In short

A Vernier calliper resolves lengths to 0.01 cm (0.1 mm). A screw gauge resolves lengths to 0.001 cm (0.01 mm). Together they cover everything from the diameter of a cricket ball to the thickness of a strand of hair.

The reading formula is the same for both, with one scale riding over another:

\text{Reading} = \text{(main scale reading, MSR)} + (\text{coinciding division, VSD/CSD})\times(\text{least count, LC}) - \text{zero error}.

Vernier calliper — least count = \dfrac{1\ \text{main-scale division}}{\text{number of Vernier divisions}} = \dfrac{1\ \text{mm}}{10} = 0.1\ \text{mm} = 0.01\ \text{cm}.

Screw gauge — least count = \dfrac{\text{pitch}}{\text{number of head-scale divisions}} = \dfrac{0.5\ \text{mm}}{50} = 0.01\ \text{mm} = 0.001\ \text{cm}.

Zero error — close the jaws (or the anvil and spindle) and record the reading. If the instrument does not read exactly zero, that offset is subtracted from every subsequent reading. Positive zero error is subtracted; negative zero error (more common for an old school screw gauge) is added back with its sign.

The result format — take n readings (n \ge 5, ideally rotating the object between readings), compute the mean \bar d, the mean absolute error \overline{\Delta d}, and quote

d = \bar d \pm \overline{\Delta d}\ \text{with percentage error}\ \ \frac{\overline{\Delta d}}{\bar d} \times 100\%.

The final number has as many decimal places as the instrument's least count — no more, no fewer. A Vernier reading for a cricket ball's diameter is reported as d = (7.12 \pm 0.03)\ \text{cm}, not d = 7.122\ \text{cm}. A screw gauge reading for a copper wire's diameter is d = (0.823 \pm 0.006)\ \text{mm}, not d = 0.8 mm.

These are the two instruments every Indian CBSE, ICSE, and state-board student uses in their first formal physics practical. The experiment appears in every class 11 practical exam and every JEE Advanced / NEET physics lab test. Doing it right sets the discipline for every measurement you will make later — energy, resistance, magnetic field, anything else. The idea is never about the diameter of the sphere; it is about reporting a number honestly.

A steel ball is sitting on a small wooden block on your lab bench. Next to it is a coil of thin copper wire — the kind used to wind electromagnets in the physics practical. A Vernier calliper and a screw gauge sit alongside, their scales polished, their jaws oiled. Your lab teacher says: tell me the diameter of the ball, and tell me the diameter of the wire. You must give numbers with uncertainties. You must show your working. You have twenty minutes.

The temptation is to close the jaws of the calliper on the ball, read "7.1 cm" off the scale, write it down, and move on. The actual procedure — the one that earns full marks in the practical and that is how real measurements get made in physics — is more deliberate. Open the jaws, slide them onto the ball, look at the Vernier scale, read the main scale and the coinciding division. Roll the ball a quarter-turn between your fingers. Measure again. Rotate and measure three more times. Average. Check for zero error. Subtract it. Report the result with a percentage uncertainty.

That is what this article walks through. The instruments themselves — what the main scale and the Vernier scale look like, what the pitch of the screw is, how the least count is derived — are covered in the earlier measurement instruments article. This article is about what you actually do with them in the lab: the procedure, the observation tables, the zero correction, the error analysis, and the final presentation.

Experiment 1: Diameter of a sphere with Vernier callipers

Apparatus

A Vernier calliper (the standard Indian school-issue instrument — steel, least count 0.01 cm, main scale in mm on the lower jaw, Vernier scale of 10 divisions on the sliding jaw). A small steel ball (or a marble, or any approximately spherical object). A wooden block to hold the ball while you measure. A clean cloth to wipe the jaws between readings.

Least count calculation

This is the first thing you write in the observation table, for every experiment, before any readings are taken.

\text{LC} = \frac{\text{1 main-scale division (MSD)}}{\text{number of Vernier-scale divisions (VSD)}}

For a standard Indian calliper, the main scale is graduated in millimetres (1 MSD = 0.1 cm), and there are 10 Vernier divisions on the sliding jaw. So

\text{LC} = \frac{0.1\ \text{cm}}{10} = 0.01\ \text{cm} = 0.1\ \text{mm}.

Why: the 10 Vernier divisions occupy 9 mm on the main scale (that is the whole point of the Vernier design), so each Vernier division is 0.9 mm. The difference between one Vernier division and one main-scale division is 1 - 0.9 = 0.1 mm — the least count.

Checking for zero error

Close the jaws of the calliper gently until they touch. Look at the Vernier scale:

Vernier zero-error casesThree panels showing a section of a Vernier scale closed jaws. Top panel: no zero error — the Vernier zero lines up with the main zero. Middle panel: positive zero error — the Vernier zero is slightly to the right of the main zero; the third Vernier division coincides, giving +0.03 cm zero error. Bottom panel: negative zero error — the Vernier zero is slightly to the left; the seventh Vernier division coincides, giving −0.03 cm zero error. 0 1 0 V No zero error (ideal) 0 0' 3 +0.03 cm zero error → subtract 0 0' 7 −0.03 cm zero error → add back
The three zero-error cases for a Vernier calliper. In the middle panel, the Vernier zero has shifted 0.03 cm to the right, so the instrument reads 0.03 cm too much — that is subtracted from each measurement. In the bottom panel, it has shifted 0.03 cm to the left, so the instrument reads too little — that is added back.

Record the zero error explicitly before taking measurements. "Zero error = +0.03 cm, zero correction = -0.03 cm." The distinction: zero error is what the instrument wrongly reports; zero correction is what you apply to fix it. They are equal and opposite.

Procedure — five readings, with rotation

Open the jaws wide enough to fit the sphere. Gently close them until they just touch the ball without deforming it. Read:

Record the observed reading. Then, critically: rotate the ball by about 60° in your fingers (around a different axis each time), re-clamp, and take another reading. Repeat five times. The rotation is not optional — a real sphere is never perfectly spherical, and if you always measure the same diameter, you get the same one-direction value every time. Rotating averages over the small deviations.

Observation table — a worked set of readings

S.No. MSR (cm) VSR (division) VSR × LC (cm) Observed d (cm) Corrected d = \text{obs} - \text{zero error} (cm)
1 7.1 5 0.05 7.15 7.15 − 0.03 = 7.12
2 7.1 4 0.04 7.14 7.11
3 7.1 6 0.06 7.16 7.13
4 7.1 5 0.05 7.15 7.12
5 7.1 4 0.04 7.14 7.11

Zero error = +0.03 cm, zero correction = -0.03 cm, LC = 0.01 cm.

Error analysis — mean, absolute error, percentage error

Step 1. Compute the mean.

\bar d = \frac{7.12 + 7.11 + 7.13 + 7.12 + 7.11}{5} = \frac{35.59}{5} = 7.118\ \text{cm}.

Round to the LC's precision: \bar d = 7.12 cm.

Why: the final answer cannot have more significant digits than the instrument can resolve. 7.118 is beyond the LC of 0.01 cm; rounding to 7.12 is honest. Do not quote 7.118 — that suggests precision the instrument does not have.

Step 2. Compute the absolute error of each reading (the distance from the mean). | S.No. | d_i | \Delta d_i = |d_i - \bar d| | |---|---|---| | 1 | 7.12 | 0.00 | | 2 | 7.11 | 0.01 | | 3 | 7.13 | 0.01 | | 4 | 7.12 | 0.00 | | 5 | 7.11 | 0.01 |

Step 3. Compute the mean absolute error.

\overline{\Delta d} = \frac{0.00 + 0.01 + 0.01 + 0.00 + 0.01}{5} = 0.006\ \text{cm} \approx 0.01\ \text{cm}.

Why: the mean absolute error cannot be smaller than the least count. If it comes out less than LC (as 0.006 cm does here), report it as LC — because no measurement with this instrument can be more precise than that.

Step 4. Percentage error.

\frac{\overline{\Delta d}}{\bar d} \times 100\% = \frac{0.01}{7.12} \times 100\% = 0.14\%.

Step 5. Report the final result.

d = (7.12 \pm 0.01)\ \text{cm}, \quad \text{percentage error} = 0.14\%.

Example 1: Full observation table for a school steel ball

You are given a steel ball roughly 1.5 cm in diameter and asked to measure it. Your Vernier calliper has LC = 0.01 cm and zero error -0.02 cm. Your five readings of MSR and VSR are as in the table below. Produce the full result.

S.No. MSR (cm) VSR Observed d Corrected d = \text{obs} - (-0.02)
1 1.5 4 1.54 1.56
2 1.5 3 1.53 1.55
3 1.5 5 1.55 1.57
4 1.5 4 1.54 1.56
5 1.5 3 1.53 1.55

Step 1. Mean: \bar d = (1.56 + 1.55 + 1.57 + 1.56 + 1.55)/5 = 7.79/5 = 1.558 cm \to 1.56 cm.

Why: rounding to the LC of 0.01 cm. Quoting 1.558 would imply precision the calliper does not have.

Step 2. Absolute errors: |1.56 - 1.56| = 0, |1.55 - 1.56| = 0.01, |1.57 - 1.56| = 0.01, |1.56 - 1.56| = 0, |1.55 - 1.56| = 0.01. Mean absolute error = (0 + 0.01 + 0.01 + 0 + 0.01)/5 = 0.006 cm \to 0.01 cm (bounded below by LC).

Step 3. Percentage error: (0.01 / 1.56) \times 100\% = 0.64\%.

Step 4. Final result:

\boxed{\;d = (1.56 \pm 0.01)\ \text{cm}.\;}
Reading a Vernier calliper: main scale 1.5 cm plus Vernier division 4A zoomed-in view of a Vernier calliper. The main scale is graduated in millimetres; the Vernier zero sits between the 1.5 cm mark and the 1.6 cm mark on the main scale. The Vernier scale of 10 divisions sits below. The fourth Vernier division coincides with a main-scale division, giving a reading of 1.5 plus 0.04 = 1.54 cm before zero correction. 1.5 1.6 0 1 2 3 4 5 6 7 8 9 10 The 4th Vernier division lines up with a main division → Reading = 1.5 + 4 × 0.01 = 1.54 cm (before zero correction)
A typical Vernier scale reading. The main scale shows the position is between 1.5 cm and 1.6 cm. The 4th Vernier division is the one that lines up with a main-scale graduation — so the reading is 1.5 + (4 × 0.01) = 1.54 cm. After subtracting the zero error of $-0.02$ cm (that is, adding 0.02 cm), the corrected reading is 1.56 cm.

What this shows: the result is a number with a precision matched to the instrument, not to the arithmetic. The calliper can resolve 0.01 cm, so the answer has two decimals and the uncertainty is \pm 0.01 cm — no better, no worse.

Experiment 2: Diameter of a wire with a screw gauge

Apparatus

A screw gauge (Indian school-issue, pitch 0.5 mm, 50 head-scale divisions, least count 0.01 mm). A piece of thin copper wire, 20–30 cm long. A clean handkerchief to wipe the spindle and anvil.

Least count calculation

\text{LC} = \frac{\text{pitch (distance moved by spindle per revolution)}}{\text{number of head-scale divisions}}

For a standard Indian screw gauge, the pitch is 0.5 mm and the head scale is divided into 50 parts, so

\text{LC} = \frac{0.5\ \text{mm}}{50} = 0.01\ \text{mm} = 0.001\ \text{cm}.

Why: each full turn of the thimble moves the spindle forward by one pitch (0.5 mm). One tick on the head scale is therefore 1/50 of one pitch, or 0.01 mm — ten times finer than the Vernier calliper. This is what makes the screw gauge the instrument of choice for thin objects like wires, metal sheets, and paper.

Checking for zero error

Gently rotate the thimble (using the ratchet at the end — never the thimble itself when the anvil and spindle are close, or you will damage the threads) until the spindle just touches the anvil. Read the head scale:

Screw gauge zero-error casesThree panels showing the head-scale view of a screw gauge with jaws closed. Top: no zero error — head-scale zero lines up with the main-scale reference line. Middle: positive zero error — head-scale zero is two divisions below the reference line; the 2nd division coincides, giving +0.02 mm zero error. Bottom: negative zero error — head-scale zero is three divisions above the reference line; the 47th division coincides, giving −0.03 mm zero error. reference line (main scale) 0 No zero error reference line 0 1 2 +0.02 mm zero error → subtract reference line 47 48 49 0 −0.03 mm zero error → add back
The three zero-error cases for a screw gauge. In the middle panel the zero is two divisions below the reference, so the gauge reads 0.02 mm too much. In the bottom panel the zero is three divisions above (equivalently, the 47th head-scale division coincides), so the gauge reads 0.03 mm too little.

Procedure — five readings, rotated

Hold the wire between the anvil and the spindle. Rotate the ratchet (never the thimble directly) until you hear two clicks — that is the spring-loaded ratchet slipping once the spindle has touched the wire with a standardised pressure. This is the most important screw-gauge rule: always tighten with the ratchet, never with the thimble. Using the thimble crushes the wire and damages the threads.

Read:

Move the wire along its length — measure at 5 different points, not the same spot five times. This catches non-uniformities in the wire's cross-section, and it is the screw-gauge analogue of rotating the ball in the Vernier experiment.

Observation table — copper wire diameter

S.No. MSR (mm) HSR (division) HSR × LC (mm) Observed d (mm) Corrected d (mm)
1 0.5 32 0.32 0.82 0.82 + 0.02 = 0.84
2 0.5 30 0.30 0.80 0.82
3 0.5 31 0.31 0.81 0.83
4 0.5 33 0.33 0.83 0.85
5 0.5 30 0.30 0.80 0.82

Zero error = -0.02 mm (negative), zero correction = +0.02 mm, LC = 0.01 mm.

Step 1. Mean: \bar d = (0.84 + 0.82 + 0.83 + 0.85 + 0.82)/5 = 4.16/5 = 0.832 mm.

Step 2. Absolute errors: |0.84 - 0.832| = 0.008, |0.82 - 0.832| = 0.012, |0.83 - 0.832| = 0.002, |0.85 - 0.832| = 0.018, |0.82 - 0.832| = 0.012. Mean: (0.008 + 0.012 + 0.002 + 0.018 + 0.012)/5 = 0.0104 mm \to 0.01 mm.

Step 3. Percentage error: (0.01/0.832) \times 100\% = 1.2\%.

Step 4. Report:

\boxed{\;d = (0.83 \pm 0.01)\ \text{mm}, \quad \text{percentage error} = 1.2\%.\;}

Note how the final answer rounds to 0.83 mm (two decimals in mm) and the error is 0.01 mm (one significant figure, matching the LC). Writing "0.832 ± 0.0104" is wrong — the extra digits imply precision the instrument cannot deliver.

Combining errors — when the wire's diameter enters another formula

A common follow-up experiment is to use the measured diameter to compute the volume or the cross-sectional area, and to propagate the error into the final result. If you use the screw-gauge diameter to compute the volume of a copper wire of length \ell = (15.0 \pm 0.1) cm:

V = \pi \left(\frac{d}{2}\right)^2 \ell = \frac{\pi d^2 \ell}{4}.

The fractional error in V is the sum of the fractional errors, weighted by their powers:

\frac{\Delta V}{V} = 2\,\frac{\Delta d}{d} + \frac{\Delta \ell}{\ell}.

Why: take the natural log of V = \pi d^2 \ell /4, differentiate term-by-term, and replace differentials with uncertainties. The factor of 2 on \Delta d / d comes from d^2. This is the standard error propagation rule and appears in every Indian physics practical.

Plug in: \Delta d / d = 0.01/0.83 = 1.2\%, \Delta \ell / \ell = 0.1/15.0 = 0.67\%, so

\frac{\Delta V}{V} = 2(1.2\%) + 0.67\% = 3.1\%.

The diameter's uncertainty dominates because it enters quadratically. The takeaway: when a measurement enters a formula as a square or a higher power, its fractional error is amplified — so it is worth spending extra time getting the diameter right. This is one of the reasons the screw gauge (with its smaller fractional error) is preferred over the Vernier calliper for thin objects.

Common sources of error and how to minimise them

Common confusions

If you just wanted the practical procedure and the error format, you can stop here — this is all that is required for CBSE class 11 practicals and the JEE/NEET practical papers. The rest is for readers who want the statistical theory behind the error bar and the historical instrument that made the Vernier scale possible.

Why "absolute error" and not "standard deviation"?

The school-level formula for the uncertainty is the mean absolute deviation from the mean:

\overline{\Delta d} = \frac{1}{n}\sum_{i=1}^n |d_i - \bar d|.

A statistician would instead report the standard deviation:

\sigma = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (d_i - \bar d)^2}.

For a small number of readings drawn from a normal distribution, \sigma \approx 1.25\,\overline{\Delta d}. The two quantities are proportional; the mean absolute error is just a simpler way of computing an uncertainty without needing squares and square roots. The Indian school syllabus uses the simpler form.

For a Gaussian distribution, the true uncertainty of the mean is the standard error \sigma / \sqrt n. With n = 5 readings this is about half the standard deviation — which is why taking more readings genuinely reduces your reported error. A full-precision laboratory measurement might take 20 readings and report an uncertainty a factor of 2 smaller than a student taking 5.

The Vernier principle in one sentence

The scale is made so that 10 divisions of the Vernier slide equal 9 divisions of the main scale. When the Vernier zero is slightly past a main-scale division, exactly one Vernier division (and only one) will line up perfectly with a main-scale division. That division's index, times the least count, is the fractional part of the reading. The principle is a geometric trick for reading the interpolation between main-scale marks without needing subdivisions — an idea from the 1631 paper by Pierre Vernier, and it is why every modern precision instrument (dial callipers, digital callipers, theodolites) carries an analog of this scale.

Parallax error, in numbers

If your eye is shifted sideways by \Delta x while looking at a Vernier scale at a viewing distance D, the apparent shift of the scale relative to the main scale is about \Delta x \cdot h / D, where h is the distance between the two scales (typically about 1 mm). For \Delta x = 2 cm and D = 30 cm, this is about 70 µm — comparable to the LC of 100 µm. So parallax can genuinely be the dominant error in a careless reading.

The fix is to look straight down (perpendicular to the scale) and use only the Vernier division that looks exactly lined up, not the one that looks close.

The screw gauge's backlash and the ratchet

The screw gauge works by a spindle threaded into the frame. Any thread pair has some slack (clearance between the threads), and when you reverse direction, the spindle does not move until the slack is taken up. This is backlash. A worn gauge has 20–50 µm of backlash — enough to swamp the 10 µm LC.

The ratchet solves a different problem: once the spindle touches the wire, further tightening would compress the wire (and damage the gauge). The ratchet has a spring-loaded pawl that slips once the applied torque exceeds a calibrated threshold — you hear the click, and you know the measurement pressure is standardised. Measurement machinists in ISRO's Liquid Propulsion Systems Centre use micrometre gauges with stiffer ratchets calibrated for measuring engine components to about 1 µm precision; the principle is the same as the ₹600 school gauge, scaled up in precision and cost.

Reporting conventions — significant figures

The rule for the final reading: the last digit of the mean is the first uncertain digit, and that digit lines up with the least significant digit of the error. So:

  • Wrong: d = 7.118 \pm 0.006 cm (the 8 is beyond the error's precision)
  • Wrong: d = 7.1 \pm 0.01 cm (the error is more precise than the mean)
  • Right: d = 7.12 \pm 0.01 cm (both have the same last decimal place)

The percentage error has no such constraint — it is a dimensionless ratio and is conventionally reported to one or two significant figures (0.14%, not 0.1403%).

The absolute error in length is usually limited by the LC below, and by the spread of readings above. When the spread is below LC, quote LC. When the spread is above LC, quote the spread. Never quote below LC — that is a claim the instrument cannot support.

Where this leads next