Physics
Units and Measurements
Errors in Measurement

Units and Measurements · JEE & NEET Physics

Errors in Measurement: notes and previous year questions

Accuracy and precision, the three kinds of error, how big an error is, how errors combine in calculations, and least count.

112 JEE Main questions (2008–2026)
20 JEE Advanced questions (2005–2025)
15 NEET questions (2008–2026)

Errors in Measurement in short

Every measurement has an error; a good result gives the value and its error.
Accuracy is closeness to the true value; precision is closeness of readings to each other.
Systematic errors push every reading the same way; they spoil accuracy and are removed by correction or calibration.
Subtract a zero error, with its sign, from every reading.

1No measurement is perfect

Measure the same rod five times and you may read 5.2, 5.4, 5.1, 5.3 and 5.5 cm. Nothing is broken: every measurement has some uncertainty. The error of a reading is how far it is from the true value.

\text{error} = \text{measured value} - \text{true value}

Knowing the error tells us how far to trust a result, lets us quote it with the right precision, and shows where to improve the experiment. A good result always gives the value and its error.

2Accuracy and precision

Accuracy	Precision
How close a measurement is to the true value	How close repeated measurements are to each other
Spoilt by systematic errors	Spoilt by random errors
Improved by calibration and correction	Improved by finer instruments and repeated readings

3Types of errors

Systematic errors push every reading the same way, by the same amount. They can be corrected once the cause is known, and they spoil accuracy. They come from:

the instrument: a zero error in vernier callipers, poor calibration, a worn metre scale;
the surroundings: temperature changing a length, air pressure in gas experiments, humidity in electrical ones;
the observer: parallax from always reading on one side of a pointer, or a steady delay in starting a stopwatch;
the theory: a simplified formula, or ignoring air resistance.

Random errors change in size and sign from one reading to the next, from small unpredictable changes: draughts, vibrations, the limits of the eye. They cannot be removed, only reduced, and they spoil precision. The cure is to take many readings and use their mean, which is closer to the true value than a single reading.

Gross errors, or blunders, are plain mistakes: misreading a scale (5 for 6), writing down a wrong value, using the wrong formula, a calculation slip. The cure is care and checking.

4Absolute, relative and percentage error

The absolute error of a reading is the size of its difference from the true value. The true value is usually unknown, so we use the mean of the readings instead.

\bar a = \frac{a_1 + a_2 + \dots + a_n}{n}

mean of n readings

\Delta a_i = |a_i - \bar a|

absolute error of each reading

\Delta \bar a = \frac{\Delta a_1 + \dots + \Delta a_n}{n}

mean absolute error

a = \bar a \pm \Delta \bar a

the result

The relative error compares the error with the value. It has no unit. The percentage error is the relative error times 100%.

\delta a = \frac{\Delta \bar a}{\bar a}

\text{percentage error} = \frac{\Delta \bar a}{\bar a} \times 100\%

5Errors in sums and differences

When you add or subtract measured values, the absolute errors add. The worst case is that both errors push the answer the same way.

Z = A \pm B \;\Rightarrow\; \Delta Z = \Delta A + \Delta B

Z = kA\ (k\ \text{exact}) \;\Rightarrow\; \Delta Z = k\,\Delta A

6Errors in products, quotients and powers

When you multiply or divide, the relative errors add, and so do the percentage errors. Dividing does not subtract them.

Z = AB \text{ or } \frac{A}{B} \;\Rightarrow\; \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}

A power multiplies the relative error by the power: squaring doubles it, cubing triples it, a square root halves it.

Z = A^n \;\Rightarrow\; \frac{\Delta Z}{Z} = n\,\frac{\Delta A}{A}

In general, multiply each relative error by its power and add them all, whether the quantity is on the top or the bottom:

Z = \frac{A^a B^b}{C^c}

\frac{\Delta Z}{Z} = a\frac{\Delta A}{A} + b\frac{\Delta B}{B} + c\frac{\Delta C}{C}

ExampleWorked examples

Density of a cube, $\rho = m/l^3$ , with $m = 100 \pm 1$ g and $l = 10 \pm 0.1$ cm: $\Delta\rho/\rho = 0.01 + 3(0.01) = 0.04$ . $\rho = 0.1$ g/cm³, so $\rho = 0.100 \pm 0.004$ g/cm³.
Resistance, $R = V/I$ , with $V = 100 \pm 5$ V and $I = 10 \pm 0.2$ A: $0.05 + 0.02 = 0.07$ , so $R = 10.0 \pm 0.7\ \Omega$ .
Pendulum, $T = 2\pi\sqrt{l/g}$ , with a 2% error in $l$ only: $\Delta T/T = \tfrac12 \times 2\% = 1\%$ .
Density of a sphere, $\rho = 6m/(\pi d^3)$ , with 3% in $m$ and 1% in $d$ : $3\% + 3(1\%) = 6\%$ .
$P = a^3b^2/\sqrt{cd}$ with 1%, 2%, 3%, 4% errors: $3(1) + 2(2) + \tfrac12(3) + \tfrac12(4) = 10.5\%$ .

7Least count

The least count (LC) of an instrument is the smallest value it can measure. A single reading is written as the value ± the least count: this is the instrumental error.

Instrument	Least count
Metre scale	1 mm = 0.1 cm
Vernier callipers	0.1 mm = 0.01 cm
Screw gauge	0.01 mm = 0.001 cm
Stopwatch (dial)	0.1 s
Stopwatch (digital)	0.01 s

Vernier callipers: 10 vernier divisions cover 9 main-scale divisions of 1 mm, so one vernier division (VSD) is 0.9 mm.

\text{LC} = 1\ \text{MSD} - 1\ \text{VSD} = 0.1\ \text{mm}

Screw gauge: divide the pitch by the number of divisions on the circular scale.

\text{LC} = \frac{\text{pitch}}{\text{circular divisions}}

1 mm ÷ 100 = 0.01 mm, or 0.5 mm ÷ 50 = 0.01 mm

Summary

Key ideas

Every measurement has an error; a good result gives the value and its error.
Accuracy is closeness to the true value; precision is closeness of readings to each other.
Systematic errors push every reading the same way; they spoil accuracy and are removed by correction or calibration.
Subtract a zero error, with its sign, from every reading.
Random errors vary in size and sign; they spoil precision and are reduced by averaging many readings.
Gross errors are blunders; avoid them with care and checking.
The result of repeated readings is the mean ± the mean absolute error.
Relative error = absolute error ÷ value, with no unit; × 100 gives the percentage error.
In sums and differences, the absolute errors add. Errors never cancel.
In products and quotients, the relative errors add.
A power multiplies the relative error by the power.
Instrumental error = ± least count.

Every equation

Error: $\text{error} = \text{measured} - \text{true}$
Mean: $\bar a = (a_1 + \dots + a_n)/n$
Absolute error: $\Delta a_i = |a_i - \bar a|$
Mean absolute error: $\Delta\bar a = (\Delta a_1 + \dots + \Delta a_n)/n$
Result: $a = \bar a \pm \Delta\bar a$
Relative error: $\delta a = \Delta\bar a / \bar a$
Percentage error: $(\Delta\bar a/\bar a) \times 100\%$
Zero-error correction: $\text{true} = \text{reading} - \text{zero error}$
Sum or difference: $\Delta Z = \Delta A + \Delta B$
Exact multiple: $Z = kA \Rightarrow \Delta Z = k\,\Delta A$
Product or quotient: $\Delta Z/Z = \Delta A/A + \Delta B/B$
Power: $Z = A^n \Rightarrow \Delta Z/Z = n\,\Delta A/A$
General rule: $\Delta Z/Z = a\,\Delta A/A + b\,\Delta B/B + c\,\Delta C/C$
Error in sin θ: $\Delta(\sin\theta)/\sin\theta = \cot\theta\,\Delta\theta$
Vernier least count: $\text{LC} = 1\ \text{MSD} - 1\ \text{VSD}$
Screw gauge least count: $\text{LC} = \text{pitch}/\text{circular divisions}$

Previous year questions with solutions

Real JEE and NEET questions on errors in measurement. Try each one before you open the solution.

Q1NEET 2026One correct option

In a vernier calliper, 20 VSD coincide with 16 MSD (each division of length 1 mm ). The least count of the vernier callipers is:

A0.2 cm
B0.01 cm
C0.02 cm
D0.1 cm

Show answer and solution

Answer: Option C

Twenty vernier divisions equal sixteen main scale divisions, so $1\ \text{VSD} = \frac{16}{20}\ \text{MSD}$ and $\text{LC} = \text{MSD}\left(1 - \frac{16}{20}\right) = \frac{4}{20}\ \text{MSD}$ . With $1\ \text{MSD} = 1\ \mathrm{mm}$ that is $\frac{1}{5} = 0.2\ \mathrm{mm}$ , and $0.2\ \mathrm{mm}$ is $0.02\ \mathrm{cm}$ .

Q2JEE Main 2026One correct option

In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm . If the circular scale has 100 divisions, the least count of screw gauge is $\_\_\_\_$ mm.

A $1\times {10}^{-2}$
B $1\times {10}^{-3}$
C $5\times {10}^{-2}$
D $5\times {10}^{-3}$

Show answer and solution

Answer: Option D

Five rotations carry the spindle $2.5\ \mathrm{mm}$ , so the pitch is $2.5 \div 5 = 0.5\ \mathrm{mm}$ . Sharing that among $100$ divisions gives $0.5 \div 100 = 0.005\ \mathrm{mm}$ , which is $5 \times 10^{-3}\ \mathrm{mm}$ .

Q3JEE Advanced 2025One or more correct options

Length, breadth and thickness of a strip having a uniform cross section are measured to be 10.5 cm, 0.05 mm, and 6.0 μm, respectively. Which of the following option(s) give(s) the volume of the strip in cm³ with correct significant figures:

A $3.2\times {10}^{-5}$
B $32.0\times {10}^{-6}$
C $3.0\times {10}^{-5}$
D $3\times {10}^{-5}$

Show answer and solution

Answer: Options D

In centimetres the three measurements are $10.5$ (three significant figures), $0.05\ \mathrm{mm} = 5\times 10^{-3}\ \mathrm{cm}$ (one figure) and $6.0\ \mu\mathrm{m} = 6.0\times 10^{-4}\ \mathrm{cm}$ (two figures). Multiplying gives $10.5 \times 5\times 10^{-3} \times 6.0\times 10^{-4} = 3.15\times 10^{-5}\ \mathrm{cm^{3}}$ . A product keeps only as many significant figures as its poorest factor, which here is one, so the volume is $3\times 10^{-5}\ \mathrm{cm^{3}}$ , option D.

Practice questions, easy to hard

Three questions from the errors in measurement practice ladder: one easy, one medium, one hard.

Q4Numerical answer

Now take the same two masses, $(120 \pm 2)\ \mathrm{g}$ and $(80 \pm 3)\ \mathrm{g}$ , and subtract the second from the first.

What is the absolute error in the difference, in $\mathrm{g}$ ?

Show answer and solution

Answer: 5

The values subtract to give $40\ \mathrm{g}$ , but the errors still add: $2 + 3 = 5\ \mathrm{g}$ . The worst case is the first mass reading $2\ \mathrm{g}$ high while the second reads $3\ \mathrm{g}$ low, which pushes the difference $5\ \mathrm{g}$ too big.

Q5Numerical answer

One full measurement, start to finish. A gauge of pitch $0.5\ \mathrm{mm}$ has $50$ circular divisions, and with its studs closed the zero of the circular scale sits $3$ divisions above the reference line. A sheet between the studs shows $2$ main scale divisions with the $14$ th circular division on the line.

What is the true thickness, in $\mathrm{mm}$ ?

Show answer and solution

Answer: 1.17

The least count is $0.5 \div 50 = 0.01\ \mathrm{mm}$ , so the observed reading is $2 \times 0.5 + 14 \times 0.01 = 1.00 + 0.14 = 1.14\ \mathrm{mm}$ . Above the line means a zero error of $-0.03\ \mathrm{mm}$ , and $1.14 - (-0.03) = 1.17\ \mathrm{mm}$ .