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Units and Measurements Formulas For JEE 2026
Units and Measurements is one of the most basic chapters in JEE Physics, and it helps students understand how physical quantities are measured and expressed. In this topic, students learn concepts like SI units, dimensional analysis, significant figures, and measurement errors. These fundamentals are important because they ensure accuracy while solving numerical problems in physics.
In JEE , this chapter usually has 1–2 questions every year, making it a simple and scoring topic if prepared properly. It also forms the base for many other chapters, as correct units and calculations are essential throughout physics. For quick revision before exams, students can also use a well-structured JEE Mains Physics Formula PDF to review important formulas and concepts easily.
What Are Units?
Whenever we measure something in physics — like the length of a table or the weight of a bag — we need two things: a number (how much) and a unit (of what). For example, "5 metres" means the number is 5 and the unit is metres. Without units, a number alone is meaningless in physics.
Scientists around the world agreed on a standard system of units called the SI system (Systeme International). This ensures that "1 metre" means the same thing everywhere.
Physical Quantity
Physical Quantity: Any quantity that can be measured and expressed in terms of units. Examples: length, mass, time, speed, force.
Physical quantities are of two types:
- Fundamental (Base) Quantities — cannot be expressed in terms of other quantities (e.g., length, mass, time)
- Derived Quantities — defined using fundamental quantities (e.g., speed = length/time, force = mass $$\times$$ acceleration)
SI Base Units and Derived Units
The SI system defines exactly seven base units. Every other unit in physics can be built from combinations of these seven.
Seven Fundamental SI Units
| Quantity | Unit | Symbol | Measures... |
|---|---|---|---|
| Length | metre | m | how long/far something is |
| Mass | kilogram | kg | how much matter is in an object |
| Time | second | s | duration of an event |
| Electric Current | ampere | A | flow of electric charge |
| Temperature | kelvin | K | hotness or coldness |
| Amount of Substance | mole | mol | number of particles ($$6.022 \times 10^{23}$$) |
| Luminous Intensity | candela | cd | brightness of a light source |
Tip: For JEE, the most commonly used base units are metre (m), kilogram (kg), and second (s). Most formulas involve only these three (called the MKS system).
SI Prefixes
Very large or very small quantities are written using prefixes that represent powers of 10.
Common SI Prefixes
| Prefix | Symbol | Factor | Example |
|---|---|---|---|
| giga | G | $$10^{9}$$ | 1 GHz = $$10^9$$ Hz |
| mega | M | $$10^{6}$$ | 1 MW = $$10^6$$ W |
| kilo | k | $$10^{3}$$ | 1 km = $$10^3$$ m |
| centi | c | $$10^{-2}$$ | 1 cm = $$10^{-2}$$ m |
| milli | m | $$10^{-3}$$ | 1 mm = $$10^{-3}$$ m |
| micro | $$\mu$$ | $$10^{-6}$$ | 1 $$\mu$$m = $$10^{-6}$$ m |
| nano | n | $$10^{-9}$$ | 1 nm = $$10^{-9}$$ m |
Dimensional Analysis Formulas
Every physical quantity can be expressed as a combination of the base quantities — mass (M), length (L), time (T), etc. The dimensional formula tells us which base quantities (and their powers) make up a given quantity.
Dimensions: The powers to which the base quantities must be raised to represent a physical quantity.
Worked Example
Find the dimensions of Force.
We know: Force = mass $$\times$$ acceleration = mass $$\times$$ (velocity / time) = mass $$\times$$ (length / time$$^2$$)
So: $$[\text{Force}] = [M] \times [L T^{-2}] = [M L T^{-2}]$$
This means force has 1 power of mass, 1 power of length, and $$-2$$ power of time.
Dimensional Formula
Any physical quantity $$Q$$ can be written as:
$$$[Q] = [M^a \, L^b \, T^c]$$$where $$a, b, c$$ are called the dimensional exponents.
(For electrical/thermal quantities, we also use A for current, K for temperature, etc., but most JEE problems use only M, L, T.)
Key Dimensional Formulas
| Quantity | Formula | Dimensions |
|---|---|---|
| Velocity | distance / time | $$[L T^{-1}]$$ |
| Acceleration | velocity / time | $$[L T^{-2}]$$ |
| Force | mass $$\times$$ acceleration | $$[M L T^{-2}]$$ |
| Energy / Work | force $$\times$$ distance | $$[M L^2 T^{-2}]$$ |
| Power | work / time | $$[M L^2 T^{-3}]$$ |
| Pressure | force / area | $$[M L^{-1} T^{-2}]$$ |
| Momentum | mass $$\times$$ velocity | $$[M L T^{-1}]$$ |
| Angular Momentum | $$r \times$$ momentum | $$[M L^2 T^{-1}]$$ |
| Gravitational Const. $$G$$ | from $$F = Gm_1m_2/r^2$$ | $$[M^{-1} L^3 T^{-2}]$$ |
| Planck's Const. $$h$$ | from $$E = h\nu$$ | $$[M L^2 T^{-1}]$$ |
Uses of Dimensional Analysis
Dimensional analysis is a powerful tool with three main uses in JEE problems:
1. Checking if a Formula is Correct
Both sides of a valid equation must have the same dimensions.
Example: Is $$v = u + at$$ dimensionally correct?
- LHS: $$[v] = [LT^{-1}]$$
- RHS: $$[u] + [a][t] = [LT^{-1}] + [LT^{-2}][T] = [LT^{-1}] + [LT^{-1}]$$ ✓
2. Deriving Relationships Between Quantities
If we know a quantity depends on certain variables, we can find the powers using dimensional consistency.
Example: Time period $$T$$ of a pendulum depends on length $$l$$ and gravity $$g$$. Find the relation.
Let $$T = k \, l^a \, g^b$$. Dimensions: $$[T] = [L^a] [L^b T^{-2b}]$$
Comparing: $$L^0 T^1 = L^{a+b} T^{-2b}$$
So $$a + b = 0$$ and $$-2b = 1$$, giving $$b = -\frac{1}{2}$$, $$a = \frac{1}{2}$$.
$$\Rightarrow T = k\sqrt{l/g}$$ (the constant $$k = 2\pi$$ cannot be found by dimensions)
3. Converting Units Between Systems
If a quantity has dimensions $$[M^a L^b T^c]$$, then:
$$$n_2 = n_1 \left(\frac{M_1}{M_2}\right)^a \left(\frac{L_1}{L_2}\right)^b \left(\frac{T_1}{T_2}\right)^c$$$where subscript 1 = old system, subscript 2 = new system.
Worked Example
Convert 1 joule to CGS units.
Energy has dimensions $$[M L^2 T^{-2}]$$.
$$n_2 = 1 \times \left(\frac{1\text{ kg}}{1\text{ g}}\right)^1 \times \left(\frac{1\text{ m}}{1\text{ cm}}\right)^2 \times \left(\frac{1\text{ s}}{1\text{ s}}\right)^{-2}$$
$$= 1 \times 1000 \times (100)^2 \times 1 = 10^7$$
So 1 joule = $$10^7$$ erg.
Note: Dimensional analysis cannot: (1) determine dimensionless constants like $$2\pi$$, (2) distinguish between quantities with the same dimensions (e.g., work and torque both have $$[ML^2T^{-2}]$$), or (3) handle logarithmic, trigonometric, or exponential functions.
Significant Figures Rules
When we measure something, our answer is only as precise as our measuring instrument allows. Significant figures (sig figs) tell us how many digits in a number are meaningful/reliable.
Significant Figures: The number of digits in a measurement that are known with certainty, plus one uncertain (estimated) digit.
Rules for Counting Significant Figures
- All non-zero digits are significant. E.g., 234 has 3 sig figs.
- Zeros between non-zero digits are significant. E.g., 1007 has 4 sig figs.
- Leading zeros (before the first non-zero digit) are not significant. E.g., 0.0052 has 2 sig figs.
- Trailing zeros after a decimal point are significant. E.g., 3.200 has 4 sig figs.
- Trailing zeros in a whole number without a decimal point are ambiguous. E.g., 1200 could have 2, 3, or 4 sig figs.
Rounding Rules in Calculations
- Addition/Subtraction: Result has the same number of decimal places as the quantity with the fewest decimal places.
- Multiplication/Division: Result has the same number of significant figures as the quantity with the fewest significant figures.
Worked Example
Multiplication: $$4.56 \times 1.4 = 6.384 \rightarrow 6.4$$ (2 sig figs, limited by 1.4)
Addition: $$12.11 + 0.3 = 12.41 \rightarrow 12.4$$ (1 decimal place, limited by 0.3)
Errors in Measurement Formulas
No measurement is perfectly accurate. Every measurement has some error (also called uncertainty). Understanding errors helps us know how reliable our results are.
Absolute Error: The difference between the measured value and the true value: $$\Delta a = |a_{\text{measured}} - a_{\text{true}}|$$
Since we usually don't know the true value, we take multiple readings and use the mean as the best estimate. The error in each reading is then its difference from the mean.
Types of Errors
- Mean Absolute Error: The average of the absolute errors from all readings. $$$\overline{\Delta a} = \frac{1}{n}\sum_{i=1}^{n} |a_i - \bar{a}|$$$ where $$\bar{a}$$ is the mean value and $$n$$ is the number of readings.
- Relative Error: How large the error is compared to the quantity itself. $$$\text{Relative Error} = \frac{\Delta a}{\bar{a}}$$$
- Percentage Error: Relative error expressed as a percentage. $$$\text{Percentage Error} = \frac{\Delta a}{\bar{a}} \times 100\%$$$
Worked Example
Five readings of a length: 2.63, 2.56, 2.42, 2.71, 2.80 m
Mean: $$\bar{a} = \frac{2.63 + 2.56 + 2.42 + 2.71 + 2.80}{5} = 2.624$$ m
Absolute errors: $$|0.006|, |0.064|, |0.204|, |0.086|, |0.176|$$
Mean absolute error: $$\overline{\Delta a} = \frac{0.006 + 0.064 + 0.204 + 0.086 + 0.176}{5} = 0.107$$ m
Percentage error: $$\frac{0.107}{2.624} \times 100 = 4.08\%$$
Propagation of Errors
When we calculate a quantity using measured values, the errors in those measurements combine (propagate) into the final result. The rules depend on the type of operation.
Error Propagation Rules
For addition/subtraction: If $$Z = A \pm B$$, the absolute errors add:
$$$\Delta Z = \Delta A + \Delta B$$$For multiplication/division: If $$Z = A^p \cdot B^q / C^r$$, the relative (fractional) errors add:
$$$\frac{\Delta Z}{Z} = |p|\frac{\Delta A}{A} + |q|\frac{\Delta B}{B} + |r|\frac{\Delta C}{C}$$$Worked Example
If $$Z = \dfrac{A^2 B}{C^{1/2}}$$ and $$\dfrac{\Delta A}{A} = 2\%$$, $$\dfrac{\Delta B}{B} = 3\%$$, $$\dfrac{\Delta C}{C} = 4\%$$, find $$\dfrac{\Delta Z}{Z}$$.
Using the rule: $$\dfrac{\Delta Z}{Z} = 2 \times 2\% + 1 \times 3\% + \frac{1}{2} \times 4\% = 4\% + 3\% + 2\% = 9\%$$
Tip: In error propagation with powers, the power becomes a multiplier of the percentage error. So a quantity raised to a higher power contributes more error. JEE often asks: "which measurement should be made most carefully?" — it is the one with the highest power.
Unit Conversion Formulas
The same physical quantity can be expressed in different units. The key principle is that the actual quantity doesn't change when we change units — only the number and unit change together.
$$$n_1 \, u_1 = n_2 \, u_2$$$
Here $$n_1, n_2$$ are the numerical values and $$u_1, u_2$$ are the units. When the unit gets smaller, the number gets larger, and vice versa.
Worked Example
Convert 36 km/h to m/s.
$$36 \;\text{km/h} = 36 \times \frac{1000 \;\text{m}}{3600 \;\text{s}} = 10 \;\text{m/s}$$
Quick shortcut: To convert km/h to m/s, multiply by $$\frac{5}{18}$$. To convert m/s to km/h, multiply by $$\frac{18}{5}$$.
Vernier Caliper and Screw Gauge Formulas
In JEE, you need to know how two precision instruments work: the Vernier caliper and the screw gauge (micrometer). Both allow measurements more precise than a simple ruler.
Vernier Caliper
A Vernier caliper has two scales: the main scale (like a ruler, marked in mm) and a smaller Vernier scale that slides along it. The Vernier scale lets us measure fractions of the smallest main scale division.
Least Count (LC): The smallest measurement an instrument can make. For a Vernier caliper, it is the difference between one main scale division (MSD) and one Vernier scale division (VSD).
Vernier Caliper Formulas
- Least Count = 1 MSD $$-$$ 1 VSD = $$\dfrac{\text{Value of 1 MSD}}{n}$$
where $$n$$ = number of divisions on the Vernier scale
Typically: LC $$= \frac{1 \text{ mm}}{10} = 0.1$$ mm = 0.01 cm - Reading = MSR + (VSR $$\times$$ LC) $$-$$ Zero Error
MSR = Main Scale Reading (the division just before the 0 of Vernier scale)
VSR = Vernier Scale Reading (the Vernier division that aligns with a main scale line)
Worked Example
MSR = 3.2 cm, VSR = 6 divisions, LC = 0.01 cm, Zero error = +0.02 cm
Reading $$= 3.2 + (6 \times 0.01) - 0.02 = 3.2 + 0.06 - 0.02 = 3.24$$ cm
Screw Gauge (Micrometer)
A screw gauge uses a screw mechanism to measure even smaller lengths (up to 0.01 mm or 0.001 cm). It has a main scale along the barrel and a circular scale on the thimble that rotates.
Pitch: The distance the screw advances in one complete rotation. Typically 0.5 mm or 1 mm.
Screw Gauge Formulas
- Least Count $$= \dfrac{\text{Pitch}}{\text{No. of circular scale divisions}}$$
Typically: LC $$= \frac{0.5 \text{ mm}}{50} = 0.01$$ mm = 0.001 cm - Reading = MSR + (CSR $$\times$$ LC) $$-$$ Zero Error
MSR = Main Scale Reading (visible divisions on the barrel)
CSR = Circular Scale Reading (the circular division aligned with the reference line)
Worked Example
A screw gauge has a pitch of 0.5 mm and 50 divisions on the circular scale. When measuring a wire, MSR = 3 mm and CSR = 27. The zero error is $$-$$0.03 mm. Find the wire's diameter.
Step 1: Least Count $$= \dfrac{0.5}{50} = 0.01$$ mm
Step 2: Observed reading $$= 3 + (27 \times 0.01) = 3 + 0.27 = 3.27$$ mm
Step 3: Corrected reading $$= 3.27 - (-0.03) = 3.27 + 0.03 = 3.30$$ mm
The wire's diameter is 3.30 mm.
Zero Error
When the jaws of a Vernier caliper (or the faces of a screw gauge) are fully closed, the zero of both scales should coincide. If they don't, there is a zero error.
Zero Error Correction
- Positive zero error: The zero of the Vernier/circular scale is ahead of the main scale zero. The instrument reads more than actual. Subtract the zero error.
- Negative zero error: The zero is behind. The instrument reads less than actual. Add the magnitude of zero error.
Corrected reading = Observed reading $$-$$ Zero error
(This works for both positive and negative zero error if you keep the sign.)
Worked Example
Complete Vernier Caliper Problem
A Vernier caliper has 10 divisions on the Vernier scale, and 1 MSD = 1 mm. When the jaws are closed, the 4th Vernier division coincides with a main scale line (zero error). When measuring an object, MSR = 2.3 cm and the 7th Vernier division coincides.
Step 1: LC $$= \dfrac{1 \text{ mm}}{10} = 0.1 \text{ mm} = 0.01$$ cm
Step 2: Zero error: The 4th division coincides when jaws are closed, meaning the Vernier zero is ahead of the main scale zero. This is a positive zero error.
Zero error $$= 4 \times 0.01 = +0.04$$ cm
Step 3: Observed reading $$= 2.3 + (7 \times 0.01) = 2.37$$ cm
Step 4: Corrected reading $$= 2.37 - 0.04 =$$ 2.33 cm
Tip: JEE frequently tests Vernier caliper and screw gauge numerical problems. Always check for zero error first — many students lose marks by forgetting to apply the correction.
Units and Measurements Formulas For JEE 2026: Conclusion
Units and Measurements is one of the most fundamental chapters in JEE Physics, and mastering it can give students a strong start in their preparation. Since this topic focuses on concepts like SI units, dimensional analysis, and measurement errors, it helps build accuracy and clarity in solving numerical problems across all chapters.
A well-structured formula sheet can make revision much easier, especially during the final days before the exam. Regular practice, clear understanding of concepts, and quick revision strategies can help students score confidently in this section and improve their overall performance in JEE
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