JEE Main & Advanced

Units, Dimensions and Measurements

Units, Dimensions and Measurements for JEE Main & Advanced

1
Module 1

SI Units and Dimensional Analysis

SI Units, Fundamental and Derived QuantitiesTopic 1

A physical quantity is any property of matter or phenomenon that can be measured. Each measurement is expressed as the product of a numerical value and a unit: $Q = n \cdot u$. If $n_1 u_1 = n_2 u_2$ for the same physical quantity expressed in two unit systems, then $n_2 = n_1 \cdot (u_1 / u_2)$. This proportionality is the basis of all unit conversion.

The International System of Units (SI) adopted in 1960 is built on seven fundamental (base) quantities along with their base units:

QuantityUnitSymbol
Lengthmetrem
Masskilogramkg
Timeseconds
Electric currentampereA
Thermodynamic temperaturekelvinK
Amount of substancemolemol
Luminous intensitycandelacd

Two supplementary units are also defined: the radian (rad) for plane angle and the steradian (sr) for solid angle. A radian is the angle subtended at the centre by an arc of length equal to the radius ($1 \text{ rad} \approx 57.2958^\circ$). A steradian is the solid angle subtended at the centre of a sphere by a surface area equal to the square of its radius; the total solid angle around a point is $4\pi$ sr.

All other quantities are derived quantities, obtained from the fundamental ones via mathematical operations. For example:

  • Velocity: $v = \dfrac{\text{displacement}}{\text{time}}$, SI unit: $\text{m/s}$.
  • Force: $F = ma$, SI unit: $\text{kg·m/s}^2 = $ newton (N).
  • Energy: $E = Fd$, SI unit: $\text{kg·m}^2/\text{s}^2 = $ joule (J).
  • Pressure: $P = F/A$, SI unit: $\text{N/m}^2 = $ pascal (Pa).
  • Power: $P = E/t$, SI unit: $\text{J/s} = $ watt (W).

Coherent and non-coherent units: A unit is coherent within a system if it is derived using a product/quotient of base units with no extra numerical factor. The SI is a coherent system. Non-SI units (litre, electron-volt, atomic mass unit, light-year) are tolerated for convenience:

  • $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$
  • $1 \text{ u} = 1.661 \times 10^{-27} \text{ kg}$
  • $1 \text{ ly} = 9.461 \times 10^{15} \text{ m}$
  • $1 \text{ parsec} = 3.086 \times 10^{16} \text{ m}$
  • $1 \text{ atm} = 1.013 \times 10^{5} \text{ Pa}$

SI Prefixes scale units by powers of 10:

PrefixSymbolFactorPrefixSymbolFactor
teraT$10^{12}$millim$10^{-3}$
gigaG$10^{9}$microμ$10^{-6}$
megaM$10^{6}$nanon$10^{-9}$
kilok$10^{3}$picop$10^{-12}$
centic$10^{-2}$femtof$10^{-15}$
Worked Examples
1

A new unit of length is chosen such that the speed of light in vacuum is unity. Express the distance from Sun to Earth (approximately $1.5 \times 10^{11}$ m) in this new unit, given $c = 3 \times 10^{8}$ m/s.

Show solution

Let the new unit of length be $L_{\text{new}}$. By definition, the speed of light $c = 1 \text{ (new unit)/s}$ requires: $$1 \cdot L_{\text{new}} / 1 \text{ s} = 3 \times 10^{8} \text{ m/s} \implies L_{\text{new}} = 3 \times 10^{8} \text{ m}$$

Convert the Sun–Earth distance: $$d = \frac{1.5 \times 10^{11} \text{ m}}{3 \times 10^{8} \text{ m per new unit}} = 500 \text{ new units}$$

This is the time light takes to travel from Sun to Earth in seconds (≈ 8 minutes 20 s). The new unit is in fact the "light-second."

Final Answer: $500$ new units (light-seconds).

2

Convert the value of $g = 9.8 \text{ m/s}^2$ into the CGS system (cm and s).

Show solution

$$g = 9.8 \frac{\text{m}}{\text{s}^2} = 9.8 \cdot \frac{100 \text{ cm}}{\text{s}^2} = 980 \text{ cm/s}^2$$

Final Answer: $g = 980 \text{ cm/s}^2$.

3

The density of a substance is $8.0 \text{ g/cm}^3$. Express it in SI units (kg/m³).

Show solution

$$8.0 \frac{\text{g}}{\text{cm}^3} = 8.0 \cdot \frac{10^{-3} \text{ kg}}{(10^{-2} \text{ m})^3} = 8.0 \cdot \frac{10^{-3}}{10^{-6}} \text{ kg/m}^3 = 8.0 \times 10^{3} \text{ kg/m}^3$$

Final Answer: $8000 \text{ kg/m}^3$.

✎ Self-Check — 5 questions0 / 5
Q1.

Which of the following is not an SI base quantity?

Q2.

One light-year is approximately equal to:

Q3.

The value of $1 \text{ N}$ in dynes (CGS) is:

Q4.

If the unit of force is taken as 100 N, the unit of mass as 10 kg, and the unit of length as 100 m, then the unit of time is:

Q5.

The number of significant figures in $0.06900$ is:

Dimensional Formulae and Their ApplicationsTopic 2

The dimension of a physical quantity expresses how it depends on the seven base quantities. We write dimensions using bracketed base symbols: $$[M], [L], [T], [A], [K], [\text{mol}], [\text{cd}]$$

The dimensional formula of a derived quantity is the algebraic combination raised to appropriate exponents. For example:

  • Area: $[L^2]$
  • Volume: $[L^3]$
  • Velocity: $[LT^{-1}]$
  • Acceleration: $[LT^{-2}]$
  • Force: $[MLT^{-2}]$
  • Work/Energy: $[ML^2T^{-2}]$
  • Power: $[ML^2T^{-3}]$
  • Pressure: $[ML^{-1}T^{-2}]$
  • Momentum: $[MLT^{-1}]$
  • Angular momentum: $[ML^2T^{-1}]$
  • Frequency: $[T^{-1}]$
  • Charge: $[AT]$
  • Electric field: $[MLT^{-3}A^{-1}]$
  • Potential: $[ML^2T^{-3}A^{-1}]$
  • Resistance: $[ML^2T^{-3}A^{-2}]$
  • Magnetic field: $[MT^{-2}A^{-1}]$
  • Planck's constant: $[ML^2T^{-1}]$
  • Gravitational constant $G$: $[M^{-1}L^3T^{-2}]$

Principle of Dimensional Homogeneity: In any physically meaningful equation, every additive term must have the same dimensions. If $A = B + C$, then $[A] = [B] = [C]$. Arguments of transcendental functions ($\sin, \cos, e^x, \ln x$) must be dimensionless.

Three Main Uses of Dimensional Analysis:

  1. Checking correctness of equations — every term must have identical dimensions.
  2. Conversion of units between systems using $n_2 = n_1 \cdot (u_1/u_2)$.
  3. Deriving relationships between quantities (up to a dimensionless constant).

Limitations:

  • Cannot determine dimensionless multiplicative constants (e.g., the $\frac{1}{2}$ in $\frac{1}{2}mv^2$).
  • Fails when a quantity depends on more than three other quantities (insufficient equations).
  • Cannot decide between $\sin\theta$ and $\cos\theta$ since both are dimensionless.
  • Cannot handle equations involving addition of quantities of the same dimension (e.g., $s = ut + \frac{1}{2}at^2$ has two terms of the same dimension — coefficients can't be found).

Buckingham π-theorem (advanced): If a physical quantity depends on $n$ variables with $k$ fundamental dimensions, then there are $n - k$ independent dimensionless groups that govern the relationship.

Worked Examples
1

The time period $T$ of a simple pendulum is suspected to depend on its length $\ell$, mass $m$, and gravitational acceleration $g$. Use dimensional analysis to derive the formula.

Show solution

Assume $T = k \cdot \ell^a \cdot m^b \cdot g^c$ where $k$ is dimensionless. Writing dimensions: $$[T] = [L]^a [M]^b [LT^{-2}]^c = [M^b L^{a+c} T^{-2c}]$$

Matching exponents of $M$, $L$, $T$ on both sides:

  • $M$: $b = 0$
  • $L$: $a + c = 0$
  • $T$: $-2c = 1 \implies c = -1/2$, so $a = 1/2$

Therefore $T = k \sqrt{\ell/g}$. (Experimentally, $k = 2\pi$.)

Final Answer: $T \propto \sqrt{\ell/g}$, independent of mass.

2

Check whether the equation $v^2 = u^2 + 2as$ is dimensionally consistent (where $v, u$ are velocities, $a$ is acceleration, $s$ is displacement).

Show solution
  • $[v^2] = [LT^{-1}]^2 = [L^2 T^{-2}]$
  • $[u^2] = [L^2 T^{-2}]$
  • $[2as] = [\text{dimensionless}] \cdot [LT^{-2}] \cdot [L] = [L^2 T^{-2}]$

All three terms have dimensions $[L^2 T^{-2}]$.

Final Answer: The equation is dimensionally consistent.

3

Stoke's law gives the viscous force on a sphere of radius $r$ moving with velocity $v$ through a fluid of viscosity coefficient $\eta$ as $F \propto \eta^a r^b v^c$. Find $a, b, c$.

Show solution

Given $[F] = [MLT^{-2}]$, $[\eta] = [ML^{-1}T^{-1}]$, $[r] = [L]$, $[v] = [LT^{-1}]$: $$[MLT^{-2}] = [ML^{-1}T^{-1}]^a [L]^b [LT^{-1}]^c = [M^a L^{-a+b+c} T^{-a-c}]$$

Matching:

  • $M$: $a = 1$
  • $L$: $-a + b + c = 1 \implies b + c = 2$
  • $T$: $-a - c = -2 \implies c = 1$, so $b = 1$

Final Answer: $F = k \eta r v$ with $k = 6\pi$ (Stokes' law: $F = 6\pi \eta r v$).

✎ Self-Check — 5 questions0 / 5
Q1.

The dimensional formula for the universal gravitational constant $G$ is:

Q2.

Which pair has the same dimensions?

Q3.

The dimensions of Planck's constant $h$ are the same as those of:

Q4.

If force $F$, length $L$, and time $T$ are taken as fundamental quantities, the dimensions of mass are:

Q5.

The dimensional formula for electrical permittivity $\varepsilon_0$ is:

2
Module 2

Errors and Measurement Instruments

Significant Figures, Error Analysis and PropagationTopic 1

Significant Figures (SF) in a measurement are the digits known with certainty plus the first uncertain digit. They reflect the precision of the measuring instrument.

Rules for counting SF:

  1. All non-zero digits are significant. (`23.45` has 4 SF)
  2. Zeros between non-zero digits are significant. (`2003` has 4 SF)
  3. Leading zeros are not significant. (`0.0023` has 2 SF)
  4. Trailing zeros in a number with a decimal point are significant. (`2.300` has 4 SF)
  5. Trailing zeros in a whole number without a decimal point are ambiguous; scientific notation removes the ambiguity. (`300` → `3 × 10²` is 1 SF; `3.00 × 10²` is 3 SF)

Rules for arithmetic operations:

  • Addition/subtraction: result has the same number of decimal places as the term with the fewest decimal places. ($2.34 + 1.2 = 3.5$)
  • Multiplication/division: result has the same number of significant figures as the term with the fewest SF. ($2.0 \times 3.14 = 6.3$)

Rounding off: If the digit to be dropped is < 5, round down; if > 5, round up; if exactly 5, round to make the last retained digit even (banker's rounding).

Types of errors:

  • Systematic errors — consistent biases (zero error, calibration error, environmental factors). Can be eliminated by correction.
  • Random errors — unpredictable fluctuations. Reduced by repeated measurements and averaging.
  • Gross errors — careless mistakes (misreading, calculation error). Avoidable.

Quantitative error definitions: For $n$ measurements $a_1, a_2, \ldots, a_n$ of a quantity:

  • Mean (best estimate): $\bar{a} = \dfrac{1}{n}\sum_{i=1}^{n} a_i$
  • Absolute error in the $i$-th measurement: $\Delta a_i = |a_i - \bar{a}|$
  • Mean absolute error: $\overline{\Delta a} = \dfrac{1}{n}\sum |\Delta a_i|$
  • Relative error: $\dfrac{\overline{\Delta a}}{\bar{a}}$
  • Percentage error: $\dfrac{\overline{\Delta a}}{\bar{a}} \times 100\%$

Propagation of errors (for small errors $\Delta x \ll x$):

OperationResultError
$Z = A + B$ or $Z = A - B$sum or difference$\Delta Z = \Delta A + \Delta B$ (max absolute)
$Z = AB$ or $Z = A/B$product or quotient$\dfrac{\Delta Z}{Z} = \dfrac{\Delta A}{A} + \dfrac{\Delta B}{B}$
$Z = A^n B^m / C^p$power law$\dfrac{\Delta Z}{Z} = \|n\|\dfrac{\Delta A}{A} + \|m\|\dfrac{\Delta B}{B} + \|p\|\dfrac{\Delta C}{C}$
$Z = \ln A$logarithm$\Delta Z = \dfrac{\Delta A}{A}$
$Z = e^A$exponential$\dfrac{\Delta Z}{Z} = \Delta A$

The absolute errors add in subtraction, which is why subtracting nearly equal numbers amplifies relative error — a key practical concern.

Worked Examples
1

The length and breadth of a rectangle are $(5.7 \pm 0.1)$ cm and $(3.4 \pm 0.2)$ cm respectively. Find the area and its percentage error.

Show solution

Area $A = \ell \times b = 5.7 \times 3.4 = 19.38 \text{ cm}^2$.

Relative errors: $$\frac{\Delta \ell}{\ell} = \frac{0.1}{5.7} \approx 0.0175, \quad \frac{\Delta b}{b} = \frac{0.2}{3.4} \approx 0.0588$$

For a product: $\dfrac{\Delta A}{A} = \dfrac{\Delta \ell}{\ell} + \dfrac{\Delta b}{b} = 0.0175 + 0.0588 = 0.0763 \approx 7.6\%$.

Absolute error: $\Delta A = 0.0763 \times 19.38 \approx 1.48 \text{ cm}^2$.

Reported to correct SF: $A = (19 \pm 1) \text{ cm}^2$ or $A = (19.4 \pm 1.5) \text{ cm}^2$.

Final Answer: Percentage error ≈ 7.6%; $A \approx (19.4 \pm 1.5) \text{ cm}^2$.

2

The period $T$ of a simple pendulum is measured from $T = 2\pi\sqrt{L/g}$, where $L = (20.0 \pm 0.1)$ cm and $T = (0.90 \pm 0.01)$ s. Find the percentage error in the value of $g$.

Show solution

Squaring: $T^2 = 4\pi^2 L/g \implies g = 4\pi^2 L / T^2$.

For $g \propto L \cdot T^{-2}$: $$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$$

Substituting: $$\frac{\Delta L}{L} = \frac{0.1}{20.0} = 0.005 = 0.5\%, \quad \frac{\Delta T}{T} = \frac{0.01}{0.90} \approx 0.0111 = 1.11\%$$

$$\frac{\Delta g}{g} = 0.5\% + 2 \times 1.11\% = 0.5\% + 2.22\% = 2.72\%$$

Final Answer: Percentage error in $g$ ≈ 2.7%.

3

A physical quantity $P$ is given by $P = \dfrac{a^3 b^2}{\sqrt{c} \cdot d}$. Percentage errors in $a, b, c, d$ are $1\%, 3\%, 4\%, 2\%$ respectively. Find the maximum percentage error in $P$.

Show solution

$$\frac{\Delta P}{P} \times 100\% = 3 \cdot \frac{\Delta a}{a} + 2 \cdot \frac{\Delta b}{b} + \frac{1}{2} \cdot \frac{\Delta c}{c} + \frac{\Delta d}{d}$$

$$= 3(1\%) + 2(3\%) + \frac{1}{2}(4\%) + 1(2\%) = 3\% + 6\% + 2\% + 2\% = 13\%$$

Final Answer: Maximum percentage error = 13%.

✎ Self-Check — 5 questions0 / 5
Q1.

The number of significant figures in $230.040$ is:

Q2.

If $X = A \times B$ and percentage errors in $A$ and $B$ are $a\%$ and $b\%$, the maximum percentage error in $X$ is:

Q3.

A force of $(2500 \pm 5)$ N is applied over an area of $(0.32 \pm 0.02)$ m². The percentage error in pressure is approximately:

Q4.

The relative error in $Z$, if $Z = A^2$, equals:

Q5.

In the addition $9.327 + 12.42 + 0.073$, the result with correct significant figures is:

Vernier Calipers, Screw Gauge and Practical MeasurementsTopic 2

Vernier Calipers measure lengths up to a precision of typically $0.1$ mm or $0.05$ mm using a secondary (vernier) scale that slides along a primary (main) scale.

Vernier Constant (Least Count): $$\text{LC} = \text{(value of 1 MSD)} - \text{(value of 1 VSD)}$$ where MSD = main scale division and VSD = vernier scale division. If $n$ VSD coincide with $(n-1)$ MSD, then: $$\text{LC} = \frac{1 \text{ MSD}}{n}$$

For a typical vernier with $10$ VSD = $9$ MSD and $1$ MSD = $1$ mm: $\text{LC} = 0.1$ mm.

Reading: Total reading = (Main scale reading) + (vernier division coinciding) × (LC) ± (zero error).

Zero error occurs when the vernier shows a non-zero reading with the jaws fully closed:

  • Positive zero error: zero of vernier is to the right of main scale zero → subtract from observed reading.
  • Negative zero error: zero of vernier is to the left of main scale zero → add to observed reading.

Screw Gauge (Micrometer) measures very small lengths (typically up to $0.001$ cm = $0.01$ mm precision) using the principle that the rotation of a screw produces a linear translation equal to the pitch per revolution.

Pitch: linear distance moved by the spindle in one complete rotation = (distance moved) / (number of rotations).

Least Count: $$\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$$

For a typical screw gauge with pitch = $1$ mm and $100$ circular divisions: $\text{LC} = 0.01$ mm.

Reading: Total = (Linear/pitch scale reading) + (circular scale reading) × (LC) ± (zero error).

Backlash error: Caused by play between screw and nut; minimized by always rotating the screw in the same direction during measurement.

Travelling Microscope is used for measuring small distances with vernier-style precision in two dimensions; useful for measuring focal lengths, refractive indices, and small displacements.

Worked Examples
1

In a vernier caliper, $10$ vernier divisions coincide with $9$ main scale divisions. If $1$ MSD = $1$ mm, find the least count.

Show solution

$$10 \text{ VSD} = 9 \text{ MSD} \implies 1 \text{ VSD} = 0.9 \text{ MSD} = 0.9 \text{ mm}$$ $$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 1 - 0.9 = 0.1 \text{ mm}$$

Final Answer: Least count = $0.1$ mm = $0.01$ cm.

2

A screw gauge has pitch $1$ mm and $50$ divisions on the circular scale. While measuring the diameter of a wire, the main scale reads $3$ mm and the $27$-th division of the circular scale coincides with the reference line. Find the diameter.

Show solution

$$\text{LC} = \frac{1 \text{ mm}}{50} = 0.02 \text{ mm}$$

Reading = MSR + CSR × LC = $3 + 27 \times 0.02 = 3 + 0.54 = 3.54$ mm.

Final Answer: Diameter of wire = $3.54$ mm.

3

A screw gauge has a positive zero error of $5$ divisions. With LC = $0.01$ mm, while measuring the thickness of a sheet, the MSR is $2$ mm and the $40$-th circular division coincides. Find the correct thickness.

Show solution

Observed reading = $2 + 40 \times 0.01 = 2.40$ mm.

Positive zero error means observed > true. Correction = $5 \times 0.01 = 0.05$ mm to be subtracted.

Correct thickness = $2.40 - 0.05 = 2.35$ mm.

Final Answer: Thickness = $2.35$ mm.

✎ Self-Check — 5 questions0 / 5
Q1.

The least count of a vernier caliper with $20$ VSD = $19$ MSD (each MSD = $1$ mm) is:

Q2.

The pitch of a screw gauge is $0.5$ mm and the head scale has $100$ divisions. Its LC is:

Q3.

In a vernier caliper, if the zero of the vernier scale lies to the left of the main scale zero by $4$ divisions, with LC = $0.01$ cm, the zero error is:

Q4.

To reduce backlash error in a screw gauge, the screw should always be rotated:

Q5.

The diameter of a small ball measured by a screw gauge (LC = $0.001$ cm) is recorded as $0.245$ cm in five trials, identical each time. The percentage error in volume is:

Ready to test yourself?

Attempt the full timed mock tests — Main & Advanced level.

Start Mock Test 1 →