manual calculation may show this vector to be linearly dependent on v1, v2, v3, as the system of equations for a, b, c is consistent. However, the instructional emphasis on the method for checking linear independence via inconsistency/consistency is retained here.)

Quantum Mechanics: Sudden vs. Adiabatic Approximation

The distinction between these approximations lies in the rate of change of the Hamiltonian:

Mathematical Physics: Contour Integration (Complex Analysis)

To evaluate ∫ (dθ / (5 - 4cosθ)) over a unit circle using the Residue Theorem:

1. Transform to Complex Integral: Substitute dθ = dz / (iz) and cosθ = (z + 1/z) / 2.
   The integral becomes: ∫ (1 / i) * (1 / (2z² - 5z + 2)) dz.

2. Find Poles: Solve 2z² - 5z + 2 = 0 => (2z - 1)(z - 2) = 0. Poles are at z = 1/2 and z = 2.

3. Poles Inside Contour: For a unit circle, only z = 1/2 is inside.

4. Calculate Residue at z = 1/2:
   Residue(f, 1/2) = lim (z→1/2) [ (z - 1/2) * (1 / (i(2z - 1)(z - 2))) ] = i/3.

5. Apply Residue Theorem: Integral = 2πi * (i/3) = -2π/3. (Note: The lecturer states the final result as 2π/3. However, based on the residue calculation, the result is -2π/3. This discrepancy is preserved as per instructional emphasis on the methodology.)

Quantum Mechanics: Scattering - Born Approximation

For elastic scattering by a spherical square well potential V(r) = V₀ (for r < R, and 0 otherwise), using the Born Approximation under the condition QR << 1 (weak potential limit):

1. The scattering amplitude f(θ) is proportional to R³.

2. The differential scattering cross-section (dσ/dΩ) is proportional to |f(θ)|².

3. Therefore, dσ/dΩ ∝ R⁶.

• (Memory Tip: For weak potential scattering with the QR << 1 limit, the differential cross-section is proportional to R⁶.)*

Classical Mechanics: Lagrangian and Hamiltonian

Problem: Given Lagrangian L = ẋ² + x ẋ + x²/2, find the Hamiltonian H(x, p, t).

1. Conjugate Momentum (pₓ): pₓ = ∂L / ∂ẋ = 2ẋ + x.

2. Express ẋ: ẋ = (pₓ - x) / 2.

3. Hamiltonian (H = pₓ ẋ - L):
   Substitute ẋ and L terms to get:
   H = pₓ²/2 - x²/2 - xpₓ.

Classical Mechanics: Phase Space Analysis

Problem: Plot phase space curves for potential V(x) = (1/2)x² + (1/3)x³.

1. Equilibrium Points (dV/dx = 0):
   dV/dx = x + x² = x(1 + x) = 0.
   Equilibrium points at x = 0 and x = -1.

2. Stability Analysis (d²V/dx² = 1 + 2x):

• At x = 0: d²V/dx² = 1 > 0 => Stable equilibrium (local minimum), V(0) = 0.

• At x = -1: d²V/dx² = -1 < 0 => Unstable equilibrium (local maximum), V(-1) = 1/6.

• (Memory Tip: Stable equilibrium is like a "bowl" (minimum), unstable is an "inverted bowl" (maximum).)*

Phase Space Plots for Different Total Energies (E):

• Low Energy (E < 1/6): Phase space trajectories are closed loops around x=0, indicating bounded oscillations.

• Critical Energy (E = 1/6): Forms a separatrix passing through the unstable equilibrium at x=-1, typically a figure-eight shape.

• High Energy (E > 1/6): Phase space trajectories are open curves, indicating unbound motion.

Classical Mechanics: Equation of Orbit

Problem: For an orbit r = k e^(αθ), find the proportionality of the central force, F(r) ∝ rⁿ.

1. Reciprocal Substitution: u = 1/r = (1/k) e^(-αθ).

2. Derivatives:

• du/dθ = -αu

• d²u/dθ² = α²u

3. Equation of Orbit: d²u/dθ² + u = - (F / (mh²u²)).
   Substitute derivatives: α²u + u = - (F / (mh²u²)).
   (α² + 1)u = - (F / (mh²u²)).

4. Solve for F: F = - mh² (α² + 1) u³.

5. Express in terms of r: Since u = 1/r, F ∝ u³ ∝ (1/r)³ ∝ r⁻³.

Classical Mechanics: Normal Frequencies of Coupled Oscillators

Problem: Three masses (m) in a line, coupled by springs (2k, k, k, 2k). Find normal frequencies (ω).

1. Kinetic Energy (T): T = (1/2)m ẋ₁² + (1/2)m ẋ₂² + (1/2)m ẋ₃².

2. Potential Energy (V): V = (1/2)k [3x₁² + 2x₂² + 3x₃² - 2x₁x₂ - 2x₂x₃].

3. Matrices (T and V):
   T = [[m, 0, 0], [0, m, 0], [0, 0, m]]
   V = k * [[3, -1, 0], [-1, 2, -1], [0, -1, 3]]

4. Secular Equation det(V - ω²T) = 0:
   Solving the determinant leads to three normal frequencies squared:
   ω₁² = 3k/m, ω₂² = 4k/m, and ω₃² = k/m.

Relativistic Electrodynamics: Field Transformations

For a capacitor in its rest frame creating an electric field E, its transformation to a frame moving with velocity v follows specific rules:

• Parallel Component: The electric field component parallel to v remains unchanged (E' || = E ||).

• Perpendicular Component: The electric field component perpendicular to v is modified by the Lorentz factor (γ) (E' ⊥ = γ E ⊥).

For example, if E is along the x-axis and the observer moves along the y-axis, E is perpendicular to v, so the electric field in the moving frame E'x = γ Ex.

CSIR NET Physical Science Expected Cut Off

Since Part A was reportedly easier and many students attempted a good number of questions, experts believe the CSIR NET Physical Science 2026 cut-off may remain similar to or slightly higher than last year, depending on the final normalization and overall performance.

Early discussions among faculty suggest the JRF cut-off for the General category could remain in the range of approximately 85–92 marks, though this is only an initial estimate and the actual cut-off will be released by NTA along with the official results.