PhysicsICSE Class 9 · Measurements and Experimentation

Vernier Callipers and Screw Gauge: Least Count, Readings and Zero Error (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

A Vernier calliper (least count 0.1 mm = 0.01 cm) and a screw gauge (pitch 1 mm, 100 circular divisions, least count 0.01 mm) measure small lengths accurately. Reading = main (or linear) scale reading + coinciding (or circular) division × least count; then subtract the zero error with its sign.

Understand Vernier Callipers and Screw Gauge: Least Count, Readings and Zero Error

Why a special instrument? A metre rule reads only to 1 mm (its least count). To measure a wire's diameter or a coin's thickness with a small percentage error you need an instrument with a much smaller least count (LC), the smallest length it can read reliably. A smaller LC means a more accurate measurement.

Vernier callipers. A fixed main scale (in cm, each cm split into 10 mm) carries a sliding Vernier scale of 10 divisions that together span 9 mm. So 1 VSD = 0.9 mm and

$L C = 1 MSD - 1 VSD = 1.0 - 0.9 = 0.1 mm = 0.01 cm .$

Reading steps:

note the main scale reading (MSR) just to the left of the Vernier zero;
find the Vernier division that coincides with a main-scale mark (call its number $n$ );
$reading = MSR + n \times L C$ .

Jaws: outer jaws for external diameters, inner (upper) jaws for internal diameters, and the thin strip for depth.

Screw gauge. The spindle advances by the pitch (here 1 mm) in one full turn of the thimble, and the thimble's edge has 100 equal divisions, so

$L C = \frac{pitch}{number of circular divisions} = \frac{1 mm}{100} = 0.01 mm .$

Reading steps:

pitch scale reading (PSR): the last visible mark on the sleeve;
circular scale reading (CSR): the thimble division level with the reference line;
$reading = PSR + CSR \times L C$ .

Turn only the ratchet to close the jaws, so the object is not squeezed.

Zero error. With the jaws closed the instrument should read 0. If it does not, the reading it gives is the zero error and every observed reading must be corrected:

$true reading = observed reading - zero error (with its sign).$

Instrument	Positive zero error	Negative zero error
Vernier callipers	Vernier zero is to the right of the main zero; error $= + n \times L C$ ( $n$ = coinciding division)	Vernier zero is to the left; error $= - (10 - n) \times L C$
Screw gauge	Zero of the circular scale is below the reference line; error $= + n \times L C$	Zero is above the reference line; error $= - (100 - n) \times L C$

A negative error means the instrument reads too little, so subtracting it adds to the observed value.

Percentage thinking. The same absolute error matters more for a thin object. A 0.03 mm slip on a 1.2 mm wire is 2.5% of its diameter, and because area goes as $d^{2}$ , the error in area is about twice that.

Vernier Callipers and Screw Gauge: Least Count, Readings and Zero Error: formulas and key facts

Vernier least count: $L C = 1 MSD - 1 VSD = \frac{1 MSD}{number of Vernier divisions}$; 0.1 mm = 0.01 cm for 10 divisions.
Vernier reading: $MSR + n \times L C$; n = coinciding Vernier division.
Screw gauge least count: $L C = \frac{pitch}{circular divisions}$; 1 mm / 100 = 0.01 mm.
Screw gauge reading: $PSR + CSR \times L C$; Then correct for zero error.
Corrected reading: $true = observed - zero error$; Keep the sign of the error.

How to solve Vernier Callipers and Screw Gauge: Least Count, Readings and Zero Error questions

Work in one unit (mm for the screw gauge, cm or mm for the Vernier) and carry the sign of the zero error.

1Write the least count first, with its working.
2Find the scale reading (MSR or PSR) from the last full mark crossed.
3Multiply the coinciding or circular division by the least count and add.
4Find the zero error from the closed-jaw reading, with its sign.
5Subtract the signed error from the observed reading; give units and the right number of decimals.

Vernier Callipers and Screw Gauge: Least Count, Readings and Zero Error in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Choosing the instrument

Suppose a workshop must check that a steel rod has a diameter of 2.50 mm, to within 1% either way.

Can a metre rule do this job? Can a screw gauge?

Try it first, then show the solutionSolution

1% of 2.50 mm = 0.025 mm is the allowed tolerance.
A metre rule has LC 1 mm, which is 40 times too coarse for a 0.025 mm tolerance.
A screw gauge has LC 0.01 mm, smaller than 0.025 mm, so it is suitable.

Answer: Only the screw gauge.

Memory hook: The tolerance must be larger than the least count.

A wrong zero in the lab

Imagine a class where every student gets a wire diameter 0.03 mm too large, because their shared screw gauge shows 3 on the circular scale with the jaws closed and nobody noted it.

What is the zero error and what should the students do?

Try it first, then show the solutionSolution

Circular zero is below the reference line: positive error, $+ 3 \times 0.01 = + 0.03$ mm.
Subtract 0.03 mm from every observed diameter.

Answer: +0.03 mm; subtract it from each reading.

Memory hook: Closed jaws first, object second.

Tricks to remember Vernier Callipers and Screw Gauge: Least Count, Readings and Zero Error

Short memory aids. Say each one aloud once, then use it on a question the same day.

LC = value of one small step
Vernier LC = 1 mm ÷ 10; screw gauge LC = 1 mm ÷ 100. Always "step size ÷ number of divisions".
Use it when: Writing LC quickly.
Observed minus error
True = observed − error, always. A negative error turns "minus" into "plus".
Use it when: Sign mistakes.
Two scales, two parts
Vernier: MSR + n × LC. Screw gauge: PSR + CSR × LC. Same shape, different names.
Use it when: Reading formulae.

Vernier Callipers and Screw Gauge: Least Count, Readings and Zero Error: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Least count of a Vernier calliper · 1 mark · MCQBoard level

On a Vernier calliper 10 Vernier divisions coincide with 9 divisions of the main scale, each of 1 mm. The least count is:
(a) 0.1 mm
(b) 0.01 mm
(c) 0.9 mm
(d) 1 mm

Try it first, then show the solutionSolution

$L C = 1 MSD - 1 VSD = 1 - 0.9 = 0.1$ mm.

Answer: (a) 0.1 mm

Example 2: A simple Vernier reading · 2 marks · Short answerBoard level

On a Vernier calliper (10 Vernier divisions = 9 main-scale divisions, 1 MSD = 1 mm) the main scale reads 2.3 cm just before the Vernier zero and the 6th Vernier division coincides with a main-scale mark. Find the reading in cm.

Try it first, then show the solutionSolution

$L C = 0.1$ mm $= 0.01$ cm. (½)
Reading $= 2.3 + 6 \times 0.01$ (1)
$= 2.36$ cm. (½)

Answer: 2.36 cm

Example 3: Positive zero error, Vernier · 3 marks · Short answerHarder board level

With the jaws of a Vernier calliper closed, the Vernier zero lies to the right of the main-scale zero and the 4th Vernier division coincides. A cylinder is then measured: MSR $= 4.5$ cm and the 7th division coincides. Find the correct diameter.

Try it first, then show the solutionSolution

Zero error $= + 4 \times 0.01 = + 0.04$ cm. (1)
Observed $= 4.5 + 7 \times 0.01 = 4.57$ cm. (1)
Correct $= 4.57 - 0.04 = 4.53$ cm. (1)

Answer: 4.53 cm

Example 4: Negative zero error, screw gauge · 3 marks · Short answerHarder board level

With the jaws closed, the 97th division of the 100-division circular scale is on the reference line. A wire is then gripped: the last visible sleeve mark is 2 mm and the 47th division is on the reference line (pitch 1 mm). Find (a) the zero error, (b) the true diameter, (c) the area of cross-section (take $π = 3.14$ ).

Try it first, then show the solutionSolution

(a) The circular zero is above the line: error $= - (100 - 97) \times 0.01 = - 0.03$ mm. (1)
(b) Observed $= 2 + 47 \times 0.01 = 2.47$ mm; true $= 2.47 - (- 0.03) = 2.50$ mm. (1)
(c) $r = 1.25$ mm; $A = π r^{2} = 3.14 \times 1.5625 \approx 4.91$ mm². (1)

Answer: (a) −0.03 mm (b) 2.50 mm (c) about 4.91 mm²

Example 5: Thickness of a sheet from a stack · 4 marks · Case-basedHOTS / competency

A student measures the thickness of a stack of 7 identical cards with a screw gauge (pitch 1 mm, 100 divisions). With the jaws closed, the reading is 4 divisions below the reference line (circular zero below the line). On the stack, the last visible sleeve mark is 4 mm and the 94th division is on the reference line.
(a) State the type and size of the zero error.
(b) Find the corrected thickness of the stack.
(c) Find the thickness of one card.
(d) A metre rule (LC 1 mm) is used for one card. Express its least count as a percentage of the card thickness and comment.

Try it first, then show the solutionSolution

(a) Positive error, $+ 4 \times 0.01 = + 0.04$ mm. (1)
(b) Observed $= 4 + 94 \times 0.01 = 4.94$ mm; corrected $= 4.94 - 0.04 = 4.90$ mm. (1)
(c) $\frac{4.90}{7} = 0.70$ mm. (1)
(d) $\frac{1}{0.70} \times 100 \approx 143%$ : the rule cannot even resolve a single card, so a screw gauge on the stack is the right choice. (1)

Answer: (a) +0.04 mm (b) 4.90 mm (c) 0.70 mm (d) about 143%; a ruler is useless for one card

Where students lose marks in Vernier Callipers and Screw Gauge: Least Count, Readings and Zero Error

The slips examiners see most often, and what to write instead.

Mistake: Zero error added instead of subtracted
True reading = observed − error.
Wrong working
Observed 4.57 cm, error +0.04 cm, so 4.61 cm.
Right working
4.57 − 0.04 = 4.53 cm.
Do this: Write the formula before substituting.
Usually costs: The final answer (often 1 of 3 marks).
Mistake: Vernier division number multiplied by the wrong unit
LC is 0.01 cm, not 0.1 cm.
Wrong working
2.3 + 6 × 0.1 = 2.9 cm.
Right working
2.3 + 6 × 0.01 = 2.36 cm.
Do this: Convert the LC to the unit of the MSR first.
Usually costs: The final answer.
Mistake: Negative error taken from the wrong end
For a negative screw gauge error, count up to 100.
Wrong working
97th division on the line, so error = −97 × 0.01.
Right working
Error = −(100 − 97) × 0.01 = −0.03 mm.
Do this: Ask: is the zero above the line? Then count back from 100.
Usually costs: 1 to 2 marks.

Exam tips

Always state the least count with its calculation before taking a reading.
Give the corrected reading to the least count (two decimals in mm for the screw gauge, two in cm for the Vernier).
Write the zero error with its sign and say "positive" or "negative".
Use the ratchet in the procedure answer; never tighten the thimble directly.

Practice questions on Vernier Callipers and Screw Gauge: Least Count, Readings and Zero Error

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The Vernier constant of a Vernier calliper is:
(a) the sum of one MSD and one VSD
(b) the difference between one MSD and one VSD
(c) the number of Vernier divisions
(d) the length of the Vernier scale
Hint
It is also the least count.
Answer
(b) the difference between one MSD and one VSD
Q2 · 1 mark · Very short answerBoard level
A screw gauge (pitch 1 mm, 100 divisions) shows PSR 3 mm and CSR 28. Write the reading.
Hint
PSR + CSR × LC.
Answer
$3 + 28 \times 0.01 = 3.28$ mm.
Q3 · 2 marks · Short answerHarder board level
A Vernier scale has 20 divisions that together equal 19 main-scale divisions (1 MSD = 1 mm). Find the least count in mm and in cm.
Hint
LC = 1 MSD ÷ number of Vernier divisions.
Answer
$L C = \frac{1}{20} = 0.05$ mm $= 0.005$ cm.
Q4 · 3 marks · Short answerHarder board level
A Vernier calliper has zero error $- 0.02$ cm. A sphere gives MSR $= 1.9$ cm with the 4th division coinciding. Find (a) the true diameter, (b) the volume of the sphere ( $π = 3.14$ ).
Hint
Subtract a negative error: add 0.02.
Answer
(a) Observed $= 1.9 + 0.04 = 1.94$ cm; true $= 1.94 + 0.02 = 1.96$ cm. (b) $r = 0.98$ cm; $V = \frac{4}{3} \times 3.14 \times 0.9 8^{3} \approx 3.94$ cm³.
Q5 · 2 marks · Short answerHarder board level
Three trials with a screw gauge on a wire give 2.48 mm, 2.47 mm and 2.49 mm. The zero error is $+ 0.02$ mm. Find the mean observed diameter and the corrected diameter.
Hint
Mean first, then correct once.
Answer
Mean $= \frac{2.48 + 2.47 + 2.49}{3} = 2.48$ mm; corrected $= 2.48 - 0.02 = 2.46$ mm.
Q6 · 3 marks · Short answerHOTS / competency
A student measures a wire of true diameter 1.17 mm but forgets to subtract a zero error of $+ 0.03$ mm, so records 1.20 mm. By what percentage does the calculated area of cross-section $\frac{π d ^{2}}{4}$ exceed the true area? (nearest 0.1%)
Hint
Area scales as the square of the diameter.
Answer
$\frac{1.2 0 ^{2}}{1.1 7 ^{2}} = \frac{1.44}{1.3689} \approx 1.052$ . The area is about 5.2% too large (more than the 2.6% error in $d$ ).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Least count of Vernier callipers (10 divisions)?Show answer
0.1 mm or 0.01 cm.
2.Least count of the screw gauge in the syllabus?Show answer
0.01 mm (pitch 1 mm, 100 divisions).
3.Corrected reading formula?Show answer
Observed reading minus zero error, with its sign.
4.What does a negative zero error do to readings?Show answer
Readings are too small; the correction adds to them.
5.Why use the ratchet?Show answer
To close the jaws with a standard light force without squeezing the object.

Vernier Callipers and Screw Gauge: Least Count, Readings and Zero Error: frequently asked questions

What is the least count of a screw gauge?

It is the pitch divided by the number of circular scale divisions. For pitch 1 mm and 100 divisions it is 0.01 mm.

How do you find zero error in a Vernier calliper?

Close the jaws and read the Vernier. If the Vernier zero is to the right of the main zero the error is positive, equal to the coinciding division times the least count; if to the left it is negative, equal to −(10 − n) × LC.

Which is more accurate, Vernier callipers or a screw gauge?

The screw gauge, because its least count (0.01 mm) is ten times smaller than that of the Vernier calliper (0.1 mm).