GSEB 12th Science Physics Important Derivations 2026 – Step-by-Step Board Exam Guide

Why Derivations Are the Key to Scoring 90+ in GSEB 12th Physics

In the GSEB 12th Science Physics paper, derivations are the highest-yielding questions. While theory questions can be subjective and numericals carry a risk of calculation errors, derivations are purely mathematical. If your diagram is correct, your assumptions are stated, and your algebraic steps are logically sound, the examiner is required to award 100% full marks.

Across Sections C and D (which make up 44 marks out of the 80-mark theory paper), derivations contribute roughly 16 to 22 marks. That is nearly one-quarter of your total score. If you lock down these derivations, you guarantee a high-scoring baseline.

📌 Practice strategy: Download the official past papers from our GSEB 12th Science Papers page and practice writing out these derivations on unruled paper. This replicates actual exam conditions.

GSEB Class 12 Physics Chapter-wise Marks & Derivation Weightage

Top 10 Most Repeated Physics Derivations

1. Electric Field due to an Electric Dipole at an Axial Point

Chapter: Electric Charges and Fields | Typical Marks: 3 | Frequency: Very High

Setup & Assumptions: Consider an electric dipole consisting of two equal and opposite charges -q and +q separated by a distance 2a. Let the dipole moment be p = q(2a) directed from -q to +q. We wish to calculate the electric field at point P situated on the axis of the dipole at a distance r from its center O.

Derivation Steps:

1. The distance of point P from the positive charge +q is (r − a). The electric field E₁ due to +q at P is:
   E₁ = q / [4πε₀(r − a)²] (directed away from +q, i.e., along OP)
2. The distance of point P from the negative charge -q is (r + a). The electric field E₂ due to -q at P is:
   E₂ = q / [4πε₀(r + a)²] (directed towards -q, i.e., along PO)
3. The net electric field E at point P is the vector sum:
   E = E₁ − E₂
   E = [q / (4πε₀)] * [1/(r − a)² − 1/(r + a)²]
4. Taking the common denominator and simplifying the numerator:
   E = [q / (4πε₀)] * [((r + a)² − (r − a)²) / (r² − a²)²]
   E = [q / (4πε₀)] * [(4ra) / (r² − a²)²]
5. We can write the numerator as 2 * r * (q * 2a). Since p = q(2a), we get:
   E = 2pr / [4πε₀(r² − a²)²]
6. For a short dipole (where the distance r is much larger than the dipole half-length a, i.e., r >> a), we can neglect a² in comparison to r²:
   E = 2p / [4πε₀r³] (along the direction of dipole moment p)

Highlight the final vector equation: E = 2p / (4πε₀r³).

2. Gauss's Law Application: Electric Field due to an Infinitely Long Straight Wire

Chapter: Electric Charges and Fields | Typical Marks: 3 | Frequency: High

Setup & Assumptions: Consider an infinitely long, thin straight wire with a uniform linear charge density λ (charge per unit length). To calculate the electric field E at a distance r from the wire, we choose a cylindrical Gaussian surface of radius r and length l coaxial with the wire.

Derivation Steps:

1. The cylindrical Gaussian surface has three parts: two flat circular end caps and one curved cylindrical surface.
2. For the flat end caps, the electric field vector E is perpendicular to the area vector dS (θ = 90°). Therefore, the electric flux through the end caps is zero:
   ∫ E · dS = ∫ E dS cos(90°) = 0
3. For the curved cylindrical surface, the electric field E is parallel to the area vector dS at every point (θ = 0°). The flux is:
   Φ = ∫ E · dS = E ∫ dS = E * (2πrl) (since the curved area of cylinder is 2πrl)
4. According to Gauss's Law, the total electric flux Φ through a closed surface is equal to the enclosed charge q divided by ε₀:
   Φ = q / ε₀
5. The charge enclosed inside the Gaussian cylinder of length l is q = λl. Substituting this:
   E * (2πrl) = λl / ε₀
6. The length l cancels out from both sides, yielding the electric field:
   E = λ / (2πε₀r)

Highlight the final equation: E = λ / (2πε₀r). Note that E is inversely proportional to r (E ∝ 1/r).

3. Relation Between Drift Velocity and Electric Current

Chapter: Current Electricity | Typical Marks: 3 | Frequency: Very High

Setup & Assumptions: Consider a conductor of length l and uniform cross-sectional area A. Let the number density of free electrons (number of free electrons per unit volume) be n. When a potential difference V is applied across the ends of the conductor, an electric field E = V/l is established, causing the free electrons to drift with velocity v_d towards the positive terminal.

Derivation Steps:

1. The total volume of the conductor is: Volume = A * l.
2. The total number of free electrons in the conductor is: N = n * A * l.
3. Since each electron carries a charge e, the total charge Q contained in the conductor is:
   Q = N * e = nAle
4. The time taken by the free electrons to cross the entire length l of the conductor is:
   t = l / v_d
5. Electric current I is defined as the rate of flow of charge through any cross-section:
   I = Q / t
6. Substitute Q and t into the current formula:
   I = nAle / (l / v_d)
   I = nAev_d

Highlight the final relation: I = nAev_d. Since current density J = I/A, we can also derive J = nev_d.

4. Magnetic Field at the Center of a Circular Current-Loop (Biot-Savart Law)

Chapter: Moving Charges and Magnetism | Typical Marks: 3 | Frequency: High

Setup & Assumptions: Consider a circular loop of radius R carrying a steady current I. We want to find the magnetic field B at the center O of the loop. According to the Biot-Savart Law, the magnetic field dB due to a small current element dl is given by:
dB = (μ₀ / 4π) * (I * dl * sinθ) / R²

Derivation Steps:

1. For any current element dl on the circular loop, the angle θ between the length element dl and the position vector r pointing to the center O is exactly 90° (since the radius is perpendicular to the tangent at every point).
2. Substituting θ = 90° (sin 90° = 1) into the Biot-Savart equation:
   dB = (μ₀ / 4π) * (I * dl) / R²
3. The direction of the magnetic field dB due to all elements is the same (directed perpendicular to the plane of the loop, pointing inward or outward depending on current direction, by the right-hand grip rule). Therefore, we can find the total magnetic field B by integrating dB over the entire loop:
   B = ∫ dB = ∫ (μ₀ / 4π) * (I * dl) / R²
4. Since μ₀, I, and R are constant, they can be pulled out of the integral:
   B = (μ₀I / 4πR²) ∫ dl
5. The integral of dl over the circular loop is equal to its circumference, 2πR:
   ∫ dl = 2πR
6. Substitute this back into the equation:
   B = (μ₀I / 4πR²) * (2πR)
   B = μ₀I / (2R)

For a coil consisting of N turns, multiply the result by N: B = μ₀NI / (2R).

5. Resonant Frequency in a Series LCR AC Circuit

Chapter: Alternating Current | Typical Marks: 3 | Frequency: High

Setup & Assumptions: Consider a resistor R, an inductor L, and a capacitor C connected in series across an alternating voltage source V = V_m sin(wt). The impedance Z of a series LCR circuit is given by:
Z = √[R² + (X_L − X_C)²]
where X_L = wL is the inductive reactance, and X_C = 1/(wC) is the capacitive reactance.

Derivation Steps:

1. Electrical resonance occurs when the amplitude of the current in the circuit reaches its maximum value. This happens when the impedance Z is at its absolute minimum.
2. For Z to be minimum, the term (X_L − X_C)² must equal zero. This yields the resonance condition:
   X_L = X_C
3. Substitute the formulas for X_L and X_C at the resonant angular frequency w_r:
   w_r * L = 1 / (w_r * C)
4. Rearranging the terms to solve for w_r:
   (w_r)² = 1 / (LC)
   w_r = 1 / √CD
5. Since angular frequency w_r = 2πf_r, where f_r is the linear resonant frequency:
   2πf_r = 1 / √CD
   f_r = 1 / (2π√CD)

Highlight the final formula: f_r = 1 / (2π√(LC)). Under this condition, Z = R, and the power factor cosφ = 1 (purely resistive circuit).

6. Lens Maker's Formula

Chapter: Ray Optics and Optical Instruments | Typical Marks: 4 | Frequency: Extremely High (Guaranteed Section D candidate)

Setup & Assumptions: Consider a thin convex lens made of a material of refractive index n₂ placed in a medium of refractive index n₁. Let R₁ and R₂ be the radii of curvature of the two spherical surfaces of the lens. Let a point object O be placed on the principal axis in front of the first surface.

Derivation Steps:

1. Refraction at the first surface (radius R₁): The light ray from O travels from medium n₁ to n₂. If the second surface were absent, it would form an image at position v₁. Using the spherical refraction formula:
   n₂/v₁ − n₁/u = (n₂ − n₁)/R₁ —— (Equation 1)
2. Refraction at the second surface (radius R₂): The refracting ray now travels from medium n₂ to n₁. The image I₁ formed by the first surface acts as a virtual object for this surface, forming the final image at v. Using the spherical refraction formula:
   n₁/v − n₂/v₁ = (n₁ − n₂)/R₂ —— (Equation 2)
3. Adding Equation 1 and Equation 2:
   (n₂/v₁ − n₁/u) + (n₁/v − n₂/v₁) = (n₂ − n₁)/R₁ + (n₁ − n₂)/R₂
4. The term n₂/v₁ cancels out. Rearranging the left side and factoring out (n₂ − n₁) on the right side:
   n₁(1/v − 1/u) = (n₂ − n₁) * [1/R₁ − 1/R₂]
5. Divide both sides by n₁:
   (1/v − 1/u) = (n₂/n₁ − 1) * [1/R₁ − 1/R₂]
6. Let n = n₂/n₁ be the relative refractive index of the lens with respect to the medium:
   (1/v − 1/u) = (n − 1) * [1/R₁ − 1/R₂]
7. By definition of the focal length, if the object is at infinity (u = ∞), the image is formed at the principal focus (v = f). Substituting u = ∞ and v = f:
   1/f − 1/∞ = (n − 1) * [1/R₁ − 1/R₂]

Highlight the final Lens Maker's Formula: 1/f = (n − 1) * (1/R₁ − 1/R₂).

7. Refractive Index of a Prism (Prism Formula)

Chapter: Ray Optics and Optical Instruments | Typical Marks: 4 | Frequency: High

Setup & Assumptions: Consider a triangular glass prism of refracting angle A. Let a ray of light PQ be incident on the first face at angle i. It gets refracted along QR at angle r₁ inside the prism, and finally emerges along RS at angle e from the second face. Let the angle of deviation (the angle between the incident ray and the emergent ray) be δ.

Derivation Steps:

1. In the quadrilateral AQNR, the sum of opposite angles is 180° (since the normals at Q and R are perpendicular to the faces):
   A + ∠QNR = 180°
2. In the triangle QNR, the sum of angles is 180°:
   r₁ + r₂ + ∠QNR = 180°
3. Comparing both equations yields the relation between the prism angle A and internal refraction angles:
   A = r₁ + r₂ —— (Equation 1)
4. The total angle of deviation δ is the sum of deviations at the two refracting faces:
   δ = (i − r₁) + (e − r₂)
   δ = (i + e) − (r₁ + r₂)
5. Substitute Equation 1 into the deviation equation:
   δ = i + e − A
   A + δ = i + e —— (Equation 2)
6. At the position of minimum deviation (where δ = D_m), the ray travels parallel to the base of the prism. Under this condition, the angle of incidence equals the angle of emergence (i = e), and the internal angles are equal (r₁ = r₂ = r).
7. Substitute these conditions into Equation 1 and Equation 2:
   A = r + r = 2r ⇒ r = A/2
   A + D_m = i + i = 2i ⇒ i = (A + D_m)/2
8. According to Snell's Law, the refractive index n of the prism material is:
   n = sin(i) / sin(r)
9. Substitute the values of i and r:
   n = sin[(A + D_m)/2] / sin(A/2)

Highlight the final Prism Formula: n = sin[(A + D_m)/2] / sin(A/2).

Common Mistakes That Cost Marks in GSEB Physics Derivations

1. Drawing Diagrams Freehand: Drawing ray diagrams or magnetic field loops without a ruler and pencil makes the vectors look sloppy. GSEB examiners are instructed to deduct 1 full mark if the diagram is not clean.
2. Omitting Vector Arrows: Forgetting to draw arrows indicating direction of electric fields (E), magnetic fields (B), or ray propagation will result in a 0.5-mark penalty. Direction is part of the physics.
3. Not Defining Assumptions: Skipping the initial statement (e.g., "Consider an electric dipole of moment p separating charge...") and diving directly into equations will result in deductions. You must define the variables.
4. Skipping Algebraic Steps: Jumping from a complex fraction directly to the simplified final version without showing the factorization or cancellation steps will look like you copied the answer, resulting in step-mark deductions.
5. Forgetting to state approximations: In dipole and slit derivations, you must explicitly state when an approximation is used (e.g., "Since r >> a, we can neglect a²").

5-Day Revision Strategy for Derivations

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Frequently Asked Questions — GSEB 12th Physics Derivations

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