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Unit & Dimension, Basic Maths & Vectors

Master the foundational language of JEE Physics — SI units, dimensional formulas, unit conversion, dimensional analysis, trigonometry shortcuts, small-angle and binomial approximations, and the full algebra of vectors.

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JEE Intro

Why This Chapter Is the Gateway to All of JEE Physics

Unit and Dimension, Basic Maths, and Vectors form the language that every JEE Physics chapter is written in. A student who skips or half-learns this chapter will find that every subsequent topic — kinematics, Newton's laws, work-energy theorem, electrostatics — becomes harder than it needs to be.

Dimensional analysis is one of the most underrated tools in JEE preparation. It can eliminate wrong options instantly, verify answers before submitting, and derive formula structures in seconds. Many JEE Main questions are solvable by dimensional reasoning alone in under 30 seconds.

Vectors are even more foundational. Once a student can resolve vectors confidently, decompose forces along inclined planes, compute torques using cross products, and apply dot products to calculate work — they have unlocked the mathematical engine that runs all of mechanics and most of electromagnetism.

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Section A

Notes: Units, Dimensions, Basic Maths & Vectors

Original teaching copy for Learn at My Place

1. The SI System — Seven Base Quantities

Physics is built on measurement. Before we can write equations, describe motion, or calculate energy, we need a shared language of units. The International System of Units (SI) is that language, and it defines seven independent base quantities from which every other physical quantity can be derived.

The seven SI base quantities and their units are:

Base QuantitySymbolSI Unit
LengthLmetre (m)
MassMkilogram (kg)
TimeTsecond (s)
Electric CurrentI or Aampere (A)
Temperatureθ or Kkelvin (K)
Amount of SubstanceNmole (mol)
Luminous IntensityJcandela (cd)

Every other physical quantity — force, energy, pressure, power — is derived from these seven. A derived unit is simply a combination of base units raised to various powers. For example, speed = distance/time gives the unit m/s, which is L/T = [LT1][LT^{-1}].

JEE tip: The SI system uses only seven base units. Radian and steradian are supplementary units, not base units. Newton, joule, pascal, watt — all derived.

2. Dimensional Formulas of Key Physical Quantities

The dimensional formula of a physical quantity expresses it in terms of the base quantities M (mass), L (length), and T (time), with each raised to some power. Writing these formulas is not about memorising a table — it is about understanding the definition of each quantity and following it through.

Here is a systematic derivation of the most important quantities:

Velocity=displacementtime[LT1]\text{Velocity} = \frac{\text{displacement}}{\text{time}} \Rightarrow [LT^{-1}]
Acceleration=velocitytime[LT2]\text{Acceleration} = \frac{\text{velocity}}{\text{time}} \Rightarrow [LT^{-2}]
Force=mass×acceleration[MLT2]\text{Force} = \text{mass} \times \text{acceleration} \Rightarrow [MLT^{-2}]
Work/Energy=Force×distance[ML2T2]\text{Work/Energy} = \text{Force} \times \text{distance} \Rightarrow [ML^2T^{-2}]
Power=WorkTime[ML2T3]\text{Power} = \frac{\text{Work}}{\text{Time}} \Rightarrow [ML^2T^{-3}]
Pressure=ForceArea[ML1T2]\text{Pressure} = \frac{\text{Force}}{\text{Area}} \Rightarrow [ML^{-1}T^{-2}]
Momentum=mass×velocity[MLT1]\text{Momentum} = \text{mass} \times \text{velocity} \Rightarrow [MLT^{-1}]
Impulse=Force×time[MLT1]\text{Impulse} = \text{Force} \times \text{time} \Rightarrow [MLT^{-1}]

Notice that momentum and impulse share the same dimensional formula. This is physically meaningful — impulse equals change in momentum (Newton's second law in integral form).

Angular momentum=mvr[ML2T1]\text{Angular momentum} = mvr \Rightarrow [ML^2T^{-1}]
Torque=F×r[ML2T2]\text{Torque} = F \times r \Rightarrow [ML^2T^{-2}]
Surface tension=FL[MT2]\text{Surface tension} = \frac{F}{L} \Rightarrow [MT^{-2}]
Coefficient of viscosity=stressvelocity gradient[ML1T1]\text{Coefficient of viscosity} = \frac{\text{stress}}{\text{velocity gradient}} \Rightarrow [ML^{-1}T^{-1}]

Notice: torque and work have the same dimensional formula [ML2T2][ML^2T^{-2}], yet they are completely different physical quantities. This is a common JEE trap — two quantities sharing dimensions does not make them identical.

Important constants:
G=[M1L3T2],h=[ML2T1],kB=[ML2T2K1],R=[ML2T2K1mol1]G = [M^{-1}L^3T^{-2}], \quad h = [ML^2T^{-1}], \quad k_B = [ML^2T^{-2}K^{-1}], \quad R = [ML^2T^{-2}K^{-1}\text{mol}^{-1}]

3. Three Applications of Dimensional Analysis

Dimensional analysis is a powerful tool that uses the fact that every valid physical equation must be dimensionally homogeneous — every term must have the same dimensional formula. This gives us three major applications.

Application 1 — Checking consistency of an equation

Consider the equation v2=u2+2asv^2 = u^2 + 2as. Left side: [v2]=[L2T2][v^2] = [L^2T^{-2}]. Right side: [u2]=[L2T2][u^2] = [L^2T^{-2}] and [2as]=[LT2][L]=[L2T2][2as] = [LT^{-2}][L] = [L^2T^{-2}]. All terms match, so the equation is dimensionally consistent.

Caution: Dimensional consistency is necessary but not sufficient. The equation s=vt+at2s = vt + at^2 is dimensionally consistent but physically wrong — it should be s=ut+12at2s = ut + \frac{1}{2}at^2. The constant 12\frac{1}{2} is invisible to dimensional analysis.

Application 2 — Converting between unit systems

The rule is: numerical value × unit = constant (the physical reality doesn't change). If a quantity has dimensions [MaLbTc][M^aL^bT^c], the conversion factor is:

n2=n1(M1M2)a(L1L2)b(T1T2)cn_2 = n_1 \left(\frac{M_1}{M_2}\right)^a \left(\frac{L_1}{L_2}\right)^b \left(\frac{T_1}{T_2}\right)^c

Example: Convert 1 joule to CGS. Energy has dimensions [ML2T2][ML^2T^{-2}]. Using M1/M2=1000M_1/M_2 = 1000 g/g, L1/L2=100L_1/L_2 = 100 cm/cm, T1/T2=1T_1/T_2 = 1: n2=1×1000×1002=107n_2 = 1 \times 1000 \times 100^2 = 10^7 ergs.

Application 3 — Deriving relationships between quantities

Suppose the time period T of a simple pendulum depends on its length L, mass m, and gravitational acceleration g. Assume T=kLambgcT = k\, L^a m^b g^c. Matching dimensions on both sides of the equation:

[T]=[LaMb(LT2)c][T] = [L^a M^b (LT^{-2})^c]
T:1=2cc=1/2T: 1 = -2c \Rightarrow c = -1/2
M:0=bb=0M: 0 = b \Rightarrow b = 0
L:0=a+ca=1/2L: 0 = a + c \Rightarrow a = 1/2

So TL/gT \propto \sqrt{L/g}. Dimensional analysis gives the correct structure, but the constant k=2πk = 2\pi must come from experiment or full derivation.

Limitations of dimensional analysis:
  • Cannot determine pure dimensionless constants (π, 2, 1/2, etc.)
  • Cannot be used when a quantity depends on two others having the same dimensions
  • Cannot handle equations involving exponential, logarithmic, or trigonometric functions
  • Cannot distinguish between scalars and vectors of the same dimension

4. Basic Maths — Trigonometry and Approximations

A working knowledge of trigonometry is indispensable for JEE Physics. Every topic from inclined planes to projectile motion to optics relies on decomposing vectors and resolving forces using trigonometric ratios.

Standard Angle Table

Anglesincostan
010
30°1/2√3/21/√3
37°3/54/53/4
45°1/√21/√21
53°4/53/54/3
60°√3/21/2√3
90°10

The pair 37°–53° appears constantly in JEE because it comes from the 3-4-5 right triangle. If the sides opposite, adjacent, and hypotenuse are 3, 4, 5, then the angle at the base is 37° (sin 37° = 3/5, cos 37° = 4/5) and the complementary angle is 53°.

Small Angle Approximation

When θ is very small (much less than 1 radian, roughly less than 10°), the first-order Taylor expansion gives:

sinθθ,cosθ1,tanθθ(θ in radians)\sin\theta \approx \theta, \quad \cos\theta \approx 1, \quad \tan\theta \approx \theta \quad (\theta \text{ in radians})

This approximation is used throughout physics — in the derivation of simple pendulum equations, in optics for paraxial rays, in waves for small oscillations.

Binomial Approximation

When x1|x| \ll 1, the binomial theorem gives:

(1+x)n1+nx(1+x)^n \approx 1 + nx

This is extremely useful when a physical quantity changes by a small fraction. For example, g=g(1+h/R)2g(12h/R)g' = g(1 + h/R)^{-2} \approx g(1 - 2h/R) when hRh \ll R. The key skill is recognising when to rewrite an expression in (1+x)n(1+x)^n form.

Common applications: 1+x1+x/2\sqrt{1+x} \approx 1+x/2, 1/1+x1x/21/\sqrt{1+x} \approx 1-x/2, ex1+xe^x \approx 1+x, ln(1+x)x\ln(1+x) \approx x, all for small x.

5. Vectors — Concepts and Addition

Physical quantities split into two categories based on whether they require a direction to be fully described. Scalars — mass, temperature, energy, speed — are fully described by magnitude alone. Vectors — displacement, velocity, force, acceleration, momentum — require both magnitude and direction.

A vector is conventionally written with an arrow overhead (A\vec{A}) or in bold (**A**). Its magnitude is written as A|\vec{A}| or simply A. A vector multiplied by a positive scalar changes magnitude; multiplied by a negative scalar, it also reverses direction.

Unit Vector

The unit vector in the direction of A\vec{A} is obtained by dividing A\vec{A} by its own magnitude:

A^=AA\hat{A} = \frac{\vec{A}}{|\vec{A}|}

The standard Cartesian unit vectors are i^\hat{i} (along +x), j^\hat{j} (along +y), and k^\hat{k} (along +z). They are mutually perpendicular and each has magnitude 1.

Triangle Law of Vector Addition

To add A\vec{A} and B\vec{B}: place the tail of B\vec{B} at the tip of A\vec{A}. The resultant R=A+B\vec{R} = \vec{A} + \vec{B} is the vector from the tail of A\vec{A} to the tip of B\vec{B}.

Parallelogram Law

Place both vectors at the same origin. The diagonal of the parallelogram they form is the resultant. If θ\theta is the angle between A\vec{A} and B\vec{B}:

R=A2+B2+2ABcosθ|\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}

The direction of the resultant with respect to A\vec{A} is given by:

tanδ=BsinθA+Bcosθ\tan\delta = \frac{B\sin\theta}{A + B\cos\theta}
Special cases:
θ = 0°: R = A + B (maximum). θ = 180°: R = |A − B| (minimum). θ = 90°: R = √(A² + B²).

6. Dot Product and Cross Product

When two vectors are multiplied, the result can be either a scalar (dot product) or a vector (cross product), depending on the physical context.

Scalar (Dot) Product

AB=ABcosθ\vec{A} \cdot \vec{B} = AB\cos\theta

The dot product is the product of the magnitudes and the cosine of the angle between them. It equals zero when the vectors are perpendicular (θ=90°\theta = 90°). It is maximum (AB) when they are parallel (θ=0°\theta = 0°) and minimum (−AB) when antiparallel (θ=180°\theta = 180°).

In Cartesian form: AB=AxBx+AyBy+AzBz\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z.

Physical examples: Work = Fd\vec{F} \cdot \vec{d}, Power = Fv\vec{F} \cdot \vec{v}.

Vector (Cross) Product

A×B=ABsinθ|\vec{A} \times \vec{B}| = AB\sin\theta

The direction of A×B\vec{A} \times \vec{B} is perpendicular to the plane containing A\vec{A} and B\vec{B}, given by the right-hand rule. The cross product is zero when vectors are parallel or antiparallel.

In Cartesian form using the determinant:

A×B=i^j^k^AxAyAzBxByBz\vec{A} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z\end{vmatrix}

Physical examples: Torque = r×F\vec{r} \times \vec{F}, Angular momentum = r×p\vec{r} \times \vec{p}.

Key difference: Dot product tells you how much of one vector lies along another. Cross product tells you how much they are perpendicular to each other.

Cartesian Resolution

Any vector A\vec{A} can be written as a sum of its Cartesian components:

A=Axi^+Ayj^+Azk^,A=Ax2+Ay2+Az2\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}, \quad |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}

For a 2D vector at angle θ with the x-axis: Ax=AcosθA_x = A\cos\theta, Ay=AsinθA_y = A\sin\theta.

Solved Practice

10 Solved Examples

Open only after you try the question yourself
JEE Exam Trap: Dimensional consistency does not guarantee a correct equation. The constant of proportionality (like π, 2, or 1/2) is invisible to dimensional analysis. Always derive the full formula or use known results for numerical answers.
Example 1. Find the dimensional formula of gravitational constant G using $F = Gm_1m_2/r^2$.
Rearrange: G=Fr2/(m1m2)G = Fr^2/(m_1m_2). Substituting dimensions: G=[MLT2][L2]/[M2]=[M1L3T2]G = [MLT^{-2}][L^2]/[M^2] = [M^{-1}L^3T^{-2}].
Example 2. Check whether the equation $v = u + at$ is dimensionally homogeneous.
LHS: [v]=[LT1][v] = [LT^{-1}]. RHS: [u]=[LT1][u] = [LT^{-1}], and [at]=[LT2][T]=[LT1][at] = [LT^{-2}][T] = [LT^{-1}]. All terms equal [LT1][LT^{-1}]. The equation is dimensionally correct.
Example 3. Convert 72 N into CGS units (dynes).
Force has dimensions [MLT2][MLT^{-2}]. SI→CGS: M ratio = 10310^3 g/kg, L ratio = 10210^2 cm/m, T ratio = 1. So 72 N =72×103×102=72×105=7.2×106= 72 \times 10^3 \times 10^2 = 72 \times 10^5 = 7.2 \times 10^6 dynes.
Example 4. Using dimensional analysis, find how the time period T of a simple pendulum depends on length L and g.
Assume T=kLagbT = kL^ag^b. Dimensions: [T]=[La(LT2)b]=[La+bT2b][T] = [L^a(LT^{-2})^b] = [L^{a+b}T^{-2b}]. Matching: 2b=1b=1/2-2b = 1 \Rightarrow b = -1/2; a+b=0a=1/2a + b = 0 \Rightarrow a = 1/2. So TL/gT \propto \sqrt{L/g}.
Example 5. Using small angle approximation, find the approximate value of sin 3°.
3°=3×π/180=π/600.05243° = 3 \times \pi/180 = \pi/60 \approx 0.0524 rad. For small angles, sinθθ\sin\theta \approx \theta, so sin3°0.0524\sin 3° \approx 0.0524.
Example 6. Approximate the value of $(1.005)^{100}$ using binomial approximation.
(1+0.005)1001+100×0.005=1+0.5=1.5(1+0.005)^{100} \approx 1 + 100 \times 0.005 = 1 + 0.5 = 1.5.
Example 7. Two vectors $\vec{A} = 3\hat{i} + 4\hat{j}$ and $\vec{B} = 4\hat{i} - 3\hat{j}$. Show that they are perpendicular.
AB=(3)(4)+(4)(3)=1212=0\vec{A} \cdot \vec{B} = (3)(4) + (4)(-3) = 12 - 12 = 0. Since the dot product is zero, the vectors are perpendicular.
Example 8. Find the angle between $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{j}$.
AB=(1)(1)+(1)(1)=0\vec{A} \cdot \vec{B} = (1)(1) + (1)(-1) = 0. So cosθ=0\cos\theta = 0, giving θ=90°\theta = 90°.
Example 9. Find the magnitude of $\vec{A} \times \vec{B}$ if A = 5, B = 4, and the angle between them is 30°.
A×B=ABsinθ=5×4×sin30°=20×0.5=10|\vec{A} \times \vec{B}| = AB\sin\theta = 5 \times 4 \times \sin 30° = 20 \times 0.5 = 10.
Example 10. A force $\vec{F} = 3\hat{i} - 4\hat{j}$ N acts through displacement $\vec{d} = 2\hat{i} + 3\hat{j}$ m. Find the work done.
W=Fd=(3)(2)+(4)(3)=612=6W = \vec{F} \cdot \vec{d} = (3)(2) + (-4)(3) = 6 - 12 = -6 J.
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Section B

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