Units and Measurements
Learn how to report a measurement, check an equation and propagate uncertainty. Measurement instruments and error analysis connect the mathematics to practical experiments.
Understand the idea → see it move → solve a question. Extra theory is there when you need it.
Textbook alignment: NCERT Class 11 Physics · chapter 1. Coaching order and separate foundation modules follow your uploaded index.
Physical quantities, SI units and conversion
A physical quantity combines a numerical value and a unit. The number changes when the unit changes, while the measured quantity does not. SI uses seven base quantities: length, mass, time, electric current, temperature, amount of substance and luminous intensity.
Derived units are combinations of base units. Convert each factor before substituting into a formula: cm² carries a squared conversion factor, and a cubic centimetre is a millionth of a cubic metre.
Other useful formulas
Convert areas and volumes with the square or cube of the length conversion, respectively.
A density is 2.7 g/cm³. Convert it to kg/m³.
Show solution
One gram is 10⁻³ kg and one cm³ is 10⁻⁶ m³. Thus the density is 2.7×10³ = 2700 kg/m³.
Express 1 kWh in joules and in electron-volts.
Show solution
- 1 kW = 1000 W = 1000 J/s and 1 h = 3600 s.
- 1 kWh = 1000 × 3600 = \(3.6\times10^{6}\) J.
- In eV: \(\dfrac{3.6\times10^{6}}{1.6\times10^{-19}}=2.25\times10^{25}\) eV.
Answer: \(3.6\times10^{6}\) J = \(2.25\times10^{25}\) eV
More solved problem patterns
Which of these is not a unit of energy: erg, electron-volt, kilowatt-hour, watt?
Show solution
- erg is the CGS energy unit; eV and kWh are energy units too.
- The watt is J/s, a rate of energy transfer. It is a unit of power.
Answer: Watt
How many astronomical units make one light year?
Show solution
- \(\dfrac{1\ \text{ly}}{1\ \text{AU}}=\dfrac{9.46\times10^{15}}{1.496\times10^{11}}\).
- This is about \(6.3\times10^{4}\).
Answer: About 63 000 AU
Go deeper · extra explanation and derivations
Base units, prefixes and practical units
| Base quantity | SI unit | Symbol |
|---|---|---|
| Length | metre | m |
| Mass | kilogram | kg |
| Time | second | s |
| Electric current | ampere | A |
| Thermodynamic temperature | kelvin | K |
| Amount of substance | mole | mol |
| Luminous intensity | candela | cd |
The radian (plane angle) and steradian (solid angle) were once called supplementary units. Both are dimensionless.
Prefixes: femto \(10^{-15}\), pico \(10^{-12}\), nano \(10^{-9}\), micro \(10^{-6}\), milli \(10^{-3}\), centi \(10^{-2}\), kilo \(10^{3}\), mega \(10^{6}\), giga \(10^{9}\), tera \(10^{12}\).
Practical length units: 1 fermi = \(10^{-15}\) m; 1 ångström = \(10^{-10}\) m; 1 micron = \(10^{-6}\) m; 1 AU = \(1.496\times10^{11}\) m; 1 light year = \(9.46\times10^{15}\) m; 1 parsec = \(3.08\times10^{16}\) m = 3.26 light years.
Energy and other everyday conversions
- 1 kWh = 1000 W × 3600 s = \(3.6\times10^{6}\) J
- 1 eV = \(1.6\times10^{-19}\) J
- 1 erg = \(10^{-7}\) J; 1 dyne = \(10^{-5}\) N
- 1 calorie ≈ 4.2 J; 1 atm ≈ \(1.013\times10^{5}\) Pa
- 1 amu ≈ \(1.66\times10^{-27}\) kg
Write symbols correctly: 5 kg, not 5 kgs or 5 Kg. A unit named after a person is lower case in words (newton) and capital in symbol (N).
- "Which of the following is not a unit of length / energy / an SI base unit?"
- "1 kWh equals … J" or "… eV".
- Match-the-column of fermi, ångström, AU and light year with their values.
- "Parsec is a unit of …" (distance).
A light year is a unit of distance, not time. So are the parsec and the astronomical unit.
The kilowatt-hour is a unit of energy, while the kilowatt is a unit of power. Electricity bills charge for energy.
Check the idea
Measuring large and small distances: parallax
Very large distances cannot be measured with a tape. Instead, view the object from two points a known distance \(b\) apart. The lines of sight meet at the object at a small angle \(\theta\), the parallax angle. Then the distance is \(D=b/\theta\).
For a small parallax angle, divide the baseline by the angle in radians to estimate distance. For apparent size, multiply distance by angular size in radians. A degree value must be converted before substitution.
Other useful formulas
The angles must be in radians. Convert first: \(1^\circ=\pi/180\) rad, \(1^{\prime}=2.91\times10^{-4}\) rad and \(1^{\prime\prime}=4.85\times10^{-6}\) rad.
The Moon is viewed from two points on opposite sides of the Earth (baseline = Earth's diameter, \(1.276\times10^{7}\) m). The parallax angle is 1°54'. Find the Moon's distance.
Show solution
\(\theta=1.9^\circ=1.9\times\pi/180=3.32\times10^{-2}\) rad. \(D=b/\theta=1.276\times10^{7}/3.32\times10^{-2}\approx3.85\times10^{8}\) m.
The Sun's angular diameter is 1920″ and its distance is \(1.496\times10^{11}\) m. Find its diameter.
Show solution
- \(\alpha=1920\times4.85\times10^{-6}=9.31\times10^{-3}\) rad.
- \(d=\alpha D=9.31\times10^{-3}\times1.496\times10^{11}\).
- \(d\approx1.39\times10^{9}\) m.
Answer: About \(1.39\times10^{9}\) m
More solved problem patterns
A RADAR pulse sent to the Moon returns after 2.56 s. Find the Moon's distance. Take \(c=3\times10^{8}\) m/s.
Show solution
- The pulse travels to the Moon and back, so \(2D=ct\).
- \(D=\dfrac{3\times10^{8}\times2.56}{2}=3.84\times10^{8}\) m.
Answer: \(3.84\times10^{8}\) m
1 cm³ of oleic acid is dissolved in alcohol to make 20 cm³. Then 1 cm³ of this is diluted to 20 cm³. One drop (1/20 cm³) of the final solution spreads into a film of area 1250 cm². Estimate the molecular size.
Show solution
- Concentration of oleic acid: \(\tfrac1{20}\times\tfrac1{20}=\tfrac1{400}\).
- Oleic acid in one drop: \(\tfrac1{20}\times\tfrac1{400}=1.25\times10^{-4}\) cm³.
- Thickness \(=V/A=1.25\times10^{-4}/1250=1\times10^{-7}\) cm \(=1\times10^{-9}\) m.
Answer: About \(10^{-9}\) m
Go deeper · extra explanation and derivations
Parallax and the parsec
Hold a pencil at arm's length and close each eye in turn. The pencil appears to jump against the background. The jump is parallax, and it is smaller for farther objects. Astronomers use two positions of the Earth, or two observatories, as the two "eyes".
A parsec is the distance at which 1 AU subtends 1 arc second: \(D=\dfrac{1.496\times10^{11}}{4.85\times10^{-6}}\approx3.08\times10^{16}\) m. A star with parallax \(p\) arc seconds (baseline 1 AU) is at \(1/p\) parsec.
Echo methods and small lengths
RADAR and SONAR: a pulse goes to the object and comes back. If the round trip takes \(t\), the distance is \(D=vt/2\). The factor 2 accounts for the return journey.
Size of a molecule (oleic acid film): a known tiny volume \(V\) of oleic acid spreads on water into a film one molecule thick, of area \(A\). The thickness is \(t=V/A\), about \(10^{-9}\) m.
Lengths in nature span from about \(10^{-15}\) m (a nucleus) to \(10^{26}\) m (the observable universe).
Order of magnitude
Write the number as \(a\times10^{b}\) with \(0.5\lt a\le5\). Then \(b\) is its order of magnitude. If \(a\) is above 5, round up to \(10^{b+1}\). So \(3\times10^8\) m/s has order 8, while the Earth's mass, \(5.97\times10^{24}\) kg, has order 25.
- "The Sun's angular diameter is 1920″ and its distance … Find its diameter."
- "A RADAR / SONAR pulse returns after t seconds. Find the distance."
- "1 parsec equals …" and "parallax of 0.5″ means …".
- "Order of magnitude of …"
In echo problems the measured time is for the round trip. Forgetting to halve it doubles the distance.
Parallax angle and distance are inversely related. A smaller parallax means a farther star.
Check the idea
Dimensions and dimensional consistency
Dimensions describe the type of a quantity rather than its chosen unit. Both metres and centimetres measure length and have dimension L. Every term added in a physical equation must have the same dimensions.
A dimensionally incorrect equation is impossible. A dimensionally correct one can still have a wrong numerical coefficient or omit physical effects. Angles, ratios of like quantities and arguments of logarithms are dimensionless.
Other useful formulas
Dimensional analysis cannot distinguish a factor of 1/2 from 2 or determine a sine dependence.
Find the dimensions of pressure.
Show solution
Pressure is force/area, so \([p]=MLT^{-2}/L^2=ML^{-1}T^{-2}\). Its SI unit is Pa = N/m².
In \(x=at+bt^2\), \(x\) is distance and \(t\) is time. Find the dimensions of \(a\), \(b\) and \(a/b\).
Show solution
- Each term must be a length. \([at]=L\Rightarrow[a]=LT^{-1}\).
- \([bt^2]=L\Rightarrow[b]=LT^{-2}\).
- \([a/b]=\dfrac{LT^{-1}}{LT^{-2}}=T\).
Answer: \([a]=LT^{-1}\), \([b]=LT^{-2}\), \([a/b]=T\)
More solved problem patterns
A wave is \(y=A\sin(\omega t-kx)\). Find the dimensions of \(\omega/k\) and name the quantity.
Show solution
- The argument of sine is dimensionless.
- So \([\omega]=T^{-1}\) and \([k]=L^{-1}\).
- \([\omega/k]=LT^{-1}\). This is the wave speed.
Answer: \(LT^{-1}\), the wave speed
Statement I: \(s=ut+\tfrac14at^2\) is dimensionally correct. Statement II: It is the correct equation of motion. Judge both.
Show solution
- Each term has dimension L, so statement I is true.
- The correct coefficient is \(\tfrac12\), so statement II is false.
- Dimensional checks cannot catch a wrong number.
Answer: I true, II false
Go deeper · extra explanation and derivations
Dimensions you must know by heart
| Quantity | Dimensions | Quantity | Dimensions |
|---|---|---|---|
| Velocity | \(LT^{-1}\) | Acceleration | \(LT^{-2}\) |
| Force | \(MLT^{-2}\) | Momentum, impulse | \(MLT^{-1}\) |
| Work, energy, torque | \(ML^2T^{-2}\) | Power | \(ML^2T^{-3}\) |
| Pressure, stress, modulus, energy density | \(ML^{-1}T^{-2}\) | Surface tension, spring constant | \(MT^{-2}\) |
| Angular momentum, Planck constant | \(ML^2T^{-1}\) | Viscosity \(\eta\) | \(ML^{-1}T^{-1}\) |
| Gravitational constant G | \(M^{-1}L^3T^{-2}\) | Frequency, angular velocity | \(T^{-1}\) |
| Specific heat | \(L^2T^{-2}K^{-1}\) | Latent heat | \(L^2T^{-2}\) |
| Thermal conductivity | \(MLT^{-3}K^{-1}\) | Strain, angle, refractive index | dimensionless |
Homogeneity and the arguments of functions
Only like quantities can be added, subtracted or equated. So every term in a correct equation has the same dimensions. Arguments of \(\sin\), \(\cos\), \(e^x\) and \(\log\) must be dimensionless.
- In \(y=A\sin(\omega t-kx)\): \([\omega]=T^{-1}\), \([k]=L^{-1}\), so \(\omega/k\) has dimensions \(LT^{-1}\), a speed.
- In van der Waals' equation \(\left(P+\dfrac{a}{V^2}\right)(V-b)=RT\): \(b\) is a volume (\(L^3\)) and \(a/V^2\) is a pressure, so \([a]=ML^{-1}T^{-2}\times L^6=ML^5T^{-2}\).
Uses and limits of dimensional analysis
Uses: checking equations, converting units between systems, and finding how a quantity depends on others.
Limits:
- Pure numbers such as \(\tfrac12\) or \(2\pi\) cannot be found.
- Relations with \(\sin\), \(\cos\), \(e^x\) or \(\log\) cannot be derived.
- Equations with added terms, like \(s=ut+\tfrac12at^2\), cannot be derived (only checked).
- In mechanics only three equations (M, L, T) exist, so at most three unknown powers can be found.
- Quantities with the same dimensions (work and torque) cannot be told apart.
- "Which pair has the same dimensions?" (energy density and pressure, Planck constant and angular momentum, work and torque).
- "In \(F=a\sqrt x+bt^2\), find the dimensions of a/b."
- "Dimensions of \(a\) and \(b\) in van der Waals' equation."
- Statement questions on what dimensional analysis can and cannot do.
Same dimensions do not mean the same quantity. Work and torque are both \(ML^2T^{-2}\); torque is written in N m, never in joules.
Dimensionless is not the same as unitless. An angle is dimensionless but has a unit, the radian. Strain is dimensionless and unitless.
Check the idea
Converting between unit systems: n₁u₁ = n₂u₂
A physical quantity does not change when you change units. Only its number changes: \(n_1u_1=n_2u_2\). A bigger unit gives a smaller number.
The physical quantity stays the same when its unit changes. Convert each base unit with the appropriate power: area needs a squared length conversion and density contains both mass and cubed length conversions.
Other useful formulas
The ratios are (old unit / new unit). Inverting them is the commonest mistake. Check: a bigger new unit must give a smaller number.
Convert 1 joule into ergs.
Show solution
For energy, \(a=1\), \(b=2\), \(c=-2\). \(n_2=1\times\left(\frac{1\ \text{kg}}{1\ \text{g}}\right)\left(\frac{1\ \text{m}}{1\ \text{cm}}\right)^2\left(\frac{1\ \text{s}}{1\ \text{s}}\right)^{-2}=10^3\times10^4=10^7\). So 1 J = \(10^7\) erg.
Convert \(G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}\) into CGS units.
Show solution
- \([G]=M^{-1}L^3T^{-2}\).
- \(n_2=6.67\times10^{-11}\times\left(\frac{1000\ \text{g}}{1\ \text{g}}\right)^{-1}\left(\frac{100\ \text{cm}}{1\ \text{cm}}\right)^{3}\left(\frac{1\ \text{s}}{1\ \text{s}}\right)^{-2}\).
- \(=6.67\times10^{-11}\times10^{-3}\times10^{6}=6.67\times10^{-8}\).
Answer: \(6.67\times10^{-8}\ \text{dyne cm}^2\text{g}^{-2}\)
More solved problem patterns
In a new system, the unit of mass is 10 kg, length 10 m and time 10 s. Express 100 J in the new unit of energy.
Show solution
- Energy is \(ML^2T^{-2}\). New unit = \(10\times10^2\times10^{-2}=10\) J.
- \(n_{\text{new}}=100/10=10\).
Answer: 10 new units
If energy E, velocity V and time T are taken as base quantities, find the dimensions of mass, length and momentum.
Show solution
- From \(E=\tfrac12mv^2\): \([M]=[EV^{-2}]\).
- From \(v=\)distance/time: \([L]=[VT]\).
- Momentum \(=mv\): \([EV^{-2}][V]=[EV^{-1}]\).
Answer: \([M]=EV^{-2}\), \([L]=VT\), \([p]=EV^{-1}\)
Go deeper · extra explanation and derivations
Why the number scales inversely with the unit
A length of 2 m is 200 cm. The centimetre is 100 times smaller, so the number is 100 times larger. In general \(Q=n_1u_1=n_2u_2\), so \(n_2/n_1=u_1/u_2\).
For a quantity of dimensions \(M^aL^bT^c\), \(u_1/u_2=(M_1/M_2)^a(L_1/L_2)^b(T_1/T_2)^c\). Each base unit enters with the power it has in the dimensional formula.
A new system with scaled base units
Suppose the new unit of mass is \(\alpha\) kg, the unit of length \(\beta\) m and the unit of time \(\gamma\) s. Then the new unit of a quantity \(M^aL^bT^c\) is \(\alpha^a\beta^b\gamma^c\) times its SI unit, and
\[n_{\text{new}}=\frac{n_{\text{SI}}}{\alpha^a\beta^b\gamma^c}.\]Example: if length and time units both double, \(g=9.8\ \text{m/s}^2\) (\(LT^{-2}\)) becomes \(9.8/(2\times2^{-2})=19.6\) new units.
Choosing different base quantities
Some questions make, say, force F, velocity V and time T the base quantities. Then express mass using a known formula: \(F=ma=m\,V/T\), so \([M]=[FV^{-1}T]\). Length is \([VT]\). Any other quantity follows by substitution. If energy E, velocity V and time T are base: \(E=\tfrac12mv^2\) gives \([M]=[EV^{-2}]\).
- "Convert G = \(6.67\times10^{-11}\) N m² kg⁻² into CGS units."
- "In a new system the unit of mass is α kg, length β m, time γ s. 1 J equals …"
- "If force, velocity and time are fundamental quantities, the dimensions of mass are …"
- "The value of g is 9.8 m/s². What is it if the units of length and time are doubled?"
If both length and time units double, acceleration in new units is larger, not smaller: \(n_2=n_1\times2^{-1}\times2^{2}=2n_1\). Check the sign of each exponent.
Changing the unit never changes the dimensions of a quantity, only its numerical value.
Check the idea
Constructing a dependence using dimensions
If a quantity depends on a few known variables, dimensional powers can constrain its form. Guess a product of powers, equate exponents of each base dimension, and solve the resulting equations.
This method works when the list of relevant variables is complete and there is no unresolved dimensionless dependence. The coefficient must be obtained from theory or experiment. For a pendulum, dimensions identify the length and gravity scaling but cannot give 2π.
Other useful formulas
A dimensionless input such as the swing angle can affect C; dimensional analysis alone cannot establish amplitude independence.
Suppose wave speed depends only on string tension F and linear density μ. Find the dependence.
Show solution
Let v ∝ Fᵃμᵇ. Matching mass gives a+b=0; time gives −2a=−1, so a=1/2 and b=−1/2. Hence v ∝ √(F/μ).
The drag force on a small sphere depends on viscosity \(\eta\), radius \(r\) and speed \(v\). Find the relation.
Show solution
- Let \(F=k\eta^ar^bv^c\). Dimensions: \(MLT^{-2}=(ML^{-1}T^{-1})^a L^b(LT^{-1})^c\).
- M: \(a=1\). T: \(-a-c=-2\Rightarrow c=1\).
- L: \(-a+b+c=1\Rightarrow b=1\).
- So \(F=k\eta rv\). Theory gives \(k=6\pi\).
Answer: \(F\propto\eta rv\)
More solved problem patterns
Find the combination of \(h\), \(c\) and \(G\) that has the dimensions of length.
Show solution
- \([h]=ML^2T^{-1}\), \([c]=LT^{-1}\), \([G]=M^{-1}L^3T^{-2}\). Let \(\ell=h^ac^bG^d\).
- M: \(a-d=0\). L: \(2a+b+3d=1\). T: \(-a-b-2d=0\).
- With \(d=a\): \(b=-3a\) and \(5a+b=1\Rightarrow2a=1\Rightarrow a=\tfrac12\).
- So \(a=d=\tfrac12\), \(b=-\tfrac32\).
Answer: \(\ell\propto\sqrt{hG/c^3}\)
Why can dimensional analysis check \(s=ut+\tfrac12at^2\) but not derive it?
Show solution
- Checking only needs every term to have dimension L, which they do.
- Deriving assumes a single product of powers. Here the answer is a sum of two different products.
- The method also cannot give the factor \(\tfrac12\).
Answer: The equation is a sum of terms with a numerical coefficient.
Go deeper · extra explanation and derivations
The method step by step
- List every quantity the result may depend on. Leave out none and add none.
- Write \(Q=k\,a^xb^yc^z\) with \(k\) a dimensionless constant.
- Write the dimensions of each side and equate the powers of M, L and T.
- Solve the three equations for \(x\), \(y\), \(z\).
- Get \(k\) from theory or experiment.
Results NEET expects you to derive
- Stokes' drag \(F\propto\eta rv\) (viscosity, radius, speed).
- Centripetal force \(F\propto mv^2/r\).
- Mass on a spring \(T\propto\sqrt{m/k}\).
- Orbital period \(T\propto r^{3/2}(GM)^{-1/2}\).
- Energy of an oscillator \(E\propto m\omega^2A^2\).
- Planck length \(\ell\propto\sqrt{hG/c^3}\).
- "The time period of a planet depends on G, M and r. Find the relation."
- "Velocity of ripples depends on surface tension, density and wavelength."
- "Using h, c and G as base quantities, find the dimension of length / time / mass."
- "Which relation cannot be derived by dimensional analysis?"
With four unknown powers and only M, L, T, the method fails. You need an extra physical argument, or the problem is badly posed.
Check the idea
Significant figures and rounding
Significant figures indicate the precision represented by a measured number. Leading zeros only locate the decimal point; zeros between nonzero digits are significant. Trailing zeros after a decimal point are significant.
For addition or subtraction, round to the least precise decimal place. For multiplication or division, round to the smallest number of significant figures among measured inputs. Keep extra digits during intermediate steps and round the final result once.
Writing 1500 without context is ambiguous; scientific notation makes intended precision explicit.
Evaluate 12.11 m + 0.3 m with appropriate precision.
Show solution
The unrounded sum is 12.41 m. The second measurement is only precise to tenths, so report 12.4 m, not 12.41 m.
A cube has edge \(1.2\times10^{-2}\) m. Find its volume and total surface area with correct significant figures.
Show solution
- Volume \(a^3=(1.2\times10^{-2})^3=1.728\times10^{-6}\) m³.
- Surface area \(6a^2=6\times1.44\times10^{-4}=8.64\times10^{-4}\) m². The 6 is exact.
- The edge has 2 significant figures, so round both to 2.
Answer: \(1.7\times10^{-6}\) m³ and \(8.6\times10^{-4}\) m²
More solved problem patterns
Round 2.745 and 2.735 to three significant figures.
Show solution
- In both, the dropped digit is exactly 5.
- 2.745: the kept digit 4 is even, so it stays: 2.74.
- 2.735: the kept digit 3 is odd, so it rises: 2.74.
Answer: 2.74 and 2.74
Evaluate \(\dfrac{4.0\times3.267}{1.20}\) to the correct significant figures.
Show solution
- Unrounded: \(13.068/1.20=10.89\).
- Inputs have 2, 4 and 3 significant figures. The least is 2.
- Round to 2 significant figures: 11.
Answer: 11
Go deeper · extra explanation and derivations
Counting rules in one place
- All non-zero digits are significant.
- Zeros between non-zero digits are significant (2.005 has 4).
- Leading zeros are never significant (0.0032 has 2).
- Trailing zeros after a decimal point are significant (3.200 has 4).
- Trailing zeros in a whole number are ambiguous (4700 m). Scientific notation removes the doubt: \(4.7\times10^3\) has 2, \(4.700\times10^3\) has 4.
- A change of unit does not change the count: 2.308 cm = 23.08 mm = 0.02308 m, all four figures.
- Exact numbers (counts, the 2 in \(2\pi r\), defined conversions) have unlimited significant figures.
Rounding off (NCERT rule)
- Dropped digit more than 5: raise the last kept digit (2.746 → 2.75).
- Dropped digit less than 5: keep it (2.743 → 2.74).
- Dropped digit exactly 5: leave the kept digit unchanged if it is even, raise it if it is odd. So 2.745 → 2.74 and 2.735 → 2.74.
Keep one extra digit in intermediate steps. Round only the final answer.
Combining measurements
Multiplication and division: the answer keeps as many significant figures as the least precise input. Addition and subtraction: the answer keeps as many decimal places as the input with the fewest decimal places.
Example: a box of mass 2.3 kg gets two gold pieces of 20.15 g and 20.17 g. Total = 2.3403 kg, reported as 2.3 kg. The difference between the gold pieces is 0.02 g, reported as 0.02 g.
- "Number of significant figures in 0.06900 / 6.320 × 10⁴ / 4700."
- "Area of a plate of 2.0 cm × 3.15 cm to correct significant figures."
- "Round 2.745 / 2.735 to three significant figures."
- Mixed operations: sum then product, or a cube's volume and surface area.
Converting 2.30 kg to grams as "2300 g" hides the precision. Write \(2.30\times10^3\) g to keep three significant figures.
For addition, do not count significant figures. 12.11 + 0.3 is limited by decimal places, giving 12.4.
Check the idea
Accuracy, precision and error estimates
Accuracy means closeness to the accepted value; precision means repeatability or narrow spread. Repeated measurements can reduce random scatter but do not automatically remove a calibration bias. A mean gives a useful best estimate when random errors dominate.
For repeated readings, the mean absolute deviation is one elementary error estimate. Instrument resolution and known systematic errors must also be considered. Report an absolute uncertainty in the same unit as the quantity; relative uncertainty is a ratio.
Other useful formulas
Closely clustered readings can all be wrong if the instrument has a zero error.
Readings are 9.8, 10.0 and 10.2 cm. Find mean and mean absolute deviation.
Show solution
The mean is 10.0 cm. Deviations are 0.2, 0 and 0.2 cm; their average is 0.133… cm, approximately 0.13 cm before choosing reporting precision.
The period of a pendulum is read as 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s. Find the mean, the mean absolute error and the percentage error.
Show solution
- Mean: \((2.63+2.56+2.42+2.71+2.80)/5=2.624\approx2.62\) s.
- Absolute errors from 2.624: 0.006, 0.064, 0.204, 0.086, 0.176 s.
- Mean absolute error: \(0.536/5=0.107\approx0.11\) s.
- Percentage error: \(0.11/2.62\times100\approx4\%\).
Answer: \(T=2.62\pm0.11\) s, about 4%
More solved problem patterns
The true length of a rod is 10.00 cm. Student A reads 9.62, 9.61, 9.63 cm. Student B reads 9.8, 10.3, 10.1 cm. Compare accuracy and precision.
Show solution
- A's readings agree within 0.01 cm, so A is precise. Their mean (9.62 cm) is 0.38 cm off, so A is not accurate. A systematic error is likely.
- B's readings spread over 0.5 cm, so B is not precise. Their mean (10.07 cm) is close to the true value, so B is fairly accurate.
Answer: A is precise but not accurate; B is accurate but not precise.
Go deeper · extra explanation and derivations
Kinds of error
- Systematic errors push every reading the same way. They come from the instrument (zero error, faulty calibration), the method (ignoring heat loss) or the observer (always reading from one side, parallax). Repeating does not remove them. They need a correction.
- Random errors scatter readings above and below the true value. Averaging many readings reduces them.
- Least count error comes from the resolution of the instrument. A finer instrument reduces it.
Absolute, mean, relative and percentage error
- Mean value: \(\bar a=\dfrac{1}{n}\sum a_i\), taken as the best estimate.
- Absolute error of each reading: \(\Delta a_i=|a_i-\bar a|\).
- Mean absolute error: \(\overline{\Delta a}=\dfrac1n\sum|\Delta a_i|\).
- Relative error: \(\overline{\Delta a}/\bar a\); percentage error: \(\times100\%\).
- Report as \(\bar a\pm\overline{\Delta a}\).
Accuracy is not precision
Accuracy is how close the result is to the true value. Precision is how close repeated readings are to each other, and how fine the instrument's resolution is. A precise instrument with a zero error gives readings that agree with one another but are all wrong.
- "Find the mean value, mean absolute error and percentage error" from a list of readings.
- "Zero error of an instrument is a … error."
- "Which can be reduced by taking more readings?" (random error).
- Statement questions on accuracy versus precision.
Averaging many readings cannot remove a systematic error. Ten readings from a ruler with a worn end are all short by the same amount.
Personal error, such as always reading a scale at an angle, is a systematic error, not a random one.
Check the idea
Propagation of measurement uncertainty
A calculated quantity inherits uncertainties from its measured inputs. For a worst-case estimate, absolute uncertainties add in sums, while fractional uncertainties add with the absolute values of the powers in products and quotients.
These are first-order bounds for small uncertainties, not a claim that independent random errors always add linearly. A squared radius therefore contributes twice its fractional uncertainty to area; this makes radius measurements especially important.
Other useful formulas
Why a radius error matters twice
Compare a circle at the measured radius with its smallest and largest possible radii. Relate the exact area spread to the first-order estimate.
The equation walkthrough below explains this experiment. Interactive controls load when JavaScript is available.
Predict before you change an input
A 2% radius uncertainty gives approximately…
The exponent of radius is 2. For the exact upper bound, (1.02)² − 1 = 4.04%, close to the first-order 4%.
A quantity is calculated from measured lengths or masses.
- Write its dependence as a product of powers.
- Multiply each percentage uncertainty by the absolute power.
- Add those contributions for the small-error worst-case bound.
Why the equation works · optional
- 1
State the geometry
\[ A=\pi r^2 \]Area contains the radius twice. The units are squared too: cm² here.
- 2
Differentiate the dependence
\[ dA=2\pi r\,dr \]A small increase in radius adds a thin ring. Its area is approximately circumference times thickness.
- 3
Divide by the original area
\[ \frac{\Delta A}{A}\approx2\frac{\Delta r}{r} \]The first-order fractional area uncertainty is twice the fractional radius uncertainty. Exact upper and lower bounds are slightly asymmetric.
Model and symbols: A = πr²; nominal r = 10 cm. Slider gives a symmetric radius bound, not a confidence interval. The linear approximation is most accurate for small fractional errors.
Uncertainties do not cancel merely because the measured quantities are subtracted.
For density m/(πr²ℓ), errors in m, r and ℓ are 1%, 2% and 1%. Estimate the maximum percentage error.
Show solution
Add the weighted fractional uncertainties: 1+2×2+1=6%. π is an exact mathematical constant and adds no measurement uncertainty.
Two resistors, \(R_1=100\pm3\ \Omega\) and \(R_2=200\pm4\ \Omega\), are joined (a) in series and (b) in parallel. Find the combined resistance with its error.
Show solution
- Series: \(R=300\ \Omega\), \(\Delta R=3+4=7\ \Omega\).
- Parallel: \(R=\dfrac{100\times200}{300}=66.7\ \Omega\).
- \(\Delta R=R^2\left(\dfrac{3}{100^2}+\dfrac{4}{200^2}\right)=4444\times(3\times10^{-4}+1\times10^{-4})=1.8\ \Omega\).
Answer: (a) \(300\pm7\ \Omega\) (b) \(66.7\pm1.8\ \Omega\)
More solved problem patterns
In a pendulum experiment, \(\ell=100.0\pm0.1\) cm and \(T=2.00\pm0.01\) s. Find the percentage error in \(g=4\pi^2\ell/T^2\).
Show solution
- \(\Delta\ell/\ell=0.1/100.0=0.1\%\).
- \(\Delta T/T=0.01/2.00=0.5\%\), counted twice: 1.0%.
- Total: \(0.1+1.0=1.1\%\).
Answer: 1.1%
Temperatures are \(\theta_1=20.0\pm0.5\)°C and \(\theta_2=50.0\pm0.5\)°C. Find the temperature rise and its percentage error.
Show solution
- Rise: \(50.0-20.0=30.0\)°C.
- Error: \(0.5+0.5=1.0\)°C. Errors add in a difference.
- Percentage: \(1.0/30.0\times100\approx3.3\%\).
Answer: \(30\pm1\)°C, about 3.3%
Go deeper · extra explanation and derivations
Where the rules come from
Sum or difference. If \(Z=A\pm B\), the worst case is \((A\pm\Delta A)\pm(B\pm\Delta B)\), so \(\Delta Z=\Delta A+\Delta B\). Errors always add.
Product. If \(Z=AB\), then \((A\pm\Delta A)(B\pm\Delta B)\approx AB\pm(B\Delta A+A\Delta B)\), dropping the tiny \(\Delta A\Delta B\). Dividing by \(Z=AB\): \(\Delta Z/Z=\Delta A/A+\Delta B/B\). Quotients give the same result.
Power. \(Z=A^n\) is a product of \(n\) copies of \(A\), so \(\Delta Z/Z=n\,\Delta A/A\). This holds for fractional and negative \(n\) using \(|n|\).
Which measurement deserves the most care
The quantity with the largest power dominates. In \(g=4\pi^2\ell/T^2\), the error in T counts twice. In the volume of a sphere, the radius error counts three times. So time the pendulum over many oscillations and measure the radius with the finest instrument.
Differences of nearly equal numbers
Subtraction keeps absolute errors but shrinks the value, so the percentage error can blow up. Temperatures 20.0 ± 0.5 °C and 50.0 ± 0.5 °C differ by 30 ± 1 °C (3.3%). If they were 49.0 and 50.0 °C, the difference 1 ± 1 °C would be 100% uncertain.
- "\(P=a^3b^2/(cd)\). Errors in a, b, c, d are 1%, 2%, 3%, 4%. Percentage error in P?"
- "Error in kinetic energy if mass and speed have errors …"
- "Two resistors with errors are joined in series / parallel. Find the error."
- "Which quantity should be measured most accurately?" (the one with the highest power).
In \(Z=A-B\), the error is \(\Delta A+\Delta B\), never \(\Delta A-\Delta B\).
A power in the denominator still adds its error. In \(\rho=m/V\), the error in V adds to the error in m.
Check the idea
Vernier calipers, screw gauge and zero correction
A fine scale subdivides what the main scale cannot resolve. For a direct vernier, the least count is the difference between one main-scale division and one vernier-scale division. A screw gauge converts rotation into a small linear motion.
Combine the main reading with the coinciding vernier or circular reading. Then subtract the zero error: an instrument that reads positive when it should read zero is reading too high. Check calibration before and after an experiment.
Other useful formulas
Read a vernier without guessing
Move the jaw, identify the main-scale reading before the vernier zero, then find the line that coincides.
The equation walkthrough below explains this experiment. Interactive controls load when JavaScript is available.
Predict before you change an input
If a caliper reads +0.2 mm when closed, correct a measured reading by…
Positive zero error makes every observed length too large, so the zero correction is negative.
You are given a caliper diagram or scale readings.
- Read the main mark just before the vernier zero.
- Add coincident division × least count.
- Subtract the zero error and report the correct precision.
Why the equation works · optional
- 1
Find the least count
\[ \mathrm{LC}=1\ \mathrm{MSD}-1\ \mathrm{VSD}=1-0.9=0.1\ \mathrm{mm} \]The small mismatch between the scales makes one line coincide for each tenth of a millimetre.
- 2
Read the complete millimetres
\[ \mathrm{MSR}=\lfloor L\rfloor\ \mathrm{mm} \]Use the main-scale mark just before the vernier zero, rather than the closest mark.
- 3
Add the coincidence
\[ L=\mathrm{MSR}+n\,\mathrm{LC} \]n is the coincident vernier division, counted from zero. A positive zero error would be subtracted from this observed reading.
Model and symbols: Ideal direct vernier: 10 vernier divisions span 9 mm; 1 main division = 1 mm; least count = 0.1 mm; zero error is absent.
Zero correction has the opposite sign to zero error. A negative zero error requires adding its magnitude.
A screw gauge has pitch 0.5 mm and 50 divisions. Observed diameter is 2.36 mm and zero error is +0.02 mm. Find corrected diameter.
Show solution
The least count is 0.5/50=0.01 mm. Corrected diameter is 2.36−0.02=2.34 mm.
A vernier has 20 divisions that equal 19 main-scale divisions, and 1 MSD = 0.5 mm. Find its least count.
Show solution
- \(1\ \text{VSD}=\tfrac{19}{20}\times0.5=0.475\) mm.
- \(LC=0.5-0.475=0.025\) mm.
- Check with the shortcut: \(LC=\text{MSD}/N=0.5/20=0.025\) mm.
Answer: 0.025 mm
More solved problem patterns
Read the vernier in the figure above. The main scale is in cm with mm divisions, and 10 vernier divisions equal 9 mm.
Show solution
- LC = 1 mm/10 = 0.1 mm = 0.01 cm.
- The vernier zero is just past the 2.3 cm mark, so MSR = 2.3 cm.
- The 7th vernier mark lines up, so add \(7\times0.01=0.07\) cm.
Answer: 2.37 cm
A screw gauge has pitch 1 mm and 100 circular divisions. The sleeve shows 3 mm and division 47 is on the reference line. There is no zero error. Find the reading.
Show solution
- LC = 1 mm/100 = 0.01 mm.
- Reading = \(3+47\times0.01=3.47\) mm.
Answer: 3.47 mm
Go deeper · extra explanation and derivations
Why a vernier works
Suppose \(N\) vernier divisions span \((N-1)\) main-scale divisions. Then
\[1\ \text{VSD}=\frac{N-1}{N}\ \text{MSD},\qquad LC=1\ \text{MSD}-1\ \text{VSD}=\frac{1\ \text{MSD}}{N}.\]With 10 vernier divisions over 9 mm, LC = 0.1 mm = 0.01 cm. Each vernier mark is 0.1 mm shorter than a main mark, so the mark that lines up tells you how many tenths of a millimetre to add.
Reading: main-scale reading (the last main mark before the vernier zero) + (number of the coinciding vernier mark × LC).
Screw gauge: pitch and least count
One full turn of the thimble moves the spindle by one pitch. With \(N\) divisions on the circular scale, LC = pitch/N. A pitch of 0.5 mm and 50 divisions give LC = 0.01 mm.
Reading: linear-scale reading (last visible mark on the sleeve) + (circular division on the reference line × LC).
Backlash error: a worn screw can turn a little without moving the spindle. Always turn in one direction while measuring. The ratchet stops you from over-tightening.
- "10 VSD = 9 MSD and 1 MSD = 1 mm. Least count?"
- "Pitch 0.5 mm, 50 circular divisions, linear reading … circular reading …. Find the diameter."
- Figure-reading questions with a vernier or screw gauge.
- "Backlash error is reduced by …"
The main-scale reading is the main mark just before the vernier zero, not the main mark that coincides with a vernier mark.
Least count of a screw gauge uses the pitch, not 1 mm. If the pitch is 0.5 mm and there are 50 divisions, LC = 0.01 mm, not 0.02 mm.
Check the idea
Zero error and instrument reading practice
An instrument has a zero error when it does not read zero with its jaws or studs closed. Find that error first, then correct every reading: true value = observed reading − zero error.
First calculate the instrument’s observed reading. Then subtract the signed zero error. A positive zero error makes readings too high; a negative one means the correction is added.
Other useful formulas
For a negative zero error, the coinciding mark \(n\) is counted from the vernier zero, but the error is \((N-n)\) divisions, not \(n\).
A vernier (LC = 0.01 cm, 10 divisions) has its zero to the left of the main zero when the jaws are closed, with the 6th vernier mark coinciding. A rod reads 2.53 cm. Find its true length.
Show solution
Zero error \(e=-(10-6)\times0.01=-0.04\) cm. True length \(=2.53-(-0.04)=2.57\) cm.
A vernier (LC = 0.1 mm) has a positive zero error with the 4th mark coinciding. A rod gives MSR = 3.2 cm and the 7th mark coinciding. Find the true length.
Show solution
- Zero error \(e=+4\times0.01=+0.04\) cm.
- Observed reading \(=3.2+7\times0.01=3.27\) cm.
- True length \(=3.27-0.04=3.23\) cm.
Answer: 3.23 cm
More solved problem patterns
A screw gauge has pitch 0.5 mm and 50 divisions. With the studs closed, the circular zero is above the reference line and division 46 is on the line. A wire gives a linear reading of 2.5 mm and circular reading 12. Find the diameter.
Show solution
- LC = 0.5/50 = 0.01 mm.
- Zero error \(=-(50-46)\times0.01=-0.04\) mm.
- Observed \(=2.5+12\times0.01=2.62\) mm.
- True \(=2.62-(-0.04)=2.66\) mm.
Answer: 2.66 mm
Read the screw gauge in the figure. Pitch 0.5 mm, 50 divisions, no zero error.
Show solution
- Whole millimetres visible on the sleeve: 4 mm. The half-millimetre mark after it is also visible, so the linear reading is 4.5 mm.
- Division 28 is on the reference line: \(28\times0.01=0.28\) mm.
- Total \(=4.5+0.28=4.78\) mm.
Answer: 4.78 mm
Go deeper · extra explanation and derivations
Positive and negative zero error on a vernier
- Positive: jaws closed, vernier zero to the right of the main zero. The instrument already reads a little. If the 4th vernier mark coincides (LC 0.1 mm), \(e=+0.4\) mm.
- Negative: vernier zero to the left of the main zero. If the 6th mark coincides on a 10-division vernier, the vernier zero is \((10-6)\times0.1=0.4\) mm short of zero, so \(e=-0.4\) mm.
Zero correction is minus the zero error. Add the correction, or subtract the error. Both give the same true value.
Zero error on a screw gauge
Close the studs gently with the ratchet. Look at the circular scale.
- If its zero has passed the reference line, so that division \(n\) sits on the line, the gauge reads \(+n\times LC\) with nothing between the studs. This is a positive error. On a standard gauge, the circular zero then lies below the reference line.
- If its zero has not reached the line, so that division \(n\) near the end of the scale sits on it, the error is \(-(N-n)\times LC\). The circular zero lies above the line.
A full reading routine
- Find the least count.
- Close the jaws or studs and find the signed zero error.
- Take the main or linear scale reading.
- Add the vernier or circular division × LC.
- Subtract the zero error.
- Repeat at different positions and take the mean.
- "Jaws closed: vernier zero to the right / left and the nth mark coincides. Find the zero error and the corrected reading."
- "Screw gauge reads … when the studs touch. Find the diameter of a wire."
- Figure-based reading of a screw gauge or vernier.
- "Zero correction for a negative zero error is …"
A negative zero error means the instrument under-reads, so the correction is added. 5.12 cm with error −0.03 cm gives 5.15 cm.
In a screw gauge with pitch 0.5 mm, check whether the half-millimetre mark below the reference line is visible. Missing it puts the answer 0.5 mm too low.
Check the idea
Explain each formula’s symbols and conditions without looking. Solve the examples before revealing their steps, then use chapter practice to check what needs revision.