JEE Main 2027 Physics: 20 Most Important Formulas (With Solved Examples & PDF)

1Projectile Motion

• Time of flight: T = 2u sinθ / g
• Maximum height: H = u² sin²θ / 2g
• Range: R = u² sin2θ / g
• Range is maximum at θ = 45°: R(max) = u² / g
• H / R = tanθ / 4
• H ∝ sin²θ and T ∝ sinθ, so the ratio of heights = the square of the ratio of times

Example: A gun fires bullets with a maximum range of 6.4 m. R(max) = u²/g = 6.4 with g = 10 gives u² = 64, so u = 8 m/s.

2Newton's Laws of Motion (Bodies in Contact and Pulleys)

• a = total net force / total mass
• To find tension, draw a free body diagram of each body and apply F = ma.

Example: On one side of a pulley hangs 4 kg, and on the other side 4 kg + 6 kg (10 kg in total). a = (100 − 40)/14 = 30/7 m/s². For the 4 kg mass, T₁ = 40 + 4a = 400/7 N, and for the 6 kg mass, T₂ = 60 − 6a = 240/7 N. So T₁ : T₂ = 5 : 3.

3Work-Energy Theorem

• W = ΔKE = ½m(v² − u²)
• When the force keeps changing, this formula is much faster than integration.

Example: m = 1 kg, velocity v = 2x². At x = 0, u = 0 and at x = 5, v = 50. So W = ½ × 1 × 50² = 1250 J.

4Young's Modulus

• Y = Stress / Strain = (F/A) / (ΔL/L) = FL / (AΔL)
• Increase in length: ΔL = FL / (AY)

Tip: In graph-based questions where the F–ΔL graph is a straight line through the origin, pick any one point (F, ΔL). You will get the same answer.

Example: A wire of length 2 m and cross-section 1 mm² (10⁻⁶ m²) has Y = 2 × 10¹¹ N/m². A force of 100 N is applied. ΔL = FL / (AY) = (100 × 2) / (10⁻⁶ × 2 × 10¹¹) = 10⁻³ m = 1 mm.

5Continuity and Bernoulli's Equation

• Continuity: A₁v₁ = A₂v₂
• Bernoulli: P + ½ρv² + ρgh = constant
• Speed of water from a small hole in a vessel (Torricelli): v = √(2gh), where h = depth of the hole below the water surface

Example: A cylindrical vessel has radius 40 cm and volume 528 litres, so h = V/(πr²) = 528000 / (22/7 × 1600) = 105 cm. The vessel stands on a 105 cm high block and the hole is 70 cm below the top, so the hole is 35 + 105 = 140 cm above the ground. Horizontal range = v × t = √(2g·70) × √(2·140/g) = 140√2 cm.

6First Law of Thermodynamics

• ΔQ = ΔU + W (W = work done by the gas)
• In an adiabatic process ΔQ = 0, so ΔU = −W
• PV^γ = constant
• W = nRΔT / (1 − γ) = (P₂V₂ − P₁V₁) / (1 − γ)

Example: The volume of a monatomic gas (γ = 5/3) increases from V to 27V. P₂ = P / 27^(5/3) = P/243. Then ΔU = (P₂V₂ − P₁V₁)/(γ − 1) = −(4/3)PV. Forgetting the minus sign is a common mistake.

7Degrees of Freedom

• U = (f/2) nRT
• Cv = fR/2, Cp = (f/2 + 1)R, γ = Cp/Cv

Example: When 2 moles of gas (temperature T) and 6 moles of gas (temperature 2T) are mixed, total energy is conserved: 8 T_f = 2T + 12T, so T_f = 7T/4.

8Electric Field for Different Shapes

• Infinitely long wire: E = 2kλ / r
• Finite wire (perpendicular): E = (kλ/r)(sinθ₁ + sinθ₂)
• Arc: E = (2kλ/r) sin(θ/2)
• On the axis of a ring: E = kQx / (R² + x²)^(3/2), and E = 0 at the centre
• Hollow sphere: 0 inside, kQ/r² outside
• Solid non-conducting sphere: kQr / R³ inside, kQ/r² outside

Example: At the centre of a half ring (θ = 180°), E = 2kλ/R, where λ = Q/(πR).

9Electric Potential and Potential Energy

• Energy of two charges: U = kq₁q₂ / r
• Potential: V = kq / r
• E = −dV/dr, i.e. ΔV = −∫E·dr
• Work done by an external agent (moving the charge slowly): W = q ΔV. Work done by the field = −q ΔV

Tip: When E is given and W is asked, first integrate E to get ΔV, then multiply by q.

Example: In a region E = 2x î N/C. An external agent slowly moves a charge q = 2 μC from x = 3 m to x = 0 (against the field). ΔV = −∫E dx = −[x²] from 3 to 0 = +9 V. So W = qΔV = 2 × 10⁻⁶ × 9 = 1.8 × 10⁻⁵ J = 18 μJ.

10Combination of Resistors

• Series: R = R₁ + R₂ + ...
• Parallel: 1/R = 1/R₁ + 1/R₂ + ...
• A Wheatstone bridge is balanced when P/Q = R/S. The middle branch can then be removed.

Example: The resistance of a wire is λ Ω/m. A and B are two points on a circle of centre O and radius r, with ∠AOB = 90°. Three wires join A and B: the minor arc (length πr/2), the path A → O → B along two radii (length r + r = 2r) and the major arc (length 3πr/2). Find the equivalent resistance between A and B.
All three wires are between the same two points A and B, so they are in parallel. Resistance of each wire = λ × length: R₁ = λπr/2, R₂ = 2λr, R₃ = 3λπr/2.
1/R = 2/(πλr) + 1/(2λr) + 2/(3πλr) = (1/λr)(2/π + 1/2 + 2/(3π)) = (1/λr)(8/(3π) + 1/2) = (16 + 3π) / (6πλr).
So R = 6πλr / (16 + 3π).

11Magnetic Field

• At the centre of a ring: B = μ₀i / 2R
• On the axis of a ring: B = μ₀iR² / [2(R² + x²)^(3/2)]
• For a straight wire: B = (μ₀i / 4πr)(sinθ₁ + sinθ₂)
• Infinite wire: B = μ₀i / 2πr

Example: B at the centre is 16 μT. At x = √3 R, (R² + x²)^(3/2) = 8R³, so B = 16/8 = 2 μT.

12Charged Particle in a Magnetic Field

• Force: F = qvB sinθ (in vector form F = q(v × B))
• Radius of circular path (when v ⊥ B): r = mv / qB = √(2mK) / qB
• F is always perpendicular to B, so a is also perpendicular to B and a·B = 0.

Example: If a = 4î − (x/2)ĵ and B = 3î + 2ĵ are perpendicular, their dot product is zero: 12 − x = 0, so x = 12.

13Magnetic Flux and Faraday's Law

• Flux: φ = BA cosθ (θ is the angle between the area vector and B)
• Induced emf: ε = −dφ/dt

Tip: First write φ as a function of time, then differentiate with respect to t. If φ = BA cos ωt, then ε = BAω sin ωt, so the emf is maximum when sin ωt = 1 (the flux is zero at that instant).

Example: A loop of area 0.02 m² rotates in a field B = 0.5 T with ω = 100 rad/s. φ = BA cos ωt = 0.01 cos(100t) Wb. So ε = −dφ/dt = 0.01 × 100 sin(100t) = sin(100t) volt, and the maximum emf is BAω = 1 V.

14LCR Series Circuit

• Impedance: Z = √[R² + (X_L − X_C)²]
• X_L = ωL, X_C = 1/ωC
• At resonance X_L = X_C, so ω = 1/√(LC) and f = 1 / (2π√(LC))

Example: For L = 1.6 H and C = 40 μF, ω = 1/√(1.6 × 40×10⁻⁶) = 125 rad/s. Then X_L = ωL = 200 Ω.

15Refraction and the Lens Maker's Formula

• Refraction at a spherical surface: μ₂/v − μ₁/u = (μ₂ − μ₁)/R
• Lens maker's formula: 1/f = (μ − 1)(1/R₁ − 1/R₂)
• Lens in a medium: f_medium / f_air = (μ_L − 1) / (μ_L/μ_m − 1)

Example: A lens with μ_L = 1.5 is placed in water (4/3). The ratio = 0.5 / (9/8 − 1) = 4, so the focal length becomes 4f. If the question asks for the "change in focal length", the answer is 3f, not 4f. Read the question carefully here.

16Einstein's Photoelectric Equation

• hν = φ + K(max) = hν₀ + K(max)
• K(max) = eV₀ (V₀ = stopping potential)
• Photon energy: E = hc/λ

Example: If λ₁ = 2λ₂ and the kinetic energies are K₁ and K₂, then hc/λ₂ = 2(φ + K₁) and hc/λ₂ = φ + K₂. Equating both gives φ = K₂ − 2K₁.

17Bohr's Atomic Model

• Radius: r = 0.529 n²/Z Å, i.e. r ∝ n²/Z
• Velocity: v = 2.2 × 10⁶ × Z/n m/s, i.e. v ∝ Z/n
• Time period: T ∝ n³/Z²
• Angular momentum: mvr = nh / 2π

Example: At the centre of the current loop formed by the moving electron, B = μ₀i/2r and i = e/T. So B ∝ 1/(rT) ∝ 1/n⁵. For n = 2 and n = 4, the ratio B₁/B₂ = (4/2)⁵ = 32.

18De Broglie Wavelength

• λ = h/p = h / √(2mK) = h / √(2mqV)
• Electron: λ = 12.27 / √V Å
• Proton: 0.286 / √V Å, deuteron: 0.202 / √V Å, alpha particle: 0.101 / √V Å

Example: For an electron and a proton at the same potential difference, λ ∝ 1/√m, so λₑ / λₚ = √(mₚ / mₑ).

19Hydrogen Spectrum

• 1/λ = RZ² (1/n₁² − 1/n₂²), where n₂ > n₁ and R ≈ 1.097 × 10⁷ m⁻¹
• Series: Lyman (n₁ = 1, UV), Balmer (n₁ = 2, visible), Paschen (3), Brackett (4), Pfund (5); the last three are infrared
• For maximum λ the energy gap is minimum: n₂ = n₁ + 1
• For minimum λ the energy gap is maximum: n₂ = ∞

Example: The maximum λ of the Lyman series = 4/(3R), and the minimum λ of the Paschen series = 9/R.

20Mass Defect and Binding Energy

• Mass defect: Δm = Z·mₚ + (A − Z)·mₙ − M(nucleus)
• Binding energy: BE = Δm c² (1 u·c² = 931.5 MeV)
• BE per nucleon is highest near Fe-56.
• Light nuclei fuse (nuclear fusion) and heavy nuclei split (nuclear fission) to become more stable.

Remember: It is wrong to say that binding energy per nucleon always increases as mass number increases. The BE per nucleon curve rises up to Fe-56 and then falls. (Total binding energy, on the other hand, keeps increasing with A.)

Example: For He-4 (Z = 2, A = 4) take mₚ = 1.0073 u, mₙ = 1.0087 u and M(nucleus) = 4.0015 u. Δm = 2(1.0073) + 2(1.0087) − 4.0015 = 0.0305 u. BE = 0.0305 × 931.5 ≈ 28.4 MeV, so BE per nucleon ≈ 7.1 MeV.