Units of Measurement & the SI System
Why Standard Units Matter
A physical quantity is fully described only by a number together with a unit — saying "the rod is 5 long" is meaningless without specifying 5 what (metres, feet, etc.). To allow scientists worldwide to compare results reliably, an internationally agreed set of units, the International System of Units (SI), is used.
| Quantity | SI Unit | Symbol |
|---|
| Length | metre | m |
| Mass | kilogram | kg |
| Time | second | s |
| Electric current | ampere | A |
| Temperature | kelvin | K |
| Amount of substance | mole | mol |
| Luminous intensity | candela | cd |
Some Practical Units Used Alongside SI
Fermi (fm): $10^{-15}$ m, used for nuclear sizes. Angstrom (Å): $10^{-10}$ m, used for atomic/molecular sizes. Astronomical unit (AU): average Earth-Sun distance, $\approx 1.496 \times 10^{11}$ m. Light year: distance light travels in one year, $\approx 9.46 \times 10^{15}$ m. Parsec: $\approx 3.08 \times 10^{16}$ m, the largest of these practical astronomical units.
Solved Example
Arrange the angstrom, fermi, light year, and astronomical unit in increasing order of size.
$\text{Fermi} (10^{-15}\text{m}) < \text{Angstrom} (10^{-10}\text{m}) < \text{Astronomical unit} (1.5\times10^{11}\text{m}) < \text{Light year} (9.46\times10^{15}\text{m})$
Q. Why is an internationally standardised system of units like SI necessary in science?
Ans: Without a common, universally agreed system, measurements taken in one country or lab (e.g., in feet or local units) couldn't be reliably compared or verified against measurements taken elsewhere. A standard system like SI ensures every scientist's numerical results mean exactly the same physical quantity, everywhere, making results reproducible and comparable worldwide.
Q. Which unit would be most appropriate for expressing the distance between two stars in our galaxy — angstrom, fermi, or light year? Explain.
Ans: Light year. Angstrom and fermi are units suited to extremely small (atomic/nuclear) scales, while distances between stars are astronomically large — using such a large practical unit (light year) avoids working with impractically huge numbers that would result from using metres or kilometres directly.
Q. A parsec is larger than a light year. Roughly how many light years does one parsec correspond to?
Ans: Approximately 3.26 light years (1 parsec ≈ 3.08 × 10¹⁶ m, and 1 light year ≈ 9.46 × 10¹⁵ m, giving a ratio of about 3.26).
Fundamental vs Derived Units
Fundamental (Base) Units
Fundamental units are independent of one another and cannot be expressed in terms of any other unit — the 7 SI base units (metre, kilogram, second, ampere, kelvin, mole, candela) are all fundamental.
Derived Units
Derived units are obtained by suitably combining (multiplying/dividing) the fundamental units, based on the physical quantity's defining relationship. E.g., speed = distance/time, so its unit is m/s; force = mass × acceleration, so its unit is $\text{kg m/s}^2$, given a special name, the newton (N).
Solved Example
Derive the SI unit of pressure from its defining relation, Pressure = Force/Area.
Unit of Force $= \text{kg m/s}^2$ (newton). Unit of Area $= \text{m}^2$.
Unit of Pressure $= \dfrac{\text{kg m/s}^2}{\text{m}^2} = \text{kg m}^{-1}\text{s}^{-2}$, given the special name pascal (Pa).
Q. Why is the kilogram considered a fundamental unit, but the newton is considered a derived unit?
Ans: The kilogram (mass) is an independently defined base quantity, not built from any other unit. The newton (force) is defined through the relation Force = mass × acceleration, so its unit is constructed by combining the fundamental units of mass (kg), length (m), and time (s) — making it a derived, not fundamental, unit.
Q. Derive the SI unit of work done, given that Work = Force × displacement.
Ans: Unit of Force = kg m/s². Unit of displacement = m. Unit of Work = kg m/s² × m = kg m²s⁻², given the special name joule (J).
Systems of Units Besides SI
Before SI became the international standard, other systems were commonly used: the CGS system (centimetre, gram, second), the FPS system (foot, pound, second), and the MKS system (metre, kilogram, second — essentially the mechanical part of SI). SI was adopted internationally in 1960 to unify scientific communication and eliminate the need for constant conversion between these differing systems.
Q. Why was the SI system introduced when other systems like CGS and FPS already existed?
Ans: Having multiple competing unit systems in simultaneous use made it cumbersome and error-prone to compare or combine results from different sources, requiring frequent unit conversions. Adopting one single, internationally agreed system (SI) for all scientific work eliminated this friction and inconsistency, streamlining global scientific communication and collaboration.
Dimensions & Dimensional Formulas
What Are Dimensions
The dimensions of a physical quantity show which fundamental quantities (and to what powers) combine to represent it — expressed using the symbols $[M]$ for mass, $[L]$ for length, $[T]$ for time, $[A]$ for current, $[K]$ for temperature, etc. A dimensional formula expresses a quantity in this power notation, e.g., force has dimensional formula $[M^1 L^1 T^{-2}]$.
| Quantity | Dimensional Formula |
|---|
| Velocity | [M⁰L¹T⁻¹] |
| Acceleration | [M⁰L¹T⁻²] |
| Force | [M¹L¹T⁻²] |
| Work / Energy | [M¹L²T⁻²] |
| Power | [M¹L²T⁻³] |
| Pressure | [M¹L⁻¹T⁻²] |
Solved Example
Derive the dimensional formula of momentum (mass × velocity).
$[\text{Momentum}] = [M] \times [L T^{-1}] = [M^1 L^1 T^{-1}]$
Q. Derive the dimensional formula of kinetic energy, given KE = ½mv².
Ans: [KE] = [M][LT⁻¹]² = [M¹L²T⁻²] (the numerical factor ½ has no dimensions and is ignored).
Q. What are the dimensions of a purely numerical constant, such as π or the number 2, in any physical formula?
Ans: Dimensionless — [M⁰L⁰T⁰]. Pure numbers and mathematical constants carry no physical dimension at all, since they don't represent a measurable physical quantity.
Q. Derive the dimensional formula of angular velocity (ω = angle/time), given that angle itself is dimensionless (a ratio of two lengths, arc/radius).
Ans: Since angle is dimensionless [M⁰L⁰T⁰], angular velocity ω = angle/time has dimensional formula [M⁰L⁰T⁻¹] — dimensionally identical to frequency, even though the two represent physically different concepts.
Dimensional Analysis — Checking Equations
Principle of Homogeneity of Dimensions
For a physical equation to be dimensionally correct, the dimensions on the left-hand side must exactly match the dimensions on the right-hand side, and every term being added or subtracted within the equation must individually have the same dimensions. This principle is used to check whether a derived or given formula could possibly be correct.
Solved Example
Check the dimensional correctness of the equation $s = ut + \tfrac{1}{2}at^2$ (equation of motion).
$[s] = [L]$
$[ut] = [LT^{-1}][T] = [L]$
$[\tfrac{1}{2}at^2] = [LT^{-2}][T^2] = [L]$
All three terms have dimension $[L]$, matching the left-hand side — the equation is dimensionally correct.
Q. Check whether the equation v² = u² + 2as is dimensionally consistent.
Ans: [v²] = [LT⁻¹]² = [L²T⁻²]. [u²] = [L²T⁻²]. [2as] = [LT⁻²][L] = [L²T⁻²]. All terms match [L²T⁻²] — the equation is dimensionally correct.
Q. A student proposes the formula T = 2π√(m/k) for the time period of a spring-mass system, where m is mass and k is the spring constant (force per unit displacement, dimension [MT⁻²]). Verify this dimensionally.
Ans: [m/k] = [M]/[MT⁻²] = [T²]. So [√(m/k)] = [T]. Since T (time period) should have dimension [T], and the numerical constant 2π is dimensionless, the formula is dimensionally consistent.
Dimensional Analysis — Deriving Relations & Limitations
Using Dimensions to Derive a Formula
If a physical quantity is known (or assumed) to depend on certain other quantities in a power-law relationship, dimensional analysis can determine the powers involved by comparing dimensions on both sides.
Solved Example
Using dimensional analysis, derive an expression for the time period (T) of a simple pendulum, assuming it depends on its length (l), mass (m), and acceleration due to gravity (g).
Assume $T = k \cdot l^a m^b g^c$, where $k$ is a dimensionless constant.
Dimensionally: $[T^1] = [L]^a [M]^b [LT^{-2}]^c = [M^b L^{a+c} T^{-2c}]$
Comparing powers: For M: $b = 0$. For T: $-2c = 1 \Rightarrow c = -\tfrac{1}{2}$. For L: $a + c = 0 \Rightarrow a = \tfrac{1}{2}$.
So $T = k\, l^{1/2} g^{-1/2} = k\sqrt{l/g}$ — matching the known formula $T = 2\pi\sqrt{l/g}$ (with $k = 2\pi$, a value dimensional analysis alone cannot determine).
Key Limitations of Dimensional Analysis
Dimensional analysis cannot determine the value of dimensionless constants (like $2\pi$ above), cannot distinguish between quantities with the same dimensions (e.g., torque and energy have identical dimensions but are physically different), cannot be used if a physical quantity depends on more than 3 independent variables in a single power-law equation (since there are only 3 fundamental dimensions M, L, T to match against), and cannot handle relationships involving trigonometric, exponential, or logarithmic functions.
Q. Why can't dimensional analysis be used to find the numerical value of the constant "2π" in the pendulum's time period formula?
Ans: Dimensional analysis only compares the dimensional (unit-based) consistency of both sides of an equation — pure numbers and mathematical constants are dimensionless ([M⁰L⁰T⁰]) and simply don't appear anywhere in this comparison. Since the method has no way to "see" or determine a dimensionless multiplying factor, its exact value must instead come from experiment or a full mathematical derivation.
Q. Torque and energy both have the dimensional formula [M¹L²T⁻²]. What does this reveal about a key limitation of dimensional analysis?
Ans: It shows that dimensional analysis cannot distinguish between two physically distinct quantities if they happen to share the same combination of fundamental dimensions — despite having identical dimensional formulas, torque (a vector, related to rotational effect of a force) and energy (a scalar, related to capacity to do work) are conceptually very different quantities, a distinction dimensional analysis alone cannot capture.
Q. A quantity is suspected to depend on 4 different independent physical variables in a single power-law relationship. Can dimensional analysis (using only M, L, T) fully determine all the powers involved? Explain.
Ans: Generally, no. Dimensional analysis using the three fundamental mechanical dimensions (M, L, T) provides only three independent equations (one each for matching powers of M, L, and T on both sides) — this is enough to solve for at most 3 unknown powers. If 4 or more independent variables (and hence 4 or more unknown powers) are involved, there aren't enough independent equations to uniquely solve for every power, making the method insufficient on its own without additional physical reasoning or information.
Accuracy, Precision & Types of Errors
Accuracy vs Precision
Accuracy refers to how close a measured value is to the true value. Precision refers to how close repeated measurements are to each other (regardless of whether they're close to the true value). A measurement set can be precise without being accurate, if all readings are consistently off by a similar amount from the true value.
| Error Type | Cause | Can Be Minimised By |
|---|
| Systematic error | Consistent flaw in instrument, method, or technique (e.g., zero error, imperfect calibration) | Correcting/calibrating the instrument, refining the method |
| Random error | Unpredictable, irregular fluctuations (e.g., observer's reaction time, environmental fluctuations) | Taking many repeated readings and averaging them |
| Gross error | Human carelessness (misreading scale, wrong recording) | Careful, attentive measurement technique |
Solved Example
A student measures the length of a rod five times and gets values very close to each other but consistently 2 cm higher than the rod's known true length. Describe this in terms of accuracy, precision, and error type.
The readings are precise (very close to each other) but not accurate (consistently far from the true value). Since the offset is consistent (always +2 cm) rather than randomly scattered, this indicates a systematic error — likely a zero error in the measuring instrument, which could be corrected by recalibrating the instrument or subtracting the known offset from every reading.
Q. Why can taking repeated measurements and averaging them reduce random error, but not systematic error?
Ans: Random errors fluctuate unpredictably in both direction and magnitude from one reading to the next, so averaging many readings tends to cancel out these fluctuations, converging closer to the true value. Systematic errors, however, consistently bias every single reading in the same direction by roughly the same amount — averaging repeated readings doesn't cancel this consistent bias at all, since every reading shares the same underlying flaw.
Q. Give one example of a systematic error that could occur in a laboratory experiment.
Ans: A zero error in a measuring instrument (e.g., a vernier callipers or screw gauge that doesn't read exactly zero when the jaws/spindle are fully closed) — this causes every subsequent measurement to be consistently off by the same fixed amount, until corrected.
Q. A set of five readings for a quantity are scattered widely around the true value, with no readings close to each other or to the true value. Describe this situation in terms of accuracy and precision.
Ans: This situation is neither accurate nor precise — the readings are not close to the true value (inaccurate) and are also not close to each other (imprecise), suggesting the presence of significant random error, and possibly also poor experimental technique or an unreliable instrument.
Absolute, Relative & Percentage Error
Relative (Fractional) Error & Percentage Error
$$\text{Relative error} = \dfrac{\Delta a_{mean}}{a_{mean}}$$
$$\text{Percentage error} = \dfrac{\Delta a_{mean}}{a_{mean}} \times 100\%$$
Solved Example
Five measurements of a length gave: 2.63, 2.56, 2.42, 2.71, 2.48 (all in cm). Calculate the mean value, mean absolute error, and percentage error.
Mean $= \dfrac{2.63+2.56+2.42+2.71+2.48}{5} = \dfrac{12.8}{5} = 2.56\ \text{cm}$
Absolute errors: $|2.56-2.63|=0.07$, $|2.56-2.56|=0.00$, $|2.56-2.42|=0.14$, $|2.56-2.71|=0.15$, $|2.56-2.48|=0.08$
Mean absolute error $= \dfrac{0.07+0.00+0.14+0.15+0.08}{5} = \dfrac{0.44}{5} = 0.088 \approx 0.09\ \text{cm}$
Percentage error $= \dfrac{0.09}{2.56} \times 100 \approx 3.5\%$
Q. A quantity's true (mean) value is 50 units, and a particular measurement gives 48 units. Calculate the absolute error and percentage error of this measurement.
Ans: Absolute error = |50 − 48| = 2 units. Percentage error = (2/50) × 100 = 4%.
Q. Why is percentage error often more useful for comparison than absolute error alone?
Ans: Absolute error just gives a raw magnitude of deviation, which by itself doesn't indicate whether that deviation is significant or negligible relative to the quantity being measured. Percentage error expresses this deviation as a fraction of the actual measured value, allowing meaningful comparison of measurement quality between quantities of very different sizes (e.g., a 1 mm absolute error is huge for a 1 cm measurement but trivial for a 10 m measurement).
Combination of Errors
Errors in Sum or Difference
If $Z = A + B$ or $Z = A - B$, the absolute errors add in both cases: $\Delta Z = \Delta A + \Delta B$ (even for subtraction, since the worst-case error still adds up).
Errors in Product or Quotient
If $Z = AB$ or $Z = A/B$, the relative (fractional) errors add: $\dfrac{\Delta Z}{Z} = \dfrac{\Delta A}{A} + \dfrac{\Delta B}{B}$.
Error in a Quantity Raised to a Power
If $Z = \dfrac{A^p B^q}{C^r}$, the relative error is: $\dfrac{\Delta Z}{Z} = p\dfrac{\Delta A}{A} + q\dfrac{\Delta B}{B} + r\dfrac{\Delta C}{C}$ — each relative error is multiplied by the magnitude of its power before adding.
Solved Example
If $A = 100 \pm 2$ and $B = 50 \pm 1$, find the maximum absolute error in $Z = A + B$ and in $Z = A - B$.
For both sum and difference: $\Delta Z = \Delta A + \Delta B = 2 + 1 = 3$
$Z = A+B = 150 \pm 3$; $Z = A - B = 50 \pm 3$
Solved Example
The radius of a sphere is measured as $r = 5.0 \pm 0.1\ \text{cm}$. Calculate the percentage error in its volume, $V = \tfrac{4}{3}\pi r^3$.
Since $V \propto r^3$, the relative error in V is 3 times the relative error in r.
Relative error in $r = 0.1/5.0 = 0.02$; Percentage error in $r = 2\%$
Percentage error in $V = 3 \times 2\% = 6\%$
Q. If the percentage errors in measuring the length and breadth of a rectangle are 2% and 3% respectively, what is the percentage error in its calculated area?
Ans: Since Area = length × breadth, relative errors add: percentage error in area = 2% + 3% = 5%.
Q. The density of a material is calculated using ρ = m/V. If the percentage error in mass is 2% and in volume is 3%, find the percentage error in density.
Ans: Since density is a quotient (m/V), relative errors add: percentage error in density = 2% + 3% = 5%.
Solved Example
The time period of a pendulum is given by $T = 2\pi\sqrt{l/g}$. If the percentage error in measuring length l is 1% and the time period T is measured with a 2% error, find the percentage error in the calculated value of g.
Rearranging: $g = \dfrac{4\pi^2 l}{T^2}$, so $g \propto l\, T^{-2}$.
Percentage error in $g = (\text{percentage error in } l) + 2\times(\text{percentage error in } T) = 1\% + 2(2\%) = 1\% + 4\% = 5\%$
Q. In an experiment, the percentage error in measuring the radius of a wire is 3%, and in measuring its length is 1%. Find the percentage error in its calculated volume (V = πr²l).
Ans: Since V ∝ r²l, percentage error in V = 2 × (percentage error in r) + (percentage error in l) = 2(3%) + 1% = 6% + 1% = 7%.
Significant Figures
Definition
Significant figures in a measured value are all the digits known with certainty, plus one final estimated (uncertain) digit — they reflect the precision of the measuring instrument used.
Key Counting Rules
- All non-zero digits are significant.
- Zeros between two non-zero digits are significant.
- Leading zeros (before the first non-zero digit) are never significant.
- Trailing zeros after a decimal point are significant.
- For a number without a decimal point, trailing zeros are ambiguous — best resolved using scientific notation.
Solved Example
State the number of significant figures in: (a) 0.00706, (b) 1.500, (c) $3.20 \times 10^4$.
(a) 3 significant figures (leading zeros don't count; the zero between 7 and 6 does).
(b) 4 significant figures (trailing zeros after a decimal point count).
(c) 3 significant figures (as clearly shown by the coefficient in scientific notation).
Q. How many significant figures are there in the measurement 6.023 × 10²³?
Ans: 4 significant figures (6, 0, 2, 3 — all part of the coefficient in scientific notation).
Q. Why is scientific notation preferred over plain decimal notation when the number of significant figures might otherwise be ambiguous?
Ans: A number like 100 written in plain form could represent 1, 2, or 3 significant figures depending on the precision actually intended, since trailing zeros without a decimal point are ambiguous. Writing it in scientific notation (e.g., 1.00 × 10²) makes the number of significant figures completely unambiguous, since every digit shown in the coefficient is understood to be significant.
Q. When adding 12.36 and 4.2 (both measured values), how many decimal places should the final answer be rounded to, and why?
Ans: The final answer should be rounded to 1 decimal place, matching the term with the fewest decimal places (4.2). In addition/subtraction, the result cannot be more precise (cannot have more decimal places) than the least precise measurement used in the calculation.
Order of Magnitude
The order of magnitude of a quantity is the power of 10 nearest to its actual value, found by writing the number in scientific notation ($N \times 10^x$) and rounding: if $N \ge 5$, the order of magnitude is $10^{x+1}$; if $N < 5$, it remains $10^x$. This gives a quick, approximate sense of a quantity's scale without needing its exact value.
Solved Example
Find the order of magnitude of the numbers 4,900 and 5,200.
$4900 = 4.9 \times 10^3$. Since $4.9 < 5$, order of magnitude $= 10^3$.
$5200 = 5.2 \times 10^3$. Since $5.2 \ge 5$, order of magnitude $= 10^4$.
Q. What is the order of magnitude of the number 720?
Ans: 720 = 7.2 × 10². Since 7.2 ≥ 5, the order of magnitude is 10³.
Least Count of Measuring Instruments
Definition
The least count of an instrument is the smallest measurement it can reliably resolve/read — every individual measurement made with that instrument therefore carries an inherent uncertainty at least as large as its least count.
Vernier Callipers
Least count $= \text{value of 1 main scale division} - \text{value of 1 vernier scale division}$, typically $0.01\ \text{cm}$ for a standard callipers with 10 vernier divisions matching 9 main scale divisions.
Screw Gauge (Micrometer)
Least count $= \dfrac{\text{Pitch}}{\text{Number of divisions on the circular scale}}$, where pitch is the linear distance moved by the spindle for one complete rotation of the circular scale — typically giving a least count of $0.001\ \text{cm}$, finer than a vernier callipers.
Solved Example
A screw gauge has a pitch of 1 mm and 100 divisions on its circular scale. Calculate its least count.
$\text{Least count} = \dfrac{\text{Pitch}}{\text{Number of circular divisions}} = \dfrac{1\ \text{mm}}{100} = 0.01\ \text{mm} = 0.001\ \text{cm}$
Q. Why does a screw gauge generally give more precise measurements than a vernier callipers?
Ans: A screw gauge typically has a much smaller least count (commonly 0.001 cm) compared to a standard vernier callipers (typically 0.01 cm) — since least count represents the smallest reliably measurable increment, a smaller least count means finer, more precise measurements are possible with the screw gauge.
Q. If a screw gauge has a pitch of 0.5 mm and 50 divisions on its circular scale, calculate its least count and compare it to a vernier callipers with a least count of 0.01 cm.
Ans: Least count of screw gauge = 0.5 mm / 50 = 0.01 mm = 0.001 cm. Comparing to the vernier callipers' least count of 0.01 cm, the screw gauge here is 10 times more precise, able to resolve measurements ten times finer than the vernier callipers.
Q. A vernier callipers has 10 vernier divisions matching 9 main scale divisions, with each main scale division equal to 1 mm. Calculate its least count.
Ans: Value of 1 vernier division = 9/10 = 0.9 mm. Least count = 1 main scale division − 1 vernier division = 1 − 0.9 = 0.1 mm = 0.01 cm.
Zero Error
Positive & Negative Zero Error
An instrument has zero error if it doesn't read exactly zero when it should (e.g., when the jaws of a vernier callipers or the spindle of a screw gauge are fully closed with nothing between them). Positive zero error occurs when the zero of the moving (vernier/circular) scale lies to the right of (past) the main scale's zero — this error must be subtracted from every subsequent raw reading. Negative zero error occurs when it lies to the left of (before) the main scale's zero — this error must be added to every subsequent raw reading.
Solved Example
A screw gauge shows a reading of 5.32 mm for an object, but has a zero error of +0.02 mm. Find the corrected reading.
Since the zero error is positive, it must be subtracted from the observed reading: Corrected reading $= 5.32 - 0.02 = 5.30\ \text{mm}$.
Q. Why must a positive zero error be subtracted from readings, while a negative zero error must be added?
Ans: A positive zero error means the instrument already reads slightly above zero even with nothing being measured — so every subsequent measurement is inflated by this same fixed amount, requiring it to be subtracted to correct back down to the true value. A negative zero error means the instrument reads slightly below zero (into negative territory) when it should read zero, so every subsequent reading is understated by this fixed amount, requiring the error to be added back to reach the true value.
⚡ Formula Cheat-Sheet
Force dimension[M¹L¹T⁻²]
Energy/Work dimension[M¹L²T⁻²]
Sum/difference errorΔZ = ΔA + ΔB
Product/quotient errorΔZ/Z = ΔA/A + ΔB/B
Power error ruleΔZ/Z = p(ΔA/A) + q(ΔB/B)
Vernier callipers L.C.1 MSD − 1 VSD (~0.01 cm)
Screw gauge L.C.Pitch ÷ circular divisions