JEE Physics — Complete Formula Sheet
Every chapter, every core formula, JEE-specific shortcuts and the mistakes examiners love to test — all in one page.
JEE Physics — Complete Formula Sheet (Main & Advanced)
All 18 major JEE Physics chapters — Mechanics to Semiconductor Electronics — with core formulas rendered in clean LaTeX, short concept notes, JEE-specific tricks, and the exact mistakes examiners exploit every year.
Quick Jump — All Chapters
Chapter-Wise Notes, Formulas & JEE Tricks
Units, Dimensions & Measurements
Every physical quantity is expressed as a combination of 7 base dimensions (M, L, T, A, K, mol, cd). Dimensional analysis lets you check the correctness of an equation and derive relationships between quantities without solving the full physics.
In JEE, dimensional analysis is used for three things: checking correctness of an equation, converting units between systems, and deriving a formula when you know which quantities it depends on. The method of dimensional homogeneity says every term added or subtracted in a valid physical equation must carry identical dimensions — this is the fastest sanity check available in the exam hall. Errors in measurement combine differently depending on the operation: for addition/subtraction, absolute errors add directly; for multiplication/division, relative (fractional) errors add; for a quantity raised to a power n, the relative error gets multiplied by n.
Key Formulas
Solved JEE-Style Examples
Solution: KE = ½mv², so the relative error in KE = relative error in m + 2×(relative error in v) = 2% + 2(3%) = 2% + 6% = 8%.
Solution: [v] = [LT⁻¹], [u] = [LT⁻¹], [at] = [LT⁻²][T] = [LT⁻¹]. All three terms have the same dimensions [LT⁻¹], so the equation is dimensionally consistent.
Use dimensional analysis to eliminate wrong-looking options in MCQs before even reading the physics — if the dimensions don't match, the option is automatically wrong, saving crucial time in JEE Main.
Forgetting that dimensional analysis cannot determine dimensionless constants (like 1/2 in KE=½mv²) — it only checks the form of the relationship, not exact numerical coefficients.
Kinematics
Kinematics describes motion without worrying about its cause. 1D, 2D (projectile) and relative motion are the three major problem types tested repeatedly in JEE.
Projectile motion is really two independent 1D motions happening simultaneously: uniform velocity horizontally (no force acts horizontally if we ignore air resistance) and uniformly accelerated motion vertically (only gravity acts). The trajectory traced is always a parabola, described by y = x tanθ − (gx²)/(2u²cos²θ). For relative motion problems (like river-boat or rain-man problems), always convert everything into vector form and subtract velocities as vectors — never subtract magnitudes directly unless the vectors happen to be along the same line.
Key Formulas
Solved JEE-Style Examples
Solution: T = 2u sinθ/g = 2(20)(0.5)/10 = 2 s. R = u²sin2θ/g = (400)(sin60°)/10 = (400)(0.866)/10 = 34.6 m.
Solution: Resultant speed = √(3²+4²) = 5 m/s. Time to cross depends only on the perpendicular component: t = 400/4 = 100 s (the current doesn't affect crossing time, only the drift distance downstream).
For projectile motion, always split velocity into horizontal (constant) and vertical (uniformly accelerated) components FIRST — nearly every projectile problem becomes a simple 1D kinematics problem in each direction once split.
Using average speed formulas (like 2v1v2/(v1+v2)) for average VELOCITY questions — average velocity depends on displacement/time, not on the harmonic mean of speeds, a very common JEE Main trap.
Laws of Motion
Newton's three laws govern how forces affect motion. Free body diagrams (FBDs) are the single most important skill here — nearly every problem is solved by correctly identifying all forces acting on a body.
When solving connected-body problems (blocks on a pulley, blocks on an incline connected by a string), always treat the string tension as the same throughout an ideal (massless, inextensible) string over a frictionless pulley. For circular motion at the top of a vertical loop, the minimum speed condition occurs when gravity alone provides the centripetal force (Normal force N=0): v_min = √(gr).
Key Formulas
Solved JEE-Style Examples
Solution: Net force = (m1−m2)g = (3−2)(10) = 10 N. Total mass = 5 kg. a = 10/5 = 2 m/s².
Solution: Friction force f = μN = 0.