1. Chapter at a Glance
Every physical quantity is a number times a unit. This chapter teaches the SI system, how to check equations with dimensions, how to write results with the correct significant figures, and how errors in measured quantities combine. It also covers the reading of vernier callipers and screw gauges, which JEE Main tests under "experimental skills". Expect one direct question every year — usually on dimensions or percentage error.
- Dimensional formula of a quantity or a constant (G, h, ε₀, μ₀, η).
- Checking or deriving a relation by dimensional analysis.
- Percentage error in a quantity calculated from measured values.
- Least count and zero error of vernier callipers and screw gauge.
2. Units and the SI System
The SI system has seven base quantities. All other quantities are derived from them.
| Base quantity | SI unit | Symbol | Dimension |
|---|---|---|---|
| Length | metre | m | [L] |
| Mass | kilogram | kg | [M] |
| Time | second | s | [T] |
| Electric current | ampere | A | [A] |
| Thermodynamic temperature | kelvin | K | [K] or [θ] |
| Amount of substance | mole | mol | [mol] |
| Luminous intensity | candela | cd | [cd] |
Plane angle (radian) and solid angle (steradian) are dimensionless. Useful practical units: 1 light-year ≈ 9.46 × 10¹⁵ m, 1 AU ≈ 1.5 × 10¹¹ m, 1 parsec ≈ 3.08 × 10¹⁶ m, 1 fermi = 10⁻¹⁵ m, 1 Å = 10⁻¹⁰ m.
3. Dimensions and Dimensional Analysis
The dimensional formula shows how a quantity depends on base quantities. For example, force = mass × acceleration = [MLT⁻²].
| Quantity | Dimensional formula |
|---|---|
| Work, energy, torque | [ML²T⁻²] |
| Power | [ML²T⁻³] |
| Pressure, stress, Young's modulus | [ML⁻¹T⁻²] |
| Impulse, momentum | [MLT⁻¹] |
| Planck's constant, angular momentum | [ML²T⁻¹] |
| Gravitational constant G | [M⁻¹L³T⁻²] |
| Coefficient of viscosity η | [ML⁻¹T⁻¹] |
| Surface tension, spring constant | [MT⁻²] |
| Permittivity ε₀ | [M⁻¹L⁻³T⁴A²] |
| Permeability μ₀ | [MLT⁻²A⁻²] |
| Magnetic field B | [MT⁻²A⁻¹] |
| Electric potential | [ML²T⁻³A⁻¹] |
Uses of dimensional analysis
- Checking correctness (principle of homogeneity): every term added or equated must have the same dimensions.
- Converting units: n₂ = n₁ (M₁/M₂)a(L₁/L₂)b(T₁/T₂)c for a quantity [MaLbTc].
- Deriving relations when a quantity depends on known variables as a product of powers.
It cannot find numerical constants (like 2π), cannot handle sums of terms, and cannot derive relations with trigonometric, logarithmic or exponential functions. A dimensionally correct equation may still be wrong.
4. Significant Figures and Errors
Significant figures
- All non-zero digits are significant; zeros between non-zero digits are significant.
- Leading zeros (0.00340) are not significant; trailing zeros after a decimal point are significant. So 0.00340 has 3 significant figures.
- Multiplication/division: keep as many significant figures as the least precise value. Addition/subtraction: keep as many decimal places as the least precise value.
Errors
- Absolute error Δa = |atrue − ameasured|; the mean absolute error is the average of these.
- Relative error = Δa/a; percentage error = (Δa/a) × 100.
- Systematic errors (instrument, zero error, method) act in one direction; random errors are reduced by taking many readings.
For Z = A ± B, the absolute errors add: ΔZ = ΔA + ΔB. Errors always add — never subtract.
5. Vernier Callipers and Screw Gauge
- Vernier callipers: LC = 1 MSD − 1 VSD = (value of 1 MSD)/(number of vernier divisions). Reading = MSR + (coinciding vernier division × LC).
- Screw gauge: pitch = distance moved in one rotation; LC = pitch/(number of circular scale divisions). Reading = MSR + (CSR × LC).
- Zero error: if the zero is on the positive side, the error is positive; corrected reading = observed reading − zero error (with sign). Zero correction is the negative of the zero error.
6. Formula Master Table
| Quantity | Formula |
|---|---|
| Unit conversion | n₂ = n₁[M₁/M₂]a[L₁/L₂]b[T₁/T₂]c |
| Error in product/power | ΔZ/Z = pΔA/A + qΔB/B + rΔC/C |
| Error in sum/difference | ΔZ = ΔA + ΔB |
| Vernier least count | LC = 1 MSD − 1 VSD |
| Screw gauge least count | LC = pitch/N |
| Corrected reading | observed − zero error |
| Speed of light relation | c = 1/√(μ₀ε₀) ⇒ [μ₀ε₀] = [L⁻²T²] |
7. Mind Map
- 7 SI base units
- Derived units
- Practical units
- Homogeneity check
- Unit conversion
- Deriving relations
- Significant figures
- Absolute, relative, %
- Propagation
- Vernier LC
- Screw gauge LC
- Zero error
8. Tips and Tricks
- Remember a few "anchor" dimensions (energy, force, pressure) and build the rest: e.g. [h] = [E]/[ν] = [ML²T⁻¹].
- Electrical dimensions: start from [ε₀] using F = q²/(4πε₀r²), with q = [AT].
- In error questions, the quantity with the highest power dominates — measure it most carefully.
- For "which pair has the same dimensions", look for classic pairs: work–torque, impulse–momentum, stress–pressure, angular momentum–Planck's constant.
- Subtracting errors when a quantity is divided — relative errors always add.
- Counting leading zeros as significant figures.
- Adding a positive zero error instead of subtracting it.
9. Important One-liners
- Torque and work have the same dimensions but different physical meaning.
- Strain, refractive index, relative density and angles are dimensionless.
- A dimensionless quantity can still have a unit (radian).
- Least count decides the maximum possible absolute error of a single reading.
- Taking the mean of many readings reduces random error, not systematic error.
- The number of significant figures does not change when units are changed.
10. Solved Numericals: Average to Advanced
Five average-level and five advanced-level problems. Try each one yourself first, then open the step-by-step solution.
Average level
In a new system, the unit of mass is 2 kg, the unit of length is 3 m and the unit of time is 4 s. What is the magnitude of a force of 100 N in this system?
Show answer & solution
Given: n₁ = 100 (SI), M₂ = 2 kg, L₂ = 3 m, T₂ = 4 s. Concept: force is [MLT⁻²], so n₂ = n₁(M₁/M₂)(L₁/L₂)(T₁/T₂)⁻².
- M₁/M₂ = 1/2 and L₁/L₂ = 1/3.
- T₁/T₂ = 1/4, so (T₁/T₂)⁻² = 16.
- n₂ = 100 × (1/2) × (1/3) × 16 = 1600/6 ≈ 266.7.
Answer: about 266.7 new units
Show by dimensional analysis that 1 joule = 10⁷ erg (erg is the CGS unit of energy: g cm² s⁻²).
Show answer & solution
Concept: energy is [ML²T⁻²]; n₂ = n₁(M₁/M₂)¹(L₁/L₂)²(T₁/T₂)⁻².
- M₁/M₂ = 1 kg/1 g = 1000.
- L₁/L₂ = 1 m/1 cm = 100, so (L₁/L₂)² = 10⁴.
- T₁/T₂ = 1 s/1 s = 1.
- n₂ = 1 × 10³ × 10⁴ × 1 = 10⁷.
Answer: 1 J = 10⁷ erg
The mass of a cube is measured with an error of 2% and its side with an error of 1%. Find the maximum percentage error in its density.
Show answer & solution
Concept: ρ = m/a³, so relative errors add with their powers.
- Δρ/ρ = Δm/m + 3(Δa/a).
- = 2% + 3 × 1% = 5%.
Answer: 5%
Give the results with correct significant figures: (a) 2.35 m + 11.2 m + 0.068 m, (b) 4.2 cm × 1.357 cm.
Show answer & solution
Concept: addition keeps the fewest decimal places; multiplication keeps the fewest significant figures.
- (a) Raw sum = 13.618 m. The least precise term, 11.2, has one decimal place.
- Round to one decimal place: 13.6 m.
- (b) Raw product = 5.6994 cm². The factor 4.2 has two significant figures.
- Round to two significant figures: 5.7 cm².
Answer: (a) 13.6 m, (b) 5.7 cm²
A vernier calliper has 1 MSD = 1 mm and 10 VSD = 9 MSD. With the jaws closed, the vernier zero is to the right of the main-scale zero and the 3rd vernier division coincides. Measuring a rod, the main-scale reading is 25 mm and the 6th vernier division coincides. Find the true length.
Show answer & solution
Concept: LC = 1 MSD − 1 VSD; corrected reading = observed − zero error.
- 1 VSD = 0.9 mm, so LC = 1 − 0.9 = 0.1 mm.
- Zero error is positive: +3 × 0.1 = +0.3 mm.
- Observed reading = 25 + 6 × 0.1 = 25.6 mm.
- True length = 25.6 − 0.3 = 25.3 mm.
Answer: 25.3 mm
Advanced level
(a) Assuming the period T of a simple pendulum depends on its length l, the bob's mass m and g, find the form of the relation. (b) If l = 100.0 cm ± 0.1 cm and T = 2.00 s ± 0.01 s, find the percentage error in g.
Show answer & solution
Concept: write T = k la mb gc and match powers of M, L and T.
- [T] = [L]a[M]b[LT⁻²]c = Mb La+c T−2c.
- M: b = 0; T: −2c = 1 ⇒ c = −½; L: a + c = 0 ⇒ a = ½.
- So T = k√(l/g); the mass does not appear.
- g = 4π²l/T² ⇒ Δg/g = Δl/l + 2ΔT/T.
- Δl/l = 0.1/100 = 0.1%; 2ΔT/T = 2 × 0.01/2.00 = 1%.
- Δg/g = 0.1% + 1% = 1.1%.
Answer: T ∝ √(l/g); error in g = 1.1%
A screw gauge has a pitch of 0.5 mm and 50 circular divisions. With the studs touching, the circular-scale zero is 3 divisions above the reference line (negative zero error). For a wire, the main scale reads 3.5 mm and the circular scale reads 27. Find the diameter.
Show answer & solution
Concept: LC = pitch/N; corrected reading = observed − zero error.
- LC = 0.5/50 = 0.01 mm.
- Zero error = −3 × 0.01 = −0.03 mm.
- Observed reading = 3.5 + 27 × 0.01 = 3.77 mm.
- Diameter = 3.77 − (−0.03) = 3.80 mm.
Answer: 3.80 mm
A force is given by F = a√x + bt², where x is distance and t is time. Find the dimensions of a, b and a/b.
Show answer & solution
Concept: each term must have the dimensions of force, [MLT⁻²].
- a√x = [MLT⁻²] ⇒ a = [MLT⁻²]/[L½] = [ML½T⁻²].
- bt² = [MLT⁻²] ⇒ b = [MLT⁻²]/[T²] = [MLT⁻⁴].
- a/b = [ML½T⁻²]/[MLT⁻⁴] = [L−½T²].
Answer: a = [ML½T⁻²], b = [MLT⁻⁴], a/b = [L−½T²]
Five readings of a pendulum's period are 3.12 s, 3.05 s, 3.20 s, 2.98 s and 3.15 s. Find the mean value, the mean absolute error and the percentage error.
Show answer & solution
Concept: mean absolute error = average of |reading − mean|.
- Mean = (3.12 + 3.05 + 3.20 + 2.98 + 3.15)/5 = 15.50/5 = 3.10 s.
- Absolute errors: 0.02, 0.05, 0.10, 0.12, 0.05 s.
- Mean absolute error = 0.34/5 = 0.068 ≈ 0.07 s.
- Percentage error = (0.068/3.10) × 100 ≈ 2.2%.
- Result: T = (3.10 ± 0.07) s.
Answer: 3.10 s, ±0.07 s, about 2.2%
Two resistors R₁ = (100 ± 3) Ω and R₂ = (200 ± 4) Ω are connected (a) in series and (b) in parallel. Find the equivalent resistance with its error in each case.
Show answer & solution
Concept: for a sum, absolute errors add. For 1/R = 1/R₁ + 1/R₂, differentiating gives ΔR/R² = ΔR₁/R₁² + ΔR₂/R₂².
- Series: R = 100 + 200 = 300 Ω; ΔR = 3 + 4 = 7 Ω.
- Parallel: R = (100 × 200)/300 = 66.7 Ω.
- ΔR₁/R₁² = 3/10⁴ = 3 × 10⁻⁴; ΔR₂/R₂² = 4/(4 × 10⁴) = 1 × 10⁻⁴.
- ΔR = R² × (4 × 10⁻⁴) = 4444.4 × 4 × 10⁻⁴ ≈ 1.8 Ω.
Answer: series (300 ± 7) Ω; parallel (66.7 ± 1.8) Ω
11. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
The dimensional formula of the universal gravitational constant G is:
- [ML²T⁻²]
- [M⁻¹L³T⁻²]
- [M⁻¹L²T⁻²]
- [ML³T⁻²]
Show solution
G = Fr²/(m₁m₂) = [MLT⁻²][L²]/[M²] = [M⁻¹L³T⁻²].
Answer: (B)
Which pair of quantities has the same dimensions?
- Force and power
- Torque and energy
- Momentum and energy
- Stress and strain
Show solution
Torque = force × distance = [ML²T⁻²] = energy. Strain is dimensionless while stress is not.
Answer: (B)
The dimensions of μ₀ε₀ are:
- [LT⁻¹]
- [L²T⁻²]
- [L⁻²T²]
- Dimensionless
Show solution
c = 1/√(μ₀ε₀) ⇒ μ₀ε₀ = 1/c² = [L⁻²T²].
Answer: (C)
The number of significant figures in 0.003400 is:
- 2
- 3
- 4
- 7
Show solution
Leading zeros are not significant; the digits 3, 4 and the two trailing zeros after the decimal point are significant: 4.
Answer: (C)
The momentum of a body is measured with an error of 2% and its mass with an error of 1%. Find the maximum percentage error in its kinetic energy (K = p²/2m).
Show solution
ΔK/K = 2(Δp/p) + Δm/m = 2 × 2% + 1% = 5%.
Answer: 5
In the equation (P + a/V²)(V − b) = RT for one mole of gas, the dimensions of a/b are:
- [ML²T⁻²]
- [ML⁻¹T⁻²]
- [ML⁵T⁻²]
- [L³]
Show solution
a/V² must have dimensions of pressure: a = [ML⁻¹T⁻²][L⁶] = [ML⁵T⁻²]. b must have dimensions of volume: [L³]. So a/b = [ML²T⁻²] (energy).
Answer: (A)
12. Quick Revision Summary
- Seven SI base units; everything else is derived.
- Homogeneity: only like dimensions can be added or equated.
- Relative errors add with their powers; absolute errors add in sums and differences.
- Vernier LC = 1 MSD − 1 VSD; screw gauge LC = pitch/N.
- Corrected reading = observed − zero error (sign included).